SOLUTIONS MANUAL for Thermodynamics An Engineering Approach, 10th Edition by Yunus Cengel by ACADEMIAMILL

1-1

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Table Of Contents Chapter 1) Introduction and Basic Concepts Chapter 2) Energy, Energy Transfer, and General Energy Analysis Chapter 3) Properties of Pure Substances Chapter 4) Energy Analysis of Closed Systems Chapter 5) Mass and Energy Analysis of Control Volumes Chapter 6) The Second Law of Thermodynamics Chapter 7) Entropy Chapter 8) Entropy Analysis Chapter 9) Exergy Chapter 10) Gas Power Cycles Chapter 11) Vapor and Combined Power Cycles Chapter 12) Refrigeration Cycles Chapter 13) Thermodynamic Property Relations Chapter 14) Gas Mixtures Chapter 15) Gas-Vapor Mixtures and Air-Conditioning Chapter 16) Chemical Reactions Chapter 17) Chemical and Phase Equilibrium Chapter 18) Compressible Flow

1-2

Chapter 1 INTRODUCTION AND BASIC CONCEPTS

PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw Hill LLC and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw Hill LLC: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw Hill LLC. CYU - Check Your Understanding

CYU 1-1 ■ Check Your Understanding CYU 1-1.1 Choose the wrong statement about the first law of thermodynamics. (a) During an interaction, energy can change from one form to another. (b) During an interaction, the total amount of energy remains constant. (c) Energy can be destroyed but it cannot be created. (d) Energy is a thermodynamic property. (e) A system with more energy input than output will gain energy. Answer: (c) Energy can be destroyed but it cannot be created. CYU 1-1.2 Which statement best expresses the second law of thermodynamics? (a) The temperature of a well-sealed room increases when a fan in the room is turned on. (b) A cup of cool coffee in a warm room never gets hot by itself. (c) Heat is generated when there is friction between two surfaces. (d) A household refrigerator supplies heat to the kitchen when operating. (e) A person who has a smaller energy input than output will lose weight. Answer: (b) A cup of cool coffee in a warm room never gets hot by itself.

CYU 1-2 ■ Check Your Understanding CYU 1-2.1 Which of the following is not a correct unit for work? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-3

(a) (b) Nm (c) (d) kJ (e) kWh Answer: (a) CYU 1-2.2 Using unit considerations, select the correct equations. I. Mass = Force  Acceleration II. Work = Force  Distance III. Work = Power  Time IV. Density = Mass  Volume (a) I and II (b) I and III (c) II and III (d) I, II, and III (e) II, III, and IV Answer: (c) II and III CYU 1-2.3 What is the weight of 10 kg? (a) 10 N (b) 10 lbf (c) 9.8 N (d) 98 N (e) 322 lbf Answer: (d) 98 N

since

CYU 1-3 ■ Check Your Understanding CYU 1-3.1 Choose the wrong statement regarding systems and control volumes. (a) The mass or region outside the system is called the surroundings. (b) The real or imaginary surface that separates the system from its surroundings is called the boundary. (c) The boundary has zero thickness. (d) The boundary can neither contain any mass nor occupy any volume in space. (e) The boundary of a system is fixed; it cannot move. Answer: (e) The boundary of a system is fixed; it cannot move. CYU 1-3.2 Choose the wrong statement regarding systems and control volumes. (a) Both mass and energy can cross the boundary of a control volume. (b) Mass cannot enter or leave a closed system but energy can. (c) Mass can enter or leave an isolated system but energy cannot. (d) The boundaries of a control volume can be real or imaginary. (e) The volume of a closed system does not have to be fixed. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-4

Answer: (c) Mass can enter or leave an isolated system but energy cannot.

CYU 1-4 ■ Check Your Understanding CYU 1-4.1 Which properties are intensive? I. Volume, II. Density, III. Temperature, C IV. Pressure, kPa (a) II and III (b) III and IV (c) I, II and III (d) II, III, and IV (e) I, II, III, and IV Answer: (d) II, III, and IV

CYU 1-5 ■ Check Your Understanding CYU 1-5.1 A fluid, respectively? (a) 1, 0.5 (b) 0.5, 1 (c) 0.5, 2 (d) 2, 0.5 (e) 0.5, 0.5

tank contains 0.5 kg of a gas. What is the density (in

) and specific volume (in

) of the

Answer: (c) 0.5, 2

CYU 1-5.2 What is the specific volume of a fluid whose specific gravity is 1.25? (a) (b) (c) (d) (e) Answer: (d)

CYU 1-5.3 What is the specific weight of a fluid whose specific gravity is 1? (a) (b) (c) (d) (e) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-5

Answer: (e) since

CYU 1-6 ■ Check Your Understanding CYU 1-6.1 Choose the wrong statement regarding state and equilibrium. (a) A set of properties completely describes the state of a system. (b) At a given state, all the properties of a system have fixed values. (c) The state of a system only changes if the values of at least two properties change. (d) In an equilibrium state there are no driving forces within the system. (e) A system in equilibrium experiences no changes when it is isolated from its surroundings. Answer: (c) The state of a system only changes if the values of at least two properties change.

CYU 1-8 ■ Check Your Understanding CYU 1-8.1 The initial and final temperatures of an object are 20C and 300 K. The temperature change of the object is (a) 27C (b) (b) 20C (c) 7C (d) 300 K (e) 280 K Answer: (c) 7C T1 = 20C T2 = 300 K = 27C dT= 27 – 20 = 7C

CYU 1-8.2 A temperature change of 10C is equal to I. 18F II. 18 R III. 10 K (a) Only I (b) Only II (c) I and II (d) I and III (e) I, II, and III Answer: (e) I, II, and III dT(K) = dT(C)  dT = 10 C = 10 K dT(F) = 1.8 dT(C)  dT = 10 C = 18 F dT(R) = 1.8 dT(C)  dT = 10 C = 18 R

1-6

III. 36 R (a) Only I (b) Only II (c) I and II (d) I and III (e) I, II, and III Answer: (d) I and III dT(F) = 1.8 dT(C)  dT = 36 F = 20 C dT(F) = 1.8 dT(K)  dT = 36 F = 20 K dT(R) = dT(F)  dT = 36 F = 36 R

CYU 1-9 ■ Check Your Understanding CYU 1-9.1 Select all of the pressure units. I. II. bar III. MPa IV. (a) I, II, III, and IV (b) II, III, and IV

(e) I and II

Answer: (b) II, III, and IV CYU 1-9.2 The pressure units Pa and psi are equivalent to (a) , (b) , (c) , (d) , (e) , Answer: (d)

CYU 1-9.3 The gage pressure of a gas in a tank is 230 kPa. If the atmospheric pressure is 100 kPa, what is the absolute pressure of the gas? I. I. 330 kPa II. 33 bar III. 130 kPa IV. 0.13 MPa V. 0.23 MPa (a) Only I (b) Only III (c) Only V (d) III and IV (e) I and II Answer: (a) Only I

CYU 1-9.4 A tank whose surface is open to the atmosphere is filled with water with a depth of 10 m. If the atmospheric pressure is 100 kPa, what is the absolute pressure at the bottom of the tank? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-7

(a) 98,100 kPa

(b) 98,000 kPa

Answer: (e) 198 kPa ( since

)(

)

(

)

and

CYU 1-10 ■ Check Your Understanding CYU 1-9.1 The pressure at the top of a mountain is 95 kPa. If the atmospheric pressure is 100 kPa, what is the height of the mountain? Take the average density of air and the gravitational acceleration to be and . (a) 50 m (b) 100 m (c) 500 m (d) 1000 m (e) 5000 m Answer: (c) 500 m ( since

)(

) (

)

and

CYU 1-11 ■ Check Your Understanding CYU 1-11.1 Which numbers have three significant digits? I. 0.005 II. 0.050 III. 0.500 IV. 50.0 (a) I and II (b) III and IV (c) I, III, and IV (d) II, III, and IV (e) I, II, III, and IV Answer: (b) III and IV

1-8

PROBLEMS Thermodynamics 1-1C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy, and thus the bicyclist picks up speed. There is no creation of energy, and thus no violation of the conservation of energy principle.

1-2C A car going uphill without the engine running would increase the energy of the car, and thus it would be a violation of the first law of thermodynamics. Therefore, this cannot happen. Using a level meter (a device with an air bubble between two marks of a horizontal water tube) it can be shown that the road that looks uphill to the eye is actually downhill.

1-3C There is no truth to his claim. It violates the second law of thermodynamics.

1-4C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles.

1-9

Mass, Force, and Units 1-5C In this unit, the word light refers to the speed of light. The light-year unit is then the product of a velocity and time. Hence, this product forms a distance dimension and unit.

1-6C Pound-mass lbm is the mass unit in English system whereas pound-force lbf is the force unit. One pound-force is the force required to accelerate a mass of 32.174 lbm by . In other words, the weight of a 1-lbm mass at sea level is 1 lbf.

1-7C There is no acceleration, thus the net force is zero in both cases.

1-8 The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be

EES (Engineering Equation Solver) SOLUTION "Given" m=200 [kg] g=9.6 [m/s^2] "Analysis" W=m*g "To Run the program, press F2 or click on the calculator icon from the Calculate menu."

1-9E The mass of an object is given. Its weight is to be determined. Analysis Applying Newton's second law, the weight is determined to be (

)

EES (Engineering Equation Solver) SOLUTION "Given" m=10 [lbm] g=32.0 [ft/s^2] "Analysis" W=m*g*Convert(lbm-ft/s^2, lbf)

1-10 The acceleration of an aircraft is given in g‟s. The net upward force acting on a man in the aircraft is to be determined. Analysis From the Newton's second law, the force applied is (

)

1-10

EES (Engineering Equation Solver) SOLUTION "Given" m=90 [kg] a=6*g "Analysis" g=9.81 [m/s^2] W=m*a

1-11 Gravitational acceleration g and thus the weight of bodies decreases with increasing elevation. The percent reduction in the weight of an airplane cruising at 13,000 m is to be determined. Properties The gravitational acceleration g is given to be at sea level and at an altitude of 13,000 m. Analysis Weight is proportional to the gravitational acceleration g, and thus the percent reduction in weight is equivalent to the percent reduction in the gravitational acceleration, which is determined from

Therefore, the airplane and the people in it will weight 0.41% less at 13,000 m altitude. Discussion Note that the weight loss at cruising altitudes is negligible.

EES (Engineering Equation Solver) SOLUTION "Given" g_sea=9.807 [m/s^2] g_plane=9.767 [m/s^2] altitude=13000 [m] "Analysis" Reduction%=(g_sea-g_plane)/g_sea*Convert(,%) "since weight is proportional to the g"

1-12 A plastic tank is filled with water. The weight of the combined system is to be determined. Assumptions The density of water is constant throughout. Properties The density of water is given to be . Analysis The mass of the water in the tank and the total mass are

Thus, (

)

1-11

EES (Engineering Equation Solver) SOLUTION "Given" m_tank=3 [kg] V=0.2 [m^3] rho=1000 [kg/m^3] "Analysis" m_water=rho*V m_total=m_water+m_tank g=9.81 [m/s^2] W=m_total*g

1-13 A rock is thrown upward with a specified force. The acceleration of the rock is to be determined. Analysis The weight of the rock is (

)

Then the net force that acts on the rock is From the Newton's second law, the acceleration of the rock becomes (

)

EES (Engineering Equation Solver) SOLUTION "Given" m=2 [kg] g=9.79 [m/s^2] F_up=200 [N] "The weight of the rock is" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-12

W=m*g "The force balance on the rock yields the net force acting on the rock as" F_net = F_up - F_down F_down=W "The acceleration of the rock is determined from Newton's second law." F_net=a*m

1-14 Problem 1-13 is reconsidered. The entire solution by appropriate software is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES (Engineering Equation Solver) software, and the solution is given below.

m=2 [kg] F_up=200 [N] g=9.79 [m/s^2] W=m*g F_net=F_up-F_down F_down=W F_net=m*a a=90.21 [m/s^2] F_down=19.58 [N] F_net=180.4 [N] F_up=200 [N] g=9.79 [m/s^2] m=2 [kg] W=19.58 [N]

m[kg]

a [m/s2]

190.2

90.21

56.88

40.21

30.21

23.54

18.78

15.21

12.43

10.21

1-13

1-15 A resistance heater is used to heat water to desired temperature. The amount of electric energy used in kWh and kJ are to be determined. Analysis The resistance heater consumes electric energy at a rate of 4 kW or energy used in 3 hours becomes

. Then the total amount of electric

Discussion Note that kW is a unit for power whereas kWh is a unit for energy.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_e=4 [kW] time=3 [h] "Analysis" Energy=W_dot_e*time Energy_kJ=Energy*Convert(kWh,kJ)

1-16E An astronaut took his scales with him to space. It is to be determined how much he will weigh on the spring and beam scales in space. Analysis (a) A spring scale measures weight, which is the local gravitational force applied on a body: (

)

(b) A beam scale compares masses and thus is not affected by the variations in gravitational acceleration. The beam scale will read what it reads on earth, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-14

EES (Engineering Equation Solver) SOLUTION "Given" m=150 [lbm] g_moon=5.48 [ft/s^2] "Analysis" W_spring=m*g_moon*Convert(lbm-ft/s^2, lbf) g_earth=32.174 [ft/s^2] W_beam=m*g_earth*Convert(lbm-ft/s^2, lbf)

1-17 During an analysis, a relation with inconsistent units is obtained. A correction is to be found, and the probable cause of the error is to be determined. Analysis The two terms on the right-hand side of the equation

do not have the same units, and therefore they cannot be added to obtain the total energy. Multiplying the last term by mass will eliminate the kilograms in the denominator, and the whole equation will become dimensionally homogeneous; that is, every term in the equation will have the same unit. Discussion Obviously this error was caused by forgetting to multiply the last term by mass at an earlier stage.

1-18 A gas tank is being filled with gasoline at a specified flow rate. Based on unit considerations alone, a relation is to be obtained for the filling time. Assumptions Gasoline is an incompressible substance and the flow rate is constant. Analysis The filling time depends on the volume of the tank and the discharge rate of gasoline. Also, we know that the unit of time is „seconds‟. Therefore, the independent quantities should be arranged such that we end up with the unit of seconds. Putting the given information into perspective, we have t [s]  V [L], and ̇ It is obvious that the only way to end up with the unit “s” for time is to divide the tank volume by the discharge rate. Therefore, the desired relation is ̇ Discussion Note that this approach may not work for cases that involve dimensionless (and thus unitless) quantities.

Systems, Properties, State, and Processes 1-19C Carbon dioxide is generated by the combustion of fuel in the engine. Any system selected for this analysis must include the fuel and air while it is undergoing combustion. The volume that contains this air-fuel mixture within pistoncylinder device can be used for this purpose. One can also place the entire engine in a control boundary and trace the system-surroundings interactions to determine the rate at which the engine generates carbon dioxide.

1-20C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system.

1-21C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the system.

1-15

1-22C When analyzing the control volume selected, we must account for all forms of water entering and leaving the control volume. This includes all streams entering or leaving the lake, any rain falling on the lake, any water evaporated to the air above the lake, any seepage to the underground earth, and any springs that may be feeding water to the lake.

1-23C In order to describe the state of the air, we need to know the value of all its properties. Pressure, temperature, and water content (i.e., relative humidity or dew point temperature) are commonly cited by weather forecasters. But, other properties like wind speed and chemical composition (i.e., pollen count and smog index, for example} are also important under certain circumstances. Assuming that the air composition and velocity do not change and that no pressure front motion occurs during the day, the warming process is one of constant pressure (i.e., isobaric).

1-24C Intensive properties do not depend on the size (extent) of the system but extensive properties do.

1-25C The original specific weight is

If we were to divide the system into two halves, each half weighs of one of these halves is

and occupies a volume of

. The specific weight

which is the same as the original specific weight. Hence, specific weight is an intensive property.

1-26C The number of moles of a substance in a system is directly proportional to the number of atomic particles contained in the system. If we divide a system into smaller portions, each portion will contain fewer atomic particles than the original system. The number of moles is therefore an extensive property.

1-27C The state of a simple compressible system is completely specified by two independent, intensive properties.

1-28C Yes, because temperature and pressure are two independent properties and the air in an isolated room is a simple compressible system.

1-29C A process during which a system remains almost in equilibrium at all times is called a quasi-equilibrium process. Many engineering processes can be approximated as being quasi-equilibrium. The work output of a device is maximum and the work input to a device is minimum when quasi-equilibrium processes are used instead of nonquasi-equilibrium processes.

1-30C A process during which the temperature remains constant is called isothermal; a process during which the pressure remains constant is called isobaric; and a process during which the volume remains constant is called isochoric.

1-16

1-31C The specific gravity, or relative density, and is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4°C, for which ). That is, When specific gravity is known, density is determined from .

1-32 The variation of density of atmospheric air with elevation is given in tabular form. A relation for the variation of density with elevation is to be obtained, the density at 7 km elevation is to be calculated, and the mass of the atmosphere using the correlation is to be estimated. Assumptions 1 Atmospheric air behaves as an ideal gas. 2 The earth is perfectly sphere with a radius of 6377 km, and the thickness of the atmosphere is 25 km. Properties The density data are given in tabular form as r, km 6377 6378 6379 6380 6381 6382 6383 6385 6387 6392 6397 6402

z, km 0 1 2 3 4 5 6 8 10 15 20 25

, 1.225 1.112 1.007 0.9093 0.8194 0.7364 0.6601 0.5258 0.4135 0.1948 0.08891 0.04008

Analysis Using EES, (1) Define a trivial function rho = a + z in equation window, (2) select new parametric table from Tables, and type the data in a two-column table, (3) select Plot and plot the data, and (4) select plot and click on “curve fit” to get curve fit window. Then specify 2nd order polynomial and enter/edit equation. The results are:

(or,

for the unit of

)

where z is the vertical distance from the earth surface at sea level. At z = 7 km, the equation would give

1-17

∫

[

]

where is the radius of the earth, h = 25 km is the thickness of the atmosphere, and a = 1.20252, b = -0.101674, and c = 0.0022375 are the constants in the density function. Substituting and multiplying by the factor 10 9 for the density unity , the mass of the atmosphere is determined to be

Performing the analysis with excel would yield exactly the same results. "Using linear regression feature of EES based on the data on parametric table, we obtain" rho=1.20251659E+00-1.01669722E-01*z+2.23747073E-03*z^2 z=7 [km] "The mass of the atmosphere is obtained by integration to be" m=4*pi*(a*r0^2*h+r0*(2*a+b*r0)*h^2/2+(a+2*b*r0+c*r0^2)*h^3/3+(b+2*c*r0)*h^4/4+c*h^5/5)*1E9 a=1.20252 b=-0.101670 c=0.0022375 r0=6377 [km] h=25 [km]

Temperature 1-33C They are Celsius (C) and kelvin (K) in the SI, and fahrenheit (F) and rankine (R) in the English system.

1-34C Probably, but not necessarily. The operation of these two thermometers is based on the thermal expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometers will always give identical readings. Otherwise, the two readings may deviate.

1-35C Two systems having different temperatures and energy contents are brought in contact. The direction of heat transfer is to be determined.

Analysis Heat transfer occurs from warmer to cooler objects. Therefore, heat will be transferred from system B to system A until both systems reach the same temperature.

1-36C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the same temperature reading, even if they are not in contact.

1-37E A temperature is given in C. It is to be expressed in F, K, and R. Analysis Using the conversion relations between the various temperature scales, T(K] = T(C) + 273 = 18C + 273 = 291 K T(F] = 1.8T(C) + 32 = (1.8)(18) + 32 = 64.4F T(R] = T(F) + 460 = 64.4 + 460 = 524.4 R

1-18

EES (Engineering Equation Solver) SOLUTION "Given" T_C=18 [C] "Analysis" T_K=T_C+273 T_F=1.8*T_C+32 T_R=T_F+460

1-38E The temperature of steam given in K unit is to be converted to F unit. Analysis Using the conversion relations between the various temperature scales,

EES (Engineering Equation Solver) SOLUTION "Given" T_K=300 [K] "Analysis" T_C=T_K-273 T_F=1.8*T_C+32

1-39 A temperature change is given in C. It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Thus, T(K] = T(C) = 130 K

EES (Engineering Equation Solver) SOLUTION "Given" DELTAT_C=130 [C] "Analysis" DELTAT_K=DELTAT_C

1-40E A temperature change is given in F. It is to be expressed in C, K, and R. Analysis This problem deals with temperature changes, which are identical in Rankine and Fahrenheit scales. Thus, T(R) = T(F) = 45 R The temperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in Fahrenheit and Rankine scales by and T(C) = T(K) = 25C

1-19

DELTAT_F=45 [F] "Analysis" DELTAT_R=DELTAT_F DELTAT_K=DELTAT_R/1.8 DELTAT_C=DELTAT_K

1-41E The temperature of oil given in F unit is to be converted to C unit. Analysis Using the conversion relation between the temperature scales,

EES (Engineering Equation Solver) SOLUTION "Given" T_F=150 [F] "Analysis" T_C=(T_F-32)/1.8

1-42E The temperature of air given in C unit is to be converted to F unit. Analysis Using the conversion relation between the temperature scales,

EES (Engineering Equation Solver) SOLUTION "Given" T_C=150 [C] "Analysis" T_F=1.8*T_C+32

1-43E During a heating process, the temperature of a system rises by 10°C. Express this rise in temperature in K, °F, and R. 1- The temperature rise of a system is to be expressed in different units. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales. Then, T(K) = T(C) = 10 K The temperature changes in Fahrenheit and Rankine scales are also identical and are related to the changes in Celsius and Kelvin scales through Eqs. 1–11 and 1–14: T(R) = 1.8 T(K) = (1.8)(10) = 18 R and T(F) = T(R) = 18F Discussion Note that the units °C and K are interchangeable when dealing with temperature differences.

1-20

Pressure, Manometer, and Barometer 1-44C The pressure relative to the atmospheric pressure is called the gage pressure, and the pressure relative to an absolute vacuum is called absolute pressure.

1-45C The atmospheric pressure, which is the external pressure exerted on the skin, decreases with increasing elevation. Therefore, the pressure is lower at higher elevations. As a result, the difference between the blood pressure in the veins and the air pressure outside increases. This pressure imbalance may cause some thin-walled veins such as the ones in the nose to burst, causing bleeding. The shortness of breath is caused by the lower air density at higher elevations, and thus lower amount of oxygen per unit volume.

1-46C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body. For a constant volume of blood to be discharged by the heart, the blood pressure must increase to overcome the increased resistance to flow.

1-47C No, the absolute pressure in a liquid of constant density does not double when the depth is doubled. It is the gage pressure that doubles when the depth is doubled.

1-48C The density of air at sea level is higher than the density of air on top of a high mountain. Therefore, the volume flow rates of the two fans running at identical speeds will be the same, but the mass flow rate of the fan at sea level will be higher.

1-49E The pressure given in kPa unit is to be converted to psia. Analysis Using the kPa to psia units conversion factor, (

)

EES (Engineering Equation Solver) SOLUTION "Given" P_kPa=200 [kPa] "Analysis" P_psia=P_kPa/(6.895 [kPa/psia])

1-50E A manometer measures a pressure difference as inches of water. This is to be expressed in psia unit. Properties The density of water is taken to be (Table A-3E). Analysis Applying the hydrostatic equation,

(

)(

)

1-21

EES (Engineering Equation Solver) SOLUTION "Given" h_in=40 [in] "or 40 inWater" "Analysis" g=32.174 [ft/s^2] "acceleration of gravity" rho_w=62.4 [lbm/ft^3] h=h_in*Convert(in, ft) DELTAP=rho_w*g*h*Convert(lbm/ft-s^2, psia)

1-51 The pressure in a vacuum chamber is measured by a vacuum gage. The absolute pressure in the chamber is to be determined. Analysis The absolute pressure in the chamber is determined from

EES (Engineering Equation Solver) SOLUTION "Given" P_vac=35 [kPa] P_atm=92 [kPa] "Analysis" P_abs=P_atm-P_vac

1-52E The maximum pressure of a tire is given in English units. It is to be converted to SI units. Assumptions The listed pressure is gage pressure. Analysis Noting that 1 atm = 101.3 kPa = 14.7 psi, the listed maximum pressure can be expressed in SI units as (

)

Discussion We could also solve this problem by using the conversion factor 1 psi = 6.895 kPa.

EES (Engineering Equation Solver) SOLUTION "Given" P_psia=35 [psia] "Analysis" P_kPa=P_psia*(101.3 [kPa])/(14.7 [psia])

1-22

1-53E A pressure gage connected to a tank reads 50 psi. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be . Analysis The atmospheric (or barometric) pressure can be expressed as (

)(

)

Then the absolute pressure in the tank is

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=50 [psi] h_in=29.1 [in] "or 29.1 inHg" rho_mercury=848.4 [lbm/ft^3] "Analysis" g=32.174 [ft/s^2] "acceleration of gravity" h=h_in*Convert(in, ft) P_atm=rho_mercury*g*h*Convert(lbm/ft-s^2, psia) P_abs=P_atm+P_gage

1-54 A pressure gage connected to a tank reads 500 kPa. The absolute pressure in the tank is to be determined. Analysis The absolute pressure in the tank is determined from

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=500 [kPa] P_atm=94 [kPa] "Analysis" P_abs=P_atm+P_gage © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-23

1-55E The weight and the foot imprint area of a person are given. The pressures this man exerts on the ground when he stands on one and on both feet are to be determined. Assumptions The weight of the person is distributed uniformly on foot imprint area. Analysis The weight of the man is given to be 200 lbf. Noting that pressure is force per unit area, the pressure this man exerts on the ground is (a) On both feet: (b) On one foot:

Discussion Note that the pressure exerted on the ground (and on the feet) is reduced by half when the person stands on both feet.

EES (Engineering Equation Solver) SOLUTION "Given" W=200 [lbf] A=1/2*72 [in^2] "Analysis" "(a)" P_a=W/A*Convert(lbf/in^2, psi) "(b)" P_b=W/(2*A)*Convert(lbf/in^2, psi)

1-56 The gage pressure in a liquid at a certain depth is given. The gage pressure in the same liquid at a different depth is to be determined. Assumptions The variation of the density of the liquid with depth is negligible. Analysis The gage pressure at two different depths of a liquid can be expressed as and Taking their ratio,

Solving for

and substituting gives

1-24

Discussion Note that the gage pressure in a given fluid is proportional to depth.

EES (Engineering Equation Solver) SOLUTION "Given" h1=3 [m] P1_gage=42 [kPa] h2=9 [m] "Analysis" P2_gage=(h2/h1)*P1_gage

1-57 The absolute pressure in water at a specified depth is given. The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined. Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.85. We take the density of water to be density of the liquid is obtained by multiplying its specific gravity by the density of water,

. Then

Analysis (a) Knowing the absolute pressure, the atmospheric pressure can be determined from (

)

(

)

(b) The absolute pressure at the same depth in the other liquid is

Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-25

EES (Engineering Equation Solver) SOLUTION "Given" h=9 [m] P=185 [kPa] rho_s=0.85 "Analysis" rho=rho_s*rho_water rho_water=1000 [kg/m^3] "taken" "(a)" P_atm=P-rho_water*g*h*Convert(kg/m-s^2, kPa) g=9.81 [m/s^2] "acceleration of gravity" "(b)" P_abs=P_atm+rho*g*h*Convert(kg/m-s^2, kPa)

1-58 A man is standing in water vertically while being completely submerged. The difference between the pressures acting on the head and on the toes is to be determined. Assumptions Water is an incompressible substance, and thus the density does not change with depth. Properties We take the density of water to be . Analysis The pressures at the head and toes of the person can be expressed as and

where h is the vertical distance of the location in water from the free surface. The pressure difference between the toes and the head is determined by subtracting the first relation above from the second, Substituting, (

)(

)

Discussion This problem can also be solved by noting that the atmospheric pressure (1 atm = 101.325 kPa) is equivalent to 10.3-m of water height, and finding the pressure that corresponds to a water height of 1.75 m.

EES (Engineering Equation Solver) SOLUTION "Given" h=1.75 [m] "Analysis" g=9.81 [m/s^2] DELTAP=rho_water*g*h*Convert(kg/m-s^2, kPa) rho_water=1000 [kg/m^3] "taken" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-26

1-59 A mountain hiker records the barometric reading before and after a hiking trip. The vertical distance climbed is to be determined. Assumptions The variation of air density and the gravitational acceleration with altitude is negligible. Properties The density of air is given to be

Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area, we obtain

(

)(

)

It yields h = 850 m which is also the distance climbed.

EES (Engineering Equation Solver) SOLUTION "Given" P_bottom=0.750 [bar] P_top=0.650 [bar] rho=1.2 [kg/m^3] "Analysis" rho*g*h*Convert(kg/m-s^2, bar)=(P_bottom-P_top) g=9.81 [m/s^2]

1-60 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building. The height of the building is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The density of air is given to be mercury is .

. The density of

Analysis Atmospheric pressures at the top and at the bottom of the building are

(

)(

)

(

)(

)

1-27

Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain

(

)(

)

It yields h = 231 m which is also the height of the building.

EES (Engineering Equation Solver) SOLUTION "Given" h_top=0.675 [m] "or 675 mmHg" h_bottom=0.695 [m] "or 695 mmHg" rho_air=1.18 [kg/m^3] rho_mercury=13600 [kg/m^3] "Analysis" g=9.81 [m/s^2] P_top=rho_mercury*g*h_top*Convert(kg/m-s^2, kPa) P_bottom=rho_mercury*g*h_bottom*Convert(kg/m-s^2, kPa) (P_bottom-P_top)=rho_air*g*h*Convert(kg/m-s^2, kPa)

1-61 Problem 1-60 is reconsidered. The entire software solution is to be printed out, including the numerical results with proper units. Analysis The problem is solved using EES, and the solution is given below. P_bottom=695 [mmHg] P_top=675 [mmHg] g=9.81 [m/s^2] rho=1.18 [kg/m^3] DELTAP_abs=(P_bottom-P_top)*CONVERT(mmHg, kPa) "Delta P reading from the barometers, converted from mmHg to kPa" DELTAP_h=rho*g*h*Convert(Pa, kPa) "Delta P due to the air fluid column height, h, between the top and bottom of the building" DELTAP_abs=DELTAP_h © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-28

DELTAP_abs=2.666 [kPa] DELTAP_h=2.666 [kPa] g=9.81 [m/s^2] h=230.3 [m] P_bottom=695 [mmHg] P_top=675 [mmHg] rho=1.18 [kg/m^3]

1-62 A gas contained in a vertical piston-cylinder device is pressurized by a spring and by the weight of the piston. The pressure of the gas is to be determined. Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield Thus,

(

)

EES (Engineering Equation Solver) SOLUTION "Given" m_piston=3.2 [kg] A=35E-4 [m^2] F_spring=150 [N] P_atm=95 [kPa] "Analysis" g=9.807 [m/s^2] W=m_piston*g P*A=P_atm*A+(W+F_spring)*Convert(N/m^2, kPa) "force balance on the piston"

1-63 Problem 1-62 is reconsidered. The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated. The pressure against the spring force is to be plotted, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-29

P_atm= 95 [kPa] m_piston=3.2 [kg] F_spring=150 [N] A=35*CONVERT(cm^2, m^2) g=9.81 [m/s^2] W_piston=m_piston*g F_atm=P_atm*A*CONVERT(kPa, N/m^2) "From the free body diagram of the piston, the balancing vertical forces yield" F_gas= F_atm+F_spring+W_piston P_gas=F_gas/A*CONVERT(N/m^2, kPa)

] 0 50 100 150 200 250 300 350 400 450 500

] 104 118.3 132.5 146.8 161.1 175.4 189.7 204 218.3 232.5 246.8

1-30

1-64 A gas is contained in a vertical cylinder with a heavy piston. The pressure inside the cylinder and the effect of volume change on pressure are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis (a) The gas pressure in the piston–cylinder device depends on the atmospheric pressure and the weight of the piston. Drawing the free-body diagram of the piston and balancing the vertical forces yield

Thus,

(

)(

)

(b) The volume change will have no effect on the free-body diagram drawn in part (a), and therefore the pressure inside the cylinder will remain the same. Discussion If the gas behaves as an ideal gas, the absolute temperature doubles when the volume is doubled at constant pressure.

EES (Engineering Equation Solver) SOLUTION "Given" m_piston=60 [kg] A=0.04 [m^2] P_atm=97 [kPa] g=9.81 [m/s^2] "Analysis" W=m_piston*g P*A=P_atm*A+W*Convert(N/m^2, kPa) "force balance on the piston"

1-65 Both a gage and a manometer are attached to a gas to measure its pressure. For a specified reading of gage pressure, the difference between the fluid levels of the two arms of the manometer is to be determined for mercury and water. Properties The densities of water and mercury are given to be and be . Analysis The gage pressure is related to the vertical distance h between the two fluid levels by → © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-31

(a) For mercury, (

)(

)

(

)(

)

(b) For water,

EES (Engineering Equation Solver) SOLUTION Fluid$='Mercury' P_atm = 101.325 [kPa] DELTAP=80 [kPa] "Note how DELTAP is displayed on the Formatted Equations Window." g=9.807 [m/s^2] "local acceleration of gravity at sea level" rho=Fluid_density(Fluid$) "Get the fluid density, either Hg or H2O, from the function." DELTAP = rho*g*h*CONVERT(Pa, kPa) "force balance" "Instead of multiplying by CONVERT('Pa','kPa') we could have divided by 1000 but then the Check Units option would not work on this equation since the constant 1000 does not have units. "

1-66 Problem 1-65 is reconsidered. The effect of the manometer fluid density in the range of 800 to on the differential fluid height of the manometer is to be investigated. Differential fluid height against the density is to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Let's modify this problem to also calculate the absolute pressure in the tank by supplying the atmospheric pressure." Function fluid_density(Fluid$) "This function is needed since if-then-else logic can only be used in functions or procedures. The underscore displays whatever follows as subscripts in the Formatted Equations Window." If fluid$='Mercury' then fluid_density=13600 else fluid_density=1000 end Fluid$='Mercury' P_atm = 101.325 [kPa] DELTAP=80 [kPa] "Note how DELTAP is displayed on the Formatted Equations Window." g=9.807 [m/s^2] "local acceleration of gravity at sea level" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-32

rho=Fluid_density(Fluid$) "Get the fluid density, either Hg or H2O, from the function" DELTAP = RHO*g*h/1000 "Instead of dividiing by 1000 Pa/kPa we could have multiplied by the EES function, CONVERT(Pa,kPa)" h_mm=h*convert(m, mm) "The fluid height in mm is found using the built-in CONVERT function." P_abs= P_atm + DELTAP

] ] 800 2156 3511 4867 6222 7578 8933 10289 11644 13000

10197 3784 2323 1676 1311 1076 913.1 792.8 700.5 627.5

1-33

1-67 The air pressure in a tank is measured by an oil manometer. For a given oil-level difference between the two columns, the absolute pressure in the tank is to be determined. Properties The density of oil is given to be . Analysis The absolute pressure in the tank is determined from (

)

EES (Engineering Equation Solver) SOLUTION "Given" h=0.80 [m] P_atm=98 [kPa] rho_oil=850 [kg/m^3] "Analysis" g=9.81 [m/s^2] P=P_atm+rho_oil*g*h*Convert(kg/m-s^2, kPa)

1-68E The pressure in a tank is measured with a manometer by measuring the differential height of the manometer fluid. The absolute pressure in the tank is to be determined for the cases of the manometer arm with the higher and lower fluid level being attached to the tank. Assumptions The fluid in the manometer is incompressible. Properties The specific gravity of the fluid is given to be SG = 1.25. The density of water at 32F is (Table A-3E)

1-34

Analysis The density of the fluid is obtained by multiplying its specific gravity by the density of water, The pressure difference corresponding to a differential height of 28 in between the two arms of the manometer is (

)(

)

Then the absolute pressures in the tank for the two cases become: (a) The fluid level in the arm attached to the tank is higher (vacuum): (b) The fluid level in the arm attached to the tank is lower: Discussion Note that we can determine whether the pressure in a tank is above or below atmospheric pressure by simply observing the side of the manometer arm with the higher fluid level.

EES (Engineering Equation Solver) SOLUTION "Given" rho_s=1.25 h=28[in]*Convert(in, ft) P_atm=12.7 [psia] "Properties" rho_water=density(water, T=80, P=14.7) g=32.174 [ft/s^2] "acceleration of gravity" "Analysis" rho=rho_s*rho_water DELTAP=rho*g*h*Convert(lbm/ft-s^2, psia) "(a)" P_abs_a=P_atm-DELTAP "(b)" P_abs_b=P_atm+DELTAP

1-69E The pressure in a natural gas pipeline is measured by a double U-tube manometer with one of the arms open to the atmosphere. The absolute pressure in the pipeline is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. 3 The pressure throughout the natural gas (including the tube) is uniform since its density is low. Properties We take the density of water to be . The specific gravity of mercury is given to be 13.6, and thus its density is .

Analysis Starting with the pressure at point 1 in the natural gas pipeline, and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to gives

1-35

Solving for Substituting, ](

)(

)

Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly. Also, it can be shown that the 15-in high air column with a density of corresponds to a pressure difference of 0.00065 psi. Therefore, its effect on the pressure difference between the two pipes is negligible.

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=14.2 [psia] h_Hg=6[in]*Convert(in,ft) h_w=27[in]*Convert(in,ft) rho_s=13.6 "Analysis" P=P_atm+g*(rho_Hg*h_Hg+rho_w*h_w)*Convert(lbm/ft-s^2, psia) rho_w=62.4 [lbm/ft^3] "taken" rho_Hg=rho_s*rho_w g=32.174 [ft/s^2] 1-70E The pressure in a natural gas pipeline is measured by a double U-tube manometer with one of the arms open to the atmosphere. The absolute pressure in the pipeline is to be determined. Assumptions 1 All the liquids are incompressible. 2 The pressure throughout the natural gas (including the tube) is uniform since its density is low. Properties We take the density of water to be . The specific gravity of mercury is given to be 13.6, and thus its density is . The specific gravity of oil is given to be 0.69, and thus its density is .

1-36

](

)(

)

Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=14.2 [psia] h_Hg=6/12 "[ft]" h_w=27/12 "[ft]" h_oil=15/12 "[ft]" rho_s=13.6 rho_s_oil=0.69 "Analysis" P=P_atm+g*(rho_Hg*h_Hg+rho_w*h_w-rho_oil*h_oil)*Convert(lbm/ft-s^2, psia) rho_w=62.4 [lbm/ft^3] "taken" rho_Hg=rho_s*rho_w rho_oil=rho_s_oil*rho_w g=32.174 [ft/s^2]

1-71E The systolic and diastolic pressures of a healthy person are given in mmHg. These pressures are to be expressed in kPa, psi, and meter water column. Assumptions Both mercury and water are incompressible substances. Properties We take the densities of water and mercury to be and , respectively. Analysis Using the relation for gage pressure, the high and low pressures are expressed as (

)(

)

(

)(

)

Noting that 1 psi = 6.895 kPa, (

)

(

)

and

1-37

For a given pressure, the relation can be expressed for mercury and water as and Setting these two relations equal to each other and solving for water height gives

Therefore,

Discussion Note that measuring blood pressure with a “water” monometer would involve differential fluid heights higher than the person, and thus it is impractical. This problem shows why mercury is a suitable fluid for blood pressure measurement devices.

EES (Engineering Equation Solver) SOLUTION "Given" h_high=0.120 [m] "or 120 mmHg" h_low=0.080 [m] "or 80 mmHg" "Properties" rho_water=1000 [kg/m^3] "taken" rho_mercury=13600 [kg/m^3] "taken" g=9.81 [m/s^2] "Analysis" P_high_kPa=rho_mercury*g*h_high*Convert(kg/m-s^2, kPa) P_low_kPa=rho_mercury*g*h_low*Convert(kg/m-s^2, kPa) P_high_psia=P_high_kPa*Convert(kPa, psia) P_low_psia=P_low_kPa*Convert(kPa, psia) h_water_high=rho_mercury/rho_water*h_high "from P = rho_water*g*h_water = rho_mercury*g*h_mercury" h_water_low=rho_mercury/rho_water*h_low

1-72 A vertical tube open to the atmosphere is connected to the vein in the arm of a person. The height that the blood will rise in the tube is to be determined. Assumptions 1 The density of blood is constant. 2 The gage pressure of blood is 120 mmHg. Properties The density of blood is given to be . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-38

Analysis For a given gage pressure, the relation for mercury and blood as and Setting these two relations equal to each other we get

can be expressed

Solving for blood height and substituting gives

Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein. This explains why IV tubes must be placed high to force a fluid into the vein of a patient.

EES (Engineering Equation Solver) SOLUTION "Given" h_mercury=0.120 [m] "or 120 mmHg" rho_blood=1050 [kg/m^3] rho_mercury=13600 [kg/m^3] "Analysis" h_blood=rho_mercury/rho_blood*h_mercury "from P = rho_blood*g*h_blood = rho_mercury*g*h_mercury"

1-73 Water is poured into the U-tube from one arm and oil from the other arm. The water column height in one arm and the ratio of the heights of the two fluids in the other arm are given. The height of each fluid in that arm is to be determined. Assumptions Both water and oil are incompressible substances. Properties The density of oil is given to be . We take the density of water to be . Analysis The height of water column in the left arm of the monometer is given to be . We let the height of water and oil in the right arm to be and , respectively. Then, . Noting that both arms are open to the atmosphere, the pressure at the bottom of the U-tube can be expressed as and

1-39

Setting them equal to each other and simplifying, Noting that

, the water and oil column heights in the second arm are determined to be

Discussion Note that the fluid height in the arm that contains oil is higher. This is expected since oil is lighter than water.

EES (Engineering Equation Solver) SOLUTION "Given" h_w1=0.70 [m] h_a/h_w2=4 rho_a=790 [kg/m^3] "Analysis" rho_w=1000 [kg/m^3] "taken" h_w1=h_w2+(rho_a/rho_w)*h_a "from P_bottom = P_atm+rho_w*g*h_w1 = P_atm+rho_w*g*h_w2+rho_a*g*h_a"

1-74 A double-fluid manometer attached to an air pipe is considered. The specific gravity of one fluid is known, and the specific gravity of the other fluid is to be determined. Assumptions 1 Densities of liquids are constant. 2 The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties The specific gravity of one fluid is given to be 13.55. We take the standard density of water to be .

1-40

Analysis Starting with the pressure of air in the tank, and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the free surface where the oil tube is exposed to the atmosphere, and setting the result equal to give Rearranging and solving for

, (

)(

)

Discussion Note that the right fluid column is higher than the left, and this would imply above atmospheric pressure in the pipe for a single-fluid manometer.

EES (Engineering Equation Solver) SOLUTION "Given" SG_1=13.55 h1=0.28 [m] h2=0.40 [m] P_air=69 [kPa] P_atm=100 [kPa] "Analysis" rho_w=1000 [kg/m^3] "taken" g=9.81 [m/s^2] SG_2=SG_1*h1/h2+(P_air-P_atm)/(rho_w*g*h2)*Convert(kPa-m^2, kg-m/s^2)

1-75 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be Analysis The absolute pressure is determined from

(

Note that

and

, respectively.

)

1-41

P_atm=758 [mmHg] h_a=0.05 [m] h_B=0.15 [m] rhog_A=10 [kN/m^3] rhog_B=8 [kN/m^3] "Analysis" P1=P_atm*Convert(mmHg,kPa)+rhog_A*h_A+rhog_B*h_B

1-76 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be Analysis The absolute pressure is determined from

Note that

and

, respectively.

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=90 [kPa] h_a=0.05 [m] h_B=0.15 [m] rhog_A=100 [kN/m^3] rhog_B=8 [kN/m^3] "Analysis" P1=P_atm+rhog_A*h_A+rhog_B*h_B

1-77 The pressure indicated by a manometer is to be determined. Properties The specific weights of fluid A and fluid B are given to be and , respectively. Analysis The absolute pressure is determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-42

æ0.1333 kPa ÷ ö ÷ = (720 mm Hg) çç ÷ çè 1 mm Hg ÷ ø + (10 kN/m3 )(0.05 m) + (20 kN/m3 )(0.15 m) Note that

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=720 [mmHg] h_a=0.05 [m] h_B=0.15 [m] rhog_A=10 [kN/m^3] rhog_B=20 [kN/m^3] "Analysis" P1=P_atm*Convert(mmHg,kPa)+rhog_A*h_A+rhog_B*h_B

1-78 The hydraulic lift in a car repair shop is to lift cars. The fluid gage pressure that must be maintained in the reservoir is to be determined. Assumptions The weight of the piston of the lift is negligible. Analysis Pressure is force per unit area, and thus the gage pressure required is simply the ratio of the weight of the car to the area of the lift,

(

)

1-43

Discussion Note that the pressure level in the reservoir can be reduced by using a piston with a larger area.

EES (Engineering Equation Solver) SOLUTION "Given" D=0.30 [m] m=2500 [kg] "Analysis" g=9.81 [m/s^2] W=m*g A=pi*D^2/4 P_gage=W/A*Convert(N/m^2, kPa)

1-79 The fluid levels in a multi-fluid U-tube manometer change as a result of a pressure drop in the trapped air space. For a given pressure drop and brine level change, the area ratio is to be determined. Assumptions 1 All the liquids are incompressible. 2 Pressure in the brine pipe remains constant. 3 The variation of pressure in the trapped air space is negligible. Properties The specific gravities are given to be 13.56 for mercury and 1.1 for brine. We take the standard density of water to be .

Analysis It is clear from the problem statement and the figure that the brine pressure is much higher than the air pressure, and when the air pressure drops by 0.7 kPa, the pressure difference between the brine and the air space increases also by the same amount. Starting with the air pressure (point A) and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the brine pipe (point B), and setting the result equal to before and after the pressure change of air give Before: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-44

After: Subtracting, (1) where and are the changes in the differential mercury and brine column heights, respectively, due to the drop in air pressure. Both of these are positive quantities since as the mercury-brine interface drops, the differential fluid heights for both mercury and brine increase. Noting also that the volume of mercury is constant, we have and

Substituting, ] It gives

EES (Engineering Equation Solver) SOLUTION "Given" DELTAP=0.7 [kPa]*Convert(kPa, kg/m-s^2) DELTAh_br=0.005 [m] "Analysis" rho_w=1000 [kg/m^3] g=9.81 [m/s^2] DELTAP/(rho_w*g)=13.56*DELTAh_br*(1+Ratio)-(1.1*DELTAh_br)

1-80 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be 0.72 and 13.6, respectively.

. The specific gravities of oil and mercury are given to be

Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to gives Rearranging, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-45

or,

Substituting, ( Solving for

)(

)

gives

Therefore, the differential height of the mercury column must be 58.2 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=80 [kPa] h_w=0.3 [m] h_oil=0.75 [m] rho_s_oil=0.72 rho_s_Hg=13.6 "Properties" rho_w=1000 [kg/m^3] "taken" rho_oil=rho_s_oil*rho_w rho_Hg=rho_s_Hg*rho_w g=9.807 [m/s^2] "Analysis" P_gage=g*(rho_oil*h_oil+rho_Hg*h_Hg-rho_w*h_w)*Convert(kg/m-s^2, kPa) 1-81 The gage pressure of air in a pressurized water tank is measured simultaneously by both a pressure gage and a manometer. The differential height h of the mercury column is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties We take the density of water to be . The specific gravities of oil and mercury are given to be 0.72 and 13.6, respectively. Analysis Starting with the pressure of air in the tank (point 1), and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to gives Rearranging,

or, Substituting, [

](

)

1-46

Solving for

gives

Therefore, the differential height of the mercury column must be 28.2 cm. Discussion Double instrumentation like this allows one to verify the measurement of one of the instruments by the measurement of another instrument.

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=40 [kPa] h_w=0.3 [m] h_oil=0.75 [m] rho_s_oil=0.72 rho_s_Hg=13.6 "Properties" rho_w=1000 [kg/m^3] "taken" rho_oil=rho_s_oil*rho_w rho_Hg=rho_s_Hg*rho_w g=9.81 [m/s^2] "Analysis" P_gage=g*(rho_oil*h_oil+rho_Hg*h_Hg-rho_w*h_w)*Convert(kg/m-s^2, kPa)

Solving Engineering Problems and Equation Solvers 1-82C Despite the convenience and capability the engineering software packages offer, they are still just tools, and they will not replace the traditional engineering courses. They will simply cause a shift in emphasis in the course material from mathematics to physics. They are of great value in engineering practice, however, as engineers today rely on software packages for solving large and complex problems in a short time, and perform optimization studies efficiently.

1-83

Determine a positive real root of the following equation using appropriate software:

Solution by EES Software (Copy the following line and paste on a blank EES screen to verify solution): 2*x^3-10*x^0.5-3*x = -3 Answer: x = 2.063 (using an initial guess of x = 2)

1-47

1-84

Solve the following system of 2 equations with 2 unknowns using appropriate software:

3xy + y = 3.5 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^3–y^2=5.9 3*x*y+y=3.5 Answer x=1.836 y=0.5378

1-85 Solve the following system of 3 equations with 3 unknowns using appropriate software: 2x – y + z = 7 xy + 2z = 4 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): 2*x-y+z=7 3*x^2+2*y=z+3 x*y+2*z=4 Answer x=1.609, y= – 0.9872, z=2.794

1-86

Solve the following system of 3 equations with 3 unknowns using appropriate software:

x+y–z=2 Solution by EES Software (Copy the following lines and paste on a blank EES screen to verify solution): x^2*y-z=1 x-3*y^0.5+x*z= – 2 x+y-z=2 Answer x=1, y=1, z=0

Review Problems 1-87E The thrust developed by the jet engine of a Boeing 777 is given to be 85,000 pounds. This thrust is to be expressed in N and kgf. Analysis Noting that 1 lbf = 4.448 N and 1 kgf = 9.81 N, the thrust developed can be expressed in two other units as Thrust in N:

(

Thrust in kgf:

) (

)

EES (Engineering Equation Solver) SOLUTION "Given" Thrust=85000 [lbf] "Analysis" Thrust_N=Thrust*Convert(lbf, N) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-48

Thrust_kgf=Thrust_N*(1/9.807) "since 1 kgf = 9.807 N"

1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation ( Sea level: Denver: Mt. Ev.:

)

(z = 0 m): (z = 1610 m): (z = 8848 m):

EES (Engineering Equation Solver) SOLUTION "Given" a=9.807 [m/s^2] b=3.32E-6 [1/s^2] m=80 [kg] z1=0 [m] z2=1610 [m] z3=8848 [m] "Analysis" g1=a-b*z1 W1=m*g1 g2=a-b*z2 W2=m*g2 g3=a-b*z3 W3=m*g3

1-89E A man is considering buying a 12-oz steak for $5.50, or a 300-g steak for $5.20. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be 12 ounce steak: (

)(

)

300 gram steak: (

)(

)

1-49

Therefore, the steak at the traditional market is a better buy.

EES (Engineering Equation Solver) SOLUTION "Given" m1=12 [oz] price1=5.50 [$] m2=300 [g] price2=5.20 [$] "Analysis" UnitCost1=price1/m1*Convert(1/oz, 1/kg) UnitCost2=price2/m2*Convert(1/g, 1/kg)

1-90E The mass of a substance is given. Its weight is to be determined in various units. Analysis Applying Newton's second law, the weight is determined in various units to be (

)

(

)

( ( (

)(

)

) )

(

)

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] "Analysis" g=9.81 [m/s^2] W_1=m*g W_2=m*g*Convert(N, kN) W_3=m*g*Convert(kg-m/s^2, N) W_4=m*g/9.81 "1 kgf = 9.81 N" W_5=m*g*Convert(kg-m/s^2, lbm-ft/s^2) W_6=m*g*Convert(kg-m/s^2, lbf)

1-50

1-91E The pressure in a steam boiler is given in

. It is to be expressed in psi, kPa, atm, and bars.

Analysis We note that , 1 atm = 14.696 psi, 1 atm = 101.325 kPa, and 1 atm = 1.01325 bar (inner cover page of text). Then the desired conversions become: In atm:

(

)

In psi:

(

)(

In kPa:

(

)(

)

In bars:

(

)(

)

Discussion Note that the units atm,

)

, and bar are almost identical to each other.

EES (Engineering Equation Solver) SOLUTION "Given" P=92 "kgf/cm^2" "Analysis" P_atm=P*1/1.03323 "since 1 atm = 1.03323 kgf/cm^2" Ppsi=P*1/1.03323*Convert(atm, psi) P_kPa=P*1/1.03323*Convert(atm, kPa) Pbars=P*1/1.03323*Convert(atm, bar)

1-92 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. Assumptions 1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded.

Analysis Noting that pressure is force per unit area, the pressure on the smaller piston is determined from (

)

From Pascal‟s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from →

(

)→

Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal‟s principle.

1-51

m_1=25 [kg] D_1=0.10 [m] "ANALYSIS" g=9.81 [m/s^2] F_1=m_1*g A_1=pi*D_1^2/4 P_1=F_1/A_1 P_2=P_1 F_2=m_2*g A_2=F_2/P_2 A_2=pi*D_2^2/4

1-93 The average atmospheric pressure is given as where z is the altitude in km. The atmospheric pressures at various locations are to be determined. Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation Atlanta: Denver:

(z = 0.306 km): (z = 1.610 km):

M. City: Mt. Ev.:

(z = 2.309 km): (z = 8.848 km):

EES (Engineering Equation Solver) SOLUTION "Given" z_Atlanta=0.306 [km] z_Denver=1.610 [km] z_MexicoCity=2.309 [km] z_Everest=8.848 [km] "Analysis" P_atm_Atlanta=101.325*(1-0.02256*z_Atlanta)^5.256 P_atm_Denver=101.325*(1-0.02256*z_Denver)^5.256 P_atm_MexicoCity=101.325*(1-0.02256*z_MexicoCity)^5.256 P_atm_Everest=101.325*(1-0.02256*z_Everest)^5.256

1-94E Hyperthermia of 5C is considered fatal. This fatal level temperature change of body temperature is to be expressed in F, K, and R. Analysis The magnitudes of 1 K and 1C are identical, so are the magnitudes of 1 R and 1F. Also, a change of 1 K or 1C in temperature corresponds to a change of 1.8 R or 1.8F. Therefore, the fatal level of hypothermia is (a) 5 K (b) 5×1.8 = 9F (c) 5×1.8 = 9 R

EES (Engineering Equation Solver) SOLUTION "Given" DELTAT_C=5 [C] "Analysis" "(a)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-52

DELTAT_K=DELTAT_C*Convert(C, K) "(b)" DELTAT_F=DELTAT_C*Convert(C, F) "(c)" DELTAT_R=DELTAT_C*Convert(C, R)

1-95E The boiling temperature of water decreases by 3C for each 1000 m rise in altitude. This decrease in temperature is to be expressed in F, K, and R. Analysis The magnitudes of 1 K and 1C are identical, so are the magnitudes of 1 R and 1F. Also, a change of 1 K or 1C in temperature corresponds to a change of 1.8 R or 1.8F. Therefore, the decrease in the boiling temperature is (a) 3 K for each 1000 m rise in altitude, and (b) (c) 31.8 = 5.4F = 5.4 R for each 1000 m rise in altitude.

EES (Engineering Equation Solver) SOLUTION "Given" DELTAT_C=3 [C] "Analysis" "(a)" DELTAT_K=DELTAT_C*Convert(C, K) "(b)" DELTAT_F=DELTAT_C*Convert(C, F) "(c)" DELTAT_R=DELTAT_C*Convert(C, R)

1-96E A house is losing heat at a rate of per C temperature difference between the indoor and the outdoor temperatures. The rate of heat loss is to be expressed per F, K, and R of temperature difference between the indoor and the outdoor temperatures. Analysis The magnitudes of 1 K and 1C are identical, so are the magnitudes of 1 R and 1F. Also, a change of 1 K or 1C in temperature corresponds to a change of 1.8 R or 1.8F. Therefore, the rate of heat loss from the house is (a) (b) (c)

per K difference in temperature, and per R or F rise in temperature.

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_loss_C=1800 [kJ/h-C] "Analysis" "(a)" Q_dot_loss_K=Q_dot_loss_C*Convert(C, K) "(b)" Q_dot_loss_F=Q_dot_loss_C*Convert(1/C, 1/F) "(c)" Q_dot_loss_R=Q_dot_loss_C*Convert(1/C, 1/R)

1-53

1-97E The average body temperature of a person rises by about 2C during strenuous exercise. This increase in temperature is to be expressed in F, K, and R. Analysis The magnitudes of 1 K and 1C are identical, so are the magnitudes of 1 R and 1F. Also, a change of 1 K or 1C in temperature corresponds to a change of 1.8 R or 1.8F. Therefore, the rise in the body temperature during strenuous exercise is (a) 2 K (b) 2×1.8 = 3.6F (c) 2×1.8 = 3.6 R

EES (Engineering Equation Solver) SOLUTION "Given" DELTAT_C=2 [C] "Analysis" "(a)" DELTAT_K=DELTAT_C*Convert(C, K) "(b)" DELTAT_F=DELTAT_C*Convert(C, F) "(c)" DELTAT_R=DELTAT_C*Convert(C, R)

1-98 The average temperature of the atmosphere is expressed as where z is altitude in km. The temperature outside an airplane cruising at 12,000 m is to be determined. Analysis Using the relation given, the average temperature of the atmosphere at an altitude of 12,000 m is determined to be

Discussion This is the “average” temperature. The actual temperature at different times can be different.

EES (Engineering Equation Solver) SOLUTION "Given" z=12000*Convert(m, km) "Analysis" T_atm_K=288.15-6.5*z T_atm_C=T_atm_K-273.15

1-99 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 180 kPa. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder. Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield

(

)

1-54

It yields m = 20.4 kg

EES (Engineering Equation Solver) SOLUTION "Given" P=180 [kPa] P_atm=100 [kPa] A=25E-4 [m^2] "Analysis" g=9.81 [m/s^2] W=m*g W=(P*A-P_atm*A)*Convert(kPa, kg/m-s^2)

1-100 A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined. Assumptions Friction between the piston and the cylinder is negligible. Analysis The gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield (

)

The force balance when the weights are placed is used to determine the mass of the weights

(

)→

1-55

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] m_p=10 [kg] D=0.14 [m] P_2=2*P_1 "ANALYSIS" A_p=pi*D^2/4 P_1=P_atm+(m_p*g)/A_p*Convert(Pa, kPa) "vertical force balance without weights" P_2=P_atm+((m_p+m_weights)*g)/A_p*Convert(Pa, kPa) "vertical force balance with weights" g=9.81 [m/s^2]

1-101 The deflection of the spring of the two-piston cylinder with a spring shown in the figure is to be determined. Analysis Summing the forces acting on the piston in the vertical direction gives

which when solved for the deflection of the spring and substituting s

gives

] ]

We expressed the spring constant k in kN/m, the pressures in kPa (i.e., kN/m2) and the diameters in m units.

EES (Engineering Equation Solver) SOLUTION "Given" k=8 [kN/cm] P1=5000 [kPa] P2=10000 [kPa] P3=1000 [kPa] D1=0.08 [m] D2=0.03 [m] "Analysis" A1=pi*D1^2/4 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-56

A2=pi*D2^2/4 A3=A1-A2 F1=P1*A1 F2=P2*A2 F3=P3*A3 F_s+F2+F3=F1 F_s=k*x

1-102 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). 2 The weight of the duct and the air in is negligible. Properties The density of air is given to be

. We take the density of water to be

Analysis Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy force exerted by water. The volume of the underground section of the duct is ] Then the buoyancy force becomes (

)

Discussion The upward force exerted by water on the duct is 6.07 kN, which is equivalent to the weight of a mass of 619 kg. Therefore, this force must be treated seriously.

EES (Engineering Equation Solver) SOLUTION "Given" L=35 [m] D=0.15 [m] rho_air=1.3 [kg/m^3] rho_water=1000 [kg/m^3] "Analysis" g=9.81 [m/s^2] V=pi*D^2/4*L F_B=rho_water*g*V*Convert(kg/m-s^2, kPa)

1-103 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be Analysis The buoyancy force acting on the balloon is

. The density of helium gas is 1/7th of this.

1-57

(

)

The total mass is (

)

The total weight is (

)

Thus the net force acting on the balloon is Then the acceleration becomes (

)

EES (Engineering Equation Solver) SOLUTION "Given" D=12 [m] N_person=2 m_person=85 [kg] rho_air=1.16 [kg/m^3] rho_He=rho_air/7 "Analysis" g=9.81 [m/s^2] V_ballon=pi*D^3/6 F_B=rho_air*g*V_ballon m_He=rho_He*V_ballon m_people=N_person*m_person m_total=m_He+m_people W=m_total*g F_net=F_B-W a=F_net/m_total © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-58

1-104 Problem 1-103 is reconsidered. The effect of the number of people carried in the balloon on acceleration is to be investigated. Acceleration is to be plotted against the number of people, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Given" D=12 [m] N_person=2 m_person=85 [kg] rho_air=1.16 [kg/m^3] rho_He=rho_air/7 "Analysis" g=9.81 [m/s^2] V_ballon=pi*D^3/6 F_B=rho_air*g*V_ballon m_He=rho_He*V_ballon m_people=N_person*m_person m_total=m_He+m_people W=m_total*g F_net=F_B-W a=F_net/m_total

1 2 3 4 5 6 7 8 9 10

] 34 22.36 15.61 11.2 8.096 5.79 4.01 2.595 1.443 0.4865

1-105 A balloon is filled with helium gas. The maximum amount of load the balloon can carry is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-59

Properties The density of air is given to be helium gas is 1/7th of this. Analysis The buoyancy force acting on the balloon is

. The density of

(

)

The mass of helium is (

)

In the limiting case, the net force acting on the balloon will be zero. That is, the buoyancy force and the weight will balance each other:

Thus,

EES (Engineering Equation Solver) SOLUTION "Given" D=12 [m] N_person=2 m_person=85 [kg] rho_air=1.16 [kg/m^3] rho_He=rho_air/7 "Analysis" g=9.81 [m/s^2] m_total_a=F_B/g m_people_a=m_total_a-m_He V_ballon=pi*D^3/6 F_B=rho_air*g*V_ballon m_He=rho_He*V_ballon © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-60

m_people=N_person*m_person m_total=m_He+m_people W=m_total*g F_net=F_B-W a=F_net/m_total

1-106 A 6-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Properties The density of water is given to be

. The specific gravity of oil is given to be 0.85.

Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water, The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, ](

)

EES (Engineering Equation Solver) SOLUTION "Given" h_water=3 [m] h_oil=3 [m] rho_water=1000 [kg/m^3] rho_s_oil=0.85 "Analysis" g=9.807 [m/s^2] rho_oil=rho_s_oil*rho_water DELTAP_oil=rho_oil*g*h_oil*Convert(kg/m-s^2, kPa) DELTAP_water=rho_water*g*h_water*Convert(kg/m-s^2, kPa) DELTAP_total=DELTAP_oil+DELTAP_water

1-107 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-61

Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock ( ) yields

(

)

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=100 [kPa] A=4E-6 [m^2] P_atm=101 [kPa] "Analysis" g=9.807 [m/s^2] W=m*g W=P_gage*A*Convert(kPa, kg/m-s^2)

1-108 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be Analysis The local atmospheric pressure is determined from

and

(

)

The atmospheric pressure may be expressed in mmHg as (

)(

)

EES (Engineering Equation Solver) SOLUTION "GIVEN" h=6400 [m] P_plane=25 [kPa] rho=1.15 [kg/m^3] rho_Hg=13600 [kg/m^3] "ANALYSIS" g=9.81 [m/s^2] P_atm-P_plane=rho*g*h*Convert(Pa, kPa) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-62

h_Hg=P_atm/(rho_Hg*g)*Convert(kPa, Pa)*Convert(m, mm)

1-109 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is measured. It is to be determined how high the water will rise in the tube. Properties The density of water is given to be . Analysis The pressure at the bottom of the tube can be expressed as Solving for h,

(

)(

)

EES (Engineering Equation Solver) SOLUTION "Given" P=107 [kPa] P_atm=99 [kPa] rho=1000 [kg/m^3] g=9.81 [m/s^2] "Analysis" (P-P_atm)*Convert(kPa, kg/m-s^2)=rho*g*h

1-110E Equal volumes of water and oil are poured into a U-tube from different arms, and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same. The excess pressure applied on the oil side is to be determined. Assumptions 1 Both water and oil are incompressible substances. 2 Oil does not mix with water. 3 The cross-sectional area of the U-tube is constant. Properties The density of oil is given to be

. We take the density of water to be

1-63

Analysis Noting that the pressure of both the water and the oil is the same at the contact surface, the pressure at this surface can be expressed as Noting that

and rearranging,

(

)(

)

= 0.227 psi Discussion When the person stops blowing, the oil will rise and some water will flow into the right arm. It can be shown that when the curvature effects of the tube are disregarded, the differential height of water will be 23.7 in to balance 30-in of oil.

EES (Engineering Equation Solver) SOLUTION "Given" h_a=(30/12) [ft] h_w=(30/12) [ft] rho_a=49.3 [lbm/ft^3] "Analysis" P_gage_blow=(rho_w*g*h_w-rho_a*g*h_a)*Convert(lbm/ft-s^2, psia) rho_w=62.4 [lbm/ft^3] "taken" g=32.174 [ft/s^2]

1-111E A water pipe is connected to a double-U manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Assumptions 1 All the liquids are incompressible. 2 The solubility of the liquids in each other is negligible. Properties The specific gravities of mercury and oil are given to be 13.6 and 0.80, respectively. We take the density of water to be .

1-64

Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to gives Solving for Substituting, Pwater pipe = 14.2 psia + (62.4 lbm/ft 3 )(32.2 ft/s 2 )[(20/12 ft) - 0.8(60/12 ft) + 13.6(25/12 ft) 2 ö æ öæ 1 lbf ç 1 ft ÷ ÷ ÷ + 0.8(30/12 ft)]´ çç ÷ç ÷ ÷çèç144 in 2 ø çè32.2 lbm ×ft/s 2 ø ÷

Therefore, the absolute pressure in the water pipe is 26.4 psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly.

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=14.2 [psia] h_w=20/12 "[ft]" h_alc=60/12 "[ft]" h_Hg=25/12 "[ft]" h_oil=30/12 "[ft]" rho_s_alc=0.79 rho_s_Hg=13.6 rho_s_oil=0.8 "Properties" rho_w=62.4 [lbm/ft^3] "taken" rho_alc=rho_s_alc*rho_w rho_Hg=rho_s_Hg*rho_w rho_oil=rho_s_oil*rho_w g=32.174 [ft/s^2] "Analysis" P_atm=P_waterpipe+(-rho_w*g*h_w+rho_alc*g*h_alc-rho_Hg*g*h_Hg-rho_oil*g*h_oil)*Convert(lbm/ft-s^2, psia)

1-65

1-112 A gasoline line is connected to a pressure gage through a double-U manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. Properties The specific gravities of oil, mercury, and gasoline are given to be 0.79, 13.6, and 0.70, respectively. We take the density of water to be .

Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the gasoline pipe, and setting the result equal to gives Rearranging, Substituting, Pgasoline = 370 kPa - (1000 kg/m3 )(9.81 m/s 2 )[(0.45 m) - 0.79(0.5 m) + 13.6(0.1 m) + 0.70(0.22 m)] æ öæ 1 kPa ö 1 kN ÷ ÷ ÷çç ´ ççç ÷ 2÷ ÷ è1000 kg ×m/s ÷ øçè1 kN/m 2 ø

Therefore, the pressure in the gasoline pipe is 15.4 kPa lower than the pressure reading of the pressure gage. Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several fluids.

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=370 [kPa] h_w=0.45 [m] h_alc=0.5 [m] h_Hg=0.1 [m] h_gas=0.22 [m] rho_s_alc=0.79 rho_s_Hg=13.6 rho_s_gas=0.70 "Properties" rho_w=1000 [kg/m^3] "taken" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-66

rho_alc=rho_s_alc*rho_w rho_Hg=rho_s_Hg*rho_w rho_gas=rho_s_gas*rho_w g=9.807 [m/s^2]

1-113 A gasoline line is connected to a pressure gage through a double-U manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. Properties The specific gravities of oil, mercury, and gasoline are given to be 0.79, 13.6, and 0.70, respectively. We take the density of water to be .

Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the terms until we reach the gasoline pipe, and setting the result equal to gives Rearranging, Substituting, Pgasoline = 180 kPa - (1000 kg/m 3 )(9.807 m/s 2 )[(0.45 m) - 0.79(0.5 m) + 13.6(0.1 m) + 0.70(0.22 m)] æ öæ 1 kPa ö 1 kN ÷ ÷ ÷çç ´ çç ÷ ÷ ÷ çè1000 kg ×m/s 2 ÷ øèç1 kN/m 2 ø

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=180 [kPa] h_w=0.45 [m] h_alc=0.5 [m] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-67

h_Hg=0.1 [m] h_gas=0.22 [m] rho_s_alc=0.79 rho_s_Hg=13.6 rho_s_gas=0.70 "Properties" rho_w=1000 [kg/m^3] "taken" rho_alc=rho_s_alc*rho_w rho_Hg=rho_s_Hg*rho_w rho_gas=rho_s_gas*rho_w g=9.807 [m/s^2]

1-114 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. Properties The density of the liquid is given to be

Analysis The gage pressure in the duct is determined from (

)(

)

The length of the differential fluid column is Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability.

EES (Engineering Equation Solver) SOLUTION "Given" theta=45 [degrees] h=0.12 [m] P=115 [kPa] P_atm=92 [kPa] rho=0.81 [kg/L] "Analysis" g=9.81 [m/s^2] P_gage=rho*Convert(kg/L, kg/m^3)*g*h*Convert(kg/m-s^2, Pa) L=h/sin(theta)*Convert(m, cm)

1-68

1-115 The flow of air through a wind turbine is considered. Based on unit considerations, a proportionality relation is to be obtained for the mass flow rate of air through the blades. Assumptions Wind approaches the turbine blades with a uniform velocity. Analysis The mass flow rate depends on the air density, average wind velocity, and the cross-sectional area which depends on hose diameter. Also, the unit of mass flow rate ̇ is kg/s. Therefore, the independent quantities should be arranged such that we end up with the proper unit. Putting the given information into perspective, we have ̇

] is a function of

], D [m], and

It is obvious that the only way to end up with the unit “kg/s” for mass flow rate is to multiply the quantities  and V with the square of D. Therefore, the desired proportionality relation is ̇ or ̇ where the constant of proportionality is C = π/4 so that ̇ Discussion Note that the dimensionless constants of proportionality cannot be determined with this approach.

1-116 A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient, the air density, the car velocity, and the frontal area of the car. Analysis The drag force depends on a dimensionless drag coefficient, the air density, the car velocity, and the frontal area. Also, the unit of force F is newton N, which is equivalent to . Therefore, the independent quantities should be arranged such that we end up with the unit for the drag force. Putting the given information into perspective, we have [

]

], and

]

It is obvious that the only way to end up with the unit “ ” for drag force is to multiply mass with the square of the velocity and the fontal area, with the drag coefficient serving as the constant of proportionality. Therefore, the desired relation is Discussion Note that this approach is not sensitive to dimensionless quantities, and thus a strong reasoning is required.

1-117E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are and T(F) = 1.8T(C) + 32. The first thought that comes to mind is to replace T(F) in the equation by its equivalent 1.8T(C) + 32, and V in mph by , which is the “regular” way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.609 to convert it to the desired unit of mph. That is, ]

√

]

or ]

√ ]

where V convection relation: ]

√ ]

1-69

EES (Engineering Equation Solver) SOLUTION "Given" T_equiv_F=91.4-(91.4-T_ambient_F)*(0.475-0.0203*Vel_mph+0.304*sqrt(Vel_mph)) "where T_ambient in F and Vel in mph" "Analysis" 1.8*T_equiv_C+32=91.4-(91.4-(1.8*T_ambient_C+32))*(0.4750.0203*Vel_kmh/1.609+0.304*sqrt(Vel_kmh/1.609)) "where T_ambient in C and Vel in km/h. Also 1 mph = 1.609 km/h and T(F) = 1.8*T(C)+32" "Let us find T_equiv using these formulas for the following values" T_ambient_F=70 [F] Vel_mph=40 [mph] T_ambient_C=(T_ambient_F-32)/1.8 Vel_kmh=Vel_mph*Convert(mph, km/h)

1-118E Problem 1-117E is reconsidered. The equivalent wind-chill temperatures in °F as a function of wind velocity in the range of 4 mph to 40 mph for the ambient temperatures of 20, 40, and 60°F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T_ambient=60 [F] V=20 [mph] T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*V + 0.304*sqrt(V))

V [mph] 4 8 12 16 20 24 28 32 36 40

] 59.94 54.59 51.07 48.5 46.54 45.02 43.82 42.88 42.16 41.61

The table is for

1-70

Fundamentals of Engineering (FE) Exam Problems 1-119 During a heating process, the temperature of an object rises by 10C. This temperature rise is equivalent to a temperature rise of (a) 10F (b) 42F (c) 18 K (d) 18 R (e) 283 K Answer (d) 18 R Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_inC=10 [C] T_inR=T_inC*1.8 T_inR_alt=T_inC*Convert(C, R) "using EES unit conversion function" "Some Wrong Solutions with Common Mistakes:" W1_TinF=T_inC "F, setting C and F equal to each other" W2_TinF=T_inC*1.8+32 "F, converting to F " W3_TinK=1.8*T_inC "K, wrong conversion from C to K" W4_TinK=T_inC+273 "K, converting to K"

1-120 An apple loses 3.6 kJ of heat as it cools per C drop in its temperature. The amount of heat loss from the apple per F drop in its temperature is (a) 0.5 kJ (b) 1.8 kJ (c) 2.0 kJ (d) 3.6 kJ (e) 6.5 kJ Answer (c) 2.0 kJ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-71

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_perC=3.6 [kJ/C] Q_perF=Q_perC/1.8 Q_perF_alt=Q_perC*Convert(kJ/C, kJ/F) "using EES unit conversion factor" "Some Wrong Solutions with Common Mistakes:" W1_Q=Q_perC*1.8 "multiplying instead of dividing" W2_Q=Q_perC "setting them equal to each other"

1-121 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is (a) 1 lbf (b) 9.81 lbf (c) 32.2 lbf (d) 0.1 lbf (e) 0.031 lbf Answer (a) 1 lbf Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=1 [lbm] g=32.2 [ft/s^2] W=m*g*Convert(lbm-ft/s^2, lbf) "Some Wrong Solutions with Common Mistakes:" g_SI=9.81 [m/s^2] W1_W= m*g_SI "Using wrong conversion" W2_W= m*g "Using wrong conversion" W3_W= m/g_SI "Using wrong conversion" W4_W= m/g "Using wrong conversion"

1-122 Consider a fish swimming 5 m below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of 25 m below the free surface is (a) 196 Pa (b) 5400 Pa (c) 30,000 Pa (d) 196,000 Pa (e) 294,000 Pa Answer (d) 196,000 Pa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). z1=5 [m] z2=25 [m] rho=1000 [kg/m^3] g=9.81 [m/s^2] DELTAP=rho*g*(z2-z1) "Some Wrong Solutions with Common Mistakes:" W1_P=rho*g*(z2-z1)/1000 "dividing by 1000" W2_P=rho*g*(z1+z2) "adding depts instead of subtracting" W3_P=rho*(z1+z2) "not using g" W4_P=rho*g*(0+z2) "ignoring z1" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-72

1-123 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 , the height of the building is (a) 17 m (b) 20 m (c) 170 m (d) 204 m (e) 252 m Answer (d) 204 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=96.0 [kPa] P2=98.0 [kPa] rho=1.0 [kg/m^3] g=9.81 [m/s^2] DELTAP=P2-P1 DELTAP=rho*g*h*Convert(Pa,kPa) "Some Wrong Solutions with Common Mistakes:" DELTAP=rho*W1_h/1000 "not using g" DELTAP=g*W2_h/1000 "not using rho" P2=rho*g*W3_h/1000 "ignoring P1" P1=rho*g*W4_h/1000 "ignoring P2"

1-124 Consider a 2.5-m deep swimming pool. The pressure difference between the top and bottom of the pool is (a) 2.5 kPa (b) 12.0 kPa (c) 19.6 kPa (d) 24.5 kPa (e) 250 kPa Answer (d) 24.5 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). z2=2.5 [m] z1=0 [m] rho=1000 [kg/m^3] g=9.81 [m/s^2] DELTAP=rho*g*(z2-z1)*Convert(Pa, kPa) "Some Wrong Solutions with Common Mistakes:" W1_P=rho*(z1+z2)/1000 "not using g" W2_P=rho*g*(z2-z1)/2000 "taking half of z" W3_P=rho*g*(z2-z1) "not dividing by 1000"

1-125, 1-126 Design and Essay Problems

1-73

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 2 ENERGY, ENERGY TRANSFER, AND GENERAL ENERGY ANALYSIS

1-74

CYU - Check Your Understanding

CYU 2-1 ■ Check Your Understanding CYU 2-1.1 A refrigerator is placed in the middle of a room. The refrigerator consumes 0.5 kW of electricity while removing heat from the refrigerated space at a rate of 0.6 kW. What is the net effect of operating this refrigerator on the room? (a) 0.1 kW cooling (b) 0.6 kW cooling (c) 0.5 kW heating (d) 1.1 kW heating (e) 1.1 kW cooling Answer: (c) 0.5 kW heating CYU 2-1.2 A fan is operating in an isolated a room. The fan consumes 300 W of electricity and converts 75 percent of this electricity to the kinetic energy of air. What is the heating rate of the room by the fan? (a) 375 W (b) 300 W (c) 225 W (d) 75 W (e) 0 Answer: (b) 300 W

CYU 2-2 ■ Check Your Understanding CYU 2-2.1 Select the correct list of energy forms which constitute internal energy: (a) Potential and kinetic (b) Sensible, chemical, and kinetic (c) Sensible and latent (d) Sensible, chemical, and nuclear (e) Sensible, latent, chemical, and nuclear Answer: (e) Sensible, latent, chemical, and nuclear CYU 2-2.2 To what velocity do we need to accelerate a car at rest to increase its kinetic energy by 1kJ / kg ? (a) (b) (c) (d)

1m / s 1.4 m / s 10 m / s 44.7 m / s

1-75

(e) 90 m / s Answer: (d) 44.7 m / s ke  V 2 / 2  1kJ / kg  V 2 / 2  1000 m2 / s2  V 2 / 2  V  44.7 m / s since 1kJ / kg  1000 m2 / s 2

CYU 2-2.3 How many meters do we need to raise a mass of 1 kg to increase its potential energy by 1 kJ? (a) 1 m (b) 9.8 m (c) 49 m (d) 98 m (e) 102 m Answer: (e) 102 m PE  mgz  1kJ  (1kg)(9.8 m / s 2 )z  1000 kg m2 / s2  (1kg)(9.8 m / s 2 )z  z  102 m since 1kJ / kg  1000 m2 / s 2 CYU 2-2.4 What is the total energy of a 5-kg object with KE = 10 kJ, PE = 15 kJ, u  20 kJ / kg (a) 20 kJ (b) 25 kJ (c) 45 kJ (d) 125 kJ (e) 225 kJ Answer: (d) 125 kJ E  U  KE  PE  (5 kg)(20 kJ / kg)  10 kJ  15 kJ  125 kJ CYU 2-2.5 What is the total mechanical energy of a fluid flowing in a horizontal pipe at a velocity is 10 m / s ? The pressure of the fluid is 200 kPa and its specific volume is 0.001m3 /kg . (a) 0.050 kJ / kg (b) 0.2 kJ / kg (c) 0.25 kJ / kg (d) 50 kJ / kg (e) 50.2 kJ / kg Answer: (c) 0.25 kJ / kg

e  Pv  ke  Pv  V 2 / 2  (200kPa)  0.001m3 / kg   (1/1000)(10 m / s)2 / 2  0.2  0.05  0.25 kJ / kg since 1kJ / kg  1000 m2 / s 2

CYU 2-3 ■ Check Your Understanding CYU 2-3.1 If an energy transfer between a closed system and its surroundings is not heat, it must be (a) Work (b) Energy transfer by mass (c) Work or energy transfer by mass (d) Mechanical energy (e) Thermal energy © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-76

Answer: (a) Work CYU 2-3.2 Select the wrong statement regarding energy transfer mechanisms. (a) Heat always flows from high temperature to low temperature. (b) There cannot be any net heat transfer between two systems that are at the same temperature. (c) If a system undergoes an adiabatic process, the temperature of the system must remain constant. (d) Heat transfer is recognized only as it crosses the boundary of a system. Answer: (c) If a system undergoes an adiabatic process, the temperature of the system must remain constant. CYU 2-3.3 A 2-kg closed system receives 6 kJ heat from a source for a period of 10 min. The rate of heat transfer and the heat transfer per unit mass are (a) 3 W, 10 kJ / kg (b) 10 W, 3 kJ / kg (c) 3 kW, 10 kJ / kg (d) 0.6 kW, 3 kJ / kg (e) 0.6 W, 3 kJ / kg Answer: (b) 10 W, 3 kJ / kg

Q.  Q / dt  (6000 J) / (10  60 s)  10 J / s  10 W q  Q / m  (6 kJ) / (2 kg)  3 kJ / kg

CYU 2-4 ■ Check Your Understanding CYU 2-4.1 Select the wrong statement regarding heat and work interactions. (a) Both heat and work are boundary phenomena. (b) Heat and work are defined at a state of a system. (c) The magnitudes of heat and work depend on the path followed during a process. (d) The magnitudes of heat and work depend on the initial and final states of a system. (e) The temperature of a well-insulated system can be changed by energy transfer as work. Answer: (b) Heat or work is defined at a state. CYU 2-4.2 Work is done on a 0.5-kg closed system by a rotating shaft in the amount of 10 kJ during a period of 20 s. The work per unit mass and the power are (a) 10 kJ / kg , 5 kW (b) 5 kJ / kg , 2 kW (c) 20 kJ / kg , 0.5 kW (d) 2000 J / kg , 50 W (e) 2 kJ / kg , 5 kW Answer: (c) 20 kJ / kg , 0.5 kW

w  W / m  (10 kJ) / (0.5 kg)  20 kJ / kg W.  W / dt  (10 kJ) / (20 s)  0.5 kJ / s  0.5 kW

CYU 2-5 ■ Check Your Understanding CYU 2-5.1 Which of the following is not mechanical work? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-77

(a) Spring work (b) Shaft work (c) Work for stretching of a liquid film (d) Work to accelerate a body (e) Electrical work Answer: (e) Electrical work CYU 2-5.2 Which of the following are mechanical work? I Electrical work II Magnetic work III Spring work IV Shaft work (a) I and II (b) I and III (c) III and IV (d) I, II, and IV (e) I, III, and IV Answer: (c) III and IV CYU 2-5.3 The velocity of a 1000-kg car is increased from rest to 72 km / h in 10 s. What is the required power for acceleration? (a) 518 kW (b) 259 kW (c) 200 kW (d) 20 kW (e) 2 kW Answer: (d) 20 kW W  0.5 m  V22  V12   0.5(1000 kg) (72 / 3.6 m / s) 2  0   200, 000 kg m2 / s 2  200 kJ since 1kJ / kg  1000 m2 / s 2 and 1m / s  3.6 km / h

W.  W / dt  (200 kJ) / (10 s)  20 kW

CYU 2-6 ■ Check Your Understanding CYU 2-6.1 Energy can be transferred to or from a system by I Mass flow II II. Work III Heat transfer (a) I, II, and III (b) I and II (c) I and III (d) II and III (e) Only III Answer: (a) I, II, and III CYU 2-6.2 Heat is lost from a system in the amount of 3 kJ. If the energy of the system is decreased by 5 kJ, what is the work interaction? (a) Win  2 kJ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-78

(b) Wout  2 kJ (c) Win  8 kJ (d) Wout  8 kJ (e) Win  3 kJ Answer: (b) Wout  2 kJ Ein  Eout  dEsys

 Qin  Win    Qout  Wout   E2  E1 (0  0)   3 kJ  Wout   5kJ Wout  2 kJ CYU 2-6.3 30 kJ of energy enters a well-insulated piston-cylinder device by mass flow while piston rises and does 40 kJ of work. What is the change in the total energy of the system? (a) 40 J (b) 70 kJ (c) 70 kJ (d) 10 kJ (e) 10 kJ Answer: (d) 10 kJ Ein  Eout  dEsys

Q  W  E in

mass,in

  Q  W  E out

out

mass,out

 E E 2

30 kJ  40 kJ  10 kJ CYU 2-6.4 Heat is transferred to a closed system in the amount of 13 kJ while 8 kJ electrical work is done on the system. If there are no kinetic and potential energy changes, what is the internal energy change of the system? (a) 5 kJ (b) 5 kJ (c) 21 kJ (d) 21 kJ (e) 8 kJ Answer: (c) 21 kJ Ein  Eout  dEsys

 Qin  Win    Qout  Wout   dEsys  dU 13kJ  8kJ  21kJ CYU 2-6.5 Using the formal sign convention, the heat and work interactions are given as W = 60 J and Q = 35 J. What is the energy change of the system E? (a) 60 J (b) 25 J (c) 25 J (d) 95 J (e) 95 J Answer: (e) 95 J Q - W = dE 35 J - (-60 J) = 95 J

1-79

CYU 2-7 ■ Check Your Understanding CYU 2-7.1 A house is heated by a natural gas heater. During a 1-h period, 0.5 kg natural gas is consumed and 20,000 kJ heat is delivered to the house. The hating value of natural gas is 50,000 kJ / kg . What is the efficiency of the heater? (a) 0% (b) 20% (c) 50% (d) 80% (e) 100% Answer: (d) 80% Qin  mHV  (0.5 kg)(50,000 kJ / kg)  25,000 kJ eta  Qsupplied /Qin  (20, 000 kJ) / (25, 000 kJ)  0.8  80% CYU 2-7.2 The overall efficiency of a natural gas power plant is 45 percent. The thermal efficiency is 50 percent and the generator efficiency is 100 percent. What is the efficiency of the natural gas furnace? (a) 100% (b) 90% (c) 45.5% (d) 22.5% (e) 10% Answer: (b) 90% eta overall  eta th eta gen eta furnace

0.45  (0.50)(1) eta furnace eta furnace  0.9 CYU 2-7.3 The motor of a pump consumes 5 kW of electrical power. If the pump efficiency is 80 percent and the motor efficiency is 90 percent, what is the shaft power input to the pump? (a) 1.4 kW (b) 3.6 kW (c) 4 kW (d) 4.5 kW (e) 5 kW Answer: (d) 4.5 kW

Shaft power  ( Electric power )  eta motor   (5kW)(0.9)  4.5kW

CYU 2-7.4 A turbine-generator unit has a combined efficiency of 72 percent. The turbine efficiency is 80 percent and the mechanical energy decrease of the fluid across the turbine is 1 MW. The shaft power output from the turbine and the electrical power output from the generator, respectively, are (a) 0.8 MW, 0.72MW (b) 0.72 MW, 0.8 MW (c) 0.9 MW, 0.72 MW (d) 0.72 MW, 0.9 MW (e) 0.9 MW, 0.8 MW Answer: (a) 0.8 MW, 0.72 MW

Shaft power   dE mech  eta turb   (1MW)(0.8)  0.8MW

eta overall  eta turb eta gen  0.72  (0.80) eta gen  eta gen  0.9

1-80

CYU 2-8 ■ Check Your Understanding CYU 2-8.1 Which of the emission listed below is not an air pollutant? (a) Hydrocarbons (b) Nitrogen oxides (c) Carbon monoxide (d) Carbon dioxide (e) Sulfur dioxide Answer: (d) Carbon dioxide CYU 2-8.2 Ground-level ozone, which is the primary component of smog, forms when _______ and _______ react in the presence of sunlight on hot, calm days. (a) CO, NO x (b) CO, HC (c) HC, NO x (d) CO, SO2 (e) SO2 , NOx Answer: (c) HC, NO x CYU 2-8.3 The primary greenhouse gas is (a) Hydrocarbons (b) Nitrogen oxides (c) Carbon monoxide (d) Carbon dioxide (e) Water vapor Answer: (d) Carbon dioxide

1-81

PROBLEMS Forms of Energy 2-1C The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame. The microscopic forms of energy, on the other hand, are those related to the molecular structure of a system and the degree of the molecular activity, and are independent of outside reference frames.

2-2C The sum of all forms of the energy a system possesses is called total energy. In the absence of magnetic, electrical and surface tension effects, the total energy of a system consists of the kinetic, potential, and internal energies.

2-3C The internal energy of a system is made up of sensible, latent, chemical and nuclear energies. The sensible internal energy is due to translational, rotational, and vibrational effects.

2-4C Thermal energy is the sensible and latent forms of internal energy, and it is referred to as heat in daily life.

2-5C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller. It differs from thermal energy in that thermal energy cannot be converted to work directly and completely. The forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.

2-6C In electric heaters, electrical energy is converted to sensible internal energy.

2-7C Hydrogen is also a fuel, since it can be burned, but it is not an energy source since there are no hydrogen reserves in the world. Hydrogen can be obtained from water by using another energy source, such as solar or nuclear energy, and then the hydrogen obtained can be used as a fuel to power cars or generators. Therefore, it is more proper to view hydrogen is an energy carrier than an energy source.

2-8C Initially, the rock possesses potential energy relative to the bottom of the sea. As the rock falls, this potential energy is converted into kinetic energy. Part of this kinetic energy is converted to thermal energy as a result of frictional heating due to air resistance, which is transferred to the air and the rock. Same thing happens in water. Assuming the impact velocity of the rock at the sea bottom is negligible, the entire potential energy of the rock is converted to thermal energy in water and air.

1-82

2-9 A hydraulic turbine-generator is to generate electricity from the water of a large reservoir. The power generation potential is to be determined. Assumptions 1. The elevation of the reservoir remains constant. 2. The mechanical energy of water at the turbine exit is negligible.

Analysis The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. æ 1 kJ/kg ö ÷= 1.177 kJ/kg emech = pe = gz = (9.81 m/s 2 )(120 m) çç ÷ çè1000 m 2 /s 2 ÷ ø

Then the power generation potential becomes æ1 kW ö ÷ Wmax = Emech = memech = (1500 kg/s)(1.177 kJ/kg) çç ÷= 1766 kW ÷ çè1 kJ/s ø

Therefore, the reservoir has the potential to generate 1766 kW of power. Discussion This problem can also be solved by considering a point at the turbine inlet, and using flow energy instead of potential energy. It would give the same result since the flow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir.

EES (Engineering Equation Solver) SOLUTION "Given" h=120 [m] m_dot=1500 [kg/s] "Analysis" g=9.81 [m/s^2] e_mech=(g*h)*Convert(m^2/s^2, kJ/kg) W_dot_max=m_dot*e_mech

2-10E The specific kinetic energy of a mass whose velocity is given is to be determined. Analysis According to the definition of the specific kinetic energy,

ke =

ö V 2 (100 ft/s) 2 æ çç 1 Btu/lbm ÷ = ÷= 0.200 Btu / lbm 2 2÷ ç 2 2 è 25, 037 ft /s ø

1-83

V=100 [ft/s] "Analysis" ke=V^2/2*(1 [Btu/lbm])/(25037 [ft^2/s^2])

2-11 The specific kinetic energy of a mass whose velocity is given is to be determined. Analysis Substitution of the given data into the expression for the specific kinetic energy gives ke =

ö V 2 (30 m/s) 2 æ çç 1 kJ/kg ÷ = = 0.45 kJ / kg 2 2÷ ÷ ç è1000 m /s ø 2 2

EES (Engineering Equation Solver) SOLUTION "Given" V=30 [m/s] "Analysis" ke=V^2/2*(1 [kJ/kg])/(1000 [m^2/s^2])

2-12E The total potential energy of an object that is below a reference level is to be determined. Analysis Substituting the given data into the potential energy expression gives

æ 1 Btu/lbm ö÷ PE = mgz = (100 lbm)(31.7 ft/s 2 )(- 20 ft) çç ÷= - 2.53 Btu èç 25, 037 ft 2 /s 2 ø÷

EES (Engineering Equation Solver) SOLUTION "Given" z=20 [ft] g=31.7 [ft/s^2] m=100 [lbm] "Analysis" PE=m*g*z*(1 [Btu/lbm])/(25037 [ft^2/s^2]) 2-13 The specific potential energy of an object is to be determined. Analysis The specific potential energy is given by æ 1 kJ/kg ÷ ö pe = gz = (9.8 m/s 2 )(50 m) çç ÷= 0.49 kJ / kg çè1000 m 2 /s 2 ÷ ø

EES (Engineering Equation Solver) SOLUTION "Given" z=50 [m] g=9.8 [m/s^2] "Analysis" pe=g*z*(1 [kJ/kg])/(1000 [m^2/s^2])

2-14 The total potential energy of an object is to be determined. Analysis Substituting the given data into the potential energy expression gives æ 1 kJ/kg ö ÷ PE = mgz = (100 kg)(9.81 m/s 2 )(20 m) çç ÷ ÷= 19.6 kJ çè1000 m 2 /s 2 ø © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-84

EES (Engineering Equation Solver) SOLUTION "Given" m=100 [kg] z=20 [m] "Analysis" g=9.81 [m/s^2] PE=m*g*z*(1 [kJ/kg])/(1000 [m^2/s^2])

2-15 A water jet strikes the buckets located on the perimeter of a wheel at a specified velocity and flow rate. The power generation potential of this system is to be determined. Assumptions Water jet flows steadily at the specified speed and flow rate. Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses, and it can be converted to work entirely. Therefore, the power potential of the water jet is its kinetic energy, which is V 2/ 2 per unit mass, and mV 2/ 2 for a given mass flow rate:

emech = ke =

ö V 2 (60 m/s) 2 æ ÷ çç 1 kJ/kg ÷ = = 1.8 kJ/kg 2 2 ÷ çè1000 m /s ø 2 2

Wmax = Emech = memech

æ1 kW ÷ ö = (120 kg/s)(1.8 kJ/kg) çç = 216 kW ÷ çè1 kJ/s ÷ ø Therefore, 216 kW of power can be generated by this water jet at the stated conditions. Discussion An actual hydroelectric turbine (such as the Pelton wheel) can convert over 90% of this potential to actual electric power.

EES (Engineering Equation Solver) SOLUTION "Given" V=60 [m/s] m_dot=120 [kg/s] "Analysis" g=9.81 [m/s^2] e_mech=V^2/2*Convert(m^2/s^2, kJ/kg) W_dot_max=m_dot*e_mech

1-85

2-16 A river is flowing at a specified velocity, flow rate, and elevation. The total mechanical energy of the river water per unit mass, and the power generation potential of the entire river are to be determined. Assumptions 1. The elevation given is the elevation of the free surface of the river. 2. The velocity given is the average velocity. 3. The mechanical energy of water at the turbine exit is negligible. Properties We take the density of water to be   1000kg / m3 . Analysis Noting that the sum of the flow energy and the potential energy is constant for a given fluid body, we can take the elevation of the entire river water to be the elevation of the free surface, and ignore the flow energy. Then the total mechanical energy of the river water per unit mass becomes

emech = pe + ke = gh +

æ 1 kJ/kg ÷ ö V2 æ (3 m/s) 2 ö ÷ çç ÷ = ççç(9.81 m/s 2 )(90 m) + ÷= 0.887 kJ / kg ÷ 2 2 ÷ ç ÷ 2 2 è øè1000 m /s ø

The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass flow rate, m = r V = (1000 kg/m3 )(500 m3 /s) = 500,000 kg/s

Wmax = Emech = memech = (500,000 kg/s)(0.887 kJ/kg) = 444,000 kW = 444 MW Therefore, 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely. Discussion Note that the kinetic energy of water is negligible compared to the potential energy, and it can be ignored in the analysis. Also, the power output of an actual turbine will be less than 444 MW because of losses and inefficiencies.

EES (Engineering Equation Solver) SOLUTION "Given" V=3 [m/s] V_dot=500 [m^3/s] h=90 [m] "Analysis" g=9.81 [m/s^2] e_mech=(g*h+V^2/2)*Convert(m^2/s^2, kJ/kg) rho=1000 [kg/m^3] m_dot=rho*V_dot W_dot_max=m_dot*e_mech

2-17 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass and the power generation potential are to be determined. Assumptions The wind is blowing steadily at a constant uniform velocity. Properties The density of air is given to be   1.25 kg / m3

1-86

Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V 2/ 2 per unit mass, and mV 2/ 2 for a given mass flow rate:

emech = ke =

ö V 2 (10 m/s) 2 æ çç 1 kJ/kg ÷ = ÷ ÷= 0.050 kJ/kg çè1000 m 2 /s 2 ø 2 2

m = r VA = r V

p D2 p (60 m)2 = (1.25 kg/m3 )(10 m/s) = 35,340 kg/s 4 4

Wmax = Emech = memech = (35,340 kg/s)(0.050 kJ/kg) = 1770 kW Therefore, 1770 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

EES (Engineering Equation Solver) SOLUTION "Given" V=10 [m/s] D=60 [m] rho=1.25 [kg/m^3] "Analysis" g=9.81 [m/s^2] e_mech=V^2/2*Convert(m^2/s^2, kJ/kg) A=pi*D^2/4 m_dot=rho*V*A W_dot_max=m_dot*e_mech

Energy Transfer by Heat and Work 2-18C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric.

2-19C Point functions depend on the state only whereas the path functions depend on the path followed during a process. Properties of substances are point functions, heat and work are path functions.

2-20C Energy can cross the boundaries of a closed system in two forms: heat and work.

1-87

2-21C An adiabatic process is a process during which there is no heat transfer. A system that does not exchange any heat with its surroundings is an adiabatic system.

2-22C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat; all other forms are work.

2-23C (a) The car's radiator transfers heat from the hot engine cooling fluid to the cooler air. No work interaction occurs in the radiator. (b) The hot engine transfers heat to cooling fluid and ambient air while delivering work to the transmission. (c) The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced. No work is produced since there is no motion of the forces acting at the interface between the tire and road. (d) There is minor amount of heat transfer between the tires and road. Presuming that the tires are hotter than the road, the heat transfer is from the tires to the road. There is no work exchange associated with the road since it cannot move. (e) Heat is being added to the atmospheric air by the hotter components of the car. Work is being done on the air as it passes over and through the car.

2-24C It is a work interaction since the electrons are crossing the system boundary, thus doing electrical work.

2-25C It is a heat interaction since it is due to the temperature difference between the sun and the room.

2-26C Compressing a gas in a piston-cylinder device is a work interaction.

2-27 The power produced by an electrical motor is to be expressed in different units. Analysis Using appropriate conversion factors, we obtain

æ1 J/s öæ ö ÷ çç1 N ×m ÷ ÷ ÷= 5 N×m / s (a) W = (5 W) ççç ÷ è1 W øèç 1 J ø÷ 2ö æ æ1 J/s öæ 1 N ×m ö ÷ ÷ç1 kg ×m/s ÷ ÷ = 5 kg ×m2 / s3 ÷ç ÷ç ç ÷ ÷ ÷ç ç 1 J øè ÷ ç 1N 1 W øè ø

(b) W = (5 W) ççèç

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=5 [W] "Analysis" W_dot_1=W_dot*Convert(W, N-m/s) W_dot_2=W_dot*Convert(W, kg-m^2/s^3)

Mechanical Forms of Work 2-28C The work done (i.e., energy transferred to the car) is the same, but the power is different. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-88

2-29E A construction crane lifting a concrete beam is considered. The amount of work is to be determined considering (a) the beam and (b) the crane as the system. Analysis (a) The work is done on the beam and it is determined from

æ ö 1 lbf ÷ W = mg D z = (3´ 2000 lbm)(32.174 ft/s 2 ) çç (24 ft) ÷ çè32.174 lbm ×ft/s 2 ÷ ø

= 144,000 lbf ×ft æ ö÷ 1 Btu = (144, 000 lbf ×ft) çç ÷= 185 Btu çè778.169 lbf ×ft ø÷

(b) Since the crane must produce the same amount of work as is required to lift the beam, the work done by the crane is

W = 144,000 lbf ×ft = 185 Btu

EES (Engineering Equation Solver) SOLUTION "Given" m=2*3000 [lbm] "1 ton = 2000 lbm" z=36 [ft] "Analysis" g=32.2 [ft/s^2] W_1=m*g*z*Convert(lbm-ft/s^2, lbf) W_2=W_1*Convert(lbf-ft, Btu)

2-30E The engine of a car develops 225 hp at 3000 rpm. The torque transmitted through the shaft is to be determined. Analysis The torque is determined from æ550 lbf ×ft/s ÷ ö W 225 hp çç ÷ T = sh = = 394 lbf ×ft ÷ ç ÷ 2p n 2p (3000/60)/s è 1 hp ø

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_sh=225 [hp] n_dot=3000 [1/min] "Analysis" T=W_dot_sh/(2*pi*n_dot)*Convert(min, s)*Convert(hp, lbf-ft/s)

2-31E The work required to compress a spring is to be determined. Analysis The force at any point during the deflection of the spring is given by F  F0  kx , where F0 is the initial force and x is the deflection as measured from the point where the initial force occurred. From the perspective of the spring, this force acts in the direction opposite to that in which the spring is deflected. Then, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-89 2

W = ò Fds = ò ( F0 + kx)dx 1

= F0 ( x2 - x1 ) +

k 2 ( x2 - x12 ) 2

= (100 lbf) [(1- 0)in ]+

200 lbf/in 2 (1 - 02 )in 2 2

= 200 lbf ×in æ öæ ö 1 Btu ÷ ç 1 ft ÷= 0.0214 Btu = (200 lbf ×in) çç ÷ ÷çç12 in ø÷ ÷ èç778.169 lbf ×ft øè

EES (Engineering Equation Solver) SOLUTION "Given" k=200 [lbf/in] F_0=100 [lbf] x_2=1 [in] "Analysis" x_1=0 [in] W=(F_0*(x_2-x_1)+k/2*(x_2^2-x_1^2))*Convert(lbf-in, Btu)

2-32 The work required to compress a spring is to be determined. Analysis Since there is no preload, F = kx. Substituting this into the work expression gives 2

W = ò Fds = ò kxdx = k ò xdx = 1

k 2 2 ( x2 - x1 ) 2

300 kN/m é 2 2ù êë(0.03 m) - 0 úû 2

= 0.135 kN ×m æ 1 kJ ö ÷= 0.135 kJ = (0.135 kN ×m) çç ÷ çè1 kN ×m ÷ ø

1-90

EES (Engineering Equation Solver) SOLUTION "Given" k=3 [kN/cm] F_0=0 [N] x_2=3 [cm] "Analysis" x_1=0 [cm] W=k/2*(x_2^2-x_1^2)*Convert(kN-cm, kJ)

2-33 A ski lift is operating steadily at 10 km / h . The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined. Assumptions 1. Air drag and friction are negligible. 2. The average mass of each loaded chair is 250 kg. 3. The mass of chairs is small relative to the mass of people, and thus the contribution of returning empty chairs to the motion is disregarded (this provides a safety factor). Analysis The lift is 1000 m long and the chairs are spaced 20 m apart. Thus at any given time there are 1000 / 20  50 chairs being lifted. Considering that the mass of each chair is 250 kg, the load of the lift at any given time is

Load  (50 chairs)(250 kg / chair)  12,500 kg Neglecting the work done on the system by the returning empty chairs, the work needed to raise this mass by 200 m is

æ ö 1 kJ ÷ ÷ Wg = mg (z2 - z1 ) = (12,500 kg)(9.81 m/s2 )(200 m) çç ÷= 24,525 kJ ÷ çè1000 kg ×m2 /s 2 ø At 10 km / h , it will take Dt =

distance 1 km = = 0.1 h = 360 s velocity 10 km/h

to do this work. Thus the power needed is Wg =

Wg Dt

24,525 kJ = 68.1 kW 360 s

The velocity of the lift during steady operation, and the acceleration during start up are æ 1 m/s ö ÷ V = (10 km/h) çç ÷ ÷= 2.778 m/s çè3.6 km/h ø

D V 2.778 m/s - 0 = = 0.556 m/s 2 Dt 5s

During acceleration, the power needed is Wa =

æ 1 kJ/kg ÷ ö 1 1 m(V22 - V12 ) / D t = (12,500 kg) ((2.778 m/s) 2 - 0)çç /(5 s) = 9.6 kW 2 2÷ ÷ ç è1000 m /s ø 2 2

Assuming the power applied is constant, the acceleration will also be constant and the vertical distance traveled during acceleration will be

1 2 1 200 m 1 at sin a = at 2 = (0.556 m/s 2 )(5 s)2 (0.2) = 1.39 m 2 2 1000 m 2

and

æ 1 kJ/kg ö ÷ ÷/(5 s) = 34.1 kW Wg = mg (z2 - z1 ) / D t = (12,500 kg)(9.81 m/s 2 )(1.39 m) çç 2 2÷ çè1000 kg ×m /s ÷ ø Thus, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-91

Wtotal = Wa + Wg = 9.6 + 34.1 = 43.7 kW

EES (Engineering Equation Solver) SOLUTION "Given" distance=1000 [m] DELTAz=200 [m] space=20 [m] N_people=3 Vel=10[km/h]*Convert(km/h, m/s) m_chair=250 [kg] t=5 [s] "Properties" g=9.807 [m/s^2] "Analysis" "Power required to operate" N_chair=distance/space m=N_chair*m_chair W_g=m*g*DELTAz*Convert(kg-m^2/s^2, kJ) DELTAt=distance/Vel W_dot_g=W_g/DELTAt "Power required to accelerate" a=Vel/t W_a=1/2*m*Vel^2*Convert(kg-m^2/s^2, kJ) W_dot_a=W_a/t sin(alpha)=DELTAz/distance h=1/2*a*t^2*sin(alpha) W_g_b=m*g*h*Convert(kg-m^2/s^2, kJ) W_dot_g_b=W_g_b/t W_dot_total=W_dot_a+W_dot_g_b

2-34 The engine of a car develops 75 kW of power. The acceleration time of this car from rest to 100 km / h on a level road is to be determined. Analysis The work needed to accelerate a body is the change in its kinetic energy,

Wa =

ææ100,000 m ö2 ö÷æ ö 1 1 1 kJ ÷ çç ÷ ÷ m (V22 - V12 ) = (1500 kg) ççççç - 0÷ = 578.7 kJ ÷ ÷ ÷ 2 2 ÷ ç ÷è1000 kg ×m /s ÷ çè 2 2 ø èç 3600 s ø ø÷

Thus the time required is W 578.7 kJ Dt = a = = 7.72 s 75 kJ/s Wa This answer is not realistic because part of the power will be used against the air drag, friction, and rolling resistance.

EES (Engineering Equation Solver) SOLUTION "Given" m=1500 [kg] W_dot_a=75 [kW] Vel1=0 Vel2=100[km/h]*Convert(km/h, m/s) "Analysis" W_a=1/2*m*(Vel2^2-Vel1^2)*Convert(kg-m^2/s^2, kJ) DELTAt=W_a/W_dot_a

1-92

2-35 A damaged car is being towed by a truck. The extra power needed is to be determined for three different cases. Assumptions Air drag, friction, and rolling resistance are negligible. Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies. That is, Wtotal = Wa + Wg

(a) Zero. (b) Wa = 0 . Thus, Wtotal = Wg = mg ( z2 - z1 ) / D t = mg

Dz = mgVz = mgV sin 30 Dt

æ50,000 m ö÷æ 1 kJ/kg ö÷ çç ÷(0.5) ÷ = (1200 kg)(9.8 1m/s 2 ) ççç ÷ ÷ç1000 m 2 /s 2 ÷ ÷ è 3600 s øè ø = 81.7 kW (c) Wg = 0 . Thus, ææ90,000 m ö öæ ö ÷ 1 1 ç ÷ ÷ç 1 kJ/kg ÷ ÷ ÷ m(V22 - V12 ) / D t = (1200 kg) çççç - 0÷ç /(12 s) = 31.3 kW ÷ ÷ 2 2 ÷ ççèç 3600 s ø ÷ ÷ ÷ 2 2 è øèç1000 m /s ø 2

Wtotal = Wa =

EES (Engineering Equation Solver) SOLUTION "Given" m=1200 [kg] Vel_b=50[km/h]*Convert(km/h, m/s) alpha=30 [degrees] Vel1_c=0 [m/s] Vel2_c=90[km/h]*Convert(km/h, m/s) t=12 [s] "Properties" g=9.807 [m/s^2] "Analysis" "(a)" W_dot_g_a=0 W_dot_a_a=0 W_dot_total_a=W_dot_a_a+W_dot_g_a "(b)" W_dot_a_b=0 W_dot_g_b=m*g*Vel_b*sin(alpha)*Convert(kg-m^2/s^2, kJ) W_dot_total_b=W_dot_a_b+W_dot_g_b "(c)" W_dot_g_c=0 W_a_c=1/2*m*(Vel2_c^2-Vel1_c^2)*Convert(kg-m^2/s^2, kJ) W_dot_a_c=W_a_c/t W_dot_total_c=W_dot_a_c+W_dot_g_c

2-36 As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter increases. The amount of work produced by this bubble is to be determined. Assumptions 1. The bubble is treated as a spherical bubble. 2. The surface tension coefficient is taken constant. Analysis Executing the work integral for a constant surface tension coefficient gives

1-93 2

W = s ò dA = s ( A2 - A1 ) = s 4p (r22 - r12 ) 1

= 4p (0.02 N/m) éëê(0.015 m) 2 - (0.005 m) 2 ù ú û

= 5.03´ 10- 5 N ×m æ 1 kJ ÷ ö = (5.03´ 10- 5 N ×m) çç ÷ ÷ çè1000 N ×m ø

= 5.03×10- 8 kJ

EES (Engineering Equation Solver) SOLUTION "Given" D1=0.01 [m] D2=0.03 [m] sigma=0.02 [N-m] "Analysis" r1=D1/2 r2=D2/2 W=sigma*4*pi*(r2^2-r1^2)*(1 [kJ])/(1000 [N-m])

2-37 The work required to stretch a steel rod in a specified length is to be determined. Assumptions The Young‟s modulus does not change as the rod is stretched. Analysis The original volume of the rod is

V0 =

p D2 p (0.005 m)2 L= (10 m) = 1.963´ 10- 4 m3 4 4

The work required to stretch the rod 3 cm is

W= =

V0 E 2 (e2 - e12 ) 2 2 ù ö (1.963´ 10- 4 m3 )(21´ 104 kN/m2 ) éêæ çç0.03 m ÷ - 02 ú ÷ ê ú ÷ ç è ø 2 ú ëê 10 m û

= 1.855´ 10- 4 kN ×m = 1.855´ 10- 4 kJ = 0.1855 J

EES (Engineering Equation Solver) SOLUTION "Given" D=0.005 [m] L=10 [m] dL=0.03 [m] E=21E4 [kN/m^2] "Analysis" V=pi*D^2/4*L epsilon_1=0 epsilon_2=dL/L W=(V*E)/2*(epsilon_2^2-epsilon_1^2)

1-94

The First Law of Thermodynamics 2-38C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass transport.

2-39C No. This is the case for adiabatic systems only.

2-40C Warmer. Because energy is added to the room air in the form of electrical work.

2-41 Water is heated in a pan on top of a range while being stirred. The energy of the water at the end of the process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the pan as our system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives Ein - Eout = D Esystem Net energy transfer by heat, work, and mass

Change in internal, kinetic, potential, etc. energies

Qin + Wsh,in - Qout = D U = U 2 - U1

30 kJ + 0.5 kJ - 5 kJ = U 2 - 12.5 kJ U 2 = 38.0 kJ Therefore, the final internal energy of the system is 38.0 kJ.

EES (Engineering Equation Solver) SOLUTION "Given" Q_in=30 [kJ] Q_out=5 [kJ] W_pw_in=500 [N-m]*Convert(N-m, kJ) E_1=12.5 [kJ] "Analysis" E_2-E_1=Q_in+W_pw_in-Q_out

2-42 The specific energy change of a system which is accelerated is to be determined. Analysis Since the only property that changes for this system is the velocity, only the kinetic energy will change. The change in the specific energy is D ke =

ö V22 - V12 (30 m/s) 2 - (0 m/s) 2 æ çç 1 kJ/kg ÷ = = 0.45 kJ / kg 2 2÷ ÷ ç è1000 m /s ø 2 2

EES (Engineering Equation Solver) SOLUTION "Given" Vel_1=0 [m/s] Vel_2=30 [m/s] "Analysis" DELTAke=(Vel_2^2-Vel_1^2)/2*Convert(m^2/s^2, kJ/kg)

1-95

2-43 A fan is to accelerate quiescent air to a specified velocity at a specified flow rate. The minimum power that must be supplied to the fan is to be determined. Assumptions The fan operates steadily. Properties The density of air is given to be   1.18 kg / m3 . Analysis A fan transmits the mechanical energy of the shaft (shaft power) to mechanical energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance can be written as

Ein - Eout

= dEsystem / dt

Rate of net energy transfer by heat, work, and mass

0 (steady)

= 0  Ein = Eout

Rate of change in internal, kinetic, potential, etc. energies

Wsh, in = mair keout = mair

Vout2 2

where mair = r V = (1.18 kg/m3 )(9 m3 /s) = 10.62 kg/s

Substituting, the minimum power input required is determined to be Wsh, in = mair

2 ö Vout (8 m/s) 2 æ çç 1 J/kg ÷ = (10.62 kg/s) ÷= 340 J/s = 340 W çè1 m 2 /s 2 ÷ ø 2 2

Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another, and it does not allow any energy to be created or destroyed during a process. In reality, the power required will be considerably higher because of the losses associated with the conversion of mechanical shaft energy to kinetic energy of air.

EES (Engineering Equation Solver) SOLUTION "Given" V=8 [m/s] V_dot=9 [m^3/s] rho=1.18 [kg/m^3] "Analysis" m_dot=rho*V_dot W_dot=m_dot*V^2/2

2-44E Water is heated in a cylinder on top of a range. The change in the energy of the water during this process is to be determined. Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible. Analysis We take the water in the cylinder as the system. This is a closed system since no mass enters or leaves. Applying the energy balance on this system gives Ein - Eout = D Esystem Net energy transfer by heat, work, and mass

Change in internal, kinetic, potential, etc. energies

Qin - Wout - Qout = D U = U 2 - U1

65 Btu - 5 Btu - 8 Btu = D U D U = U 2 - U1 = 52 Btu Therefore, the energy content of the system increases by 52 Btu during this process. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-96

EES (Engineering Equation Solver) SOLUTION "Given" Q_in=65 [Btu] Q_out=8 [Btu] W_b_out=5 [Btu] "Analysis" Q_in-Q_out-W_b_out=DELTAU

2-45E The heat loss from a house is to be made up by heat gain from people, lights, appliances, and resistance heaters. For a specified rate of heat loss, the required rated power of resistance heaters is to be determined. Assumptions 1 The house is well-sealed, so no air enters or heaves the house. 2 All the lights and appliances are kept on. 3 The house temperature remains constant.

Analysis Taking the house as the system, the energy balance can be written as

Ein - Eout Rate of net energy transfer by heat, work, and mass

= dEsystem / dt ¿ 0 (steady) = 0  Ein = Eout Rate of change in internal, kinetic, potential, etc. energies

where Eout = Qout = 60, 000 Btu/h

and Ein = Epeople + Elights + Eappliance + Eheater = 6000 Btu/h + Eheater

Substituting, the required power rating of the heaters becomes æ 1 kW ö ÷ Eheater = 60, 000 - 6000 = 54, 000 Btu/h çç ÷ ÷= 15.8 kW çè3412 Btu/h ø Discussion When the energy gain of the house equals the energy loss, the temperature of the house remains constant. But when the energy supplied drops below the heat loss, the house temperature starts dropping.

EES (Engineering Equation Solver) SOLUTION "Given" E_dot_out=60000 [Btu/h] E_dot_gen=6000 [Btu/h] "Analysis" E_dot_heater=(E_dot_out-E_dot_gen)*Convert(Btu/h, kW)

2-46E A water pump increases water pressure. The power input is to be determined.

1-97

Analysis The power input is determined from W = V ( P2 - P1 )

æ 1 Btu öæ 1 hp ö ÷ ÷ ç ÷ = (0.8 ft 3 /s)(70 - 15)psia çç ÷ççè0.7068 Btu/s ÷ ÷ çè5.404 psia ×ft 3 ÷ ø ø

= 11.5 hp The water temperature at the inlet does not have any significant effect on the required power.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=15 [psia] P_2=70 [psia] V_dot=0.8 [ft^3/s] "Analysis" W_dot=V_dot*(P_2-P_1)*Convert(psia-ft^3, Btu)*Convert(Btu/s, hp)

2-47 The lighting energy consumption of a storage room is to be reduced by installing motion sensors. The amount of energy and money that will be saved as well as the simple payback period are to be determined. Assumptions The electrical energy consumed by the ballasts is negligible. Analysis The plant operates 12 hours a day, and thus currently the lights are on for the entire 12 hour period. The motion sensors installed will keep the lights on for 3 hours, and off for the remaining 9 hours every day. This corresponds to a total of 9365 = 3285 off hours per year. Disregarding the ballast factor, the annual energy and cost savings become

Energy Savings  (Number of lamps )( Lamp wattage ) (Reduction of annual operating hours)  (24 lamps)(60 W / lamp)(3285 hours / year)  4730kWh / year Cost Savings  (Energy Savings) (Unit cost of energy)

 (4730kWh / year)($0.11/ kWh)  $520 / year The implementation cost of this measure is the sum of the purchase price of the sensor plus the labor, Implementation Cost = Material + Labor = $32 + $40 = $72 This gives a simple payback period of Simple payback period =

Implementation cost $72 = = 0.138 year (1.66 months) Annual cost savings $520 / year

Therefore, the motion sensor will pay for itself in less than 2 months.

EES (Engineering Equation Solver) SOLUTION "Given" N_lamp=6*4 E_dot_lamp=60 [W] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-98

CurrentHours=12*365 "[h/year]" NewHours=3*365 "[h/year]" Cost_electricity=0.11 [$/kWh] MaterialCost=32 [$] LaborCost=40 [$] "Analysis" EnergySavings=N_lamp*E_dot_lamp*(CurrentHours-NewHours)*Convert(W, kW) CostSavings=EnergySavings*Cost_electricity ImplementationCost=MaterialCost+LaborCost PaybackPeriod=ImplementationCost/CostSavings

2-48 The classrooms and faculty offices of a university campus are not occupied an average of 4 hours a day, but the lights are kept on. The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined. Analysis The total electric power consumed by the lights in the classrooms and faculty offices is Elighting, classroom = (Power consumed per lamp) ´ (No. of lamps) = (200´ 12´ 110 W) = 264,000 = 264 kW Elighting, offices = (Power consumed per lamp)´ (No. of lamps) = (400´ 6´ 110 W) = 264,000 = 264 kW Elighting, total = Elighting, classroom + Elighting, offices = 264 + 264 = 528 kW

Noting that the campus is open 240 days a year, the total number of unoccupied work hours per year is

Unoccupied hours  (4 hours / day)(240 days / year)  960 h / yr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are Energy savings = ( Elighting, total )(Unoccupied hours) = (528 kW)(960 h/yr) = 506,880 kWh

Cost savings = (Energy savings)(Unit cost of energy) = (506,880 kWh/yr)($0.11/kWh) = $55,757 / yr Discussion Note that simple conservation measures can result in significant energy and cost savings.

EES (Engineering Equation Solver) SOLUTION "Given" N_classroom=200 N_office=400 N_tube_c=12 "tubes/classroom" N_tube_o=6 "tubes/office" E_dot_tube=110 [W] Days=240 [1/year] Hours=4 [h] Cost_electricity=0.11 [$/kWh] "Analysis" E_dot_classrooms=N_classroom*N_tube_c*E_dot_tube*Convert(W, kW) E_dot_offices=N_office*N_tube_o*E_dot_tube*Convert(W, kW) E_dot_total=E_dot_classrooms+E_dot_offices UnoccupiedHours=Hours*Days EnergySavings=E_dot_total*UnoccupiedHours CostSavings=EnergySavings*Cost_electricity

2-49 A room contains a light bulb, a TV set, a refrigerator, and an iron. The rate of increase of the energy content of the room when all of these electric devices are on is to be determined. Assumptions 1. The room is well sealed, and heat loss from the room is negligible. 2. All the appliances are kept on. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-99

Analysis Taking the room as the system, the rate form of the energy balance can be written as

Ein - Eout Rate of net energy transfer by heat, work, and mass

dEsystem / dt

 dEroom / dt = Ein

Rate of change in internal, kinetic, potential, etc. energies

since no energy is leaving the room in any form, and thus Eout = 0. Also, Ein = Elights + ETV + Erefrig + Eiron

= 40 + 110 + 300 + 1200 W = 1650 W

Substituting, the rate of increase in the energy content of the room becomes

dEroom / dt = Ein = 1650 W Discussion Note that some appliances such as refrigerators and irons operate intermittently, switching on and off as controlled by a thermostat. Therefore, the rate of energy transfer to the room, in general, will be less.

EES (Engineering Equation Solver) SOLUTION "Given" T_i=20 [C] E_dot_lights=40 [W] E_dot_TV=110 [W] E_dot_ref=300 [W] E_dot_iron=1200 [W] "Analysis" E_dot_total=E_dot_lights+E_dot_TV+E_dot_ref+E_dot_iron

2-50 An inclined escalator is to move a certain number of people upstairs at a constant velocity. The minimum power required to drive this escalator is to be determined. Assumptions 1. Air drag and friction are negligible. 2. The average mass of each person is 75 kg. 3. The escalator operates steadily, with no acceleration or breaking. 4. The mass of escalator itself is negligible. Analysis At design conditions, the total mass moved by the escalator at any given time is

Mass  (50 persons)(75 kg / person)  3750 kg The vertical component of escalator velocity is

Vvert = V sin 45° = (0.6 m/s)sin45° Under stated assumptions, the power supplied is used to increase the potential energy of people. Taking the people on elevator as the closed system, the energy balance in the rate form can be written as

1-100

Ein - Eout Rate of net energy transfer by heat, work, and mass

Win =

dEsystem / dt

= 0  Ein = dEsys / dt @

D Esys Dt

Rate of change in internal, kinetic, potential, etc. energies

D PE mgD z = = mgVvert Dt Dt

That is, under stated assumptions, the power input to the escalator must be equal to the rate of increase of the potential energy of people. Substituting, the required power input becomes æ 1 kJ/kg ÷ ö Win = mgVvert = (3750 kg)(9.81 m/s 2 )(0.6 m/s)sin45° çç = 12.5 kJ/s = 15.6 kW ÷ çè1000 m 2 /s 2 ÷ ø When the escalator velocity is doubled to V  1.2 m / s , the power needed to drive the escalator becomes æ 1 kJ/kg ÷ ö Win = mgVvert = (3750 kg)(9.81 m/s 2 )(1.2 m/s)sin45° çç = 25.0 kJ/s = 31.2 kW ÷ çè1000 m 2 /s 2 ÷ ø

Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity.

EES (Engineering Equation Solver) SOLUTION "Given" N_people=50 m_person=75 [kg] V=0.6 [m/s] "or 1.2 m/s" theta=45 [degrees] "Analysis" m_people=N_people*m_person V_vert=V*sin(theta) W_dot_in=m_people*g*V_vert g=9.81 [m/s^2]

2-51 A car cruising at a constant speed to accelerate to a specified speed within a specified time. The additional power needed to achieve this acceleration is to be determined. Assumptions 1. The additional air drag, friction, and rolling resistance are not considered. 2. The road is a level road. Analysis We consider the entire car as the system, except that let‟s assume the power is supplied to the engine externally for simplicity (rather that internally by the combustion of a fuel and the associated energy conversion processes). The energy balance for the entire mass of the car can be written in the rate form as D Esys Ein - Eout = dEsystem / dt = 0  Ein = dEsys / dt @ Dt Rate of net energy transfer by heat, work, and mass

Win =

Rate of change in internal, kinetic, potential, etc. energies

D KE m(V22 - V12 ) / 2 = Dt Dt

since we are considering the change in the energy content of the car due to a change in its kinetic energy (acceleration). Substituting, the required additional power input to achieve the indicated acceleration becomes

ö V 2 - V12 (110/3.6 m/s)2 - (70/3.6 m/s)2 æ çç 1 kJ/kg ÷ Win = m 2 = (2100 kg) 2 2÷ ÷= 117 kJ/s = 117 kW ç è1000 m /s ø 2D t 2(5 s) since 1m / s  3.6 km / h . If the total mass of the car were 700 kg only, the power needed would be

ö V 2 - V12 (110/3.6 m/s)2 - (70/3.6 m/s)2 æ çç 1 kJ/kg ÷ Win = m 2 = (700 kg) ÷= 38.9 kW çè1000 m2 /s 2 ÷ ø 2D t 2(5 s) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-101

Discussion Note that the power needed to accelerate a car is inversely proportional to the acceleration time. Therefore, the short acceleration times are indicative of powerful engines.

EES (Engineering Equation Solver) SOLUTION "Given" m=2100 [kg] V_1=70 [km/h]*Convert(km/h, m/s) V_2=110 [km/h]*Convert(km/h, m/s) time=5 [s] "Analysis" W_dot_in=m*(V_2^2-V_1^2)/(2*time)*Convert(m^2/s^2, kJ/kg)

2-52E The high rolling resistance tires of a car are replaced by low rolling resistance ones. For a specified unit fuel cost, the money saved by switching to low resistance tires is to be determined. Assumptions 1. The low rolling resistance tires deliver 2 mpg over all velocities. 2. The car is driven 15,000 miles per year. Analysis The annual amount of fuel consumed by this car on high- and low-rolling resistance tires are Annual Fuel Consumption High =

Miles driven per year 15,000 miles/year = = 428.6 gal/year Miles per gallon 35 miles/gal

Annual Fuel Consumption Low =

Miles driven per year 15,000 miles/year = = 405.4 gal/year Miles per gallon 37 miles/gal

Then the fuel and money saved per year become

Fuel Savings = Annual Fuel Consumption High - Annual Fuel Consumption Low = 428.6 gal/year - 405.4 gal/year = 23.2 gal/year Cost savings = (Fuel savings)(Unit cost of fuel) = (23.2 gal/year)($3.5/gal) = $81.1/ year Discussion A typical tire lasts about 3 years, and thus the low rolling resistance tires have the potential to save about $150 to the car owner over the life of the tires, which is comparable to the installation cost of the tires.

EES (Engineering Equation Solver) SOLUTION "Given" MPG_1=35 [mpg] MPG_2=37 [mpg] Miles=15000 [miles] UnitCost=3.5 [$/gal] "Analysis" AFC_high=Miles/MPG_1 AFC_low=Miles/MPG_2 FuelSavings=AFC_high-AFC_low CostSavings=FuelSavings*UnitCost

Energy Conversion Efficiencies 2-53C Mechanical efficiency is defined as the ratio of the mechanical energy output to the mechanical energy input. A mechanical efficiency of 100% for a hydraulic turbine means that the entire mechanical energy of the fluid is converted to mechanical (shaft) work.

1-102

2-54C The combined pump-motor efficiency of a pump/motor system is defined as the ratio of the increase in the mechanical energy of the fluid to the electrical power consumption of the motor, hpump-motor = hpump hmotor =

Emech,out - Emech,in Welect,in

D Emech,fluid Welect,in

Wpump Welect,in

The combined pump-motor efficiency cannot be greater than either of the pump or motor efficiency since both pump and motor efficiencies are less than 1, and the product of two numbers that are less than one is less than either of the numbers.

2-55C The turbine efficiency, generator efficiency, and combined turbine-generator efficiency are defined as follows: W shaft, out Mechanical energy output  turbine   Mechanical energy extracted from the fluid | E | mech, fluid

Electrical power output W elect, out  generator   Mechanical power input W shaft, in

 turbine -gen   turbine generator  

Welect, out  E

Emech, in

mech, out



Welect, out | E mech, fluid |

2-56C No, the combined pump-motor efficiency cannot be greater that either of the pump efficiency of the motor efficiency. This is because hpump-motor = hpump hmotor , and both hpump and hmotor are less than one, and a number gets smaller when multiplied by a number smaller than one.

2-57 A hooded electric open burner and a gas burner are considered. The amount of the electrical energy used directly for cooking and the cost of energy per “utilized” kWh are to be determined. Analysis The electric heater is rated at 2.4 kW. Therefore, the rate of energy consumption by the electric heater is Qinput, electric = 2.4 kW

The efficiency of the electric heater is given to be 73 percent. Therefore, a burner that consumes 2.4-kW of electrical energy will supply Qutilized = (Energy input) ´ (Efficiency) = (2.4 kW)(0.73) = 1.75 kW

of useful energy. The unit cost of utilized energy is inversely proportional to the efficiency, and is determined from Cost of utilized energy =

Cost of energy input $0.10 / kWh = = $0.137 / kWh Efficiency 0.73

Noting that the efficiency of a gas burner is 38 percent, the energy input to a gas burner that supplies utilized energy at the same rate (1.75 kW) is

Qinput, gas =

Qutilized 1.75 kW = = 4.61kW (= 15,700 Btu/h) Efficiency 0.38

since 1kW  3412Btu / h . Therefore, a gas burner should have a rating of at least 15, 700 Btu / h to perform as well as the electric unit. Noting that 1 therm = 29.3 kWh, the unit cost of utilized energy in the case of gas burner is determined the same way to be Cost of utilized energy =

Cost of energy input $1.20 / (29.3 kWh) = = $0.108 / kWh Efficiency 0.38

1-103

UnitCost_electricity=0.10 [$/kWh] UnitCost_gas=1.20 [$/therm] eta_electricity=0.73 eta_gas=0.38 "Analysis" Q_dot_utilized_electricity=E_dot*eta_electricity Cost_electricity=UnitCost_electricity/eta_electricity Q_dot_input_gas=Q_dot_utilized_electricity/eta_gas Cost_gas=UnitCost_gas/eta_gas*Convert(kWh, therm)

2-58E The combustion efficiency of a furnace is raised from 0.7 to 0.8 by tuning it up. The annual energy and cost savings as a result of tuning up the boiler are to be determined. Assumptions The boiler operates at full load while operating. Analysis The heat output of boiler is related to the fuel energy input to the boiler by

Boiler output = (Boiler input)(Combustion efficiency) or

Qout = Qin hfurnace

6 The current rate of heat input to the boiler is given to be Qin, current = 5.5´ 10 Btu/h.

Then the rate of useful heat output of the boiler becomes

Qout = (Qin hfurnace )current = (5.5´ 106 Btu/h)(0.7) = 3.85´ 106 Btu/h The boiler must supply useful heat at the same rate after the tune up. Therefore, the rate of heat input to the boiler after the tune up and the rate of energy savings become

Qin, new = Qout / hfurnace, new = (3.85´ 106 Btu/h)/0.8 = 4.81´ 106 Btu/h Qin, saved = Qin, current - Qin, new = 5.5´ 106 - 4.81´ 106 = 0.69´ 106 Btu/h Then the annual energy and cost savings associated with tuning up the boiler become Energy Savings  Qin saved (Operation hours)  (0.69 106 Btu / h)(4200 h / year )  2.89  109 Btu / yr

Cost Savings  (Energy Savings)(Unit cost of energy)  (2.89 109 Btu / yr)($13 /106 Btu)  $37, 500 / year

Discussion Notice that tuning up the boiler will save $37,500 a year, which is a significant amount. The implementation cost of this measure is negligible if the adjustment can be made by in-house personnel. Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year.

1-104

Q_dot_in_current=5.5E6 [Btu/h] eta_furnace_current=0.7 eta_furnace_new=0.8 Hours=4200 [h/year] UnitCost=13E-6 [$/Btu] "Analysis" Q_dot_out=Q_dot_in_current*eta_furnace_current Q_dot_in_new=Q_dot_out/eta_furnace_new Q_dot_in_saved=Q_dot_in_current-Q_dot_in_new Energysavings=Q_dot_in_saved*Hours CostSavings=EnergySavings*UnitCost 2-59E Problem 2-58E is reconsidered. The effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings as the efficiency varies from 0.7 to 0.9 and the unit cost varies from $12 to $14 per million Btu are the investigated. The annual energy saved and the cost savings are to be plotted against the efficiency for unit costs of $12, $13, and $14 per million Btu. Analysis The problem is solved using EES, and the solution is given below. "Given" Q_dot_in_current=5.5E6 [Btu/h] eta_furnace_current=0.7 eta_furnace_new=0.8 Hours=4200 [h/year] UnitCost=13E-6 [$/Btu] "Analysis" Q_dot_out=Q_dot_in_current*eta_furnace_current Q_dot_in_new=Q_dot_out/eta_furnace_new Q_dot_in_saved=Q_dot_in_current-Q_dot_in_new Energysavings=Q_dot_in_saved*Hours CostSavings=EnergySavings*UnitCost ηfurnace,new

EnergySavings [Btu/year]

CostSavings [$/year]

0.7 0.72 0.74 0.76 0.78 0.8 0.82 0.84 0.86 0.88 0.9

0.00E+00 6.42E+08 1.25E+09 1.82E+09 2.37E+09 2.89E+09 3.38E+09 3.85E+09 4.30E+09 4.73E+09 5.13E+09

0 8342 16232 23708 30800 37538 43946 50050 55870 61425 66733

Table values are for UnitCost=13E-5 [$/Btu]

1-105

CostSavings ($/year)

70000 60000

14x10-6 $/Btu

50000

13x10-6 $/Btu

40000

12x10-6 $/Btu

30000 20000 10000 0 0.7

0.74

0.78

0.82

0.86

0.9

0.86

0.9

h furnace,new

Energysavings (Btu/year)

6x109 5x109 4x109 3x109 2x109 1x109 0 0.7

0.74

0.78

0.82

h furnace,new

2-60 A worn out standard motor is replaced by a high efficiency one. The reduction in the internal heat gain due to the higher efficiency under full load conditions is to be determined. Assumptions 1. The motor and the equipment driven by the motor are in the same room. 2. The motor operates at full load so that f load  1 . Analysis The heat generated by a motor is due to its inefficiency, and the difference between the heat generated by two motors that deliver the same shaft power is simply the difference between the electric power drawn by the motors, Win, electric, standard = Wshaft / hmotor = (75´ 746 W)/0.91=61,484 W Win, electric, efficient = Wshaft / hmotor = (75´ 746 W)/0.954=58,648 W

Then the reduction in heat generation becomes

Qreduction = Win, electric, standard - Win, electric, efficient = 61,484 - 58,648 = 2836 W

1-106

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_shaft=75 [hp] eta_motor_standard=0.91 eta_motor_efficient=0.954 "Analysis" W_dot_in_standard=W_dot_shaft/eta_motor_standard*Convert(hp, W) W_dot_in_efficient=W_dot_shaft/eta_motor_efficient*Convert(hp, W) Q_dot_reduction=W_dot_in_standard-W_dot_in_efficient

2-61 An electric car is powered by an electric motor mounted in the engine compartment. The rate of heat supply by the motor to the engine compartment at full load conditions is to be determined. Assumptions The motor operates at full load so that the load factor is 1. Analysis The heat generated by a motor is due to its inefficiency, and is equal to the difference between the electrical energy it consumes and the shaft power it delivers, Win, electric = Wshaft / hmotor = (90 hp)/0.91 = 98.90 hp

Qgeneration = Win, electric - Wshaft out = 98.90 - 90 = 8.90 hp = 6.64 kW

since 1 hp = 0.746 kW. Discussion Note that the electrical energy not converted to mechanical power is converted to heat.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_shaft_out=90 [hp] eta_motor=0.91 "Analysis" W_dot_in_electric=W_dot_shaft_out/eta_motor Q_dot_generation=(W_dot_in_electric-W_dot_shaft_out)*Convert(hp, W)

2-62 Several people are working out in an exercise room. The rate of heat gain from people and the equipment is to be determined. Assumptions The average rate of heat dissipated by people in an exercise room is 600 W. Analysis The 6 weight lifting machines do not have any motors, and thus they do not contribute to the internal heat gain directly. The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods. Noting that 1 hp = 745.7 W, the total heat generated by the motors is

Qmotors = (No. of motors)´ Wmotor ´ fload ´ f usage / hmotor = 7´ (2.5´ 746 W)´ 0.70´ 1.0/0.77 = 11,870 W © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-107

The heat gain from 14 people is

Qpeople = 14´ (600 W) = 8400 W Then the total rate of heat gain of the exercise room during peak period becomes

Qtotal = Qmotors + Qpeople = 11,870 + 8400 = 20, 270 W

EES (Engineering Equation Solver) SOLUTION "Given" N_motor=7 W_dot_motor=2.5 [hp] f_load=0.7 eta_motor=0.77 N_people=14 Q_dot_person=600 [W] "Analysis" Q_dot_motors=N_motor*W_dot_motor*f_load/eta_motor*Convert(hp, W) Q_dot_people=N_people*Q_dot_person Q_dot_total=Q_dot_motors+Q_dot_people

2-63 A room is cooled by circulating chilled water through a heat exchanger, and the air is circulated through the heat exchanger by a fan. The contribution of the fan-motor assembly to the cooling load of the room is to be determined. Assumptions The fan motor operates at full load so that f load  1 . Analysis The entire electrical energy consumed by the motor, including the shaft power delivered to the fan, is eventually dissipated as heat. Therefore, the contribution of the fan-motor assembly to the cooling load of the room is equal to the electrical energy it consumes,

Qinternal generation = Win, electric = Wshaft / hmotor = (0.25 hp)/0.60 = 0.463 hp = 311 W

since 1 hp = 746 W.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_shaft=0.25 [hp] eta_motor=0.60 "Analysis" Q_dot_motor=W_dot_shaft/eta_motor*Convert(hp, W)

1-108

2-64 A hydraulic turbine-generator is to generate electricity from the water of a lake. The overall efficiency, the turbine efficiency, and the shaft power are to be determined. Assumptions 1. The elevation of the lake and that of the discharge site remains constant. 2. Irreversible losses in the pipes are negligible. Properties The density of water can be taken to be   1000 kg / m3 . The gravitational acceleration is g  9.81 m / s 2 Analysis (a) We take the bottom of the lake as the reference level for convenience. Then kinetic and potential energies of water are zero, and the mechanical energy of water consists of pressure energy only which is

emech,in - emech,out =

P = gh r

æ 1 kJ/kg ö÷ = (9.81 m/s 2 )(50 m) çç ÷ çè1000 m 2 /s 2 ø÷ = 0.491 kJ/kg Then the rate at which mechanical energy of fluid supplied to the turbine and the overall efficiency become

D Emech,fluid = m(emech,in - emech,in ) = (5000 kg/s)(0.491 kJ/kg) = 2455 kW hoverall = hturbine-gen =

Welect,out | D Emech,fluid |

1862 kW = 0.760 2455 kW

(b) Knowing the overall and generator efficiencies, the mechanical efficiency of the turbine is determined from hturbine-gen = hturbine hgenerator ® hturbine =

hturbine-gen hgenerator

0.76 = 0.800 0.95

Wshaft,out = hturbine D Emech,fluid = (0.800)(2455 kW) = 1964 kW » 1960 kW Therefore, the lake supplies 2455 kW of mechanical energy to the turbine, which converts 1964 kW of it to shaft work that drives the generator, which generates 1862 kW of electric power.

1-109

h=50 [m] m_dot=5000 [kg/s] W_dot_elect=1862 [kW] eta_gen=0.95 "Analysis" "(a)" g=9.81 [m/s^2] DELTAe_mech=g*h*Convert(m^2/s^2, kJ/kg) DELTAE_dot_mech_fluid=m_dot*DELTAe_mech eta_turbine_gen=W_dot_elect/DELTAE_dot_mech_fluid "(b)" eta_turbine=eta_turbine_gen/eta_gen "(c)" W_dot_shaft=eta_turbine*DELTAE_dot_mech_fluid

2-65 A pump with a specified shaft power and efficiency is used to raise water to a higher elevation. The maximum flow rate of water is to be determined. Assumptions 1. The flow is steady and incompressible. 2. The elevation difference between the reservoirs is constant. 3. We assume the flow in the pipes to be frictionless since the maximum flow rate is to be determined,

Properties We take the density of water to be ρ  1000 kg / m3 . Analysis The useful pumping power (the part converted to mechanical energy of water) is

Wpump,u = hpumpWpump, shaft = (0.82)(7 hp) = 5.74 hp The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. That is,

1-110

Noting that D Emech = Wpump, u , the volume flow rate of water is determined to be V=

Wpump,u r gz2

æ745.7 W ÷ öæ1 N ×m/s öæ1 kg ×m/s 2 ö 5.74 hp ÷= 0.0291 m3 / s çç ÷ çç ÷ ÷ ÷çç ÷ 2 ÷ øçè 1 N ÷ (1000 kg/m )(9.81 m/s )(15 m) èç 1 hp ÷ øèç 1 W ÷ç ø 3

Discussion This is the maximum flow rate since the frictional effects are ignored. In an actual system, the flow rate of water will be less because of friction in pipes.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_shaft=7 [hp] z=15 [m] eta_pump=0.82 "Analysis" rho=1000 [kg/m^3] g=9.81 [m/s^2] W_dot_pump_u=eta_pump*W_dot_shaft*Convert(hp, W) V_dot=W_dot_pump_u/(rho*g*z)

2-66 Geothermal water is raised from a given depth by a pump at a specified rate. For a given pump efficiency, the required power input to the pump is to be determined.

Assumptions 1. The pump operates steadily. 2. Frictional losses in the pipes are negligible. 3. The changes in kinetic energy are negligible. 4. The geothermal water is exposed to the atmosphere and thus its free surface is at atmospheric pressure. Properties The density of geothermal water is given to be   1050 kg / m3 . Analysis The elevation of geothermal water and thus its potential energy changes, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of geothermal water is equal to the change in its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. That is, D Emech = mD emech = mD pe = mg D z = r V g D z

1-111

æ 1N ÷ öæ 1 kW ö ÷ ÷çç = (1050 kg/m3 )(0.3 m3 /s)(9.81 m/s 2 )(200 m) ççç ÷ 2÷ ÷ è1 kg ×m/s ÷ øèç1000 N ×m/s ø

= 618.0 kW Then the required power input to the pump becomes Wpump, elect =

D Emech 618 kW = = 835 kW hpump-motor 0.74

Discussion The frictional losses in piping systems are usually significant, and thus a larger pump will be needed to overcome these frictional losses.

EES (Engineering Equation Solver) SOLUTION "Given" rho=1050 [kg/m^3] V_dot=0.3 [m^3/s] z=200 [m] eta=0.74 "Analysis" g=9.81 [m/s^2] m_dot=rho*V_dot DELTAE_dot_mech_fluid=m_dot*g*z*Convert(m^2/s^2, kJ/kg) W_dot_pump_elect=DELTAE_dot_mech_fluid/eta

2-67 Wind is blowing steadily at a certain velocity. The mechanical energy of air per unit mass, the power generation potential, and the actual electric power generation are to be determined.

Assumptions 1. The wind is blowing steadily at a constant uniform velocity. 2. The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be   1.25 kg / m3 . Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. 2 Therefore, the power potential of the wind is its kinetic energy, which is V / 2 per unit mass, and mV 2 / 2 for a given mass flow rate: emech = ke =

ö V 2 (7 m/s) 2 æ çç 1 kJ/kg ÷ = = 0.0245 kJ/kg 2 2÷ ÷ ç è1000 m /s ø 2 2

1-112

m = r VA = r V

p D2 p (80 m)2 = (1.25 kg/m3 )(7 m/s) = 43,982 kg/s 4 4

Wmax = Emech = memech = (43,982 kg/s)(0.0245 kJ/kg) = 1078 kW The actual electric power generation is determined by multiplying the power generation potential by the efficiency,

Welect = hwind turbineWmax = (0.30)(1078 kW) = 323 kW Therefore, 323 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions.

EES (Engineering Equation Solver) SOLUTION "Given" V=7 [m/s] D=80 [m] eta_overall=0.30 rho=1.25 [kg/m^3] "Analysis" g=9.81 [m/s^2] A=pi*D^2/4 m_dot=rho*A*V W_dot_max=m_dot*V^2/2*Convert(m^2/s^2, kJ/kg) W_dot_elect=eta_overall*W_dot_max

2-68 Problem 2-67 is reconsidered. The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 m / s to 20 m / s in increments of 5 m / s , and the diameter varies from 20 m to 120 m in increments of 20 m is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Given" V=7 [m/s] D=80 [m] eta_overall=0.30 rho=1.25 [kg/m^3] "Analysis" g=9.81 [m/s^2] A=pi*D^2/4 m_dot=rho*A*V W_dot_max=m_dot*V^2/2*Convert(m^2/s^2, kJ/kg) W_dot_elect=eta_overall*W_dot_max D (m)

V (m/s)

m (kg/s)

Welect (kW)

1-113

20 20 20 20 40 40 40 40 60 60 60 60 80 80 80 80 100 100 100 100 120 120 120 120

5 10 15 20 5 10 15 20 5 10 15 20 5 10 15 20 5 10 15 20 5 10 15 20

1963 3927 5890 7854 7854 15708 23562 31416 17671 35343 53014 70686 31416 62832 94248 125664 49087 98175 147262 196350 70686 141372 212058 282743

7.363 58.9 198.8 471.2 29.45 235.6 795.2 1885 66.27 530.1 1789 4241 117.8 942.5 3181 7540 184.1 1473 4970 11781 265.1 2121 7157 16965

1-114

2-69 Water is pumped from a lower reservoir to a higher reservoir at a specified rate. For a specified shaft power input, the power that is converted to thermal energy is to be determined.

Assumptions 1. The pump operates steadily. 2. The elevations of the reservoirs remain constant. 3. The changes in kinetic energy are negligible. Properties We take the density of water to be   1000 kg / m3 Analysis The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. That is, D Emech = mD emech = mD pe = mg D z = r V g D z

æ 1N ö æ 1 kW ÷ ö ÷ ç ÷ = (1000 kg/m3 )(0.03 m3 /s)(9.81 m/s 2 )(45 m) çç ÷ ÷ççè1000 N ×m/s ø ÷= 13.2 kW ÷ çè1 kg ×m/s 2 ø Then the mechanical power lost because of frictional effects becomes Wfrict = Wpump, in - D Emech = 20 - 13.2 kW = 6.8 kW

Discussion The 6.8 kW of power is used to overcome the friction in the piping system. The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy, which results in a slight rise in fluid temperature. Note that this pumping process could be accomplished by a 13.2 kW pump (rather than 20 kW) if there were no frictional losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 13.2 kW of power from the water.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_shaft=20 [kW] z=45 [m] V_dot=0.03 [m^3/s] "Analysis" rho=1000 [kg/m^3] g=9.81 [m/s^2] m_dot=rho*V_dot DELTAE_dot_mech_fluid=m_dot*g*z*Convert(m^2/s^2, kJ/kg) W_dot_fric=W_dot_shaft-DELTAE_dot_mech_fluid

1-115

2-70E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The mechanical power used to overcome frictional effects is to be determined. Assumptions 1. The flow is steady and incompressible. 2. The elevation difference between the lake and the free surface of the pool is constant. 3. The average flow velocity is constant since pipe diameter is constant. Properties We take the density of water to be   62.4lbm / ft 3 .

Analysis The useful mechanical pumping power delivered to water is Wpump,u = hpumpWpump = (0.80)(20 hp) = 16 hp

The elevation of water and thus its potential energy changes during pumping, but it experiences no changes in its velocity and pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. That is, D Emech = mD emech = mD pe = mg D z = r V g D z

Substituting, the rate of change of mechanical energy of water becomes æ öæ 1 hp ö 1 lbf ÷ ÷ D Emech = (62.4 lbm/ft 3 )(1.5 ft 3 /s)(32.2 ft/s 2 )(80 ft) çç ÷ççç ÷= 13.63 hp ÷ èç32.2 lbm ×ft/s 2 ÷ øè550 lbf ×ft/s ø Then the mechanical power lost in piping because of frictional effects becomes Wfrict = Wpump, u - D Emech = 16 - 13.63 hp = 2.37 hp

Discussion Note that the pump must supply to the water an additional useful mechanical power of 2.37 hp to overcome the frictional losses in pipes.

EES (Engineering Equation Solver) SOLUTION "Given" eta_pump=0.80 W_dot_pump=20 [hp] V_dot=1.5 [ft^3/s] z=80 [ft] "Analysis" rho=62.4 [lbm/ft^3] g=32.2 [ft/s^2] W_dot_pump_u=eta_pump*W_dot_pump m_dot=rho*V_dot DELTAE_dot_mech_fluid=m_dot*g*z*Convert(lbm-ft/s^2, lbf)*Convert(lbf-ft/s, hp) W_dot_fric=W_dot_pump_u-DELTAE_dot_mech_fluid

1-116

2-71 Water is pumped from a lake to a storage tank at a specified rate. The overall efficiency of the pump-motor unit and the pressure difference between the inlet and the exit of the pump are to be determined. Assumptions 1. The elevations of the tank and the lake remain constant. 2. Frictional losses in the pipes are negligible. 3. The changes in kinetic energy are negligible. 4. The elevation difference across the pump is negligible. Properties We take the density of water to be   1000 kg / m3 .

Analysis (a) We take the free surface of the lake to be point 1 and the free surfaces of the storage tank to be point 2. We also take the lake surface as the reference level  z1  0  , and thus the potential energy at points 1 and 2 are pe1  0 and pe2  gz2 . The flow energy at both points is zero since both 1 and 2 are open to the atmosphere  P1  P2  Patm  . Further, the kinetic energy at both points is zero  ke1  ke 2  0  since the water at both locations is essentially stationary. The mass flow rate of water and its potential energy at point 2 are m = r V = (1000 kg/m3 )(0.070 m 3 /s) = 70 kg/s æ 1 kJ/kg ÷ ö pe2 = gz2 = (9.81 m/s 2 )(15 m) çç = 0.1472 kJ/kg ÷ çè1000 m 2 /s 2 ÷ ø

Then the rate of increase of the mechanical energy of water becomes D Emech,fluid = m(emech,out - emech,in ) = m( pe2 - 0) = mpe2 = (70 kg/s)(0.1472 kJ/kg) = 10.3 kW

The overall efficiency of the combined pump-motor unit is determined from its definition, hpump-motor =

D Emech,fluid Welect,in

10.3 kW = 0.669 15.4 kW

or 66.9%

(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pump consists of the change in the flow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energy supplied by the pump, which is 10.3 kW: D Emech,fluid = m(emech,out - emech,in ) = m

P2 - P1 = VDP r

Solving for P and substituting, DP =

D Emech,fluid V

ö 10.3 kJ/s æ 1 kPa ×m3 ÷ ÷ = 147 kPa ççç ÷ 3 0.070 m /s è 1 kJ ÷ ø

Therefore, the pump must boost the pressure of water by 147 kPa in order to raise its elevation by 15 m. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-117

Discussion Note that only two-thirds of the electric energy consumed by the pump-motor is converted to the mechanical energy of water; the remaining one-third is wasted because of the inefficiencies of the pump and the motor.

EES (Engineering Equation Solver) SOLUTION "Given" z=15 [m] V_dot=0.070 [m^3/s] W_dot_elect_in=15.4 [kW] "Analysis" rho=1000 [kg/m^3] g=9.81 [m/s^2] m_dot=rho*V_dot DELTAE_dot_mech_fluid=m_dot*g*z*Convert(m^2/s^2, kJ/kg) eta_pump_motor=DELTAE_dot_mech_fluid/W_dot_elect_in DELTAP=DELTAE_dot_mech_fluid/V_dot

2-72 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity. The electric power generation, the daily electricity production, and the monetary value of this electricity are to be determined. Assumptions 1. The wind is blowing steadily at a constant uniform velocity. 2. The efficiency of the wind turbine is independent of the wind speed. Properties The density of air is given to be   1.25 kg / m3 .

ö V 2 (8 m/s) 2 æ çç 1 kJ/kg ÷ = ÷ ÷= 0.032 kJ/kg çè1000 m 2 /s 2 ø 2 2

m = r VA = r V

p D2 p (100 m)2 = (1.25 kg/m3 )(8 m/s) = 78,540 kg/s 4 4

Wmax = Emech = memech = (78,540 kg/s)(0.032 kJ/kg) = 2513 kW The actual electric power generation is determined from

Welect = hwind turbineWmax = (0.32)(2513 kW) = 804.2 kW Then the amount of electricity generated per day and its monetary value become Amount of electricity = (Wind power)(Operating hours) = (804.2 kW)(24 h) = 19,300 kWh

Revenues  (Amount of electricity)(Unit price)  (19,300kWh)($0.09 / kWh)  $1737 (per day) Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a reasonable cost, which explains the overwhelming popularity of wind turbines in recent years.

1-118

EES (Engineering Equation Solver) SOLUTION "Given" D=100 [m] V=8 [m/s] eta=0.32 rho=1.25 [kg/m^3] UnitPrice=0.09 [$/kWh] time=24 [h] "Analysis" A=pi*D^2/4 m_dot=rho*A*V W_dot_max=1/2*m_dot*V^2*Convert(m^2/s^2, kJ/kg) W_dot_elect=eta*W_dot_max Amount_Electricity=W_dot_elect*time Revenues=Amount_Electricity*UnitPrice

2-73 The available head of a hydraulic turbine and its overall efficiency are given. The electric power output of this turbine is to be determined. Assumptions 1. The flow is steady and incompressible. 2. The elevation of the reservoir remains constant. Properties We take the density of water to be ρ  1000 kg / m3 .

Analysis The total mechanical energy the water in a reservoir possesses is equivalent to the potential energy of water at the free surface, and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. Therefore, the actual power produced by the turbine can be expressed as Wturbine = hturbine mghturbine = hturbine r V ghturbine

Substituting,

æ 1N ÷ öæ 1 kW ö ÷ ÷çç Wturbine = (0.91)(1000 kg/m3 )(0.25 m3 /s)(9.81 m/s 2 )(85 m) çç ÷= 190 kW 2÷ çè1 kg ×m/s ÷ ø øèç1000 N ×m/s ÷ Discussion Note that the power output of a hydraulic turbine is proportional to the available elevation difference (turbine head) and the flow rate.

1-119

z=85 [m] V_dot=0.25 [m^3/s] eta_turb=0.91 "Analysis" rho=1000 [kg/m^3] g=9.81 [m/s^2] m_dot=rho*V_dot W_dot_turb=eta_turb*m_dot*g*z*Convert(m^2/s^2, kJ/kg)

2-74 The mass flow rate of water through the hydraulic turbines of a dam is to be determined. Analysis The mass flow rate is determined from

W = mg ( z2 - z1 ) ¾ ¾ ® m=

W = g ( z2 - z1 )

50, 000 kJ/s = 24,700 kg / s æ 1 kJ/kg ö÷ (9.8 m/s 2 )(206 - 0) m çç ÷ çè1000 m 2 /s 2 ø÷

EES (Engineering Equation Solver) SOLUTION "Given" "Analysis" z_1=0 [m] z_2=206 [m] W_dot=50000 [kW] eta_turbine=1 g=9.81 [m/s^2] W_dot=eta_turbine*m_dot*g*(z_2-z_1)*Convert(m^2/s^2, kJ/kg)

2-75 A pump is pumping oil at a specified rate. The pressure rise of oil in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1. The flow is steady and incompressible. 2. The elevation difference across the pump is negligible. Properties The density of oil is given to be   860 kg / m3 .

Analysis Then the total mechanical energy of a fluid is the sum of the potential, flow, and kinetic energies, and is expressed per unit mass as emech = gh + Pv + V 2 /2. To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is æ æ V2 V2ö V 2 - V12 ö ÷ ÷= V çç( P2 - P1 ) + r 2 ÷ D Emech, fluid = m(emech, out - emech, in ) = m ççç( Pv ) 2 + 2 - ( Pv)1 - 1 ÷ ÷ ÷ ÷ ÷ çè 2 2ø 2 èçç ø

since m = r V = V / v , and there is no change in the potential energy of the fluid. Also, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-120

V1 =

V V 0.1 m3 /s = = = 19.9 m/s 2 A1 p D1 / 4 p (0.08 m) 2 / 4

V2 =

V V 0.1 m3 /s = = = 8.84 m/s A2 p D22 / 4 p (0.12 m) 2 / 4

Substituting, the useful pumping power is determined to be Wpump,u = D Emech,fluid 2 2 æ öæ 1 kW ö ö 1 kN ç500 kN/m 2 + (860 kg/m 3 ) (8.84 m/s) - (19.9 m/s) æ ÷ ç ÷ ç ÷ ÷÷ = (0.1 m3 /s) ç ÷ ç ç ÷ 2÷ ç ÷ ç ç1 kN ×m/s ø ç 2 è1000 kg ×m/s ÷ ø÷ è øè

= 36.3 kW Then the shaft power and the mechanical efficiency of the pump become Wpump,shaft = hmotorWelectric = (0.90)(44 kW) = 39.6 kW

hpump =

Wpump, u Wpump, shaft

36.3 kW = 0.918 = 91.8% 39.6 kW

Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.9  0.918 = 0.826.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_elect=44 [kW] rho=860 [kg/m^3] V_dot=0.1 [m^3/s] D1=0.08 [m] D2=0.12 [m] DELTAP=500 [kPa] eta_motor=0.90 "Analysis" A_1=pi*D1^2/4 V_1=V_dot/A_1 A_2=pi*D2^2/4 V_2=V_dot/A_2 W_dot_pump_u=V_dot*(DELTAP+rho*1/2*(V_2^2-V_1^2)*Convert(kg-m/s^2, kN)) W_dot_pump_shaft=eta_motor*W_dot_elect eta_pump=W_dot_pump_u/W_dot_pump_shaft

2-76 A wind turbine produces 180 kW of power. The average velocity of the air and the conversion efficiency of the turbine are to be determined. Assumptions The wind turbine operates steadily. Properties The density of air is given to be 1.31kg / m3 . Analysis (a) The blade diameter and the blade span area are

æ 1 m/s ö ÷ (250 km/h) çç ÷ ÷ çè3.6 km/h ø D= = = 88.42 m æ1 min ÷ ö pn ç p (15/min) ç ÷ çè 60 s ÷ ø Vtip

p D 2 p (88.42 m)2 = = 6140 m2 4 4

1-121

Then the average velocity of air through the wind turbine becomes V=

m 42, 000 kg/s = = 5.23 m / s r A (1.31 kg/m3 )(6140 m 2 )

(b) The kinetic energy of the air flowing through the turbine is 1 1 mV 2 = (42, 000 kg/s)(5.23 m/s) 2 = 574.3 kW 2 2 Then the conversion efficiency of the turbine becomes KE =

W 180 kW = = 0.313 = 31.3% KE 574.3 kW

Discussion Note that about one-third of the kinetic energy of the wind is converted to power by the wind turbine, which is typical of actual turbines.

EES (Engineering Equation Solver) SOLUTION "Given" n_dot=15 [1/min]*Convert(1/min, 1/s) m_dot=42000 [kg/s] V_tip=250 [km/h]*Convert(km/h, m/s) W_dot=180 [kW] rho=1.31 [kg/m^3] "Analysis" D=V_tip/(pi*n_dot) A=pi*D^2/4 V=m_dot/(rho*A) KE_dot=1/2*m_dot*V^2*Convert(m^2/s^2, kJ/kg) eta=W_dot/KE_dot

2-77 Liquefied natural gas (LNG) enters a cryogenic turbine at 5000 kPa and 160C at a rate of 27 kg/s and leaves at 600 kPa. The actual power produced by the turbine is measured to be 240 kW. If the density of LNG is 423.8 kg / m3 , determine the efficiency of the cryogenic turbine. 2-77 Liquefied natural gas (LNG) is expanded in a cryogenic turbine to produce power. The efficiency of the turbine is to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. Analysis Noting that LNG remains as liquid across the turbine, the ideal or maximum power produced by this cryogenic or hydraulic turbine can be determined from P P (5000  600)kPa Wmax  m 1 2  (27 kg/s)  279.7 kW  423.8 kg/m3 The efficiency of the turbine is defined as the actual power produced divided by the ideal power produced: W 240 kW turb  act   0.858  85.8% Wmax 279.1 kW

EES (Engineering Equation Solver) SOLUTION T_1=-160 [C] P_1=4000 [kPa] P_2=400 [kPa] m_dot=28 [kg/s] W_dot_act=185 [kW] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-122

rho_1=density(Methane,T=T_1,P=P_1) W_dot_rev=m_dot*(P_1-P_2)/rho_1 eta_turbine=W_dot_act/W_dot_rev

2-78 A well-established electrical energy storage technology is pumped hydroelectric storage (PHS). The electricity generated from a renewable energy system such as solar panels or wind turbines is used to pump water from a lower reservoir to a higher reservoir during off-peak hours. During peak hours, the water in the higher reservoir is used to drive a water turbine to generate electricity. Consider a PHS system in which 25,000 m3 of water is pumped to an average height of 38 m. Determine the amounts of electricity consumed by the pump and produced by the turbine if the overall efficiency of the pump-motor unit and that of the turbine-generator unit are 75 percent. What is the overall efficiency of this PHS system? 2-78 A pumped hydroelectric storage (PHS) system is considered. The amounts of electricity consumed by the pump and produced by the turbine are to be determined. Assumptions 1. Frictional losses in piping are negligible. Properties We take the density of water to be   1000 kg / m3  1kg / L . Analysis The total mass of water that is pumped is m  V  (1000 kg/m3 )(25,000 m3 )  2.50 107 kg

The electricity consumed by the pump is

Wpump,in 



mgz

pump-motor (2.50  107 kg)(9.81 m/s 2 )(38 m)  1 kJ/kg  1 kWh    2 2  0.75  1000 m /s  3600 kJ 

 3452 kWh The electricity produced by the turbine is

Wturb,out  turbine-gen mgz  1 kJ/kg  1 kWh   (0.75)(2.50  107 kg)(9.81 m/s 2 )(38 m)   2 2   1000 m /s  3600 kJ   1942 kWh The overall efficiency of this PHS system is determined from

overall 

Wturb,out Wpump,in



1942 kWh  0.5625  56.3% 3452 kWh

The same result could be obtained by the product of the efficiencies of the pump-motor unit and the turbine-generator unit: 0.75 x 0.75 = 0.5625.

EES (Engineering Equation Solver) SOLUTION V=25000 [m^3] h=38 [m] eta_pm=0.75 eta_tg=0.75 rho=1000 [kg/m^3] g=9.81 [m/s^2] m=rho*V W_in=(m*g*h)/eta_pm*convert (J, kWh) W_out=eta_tg*(m*g*h)*convert (J, kWh) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-123

Eta_overall=W_out/W_in

Energy and Environment 2-79C Energy conversion pollutes the soil, the water, and the air, and the environmental pollution is a serious threat to vegetation, wild life, and human health. The emissions emitted during the combustion of fossil fuels are responsible for smog, acid rain, and global warming and climate change. The primary chemicals that pollute the air are hydrocarbons (HC, also referred to as volatile organic compounds, VOC), nitrogen oxides (NOx), and carbon monoxide (CO). The primary source of these pollutants is the motor vehicles.

2-80C Fossil fuels include small amounts of sulfur. The sulfur in the fuel reacts with oxygen to form sulfur dioxide ( SO2 ), which is an air pollutant. The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids. The acids formed usually dissolve in the suspended water droplets in clouds or fog. These acid-laden droplets are washed from the air on to the soil by rain or snow. This is known as acid rain. It is called “rain” since it comes down with rain droplets. As a result of acid rain, many lakes and rivers in industrial areas have become too acidic for fish to grow. Forests in those areas also experience a slow death due to absorbing the acids through their leaves, needles, and roots. Even marble structures deteriorate due to acid rain.

2-81C Carbon monoxide, which is a colorless, odorless, poisonous gas that deprives the body's organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen. At low levels, carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles, slows body reactions and reflexes, and impairs judgment. It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain. At high levels, it can be fatal, as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars.

2-82C Carbon dioxide ( CO2 ), water vapor, and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth. This is known as the greenhouse effect. The greenhouse effect makes life on earth possible by keeping the earth warm. But excessive amounts of these gases disturb the delicate balance by trapping too much energy, which causes the average temperature of the earth to rise and the climate at some localities to change. These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change. The greenhouse effect can be reduced by reducing the net production of CO2 by consuming less energy (for example, by buying energy efficient cars and appliances) and planting trees.

2-83C Smog is the brown haze that builds up in a large stagnant air mass, and hangs over populated areas on calm hot summer days. Smog is made up mostly of ground-level ozone ( O3 ), but it also contains numerous other chemicals, including carbon monoxide (CO), particulate matter such as soot and dust, volatile organic compounds (VOC) such as benzene, butane, and other hydrocarbons. Ground-level ozone is formed when hydrocarbons and nitrogen oxides react in the presence of sunlight in hot calm days. Ozone irritates eyes and damage the air sacs in the lungs where oxygen and carbon dioxide are exchanged, causing eventual hardening of this soft and spongy tissue. It also causes shortness of breath, wheezing, fatigue, headaches, nausea, and aggravate respiratory problems such as asthma.

2-84E A household uses fuel oil for heating, and electricity for other energy needs. Now the household reduces its energy use by 15%. The reduction in the CO2 production this household is responsible for is to be determined. Properties The amount of CO2 produced is 1.54 lbm per kWh and 26.4 lbm per gallon of fuel oil (given).

1-124

Analysis Noting that this household consumes 14,000 kWh of electricity and 900 gallons of fuel oil per year, the amount of CO2 production this household is responsible for is

Amount of CO2 produced = (Amount of electricity consumed)(Amount of CO2 per kWh) + (Amount of fuel oil consumed)(Amount of CO2 per gallon) = (14,000 kWh/yr)(1.54 lbm/kWh) + (900 gal/yr)(26.4 lbm/gal) = 45,320 CO2 lbm/year Then reducing the electricity and fuel oil usage by 15% will reduce the annual amount of CO2 production by this household by

Reduction in CO2 produced = (0.15)(Current amount of CO2 production) = (0.15)(45,320 CO2 kg/year)

= 6798 CO2 lbm / year Therefore, any measure that saves energy also reduces the amount of pollution emitted to the environment.

EES (Engineering Equation Solver) SOLUTION "Given" E=14000 [kWh/year] V=900 [gal/year] m_CO2_fueloil=26.4 [lbm/gal] m_CO2_electricity=1.54 [lbm/kWh] f_reduction=0.15 "Analysis" CO2_produced=E*m_CO2_electricity+V*m_CO2_fueloil CO2_reduced=f_reduction*CO2_produced

2-85 A power plant that burns natural gas produces 0.59 kg of carbon dioxide ( CO2 ) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a natural gas power plant. Properties 0.59 kg of CO2 is produced per kWh of electricity generated (given). Analysis Noting that there are 300,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO2 produced = (Amount of electricity consumed)(Amount of CO2 per kWh)

= (300,000 household)(700 kWh/year household)(0.59 kg/kWh) = 1.23´ 108 CO2 kg/year

= 123,000 CO2 ton / year Therefore, the refrigerators in this city are responsible for the production of 123,000 tons of CO2 .

EES (Engineering Equation Solver) SOLUTION "Given" m_CO2=0.59 [kg/kWh] E=700 [kWh/year] N_household=300000 "Analysis" CO2=N_household*E*m_CO2

1-125

2-86 A power plant that burns coal, produces 1.1 kg of carbon dioxide ( CO2 ) per kWh. The amount of CO2 production that is due to the refrigerators in a city is to be determined. Assumptions The city uses electricity produced by a coal power plant. Properties 1.1 kg of CO2 is produced per kWh of electricity generated (given).

Analysis Noting that there are 300,000 households in the city and each household consumes 700 kWh of electricity for refrigeration, the total amount of CO2 produced is Amount of CO2 produced = (Amount of electricity consumed)(Amount of CO2 per kWh)

= (300,000 household)(700 kWh/household)(1.1 kg/kWh) = 2.31´ 108 CO2 kg/year

= 231,000 CO2 ton / year Therefore, the refrigerators in this city are responsible for the production of 231,000 tons of CO2 .

EES (Engineering Equation Solver) SOLUTION "Given" m_CO2=1.1 [kg/kWh] E=700 [kWh/year] N_household=300000 "Analysis" CO2=N_household*E*m_CO2

2-87 A household has 2 cars, a natural gas furnace for heating, and uses electricity for other energy needs. The annual amount of NOx emission to the atmosphere this household is responsible for is to be determined. Properties The amount of NOx produced is 7.1 g per kWh, 4.3 g per therm of natural gas, and 11 kg per car (given).

Analysis Noting that this household has 2 cars, consumes 1200 therms of natural gas, and 9,000 kWh of electricity per year, the amount of NOx production this household is responsible for is Amount of NOx produced = (No. of cars)(Amount of NOx produced per car) + (Amount of electricity consumed)(Amount of NOx per kWh)

+ (Amount of gas consumed)(Amount of NOx per gallon) = (2 cars)(11 kg/car) + (9000 kWh/yr)(0.0071 kg/kWh)

1-126

= 91.06 NOx kg / year Discussion Any measure that saves energy will also reduce the amount of pollution emitted to the atmosphere.

EES (Engineering Equation Solver) SOLUTION "Given" mileage=20000 [km/year] m_NOx_car=11 [kg/year] m_NOx_naturalgas=0.0043 [kg/therm] m_NOx_powerplant=0.0071 [kg/kWh] N_car=2 E=9000 [kWh/year] Gas=1200 [therm/year] "Analysis" NOx=N_car*m_NOx_car+E*m_NOx_powerplant+Gas*m_NOx_naturalgas

2-88E A person trades in his Ford Taurus for a Ford Explorer. The extra amount of CO2 emitted by the Explorer within 5 years is to be determined. Assumptions The Explorer is assumed to use 850 gallons of gasoline a year compared to 650 gallons for Taurus. Analysis The extra amount of gasoline the Explorer will use within 5 years is

Extra Gasoline

 (Extra per year)(No. of years)  (850  650gal / yr)(5yr)

 1000gal Extra CO2 produced  (Extra gallons of gasoline used)(CO2 emission per gallon)

 (1000gal)(19.7lbm / gal)  19, 700lbm CO2

Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced.

EES (Engineering Equation Solver) SOLUTION "Given" V_Taurus=650 [gal/year] V_Explorer=850 [gal/year] m_CO2=19.7 [lbm/gal] DELTAt=5 [year] "Analysis" ExtraGasoline=(V_Explorer-V_Taurus)*DELTAt ExtraCO2=ExtraGasoline*m_CO2

Special Topic: Mechanisms of Heat Transfer 2-89C The three mechanisms of heat transfer are conduction, convection, and radiation.

2-90C Diamond has a higher thermal conductivity than silver, and thus diamond is a better conductor of heat.

1-127

2-91C In forced convection, the fluid is forced to move by external means such as a fan, pump, or the wind. The fluid motion in natural convection is due to buoyancy effects only.

2-92C A blackbody is an idealized body that emits the maximum amount of radiation at a given temperature, and that absorbs all the radiation incident on it. Real bodies emit and absorb less radiation than a blackbody at the same temperature.

2-93C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature. Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface. The Kirchhoff's law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength.

2-94 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of heat transfer through the wall is to be determined.

Assumptions 1. Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2. Thermal properties of the wall are constant.

Properties The thermal conductivity of the wall is given to be k  0.69 W / m C . Analysis Under steady conditions, the rate of heat transfer through the wall is

Qcond = kA

DT (20 - 5)°C = (0.69 W/m ×°C)(5´ 6 m2 ) = 1035 W L 0.3 m

EES (Engineering Equation Solver) SOLUTION "Given" A=5*6 [m^2] L=0.30 [m] k=0.69 [W/m-C] T1=20 [C] T2=5 [C] "Analysis" Q_dot_cond=k*A*(T1-T2)/L

1-128

2-95 Heat is transferred steadily to boiling water in the pan through its bottom. The inner surface temperature of the bottom of the pan is given. The temperature of the outer surface is to be determined. Assumptions 1. Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. 2. Thermal properties of the aluminum pan are constant. Properties The thermal conductivity of the aluminum is given to be k  237 W / m C . Analysis The heat transfer surface area is A   r 2   (0.1m)2  0.0314 m2

Under steady conditions, the rate of heat transfer through the bottom of the pan by conduction is

Q = kA

T - T1 DT = kA 2 L L

Substituting,

T - 105°C 700 W = (237 W/m ×°C)(0.0314 m2 ) 2 0.006 m which gives

T2  105.6C EES SOLUTION "Given" k=237 [W/m-C] D=0.20 [m] L=0.006 [m] Q_dot_cond=700 [W] T1=105 [C] "Analysis" A=pi*D^2/4 Q_dot_cond=k*A*(T2-T1)/L

2-96 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is measured. The thermal conductivity of the plate material is to be determined. Assumptions 1. Steady operating conditions exist since the surface temperatures of the plate remain constant at the specified values. 2. Heat transfer through the plate is one-dimensional. 3. Thermal properties of the plate are constant. Analysis The thermal conductivity is determined directly from the steady one-dimensional heat conduction relation to be

1-129

T- T Q = kA 1 2 L

(Q / A) L (500 W/m2 )(0.02 m) = = 0.1 W / m.°C T1 - T2 (100-0)°C

EES (Engineering Equation Solver) SOLUTION "Given" L=0.02 [m] T1=0 [C] T2=100 [C] q_cond=500 [W/m^2] "Analysis" q_cond=k*(T2-T1)/L

2-97 Hot air is blown over a flat surface at a specified temperature. The rate of heat transfer from the air to the plate is to be determined.

Assumptions 1. Steady operating conditions exist. 2. Heat transfer by radiation is not considered. 3. The convection heat transfer coefficient is constant and uniform over the surface. Analysis Under steady conditions, the rate of heat transfer by convection is Qconv = hAD T

= (55 W/m2 ×°C)(2´ 4 m2 )(80 - 30)°C

= 22,000 W = 22 kW

EES (Engineering Equation Solver) SOLUTION "Given" T_f=80 [C] A=2[m]*4[m] T_s=30 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-130

h=55 [W/m^2-C] "Analysis" Q_dot_conv=h*A*(T_f-T_s)

2-98 A person is standing in a room at a specified temperature. The rate of heat transfer between a person and the surrounding air by convection is to be determined. Assumptions 1. Steady operating conditions exist. 2. Heat transfer by radiation is not considered. 3. The environment is at a uniform temperature. Analysis The heat transfer surface area of the person is A   DL   (0.3 m)(1.75 m)  1.649 m2

Under steady conditions, the rate of heat transfer by convection is Qconv = hAD T = (10 W/m 2 ×°C)(1.649 m 2 )(34 - 20)°C = 231 W

EES (Engineering Equation Solver) SOLUTION "Given" D=0.30 [m] height=1.75 [m] T_s=34 [C] h=10 [W/m^2-C] T_f=20 [C] "Analysis" A=pi*D*height Q_dot_conv=h*A*(T_s-T_f)

2-99 A 1000-W iron is left on the iron board with its base exposed to the air at 23°C. The temperature of the base of the iron is to be determined in steady operation. Assumptions 1. Steady operating conditions exist. 2. The thermal properties of the iron base and the convection heat transfer coefficient are constant and uniform. 3. The temperature of the surrounding surfaces is the same as the temperature of the surrounding air. Properties The emissivity of the base surface is given to be  = 0.4.

1-131

Analysis At steady conditions, the 1000 W of energy supplied to the iron will be dissipated to the surroundings by convection and radiation heat transfer. Therefore, Qtotal = Qconv + Qrad = 1000 W

where Qconv = hAD T = (20 W/m 2 ×K)(0.02 m 2 )(Ts - 296 K) = 0.4(Ts - 296 K) W

and Qrad = es A(Ts4 - To4 ) = 0.4(0.02 m 2 )(5.67 ´ 10- 8 W/m 2 ×K 4 )[Ts 4 - (296 K) 4 ]

= 0.04536´ 10- 8 [Ts4 - (296 K) 4 ]W

Substituting,

1000 W = 0.4(Ts - 296 K) + 0.04536´ 10- 8 [Ts4 - (296 K) 4 ] Solving by trial and error gives

Ts = 1106 K = 833°C Discussion We note that the iron will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 1106 K.

EES (Engineering Equation Solver) SOLUTION "Given" E_dot_iron=1000 [W] T_f=23[C]+273 [K] h=20 [W/m^2-C] epsilon=0.4 A=0.02 [m^2] "Properties" sigma=5.67E-8 [W/m^2-K^4] "Analysis" Q_dot_conv=h*A*(T_s-T_f) Q_dot_rad=epsilon*sigma*A*(T_s^4-T_f^4) Q_dot_total=Q_dot_conv+Q_dot_rad Q_dot_total=E_dot_iron T_s_C=T_s-273

2-100 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation. The surface temperature of the plate is to be determined when it stabilizes. Assumptions 1. Steady operating conditions exist. 2. Heat transfer through the insulated side of the plate is negligible. 3. The heat transfer coefficient is constant and uniform over the plate. 4. Heat loss by radiation is negligible. Properties The solar absorptivity of the plate is given to be  = 0.8. Analysis When the heat loss from the plate by convection equals the solar radiation absorbed, the surface temperature of the plate can be determined from

1-132

Qsolarabsorbed = Qconv

a Qsolar = hA(Ts - To ) 0.8´ A´ 450 W/m2 = (50 W/m 2 ×°C)A(Ts - 25)

Canceling the surface area A and solving for Ts gives

Ts = 32.2°C

EES (Engineering Equation Solver) SOLUTION "Given" alpha=0.8 q_dot_solar=450 [W/m^2] T_f=25 [C] h=50 [W/m^2-C] "Analysis" q_dot_solarabsorbed=alpha*q_dot_solar q_dot_conv=h*(T_s-T_f) q_dot_solarabsorbed=q_dot_conv

2-101 Prob. 2-100 is reconsidered. Effect of convection heat transfer coefficient on the surface temperature of the plate is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Given" alpha=0.8 q_dot_solar=450 [W/m^2] T_f=25 [C] h=50 [W/m^2-C] "Analysis" q_dot_solarabsorbed=alpha*q_dot_solar q_dot_conv=h*(T_s-T_f) q_dot_solarabsorbed=q_dot_conv h[W / m2  C] 10 15 20 25 30 35 40 45

Ts [C] 61 49 43 39.4 37 35.29 34 33

1-133

50 55 60 65 70 75 80 85 90

32.2 31.55 31 30.54 30.14 29.8 29.5 29.24 29

1-134

2-102 A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined. Assumptions 1. Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2. Heat transfer through the shell is one-dimensional. 3. Thermal properties of the iron shell are constant. 4. The inner surface of the shell is at the same temperature as the iced water, 0°C. Properties The thermal conductivity of iron is k  80.2 W / m· C (Table 2-3). The heat of fusion of water is at 1 atm is 333.7 kJ / kg . Analysis This spherical shell can be approximated as a plate of thickness 0.4 cm and surface area A   D 2   (0.4 m)2  0.5027 m2

Then the rate of heat transfer through the shell by conduction is

Qcond = kA

DT (3 - 0)°C = (80.2 W/m ×°C)(0.5027 m2 ) = 30, 235 W L 0.004 m

Considering that it takes 333.7 kJ of energy to melt 1 kg of ice at 0°C, the rate at which ice melts in the container can be determined from

mice =

Q 30.235 kJ/s = = 0.0906 kg / s hif 333.7 kJ/kg

Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall. The error in this case is very small because of the large diameter to thickness ratio. For better accuracy, we could use the inner surface area (D = 39.2 cm) or the mean surface area (D = 39.6 cm) in the calculations.

EES (Engineering Equation Solver) SOLUTION "Given" D=0.40 [m] L=0.004 [m] T1=0 [C] T2=3 [C] "Properties" k=80.2 [W/m-K] "from Table A-3" k_EES=k_('Iron', 2.5) "EES function returns thermal conductivity of iron at 2.5 C" h_if=333.7 [kJ/kg] "for water from the text" "Analysis" A=pi*D^2 Q_dot_cond=k*A*(T2-T1)/L "The curved wall is approximated as a plane wall because of the large diameter to thickness ratio" m_dot_ice=Q_dot_cond/h_if*Convert(W, kW)

1-135

Review Problems 2-103 A gravity driven lighting device requires that a sand bag is raised in every 20 minutes. The velocity of the sand bag as it descends and the overall efficiency of the device are to be determined. Analysis (a) It takes 20 min for the sand bag to descend from a 2-m height. Then, the velocity is V=

ö Dz 2m æ çç1000 mm ÷ = ÷ ÷= 1.67 mm / s ç D t (20´ 60) s è 1 m ø

(b) The energy stored in the sand bag at a higher elevation is stored in the form of potential energy, which is equal to energy input to the device. Therefore, æ 1 J/kg ÷ ö Ein = D PE = mg D z = (10 kg)(9.81 m/s 2 )(2 m) çç = 196.2 J ÷ çè1 m 2 /s 2 ÷ ø For a time period of 20 min, the input power to the device is Ein =

ö Ein 196.2 J æ ÷ çç1 W ÷ = = 0.1635 W ÷ D t (20´ 60) s çè1 J/s ø

The LED bulb provides 16 lumens of lighting. For an efficacy of 150 lumens per watt, the corresponding output electric power is Eout =

Elighting Efficacy

16 lumens = 0.1067 W 150 lumens/W

The overall efficiency of the device is the ratio of the output power to the input power:

hoverall =

Eout 0.1067 W = = 0.652 = 65.2 percent 0.1635 W Ein

Discussion One can double the amount of lighting from 16 lumens to 32 lumens by doubling the weight of the sand bag. However, this requires lifting of a 20-kg weight by a strong person available on the site.

EES (Engineering Equation Solver) SOLUTION "Given" E_dot_lighting=16 [lumen] m=10 [kg] z=2 [m] t=(20*60) [s] Efficacy=150 [lumen/W] "Analysis" V=z/t E_in=m*g*z g=9.81 [m/s^2] E_dot_in=E_in/t E_dot_out=E_dot_lighting/Efficacy eta=E_dot_out/E_dot_in

2-104 A classroom has a specified number of students, instructors, and fluorescent light bulbs. The rate of internal heat generation in this classroom is to be determined. Assumptions 1. There is a mix of men, women, and children in the classroom. 2. The amount of light (and thus energy) leaving the room through the windows is negligible. Properties The average rate of heat generation from people seated in a room/office is given to be 100 W. Analysis The amount of heat dissipated by the lamps is equal to the amount of electrical energy consumed by the lamps, including the 10% additional electricity consumed by the ballasts. Therefore, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-136

Qlighting = (Energy consumed per lamp)´ (No. of lamps) =(40 W)(1.1)(18)=792 W

Qpeople = (No. of people)´ Qperson = 56´ (100 W) = 5600 W Then the total rate of heat gain (or the internal heat load) of the classroom from the lights and people become Qtotal = Qlighting + Qpeople = 792 + 5600 = 6392 W

EES (Engineering Equation Solver) SOLUTION "Given" N_people=56 Q_dot_person=100 [W] N_lamps=18 Q_dot_lamp=40 [W] f=1.1 "Analysis" Q_dot_lighting=f*N_lamps*Q_dot_lamp Q_dot_people=N_people*Q_dot_person Q_dot_total=Q_dot_lighting+Q_dot_people

2-105 A decision is to be made between a cheaper but inefficient natural gas heater and an expensive but efficient natural gas heater for a house. Assumptions The two heaters are comparable in all aspects other than the initial cost and efficiency. Analysis Other things being equal, the logical choice is the heater that will cost less during its lifetime. The total cost of a system during its lifetime (the initial, operation, maintenance, etc.) can be determined by performing a life cycle cost analysis. A simpler alternative is to determine the simple payback period. The annual heating cost is given to be $1200. Noting that the existing heater is 55% efficient, only 55% of that energy (and thus money) is delivered to the house, and the rest is wasted due to the inefficiency of the heater. Therefore, the monetary value of the heating load of the house is

Cost of useful heat  (55%)(Current annual heating cost)

 0.55  ($1200 / yr)  $660 / yr This is how much it would cost to heat this house with a heater that is 100% efficient. For heaters that are less efficient, the annual heating cost is determined by dividing $660 by the efficiency: 82% heater: Annual cost of heating  (Cost of useful heat) / Efficiency  ($660 / yr) / 0.82  $805 / yr 95% heater:

Annual cost of heating  (Cost of useful heat) / Efficiency  ($660 / yr) / 0.95  $695 / yr

Annual cost savings with the efficient heater = 805 - 695 = $110 Excess initial cost of the efficient heater = 2700 - 1600 = $1100 The simple payback period becomes Simple payback period =

Excess initial cost $1100 = = 10 years Annaul cost savings $110 / yr

Therefore, the more efficient heater will pay for the $1100 cost differential in this case in 10 years, which is more than the 8-year limit. Therefore, the purchase of the cheaper and less efficient heater is a better buy in this case. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-137

EES (Engineering Equation Solver) SOLUTION "Given" eta_old=0.55 eta_conventional=0.82 eta_highefficiency=0.95 Cost_conventional=1600 [$] Cost_highefficiency=2700 [$] CurrentCost=1200 [$/year] "Analysis" CostHeat=eta_old*CurrentCost AnnualCost_conventional=CostHeat/eta_conventional AnnualCost_highefficiency=CostHeat/eta_highefficiency PaybackPeriod=(Cost_highefficiency-Cost_conventional)/(AnnualCost_conventional-AnnualCost_highefficiency)

2-106 A homeowner is considering three different heating systems for heating his house. The system with the lowest energy cost is to be determined. Assumptions The differences in installation costs of different heating systems are not considered. Properties The energy contents, unit costs, and typical conversion efficiencies of different systems are given in the problem statement. Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from Unit cost of useful energy =

Unit cost of energy supplied Conversion efficiency

Substituting, Natural gas heater :

Unit cost of useful energy =

ö $1.24/therm æ çç 1 therm ÷ ÷= $13.5´ 10- 6 / kJ 0.87 èç105,500 kJ ø÷

Heating oil heater :

Unit cost of useful energy =

ö $2.3/gal æ çç 1 gal ÷ ÷= $19.1´ 10- 6 / kJ ç ÷ 0.87 è138,500 kJ ø

Electric heater:

Unit cost of useful energy =

ö $0.12/kWh) æ - 6 çç 1 kWh ÷ ÷ ÷= $33.3´ 10 / kJ ç è3600 kJ ø 1.0

Therefore, the system with the lowest energy cost for heating the house is the natural gas heater.

EES (Engineering Equation Solver) SOLUTION "Given" UnitCost_electric=0.12 [$/kWh] UnitCost_gas=1.24 [$/therm] UnitCost_oil=2.3 [$/gal] eta_electric=1.0 eta_gas=0.87 eta_oil=0.87 "Analysis" Cost_UsefulEnergy_gas=UnitCost_gas/eta_gas*1/Convert(therm, kJ) Cost_UsefulEnergy_oil=UnitCost_oil/eta_oil*1/138500 "[$/kJ], since 1 gal = 138500 kJ" Cost_UsefulEnergy_electric=UnitCost_electric/eta_electric*1/Convert(kWh, kJ)

2-107 It is estimated that 570,000 barrels of oil would be saved per day if the thermostat setting in residences in winter were lowered by 6F (3.3C). The amount of money that would be saved per year is to be determined. Assumptions The average heating season is given to be 180 days, and the cost of oil to be $40 / barrel . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-138

Analysis The amount of money that would be saved per year is determined directly from

(570,000 barrel/day)(180 days/year)($55/barrel) = $5.64 ×109 Therefore, the proposed measure will save more than 5.6-billion dollars a year in energy costs.

EES (Engineering Equation Solver) SOLUTION "Given" OilSavings=570000 "[barrel/day]" DELTAT=3.3 [C] HeatingSeason=180 [day/year] UnitCost=55 "[$/barrel]" "Analysis" MoneySaved=OilSavings*HeatingSeason*UnitCost

2-108 The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by adding adequate insulation. The time it will take for the added insulation to pay for itself from the energy it saves is to be determined. Assumptions It is given that the annual energy usage of a house is $1200 a year, and 46% of it is used for heating and cooling. The cost of added insulation is given to be $200. Analysis The amount of money that would be saved per year is determined directly from

Money saved =($1200 / year)(0.46)(0.30) = $166/yr

Then the simple payback period becomes Payback period =

Cost $200 = = 1.2 yr Money saved $166/yr

Therefore, the proposed measure will pay for itself in less than one and a half year.

EES (Engineering Equation Solver) SOLUTION "Given" Cost_energy=1200 [$/year] f_heat_cool=0.46 reduction_heat_cool=0.30 Cost_insulation=200 [$] "Analysis" MoneySaved=Cost_energy*f_heat_cool*reduction_heat_cool Payback=Cost_insulation/MoneySaved

2-109 A diesel engine burning light diesel fuel that contains sulfur is considered. The rate of sulfur that ends up in the exhaust and the rate of sulfurous acid given off to the environment are to be determined. Assumptions 1. All of the sulfur in the fuel ends up in the exhaust. 2. For one kmol of sulfur in the exhaust, one kmol of sulfurous acid is added to the environment. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-139

Properties The molar mass of sulfur is 32·kg / kmol . Analysis The mass flow rates of fuel and the sulfur in the exhaust are mfuel =

mair (336 kg air/h) = = 18.67 kg fuel/h AF (18 kg air/kg fuel)

mSulfur = (750´ 10-6 )mfuel = (500´ 10-6 )(18.67 kg/h) = 0.00933 kg / h The rate of sulfurous acid given off to the environment is mH2SO3 =

M H2SO3 2´ 1 + 32 + 3´ 16 mSulfur = (0.00933 kg/h) = 0.0239 kg / h M Sulfur 32

Discussion This problem shows why the sulfur percentage in diesel fuel must be below certain value to satisfy regulations.

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol_engine=0.004 [m^3] N_dot=2500[1/min]*Convert(1/min, 1/s) AF=18 Sulfur_PPM=500E-6 m_dot_air=336 [kg/h] MM_sulfur=32 [kg/kmol] "PROPERTIES" MM_O2=MolarMass(O2) MM_H2=MolarMass(H2) MM_H2SO3=MM_H2+MM_sulfur+1.5*MM_O2 "ANALYSIS" m_dot_fuel=m_dot_air/AF m_dot_sulfur=Sulfur_PPM*m_dot_fuel m_dot_H2SO3=MM_H2SO3/MM_sulfur*m_dot_sulfur

2-110 The work required to compress a spring is to be determined. Analysis (a) With the preload, F  F0  kx . Substituting this into the work integral gives

W = ò Fds = ò (kx + F0 )dx 1

k 2 ( x2 - x12 ) + F0 ( x2 - x1 ) 2

300 N/cm é 2 2ù êë(1 cm) - 0 ú û+ (100 N) [(1 cm) - 0] 2

1-140

EES (Engineering Equation Solver) SOLUTION "Given" x2=1 [cm] k=300 [N/cm] F0=100 [N] "Analysis" x1=0 [cm] W=k/2*(x2^2-x1^2)+F0*(x2-x1) W_kJ=W*Convert(N-cm,kJ)

2-111 The work required to compress a gas in a gas spring is to be determined. Assumptions All forces except that generated by the gas spring will be neglected. Analysis When the expression given in the problem statement is substituted into the work integral relation, and advantage is taken of the fact that the force and displacement vectors are collinear, the result is

Constant dx xk 1

W = ò Fds = ò 1

Constant 1- k ( x2 - x11- k ) 1- k

1000 N ×m1.3 é (0.3 m)- 0.3 - (0.1 m)- 0.3 ùûú 1- 1.3 ëê

= 1867 N ×m = 1867 J = 1.87 kJ

EES (Engineering Equation Solver) SOLUTION "Given" C=1000 [N-m^1.3] k=1.3 x1=0.1 [m] x2=0.3 [m] "Analysis" W=C/(1-k)*(x2^(1-k)-x1^(1-k))

2-112 A TV set is kept on a specified number of hours per day. The cost of electricity this TV set consumes per month is to be determined. Assumptions 1. The month is 30 days. 2. The TV set consumes its rated power when on. Analysis The total number of hours the TV is on per month is Operating hours  (6 h / day )(30 days )  180 h Then the amount of electricity consumed per month and its cost become © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-141

Amount of electricity = (Power consumed)(Operating hours) = (0.120 kW)(180 h) = 21.6 kWh Cost of electricity  (Amount of electricity)(Unit cost)  (21.6kWh)($0.12 / kWh)  $2.59(per month) Properties Note that an ordinary TV consumes more electricity that a large light bulb, and there should be a conscious effort to turn it off when not in use to save energy.

EES (Engineering Equation Solver) SOLUTION "Given" E_dot_TV=0.120 [kW] time=30*6 [h] UnitCost=0.12 [$/kWh] "Analysis" Cost=E_dot_TV*time*UnitCost

2-113E The power required to pump a specified rate of water to a specified elevation is to be determined. Properties The density of water is taken to be 62.4 lbm / ft 3 (Table A-3E). Analysis The required power is determined from W = mg ( z2 - z1 ) = r V g ( z2 - z1 )

æ 35.315 ft 3 /s ö÷ æ ö 1 lbf ÷ ÷ = (62.4 lbm/ft 3 )(200 gal/min) ççç (32.174 ft/s 2 )(300 ft) çç ÷ ÷ 2 ÷ ç è32.174 lbm ×ft/s ø è15,850 gal/min ø÷ æ ö÷ 1 kW = 8342 lbf ×ft/s = (8342 lbf ×ft/s) çç ÷= 11.3 kW çè737.56 lbf ×ft/s ø÷

EES (Engineering Equation Solver) SOLUTION "Given" z_1=-200 [ft] z_2=100 [ft] V_dot=200 [gal/min]*Convert(gal/min, ft^3/s) "Analysis" rho=62.4 [lbm/ft^3] g=32.174 [ft/s^2] W_dot=rho*V_dot*g*(z_2-z_1)*Convert(lbm-ft/s^2, lbf) W_dot_kW=W_dot*Convert(lbf-ft/s, kW)

2-114 The weight of the cabin of an elevator is balanced by a counterweight. The power needed when the fully loaded cabin is rising, and when the empty cabin is descending at a constant speed are to be determined. Assumptions 1. The weight of the cables is negligible. 2. The guide rails and pulleys are frictionless. 3. Air drag is negligible. Analysis (a) When the cabin is fully loaded, half of the weight is balanced by the counterweight. The power required to raise the cabin at a constant speed of 1.2 m / s is æ 1N ÷ öæ 1 kW ö mgz ÷= 4.71kW ÷çç W= = mgV = (400 kg)(9.81 m/s 2 )(1.2 m/s) çç ÷ 2÷ ø Dt èç1 kg ×m/s ÷ øèç1000 N ×m/s ÷ If no counterweight is used, the mass would double to 800 kg and the power would be 24.71 = 9.42 kW. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-142

(b) When the empty cabin is descending (and the counterweight is ascending) there is mass imbalance of 400-150 = 250 kg. The power required to raise this mass at a constant speed of 1.2 m / s is æ 1N ÷ öæ 1 kW ö mgz ÷ ÷çç W= = mgV = (250 kg)(9.81 m/s 2 )(1.2 m/s) çç ÷= 2.94 kW 2÷ ø Dt èç1 kg ×m/s ÷ øèç1000 N ×m/s ÷ If a friction force of 800 N develops between the cabin and the guide rails, we will need æ 1 kW ö F z ÷ Wfriction = friction = FfrictionV = (800 N)(1.2 m/s)çç = 0.96 kW ÷ çè1000 N ×m/s ÷ ø Dt of additional power to combat friction which always acts in the opposite direction to motion. Therefore, the total power needed in this case is

Wtotal = W + Wfriction = 2.94 + 0.96 = 3.90 kW

EES (Engineering Equation Solver) SOLUTION "Given" m_loaded=800 [kg] m_empty=150 [kg] m_counter=400 [kg] Vel_rise=1.2 [m/s] Vel_descend=1.2 [m/s] F_friction=800 [N] "Properties" g=9.81 [m/s^2] "Analysis" "(a)" m_a=m_loaded-m_counter W1_dot_a=m_a*g*Vel_rise*Convert(N-m/s, kW) W2_dot_a=m_loaded*g*Vel_rise*Convert(N-m/s, kW) "(b)" m_b=m_counter-m_empty W1_dot_b=m_b*g*Vel_descend*Convert(N-m/s, kW) W_dot_friction=F_friction*Vel_descend*Convert(N-m/s, kW) W2_dot_b=W1_dot_b+W_dot_friction

2-115 The power that could be produced by a water wheel is to be determined. Properties The density of water is taken to be 1000 m3 / kg (Table A-3). Analysis The power production is determined from W = mg ( z2 - z1 ) = r V g ( z2 - z1 )

æ 1 kJ/kg ÷ ö = (1000 kg/m3 )(0.480/60 m3 /s)(9.81 m/s 2 )(14 m) çç ÷ çè1000 m 2 /s 2 ÷ ø = 1.10 kW

1-143

EES (Engineering Equation Solver) SOLUTION "Given" z_1=0 [m] z_2=14 [m] V_dot=480 [L/min]*Convert(L/min, m^3/s) "Analysis" rho=1000 [kg/m^3] g=9.81 [m/s^2] W_dot=rho*V_dot*g*(z_2-z_1)*Convert(m^2/s^2, kJ/kg)

2-116 The available head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determined. Assumptions 1. The flow is steady. 2. Water levels at the reservoir and the discharge site remain constant. 3. Frictional losses in piping are negligible. Properties We take the density of water to be   1000 kg / m3  1kg / L .

Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam (relative to free surface of discharge water), and it can be converted to work entirely. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and mgz for a given mass flow rate.

æ 1 kJ/kg ÷ ö emech = pe = gz = (9.81 m/s 2 )(90 m) çç = 0.8829 kJ/kg ÷ çè1000 m 2 /s 2 ÷ ø The mass flow rate is m = r V = (1000 kg/m3 )(65 m3 /s) = 65,000 kg/s

Then the maximum and actual electric power generation become

æ 1 MW ÷ ö Wmax = Emech = memech = (65, 000 kg/s)(0.8829 kJ/kg) çç = 57.39 MW ÷ çè1000 kJ/s ÷ ø Welectric = hoverallWmax = 0.84(57.39 MW) = 48.2 MW

Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the efficiency of the turbine–generator unit.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=65 [m^3/s] z=90 [m] eta=0.84 "Analysis" rho=1000 [kg/m^3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-144

g=9.81 [m/s^2] m_dot=rho*V_dot W_dot_max=m_dot*g*z*Convert(m^2/s^2, kJ/kg) W_dot_elect=eta*W_dot_max

2-117 An entrepreneur is to build a large reservoir above the lake level, and pump water from the lake to the reservoir at night using cheap power, and let the water flow from the reservoir back to the lake during the day, producing power. The potential revenue this system can generate per year is to be determined. Assumptions 1. The flow in each direction is steady and incompressible. 2. The elevation difference between the lake and the reservoir can be taken to be constant, and the elevation change of reservoir during charging and discharging is disregarded. 3. Frictional losses in piping are negligible. 4. The system operates every day of the year for 10 hours in each mode. Properties We take the density of water to be ρ  1000 kg / m3 .

Analysis The total mechanical energy of water in an upper reservoir relative to water in a lower reservoir is equivalent to the potential energy of water at the free surface of this reservoir relative to free surface of the lower reservoir. Therefore, the power potential of water is its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. This also represents the minimum power required to pump water from the lower reservoir to the higher reservoir.

Wmax, turbine = Wmin, pump = Wideal = D Emech = mD emech = mD pe = mgD z = rV gD z æ 1N ÷ öæ 1 kW ö ÷ çç ÷ = (1000 kg/m3 )(2 m3 /s)(9.81 m/s 2 )(40 m) ççç ÷ ÷ 2÷ ÷ è1 kg ×m/s øçè1000 N ×m/s ø = 784.8 kW The actual pump and turbine electric powers are Wpump, elect =

Wideal 784.8 kW = = 1046 kW hpump-motor 0.75

Wturbine = hturbine-genWideal = 0.75(784.8 kW) = 588.6 kW

Then the power consumption cost of the pump, the revenue generated by the turbine, and the net income (revenue minus cost) per year become Cost = Wpump, elect D t ´ Unit price = (1046 kW)(365´ 10 h/year)($0.05/kWh) = $190,968/year

Revenue = WturbineD t ´ Unit price = (588.6 kW)(365´ 10 h/year)($0.12/kWh) = $257,807/year Net income  Revenue  Cost  257,807  190,968  $66, 839 / year Discussion It appears that this pump-turbine system has a potential to generate net revenues of about $67,000 per year. A decision on such a system will depend on the initial cost of the system, its life, the operating and maintenance costs, the interest rate, and the length of the contract period, among other things.

1-145

z=40 [m] V_dot=2 [m^3/s] eta_pump_motor=0.75 eta_turb_gen=0.75 time=365*10 [h/year] UnitPrice_night=0.05 [$/kWh] UnitPrice_day=0.12 [$/kWh] "Analysis" rho=1000 [kg/m^3] g=9.81 [m/s^2] m_dot=rho*V_dot W_dot_max=m_dot*g*z*Convert(m^2/s^2, kJ/kg) W_dot_pump_elect=W_dot_max/eta_pump_motor W_dot_turb_elect=W_dot_max*eta_turb_gen Cost=W_dot_pump_elect*time*UnitPrice_night Revenue=W_dot_turb_elect*time*UnitPrice_day NetIncome=Revenue-Cost

2-118 The pump of a water distribution system is pumping water at a specified flow rate. The pressure rise of water in the pump is measured, and the motor efficiency is specified. The mechanical efficiency of the pump is to be determined. Assumptions 1. The flow is steady. 2. The elevation difference across the pump is negligible. 3. Water is incompressible.

Analysis From the definition of motor efficiency, the mechanical (shaft) power delivered by the he motor is Wpump, shaft = hmotorWelectric = (0.90)(15 kW) = 13.5 kW

To determine the mechanical efficiency of the pump, we need to know the increase in the mechanical energy of the fluid as it flows through the pump, which is

D Emech, fluid = m(emech, out - emech, in ) = m[( Pv)2 - ( Pv)1 ] = m( P2 - P1 )v = V ( P2 - P1 ) æ 1 kJ ö÷ = (0.050 m3 /s)(300-100 kPa) çç ÷= 10 kJ/s = 10 kW çè1 kPa ×m3 ø÷ since m = r V = V / v and there is no change in kinetic and potential energies of the fluid. Then the pump efficiency becomes hpump =

D Emech, fluid Wpump, shaft

10 kW = 0.741 13.5 kW

or 74.1%

Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor efficiencies, which is 0.90.741 = 0.667.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_elect=15 [kW] eta_motor=0.90 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-146

V_dot=0.050 [m^3/s] P1=100 [kPa] P2=300 [kPa] "Analysis" rho=1000 [kg/m^3] W_dot_pump_shaft=eta_motor*W_dot_elect DELTAE_dot_mech_fluid=V_dot*(P2-P1) eta_pump=DELTAE_dot_mech_fluid/W_dot_pump_shaft

2-119 An automobile moving at a given velocity is considered. The power required to move the car and the area of the effective flow channel behind the car are to be determined. Analysis The absolute pressure of the air is

æ0.1333 kPa ÷ ö ÷ P = (700 mm Hg) ççç = 93.31 kPa ÷ è 1 mm Hg ÷ ø

and the specific volume of the air is v=

RT (0.287 kPa ×m3 /kg ×K)(293 K) = = 0.9012 m3 /kg P 93.31 kPa

The mass flow rate through the control volume is

AV (3 m2 )(90/3.6 m/s) 1 1 = = 83.22 kg/s v 0.9012 m3 /kg

The power requirement is ö V 2 - V22 (90 / 3.6 m/s) 2 - (82 / 3.6 m/s) 2 æ çç 1 kJ/kg ÷ W= m 1 = (83.22 kg/s) = 4.42 kW 2 2÷ ÷ ç è1000 m /s ø 2 2

The outlet area is

A2V2 mv (83.22 kg/s)(0.9012 m3 /kg) ¾¾ ® A2 = = = 3.29 m 2 v V2 (82/3.6) m/s

EES (Engineering Equation Solver) SOLUTION "Given" A_1=3 [m^2] Vel_1=90 [km/h]*Convert(km/h, m/s) P=700 [mmHg]*Convert(mmHg,kPa) T=(20+273) [K] Vel_2=82 [km/h]*Convert(km/h, m/s) "Analysis" R=0.287 [kJ/kg-K] v=(R*T)/P m_dot=(A_1*Vel_1)/v W_dot=m_dot*(Vel_1^2-Vel_2^2)/2*Convert(m^2/s^2, kJ/kg) A_2=(m_dot*v)/Vel_2

1-147

2-120 Wind energy has been used since 4000 BC to power sailboats, grind grain, pump water for farms, and generate electricity. In the United States alone, more than 6 million small windmills, most of them under 5 hp, have been used since the 1850s to pump water. Small windmills have been used to generate electricity since 1900, but the development of modern wind turbines occurred towards the end of the twentieth century in response to the energy crises in the early 1970s. Regions with an average wind speed of above 5 m / s are potential sites for economical wind power generation. A typical wind turbine has a blade span (or rotor) diameter of about 100 m and generates about 3 MW of power. Recent wind turbines have power capacities above 10 MW and rotor diameters of over 150 m. Consider a wind turbine with an 80-m-diameter rotor that is rotating at 20 rpm (revolutions per minute) under steady winds at an average velocity of 30 km/h. Assuming the wind turbine has an efficiency of 35 percent (i.e., it converts 35 percent of the kinetic energy of the wind to electricity), determine (a) the power produced, in kW; (b) the tip speed of the blade, in km/h; and (c) the revenue generated by the wind turbine per year if the electric power produced is sold to the utility at $1.15 / kWh . Take the density of air to be 1.20 kg / m3 . 2-120 A wind turbine is rotating at 20 rpm under steady winds of 30 km / h . The power produced, the tip speed of the blade, and the revenue generated by the wind turbine per year are to be determined. Assumptions 1. 2.

Steady operating conditions exist. The wind turbine operates continuously during the entire year at the specified conditions.

Properties The density of air is given to be   1.20 kg / m3 . Analysis (a) The blade span area and the mass flow rate of air through the turbine are A   D2 / 4   (80 m)2 / 4  5027 m 2

 1000 m  1 h  V  (30 km/h)     8.333 m/s  1 km  3600 s  m   AV  (1.2 kg/m3 )(5027 m2 )(8.333 m/s)=50,270 kg/s

Noting that the kinetic energy of a unit mass is V 2 / 2 and the wind turbine captures 35% of this energy, the power generated by this wind turbine becomes

1  1 kJ/kg  1  W    m V 2   (0.35) (50,270 kg/s )(8.333 m/s ) 2    610.9 kW 2 2   1000 m 2 /s 2  (b) Noting that the tip of blade travels a distance of D per revolution, the tip velocity of the turbine blade for an rpm of n becomes V tip  Dn   (80 m)(20 / min)  5027 m/min = 83.8 m/s  302 km/h

(c) The amount of electricity produced and the revenue generated per year are Electricit y produced  W t  (610 .9 kW)(365  24 h/yr) = 5.351  10 6 kWh/yr

Revenue generated  (Electricity produced)(Unit price)  (5.351106 kWh/year)($1.15/kWh)

 $6,154,000 / yr Discussion (a) If we repeat this problem at a wind velocity of 20 km / h , the power generated, the tip velocity, and the revenue generated become 181 kW, 302 km / h , and $1,823,000 /yr , respectively.

1-148

Fundamentals of Engineering (FE) Exam Problems 2-121 A 2-kW electric resistance heater in a room is turned on and kept on for 50 min. The amount of energy transferred to the room by the heater is (a) 2 Kj (b) 100 kJ (c) 3000 kJ (d) 6000 kJ (e) 12,000 kJ Answer (d) 6000 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). We= 2 [kJ/s] time=50*60 [s] We_total=We*time "Some Wrong Solutions with Common Mistakes:" W1_Etotal=We*time/60 "using minutes instead of s" W2_Etotal=We "ignoring time"

2-122 Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.13 / kWh , the electricity cost of this refrigerator per month (30 days) is (a) $4.9 (b) $5.8 (c) $7.5 (d) $8.3 (e) $9.7 Answer (c) $7.5 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W_e=0.320 [kW] Hours=0.25*(24*30) [h] price=0.13 [$/kWh] Cost=W_e*Hours*price "Some Wrong Solutions with Common Mistakes:" W1_cost= W_e*24*30*price "running continuously"

2-123 A 75 hp (shaft) compressor in a facility that operates at full load for 2500 hours a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is $0.11/ kWh , the annual electricity cost of this compressor is (a) $14,300 (b) $15,380 (c) $16,540 (d) $19,180 (e) $22,180 Answer (c) $16,540 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-149

W_comp=75 [hp] Hours=2500 [h/yr] Eff=0.93 price=0.11 [$/kWh] W_e=W_comp*Convert(hp, kW)*Hours/Eff Cost=W_e*price "Some Wrong Solutions with Common Mistakes:" W1_cost= W_comp*0.7457*Hours*price*Eff "multiplying by efficiency" W2_cost= W_comp*Hours*price/Eff "not using conversion" W3_cost= W_comp*Hours*price*Eff "multiplying by efficiency and not using conversion" W4_cost= W_comp*0.7457*Hours*price "Not using efficiency"

2-124 In a hot summer day, the air in a well-sealed room is circulated by a 0.50-hp (shaft) fan driven by a 65% efficient motor. (Note that the motor delivers 0.50 hp of net shaft power to the fan). The rate of energy supply from the fan-motor assembly to the room is (a) (b) (c) (d) (e)

0.769 kJ / s 0.325 kJ / s 0.574 kJ / s 0.373 kJ / s 0.242 kJ / s

Answer (c) 0.574 kJ / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W_fan=0.50*0.7457 [kW] Eff=0.65 E=W_fan/Eff "Some Wrong Solutions with Common Mistakes:" W1_E=W_fan*Eff "Multiplying by efficiency" W2_E=W_fan "Ignoring efficiency" W3_E=W_fan/Eff/0.7457 "Using hp instead of kW"

2-125 A fan is to accelerate quiescent air to a velocity to 9 m / s at a rate of 3 m3 / min . If the density of air is 1.15 kg / m3 , the minimum power that must be supplied to the fan is (a) 41 W (b) 122 W (c) 140 W (d) 206 W (e) 280 W Answer (c) 140 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=9 [m/s] V_dot=3 [m^3/s] rho=1.15 [kg/m^3] m_dot=rho*V_dot W_e=m_dot*V^2/2 "Some Wrong Solutions with Common Mistakes:" W1_W_e=V_dot*V^2/2 "Using volume flow rate" W2_W_e=m_dot*V^2 "forgetting the 2" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-150

W3_W_e=V^2/2 "not using mass flow rate"

2-126 A 900-kg car cruising at a constant speed of 60 km / h is to accelerate to 100 km / h in 4 s. The additional power needed to achieve this acceleration is (a) 56 kW (b) 222 kW (c) 2.5 kW (d) 62 kW (e) 90 kW Answer (a) 56 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=900 [kg] V1=60 [km/h] V2=100 [km/h] t=4 [s] Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000/t "Some Wrong Solutions with Common Mistakes:" W1_Wa=((V2/3.6)^2-(V1/3.6)^2)/2/t "Not using mass" W2_Wa=m*((V2)^2-(V1)^2)/2000/t "Not using conversion factor" W3_Wa=m*((V2/3.6)^2-(V1/3.6)^2)/2000 "Not using time interval" W4_Wa=m*((V2/3.6)-(V1/3.6))/1000/t "Using velocities"

2-127 The elevator of a large building is to raise a net mass of 550 kg at a constant speed of 12 m / s using an electric motor. Minimum power rating of the motor should be (a) 0 kW (b) 4.8 kW (c) 12 kW (d) 45 kW (e) (65 kW Answer (e) 65 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=550 [kg] V=12 [m/s] g=9.81 [m/s^2] W=m*g*V*Convert(kg-m^2/s^3, kW) "Some Wrong Solutions with Common Mistakes:" W1_W=m*V "Not using g" W2_W=m*g*V^2/2000 "Using kinetic energy" W3_W=m*g/V "Using wrong relation"

2-128 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3 / s from an elevation of 65 m using a turbine–generator with an efficiency of 85 percent. When frictional losses in piping are disregarded, the electric power output of this plant is (a) 3.9 MW (b) 38 MW (c) 45 MW © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-151

(d) 53 MW (e) 65 MW Answer (b) 38 MW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_dot=70 [m^3/s] z=65 [m] Eff=0.85 g=9.81 [m/s^2] rho=1000 [kg/m^3] We=rho*V_dot*g*z*Eff*Convert(W, MW) "Some Wrong Solutions with Common Mistakes:" W1_We=rho*V_dot*z*Eff/10^6 "Not using g" W2_We=rho*V_dot*g*z/Eff/10^6 "Dividing by efficiency" W3_We=rho*V_dot*g*z/10^6 "Not using efficiency"

2-129 A 2-kW pump is used to pump kerosene (   0.820 kg / L ) from a tank on the ground to a tank at a higher elevation. Both tanks are open to the atmosphere, and the elevation difference between the free surfaces of the tanks is 30 m. The maximum volume flow rate of kerosene is (a) (b) (c) (d) (e)

8.3 L / s 7.2 L / s 6.8 L / s 12.1L / s 17.8 L / s

Answer (a) 8.3 L / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W=2 [kW] rho=0.820 [kg/L] z=30 [m] g=9.81 [m/s^2] W=rho*Vdot*g*z*Convert(W, kW) "Some Wrong Solutions with Common Mistakes:" W=W1_Vdot*g*z/1000 "Not using density"

2-130 A glycerin pump is powered by a 5-kW electric motor. The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa. If the flow rate through the pump is 18 L / s and the changes in elevation and the flow velocity across the pump are negligible, the overall efficiency of the pump is (a) 69% (b) 72% (c) 76% (d) 79% (e) 82% Answer (c) 76% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-152

We=5 [kW] Vdot= 0.018 [m^3/s] DP=211 [kPa] Emech=Vdot*DP Eff=Emech/We

The following problems are based on the optional special topic of heat transfer 2-131 A 10-cm high and 20-cm wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of 0.08 W and transferring it by convection to the surrounding air at 25C. Heat transfer from the back surface of the board is negligible. If the convection heat transfer coefficient on the surface of the board is 10 W / m 2  C and radiation heat transfer is negligible, the average surface temperature of the chips is (a) 26C (b) 45C (c) 15C (d) 80C (e) 65C Answer (e) 65C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=0.10*0.20 [m^2] Q= 100*0.08 [W] Tair=25 [C] h=10 [W/m^2-C] Q= h*A*(Ts-Tair) "Some Wrong Solutions with Common Mistakes:" Q= h*(W1_Ts-Tair) "Not using area" Q= h*2*A*(W2_Ts-Tair) "Using both sides of surfaces" Q= h*A*(W3_Ts+Tair) "Adding temperatures instead of subtracting" Q/100= h*A*(W4_Ts-Tair) "Considering 1 chip only"

2-132 A 50-cm-long, 0.2-cm-diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coefficient in water at 1 atm experimentally. The surface temperature of the wire is measured to be 130C when a wattmeter indicates the electric power consumption to be 4.1 kW. Then the heat transfer coefficient is (a) 43,500 W / m2·C (b) 137 W / m 2 ·C (c) 68,330 W / m2·C (d) 10,038 W / m2 ·C (e) 37,540 W / m2·C Answer (a) 43,500 W / m2·C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). L=0.5 [m] D=0.002 [m] We=4100 [W] Ts=130 [C] Tf=100 [C] "Boiling temperature of water at 1 atm" A=pi*D*L © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-153

We= h*A*(Ts-Tf) "Some Wrong Solutions with Common Mistakes:" We= W1_h*(Ts-Tf) "Not using area" We= W2_h*(L*pi*D^2/4)*(Ts-Tf) "Using volume instead of area" We= W3_h*A*Ts "Using Ts instead of temp difference"

2-133 A 3  m 2 hot black surface at 80C is losing heat to the surrounding air at 25C by convection with a convection heat transfer coefficient of 12 W / m 2 .C , and by radiation to the surrounding surfaces at 15C. The total rate of heat loss from the surface is (a) 1987 W (b) 2239 W (c) 2348 W (d) 3451 W (e) 3811 W Answer (d) 3451 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=3 [m^2] Ts=80 [C] Tf=25 [C] h_conv=12 [W/m^2-C] Tsurr=15 [C] sigma=5.67E-8 [W/m^2-K^4] eps=1 Q_conv=h_conv*A*(Ts-Tf) Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) Q_total=Q_conv+Q_rad "Some Wrong Solutions with Common Mistakes:" W1_Ql=Q_conv "Ignoring radiation" W2_Q=Q_rad "ignoring convection" W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) "Using C in radiation calculations" W4_Q=Q_total/A "not using area"

2-134 Heat is transferred steadily through a 0.2-m thick 8 m by 4 m wall at a rate of 2.4 kW. The inner and outer surface temperatures of the wall are measured to be 15C to 5C. The average thermal conductivity of the wall is (a) (b) (c) (d) (e)

0.002 W / m·C 0.75 W / m·C 1.0 W / m·C 1.5 W / m·C 3.0 W / m·C

Answer (d) 1.5 W / m·C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=8*4 [m^2] L=0.2 [m] T1=15 [C] T2=5 [C] Q=2400 [W] Q=k*A*(T1-T2)/L © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-154

"Some Wrong Solutions with Common Mistakes:" Q=W1_k*(T1-T2)/L "Not using area" Q=W2_k*2*A*(T1-T2)/L "Using areas of both surfaces" Q=W3_k*A*(T1+T2)/L "Adding temperatures instead of subtracting" Q=W4_k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"

2-135 The roof of an electrically heated house is 7 m long, 10 m wide, and 0.25 m thick. It is made of a flat layer of concrete whose thermal conductivity is 0.92 W / m·C . During a certain winter night, the temperatures of the inner and outer surfaces of the roof are measured to be 15C and 4C, respectively. The average rate of heat loss through the roof that night was (a) 41 W (b) 177 W (c) 4894 W (d) 5567 W (e) 2834 W Answer (e) 2834 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). A=7*10 [m^2] L=0.25 [m] k=0.92 [W/m-C] T1=15 [C] T2=4 [C] Q_cond=k*A*(T1-T2)/L "Some Wrong Solutions with Common Mistakes:" W1_Q=k*(T1-T2)/L "Not using area" W2_Q=k*2*A*(T1-T2)/L "Using areas of both surfaces" W3_Q=k*A*(T1+T2)/L "Adding temperatures instead of subtracting" W4_Q=k*A*L*(T1-T2) "Multiplying by thickness instead of dividing by it"

2-136 … 2-144 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

1-155

Chapter 3 PROPERTIES OF PURE SUBSTANCES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw Hill LLC and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw Hill LLC: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw Hill LLC.

1-156

CYU - Check Your Understanding CYU 3-3 ■ Check Your Understanding CYU 3-3.1 A vapor that is not about to condense is called a _______ and a vapor that is about to condense is called a _______. (a) Subcooled vapor, saturated vapor (b) Superheated vapor, Saturated vapor (c) Saturated vapor, superheated vapor (d) Saturated vapor, compressed vapor (e) Superheated vapor, subcooled vapor Answer: (b) Superheated vapor, saturated vapor CYU 3-3.2 An ordinary glass of water at room temperature is at a state of I. Saturated liquid II. Compressed liquid III. Subcooled liquid (a) Only I (b) Only II (c) I and II (d) I and III (e) II and III Answer: (e) II and III CYU 3-3.3 At what state is the temperature of water at 1 atm pressure the lowest? I. Saturated liquid II. Compressed liquid III. Saturated vapor IV. Superheated vapor (a) Only I (b) Only II (c) Only III (d) I and II (e) I and III Answer: (b) Only II CYU 3-3.4 At what state is the pressure of water at 30ºC the lowest? I. Saturated liquid II. Compressed liquid III. Saturated vapor IV. Superheated vapor (a) Only I (b) Only II (c) Only III (d) Only IV (e) I and III Answer: (d) Only IV © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-157

CYU 3-3.5 For which case does the cooking of a certain stew take the longest time? (a) Sea level, regular pan (b) Sea level, pressure cooker (c) 1500-m elevation, regular pan (d) 3000-m elevation, regular pan (e) 1500-m elevation, pressure cooker Answer: (d) 3000-m elevation, regular pan

CYU 3-4 ■ Check Your Understanding CYU 3-4.1 For which case is the value of the specific volume of water the highest? (a) Saturated liquid, 100 kPa (b) Saturated liquid, 500 kPa (c) Saturated vapor, 100 kPa (d) Saturated vapor, 500 kPa (e) Saturated vapor, 800 kPa Answer: (c) Saturated vapor, 100 kPa CYU 3-4.2 At which two states are the temperatures of a pure substance at a specified pressure equal? I. Saturated liquid II. Compressed liquid III. Saturated vapor IV. Superheated vapor (a) I and II (b) I and III (c) II and III (d) I and IV (e) II and IV Answer: (b) I and III CYU 3-4.3 The triple point pressures of some substances are given below. Which substance do you think will experience sublimation? (a) Water, 0.61 kPa (b) Ammonia, 6.08 kPa (c) Carbon dioxide, 517 kPa (d) Xenon, 81.5 kPa Answer: (c) Carbon dioxide, 517 kPa

CYU 3-5 ■ Check Your Understanding CYU 3-5.1 Select the wrong description associated with a compressed liquid or a superheated vapor: (a) Superheated vapor: P is less than Psat at a given T. (b) Compressed liquid: T is greater than Tsat at a given P. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-158

(c) Superheated vapor: v is greater than v g at a given T or P. (d) Compressed liquid: v is less than v f at a given T or P. (e) Compressed liquid: h is less than h f at a given T or P. Answer: (b) Compressed liquid: T is greater than Tsat at a given P. CYU 3-5.2 What is the internal energy of a fluid whose enthalpy, pressure, and specific volume are given to be 2500 kJ / kg , 2 MPa, and 0.1m3 / kg , respectively. (a) 2700kJ / kg (b) 2520kJ / kg (c) 2500kJ / kg (d) 2480kJ / kg (e) 2300kJ / kg Answer: (e) 2300kJ / kg

u  h  Pv  2500kJ / kg  (2000kPa)  0.1m3 / kg   2300kJ / kg

CYU 3-5.3 Water is boiled at a constant pressure of 90 kPa. How much heat need to be transferred to one kg water to turn it into vapor? (a) q  u fg @ 90 kPa (b) q  h fg @ 90 kPa (c) q  hg @ 90 kPa (d)

q  h fg @ 100 C

(e)

q  u fg @ 100 C s

Answer: (b) q  h fg @ 90 kPa

CYU 3-5.4 What is the specific volume and entropy of water at P = 300 kPa and T = 133.52C? See water tables. (a) v  0.001073 m3 / kg, s  1.6717kJ / kg  K (b) v  0.001073 m3 / kg, s  6.9917kJ / kg  K (c) v  0.60582 m3 / kg, s  1.6717kJ / kg  K (d) v  0.60582 m3 / kg, s  6.9917kJ / kg  K (e) Insufficient information Answer: (e) Insufficient information Water is saturated mixture. Temperature and pressure are dependent in this region. CYU 3-5.5 What is the internal energy and enthalpy of water at P = 300 kPa and T = 110C? See water tables. (a) u  561.11 kJ / kg, h  561.43 kJ / kg (b) u  2543.2 kJ / kg, h  2724.9 kJ / kg (c) u  461.27 kJ / kg, h  461.42 kJ / kg (d) u  2517.7 kJ / kg, h  2691.1 kJ / kg (e) u  1982.1 kJ / kg, h  2163.5 kJ / kg Answer: u  461.27 kJ / kg, h  461.42 kJ / kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-159 u  u f @110 C  461.27 kJ / kg (Table A-4) h  u f @110 C  461.42 kJ / kg (Table A-4)

CYU 3-5.6 What is the entropy of water at P = 0.6 MPa and T = 200C? See water tables. (a) 6.9683 kJ / kg  K (b) 6.7593 kJ / kg  K (c) 1.9308 kJ / kg  K (d) 6.4302 kJ / kg  K (e) 2.3305 kJ / kg  K Answer: (a) 6.9683 kJ / kg  K Superheated vapor table, Table A-6 CYU 3-5.7 What is the pressure of water at T = 170C and u  2000 kJ / kg ? See water tables. (a) Less than 792.18 kPa (b) 792.18 kPa (c) Greater than 792.18 kPa (d) 100 kPa (e) 1000 kPa Answer: (b) 792.18 kPa u  u f @170 C  718.20 kJ / kg (Table A-4) h  u f @170 C  2575.7 kJ / kg (Table A-4)

Saturated mixture P  Psat @ 170 C  792.18 kPa (Table A-4)

CYU 3-6 ■ Check Your Understanding CYU 3-6.1 A fixed amount of ideal gas in a rigid tank at T1 and P1 undergoes a process until the pressure increases to P2 . The temperature at the final state T2 is (a) T2  T1  P2 / P1  (b)

T2  T1  P1 / P2 

T2  T1  P2 / P1 

CYU 3-6.2 A piston-cylinder device contains a fixed mass of air at a given pressure and temperature. If both the temperature and pressure are doubled, the volume of the air will © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-160

(a) Double (b) Decrease by half (c) Remain constant (d) Insufficient information Answer: (c) Remain constant P1 V1  mRT1

 2 P1  V2  mR  2T1  V2  V1

CYU 3-6.3 A rigid tank contains a fixed mass of air at 10C and 100 kPa. If the temperature is increased to 20C, the air pressure will be (a) 100 kPa (b) Between 100 and 200 kPa (c) 200 kPa (d) More than 200 kPa (e) Insufficient information Answer: (b) Between 100 and 200 kPa P1 v  R(283 K) P2 v  R(293 K) P2  (293 / 283) P1  (293 / 283)(100kPa)

CYU 3-6.4 A rigid tank contains a fixed mass of nitrogen gas at 400 K and 300 kPa. If half of the nitrogen mass is withdrawn from the tank and the pressure is decreased to 150 kPa, the nitrogen temperature will be (a) More than 400 K (b) 400 K (c) Between 200 and 400 K (d) 200 K (e) Less than 200 K Answer: (b) 400 K P1 V1  m1 RT1 0.5P1 V1  0.5m1 RT2 RT2 T2  T1

CYU 3-7 ■ Check Your Understanding CYU 3-7.1 Gases typically behave as an ideal gas at (a) Low pressures and low temperatures (b) Low pressures and high temperatures (c) High pressures and high temperatures (d) High pressures and low temperatures (e) Around critical pressure and temperature Answer: (b) Low pressures and high temperatures

1-161

CYU 3-7.2 The compressibility factor is defined as (a) Z  RT / PV (b) Z  mRT / PV (c) Z  mPv / RT (d) Z  videal / vactual (e) Z  vactual / videal Answer: (e) Z  vactual / videal CYU 3-7.3 A gas deviates from the ideal gas behavior most when the compressibility factor Z is (a) 0.3 (b) 0.5 (c) 0.7 (d) 1 (e) 1.1 Answer: (a) 0.3

1-162

PROBLEMS Pure Substances, Phase Change Processes, Property Diagrams 3-1C Yes, since the chemical composition throughout the tank remain the same.

3-2C Yes. Because it has the same chemical composition throughout.

3-3C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.

3-4C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid.

3-5C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure.

3-6C Yes. The saturation temperature of a pure substance depends on pressure. The higher the pressure, the higher the saturation or boiling temperature.

3-7C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a pure substance coexist in equilibrium.

3-8C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and thus the greater the cooking temperature.

3-9C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases.

3-10C No.

3-11C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other one.

3-12C Yes.

1-163

Property Tables 3-13C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor region.

3-14C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.

3-15C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent of the selected reference state.

3-16C The term h fg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure. It can be determined from h fg  hg  h f .

3-17C Yes. It decreases with increasing pressure and becomes zero at the critical pressure.

3-18C Yes; the higher the temperature the lower the h fg value.

3-19C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the h fg .

3-20C A given volume of water will boil at a higher temperature in a tall and narrow pot since the pressure at the bottom (and thus the corresponding saturation pressure) will be higher in that case.

3-21C The molar mass of gasoline  C8 H18  is 114 kg / kmol , which is much larger than the molar mass of air that is 29 kg / kmol . Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature

than the surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment

3-22C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus v T , P @v f @ T .

3-23C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.

3-24 Complete the following table for H 2 O :

1-164

T, C

P, kPa

u, kJ / kg

Phase description

143.61

400

1450

Saturated mixture

220

2319.6

2601.3

Saturated vapor

190

2500

805.15

Compressed liquid

466.21

4000

3040

Superheated vapor

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=400 [kPa] u[1]=1450 [kJ/kg] T[2]=220 [C] x[2]=1 T[3]=190 [C] P[3]=2500 [kPa] P[4]=4000 [kPa] u[4]=3040 [kJ/kg] "Analysis" Fluid$='steam_iapws' T[1]=temperature(Fluid$, P=P[1], u=u[1]) x[1]=quality(Fluid$, P=P[1], u=u[1]) P[2]=pressure(Fluid$, T=T[2], x=x[2]) u[2]=intenergy(Fluid$, T=T[2], x=x[2]) u[3]=intenergy(Fluid$, T=T[3], P=P[3]) x[3]=quality(Fluid$, T=T[3], P=P[3]) T[4]=temperature(Fluid$, P=P[4], u=u[4]) x[4]=quality(Fluid$, P=P[4], u=u[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

3-25 Complete the following table for H 2 O : T, C

P, kPa

v, m3 / kg

Phase description

140

361.53

0.035

Saturated mixture

155.46

550

0.001097

Saturated liquid

125

750

0.001065

Compressed liquid

300

1803

0.140

Superheated vapor

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=140 [C] v[1]=0.035 [m^3/kg] P[2]=550 [kPa] x[2]=0 T[3]=125 [C] P[3]=750 [kPa] T[4]=300 [C] v[4]=0.140 [m^3/kg] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-165

Fluid$='steam_iapws' P[1]=pressure(Fluid$, T=T[1], v=v[1]) x[1]=quality(Fluid$, T=T[1], v=v[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) v[2]=volume(Fluid$, P=P[2], x=x[2]) v[3]=volume(Fluid$, T=T[3], P=P[3]) x[3]=quality(Fluid$, T=T[3], P=P[3]) P[4]=pressure(Fluid$, T=T[4], v=v[4]) x[4]=quality(Fluid$, T=T[4], v=v[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

3-26E Complete the following table for H 2 O : T, C

P, psia

u, Btu / lbm

Phase description

300

67.03

782

Saturated mixture

267.22

236.02

Saturated liquid

500

120

1174.4

Superheated vapor

400

373.84

Compressed liquid

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=300 [F] u[1]=782 [Btu/lbm] P[2]=40 [psia] x[2]=0 T[3]=500 [F] P[3]=120 [psia] T[4]=400 [F] P[4]=400 [psia] "Analysis" Fluid$='steam_iapws' P[1]=pressure(Fluid$, T=T[1], u=u[1]) x[1]=quality(Fluid$, T=T[1], u=u[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) u[2]=intenergy(Fluid$, P=P[2], x=x[2]) u[3]=intenergy(Fluid$, T=T[3], P=P[3]) x[3]=quality(Fluid$, T=T[3], P=P[3]) u[4]=intenergy(Fluid$, T=T[4], P=P[4]) x[4]=quality(Fluid$, T=T[4], P=P[4]) "x = 100 for superheated vapor and x = −100 for compressed liquid"

3-27E Problem 3-26E is reconsidered. The missing properties of water are to be determined using appropriate software, and the solution is to be repeated for refrigerant-134a, refrigerant-22, and ammonia. Analysis The problem is solved using EES, and the solution is given below. "Given" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-166

T[1]=300 [F] u[1]=782 [Btu/lbm] P[2]=40 [psia] x[2]=0 T[3]=500 [F] P[3]=120 [psia] T[4]=400 [F] P[4]=420 [psia] "Analysis" Fluid$='steam_iapws' "Change the fluid to 'R-134a', 'R-22' and 'Ammonia' and Solve" P[1]=pressure(Fluid$, T=T[1], u=u[1]) x[1]=quality(Fluid$, T=T[1], u=u[1]) T[2]=temperature(Fluid$, P=P[2], x=x[2]) u[2]=intenergy(Fluid$, P=P[2], x=x[2]) u[3]=intenergy(Fluid$, P=P[3], T=T[3]) x[3]=quality(Fluid$, P=P[3], T=T[3]) u[4]=intenergy(Fluid$, P=P[4], T=T[4]) x[4]=quality(Fluid$, P=P[4], T=T[4]) "x=100 for superheated vapor and x=−100 for compressed liquid" Solution for steam P, psia T, F 300 67.028 267.2 40 500 120 400 400

x 0.6173 0 100 −100

u, Btu / lbm 782 236 1174 373.8

Solution for R134a P, psia T, F 300 29.01 40 500 120 400 400

x 0 100 100

u, Btu / lbm 782 21.24 203.5 173.6

Solution for R22 P, psia T, F 300 1.534 40 500 120 400 400

x 0 100 100

u, Btu / lbm 782 78.08 240.1 218.7

Solution for ammonia P, psia T, F 300 11.67 40 500 120 400 400

x 0 100 100

u, Btu / lbm 782 63.47 785.5 727

3-28 Complete the following table for H 2 O : T, C

P, kPa

h, kJ / kg

Phase description

1-167

120.21

200

2045.8

0.7

Saturated mixture

140

361.53

1800

0.565

Saturated mixture

177.66

950

752.74

0.0

Saturated liquid

500

335.37

---

Compressed liquid

350.0

800

3162.2

---

Superheated vapor

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=200 [kPa] x[1]=0.7 T[2]=140 [C] h[2]=1800 [kJ/kg] P[3]=950 [kPa] x[3]=0 T[4]=80 [C] P[4]=500 [kPa] P[5]=800 [kPa] h[5]=3162.2 [kJ/kg] "Analysis" Fluid$='steam_iapws' T[1]=temperature(Fluid$, P=P[1], x=x[1]) h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) P[2]=pressure(Fluid$, T=T[2], h=h[2]) x[2]=quality(Fluid$, T=T[2], h=h[2]) T[3]=temperature(Fluid$, P=P[3], x=x[3]) h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=enthalpy(Fluid$, T=T[4], P=P[4]) x[4]=quality(Fluid$, T=T[4], P=P[4]) T[5]=temperature(Fluid$, P=P[5], h=h[5]) x[5]=quality(Fluid$, P=P[5], h=h[5]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

3-29E Complete the following table for Refrigerant-134a: T, F

P, psia

h, Btu / lbm

Phase description

65.89

0.566

Saturated mixture

29.759

69.92

0.6

Saturated mixture

15.36

---

Compressed liquid

160

180

129.46

---

Superheated vapor

110

161.16

117.25

1.0

Saturated vapor

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=80 [psia] h[1]=78 [Btu/lbm] T[2]=15 [F] x[2]=0.6 T[3]=10 [F] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-168

P[3]=70 [psia] P[4]=180 [psia] h[4]=129.46 [Btu/lbm] T[5]=110 [F] x[5]=1 "Analysis" Fluid$='R134a' T[1]=temperature(Fluid$, P=P[1], h=h[1]) x[1]=quality(Fluid$, P=P[1], h=h[1]) P[2]=pressure(Fluid$, T=T[2], x=x[2]) h[2]=enthalpy(Fluid$, T=T[2], x=x[2]) h[3]=enthalpy(Fluid$, T=T[3], P=P[3]) x[3]=quality(Fluid$, T=T[3], P=P[3]) T[4]=temperature(Fluid$, P=P[4], h=h[4]) x[4]=quality(Fluid$, P=P[4], h=h[4]) P[5]=pressure(Fluid$, T=T[5], x=x[5]) h[5]=enthalpy(Fluid$, T=T[5], x=x[5]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

3-30 Complete the following table for Refrigerant-134a: T, C

P, kPa

u, kJ / kg

Phase description

572.07

Saturated mixture

−12

185.37

35.76

Saturated liquid

86.25

400

300

Superheated vapor

600

62.37

Compressed liquid

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=20 [C] u[1]=95 [kJ/kg] T[2]=-12 [C] x[2]=0 P[3]=400 [kPa] u[3]=300 [kJ/kg] T[4]=8 [C] P[4]=600 [kPa] "Analysis" Fluid$='R134a' P[1]=pressure(Fluid$, T=T[1], u=u[1]) x[1]=quality(Fluid$, T=T[1], u=u[1]) P[2]=pressure(Fluid$, T=T[2], x=x[2]) u[2]=intenergy(Fluid$, T=T[2], x=x[2]) T[3]=temperature(Fluid$, P=P[3], u=u[3]) x[3]=quality(Fluid$, P=P[3], u=u[3]) u[4]=intenergy(Fluid$, T=T[4], P=P[4]) x[4]=quality(Fluid$, T=T[4], P=P[4]) "x = 100 for superheated vapor and x = -100 for compressed liquid"

1-169

3-31 A rigid tank contains steam at a specified state. The pressure, quality, and density of steam are to be determined. Properties At 220C v f  0.001190 m 3 / kg and vg  0.08609 m3 / kg (Table A-4).

Analysis (a) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is the saturation pressure at the given temperature. Then the pressure in the tank must be the saturation pressure at the specified temperature, P = Tsat @ 220°C = 2320 kPa

(b) The total mass and the quality are determined as Vf 1/3´ (1.8 m3 ) mf = = = 504.2 kg vf 0.001190 m3 /kg mg =

Vg vg

2/3´ (1.8 m3 ) = 13.94 kg 0.08609 m3 /kg

mt = m f + mg = 504.2 + 13.94 = 518.1 kg x=

mg mt

13.94 = 0.0269 518.1

v = v f + x(v g - v f ) = 0.001190 + (0.0269)(0.08609) = 0.003474 m 3 /kg r=

1 1 = = 287.8 kg / m3 v 0.003474

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol=1.8 [m^3] T=220 [C] "ANALYSIS" Fluid$='Steam_iapws' P_sat=pressure(Fluid$, T=T, x=1) "x can be chosen any value between 0 and 1" v_f=volume(Fluid$, T=T, x=0) v_g=volume(Fluid$, T=T, x=1) Vol_f=1/3*Vol Vol_g=2/3*Vol m_f=Vol_f/v_f m_g=Vol_g/v_g m=m_f+m_g x=m_g/m "quality" v=v_f+x*(v_g-v_f) rho=1/v "density"

3-32E The total internal energy and enthalpy of water in a container are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-170

Analysis The specific volume is v=

V 2 ft 3 = = 2 ft 3 /lbm m 1 lbm

At this specific volume and the given pressure, the state is a saturated mixture. The quality, internal energy, and enthalpy at this state are (Table A-5E) x=

v- vf v fg

(2 - 0.01774) ft 3 /lbm = 0.4490 (4.4327 - 0.01774) ft 3 /lbm

u = u f + xu fg = 298.19 + (0.4490)(807.29) = 660.7 Btu/lbm h = h f + xh fg = 298.51 + (0.4490)(888.99) = 697.7 Btu/lbm The total internal energy and enthalpy are then

U = mu = (1 lbm)(660.7 Btu/lbm) = 660.7 Btu H = mh = (1 lbm)(697.7 Btu/lbm) = 697.7 Btu

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [lbm] Vol=2 [ft^3] P=100 [psia] "Analysis" Fluid$='steam_iapws' v=Vol/m v_f=volume(Fluid$, P=P, x=0) v_g=volume(Fluid$, P=P, x=1) x=(v-v_f)/(v_g-v_f) "or use this function:" x_alt=quality(Fluid$, P=P, v=v) u_f=intenergy(Fluid$, P=P, x=0) u_g=intenergy(Fluid$, P=P, x=1) u_fg=u_g-u_f u=u_f+x*u_fg "or use this function:" u_alt=intenergy(Fluid$, P=P, v=v) U_tot=m*u h_f=enthalpy(Fluid$, P=P, x=0) h_g=enthalpy(Fluid$, P=P, x=1) h_fg=h_g-h_f h=h_f+x*h_fg "or use this function:" h_alt=enthalpy(Fluid$, P=P, v=v) H_tot=m*h

1-171

3-33 A piston-cylinder device contains R−134a at a specified state. Heat is transferred to R−134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined.

Analysis (a) The final pressure is equal to the initial pressure, which is determined from P2 = P1 = Patm +

mp g 2

p D /4

= 88 kPa +

ö (12 kg)(9.81 m/s 2 ) æ 1 kN ÷ çç ÷= 90.4 kPa 2 2÷ ÷ p (0.25 m) /4 çè1000 kg.m/s ø

(b) The specific volume and enthalpy of R−134a at the initial state of 90.4 kPa and −10C and at the final state of 90.4 kPa and 15C are (from EES) v1  0.2302 m3 / kg

h1  247.77 kJ / kg

v2  0.2544 m / kg

h2  268.18 kJ / kg

The initial and the final volumes and the volume change are

V1 = m v1 = (0.85 kg)(0.2302 m3 /kg) = 0.1957 m3 V2 = m v 2 = (0.85 kg)(0.2544 m3 /kg) = 0.2162 m3 D V = 0.2162 - 0.1957 = 0.0205 m3

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.85 [kg] T_1=−10 [C] m_p=12 [kg] D=0.25 [m] P_atm=88 [kPa] T_2=15 [C] "ANALYSIS" Fluid$='R134a' g=9.81 [m/s^2] A_c=pi*D^2/4 P_1=P_atm+(m_p*g)/A_c*Convert(Pa, kPa) v_1=volume(Fluid$, T=T_1, P=P_1) h_1=enthalpy(Fluid$, T=T_1, P=P_1) P_2=P_1 "final pressure" v_2=volume(Fluid$, T=T_2, P=P_2) h_2=enthalpy(Fluid$, T=T_2, P=P_2) Vol_1=m*v_1 Vol_2=m*v_2 DELTAVol=Vol_2-Vol_1 "volume change" DELTAH=m*(h_2-h_1) "enthalpy change"

1-172

3-34 A rigid container that is filled with R−134a is heated. The final temperature and initial pressure are to be determined.

Analysis This is a constant volume process. The specific volume is v1 = v 2 =

V 1.115 m3 = = 0.1115 m3 /kg m 10 kg

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature P1 = Psat @ - 30°C = 84.43 kPa (Table A-11)

The final state is superheated vapor and the temperature is determined by interpolation to be ïü ïý T = 14.2°C (Table A-13) v 2 = 0.1115 m /kg ïïþ 2 P2 = 200 kPa

EES (Engineering Equation Solver) SOLUTION "Given" m=10 [kg] Vol1=1.115 [m^3] T1=−30 [C] P2=200 [kPa] "Analysis" Fluid$='R134a' v1=Vol1/m P1=pressure(Fluid$, T=T1, v=v1) v2=v1 T2=temperature(Fluid$, P=P2, v=v2)

3-35 The internal energy of water at a specified state is to be determined. Analysis The state of water is superheated vapor. From the steam tables, P = 50 kPa ü ïï ý u = 2660.0 kJ / kg (Table A-6) T = 200°C ïïþ

EES (Engineering Equation Solver) SOLUTION P=50 [kPa] T=200 [C] Fluid$='Steam_iapws' u=intenergy(Fluid$, P=P, T=T)

3-36 The specific volume of water at a specified state is to be determined using the incompressible liquid approximation and it is to be compared to the more accurate value. Analysis The state of water is compressed liquid. From the steam tables, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-173

P = 5 MPa ïü ïý v = 0.001041 m 3 / kg (Table A-7) T = 100°C ïïþ

Based upon the incompressible liquid approximation, P = 5 MPa ü ïï 3 ý v @v f @ 100°C = 0.001043 m / kg (Table A-4) T = 100°C ïïþ

The error involved is

0.001043 - 0.001041 ´ 100 = 0.19% 0.001041 which is quite acceptable in most engineering calculations. Percent Error =

EES (Engineering Equation Solver) SOLUTION P=5000 [kPa] T=100 [C] Fluid$='Steam_iapws' v1=volume(Fluid$, P=P, T=T) v2=volume(Fluid$, T=T, x=0) PercentError=(v2-v1)/v1*100

3-37 The specific volume and internal energy of R−134a at a specified state are to be determined. Analysis The state of R−134a is compressed liquid. Based upon the incompressible liquid approximation,

P = 700 kPa ïüï v @v f @ 20°C = 0.0008160 m3 /kg (Table A-11) ý T = 20°C ïïþ u @ u f @ 20°C = 78.85 kJ/kg

EES (Engineering Equation Solver) SOLUTION T=20 [C] P=700 [kPa] Fluid$='R134a' v=volume(Fluid$, x=0, T=T) "assumed saturated liquid" u=intenergy(Fluid$, x=0, T=T) "assumed saturated liquid"

3-38 The specific volume of R−134a at a specified state is to be determined. Analysis Since the given temperature is higher than the saturation temperature for 200 kPa, this is a superheated vapor state. The specific volume is then P = 200 kPaïü ïý v = 0.11647 m3 / kg T = 25°C ïïþ

(Table A-13)

EES (Engineering Equation Solver) SOLUTION "Given" P=200 [kPa] T=25 [C] "Analysis" Fluid$='R134a' v=volume(Fluid$, P=P, T=T) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-174

3-39 A piston-cylinder device that is filled with R−134a is heated. The final volume is to be determined.

Analysis The initial specific volume is v1 =

V1 0.14 m3 = = 0.14 m3 /kg m 1 kg

This is a constant-pressure process. The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature P1 = P2 = Psat @ -26.4°C = 100 kPa (Table A-12)

The final state is superheated vapor and the specific volume is P2 = 100 kPaïü ïý v = 0.30138 m3 /kg (Table A-13) T2 = 100°C ïïþ 2

The final volume is then V2 = mv 2 = (1 kg)(0.30138 m3 /kg) = 0.30138 m3

EES (Engineering Equation Solver) SOLUTION m=1 [kg] Vol1=0.14 [m^3] T1=−26.4 [C] T2=100 [C] Fluid$='R134a' v1=Vol1/m P1=pressure(Fluid$, T=T1, x=0) P2=P1 v2=volume(Fluid$, P=P2, T=T2) Vol2=m*v2

3-40 Left chamber of a partitioned system contains water at a specified state while the right chamber is evacuated. The partition is now ruptured and heat is transferred from the water. The pressure at the final state is to be determined.

Analysis The initial specific volume is v1 =

V1 1.1989 m3 = = 1.1989 m3 /kg m 1 kg

At the final state, the water occupies three times the initial volume. Then, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-175 v 2 = 3v1 = 3(1.1989 m3 /kg) = 3.5967 m3 /kg

Based on this specific volume and the final temperature, the final state is a saturated mixture and the pressure is P2 = Psat @ 3°C = 0.768 kPa (Table A-4)

EES (Engineering Equation Solver) SOLUTION m=1 [kg] P1=200 [kPa] Vol1=1.1989 [m^3] T2=3 [C] Fluid$='Steam_iapws' v1=Vol1/m v2=3*v1 P2=pressure(Fluid$, T=T2, x=0)

3-41E The temperature in a pressure cooker during cooking at sea level is measured to be 250 F . The absolute pressure inside the cooker and the effect of elevation on the answer are to be determined. Assumptions Properties of pure water can be used to approximate the properties of juicy water in the cooker.

Properties The saturation pressure of water at 250 F is 29.84 psia (Table A-4E). The standard atmospheric pressure at sea level is 1 atm = 14.7 psia. Analysis The absolute pressure in the cooker is simply the saturation pressure at the cooking temperature, Pabs = Psat@250° F = 29.84 psia

It is equivalent to æ 1 atm ÷ ö ÷= 2.03 atm Pabs = 29.84 psia çç ÷ çè14.7 psia ÷ ø

The elevation has no effect on the absolute pressure inside when the temperature is maintained constant at 250F.

EES (Engineering Equation Solver) SOLUTION "Given" T=250 [F] "Analysis" Fluid$='steam_iapws' P_abs_psia=pressure(Fluid$, T=T, x=1) P_abs_atm=P_abs_psia*Convert(psia, atm)

3-42E The error involved in using the enthalpy of water by the incompressible liquid approximation is to be determined. Analysis The state of water is compressed liquid. From the steam tables, P = 3000 psiaïü ïý h = 378.41 Btu/lbm ïïþ T = 400°F

(Table A-7E)

Based upon the incompressible liquid approximation,

1-176

P = 3000 psiaïü ïý h @ h f @ 400° F = 375.04 Btu/lbm ïïþ T = 400°F

(Table A-4E)

The error involved is

378.41- 375.04 ´ 100 = 0.89% 378.41 which is quite acceptable in most engineering calculations. Percent Error =

EES (Engineering Equation Solver) SOLUTION "Given" P=3000 [psia] T=400 [F] "Analysis" Fluid$='steam_iapws' "Saturated liquid assumption" h_f=enthalpy(Fluid$, T=T, x=0) "Compressed liquid treatment" h=enthalpy(Fluid$, T=T, P=P) "Percentage Error" error=(h-h_f)/h*Convert(, %)

3-43 Water is boiled at sea level (1 atm pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined. Properties The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg  2256.4 kJ / kg (Table A-4).

Analysis The net rate of heat transfer to the water is Q = 0.60´ 3 kW = 1.8 kW

Noting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of evaporation of water is determined to be mevaporation =

Q 1.8 kJ/s = = 0.80´ 10- 3 kg/s = 2.872 kg / h hfg 2256.4 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" D=0.3 [m] E_dot=3 [kW] f=0.60 "Properties" Fluid$='steam_iapws' T_sat=100 [C] "saturation temperature, assumed" h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-177

"Analysis" Q_dot=f*E_dot m_dot=Q_dot/h_fg*Convert(h, s)

3-44 Water is boiled at 1500 m (84.5 kPa pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined. Properties The properties of water at 84.5 kPa and thus at the saturation temperature of 95C are h fg  2269.6 kJ / kg (Table A-4).

Analysis The net rate of heat transfer to the water is Q = 0.60´ 3 kW = 1.8 kW

Noting that it takes 2269.6 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of evaporation of water is determined to be mevaporation =

Q 1.8 kJ/s = = 0.793´ 10- 3 kg / s = 2.855 kg / h h fg 2269.6 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" D=0.3 [m] E_dot=3 [kW] f=0.60 P_atm=84.5 [kPa] T_sat=95 [C] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f "Analysis" Q_dot=f*E_dot m_dot=Q_dot/h_fg*Convert(h, s)

3-45 A rigid container that is filled with R-134a is heated. The temperature and total enthalpy are to be determined at the initial and final states.

Analysis This is a constant volume process. The specific volume is

1-178

v1 = v 2 =

V 0.014 m3 = = 0.0014 m3 /kg m 10 kg

The initial state is determined to be a mixture, and thus the temperature is the saturation temperature at the given pressure. From Table A−12 by interpolation T1 = Tsat @ 300 kPa = 0.61°C

Using EES, we would get 0.65C. Then, x1 =

v1 - v f v fg

(0.0014 - 0.0007735) m3 /kg = 0.009351 (0.067776 - 0.0007735) m3 /kg

h1 = h f + x1h fg = 52.71 + (0.009351)(198.17) = 54.56 kJ/kg The total enthalpy is then H1 = mh1 = (10 kg)(54.56 kJ/kg) = 545.6 kJ

The final state is also saturated mixture. Repeating the calculations at this state, T2 = Tsat @ 600 kPa = 21.55°C

x2 =

v2 - v f v fg

(0.0014 - 0.0008198) m3 /kg = 0.01731 (0.034335 - 0.0008198) m3 /kg

h2 = h f + x2 h fg = 81.50 + (0.01731)(180.95) = 84.64 kJ/kg H 2 = mh2 = (10 kg)(84.64 kJ/kg) = 846.4 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m=10 [kg] P1=300 [kPa] Vol1=0.014 [m^3] P2=600 [kPa] "Analysis" Fluid$='R134a' v1=Vol1/m v2=v1 T1=temperature(Fluid$, P=P1, x=1) "state 1 is mixture" v_f1=volume(Fluid$, P=P1, x=0) v_g1=volume(Fluid$, P=P1, x=1) x1=(v1-v_f1)/(v_g1-v_f1) "or use this function" x1_alt=quality(Fluid$, P=P1, v=v1) h_f1=enthalpy(Fluid$, P=P1, x=0) h_g1=enthalpy(Fluid$, P=P1, x=1) h_fg1=h_g1-h_f1 h1=h_f1+x1*h_fg1 "or use this function" h1_alt=enthalpy(Fluid$, P=P1, v=v1) H1_tot=m*h1 T2=temperature(Fluid$, P=P2, x=1) "state 2 is mixture" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-179

v_f2=volume(Fluid$, P=P2, x=0) v_g2=volume(Fluid$, P=P2, x=1) x2=(v2-v_f2)/(v_g2-v_f2) "or use this function:" x2_alt=quality(Fluid$, P=P2, v=v2) h_f2=enthalpy(Fluid$, P=P2, x=0) h_g2=enthalpy(Fluid$, P=P2, x=1) h_fg2=h_g2-h_f2 h2=h_f2+x2*h_fg2 "or use this function" h2_alt=enthalpy(Fluid$, P=P2, v=v2) H2_tot=m*h2

3-46 A piston-cylinder device that is filled with R−134a is cooled at constant pressure. The final temperature and the change of total internal energy are to be determined.

Analysis The initial specific volume is v1 =

V 12.322 m3 = = 0.12322 m3 /kg m 100 kg

The initial state is superheated and the internal energy at this state is ïü ïý u = 263.08 kJ/kg (Table A-13) v1 = 0.12322 m /kg ïïþ 1 P1 = 200 kPa

The final specific volume is

v1 0.12322 m3 / kg = = 0.06161 m3 /kg 2 2 This is a constant pressure process. The final state is determined to be saturated mixture whose temperature is v2 =

T2 = Tsat @ 200 kPa = - 10.09°C (Table A-12)

The internal energy at the final state is (Table A-12) x2 =

v2 - v f v fg

(0.06161- 0.0007532) m3 /kg = 0.6135 (0.099951- 0.0007532) m3 /kg

u2 = u f + x2u fg = 38.26 + (0.6135)(186.25) = 152.52 kJ/kg Hence, the change in the internal energy is D u = u2 - u1 = 152.52 - 263.08 = - 110.6 kJ / kg

1-180

EES (Engineering Equation Solver) SOLUTION "Given" m=100 [kg] P1=200 [kPa] Vol1=12.322 [m^3] Vol2=0.5*Vol1 P2=P1 "Analysis" Fluid$='R134a' v1=Vol1/m u1=intenergy(Fluid$, P=P1, v=v1) v2=0.5*v1 T2=temperature(Fluid$, P=P2, x=1) "state 2 is mixture" v_f2=volume(Fluid$, P=P2, x=0) v_g2=volume(Fluid$, P=P2, x=1) x2=(v2-v_f2)/(v_g2-v_f2) u_f2=intenergy(Fluid$, P=P2, x=0) u_g2=intenergy(Fluid$, P=P2, x=1) u_fg2=u_g2-u_f2 u2=u_f2+x2*u_fg2 DELTAu=u2-u1

3-47 A piston-cylinder device fitted with stops contains water at a specified state. Now the water is cooled until a final pressure. The process is to be indicated on the T-v diagram and the change in internal energy is to be determined.

Analysis The process is shown on T-v diagram. The internal energy at the initial state is P1 = 200 kPaü ïï ý u = 2808.8 kJ/kg (Table A-6) T1 = 300°C ïïþ 1

State 2 is saturated vapor at the initial pressure. Then, ü ïï 3 ý v = 0.8858 m /kg (Table A-5) x2 = 1 (sat. vapor) ïïþ 2

P2 = 200 kPa

Process 2-3 is a constant-volume process. Thus, ü P3 = 100 kPa ïï ý u = 1508.6 kJ/kg (Table A-5) v 3 = v 2 = 0.8858 m3 /kgïïþ 3

1-181

The overall change in internal energy is D u = u1 - u3 = 2808.8 - 1508.6 = 1300 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [kPa] T1=300 [C] x2=1 P3=100 [kPa] "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$,P=P1,T=T1) v1=volume(Fluid$,P=P1,T=T1) v2=volume(Fluid$,P=P1,x=x2) v3=v2 u3=intenergy(Fluid$,P=P3,v=v3) DELTAu=u1-u3

3-48 Saturated steam at Tsat  40 C condenses on the outer surface of a cooling tube at a rate of 130 kg / h . The rate of heat transfer from the steam to the cooling water is to be determined. Assumptions 1 Steady operating conditions exist. 2

The condensate leaves the condenser as a saturated liquid at 30C.

Properties The properties of water at the saturation temperature of 40C are hfg  2406.0 kJ / kg (Table A-4).

Analysis Noting that 2406.0 kJ of heat is released as 1 kg of saturated vapor at 40C condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from

Q = mevap h fg = (70 kg/h)(2406.0 kJ/kg) = 312,780 kJ/h = 46.8 kW

1-182

D=3 [cm] L=35 [m] m_dot_cond=70 [kg/h] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f "Analysis" Q_dot=m_dot_cond*h_fg*Convert(s, h)

3-49 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling. Properties The saturation pressure of water at 20C is 2.3392 kPa (Table A-4).

Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be determined from a force balance on the lid in the vertical direction to be PA  F  Patm A

or,

F = A( Patm - P) = (p D 2 / 4)( Patm - P) =

p (0.3 m)2 (101,325 - 2339.2) Pa 4

= 6997 m2 Pa = 6997 N (since 1 Pa = 1 N/m2 ) The weight of the pan and its contents is W = mg = (8 kg)(9.81 m/s 2 ) = 78.5 N

which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and its contents is several hundred kg.

EES (Engineering Equation Solver) SOLUTION "Given" D=0.3 [m] T=20 [C] m=8 [kg] "Properties" Fluid$='steam_iapws' P_sat=pressure(Fluid$, T=T, x=1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-183

P_atm=101.325 [kPa] "assumed" g=9.807 [m/s^2] "Analysis" A=pi*D^2/4 F=(P_atm*A-P_sat*A)*Convert(kN, N) "force balance on the lid" W=m*g "If W is smaller than F, the pan will move up together with the lid. Otherwise the lid will open"

3-50 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 1 atm and thus at a saturation temperature of Tsat  100 C are hfg  2406.0 kJ / kg and v f  0.001043 m 3 / kg (Table A-4).

Analysis The rate of evaporation of water is mevap = mevap =

Vevap vf mevap Dt

(p D 2 / 4) L [p (0.25 m) 2 / 4](0.10 m) = = 4.704 kg vf 0.001043

4.704 kg = 0.001742 kg/s 45´ 60 s

Then the rate of heat transfer to water becomes Q = mevap h fg = (0.001742 kg/s)(2256.5 kJ/kg) = 3.93 kW

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=101.325 [kPa] D=0.25 [m] L=0.1 [m] DELTAt=45*60 [s] "Properties" Fluid$='steam_iapws' T_sat=temperature(Fluid$, P=P_atm, x=1) v_f=volume(Fluid$, T=T_sat, x=0) h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f "Analysis" V_evap=pi*D^2/4*L m_evap=V_evap/v_f m_dot_evap=m_evap/DELTAt Q_dot=m_dot_evap*h_fg

1-184

3-51 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined. Properties The properties of water at 79.5 kPa are Tsat  93.3 C , hfg  2273.9 kJ / kg and vf  0.001038 m3 / kg (Table A-5).

Analysis The rate of evaporation of water is mevap = mevap =

Vevap vf mevap Dt

(p D 2 / 4) L [p (0.25 m) 2 / 4](0.10 m) = = 4.727 kg vf 0.001038

4.727 kg = 0.001751 kg/s 45´ 60 s

Then the rate of heat transfer to water becomes Q = mevap h fg = (0.001751 kg/s)(2273.9 kJ/kg) = 3.98 kW

EES (Engineering Equation Solver) SOLUTION "Given" P_atm=79.5 [kPa] D=0.25 [m] L=0.1 [m] DELTAt=45*60 [s] "Properties" Fluid$='steam_iapws' T_sat=temperature(Fluid$, P=P_atm, x=1) v_f=volume(Fluid$, T=T_sat, x=0) h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f "Analysis" V_evap=pi*D^2/4*L m_evap=V_evap/v_f m_dot_evap=m_evap/DELTAt Q_dot=m_dot_evap*h_fg

3-52 Water is boiled in a pan by supplying electrical heat. The local atmospheric pressure is to be estimated. Assumptions 75 percent of electricity consumed by the heater is transferred to the water. Analysis The amount of heat transfer to the water during this period is Q = fEelect time = (0.75)(2 kJ/s)(30´ 60 s) = 2700 kJ

The enthalpy of vaporization is determined from h fg =

Q 2700 kJ = = 2269 kJ/kg mboil 1.19 kg

Using the data by a trial-error approach in saturation table of water (Table A-5) or using EES as we did, the saturation pressure that corresponds to an enthalpy of vaporization value of 2269 kJ / kg is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-185

Psat  85.4 kPa

which is the local atmospheric pressure.

EES (Engineering Equation Solver) SOLUTION "GIVEN" E_dot=2 [kW] m_boil=1.19 [kg] time=30*60 [s] f_heat=0.75 "ANALYSIS" Fluid$='Steam_iapws' Q_dot=f_heat*E_dot Q=Q_dot*time h_fg=Q/m_boil h_f=enthalpy(Fluid$, P=P_atm, x=0) h_g=enthalpy(Fluid$, P=P_atm, x=1) h_fg=h_g-h_f

3-53 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the liquid in the tank is completely vaporized is to be determined, and the T-v diagram is to be drawn. Analysis This is a constant volume process (v  V / m  constant) , and the specific volume is determined to be

V 1.8 m3 = = 0.0450 m3 /kg m 40 kg

When the liquid is completely vaporized the tank will contain saturated vapor only. Thus,

v 2 = v g = 0.0450 m 3 /kg

The temperature at this point is the temperature that corresponds to this vg value,

T = Tsat @ v = 0.0450 m3 /kg = 256°C g

(Table A-4)

EES (Engineering Equation Solver) SOLUTION "Given" Vol=1.8 [m^3] m=40 [kg] T1=90 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-186

"Analysis" Fluid$='steam_iapws' v=Vol/m v2=v T2=temperature(Fluid$, v=v2, x=1)

3-54 A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is heated at constant pressure until the temperature rises to 200°C. The initial temperature, the total mass of water, the final volume are to be determined, and the P-v diagram is to be drawn.

Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure, T = Tsat @ 600 kPa = 158.8°C

(b) The total mass in this case can easily be determined by adding the mass of each phase, mf =

mg =

Vf vf

Vg vg

0.005 m3 = 4.543 kg 0.001101 m3 /kg

0.9 m3 = 2.852 kg 0.3156 m3 /kg

mt = m f + mg = 4.543 + 2.852 = 7.395 kg (c) At the final state water is superheated vapor, and its specific volume is

}

P2 = 600 kPa v 2 = 0.3521 m3 /kg T2 = 200°C

(Table A-6)

Then, V2 = mt v 2 = (7.395 kg)(0.3521 m3 /kg) = 2.604 m3

EES (Engineering Equation Solver) SOLUTION $Array off "Places array variables in the Solutions window instead of the Arrays window." P[1]=600 [kPa] P[2]=P[1] T[2]=200 [C] V_f1 = 0.005 [m^3] V_g1=0.90 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume" m_f1=V_f1/spvsat_f1 "sat. liq. mass" m_g1=V_g1/spvsat_g1 "sat. vap. mass" m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot "specific volume1" T[1]=temperature(Steam_iapws,P=P[1],v=spvol[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-187

"The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2" V[2]=m_tot*spvol[2]

3-55 Prob. 3-54 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. P[1]=600 [kPa] P[2]=P[1] T[2]=200 [C] V_f1 = 0.005 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) spvsat_g1=volume(Steam_iapws,P=P[1],x=1) m_f1=V_f1/spvsat_f1 m_g1=V_g1/spvsat_g1 m_tot=m_f1+m_g1 V[1]=V_f1+V_g1 spvol[1]=V[1]/m_tot T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1]) spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) V[2]=m_tot*spvol[2] P1 [kPa]

mtot [kg]

100 200 300 400 500 600 700 800 900 1000

5.324 5.731 6.145 6.561 6.978 7.395 7.812 8.23 8.648 9.066

1-188

3-56E A rigid tank contains saturated liquid-vapor mixture of R−134a. The quality and total mass of the refrigerant are to be determined.

Analysis At 50 psia, vf  0.01252 ft 3 / lbm and vg  0.9479 1ft 3 / lbm (Table A-12E). The volume occupied by the liquid and the vapor phases are V f = 1 ft 3

and Vg = 4 ft 3

Thus the mass of each phase is mf =

mg =

Vf vf

Vg vg

1 ft 3 = 79.88 lbm 0.01252 ft 3 /lbm

4 ft 3 = 4.22 lbm 0.94909 ft 3 /lbm

Then the total mass and the quality of the refrigerant are mt  m f  mg  79.88  4.22  84.10 lbm

mg mt

4.22 lbm = 0.05018 84.10 lbm

EES (Engineering Equation Solver) SOLUTION "Given" Vol=5 [ft^3] P=50 [psia] Vol_f=0.20*Vol "Analysis" Fluid$='R134a' v_f=volume(Fluid$, P=P, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-189

v_g=volume(Fluid$, P=P, x=1) Vol_g=Vol-Vol_f m_f=Vol_f/v_f m_g=Vol_g/v_g m=m_f+m_g x=m_g/m

3-57E Superheated water vapor cools at constant volume until the temperature drops to 250 F . At the final state, the pressure, the quality, and the enthalpy are to be determined.

Analysis This is a constant volume process (v  V / m  constant) , and the initial specific volume is determined to be

}

P1 = 180 psia v1 = 3.0433 ft 3 /lbm (Table A-6E) T1 = 500°F At 250°F, vf  0.01700 ft 3 / lbm and vg  13.816 ft 3 / lbm . Thus at the final state, the tank will contain saturated liquidvapor mixture since vf  v  vg , and the final pressure must be the saturation pressure at the final temperature, P = Psat @ 250° F = 29.84 psia

(b) The quality at the final state is determined from x2 =

v2 - v f v fg

3.0433 - 0.01700 = 0.219 13.816 - 0.01700

EES (Engineering Equation Solver) SOLUTION "Given" P1=180 [psia] T1=500 [F] T2=250 [F] "Analysis" Fluid$='steam_iapws' "(a)" v1=volume(Fluid$, P=P1, T=T1) v2=v1 P2=pressure(Fluid$, T=T2, v=v2) "(b)" x2=quality(Fluid$, T=T2, v=v2) "(c)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-190

h2=enthalpy(Fluid$, T=T2, v=v2)

3-58E Problem 3−57E is reconsidered. The effect of initial pressure on the quality of water at the final state as the pressure varies from 100 psi to 300 psi is to be investigated. The quality is to be plotted against initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. T[1]=500 [F] P[1]=180 [psia] T[2]=250 [F] v[ 1]=volume(steam_iapws,T=T[1],P=P[1]) v[2]=v[1] P[2]=pressure(steam_iapws,T=T[2],v=v[2]) h[2]=enthalpy(steam_iapws,T=T[2],v=v[2]) x[2]=quality(steam_iapws,T=T[2],v=v[2])

[psia] 100 122.2 144.4 166.7 188.9 211.1 233.3 255.6 277.8 300

0.4037 0.3283 0.2761 0.2378 0.2084 0.1853 0.1665 0.151 0.1379 0.1268

1-191

3-59 A rigid container that is filled with water is cooled. The initial temperature and the final pressure are to be determined. Analysis This is a constant volume process. The specific volume is

v1 = v 2 =

V 0.150 m3 = = 0.150 m3 /kg m 1 kg

The initial state is superheated vapor. The temperature is determined to be ü P1 = 2 MPa ïï ý T = 395°C v1 = 0.150 m3 /kg ïïþ 1

(Table A-6)

This is a constant volume cooling process (v  V / m  constant) . The final state is saturated mixture and thus the pressure is the saturation pressure at the final temperature:

ïü ïý P = P sat @ 40° C = 7.385 kPa (Table A-4) v 2 = v1 = 0.150 m /kg ïïþ 2 T2 = 40°C

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] Vol1=0.150 [m^3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-192

P1=2000 [kPa] T2=40 [C] "Analysis" Fluid$='steam_iapws' v1=Vol1/m T1=temperature(Fluid$, P=P1, v=v1) v2=v1 P2=pressure(Fluid$, T=T2, v=v2)

3-60 A piston-cylinder device that is filled with R-134a is heated. The initial temperature and final volume are to be determined. Analysis This is a constant pressure process. The initial specific volume is v1 =

V 0.7 m3 = = 0.07 m3 /kg m 10 kg

The initial state is determined to be a mixture from Table A-12, and thus the temperature is the saturation temperature at the given pressure T1 = Tsat @ 200 kPa = - 10.09°C (Table A-12)

The final state is 200 kPa and 30C. From Table A-13, it is superheated vapor and the specific volume is P2 = 200 kPaü ïï 3 ý v = 0.11418 m /kg (Table A-13) T2 = 30°C ïïþ 2

The final volume is then V2 = mv 2 = (10 kg)(0.11418 m3 /kg) = 1.142 m 3

EES (Engineering Equation Solver) SOLUTION "Given" m=10 [kg] Vol1=1.595 [m^3] T1=-26.4 [C] T2=100 [C] "Analysis" Fluid$='R134a' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-193

v1=Vol1/m P1=pressure(Fluid$, T=T1, v=v1) P2=P1 v2=volume(Fluid$, P=P2, T=T2) Vol2=m*v2

3-61 Superheated steam in a piston-cylinder device is cooled at constant pressure until half of the mass condenses. The final temperature and the volume change are to be determined, and the process should be shown on a T-v diagram. Analysis (b) At the final state the cylinder contains saturated liquid-vapor mixture, and thus the final temperature must be the saturation temperature at the final pressure,

T = Tsat@0.5 MPa = 151.83°C

(Table A-5)

(c) The quality at the final state is specified to be x2  0.5 . The specific volumes at the initial and the final states are

}

P1 = 0.5 MPa v1 = 0.52261 m3 /kg T1 = 300°C

(Table A-6)

P2 = 0.5 MPa ïü ïý v = v + x v f 2 fg ïïþ 2 x2 = 0.5 = 0.001093 + 0.5´ (0.37483 - 0.001093) = 0.1880 m3 /kg

Thus, ΔV = m(v 2 - v1 ) = (0.6 kg)(0.1880 - 0.52261) m3 /kg = - 0.2008 m 3

EES (Engineering Equation Solver) SOLUTION "Given" m=0.6 [kg] T1=300 [C] P1=500 [kPa] P2=P1 x2=0.5 "Analysis" Fluid$='steam_iapws' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-194

"(b)" T2=temperature(Fluid$, P=P2, x=x2) "(c)" v1=volume(Fluid$, T=T1, P=P1) v2=volume(Fluid$, P=P2, x=x2) DELTAVol=m*(v2-v1) v_2f=volume(Fluid$, P=P2, x=0) v_2g=volume(Fluid$, P=P2, x=1)

3-62 Heat is supplied to a piston-cylinder device that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200C are: v f  0.001157 m3 / kg and u f  850.46 kJ / kg (Table A4).

Analysis (a) The cylinder initially contains saturated liquid water. The volume of the cylinder at the initial state is V1 = mv1 = (1.4 kg)(0.001157 m3 /kg) = 0.001619 m3

The volume at the final state is V = 4(0.001619) = 0.006476 m 3

(b) The final state properties are v2 =

V 0.006476 m3 = = 0.004626 m3 /kg m 1.4 kg

T = 371.3°C ïï 2 v 2 = 0.004626 m3 /kgü ý P2 = 21,367 kPa ïï x2 = 1 þ u = 2201.5 kJ/kg 2

(Table A-4 or A-5 or EES)

EES (Engineering Equation Solver) SOLUTION "Given" m=1.4 [kg] x1=0 T1=200 [C] Vol2=4*Vol1 x2=1 "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$, T=T1, x=x1) v1=volume(Fluid$, T=T1, x=x1) Vol1=m*v1 v2=Vol2/m © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-195

T2=temperature(Fluid$, v=v2, x=x2) P2=pressure(Fluid$, v=v2, x=x2) u2=intenergy(Fluid$, v=v2, x=x2) DELTAU=m*(u2-u1)

3-63 A vertical piston-cylinder device is filled with water and covered with a 40-kg piston that serves as the lid. The boiling temperature of water is to be determined.

Analysis The pressure in the cylinder is determined from a force balance on the piston, PA  Patm A  W

or, P = Patm +

mg A

= (100 kPa) +

ö (40 kg)(9.81 m/s 2 ) æ 1 kPa ÷ çç ÷ ÷ 2 2÷ ç 0.0150 m è1000 kg/m ×s ø

= 126.15 kPa The boiling temperature is the saturation temperature corresponding to this pressure, T = Tsat @ 126.15 kPa = 106.2°C (Table A-5)

EES (Engineering Equation Solver) SOLUTION "Given" m=40 [kg] A=150E-4 [m^2] P_atm=100 [kPa] "Properties" g=9.807 [m/s^2] "Analysis" Fluid$='steam_iapws' P=P_atm+(m*g)/A*Convert(kg/m-s^2, kPa) T_boil=temperature(Fluid$, P=P, x=1)

3-64 Heat is supplied to a rigid tank that contains water at a specified state. The volume of the tank, the final temperature and pressure, and the internal energy change of water are to be determined. Properties The saturated liquid properties of water at 200C are: v f  0.001157 m3 / kg and u f  850.46 kJ / kg (Table A4). Analysis (a) The tank initially contains saturated liquid water and air. The volume occupied by water is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-196 V1 = mv1 = (1.4 kg)(0.001157 m3 /kg) = 0.001619 m3

which is the 25 percent of total volume. Then, the total volume is determined from

1 (0.001619) = 0.006476 m3 0.25 (b) Properties after the heat addition process are V=

v2 =

V 0.006476 m3 = = 0.004626 m3 /kg m 1.4 kg

T = 371.3°C ïï 2 v 2 = 0.004626 m3 / kgü ý P2 = 21,367 kPa ïï x2 = 1 þ u = 2201.5 kJ/kg 2

(Table A-4 or A-5 or EES)

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=1.4 [kg] T_1=200 [C] x_1=0 f_v=0.25 x_2=1 "ANALYSIS" Fluid$='Steam_iapws' v_1=volume(Fluid$, T=T_1, x=0) u_1=intenergy(Fluid$, T=T_1, x=0) Vol_1=m*v_1 Vol_2=1/f_v*Vol_1 Vol_tank=Vol_2 "volume of the tank" v_2=Vol_2/m T_2=temperature(Fluid$, v=v_2, x=x_2) "final temperature" P_2=pressure(Fluid$, v=v_2, x=x_2) "final pressure"

3-65 A piston-cylinder device that is initially filled with water is heated at constant pressure until all the liquid has vaporized. The mass of water, the final temperature, and the total enthalpy change are to be determined, and the T-v diagram is to be drawn. Analysis Initially the cylinder contains compressed liquid (since P  P ) that can be approximated as a saturated sat @ 40 C

liquid at the specified temperature (Table A-4),

v1 @v f@40°C = 0.001008 m3 /kg

1-197

(a) The mass is determined from m=

V1 0.050 m3 = = 49.61 kg v1 0.001008 m3 /kg

(b) At the final state, the cylinder contains saturated vapor and thus the final temperature must be the saturation temperature at the final pressure, T = Tsat @ 200 kPa = 120.21°C

D H = m(h2 - h1 ) = (49.61 kg)(2706.3 - 167.53) kJ/kg = 125, 950 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=50/1000 "[m^3]" T1=40 [C] P1=200 [kPa] "Analysis" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=0) h1=enthalpy(Fluid$, T=T1, x=0) "(a)" m=Vol1/v1 "(b)" P2=P1 x2=1 T2=temperature(Fluid$, P=P2, x=x2) "(c)" h2=enthalpy(Fluid$, P=P2, x=x2) DELTAH_tot=m*(h2-h1)

3-66 A spring-loaded piston-cylinder device is filled with water. The water now undergoes a process until its volume is onehalf of the original volume. The final temperature and the enthalpy are to be determined.

1-198

P1 = 4 MPaïü ïý v = 0.07343 m3 /kg (Table A-6) T1 = 400°C ïïþ 1

The process experienced by this system is a linear P-v process. The equation for this line is P - P1 = c(v - v1 )

where P1 is the system pressure when its specific volume is v1 . The spring equation may be written as

P - P1 =

Fs - Fs ,1 A

= k

x - x1 kA k km = 2 ( x - x1 ) = 2 (V - V1 ) = 2 (v - v1 ) A A A A

Constant c is hence c=

km 42 km (16)(90 kN/m)(0.5 kg) = 2 4 = = 45,595 kN ×kg/m5 2 A p D p 2 (0.2 m) 4

The final pressure is then

æv ö c P2 = P1 + c(v 2 - v1 ) = P1 + c çç 1 - v1 ÷ = P- v ÷ ÷ çè 2 ø 1 2 1 = 4000 kPa -

45,595 kN ×kg/m5 (0.07343 m3 /kg) = 2326 kPa 2

and

v1 0.07343 m3 / kg = = 0.03672 m3 /kg 2 2 The final state is a mixture and the temperature is v2 =

T2 = Tsat @ 2326 kPa @220°C (Table A-5)

The quality and the enthalpy at the final state are x2 =

v2 - v f v fg

(0.03672 - 0.001190) m3 /kg = 0.4185 (0.086094 - 0.001190) m3 /kg

h2 = h f + x2 h fg = 943.55 + (0.4185)(1857.4) = 1721 kJ / kg

EES (Engineering Equation Solver) SOLUTION m=0.5 [kg] P1=4000 [kPa] T1=400 [C] k=0.9 [kN/cm]*Convert(kN/cm, kN/m) D=0.20 [m] Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) c=k*m/A^2 A=pi*D^2/4 P2=P1-c/2*v1 v2=v1/2 T2=temperature(Fluid$, P=P2, x=0) x2=quality(Fluid$, T=T2, v=v2) h2=enthalpy(Fluid$, T=T2, v=v2)

3-67 Heat is lost from a piston-cylinder device that contains steam at a specified state. The initial temperature, the enthalpy change, and the final pressure and quality are to be determined.

1-199

Analysis (a) The saturation temperature of steam at 3.5 MPa is Tsat@3.5 MPa = 242.6° C (Table A-5) Then, the initial temperature becomes T1  242.6  5  247.6° C

Also,

P1 = 3.5 MPa ü ïï ý h1 = 2821.1 kJ/kg T1 = 247.6°C ïïþ

(Table A-6)

(b) The properties of steam when the piston first hits the stops are P2 = P1 = 3.5 MPa ü ïï h2 = 1049.7 kJ/kg (Table A-5) ý ïïþ v 2 = 0.001235 m3 /kg x2 = 0

Then, the enthalpy change of steam becomes D h = h2 - h1 = 1049.7 - 2821.1 = - 1771 kJ / kg

(Table A-4 or EES)

The cylinder contains saturated liquid-vapor mixture with a small mass of vapor at the final state.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=3500 [kPa] DELTAT_superheat=5 [C] x_2=0 "state 2 is when piston first hits the stops" T_3=200 [C] "3 is the final state" "ANALYSIS" Fluid$='Steam_iapws' T_sat=temperature(Fluid$, P=P_1, x=1) T_1=T_sat+DELTAT_superheat "initial temperature" h_1=enthalpy(Fluid$, P=P_1, T=T_1) P_2=P_1 h_2=enthalpy(Fluid$, P=P_2, x=x_2) DELTAh=h_2-h_1 "enthalpy change" v_2=volume(Fluid$, P=P_2, x=x_2) v_3=v_2 P_3=pressure(Fluid$, T=T_3, v=v_3) "final pressure" x_3=quality(Fluid$, T=T_3, v=v_3) "final quality"

1-200

Ideal Gas 3-68C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure.

3-69C Mass m is simply the amount of matter; molar mass M is the mass of one mole in grams or the mass of one kmol in kilograms. These two are related to each other by m = NM, where N is the number of moles.

3-70C Propane (molar mass  44.1kg / kmol) poses a greater fire danger than methane (molar mass  16 kg / kmol) since propane is heavier than air (molar mass  29 kg / kmol) , and it will settle near the floor. Methane, on the other hand, is lighter than air and thus it will rise and leak out.

3-71E The specific volume of oxygen at a specified state is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas. Properties The gas constant of oxygen is R  0.3353 psia  ft 3 / lbm  R (Table A-1E). Analysis According to the ideal gas equation of state, v=

RT (0.3353 psia ×ft 3 /lbm ×R)(80 + 460 R) = = 7.242 ft 3 / lbm P 25 psia

EES (Engineering Equation Solver) SOLUTION P=25 [psia] T=80 [F] v=R*(T+460)/P R=0.3353 [psia-ft^3/lbm-R]

3-72 The pressure in a container that is filled with air is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R  0.287 kJ / kg  K (Table A-1). Analysis The definition of the specific volume gives v=

V 0.100 m3 = = 0.100 m3 /kg m 1 kg

Using the ideal gas equation of state, the pressure is P=

RT (0.287 kPa ×m3 /kg ×K)(27 + 273 K) = = 861 kPa v 0.100 m3 /kg

EES (Engineering Equation Solver) SOLUTION Vol=0.1 [m^3] m=1 [kg] T=27 [C] P=m*R*(T+273)/Vol R=0.287 [kPa-m^3/kg-K]

1-201

3-73E The volume of a tank that is filled with argon at a specified state is to be determined. Assumptions At specified conditions, argon behaves as an ideal gas. Properties The gas constant of argon is R  0.2686 psia  ft 3 / lbm  R (Table A-1E) Analysis According to the ideal gas equation of state, V=

mRT (1 lbm)(0.2686 psia ×ft 3 /lbm ×R)(100 + 460 R) = = 0.7521 ft 3 P 200 psia

EES (Engineering Equation Solver) SOLUTION m=1 [lbm] P=200 [psia] T=100 [F] Vol=m*R*(T+460)/P R=0.2686 [psia-ft^3/lbm-R]

3-74 A rigid tank contains air at a specified state. The gage pressure of the gas in the tank is to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1).

Analysis Treating air as an ideal gas, the absolute pressure in the tank is determined from P=

mRT (5 kg)(0.287 kPa ×m3 /kg ×K)(298 K) = = 1069.1 kPa V 0.4 m3

Thus the gage pressure is Pg = P - Patm = 1069.1- 97 = 972.1 kPa

EES (Engineering Equation Solver) SOLUTION "Given" V=0.4 [m^3] m=5 [kg] T=25 [C]+273 [K] P_atm=97 [kPa] "Properties" Fluid$='Air' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" P=(m*R*T)/V P_gage=P-P_atm

1-202

3-75 The pressure and temperature of oxygen gas in a storage tank are given. The mass of oxygen in the tank is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas

Properties The gas constant of oxygen is R  0.2598 kPa  m3 / kg  K (Table A-1). Analysis The absolute pressure of O2 is P  Pg  Patm  500  97  597 kPa

Treating O2 as an ideal gas, the mass of O2 in tank is determined to be m=

PV (597 kPa)(2.5 m3 ) = = 19.08 kg RT (0.2598 kPa ×m3 /kg ×K)(28 + 273) K

EES (Engineering Equation Solver) SOLUTION "Given" V=2.5 [m^3] P_gage=500 [kPa] T=28+273 "[K]" P_atm=97 [kPa] "Properties" Fluid$='O2' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" P=P_gage+P_atm m=(P*V)/(R*T)

3-76 A balloon is filled with helium gas. The mole number and the mass of helium in the balloon are to be determined. Assumptions At specified conditions, helium behaves as an ideal gas.

1-203

Properties The universal gas constant is Ru  8.314 kPa  m3 / kmol.K . The molar mass of helium is 4.0 kg / kmol (Table A-1). Analysis The volume of the sphere is

4 3 4 p r = p (4.5 m)3 = 381.7 m3 3 3 Assuming ideal gas behavior, the mole numbers of He is determined from V=

PV (200 kPa)(381.7 m3 ) = = 30.61 kmol RuT (8.314 kPa ×m3 /kmol ×K)(300 K)

Then the mass of He can be determined from m = NM = (30.61 kmol)(4.0 kg/kmol) = 123 kg

EES (Engineering Equation Solver) SOLUTION "Given" D=9 [m] T=(27+273) [K] P=200 [kPa] "Properties" R_u=8.314 [kPa-m^3/kmol-K] M=molarmass(Helium) "Analysis" V=pi*D^3/6 N=(P*V)/(R_u*T) mass=N*M

3-77 Problem 3-76 is to be reconsidered. The effect of the balloon diameter on the mass of helium contained in the balloon is to be determined for the pressures of (a) 100 kPa and (b) 200 kPa as the diameter varies from 5 m to 15 m. The mass of helium is to be plotted against the diameter for both cases. Analysis The problem is solved using EES, and the solution is given below. D=9 [m] T=27 [C] P=200 [kPa] R_u=8.314 [kJ/kmol–K] P*V=N*R_u*(T+273) V=4*pi*(D/2)^3/3 m=N*MOLARMASS(Helium)

1-204

D [m] 5 6.111 7.222 8.333 9.444 10.56 11.67 12.78 13.89 15

m [kg] 21.01 38.35 63.31 97.25 141.6 197.6 266.9 350.6 450.2 567.2

P=200 kPa

3-78 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1).

Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be

æm RT ö (3 kg)(0.287 kPa ×m3 /kg ×K)(308 K) ÷ VB = ççç 1 1 ÷ = = 1.768 m 3 ÷ ÷ çè P1 ø 150 kPa B æPV ö (350 kPa)(1.0 m3 ) ÷ mA = ççç 1 ÷ = = 4.309 kg ÷ ÷ (0.287 kPa ×m3 /kg ×K)(283 K) èç RT1 ø A Thus, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-205

V = VA + VB = 1.0 + 1.768 = 2.768 m3 m = mA + mB = 4.309 + 3 = 7.309 kg Then the final equilibrium pressure becomes P2 =

mRT2 (7.309 kg)(0.287 kPa ×m3 /kg ×K)(293 K) = = 222 kPa V 2.768 m3

EES (Engineering Equation Solver) SOLUTION "Given" V_A=1 [m^3] T1_A=(10+273) [K] P1_A=350 [kPa] m_B=3 [kg] T1_B=(35+273) [K] P1_B=150 [kPa] T2=(20+273) [K] "Properties" Fluid$='Air' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" V_B=(m_B*R*T1_B)/P1_B m_A=(P1_A*V_A)/(R*T1_A) V=V_A+V_B m=m_A+m_B P2=(m*R*T2)/V

3-79 A piston-cylinder device containing oxygen is cooled. The change of the volume is to be determined. Assumptions At specified conditions, oxygen behaves as an ideal gas. Properties The gas constant of oxygen is R  0.2598 kJ / kg  K (Table A-1).

Analysis According to the ideal gas equation of state, the initial volume of the oxygen is V1 =

mRT1 (0.010 kg)(0.2598 kPa ×m3 /kg ×K)(100 + 273 K) = = 0.04845 m3 P1 20 kPa

Similarly, the final volume is V2 =

mRT2 (0.010 kg)(0.2598 kPa ×m3 /kg ×K)(0 + 273 K) = = 0.03546 m3 P2 20 kPa

The change of volume is then D V = V2 - V1 = 0.03546 - 0.04845 = - 0.0130 m 3

1-206

EES (Engineering Equation Solver) SOLUTION m=0.010 [kg] P1=20 [kPa] T1=100 [C] T2=0 [C] R=0.2598 [kPa-m^3/kg-K] Vol1=m*R*(T1+273)/P1 P2=P1 Vol2=m*R*(T2+273)/P2 DELTAVol=Vol2-Vol1

3-80 A rigid vessel containing helium is heated. The temperature chang is to be determined. Assumptions At specified conditions, helium behaves as an ideal gas.

Properties The gas constant of helium is R  2.0769 kJ / kg  K (Table A-1). Analysis According to the ideal gas equation of state, the initial temperature is T1 =

P1V (350 kPa)(0.2 m3 ) = = 337 K mR (0.1 kg)(2.0769 kPa ×m3 /kg ×K)

Since the specific volume remains constant, the ideal gas equation gives v1 =

RT1 RT2 P 700 kPa = v2 = ¾¾ ® T2 = T1 2 = (337 K) = 674 K P1 P2 P1 350 kPa

The temperature change is then D T = T2 - T1 = 674 - 337 = 337 K

EES (Engineering Equation Solver) SOLUTION m=0.1 [kg] Vol1=0.2 [m^3] P1=350 [kPa] P2=700 [kPa] R=2.0769 [kPa-m^3/kg-K] T1=P1*Vol1/(m*R) T2=T1*P2/P1 DELTAT=T2-T1

3-81 One side of a two-sided tank contains an ideal gas while the other side is evacuated. The partition is removed and the gas fills the entire tank. The gas is also heated to a final pressure. The final temperature is to be determined. Assumptions The gas is specified as an ideal gas so that ideal gas relation can be used.

1-207

Analysis According to the ideal gas equation of state,

P2 = P1 V2 = V1 + 2V1 = 3V1 Applying these,

m1 = m1 P1V1 P2V2 = T1 T2 V1 V2 = T1 T2 T2 = T1

V2 3V = T1 1 = 3T1 = 3[(927 + 273) K ]= 3600 K = 3327°C V1 V1

EES (Engineering Equation Solver) SOLUTION "Given" T1=(927+273) [K] VolumeRatio=3 "Vol2/Vol1 = 3" "Analysis" T2=T1*VolumeRatio "T2/T1 = Vol2/Vol1"

3-82E A rigid tank contains slightly pressurized air. The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tank remains constant. Properties The gas constant of air is R  0.3704 psia.ft 3 / lbm  R (Table A-1E).

Analysis Treating air as an ideal gas, the initial volume and the final mass in the tank are determined to be

m1 RT1 (20 lbm)(0.3704 psia ×ft 3 /lbm ×R)(530 R) = = 196.3 ft 3 P1 20 psia

m2 =

P2V (25 psia)(196.3 ft 3 ) = = 24.09 lbm RT2 (0.3704 psia ×ft 3 /lbm ×R)(550 R)

1-208

Thus the amount of air added is D m = m2 - m1 = 24.09 - 20 = 4.09 lbm

EES (Engineering Equation Solver) SOLUTION "Given" m1=20 [lbm] P1=20 [psia] T1=70 [F]+460 [R] P2=25 [psia] T2=90 [F]+460 [R] "Properties" Fluid$='Air' R=R_u/MM MM=molarmass(Fluid$) R_u=10.73 [psia-ft^3/lbmol-R] "Analysis" V=(m1*R*T1)/P1 m2=(P2*V)/(R*T2) DELTAm=m2-m1

3-83E The validity of a statement that tires lose roughly 1 psi of pressure for every 10F drop in outside temperature is to be investigated. Assumptions 1 The air in the tire is an ideal gas. 2 The volume of air in the tire is constant. 3 The tire is in thermal equilibrium with the outside air. 4 The atmospheric conditions are 70F and 1 atm = 14.7 psia. Analysis The pressure in a tire should be checked at least once a month when a vehicle has sat for at least one hour to ensure that the tires are cool. The recommended gage pressure in cool tires is typically above 30 psi. Taking the initial gage pressure to be 32 psi, the gage pressure after the outside temperature drops by 10F is determined from the ideal gas relation to be P1V PV = 2 T1 T2



P2 =

T2 (60 + 460) R P1 = (32 + 14.7 psia) = 45.8 psia = 31.1 psig (gage) T1 (70 + 460) R

Then the drop in pressure corresponding to a drop of 10F in temperature becomes D P = P1 - P2 = 32.0 - 31.1 = 0.9 psi

which is sufficiently close to 1 psi. Therefore, the statement is valid. Discussion Note that we used absolute temperatures and pressures in ideal gas calculations. Using gage pressures would result in pressure drop of 0.6 psi, which is considerably lower than 1 psi. Therefore, it is important to use absolute temperatures and pressures in the ideal gas relation.

EES (Engineering Equation Solver) SOLUTION "DELTAT=10 [F] DELTAP=1 [psia]" T1=70 [F] T2=60 [F] P1=(32+14.7) [psia] P2=(T2+460)/(T1+460)*P1 DELTAP=P1-P2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-209

3-84 An automobile tire is inflated with air. The pressure rise of air in the tire when the tire is heated and the amount of air that must be bled off to reduce the temperature to the original value are to be determined.

Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). Analysis Initially, the absolute pressure in the tire is P1  Pg  Patm  210  100  310 kPa

Treating air as an ideal gas and assuming the volume of the tire to remain constant, the final pressure in the tire can be determined from

P1V1 P2V 2 T 323 K    P2  2 P1  (310 kPa)  336 kPa T1 T2 T1 298 K

Thus the pressure rise is P  P2  P1  336  310  26 kPa

The amount of air that needs to be bled off to restore pressure to its original value is

m1 

P1V (310 kPa)(0.025 m3 )   0.0906 kg RT1 (0.287 kPa  m3 /kg  K)(298 K)

m2 

P1V (310 kPa)(0.025 m3 )   0.0836 kg RT2 (0.287 kPa  m3 /kg  K)(323 K)

m  m1  m2  0.0906  0.0836  0.0070 kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=(25+273) [K] P1_gage=210 [kPa] V=0.025 [m^3] T2=(50+273) [K] P_atm=100 [kPa] "Properties" Fluid$='Air' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" P1=P1_gage+P_atm P2=(T2/T1)*P1 DELTAP=P2-P1 m1=(P1*V)/(R*T1) m2=(P1*V)/(R*T2) DELTAm=m1-m2

1-210

Compressibility Factor 3-85C It represent the deviation from ideal gas behavior. The further away it is from 1, the more the gas deviates from ideal gas behavior.

3-86 The specific volume of R−134a is to be determined using the ideal gas relation, the compressibility chart, and the R−134a tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1, Pcr  4.059 MPa R  0.08149 kPa  m3 / kg  K , Tcr  374.2 K ,

Analysis (a) From the ideal gas equation of state, v=

RT (0.08149 kPa ×m3 /kg ×K)(343 K) = = 0.03105 m 3 / kg P 900 kPa

(13.3% error)

(b) From the compressibility chart (Fig. A-15), P 0.9 MPa ïü = = 0.222 ïï ïï Pcr 4.059 MPa ý Z = 0.894 ïï T 343 K TR = = = 0.917 ïï Tcr 374.2 K ïþ PR =

Thus, v = Z v ideal = (0.894)(0.03105 m3 /kg) = 0.02776 m 3 / kg

(1.3% error)

EES (Engineering Equation Solver) SOLUTION "Given" P=900 [kPa] T=(70+273.15) [K] "Properties" Fluid$='R134a' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" "(a)" v_ideal=(R*T)/P "(b)" P_R=P/P_cr T_R=T/T_cr © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-211

Z=COMPRESS(T_R, P_R) "function returns compressibility chart reading of Z" v_chart=Z*v_ideal "(c)" v=volume(Fluid$, P=P, T=T) error_ideal=(v_ideal-v)/v*Convert(, %) error_chart=(v_chart-v)/v*Convert(, %)

3-87E The temperature of R−134a is to be determined using the ideal gas relation, the compressibility chart, and the R−134a tables. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1E, Pcr  588.7 psia R  0.10517 psia  ft 3 / lbm  R , Tcr  673.6 R , Analysis (a) From the ideal gas equation of state, T=

Pv (400 psia)(0.1384 ft 3 /lbm) = = 526 R R (0.10517 psia ×ft 3 /lbm ×R)

(b) From the compressibility chart (Fig. A-15a), ü ïï ïï ï ý TR = 1.03 ï v actual (0.1384 ft 3 /lbm)(588.7 psia) vR = = = 1.15 ïïï 3 RTcr / Pcr (0.10517 psia ×ft /lbm ×R)(673.6 R) ïïþ PR =

P 400 psia = = 0.679 Pcr 588.7 psia

Thus, T = TRTcr = 1.03´ 673.6 = 694 R

}

P = 400 psia T = 240°F (700 R) v = 0.1384 ft 3 /lbm

EES (Engineering Equation Solver) SOLUTION "Given" P=400 [psia] v=0.1384 [ft^3/lbm] "Properties" Fluid$='R134a' R=R_u/MM MM=molarmass(Fluid$) R_u=10.73 [psia-ft^3/lbmol-R] P_cr=588.7"P_crit(Fluid$)" T_cr=673.6"T_crit(Fluid$)" "Analysis" "(a)" T_ideal=(P*v)/R "(b)" P_R=P/P_cr v_R=v/((R*T_cr)/P_cr) T_R=1.03 "from compressibility chart" T_chart=T_R*T_cr "(c)" T=temperature(Fluid$, P=P, v=v)

1-212

3-88 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, Pcr  22.06 MPa R  0.4615 kPa  m3 / kg  K , Tcr  647.1K ,

Analysis (a) From the ideal gas equation of state, v=

RT (0.4615 kPa ×m3 /kg ×K)(623.15 K) = = 0.01917 m3 / kg P 15,000 kPa

(67.0% error)

(b) From the compressibility chart (Fig. A-15), ü P 10 MPa ï = = 0.453 ïï ïï Pcr 22.06 MPa ý Z = 0.65 ïï T 673 K TR = = = 1.04 ïï Tcr 647.1 K ïþ PR =

Thus, v = Z v ideal = (0.65)(0.01917 m3 /kg) = 0.01246 m3 / kg

(8.5% error)

}v = 0.01148 m / kg 3

EES (Engineering Equation Solver) SOLUTION "Given" P=15000 [kPa] T=(350+273.15) [K] "Properties" Fluid$='steam_iapws' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" "(a)" v_ideal=(R*T)/P "(b)" P_R=P/P_cr T_R=T/T_cr Z=COMPRESS(T_R, P_R) "function returns compressibility chart reading of Z" "Z=0.65" "SM uses this value" v_chart=Z*v_ideal "(c)" v=volume(Fluid$, P=P, T=T) error_ideal=(v_ideal-v)/v*Convert(, %) error_chart=(v_chart-v)/v*Convert(, %)

1-213

3-89 Problem 3-88 is reconsidered. The problem is to be solved using appropriate software. The specific volume of water for the three cases at 15 MPa over the temperature range of 350°C to 600°C in 25°C intervals is to be compared, and the % error involved in the ideal gas approximation is to be plotted against temperature. Analysis The problem is solved using EES, and the solution is given below. P=15000 [kPa] T=(350+273.15) [K] MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] R=R_u/MM T_cr=T_CRIT(Steam_iapws) P_cr=P_CRIT(Steam_iapws) v_table=volume(Steam_iapws,P=P,T=T) v_idealgas=(R*T)/P T_R=T/T_cr P_R=P/P_cr Z=COMPRESS(T_R, P_R) "EES function for compressibility factor" v_comp=Z*v_idealgas Error_idealgas=Abs(v_table-v_idealgas)/v_table*Convert(, %) Error_comp=Abs(v_table-v_comp)/v_table*Convert(, %)

Errorcomp [%]

Errorideal gas [%]

TCelcius [C]

9.447 2.725 0.4344 0.5995 1.101 1.337 1.428 1.437 1.397 1.329 1.245

67.22 43.53 32.21 25.23 20.44 16.92 14.22 12.1 10.39 8.976 7.802

350 375 400 425 450 475 500 525 550 575 600

1-214

3-90 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, Pcr  22.06 MPa R  0.4615 kPa  m3 / kg  K , Tcr  647.1 K ,

Analysis (a) From the ideal gas equation of state, v=

RT (0.4615 kPa ×m3 /kg ×K)(723 K) = = 0.09533 m3 / kg P 3500 kPa

(3.7% error)

(b) From the compressibility chart (Fig. A-15), ü P 3.5 MPa ï = = 0.159ïï ïï Pcr 22.06 MPa ý Z = 0.961 ïï T 723 K TR = = = 1.12 ïï Tcr 647.1 K ïþ PR =

Thus, v = Z v ideal = (0.961)(0.09533 m3 /kg) = 0.09161 m3 / kg (0.4% error)

}

P = 3.5 MPa v = 0.09196 m3 / kg T = 450°C

EES (Engineering Equation Solver) SOLUTION "Given" P=3500 [kPa] T=(450+273.15) [K] "Properties" Fluid$='steam_iapws' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" "(a)" v_ideal=(R*T)/P "(b)" P_R=P/P_cr T_R=T/T_cr Z=COMPRESS(T_R, P_R) "function returns compressibility chart reading of Z" v_chart=Z*v_ideal "(c)" v=volume(Fluid$, P=P, T=T) error_ideal=(v_ideal-v)/v*Convert(, %) error_chart=(v_chart-v)/v*Convert(, %)

1-215

3-91 The specific volume of nitrogen gas is to be determined using the ideal gas relation and the compressibility chart. The errors involved in these two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of nitrogen are, from Table A-1, Pcr  3.39 MPa R  0.2968 kPa  m3 / kg  K , Tcr  126.2 K , Analysis

(a) From the ideal gas equation of state, v=

RT (0.2968 kPa ×m3 /kg ×K)(150 K) = = 0.004452 m3 / kg P 10,000 kPa

(86.4% error)

(b) From the compressibility chart (Fig. A-15), P 10 MPa ïü = = 2.95 ïï ïï Pcr 3.39 MPa ý Z = 0.54 ï T 150 K TR = = = 1.19 ïï ïï Tcr 126.2 K þ PR =

Thus, v = Z v ideal = (0.54)(0.004452 m3 /kg) = 0.002404 m 3 / kg

(0.7% error)

EES (Engineering Equation Solver) SOLUTION "Given" P=10000 [kPa] T=150 [K] v=0.002388 [m^3/kg] "Properties" Fluid$='N2' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" "(a)" v_ideal=(R*T)/P "(b)" P_R=P/P_cr T_R=T/T_cr Z=COMPRESS(T_R, P_R) "function returns compressibility chart reading of Z" v_chart=Z*v_ideal error_ideal=(v_ideal-v)/v*Convert(, %) error_chart=(v_chart-v)/v*Convert(, %)

1-216

3-92 Ethylene is heated at constant pressure. The specific volume change of ethylene is to be determined using the compressibility chart. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1, Pcr  5.12 MPa R  0.2964 kPa  m3 / kg  K , Tcr  282.4 K ,

Analysis From the compressibility chart at the initial and final states (Fig. A-15), ü ïï ïï ï Z = 0.56 ý 1 ï P1 5 MPa PR1 = = = 0.977 ïïï Pcr 5.12 MPa ïþ

TR1 =

T1 293 K = = 1.038 Tcr 282.4 K

üï T2 473 K = = 1.675ïï Tcr 282.4 KR ý Z1 = 0.961 ïï PR 2 = PR1 = 0.977 ïïþ

TR 2 =

The specific volume change is Dv = =

R ( Z 2T2 - Z1T1 ) P 0.2964 kPa ×m3 /kg ×K [(0.961)(473 K) - (0.56)(293 K) ] 5000 kPa

= 0.0172 m3 / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=5000 [kPa] P2=P1 T1=(20+273.15) [K] T2=(200+273.15) [K] "Properties" Fluid$='ethylene' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" P_R1=P1/P_cr T_R1=T1/T_cr Z1=COMPRESS(T_R1, P_R1) "function returns compressibility chart reading of Z" P_R2=P2/P_cr T_R2=T2/T_cr Z2=COMPRESS(T_R2, P_R2) "function returns compressibility chart reading of Z" DELTAv=R/P1*(Z2*T2-Z1*T1)

1-217

3-93 CO2 gas flows through a pipe. The volume flow rate and the density at the inlet and the volume flow rate at the exit of the pipe are to be determined. Properties The gas constant, the critical pressure, and the critical temperature of CO2 are (Table A-1) R  0.1889 kPa  m3 / kg  K ,

Tcr  304.2 K ,

Pcr  7.39 Mpa

Analysis (a) From the ideal gas equation of state, V1 =

mRT1 (2 kg/s)(0.1889 kPa ×m3 /kg ×K)(500 K) = = 0.06297 m3 / kg P1 (3000 kPa)

r1 =

P1 (3000 kPa) = = 31.76 kg / m3 RT1 (0.1889 kPa ×m3 /kg ×K)(500 K)

V2 =

mRT2 (2 kg/s)(0.1889 kPa ×m3 /kg ×K)(450 K) = = 0.05667 m3 / kg P2 (3000 kPa)

(2.1% error)

(2.1% error) (3.6% error)

(b) From the compressibility chart (EES function for compressibility factor is used) P1 3 MPa ïü = = 0.407 ïï ïï Pcr 7.39 MPa ý Z1 = 0.9791 ï T1 500 K TR ,1 = = = 1.64 ïïï Tcr 304.2 K ïþ PR =

P2 3 MPa ïü = = 0.407 ïï ïï Pcr 7.39 MPa ý Z 2 = 0.9656 ï T 450 K TR ,2 = 2 = = 1.48 ïïï Tcr 304.2 K ïþ PR =

Thus,

V1 =

Z1mRT1 (0.9791)(2 kg/s)(0.1889 kPa ×m3 /kg ×K)(500 K) = = 0.06165 m3 / kg P1 (3000 kPa)

r1 =

P1 (3000 kPa) = = 32.44 kg / m3 Z1 RT1 (0.9791)(0.1889 kPa ×m3 /kg ×K)(500 K)

V2 =

Z 2 mRT2 (0.9656)(2 kg/s)(0.1889 kPa ×m3 /kg ×K)(450 K) = = 0.05472 m3 / kg P2 (3000 kPa)

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=3000 [kPa] T_1=500 [K] m_dot=2 [kg/s] T_2=450 [K] "PROPERTIES" Fluid$='CO2' R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Fluid$) P_crit=P_crit(Fluid$) T_crit=T_crit(Fluid$) "ANALYSIS" "(a) Ideal gas solution" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-218

V_dot_1_a=(m_dot*R*T_1)/P_1 rho_1_a=P_1/(R*T_1) P_2=P_1 V_dot_2_a=(m_dot*R*T_2)/P_2 "(b) Compressibility chart solution" P_R=P_1/P_crit T_R_1=T_1/T_crit T_R_2=T_2/T_crit Z_1=COMPRESS(T_R_1, P_R) "EES function for compressibility factor" Z_2=COMPRESS(T_R_2, P_R) V_dot_1_b=(Z_1*m_dot*R*T_1)/P_1 rho_1_b=P_1/(Z_1*R*T_1) V_dot_2_b=(Z_2*m_dot*R*T_2)/P_2 "Percent errors" Error_V1=(V_dot_1_a-V_dot_1_b)/V_dot_1_b*Convert(, %) Error_rho_1=(rho_1_b-rho_1_a)/rho_1_b*Convert(, %) Error_V2=(V_dot_2_a-V_dot_2_b)/V_dot_2_b*Convert(, %)

3-94E Ethane in a rigid vessel is heated. The final pressure is to be determined using the compressibility chart.

Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1E, Pcr  708 psia R  0.3574 psia  ft 3 / lbm  R , Tcr  549.8 R , Analysis From the compressibility chart at the initial state (from Fig. A-15 or EES. We used EES throughout the solution.), ü ïï ïï ï Z = 0.977 ý 1 ï P1 50 psia PR1 = = = 0.0706 ïïï Pcr 708 psia ïþ The specific volume does not change during the process. Then, TR1 =

T1 560 R = = 1.019 Tcr 549.8 R

v1 = v 2 =

Z1 RT1 (0.977)(0.3574 psia ×ft 3 /lbm ×R)(560 R) = = 3.91 ft 3 /lbm P1 50 psia

At the final state,

üï ïï ïï ý Z 2 = 1.0 3 ï v 2, actual 3.911 ft /lbm v R2 = = = 14.1 ïïï 3 RTcr / Pcr (0.3574 psia ×ft /lbm ×R)(549.8 R)/(708 psia) ïïþ TR 2 =

T2 1000 R = = 1.819 Tcr 549.8 R

P2 =

Z 2 RT2 (1.0)(0.3574 psia ×ft 3 /lbm ×R)(1000 R) = = 91.4 psia v2 3.91 ft 3 /lbm

Thus,

1-219

EES (Engineering Equation Solver) SOLUTION "Given" P1=50 [psia] T1=(100+460) [R] T2=(540+460) [R] "Properties" Fluid$='ethane' R=0.3574 [psia-ft^3/lbm-R] P_cr=708"P_crit(Fluid$)" T_cr=549.8"T_crit(Fluid$)" "Analysis" P_R1=P1/P_cr T_R1=T1/T_cr Z1=COMPRESS(T_R1, P_R1) "function returns compressibility chart reading of Z" v1=(Z1*R*T1)/P1 v2=v1 T_R2=T2/T_cr v_R2=v2/(R*T_cr/P_cr) Z2=1 "from the compressibility chart" P2=(Z2*R*T2)/v2

3-95 The pressure of R-134a is to be determined using the ideal gas relation, the compressibility chart, and the R-134a tables. Properties The gas constant, the critical pressure, and the critical temperature of refrigerant-134a are, from Table A-1, Pcr  4.059 MPa R  0.08149 kPa  m3 / kg  K , Tcr  374.2 K ,

Analysis The specific volume of the refrigerant is v=

V 0.016773 m3 = = 0.016773 m3 /kg m 1 kg

(a) From the ideal gas equation of state, P=

RT (0.08149 kPa ×m3 /kg ×K)(383 K) = = 1861 kPa v 0.016773 m3 /kg

(b) From the compressibility chart (Fig. A-15), ü ïï ïï ï ý PR = 0.39 ï v actual 0.016773 m3 /kg vR = = = 2.24 ïïï 3 RTcr / Pcr (0.08149 kPa ×m /kg ×K)(374.2 K)/(4059 kPa) ïïþ TR =

T 383 K = = 1.023 Tcr 374.2 K

Thus, P = PR Pcr = (0.39)(4059 kPa) = 1583 kPa

1-220

T = 110°C P = 1600 kPa v = 0.016773 m3 /kg

}

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.016773 [m^3] m=1 [kg] T=110 [C]+273 [K] "Properties" Fluid$='R134a' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" v=Vol/m "(a)" P_ideal=(R*T)/v "(b)" T_R=T/T_cr v_R=v/((R*T_cr)/P_cr) P_R=0.39 "from compressibility chart" P_chart=P_R*P_cr "(c)" P=pressure(Fluid$, T=T, v=v)

3-96E Water vapor is heated at constant pressure. The final temperature is to be determined using ideal gas equation, the compressibility charts, and the steam tables.

Properties The critical pressure and the critical temperature of water are, from Table A-1E, Pcr  3200 psia R  0.5956 psia  ft 3 / lbm  R , Tcr  1164.8 R , Analysis (a) From the ideal gas equation, T2 = T1

v2 = (400 + 460 R)(2) = 1720 R v1

(b) The properties of steam are (Table A-4E)

P1 = P2 = Psat@400° F = 247.26 psia v1 = v g@400° F = 1.8639 ft 3 /lbm

1-221

At the final state, from the compressibility chart (Fig. A-15),

PR 2 = v R2 =

P2 247.26 psia = = 0.0773 Pcr 3200 psia v 2,actual RTcr / Pcr

3.7278 ft 3 /lbm = 17.19 (0.5956 psia ×ft 3 /lbm ×R)(1164.8 R)/(3200 psia)

üï ïï ïï ý Z 2 = 0.985 ïï ïï ïïþ

Thus, T2 =

P2v 2 (247.26 psia)(3.7278 ft 3 /lbm) = = 1571 R Z2 R (0.985)(0.5956 psia ×ft 3 /lbm ×R)

(from Table A-6E or EES)

EES (Engineering Equation Solver) SOLUTION "Given" T1=(400+459.67) [R] x1=1 P2=P1 VolumeRatio=2 "Vol2/Vol1=2" "Properties" Fluid$='steam_iapws' R=0.5956 [psia-ft^3/lbm-R] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" "(a)" T2_a=T1*VolumeRatio "(b)" P1=pressure(Fluid$, T=T1, x=x1) v1=volume(Fluid$, T=T1, x=x1) v2=VolumeRatio*v1 P_R2=P2/P_cr v_R2=v2/(R*T_cr/P_cr) Z2=0.985 "from compressibility chart" T2_b=(P2*v2)/(Z2*R) "(c)" T2_c=temperature(Fluid$, P=P2, v=v2)

3-97 The percent error involved in treating CO2 at a specified state as an ideal gas is to be determined. Properties The critical pressure, and the critical temperature of CO2 are, from Table A-1, Tcr = 304.2 K

and

Pcr = 7.39 MPa

1-222

PR =

P 5 MPa = = 0.677 Pcr 7.39 MPa

TR =

T 298 K = = 0.980 Tcr 304.2 K

ü ïï ïï ï Z = 0.69 ý ïï ïï ïþ

Then the error involved in treating CO2 as an ideal gas is

v - v ideal 1 1 = 1= 1= - 0.44.9 or 44.9% v Z 0.69

Error =

EES (Engineering Equation Solver) SOLUTION "Given" P=5000 [kPa] T=(25+273) [K] "Properties" Fluid$='CO2' P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" T_R=T/T_cr P_R=P/P_cr Z=COMPRESS(T_R, P_R) "function returns compressibility chart reading of Z" Error=(1-1/Z)*Convert(, %)

3-98 Methane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of methane are, from Table A-1, Pcr  4.64 MPa R  0.5182 kPa  m3 / kg  K , Tcr  191.1K ,

Analysis From the ideal gas equation, T2  T1

v2  (300 K )(1.8)  540 K v1

From the compressibility chart at the initial state (Fig. A-15),

    Z1  0.86, v R1  0.63 P 10 MPa PR1  1   2.16   Pcr 4.64 MPa

TR1 

T1 300 K   1.57 Tcr 191.1 K

At the final state, PR 2  PR1  2.16

  Z2 1

v R 2  1.8v R1  1.8(0.63)  1.134 

(Fig. A-28)

1-223

T2 

P2v 2 P v T 10,000 kPa (1.134)(19 1.1 K)  2 R 2 cr   467 K Z 2 R Z 2 Pcr 1 4640 kPa

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate.

EES (Engineering Equation Solver) SOLUTION "Given" P1=10000 [kPa] T1=300 [K] P2=P1 VolumeRatio=1.8 "Vol2/Vol1=1.5" "Properties" Fluid$='methane' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=4640 [kPa] "P_cr = P_crit(Fluid$)" T_cr=191.1 [K] "T_cr = T_crit(Fluid$)" "Analysis" T2_a=T1*VolumeRatio P_R1=P1/P_cr T_R1=T1/T_cr Z1=0.86 "Z1_alt= COMPRESS(T_R1, P_R1)" "function returns compressibility chart reading of Z" v_R1=0.63 "from compressibility chart" P_R2=P2/P_cr v_R2=VolumeRatio*v_R1 Z2=1 "from compressibility chart" v2=v_R2*(R*T_cr/P_cr) T2_b=(P2*v2)/(Z2*R)

Other Equations of State

3-99C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection point at the critical point.

3-100E The temperature of R-134a in a tank at a specified state is to be determined using the ideal gas relation, the van der Waals equation, and the refrigerant tables. Properties The gas constant, critical pressure, and critical temperature of R-134a are (Table A-1E) Tcr  673.6 R , Pcr  588.7 psia R  0.1052 psia  ft 3 / lbm  R , Analysis (a) From the ideal gas equation of state, T=

Pv (400 psia)(0.1144 ft 3 /lbm) = = 435 R R 0.1052 psia ×ft 3 /lbm ×R

(b) The van der Waals constants for the refrigerant are determined from

27 R 2Tcr2 (27)(0.1052 psia ×ft 3 /lbm ×R) 2 (673.6 R) 2 = = 3.596 ft 6 ×psia/lbm 2 64 Pcr (64)(588.7 psia)

1-224

RTcr (0.1052 psia ×ft 3 /lbm ×R)(673.6 R) = = 0.01504 ft 3 /lbm 8Pcr 8´ 588.7 psia

ö 1æ aö 1 æ çç400 + 3.591 ÷ ÷ (v - b) = ÷ çççP + 2 ÷ ÷(0.1144 - 0.01504) = 638 R 2÷ ÷ ç Rè 0.1052 è v ø (0.3479) ø

Then,

}

P = 400 psia T = 100°F (660 R ) v = 0.1144 ft 3 /lbm

EES (Engineering Equation Solver) SOLUTION "Given" P=400 [psia] v=0.1144 [ft^3/lbm] "Properties" Fluid$='R134a' M=molarmass(Fluid$) R_u=10.73 [psia-ft^3/lbmol-R] R=R_u/M P_cr=588.7"P_crit(Fluid$)" T_cr=673.6"T_crit(Fluid$)" "Analysis" "(a)" T_ideal=(P*v)/R "(b)" a=(27*R^2*T_cr^2)/(64*P_cr) b=(R*T_cr)/(8*P_cr) T_Waals=1/R*(P+a/v^2)*(v-b) "(c)" T_actual=temperature(Fluid$, P=P, v=v)

3-101 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas, van der Waals, and Beattie-Bridgeman equations. The error involved in each case is to be determined. Properties The gas constant, molar mass, critical pressure, and critical temperature of nitrogen are (Table A-1) Tcr  126.2 K , Pcr  3.39 MPa R  0.2968 kPa  m3 / kg  K , M  28.013 kg / kmol ,

Analysis The specific volume of nitrogen is v=

V 3.27 m3 = = 0.0327 m3 /kg m 100 kg

(a) From the ideal gas equation of state, P=

RT (0.2968 kPa ×m3 /kg ×K)(175 K) = = 1588 kPa (5.5% error) v 0.0327 m3 /kg

1-225

(b) The van der Waals constants for nitrogen are determined from

27 R 2Tcr2 (27)(0.2968 kPa ×m3 /kg ×K) 2 (126.2 K) 2 = = 0.175m6 ×kPa/kg 2 64 Pcr (64)(3390 kPa)

RTcr (0.2968 kPa ×m3 /kg ×K)(126.2 K) = = 0.00138 m3 /kg 8Pcr 8´ 3390 kPa

RT a 0.2968´ 175 0.175 - 2 = = 1495 kPa (0.7% error) v- b v 0.0327 - 0.00138 (0.0327) 2

Then,

(c) The constants in the Beattie-Bridgeman equation are æ aö æ 0.02617 ö ÷ ç A = Ao çç1- ÷ ÷ ÷ ÷= 136.2315èçç1- 0.9160 ø ÷= 132.339 èç v ø æ bö æ - 0.00691ö ÷ ç B = Bo çç1- ÷ ÷ ÷ ÷= 0.05046 èçç1- 0.9160 ø ÷= 0.05084 èç v ø c = 4.2´ 104 m3 ×K 3 /kmol

since v = M v = (28.013 kg/kmol)(0.0327 m3 /kg) = 0.9160 m3 /kmol . Substituting, P=

ö Ru T æ A ç1- c ÷ (v + B)- 2 2 ç 3÷ ÷ ç v è vT ø v

8.314´ 175 æ 4.2´ 10 4 ö 132.339 ÷ ç1÷ (0.9160 + 0.05084)÷ 2 ç 2 ç ÷ (0.9160) è 0.9160´ 1753 ø (0.9160)

= 1504 kPa (0.07% error)

EES (Engineering Equation Solver) SOLUTION "Given" Vol=3.27 [m^3] mass=100 [kg] T=175 [K] P_actual=1505 [kPa] "Properties" Fluid$='Nitrogen' M=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] R=R_u/M P_cr=3390 [kPa] T_cr=126.2 [K] "Analysis" v=Vol/mass "(a)" P_ideal=(R*T)/v Error_ideal=(P_ideal-P_actual)/P_actual*Convert(, %) "(b)" a_Waals=(27*R^2*T_cr^2)/(64*P_cr) b_Waals=(R*T_cr)/(8*P_cr) P_Waals=(R*T)/(v-b_Waals)-a_Waals/v^2 Error_Waals=(P_Waals-P_actual)/P_actual*Convert(, %) "(c)" v_bar=M*v A_0=136.2315 B_0=0.05046 a_Beattie=0.02617 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-226

b_Beattie=-0.00691 c=4.2E4 "Coefficients from the text" A=A_0*(1-a_Beattie/v_bar) B=B_0*(1-b_Beattie/v_bar) P_Beattie=(R_u*T)/v_bar^2*(1-c/(v_bar*T^3))*(v_bar+B)-A/v_bar^2 Error_Beattie=(P_Beattie-P_actual)/P_actual*Convert(, %)

3-102 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas relation and the BeattieBridgeman equation. The error involved in each case is to be determined. Properties The gas constant and molar mass of nitrogen are (Table A-1) R  0.2968 kPa  m3 / kg  K and M  28.013 kg / kmol

Analysis (a) From the ideal gas equation of state, P=

RT (0.2968 kPa ×m3 /kg ×K)(150 K) = = 1063 kPa (6.3% error) v 0.041884 m3 /kg

(b) The constants in the Beattie-Bridgeman equation are æ aö æ 0.02617 ö ÷ ç A = Ao çç1- ÷ ÷ ÷ ÷= 136.2315èçç1- 1.1733 ø ÷= 133.193 çè v ø æ bö æ - 0.00691ö ÷ ç B = Bo çç1- ÷ ÷ ÷ ÷= 0.05046 èçç1- 1.1733 ø ÷= 0.05076 çè v ø c = 4.2´ 104 m3×K 3 /kmol

since v = M v = (28.013 kg/kmol)(0.041884 m3 /kg) = 1.1733 m3 /kmol.

Substituting, P=

ö ö Ru T æ A 8.314´ 150 æ 4.2 ´ 10 4 ÷ 133.193 ç1çç1- c ÷ ÷ v + B = ( ) (1.1733 + 0.05076)÷ ç ÷ 2 ç 3÷ 2 2 ç 3÷ 2 v è vT ø v (1.1733) è 1.1733´ 150 ø (1.1733)

= 1000.4 kPa (negligible error)

EES (Engineering Equation Solver) SOLUTION "An EES function is used to evaluate teh Beattie-Bridgeman equation of state." Function BeattBridg(T, v, M, R_u) v_bar=v*M "Conversion from m^3/kg to m^3/kmol" "The constants for the Beattie-Bridgeman equation of state are found in Table A-29" Ao=136.2315; aa=0.02617; Bo=0.05046; bb=-0.00691; cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) "The Beattie-Bridgeman equation of state is" BeattBridg:=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-227

End M=MOLARMASS(Nitrogen) R_u=8.314 "[kJ/kmol-K] Universal gas constant" R=R_u/M "Particular gas constant" P_EES=PRESSURE(Nitrogen,T=T_table,v=v) "[kPa] EES data for nitrogen as a real gas" P_idealgas=R*T_idealgas/v "Ideal gas equation" P_BB=BeattBridg(T_BB,v,M,R_u) "Beattie-Bridgeman equation of state Function" "Data for problem" T=150 [K] T_table=T; T_BB=T;T_idealgas=T v=0.041884 [m3/kg] P_exper=1000 [kPa]

3-103 Problem 3-102 is reconsidered. Using appropriate software, the specific volume is to be plotted versus temperature for a pressure of 1000 kPa over the range of 110 K < T < 150 K. Analysis The problem is solved using EES, and the solution is given below. P=1000 [kPa] T=150 [K] M=molarmass(Nitrogen) R_u=8.314 [kJ/kmol-K] R=R_u/M v_table=volume(Nitrogen, T=T, P=P) "EES data for nitrogen as a real gas" v_ideal=R*T/P "Ideal gas equation" v_BB=v_bar/M Ao=136.2315; aa=0.02617; Bo=0.05046;bb=-0.00691;cc=4.20*1E4 B=Bo*(1-bb/v_bar) A=Ao*(1-aa/v_bar) P=R_u*T/(v_bar**2)*(1-cc/(v_bar*T**3))*(v_bar+B)-A/v_bar**2 "Beattie-Bridgeman equation of state function"

T [K] 110 115 120 125 130 135 140 145 150

vtable [m3 / kg]

VBB [m3 / kg]

Videal [m3 / kg]

0.0271 0.02919 0.03117 0.03306 0.0349 0.03669 0.03845 0.04017 0.04188

0.02723 0.02929 0.03125 0.03313 0.03496 0.03674 0.03849 0.04021 0.0419

0.03265 0.03413 0.03561 0.0371 0.03858 0.04007 0.04155 0.04303 0.04452

1-228

0.046 0.044 0.042 ideal gas

0.038

v [m /kg]

0.04

0.036

Beattie-Bridgeman

0.034

Table

0.032 0.03 0.028 110

115

120

125

130

135

140

145

150

T [K]

3-104 Carbon monoxide is heated in a piston-cylinder device. The final volume of the carbon monoxide is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state.

Properties The gas constant and molar mass of CO are (Table A-1) R  0.2968 kPa  m3 / kg  K , M  28.011 kg / kmol Analysis (a) From the ideal gas equation of state, V2 =

mRT2 (0.100 kg)(0.2968 kPa ×m3 /kg ×K)(773 K) = = 0.02294 m 3 P 1000 kPa

(b) Using the coefficients of Table 3-4 for carbon monoxide and the given data, the Benedict-Webb-Rubin equation of state for state 2 is

P2 = 1000 = +

RuT2 æ C ö 1 bRuT2 - a aa c æ gö 2 ÷ + çççB0 RuT2 - A0 - 02 ÷ + + 6 + 3 2 çç1 + 2 ÷ ÷ ÷ 2 3 ÷exp(- g / v ) ÷v v2 T2 ø v v v T2 èç v ø èç (8.314)(773) æ 8.673´ 105 ö 1 0.002632 ´ 8.314 ´ 773 - 3.71 ÷ ÷ + ççç0.05454 ´ 8.314 ´ 773 - 135.9 + ÷ 2 2 ÷v v2 773 v3 è ø ö 3.71´ 0.000135 1.054 ´ 105 æ 2 ç1 + 0.0060 ÷ + 3 ÷ 6 2 ç 2 ÷exp(- 0.0060 / v ) ç ø v v (773) è v

1-229 v 2 = 6.460 m3 /kmol

Then,

v2 =

v2 6.460 m3 /kmol = = 0.2306 m3 /kg M 28.011 kg/kmol

V2 = mv 2 = (0.100 kg)(0.2306 m3 /kg) = 0.02306 m3

EES (Engineering Equation Solver) SOLUTION "Given" m=0.100 [kg] P=1000 [kPa] T1=(200+273) [K] T2=(500+273) [K] "Properties" Fluid$='CO' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" Vol2_ideal=(m*R*T2/P) P=(R_u*T)/v+(B_0*R_u*T-A_0-C_0/T^2)*1/v^2+(b*R_u*Ta)/v^3+(a*alpha)/v^6+c/(v^3*T^2)*(1+gamma/v^2)*exp(-gamma/v^2) T=T2 a=3.71 A_0=135.87 b=0.002632 B_0=0.05454 c=1.054E5 C_0=8.673E5 alpha=1.350E-4 gamma=0.0060 v2=v/MM Vol2=m*v2

3-105 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, van der Waals equation, and the steam tables. Properties The gas constant, critical pressure, and critical temperature of steam are (Table A-1) R  0.4615 kPa  m3 / kg  K , Tcr  647.1K , Pcr  22.06 MPa

Analysis The specific volume of steam is v=

V 1 m3 = = 0.3520 m3 /kg m 2.841 kg

1-230

Pv (600 kPa)(0.352 m3 /kg) = = 457.6 K R 0.4615 kPa ×m3 /kg ×K

(b) The van der Waals constants for steam are determined from

27 R 2Tcr2 (27)(0.4615 kPa ×m3 /kg ×K) 2 (647.1 K) 2 = = 1.705 m6 ×kPa/kg 2 64 Pcr (64)(22,060 kPa)

RTcr (0.4615 kPa ×m3 /kg ×K)(647.1 K) = = 0.00169 m3 /kg 8Pcr 8´ 22,060 kPa

ö 1æ aö 1 æ çç600 + 1.705 ÷ ÷ (v - b) = (0.352 - 0.00169) = 465.9 K ÷ çççP + 2 ÷ ÷ 2÷ ÷ ç Rè 0.4615 è v ø (0.3520) ø

Then,

}

P = 0.6 MPa T = 200°C v = 0.3520 m3 /kg

( = 473 K)

EES (Engineering Equation Solver) SOLUTION "Given" Vol=1 [m^3] mass=2.841 [kg] P=600 [kPa] "Properties" Fluid$='steam_iapws' M=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] R=R_u/M P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" v=Vol/mass "(a)" T_ideal=(P*v)/R "(b)" a=(27*R^2*T_cr^2)/(64*P_cr) b=(R*T_cr)/(8*P_cr) T_Waals=1/R*(P+a/v^2)*(v-b) "(c)" T_actual=temperature(Fluid$, P=P, v=v)

3-106 Problem 3-105 is reconsidered. The problem is to be solved using appropriate software. The temperature of water is to be compared for the three cases at constant specific volume over the pressure range of 0.1 MPa to 1 MPa in 0.1 MPa increments. The %error involved in the ideal gas approximation is to be plotted against pressure. Analysis The problem is solved using EES, and the solution is given below. Function vanderWaals(T,v,M,R_u,T_cr,P_cr) v_bar=v*M "The constants for the van der Waals equation of state:" a=27*R_u^2*T_cr^2/(64*P_cr) b=R_u*T_cr/(8*P_cr) vanderWaals:=R_u*T/(v_bar-b)-a/v_bar**2 "The van der Waals equation of state" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-231

End m=2.841 [kg] Vol=1 [m^3] {P=6*convert(MPa,kPa)} T_cr=T_CRIT(Steam_iapws) P_cr=P_CRIT(Steam_iapws) v=Vol/m P_table=P P_vdW=P P_idealgas=P T_table=temperature(Steam_iapws,P=P_table,v=v) "EES data for steam as a real gas" {P_table=pressure(Steam_iapws, T=T_table,v=v)} {T_sat=temperature(Steam_iapws,P=P_table,v=v)} MM=MOLARMASS(water) R_u=8.314 [kJ/kmol-K] R=R_u/MM P_idealgas=R*T_idealgas/v "Ideal gas equation" "The value of P_vdW is found from van der Waals equation of state function" P_vdW=vanderWaals(T_vdW,v,MM,R_u,T_cr,P_cr) Error_idealgas=Abs(T_table-T_idealgas)/T_table*Convert(, %) Error_vdW=Abs(T_table-T_vdW)/T_table*Convert(, %)

P [kPa]

Tideal gas [K]

Ttable [K]

TvdW [K]

Errorideal gas [K]

100 200 300 400 500 600 700 800 900 1000

76.27 152.5 228.8 305.1 381.4 457.6 533.9 610.2 686.4 762.7

372.8 393.4 406.7 416.8 425 473 545.3 619.1 693.7 768.6

86.35 162.3 238.2 314.1 390 465.9 541.8 617.7 693.6 769.5

79.54 61.22 43.74 26.8 10.27 3.249 2.087 1.442 1.041 0.7725

1-232

1-233

3-107E Carbon dioxide is heated in a constant pressure apparatus. The final volume of the carbon dioxide is to be determined using the ideal gas equation and the Benedict-Webb-Rubin equation of state.

Properties The gas constant and molar mass of CO2 are (Table A-1E) R  0.2438 psia  ft 3 / lbm  R , M  44.01 lbm / lbmol

Analysis (a) From the ideal gas equation of state, V2 =

mRT2 (1 lbm)(0.2438 psia ×ft 3 /lbm ×R)(1260 R) = = 0.3072 ft 3 P 1000 psia

(b) Using the coefficients of Table 3-4 for carbon dioxide and the given data in SI units, the Benedict-Webb-Rubin equation of state for state 2 is

P2 = 6895 = +

RuT2 æ C ö 1 bRuT2 - a aa c æ gö 2 ÷ + çççB0 RuT2 - A0 - 02 ÷ + + 6 + 3 2 çç1 + 2 ÷ ÷ ÷ 2 3 ÷exp(- g / v ) ÷ v2 T2 øv v v v T2 çè v ø èç (8.314)(700) æ 1.404´ 107 ö 1 0.007210 ´ 8.314 ´ 700 - 13.86 ÷ ÷ 2+ + çç0.04991´ 8.314´ 700 - 277.30 ÷ 2 ÷ ç v2 700 v3 è øv ö 13.86´ 8.470´ 10- 5 1.511´ 106 æ 2 çç1 + 0.00539 ÷ + 3 ÷exp(- 0.00539 / v ) ÷ v6 v (700) 2 çè v2 ø

The solution of this equation by an equation solver such as EES gives v 2 = 0.8477 m3 /kmol

Then,

v2 =

v 2 0.8477 m3 /kmol = = 0.01926 m3 /kg M 44.01 kg/kmol

æ35.315 ft 3 ÷ ö ÷ V2 = mv 2 = (1/ 2.2046 kg)(0.01926 m 3 /kg) ççç = 0.3087 ft 3 ÷ 3 ÷ è 1m ø

EES (Engineering Equation Solver) SOLUTION m=1 [lbm] P1=1000 [psia] T1=(200+460) [R] T2=(800+460) [R] MM=44.01 R=0.2438 [psia-ft^3/lbmol-R] Vol2_ideal=m*R*T2/P1 R_u=8.314 P=6895 T=700 P=(R_u*T)/v+(B_0*R_u*T-A_0-C_0/T^2)*1/v^2+(b*R_u*Ta)/v^3+(a*alpha)/v^6+c/(v^3*T^2)*(1+gamma/v^2)*exp(-gamma/v^2) a=13.86 A_0=277.30 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-234

b=0.007210 B_0=0.04991 c=1.511E6 C_0=1.404E7 alpha=8.470E-5 gamma=0.00539 v2=v/MM Vol2=m*Convert(lbm, kg)*v2*Convert(m^3,ft^3)

Special Topic: Vapor Pressure and Phase Equilibrium 3-108 The vapor pressure in the air at the beach when the air temperature is 30C is claimed to be 5.2 kPa. The validity of this claim is to be evaluated. Properties The saturation pressure of water at 30C is 4.247 kPa (Table A-4).

Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature, which is Pv , max = Psat @ Tair = Psat @ 30° C = 4.247 kPa

which is less than the claimed value of 5.2 kPa. Therefore, the claim is false.

EES (Engineering Equation Solver) SOLUTION "Given" T_air=30 [C] P_v_claimed=5.2 [kPa] "Analysis" Fluid$='steam_iapws' P_sat=pressure(Fluid$, T=T_air, x=1) P_v_max=P_sat "If P_v-max is smaller than P_v_claimed, the claim is false."

3-109 A glass of water is left in a room. The vapor pressures at the free surface of the water and in the room far from the glass are to be determined.

Assumptions The water in the glass is at a uniform temperature. Properties The saturation pressure of water is 2.339 kPa at 20C, and 1.706 kPa at 15C (Table A-4). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-235

Analysis The vapor pressure at the water surface is the saturation pressure of water at the water temperature, Pv , water surface = Psat @ Twater = Psat @ 15 °C = 1.706 kPa

Noting that the air in the room is not saturated, the vapor pressure in the room far from the glass is Pv , air = f Psat @ Tair = f Psat @ 20 °C = (0.4)(2.339 kPa) = 0.936 kPa

EES (Engineering Equation Solver) SOLUTION "Given" T_air=20 [C] phi=0.40 T_water=15 [C] "Properties" Fluid$='steam_iapws' P1_sat=pressure(Fluid$, T=T_air, x=1) P2_sat=pressure(Fluid$, T=T_water, x=1) "Analysis" "(a)" P_v_watersurface=P2_sat "(b)" P_v_air=phi*P1_sat

3-110 A person buys a supposedly cold drink in a hot and humid summer day, yet no condensation occurs on the drink. The claim that the temperature of the drink is below 10C is to be evaluated.

Properties The saturation pressure of water at 35C is 5.629 kPa (Table A-4). Analysis The vapor pressure of air is Pv , air = f Psat @ Tair = f Psat@35°C = (0.7)(5.629 kPa) = 3.940 kPa

The saturation temperature corresponding to this pressure (called the dew-point temperature) is Tsat = Tsat @ Pv = Tsat@3.940 kPa = 28.7°C

That is, the vapor in the air will condense at temperatures below 28.7C. Noting that no condensation is observed on the can, the claim that the drink is at 10C is false.

EES (Engineering Equation Solver) SOLUTION "Given" T_air=35 [C] phi=0.70 T_drink_claimed=10 [C] "Properties" Fluid$='steam_iapws' P_sat=pressure(Fluid$, T=T_air, x=1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-236

"Analysis" P_v_air=phi*P_sat T_sat=temperature(Fluid$, P=P_v_air, x=1) "If T_sat is greater than T_drink_claimed, the claim is false."

3-111 Two rooms are identical except that they are maintained at different temperatures and relative humidities. The room that contains more moisture is to be determined. Properties The saturation pressure of water is 2.339 kPa at 20C, and 3.17 kPa at 25C (Table A-4). Analysis The vapor pressures in the two rooms are Room 1: Pv1 = f 1 Psat @ T = f 1 Psat @ 25°C = (0.4)(3.17 kPa) = 1.27 kPa 1

Room 2:

Pv 2 = f 2 Psat @ T2 = f 2 Psat @ 20°C = (0.55)(2.339 kPa) = 1.29 kPa

Therefore, room 1 at 30C and 40% relative humidity contains more moisture.

EES (Engineering Equation Solver) SOLUTION "Given" T1_air=25 [C] phi_1=0.40 T2_air=20 [C] phi_2=0.55 "Properties" Fluid$='steam_iapws' P1_sat=pressure(Fluid$, T=T1_air, x=1) P2_sat=pressure(Fluid$, T=T2_air, x=1) "Analysis" P1_v_air=phi_1*P1_sat P2_v_air=phi_2*P2_sat "The room with the higher P_v_air contains more moisture."

3-112E A thermos bottle half-filled with water is left open to air in a room at a specified temperature and pressure. The temperature of water when phase equilibrium is established is to be determined.

Assumptions The temperature and relative humidity of air over the bottle remain constant. Properties The saturation pressure of water at 60F is 0.2564 psia (Table A-4E). Analysis The vapor pressure of air in the room is Pv , air = f Psat @ Tair = f Psat @ 70° F = (0.35)(0.2564 psia) = 0.08973 psia

Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface. Therefore, Pv , water surface = Pv , air = 0.08973 psia

1-237

and Twater = Tsat @ Pv = Tsat @ 0.08973 psia = 32.3°F

Discussion Note that the water temperature drops to 32.3F in an environment at 60F when phase equilibrium is established.

EES (Engineering Equation Solver) SOLUTION "Given" T_air=60 [F] phi=0.35 "Properties" Fluid$='steam_iapws' P_sat=pressure(Fluid$, T=T_air, x=1) "Analysis" P_v_air=phi*P_sat P_v_watersurface=P_v_air T_water=temperature(Fluid$, P=P_v_watersurface, x=1)

Review Problems 3-113 Complete the following table for H 2 O : P, kPa

T,  C

v, m3 / kg

u, kJ / kg

Phase description

200

0.001004

125.71

Compressed liquid

270.3

130

Insufficient information

200

400

1.5493

2967.2

Superheated steam

300

133.52

0.500

2196.4

Saturated mixture, x=0.825

500

473.1

0.6858

3084

Superheated steam

u, kJ / kg

Phase description

EES (Engineering Equation Solver) SOLUTION Fluid$='steam_iapws' v1=volume(Fluid$, P=200, T=30) u1=intenergy(Fluid$, P=200, T=30) T_sat1=temperature(Fluid$, P=200, x=0) T_sat2=temperature(Fluid$, P=270.3, x=0) P3=pressure(Fluid$, T=400, v=1.5493) u3=intenergy(Fluid$, T=400, v=1.5493) T4=temperature(Fluid$, P=300, v=0.500) u4=intenergy(Fluid$, P=300, v=0.500) x4=quality(Fluid$, P=300, v=0.500) T5=temperature(Fluid$, P=500, u=3084) v5=volume(Fluid$, P=500, u=3084)

3-114 Complete the following table for R-134a: P, kPa

T, C

v, m3 / kg

1-238

320

−12

0.0007498

35.76

Compressed liquid

1000

39.37

Insufficient information

140

0.17794

263.80

Superheated vapor

180

−12.73

0.0700

153.57

Saturated mixture, x=0.6315

200

22.10

0.1197

249

Superheated vapor

EES (Engineering Equation Solver) SOLUTION Fluid$='R134a' v1=volume(Fluid$, P=320, T=−12) u1=intenergy(Fluid$, P=320, T=−12) T_sat1=temperature(Fluid$, P=320, x=0) T_sat2=temperature(Fluid$, P=1000, x=0) P3=pressure(Fluid$, T=40, v=0.17794) u3=intenergy(Fluid$, T=40, v=0.17794) T4=temperature(Fluid$, P=180, v=0.0700) u4=intenergy(Fluid$, P=180, v=0.0700) x4=quality(Fluid$, P=180, v=0.0700) T5=temperature(Fluid$, P=200, u=249) v5=volume(Fluid$, P=200, u=249)

3-115 A rigid tank contains an ideal gas at a specified state. The final temperature is to be determined for two different processes.

Analysis (a) The first case is a constant volume process. When half of the gas is withdrawn from the tank, the final temperature may be determined from the ideal gas relation as æ100 kPa ÷ ö m P T2 = 1 2 T1 = (2)çç ÷(600 K) = 400 K çè300 kPa ÷ ø m2 P1 (b) The second case is a constant volume and constant mass process. The ideal gas relation for this case yields P2 =

æ400 K ÷ ö T2 P1 = çç (300 kPa) = 200 kPa ÷ ÷ ç è 600 K ø T1

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=300 [kPa] T_1=600 [K] P_2=100 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-239

m_1=1 "assumed, it does not matter what value is chosen" m_2=1/2*m_1 "ANALYSIS" (m_1*T_1)/P_1=(m_2*T_2)/P_2 "ideal gas equation, volume is constant" T_1/P_1=T_2/P_2_b "ideal gas equation, both volume and mass are constant"

3-116 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density of CO2 at the given state and the volume flow rate at the exit of the pipe are to be determined.

Analysis (a) The volume and mass flow rates may be determined from ideal gas relation as V1 =

NRu T1 (0.4 kmol/s)(8.314 kPa.m3 /kmol.K)(500 K) = = 0.5543 m3 / s P 3000 kPa

m1 =

P1V1 (3000 kPa)(0.5543 m3 / s) = = 17.60 kg / s RT1 (0.1889 kPa.m3 /kg.K)(500 K)

The density is r1 =

m1 V1

(17.60 kg/s) = 31.76 kg / m3 (0.5543 m 3 /s)

(b) The volume flow rate at the exit is V2 =

NRuT2 (0.4 kmol/s)(8.314 kPa.m3 /kmol.K)(450 K) = = 0.4988 m3 / s P 3000 kPa

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=3000 [kPa] T_1=500 [K] N_dot=0.4 [kmol/s] T_2=450 [K] "PROPERTIES" R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(CO2) "ANALYSIS" V_dot_1=(N_dot*R_u*T_1)/P_1 m_dot=(P_1*V_dot_1)/(R*T_1) rho_1=m_dot/V_dot_1 P_2=P_1 V_dot_2=(N_dot*R_u*T_2)/P_2

3-117 The pressure in an automobile tire increases during a trip while its volume remains constant. The percent increase in the absolute temperature of the air in the tire is to be determined.

1-240

Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas. Properties The local atmospheric pressure is 90 kPa. Analysis The absolute pressures in the tire before and after the trip are

P1 = Pgage,1 + Patm = 200 + 90 = 290 kPa P2 = Pgage,2 + Patm = 220 + 90 = 310 kPa Noting that air is an ideal gas and the volume is constant, the ratio of absolute temperatures after and before the trip are P1V1 PV T2 P 310 kPa = 2 2 ® = 2= =1.069 T1 T2 T1 P1 290 kPa

Therefore, the absolute temperature of air in the tire will increase by 6.9% during this trip.

EES (Engineering Equation Solver) SOLUTION "Given" P1_gage=200 [kPa] P2_gage=220 [kPa] P_atm=90 [kPa] V=0.035 [m^3] "Analysis" P1=P1_gage+P_atm P2=P2_gage+P_atm RatioP=P2/P1 RatioT=RatioP "from ideal gas relation (P1*V/T1)=(P2*V/T2)" Increase=(RatioT-1)*Convert(, %)

3-118 A tank contains argon at a specified state. Heat is transferred from argon until it reaches a specified temperature. The final gage pressure of the argon is to be determined. Assumptions 1 Argon is an ideal gas.

1-241

Properties The local atmospheric pressure is given to be 100 kPa. Analysis Noting that the specific volume of argon in the tank remains constant, from ideal gas relation, we have P2 = P1

T2 (300 + 273)K = (200 + 100 kPa) = 196.9 kPa T1 (600 + 273)K

Then the gage pressure becomes Pgage, 2 = P2 - Patm = 196.9 - 100 = 96.9 kPa

EES (Engineering Equation Solver) SOLUTION "Given" T1=600[C]+273 [K] P1_gage=200 [kPa] T2=300[C]+273 [K] P_atm=100 [kPa] "Analysis" P1=P1_gage+P_atm P2=P1*(T2/T1) P2_gage=P2-P_atm

3-119 The cylinder conditions before the heat addition process is specified. The pressure after the heat addition process is to be determined.

Assumptions 1 The contents of cylinder are approximated by the air properties. 2 Air is an ideal gas. Analysis The final pressure may be determined from the ideal gas relation P2 =

æ1900 + 273 K ÷ ö T2 P1 = çç (1200 kPa) = 3607 kPa @ 3.61 MPa ÷ çè 450 + 273 K ÷ ø T1

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=1200 [kPa] T_1=(450+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-242

T_2=(1900+273) [K] "ANALYSIS" T_1/P_1=T_2/P_2 "ideal gas relation, mass and volume are constant"

3-120 Nitrogen gas in a rigid tank is heated to a final gage pressure. The final temperature is to be determined.

Assumptions At specified conditions, nitrogen behaves as an ideal gas. Analysis According to the ideal gas equation of state at constant volume,

m1 = m1 P1V1 P2V2 = T1 T2 Since

V1 = V2 P1 P = 2 T1 T2 T2 = T1

P2 (250 + 100) kPa = [(227 + 273) K ] = 875 K = 602°C P2 (100 + 100) kPa

EES (Engineering Equation Solver) SOLUTION "Given" T1=(227+273) [K] P1_gage=100 [kPa] P2_gage=250 [kPa] P_atm=100 [kPa] "Analysis" P1=P1_gage+P_atm P2=P2_gage+P_atm T2=T1*(P2/P1)

3-121 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are to be determined.

1-243

Analysis The initial specific volume is 0.090 m3 / kg . Using this with the initial temperature reveals that the initial state is a mixture. The initial pressure is then the saturation pressure, ü ïï ý P = Psat @ -40°C = 51.25 kPa v1 = 0.090 m /kg ïïþ 1 T1 = - 40°C

(Table A-11)

This is a constant volume cooling process (v  V / m  constant) . The final state is superheated vapor and the final temperature is then ïü ïý T = 50°C v 2 = v1 = 0.090 m /kg ïïþ 2 P2 = 280 kPa

(Table A-13)

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] Vol=0.090 [m^3] T1=-40 [C] P2=280 [kPa] "Analysis" Fluid$='R134a' v1=Vol/m P1=pressure(Fluid$, T=T1, v=v1) v2=v1 T2=temperature(Fluid$, P=P2, v=v2)

3-122 The refrigerant in a rigid tank is allowed to cool. The pressure at which the refrigerant starts condensing is to be determined, and the process is to be shown on a P-v diagram.

Analysis This is a constant volume process (v  V / m  constant) , and the specific volume is determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-244

V 0.117 m3 = = 0.117 m3 /kg m 1 kg

When the refrigerant starts condensing, the tank will contain saturated vapor only. Thus, v 2 = v g = 0.117 m3 /kg

The pressure at this point is the pressure that corresponds to this vg value,

P2 = Psat @ v = 0.117 m3 /kg = 170 kPa g

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.117 [m^3] m=1 [kg] P1=240 [kPa] x2=1 "Analysis" Fluid$='R134a' v=Vol/m v2=v P2=pressure(Fluid$, v=v2, x=x2)

3-123 Water at a specified state is contained in a tank. It is now cooled. The process will be indicated on the P-v and T- v diagrams. Analysis The properties at the initial and final states are P1 = 300 kPaü ïï 3 ý v = 0.7964 m /kg (Table A-6) T1 = 250°C ïïþ 1 ïü ïý T = 111.35°C (Table A-5) v 2 = v1 = 0.7964 m /kg ïïþ 2 P2 = 150 kPa

Using Property Plot feature of EES, and by adding state points we obtain following diagrams.

1-245 Steam IAPWS

106 105

P [kPa]

104 103

300 kPa

102

250°C 150 kPa

111,4°C

101 100 10-4

10-3

10-2

10-1 v [m3/kg]

100

101

102

Steam IAPWS

700 600

T [°C]

500 400 300 200 100 0 10-3

250 C 300 kPa 111.35 C

150 kPa

10-2

10-1

2 100

101

102

v [m3/kg]

EES (Engineering Equation Solver) SOLUTION "Given" P1=300 [kPa] T1=250 [C] P2=150 [kPa] "Analysis" Fluid$='steam_iapws' v1=volume(Fluid$,P=P1,T=T1) v2=v1 T2=temperature(Fluid$,P=P2,v=v2)

3-124 A large tank contains nitrogen at a specified temperature and pressure. Now some nitrogen is allowed to escape, and the temperature and pressure of nitrogen drop to new values. The amount of nitrogen that has escaped is to be determined.

1-246

Properties The gas constant for nitrogen is 0.2968 kPa  m3 / kg  K (Table A-1). Analysis Treating N 2 as an ideal gas, the initial and the final masses in the tank are determined to be

m1 =

P1V (600 kPa)(9 m3 ) = = 62.74 kg RT1 (0.2968 kPa ×m3 /kg ×K)(290 K)

m2 =

P2V (400 kPa)(9 m3 ) = = 42.12 kg RT2 (0.2968 kPa ×m3 /kg ×K)(278 K)

Thus the amount of N 2 that escaped is D m = m1 - m2 = 62.74 - 42.12 = 20.6 kg

EES (Engineering Equation Solver) SOLUTION "Given" V=9 [m^3] T1=(17+273) [K] P1=600 [kPa] P2=400 [kPa] T2=(15+273) [K] "Properties" R=R_u/M M=molarmass(N2) R_u=8.314 [kJ/kmol-K] "Analysis" m1=(P1*V)/(R*T1) m2=(P2*V)/(R*T2) DELTAm=m1-m2

3-125 Superheated refrigerant-134a is cooled at constant pressure until it exists as a compressed liquid. The changes in total volume and internal energy are to be determined, and the process is to be shown on a T-v diagram.

1-247

Analysis The refrigerant is a superheated vapor at the initial state and a compressed liquid at the final state. From Tables A-13 and A-11,

P1 = 1.2 MPa ïü ï u1 = 277.23 kJ/kg ý v = 0.019502 m3 /kg ïï 1 T1 = 70°C þ P2 = 1.2 MPa ïü ï u2 @ u f @ 20°C = 78.85 kJ/kg 3 ý v @v ïï 2 f @ 20 ° C = 0.0008160 m /kg T2 = 20°C þ

Thus, (b) D V = m(v 2 - v1 ) = (10 kg)(0.0008160 - 0.019502) m3 /kg = - 0.187 m3 (c) D U = m(u2 - u1 ) = (10 kg)(78.85 - 277.23) kJ/kg = - 1984 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m=10 [kg] P1=1200 [kPa] T1=70 [C] P2=P1 T2=20 [C] "Analysis" Fluid$='R134a' u1=intenergy(Fluid$, T=T1, P=P1) v1=volume(Fluid$, T=T1, P=P1) u2=intenergy(Fluid$, T=T2, x=0) v2=volume(Fluid$, T=T2, x=0) DELTAVol=m*(v2-v1) DELTAU=m*(u2-u1)

3-126 The rigid tank contains saturated liquid-vapor mixture of water. The mixture is heated until it exists in a single phase. For a given tank volume, it is to be determined if the final phase is a liquid or a vapor.

1-248

Analysis This is a constant volume process (v  V / m  constant) , and thus the final specific volume will be equal to the initial specific volume, v 2 = v1

The critical specific volume of water is 0.003106 m3 / kg . Thus if the final specific volume is smaller than this value, the water will exist as a liquid, otherwise as a vapor. V = 4L ¾¾ ® v=

V 0.004 m3 = = 0.002 m3 /kg < v cr Thus, liquid. m 2 kg

V = 400 L ¾ ¾ ® v=

V 0.4 m3 = = 0.2 m3 /kg > v cr . m 2 kg

Thus, vapor.

EES (Engineering Equation Solver) SOLUTION "Given" Vol_a=4 [L]*Convert(L, m^3) Vol_b=400 [L]*Convert(L, m^3) m=2 [kg] T=50 [C] "Analysis" Fluid$='steam_iapws' v_cr=v_crit(Fluid$) v_a=Vol_a/m v_b=Vol_b/m "If v_a or v_b is smaller than v_cr, the water will exist as a liquid, otherwise as a vapor"

3-127 A piston-cylinder device contains steam at a specified state. Steam is cooled at constant pressure. The volume change is to be determined using compressibility factor.

Properties The gas constant, the critical pressure, and the critical temperature of steam are Pcr  22.06 MPa R  0.4615 kPa  m3 / kg  K , Tcr  647.1K , Analysis The exact solution is given by the following:

1-249

P = 200 kPa üïï 3 ý v = 1.31623 m /kg T1 = 300°C ïïþ 1 P = 200 kPa üïï 3 ý v = 0.95986 m /kg T2 = 150°C ïïþ 2

(Table A-6)

D Vexact = m(v1 - v 2 ) = (0.2 kg)(1.31623 - 0.95986)m3 /kg = 0.07128 m 3

Using compressibility chart (EES function for compressibility factor is used) P1 0.2 MPa ïü = = 0.0091 ïï ïï Pcr 22.06 MPa ý Z1 = 0.9956 ï T1 300 + 273 K TR ,1 = = = 0.886ïïï Tcr 647.1 K ïþ PR =

ü P2 0.2 MPa ï = = 0.0091ïï ïï Pcr 22.06 MPa ý Z 2 = 0.9897 ï T2 150 + 273 K TR ,2 = = = 0.65ïïï Tcr 647.1 K ïþ PR =

V1 =

Z1mRT1 (0.9956)(0.2 kg)(0.4615 kPa ×m3 /kg ×K)(300 + 273 K) = = 0.2633 m3 P1 (200 kPa)

V2 =

Z 2 mRT2 (0.9897)(0.2 kg)(0.4615 kPa ×m3 /kg ×K)(150 + 273 K) = = 0.1932 m3 P2 (200 kPa)

D Vchart = V1 - V2 = 0.2633 - 0.1932 = 0.07006 m3 ,

Error: 1.7%

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.2 [kg] P_1=200 [kPa] T_1=300[C]+273 [K] P_2=P_1 T_2=150[C]+273 [K] "PROPERTIES" Fluid1$='H2O' Fluid2$='Steam_iapws' R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Fluid1$) P_crit=P_crit(Fluid1$) T_crit=T_crit(Fluid1$) "ANALYSIS" "Exact solution" v_1=volume(Fluid2$, P=P_1, T=T_1) v_2=volume(Fluid2$, P=P_2, T=T_2) Vol_1_a=m*v_1 Vol_2_a=m*v_2 DELTAVol_a=Vol_1_a-Vol_2_a "(b) Compressibility factor solution" P_R_1=P_1/P_crit T_R_1=T_1/T_crit P_R_2=P_2/P_crit T_R_2=T_2/T_crit Z_1=COMPRESS(T_R_1, P_R_1) "EES function for compressibility factor" Z_2=COMPRESS(T_R_2, P_R_2) Vol_1_b=(Z_1*m*R*T_1)/P_1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-250

Vol_2_b=(Z_2*m*R*T_2)/P_2 DELTAVol_b=Vol_1_b-Vol_2_b Error=(DELTAVol_a-DELTAVol_b)/DELTAVol_a*Convert(, %) "percent error"

3-128 One section of a tank is filled with saturated liquid R-134a while the other side is evacuated. The partition is removed, and the temperature and pressure in the tank are measured. The volume of the tank is to be determined.

Analysis The mass of the refrigerant contained in the tank is m=

V1 0.03 m3 = = 34.96 kg v1 0.0008580 m3 /kg

since v1 = v f @ 0.9 MPa = 0.0008580 m3 /kg

At the final state (Table A-13), P2 = 280 kPa ïü ïý v = 0.07997 m3 /kg ïïþ 2

T2 = 20°C

Thus, Vtank = V2 = mv 2 = (34.96 kg)(0.07997 m3 /kg) = 2.80 m3

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.03 [m^3] P1=900 [kPa] x1=0 T2=20 [C] P2=280 [kPa] "Analysis" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) m=Vol1/v1 v2=volume(Fluid$, T=T2, P=P2) Vol_tank=m*v2

3-129 Problem 3-128 is reconsidered. The effect of the initial pressure of refrigerant-134 on the volume of the tank is to be investigated as the initial pressure varies from 0.5 MPa to 1.5 MPa. The volume of the tank is to be plotted versus the initial pressure, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. x_1=0.0 Vol_1=0.03 [m^3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-251

P_1=900 [kPa] T_2=20 [C] P_2=280 [kPa] v_1=volume(R134a,P=P_1,x=x_1) Vol_1=m*v_1 v_2=volume(R134a,P=P_2,T=T_2) Vol_2=m*v_2

Vol2

[kPa]

 m3 

500 600 700 800 900 1000 1100 1200 1300 1400 1500

2.977 2.926 2.88 2.837 2.796 2.757 2.721 2.685 2.651 2.617 2.584

3-130 A tank contains helium at a specified state. Heat is transferred to helium until it reaches a specified temperature. The final gage pressure of the helium is to be determined. Assumptions 1 Helium is an ideal gas.

1-252

Properties The local atmospheric pressure is given to be 100 kPa. Analysis Noting that the specific volume of helium in the tank remains constant, from ideal gas relation, we have P2 = P1

T2 (300 + 273)K = (140 + 100 kPa) = 366.2 kPa T1 (37 + 273)K

Then the gage pressure becomes Pgage,2 = P2 - Patm = 366.2 - 100 = 266 kPa

EES (Engineering Equation Solver) SOLUTION "Given" T1=(37+273) [K] P1_gage=140 [kPa] T2=(200+273) [K] P_atm=100 [kPa] "Analysis" P1=P1_gage+P_atm P2=P1*(T2/T1) P2_gage=P2-P_atm

3-131 (a) On the P-v diagram, the constant temperature process through the state P = 300 kPa, v  0.525 m3 / kg as pressure changes from P1  200 kPa to P2  400 kPa is to be sketched. The value of the temperature on the process curve on the P-v diagram is to be placed. SteamIAPWS

106

105

P [kPa]

104

103

102

400 300 200

2 133.5°C

101

0.525 100 10-4

10-3

10-2

10-1

100

101

102

v [m /kg]

1-253

(b) On the T-v diagram the constant specific vol-ume process through the state T  120 C , v  0.7163 m3 / kg from P1  100 kPa to P2  300 kPa is to be sketched. For this data set, the temperature values at states 1 and 2 on its axis is to be placed. The value of the specific volume on its axis is also to be placed. SteamIAPWS

700

600

300 kPa

T [°C]

500

198.7 kPa

400

100 kPa 300

200

100

1 0.7163

0 10-3

10-2

10-1

100

101

102

v [m /kg]

3-132 Water at a specified state is contained in a piston-cylinder device fitted with stops. Water is now heated until a final pressure. The process will be indicated on the P-v and T- v diagrams.

Analysis The properties at the three states are P1 = 300 kPa ü ïï ý T = 133.5°C v1 = 0.5 m3 /kg ïïþ 1

(Table A-5)

ïü ïý v = 0.6058 m3 /kg, T = 133.5°C 2 x2 = 1 (sat. vap.) ïïþ 2

P2 = 300 kPa

ü P2 = 600 kPa ïï ý T = 517.8°C v 3 = 0.6058 m3 /kg ïïþ 2

(Table A-5)

(Table A-6)

Using Property Plot feature of EES, and by adding state points we obtain following diagrams.

1-254

EES (Engineering Equation Solver) SOLUTION "Given" P1=300 [kPa] v1=0.5 [m^3/kg] x2=1 P2=P1 P3=600 [kPa] v3=v2 "Analysis" Fluid$='steam_iapws' T1=temperature(Fluid$,P=P1,v=v1) T2=temperature(Fluid$,P=P2,x=x2) v2=volume(Fluid$,P=P2,x=x2) T3=temperature(Fluid$,P=P3,v=v3)

1-255

3-133 Ethane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1, Pcr  4.48 MPa R  0.2765 kPa  m3 / kg  K , Tcr  305.5 K ,

Analysis From the ideal gas equation, T2 = T1

v2 = (373 K)(1.6) = 596.8 K v1

From the compressibility chart at the initial state (Fig. A-15), ü ïï ïï ï Z = 0.61, v = 0.35 ý 1 R1 ï P1 10 MPa PR1 = = = 2.232 ïïï Pcr 4.48 MPa ïþ At the final state, TR1 =

T1 373 K = = 1.221 Tcr 305.5 K

ü PR 2 = PR1 = 2.232 ïï ý Z 2 = 0.83 v R 2 = 1.6v R1 = 1.6(0.35) = 0.56 ïïþ

Thus, T2 =

P2v 2 P v T 10,000 kPa (0.56)(305.5 K) = 2 R 2 cr = = 460 K Z2 R Z 2 Pcr 0.83 4480 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P1=10000 [kPa] T1=(100+273.15) [K] P2=P1 VolumeRatio=1.6 "Vol2/Vol1=1.6" "Properties" Fluid$='ethane' R=R_u/MM MM=molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] P_cr=4480 [kPa] "using EES: P_cr = P_crit(Fluid$)" T_cr=305.5 [K] "using EES: T_cr = T_crit(Fluid$)" "Analysis" "(a)" T2_a=T1*VolumeRatio "(b)" T_R1=T1/T_cr P_R1=P1/P_cr Z1=COMPRESS(T_R1, P_R1) "function returns compressibility chart reading of Z" v_R1=0.35 "from compressibility chart" P_R2=P2/P_cr © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-256

v_R2=VolumeRatio*v_R1 Z2=0.83 "from compressibility chart" v2_b=v_R2*(R*T_cr/P_cr) T2_b=(P2*v2_b)/(Z2*R)

3-134 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation, the generalized chart, and the steam tables. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1,

R = 0.4615 kPa ×m3 /kg ×K ,

Tcr = 647.1 K ,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state, P=

RT (0.4615 kPa ×m3 /kg ×K)(673 K) = = 15,529 kPa v 0.02 m3 /kg

(b) From the compressibility chart (Fig. A-15a), ü ïï ïï ï ý PR = 0.57 ï v actual (0.02 m3 /kg)(22,060 kPa) vR = = = 1.48 ïïï 3 RTcr / Pcr (0.4615 kPa ×m /kg ×K)(647.1 K) ïïþ TR =

T 673 K = = 1.040 Tcr 647.1 K

Thus, P = PR Pcr = 0.57´ 22, 060 = 12,574 kPa

T = 400°C P = 12,515 kPa (from EES) v = 0.02 m3 /kg

}

EES (Engineering Equation Solver) SOLUTION "Given" T=(400+273.15) [K] v=0.02 [m^3/kg] "Properties" Fluid$='steam_iapws' R=R_u/M M=molarmass(Fluid$) R_u=8.314 [kJ/kmol-K] P_cr=P_crit(Fluid$) T_cr=T_crit(Fluid$) "Analysis" "(a)" P_ideal=(R*T)/v "(b)" T_R=T/T_cr © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-257

v_R=v/((R*T_cr)/P_cr) P_R=0.57 "from compressibility chart" P_chart=P_R*P_cr "(c)" P=pressure(Fluid$, T=T, v=v)

3-135E The specific volume of nitrogen at a given state is to be determined using the ideal gas relation, the Benedict-WebbRubin equation, and the compressibility factor. Properties The properties of nitrogen are (Table A-1E) R  0.3830 psia  ft 3 / lbm  R , M  28.013 lbm / lbmol , Tcr  227.1 R , Pcr  492 psia

Analysis (a) From the ideal gas equation of state, RT (0.3830 psia ×ft 3 /lbm ×R)(360 R) = = 0.3447 ft 3 / lbm P 400 psia

(b) Using the coefficients of Table 3-4 for nitrogen and the given data in SI units, the Benedict-Webb-Rubin equation of state is

P= 2758 = +

RuT æ C ö 1 bRuT - a aa c æ gö ÷ + çççB0 RuT - A0 - 02 ÷ + + 6 + 3 2 çç1 + 2 ÷ exp(- g / v 2 ) ÷ ÷ 3 ÷ ç ÷ ç è ø v T øv v v v T v è ö1 (8.314)(200) æ 8.164´ 105 ÷ 0.002328´ 8.314 ´ 200 - 2.54 ÷ + çç0.04074´ 8.314´ 200 - 106.73 + ÷ 2 ÷v 2 çè v2 200 v3 ø ö 2.54´ 1.272´ 10- 4 7.379´ 10 4 æ 2 çç1 + 0.0053 ÷ + 3 ÷ 6 2 ç 2 ÷exp(- 0.0053 / v ) ø v v (200) è v

The solution of this equation by an equation solver such as EES gives v = 0.5666 m3 /kmol

Then, v=

v 0.5666 m3 /kmol æ 16.02 ft 3 /lbm ö ÷= 0.3240 ft 3 / lbm ç ÷ = ç ÷ 3 ÷ M 28.013 kg/kmol ç è 1 m /kg ø

(c) From the compressibility chart (Fig. A-15), ü T 360 R ï = = 1.585 ïï ïï Tcr 227.1 R ý Z = 0.94 ï P 400 psia PR = = = 0.813 ïï ïï Pcr 492 psia þ

TR =

Thus, v = Z v ideal = (0.94)(0.3447 ft 3 /lbm) = 0.3240 ft 3 / lbm

EES (Engineering Equation Solver) SOLUTION "Given" P_psia=400 [psia] T_Ran=(-100+460) [R] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-258

Fluid$='N2' R=0.3830 [psia-ft^3/lbm-R] MM=28.013 [lbm/lbmol] T_cr=227.1 [R] P_cr=492 [psia] "Analysis" "(a)" v_ideal=(R*T_Ran)/P_psia "(b)" P=(R_u*T)/v+(B_0*R_u*T-A_0-C_0/T^2)*1/v^2+(b*R_u*Ta)/v^3+(a*alpha)/v^6+c/(v^3*T^2)*(1+gamma/v^2)*exp(-gamma/v^2) P=P_psia*Convert(psia, kPa) T=T_Ran*Convert(R, K) R_u=8.314 [kJ/kmol-K] a=3.71 A_0=135.87 b=0.002632 B_0=0.05454 c=1.054E5 C_0=8.673E5 alpha=1.350E-4 gamma=0.0060 v_BWR=v/MM*Convert(m^3/kg, ft^3/lbm) "(c)" T_R=T_Ran/T_cr P_R=P_psia/P_cr Z=COMPRESS(T_R, P_R) "function returns compressibility factor at T_R and P_R" v_chart=Z*v_ideal

3-136 A hot air balloon with 2 people in its cage is about to take off. The average temperature of the air in the balloon for two environment temperatures is to be determined.

Assumptions Air is an ideal gas. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K . Analysis The buoyancy force acting on the balloon is Vballoon = 4p r 3 / 3 = 4p (9 m)3 /3 = 3054 m3

r cool air =

P 93 kPa = = 1.137 kg/m3 RT (0.287 kPa ×m3 /kg ×K)(285 K)

FB = r cool air gVballoon

1-259

æ 1N ö ÷ ÷ = (1.137 kg/m3 )(9.81 m/s2 )(3054 m3 ) ççç = 34,056 N ÷ ÷ è1 kg ×m/s 2 ø The vertical force balance on the balloon gives

FB = Whot air + Wcage + Wpeople = (mhot air + mcage + mpeople ) g Substituting, æ 1N ö ÷ ÷ 34,056 N = (mhot air + 120 kg + 170 kg)(9.81 m/s2 ) çç 2÷ èç1 kg ×m/s ÷ ø

which gives mhot air = 3182 kg

Therefore, the average temperature of the air in the balloon is T=

PV (93 kPa)(3054 m3 ) = = 311 K mR (3182 kg)(0.287 kPa ×m3 /kg ×K)

Repeating the solution above for an atmospheric air temperature of 25C gives 327 K for the average air temperature in the balloon.

EES (Engineering Equation Solver) SOLUTION "Given" D=18 [m] m_cage=120 [kg] m_person=85 [kg] N_people=2 P_atm=93 [kPa] T_atm=12+273 "[K]" "Properties" R=R_u/M M=molarmass(air) R_u=8.314 [kJ/kmol-K] g=9.81 [m/s^2] "Analysis" V_balloon=pi*D^3/6 m_people=N_people*m_person W_hotair=m_hotair*g W_cage=m_cage*g W_people=m_people*g rho_coolair=P_atm/(R*T_atm) F_B=rho_coolair*g*V_balloon F_B=W_hotair+W_cage+W_people T=(P_atm*V_balloon)/(m_hotair*R)

3-137 The specific volume of oxygen at a given state is to be determined using the ideal gas relation, the Beattie-Bridgeman equation, and the compressibility factor. Properties The properties of oxygen are (Table A-1) R  0.2598 kPa  m3 / kg  K , M  31.999 kg / kmol , Tcr  154.8 K , Pcr  5.08 MPa

1-260

Analysis (a) From the ideal gas equation of state, v=

RT (0.2598 kPa ×m3 /kg ×K)(293 K) = = 0.01903 m 3 / kg P 4000 kPa

(b) The constants in the Beattie-Bridgeman equation are expressed as æ aö æ ö ç 0.02562 ÷ A = Ao çç1- ÷ ÷ ÷ ÷= 151.0857 ççè1÷ èç v ø ø v æ bö æ 0.004208 ö ÷ B = Bo çç1- ÷ ÷= 0.04624 ççç1÷ ÷ ÷ èç v ø è ø v c = 4.80´ 104 m3 ×K 3 /kmol

Substituting these coefficients into the Beattie-Bridgeman equation P=

ö Ru T æ A çç1- c ÷ ÷(v + B)- 2 ø v 2 çè v T 3 ÷ v

and solving using an equation solver such as EES gives v = 0.5931 m3 /kmol

and v=

v 0.5931 m3 /kmol = = 0.01853 m 3 / kg M 31.999 kg/kmol

TR =

T 293 K = = 1.893 Tcr 154.8 K

Z = 0.975

Thus, v = Z v ideal = (0.975)(0.01903 m3 /kg) = 0.01855 m 3 / kg

EES (Engineering Equation Solver) SOLUTION P=4000 [kPa] T=(20+273) [K] R=0.2598 M=31.999 T_cr=154.8 P_cr=5080 R_u=8.314 [kPa-m^3/kmol-K] v_ideal=R*T/P v_bar=M*v A_0=151.0857 B_0=0.04624 a_Beattie=0.02562 b_Beattie=-0.004208 c=4.80E4 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-261

A=A_0*(1-a_Beattie/v_bar) B=B_0*(1-b_Beattie/v_bar) P=(R_u*T)/v_bar^2*(1-c/(v_bar*T^3))*(v_bar+B)-A/v_bar^2 P_R=P/P_cr T_R=T/T_cr Z=COMPRESS(T_R, P_R) "function returns compressibility chart reading of Z" "Z=0.975" "SM uses this value" v_chart=Z*v_ideal

Fundamentals of Engineering (FE) Exam Problems 3-138 A 1  m3 rigid tank contains 10 kg of water (in any phase or phases) at 160C. The pressure in the tank is (a) 738 kPa (b) 618 kPa (c) 370 kPa (d) 2000 kPa (e) 1618 kPa Answer (b) 618 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=1 [m^3] m=10 [kg] T=160 [C] v=V_tank/m P=PRESSURE(Steam_IAPWS,v=v,T=T) "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] W1_P*V_tank=m*R*(T+273) "Treating steam as ideal gas" W2_P*V_tank=m*R*T "Treating steam as ideal gas and using deg.C"

3-139 A rigid 3  m 3 rigid vessel contains steam at 2 MPa and 500C. The mass of the steam is (a) 13 kg (b) 17 kg (c) 22 kg (d) 28 kg (e) 35 kg Answer (b) 17 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=3 [m^3] P1=2000 [kPa] T1=500 [C] v1=VOLUME(Steam_IAPWS,T=T1,P=P1) m=V/v1 "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] P1*V=W1_m*R*(T1+273) "Treating steam as ideal gas" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-262

P1*V=W2_m*R*T1 "Treating steam as ideal gas and using deg.C"

3-140 A 240  m 3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 25% of the mass is liquid and the 75% of the mass is vapor, the total mass in the tank is (a) 240 kg (b) 265 kg (c) 307 kg (d) 361 kg (e) 450 kg Answer (d) 361 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V_tank=240 [m^3] P1=200 [kPa] x=0.75 v_f=VOLUME(Steam_IAPWS, x=0,P=P1) v_g=VOLUME(Steam_IAPWS, x=1,P=P1) v=v_f+x*(v_g-v_f) m=V_tank/v "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] T=TEMPERATURE(Steam_IAPWS,x=0,P=P1) P1*V_tank=W1_m*R*(T+273) "Treating steam as ideal gas" P1*V_tank=W2_m*R*T "Treating steam as ideal gas and using deg.C" W3_m=V_tank "Taking the density to be 1 kg/m^3"

3-141 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 10 minutes. If the heat loss from the coffee maker is negligible, the power rating of the heating element is (a) 3.8 kW (b) 2.2 kW (c) 1.9 kW (d) 1.6 kW (e) 0.8 kW Answer 1.9 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m1=1 [kg] time=10*60 [s] m_evap=0.5*m1 P=101.325 [kPa] Power*time=m_evap*h_fg h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" W1_Power*time=m_evap*h_g "Using h_g" W2_Power*time/60=m_evap*h_g "Using minutes instead of seconds for time" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-263

W3_Power=2*Power "Assuming all the water evaporates"

3-142 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 1.25 kg of liquid water evaporates in 30 minutes. The rate of heat transfer to the water is (a) 1.57 kW (b) 1.86 kW (c) 2.09 kW (d) 2.43 kW (e) 2.51 kW Answer (a) 1.57 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=1.25 [kg] time=30*60 [s] P=101.325 [kPa] h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f Q*time=m_evap*h_fg "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time/60=m_evap*h_g "Using minutes instead of seconds for time" W3_Q*time=m_evap*h_f "Using h_f"

3-143 Water is boiled in a pan on a stove at sea level. During 10 min of boiling, its is observed that 200 g of water has evaporated. Then the rate of heat transfer to the water is (a) 0.84 kJ / min (b) 45.1 kJ / min (c) 41.8 kJ / min (d) 53.5 kJ / min (e) 225.7 kJ / min Answer (b) 45.1 kJ / min Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_evap=0.2 [kg] time=10 [min] P=101.325 [kPa] h_f=ENTHALPY(Steam_IAPWS, x=0,P=P) h_g=ENTHALPY(Steam_IAPWS, x=1,P=P) h_fg=h_g-h_f Q*time=m_evap*h_fg "Some Wrong Solutions with Common Mistakes:" W1_Q*time=m_evap*h_g "Using h_g" W2_Q*time*60=m_evap*h_g "Using seconds instead of minutes for time" W3_Q*time=m_evap*h_f "Using h_f" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-264

3-144 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40C. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 2.2 atm, the final temperature in the tank is (a) 71C (b) 44C (c) −100C (d) 20C (e) 172C Answer (a) 71C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m1=2 [kg] P1=4 [atm] P2=2.2 [atm] T1=(40+273) [K] m2=0.5*m1 P1/P2=(m1*T1)/(m2*T2) "since R=constant and V= constant" T2_C=T2-273 "Some Wrong Solutions with Common Mistakes:" P1/P2=m1*(T1-273)/(m2*W1_T2) "Using C instead of K" P1/P2=m1*T1/(m1*(W2_T2+273)) "Disregarding the decrease in mass" P1/P2=m1*T1/(m1*W3_T2) "Disregarding the decrease in mass, and not converting to deg. C" W4_T2=(T1-273)/2 "Taking T2 to be half of T1 since half of the mass is discharged"

3-145 The pressure of an automobile tire is measured to be 190 kPa (gage) before a trip and 215 kPa (gage) after the trip at a location where the atmospheric pressure is 95 kPa. If the temperature of air in the tire before the trip is 25C, the air temperature after the trip is (a) 51.1C (b) 64.2C (c) 27.2C (d) 28.3C (e) 25.0C Answer (a) 51.1C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Patm=95 [kPa] P1=190 [kPa]+Patm P2=215 [kPa]+Patm T1=(25+273) [K] P1/P2=T1/T2 "since R, V, and m are constant" T2_C=T2-273 "Some Wrong Solutions with Common Mistakes:" P1/P2=(T1-273)/W1_T2 "Using C instead of K" (P1-Patm)/(P2-Patm)=T1/(W2_T2+273) "Using gage pressure instead of absolute pressure" (P1-Patm)/(P2-Patm)=(T1-273)/W3_T2 "Making both of the mistakes above" W4_T2=T1-273 "Assuming the temperature to remain constant"

1-265

3-146 Consider a sealed can that is filled with refrigerant-134a. The contents of the can are at the room temperature of 25C. Now a leak developes, and the pressure in the can drops to the local atmospheric pressure of 90 kPa. The temperature of the refrigerant in the can is expected to drop to (rounded to the nearest integer) (a) 0C (b) −29C (c) −16C (d) 5C (e) 25C Answer (b) −29C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 [C] P2=90 [kPa] T2=TEMPERATURE(R134a,x=0,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant"

3-147 … 3-149 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

1-266

Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS

1-267

CYU - Check Your Understanding CYU 4-1 ■ Check Your Understanding CYU 4-1.1 Select the wrong statement regarding the boundary work during a quasi-equilibrium process. (a) The boundary work during an isothermal process is (b) The boundary work during a constant volume process is zero. (c) The boundary work during a constant pressure process is (d) The boundary work during a polytropic process is (e) The boundary work during an isothermal process is Answer: (c) The boundary work during a constant pressure process is (There should be no ‘m’). CYU 4-1.2 A vertical piston-cylinder device with a piston which is free to move contains air at 100 kPa and 400 K. If 60 kJ of heat is lost from the air to the surroundings, what is the final pressure of the air in the cylinder? (a) 100 kPa (b) 50 kPa (c) 25 kPa (d) 200 kPa (e) 400 kPa Answer: (a) 100 kPa , pressure is constant during the process. CYU 4-1.3 A piston-cylinder device contains of air at 100 kPa and 300 K. At this state, the piston is free to move. Now, heat is transferred to the air and the volume of the air triples. What is the boundary work done during this process? (a) 100 kJ (b) 50 kJ (c) 20 kJ (d) 10 k (e) 0 kJ Answer: (c) 20 kJ

CYU 4-2 ■ Check Your Understanding CYU 4-2.1 Select the quantities which are part of the total energy of a closed system. I. Heat transfer II. Work III. Internal energy IV. Kinetic energy (a) I and II (b) I and III (c) III and IV (d) I, II, and III © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-268

(e) I, III, and IV Answer: (c) III and IV CYU 4-2.2 Steam in a stationary adiabatic piston-cylinder device is being stirred by mixer driven by a shaft. The piston is free to move. The energy balance relation for this process in its simplest form is (a) (b) (c) (d) (e) Answer: (c) CYU 4-2.3 A well-insulated piston-cylinder device contains m kg of saturated liquid water at T and P. Now, an electrical resistance heater submerged in water is turned on. The piston is free to rise at constant pressure. What is the amount of electrical energy consumed when the last drop of liquid in the cylinder is vaporized? (a) (b) (c) (d) (e) Answer: (d)

(

)

CYU 4-2.4 A rigid tank is divided into two equal parts by a partition. Initially, one side of the tank contains water, and the other side is evacuated. The partition is then removed, and the water expands into the entire tank. 180 J of heat is transferred to the water during the process. The internal energy change of water and the work done during this process are, respectively (a) 0, 180 J (b) 180 J, 0 (c) 180 J, 180 J (d) 0, 0 (e) Insufficient information Answer: (b) 180 J, 0

CYU 4-2.5 During a process, 8 kJ of heat is lost from a stationary closed system while 5 kJ of work is done by the system. The energy change E of the system is (a) 13 kJ (b) 13 kJ (c) 3 kJ (d) 3 kJ (e) 5 kJ Answer: (b) 13 kJ

- (8 kJ) - (5 kJ) = -13 kJ

1-269

CYU 4-3 ■ Check Your Understanding CYU 4-3.1 Select the wrong statement regarding specific heats. (a) The specific heat at constant volume is a measure of the variation of internal energy of a substance with temperature. (b) The specific heat units kJ/kg°C and kJ/kgK are identical and can be used interchangeably. (c) The specific heat at constant pressure can be less than the specific heat at constant volume. (d) The energy needed to raise the temperature of a unit mass of a substance by 1ºC depends on how the process is executed. (e) The specific heat at constant pressure is a measure of the variation of enthalpy of a substance with temperature. Answer: (c) The specific heat at constant pressure can be less than the specific heat at constant volume. CYU 4-3.2 A piston-cylinder device contains 1 kg of air at 10C. As a result of heat transfer from the surroundings, the air temperature rises to 20C at constant pressure. What is the internal energy change of the air during this process? For air, take and . (a) 3 kJ (b) 7 kJ (c) 10 kJ (d) 13 kJ (e) 0 kJ Answer: (b) 7 kJ

CYU 4-3.3 An adiabatic piston-cylinder device contains 1 kg air at 10C. Now, an electric heater in the cylinder is turned on and kept on until the air temperature rises by 1C while the pressure remains constant. What is the amount of boundary work done during this process? For air, take and . (a) 0.3 kJ (b) 0.7 kJ (c) 1 kJ (d) 1.3 kJ (e) 0 kJ Answer: (a) 0.3 kJ

(

)

CYU 4-4 ■ Check Your Understanding CYU 4-4,1 Select the correct equations for ideal gases. I. II. III. (a) I and II (b) I and III (c) II and III (d) I, II, and III © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-270

(e) Only I Answer: (a) I and II CYU 4-4.2 A piston-cylinder device contains m kg of oxygen at , , and . As a result of heat transfer from an external source, the state of oxygen changes to and while the pressure remains constant at . The amount of heat transfer during this process (in kJ) can be determined from (a) (b) (c) (d) (e) ∫ Answer: (a) CYU 4-4.3 A rigid tank contains m kg of an ideal gas at heat Q lost from the gas, the state of the gas changes to (a) (b) (c) (d) (e)

and . As a result of electrical work done on the gas and and . An energy balance on the system can be written as

Answer: (d) CYU 4-4.4 A piston-cylinder device contains m kg of air at , , and . Now, heat in the amount of Q is transferred to the air and the state of air changes to and while the pressure remains constant at . An energy balance on the system can be written as (a) (b) (c) (d) (e) Answer: (c) CYU 4-4.5 A rigid tank contains 0.2 kg of air at 90C. As a result of heat loss to the surroundings, the air temperature is decreased to 70C. What is the amount of heat transfer during this process? For air, take and . (a) 1.4 kJ (b) 2.8 kJ (c) 4 kJ (d) 14 kJ (e) 20 kJ Answer: (b) 2.8 kJ

CYU 4-4.6 A piston-cylinder device contains 1 kg of an ideal gas at 10C. Now, 11 kJ of electrical work is done on the gas, increasing its temperature to 20C at constant pressure. What is the amount of boundary work done during this process? For the ideal gas, take . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-271

(a) 0 kJ (b) 3 kJ (c) 8 kJ (d) 11 kJ (e) 14 kJ Answer: (b) 3 kJ

CYU 4-5 ■ Check Your Understanding CYU 4-5.1 Liquid water at mass is (a) (b) (c) (d) (e)

, and

is compressed by a pump to a pressure of

. The work input to the pump per unit

Answer: (a) CYU 4-5.2 A hot copper block at with a mass of and specific heat of is placed in a tank containing cold water at with a mass of and specific heat of . Heat is lost from the tank to the surroundings in the amount of Q. Copper block and water reach thermal equilibrium at the end of the process at a temperature of . The energy balance for this system (copper block + water) can be written as (a) (b) (c) (d) (e) Answer: (e) (

)

(

)

CYU 4-5.3 A tank contains 1 kg water at 50C. Now, 20,000 J of heat is lost from the water to its surroundings. What is the final temperature of the water in the tank? For water, take . (a) 45C (b) 40C (c) 30C (d) 20C (e) 0C Answer: (a) 45C

1-272

1-273

PROBLEMS Moving Boundary Work 4-1C Yes.

4-2C It represents the boundary work for quasi-equilibrium processes.

4-3C The area under the process curve, and thus the boundary work done, is greater in the constant pressure case.

4-4 1 kPa  m 3  1 k(N / m 2 )  m 3  1 kN  m  1 kJ

EES (Engineering Equation Solver) SOLUTION "Given" "Show that 1 kPa-m^3 = 1 kJ" "Analysis" A=1 [kPa-m^3] B=A*Convert(kPa-m^3, kJ)

4-5 The boundary work done during the process shown in the figure is to be determined. Assumptions The process is quasi-equilibrium. Analysis No work is done during the process 2-3 since the area under process line is zero. Then the work done is equal to the area under the process line 1-2:

(

)

1-274

P1=500 [kPa] P2=100 [kPa] P3=400 [kPa] v1=0.5 [m^3/kg] v2=1 [m^3/kg] W_b_out=m*(P1+P2)/2*(v2-v1)

4-6 Nitrogen gas in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1. The process is quasi-equilibrium. 2. Nitrogen is an ideal gas. Analysis The boundary work is determined from its definition to be

∫ (

)(

)

Discussion The negative sign indicates that work is done on the system (work input).

EES (Engineering Equation Solver) SOLUTION "Given" T1=300 [K] P1=150 [kPa] V1=0.2 [m^3] T2=T1 P2=800 [kPa] "Analysis" W_b_out=P1*V1*ln(P1/P2)

4-7 Helium is compressed in a piston-cylinder device. The initial and final temperatures of helium and the work required to compress it are to be determined. Assumptions The process is quasi-equilibrium. Properties The gas constant of helium is Analysis The initial specific volume is

(Table A-1).

1-275

Using the ideal gas equation,

Since the pressure stays constant,

and the work integral expression gives ∫

(

)

That is,

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=1 [kg] Vol1=5 [m^3] Vol2=2 [m^3] P=130 [kPa] "PROPERTIES" R=2.0769 [kJ/kg-K] "ANALYSIS" v1=Vol1/m T1=(P*v1/R) T2=T1*(Vol2/Vol1) W_b_out=P*(Vol2-Vol1) W_b_in=-W_b_out

4-8 A piston-cylinder device with a set of stops contains steam at a specified state. Now, the steam is cooled. The compression work for two cases and the final temperature are to be determined. Analysis (a) The specific volumes for the initial and final states are (Table A-6)

1-276

}

Noting that pressure is constant during the process, the boundary work is determined from (b) The volume of the cylinder at the final state is 40% of initial volume. Then, the boundary work becomes The temperature at the final state is }

(Table A-5)

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.6 [kg] P_1=1000 [kPa] T_1=400 [C] P_2_a=P_1 T_2_a=250 [C] P_2_b=500 [kPa] f=0.40 "ANALYSIS" Fluid$='Steam_iapws' "(a)" v_1=volume(Fluid$, P=P_1, T=T_1) v_2=volume(Fluid$, P=P_2_a, T=T_2_a) Vol_1=m*v_1 Vol_2=m*v_2 W_b_a=P_1*(Vol_1-Vol_2) "(b)" Vol_stop=f*Vol_1 W_b_b=P_1*(Vol_1-Vol_stop) v_stop=Vol_stop/m T_2_b=temperature(Fluid$, P=P_2_b, v=v_stop)

4-9 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4 through A-6) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-277

} } Analysis The boundary work is determined from its definition to be ∫ (

)

Discussion The positive sign indicates that work is done by the system (work output). We obtained the specific volume at the final state by linear interpolation in Table A-6. A more accurate value can be obtained using EES, and it gives 1.44449 . The boundary work in this case becomes 214 kJ. It turns out that obtaining the specific value in Table A-6 by interpolation in this case is not a good idea.

EES (Engineering Equation Solver) SOLUTION "Given" m=5 [kg] x1=1 "saturated vapor" P1=150 [kPa] P2=P1 T2=200 [C] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, x=x1) v2=volume(Fluid$, P=P2, T=T2) "Analysis" W_b_out=m*P1*(v2-v1)

4-10E Superheated water vapor in a cylinder is cooled at constant pressure until 70% of it condenses. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Properties Noting that the pressure remains constant during this process, the specific volumes at the initial and the final states are (Table A-4E through A-6E)

1-278

} }

Analysis The boundary work is determined from its definition to be ∫ (

)

Discussion The negative sign indicates that work is done on the system (work input).

EES (Engineering Equation Solver) SOLUTION "Given" m=16 [lbm] P1=40 [psia] T1=600 [F] P2=P1 x2=1-0.70 "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) v2=volume(Fluid$, P=P2, x=x2) "Analysis" W_b_out=m*P1*(v2-v1)*Convert(psia-ft^3, Btu)

4-11 Water is expanded isothermally in a closed system. The work produced is to be determined. Assumptions The process is quasi-equilibrium. Analysis From water table

1-279

P1 = P2 = Psat @ 200°C = 1554.9 kPa v1 = v f @ 200°C = 0.001157 m 3 /kg v 2 = v f + xv fg = 0.001157 + 0.80(0.12721 - 0.001157) = 0.10200 m 3 /kg

The definition of specific volume gives

The work done during the process is determined from ∫

(

)

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=1 [m^3] x1=0 T1=200 [C] T2=T1 x2=0.80 "Properties" Fluid$='steam_iapws' P1=pressure(Fluid$, T=T1, x=x1) P2=P1 v1=volume(Fluid$, T=T1, x=x1) v2_f=volume(Fluid$, T=T2, x=0) v2_g=volume(Fluid$, T=T2, x=1) v2=v2_f+x2*(v2_g-v2_f) "Analysis" Vol2=Vol1*(v2/v1) W_b_out=P1*(Vol2-Vol1)

4-12 Argon is compressed in a polytropic process. The final temperature is to be determined. Assumptions The process is quasi-equilibrium. Analysis For a polytropic expansion or compression process, For an ideal gas, Combining these equations produces ( )

(

)

1-280

P2=1200 [kPa] T2=(T1+273)*(P2/P1)^((n-1)/n)

4-13 A gas in a cylinder is compressed to a specified volume in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined by plotting the process on a P-V diagram and also by integration. Assumptions The process is quasi-equilibrium. Analysis (a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus,

and

(

)

(b) The boundary work can also be determined by integration to be

∫

Discussion The negative sign indicates that work is done on the system (work input).

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.42 [m^3] V2=0.12 [m^3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-281

a=-1200 [kPa/m^3] b=600 [kPa] "Analysis (a) Finding the area under the process curve" P1=a*V1+b P2=a*V2+b W_b_out_area=(P1+P2)/2*(V2-V1) "(b) Having EES to perform integration" P=a*V+b W_b_out_integration=integral(P, V, V2, V1) "Integrating by hand gives" W_b_out_hand=a*(V2^2-V1^2)/2+b*(V2-V1)

4-14 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1. 2.

The process is quasi-equilibrium. Air is an ideal gas.

Properties The gas constant of air is

(Table A-1).

Analysis The boundary work is determined from its definition to be ∫

Discussion The negative sign indicates that work is done on the system (work input).

EES (Engineering Equation Solver) SOLUTION "Given" m=1.5 [kg] P1=120 [kPa] T1=(24+273) [K] P2=600 [kPa] T2=T1 "Properties" R=R_u/MM MM=Molarmass(Air) R_u=8.314 [kJ/kmol-K] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-282

W_b_out=m*R*T1*ln(P1/P2)

4-15 A gas in a cylinder expands polytropically to a specified volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The boundary work for this polytropic process can be determined directly from (

( )

)

and

∫ (

)

Discussion The positive sign indicates that work is done by the system (work output).

EES (Engineering Equation Solver) SOLUTION Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. A function is convenient if n is considered to be a variable since the expression for boundary work depens on whether n=1 or n<>1 and logic is not allowed in EES equations." If n<>1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n) else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) endif end $Array off "This command forces array variables to show up in the Solutions Window and not in the Array Window - useful if there are only a few array variables." "Input Data" n=1.5 P[1] = 350 [kPa] V[1] = 0.03 [m^3] V[2] = 0.2 [m^3] Gas$='AIR' "System: The gas enclosed in the piston-cylinder device." "Process: Polytropic expansion or compression, P*V^n = C" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-283

P[2]*V[2]^n=P[1]*V[1]^n "Find work using the function" W_b = BoundWork(P[1],V[1],P[2],V[2],n) "If we modify this problem and specify the mass, then we can calculate the final temperature of the fluid for compression or expansion" m[1] = m[2] "Conservation of mass for the closed system" "Let's solve the problem for m[1] = 0.05 kg" m[1] = 0.05 [kg] "Find the temperatures from the pressure and specific volume." T[1]=temperature(gas$,P=P[1],v=V[1]/m[1]) T[2]=temperature(gas$,P=P[2],v=V[2]/m[2])

4-16 Problem 4-15 is reconsidered. The process described in the problem is to be plotted on a P-V diagram, and the effect of the polytropic exponent n on the boundary work as the polytropic exponent varies from 1.1 to 1.6 is to be plotted. Analysis The problem is solved using EES, and the solution is given below. Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. This function is required since the expression for boundary work depens on whether n=1 or n<>1" If n<>1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n) "when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "when n=1" endif end "n=1.5" P[1] = 350 [kPa] V[1] = 0.03 [m^3] V[2] = 0.2 [m^3] Gas$='AIR' P[2]*V[2]^n=P[1]*V[1]^n W_b = BoundWork(P[1],V[1],P[2],V[2],n)

1-284

1.267 1.322 1.378 1.433 1.489 1.544 1.6

15.63 14.9 14.22 13.58 12.98 12.42 11.89

1-285

4-17 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a specified value. The boundary work done during this process is to be determined. Assumptions 1. The process is quasi-equilibrium. 2. Nitrogen is an ideal gas. Properties The gas constant for nitrogen is Analysis The boundary work for this polytropic process can be determined from

(Table A-2a)

∫

Discussion The negative sign indicates that work is done on the system (work input).

EES (Engineering Equation Solver) SOLUTION "Given" m=5 [kg] P1=100 [kPa] T1=250 [K] T2=450 [K] n=1.4 "Properties" R=R_u/MM MM=Molarmass(N2) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" "Integrating by hand gives" W_b_out=(m*R*(T2-T1))/(1-n) "Having EES to perform integration" V1=(m*R*T1)/P1 V2=(m*R*T2)/P2 P2=C/(V2^n) C=P1*V1^n P*V^n=C W_b_out_integration=integral(P, V, V2, V1)

4-18E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume. The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-286

(a) The pressure of the gas changes linearly with volume, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus, At state 1:

At state 2:

and, (

)

Discussion The positive sign indicates that work is done by the system (work output).

EES (Engineering Equation Solver) SOLUTION "Given" P1=15 [psia] P2=100 [psia] a=5 [psia/ft^3] V1=7 [ft^3] "Analysis" "Finding the area under the process curve" P1=a*V1+b P2=a*V2+b W_b_out_area=(P1+P2)/2*(V2-V1)*Convert(psia-ft^3, Btu) "Having EES to perform integration" P=a*V+b W_b_out_integration=integral(P, V, V1, V2)*Convert(psia-ft^3, Btu) "Integrating by hand gives" W_b_out_hand=(a*(V2^2-V1^2)/2+b*(V2-V1))*Convert(psia-ft^3, Btu)

4-19 A piston-cylinder device contains nitrogen gas at a specified state. The boundary work is to be determined for the isothermal expansion of nitrogen. Properties The properties of nitrogen are

, k = 1.4 (Table A-2a).

1-287

Analysis We first determine initial and final volumes from ideal gas relation, and find the boundary work using the relation for isothermal expansion of an ideal gas

( )

(

)

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.4 [kg] P_1=160 [kPa] T_1=(140+273) [K] P_2=100 [kPa] T_2=T_1 "PROPERTIES" R=0.2968 [kPa-m^3/kg-K] "ANALYSIS" V_1=(m*R*T_1)/P_1 V_2=(m*R*T_2)/P_2 W_b=P_1*V_1*ln(V_2/V_1)

4-20 A piston-cylinder device contains air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determined. Properties The properties of air are Analysis For the isothermal expansion process:

( )

, k = 1.4 (Table A-2a).

(

)

1-288

For the polytropic compression process: →

→

For the constant pressure compression process: The net work for the cycle is the sum of the works for each process

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.15 [kg] P_1=2000 [kPa] T_1=(350+273) [K] P_2=500 [kPa] T_2=T_1 P_3=P_1 n=1.2 "PROPERTIES" Fluid$='Air' R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Fluid$) "ANALYSIS" V_1=(m*R*T_1)/P_1 V_2=(m*R*T_2)/P_2 W_b_12=P_1*V_1*ln(V_2/V_1) "Process 1-2: isothermal expansion" P_2*V_2^n=P_3*V_3^n "get V_3 from this polytropic relation" W_b_23=(P_3*V_3-P_2*V_2)/(1-n) "Process 2-3: polytropic compression" W_b_31=P_3*(V_1-V_3) "Process 3-1: constant pressure compression" W_net=W_b_12+W_b_23+W_b_31 "Compression work is negative and expansion work is positive"

4-21 Several sets of pressure and volume data are taken as a gas expands. The boundary work done during this process is to be determined using the experimental data. Assumptions The process is quasi-equilibrium. Analysis Plotting the given data on a P-V diagram on a graph paper and evaluating the area under the process curve, the work done is determined to be 0.25 kJ.

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.001 [m^3] V2=0.002 [m^3] "Analysis" P=498.77-246008*V+4.8306E+07*V^2 "This function is obtained using regression feature of EES under Tables menu based on P and V data from Parametric Table" W_b_out_integral=integral(P, V, V1, V2)

1-289

4-22 A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and temperature rises to specified values. The work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The initial state is saturated mixture at 90C. The pressure and the specific volume at this state are (Table A-4),

The final specific volume at 800 kPa and 250°C is (Table A-6) Since this is a linear process, the work done is equal to the area under the process line 1-2:

(

)

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] T1=90 [C] x1=0.10 P2=800 [kPa] T2=250 [C] "Properties" Fluid$='steam_iapws' P1=pressure(Fluid$, T=T1, x=x1) v1_f=volume(Fluid$, T=T1, x=0) v1_g=volume(Fluid$, T=T1, x=1) v1=v1_f+x1*(v1_g-v1_f) v2=volume(Fluid$, P=P2, T=T2) "Analysis" W_b_out=(P1+P2)/2*m*(v2-v1)

4-23 An ideal gas undergoes two processes in a piston-cylinder device. The process is to be sketched on a P-V diagram; an expression for the ratio of the compression to expansion work is to be obtained and this ratio is to be calculated for given values of n and r. Assumptions The process is quasi-equilibrium. Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-290

(a) The processes on a P-V diagram is as follows: (b) The ratio of the compression-to-expansion work is called the back-work ratio BWR. Process 1-2:

∫ The process is

= constant,

, and the integration results in

where the ideal gas law has been used. However, the compression work is

Process 2-3: ∫ The process is P = constant and the integration gives where

. Using the ideal gas law, the expansion work is

The back-work ratio is defined as

Since process 1-2 is polytropic, the temperature-volume relation for the ideal gas is ( ) where r is the compression ratio

( )

. Since process 2-3 is constant pressure, the combined ideal gas law gives

The back-work ratio becomes

1-291

EES (Engineering Equation Solver) SOLUTION "GIVEN" n=1.4 r=6 "(c)" BWR=1/(n-1)*(1-r^(1-n))/(r-1)

4-24 Water in a cylinder equipped with a spring is heated and evaporated. The vapor expands until it compresses the spring 20 cm. The final pressure and temperature, and the boundary work done are to be determined, and the process is to be shown on a P-V diagram. Assumptions The process is quasi-equilibrium. Analysis (a) The final pressure is determined from (

)

The specific and total volumes at the three states are }

At 450 kPa, and mixture and thus the final temperature is

. Noting that

, the final state is a saturated

(b) The pressure remains constant during process 1-2 and changes linearly (a straight line) during process 2-3. Then the boundary work during this process is simply the total area under the process curve,

(

)(

)

1-292

Discussion The positive sign indicates that work is done by the system (work output).

EES (Engineering Equation Solver) SOLUTION "Given" m=50 [kg] P1=250 [kPa] T1=25 [C] A=0.1 [m^2] Vol2=0.2 [m^3] k=100 [kN/m] x=0.20 [m] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) "Analysis" "(a)" P2=P1 F_s=k*x P3=P2+F_s/A Vol1=m*v1 Vol3=Vol2+x*A v3=Vol3/m T3=temperature(Fluid$, P=P3, v=v3) "(b)" W_b_out=P1*(Vol2-Vol1)+(P2+P3)/2*(Vol3-Vol2)

4-25 Problem 4-24 is reconsidered. The effect of the spring constant on the final pressure in the cylinder and the boundary work done as the spring constant varies from to is to be investigated. The final pressure and the boundary work are to be plotted against the spring constant. Analysis The problem is solved using EES, and the solution is given below. P[1]=250 [kPa] m=50 [kg] T[1]=25 [C] P[2]=P[1] V[2]=0.2 [m^3] A=0.1[m^2] k=100 [kN/m] DELTAx=20 [cm] Spring_const=k/A^2 V[1]=m*spvol[1] spvol[1]=volume(Steam_iapws,P=P[1],T=T[1]) V[2]=m*spvol[2] V[3]=V[2]+A*DELTAx*convert(cm,m) V[3]=m*spvol[3] T[2]=temperature(Steam_iapws,P=P[2],v=spvol[2]) T[3]=temperature(Steam_iapws,P=P[3],v=spvol[3]) Wnet_other = 0 W_out=Wnet_other + W_b12+W_b23 W_b12=P[1]*(V[2]-V[1]) W_b23=P[2]*(V[3]-V[2])+Spring_const/2*(V[3]-V[2])^2 P[3]=P[2]+(Spring_const)*(V[3]-V[2]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-293

] ] 50 100 150 200 250 300 350 400 450 500

350 450 550 650 750 850 950 1050 1150 1250

] 43.46 44.46 45.46 46.46 47.46 48.46 49.46 50.46 51.46 52.46

1-294

4-26 gas in a cylinder is compressed until the volume drops to a specified value. The pressure changes during the process with volume as . The boundary work done during this process is to be determined. Assumptions The process is quasi-equilibrium. Analysis The boundary work done during this process is determined from ∫

∫ ( (

)

( )(

) )

Discussion The negative sign indicates that work is done on the system (work input).

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.3 [m^3] V2=0.1 [m^3] a=8 [kPa-m^6] "Analysis" "Integrating by hand gives" W_b_out=-a*(1/V2-1/V1) "Having EES to perform integration" P=a/V^2 W_b_out_integration=integral(P, V, V2, V1)

Closed System Energy Analysis 4-27E A piston-cylinder device involves expansion work and work input by a stirring device. The net change of internal energy is to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-295

1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved 3. The thermal energy stored in the cylinder itself is negligible. Analysis This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟

where and

Substituting,

EES (Engineering Equation Solver) SOLUTION "GIVEN" W_out=15000 [lbf-ft] W_in=10.228 [Btu] "ANALYSIS" W_in-W_out*Convert(lbf-ft, Btu)=DELTAU

4-28 Motor oil is contained in a rigid container that is equipped with a stirring device. The rate of specific energy increase is to be determined. Analysis This is a closed system since no mass enters or leaves. The energy balance for closed system can be expressed as ⏟

⏟ ̇

Then,

Dividing this by the mass in the system gives ̇

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=2.5 [kg] Q_dot_in=1 [W] W_dot_in=1.5 [W] "ANALYSIS" Q_dot_in+W_dot_in=m*DELTAe_dot

4-29 The table is to be completed using conservation of energy principle for a closed system. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-296

Analysis The energy balance for a closed system can be expressed as ⏟

⏟

Application of this equation gives the following completed table: m (kg) 280

440

1020

860

-53.3

-350

130

550

-96

-40

260

300

-150

300

550

750

500

-250

-400

-200

500

300

-100

4-30 A rigid tank is initially filled with superheated R-134a. Heat is transferred to the tank 4-31 until the pressure inside rises to a specified value. The mass of the refrigerant and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions. Analysis (a) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ⏟

⏟

Using data from the refrigerant tables (Tables A-11 through A-13), the properties of R-134a are determined to be 3 P1 = 160 kPa ïü ïý v f = 0.0007435, v g = 0.12355 m /kg ïïþ u f = 31.06, u fg = 190.31 kJ/kg x1 = 0.4

}

1-297

(b) Then the heat transfer to the tank becomes

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.5 [m^3] P1=160 [kPa] x1=0.40 P2=700 [kPa] "Analysis" Fluid$='R134a' "(a)" v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) v2=v1 "fixed mass, rigid tank" u2=intenergy(Fluid$, P=P2, v=v2) m=Vol/v1 "(b)" Q_in=m*(u2-u1) "energy balance" v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) u1_f=intenergy(Fluid$, P=P1, x=0) u1_g=intenergy(Fluid$, P=P1, x=1) u1_fg=u1_g-u1_f

4-31E A rigid tank is initially filled with saturated R-134a vapor. Heat is transferred from the refrigerant until the pressure inside drops to a specified value. The final temperature, the mass of the refrigerant that has condensed, and the amount of heat transfer are to be determined. Also, the process is to be shown on a P-v diagram. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions. Analysis (a) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ⏟ ⏟

Using data from the refrigerant tables (Tables A-11E through A-13E), the properties of R-134a are determined to be } }

1-298

The final state is saturated mixture. Thus, (b) The total mass and the amount of refrigerant that has condensed are

Also, (c) Substituting,

= 4167 Btu

EES (Engineering Equation Solver) SOLUTION "Given" Vol=20 [ft^3] P1=160 [psia] x1=1 P2=50 [psia] "Analysis" Fluid$='R134a' "(a)" v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) v2=v1 "fixed mass, rigid tank" u2=intenergy(Fluid$, P=P2, v=v2) T2=temperature(Fluid$, P=P2, v=v2) "(b)" x2=quality(Fluid$, P=P2, v=v2) m=Vol/v1 m_f=(1-x2)*m "(b)" -Q_out=m*(u2-u1) "energy balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-299

v2_f=volume(Fluid$, P=P2, x=0) v2_g=volume(Fluid$, P=P2, x=1) u2_f=intenergy(Fluid$, P=P2, x=0) u2_g=intenergy(Fluid$, P=P2, x=1) u2_fg=u2_g-u2_f

4-32 Water contained in a rigid vessel is heated. The heat transfer is to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved 3. The thermal energy stored in the vessel itself is negligible. Analysis We take water as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

The properties at the initial and final states are (Tables A-4 and A-6, or from EES) } }

Note that it is difficult to obtain internal energy at the final state Instead, we used EES. The mass in the system is

from Table A-6.

Substituting,

1-300

V=0.010 [m^3] T1=100 [C] x1=0.123 T2=180 [C] "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$,T=T1,x=x1) v1=volume(Fluid$,T=T1,x=x1) v2=v1 u2=intenergy(Fluid$,T=T2,v=v2) x2=quality(Fluid$,T=T2,v=v2) m=V/v1 Q_in=m*(u2-u1)

4-33E R-134a contained in a rigid vessel is heated. The heat transfer is to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved 3. The thermal energy stored in the vessel itself is negligible.

Analysis We take R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟

The properties at the initial and final states are (Tables A-11E, A-13E) } } Note that the final state is superheated vapor and the internal energy at this state should be obtained by interpolation using 50 psia and 60 psia mini tables (100F line) in Table A-13E. The mass in the system is

Substituting,

1-301

EES (Engineering Equation Solver) SOLUTION Vol=1 [ft^3] T1=-20 [F] x1=0.277 T2=100 [F] Fluid$='R134a' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) v2=v1 u2=intenergy(Fluid$, T=T2, v=v2) m=Vol/v1 Q_in=m*(u2-u1)

4-34 A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled at constant pressure. The amount of heat loss is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

since during a constant pressure quasi-equilibrium process. The properties of R-134a are (Tables A-11 through A-13):

1-302

} } Substituting,

EES (Engineering Equation Solver) SOLUTION "Given" m=5 [kg] P1=800 [kPa] T1=70 [C] P2=P1 T2=15 [C] "Analysis" Fluid$='R134a' h1=enthalpy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, T=T2, x=0) "use saturated liquid state" -Q_out=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process"

4-35E A cylinder contains water initially at a specified state. The water is heated at constant pressure. The final temperature of the water is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. The thermal energy stored in the cylinder itself is negligible. 3. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

since during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-6E)

1-303

} Substituting, Then,

}

EES (Engineering Equation Solver) SOLUTION "Given" m=0.5 [lbm] P1=120 [psia] Vol1=2 [ft^3] Q_in=200 [Btu] P2=P1 "Analysis" Fluid$='steam_iapws' v1=Vol1/m h1=enthalpy(Fluid$, P=P1, v=v1) Q_in=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process" T2=temperature(Fluid$, P=P2, h=h2)

4-36 Saturated liquid water is heated at constant pressure to a saturated vapor in a piston-cylinder device. The heat transfer is to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

1-304

⏟

since during a constant pressure quasi-equilibrium process. Since water changes from saturated liquid to saturated vapor, we have

since

EES (Engineering Equation Solver) SOLUTION "Given" m=2 [kg] T=150 [C] "Analysis" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T, x=0) h_g=enthalpy(Fluid$, T=T, x=1) h_fg=h_g-h_f Q_in=m*h_fg

4-37 A cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically as it is stirred by a paddle-wheel at constant pressure. The voltage of the current source is to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-305

2. The cylinder is well-insulated and thus heat transfer is negligible. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟

since during a constant pressure quasi-equilibrium process. The properties of water are (Tables A-4 through A-6)

P1 = 175 kPa ïü ïý h1 = h f @175 kPa = 487.01 kJ/kg ïïþ v1 = v f @175 kPa = 0.001057 m3 /kg sat.liquid }

Substituting,

(

)

1-306

Vol1=5 [L] x1=0 P1=175 [kPa] P2=P1 I=8 [Amp] DELTAt=45*60 [s] x2=0.5 W_pw_in=400 [kJ] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, x=x1) v1=volume(Fluid$, P=P1, x=x1) h2=enthalpy(Fluid$, P=P2, x=x2) m=Vol1/v1*Convert(L, m^3) W_e_in+W_pw_in=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process" W_e_in*Convert(kJ, Volt-Amp-s)=Voltage*I*DELTAt

4-38 A room is heated by an electrical radiator containing heating oil. Heat is lost from the room. The time period during which the heater is on is to be determined. Assumptions 1. 2. 3. 4.

Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa. The kinetic and potential energy changes are negligible, Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. The local atmospheric pressure is 100 kPa. 5 The room is air-tight so that no air leaks in and out during the process.

Properties The gas constant of air is temperature (Table A-2). Oil properties are given to be

(Table A-1). Also, and

for air at room .

Analysis We take the air in the room and the oil in the radiator to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary constant-volume closed system can be expressed as

⏟

⏟ ̇ ̇ ]

]

The mass of air and oil are

1-307

→ Discussion In practice, the pressure in the room will remain constant during this process rather than the volume, and some air will leak out as the air expands. As a result, the air in the room will undergo a constant pressure expansion process. Therefore, it is more proper to be conservative and to using H instead of use U in heating and air-conditioning applications.

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol_oil=0.040 [m^3] Vol_room=50 [m^3] W_dot_in=2.4 [kW] Q_dot_out=0.35 [kW] T_1=10 [C] T_oil_2=50 [C] T_air_2=20 [C] rho_oil=950 [kg/m^3] c_p_oil=2.2 [kJ/kg-C] "PROPERTIES" c_v_air=0.718 [kJ/kg-K] R=0.287 [kJ/kg-K] "Table A-2" "ANALYSIS" P=101.3 [kPa] "assumed" m_air=(P*Vol_room)/(R*(T_1+273)) m_oil=rho_oil*Vol_oil (W_dot_in-Q_dot_out)*time=m_air*c_v_air*(T_air_2-T_1)+m_oil*C_p_oil*(T_oil_2-T_1) "energy balance"

4-39 A saturated water mixture contained in a spring-loaded piston-cylinder device is heated until the pressure and volume rise to specified values. The heat transfer and the work done are to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟

1-308

The initial state is saturated mixture at 75 kPa. The specific volume and internal energy at this state are (Table A-5),

The mass of water is

The final specific volume is

The final state is now fixed. The internal energy at this specific volume and 225 kPa pressure is (Table A-6) Since this is a linear process, the work done is equal to the area under the process line 1-2: (

)

Substituting into energy balance equation gives

EES (Engineering Equation Solver) SOLUTION "Given" P1=75 [kPa] x1=0.08 Vol1=2 [m^3] Vol2=5 [m^3] P2=225 [kPa] "Analysis" Fluid$='Steam_IAPWS' v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) u1_f=intenergy(Fluid$, P=P1, x=0) u1_g=intenergy(Fluid$, P=P1, x=1) u1_fg=u1_g-u1_f v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) m=Vol1/v1 v2=Vol2/m u2=intenergy(Fluid$, P=P2, v=v2) W_b_out=(P1+P2)/2*(Vol2-Vol1) Q_in=W_b_out+m*(u2-u1)

1-309

4-40 A cylinder equipped with a set of stops for the piston to rest on is initially filled with saturated water vapor at a specified pressure. Heat is transferred to water until the volume doubles. The final temperature, the boundary work done by the steam, and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟

The properties of steam are (Tables A-4 through A-6)

P1 = 250 kPa ïüï v1 = v g @ 250 kPa = 0.71873 m3 /kg ý ïïþ u1 = u g @ 250 kPa = 2536.8 kJ/kg sat.vapor

üï T3 = 662°C ïý v 3 = 1.4375 m3 /kg ïïþ u3 = 3411.4 kJ/kg P3 = 300 kPa

(b) The work done during process 1-2 is zero (since V = const) and the work done during the constant pressure process 2-3 is ∫

(

)

1-310

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.6 [m^3] P1=250 [kPa] x1=1 P2=300 [kPa] "state 2 is when the piston starts rising" Vol3=2*Vol1 "state 3 is final state" "Analysis" Fluid$='steam_iapws' "(a)" v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) m=Vol1/v1 v3=Vol3/m P3=P2 u3=intenergy(Fluid$, P=P3, v=v3) T3=temperature(Fluid$, P=P3, v=v3) "(b)" Vol2=Vol1 W_b_out=P2*(Vol3-Vol2) "(c)" Q_in-W_b_out=m*(u3-u1) "energy balance"

4-41 One part of an insulated tank contains compressed liquid while the other side is evacuated. The partition is then removed, and water is allowed to expand into the entire tank. The final temperature and the volume of the tank are to be determined. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. The tank is insulated and thus heat transfer is negligible. 3. There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ⏟ ⏟

1-311

The properties of water are (Tables A-4 through A-6)

P1 = 600 kPa ïüï v1 @v f @ 60°C = 0.001017 m3 /kg ý ïïþ u1 @ u f @ 60°C = 251.16 kJ/kg T1 = 60°C We now assume the final state in the tank is saturated liquid-vapor mixture and determine quality. This assumption will be verified if we get a quality between 0 and 1.

P2 = 10 kPa ïüï v f = 0.001010, v g = 14.670 m3 /kg ý ïïþ u f = 191.79, u fg = 2245.4 kJ/kg (u2 = u1 )

Thus, ] and,

EES (Engineering Equation Solver) SOLUTION "Given" m=2.5 [kg] T1=60 [C] P1=600 [kPa] P2=10 [kPa] "Analysis" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=0) u1=intenergy(Fluid$, T=T1, x=0) u2=u1 "from energy balance" x2=quality(Fluid$, P=P2, u=u2) T2=temperature(Fluid$, P=P2, u=u2) v2=volume(Fluid$, P=P2, u=u2) Vol2=m*v2

4-42 Problem 4-41 is reconsidered. The effect of the initial pressure of water on the final temperature in the tank as the initial pressure varies from 100 kPa to 600 kPa is to be investigated. The final temperature is to be plotted against the initial pressure. Analysis The problem is solved using EES, and the solution is given below. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-312

m=2.5 [kg] P[1]=600 [kPa] T[1]=60 [C] P[2]=10 [kPa] Fluid$='Steam_IAPWS' E_in-E_out=DELTAE E_in=0 E_out=0 DELTAE=m*(u[2]-u[1]) u[1]=INTENERGY(Fluid$,P=P[1], T=T[1]) v[1]=volume(Fluid$,P=P[1], T=T[1]) T[2]=temperature(Fluid$,P=P[2], u=u[2])

T_2=T[2] v[2]=volume(Fluid$,P=P[2], u=u[2]) V_total=m*v[2]

] 100 200 300 400 500 600

] 45.79 45.79 45.79 45.79 45.79 45.79

1-313

4-43 Two tanks initially separated by a partition contain steam at different states. Now the partition is removed and they are allowed to mix until equilibrium is established. The temperature and quality of the steam at the final state and the amount of heat lost from the tanks are to be determined. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions.

Analysis (a) We take the contents of both tanks as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as ⏟

⏟

] ] The properties of steam in both tanks at the initial state are (Tables A-4 through A-6) } © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-314

} ] The total volume and total mass of the system are

Now, the specific volume at the final state may be determined

which fixes the final state and we can determine other properties }

(b) Substituting, ]

]

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_A=2 [kg] T_1A=300 [C] P_1A=1000 [kPa] m_B=3 [kg] T_1B=150 [C] x_1B=0.5 P_2=300 [kPa] "PROPERTIES" Fluid$='Steam_iapws' u_1A=intenergy(Fluid$, T=T_1A, P=P_1A) v_1A=volume(Fluid$, T=T_1A, P=P_1A) u_1B=intenergy(Fluid$, T=T_1B, x=x_1B) v_1B=volume(Fluid$, T=T_1B, x=x_1B) "ANALYSIS" Vol_1A=m_A*v_1A "volume of Tank A" Vol_1B=m_B*v_1B "volume of Tank B" Vol_t=Vol_1A+Vol_1B "total volume" m_t=m_A+m_B "total mass" v_2=Vol_t/m_t "final specific volume" T_2=temperature(Fluid$, P=P_2, v=v_2) "final temperature" x_2=quality(Fluid$, P=P_2, v=v_2) "final quality" u_2=intenergy(Fluid$, P=P_2, v=v_2) "final internal energy" -Q_out=m_A*(u_2-u_1A)+m_B*(u_2-u_1B) "energy balance"

1-315

Specific Heats, u and h of Ideal Gases 4-44C Very close, but no. Because the heat transfer during this process is

, and

varies with temperature.

4-45C The energy required is vary with pressure.

, which will be the same in both cases. This is because the

of an ideal gas does not

4-46C The energy required is vary with volume.

, which will be the same in both cases. This is because the

of an ideal gas does not

4-47C For the constant pressure case. This is because the heat transfer to an ideal gas is at constant volume, and is always greater than .

at constant pressure,

4-48C It can be used for any kind of process of an ideal gas.

4-49C It can be used for any kind of process of an ideal gas.

4-50C Modeling both gases as ideal gases with constant specific heats, we have

Since both gases undergo the same temperature change, the gas with the greatest will experience the largest change in internal energy. Similarly, the gas with the largest will have the greatest enthalpy change. Inspection of Table A-2a indicates that air will experience the greatest change in both cases.

4-51 The desired result is obtained by multiplying the first relation by the molar mass M, or

4-52E The internal energy change of air is to be determined for two cases of specified temperature changes. Assumptions At specified conditions, air behaves as an ideal gas. Properties The constant-volume specific heat of air at room temperature is (Table A-2Ea). Analysis Using the specific heat at constant volume, If we use the same room temperature specific heat value, the internal energy change will be the same for the second case. However, if we consider the variation of specific heat with temperature and use the specific heat values from Table A-2Eb, we have at 150F and at 50F. Then,

The two results differ from each other by about 0.8%.

1-316

EES (Engineering Equation Solver) SOLUTION T1_a=100 [F] T2_a=200 [F] T1_b=0 [F] T2_b=100 [F] c_v=0.171 [Btu/lbm-R] "at room temperature, Table A-2a" c_v_a=0.1725 [Btu/lbm-R] "at 150 F, Table A-2b" c_v_b=0.1712 [Btu/lbm-R] "at 50 F, Table A-2b" "Analysis" DELTAu=c_v*(T2_a-T1_a) DELTAu_a=c_v_a*(T2_a-T1_a) DELTAu_b=c_v_b*(T2_a-T1_a)

4-53 Neon is compressed isothermally in a compressor. The specific volume and enthalpy changes are to be determined. Assumptions At specified conditions, neon behaves as an ideal gas. Properties The gas constant of neon is

and the constant-pressure specific heat of neon is

(Table A-2a). Analysis At the compressor inlet, the specific volume is

Similarly, at the compressor exit,

The change in the specific volume caused by the compressor is Since the process is isothermal,

EES (Engineering Equation Solver) SOLUTION P1=100 [kPa] T=20 [C] P2=500 [kPa] R=0.4119 [kJ/kg-K] v1=R*(T+273)/P1 v2=R*(T+273)/P2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-317

DELTAv=v2-v1 DELTAh=0 "since DELTAh=c_p*DELTAT and DELTAT=0"

4-54 The enthalpy change of oxygen is to be determined for two cases of specified temperature changes. Assumptions At specified conditions, oxygen behaves as an ideal gas. Properties The constant-pressure specific heat of oxygen at room temperature is

(Table A-2a).

Analysis Using the specific heat at constant pressure, If we use the same room temperature specific heat value, the enthalpy change will be the same for the second case. However, if we consider the variation of specific heat with temperature and use the specific heat values from Table A-2b, we have at 200C (= 473 K) and at 100C ( 373 K). Then,

The two results differ from each other by about 3%. The pressure has no influence on the enthalpy of an ideal gas.

EES (Engineering Equation Solver) SOLUTION "Given" T1_a=150 [C] T2_a=250 [C] T1_b=0 [C] T2_b=100 [C] "Properties" c_p_a=0.918 [kJ/kg-K] "at room temperature, Table A-2a" c_p_b=0.964 [kJ/kg-K] "at 473 K, Table A-2b" c_p_c=0.934 [kJ/kg-K] "at 373 K, Table A-2b" "Analysis" DELTAh_a=c_p_a*(T2_a-T1_a) DELTAh_b=c_p_b*(T2_a-T1_a) DELTAh_c=c_p_c*(T2_b-T1_b)

4-55 A spring-loaded piston-cylinder device is filled with nitrogen. Nitrogen is now heated until its volume increases by 10%. The changes in the internal energy and enthalpy of the nitrogen are to be determined. Properties The gas constant of nitrogen is . The specific heats of nitrogen at room temperature are and (Table A-2a). Analysis The initial volume of nitrogen is

The process experienced by this system is a linear P-v process. The equation for this line is where

is the system pressure when its specific volume is

. The spring equation may be written as

Constant c is hence

1-318

The final pressure is then

The final temperature is

Using the specific heats,

EES (Engineering Equation Solver) SOLUTION m=0.010 [kg] k=1 [kN/m] D=0.10 [m] P1=120 [kPa] T1=(27+273) [K] Vol2=1.10*Vol1 R=0.2968 [kJ/kg-K] c_v=0.743 [kJ/kg-K] c_p=1.039 [kJ/kg-K] Vol1=m*R*T1/P1 c=k/A^2 A=pi*D^2/4 P2=P1+c*(Vol2-Vol1) T2=P2*Vol2/(m*R) DELTAu=c_v*(T2-T1) DELTAh=c_p*(T2-T1)

4-56 The internal energy change of hydrogen gas during a heating process is to be determined using an empirical specific heat relation, constant specific heat at average temperature, and constant specific heat at room temperature. Analysis (a) Using the empirical relation for ̄ ̄

from Table A-2c and relating it to ̄

1-319

where a = 29.11,

, ̄

∫

, and ̄

∫

. Then, ]

= (29.11- 8.314)(800 - 200) - 12 (0.1916 ´ 10- 2 )(800 2 - 200 2 ) + 13 (0.4003´ 10- 5 )(8003 - 2003 ) - 14 (0.8704 ´ 10- 9 )(800 4 - 200 4 ) ̄

(b) Using a constant

value from Table A-2b at the average temperature of 500 K,

value from Table A-2a at room temperature,

EES (Engineering Equation Solver) SOLUTION "Given" T1=200 [K] T2=800 [K] "Properties" Fluid$='H2' M=molarmass(Fluid$) R_u=8.314 [kJ/kmol-K] c_v_avg=10.389 [kJ/kg-K] "at 500 K, Table A-2b" c_v_room=10.183 [kJ/kg-K] "at 300 K, Table A-2a" "Analysis" "(a)" a=29.11 b=-0.1916E-2 c=0.4003E-5 d=-0.8704E-9 "from Table A-2c of the text" DELTAu_bar=(a-R_u)*(T2-T1)+1/2*b*(T2^2-T1^2)+1/3*c*(T2^3-T1^3)+1/4*d*(T2^4-T1^4) DELTAu_empirical=DELTAu_bar/M "(b)" DELTAu_average=c_v_avg*(T2-T1) "(c)" DELTAu_room=c_v_room*(T2-T1)

4-57E The enthalpy change of oxygen gas during a heating process is to be determined using an empirical specific heat relation, constant specific heat at average temperature, and constant specific heat at room temperature. Analysis (a) Using the empirical relation for ̄ from Table A-2Ec, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-320

̄ where a = 6.085,

, ̄

, and ∫

∫

. Then, ]

= 6.085(1500 - 800) + 12 (0.2017 ´ 10- 2 )(1500 2 - 800 2 ) - 13 (0.05275´ 10- 5 )(15003 - 8003 ) + 14 (0.05372 ´ 10- 9 )(1500 4 - 800 4 ) ̄ (b) Using the constant

value from Table A-2Eb at the average temperature of 1150 R,

value from Table A-2Ea at room temperature,

EES (Engineering Equation Solver) SOLUTION "Given" T1=800 [R] T2=1500 [R] "Properties" c_p=0.219 [Btu/lbm-R] "at 537 R, Table A-2Eb" c_p_avg=0.242 [Btu/lbm-R] "at 1150 R, Table A-2Ea" MM=31.999 [lbm/lbmol] "Analysis" "(a)" a=6.085 b=0.2017E-2 c=-0.05275E-5 d=0.05372E-9 "constants are from Table A-2Ec of the text" DELTAh_bar=a*(T2-T1)+1/2*b*(T2^2-T1^2)+1/3*c*(T2^3-T1^3)+1/4*d*(T2^4-T1^4) DELTAh_empirical=DELTAH_bar/MM "(b)" DELTAh_average=c_p_avg*(T2-T1) "(c)" DELTAh_room=c_p*(T2-T1)

Closed System Energy Analysis: Ideal Gases 4-58C No, it isn't. This is because the first law relation Q - W = U reduces to W = 0 in this case since the system is adiabatic (Q = 0) and U = 0 for the isothermal processes of ideal gases. Therefore, this adiabatic system cannot receive any net work at constant temperature.

1-321

4-59 Nitrogen in a rigid vessel is cooled by rejecting heat. The internal energy change of the nitrogen is to be determined. Assumptions 1. Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 126.2 K and 3.39 MPa. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats at room temperature can be used for nitrogen. Analysis We take the nitrogen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

Thus,

EES (Engineering Equation Solver) SOLUTION q_out=100 [kJ/kg] DELTAu=-q_out

4-60E Nitrogen in a rigid vessel is cooled. The work done and heat transfer are to be determined. Assumptions 1. Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 126.2 K (227.1 R) and 3.39 MPa (492 psia). 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats at room temperature can be used for nitrogen. Properties For nitrogen, at room temperature (Table A-2Ea). Analysis We take the nitrogen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

1-322

There is no work done since the vessel is rigid: Since the specific volume is constant during the process, the final temperature is determined from ideal gas equation to be

Substituting,

EES (Engineering Equation Solver) SOLUTION P1=100 [psia] T1=(300+460) [R] P2=50 [psia] w=0 "rigid vessel" T2=T1*(P2/P1) q_out=c_v*(T1-T2) c_v=0.177 [Btu/lbm-R]

4-61 Air in a closed system undergoes an isothermal process. The initial volume, the work done, and the heat transfer are to be determined. Assumptions 1. Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 132.5 K and 3.77 MPa. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats can be used for air. Properties The gas constant of air is (Table A-1). Analysis We take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

Ein  Eout



Net energy t ransfer by heat , w ork, and mass

Esyst em Change in int ernal, kinet ic, pot ent ial, et c. energies

Qin  Wb ,out  U  mcv (T2  T1)

Qin  Wb ,out  0

(since T1  T2 )

Qin  W b ,out The initial volume is

1-323

Using the boundary work relation for the isothermal process of an ideal gas gives 2

W b ,out  m  Pd v  m RT  1

 m RT ln

v2 P  m RT ln 1 v1 P2

 (2 kg)(0.287 kPa  m 3 / kg  K)(473 K)ln

600 kPa  547.1 kJ 80 kPa

From energy balance equation,

EES (Engineering Equation Solver) SOLUTION "Given" m=2 [kg] P1=600 [kPa] T=(200+273) [K] P2=80 [kPa] "Properties" R=0.287 [kJ/kg-K] "Analysis" Vol1=(m*R*T)/P1 W_b_out=m*R*T*ln(P1/P2) Q_in-W_b_out=DELTAU DELTAU=0 "temperature is constant"

4-62E Carbon dioxide contained in a piston-cylinder device undergoes a constant-pressure process. The work done and the heat transfer are to be determined. Assumptions 1. Carbon dioxide is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 547.5 R and 1071 psia. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats at room temperature can be used for . Properties The properties of at room temperature are and (Table A-2Ea). Analysis We take as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

1-324

Using the boundary work relation for the isobaric process of an ideal gas gives Substituting into energy balance equation,

EES (Engineering Equation Solver) SOLUTION P1=15 [psia] T1=(80+460) [R] T2=(200+460) [R] R=0.04513 [Btu/lbm-R] c_v=0.158 [Btu/lbm-R] w_b_out=R*(T2-T1) q_in=w_b_out+c_v*(T2-T1)

4-63 The hydrogen gas in a rigid tank is cooled until its temperature drops to 300 K. The final pressure in the tank and the amount of heat transfer are to be determined. Assumptions 1. Hydrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -240C and 1.30 MPa. 2. The tank is stationary, and thus the kinetic and potential energy changes are negligible, Properties The gas constant of hydrogen is (Table A-1). The constant volume specific heat of hydrogen at the average temperature of 450 K is, (Table A-2). Analysis (a) The final pressure of hydrogen can be determined from the ideal gas relation, → (b) We take the hydrogen in the tank as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

where

Substituting into the energy balance,

1-325

EES (Engineering Equation Solver) SOLUTION "Given" V=3 [m^3] P1=250 [kPa] T1=550 [K] T2=350 [K] "Properties" Fluid$='H2' u1=intenergy(Fluid$, T=T1) u2=intenergy(Fluid$, T=T2) c_p=CP(Fluid$,T=298) c_v=CV(Fluid$,T=298) R=c_p-c_v "Analysis" "(a)" P2=(T2/T1)*P1 "(b)" m=(P1*V)/(R*T1) -Q_out=m*(u2-u1) "energy balance"

4-64 Oxygen is heated to experience a specified temperature change. The heat transfer is to be determined for two cases. Assumptions 1. Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats can be used for oxygen. Properties The specific heats of oxygen at the average temperature of (20+120)/2=70C=343 K are and (Table A-2b). Analysis We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for a constant-volume process can be expressed as ⏟

⏟

The energy balance during a constant-pressure process (such as in a pistoncylinder device) can be expressed as ⏟

⏟

1-326

since during a constant pressure quasi-equilibrium process. Substituting for both cases,

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] T1=20 [C] T2=120 [C] "Properties" c_p=0.927 [kJ/kg-K] c_v=0.667 [kJ/kg-K] "at 343 K, Table A-2b" "Analysis" Q_in_V=m*c_v*(T2-T1) "at constant volume" Q_in_P=m*c_p*(T2-T1) "at constant pressure"

4-65E A paddle wheel in an oxygen tank is rotated until the pressure inside rises to 20 psia while some heat is lost to the surroundings. The paddle wheel work done is to be determined. Assumptions 1. Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -181F and 736 psia. 2. The kinetic and potential energy changes are negligible, 3. The energy stored in the paddle wheel is negligible. 4. This is a rigid tank and thus its volume remains constant. Properties The gas constant and molar mass of oxygen are and (Table A1E). The specific heat of oxygen at the average temperature of is (Table A-2E). Analysis We take the oxygen in the tank as our system. This is a closed system since no mass enters or leaves. The energy balance for this system can be expressed as

⏟

1-327

The final temperature and the mass of oxygen are →

Substituting,

EES (Engineering Equation Solver) SOLUTION "Given" V=10 [ft^3] P1=14.7 [psia] T1=(80+460) [R] P2=20 [psia] Q_out=20 [Btu] "Properties" Fluid$='O2' c_p=CP(Fluid$,T=540) c_v=CV(Fluid$,T=540) R=(c_p-c_v)*Convert(Btu/lbm-R, psia-ft^3/lbm-R) u1=intenergy(Fluid$, T=T1) u2=intenergy(Fluid$, T=T2) "Analysis" T2=(P2/P1)*T1 m=(P1*V)/(R*T1) W_pw_in-Q_out=m*(u2-u1) "energy balance"

4-66 A room is heated by a radiator, and the warm air is distributed by a fan. Heat is lost from the room. The time it takes for the air temperature to rise to 20°C is to be determined. Assumptions 1. Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -141C and 3.77 MPa. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 4. The local atmospheric pressure is 100 kPa. 5 The room is air-tight so that no air leaks in and out during the process. Properties The gas constant of air is (Table A-1). Also, for air at room temperature (Table A-2). Analysis We take the air in the room to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary constant-volume closed system can be expressed as

1-328

⏟

or, ̇

̇ ̇

The mass of air is

Using the

value at room temperature, ]

It yields t = 831 s Discussion In practice, the pressure in the room will remain constant during this process rather than the volume, and some air will leak out as the air expands. As a result, the air in the room will undergo a constant pressure expansion process. Therefore, it is more proper to be conservative and to using H instead of use U in heating and air-conditioning applications.

EES (Engineering Equation Solver) SOLUTION "Given" V=4*5*7 [m^3] Q_dot_in=10000[kJ/h]*Convert(kJ/h, kW) W_dot_fan_in=100[W]*Convert(W, kW) Q_dot_out=5000[kJ/h]*Convert(kJ/h, kW) T1=(10+273) [K] T2=(20+273) [K] "Properties" Fluid$='Air' c_p=CP(Fluid$,T=298) c_v=CV(Fluid$,T=298) R=c_p-c_v "Analysis" P=100 [kPa] “assumed local atmospheric pressure" m=(P*V)/(R*T1) (Q_dot_in+W_dot_fan_in-Q_dot_out)*DELTAt=m*C_v*(T2-T1) "energy balance"

1-329

4-67 One part of an insulated rigid tank contains an ideal gas while the other side is evacuated. The final temperature and pressure in the tank are to be determined when the partition is removed. Assumptions 1. The kinetic and potential energy changes are negligible, 2. The tank is insulated and thus heat transfer is negligible. Analysis We take the entire tank as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

Therefore, Since u = u(T) for an ideal gas. Then, →

EES (Engineering Equation Solver) SOLUTION "Given" m=4 [kg] P1=800 [kPa] T1=(50+273) [K] RatioV=1/2 "RatioV=V1/V2" "Analysis" T2=T1 "from energy balance" T2_C=T2-273 P2=RatioV*P1 "from ideal gas relation"

4-68 An ideal gas at a specified state contained in a piston-cylinder device undergoes an isothermal process during which heat is lost from the gas. The final pressure and volume are to be determined. Assumptions 1. The kinetic and potential energy changes are negligible, .

1-330

Properties The properties of air are and (Table A-2a). Analysis We take the ideal gas as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system fort he process 1-3 can be expressed as ⏟

⏟

The boundary work for an isothermal process is determined from

∫

Thus, (

)

(

)

(

)

and The final pressure is found from the combined ideal gas law: →

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] Vol1=0.6 [m^3] Q_out=60 [kJ] "Properties" R=0.287 [kJ/kg-K] c_v=0.718 [kJ/kg-K] "Table A-2a" "Analysis" DELTAU=0 "temperature is constant, T1=T2" -Q_out-W_b_out=DELTAU © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-331

W_b_out=P1*Vol1*ln(VolumeRatio) "VolumeRatio=Vol2/Vol1" Vol2=Vol1*VolumeRatio P2=P1*(1/VolumeRatio)

4-69 A resistance heater is to raise the air temperature in the room from 5 to 25°C within 11 min. The required power rating of the resistance heater is to be determined. Assumptions 1. 2. 3. 4. 5.

Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of –141C and 3.77 MPa. The kinetic and potential energy changes are negligible, Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. Heat losses from the room are negligible. The room is air-tight so that no air leaks in and out during the process.

Properties The gas constant of air is (Table A-1). Also, for air at room temperature (Table A-2). Analysis We take the air in the room to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary constant-volume closed system can be expressed as

⏟

or, ̇ The mass of air is

Substituting, the power rating of the heater becomes ̇ Discussion In practice, the pressure in the room will remain constant during this process rather than the volume, and some air will leak out as the air expands. As a result, the air in the room will undergo a constant pressure expansion process. Therefore, it is more proper to be conservative and to use H instead of using U in heating and air-conditioning applications.

1-332

EES (Engineering Equation Solver) SOLUTION "Given" V=4*5*6 [m^3] T1=(5+273) [K] T2=(25+273) [K] DELTAt=17*60 [s] P=100 [kPa] "Properties" R=0.287 [kJ/kg-K] c_v=0.718 [kJ/kg-K] "Table A-2a" "Analysis" m=(P*V)/(R*T1) W_dot_e_in=(m*c_v*(T2-T1))/DELTAt "energy balance"

4-70 An insulated cylinder initially contains at a specified state. The is heated electrically for 10 min at constant pressure until the volume doubles. The electric current is to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. The is an ideal gas with constant specific heats. 3. The thermal energy stored in the cylinder itself and the resistance wires is negligible. 4. The compression or expansion process is quasi-equilibrium. Properties The gas constant and molar mass of are and (Table A-1). The specific heat of at the average temperature of is (Table A-2b). Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as ⏟

since temperature of

⏟

during a constant pressure quasi-equilibrium process. The final is

→ The mass of

1-333

Substituting, Then, (

)

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.3 [m^3] P1=200 [kPa] T1=(27+273) [K] Voltage=110 [Volt] DELTAt=(10*60) [s] P2=P1 V2=2*V1 "Properties" Fluid$='CO2' C_p=CP(Fluid$,T=298) C_v=CV(Fluid$,T=298) R=C_p-C_v h1=enthalpy(Fluid$, T=T1) "Analysis" T2=(P2/P1)*(V2/V1)*T1 h2=enthalpy(Fluid$, T=T2) m=(P1*V1)/(R*T1) W_e_in=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process" I=W_e_in/(Voltage*DELTAt)*Convert(kJ/s, Volt-Amp)

4-71 Argon is compressed in a polytropic process. The work done and the heat transfer are to be determined. Assumptions 1. Argon is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 151 K and 4.86 MPa. 2. The kinetic and potential energy changes are negligible, Properties The properties of argon are and (Table A-2a). Analysis We take argon as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

Using the boundary work relation for the polytropic process of an ideal gas gives © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-334

*( )

)

Thus, The temperature at the final state is ( )

(

)

From the energy balance equation, Thus,

EES (Engineering Equation Solver) SOLUTION "Given" n=1.2 P1=120 [kPa] T1=(10+273) [K] P2=800 [kPa] "Properties" R=0.2081 [kJ/kg-K] c_v=0.3122 [kJ/kg-K] "Table A-2a" "Analysis" w_b_out=R*T1/(1-n)*((P2/P1)^((n-1)/n)-1) T2=T1*(P2/P1)^((n-1)/n) q_in=w_b_out+c_v*(T2-T1)

4-72 An insulated cylinder is initially filled with air at a specified state. A paddle-wheel in the cylinder stirs the air at constant pressure. The final temperature of air is to be determined. Assumptions 1. Air is an ideal gas with variable specific heats. 2. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 3. There are no work interactions involved other than the boundary work. 4. The cylinder is well-insulated and thus heat transfer is negligible. 5. The thermal energy stored in the cylinder itself and the paddle-wheel is negligible. 6. The compression or expansion process is quasi-equilibrium. Properties The gas constant of air is (Table A-1). Also, temperature (Table A-2). The enthalpy of air at the initial temperature is

for air at room

(Table A-17) Analysis We take the air in the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟ →

since The mass of air is

during a constant pressure quasi-equilibrium process.

1-335

Substituting into the energy balance, From Table A-17, Alternative solution Using specific heats at room temperature, be

, the final temperature is determined to

which gives

EES (Engineering Equation Solver) SOLUTION "Given" V=0.100 [m^3] P1=400 [kPa] T1=(25+273) [K] W_pw_in=15 [kJ] P2=P1 "Properties" Fluid$='Air' C_p=CP(Fluid$,T=298) C_v=CV(Fluid$,T=298) R=C_p-C_v h1=enthalpy(Fluid$, T=T1) "Analysis" m=(P1*V)/(R*T1) W_pw_in=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process" T2=temperature(Fluid$, h=h2)

4-73 A piston-cylinder device contains air. A paddle wheel supplies a given amount of work to the air. The heat transfer is to be determined. Assumptions 1. Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 132.5 K and 3.77 MPa. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats can be used for air. Analysis We take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-336

⏟

Using the boundary work relation on a unit mass basis for the isothermal process of an ideal gas gives

Substituting into the energy balance equation (expressed on a unit mass basis) gives Discussion Note that the energy content of the system remains constant in this case, and thus the total energy transfer output via boundary work must equal the total energy input via shaft work and heat transfer.

EES (Engineering Equation Solver) SOLUTION "Given" P1=400 [kPa] T1=(17+273) [K] w_in=75 [kJ/kg] T2=T1 VolumeRatio=3 "Properties" R=0.287 [kJ/kg-K] "Analysis" w_b_out=R*T1*ln(VolumeRatio) DELTAu=0 "temperature is constant" q_in+w_in-w_b_out=DELTAu

4-74 A cylinder is initially filled with air at a specified state. Air is heated electrically at constant pressure, and some heat is lost in the process. The amount of electrical energy supplied is to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. Air is an ideal gas with variable specific heats. 3. The thermal energy stored in the cylinder itself and the resistance wires is negligible. 4. The compression or expansion process is quasi-equilibrium. Properties The initial and final enthalpies of air are (Table A-17)

1-337

Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

⏟

⏟ →

since

during a constant pressure quasi-equilibrium process. Substituting,

or, (

)

Alternative solution The specific heat of air at the average temperature of Table A-2b, . Substituting,

is, from

or, (

)

Discussion Note that for small temperature differences, both approaches give the same result.

EES (Engineering Equation Solver) SOLUTION "Given" m=15 [kg] T1=(25+273) [K] T2=(95+273) [K] P1=300 [kPa] P2=P1 Q_out=60 [kJ] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) h2=enthalpy(Fluid$, T=T2) "Analysis" W_e_in-Q_out=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process" W_e_in_kWh=W_e_in*Convert(kJ, kWh) "Alternative Solution" c_p=1.0065 [kJ/kg-K] "at 324 K, Table A-2b" W_e_in_alt=(m*c_p*(T2-T1)+Q_out)*Convert(kJ, kWh)

4-75E One part of an insulated rigid tank contains air while the other side is evacuated. The internal energy change of the air and the final air pressure are to be determined when the partition is removed. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-338

Assumptions Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 221.5F and 547 psia. 2. The kinetic and potential energy changes are negligible, .3 3. Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 4. The tank is insulated and thus heat transfer is negligible. Analysis We take the entire tank as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as 1.

⏟

Since the internal energy does not change, the temperature of the air will also not change. Applying the ideal gas equation gives →

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=1.5 [ft^3] P1=100 [psia] T1=(100+460) [R] "Analysis" DELTAU=0 "since no energy enters or leaves" T2=T1 "since internal energy dsoes not change DELTAu=c_v*(T2-T1)" Vol2=2*Vol1 P2=P1*(Vol1/Vol2)

4-76 A cylinder initially contains nitrogen gas at a specified state. The gas is compressed polytropically until the volume is reduced by one-half. The work done and the heat transfer are to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. The is an ideal gas with constant specific heats. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Properties The gas constant of is (Table A-1). The value of at the anticipated average temperature of 350 K is (Table A-2b). Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The energy balance for this closed system can be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-339

⏟

The final pressure and temperature of nitrogen are

→

( )

→ Then the boundary work for this polytropic process can be determined from ∫

Substituting into the energy balance gives

EES (Engineering Equation Solver) SOLUTION "Given" m=2.2 [kg] P1=100 [kPa] T1=(25+273) [K] n=1.3 RatioV=2 "RatioV=V1/V2" "Properties" R=0.2968 [kJ/kg-K] c_v=0.744 [kJ/kg-K] "at 350 K, Table A-2b" "Analysis" P2=ratioV^n*P1 T2=(P2/P1)*(1/RatioV)*T1 W_b_in=(-m*R*(T2-T1))/(1-n) W_b_in-Q_out=m*c_v*(T2-T1) "energy balance"

4-77 Problem 4-76 is reconsidered. The process is to be plotted on a P-V diagram, and the effect of the polytropic exponent n on the boundary work and heat transfer as the polytropic exponent varies from 1.0 to 1.4 is to be investigated. The boundary work and the heat transfer are to be plotted versus the polytropic exponent. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-340

Analysis The problem is solved using EES, and the solution is given below. Procedure Work(P[2],V[2],P[1],V[1],n:W_in) If n=1 then W_in=-P[1]*V[1]*ln(V[2]/V[1]) Else W_in=-(P[2]*V[2]-P[1]*V[1])/(1-n) endif End Vratio=0.5 "V[2]/V[1] = Vratio" "n=1.3" "Polytropic exponent" P[1] = 100 [kPa] T[1] = (25+273) [K] m=2.2 [kg] MM=molarmass(nitrogen) R_u=8.314 [kJ/kmol-K] R=R_u/MM V[1]=m*R*T[1]/P[1] V[2]=Vratio*V[1] P[2]*V[2]/T[2]=P[1]*V[1]/T[1] P[2]*V[2]^n=P[1]*V[1]^n Q_out= W_in-m*(u[2]-u[1]) "energy balance" u[1]=intenergy(N2, T=T[1]) u[2]=intenergy(N2, T=T[2]) Call Work(P[2],V[2],P[1],V[1],n:W_in) "The following is required for the P-v plots" {P_plot*spv_plot/T_plot=P[1]*V[1]/m/T[1] "The combined ideal gas law for states 1 and 2 plus the polytropic process relation give P[2] and T[2]" P_plot*spv_plot^n=P[1]*(V[1]/m)^n} {spV_plot=R*T_plot/P_plot"[m^3]"}

n 1 1.044 1.089

] 134.9 121.8 108.2

] 134.9 137 139.1

1-341

1.133 1.178 1.222 1.267 1.311 1.356 1.4

94.25 79.8 64.86 49.42 33.45 16.93 –0.1588

141.3 143.5 145.8 148.1 150.5 152.9 155.4

4-78 A cylinder initially contains argon gas at a specified state. The gas is stirred while being heated and expanding isothermally. The amount of heat transfer is to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. The air is an ideal gas with constant specific heats. 3. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass crosses the system boundary. The energy balance for this closed system can be expressed as

⏟

since u = u(T) for ideal gases, and thus

when

⏟

. Therefore,

1-342

EES (Engineering Equation Solver) SOLUTION "Given" m=4 [kg] P1=250 [kPa] T1=(35+273) [K] W_b_out=15 [kJ] W_pw_in=3 [kJ] "Analysis" Q_in+W_pw_in-W_b_out=0 "since U2=U1 for isothermal process of an ideal gas"

4-79 Helium contained in a spring-loaded piston-cylinder device is heated. The work done and the heat transfer are to be determined. Assumptions 1. Helium is an ideal gas since it is at a high temperature relative to its critical temperature of 5.3 K. 2. The kinetic and potential energy changes are negligible, . Properties The properties of helium are and (Table A-2a). Analysis We take helium as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

The initial and final specific volumes are

Pressure changes linearly with volume and the work done is equal to the area under the process line 1-2:

(

)

1-343

EES (Engineering Equation Solver) SOLUTION m=5 [kg] P1=100 [kPa] T1=(20+273) [K] P2=800 [kPa] T2=(160+273) [K] R=2.0769 [kJ/kg-K] c_v=3.1156 [kJ/kg-K] Vol1=m*R*T1/P1 Vol2=m*R*T2/P2 W_b_out=(P1+P2)/2*(Vol2-Vol1) W_b_in=-W_b_out Q_in=W_b_out+m*c_v*(T2-T1) Q_out=-Q_in

4-80 A cylinder equipped with a set of stops for the piston to rest on is initially filled with helium gas at a specified state. The amount of heat that must be transferred to raise the piston is to be determined. Assumptions 1. Helium is an ideal gas with constant specific heats. 2. The kinetic and potential energy changes are negligible, . 3. There are no work interactions involved. 4. The thermal energy stored in the cylinder itself is negligible. Properties The specific heat of helium at room temperature is (Table A-2). Analysis We take the helium gas in the cylinder as the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this constant volume closed system can be expressed as ⏟

⏟

The final temperature of helium can be determined from the ideal gas relation to be

→ Substituting into the energy balance relation gives

1-344

m=0.5 [kg] P1=100 [kPa] T1=(25+273) [K] P2=500 [kPa] "Properties" c_v=CV(Helium, T=298, P=101) "Analysis" T2=(P2/P1)*T1 Q_in=m*c_v*(T2-T1) "energy balance"

4-81 A cylinder equipped with a set of stops for the piston is initially filled with air at a specified state. Heat is transferred to the air until the volume doubled. The work done by the air and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. Air is an ideal gas with variable specific heats. 2. The kinetic and potential energy changes are negligible, . 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Properties The gas constant of air is (Table A-1). Analysis We take the air in the cylinder as the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed as ⏟

⏟

The initial and the final volumes and the final temperature of air are

→ No work is done during process 1-2 since . The pressure remains constant during process 2-3 and the work done during this process is

1-345

∫ The initial and final internal energies of air are (Table A-17)

Then from the energy balance, Alternative solution The specific heat of air at the average temperature of A-2b, . Substituting,

is, from Table

EES (Engineering Equation Solver) SOLUTION "Given" m=3 [kg] P1=200 [kPa] T1=(27+273) [K] P2=400 [kPa] “state 2 is when the piston starts to move" V3=2*V1 "state 3 is final state" "Properties" Fluid$='Air' c_p=CP(Fluid$,T=298) c_v=CV(Fluid$,T=298) R=c_p-c_v u1=intenergy(Fluid$, T=T1) "Analysis" V1=(m*R*T1)/P1 P3=P2 V2=V1 W_b_out=P2*(V3-V2) T3=(P3/P1)*(V3/V1)*T1 u3=intenergy(Fluid$, T=T3) Q_in-W_b_out=m*(u3-u1) "energy balance"

4-82 A cylinder equipped with a set of stops on the top is initially filled with air at a specified state. Heat is transferred to the air until the piston hits the stops, and then the pressure doubles. The work done by the air and the amount of heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. Air is an ideal gas with variable specific heats. 2. The kinetic and potential energy changes are negligible, ke  pe  0 . 3. The thermal energy stored in the cylinder itself is negligible. Properties The gas constant of air is (Table A-1). Analysis We take the air in the cylinder to be the system. This is a closed system since no mass crosses the boundary of the system. The energy balance for this closed system can be expressed as

1-346

⏟

The initial and the final volumes and the final temperature of air are determined from

→ No work is done during process 2-3 since . The pressure remains constant during process 1-2 and the work done during this process is ∫ (

)

The initial and final internal energies of air are (Table A-17)

Substituting,

Alternative solution The specific heat of air at the average temperature of A-2b, . Substituting

is, from Table

EES (Engineering Equation Solver) SOLUTION "Input Data" P[1] = 200 [kPa] T[1] = (27+273) [K] m=3 [kg] "constant mass, closed system" R_u=8.314 [kJ/kmol-K] R=R_u/molarmass(air) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-347

V[1]=m"kg"*R"kJ/kg-K"*T[1]"K"/P[1 ]"kPa" "[m^3] note that comments (here units) can be placed within an equation." "For process 1 to 2, P=constant until the piston hits the stops and V doubles " "Process equations" V[2]=2*V[1] P[2]=P[1] P[2]*V[2]/T[2]=P[1]*V[1]/T[1]"The combined ideal gas law for states 1 and 2 gives T[2]" "For process 2 to 3, the piston is against the stops and V=const. and P doubles" V[3]=V[2] P[3]=2*P[2] P[3]*V[3]/T[3]=P[2]*V[2]/T[2]"The combined ideal gas law for states 2 and 3 gives T[3]" spvol[1]=V[1]/m; spvol[2]=V[2]/m; spvol[3]=V[3]/m"specific volumes for plots" "Conservation of Energy for the air, a closed system, we neglect Delta KE and Delta PE." Q_in - W_out = DELTAE DELTAE=m*(u[3]-u[1]) u[1]=intenergy(air, T=T[1]) "internal energy for air as an ideal gas" u[3]=intenergy(air, T=T[3]) W_out = W_12_out+W_23_out "total work" W_12_out=P[1]*(V[2]-V[1])"constant pressure process" W_23_out=0 "constant volume process"

Closed System Energy Analysis: Solids and Liquids 4-83 An iron block is heated. The internal energy and enthalpy changes are to be determined for a given temperature change. Assumptions Iron is an incompressible substance with a constant specific heat. Properties The specific heat of iron is (Table A-3b). Analysis The internal energy and enthalpy changes are equal for a solid. Then,

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] T1=25 [C] T2=75 [C] "Properties" c=0.45 [kJ/kg-K] "Table A-3b" "Analysis" DELTAH=m*c*(T2-T1) DELTAU=DELTAH

4-84E Liquid water experiences a process from one state to another. The internal energy and enthalpy changes are to be determined under different assumptions. Analysis (a) Using the properties from compressed liquid tables

1-348

æ 1 Btu ö ÷ ÷= 18.21 Btu/lbm = 18.07 Btu/lbm + (0.01602 ft 3 /lbm)(50 - 0.17812) psia ççç 3÷ ÷ è5.404 psia ×ft ø }

(b) Using incompressible substance approximation and property tables (Table A-4E),

(Table A-3Ea),

EES (Engineering Equation Solver) SOLUTION "Given" P1=50 [psia] T1=50 [F] P2=2000 [psia] T2=100 [F] "Analysis" Fluid$='steam_iapws' "(a)" u1_a=intenergy(Fluid$, T=T1, x=0) h_f1=enthalpy(Fluid$, T=T1, x=0) v_f1=volume(Fluid$, T=T1, x=0) P_sat1=pressure(Fluid$, T=T1, x=0) h1_a=h_f1+v_f1*(P1-P_sat1)*Convert(psia-ft^3, Btu) u2_a=intenergy(Fluid$, P=P2, T=T2) h2_a=enthalpy(Fluid$, P=P2, T=T2) DELTAu_a=u2_a-u1_a DELTAh_a=h2_a-h1_a "(b)" u1_b=intenergy(Fluid$, T=T1, x=0) h1_b=enthalpy(Fluid$, T=T1, x=0) u2_b=intenergy(Fluid$, T=T2, x=0) h2_b=enthalpy(Fluid$, T=T2, x=0) DELTAu_b=u2_b-u1_b DELTAh_b=h2_b-h1_b "(c)" c=1.0 [Btu/lbm-R] "Table A-3Ea" DELTAu_c=c*(T2-T1) DELTAh_c=DELTAu_c

4-85E A person shakes a canned of drink in a iced water to cool it. The mass of the ice that will melt by the time the canned drink is cooled to a specified temperature is to be determined. Assumptions 1. The thermal properties of the drink are constant, and are taken to be the same as those of water. 2. The effect of agitation on the amount of ice melting is negligible. 3. The thermal energy capacity of the can itself is negligible, and thus it does not need to be considered in the analysis. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-349

Properties The density and specific heat of water at the average temperature of and (Table A-3E). The heat of fusion of water is

are

Analysis We take a canned drink as the system. The energy balance for this closed system can be expressed as

⏟

Noting that 1 gal = 128 oz and heat transfer from a ball is

⏟ → = 957.5 oz, the total amount of (

Noting that the heat of fusion of water is

)(

)

, the amount of ice that will melt to cool the drink is

since heat transfer to the ice must be equal to heat transfer from the can. Discussion The actual amount of ice melted will be greater since agitation will also cause some ice to melt.

EES (Engineering Equation Solver) SOLUTION "Given" T1=85 [F] V=12 [oz] T_icedwater=32 [F] T2=37 [F] "Properties" rho=62.3 [lbm/ft^3] "at (75+45)/2=60 F, Table A-3E" c_p=1.0 [Btu/lbm-R] "at (75+45)/2=60 F, Table A-3E" h_if=143.5 [Btu/lbm] "from the text" "Analysis" m=rho*V*Convert(oz, ft^3) -Q_out=m*c_p*(T2-T1) "energy balance" m_ice=Q_out/h_if

4-86 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked is to be determined. Assumptions 1. 2. 3. 4.

The egg is spherical in shape with a radius of . The thermal properties of the egg are constant. Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible. There are no changes in kinetic and potential energies.

Properties The density and specific heat of the egg are given to be

and

1-350

Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as

⏟

Then the mass of the egg and the amount of heat transfer become

EES (Engineering Equation Solver) SOLUTION "Given" D=0.055 [m] T1=8 [C] T_water=97 [C] T2=80 [C] rho=1020 [kg/m^3] c_p=3.32 [kJ/kg-C] "Analysis" V=pi*D^3/6 m=rho*V Q_in=m*c_p*(T2-T1) "energy balance"

4-87 An iron whose base plate is made of an aluminum alloy is turned on. The minimum time for the plate to reach a specified temperature is to be determined. Assumptions 1. It is given that 85 percent of the heat generated in the resistance wires is transferred to the plate. 2. The thermal properties of the plate are constant. 3. Heat loss from the plate during heating is disregarded since the minimum heating time is to be determined. 4. There are no changes in kinetic and potential energies. 5. The plate is at a uniform temperature at the end of the process. Properties The density and specific heat of the aluminum alloy plate are given to be

and

. Analysis The mass of the iron's base plate is Noting that only 85 percent of the heat generated is transferred to the plate, the rate of heat transfer to the iron's base plate is

1-351

̇ We take plate to be the system. The energy balance for this closed system can be expressed as ⏟

⏟ →

Solving for t and substituting, ̇ which is the time required for the plate temperature to reach the specified temperature.

EES (Engineering Equation Solver) SOLUTION "Given" E_dot=1000 [W] L=0.005 [m] A=0.03 [m^2] T1=22 [C] T2=200 [C] f_heat=0.90 "Properties" rho=2770 [kg/m^3] "given" c_p=875 [J/kg-C] "given" "Analysis" V=L*A m=rho*V Q_dot_in=f_heat*E_dot Q_dot_in*DELTAt=m*c_p*(T2-T1) "energy balance"

4-88 Stainless steel ball bearings leaving the oven at a specified uniform temperature at a specified rate are exposed to air and are cooled before they are dropped into the water for quenching. The rate of heat transfer from the ball bearing to the air is to be determined. Assumptions 1. 2. 3.

The thermal properties of the bearing balls are constant. The kinetic and potential energy changes of the balls are negligible. The balls are at a uniform temperature at the end of the process

Properties The density and specific heat of the ball bearings are given to be

and

Analysis We take a single bearing ball as the system. The energy balance for this closed system can be expressed as

1-352

⏟

The total amount of heat transfer from a ball is

Then the rate of heat transfer from the balls to the air becomes ̇ ̇ Therefore, heat is lost to the air at a rate of 2.34 kW.

EES (Engineering Equation Solver) SOLUTION "Given" D=0.012 [m] n_dot_ball=800 [1/min] T1=900 [C] T_air=25 [C] T2=850 [C] rho=8085 [kg/m^3] c_p=0.480 [kJ/kg-C] "Analysis" V=pi*D^3/6 m=rho*V -Q_out=m*c_p*(T2-T1) "energy balance" Q_dot_total=n_dot_ball*Q_out/Convert(kW, kJ/min)

4-89E Large brass plates are heated in an oven at a rate of be determined.

. The rate of heat transfer to the plates in the oven is to

Assumptions 1. 2.

The thermal properties of the plates are constant. The changes in kinetic and potential energies are negligible.

Properties The density and specific heat of the brass are given to be

and

Analysis We take the plate to be the system. The energy balance for this closed system can be expressed as

1-353

⏟

The mass of each plate and the amount of heat transfer to each plate is ] Then the total rate of heat transfer to the plates becomes ̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" L=(1.6/12) [ft] A=2[ft]*2[ft] T1=75 [F] T_oven=1500 [F] n_dot_plate=300 [1/min] T2=900 [F] "Properties" rho=532.5 [lbm/ft^3] "given" c_p=0.091 [Btu/lbm-F] "given" "Analysis" V=L*A m=rho*V Q_in=m*c_p*(T2-T1) "energy balance" Q_dot_total=n_dot_plate*Q_in/Convert(min, s)

4-90 Long cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in the oven is to be determined. Assumptions 1. The thermal properties of the rods are constant. 2. The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the steel rods are given to be

and

Analysis Noting that the rods enter the oven at a velocity of 2 m/min and exit at the same velocity, we can say that a 3-m long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is ] We take the 2-m section of the rod in the oven as the system. The energy balance for this closed system can be expressed as

1-354

⏟

Substituting, Noting that this much heat is transferred in 1 min, the rate of heat transfer to the rod becomes ̇

EES (Engineering Equation Solver) SOLUTION "Given" D=0.08 [m] Vel=2 [m/min] T_oven=900 [C] T1=30 [C] T2=500 [C] rho=7833 [kg/m^3] c_p=0.465 [kJ/kg-C] "Analysis" A=pi*D^2/4 DELTAt=1 [min] L=Vel*DELTAt "since 3 m long section of the rod is heated in 1 min" V=L*A m=rho*V "mass of the rod heated in 1 minute" Q_in=m*c_p*(T2-T1) "energy balance" Q_dot_in=Q_in/DELTAt*Convert(kJ/min, kW)

4-91 An electronic device is on for 5 minutes, and off for several hours. The temperature of the device at the end of the 5min operating period is to be determined for the cases of operation with and without a heat sink. Assumptions 1. The device and the heat sink are isothermal. 2. The thermal properties of the device and of the sink are constant. 3. Heat loss from the device during on time is disregarded since the highest possible temperature is to be determined. Properties The specific heat of the device is given to be . The specific heat of aluminum at room temperature of 300 K is (Table A-3). Analysis We take the device to be the system. Noting that electrical energy is supplied, the energy balance for this closed system can be expressed as

1-355

⏟

̇ Substituting, the temperature of the device at the end of the process is determined to be (without the heat sink) Case 2 When a heat sink is attached, the energy balance can be expressed as ̇ Substituting, the temperature of the device-heat sink combination is determined to be

(25 J/s)(5´ 60 s) = (0.020 kg)(850 J/kg.°C)(T2 - 25)°C + (0.500 kg)(902 J/kg.°C)(T2 - 25)°C T2 = 41.0°C (with heat sink) Discussion These are the maximum temperatures. In reality, the temperatures will be lower because of the heat losses to the surroundings.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_e_in=25 [W] m=0.020 [kg] DELTAt=5[min]*Convert(min, s) T1=25 [C] m_HeatSink=0.5 [kg] c_p_device=850 [J/kg-C] "Properties" c_p_Al=902 [J/kg-C] "at 300 K, Table A-3" "Analysis" "Without heat sink" W_dot_e_in*DELTAt=m*c_p_device*(T2_a-T1) "With heat sink" W_dot_e_in*DELTAt=m*c_p_device*(T2_b-T1)+m_HeatSink*C_p_Al*(T2_b-T1)

4-92 Problem 4-91 is reconsidered. The effect of the mass of the heat sink on the maximum device temperature as the mass of heat sink varies from 0 kg to 1 kg is to be investigated. The maximum temperature is to be plotted against the mass of heat sink. Analysis The problem is solved using EES, and the solution is given below. Q_dot_out = 25 [W] m_device=20 [g] cp_device=850 [J/kg-C] A=5 [cm^2] DELTAt=5 [min] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-356

T_amb=25 [C] m_sink=0.5 [kg] cp_al=0.902 [kJ/kg-C] T_2=T_amb E_dot_in - E_dot_out = DELTAE_dot E_dot_in =0 E_dot_out = Q_dot_out DELTAE_dot= m_device*convert(g,kg)*cp_device*(T_2-T_1_device)/(DELTAt*convert(min,s)) E_dot_in - E_dot_out = DELTAE_dot_combined DELTAE_dot_combined= (m_device*convert(g,kg)*Cp_device*(T_2-T_1_device&sink)+m_sink*cp_al*(T_2T_1_device&sink)*convert(kJ,J))/(DELTAt*convert(min,s)) ] 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

] 466.2 94.96 62.99 51.08 44.85 41.03 38.44 36.57 35.15 34.05 33.16

1-357

4-93 The face of a person is slapped. For the specified temperature rise of the affected part, the impact velocity of the hand is to be determined. Assumptions 1. The hand is brought to a complete stop after the impact. 2. The face takes the blow without significant movement. 3. No heat is transferred from the affected area to the surroundings, and thus the process is adiabatic. 4. No work is done on or by the system. 5. The potential energy change is zero, PE = 0 and E = U + KE. Analysis We analyze this incident in a professional manner without involving any emotions. First, we identify the system, draw a sketch of it, and state our observations about the specifics of the problem. We take the hand and the affected portion of the face as the system. This is a closed system since it involves a fixed amount of mass (no mass transfer). We observe that the kinetic energy of the hand decreases during the process, as evidenced by a decrease in velocity from initial value to zero, while the internal energy of the affected area increases, as evidenced by an increase in the temperature. There seems to be no significant energy transfer between the system and its surroundings during this process. Under the stated assumptions and observations, the energy balance on the system can be expressed as ⏟

⏟

] That is, the decrease in the kinetic energy of the hand must be equal to the increase in the internal energy of the affected area. Solving for the velocity and substituting the given quantities, the impact velocity of the hand is determined to be √

√

(

)

Discussion Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering.

EES (Engineering Equation Solver) SOLUTION "Given" DELTAT_tissue=2.4 [C] m_hand=0.9 [kg] m_tissue=0.150 [kg] c_p_tissue=3.8 [kJ/kg-C] "Analysis" DELTAU_tissue+DELTAKE_hand=0 "energy balance" DELTAU_tissue=m_tissue*c_p_tissue*DELTAT_tissue DELTAKE_hand=-m_hand*V^2/2*Convert(m^2/s^2, kJ/kg)

Special Topic: Biological Systems 4-94C The energy released during metabolism in humans is used to maintain the body temperature at 37C.

1-358

4-95C The food we eat is not entirely metabolized in the human body. The fraction of metabolizable energy contents are 95.5% for carbohydrates, 77.5% for proteins, and 97.7% for fats. Therefore, the metabolizable energy content of a food is not the same as the energy released when it is burned in a bomb calorimeter.

4-96C Yes. Each body rejects the heat generated during metabolism, and thus serves as a heat source. For an average adult male it ranges from 84 W at rest to over 1000 W during heavy physical activity. Classrooms are designed for a large number of occupants, and thus the total heat dissipated by the occupants must be considered in the design of heating and cooling systems of classrooms.

4-97 Six people are fast dancing in a room, and there is a resistance heater in another identical room. The room that will heat up faster is to be determined. Assumptions 1. The rooms are identical in every other aspect. 2. Half of the heat dissipated by people is in sensible form. 3. The people are of average size. Properties An average fast dancing person dissipates of energy (sensible and latent) (Table 4-2). Analysis Three couples will dissipate E = (6 persons)( .person)( )= = 4190 W of energy. (About half of this is sensible heat). Therefore, the room with the people dancing will warm up much faster than the room with a 2-kW resistance heater.

EES (Engineering Equation Solver) SOLUTION "Given" E_dot_electric=2 [kW] N_persons=6 "Properties" e_dancing=600 [kcal/h] "from Table 4-2 of the text" "Analysis" E_dot_dancing=N_persons*e_dancing*Convert(kcal/h, W)

4-98 The average body temperature of the human body rises by 2C during strenuous exercise. The increase in the thermal energy content of the body as a result is to be determined. Properties The average specific heat of the human body is given to be . Analysis The change in the sensible internal energy of the body is U = mcT = (80 kg)( )(2C) = 576 kJ as a result of body temperature rising 2C during strenuous exercise.

EES (Engineering Equation Solver) SOLUTION "Given" m_man=80 [kg] T1=37 [C] T2=39 [C] C_body=3.6 [kJ/kg-C] "Analysis" DELTAU=m_man*C_body*(T2-T1)

1-359

4-99 Two people are identical except one jogs for 30 min while the other watches TV. The weight difference between these two people in one month is to be determined. Assumptions The two people have identical metabolism rates, and are identical in every other aspect. Properties An average 68-kg person consumes while jogging, and while watching TV (Table 4-2). Analysis An 80-kg person who jogs 0.5 h a day will have jogged a total of 15 h a month, and will consume ]

(

)(

)

more calories than the person watching TV. The metabolizable energy content of 1 kg of fat is 33,100 kJ. Therefore, the weight difference between these two people in 1-month will be

EES (Engineering Equation Solver) SOLUTION "Given" m_person=80 [kg] t_jogging_day=(30/60) [h] "Properties" e_jogging=540 [kcal/h] e_TV=72 [kcal/h] "from Table 4-2 of the text for an average 68-kg person" m_average=68 [kg] e_fat=33100 [kJ/kg] "from the text" "Analysis" t_jogging_month=t_jogging_day*30 "1 month = 30 days" DELTAE_consumed=(e_jogging-e_TV)*t_jogging_month*m_person/m_average*Convert(kcal, kJ) DELTAm_fat=DELTAE_consumed/e_fat 4-100 A 90-kg person eats 1-L of ice cream. The length of time this person needs to jog to burn off these calories is to be determined.

Assumptions The person meets his entire calorie needs from the ice cream while jogging. Properties An average 68-kg person consumes while jogging, and the energy content of a 100-ml of ice cream is 110 Cal (Tables 4-1 and 4-2). Analysis The rate of energy consumption of a 55-kg person while jogging is ̇

(

)

Noting that a 100-ml serving of ice cream has 110 Cal of metabolizable energy, a 1-liter box of ice cream will have 1100 Calories. Therefore, it will take

of jogging to burn off the calories from the ice cream.

EES (Engineering Equation Solver) SOLUTION "Given" m_person=90 [kg] V_eaten=1 [L] t_bicycling=1 [h] "Properties" e_jogging=540 [kcal/h] "from Table 1-4 of the text for an average 68-kg person" m_average=68 [kg] e_icecream=110 [kcal] "from Table 1-3 of the text for a 100-mL ice cream" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-360

V=(100/1000) [L] "Analysis" E_dot_consumed=e_jogging*m_person/m_average E_icecream_eaten=e_icecream*V_eaten/V DELTAt=E_icecream_eaten/E_dot_consumed

4-101 A person with 20-kg of body fat goes on a hunger strike. The number of days this person can survive on the body fat alone is to be determined. Assumptions 1. The person is an average male who remains in resting position at all times. 2. The person meets the entire calorie needs from the body fat alone. Properties The metabolizable energy content of fat is . An average resting person burns calories at a rate of (Table 4-2). Analysis The metabolizable energy content of 20 kg of body fat is The person will consume (

)

Therefore, this person can survive

on the body fat alone. This result is not surprising since people are known to survive over 100 days without any food intake.

EES (Engineering Equation Solver) SOLUTION "Given" m_fat=20 [kg] "Properties" e_fat=33100 [kJ/kg] "from the text" e_resting=72 [kcal/h] "from Table 4-2 of the text" "Analysis" E_fat_total=e_fat*m_fat E_consumed=e_resting*Convert(kcal/h, kJ/day) DELTAt=E_fat_total/E_consumed

4-102 Two 50-kg people are identical except one eats their baked potato with 4 teaspoons of butter while the other eats theirs plain every evening. The weight difference between these two people in one year is to be determined. Assumptions 1. These two people have identical metabolism rates, and are identical in every other aspect. 2. All the calories from the butter are converted to body fat. Properties The metabolizable energy content of 1 kg of body fat is 33,100 kJ. The metabolizable energy content of 1 teaspoon of butter is 35 Calories (Table 4-1). Analysis A person who eats 4 teaspoons of butter a day will consume (

)

1-361

(

)

of additional body fat that year.

EES (Engineering Equation Solver) SOLUTION "Given" m_person=50 [kg] N_teaspoon=4 [1/day] "Properties" e_fat=33100 [kJ/kg] "from the text" e_butter=35 [kcal] "from Table 4-1 of the text" "Analysis" E_consumed=N_teaspoon*e_butter*Convert(kcal/day, kcal/year) m_fat=E_consumed/e_fat*Convert(kcal, kJ)

4-103 A person changes his/her diet to lose weight. The time it will take for the body mass index (BMI) of the person to drop from 30 to 20 is to be determined. Assumptions The deficit in the calori intake is made up by burning body fat. Properties The metabolizable energy contents are 350 Cal for a slice of pizza and 87 Cal for a 200-ml regular cola. The metabolizable energy content of 1 kg of body fat is 33,100 kJ. Analysis The lunch calories before the diet is The lunch calories after the diet is The calorie reduction is The corresponding reduction in the body fat mass is (

)

The weight of the person before and after the diet is

Then it will take

for the BMI of this person to drop to 20.

EES (Engineering Equation Solver) SOLUTION "GIVEN" BMI_1=30 H=1.6 [m] e_pizza=350 [kcal] e_Coke=87 [kcal] E_fat=33100 [kJ/kg] BMI_2=20 "ANALYSIS" E_old=3*e_pizza+2*e_Coke "initial lunch clories" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-362

E_new=2*e_pizza+1*e_Coke "final lunch calories" E_reduced=E_old-e_new "calorie reduction" m_fat=E_reduced/E_fat*Convert(kcal, kJ) "lost fat" W_1=BMI_1*H^2 "initial weight" W_2=BMI_2*H^2 "final weight" DELTAW=W_1-W_2 "weight difference" Day=DELTAW/m_fat "number of days"

Review Problems 4-104 The compression work from to using a polytropic process is to be compared for neon and air. Assumptions The process is quasi-equilibrium. Properties The gas constants for neon and air R = 0.4119 and , respectively (Table A-2a). Analysis For a polytropic process, The boundary work during a polytropic process of an ideal gas is ∫

∫

*( )

The negative of this expression gives the compression work during a polytropic process. Inspection of this equation reveals that the gas with the smallest gas constant (i.e., largest molecular weight) requires the least work for compression. In this problem, air will require the least amount of work.

4-105 The expansion work from to in a closed system polytropic process is to be compared for neon and air. Assumptions The process is quasi-equilibrium. Properties The gas constants for neon and air are R = 0.4119 and , respectively (Table A-2a). Analysis For a polytropic process, The boundary work during a polytropic process of an ideal gas is ∫

∫

*( )

Inspection of this equation reveals that the gas with the largest gas constant (i.e., smallest molecular weight) will produce the greatest amount of work. Hence, the neon gas will produce the greatest amount of work.

4-106 A classroom has 40 students, each dissipating 84 W of sensible heat. It is to be determined if it is necessary to turn the heater on in the room to avoid cooling of the room. Properties Each person is said to be losing sensible heat to the room air at a rate of 84 W. Analysis We take the room is losing heat to the outdoors at a rate of ̇

(

)

The rate of sensible heat gain from the students is ̇ which is greater than the rate of heat loss from the room. Therefore, it is not necessary to turn the heater. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-363

EES (Engineering Equation Solver) SOLUTION Q_dot_out=12000 [kJ/h] n_student=40 q=84 [W] Q_dot_loss=Q_dot_out*Convert(kJ/h, W) Q_dot_gain=n_student*q

4-107 For a 10C temperature change of air, the final velocity and final elevation of air are to be determined so that the internal, kinetic and potential energy changes are equal. Properties The constant-volume specific heat of air at room temperature is

(Table A-2).

Analysis The internal energy change is determined from

Equating kinetic and potential energy changes to internal energy change, the final velocity and elevation are determined from → →

(

)→ (

)→

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_1=0 [C] T_2=10 [C] Vel_1=0 [m/s] z_1=0 [m] "PROPERTIES" C_v=CV(Air, T=T_ave) T_ave=1/2*(T_1+T_2) "ANALYSIS" DELTAu=C_v*(T_2-T_1) DELTAu=DELTAke DELTAu=DELTApe DELTAke=1/2*(Vel_2^2-Vel_1^2)*Convert(m^2/s^2, kJ/kg) DELTApe=g*(z_2-z_1)*Convert(m^2/s^2, kJ/kg) g=9.81 [m/s^2]

1-364

4-108 A gas mixture contained in a rigid tank is cooled. The heat transfer is to be determined. Assumptions 1. The gas mixture is an ideal gas. 2. The kinetic and potential energy changes are negligible, . 3 Constant specific heats can be used. Properties The specific heat of gas mixture is given to be . Analysis We take the contents of the tank as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

As the specific volume remains constant during this process, the ideal gas equation gives

Substituting,

EES (Engineering Equation Solver) SOLUTION c_v=0.748 [kJ/kg-K] P1=200 [kPa] T1=(200+273) [K] P2=100 [kPa] T2=T1*P2/P1 q_out=c_v*(T1-T2)

4-109 Heat is transferred to a piston-cylinder device containing air. The expansion work is to be determined. Assumptions 1. There is no friction between piston and cylinder. 2. Air is an ideal gas. Properties The gas contant for air is

(Table A-2a).

1-365

Analysis Noting that the gas constant represents the boundary work for a unit mass and a unit temperature change, the expansion work is simply determined from

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.5 [kg] DELTAT=5 [C] "ANALYSIS" R_u=8.314 [kJ/kmol-K] MM=MolarMass(air) R=R_u/MM W_b=m*DELTAT*R "Gas constant, R, represents the boundary work for a unit mass and a unit temperature change"

4-110 A cylinder is initially filled with saturated liquid-vapor mixture of R-134a at a specified pressure. Heat is transferred to the cylinder until the refrigerant vaporizes completely at constant pressure. The initial volume, the work done, and the total heat transfer are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2. The thermal energy stored in the cylinder itself is negligible. 3. The compression or expansion process is quasi-equilibrium. Analysis (a) Using property data from R-134a tables (Tables A-11 through A-13), the initial volume of the refrigerant is determined to be

}

1-366

} ∫ (

)

(c) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

Substituting,

EES (Engineering Equation Solver) SOLUTION "Given" m=0.2 [kg] P1=200 [kPa] x1=1-0.75 P2=P1 x2=1 "Analysis" Fluid$='R134a' "(a)" v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) Vol1=m*v1 "(b)" v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) W_b_out=m*P1*(v2-v1) "(c)" Q_in-W_b_out=m*(u2-u1) "energy balance"

4-111E An insulated rigid vessel contains air. A paddle wheel supplies work to the air. The work supplied and final temperature are to be determined. Assumptions

1-367

Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 238.5 R and 547 psia. 2. The kinetic and potential energy changes are negligible, 3. Constant specific heats can be used for air. Properties The specific heats of air at room temperature are and (Table A-2Ea). Analysis We take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

As the specific volume remains constant during this process, the ideal gas equation gives

Substituting,

EES (Engineering Equation Solver) SOLUTION "Given" m=2 [lbm] P1=30 [psia] T1=(60+460) [R] P2=40 [psia] "Properties" c_v=0.171 [Btu/lbm-R] "Table A-2Ea" "Analysis" T2=T1*(P2/P1) T2_F=ConvertTemp(R, F, T2) W_in=m*c_v*(T2-T1) "energy balance"

4-112 Air at a given state is expanded polytropically in a piston-cylinder device to a specified pressure. The final temperature is to be determined.

1-368

Analysis The polytropic relations for an ideal gas give ( )

(

)

EES (Engineering Equation Solver) SOLUTION "Given" n=1.2 P1=1000 [kPa] T1=(400+273) [K] P2=110 [kPa] "Analysis" T2=T1*(P2/P1)^((n-1)/n) T2_C=ConvertTemp(K, C, T2)

4-113 Nitrogen in a rigid vessel is heated. The work done and heat transfer are to be determined. Assumptions 1. Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 126.2 K and 3.39 MPa. 2. The kinetic and potential energy changes are negligible, 3. Constant specific heats at room temperature can be used for nitrogen. Properties For nitrogen, at room temperature (Table A-2a). Analysis We take the nitrogen as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

⏟

There is no work done since the vessel is rigid: Since the specific volume is constant during the process, the final temperature is determined from ideal gas equation to be

Substituting,

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(25+273) [K] P2=300 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-369

"Properties" c_v=0.743 [kJ/kg-K] "Table A-2a" "Analysis" w=0 "rigid tank, no work is done" T2=T1*(P2/P1) q_in=c_v*(T2-T1)

4-114 A well-insulated rigid vessel contains saturated liquid water. Electrical work is done on water. The final temperature is to be determined. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. The thermal energy stored in the tank itself is negligible. Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as ⏟

⏟

The amount of electrical work added during 30 minutes period is

(

)

The properties at the initial state are (Table A-4)

Substituting, → The final state is compressed liquid. Noting that the specific volume is constant during the process, the final temperature is determined using EES to be } Discussion We cannot find this temperature directly from steam tables at the end of the book. Approximating the final compressed liquid state as saturated liquid at the given internal energy, the final temperature is determined from Table A-4 to be Note that this approximation resulted in an answer, which is not very close to the exact answer determined using EES.

1-370

x1=0 T1=40 [C] I=10 [Amp] Voltage=50 [Volt] DELTAt=(30*60) [s] "Analysis" Fluid$='steam_iapws' W_e_in=Voltage*I*DELTAt*Convert(Volt-Amp, W)*Convert(J, kJ) u1=intenergy(Fluid$, T=T1, x=x1) v1=volume(Fluid$, T=T1, x=x1) W_e_in=m*(u2-u1) "solved for u2" v2=v1 T2=temperature(Fluid$, v=v2, u=u2) "compressed liquid state" T2_alt=temperature(Fluid$, u=u2, x=0) "approximating state as saturated liquid at u2"

4-115 A cylinder equipped with a set of stops for the piston is initially filled with saturated liquid-vapor mixture of water a specified pressure. Heat is transferred to the water until the volume increases by 20%. The initial and final temperature, the mass of the liquid when the piston starts moving, and the work done during the process are to be determined, and the process is to be shown on a P-v diagram. Assumptions The process is quasi-equilibrium. Analysis (a) Initially the system is a saturated mixture at 160 kPa pressure, and thus the initial temperature is

The total initial volume is Then the total and specific volumes at the final state are

1-371

} (b) When the piston first starts moving,

and

. The specific volume at this state is

which is greater than at 500 kPa. Thus no liquid is left in the cylinder when the piston starts moving. (c) No work is done during process 1-2 since . The pressure remains constant during process 2-3 and the work done during this process is ∫

(

)

EES (Engineering Equation Solver) SOLUTION "Given" m=3 [kg] P1=160 [kPa] m1_f=1 [kg] P2=500 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" x1=1-(m1_f/m) T1=temperature(Fluid$, P=P1, x=x1) v1=volume(Fluid$, P=P1, x=x1) Vol1=m*v1 Vol3=1.2*Vol1 "state 3 is the final state" v3=Vol3/m P3=P2 T3=temperature(Fluid$, P=P3, v=v3) "(b)" Vol2=Vol1 "state 2 is when the piston starts moving" v2=Vol2/m x2=quality(Fluid$, P=P2, v=v2) "x2=100 meaning that no liquid is left" m2_f=(1-x2)*m "(c)" W_b_out=P2*(Vol3-Vol2) v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) v2_g=volume(Fluid$, P=P2, x=1)

4-116 A cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant is heated both electrically and by heat transfer at constant pressure for 6 min. The electric current is to be determined, and the process is to be shown on a T-v diagram. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2. The thermal energy stored in the cylinder itself and the wires is negligible. 3. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

1-372

⏟

since through A-13)

⏟

during a constant pressure quasi-equilibrium process. The properties of R-134a are (Tables A-11

} } Substituting, (

)

EES (Engineering Equation Solver) SOLUTION "Given" m=12 [kg] P1=240 [kPa] x1=1 Q_in=300 [kJ] P2=P1 Voltage=110 [Volt] time=6*60 [s] T2=70 [C] "Analysis" Fluid$='R134a' h1=enthalpy(Fluid$, P=P1, x=x1) h2=enthalpy(Fluid$, P=P2, T=T2) Q_in+W_e_in=m*(h2-h1) "since (U2-U1)+W_b = H2-H1 for a constant-pressure process" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-373

W_e_in*Convert(kJ/s, Volt-Amp)=Voltage*I*time

4-117 Saturated water vapor contained in a spring-loaded piston-cylinder device is condensed until it is saturated liquid at a specified temperature. The heat transfer is to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

The properties at the initial and final states are (Table A-4),

Since this is a linear process, the work done is equal to the area under the process line 1-2:

(

)

Thus, Substituting into energy balance equation gives

1-374

T2=50 [C] x2=0 "Analysis" Fluid$='steam_iapws' P1=pressure(Fluid$,T=T1,x=x1) u1=intenergy(Fluid$,T=T1,x=x1) v1=volume(Fluid$,T=T1,x=x1) P2=pressure(Fluid$,T=T2,x=x2) u2=intenergy(Fluid$,T=T2,x=x2) v2=volume(Fluid$,T=T2,x=x2) w_in=(P1+P2)/2*(v1-v2) q_out=w_in+(u1-u2)

4-118 A cylinder is initially filled with helium gas at a specified state. Helium is compressed polytropically to a specified temperature and pressure. The heat transfer during the process is to be determined. Assumptions 1. Helium is an ideal gas with constant specific heats. 2. The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Properties The gas constant of helium is (Table A-1). Also, 2). Analysis The mass of helium and the exponent n are determined to be

(Table A-

→ →

( )

→

(

)

→

Then the boundary work for this polytropic process can be determined from

∫

We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Taking the direction of heat transfer to be to the cylinder, the energy balance for this stationary closed system can be expressed as ⏟

⏟

1-375

Substituting, The negative sign indicates that heat is lost from the system.

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(10+273) [K] V1=0.2 [m^3] P2=700 [kPa] T2=(290+273) [K] "Properties" c_v=3.1156 [kJ/kg-K] "Table A-2a" R=2.0769 [kJ/kg-K] "Table A-2a" "Analysis" m=(P1*V1)/(R*T1) V2=(T2/T1)*(P1/P2)*V1 P1*V1^n=P2*V2^n C=P1*V1^n P*V^n=C W_b_in=integral(P, V, V2, V1) "EES function that performs integration" W_b_in_alt=-m*R*(T2-T1)/(1-n) Q_in+W_b_in=m*c_v*(T2-T1)

4-119 A cylinder equipped with an external spring is initially filled with air at a specified state. Heat is transferred to the air, and both the temperature and pressure rise. The total boundary work done by the air, and the amount of work done against the spring are to be determined, and the process is to be shown on a P-v diagram. Assumptions 1. The process is quasi-equilibrium. 2. The spring is a linear spring. Analysis (a) The pressure of the gas changes linearly with volume during this process, and thus the process curve on a P-V diagram will be a straight line. Then the boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus,

(

)

1-376

(b) If there were no spring, we would have a constant pressure process at P = 100 kPa. The work done during this process is

∫ (

)

Thus,

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] V1=0.15 [m^3] P2=800 [kPa] V2=0.45 [m^3] "Analysis" "(a)" W_b_out=(P1+P2)/2*(V2-V1) "Area under the process curve" "(b)" W_b_out_nospring=P1*(V2-V1) W_spring=W_b_out-W_b_out_nospring

4-120 A cylinder and a rigid tank initially contain the same amount of an ideal gas at the same state. The temperature of both systems is to be raised by the same amount. The amount of extra heat that must be transferred to the cylinder is to be determined. Analysis In the absence of any work interactions, other than the boundary work, the H and U represent the heat transfer for ideal gases for constant pressure and constant volume processes, respectively. Thus the extra heat that must be supplied to the air maintained at constant pressure is where

1-377

Substituting,

EES (Engineering Equation Solver) SOLUTION "Given" m=12 [kg] DELTAT=15 [K] MM=25 [kg/kmol] "Properties" R=R_u/MM R_u=8.314 [kJ/kmol-C] "Analysis" Q_in_extra=m*R*DELTAT " from Q_in_extra=DELTAH-DELTAU=m*c_p*DELTAT-m*c_v*DELTAT=m*(c_p-c_v)*DELTAT=m*R*DELTAT"

4-121 An insulated cylinder initially contains a saturated liquid-vapor mixture of water at a specified temperature. The entire vapor in the cylinder is to be condensed isothermally by adding ice inside the cylinder. The amount of ice that needs to be added is to be determined. Assumptions 1. Thermal properties of the ice are constant. 2. The cylinder is well-insulated and thus heat transfer is negligible. 3. There is no stirring by hand or a mechanical device (it will add energy). Properties The specific heat of ice at about 0C is (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are given to be 0C and . Analysis We take the contents of the cylinder (ice and saturated water) as our system, which is a closed system. Noting that the temperature and thus the pressure remains constant during this phase change process and thus , the energy balance for this system can be written as

⏟

]

The properties of water at 120C are (Table A-4)

1-378

Then,

Noting that

and

and substituting gives ]

m = 0.0294 kg = 29.4 g ice

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.01 [m^3] x1=0.2 T1=120 [C] T_ice=0 [C] x2=0 T2=120 [C] T_melting=0 [C] h_if=333.7 [kJ/kg] "Properties" C_w=4.18 [kJ/kg-C] C_ice=2.11 [kJ/kg-C] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=x1) v1=volume(Fluid$, T=T1, x=x1) h2=enthalpy(Fluid$, T=T2, x=x2) m_w=Vol1/v1 0=m_ice*C_ice*(T_melting-T_ice)+m_ice*h_if+m_ice*C_w*(T2-T_melting)+m_w*(h2-h1) "energy balance on ice and water"

4-122 The heating of a passive solar house at night is to be assisted by solar heated water. The length of time that the electric heating system would run that night with or without solar heating are to be determined. Assumptions 1. Water is an incompressible substance with constant specific heats. 2. The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3. The house is maintained at 22°C at all times. Properties The density and specific heat of water at room temperature are Analysis (a) The total mass of water is

and

(Table A-3).

Taking the contents of the house, including the water as our system, the energy balance relation can be written as

1-379

⏟

or, ̇

]

Substituting, It gives t = 17,170 s = 4.77 h (b) If the house incorporated no solar heating, the energy balance relation above would simplify further to ̇ Substituting, It gives t = 33,333 s = 9.26 h

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_out=50000[kJ/h]*Convert(kJ/h, kW) T1=80 [C] T2=22 [C] time=10*3600 [s] V=50*20[L]*Convert(L, m^3) W_dot_e_in=15 [kW] "Properties" rho=1000 [kg/m^3] "Table A-3" c_p=4.18 [kJ/kg-C] "Table A-3" "Analysis" "(a)" m=rho*V W_dot_e_in*time_a-Q_dot_out*time=m*C_p*(T2-T1) "energy balance on house" time_a_h=time_a*Convert(s, h) "(b)" W_dot_e_in*time_b-Q_dot_out*time=0 "energy balance" time_b_h=time_b*Convert(s, h)

1-380

4-123 Water is boiled at sea level (1 atm pressure) in a coffee maker, and half of the water evaporates in 25 min. The power rating of the electric heating element and the time it takes to heat the cold water to the boiling temperature are to be determined.

Assumptions 1. The electric power consumption by the heater is constant. 2. Heat losses from the coffee maker are negligible. Properties The enthalpy of vaporization of water at the saturation temperature of 100C is At an average temperature of , the specific heat of water is about (Table A-3).

(Table A-4). , and the density is

Analysis The density of water at room temperature is very nearly , and thus the mass of 1 L water at 18C is nearly 1 kg. Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a liquid at a specified temperature, the amount of electrical energy needed to vaporize 0.5 kg of water in 25 min is ̇

Therefore, the electric heater consumes (and transfers to water) 0.752 kW of electric power. The time it takes for the entire water to be heated from 18C to 100C is determined to be ̇

̇ Discussion We can also solve this problem using 100C, we have and results will be practically the same.

data (instead of density), and data instead of specific heat. At . At 18C, we have (Table A-4). The two

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.001 [m^3] time_evap=13*60 [s] m_evap=1/2*m T1=18 [C] T2=T_sat T_sat=100 [C] "at sea level" "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f rho=1000 "density(water, T=T_ave, P=P)" C_w=4.18 "CP(water, T=T_ave, P=P)" T_ave=1/2*(T1+T2) P=101.325 [kPa] "at sea level" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-381

"Analysis" m=rho*Vol W_dot_e=(m_evap*h_fg)/time_evap W_dot_e=(m*C_w*(T2-T1))/time time_min=time*Convert(s, min)

4-124 A sample of a food is burned in a bomb calorimeter, and the water temperature rises by 3.2°C when equilibrium is established. The energy content of the food is to be determined. Assumptions 1. Water is an incompressible substance with constant specific heats. 2. Air is an ideal gas with constant specific heats. 3. The energy stored in the reaction chamber is negligible relative to the energy stored in water. 4. The energy supplied by the mixer is negligible. Properties The specific heat of water at room temperature is (Table A-3). The constant volume specific heat of air at room temperature is (Table A-2). Analysis The chemical energy released during the combustion of the sample is transferred to the water as heat. Therefore, disregarding the change in the sensible energy of the reaction chamber, the energy content of the food is simply the heat transferred to the water. Taking the water as our system, the energy balance can be written as ⏟ ⏟ or

] Substituting, for a 2-g sample. Then the energy content of the food per unit mass is (

)

To make a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber, we treat the entire mass within the chamber as air and determine the change in sensible internal energy: ] which is less than 1% of the internal energy change of water. Therefore, it is reasonable to disregard the change in the sensible energy content of the reaction chamber in the analysis.

EES (Engineering Equation Solver) SOLUTION "Given" m_w=3 [kg] m_sample=0.002 [kg] m_a=0.100 [kg] DELTAT=3.2 [C] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-382

c_v_a=0.718 [kJ/kg-C] "Table A-2a" c_p_w=4.18 [kJ/kg-C] "Table A-3" "Analysis" Q_in=m_w*c_p_w*DELTAT "energy balance on water" e_food=Q_in/m_sample DELTAU_chamber=m_a*c_v_a*DELTAT

4-125 A man drinks one liter of cold water at 3°C in an effort to cool down. The drop in the average body temperature of the person under the influence of this cold water is to be determined. Assumptions 1. Thermal properties of the body and water are constant. 2. The effect of metabolic heat generation and the heat loss from the body during that time period are negligible. Properties The density of water is very nearly

, and the specific heat of water at room temperature is

(Table A-3). The average specific heat of human body is given to be Analysis. The mass of the water is

We take the man and the water as our system, and disregard any heat and mass transfer and chemical reactions. Of course these assumptions may be acceptable only for very short time periods, such as the time it takes to drink the water. Then the energy balance can be written as ⏟

⏟

or ]

]

Substituting It gives Then T=39  38.4 = 0.6C Therefore, the average body temperature of this person should drop about half a degree celsius.

EES (Engineering Equation Solver) SOLUTION "Given" m_body=68 [kg] T1_body=39 [C] V_w=0.001 [m^3] T1_w=3 [C] c_p_body=3.6 [kJ/kg-C] "Properties" rho_w=1000 [kg/m^3] "Table A-3" c_p_w=4.18 [kJ/kg-C] "Table A-3" "Analysis" m_w=rho_w*V_w 0=m_body*c_p_body*(T2-T1_body)+m_w*c_p_w*(T2-T1_w) "energy balance on the body and water" DELTAT=T1_body-T2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-383

4-126 A piston-cylinder device initially contains saturated liquid water. An electric resistor placed in the tank is turned on until the tank contains saturated water vapor. The volume of the tank, the final temperature, and the power rating of the resistor are to be determined. Assumptions 1. The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions. Properties The initial properties of steam are (Table A-4) } Analysis (a) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as

⏟

since for a constant-pressure process. The initial and final volumes are

(b) Now, the final state can be fixed by calculating specific volume

The final state is saturated mixture and both pressure and temperature remain constant during the process. Other properties are }

(Table A-4 or EES)

EES (Engineering Equation Solver) SOLUTION "Given" m=1.8 [kg] x1=0 T1=120 [C] time=10*60 [s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-384

VolumeRatio=4 "Analysis" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=x1) h1=enthalpy(Fluid$, T=T1, x=x1) P1=pressure(Fluid$, T=T1, x=x1) Vol1=m*v1 Vol2=VolumeRatio*Vol1 P2=P1 v2=Vol2/m T2=temperature(Fluid$, P=P2, v=v2) h2=enthalpy(Fluid$, P=P2, v=v2) x2=quality(Fluid$, P=P2, v=v2) W_e_in=m*(h2-h1) W_dot_e_in=W_e_in/time

4-127 An insulated rigid tank initially contains saturated liquid water and air. An electric resistor placed in the tank is turned on until the tank contains saturated water vapor. The volume of the tank, the final temperature, and the power rating of the resistor are to be determined. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions. 3 Energy added to the air is neglected. Properties The initial properties of steam are (Table A-4) } Analysis (a) We take the contents of the tank as the system. We neglect energy added to the air in our analysis based on the problem statement. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as

⏟

The initial water volume and the tank volume are

(b) Now, the final state can be fixed by calculating specific volume

1-385

} (c) Substituting, Finally, the power rating of the resistor is ̇

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=1.4 [kg] T_1=200 [C] x_1=0 f_v=0.25 x_2=1 time=20[min]*Convert(min, s) "ANALYSIS" Fluid$='Steam_iapws' v_1=volume(Fluid$, T=T_1, x=0) u_1=intenergy(Fluid$, T=T_1, x=0) Vol_1=m*v_1 Vol_2=1/f_v*Vol_1 Vol_tank=Vol_2 "volume of the tank" v_2=Vol_2/m T_2=temperature(Fluid$, v=v_2, x=x_2) "final temperature" u_2=intenergy(Fluid$, v=v_2, x=x_2) W_e_in=m*(u_2-u_1) "energy balance" W_dot_e_in=W_e_in/time "power rating of the resistor"

4-128 A 1- ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibrium temperature in the tank is to be determined. Assumptions 1. Thermal properties of the ice and water are constant. 2. Heat transfer to the water tank is negligible. 3. There is no stirring by hand or a mechanical device (it will add energy).

Properties The specific heat of water at room temperature is

, and the specific heat of ice at about 0C is

(Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and .

1-386

Analysis We take the ice and the water as our system, and disregard any heat transfer between the system and the surroundings. Then the energy balance for this process can be written as ⏟

⏟

]

Substituting, ] It gives which is the final equilibrium temperature in the tank.

EES (Engineering Equation Solver) SOLUTION "Given" m_w=1*1000 [kg] T1_w=20 [C] m_ice=130 [kg] T1_ice=-5 [C] T_melting=0 [C] "Properties" C_w=4.18 [kJ/kg-K] C_ice=2.11 [kJ/kg-K] h_if=333.7 [kJ/kg] "given" "Analysis" 0=m_ice*C_ice*(T_melting-T1_ice)+m_ice*h_if+m_ice*C_w*(T2-T_melting)+m_w*C_w*(T2-T1_w) "energy balance on ice and water"

4-129 A 0.3-L glass of water at 20°C is to be cooled with ice to 5°C. The amount of ice or cold water that needs to be added to the water is to be determined. Assumptions 1. Thermal properties of the ice and water are constant. 2. Heat transfer to the glass is negligible. 3. There is no stirring by hand or a mechanical device (it will add energy). Properties The density of water is , and the specific heat of water at room temperature is (Table A-3). The specific heat of ice at about 0C is (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and .

1-387

Analysis (a) The mass of the water is We take the ice and the water as our system, and disregard any heat and mass transfer. This is a reasonable assumption since the time period of the process is very short. Then the energy balance can be written as ⏟

⏟

] Noting that

and

]

and substituting gives ]

m = 0.0546 kg = 54.6 g (b) When

instead of 0C, substituting gives ]

]

m = 0.0487 kg = 48.7 g Cooling with cold water can be handled the same way. All we need to do is replace the terms for ice by a term for cold water at 0C: ]

]

Substituting, [

(

)

]

(

)

It gives m = 0.9 kg = 900 g Discussion Note that this is about 16 times the amount of ice needed, and it explains why we use ice instead of water to cool drinks. Also, the temperature of ice does not seem to make a significant difference.

EES (Engineering Equation Solver) SOLUTION "Given" V_w=0.3[L]*Convert(L, m^3) T1=20 [C] T2=5 [C] T1_ice_a=0 [C] T1_ice_b=-20 [C] T_cw=0 [C] T_melting=0 [C] h_if=333.7 [kJ/kg] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-388

"Properties" C_w=4.18 [kJ/kg-C] C_ice=2.11 [kJ/kg-C] rho_w=1000 [kg/m^3] "Analysis" "(a)" m_w=rho_w*V_w 0=m_ice_a*C_ice*(T_melting-T1_ice_a)+m_ice_a*h_if+m_ice_a*C_ice*(T2-T_melting)+m_w*C_w*(T2-T1) "energy balance on ice and water" "(b)" 0=m_ice_b*C_ice*(T_melting-T1_ice_b)+m_ice_b*h_if+m_ice_b*C_ice*(T2-T_melting)+m_w*C_w*(T2-T1) 0=m_cw*C_w*(T2-T_cw)+m_w*C_w*(T2-T1) "cold water case"

4-130 Problem 4-129 is reconsidered. The effect of the initial temperature of the ice on the final mass of ice required as the ice temperature varies from -26°C to 0°C is to be investigated. The mass of ice is to be plotted against the initial temperature of ice. Analysis The problem is solved using EES, and the solution is given below. V = 0.3 [L] T_1_ice = 0 [C] T_1 = 20 [C] T_2 = 5 [C] c_ice = 2.11 [kJ/kg-C] c_water = 4.18 [kJ/kg-C] h_if = 333.7 [kJ/kg] T_1_ColdWater = 0 [C] rho_water = 1 [kg/L] m_water = rho_water*V E_in - E_out = DELTAE_sys E_in = 0 E_out = 0 DELTAE_sys = DELTAU_water+DELTAU_ice "The system is water+ice" DELTAU_water = m_water*C_water*(T_2 - T_1) T_melting = 0 [C] DELTAU_ice = DELTAU_solid_ice+DELTAU_melted_ice DELTAU_solid_ice =m_ice*c_ice*(T_melting-T_1_ice) + m_ice*h_if DELTAU_melted_ice=m_ice*c_water*(T_2 - T_melting) m_ice_grams=m_ice*convert(kg,g) DELTAE_sys = DELTAU_water+DELTAU_ColdWater "The system is water+cold water" DELTAU_water = m_water*c_water*(T_2_ColdWater - T_1) DELTAU_ColdWater = m_ColdWater*c_water*(T_2_ColdWater - T_1_ColdWater) m_ColdWater_grams=m_ColdWater*convert(kg,g)

] -26 -24 -22

] 45.94 46.42 46.91

1-389

-20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0

47.4 47.91 48.43 48.97 49.51 50.07 50.64 51.22 51.81 52.42 53.05

1-390

4-131 A well-insulated room is heated by a steam radiator, and the warm air is distributed by a fan. The average temperature in the room after 45 min is to be determined. Assumptions 1. Air is an ideal gas with constant specific heats at room temperature. 2. The kinetic and potential energy changes are negligible. 3. The air pressure in the room remains constant and thus the air expands as it is heated, and some warm air escapes. Properties The gas constant of air is (Table A-1). Also, for air at room temperature (Table A-2). Analysis We first take the radiator as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as ⏟

⏟

Using data from the steam tables (Tables A-4 through A-6), some properties are determined to be } }

Substituting, The volume and the mass of the air in the room are

and

The amount of fan work done in 45 min is ̇ We now take the air in the room as the system. The energy balance for this closed system is expressed as

since the boundary work and U combine into H for a constant pressure expansion or compression process. It can also be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-391

̇ ̇

Substituting, ( ) which yields Therefore, the air temperature in the room rises from 7C to 10.7C in 45 min.

EES (Engineering Equation Solver) SOLUTION "Given" Vol_room=3*4*6 [m^3] T1_air=7 [C] P_air=100 [kPa] Vol_rad=0.015 P1=200 [kPa] T1=200 [C] P2=100 [kPa] time=45*60 [s] W_dot_fan_in=0.120 [kW] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) u1=intenergy(Fluid$, P=P1, T=T1) v2=v1 "fixed mass, constant volume" u2=intenergy(Fluid$, P=P2, v=v2) c_p_air=1.005 [kJ/kg-K] "Table A-2a" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" m_s=Vol_rad/v1 -Q_out_steam=m_s*(u2-u1) "energy balance on steam" m_air=(P_air*Vol_room)/(R*(T1_air+273)) W_fan_in=W_dot_fan_in*time Q_in_air=Q_out_steam Q_in_air+W_fan_in=m_air*C_p_air*(T2_air-T1_air) "energy balance on room air" "DELTAU+W_b = DELTAH for a constant pressure process"

4-132 Two adiabatic chambers are connected by a valve. One chamber contains oxygen while the other one is evacuated. The valve is now opened until the oxygen fills both chambers and both tanks have the same pressure. The total internal energy change and the final pressure in the tanks are to be determined. Assumptions 1. Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2. The kinetic and potential energy changes are negligible, . 3. Constant specific heats at room temperature can be used. 4. Both chambers are insulated and thus heat transfer is negligible. Analysis We take both chambers as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

1-392

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = D U = mcv (T2 - T1 )

Since the internal energy does not change, the temperature of the air will also not change. Applying the ideal gas equation gives →

EES (Engineering Equation Solver) SOLUTION Vol1=2 [m^3] P1=1000 [kPa] T1=(127+273) [K] Vol2=2*Vol1 P2=P1*Vol1/Vol2

4-133 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and the two tanks come to the same state at the temperature of the surroundings. The final pressure and the amount of heat transfer are to be determined. Assumptions 1. The tanks are stationary and thus the kinetic and potential energy changes are zero. 2. The tank is insulated and thus heat transfer is negligible. 3. There are no work interactions. Analysis We take the entire contents of the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as

⏟

] ] The properties of water in each tank are (Tables A-4 through A-6) Tank A: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-393

P1 = 400 kPa üïï v f = 0.001084, v g = 0.46242 m3 /kg ý ïïþ u f = 604.22, u fg = 1948.9 kJ/kg x1 = 0.80 ] Tank B: }

ü ïï v f = 0.001003, v g = 43.340 m 3 /kg ý u fg = 2304.3 kJ/kg v 2 = 0.73117 m3 /kg ïïþ u f = 104.83, Thus at the final state the system will be a saturated liquid-vapor mixture since T2 = 25°C

. The final pressure must be

Also,

Substituting, ]

EES (Engineering Equation Solver) SOLUTION "Given" Vol_A=0.2 [m^3] P1_A=400 [kPa] x1_A=0.80 Vol_B=0.5 [m^3] P1_B=200 [kPa] T1_B=250 [C] T2=25 [C] "Properties" Fluid$='steam_iapws' v1_A=volume(Fluid$, P=P1_A, x=x1_A) u1_A=intenergy(Fluid$, P=P1_A, x=x1_A) v1_B=volume(Fluid$, P=P1_B, T=T1_B) u1_B=intenergy(Fluid$, P=P1_B, T=T1_B) "Analysis" m1_A=Vol_A/v1_A m1_B=Vol_B/v1_B m_t=m1_A+m1_B Vol_t=Vol_A+Vol_B v2=Vol_t/m_t P2=pressure(Fluid$, T=T2, v=v2) u2=intenergy(Fluid$, T=T2, v=v2) -Q_out=m_t*u2-m1_A*u1_A-m1_B*u1_B "energy balance"

1-394

4-134 Problem 4-133 is reconsidered. The effect of the environment temperature on the final pressure and the heat transfer as the environment temperature varies from 0°C to 50°C is to be investigated. The final results are to be plotted against the environment temperature. Analysis The problem is solved using EES, and the solution is given below. Vol_A=0.2 [m^3] P_A[1]=400 [kPa] x_A[1]=0.8 T_B[1]=250 [C] P_B[1]=200 [kPa] Vol_B=0.5 [m^3] T_final=25 [C] E_in-E_out=DELTAE E_in=0 E_out=Q_out DELTAE=m_A*(u_A[2]-u_A[1])+m_B*(u_B[2]-u_B[1]) m_A=Vol_A/v_A[1] m_B=Vol_B/v_B[1] Fluid$='Steam_IAPWS' u_A[1]=INTENERGY(Fluid$,P=P_A[1], x=x_A[1]) v_A[1]=volume(Fluid$,P=P_A[1], x=x_A[1]) T_A[1]=temperature(Fluid$,P=P_A[1], x=x_A[1]) u_B[1]=INTENERGY(Fluid$,P=P_B[1],T=T_B[1]) v_B[1]=volume(Fluid$,P=P_B[1],T=T_B[1]) u_B[2]=u_final u_A[2]=u_final m_final=m_A+m_B Vol_final=Vol_A+Vol_B v_final=Vol_final/m_final u_final=INTENERGY(Fluid$,T=T_final, v=v_final) P_final=pressure(Fluid$,T=T_final, v=v_final) ] 0.6112 0.9069 1.323 1.898 2.681 3.734 5.13 6.959 9.325 12.35

] 2300 2274 2247 2218 2187 2153 2116 2075 2030 1978

] 0 5.556 11.11 16.67 22.22 27.78 33.33 38.89 44.44 50

1-395

4-135 A piston-cylinder device contains an ideal gas. An external shaft connected to the piston exerts a force. For an isothermal process of the ideal gas, the amount of heat transfer, the final pressure, and the distance that the piston is displaced are to be determined.

Assumptions 1. The kinetic and potential energy changes are negligible, . 2. The friction between the piston and the cylinder is negligible. Analysis (a) We take the ideal gas in the cylinder to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary closed system can be expressed as ⏟

⏟

Thus, the amount of heat transfer is equal to the boundary work input (b) The relation for the isothermal work of an ideal gas may be used to determine the final volume in the cylinder. But we first calculate initial volume

Then, ( ) (

)→

The final pressure can be determined from ideal gas relation applied for an isothermal process → → (c) The final position of the piston and the distance that the piston is displaced are © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-396

→

→ D

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] T_1=24[C]+273 [K] D=0.10 [m] L_1=0.20 [m] W_in=0.1 [kJ] "ANALYSIS" W_in-Q_out=0 "energy balance, isothermal process of an ideal gas" -W_in=P_1*V_1*ln(V_2/V_1) "isothermal work in a closed system, compression is negative" P_1*V_1=P_2*V_2 V_1=pi*D^2/4*L_1 V_2=pi*D^2/4*L_2 DELTAL=L_1-L_2

4-136 A piston-cylinder device with a set of stops contains superheated steam. Heat is lost from the steam. The pressure and quality (if mixture), the boundary work, and the heat transfer until the piston first hits the stops and the total heat transfer are to be determined. Assumptions 1. The kinetic and potential energy changes are negligible, . 2. The friction between the piston and the cylinder is negligible. Analysis (a) We take the steam in the cylinder to be the system. This is a closed system since no mass crosses the system boundary. The energy balance for this stationary closed system can be expressed as ⏟

⏟

Denoting when piston first hits the stops as state (2) and the final state as (3), the energy balance relations may be written as

The properties of steam at various states are (Tables A-4 through A-6)

1-397

} } } (b) Noting that the pressure is constant until the piston hits the stops during which the boundary work is done, it can be determined from its definition as (c) Substituting into energy balance relations, (d)

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=0.35 [kg] P_1=3500 [kPa] DELTAT_superheat=7.4 [C] x_2=0 "state 2 is when piston first hits the stops" T_3=200 [C] "3 is the final state" "ANALYSIS" Fluid$='Steam_iapws' T_sat=temperature(Fluid$, P=P_1, x=1) T_1=T_sat+DELTAT_superheat u_1=intenergy(Fluid$, P=P_1, T=T_1) v_1=volume(Fluid$, P=P_1, T=T_1) P_2=P_1 v_2=volume(Fluid$, P=P_2, x=x_2) u_2=intenergy(Fluid$, P=P_2, x=x_2) v_3=v_2 P_3=pressure(Fluid$, T=T_3, v=v_3) "final pressure" x_3=quality(Fluid$, T=T_3, v=v_3) "final quality" u_3=intenergy(Fluid$, T=T_3, v=v_3) W_b_in-Q_out_12=m*(u_2-u_1) "energy balance, process 1-2" W_b_in-Q_out_13=m*(u_3-u_1) "energy balance, process 1-3" W_b_in=m*P_1*(v_1-v_2) "boundary work"

4-137E Two adiabatic tanks containing water at different states are connected by a valve. The valve is now opened, allowing the water vapor from the high-pressure tank to move to the low-pressure tank until the pressure in the two becomes equal. The final pressure and the final mass in each tank are to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no heat or work interactions involved Analysis We take both tanks as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

1-398

⏟

where the high-pressure and low-pressure tanks are denoted by A and B, respectively. The specific volume in each tank is } } The total mass contained in the tanks is

Similarly, the initial total internal energy contained in both tanks is The internal energy and the specific volume are

Now, the final state is fixed. The pressure in this state may be obtained by iteration in water tables (Table A-5E). We used EES to get the exact result: Since the specific volume is now the same in both tanks, and both tanks have the same volume, the mass is equally divided between the tanks at the end of this process,

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol_A=10 [ft^3] P_1A=450 [psia] x_1A=0.10 Vol_B=10 [ft^3] P_1B=15 [psia] x_1B=0.75 "PROPERTIES" Fluid$='Steam_iapws' v_1A_f=volume(Fluid$, P=P_1A, x=0) v_1A_g=volume(Fluid$, P=P_1A, x=1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-399

v_1A=v_1A_f+x_1A*(v_1A_g-v_1A_f) u_1A_f=intenergy(Fluid$, P=P_1A, x=0) u_1A_g=intenergy(Fluid$, P=P_1A, x=1) u_1A_fg=u_1A_g-u_1A_f u_1A=u_1A_f+x_1A*u_1A_fg v_1B_f=volume(Fluid$, P=P_1B, x=0) v_1B_g=volume(Fluid$, P=P_1B, x=1) v_1B=v_1B_f+x_1B*(v_1B_g-v_1B_f) u_1B_f=intenergy(Fluid$, P=P_1B, x=0) u_1B_g=intenergy(Fluid$, P=P_1B, x=1) u_1B_fg=u_1B_g-u_1B_f u_1B=u_1B_f+x_1B*u_1B_fg m_1A=Vol_A/v_1A m_1B=Vol_B/v_1B m=m_1A+m_1B U_tot_1=m_1A*u_1A+m_1B*u_1B u_1=U_tot_1/m u_2=u_1 v_1=(Vol_A+Vol_B)/m v_2=v_1 P_2=pressure(Fluid$, u=u_2, v=v_2) m_2A=m/2 m_2B=m/2

4-138 An insulated rigid tank is divided into two compartments, each compartment containing the same ideal gas at different states. The two gases are allowed to mix. The simplest expression for the mixture temperature in a specified format is to be obtained. Analysis We take the both compartments together as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

⏟

and Solving for final temperature, we find

1-400

4-139 Solar energy is to be stored as sensible heat using phase-change materials, granite rocks, and water. The amount of heat that can be stored in a space using these materials as the storage medium is to be determined. Assumptions 1. The materials have constant properties at the specified values. 2. No allowance is made for voids, and thus the values calculated are the upper limits. Analysis The amount of energy stored in a medium is simply equal to the increase in its internal energy, which, for incompressible substances, can be determined from (a) The latent heat of glaubers salts is given to be of energy stored is becomes

. Disregarding the sensible heat storage in this case, the amount

This value would be even larger if the sensible heat storage due to temperature rise is considered. (b) The density of granite is (Table A-3), and its specific heat is given to be amount of energy that can be stored in the rocks when the temperature rises by 20C becomes (c) The density of water is about (Table A-3), and its specific heat is given to be amount of energy that can be stored in the water when the temperature rises by 20C becomes

. Then the

Discussion Note that the greatest amount of heat can be stored in phase-change materials essentially at constant temperature. Such materials are not without problems, however, and thus they are not widely used.

EES (Engineering Equation Solver) SOLUTION "Given" T_melting_salt=32 [C] h_if_salt=329*1000 [kJ/m^3] V=5 [m^3] C_rock=2.32 [kJ/kg-C] DELTAT_rock=20 [C] C_water=4.00 [kJ/kg-C] DELTAT_water=20 [C] "Properties" rho_rock=2700 [kg/m^3] "from Table A-3 of the text" rho_water=1000 [kg/m^3] "from Table A-3 of the text" "Analysis" "(a)" DELTAU_salt=V*h_if_salt "(b)" DELTAU_rock=rho_rock*V*C_rock*DELTAT_rock "(c)" DELTAU_water=rho_water*V*C_water*DELTAT_water

Fundamentals of Engineering (FE) Exam Problems 4-140 A rigid tank contains nitrogen gas at 500 kPa and 300 K. Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800 kPa. The work done during this process is (a) 500 kJ (b) 1500 kJ (c) 0 kJ (d) 900 kJ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-401

(e) 2400 kJ Answer (b) 0 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=3 [m^3] P1=500 [kPa] T1=300 [K] P2=800 [kPa] W=0 "since constant volume" "Some Wrong Solutions with Common Mistakes:" R=0.297 W1_W=V*(P2-P1) "Using W=V*DELTAP" W2_W=V*P1 W3_W=V*P2 W4_W=R*T1*ln(P1/P2)

4-141 A volume of (a) 82 kJ (b) 180 kJ (c) 240 kJ (d) 275 kJ (e) 315 kJ

cylinder contains nitrogen gas at 600 kPa and 300 K. Now the gas is compressed isothermally to a . The work done on the gas during this compression process is

Answer (d) 275 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=0.5 [m^3] V2=0.2 [m^3] P1=600 [kPa] T1=300 [K] R=8.314/28 P1*V1=m*R*T1 W=m*R*T1* ln(V1/V2) "constant temperature process" "Some Wrong Solutions with Common Mistakes:" W1_W=R*T1* ln(V2/V1)"Forgetting m" W2_W=P1*(V1-V2) "Using V*DeltaP" P1*V1/T1=P2*V2/T1 W3_W=(V1-V2)*(P1+P2)/2 "Using P_ave*Delta V" W4_W=P1*V1-P2*V2 "Using W=P1V1-P2V2"

4-142 A well-sealed room contains 60 kg of air at 200 kPa and 25C. Now solar energy enters the room at an average rate of while a 120-W fan is turned on to circulate the air in the room. If heat transfer through the walls is negligible, the air temperature in the room in 30 min will be (a) 25.6C (b) 49.8C (c) 53.4C © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-402

(d) 52.5C (e) 63.4C Answer (e) 63.4C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=60 [kg] P1=200 [kPa] T1=25 [C] Qsol=0.8 [kJ/s] time=30*60 [s] Wfan=0.12 [kJ/s] R=0.287 [kJ/kg-K] cv=0.718 [kJ/kg-K] time*(Wfan+Qsol)=m*cv*(T2-T1) "from energy balance E_in-E_out=dE_system" "Some Wrong Solutions with Common Mistakes:" cp=1.005 [kJ/kg-K] time*(Wfan+Qsol)=m*cp*(W1_T2-T1) "Using cp instead of cv " time*(-Wfan+Qsol)=m*cv*(W2_T2-T1) "Subtracting Wfan instead of adding" time*Qsol=m*cv*(W3_T2-T1) "Ignoring Wfan" time*(Wfan+Qsol)/60=m*cv*(W4_T2-T1) "Using min for time instead of s"

4-143 A room contains 75 kg of air at 100 kPa and 15C. The room has a 250-W refrigerator (the refrigerator consumes 250 W of electricity when running), a 120-W TV, a 1.8-kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is (a) (b) (c) (d) (e) Answer (e) Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=75 [kg] P_1=100 [kPa] T_1=15 [C] time=30*60 [s] W_ref=0.250 [kJ/s] W_TV=0.120 [kJ/s] W_heater=1.8 [kJ/s] W_fan=0.05 [kJ/s] R=0.287 [kJ/kg-K] "Applying energy balance E_in-E_out=dE_system gives E_out=E_in since T=constant and dE=0" E_gain=W_ref+W_TV+W_heater+W_fan Q_loss=E_gain*Convert(kJ/s, kJ/h) "Some Wrong Solutions with Common Mistakes:" E_gain1=-W_ref+W_TV+W_heater+W_fan "Subtracting Wrefrig instead of adding" W1_Qloss=E_gain1*3600 "kJ/h" E_gain2=W_ref+W_TV+W_heater-W_fan "Subtracting Wfan instead of adding" W2_Qloss=E_gain2*3600 "kJ/h" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-403

E_gain3=-W_ref+W_TV+W_heater-W_fan "Subtracting Wrefrig and Wfan instead of adding" W3_Qloss=E_gain3*3600 "kJ/h" E_gain4=W_ref+W_heater+W_fan "Ignoring the TV" W4_Qloss=E_gain4*3600 "kJ/h"

4-144 A frictionless piston-cylinder device and a rigid tank contain 3 kmol of an ideal gas at the same temperature, pressure and volume. Now heat is transferred, and the temperature of both systems is raised by 10C. The amount of extra heat that must be supplied to the gas in the cylinder that is maintained at constant pressure is (a) 0 kJ (b) 27 kJ (c) 83 kJ (d) 249 kJ (e) 300 kJ Answer (d) 249 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N=3 [kmol] T_change=10 [C] Ru=8.314 [kJ/kmol-K] Q_diff=N*Ru*T_change "cp-cv=R, and thus Q_diff=m*R*dT=N*Ru*dT" "Some Wrong Solutions with Common Mistakes:" W1_Qdiff=0 "Assuming they are the same" W2_Qdiff=Ru*T_change "Not using mole numbers" W3_Qdiff=Ru*T_change/N "Dividing by N instead of multiplying" W4_Qdiff=N*Rair*T_change; Rair=0.287 "using Ru instead of R"

4-145 A piston-cylinder device contains 5 kg of air at 400 kPa and 30C. During a quasi-equilibrium isothermal expansion process, 15 kJ of boundary work is done by the system, and 3 kJ of paddle-wheel work is done on the system. The heat transfer during this process is (a) 12 kJ (b) 18 kJ (c) 2.4 kJ (d) 3.5 kJ (e) 60 kJ Answer (a) 12 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=5 [kg] P_1=400 [kPa] T=30 [C] Wout_b=15 [kJ] Win_pw=3 [kJ] R=0.287 [kJ/kg-K] cv=0.718 [kJ/kg-K] Q_in+Win_pw-Wout_b=0 "From energy balance E_in-E_out=dE_system, T=constant and thus dE_system=0" "Some Wrong Solutions with Common Mistakes:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-404

W1_Qin=Q_in/cv "Dividing by cv" W2_Qin=Win_pw+Wout_b "Adding both quantities" W3_Qin=Win_pw "Setting it equal to paddle-wheel work" W4_Qin=Wout_b "Setting it equal to boundaru work"

4-146 A glass of water with a mass of 0.32 kg at 20C is to be cooled to 0C by dropping ice cubes at 0C into it. The latent heat of fusion of ice is , and the specific heat of water is . The amount of ice that needs to be added is (a) 32 g (b) 40 g (c) 80 g (d) 93 g (e) 110 g Answer (c) 80 g Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_w=0.32 [kg] T1=20 [C] T2=0 [C] c=4.18 [kJ/kg-K] h_melting=334 [kJ/kg-K] DELTAT=T2-T1 "Noting that there is no energy transfer with the surroundings and the latent heat of melting of ice is transferred form the water, and applying energy balance E_in-E_out=dE_system to ice+water gives" dE_ice+dE_w=0 dE_ice=m_ice*h_melting dE_w=m_w*c*DELTAT "Some Wrong Solutions with Common Mistakes:" W1_mice*h_melting*(T1-T2)+m_w*c*DELTAT=0 "Multiplying h_latent by temperature difference" W2_mice=m_w "taking mass of water to be equal to the mass of ice"

4-147 A 2-kW electric resistance heater submerged in 5-kg water is turned on and kept on for 10 min. During the process, 300 kJ of heat is lost from the water. The temperature rise of water is (a) 0.4C (b) 43.1C (c) 57.4C (d) 71.8C (e) 180.0C Answer (b) 43.1C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=5 [kg] Q_loss=300 [kJ] time=10*60 [s] W_e=2 [kJ/s] c=4.18 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-405

time*W_e-Q_loss = dU_system "from energy balance E_in-E_out=dE_system" dU_system=m*c*DELTAT "Some Wrong Solutions with Common Mistakes:" time*W_e = m*c*W1_T "Ignoring heat loss" time*W_e+Q_loss = m*c*W2_T "Adding heat loss instead of subtracting" time*W_e-Q_loss = m*1.0*W3_T "Using specific heat of air or not using specific heat"

4-148 A 2-kW baseboard electric resistance heater in a vacant room is turned on and kept on for 15 min. The mass of the air in the room is 75 kg, and the room is tightly sealed so that no air can leak in or out. The temperature rise of air at the end of 15 min is (a) 8.5C (b) 12.4C (c) 24.0C (d) 33.4C (e) 54.8C Answer (d) 33.4C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=75 [kg] time=15*60 [s] W_e=2 [kJ/s] R=0.287 [kJ/kg-K] cv=0.718 [kJ/kg-K] time*W_e=m*cv*DELTAT "from energy balance E_in-E_out=dE_system" "Some Wrong Solutions with Common Mistakes:" cp=1.005 [kJ/kg-K] time*W_e=m*cp*W1_DELTAT "Using cp instead of cv" time*W_e/60=m*cv*W2_DELTAT "Using min for time instead of s"

4-149 1.5 kg of liquid water initially at 12C is to be heated to 95C in a teapot equipped with a 800 W electric heating element inside. The specific heat of water can be taken to be , and the heat loss from the water during heating can be neglected. The time it takes to heat the water to the desired temperature is (a) 5.9 min (b) 7.3 min (c) 10.8 min (d) 14.0 min (e) 17.0 min Answer (c) 10.8 min Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=1.5 [kg] T1=12 [C] T2=95 [C] Q_loss=0 [kJ] W_e=0.8 [kJ/s] c=4.18 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-406

time*W_e-Q_loss = dU_system "time in minutes" "from energy balance E_in-E_out=dE_system" dU_system=m*c*(T2-T1) time_min=time/60 "Some Wrong Solutions with Common Mistakes:" W1_time*60*W_e-Q_loss = m*c*(T2+T1) "Adding temperatures instead of subtracting" W2_time*60*W_e-Q_loss = c*(T2-T1) "Not using mass"

4-150 A container equipped with a resistance heater and a mixer is initially filled with 3.6 kg of saturated water vapor at 120C. Now the heater and the mixer are turned on; the steam is compressed, and there is heat loss to the surrounding air. At the end of the process, the temperature and pressure of steam in the container are measured to be 300C and 0.5 MPa. The net energy transfer to the steam during this process is (a) 274 kJ (b) 914 kJ (c) 1213 kJ (d) 988 kJ (e) 1291 kJ Answer (d) 988 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=3.6 [kg] T1=120 [C] x1=1 P2=500 [kPa] T2=300 [C] u1=INTENERGY(Steam_IAPWS,T=T1,x=x1) u2=INTENERGY(Steam_IAPWS,T=T2,P=P2) "Noting that Eout=0 and dU_system=m*(u2-u1), applying energy balance E_in-E_out=dE_system gives" E_out=0 E_in=m*(u2-u1) "Some Wrong Solutions with Common Mistakes:" cp_steam=1.8723 [kJ/kg-K] cv_steam=1.4108 [kJ/kg-K] W1_Ein=m*cp_Steam*(T2-T1) "Assuming ideal gas and using cp" W2_Ein=m*cv_steam*(T2-T1) "Assuming ideal gas and using cv" W3_Ein=u2-u1 "Not using mass" h1=ENTHALPY(Steam_IAPWS,T=T1,x=x1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) W4_Ein=m*(h2-h1) "Using enthalpy"

4-151 An ordinary egg with a mass of 0.1 kg and a specific heat of is dropped into boiling water at 95C. If the initial temperature of the egg is 5C, the maximum amount of heat transfer to the egg is (a) 12 kJ (b) 30 kJ (c) 24 kJ (d) 18 kJ (e) infinity Answer (b) 30 kJ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-407

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.1 [kg] T1=5 [C] T2=95 [C] c=3.32 [kJ/kg-K] E_in = dU_system "from energy balance E_in-E_out=dE_system" dU_system=m*c*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Ein = m*c*T2 "Using T2 only" W2_Ein=m*(ENTHALPY(Steam_IAPWS,T=T2,x=1)-ENTHALPY(Steam_IAPWS,T=T2,x=0)) "Using h_fg"

4-152 An apple with an average mass of 0.18 kg and average specific heat of The amount of heat transferred from the apple is (a) 7.9 kJ (b) 11.2 kJ (c) 14.5 kJ (d) 17.6 kJ (e) 19.1 kJ

is cooled from 17C to 5C.

Answer (a) 7.9 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.18 [kg] T1=17 [C] T2=5 [C] c=3.65 [kJ/kg-K] -Q_out = dU_system "from energy balance E_in-E_out=dE_system" dU_system=m*c*(T2-T1) "Some Wrong Solutions with Common Mistakes:" -W1_Qout =c*(T2-T1) "Not using mass" -W2_Qout =m*c*(T2+T1) "adding temperatures"

4-153 A 6-pack canned drink is to be cooled from 18C to 3C. The mass of each canned drink is 0.355 kg. The drinks can be treated as water, and the energy stored in the aluminum can itself is negligible. The amount of heat transfer from the 6 canned drinks is (a) 22 kJ (b) 32 kJ (c) 134 kJ (d) 187 kJ (e) 223 kJ Answer (c) 134 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=6*0.355 [kg] T1=18 [C] T2=3 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-408

c=4.18 [kJ/kg-K] DELTAT=T2-T1 "Applying energy balance E_in-E_out=dE_system and noting that dU_system=m*c*DELTAT gives" -Q_out=m*c*DELTAT "Some Wrong Solutions with Common Mistakes:" -W1_Qout=m*c*DELTAT/6 "Using one can only" -W2_Qout=m*c*(T1+T2) "Adding temperatures instead of subtracting" -W3_Qout=m*1.0*DELTAT "Using specific heat of air or forgetting specific heat"

4-154 A room is filled with saturated steam at 100C. Now a 5-kg bowling ball at 25C is brought to the room. Heat is transferred to the ball from the steam, and the temperature of the ball rises to 100C while some steam condenses on the ball as it loses heat (but it still remains at 100C). The specific heat of the ball can be taken to be . The mass of steam that condensed during this process is (a) 80 g (b) 128 g (c) 299 g (d) 351 g (e) 405 g Answer (c) 299 g Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_ball=5 [kg] T=100 [C] T1=25 [C] T2=100 [C] cp=1.8 [kJ/kg-C] Q=m_ball*cp*(T2-T1) Q=m_steam*h_fg h_f=ENTHALPY(Steam_IAPWS, x=0,T=T) h_g=ENTHALPY(Steam_IAPWS, x=1,T=T) h_fg=h_g-h_f "Some Wrong Solutions with Common Mistakes:" Q=W1m_steam*h_g "Using h_g" Q=W2m_steam*cp*(T2-T1) "Using m*c*DeltaT = Q for water" Q=W3m_steam*h_f "Using h_f"

4-155 An ideal gas has a gas constant and a constant-volume specific heat has a temperature change of 100C, choose the correct answer for each of the following: 1.

. If the gas

The change in enthalpy is, in kJ/kg (a) 30 (b) 70 (c) 100 (d) insufficient information to determine

Answer (c) 100

1-409

The change in internal energy is, in kJ/kg (a) 30 (b) 70 (c) 100 (d) insufficient information to determine

Answer (b) 70 3.

The work done is, in kJ/kg (a) 30 (b) 70 (c) 100 (d) insufficient information to determine

Answer (d) insufficient information to determine 4.

The heat transfer is, in kJ/kg (a) 30 (b) 70 (c) 100 (d) insufficient information to determine

Answer (d) insufficient information to determine 5.

The change in the pressure-volume product is, in kJ/kg (a) 30 (b) 70 (c) 100 (d) insufficient information to determine

Answer (a) 30 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). R=0.3 [kJ/kg-K] c_v=0.7 [kJ/kg-K] DELTAT=100 [K] "(1)" c_p=R+c_v DELTAh=c_p*DELTAT "(2)" DELTAu=c_v*DELTAT "(5)" PV=R*DELTAT

4-156 Saturated steam vapor is contained in a piston–cylinder device. While heat is added to the steam, the piston is held stationary, and the pressure and temperature become 1.2 MPa and 700°C, respectively. Additional heat is added to the steam until the temperature rises to 1200°C, and the piston moves to maintain a constant pressure. 1. The initial pressure of the steam is most nearly © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-410

(a) 250 kPa (b) 500 kPa (c) 750 kPa (d) 1000 kPa (e) 1250 kPa Answer (b) 500 kPa 2.

The work done by the steam on the piston is most nearly (a) (b) (c) (d) (e)

Answer (a) 3.

The total heat transferred to the steam is most nearly (a) (b) (c) (d) (e)

Answer (c) Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P2=1200 [kPa] T2=700 [C] T3=1200 [C] P3=P2 "1" v2=volume(steam_iapws, P=P2, T=T2) v1=v2 P1=pressure(steam_iapws, x=1, v=v1) "2" v3=volume(steam_iapws, P=P3, T=T3) w_b=P2*(v3-v2) "3" u1=intenergy(steam_iapws, x=1, v=v1) u3=intenergy(steam_iapws, P=P3, T=T3) q=u3-u1+w_b

4-157 A piston–cylinder device contains an ideal gas. The gas undergoes two successive cooling processes by rejecting heat to the surroundings. First the gas is cooled at constant pressure until . Then the piston is held stationary while the gas is further cooled to , where all temperatures are in K. 1.

The ratio of the final volume to the initial volume of the gas is

1-411

(a) 0.25 (b) 0.50 (c) 0.67 (d) 0.75 (e) 1.0 Answer (d) 0.75 Solution From the ideal gas equation

The work done on the gas by the piston is (a) (b) (c) (d) ( (e)

)

Answer (a) Solution From boundary work relation (per unit mass) ∫ 3.

→

The total heat transferred from the gas is (a) (b) (c) (d) ( (e)

)

Answer (d) (

)

Solution From an energy balance

qin = c p (T2 - T1 ) + cv (T3 - T2 ) = c p (3 / 4T1 - T1 ) + cv (1/ 2T1 - 3 / 4T1 ) = qout =

- (c p + cv )T1 4

(c p + cv )T1 4

1-412

4-158 … 4-164 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

1-413

1-414

CYU - Check Your Understanding CYU 5-1 ■ Check Your Understanding CYU 5-1.1 A steady-flow system with gas flow has one inlet (state 1) and two exits (states 2 and 3). Select the correct expressions for mass and volume flow rates. I.

m1 = m 2 + m 3

II. V1 = V2 + V3 (a) Only I (b) Only II (c) I and II (d) None of these Answer: (a) Only I CYU 5-1.2 During a process, 5 kg of mass enters a control volume. If the change in the mass content of the control volume is 3 kg at the end of the process, the amount of mass that leaves the control volume is (a) 0 kg (b) 2 kg (c) 3 kg (d) 5 kg (e) 8 kg Answer: (e) 8 kg min  mout  dm

5 kg  mout  3 kg mout  8 kg CYU 5-1.3 Liquid water flows through a 30-cm 2 cross-section pipe at a velocity of 2 m / s . If the density of water is 1 Kg / L , the mass and volume flow rates of water, respectively, are (a) 0.6 kg / s, 0.6 m3 / s (b) 6 kg / s, 6 m3 / s (c) 6 kg / s, 0.006 m3 / s (d) 60 kg / s, 0.06 m3 / s (e) 0.06 kg / s, 60 m3 / s Answer: (c) 6 kg / s, 0.006 m3 / s

m.  rho A V  1000 kg / m3  0.0030 m 2  (2 m / s)  6 kg / s

V.  AV   0.0030 m2  (2 m / s)  0.006 m3 / s

CYU 5-1.4 Consider the steady flow of a liquid in a converging pipe. The mass and volume flow rates of the liquid at the inlet are 2 kg / s and 2 L / s , respectively. If the inlet cross-sectional area of the pipe is twice that at the exit, the mass and volume flow rates of the fluid at the exit, respectively, are (a) 2 kg / s, 1L / s © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-415

(b) 1kg / s, 2 L / s (c) 1kg / s, 1L / s (d) 2 kg / s, 2 L / s (e) 1kg / s, 0.5 L / s Answer: (d) 2 kg / s, 2 L / s For a single-stream, steady-flow piping system, and liquid flow, m1  m2  2 kg / s

V1  V2  2 L / s

CYU 5-2 ■ Check Your Understanding CYU 5-2.1 A fluid of mass m, enthalpy h, internal energy u, kinetic energy ke, and potential energy pe enters a tank. Choose the expression that represents the total amount of energy that has entered the tank. (a) mh (b) m(ke) (c) m(u + h + ke + pe) (d) m(u + ke + pe) (e) m(h + ke + pe) Answer: (e) m(h + ke + pe) CYU 5-2.2 What is the flow work of a fluid flowing in a pipe at 150 kPa and 30C with a velocity of 4 m / s ? The density of the fluid is 2 kg / m3 . (a) 0.075 kJ / kg (b) 75 kJ / kg (c) 0.3 kJ / kg (d) 300 kJ / kg (e) 8 kJ / kg Answer: (b) 75 kJ / kg

w flow  Pv  P / rho  (150 kPa) /  2 kg / m3   75 kJ / kg

CYU 5-2.3 Consider a fluid flowing in a pipe at 150 kPa and 30C. Various specific energy terms of this fluid are given as h  13 kJ / kg, u  10 kJ / kg, ke  1.5 kJ / kg, pe  0.5 kJ / kg. What is the flow energy of this fluid? (a) 13 kJ / kg (b) 10 kJ / kg (c) 5 kJ / kg (d) 3 kJ / kg (e) 1.5 kJ / kg Answer: (d) 3 kJ / kg

w flow  h  u  13  10  3 kJ / kg

1-416

CYU 5-3 ■ Check Your Understanding CYU 5-3.1 Select the wrong statement regarding energy analysis of steady-flow systems. (a) The mass content of control volume during a steady-state process is constant. (b) The energy content of control volume during a steady-state process is constant. (c) A steady-flow device cannot involve more than one inlet and one outlet. (d) The fluid properties may be different at different inlets and exits during a steady-flow process. (e) The fluid properties at an inlet or exit remain constant during a steady-flow process. Answer: (c) A steady-flow device cannot involve more than one inlet and one outlet. CYU 5-3.2 A steady-flow device with one inlet (state 1), one exit (state 2) involves work input and no heat transfer. If the kinetic and potential energy changes are negligible, the energy balance relation for this device is (a) win  h1  h2 (b) win  h1  h2 (c) win  u1  u2 (d) win  u2  u1 (e) win  qout  h1  h2 Answer: (a) win  h1  h2 CYU 5-3.3 A steady-flow device with one inlet (state 1), one exit (state 2) involves work input and heat output with negligible potential energy change. The energy balance relation for this device is (a) qout  win  h2  h1 (b) win   h1  V12 / 2   qout   h2  V22 / 2 

(d) win   h1  V12 / 2   qout   h2  V22 / 2  (e) qout  win  u2  u1 Answer: (b) win   h1  V12 / 2   qout   h2  V22 / 2 

CYU 5-4 ■ Check Your Understanding CYU 5-4.1 Which device below handles a liquid and requires work input, usually with negligible heat transfer and negligible changes in kinetic and potential energies of fluid streams? (a) Nozzle (b) Turbine (c) Compressor (d) Pump (e) Heat exchanger Answer: (d) Pump © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-417

CYU 5-4.2 Which device below increases the pressure of gas usually with negligible changes in kinetic and potential energies of fluid streams? (a) Nozzle (b) Turbine (c) Compressor (d) Pump (e) Heat exchanger Answer: (c) Compressor CYU 5-4.3 Select the correct statements for the operation of a throttling valve with negligible heat transfer and negligible changes in kinetic and potential energies. I. Throttling valves produce a pressure drop without involving any work. II. For an ideal gas, the temperature decreases during a throttling process. III. Internal energy values at the inlet and exit of a throttling valve are the same. (a) Only I (b) Only II (c) I and II (d) II and III (e) I, II, and III Answer: (a) Only I CYU 5-4.4 Air is compressed steadily by an adiabatic compressor. The energy balance relation for this process on a unit mass basis is (a) u1  u2  wout (b) u1  win  qout  u2 (c) win  h1  h2 (d) h1  wout  h2 (e) h1  win  qout  h2 Answer: (c) win  h1  h2 CYU 5-4.5 Air incoming with a low velocity is accelerated by an adiabatic nozzle steadily to a high velocity. The simplest form of the energy balance relation for this device is (a) u1   u2  V22 / 2  )

u  V / 2  q  u  V / 2 (c)  h  V / 2   q   h  V / 2  (d) w   h  V / 2   q   h  V / 2  (b)

2 1

out

2 2

2 1

out

2 2

2 1

out

2 2

(e) h1  V22 / 2  h2 Answer: (e) h1  V22 / 2  h2 CYU 5-4.6 Steam enters a turbine which is not adequately insulated steadily at a mass flow rate of m . If the kinetic and potential energy changes are negligible, the power output of the turbine can be expressed as (a) Wout = m(h2 - h1 ) - Qin (b) W out = m (h2 - h1) (c) W out = m (h2 - h1)- Q out © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-418

(d)

W out = m (h1 - h2 )- Q out

(e)

W out = m (h2 - h1)+ Q out

Answer: (d) W out = m (h1 - h2 )- Q out CYU 5-4.7 An uninsulated mixing chamber is used to mix a cold fluid stream entering at a rate of m c by a hot fluid stream entering at a rate of m h . If the rate of heat loss from the mixing chamber is ̇

, the energy balance relation for this mixing

chamber is (a) m chc + m hhh = Q out (b) m chc + m hhh = (m c + m h )hm ixture (c) m chc + m hhh = (m c + m h )hm ixture + Q out (d) m chc + m hhh + Q out = (m c + m h )hm ixture (e) m chc + m hhh + Q in = (m c + m h )hm ixture Answer: (c) m chc + m hhh = (m c + m h )hm ixture + Q out

CYU 5-5 ■ Check Your Understanding CYU 5-5.1 An initially evacuated adiabatic tank is charged by air flowing in a line at Pi (in kPa) and Ti (in K). The valve at the inlet of the tank is closed when the air pressure inside the tank becomes Pi . The temperature of the air in the tank at this point is (a) Ti (b) cvTi (c) c pTi (d) (c p  cv )Ti (e) kTi Answer: (e) kTi Ein  Eout  dEsys

mi hi  m2u2 hi  u2 c p Ti  c v T2

T2   cp / c v  Ti

T2  kTi

1-419

CYU 5-5.2 An adiabatic tank initially containing steam at P1 and T1 with a mass of m1 is charged by steam flowing in a line at Pi and Ti . The valve at the inlet of the tank is closed when the air pressure and temperature inside the tank become

P2 and T2 . Taking the tank as the system, the energy for this process can be expressed as (a) (m2  m1 ) hi  m2u2  m1u1 (b) (m2  m1 ) hi  m2 h2  m1h1 (c) (m2  m1 ) ui  m2u2  m1u1 (d) (m2  m1 ) hi  m2u2 (e) (m2  m1 ) hi  m1u1 Answer: (a) (m2  m1 ) hi  m2u2  m1u1

Ein - Eout = dEsys mi hi = m2u2 - m1u1

(m2 - m1 )hi = m2u2 - m1u1 CYU 5-5.3 A tank initially contains air at P1 and T1 with a mass of m1 . Now, a valve is opened and air is allowed to leave the tank until the temperature and pressure in the tank drop to P2 and T2 . Heat is also lost from the tank during this process. Taking the tank as the system, the energy balance for this process can be expressed as (a) (m1 - m2 )he - Qout = m2 h2 - m1h1 (b) (m1 - m2 )hi - (m1 - m2 )he - Qout = m2u2 - m1u1 (c) (m1 - m2 )he = m2u2 - m1u1 - Qout (d) - (m1 - m2 )he - Qout = m2u2 - m1u1 (e) (m1 - m2 )he = m2u2 - m1u1 - Qout Answer: (d) - (m1 - m2 )he - Qout = m2u2 - m1u1

Ein - Eout = dEsys - me he - Qout = m2u2 - m1u1

- (m1 - m2 )he - Qout = m2u2 - m1u1 CYU 5-5.4 An initially evacuated adiabatic tank is charged by a fluid flowing steadily in a line at Pi and Ti . The valve of the tank is closed when the fluid pressure inside the tank reaches the line pressure Pi . The enthalpy and internal energy of the fluid in the line are 3000 kJ / kg and 2800 kJ / kg , respectively. The internal energy of the fluid in the tank at the final state is (a) 5800 kJ / kg (b) 3000 kJ / kg (c) 2800 kJ / kg (d) 200 kJ / kg (e) Insufficient information Answer: (b) 3000 kJ / kg Ein  Eout  dEsys

mi hi  m2u2 hi  u2  3000 kJ / kg

1-420

PROBLEMS Conservation of Mass 5-1C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process.

5-2C Mass flow rate is the amount of mass flowing through a cross-section per unit time whereas the volume flow rate is the amount of volume flowing through a cross-section per unit time.

5-3C The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady-flow process.

5-4C No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady (unless the density is constant). To be steady, the mass flow rate through the device must remain constant.

1-421

5-5 The ventilating fan of the bathroom of a building runs continuously. The mass of air “vented out” per day is to be determined.

Assumptions Flow through the fan is steady. Properties The density of air in the building is given to be 1.20 kg / m3 . Analysis The mass flow rate of air vented out is

mair = r Vair = (1.20 kg/m3 )(0.030 m3 /s) = 0.036 kg/s Then the mass of air vented out in 24 h becomes

m = mair D t = (0.036 kg/s)(24´ 3600 s) = 3110 kg Discussion Note that more than 3 tons of air is vented out by a bathroom fan in one day.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=30[L/s]*Convert(L/s, m^3/s) rho=1.20 [kg/m^3] "Analysis" DELTAt=24 [h]*Convert(h, s) "seconds in one day" m_dot=rho*V_dot m=m_dot*DELTAt

5-6 Air flows through a pipe. Heat is supplied to the air. The volume flow rates of air at the inlet and exit, the velocity at the exit, and the mass flow rate are to be determined.

Properties The gas constant for air is 0.287 kJ / kg.K (Table A-2). Analysis (a) (b) The volume flow rate at the inlet and the mass flow rate are V1 = AcV1 =

p D2 p (0.16 m) 2 V1 = (5 m/s) = 0.1005 m 3 / s 4 4

m = r 1 AcV1 =

P1 p D 2 (200 kPa) p (0.16 m) 2 V1 = (5 m/s) = 0.2391 kg / s RT1 4 (0.287 kJ/kg.K)(20 + 273 K) 4

1-422

(c) Noting that mass flow rate is constant, the volume flow rate and the velocity at the exit of the pipe are determined from

V2 =

m m 0.2391 kg/s = = = 0.1193 m 3 / s P 180 kPa r2 2 (0.287 kJ/kg.K)(40 + 273 K) RT2 V2 0.1193 m3 / s = = 5.935 m / s Ac p (0.16 m) 2 4

EES (Engineering Equation Solver) SOLUTION "GIVEN" D=0.16 [m] P_1=200 [kPa] T_1=20+273 [K] Vel_1=5 [m/s] P_2=180 [kPa] T_2=40+273 [K] "PROPERTIES" R=0.287 "R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Air)" "ANALYSIS" A=pi*D^2/4 rho_1=P_1/(R*T_1) Vol_dot_1=A*Vel_1 m_dot=rho_1*A*Vel_1 rho_2=P_2/(R*T_2) Vol_dot_2=m_dot/rho_2 Vel_2=Vol_dot_2/A

5-7E Steam flows in a pipe. The minimum diameter of the pipe for a given steam velocity is to be determined. Assumptions Flow through the pipe is steady.

Properties The specific volume of steam at the given state is (Table A-6E)

P = 200 psia ïü ïý v = 3.0586 ft 3 /lbm 1 T = 600°F ïïþ Analysis The cross-sectional area of the pipe is m=

1 mv (200 lbm/s)(3.0586 ft 3 /lbm) AcV ¾ ¾ ® A= = = 10.37 ft 2 v V 59 ft/s

Solving for the pipe diameter gives

p D2 ¾¾ ® D= 4

4A = p

4(10.37 ft 2 ) = 3.63 ft p

Therefore, the diameter of the pipe must be at least 3.63 ft to ensure that the velocity does not exceed 59 ft / s . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-423

EES (Engineering Equation Solver) SOLUTION m_dot=200 [lbm/s] P=200 [psia] T=600 [F] Vel=59 [ft/s] Fluid$='Steam_iapws' v=volume(Fluid$, P=P, T=T) A=m_dot*v/Vel A=pi*D^2/4

5-8E A garden hose is used to fill a water bucket. The volume and mass flow rates of water, the filling time, and the discharge velocity are to be determined. Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is no waste of water by splashing. Properties We take the density of water to be 62.4 lbm / ft 3 (Table A-3E). Analysis (a) The volume and mass flow rates of water are

V = AV = (p D 2 / 4)V = [p (1/12 ft) 2 / 4](8 ft/s) = 0.04363 ft 3 / s m = r V = (62.4 lbm/ft 3 )(0.04363 ft 3 /s) = 2.72 lbm / s (b) The time it takes to fill a 20-gallon bucket is Dt =

3 ö V 20 gal æ ÷ çç 1 ft ÷= 61.3 s = ÷ 3 ÷ ç 0.04363 ft /s è 7.4804 gal ø V

Ve =

V V 0.04363 ft 3 /s = = = 32 ft / s Ae p De2 / 4 [p (0.5 /12 ft) 2 / 4]

Discussion Note that for a given flow rate, the average velocity is inversely proportional to the square of the velocity. Therefore, when the diameter is reduced by half, the velocity quadruples.

EES (Engineering Equation Solver) SOLUTION "Given" V=20 [gal] D=1[in]*Convert(in, ft) D_e=0.5[in]*Convert(in, ft) Vel=8 [ft/s] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-424

rho=density(water, T=25, P=100) "Analysis" "(a)" A=pi*D^2/4 V_dot=A*Vel m_dot=rho*V_dot "(b)" DELTAt=V/V_dot*Convert(gal, ft^3) "c" A_e=pi*D_e^2/4 Vel_e=V_dot/A_e

5-9E Helium at a specified state is compressed to another specified state. The mass flow rate and the inlet area are to be determined. Assumptions Flow through the compressor is steady. Properties The gas cosntant of helium is R  2.6809 psia  ft 3 / lbm  R (Table A-1E)

Analysis The mass flow rate is determined from

A2V2 AV P (0.01 ft 2 )(100 ft/s)(200 psia) = 2 2 2= = 0.07038 lbm / s v2 RT2 (2.6809 psia ×ft 3 /lbm ×R)(1060 R)

The inlet area is determined from

A1 =

mv1 mRT1 (0.07038 lbm/s)(2.6809 psia ×ft 3 /lbm ×R)(530 R) = = = 0.1333 ft 2 V1 V1 P1 (50 ft/s)(15 psia)

EES (Engineering Equation Solver) SOLUTION "Given" P1=15 [psia] T1=(70+460) [R] P2=200 [psia] T2=(600+460) [R] A2=0.01 [ft^2] Vel2=100 [ft/s] Vel1=50 [ft/s] "Properties" R=2.6809 [psia-ft^3/lbm-R] "Table A-1E" "Analysis" v2=(R*T2/P2) m_dot=(A2*Vel2)/v2 v1=(R*T1/P1) A1=(m_dot*v1)/Vel1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-425

5-10 Air flows through an aircraft engine. The volume flow rate at the inlet and the mass flow rate at the exit are to be determined. Assumptions 1 Air is an ideal gas. 2 The flow is steady. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). Analysis The inlet volume flow rate is 2 3 V1 = AV 1 1 = (1 m )(180 m/s) = 180 m / s

The specific volume at the inlet is

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(20 + 273 K) = = 0.8409 m3 /kg P1 100 kPa

Since the flow is steady, the mass flow rate remains constant during the flow. Then,

V1 180 m3 /s = = 214.1 kg / s v1 0.8409 m3 /kg

EES (Engineering Equation Solver) SOLUTION "Given" A1=1 [m^2] P1=100 [kPa] T1=(20+273) [K] Vel1=180 [m/s] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" Vol1_dot=A1*Vel1 v1=(R*T1)/P1 m_dot=Vol1_dot/v1

5-11 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until the density rises to a specified level. The mass of air that entered the tank is to be determined. Properties The density of air is given to be 1.18 kg / m3 at the beginning, and 5.30 kg / m3 at the end.

Analysis We take the tank as the system, which is a control volume since mass. crosses the boundary The mass balance for this system can be expressed as Mass balance:

min - mout = D msystem ® mi = m2 - m1 = r 2V - r 1V Substituting,

1-426

Therefore, 8.24 kg of mass entered the tank.

EES (Engineering Equation Solver) SOLUTION "Given" V=2 [m^3] rho_1=1.18 [kg/m^3] rho_2=5.30 [kg/m^3] "Analysis" m_i=(rho_2-rho_1)*V

5-12 Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 2.21kg / m3 at the inlet, and 0.762 kg / m3 at the exit.

Analysis (a) The mass flow rate of air is determined from the inlet conditions to be 3 2 m = r 1 AV 1 1 = (2.21 kg/m )(0.009 m )(40 m/s) = 0.796 kg / s

(b) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit area of the nozzle is determined to be m = r 2 A2V2

¾¾ ®

A2 =

m 0.796 kg/s = = 0.00580 m 2 = 58.0 cm 2 r 2V2 (0.762 kg/ m3 )(180 m/s)

EES (Engineering Equation Solver) SOLUTION "Given" rho_1=2.21 [kg/m^3] Vel1=40 [m/s] rho_2=0.762 [kg/m^3] Vel2=180 [m/s] A1=90E-4 [m^2] "Analysis" "(a)" m_dot=rho_1*A1*Vel1 "(b)" A2=m_dot/(rho_2*Vel2)*Convert(m^2, cm^2)

5-13 A spherical hot-air balloon is considered. The time it takes to inflate the balloon is to be determined. Assumptions 1 Air is an ideal gas. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). Analysis The specific volume of air entering the balloon is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-427

RT (0.287 kPa ×m3 /kg ×K)(20 + 273 K) = = 0.7008 m3 /kg P 120 kPa

The mass flow rate at this entrance is

AcV p D 2 V p (1.0 m) 2 3 m/s = = = 3.362 kg/s v 4 v 4 0.7008 m3 /kg

The initial mass of the air in the balloon is

mi =

Vi p D3 p (5 m)3 = = = 93.39 kg v 6v 6(0.7008 m3 /kg)

Similarly, the final mass of air in the balloon is

mf =

Vf v

p D3 p (17 m)3 = = 3671 kg 6v 6(0.7008 m3 /kg)

The time it takes to inflate the balloon is determined from

Dt =

m f - mi m

(3671- 93.39) kg = 1064 s = 17.7 min 3.362 kg/s

EES (Engineering Equation Solver) SOLUTION "Given" T=(20+273) [K] P=120 [kPa] D1=5 [m] Vel=3 [m/s] D_opening=1 [m] D2=17 [m] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" v=R*T/P A_c=pi*D_opening^2/4 m_dot=A_c*Vel/v Vol_i=pi*D1^3/6 m_i=Vol_i/v Vol_f=pi*D2^3/6 m_f=Vol_f/v DELTAt=(m_f-m_i)/m_dot DELTAT_min=DELTAT*Convert(s, min)

5-14 Water flows through the tubes of a boiler. The velocity and volume flow rate of the water at the inlet are to be determined. Assumptions Flow through the boiler is steady. Properties The specific volumes of water at the inlet and exit are (Tables A-6 and A-7) P1 = 7 MPa ü ïï 3 ý v1 = 0.001017 m /kg ïïþ T1 = 65°C P2 = 6 MPa ïü ïý v = 0.05217 m3 /kg 2 T2 = 450°C ïïþ

1-428

Analysis The cross-sectional area of the tube is

p D 2 p (0.13 m)2 = = 0.01327 m2 4 4 The mass flow rate through the tube is same at the inlet and exit. It may be determined from exit data to be Ac =

AcV2 (0.01327 m 2 )(80 m/s) = = 20.35 kg/s v2 0.05217 m3 /kg

The water velocity at the inlet is then

V1 =

mv1 (20.35 kg/s)(0.001017 m3 /kg) = = 1.560 m / s Ac 0.01327 m2

The volumetric flow rate at the inlet is

V1 = AcV1 = (0.01327 m2 )(1.560 m/s) = 0.0207 m 3 / s EES (Engineering Equation Solver) SOLUTION "GIVEN" D=0.130 [m] P1=7000 [kPa] T1=65 [C] P2=6000 [kPa] T2=450 [C] Vel2=80 [m/s] "PROPERTIES" Fluid$='Steam_iapws' v1=volume(Fluid$, P=P1, T=T1) v2=volume(Fluid$, P=P2, T=T2) "ANALYSIS" A_c=pi*D^2/4 m_dot=(A_c*Vel2)/v2 Vel1=(m_dot*v1)/A_c Vol1_dot=A_c*Vel1

5-15 A desktop computer is to be cooled by a fan at a high elevation where the air density is low. The mass flow rate of air through the fan and the diameter of the casing for a given velocity are to be determined. Assumptions Flow through the fan is steady. Properties The density of air at a high elevation is given to be 0.7 kg / m3 . Analysis The mass flow rate of air is

mair = r Vair = (0.7 kg/m3 )(0.34 m3 /min) = 0.238 kg / min If the mean velocity is 110 m / min , the diameter of the casing is V = AV =

p D2 V ® D= 4

4V = pV

4(0.34 m 3 /min) = 0.0627 m = 6.3 cm p (110 m/min)

Therefore, the diameter of the casing must be at least 6.3 cm to ensure that the mean velocity does not exceed 110 m / min . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-429

Discussion This problem shows that engineering systems are sized to satisfy certain constraints imposed by certain considerations.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=0.34 [m^3/min] Elevation=3400 [m] rho=0.70 [kg/m^3] Vel=110 [m/min] "Analysis" m_dot=rho*V_dot*Convert(s, min) V_dot=A*Vel A=pi*D^2/4

5-16 Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percent increase in the velocity of air as it flows through the drier is to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 1.20 kg / m3 at the inlet, and 1.05 kg / m3 at the exit.

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . Then,

m1 = m2

r 1 AV1 = r 2 AV2 V2 r 1.20 kg/m3 = 1= = 1.263 V1 r 2 0.95 kg/m3

(or, and increase of 26.3%)

Therefore, the air velocity increases 26.3% as it flows through the hair drier.

EES (Engineering Equation Solver) SOLUTION "Given" rho_1=1.20 [kg/m^3] rho_2=0.95 [kg/m^3] "Analysis" Ratio_Vel=rho_1/rho_2 "from rho_1*A*Vel1=rho_2*A*Vel2" Increase_Vel=(Ratio_Vel-1)*Convert(, %)

5-17 Refrigerant-134a flows through a pipe. Heat is supplied to R-134a. The volume flow rates of air at the inlet and exit, the mass flow rate, and the velocity at the exit are to be determined.

1-430

Properties The specific volumes of R-134a at the inlet and exit are (Table A-13) P1 = 200 kPa ïü ïýv = 0.1142 m3 /kg 1 T1 = 20°C ïïþ

P1 = 180 kPa ïü ïýv = 0.1374 m3 /kg 2 T1 = 40°C ïïþ

Analysis (a) (b) The volume flow rate at the inlet and the mass flow rate are

V1 = AcV1 = m=

p D2 p (0.28 m)2 V1 = (5 m/s) = 0.3079 m3 / s 4 4

1 1 p D2 1 p (0.28 m) 2 AcV1 = V1 = (5 m/s) = 2.696 kg / s 3 v1 v1 4 4 0.1142 m /kg

(c) Noting that mass flow rate is constant, the volume flow rate and the velocity at the exit of the pipe are determined from

V2 = mv 2 = (2.696 kg/s)(0.1374 m3 /kg) = 0.3705 m3 / s V2 =

V2 0.3705 m3 / s = = 6.02 m / s Ac p (0.28 m) 2 4

EES (Engineering Equation Solver) SOLUTION "GIVEN" D=0.28 [m] P_1=200 [kPa] T_1=20 [C] Vel_1=5 [m/s] P_2=180 [kPa] T_2=40 [C] "PROPERTIES" Fluid$='R134a' v_1=volume(Fluid$, P=P_1, T=T_1) v_2=volume(Fluid$, P=P_2, T=T_2) "ANALYSIS" A=pi*D^2/4 Vol_dot_1=A*Vel_1 m_dot=1/v_1*A*Vel_1 Vol_dot_2=m_dot*v_2 Vel_2=Vol_dot_2/A

5-18 The minimum fresh air requirements of a residential building is specified to be 0.35 air changes per hour. The size of the fan that needs to be installed and the diameter of the duct are to be determined. Analysis The volume of the building and the required minimum volume flow rate of fresh air are

(

)

Vroom = (3.0m) 200m2 = 600m3 V = Vroom ´ ACH = 600m3 (0.35 / h) = 210m3 / h = 210, 000L / h = 3500 L / min

(

)

1-431

The volume flow rate of fresh air can be expressed as

V  VA  V (D 2 / 4) Solving for the diameter D and substituting, D

4V  V

4(210 / 3600 m 3 /s )  0.136 m  (4 m/s)

Therefore, the diameter of the fresh air duct should be at least 13.6 cm if the velocity of air is not to exceed 4 m / s .

EES (Engineering Equation Solver) SOLUTION "Given" ACH=0.35 [1/h] h=3.0 [m] A_room=200 [m^2] Vel=4 [m/s] "Analysis" V_room=h*A_room V_dot=V_room*ACH*Convert(m^3, L)*Convert(min, h) V_dot=Vel*A_duct*Convert(m^3, L)*Convert(min, s) A_duct=pi*D^2/4

Flow Work and Energy Transfer by Mass 5-19C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass.

5-20C Flowing fluids possess flow energy in addition to the forms of energy a fluid at rest possesses. The total energy of a fluid at rest consists of internal, kinetic, and potential energies. The total energy of a flowing fluid consists of internal, kinetic, potential, and flow energies.

5-21C Flow energy or flow work is the energy needed to push a fluid into or out of a control volume. Fluids at rest do not possess any flow energy.

5-22E A water pump increases water pressure. The flow work required by the pump is to be determined. Assumptions 1 Flow through the pump is steady. 2 The state of water at the pump inlet is saturated liquid. 3 The specific volume remains constant. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-432

Properties The specific volume of saturated liquid water at 15 psia is v  v f@ 15 psia  0.01672 ft 3 /lbm (Table A-5E)

Then the flow work relation gives

wflow = P2 v2 - Pv 1 1 = v ( P2 - P1 ) æ ö 1Btu ÷ = 0.01672ft 3 / lbm (80 - 15)psia çç 3 ÷ è 5.404psia ×ft ø = 0.201Btu / lbm

(

)

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=15 [psia] P2=80 [psia] "PROPERTIES" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, x=0) "ANALYSIS" w_flow=v1*(P2-P1)*Convert(psia-ft^3, Btu)

5-23 An air compressor compresses air. The flow work required by the compressor is to be determined. Assumptions 1 Flow through the compressor is steady. 2 Air is an ideal gas.

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). Analysis Combining the flow work expression with the ideal gas equation of state gives

wflow = P2v 2 - P1v1

= R(T2 - T1 ) = (0.287 kJ/kg ×K)(400 - 20)K = 109 kJ / kg

1-433

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.006 [m^3] P1=120 [kPa] T1=20 [C] P2=1000 [kPa] T2=400 [C] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" w_flow=R*(T2-T1) "w_flow = P2*v2 - P1*v1 = R(T2-T1)"

5-24 Warm air in a house is forced to leave by the infiltrating cold outside air at a specified rate. The net energy loss due to mass transfer is to be determined. Assumptions 1 The flow of the air into and out of the house through the cracks is steady. 2 The kinetic and potential energies are negligible. 3 Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2).

Analysis The density of air at the indoor conditions and its mass flow rate are r=

P 101.325 kPa = = 1.189 kg/m 3 RT (0.287 kPa ×m 3 /kg ×K)(24 + 273)K

æ 1h ÷ ö m = r V = (1.189 kg/m3 )(90 m3 /h) çç = 0.02971 kg/s ÷ ÷ çè3600 s ø

Noting that the total energy of a flowing fluid is equal to its enthalpy when the kinetic and potential energies are negligible, and that the rate of energy transfer by mass is equal to the product of the mass flow rate and the total energy of the fluid per unit mass, the rates of energy transfer by mass into and out of the house by air are

Emass, in = mqin = mh1 Emass, out = mqout = mh2 The net energy loss by air infiltration is equal to the difference between the outgoing and incoming energy flow rates, which is

D Emass = Emass, out - Emass, in = m(h2 - h1 ) = mc p (T2 - T1 ) = (0.02971 kg/s)(1.005 kJ/kg ×°C)(24 - 5)°C

1-434

This quantity represents the rate of energy transfer to the refrigerant in the compressor. Discussion The rate of energy loss by infiltration will be less in reality since some air will leave the house before it is fully heated to 24C.

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_house=(24+273) [K] V_dot=90 [m^3/h] T_outdoor=(5+273) [K] "Properties" c_p=CP(air, T=300) c_v=CV(air, T=300) R=c_p-c_v "Analysis" rho=P/(R*T_house) m_dot=rho*V_dot*Convert(s, h) DELTAE_dot_mass=m_dot*c_p*(T_house-T_outdoor)

5-25 Refrigerant-134a enters a compressor as a saturated vapor at a specified pressure, and leaves as superheated vapor at a specified rate. The rates of energy transfer by mass into and out of the compressor are to be determined. Assumptions 1 The flow of the refrigerant through the compressor is steady. 2 The kinetic and potential energies are negligible, and thus they are not considered.

Properties The enthalpy of refrigerant-134a at the inlet and the exit are (Tables A-12 and A-13)

h1 = hg @ 0.14MPa = 239.19kJ / kg

P2 = 0.8MPaïü ïý h = 296.82kJ / kg T2 = 60 C ïïþ 2

Analysis Noting that the total energy of a flowing fluid is equal to its enthalpy when the kinetic and potential energies are negligible, and that the rate of energy transfer by mass is equal to the product of the mass flow rate and the total energy of the fluid per unit mass, the rates of energy transfer by mass into and out of the compressor are

Emass, in = mqin = mh1 = (0.06 kg/s)(239.19 kJ/kg) = 14.35 kJ/s = 14.35 kW Emass, out = mqout = mh2 = (0.06 kg/s)(296.82 kJ/kg) = 17.81 kJ/s = 17.81kW Discussion The numerical values of the energy entering or leaving a device by mass alone does not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity here is the difference between the outgoing and incoming energy flow rates, which is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-435

D Emass = Emass, out - Emass, in = 17.81- 14.35 = 3.46 kW This quantity represents the rate of energy transfer to the refrigerant in the compressor.

EES (Engineering Equation Solver) SOLUTION "Given" P1=140 [kPa] x1=1 P2=800 [kPa] T2=60 [C] m_dot=0.06 [kg/s] "Properties" Fluid$='R134a' h1=enthalpy(Fluid$, P=P1, x=x1) h2=enthalpy(Fluid$, P=P2, T=T2) "Analysis" theta_in=h1 E_dot_mass_in=m_dot*theta_in theta_out=h2 E_dot_mass_out=m_dot*theta_out

5-26E Steam is leaving a pressure cooker at a specified pressure. The velocity, flow rate, the total and flow energies, and the rate of energy transfer by mass are to be determined. Assumptions 1 The flow is steady, and the initial start-up period is disregarded. 2 The kinetic and potential energies are negligible, and thus they are not considered. 3 Saturation conditions exist within the cooker at all times so that steam leaves the cooker as a saturated vapor at 20 psia. Properties The properties of saturated liquid water and water vapor at 20 psia are v f  0.01683 ft 3 / lbm, vg  20.093 ft 3 / lbm, u g  1081.8 Btu / lbm, and hg  1156.2 Btu / lbm (Table A-5E). Analysis (a) Saturation conditions exist in a pressure cooker at all times after the steady operating conditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steam has the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated, the mass flow rate of the exiting steam, and the exit velocity are

D Vliquid vf

æ0.13368 ft 3 ö÷ 0.6 gal çç ÷ ÷= 4.766 lbm 0.01683 ft 3 /lbm çè 1 gal ÷ ø

m 4.766 lbm = = 0.1059 lbm/min = 1.765 ×10-3 lbm / s Dt 45 min

2ö mv g (1.765´ 10-3 lbm/s)(20.093 ft 3 /lbm) æ m çç144 in ÷ ÷= 34.1 ft / s = = çè 1 ft 2 ø÷ ÷ r g Ac Ac 0.15 in 2

1-436

(b) Noting that h = u + Pv and that the kinetic and potential energies are disregarded, the flow and total energies of the exiting steam are

eflow = Pv = h - u = 1156.2 - 1081.8 = 74.4 Btu / lbm q = h + ke + pe @ h = 1156.2 Btu / lbm Note that the kinetic energy in this case is ke  V 2 / 2  (34.1 ft / s) 2 / 2  581 ft 2 / s 2  0.0232 Btu/lbm , which is very small compared to enthalpy. (c) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate and the total energy of the exiting steam per unit mass,

Emass = mq = (1.765´ 10- 3 lbm/s)(1156.2 Btu/lbm) = 2.04 Btu / s Discussion The numerical value of the energy leaving the cooker with steam alone does not mean much since this value depends on the reference point selected for enthalpy (it could even be negative). The significant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside (which is h fg ) since it relates directly to the amount of energy supplied to the cooker.

EES (Engineering Equation Solver) SOLUTION "Given" P=20 [psia] DELTAV_liquid=0.6 [gal] DELTAt=45 [min] A_c=0.15 [in^2] "Properties" Fluid$='steam_iapws' v_f=volume(Fluid$, P=P, x=0) v_g=volume(Fluid$, P=P, x=1) u_g=intenergy(Fluid$, P=P, x=1) h_g=enthalpy(Fluid$, P=P, x=1) "Analysis" "(a)" m=DELTAV_liquid/v_f*Convert(gal, ft^3) m_dot=m/DELTAt*Convert(s, min) Vel=(m_dot*v_g)/A_c*Convert(ft^2, in^2) "(b)" e_flow=h_g-u_g theta=h_g "(c)" E_dot_mass=m_dot*theta

Steady Flow Energy Balance: Nozzles and Diffusers 5-27C A steady-flow system involves no changes with time anywhere within the system or at the system boundaries

5-28C No.

5-29C It is mostly converted to internal energy as shown by a rise in the fluid temperature.

1-437

5-30C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a decrease in the fluid temperature.

5-31E Air is accelerated in an adiabatic nozzle. The velocity at the exit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 The nozzle is adiabatic. Properties The specific heat of air at the average temperature of (700  645) / 2  672.5 F is c p  0.253 Btu / lbm  R (Table A-2Eb).

Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 /2) = m(h2 +V22 /2) h1 + V12 /2 = h2 +V22 /2

Solving for exit velocity, 0.5

0.5

é 2 ù V2 = éêëV12 + 2(h1 - h2 )ù ú û = êëV1 + 2c p (T1 - T2 )ú û

0.5

é æ25,037 ft 2 /s 2 ÷ öù ú ÷ = êê(80 ft/s) 2 + 2(0.253 Btu/lbm ×R)(700 - 645)R çç ÷ çè 1 Btu/lbm ÷ øúúû êë

= 838.6 ft / s

EES (Engineering Equation Solver) SOLUTION "Given" P1=300 [psia] T1=700 [F] Vel1=80 [ft/s] P2=250 [psia] T2=645 [F] "Properties" c_p=0.253 [Btu/lbm-R] "at (700+645)/2= 672.5 F, Table A-2Eb" "Analysis" 0=c_p*(T2-T1)+(Vel2^2-Vel1^2)/2*Convert(ft^2/s^2, Btu/lbm) "energy balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-438

5-32 Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 The diffuser is adiabatic. Properties The specific heat of air at the average temperature of (30  90) / 2  60 C  333 K is c p  1.007 kJ / kg  K (Table A2b). Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 /2) = m(h2 +V22 /2)

h1 + V12 /2 = h2 +V22 /2 Solving for exit velocity, 0.5

0.5

é 2 ù V2 = éëêV12 + 2(h1 - h2 )ù ú = êëV1 + 2c p (T1 - T2 )û ú û

0.5

é æ1000 m 2 /s 2 ÷ öù ú ÷ = êê(350 m/s)2 + 2(1.007 kJ/kg ×K)(30 - 90)K ççç ÷ è 1 kJ/kg ø÷úûú ëê

= 40.7 m / s

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=30 [C] Vel1=350 [m/s] P2=200 [kPa] T2=90 [C] "Properties" c_p=1.007 [kJ/kg-K] "at (20+90)/2= 55 C = 328 K, Table A-2b" "Analysis" 0=c_p*(T2-T1)+(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) "energy balance"

5-33E Air is accelerated in a nozzle from 150 ft / s to 900 ft / s . The exit temperature of air and the exit area of the nozzle are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-439

1 2 3 4

This is a steady-flow process since there is no change with time. Air is an ideal gas with variable specific heats. Potential energy changes are negligible. There are no work interactions.

Properties The enthalpy of air at the inlet is h1  143.47 Btu / lbm (Table A-17E). Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V /2) = Qout + m(h2 +V22 /2) (since W @D pe @ 0) æ V 2 - V12 ö ÷ ÷ - Qout = m çççh2 - h1 + 2 ÷ ÷ çè 2 ø

h2 = - qout + h1 -

V22 - V12 2

= - 6.5 Btu/lbm + 143.47 Btu/lbm -

ö (900 ft/s)2 - (150 ft/s) 2 æ çç 1 Btu/lbm ÷ ÷ 2 2÷ ç 2 è 25,037 ft /s ø

= 121.2 Btu/lbm Thus, from Table A-17E, T2 = 507 R (b) The exit area is determined from the conservation of mass relation,

æRT / P ÷ öV1 v V 1 1 A2V2 = AV ® A2 = 2 1 A1 = ççç 2 2 ÷ A1 1 1 ¾ ¾ ÷ çè RT1 / P1 ÷ v2 v1 v1 V2 øV2 A2 =

(508/14.7)(150 ft/s) (0.1 ft 2 ) = 0.0480 ft 2 (600/50)(900 ft/s)

EES (Engineering Equation Solver) SOLUTION "Given" P1=50 [psia] T1=140[F]+460 [R] Vel1=150 [ft/s] A1=0.1 [ft^2] P2=14.7 [psia] Vel2=900 [ft/s] q_out=6.5 [Btu/lbm] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-440

Fluid$='Air' C_p=CP(Fluid$,T=540) C_v=CV(Fluid$,T=540) R=(C_p-C_v)*Convert(Btu/lbm-R, psia-ft^3/lbm-R) h1=enthalpy(Fluid$, T=T1) "Analysis" "(a)" h1+Vel1^2/2*Convert(ft^2/s^2, Btu/lbm)=q_out+h2+Vel2^2/2*Convert(ft^2/s^2, Btu/lbm) "energy balance" T2=temperature(Fluid$, h=h2) "(b)" v1=(R*T1)/P1 v2=(R*T2)/P2 1/v1*A1*Vel1=1/v2*A2*Vel2 "mass balance"

5-34 Air is accelerated in a nozzle from 120 m / s to 380 m / s . The exit temperature and pressure of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The enthalpy of air at the inlet temperature of 500 K is h1  503.02 kJ / kg (Table A-17). Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 / 2) = m(h2 +V22 /2) (since Q @ W @D pe @ 0) 0 = h2 - h1 +

V22 - V12 2

or, 2

ö (380 m/s) - (120 m/s) æ V 2 - V12 çç 1 kJ/kg ÷ h2 = h1 - 2 = 503.02 kJ/kg = 438.02 kJ/kg 2 2÷ ÷ ç è ø 2 2 1000 m /s

Then from Table A-17 we read T2 = 436.5 K @437 K (b) The exit pressure is determined from the conservation of mass relation,

1-441

1 1 1 1 A2V2 = AV ® A2V2 = AV 1 1 ¾ ¾ 1 1 v2 v1 RT2 / P2 RT1 / P1

Thus, P2 =

A1T2V1 2 (436.5 K)(120 m/s) P1 = (600 kPa) = 331 kPa A2TV 1 (500 K)(380 m/s) 1 2

EES (Engineering Equation Solver) SOLUTION "Given" P1=600 [kPa] T1=500 [K] Vel1=120 [m/s] Vel2=380 [m/s] A1=2*A2 "Properties" Fluid$='Air' C_p=CP(Fluid$,T=298) C_v=CV(Fluid$,T=298) R=C_p-C_v h1=enthalpy(Fluid$, T=T1) "Analysis" "(a)" h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg) "energy balance" T2=temperature(Fluid$, h=h2) "(b)" v1=(R*T1)/P1 v2=(R*T2)/P2 A1=1 "assumed, it does not matter what value is assumed" 1/v1*A1*Vel1=1/v2*A2*Vel2 "mass balance"

5-35 CO2 gas is accelerated in a nozzle to 450 m / s . The inlet velocity and the exit temperature are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2

CO2 is an ideal gas with variable specific heats.

3 4 5

Potential energy changes are negligible. The device is adiabatic and thus heat transfer is negligible. There are no work interactions.

Properties The gas constant and molar mass of CO2 are 0.1889 kPa  m3 / kg  K and 44 kg / kmol (Table A-1). The enthalpy of CO2 at 500C is h1  30, 797 kJ / kmol (Table A-20). Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. Using the ideal gas relation, the specific volume is determined to be

1-442

v1 =

RT1 (0.1889 kPa ×m /kg ×K )(773 K ) = = 0.146 m3 /kg P1 1000 kPa

mv1 (6000/3600 kg/s)(0.146 m /kg ) 1 AV ® V1 = = = 60.8 m / s 1 1 ¾ ¾ v1 A1 40´ 10- 4 m 2

Thus, 3

(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m(h1 + V12 / 2) = m(h2 +V22 /2) (since Q @ W @D pe @ 0)

0 = h2 - h1 +

V22 - V12 2

Substituting,

h2 = h1 -

V22 - V12 M 2 2

= 30,797 kJ/kmol -

ö (450 m/s) - (60.8 m/s) æ çç 1 kJ/kg ÷ 2 2÷ ÷(44 kg/kmol) ç è1000 m /s ø

= 26,423 kJ/kmol

Then the exit temperature of CO2 from Table A-20 is obtained to be T2 = 685.8 K @686 K Alternative Solution Using constant specific heats for CO2 , at 700 K, c p  1.126 kJ / kg  K (Table A-2b)

V22 - V12 2 V 2 - V12 0 = c p (T2 - T1 ) + 2 2

0 = h2 - h1 +

(1.126 kJ/kgK)(T2 - 773)K = -

ö (450 m/s) - (60.8 m/s) æ ç 1 kJ/kg ÷ 2

÷ çç è1000 m 2 /s 2 ÷ ø

T2 = 685 K

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [kPa] T1=(500+273) [K] A1=40E-4 [m^2] P2=100 [kPa] Vel2=450 [m/s] m_dot=6000[kg/h]*Convert(kg/h, kg/s) "Properties" Fluid$='CO2' R=0.1889 [kJ/kg-K] "Table A-2a" h1=enthalpy(Fluid$, T=T1) "Analysis" "(a)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-443

v1=(R*T1)/P1 Vel1=(m_dot*v1)/A1 "(b)" h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg) "energy balance" T2=temperature(Fluid$, h=h2)

5-36 Heat is lost from the steam flowing in a nozzle. The velocity and the volume flow rate at the nozzle exit are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy change is negligible. 3 There are no work interactions. Analysis We take the steam as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Energy balance:

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2ö æ æ V2ö ççh + V2 ÷ ÷ ÷ m çççh1 + 1 ÷ = m 2 ÷ ÷+ Qout 2 ø÷ 2 ø÷ èç èçç

h1 +

since W @D pe @ 0)

V12 V2 Q = h2 + 2 + out 2 2 m

The properties of steam at the inlet and exit are (Table A-6) 3 P1 = 800 kPa ïü ïý v1 = 0.38429 m /kg T1 = 400°C ïïþ h1 = 3267.7 kJ/kg

3 P2 = 400 kPa ü ïïý v 2 = 0.74334 m /kg T1 = 375°C ïïþ h2 = 3221.8 kJ/kg

The mass flow rate of the steam is m=

1 1 AV (0.08 m 2 )(10 m/s) = 2.082 kg/s 1 1 = v1 0.38429 m3 /s

Substituting,

3267.7 kJ/kg +

ö V 2 æ 1 kJ/kg ö÷ (10 m/s)2 æ 25 kJ/s çç 1 kJ/kg ÷ = 3221.8 kJ/kg + 2 çç ÷ ÷+ 2 2 ÷ èç1000 m /s ø 2 2 èç1000 m 2 /s 2 ø÷ 2.082 kg/s V2 = 260 m / s

The volume flow rate at the exit of the nozzle is

1-444

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_1=400 [C] P_1=800 [kPa] Vel_1=10 [m/s] A_1=800E-4 [m^2] T_2=375 [C] P_2=400 [kPa] Q_dot_out=25 [kW] "ANALYSIS" Fluid$='Steam_iapws' v_1=volume(Fluid$, T=T_1, P=P_1) h_1=enthalpy(Fluid$, T=T_1, P=P_1) v_2=volume(Fluid$, T=T_2, h=h_2) h_2=enthalpy(Fluid$, T=T_2, P=P_2) m_dot*(h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg))=m_dot*(h_2+Vel_2^2/2*Convert(m^2/s^2, kJ/kg))+Q_dot_out "energy balance on the nozzle" m_dot=1/v_1*A_1*Vel_1 "mass flow rate of steam" Vol_dot_2=m_dot*v_2 "Volume flow rate at the exit"

5-37 Air is decelerated in a diffuser from 230 m / s to 30 m / s . The exit temperature of air and the exit area of the diffuser are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The enthalpy of air at the inlet temperature of 400 K is h1  400.98 kJ/Kg (Table A-17). Analysis

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 / 2) = m(h2 +V22 /2) (since Q @ W @D pe @ 0) , 0 = h2 - h1 +

V22 - V12 2

1-445

or, 2

h2 = h1 -

ö (30 m/s) - (230 m/s) æ V22 - V12 ÷= 426.98 kJ/kg çç 1 kJ/kg ÷ = 400.98 kJ/kg 2 2 ÷ çè1000 m /s ø 2 2

From Table A-17,

T2 = 425.6 K (b) The specific volume of air at the diffuser exit is

RT2 (0.287 kPa ×m /kg ×K )(425.6 K ) = = 1.221 m3 /kg P2 (100 kPa ) 3

v2 =

From conservation of mass,

1 A2V2 v2

¾¾ ®

A2 =

mv 2 (6000 3600 kg/s)(1.221 m3 /kg) = = 0.0678 m2 V2 30 m/s

EES (Engineering Equation Solver) SOLUTION "Given" P1=80 [kPa] T1=(127+273) [K] Vel1=230 [m/s] P2=100 [kPa] Vel2=30 [m/s] m_dot=6000[kg/h]*Convert(kg/h, kg/s) "Properties" Fluid$='Air' R=0.287 [kJ/kg-K] "Table A-2a" h1=enthalpy(Fluid$, T=T1) "Analysis" "(a)" h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg) "energy balance" T2=temperature(Fluid$, h=h2) "(b)" v2=(R*T2)/P2 m_dot=1/v2*A2*Vel2

5-38E Air is decelerated in a diffuser from 750 ft / s to a low velocity. The exit temperature and the exit velocity of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions. Properties The enthalpy of air at the inlet temperature of 65F (or 520 R) is h1  125.40 Btu / lbm (Table A-17E). Analysis

1-446

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 /2) = m(h2 +V22 /2) (since Q @W @D pe @ 0) ,

0 = h2 - h1 +

V22 - V12 2

or, 2

h2 = h1 -

ö 0 - (750 ft/s ) æ V22 - V12 çç 1 Btu/lbm ÷ = 125.40 Btu/lbm ÷= 136.63 Btu/lbm 2 2÷ ç 2 2 è 25,037 ft /s ø

From Table A-17E, T2 = 571.6 R = 112 F

(b) The exit velocity of air is determined from the conservation of mass relation, 1 1 1 1 A2V2 = AV ® A2V2 = AV 1 1 ¾ ¾ 1 1 v2 v1 RT2 / P2 RT1 / P1

Thus, V2 =

A1T2 P1 1 (571.6 R)(13 psia) V1 = (750 ft/s) = 244 ft / s A2T1 P2 3 (525 R)(14.5 psia)

EES (Engineering Equation Solver) SOLUTION "Given" P1=13 [psia] T1=(65+460) [R] Vel1=750 [ft/s] P2=14.5 [psia] Vel2=0 "assumed negligible in energy balance equation" A2=3*A1 "Properties" Fluid$='Air' R=0.3704 [psia-ft^3/lbm-R] "Table A-2a" h1=enthalpy(Fluid$, T=T1) "Analysis" "(a)" h1+Vel1^2/2*Convert(ft^2/s^2, Btu/lbm)=h2+Vel2^2/2*Convert(ft^2/s^2, Btu/lbm) "energy balance" T2=temperature(Fluid$, h=h2) "(b)" v1=(R*T1)/P1 v2=(R*T2)/P2 A1=1 "assumed, it does not matter what value is assumed" 1/v1*A1*Vel1=1/v2*A2*Vel2_actual "mass balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-447

5-39 R-134a is accelerated in a nozzle from a velocity of 20 m / s . The exit velocity of the refrigerant and the ratio of the inlet-to-exit area of the nozzle are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

Properties From the refrigerant tables (Table A-13) 3 P1 = 700 kPa ü ïïý v1 = 0.043358 m /kg T1 = 120°C ïïþ h1 = 358.92 kJ/kg

and P2 = 400 kPa ü ïï v 2 = 0.056796 m3 /kg ý ïïþ h2 = 275.09 kJ/kg T2 = 30°C

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 / 2) = m(h2 +V22 /2) (since Q @ W @D pe @ 0)

0 = h2 - h1 +

V22 - V12 2

Substituting, 2

V 2 - (20 m/s) æ 1 kJ/kg ÷ ö çç 0 = (275.09 - 358.92)kJ/kg + 2 2 2÷ ÷ ç è1000 m /s ø 2

It yields

V2  409.9 m / s (b) The ratio of the inlet to exit area is determined from the conservation of mass relation,

(0.043358 m /kg )(409.9 m/s) A v V 1 1 A2V2 = AV ® 1 = 1 2 = = 15.65 1 1 ¾ ¾ v2 v1 A2 v 2 V1 (0.056796 m3 /kg )(20 m/s) 3

EES (Engineering Equation Solver) SOLUTION "Given" P1=700 [kPa] T1=120 [C] Vel1=20 [m/s] P2=400 [kPa] T2=30 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-448

"Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) v2=volume(Fluid$, P=P2, T=T2) h2=enthalpy(Fluid$, P=P2, T=T2) "Analysis" "(a)" h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg) "energy balance" "(b)" RatioA=(v1/v2)*(Vel2/Vel1) "RatioA=A1/A2, from mass balance 1/v1*A1*Vel1=1/v2*A2*Vel2"

5-40 R-134a is decelerated in a diffuser from a velocity of 160 m / s . The exit velocity of R-134a and the mass flow rate of the R-134a are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 There are no work interactions.

Properties From the R-134a tables (Tables A-11 through A-13) 3 P1 = 600 kPa ïü ïý v1 = 0.034335 m /kg ïïþ h1 = 262.46 kJ/kg sat. vapor

and 3 P2 = 700 kPa ïü ïý v 2 = 0.031696 m /kg ïïþ h2 = 278.59 kJ/kg T2 = 40°C

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. Then the exit velocity of R-134a is determined from the steady-flow mass balance to be

v A 1 1 1 (0.031696 m3 /kg) A2V2 = AV ® V2 = 2 1 V1 = (160 m/s) = 82.06 m/s @82.1 m / s 1 1 ¾¾ v2 v1 v1 A2 1.8 (0.034335 m3 /kg) (b) We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Qin + m(h1 + V12 /2) = m(h2 +V22 /2) (since W @D pe @ 0)

æ V 2 - V12 ö÷ ÷ Qin = m çççh2 - h1 + 2 ÷ çè 2 ø÷ Substituting, the mass flow rate of the refrigerant is determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-449 2 é öù (82.06 m/s) - (160 m/s) 2 æ ú çç 1 kJ/kg ÷ 2 kJ/s = m êê(278.59 - 262.46)kJ/kg + ÷ çè1000 m 2 /s 2 ÷ øú 2 êë ú û

It yields

m = 0.298 kg / s

EES (Engineering Equation Solver) SOLUTION "Given" P1=600 [kPa] x1=1 Vel1=160 [m/s] P2=700 [kPa] T2=40 [C] Q_dot_in=2 [kJ/s] A2=1.8*A1 "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) h1=enthalpy(Fluid$, P=P1, x=x1) v2=volume(Fluid$, P=P2, T=T2) h2=enthalpy(Fluid$, P=P2, T=T2) "Analysis" "(a)" A1=1 "assumed, it does not matter what value is assumed" 1/v1*A1*Vel1=1/v2*A2*Vel2 "mass balance" "(b)" Q_dot_in+m_dot*(h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg))=m_dot*(h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)) "energy balance"

5-41 Air is decelerated in a diffuser from 220 m / s . The exit velocity and the exit pressure of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with variable specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. Properties The gas constant of air is 0.287 kPa.m3 / kg  K (Table A-1). The enthalpies are (Table A-17)

T1 = 27°C = 300 K ® h1 = 300.19 kJ/kg T2 = 42°C = 315 K ® h2 = 315.27 kJ/kg Analysis

(a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-450

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m(h1 + V12 /2) = Qout + m(h2 +V22 /2) (since W @D pe @ 0)

æ V 2 - V12 ö÷ ÷ - Qout = m çççh2 - h1 + 2 ÷ çè 2 ÷ ø Substituting, the exit velocity of the air is determined to be æ öö V 2 - (220 m/s) 2 æ ÷ çç 1 kJ/kg ÷ ÷ - 18 kJ/s = (2.5 kg/s)ççç(315.27 - 300.19) kJ/kg + 2 ÷ 2 2÷ ÷ ç ÷ è1000 m /s øø 2 èç

It yields

V2  62.0 m / s (b) The exit pressure of air is determined from the conservation of mass and the ideal gas relations,

(0.04 m )(62.0 m/s) AV 1 A2V2 ¾ ¾ ® v2 = 2 2 = = 0.992 m3 /kg v2 m 2.5 kg/s 2

and RT2 (0.287 kPa ×m /kg ×K )(315 K ) = = 91.1 kPa v2 0.992 m 3 /kg 3

P2v 2 = RT2 ¾ ¾ ® P2 =

EES (Engineering Equation Solver) SOLUTION "Given" P1=80 [kPa] T1=27[C]+273 [K] Vel1=220 [m/s] T2=42[C]+273 [K] A2=400E-4 [m^2] m_dot=2.5 [kg/s] Q_dot_out=18 [kJ/s] "Properties" Fluid$='Air' C_p=CP(Fluid$,T=298) C_v=CV(Fluid$,T=298) R=C_p-C_v h1=enthalpy(Fluid$, T=T1) h2=enthalpy(Fluid$, T=T2) "Analysis" "(a)" m_dot*(h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg))=Q_dot_out+m_dot*(h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)) "energy balance" "(b)" v2=(A2*Vel2)/m_dot P2=(R*T2)/v2

5-42 Air is accelerated in a nozzle from 45 m / s to 180 m / s . The mass flow rate, the exit temperature, and the exit area of the nozzle are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-451

1 2 3 4 5

This is a steady-flow process since there is no change with time. Air is an ideal gas with constant specific heats. Potential energy changes are negligible. The device is adiabatic and thus heat transfer is negligible. There are no work interactions.

Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The specific heat of air at the anticipated average temperature of 450 K is c p  1.02 kJ / kg  C (Table A-2).

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . Using the ideal gas relation, the specific volume and the mass flow rate of air are determined to be

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(473 K) = = 0.4525 m3 /kg P1 300 kPa

1 1 AV (0.0110 m 2 )(45 m/s) = 1.094 kg / s 1 1 = v1 0.4525 m3 /kg

(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m(h1 + V12 / 2) = m(h2 +V22 /2) (since Q @ W @D pe @ 0)

0 = h2 - h1 +

V22 - V12 2

¾¾ ®

0 = c p , ave (T2 - T1 )+

V22 - V12 2

Substituting, 0 = (1.02 kJ/kg ×K)(T2 - 200°C) +

ö (180 m/s) 2 - (45 m/s) 2 æ çç 1 kJ/kg ÷ 2 2÷ ç è ø 2 1000 m /s ÷

It yields T2 = 185.2 C

 0.287 kPa  m / kg  K  (185.2  273 K)  1.315m3 / kg RT v2  2  P2 100kPa 3

m

1 A2V2 v2

 1.094 kg / s 

1 A2 (180 m / s)  A2  0.00799 m 2  79.9 cm 2 1.315 m3 / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=300 [kPa] T1=(200+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-452

Vel1=45 [m/s] A1=110E-4 [m^2] P2=100 [kPa] Vel2=180 [m/s] "Properties" Fluid$='Air' R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.02 [kJ/kg-K] "at 450 K, Table A-2b" "Analysis" "(a)" v1=(R*T1)/P1 m_dot=1/v1*A1*Vel1 "(b)" h1=enthalpy(Fluid$, T=T1) 0=c_p*(T2-T1)+(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) "energy balance" T2_C=ConvertTemp(K, C, T2) "(c)" v2=(R*T2)/P2 m_dot=1/v2*A2*Vel2 A2_cm=A2*Convert(m^2, cm^2)

5-43 Problem 5-42 is reconsidered. The effect of the inlet area on the mass flow rate, exit velocity, and the exit area as the inlet area varies from 50 cm 2 to 150 cm2 is to be investigated, and the final results are to be plotted against the inlet area. Analysis The problem is solved using EES, and the solution is given below. Function HCal(WorkFluid$, Tx, Px) "Function to calculate the enthalpy of an ideal gas or real gas" If 'Air' = WorkFluid$ then HCal:=ENTHALPY(Air,T=Tx) "for idal gas" else HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px) "for real gas" endif end HCal WorkFluid$ = 'Air' T[1] = 200 [C] P[1] = 300 [kPa] Vel[1] = 45 [m/s] P[2] = 100 [kPa] Vel[2] = 180 [m/s] A[1]=110 [cm^2] Am[1]=A[1]*convert(cm^2,m^2) h[1]=HCal(WorkFluid$,T[1],P[1]) h[2]=HCal(WorkFluid$,T[2],P[2]) v[1]=volume(workFluid$,T=T[1],p=P[1]) "The Volume function has the same form for an ideal gas as for a real fluid." v[2]=volume(WorkFluid$,T=T[2],p=P[2]) m_dot[1]= m_dot[2] "Conservation of mass" m_dot[1]=Am[1]*Vel[1]/v[1] m_dot[2]= Am[2]*Vel[2]/v[2] h[1]+Vel[1]^2/(2*1000) = h[2]+Vel[2]^2/(2*1000) "Energy balance" A_ratio=A[1]/A[2] A[2]=Am[2]*convert(m^2,cm^2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-453

écm 2 ù ëê ûú 50 60 70 80 90 100 110 120 130 140 150

écm 2 ù ëê ûú 36.32 43.59 50.85 58.12 65.38 72.65 79.91 87.18 94.44 101.7 109

[kg/s ]

[C ]

0.497 0.5964 0.6958 0.7952 0.8946 0.9941 1.093 1.193 1.292 1.392 1.491

185.2 185.2 185.2 185.2 185.2 185.2 185.2 185.2 185.2 185.2 185.2

110

1.6

100

1.4

1.2

80 70

0.8

m[1] [kg/s]

A[2] [cm ]

50 0.6

40 30 50

110

130

0.4 150

A[1] [cm ]

Turbines and Compressors 5-44C Yes.

5-45C Yes. Because energy (in the form of shaft work) is being added to the air.

5-46C No.

5-47 Air is expanded in a turbine. The mass flow rate and outlet area are to be determined. Assumptions 1 Air is an ideal gas. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-454

The flow is steady.

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1).

Analysis The specific volumes of air at the inlet and outlet are

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(600 + 273 K) = = 0.2506 m3 /kg P1 1000 kPa

v2 =

RT2 (0.287 kPa ×m3 /kg ×K)(200 + 273 K) = = 1.3575 m3 /kg P2 100 kPa

The mass flow rate is

AV (0.1 m2 )(30 m/s) 1 1 = = 11.97 kg / s v1 0.2506 m3 /kg

The outlet area is

A2 =

mv 2 (11.97 kg/s)(1.3575 m3 /kg) = = 1.625 m 2 V2 10 m/s

EES (Engineering Equation Solver) SOLUTION P1=1000 [kPa] T1=(600+273) [K] P2=100 [kPa] T2=(200+273) [K] A1=0.1 [m^2] Vel1=30 [m/s] Vel2=10 [m/s] R=0.287 [kJ/kg-K] v1=(R*T1)/P1 v2=(R*T2)/P2 m_dot=1/v1*A1*Vel1 A2=m_dot*v2/Vel2

5-48E Air is expanded in a gas turbine. The inlet and outlet mass flow rates are to be determined.

1-455

Assumptions 1 Air is an ideal gas. 2 The flow is steady. Properties The gas constant of air is R  0.3704 psia  ft 3 / lbm  R (Table A-1E). Analysis The specific volumes of air at the inlet and outlet are

v1 =

RT1 (0.3704 psia ×ft 3 /lbm ×R)(700 + 460 R) = = 2.864 ft 3 /lbm P1 150 psia

v2 =

RT2 (0.3704 psia ×ft 3 /lbm ×R)(100 + 460 R) = = 13.83 ft 3 /lbm P2 15 psia

The volume flow rates at the inlet and exit are then

V1 = mv1 = (5 lbm/s)(2.864 ft 3 /lbm) = 14.32 ft 3 / s V2 = mv 2 = (5 lbm/s)(13.83 ft 3 /lbm) = 69.15 ft 3 / s EES (Engineering Equation Solver) SOLUTION P1=150 [psia] T1=(700+460) [R] P2=15 [psia] T2=(100+460) [R] m_dot=5 [lbm/s] R=0.3704 [psia-ft^3/lbm-R] v1=(R*T1)/P1 v2=(R*T2)/P2 Vol_dot_1=m_dot*v1 Vol_dot_2=m_dot*v2

5-49 R-134a at a given state is compressed to a specified state. The mass flow rate and the power input are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer with the surroundings is negligible.

1-456

Analysis We take the compressor as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the compressor, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Win + mh1 = mh2

(since D ke @D pe @ 0)

Win = m(h2 - h2 )

From R134a tables (Tables A-11, A-12, A-13) P1 = 100 kPa ïü ïý h1 = 236.34 kJ/kg T1 = - 24°C ïïþ v1 = 0.1947 m3 /kg

P2 = 800 kPa ü ïï ý h2 = 296.82 kJ/kg T2 = 60°C ïïþ

The mass flow rate is

V1 (1.35 / 60) m3 /s = = 0.1155 kg / s v1 0.1947 m3 /kg

Substituting,

Win = m(h2 - h1 ) = (0.1155 kg/s)(296.82 - 236.34)kJ/kg = 6.99 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=100 [kPa] T1=-24 [C] V_dot_1=(1.35/60) [m^3/s] P2=800 [kPa] T2=60 [C] "ANALYSIS" Fluid$='R134a' h1=enthalpy(Fluid$, P=P1, T=T1) v1=volume(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, T=T2) m_dot=V_dot_1/v1 W_dot_in=m_dot*(h2-h1)

5-50 Saturated R-134a vapor is compressed to a specified state. The power input is given. The exit temperature is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-457

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer with the surroundings is negligible.

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Win + mh1 = mh2

(since D ke @D pe @ 0)

Win = m(h2 - h2 )

From R134a tables (Table A-12) P1 = 180 kPa ïü ïý h1 = 242.86 kJ/kg ïïþ v1 = 0.1104 m3 /kg x1 = 0

The mass flow rate is

V1 (0.35 / 60) m3 /s = = 0.05283 kg/s v1 0.1104 m3 /kg

Substituting for the exit enthalpy, Win = m(h2 - h1) 2.35 kW = (0.05283 kg/s)(h2 - 242.86)kJ/kg ¾ ¾ ® h2 = 287.34 kJ/kg

From Table A-13, ü ïï ý T2 = 52.5°C h2 = 287.34 kJ/kg ïïþ P2 = 900 kPa

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=180 [kPa] x1=1 V_dot_1=(0.35/60) [m^3/s] P2=900 [kPa] W_dot_actual=2.35 "ANALYSIS" h1=enthalpy(R134a, P=P1, x=x1) v1=volume(R134a, P=P1, x=x1) m_dot=V_dot_1/v1 W_dot_actual=m_dot*(h2-h1) T2=temperature(R134a, P=P2, h=h2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-458

5-51 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A-4 through 6) 3 P1 = 4 MPa ïü ïý v1 = 0.086442 m /kg T1 = 500°C ïïþ h1 = 3446.0 kJ/kg

and P2 = 30 kPa ïü ïý h = h + x h = 289.27 + 0.92´ 2335.3 = 2437.7 kJ/kg 2 f 2 fg x2 = 0.92 ïïþ

Analysis

(a) The change in kinetic energy is determined from 2

2 ö V 2 - V12 (50 m/s) - (80 m/s) æ çç 1 kJ/kg ÷ D ke = 2 = = - 1.95 kJ / kg ÷ çè1000 m 2 /s 2 ÷ ø 2 2

(b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m(h1 + V12 / 2) = Wout + m(h2 +V22 /2) (since Q @D pe @ 0)

æ V 2 - V12 ö÷ ÷ Wout = - m çççh2 - h1 + 2 ÷ çè 2 ÷ ø Then the power output of the turbine is determined by substitution to be

Wout = - (12 kg/s)(2437.7 - 3446.0 - 1.95) kJ/kg = 12,123 kW = 12.1 MW (c) The inlet area of the turbine is determined from the mass flow rate relation,

mv1 (12 kg/s)(0.086442 m3 /kg) 1 AV ¾ ¾ ® A = = = 0.0130 m2 1 1 1 v1 V1 80 m/s

1-459

EES (Engineering Equation Solver) SOLUTION "Given" P1=4000 [kPa] T1=500 [C] Vel1=80 [m/s] P2=30 [kPa] x2=0.92 Vel2=50 [m/s] m_dot=12 [kg/s] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, x=x2) h2_f=enthalpy(Fluid$, P=P2, x=0) h2_g=enthalpy(Fluid$, P=P2, x=1) h2_fg=h2_g-h2_f "Analysis" "(a)" DELTAKE=(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) "(b)" m_dot*(h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg))=W_dot_out+m_dot*(h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)) "energy balance" "(c)" A1=(m_dot*v1)/Vel1

5-52 Problem 5-51 is reconsidered. The effect of the turbine exit pressure on the power output of the turbine as the exit pressure varies from 10 kPa to 200 kPa is to be investigated. The power output is to be plotted against the exit pressure. Analysis The problem is solved using EES, and the solution is given below.

T[1] = 500 [C] P[1] = 4000 [kPa] Vel[1] = 80 [m/s] P[2] = 30 [kPa] X_2=0.92 Vel[2] = 50 [m/s] m_dot[1]=12 [kg/s] Fluid$='Steam_IAPWS' h[1]=enthalpy(Fluid$,T=T[1],P=P[1]) h[2]=enthalpy(Fluid$,P=P[2],x=x_2) T[2]=temperature(Fluid$,P=P[2],x=x_2) v[1]=volume(Fluid$,T=T[1],P=P[1]) v[2]=volume(Fluid$,P=P[2],x=x_2) m_dot[1]= m_dot[2] m_dot[1]=A[1]*Vel[1]/v[1] m_dot[2]= A[2]*Vel[2]/v[2] m_dot[1]*(h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) = m_dot[2]*(h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg))+W_dot_turb*convert(MW,kJ/s) DELTAke=Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)-Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)

1-460

Wturb

[kPa ]

[MW ]

[C ]

10 31.11 52.22 73.33 94.44 115.6 136.7 157.8 178.9 200

12.67 12.1 11.82 11.63 11.48 11.36 11.25 11.16 11.08 11.01

45.81 69.93 82.4 91.16 98.02 103.7 108.6 112.9 116.7 120.2

5-53E Steam expands in a turbine. The rate of heat loss from the steam for a power output of 4 MW is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-461

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Properties From the steam tables (Tables A-4E through 6E) P1 = 1000 psia ïü ïý h = 1448.6 Btu/lbm ïïþ 1

T1 = 900°F

P2 = 5 psia ü ïï ý h2 = 1130.7 Btu/lbm sat.vapor ïïþ Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + Wout + mh2

(since D ke @D pe @ 0)

Qout = - m(h2 - h1 ) - Wout

Substituting, æ 1 Btu ö ÷ Qout = - (45000/3600 lbm/s)(1130.7 - 1448.6) Btu/lbm - 4000 kJ/s çç ÷ ÷= 182.0 Btu / s çè1.055 kJ ø

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [psia] T1=900 [F] P2=5 [psia] x2=1 m_dot=45000[lbm/h]*Convert(1/h, 1/s) W_dot_out=4000 [kW] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, x=x2) "Analysis" m_dot*h1=Q_dot_out+W_dot_out*Convert(kW, Btu/s)+m_dot*h2 "energy balance"

5-54 Steam expands in a turbine. The exit temperature of the steam for a power output of 2 MW is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-462

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A-4 through 6) P1 = 8 MPa ü ïï ý h1 = 3399.5 kJ/kg T1 = 500°C ïïþ

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Wout + mh2

(since Q @D ke @D pe @ 0)

Wout = m(h1 - h2 )

Substituting, 2500 kJ/s = (3 kg/s)(3399.5 - h2 ) kJ/kg

h2 = 2566.2 kJ/kg Then the exit temperature becomes ïü ïý T = 60.1°C 2 h2 = 2566.2 kJ/kg ïïþ P2 = 20 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P1=8000 [kPa] T1=500 [C] P2=20 [kPa] m_dot=3 [kg/s] W_dot_out=2500 [kW] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) "Analysis" m_dot*h1=W_dot_out+m_dot*h2 "energy balance" T2=temperature(Fluid$, P=P2, h=h2)

1-463

5-55 Air is compressed at a rate of 10 L / s by a compressor. The work required per unit mass and the power required are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The constant pressure specific heat of air at the average temperature of (20  300) / 2  160 C  433 K is c p  1.018 kJ / kg  K (Table A-2b). The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1).

Analysis

(a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Win + mh1 = mh2 (since D ke @D pe @ 0)

Win = m(h2 - h1 ) = mc p (T2 - T1 )

Thus,

win = c p (T2 - T1 ) = (1.018 kJ/kg ×K)(300 - 20)K = 285.0 kJ / kg (b) The specific volume of air at the inlet and the mass flow rate are

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(20 + 273 K) = = 0.7008 m3 /kg P1 120 kPa

V1 0.010 m3 /s = = 0.01427 kg/s v1 0.7008 m3 /kg

Then the power input is determined from the energy balance equation to be

Win = mc p (T2 - T1 ) = (0.01427 kg/s)(1.018 kJ/kg ×K)(300 - 20)K = 4.068 kW EES (Engineering Equation Solver) SOLUTION "Given" Vol1_dot=0.010 [m^3/s] P1=120 [kPa] T1=(20+273) [K] P2=1000 [kPa] T2=(300+273) [K] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.018 [kJ/kg-K] "at (20+300)/2=160 C=433 K, Table A-2b" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-464

"Analysis" "(a)" w_in=c_p*(T2-T1) "(b)" v1=(R*T1)/P1 m_dot=Vol1_dot/v1 W_dot_in=m_dot*c_p*(T2-T1) "energy balance"

5-56 CO2 is compressed by a compressor. The volume flow rate of CO2 at the compressor inlet and the power input to the compressor are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with variable specific heats. 4 The device is adiabatic and thus heat transfer is negligible. Properties The gas constant of CO2 is R  0.1889 kPa  m3 / kg  K , and its molar mass is M  44 kg / kmol (Table A-1). The inlet and exit enthalpies of CO2 are (Table A-20)

T1 = 300 K ® h1 = 9, 431 kJ/kmol T2 = 450 K ® h2 = 15, 483 kJ/kmol

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. The inlet specific volume of air and its volume flow rate are 3 RT1 (0.1889 kPa ×m /kg ×K )(300 K ) v1 = = = 0.5667 m 3 /kg P1 100 kPa

V = mv = (0.5 kg/s)(0.5667 m3 /kg) = 0.283 m3 / s

1 ` (b) We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Win + mh1 = mh2 (since Q @D ke @D pe @ 0) Win = m(h2 - h1 ) = m(h2 - h1 ) / M

Substituting

1-465

Win =

(0.5 kg/s)(15,483 - 9,431 kJ/kmol) 44 kg/kmol

= 68.8 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=300 [K] P2=600 [kPa] T2=450 [K] m_dot=0.5 [kg/s] "Properties" Fluid$='CO2' R=0.1889 [kJ/kg-K] "Table A-2a" h1=enthalpy(Fluid$, T=T1) h2=enthalpy(Fluid$, T=T2) "Analysis" v1=(R*T1)/P1 Vol1_dot=m_dot*v1 W_dot_in+m_dot*h1=m_dot*h2 "energy balance"

5-57 Steam is expanded in a turbine. The power output is given. The rate of heat transfer is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Properties From the steam tables (Table A-4, A5, A-6)

P1 = 6 MPa ïüï ý h1 = 3658.8 kJ/kg T1 = 600°C ïïþ

P2 = 0.5 MPa üïï ý h2 = 2855.8 kJ/kg T2 = 200°C ïïþ Analysis We take the turbine as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the compressor, the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

1-466

æ æ V 2 ö÷ V 2 ö÷ m çççh1 + 1 ÷ = m çççh2 + 2 ÷ ÷ ÷+ Wout + Qout çè 2 ø÷ 2 ø÷ èç

(since D pe @ 0)

æ V 2 - V22 ö÷ ÷ Qout = - Wout + m çççh1 - h2 + 1 ÷ çè 2 ÷ ø Substituting,

æ V 2 - V22 ö÷ ÷ Qout = - Wout + m çççh1 - h2 + 1 ÷ çè 2 ÷ ø é öù (0 - 180 m/s) 2 æ çç 1 kJ/kg ÷ ú = 20,350 kW + (26 kg/s) ê(3658.8 - 2855.8)kJ/kg + 2 2÷ ç ê è1000 m /s ÷ øúû 2 ë

= 105 kW

EES (Engineering Equation Solver) SOLUTION "Given" m_dot=26 [kg/s] Vel1=0 [m/s] P1=6000 [kPa] T1=600 [C] P2=500 [kPa] T2=200 [C] Vel2=180 [m/s] W_dot_out=20350 [kW] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$,T=T1,P=P1) h2=enthalpy(Fluid$,T=T2,P=P2) Q_dot_out=-W_dot_out+m_dot*(h1-h2+(Vel1^2-Vel2^2)/2*Convert(m^2/s^2, kJ/kg))

5-58 Air is compressed in an adiabatic compressor. The mass flow rate of the air and the power input are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The compressor is adiabatic. 3 Air is an ideal gas with constant specific heats. Properties The constant pressure specific heat of air at the average temperature of (20  400) / 2  210 C  483 K is c p  1.026 kJ / kg  K (Table A-2b). The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1).

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

1-467

æ æ V 2 - V12 ö÷ V 2 - V12 ö÷ ÷ ÷ Win = m çççh2 - h1 + 2 = m çççc p (T2 - T1 ) + 2 ÷ ÷ çè 2 ø÷ 2 ø÷ èç The specific volume of air at the inlet and the mass flow rate are

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(20 + 273 K) = = 0.8409 m3 /kg P1 100 kPa

AV (0.15 m2 )(30 m/s) 1 1 = = 5.351 kg / s v1 0.8409 m3 /kg

Similarly at the outlet,

v2 =

RT2 (0.287 kPa ×m3 /kg ×K)(400 + 273 K) = = 0.1073 m3 /kg P2 1800 kPa

V2 =

mv 2 (5.351 kg/s)(0.1073 m3 /kg) = = 7.177 m/s A2 0.08 m2

(b) Substituting into the energy balance equation gives

æ V 2 - V12 ö÷ ÷ Win = m çççc p (T2 - T1 ) + 2 ÷ çè 2 ø÷ é öù (7.177 m/s) 2 - (30 m/s) 2 æ çç 1 kJ/kg ÷ ú = (5.351 kg/s) ê(1.026 kJ/kg ×K)(400 - 20)K + 2 2÷ ÷ ç ê ú è ø 2 1000 m /s ë û

= 2084 kW

EES (Engineering Equation Solver) SOLUTION P1=100 [kPa] T1=(20+273) [K] P2=1800 [kPa] T2=(400+273) [K] A1=0.15 [m^2] Vel1=30 [m/s] A2=0.08 [m^2] c_p=1.026 [kJ/kg-K] "at (20+400)/2=210 C = 483 K, Table A-2b" R=0.287 [kJ/kg-C] v1=R*T1/P1 m_dot=A1*Vel1/v1 v2=R*T2/P2 Vel2=m_dot*v2/A2 W_dot_in=m_dot*(c_p*(T2-T1)+(Vel2^2-Vel1^2)/2000)

1-468

5-59 Air is compressed by a compressor. The mass flow rate of air through the compressor is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

Properties The inlet and exit enthalpies of air are (Table A-17) T1  25 C  298K  h1  h @ 298 K  298.2kJ / kg T2  347 C  620K  h2  h @ 620 K  628.07kJ / kg

Analysis We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Win + m(h1 + V12 / 2) = Qout + m(h2 +V22 /2) (since D pe @ 0)

æ ö V 2 - V12 ÷ ÷ Win - Qout = m çççh2 - h1 + 2 ÷ çè 2 ÷ ø Substituting, the mass flow rate is determined to be 2 é öù (90 m/s) - 0 æ ú çç 1 kJ/kg ÷ 250 kJ/s - (1500/60 kJ/s) = m êê628.07 - 298.2 + ÷ ÷ú® m = 0.674 kg / s çè1000 m 2 /s 2 ø 2 êë ú û

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=25 [C] Vel1=0 P2=1000 [kPa] T2=347 [C] Vel2=90 [m/s] Q_dot_out=1500[kJ/min]*Convert(kJ/min, kJ/s) W_dot_in=250 [kW] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) h2=enthalpy(Fluid$, T=T2) "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-469

W_dot_in+m_dot*(h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg))=Q_dot_out+m_dot*(h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)) "energy balance"

5-60 Steam expands in a two-stage adiabatic turbine from a specified state to another state. Some steam is extracted at the end of the first stage. The power output of the turbine is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible.

Properties From the steam tables (Table A-6)

P1 = 12.5 MPa ïüï ý h1 = 3476.5 kJ/kg ïïþ T1 = 550°C P2 = 1 MPa ïüï ý h2 = 2828.3 kJ/kg T2 = 200°C ïïþ P3 = 100 kPa üïï ý h3 = 2675.8 kJ/kg ïïþ T3 = 100°C Analysis The mass flow rate through the second stage is

m3 = m1 - m2 = 20 - 1 = 19 kg/s We take the entire turbine, including the connection part between the two stages, as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters the turbine and two fluid streams leave, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 = m2 h2 + m3 h3 + Wout

Wout = m1h1 - m2 h2 - m3 h3

Substituting, the power output of the turbine is

Wout = (20 kg/s)(3476.5 kJ/kg) - (1 kg/s)(2828.3 kJ/kg) - (19 kg/s)(2675.8 kJ/kg)

= 15,860 kW

1-470

EES (Engineering Equation Solver) SOLUTION P1=12500 [kPa] T1=550 [C] m_dot_1=20 [kg/s] P2=1000 [kPa] T2=200 [C] m_dot_2=1 [kg/s] P3=100 [kPa] T3=100 [C] Fluid$='steam_iapws' m_dot_3=m_dot_1-m_dot_2 h1=enthalpy(Fluid$,T=T1,P=P1) h2=enthalpy(Fluid$,T=T2,P=P2) h3=enthalpy(Fluid$,T=T3,P=P3) W_dot_out=m_dot_1*h1-m_dot_2*h2-m_dot_3*h3

Throttling Valves 5-61C Because usually there is a large temperature drop associated with the throttling process.

5-62C No. Because air is an ideal gas and h = h(T) for ideal gases. Thus if h remains constant, so does the temperature.

5-63C If it remains in the liquid phase, no. But if some of the liquid vaporizes during throttling, then yes.

5-64C Yes.

5-65C The temperature of a fluid can increase, decrease, or remain the same during a throttling process. Therefore, this claim is valid since no thermodynamic laws are violated.

5-66 Refrigerant-134a is throttled by a valve. The temperature drop of the refrigerant and the specific volume after expansion are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of R-134a is, from the refrigerant tables (Tables A-11 through 13), o P1 = 0.7 MPa ïü ïý T1 = Tsat = 26.69 C ïïþ h1 = h f = 88.82 kJ/kg sat. liquid

1-471

Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout = D EsystemZ 0 (steady) = 0 ® Ein = Eout ® mh1 = mh2 ® h1 = h2

since Q @W = D ke @D pe @ 0. Then, P2 = 160 kPa ïü ïý h f = 31.18 kJ/kg, Tsat = - 15.60°C ïï hg = 241.14 kJ/kg (h2 = h1 ) þ

Obviously h f < h2 < hg , thus the refrigerant exists as a saturated mixture at the exit state and thus T2 = Tsat = - 15.60° C. Then the temperature drop becomes

D T = T2 - T1 = - 15.60 - 26.69 = - 42.3°C The quality at this state is determined from

x2 =

h2 - h f h fg

88.82 - 31.18 = 0.2745 209.96

Thus,

v 2 = v f + x2v fg = 0.0007435 + 0.2745´ (0.12355 - 0.0007435) = 0.0345 m3 / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=700 [kPa] x1=0 P2=160 [kPa] "Properties" Fluid$='R134a' T1=temperature(Fluid$, P=P1, x=x1) h1=enthalpy(Fluid$, P=P1, x=x1) "Analysis" h1=h2 "energy balance" T2=temperature(Fluid$, P=P2, h=h2) DELTAT=T1-T2 v2=volume(Fluid$, P=P2, h=h2)

5-67 Steam is throttled from a specified pressure to a specified state. The quality at the inlet is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-472

3 4

Heat transfer to or from the fluid is negligible. There are no work interactions involved.

mh1 = mh2 h1 = h2 since

Q @W = D ke @ Δpe @ 0.

The enthalpy of steam at the exit is (Table A-6), P2 = 50 kPa ü ïï ý h2 = 2682.4 kJ/kg T2 = 100°C ïïþ

The quality of the steam at the inlet is (Table A-5) ïü ïý x = h2 - h f = 2682.4 - 844.55 = 0.944 1 h1 = h2 = 2682.4 kJ/kg ïïþ h fg 1946.4 P1 = 1500 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P1=1500 [kPa] P2=50 [kPa] T2=100 [C] "Analysis" Fluid$='steam_iapws' h2=enthalpy(Fluid$, P=P2, T=T2) h1=h2 "energy balance" h1_f=enthalpy(Fluid$, P=P1, x=0) h1_g=enthalpy(Fluid$, P=P1, x=1) h1_fg=h1_g-h1_f x1=(h1-h1_f)/h1_fg

5-68 Refrigerant-134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved.

1-473

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout

mh1 = mh2 h1 = h2 since

Q @W = D ke @ Δpe @ 0 .

The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-11),

T1 = 50°C sat. liquid

ïü ïý h = h = 123.52 kJ/kg f ïïþ 1

The exit quality is T2 = - 20°C h2 = h1

ïü ïý x = h2 - h f = 123.52 - 25.47 = 0.460 ïïþ 2 h fg 212.96

EES (Engineering Equation Solver) SOLUTION "Given" T1=50 [C] x1=0 T2=-20 [C] "Analysis" Fluid$='R134a' h1=enthalpy(Fluid$, T=T1, x=x1) h2=h1 "energy balance" h2_f=enthalpy(Fluid$, T=T2, x=0) h2_g=enthalpy(Fluid$, T=T2, x=1) h2_fg=h2_g-h2_f x2=(h2-h2_f)/h2_fg

5-69 Steam is throttled by a well-insulated valve. The temperature drop of the steam after the expansion is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of steam is (Tables A-6),

1-474

P1 = 8 MPa ü ïï ý h1 = 2988.1 kJ/kg T1 = 350°C ïïþ

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout = D EsystemZ 0 (steady) = 0

Ein = Eout

mh1 = mh2 h1 = h2 since Q @W = D ke @D pe @ 0. Then the exit temperature of steam becomes P2 = 2 MPa ü ïï ý T2 = 285°C ïï h = h (2 1) þ

EES (Engineering Equation Solver) SOLUTION "Given" P1=8000 [kPa] T1=350 [C] P2=2000 [kPa] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) "Analysis" h1=h2 "energy balance" T2=temperature(Fluid$, P=P2, h=h2)

5-70 Problem 5-69 is reconsidered. The effect of the exit pressure of steam on the exit temperature after throttling as the exit pressure varies from 6 MPa to 1 MPa is to be investigated. The exit temperature of steam is to be plotted against the exit pressure. Analysis The problem is solved using EES, and the solution is given below. P_in=8000 [kPa] T_in=350 [C] P_out=2000 [kPa] WorkingFluid$='Steam_iapws' "WorkingFluid: can be changed to other fluids" m_dot_in=m_dot_out "steady-state mass balance" m_dot_in=1 "mass flow rate is arbitrary" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-475

m_dot_in*h_in+Q_dot-W_dot-m_dot_out*h_out=0 "steady-state energy balance" Q_dot=0 "assume the throttle to operate adiabatically" W_dot=0 "throttles do not have any means of producing power" h_in=enthalpy(WorkingFluid$,T=T_in,P=P_in) "property table lookup" T_out=temperature(WorkingFluid$,P=P_out,h=h_out) "property table lookup" x_out=quality(WorkingFluid$,P=P_out,h=h_out) "x_out is the quality at the outlet" P[1]=P_in; P[2]=P_out; h[1]=h_in; h[2]=h_out "use arrays to place points on property plot"

Pout

Tout

[kPa] 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000

[C] 270.5 277.7 284.6 291.2 297.6 303.7 309.5 315.2 320.7 325.9 331

5-71E Refrigerant-134a is throttled by a valve. The temperature and internal energy change are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-476

2 3 4

Kinetic and potential energy changes are negligible. Heat transfer to or from the fluid is negligible. There are no work interactions involved.

Ein = Eout

mh1 = mh2 h1 = h2 since Q @W = D ke @D pe @ 0. The properties are (Tables A-11E through 13E), h1 = 41.79 Btu/lbm P1 = 120 psia ü ïï ý u1 = 41.49 Btu/lbm ïïþ x1 = 0 T1 = 90.49°F ü ïï T2 = - 2.43°F ý h2 = h1 = 41.79 Btu/lbm ïïþ u2 = 38.96 Btu/lbm P2 = 20 psia

D T = T2 - T1 = - 2.43 - 90.49 = - 92.9°F D u = u2 - u1 = 38.96 - 41.49 = - 2.53 Btu / lbm That is, the temperature drops by 92.9F and internal energy drops by 2.53 Btu / lbm .

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=120 [psia] x1=0 P2=20 [psia] "SOLUTION" h1=enthalpy(R134a, P=P1, x=x1) u1=intenergy(R134a, P=P1, x=x1) T1=temperature(R134a, P=P1, x=x1) h2=h1 T2=temperature(R134a, P=P2, h=h2) u2=intenergy(R134a, P=P2, h=h2) DELTAT=T2-T1 DELTAu=u2-u1

5-72E High-pressure air is throttled to atmospheric pressure. The temperature of air after the expansion is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-477

4 5

There are no work interactions involved. Air is an ideal gas.

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout

mh1 = mh2 h1 = h2 since Q @W = D ke @D pe @ 0 .

For an ideal gas, h = h(T) Therefore, T2 = T1 = 90 F

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [psia] T1=90 [F] P2=35 [psia] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) "Analysis" h1=h2 "energy balance" T2=temperature(Fluid$, h=h2)

5-73 Refrigerant-134a is throttled by a valve. The pressure and internal energy after expansion are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no work interactions involved. Properties The inlet enthalpy of R-134a is, from the refrigerant tables (Tables A-11 through 13), © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-478

P1  0.8 MPa   h1  h f @ 25 C  86.41 kJ/kg T1  25C 

1  m 2  m  . We take the throttling valve as the system, which is a Analysis There is only one inlet and one exit, and thus m control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout = ΔEsystem 0 (steady) = 0 Ein = Eout

mh1 = mh2 h1 = h2 since Q  W  ke  pe  0 . Then, T2  20C  h f  25.47 kJ/kg, u f  25.37 kJ/kg h2  h1   hg  238.43 kJ/kg u g  218.86 kJ/kg

Obviously h f < h2 < hg , thus the refrigerant exists as a saturated mixture at the exit state, and thus

P2 = Psat @- 20 C = 132.82 kPa Also, x2 

h2  h f h fg



86.41  25.47  0.2862 212 .96

Thus, u 2  u f  x 2 u fg  25.37  0.2862  193 .49  80.7 kJ/kg

EES (Engineering Equation Solver) SOLUTION Refrig$='R134a' "refrigerant: can be changed to R12 or other fluids" P_in=800 [kPa] T_in=25 [C] T_out=-20 [C] "Analysis" m_dot_in=m_dot_out "steady-state mass balance" m_dot_in=1 "mass flow rate is arbitrary" m_dot_in*h_in-m_dot_out*h_out+Q_dot-W_dot=0 "steady-state energy balance" Q_dot=0 "assume the throttle to operate adiabatically" W_dot=0 "throttles do not have any means of producing power" h_in=enthalpy(Refrig$,T=T_in,P=P_in) "property evaluation" P_out=pressure(Refrig$,T=T_out,h=h_out) "property evaluation" u_out=intEnergy(Refrig$,T=T_out,h=h_out) "property evaluation" x_out=quality(Refrig$,T=T_out,h=h_out) "x_out is the quality at the outlet" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-479

P[1]=P_in; P[2]=P_out; h[1]=h_in; h[2]=h_out "use arrays to place points on property plot"

Mixing Chambers and Heat Exchangers 5-74C Under the conditions of no heat and work interactions between the mixing chamber and the surrounding medium.

5-75C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium.

5-76C Yes, if the mixing chamber is losing heat to the surrounding medium.

5-77 Refrigerant-134a is condensed in a water-cooled condenser. The mass flow rate of the cooling water required is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties The enthalpies of R-134a at the inlet and the exit states are (Tables A-11 through A-13)

P3 = 700 kPa T3 = 70°C P4 = 700 kPa sat. liquid

ïüï ý h3 = 308.34 kJ/kg ïïþ ïü ïý h = h f @ 700 kPa = 88.82 kJ/kg ïïþ 4

Water exists as compressed liquid at both states, and thus (Table A-4) h1  h f @15 C  62.98kJ / kg h2  h f @ 25 C  104.83kJ / kg

Analysis We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance (for each fluid stream): © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-480 min - mout = D msystem

0 (steady)

= 0 ®

min = mout ®

m1 = m2 = mw

and

m3 = m4 = mR

Energy balance (for the heat exchanger):

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 + m3 h3 = m2 h2 + m4 h4

(since Q = W = D ke @D pe @ 0)

Combining the two,

mw (h2 - h1 ) = mR (h3 - h4 ) Solving for mw : mw =

h3 - h4 mR h2 - h1

Substituting, mw =

(308.34 - 88.82) kJ/kg (8 kg/min) = 42.0 kg / min (104.83 - 62.98) kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=700 [kPa] T1=70 [C] m_dot_1=8 [kg/min] P2=P1 x2=0 P3=300 [kPa] T3=15 [C] P4=P3 T4=25 [C] "Properties" h1=enthalpy(R134a, T=T1, P=P1) h2=enthalpy(R134a, P=P2, x=x2) h3=enthalpy(steam_iapws, T=T3, x=0) h4=enthalpy(steam_iapws, T=T4, x=0) "Analysis" m_dot_1=m_dot_2 "mass balance on R134a" m_dot_3=m_dot_4 "mass balance on water" m_dot_1*h1+m_dot_3*h3=m_dot_2*h2+m_dot_4*h4 "energy balance" m_dot_w=m_dot_3

5-78 Hot and cold streams of a fluid are mixed in a mixing chamber. Heat is lost from the chamber. The energy carried from the mixing chamber is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work and heat interactions. Analysis We take the mixing device as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

1-481

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1e1 + m2 e2 = m3 e3 + Qout From a mass balance m3 = m1 + m2 = 5 + 15 = 20 kg/s

Substituting into the energy balance equation solving for the exit enthalpy gives m1e1 + m2 e2 = m3 e3 + Qout

e3 =

m1e1 + m2 e2 - Qout (5 kg/s)(150 kJ/kg + (15 kg/s)(50 kJ/kg) - 5.5 kW = = 74.7 kJ / kg m3 20 kg/s

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_1=5 [kg/s] e1=150 [kJ/kg] m_dot_2=15 [kg/s] e2=50 [kJ/kg] Q_dot_out=5.5 [kW] "Analysis" m_dot_1+m_dot_2=m_dot_3 "mass balance" m_dot_1*e1+m_dot_2*e2=m_dot_3*e3+Q_dot_out "energy balance"

5-79 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions. Properties Noting that T < Tsat @ 250 kPa = 127.41 C , the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus, h1  h f @80 C  335.02kJ / kg

h2  h f @ 20 C  83.915kJ / kg h3  h f @ 42 C  175.90kJ / kg

Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-482

Mass balance: min - mout = D msystem

0 (steady)

= 0

¾¾ ®

m1 + m2 = m3

Energy balance:

Ein - Eout Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m2 h2 = m3 h3 (since Q = W = D ke @D pe @ 0) Combining the two relations and solving for m2 gives

m1h1 + m2 h2 = (m1 + m2 )h3 m2 =

h1 - h3 m1 h3 - h2

Substituting, the mass flow rate of cold water stream is determined to be m2 =

(335.02 - 175.90) kJ/kg (0.5 kg/s) = 0.865 kg / s (175.90 - 83.915) kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=80 [C] m_dot_1=0.5 [kg/s] T2=20 [C] T3=42 [C] P=250 [kPa] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=0) h2=enthalpy(Fluid$, T=T2, x=0) h3=enthalpy(Fluid$, T=T3, x=0) "Analysis" m_dot_1+m_dot_2=m_dot_3 "mass balance" m_dot_1*h1+m_dot_2*h2=m_dot_3*h3 "energy balance"

5-80E Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter at the same rate, the temperature and quality (if saturated) of the exit stream is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-483

3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From steam tables (Tables A-5E through A-6E), h1  h f @80 F  48.065 Btu / lbm

h2  hg @ 20 psia  1156.2 Btu / lbm

Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance:

min - mout = D msystemZ 0 (steady) = 0 min = mout m1 + m2 = m3 = 2m

m1 = m2 = m Energy balance:

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m2 h2 = m3 h3 (since Q = W = D ke @D pe @ 0) Combining the two gives

mh1 + mh2 = 2mh3 or h3 = (h1 + h2 ) / 2 Substituting,

h3  (48.065  1156.2) / 2  602.13 Btu / lbm At 20 psia, h f  196.27 Btu / lbm and hg  1156.2 Btu / lbm . Thus the exit stream is a saturated mixture since h f < h3 < hg Therefore,

T3 = Tsat @ 20 psia = 228 F and x3 =

h3 - h f h fg

602.13 - 196.27 = 0.423 1156.2 - 196.27

EES (Engineering Equation Solver) SOLUTION "Given" T1=80 [F] P1=20 [psia] P2=20 [psia] x2=1 m_dot_1=m_dot_2 "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-484

Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=0) h2=enthalpy(Fluid$, P=P2, x=x2) h2_f=enthalpy(Fluid$, P=P2, x=0) h2_g=enthalpy(Fluid$, P=P2, x=1) "Analysis" m_dot_1=1 "assumed, it does not matter what value is assumed" m_dot_1+m_dot_2=m_dot_3 "mass balance" m_dot_1*h1+m_dot_2*h2=m_dot_3*h3 "energy balance" P3=P1 T3=temperature(Fluid$, P=P3, h=h3) x3=quality(Fluid$, P=P3, h=h3)

5-81 An adiabatic open feedwater heater mixes steam with feedwater. The outlet mass flow rate and the outlet velocity are to be determined.

Assumptions Steady operating conditions exist. Properties From a mass balance

m3 = m1 + m2 = 0.2 + 10 = 10.2 kg / s The specific volume at the exit is (Table A-4) P3 = 100 kPa ü ïï 3 ý v 3 @v f @ 60°C = 0.001017 m /kg ïïþ T3 = 60°C

The exit velocity is then

V3 = =

m3v 3 4m3v 3 = A3 p D2 4(10.2 kg/s)(0.001017 m3 /kg) p (0.03 m)2

= 14.7 m / s

EES (Engineering Equation Solver) SOLUTION m_dot_1=0.2 [kg/s] P1=100 [kPa] T1=160 [C] m_dot_2=10 [kg/s] P2=100 [kPa] T2=50 [C] P3=100 [kPa] T3=60 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-485

D3=0.03 [m] Fluid$='steam_iapws' v3=volume(Fluid$, T=T3, x=0) m_dot_3=m_dot_1+m_dot_2 Vel3=m_dot_3*v3/A3 A3=pi*D3^2/4

5-82 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the exit temperature of hot water are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ / kg  C , respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = mc p (T2 - T1 )

Then the rate of heat transfer to the cold water in this heat exchanger becomes Q = [mc p (Tout - Tin )]cold water = (0.60 kg/s)(4.18 kJ/kg.°C)(45°C - 15°C) = 75.24 kW

Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of the hot water is determined to be

Q = [mc p (Tin - Tout )]hot water ¾ ¾ ® Tout = Tin -

Q 75.24 kW = 100°C = 94.0°C mc p (3 kg/s)(4.19 kJ/kg.°C)

1-486

EES (Engineering Equation Solver) SOLUTION "Given" T1=15 [C] T2=45 [C] m_dot_cw=0.60 [kg/s] T3=100 [C] m_dot_hw=3 [kg/s] c_p_cw=4.18 [kJ/kg-C] c_p_hw=4.19 [kJ/kg-C] "Analysis" Q_dot=m_dot_cw*c_p_cw*(T2-T1) "energy balance on cold water" Q_dot=m_dot_hw*c_p_hw*(T3-T4) "energy balance on hot water"

5-83E Steam is condensed by cooling water in a condenser. The rate of heat transfer in the heat exchanger and the rate of condensation of steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.

Properties The specific heat of water is 1.0 Btu / lbm. F (Table A-3E). The enthalpy of vaporization of water at 75F is

1050.9 Btu / lbm (Table A-4E). Analysis We take the tube-side of the heat exchanger where cold water is flowing as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = mc p (T2 - T1 )

Then the rate of heat transfer to the cold water in this heat exchanger becomes Q = [mc p (Tout - Tin )]water = (45 lbm/s)(1.0 Btu/lbm.°F)(65°F - 50°F) = 675 Btu / s

Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate of condensation of the steam in the heat exchanger is determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-487

Q = (mh fg )steam = ¾ ¾ ® msteam =

Q 675 Btu/s = = 0.642 lbm / s h fg 1050.9 Btu/lbm

EES (Engineering Equation Solver) SOLUTION "Given" T_steam=75 [F] T1=50 [F] T2=65 [F] m_dot_w=45 [lbm/s] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_steam, x=0) h_g=enthalpy(Fluid$, T=T_steam, x=1) h_fg=h_g-h_f c_p=1.0 [Btu/lbm-F] "Table A-3E" "Analysis" Q_dot=m_dot_w*c_p*(T2-T1) "energy balance on water" m_dot_steam=Q_dot/h_fg

5-84 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the outlet temperature of the air are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ / kg. C , respectively.

Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0 ® Ein = Eout

Rate of change in internal, kinetic, potential, etc. energies

mh1 = Qout + mh2 (since D ke @D pe @ 0)

Qout = mc p (T1 - T2 ) Then the rate of heat transfer from the exhaust gases becomes © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-488

Q = [mc p (Tin - Tout )]gas = (0.95 kg/s)(1.1 kJ/kg.°C)(160°C - 95°C)

= 67.93 kW The mass flow rate of air is m=

PV (95 kPa)(0.6 m3 /s) = = 0.6778 kg/s RT (0.287 kPa.m3 /kg.K)´ 293 K

Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes Q = mc p (Tc, out - Tc, in ) ¾ ¾ ® Tc, out = Tc, in +

Q 67.93 kW = 20°C + = 120°C mc p (0.6778 kg/s)(1.005 kJ/kg.°C)

EES (Engineering Equation Solver) SOLUTION "Given" P1=95 [kPa] T1=20 [C] Vol_dot_air=0.6 [m^3/s] T3=160 [C] T4=95 [C] m_dot_gases=0.95 [kg/s] c_p_air=1.005 [kJ/kg-C] c_p_gases=1.10 [kJ/kg-C] "Analysis" R=0.287 [kPa-m^3/kg-K] "Table A-2a" m_dot_air=(P1*Vol_dot_air)/(R*(T1+273)) Q_dot=m_dot_gases*c_p_gases*(T3-T4) "energy balance on combustion gases" Q_dot=m_dot_air*c_p_air*(T2-T1) "energy balance on air"

5-85E An adiabatic open feedwater heater mixes steam with feedwater. The outlet mass flow rate and the outlet velocity are to be determined for two exit temperatures.

Assumptions Steady operating conditions exist. Analysis From a mass balance

m3 = m1 + m2 = 0.1+ 2 = 2.1 lbm / s The specific volume at the exit is (Table A-4E) P3 = 10 psia ü ïï 3 ý v 3 @v f @120° F = 0.01620 ft /lbm T3 = 120°F ïïþ

The exit velocity is then

1-489

V3 = =

m3v 3 4m3v 3 = A3 p D2

4(2.1 lbm/s)(0.01620 ft 3 /lbm) p (0.5 ft)2

= 0.1733 ft / s When the temperature at the exit is 180F, we have

m3 = m1 + m2 = 0.1+ 2 = 2.1 lbm / s P3 = 10 psia ü ïï 3 ý v 3 @v f @180° F = 0.01651 ft /lbm T3 = 180°F ïïþ

V3 =

m3v 3 4m3v 3 4(2.1 lbm/s)(0.01651 ft 3 /lbm) = = = 0.177 ft / s A3 p D2 p (0.5 ft)2

The mass flow rate at the exit is same while the exit velocity slightly increases when the exit temperature is 180F instead of 120F.

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_1=0.1 [lbm/s] P1=10 [psia] T1=200 [F] m_dot_2=2 [lbm/s] P2=10 [psia] T2=100 [F] P3=10 [psia] T3=120 [F] D3=0.5 [ft] T3_a=180 [F] "Analysis" Fluid$='steam_iapws' m_dot_1+m_dot_2=m_dot_3 v3=volume(Fluid$, T=T3, x=0) A3=pi*D3^2/4 Vel3=(m_dot_3*v3)/A3 "If T3=180 F:" v3_a=volume(Fluid$, T=T3_a, x=0) Vel3_a=(m_dot_3*v3_a)/A3

5-86 Refrigerant-134a is to be cooled by air in the condenser. For a specified volume flow rate of air, the mass flow rate of the refrigerant is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Air is an ideal gas with constant specific heats at room temperature.

1-490

Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The constant pressure specific heat of air is c p  1.005 kJ / kg  C (Table A-2). The enthalpies of the R-134a at the inlet and the exit states are (Tables A-11 through A-

13)

P3 = 1 MPa ïüï ý h3 = 324.66 kJ/kg T3 = 90°C ïïþ

P4 = 1 MPa ïüï ý h4 @ h f @ 30 oC = 93.58 kJ/kg T4 = 30°C ïïþ Analysis The inlet specific volume and the mass flow rate of air are v1 =

RT1 (0.287 kPa ×m /kg ×K )(300 K ) = = 0.861 m 3 /kg P1 100 kPa

V1 600 m3 /min = = 696.9 kg/min v1 0.861 m3 /kg

and

We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream): min - mout = D msystemZ 0 (steady) = 0 ® min = mout ® m1 = m2 = ma and m3 = m4 = mR

Energy balance (for the entire heat exchanger):

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m3 h3 = m2 h2 + m4 h4 (since Q = W = D ke @D pe @ 0)

Combining the two, ma (h2 - h1 ) = mR (h3 - h4 )

Solving for mR : mR =

c p (T2 - T1 ) h2 - h1 ma @ ma h3 - h4 h3 - h4

Substituting, mR =

(1.005 kJ/kg ×°C)(60 - 27)°C (696.9 kg/min) = 100.0 kg / min (324.66 - 93.58) kJ/kg

1-491

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [kPa] T1=90 [C] P2=1000 [kPa] T2=30 [C] P3=100 [kPa] T3=27 [C] Vol_dot_3=600 [m^3/min] P4=95 [kPa] T4=60 [C] "Properties" Fluid$='R134a' h1=enthalpy(Fluid$, T=T1, P=P1) h2=enthalpy(Fluid$, T=T2, x=0) C_p_air=1.005 R_air=0.287 "Analysis" v3=(R_air*(T3+273))/P3 m_dot_3=Vol_dot_3/v3 m_dot_1=m_dot_2 "mass balance on R134a" m_dot_3=m_dot_4 "mass balance on air" m_dot_1*h1-m_dot_2*h2=m_dot_4*C_p_air*(T4-T3) "energy balance" m_dot_R=m_dot_1

5-87 Refrigerant-22 is evaporated in an evaporator by air. The rate of heat transfer from the air and the temperature change of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work and heat interactions between the evaporator and the surroundings.

Analysis We take the condenser as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mR h1 + ma h3 = mR h2 + ma h4 mR (h2 - h1 ) = ma (h3 - h4 ) = ma c p D Ta

If we take the refrigerant as the system, the energy balance can be written as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-492

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mR h1 + Qin = mR h2

Qin = mR (h2 - h1 )

(a) The mass flow rate of the refrigerant is

mR =

V1 (2.65 / 3600) m3 /s = = 0.02910 kg/s v1 0.0253 m3 /kg

The rate of heat absorbed from the air is Qin = mR (h2 - h1 ) = (0.02910 kg/s)(398.0 - 220.2)kJ/kg = 5.17 kW

(b) The temperature change of air can be determined from an energy balance on the evaporator:

QL = mR (h3 - h2 ) = ma c p (Ta1 - Ta 2 ) 5.17 kW = (0.75 kg/s)(1.005 kJ/kg ×°C)D Ta D Ta = 6.9°C The specific heat of air is taken as 1.005 kJ / kg  C (Table A-2).

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=200 [kPa] x1=0.22 P2=P1 V_dot_1=(2.65/3600) [m^3/s] DELTAT_superheat=5 [C] m_dot_a=0.75 "PROPERTIES" c_p_a=1.005 [kJ/kg-C] "Table A-3" "SOLUTION" "h1=enthalpy(R22, P=P1, x=x1) v1=volume(R22, P=P1, x=x1) T2=T_sat+DELTAT_superheat T_sat=temperature(R22, P=P2, x=0) h2=enthalpy(R22, P=P2, T=T2)" h1=220.2 v1=0.0253 h2=398 m_dot_R=V_dot_1/v1 Q_dot_L=m_dot_R*(h2-h1) Q_dot_L=m_dot_a*c_p_a*DELTAT_a

5-88 Two streams of cold and warm air are mixed in a chamber. If the ratio of hot to cold air is 1.6, the mixture temperature and the rate of heat gain of the room are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-493

The device is adiabatic and thus heat transfer is negligible.

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K . The enthalpies of air are obtained from air table (Table A17) as h1  h@ 280K  280.13kJ / kg h2  h@307K  307.23kJ / kg

hroom  h@ 297K  297.18kJ / kg

Analysis (a) We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: min - mout = D msystemZ 0 (steady) = 0 ® min = mout ® m1 + 1.6m1 = m3 = 2.6m1 since m2 = 1.6m1

Energy balance:

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m2 h2 = m3 h3

(since Q @W @D ke @D pe @ 0)

Combining the two gives m1h1 + 1.6m1h2 = 2.6m1h3 or h3 = (h1 + 1.6h2 ) / 2.6

Substituting,

h3  (280.13  1.6  307.23) / 2.6  296.81 kJ / kg From air table at this enthalpy, the mixture temperature is T3  T @ h  296.81 kJ / kg  296.6 K  23.6 C

(b) The mass flow rates are determined as follows v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(7 + 273 K) = = 0.7654 m3 / kg P 105 kPa

m1 =

V1 0.55 m3 /s = = 0.7186 kg/s v1 0.7654 m3 /kg

m3 = 2.6m1 = 2.6(0.7186 kg/s) = 1.868 kg/s The rate of heat gain of the room is determined from

Qgain = m3 (hroom - h3 ) = (1.868 kg/s)(297.18 - 296.81) kJ/kg = 0.691kW Therefore, the room gains heat at a rate of 0.691 kW.

1-494

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_1=7 [C] T_2=34 [C] Vol_dot_1=0.55 [m^3/s] m_dot_2/m_dot_1=1.6 T_room=24 [C] P=105 [kPa] "PROPERTIES" Fluid$='Air' h_1=enthalpy(Fluid$, T=T_1) h_2=enthalpy(Fluid$, T=T_2) h_room=enthalpy(Fluid$, T=T_room) R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Fluid$) "ANALYSIS" v_1=(R*(T_1+273))/P m_dot_1=Vol_dot_1/v_1 m_dot_3=m_dot_1+m_dot_2 "mass balance" m_dot_1*h_1+m_dot_2*h_2=m_dot_3*h_3 "energy balance" T_3=temperature(Fluid$, h=h_3) "mixture temperature" Q_dot_gain=m_dot_3*(h_room-h_3) "rate of cooling to room"

5-89 A heat exchanger that is not insulated is used to produce steam from the heat given up by the exhaust gases of an internal combustion engine. The temperature of exhaust gases at the heat exchanger exit and the rate of heat transfer to the water are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Exhaust gases are assumed to have air properties with constant specific heats. Properties The constant pressure specific heat of the exhaust gases is taken to be c p  1.045 kJ / kg  C (Table A-2). The inlet and exit enthalpies of water are (Tables A-4 and A-5)

ïüï ý hw,in = 62.98 kJ/kg x = 0 (sat. liq.)ïïþ

Tw,in = 15°C

1-495

Pw,out = 2 MPa üïï ý hw,out = 2798.3 kJ/kg x = 1 (sat. vap.)ïïþ Analysis We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance (for each fluid stream): min - mout = D msystemZ 0 (steady) = 0 ¾ ¾ ® min = mout

Energy balance (for the entire heat exchanger): Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mexh hexh,in + mw hw,in = mexh hexh,out + mw hw,out + Qout (since W = D ke @D pe @0) or

mexh c pTexh,in + mw hw,in = mexh c pTexh,out + mw hw,out + Qout

Noting that the mass flow rate of exhaust gases is 15 times that of the water, substituting gives

15mw (1.045 kJ/kg ×°C)(400°C) + mw (62.98 kJ/kg)

(1)

= 15mw (1.045 kJ/kg.°C)Texh,out + mw (2798.3 kJ/kg) + Qout The heat given up by the exhaust gases and heat picked up by the water are

Qexh = mexh c p (Texh,in - Texh,out ) = 15mw (1.045 kJ/kg.°C)(400 - Texh,out )°C Qw = mw (hw,out - hw,in ) = mw (2798.3 - 62.98)kJ/kg

(2)

(3)

The heat loss is Qout = f heat loss Qexh = 0.1Qexh

(4)

The solution may be obtained by a trial-error approach. Or, solving the above equations simultaneously using EES software, we obtain

Texh,out = 206.1°C, Qw = 97.26 kW, mw = 0.03556 kg/s, mexh = 0.5333 kg/s

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_w_out=2000 [kPa] T_exh_in=400 [C] m_dot_exh=32[kg/min]*Convert(kg/min, kg/s) m_dot_w=1/15*m_dot_exh T_w_in=15 [C] f_heatloss=0.10 "PROPERTIES" c_p_exh=1.045 [kJ/kg-K] "Table A-2a" "ANALYSIS" Fluid$='Steam_iapws' h_w_in=enthalpy(Fluid$, T=T_w_in, x=0) h_w_out=enthalpy(Fluid$, P=P_w_out, x=1) m_dot_exh*c_p_exh*T_exh_in+m_dot_w*h_w_in=m_dot_exh*c_p_exh*T_exh_out+m_dot_w*h_w_out+Q_dot_ out "energy balance" Q_dot_w=m_dot_w*(h_w_out-h_w_in) Q_dot_exh=m_dot_exh*c_p_exh*(T_exh_in-T_exh_out) Q_dot_out=f_heatloss*Q_dot_exh

1-496

5-90 Water is heated by hot oil in a heat exchanger. The rate of heat transfer in the heat exchanger and the outlet temperature of oil are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.

Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ / kg  C , respectively. Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the rate of heat transfer to the cold water in this heat exchanger becomes

Q = [mc p (Tout - Tin )]water = (4.5 kg/s)(4.18 kJ/kg.°C)(70°C - 20°C) = 940.5 kW Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of oil is determined from Q = [mc p (Tin - Tout )]oil ¾ ¾ ® Tout = Tin -

Q 940.5 kW = 170°C = 129.1°C mc p (10 kg/s)(2.3 kJ/kg.°C)

EES (Engineering Equation Solver) SOLUTION "Given" T1=20 [C] T2=70 [C] m_dot_w=4.5 [kg/s] T3=170 [C] m_dot_oil=10 [kg/s] c_p_w=4.18 [kJ/kg-C] c_p_oil=2.30 [kJ/kg-C] "Analysis" Q_dot=m_dot_w*c_p_w*(T2-T1) "energy balance on water" Q_dot=m_dot_oil*c_p_oil*(T3-T4) "energy balance on oil"

1-497

5-91 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The heat of vaporization of water at 50C is hfg  2382.0 kJ / kg and specific heat of cold water is c p  4.18 kJ / kg  C (Tables A-3 and A-4).

Analysis We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the heat transfer rate to the cooling water in the condenser becomes

Q = [mc p (Tout - Tin )]cooling water = (101 kg/s)(4.18 kJ/kg.°C)(27°C - 18°C) = 3800 kJ/s The rate of condensation of steam is determined to be Q = (mh fg )steam ¾ ¾ ® msteam =

Q 3800 kJ/s = = 1.60 kg / s h fg 2382.0 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" T_steam=50 [C] T1=18 [C] T2=27 [C] m_dot_w=101 [kg/s] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_steam, x=0) h_g=enthalpy(Fluid$, T=T_steam, x=1) h_fg=h_g-h_f © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-498

c_p=4.18 [kJ/kg-C] "taken" "Analysis" Q_dot=m_dot_w*c_p*(T2-T1) "energy balance on water" m_dot_steam=Q_dot/h_fg

5-92 Problem 5-91 is reconsidered. The effect of the inlet temperature of cooling water on the rate of condensation of steam as the inlet temperature varies from 10°C to 20°C at constant exit temperature is to be investigated. The rate of condensation of steam is to be plotted against the inlet temperature of the cooling water. Analysis The problem is solved using EES, and the solution is given below. T_s[1]=50 [C] T_s[2]=50 [C] m_dot_water=101 [kg/s] T_water[1]=18 [C] T_water[2]=27 [C] c_p_water = 4.20 [kJ/kg-°C] E_dot_in - E_dot_out = DELTAE_dot_cv DELTAE_dot_cv=0 E_dot_in=m_dot_s*h_s[1] + m_dot_water*h_water[1] E_dot_out=m_dot_s*h_s[2] + m_dot_water*h_water[2] h_s[1] =enthalpy(steam_iapws,T=T_s[1],x=1) h_s[2] =enthalpy(steam_iapws,T=T_s[2],x=0) h_fg_s=h_s[1]-h_s[2] h_water[1] =c_p_water*T_water[1] h_water[2] =c_p_water*T_water[2]

Twater,1

 kg/s

[C]

3.028 2.671 2.315 1.959 1.603 1.247

10 12 14 16 18 20

1-499

5-93 Two streams of water are mixed in an insulated chamber. The temperature of the exit stream is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

Properties From the steam tables (Table A-4),

h1  h f @90 C  377.04 kJ / kg h2  h f @50 C  209.34 kJ / kg Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: min - mout = D msystem 0 (steady) = 0

min = mout

m1 + m2 = m3 Energy balance:

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 + m2 h2 = m3 h3 (since Q = W = D ke @D pe @ 0)

1-500

Solving for the exit enthalpy, h3 =

m1h1 + m2 h2 (30)(377.04) + (200)(209.34) = = 231.21 kJ/kg m1 + m2 30 + 200

The temperature corresponding to this enthalpy is T3 = 55.2 C (Table A-4)

EES (Engineering Equation Solver) SOLUTION m_dot_1=30 [kg/s] T1=90 [C] m_dot_2=200 [kg/s] T2=50 [C] Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=0) h2=enthalpy(Fluid$, T=T2, x=0) h3=(m_dot_1*h1+m_dot_2*h2)/(m_dot_1+m_dot_2) T3=temperature(Fluid$, h=h3, x=0)

5-94 Two mass streams of the same idela gas are mixed in a mixing chamber. Heat is transferred to the chamber. Three expressions as functions of other parameters are to be obtained. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions.

Analysis (a) We take the mixing device as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m2 h2 + Qin = m3 h3

From a mass balance,

m3 = m1 + m2 Since

h = c pT ,

1-501

m1c pT1 + m2 c pT2 + Qin = m3c pT3 m1 m Q T1 + 2 T2 + in m3 m3 m3 c p

T3 =

(b) Expression for volume flow rate: V3 = m3v 3 = m3 V3 =

RT3 P3

ö m3 R æ çç m1 T + m2 T + Qin ÷ ÷ ÷ 1 2 ç ÷ ÷ P3 çè m3 m3 m3 c p ø

P3 = P1 = P2 = P V3 =

m1 RT1 m2 RT2 RQin + + P1 P2 P3 c p

V3 = V1 + V2 +

RQin Pc p

Pipe and duct Flow 5-95 Water is heated by a 7-kW resistance heater as it flows through an insulated tube. The mass flow rate of water is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Water is an incompressible substance with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The tube is adiabatic and thus heat losses are negligible. Properties The specific heat of water at room temperature is c p  4.18 kJ / kg  C (Table A-3).

Analysis We take the water pipe as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

We,in + mh1 = mh2 (since Qout @D ke @D pe @0) We,in = m(h2 - h1 ) = m[c(T2 - T1 ) + v D P ¿ 0 ] = mc (T2 - T1 )

1-502

Substituting, the mass flow rates of water is determined to be m=

We,in c(T2 - T1 )

7 kJ/s = 0.0304 kg / s (4.184 kJ/kg ×o C)(75 - 20) o C

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_e_in=7 [kW] T1=20 [C] T2=75 [C] "Properties" C_p=CP(water, T=T_ave, P=P) T_ave=1/2*(T1+T2) P=100 [kPa] "assumed" "Analysis" W_dot_e_in=m_dot*C_p*(T2-T1) "energy balance on water pipe"

5-96 Air at a specified rate is heated by an electrical heater. The current is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The heat losses from the air is negligible. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2a).

Analysis We take the pipe in which the air is heated as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 + We,in = mh2

We,in = m(h2 - h1 ) VI = mc p (T2 - T1 )

The inlet specific volume and the mass flow rate of air are

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(288 K) = = 0.8266 m3 /kg P1 100 kPa

V1 0.3 m3 /s = = 0.3629 kg/s v1 0.8266 m3 /kg

Substituting into the energy balance equation and solving for the current gives © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-503

mc p (T2 - T1 ) V

ö (0.3629 kg/s)(1.005 kJ/kg ×K)(30 - 15)K æ ÷ çç1000 VI ÷ = 49.7 Amperes ÷ çè 1 kJ/s ø 110 V

EES (Engineering Equation Solver) SOLUTION "Given" Voltage=110 [volt] Vol1_dot=0.3 [m^3/s] P1=100 [kPa] T1=(15+273) [K] P2=100 [kPa] T2=(30+273) [K] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" "Analysis" v1=(R*T1)/P1 m_dot=Vol1_dot/v1 I=(m_dot*c_p*(T2-T1))/Voltage*Convert(kJ/s, Volt-Amp)

5-97 The ducts of a heating system pass through an unheated area. The rate of heat loss from the air in the ducts is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 There are no work interactions involved. Properties The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2)

Analysis We take the heating duct as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2

(since W @D ke @D pe @ 0)

Qout = m(h1 - h2 ) = mc p (T1 - T2 ) Substituting, Qout = (120 kg/min)(1.005 kJ/kg ×°C)(4°C) = 482 kJ / min

1-504

DELTAT=4 [C] m_dot=120 [kg/min] "Properties" C_p=CP(air, T=25[C]) "Analysis" Q_dot_out=m_dot*C_p*DELTAT "energy balance on heating duct"

5-98E The cooling fan of a computer draws air, which is heated in the computer by absorbing the heat of PC circuits. The electrical power dissipated by the circuits is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 All the heat dissipated by the circuits are picked up by the air drawn by the fan. Properties The gas constant of air is 0.3704 psia  ft 3 / lbm  R (Table A-1E). The constant pressure specific heat of air at room temperature is c p  0.240 Btu / lbm  F (Table A-2Ea).

Analysis We take the pipe in which the air is heated as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 + We,in = mh2

We,in = m(h2 - h1 ) We,in = mc p (T2 - T1 )

The inlet specific volume and the mass flow rate of air are

v1 =

RT1 (0.3704 psia ×ft 3 /lbm ×R)(530 R) = = 13.35 ft 3 /lbm P1 14.7 psia

V1 0.3 ft 3 /s = = 0.02246 lbm/s v1 13.35 ft 3 /lbm

Substituting, æ 1 kW ö ÷ We,out = (0.02246 lbm/s)(0.240 Btu/lbm ×R)(83 - 70)Btu/lbm çç = 0.0740 kW ÷ çè0.94782 Btu/s ÷ ø

1-505

P1=14.7 [psia] T1=(70+460) [R] P2=14.7 [psia] T2=(83+460) [R] "Properties" R=0.3704 [psia-ft^3/lbm-R] "Table A-1E" c_p=0.240 [Btu/lbm-R] "Table A-2Ea" "Analysis" v1=(R*T1)/P1 m_dot=Vol1_dot/v1 W_dot_e_out=m_dot*c_p*(T2-T1)*Convert(Btu/s, kW)

5-99 Saturated liquid water is heated in a steam boiler at a specified rate. The rate of heat transfer in the boiler is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions.

Analysis We take the pipe in which the water is heated as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 + Qin = mh2 Qin = m(h2 - h1 )

The enthalpies of water at the inlet and exit of the boiler are (Table A-5, A-6).

P1 = 2 MPa x= 0

P2 = 2 MPa T2 = 250°C

ü ïï ý h1 @ h f @ 2 MPa = 908.47 kJ/kg ïïþ

ïüï ý ïïþ

h2 = 2903.3 kJ/kg

Substituting, Qin = (4 kg/s)(2903.3 - 908.47)kJ/kg = 7980 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P=2000 [kPa] m_dot=4 [kg/s] T2=250 [C] "ANALYSIS" Fluid$='steam_iapws' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-506

h1=enthalpy(Fluid$, P=P, x=0) h2=enthalpy(Fluid$, P=P, T=T2) Q_dot_in=m_dot*(h2-h1)

5-100E Electronic devices mounted on a cold plate are cooled by water. The amount of heat generated by the electronic devices is to be determined. Assumptions 1 Steady operating conditions exist. 2 About 15 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation. 3 Kinetic and potential energy changes are negligible. Properties The properties of water at room temperature are ρ  62.1 lbm / ft 3 and c p  1.00 Btu / lbm. F (Table A-3E).

Analysis We take the tubes of the cold plate to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then mass flow rate of water and the rate of heat removal by the water are determined to be

m = r AV = r

p D2 p (0.25 /12 ft)2 V = (62.1 lbm/ft 3 ) (40 ft/min) = 0.8483 lbm/min = 50.9 lbm/h 4 4

Q = mc p (Tout - Tin ) = (50.9 lbm/h)(1.00 Btu/lbm.°F)(105 - 70)°F = 1781 Btu/h which is 85 percent of the heat generated by the electronic devices. Then the total amount of heat generated by the electronic devices becomes

1781 Btu/h = 2096 Btu / h = 614 W 0.85

EES (Engineering Equation Solver) SOLUTION "Given" T1=70 [F] T2=105 [F] Vel=40 [ft/min] D=0.25[in]*Convert(in, ft) f_water=0.85 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-507

"Properties" Fluid$='Steam_iapws' rho=62.1 [lbm/ft^3] "Table A-3" c_p=1.0 [Btu/lbm-F] "Table A-3" "Analysis" A=pi*D^2/4 m_dot=rho*A*Vel Q_dot_in=m_dot*c_p*(T2-T1) "energy balance on water" Q_dot=Q_dot_in/f_water*Convert(Btu/min, W)

5-101 Air enters a hollow-core printed circuit board. The exit temperature of the air is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 The local atmospheric pressure is 1 atm. 4 Kinetic and potential energy changes are negligible. Properties The gas constant of air is R  0.287 kJ / kg  C (Table A-1). The specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2).

Analysis The density of air entering the duct and the mass flow rate are r=

P 101.325 kPa = = 1.185 kg/m3 RT (0.287 kPa.m3 /kg.K)(25+273)K

m = r V = (1.185 kg/m3 )(0.0008 m3 / s) = 0.0009477 kg/s

We take the hollow core to be the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the exit temperature of air leaving the hollow core becomes Qin = mc p (T2 - T1 ) ® T2 = T1 +

Qin 15 J/s = 25°C + = 46.0°C mc p (0.0009477 kg/s)(1005 J/kg.°C)

1-508

w=0.0015 [m] Q_dot=20 [W] T1=25 [C] V_dot=0.8[L/s]*Convert(L/s, m^3/s) "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" "Analysis" P=101.325 [kPa] "standard atmospheric pressure is assumed" rho=P/(R*(T1+273)) m_dot=rho*V_dot Q_dot*Convert(W, kW)=m_dot*c_p*(T2-T1) "energy balance on air"

5-102 A computer is cooled by a fan blowing air through the case of the computer. The required flow rate of the air and the fraction of the temperature rise of air that is due to heat generated by the fan are to be determined. Assumptions 1 Steady flow conditions exist. 2 Air is an ideal gas with constant specific heats. 3 The pressure of air is 1 atm. 4 Kinetic and potential energy changes are negligible Properties The specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2). Analysis (a) We take the air space in the computer as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + Win + mh1 = mh2 (since D ke @D pe @ 0)

Qin + Win = mc p (T2 - T1 ) Noting that the fan power is 25 W and the 8 PCBs transfer a total of 80 W of heat to air, the mass flow rate of air is determined to be Qin + Win = mc p (Te - Ti ) ® m =

Qin + Win (8´ 10) W + 25 W = = 0.0104 kg / s c p (Te - Ti ) (1005 J/kg.°C)(10°C)

(b) The fraction of temperature rise of air that is due to the heat generated by the fan and its motor can be determined from Q = mc p D T ® D T =

f =

Q 25 W = = 2.4°C mc p (0.0104 kg/s)(1005 J/kg.°C)

2.4°C = 0.24 = 24% 10°C

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_PCB=8*10 [W] h=0.12 [m] L=0.18 [m] W_dot_fan=25 [W] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-509

DELTAT=10 [C] "Properties" c_p=1005 [J/kg-C] "Table A-2a" "Analysis" "(a)" Q_dot_total=W_dot_PCB+W_dot_fan Q_dot_total=m_dot*c_p*DELTAT "energy balance on air" "(b)" W_dot_fan=m_dot*c_p*DELTAT_fan f=DELTAT_fan/DELTAT*Convert(, %)

5-103 A desktop computer is to be cooled safely by a fan in hot environments and high elevations. The air flow rate of the fan and the diameter of the casing are to be determined. Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The specific heat of air at the average temperature of Tavg  (45  60) / 2  52.5 C  325.5 K is c p  1.0065 kJ / kg  C . The gas constant for air is R  0.287 kJ / kg  K (Table A-2).

Analysis The fan selected must be able to meet the cooling requirements of the computer at worst conditions. Therefore, we assume air to enter the computer at 66.63 kPa and 45C, and leave at 60C. We take the air space in the computer as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the required mass flow rate of air to absorb heat at a rate of 60 W is determined to be Q = mc p (Tout - Tin ) ® m =

Q 60 W = = 0.00397 kg/s = 0.238 kg/min c p (Tout - Tin ) (1006.5 J/kg.°C)(60-45)°C

The density of air entering the fan at the exit and its volume flow rate are

P 66.63 kPa = = 0.6972 kg/m 3 3 RT (0.287 kPa.m /kg ×K)(60+273)K

m 0.238 kg/min = = 0.341 m 3 / min r 0.6972 kg/m3

For an average exit velocity of 110 m / min , the diameter of the casing of the fan is determined from V = AcV =

p D2 V ® D= 4

4V = pV

(4)(0.341 m3 /min) = 0.063 m = 6.3 cm p (110 m/min)

1-510

T_in=45 [C] T_out=60 [C] elevation=3400 [m] P=66.63 [kPa] Vel=110 [m/min] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.0065 [kJ/kg-K] "at (45+60)/2=52.5 C=325.5 K, Table A-2b" "Analysis" Q_dot=m_dot*c_p*(T_out-T_in) "energy balance on air" rho=P/(R*(T_out+273)) Vol_dot=m_dot/rho*Convert(kg/s, kg/min) Vol_dot=A_c*Vel A_c=pi*D^2/4

5-104 A desktop computer is to be cooled safely by a fan in hot environments and high elevations. The air flow rate of the fan and the diameter of the casing are to be determined. Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The specific heat of air at the average temperature of Tave  (45  60) / 2  52.5 C is c p  1.0065 kJ / kg  C The gas constant for air is R  0.287 kJ / kg  K (Table A-2). Analysis The fan selected must be able to meet the cooling requirements of the computer at worst conditions. Therefore, we assume air to enter the computer at 66.63 kPa and 45C, and leave at 60C. We take the air space in the computer as the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the required mass flow rate of air to absorb heat at a rate of 100 W is determined to be Q = mc p (Tout - Tin )

¾¾ ®

Q 100 W = = 0.006624 kg/s = 0.397 kg/min c p (Tout - Tin ) (1006.5 J/kg ×°C)(60-45)°C

The density of air entering the fan at the exit and its volume flow rate are

P 66.63 kPa = = 0.6972 kg/m 3 RT (0.287 kPa ×m 3 /kg ×K)(60+273)K

m 0.397 kg/min = = 0.57 m 3 / min r 0.6972 kg/m 3

For an average exit velocity of 110 m / min , the diameter of the casing of the fan is determined from V = AcV =

p D2 V ¾¾ ® D= 4

4V = pV

(4)(0.57 m3 /min) = 0.081 m = 8.1 cm p (110 m/min)

1-511

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot=0.100 [kW] T_in=45 [C] T_out=60 [C] elevation=3400 [m] P=66.63 [kPa] Vel=110 [m/min] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.0065 [kJ/kg-K] "at (45+60)/2=52.5 C=325.5 K, Table A-2b" "Analysis" Q_dot=m_dot*c_p*(T_out-T_in) "energy balance on air" rho=P/(R*(T_out+273)) Vol_dot=m_dot/rho*Convert(kg/s, kg/min) Vol_dot=A_c*Vel A_c=pi*D^2/4

5-105 A room is to be heated by an electric resistance heater placed in a duct in the room. The power rating of the electric heater and the temperature rise of air as it passes through the heater are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, and thus heat transfer through it is negligible. 5 No air leaks in and out of the room. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The specific heats of air at room temperature are c p  1.005 and cv  0.718 kJ / kg  K (Table A-2).

Analysis (a) The total mass of air in the room is

V = 4´ 5´ 6 m3 = 120 m3 m=

P1V (98 kPa)(120 m3 ) = = 142.3 kg RT1 (0.287 kPa ×m3 /kg ×K)(288 K)

We first take the entire room as our system, which is a closed system since no mass leaks in or out. The power rating of the electric heater is determined by applying the conservation of energy relation to this constant volume closed system: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in + Wfan,in - Qout = D U

(since D KE = D PE = 0)

1-512

D t (We,in + Wfan,in - Qout ) = mcv ,avg (T2 - T1 ) Solving for the electrical work input gives

We,in = Qout - Wfan,in + mcv (T2 - T1 ) / D t = (150/60 kJ/s) - (0.2 kJ/s) + (142.3 kg)(0.718 kJ/kg ×°C)(25 - 15)°C/(25´ 60 s)

= 2.981 kW (b) We now take the heating duct as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m. The energy balance for this adiabatic steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

We,in + Wfan,in + mh1 = mh2 (since Q = D ke @D pe @0) We,in + Wfan,in = m(h2 - h1 ) = mc p (T2 - T1 )

Thus, D T = T2 - T1 =

We,in + Wfan,in mc p

(2.981 + 0.2) kJ/s = 4.75°C 40/60 kg/s )(1.005 kJ/kg ×K ) (

EES (Engineering Equation Solver) SOLUTION "Given" V=4[m]*5[m]*6[m] T1=15 [C] P1=98 [kPa] Q_dot_out=150[kJ/min]*Convert(kJ/min, kJ/s) W_dot_fan_in=0.200 [kW] m_dot=40[kg/min]*Convert(kg/min, kg/s) time=25*60 [s] T2=25 [C] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-C] "Table A-2a" c_v=0.718 [kJ/kg-C] "Table A-2a" "Analysis" "(a)" m=(P1*V)/(R*(T1+273)) (W_dot_e_in+W_dot_fan_in-Q_dot_out)*time=m*c_v*(T2-T1) "energy balance on the room" "(b)" W_dot_e_in+W_dot_fan_in=m_dot*c_p*DELTAT "energy balance on the heating duct"

5-106 A house is heated by an electric resistance heater placed in a duct. The power rating of the electric heater is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-513

Properties The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  K (Table A-2)

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

We,in + Wfan,in + mh1 = Qout + mh2 (since D ke @D pe @0) We,in + Wfan,in = Qout + m(h2 - h1 ) = Qout + mc p (T2 - T1 ) Substituting, the power rating of the heating element is determined to be

We,in = Qout + mc p D T - Wfan,in = (0.3 kJ/s) + (0.6 kg/s)(1.005 kJ/kg ×°C)(7°C) - 0.3 kW

= 4.22 kW

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_fan_in=0.300 [kW] m_dot=0.6 [kg/s] DELTAT=7 [C] Q_dot_out=0.300 [kW] "Properties" c_p=1.005 [kJ/kg-C] "Table A-2a" "Analysis" W_dot_e_in+W_dot_fan_in-Q_dot_out=m_dot*c_p*DELTAT "energy balance on the heating duct"

5-107 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The rate at which heat needs to be removed from the oil to keep its temperature constant is to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the roll are constant. 3 Kinetic and potential energy changes are negligible

1-514

Properties The properties of the steel plate are given to be   7854 kg / m3 and c p  0.434 kJ / kg  C . Analysis The mass flow rate of the sheet metal through the oil bath is m = r V = r wtV = (7854 kg/m3 )(2 m)(0.005 m)(10 m/min) = 785.4 kg/min

We take the volume occupied by the sheet metal in the oil bath to be the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2 (since D ke @D pe @ 0)

Qout = mc p (T1 - T2 ) Then the rate of heat transfer from the sheet metal to the oil bath becomes

Qout = mc p (Tin - Tout )metal = (785.4 kg/min)(0.434 kJ/kg.°C)(820 - 51.1)°C = 262,090 kJ/min = 4368 kW This is the rate of heat transfer from the metal sheet to the oil, which is equal to the rate of heat removal from the oil since the oil temperature is maintained constant.

EES (Engineering Equation Solver) SOLUTION "Given" w=2 [m] t=0.005 [m] T1=820 [C] T2=51.1 [C] T_oil=45 [C] Vel=10[m/min]*Convert(m/min, m/s) rho=7854 [kg/m^3] c_p=0.434 [kJ/kg-C] "Analysis" V_dot=w*t*Vel m_dot=rho*V_dot Q_dot=m_dot*c_p*(T1-T2) "energy balance on air"

5-108 Problem 5-107 is reconsidered. The effect of the moving velocity of the steel plate on the rate of heat transfer from the oil bath as the velocity varies from 5 to 50 m/min is to be investigated. Tate of heat transfer is to be plotted against the plate velocity. Analysis The problem is solved using EES, and the solution is given below. Vel = 10 [m/min] T_bath = 45 [C] T_1 = 820 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-515

T_2 = 51.1 [C] rho = 7854 [kg/m^3] c_p = 0.434 [kJ/kg-C] width = 2 [m] thick = 0.5 [cm] Vol_dot = width*thick*convert(cm,m)*Vel/convert(min,s) m_dot = rho*Vol_dot E_dot_metal_in = E_dot_metal_out E_dot_metal_in=m_dot*h_1 E_dot_metal_out=m_dot*h_2+Q_dot_metal_out h_1 = c_p*T_1 h_2 = c_p*T_2 Q_dot_oil_out = Q_dot_metal_out

Qoilout

Vel

[kW] 218.3 436.6 654.9 873.2 1091 1310 1528 1746 1965 2183

[m / min] 5 10 15 20 25 30 35 40 45 50

5-109E Water is heated in a parabolic solar collector. The required length of parabolic collector is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat loss from the tube is negligible so that the entire solar energy incident on the tube is transferred to the water. 3 Kinetic and potential energy changes are negligible Properties The specific heat of water at room temperature is c p  1.00 Btu / lbm. F (Table A-3E).

Analysis We take the thin aluminum tube to be the system, which is a control volume. The energy balance for this steadyflow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0)

1-516

Then the total rate of heat transfer to the water flowing through the tube becomes

Qtotal = mc p (Te - Ti ) = (4 lbm/s)(1.00 Btu/lbm.°F)(180 - 55)°F = 500 Btu/s = 1,800,000 Btu/h The length of the tube required is

EES (Engineering Equation Solver) SOLUTION "Given" T1=55 [F] T2=180 [F] m_dot=4 [lbm/s] D=1.25[in]*Convert(in, ft) Q_dot=400 [Btu/h-ft] "Properties" c_p=1.0 [Btu/lbm-F] "Table A-3E" "Analysis" Q_dot_total=m_dot*c_p*(T2-T1)*Convert(Btu/s, Btu/h) L=Q_dot_total/Q_dot

5-110 Heat is supplied to the argon as it flows in a heater. The exit temperature of argon and the volume flow rate at the exit are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Properties The gas constant of argon is 0.2081 kPa  m3 / kg  K . The constant pressure specific heat of air at room temperature is c p  0.5203 kJ / kg  C (Table A-2a).

Analysis (a) We take the pipe(heater) in which the argon is heated as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 + Qin = mh2

Qin = m(h2 - h1 )

Qin = mc p (T2 - T1 ) Substituting and solving for the exit temperature,

1-517

T2 = T1 +

Qin 150 kW = 300 K + = 346.2 K = 73.2°C mc p (6.24 kg/s)(0.5203 kJ/kg ×K)

(b) The exit specific volume and the volume flow rate are

v2 =

RT2 (0.2081 kPa ×m3 /kg ×K)(346.2 K) = = 0.7204 m3 /kg P2 100 kPa

V2 = mv 2 = (6.24 kg/s)(0.8266 m3 /kg) = 4.50 m 3 / s

EES (Engineering Equation Solver) SOLUTION "Given" T1=300 [K] P=100 [kPa] m_dot=6.24 [kg/s] Q_dot_in=150 [kW] "Properties" R=0.2081 [kJ/kg-K] "Table A-2a" c_p=0.5203 [kJ/kg-K] "Table A-2a" "Analysis" T2=T1+Q_dot_in/(m_dot*c_p) v2=(R*T2)/P Vol2_dot=m_dot*v2

5-111 Steam pipes pass through an unheated area, and the temperature of steam drops as a result of heat losses. The mass flow rate of steam and the rate of heat loss from are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions involved.

Properties From the steam tables (Table A-6),

P1 = 2 MPa ïüï v1 = 0.12551 m3 /kg ý T1 = 300°C ïïþ h1 = 3024.2 kJ/kg P2 = 1800 kPa ïüï ý h2 = 2911.7 kJ/kg T2 = 250°C ïïþ Analysis (a) The mass flow rate of steam is determined directly from m=

1 1 ép (0.08 m)2 ù(2.5 m/s ) = 0.4005 kg / s AV 1 1 = ú û v1 0.12551 m 3 /kg êë

(b) We take the steam pipe as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m. The energy balance for this steady-flow system can be expressed in the rate form as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-518

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2

(since W @D ke @D pe @ 0)

Qout = m(h1 - h2 )

Substituting, the rate of heat loss is determined to be Qloss = (0.4005 kg/s)(3024.2 - 2911.7) kJ/kg = 45.1 kJ / s

EES (Engineering Equation Solver) SOLUTION "Given" D1=0.16 [m] P1=2000 [kPa] T1=300 [C] Vel1=2.5 [m/s] D2=0.14 [m] P2=1800 [kPa] T2=250 [C] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, P=P1) h1=enthalpy(Fluid$, T=T1, P=P1) h2=enthalpy(Fluid$, T=T2, P=P2) "Analysis" "(a)" A1=pi*D1^2/4 m_dot=1/v1*A1*Vel1 "(b)" Q_dot_out=m_dot*(h1-h2) "energy balance on steam pipe"

5-112 R-134a is condensed in a condenser. The heat transfer per unit mass is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions.

Analysis We take the pipe in which R-134a is condensed as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

1-519

Qout = m(h1 - h2 )

qout = h1 - h2 The enthalpies of R-134a at the inlet and exit of the condenser are (Table A-12, A-13).

P1 = 900 kPa üïï ý h1 = 295.15 kJ/kg ïïþ T1 = 60°C P2 = 900 kPa ü ïï ý h2 = h f @900 kPa = 101.62 kJ/kg ïïþ x= 0 Substituting,

qout = 295.15 - 101.62 = 193.5 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P=900 [kPa] T1=60 [C] x2=0 "Properties" Fluid$='R134a' h1=enthalpy(Fluid$, P=P, T=T1) h2=enthalpy(Fluid$, P=P, x=x2) "Analysis" q_out=h1-h2 "energy balance on steam pipe"

5-113 A hair dryer consumes 1200 W of electric power when running. The inlet volume flow rate and the exit velocity of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The power consumed by the fan and the heat losses are negligible. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  K (Table A-2)

Analysis We take the hair dryer as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as

1-520

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

We,in + mh1 = mh2 (since Qout @D ke @D pe @0) We,in = m(h2 - h1 ) = mc p (T2 - T1 )

Substituting, the mass and volume flow rates of air are determined to be m=

v1 =

We,in c p (T2 - T1 )

1.2 kJ/s

(1.005 kJ/kg ×°C)(47 - 22)°C

= 0.04776 kg/s

3 RT1 (0.287 kPa ×m /kg ×K )(295 K ) = = 0.8467 m3 /kg P1 (100 kPa )

V1 = mv1 = (0.04776 kg/s)(0.8467 m3 /kg ) = 0.0404 m 3 / s (b) The exit velocity of air is determined from the mass balance m1 = m2 = m to be

v2 =

RT2 (0.287 kPa ×m3 /kg ×K)(320 K) = = 0.9184 m3 /kg P2 100 kPa

1 A2V2 v2

¾¾ ®

V2 =

mv 2 (0.04776 kg/s)(0.9184 m3 /kg) = = 7.31 m / s A2 60´ 10- 4 m2

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_e_in=1.2 [kW] P1=100 [kPa] T1=22 [C] T2=47 [C] A2=60E-4 [m^2] "Properties" C_p=CP(air, T=T_ave) C_v=CV(air, T=T_ave) T_ave=1/2*(T1+T2) R=C_p-C_v "Analysis" "(a)" W_dot_e_in=m_dot*C_p*(T2-T1) "energy balance on hair dryer" v1=(R*(T1+273))/P1 Vol_dot_1=m_dot*v1 "(b)" P2=P1 v2=(R*(T2+273))/P2 Vel2=(m_dot*v2)/A2

5-114 Problem 5-113 is reconsidered. The effect of the exit cross-sectional area of the hair drier on the exit velocity as the exit area varies from 25 cm 2 to 75 cm2 is to be investigated. The exit velocity is to be plotted against the exit crosssectional area.

1-521

Analysis The problem is solved using EES, and the results are tabulated and plotted below. R=0.287 [kPa-m^3/kg-K] P= 100 [kPa] T_1 = 22 [C] T_2= 47 [C] A_2 = 60 [cm^2] A_1 = 53.35 [cm^2]

W_dot_ele=1200 [W] E_dot_in = E_dot_out E_dot_in = W_dot_ele*convert(W,kW) + m_dot_1*(h_1+Vel_1^2/2*convert(m^2/s^2,kJ/kg)) E_dot_out = m_dot_2*(h_2+Vel_2^2/2*convert(m^2/s^2,kJ/kg)) h_2 = enthalpy(air, T = T_2) h_1= enthalpy(air, T = T_1) V_dot_1 = m_dot_1*v_1 P*v_1=R*(T_1+273) V_dot_2 = m_dot_2*v_2 P*v_2=R*(T_2+273) m_dot_1 = m_dot_2 Vel_1=V_dot_1/(A_1*convert(cm^2,m^2)) Vel_2=V_dot_2/(A_2*convert(cm^2,m^2))

A 2 [cm 2 ] 25 30 35 40 45 50 55 60 65 70 75

Vel2 [m / s] 17.46 14.57 12.51 10.95 9.738 8.768 7.973 7.31 6.749 6.267 5.85

1-522 18

Vel2 [m/s]

Set A1 = 53.35 cm2

14 12 10 8 6 4 20

A2 [cm2]

5-115E The ducts of an air-conditioning system pass through an unconditioned area. The inlet velocity and the exit temperature of air are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 There are no work interactions involved. Properties The gas constant of air is 0.3704 psia. ft 3 / lbm.R (Table A-1E). The constant pressure specific heat of air at room temperature is c p  0.240 Btu / lbm  R (Table A-2E)

Analysis (a) The inlet velocity of air through the duct is

V1 =

V1 V 450 ft 3 /min = 12 = = 825 ft / min A1 p r p (5/12 ft)2

Then the mass flow rate of air becomes v1 =

RT1 (0.3704 psia ×ft /lbm ×R )(510 R ) = = 12.6 ft 3 /lbm P1 (15 psia )

V1 450 ft 3 /min = = 35.7 lbm/min = 0.595 lbm / s v1 12.6 ft 3 /lbm

(b) We take the air-conditioning duct as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

1-523

T2 = T1 +

Qin 2 Btu/s = 50 o F + = 64.0 o F mc p (0.595 lbm/s)(0.24 Btu/lbm ×o F)

Ein = Eout

Qin + mh1 = mh2

(since W @D ke @D pe @ 0)

Qin = m(h2 - h1 ) = mc p (T2 - T1 )

EES (Engineering Equation Solver) SOLUTION "Given" P1=15 [psia] T1=50 [F] Vol_dot_1=450 [ft^3/min] D=10[in]*Convert(in, ft) Q_dot_in=2 [Btu/s] "Properties" Fluid$='air' C_p=CP(Fluid$, T=80[F]) C_v=CV(Fluid$, T=80[F]) R=(C_p-C_v)*Convert(Btu/lbm-R, psia-ft^3/lbm-R) h1=enthalpy(air, T=T1) "Analysis" "(a)" A=pi*D^2/4 Vel1=Vol_dot_1/A v1=(R*(T1+460))/P1 m_dot=Vol_dot_1/v1*Convert(lbm/min, lbm/s) "(b)" Q_dot_in=m_dot*(h2-h1) "energy balance on hair dryer" T2=temperature(Fluid$, h=h2)

5-116 Steam flows through a non-constant cross-section pipe. The inlet and exit velocities of the steam are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy change is negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

Analysis We take the pipe as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as Mass balance: min - mout = D msystem 0 (steady) = 0

1-524

min = mout ¾ ¾ ® A1

V1 V p D12 V1 p D22 V2 = A1 1 ¾ ¾ ® = v1 v1 4 v1 4 v2

Energy balance: Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

V12 V2 = h2 + 2 (since Q @W @D ke @D pe @ 0) 2 2 The properties of steam at the inlet and exit are (Table A-6) h1 +

3 P1 = 200 kPa ïü ïý v1 = 1.0805 m /kg T1 = 200°C ïïþ h1 = 2870.7 kJ/kg

P2 = 150 kPa ü ïï v 2 = 1.2855 m3 /kg ý ïïþ h2 = 2772.9 kJ/kg T1 = 150°C

Assuming inlet diameter to be 1.8 m and the exit diameter to be 1.0 m, and substituting,

V1 V2 p (1.8 m)2 p (1.0 m)2 = 3 4 4 (1.0805 m /kg) (1.2855 m3 /kg) 2870.7 kJ/kg +

(1)

ö ö V12 æ V22 æ çç 1 kJ/kg ÷ çç 1 kJ/kg ÷ = 2772.9 kJ/kg + ÷ 2 2÷ 2 2÷ ÷ ç ç 2 è1000 m /s ø 2 è1000 m /s ø

(2)

There are two equations and two unknowns. Solving equations (1) and (2) simultaneously using an equation solver such as EES, the velocities are determined to be

V1  118.8 m / s V2  458.0 m / s

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=200 [kPa] T_1=200 [C] P_2=150 [kPa] T_2=150 [C] D_1=1.8 [m] "assumed, it does not matter what value is assumed" D_2=1 [m] "PROPERTIES" Fluid$='Steam_iapws' h_1=enthalpy(Fluid$, T=T_1, P=P_1) v_1=volume(Fluid$, T=T_1, P=P_1) h_2=enthalpy(Fluid$, T=T_2, P=P_2) v_2=volume(Fluid$, T=T_2, P=P_2) "ANALYSIS" h_1+(Vel_1^2/2)*Convert(m^2/s^2, kJ/kg)=h_2+(Vel_2^2/2)*Convert(m^2/s^2, kJ/kg) "energy balance" Vel_1/Vel_2=(D_2/D_1)^2*(v_1/v_2) "from the mass balance"

Charging and Discharging Processes 5-117 An insulated rigid tank is evacuated. A valve is opened, and air is allowed to fill the tank until mechanical equilibrium is established. The final temperature in the tank is to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-525

This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The device is adiabatic and thus heat transfer is negligible. Properties The specific heat ratio for air at room temperature is k = 1.4 (Table A-2).

Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

mi = m2

(since mout = minitial = 0)

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

mi hi = m2u2 (since Q @W @ Eout = Einitial = ke @ pe @0) Combining the two balances:

u2 = hi

cv T2 = c pTi

T2 = (c p / cv )Ti = kTi

Substituting,

T2 = 1.4´ 290 K = 406 K = 133°C

EES (Engineering Equation Solver) SOLUTION "Given" P_i=95 [kPa] T_i=(17+273) [K] P2=95 [kPa] "Properties" k=1.4 "Table A-2a" "Analysis" T2=k*T_i T2_C=ConvertTemp(K, C, T2)

5-118 Steam in a supply line is allowed to enter an initially evacuated tank. The temperature of the steam in the supply line and the flow work are to be determined.

1-526

Analysis Flow work of the steam in the supply line is converted to sensible internal energy in the tank. That is,

hline = utank where Ptank = 4 MPa ïü ïý u = 3189.5 kJ/kg (Table A-6) tank Ttank = 550°C ïïþ

Now, the properties of steam in the line can be calculated ïü ïý Tline = 389.5°C (Table A-6) hline = 3189.5 kJ/kgïïþ uline = 2901.5 kJ/kg Pline = 4 MPa

The flow work per unit mass is the difference between enthalpy and internal energy of the steam in the line

wflow = hline - uline = 3189.5 - 2901.5 = 288 kJ / kg

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_line=4000 [kPa] P_tank=4000 [kPa] T_tank=550 [C] "ANALYSIS" Fluid$='Steam_iapws' u_tank=intenergy(Fluid$, T=T_tank, P=P_tank) h_line=u_tank T_line=temperature(Fluid$, P=P_line, h=h_line) u_line=intenergy(Fluid$, P=P_line, h=h_line) w_flow=h_line-u_line

5-119 Steam flowing in a supply line is allowed to enter into an insulated tank until a specified state is achieved in the tank. The mass of the steam that has entered and the pressure of the steam in the supply line are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid entering the tank remains constant. 2 Kinetic and potential energies are negligible.

1-527

Properties The initial and final properties of steam in the tank are (Tables A-5 and A-6) 3 ïü ïý v1 = 0.19436 m /kg x1 = 1 (sat. vap.)ïïþ u1 = 2582.8 kJ/kg

P1 = 1 MPa

3 P2 = 2 MPa ü ïïý v 2 = 0.12551 m /kg T1 = 300°C ïïþ u2 = 2773.2 kJ/kg

Ein - Eout

mi = m2 - m1 =

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

mi hi = m2u2 - m1u1 (since Q @ ke @ pe @0) The initial and final masses and the mass that has entered are

m1 =

V 2 m3 = = 10.29 kg v1 0.19436 m3 /kg

m2 =

V 2 m3 = = 15.94 kg v 2 0.12551 m3 /kg

mi = m2 - m1 = 15.94 - 10.29 = 5.645 kg Substituting,

(5.645 kg)hi = (15.94 kg)(2773.2 kJ/kg) - (10.29 kg)(2582.8 kJ/kg) ¾ ¾ ® hi = 3120.3 kJ/kg The pressure in the supply line is hi = 3120.3 kJ/kgü ïï ý Pi = 8931 kPa (determined from EES) ïïþ Ti = 400°C

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol=2 [m^3] P_1=1000 [kPa] x_1=1 P_2=2000 [kPa] T_2=300 [C] T_i=400 [C] "PROPERTIES" Fluid$='Steam_iapws' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-528

u_1=intenergy(Fluid$, P=P_1, x=x_1) v_1=volume(Fluid$, P=P_1, x=x_1) u_2=intenergy(Fluid$, P=P_2, T=T_2) v_2=volume(Fluid$, P=P_2, T=T_2) "ANALYSIS" m_i=m_2-m_1 "mass balance" m_i*h_i=m_2*u_2-m_1*u_1 "energy balance" m_1=Vol/v_1 m_2=Vol/v_2 P_i=pressure(Fluid$, T=T_i, h=h_i) "pressure of steam in supply line"

5-120 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1).

Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® mi = m2

(since mout = minitial = 0)

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 (since W @ Eout = Einitial = ke @ pe @0) Combining the two balances: Qin = m2 (u2 - hi )

where

m2 =

P2V (100 kPa)(0.035 m3 ) = = 0.04134 kg RT2 (0.287 kPa ×m3 /kg ×K)(295 K)

1-529 A-17 Ti = T2 = 295 K ¾ Table ¾¾ ¾®

hi = 295.17 kJ/kg u2 = 210.49 kJ/kg

Substituting,

Qin  (0.04134 kg)(210.49  295.17)kJ / kg  3.50 kJ or

Qout = 3.50 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the direction.

EES (Engineering Equation Solver) SOLUTION "Given" V=35[L]*Convert(L, m^3) P_i=100 [kPa] T_i=22 [C] T2=T_i P2=P_i "Properties" Fluid$='air' R=0.287 [kJ/kg-K] "Table A-2a" h_i=enthalpy(Fluid$, T=T_i) u2=intenergy(Fluid$, T=T2) "Analysis" m_i=m2 "mass balance" Q_in+m_i*h_i=m2*u2 "energy balance" m2=(P2*V)/(R*(T2+273))

5-121 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line, and air is allowed to enter the tank until mechanical equilibrium is established. The mass of air that entered and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the tank (will be verified).

Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The properties of air are (Table A-17) Ti = 295 K

¾¾ ®

hi = 295.17 kJ/kg

T1 = 295 K

¾¾ ®

u1 = 210.49 kJ/kg

1-530

T2 = 350 K

¾¾ ®

u2 = 250.02 kJ/kg

Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

mi = m2 - m1

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 - m1u1 (since W @ ke @ pe @0) The initial and the final masses in the tank are

m1 =

P1V (100 kPa)(2 m3 ) = = 2.362 kg RT1 (0.287 kPa ×m3 /kg ×K)(295 K)

m2 =

P2V (600 kPa)(2 m3 ) = = 11.946 kg RT2 (0.287 kPa ×m3 /kg ×K)(350 K)

Then from the mass balance,

mi = m2 - m1 = 11.946 - 2.362 = 9.584 kg (b) The heat transfer during this process is determined from

Qin = - mi hi + m2u2 - m1u1 = - (9.584 kg )(295.17 kJ/kg )+ (11.946 kg )(250.02 kJ/kg )- (2.362 kg )(210.49 kJ/kg )

= - 339 kJ ®

Qout = 339 kJ

Discussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction.

EES (Engineering Equation Solver) SOLUTION "Given" V=2 [m^3] P1=100 [kPa] T1=22 [C] P_i=600 [kPa] T_i=22 [C] P2=P_i T2=77 [C] "Properties" Fluid$='air' R=0.287 [kJ/kg-K] "Table A-2a" h_i=enthalpy(Fluid$, T=T_i) u1=intenergy(Fluid$, T=T1) u2=intenergy(Fluid$, T=T2) "Analysis" "(a)" m_i=m2-m1 "mass balance" m1=(P1*V)/(R*(T1+273)) m2=(P2*V)/(R*(T2+273)) "(b)" Q_in+m_i*h_i=m2*u2-m1*u1 "energy balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-531

5-122 A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vapor is allowed to escape at constant pressure until the temperature rises to 500C. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the steam leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

Properties The properties of water are (Tables A-4 through A-6)

P1 = 2 MPa ïüï v1 = 0.12551 m3 /kg ý T1 = 300°C ïïþ u1 = 2773.2 kJ/kg, h1 = 3024.2 kJ/kg

P2 = 2 MPa üïï v 2 = 0.17568 m3 /kg ý T2 = 500°C ïïþ u2 = 3116.9 kJ/kg, h2 = 3468.3 kJ/kg Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 (since W @ ke @ pe @0) The state and thus the enthalpy of the steam leaving the tank is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values. Thus,

h + h2 3024.2 + 3468.3 kJ/kg he @ 1 = = 3246.2 kJ/kg 2 2 The initial and the final masses in the tank are m1 =

V1 0.2 m3 = = 1.594 kg v1 0.12551 m3 /kg

m2 =

V2 0.2 m3 = = 1.138 kg v 2 0.17568 m3 /kg

Then from the mass and energy balance relations,

me = m1 - m2 = 1.594 - 1.138 = 0.456 kg Qin = me he + m2u2 - m1u1 = (0.456 kg )(3246.2 kJ/kg )+ (1.138 kg )(3116.9 kJ/kg )- (1.594 kg )(2773.2 kJ/kg )

1-532

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.2 [m^3] P1=2000 [kPa] T1=300 [C] P2=P1 T2=500 [C] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) u1=intenergy(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) v2=volume(Fluid$, P=P2, T=T2) u2=intenergy(Fluid$, P=P2, T=T2) h2=enthalpy(Fluid$, P=P2, T=T2) h_e=1/2*(h1+h2) "assumed" "Analysis" m_e=m1-m2 "mass balance" m1=Vol/v1 m2=Vol/v2 Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance"

5-123E A rigid tank initially contains saturated water vapor. The tank is connected to a supply line, and water vapor is allowed to enter the tank until one-half of the tank is filled with liquid water. The final pressure in the tank, the mass of steam that entered, and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

Properties The properties of water are (Tables A-4E through A-6E)

T1 = 300°F ïüï v1 = v g @ 300° F = 6.4663 ft 3 /lbm ý sat. vapor ïïþ u1 = u g @ 300° F = 1099.8 Btu/lbm T2 = 300° F üïï v f = 0.01745, v g = 6.4663 ft 3 /lbm ý sat. mixture ïïþ u f = 269.51, u g = 1099.8 Btu/lbm Pi = 200 psia üïï ý hi = 1210.9 Btu/lbm Ti = 400°F ïïþ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-533

Mass balance: min - mout = Δmsystem ® mi = m2 - m1 Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 - m1u1 (since W @ ke @ pe @0) (a) The tank contains saturated mixture at the final state at 250F, and thus the exit pressure is the saturation pressure at this temperature,

P2 = Psat @ 300° F = 67.03 psia (b) The initial and the final masses in the tank are

m1 =

V 3 ft 3 = = 0.464 lbm v1 6.4663 ft 3 /lbm

m2 = m f + mg =

Vf vf

Vg vg

1.5 ft 3 1.5 ft 3 . + = 85.97 + 0.232 = 86.20 lbm 0.01745 ft 3 /lbm 6.4663 ft 3 /lbm

Then from the mass balance

mi = m2 - m1 = 86.20 - 0.464 = 85.74 lbm (c) The heat transfer during this process is determined from the energy balance to be

Qin = - mi hi + m2u2 - m1u1 = - (85.74 lbm)(1210.9 Btu/lbm )+ 23,425 Btu - (0.464 lbm)(1099.8 Btu/lbm)

= - 80,900 Btu ® Qout = 80,900 Btu since

U 2 = m2u2 = m f u f + mg ug = 85.97´ 269.51 + 0.232´ 1099.8 = 23,425 Btu

Discussion A negative result for heat transfer indicates that the assumed direction is wrong, and should be reversed.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=3 [ft^3] T1=300 [F] x1=1 P_i=200 [psia] T_i=400 [F] T2=300 [F] Vol2_f=Vol/2 "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) v2_f=volume(Fluid$, T=T2, x=0) v2_g=volume(Fluid$, T=T2, x=1) h_i=enthalpy(Fluid$, T=T_i, P=P_i) "Analysis" "(a)" P2=pressure(Fluid$, T=T2, x=1) "mixture" "(b)" m1=Vol/v1 m2_f=Vol2_f/v2_f m2_g=(Vol-Vol2_f)/v2_g © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-534

m2=m2_f+m2_g m_i=m2-m1 "mass balance" "(c)" x2=m2_g/m2 u2=intenergy(Fluid$, T=T2, x=x2) Q_in+m_i*h_i=m2*u2-m1*u1 "energy balance"

5-124E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is opened, and air is allowed to escape at constant temperature until the pressure inside drops to 25 psia. The amount of electrical work transferred is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats. Properties The gas constant of air is R  0.3704 psia.ft 3 / lbm  R (Table A-1E). The properties of air are (Table A-17E) Ti = 580 R ¾ ¾ ® hi = 138.66 Btu/lbm T1 = 580 R ¾ ¾ ® u1 = 98.90 Btu/lbm T2 = 580 R ¾ ¾ ® u2 = 98.90 Btu/lbm

min - mout = D msystem ® me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in - me he = m2u2 - m1u1 (since Q @ ke @ pe @ 0)

The initial and the final masses of air in the tank are m1 =

(50 psia )(40 ft 3 ) P1V = = 9.310 lbm RT1 (0.3704 psia ×ft 3 /lbm ×R )(580 R )

m2 =

(25 psia )(40 ft ) P2V = = 4.655 lbm RT2 (0.3704 psia ×ft 3 /lbm ×R )(580 R ) 3

1-535

Then from the mass and energy balances,

me = m1 - m2 = 9.310 - 4.655 = 4.655 kg We,in = me he + m2u2 - m1u1 = (4.655 lbm)(138.66 Btu/lbm)+ (4.655 lbm)(98.90 Btu/lbm)- (9.310 lbm)(98.90 Btu/lbm)

= 185 Btu

EES (Engineering Equation Solver) SOLUTION "Given" V=40 [ft^3] P1=50 [psia] T1=(120+460) [R] P2=25 [psia] T2=T1 T_e=T1 "Properties" Fluid$='air' h_e=enthalpy(Fluid$, T=T_e) u1=intenergy(Fluid$, T=T1) u2=intenergy(Fluid$, T=T2) R=0.3704 [psia-ft^3/lbm-R] "from Table A-1E of the text" "Analysis" m_e=m1-m2 "mass balance" m1=(P1*V)/(R*T1) m2=(P2*V)/(R*T2) W_e_in=m_e*h_e+m2*u2-m1*u1 "energy balance"

5-125 A pressure cooker is initially half-filled with liquid water. If the pressure cooker is not to run out of liquid water for 1 h, the highest rate of heat transfer allowed is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.

Properties The properties of water are (Tables A-4 through A-6) P1 = 175 kPa

v f = 0.001057 m3 /kg, v g = 1.0037 m3 /kg u f = 486.82 kJ/kg, u g = 2524.5 kJ/kg

1-536

P2 = 175 kPa üïï v 2 = v g @ 175 kPa = 1.0036 m3 /kg ý ïïþ u2 = u g @ 175 kPa = 2524.5 kJ/kg sat. vapor Pe = 175 kPa ü ïï ý he = hg @ 175 kPa = 2700.2 kJ/kg ïïþ sat. vapor Analysis We take the cooker as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniformflow system can be expressed as

Mass balance: min - mout = Δmsystem ® me = m1 - m2 Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 (since W @ ke @ pe @0) The initial mass, initial internal energy, and final mass in the tank are m1 = m f + mg =

Vf vf

Vg vg

0.002 m3 0.002 m3 + = 1.893 + 0.002 = 1.895 kg 3 0.001057 m /kg 1.0036 m3 /kg

U1 = m1u1 = m f u f + mg u g = (1.893)(486.82)+ (0.002)(2524.5) = 926.6 kJ

m2 =

V 0.004 m3 = = 0.003985 kg v 2 1.0037 m3 /kg

Then from the mass and energy balances,

me = m1 - m2 = 1.895 - 0.003985 = 1.891 kg Qin = me he + m2u2 - m1u1 = (1.891 kg)(2700.2 kJ/kg)+ (0.003985 kg)(2524.5 kJ/kg )- 926.6 kJ = 4188 kJ

Thus, Q=

Q 4188 kJ = = 0.931 kW D t (75´ 60) s

EES (Engineering Equation Solver) SOLUTION "Given" Vol=4[L]*Convert(L, m^3) P1=175 [kPa] Vol1_f=0.50*Vol P_e=P1 x_e=1 P2=P1 x2=1 DELTAt=(75*60) [s] "Properties" Fluid$='steam_iapws' v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) h_e=enthalpy(Fluid$, P=P_e, x=x_e) "Analysis" m1_f=Vol1_f/v1_f © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-537

m1_g=(Vol-Vol1_f)/v1_g m1=m1_f+m1_g m2=Vol/v2 m_e=m1-m2 "mass balance" x1=m1_g/m1 u1=intenergy(Fluid$, P=P1, x=x1) Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance" Q_dot=Q_in/DELTAt

5-126 R-134a from a tank is discharged to an air-conditioning line in an isothermal process. The final quality of the R-134a in the tank and the total heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the exit remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.

- me = m2 - m1 me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 Qin = m2u2 - m1u1 + me he Combining the two balances:

Qin = m2u2 - m1u1 + (m1 - m2 )he The initial state properties of R-134a in the tank are

v = 0.0008260 m3 /kg T1 = 24°C üïï 1 ý u1 = 84.44 kJ/kg ïïþ x= 0 he = 84.98 kJ/kg

(Table A-11)

Note that we assumed that the refrigerant leaving the tank is at saturated liquid state, and found the exiting enthalpy accordingly. The volume of the tank is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-538

V = m1v1 = (5 kg)(0.0008260 m3 /kg) = 0.004130 m3 The final specific volume in the container is

v2 =

V 0.004130 m3 = = 0.01652 m3 /kg m2 0.25 kg

The final state is now fixed. The properties at this state are (Table A-11) v2 - v f 0.01652 - 0.0008260 x = = = 0.5056 ïü ïý 2 v fg 0.031869 - 0.0008260 3 v 2 = 0.01652 m /kg ïïþ u2 = u f + x2 u fg = 84.44 kJ/kg + (0.5056)(158.68 kJ/kg) = 164.67 kJ/kg T2 = 24°C

Substituting into the energy balance equation,

Qin = m2u2 - m1u1 + (m1 - m2 )he = (0.25 kg)(164.67 kJ/kg) - (5 kg)(84.44 kJ/kg) + (4.75 kg)(84.98 kJ/kg)

= 22.6 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m1=5 [kg] T1=24 [C] x1=0 m2=0.25 [kg] T2=T1 "Analysis" Fluid$='R134a' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) h_e=enthalpy(Fluid$, T=T1, x=x1) Vol=m1*v1 v2=Vol/m2 v2_f=volume(Fluid$, T=T2, x=0) v2_g=volume(Fluid$, T=T2, x=1) x2=(v2-v2_f)/(v2_g-v2_f) u2_f=intenergy(Fluid$, T=T2, x=0) u2_g=intenergy(Fluid$, T=T2, x=1) u2_fg=u2_g-u2_f u2=u2_f+x2*u2_fg m_e=m1-m2 "mass balance" Q_in=m2*u2-m1*u1+m_e*h_e "energy balance"

5-127E Oxygen is supplied to a medical facility from 10 compressed oxygen tanks in an isothermal process. The mass of oxygen used and the total heat transfer to the tanks are to be determined. Assumptions 1 This is an unsteady process but it can be analyzed as a uniform-flow process. 2 Oxygen is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved.

1-539

Properties The gas constant of oxygen is R  0.3353 psia.ft 3 / lbm  R (Table A-1E). The specific heats of oxygen at room temperature are c p  0.219 Btu / lbm  R and cv  0.157 Btu / lbm  R (Table A-2Ea). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as

Mass balance: min - mout = D msystem

- me = m2 - m1

me = m1 - m2 Energy balance: Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 Qin = m2u2 - m1u1 + me he Qin = m2 cv T2 - m1cv T1 + me c pTe

Combining the two balances:

Qin = m2 cv T2 - m1cv T1 + (m1 - m2 )c pTe The initial and final masses, and the mass used are

m1 =

P1V (1500 psia)(15 ft 3 ) = = 124.3 lbm RT1 (0.3353 psia ×ft 3 /lbm ×R)(80 + 460 R)

m2 =

P2V (300 psia)(15 ft 3 ) = = 24.85 lbm RT2 (0.3353 psia ×ft 3 /lbm ×R)(80 + 460 R)

me = m1 - m2 = 124.3 - 24.85 = 99.41lbm Substituting into the energy balance equation, Qin = m2 cv T2 - m1cv T1 + me c pTe

= (24.85)(0.157)(540) - (124.3)(0.157)(540) + (99.41)(0.219)(540)

= 3328 Btu

EES (Engineering Equation Solver) SOLUTION "Given" Vol=10*1.5 [ft^3] P1=1500 [psia] T=(80+460) [R] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-540

P2=300 [psia] "Properties" R=0.3353 [psia-ft^3/lbm-R] "Table A-1E" c_p=0.219 [Btu/lbm-R] "Table A-2a" c_v=0.157 [Btu/lbm-R] "Table A-2a" "Analysis" m1=P1*Vol/(R*T) m2=P2*Vol/(R*T) m_e=m1-m2 Q_in=m2*c_v*T-m1*c_v*T+m_e*c_p*T

5-128 A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The mass of the R-134a that entered and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

Properties The properties of refrigerant are (Tables A-11 through A-13)

P1 = 0.8 MPa üïï v1 = v g @ 0.8 MPa = 0.02565 m 3 /kg ý ïïþ u1 = u g @ 0.8 MPa = 246.82 kJ/kg sat.vapor P2 = 1.2 MPa üïï v 2 = v f @ 1.2 MPa = 0.0008935 m3 /kg ý ïïþ u2 = u f @ 1.2 MPa = 116.72 kJ/kg sat. liquid

Pi = 1.2 MPa ïüï ý hi = h f @ 40°C = 108.28 kJ/kg ïïþ Ti = 40°C Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® mi = m2 - m1 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 - m1u1 (since W @ ke @ pe @0) (a) The initial and the final masses in the tank are © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-541

m1 =

V1 0.05 m3 = = 1.952 kg v1 0.02565 m3 /kg

m2 =

V2 0.05 m3 = = 55.96 kg v 2 0.0008935 m3 /kg

Then from the mass balance

mi = m2 - m1 = 55.96 - 1.952 = 54.01 kg (c) The heat transfer during this process is determined from the energy balance to be

Qin = - mi hi + m2u2 - m1u1 = - (54.01 kg )(108.28 kJ/kg )+ (55.96 kg )(116.72 kJ/kg )- (1.952 kg )(246.82 kJ/kg )

= 202 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.05 [m^3] P1=800 [kPa] x1=1 P_i=1200 [kPa] T_i=40 [C] P2=1200 [kPa] x2=0 "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) h_i=enthalpy(Fluid$, T=T_i, x=0) "approximated as saturated liquid at the given temperature" "Analysis" "(a)" m_i=m2-m1 "mass balance" m1=Vol/v1 m2=Vol/v2 "(b)" Q_in+m_i*h_i=m2*u2-m1*u1 "energy balance"

5-129 A rigid tank initially contains saturated liquid-vapor mixture of refrigerant-134a. A valve at the bottom of the tank is opened, and liquid is withdrawn from the tank at constant pressure until no liquid remains inside. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

1-542

Properties The properties of R-134a are (Tables A-11 through A-13) P1 = 800 kPa ®

v f = 0.0008457 m3 /kg, v g =0.025645 m 3 /kg u f = 94.80 kJ/kg, ug =246.82 kJ/kg

P2 = 800 kPa ïüï v 2 = v g @ 800 kPa = 0.025645 m3 /kg ý ïïþ u2 = u g @ 800 kPa = 246.82 kJ/kg sat. vapor Pe = 800 kPa ïü ïý h = h f @ 800 kPa = 95.48 kJ/kg ïïþ e sat. liquid Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

me = m1 - m2

Energy balance: Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = me he + m2u2 - m1u1 (since W @ke @ pe @0) The initial mass, initial internal energy, and final mass in the tank are m1 = m f + mg =

Vf vf

Vg vg

0.12´ 0.25 m3 0.12´ 0.75 m3 + = 35.474 + 3.509 = 38.98 kg 3 0.0008457 m /kg 0.025645 m 3 /kg

U1 = m1u1 = m f u f + mg u g = (35.474)(94.80)+ (3.509)(246.82) = 4229.0 kJ

m2 =

V 0.1 2 m3 = = 4.679 kg v 2 0.025645 m3 /kg

Then from the mass and energy balances,

me = m1 - m2 = 38.98 - 4.679 = 34.30 kg Qin = (34.30 kg )(95.48 kJ/kg )+ (4.679 kg )(246.82 kJ/kg )- 4229.0 kJ = 201 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.12 [m^3] P1=800 [kPa] Vol1_f=0.25*Vol P_e=P1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-543

x_e=0 P2=P1 x2=1 "Properties" Fluid$='R134a' v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) h_e=enthalpy(Fluid$, P=P_e, x=x_e) "Analysis" m1_f=Vol1_f/v1_f m1_g=(Vol-Vol1_f)/v1_g m1=m1_f+m1_g m2=Vol/v2 m_e=m1-m2 "mass balance" x1=m1_g/m1 u1=intenergy(Fluid$, P=P1, x=x1) Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance"

5-130 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of the mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

Properties The properties of water are (Tables A-4 through A-6) v = v f @ 200 o C = 0.001157 m 3 /kg T1 = 200°C ïü ïý 1 sat. liquid ïïþ u1 = u = 850.46 kJ/kg o f @ 200 C

Te = 200 C üïï ý he = h f @ 200 o C = 852.26 kJ/kg sat. liquid ïïþ Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® me = m1 - m2

1-544

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = me he + m2u2 - m1u1 (since W @ke @ pe @0) The initial and the final masses in the tank are

m1 =

V1 0.3 m3 = = 259.4 kg v1 0.001157 m3 /kg

m2 = 12 m1 = 12 (259.4 kg ) = 129.7 kg

Then from the mass balance,

me = m1 - m2 = 259.4 - 129.7 = 129.7 kg Now we determine the final internal energy,

v2 = x2 =

V 0.3 m3 = = 0.002313 m3 /kg m2 129.7 kg v2 - v f v fg

0.002313 - 0.001157 = 0.009171 0.12721- 0.001157

ïüï ý u2 = u f + x2 u fg = 850.46 + (0.009171)(1743.7 ) = 866.46 kJ/kg x2 = 0.009171 ïïþ

T2 = 200°C

Then the heat transfer during this process is determined from the energy balance by substitution to be Q = (129.7 kg )(852.26 kJ/kg )+ (129.7 kg )(866.46 kJ/kg )- (259.4 kg )(850.46 kJ/kg )

= 2308 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.3 [m^3] T1=200 [C] x1=0 T_e=T1 x_e=x1 T2=T1 m2=m1/2 "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) h_e=enthalpy(Fluid$, T=T_e, x=x_e) "Analysis" m_e=m1-m2 "mass balance" m1=Vol/v1 v2=Vol/m2 u2=intenergy(Fluid$, T=T2, v=v2) Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance"

5-131 A hot-air balloon is considered. The final volume of the balloon and work produced by the air inside the balloon as it expands the balloon skin are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-545

1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There is no heat transfer. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). Analysis The specific volume of the air at the entrance and exit, and in the balloon is v=

RT (0.287 kPa ×m3 /kg ×K)(35 + 273 K) = = 0.8840 m3 /kg P 100 kPa

The mass flow rate at the entrance is then

mi =

AV (1 m2 )(2 m/s) i i = = 2.262 kg/s v 0.8840 m3 /kg

while that at the outlet is

me =

AeVe (0.5 m2 )(1 m/s) = = 0.5656 kg/s v 0.8840 m3 /kg

Applying a mass balance to the balloon, min - mout = D msystem

mi - me = m2 - m1 m2 - m1 = (mi - me )D t = [(2.262 - 0.5656) kg/s](2´ 60 s) = 203.6 kg The volume in the balloon then changes by the amount

D V = (m2 - m1 )v = (203.6 kg)(0.8840 m3 /kg) = 180 m 3 and the final volume of the balloon is

V2 = V1 + D V = 75 + 180 = 255 m3 In order to push back the boundary of the balloon against the surrounding atmosphere, the amount of work that must be done is æ 1 kJ ö ÷ Wb,out = PD V = (100 kPa)(180 m3 ) çç ÷= 18,000 kJ ÷ çè1 kPa ×m3 ø

EES (Engineering Equation Solver) SOLUTION "Given" A_e=0.5 [m^2] A_i=1 [m^2] DELTAt=2*60 [s] P_i=100 [kPa] T_i=(35+273) [K] Vel_i=2 [m/s] Vel_e=1 [m/s] Vol1=75 [m^3] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" v_i=(R*T_i)/P_i m_dot_i=(A_i*Vel_i)/v_i v_e=v_i m_dot_e=(A_e*Vel_e)/v_e DELTAm=(m_dot_i-m_dot_e)*DELTAt "DELTAm = m2-m1" DELTAVol=DELTAm*v_i © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-546

Vol2=Vol1+DELTAVol W_b_out=P_i*DELTAVol

5-132 A balloon is initially filled with helium gas at atmospheric conditions. The tank is connected to a supply line, and helium is allowed to enter the balloon until the pressure rises from 100 to 125 kPa. The final temperature in the balloon is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Helium is an ideal gas with constant specific heats. 3 The expansion process is quasi-equilibrium. 4 Kinetic and potential energies are negligible. 5 There are no work interactions involved other than boundary work. 6 Heat transfer is negligible. Properties The gas constant of helium is R  2.0769 kJ / kg  K (Table A-1). The specific heats of helium are c p = 5.1926 and cv  3.1156 kJ / kg  K (Table A-2a). Analysis We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

mi = m2 - m1

Energy balance: Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

mi hi = Wb,out + m2u2 - m1u1 (since Q @ ke @ pe @ 0)

m1 =

(100 kPa )(40 m 3 ) P1V1 = = 6.641 kg RT1 (2.0769 kPa ×m 3 /kg ×K )(290 K )

P1 V1 = P2 V2 m2 =

¾¾ ®

V2 =

P2 125 kPa V1 = (40 m3 ) = 50 m3 P1 100 kPa

(125 kPa )(50 m3 ) P2V2 3009.3 = = kg 3 RT2 (2.0769 kPa ×m /kg ×K )(T2 K ) T2

Then from the mass balance, mi = m2 - m1 =

3009.3 - 6.641 kg T2

1-547

Noting that P varies linearly with V, the boundary work done during this process is

(100 + 125) kPa P1 + P2 (V2 - V1 ) = (50 - 40) m3 = 1125 kJ 2 2 Using specific heats, the energy balance relation reduces to Wb =

Wb,out = mi c pTi - m2 cv T2 + m1cv T1 Substituting, æ3009.3 ö 3009.3 ÷(5.1926)(298)1125 = ççç - 6.641÷ (3.1156 )T2 + (6.641)(3.1156 )(290 ) ÷ ÷ çè T2 T2 ø

It yields

T2 = 315 K

EES (Engineering Equation Solver) SOLUTION "Given" V1=40 [m^3] P1=100 [kPa] T1=(17+273) [K] P_i=125 [kPa] T_i=(25+273) [K] P2=P_i "Properties" R=2.0769 [kJ/kg-K] "Table A-2a" c_p=5.1926 [kJ/kg-K] "Table A-2a" c_v=3.1156 [kJ/kg-K] "Table A-2a" "Analysis" m_i=m2-m1 "mass balance" m1=(P1*V1)/(R*T1) V2=(P2/P1)*V1 m2=(P2*V2)/(R*T2) W_b_out=(P1+P2)/2*(V2-V1) "area under the process curve since P varies linearly with V" m_i*c_p*T_i=W_b_out+m2*c_v*T2-m1*c_v*T1 "energy balance"

5-133 An insulated rigid tank initially contains helium gas at high pressure. A valve is opened, and half of the mass of helium is allowed to escape. The final temperature and pressure in the tank are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the helium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats. Properties The specific heat ratio of helium is k = 1.667 (Table A-2). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as

1-548

Mass balance:

min - mout = D msystem ® me = m1 - m2 m2 = 12 m1 (given) ¾ ¾ ® me = m2 = 12 m1

Energy balance: Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- me he = m2u2 - m1u1 (since W @Q @ ke @ pe @0) Note that the state and thus the enthalpy of helium leaving the tank is changing during this process. But for simplicity, we assume constant properties for the exiting steam at the average values. Combining the mass and energy balances:

0 = 12 m1he + 12 m1u2 - m1u1

0 = he + u2 - 2u1 or 0 = c p

Dividing by m1 / 2 Dividing by cv :

0 = k (T1 + T2 )+ 2T2 - 4T1

Solving for T2 :

T2 =

T1 + T2 + cv T2 - 2cv T1 2 since k = c p / cv

(4 - k ) (4 - 1.667) T1 = (403 K ) = 257 K (2 + k ) (2 + 1.667)

The final pressure in the tank is P1V m RT mT 1 257 = 1 1 ¾¾ ® P2 = 2 2 P1 = (3000 kPa ) = 956 kPa P2V m2 RT2 m1T1 2 403

EES (Engineering Equation Solver) SOLUTION "Given" V=0.15 [m^3] P1=3000 [kPa] T1=(130+273) [K] Ratio_m=0.5 "m2/m1=0.5" "Properties" k=1.667 "Table A-2a" "Analysis" T2=(4-k)/(2+k)*T1 P2=Ratio_m*T2/T1*P1

5-134 A vertical cylinder initially contains air at room temperature. Now a valve is opened, and air is allowed to escape at constant pressure and temperature until the volume of the cylinder goes down by half. The amount air that left the cylinder and the amount of heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-549

3 4 5

There are no work interactions other than boundary work. Air is an ideal gas with constant specific heats. The direction of heat transfer is to the cylinder (will be verified).

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1).

Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® me = m1 - m2 Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + Wb,in - me he = m2u2 - m1u1 (since ke @ pe @ 0) The initial and the final masses of air in the cylinder are

(300 kPa )(0.2 m3 ) P1V1 m1 = = = 0.714 kg RT1 (0.287 kPa ×m3 /kg ×K )(293 K ) m2 =

(300 kPa )(0.1 m3 ) P2V2 = = 0.357 kg = 12 m1 3 RT2 (0.287 kPa ×m /kg ×K )(293 K )

Then from the mass balance,

me = m1 - m2 = 0.714 - 0.357 = 0.357 kg (b) This is a constant pressure process, and thus the Wb and the U terms can be combined into H to yield

Q = me he + m2 h2 - m1h1 Noting that the temperature of the air remains constant during this process, we have

hi = h1 = h2 = h. Also, me = m2 = 12 m1 .

Thus, Q = (12 m1 + 12 m1 - m1 )h = 0

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.2 [m^3] T1=(20+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-550

P1=300 [kPa] V2=0.5*V1 T2=T1 P2=P1 T_e=T1 "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) h2=enthalpy(Fluid$, T=T2) h_e=enthalpy(Fluid$, T=T_e) R=0.287 [kPa-m^3/kg-K] "Analysis" "(a)" m_e=m1-m2 "mass balance" m1=(P1*V1)/(R*T1) m2=(P2*V2)/(R*T2) "(b)" Q_in=m_e*h_e+m2*h2-m1*h1 "energy balance, since U2-U1+W_b_out=H2-H1 for a constant pressure process"

5-135 A vertical piston-cylinder device contains air at a specified state. Air is allowed to escape from the cylinder by a valve connected to the cylinder. The final temperature and the boundary work are to be determined. Properties The gas constant of air is R  0.287 kJ / kg  K (Table A-1).

Analysis The initial and final masses in the cylinder are

m1 =

PV1 (600 kPa)(0.25 m3 ) = = 0.9121 m3 RT1 (0.287 kJ/kg.K)(300 + 273 K)

m2 = 0.25m1 = 0.25(0.9121 kg) = 0.2280 kg Then the final temperature becomes

T2 =

PV2 (600 kPa)(0.05 m3 ) = = 458 K m2 R (0.2280 kg)(0.287 kJ/kg.K)

Noting that pressure remains constant during the process, the boundary work is determined from

Wb = P(V1 - V2 ) = (600 kPa)(0.25 - 0.05)m3 = 120 kJ

EES (Engineering Equation Solver) SOLUTION "GIVEN" V_1=0.25 [m^3] P_1=600 [kPa] T_1=300[c]+273 [K] P_2=P_1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-551

V_2=0.05 [m^3] "PROPERTIES" R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Air) "ANALYSIS" m_1=(P_1*V_1)/(R*T_1) m_2=0.25*m_1 T_2=(P_2*V_2)/(m_2*R) W_b=P_1*(V_1-V_2)

5-136 A cylinder initially contains superheated steam. The cylinder is connected to a supply line, and is superheated steam is allowed to enter the cylinder until the volume doubles at constant pressure. The final temperature in the cylinder and the mass of the steam that entered are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved other than boundary work. 5 The device is insulated and thus heat transfer is negligible.

Properties The properties of steam are (Tables A-4 through A-6)

P1 = 500 kPa üïï v1 = 0.42503 m3 /kg ý T1 = 200°C ïïþ u1 = 2643.3 kJ/kg Pi = 1 MPa ïüï ý hi = 3158.2 kJ/kg Ti = 350°C ïïþ Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: Energy balance:

min - mout = D msystem ® mi = m2 - m1 Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

mi hi = Wb,out + m2u2 - m1u1 (since Q @ ke @ pe @ 0)

1-552

0 = Wb,out - (m2 - m1 )hi + m2u2 - m1u1

The boundary work done during this process is 2 æ 1 kJ ÷ ö Wb, out = ò P dV = P (V2 - V1 ) = (500 kPa )(0.02 - 0.01)m 3 çç ÷= 5 kJ çè1 kPa ×m3 ÷ 1 ø

The initial and the final masses in the cylinder are

m1 =

V1 0.01 m3 = = 0.0235 kg v1 0.42503 m3 /kg

m2 =

V2 0.02 m3 = v2 v2

Substituting, æ0.02 ö 0.02 ÷(3158.2)+ 0 = 5 - ççç - 0.0235÷ u2 - (0.0235)(2643.3) ÷ ÷ çè v 2 v2 ø

Then by trial and error (or using EES program), T2  262 C and v2  0.4858 m3 / kg

(b) The final mass in the cylinder is

m2 =

V2 0.02 m3 = = 0.0412 kg v 2 0.4858 m3 /kg

Then,

mi  m2  m1  0.0412  0.0235  0.0176 kg

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.01 [m^3] T1=200 [C] P1=500 [kPa] P_i=1000 [kPa] T_i=350 [C] Vol2=2*Vol1 "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, P=P1) u1=intenergy(Fluid$, T=T1, P=P1) h_i=enthalpy(Fluid$, T=T_i, P=P_i) "Analysis" "(a) and (b)" m_i=m2-m1 "mass balance" m_i*h_i=W_b_out+m2*u2-m1*u1 "energy balance" W_b_out=P1*(Vol2-Vol1) m1=Vol1/v1 m2=Vol2/v2 v2=volume(Fluid$, T=T2, P=P2) u2=intenergy(Fluid$, T=T2, P=P2) P2=P1

5-137 Steam at a specified state is allowed to enter a piston-cylinder device in which steam undergoes a constant pressure expansion process. The amount of mass that enters and the amount of heat transfer are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-553

Properties The properties of steam at various states are (Tables A-4 through A-6) v1 =

V1 0.1 m3 = = 0.16667 m3 /kg m1 0.6 kg

P2 = P1 ïü ïý u = 2004.4 kJ/kg 1 v1 = 0.16667 m /kgïïþ P1 = 800 kPa

3 P2 = 800 kPa ü ïïý v 2 = 0.29321 m /kg T2 = 250 °C ïïþ u2 = 2715.9 kJ/kg

Pi = 5 MPa ïü ïý h = 3434.7 kJ/kg i Ti = 500 °C ïïþ

min - mout = D msystem

mi = m2 - m1

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - Wb,out + mi hi = m2u2 - m1u1

(since ke @ pe @ 0)

Noting that the pressure remains constant, the boundary work is determined from Wb,out = P(V2 - V1 ) = (800 kPa)(2´ 0.1- 0.1) m 3 = 80 kJ

The final mass and the mass that has entered are

m2 =

V2 0.2 m3 = = 0.6821 kg v 2 0.29321 m3 /kg

mi = m2 - m1 = 0.6821- 0.6 = 0.08210 kg (b) Finally, substituting into energy balance equation

1-554

Qin - 80 kJ + (0.08210 kg)(3434.7 kJ/kg) = (0.6821 kg)(2715.9 kJ/kg) - (0.6 kg)(2004.4 kJ/kg) Qin = 448 kJ

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_1=0.6 [kg] Vol_1=0.1 [m^3] Vol_2=2*Vol_1 P_1=800 [kPa] P_i=5000 [kPa] T_i=500 [C] T_2=250 [C] "PROPERTIES" Fluid$='Steam_iapws' v_1=Vol_1/m_1 P_2=P_1 u_1=intenergy(Fluid$, P=P_1, v=v_1) u_2=intenergy(Fluid$, P=P_2, T=T_2) v_2=volume(Fluid$, P=P_2, T=T_2) h_i=enthalpy(Fluid$, P=P_i, T=T_i) "ANALYSIS" m_i=m_2-m_1 "mass balance" Q_in+m_i*h_i-W_b_out=m_2*u_2-m_1*u_1 "energy balance" W_b_out=P_1*(Vol_2-Vol_1) "boundary work" Vol_2=m_2*v_2

5-138 The air in an insulated, rigid compressed-air tank is released until the pressure in the tank reduces to a specified value. The final temperature of the air in the tank is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The tank is well-insulated, and thus there is no heat transfer. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The specific heats of air at room temperature are c p  1.005 kJ / kg  K and cv  0.718 kJ / kg  K (Table A-2a).

1-555

Mass balance:

min - mout = D msystem - me = m2 - m1 me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

me he = m2u2 - m1u1 0 = m2u2 - m1u1 + me he 0 = m2 cv T2 - m1cv T1 + me c pTe

Combining the two balances:

0 = m2 cv T2 - m1cv T1 + (m1 - m2 )c pTe The initial and final masses are given by

m1 =

P1V (2400 kPa)(0.5 m3 ) = = 14.27 kg RT1 (0.287 kPa ×m3 /kg ×K)(20 + 273 K)

m2 =

P2V (2000 kPa)(0.5 m3 ) 3484 = = 3 RT2 (0.287 kPa ×m /kg ×K)T2 T2

The temperature of air leaving the tank changes from the initial temperature in the tank to the final temperature during the discharging process. We assume that the temperature of the air leaving the tank is the average of initial and final temperatures in the tank. Substituting into the energy balance equation gives 0 = m2 cv T2 - m1cv T1 + (m1 - m2 )c pTe

æ æ293 + T2 ö÷ 3484 3484 ö÷ ÷(1.005) çç (0.718)T2 - (14.27)(0.718)(293) + ççç14.27 ÷ ÷ çè 2 ÷ ÷ ç ø T2 T2 ø è

whose solution by trial-error or by an equation solver such as EES is

T2 = 278 K = 5.0°C

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.5 [m^3] P1=2400 [kPa] T1=(20+273) [K] P2=2000 [kPa] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_v=0.718 [kJ/kg-K] "Table A-2a" "Analysis" m_e=m1-m2 "mass balance" m1=(P1*Vol)/(R*T1) m2=(P2*Vol)/(R*T2) T_e=(T1+T2)/2 0=m2*c_v*T2-m1*c_v*T1+m_e*c_p*T_e "energy balance" T2_C=ConvertTemp(K, C, T2)

1-556

Review Problems 5-139 The rate of accumulation of water in a pool and the rate of discharge are given. The rate supply of water to the pool is to be determined. Assumptions 1 Water is supplied and discharged steadily. 2 The rate of evaporation of water is negligible. 3 No water is supplied or removed through other means. Analysis The conservation of mass principle applied to the pool requires that the rate of increase in the amount of water in the pool be equal to the difference between the rate of supply of water and the rate of discharge. That is, dmpool dt

= mi - me

mi =

dmpool dt

+ me

Vi =

dVpool dt

+ Ve

since the density of water is constant and thus the conservation of mass is equivalent to conservation of volume. The rate of discharge of water is Ve = AeVe = (p D 2 /4)Ve = [p (0.07 m) 2 /4](4 m/s) = 0.01539 m 3 /s

The rate of accumulation of water in the pool is equal to the cross-section of the pool times the rate at which the water level rises, dVpool dt

= Across-sectionVlevel = (6 m ´ 9 m)(0.025 m/min) = 1.35 m3 /min = 0.0225 m3 /s

Substituting, the rate at which water is supplied to the pool is determined to be Vi =

dVpool dt

+ Ve = 0.0225 + 0.01539 = 0.0379 m 3 / s

Therefore, water is supplied at a rate of 0.0379 m3 / s  37.9 L / s .

EES (Engineering Equation Solver) SOLUTION "Given" A_CrossSection=6[m]*9[m] D=0.07 [m] Vel_e=4 [m/s] Vel_level=2.5[cm/min]*Convert(cm/min, m/s) "Analysis" A_e=pi*D^2/4 V_dot_e=A_e*Vel_e V_dot=A_CrossSection*Vel_level V_dot_i=V_dot+V_dot_e

5-140 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. The mass flow rate of the plate is to be determined.

1-557

Assumptions The plate moves through the bath steadily. Properties The density of steel plate is given to be   7854 kg / m3 . Analysis The mass flow rate of the sheet metal through the oil bath is m = r V = r wtV = (7854 kg/m3 )(1 m)(0.005 m)(10 m/min) = 393 kg/min = 6.55 kg / s

Therefore, steel plate can be treated conveniently as a “flowing fluid” in calculations.

EES (Engineering Equation Solver) SOLUTION "Given" w=1 [m] t=0.005 [m] rho=7854 [kg/m^3] Vel=10[m/min]*Convert(m/min, m/s) "Analysis" m_dot=rho*w*t*Vel

5-141 Helium flows steadily in a pipe and heat is lost from the helium during this process. The heat transfer and the volume flow rate at the exit are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Properties The gas constant of helium is 2.0769 kPa  m3 / kg  K . The constant pressure specific heat of air at room temperature is c p  5.1926 kJ / kg  C (Table A-2a).

Analysis (a) We take the pipe in which the argon is heated as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = mh2 + Qout Qout = m(h1 - h2 )

1-558

Substituting, Qout = mc p (T1 - T2 ) = (8 kg/s)(5.1926 kJ/kg ×K)(427 - 27)K = 16, 620 kW

(b) The exit specific volume and the volume flow rate are

v2 =

RT2 (2.0769 kPa ×m3 /kg ×K)(300 K) = = 6.231 m3 /kg P2 100 kPa

V2 = mv 2 = (8 kg/s)(6.231 m3 /kg) = 49.85 m 3 / s

EES (Engineering Equation Solver) SOLUTION "Given" m_dot=8 [kg/s] T1=(427+273) [K] P1=100 [kPa] T2=(27+273) [K] P2=P1 "Properties" R=2.0769 [kJ/kg-K] "Table A-2a" c_p=5.1926 [kJ/kg-K] "Table A-2a" "Analysis" Q_dot_out=m_dot*c_p*(T1-T2) v2=(R*T2)/P2 Vol2_dot=m_dot*v2

5-142 Air is accelerated in a nozzle. The density of air at the nozzle exit is to be determined. Assumptions Flow through the nozzle is steady. Properties The density of air is given to be 4.18 kg / m3 at the inlet.

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . Then,

m1 = m2 r 1 AV 1 1 = r 2 A2V2 r2 =

A1 V1 120 m/s r1 = 2 (4.18 kg/m3 ) = 2.64 kg / m 3 A2 V2 380 m/s

Discussion Note that the density of air decreases considerably despite a decrease in the cross-sectional area of the nozzle.

EES (Engineering Equation Solver) SOLUTION "Given" rho_1=4.18 [kg/m^3] RatioA=2 "RatioA=A1/A2" Vel1=120 [m/s] Vel2=380 [m/s] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-559

rho_2=RatioA*(Vel1/Vel2)*rho_1

5-143 Water is boiled at Tsat = 100 C by an electric heater. The rate of evaporation of water is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the outer surfaces of the water tank are negligible. Properties The enthalpy of vaporization of water at 100C is hfg  2256.4 kJ / kg (Table A-4).

Analysis Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a unit mass of a liquid at a specified temperature, the rate of evaporation of water is determined to be

mevaporation =

We,boiling h fg

3 kJ/s 2256.4 kJ/kg

= 0.00133 kg / s = 4.79 kg / h

EES (Engineering Equation Solver) SOLUTION "Given" T_sat=100 [C] W_dot_e=3 [kJ/s] "Analysis" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_sat, x=0) h_g=enthalpy(Fluid$, T=T_sat, x=1) h_fg=h_g-h_f m_dot_evaporation=W_dot_e/h_fg*Convert(kg/s, kg/h)

5-144 An air compressor consumes 6.2 kW of power to compress a specified rate of air. The flow work required by the compressor is to be compared to the power used to increase the pressure of the air. Assumptions 1 Flow through the compressor is steady. 2 Air is an ideal gas.

1-560

Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). Analysis The specific volume of the air at the inlet is

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(20 + 273 K) = = 0.7008 m3 /kg P1 120 kPa

The mass flow rate of the air is

V1 0.015 m3 /s = = 0.02140 kg/s v1 0.7008 m3 /kg

Combining the flow work expression with the ideal gas equation of state gives the flow work as

wflow = P2v 2 - P1v1 = R(T2 - T1 ) = (0.287 kJ/kg ×K)(300 - 20)K = 80.36 kJ/kg The flow power is

Wflow = mwflow = (0.02140 kg/s)(80.36 kJ/kg) = 1.72 kW The remainder of compressor power input is used to increase the pressure of the air:

W = Wtotal,in - Wflow = 6.2 - 1.72 = 4.48 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol1_dot=0.015 [m^3/s] P1=120 [kPa] T1=(20+273) [K] P2=800 [kPa] T2=(300+273) [K] W_dot_in=6.2 [kW] "PROPERTIES" R=0.287 [kJ/kg-K] "Table A-2a" "ANALYSIS" v1=(R*T1)/P1 m_dot=Vol1_dot/v1 w_flow=R*(T2-T1) W_dot_flow=m_dot*w_flow W_dot_pressure_increase=W_dot_in-W_dot_flow

5-145 Steam expands in a turbine whose power production is 12,350 kW. The rate of heat lost from the turbine is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Properties From the steam tables (Tables A-6 and A-4) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-561

P1 = 1.6 MPa ü ïï ý h1 = 3146.0 kJ/kg T1 = 350°C ïïþ T2 = 30°C ü ïï ý h2 = 2555.6 kJ/ kg ïïþ x2 = 1

Analysis We take the turbine as the system, which is a control volume since mass crosses the boundary. Noting that there is one inlet and one exiti the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 = m2 h2 + Wout + Qout Qout = m(h1 - h2 ) - Wout Substituting,

Qout = (22 kg/s)(3146.0 - 2555.6) kJ/kg - 12,350 kW = 640 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=1600 [kPa] T1=350 [C] T2=30 [C] x2=1 m_dot=22 [kg/s] W_dot_out=12350 [kW] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$,T=T1,P=P1) h2=enthalpy(Fluid$,T=T2,x=x2) Q_dot_out=m_dot*(h1-h2)-W_dot_out

5-146E Refrigerant-134a is compressed steadily by a compressor. The mass flow rate of the refrigerant and the exit temperature are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties From the refrigerant tables (Tables A-11E through A-13E) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-562

P1 = 15 psia ü ïï v1 = 3.2551 ft 3 /lbm ý ïïþ h1 = 107.52 Btu/lbm T1 = 20°F

Analysis (a) The mass flow rate of refrigerant is

V1 10 ft 3 /s = = 3.072 lbm / s v1 3.2551 ft 3 /lbm

(b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Win + mh1 = mh2

(since Q @D ke @D pe @ 0)

Win = m(h2 - h1 )

Substituting,

æ0.7068 Btu/s ÷ ö ÷= (45 hp) çç (3.072 lbm/s)(h2 - 107.52) Btu/lbm ÷ ÷ çè 1 hp ø

h2 = 117.87 Btu/lbm Then the exit temperature becomes ü ïï ý h2 = 117.87 Btu/lbm ïïþ P2 = 100 psia

T2 = 95.7°F (Table A-13E)

EES (Engineering Equation Solver) SOLUTION "Given" P1=15 [psia] T1=20 [F] Vol_dot=10 [ft^3/s] P2=100 [psia] W_dot_in=45 [hp] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-563

"Analysis" "(a)" m_dot=Vol_dot/v1 "(b)" W_dot_in*Convert(hp, Btu/s)+m_dot*h1=m_dot*h2 "energy balance" T2=temperature(Fluid$, P=P2, h=h2)

5-147E Nitrogen gas flows through a long, constant-diameter adiabatic pipe. The velocities at the inlet and exit are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Nitrogen is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 The re is no heat transfer from the nitrogen. Properties The specific heat of nitrogen at the room temperature is c p  0.248 Btu / lbm  R (Table A-2Ea).

Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take the pipe as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = m(h2 +V22 /2) h1 + V12 / 2 = h2 +V22 /2

V12 - V22 = c p (T2 - T1 ) 2 Combining the mass balance and ideal gas equation of state yields m1 = m2 AV AV 1 1 = 2 2 v1 v2 V2 =

A1 v 2 v T P V1 = 2 V1 = 2 1 V1 A2 v1 v1 T1 P2

Substituting this expression for V2 into the energy balance equation gives 0.5

0.5 é ù é ù ê ú ê ú ê2c (T - T ) ú ê2(0.248)(70 - 120) æ25, 037 ft 2 /s 2 öú ê p 2 1 ú ÷ çç ê ÷ú = 515 ft / s V1 = ê = ú ÷ 2 2 ê ÷ú ç 1 Btu/lbm æ ö è ø ê æT2 P1 ö÷ ú 530 100 ê 1- ç ú ÷ ê1- çç ú ÷ ÷ ç ê ú çè580 50 ø÷ ê çè T1 P2 ø÷ ÷ú ê ë ûú ë û The velocity at the exit is

1-564

V2 =

T2 P1 530 100 V1 = 515 = 941 ft / s T1 P2 580 50

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=100 [psia] T1=(120+460) [R] P2=50 [psia] T2=(70+460) [R] "PROPERTIES" c_p=0.248 [Btu/lbm-R] "Table A-2Ea" "ANALYSIS" (Vel1^2-Vel2^2)/2*Convert(ft^2/s^2, Btu/lbm)=c_p*(T2-T1) "energy balance" Vel2=T2/T1*P1/P2*Vel1 "from mass balance"

5-148 Water at a specified rate is heated by an electrical heater. The current is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The heat losses from the water is negligible. Properties The specific heat and the density of water are taken to be c p  4.18 kJ / kg  C and   1 kg / L (Table A-3).

Analysis We take the pipe in which water is heated as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 + We,in = mh2 We,in = m(h2 - h1 ) VI = mc p (T2 - T1 )

The mass flow rate of the water is

m = r V = (1 kg/L)(0.1 L/s) = 0.1 kg/s Substituting into the energy balance equation and solving for the current gives I=

mc p (T2 - T1 ) V

ö (0.1 kg/s)(4.18 kJ/kg ×K)(30 - 18)K æ çç1000 VI ÷ ÷ ÷= 45.6 A ç è 1 kJ/s ø 110 V

1-565

EES (Engineering Equation Solver) SOLUTION "GIVEN" Voltage =110 [Volt] Vol_dot=0.1 [L/s] T1=18 [C] T2=30 [C] "PROPERTIES" rho=1 [kg/L] "Table A-3" c_p=4180 [J/kg-K] "Table A-3" "ANALYSIS" m_dot=rho*Vol_dot I=(m_dot*c_p*(T2-T1))/Voltage

5-149 A fan is powered by a 0.5 hp motor, and delivers air at a rate of 85 m3 / min . The highest possible air velocity at the fan exit is to be determined. Assumptions 1

This is a steady-flow process since there is no change with time at any point and thus D mCV = 0 and D ECV = 0 .

The inlet velocity and the change in potential energy are negligible, V1 @0 and D pe @0 .

3 4 5

There are no heat and work interactions other than the electrical power consumed by the fan motor. The efficiencies of the motor and the fan are 100% since best possible operation is assumed. Air is an ideal gas with constant specific heats at room temperature.

Properties The density of air is given to be   1.18 kg / m3 . The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2).

Analysis We take the fan-motor assembly as the system. This is a control volume since mass crosses the system boundary during the process. We note that there is only one inlet and one exit, and thus m1 = m2 = m . The velocity of air leaving the fan will be highest when all of the entire electrical energy drawn by the motor is converted to kinetic energy, and the friction between the air layers is zero. In this best possible case, no energy will be converted to thermal energy, and thus the temperature change of air will be zero, T2 = T1 . Then the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

1-566

We,in + mh1 = m(h2 +V22 /2) (since V1 @ 0 and D pe @ 0) Noting that the temperature and thus enthalpy remains constant, the relation above simplifies further to

We,in = mV22 /2 where

m = r V = (1.18 kg/m3 )(85 m3 /min)=100.3 kg/min = 1.672 kg/s Solving for V2 and substituting gives

V2 =

2We,in m

ö öæ1 m 2 / s 2 ÷ 2(0.5 hp) æ çç çç745.7 W ÷ ÷ ÷ = 21.1 m / s ÷ ÷ ç ÷ 1W ÷ 1.672 kg/s èç 1 hp øè ø

Discussion In reality, the velocity will be less because of the inefficiencies of the motor and the fan.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_e_in=0.5 [hp] V_dot=85 [m^3/min] rho=1.18 [kg/m^3] "Analysis" m_dot=rho*V_dot*Convert(kg/min, kg/s) W_dot_e_in*Convert(hp, W)=m_dot*Vel^2/2 "energy balance, since temperature and thus enthalpy remains constant"

5-150 Steam flows in an insulated pipe. The mass flow rate of the steam and the speed of the steam at the pipe outlet are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work and heat interactions.

Analysis We take the pipe in which steam flows as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = mh2 h1 = h2 The properties of the steam at the inlet and exit are (Table A-6)

1-567

ü ïï 3 ý v 2 = 0.23099 m /kg h2 = h1 = 2935.6 kJ/kgïïþ P2 = 1000 kPa

The mass flow rate is

AV p D12 V1 p (0.15 m)2 4 m/s 1 1 = = = 0.3674 kg / s v1 4 v1 4 0.19241 m3 /kg

The outlet velocity will then be

V2 =

mv 2 4mv 2 4(0.3674 kg/s)(0.23099 m3 /kg) = = = 10.8 m / s A2 p D22 p (0.10 m)2

EES (Engineering Equation Solver) SOLUTION "Given" P1=1200 [kPa] T1=250 [C] Vel1=4 [m/s] P2=1000 [kPa] D1=0.15 [m] D2=0.1 [m] "Analysis" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) h2=h1 v2=volume(Fluid$, P=P2, h=h2) A1=pi*D1^2/4 m_dot=(A1*Vel1)/v1 A2=pi*D2^2/4 Vel2=(m_dot*v2)/A2

5-151 Heat is lost from the steam flowing in a nozzle. The exit velocity and the mass flow rate are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy change is negligible. 3 There are no work interactions.

Analysis (a) We take the steam as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Energy balance:

Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

1-568

Ein = Eout 2ö æ æ V2ö ççh + V2 ÷ ÷ ÷ m çççh1 + 1 ÷ = m 2 ÷ ÷+ Qout ÷ 2ø 2 ø÷ èç èçç

since W @D pe @ 0)

or V2 =

2(h1 - h2 - qout )

The properties of steam at the inlet and exit are (Table A-6) P1 = 200 kPa ïü ïý h = 2769.1 kJ/kg 1 T1 = 150°C ïïþ

P2 = 75 kPa ïüï v 2 = 2.2172 m3 /kg ý ïïþ h2 = 2662.4 kJ/kg sat. vap. Substituting, V2 =

2(h1 - h2 - qout ) =

æ 1 kJ/kg ö ÷ 2(2769.1- 2662.4 - 26)kJ/kg çç ÷ ÷ = 401.7 m / s çè1000 m 2 /s 2 ø

(b) The mass flow rate of the steam is m=

1 1 A2V2 = (0.001 m 2 )(401.7 m/s) = 0.181 kg / s v2 2.2172 m3 /kg

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vel_1=0 [m/s] T_1=150 [C] P_1=200 [kPa] P_2=75 [kPa] x_2=1 q_out=26 [kJ/kg] A_2=0.001 [m^2] "ANALYSIS" Fluid$='Steam_iapws' h_1=enthalpy(Fluid$, T=T_1, P=P_1) v_2=volume(Fluid$, P=P_2, x=x_2) h_2=enthalpy(Fluid$, P=P_2, x=x_2) h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg)=h_2+Vel_2^2/2*Convert(m^2/s^2, kJ/kg)+q_out "energy balance on the nozzle" m_dot=1/v_2*A_2*Vel_2 "mass flow rate of steam"

5-152 Saturated steam at 1 atm pressure and thus at a saturation temperature of Tsat = 100 C condenses on a vertical plate maintained at 90C by circulating cooling water through the other side. The rate of condensation of steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 The steam condenses and the condensate drips off at 100C. (In reality, the condensate temperature will be between 90 and 100, and the cooling of the condensate a few C should be considered if better accuracy is desired).

1-569

Properties The enthalpy of vaporization of water at 1 atm (101.325 kPa) is hfg  2256.5 kJ / kg (Table A-5). Analysis The rate of heat transfer during this condensation process is given to be 180 kJ / s . Noting that the heat of vaporization of water represents the amount of heat released as a unit mass of vapor at a specified temperature condenses, the rate of condensation of steam is determined from mcondensation =

Q 180 kJ/s = = 0.0798 kg / s 2256.5 kJ/kg h fg

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_plate=90 [C] Q_dot=180 [kJ/s] "Analysis" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, P=P, x=0) h_g=enthalpy(Fluid$, P=P, x=1) h_fg=h_g-h_f m_dot_condensation=Q_dot/h_fg

5-153 Steam at a saturation temperature of Tsat = 40 C condenses on the outside of a thin horizontal tube. Heat is transferred to the cooling water that enters the tube at 25C and exits at 35C. The rate of condensation of steam is to be determined. Assumptions 1 Steady operating conditions exist. 2 Water is an incompressible substance with constant properties at room temperature. 3 The changes in kinetic and potential energies are negligible. Properties The properties of water at room temperature are   997 kg / m3 and c p  4.18 kJ / kg  C (Table A-3). The enthalpy of vaporization of water at 40C is hfg  2406.0 kJ / kg (Table A-4).

1-570

Analysis The mass flow rate of water through the tube is

mwater = r VAc = (997 kg/m3 )(2 m/s)[p (0.03 m) 2 / 4] = 1.409 kg/s Taking the volume occupied by the cold water in the tube as the system, which is a steady-flow control volume, the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = Qwater = mwater c p (T2 - T1 )

Then the rate of heat transfer to the water and the rate of condensation become Q = mc p (Tout - Tin ) = (1.409 kg/s)(4.18 kJ/kg ×°C)(35 - 25)°C = 58.9 kW

Q = mcond h fg ® mcond =

Q 58.9 kJ/s = = 0.0245 kg / s h fg 2406.0 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" T_steam=40 [C] L=5 [m] D=0.03 [m] T1=25 [C] T2=35 [C] Vel=2 [m/s] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_steam, x=0) h_g=enthalpy(Fluid$, T=T_steam, x=1) h_fg=h_g-h_f rho=997 [kg/m^3] "Table A-3" c_p=4.18 [kJ/kg-C] "Table A-3" "Analysis" A_c=pi*D^2/4 m_dot_water=rho*A_c*Vel Q_dot_water=m_dot_water*c_p*(T2-T1) Q_dot_condensation=Q_dot_water Q_dot_condensation=m_dot_cond*h_fg

1-571

5-154 The feedwater of a steam power plant is preheated using steam extracted from the turbine. The ratio of the mass flow rates of the extracted seam the feedwater are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.

Properties The enthalpies of steam and feedwater at are (Tables A-4 through A-6)

P1 = 1 MPa ïü ïý h = 2828.3 kJ/kg 1 T1 = 200°C ïïþ

P1 = 1 MPa ïü ïý h2 = h f @ 1 MPa = 762.51 kJ/kg T2 = 179.9°C sat. liquid ïïþ and

P3 = 2.5 MPa ïü ïý h @ h = 209.34 kJ/kg f @ 50 o C ïïþ 3 T3 = 50°C ïü ïý h @ h 4 f @ 170 ° C = 718.55 kJ/kg T4 = T2 - 10 @170°C ïïþ P4 = 2.5 MPa

min - mout = D msystemZ 0 (steady) = 0 ® min = mout ® m1 = m2 = ms and m3 = m4 = mfw Energy balance (for the heat exchanger):

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m3 h3 = m2 h2 + m4 h4 (since Q = W = D ke @D pe @ 0) Combining the two,

ms (h2 - h1 ) = m fw (h3 - h4 ) Dividing by m fw and substituting, ms h - h4 (718.55 - 209.34)kJ/kg = 3 = = 0.246 m fw h2 - h1 (2828.3 - 762.51)kJ/kg

1-572

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [kPa] T1=200 [C] P2=P1 x2=0 P3=2500 [kPa] T3=50 [C] T4=T2-10 P4=P3 "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, P=P1) h2=enthalpy(Fluid$, P=P2, x=x2) T2=temperature(Fluid$, P=P2, x=x2) h3=enthalpy(Fluid$, T=T3, x=0) h4=enthalpy(Fluid$, T=T4, x=0) "Analysis" m_dot_1=m_dot_2 "mass balance on steam" m_dot_3=m_dot_4 "mass balance on feedwater" m_dot_1*h1+m_dot_3*h3=m_dot_2*h2+m_dot_4*h4 "energy balance on heat exchanger" y=m_dot_1/m_dot_3 "where" m_dot_3=1 "assumed, it does not matter what value is assumed when calculating y"

5-155 Air is preheated by the exhaust gases of a gas turbine in a regenerator. For a specified heat transfer rate, the exit temperature of air and the mass flow rate of exhaust gases are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Heat loss from the regenerator to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 5 Exhaust gases can be treated as air. 6 Air is an ideal gas with variable specific heats. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The enthalpies of air are (Table A-17)

T1 = 550 K ® h1 = 555.74 kJ/kg T3 = 800 K ® h3 = 821.95 kJ/kg T4 = 600 K ® h4 = 607.02 kJ/kg

1-573

Analysis (a) We take the air side of the heat exchanger as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mair h1 = mair h2 (since W = D ke @D pe @ 0)

Qin = mair (h2 - h1 ) Substituting,

2700 kJ / s  (800 / 60 kg / s)  h2  554.71 kJ / kg   h2  757.58 kJ / kg Then from Table A-17 we read

T2 = 741 K (b) Treating the exhaust gases as an ideal gas, the mass flow rate of the exhaust gases is determined from the steady-flow energy relation applied only to the exhaust gases, Ein = Eout

mexhaust h3 = Qout + mexhaust h4 (since W = D ke @D pe @ 0)

Qout = mexhaust (h3 - h4 ) 2700 kJ/s = mexhaust (821.95 - 607.02) kJ/kg It yields

mexhaust = 12.6 kg / s

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [kPa] T1=550 [K] m_dot_air=800[kg/min]*Convert(kg/min, kg/s) Q_dot_air_in=2700 [kJ/s] P3=140 [kPa] T3=800 [K] P4=130 [kPa] T4=600 [K] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) h3=enthalpy(Fluid$, T=T3) h4=enthalpy(Fluid$, T=T4) "Analysis" "(a)" Q_dot_air_in+m_dot_air*h1=m_dot_air*h2 "energy balance on air" T2=temperature(Fluid$, h=h2) "(b)" Q_dot_exhaust_out=Q_dot_air_in m_dot_exhaust*h3=Q_dot_exhaust_out+m_dot_exhaust*h4 "energy balance on exhaust gases"

1-574

5-156 Cold water enters a steam generator at 20C, and leaves as saturated vapor at Tsat = 200 C . The fraction of heat used to preheat the liquid water from 20C to saturation temperature of 200C is to be determined. Assumptions 1 Steady operating conditions exist. 2 Heat losses from the steam generator are negligible. 3 The specific heat of water is constant at the average temperature. Properties The heat of vaporization of water at 200C is hfg  1939.8 kJ / kg (Table A-4), and the specific heat of liquid water is c  4.18 kJ / kg. C (Table A-3).

Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a specified temperature. Using the average specific heat, the amount of heat transfer needed to preheat a unit mass of water from 20C to 200C is qpreheating = cD T

= (4.18 kJ/kg ×°C)(200 - 20)°C

=752.4 kJ/kg and

qtotal = qboiling + qpreheating = 1939.8 + 364.1 = 2692.2 kJ/kg Therefore, the fraction of heat used to preheat the water is

Fraction to preheat =

qpreheating qtotal

752.4 = 0.2795 (or 28.0%) 2692.2

EES (Engineering Equation Solver) SOLUTION "Given" T1=20 [C] T2=200 [C] x2=1 "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T2, x=0) h_g=enthalpy(Fluid$, T=T2, x=1) h_fg=h_g-h_f c_w=4.18 [kJ/kg-K] "Table A-3" "Analysis" q_preheating=c_w*(T2-T1) q_boiling=h_fg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-575

q_total=q_boiling+q_preheating f_preheating=q_preheating/q_total

5-157 An ideal gas expands in a turbine. The volume flow rate at the inlet for a power output of 650 kW is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Properties The properties of the ideal gas are given as R  0.30 kPa  m3 / kg  K, c p  1.13 kJ / kg  C, cv  0.83 kJ / kg  C .

Analysis We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout ¾ ¾ ® mh1 = Wout + mh2 (since Q @D ke = D pe @ 0) which can be rearranged to solve for mass flow rate m=

Wout Wout 650 kW = = = 1.438 kg/s h1 - h2 c p (T1 - T2 ) (1.13 kJ/kg.K)(1200 - 800)K

The inlet specific volume and the volume flow rate are

v1 =

3 RT1 (0.3 kPa ×m /kg ×K )(1200 K ) = = 0.4 m3 /kg P1 900 kPa

Thus,

V = mv1 = (1.438 kg/s)(0.4 m3 /kg) = 0.575 m3 / s

EES (Engineering Equation Solver) SOLUTION "Given" T1=1200 [K] P1=900 [kPa] T2=800 [K] W_dot=650 [kW] c_p=1.13 [kJ/kg-K] c_v=0.83 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-576

R=0.30 [kJ/kg-K] "Analysis" W_dot=m_dot*c_p*(T1-T2) v_1=(R*T1)/P1 V_dot_1=m_dot*v_1

5-158 Helium is compressed by a compressor. The power required is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats. 4 The compressor is adiabatic. Properties The constant pressure specific heat of helium is c p  5.1926 kJ / kg  K . The gas constant is

R  2.0769 kJ / kg  K (Table A-2a).

Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Win + mh1 = mh2 (since D ke @D pe @ 0) Win = m(h2 - h1 ) = mc p (T2 - T1 ) The specific volume of air at the inlet and the mass flow rate are

v1 =

RT1 (2.0769 kPa ×m3 /kg ×K)(20 + 273 K) = = 5.532 m3 /kg P1 110 kPa

AV (0.1 m2 )(7 m/s) 1 1 = = 0.1265 kg/s v1 5.532 m3 /kg

Then the power input is determined from the energy balance equation to be

Win = mc p (T2 - T1 ) = (0.1265 kg/s)(5.1926 kJ/kg ×K)(200 - 20)K = 118 kW

1-577

T1=(20+273) [K] P2=400 [kPa] T2=(200+273) [K] A1=0.1 [m^2] Vel1=7 [m/s] "Properties" R=2.0769 [kJ/kg-K] "Table A-2a" c_p=5.1926 [kJ/kg-K] "Table A-2a" "Analysis" v1=(R*T1)/P1 m_dot=(A1*Vel1)/v1 W_dot_in=m_dot*c_p*(T2-T1)

5-159 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the mass flow rate of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water are constant. Properties The specific heat of chicken are given to be 3.54 kJ / kg. C . The specific heat of water is 4.18 kJ / kg. C (Table A-3).

Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of

mchicken = (500 chicken / h)(2.2 kg / chicken) = 1100 kg / h = 0.3056 kg / s Taking the chicken flow stream in the chiller as the system, the energy balance for steadily flowing chickens can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2 (since D ke @D pe @ 0) Qout = Qchicken = mchicken c p (T1 - T2 )

Then the rate of heat removal from the chickens as they are cooled from 15C to 3C becomes Qchicken = (mc p D T )chicken = (0.3056 kg/s)(3.54 kJ/kg.°C)(15 - 3)°C = 13.0 kW

1-578

The chiller gains heat from the surroundings at a rate of 200 kJ / h  0.0556 kJ / s . Then the total rate of heat gain by the water is Qwater = Qchicken + Qheat gain = 13.0 + 0.056 = 13.056 kW

Noting that the temperature rise of water is not to exceed 2C as it flows through the chiller, the mass flow rate of water must be at least mwater =

Qwater 13.056 kW = = 1.56 kg / s (c p D T ) water (4.18 kJ/kg.°C)(2°C)

If the mass flow rate of water is less than this value, then the temperature rise of water will have to be more than 2C.

EES (Engineering Equation Solver) SOLUTION "Given" m_chicken=2.2 [kg] T_chilledwater=0.5 [C] T1=15 [C] T2=3 [C] n_dot_chicken=500[1/h]*Convert(1/h, 1/s) Q_dot_heatgain=200[kJ/h]*Convert(kJ/h, kJ/s) DELTAT_water=2 [C] c_chicken=3.54 [kJ/kg-C] "Properties" c_water=4.18 [kJ/kg-C] "Table A-3" "Analysis" "(a)" m_dot_chicken=n_dot_chicken*m_chicken Q_dot_chicken=m_dot_chicken*c_chicken*(T1-T2) Q_dot_water=Q_dot_chicken+Q_dot_heatgain "(b)" Q_dot_water=m_dot_water*c_water*DELTAT_water 5-160 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the mass flow rate of water are to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water are constant. 3 Heat gain of the chiller is negligible. Properties The specific heat of chicken are given to be 3.54 kJ / kg. C . The specific heat of water is 4.18 kJ / kg. C (Table A3).

1-579

(a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered to flow steadily through the chiller at a mass flow rate of

mchicken = (500 chicken/h)(2.2 kg/chicken) = 1100 kg/h = 0.3056kg/s Taking the chicken flow stream in the chiller as the system, the energy balance for steadily flowing chickens can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2 (since D ke @D pe @ 0) Qout = Qchicken = mchicken c p (T1 - T2 )

Then the rate of heat removal from the chickens as they are cooled from 15C to 3C becomes Qchicken = (mc p D T )chicken = (0.3056 kg/s)(3.54 kJ/kg.°C)(15 - 3)°C = 13.0 kW

Heat gain of the chiller from the surroundings is negligible. Then the total rate of heat gain by the water is

Qwater = Qchicken = 13.0 kW Noting that the temperature rise of water is not to exceed 2C as it flows through the chiller, the mass flow rate of water must be at least mwater =

Qwater 13.0 kW = = 1.56 kg / s (c p D T ) water (4.18 kJ/kg.°C)(2°C)

If the mass flow rate of water is less than this value, then the temperature rise of water will have to be more than 2C.

EES (Engineering Equation Solver) SOLUTION "Given" m_chicken=2.2 [kg] T_chilledwater=0.5 [C] T1=15 [C] T2=3 [C] n_dot_chicken=500[1/h]*Convert(1/h, 1/s) DELTAT_water=2 [C] c_chicken=3.54 [kJ/kg-C] "Properties" c_water=4.18 [kJ/kg-C] "Table A-3" "Analysis" "(a)" m_dot_chicken=n_dot_chicken*m_chicken Q_dot_chicken=m_dot_chicken*c_chicken*(T1-T2) Q_dot_water=Q_dot_chicken "(b)" Q_dot_water=m_dot_water*c_water*DELTAT_water

5-161E A refrigeration system is to cool eggs by chilled air at a rate of 10,000 eggs per hour. The rate of heat removal from the eggs, the required volume flow rate of air, and the size of the compressor of the refrigeration system are to be determined. Assumptions 1 Steady operating conditions exist. 2 The eggs are at uniform temperatures before and after cooling. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-580

3 4 5

The cooling section is well-insulated. The properties of eggs are constant. The local atmospheric pressure is 1 atm.

Properties The properties of the eggs are given to   67.4 lbm / ft 3 and c p  0.80 Btu / lbm. F . The specific heat of air at room temperature c p  0.24 Btu / lbm. F (Table A-2E). The gas constant of air is R  0.3704 psia.ft 3 / lbm  R (Table A1E). Analysis (a) Noting that eggs are cooled at a rate of 10,000 eggs per hour, eggs can be considered to flow steadily through the cooling section at a mass flow rate of

megg = (10,000 eggs/h)(0.14 lbm/egg) = 1400 lbm/h = 0.3889 lbm/s Taking the egg flow stream in the cooler as the system, the energy balance for steadily flowing eggs can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2 (since D ke @D pe @ 0) Qout = Qegg = megg c p (T1 - T2 )

Then the rate of heat removal from the eggs as they are cooled from 90F to 50F at this rate becomes Qegg = (mc p D T )egg = (0.14 lbm/egg)(3000 egg/h)(0.80 Btu/lbm.°F)(90 - 50)°F = 13, 440 Btu / h

(b) All the heat released by the eggs is absorbed by the refrigerated air since heat transfer through he walls of cooler is negligible, and the temperature rise of air is not to exceed 10F. The minimum mass flow and volume flow rates of air are determined to be mair =

Qair 13,440 Btu/h = = 5600 lbm/h (c p D T )air (0.24 Btu/lbm.°F)(10°F)

r air =

P 14.7 psia = = 0.08034 lbm/ft 3 RT (0.3704 psia.ft 3 /lbm.R)(34 + 460)R

Vair =

mair 5600 lbm/h = = 69, 700 ft 3 / h r air 0.08034 lbm/ft 3 ³

EES (Engineering Equation Solver) SOLUTION "Given" m_egg=0.14 [lbm] T1_egg=90 [F] T2_egg=50 [F] T1_air=34 [F] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-581

n_dot_egg=3000 [1/h] DELTAT_air=10 [F] rho_egg=67.4 [lbm/ft^3] c_egg=0.80 [Btu/lbm-F] "Properties" c_p_air=0.240 [Btu/lbm-F] "Table A-2Ea" R=0.3704 [psia-ft^3/lbm-R] "Table A-2Ea" "Analysis" "(a)" m_dot_egg=n_dot_egg*m_egg Q_dot_egg=m_dot_egg*c_egg*(T1_egg-T2_egg) "(b)" Q_dot_air=Q_dot_egg m_dot_air=Q_dot_air/(c_p_air*DELTAT_air) rho_air=P/(R*(T1_air+460)) P=14.7 [psia] "assumed" V_dot_air=m_dot_air/rho_air

5-162 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates of heat and water mass that need to be supplied to the water are to be determined. Assumptions 1 Steady operating conditions exist. 2 3 4

The entire water body is maintained at a uniform temperature of 50C. Heat losses from the outer surfaces of the bath are negligible. Water is an incompressible substance with constant properties.

Properties The specific heat of water at room temperature is c p  4.18 kJ / kg  C . Also, the specific heat of glass is 0.80 kJ / kg  C (Table A-3).

Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is

mbottle = mbottle ´ Bottle flow rate = (0.150 kg/bottle)(450/60 bottles/s) = 1.125 kg/s Taking the bottle flow section as the system, which is a steady-flow control volume, the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

1-582

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = Qbottle = mwater c p (T2 - T1 )

Then the rate of heat removal by the bottles as they are heated from 20 to 50C is Qbottle = mbottle c p D T = (1.125 kg/s)(0.8 kJ/kg.°C)(50 - 20)°C = 27 kW

The amount of water removed by the bottles is

mwater, out = (Flow rate of bottles)(Water removed per bottle)

= (450 bottles / min )(0.0002 g/bottle ) = 90 g/min = 0.0015 kg / s Noting that the water removed by the bottles is made up by fresh water entering at 15C, the rate of heat removal by the water that sticks to the bottles is Qwater removed = mwater removed c p D T = (0.0015 kg/s)(4.18 kJ/kg ×°C)(50 - 15)°C = 0.2195 kW

Therefore, the total amount of heat removed by the wet bottles is Qtotal, removed = Qglass removed + Qwater removed = 27 + 0.2195 = 27.22 kW

Discussion In practice, the rates of heat and water removal will be much larger since the heat losses from the tank and the moisture loss from the open surface are not considered.

EES (Engineering Equation Solver) SOLUTION "Given" T1=20 [C] T2=50 [C] n_dot_bottle=450 [1/min] m_bottle=0.150 [kg] m_WaterRemoved=0.0002 [kg] T_water=15 [C] "Properties" c_glass=0.80 [kJ/kg-C] "Table A-3" c_water=4.18 [kJ/kg-C] "Table A-3" "Analysis" "(a)" m_dot_bottle=n_dot_bottle*m_bottle*Convert(kg/min, kg/s) Q_dot_bottle=m_dot_bottle*c_glass*(T2-T1) m_dot_water_removed=n_dot_bottle*m_WaterRemoved*Convert(kg/min, kg/s) "(b)" Q_dot_water_removed=m_dot_water_removed*c_water*(T2-T_water) Q_dot_total_removed=Q_dot_bottle+Q_dot_water_removed

5-163 A regenerator is considered to save heat during the cooling of milk in a dairy plant. The amounts of fuel and money such a generator will save per year are to be determined. Assumptions 1 Steady operating conditions exist. 2 The properties of the milk are constant.

1-583

Properties The average density and specific heat of milk can be taken to be milk  water  1kg / L and

c p , milk  3.79 kJ / kg. C (Table A-3). Analysis The mass flow rate of the milk is

mmilk = r Vmilk = (1 kg/L)(20 L/s) = 20 kg/s =72,000 kg/h Taking the pasteurizing section as the system, the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = mmilk c p (T2 - T1 )

Therefore, to heat the milk from 4 to 72C as being done currently, heat must be transferred to the milk at a rate of Qcurrent = [mcp (Tpasturization - Trefrigeration )]milk

= (20 kg/s)(3.79 kJ/kg.ëC)(72 - 4)°C

= 5154 kJ/s The proposed regenerator has an effectiveness of  = 0.82, and thus it will save 82 percent of this energy. Therefore,

Qsaved = eQcurrent = (0.82)(5154 kJ/s) = 4227 kJ/s Noting that the boiler has an efficiency of ηboiler = 0.90 , the energy savings above correspond to fuel savings of

Fuel Saved =

Qsaved 4227 kJ/s 1 therm = = 0.04452 therm/s hboiler 0.90 105,500 kJ

Noting that 1 year = 365  24 = 8760 h and unit cost of natural gas is $1.10 / therm , the annual fuel and money savings will be Fuel Saved  (0.04452 therms / s)(8760  3600 s)  1.404  106 therms / yr

Money saved =(Fuel saved)(Unit cost of fuel) =(1.404´ 106 therm/yr)($1.10/therm)

=$1.544×106 / yr

1-584

EES (Engineering Equation Solver) SOLUTION "Given" T_refrig=4 [C] T_pastur=72 [C] V_dot_milk=20[L/s]*Convert(L/s, m^3/s) time=365*24*3600 [s] eta_boiler=0.90 T_coldwater=18 [C] epsilon_regen=0.82 Cost_gas=1.10 [$/therm] "Properties" rho_milk=1000 [kg/m^3] "water density is used" c_milk=3.79 [kJ/kg-C] "Table A-3" "Analysis" m_dot_milk=rho_milk*V_dot_milk Q_dot_current=m_dot_milk*c_milk*(T_pastur-T_refrig) Q_dot_saved=epsilon_regen*Q_dot_current FuelSaved=(Q_dot_saved/eta_boiler)*time*Convert(kJ, therm) MoneySaved=FuelSaved*Cost_gas

5-164 Long aluminum wires are extruded at a velocity of 8 m / min , and are exposed to atmospheric air. The rate of heat transfer from the wire is to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant.

Properties The properties of aluminum are given to be   2702 kg / m3 and c p  0.896 kJ / kg  C . Analysis The mass flow rate of the extruded wire through the air is

m = r V = r (p r02 )V = (2702 kg/m3 )p (0.0025 m) 2 (8 m/min) = 0.4244 kg/min = 0.007074 kg/s Taking the volume occupied by the extruded wire as the system, which is a steady-flow control volume, the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2 (since D ke @D pe @ 0) Qout = Qwire = mwire c p (T1 - T2 )

Then the rate of heat transfer from the wire to the air becomes Q = mc p [T (t ) - T¥ ] = (0.007074 kg/s)(0.896 kJ/kg.°C)(350 - 50)°C = 1.90 kW

1-585

EES (Engineering Equation Solver) SOLUTION "Given" D=0.005 [m] T1=350 [C] T2=50 [C] T_air=25 [C] Vel=8 [m/min]*Convert(m/min, m/s) rho=2702 [kg/m^3] c_p=0.896 [kJ/kg-C] "Analysis" A=pi*D^2/4 m_dot=rho*A*Vel Q_dot_wire=m_dot*c_p*(T1-T2)

5-165 Long copper wires are extruded at a velocity of 8 m / min , and are exposed to atmospheric air. The rate of heat transfer from the wire is to be determined. Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant. Properties The properties of copper are given to be   8950 kg / m3 and c p  0.383 kJ / kg. C . Analysis The mass flow rate of the extruded wire through the air is

m = r V = r (p r02 )V = (8950 kg/m3 )p (0.0025 m) 2 (8 / 60 m/s) = 0.02343 kg/s

Taking the volume occupied by the extruded wire as the system, which is a steady-flow control volume, the energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Qout + mh2 (since D ke @D pe @ 0) Qout = Qwire = mwire c p (T1 - T2 )

Then the rate of heat transfer from the wire to the air becomes Q = mc p [T (t ) - T¥ ] = (0.02343 kg/s)(0.383 kJ/kg.°C)(350 - 50)°C = 2.69 kW

EES (Engineering Equation Solver) SOLUTION "Given" D=0.005 [m] T1=350 [C] T2=50 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-586

T_air=30 [C] Vel=8[m/min]*Convert(m/min, m/s) rho=8950 [kg/m^3] c_p=0.383 [kJ/kg-C] "Analysis" A=pi*D^2/4 m_dot=rho*A*Vel Q_dot_wire=m_dot*c_p*(T1-T2)

5-166E Steam is mixed with water steadily in an adiabatic device. The temperature of the water leaving this device is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work and heat interactions. 4 There is no heat transfer between the mixing device and the surroundings.

Analysis We take the mixing device as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m2 h2 = m3 h3 From a mass balance

m3 = m1 + m2 = 0.05 + 1 = 1.05 lbm/s The enthalpies of steam and water are (Table A-6E and A-4E) P1 = 80 psia ïü ïý h = 1230.8 Btu/lbm 1 T1 = 400°F ïïþ P2 = 80 psia ü ïï ý h2 @ h f @ 60° F = 28.08 Btu/lbm T2 = 60°F ïïþ

Substituting into the energy balance equation solving for the exit enthalpy gives h3 =

m1h1 + m2 h2 (0.05 lbm/s)(1230.8 Btu/lbm) + (1 lbm/s)(28.08 Btu/lbm) = = 85.35 Btu/lbm m3 1.05 lbm/s

At the exit state P3 = 80 psia and h3  85.35 kJ / kg . An investigation of Table A-5E reveals that this is compressed liquid state. Approximating this state as saturated liquid at the specified temperature, the temperature may be determined from Table A-4E to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-587

ü ïï ý T3 @T f @ h= 85.35 Btu/lbm = 117°F h3 = 85.35 Btu/lbm ïïþ P3 = 80 psia

Discussion The exact answer is determined at the compressed liquid state using EES to be 117.1F, indicating that the saturated liquid approximation is a reasonable one.

EES (Engineering Equation Solver) SOLUTION "Given" P1=80 [psia] T1=400 [F] P2=80 [psia] T2=60 [F] m_dot_1=0.05 [lbm/s] m_dot_2=1 [lbm/s] P3=80 [psia] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, T=T2, x=0) "assumed as saturated liquid" m_dot_3=m_dot_1+m_dot_2 h3=(m_dot_1*h1+m_dot_2*h2)/m_dot_3 T3_approximate=temperature(Fluid$, h=h3, x=0) "saturated liquid assumption" T3_exact=temperature(Fluid$, P=P3, h=h3) "compressed liquid"

5-167 A constant-pressure R-134a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams. The flow power needed to operate this unit and the mass flow rate of the two outlet streams are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work and heat interactions. Analysis The specific volume at the inlet is (Table A-12) P1 = 320 kPa ïü ïý v = v + x (v - v ) = 0.0007771 + (0.55)(0.06368 - 0.0007771) = 0.03537 m3 /kg f 1 g f ïïþ 1 x1 = 0.55

The mass flow rate at the inlet is then

m1 =

V1 0.006 m3 /s = = 0.1696 kg/s v1 0.03537 m3 /kg

For each kg of mixture processed, 0.55 kg of vapor are processed. Therefore,

m2 = 0.7m1 = 0.55´ 0.1696 = 0.09329 kg / s m3 = m1 - m2 = 0.45m1 = 0.45´ 0.1696 = 0.07633 kg / s The flow power for this unit is Wflow = m2 P2v 2 + m3 P3v 3 - m1 P1v1 = (0.09329 kg/s)(320 kPa)(0.06368 m3 /kg) + (0.07633 kg/s)(320 kPa)(0.0007771 m3 /kg) - (0.1696 kg/s)(320 kPa)(0.03537 m3 /kg)

= 0 kW

1-588

EES (Engineering Equation Solver) SOLUTION "Given" Vol_dot_1=0.006 [m^3/s] P=320 [kPa] x1=0.55 "Analysis" Fluid$='R134a' v1=volume(Fluid$,P=P, x=x1) v_f=volume(Fluid$,P=P, x=0) v_g=volume(Fluid$,P=P, x=1) m_dot_1=Vol_dot_1/v1 m_dot_2=x1*m_dot_1 m_dot_3=m_dot_1-m_dot_2 W_dot_flow=m_dot_2*P*v_g+m_dot_3*P*v_f-m_dot_1*P*v1

5-168E A study quantifies the cost and benefits of enhancing IAQ by increasing the building ventilation. The net monetary benefit of installing an enhanced IAQ system to the employer per year is to be determined. Assumptions The analysis in the report is applicable to this work place. Analysis The report states that enhancing IAQ increases the productivity of a person by $90 per year, and decreases the cost of the respiratory illnesses by $39 a year while increasing the annual energy consumption by $6 and the equipment cost by about $4 a year. The net monetary benefit of installing an enhanced IAQ system to the employer per year is determined by adding the benefits and subtracting the costs to be

Net benefit  Total benefits  total cost  (90  39)  (6  4)  $119 / year (per person) The total benefit is determined by multiplying the benefit per person by the number of employees,

Total net benefit  No. of employees  Net benefit per person  120  $119 / year  $14, 280 / year Discussion Note that the unseen savings in productivity and reduced illnesses can be very significant when they are properly quantified.

EES (Engineering Equation Solver) SOLUTION "Given" V1_dot=5 [cfm] V2_dot=20 [cfm] Increase_Productivity=0.25 [%] Value_Productivity=90 [$/year] Decrease_Illness=10 [%] Saving_Illness=39 [$/year] Consumption_Energy=6 [$/year] Consumption_Equipment=4 [$/year] N_employee=120 "Analysis" TotalBenefit=Value_Productivity+Saving_Illness TotalCost=Consumption_Energy+Consumption_Equipment NetBenefit=TotalBenefit-TotalCost TotalNetBenefit=N_employee*NetBenefit

5-169E A winterizing project is to reduce the infiltration rate of a house from 2.2 ACH to 1.1 ACH. The resulting cost savings are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-589

1 2

The house is maintained at 72F at all times. The latent heat load during the heating season is negligible.

3 4 5 6

The infiltrating air is heated to 72F before it exfiltrates. Air is an ideal gas with constant specific heats at room temperature. The changes in kinetic and potential energies are negligible. Steady flow conditions exist.

Properties The gas constant of air is 0.3704 psia. ft 3 / lbm  R (Table A-1E). The specific heat of air at room temperature is 0.24 Btu / lbm  F (Table A-2E).

Analysis The density of air at the outdoor conditions is ro =

Po 13.5 psia = = 0.0734 lbm/ft 3 RTo (0.3704 psia ×ft 3 /lbm ×R)(496.5 R)

The volume of the house is

Vbuilding = (Floor area)(Height) = (4500 ft 2 )(9 ft) = 40,500 ft 3 We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the house. The energy balance for this imaginary steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = mc p (T2 - T1 ) = rV c p (T2 - T1 )

The reduction in the infiltration rate is 2.2 – 1.1 = 1.1 ACH. The reduction in the sensible infiltration heat load corresponding to it is Qinfiltration, saved = r o c p (ACHsaved )(Vbuilding )(Ti - To )

= (0.0734 lbm/ft 3 )(0.24 Btu/lbm.°F)(1.1/h)(40,500 ft 3 )(72 - 36.5)°F =27,860 Btu/h = 0.2786 therm/h

since 1 therm = 100,000 Btu. The number of hours during a six month period is 63024 = 4320 h. Noting that the furnace efficiency is 0.92 and the unit cost of natural gas is $1.24 / therm , the energy and money saved during the 6-month period are Energy savings = (Qinfiltration, saved )(No. of hours per year)/Efficiency

= (0.2786 therm/h)(4320 h/year)/0.92

= 1308 therms/year Cost savings = (Energy savings)(Unit cost of energy) = (1308 therms/year)($1.24/therm) = $1622 / year

1-590

Therefore, reducing the infiltration rate by one-half will reduce the heating costs of this homeowner by $1622 per year.

EES (Engineering Equation Solver) SOLUTION "Given" P_o=13.5 [psia] T_o=36.5 [F] Height=9 [ft] Area=4500 [ft^2] ACH_old=2.2 [1/h] ACH_new=1.1 [1/h] UnitCost=1.24 [$/therm] time=6*30*24 [h/year] T_i=72 [F] eta_furnace=0.92 "Properties" R=0.3704 [psia-ft^3/lbm-R] "Table A-1E" c_p=0.24 [Btu/lbm-R] "Table A-2Ea" "Analysis" rho_o=P_o/(R*(T_o+460)) V_building=Area*Height ACH_saved=ACH_old-ACH_new Q_dot_saved=rho_o*C_p*ACH_saved*V_building*(T_i-T_o)*Convert(Btu, therm) E_saved=(Q_dot_saved*time)/eta_furnace CostSaved=E_saved*UnitCost

5-170 The ventilating fan of the bathroom of an electrically heated building in San Francisco runs continuously. The amount and cost of the heat “vented out” per month in winter are to be determined. Assumptions 1 We take the atmospheric pressure to be 1 atm = 101.3 kPa since San Francisco is at sea level. 2

The building is maintained at 22C at all times.

3 4 5

The infiltrating air is heated to 22C before it exfiltrates. Air is an ideal gas with constant specific heats at room temperature. Steady flow conditions exist.

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). The specific heat of air at room temperature is

c p  1.005 kJ / kg  C (Table A-2).

Analysis The density of air at the indoor conditions of 1 atm and 22C is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-591

ro =

Po (101.3 kPa) = = 1.197 kg/m3 RTo (0.287 kPa.m3 /kg.K)(22 + 273 K)

Then the mass flow rate of air vented out becomes

mair = r Vair = (1.197 kg/m3 )(0.030 m3 /s) = 0.03590 kg/s We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the house. The energy balance for this imaginary steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = mc p (T2 - T1 )

Noting that the indoor air vented out at 22C is replaced by infiltrating outdoor air at 12.2C, the sensible infiltration heat loss (heat gain for the infiltrating air) due to venting by fans can be expressed Qloss by fan = mair cp (Tindoors - Toutdoors )

= (0.0359 kg/s)(1.005 kJ/kg.°C)(22 - 12.2)°C = 0.3536 kJ/s = 0.3536 kW Then the amount and cost of the heat “vented out” per month (1 month = 30  24 = 720 h) becomes Energy loss = Qloss by fan D t = (0.3536 kW)(720 h/month) = 254.6 kWh / month

Money loss = (Energy loss)(Unit cost of energy) = (254.6 kWh/month)($0.12/kWh) = $22.9 / month Discussion Note that the energy and money loss associated with ventilating fans can be very significant. Therefore, ventilating fans should be used with care.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=30[L/s]*Convert(L/s, m^3/s) T_o=12.2 [C] T_i=22 [C] cost=0.09 [$/kWh] time=30*24 [h] "Properties" "C_p=CP(air, T=T_ave) C_v=CV(air, T=T_ave) T_ave=1/2*(T_i+T_o) R=C_p-C_v" c_p=1.005 R=0.287 "Analysis" rho=P_o/(R*(T_i+273)) P_o=101.325 [kPa] "sea level atmospheric pressure" m_dot=rho*V_dot Q_dot_out=m_dot*C_p*(T_i-T_o) E_loss=Q_dot_out*time MoneyLoss=E_loss*cost

5-171E A small positioning control rocket in a satellite is driven by a container filled with R-134a at saturated liquid state. The number of bursts this rocket experience before the quality in the container is 90% or more is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-592

Analysis The initial and final specific volumes are T1 = - 10°F ïü ïý v = 0.01170 ft 3 /lbm (Table A-11E) ïïþ 1 x1 = 0 T2 = - 10°F ü ïï 3 ý v 2 = v f + x2v fg = 0.01170 + (0.90)(2.7097 - 0.01170) = 2.4399 ft /lbm ïïþ x2 = 0.90

The initial and final masses in the container are

m1 =

V 2 ft 3 = = 170.9 lbm v1 0.01170 m3 /kg

m2 =

V 2 ft 3 = = 0.8197 lbm v 2 2.4399 ft 3 /lbm

Then,

D m = m1 - m2 = 170.9 - 0.8197 = 170.1 lbm The amount of mass released during each control burst is

D mb = mD t = (0.05 lbm/s)(5 s) = 0.25 lbm The number of bursts that can be executed is then Nb =

Dm 170.1 lbm = = 680 bursts D mb 0.25 lbm/burst

EES (Engineering Equation Solver) SOLUTION "Given" Vol=2 [ft^3] T1=-10 [F] x1=0 DELTAt=5 [s] m_dot=0.05 [lbm/s] x2=0.90 T2=-10 [F] "Analysis" Fluid$='R134a' v1=volume(Fluid$, T=T1, x=x1) v2_f=volume(Fluid$, T=T2, x=0) v2_g=volume(Fluid$, T=T2, x=1) v2=v2_f+x2*(v2_g-v2_f) m1=Vol/v1 m2=Vol/v2 DELTAm=m1-m2 DELTAm_b=m_dot*DELTAt N_b=DELTAm/DELTAM_b

5-172 Outdoors air at -5C and 95 kPa enters the building at a rate of 60 L / s while the indoors is maintained at 25C. The rate of sensible heat loss from the building due to infiltration is to be determined. Assumptions 1 2

The house is maintained at 25C at all times. The latent heat load is negligible.

The infiltrating air is heated to 25C before it exfiltrates.

1-593

4 5 6

Air is an ideal gas with constant specific heats at room temperature. The changes in kinetic and potential energies are negligible. Steady flow conditions exist.

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K . The specific heat of air at room temperature is

c p  1.005 kJ / kg  C (Table A-2).

Analysis The density of air at the outdoor conditions is ro =

Po 95 kPa = = 1.235 kg/m 3 RTo (0.287 kPa.m 3 /kg.K)(-5 + 273 K)

We can view infiltration as a steady stream of air that is heated as it flows in an imaginary duct passing through the building. The energy balance for this imaginary steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = mc p (T2 - T1 )

Then the sensible infiltration heat load corresponding to an infiltration rate of 60 L / s becomes Qinfiltration = r oVair c p (Ti - To )

= (1.235 kg/m3 )(0.060 m3 /s)(1.005 kJ/kg.°C)[25 - ( - 5)]°C

= 2.23 kW Therefore, sensible heat will be lost at a rate of 2.23 kJ / s due to infiltration.

EES (Engineering Equation Solver) SOLUTION "Given" T_o=-5 [C] P_o=95 [kPa] V_dot=60[L/s]*Convert(L/s, m^3/s) T_i=25 [C] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" "Analysis" rho_o=P_o/(R*(T_o+273)) Q_dot_infiltration=rho_o*V_dot*c_p*(T_i-T_o)

1-594

5-173 Chilled air is to cool a room by removing the heat generated in a large insulated classroom by lights and students. The required flow rate of air that needs to be supplied to the room is to be determined. Assumptions 1 The moisture produced by the bodies leave the room as vapor without any condensing, and thus the classroom has no latent heat load. 2 Heat gain through the walls and the roof is negligible. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Steady operating conditions exist. Properties The specific heat of air at room temperature is 1.005 kJ / kg  C (Table A-2). The average rate of sensible heat generation by a person is given to be 60 W. Analysis The rate of sensible heat generation by the people in the room and the total rate of sensible internal heat generation are

Qgen, sensible = qgen, sensible (No. of people) = (60 W/person)(150 persons) = 9000 W Qtotal, sensible = Qgen, sensible + Qlighting = 9000 + 6000 = 15, 000 W

Both of these effects can be viewed as heat gain for the chilled air stream, which can be viewed as a steady stream of cool air that is heated as it flows in an imaginary duct passing through the room. The energy balance for this imaginary steadyflow system can be expressed in the rate form as Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Qin + mh1 = mh2 (since D ke @D pe @ 0) Qin = Qtotal, sensible = mc p (T2 - T1 )

Then the required mass flow rate of chilled air becomes mair =

Qtotal, sensible cpD T

15 kJ/s = 1.49 kg / s (1.005 kJ/kg ×°C)(25 - 15)°C

Discussion The latent heat will be removed by the air-conditioning system as the moisture condenses outside the cooling coils.

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_gen_sensible=150*60[W]*Convert(W, kW) Q_dot_lighting=6 [kW] T1=15 [C] T2=25 [C] "Properties" c_p=1.005 [kJ/kg-K] "Table A-2a" "Analysis" Q_dot_total=Q_dot_gen_sensible+Q_dot_lighting Q_dot_total=m_dot_air*c_p*(T2-T1)

5-174 The average air velocity in the circular duct of an air-conditioning system is not to exceed 8 m / s . If the fan converts 80 percent of the electrical energy into kinetic energy, the size of the fan motor needed and the diameter of the main duct are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-595

This is a steady-flow process since there is no change with time at any point and thus D mCV = 0 and D ECV = 0.

The inlet velocity is negligible, V1 @ 0.

3 4

There are no heat and work interactions other than the electrical power consumed by the fan motor. Air is an ideal gas with constant specific heats at room temperature.

Properties The density of air is given to be   1.20 kg / m3 . The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  C (Table A-2).

m = r V = (1.20 kg/m3 )(130 m3 /min) = 156 kg/min = 2.6 kg/s D KE = m

ö V22 - V12 (8 m/s) 2 - 0 æ çç 1 kJ/kg ÷ = (2.6 kg/s) = 0.0832 kW 2 2÷ ÷ ç è1000 m /s ø 2 2

It is stated that this represents 80% of the electrical energy consumed by the motor. Then the total electrical power consumed by the motor is determined to be 0.7Wmotor = D KE ® Wmotor =

D KE 0.0832 kW = = 0.104 kW 0.8 0.8

The diameter of the main duct is V = VA = V (p D 2 / 4) ®

4V = pV

ö 4(130 m 3 / min) æ çç1 min ÷ ÷ = 0.587 m çè 60 s ÷ ø p (8 m/s)

Therefore, the motor should have a rated power of at least 0.104 kW, and the diameter of the duct should be at least 58.7 cm

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=130[m^3/min]*Convert(m^3/min, m^3/s) Vel2=8 [m/s] f=0.8 rho=1.20 [kg/m^3] "Analysis" m_dot=rho*V_dot DELTAKE=m_dot*(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) Vel1=0 DELTAKE=f*W_dot_motor V_dot=Vel2*A A=pi*D^2/4 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-596

5-175 A building is to be heated by a 30-kW electric resistance heater placed in a duct inside. The time it takes to raise the interior temperature from 14C to 24C, and the average mass flow rate of air as it passes through the heater in the duct are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, and thus heat transfer through it is negligible. 5 No air leaks in and out of the building. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The specific heats of air at room temperature are c p  1.005 and cv  0.718 kJ / kg  K (Table A-2).

Analysis (a) The total mass of air in the building is m=

(95 kPa )(400 m 3 ) P1V1 = = 461.3 kg. RT1 (0.287 kPa ×m 3 /kg ×K )(287 K )

We first take the entire building as our system, which is a closed system since no mass leaks in or out. The time required to raise the air temperature to 24°C is determined by applying the energy balance to this constant volume closed system: Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We, in + Wfan, in - Qout = D U

(since D KE = D PE = 0)

D t (We, in + Wfan, in - Qout ) = mcv , avg (T2 - T1 ) Solving for t gives Dt =

mcv , avg (T2 - T1 ) We, in + Wfan, in - Qout

(461.3 kg)(0.718 kJ/kg ×o C)(24 - 14) o C = 146 s (30 kJ/s) + (0.25 kJ/s) - (450/60 kJ/s)

(b) We now take the heating duct as the system, which is a control volume since mass crosses the boundary. There is only one inlet and one exit, and thus m1 = m2 = m. The energy balance for this adiabatic steady-flow system can be expressed in the rate form as Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

1-597

We, in + Wfan, in cpD T

(30 + 0.25) kJ/s = 6.02 kg / s (1.005 kJ/kg ×o C)(5 o C)

Ein = Eout We,in + Wfan, in + mh1 = mh2 (since Q = D ke @D pe @0)

We,in + Wfan, in = m(h2 - h1 ) = mc p (T2 - T1 )

EES (Engineering Equation Solver) SOLUTION "Given" Vol_building=400 [m^3] W_dot_e_in=30 [kW] T1_air=14 [C] T2_air=24 [C] P1=95 [kPa] Q_dot_out=450[kJ/min]*Convert(kJ/min, kJ/s) W_dot_fan_in=0.250 [kW] DELTAT=5 [C] "Properties" Fluid$='air' u1=intenergy(Fluid$, T=T1_air) u2=intenergy(Fluid$, T=T2_air) C_p=CP(Fluid$, T=25[C]) C_v=CV(Fluid$, T=25[C]) R=C_p-C_v "Analysis" "(a)" m=(P1*Vol_building)/(R*(T1_air+273)) (W_dot_e_in+W_dot_fan_in-Q_dot_out)*time=m*(u2-u1) "energy balance on building" "(b)" W_dot_e_in+W_dot_fan_in=m_dot*C_p*DELTAT "energy balance on heating duct"

5-176 The maximum flow rate of a standard shower head can be reduced from 13.3 to 10.5 L / min by switching to lowflow shower heads. The ratio of the hot-to-cold water flow rates and the amount of electricity saved by a family of four per year by replacing the standard shower heads by the low-flow ones are to be determined. Assumptions 1

This is a steady-flow process since there is no change with time at any point and thus D mCV = 0 and D ECV = 0.

The kinetic and potential energies are negligible, ke @ pe @0.

Heat losses from the system are negligible and thus Q @ 0.

4 5 6 7 8

There are no work interactions involved. Showers operate at maximum flow conditions during the entire shower. Each member of the household takes a 5-min shower every day. Water is an incompressible substance with constant properties. The efficiency of the electric water heater is 100%.

Properties The density and specific heat of water at room temperature are ρ  1 kg / L and c  4.18 kJ / kg. C (Table A-3).

1-598

Analysis (a) We take the mixing chamber as the system. This is a control volume since mass crosses the system boundary during the process. We note that there are two inlets and one exit. The mass and energy balances for this steady-flow system can be expressed in the rate form as follows: Mass balance:

min - mout = D msystemZ 0 (steady) = 0 min = mout ® m1 + m2 = m3 Energy balance:

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m1h1 + m2 h2 = m3 h3 (since Q @ 0, W = 0, ke @ pe @ 0) Combining the mass and energy balances and rearranging,

m1h1 + m2 h2 = (m1 + m2 )h3 m2 (h2 - h3 ) = m1 (h3 - h1 ) Then the ratio of the mass flow rates of the hot water to cold water becomes h - h1 c(T3 - T1 ) T3 - T1 m2 (42 - 15)°C = 3 = = = = 2.08 m1 h2 - h3 c(T2 - T3 ) T2 - T3 (55 - 42)°C

(b) The low-flow heads will save water at a rate of

Vsaved = [(13.3-10.5) L/min](5 min/person.day)(4 persons)(365 days/yr) = 20,440 L/year

msaved = r Vsaved = (1 kg/L)(20,440 L/year) = 20,440 kg/year Then the energy saved per year becomes

Energy saved = msaved cD T = (20,440 kg/year)(4.18 kJ/kg.°C)(42 - 15)°C

= 2,307,000 kJ/year = 641 kWh

(since 1 kWh = 3600 kJ)

Therefore, switching to low-flow shower heads will save about 641 kWh of electricity per year.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot_old=13.3[L/min]*Convert(L/min, m^3/min) V_dot_new=10.5[L/min]*Convert(L/min, m^3/min) time=365*5 [min] N_person=4 T1=15 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-599

T2=55 [C] T3=42 [C] C_p=4.18 [kJ/kg-C] "Properties" rho=1000 [kg/m^3] "Analysis" "(a)" m_dot_1+m_dot_2=m_dot_3 "mass balance" m_dot_1*C_p*(T3-T1)=m_dot_2*C_p*(T2-T3) "energy balance" m_dot_1=1 "assumed, it does not matter what value is assumed" Ratio_m=m_dot_2/m_dot_1 "(b)" V_dot_saved=(V_dot_old-V_dot_new)*time*N_person m_dot_saved=rho*V_dot_saved E_saved=m_dot_saved*C_p*(T3-T1)*Convert(kJ, kWh)

5-177 Problem 5-176 is reconsidered. The effect of the inlet temperature of cold water on the energy saved by using the low-flow showerhead as the inlet temperature varies from 10°C to 20°C is to be investigated. The electric energy savings is to be plotted against the water inlet temperature. Analysis The problem is solved using EES, and the results are tabulated and plotted below. T_1 = 15 [C] T_2 = 55 [C] T_3 = 42 [C] V_dot_old = 13.3 [L/min] V_dot_new = 10.5 [L/min] c_p = 4.18 [kJ/kg-K ] density=1[kg/L] m_dot_1=1[kg/s] "We can set m_dot_1 = 1 without loss of generality." m_dot_in - m_dot_out = DELTAm_dot_sys DELTAm_dot_sys=0 m_dot_in =m_dot_1 + m_dot_2 m_dot_out = m_dot_3 E_dot_in - E_dot_out = DELTAE_dot_sys DELTAE_dot_sys=0 E_dot_in = m_dot_1*h_1 + m_dot_2*h_2 E_dot_out = m_dot_3*h_3 h_1 = c_p*T_1 h_2 = c_p*T_2 h_3 = c_p*T_3 V_dot_saved=(V_dot_old-V_dot_new)*5*4*365 m_dot_saved=density*V_dot_saved E_dot_saved=m_dot_saved*c_p*(T_3 - T_1)*convert(kJ,kWh) m_ratio = m_dot_2/m_dot_1

1-600

Esaved [kWh / year ]

T1[C]

759.5 712 664.5 617.1 569.6 522.1

10 12 14 16 18 20

5-178 A submarine that has an air-ballast tank originally partially filled with air is considered. Air is pumped into the ballast tank until it is entirely filled with air. The final temperature and mass of the air in the ballast tank are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 The process is adiabatic. 3 There are no work interactions. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The specific volume of water is taken 0.001 m3 / kg . Analysis The conservation of mass principle applied to the air gives

dma = min dt and as applied to the water becomes dmw = - mout dt The first law for the ballast tank produces d (mu ) a d (mu ) w + + hw mw - ha ma dt dt Combining this with the conservation of mass expressions, rearranging and canceling the common dt term produces 0=

d (mu)a + d (mu)w = ha dma + hw dmw Integrating this result from the beginning to the end of the process gives © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-601

[(mu )2 - (mu )1 ]a + [(mu )2 - (mu )1 ]w = ha (m2 - m1 ) a + hw (m2 - m1 ) w Substituting the ideal gas equation of state and the specific heat models for the enthalpies and internal energies expands this to æPV ö PV2 PV1 PV1 ÷ ÷ cv T2 cv T1 - mw,1cwTw = c pTin ççç 2 ÷- mw,1cwTw ÷ ç RT2 RT1 è RT2 RT1 ø

When the common terms are cancelled, this result becomes

T2 =

V2 700 = = 386.8 K V1 100 1 1 + (700 - 100) + (V2 - V1 ) 288 (1.4)(293) T1 kTin

The final mass from the ideal gas relation is

m2 =

PV2 (1500 kPa)(700 m3 ) = = 9460 kg RT2 (0.287 kPa ×m3 /kg ×K)(386.8 K)

EES (Engineering Equation Solver) SOLUTION "Given" Vol2=700 [m^3] Vol1=100 [m^3] P1=1500 [kPa] T1=(15+273) [K] P_in=1500 [kPa] T_in=(20+273) [K] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" T2=Vol2/(Vol1/T1+1/(k*T_in)*(Vol2-Vol1)) P2=P1 m2=(P2*Vol2)/(R*T2)

5-179 A submarine that has an air-ballast tank originally partially filled with air is considered. Air is pumped into the ballast tank in an isothermal manner until it is entirely filled with air. The final mass of the air in the ballast tank and the total heat transfer are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 There are no work interactions. Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). The specific heats of air at room temperature are c p  1.005 kJ / kg  K and cv  0.718 kJ / kg  K (Table A-2a). The specific volume of water is taken 0.001 m3 / kg . Analysis The initial air mass is

m1 =

P1V1 (1500 kPa)(100 m3 ) = = 1814 kg RT1 (0.287 kPa ×m3 /kg ×K)(288.15 K)

and the initial water mass is

mw =

V1 600 m3 = = 600, 000 kg v1 0.001 m3 /kg

1-602

m2 =

P2V2 (1500 kPa)(700 m3 ) = = 12,697 kg RT2 (0.287 kPa ×m3 /kg ×K)(288.15 K)

The first law when adapted to this system gives

Qin + mi hi - me he = m2u2 - m1u1 Qin = m2u2 - m1u1 + me he - mi hi Qin = m2 cv T - (ma ,1cv T + mwuw ) + mw hw - (m2 - m1 )c pT

Noting that

uw @ hw = 62.98 kJ/kg Substituting,

Qin = 12,697´ 0.718´ 288 - (1814´ 0.718´ 288 + 600,000´ 62.98)+ 600,000´ 62.98 - (12,697 - 1814)´ 1.005´ 288

= 0 kJ The process is adiabatic.

EES (Engineering Equation Solver) SOLUTION "Given" Vol2=700 [m^3] Vol1=100 [m^3] P1=1500 [kPa] T1=(15+273.15) [K] P_in=1500 [kPa] T_in=(20+273.15) [K] T2=T1 P2=P1 "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_v=1.005 [kJ/kg-K] "Table A-2a" v_w=0.001 [m^3/kg] "for water" "Analysis" m1=(P1*Vol1)/(R*T1) Vol_w1=Vol2-Vol1 m_w=Vol_w1/v_w m2=(P2*Vol2)/(R*T2) h_w=enthalpy(steam_iapws, T=T1, x=0) u_w=h_w "approximated" Q_in=m2*c_v*T2-(m1*c_v*T1+m_w*u_w)+m_w*h_w-(m2-m1)*c_p*T1

5-180 Steam expands in a turbine steadily. The mass flow rate of the steam, the exit velocity, and the power output are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible.

1-603

Properties From the steam tables (Tables A-4 through 6) and

3 P1 = 7 MPa ïü ïý v1 = 0.05567 m /kg T1 = 600°C ïïþ h1 = 3650.6 kJ/kg

3 P2 = 25 kPa ïü ïý v 2 = v f + x2v fg = 0.00102 + (0.95)(6.2034 - 0.00102 ) = 5.8933 m /kg h2 = h f + x2 h fg = 271.96 + (0.95)(2345.5) = 2500.2 kJ/kg x2 = 0.95 ïïþ

Analysis (a) The mass flow rate of the steam is m=

1 1 V1 A1 = (60 m/s)(0.015 m 2 ) = 16.17 kg / s v1 0.05567 m3 /kg

(b) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit velocity is determined from

mv 2 (16.17 kg/s)(5.8933 m3 /kg) 1 V2 A2 ¾ ¾ ® V2 = = = 680.6 m / s v2 A2 0.14 m2

(c) We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = Wout + Qout + m(h2 +V22 /2) (since D pe @ 0) æ V 2 - V12 ö ÷ ÷ Wout = - Qout - m çççh2 - h1 + 2 ÷ ÷ çè 2 ø Then the power output of the turbine is determined by substituting to be 2 æ (680.6 m/s) 2 - (60 m/s ) æ 1 kJ/kg ÷ öö ÷ ç ÷ çç Wout = - (16.17 ´ 20) kJ/s - (16.17 kg/s )çç2500.2 - 3650.6 + ÷ ÷ 2 2 ÷ ÷ ççè èç1000 m /s ø÷ 2 ÷ ø

= 14,560 kW

EES (Engineering Equation Solver) SOLUTION $Array off "places array variables in the Solutions Window instead of the Arrays Window." A[1]=150 [cm^2] T[1]=600 [C] P[1]=7000 [kPa] Vel[1]= 60 [m/s] A[2]=1400 [cm^2] P[2]=25 [kPa] q_out = 20 [kJ/kg] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-604

m_dot = A[1]*convert(cm^2, m^2)*Vel[1]/v[1] v[1]=volume(steam_iapws, T=T[1], P=P[1]) "specific volume of steam at state 1" Vel[2]=m_dot*v[2]/(A[2]*Convert(cm^2, m^2)) v[2]=volume(steam_iapws, x=0.95, P=P[2]) "specific volume of steam at state 2" T[2]=temperature(steam_iapws, P=P[2], v=v[2]) "not required, but good to know" "Conservation of Energy for steady-flow: Ein_dot - Eout_dot = DeltaE_dot - For steady-flow, DeltaE_dot = 0" DELTAE_dot=0 "For the turbine as the control volume, neglecting the PE of each flow steam:" E_dot_in=E_dot_out h[1]=enthalpy(steam_iapws,T=T[1], P=P[1]) E_dot_in=m_dot*(h[1]+ Vel[1]^2/2*convert(J, kJ) ) h[2]=enthalpy(steam_iapws, x=0.95, P=P[2]) E_dot_out=m_dot*(h[2]+ Vel[2]^2/2*convert(J, kJ)) + m_dot *q_out+ W_dot_out Power=W_dot_out*Convert(kW, MW) Q_dot_out=m_dot*q_out

5-181 Problem 5-180 is reconsidered. The effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine as the exit pressure varies from 10 kPa to 50 kPa (with the same quality), and the exit area to varies from 1000 cm 2 to 3000 cm 2 is to be investigated. The exit velocity and the power output are to be plotted against the exit pressure for the exit areas of 1000, 2000, and 3000 cm 2 . Analysis The problem is solved using EES, and the results are tabulated and plotted below. A[1]=150 [cm^2] T[1]=600 [C] P[1]=7000 [kPa] Vel[1]= 60 [m/s] A[2]=1400 [cm^2] P[2]=25 [kPa] q_out = 20 [kJ/kg] Fluid$='Steam_IAPWS' m_dot = A[1]*Vel[1]/v[1]*convert(cm^2,m^2) v[1]=volume(Fluid$, T=T[1], P=P[1]) Vel[2]=m_dot*v[2]/(A[2]*convert(cm^2,m^2)) v[2]=volume(Fluid$, x=0.95, P=P[2]) T[2]=temperature(Fluid$, P=P[2], v=v[2]) E_dot_in=E_dot_out h[1]=enthalpy(Fluid$,T=T[1], P=P[1]) E_dot_in=m_dot*(h[1]+ Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) h[2]=enthalpy(Fluid$,x=0.95, P=P[2]) E_dot_out=m_dot*(h[2]+ Vel[2]^2/2*Convert(m^2/s^2, kJ/kg))+ m_dot *q_out+ W_dot_out Power=W_dot_out Q_dot_out=m_dot*q_out

Power Vel2 [m/ s] [kW] 10 -22158 2253 14.44 -1895 1595 18.89 6071 1239 23.33 9998 1017 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

P2 [kPa]

1-605

27.78 32.22 36.67 41.11 45.56 50

12212 13573 14464 15075 15507 15821

863.2 751.1 665.4 597.8 543 497.7

Table values are for A[2]  1000 [cm 2 ]

5-182 Water is to be heated steadily from 20°C to 48°C by an electrical resistor inside an insulated pipe. The power rating of the resistance heater and the average velocity of the water are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time at any point within the system and thus D mCV = 0 and D ECV = 0. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-606

Water is an incompressible substance with constant specific heats.

The kinetic and potential energy changes are negligible, D ke @D pe @0.

The pipe is insulated and thus the heat losses are negligible.

Properties The density and specific heat of water at room temperature are ρ  1000 kg / m3 and c  4.18 kJ / kg. C (Table A-3).

Analysis (a) We take the pipe as the system. This is a control volume since mass crosses the system boundary during the process. Also, there is only one inlet and one exit and thus m1 = m2 = m. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout We,in + mh1 = mh2 (since Qout @D ke @D pe @0)

We, in = m(h2 - h1 ) = m[c(T2 - T1 ) + v D P ¿ 0 ] = mc (T2 - T1 ) The mass flow rate of water through the pipe is

m = r V1 = (1000 kg/m3 )(0.024 m3 /min) = 24 kg/min Therefore,

We, in = mc(T2 - T1 ) = (24/60 kg/s)(4.18 kJ/kg ×o C)(48 - 20)°C = 46.8 kW (b) The average velocity of water through the pipe is determined from

V V 0.024 m3 /min = = = 5.43 m / min A p r 2 π (0.0375 m) 2

EES (Engineering Equation Solver) SOLUTION "Given" D=0.075 [m] T1=20 [C] T2=48 [C] Vol_dot=24[L/min]*Convert(L/min, m^3/s) "Properties" rho=1000 [kg/m^3] "Table A-3" c_p=4.18 [kJ/kg-C] "Table A-3" "Analysis" "(a)" m_dot=rho*Vol_dot W_dot_e_in=m_dot*C_p*(T2-T1) "(b)" A=pi*D^2/4 Vol_dot=Vel*A*Convert(m/min, m/s)

1-607

5-183 A tank initially contains saturated mixture of R-134a. A valve is opened and R-134a vapor only is allowed to escape slowly such that temperature remains constant. The heat transfer necessary with the surroundings to maintain the temperature and pressure of the R-134a constant is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the exit remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.

- me = m2 - m1 me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 Qin = m2u2 - m1u1 + me he Combining the two balances:

Qin = m2u2 - m1u1 + (m1 - m2 )he The specific volume at the initial state is

v1 =

V 0.0015 m3 = = 0.002727 m3 /kg m1 0.55 kg

The initial state properties of R-134a in the tank are

v1 - v f 0.002727 - 0.0008312 x = = = 0.06499 ïü ïý 1 v fg 0.030008 - 0.0008312 (Table A-11) 3 v1 = 0.002727 m /kg ïïþ u1 = u f + x1u fg = 87.26 + (0.06499)(156.89) = 97.45 kJ/kg T1 = 26°C

The enthalpy of saturated vapor refrigerant leaving the bottle is

he = hg @ 26°C = 264.73 kJ/kg The specific volume at the final state is

v2 =

V 0.0015 m3 = = 0.01 m3 /kg m2 0.15 kg

1-608

v2 - v f 0.01- 0.0008312 ü = = 0.3143 ïï x2 = v 0.030008 - 0.0008312 (Table A-11) ý fg 3 v 2 = 0.01 m /kg ïïþ u2 = u f + x1u fg = 87.26 + (0.3143)(156.89) = 136.56 kJ/kg T2 = 26°C

Substituting into the energy balance equation,

Qin = m2u2 - m1u1 + (m1 - m2 )he = (0.15 kg)(136.56 kJ/kg) - (0.55 kg)(97.45 kJ/kg) + (0.55 - 0.15 kg)(264.73 kJ/kg) = 72.8 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.0015 [m^3] m1=0.55 [kg] T1=26 [C] T2=T1 m2=0.15 [kg] "Analysis" Fluid$='R134a' v1=Vol/m1 v1_f=volume(Fluid$, T=T1, x=0) v1_g=volume(Fluid$, T=T1, x=1) x1=(v1-v1_f)/(v1_g-v1_f) u1_f=intenergy(Fluid$, T=T1, x=0) u1_g=intenergy(Fluid$, T=T1, x=1) u1_fg=u1_g-u1_f u1=u1_f+x1*u1_fg h_e=enthalpy(Fluid$, T=T1, x=1) v2=Vol/m2 v2_f=volume(Fluid$, T=T2, x=0) v2_g=volume(Fluid$, T=T2, x=1) x2=(v2-v2_f)/(v2_g-v2_f) u2_f=intenergy(Fluid$, T=T2, x=0) u2_g=intenergy(Fluid$, T=T2, x=1) u2_fg=u2_g-u2_f u2=u2_f+x2*u2_fg m_e=m1-m2 "mass balance" Q_in=m2*u2-m1*u1+m_e*h_e "energy balance"

5-184 R-134a is allowed to leave a piston-cylinder device with a pair of stops. The work done and the heat transfer are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device is assumed to be constant. 2 Kinetic and potential energies are negligible.

1-609

Properties The properties of R-134a at various states are (Tables A-11 through A-13)

v = 0.032659 m3 /kg P1 = 800 kPa ïüï 1 ý u = 290.86 kJ/kg T1 = 80 °C ïïþ 1 h1 = 316.99 kJ/kg v = 0.042115 m3 /kg P2 = 500 kPa ïüï 2 ý u = 242.42 kJ/kg T2 = 20 °C ïïþ 2 h2 = 263.48 kJ/kg Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wb, in - Qout - me he = m2u2 - m1u1

(since ke @ pe @ 0)

The volumes at the initial and final states and the mass that has left the cylinder are V1 = m1v1 = (2 kg)(0.032659 m3 /kg) = 0.06532 m3 V2 = m2v 2 = (1/ 2)m1v 2 = (1/2)(2 kg)(0.042115 m3 /kg) = 0.04212 m 3

me = m1 - m2 = 2 - 1 = 1 kg The enthalpy of the refrigerant withdrawn from the cylinder is assumed to be the average of initial and final enthalpies of the refrigerant in the cylinder

he = (1/ 2)(h1 + h2 ) = (1/ 2)(316.99 + 263.48) = 290.23 kJ/kg Noting that the pressure remains constant after the piston starts moving, the boundary work is determined from

Wb, in = P2 (V1 - V2 ) = (500 kPa)(0.06532 - 0.04212) m3 = 11.6 kJ (b) Substituting,

11.6 kJ - Qout - (1 kg)(290.23 kJ/kg) = (1 kg)(242.42 kJ/kg) - (2 kg)(290.86 kJ/kg) Qout = 60.7 kJ

1-610

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_1=2 [kg] P_1=800 [kPa] T_1=80 [C] P_2=500 [kPa] T_2=20 [C] m_2=1/2*m_1 "PROPERTIES" Fluid$='R134a' u_1=intenergy(Fluid$, P=P_1, T=T_1) h_1=enthalpy(Fluid$, P=P_1, T=T_1) v_1=volume(Fluid$, P=P_1, T=T_1) Vol_1=m_1*v_1 u_2=intenergy(Fluid$, P=P_2, T=T_2) h_2=enthalpy(Fluid$, P=P_2, T=T_2) v_2=volume(Fluid$, P=P_2, T=T_2) Vol_2=m_2*v_2 h_e=1/2*(h_1+h_2) "ANALYSIS" m_e=m_1-m_2 "mass balance" W_b_in-m_e*h_e-Q_out=m_2*u_2-m_1*u_1 "energy balance" W_b_in=P_2*(Vol_1-Vol_2) "boundary work"

5-185 Air is allowed to leave a piston-cylinder device with a pair of stops. Heat is lost from the cylinder. The amount of mass that has escaped and the work done are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device is assumed to be constant. 2 Kinetic and potential energies are negligible. 3 Air is an ideal gas with constant specific heats at the average temperature. Properties The properties of air are R  0.287 kPa  m3 / kg  K (Table A-1), cv  0.733 kJ / kg  K , cv  1.020 kJ / kg  K at the anticipated average temperature of 450 K (Table A-2b).

1-611

Energy balance:

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wb, in - Qout - me he = m2u2 - m1u1 (since ke @ pe @ 0)

Wb, in - Qout - me C pTe = m2 cv T2 - m1cv T1

The temperature of the air withdrawn from the cylinder is assumed to be the average of initial and final temperatures of the air in the cylinder. That is,

Te = (1/ 2)(T1 + T2 ) = (1/ 2)(473 + T2 ) The volumes and the masses at the initial and final states and the mass that has escaped from the cylinder are given by

V1 =

m1 RT1 (1.2 kg)(0.287 kPa. m3 /kg.K)(200 + 273 K) = = 0.2327 m3 P1 (700 kPa)

V2 = 0.80V1 = (0.80)(0.2327) = 0.1862 m3

m2 =

P2V2 (600 kPa)(0.1862 m3 ) 389.18 = = kg RT2 T2 (0.287 kPa. m3 /kg.K)T2

æ ö 389.18 ÷ ÷ me = m1 - m2 = ççç1.2 kg ÷ çè T2 ÷ ø Noting that the pressure remains constant after the piston starts moving, the boundary work is determined from

Wb, in = P2 (V1 - V2 ) = (600 kPa)(0.2327 - 0.1862) m3 = 27.9 kJ Substituting,

æ 389.18 ö ÷ ÷ 27.9 kJ - 40 kJ - ççç1.2 (1.020 kJ/kg.K)(1/2)(473 + T2 ) ÷ ÷ çè T2 ø æ389.18 ÷ ö ÷ = ççç (0.733 kJ/kg.K)T2 - (1.2 kg)(0.733 kJ/kg.K)(473 K) ÷ ÷ çè T2 ø The final temperature may be obtained from this equation by a trial-error approach or using EES to be

T2 = 415.0 K Then, the amount of mass that has escaped becomes

me = 1.2 -

389.18 = 0.262 kg 415.0 K

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_1=1.2 [kg] P_1=700 [kPa] T_1=200[C]+273 [K] P_2=600 [kPa] Vol_2=0.80*Vol_1 Q_out=40 [kJ] "PROPERTIES" Fluid$='Air' C_p=CP(Fluid$, T=T_ave) C_v=CV(Fluid$, T=T_ave) T_ave=1/2*(T_1+T_2) R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Fluid$) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-612

"ANALYSIS" m_e=m_1-m_2 "mass balance" W_b_in-m_e*C_p*T_e-Q_out=m_2*C_v*T_2-m_1*C_v*T_1 "energy balance" T_e=1/2*(T_1+T_2) "approximate temperature of air escaping" W_b_in=P_2*(Vol_1-Vol_2) "boundary work" Vol_1=(m_1*R*T_1)/P_1 m_2=(P_2*Vol_2)/(R*T_2)

5-186 Heat is transferred to a pressure cooker at a specified rate for a specified time period. The cooking temperature and the water remaining in the cooker are to be determined. Assumptions 1 This process can be analyzed as a uniform-flow process since the properties of the steam leaving the control volume remain constant during the entire cooking process. 2

The kinetic and potential energies of the streams are negligible, ke  pe  0.

3 The pressure cooker is stationary and thus its kinetic and potential energy changes are zero; that is, KE = PE = 0 and ΔEsystem = ΔU system . 4 The pressure (and thus temperature) in the pressure cooker remains constant. 5 Steam leaves as a saturated vapor at the cooker pressure. 6 There are no boundary, electrical, or shaft work interactions involved. 7 Heat is transferred to the cooker at a constant rate. Analysis We take the pressure cooker as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is an unsteady-flow process since changes occur within the control volume. Also, there is one exit and no inlets for mass flow. (a) The absolute pressure within the cooker is

Pabs = Pgage + Patm = 75 + 100 = 175 kPa Since saturation conditions exist in the cooker at all times, the cooking temperature must be the saturation temperature corresponding to this pressure. From Table A–5, it is

T2 = Tsat @ 175 kPa = 116.04°C which is about 16°C higher than the ordinary cooking temperature. (b) Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

- me = (m2 - m1 )CV or me = (m1 - m2 )CV

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 (since W @ ke @ pe @0)

1-613

Combining the mass and energy balances gives

Qin = (m1 - m2 )he + m2u2 - m1u1 The amount of heat transfer during this process is found from

Qin = Qin D T = (0.5 kJ/s)(30´ 60 s) = 900 kJ Steam leaves the pressure cooker as saturated vapor at 175 kPa at all times. Thus,

he = hg @ 175 kPa = 2700.2 kJ/kg The initial internal energy is found after the quality is determined:

v1 = x1 =

V 0.006 m3 = = 0.006 m3 /kg m1 1 kg v1 - v f v fg

0.006 - 0.001 = 0.00499 1.004 - 0.001

Thus,

u1 = u f + x1u fg = 486.82 + (0.00499)(2037.7) kJ/kg = 497 kJ/kg and

U1 = m1u1 = (1 kg)(497 kJ/kg) = 497 kJ The mass of the system at the final state is m2  V / v2 . Substituting this into the energy equation yields æ æV ö Vö ÷ ÷ ÷ Qin = çççm1 he + ççç u2 - m1u1 ÷ ÷ ÷ ÷ ÷ v2 ø èç èçv 2 ø

There are two unknowns in this equation, u2 and v2 . Thus we need to relate them to a single unknown before we can determine these unknowns. Assuming there is still some liquid water left in the cooker at the final state (i.e., saturation conditions exist), v2 and u2 can be expressed as

v 2 = v f + x2v fg = 0.001 + x2 (1.004 - 0.001) m3 /kg u2 = u f + x2u fg = 486.82 + x2 (2037.7) kJ/kg

Recall that during a boiling process at constant pressure, the properties of each phase remain constant (only the amounts change). When these expressions are substituted into the above energy equation, x2 becomes the only unknown, and it is determined to be

x2 = 0.009 Thus,

v 2 = 0.001 + (0.009)(1.004 - 0.001) m3 /kg = 0.010 m3 /kg and

m2 =

V 0.006 m3 = = 0.6 kg v 2 0.01 m3 /kg

Therefore, after 30 min there is 0.6 kg water (liquid + vapor) left in the pressure cooker. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-614

Discussion Note that almost half of the water in the pressure cooker has evaporated during cooking.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.006 [m^3] P_gage=75 [kPa] m1=1 [kg] Q_dot_in=0.500 [kW] DELTAt=(30*60) [s] P_atm=100 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" P=P_gage+P_atm T2=temperature(Fluid$, P=P, x=1) "saturation temperature" "(b)" Q_in=Q_dot_in*DELTAt h_e=enthalpy(Fluid$, P=P, x=1) v1=Vol/m1 v_f=volume(Fluid$, P=P, x=0) v_g=volume(Fluid$, P=P, x=1) x1=(v1-v_f)/(v_g-v_f) u_f=intenergy(Fluid$, P=P, x=0) u_g=intenergy(Fluid$, P=P, x=1) u_fg=u_g-u_f u1=u_f+x1*u_fg m2=Vol/v2 v2=v_f+x2*(v_g-v_f) u2=u_f+x2*u_fg m_e=m1-m2 "mass balance" Q_in=m2*u2-m1*u1+m_e*h_e "energy balance"

5-187 The air in a tank is released until the pressure in the tank reduces to a specified value. The mass withdrawn from the tank is to be determined for three methods of analysis. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are no work or heat interactions involved. Properties The gas constant of air is 0.287 kPa  m3 / kg  K (Table A-1). The specific heats of air at room temperature are c p  1.005 kJ / kg  K and cv  0.718 kJ / kg  K . Also k = 1.4 (Table A-2a).

1-615

- me = m2 - m1 me = m1 - m2 Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- me he = m2u2 - m1u1 0 = m2u2 - m1u1 + me he 0 = m2 cv T2 - m1cv T1 + me c pTe

Combining the two balances:

0 = m2 cv T2 - m1cv T1 + (m1 - m2 )c pTe The initial and final masses are given by

m1 =

P1V (800 kPa)(1 m3 ) = = 9.354 kg RT1 (0.287 kPa ×m3 /kg ×K)(25 + 273 K)

m2 =

P2V (150 kPa)(1 m3 ) 522.6 = = 3 RT2 (0.287 kPa ×m /kg ×K)T2 T2

æ ö æ298 + T2 ö 522.6 522.6 ÷ ÷ ÷(1.005) çç (0.718)T2 - (9.354)(0.718)(298) + ççç9.354 ÷ ÷ ÷ çè 2 ÷ ç ø T2 T2 ø è

whose solution is

T2 = 191.0 K Substituting, the final mass is

522.6 = 2.736 kg 191 and the mass withdrawn is m2 =

me = m1 - m2 = 9.354 - 2.736 = 6.618 kg (b) Considering the process in two parts, first from 800 kPa to 400 kPa and from 400 kPa to 150 kPa, the solution will be as follows: From 800 kPa to 400 kPa:

m2 = 0=

P2V (400 kPa)(1 m3 ) 1394 = = RT2 (0.287 kPa ×m3 /kg ×K)T2 T2

æ ö æ298 + T2 ÷ ö 1394 1394 ÷ ÷ (0.718)T2 - (9.354)(0.718)(298) + ççç9.354 (1.005) çç ÷ ÷ ÷ ç è ø T2 T2 ÷ 2 èç ø

1-616

m2 =

1394 = 5.687 kg 245.1

me,1 = m1 - m2 = 9.354 - 5.687 = 3.667 kg From 400 kPa to 150 kPa: 0=

æ æ245.1 + T2 ÷ ö 522.6 522.6 ö ÷ ÷(1.005) çç (0.718)T2 - (5.687)(0.718)(245.1) + ççç5.687 ÷ ÷ ÷ ç ÷ ç è ø T2 T2 ø 2 è

T2 = 186.5 K

m2 =

522.6 = 2.803 kg 186.5

me,2 = m1 - m2 = 5.687 - 2.803 = 2.884 kg The total mass withdrawn is

me = me,1 + me,2 = 3.667 + 2.884 = 6.551 kg (c) The mass balance may be written as

dm = - me dt When this is combined with the ideal gas equation of state, it becomes V d (P / T ) = - me R dt since the tank volume remains constant during the process. An energy balance on the tank gives d (mu ) = - he me dt cv

d (mT ) dm = c pT dt dt

V dP V d (P / T ) = c pT R dt R dt

ædP P dT ö dP ÷ = c p çç ÷ ÷ çè dt T dt ø dt

(c p - cv )

dP dT = cp P dt

When this result is integrated, it gives ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P ÷ ø 1

0.4/1.4

æ150 kPa ÷ ö = (298 K) çç ÷ çè800 kPa ø÷

= 184.7 K

The final mass is

m2 =

P2V (150 kPa)(1 m3 ) = = 2.830 kg RT2 (0.287 kPa ×m3 /kg ×K)(184.7 K)

and the mass withdrawn is

me = m1 - m2 = 9.354 - 2.830 = 6.524 kg Discussion The result in first method is in error by 1.4% while that in the second method is in error by 0.4%.

1-617

P1=800 [kPa] T1=(25+273) [K] P2=150 [kPa] P2_int=400 [kPa] "input for part (b)" "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_v=0.718 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" "(a)" m_e=m1-m2 "mass balance" m1=(P1*Vol)/(R*T1) m2=(P2*Vol)/(R*T2) T_e=0.5*(T1+T2) 0=m2*c_v*T2-m1*c_v*T1+m_e*c_p*T_e "energy balance" "(b)" m_e_1=m1-m2_int m2_int=(P2_int*Vol)/(R*T2_int) T_e_int=0.5*(T1+T2_int) 0=m2_int*c_v*T2_int-m1*c_v*T1+m_e_1*c_p*T_e_int m_e_2=m2_int-m2_end m2_end=(P2*Vol)/(R*T2_end) T_e_end=0.5*(T2_int+T2_end) 0=m2_end*c_v*T2_end-m2_int*c_v*T2_int+m_e_2*c_p*T_e_end m_e_b=m_e_1+m_e_2 "(c)" T2_c=T1*(P2/P1)^((k-1)/k) m2_c=(P2*Vol)/(R*T2_c) m_e_c=m1-m2_c "Errors" Error_a=(m_e_c-m_e)/m_e_c*Convert(, %) Error_b=(m_e_c-m_e_b)/m_e_c*Convert(, %)

5-188 Geothermal water flows through a flash chamber, a separator, and a turbine in a geothermal power plant. The temperature of the steam after the flashing process and the power output from the turbine are to be determined for different flash chamber exit pressures. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The devices are insulated so that there are no heat losses to the surroundings. 4 Properties of steam are used for geothermal water.

Analysis For all components, we take the steam as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-618

Energy balance: Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

For each component, the energy balance reduces to Flash chamber:

h1 = h2

Separator:

m2 h2 = m3 h3 + mliquid hliquid

Turbine:

WT = m3 (h3 - h4 )

(a) For a flash chamber exit pressure of P2 = 1 MPa The properties of geothermal water are h1 = hsat @ 230 °C = 990.14 kJ/kg

h2 = h1 x2 =

h2 - h f @1000 kPa h fg @1000 kPa

990.14 - 762.51 = 0.113 2014.6

T2 = Tsat @1000 kPa = 179.9°C

P3 = 1000 kPa ïü ïý h = 2777.1 kJ/kg ïïþ 3 x3 = 1 P4 = 20 kPa ïü ïý h = h + x h = 251.42 + (0.05)(2357.5 kJ/kg) = 2491.1 kJ/kg 4 f 4 fg x4 = 0.95 ïïþ The mass flow rate of vapor after the flashing process is

m3 = x2 m2 = (0.113)(50 kg/s) = 5.649 kg/s Then, the power output from the turbine becomes

WT = (5.649 kg/s)(2777.1- 2491.1) = 1616 kW Repeating the similar calculations for other pressures, we obtain (b) For P2 = 500 MPa ,

T2 = 151.8°C,

WT = 2134 kW

T2 = 99.6°C,

WT = 2333 kW

(d) For P2 = 50 MPa ,

T2 = 81.3°C,

WT = 2173 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_1=230 [C] x_1=0 m_dot_1=50 [kg/s] P_4=20 [kPa] x_4=0.95 P_2=1000 [kPa] "ANALYSIS" Fluid$='Steam_iapws' h_1=enthalpy(Fluid$, T=T_1, x=x_1) h_2=h_1 x_2=quality(Fluid$, P=P_2, h=h_2) T_2=temperature(Fluid$, P=P_2, h=h_2) "temperature after flashing process" m_dot_3=x_2*m_dot_1 "mass flow rate of vapor after flashing" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-619

h_3=enthalpy(Fluid$, P=P_2, x=1) h_4=enthalpy(Fluid$, P=P_4, x=x_4) W_dot_T=m_dot_3*(h_3-h_4) "turbine power"

5-189 An adiabatic air compressor is powered by a direct-coupled steam turbine, which is also driving a generator. The net power delivered to the generator is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The devices are adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats.

Properties From the steam tables (Tables A-4 through 6) and

P3 = 12.5 MPa ïü ïý h = 3343.6 kJ/kg ïïþ 3 T3 = 500°C P4 = 10 kPa ïü ïý h = h + x h = 191.81 + (0.92)(2392.1) = 2392.5 kJ/kg 4 f 4 fg x4 = 0.92 ïïþ

From the air table (Table A-17), T1 = 295 K ¾ ¾ ® h1 = 295.17 kJ/kg T2 = 620 K ¾ ¾ ® h2 = 628.07 kJ/kg

Analysis There is only one inlet and one exit for either device, and thus min = mout = m. We take either the turbine or the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for either steadyflow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

For the turbine and the compressor it becomes Compressor:

Wcomp, in + mair h1 = mair h2 ® Wcomp, in = mair (h2 - h1 ) Turbine:

msteam h3 = Wturb, out + msteam h4 ® Wturb, out = msteam (h3 - h4 ) Substituting, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-620

Wcomp, in = (10 kg/s)(628.07 - 295.17 ) kJ/kg = 3329 kW Wturb, out = (25 kg/s)(3343.6 - 2392.5) kJ/kg = 23,777 kW Therefore,

Wnet,out = Wturb,out - Wcomp,in = 23, 777 - 3329 = 20,448 kW

EES (Engineering Equation Solver) SOLUTION "Given" P3=12500 [kPa] T3=500 [C] P4=10 [kPa] x4=0.92 m_dot_steam=25 [kg/s] P1=98 [kPa] T1=(295-273) [C] P2=1000 [kPa] T2=(620-273) [C] m_dot_air=10 [kg/s] "Properties" Fluid$='steam_iapws' h3=enthalpy(Fluid$, P=P3, T=T3) h4=enthalpy(Fluid$, P=P4, x=x4) h1=enthalpy(air, T=T1) h2=enthalpy(air, T=T2) "Analysis" W_dot_comp_in=m_dot_air*(h2-h1) W_dot_turb_out=m_dot_steam*(h3-h4) W_dot_net=W_dot_turb_out-W_dot_comp_in

5-190 The turbocharger of an internal combustion engine consisting of a turbine, a compressor, and an aftercooler is considered. The temperature of the air at the compressor outlet and the minimum flow rate of ambient air are to be determined. Assumptions 1 All processes are steady since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Air properties are used for exhaust gases. 4 Air is an ideal gas with constant specific heats. 5 The mechanical efficiency between the turbine and the compressor is 100%. 6 All devices are adiabatic. 7 The local atmospheric pressure is 100 kPa.

1-621

Properties The constant pressure specific heats of exhaust gases, warm air, and cold ambient air are taken to be c p = 1.063,1.008 , and 1.005 kJ / kg  K , respectively (Table A-2b). Analysis (a) An energy balance on turbine gives

WT = mexh c p ,exh (Texh,1 - Texh,2 )

= (0.02 kg/s)(1.063 kJ/kg ×K)(400 - 350)K

= 1.063 kW This is also the power input to the compressor since the mechanical efficiency between the turbine and the compressor is assumed to be 100%. An energy balance on the compressor gives the air temperature at the compressor outlet WC = ma c p ,a (Ta, 2 - Ta, 1 ) 1.063 kW = (0.018 kg/s)(1.008 kJ/kg ×K)(Ta, 2 - 50)K ¾ ¾ ® Ta, 2 = 108.6°C (b) An energy balance on the aftercooler gives the mass flow rate of cold ambient air ma c p , a (Ta, 2 - Ta, 3 ) = mca c p, ca (Tca, 2 - Tca, 1 )

(0.018 kg/s)(1.008 kJ/kg ×°C)(108.6 - 80)°C = mca (1.005 kJ/kg ×°C)(40 - 30)°C mca = 0.05161 kg/s The volume flow rate may be determined if we first calculate specific volume of cold ambient air at the inlet of aftercooler. That is,

v ca =

RT (0.287 kJ/kg ×K)(30 + 273 K) = = 0.8696 m3 /kg P 100 kPa

Vca = mv ca = (0.05161 kg/s)(0.8696 m3 /kg) = 0.0449 m3 / s = 44.9 L / s

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_exh[1]=400 [C] P_exh_1=120 [kPa] T_exh[2]=350 [C] m_dot_exh=0.02 [kg/s] T_air[1]=50 [C] P_air[1]=100 [kPa] P_air[2]=130 [kPa] m_dot_air=0.018 [kg/s] T_air[3]=80 [C] T_coldair[1]=30 [C] T_coldair[2]=40 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-622

eta_mec=1.0 "PROPERTIES" c_p_exh=1.063 [kJ/kg-K] "Table A-2b" c_p_air=1.008 [kJ/kg-K] "Table A-2b" c_p_coldair=1.005 [kJ/kg-K] "Table A-2b" R_air=0.287 [kJ/kg-K] "Table A-2b" "ANALYSIS" W_dot_T=m_dot_exh*c_p_exh*(T_exh[1]-T_exh[2]) "turbine power output" W_dot_C=eta_mec*W_dot_T "compressor power input" W_dot_C=m_dot_air*c_p_air*(T_air[2]-T_air[1]) m_dot_air*c_p_air*(T_air[2]-T_air[3])=m_dot_coldair*c_p_coldair*(T_coldair[2]-T_coldair[1]) "energy balance on aftercooler" Vol_dot_coldair=m_dot_coldair*v_coldair "Volume flow rate of ambient air at the inlet of aftercooler" P_air[1]*v_coldair=R_air*(T_coldair[1]+273) "ideal gas equation for the ambient air at the inlet of aftercooler" T_air_2=T_air[2]

5-191 A newly developed electrical energy storage technology is cryogenic energy storage (CES). The electricity generated from a renewable energy source such as solar or wind is used to liquefy air or nitrogen at a very low temperature during off-peak hours. The liquefied gas is then stored in cryogenic tanks. During peak hours, liquid nitrogen is pumped to a higher pressure and it is evaporated and superheated by the heat of ambient air. The resulting high-pressure gas is used to drive a gas turbine and produce electricity. Consider a CES system in which nitrogen is liquefied and stored in a 500-m3 cryogenic tank during offpeak hours. When needed later during peak hours, the entire liquid nitrogen is withdrawn from the tank and it enters an ideal pump at T1 = - 200 C and P1 = 0.12 MPa . It is pumped to a pressure of P2 = 15 MPa and is heated by ambient air to T3 = 20 C at constant pressure. It is expanded in a gas turbine to P4 = 100 MPa and T4 = - 177 C . Determine the amount of heat input in the evaporator and the amounts of work consumed by the pump and produced by the turbine, all in kWh. The properties of nitrogen are, h1  440.1 kJ / kg, 1  825.9 kg / m3 , h3  33.36 kJ / kg, h4  211.6 kJ / kg . 5-191 A cryogenic energy storage (CES) system is considered. The amount of heat input in the evaporator and the amounts of work consumed by the pump and produced by the turbine are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 The losses in the pump are negligible. 3 The devices are adiabatic.

1-623

h1  440.1 kJ / kg, ρ1  825.9 kg / m3 , h3  33.36 kJ / kg, h4  211.6 kJ / kg

Analysis Using the density value at pump inlet, the mass of nitrogen in the storage tank is determined from

m  V  (825.9 kg/m3 )(500 m3 )  412,962 kg Neglecting elevation and velocity changes across the pump, the work input to the pump per unit mass is determined as

wpump,in 

P2  P1





(15,000  120) kPa  18.02 kJ/kg 825.9 kg/m3

We take the pump as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the turbine, the energy balance for this control volume can be expressed per unit mass basis as

ein  eout

wpump,in  h1  h2

(since Q  0, ke  0, pe  0)

Therefore, the enthalpy of nitrogen at the exit of the pump is

h2  h1  wpump,in  440.1  18.02  422.0 kJ/kg We take the evaporator as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the turbine, the energy balance for this control volume can be expressed per unit mass basis as

ein  eout

qin  h2  h3

(since Q  0, ke  0, pe  0)

Then, the amount of heat input in the evaporator is

 1 kWh  Qin  m(h3  h2 )  (412,962 kg)[(33.36)  (422.0)] kJ/kg    44,586 kW h  3600 kJ  The total work input to the pump is

 1 kWh  Wpump,in  mwpump,in  (412,962 kg)(18.02 kJ/kg)    2067 kW h  3600 kJ  We take the turbine as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the turbine, the energy balance for this control volume can be expressed per unit mass basis as

ein  eout

h3  h4  wturb,out

(since Q  0, ke  0, pe  0)

Then, the work output from the turbine becomes

 1 kWh  Wturb,out  m(h3  h4 )  (412,962 kg)[(33.36)  (211.6)] kJ/kg    20,442 kW h  3600 kJ 

EES (Engineering Equation Solver) SOLUTION T1=-200 [C] P1=120 [kPa] V=500 [m^3] P2=15000 [kPa] T3=20 [C] P3=P2 P4=100 [kPa] T4=-177 [C] rho=density(Nitrogen, P=P1, T=T1) m=rho*V h1=enthalpy(Nitrogen, P=P1, T=T1) w_pump=(P2-P1)/rho h2=h1+w_pump © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-624

h3=enthalpy(Nitrogen, P=P3, T=T3) h4=enthalpy(Nitrogen, P=P4, T=T4) w_out_permass=(h3-h4) Q_in=m*(h3-h2)*convert(kJ,kWh) W_in_pump=m*w_pump*convert(kJ,kWh) W_out_turb=m*(h3-h4)*convert(kJ,kWh)

5-192 A water tank open to the atmosphere is initially filled with water. The tank discharges to the atmosphere through a long pipe connected to a valve. The initial discharge velocity from the tank and the time required to empty the tank are to be determined. Assumptions 1 The flow is incompressible. 2 The draining pipe is horizontal. 3 The tank is considered to be empty when the water level drops to the center of the valve.

Analysis (a) Substituting the known quantities, the discharge velocity can be expressed as V=

2 gz = 1.5 + fL / D

2 gz = 1.5 + 0.015(100 m)/(0.10 m)

0.1212 gz

Then the initial discharge velocity becomes

V1 =

0.1212 gz1 =

0.1212(9.81 m/s 2 )(2 m) = 1.54 m / s

where z is the water height relative to the center of the orifice at that time. (b) The flow rate of water from the tank can be obtained by multiplying the discharge velocity by the pipe cross-sectional area,

p D2 0.1212 gz 4 Then the amount of water that flows through the pipe during a differential time interval dt is V = ApipeV2 =

p D2 0.1212 gzdt (1) 4 which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, dV = V dt =

p D02 dz (2) 4 where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since the positive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting Eqs. (1) and (2) equal to each other and rearranging, dV = Atank (- dz ) = -

1-625

p D2 4

0.1212 gzdt = -

p D02 D02 dz ® dt = 4 D2

dz 0.1212 gz

D02 D

- 1

0.1212 g

z 2 dz

The last relation can be integrated easily since the variables are separated. Letting t f be the discharge time and integrating it from t = 0 when z = z1 to t = t f when z = 0 (completely drained tank) gives

dt = -

t= 0

D02

ò 0.1212 g

z = z1

- 1/ 2

dz ® t f = -

D02

0.1212 g

1 2

= z1

2 D02 D

0.1212 g

z12

Simplifying and substituting the values given, the draining time is determined to be tf =

2 D02 D2

z1 2(10 m) 2 = 0.1212 g (0.1 m) 2

2m = 25,940 s = 7.21 h 0.1212(9.81 m/s 2 )

Discussion The draining time can be shortened considerably by installing a pump in the pipe.

EES (Engineering Equation Solver) SOLUTION "Given" D_0=10 [m] z1=2 [m] D=0.10 [m] L=100 [m] f=0.015 "Analysis" "(a)" g=9.807 [m/s^2] Vel1=sqrt((2*g*z1)/(1.5+(f*l)/D)) "(b)" t_f=(2*D_0^2)/D^2*sqrt(z1/(0.1212*g))*Convert(s, h) "Obtained after some manipulations and integration"

5-193 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified).

1-626

Analysis We take the bottle as the system. It is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

mi = m2

(since mout = minitial = 0)

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 (since W @ Eout = Einitial = ke @ pe @0) Combining the two balances:

Qin = m2 (u2 - hi ) = m2 (cv T2 - c pTi ) but

Ti = T2 = T0 and c p - cv = R.

Substituting, Qin = m2 (cv - c p )T0 = - m2 RT0 = -

P0V RT0 = - P0V RT0

Therefore,

Qout = P0V (Heat is lost from the tank)

Fundamentals of Engineering (FE) Exam Problems 5-194 An adiabatic heat exchanger is used to heat cold water at 15C entering at a rate of 5 kg / s by hot air at 90C entering also at rate of 5 kg / s . If the exit temperature of hot air is 20C, the exit temperature of cold water is (a) 27C (b) 32C (c) 52C (d) 85C (e) 90C © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-627

Answer (b) 32C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Tw1=15 [C] m_dot_w=5 [kg/s] Tair1=90 [C] Tair2=20 [C] m_dot_air=5 [kg/s] c_w=4.18 [kJ/kg-C] cp_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(Tw2-Tw1) "Some Wrong Solutions with Common Mistakes:" (Tair1-Tair2)=(W1_Tw2-Tw1) "Equating temperature changes of fluids" cv_air=0.718 [kJ/kg-K] m_dot_air*cv_air*(Tair1-Tair2)=m_dot_w*c_w*(W2_Tw2-Tw1) "Using cv for air" W3_Tw2=Tair1 "Setting inlet temperature of hot fluid = exit temperature of cold fluid" W4_Tw2=Tair2 "Setting exit temperature of hot fluid = exit temperature of cold fluid"

5-195 A heat exchanger is used to heat cold water at 15C entering at a rate of 2 kg / s by hot air at 85C entering at rate of

3 kg / s . The heat exchanger is not insulated, and is loosing heat at a rate of 25 kJ / s . If the exit temperature of hot air is 20C, the exit temperature of cold water is (a) 28C (b) 35C (c) 38C (d) 41C (e) 80C Answer (b) 35C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Tw1=15 [C] m_dot_w=2 [kg/s] Tair1=85 [C] Tair2=20 [C] m_dot_air=3 [kg/s] Q_loss=25 [kJ/s] c_w=4.18 [kJ/kg-C] cp_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(Tw2-Tw1)+Q_loss "Some Wrong Solutions with Common Mistakes:" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(W1_Tw2-Tw1) "Not considering Q_loss" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(W2_Tw2-Tw1)-Q_loss "Taking heat loss as heat gain" (Tair1-Tair2)=(W3_Tw2-Tw1) "Equating temperature changes of fluids" cv_air=0.718 [kJ/kg-K] m_dot_air*cv_air*(Tair1-Tair2)=m_dot_w*c_w*(W4_Tw2-Tw1)+Q_loss "Using cv for air"

1-628

5-196 An adiabatic heat exchanger is used to heat cold water at 15C entering at a rate of 5 kg / s by hot water at 90C entering at rate of 4 kg / s . If the exit temperature of hot water is 50C, the exit temperature of cold water is (a) 42C (b) 47C (c) 55C (d) 78C (e) 90C Answer (b) 47C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Tcold_1=15 [C] m_dot_cold=5 [kg/s] Thot_1=90 [C] Thot_2=50 [C] m_dot_hot=4 [kg/s] c_w=4.18 [kJ/kg-C] Q_loss=0 [kJ/s] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_hot*c_w*(Thot_1-Thot_2)=m_dot_cold*c_w*(Tcold_2-Tcold_1)+Q_loss "Some Wrong Solutions with Common Mistakes:" Thot_1-Thot_2=W1_Tcold_2-Tcold_1 "Equating temperature changes of fluids" W2_Tcold_2=90 "Taking exit temp of cold fluid=inlet temp of hot fluid"

5-197 In a shower, cold water at 10C flowing at a rate of 5 kg / min is mixed with hot water at 60C flowing at a rate of

2 kg / min. The exit temperature of the mixture will be (a) 24.3C (b) 35.0C (c) 40.0C (d) 44.3C (e) 55.2C Answer (a) 24.3C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Tcold_1=10 [C] m_dot_cold=5 [kg/min] Thot_1=60 [C] m_dot_hot=2 [kg/min] c_w=4.18 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_hot*c_w*Thot_1+m_dot_cold*c_w*Tcold_1=(m_dot_hot+m_dot_cold)*c_w*Tmix "Some Wrong Solutions with Common Mistakes:" W1_Tmix=(Tcold_1+Thot_1)/2 "Taking the average temperature of inlet fluids"

1-629

5-198 In a heating system, cold outdoor air at 7C flowing at a rate of 4 kg / min is mixed adiabatically with heated air at 70C flowing at a rate of 5 kg / min . The exit temperature of the mixture is (a) 34C (b) 39C (c) 42C (d) 57C (e) 70C Answer (c) 42C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Tcold_1=7 [C] m_dot_cold=4 [kg/min] Thot_1=70 [C] m_dot_hot=5 [kg/min] c_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_hot*c_air*Thot_1+m_dot_cold*c_air*Tcold_1=(m_dot_hot+m_dot_cold)*c_air*Tmix "Some Wrong Solutions with Common Mistakes:" W1_Tmix=(Tcold_1+Thot_1)/2 "Taking the average temperature of inlet fluids"

5-199 Refrigerant-134a expands in an adiabatic turbine from 1.2 MPa and 100C to 0.18 MPa and 50C at a rate of 1.25 kg / s . The power output of the turbine is (a) 44.7 kW (b) 66.4 kW (c) 72.7 kW (d) 89.2 kW (e) 112.0 kW Answer (a) 44.7 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1200 [kPa] T1=100 [C] P2=180 [kPa] T2=50 [C] m_dot=1.25 [kg/s] Q_dot_loss=0 [kJ/s] h1=ENTHALPY(R134a,T=T1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" -W_dot_out-Q_dot_loss=m_dot*(h2-h1) "Some Wrong Solutions with Common Mistakes:" -W1_Wout-Q_dot_loss=(h2-h1)/m_dot "Dividing by mass flow rate instead of multiplying" -W2_Wout-Q_dot_loss=h2-h1 "Not considering mass flow rate" u1=INTENERGY(R134a,T=T1,P=P1) u2=INTENERGY(R134a,T=T2,P=P2) -W3_Wout-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy" -W4_Wout-Q_dot_loss=u2-u1 "Using internal energy and ignoring mass flow rate" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-630

5-200 Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbine at 1 MPa and 1500 K at a rate of 0.1 kg / s , and exit at 0.2 MPa and 900 K. If heat is lost from the turbine to the surroundings at a rate of

15 kJ / s , the power output of the gas turbine is (a) 15 kW (b) 30 kW (c) 45 kW (d) 60 kW (e) 75 kW Answer (c) 45 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1000 [kPa] T1=1500 [K] P2=200 [kPa] T2=900 [K] m_dot=0.1 [kg/s] Q_dot_loss=15 [kJ/s] cp_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_out+Q_dot_loss=m_dot*cp_air*(T1-T2) "Alternative: Variable specific heats - using EES data" W_dot_out_variable+Q_dot_loss=m_dot*(ENTHALPY(Air,T=T1)-ENTHALPY(Air,T=T2)) "Some Wrong Solutions with Common Mistakes:" W1_Wout=m_dot*cp_air*(T1-T2) "Disregarding heat loss" W2_Wout-Q_dot_loss=m_dot*cp_air*(T1-T2) "Assuming heat gain instead of loss"

5-201 Steam expands in a turbine from 4 MPa and 500C to 0.5 MPa and 250C at a rate of 1350 kg / h . Heat is lost from the turbine at a rate of 25 kJ / s during the process. The power output of the turbine is (a) 157 kW (b) 207 kW (c) 182 kW (d) 287 kW (e) 246 kW Answer (a) 157 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=500 [C] P1=4000 [kPa] T2=250 [C] P2=500 [kPa] m_dot=(1350/3600) [kg/s] Q_dot_loss=25 [kJ/s] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_out+Q_dot_loss=m_dot*(h1-h2) "Some Wrong Solutions with Common Mistakes:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-631

W1_Wout=m_dot*(h1-h2) "Disregarding heat loss" W2_Wout-Q_dot_loss=m_dot*(h1-h2) "Assuming heat gain instead of loss" u1=INTENERGY(Steam_IAPWS,T=T1,P=P1) u2=INTENERGY(Steam_IAPWS,T=T2,P=P2) W3_Wout+Q_dot_loss=m_dot*(u1-u2) "Using internal energy instead of enthalpy" W4_Wout-Q_dot_loss=m_dot*(u1-u2) "Using internal energy and wrong direction for heat"

5-202 Steam is compressed by an adiabatic compressor from 0.2 MPa and 150C to 0.8 MPa and 350C at a rate of 1.30 kg / s . The power input to the compressor is (a) 511 kW (b) 393 kW (c) 302 kW (d) 717 kW (e) 901 kW Answer (a) 511 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=200 [kPa] T1=150 [C] P2=800 [kPa] T2=350 [C] m_dot=1.30 [kg/s] Q_dot_loss=0 [kJ/s] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_in-Q_dot_loss=m_dot*(h2-h1) "Some Wrong Solutions with Common Mistakes:" W1_Win-Q_dot_loss=(h2-h1)/m_dot "Dividing by mass flow rate instead of multiplying" W2_Win-Q_dot_loss=h2-h1 "Not considering mass flow rate" u1=INTENERGY(Steam_IAPWS,T=T1,P=P1) u2=INTENERGY(Steam_IAPWS,T=T2,P=P2) W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy" W4_Win-Q_dot_loss=u2-u1 "Using internal energy and ignoring mass flow rate"

5-203 Refrigerant-134a is compressed by a compressor from the saturated vapor state at 0.14 MPa to 0.9 MPa and 60C at a rate of 0.108 kg / s . The refrigerant is cooled at a rate of 1.10 kJ / s during compression. The power input to the compressor is (a) 4.94 kW (b) 6.04 kW (c) 7.14 kW (d) 7.50 kW (e) 8.13 kW Answer (c) 7.14 kW

1-632

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=140 [kPa] x1=1 P2=900 [kPa] T2=60 [C] m_dot=0.108 [kg/s] Q_dot_loss=1.10 [kJ/s] h1=ENTHALPY(R134a,x=x1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_in-Q_dot_loss=m_dot*(h2-h1) "Some Wrong Solutions with Common Mistakes:" W1_Win+Q_dot_loss=m_dot*(h2-h1) "Wrong direction for heat transfer" W2_Win =m_dot*(h2-h1) "Not considering heat loss" u1=INTENERGY(R134a,x=x1,P=P1) u2=INTENERGY(R134a,T=T2,P=P2) W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy" W4_Win+Q_dot_loss=u2-u1 "Using internal energy and wrong direction for heat transfer"

5-204 Refrigerant-134a at 1.4 MPa and 70C is throttled to a pressure of 0.6 MPa. The temperature of the refrigerant after throttling is (a) 70C (b) 66C (c) 57C (d) 49C (e) 22C Answer (c) 57C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1400 [kPa] T1=70 [C] P2=600 [kPa] h1=ENTHALPY(R134a,T=T1,P=P1) T2=TEMPERATURE(R134a,h=h1,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming the temperature to remain constant" W2_T2=TEMPERATURE(R134a,x=0,P=P2) "Taking the temperature to be the saturation temperature at P2" u1=INTENERGY(R134a,T=T1,P=P1) W3_T2=TEMPERATURE(R134a,u=u1,P=P2) "Assuming u=constant" v1=VOLUME(R134a,T=T1,P=P1) W4_T2=TEMPERATURE(R134a,v=v1,P=P2) "Assuming v=constant"

5-205 Steam enters a diffuser steadily at 0.5 MPa, 300C, and 90 m / s at a rate of 3.5 kg / s . The inlet area of the diffuser is (a) 22 cm 2 (b) 53 cm 2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-633

(c) 126 cm2 (d) 175 cm 2 (e) 203 cm2 Answer (e) 203 cm2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel_1=90 [m/s] m=3.5 [kg/s] T1=300 [C] P1=500 [kPa] "The rate form of energy balance is E_dot_in - E_dot_out = DELTAE_dot_cv" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) m=(1/v1)*A*Vel_1 "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] P1*v1ideal=R*(T1+273) m=(1/v1ideal)*W1_A*Vel_1 "assuming ideal gas" P1*v2ideal=R*T1 m=(1/v2ideal)*W2_A*Vel_1 "assuming ideal gas and using C" m=W3_A*Vel_1 "not using specific volume"

5-206 Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 280 m / s at a rate of 2.5 kg / s . If the temperature and pressure of the steam at the nozzle exit are 400C and 2 MPa, the exit area of the nozzle is (a) 8.4 cm 2 (b) 10.7 cm 2 (c) 13.5 cm 2 (d) 19.6 cm 2 (e) 23.0 cm2 Answer (c) 13.5 cm 2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel_1=0 [m/s] Vel_2=280 [m/s] m=2.5 [kg/s] T2=400 [C] P2=2000 [kPa] "The rate form of energy balance is E_dot_in - E_dot_out = DELTAE_dot_cv" v2=VOLUME(Steam_IAPWS,T=T2,P=P2) m=(1/v2)*A2*Vel_2 "A2 in m^2" "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] P2*v2ideal=R*(T2+273) m=(1/v2ideal)*W1_A2*Vel_2 "assuming ideal gas" P1*v2ideal=R*T2 m=(1/v2ideal)*W2_A2*Vel_2 "assuming ideal gas and using C" m=W3_A2*Vel_2 "not using specific volume" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

1-634

5-207 Air at 27C and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligible, the exit temperature of air will be (a) 10C (b) 15C (c) 20C (d) 23C (e) 27C Answer (e) 27C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=27 [C] P1=5 [atm] P2=1 [atm] T2=T1 "The temperature of an ideal gas remains constant during throttling, and thus T2=T1" "Some Wrong Solutions with Common Mistakes:" W1_T2=T1*P1/P2 "Assuming v=constant and using C" W2_T2=(T1+273)*P1/P2-273 "Assuming v=constant and using K" W3_T2=T1*P2/P1 "Assuming v=constant and pressures backwards and using C" W4_T2=(T1+273)*P2/P1 "Assuming v=constant and pressures backwards and using K"

5-208 Steam at 1 MPa and 300C is throttled adiabatically to a pressure of 0.4 MPa. If the change in kinetic energy is negligible, the specific volume of the steam after throttling will be (a) 0.358 m3 / kg (b) 0.233 m3 / kg (c) 0.375 m3 / kg (d) 0.646 m3 / kg (e) 0.646 m3 / kg Answer (d) 0.646 m3 / kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1000 [kPa] T1=300 [C] P2=400 [kPa] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=h1 v2=VOLUME(Steam_IAPWS,h=h2,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) "Assuming the volume to remain constant" u1=INTENERGY(Steam,T=T1,P=P1) W2_v2=VOLUME(Steam_IAPWS,u=u1,P=P2) "Assuming u=constant" W3_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) "Assuming T=constant"

1-635

5-209 Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 50C at a rate of 2 kg / s , the exit temperature of air will be (a) 46.0C (b) 50.0C (c) 54.0C (d) 55.4C (e) 58.0C Answer (c) 54.0C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=50 [C] m_dot=2 [kg/s] W_dot_e=8 [kJ/s] cp=1.005 [kJ/kg-K] W_dot_e=m_dot*cp*(T2-T1) "Some Wrong Solutions with Common Mistakes:" cv=0.718 [kJ/kg-K] W_dot_e=cp*(W1_T2-T1) "Not using mass flow rate" W_dot_e=m_dot*cv*(W2_T2-T1) "Using cv" W_dot_e=m_dot*cp*W3_T2 "Ignoring T1"

5-210 Saturated water vapor at 40C is to be condensed as it flows through a tube at a rate of 0.20 kg/s. The condensate leaves the tube as a saturated liquid at 40C. The rate of heat transfer from the tube is (a) 34 kJ / s (b) 481 kJ / s (c) 2406 kJ / s (d) 514 kJ / s (e)

548 kJ / s

Answer (b) 481 kJ / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=40 [C] m_dot=0.20 [kg/s] h_f=ENTHALPY(Steam_IAPWS,T=T1,x=0) h_g=ENTHALPY(Steam_IAPWS,T=T1,x=1) h_fg=h_g-h_f Q_dot=m_dot*h_fg "Some Wrong Solutions with Common Mistakes:" W1_Q=m_dot*h_f "Using hf" W2_Q=m_dot*h_g "Using hg" W3_Q=h_fg "not using mass flow rate" W4_Q=m_dot*(h_f+h_g) "Adding hf and hg"

1-636

6-1

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 6 THE SECOND LAW OF THERMODYNAMICS

6-2

CYU - Check Your Understanding CYU 6-1 ■ Check Your Understanding CYU 6-1.1 Select the wrong statement about the laws of thermodynamics. (a) A process cannot occur unless it satisfies both the first and the second laws of thermodynamics. (b) The second law asserts that energy has quality. (c) The first law asserts that the higher the quality energy, the more of it can be converted to work. (d) The second law is used in determining the theoretical limits for the performance of engineering systems. (e) The first law is an expression of the conservation of energy principle. Answer: (c) The first law asserts that the higher the quality energy, the more of it can be converted to work. CYU 6-1.2 A cup of coffee at 90C is placed in a room at 20C. Someone claims that as a result of heat transfer from the room air, the temperature of the tea increases to 95C. Select the correct statement for this process. (a) This process cannot take place because it violates the first law. (b) This process cannot take place because it violates the second law. (c) This process cannot take place because it violates both the first and second laws. (d) This process can take place. Answer: (b) It cannot take place because it violates the second law.

CYU 6-2 ■ Check Your Understanding CYU 6-2.1 Which of the following can reasonably be modeled as a thermal energy reservoir? (a) Large lakes (b) Atmospheric air (c) Large rivers (d) Industrial furnaces (e) All of them Answer: (e) All of them CYU 6-2.2 Consider the air-conditioning process during which heat is absorbed from the indoors at 23C and is discarded to the outdoors at 40C. Identify the heat source and the heat sink for this process. (a) Indoors: Sink, Outdoors: Source (b) Indoors: Source, Outdoors: Sink (c) Indoors: Sink, Outdoors: Sink (d) Indoors: Source, Outdoors: Source Answer: (b) Indoors: Source, Outdoors: Sink

CYU 6-3 ■ Check Your Understanding CYU 6-3.1 Select the correct statements regarding the operation of heat engines. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-3

I. Actual heat engines cannot have a thermal efficiency of 100 percent but ideal heat engines can. II. No heat engine can convert all the heat it receives to useful work. III. The requirement that a heat engine must operate on a cycle is the expression of the Kelvin-Planck statement of the second law. (a) Only I (b) Only II (c) Only III (d) I and II (e) I and III Answer: (b) Only II CYU 6-3.2 Of the net work output and thermal efficiency relations below for heat engines, select the correct ones? I. Wnet = Wout - Win II. Wnet = Qin - Qout III. hth = Wnet / Qin IV. hth = 1- Qout / Qin (a) I and II (b) III and IV (c) I and III (d) II and IV (e) I, II, III, and IV Answer: (e) I, II, III, and IV CYU 6-3.3 A heat engine absorbs heat from a source at a rate of 200 kW and rejects 120 kW of it to a sink. What are the net power output and the thermal efficiency for this engine? (a) 80 kW, 40% (b) 120 kW, 60% (c) 80 kW, 60% (d) 120 kW, 40% (e) 120 kW, 20% Answer: (a) 80 kW, 40% Wnet = Qin - Qout = 200 - 120 = 80 kW eta th = Wnet / Qin = 80 / 200 = 0.40 CYU 6-3.4 A steam power plant is supplied heat from a source at a rate of 1000 kW and produces a net power output of 300 kW. If the pump power input is 50 kW, what are the turbine power output and the rate of heat rejected from the plant, respectively? (a) 950 kW, 350 kW (b) 700 kW, 300 kW (c) 300 kW, 700 kW (d) 350 kW, 950 kW (e) 350 kW, 700 kW Answer: (e) 350 kW, 700 kW Wturb = Wnet + Wpump = 300 + 50 = 350 kW

Qout = Qin - Wnet = 1000 - 300 = 700 kW

6-4

CYU 6-4 ■ Check Your Understanding CYU 6-4.1 The state of the refrigerant at various locations of an ideal vapor-compression refrigeration cycle are given below. Select the wrong expression. (a) Saturated liquid at the condenser exit (b) Saturated vapor at the compressor exit (c) Superheated vapor at the condenser inlet (d) Saturated mixture at the evaporator inlet (e) Saturated liquid at the expansion valve inlet Answer: (b) Saturated vapor at the compressor exit CYU 6-4.2 Which statements are correct regarding refrigerators and heat pumps? I. The COP of a refrigerator is equal to the COP of a heat pump for fixed values of QL and QH . II. The minimum COP value of a heat pump is zero. III. The COP of a refrigerator cannot be less than 1. (a) I, II, and III (b) I and II (c) I and III (d) II and III (e) None of these Answer: (e) None of these CYU 6-4.3 Select the correct statements below regarding refrigerators and heat pumps. I. The COP of a refrigerator decreases with decreasing refrigerated space temperature. II. The Clausius statement asserts that a refrigerator cannot operate without energy input from an external source, usually in the form of work. III. In the evaporator of a vapor-compression refrigeration cycle, heat flows from a low-temperature medium to hightemperature medium. (a) Only III (b) I and II (c) I and III (d) II and III (e) II, II, and III Answer: (b) I and II CYU 6-4.4 A device running on a refrigeration cycle absorbs heat from a cool space at a rate of 4 kW and rejects 10 kW of heat to a warm space. What are the COP values of this device for the cases of operating as a refrigerator or a heat pump? (a) COPR = 1.5, COPHP = 2.5 (b) COPR = 2 / 5, COPHP = 3 / 5 (c) COPR = 2 / 3, COPHP = 5 / 3 (d) COPR = 3 / 5, COPHP = 7 / 3 (e) COPR = 1.5, COPHP = 3 / 5 Answer: (c) COPR = 2 / 3, COPHP = 5 / 3

Wnet = QH - QL = 10 - 4 = 6 kW COPR = QL / Wnet = 4 / 6 = 2 / 3 COPHP = QH / Wnet = 10 / 6 = 5 / 3

6-5

CYU 6-4.5 A heat pump absorbs heat from the cold outdoors at a rate of 3 kW and transfers 8 kW of heat to the warm indoors. What is the COP of this heat pump and the power input, respectively? (a) 8 / 3, 5 kW (b) 5 / 8, 3 kW (c) 3 / 8, 3 kW (d) 8 / 5, 8 kW (e) 8 / 5, 5 kW Answer: (e) 8 / 5, 5 kW

Wnet = QH - QL = 8 - 3 = 5 kW COPHP = QH / Wnet = 8 / 5

CYU 6-5 ■ Check Your Understanding CYU 6-5.1 Which of the following is a violation of the second law of thermodynamics? (a) A coal power plant with a thermal efficiency of 100 percent (b) A hydroelectric power plant with an efficiency of 100 percent (c) A generator with an efficiency of 100 percent. (d) An electric motor with an efficiency of 100 percent. (e) A resistance heater with an efficiency of 100 percent (supplying as much heat as the electricity it consumes) Answer: (a) A coal power plant with a thermal efficiency of 100 percent

CYU 6-6 ■ Check Your Understanding CYU 6-6.1 Which of the following renders a process irreversible? (a) Mixing of two fluids (b) Friction (c) Unrestrained expansion (d) Heat transfer across a finite temperature difference (e) All of them Answer: (e) All of them CYU 6-6.2 Of the statements below about reversible processes, select the correct ones. I. Reversible processes can be viewed as theoretical limits for the corresponding irreversible ones. II. A turbine delivers the most power when the process is reversible. III. A pump consumes the least power when the process is reversible. IV. Reversible processes serve as idealized models to which actual processes can be compared. (a) Only I and IV (b) Only I and II (c) Only II (d) Only I, II and III (e) All of them Answer: (e) All of them CYU 6-6.3 Which statement is wrong regarding irreversible processes? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-6

(a) Expansion of a fluid in a throttling valve is an irreversible process. (b) The larger the temperature difference during a heat transfer process, the smaller the irreversibility. (c) The larger the irreversibilities in a power plant, the smaller the thermal efficiency. (d) The smaller the electrical resistance during current flow in a wire, the smaller the irreversibility. (e) None of these Answer: (b) The larger the temperature difference during a heat transfer process, the smaller the irreversibility.

CYU 6-7 ■ Check Your Understanding CYU 6-7.1 Which statements below are correct regarding the Carnot cycle? I. The Carnot cycle is made up of reversible processes. II. The Carnot cycle cannot be executed in a steady-flow system. III. The Carnot cycle is the most efficient cycle operating between two specified temperature limits. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III

CYU 6-9 ■ Check Your Understanding CYU 6-9.1 Which statements below are correct regarding the Kelvin scale? I. On the Kelvin scale, the temperature ratios are related to the heat transfer ratios of reversible heat engines. II. On the Kelvin scale, temperatures vary between zero and infinity. III. The magnitudes of temperature change units on the Kelvin and Celsius scales are identical. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (e) I, II, and III

CYU 6-10 ■ Check Your Understanding CYU 6-10.1 Select the wrong statement below regarding the quality of energy. (a) Up to 30 percent of heat from a geothermal heat reservoir at 127C can be possibly converted to work in a heat engine when the sink temperature is 27C. (b) 100 percent of work can be converted to heat. (c) Only a fraction of heat can be converted to work. (d) A higher fraction of the high-temperature heat can be converted to work.

6-7

Answer: (a) Up to 30 percent of heat from a geothermal heat reservoir at 127C can be possibly converted to work in a heat engine when the sink temperature is 27C. CYU 6-10.2 Which equations can be used to determine the thermal efficiency of irreversible heat engines? I. hth = 1- QL / QH II. hth = Wnet / QH III. hth = 1- TL / TH (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (b) I and II CYU 6-10.3 A heat engine is said to absorb heat from a source at 1000 K at a rate of 250 kW and rejects 50 kW of it to a sink at 300 K. This heat engine is (a) Reversible (b) Irreversible (c) Impossible (d) Insufficient information Answer: (c) Impossible etath,rev = 1- TL / TH = 1- 300 /1000 = 0.70 etath = 1- Qth / QH = 1- 50 / 250 = 0.80 CYU 6-10.4 A heat engine absorbs heat from a source at 1000 K at a rate of 250 kW and rejects 100 kW of it to a sink at 300 K. The thermal efficiency of this heat engine is (a) 0.40 (b) 0.50 (c) 0.60 (d) 0.75 (e) 1 Answer: (c) 0.60 etath,rev = 1- TL / TH = 1- 300 /1000 = 0.70 etath, = 1- QL / QH = 1- 100 / 250 = 0.60 CYU 6-10.5 A heat engine absorbs heat from a source at 1000 K at a rate of 250 kW and rejects 100 kW of it to a sink at 300 K. The maximum possible thermal efficiency this heat engine can have is (a) 0.40 (b) 0.50 (c) 0.60 (d) 0.70 (e) 1 Answer: (d) 0.70 etath,rev = 1- TL / TH = 1- 300 /1000 = 0.70

etath, = 1- QL / QH = 1- 100 / 250 = 0.60

6-8

CYU 6-11 ■ Check Your Understanding CYU 6-11.1 Of the COP relations below, which one cannot be used to determine the COP of a Carnot heat pump? (a) COPHP,rev = QL / (QH - Win ) (b) COPHP,rev = 1/ (1- TL / TH ) (c) COPHP,rev = TH / (TH - TL ) (d) COPHP,rev = QH / Win (e) COPHP,rev = QH / (QH - QL ) Answer: (a) COPHP,rev = QL / (QH - Win ) CYU 6-11.2 A refrigerator is claimed to absorb 5 kJ of heat from the refrigerated space at 200 K and rejects 10 kJ of heat to a medium at 300 K. This refrigerator is (a) Reversible (b) Irreversible (c) Impossible (d) Insufficient information Answer: (b) Irreversible COPR ,rev = TL / (TH - TL ) = 200 / (300 - 200) = 2

COPR = Q L / (QH - Q L ) = 5 / (10 - 5) = 1 CYU 6-11.3 A heat pump absorbs heat at a rate of 5 kW from a cold medium at 250 K and supplies 10 kW of heat to a warm medium at 300 K. The COP of this heat pump is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer: (b) 2 COPHP = QH / (Q H - Q L ) = 10 / (10 - 5) = 2

COPHP,rev = TH / (TH - TL ) = 250 / (300 - 250) = 5 CYU 6-11.4 A refrigerator absorbs heat at a rate of 5 kW from a cold medium at 250 K and rejects 10 kW of heat to a warm medium at 300 K. The maximum possible COP a refrigerator operating between these temperature limits is (a) 0.5 (b) 1 (c) 2 (d) 3 (e) 5 Answer: (e) 5 COPR ,rev = TL / (TH - TL ) = 250 / (300 - 250) = 5

COPR = Q L / (QH - Q L ) = 5 / (10 - 5) = 1

6-9

PROBLEMS Second Law of Thermodynamics and Thermal Energy Reservoirs 6-1C Water is not a fuel; thus the claim is false.

6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 6 kWh of electricity.

6-3C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity.

6-4C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room.

6-5C No. Heat cannot flow from a low-temperature medium to a higher temperature medium.

6-6C A thermal-energy reservoir is a body that can supply or absorb finite quantities of heat isothermally. Some examples are the oceans, the lakes, and the atmosphere.

6-7C Yes. Because the temperature of the oven remains constant no matter how much heat is transferred to the potatoes.

6-8C The surrounding air in the room that houses the TV set.

Heat Engines and Thermal Efficiency 6-9C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, and reject the rest to a sink.

6-10C It is expressed as "No heat engine can exchange heat with a single reservoir, and produce an equivalent amount of work".

6-11C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.

6-12C (a) No, (b) Yes. According to the second law, no heat engine can have and efficiency of 100%.

6-10

6-13C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.

6-14C No. The Kelvin-Plank limitation applies only to heat engines; engines that receive heat and convert some of it to work.

6-15C No. Because 100% of the work can be converted to heat.

6-16C Method (b). With the heating element in the water, heat losses to the surrounding air are minimized, and thus the desired heating can be achieved with less electrical energy input.

6-17 The heat input and thermal efficiency of a heat engine are given. The work output of the heat engine is to be determined. Assumptions 1. The plant operates steadily. 2. Heat losses from the working fluid at the pipes and other components are negligible. Analysis Applying the definition of the thermal efficiency to the heat engine,

Wnet = hth QH = (0.35)(1.3 kJ) = 0.455 kJ

EES (Engineering Equation Solver) SOLUTION Q_H=1.3 [kJ] eta_th=0.35 W_net=eta_th*Q_H

6-18 The rates of heat supply and heat rejection of a power plant are given. The power output and the thermal efficiency of this power plant are to be determined. Assumptions 1. The plant operates steadily. 2. Heat losses from the working fluid at the pipes and other components are taken into consideration. Analysis

(a) The total heat rejected by this power plant is

6-11

QL = 165 + 8 = 173 GJ/h Then the net power output of the plant becomes

Wnet,out = QH - QL = 280 - 173 = 107 GJ/h = 29.7 MW (b) The thermal efficiency of the plant is determined from its definition,

hth =

Wnet, out QH

107 GJ/h = 0.382 = 38.2% 280 GJ/h

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_H=280 [GJ/h] Q_dot_losses=8 [GJ/h] Q_dot_WasteHeat=165 [GJ/h] "Analysis" "(a)" Q_dot_L=Q_dot_losses+Q_dot_WasteHeat W_dot_net_out=Q_dot_H-Q_dot_L W_dot_net_out_MW=W_dot_net_out*Convert(GJ/h, MW) "(b)" eta_th=W_dot_net_out/Q_dot_H

6-19E The rate of heat input and thermal efficiency of a heat engine are given. The power output of the heat engine is to be determined. Assumptions 1. 2.

The plant operates steadily. Heat losses from the working fluid at the pipes and other components are negligible.

Analysis Applying the definition of the thermal efficiency to the heat engine,

Wnet = hth QH æ ö 1 hp ÷ = (0.4)(3´ 104 Btu/h) çç ÷ ÷ çè 2544.5 Btu/h ø

= 4.72 hp

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_H=3E4 [Btu/h] eta_th=0.40 "Analysis" W_dot_net=eta_th*Q_dot_H*Convert(Btu/h, hp) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-12

6-20 The power output and thermal efficiency of a power plant are given. The rate of heat rejection is to be determined, and the result is to be compared to the actual case in practice. Assumptions 1. The plant operates steadily. 2. Heat losses from the working fluid at the pipes and other components are negligible. Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation,

QH =

Wnet, out hth

600 MW = 1500 MW 0.4

The rate of heat transfer to the river water is determined from the first law relation for a heat engine,

QL = QH - Wnet, out = 1500 - 600 = 900 MW In reality the amount of heat rejected to the river will be lower since part of the heat will be lost to the surrounding air from the working fluid as it passes through the pipes and other components.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net_out=600 [MW] eta_th=0.40 "Analysis" Q_dot_H=W_dot_net_out/eta_th Q_dot_L=Q_dot_H-W_dot_net_out

6-21 The heat rejection and thermal efficiency of a heat engine are given. The heat input to the engine is to be determined. Assumptions 1. The plant operates steadily. 2. Heat losses from the working fluid at the pipes and other components are negligible. Analysis According to the definition of the thermal efficiency as applied to the heat engine,

6-13

qH - qL = h th qH which when rearranged gives qH =

qL 500 kJ/kg = = 909 kJ / kg 1- hth 1- 0.45

EES (Engineering Equation Solver) SOLUTION "Given" eta_th=0.45 q_L=500 [kJ/kg] "Analysis" w_net=q_H-q_L w_net=eta_th*q_H

6-22E The work output and heat rejection of a heat engine that propels a ship are given. The thermal efficiency of the engine is to be determined.

Assumptions 1. The plant operates steadily. 2. Heat losses from the working fluid at the pipes and other components are negligible. Analysis Applying the first law to the heat engine gives

qH = wnet + qL = 500 Btu/lbm + 300 Btu/lbm = 800 Btu/lbm Substituting this result into the definition of the thermal efficiency, hth =

wnet 500 Btu/lbm = = 0.625 or 62.5% qH 800 Btu/lbm

EES (Engineering Equation Solver) SOLUTION w_net=500 [Btu/lbm] q_L=300 [Btu/lbm] q_H=w_net+q_L eta_th=w_net/q_H

6-23 The power output and fuel consumption rate of a power plant are given. The thermal efficiency is to be determined.

6-14

Assumptions The plant operates steadily. Properties The heating value of coal is given to be 30,000 kJ / kg . Analysis The rate of heat supply to this power plant is

QH = mcoal qHV,coal = (60,000 kg/h )(30,000 kJ/kg ) = 1.8´ 109 kJ/h = 500 MW Then the thermal efficiency of the plant becomes

hth =

Wnet, out QH

150 MW = 0.300 = 30.0% 500 MW

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net_out=150 [MW] m_dot_coal=60*1000 [kg/h] HV_coal=30000 [kJ/kg] "Analysis" Q_dot_H=m_dot_coal*HV_coal*Convert(kJ/h, MW) eta_th=W_dot_net_out/Q_dot_H

6-24 The power output and fuel consumption rate of a car engine are given. The thermal efficiency of the engine is to be determined.

Assumptions The car operates steadily. Properties The heating value of the fuel is given to be 44,000 kJ / kg . Analysis The mass consumption rate of the fuel is

mfuel = (r V )fuel = (0.8 kg/L)(22 L/h) = 17.6 kg/h The rate of heat supply to the car is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-15

QH = mcoal qHV,coal = (17.6 kg/h)(44,000 kJ/kg) = 774,400 kJ/h = 215.1 kW Then the thermal efficiency of the car becomes

hth =

Wnet, out QH

55 kW = 0.256 = 25.6% 215.1 kW

EES (Engineering Equation Solver) SOLUTION "Given" V_dot_fuel=22 [L/h] W_dot_net_out=55 [kW] HV_fuel=44000 [kJ/kg] rho_fuel=0.8 [g/cm^3] "Analysis" m_dot_fuel=rho_fuel*V_dot_fuel*Convert(g/cm^3, kg/L) Q_dot_H=m_dot_fuel*HV_fuel*Convert(kJ/h, kW) eta_th=W_dot_net_out/Q_dot_H

6-25E The power output and thermal efficiency of a solar pond power plant are given. The rate of solar energy collection is to be determined.

Assumptions The plant operates steadily. Analysis The rate of solar energy collection or the rate of heat supply to the power plant is determined from the thermal efficiency relation to be

QH =

Wnet, out hth

öæ3600 s ö÷ 150 kW æ 7 çç 1 Btu ÷ ç ÷ ÷çç 1 h ø÷ ÷= 1.71×10 Btu / h ç 0.03 è1.055 kJ øè

EES (Engineering Equation Solver) SOLUTION "Given" eta_th=0.03 W_dot_net_out=150 [kW] "Analysis" Q_dot_H=W_dot_net_out/eta_th*Convert(kW, Btu/h)

6-26 A coal-burning power plant produces 300 MW of power. The amount of coal consumed during a one-day period and the rate of air flowing through the furnace are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-16

Assumptions 1. The power plant operates steadily. 2. The kinetic and potential energy changes are zero. Properties The heating value of the coal is given to be 28,000 kJ / kg . Analysis (a) The rate and the amount of heat inputs to the power plant are

Qin =

Wnet,out hth

300 MW = 937.5 MW 0.32

Qin = Qin D t = (937.5 MJ/s)(24´ 3600 s) = 8.1´ 10 7 MJ The amount and rate of coal consumed during this period are

mcoal =

Qin 8.1´ 107 MJ = = 2.893´ 106 kg qHV 28 MJ/kg

mcoal =

mcoal 2.893´ 106 kg = = 33.48 kg/s Dt 24´ 3600 s

(b) Noting that the air-fuel ratio is 12, the rate of air flowing through the furnace is

mair = (AF)mcoal = (12 kg air/kg fuel)(33.48 kg/s) = 401.8 kg / s

EES (Engineering Equation Solver) SOLUTION "GIVEN" W_dot_net=3E5 [kW] eta_th=0.32 AF=12 HV_coal=28000 [kJ/kg] time=24*3600 [s] "ANALYSIS" Q_dot_in=W_dot_net/eta_th Q_in=Q_dot_in*time m_coal=Q_in/HV_coal "mass of fuel consumed in one day" m_dot_coal=m_coal/time m_dot_air=AF*m_dot_coal "air rate through the furnace"

6-27E An OTEC power plant operates between the temperature limits of 86F and 41F. The cooling water experiences a temperature rise of 6F in the condenser. The amount of power that can be generated by this OTEC plans is to be determined.

Assumptions 1. Steady operating conditions exist. 2. Water is an incompressible substance with constant properties. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-17

Properties The density and specific heat of water are taken r = 64.0 lbm / ft 3 and c = 1.0 Btu / lbm.° F , respectively.

Analysis The mass flow rate of the cooling water is mwater = r Vwater

æ 1 ft 3 ö ÷ ÷ = (64.0 lbm/ft 3 )(13,300 gal/min) ççç ÷ ÷ è7.4804 gal ø

= 113,790 lbm/min = 1897 lbm/s The rate of heat rejection to the cooling water is

Qout = mwater c(Tout - Tin ) = (1897 lbm/s)(1.0 Btu/lbm.°F)(6°F) = 11,380 Btu/s Noting that the thermal efficiency of this plant is 2.5%, the power generation is determined to be

W W W = ® 0.025 = ® W = 292 Btu/s = 308 kW W + (11,380 Btu/s) Qin W + Qout

since 1kW = 0.9478 Btu / s .

EES (Engineering Equation Solver) SOLUTION "Given" T_H=86 [F] T_L=41 [F] V_dot=13300 [gal/min] D=40 [in] DELTAT=6 [F] eta_th=0.025 "Properties" rho=64 [lbm/ft^3] "given" c=1.0 [Btu/lbm-F] "Analysis" m_dot=rho*V_dot*Convert(gal, ft^3)*Convert(lbm/min, lbm/s) Q_dot_out=m_dot*C*DELTAT eta_th=W_dot/Q_dot_in Q_dot_in=W_dot+Q_dot_out W_dot_kW=W_dot*Convert(Btu/s, kW)

6-28 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants. The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 5 years is to be determined.

Assumptions 1. Power is generated continuously by either plant at full capacity. 2. The time value of money (interest, inflation, etc.) is not considered. Properties The heating value of the coal is given to be 28´ 106 kJ / ton . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-18

Analysis For a power generation capacity of 150,000 MW, the construction costs of coal and IGCC plants and their difference are Construction cost coal = (150, 000, 000 kW)($1300/kW)=$195´ 109 Construction cost IGCC = (150, 000, 000 kW)($1500/kW)=$225´ 109

Construction cost difference = $225´ 109 - $195´ 109 = $30´ 109 The amount of electricity produced by either plant in 5 years is

We = W D t = (150,000,000 kW)(5´ 365´ 24 h)=6.570´ 1012 kWh The amount of fuel needed to generate a specified amount of power can be determined from h=

We Qin

® Qin =

We h

or mfuel =

We Qin = Heating value h (Heating value)

Then the amount of coal needed to generate this much electricity by each plant and their difference are

mcoal, coal plant =

ö We 6.570´ 1012 kWh æ 9 çç3600 kJ ÷ = ÷= 2.112´ 10 tons ÷ h (Heating value) (0.40)(28´ 106 kJ/ton) çè 1 kWh ø

mcoal, IGCC plant =

ö We 6.570´ 1012 kWh æ çç3600 kJ ÷ = = 1.760´ 109 tons ÷ 6 ÷ ç h (Heating value) (0.48)(28´ 10 kJ/ton) è 1 kWh ø

D mcoal = mcoal, coal plant - mcoal, IGCC plant = 2.112´ 109 - 1.760´ 109 =0.352 ´ 109 tons For D mcoal to pay for the construction cost difference of $30 billion, the price of coal should be

Unit cost of coal =

Construction cost difference $30´ 109 = = $85.2 / ton D mcoal 0.352´ 109 tons

Therefore, the IGCC plant becomes attractive when the price of coal is above $85.2 per ton.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=15E7 [kW] Cost_coal=1300 [$/kW] eta_coal=0.40 Cost_IGCC=1500 [$/kW] eta_IGCC=0.48 HV_coal=28000 [kJ/kg] PaybackYears=5 [yr] "Analysis" time=PaybackYears*Convert(yr, h) ConstCost_coal=W_dot*Cost_coal ConstCost_IGCC=W_dot*Cost_IGCC ConstCostDif=ConstCost_IGCC-ConstCost_coal W_e=W_dot*time m_coal_coal=W_e/(eta_coal*HV_coal)*Convert(kWh, kJ) m_coal_IGCC=W_e/(eta_IGCC*HV_coal)*Convert(kWh, kJ) DELTAm_coal=m_coal_coal-m_coal_IGCC UnitCost_coal=ConstCostDif/DELTAm_coal

6-29 Problem 6-28 is reconsidered. The price of coal is to be investigated for varying simple payback periods, plant construction costs, and operating efficiency. Analysis The problem is solved using EES, and the solution is given below. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-19

"Given" W_dot=15E7 [kW] Cost_coal=1300 [$/kW] eta_coal=0.40 Cost_IGCC=1500 [$/kW] eta_IGCC=0.48 HV_coal=28000 [kJ/kg] PaybackYears=5 [yr] "Analysis" time=PaybackYears*Convert(yr, h) ConstCost_coal=W_dot*Cost_coal ConstCost_IGCC=W_dot*Cost_IGCC ConstCostDif=ConstCost_IGCC-ConstCost_coal W_e=W_dot*time m_coal_coal=W_e/(eta_coal*HV_coal)*Convert(kWh, kJ) m_coal_IGCC=W_e/(eta_IGCC*HV_coal)*Convert(kWh, kJ) DELTAm_coal=m_coal_coal-m_coal_IGCC UnitCost_coal=ConstCostDif/DELTAm_coal*1000

PaybackYears [yr] 1 2 3 4 5 6 7 8 9 10

UnitCost coal [$ / ton ] 426.2 213.1 142.1 106.5 85.24 71.03 60.88 53.27 47.35 42.62 450

UnitCostcoal [$/ton]

400 350 300 250 200 150 100 50 0 1

PaybackYears [yr]

6-20

ηcoal

UnitCost coal

0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45

[$ / ton ] 28.41 31.09 34.09 37.5 41.4 45.9 51.14 57.34 64.78 73.87 85.24 99.85 119.3 146.6 187.5 255.7 300

UnitCostcoal [$/ton]

250

200

150

100

0 0.3

0.32

0.34

0.36

0.38

0.4

0.42

0.44

0.46

coal

Cost IGCC

UnitCost coal

[$ / kW] 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200

[$ / ton ] 0 42.62 85.24 127.9 170.5 213.1 255.7 298.3 340.9 383.6

6-21 400

UnitCostcoal [$/ton]

350 300 250 200 150 100 50 0 1300

1400

1500

1600

1700

1800

1900

2000

2100

2200

CostIGCC [$/kW]

6-30 The projected power needs of the United States is to be met by building inexpensive but inefficient coal plants or by building expensive but efficient IGCC plants. The price of coal that will enable the IGCC plants to recover their cost difference from fuel savings in 3 years is to be determined.

Assumptions 1. Power is generated continuously by either plant at full capacity. 2. The time value of money (interest, inflation, etc.) is not considered. Properties The heating value of the coal is given to be 28´ 106 kJ / ton . Analysis For a power generation capacity of 150,000 MW, the construction costs of coal and IGCC plants and their difference are Construction cost coal = (150, 000, 000 kW)($1300/kW)=$195´ 109 Construction cost IGCC = (150, 000, 000 kW)($1500/kW)=$225´ 109

Construction cost difference = $225´ 109 - $195´ 109 = $30´ 109 The amount of electricity produced by either plant in 3 years is

We = W D t = (150,000,000 kW)(3´ 365´ 24 h)=3.942´ 1012 kWh The amount of fuel needed to generate a specified amount of power can be determined from h=

We Qin

® Qin =

We h

or mfuel =

We Qin = Heating value h (Heating value)

Then the amount of coal needed to generate this much electricity by each plant and their difference are

mcoal, coal plant =

ö We 3.942´ 1012 kWh æ 9 çç3600 kJ ÷ = ÷ 6 ÷= 1.267´ 10 tons ç h (Heating value) (0.40(28´ 10 kJ/ton) è 1 kWh ø

6-22

mcoal, IGCC plant =

ö We 3.942´ 1012 kWh æ ÷ çç3600 kJ ÷ = = 1.055´ 109 tons 6 ÷ h (Heating value) (0.48)(28´ 10 kJ/ton) çè 1 kWh ø

D mcoal = mcoal, coal plant - mcoal, IGCC plant = 1.267´ 109 - 1.055´ 109 =0.211´ 109 tons For D mcoal to pay for the construction cost difference of $30 billion, the price of coal should be

Unit cost of coal =

Construction cost difference $30´ 109 = = $142 / ton D mcoal 0.211´ 109 tons

Therefore, the IGCC plant becomes attractive when the price of coal is above $142 per ton.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=15E7 [kW] Cost_coal=1300 [$/kW] eta_coal=0.40 Cost_IGCC=1500 [$/kW] eta_IGCC=0.48 HV_coal=28000 [kJ/kg] PaybackYears=3 [yr] "Analysis" time=PaybackYears*Convert(yr, h) ConstCost_coal=W_dot*Cost_coal ConstCost_IGCC=W_dot*Cost_IGCC ConstCostDif=ConstCost_IGCC-ConstCost_coal W_e=W_dot*time m_coal_coal=W_e/(eta_coal*HV_coal)*Convert(kWh, kJ) m_coal_IGCC=W_e/(eta_IGCC*HV_coal)*Convert(kWh, kJ) DELTAm_coal=m_coal_coal-m_coal_IGCC UnitCost_coal=ConstCostDif/DELTAm_coal

Refrigerators and Heat Pumps 6-31C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium.

6-32C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from a refrigerated space whereas the purpose of an air-conditioner is remove heat from a living space.

6-33C The coefficient of performance of a refrigerator represents the amount of heat removed from the refrigerated space for each unit of work supplied. It can be greater than unity.

6-34C The coefficient of performance of a heat pump represents the amount of heat supplied to the heated space for each unit of work supplied. It can be greater than unity.

6-35C No. The heat pump captures energy from a cold medium and carries it to a warm medium. It does not create it. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-23

6-36C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It does not create it.

6-37C No. Because the refrigerator consumes work to accomplish this task.

6-38C No. Because the heat pump consumes work to accomplish this task.

6-39C No device can transfer heat from a cold medium to a warm medium without requiring a heat or work input from the surroundings.

6-40C The violation of one statement leads to the violation of the other one, as shown in Sec. 6-4, and thus we conclude that the two statements are equivalent.

6-41 The COP and the power input of a residential heat pump are given. The rate of heating effect is to be determined. Assumptions The heat pump operates steadily. Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives

QH = COPHPWnet,in = (1.6)(2 kW) = 3.2 kW = 3.2 kJ / s

EES (Engineering Equation Solver) SOLUTION COP_HP=1.6 W_dot_net_in=2 [kW] Q_dot_H=COP_HP*W_dot_net_in

6-42 The cooling effect and the COP of a refrigerator are given. The power input to the refrigerator is to be determined.

Assumptions The refrigerator operates steadily. Analysis Rearranging the definition of the refrigerator coefficient of performance and applying the result to this refrigerator gives

Wnet,in =

QL 5 kW = = 3.85 kW COPR 1.3

6-24

COP_R=1.3 W_dot_net_in=Q_dot_L/COP_R

6-43 The cooling effect and the power consumption of an air conditioner are given. The rate of heat rejection from this air conditioner is to be determined.

Assumptions The air conditioner operates steadily. Analysis Applying the first law to the air conditioner gives

QH = Wnet,in + QL = 0.75 + 1 = 1.75 kW

EES (Engineering Equation Solver) SOLUTION Q_dot_L=1 [kW] W_dot_net_in=0.75 [kW] Q_dot_H=W_dot_net_in+Q_dot_L

6-44 The cooling effect and the rate of heat rejection of a refrigerator are given. The COP of the refrigerator is to be determined. Assumptions The refrigerator operates steadily. Analysis Applying the first law to the refrigerator gives

Wnet,in = QH - QL = 22,000 - 15,000 = 7000 kJ/h Applying the definition of the coefficient of performance,

COPR =

QL 15,000 kJ/h = = 2.14 7000 Wnet,in

EES (Engineering Equation Solver) SOLUTION Q_dot_L=15000 [kJ/h] Q_dot_H=22000 [kJ/h] W_dot_net_in=Q_dot_H-Q_dot_L COP_R=Q_dot_L/W_dot_net_in

6-45 The heat removal rate of a refrigerator per kW of power it consumes is given. The COP and the rate of heat rejection are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-25

Assumptions The refrigerator operates steadily. Analysis The coefficient of performance of the refrigerator is determined from its definition,

COPR =

ö QL 5040 kJ/h æ çç 1 kW ÷ = = 1.4 ÷ ÷ ç 1 kW è3600 kJ/h ø Wnet, in

The rate of heat rejection to the surrounding air, per kW of power consumed, is determined from the energy balance,

QH = QL + Wnet, in = (5040 kJ/h) + (1´ 3600 kJ/h ) = 8640 kJ / h

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_L=5040 [kJ/h] W_dot_net_in=1 [kW] "Analysis" COP_R=Q_dot_L/W_dot_net_in*1/Convert(kW, kJ/h) Q_dot_H=Q_dot_L+W_dot_net_in*Convert(kW, kJ/h)

6-46 The rate of heat supply of a heat pump per kW of power it consumes is given. The COP and the rate of heat absorption from the cold environment are to be determined.

Assumptions The heat pump operates steadily. Analysis The coefficient of performance of the refrigerator is determined from its definition,

COPHP =

ö QH 8000 kJ/h æ ÷= 2.22 çç 1 kW ÷ = ÷ 1 kW çè3600 kJ/h ø Wnet, in

The rate of heat absorption from the surrounding air, per kW of power consumed, is determined from the energy balance,

QL = QH - Wnet, in = (8,000 kJ/h) - (1)(3600 kJ/h ) = 4400 kJ / h

6-26

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_H=8000 [kJ/h] W_dot_net_in=1 [kW] "Analysis" COP_HP=Q_dot_H/W_dot_net_in*1/Convert(kW, kJ/h) Q_dot_L=Q_dot_H-W_dot_net_in*Convert(kW, kJ/h)

6-47 The COP and the work input of a heat pump are given. The heat transferred to and from this heat pump are to be determined. Assumptions The heat pump operates steadily. Analysis Applying the definition of the heat pump coefficient of performance,

QH = COPHPWnet,in = (1.7)(50 kJ) = 85 kJ Adapting the first law to this heat pump produces

QL = QH - Wnet,in = 85 kJ - 50 kJ = 35 kJ

EES (Engineering Equation Solver) SOLUTION COP_HP=1.7 W_net_in=50 [kJ] Q_H=COP_HP*W_net_in Q_L=Q_H-W_net_in

6-48E The COP and the heating effect of a heat pump are given. The power input to the heat pump is to be determined. Assumptions The heat pump operates steadily. Analysis Applying the definition of the coefficient of performance,

Wnet,in =

ö QH 100,000 Btu/h æ 1 hp ÷ çç = = 28.1 hp ÷ ÷ ç è 2544.5 Btu/h ø COPHP 1.4

6-27

EES (Engineering Equation Solver) SOLUTION COP_HP=1.4 Q_dot_H=100000 [Btu/h] W_net_in=Q_dot_H/COP_HP*Convert(Btu/h,hp)

6-49 The power consumption and the cooling rate of an air conditioner are given. The COP and the rate of heat rejection are to be determined. Assumptions The air conditioner operates steadily.

Analysis (a) The coefficient of performance of the air-conditioner (or refrigerator) is determined from its definition,

COPR =

ö QL 750 kJ/min æ çç 1 kW ÷ = = 2.38 ÷ ÷ ç è 5.25 kW 60 kJ/min ø Wnet, in

(b) The rate of heat discharge to the outside air is determined from the energy balance,

QH = QL + Wnet, in = (750 kJ/min)+ (5.25´ 60 kJ/min) = 1065 kJ / min

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_L=750 [kJ/min] W_dot_net_in=5.25 [kW] "Analysis" "(a)" COP_R=Q_dot_L/W_dot_net_in*Convert(kJ/min, kW) "(b)" Q_dot_H=Q_dot_L+W_dot_net_in*Convert(kW, kJ/min)

6-50 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5 watermelons is to be determined. Assumptions 1. 2. 3.

The refrigerator operates steadily. The heat gain of the refrigerator through its walls, door, etc. is negligible. The watermelons are the only items in the refrigerator to be cooled.

Properties The specific heat of watermelons is given to be c = 4.2 kJ / kg.° C . Analysis The total amount of heat that needs to be removed from the watermelons is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-28

QL = (mcD T )watermelons = 5´ (10 kg )(4.2 kJ/kg ×°C)(28 - 8)°C = 4200 kJ The rate at which this refrigerator removes heat is

QL = (COPR )(Wnet, in ) = (1.5)(0.45 kW) = 0.675 kW

That is, this refrigerator can remove 0.675 kJ of heat per second. Thus the time required to remove 4200 kJ of heat is Dt =

QL 4200 kJ = = 6222 s = 104 min QL 0.675 kJ/s

This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the watermelons.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net_in=0.45 [kW] COP_R=1.5 m=5*10 [kg] T1=28 [C] T2=8 [C] c=4.2 [kJ/kg-C] "Analysis" Q_L=m*c*(T1-T2) Q_dot_L=COP_R*W_dot_net_in time=Q_L/Q_dot_L*1/Convert(min, s)

6-51 An air conditioner with a known COP cools a house to desired temperature in 15 min. The power consumption of the air conditioner is to be determined. Assumptions 1. 2. 3.

The air conditioner operates steadily. The house is well-sealed so that no air leaks in or out during cooling. Air is an ideal gas with constant specific heats at room temperature.

6-29

Properties The constant volume specific heat of air is given to be cv = 0.72 kJ / kg ×° C . Analysis Since the house is well-sealed (constant volume), the total amount of heat that needs to be removed from the house is 0

QL = (mcv D T )House = (800 kg )(0.72 kJ/kg ×°C)(35 - 20)°C = 8640 kJ This heat is removed in 30 minutes. Thus the average rate of heat removal from the house is

QL 8640 kJ = = 4.8 kW D t 30´ 60 s Using the definition of the coefficient of performance, the power input to the air-conditioner is determined to be QL =

Wnet, in =

QL 4.8 kW = = 1.71 kW COPR 2.8

EES (Engineering Equation Solver) SOLUTION "Input Data" T_1=35 [C] T_2=20 [C] c_v = 0.72 [kJ/kg-C] m_house=800 [kg] DELTAtime=30 [min] COP=2.8 "Assuming no work done on the house and no heat energy added to the house in the time period with no change in KE and PE, the first law applied to the house is:" E_dot_in - E_dot_out = DELTAE_dot E_dot_in = 0 E_dot_out = Q_dot_L DELTAE_dot = m_house*DELTAu_house/DELTAtime DELTAu_house = c_v*(T_2-T_1) "Using the definition of the coefficient of performance of the A/C:" W_dot_in = Q_dot_L/COP *convert(kJ/min, kW) Q_dot_H= W_dot_in*convert(kW, kJ/min) + Q_dot_L

6-52 Problem 6-51 is reconsidered. The rate of power drawn by the air conditioner required to cool the house as a function for air conditioner EER ratings in the range 5 to 15 is to be investigated. Representative costs of air conditioning units in the EER rating range are to be included. Analysis The problem is solved using EES, and the results are tabulated and plotted below. T_1=35 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-30

T_2=20 [C] c_v = 0.72 [kJ/kg-C] m_house=800 [kg] DELTAtime=30 [min] EER=5 COP=EER/3.412 E_dot_in - E_dot_out = DELTAE_dot E_dot_in = 0 E_dot_out = Q_dot_L DELTAE_dot = m_house*DELTAu_house/DELTAtime DELTAu_house = c_v*(T_2-T_1) W_dot_in = Q_dot_L/COP*convert('kJ/min','kW') Q_dot_H= W_dot_in*convert('KW','kJ/min') + Q_dot_L

EER

Win

5 6 7 8 9 10 11 12 13 14 15

[kW] 3.276 2.73 2.34 2.047 1.82 1.638 1.489 1.365 1.26 1.17 1.092

6-53E The COP and the refrigeration rate of an ice machine are given. The power consumption is to be determined. Assumptions The ice machine operates steadily. Analysis The cooling load of this ice machine is

QL = mqL = (28 lbm/h )(169 Btu/lbm) = 4732 Btu/h Using the definition of the coefficient of performance, the power input to the ice machine system is determined to be

Wnet, in =

ö QL 4732 Btu/h æ çç 1 hp ÷ = = 0.775 hp ÷ ÷ ç è ø COPR 2.4 2545 Btu/h

6-31

EES (Engineering Equation Solver) SOLUTION "Given" T1=55 [F] T2=25 [F] COP_R=2.4 m_dot_ice=28 [lbm/h] q_L=169 [Btu/lbm] "Analysis" Q_dot_L=m_dot_ice*q_L W_dot_net_in=Q_dot_L/COP_R*Convert(Btu/h, hp)

6-54 A refrigerator is used to cool water to a specified temperature. The power input is given. The flow rate of water and the COP of the refrigerator are to be determined. Assumptions The refrigerator operates steadily. Properties The specific heat of water is 4.18 kJ / kg.° C and its density is 1 kg / L . Analysis The rate of cooling is determined from

QL = QH - Win = (570 / 60) kW - 2.65 kW = 6.85 kW The mass flow rate of water is

QL = mc p (T1 - T2 ) ¾ ¾ ® m=

QL 6.85 kW = = 0.09104 kg/s c p (T1 - T2 ) (4.18 kJ/kg ×°C)(23 - 5) °C

The volume flow rate is V=

ö m 0.09104 kg/s æ çç 60 s ÷ = ÷= 5.46 L / min çè1 min ÷ ø r 1 kg/L

The COP is

COP =

QL 6.85 kW = = 2.58 Win 2.65 kW

EES (Engineering Equation Solver) SOLUTION "Given" T1=23 [C] T2=5 [C] Q_dot_H=570 [kJ/min] W_dot_in=2.65 [kW] c_p=4.18 [kJ/kg-C] rho=1 [kg/L] "Analysis" Q_dot_L=q_dot_H*Convert(kJ/min, kW)-W_dot_in m_dot=Q_dot_L/(c_p*(T1-T2)) Vol_dot=m_dot/rho*Convert(L/s, L/min) COP=Q_dot_L/W_dot_in

6-55 The COP and the refrigeration rate of a refrigerator are given. The power consumption of the refrigerator is to be determined. Assumptions The refrigerator operates steadily. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-32

Analysis Since the refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ / h , the refrigerator removes heat at a rate of

QL = 4´ (800 kJ/h) = 3200 kJ/h when running. Thus the power the refrigerator draws when it is running is

Wnet, in =

QL 3200 kJ/h = = 1455 kJ/h = 0.40 kW COPR 2.2

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_L=4*800 [kJ/h] COP_R=2.2 "Analysis" W_dot_net_in=Q_dot_L/COP_R*Convert(kJ/h, kW)

6-56E An office that is being cooled adequately by a 12,000 Btu / h window air-conditioner is converted to a computer room. The number of additional air-conditioners that need to be installed is to be determined. Assumptions 1. The computers are operated by 7 adult men. 2. The computers consume 40 percent of their rated power at any given time. Properties The average rate of heat generation from a person seated in a room/office is 100 W (given). Analysis The amount of heat dissipated by the computers is equal to the amount of electrical energy they consume. Therefore,

Qcomputers = (Rated power)´ (Usage factor) = (8.4 kW)(0.4) = 3.36 kW Qpeople = (No. of people)´ Qperson = 7´ (100 W) = 700 W

6-33

since 1 W = 3.412 Btu / h . Then noting that each available air conditioner provides 7000 Btu / h cooling, the number of air-conditioners needed becomes No. of air conditioners =

Cooling load 13,853 Btu/h = Cooling capacity of A/C 7000 Btu/h

= 1.98 » 2 Air conditioners

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_L_AC=7000 [Btu/h] Q_dot_computers=0.4*8.4 [kW] Q_dot_people=7*0.100 [kW] "Analysis" Q_dot_total=(Q_dot_computers+Q_dot_people)*Convert(kW, Btu/h) N_AC=Q_dot_total/Q_dot_L_Ac

6-57 A house is heated by resistance heaters, and the amount of electricity consumed during a winter month is given. The amount of money that would be saved if this house were heated by a heat pump with a known COP is to be determined. Assumptions The heat pump operates steadily. Analysis The amount of heat the resistance heaters supply to the house is equal to he amount of electricity they consume. Therefore, to achieve the same heating effect, the house must be supplied with 1200 kWh of energy. A heat pump that supplied this much heat will consume electrical power in the amount of

Wnet, in =

QH 1200 kWh = = 500 kWh COPHP 2.4

which represent a savings of 1200 − 500 = 700 kWh. Thus the homeowner would have saved

(700 kWh)(0.12$ / kWh) = $84.0

EES (Engineering Equation Solver) SOLUTION "Given" W_heater=1200 [kWh] COP_HP=2.4 UnitCost=8.5[cents/kWh]*Convert(cents/kWh, $/kWh) "Analysis" Q_H=W_heater W_net_in=Q_H/COP_HP E_savings=W_heater-W_net_in CostSavings=E_savings*UnitCost

6-58 Refrigerant-134a flows through the condenser of a residential heat pump unit. For a given compressor power consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined. Assumptions 1. The heat pump operates steadily. 2. The kinetic and potential energy changes are zero. Properties The enthalpies of R-134a at the condenser inlet and exit are

6-34

P1 = 800 kPa ü ïï ý h1 = 271.24 kJ/kg ïïþ T1 = 35°C P2 = 800 kPa ü ïï ý h2 = 95.48 kJ/kg ïïþ x2 = 0 Analysis (a) An energy balance on the condenser gives the heat rejected in the condenser

QH = m(h1 - h2 )

= (0.018 kg/s)(271.24 - 95.48) kJ/kg = 3.164 kW

The COP of the heat pump is

COP =

QH 3.164 kW = = 2.64 1.2 kW Win

(b) The rate of heat absorbed from the outside air

QL = QH - Win = 3.164 - 1.2 = 1.96 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=800 [kPa] T_1=35 [C] m_dot=0.018 [kg/s] P_2=800 [kPa] x_2=0 W_dot_in=1.2 [kW] "ANALYSIS" Fluid$='R134a' h_1=enthalpy(Fluid$, P=P_1, T=T_1) h_2=enthalpy(Fluid$, P=P_2, x=x_2) Q_dot_H=m_dot*(h_1-h_2) COP=Q_dot_H/W_dot_in Q_dot_L=Q_dot_H-W_dot_in "heat absorbed from outside air"

6-59 A commercial refrigerator with R-134a as the working fluid is considered. The evaporator inlet and exit states are specified. The mass flow rate of the refrigerant and the rate of heat rejected are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-35

1. The refrigerator operates steadily. 2. The kinetic and potential energy changes are zero. Properties The properties of R-134a at the evaporator inlet and exit states are (Tables A-11 through A-13)

P1 = 100 kPa ü ïï ý h1 = 60.71 kJ/kg ïïþ x1 = 0.2 P2 = 100 kPa ü ïï ý h2 = 234.75 kJ/kg T2 = - 26 °C ïïþ Analysis (a) The refrigeration load is

QL = (COP)Win = (1.2)(0.600 kW) = 0.72 kW The mass flow rate of the refrigerant is determined from

mR =

QL 0.72 kW = = 0.00414 kg / s h2 - h1 (234.75 - 60.71) kJ/kg

(b) The rate of heat rejected from the refrigerator is

QH = QL + Win = 0.72 + 0.60 = 1.32 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] x_1=0.20 P_2=100 [kPa] T_2=-26 [C] W_dot_in=0.60 [kW] COP=1.2 "ANALYSIS" Fluid$='R134a' h_1=enthalpy(Fluid$, P=P_1, x=x_1) h_2=enthalpy(Fluid$, P=P_2, T=T_2) Q_dot_L=COP*W_dot_in "refrigeration load" m_dot=Q_dot_L/(h_2-h_1) "mass flow rate of refrigerant" Q_dot_H=Q_dot_L+W_dot_in "heat rejected to kitchen air"

6-36

Perpetual-Motion Machines 6-60C This device creates energy, and thus it is a PMM1.

6-61C This device creates energy, and thus it is a PMM1.

Reversible and Irreversible Processes 6-62C Because reversible processes can be approached in reality, and they form the limiting cases. Work producing devices that operate on reversible processes deliver the most work, and work consuming devices that operate on reversible processes consume the least work.

6-63C No. Because it involves heat transfer through a finite temperature difference.

6-64C This process is irreversible. As the block slides down the plane, two things happen, (a) the potential energy of the block decreases, and (b) the block and plane warm up because of the friction between them. The potential energy that has been released can be stored in some form in the surroundings (e.g., perhaps in a spring). When we restore the system to its original condition, we must (a) restore the potential energy by lifting the block back to its original elevation, and (b) cool the block and plane back to their original temperatures. The potential energy may be restored by returning the energy that was stored during the original process as the block decreased its elevation and released potential energy. The portion of the surroundings in which this energy had been stored would then return to its original condition as the elevation of the block is restored to its original condition. In order to cool the block and plane to their original temperatures, we have to remove heat from the block and plane. When this heat is transferred to the surroundings, something in the surroundings has to change its state (e.g., perhaps we warm up some water in the surroundings). This change in the surroundings is permanent and cannot be undone. Hence, the original process is irreversible.

6-65C The irreversibilities that occur within the system boundaries are internal irreversibilities; those which occur outside the system boundaries are external irreversibilities.

6-66C The chemical reactions of combustion processes of a natural gas and air mixture will generate carbon dioxide, water, and other compounds and will release heat energy to a lower temperature surroundings. It is unlikely that the surroundings will return this energy to the reacting system and the products of combustion react spontaneously to reproduce the natural gas and air mixture.

6-67C Adiabatic stirring processes are irreversible because the energy stored within the system can not be spontaneously released in a manor to cause the mass of the system to turn the paddle wheel in the opposite direction to do work on the surroundings.

6-68C When the compression process is non-quasi equilibrium, the molecules before the piston face cannot escape fast enough, forming a high pressure region in front of the piston. It takes more work to move the piston against this highpressure region.

6-37

6-69C When an expansion process is non-quasi equilibrium, the molecules before the piston face cannot follow the piston fast enough, forming a low pressure region behind the piston. The lower pressure that pushes the piston produces less work.

6-70C A reversible expansion or compression process cannot involve unrestrained expansion or sudden compression, and thus it is quasi-equilibrium. A quasi-equilibrium expansion or compression process, on the other hand, may involve external irreversibilities (such as heat transfer through a finite temperature difference), and thus is not necessarily reversible.

The Carnot Cycle and Carnot's Principle 6-71C The four processes that make up the Carnot cycle are isothermal expansion, reversible adiabatic expansion, isothermal compression, and reversible adiabatic compression.

6-72C They are (1) the thermal efficiency of an irreversible heat engine is lower than the efficiency of a reversible heat engine operating between the same two reservoirs, and (2) the thermal efficiency of all the reversible heat engines operating between the same two reservoirs are equal.

6-73C (a) No, (b) No. They would violate the Carnot principle.

6-74C False. The second Carnot principle states that no heat engine cycle can have a higher thermal efficiency than the Carnot cycle operating between the same temperature limits.

6-75C Yes. The second Carnot principle states that all reversible heat engine cycles operating between the same temperature limits have the same thermal efficiency.

Carnot Heat Engines 6-76C No.

6-77C The one that has a source temperature of 600°C. This is true because the higher the temperature at which heat is supplied to the working fluid of a heat engine, the higher the thermal efficiency.

6-78 The source and sink temperatures of a power plant are given. The maximum efficiency of the plant is to be determined. Assumptions The plant operates steadily. Analysis The maximum efficiency this plant can have is the Carnot efficiency, which is determined from

6-38

hth,max = 1-

TL 300 K = 1= 0.75 or 75% TH 1200 K

EES (Engineering Equation Solver) SOLUTION T_H=1200 [K] T_L=300 [K] eta_th_max=1-T_L/T_H

6-79 The source and sink temperatures of a power plant are given. The maximum efficiency of the plant is to be determined. Assumptions The plant operates steadily. Analysis The maximum efficiency this plant can have is the Carnot efficiency, which is determined from

hth,max = 1-

TL 300 K = 1= 0.70 or 70% TH 1000 K

EES (Engineering Equation Solver) SOLUTION T_H=1000 [K] T_L=300 [K] eta_th_max=1-T_L/T_H

6-80E The claim of a thermodynamicist regarding to the thermal efficiency of a heat engine is to be evaluated. Assumptions The plant operates steadily. Analysis The maximum thermal efficiency would be achieved when this engine is completely reversible. When this is the case,

6-39

hth,max = 1-

TL 510 R = 1= 0.595 TH 1260 R

Since the thermal efficiency of the actual engine is less than this, this engine is possible.

EES (Engineering Equation Solver) SOLUTION T_H=1260 [R] T_L=510 [R] eta_th_max=1-T_L/T_H

6-81E The sink temperature of a Carnot heat engine, the rate of heat rejection, and the thermal efficiency are given. The power output of the engine and the source temperature are to be determined. Assumptions The Carnot heat engine operates steadily. Analysis (a) The rate of heat input to this heat engine is determined from the definition of thermal efficiency,

hth = 1-

QL 800 Btu/min ¾¾ ® 0.47 = 1¾¾ ® QH = 1509 Btu/min QH QH

Then the power output of this heat engine can be determined from

Wnet, out = hth QH = (0.47)(1509 Btu/min ) = 709.2 Btu/min = 16.7 hp (b) For reversible cyclic devices we have æQH ÷ ö æ ö çç ÷ = ççTH ÷ ÷ ÷ ÷ çèç Q ø ççè T ø ÷ ÷ L L rev

Thus the temperature of the source TH must be æQ ö æ1509 Btu/min ÷ ö ÷ TH = ççç H ÷ ÷(520 R ) = 981 R ÷ TL = ççèç ø 800 Btu/min ÷ èç QL ÷ ørev

6-40

EES (Engineering Equation Solver) SOLUTION "Given" eta_th_C=0.47 T_L=60[F]+460 [R] Q_dot_L=800 [Btu/min] "Analysis" "(a)" eta_th_C=1-Q_dot_L/Q_dot_H W_dot_net_out=eta_th_C*Q_dot_H*1/Convert(hp, Btu/min) "(b)" T_H=(Q_dot_H/Q_dot_L)*T_L

6-82E The source and sink temperatures and the power output of a reversible heat engine are given. The rate of heat input to the engine is to be determined. Assumptions The heat engine operates steadily. Analysis The thermal efficiency of this reversible heat engine is determined from

hth,max = 1-

TL 500 R = 1= 0.6667 TH 1500 R

A rearrangement of the definition of the thermal efficiency produces

hth,max =

ö Wnet Wnet 5 hp æ ÷= 19, 080 Btu / h çç 2544.5 Btu/h ÷ ¾¾ ® QH = = ÷ ÷ hth,max 0.6667 çè 1 hp QH ø

EES (Engineering Equation Solver) SOLUTION T_H=1500 [R] T_L=500 [R] W_dot_net=5 [hp] eta_th_max=1-T_L/T_H Q_dot_H=W_dot_net/eta_th_max*Convert(hp, Btu/h)

6-83 An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from

6-41

hth, max = hth, C = 1-

TL 290 K = 1= 0.420 or 42.0% TH 500 K

The actual thermal efficiency of the heat engine in question is hth =

Wnet 300 kJ = = 0.429 or 42.9% QH 700 kJ

which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false.

EES (Engineering Equation Solver) SOLUTION "Given" Q_H=700 [kJ] W_net_out=300 [kJ] T_H=500 [K] T_L=290 [K] "Analysis" eta_th_C=1-T_L/T_H eta_th_actual=W_net_out/Q_H "The claim is false if eta_th_actual is greater than eta_th_C"

6-84 The source and sink temperatures of a Carnot heat engine and the rate of heat supply are given. The thermal efficiency and the power output are to be determined. Assumptions The Carnot heat engine operates steadily. Analysis (a) The thermal efficiency of a Carnot heat engine depends on the source and the sink temperatures only, and is determined from

hth, C = 1-

TL 300 K = 1= 0.70 or 70% TH 1000 K

(b) The power output of this heat engine is determined from the definition of thermal efficiency,

Wnet, out = hth QH = (0.70)(800 kJ/min ) = 560 kJ/min = 9.33 kW

6-42

EES (Engineering Equation Solver) SOLUTION "Given" T_H=1000 [K] T_L=300 [K] Q_dot_H=800 [kJ/min] "Analysis" "(a)" eta_th_C=1-T_L/T_H "(b)" W_dot_net_out=eta_th_C*Q_dot_H*1/Convert(kW, kJ/min)

6-85 The source and sink temperatures of a heat engine and the rate of heat supply are given. The maximum possible power output of this engine is to be determined. Assumptions The heat engine operates steadily. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from

hth, max = hth, C = 1-

TL 298 K = 1= 0.600 or 60.0% TH (477 + 273) K

Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be

Wnet, out = hth QH = (0.600)(65,000 kJ/min ) = 39,000 kJ/min = 653 kW

EES (Engineering Equation Solver) SOLUTION T_H = 477 [C] T_L = 25 [C] Q_H = 65000 [kJ/min] "First Law applied to the heat engine" Q_H - Q_L- W_net = 0 W_net_kW=W_net*convert(kJ/min,kW) "Cycle Thermal Efficiency - Temperatures must be absolute" eta_th = 1 - (T_L + 273)/(T_H + 273) "Definition of cycle efficiency" eta_th=W_net / Q_H

6-86 Problem 6-85 is reconsidered. The effects of the temperatures of the heat source and the heat sink on the power produced and the cycle thermal efficiency as the source temperature varies from 300°C to 1000°C and the sink temperature varies from 0°C to 50°C are to be studied. The power produced and the cycle efficiency against the source temperature for sink temperatures of 0°C, 25°C, and 50°C are to be plotted. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-43

Analysis The problem is solved using EES, and the results are tabulated and plotted below. T_H = 477 [C] T_L =25 [C] Q_dot_H = 65000 [kJ/min] Q_dot_H - Q_dot_L- W_dot_net = 0 W_dot_net_KW=W_dot_net*convert(kJ/min,kW) eta_th = 1 - (T_L + 273)/(T_H + 273) eta_th=W_dot_net / Q_dot_H

WnetkW

[C] [kW] 300 567.2 400 643.9 500 700.7 600 744.6 700 779.4 800 807.7 900 831.2 1000 851 Values for TL = 0° C

ηth 0.5236 0.5944 0.6468 0.6873 0.7194 0.7456 0.7673 0.7855

6-44

6-87E An inventor claims to have developed a heat engine. The inventor reports temperature, heat transfer, and work output measurements. The claim is to be evaluated. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from

hth, max = hth, C = 1-

TL 540 R = 1= 0.400 or 40.0% TH 900 R

The actual thermal efficiency of the heat engine in question is hth =

Wnet 160 Btu = = 0.533 or 53.3% QH 300 Btu

which is greater than the maximum possible thermal efficiency. Therefore, this heat engine is a PMM2 and the claim is false.

EES (Engineering Equation Solver) SOLUTION "Given" Q_H=300 [Btu] W_net_out=160 [Btu] T_H=900 [R] T_L=540 [R] "Analysis" eta_th_C=1-T_L/T_H eta_th_actual=W_net_out/Q_H "The claim is false if eta_th_actual is greater than eta_th_C"

6-45

6-88 The source and sink temperatures of an OTEC (Ocean Thermal Energy Conversion) power plant are given. The maximum thermal efficiency is to be determined. Assumptions The power plant operates steadily. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from

hth, max = hth, C = 1-

TL 276 K = 1= 0.071 or 7.1% TH 297 K

EES (Engineering Equation Solver) SOLUTION "Given" T_H=24[C]+273 [K] T_L=3[C]+273 [K] "Analysis" eta_th_C=1-T_L/T_H

6-89 The claim that the efficiency of a completely reversible heat engine can be doubled by doubling the temperature of the energy source is to be evaluated. Assumptions The heat engine operates steadily. Analysis The upper limit for the thermal efficiency of any heat engine occurs when a completely reversible engine operates between the same energy reservoirs. The thermal efficiency of this completely reversible engine is given by

hth,rev = 1-

TL T - TL = H TH TH

If we were to double the absolute temperature of the high temperature energy reservoir, the new thermal efficiency would be hth,rev = 1-

TL 2T - TL T - TL = H < 2 H 2TH 2TH TH

The thermal efficiency is then not doubled as the temperature of the high temperature reservoir is doubled.

6-90 A geothermal power plant uses geothermal liquid water at 150ºC at a specified rate as the heat source. The actual and maximum possible thermal efficiencies and the rate of heat rejected from this power plant are to be determined. Assumptions 1. The power plant operates steadily. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-46

2. The kinetic and potential energy changes are zero. 3. Steam properties are used for geothermal water. Properties Using saturated liquid properties, (Table A-4)

Tsource,1  150 C   hgeo,1  632.18 kJ/kg x source  0  Tsource,2  90 C   hsource  377.04 kJ/kg x source  0  Tsink  25 C   hsink  104.83 kJ/kg x sink  0  Analysis (a) The rate of heat input to the plant is  (h Q  m h )  (210 kg/s) (632 .18  377 .04) kJ/kg  53,580 kW in

geo

geo,1

geo,2

The actual thermal efficiency is W net, out 8000 kW  th    0.1493  14.9%  53,580 kW Qin (b) The maximum thermal efficiency is the thermal efficiency of a reversible heat engine operating between the source and sink temperatures

 th, max  1 

TL (25  273) K  1  0.2955  29.6% TH (150  273) K

Q out  Q in  W net, out  53,580  8000  45,580 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_geo_1=150 [C] T_geo_2=90 [C] m_dot_geo=210 [kg/s] W_dot_net=8000 [kW] T_ambient=25 [C] "ANALYSIS" Fluid$='Steam_iapws' h_geo_1=enthalpy(Fluid$, T=T_geo_1, x=0) "assumed saturated liquid" h_geo_2=enthalpy(Fluid$, T=T_geo_2, x=0) h_dead=enthalpy(Fluid$, T=T_ambient, x=0) "dead state enthalpy" Q_dot_in=m_dot_geo*(h_geo_1-h_geo_2) "energy of geothermal source" eta_th=W_dot_net/Q_dot_in eta_th_max=1-(T_ambient+273)/(T_geo_1+273) Q_dot_out=Q_dot_in-W_dot_net

Carnot Refrigerators and Heat Pumps 6-91C It is the COP that a Carnot refrigerator would have, COPR =

1 . TH / TL - 1

6-47

6-92C The difference between the temperature limits is typically much higher for a refrigerator than it is for an air conditioner. The smaller the difference between the temperature limits a refrigerator operates on, the higher is the COP. Therefore, an air-conditioner should have a higher COP.

6-93C The deep freezer should have a lower COP since it operates at a much lower temperature, and in a given environment, the COP decreases with decreasing refrigeration temperature.

6-94C By increasing TL or by decreasing TH .

6-95C No. At best (when everything is reversible), the increase in the work produced will be equal to the work consumed by the refrigerator. In reality, the work consumed by the refrigerator will always be greater than the additional work produced, resulting in a decrease in the thermal efficiency of the power plant.

6-96C No. At best (when everything is reversible), the increase in the work produced will be equal to the work consumed by the refrigerator. In reality, the work consumed by the refrigerator will always be greater than the additional work produced, resulting in a decrease in the thermal efficiency of the power plant.

6-97C Bad idea. At best (when everything is reversible), the increase in the work produced will be equal to the work consumed by the heat pump. In reality, the work consumed by the heat pump will always be greater than the additional work produced, resulting in a decrease in the thermal efficiency of the power plant.

6-98 The claim of a thermodynamicist regarding the COP of a heat pump is to be evaluated. Assumptions The heat pump operates steadily. Analysis The maximum heat pump coefficient of performance would occur if the heat pump were completely reversible,

COPHP,max =

TH 293 K = = 14.7 TH - TL 293 K - 273 K

Since the claimed COP is less than this maximum, the claim is valid.

EES (Engineering Equation Solver) SOLUTION T_L=273 [K] T_H=293 [K] COP_HP_max=T_H/(T_H-T_L)

6-48

6-99 The minimum work per unit of heat transfer from the low-temperature source for a heat pump is to be determined. Assumptions The heat pump operates steadily. Analysis Application of the first law gives

Wnet,in QL

QH - QL QH = - 1 QL QL

For the minimum work input, this heat pump would be completely reversible and the thermodynamic definition of temperature would reduce the preceding expression to

Wnet,in QL

TH 535 K - 1= - 1 = 0.163 TL 460 K

EES (Engineering Equation Solver) SOLUTION T_L=460 [K] T_H=535 [K] W\Q_L=T_H/T_L-1

6-100 The rate of cooling provided by a reversible refrigerator with specified reservoir temperatures is to be determined. Assumptions The refrigerator operates steadily. Analysis The COP of this reversible refrigerator is

COPR,max =

TL 250 K = = 5 TH - TL 300 K - 250 K

Rearranging the definition of the refrigerator coefficient of performance gives

QL = COPR,maxWnet,in = (5)(10 kW) = 50 kW

EES (Engineering Equation Solver) SOLUTION T_L=250 [K] T_H=300 [K] W_dot_net_in=10 [kW] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-49

COP_R_max=T_L/(T_H-T_L) Q_dot_L=COP_R_max*W_dot_net_in

6-101 The cooled space and the outdoors temperatures for a Carnot air-conditioner and the rate of heat removal from the airconditioned room are given. The power input required is to be determined. Assumptions The air-conditioner operates steadily. Analysis The COP of a Carnot air conditioner (or Carnot refrigerator) depends on the temperature limits in the cycle only, and is determined from

COPR, C =

1 1 = = 27.0 (TH / TL )- 1 (35 + 273 K )/ (24 + 273 K )- 1

The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator,

Wnet, in =

QL 750 kJ/min = = 27.8 kJ/min = 0.463 kW COPR, max 27.0

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_L=750 [kJ/min] T_L=24[C]+273 [K] T_H=35[C]+273 [K] "Analysis" COP_R_C=1/((T_H/T_L)-1) W_dot_net_in_min=Q_dot_L/COP_R_C*1/Convert(kW, kJ/min)

6-102 The power input and the COP of a Carnot heat pump are given. The temperature of the low-temperature reservoir and the heating load are to be determined. Assumptions The heat pump operates steadily. Analysis The temperature of the low-temperature reservoir is

6-50

COPHP,max =

TH 297 K ¾¾ ® 12.5 = ¾¾ ® TL = 273 K TH - TL (297 - TL ) K

The heating load is

COPHP,max =

QH QH ¾¾ ® 12.5 = ¾¾ ® QH = 26.9 kW 2.15 kW Win

EES (Engineering Equation Solver) SOLUTION "Given" COP_HP_max=12.5 T_H=(24+273) [K] W_dot_in=2.15 [kW] "Analysis" COP_HP_max=T_H/(T_H-T_L) Q_dot_H=COP_HP_max*W_dot_in

6-103E An air-conditioning system maintains a house at a specified temperature. The rate of heat gain of the house and the rate of internal heat generation are given. The maximum power input required is to be determined. Assumptions The air-conditioner operates steadily. Analysis The power input to an air-conditioning system will be a minimum when the air-conditioner operates in a reversible manner. The coefficient of performance of a reversible air-conditioner (or refrigerator) depends on the temperature limits in the cycle only, and is determined from

COPR, rev =

1 1 = = 17.67 (TH / TL )- 1 (100 + 460 R )/ (70 + 460 R )- 1

The cooling load of this air-conditioning system is the sum of the heat gain from the outside and the heat generated within the house,

QL = 800 + 100 = 900 Btu/min The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator,

Wnet, in, min =

QL 900 Btu/min = = 50.93 Btu/min = 1.20 hp COPR, max 17.67

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(70+460) [R] T_H=(100+460) [R] Q_dot_heatgain=800 [Btu/min] Q_dot_generation=100 [Btu/min] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-51

COP_R_C=1/((T_H/T_L)-1) Q_dot_L=Q_dot_heatgain+Q_dot_generation W_dot_net_in_min=Q_dot_L/COP_R_C*1/Convert(hp, Btu/min)

6-104 The rate of heat removal and the source and sink temperatures are given for a Carnot refrigerator. The COP of the refrigerator and the power input are to be determined. Assumptions The refrigerator operates steadily. Analysis The COP of the Carnot refrigerator is determined from

COPR,max =

TL 288 K = = 13.71 TH - TL (36 - 15) K

The power input is

COPR,max =

QL 16, 000 kJ/h ¾¾ ® 13.71 = ¾¾ ® Win = 1167 kJ/h = 0.324 kW Win Win

The rate of heat rejected is

QH = QL + Wnet,in = 16,000 kJ/h + 1167 kJ/h = 17,167 kJ / h

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(15+273) [K] Q_dot_L=16000 [kJ/h] T_H=(36+273) [K] "Analysis" COP_R_max=T_L/(T_H-T_L) W_dot_in=Q_dot_L/COP_R_max W_dot_in_kW=W_dot_in*Convert(kJ/h, kW) Q_dot_H=Q_dot_L+W_dot_in

6-105E The power required by a reversible refrigerator with specified reservoir temperatures is to be determined. Assumptions The refrigerator operates steadily. Analysis The COP of this reversible refrigerator is

6-52

COPR, max =

TL 450 R = = 5 TH - TL 540 R - 450 R

Using this result in the coefficient of performance expression yields

Wnet, in =

ö QL 15, 000 Btu/h æ 1 kW ÷ çç = ÷= 0.879 kW çè3412.14 Btu/h ÷ ø COPR, max 5

EES (Engineering Equation Solver) SOLUTION "Given" T_L=450 [R] T_H=540 [R] Q_dot_L=15000 [Btu/h] "Analysis" COP_R_max=T_L/(T_H-T_L) W_dot_net=Q_dot_L/COP_R_max*Convert(Btu/h, kW)

6-106 The refrigerated space temperature, the COP, and the power input of a Carnot refrigerator are given. The rate of heat removal from the refrigerated space and its temperature are to be determined. Assumptions The refrigerator operates steadily. Analysis (a) The rate of heat removal from the refrigerated space is determined from the definition of the COP of a refrigerator,

QL = COPRWnet, in = (4.5)(0.5 kW) = 2.25 kW = 135 kJ / min (b) The temperature of the refrigerated space TL is determined from the coefficient of performance relation for a Carnot refrigerator,

COPR, rev =

1 (TH / TL )- 1

¾¾ ®

4.5 =

1 (25 + 273 K )/TL - 1

It yields

TL = 243.8 K = 29.2°C

6-53

EES (Engineering Equation Solver) SOLUTION "Given" T_H=25[C]+273 [K] W_dot_net_in_min=0.500 [kW] COP_R_C=4.5 "Analysis" "(a)" Q_dot_L=COP_R_C*W_dot_net_in_min*Convert(kW, kJ/min) "(b)" COP_R_C=1/((T_H/T_L)-1)

6-107 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house is given. The minimum power input required is to be determined. Assumptions The heat pump operates steadily. Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from

COPHP, rev =

1 1 = = 10.2 1- (TL / TH ) 1- (- 5 + 273 K )/ (24 + 273 K )

The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be

Wnet, in, min =

ö QH 80,000 kJ/h æ ÷ çç 1 h ÷ = = 2.18 kW ÷ çè3600 s ø COPHP 10.2

which is the minimum power input required.

EES (Engineering Equation Solver) SOLUTION "Given" T_H=24[C]+273 [K] T_L=-5[C]+273 [K] Q_dot_H=80000 [kJ/h] "Analysis" COP_HP_C=1/(1-(T_L/T_H)) W_dot_net_in_min=Q_dot_H/COP_HP_C*1/Convert(kW, kJ/h)

6-108 A commercial refrigerator with R-134a as the working fluid is considered. The condenser inlet and exit states are specified. The mass flow rate of the refrigerant, the refrigeration load, the COP, and the minimum power input to the compressor are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-54

1. The refrigerator operates steadily. 2. The kinetic and potential energy changes are zero. Properties The properties of R-134a and water are (Steam and R-134a tables)

P1 = 1.2 MPa ü ïï ý h1 = 278.28 kJ/kg ïïþ T1 = 50°C T2 = Tsat@1.2 MPa + D Tsubcool = 46.3 - 5 = 41.3°C

P2 = 1.2 MPa ïü ïý h = 110.19 kJ/kg 2 T2 = 41.3°C ïïþ Tw,1 = 18°Cü ïï ý hw,1 = 75.54 kJ/kg xw,1 = 0 ïïþ

Tw,2 = 26°Cü ïï ý hw,2 = 109.01 kJ/kg ïï xw,2 = 0 þ Analysis (a) The rate of heat transferred to the water is the energy change of the water from inlet to exit

QH = mw (hw,2 - hw,1 ) = (0.25 kg/s)(109.01- 75.54) kJ/kg = 8.367 kW The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser. That is,

QH = mR (h1 - h2 ) ¾ ¾ ® mR =

QH 8.367 kW = = 0.0498 kg / s h1 - h2 (278.28 - 110.19) kJ/kg

(b) The refrigeration load is

QL = QH - Win = 8.37 - 3.30 = 5.07 kW (c) The COP of the refrigerator is determined from its definition,

COP =

QL 5.07 kW = = 1.54 3.3 kW Win

(d) The COP of a reversible refrigerator operating between the same temperature limits is COPmax =

1 1 = = 4.49 TH / TL - 1 (18 + 273) / (- 35 + 273) - 1

Then, the minimum power input to the compressor for the same refrigeration load would be

Win,min =

QL 5.07 kW = = 1.13 kW COPmax 4.49

6-55

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=-35[C]+273 [K] T_H=18[C]+273 [K] T_w1=18 [C] m_dot_w=0.25 [kg/s] T_w2=26 [C] P_1=1200 [kPa] T_1=50 [C] P_2=1200 [kPa] DELTAT_subcool=5 [C] W_dot_in=3.3 [kW] "PROPERTIES" Fluid1$='R134a' Fluid2$='Steam_iapws' h_1=enthalpy(Fluid1$, P=P_1, T=T_1) T_sat=temperature(Fluid1$, P=P_2, x=0) T_2=T_sat-DELTAT_subcool h_2=enthalpy(Fluid1$, P=P_2, T=T_2) h_w1=enthalpy(Fluid2$, T=T_w1, x=0) h_w2=enthalpy(Fluid2$, T=T_w2, x=0) "ANALYSIS" Q_dot_H=m_dot_w*(h_w2-h_w1) m_dot_R=Q_dot_H/(h_1-h_2) "mass flow rate of refrigerant" Q_dot_L=Q_dot_H-W_dot_in "refrigeration load" COP=Q_dot_L/W_dot_in "actual COP" COP_max=1/(T_H/T_L-1) "maximum COP" W_dot_in_min=Q_dot_L/COP_max "minimum power input"

6-109 A heat pump maintains a house at a specified temperature in winter. The maximum COPs of the heat pump for different outdoor temperatures are to be determined. Analysis The coefficient of performance of a heat pump will be a maximum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined for all three cases above to be

COPHP, rev =

1 1 = = 29.3 1- (TL / TH ) 1- (10 + 273 K )/ (20 + 273 K )

COPHP, rev =

1 1 = = 11.7 1- (TL / TH ) 1- (- 5 + 273 K )/ (20 + 273 K )

COPHP, rev =

1 1 = = 5.86 1- (TL / TH ) 1- (- 30 + 273 K )/ (20 + 273 K )

6-56

EES (Engineering Equation Solver) SOLUTION "Given" T_H=(20+273) [K] T_L_a=(10+273) [K] T_L_b=(-5+273) [K] T_L_c=(-30+273) [K] "Analysis" "(a)" COP_HP_C_a=1/(1-(T_L_a/T_H)) "(b)" COP_HP_C_b=1/(1-(T_L_b/T_H)) "(c)" COP_HP_C_c=1/(1-(T_L_c/T_H))

6-110E A heat pump maintains a house at a specified temperature. The rate of heat loss of the house is given. The minimum power inputs required for different source temperatures are to be determined. Assumptions The heat pump operates steadily. Analysis (a) The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. If the outdoor air at 25°F is used as the heat source, the COP of the heat pump and the required power input are determined to be

COPHP, max = COPHP, rev = =

1 1- (TL / TH )

1 = 10.15 1- (25 + 460 R )/ (78 + 460 R )

and

Wnet, in, min =

ö QH 70,000 Btu/h æ çç 1 hp ÷ = = 2.71 hp ÷ ÷ ç è 2545 Btu/h ø COPHP, max 10.15

(b) If the well-water at 50°F is used as the heat source, the COP of the heat pump and the required power input are determined to be

COPHP, max = COPHP, rev =

1 1 = = 19.2 1- (TL / TH ) 1- (50 + 460 R )/ (78 + 460 R )

and

Wnet, in, min =

ö QH 70,000 Btu/h æ ÷ çç 1 hp ÷ = = 1.43 hp ÷ çè 2545 Btu/h ø COPHP, max 19.2

6-57

EES (Engineering Equation Solver) SOLUTION "Given" T_H=(78+460) [R] T_L_air=(25+460) [R] T_L_water=(50+460) [R] Q_dot_H=70000 [Btu/h] "Analysis" "(a)" COP_HP_C_a=1/(1-(T_L_air/T_H)) W_dot_net_in_min_a=Q_dot_H/COP_HP_C_a*1/Convert(hp, Btu/h) "(a)" COP_HP_C_b=1/(1-(T_L_water/T_H)) W_dot_net_in_min_b=Q_dot_H/COP_HP_C_b*1/Convert(hp, Btu/h)

6-111 A reversible heat pump is considered. The temperature of the source and the rate of heat transfer to the sink are to be determined. Assumptions The heat pump operates steadily. Analysis Combining the first law, the expression for the coefficient of performance, and the thermodynamic temperature scale gives

COPHP,max =

TH TH - TL

which upon rearrangement becomes

æ ö÷ æ 1÷ ö 1 ÷ TL = TH ççç1÷ ÷= (300 K) ççç1÷= 112.5 K è ø çè COPHP,max ÷ 1.6 ø Based upon the definition of the heat pump coefficient of performance,

QH = COPHP,maxWnet,in = (1.6)(1.5 kW) = 2.4 kW

EES (Engineering Equation Solver) SOLUTION T_H=300 [K] W_dot_net_in=1.5 [kW] COP_HP_max=1.6 COP_HP_max=T_H/(T_H-T_L) Q_dot_H=COP_HP_max*W_dot_net_in

6-112 A Carnot heat pump consumes 4.8-kW of power when operating, and maintains a house at a specified temperature. The average rate of heat loss of the house in a particular day is given. The actual running time of the heat pump that day, the heating cost, and the cost if resistance heating is used instead are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-58

Analysis (a) The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only, and is determined from

COPHP, rev =

1 1 = = 12.96 1- (TL / TH ) 1- (2 + 273 K )/ (25 + 273 K )

The amount of heat the house lost that day is

QH = QH (1 day) = (55,000 kJ/h )(24 h ) = 1,320,000 kJ Then the required work input to this Carnot heat pump is determined from the definition of the coefficient of performance to be Wnet, in =

QH 1,320,000 kJ = = 101,880 kJ COPHP 12.96

Thus the length of time the heat pump ran that day is

Dt =

Wnet, in Wnet, in

101,880 kJ = 21,225 s = 5.90 h 4.8 kJ/s

(b) The total heating cost that day is Cost = W ´ price = (Wnet, in ´ D t )(price) = (4.8 kW)(5.90 h )(0.11 $/kWh ) = $3.11

(c) If resistance heating were used, the entire heating load for that day would have to be met by electrical energy. Therefore, the heating system would consume 1,320,000 kJ of electricity that would cost æ1 kWh ÷ ö New Cost = QH ´ price = (1,320,000 kJ )çç (0.11 $/kWh ) = $40.3 ÷ çè3600 kJ ÷ ø

EES (Engineering Equation Solver) SOLUTION "Given" T_H=(25+273) [K] T_L=(2+273) [K] Q_dot_H=55000 [kJ/h] W_dot_net_in=4.8 [kW] UnitPrice=0.11 [$/kWh] "Analysis" "(a)" COP_HP_C=1/(1-(T_L/T_H)) Q_H=Q_dot_H*daytime daytime=24 [h] W_net_in=Q_H/COP_HP_C time=W_net_in/W_dot_net_in*1/Convert(h, s) "(b)" Cost=W_dot_net_in*time*UnitPrice © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-59

"(c)" NewCost=Q_H*UnitPrice*1/Convert(kWh, kJ)

6-113 A Carnot heat engine is used to drive a Carnot refrigerator. The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined. Assumptions The heat engine and the refrigerator operate steadily. Analysis (a) The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from

hth, max = hth, C = 1-

TL 300 K = 1= 0.744 TH 1173 K

Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be

Wnet, out = hth QH = (0.744)(800 kJ/min ) = 595.2 kJ/min which is also the power input to the refrigerator, Wnet, in . The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used. The COP of the Carnot refrigerator is

COPR, rev =

1 1 = = 8.37 (TH / TL )- 1 (27 + 273 K)/ (- 5 + 273 K )- 1

Then the rate of heat removal from the refrigerated space becomes QL , R = (COPR, rev )(Wnet, in ) = (8.37)(595.2 kJ/min ) = 4982 kJ / min

(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine ( QL , HE ) and the heat discarded by the refrigerator ( QH ,R ),

QL, HE = QH , HE - Wnet, out = 800 - 595.2 = 204.8 kJ/min QH , R = QL, R + Wnet, in = 4982 + 595.2 = 5577.2 kJ/min and

Qambient = QL, HE + QH , R = 204.8 + 5577.2 = 5782 kJ / min

EES (Engineering Equation Solver) SOLUTION "Given" T_H_HE=(900+273) [K] T_L_HE=(27+273) [K] Q_dot_H_HE=800 [kJ/min] T_L_R=(-5+273) [K] T_H_R=(27+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-60

"Analysis" "(a)" eta_th_C=1-T_L_HE/T_H_HE W_dot_net_out_HE=eta_th_C*Q_dot_H_HE COP_R_C=1/((T_H_R/T_L_R)-1) W_dot_net_in_R=W_dot_net_out_HE Q_dot_L_R=COP_R_C*W_dot_net_in_R "(b)" Q_dot_L_HE=Q_dot_H_HE-W_dot_net_out_HE Q_dot_H_R=Q_dot_L_R+W_dot_net_in_R Q_dot_ambient=Q_dot_L_HE+Q_dot_H_R

6-114 A heat pump that consumes 4-kW of power when operating maintains a house at a specified temperature. The house is losing heat in proportion to the temperature difference between the indoors and the outdoors. The lowest outdoor temperature for which this heat pump can do the job is to be determined. Assumptions The heat pump operates steadily. Analysis Denoting the outdoor temperature by TL , the heating load of this house can be expressed as

QH = (4500 kJ/h ×K)(297 - TL ) = (1.25 kW/K)(297 - TL )K

The coefficient of performance of a Carnot heat pump depends on the temperature limits in the cycle only, and can be expressed as

COPHP =

1 1 = 1- (TL / TH ) 1- TL /(297 K)

COPHP =

(1.25 kW/K )(297 - TL )K QH = 4 kW Wnet, in

Equating the two relations above and solving for TL , we obtain

TL = 266.2 K = −6.8°C

EES (Engineering Equation Solver) SOLUTION "Given" q_H=4500 [kJ/h-K] W_dot_net_in=4 [kW] T_H=(24+273) [K] "Analysis" Q_dot_H=q_H*(T_H-T_L)*1/Convert(kW, kJ/h) COP_HP=1/(1-(T_L/T_H)) COP_HP=Q_dot_H/W_dot_net_in T_L_C=T_L-273 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-61

6-115 An expression for the COP of a completely reversible refrigerator in terms of the thermal-energy reservoir temperatures, TL and TH is to be derived. Assumptions The refrigerator operates steadily. Analysis Application of the first law to the completely reversible refrigerator yields

Wnet,in = QH - QL

This result may be used to reduce the coefficient of performance,

COPR,rev =

QL QL 1 = = Wnet,in QH - QL QH / QL - 1

Since this refrigerator is completely reversible, the thermodynamic definition of temperature tells us that,

QH T = H QL TL When this is substituted into the COP expression, the result is

COPR,rev =

TL 1 = TH / TL - 1 TH - TL

6-116 The COP of a completely reversible refrigerator as a function of the temperature of the sink is to be calculated and plotted. Assumptions The refrigerator operates steadily. Analysis The coefficient of performance for this completely reversible refrigerator is given by

COPR,max =

TL 250 K = TH - TL TH - 250 K

Using EES, we tabulate and plot the variation of COP with the sink temperature as follows: T_L=250 [K] COP_R_max=T_L/(T_H-T_L)

6-62

TH [ K]

COPR,max

300 320 340 360 380 400 420 440 460 480 500

5 3.571 2.778 2.273 1.923 1.667 1.471 1.316 1.19 1.087 1 5

COPR,max

0 300

340

380

420

460

500

TH [K]

Special Topic: Household Refrigerators 6-117C Today’s refrigerators are much more efficient than those built in the past as a result of using smaller and higher efficiency motors and compressors, better insulation materials, larger coil surface areas, and better door seals.

6-118C It is important to clean the condenser coils of a household refrigerator a few times a year since the dust that collects on them serves as insulation and slows down heat transfer. Also, it is important not to block air flow through the condenser coils since heat is rejected through them by natural convection, and blocking the air flow will interfere with this heat rejection process. A refrigerator cannot work unless it can reject the waste heat.

6-119C It is a bad idea to overdesign the refrigeration system of a supermarket so that the entire air-conditioning needs of the store can be met by refrigerated air without installing any air-conditioning system. This is because the refrigerators cool the air to a much lower temperature than needed for air conditioning, and thus their efficiency is much lower, and their operating cost is much higher.

6-120C It is a bad idea to meet the entire refrigerator/freezer requirements of a store by using a large freezer that supplies sufficient cold air at -20°C instead of installing separate refrigerators and freezers . This is because the freezers cool the air to a much lower temperature than needed for refrigeration, and thus their efficiency is much lower, and their operating cost is much higher. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-63

6-121 A refrigerator consumes 400 W when running, and $170 worth of electricity per year under normal use. The fraction of the time the refrigerator will run in a year is to be determined. Assumptions The electricity consumed by the light bulb is negligible. Analysis The total amount of electricity the refrigerator uses a year is Total electric energy used = We,total =

Total cost of energy $170/year = = 1360 kWh/year Unit cost of energy $0.125/kWh

The number of hours the refrigerator is on per year is

Total operating hours = D t =

We,total We

1360 kWh/year = 3400 h/year 0.400 kW

Noting that there are 365 x 24 = 8760 hours in a year, the fraction of the time the refrigerator is on during a year is determined to be Time fraction on =

Total operating hours 3400/year = = 0.388 Total hours per year 8760 h/year

Therefore, the refrigerator remained on 38.8% of the time.

EES (Engineering Equation Solver) SOLUTION "Given" Cost=170 [$/year] UnitCost=0.125 [$/kWh] W_dot_e=0.400 [kW] "Analysis" W_e_total=Cost/UnitCost Hours=W_e_total/W_dot_e TimeFraction=Hours/time time=365*24 [h/year]

6-122 The light bulb of a refrigerator is to be replaced by a $25 energy efficient bulb that consumes less than half the electricity. It is to be determined if the energy savings of the efficient light bulb justify its cost. Assumptions The new light bulb remains on the same number of hours a year. Analysis The lighting energy saved a year by the energy efficient bulb is

6-64

Lighting energy saved = (Lighting power saved)(Operating hours) = [(40 - 18)W](60 h/year) = 1320 Wh = 1.32 kWh

This means 1.32 kWh less heat is supplied to the refrigerated space by the light bulb, which must be removed from the refrigerated space. This corresponds to a refrigeration savings of

Lighting energy saved 1.32 kWh = = 1.02 kWh COP 1.3 Then the total electrical energy and money saved by the energy efficient light bulb become Refrigeration energy saved =

Total energy saved = (Lighting + Refrigeration) energy saved = 1.32 + 1.02 = 2.34 kWh/year Money saved = (Total energy saved)(Unit cost of energy) = (2.34 kWh/year)($0.13/kWh) = $0.30 / year That is, the light bulb will save only 30 cents a year in energy costs, and it will take $ 25/$ 0.19 = 83 years for it to pay for itself from the energy it saves. Therefore, it is not justified in this case.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_current=0.040 [kW] W_dot_efficient=0.018 [kW] Hours=60 [h/year] COP_R=1.3 Cost=25 [$] UnitCost=0.13 [$/kWh] "Analysis" EnergySaved_lighting=(W_dot_current-W_dot_efficient)*Hours EnergySaved_refrigeration=EnergySaved_lighting/COP_R EnergySaved_total=EnergySaved_lighting+EnergySaved_refrigeration MoneySaved=EnergySaved_total*UnitCost Payback=Cost/MoneySaved

6-123 The door of a refrigerator is opened 8 times a day, and half of the cool air inside is replaced by the warmer room air. The cost of the energy wasted per year as a result of opening the refrigerator door is to be determined for the cases of moist and dry air in the room. Assumptions 1. The room is maintained at 20C and 95 kPa at all times. 2. Air is an ideal gas with constant specific heats at room temperature. 3. The moisture is condensed at an average temperature of 4°C. 4. Half of the air volume in the refrigerator is replaced by the warmer kitchen air each time the door is opened. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-65

Properties The gas constant of air is R = 0.287 kPa ×m3 / kg ×K (Table A-1). The specific heat of air at room temperature is

c p = 1.005 kJ / kg.° C (Table A-2a). The heat of vaporization of water at 4°C is h fg = 2491.4 kJ / kg (Table A-4). Analysis The volume of the refrigerated air replaced each time the refrigerator is opened is 0.3 m3 (half of the 0.6 m3 air volume in the refrigerator). Then the total volume of refrigerated air replaced by room air per year is

Vair, replaced = (0.3 m3 )(20/day)(365 days/year) = 2190 m3 /year

ro =

Po 95 kPa = = 1.195 kg/m3 RTo (0.287 kPa.m 3 /kg.K)(4 + 273 K)

mair = r Vair = (1.195 kg/m3 )(2190 m3 /year) = 2617 kg/year The amount of moisture condensed and removed by the refrigerator is

mmoisture = mair (moisture removed per kg air) = (2617 kg air/year)(0.006 kg/kg air) = 15.70 kg/year The sensible, latent, and total heat gains of the refrigerated space become Qgain, sensible = mair c p (Troom - Trefrig )

= (2617 kg/year)(1.005 kJ/kg.°C)(20 - 4)°C = 42,082 kJ/year Qgain, latent = mmoisture hfg

= (15.70 kg/year)(2491.4 kJ/kg) = 39,121 kJ/year Qgain, total = Qgain, sensible + Qgain, latent = 42, 082 + 39,121 = 81, 202 kJ/year

For a COP of 1.4, the amount of electrical energy the refrigerator will consume to remove this heat from the refrigerated space and its cost are Electrical energy used (total) =

Qgain, total COP

ö 81,202 kJ/year æ çç 1 kWh ÷ = 16.11 kWh/year ÷ ÷ ç è3600 kJ ø 1.4

Cost of energy used (total) = (Energy used)(Unit cost of energy) = (16.11 kWh/year)($0.115/kWh)

= $1.85 / year If the room air is very dry and thus latent heat gain is negligible, then the amount of electrical energy the refrigerator will consume to remove the sensible heat from the refrigerated space and its cost become Electrical energy used (sensible)=

Qgain, sensible COP

ö 42,082 kJ/year æ çç 1 kWh ÷ ÷ ÷= 8.350 kWh/year çè3600 kJ ø 1.4

Cost of energy used (sensible) = (Energy used)(Unit cost of energy) = (8.350 kWh/year)($0.115/kWh)

6-66

EES (Engineering Equation Solver) SOLUTION "Given" V=0.9 [m^3] V_air=0.6 [m^3] T_R=4 [C] T_o=20 [C] P_o=95 [kPa] MoistureRemoved=0.006 [kg/kg] N_opening=20 [1/day] COP_R=1.4 UnitCost=0.115 [$/kWh] "Properties" Fluid$= 'steam_iapws' h_f=enthalpy(Fluid$, T=T_R, x=0) h_g=enthalpy(Fluid$, T=T_R, x=1) h_fg=h_g-h_f R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" "Analysis" V_air_replaced=1/2*V_air*N_opening*Convert(year, day) rho=P_o/(R*(T_R+273)) m_air=rho*V_air_replaced m_moisture=m_air*MoistureRemoved Q_gain_sensible=m_air*C_p*(T_o-T_R) Q_gain_latent=m_moisture*h_fg Q_gain_total=Q_gain_sensible+Q_gain_latent E_electric=Q_gain_total/COP_R*1/Convert(kWh, kJ) Cost=E_electric*UnitCost "If the room is very dry" E_electric_dry=Q_gain_sensible/COP_R*1/Convert(kWh, kJ) Cost_dry=E_electric_dry*UnitCost

Review Problems 6-124 The claim of a manufacturer of ice cream freezers regarding the COP of these freezers is to be evaluated. Assumptions The refrigerator operates steadily. Analysis The maximum refrigerator coefficient of performance would occur if the refrigerator were completely reversible,

COPR,max =

TL 250 K = = 5 TH - TL 300 K - 250 K

Since the claimed COP is less than this maximum, this refrigerator is possible.

6-67

EES (Engineering Equation Solver) SOLUTION T_H=300 [K] T_L=250 [K] COP_R_max=T_L/(T_H-T_L)

6-125 The claim of a heat pump designer regarding the COP of the heat pump is to be evaluated. Assumptions The heat pump operates steadily. Analysis The maximum heat pump coefficient of performance would occur if the heat pump were completely reversible,

COPHP,max =

TH 300 K = = 7.5 TH - TL 300 K - 260 K

Since the claimed COP is less than this maximum, this heat pump is possible.

EES (Engineering Equation Solver) SOLUTION T_H=300 [K] T_L=260 [K] COP_HP_max=T_H/(T_H-T_L)

6-126 An air-conditioning system maintains a house at a specified temperature. The rate of heat gain of the house, the rate of internal heat generation, and the COP are given. The required power input is to be determined. Assumptions Steady operating conditions exist. Analysis The cooling load of this air-conditioning system is the sum of the heat gain from the outdoors and the heat generated in the house from the people, lights, and appliances:

QL = 20, 000 + 8, 000 = 28,000 kJ/h Using the definition of the coefficient of performance, the power input to the air-conditioning system is determined to be

Wnet, in =

ö QL 28,000 kJ/h æ çç 1 kW ÷ = = 3.11 kW ÷ ÷ ç è3600 kJ/h ø COPR 2.5

6-68

EES (Engineering Equation Solver) SOLUTION "Given" T=20 [C] Q_dot_heatgain=20000 [kJ/h] Q_dot_generated=8000 [kJ/h] COP_R=2.5 "Analysis" Q_dot_L=Q_dot_heatgain+Q_dot_generated W_dot_net_in=Q_dot_L/COP_R*1/Convert(kW, kJ/h)

6-127E A Carnot heat pump maintains a house at a specified temperature. The rate of heat loss from the house and the outdoor temperature are given. The COP and the power input are to be determined. Analysis (a) The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only, and is determined from

COPHP, rev =

1 1 = = 13.4 1- (TL / TH ) 1- (35 + 460 R )/ (75 + 460 R )

(b) The heating load of the house is

QH = (2500 Btu/h ×°F)(75 - 35)°F = 100,000 Btu/h Then the required power input to this Carnot heat pump is determined from the definition of the coefficient of performance to be

Wnet, in =

ö QH 100,000 Btu/h æ ÷= 2.93 hp çç 1 hp ÷ = ÷ çè 2545 Btu/h ø COPHP 13.4

EES (Engineering Equation Solver) SOLUTION "Given" T_H=75[F]+460 [R] T_L=35[F]+460 [R] q_H=2500 [Btu/h-F] "Analysis" "(a)" COP_HP_C=1/(1-(T_L/T_H)) "(b)" Q_dot_H=q_H*(T_H-T_L) W_dot_net_in_min=Q_dot_H/COP_HP_C*1/Convert(hp, Btu/h)

6-69

6-128E A refrigerator with a water-cooled condenser is considered. The cooling load and the COP of a refrigerator are given. The power input, the exit temperature of water, and the maximum possible COP of the refrigerator are to be determined. Assumptions The refrigerator operates steadily.

Analysis (a) The power input is Win =

öæ 1 h ö QL 21, 000 Btu/h æ ÷ çç1.055 kJ ÷ çç = ÷ ÷ ÷ ÷= 3.239 kW ç è 1 Btu øèç3600 s ø COP 1.9

(b) The rate of heat rejected in the condenser is

QH = QL + Win æ 1 Btu öæ 3600 s ö ÷ ÷ = 21, 000 Btu/h + 3.239 kW çç ÷ççç ÷ ÷ ÷ èç1.055 kJ øè 1h ø

= 32, 053 Btu/h The exit temperature of the water is

QH = mc p (T2 - T1 ) T2 = T1 +

QH mc p

= 65°F +

32, 053 Btu/h = 71.14°F æ3600 s ö÷ ç (1.45 lbm/s) ç ÷(1.0 Btu/lbm ×°F) çè 1 h ø÷

TL 25 + 460 = = 11.26 TH - TL 0.5(65 + 71.14) - 25

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(25+460) [R] Q_dot_L=21000 [Btu/h] T1=65 [F] m_dot=1.45 [lbm/s] COP_HP=1.9 "Properties" c_p=1.0 [Btu/lbm-R] "Table A-3E" "Analysis" "(a)" W_dot_in=Q_dot_L/COP_HP*Convert(Btu/h, kW) "(b)" Q_dot_H=Q_dot_L+W_dot_in*Convert(kW, Btu/h) Q_dot_H=m_dot*Convert(lbm/s, lbm/h)*c_p*(T2-T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-70

"(c)" T_H=0.5*(T1+T2)+460 "average temperature of water is taken as T_H" COP_HP_max=T_L/(T_H-T_L)

6-129 A refrigeration system is to cool bread loaves at a rate of 1200 per hour by refrigerated air at − 30C. The rate of heat removal from the breads, the required volume flow rate of air, and the size of the compressor of the refrigeration system are to be determined. Assumptions 1. Steady operating conditions exist. 2. The thermal properties of the bread loaves are constant. 3. The cooling section is well-insulated so that heat gain through its walls is negligible.

Properties The average specific and latent heats of bread are given to be 2.93 kJ / kg.° C and 109.3 kJ / kg , respectively. The gas constant of air is 0.287 kPa.m3 / kg.K (Table A-1), and the specific heat of air at the average temperature of (- 30 + - 22) / 2 = - 26° C = 250 K is c p = 1.0 kJ / kg.° C (Table A-2).

Analysis (a) Noting that the breads are cooled at a rate of 500 loaves per hour, breads can be considered to flow steadily through the cooling section at a mass flow rate of

mbread = (1200 breads/h)(0.350 kg/bread) = 420 kg/h = 0.1167 kg/s − 10C and frozen becomes

Then the rate of heat

Qbread = (mc p D T )bread = (420 kg/h)(2.93 kJ/kg.°C)[(30 - ( - 10)]°C = 49,224 kJ/h Qfreezing = (mhlatent )bread = (420 kg/h )(109.3 kJ/kg) = 45,906 kJ/h and

Qtotal = Qbread + Qfreezing = 49, 224 + 45,906 = 95,130 kJ / h (b) All the heat released by the breads is absorbed by the refrigerated air, and the temperature rise of air is not to exceed 8C. The minimum mass flow and volume flow rates of air are determined to be

mair = r=

Qair 95,130 kJ/h = = 11,891 kg/h (c p D T )air (1.0 kJ/kg.°C)(8°C)

P 101.3 kPa = = 1.453 kg/m3 RT (0.287 kPa.m3 /kg.K)( - 30 + 273) K

Vair =

mair 11,891 kg/h = = 8185 m 3 / h r air 1.453 kg/m3

Qrefrig COP

95,130 kJ/h = 79,275 kJ/h = 22.02 kW 1.2

6-71

EES (Engineering Equation Solver) SOLUTION "Given" m_bread=0.350 [kg] N_dot_bread=1200 [1/h] T1_bread=30 [C] T2_bread=-10 [C] T_air=-30 [C] DELTAT_air=8 [C] COP=1.2 c_p_bread=2.93 [kJ/kg-C] h_fg_bread=109.3 [kJ/kg] "Properties" c_p_air=1.0 [kJ/kg-C] "at (-30-22)/2=-26 C = 250 K, Table A-2b" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" "(a)" m_dot_bread=N_dot_bread*m_bread Q_dot_bread=m_dot_bread*c_p_bread*(T1_bread-T2_bread) Q_dot_freezing=m_dot_bread*h_fg_bread Q_dot_total=Q_dot_bread+Q_dot_freezing "(b)" m_dot_air=Q_dot_total/(c_p_air*DELTAT_air) rho_air=P/(R*(T_air+273)) P=101.325 [kPa] "assumed" V_dot_air=m_dot_air/rho_air "(c)" W_dot_refrig=Q_dot_total/COP*1/Convert(kW, kJ/h)

6-130 A heat pump with a specified COP and power consumption is used to heat a house. The time it takes for this heat pump to raise the temperature of a cold house to the desired level is to be determined. Assumptions 1. Air is an ideal gas with constant specific heats at room temperature. 2. The heat loss of the house during the warp-up period is negligible. 3. The house is well-sealed so that no air leaks in or out. Properties The constant volume specific heat of air at room temperature is cv = 0.718 kJ / kg ×° C . Analysis Since the house is well-sealed (constant volume), the total amount of heat that needs to be supplied to the house is

QH = (mcv D T )house = (1500 kg )(0.718 kJ/kg ×°C)(22 - 7 )°C = 16,155 kJ

The rate at which this heat pump supplies heat is

QH = COPHPWnet, in = (2.8)(5 kW) = 14 kW That is, this heat pump can supply 14 kJ of heat per second. Thus the time required to supply 16,155 kJ of heat is Dt =

QH 16,155 kJ = = 1154 s = 19.2 min 14 kJ/s QH

6-72

EES (Engineering Equation Solver) SOLUTION "Given" COP_HP=2.8 W_dot_net_in=5 [kW] T1=7 [C] T2=22 [C] m=1500 [kg] "Properties" c_v=0.718 [kJ/kg-C] "Table A-2a" "Analysis" Q_H=m*c_v*(T2-T1) Q_dot_H=COP_HP*W_dot_net_in time=Q_H/Q_dot_H

6-131 A solar pond power plant operates by absorbing heat from the hot region near the bottom, and rejecting waste heat to the cold region near the top. The maximum thermal efficiency that the power plant can have is to be determined. Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot efficiency, which is determined from

hth, max = hth, C = 1-

TL 308 K = 1= 0.127 or 12.7% TH 353 K

In reality, the temperature of the working fluid must be above 35C in the condenser, and below 80C in the boiler to allow for any effective heat transfer. Therefore, the maximum efficiency of the actual heat engine will be lower than the value calculated above.

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(35+273) [K] T_H=(80+273) [K] "Analysis" eta_th=1-T_L/T_H

6-132 A Carnot heat engine cycle is executed in a closed system with a fixed mass of steam. The net work output of the cycle and the ratio of sink and source temperatures are given. The low temperature in the cycle is to be determined. Assumptions The engine is said to operate on the Carnot cycle, which is totally reversible. Analysis The thermal efficiency of the cycle is

6-73

hth = 1-

TL 1 = 1- = 0.5 TH 2

Also, hth =

W QH

¾¾ ®

QH =

W 60 kJ = = 120 kJ hth 0.5

QL = QH - W = 120 - 60 = 60 kJ and qL =

QL 60 kJ = = 2400 kJ/kg = h fg @ TL m 0.025 kg

since the enthalpy of vaporization h fg at a given T or P represents the amount of heat transfer as 1 kg of a substance is converted from saturated liquid to saturated vapor at that T or P. Therefore, TL is the temperature that corresponds to the h fg value of 2400 kJ / kg , and is determined from the steam tables (Table A-4) to be TL = 42.5° C

EES (Engineering Equation Solver) SOLUTION "Given" m=0.025 [kg] RatioT=0.5 "RatioT=T_L/T_H" W_net_out=60 [kJ] "Properties" Fluid$= 'steam_iapws' h_f=enthalpy(Fluid$, T=T_L, x=0) h_g=enthalpy(Fluid$, T=T_L, x=1) h_fg=h_g-h_f "Analysis" eta_th=1-RatioT eta_th=W_net_out/Q_H Q_L=Q_H-W_net_out Q_L=m*h_fg

6-133 Problem 6-132 is reconsidered. The effect of the net work output on the required temperature of the steam during the heat rejection process as the work output varies from 40 kJ to 60 kJ is to be investigated. Analysis The problem is solved using EES, and the results are tabulated and plotted below. "Given" m=0.025 [kg] RatioT=0.5 "RatioT=T_L/T_H" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-74

W_net_out=60 [kJ] "Properties" Fluid$= 'steam_iapws' h_f=enthalpy(Fluid$, T=T_L, x=0) h_g=enthalpy(Fluid$, T=T_L, x=1) h_fg=h_g-h_f "Analysis" eta_th=1-RatioT eta_th=W_net_out/Q_H Q_L=Q_H-W_net_out Q_L=m*h_fg

Wout

TL,C

[kJ]

[C] 270.8 252.9 232.8 209.9 184 154.4 120.8 83.17 42.5

40 42.5 45 47.5 50 52.5 55 57.5 60

300

250

TL [C]

200

150

100

0 40

W net,out [kJ]

6-134 A Carnot refrigeration cycle is executed in a closed system with a fixed mass of R-134a. The net work input and the ratio of maximum-to-minimum temperatures are given. The minimum pressure in the cycle is to be determined. Assumptions The refrigerator is said to operate on the reversed Carnot cycle, which is totally reversible. Analysis The coefficient of performance of the cycle is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-75

COPR =

1 1 = = 5 TH / TL - 1 1.2 - 1

COPR =

QL ¾¾ ® QL = COPR ´ Win = (5)(22 kJ ) = 110 kJ Win

Also,

QH = QL + W = 110 + 22 = 132 kJ and qH =

QH 132 kJ = = 137.5 kJ/kg = h fg @ TH m 0.96 kg

since the enthalpy of vaporization h fg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P. Therefore, TH is the temperature that corresponds to the h fg value of 137.5 kJ / kg , and is determined from the R-134a tables to be

TH @61.3°C = 334.3 K Then,

TL =

TH 334.3 K = = 278.6 K @ 5.6°C 1.2 1.2

Therefore,

Pmin = Psat @ 5.6°C = 356 kPa

EES (Engineering Equation Solver) SOLUTION "Given" m=0.96 [kg] RatioT=1.2 "RatioT=T_H/T_L" W_net_in=22 [kJ] "Properties" Fluid$= 'R134a' h_f=enthalpy(Fluid$, T=T_H, x=0) h_g=enthalpy(Fluid$, T=T_H, x=1) h_fg=h_g-h_f "Analysis" COP_R=1/(RatioT-1) COP_R=Q_L/W_net_in Q_H=Q_L+W_net_in Q_H=m*h_fg T_H=RatioT*T_L P_min=pressure(Fluid$, T=T_L, x=1)

6-76

6-135 Problem 6-134 is reconsidered. The effect of the net work input on the minimum pressure as the work input varies from 10 kJ to 30 kJ is to be investigated. The minimum pressure in the refrigeration cycle is to be plotted as a function of net work input. Analysis The problem is solved using EES, and the results are tabulated and plotted below. m_R134a = 0.96 [kg] THtoTLRatio = 1.2 "T_H = 1.2T_L" W_in = 22 [kJ] "Depending on the value of W_in, adjust the guess value of T_H." COP_R = 1/( THtoTLRatio- 1) Q_L = W_in*COP_R Q_L + W_in = Q_H Q_H=m_R134a*h_fg h_fg=enthalpy(R134a,T=T_H,x=1) - enthalpy(R134a,T=T_H,x=0) T_H=THtoTLRatio*T_L P_min = pressure(R134a,T=T_L,x=0)*convert(kPa,MPa) T_L_C = T_L - 273 Win

Pmin

TL,C

[kW]

[MPa]

[K]

10 15 20 25 30

0.8673 0.6837 0.45 0.2251 0.06978

368.8 358.9 342.7 319.3 287.1

307.3 299 285.6 266.1 239.2

[C] 34.32 26.05 12.61 -6.907 -33.78

6-136 A Carnot heat engine cycle is executed in a steady-flow system with steam. The thermal efficiency and the mass flow rate of steam are given. The net power output of the engine is to be determined. Assumptions All components operate steadily. Properties The enthalpy of vaporization hfg of water at 275C is 1574.5 kJ / kg (Table A-4).

6-77

Analysis The enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer as 1 kg of a substance is converted from saturated liquid to saturated vapor at that T or P. Therefore, the rate of heat transfer to the steam during heat addition process is

QH = mh fg @ 275°C = (3 kg/s)(1574.5 kJ/kg) = 4723 kJ/s Then the power output of this heat engine becomes

Wnet, out = hth QH = (0.30)(4723 kW) = 1417 kW

EES (Engineering Equation Solver) SOLUTION "Given" eta_th=0.30 T_H=275 [C] m_dot=3 [kg/s] "Properties" Fluid$= 'steam_iapws' h_f=enthalpy(Fluid$, T=T_H, x=0) h_g=enthalpy(Fluid$, T=T_H, x=1) h_fg=h_g-h_f "Analysis" Q_dot_H=m_dot*h_fg W_dot_net_out=eta_th*Q_dot_H

6-137 Two Carnot heat engines operate in series between specified temperature limits. If the thermal efficiencies of both engines are the same, the temperature of the intermediate medium between the two engines is to be determined. Assumptions The engines are said to operate on the Carnot cycle, which is totally reversible.

Analysis The thermal efficiency of the two Carnot heat engines can be expressed as

6-78

hth, I = 1-

T T and hth, II = 1- L TH T

Equating, 1-

T T = 1- L TH T

Solving for T,

T = TH TL =

(1400 K)(300 K) = 648 K

EES (Engineering Equation Solver) SOLUTION "Given" T_H=1400 [K] T_L=300 [K] "Analysis" eta_th=1-T/T_H eta_th=1-T_L/T

6-138 Half of the work output of a Carnot heat engine is used to drive a Carnot heat pump that is heating a house. The minimum rate of heat supply to the heat engine is to be determined. Assumptions Steady operating conditions exist. Analysis The coefficient of performance of the Carnot heat pump is

COPHP, C =

1 1 = = 14.75 1- (TL / TH ) 1- (2 + 273 K )/ (22 + 273 K )

Then power input to the heat pump, which is supplying heat to the house at the same rate as the rate of heat loss, becomes

Wnet, in =

QH 62,000 kJ/h = = 4203 kJ/h COPHP, C 14.75

which is half the power produced by the heat engine. Thus the power output of the heat engine is

Wnet, out = 2Wnet, in = 2(4203 kJ/h) = 8406 kJ/h To minimize the rate of heat supply, we must use a Carnot heat engine whose thermal efficiency is determined from

6-79

TL 293 K = 1= 0.727 TH 1073 K

hth, C = 1-

Then the rate of heat supply to this heat engine is determined from the definition of thermal efficiency to be QH , HE =

Wnet, out hth, HE

8406 kJ/h = 11, 560 kJ / h 0.727

EES (Engineering Equation Solver) SOLUTION "Given" T_H_HE=(800+273) [K] T_L_HE=(20+273) [K] W_dot_net_out_HE=2*W_dot_net_in_HP T_L_HP=(2+273) [K] T_H_HP=(22+273) [K] Q_dot_H_HP=62000 [kJ/h] "Analysis" COP_HP_C=1/(1-(T_L_HP/T_H_HP)) W_dot_net_in_HP=Q_dot_H_HP/COP_HP_C eta_th_C=1-T_L_HE/T_H_HE Q_dot_H_HE=W_dot_net_out_HE/eta_th_C

6-139 The thermal efficiency and power output of a gas turbine are given. The rate of fuel consumption of the gas turbine is to be determined. Assumptions Steady operating conditions exist. Properties The density and heating value of the fuel are given to be 0.8 g / cm3 and 42,000 kJ / kg , respectively. Analysis This gas turbine is converting 21% of the chemical energy released during the combustion process into work. The amount of energy input required to produce a power output of 6,000 kW is determined from the definition of thermal efficiency to be

QH =

Wnet, out hth

6000 kJ/s = 28,570 kJ/s 0.21

To supply energy at this rate, the engine must burn fuel at a rate of m=

28,570 kJ/s = 0.6803 kg/s 42,000 kJ/kg

since 42,000 kJ of thermal energy is released for each kg of fuel burned. Then the volume flow rate of the fuel becomes

6-80

m 0.6803 kg/s = = 0.850 L / s r 0.8 kg/L

EES (Engineering Equation Solver) SOLUTION "Given" eta_th=0.21 W_dot_net_out=6000 [kW] HV=42000 [kJ/kg] rho=0.8 [g/cm^3] "Analysis" Q_dot_H=W_dot_net_out/eta_th m_dot=Q_dot_H/HV V_dot=m_dot/rho

6-140 A Carnot heat pump cycle is executed in a steady-flow system with R-134a flowing at a specified rate. The net power input and the ratio of the maximum-to-minimum temperatures are given. The ratio of the maximum to minimum pressures is to be determined. Analysis The coefficient of performance of the cycle is

COPHP =

1 1 = = 6.0 1- TL / TH 1- 1/ 1.2

and

QH = COPHP ´ Win = (6.0)(5 kW) = 30.0 kJ/s qH =

QH 30.0 kJ/s = = 166.67 kJ/kg = h fg @ TH m 0.18 kg/s

since the enthalpy of vaporization h fg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P. Therefore, TH is the temperature that corresponds to the h fg value of 166.67 kJ / kg , and is determined from the R-134a tables to be

TH @36.5°C = 309.6 K and, Pmax = Psat @ 36.5°C = 924.5 kPa

TL =

TH 309.6 K = = 258.0 K @- 15.15°C 1.2 1.2

Also,

6-81

Pmin = Psat @ - 15.15°C = 163.2 kPa Then the ratio of the maximum to minimum pressures in the cycle is Pmax 924.5 kPa = = 5.67 Pmin 163.2 kPa

EES (Engineering Equation Solver) SOLUTION "Given" m_dot=0.18 [kg/s] RatioT=1/1.2 "RatioT=T_L/T_H" W_dot_net_in=5 [kW] "Properties" Fluid$= 'R134a' h_f=enthalpy(Fluid$, T=T_H, x=0) h_g=enthalpy(Fluid$, T=T_H, x=1) h_fg=h_g-h_f "Analysis" COP_HP=1/(1-RatioT) Q_dot_H=COP_HP*W_dot_net_in Q_dot_H=m_dot*h_fg P_max=pressure(Fluid$, T=T_H, x=1) T_L=T_H*RatioT P_min=pressure(Fluid$, T=T_L, x=1) RatioP=P_max/P_min

6-141 The cargo space of a refrigerated truck is to be cooled from 25C to an average temperature of 5C. The time it will take for an 11-kW refrigeration system to precool the truck is to be determined. Assumptions 1. The ambient conditions remain constant during precooling. 2. The doors of the truck are tightly closed so that the infiltration heat gain is negligible. 3. The air inside is sufficiently dry so that the latent heat load on the refrigeration system is negligible. 4. Air is an ideal gas with constant specific heats.

Properties The density of air is taken 1.2 kg / m3 , and its specific heat at the average temperature of 15C is

c p = 1.0 kJ / kg.° C (Table A-2). Analysis The mass of air in the truck is

mair = r airVtruck = (1.2 kg/m3 )(12 m´ 2.3 m´ 3.5 m) = 116 kg The amount of heat removed as the air is cooled from 25 to 5C

6-82

Qcooling, air = (mc p D T )air = (116 kg)(1.0 kJ/kg.°C)(25 - 5)°C = 2,320 kJ Noting that UA is given to be 80 W / ° C and the average air temperature in the truck during precooling is (25 + 5) / 2 = 15° C , the average rate of heat gain by transmission is determined to be

Qtransmission,avg = UAD T = (120 W/°C)(25 - 15)°C = 1200 W = 1.2 kJ/s Therefore, the time required to cool the truck from 25 to 5C is determined to be

Qrefrig.D t = Qcooling, air + Qtransmission D t ¾ ¾ ® Dt =

Qcooling, air Qrefrig. - Qtransmission

2320 kJ = 237 s = 3.95 min (11- 1.2) kJ/s

EES (Engineering Equation Solver) SOLUTION "Given" V_truck=12*2.3*3.5 [m^3] T1=25 [C] T2=5 [C] UA=0.120 [kW/C] T_amb=25 [C] Q_dot_refrig=11 [kW] "Properties" rho=1.2 [kg/m^3] "Table A-2" c_p=1.0 [kJ/kg-C] "Table A-2" "Analysis" m=rho*V_truck Q_cooling_air=m*c_p*(T1-T2) T_ave=1/2*(T1+T2) Q_dot_transmission=UA*(T_amb-T_ave) Q_dot_refrig*time=Q_cooling_air+Q_dot_transmission*time

6-142 The maximum flow rate of a standard shower head can be reduced from 13.3 to 10.5 L / min by switching to lowflow shower heads. The amount of oil and money a family of four will save per year by replacing the standard shower heads by the low-flow ones are to be determined. Assumptions 1. Steady operating conditions exist. 2. Showers operate at maximum flow conditions during the entire shower. 3. Each member of the household takes a 5-min shower every day.

Properties The specific heat of water is c = 4.18 kJ / kg.° C and heating value of heating oil is 146,300 kJ / gal (given). The density of water is r = 1 kg / L . Analysis The low-flow heads will save water at a rate of © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-83

Vsaved = [(13.3 - 10.5) L/min](6 min/person.day)(4 persons)(365 days/yr) = 24,528 L/year msaved = rVsaved = (1 kg/L)(24,528 L/year) = 24,528 kg/year Then the energy, fuel, and money saved per year becomes

Energy saved = msaved cD T = (24,528 kg/year)(4.18 kJ/kg.°C)(42-15)°C = 2,768,000 kJ/year Fuel saved =

Energy saved 2,768,000 kJ/year = = 29.1 gal / year (Efficiency)(Heating value of fuel) (0.65)(146,300 kJ/gal)

Money saved = (Fuel saved)(Unit cost of fuel) = (29.1gal/year)($2.80/gal) = $81.5 / year Therefore, switching to low-flow shower heads will save about $80 per year in energy costs.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot_old=13.3 [L/min] V_dot_new=10.5 [L/min] N_people=4 time_shower=6 [min/day] T1=15 [C] T2=55 [C] T3=42 [C] eta=0.65 UnitCost=1.20 [$/gal] C_p_w=4.18 [kJ/kg-C] HV_oil=146300 [kJ/gal] "Properties" rho_w=1000 [kg/m^3] "Analysis" V_dot_saved=(V_dot_old-V_dot_new)*time_shower*N_people*time*1/Convert(m^3, L) time=365 [day/year] m_dot_saved=rho_w*V_dot_saved EnergySaved=m_dot_saved*C_p_w*(T3-T1) FuelSaved=EnergySaved/(eta*HV_oil) MoneySaved=FuelSaved*UnitCost

6-143 The drinking water needs of a production facility with 20 employees is to be met by a bobbler type water fountain. The size of compressor of the refrigeration system of this water cooler is to be determined. Assumptions 1. Steady operating conditions exist. 2. Water is an incompressible substance with constant properties at room temperature. 3. The cold water requirement is 0.4 L / h per person. Properties The density and specific heat of water at room temperature are r = 1.0 kg / L and c = 4.18 kJ / kg.° C (Table A-3). Analysis The refrigeration load in this case consists of the heat gain of the reservoir and the cooling of the incoming water. The water fountain must be able to provide water at a rate of

mwater = r Vwater = (1 kg/L)(0.4 L/h ×person)(20 persons) = 8.0 kg/h To cool this water from 22C to 8C, heat must removed from the water at a rate of

6-84

Qcooling = mcp (Tin - Tout )

= (8.0 kg/h)(4.18 kJ/kg.°C)(22 - 8)°C = 468 kJ/h = 130 W

(since 1 W = 3.6 kJ/h)

Then total refrigeration load becomes

Qrefrig, total = Qcooling + Qtransfer = 130 + 45 = 175 W Noting that the coefficient of performance of the refrigeration system is 2.9, the required power input is Wrefrig =

Qrefrig COP

175 W = 60.3 W 2.9

Therefore, the power rating of the compressor of this refrigeration system must be at least 60.3 W to meet the cold water requirements of this office.

EES (Engineering Equation Solver) SOLUTION "Given" N_people=20 T1=22 [C] T2=8 [C] V_dot=0.4E-3 [m^3/h] T_amb=25 [C] Q_dot_HeatTransfer=45 [W] COP=2.9 "Properties" rho=1000 [kg/m^3] "Table A-3" c_p=4.18 [kJ/kg-C] "Table A-3" "Analysis" m_dot=rho*V_dot*N_people Q_dot_cooling=m_dot*c_p*(T1-T2)*1/Convert(W, kJ/h) Q_dot_refrig=Q_dot_cooling+Q_dot_HeatTransfer W_dot_refrig=Q_dot_refrig/COP

6-144 A typical heat pump powered water heater costs about $800 more to install than a typical electric water heater. The number of years it will take for the heat pump water heater to pay for its cost differential from the energy it saves is to be determined. Assumptions 1. The price of electricity remains constant. 2. Water is an incompressible substance with constant properties at room temperature. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-85

Time value of money (interest, inflation) is not considered.

Analysis The amount of electricity used to heat the water and the net amount transferred to water are Total energy used (electrical) =

Total cost of energy Unit cost of energy

$350/year $0.11/kWh

= 3182 kWh/year Total energy transfer to water = Ein = (Efficiency)(Total energy used)

= 0.95´ 3182 kWh/year

= 3023 kWh/year The amount of electricity consumed by the heat pump and its cost are Energy usage (of heat pump) =

Energy transfer to water 3023 kWh/year = = 916.0 kWh/year COPHP 3.3

Energy cost (of heat pump) = (Energy usage)(Unit cost of energy) = (916.0 kWh/year)($0.11/kWh)

= $100.8/year Then the money saved per year by the heat pump and the simple payback period become

Money saved = (Energy cost of electric heater) - (Energy cost of heat pump) = $350 - $100.8 = $249.2

Simple payback period =

Additional installation cost $800 = = 3.21 years Money saved $249.2/year

Discussion The economics of heat pump water heater will be even better if the air in the house is used as the heat source for the heat pump in summer, and thus also serving as an air-conditioner.

EES (Engineering Equation Solver) SOLUTION "Knowns" Cost_ElectHeat = 350[$/year] Cost_KWh=0.11 [$/kWh] eta=0.95 Cost_extra_HP=800 [$] "The energy supplied to the water by the electric heater is 90% of energy purchased. The energy purchased, in kW, is the yearly cost of the electric heat divided by the cost per KWh." E_ElectHeater = eta*Cost_ElectHeat/Cost_KWh "For the same amont of heated water and assuming that all the heat energy leaving the heat pump goes into the water, then Energy supplied by heat pump heater = Energy supplied by electric heater" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-86

E_HeatPump = E_ElectHeater "Electrical Work enegy supplied to heat pump = Heat added to water/COP" COP=3.3 COP=E_HeatPump/W_HeatPump "the definition of COP" "Cost per year to operate the heat pump is" Cost_HeatPump=W_HeatPump*Cost_KWh "Let N_BrkEven be the number of years to break even. At the break even point, the total cost difference between the two water heaters is zero. The years to break even, neglecting the cost to borrow the extra $800 to install heat pump is then:" CostDiff_total = 0 CostDiff_total=Cost_extra_HP+N_BrkEven*(Cost_HeatPump-Cost_ElectHeat)

6-145 Problem 6-144 is reconsidered. The effect of the heat pump COP on the yearly operation costs and the number of years required to break even are to be considered. Analysis The problem is solved using EES, and the results are tabulated and plotted below. Cost_ElectHeater = 350 [$/year] eta=0.95 E_ElectHeater = eta*Cost_ElectHeater /UnitCost UnitCost=0.11 [$/kWh] COP=3.3 E_HeatPump = E_ElectHeater W_HeatPump = E_HeatPump/COP Cost_HeatPump=W_HeatPump*UnitCost CostDiff_total = 0 [$] CostDiff_total=AddCost+N_BrkEven*(Cost_HeatPump-Cost_ElectHeater) "N_BrkEven: the number of years to break even" AddCost=800 [$] COP

BBrkEven

Cost HeatPump

Cost ElektHeater

2 2.3 2.6 2.9 3.2 3.5 3.8 4.1 4.4 4.7 5

[years] 4.354 3.894 3.602 3.399 3.251 3.137 3.048 2.975 2.915 2.865 2.822

[$ / year] 166.3 144.6 127.9 114.7 103.9 95 87.5 81.1 75.57 70.74 66.5

[$ / year] 350 350 350 350 350 350 350 350 350 350 350

6-87

6-146 A home owner is to choose between a high-efficiency natural gas furnace and a ground-source heat pump. The system with the lower energy cost is to be determined. Assumptions The two heaters are comparable in all aspects other than the cost of energy. Analysis The unit cost of each kJ of useful energy supplied to the house by each system is Natural gas furnace: Unit cost of useful energy = Heat Pump System:

Unit cost of useful energy =

ö ($0.75/therm) æ çç 1 therm ÷ ÷= $7.33´ 10- 6 /kJ ÷ ç 0.97 è105,500 kJ ø ö ($0.115/kWh) æ çç 1 kWh ÷ = $9.13´ 10- 6 /kJ ÷ ÷ ç è ø 3.5 3600 kJ

The energy cost of high-efficiency natural gas furnace will be lower due to low cost of natural gas.

EES (Engineering Equation Solver) SOLUTION "Given" eta_gas=0.97 COP=3.5 UnitCost_electric=0.115 [$/kWh] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-88

UnitCost_gas=0.75 [$/therm] "Analysis" Cost_UsefulEnergy_gas=UnitCost_gas/eta_gas*1/Convert(therm, kJ) Cost_UsefulEnergy_electric=UnitCost_electric/COP*1/Convert(kWh, kJ)

6-147 Switching to energy efficient lighting reduces the electricity consumed for lighting as well as the cooling load in summer, but increases the heating load in winter. It is to be determined if switching to efficient lighting will increase or decrease the total energy cost of a building. Assumptions The light escaping through the windows is negligible so that the entire lighting energy becomes part of the internal heat generation. Analysis (a) Efficient lighting reduces the amount of electrical energy used for lighting year-around as well as the amount of heat generation in the house since light is eventually converted to heat. As a result, the electrical energy needed to air condition the house is also reduced. Therefore, in summer, the total cost of energy use of the household definitely decreases. (b) In winter, the heating system must make up for the reduction in the heat generation due to reduced energy used for lighting. The total cost of energy used in this case will still decrease if the cost of unit heat energy supplied by the heating system is less than the cost of unit energy provided by lighting. The cost of 1 kWh heat supplied from lighting is $0.12 since all the energy consumed by lamps is eventually converted to thermal energy. Noting that 1 therm = 105,500 kJ = 29.3 kWh and the furnace is 80% efficient, the cost of 1 kWh heat supplied by the heater is

Cost of 1 kWh heat supplied by furnace = (Amount of useful energy/hfurnace )(Price) æ 1 therm ö ÷ = [(1 kWh)/0.80]($1.40/therm) çç ÷ çè 29.3 kWh ÷ ø

= $0.060 (per kWh heat) which is less than $0.12. Thus we conclude that switching to energy efficient lighting will reduce the total energy cost of this building both in summer and in winter. Discussion To determine the amount of cost savings due to switching to energy efficient lighting, consider 10 h of operation of lighting in summer and in winter for 1 kW rated power for lighting. Current lighting: Lighting cost: (Energy used)(Unit cost) = (1 kW)(10 h)( $0.12 / kWh ) = $1.20 Increase in air conditioning cost: (Heat from lighting/COP)(unit cost) = ( 10 kWh/3.5 )( $0.12 / kWh ) = $0.34 Decrease in the heating cost = [Heat from lighting/Eff](unit cost) = ( 10 / 0.8 kWh )( $1.40 / 29.3 / kWh ) = $0.60

Energy efficient lighting: Lighting cost: (Energy used)(Unit cost) = (0.25 kW)(10 h)( $0.12 / kWh ) = $0.30 Increase in air conditioning cost: (Heat from lighting/COP)(unit cost) = ( 2.5 kWh / 3.5 )( $0.12 / kWh ) = $0.086 Decrease in the heating cost = [Heat from lighting/Eff](unit cost) = ( 2.5 / 0.8 kWh )( $1.40 / 29.3 / kWh ) = $0.15 Note that during a day with 10 h of operation, the total energy cost decreases from $1.54 to $0.39 in summer, and from $0.60 to $0.15 in winter when efficient lighting is used.

6-89

COP=3.5 Cost_electricity=0.12 [$/kWh] Cost_natgas=1.40 [$/therm] "Analysis" Cost_kWh=1/eta_furnace*Cost_natgas*1/Convert(therm, kWh)

6-148 A heat pump is used to heat a house. The maximum money saved by using the lake water instead of outside air as the heat source is to be determined. Assumptions 1. Steady operating conditions exist. 2. The kinetic and potential energy changes are zero. Analysis When outside air is used as the heat source, the cost of energy is calculated considering a reversible heat pump as follows: COPmax =

Win,min =

1 1 = = 11.92 1- TL / TH 1- (0 + 273) / (25 + 273)

QH (140, 000 / 3600) kW = = 3.262 kW COPmax 11.92

Cost air = (3.262 kW)(100 h)($0.12/kWh) = $39.15 Repeating calculations for lake water, COPmax =

Win,min =

1 1 = = 19.87 1- TL / TH 1- (10 + 273) / (25 + 273)

QH (140, 000 / 3600) kW = = 1.957 kW COPmax 19.87

Cost lake = (1.957 kW)(100 h)($0.12/kWh) = $23.49 Then the money saved becomes

Money Saved = Cost air - Cost lake = $39.15 - $23.49 = $15.7

EES (Engineering Equation Solver) SOLUTION "GIVEN" Q_dot_H=140000 [kJ/h]*Convert(kJ/h, kW) T_H=(25+273) [K] T_L_air=(0+273) [K] T_L_w=(10+273) [K] time=100 [h] UnitCost=0.12 [$/kWh] "ANALYSIS" COP_max_air=1/(1-T_L_air/T_H) "maximum COP" W_dot_in_min_air=Q_dot_H/COP_max_air "minimum power input" Cost_air=W_dot_in_min_air*time*UnitCost COP_max_w=1/(1-T_L_w/T_H) "maximum COP" W_dot_in_min_w=Q_dot_H/COP_max_w "minimum power input" Cost_water=W_dot_in_min_w*time*UnitCost Saving=Cost_air-Cost_water

6-90

6-149 The ventilating fans of a house discharge a houseful of warmed air in one hour (ACH = 1). For an average outdoor temperature of 5C during the heating season, the cost of energy “vented out” by the fans in 1 h is to be determined. Assumptions 1. Steady operating conditions exist. 2. The house is maintained at 22C and 92 kPa at all times. 3. The infiltrating air is heated to 22C before it is vented out. 4. Air is an ideal gas with constant specific heats at room temperature. 5. The volume occupied by the people, furniture, etc. is negligible. Properties The gas constant of air is R = 0.287 kPa.m3 / kg.K (Table A-1). The specific heat of air at room temperature is

c p = 1.0 kJ / kg.° C (Table A-2a).

Analysis The density of air at the indoor conditions of 92 kPa and 22C is ro =

Po 92 kPa = = 1.087 kg/m 3 3 RTo (0.287 kPa.m /kg.K)(22+273 K)

Noting that the interior volume of the house is 200 2.8 = 560 m3 , the mass flow rate of air vented out becomes

mair = r Vair = (1.087 kg/m3 )(560 m3 /h) = 608.7 kg/h = 0.169 kg/s Noting that the indoor air vented out at 22 a rate of Qloss,fan = mair c p (Tindoors - Toutdoors )

= (0.169 kg/s)(1.0 kJ/kg.°C)(22 - 5)°C = 2.874 kJ/s =2.874 kW Then the amount and cost of the heat “vented out” per hour becomes

Fuel energy loss = Qloss,fan D t / hfurnace = (2.874 kW)(1 h)/0.96 = 2.994 kWh Money loss = (Fuel energy loss)(Unit cost of energy) æ 1 therm ÷ ö = (2.994 kWh)($1.20/therm) çç = $0.123 ÷ çè 29.3 kWh ÷ ø

Discussion Note that the energy and money loss associated with ventilating fans can be very significant. Therefore, ventilating fans should be used sparingly.

EES (Engineering Equation Solver) SOLUTION "Given" A=200 [m^2] H=2.8 [m] eta_heater=0.96 T_i=22 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-91

P=92 [kPa] UnitCost=1.20 [$/therm] time=1 [h] T_o=5 [C] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.0 [kJ/kg-C] "Table A-2a" "Analysis" V=A*H rho=P/(R*(T_i+273)) m_dot=rho*V/time*1/Convert(kg/s, kg/h) Q_dot_loss_fan=m_dot*c_p*(T_i-T_o) EnergyLoss=Q_dot_loss_fan*time/eta_heater MoneyLoss=EnergyLoss*UnitCost*1/Convert(therm, kWh)

6-150 The ventilating fans of a house discharge a houseful of air-conditioned air in one hour (ACH = 1). For an average outdoor temperature of 28C during the cooling season, the cost of energy “vented out” by the fans in 1 h is to be determined. Assumptions 1. Steady operating conditions exist. 2. The house is maintained at 22C and 92 kPa at all times. 3. The infiltrating air is cooled to 22C before it is vented out. 4. Air is an ideal gas with constant specific heats at room temperature. 5. The volume occupied by the people, furniture, etc. is negligible. 6. Latent heat load is negligible. Properties The gas constant of air is R = 0.287 kPa.m3 / kg.K (Table A-1). The specific heat of air at room temperature is

c p = 1.0 kJ / kg.° C (Table A-2a). Analysis The density of air at the indoor conditions of 92 kPa and 22C is ro =

Po 92 kPa = = 1.087 kg/m 3 RTo (0.287 kPa.m 3 /kg.K)(22+273 K)

Noting that the interior volume of the house is 200´ 2.8 = 560 m3 , the mass flow rate of air vented out becomes

mair = r Vair = (1.087 kg/m3 )(560 m3 /h) = 608.7 kg/h = 0.1690 kg/s Noting that the indoor air vented out at 22C is replaced by infiltrating outdoor air at 33C, this corresponds to energy loss at a rate of Qloss,fan = mair c p (Toutdoors - Tindoors )

= (0.1690 kg/s)(1.0 kJ/kg.°C)(33 - 22)°C = 1.859 kJ/s =1.859 kW Then the amount and cost of the electric energy “vented out” per hour becomes © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-92

Electric energy loss = Qloss,fan D t / COP = (1.859 kW)(1 h)/2.1 = 0.8854 kWh Money loss = (Fuel energy loss)(Unit cost of energy) = (0.8854 kWh)($0.12 / kWh) = $0.106 Discussion Note that the energy and money loss associated with ventilating fans can be very significant. Therefore, ventilating fans should be used sparingly.

EES (Engineering Equation Solver) SOLUTION "Given" A=200 [m^2] H=2.8 [m] eta_heater=0.96 T_i=22 [C] P=92 [kPa] UnitCost=0.12 [$/kWh] time=1 [h] T_o=33 [C] COP=2.1 "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.0 [kJ/kg-C] "Table A-2a" "Analysis" V=A*H rho=P/(R*(T_i+273)) m_dot=rho*V/time*1/Convert(kg/s, kg/h) Q_dot_loss_fan=m_dot*c_p*(T_o-T_i) EnergyLoss=Q_dot_loss_fan*time/COP MoneyLoss=EnergyLoss*UnitCost

6-151 A geothermal heat pump with R-134a as the working fluid is considered. The evaporator inlet and exit states are specified. The mass flow rate of the refrigerant, the heating load, the COP, and the minimum power input to the compressor are to be determined. Assumptions 1. The heat pump operates steadily. 2. The kinetic and potential energy changes are zero. 3. Steam properties are used for geothermal water. Properties The properties of R-134a and water are (Steam and R-134a tables)

6-93

T1 = 12°Cü ïï h1 = 96.54 kJ/kg ý x1 = 0.15 ïïþ P1 = 443.3 kPa P2 = P1 = 443.3 kPa ü ïï ý h2 = 257.33 kJ/kg ïïþ x2 = 1 Tw,1 = 60°Cïü ïý h = 251.18 kJ/kg ïï w,1 xw,1 = 0 þ Tw,2 = 40°Cïü ïý h = 167.53 kJ/kg ïï w,2 xw,2 = 0 þ

Analysis (a) The rate of heat transferred from the water is the energy change of the water from inlet to exit

QL = mw (hw,1 - hw,2 ) = (0.065 kg/s)(251.18 - 167.53) kJ/kg = 5.437 kW The energy increase of the refrigerant is equal to the energy decrease of the water in the evaporator. That is,

QL = mR (h2 - h1 ) ¾ ¾ ® mR =

QL 5.437 kW = = 0.0338 kg / s h2 - h1 (257.33 - 96.54) kJ/kg

(b) The heating load is

QH = QL + Win = 5.437 + 1.6 = 7.04 kW (c) The COP of the heat pump is determined from its definition,

COP =

QH 7.04 kW = = 4.40 1.6 kW Win

(d) The COP of a reversible heat pump operating between the same temperature limits is COPmax =

1 1 = = 9.51 1- TL / TH 1- (25 + 273) / (60 + 273)

Then, the minimum power input to the compressor for the same refrigeration load would be

Win,min =

QH 7.04 kW = = 0.740 kW COPmax 9.51

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=(25+273) [K] T_H=(60+273) [K] T_w1=60 [C] m_dot_w=0.065 [kg/s] T_w2=40 [C] T_1=12 [C] x_1=0.15 x_2=1 P_2=P_1 W_dot_in=1.6 [kW] "PROPERTIES" Fluid1$='R134a' Fluid2$='Steam_iapws' P_1=pressure(Fluid1$, T=T_1, x=x_1) h_1=enthalpy(Fluid1$, T=T_1, x=x_1) h_2=enthalpy(Fluid1$, P=P_2, x=x_2) h_w1=enthalpy(Fluid2$, T=T_w1, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-94

h_w2=enthalpy(Fluid2$, T=T_w2, x=0) "ANALYSIS" Q_dot_L=m_dot_w*(h_w1-h_w2) m_dot_R=Q_dot_L/(h_2-h_1) "mass flow rate of refrigerant" Q_dot_H=Q_dot_L+W_dot_in "heating load" COP=Q_dot_H/W_dot_in "actual COP" COP_max=1/(1-T_L/T_H) "maximum COP" W_dot_in_min=Q_dot_H/COP_max "minimum power input"

6-152 A heat pump is used as the heat source for a water heater. The rate of heat supplied to the water and the minimum power supplied to the heat pump are to be determined. Assumptions 1. Steady operating conditions exist. 2. The kinetic and potential energy changes are zero. Properties The specific heat and specific volume of water at room temperature are c p = 4.18 kJ / kg.K and v = 0.001m3 / kg (Table A-3).

Analysis (a) An energy balance on the water heater gives the rate of heat supplied to the water

QH = mc p (T2 - T1 ) = =

V c p (T2 - T1 ) v

(0.02 / 60) m3 /s (4.18 kJ/kg ×°C)(50 - 10)°C = 55.73 kW 0.001 m3 /kg

(b) The COP of a reversible heat pump operating between the specified temperature limits is COPmax =

1 1 = = 10.1 1- TL / TH 1- (0 + 273) / (30 + 273)

Then, the minimum power input would be

Win,min =

QH 55.73 kW = = 5.52 kW COPmax 10.1

EES (Engineering Equation Solver) SOLUTION "GIVEN" T1=10 [C] T2=50 [C] V_dot=0.02[m^3/min]*Convert(m^3/min, m^3/s) T_L=(0+273) [K] T_H=(30+273) [K] "Properties" c_p=4.18 [kJ/kg-C] "Table A-3" rho=1000 [kg/m^3] "Table A-3" "ANALYSIS" Q_dot_H=rho*V_dot*c_p*(T2-T1) COP_max=1/(1-T_L/T_H) "maximum COP" W_dot_in_min=Q_dot_H/COP_max "minimum power input"

6-95

6-153 The maximum work that can be extracted from a pond containing 105 kg of water at 350 K when the temperature of the surroundings is 300 K is to be determined. Temperature intervals of (a) 5 K, (b) 2 K, and (c) 1 K until the pond temperature drops to 300 K are to be used. Analysis The problem is solved using EES, and the solution is given below. T_L = 300 [K] m_pond = 1E+5 [kg] c_pond = 4.18 [kJ/kg-K] T_H_high = 350 [K] T_H_low = 300 [K] deltaT_H = 1 [K] "deltaT_H is the stepsize for the EES integral function." eta_th_C = 1 - T_L/T_H deltaQ_pond = m_pond*C_pond*deltaT_H deltaQ_H = -deltaQ_pond IntegrandW_out = eta_th_C*m_pond*c_pond "Exact Solution by integration from T_H = 350 K to 300 K:" W_out_exact = -m_pond*c_pond*(T_H_low - T_H_high -T_L*ln(T_H_low/T_H_high)) "EES integral function where the stepsize is an input to the solution." W_EES_1 = integral(IntegrandW_out,T_H, T_H_low, T_H_high,deltaT_H) W_EES_2 = integral(IntegrandW_out,T_H, T_H_low, T_H_high,2*deltaT_H) W_EES_5 = integral(integrandW_out,T_H, T_H_low, T_H_high,5*deltaT_H) SOLUTION c_pond= 4.18 [ kJ / kg-K] deltaQ_H= − 418000 [kJ] deltaQ_pond=418000 [kJ] deltaT_H=1 [K] eta_th_C=0.1429 IntegrandW_out=59714 [kJ] m_pond=100000 [kg] T_H=350 [K] T_H_high=350 [K] T_H_low=300 [K] T_L=300 [K] W_EES_1=1.569E+06 [kJ] W_EES_2=1.569E+06 [kJ] W_EES_5=1.569E+06 [kJ] W_out_exact=1.570E+06 [kJ] Continued in the next page:

This problem can also be solved exactly by integration as follows: The maximum work will be obtained if a Carnot heat engine is used. The sink temperature of this heat engine will remain constant at 300 K but the source temperature will be decreasing from 350 K to 300 K. Then the thermal efficiency of the Carnot heat engine operating between pond and the ambient air can be expressed as

hth, C = 1-

TL 300 = 1TH TH

where TH is a variable. The conservation of energy relation for the pond can be written in the differential form as ¶ Qpond = mcdTH

6-96

¶ QH = - ¶ Qpond = - mcdTH = - (105 kg)(4.18 kJ/kg ×K )dTH Also, æ 300 ö ÷ 105 kg (4.18 kJ/kg ×K )¶ T ÷ ¶Wnet = hth, C ¶ QH = - ççç1( ) H ÷ ÷ çè TH ø

The total work output is obtained by integration,

Wnet = ò

350 300

hth,C ¶ QH = ò

= 4.18´ 105 ò

æ 300 ö ÷ çç1÷(105 kg )(4.18 kJ/kg ×K )dTH 300 ç ÷ çè T ÷ ø 350

æ 300 ö ÷ çç1÷dTH = 15.7´ 105 kJ 300 ç ÷ çè T ÷ ø 350

which is the exact result. The values obtained by computer solution will approach this value as the temperature interval is decreased.

6-154 A Carnot heat engine is operating between specified temperature limits. The source temperature that will double the efficiency is to be determined. Analysis Denoting the new source temperature by TH * , the thermal efficiency of the Carnot heat engine for both cases can be expressed as

hth, C = 1-

TL TL * and hth, = 2hth, C C = 1TH TH*

Substituting, 1-

æ T ö TL ÷ = 2 ççç1- L ÷ * ÷ çè TH ÷ TH ø

Solving for TH* ,

TH* =

TH TL TH - 2TL

6-97

6-155E The thermal efficiency of a completely reversible heat engine as a function of the source temperature is to be calculated and plotted. Assumptions The heat engine operates steadily. Analysis With the specified sink temperature, the thermal efficiency of this completely reversible heat engine is hth,rev = 1-

TL 500 R = 1TH TH

Using EES, we tabulate and plot the variation of thermal efficiency with the source temperature: T_L=500 [R] eta_th_rev=1-T_L/T_H

TH [R] 500 650 800 950 1100 1250 1400 1550 1700 1850 2000

ηth,rev

0 0.2308 0.375 0.4737 0.5455 0.6 0.6429 0.6774 0.7059 0.7297 0.75

6-156 It is to be shown that COPHP = COPR + 1 for the same temperature and heat transfer terms. Analysis Using the definitions of COPs, the desired relation is obtained to be

COPHP =

Q + Wnet, in QH QL = L = + 1 = COPR + 1 Wnet, in Wnet, in Wnet, in

6-157 It is to be proven that a refrigerator's COP cannot exceed that of a completely reversible refrigerator that shares the same thermal-energy reservoirs. Assumptions The refrigerator operates steadily. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-98

Analysis We begin by assuming that the COP of the general refrigerator B is greater than that of the completely reversible refrigerator A, COPB > COPA . When this is the case, a rearrangement of the coefficient of performance expression yields

WB =

QL QL < = WA COPB COPA

That is, the magnitude of the work required to drive refrigerator B is less than that needed to drive completely reversible refrigerator A. Applying the first law to both refrigerators yields

QH , B < QH , A since the work supplied to refrigerator B is less than that supplied to refrigerator A, and both have the same cooling effect, QL .

Since A is a completely reversible refrigerator, we can reverse it without changing the magnitude of the heat and work transfers. This is illustrated in the figure below. The heat, QL , which is rejected by the reversed refrigerator A can now be routed directly to refrigerator B. The net effect when this is done is that no heat is exchanged with the TL reservoir. The magnitude of the heat supplied to the reversed refrigerator A, QH , A has been shown to be larger than that rejected by refrigerator B. There is then a net heat transfer from the TH reservoir to the combined device in the dashed lines of the figure whose magnitude is given by QH , A - QH , B . Similarly, there is a net work production by the combined device whose magnitude is given by WA - WB . The combined cyclic device then exchanges heat with a reservoir at a single temperature and produces work which is clearly a violation of the Kelvin-Planck statement of the second law. Our assumption the COPB > COPA must then be wrong.

Fundamentals of Engineering (FE) Exam Problems 6-158 Consider a Carnot refrigerator and a Carnot heat pump operating between the same two thermal energy reservoirs. If the COP of the refrigerator is 3.4, the COP of the heat pump is (a) 1.7 (b) 2.4 (c) 3.4 (d) 4.4 (e) 5.0 Answer (d) 4.4 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-99

COP_R=3.4 COP_HP=COP_R+1 "Some Wrong Solutions with Common Mistakes:" W1_COP=COP_R-1 "Subtracting 1 instead of adding 1" W2_COP=COP_R "Setting COPs equal to each other"

6-159 A 2.4−m high 200 - m 2 -conditioning system whose COP is 3.2. It is estimated that the kitchen, bath, and other ventilating fans of the house discharge a houseful of conditioned air once every hour. If the average outdoor temperature is 32C, the density of air is 1.20 kg / m3 , and the unit cost of electricity is $0.10 / kWh , the amount of money “vented out” by the fans in 10 hours is (a) $0.50 (b) $1.60 (c) $5.00 (d) $11.00 (e) $16.00 Answer (a) $0.50 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). COP=3.2 T1=22 [C] T2=32 [C] Price=0.10 [$/kWh] V=2.4*200 [m^3] cp=1.005 [kJ/kg-C] rho=1.20 [kg/m^3] m=rho*V m_total=m*10 Ein=m_total*cp*(T2-T1)/COP Cost=(Ein/3600)*Price "Some Wrong Solutions with Common Mistakes:" W1_Cost=(Price/3600)*m_total*cp*(T2-T1)*COP "Multiplying by Eff instead of dividing" W2_Cost=(Price/3600)*m_total*cp*(T2-T1) "Ignoring efficiency" W3_Cost=(Price/3600)*m*cp*(T2-T1)/COP "Using m instead of m_total" W4_Cost=(Price/3600)*m_total*cp*(T2+T1)/COP "Adding temperatures"

6-160 A window air conditioner that consumes 1 kW of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room, and is plugged in. The rate of cooling or heating this air conditioner will provide to the air in the room when running is (a) 3 kJ / s , cooling (b) 1 kJ / s , cooling (c) 0.33 kJ / s , heating (d) 1 kJ / s , heating (e) 3 kJ / s , heating Answer (d) 1 kJ / s , heating © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-100

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W_in=1 [kW] COP=3 "From energy balance, heat supplied to the room is equal to electricity consumed," E_supplied=W_in "heating" "Some Wrong Solutions with Common Mistakes:" W1_E=-W_in "kJ/s, cooling" W2_E=-COP*W_in "kJ/s, cooling" W3_E=W_in/COP "kJ/s, heating" W4_E=COP*W_in "kJ/s, heating"

6-161 The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23C to 6C at an average rate of 18 kg / h . If the COP of this refrigerator is 3.1, the required power input to this refrigerator is (a) 1100 W (b) 355 W (c) 195 W (d) 115 W (e) 35 W Answer (d) 115 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). COP=3.1 T1=23 [C] T2=6 [C] m_dot=(18/3600) [kg/s] cp=4.18 [kJ/kg-C] Q_L=m_dot*cp*(T1-T2) W_in=Q_L*1000/COP "Some Wrong Solutions with Common Mistakes:" W1_Win=m_dot*cp*(T1-T2) *1000*COP "Multiplying by COP instead of dividing" W2_Win=m_dot*cp*(T1-T2) *1000 "Not using COP" W3_Win=m_dot*(T1-T2) *1000/COP "Not using specific heat" W4_Win=m_dot*cp*(T1+T2) *1000/COP "Adding temperatures"

6-162 The label on a washing machine indicates that the washer will use $85 worth of hot water if the water is heated by a 90% efficiency electric heater at an electricity rate of $0.125 / kWh . If the water is heated from 18C to 45C, the amount of hot water an average family uses per year, in metric tons, is (a) 19.5 tons (b) 21.7 tons (c) 24.1 tons (d) 27.2 tons (e) 30.4 tons Answer (a) 19.5 tons Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-101

Eff=0.90 T1=18 [C] T2=45 [C] Cost=85 [$] Price=0.125 [$/kWh] c=4.18 [kJ/kg-C] Ein=(Cost/Price)*3600 Ein=m*c*(T2-T1)/Eff "Some Wrong Solutions with Common Mistakes:" Ein=W1_m*c*(T2-T1)*Eff "Multiplying by Eff instead of dividing" Ein=W2_m*c*(T2-T1) "Ignoring efficiency" Ein=W3_m*(T2-T1)/Eff "Not using specific heat" Ein=W4_m*c*(T2+T1)/Eff "Adding temperatures"

6-163 A heat pump is absorbing heat from the cold outdoors at 5C and supplying heat to a house at 25C at a rate of 18,000 kg / h . If the power consumed by the heat pump is 1.9 kW, the coefficient of performance of the heat pump is (a) 1.3 (b) 2.6 (c) 3.0 (d) 3.8 (e) 13.9 Answer (b) 2.6 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=5 [C] TH=25 [C] QH=(18000/3600) [kW] Win=1.9 [kW] COP=QH/Win "Some Wrong Solutions with Common Mistakes:" W1_COP=Win/QH "Doing it backwards" W2_COP=TH/(TH-TL) "Using temperatures in C" W3_COP=(TH+273)/(TH-TL) "Using temperatures in K" W4_COP=(TL+273)/(TH-TL) "Finding COP of refrigerator using temperatures in K"

6-164 A heat engine cycle is executed with steam in the saturation dome. The pressure of steam is 1 MPa during heat addition, and 0.4 MPa during heat rejection. The highest possible efficiency of this heat engine is (a) 8.0% (b) 15.6% (c) 20.2% (d) 79.8% (e) 100% Answer (a) 8.0% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P_H=1000 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-102

P_L=400 [kPa] T_H=TEMPERATURE(Steam_IAPWS,x=0,P=P_H) T_L=TEMPERATURE(Steam_IAPWS,x=0,P=P_L) Eta_Carnot=1-(T_L+273)/(T_H+273) "Some Wrong Solutions with Common Mistakes:" W1_Eta_Carnot=1-P_L/P_H "Using pressures" W2_Eta_Carnot=1-T_L/T_H "Using temperatures in C" W3_Eta_Carnot=T_L/T_H "Using temperatures ratio"

6-165 A heat pump cycle is executed with R-134a under the saturation dome between the pressure limits of 1.2 MPa and 0.16 MPa. The maximum coefficient of performance of this heat pump is (a) 5.7 (b) 5.2 (c) 4.8 (d) 4.5 (e) 4.1 Answer (b) 5.2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P_H=1200 [kPa] P_L=160 [kPa] T_H=TEMPERATURE(R134a,x=0,P=P_H) T_L=TEMPERATURE(R134a,x=0,P=P_L) COP_HP=(T_H+273)/(T_H-T_L) "Some Wrong Solutions with Common Mistakes:" W1_COP=P_H/(P_H-P_L) "Using pressures" W2_COP=T_H/(T_H-T_L) "Using temperatures in C" W3_COP=T_L/(T_H-T_L) "Refrigeration COP using temperatures in C" W4_COP=(T_L+273)/(T_H-T_L)"Refrigeration COP using temperatures in K"

6-166 A refrigeration cycle is executed with R-134a under the saturation dome between the pressure limits of 1.6 MPa and 0.2 MPa. If the power consumption of the refrigerator is 3 kW, the maximum rate of heat removal from the cooled space of this refrigerator is (a) 0.45 kJ / s (b) 0.78 kJ / s (c) 3.0 kJ / s (d) 11.6 kJ / s (e) 14.6 kJ / s Answer (d) 11.6 kJ / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P_H=1600 [kPa] P_L=200 [kPa] W_in=3 [kW] T_H=TEMPERATURE(R134a,x=0,P=P_H) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-103

T_L=TEMPERATURE(R134a,x=0,P=P_L) COP=(T_L+273)/(T_H-T_L) Q_L=W_in*COP "Some Wrong Solutions with Common Mistakes:" W1_QL=W_in*T_L/(T_H-T_L) "Using temperatures in C" W2_QL=W_in "Setting heat removal equal to power input" W3_QL=W_in/COP "Dividing by COP instead of multiplying" W4_QL=W_in*(T_H+273)/(T_H-T_L) "Using COP definition for Heat pump"

6-167 A heat pump with a COP of 3.2 is used to heat a perfectly sealed house (no air leaks). The entire mass within the house (air, furniture, etc.) is equivalent to 1200 kg of air. When running, the heat pump consumes electric power at a rate of 5 kW. The temperature of the house was 7C when the heat pump was turned on. If heat transfer through the envelope of the house (walls, roof, etc.) is negligible, the length of time the heat pump must run to raise the temperature of the entire contents of the house to 22C is (a) 13.5 min (b) 43.1 min (c) 138 min (d) 18.8 min (e) 808 min Answer (a) 13.5 min Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). COP=3.2 m=1200 [kg] T1=7 [C] T2=22 [C] Win=5 [kW] cv=0.718 [kJ/kg-C] QH=m*cv*(T2-T1) Win*time=QH/COP/60 "Some Wrong Solutions with Common Mistakes:" Win*W1_time*60=m*cv*(T2-T1) *COP "Multiplying by COP instead of dividing" Win*W2_time*60=m*cv*(T2-T1) "Ignoring COP" Win*W3_time=m*cv*(T2-T1) /COP "Finding time in seconds instead of minutes" Win*W4_time*60=m*cp*(T2-T1) /COP "Using cp instead of cv" cp=1.005 [kJ/kg-K]

6-168 A heat engine cycle is executed with steam in the saturation dome between the pressure limits of 7 MPa and 2 MPa. If heat is supplied to the heat engine at a rate of 150 kJ / s , the maximum power output of this heat engine is (a) 8.1 kW (b) 19.7 kW (c) 38.6 kW (d) 107 kW (e) 130 kW Answer (b) 19.7 kW

6-104

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). PH=7000 [kPa] PL=2000 [kPa] Q_in=150 [kW] TH=TEMPERATURE(Steam_IAPWS,x=0,P=PH) TL=TEMPERATURE(Steam_IAPWS,x=0,P=PL) Eta=1-(TL+273)/(TH+273) W_out=Eta*Q_in "Some Wrong Solutions with Common Mistakes:" W1_W_out=(1-TL/TH)*Q_in "Using temperatures in C" W2_W_out=(1-PL/PH)*Q_in "Using pressures" W3_W_out=Q_in/Eta "Dividing by efficiency instead of multiplying" W4_W_out=(TL+273)/(TH+273)*Q_in "Using temperature ratio"

6-169 A heat engine receives heat from a source at 1000C and rejects the waste heat to a sink at 50C. If heat is supplied to this engine at a rate of 100 kJ / s , the maximum power this heat engine can produce is (a) 25.4 kW (b) 55.4 kW (c) 74.6 kW (d) 95.0 kW (e) 100.0 kW Answer (c) 74.6 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TH=1000 [C] TL=50 [C] Q_in=100 [kW] Eta=1-(TL+273)/(TH+273) W_out=Eta*Q_in "Some Wrong Solutions with Common Mistakes:" W1_W_out=(1-TL/TH)*Q_in "Using temperatures in C" W2_W_out=Q_in "Setting work equal to heat input" W3_W_out=Q_in/Eta "Dividing by efficiency instead of multiplying" W4_W_out=(TL+273)/(TH+273)*Q_in "Using temperature ratio"

6-170 An air-conditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of 32 kJ / s to maintain its temperature constant at 20C. If the temperature of the outdoors is 35C, the power required to operate this air-conditioning system is (a) 0.58 kW (b) 3.20 kW (c) 1.56 kW (d) 2.26 kW (e) 1.64 kW Answer (e) 1.64 kW

6-105

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=20 [C] TH=35 [C] QL=32 [kJ/s] COP=(TL+273)/(TH-TL) COP=QL/Win "Some Wrong Solutions with Common Mistakes:" QL=W1_Win*TL/(TH-TL) "Using temperatures in C" QL=W2_Win "Setting work equal to heat input" QL=W3_Win/COP "Dividing by COP instead of multiplying" QL=W4_Win*(TH+273)/(TH-TL) "Using COP of HP"

6-171 A refrigerator is removing heat from a cold medium at 3C at a rate of 5400 kJ / h and rejecting the waste heat to a medium at 30C. If the coefficient of performance of the refrigerator is 2, the power consumed by the refrigerator is (a) 0.5 kW (b) 0.75 kW (c) 1.0 kW (d) 1.5 kW (e) 3.0 kW Answer (b) 0.75 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=3 [C] TH=30 [C] QL=(5400/3600) [kJ/s] COP=2 QL=Win*COP "Some Wrong Solutions with Common Mistakes:" QL=W1_Win*(TL+273)/(TH-TL) "Using Carnot COP" QL=W2_Win "Setting work equal to heat input" QL=W3_Win/COP "Dividing by COP instead of multiplying" QL=W4_Win*TL/(TH-TL) "Using Carnot COP using C"

6-172 Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one. If the source temperature of the first engine is 130 0 K and the sink temperature of the second engine is 300 K and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is (a) 625 K (b) 800 K (c) 860 K (d) 453 K (e) 758 K Answer (a) 625 K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-106

TH=1300 [K] TL=300 [K] eta_1=1-Tmid/TH eta_2=1-TL/Tmid eta_1=eta_2 "Some Wrong Solutions with Common Mistakes:" W1_Tmid=(TL+TH)/2 "Using average temperature" W2_Tmid=SQRT(TL*TH) "Using average temperature"

6-173 A typical new household refrigerator consumes about 680 kWh of electricity per year, and has a coefficient of performance of 1.4. The amount of heat removed by this refrigerator from the refrigerated space per year is (a) 952 MJ / yr (b) 1749 MJ / yr (c) 2448 MJ / yr (d) 3427 MJ / yr (e) 4048 MJ / yr Answer (d) 3427 MJ / yr Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). W_in=(680*3.6) [MJ/yr] COP_R=1.4 QL=W_in*COP_R "Some Wrong Solutions with Common Mistakes:" W1_QL=W_in*COP_R/3.6 "Not using the conversion factor" W2_QL=W_in "Ignoring COP" W3_QL=W_in/COP_R "Dividing by COP instead of multiplying"

6-174 … 6-181 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-107

Chapter 7 ENTROPY

6-108

CYU - Check Your Understanding CYU 7-1 ■ Check Your Understanding CYU 7-1.1 Select the incorrect statement regarding entropy and Clausius inequality: (a) The integral of  Q / T gives the value of entropy change only if the integration is carried out along an internally reversible path between the two states. (b) The Clausius inequality is valid for all thermodynamic cycles, reversible or irreversible. (c) The equality in the Clausius inequality holds for internally reversible cycles and the inequality for the irreversible ones. (d) The entropy change between two specified states depends on the path followed during a process. (e) Entropy is a property. Answer: (d) The entropy change between two specified states depends on the path followed during a process. CYU 7-1.2 Heat at a rate of 70 kW is transferred to a thermal energy reservoir at 25C. What is the entropy change of the reservoir during this process? (a) 70 / 25 kW / K (b) 70 / 25 kW / K (c) 70 / 298 kW / K (d) 70 / 298 kW / K (e) 0 Answer: (c) 70 / 298 kW / K

dS  Q / T  70 kW / 298 K Heat transfer is to the reservoir.

CYU 7-2 ■ Check Your Understanding CYU 7-2.1 Select the incorrect statement below: (a) The entropy change of a closed system during a reversible process is equal to the integral of  Q / T for that process. (b) The entropy generation is always a positive quantity or zero. (c) In the absence of any entropy transfer, the entropy change of a system is equal to the entropy generation. (d) The entropy of an isolated system during an irreversible process always increases. (e) The entropy of a closed system can only be changed by heat transfer. Answer: (e) The entropy of a closed system can only be changed by heat transfer. CYU 7-2.2 Select the incorrect statement below: (a) No entropy is generated during reversible processes. (b) Entropy is a conserved during a process, just like energy. (c) The more irreversible a process is, the larger the entropy generated during that process. (d) The entropy change of a system can be negative during a process. (e) The entropy generation cannot be negative during a process. Answer: (b) Entropy is a conserved during a process, just like energy.

6-109

CYU 7-2.3 Heat is transferred to a closed system from a thermal energy reservoir. If the entropy change of the reservoir is 0.8 kJ / K and that of the system is 0.5 kJ / K , this process is (a) Impossible (b) Irreversible (c) Reversible (d) Insufficient information Answer: (a) Impossible Sgen  dStotal  dSreservoir  dSsystem  (0.8 kJ / K)  (0.5 kJ / K)  0.3 kJ / K CYU 7-2.4 1500 kJ of heat is transferred K to a closed system from a thermal energy reservoir at 500. If the entropy generation during this process is 2 kJ / K , what is the entropy change of the system? (a) 3 kJ / K (b) 3 kJ / K (c) 2 kJ / K (d) 2 kJ / K (e) 5 kJ / K Answer: (e) 5 kJ / K dSreservoir  Q / T  1500 kJ / 500 K  3 kJ / K Sgen  dStotal  dSreservoir  dSsystem  (3 kJ / K)  dSsystem  2 kJ / K dSsystem  5 kJ / K

CYU 7-3 ■ Check Your Understanding CYU 7-3.1 The entropy of a compressed liquid at a temperature T and pressure P can be determined approximately as (a) s @ T , P  sf @ P (b) s @ T , P  sg @ P (c) s @ T , P  sf @ T (d) s @ T , P  sg @ T (e) s @ T , P  sfg @ P Answer: (c) s @ T , P  sf @ T CYU 7-3.2 The entropy change of a pure substance can be determined from (a) s  s2  s1  sgen (b) s  sgen (c) s  sgen   s2  s1  (d) s  q / T  s2  s1 (e) s  s2  s1 Answer: (e) s  s2  s1

6-110

CYU 7-4.1 Which process is called an isentropic process? (a) Reversible and isothermal (b) Adiabatic and reversible (c) Isenthalpic and adiabatic (d) Irreversible and isothermal (e) Isothermal and adiabatic Answer: (b) Adiabatic and reversible CYU 7-4.2 For which processes can an isentropic process serve as an appropriate model? I. Steam expanding in an insulated turbine II. Acceleration of air in an adiabatic nozzle III. Compression of air in a compressor which is intentionally cooled (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (b) I and II CYU 7-4.3 Steam at 4 MPa and 700 K expands in an adiabatic turbine to a pressure of 20 kPa and producing 90 kW of power. What is the entropy change of steam across the turbine if this process is internally reversible? (a) 0 (b) 90 / 700 kW / K (c) 700 / 90 kW / K (d) 90 / 700 kW / K (e) Insufficient information Answer: (a) 0

CYU 7-5 ■ Check Your Understanding CYU 7-5.1 For a process curve on a T-s diagram, which of statements below are correct? I. The area under the process curve for an isentropic process is zero. II. The area under the process curve on a T-S diagram represents heat transfer during an internally reversible process. III. An isentropic process on a T-s diagram appears as a vertical line. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (e) I, II, and III

CYU 7-6 ■ Check Your Understanding CYU 7-6.1 Select the correct statement regarding entropy? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-111

(a) During a heat transfer process between a hot body at 50ºC and a cold body at 10ºC, the increase in the entropy of the cold body is equal to the decrease in the entropy of the hot body. (b) In the absence of friction, the pendulum of a grandfather clock does not create any entropy. (c) The mixing of a liquid in by ana electric mixer does not increase the entropy of the gas. (d) No entropy is generated during the complete combustion of a fuel in an adiabatic test chamber. Answer: (b) In the absence of friction, the pendulum of a grandfather clock does not create any entropy. CYU 7-6.2 An electric resistance heater in a well-insulated room is turned on and kept on until 5 kWh of electricity is consumed. If the entropy change of the room is 0.5 kWh / K , what is the entropy generated during this process? (a) 0 (b) 0.5 kWh / K (c) 0.5 kWh / K (d) 4.5 kWh / K (e) 5 kWh / K Answer: (c) 0.5 kWh / K No entropy is transferred by work. Entropy of the system increases. dSsys  Sgen  0.5 kWh / K

CYU 7-9 ■ Check Your Understanding CYU 7-9.1 Select the correct statement regarding entropy of ideal gases: (a) The entropy of an ideal gas varies with specific volume or pressure as well as the temperature. (b) The s  value is functions of temperature and pressure. (c) The s  value is functions of temperature and specific volume. (d) For monatomic ideal gases such as helium, entropy is a function of temperature only. Answer: (a) The entropy of an ideal gas varies with specific volume or pressure as well as the temperature.

P P1

CYU 7-9.2 The relation s2 = s1 + R ln 2 is applicable to I. Ideal gases with variable specific heats II. Isentropic processes of ideal gases III. Ideal gases with constant specific heats IV. Ideal gases undergoing a constant specific volume process (a) Only I (b) I and II (c) I, III and IV (d) II and III (e) I and IV Answer: (b) I and II CYU 7-9.3 An ideal gas undergoes a process during which its state changes from T1 , v1 to T2 , v 2 . Considering variation of specific heats with temperature, the most appropriate relation to calculate the entropy change of the gas is

6-112

v v1

+ + (a) s2 - s1 = s2 - s1 + R ln 2

T T1

(b) s2 - s1 = cp ln 2

P P1

o o (c) s2 - s1 = s2 - s1 - R ln 2

T T1

P P1

T T1

v v1

(d) s2 - s1 = cp ln 2 - R ln 2 (e) s2 - s1 = cv ln 2 + R ln 2

v v1

Answer: (a) s2 - s1 = s2 - s1 + R ln 2 +

CYU 7-9.4 An ideal gas undergoes an isothermal process during which its state changes from P1 , v1 to P2 , v 2 . The entropy change of the gas can be determined from

P P1

(a) D s= - R ln 2

v v1

(b) D s= cp ln 2

v v1

P P1

(d) D s= cp ln 2

v v1

(e) D s= cv ln 2

P P1

Answer: (a) D s= - R ln 2

6-113

PROBLEMS Entropy and the Increase of Entropy Principle 7-1C No. The ò dQ represents the net heat transfer during a cycle, which could be positive.

7-2C No. A system may produce more (or less) work than it receives during a cycle. A steam power plant, for example, produces more work than it receives during a cycle, the difference being the net work output.

7-3C The entropy change will be the same for both cases since entropy is a property and it has a fixed value at a fixed state.

7-4C No. In general, that integral will have a different value for different processes. However, it will have the same value for all reversible processes.

7-5C No. A system may reject more (or less) heat than it receives during a cycle. The steam in a steam power plant, for example, receives more heat than it rejects during a cycle. 7-6C Yes.

7-7C No. An isothermal process can be irreversible. Example: A system that involves paddle-wheel work while losing an equivalent amount of heat.

7-8C Yes.

7-9C The value of this integral is always larger for reversible processes.

7-10C It is possible to create entropy, but it is not possible to destroy it.

7-11C No. Because the entropy of the surrounding air increases even more during that process, making the total entropy change positive.

7-12C If the system undergoes a reversible process, the entropy of the system cannot change without a heat transfer. Otherwise, the entropy must increase since there are no offsetting entropy changes associated with reservoirs exchanging heat with the system.

6-114

7-13C The claim that work will not change the entropy of a fluid passing through an adiabatic steady-flow system with a single inlet and outlet is true only if the process is also reversible. Since no real process is reversible, there will be an entropy increase in the fluid during the adiabatic process in devices such as pumps, compressors, and turbines.

7-14C Yes. This will happen when the system is losing heat, and the decrease in entropy as a result of this heat loss is equal to the increase in entropy as a result of irreversibilities.

7-15C Sometimes.

7-16C Never.

7-17C Always.

7-18C Increase.

7-19C Sometimes.

7-20C Greater than.

7-21 Heat is transferred from a hot reservoir to a cold reservoir. The entropy change of the two reservoirs is to be calculated and it is to be determined if the second law is satisfied. Assumptions The reservoirs operate steadily.

Analysis The rate of entropy change of everything involved in this transfer of heat is given by

DS =

QH QL - 2 kW 2 kW + = + = 0.00417 kW / K TH TL 800 K 300 K

Since this rate is positive (i.e., the entropy increases as time passes), this transfer of heat is possible.

EES (Engineering Equation Solver) SOLUTION Q_dot_H=2 [kW] T_H=800 [K] T_L=300 [K] Q_dot_L=Q_dot_H DELTAS_dot=-Q_dot_H/T_H+Q_dot_L/T_L © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-115

7-22E A reversible air conditioner with specified reservoir temperatures is considered. The entropy change of two reservoirs is to be calculated and it is to be determined if this air conditioner satisfies the increase in entropy principle. Assumptions The air conditioner operates steadily. Analysis According to the thermodynamic temperature scale,

QH = QL

TH 570 R = (36, 000 Btu/h) = 38, 720 Btu/h TL 530 R

The rate of entropy change of the hot reservoir is then

D SH =

QH - 38, 720 Btu/h = = - 67.92 Btu/h ×R TH 570 R

Similarly, the rate of entropy change of the cold reservoir is

D SL =

QL 36, 000 Btu/h = = 67.92 Btu/h ×R TL 530 R

The net rate of entropy change of everything in this system is

D S total = D S H + D S L = - 67.92 + 67.92 = 0 Btu / h ×R The net rate of entropy change is zero as it must be in order to satisfy the second law.

EES (Engineering Equation Solver) SOLUTION Q_dot_L=36000 [Btu/h] T_H=570 [R] T_L=530 [R] Q_dot_H=Q_dot_L*T_H/T_L DELTAS_dot_H=-Q_dot_H/T_H DELTAS_dot_L=Q_dot_L/T_L DELTAS_dot_total=DELTAS_dot_H+DELTAS_dot_L

7-23 Heat is transferred directly from an energy-source reservoir to an energy-sink. The entropy change of the two reservoirs is to be calculated and it is to be determined if the increase of entropy principle is satisfied. Assumptions The reservoirs operate steadily. Analysis The entropy change of the source and sink is given by

6-116

DS =

QH QL - 100 kJ 100 kJ + = + = 0.0833 kJ / K TH TL 1200 K 600 K

Since the entropy of everything involved in this process has increased, this transfer of heat is possible.

EES (Engineering Equation Solver) SOLUTION "Given" Q_H=100 [kJ] T_H=1200 [K] T_L=600 [K] "Analysis" Q_L=Q_H DELTAS=-Q_H/T_H+Q_L/T_L

7-24 It is assumed that heat is transferred from a cold reservoir to the hot reservoir contrary to the Clausius statement of the second law. It is to be proven that this violates the increase in entropy principle. Assumptions The reservoirs operate steadily. Analysis According to the definition of the entropy, the entropy change of the high-temperature reservoir shown below is

D SH =

Q 100 kJ = = 0.08333 kJ/K TH 1200 K

and the entropy change of the low-temperature reservoir is

D SL =

Q - 100 kJ = = - 0.1667 kJ/K TL 600 K

The total entropy change of everything involved with this system is then

D Stotal = D SH + D SL = 0.08333 - 0.1667 = - 0.0833 kJ / K which violates the increase in entropy principle since the total entropy change is negative.

6-117

Q_H=100 [kJ] T_H=1200 [K] T_L=600 [K] "Analysis" Q_L=Q_H DELTAS_H=Q_H/T_H DELTAS_L=-Q_L/T_L DELTAS_total=DELTAS_H+DELTAS_L

7-25 Heat is transferred isothermally from a source to the working fluid of a Carnot engine. The entropy change of the working fluid, the entropy change of the source, and the total entropy change during this process are to be determined. Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is given by

DS =

Q T

Noting that heat transferred from the source is equal to the heat transferred to the working fluid, the entropy changes of the fluid and of the source become

(b)

D Sfluid =

Qfluid Qin,fluid 900 kJ = = = 1.337 kJ / K Tfluid Tfluid 673 K

D Ssource =

Q Qsource 900 kJ = - out, source = = - 1.337 kJ / K Tsource Tsource 673 K

Sgen = D Stotal = D Sfluid + D Ssource = 1.337 - 1.337 = 0

EES (Engineering Equation Solver) SOLUTION "Given" Q_in_fluid=900 [kJ] T_source=ConvertTemp(C, K, 400) "Analysis" "(a)" DELTAS_fluid=Q_in_fluid/T_fluid T_fluid=T_source "(b)" DELTAS_source=-Q_out_source/T_source Q_out_source=Q_in_fluid "(c)" DELTAS_total=DELTAS_fluid+DELTAS_source

6-118

7-26 Problem 7-25 is reconsidered. The effects of the varying the heat transferred to the working fluid and the source temperature on the entropy change of the working fluid, the entropy change of the source, and the total entropy change for the process as the source temperature varies from 100°C to 1000°C are to be investigated. The entropy changes of the source and of the working fluid are to be plotted against the source temperature for heat transfer amounts of 500 kJ, 900 kJ, and 1300 kJ. Analysis The problem is solved using EES, and the results are tabulated and plotted below. T_H = 400 [C] Q_H = 900 [kJ] T_Sys = T_H DELTAS_source = -Q_H/(T_H+273) DELTAS_fluid = +Q_H/(T_Sys+273) S_gen = DELTAS_source + DELTAS_fluid

Sfluid

Ssource

Sgen

[kJ / K]

[kJ / K] 0 0 0 0 0 0 0 0 0 0

[C]

3.485 2.748 2.269 1.932 1.682 1.489 1.336 1.212 1.108 1.021

100 200 300 400 500 600 700 800 900 1000

7-27E Heat is transferred isothermally from the working fluid of a Carnot engine to a heat sink. The entropy change of the working fluid is given. The amount of heat transfer, the entropy change of the sink, and the total entropy change during the process are to be determined.

6-119

Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is given by

Q T Noting that heat transferred from the working fluid is equal to the heat transferred to the sink, the heat transfer become DS =

Qfluid = Tfluid D Sfluid = (555 R )(- 0.7 Btu/R ) = - 388.5 Btu ®

Qfluid,out = 388.5 Btu

(b) The entropy change of the sink is determined from

D Ssink =

Qsink, in Tsink

388.5 Btu = 0.7 Btu / R 555 R

Sgen = D Stotal = D Sfluid + D Ssink = - 0.7 + 0.7 = 0 This is expected since all processes of the Carnot cycle are reversible processes, and no entropy is generated during a reversible process.

EES (Engineering Equation Solver) SOLUTION "Given" DELTAS_fluid=-0.7 [Btu/R] T_sink=(95+460) [R] "Analysis" "(a)" Q_fluid_out=-T_fluid*DELTAS_fluid T_fluid=T_sink "(b)" DELTAS_sink=Q_sink_in/T_sink Q_sink_in=Q_fluid_out "(c)" DELTAS_total=DELTAS_fluid+DELTAS_sink

7-28 Air is compressed steadily by a compressor. The air temperature is maintained constant by heat rejection to the surroundings. The rate of entropy change of air is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas. 4. The process involves no internal irreversibilities such as friction, and thus it is an isothermal, internally reversible process. Properties Noting that h = h(T) for ideal gases, we have h1  h2 since T1  T2  25 C .

6-120

Analysis We take the compressor as the system. Noting that the enthalpy of air remains constant, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Win = Qout

Therefore,

Qout = Win = 40 kW Noting that the process is assumed to be an isothermal and internally reversible process, the rate of entropy change of air is determined to be

D Sair = -

Qout,air Tsys

40 kW = - 0.134 kW / K 298 K

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_in=40 [kW] T1=(25+273) [K] T2=T1 T_surr=(20+273) [K] "Analysis" W_dot_in=Q_dot_out "energy balance" DELTAS_dot_air=-Q_dot_out/T1

7-29 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat is transferred to the refrigerant from the cooled space, and the liquid is vaporized. The entropy change of the refrigerant, the entropy change of the cooled space, and the total entropy change for this process are to be determined. Assumptions 1. Both the refrigerant and the cooled space involve no internal irreversibilities such as friction. 2. Any temperature change occurs within the wall of the tube, and thus both the refrigerant and the cooled space remain isothermal during this process. Thus it is an isothermal, internally reversible process. Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes, the entropy change for them can be determined from

DS =

Q T

6-121

(a) The pressure of the refrigerant is maintained constant. Therefore, the temperature of the refrigerant also remains constant at the saturation value,

T = Tsat@140 kPa = - 18.77°C = 254.4 K (Table A-12) Then,

D Srefrigerant =

Qrefrigerant, in Trefrigerant

180 kJ = 0.7076 kJ / K 254.4 K

(b) Similarly, D Sspace = -

Qspace, out Tspace

180 kJ = - 0.6844 kJ / K 263 K

Sgen = D Stotal = D Srefrigerant + D Sspace = 0.7076 - 0.6844 = 0.0232 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" P1=140 [kPa] Q_refrigerant_in=180 [kJ] T_space=(-10+273) [K] "Analysis" "(a)" T_refrigerant=temperature(R134a, P=P1, x=1) DELTAS_refrigerant=Q_refrigerant_in/T_refrigerant "(b)" DELTAS_space=-Q_space_out/T_space Q_space_out=Q_refrigerant_in "(c)" DELTAS_total=DELTAS_refrigerant+DELTAS_space

7-30 A rigid tank contains an ideal gas that is being stirred by a paddle wheel. The temperature of the gas remains constant as a result of heat transfer out. The entropy change of the gas is to be determined. Assumptions The gas in the tank is given to be an ideal gas. Analysis The temperature and the specific volume of the gas remain constant during this process. Therefore, the initial and the final states of the gas are the same. Then s 2  s1 since entropy is a property. Therefore,

6-122

D Ssys = 0

EES (Engineering Equation Solver) SOLUTION "Given" T1=40 [C] W_pw_in=200 [kJ] T2=T1 T_surr=30 [C] "Analysis" DELTAS_sys=0 "since the temperature and specific volume of the gas remain constant and the initial and the final states are the same. Then s2=s1."

7-31 According to the conservation of mass principle,

dmCV = å m- å m dt in out dm = min dt

An entropy balance adapted to this system becomes

dSsurr d (ms) + - min sin ³ 0 dt dt When this is combined with the mass balance, and the constant entropies are removed from the derivatives, it becomes

dSsurr dm dm +s - sin ³ 0 dt dt dt Multiplying by dt and integrating the result yields

D Ssurr + (s - sin )D m ³ 0 or

D Ssurr ³ (s - sin )D m

7-32 According to the conservation of mass principle,

dmCV = å m- å m dt in out dm = - mout dt

An entropy balance adapted to this system becomes

dSsurr d (ms) + + mout s ³ 0 dt dt When this is combined with the mass balance, it becomes

dSsurr dm dm +s - s ³ 0 dt dt dt Multiplying by dt and integrating the result yields

6-123

Since all the entropies are same, this reduces to

D Ssurr ³ 0 Hence, the entropy of the surroundings can only increase or remain fixed.

Entropy Changes of Pure Substances 7-33C Yes, because an internally reversible, adiabatic process involves no irreversibilities or heat transfer.

7-34E R-134a is expanded isentropically in a closed system. The heat transfer and work production are to be determined.

Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis As there is no area under the process line shown on the T-s diagram and this process is reversible,

Q = 0 Btu

The energy balance for this system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wout = D U = m(u2 - u1 ) Wout = m(u1 - u2 ) The initial state properties are

ïü ïý u1 = 109.46 Btu/lbm (Table A-13E) P1 = 100 psia ïïþ s1 = 0.22902 Btu/lbm ×R

T1 = 100°F

6-124

The final state properties for this isentropic process are (Table A-12E) ïü ïý s2 = s1 = 0.22902 Btu/lbm ×R ïïþ

P2 = 10 psia

x2 =

s2 - s f s fg

0.22902 - 0.00741 = 0.99788 0.22208

u2 = u f + x2 u fg = 3.132 + (0.99788)(87.463) = 90.41 Btu/lbm

Substituting,

Wout = m(u1 - u2 ) = (1 lbm)(109.46 - 90.41) Btu/lbm = 19.05 Btu

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [lbm] P1=100 [psia] T1=100 [F] P2=10 [psia] "Analysis" Fluid$='R134a' Q=0 u1=intenergy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2=s1 x2=quality(Fluid$, P=P2, s=s2) u2=intenergy(Fluid$, P=P2, s=s2) W_out=m*(u1-u2)

7-35E A piston-cylinder device that is filled with water is heated. The total entropy change is to be determined. Analysis The initial specific volume is

v1 =

v1 2.5 ft 3 = = 1.25 ft 3 /lbm m 2 lbm

which is between v f and vg for 300 psia. The initial quality and the entropy are then (Table A-5E) x1 =

v1 - v f v fg

(1.25 - 0.01890) ft 3 /lbm = 0.8075 (1.5435 - 0.01890) ft 3 /lbm

s1 = s f + x1s fg = 0.58818 Btu/lbm ×R + (0.8075)(0.92289 Btu/lbm ×R) = 1.3334 Btu/lbm ×R

The final state is superheated vapor and

6-125

ü ïï ý s = 1.5706 Btu/lbm ×R (Table A-6E) P2 = P1 = 300 psia ïïþ 2 Hence, the change in the total entropy is T2 = 500°F

D S = m(s2 - s1 ) = (2 lbm)(1.5706 - 1.3334) Btu/lbm ×R = 0.4744 Btu / R

EES (Engineering Equation Solver) SOLUTION "Given" m=2 [lbm] P1=300 [psia] Vol1=2.5 [ft^3] P2=P1 T2=500 [F] "Analysis" Fluid$='steam_iapws' v1=Vol1/m v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) x1=(v1-v1_f)/(v1_g-v1_f) s1_f=entropy(Fluid$, P=P1, x=0) s1_g=entropy(Fluid$, P=P1, x=1) s1_fg=s1_g-s1_f s1=s1_f+x1*(s1_fg) s2=entropy(Fluid$, P=P2, T=T2) DELTAS=m*(s2-s1)

7-36 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. An electric heater inside is turned on and kept on until all the liquid vaporized. The entropy change of the water during this process is to be determined. Analysis From the steam tables (Tables A-4 through A-6)

P1 = 200 kPa üïï v1 = v f + x1v fg = 0.001061 + (0.25)(0.88578 - 0.001061) = 0.22224 m 3 /kg ý ïïþ s1 = s f + x1 s fg = 1.5302 + (0.25)(5.5968) = 2.9294 kJ/kg ×K x1 = 0.25 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-126

Then the entropy change of the steam becomes

D S = m (s2 - s1 ) = (3 kg)(6.6335 - 2.9294) kJ/kg ×K = 11.1 kJ / K v 2 = v1 = 0.22224 m3 /kgü ïïý s = 6.6335 kJ/kg ×K ïï 2 x2 = 1 (sat. vapor) þ

EES (Engineering Equation Solver) SOLUTION "Given" m=3 [kg] P1=200 [kPa] x1=1-0.75 x2=1 "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, x=x1) v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) s1=entropy(Fluid$, P=P1, x=x1) s1_f=entropy(Fluid$, P=P1, x=0) s1_g=entropy(Fluid$, P=P1, x=1) s1_fg=s1_g-s1_f v2=v1 "fixed mass, constant volume" s2=entropy(Fluid$, v=v2, x=x2) "Analysis" DELTAS=m*(s2-s1)

7-37 The change in the entropy of R-134a as it is heated at constant pressure is to be calculated using the relation dS  ( Q / T )int rev , and it is to be verified by using R-134a tables. Analysis As R-134a is converted from a saturated liquid to a saturated vapor, both the pressure and temperature remains constant. Then, the relation dS  ( Q / T )int rev reduces to

ds =

dh T

When this result is integrated between the saturated liquid and saturated vapor states, the result is (Table A-12) sg - s f =

hg - h f T

h fg @ 200 kPa Tsat @ 200 kPa

206.09 kJ/kg = 0.78343 kJ / kg ×K (- 10.09 + 273.15) K

Finding the result directly from the R-134a tables

sg - s f = s fg @ 200 kPa = 0.78339 kJ / kg ×K

(Table A-12)

The two results are practically identical.

6-127

EES (Engineering Equation Solver) SOLUTION P=200 [kPa] x1=0 x2=1 Fluid$='R134a' h_f=enthalpy(Fluid$, P=P, x=x1) h_g=enthalpy(Fluid$, P=P, x=x2) T=temperature(Fluid$, P=P, x=x1) DELTAs=(h_g-h_f)/(T+273.15) s_f=entropy(Fluid$, P=P, x=x1) s_g=entropy(Fluid$, P=P, x=x2) s_fg=s_g-s_f

7-38 The radiator of a steam heating system is initially filled with superheated steam. The valves are closed, and steam is allowed to cool until the temperature drops to a specified value by transferring heat to the room. The entropy change of the steam during this process is to be determined. Analysis From the steam tables (Tables A-4 through A-6),

3 P1 = 200 kPa ïü ïý v1 = 0.95986 m /kg ïïþ s1 = 7.2810 kJ/kg ×K T1 = 150°C

T2 = 40°C v 2 = v1

ïü ïý x = v 2 - v f = 0.95986 - 0.001008 = 0.04914 ïïþ 2 v fg 19.515 - 0.001008

s2 = s f + x2 s fg = 0.5724 + (0.04914)(7.6832) = 0.9499 kJ/kg ×K The mass of the steam is

V 0.020 m3 = = 0.02084 kg v1 0.95986 m3 /kg

Then the entropy change of the steam during this process becomes

D S = m (s2 - s1 ) = (0.02084 kg )(0.9499 - 7.2810) kJ/kg ×K = - 0.132 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.020 [m^3] P1=200 [kPa] T1=150 [C] T2=40 [C] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-128

v2=v1 "fixed mass, constant volume" s2=entropy(Fluid$, T=T2, v=v2) "Analysis" m=Vol/v1 DELTAS=m*(s2-s1)

7-39 A rigid tank is divided into two equal parts by a partition. One part is filled with compressed liquid water while the other side is evacuated. The partition is removed and water expands into the entire tank. The entropy change of the water during this process is to be determined.

Analysis The properties of the water are (Table A-4) 3 P1 = 400 kPa ü ïï v1 @v f@60°C = 0.001017 m /kg ý ïïþ s1 = s f @ 60°C = 0.8313 kJ/kg ×K T1 = 60°C

Noting that

v 2 = 2v1 = (2)(0.001017) = 0.002034 m3 /kg v2 - v f 0.002034 - 0.001026 x = = = 0.0002524 ïü ïý 2 v fg 3.993 - 0.001026 3 v 2 = 0.002034 m /kg ïïþ s2 = s f + x2 s fg = 1.0261 + (0.0002524)(6.6430) = 1.0278 kJ/kg ×K P2 = 40 kPa

Then the entropy change of the water becomes

D S = m (s2 - s1 ) = (2.5 kg )(1.0278 - 0.8313) kJ/kg ×K = 0.492 kJ / K

EES (Engineering Equation Solver) SOLUTION $Array off "Places array variables in the solutions window rather than the arrays window." "Input Data" P[1]=400 [kPa] T[1]=60 [C] m=2.5 [kg] P[2]=40 [kPa] V[1]=m*spv[1] spv[1]=volume(steam_iapws,T=T[1], P=P[1]) "specific volume of steam at state 1" s[1]=entropy(steam_iapws,T=T[1],P=P[1]) "entropy of steam at state 1" V[2]=2*V[1] "Steam expands to fill entire volume at state 2" "State 2 is identified by P[2] and spv[2]" spv[2]=V[2]/m "specific volume of steam at state 2" s[2]=entropy(steam_iapws,P=P[2],v=spv[2]) "entropy of steam at state 2" T[2]=temperature(steam_iapws,P=P[2],v=spv[2]) DELTAS_sys=m*(s[2]-s[1]) "Total entopy change of steam" v_2f=volume(steam_iapws, P=P[2], x=0) v_2g=volume(steam_iapws, P=P[2], x=1) x2=(spv[2]-v_2f)/(v_2g-v_2f) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-129

s_2f=entropy(steam_iapws, P=P[2], x=0) s_2g=entropy(steam_iapws, P=P[2], x=1) s_fg=s_2g-s_2f

7-40 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant expands in a reversible manner until the pressure drops to a specified value. The final temperature in the cylinder and the work done by the refrigerant are to be determined. Assumptions 1. The kinetic and potential energy changes are negligible. 2. The cylinder is well-insulated and thus heat transfer is negligible. 3. The thermal energy stored in the cylinder itself is negligible. 4. The process is stated to be reversible. Analysis (a) This is a reversible adiabatic (i.e., isentropic) process, and thus s 2  s1 . From the refrigerant tables (Tables A-11 through A-13),

P1 = 0.8 MPa üïï ý ïïþ sat. vapor

v1 = v g @ 0.8 MPa = 0.025645 m3 /kg u1 = ug @ 0.8 MPa = 246.82 kJ/kg s1 = sg @ 0.8 MPa = 0.91853 kJ/kg ×K

Also,

V 0.05 m3 = = 1.950 kg v1 0.025645 m3 /kg

and s2 - s f 0.91853 - 0.24757 = = 0.9874 P2 = 0.4 MPa ïüï x2 = s fg 0.67954 ý ïïþ s2 = s1 u2 = u f + x2 u fg = 63.61 + (0.9874)(171.49) = 232.94 kJ/kg

T2 = Tsat @ 0.4 MPa = 8.91°C (b) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this adiabatic closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wb,out = D U Wb,out = m(u1 - u2 )

Substituting, the work done during this isentropic process is determined to be

6-130

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.05 [m^3] P1=800 [kPa] x1=1 P2=400 [kPa] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) "Analysis" "(a)" s2=s1 "reversible-adiabatic (i.e., isentropic) process" u2=intenergy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) x2=quality(Fluid$, P=P2, s=s2) "(b)" m=Vol/v1 -W_b_out=m*(u2-u1) "energy balance"

7-41 Problem 7-40 is reconsidered. The work done by the refrigerant is to be calculated and plotted as a function of final pressure as the pressure varies from 0.8 MPa to 0.4 MPa. The work done for this process is to be compared to one for which the temperature is constant over the same pressure range. Analysis The problem is solved using EES, and the results are tabulated and plotted below. Procedure IsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isotherm) T_isotherm=Temperature(R134a,P=P_1,x=x_1) T=T_isotherm u_1 = INTENERGY(R134a,P=P_1,x=x_1) v_1 = volume(R134a,P=P_1,x=x_1) s_1 = entropy(R134a,P=P_1,x=x_1) u_2 = INTENERGY(R134a,P=P_2,T=T) s_2 = entropy(R134a,P=P_2,T=T) Q_isotherm = (T+273)*m_sys*(s_2 - s_1) DELTAE_isotherm = m_sys*(u_2 - u_1) E_in = Q_isotherm E_out = DELTAE_isotherm+E_in Work_out_isotherm=E_out END P_1 = 800 [kPa] x_1 = 1.0 V_sys = 0.05[m^3] P_2 = 400 [kPa] E_in - E_out = DELTAE_sys E_out = Work_out_isen E_in = 0 DELTAE_sys = m_sys*(u_2 - u_1) u_1 = INTENERGY(R134a,P=P_1,x=x_1) v_1 = volume(R134a,P=P_1,x=x_1) s_1 = entropy(R134a,P=P_1,x=x_1) V_sys = m_sys*v_1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-131

s_2 = s_1 u_2 = INTENERGY(R134a,P=P_2,s=s_2) T_2_isen = temperature(R134a,P=P_2,s=s_2) Call IsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isotherm)

Work out,isen

Work out,isotherm

Qisotherm

[kPa]

[kJ] 27.09 18.55 11.44 5.347 0

[kJ] 60.02 43.33 28.2 13.93 0

[kJ]

400 500 600 700 800

47.08 33.29 21.25 10.3 0

7-42 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically at constant pressure. The entropy change of the water during this process is to be determined. Assumptions 1. The kinetic and potential energy changes are negligible. 2. The cylinder is well-insulated and thus heat transfer is negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-132

3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. AnalysisFrom the steam tables (Tables A-4 through A-6),

v = v f @150 kPa = 0.001053 m3 /kg P1 = 150 kPa ïüï 1 ý h1 = h f @150 kPa = 467.13 kJ/kg sat. liquid ïïþ s1 = s f @150 kPa = 1.4337 kJ/kg ×K Also,

V 0.005 m3 = = 4.750 kg v1 0.001053 m3 /kg

We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in - Wb, out = D U We, in = m(h2 - h1 )

since U  Wb  H during a constant pressure quasi-equilibrium process. Solving for h2 ,

h2 = h1 +

We,in m

= 467.13 +

1700 kJ = 825.06 kJ/kg 4.75 kg

Thus, h2 - h f 825.06 - 467.13 x = = = 0.1608 ïü ïý 2 h fg 2226.0 ï h2 = 825.06 kJ/kg ïþ s2 = s f + x2 s fg = 1.4337 + (0.1608)(5.7894) = 2.3646 kJ/kg ×K P2 = 150 kPa

Then the entropy change of the water becomes

D S = m (s2 - s1 ) = (4.750 kg )(2.3646 - 1.4337)kJ/kg ×K = 4.42 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.005 [m^3] P1=150 [kPa] x1=0 P2=P1 W_e_in=1700 [kJ] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, x=x1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-133

h1=enthalpy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) "Analysis" m=Vol/v1 W_e_in=m*(h2-h1) "since U2-U1+W_b = H2-H1 for a constant pressure process" s2=entropy(Fluid$, P=P2, h=h2) x2=quality(Fluid$, P=P2, h=h2) DELTAS=m*(s2-s1)

7-43 R-134a undergoes a process during which the entropy is kept constant. The final temperature and internal energy are to be determined. Analysis The initial entropy is

ïü ïý s = 0.9335 kJ/kg ×K (Table A-13) P1 = 600 kPa ïïþ 1

T1 = 25°C

The entropy is constant during the process. The final state is a mixture since the entropy is between s f and s g for 100 kPa. The properties at this state are (Table A-12) T2 = Tsat @ 100 kPa = - 26.37°C

x2 =

s2 - s f s fg

(0.9335 - 0.07182) kJ/kg ×K = 0.9791 0.8801 kJ/kg ×K

u2 = u f + x2u fg = 17.19 + (0.9791)(198.01) = 211.1kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] P1=600 [kPa] T1=25 [C] P2=100 [kPa] s2=s1 "Analysis" Fluid$='R134a' s1=entropy(Fluid$, P=P1, T=T1) T2=temperature(Fluid$, P=P2, x=1) "state 2 is mixture" s2_f=entropy(Fluid$, P=P2, x=0) s2_g=entropy(Fluid$, P=P2, x=1) s2_fg=s2_g-s2_f x2=(s2-s2_f)/s2_fg u2_f=intenergy(Fluid$, P=P2, x=0) u2_g=intenergy(Fluid$, P=P2, x=1) u2_fg=u2_g-u2_f u2=u2_f+x2*u2_fg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-134

7-44 Refrigerant-134a is expanded in a turbine during which the entropy remains constant. The inlet and outlet velocities are to be determined. Analysis The initial state is superheated vapor and thus

P1 = 600 kPa ü ïï v1 = 0.04307 m3 /kg (Table A-13) ý T1 = 70°C ïïþ s1 = 1.0706 kJ/kg ×K The entropy is constant during the process. The properties at the exit state are

ü ïï 3 ý v 2 = 0.2274 m /kg (Table A-13) ï s2 = s1 = 1.0706 kJ/kg ×K ïþ The inlet and outlet veloicites are P2 = 100 kPa

V1 =

mv1 (0.75 kg/s)(0.04307 m3 /kg) = = 0.0646 m / s A1 0.5 m2

V2 =

mv 2 (0.75 kg/s)(0.2274 m3 /kg) = = 0.171 m / s A2 1.0 m2

EES (Engineering Equation Solver) SOLUTION "Given" P1=600 [kPa] T1=70 [C] P2=100 [kPa] A1=0.5 [m^2] A2=1 [m^2] m_dot=0.75 [kg/s] "Analysis" Fluid$='R134a' v1=volume(Fluid$,T=T1,P=P1) s1=entropy(Fluid$,T=T1,P=P1) s2=s1 v2=volume(Fluid$,P=P2,s=s2) x2=quality(Fluid$,P=P2,s=s2) Vel1=m_dot*v1/A1 Vel2=m_dot*v2/A2

7-45 R-134a undergoes an isothermal process in a closed system. The work and heat transfer are to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis The energy balance for this system can be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-135

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win - Qout = D U = m(u2 - u1 ) The initial state properties are

P1 = 320 kPa ïü ïý u1 = 261.62 kJ/kg (Table A-13) ïïþ s1 = 1.0452 kJ/kg ×K T1 = 40°C For this isothermal process, the final state properties are (Table A-11)

T2 = T1 = 40°C ü ïï u2 = u f + x2u fg = 107.39 + (0.45)(143.61) = 172.02 kJ/kg ý ïïþ s2 = s f + x2 s fg = 0.39493 + (0.45)(0.52059) = 0.62920 kJ/kg ×K x2 = 0.45 The heat transfer is determined from

qin = T0 (s2 - s1 ) = (313 K)(0.62920 - 1.0452) kJ/kg ×K = - 130.2 kJ/kg The negative sign shows that the heat is actually transferred from the system. That is,

qout = 130.2 kJ / kg The work required is determined from the energy balance to be

win = qout + (u2 - u1 ) = 130.2 kJ/kg + (172.02 - 261.62) kJ/kg = 40.6 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=320 [kPa] T1=40 [C] T2=T1 x2=0.45 "Analysis" Fluid$='R134a' u1=intenergy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) u2_f=intenergy(Fluid$, T=T2, x=0) u2_g=intenergy(Fluid$, T=T2, x=1) u2_fg=u2_g-u2_f u2=u2_f+x2*u2_fg s2_f=entropy(Fluid$, T=T2, x=0) s2_g=entropy(Fluid$, T=T2, x=1) s2_fg=s2_g-s2_f s2=s2_f+x2*s2_fg q_in=(T1+273)*(s2-s1) q_out=-q_in w_in=q_out+(u2-u1)

6-136

7-46 A rigid tank contains saturated water vapor at a specified temperature. Steam is cooled to ambient temperature. The process is to be sketched and entropy changes for the steam and for the process are to be determined. Assumptions 1. The kinetic and potential energy changes are negligible. Analysis (b) From the steam tables (Tables A-4 through A-6),

v1 = v g = 1.6720 kJ/kg T1 = 100°Cïüï ý u1 = u g = 2506.0 kJ/kg ïïþ x= 1 s1 = 7.3542 kJ/kg ×K

x2 = 0.03856 T2 = 25°Cüïï ý u2 = 193.67 kJ/kg v 2 = v1 ïïþ s2 = 0.6830 kJ/kg ×K The entropy change of steam is determined from

D Sw = m(s2 - s1 ) = (5 kg)(0.6830-7.3542)kJ/kg ×K = - 33.36 kJ / K (c) We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = m(u2 - u1 ) That is,

Qout = m(u1 - u2 ) = (5 kg)(2506.0-193.67)kJ/kg = 11,562 kJ The total entropy change for the process is

Sgen = D S w +

Qout 11,562 kJ = - 33.36 kJ/K + = 5.44 kJ / K Tsurr 298 K

EES (Engineering Equation Solver) SOLUTION "Given" m=5 [kg] T1=100 [C] x1=1 T2=25 [C] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) s1=entropy(Fluid$, T=T1, x=x1) v2=v1 x2=quality(Fluid$, T=T2, v=v2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-137

u2=intenergy(Fluid$, T=T2, v=v2) s2=entropy(Fluid$, T=T2, v=v2) "Analysis" DELTAS_w=m*(s2-s1) Q_out=m*(u1-u2) S_gen=DELTAS_w+Q_out/T_surr T_surr=T2+273

7-47 A rigid tank is initially filled with a saturated mixture of R-134a. Heat is transferred to the tank from a source until the pressure inside rises to a specified value. The entropy change of the refrigerant, entropy change of the source, and the total entropy change for this process are to be determined. Assumptions 1. The tank is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions.

Analysis (a) From the refrigerant tables (Tables A-11 through A-13),

u1 = u f + x1u fg = 38.26 + (0.4)(186.25) = 112.76 kJ/kg

P1 = 200 kPa ïüï ý ïïþ x1 = 0.4

s1 = s f + x1s fg = 0.15449 + (0.4)(0.78339) = 0.4678 kJ/kg ×K v1 = v f + x1v fg = 0.0007532 + (0.4)(0.099951- 0.0007532) = 0.04043 m3 /kg x2 =

P2 = 400 kPa ïüï ý ïïþ v 2 = v1

v2 - v f v fg

0.04043 - 0.0007905 = 0.7853 0.051266 - 0.0007905

u2 = u f + x2 u fg = 63.61 + (0.7853)(171.49) = 198.29 kJ/kg s2 = s f + x2 s fg = 0.24757 + (0.7853)(0.67954) = 0.7813 kJ/kg ×K

The mass of the refrigerant is

V 0.5 m3 = = 12.37 kg v1 0.04043 m3 /kg

Then the entropy change of the refrigerant becomes

D Ssystem = m (s2 - s1 ) = (12.37 kg )(0.7813 - 0.4678) kJ/kg ×K = 3.876 kJ / K (b) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationary closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = D U = m(u2 - u1 ) Substituting,

6-138

The heat transfer for the source is equal in magnitude but opposite in direction. Therefore, Qsource, out  Qtank, in  1058 kJ

and

D Ssource = -

Qsource,out Tsource

1058 kJ = - 3.434 kJ / K 308 K

D S total = D Ssystem + D Ssource = 3.876 + (- 3.434) = 0.441 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.5 [m^3] P1=200 [kPa] x1=0.40 P2=400 [kPa] T_source=35 [C] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) v2=v1 "fixed mass, constant volume" u2=intenergy(Fluid$, P=P2, v=V2) s2=entropy(Fluid$, P=P2, v=V2) "Analysis" "(a)" m=Vol/v1 DELTAS_sys=m*(s2-s1) "(b)" Q_sys_in=m*(u2-u1) "energy balance" Q_source_out=Q_sys_in DELTAS_source=-Q_source_out/(T_source+273) "(c)" DELTAS_total=DELTAS_sys+DELTAS_source

7-48 Problem 7-47 is reconsidered. The effects of the source temperature and final pressure on the total entropy change for the process as the source temperature varies from 30°C to 210°C, and the final pressure varies from 250 kPa to 500 kPa are to be investigated. The total entropy change for the process is to be plotted as a function of the source temperature for final pressures of 250 kPa, 400 kPa, and 500 kPa. Analysis The problem is solved using EES, and the results are tabulated and plotted below. P_1 = 200 [kPa] x_1 = 0.4 V_sys = 0.5 [m^3] P_2 = 400 [kPa] T_source = 35 [C] E_in - E_out = DELTAE_sys E_out = 0 [kJ] E_in = Q DELTAE_sys = m_sys*(u_2 - u_1) u_1 = INTENERGY(R134a,P=P_1,x=x_1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-139

v_1 = volume(R134a,P=P_1,x=x_1) V_sys = m_sys*v_1 v_2 = v_1 u_2 = INTENERGY(R134a,P=P_2,v=v_2) s_1 = entropy(R134a,P=P_1,x=x_1) s_2 = entropY(R134a,P=P_2,v=v_2) DELTAS_sys = m_sys*(s_2 - s_1) DELTAS_source = -Q/(T_source + 273) DELTAS_total = DELTAS_source + DELTAS_sys

Stotal

Tsource

[kJ / K] 0.3848 0.6997 0.9626 1.185 1.376 1.542 1.687

[C] 30 60 90 120 150 180 210

7-49 Steam enters a nozzle at a specified state and leaves at a specified pressure. The process is to be sketched on the T-s diagram and the maximum outlet velocity is to be determined. Analysis (b) The inlet state properties are

P1 = 6000 kPa ü ïï ý ïïþ x1 = 1

h1 = 2784.6 kJ/kg s1 = 5.8902 kJ/kg ×K

(Table A-5)

For the maximum velocity at the exit, the entropy will be constant during the process. The exit state enthalpy is (Table A-6) s2 - s f 5.8902 - 2.2159 x2 = = = 0.8533 ü ïï s 4.3058 ý fg s2 = s1 = 5.8902 kJ/kg ×K ïïþ h2 = h f + xh fg = 798.33 + 0.8533´ 1985.4 = 2492.5 kJ/kg

P2 = 1200 kPa

We take the nozzle as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the nozzle, the energy balance for this steady-flow system can be expressed in the rate form as

6-140

Ein - Eout

D Esystem ¿ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2ö æ æ V2ö ççh + V2 ÷ ÷ ÷ m çççh1 + 1 ÷ = m 2 ÷ ÷ 2 ø÷ 2 ø÷ èç èçç

(since W @ Q @D pe @ 0)

æV 2 - V12 ÷ ö ÷ h1 - h2 = ççç 2 ÷ ÷ çè 2 ø Solving for the exit velocity and substituting,

æV 2 - V12 ÷ ö ÷ h1 - h2 = ççç 2 ÷ çè 2 ø÷ 0.5

2 2 öù é æ ê(0 m/s) 2 + 2(2784.6 - 2492.5) kJ/kg çç1000 m /s ÷ ÷ú V2 = éëêV + 2(h1 - h2 )ù = ú ê û çè 1 kJ/kg ÷ ÷ øú ú ëê û 2 1

0.5

= 764.3 m / s

EES (Engineering Equation Solver) SOLUTION "Given" Vel1=0 P1=6000 [kPa] x1=1 P2=1200 [kPa] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$,P=P1,x=x1) s1=entropy(Fluid$,P=P1,x=x1) s2=s1 h2=enthalpy(Fluid$,P=P2,s=s2) x2=quality(Fluid$,P=P2,s=s2) T2=temperature(Fluid$,P=P2,s=s2) h1-h2=(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg)

7-50 Steam enters a diffuser at a specified state and leaves at a specified pressure. The minimum outlet velocity is to be determined.

6-141

Analysis The inlet state is superheated vapor and thus

P1 = 150 kPa ïü ïý h1 = 2711.4 kJ/kg (Table A-6) T1 = 120°C ïïþ s1 = 7.2699 kJ/kg ×K For minimum velocity at the exit, the entropy will be constant during the process. The exit state enthalpy is (Table A-6)

ïü ïý h = 2845.7 kJ/kg (Table A-6) s2 = s1 = 7.2699 kJ/kg ×K ïïþ 2

P2 = 300 kPa

We take the diffuser as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the diffuser, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2ö æ æ V2ö ççh + V2 ÷ ÷ ÷ m çççh1 + 1 ÷ = m 2 ÷ ÷ 2 ø÷ 2 ø÷ èç èçç

(since W @ Q @D pe @ 0)

æV 2 - V12 ÷ ö ÷ h1 - h2 = ççç 2 ÷ çè 2 ø÷ Solving for the exit velocity and substituting, æV 2 - V12 ÷ ö ÷ h1 - h2 = ççç 2 ÷ ÷ çè 2 ø 0.5

2 2 öù é æ ê(550 m/s) 2 + 2(2711.4 - 2845.7) kJ/kg çç1000 m /s ÷ ÷ú V2 = éëêV + 2(h1 - h2 )ù = ú ê û çè 1 kJ/kg ÷ ÷ øú ú ëê û = 184.1 m / s 2 1

0.5

EES (Engineering Equation Solver) SOLUTION "Given" P1=150 [kPa] T1=120 [C] Vel1=550 [m/s] P2=300 [kPa] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$,P=P1, T=T1) s1=entropy(Fluid$,P=P1, T=T1) s2=s1 h2=enthalpy(Fluid$,P=P2,s=s2) h1-h2=(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg)

6-142

7-51E R-134a is expanded in a turbine during which the entropy remains constant. The enthalpy difference is to be determined.

Analysis The initial state is superheated vapor and thus

P1 = 250 psia ïü ïý ïïþ T1 = 175°F

h1 = 129.94 Btu/lbm s1 = 0.23279 Btu/lbm ×R

(Table A-13E or EES)

The entropy is constant during the process. The final state is also superheated vapor and the enthalpy at this state is

ïü ïý h = 106.94 Btu/lbm (Table A-13E or EES) s2 = s1 = 0.23281 Btu/lbm ×Rïïþ 2

T2 = 20°F

Note that the properties at the inlet and exit states can also be determined from Table A-13E by interpolation but the values will not be as accurate as those by EES. The change in the enthalpy across the turbine is then

D h = h2 - h1 = 106.94 - 129.94 = - 23.0 Btu / lbm

EES (Engineering Equation Solver) SOLUTION Fluid$='R134a' P1=250 [psia] T1=175 [F] h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2=s1 T2=20 [F] h2=enthalpy(Fluid$, s=s2, T=T2) DELTAh=h1-h2

7-52 Water is compressed in a compressor during which the entropy remains constant. The final temperature and enthalpy are to be determined. Analysis The initial state is superheated vapor and the entropy is

T1 = 160°C ïü ïý h1 = 2800.7 kJ/kg (from EES) P1 = 35 kPa ïïþ s1 = 8.1531 kJ/kg ×K

6-143

Note that the properties can also be determined from Table A-6 by interpolation but the values will not be as accurate as those by EES. The final state is superheated vapor and the properties are (Table A-6)

ïü ïý T2 = 440.5°C s2 = s1 = 8.1531 kJ/kg ×K ïïþ h2 = 3361.0 kJ / kg

P2 = 300 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P1=35 [kPa] T1=160 [C] P2=300 [kPa] s2=s1 "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$,P=P1, T=T1) s1=entropy(Fluid$,P=P1, T=T1) T2=temperature(Fluid$,P=P2,s=s2) h2=enthalpy(Fluid$,P=P2,s=s2)

7-53 Saturated Refrigerant-134a vapor at 160 kPa is compressed steadily by an adiabatic compressor. The minimum power input to the compressor is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. Analysis The power input to an adiabatic compressor will be a minimum when the compression process is reversible. For the reversible adiabatic process we have s2  s1 . From the refrigerant tables (Tables A-11 through A-13),

P1 = 160 kPa üïï ý ïïþ sat. vapor

v1 = v g @ 160 kPa = 0.12355 m3 /kg h1 = hg @ 160 kPa = 241.14 kJ/kg s1 = sg @ 160 kPa = 0.94202 kJ/kg ×K

P2 = 900 kPa ü ïï ý h2 = 277.12 kJ/kg ïïþ s2 = s1 Also,

V1 2 m3 /min = = 16.19 kg/min = 0.2698 kg/s v1 0.12355 m3 /kg

6-144

There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Win + mh1 = mh2

(since Q @D ke @D pe @ 0)

Win = m(h2 - h1 )

Substituting, the minimum power supplied to the compressor is determined to be

Win = (0.2698 kg/s)(277.12 - 241.14) kJ/kg = 9.71 kW

EES (Engineering Equation Solver) SOLUTION "Given" Vol_dot=2 [m^3/min] P1=160 [kPa] x1=1 P2=900 [kPa] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) h1=enthalpy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) s2=s1 "for power input to be minimum" h2=enthalpy(Fluid$, P=P2, s=s2) "Analysis" m_dot=Vol_dot/v1*1/Convert(kg/s, kg/min) W_dot_in=m_dot*(h2-h1) "energy balance"

7-54E R-134a is compressed in an isentropic compressor. The work required is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. The process is isentropic (i.e., reversible-adiabatic). Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + Win = mh2 Win = m(h2 - h1 )

6-145

The inlet state properties are

T1 = 0°Fïü ïý x1 = 1 ïïþ

h1 = 103.10 Btu/lbm s1 = 0.22542 Btu/lbm ×R

(Table A-11E)

For this isentropic process, the final state enthalpy is

ü ïï ý h = 123.30 Btu/lbm (Table A-13E) s2 = s1 = 0.22542 Btu/lbm ×R ïïþ 2 Substituting, P2 = 200 psia

win = h2 - h1 = (123.30 - 103.10) Btu/lbm = 20.2 Btu / lbm

EES (Engineering Equation Solver) SOLUTION Fluid$='R134a' x1=1 T1=0 [F] P2=200 [psia] h1=enthalpy(Fluid$, T=T1, x=x1) s1=entropy(Fluid$, T=T1, x=x1) s2=s1 h2=enthalpy(Fluid$, s=s2, P=P2) w_in=h2-h1

7-55 Steam is expanded in an isentropic turbine. The work produced is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. The process is isentropic (i.e., reversible-adiabatic). Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

6-146

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 = m2 h2 + m3 h3 + Wout Wout = m1h1 - m2 h2 - m3 h3

From a mass balance,

m2 = 0.05m1 = (0.05)(5 kg/s) = 0.25 kg/s m3 = 0.95m1 = (0.95)(5 kg/s) = 4.75 kg/s Noting that the expansion process is isentropic, the enthalpies at three states are determined as follows:

P3 = 50 kPa ïü ïý T3 = 100°C ïïþ

h3 = 2682.4 kJ/kg s3 = 7.6953 kJ/kg ×K

(Table A-6)

ïü ïý h = 3851.2 kJ/kg (Table A-6) s1 = s3 = 7.6953 kJ/kg ×K ïïþ 1

P1 = 3 MPa

ü ïï ý h = 3206.5 kJ/kg s2 = s3 = 7.6953 kJ/kg ×K ïïþ 2 Substituting, P2 = 500 kPa

(Table A-6)

Wout = m1h1 - m2 h2 - m3 h3

= (2 kg/s)(3851.2 kJ/kg) - (0.1 kg/s)(3206.5 kJ/kg) - (1.9 kg/s)(2682.4 kJ/kg) = 2285 kW

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_1=2 [kg/s] P1=3000 [kPa] P3=50 [kPa] T3=100 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-147

P2=500 [kPa] f=0.05 "fraction diverted fro feedwater heating" "Analysis" Fluid$='steam_iapws' m_dot_2=f*m_dot_1 m_dot_3=(1-f)*m_dot_1 h3=enthalpy(Fluid$, P=P3, T=T3) s3=entropy(Fluid$, P=P3, T=T3) s1=s3 h1=enthalpy(Fluid$, P=P1, s=s1) s2=s3 h2=enthalpy(Fluid$, P=P2, s=s2) W_dot_out=m_dot_1*h1-m_dot_2*h2-m_dot_3*h3

7-56 Water is compressed isentropically in a closed system. The work required is to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved other than the boundary work. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Analysis The energy balance for this system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win = D U = m(u2 - u1 ) The initial state properties are u1 = u f + x1u fg = 42.020 + (0.814)(2346.6) = 1952.2 kJ/kg s1 = s f + x1s fg = 0.1511 + (0.814)(8.7488) = 7.2726 kJ/kg ×K

Since the entropy is constant during this process,

ïü ïý u = 3132.6 kJ/kg (Table A-6) s2 = s1 = 7.2726 kJ/kg ×K ïïþ 2

P2 = 3 MPa Substituting,

win = u2 - u1 = (3132.6 - 1952.2) kJ/kg = 1180.4 kJ / kg

6-148

EES (Engineering Equation Solver) SOLUTION "Given" T1=10 [C] x1=0.814 P2=3000 [kPa] "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$,T=T1, x=x1) s1=entropy(Fluid$,T=T1, x=x1) s2=s1 u2=intenergy(Fluid$,P=P2,s=s2) w_in=u2-u1 7-57 Water vapor is expanded adiabatically in a piston-cylinder device. The entropy change is to be determined and it is to be discussed if this process is realistic. Analysis (a) The properties at the initial state are

P1 = 600 kPa ïü ïý u1 = 2566.8 kJ/kg (Table A-5) ïïþ s1 = 6.7593 kJ/kg ×K x1 = 1 We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wout = D U = m(u2 - u1 )

(since Q = KE = PE = 0)

Solving for the final state internal energy,

u2 = u1 +

Wout 700 kJ = 2566.8 kJ/kg + = 2216.8 kJ/kg m 2 kg

The entropy at the final state is (from Table A-5)

6-149

u2 - u f 2216.8 - 417.40 = = 0.8617 ü ïï x2 = u 2088.2 ý fg u2 = 2216.8 kJ/kgïïþ s2 = s f + xs fg = 1.3028 + 0.8617 ´ 6.0562 = 6.5215 kJ/kg ×K P2 = 100 kPa

The entropy change is

D s = s2 - s1 = 6.5215 - 6.7593 = - 0.238 kJ / kg×K (b) The process is not realistic since entropy cannot decrease during an adiabatic process. In the limiting case of a reversible (and adiabatic) process, the entropy would remain constant.

EES (Engineering Equation Solver) SOLUTION "Given" m=2 [kg] P1=600 [kPa] x1=1 P2=100 [kPa] W_out=700 [kJ] "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$,P=P1,x=x1) s1=entropy(Fluid$,P=P1,x=x1) W_out=m*(u1-u2) u2_f=intenergy(Fluid$, P=P2, x=0) u2_g=intenergy(Fluid$, P=P2, x=1) u2_fg=u2_g-u2_f x2=(u2-u2_f)/u2_fg s2_f=entropy(Fluid$, P=P2, x=0) s2_g=entropy(Fluid$, P=P2, x=1) s2_fg=s2_g-s2_f s2=s2_f+x2*s2_fg DELTAs=s2-s1

7-58 A cylinder is initially filled with saturated water vapor mixture at a specified temperature. Steam undergoes a reversible heat addition and an isentropic process. The processes are to be sketched and heat transfer for the first process and work done during the second process are to be determined. Assumptions 1. The kinetic and potential energy changes are negligible. 2. The thermal energy stored in the cylinder itself is negligible. 3. Both processes are reversible.

6-150

T1 = 100°Cü ïï ý h1 = h f + xh fg = 419.17 + (0.5)(2256.4) = 1547.4 kJ/kg x = 0.5 ïïþ T2 = 100°Cïüï ý ïïþ x2 = 1

h2 = hg = 2675.6 kJ/kg u2 = u g = 2506.0 kJ/kg s2 = sg = 7.3542 kJ/kg ×K

P3 = 15 kPa ïü ïý u = 2247.9 kJ/kg ïïþ 3 s3 = s2 We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - Wb,out = D U = m(u2 - u1 )

For process 1-2, it reduces to Q12,in = m(h2 - h1 ) = (5 kg)(2675.6 - 1547.4)kJ/kg = 5641kJ

EES (Engineering Equation Solver) SOLUTION "Given" m=5 [kg] T1=100 [C] x1=0.5 T2=T1 x2=1 P3=15 [kPa] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=x1) h2=enthalpy(Fluid$, T=T2, x=x2) u2=intenergy(Fluid$, T=T2, x=x2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-151

s2=entropy(Fluid$, T=T2, x=x2) s3=s2 u3=intenergy(Fluid$, P=P3, s=s3) "Analysis" Q_12_in=m*(h2-h1) W_23_b_out=m*(u2-u3)

7-59 Heat is added to a pressure cooker that is maintained at a specified pressure. The minimum entropy change of the thermal-energy reservoir supplying this heat is to be determined. Assumptions 1. Only water vapor escapes through the pressure relief valve. Analysis According to the conservation of mass principle,

dmCV = å m- å m dt in out dm = - mout dt

An entropy balance adapted to this system becomes

dSsurr d (ms) + + mout s ³ 0 dt dt When this is combined with the mass balance, it becomes

dSsurr d (ms) dm + - s ³ 0 dt dt dt Multiplying by dt and integrating the result yields

D Ssurr + m2 s2 - m1s1 - sout (m2 - m1 ) ³ 0 The properties at the initial and final states are (from Table A-5 at P1  175 kPa and P2  150 kPa ) v1 = v f + xv fg = 0.001057 + (0.10)(1.0037 - 0.001057) = 0.1013 m 3 /kg

s1 = s f + xs fg = 1.4850 + (0.10)(5.6865) = 2.0537 kJ/kg ×K v 2 = v f + xv fg = 0.001053 + (0.40)(1.1594 - 0.001053) = 0.4644 m 3 /kg

s2 = s f + xs fg = 1.4337 + (0.40)(5.7894) = 3.7494 kJ/kg ×K

The initial and final masses are

m1 =

V 0.020 m3 = = 0.1974 kg v1 0.01013 m3 /kg

m2 =

V 0.020 m3 = = 0.04307 kg v 2 0.4644 m3 /kg

The entropy of escaping water vapor is sout = sg @ 150 kPa = 7.2231 kJ/kg ×K

Substituting, D Ssurr + m2 s2 - m1 s1 - sout (m2 - m1 ) ³ 0 D Ssurr + (0.04307)(3.7494) - (0.1974)(2.0537) - (7.2231)(0.04307 - 0.1974) ³ 0 D Ssurr + 0.8708 ³ 0

The entropy change of the thermal energy reservoir must then satisfy

6-152

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.020 P1=175 P2=150 x1=0.10 x2=0.40 "Analysis: " Fluid$='Steam_IAPWS' v1_f=volume(Fluid$,P=P1,x=0) v1_g=volume(Fluid$,P=P1,x=1) s1_f=entropy(Fluid$,P=P1,x=0) s1_g=entropy(Fluid$,P=P1,x=1) s1_fg=s1_g-s1_f v2_f=volume(Fluid$,P=P2,x=0) v2_g=volume(Fluid$,P=P2,x=1) s2_f=entropy(Fluid$,P=P2,x=0) s2_g=entropy(Fluid$,P=P2,x=1) s2_fg=s2_g-s2_f v1=volume(Fluid$,P=P1,x=x1) s1=entropy(Fluid$,P=P1,x=x1) v2=volume(Fluid$,P=P2,x=x2) s2=entropy(Fluid$,P=P2,x=x2) s_out=s2_g m1=Vol/v1 m2=Vol/v2 DELTAS_surr+m2*s2-m1*s1-s_out*(m2-m1)=0

7-60 Heat is added to a pressure cooker that is maintained at a specified pressure. Work is also done on water. The minimum entropy change of the thermal-energy reservoir supplying this heat is to be determined. Assumptions 1. Only water vapor escapes through the pressure relief valve. Analysis According to the conservation of mass principle,

dmCV = å m- å m dt in out dm = - mout dt

An entropy balance adapted to this system becomes

dSsurr d (ms) + + mout s ³ 0 dt dt When this is combined with the mass balance, it becomes

dSsurr d (ms) dm + - s ³ 0 dt dt dt Multiplying by dt and integrating the result yields

D Ssurr + m2 s2 - m1s1 - sout (m2 - m1 ) ³ 0 Note that work done on the water has no effect on this entropy balance since work transfer does not involve any entropy transfer. The properties at the initial and final states are (from Table A-5 at P1  175 kPa and P2  150 kPa ) v1 = v f + xv fg = 0.001057 + (0.10)(1.0037 - 0.001057) = 0.1013 m 3 /kg

6-153

s1 = s f + xs fg = 1.4850 + (0.10)(5.6865) = 2.0537 kJ/kg ×K v 2 = v f + xv fg = 0.001053 + (0.40)(1.1594 - 0.001053) = 0.4644 m 3 /kg

s2 = s f + xs fg = 1.4337 + (0.40)(5.7894) = 3.7494 kJ/kg ×K

The initial and final masses are

m1 =

V 0.020 m3 = = 0.1974 kg v1 0.01013 m3 /kg

m2 =

V 0.020 m3 = = 0.04307 kg v 2 0.4644 m3 /kg

The entropy of escaping water vapor is sout = sg@150 kPa = 7.2231 kJ/kg ×K

Substituting, D Ssurr + m2 s2 - m1 s1 - sout (m2 - m1 ) ³ 0 D Ssurr + (0.04307)(3.7494) - (0.1974)(2.0537) - (7.2231)(0.04307 - 0.1974) ³ 0 D Ssurr + 0.8708 ³ 0

The entropy change of the thermal energy reservoir must then satisfy

D Ssurr ³ - 0.8708 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.020 P1=175 P2=150 x1=0.10 x2=0.40 W_in=100 [kJ] "Analysis: " Fluid$='Steam_IAPWS' v1_f=volume(Fluid$,P=P1,x=0) v1_g=volume(Fluid$,P=P1,x=1) s1_f=entropy(Fluid$,P=P1,x=0) s1_g=entropy(Fluid$,P=P1,x=1) s1_fg=s1_g-s1_f v2_f=volume(Fluid$,P=P2,x=0) v2_g=volume(Fluid$,P=P2,x=1) s2_f=entropy(Fluid$,P=P2,x=0) s2_g=entropy(Fluid$,P=P2,x=1) s2_fg=s2_g-s2_f v1=volume(Fluid$,P=P1,x=x1) s1=entropy(Fluid$,P=P1,x=x1) v2=volume(Fluid$,P=P2,x=x2) s2=entropy(Fluid$,P=P2,x=x2) s_out=s2_g m1=Vol/v1 m2=Vol/v2 DELTAS_surr+m2*s2-m1*s1-s_out*(m2-m1)=0

6-154

7-61E An insulated rigid can initially contains R-134a at a specified state. A crack develops, and refrigerant escapes slowly. The final mass in the can is to be determined when the pressure inside drops to a specified value.

Assumptions 1. The can is well-insulated and thus heat transfer is negligible. 2. The refrigerant that remains in the can underwent a reversible adiabatic process. Analysis Noting that for a reversible adiabatic (i.e., isentropic) process, s1  s2 , the properties of the refrigerant in the can are (Tables A-11E through A-13E)

P1 = 90 psia ïü ïý s @ s f @ 30 ° F = 0.04750 Btu/lbm ×R ïïþ 1 T1 = 30°F

P2 = 20 psia ïüï ý ïïþ s2 = s1

x2 =

s2 - s f s fg

0.04750 - 0.02603 = 0.1075 0.1997

v 2 = v f + x2v fg = 0.01181 + (0.1075)(2.2781- 0.01181) = 0.2555 ft 3 /lbm

Thus the final mass of the refrigerant in the can is

V 0.55 ft 3 = = 2.15 lbm v 2 0.2555 ft 3 /lbm

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.55 [ft^3] P1=90 [psia] T1=30 [F] P2=20 [psia] "Properties" Fluid$='R134a' s1=entropy(Fluid$, T=T1, x=0) "use saturated liquid state at given temperature" s2=s1 "reversible-adiabatic (i.e., isentropic) process" v2=volume(Fluid$, P=P2, s=s2) x2=quality(Fluid$, P=P2, s=s2) v2_f=volume(Fluid$, P=P2, x=0) v2_g=volume(Fluid$, P=P2, x=1) s2_f=entropy(Fluid$, P=P2, x=0) s2_g=entropy(Fluid$, P=P2, x=1) s2_fg=s2_g-s2_f "Analysis" m=Vol/v2

7-62 The total heat transfer for the process 1-3 shown in the figure is to be determined. Analysis For a reversible process, the area under the process line in T-s diagram is equal to the heat transfer during that process. Then, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-155

Q1-3 = Q1-2 + Q2-3 2

= ò TdS + ò TdS 1

T1 + T2 ( S 2 - S1 ) + T2 ( S3 - S 2 ) 2

(360 + 273) + (55 + 273) K (3 - 1)kJ/K + (360 + 273 K)(2 - 3)kJ/K 2 = 328 kJ =

EES (Engineering Equation Solver) SOLUTION "Given" T1=(55+273) [K] T2=(360+273) [K] T3=T2 S1=1 [kJ/K] S2=3 [kJ/K] S3=2 [kJ/K] "Analysis" Q_12=(T1+T2)/2*(S2-S1) Q_23=T2*(S3-S2) Q_13=Q_12+Q_23

7-63 The total heat transfer for the process 1-2 shown in the figure is to be determined. Analysis For a reversible process, the area under the process line in T-s diagram is equal to the heat transfer during that process. Then,

Q1-2 = ò TdS 1

6-156

T1 + T2 ( S2 - S1 ) 2

(500 + 273) + (100 + 273) K (1.0 - 0.2)kJ/K 2 = 458 kJ =

EES (Engineering Equation Solver) SOLUTION "Given" T1=(100+273) [K] T2=(500+273) [K] S1=0.2 [kJ/K] S2=1 [kJ/K] "Analysis" Q_12=(T1+T2)/2*(S2-S1)

7-64 The heat transfer for the process 1-3 shown in the figure is to be determined. Analysis For a reversible process, the area under the process line in T-s diagram is equal to the heat transfer during that process. Then,

q1-3 = ò Tds + ò Tds 1

T1 + T2 ( s2 - s1 ) + 0 2

(120 + 273) + (30 + 273) K (1.0 - 0.02)kJ/kg ×K 2

= 341 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=(120+273) [K] T2=(30+273) [K] T3=(100+273) [K] s1=0.02 [kJ/kg-K] s2=1 [kJ/kg-K] s3=1 [kJ/kg-K] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-157

q_12=(T1+T2)/2*(s2-s1) q_23=0 q_13=q_12+q_23

Entropy Change of Incompressible Substances 7-65C No, because entropy is not a conserved property.

7-66 An adiabatic pump is used to compress saturated liquid water in a reversible manner. The work input is to be determined by different approaches. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Heat transfer to or from the fluid is negligible. Analysis The properties of water at the inlet and exit of the pump are (Tables A-4 through A-6) h1 = 191.81 kJ/kg P1 = 10 kPa ïü ïý s = 0.6492 kJ/kg ïïþ 1 x1 = 0 v1 = 0.001010 m3 /kg

P2 = 15 MPa üïï h2 = 206.90 kJ/kg ý ïïþ v 2 = 0.001004 m3 /kg s2 = s1

(a) Using the entropy data from the compressed liquid water table wP = h2 - h1 = 206.90 - 191.81 = 15.10 kJ / kg

(b) Using inlet specific volume and pressure values wP = v1 ( P2 - P1 ) = (0.001010 m3 /kg)(15,000 - 10)kPa = 15.14 kJ / kg

Error = 0.3% (c) Using average specific volume and pressure values

wP = v avg ( P2 - P1 ) = éëê1/ 2(0.001010 + 0.001004) m3 /kg ùûú(15,000 - 10)kPa = 15.10 kJ / kg Error = 0% Discussion The results show that any of the method may be used to calculate reversible pump work.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=10 [kPa] x_1=0 P_2=15000 [kPa] "ANALYSIS" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-158

Fluid$='Steam_iapws' "(a)" h_1=enthalpy(Fluid$, P=P_1, x=x_1) s_1=entropy(Fluid$, P=P_1, x=x_1) s_2=s_1 "isentropic process" h_2=enthalpy(Fluid$, P=P_2, s=s_2) w_p_a=h_2-h_1 "(b)" v_1=volume(Fluid$, P=P_1, x=x_1) w_p_b=v_1*(P_2-P_1) "(c)" v_2=volume(Fluid$, P=P_2, s=s_2) v_ave=1/2*(v_1+v_2) w_p_c=v_ave*(P_2-P_1) Error_b=(w_p_b-w_p_a)/w_p_a*Convert(, %) Error_c=(w_p_c-w_p_a)/w_p_a*Convert(, %)

7-67 Computer chips are cooled by placing them in saturated liquid R-134a. The entropy changes of the chips, R-134a, and the entire system are to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved. 3. There is no heat transfer between the system and the surroundings. Analysis (a) The energy balance for this system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = D U = [m(u2 - u1 )]chips + [m(u2 - u1 ) ]R-134a

[m(u1 - u2 )]chips = [m(u2 - u1 )]R-134a The heat released by the chips is

Qchips = mc(T1 - T2 ) = (0.010 kg)(0.3 kJ/kg ×K) [20 - (- 40)] K = 0.18 kJ The mass of the refrigerant vaporized during this heat exchange process is

mg, 2 =

QR - 134a QR- 134a 0.18 kJ = = = 0.0007970 kg hg - h f h fg @- 40°C 225.86 kJ/kg

Only a small fraction of R-134a is vaporized during the process. Therefore, the temperature of R-134a remains constant during the process. The change in the entropy of the R-134a is (at -11) D SR - 134a = mg , 2 sg , 2 + m f , 2 s f , 2 - m f , 1 s f , 1

= (0.0007970)(0.96866) + (0.005 - 0.0007970)(0) - (0.005)(0) = 0.000772 kJ / K

(b) The entropy change of the chips is D Schips = mc ln

T2 (- 40 + 273)K = (0.010 kg)(0.3 kJ/kg ×K)ln = - 0.000687 kJ / K T1 (20 + 273)K

6-159

The positive result for the total entropy change (i.e., entropy generation) indicates that this process is possible.

EES (Engineering Equation Solver) SOLUTION "Given" m_chips=0.010 [kg] c_p_chips=0.3 [kJ/kg-K] T1=20 [C] m_R=0.005 [kg] T2=-40 [C] "Analysis" Fluid$='R134a' "(a)" Q_chips=m_chips*c_p_chips*(T1-T2) Q_R=Q_chips u_f=intenergy(Fluid$, T=T2, x=0) u_g=intenergy(Fluid$, T=T2, x=1) u_fg=u_g-u_f m_g2=Q_R/u_fg s_g2=entropy(Fluid$, T=T2, x=1) s_f2=entropy(Fluid$, T=T2, x=0) s_f1=s_f2 m_f2=m_R-m_g2 m_f1=m_R DELTAS_R=m_g2*s_g2+m_f2*s_f2-m_f1*s_f1 "(b)" DELTAS_chips=m_chips*c_p_chips*ln((T2+273)/(T1+273)) "(c)" DELTAS_total=DELTAS_R+DELTAS_chips

7-68 A hot iron block is dropped into water in an insulated tank. The total entropy change during this process is to be determined. Assumptions 1. Both the water and the iron block are incompressible substances with constant specific heats at room temperature. 2. The system is stationary and thus the kinetic and potential energies are negligible. 3. The tank is well-insulated and thus there is no heat transfer. 4. The water that evaporates, condenses back. Properties The specific heat of water at 25C is c p  4.18 kJ / kg  C . The specific heat of iron at room temperature is c p  0.45 kJ / kg  C (Table A-3). Analysis We take the entire contents of the tank, water + iron block, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = DU

6-160

or,

D Uiron + D U water = 0 [mc(T2 - T1 )]iron + [mc(T2 - T1 )]water = 0

Substituting,

(25 kg)(0.45 kJ/kg ×K)(T2 - 280°C) + (100 kg)(4.18 kJ/kg ×K)(T2 - 18°C) = 0 ® T2 = 24.9°C = 297.9 K

The entropy generated during this process is determined from

æT ö æ297.9 K ÷ ö ÷= (25 kg )(0.45 kJ/kg ×K ) ln çç D Siron = mcavg ln ççç 2 ÷ = - 6.961 kJ/K ÷ ÷ ÷ ç ÷ è 553 K ø èç T1 ø æT ö æ297.9 K ö ÷= 9.749 kJ/K ÷ D S water = mcavg ln ççç 2 ÷ = (100 kg)(4.18 kJ/kg ×K ) ln çç ÷ ÷ çè 291 K ÷ ø èç T1 ÷ ø Thus, Sgen = D Stotal = D Siron + D S water = - 6.961 + 9.749 = 2.79 kJ / K

Discussion The results can be improved somewhat by using specific heats at average temperature.

EES (Engineering Equation Solver) SOLUTION "Given" m_iron=25 [kg] T1_iron=(280+273) [K] m_w=100 [kg] T1_w=(18+273) [K] "Properties" c_w=4.18 [kJ/kg-K] "Table A-3" c_iron=0.45 [kJ/kg-K] "Table A-3" "Analysis" m_w*c_w*(T2-T1_w)+m_iron*c_iron*(T2-T1_iron)=0 "energy balance" DELTAS_iron=m_iron*c_iron*ln(T2/T1_iron) T2_C=T2-273 DELTAS_w=m_w*c_w*ln(T2/T1_w) DELTAS_total=DELTAS_iron+DELTAS_w

7-69 An aluminum block is brought into contact with an iron block in an insulated enclosure. The final equilibrium temperature and the total entropy change for this process are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-161

1. 2. 3.

Both the aluminum and the iron block are incompressible substances with constant specific heats. The system is stationary and thus the kinetic and potential energies are negligible. The system is well-insulated and thus there is no heat transfer.

Properties The specific heat of aluminum at the anticipated average temperature of 400 K is c p  0.949 kJ / kg  C . The specific heat of iron at room temperature (the only value available in the tables) is c p  0.45 kJ / kg  C (Table A-3). Analysis We take the iron+aluminum blocks as the system, which is a closed system. The energy balance for this system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = DU

or, D U alum + D Uiron = 0

[mc(T2 - T1 )]alum + [mc(T2 - T1 )]iron = 0 Substituting, (30 kg)(0.949 kJ/kg ×K)(T2 - 140°C) + (40 kg)(0.45 kJ/kg ×K)(T2 - 60°C) = 0

T2 = 109°C = 382 K The total entropy change for this process is determined from

æT ö æ382 K ö÷ ÷ D Siron = mcavg ln ççç 2 ÷ = (40 kg )(0.45 kJ/kg ×K ) ln çç ÷= 2.472 kJ/K ÷ çè T1 ÷ èç333 K ÷ ø ø æT ö æ382 K ö÷ ÷ D Salum = mcavg ln ççç 2 ÷ = (30 kg )(0.949 kJ/kg ×K ) ln çç ÷= - 2.221 kJ/K ÷ çè T1 ø÷ èç 413 K ø÷ Thus,

D Stotal = D Siron + D Salum = 2.472 - 2.221 = 0.251 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" m_Al=30 [kg] T1_Al=(140+273) [K] m_iron=40 [kg] T1_iron=(60+273) [K] "Properties" c_Al=0.949 [kJ/kg-K] "at 450 K, Table A-3" c_iron=0.45 [kJ/kg-K] "at room temperature, Table A-3" "Analysis" m_Al*c_Al*(T2-T1_Al)+m_iron*c_iron*(T2-T1_iron)=0 "energy balance" T2_C=T2-273 DELTAS_iron=m_iron*c_iron*ln(T2/T1_iron) DELTAS_Al=m_Al*c_Al*ln(T2/T1_Al) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-162

DELTAS_total=DELTAS_iron+DELTAS_Al

7-70 Problem 7-69 is reconsidered. The effect of the mass of the iron block on the final equilibrium temperature and the total entropy change for the process is to be studied. The mass of the iron is to vary from 10 to 100 kg. The equilibrium temperature and the total entropy change are to be plotted as a function of iron mass. Analysis The problem is solved using EES, and the results are tabulated and plotted below. T_1_iron = 60 [C] m_iron = 40 [kg] T_1_al = 140 [C] m_al = 30 [kg] c_al = 0.949 [kJ/kg-K] "FromTable A-3 at the anticipated average temperature of 450 K." c_iron= 0.45 [kJ/kg-K] "FromTable A-3 at room temperature, the only value available." E_in - E_out = DELTAE_sys E_out = 0 E_in = 0 DELTAE_sys = m_iron*DELTAu_iron + m_al*DELTAu_al DELTAu_iron = c_iron*(T_2_iron - T_1_iron) DELTAu_al = c_al*(T_2_al - T_1_al) T_2_iron = T_2 T_2_al = T_2 DELTAS_iron = m_iron*c_iron*ln((T_2_iron+273) / (T_1_iron+273)) DELTAS_al = m_al*c_al*ln((T_2_al+273) / (T_1_al+273)) DELTAS_total = DELTAS_iron + DELTAS_al

Stotal kJ / kg 

0.08547 0.1525 0.2066 0.2511 0.2883 0.32 0.3472 0.3709 0.3916 0.41

miron [kg] 10 20 30 40 50 60 70 80 90 100

T2 [C] 129.1 120.8 114.3 109 104.7 101.1 97.98 95.33 93.02 91

6-163

7-71 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the total entropy change are to be determined.

Assumptions 1. Both the water and the copper block are incompressible substances with constant specific heats at room temperature. 2. The system is stationary and thus the kinetic and potential energies are negligible. 3. The tank is well-insulated and thus there is no heat transfer. Properties The density and specific heat of water at 25C are   997 kg / m3 and c p  4.18 kJ / kg. C . The specific heat of copper at 27C is c p  0.386 kJ / kg  C (Table A-3). Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = DU

or,

D U Cu + D U water = 0 [mc(T2 - T1 )]Cu + [mc(T2 - T1 )]water = 0 where mwater = r V = (997 kg/m3 )(0.090 m3 ) = 89.73 kg

Using specific heat values for copper and liquid water at room temperature and substituting, (50 kg)(0.386 kJ/kg ×°C)(T2 - 140)°C + (89.73 kg)(4.18 kJ/kg ×°C)(T2 - 10)°C = 0

6-164

T2  16.4 C  289.4 K The entropy generated during this process is determined from

æT ö ö ÷= (50 kg )(0.386 kJ/kg ×K ) ln æ çç 289.4 K ÷ D Scopper = mcavg ln ççç 2 ÷ ÷ ÷= - 6.864 kJ/K ç ÷ çè T1 ÷ è 413 K ø ø

æT ÷ ö æ289.4 K ÷ ö D S water = mcavg ln ççç 2 ÷ = (89.73 kg )(4.18 kJ/kg ×K ) ln çç ÷= 8.388 kJ/K ÷ ÷ çè T1 ø èç 283 K ÷ ø Thus, D Stotal = D Scopper + D Swater = - 6.864 + 8.388 = 1.52 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" m_Cu=50 [kg] T1_Cu=(140+273) [K] Vol_w=0.090 [m^3] T1_w=(10+273) [K] "Properties" rho_w=997 [kg/m^3] "Table A-3" c_w=4.18 [kJ/kg-C] "Table A-3" c_Cu=0.386 [kJ/kg-C] "Table A-3" "Analysis" m_w=rho_w*Vol_w m_w*c_w*(T2-T1_w)+m_Cu*c_Cu*(T2-T1_Cu)=0 "energy balance" T2_C=T2-273 DELTAS_Cu=m_Cu*c_Cu*ln(T2/T1_Cu) DELTAS_w=m_w*c_w*ln(T2/T1_w) DELTAS_total=DELTAS_Cu+DELTAS_w

7-72 An iron block and a copper block are dropped into a large lake. The total amount of entropy change when both blocks cool to the lake temperature is to be determined. Assumptions 1. The water, the iron block and the copper block are incompressible substances with constant specific heats at room temperature. 2. Kinetic and potential energies are negligible. Properties The specific heats of iron and copper at room temperature are ciron  0.45 kJ / kg. C and ccopper  0.386 kJ / kg. C (Table A-3). Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature (15C) when the thermal equilibrium is established. Then the entropy changes of the blocks become

æT ö æ288 K ÷ö D Siron = mcavg ln ççç 2 ÷÷÷= (30 kg )(0.45 kJ/kg ×K )ln çç = - 2.746 kJ/K çè353 K ÷÷ø èç T1 ø÷ æT ö æ288 K ÷ ö ÷= (40 kg )(0.386 kJ/kg ×K )ln çç D Scopper = mcavg ln ççç 2 ÷ = - 3.141 kJ/K ÷ ÷ ÷ ç ÷ è353 K ø èç T1 ø We take both the iron and the copper blocks, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance for this system can be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-165

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = D U iron + D U copper

or, Qout = [mc(T1 - T2 )]iron + [mc(T1 - T2 )]copper

Substituting,

Qout = (30 kg)(0.45 kJ/kg ×K )(353 - 288)K + (40 kg)(0.386 kJ/kg ×K )(353 - 288)K = 1881 kJ Thus, D Slake =

Qlake, in Tlake

1881 kJ = 6.528 kJ/K 288 K

Then the total entropy change for this process is D Stotal = D Siron + D Scopper + D Slake = (- 2.746) + (- 3.141) + 6.528 = 0.642 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" m_iron=30 [kg] m_Cu=40 [kg] T1=ConvertTemp(C, K, 80[C]) T_lake=ConvertTemp(C, K, 15[C]) "Properties" c_Cu=0.386 [kJ/kg-K] "Table A-3" c_iron=0.45 [kJ/kg-K] "Table A-3" "Analysis" T2=T_lake "since the thermal energy capacity of the lake is very large" -Q_out=m_Cu*c_Cu*(T2-T1)+m_iron*c_iron*(T2-T1) "energy balance" DELTAS_iron=m_iron*c_iron*ln(T2/T1) DELTAS_Cu=m_Cu*c_Cu*ln(T2/T1) DELTAS_lake=Q_lake_in/T_lake Q_lake_in=Q_out DELTAS_total=DELTAS_iron+DELTAS_Cu+DELTAS_lake

6-166

Entropy Changes of Ideal Gases 7-73C The Pr and vr are called relative pressure and relative specific volume, respectively. They are derived for isentropic processes of ideal gases, and thus their use is limited to isentropic processes only.

7-74C No. The entropy of an ideal gas depends on the pressure as well as the temperature.

7-75C The entropy of a gas can change during an isothermal process since entropy of an ideal gas depends on the pressure as well as the temperature.

7-76C The entropy change relations of an ideal gas simplify to s  c p ln T2 / T1  for a constant pressure process and

s  cv ln T2 / T1  for a constant volume process.

Noting that c p  cv , the entropy change will be larger for a constant pressure process.

7-77 The entropy difference between the two states of oxygen is to be determined. Assumptions Oxygen is an ideal gas with constant specific heats. Properties The specific heat of oxygen at the average temperature of (39  337) / 2  188 C  461 K is c p  0.960 kJ / kg  K (Table A-2b). Analysis From the entropy change relation of an ideal gas, D soxygen = c p ln

T2 P (337 + 273)K - R ln 2 = (0.960 kJ/kg ×K)ln - 0 = 0.6436 kJ / kg ×K T1 P1 (39 + 273)K

since the pressure is same at the initial and final states.

EES (Engineering Equation Solver) SOLUTION T1=(39+273) [K] T2=(337+273) [K] c_p=0.960 [kJ/kg-K] "at 188 C = 461 K for oxygen" DELTAs=c_p*ln(T2/T1)

7-78E The entropy change of air during an expansion process is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The specific heat of air at the average temperature of (500  50) / 2  275 F is c p  0.243 Btu / lbm  R (Table A-2Eb). The gas constant of air is R  0.06855 Btu / lbm  R (Table A-2Ea). Analysis From the entropy change relation of an ideal gas, T P D sair = c p ln 2 - R ln 2 T1 P1 = (0.243 Btu/lbm ×R)ln

(50 + 460)R 100 psia - (0.06855 Btu/lbm ×R)ln (500 + 460)R 200 psia

6-167

= - 0.1062 Btu / lbm ×R EES (Engineering Equation Solver) SOLUTION T1=(500+460) [R] T2=(50+460) [R] P1=200 [psia] P2=100 [psia] R=0.06855 [Btu/lbm-R] c_p=0.243 [Btu/lbm-R] "at 275 F for air" DELTAs=c_p*ln(T2/T1)-R*ln(P2/P1)

7-79 The final temperature of air when it is expanded isentropically is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air at an anticipated average temperature of 550 K is k = 1.381 (Table A-2b). Analysis From the isentropic relation of an ideal gas under constant specific heat assumption, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P ø÷ 1

0.381/1.381

æ100 kPa ÷ ö = (477 + 273 K) çç ÷ çè1000 kPa ÷ ø

= 397 K

Discussion The average air temperature is (750  397.4) / 2  573.7 K , which is sufficiently close to the assumed average temperature of 550 K.

EES (Engineering Equation Solver) SOLUTION T1=(477+273) [K] P1=1000 [kPa] P2=100 [kPa] k=1.381 "at 550 K fo air" T2=T1*(P2/P1)^((k-1)/k)

7-80E The final temperature of air when it is expanded isentropically is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air at an anticipated average temperature of 300F is k = 1.394 (Table A-2Eb). Analysis From the isentropic relation of an ideal gas under constant specific heat assumption, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P1 ÷ ø

0.394/1.394

æ20 psia ÷ ö ÷ = (500 + 460 R) ççç ÷ è100 psia ø÷

= 609 R

Discussion The average air temperature is (960  609) / 2  785 R  325 F , which is sufficiently close to the assumed average temperature of 300F.

EES (Engineering Equation Solver) SOLUTION T1=(500+460) [R] P1=100 [psia] P2=20 [psia] k=1.394 "at 300 F fo air" T2=T1*(P2/P1)^((k-1)/k) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-168

7-81 The final temperatures of helium and nitrogen when they are compressed isentropically are to be compared. Assumptions Helium and nitrogen are ideal gases with constant specific heats. Properties The specific heat ratios of helium and nitrogen at room temperature are k = 1.667 and k = 1.4, respectively (Table A-2a). Analysis From the isentropic relation of an ideal gas under constant specific heat assumption, ( k - 1)/ k

æP ö ÷ T2,He = T1 ççç 2 ÷ ÷ çè P ø÷ 1

( k - 1)/ k

æP ö ÷ T2,N2 = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

0.667/1.667

æ1000 kPa ÷ ö = (298 K) çç ÷ èç 100 kPa ÷ ø

= 749 K

0.4/1.4

æ1000 kPa ö÷ = (298 K) çç ÷ èç 100 kPa ø÷

= 575 K

Hence, the helium produces the greater temperature when it is compressed.

EES (Engineering Equation Solver) SOLUTION T1=(25+273) [K] P1=100 [kPa] P2=1000 [kPa] k_He=1.667 k_N2=1.4 T2_He=T1*(P2/P1)^((k_He-1)/k_He) T2_N2=T1*(P2/P1)^((k_N2-1)/k_N2)

7-82 The final temperatures of neon and air when they are expanded isentropically are to be compared. Assumptions Neon and air are ideal gases with constant specific heats. Properties The specific heat ratios of neon and air at room temperature are k = 1.667 and k = 1.4, respectively (Table A-2a). Analysis From the isentropic relation of an ideal gas under constant specific heat assumption, ( k - 1)/ k

æP ö ÷ T2,Ne = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

( k - 1)/ k

æP ö ÷ T2,air = T1 ççç 2 ÷ ÷ èç P1 ÷ ø

0.667/1.667

æ100 kPa ÷ ö = (773 K) çç ÷ ÷ èç1000 kPa ø

= 308 K

0.4/1.4

æ100 kPa ö÷ = (773 K) çç ÷ èç1000 kPa ø÷

= 400 K

Hence, the neon produces the smaller temperature when it is expanded.

EES (Engineering Equation Solver) SOLUTION T1=(500+273) [K] P1=1000 [kPa] P2=100 [kPa] k_Ne=1.667 k_air=1.4 T2_Ne=T1*(P2/P1)^((k_Ne-1)/k_Ne) T2_air=T1*(P2/P1)^((k_air-1)/k_air)

7-83 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tank stirs the gas, and the pressure and temperature of CO2 rises. The entropy change of CO2 during this process is to be determined using constant specific heats. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-169

Assumptions At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at room temperature. Properties The specific heat of CO2 is cv  0.657 kJ / kg.K (Table A-2). Analysis Using the ideal gas relation, the entropy change is determined to be

P2V PV T P 150 kPa = 1 ¾¾ ® 2 = 2 = = 1.5 T2 T1 T1 P1 100 kPa

Thus, ¿0 æ T V ö÷ T D S = m (s2 - s1 ) = m çççcv , avg ln 2 + R ln 2 ÷ = mcv , avg ln 2 ÷ çè T1 V1 ø÷ T1

= (2.7 kg )(0.657 kJ/kg ×K ) ln (1.5)

= 0.719 kJ / K EES (Engineering Equation Solver) SOLUTION "Given" Vol=1.5 [m^3] m=2.7 [kg] P1=100 [kPa] P2=150 [kPa] "Properties" c_v=0.657 [kJ/kg-K] "Analysis" RatioT=P2/P1 "from ideal gas relation, RatioT=T2/T1" DELTAS=m*c_v*ln(RatioT) "since V2=V1"

7-84 A tank initially contains air at 100 kPa and 305 K. The electrical resistance heater in the tank is turned on until the air is at 375 kPa and 600 K. Accounting for the variation of specific heats with temperature, determine the entropy change of the air. 7-84 Air in a tank is heated by a electric heater. The entropy change of air is to be determined. Assumptions Air is an ideal gas with variable specific heats. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-2a). Analysis Since the pressure information is available, we use the entropy change relation involving the entropy term s  . From Table A-17: T1 = 305 K ¾ ¾ ® s1o = 1.71865 kJ/kg ×K

T2 = 600 K ¾ ¾ ® s2o = 2.40902 kJ/kg ×K

6-170

Then,

s2  s1  s2o  s1o  R ln

P2 P1

 375 kPa   (2.40902  1.71865) kJ/kg  K  (0.287 kJ/kg  K) ln    100 kPa 

 0.311 kJ / kg  K

7-85 A piston-cylinder device initially contains air at 100 kPa and 305 K. The gas is now compressed isentropically until its pressure rises to 375 kPa. Accounting for the variation of specific heats with temperature, determine the final air temperature. 7-85 Air in a piston-cylinder device is compressed isentropically. The final air temperature is to be determined. Assumptions Air is an ideal gas with variable specific heats. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-2a). Analysis Since the pressure information is available, we use the isentropic relation involving the entropy term s  . From Table A-17: T1 = 305 K ¾ ¾ ® s1o = 1.71865 kJ/kg ×K

The entropy term at the final state for the isentropic process is s2o  s1o  R ln

P2  375 kPa   1.71865 kJ/kg  K  (0.287 kJ/kg  K) ln    2.0980 kJ/kg  K P1  100 kPa 

The corresponding temperature from Table A-17 is s2o  2.0980 kJ/kg  K   T2  444.1K

7-86 A piston-cylinder device initially contains air at 12°C. Heat is transferred and air expands until its temperature rises to 450 K while its volume triples. Accounting for the variation of specific heats with temperature, determine the entropy change of the air. 7-86 Air is expanded in a piston-cylinder device until its volume triples. The entropy change of air is to be determined. Assumptions Air is an ideal gas with variable specific heats. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-2a). Analysis Since the volume ratio is available, we use the entropy change relation involving the entropy term s  . From Table A-17: T1 = 285 K ¾ ¾ ® s1+ = 0.02829 kJ/kg ×K T2 = 450 K ¾ ¾ ® s2+ = 0.35826 kJ/kg ×K

Then,

s2  s1  s2  s1  R ln

V2 V1

 (0.35826  0.02829) kJ/kg  K  (0.287 kJ/kg  K)ln(3)

 0.645 kJ / kg  K

6-171

7-87 Air initially at 450 K in a piston-cylinder device is allowed to expand isentropically until its volume triples. Accounting for the variation of specific heats with temperature, determine the final temperature of the air. 7-87 Air is expanded in a piston-cylinder device until its volume triples. The final temperature of the air is to be determined. Assumptions Air is an ideal gas with variable specific heats. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-2a). Analysis Since the volume ratio is available, we use the isentropic relation involving the entropy term s  . From Table A17: T1 = 450 K ¾ ¾ ®

s1+ = 0.35826 kJ/kg ×K

The entropy term at the final state for the isentropic process is s2  s1+  R ln

V2  0.35826 kJ/kg  K  (0.287 kJ/kg  K) ln(3)  0.04296 kJ/kg  K V1

The corresponding temperature from Table A-17 is s2  0.04296 kJ/kg  K   T2  290.9 K

7-88 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinder is turned on, and air is heated for 15 min at constant pressure. The entropy change of air during this process is to be determined for the cases of constant and variable specific heats. Assumptions At specified conditions, air can be treated as an ideal gas. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-1). Analysis The mass of the air and the electrical work done during this process are

(120 kPa )(0.3 m ) P1V1 = = 0.4325 kg RT1 (0.287 kPa ×m 3 /kg ×K )(290 K ) 3

We,in = We,in D t = (0.2 kJ/s)(15´ 60 s) = 180 kJ

The energy balance for this stationary closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We, in - Wb, out = D U ¾ ¾ ® We, in = m(h2 - h1 ) @ c p (T2 - T1 )

since U  Wb  H during a constant pressure quasi-equilibrium process. (a) Using a constant c p value at the anticipated average temperature of 450 K, the final temperature becomes © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-172

Thus, T2 = T1 +

We, in mc p

= 290 K +

180 kJ

(0.4325 kg)(1.02 kJ/kg ×K )

= 698.0 K

Then the entropy change becomes ¿0 æ T2 P2 ö÷ T ç ÷÷= mc p , avg ln 2 ç D Ssys = m (s2 - s1 ) = m çc p , avg ln - R ln ÷ T1 P1 ø T1 èç

æ698 K ÷ ö = (0.4325 kg )(1.020 kJ/kg ×K ) ln çç = 0.387 kJ / K ÷ ÷ çè 290 K ø (b) The final temperature for variable specific heats case is determined as follows: T1 = 290 K ¾ ¾ ® h1 = 290.16 kJ/kg (Table A-17) We, in = m(h2 - h1 ) 180 kJ = (0.4325 kg)(h2 - 290.16)kJ/kg h2 = 706.34 kJ/kg h2 = 706.34 kJ/kg ¾ ¾ ® T2 = 693.6 K

(Table A-17)

Assuming variable specific heats using the entropy term so, and reading values from Table A-17: T1 = 290 K ¾ ¾ ® s1o = 1.66802 kJ/kg ×K T2 = 693.6 K ¾ ¾ ® s2o = 2.56288 kJ/kg ×K ¿0 æ P ö o o ÷ D Ssys = m çççs2o - s1o - R ln 2 ÷ ÷= m (s2 - s1 ) = (0.4325 kg )(2.56288 - 1.66802)kJ/kg ×K = 0.387 kJ / K çè P1 ø÷

(c) Assuming variable specific heats using the entropy term s  , and reading values from Table A-17: T1 = 290 K ¾ ¾ ® s1+ = 0.04076 kJ/kg ×K T2 = 693.6 K ¾ ¾ ® s2+ = 0.68536 kJ/kg ×K

V2 =

mRT2 (0.4325 kg)(0.287 kPa ×m3 /kg ×K)(698 K) = = 0.7220 m3 P2 120 kPa

 V  S sys  m  s2  s1  R ln 2  V1     0.7220 m3    (0.4325 kg) (0.68536  0.04076) kJ/kg  K  (0.287 kJ/kg  K)ln  3    0.3 m     0.388 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.300 [m^3] P1=120 [kPa] T1=(17+273) [K] W_dot_e_in=0.200 [kW] time=15*60 [s] P2=P1 "Properties" R=0.287 [kJ/kg-K] "Table A-2a" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-173

c_p=1.02 [kJ/kg-K] "at 450 K, Table A-2b" "Analysis" m=(P1*Vol1)/(R*T1) W_e_in=W_dot_e_in*time "(a)" W_e_in=m*c_p*(T2-T1) "energy balance, since U2-U1+W_b = H2-H1 for a constant pressure process" DELTAS_a=m*(c_p*ln(T2/T1)-R*ln(P2/P1)) "(b)" Fluid$='air' h1=enthalpy(Fluid$, T=T1) W_e_in=m*(h2-h1) "energy balance" s1=entropy(Fluid$, T=T1, P=P1) s2=entropy(Fluid$, h=h2, P=P2) DELTAS_b=m*(s2-s1)

7-89 A cylinder contains N 2 gas at a specified pressure and temperature. The gas is compressed polytropically until the volume is reduced by half. The entropy change of nitrogen during this process is to be determined using constant specific heats. Assumption 1.

At specified conditions, N 2 can be treated as an ideal gas.

Nitrogen has constant specific heats at room temperature.

Properties The gas constant of nitrogen is R  0.2968 kJ / kg.K (Table A-1). The constant volume specific heat of nitrogen at room temperature is cv  0.743 kJ / kg.K (Table A-2).

Analysis From the polytropic relation, n- 1

T2 æ v ö ÷ = ççç 1 ÷ ÷ T1 èçv 2 ø÷

n- 1

æv ö ÷ ¾¾ ® T2 = T1 ççç 1 ÷ ÷ èçv 2 ø÷

1.3- 1

= (310 K )(2)

= 381.7 K

Then the entropy change of nitrogen becomes

æ T Vö D S N2 = m çççcv , avg ln 2 + R ln 2 ÷÷÷ çè T1 V1 ÷ø æ ö 381.7 K ÷ = (0.75 kg )çç(0.743 kJ/kg ×K ) ln + (0.2968 kJ/kg ×K ) ln (0.5)÷ ÷ çè ø 310 K

= - 0.0385 kJ / K EES (Engineering Equation Solver) SOLUTION "Given" m=0.75 [kg] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-174

P1=140 [kPa] T1=(37+273) [K] n=1.3 RatioV=0.5 "RatioV=V2/V1" "Properties" c_v=0.743 [kJ/kg-K] "Table A-2a" R=0.2968 [kJ/kg-K] "Table A-2a" "Analysis" T2=T1*(1/RatioV)^(n-1) "from polytropic relation" DELTAS=m*(c_v*ln(T2/T1)+R*ln(RatioV))

7-90 Problem 7-89 is reconsidered. The effect of varying the polytropic exponent from 1 to 1.4 on the entropy change of the nitrogen is to be investigated, and the processes are to be shown on a common P-v diagram. Analysis The problem is solved using EES, and the results are tabulated and plotted below. m=0.75 [kg] P1=140 [kPa] T1=(37+273) [K] n=1.3 RatioV=0.5 "RatioV=V2/V1" c_v=0.743 [kJ/kg-K] R=0.297 [kJ/kg-K] T2=T1*(1/RatioV)^(n–1) "from polytropic relation" DELTAS=m*(c_v*ln(T2/T1)+R*ln(RatioV)) P1*V1=m*R*T1 n

S [kJ / kg]

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4

–0.1544 –0.1351 –0.1158 –0.09646 –0.07715 –0.05783 –0.03852 –0.01921 0.000104

6-175

7-91E A fixed mass of helium undergoes a process from one specified state to another specified state. The entropy change of helium is to be determined for the cases of reversible and irreversible processes. Assumptions 1. At specified conditions, helium can be treated as an ideal gas. 2. Helium has constant specific heats at room temperature. Properties The gas constant of helium is R  0.4961 Btu / lbm  R (Table A-1E). The constant volume specific heat of helium is cv  0.753 Btu / lbm  R (Table A-2E).

Analysis From the ideal-gas entropy change relation,

æ T v ö ÷ D SHe = m çççcv , ave ln 2 + R ln 2 ÷ ÷ çè T1 v1 ÷ ø æ ö æ20 ft 3 /lbm ö 700 R ÷÷ ÷ ÷ = (25 lbm) ççç(0.753 Btu/lbm ×R) ln + (0.4961 Btu/lbm ×R ) ln ççç ÷ ÷ 3 520 R è50 ft /lbm ÷ ø÷ èç ø = - 5.77 Btu / R The entropy change will be the same for both cases.

EES (Engineering Equation Solver) SOLUTION "Given" m=25 [lbm] v1=50 [ft^3/lbm] T1=(60+460) [R] v2=20 [ft^3/lbm] T2=(240+460) [R] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-176

"Properties" c_v=0.753 [Btu/lbm-R] "Table A-2a" R=0.4961 [Btu/lbm-R] "Table A-2a" "Analysis" DELTAS=m*(c_v*ln(T2/T1)+R*ln(v2/v1))

7-92 Air is expanded in a piston-cylinder device isothermally until a final pressure. The amount of heat transfer is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The specific heat of air at the given temperature of c p  1.013 kJ / kg  K (Table A-2b). The gas constant of air is R  0.287 kJ / kg.K (Table A-2a). Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - Wout = D U = m(u2 - u1 )

(since KE = PE = 0)

Qin - Wout = D U = m(u2 - u1 ) = 0 since T1 = T1

Qin = Wout The boundary work output during this isothermal process is Wout = mRT ln

P1 200 kPa = (1 kg)(0.287 kJ/kg ×K)(127 + 273 K)ln = 79.6 kJ P2 100 kPa

Thus, Qin = Wout = 79.6 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] P1=200 [kPa] T1=(127+273) [K] T2=T1 P2=100 [kPa] "Properties" c_p=1.013 [kJ/kg-K] "at 127 C = 400 K, Table A-2b" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" W_out=m*R*T1*ln(P1/P2) Q_in=W_out

7-93 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature and pressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. The total entropy change during this process is to be determined. Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations apply. Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-177

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = D U = m(u2 - u1 )

u2 = u1 T2 = T1

since u = u(T) for an ideal gas. Then the entropy change of the gas becomes ¿0 æ T V ÷ö V D S = N çççcv , avg ln 2 + Ru ln 2 ÷÷= NRu ln 2 çè T1 V1 ÷ø V1

= (12 kmol)(8.314 kJ/kmol ×K ) ln (2) = 69.2 kJ / K

This also represents the total entropy change since the tank does not contain anything else, and there are no interactions with the surroundings.

EES (Engineering Equation Solver) SOLUTION "Given" N=12 [kmol] P1=330 [kPa] T1=(50+273) [K] V2\V1=2 "Properties" R_u=8.314 "[kJ/kmol-K]" "Analysis" DELTAS=N*R_u*ln(V2\V1) "since T2=T1 from the energy balance"

7-94 Air is compressed in a piston-cylinder device until a final pressure. The minimum work input is given. The mass of air in the device is to be determined using constant and variable specific heats. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at 300 K is cv  0.718 kJ / kg.K and k = 1.4 (Table A-2a). Analysis (a) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win = D U = m(u2 - u1 )

(since Q = KE = PE = 0)

6-178

For the minimum work input to the compressor, the process must be reversible as well as adiabatic (i.e., isentropic). This being the case, the exit temperature will be ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P ÷ ø 1

0.4/1.4

æ600 kPa ö÷ = (27 + 273 K) çç ÷ çè100 kPa ø÷

= 500.5 K

Substituting into the energy balance equation gives Win = m(u2 - u1 ) = mc p (T2 - T1 ) ¾ ¾ ® m=

Win 1000 kJ = = 6.94 kJ / kg cv (T2 - T1 ) (0.718 kJ/kg ×K)(500.5 - 300)K

(b) Since the pressure information is available, we use the isentropic relation involving the entropy term s  . From Table A-17: T1 = 300 K ¾ ¾ ® u1 = 214.07 kJ/kg, s1o = 1.70203 kJ/kg ×K

The entropy term at the final state for the isentropic process is s2o  s1o  R ln

P2  600 kPa   1.70203 kJ/kg  K  (0.287 kJ/kg  K) ln    2.21626 kJ/kg  K P1  100 kPa 

The corresponding internal energy from Table A-17 is s2o  2.21626 kJ/kg  K   u2  358.33 kJ/kg

Then, Win = m(u2 - u1 ) ¾ ¾ ® m=

Win 1000 kJ = = 6.93 kg u2 - u1 (358.33 - 214.07) kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=(27+273) [K] P1=100 [kPa] W_in=1000 [kJ] P2=600 [kPa] "Properties" c_v=0.718 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" T2=T1*(P2/P1)^((k-1)/k) m=W_in/(c_v*(T2-T1))

7-95 Air is expanded in an adiabatic turbine. The maximum work output is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-179

2. 3.

The process is adiabatic, and thus there is no heat transfer. Air is an ideal gas with constant specific heats.

Properties The properties of air at an anticipated average temperature of 550 K are c p  1.040 kJ / kg  K and k = 1.381 (Table A-2b). Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 = mh2 + Wout Wout = m(h1 - h2 )

wout = h1 - h2 For the minimum work input to the compressor, the process must be reversible as well as adiabatic (i.e., isentropic). This being the case, the exit temperature will be

( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P1 ø÷

0.381/1.381

æ 200 kPa ÷ ö = (500 + 273 K) çç ÷ ÷ èç3500 kPa ø

= 351 K

Substituting into the energy balance equation gives wout = h1 - h2 = c p (T1 - T2 ) = (1.040 kJ/kg ×K)(773 - 351)K = 439 kJ / kg

Discussion The average air temperature is (773  351) / 2  562 K , which is sufficiently close to the assumed average temperature of 550 K.

EES (Engineering Equation Solver) SOLUTION "Given" T1=(500+273) [K] P1=3500 [kPa] P2=200 [kPa] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-180

"at 500 K for air" c_p=1.040 [kJ/kg-K] "Table A-2a" k=1.381 "Table A-2a" "Analysis" T2=T1*(P2/P1)^((k-1)/k) w_out=c_p*(T1-T2)

7-96 Air is compressed in a piston-cylinder device in a reversible and isothermal manner. The entropy change of air and the work done are to be determined. Assumptions 1. At specified conditions, air can be treated as an ideal gas. 2. The process is specified to be reversible. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-1). Analysis (a) Noting that the temperature remains constant, the entropy change of air is determined from

T D Sair = c p , avg ln 2 T1

P P - R ln 2 = - R ln 2 P1 P1

æ600 kPa ÷ ö = - (0.287 kJ/kg ×K )ln çç = - 0.545 kJ / kg ×K ÷ çè 90 kPa ÷ ø Also, for a reversible isothermal process, q = T D s = (293 K )(- 0.545 kJ/kg ×K ) = - 160 kJ/kg ® qout = 160 kJ/kg

(b) The work done during this process is determined from the closed system energy balance,

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win - Qout = D U = mcv (T2 - T1 ) = 0 win = qout = 160 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=90 [kPa] T1=(20+273) [K] P2=600 [kPa] T2=T1 "Properties" "c_v=CV(air, T=T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-181

c_p=CP(air, T=T1) R=c_p-c_v" R=0.287 "Analysis" "(a)" DELTAs=-R*ln(P2/P1) "since T2=T1" "(b)" q_out=-T1*DELTAs w_in-q_out=0 "energy balance, since T2=T1"

7-97 Helium gas is compressed in a piston-cylinder device in a reversible and adiabatic manner. The final temperature and the work are to be determined for the cases of the process taking place in a piston-cylinder device and a steady-flow compressor. Assumptions 1. Helium is an ideal gas with constant specific heats. 2. The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. Properties The specific heats and the specific heat ratio of helium are cv  3.1156 kJ / kg.K , cp  5.1926 kJ / kg.K, and k = 1.667 (Table A-2). Analysis (a) From the ideal gas isentropic relations,

æP2 ö÷( T2 = T1 ççç ÷ ÷ çè P ÷ ø

k - 1) k

0.667 1.667

æ450 kPa ÷ ö = (303 K )çç ÷ çè 90 kPa ø÷

= 576.9 K

(a) We take the air in the cylinder as the system. The energy balance for this stationary closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win = D U = m(u2 - u1 ) @ mcv (T2 - T1 )

Thus, win = cv (T2 - T1 ) = (3.1156 kJ/kg ×K)(576.9 - 303)K = 853.4 kJ / kg

(b) If the process takes place in a steady-flow device, the final temperature will remain the same but the work done should be determined from an energy balance on this steady-flow device,

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Win + mh1 = mh2 Win = m(h2 - h1 ) @ mc p (T2 - T1 )

6-182

win = c p (T2 - T1 ) = (5.1926 kJ/kg ×K)(576.9 - 303)K = 1422.3 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=90 [kPa] T1=30[C]+273 [K] P2=450 [kPa] "Properties" Fluid$='helium' u1=intenergy(Fluid$, T=T1, P=P1) h1=enthalpy(Fluid$, T=T1, P=P1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" u2=intenergy(Fluid$, P=P2, s=s2) h2=enthalpy(Fluid$, P=P2, s=s2) "Analysis" T2=T1*(P2/P1)^((k-1)/k) k=1.667 "(a)" w_in_a=u2-u1 "(b)" w_in_b=h2-h1

7-98 Nitrogen is compressed in an adiabatic compressor. The minimum work input is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. The process is adiabatic, and thus there is no heat transfer. 3. Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at an anticipated average temperature of 400 K are c p  1.044 kJ / kg  K and k = 1.397 (Table A-2b). Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + Win = mh2 Win = m(h2 - h1 )

6-183

For the minimum work input to the compressor, the process must be reversible as well as adiabatic (i.e., isentropic). This being the case, the exit temperature will be

( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P ÷ ø 1

0.397/1.397

æ600 kPa ÷ ö = (303 K) çç ÷ èç120 kPa ÷ ø

= 479 K

Substituting into the energy balance equation gives win = h2 - h1 = c p (T2 - T1 ) = (1.044 kJ/kg ×K)(479 - 303)K = 184 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=(30+273) [K] P1=120 [kPa] P2=600 [kPa] "Properties" "at 400 K for nitrogen" c_p=1.044 [kJ/kg-K] "Table A-2a" k=1.397 "Table A-2a" "Analysis" T2=T1*(P2/P1)^((k-1)/k) w_in=c_p*(T2-T1)

7-99 Air is expanded adiabatically in a piston-cylinder device. The entropy change is to be determined and it is to be discussed if this process is realistic. Assumptions 1. Air is an ideal gas with constant specific heats. Properties The properties of air at 300 K are c p  1.005 kJ / kg  K , cv  0.718 kJ / kg  K and k = 1.4. Also, R  0.287 kJ / kg.K (Table A-2a). Analysis (a) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wout = D U = m(u2 - u1 )

(since Q = KE = PE = 0)

- Wout = mcv (T2 - T1 ) Solving for the final temperature, Wout = mcv (T1 - T2 ) ¾ ¾ ® T2 = T1 -

Wout 600 kJ = (427 + 273 K) = 532.9 K mcv (5 kg)(0.718 kJ/kg ×K)

6-184

From the entropy change relation of an ideal gas,

T P D sair = c p ln 2 - R ln 2 T1 P1 = (1.005 kJ/kg ×K)ln

532.9 K 100 kPa - (0.287 kJ/kg ×K)ln 700 K 600 kPa

= 0.240 kJ / kg×K (b) Since the entropy change is positive for this adiabatic process, the process is irreversible and realistic.

EES (Engineering Equation Solver) SOLUTION "Given" m=5 [kg] T1=(427+273) [K] P1=600 [kPa] P2=100 [kPa] W_out=600 [kJ] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_v=0.718 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" T2=T1-W_out/(m*c_v) DELTAs_air=c_p*ln(T2/T1)-R*ln(P2/P1)

7-100 A container filled with liquid water is placed in a room and heat transfer takes place between the container and the air in the room until the thermal equilibrium is established. The final temperature, the amount of heat transfer between the water and the air, and the entropy generation are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. Air is an ideal gas with constant specific heats. 3. The room is well-sealed and there is no heat transfer from the room to the surroundings. 4. Sea level atmospheric pressure is assumed. P = 101.3 kPa. Properties The properties of air at room temperature are R  0.287 kPa  m3 / kg  K , c p  1.005 kJ / kg  K , cv  0.718 kJ / kg  K . The specific heat of water at room temperature is cw  4.18 kJ / kg  K (Tables A-2, A-3).

Analysis (a) The mass of the air in the room is ma =

PV (101.3 kPa)(90 m3 ) = = 111.5 kg RTa1 (0.287 kPa ×m3 /kg ×K)(12 + 273 K)

6-185

An energy balance on the system that consists of the water in the container and the air in the room gives the final equilibrium temperature 0 = mwcw (T2 - Tw1 ) + ma cv (T2 - Ta1 ) 0 = (45 kg)(4.18 kJ/kg ×K)(T2 - 95) + (111.5 kg)(0.718 kJ/kg ×K)(T2 - 12) ¾ ¾ ® T2 = 70.2°C

(b) The heat transfer to the air is

Q = ma cv (T2 - Ta1 ) = (111.5 kg)(0.718 kJ/kg.K)(70.2 - 12) = 4660 kJ (c) The entropy generation associated with this heat transfer process may be obtained by calculating total entropy change, which is the sum of the entropy changes of water and the air. D S w = mw cw ln

P2 =

T2 (70.2 + 273) K = (45 kg)(4.18 kJ/kg.K)ln = - 13.11 kJ/K Tw1 (95 + 273) K

ma RT2 (111.5 kg)(0.287 kPa ×m3 /kg ×K)(70.2 + 273 K) = = 122 kPa V (90 m3 )

é (70.2 + 273) K 122 kPa ù ú= 14.88 kJ/K = (111.5 kg) ê(1.005 kJ/kg.K)ln - (0.287 kJ/kg.K)ln ê (12 + 273) K 101.3 kPa ú ë û Sgen = D Stotal = D S w + D Sa = - 13.11 + 14.88 = 1.77 kJ / K

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_w=45 [kg] T_w1=(95+273) [K] V_room=90 [m^3] T_a1=(12+273) [K] P_1=101.3 [kPa] "assume atmospheric pressure" "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_v=0.718 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" c_w=4.18 [kJ/kg-K] "Table A-3" "ANALYSIS" "(a)" m_a=(P_1*V_room)/(R*T_a1) 0=m_w*c_w*(T_2-T_w1)+m_a*c_v*(T_2-T_a1) "energy balance" T2_C=T_2-273 "(b)" Q=m_a*c_v*(T_2-T_a1) "heat transfer" "(c)" DELTAS_w=m_w*c_w*ln(T_2/T_w1) DELTAS_a=m_a*(c_p*ln(T_2/T_a1)-R*ln(P_2/P_1)) P_2=(m_a*R*T_2)/V_room S_gen=DELTAS_w+DELTAS_a "entropy generation"

7-101 Oxygen is expanded in an adiabatic nozzle. The maximum velocity at the exit is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. The process is adiabatic, and thus there is no heat transfer. 3. Oxygen is an ideal gas with constant specific heats. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-186

Properties The properties of oxygen at room temperature are c p  0.918 kJ / kg  K and k = 1.395 (Table A-2a). Analysis For the maximum velocity at the exit, the process must be reversible as well as adiabatic (i.e., isentropic). This being the case, the exit temperature will be ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

0.395/1.395

æ120 kPa ÷ ö = (363 K) çç ÷ èç300 kPa ø÷

= 280.0 K

There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

æ V2ö æ V2ö m çççh1 + 1 ÷÷÷= m çççh2 + 2 ÷÷÷ çè 2 ø÷ 2 ø÷ èç V12 V2 = h2 + 2 2 2 Solving for the exit velocity, h1 +

0.5

V2 = éëêV12 + 2(h1 - h2 )ù ú û

0.5

= éëêV12 + 2c p (T1 - T2 )ù ú û

0.5

é æ1000 m 2 /s 2 ÷ öù ú ÷ = êê(3 m/s) 2 + 2(0.918 kJ/kg ×K)(363 - 280)K çç ÷ ÷ú çè 1 kJ/kg ø êë ú û = 390 m / s

EES (Engineering Equation Solver) SOLUTION "Given" T1=(90+273) [K] P1=300 [kPa] Vel1=3 [m/s] P2=120 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-187

"Properties" "at room temperature for oxygen" c_p=0.918 [kJ/kg-K] "Table A-2a" k=1.395 "Table A-2a" "Analysis" T2=T1*(P2/P1)^((k-1)/k) Vel2^2=Vel1^2+2*c_p*(T1-T2)*Convert(kJ/kg, m^2/s^2)

7-102 Air is accelerated in an isentropic nozzle. The maximum velocity at the exit is to be determined. Assumptions 1. Air is an ideal gas with constant specific heats. 2. The nozzle operates steadily. Properties The properties of air at room temperature are c p  1.005 kJ / kg  K , k = 1.4 (Table A-2a). Analysis The exit temperature is determined from ideal gas isentropic relation to be, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P ÷ ø 1

0.4/1.4

æ100 kPa ÷ ö = (400 + 273 K )çç ÷ èç800 kPa ÷ ø

= 371.5 K

We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = m(h2 +V22 /2)

0 = h2 - h1 +

V22 - 0 2

0 = c p (T2 - T1 ) +

V22 2

Therefore,

V2 =

2c p (T1 - T2 ) =

æ1000 m2 /s 2 ÷ ö ÷ 2(1.005 kJ/kg.K)(673-371.5)K ççç = 778.5 m / s ÷ è 1 kJ/kg ÷ ø

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=800 [kPa] T1=400 [C]+273 [K] Vel_1=0 [m/s] P2=100 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-188

"PROPERTIES" c_p=1.005 [kJ/kg-K] k=1.4 "ANALYSIS" T2=T1*(P2/P1)^((k-1)/k) Vel_2^2=2*c_p*(T1-T2)*Convert(kJ/kg, m^2/s^2)

7-103E Air is charged to an initially evacuated container from a supply line. The minimum temperature of the air in the container after it is filled is to be determined. Assumptions 1. This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2. Air is an ideal gas with constant specific heats. 3. Kinetic and potential energies are negligible. 4. The tank is well-insulated, and thus there is no heat transfer. Properties The specific heat of air at room temperature is c p  0.240 Btu / lbm  R (Table A-2Ea). Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and entropy balances for this uniform-flow system can be expressed as Mass balance: min - mout = D msystem

mi = m2 - m1

mi = m2 Entropy balance:

m2 s2 - m1s1 + me se - mi si ³ 0 m2 s2 - mi si ³ 0 Combining the two balances,

m2 s2 - m2 si ³ 0 s2 - si ³ 0

The minimum temperature will result when the equal sign applies. Noting that P2  Pi , we have s2 - si = c p ln

T2 P T - R ln 2 = 0 ¾ ¾ ® c p ln 2 = 0 Ti Pi Ti

Then, T2 = Ti = 140°F

6-189

Alternative Solution The reversible process requires a reversible process with work extraction during the filling-up the vessel. The work will be due to an imaginary piston doing boundary work with pressure equal to line pressure: Wout = Pi (V2 - V1 ) = P2V2 = m2 RT2

An energy balance for the reversible process (no entropy generation) can be written as mi hi = m2u2 - m1u1 + Wout

m2 hi = m2u2 + m2 RT2 hi = u2 + RT2 c pTi = cv T2 + RT2 c pTi = (cv + R)T2 c pTi = c pT2

Ti = T2 = 140°F

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_i=150 [psia] T_i=(140+460) [R] P2=P_i "Properties" c_p=0.240 [Btu/lbm-R] "Table A-2Ea" R=0.287 [Btu/lbm-R] "Table A-2Ea" "ANALYSIS" DELTAs=c_p*ln(T2/T_i)-R*ln(P2/P_i) DELTAs=0 "for minimum temperature in the tank" T2_C=T2-460

7-104 An insulated rigid tank contains argon gas at a specified pressure and temperature. A valve is opened, and argon escapes until the pressure drops to a specified value. The final mass in the tank is to be determined. Assumptions 1. At specified conditions, argon can be treated as an ideal gas. 2. The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. Properties The specific heat ratio of argon is k = 1.667 (Table A-2). Analysis From the ideal gas isentropic relations,

æP2 ö÷( T2 = T1 ççç ÷ ÷ èç P ÷ ø

k - 1) k

0.667 1.667

æ200 kPa ÷ ö = (303 K )çç ÷ çè 450 kPa ø÷

= 219.0 K

6-190

The final mass in the tank is determined from the ideal gas relation, (200 kPa )(303 K ) P1V m RT PT = 1 1 ¾¾ ® m2 = 2 1 m1 = (4 kg) = 2.46 kg P2V m2 RT2 PT (450 kPa )(219 K) 1 2

EES (Engineering Equation Solver) SOLUTION "Given" m1=4 [kg] P1=450 [kPa] T1=(30+273) [K] P2=200 [kPa] "Properties" k=1.667 "Table A-2a" "Analysis" T2=T1*(P2/P1)^((k-1)/k) m2=(P2/P1)*(T1/T2)*m1

7-105 Problem 7-104 is reconsidered. The effect of the final pressure on the final mass in the tank is to be investigated as the pressure varies from 450 kPa to 150 kPa, and the results are to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. c_p = 0.5203 [kJ/kg-K ] c_v = 0.3122 [kJ/kg-K ] R=0.2081 [kPa-m^3/kg-K] P_1= 450 [kPa] T_1 = 30 [C] m_1 = 4 [kg] P_2= 200 [kPa] k = c_p/c_v T_2 = ((T_1+273)*(P_2/P_1)^((k-1)/k)-273) V_2 = V_1 P_1*V_1=m_1*R*(T_1+273) P_2*V_2=m_2*R*(T_2+273)

[kPa]

[kg] 2.069 2.459 2.811 3.136 3.44 3.727 4

150 200 250 300 350 400 450

6-191

7-106 For ideal gases, c p  cv  R and P2V 2 P1V1  T2 T1

 

V 2 T2 P1  V1 T1 P2

Thus,

T  V  s2  s1  cv ln  2   R ln  2   T1   V1  T  T P   cv ln  2   R ln  2 1   T1   T1 P2 

T  T  P   cv ln  2   R ln  2   R ln  2   T1   T1   P1  T  P   c p ln  2   R ln  2   T1   P1 

7-107 For an ideal gas, dh  c p dT and v  RT / P . From the second Tds relation,

ds 

dh v dP c p dP RT dP dT dP     cp R T T T P T T P

Integrating,

T  P  s 2  s1  c p ln  2   R ln  2   T1   P1  Since c p is assumed to be constant.

Review Problems 7-108 The operating conditions and thermal reservoir temperatures of a heat pump are given. It is to be determined if the increase of entropy principle is satisfied. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-192

Assumptions The heat pump operates steadily. Analysis Applying the first law to the cyclic heat pump gives

QL = QH - Wnet,in = 25 kW - 5 kW = 20 kW According to the definition of the entropy, the rate at which the entropy of the high-temperature reservoir increases is

D SH =

QH 25 kW = = 0.0833 kW/K TH 300 K

Similarly, the rate at which the entropy of the low-temperature reservoir decreases is D SL =

QL - 20 kW = = - 0.0769 kW/K TL 260 K

The rate at which the entropy of everything changes is then D S total = D S H + D S L = 0.0833 - 0.0769 = 0.0064 kW/K

which is positive and therefore it satisfies the increase in entropy principle.

EES (Engineering Equation Solver) SOLUTION "Given" T_L=260 [K] T_H=300 [K] Q_dot_H=25 [kW] W_dot_net_in=5 [kW] "Analysis" Q_dot_L=Q_dot_H-W_dot_net_in DELTAS_dot_H=Q_dot_H/T_H DELTAS_dot_L=-Q_dot_L/T_L DELTAS_dot_total=DELTAS_dot_H+DELTAS_dot_L

7-109 The source and sink temperatures and the COP of a refrigerator are given. The total entropy change of the two reservoirs is to be calculated and it is to be determined if this refrigerator satisfies the second law. Assumptions The refrigerator operates steadily. Analysis Combining the first law and the definition of the coefficient of performance produces

6-193

æ ö æ 1ö 1 ÷ ÷ QH = QL ççç1 + = (1 kJ) çç1 + ÷ ÷ ÷= 1.25 kJ çè 4 ø÷ çè COPR ÷ ø when COP = 4. The entropy change of everything is then D Stotal = D SH + D SL

QH QL 1.25 kJ - 1 kJ + = + = 0.000173 kJ / K TH TL 303 K 253 K

Since the entropy increases, a refrigerator with COP = 4 is possible. When the coefficient of performance is increased to 6,

æ æ 1ö 1 ÷ö ÷÷= (1 kJ) çç1 + ÷÷÷= 1.167 kJ QH = QL ççç1 + èç 6 ø èç COPR ÷ø and the net entropy change is

D Stotal = D SH + D SL =

QH QL 1.167 kJ - 1 kJ + = + = - 0.000101 kJ / K TH TL 303 K 253 K

and the refrigerator can no longer be possible.

EES (Engineering Equation Solver) SOLUTION "Given" COP_a=4 T_L=(-20+273) [K] T_H=(30+273) [K] Q_L=1 [kJ] COP_b=6 "Analysis" Q_H_a=Q_L*(1+1/COP_a) DELTAS_total_a=Q_H_a/T_H-Q_L/T_L Q_H_b=Q_L*(1+1/COP_b) DELTAS_total_b=Q_H_b/T_H-Q_L/T_L

7-110 Steam is expanded adiabatically in a closed system. The minimum internal energy that can be achieved during this process is to be determined.

6-194

Analysis The entropy at the initial state is P1 = 1500 kPa ïü ïý s = 6.9957 kJ/kg ×K (from Table A-6 or EES) ïïþ 1 T1 = 320°C The internal energy will be minimum if the process is isentropic. Then, ïü ïý s2 = s1 = 6.9957 kJ/kg ×Kïïþ

P2 = 100 kPa

x2 =

s2 - s f s fg

6.9957 - 1.3028 = 0.9400 6.0562

u2 = u f + x2 u fg = 417.40 + (0.9400)(2088.2) = 2380.3 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=1500 [kPa] T1=320 [C] P2=100 [kPa] "Analysis" Fluid$='steam_iapws' s1=entropy(Fluid$,T=T1,P=P1) s2=s1 u2=intenergy(Fluid$,P=P2,s=s2) x2=quality(Fluid$,P=P2,s=s2)

7-111E Water is expanded in an isothermal, reversible process. It is to be determined if the process is possible. Analysis The entropies at the initial and final states are (Tables A-5E and A-6E) P1 = 30 psia ü ïï T1 = 250.3°F ý x1 = 0.70 ïïþ s1 = s f + x1 s fg = 0.36821 + (0.70)(1.33132) = 1.3001 Btu/lbm ×R ü ïï ý s = 1.8277 Btu/lbm ×R T2 = T1 = 250.3°F ïïþ 2 The heat transfer during this isothermal, reversible process is the area under the process line: P2 = 10 psia

6-195

= (250 + 460 K)(1.8277 - 1.3001) Btu/lbm ×R = 374.6 Btu/lbm The total entropy change (i.e., entropy generation) is the sum of the entropy changes of water and the reservoir:

D stotal = D swater + D sR = s2 - s1 +

- q TR

= (1.8277 - 1.3001) Btu/lbm ×R +

- 374.6 Btu/lbm (300 + 460) R

= 0.0347 Btu / lbm× R Note that the sign of heat transfer is with respect to the reservoir.

EES (Engineering Equation Solver) SOLUTION "Given" P1=30 [psia] x1=0.70 P2=10 [psia] T_R=300 [F] "Analysis" Fluid$='steam_iapws' T1=temperature(Fluid$,P=P1,x=x1) s1=entropy(Fluid$,P=P1,x=x1) T2=T1 s2=entropy(Fluid$,P=P2,T=T2) q=(T1+460)*(s2-s1) DELTAs_total=s2-s1-q/(T_R+460)

7-112 Oxygen is expanded adiabatically in a piston-cylinder device. The maximum volume is to be determined. Assumptions 1. Changes in the kinetic and potential energies are negligible. 2. Oxygen is an ideal gas with constant specific heats. Properties The gas constant of oxygen is R  0.2598 kPa  m3 / kg  K . The specific heat ratio at the room temperature is k = 1.395 (Table A-2a). Analysis The volume of oxygen at the initial state is

V1 =

mRT1 (3 kg)(0.2598 kPa ×m3 /kg ×K)(373 + 273 K) = = 0.5300 m3 P1 950 kPa

The volume at the final state will be maximum if the process is isentropic. The volume for this case is determined from the isentropic relation of an ideal gas to be 1/ k

1/1.395 æP ö ö 950 kPa ÷ 3 æ ç ÷ V2,max = V1 ççç 1 ÷ = (0.5300 m ) = 2.66 m 3 ÷ çç ÷ ÷ çè P2 ÷ è ø 100 kPa ø

EES (Engineering Equation Solver) SOLUTION "Given" m=3 [kg] P1=950 [kPa] T1=(373+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-196

P2=100 [kPa] "Analysis" R=0.2598 [kPa-m^3/kg-K] k=1.395 V1=(m*R*T1)/P1 V2_max=V1*(P1/P2)^(1/k)

7-113E A solid block is heated with saturated water vapor. The final temperature of the block and water, and the entropy changes of the block, water, and the entire system are to be determined. Assumptions 1. The system is stationary and thus the kinetic and potential energy changes are zero. 2. There are no work interactions involved. 3. There is no heat transfer between the system and the surroundings. Analysis (a) As the block is heated, some of the water vapor will be condensed. We will assume (will be checked later) that the water is a mixture of liquid and vapor at the end of the process. Based upon this assumption, the final temperature of the water and solid block is 227.9F (The saturation temperature at 20 psia). The heat picked up by the block is Qblock = mc(T2 - T1 ) = (100 lbm)(0.5 Btu/lbm ×R)(227.9 - 80)R = 7396 Btu

The water properties at the initial state are P1 = 20 psia ü ïï ý ïïþ x1 = 1

T1 = 227.9°F

(Table A-5E)

h1 = 1156.2 Btu/lbm s1 = 1.7319 Btu/lbm ×R

The heat released by the water is equal to the heat picked up by the block. Also noting that the pressure of water remains constant, the enthalpy of water at the end of the heat exchange process is determined from h2 = h1 -

Qwater 7396 Btu = 1156.2 Btu/lbm = 416.61 Btu/lbm mw 10 lbm

The state of water at the final state is saturated mixture. Thus, our initial assumption was correct. The properties of water at the final state are ü ïï ý h2 = 416.61 Btu/lbm ïïþ

P2 = 120 psia

x2 =

h2 - h f h fg

416.61- 196.27 = 0.2295 959.93

s2 = s f + x2 s fg = 0.33582 + (0.2295)(1.3961) = 0.65628 Btu/lbm ×R

The entropy change of the water is then D Swater = mw (s2 - s1 ) = (10 lbm)(0.65628 - 1.7319)Btu/lbm = - 10.76 Btu / R

(b) The entropy change of the block is

T (227.9 + 460)R D Sblock = mc ln 2 = (100 lbm)(0.5 Btu/lbm ×R)ln = 12.11 Btu / R T1 (80 + 460)R (c) The total entropy change is D Stotal = Sgen = D Swater + D Sblock = - 10.76 + 12.11 = 1.35 Btu / R

The positive result for the total entropy change (i.e., entropy generation) indicates that this process is possible.

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_block=100 [lbm] c_p_block=0.5 [Btu/lbm-R] T1_block=80 [F] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-197

m_w=10 [lbm] P1=20 [psia] x1=1 "ANALYSIS" Fluid$='Steam_iapws' "(a)" T2=temperature(Fluid$, P=P1, x=1) "assumed final temperature" Q_block=m_block*c_p_block*(T2-T1_block) h1=enthalpy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) Q_water=Q_block h2=h1-Q_water/m_w P2=P1 h2_f=enthalpy(Fluid$, P=P2, x=0) h2_g=enthalpy(Fluid$, P=P2, x=1) h2_fg=h2_g-h2_f x2=(h2-h2_f)/h2_fg s2_f=entropy(Fluid$, P=P2, x=0) s2_g=entropy(Fluid$, P=P2, x=1) s2_fg=s2_g-s2_f s2=s2_f+x2*s2_fg DELTAS_w=m_w*(s2-s1) "(b)" DELTAS_block=m_block*c_p_block*ln((T2+460)/(T1_block+460)) "(c)" DELTAS_total=DELTAS_w+DELTAS_block

7-114 A piston-cylinder device contains air that undergoes a reversible thermodynamic cycle composed of three processes. The work and heat transfer for each process are to be determined. Assumptions 1. All processes are reversible. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats.

Properties The gas constant of air is R  0.287 kPa  m3 / kg  K (Table A-1). Analysis Using variable specific heats, the properties can be determined using the air table (Table A-17) as follows

T1 = T2 = 300 K ¾ ¾ ® s3o  s2o  R ln

u1 = u2 = 214.07 kJ/kg s10 = s20 = 1.70203 kJ/kg.K

P3  400 kPa   1.70203 kJ/kg  K  (0.287 kJ/kg  K) ln    1.98353 kJ/kg  K P2  150 kPa 

6-198

 

u3  283.79 kJ/kg T3  396.7 K

The mass of the air and the volumes at the various states are m=

P1V1 (400 kPa)(0.3 m3 ) = = 1.394 kg RT1 (0.287 kPa ×m3 /kg ×K)(300 K)

V2 =

mRT2 (1.394 kg)(0.287 kPa ×m3 /kg ×K)(300 K) = = 0.8 m3 P2 150 kPa

V3 =

mRT3 (1.394 kg)(0.287 kPa ×m3 /kg ×K)(396.7 K) = = 0.3968 m3 P3 400 kPa

Process 1-2: Isothermal expansion ( T2  T1 )

D S1- 2 = - mR ln

P2 150 kPa = (1.394 kg)(0.287 kJ/kg.K)ln = 0.3924 kJ/kg.K P1 400 kPa

Qin, 1- 2 = T1D S1- 2 = (300 K)(0.3924 kJ/K) = 117.7 kJ

Wout, 1- 2 = Qin,1- 2 = 117.7 kJ

Process 2-3: Isentropic (reversible-adiabatic) compression ( s2  s1 ) Win, 2- 3 = m(u3 - u2 ) = (1.394 kg)(283.79 - 214.07) kJ/kg = 97.2 kJ

Q23  0 kJ Process 3-1: Constant pressure compression process ( P1  P3 ) Win, 3- 1 = P3 (V3 - V1 ) = (400 kg)(0.3968 - 0.3) kJ/kg = 38.7 kJ Qout, 3- 1 = Win,3- 1 - m(u1 - u3 ) = 38.7 kJ - (1.394 kg)(214.07 - 283.79) kJ/kg = 135.9 kJ

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[1]=400 [kPa] T[1]=300 [K] V[1]=0.3 [m^3] P[2]=150 [kPa] T[2]=T[1] P[3]=P[1] s[3]=s[2] "PROPERTIES" Fluid$='Air' R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=Molarmass(Fluid$) u[1]=intenergy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) u[2]=intenergy(Fluid$, T=T[2]) s[2]=entropy(Fluid$, T=T[2], P=P[2]) u[3]=intenergy(Fluid$, P=P[3], s=s[3]) T[3]=temperature(Fluid$, P=P[3], s=s[3]) "ANALYSIS" m=(P[1]*V[1])/(R*T[1]) V[2]=(m*R*T[2])/P[2] V[3]=(m*R*T[3])/P[3] Q_12_in-W_12_out=m*(u[2]-u[1]) "energy balance, isothermal expansion, process 1-2" Q_12_in=T[1]*DELTAS_12 "heat transfer , process 1-2" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-199

DELTAS_12=m*(s[2]-s[1]) W_23_in=m*(u[3]-u[2]) "energy balance, isentropic compression, process 2-3" Q_23=0 "adiabatic process" W_31_in-Q_31_out=m*(u[1]-u[3]) "energy balance, constant pressure compression, process 3-1" W_31_in=P[1]*(V[3]-V[1])

7-115E A cylinder initially filled with helium gas at a specified state is compressed polytropically to a specified temperature and pressure. The entropy changes of the helium and the surroundings are to be determined, and it is to be assessed if the process is reversible, irreversible, or impossible. Assumptions 1. Helium is an ideal gas with constant specific heats. 2. The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 3. The thermal energy stored in the cylinder itself is negligible. 4. The compression or expansion process is quasi-equilibrium. Properties The gas constant of helium is R  2.6805 psia  ft 3 / lbm  R  0.4961 Btu / lbm  R . The specific heats of helium are cv  0.753 and c p  1.25 Btu / lbm  R (Table A-2E). Analysis (a) The mass of helium is

(25 psia )(15 ft ) P1V1 = = 0.264 lbm RT1 (2.6805 psia ×ft 3 /lbm ×R )(530 R ) 3

Then the entropy change of helium becomes

æ T Pö ÷ D Ssys = D S helium = m çççc p , avg ln 2 - R ln 2 ÷ ÷ ÷ çè T1 P1 ø

é 760 R 70 psia ù ú= - 0.016 Btu / R = (0.264 lbm) ê(1.25 Btu/lbm ×R) ln - (0.4961 Btu/lbm ×R) ln ê 530 R 25 psia ú ë û (b) The exponent n and the boundary work for this polytropic process are determined to be

(760 R )(25 psia ) P1V1 PV T P = 2 2 ¾¾ ® V2 = 2 1 V1 = (15 ft 3 ) = 7.682 ft 3 T1 T2 T1 P2 (530 R )(70 psia ) n

n æP ö æV1 ö æ70 ö æ 15 ö çç ÷ ÷ ÷ ç ç ÷ ÷ P2V2n = P1V1n ¾ ¾ ® ççç 2 ÷ = ¾ ¾ ® = ® n = 1.539 ÷ ÷ çç ÷ çç ÷ ÷ ÷ ¾¾ ÷ èççV2 ø ÷ è 25 ø è7.682 ø èç P1 ø

Then the boundary work for this polytropic process can be determined from 2

Wb,in = - ò P dV = 1

mR (T2 - T1 ) P2V2 - P1V1 =1- n 1- n

(0.264 lbm)(0.4961 Btu/lbm ×R )(760 - 530)R 1- 1.539

= 55.9 Btu

6-200

We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heat transfer to be from the cylinder, the energy balance for this stationary closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout + Wb, in = D U = m(u2 - u1 ) - Qout = m(u2 - u1 ) - Wb, in Qout = Wb, in - mcv (T2 - T1 )

Substituting,

Qout = 55.9 Btu - (0.264 lbm)(0.753 Btu/lbm ×R )(760 - 530)R = 10.2 Btu Noting that the surroundings undergo a reversible isothermal process, its entropy change becomes

D Ssurr =

Qsurr, in Tsurr

10.2 Btu = 0.019 Btu / R 530 R

(c) Noting that the system + surroundings combination can be treated as an isolated system, D Stotal = D Ssys + D Ssurr = - 0.016 + 0.019 = 0.003 Btu/R > 0

Therefore, the process is irreversible.

EES (Engineering Equation Solver) SOLUTION "Given" V1=15 [ft^3] P1=25 [psia] T1=(70+460) [R] P2=70 [psia] T2=(300+460) [R] T_surr=(70+460) [R] "Properties" Fluid$='helium' u1=intenergy(Fluid$, T=T1, P=P1) s1=entropy(Fluid$, T=T1, P=P1) u2=intenergy(Fluid$, T=T2, P=P2) s2=entropy(Fluid$, T=T2, P=P2) R=2.6805 [psia-ft^3/lbm-R] "from Table A-1E of the text" "Analysis" "(a)" m=(P1*V1)/(R*T1) DELTAS_sys=m*(s2-s1) "(b)" (P1*V1)/T1=(P2*V2)/T2 "from ideal gas relation" P1*V1^n=P2*V2^n P1*V1^n=C P*V^n=C W_b_out=integral(P, V, V1, V2)*Convert(psia-ft^3, Btu) "EES function that performs integration" W_b_in=-W_b_out -Q_out+W_b_in=m*(u2-u1) "energy balance" DELTAS_surr=Q_surr_in/T_surr Q_surr_in=Q_out "(c)" DELTAS_total=DELTAS_sys+DELTAS_surr

6-201

7-116 A piston-cylinder device contains steam that undergoes a reversible thermodynamic cycle composed of three processes. The work and heat transfer for each process and for the entire cycle are to be determined. Assumptions 1. All processes are reversible. 2. Kinetic and potential energy changes are negligible. Analysis The properties of the steam at various states are (Tables A-4 through A-6)

u1 = 2884.5 kJ/kg P1 = 400 kPaüïï 3 ýv = 0.71396 m /kg T1 = 350°C ïïþ 1 s1 = 7.7399 kJ/kg.K

P2 = 150 kPa ïüï u2 = 2888.0 kJ/kg ý T2 = 350°C ïïþ s2 = 8.1983 kJ/kg.K ïüï u3 = 3132.9 kJ/kg ý s3 = s2 = 8.1983 kJ/kg.Kïïþv 3 = 0.89148 m3 /kg P3 = 400 kPa

The mass of the steam in the cylinder and the volume at state 3 are

V1 0.5 m3 = = 0.7003 kg v1 0.71396 m3 /kg

V3 = mv 3 = (0.7003 kg)(0.89148 m3 /kg) = 0.6243 m3

Process 1-2: Isothermal expansion ( T2  T1 ) D S1- 2 = m(s2 - s1 ) = (0.7003 kg)(8.1983 - 7.7399)kJ/kg ×K = 0.3210 kJ/kg ×K

Qin, 1- 2 = T1D S1- 2 = (350 + 273 K)(0.3210 kJ/K) = 200.0 kJ Wout, 1- 2 = Qin, 1- 2 - m(u2 - u1 ) = 120.0 kJ - (0.7003 kg)(2888.0 - 2884.5)kJ/kg = 197.5 kJ

Process 2-3: Isentropic (reversible-adiabatic) compression ( s3  s2 ) Win, 2- 3 = m(u3 - u2 ) = (0.7003 kg)(3132.9 - 2888.0)kJ/kg = 171.5 kJ

Q23  0 kJ Process 3-1: Constant pressure compression process ( P1  P3 ) Win, 3- 1 = P3 (V3 - V1 ) = (400 kPa)(0.6243 - 0.5) = 49.73 kJ Qout, 3- 1 = Win, 3- 1 - m(u1 - u3 ) = 49.73 kJ - (0.7003 kg)(2884.5 - 3132.9) = 223.7 kJ

The net work and net heat transfer are Wnet, in = Win, 3- 1 + Win, 2- 3 - Wout, 1- 2 = 49.73 + 171.5 - 197.5 = 23.7 kJ Qnet, in = Qin, 1- 2 - Qout, 3- 1 = 200.0 - 223.7 = - 23.7 kJ ¾ ¾ ® Qnet, out = 23.7 kJ

Discussion The results are not surprising since for a cycle, the net work and heat transfers must be equal to each other.

6-202

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[1]=400 [kPa] T[1]=350 [C] Vol[1]=0.5 [m^3] P[2]=150 [kPa] T[2]=T[1] P[3]=P[1] s[3]=s[2] "PROPERTIES" Fluid$='Steam_iapws' u[1]=intenergy(Fluid$, T=T[1], P=P[1]) v[1]=volume(Fluid$, T=T[1], P=P[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) u[2]=intenergy(Fluid$, T=T[2], P=P[2]) s[2]=entropy(Fluid$, T=T[2], P=P[2]) u[3]=intenergy(Fluid$, P=P[3], s=s[3]) v[3]=volume(Fluid$, P=P[3], s=s[3]) "ANALYSIS" m=Vol[1]/v[1] Vol[3]=m*v[3] Q_12_in-W_12_out=m*(u[2]-u[1]) "energy balance, isothermal expansion, process 1-2" Q_12_in=(T[1]+273)*DELTAS_12 "heat transfer , process 1-2" DELTAS_12=m*(s[2]-s[1]) W_23_in=m*(u[3]-u[2]) "energy balance, isentropic compression, process 2-3" W_31_in-Q_31_out=m*(u[1]-u[3]) "energy balance, constant pressure compression, process 3-1" W_31_in=P[1]*(Vol[3]-Vol[1]) W_net_in=W_23_in+W_31_in-W_12_out "net work input" Q_net_in=Q_12_in-Q_31_out "net heat input"

7-117 An electric resistance heater is doing work on carbon dioxide contained an a rigid tank. The final temperature in the tank, the amount of heat transfer, and the entropy generation are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. Carbon dioxide is ideal gas with constant specific heats at room temperature. Properties The properties of CO2 at an anticipated average temperature of 350 K are R  0.1889 kPa  m3 / kg.K , c p  0.895 kJ / kg.K, cv  0.706 kJ / kg.K(Table A-2b). Analysis

(a) The mass and the final temperature of CO2 may be determined from ideal gas equation m=

P1V (100 kPa)(0.8 m3 ) = = 1.694 kg RT1 (0.1889 kPa ×m3 /kg ×K)(250 K)

6-203

T2 =

P2V (175 kPa)(0.8 m3 ) = = 437.5 K mR (1.694 kg)(0.1889 kPa ×m3 /kg ×K)

(b) The amount of heat transfer may be determined from an energy balance on the system Qout = Ee, in D t - mcv (T2 - T1 )

= (0.5 kW)(40´ 60 s) - (1.694 kg)(0.706 kJ/kg.K)(437.5 - 250)K = 975.8 kJ

(c) The entropy generation associated with this process may be obtained by calculating total entropy change, which is the sum of the entropy changes of CO2 and the surroundings

æ T P ö Qout ÷ Sgen = D SCO2 + D Ssurr = m çççc p ln 2 - R ln 2 ÷ ÷+ çè T1 P1 ÷ ø Tsurr

é 437.5 K 175 kPa ù 975.8 kJ ú+ = (1.694 kg) ê(0.895 kJ/kg.K)ln - (0.1889 kJ/kg.K)ln 250 K 100 kPa ûú 300 K ëê = 3.92 kJ / K

EES (Engineering Equation Solver) SOLUTION "GIVEN" V=0.8 [m^3] T_1=250 [K] P_1=100 [kPa] E_dot_in=0.5 [kW] time=40[min]*Convert(min, s) P_2=175 [kPa] T_surr=300 [K] "PROPERTIES" R=0.1889 [kJ/kg-K] "Table A-2a" c_v=0.706 [kJ/kg-K] "at 350 K, Table A-2b" c_p=0.895 [kJ/kg-K] "at 350 K, Table A-2b" "ANALYSIS" m=(P_1*V)/(R*T_1) T_2=(P_2*V)/(m*R) "final temperature" E_dot_in*time-Q_out=m*c_v*(T_2-T_1) "energy balance" DELTAS_CO2=m*(c_p*ln(T_2/T_1)-R*ln(P_2/P_1)) DELTAS_surr=Q_out/T_surr S_gen=DELTAS_CO2+DELTAS_surr "entropy generation"

7-118 The claim of an inventor that an adiabatic steady-flow device produces a specified amount of power is to be evaluated. Assumptions 1. Kinetic and potential energy changes are negligible. 2. The device is adiabatic and thus heat transfer is negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at the anticipated average temperature of 400 K are c p  1.013 kJ / kg.K and k = 1.395 (Table A-2b). Analysis We take the steady-flow device as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the turbine, the energy balance for this steady-flow system can be expressed in the rate form as

6-204

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 = Wout + mh2 (since Q @D ke @ Δpe @ 0) Wout = m(h1 - h2 ) @ mc p (T1 - T2 )

The adiabatic device would produce the maximum power if the process is isentropic. The isentropic exit temperature is

( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

0.395/1.395

æ100 kPa ÷ ö = (573 K) çç ÷ ÷ èç1200 kPa ø

= 283.5 K

The maximum power this device can produce is then Ws,out = mc p (T1 - T2 s ) = (1 kg/s)(1.013 kJ/kg ×K)(573 - 283.5)K = 293 kW

This is greater than the power claim of the inventor, and thus the claim is valid.

EES (Engineering Equation Solver) SOLUTION "GIVEN" W_dot_out=230 [kW] m_dot=1 [kg/s] P1=1200 [kPa] T1=(300+273) [K] P2=100 [kPa] "PROPERTIES" c_p=1.013 [kJ/kg-K] "at 400 K, Table A-2b" k=1.395 "at 400 K, Table A-2a" "ANALYSIS" T2_s=T1*(P2/P1)^((k-1)/k) W_s_out=m_dot*c_p*(T1-T2_s)

6-205

7-119 Heat is lost from the helium as it is throttled in a throttling valve. The exit pressure and temperature of helium and the entropy generation are to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Helium is an ideal gas with constant specific heats. Properties The properties of helium are R  2.0769 kPa  m3 / kg.K , c p  5.1926 kJ / kg  K(Table A-2a).

Analysis (a) The final temperature of helium may be determined from an energy balance on the control volume qout = c p (T1 - T2 ) ¾ ¾ ® T2 = T1 -

qout 1.75 kJ/kg = 60°C = 59.7°C = 332.7 K cp 5.1926 kJ/kg.°C

The final pressure may be determined from the relation for the entropy change of helium T P D sHe = c p ln 2 - R ln 2 T1 P1

0.34 kJ/kg.K = (5.1926 kJ/kg.K)ln

P2 332.7 K - (2.0769 kJ/kg.K)ln 333 K 400 kPa

P2 = 339 kPa (b) The entropy generation associated with this process may be obtained by adding the entropy change of helium as it flows in the valve and the entropy change of the surroundings

sgen = D sHe + D ssurr = D sHe +

qout 1.75 kJ/kg = 0.34 kJ/kg.K + = 0.346 kJ / kg.K Tsurr (25 + 273) K

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=400 [kPa] T_1=(60+273) [K] q_out=1.75 [kJ/kg] T_0=(25+273) [K] P_0=100 [kPa] DELTAs_He=0.34 [kJ/kg-K] "PROPERTIES" R=2.0769 [kJ/kg-K] "Table A-2a" c_p=5.1926 [kJ/kg-K] "Table A-2a" "ANALYSIS" q_out=c_p*(T_1-T_2) "energy balance" T_2_C=T_2-273 s_gen=DELTAs_He+DELTAs_surr "entropy generation" DELTAs_He=c_p*ln(T_2/T_1)-R*ln(P_2/P_1) DELTAs_surr=q_out/T_0

7-120 R-134a is compressed in an isentropic compressor. The work required is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-206

Assumptions 1. This is a steady-flow process since there is no change with time. 2. The process is isentropic (i.e., reversible-adiabatic). 3. Kinetic and potential energy changes are negligible. Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + Win = mh2 Win = m(h2 - h1 )

The inlet state properties are

T1 = - 10°Cïü ïý h1 = 244.55 kJ/kg (Table A-11) ïïþ s1 = 0.93782 kJ/kg ×K x1 = 1 For this isentropic process, the final state enthalpy is P2 = 800 kPa ïü ïý h = 273.29 kJ/kg (Table A-13) s2 = s1 = 0.93782 kJ/kg ×K ïïþ 2 Substituting,

win = h2 - h1 = (273.29 - 244.55) kJ/kg = 28.7 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=-10 [C] x1=1 P2=800 [kPa] "Analysis" Fluid$='R134a' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-207

h1=enthalpy(Fluid$,x=x1,T=T1) s1=entropy(Fluid$,x=x1,T=T1) s2=s1 h2=enthalpy(Fluid$,P=P2,s=s2) w_in=h2-h1

7-121 Refrigerant-134a is compressed in a compressor. The rate of heat loss from the compressor, the exit temperature of R-134a, and the rate of entropy generation are to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. Analysis (a) The properties of R-134a at the inlet of the compressor are (Table A-12)

v = 0.12355 m3 /kg P1 = 160 kPa üïï 1 ý h1 = 241.14 kJ/kg ïïþ x1 = 1 s1 = 0.94202 kJ/kg.K The mass flow rate of the refrigerant is

V1 0.03 m3 /s = = 0.2428 kg/s v1 0.12355 m3 /kg

Given the entropy increase of the surroundings, the heat lost from the compressor is

D Ssurr =

Qout ¾¾ ® Qout = Tsurr D Ssurr = (20 + 273 K)(0.008 kW/K) = 2.344 kW Tsurr

(b) An energy balance on the compressor gives Win - Qout = m(h2 - h1 )

10 kW - 2.344 kW = (0.2428 kg/s)(h2 - 241.14) kJ/kg ¾ ¾ ® h2 = 272.67 kJ/kg

The exit state is now fixed. Then,

ïüï T2 = 36.4°C ý h2 = 272.67 kJ/kgïïþ s2 = 0.93589 kJ/kg.K P2 = 800 kPa

(c) The entropy generation associated with this process may be obtained by adding the entropy change of R-134a as it flows in the compressor and the entropy change of the surroundings

Sgen = D SR + D Ssurr = m(s2 - s1 ) + D Ssurr = (0.2428 kg/s)(0.93589 - 0.94202) kJ/kg.K + 0.008 kW/K

= 0.00651kJ / K

6-208

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=160 [kPa] x_1=1 Vol_dot=0.03 [m^3/s] P_2=800 [kPa] W_dot_in=10 [kW] DELTAS_dot_surr=0.008 [kW/K] T_0=(20+273) [K] "ANALYSIS" Fluid$='R134a' h_1=enthalpy(Fluid$, P=P_1, x=x_1) v_1=volume(Fluid$, P=P_1, x=x_1) s_1=entropy(Fluid$, P=P_1, x=x_1) m_dot=Vol_dot/v_1 m_dot*h_1+W_dot_in=m_dot*h_2+Q_dot_out "energy balance" Q_dot_out=T_0*DELTAS_dot_surr "rate of heat loss" T_2=temperature(Fluid$, P=P_2, s=s_2) "exit temperature" s_2=entropy(Fluid$, P=P_2, h=h_2) DELTAS_dot_R=m_dot*(s_2-s_1) "entropy change of R-134a" S_dot_gen=DELTAS_dot_R+DELTAS_dot_surr "entropy generation"

7-122 One ton of liquid water at 80°C is brought into a room. The final equilibrium temperature in the room and the entropy change during this process are to be determined. Assumptions 1. The room is well insulated and well sealed. 2. The thermal properties of water and air are constant at room temperature. 3. The system is stationary and thus the kinetic and potential energy changes are zero. 4. There are no work interactions involved. Properties The gas constant of air is R  0.287 kPa  m3 / kg.K (Table A-1). The specific heat of water at room temperature is c  4.18 kJ / kg  C (Table A-3). For air is cv  0.718 kJ / kg  C at room temperature.

Analysis (a) The volume and the mass of the air in the room are V  4  5  7  140 m3

6-209

mair =

(100 kPa )(140 m3 ) P1V1 = = 165.4 kg RT1 (0.2870 kPa ×m 3 /kg ×K )(295 K )

Taking the contents of the room, including the water, as our system, the energy balance can be written as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem

® 0 = D U = (D U )water + (D U )air

Change in internal, kinetic, potential, etc. energies

or émc (T2 - T1 )ù + émcv (T2 - T1 )ù = 0 ë ûwater ë ûair

Substituting,

(1000 kg)(4.18 kJ/kg ×oC)(Tf - 80) oC + (165.4 kg )(0.718 kJ/kg ×o C)(Tf - 22) o C = 0 It gives the final equilibrium temperature in the room to be T f  78.4 C

(b) Considering that the system is well-insulated and no mass is entering and leaving, the total entropy change during this process is the sum of the entropy changes of water and the room air, D Stotal = Sgen = D Sair + D Swater

where D Sair = mcv ln

T2 V + mR ln 2 T1 V1

¿0

= (165.4 kg )(0.718 kJ/kg ×K ) ln

351.4 K = 20.78 kJ / K 295 K

T 351.4 K D S water = mc ln 2 = (1000 kg )(4.18 kJ/kg ×K ) ln = - 18.99 kJ / K T1 353 K Substituting, the total entropy change is determined to be

Stotal  20.78 ð 18.99  1.79 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" m_w=1000 [kg] T1_w=(80+273) [K] Vol=4*5*7 [m^3] T1_air=(22+273) [K] P1_air=100 [kPa] "Properties" c_p_w=CP(water, T=T_w_ave, P=100) T_w_ave=1/2*(T1_w+T2) c_v_air=CV(air, T=T_air_ave) c_p_air=CP(air, T=T_air_ave) T_air_ave=1/2*(T1_air+T2) R=c_p_air-c_v_air "Analysis" "(a)" m_air=(P1_air*Vol)/(R*T1_air) m_w*c_p_w*(T2-T1_w)+m_air*c_v_air*(T2-T1_air)=0 "energy balance" "(b)" DELTAS_w=m_w*c_p_w*ln(T2/T1_w) DELTAS_air=m_air*c_v_air*ln(T2/T1_air) DELTAS_total=DELTAS_w+DELTAS_air

6-210

7-123 An expression for the enthalpy change of an ideal gas during the isentropic process is to be obtained. Analysis The expression is obtained as follows:

dh  Tds 0  vdP  vdP

h   vdP, where v  const P 1/ k 2

h   const P

1/ k

P11/ k vP1/ k P11/ k dP  const  1  1/ k 1 1  1/ k 1

k k k h  Pv  RT  R T2  T1  k 1 1 k 1 k 1 1

h  c p T2  T1  7-124 It is to be shown that for thermal energy reservoirs, the entropy change relation D S = mc ln(T2 / T1 ) reduces to

D S = Q / T as T2 ® T1 . Analysis Consider a thermal energy reservoir of mass m, specific heat c, and initial temperature T1 . Now heat, in the amount of Q, is transferred to this reservoir. The first law and the entropy change relations for this reservoir can be written as

and

Q = mc (T2 - T1 )

¾¾ ®

mc =

Q T2 - T1

ln (T2 / T1 ) T D S = mc ln 2 = Q T1 T2 - T1 Taking the limit as T2  T1 by applying the L'Hospital's rule,

DS = Q

1/ T1 Q = 1 T1

which is the desired result.

7-125 The validity of the Clausius inequality is to be demonstrated using a reversible and an irreversible heat engine operating between the same temperature limits. Analysis Consider two heat engines, one reversible and one irreversible, both operating between a high-temperature reservoir at TH and a low-temperature reservoir at TL . Both heat engines receive the same amount of heat, QH . The © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-211

reversible heat engine rejects heat in the amount of QL , and the irreversible one in the amount of QL , irrev  QL  Qdiff , where

Qdiff is a positive quantity since the irreversible heat engine produces less work. Noting that QH and QL are transferred at constant temperatures of TH and TL , respectively, the cyclic integral of  Q / T for the reversible and irreversible heat engine cycles become æd Q ö dQ dQ Q Q 1 1 ò çççè T ÷÷÷ørev = ò THH - ò TL L = TH ò d QH - TL ò d QL = THH - TLL = 0 since  QH / TH    QL / TL  for reversible cycles. Also,

æd Q ö

ò çççè T ø÷÷÷

irrev

Q QH QL, irrev QH QL Qdiff = = - diff < 0 TH TL TH TL TL TL

æd Q ö since Qdiff is a positive quantity. Thus, ò çç ÷ ÷£ 0 . çè T ø÷

7-126 Two identical bodies at different temperatures are connected to each other through a heat engine. It is to be shown that the final common temperature of the two bodies will be T f = T1T2 when the work output of the heat engine is maximum. Analysis For maximum power production, the entropy generation must be zero. Taking the source, the sink, and the heat engine as our system, which is adiabatic, and noting that the entropy change for cyclic devices is zero, the entropy generation for this system can be expressed as

Sgen = (D S )source + (D S )engine

¿0

+ (D S )sink = 0

6-212

mc ln ln

Tf T1

+ ln

Tf T2

Tf T1

+ 0 + mc ln

= 0 ¾¾ ® ln

Tf T2

= 0

Tf Tf T1 T2

= 0 ¾¾ ® T f2 = T1T2

and thus Tf =

T1T2

for maximum power production.

7-127 The temperature of an ideal gas is given as functions of entropy and pressure. The T-P relation for the gas undergoing an isentropic process is to be obtained. Analysis The expressions for temperatures T1 and T2 are

T1 = T (s1 , P1 ) and T2 = T (s2 , P2 ) T1 = AP1( k - 1)/ k exp( s1 / c p )

T2 = AP2( k - 1)/ k exp( s2 / c p )

Now divide T2 by T1

AP2( k - 1)/ k exp( s2 / c p ) T2 = T1 AP1( k - 1)/ k exp( s1 / c p ) Since A and c p are constants and the process is isentropic s2  s1 , the temperature ratio reduces to ( k - 1)/ k

T2 æ Pö ÷ = ççç 2 ÷ ÷ T1 çè P1 ÷ ø

Fundamentals of Engineering (FE) Exam Problems

7-128 Steam is condensed at a constant temperature of 30C as it flows through the condenser of a power plant by rejecting heat at a rate of 55 MW. The rate of entropy change of steam as it flows through the condenser is (a) 1.83 MW / K (b) 0.18 MW / K (c) 0 MW / K (d) 0.56 MW / K (e) 1.22 MW / K Answer (b) 0.18 MW / K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=30 [C] Q_out=55 [MW] S_change=-Q_out/(T1+273) "Some Wrong Solutions with Common Mistakes:" W1_S_change=0 "Assuming no change" W2_S_change=Q_out/T1 "Using temperature in C" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-213

W3_S_change=Q_out/(T1+273) "Wrong sign" W4_S_change=-s_fg "Taking entropy of vaporization" s_fg=(ENTROPY(Steam_IAPWS,T=T1,x=1)-ENTROPY(Steam_IAPWS,T=T1,x=0))

7-129 Steam is compressed from 6 MPa and 300 C to 10 MPa isentropically. The final temperature of the steam is (a) 290 C (b) 300 C (c) 311 C (d) 371 C (e) 422 C Answer (d) 371 C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=6000 [kPa] T1=300 [C] P2=10000 [kPa] s2=s1 s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) T2=TEMPERATURE(Steam_IAPWS,s=s2,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant" W2_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P2" W3_T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2) "Saturation temperature at P1"

7-130 An apple with an average mass of 0.12 kg and average specific heat of 3.65 kJ / kg  C is cooled from 25C to 5C. The entropy change of the apple is (a) 0.705 kJ / K (b) 0.254 kJ / K (c) 0.0304 kJ / K (d) 0 kJ / K (e) 0.348 kJ / K Answer (c) 0.0304 kJ / K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.12 [kg] c=3.65 [kJ/kg-K] T1=25 [C] T2=5 [C] S_change=m*c*ln((T2+273)/(T1+273)) "Some Wrong Solutions with Common Mistakes:" W1_S_change=c*ln((T2+273)/(T1+273)) "Not using mass" W2_S_change=m*c*ln(T2/T1) "Using C" W3_S_change=m*c*(T2-T1) "Using Wrong relation" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-214

7-131 A piston-cylinder device contains 5 kg of saturated water vapor at 3 MPa. Now heat is rejected from the cylinder at constant pressure until the water vapor completely condenses so that the cylinder contains saturated liquid at 3 MPa at the end of the process. The entropy change of the system during this process is (a) 0 kJ / K (b) 3.5 kJ / K (c) 12.5 kJ / K (d) 17.7 kJ / K (e) 19.5 kJ / K Answer (d) 17.7 kJ / K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=5 [kg] P1=3000 [kPa] s_fg=(ENTROPY(Steam_IAPWS,P=P1,x=1)-ENTROPY(Steam_IAPWS,P=P1,x=0)) S_change=-m*s_fg

7-132 Argon gas expands in an adiabatic turbine from 3 MPa and 750C to 0.3 MPa at a rate of 5 kg / s . The maximum power output of the turbine is (a) 0.64 MW (b) 1.12 MW (c) 1.60 MW (d) 1.95 MW (e) 2.40 MW Answer (c) 1.60 MW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=3000 [kPa] T1=(750+273) [K] P2=300 [kPa] m=5 [kg/s] cp=0.5203 [kJ/kg-K] k=1.667 T2=T1*(P2/P1)^((k-1)/k) W_max=m*cp*(T1-T2) "Some Wrong Solutions with Common Mistakes:" cv=0.2081[kJ/kg-K] W1_Wmax=m*cv*(T1-T2) "Using cv" T22=T1*(P2/P1)^((k-1)/k) "Using C instead of K" W2_Wmax=m*cp*(T1-T22) W3_Wmax=cp*(T1-T2) "Not using mass flow rate" T24=T1*(P2/P1) "Assuming T is proportional to P, using C" W4_Wmax=m*cp*(T1-T24)

6-215

7-133 A unit mass of a substance undergoes an irreversible process from state 1 to state 2 while gaining heat from the surroundings at temperature T in the amount of q. If the entropy of the substance is s1 at state 1, and s2 at state 2, the entropy change of the substance s during this process is (a) s  s2  s1 (b) s  s2  s1 (c) s  s2  s1 (d) s  s2  s1  q / T (e) s  s2  s1  q / T Answer (c) s  s2  s1

7-134 A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while loosing heat to the surroundings at temperature T in the amount of q. If the gas constant of the gas is R, the entropy change of the gas s during this process is (a) s  R ln  P2 / P1  (b) s  R ln  P2 / P1   q / T (c) s  R ln  P1 / P2  (d) s  R ln  P1 / P2   q / T (e) s = 0 Answer (c) s  R ln  P1 / P2 

7-135 Helium gas enters an adiabatic nozzle steadily at 500C and 600 kPa with a low velocity, and exits at a pressure of 90 kPa. The highest possible velocity of helium gas at the nozzle exit is (a) 1475 m / s (b) 1662 m / s (c) 1839 m / s (d) 2066 m / s (e) 3040 m / s Answer (d) 2066 m / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=500 [C] P1=600 [kPa] Vel1=0 P2=90 [kPa] k=1.667 cp=5.1926 [kJ/kg-K] cv=3.1156 [kJ/kg-K] T2=(T1+273)*(P2/P1)^((k-1)/k)-273 "The exit velocity will be highest for isentropic expansion" "Energy balance for this case is h+ke=constant for the fluid stream (Q=W=pe=0)" (0.5*Vel1^2)/1000+cp*T1=(0.5*Vel2^2)/1000+cp*T2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-216

"Some Wrong Solutions with Common Mistakes:" T2a=T1*(P2/P1)^((k-1)/k) "Using C for temperature" (0.5*Vel1^2)/1000+cp*T1=(0.5*W1_Vel2^2)/1000+cp*T2a T2b=T1*(P2/P1)^((k-1)/k) "Using cv" (0.5*Vel1^2)/1000+cv*T1=(0.5*W2_Vel2^2)/1000+cv*T2b T2c=T1*(P2/P1)^k "Using wrong relation" (0.5*Vel1^2)/1000+cp*T1=(0.5*W3_Vel2^2)/1000+cp*T2c

7-136 Combustion gases with a specific heat ratio of 1.3 enter an adiabatic nozzle steadily at 800C and 800 kPa with a low velocity, and exit at a pressure of 85 kPa. The lowest possible temperature of combustion gases at the nozzle exit is (a) 43 C (b) 237 C (c) 367 C (d) 477 C (e) 640 C Answer (c) 367 C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 T1=800 [C] P1=800 [kPa] P2=85 [kPa] T2=(T1+273)*(P2/P1)^((k-1)/k)-273 "Nozzle exit temperature will be lowest for isentropic operation" "Some Wrong Solutions with Common Mistakes:" W1_T2=T1*(P2/P1)^((k-1)/k) "Using C for temperature" W2_T2=(T1+273)*(P2/P1)^((k-1)/k) "Not converting the answer to C" W3_T2=T1*(P2/P1)^k "Using wrong relation"

7-137 Steam enters an adiabatic turbine steadily at 400C and 5 MPa, and leaves at 20 kPa. The highest possible percentage of mass of steam that condenses at the turbine exit and leaves the turbine as a liquid is (a) 4% (b) 8% (c) 12% (d) 18% (e) 0% Answer (d) 18% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=5000 [kPa] T1=400 [C] P2=20 [kPa] s2=s1 s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) x2=QUALITY(Steam_IAPWS,s=s2,P=P2) moisture=1-x2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-217

7-138 Liquid water enters an adiabatic piping system at 15C at a rate of 8 kg / s . If the water temperature rises by 0.2C during flow due to friction, the rate of entropy generation in the pipe is (a) 23 W / K (b) 55 W / K (c) 68 W / K (d) 220 W / K (e) 443 W / K Answer (a) 23 W / K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=8 [kg/s] T1=15 [C] DELTAT=0.2 [C] cp=4180 [J/kg-K] T2=T1+DELTAT S_gen=m*cp*ln((T2+273)/(T1+273)) "Some Wrong Solutions with Common Mistakes:" W1_Sgen=m*cp*ln(T2/T1) "Using deg. C" W2_Sgen=cp*ln(T2/T1) "Not using mass flow rate with deg. C" W3_Sgen=cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with deg. C"

7-139 Helium gas is compressed from 1 atm and 25C to a pressure of 10 atm adiabatically. The lowest temperature of helium after compression is (a) 25C (b) 63C (c) 250C (d) 384C (e) 476C Answer (e) 476C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=(1*101.325) [kPa] T1=25 [C] P2=(10*101.325) [kPa] k=1.667 T2=(T1+273)*(P2/P1)^((k-1)/k) "The exit temperature will be lowest for isentropic compression" T2_C= T2-273 "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming temperature remains constant" W2_T2=T1*(P2/P1)^((k-1)/k) "Using C instead of K" W3_T2=(T1+273)*(P2/P1)-273 "Assuming T is proportional to P" W4_T2=T1*(P2/P1) "Assuming T is proportional to P, using C"

6-218

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 8 ENTROPY ANALYSIS

6-219

CYU - Check Your Understanding CYU 8-1 ■ Check Your Understanding CYU 8-1.1 Which process involves the smallest amount of work on a unit mass basis for the same pressure change? (a) Compressing air in a compressor (b) Compressing R-134a vapor in a compressor (c) Compressing water in a pump (d) Expanding steam in a turbine (e) Expanding combustion gases in a turbine Answer: (c) Compressing water in a pump CYU 8-1.2 For a pump, select the correct relation for the reversible work input. (a) w rev = cv (T2 - T1) (b) w rev = cp (P2 - P1) (c) w rev = (P2 - P1)/v (d) w rev = (P2 - P1)/r (e) w rev = v (T2 - T1) Answer: (d) w rev = (P2 - P1)/r

CYU 8-2 ■ Check Your Understanding CYU 8-2.1 Which compression process below requires the least amount of work for the same pressure change? (a) Isentropic (b) Isothermal (c) Polytropic (d) Two-stage compression with intercooling (e) Three-stage compression with intercooling Answer: (b) Isothermal CYU 8-2.2 Which of the following is the correct approach to minimize the work input to an air compressor? (a) Insulating the compressor to avoid heat losses (b) Preheating the air before it enters the compressor (c) Cooling the air as it is compressed in the compressor (d) Compressing the air to a higher pressure than the process requires and then to bleed air to a lower pressure (e) Using multistage compression with heating between stages Answer: (c) Cooling the air as it is compressed in the compressor CYU 8-2.3 Air is compressed from 100 kPa to 1600 kPa in two stages with intercooling. Which intermediate pressure should be selected to minimize the work input? (a) 200 kPa © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-220

(b) 300 kPa (c) 400 kPa (d) 850 kPa (e) 1000 kPa Answer: (c) 400 kPa Px / P1 = P2 / Px ® Px /100 = 1600 / Px ® Px = 400 kPa

CYU 8-3 ■ Check Your Understanding CYU 8-3.1 For which device is it not appropriate to define the isentropic efficiency? (a) Nozzle (b) Pump (c) Turbine (d) Single stage compressor (e) Multi-stage compressor with intercooling Answer: (e) Multi-stage compressor with intercooling CYU 8-3.2 Which property is not held at the same value for the actual and the isentropic operation of a turbine? (a) Inlet temperature (b) Exit temperature (c) Inlet pressure (d) Exit pressure (e) Inlet velocity Answer: (b) Exit temperature CYU 8-3.3 Which relations can be used to determine the isentropic efficiency of a steam turbine? 1: inlet state, 2: actual exit state, 2s: isentropic exit state IV. hturb = wactual

wisentropic

hturb =

VI. hturb =

c p (T1 - T2 ) c p (T1 - T2 s ) h1 - h2 h1 - h2s

(a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III CYU 8-3.4 Which relation(s) can be used to determine the isentropic efficiency of an air compressor if the variation of specific heats with temperature is to be accounted for? 1: inlet state, 2: actual exit state, 2s: isentropic exit state I.

hcomp =

wisentropic wactual

6-221

II.

hcomp =

III. hcomp =

c p (T2 s - T1 ) c p (T2 - T1 ) h2 s - h1 h2 - h1

(a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III CYU 8-3.5 Which relation(s) can be used to determine the isentropic efficiency of a water pump? 1: inlet state, 2: actual exit state, 2s: isentropic exit state I.

hpump =

h2 s - h1 h2 - h1

II.

hpump =

h2 - h1 v1 (P2 - P1 )

III. hcomp =

wactual wisentropic

(a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (a) Only I

CYU 8-4 ■ Check Your Understanding CYU 8-4.1 By which mechanism(s) can entropy be transferred to or from a system? I. Heat transfer II. Work III. Mass flow (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III CYU 8-4.2 The magnitude and direction of entropy transfer that accompanies heat transfer Q through a section at temperature T are (a) Q / T ; same direction as Q (b) Q / T ; opposite direction of Q (c) T / Q ; same direction as Q (d) QT; opposite direction of Q © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-222

(e) ln Q / T ; same direction as Q Answer: (a) Q / T ; same direction as Q CYU 8-4.3 Work in the amount of W is done on water boiling at constant temperature T by a mixer whose shaft penetrates the system boundary. The entropy transfer to the water by this work interaction is (a) W (b) W / T (c) T / W (d) 1 / W (e) 0 Answer: (e) 0

CYU 8-5 ■ Check Your Understanding CYU 8-5.1 For an adiabatic closed system, the entropy generation expression simplifies to (a) Sgen  ΔSsystem  Sout  Sin (b) Sgen  ΔSsystem (c) Sgen  Sout  Sin (d) Sgen  Sin  Sout (e) Sgen  0 Answer: (b) Sgen  ΔSsystem

CYU 8-5.2 A rigid tank containing steam loses 300 kJ of heat to the surrounding air at 27C. If the entropy of steam decreases by 0.75 kJ / K during this process, what is the entropy generated during this process? (a) 0 (b) 0.25 kJ / K (c) 0.50 kJ / K (d) 0.75 kJ / K (e) 1.75 kJ / K Answer: (b) 0.25 kJ / K Sin - Sout + Sgen = D Ssystem

- Q / T + Sgen = D Ssystem

(- 300 kJ) / (300 K) + Sgen = - 0.75 kJ / K Sgen = 0.25 kJ / K

CYU 8-5.3 Air expands in an insulated piston-cylinder device, doing 100 kJ of work. The surroundings are at 27C. What is the entropy generation during this process if the increase in the entropy of air is 0.50 kJ / K ? (a) 0 (b) 0.25 kJ / K (c) 0.50 kJ / K (d) 1.0 kJ / K © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-223

(e) 1.5 kJ / K Answer: (c) 0.50 kJ / K Sin - Sout + Sgen = D Ssystem 0 + Sgen = D Ssystem

Sgen = 0.5 kJ / K

CYU 8-6 ■ Check Your Understanding CYU 8-6.1 For an adiabatic single-stream steady-flow device, the simplest form for the rate of entropy generation is (i: inlet, e: exit) (a)

Sgen = m (se - si)- Q /T

(b)

Sgen = m (si - se)+ Q /T

(c) Sgen = m (se - si) (d) Sgen = m (si - se) (e) Sgen = 0 Answer: (c) Sgen = m (se - si) CYU 8-6.2 An air compressor consumes 60 kW of power while losing heat at a rate of 30 kW to the surroundings at 27C during steady operation. If the entropy of air increases by 0.3 kW / K during compression, what is the rate of entropy generation during this process? (a) 0.4 kW / K (b) 0.3 kW / K (c) 0.1 kW / K (d) 1 kW / K (e) 0 Answer: (a) 0.4 kW / K S . in - S . out + S . gen = 0

- Q. / T - 0.3 kW / K + S .gen = 0 (- 30 kW) / (300 K) - 0.3 kW / K + S. gen = 0

S .gen = 0.4 kW / K

CYU 8-6.3 Steam expands in an adiabatic turbine, producing 150 kW of power. The surroundings are at 27C. What is the rate of entropy generation during this process if the increase of the entropy of steam is 0.2 kW / K ? (a) 0.7 kW / K (b) 0.5 kW / K (c) 0.3 kW / K (d) 0.2 kW / K (e) 0 Answer: (d) 0.2 kW / K S . in - S . out + S . gen = 0

6-224

- 0.2 kW / K + S. gen = 0 S .gen = 0.2 kW / K

6-225

PROBLEMS Reversible Steady-Flow Work 8-1C The work associated with steady-flow devices is proportional to the specific volume of the gas. Cooling a gas during compression will reduce its specific volume, and thus the power consumed by the compressor.

8-2C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the work output of the turbine. Therefore, this is not a good proposal.

8-3C We would not support this proposal since the steady-flow work input to the pump is proportional to the specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.

8-4 The work produced for the process 1-3 shown in the figure is to be determined. Assumptions Kinetic and potential energy changes are negligible.

Analysis The work integral represents the area to the left of the reversible process line. Then, 2

win,1-3 = ò v dP + ò v dP 1

v1 + v 2 ( P2 - P1 ) + v 2 ( P3 - P2 ) 2

(0.5 + 1.0)m3 /kg (100 - 500)kPa + (1.0 m3 /kg)(400-100)kPa 2 = 0 kJ / kg =

EES (Engineering Equation Solver) SOLUTION P1=500 [kPa] P2=100 [kPa] P3=400 [kPa] v1=0.5 [m^3/kg] v2=1.0 [m^3/kg] v3=1.0 [m^3/kg] w_in_13=(v1+v2)/2*(P2-P1)+v2*(P3-P2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-226

8-5E The work produced for the process 1-2 shown in the figure is to be determined. Assumptions Kinetic and potential energy changes are negligible. Analysis The work integral represents the area to the left of the reversible process line. Then,

win,1-2 = ò v dP 1

v1 + v 2 ( P2 - P1 ) 2

æ ö (0.1 + 1.7)ft 3 /lbm 1 Btu ÷ ÷ (500 - 100)psia çç ÷ çè5.404 psia ×ft 3 ÷ 2 ø

= 66.6 Btu / lbm EES (Engineering Equation Solver) SOLUTION P1=100 [psia] P2=500 [psia] v1=0.1 [ft^3/lbm] v2=1.7 [ft^3/lbm] w_in_12=(v1+v2)/2*(P2-P1)*Convert(psia-ft^3,Btu)

8-6E Air is compressed isothermally in a reversible steady-flow device. The work required is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. There is no heat transfer associated with the process. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.06855 Btu / lbm ×R (Table A-1E).

6-227

Analysis Substituting the ideal gas equation of state into the reversible steady-flow work expression gives 2

P dP = RT ln 2 P P1 1

win = ò v dP = RT ò 1

æ80 psia ö ÷ ÷ = (0.06855 Btu/lbm ×R)(55 + 460 K)ln çç ÷ ÷ çè13 psia ø

= 64.2 Btu / lbm

EES (Engineering Equation Solver) SOLUTION "Given" P1=13 [psia] T1=(55+460) [R] P2=80 [psia] T2=T1 "Properties" R=0.06855 [Btu/lbm-R] "Table A-2Ea" "Analysis" w_in=R*T1*ln(P2/P1)

8-7 Saturated water vapor is compressed in a reversible steady-flow device. The work required is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. There is no heat transfer associated with the process. 3. Kinetic and potential energy changes are negligible. Analysis The properties of water at the inlet state are T1 = 150°C ü ïï P1 = 476.16 kPa (Table A-4) ý ïïþ v1 = 0.39248 m3 /kg x1 = 1 Noting that the specific volume remains constant, the reversible steady-flow work expression gives 2

win = ò v dP = v1 ( P2 - P1 ) 1

6-228

 205.6 kJ / kg EES (Engineering Equation Solver) SOLUTION "Given" T1=150 [C] x1=1 P2=1000 [kPa] v2=v1 "Analysis" Fluid$='steam_iapws' P1=pressure(Fluid$, T=T1, x=x1) v1=volume(Fluid$, T=T1, x=x1) w_in=v1*(P2-P1)

8-8 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highest possible mass flow rate of water is to be determined.

Assumptions 1. Liquid water is an incompressible substance. 2. Kinetic energy changes are negligible, but potential energy changes may be significant. 3. The process is assumed to be reversible since we will determine the limiting case. Properties The specific volume of liquid water is given to be v1 = 0.001m3 / kg . Analysis The highest mass flow rate will be realized when the entire process is reversible. Thus it is determined from the reversible steady-flow work relation for a liquid, æ 2 ö Win = m ççò v dP + D ke 0 + D pe÷ ÷= m {v (P2 - P1 )+ g (z 2 - z1 )} è 1 ø

  1 kJ   1 kJ/kg   7 kJ/s  m (0.001 m3 /kg)(5000  120) kPa   (9.81 m/s 2 )(10 m)  3  2 2   1 kPa  m   1000 m /s    It yields m = 1.41 kg / s

6-229

P1=120 [kPa] P2=5000 [kPa] DELTAz=10 [m] v=0.001 [m^3/kg] "Analysis" W_dot_in=m_dot*(v*(P2-P1)+g*DELTAz*Convert(m^2/s^2, kJ/kg)) g=9.81 [m/s^2]

8-9 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to the pump is to be determined. Assumptions 1. Liquid water is an incompressible substance. 2. Kinetic and potential energy changes are negligible. 3. The process is reversible. Properties The specific volume of saturated liquid water at 20 kPa is v1 = v f @ 20kPa = 0.001017 m3 / kg (Table A-5). Analysis The power input to the pump can be determined directly from the steady-flow work relation for a liquid,

æ 2 ö Win = m ççò v dP + D ke ¿ 0 + D pe ¿ 0 ÷ = mv1 (P2 - P1 ) è 1 ø÷ Substituting, æ 1 kJ ö ÷ Win = (45 kg/s)(0.001017 m3 /kg)(6000 - 20)kPa çç ÷= 274 kW ÷ çè1 kPa ×m3 ø

EES (Engineering Equation Solver) SOLUTION "Given" P1=20 [kPa] x1=0 m_dot=45 [kg/s] P2=6000 [kPa] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, x=x1) "Analysis" W_dot_in=m_dot*v1*(P2-P1)

8-10 A steam power plant operates between the pressure limits of 5 MPa and 10 kPa. The ratio of the turbine work to the pump work is to be determined. Assumptions 1. Liquid water is an incompressible substance. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-230

2. Kinetic and potential energy changes are negligible. 3. The process is reversible. 4. The pump and the turbine are adiabatic. Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic, s1  s2 and s3  s4 . Then the properties of the steam are

P4 = 10 kPa ïü ïý h4 = hg @10 kPa = 2583.9 kJ/kg sat. vapor ïïþ s4 = sg @10 kPa = 8.1488 kJ/kg ×K

P3  5 MPa s3  s4

} h3  4608.1 kJ/kg

Also, v1 = v f @10kPa = 0.00101m3 / kg .

The work output to this isentropic turbine is determined from the steady-flow energy balance to be

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout

mh3  mh4  Wout Wout  m(h3  h4 ) Substituting, wturb, out = h3 - h4 = 4608.1- 2583.9 = 2024.2 kJ/kg

The pump work input is determined from the steady-flow work relation to be 2

wpump, in = ò v dP + D ke ¿ 0 + D pe ¿ 0 = v1 (P2 - P1 ) 1

 1 kJ   (0.00101 m 3 /kg)(5000  10)kPa  3   1 kPa  m   5.041 kJ/kg Thus, wturb, out wpump, in

2024.2 = 402 5.041

6-231

P3=5000 [kPa] P4=10 [kPa] x1=0 x4=1 "Properties" Fluid$='steam_iapws' P1=P4 P2=P3 v1=volume(Fluid$, P=P1, x=x1) h4=enthalpy(Fluid$, P=P4, x=x4) s4=entropy(Fluid$, P=P4, x=x4) s3=s4 "isentropic process" h3=enthalpy(Fluid$, P=P3, s=s3) "Analysis" w_turb_out=h3-h4 w_pump_in=v1*(P2-P1) Ratio=w_turb_out/w_pump_in

8-11 Problem 8-10 is reconsidered. The effect of the quality of the steam at the turbine exit on the net work output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to be plotted as a function of this quality. Analysis The problem is solved using EES, and the results are tabulated and plotted below. P[1] = 10 [kPa] x[1] = 0 P[2] = 5000 [kPa] x[4] = 1.0 WorkFluid$ = 'Steam_IAPWS' "Pump Analysis" T[1]=temperature(WorkFluid$,P=P[1],x=0) v[1]=volume(workFluid$,P=P[1],x=0) h[1]=enthalpy(WorkFluid$,P=P[1],x=0) s[1]=entropy(WorkFluid$,P=P[1],x=0) s[2] = s[1] h[2]=enthalpy(WorkFluid$,P=P[2],s=s[2]) T[2]=temperature(WorkFluid$,P=P[2],s=s[2]) v[2]=volume(WorkFluid$,T=T[2],p=P[2]) h[1]+W_pump = h[2] W_pump_incomp = v[1]*(P[2] - P[1]) "Turbine analaysis" P[4] = P[1] P[3] = P[2] h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4]) s[4]=entropy(WorkFluid$,P=P[4],x=x[4]) T[4]=temperature(WorkFluid$,P=P[4],x=x[4]) s[3] = s[4] h[3]=enthalpy(WorkFluid$,P=P[3],s=s[3]) T[3]=temperature(WorkFluid$,P=P[3],s=s[3]) h[3] = h[4] + W_turb W_net_out = W_turb - W_pump

x4 0.5

Wnet, out [kJ / kg]

555.6

6-232

0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

637.4 719.2 801 882.8 971.2 1087 1240 1442 1699 2019

8-12E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The power input to the compressor is to be determined, and it is also to be compared to the work input for the liquid case. Assumptions 1. Liquid refrigerant is an incompressible substance. 2. Kinetic and potential energy changes are negligible. 3. The process is reversible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-233

4. The compressor is adiabatic. Analysis The compression process is reversible and adiabatic, and thus isentropic, s1  s 2 . Then the properties of the refrigerant are (Tables A-11E through A-13E)

P1 = 15 psia ü ïï h1 = 101.00 Btu/lbm ý sat. vapor ïïþ s1 = 0.22717 Btu/lbm ×R

P1  80 psia s2  s1

} h2  115.81 Btu/lbm

The work input to this isentropic compressor is determined from the steady-flow energy balance to be

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout Win  mh1  mh2

Win  m(h2  h1 ) Thus, win = h2 - h1 = 115.81- 101.00 = 14.8 Btu / lbm

If the refrigerant were first condensed at constant pressure before it was compressed, we would use a pump to compress the liquid. In this case, the pump work input could be determined from the steady-flow work relation to be 2

win = ò v dP + D ke ¿ 0 + D pe ¿ 0 = v1 (P2 - P1 ) 1

where v3 = v f @15 psia = 0.01164 ft 3 / lbm . Substituting,

æ ö 1 Btu ÷ ÷ win = (0.01164 ft 3 /lbm) (80 - 15) psia çç ÷= 0.140 Btu / lbm ÷ çè5.4039 psia ×ft 3 ø

EES (Engineering Equation Solver) SOLUTION "Given" P1=15 [psia] x1=1 P2=80 [psia] "Properties" Fluid$='R134a' h1=enthalpy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-234

s2=s1 "isentropic process" h2=enthalpy(Fluid$, P=P2, s=s2) "Analysis" w_in_C=h2-h1 "work input to compressor" v3=volume(Fluid$, P=P1, x=0) w_in_P=v3*(P2-P1)*Convert(psia-ft^3, Btu)

8-13E Helium gas is compressed from a specified state to a specified pressure at a specified rate. The power input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, and two-stage compression. Assumptions 1. 2. 3.

Helium is an ideal gas with constant specific heats. The process is reversible. Kinetic and potential energy changes are negligible.

Properties The gas constant of helium is R = 2.6805 psia.ft 3 / lbm .

R = 0.4961 Btu / lbm ×R . The specific heat ratio of helium is k = 1.667 (Table A-2E).

Analysis The mass flow rate of helium is

(16 psia )(10 ft 3 /s) P1V1 m= = = 0.1095 lbm/s RT1 (2.6805 psia ×ft 3 /lbm ×R )(545 R ) (a) Isentropic compression with k = 1.667:

Wcomp, in = m

k- 1 / k í ü ïï ö( ) kRT1 ïïï æ P2 ÷ ïý ç ÷ 1 ì çç ÷ ÷ ïï k - 1 ïï çè P1 ø ï îï þ

1.667  0.4961 Btu/lbm  R  545 R   120 psia    0.1095 lbm/s    1.667  1

0.667/1.667

 16 psia  

   1  

 91.74 Btu/s  129.8 hp

since 1 hp = 0.7068 Btu/s

(b) Polytropic compression with n = 1.2:

Wcomp, in = m

n- 1 / n í ü ïï ö( ) nRT1 ïïï æ P2 ÷ ïý ç ÷ 1 ì çç ÷ ïï n - 1 ïï çè P1 ÷ ø ï îï þ

1.2  0.4961 Btu/lbm  R  545 R   120 psia    0.1095 lbm/s    1.2  1

 16 psia  

0.2/1.2

   1  

6-235

P2 120 psia = (0.1095 lbm/s)(0.4961 Btu/lbm ×R )(545 R )ln = 59.67 Btu/s = 84.42 hp P1 16 psia

(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across each stage is the same, and its value is determined from

Px =

P1 P2 = (16 psia )(120 psia ) = 43.82 psia

The compressor work across each stage is also the same, thus total compressor work is twice the compression work for a single stage: ( í Px ö nRT1 ïïï æ ÷ ç ÷ ç ì ÷ ÷ n - 1 ïï çèç P1 ø îï

n- 1)/ n

Wcomp, in = 2mwcomp, I = 2m

ïü ï - 1ïý ïï ï þ

1.2  0.4961 Btu/lbm  R  545 R   43 82 psia   2  0.1095 lbm/s    1.2  1

 14 psia  

0.2/1.2

   1  

 64.97 Btu/s  91.92 hp since 1 hp = 0.7068 Btu/s

EES (Engineering Equation Solver) SOLUTION "Given" P1=16 [psia] T1=(85+460) [R] V1_dot=10 [ft^3/s] P2=120 [psia] n=1.2 "Properties" R=0.4961 [Btu/lbm-R] "Table A-2Ea" R1=2.6805 [psia-ft^3/lbm-R] "Table A-2Ea" k=1.667 "Table A-2Ea" "Analysis" m_dot=(P1*V1_dot)/(R1*T1) "(a)" W_dot_comp_in_a=m_dot*(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1)*Convert(Btu/s, hp) "(b)" W_dot_comp_in_b=m_dot*(n*R*T1)/(n-1)*((P2/P1)^((n-1)/n)-1)*Convert(Btu/s, hp) "(c)" W_dot_comp_in_c=m_dot*R*T1*ln(P2/P1)*Convert(Btu/s, hp) "(d)" P_x=sqrt(P1*P2) W_dot_comp_in_d=2*m_dot*(n*R*T1)/(n-1)*((P_x/P1)^((n-1)/n)-1)*Convert(Btu/s, hp)

8-14E Problem 8-13E is reconsidered. The work of compression and entropy change of the helium is to be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667. Analysis The problem is solved using EES, and the results are tabulated and plotted below. "Given" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-236

P1=16 [psia] T1=(85+460) [R] V1_dot=10 [ft^3/s] P2=120 [psia] n=1.2 "Properties from Table A-2Ea" R=0.4961 [Btu/lbm-R] R1=2.6805 [psia-ft^3/lbm-R] k=1.667 c_p=1.25 [Btu/lbm-R] "Analysis" m_dot=(P1*V1_dot)/(R1*T1) W_dot_comp_in_a=m_dot*(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1)*Convert(Btu/s, hp) W_dot_comp_in_b=m_dot*(n*R*T1)/(n-1)*((P2/P1)^((n-1)/n)-1)*Convert(Btu/s, hp) W_dot_comp_in_c=m_dot*R*T1*ln(P2/P1)*Convert(Btu/s, hp) P_x=sqrt(P1*P2) W_dot_comp_in_d=2*m_dot*(n*R*T1)/(n-1)*((P_x/P1)^((n-1)/n)-1)*Convert(Btu/s, hp) T2/T1=(P2/P1)^((n-1)/n) DELTAS_He=m_dot*(c_p*ln(T2/T1)-R*ln(P2/P1))

1 1.1 1.2 1.3 1.4 1.5 1.6 1.667

Wcomp, in, a

Wcomp, in, b

Wcomp, in, c

Wcomp, in, d

D SHe

[hp]

[Btu / s - R]

129.8 129.8 129.8 129.8 129.8 129.8 129.8 129.8

84.42 92.64 100.3 107.5 114.1 120.3 126.1 129.8

84.42 84.42 84.42 84.42 84.42 84.42 84.42 84.42

84.42 88.41 91.92 95.04 97.82 100.3 102.6 104

–0.1095 –0.0844 –0.0635 –0.04582 –0.03066 –0.01753 –0.006036 0.0008937

6-237

8-15 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. The mass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic, polytropic, isothermal, and two-stage compression. Assumptions 1. Nitrogen is an ideal gas with constant specific heats. 2. The process is reversible. 3. Kinetic and potential energy changes are negligible. Properties The gas constant of nitrogen is R = 0.297 kJ / kg.K (Table A-1). The specific heat ratio of nitrogen is k = 1.4 (Table A-2).

Analysis (a) Isentropic compression: or,

Wcomp, in = m

kRT1 (k - 1)/ k - 1 (P2 P1 ) k- 1

10 kJ/s  m

{

}

1.4  0.297 kJ/kg  K  300 K  1.4  1

 480 kPa / 80 kPa 

0.4/1.4



1

It yields

m = 0.048 kg / s (b) Polytropic compression with n = 1.3: or,

Wcomp, in = m

nRT1 (n- 1)/ n - 1 (P2 P1 ) n- 1

{

}

6-238

10 kJ/s  m

1.3 0.297 kJ/kg  K  300 K  1.3  1

 480 kPa / 80 kPa 

0.3/1.3



1

It yields

m = 0.051 kg / s (c) Isothermal compression: Wcomp, in = mRT ln

æ480 kPa ÷ ö P1 ¾¾ ® 10 kJ/s = m (0.297 kJ/kg ×K )(300 K ) ln çç ÷ çè 80 kPa ÷ ø P2

It yields m = 0.063 kg / s

(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each stage is the same, and its value is determined to be

Px =

P1 P2 = (80 kPa )(480 kPa ) = 196 kPa

The compressor work across each stage is also the same, thus total compressor work is twice the compression work for a single stage: or,

Wcomp,in = 2mwcomp, I = 2m

10 kJ/s  2m

nRT1 (n- 1)/ n - 1 (Px P1 ) n- 1

{

}

1.3 0.297 kJ/kg  K  300 K  196 kPa / 80 kPa 0.3/1.3  1    1.3  1

It yields m = 0.056 kg / s

EES (Engineering Equation Solver) SOLUTION "Given" P1=80 [kPa] T1=ConvertTemp(C, K, 27[C]) P2=480 [kPa] W_dot_comp_in=10 [kW] n=1.3 "Properties" c_v=CV(N2, T=T1) c_p=CP(N2, T=T1) k=c_p/c_v R=c_p-c_v "Analysis" "(a)" W_dot_comp_in=m_dot_a*(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1) "(b)" W_dot_comp_in=m_dot_b*(n*R*T1)/(n-1)*((P2/P1)^((n-1)/n)-1) "(c)" W_dot_comp_in=m_dot_c*R*T1*ln(P2/P1) "(d)" P_x=sqrt(P1*P2) W_dot_comp_in=2*m_dot_d*(n*R*T1)/(n-1)*((P_x/P1)^((n-1)/n)-1)

8-16 Air is compressed in an adiabatic compressor reversibly from 100 kPa and 20°C to 750 kPa. What percentage of work input is transferred to the air as thermal energy or heat? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-239

8-16 Air is compressed isentropically from a specified state to a specified pressure. The percentage of work input that is transferred to the air as heat is to be determined. Assumptions 1. Air is an ideal gas with constant specific heats. 2. Kinetic and potential energy changes are negligible. Properties The gas constant of air is R = 0.287 kJ / kg.K (Table A-1). The specific heat ratio of air is k = 1.4 (Table A-2). Analysis The work input during the isentropic compression process is

wcomp, isen =



kRT1 (k - 1)/ k - 1 (P2 P1 ) k- 1

{

}

1.4  0.287 kJ/kg  K  293 K  1.4  1

 750 kPa /100 kPa 

0.4/1.4



1

 229.1 kJ/kg No thermal energy is added to the air if the compression process is done in a reversible and isothermal manner: wcomp, isoth = RT ln

æ750 kPa ÷ ö P2 = (0.287 kJ/kg ×K )(293 K ) ln çç = 169.4 kJ/kg ÷ ÷ ç è100 kPa ø P1

The difference between the two work values is Thermal energy = wcomp, isen - wcomp, isoth = 229.1- 169.4 = 59.7 kJ/kg

That is, during the isothermal compression process, 59.7 kJ / kg of heat should be rejected from the air as it is compressed. Then the percentage of work input that is transferred to the air as heat is

Percentageheat =

Thermal energy 59.7 kJ/kg = = 0.260 = 26.0% wcomp, isen 229.1 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=293 [K] P2=750 [kPa] "Properties" k=1.4 R=0.287 [kJ/kg-K] "Analysis" w_comp_isentropic=(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1) w_comp_isothermal=R*T1*ln(P2/P1) ThermalEnergy=w_comp_isentropic-w_comp_isothermal Percentage=ThermalEnergy/w_comp_isentropic*100

Isentropic Efficiencies of Steady-Flow Devices 8-17C The ideal process for all three devices is the reversible adiabatic (i.e., isentropic) process. The adiabatic efficiencies of these devices are defined as hT =

actual work output insentropic work input actual exit kinetic energy , hC = , and h N = insentropic work output actual work input insentropic exit kinetic energy

6-240

8-18C No, because the isentropic process is not the model or ideal process for compressors that are cooled intentionally.

8-19C Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a result of irreversibilities. Therefore, the actual exit state has to be on the right-hand side of the isentropic exit state

8-20 Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at a specified pressure. The isentropic efficiency of the turbine is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Argon is an ideal gas with constant specific heats. Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is c p = 0.5203 kJ / kg.K (Table A-2). Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the isentropic turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein = Eout

mh1  Ws ,out  mh2 s

(since Q  Δke  Δpe  0)

Ws,out  m(h1  h2 s ) From the isentropic relations,

æP2 s ÷ ö( ç ÷ T2 s = T1 çç ÷ çè P ø÷

k - 1)/ k

0.667/1.667

æ200 kPa ÷ ö = (1073 K )çç ÷ èç1500 kPa ÷ ø

= 479 K

Then the power output of the isentropic turbine becomes Ws,out = mc p (T1 - T2 s ) = (80/60 kg/min )(0.5203 kJ/kg ×K )(1073 - 479) = 412.1 kW

Then the isentropic efficiency of the turbine is determined from

hT =

Wa 370 kW = = 0.898 = 89.8% 412.1 kW Ws

EES (Engineering Equation Solver) SOLUTION "Given" T1=ConvertTemp(C, K, 800[C]) P1=1500 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-241

P2=200 [kPa] m_dot=80 [kg/min]*Convert(kg/min, kg/s) W_dot_a_out=370 [kW] "Properties" C_p=CP(Argon, T=300, P=100) C_v=CV(Argon, T=300, P=100) k=C_p/C_v "Analysis" T2_s=T1*(P2/P1)^((k-1)/k) W_dot_s_out=m_dot*C_p*(T1-T2_s) eta_T=W_dot_a_out/W_dot_s_out

8-21E Steam is compressed in an adiabatic closed system with an isentropic efficiency of 80%. The work produced and the final temperature are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. The device is adiabatic and thus heat transfer is negligible. Analysis We take the steam as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wout  U  m(u2  u1 ) wout  u1  u2 From the steam tables (Tables A-5 and A-6),

P1 = 100 psia üïï u1 = 1233.7 Btu/lbm ý T1 = 650°F ïïþ s1 = 1.7816 Btu/lbm ×R

P2  1 MPa   s2 s  s1 

x2 s 

s2 s  s f s fg



1.7816  0.47427  0.9961 1.12888

u2 s  u f  x2 s  u fg  298.19  0.9961 807.29  1068.4 Btu/lbm

The work input during the isentropic process is ws , out = u1 - u2 s = (1233.7 - 1068.4)Btu/lbm = 165.3 Btu/lbm

The actual work input is then

wa , out = hisen ws , out = (0.80)(165.3 Btu/lbm) = 132.2 Btu / lbm The internal energy at the final state is determined from

wout = u1 - u2 ¾ ¾ ® u2 = u1 - wout = (1233.7 - 132.2)Btu/lbm = 1101.4 Btu/lbm Using this internal energy and the pressure at the final state, the temperature is determined from Table A-6 to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-242

ü ïï ý T2 = 274.6°F u2 = 1101.4 Btu/lbm ïïþ P2 = 10 psia

EES (Engineering Equation Solver) SOLUTION P1=100 [psia] T1=650 [F] P2=10 [psia] eta=0.80 "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$,P=P1,T=T1) s1=entropy(Fluid$,P=P1,T=T1) s2s=s1 u2s=intenergy(Fluid$,P=P2,s=s2s) x2s=quality(Fluid$,P=P2,s=s2s) w_out_isen=u1-u2s w_out_act=eta*w_out_isen w_out_act=u1-u2 x2=quality(Fluid$,P=P2,u=u2) T2=temperature(Fluid$,P=P2,u=u2)

8-22E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82% at a specified state, and leave at a specified pressure. The work output of the turbine is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Combustion gases can be treated as air that is an ideal gas with variable specific heats. Analysis From the air table (Table A-17E) and isentropic relations, T1 = 2000 R ¾ ¾ ®

h1 = 504.71 Btu/lbm s1o = 0.93205 Btu/lbm ×R

æ60 psia ÷ ö P2 ÷= 0.88453 Btu/lbm ×R ¾ ¾ = 0.93205 + (0.06855 Btu/lbm ×R) ln çç ® h2 s = 417.48 Btu/lbm ÷ ÷ ç P1 è120 psia ø There is only one inlet and one exit, and thus m1 = m2 = m. We take the actual turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as s2o = s1o + R ln

Ein = Eout

mh1  Wa,out  mh2

(since Q  Δke  Δpe  0)

Wa,out  m(h1  h2 ) Noting that wa  Tws , the work output of the turbine per unit mass is determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-243 wa = (0.82)(504.71- 417.48)Btu/lbm = 71.5 Btu / lbm

EES (Engineering Equation Solver) SOLUTION "Given" T1=1540 [F] P1=120 [psia] P2=60 [psia] eta_T=0.82 "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2_s=s1 h2_s=enthalpy(Fluid$, P=P2, s=s2_s) "Analysis" w_s=h1-h2_s w_a=eta_T*w_s

8-23 Steam is expanded in an adiabatic turbine. The isentropic efficiency is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take the actual turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout mh1  Wa ,out  mh2

(since Q  Δke  Δpe  0)

Wa ,out  m(h1  h2 )

From the steam tables (Tables A-4 through A-6),

P1 = 4 MPa T1 = 350°C

ïü ïý h1 = 3093.3 kJ/kg ïïþ s1 = 6.5843 kJ/kg ×K

6-244

P2 s  120 kPa s2 s  s1

}

x2 s  0.8798 h2 s  2413.4 kJ/kg

From the definition of the isentropic efficiency,

hT =

Wa ,out Ws ,out

m(h1 - h2 ) h - h2 3093.3 - 2683.1 = 1 = = 0.603 = 60.3% m(h1 - h2 s ) h1 - h2 s 3093.3 - 2413.4

EES (Engineering Equation Solver) SOLUTION "Given" P1=4000 [kPa] T1=350 [C] P2=120 [kPa] x2=1 "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$,P=P1,T=T1) s1=entropy(Fluid$,P=P1,T=T1) s2s=s1 h2s=enthalpy(Fluid$,P=P2,s=s2s) x2s=quality(Fluid$,P=P2,s=s2s) h2=enthalpy(Fluid$,P=P2,x=x2) w_out_isen=h1-h2s w_out_act=h1-h2 eta_T=w_out_act/w_out_isen

8-24 Steam is expanded in an adiabatic turbine with an isentropic efficiency of 0.92. The power output of the turbine is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

6-245

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout mh1 = Wa,out + mh2

(since Q @Δke @Δpe @ 0)

Wa ,out  m(h1  h2 ) From the steam tables (Tables A-4 through A-6),

P1 = 3 MPa T1 = 400°C

ïüï ý ïïþ

P2 s  30 kPa   s2 s  s1 

h1 = 3231.7 kJ/kg s1 = 6.9235 kJ/kg ×K

x2 s 

s2 s  s f s fg



6.9235  0.9441  0.8763 6.8234

h2 s  h f  x2 s h fg  289.27  (0.8763)(2335.3)  2335.7 kJ/kg

The actual power output may be determined by multiplying the isentropic power output with the isentropic efficiency. Then, Wa ,out = hT Ws ,out = hT m(h1 - h2 s ) = (0.92)(2 kg/s)(3231.7 - 2335.7)kJ/kg = 1649 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=3000 [kPa] T1=400 [C] P2=30 [kPa] eta_T=0.92 m_dot=2 [kg/s] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2_s=s1 s2_f=entropy(Fluid$, P=P2, x=0) s2_g=entropy(Fluid$, P=P2, x=1) s2_fg=s2_g-s2_f x2_s=(s2_s-s2_f)/s2_fg h2_f=enthalpy(Fluid$, P=P2, x=0) h2_g=enthalpy(Fluid$, P=P2, x=1) h2_fg=h2_g-h2_f h2_s=h2_f+x2_s*h2_fg W_dot_s_out=m_dot*(h1-h2_s) W_dot_a_out=eta_T*W_dot_s_out

6-246

8-25 Steam is expanded in an adiabatic turbine with an isentropic efficiency of 0.85. The power output of the turbine is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout mh1 = Wa,out + mh2

(since Q @Δke @Δpe @ 0)

Wa ,out  m(h1  h2 ) From the steam tables (Tables A-4 through A-6),

P1 = 3 MPa T1 = 400°C

ïüï ý ïïþ

P2 s  30 kPa   s2 s  s1 

h1 = 3231.7 kJ/kg s1 = 6.9235 kJ/kg ×K

x2 s 

s2 s  s f s fg



6.9235  0.9441  0.8763 6.8234

h2 s  h f  x2 s h fg  289.27  (0.8763)(2335.3)  2335.7 kJ/kg

The actual power output may be determined by multiplying the isentropic power output with the isentropic efficiency. Then, Wa ,out = hT Ws ,out = hT m(h1 - h2 s ) = (0.85)(2 kg/s)(3231.7 - 2335.7)kJ/kg = 1523 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=3000 [kPa] T1=400 [C] P2=30 [kPa] eta_T=0.85 m_dot=2 [kg/s] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2_s=s1 s2_f=entropy(Fluid$, P=P2, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-247

s2_g=entropy(Fluid$, P=P2, x=1) s2_fg=s2_g-s2_f x2_s=(s2_s-s2_f)/s2_fg h2_f=enthalpy(Fluid$, P=P2, x=0) h2_g=enthalpy(Fluid$, P=P2, x=1) h2_fg=h2_g-h2_f h2_s=h2_f+x2_s*h2_fg W_dot_s_out=m_dot*(h1-h2_s) W_dot_a_out=eta_T*W_dot_s_out

8-26 Refrigerant-134a enters an adiabatic compressor with an isentropic efficiency of 0.87 at a specified state with a specified volume flow rate, and leaves at a specified pressure. The compressor exit temperature and power input to the compressor are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Analysis

(a) From the refrigerant tables (Tables A-11E through A-13E),

h1 = hg @ 100 kPa = 234.46 kJ/kg P1 = 100 kPa ïüï ý s1 = sg @ 100 kPa = 0.95191 kJ/kg ×K ïïþ sat. vapor v1 = v g @ 100 kPa = 0.19255 m3 /kg P2  1 MPa   h2 s  282.53 kJ/kg s2 s  s1  From the isentropic efficiency relation, Thus,

hC =

h2 s - h1 ¾¾ ® h2 a = h1 + (h2 s - h1 ) / h C = 234.46 + (282.53 - 234.46)/0.87 = 289.71 kJ/kg h2 a - h1

P2 a  1 MPa   T  56.5C h2 a  289.71 kJ/kg  2 a (b) The mass flow rate of the refrigerant is determined from

V1 0.7/60 m3 /s = = 0.06059 kg/s v1 0.19255 m3 /kg

There is only one inlet and one exit, and thus m1 = m2 = m. We take the actual compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as

6-248

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout Wa, in + mh1 = mh2

(since Q @Δke @Δpe @ 0)

Wa,in  m(h2  h1 ) Substituting, the power input to the compressor becomes, Wa, in = (0.06059 kg/s)(289.71- 234.46)kJ/kg = 3.35 kW

EES (Engineering Equation Solver) SOLUTION P[1] = 100 [kPa] P[2] = 1000 [kPa] Vol_dot_1 = 0.7 [m^3/min] Eta_c = 0.87 "Compressor adiabatic efficiency" A_ratio = 1.5 d_2 = 2/100 "[m]" "System: Control volume containing the compressor, see the diagram window. Property Relation: Use the real fluid properties for R134a. Process: Steady-state, steay-flow, adiabatic process." Fluid$='R134a' "Property Data for state 1" T[1]=temperature(Fluid$,P=P[1],x=1) "Real fluid equ. at the sat. vapor state" h[1]=enthalpy(Fluid$, P=P[1], x=1) "Real fluid equ. at the sat. vapor state" s[1]=entropy(Fluid$, P=P[1], x=1) "Real fluid equ. at the sat. vapor state" v[1]=volume(Fluid$, P=P[1], x=1) "Real fluid equ. at the sat. vapor state" "Property Data for state 2" s_s[1]=s[1]; T_s[1]=T[1] "needed for plot" s_s[2]=s[1] "for the ideal, isentropic process across the compressor" h_s[2]=ENTHALPY(Fluid$, P=P[2], s=s_s[2]) "Enthalpy 2 at the isentropic state 2s and pressure P[2]" T_s[2]=Temperature(Fluid$, P=P[2], s=s_s[2]) "Temperature of ideal state - needed only for plot." "Steady-state, steady-flow conservation of mass" m_dot_1 = m_dot_2 m_dot_1 = Vol_dot_1/v[1]*Convert(1/min, 1/s) Vol_dot_1/v[1]=Vol_dot_2/v[2] Vel[2]=Vol_dot_2/A[2]*convert(1/min, 1/s) A[2] = pi*(d_2)^2/4 A_ratio*Vel[1]/v[1] = Vel[2]/v[2] "Mass flow rate: = A*Vel/v, A_ratio = A[1]/A[2]" A_ratio=A[1]/A[2] "Steady-state, steady-flow conservation of energy, adiabatic compressor, see diagram window" m_dot_1*(h[1]+Vel[1]^2/2*convert(J, kJ)) + W_dot_c= m_dot_2*(h[2]+(Vel[2])^2/2*convert(J, kJ)) "Definition of the compressor adiabatic efficiency, Eta_c=W_isen/W_act" Eta_c = (h_s[2]-h[1])/(h[2]-h[1]) "Knowing h[2], the other properties at state 2 can be found." v[2]=volume(Fluid$, P=P[2], h=h[2]) "v[2] is found at the actual state 2, knowing P and h." T[2]=temperature(Fluid$, P=P[2],h=h[2]) "Real fluid equ. for T at the known outlet h and P." s[2]=entropy(Fluid$, P=P[2], h=h[2]) "Real fluid equ. at the known outlet h and P." T_exit=T[2] "Neglecting the kinetic energies, the work is:" m_dot_1*h[1] + W_dot_c_noKE= m_dot_2*h[2] "Comparing W_dot with W_dot_noKE, the assumption to neglect KE is valid."

6-249

8-27 Problem 8-26 is reconsidered. The problem is to be solved by considering the kinetic energy and by assuming an inlet-to-exit area ratio of 1.5 for the compressor when the compressor exit pipe inside diameter is 2 cm. Analysis The problem is solved using EES, and the solution is given below. P[1] = 100 [kPa] P[2] = 1000 [kPa] Vol_dot_1 = 0.7 [m^3/min] Eta_c = 0.87 A_ratio = 1.5 d_2 = 0.02 [m] Fluid$='R134a' T[1]=temperature(Fluid$,P=P[1],x=1) h[1]=enthalpy(Fluid$, P=P[1], x=1) s[1]=entropy(Fluid$, P=P[1], x=1) v[1]=volume(Fluid$, P=P[1], x=1) s_s[1]=s[1]; T_s[1]=T[1] "needed for plot" s_s[2]=s[1] h_s[2]=ENTHALPY(Fluid$, P=P[2], s=s_s[2]) T_s[2]=Temperature(Fluid$, P=P[2], s=s_s[2]) "needed for plot" m_dot_1 = m_dot_2 m_dot_1 = Vol_dot_1/v[1]*Convert(1/min, 1/s) Vol_dot_1/v[1]=Vol_dot_2/v[2] Vel[2]=Vol_dot_2/A[2]*Convert(1/min, 1/s) A[2] = pi*(d_2)^2/4 A_ratio*Vel[1]/v[1] = Vel[2]/v[2] A_ratio=A[1]/A[2] m_dot_1*(h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) + W_dot_c= m_dot_2*(h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)) Eta_c = (h_s[2]-h[1])/(h[2]-h[1]) v[2]=volume(Fluid$, P=P[2], h=h[2]) T[2]=temperature(Fluid$, P=P[2],h=h[2]) s[2]=entropy(Fluid$, P=P[2], h=h[2]) T_exit=T[2] m_dot_1*h[1] + W_dot_c_noke= m_dot_2*h[2] SOLUTION A_ratio=1.5 d_2=0.02 [m] Eta_c=0.87 Fluid$='R134a' m_dot_1 = 0.06059 [kg / s]

m_dot_2 = 0.06059 [kg / s] T_exit=56.51 [C] Vol_dot_1=0.7 [m^3 /min] Vol_dot_2=0.08229 [m^3 /min] W_dot_c=3.33 [kW] W_dot_c_noke=3.348 [kW]

6-250

8-28 R-134a is compressed by an adiabatic compressor. The isentropic efficiency of the compressor is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Analysis From the R-134a tables (Tables A-11 through A-13),

ü ïï ý x = 1 (sat. vap.) ïïþ

T1 = 0°C

h1 = 250.50 kJ/kg s1 = 0.93158 kJ/kg ×K

P2  600 kPa   h  290.30 kJ/kg T2  50C  2 a P2  600 kPa   h  265.32 kJ/kg s2  s1  0.93158 kJ/kg  K  2 s From the isentropic efficiency relation, hC =

ws h - h1 265.32 - 250.50 = 2s = = 0.372 = 37.2% wa h2 a - h1 290.30 - 250.50

EES (Engineering Equation Solver) SOLUTION "Given" T1=0 [C] x1=1 T2=50 [C] P2=600 [kPa] "Analysis" Fluid$='R134a' h1=enthalpy(Fluid$,T=T1,x=x1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-251

s1=entropy(Fluid$,T=T1,x=x1) h2=enthalpy(Fluid$,P=P2,T=T2) s2s=s1 h2s=enthalpy(Fluid$,P=P2,s=s2s) eta_C=(h2s-h1)/(h2-h1)

8-29 Air is compressed by an adiabatic compressor from a specified state to another specified state. The isentropic efficiency of the compressor and the exit temperature of air for the isentropic case are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Air is an ideal gas with variable specific heats. Analysis (a) From the air table (Table A-17), T1 = 300 K ¾ ¾ ® h1 = 300.19 kJ/kg, s1o = 1.70203 kJ/kg ×K

T2  550 K  h2 a  554.74 kJ/kg From the isentropic relation, s2o  s1o  R ln

P2  600 kPa   1.70203  (0.287) ln   h2 s  508.81 kJ/kg   2.23099 kJ/kg  K  P1  95 kPa 

Then the isentropic efficiency becomes hC =

h2 s - h1 508.81- 300.19 = = 0.820 = 82.0% h2 a - h1 554.74 - 300.19

(b) If the process were isentropic, the exit temperature would be

h2 s = 508.81 kJ/kg ¾ ¾ ® T2 s = 505.6 K @506 K

EES (Engineering Equation Solver) SOLUTION "Given" P1=95 [kPa] T1=27 [C] P2=600 [kPa] T2=277 [C] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) h2_a=enthalpy(Fluid$, T=T2) s2_s=s1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-252

h2_s=enthalpy(Fluid$, P=P2, s=s2_s) "Analysis" "(a)" eta_C=(h2_s-h1)/(h2_a-h1) "(b)" T2_s=temperature(Fluid$, P=P2, s=s2_s)

8-30E Argon enters an adiabatic compressor with an isentropic efficiency of 80% at a specified state, and leaves at a specified pressure. The exit temperature of argon and the work input to the compressor are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Argon is an ideal gas with constant specific heats. Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is c p = 0.1253 Btu / lbm ×R (Table A-2E). Analysis (a) The isentropic exit temperature T2s is determined from

æP ö( ÷ T2 s = T1 ççç 2 s ÷ ÷ çè P1 ø÷

k - 1)/ k

0.667/1.667

æ200 psia ÷ ö ÷ = (535 R )çç ÷ çè 14 psia ÷ ø

= 1550.4 R

The actual kinetic energy change during this process is 2

D kea =

ö V22 - V12 (240 ft/s) - (60 ft/s) æ çç 1 Btu/lbm ÷ = ÷= 1.078 Btu/lbm 2 2÷ ç 2 2 è 25,037 ft /s ø

The effect of kinetic energy on isentropic efficiency is very small. Therefore, we can take the kinetic energy changes for the actual and isentropic cases to be same in efficiency calculations. From the isentropic efficiency relation, including the effect of kinetic energy, hC =

c p (T2 s - T1 )+ D kes 0.1253(1550.4 - 535)+ 1.078 ws (h - h1 ) + D ke = 2s = ¾¾ ® 0.87 = wa (h2 a - h1 ) + D ke c p (T2 a - T1 )+ D kea 0.1253(T2 a - 535)+ 1.078

It yields

T2a  1703 R (b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as

Ein - Eout Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

6-253

Wa, in + m(h1 + V12 / 2) = m(h2 + V22 /2) (since Q @D pe @0)

 V22  V12  Wa,in  m  h2  h1    wa,in  h2  h1  Δkea  c p (T2 a  T1 )  Δkea 2   Substituting, the work input to the compressor is determined to be wa, in = (0.1253 Btu/lbm ×R )(1703 - 535)R + 1.078 Btu/lbm = 147 Btu / lbm

EES (Engineering Equation Solver) SOLUTION "Given" P1=14 [psia] T1=(75+460) [R] Vel1=60 [ft/s] P2=200 [psia] Vel2=240 [ft/s] eta_C=0.87 "Properties" c_p=0.1253 [Btu/lbm-R] "Table A-2Ea" k=1.667 "Table A-2Ea" "Analysis" "(a)" T2_s=T1*(P2/P1)^((k-1)/k) DELTAke_a=(Vel2^2-Vel1^2)/2*1/Convert(Btu/lbm, ft^2/s^2) eta_C=(C_p*(T2_s-T1)+DELTAke_s)/(C_p*(T2_a-T1)+DELTAke_a) DELTAke_s=DELTAke_a "assumed, since the difference will have negligible effect on efficiency" "(b)" w_a_in=c_p*(T2_a-T1)+DELTAke_a

8-31 Argon gas is compressed by an adiabatic compressor. The isentropic efficiency of the compressor is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Argon is an ideal gas with constant specific heats. Properties The specific heat ratio of argon is k = 1.667 (Table A-2). Analysis The isentropic exit temperature is ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P ø÷ 1

0.667/1.667

æ2000 kPa ÷ ö = (300 K) çç ÷ çè 200 kPa ÷ ø

= 753.8 K

From the isentropic efficiency relation, hC =



ws h - h1 c p (T2 s - T1 ) T2 s - T1 = 2s = = wa h2 a - h1 c p (T2 a - T1 ) T2 a - T1

753.8  300  0.867  86.7% 823  300

6-254

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [kPa] T1=(27+273) [K] T2=(550+273) [K] P2=2000 [kPa] "Analysis" k=1.667 T_2s=T1*(P2/P1)^((k-1)/k) eta_C=(T_2s-T1)/(T2-T1)

8-32E Air is accelerated in a 85% efficient adiabatic nozzle from low velocity to a specified velocity. The exit temperature and pressure of the air are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Air is an ideal gas with variable specific heats. Analysis From the air table (Table A-17E),

T1 = 1400 R ¾ ¾ ® h1 = 342.90 Btu/lbm, s1o = 0.83604 Btu/lbm ×R There is only one inlet and one exit, and thus m1 = m2 = m . We take the nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout m(h1 + V12 /2) = m(h2 + V22 /2) (since W = Q @D pe @ 0)

h2  h1 

V22  V12 0 2

6-255

Substituting, the exit temperature of air is determined to be 2

h2 = 342.90 kJ/kg -

ö (650 ft/s) - 0 æ ç 1 Btu/lbm ÷ 2

÷= 334.46 Btu/lbm çç ÷ è 25,037 ft 2 /s 2 ø

From the air table we read T2a  1368 R  908o F

From the isentropic efficiency relation hN =

h2 a - h1 h2 s - h1

h2 s = h1 + (h2 a - h1 )/ h N = 342.90 + (334.46 - 342.90)/ (0.85) = 332.97 Btu/lbm ¾ ¾ ® s2o = 0.82884 Btu/lbm ×R Then the exit pressure is determined from the isentropic relation to be s2o  s1o  R ln

P2 P1

 P2  0.82884 Btu/lbm  R  0.83604 Btu/lbm  R  (0.06855 Btu/lbm  R) ln    45 psia 

P2  40.5 psia EES (Engineering Equation Solver) SOLUTION "Given" P1=45 [psia] T1=940 [F] Vel1=0 [ft/s] Vel2=650 [ft/s] eta_N=0.85 "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) "Analysis" "(a)" h1+Vel1^2/2*1/Convert(Btu/lbm, ft^2/s^2)=h2_a+Vel2^2/2*1/Convert(Btu/lbm, ft^2/s^2) T2_a=temperature(Fluid$, h=h2_a) "(b)" eta_N=(h2_a-h1)/(h2_s-h1) s2_s=s1 P2=pressure(Fluid$, h=h2_s, s=s2_s)

8-33E Problem 8-32E is reconsidered. The effect of varying the nozzle isentropic efficiency from 0.8 to 1.0 on the exit temperature and pressure of the air is to be investigated, and the results are to be plotted. Analysis The problem is solved using EES, and the results are tabulated and plotted below. P[1] = 45 [psia] T[1] = 940 [F] Vel[2] = 650 [ft/s] Vel[1] = 0 [ft/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-256

eta_nozzle = 0.85 WorkFluid$ = 'Air' h[1]=enthalpy(WorkFluid$,T=T[1]) s[1]=entropy(WorkFluid$,P=P[1],T=T[1]) T_s[1] = T[1] s[2] =s[1] s_s[2] = s[1] h_s[2]=enthalpy(WorkFluid$,T=T_s[2]) T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2]) eta_nozzle = ke[2]/ke_s[2] ke[1] = Vel[1]^2/2 ke[2]=Vel[2]^2/2 h[1]+ke[1]*convert(ft^2/s^2,Btu/lbm) = h[2] + ke[2]*convert(ft^2/s^2,Btu/lbm) h[1] +ke[1]*convert(ft^2/s^2,Btu/lbm) = h_s[2] + ke_s[2]*convert(ft^2/s^2,Btu/lbm) T[2]=temperature(WorkFluid$,h=h[2]) P_2_answer = P[2] T_2_answer = T[2]

ηnozzle

Ts,2

[psia] 40.25 40.36 40.47 40.57 40.67 40.76 40.85 40.93 41.01 41.09 41.17

[F]

0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1

907.6 907.6 907.6 907.6 907.6 907.6 907.6 907.6 907.6 907.6 907.6

[F] 899.5 900.5 901.4 902.3 903.2 904 904.8 905.6 906.3 907 907.6

6-257

8-34E Air is decelerated in an adiabatic diffuser with an isentropic efficiency of 0.82. The air velocity at the exit is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. There is no heat transfer or shaft work associated with the process. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 0.240 Btu / lbm ×R and k = 1.4 (Table A-2Ea). Analysis For the isentropic process of an ideal gas, the exit temperature is determined from ( k - 1)/ k

æP ö÷ T2 s = T1 ççç 2 ÷ ÷ çè P ø÷ 1

0.4/1.4

æ20 psia ÷ ö ÷ = (30 + 460 R) çç ÷ çè11 psia ø÷

= 581.3 R

There is only one inlet and one exit, and thus m1 = m2 = m. We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout æ æ V2ö V2ö ÷ ÷ m çççh1 + 1 ÷ = m çççh2 + 2 ÷ ÷ ÷ ÷ ÷ çè 2ø 2ø èç

V2 V2 h1  1  h2  2 2 2 h1  h2 

V22  V12 2

6-258

c p (T1  T2 ) 

V22  V12  ke 2

The kinetic energy change for the isentropic case is

D ke s = c p (T2 s - T1 ) = (0.240 Btu/lbm ×R)(581.3 - 490)R = 21.90 Btu/lbm The kinetic energy change for the actual process is

D kea = hN D kes = (0.82)(21.90 Btu/lbm) = 17.96 Btu/lbm Substituting into the energy balance and solving for the exit velocity gives 0.5

2 1

V2 = (V - 2D ke a )

0.5

é æ25,037 ft 2 /s 2 ö÷ù ú = 735 ft / s ÷ = êê(1200 ft/s) 2 - 2(17.96 Btu/lbm) ççç ÷ ú ÷ 1 Btu/lbm è ø ëê ûú

EES (Engineering Equation Solver) SOLUTION "Given" P1=11 [psia] T1=(30+460) [R] P2=20 [psia] eta_D=0.82 Vel1=1200 [ft/s] "Properties" c_p=0.240 [Btu/lbm-R] "Table A-2Ea" k=1.4 "Table A-2Ea" "Analysis" T2_s=T1*(P2/P1)^((k-1)/k) DELTAke_s=c_p*(T2_s-T1) DELTAke_a=eta_D*DELTAke_s DELTAke_a=(Vel1^2-Vel2^2)/2*Convert(ft^2/s^2, Btu/lbm)

8-35 Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temperature are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Combustion gases can be treated as air that is an ideal gas with variable specific heats.

Analysis From the air table (Table A-17),

T1 = 1020 K ¾ ¾ ® h1 = 1068.89 kJ/kg, s1o = 2.99034 kJ/kg ×K From the isentropic relation, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-259

s2o  s1o  R ln

P2  85 kPa   2.99034 kJ/kg  K  (0.287 kJ/kg  K) ln   h2 s  784.11 kJ/kg   2.66947 kJ/kg  K  P1  260 kPa 

There is only one inlet and one exit, and thus m1 = m2 = m. We take the nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system for the isentropic process can be expressed as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout m(h1 + V12 / 2) = m(h2 s +V22s /2) (since W = Q @D pe @ 0)

h2 s  h1 

V22s  V12 2

Then the isentropic exit velocity becomes

V2 s = V12 + 2 (h1 - h2 s ) =

æ1000 m2 /s 2 ÷ ö ÷ = 758.9 m/s ÷ çè 1 kJ/kg ÷ ø

(80 m/s) + 2 (1068.89 - 784.11)kJ/kg ççç 2

Therefore, V2 a =

h N V2 s =

0.92 (758.9 m/s) = 727.9 m/s @728 m / s

The exit temperature of air is determined from the steady-flow energy equation, 2

h2 a = 1068.89 kJ/kg -

ö (727.9 m/s) - (80 m/s) æ ç 1 kJ/kg ÷ 2

÷ ÷= 807.15 kJ/kg ççè1000 m 2 /s 2 ø

From the air table we read

T2a  786 K

EES (Engineering Equation Solver) SOLUTION "Given" P1=260 [kPa] T1=747 [C] Vel1=80 [m/s] P2=85 [kPa] eta_N=0.92 "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) "Analysis" "(a)" s2_s=s1 h2_s=enthalpy(Fluid$, P=P2, s=s2_s) h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2_s+Vel2_s^2/2*Convert(m^2/s^2, kJ/kg) Vel2_a=sqrt(eta_N)*Vel2_s "(b)" h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2_a+Vel2_a^2/2*Convert(m^2/s^2, kJ/kg) T2_a=temperature(Fluid$, h=h2_a)

6-260

8-36 Air is expanded in an adiabatic nozzle with an isentropic efficiency of 0.96. The air velocity at the exit is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. There is no heat transfer or shaft work associated with the process. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ / kg.K and k = 1.4 (Table A-2a). Analysis For the isentropic process of an ideal gas, the exit temperature is determined from ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

0.4/1.4

æ100 kPa ö÷ = (180 + 273 K) çç ÷ èç300 kPa ø÷

= 331.0 K

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout æ æ V2ö V2ö ÷ ÷ m çççh1 + 1 ÷ = m çççh2 + 2 ÷ ÷ ÷ ÷ ÷ çè 2ø 2ø èç

V2 V2 h1  1  h2  2 2 2 h1  h2 

V22  V12 2

c p (T1  T2 ) 

V22  V12  ke 2

The kinetic energy change for the isentropic case is

D kes = c p (T1 - T2 s ) = (1.005 kJ/kg ×K)(453 - 331)K = 122.6 kJ/kg The kinetic energy change for the actual process is

D kea = hN D kes = (0.93)(122.6 kJ/kg) = 114.1 kJ/kg Substituting into the energy balance and solving for the exit velocity gives 0.5

V2 = (2D kea )

0.5

é æ1000 m2 /s 2 ÷ öù ú = 478 m / s ÷ = êê2(114.1 kJ/kg) ççç ÷ ú ÷ 1 kJ/kg è ø ëê ûú

EES (Engineering Equation Solver) SOLUTION "Given" P1=300 [kPa] T1=(180+273) [K] P2=100 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-261

Vel1=0 [m/s] eta_N=0.93 "Properties" c_p=1.005 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" T2_s=T1*(P2/P1)^((k-1)/k) DELTAke_s=c_p*(T1-T2_s) DELTAke_a=eta_N*DELTAke_s DELTAke_a=(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg)

Entropy Balance 8-37E An unknown mass of iron is dropped into water in an insulated tank while being stirred by a 200-W paddle wheel. Thermal equilibrium is established after 10 min. The mass of the iron block and the entropy generated during this process are to be determined. Assumptions 1. Both the water and the iron block are incompressible substances with constant specific heats at room temperature. 2. The system is stationary and thus the kinetic and potential energy changes are zero. 3. The system is well-insulated and thus there is no heat transfer. Properties The specific heats of water and the iron block at room temperature are c p , water = 1.00 Btu / lbm.° F and c p , iron = 0.107 Btu / lbm.° F (Table A-3E). The density of water at room temperature is 62.1 lbm / ft 3 .

Analysis (a) We take the entire contents of the tank, water + iron block, as the system. This is a closed system since no mass crosses the system boundary during the process. The energy balance on the system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wpw,in  U or,

Wpw, in = D U iron + D U water Wpw, in = [mc (T2 - T1 )]iron + [mc (T2 - T1 )]water

where

mwater = r V = (62.1 lbm/ft 3 )(0.8 ft 3 ) = 49.7 lbm

 1 Btu  Wpw  Wpw t  (0.2 kJ/s)(10  60 s)    113.7 Btu  1.055 kJ  Using specific heat values for iron and liquid water and substituting, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-262

113.7 Btu = miron (0.107 Btu/lbm ×°F)(75 - 185)°F + (49.7 lbm)(1.00 Btu/lbm ×°F)(75 - 70)°F

miron  11.4 lbm (b) Again we take the iron + water in the tank to be the system. Noting that no heat or mass crosses the boundaries of this combined system, the entropy balance for it can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0  Sgen, total  Siron  Swater where

æT ö æ535 R ö÷ ÷ D Siron = mcavg ln ççç 2 ÷ = (11.4 lbm)(0.107 Btu/lbm ×R )ln çç ÷ ÷= - 0.228 Btu/R çè645 R ø÷ èç T1 ø÷

T   535 R  S water  mcavg ln  2    49.6 lbm 1.0 Btu/lbm  R  ln    0.466 Btu/R  530 R   T1  Therefore, the entropy generated during this process is

D Stotal = Sgen = D Siron + D Swater = - 0.228 + 0.466 = 0.238 Btu / R

EES (Engineering Equation Solver) SOLUTION "Given" T1_iron=(185+460) [R] T1_w=(70+460) [R] T2=(75+460) [R] V_w=0.8 [ft^3] W_dot_pw_in=0.200 [kW] time=10*60 [s] "Properties" rho_w=62.1 [lbm/ft^3] "Table A-3E" c_w=1.00 [Btu/lbm-R] "Table A-3E" c_iron=0.107 [Btu/lbm-R] "Table A-3E" "Analysis" W_pw_in=m_w*c_w*(T2-T1_w)+m_iron*c_iron*(T2-T1_iron) "energy balance" m_w=rho_w*V_w W_pw_in=W_dot_pw_in*time*Convert(kJ, Btu) S_gen_total=DELTAS_iron+DELTAS_w "entropy balance" DELTAS_iron=m_iron*c_iron*ln(T2/T1_iron) DELTAS_w=m_w*c_w*ln(T2/T1_w)

8-38E Refrigerant-134a is expanded adiabatically from a specified state to a given pressure. The entropy generation is to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. Analysis The rate of entropy generation within the expansion device during this process can be determined by applying the rate form of the entropy balance on the system. Noting that the system is adiabatic and thus there is no heat transfer, the entropy balance for this steady-flow system can be expressed as

6-263

Sin - Sout

Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 (steady)

Rate of entropy generation

Rate of change of entropy

m1s1  m2 s2  Sgen  0 Sgen  m( s2  s1 )

sgen  s2  s1 The properties of the refrigerant at the inlet and exit states are (Tables A-11E through A-13E) P1 = 100 psia ü ïï ý h1 = 118.90 Btu/lbm, s1 = 0.22900 Btu/lbm ×R ïïþ T1 = 100°F

h2 = h1 = 118.90 Btu/lbm ü P2 = 10 psia ïï ý s2 = 0.27143 Btu/lbm ×R h2 = 118.90 Btu/lbmïïþ

Substituting,

sgen = s2 - s1 = 0.27143 - 0.22900 = 0.0424 Btu / lbm×R

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [psia] T1=100 [F] P2=10 [psia] "Analysis" Fluid$='R134a' h1=enthalpy(Fluid$, P=P1, T=T1) h2=h1 s1=entropy(Fluid$, P=P1, T=T1) s2=entropy(Fluid$, P=P2, h=h2) s_gen=s2-s1

8-39E A cylinder contains saturated liquid water at a specified pressure. Heat is transferred to liquid from a source and some liquid evaporates. The total entropy generation during this process is to be determined. Assumptions 1. No heat loss occurs from the water to the surroundings during the process. 2. The pressure inside the cylinder and thus the water temperature remains constant during the process. 3. No irreversibilities occur within the cylinder during the process. Analysis The pressure of the steam is maintained constant. Therefore, the temperature of the steam remains constant also at

T = Tsat@40 psia = 267.2°F = 727.2 R (Table A-5E) Taking the contents of the cylinder as the system and noting that the temperature of water remains constant, the entropy change of the system during this isothermal, internally reversible process becomes

6-264

D Ssystem =

Qsys,in Tsys

600 Btu = 0.8251 Btu/R 727.2 R

Similarly, the entropy change of the heat source is determined from

D Ssource = -

Qsource, out Tsource

600 Btu = - 0.4110 Btu/R 1000 + 460 R

Now consider a combined system that includes the cylinder and the source. Noting that no heat or mass crosses the boundaries of this combined system, the entropy balance for it can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0  Sgen,total  Swater  Ssource Therefore, the total entropy generated during this process is

Sgen, total = D Swater + D Ssource = 0.8251- 0.4110 = 0.414 Btu / R Discussion The entropy generation in this case is entirely due to the irreversible heat transfer through a finite temperature difference. We could also determine the total entropy generation by writing an energy balance on an extended system that includes the system and its immediate surroundings so that part of the boundary of the extended system, where heat transfer occurs, is at the source temperature.

EES (Engineering Equation Solver) SOLUTION "Given" P1=40 [psia] x1=0 P2=P1 T_source=(1000+460) [R] Q_sys_in=600 [Btu] "Analysis" T1=temperature(steam_iapws, P=P1, x=x1) DELTAS_sys=Q_sys_in/(T1+460) DELTAS_source=-Q_source_out/T_source Q_source_out=Q_sys_in S_gen_total=DELTAS_sys+DELTAS_source "entropy balance on combined system"

8-40 Heat is lost from the air that is being compressed in a compressor. The exit temperature of the air, the work input to the compressor, and the entropy generation during this process are to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.287 kPa ×m3 / kg ×K . Also, the specific heat at room temperature is c p = 1.005 kJ / kg.K (Table A-2). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-265

Analysis (a) The exit temperature of the air may be determined from the relation for the entropy change of air

D Sair = c p ln

T2 P - R ln 2 Ta1 P1

0.40 kJ/kg.K  (1.005 kJ/kg.K)ln

T2 800 kPa  (0.287 kJ/kg.K)ln (22  273) K 100 kPa

T2  358.8 K  85.8C (b) The work input to the compressor is obtained from an energy balance on the compressor

win = c p (T2 - T1 ) + qout = (1.005 kJ/kg.°C)(85.8 - 22)°C + 120 kJ/kg = 184.1kJ / kg (c) The entropy generation associated with this process may be obtained by adding the entropy change of air as it is compressed in the compressor and the entropy change of the surroundings D ssurr =

qout 120 kJ/kg = = 0.4068 kJ/kg.K Tsurr (22 + 273) K

sgen = D stotal = D sair + D ssurr = - 0.40 + 0.4068 = 0.0068 kJ / kg.K

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] T_1=22[C]+273 [K] P_2=800 [kPa] q_out=120 [kJ/kg] DELTAs_air=-0.40 [kJ/kg-K] "PROPERTIES" Fluid$='Air' C_p=CP(Fluid$, T=T_ave) T_ave=1/2*(T_1+T_2) R=R_u/MM R_u=8.314 [kJ/kmol-K] MM=MolarMass(Fluid$) "ANALYSIS" DELTAs_air=C_p*ln(T_2/T_1)-R*ln(P_2/P_1) C_p*T_1+w_in=C_p*T_2+q_out "energy balance" DELTAs_surr=q_out/T_1 "entropy change of surroundings" s_gen=DELTAs_air+DELTAs_surr "entropy generation"

8-41 A steam turbine for which the power output and the isentropic efficiency are given is considered. The mass flow rate of the steam, the temperature of the steam at the turbine exit, and the entropy generation are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-266

1. Steady operating conditions exist. 2. Potential energy changes are negligible. Analysis

(a) The properties of the steam at the inlet of the turbine and the enthalpy at the exit for the isentropic case are (Table A-6)

P1 = 7 MPa ïü ïý h1 = 3411.4 kJ/kg T1 = 500°C ïïþ s1 = 6.8000 kJ/kg.K P2  100 kPa

  h  2466.6 kJ/kg s2  s1  6.8000 kJ/kg.K  2 s The power output if the expansion was isentropic would be

Ws =

Wa 5000 kW = = 6494 kW hT 0.77

An energy balance on the turbine for the isentropic process may be used to determine the mass flow rate of the steam

  V2  V2  m  h1  1   m  h2 s  2   Ws 2  2      (45 m/s)2  1 kJ/kg   (75 m/s)2  1 kJ/kg   m 3411.4 kJ/kg   m 2466.6 kJ/kg   6494 kW    2 2  2 2  2 2  1000 m /s    1000 m /s     m  6.886 kg / s (b) An energy balance on the turbine for the actual process may be used to determine actual enthalpy at the exit

æ æ V2ö V2ö ÷= m ççh2 + 2 ÷ ÷+ Wa m çççh1 + 1 ÷ ÷ ÷ çè 2 ø÷ 2 ø÷ èçç

  (45 m/s) 2  1 kJ/kg   (75 m/s) 2  1 kJ/kg   (6.886 kg/s) 3411.4 kJ/kg   (6.886 kg/s) h   5000 kW   2  2 2  2 2  2 2  1000 m /s    1000 m /s    

h2  2683.5 kJ/kg Now, other properties at the exit state may be obtained ïü ïýT2 = 103.7°C h2 = 2683.5 kJ/kgïïþ s2 = 7.3817 kJ/kg.K P2 = 100 kPa

Sgen = m(s2 - s1 ) = (6.886 kg/s)(7.3817 - 6.8000) kJ/kg.K = 4.01kW / K

6-267

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=7000 [kPa] T_1=500 [C] Vel_1=45 [m/s] P_2=100 [kPa] Vel_2=75 [m/s] W_dot_actual=5000 [kW] eta_T=0.77 "ANALYSIS" Fluid$='Steam_iapws' h_1=enthalpy(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) m_dot*(h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg))=m_dot*(h_2+Vel_2^2/2*Convert(m^2/s^2, kJ/kg))+W_dot_actual "energy balance for actual process" m_dot*(h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg))=m_dot*(h_2s+Vel_2^2/2*Convert(m^2/s^2, kJ/kg))+W_dot_isentropic "energy balance for isentropic process" eta_T=W_dot_actual/W_dot_isentropic T_2=temperature(Fluid$, P=P_2, h=h_2) s_2=entropy(Fluid$, P=P_2, h=h_2) S_dot_gen=m_dot*(s_2-s_1)

8-42 In an ice-making plant, water is frozen by evaporating saturated R-134a liquid. The rate of entropy generation is to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. Analysis We take the control volume formed by the R-134a evaporator with a single inlet and single exit as the system. The rate of entropy generation within this evaporator during this process can be determined by applying the rate form of the entropy balance on the system. The entropy balance for this steady-flow system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 (steady)

Rate of entropy generation

m1s1  m2 s2 

Rate of change of entropy

Qin  Sgen  0 Tw

Sgen  mR ( s2  s1 )  Sgen  mR s fg 

Qin Tw

6-268

The properties of the refrigerant are (Table A-11) h fg@- 16°C = 210.18 kJ/kg

s fg@16C  0.81729 kJ/kg  K The rate of that must be removed from the water in order to freeze it at a rate of 5500 kg / h is

Qin = mw hif = (5500 / 3600 kg/s)(333.7 kJ/kg) = 509.8 kW where the heat of fusion of water at 1 atm is 333.7 kJ / kg . The mass flow rate of R-134a is mR =

Qin 509.8 kJ/s = = 2.426 kg/s h fg 210.18 kJ/kg

Substituting,

Sgen = mR s fg -

Qin 509.8 kW = (2.426 kg/s)(0.81729 kJ/kg ×K) = 0.115 kW / K Tw 273 K

EES (Engineering Equation Solver) SOLUTION "Given" T_w=(0+273) [K] T=-16 [C] x1=0 x2=1 m_dot_w=(5500/3600) [kg/s] "Properties" h_if=333.7 [kJ/kg] "heat of fusion of water at 1 atm" "Analysis" Fluid$='R134a' h_f=enthalpy(Fluid$, T=T, x=0) h_g=enthalpy(Fluid$, T=T, x=1) h_fg=h_g-h_f s_f=entropy(Fluid$, T=T, x=0) s_g=entropy(Fluid$, T=T, x=1) s_fg=s_g-s_f Q_dot_in=m_dot_w*h_if m_dot_R=Q_dot_in/h_fg S_dot_gen=m_dot_R*s_fg-Q_dot_in/T_w

8-43 Oxygen is cooled as it flows in an insulated pipe. The rate of entropy generation in the pipe is to be determined. Assumptions 1. Steady operating conditions exist. 2. The pipe is well-insulated so that heat loss to the surroundings is negligible. 3. Changes in the kinetic and potential energies are negligible. 4. Oxygen is an ideal gas with constant specific heats. Properties The properties of oxygen at room temperature are R = 0.2598 kJ / kg.K , c p = 0.918 kJ / kg.K (Table A-2a). Analysis The rate of entropy generation in the pipe is determined by applying the rate form of the entropy balance on the pipe:

6-269

Sin - Sout

Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 (steady)

Rate of entropy generation

Rate of change of entropy

m1s1  m2 s2  Sgen  0 (since Q  0)

Sgen  m( s2  s1 ) The specific volume of oxygen at the inlet and the mass flow rate are

v1 =

RT1 (0.2598 kPa ×m3 /kg ×K)(293 K) = = 0.3172 m3 /kg P1 240 kPa

AV p D 2V1 p (0.12 m) 2 (70 m/s) 1 1 = = = 2.496 kg/s v1 4v1 4(0.3172 m3 /kg)

Substituting into the entropy balance relation,

Sgen = m( s2 - s1 )

 T P   m  c p ln 2  R ln 2  T1 P1  

291 K 200 kPa    (2.496 kg/s) (0.918 kJ/kg  K)ln  (0.2598 kJ/kg  K)ln 293 K 240 kPa    0.1025 kW / K EES (Engineering Equation Solver) SOLUTION "Given" D=0.12 [m] Vel1=70 [m/s] P1=240 [kPa] T1=(20+273) [K] P2=200 [kPa] T2=(18+273) [K] "Properties" R=0.2598 [kJ/kg-K] "Table A-2a" c_p=0.918 [kJ/kg-K] "Table A-2a" "Analysis" v1=(R*T1)/P1 A=pi*D^2/4 m_dot=(A*Vel1)/v1 DELTAs=c_p*ln(T2/T1)-R*ln(P2/P1) S_dot_gen=m_dot*DELTAs

8-44E Water and steam are mixed in a chamber that is losing heat at a specified rate. The rate of entropy generation during this process is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-270

Assumptions 1.

This is a steady-flow process since there is no change with time at any point and thus ΔmCV  0 , ΔECV  0 , and

ΔSCV  0 . 2. There are no work interactions involved. 3. The kinetic and potential energies are negligible, ke = pe = 0. Analysis We take the mixing chamber as the system. This is a control volume since mass crosses the system boundary during the process. We note that there are two inlets and one exit. Under the stated assumptions and observations, the mass and energy balances for this steady-flow system can be expressed in the rate form as follows: Mass balance:

min - mout = D msystemZ 0 (steady) = 0

min  mout  m1  m2  m3 Energy balance:

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout m1h1  m2 h2  m3h3  Qout Combining the mass and energy balances gives Qout = m1h1 + m2 h2 - (m1 + m2 )h3

The desired properties at the specified states are determined from the steam tables to be P1 = 20 psia ü ïï h1 = h f @ 50° F = 18.07 Btu/lbm ý ïïþ s1 = s f @ 50° F = 0.03609 Btu/lbm ×R T1 = 50°F P2 = 20 psia ïü ïý h2 = 1162.3 Btu/lbm T2 = 240°F ïïþ s2 = 1.7406 Btu/lbm ×R

P3 = 20 psia ïü ïý h3 = h f @130° F = 97.99 Btu/lbm ïïþ s3 = s f @130° F = 0.18174 Btu/lbm ×R

T3 = 130°F

Substituting, 180 Btu/min = [300´ 18.07 + m2 ´ 1162.3 - (300 + m2 ) ´ 97.99]Btu/min ¾ ¾ ® m2 = 22.7 lbm/min

The rate of entropy generation during this process can be determined by applying the rate form of the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature of the

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 (steady)

Rate of entropy generation

m1s1  m2 s2  m3 s3 

Rate of change of entropy

Qout  Sgen  0 Tb

Substituting, the rate of entropy generation is determined to be

Sgen = m3 s3 - m1 s1 - m2 s2 +

Qout Tb

6-271

 (322.7  0.18174  300  0.03609  22.7 1.7406) Btu/min  R) 

180 Btu/min 530 R

 8.65 Btu / min  R Discussion Note that entropy is generated during this process at a rate of 8.65 Btu / min · R. This entropy generation is caused by the mixing of two fluid streams (an irreversible process) and the heat transfer between the mixing chamber and the surroundings through a finite temperature difference (another irreversible process).

EES (Engineering Equation Solver) SOLUTION "Given" P1=20 [psia] T1=50 [F] m_dot_1=300 [lbm/min] P2=20 [psia] T2=240 [F] P3=20 [psia] T3=130 [F] T_surr=(70+460) [R] Q_dot_out=180 [Btu/min] "Analysis" Fluid$='steam_iapws' "state 1 is approximated as saturated liquid" h1=enthalpy(Fluid$, T=T1, x=0) s1=entropy(Fluid$, T=T1, x=0) h2=enthalpy(Fluid$, P=P2, T=T2) s2=entropy(Fluid$, P=P2, T=T2) "state 3 is approximated as saturated liquid" h3=enthalpy(Fluid$, T=T3, x=0) s3=entropy(Fluid$, T=T3, x=0) m_dot_3=m_dot_1+m_dot_2 Q_dot_out=m_dot_1*h1+m_dot_2*h2-m_dot_3*h3 S_dot_gen=m_dot_3*s3-m_dot_1*s1-m_dot_2*s2+Q_dot_out/T_surr

8-45 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the rate of entropy generation within the heat exchanger are to be determined. Assumptions 1. Steady operating conditions exist. 2. The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the kinetic and potential energies of fluid streams are negligible. 4. Fluid properties are constant.

6-272

Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ / kg.° C , respectively. Analysis (a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout Qin  mh1  mh2 (since ke  pe  0) Qin  mc p (T2  T1 ) Then the rate of heat transfer to the cold water in this heat exchanger becomes

Qin = [mc p (Tout - Tin )]cold water = (0.95 kg/s)(4.18 kJ/kg.°C)(70°C - 10°C)=238.3 kW Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of the hot water is determined to be Q = [mc p (Tin - Tout )]hot water ¾ ¾ ® Tout = Tin -

Q 238.3 kW = 85°C = 49.5°C mc p (1.6 kg/s)(4.19 kJ/kg.°C)

(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem 0 (steady)

Rate of entropy generation

Rate of change of entropy

m1s1  m3 s3  m2 s2  m3 s4  Sgen  0 (since Q  0)

mcold s1  mhot s3  mcold s2  mhot s4  Sgen  0 Sgen  mcold ( s2  s1 )  mhot ( s4  s3 ) Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be T T Sgen = mcold c p ln 2 + mhot c p ln 4 T1 T3

 (0.95 kg/s)(4.18 kJ/kg.K)ln

70+273 49.5+273  (1.6 kg/s)(4.19 kJ/kg.K)ln 10+273 85+273

 0.06263 kW / K EES (Engineering Equation Solver) SOLUTION "Given" T1=(10+273) [K] T2=(70+273) [K] m_dot_cw=0.95 [kg/s] T3=(85+273) [K] m_dot_hw=1.6 [kg/s] c_p_cw=4.18 [kJ/kg-C] c_p_hw=4.19 [kJ/kg-C] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-273

"(a)" Q_dot=m_dot_cw*c_p_cw*(T2-T1) Q_dot=m_dot_hw*c_p_hw*(T3-T4) "(b)" S_dot_gen=m_dot_cw*c_p_cw*ln(T2/T1)+m_dot_hw*c_p_hw*ln(T4/T3) "entropy balance"

8-46 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer, the outlet temperature of the air, and the rate of entropy generation are to be determined. Assumptions 1. Steady operating conditions exist. 2. The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the kinetic and potential energies of fluid streams are negligible. 4. Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ / kg.° C , respectively. The gas constant of air is R = 0.287 kJ / kg.K (Table A-1).

Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout mh1  Qout  mh2 (since ke  pe  0) Qout  mc p (T1  T2 ) Then the rate of heat transfer from the exhaust gases becomes

Q = [mc p (Tin - Tout )]gas. = (2.2 kg/s)(1.1 kJ/kg.°C)(180°C - 95°C)= 205.7 kW The mass flow rate of air is

PV (95 kPa)(1.6 m3 /s) = = 1.808 kg/s RT (0.287 kPa ×m3 /kg ×K)(293 K)

Noting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of the air becomes Q = éêëmc p (Tout - Tin )ù ú û

air

¾¾ ® Tout = Tin +

Q 205.7 kW = 20°C + = 133.2°C mc p (1.808 kg/s)(1.005 kJ/kg ×°C)

6-274

The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem 0 (steady) Rate of change of entropy

m1s1  m3 s3  m2 s2  m3 s4  Sgen  0 (since Q  0) mexhaust s1  mair s3  mexhaust s2  mair s4  Sgen  0 Sgen  mexhaust ( s2  s1 )  mair ( s4  s3 ) Then the rate of entropy generation is determined to be T T Sgen = mexhaust c p ln 2 + mair c p ln 4 T1 T3

 (2.2 kg/s)(1.1 kJ/kg  K)ln

(95+273)K (133.2+273)K  (1.808 kg/s)(1.005 kJ/kg  K)ln (180+273)K (20+273)K

\  0.091 kW / K

EES (Engineering Equation Solver) SOLUTION "Given" P1=95 [kPa] T1=(20+273) [K] V1_dot_air=1.6 [m^3/s] T3=(180+273) [K] T4=(95+273) [K] m_dot_gas=2.2 [kg/s] c_p_air=1.005 [kJ/kg-C] c_p_gas=1.10 [kJ/kg-C] "Properties" R=0.287 [kJ/kg-K] "Table A-1" "Analysis" Q_dot=m_dot_gas*c_p_gas*(T3-T4) m_dot_air=(P1*V1_dot_air)/(R*T1) Q_dot=m_dot_air*c_p_air*(T2-T1) T2_C=T2-273 S_dot_gen=m_dot_air*c_p_air*ln(T2/T1)+m_dot_gas*c_p_gas*ln(T4/T3) "entropy balance"

8-47 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam and the rate of entropy generation are to be determined. Assumptions 1. Steady operating conditions exist. 2. The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the kinetic and potential energies of fluid streams are negligible. 4. Fluid properties are constant. Properties The enthalpy and entropy of vaporization of water at 60C are h fg = 2357.7 kJ / kg and s fg = 7.0769 kJ / kg.K (Table A-4). The specific heat of water at room temperature is c p = 4.18 kJ / kg.° C (Table A-3). Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-275

(a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout Qin  mh1  mh2 (since ke  pe  0) Qin  mc p (T2  T1 ) Then the heat transfer rate to the cooling water in the condenser becomes Q = [mc p (Tout - Tin )]cooling water

 (75 kg/s)(4.18 kJ/kg.C)(27C  18C)=2822 kJ/s The rate of condensation of steam is determined to be Q = (mh fg )steam ¾ ¾ ® msteam =

Q 2822 kJ/s = = 1.20 kg / s h fg 2357.7 kJ/kg

(b) The rate of entropy generation within the condenser during this process can be determined by applying the rate form of the entropy balance on the entire condenser. Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem 0 (steady) Rate of change of entropy

m1s1  m3 s3  m2 s2  m4 s4  Sgen  0 (since Q  0) mwater s1  msteam s3  mwater s2  msteam s4  Sgen  0 Sgen  mwater ( s2  s1 )  msteam ( s4  s3 ) m1s1  m3 s3  m2 s2  m4 s4  Sgen  0 (since Q  0) mwater s1  msteam s3  mwater s2  msteam s4  Sgen  0 Sgen  mwater ( s2  s1 )  msteam ( s4  s3 ) Noting that water is an incompressible substance and steam changes from saturated vapor to saturated liquid, the rate of entropy generation is determined to be T T Sgen = mwater c p ln 2 + msteam ( s f - sg ) = mwater c p ln 2 - msteam s fg T1 T1

6-276

 (75 kg/s)(4.18 kJ/kg  K)ln

27+273  (1.20 kg/s)(7.0769 kJ/kg  K) 18+273

 1.06 kW / K EES (Engineering Equation Solver) SOLUTION "Given" T_s=60[C] T1=18[C] T2=27[C] m_dot_w=75 [kg/s] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_s, x=0) h_g=enthalpy(Fluid$, T=T_s, x=1) h_fg=h_g-h_f s_f=entropy(Fluid$, T=T_s, x=0) s_g=entropy(Fluid$, T=T_s, x=1) s_fg=s_g-s_f C_p=4.18 [kJ/kg-K] "Analysis" "(a)" Q_dot=m_dot_w*C_p*(T2-T1) m_dot_s=Q_dot/h_fg "(b)" S_dot_gen=m_dot_w*C_p*ln((T2+273)/(T1+273))-m_dot_s*s_fg "entropy balance"

8-48 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked and the amount of entropy generation associated with this heat transfer process are to be determined. Assumptions 1.

The egg is spherical in shape with a radius of r0  2.75 cm .

The thermal properties of the egg are constant.

Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible.

There are no changes in kinetic and potential energies.

Properties The density and specific heat of the egg are given to be ρ = 1020 kg / m3 and c p = 3.32 kJ / kg.° C . Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

6-277

Qin  U egg  m(u2  u1 )  mc(T2  T1 ) Then the mass of the egg and the amount of heat transfer become m = rV = r

p D3 p (0.055 m)3 = (1020 kg/m3 ) = 0.0889 kg 6 6

Qin  mc p (T2  T1 )  (0.0889 kg)(3.32 kJ/kg.C)(70  8)C  18.3 kJ We again take a single egg as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the egg and its immediate surroundings so that the boundary temperature of the extended Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin Q  Sgen  Ssystem  Sgen   in  Ssystem Tb Tb where T 70 + 273 D Ssystem = m( s2 - s1 ) = mcavg ln 2 = (0.0889 kg)(3.32 kJ/kg.K) ln = 0.0588 kJ/K T1 8 + 273

Substituting, Sgen = -

Qin 18.3 kJ + D Ssystem = + 0.0588 kJ/K = 0.00934 kJ / K (per egg) Tb 370 K

EES (Engineering Equation Solver) SOLUTION "Given" D=0.055 [m] T1=(8+273) [K] T2=(70+273) [K] T_w=(97+273) [K] rho=1020 [kg/m^3] c_p=3.32 [kJ/kg-C] "Analysis" V=pi*D^3/6 m=rho*V Q_in=m*c_p*(T2-T1) Q_in/T_w+S_gen=DELTAS_sys "entropy balance" DELTAS_sys=m*c_p*ln(T2/T1)

8-49 Chickens are to be cooled by chilled water in an immersion chiller that is also gaining heat from the surroundings. The rate of heat removal from the chicken and the rate of entropy generation during this process are to be determined. Assumptions 1. Steady operating conditions exist. 2. Thermal properties of chickens and water are constant. 3.

The temperature of the surrounding medium is given to be 25C.

Properties The specific heat of chicken is given to be 3.54 kJ / kg.° C . The specific heat of water at room temperature is 4.18 kJ / kg.° C (Table A-3). Analysis

6-278

(a) Chickens can be considered to flow steadily through the chiller at a mass flow rate of

mchicken = (250 chicken/h)(2.2 kg/chicken) = 550 kg/h = 0.1528 kg/s Taking the chicken flow stream in the chiller as the system, the energy balance for steadily flowing chickens can be expressed in the rate form as Ein - Eout = D EsystemZ 0 (steady) = 0 ® Ein = Eout Rate of net energy transfer by heat, work, and mass

Rate of change in internal, kinetic, potential, etc. energies

mh1  Qout  mh2 (since ke  pe  0) Qout  Qchicken  mchicken c p (T1  T2 ) Then the rate of heat removal from the chickens as they are cooled from 15C to 3C becomes

Qchicken = (mc p D T )chicken = (0.1528 kg/s)(3.54 kJ/kg.º°C)(15 - 3)°C = 6.49 kW The chiller gains heat from the surroundings as a rate of 150 kJ / h = 0.0417 kJ / s . Then the total rate of heat gain by the water is

Qwater = Qchicken + Qheat gain = 6.49 + 0.0417 = 6.532 kW chiller, the mass flow rate of water must be at least mwater =

Qwater 6.532 kW = = 0.781 kg/s (c p D T ) water (4.18 kJ/kg.°C)(2°C)

(b) The rate of entropy generation is determined by applying the entropy balance on an extended system that includes the chiller and the immediate surroundings so that boundary temperature is the surroundings temperature:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem ¿ 0 (steady) Rate of change of entropy

T T Q Sgen  mchicken c p ln 2  mwater c p ln 4  in T1 T3 Tsurr 276 275.5 0.0417 kW  (0.1528 kg/s)(3.54 kJ/kg.K)ln  (0.781 kg/s)(4.18 kJ/kg.K)ln  288 273.5 298 K  0.000625 kW / K Noting that both streams are incompressible substances, the rate of entropy generation is determined to be

EES (Engineering Equation Solver) SOLUTION "Given" T1=ConvertTemp(C, K, 0.5[C]) T2=ConvertTemp(C, K, 2.5[C]) T3=ConvertTemp(C, K, 15[C]) T4=ConvertTemp(C, K, 3[C]) m_chicken=2.2 [kg] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-279

N_dot_chicken=250 [1/h] T_surr=ConvertTemp(C, K, 25[C]) Q_dot_HeatGain=150[kJ/h]*Convert(kJ/h, kW) "Properties" C_p_chicken=3.54 [kJ/kg-C] "given" C_p_w=4.18 [kJ/kg-C] "Analysis" "(a)" m_dot_chicken=N_dot_chicken*m_chicken*1/Convert(kg/s, kg/h) Q_dot_chicken=m_dot_chicken*C_p_chicken*(T3-T4) Q_dot_w=Q_dot_chicken+q_dot_HeatGain Q_dot_w=m_dot_w*C_p_w*(T2-T1) "(b)" S_dot_gen=m_dot_w*C_p_w*ln(T2/T1)+m_dot_chicken*C_p_chicken*ln(T4/T3)-Q_dot_HeatGain/T_surr "entropy balance"

8-50E Large brass plates are heated in an oven at a specified rate. The rate of heat transfer to the plates in the oven and the rate of entropy generation associated with this heat transfer process are to be determined. Assumptions 1. The thermal properties of the plates are constant. 2. The changes in kinetic and potential energies are negligible.

Properties The density and specific heat of the brass are given to be ρ = 532.5 lbm / ft 3 and c p = 0.091 Btu / lbm.° F . Analysis We take the plate to be the system. The energy balance for this closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin  U plate  m(u2  u1 )  mc(T2  T1 ) The mass of each plate and the amount of heat transfer to each plate is m = r V = r LA = (532.5 lbm/ft 3 )[(1.2 /12 ft)(2 ft)(2 ft)] = 213 lbm

Qin = mc(T2 - T1 ) = (213 lbm/plate)(0.091 Btu/lbm.°F)(1000 - 75)°F = 17,930 Btu/plate Then the total rate of heat transfer to the plates becomes

Qtotal = nplateQin, per plate = (450 plates/min)´ (17,930 Btu/plate) = 8,069,000 Btu / min = 134,500 Btu / s We again take a single plate as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the plate and its immediate surroundings so that the boundary © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-280

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin Q  Sgen  Ssystem  Sgen   in  Ssystem Tb Tb where T 1000 + 460 D Ssystem = m( s2 - s1 ) = mcavg ln 2 = (213 lbm)(0.091 Btu/lbm.R)ln = 19.46 Btu/R T1 75 + 460

Substituting, Sgen = -

Qin 17,930 Btu + D Ssystem = + 19.46 Btu/R = 9.272 Btu/R (per plate) Tb 1300 + 460 R

Then the rate of entropy generation becomes

Sgen = Sgen nball = (9.272 Btu/R ×plate)(450 plates/min) = 4172 Btu/min ×R = 69.5 Btu / s.R EES (Engineering Equation Solver) SOLUTION "Given" L=1.2[in]*Convert(in, ft) A=2*2 [ft^2] n_dot_plate=450 [1/min] T1=(75+460) [R] T2=(1000+460) [R] T_oven=(1300+460) [R] rho=532.5 [lbm/ft^3] c_p=0.091 [Btu/lbm-F] "Analysis" V=L*A m=rho*V Q_in=m*c_p*(T2-T1) Q_dot_total=n_dot_plate*Q_in*1/Convert(min, s) Q_in/T_oven+S_gen=DELTAS_sys "entropy balance" DELTAS_sys=m*C_p*ln(T2/T1) S_dot_gen=S_gen*n_dot_plate*1/Convert(min, s)

8-51 Long cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in the oven and the rate of entropy generation associated with this heat transfer process are to be determined. Assumptions 1. The thermal properties of the rods are constant. 2. The changes in kinetic and potential energies are negligible. Properties The density and specific heat of the steel rods are given to be ρ = 7833 kg / m3 and c p = 0.465 kJ / kg. ° C . Analysis (a) Noting that the rods enter the oven at a velocity of 3 m / min and exit at the same velocity, we can say that a 3-m long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is

6-281

m = r V = r LA = r L(p D 2 /4)

 (7833 kg/m3 )(3 m)[ ( 0.1 m)2 /4]  184.6 kg We take the 3-m section of the rod in the oven as the system. The energy balance for this closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin  U rod  m(u2  u1 )  mc(T2  T1 ) Substituting,

Qin = mc(T2 - T1 ) = (184.6 kg)(0.465 kJ/kg.°C)(700 - 30)°C = 57,512 kJ Noting that this much heat is transferred in 1 min, the rate of heat transfer to the rod becomes Qin = Qin /D t = (57,512 kJ)/(1 min)=57,512 kJ/min = 958.5 kW

(b) We again take the 3-m long section of the rod as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the rod and its immediate surroundings so that the boundary temperature of the extended system is at 900C at all times: Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin Q  Sgen  Ssystem  Sgen   in  Ssystem Tb Tb where T 700 + 273 D Ssystem = m( s2 - s1 ) = mcavg ln 2 = (184.6 kg)(0.465 kJ/kg.K)ln = 100.1 kJ/K T1 30 + 273

Substituting, Sgen = -

Qin 57,512 kJ + D Ssystem = + 100.1 kJ/K = 51.1 kJ/K Tb 900 + 273 K

Noting that this much entropy is generated in 1 min, the rate of entropy generation becomes Sgen =

Sgen Dt

51.1 kJ/K = 51.1 kJ/min.K = 0.85 kW / K 1 min

EES (Engineering Equation Solver) SOLUTION "Given" D=0.10 [m] L=6 [m] Vel=3 [m/min] T_oven=(900+273) [K] T1=(30+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-282

T2=(700+273) [K] rho=7833 [kg/m^3] c_p=0.465 [kJ/kg-C] "Analysis" "(a)" A=pi*D^2/4 V_dot=Vel*A m_dot=rho*V_dot Q_dot_in=m_dot*c_p*(T2-T1)*1/Convert(kW, kJ/min) "(b)" Q_dot_in/T_oven+S_dot_gen=DELTAS_dot_sys "entropy balance" DELTAS_dot_sys=m_dot*c_p*ln(T2/T1)*1/Convert(kW, kJ/min)

8-52 Stainless steel ball bearings leaving the oven at a uniform temperature of 900C at a rate of 1100 / min are exposed to air and are cooled to 850C before they are dropped into the water for quenching. The rate of heat transfer from the ball to the air and the rate of entropy generation due to this heat transfer are to be determined. Assumptions 1. The thermal properties of the bearing balls are constant. 2. The kinetic and potential energy changes of the balls are negligible. 3. The balls are at a uniform temperature at the end of the process Properties The density and specific heat of the ball bearings are given to be ρ = 8085 kg / m3 and c p = 0.480 kJ / kg. ° C . Analysis (a) We take a single bearing ball as the system. The energy balance for this closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qout  U ball  m(u2  u1 ) Qout  mc(T1  T2 )

The total amount of heat transfer from a ball is m = rV = r

p D3 p (0.018 m)3 = (8085 kg/m3 ) = 0.02469 kg 6 6

Qout  mc(T1  T2 )  (0.02469 kg)(0.480 kJ/kg.C)(900  850)C  0.5925 kJ/ball Then the rate of heat transfer from the balls to the air becomes

Qtotal = nballQout (per ball) = (1100 balls/min)´ (0.5925 kJ/ball) = 651.8 kJ/min = 10.86 kW Therefore, heat is lost to the air at a rate of 10.86 kW. (b) We again take a single bearing ball as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the ball and its immediate surroundings so that the boundary temperature of the extended system is at 20C at all times: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-283

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass



Entropy generation

Change in entropy

Qout Q  Sgen  Ssystem  Sgen  out  Ssystem Tb Tb

where T (850+273)K D Ssystem = m( s2 - s1 ) = mcavg ln 2 = (0.02469 kg)(0.480 kJ/kg ×K)ln = - 0.0005162 kJ/K T1 (900+273)K

Substituting, Sgen =

Qout 0.5925 kJ + D Ssystem = - 0.0005162 kJ/K = 0.001506 kJ/K (per ball) Tb 293 K

Then the rate of entropy generation becomes

Sgen = Sgen nball = (0.001506 kJ/K ×ball)(1100 balls/min)=1.657 kJ/min ×K=0.02761 kW / K

EES (Engineering Equation Solver) SOLUTION "Given" D=0.018 [m] n_dot_ball=1100 [1/min] T1=(900+273) [K] T2=(850+273) [K] T_air=(20+273) [K] rho=8085 [kg/m^3] c_p=0.480 [kJ/kg-C] "Analysis" "(a)" V=pi*D^3/6 m=rho*V Q_out=m*c_p*(T1-T2) Q_dot_total=n_dot_ball*Q_out*1/Convert(kW, kJ/min) "(b)" -Q_out/T_air+S_gen=DELTAS_sys "entropy balance" DELTAS_sys=m*c_p*ln(T2/T1) S_dot_gen=S_gen*n_dot_ball*1/Convert(kW, kJ/min)

8-53 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of entropy generation within the wall is to be determined. Assumptions Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. AnalysisWe take the wall to be the system, which is a closed system. Under steady conditions, the rate form of the entropy balance for the wall simplifies to

6-284

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem

Rate of entropy generation

= 0

Rate of change of entropy

Qin Qout   Sgen, wall  0 Tb,in Tb,out

1800 W 1800 W   Sgen, wall  0 289 K 277 K

Sgen, wall  0.270 W / K Therefore, the rate of entropy generation in the wall is 0.270 W / K .

EES (Engineering Equation Solver) SOLUTION "Given" A=4*10 [m^2] L=0.2 [m] T1=(16+273) [K] T2=(4+273) [K] Q_dot=1800 [W] "Analysis" Q_dot_in/T1-Q_dot_out/T2+S_dot_gen=0 "entropy balance" Q_dot_in=Q_dot Q_dot_out=Q_dot

8-54E Steam is decelerated in a diffuser from a velocity of 900 ft / s to 100 ft / s . The mass flow rate of steam and the rate of entropy generation are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Potential energy changes are negligible. 3. There are no work interactions. Properties The properties of steam at the inlet and the exit of the diffuser are (Tables A-4E through A-6E)

6-285

P1 = 20 psiaüïï h1 = 1162.3 Btu/lbm ý T1 = 240°F ïïþ s1 = 1.7406 Btu/lbm ×R h2  1160.5 Btu/lbm T2  240F  s2  1.7141 Btu/lbm  R sat. vapor  v2  16.316 ft 3 /lbm Analysis (a) The mass flow rate of the steam can be determined from its definition to be m=

1 1 A2V2 = (1 ft 2 )(100 ft/s) = 6.129 lbm / s v2 16.316 ft 3 /lbm

(b) We take diffuser as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D Esystem ¿ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout m(h1 + V12 /2) - Qout = m(h2 + V22 /2) (since W @D pe @ 0)

 V 2  V12  Qout  m  h2  h1  2  2   Substituting, the rate of heat loss from the diffuser is determined to be 2 2 æ ö ö÷ (100 ft/s) - (900 ft/s ) æ çç ÷ çç 1 Btu/lbm ÷ Qout = - (6.129 lbm/s)ç1160.5 - 1162.3 + = 108.42 Btu/s ÷ ÷ çè 25,037 ft 2 /s 2 ø÷÷ ççè 2 ø÷

The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the diffuser and its immediate surroundings so that the boundary temperature of the extended system is 77F at all times. It gives

Sin - Sout

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

Rate of net entropy transfer by heat and mass

ms1  ms2 

Qout  Sgen  0 Tb,surr

Substituting, the total rate of entropy generation during this process becomes Sgen = m (s2 - s1 )+

Qout 108.42 Btu/s = (6.129 lbm/s)(1.7141- 1.7406)Btu/lbm ×R + = 0.0395 Btu / s ×R Tb, surr 537 R

EES (Engineering Equation Solver) SOLUTION "Given" P1=20 [psia] T1=240 [F] Vel1=900 [ft/s] T2=240 [F] x2=1 Vel2=100 [ft/s] A2=1 [ft^2] T_surr=77 [F] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-286

Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, T=T2, x=x2) s2=entropy(Fluid$, T=T2, x=x2) v2=volume(Fluid$, T=T2, x=x2) "Analysis" "(a)" m_dot=1/v2*A2*Vel2 "(b)" m_dot*(h1+Vel1^2/2*1/Convert(Btu/lbm, ft^2/s^2))-Q_dot_out=m_dot*(h2+Vel2^2/2*1/Convert(Btu/lbm, ft^2/s^2)) "energy balance" m_dot*s1-m_dot*s2-Q_dot_out/(T_surr+460)+S_dot_gen=0 "entropy balance on extended stystem"

8-55 Steam is accelerated in a nozzle from a velocity of 55 m / s to 390 m / s . The exit temperature and the rate of entropy generation are to be determined. Assumptions 1. 2. 3. 4.

This is a steady-flow process since there is no change with time. Potential energy changes are negligible. There are no work interactions. The device is adiabatic and thus heat transfer is negligible.

Properties From the steam tables (Table A-6),

h1 = 3137.7 kJ/kg P1 = 2 MPa ïü ïý s = 6.9583 kJ/kg ×K 1 T1 = 350 °C ïïþ v1 = 0.13860 m3 /kg

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout m(h1 + V12 /2) = m(h2 +V22 /2) (since Q @ W @D pe @ 0)

0  h2  h1 

V22  V12 2

Substituting, 2

h2 = 3137.7 kJ/kg -

ö (390 m/s) - (55 m/s) æ ç 1 kJ/kg ÷ 2

÷= 3063.1 kJ/kg çç è1000 m 2 /s 2 ÷ ø

6-287

ü T2 = 303°C ïï ý h2 a = 3063.1 kJ/kg ïïþ s2 = 7.2454 kJ/kg ×K P2 = 0.8 MPa

The mass flow rate of steam is m=

1 1 AV (7.5´ 10- 4 m 2 )(55 m/s) = 0.2976 kg/s 1 1 = v1 0.13860 m 3 /kg

(b) Again we take the nozzle to be the system. Noting that no heat crosses the boundaries of this combined system, the entropy balance for it can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

ms1  ms2  Sgen  0 Sgen  m  s2  s1  Substituting, the rate of entropy generation during this process is determined to be

Sgen = m(s2 - s1 ) = (0.2976 kg/s)(7.2454 - 6.9583)kJ/kg ×K = 0.0854 kW / K

EES (Engineering Equation Solver) SOLUTION "Given" P1=2000 [kPa] T1=350 [C] Vel1=55 [m/s] P2=800 [kPa] Vel2=390 [m/s] A1=7.5E-4 [m^2] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) v1=volume(Fluid$, P=P1, T=T1) "Analysis" "(a)" m_dot*(h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg))=m_dot*(h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)) "energy balance" T2=temperature(Fluid$, P=P2, h=h2) s2=entropy(Fluid$, P=P2, h=h2) "(b)" m_dot=1/v1*A1*Vel1 "(b)" m_dot*s1-m_dot*s2+S_dot_gen=0 "entropy balance"

8-56 Steam expands in a turbine from a specified state to another specified state. The rate of entropy generation during this process is to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. Properties From the steam tables (Tables A-4 through 6)

6-288

P1 = 8 MPa üïï h1 = 3399.5 kJ/kg ý T1 = 500°C ïïþ s1 = 6.7266 kJ/kg ×K

P2  40 kPa  h2  2636.1 kJ/kg  sat. vapor  s2  7.6691 kJ/kg  K Analysis There is only one inlet and one exit, and thus m1 = m2 = m. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein  Eout mh1 = Qout + Wout + mh2

Qout  m(h1  h2 )  Wout Substituting, Qout = (40,000/3600 kg/s)(3399.5 - 2636.1)kJ/kg - 8200 kJ/s = 282.6 kJ/s

The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the turbine and its immediate surroundings so that the boundary temperature of the extended system is

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

ms1  ms2 

Qout  Sgen  0 Tb,surr

Substituting, the rate of entropy generation during this process is determined to be Sgen = m (s2 - s1 )+

Qout 282.6 kW = (40,000/3600 kg/s)(7.6691- 6.7266)kJ/kg ×K + = 11.4 kW / K Tb, surr 298 K

EES (Engineering Equation Solver) SOLUTION "Given" P1=8000 [kPa] T1=500 [C] P2=40 [kPa] x2=1 m_dot=40000[kg/h]*Convert(kg/h, kg/s) W_dot_out=8200 [kW] T_surr=25 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-289

"Properties" Fluid$='Steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) "Analysis" m_dot*h1=Q_dot_out+W_dot_out+m_dot*h2 "energy balance" m_dot*s1-m_dot*s2-Q_dot_out/(T_surr+273)+S_dot_gen=0 "entropy balance on extended stystem"

8-57 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam and the rate of entropy generation are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. There are no work interactions. Properties Noting that T  Tsat @ 200 kPa  120.21 C , the cold water and the exit mixture streams exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. From Tables A-4 through A-6,

h1 @ h f @15°C = 62.98 kJ/kg P1 = 200 kPa ü ïï ý ïïþ s1 @ s f @15°C = 0.2245 kJ/kg ×K T1 = 15°C

P2  200 kPa  h2  2769.1 kJ/kg  T2  150C  s2  7.2810 kJ/kg  K P3  200 kPa  h3  h f @80C  335.02 kJ/kg  T3  80C  s3  s f @80C  1.0756 kJ/kg  K Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance: min - mout = D msystemZ 0 (steady) = 0 ¾ ¾ ® m1 + m2 = m3

Energy balance:

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

6-290

Qout = m1h1 + m2 h2 - (m1 + m2 )h3 = m1 (h1 - h3 )+ m2 (h2 - h3 ) Solving for m2 and substituting, the mass flow rate of the superheated steam is determined to be m2 =

Also,

Qout - m1 (h1 - h3 ) h2 - h3

(1200/60 kJ/s) - (4.3 kg/s)(62.98 - 335.02)kJ/kg (2769.1- 335.02)kJ/kg

= 0.4806 kg / s

m3 = m1 + m2 = 4.3 + 0.4806 = 4.781 kg/s

(b) The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature of the extended all times. It gives

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

m1s1  m2 s2  m3 s3 

Qout  Sgen  0 Tb,surr

Substituting, the rate of entropy generation during this process is determined to be Sgen = m3 s3 - m2 s2 - m1 s1 +

Qout Tb, surr

  4.781 kg/s 1.0756 kJ/kg  K    0.4806 kg/s  7.2810 kJ/kg  K    4.3 kg/s  0.2245 kJ/kg  K  

(1200 / 60 kJ/s) 293 K

 0.746 kW / K

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [kPa] T1=15 [C] P2=200 [kPa] T2=150 [C] P3=200 [kPa] T3=80 [C] m1_dot=4.3 [kg/s] T_surr=20 [C] Q_dot_out=1200[kJ/min]*Convert(kJ/min, kW) "Properties" Fluid$='steam_iapws' "For state 1 and 3 we use saturated liquid state at given temperature" h1=enthalpy(Fluid$, x=0, T=T1) s1=entropy(Fluid$, x=0, T=T1) h2=enthalpy(Fluid$, P=P2, T=T2) s2=entropy(Fluid$, P=P2, T=T2) h3=enthalpy(Fluid$, x=0, T=T3) s3=entropy(Fluid$, x=0, T=T3) "Analysis" "(a)" m1_dot+m2_dot=m3_dot "mass balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-291

m1_dot*h1+m2_dot*h2=m3_dot*h3 "energy balance" "(b)" m1_dot*s1+m2_dot*s2-m3_dot*s3-Q_dot_out/(T_surr+273)+S_dot_gen=0 "entropy balance on extended system"

8-58 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer and the entropy generation during this process are to be determined. Assumptions 1. This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2. Kinetic and potential energies are negligible. 3. There are no work interactions involved. 4. The direction of heat transfer is to the tank (will be verified). Properties The properties of water are (Tables A-4 through A-6)

v1 = v f @ 120°C = 0.001060 m3 /kg T1 = 120°C ïü ïý u = u 1 f @ 120 ° C = 503.60 kJ/kg sat. liquid ïïþ s1 = s f @ 120°C = 1.5279 kJ/kg ×K Te  120C  he  h f @120C  503.81 kJ/kg  sat. liquid  se  s f @120C  1.5279 kJ/kg  K Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - mout = D msystem ® me = m1 - m2

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin  me he  m2u2  m1u1 (since W  ke  pe  0) The initial and the final masses in the tank are

m1 =

V 0.18 m3 = = 169.76 kg v1 0.001060 m3 /kg

m2 

1 1 m1  169.76 kg   84.88 kg  me 2 2

6-292

v2  x2 

V 0.18 m3   0.002121 m3 /kg m2 84.88 kg v2  v f v fg



0.002121  0.001060  0.001191 0.8913  0.001060

 u2  u f  x2u fg  503.60   0.001191 2025.3  506.01 kJ/kg  x2  0.001191  s2  s f  x2 s fg  1.5279   0.001191 5.6013  1.5346 kJ/kg  K

T2  120C

The heat transfer during this process is determined by substituting these values into the energy balance equation,

Qin = me he + m2 u2 - m1u1 = (84.88 kg )(503.81 kJ/kg )+ (84.88 kg )(506.01 kJ/kg )- (169.76 kg )(503.60 kJ/kg ) = 222.6 kJ (b) The total entropy generation is determined by considering a combined system that includes the tank and the heat source. Noting that no heat crosses the boundaries of this combined system and no mass enters, the entropy balance for it can be expressed as

Sin - Sout + Sgen = D Ssystem ¾ ¾ ® - me se + Sgen = D S tank + D Ssource Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Therefore, the total entropy generated during this process is

Sgen = me se + D S tank + D Ssource = me se + (m2 s2 - m1s1 ) -

Qsource,out Tsource

  84.88 kg 1.5279 kJ/kg  K    84.88 kg 1.5346 kJ/kg  K   169.76 kg 1.5279 kJ/kg  K  

222.6 kJ (230  273) K

 0.1237 kJ / K EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.18 [m^3] T1=120 [C] x1=0 T2=T1 T_e=120 [C] x_e=0 T_source=230 [C] m2=1/2*m1 "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) s1=entropy(Fluid$, T=T1, x=x1) h_e=enthalpy(Fluid$, T=T_e, x=x_e) s_e=entropy(Fluid$, T=T_e, x=x_e) "Analysis" "(a)" m_e=m1-m2 "mass balance" Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance" m1=Vol/v1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-293

v2=Vol/m2 u2=intenergy(Fluid$, T=T2, v=v2) s2=entropy(Fluid$, T=T2, v=v2) x2=quality(Fluid$, T=T2, v=v2) v2_f=volume(Fluid$, T=T2, x=0) v2_g=volume(Fluid$, T=T2, x=1) u2_f=intenergy(Fluid$, T=T2, x=0) u2_g=intenergy(Fluid$, T=T2, x=1) u2_fg=u2_g-u2_f s2_f=entropy(Fluid$, T=T2, x=0) s2_g=entropy(Fluid$, T=T2, x=1) s2_fg=s2_g-s2_f "(b)" -m_e*s_e+S_gen=m2*s2-m1*s1-Q_source_out/(T_source+273) "entropy balance on exetended system" Q_source_out=Q_in

8-59 Liquid water is withdrawn from a rigid tank that initially contains saturated water mixture until no liquid is left in the tank. The quality of steam in the tank at the initial state, the amount of mass that has escaped, and the entropy generation during this process are to be determined. Assumptions 1. Kinetic and potential energy changes are zero. 2. There are no work interactions. Analysis (a) The properties of the steam in the tank at the final state and the properties of exiting steam are (Tables A-4 through A6)

u2 = 2553.1 kJ/kg ïü ïýv = 0.46242 m3 /kg x2 = 1 (sat. vap.)ïïþ 2 s2 = 6.8955 kJ/kg.K

P2 = 400 kPa

P2e  400 kPa  he  604.66 kJ/kg  xe  0 (sat. liq.)  se  1.7765 kJ/kg.K The relations for the volume of the tank and the final mass in the tank are

V = m1v1 = (7.5 kg)v1

m2 

(7.5 kg)v1 V   16.219v1 v2 0.46242 m3 /kg

The mass, energy, and entropy balances may be written as

6-294

Qin - me se + S gen = m2 s2 - m1 s1 Tsource

Substituting,

me = 7.5 - 16.219v1

(1)

5 - (7.5 - 16.219v1 )(604.66) = 16.219v1 (2553.1) - 7.5u1

(2)

5 - (7.5 - 16.219v1 )(1.7765) + Sgen = 16.219v1 (6.8955) - 7.5s1 (3) 500 + 273 Eq. (2) may be solved by a trial-error approach by trying different qualities at the inlet state. Or, we can use EES to solve the equations to find x1  0.8666 Other properties at the initial state are

u1 = 2293.2 kJ/kg P1 = 400 kPa ïüï 3 ýv = 0.40089 m /kg x1 = 0.8666 ïïþ 1 s1 = 6.2129 kJ/kg.K Substituting into Eqs (1) and (3), (b) me = 7.5 - 16.219(0.40089) = 0.998 kg (c)

5 - [7.5 - 16.219(0.40089) ](1.7765) + Sgen = 16.219(0.40089)(6.8955) - 7.5(6.2129) 500 + 273 Sgen = 0.00553 kJ / K

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_1=7.5 [kg] P_1=400 [kPa] x_2=1 P_2=P_1 P_e=P_1 x_e=0 Q_in=5 [kJ] T_s=500 [C] "PROPERTIES" Fluid$='Steam_iapws' v_2=volume(Fluid$, P=P_2, x=x_2) u_2=intenergy(Fluid$, P=P_2, x=x_2) s_2=entropy(Fluid$, P=P_2, x=x_2) h_e=enthalpy(Fluid$, P=P_e, x=x_e) s_e=entropy(Fluid$, P=P_e, x=x_e) "ANALYSIS" v_1=volume(Fluid$, P=P_1, u=u_1) Vol=m_1*v_1 m_2=Vol/v_2 m_e=m_1-m_2 "mass balance" Q_in-m_e*h_e=m_2*u_2-m_1*u_1 "energy balance" x_1=quality(Fluid$, P=P_1, u=u_1) s_1=entropy(Fluid$, P=P_1, u=u_1) Q_in/(T_s+273)-m_e*s_e+S_gen=m_2*s_2-m_1*s_1 "entropy balance"

6-295

Special Topic: Reducing the Cost of Compressed Air 8-60 The compressed air requirements of a plant is being met by a 90 hp compressor that compresses air from 101.3 kPa to 1100 kPa. The amount of energy and money saved by reducing the pressure setting of compressed air to 750 kPa is to be determined. Assumptions 1. Air is an ideal gas with constant specific heats. 2. Kinetic and potential energy changes are negligible. 3. The load factor of the compressor is given to be 0.75. 4. The pressures given are absolute pressure rather than gage pressure. Properties The specific heat ratio of air is k = 1.4 (Table A-2). Analysis The electrical energy consumed by this compressor per year is Energy consumed = (Power rating)(Load factor)(Annual Operating Hours)/Motor efficiency = (90 hp)(0.746 kW / hp)(0.75)(3500 hours / year ) / 0.94 = 187, 420 kWh / year The fraction of energy saved as a result of reducing the pressure setting of the compressor is

Power Reduction Factor = 1 -

 1

( P2, reduced / P1 )( k - 1)/ k - 1 ( P2 / P1 )( k - 1)/ k - 1

(750 /101.3)(1.41)/1,4  1 (1100 /101.3)(1.41)/1,4  1

 0.2098 That is, reducing the pressure setting will result in about 11 percent savings from the energy consumed by the compressor and the associated cost. Therefore, the energy and cost savings in this case become Energy Savings = (Energy consumed)(Power reduction factor) = (187, 420 kWh / year )(0.2098) = 39, 320 kWh / year Cost Savings = (Energy savings)(Unit cost of energy) = (39,320 kWh / year )($0.105 / kWh) = $4128 / year

6-296

Therefore, reducing the pressure setting by 250 kPa will result in annual savings of 39,320 kWh that is worth $4128 in this case. Discussion Some applications require very low pressure compressed air. In such cases the need can be met by a blower instead of a compressor. Considerable energy can be saved in this manner, since a blower requires a small fraction of the power needed by a compressor for a specified mass flow rate.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=90 [hp] P1=101.3 [kPa] T1=15 [C] P2=1100 [kPa] P2_reduced=750 [kPa] Hours=3500 [h/year] f_load=0.75 eta_motor=0.94 UnitCost=0.105 [$/kWh] "Properties" k=1.4 "Table A-2a" "Analysis" EnergyConsumed=(W_dot*f_load*Hours)/eta_motor*Convert(hp, kW) f_PowerReduction=1-((P2_reduced/P1)^((k-1)/k)-1)/((P2/P1)^((k-1)/k)-1) EnergySavings=f_PowerReduction*EnergyConsumed CostSavings=EnergySavings*UnitCost

8-61 The compressed air requirements of the facility during 60 percent of the time can be met by a 25 hp reciprocating compressor instead of the existing 100 hp compressor. The amounts of energy and money saved as a result of switching to the 25 hp compressor during 60 percent of the time are to be determined. Analysis Noting that 1 hp = 0.746 kW, the electrical energy consumed by each compressor per year is determined from (Energy consumed)Large = (Power)(Hours)  (LFxTF / ηmotor ) Unloaded  (LFxTF / ηmotor ) Loaded 

 (100hp)(0.746 kW / hp)(3800 hours / year)[0.35  0.6 / 0.82  0.90  0.4 / 0.9]

185,990 kWh / year (Energy consumed)Small = (Power)(Hours)  (LFxTF / ηmotor ) Unloaded  (LFxTF / ηmotor ) Loaded 

 (25hp)(0.746 kW / hp)(3800 hours / year)[0.0  0.15  0.95  0.85] / 0.88  65,031kWh / year

Therefore, the energy and cost savings in this case become Energy Savings  ( Energy consumed )Large  ( Energy consumed )Small

6-297

 120,959 kWh / year

Cost Savings  (Energy savings)(Unit cost of energy)

 (120,959 kWh / year )($0.083 / kWh)  $10,040 / year Discussion Note that utilizing a small compressor during the times of reduced compressed air requirements and shutting down the large compressor will result in annual savings of 120,959 kWh, which is worth $10,040 in this case.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_l=100 [hp] W_dot_s=25 [hp] Hours=3800 [h/year] UnitCost=0.083 [$/kWh] LF_unloaded_l=0.35 TF_unloaded_l=0.6 eta_motor_unloaded_l=0.82 LF_loaded_l=0.90 TF_loaded_l=0.4 eta_motor_loaded_l=0.90 LF_unloaded_s=0 TF_unloaded_s=0.15 eta_motor_unloaded_s=0.88 LF_loaded_s=0.95 TF_loaded_s=0.85 eta_motor_loaded_s=0.88 "Analysis" EnergyConsumed_large=W_dot_l*Hours*(LF_unloaded_l*TF_unloaded_l/eta_motor_unloaded_l+LF_loaded_l*T F_loaded_l/eta_motor_loaded_l)*Convert(hp, kW) EnergyConsumed_small=W_dot_s*Hours*(LF_unloaded_s*TF_unloaded_s/eta_motor_unloaded_s+LF_loaded_ s*TF_loaded_s/eta_motor_loaded_s)*Convert(hp, kW) EnergySavings=EnergyConsumed_large-EnergyConsumed_small CostSavings=EnergySavings*UnitCost

8-62 A facility stops production for one hour every day, including weekends, for lunch break, but the 90 hp compressor is kept operating. If the compressor consumes 35 percent of the rated power when idling, the amounts of energy and money saved per year as a result of turning the compressor off during lunch break are to be determined. Analysis It seems like the compressor in this facility is kept on unnecessarily for one hour a day and thus 365 hours a year, and the idle factor is 0.35. Then the energy and cost savings associated with turning the compressor off during lunch break are determined to be

6-298

Energy Savings  (Power Rating)(Turned Off Hours)(Idle Factor ) / ηmotor  (90hp)(0.7457 kW / hp)(365 hours / year)(0.35) / 0.84

 10, 207 kWh / year Cost Savings  (Energy savings)(Unit cost of energy)  (10, 207 kWh / year )($0.09 / kWh)  $919 / year Discussion Note that the simple practice of turning the compressor off during lunch break will save this facility $1123 a year in energy costs. There are also side benefits such as extending the life of the motor and the compressor, and reducing the maintenance costs.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=90 [hp] OffHours=365 [h/year] f_idle=0.35 eta_motor=0.84 UnitCost=0.09 [$/kWh] "Analysis" EnergySavings=(W_dot*OffHours*f_idle)/eta_motor*Convert(hp, kW) CostSavings=EnergySavings*UnitCost

8-63 It is determined that 25 percent of the energy input to the compressor is removed from the compressed air as heat in the aftercooler with a refrigeration unit whose COP is 2.5. The amounts of the energy and money saved per year as a result of cooling the compressed air before it enters the refrigerated dryer are to be determined. Assumptions The compressor operates at full load when operating. Analysis Noting that 25 percent of the energy input to the compressor is removed by the aftercooler, the rate of heat removal from the compressed air in the aftercooler under full load conditions is Qaftercooling = (Rated Power of Compressor)(Load Factor)(Aftercooling Fraction)

= (150 hp)(0.746 kW/hp)(1.0)(0.25)=27.96 kW

6-299

The compressor is said to operate at full load for 2100 hours a year, and the COP of the refrigeration unit is 2.5. Then the energy and cost savings associated with this measure become

Energy Savings  (Qaftercoling )(Annual Operating Hours) / COP  (27.96 kW)(2100 hours / year ) / 2.5  23, 490 kWh / year

Cost Savings  (Energy savings)(Unit cost of energy saved)

 (23, 490 kWh / year)($0.065 / kWh)  $1527 / year

Discussion Note that the aftercooler will save this facility 23,490 kWh of electrical energy worth $1527 per year. The actual savings will be less than indicated above since we have not considered the power consumed by the fans and/or pumps of the aftercooler. However, if the heat removed by the aftercooler is utilized for some useful purpose such as space heating or process heating, then the actual savings will be much more.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=150 [hp] Hours=2100 [h/year] f_aftercooling=0.25 COP=2.5 UnitCost=0.065 [$/kWh] "Analysis" Q_dot_aftercooling=W_dot*f_aftercooling*Convert(hp, kW) EnergySavings=(Q_dot_aftercooling*Hours)/COP CostSavings=EnergySavings*UnitCost

8-64 The motor of a 150 hp compressor is burned out and is to be replaced by either a 93% efficient standard motor or a 96.2% efficient high efficiency motor. The amount of energy and money the facility will save by purchasing the highefficiency motor instead of standard motor are to be determined. It is also to be determined if the savings from the high efficiency motor justify the price differential. Assumptions 1. The compressor operates at full load when operating. 2. The life of the motors is 10 years. 3. There are no rebates involved. 4. The price of electricity remains constant. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-300

Analysis The energy and cost savings associated with the installation of the high efficiency motor in this case are determined to be

Energy Savings  (Power Rating)(Operating Hours)(Load Factor) 1/ ηstandard  1/ ηefficient   (150 hp)(0.7457 kW / hp)(4,368 hours / year)(1.0)(1/ 0.930  1/ 0.962)  17, 476 kWh / year

Cost Savings  (Energy savings) (Unit cost of energy)  (17, 476 kWh / year )($0.095 / kWh)  $1660 / year

The additional cost of the energy efficient motor is Cost Differential = $10,942  $9,031 = $1,911 Discussion The money saved by the high efficiency motor will pay for this cost difference in $1,911/ $1660  1.15 year , and will continue saving the facility money for the rest of the 10 years of its lifetime. Therefore, the use of the high efficiency motor is recommended in this case even in the absence of any incentives from the local utility company.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=150 [hp] eta_standard=0.93 eta_efficient=0.962 Cost_standard=9031 [$] Cost_efficient=10942 [$] Hours=4368 [h/year] UnitCost=0.095 [$/kWh] "Analysis" EnergySavings=W_dot*Hours*(1/eta_standard-1/eta_efficient)*Convert(hp, kW) CostSavings=EnergySavings*UnitCost CostDifferential=Cost_efficient-Cost_standard Payback=CostDifferential/CostSavings

8-65 The compressor of a facility is being cooled by air in a heat-exchanger. This air is to be used to heat the facility in winter. The amount of money that will be saved by diverting the compressor waste heat into the facility during the heating season is to be determined. Assumptions The compressor operates at full load when operating. Analysis Assuming c p  1.0 kJ / kg. C and operation at sea level and taking the density of air to be 1.2 kg / m3 , the mass flow rate of air through the liquid-to-air heat exchanger is determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-301

Mass flow rate of air  (Density of air)(Average velocity)(Flow area)

 (1.2 kg / m3 )(3 m / s)(1.0 m2 )  3.6 kg / s  12,960 kg / h Noting that the temperature rise of air is 32C, the rate at which heat can be recovered (or the rate at which heat is transferred to air) is

Rate of Heat Recovery  (Mass flow rate of air )(Specific heat of air) (Temperature rise )

 (12,960 kg / h)(1.0 kJ / kg  C)(32 C)  414,720 kJ / h The number of operating hours of this compressor during the heating season is

Operating hours  (20 hours/day)(5 days/week)(26 weeks/year)  2600 hours/year Then the annual energy and cost savings become

Energy Savings  (Rate of Heat Recovery)(Annual Operating Hours ) / Efficiency  (414,720 kJ / h)(2600 hours / year) ) / 0.85  1, 268,555,000 kJ / year  12,024 therms / year Cost Savings  (Energy savings )(Unit cost of energy saved)  (12,024 therms / year )($1.25 / therm)  $15, 030 / year Therefore, utilizing the waste heat from the compressor will save $15,030 per year from the heating costs. Discussion The implementation of this measure requires the installation of an ordinary sheet metal duct from the outlet of the heat exchanger into the building. The installation cost associated with this measure is relatively low. A few of the manufacturing facilities we have visited already have this conservation system in place. A damper is used to direct the air into the building in winter and to the ambient in summer. Combined compressor/heat-recovery systems are available in the market for both air-cooled (greater than 50 hp) and water cooled (greater than 125 hp) systems.

EES (Engineering Equation Solver) SOLUTION "Given" eta_heater=0.85 A=1*1 [m^2] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-302

T1=20 [C] T2=52 [C] Vel=3 [m/s] Hours=20*5*26 [h/year] UnitCost=1.25 [$/therm] "Properties" rho=1.2 [kg/m^3] "assumed" c_p=1.0 [kJ/kg-C] "assumed" "Analysis" m_dot_air=rho*Vel*A*Convert(kg/s, kg/h) Q_dot_recovered=m_dot_air*c_p*(T2-T1) EnergySavings=Q_dot_recovered*Hours/eta_heater*1/Convert(therm, kJ) CostSavings=EnergySavings*UnitCost

8-66 The compressed air lines in a facility are maintained at a gage pressure of 700 kPa at a location where the atmospheric pressure is 85.6 kPa. There is a 3-mm diameter hole on the compressed air line. The energy and money saved per year by sealing the hole on the compressed air line. Assumptions 1. Air is an ideal gas with constant specific heats. 2. Kinetic and potential energy changes are negligible. Properties The gas constant of air is R  0.287 kJ / kg.K . The specific heat ratio of air is k = 1.4 (Table A-2). Analysis Disregarding any pressure losses and noting that the absolute pressure is the sum of the gage pressure and the atmospheric pressure, the work needed to compress a unit mass of air at 15C from the atmospheric pressure of 85.6 kPa to 700 + 85.6 = 785.6 kPa is determined to be éæP ö( k - 1)/ k ù kRT1 êç 2 ÷ ú wcomp, in = 1 êççç ÷ ú ÷ ÷ hcomp (k - 1) êè P1 ø ú ë û

(1.4- 1)/1.4 ù ö (1.4)(0.287 kJ/kg.K)(288 K) éêæ ú çç785.6 kPa ÷ 1 ÷ êçè 85.6 kPa ÷ ú ø (0.8)(1.4 - 1) êë ú û

= 319.6 kJ/kg The cross-sectional area of the 5-mm diameter hole is

A = p D 2 /4 = p (3´ 10- 3 m)2 /4 = 7.069´ 10- 6 m 2 Noting that the line conditions are T0  298 K and

P0  785.6 kPa , the mass flow rate of the air leaking through the hole is determined to be 1/ ( k - 1)

æ 2 ö÷ mair = Closs çç ÷ çè k + 1ø÷

æ 2 ö÷ P0 A kR çç ÷T èç k + 1ø÷ 0 RT0

6-303 1/(1.4- 1)

æ 2 ÷ ö = (0.65) çç ÷ çè1.4 + 1÷ ø ´

785.6 kPa (7.069´ 10- 6 m 2 ) 3 (0.287 kPa.m /kg.K)(298 K)

æ1000 m 2 /s 2 ÷ öæ 2 ö ÷ ç ÷ (1.4)(0.287 kJ/kg.K) çç ÷(298 K) ÷ççè1.4 + 1÷ çè 1 kJ/kg ÷ ø ø

= 0.008451 kg/s Then the power wasted by the leaking compressed air becomes

Power wasted = mair wcomp, in = (0.008451 kg/s)(319.6 kJ/kg) = 2.701 kW Noting that the compressor operates 4200 hours a year and the motor efficiency is 0.93, the annual energy and cost savings resulting from repairing this leak are determined to be

Energy Savings  (Power wasted )(Annual operating hours ) / Motor efficiency  (2.701 kW)(4200 hours / year ) / 0.93  12, 200 kWh / year

Cost Savings  (Energy savings )(Unit cost of energy)  (12, 200 kWh / year )($0.10 / kWh)  $1220 / year

Therefore, the facility will save 12,200 kWh of electricity that is worth $1220 a year when this air leak is sealed.

EES (Engineering Equation Solver) SOLUTION "Given" P_gage=700 [kPa] P_atm=85.6 [kPa] T1=(15+273) [K] T_o=(25+273) [K] Hours=4200 [h/year] UnitCost=0.10 [$/kWh] eta_C=0.8 eta_motor=0.93 C_loss=0.65 D=0.003 [m] "Properties" k=1.4 "Table A-2a" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" P_o=P_gage+P_atm P2=P_o P1=P_atm w_comp_in=1/eta_C*(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1) A=pi*D^2/4 m_dot_air=C_loss*(2/(k+1))^(1/(k-1))*P_o/(R*T_o)*A*sqrt(k*R*Convert(kJ/kg, m^2/s^2)*(2/(k+1))*T_o) PowerWasted=m_dot_air*w_comp_in EnergySavings=PowerWasted*Hours/eta_motor CostSavings=EnergySavings*UnitCost

6-304

8-67 The total energy used to compress air in the US is estimated to be 0.5 1015 kJ per year. About 20% of the compressed air is estimated to be lost by air leaks. The amount and cost of electricity wasted per year due to air leaks is to be determined. Assumptions About 20% of the compressed air is lost by air leaks.

Analysis The electrical energy and money wasted by air leaks are

Energy wasted  (Energy consumed )(Fraction wasted)

 (0.5 1015 kJ)(1 kWh / 3600 kJ)(0.20)  27.78×109 kWh / year Money wasted  (Energy wasted )(Unit cost of energy)

 (27.78 109 kWh / year )($0.11/ kWh)  $3.056×109 / year Therefore, air leaks are costing almost $3.1 billion a year in electricity costs. The environment also suffers from this because of the pollution associated with the generation of this much electricity.

EES (Engineering Equation Solver) SOLUTION "Given" E=0.5E+15 [kJ/year] f_wasted=0.20 UnitCost=0.11 [$/kWh] "Analysis" EnergyWasted=E*f_wasted*1/Convert(kWh, kJ) MoneyWasted=EnergyWasted*UnitCost

8-68 A 150 hp compressor in an industrial facility is housed inside the production area where the average temperature compressor instead of using the inside air are to be determined. Assumptions 1. 2.

Air is an ideal gas with constant specific heats. Kinetic and potential energy changes are negligible.

Analysis The electrical energy consumed by this compressor per year is

Energy consumed  (Power rating)(Load factor)(Annual Operating Hours) / Motor efficiency  (150 hp)(0.746 kW / hp)(0.85)(4500 hours / year ) / 0.9  475,384 kWh / year

6-305

Also,

Cost of Energy  (Energy consumed )(Unit cost of energy)  (475,384 kWh / year )($0.07 / kWh)  $33, 277 / year

The fraction of energy saved as a result of drawing in cooler outside air is Power Reduction Factor = 1 -

Toutside 10 + 273 = 1= 0.0503 Tinside 25 + 273

5.03 percent savings from the energy consumed by the compressor and the associated cost. Therefore, the energy and cost savings in this case become

Energy Savings  (Energy consumed )(Power reduction factor)  (475,384kWh / year )(0.0503)  23, 929kWh / year

Therefore, drawing air in from the outside will result in annual savings of 23,929 kWh, which is worth $1795 in this case. Discussion The price of a typical 150 hp compressor is much lower than $50,000. Therefore, it is interesting to note that the cost of energy a compressor uses a year may be more than the cost of the compressor itself. The implementation of this measure requires the installation of an ordinary sheet metal or PVC duct from the compressor intake to the outside. The installation cost associated with this measure is relatively low, and the pressure drop in the duct in most cases is negligible. About half of the manufacturing facilities we have visited, especially the newer ones, have the duct from the compressor intake to the outside in place, and they are already taking advantage of the savings associated with this measure.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot=150 [hp] T_inside=25 [C] T_outside=10 [C] Hours=4500 [h/year] f_load=0.85 eta_motor=0.90 UnitCost=0.075 [$/kWh] "Analysis" EnergyConsumed=(W_dot*f_load*Hours)/eta_motor*Convert(hp, kW) f_PowerReduction=1-(T_outside+273)/(T_inside+273) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-306

EnergySavings=f_PowerReduction*EnergyConsumed CostSavings=EnergySavings*UnitCost

Review Problems 8-69 Saturated steam is compressed in an adiabatic process with an isentropic efficiency of 0.90. The work required is to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. The device is adiabatic and thus heat transfer is negligible. Analysis We take the steam as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win = D U = m(u2 - u1 ) From the steam tables (Tables A-5 and A-6),

P1 = 100 kPa ïü ïý u1 = ug @ 100 kPa = 2505.6 kJ/kg ïïþ s1 = sg @ 100 kPa = 7.3589 kJ/kg ×K sat. vapor

üïï ý u2 s = 2903.0 kJ/kg ïïþ s2 s = s1 The work input during the isentropic process is P2 = 1 MPa

Ws ,in = m(u2 s - u1 ) = (100 kg)(2903.0 - 2505.6)kJ/kg = 39,740 kJ The actual work input is then

Wa ,in =

Ws ,in hC

39, 740 kJ = 44,160 kJ 0.90

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] x1=1 P2=1000 [kPa] m=100 [kg] eta_C=0.90 "Analysis" Fluid$='steam_iapws' u1=intenergy(Fluid$,P=P1,x=1) s1=entropy(Fluid$,P=P1,x=x1) s2s=s1 u2s=intenergy(Fluid$,P=P2,s=s2s) W_s_in=m*(u2s-u1) W_a_in=W_s_in/eta_C

8-70 R-134a is expanded in an adiabatic process with an isentropic efficiency of 0.95. The final volume is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-307

Assumptions 1. Kinetic and potential energy changes are negligible. 2. The device is adiabatic and thus heat transfer is negligible. Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wout = D U = m(u2 - u1 ) From the R-134a tables (Tables A-11E through A-13E),

P1 = 120 psia ïüï u1 = 108.49 Btu/lbm ý ïïþ s1 = 0.22364 Btu/lbm ×R T1 = 100°F P2 = 20 psia s2 s = s1

s2 s - s f 0.22364 - 0.02603 = = 0.9897 ü ïï x2 s = s fg 0.19967 ý ïïþ u2 s = u f + x2 s u fg = 11.393 + (0.9897)(82.915) = 93.454 Btu/lbm

The actual work input is

wa ,out = hT ws ,out = hT (u1 - u2 s ) = (0.95)(108.49 - 93.454)Btu/lbm = 14.28 Btu/lbm The actual internal energy at the end of the expansion process is wa ,out = (u1 - u2 a ) ¾ ¾ ® u2 a = u1 - wa ,out = 108.49 - 14.28 = 94.21 Btu/lbm

The specific volume at the final state is (Table A-12E) u2 a - u f 94.21- 11.393 x = = = 0.9988 ïü ïý 2 u fg 82.915 ï u2 a = 94.21 Btu/lbm ïþ v 2 = v f + x2v fg = 0.01181 + (0.9988)(2.2781- 0.01181) = 2.2754 ft 3 /lbm P2 = 20 psia

The final volume is then

V2 = mv 2 = (10 lbm)(2.2754 ft 3 /lbm) = 22.75 ft 3

EES (Engineering Equation Solver) SOLUTION "Given" m=10 [lbm] P1=120 [psia] T1=100 [F] P2=20 [psia] eta_T=0.95 "Analysis" Fluid$='R134a' u1=intenergy(Fluid$,P=P1,T=T1) s1=entropy(Fluid$,P=P1,T=T1) s2s=s1 x2s=quality(Fluid$,P=P2,s=s2s) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-308

u2s=intenergy(Fluid$,P=P2,s=s2s) w_a_out=eta_T*(u1-u2s) u2=u1-w_a_out x2=quality(Fluid$,P=P2,u=u2) v2=volume(Fluid$,P=P2,u=u2) Vol2=m*v2

8-71 R-134a is expanded in an adiabatic process with an isentropic efficiency of 0.80. The work produced and final enthalpy are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. The device is adiabatic and thus heat transfer is negligible. Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wout = D U = m(u2 - u1 ) From the R-134a tables (Tables A-11 through A-13),

P1 = 700 kPa ïüï u1 = 256.41 kJ/kg ý ïïþ s1 = 0.9642 kJ/kg ×K T1 = 40°C P2 = 60 kPa s2 = s1

s2 - s f 0.9642 - 0.01633 = = 0.9997 üïï x2 = s 0.94812 ý fg ïïþ u2 s = u f + x2 s u fg = 3.795 + (0.9997)(205.34) = 209.07 kJ/kg

The actual work input is

wa ,out = hs ws ,out = hT (u1 - u2 s ) = (0.80)(256.41- 209.07)kJ/kg = 37.87 kJ / kg The actual internal energy at the end of the expansion process is wa ,out = (u1 - u2 a ) ¾ ¾ ® u2 = u1 - wa ,out = 256.41- 37.87 = 218.54 kJ/kg

The enthalpy at the final state is ïü ïý h = 238.43 kJ / kg (Table A-13) 2 u2 = 218.54 kJ/kg ïïþ P2 = 60 kPa

6-309

P2=60 [kPa] eta_T=0.80 "Analysis" Fluid$='R134a' u1=intenergy(Fluid$,P=P1,T=T1) s1=entropy(Fluid$,P=P1,T=T1) s2s=s1 x2s=quality(Fluid$,P=P2,s=s2s) u2s=intenergy(Fluid$,P=P2,s=s2s) w_a_out=eta_T*(u1-u2s) u2=u1-w_a_out h2=enthalpy(Fluid$,P=P2,u=u2)

8-72 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit temperature of air and the rate of entropy generation are to be determined for the cases of an insulated and uninsulated evaporator. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. There are no work interactions. 4. Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.287 kPa  m3 / kg.K (Table A-1). The constant pressure specific heat of air at room temperature is c p  1.005 kJ / kg  K (Table A-2). The properties of R-134a at the inlet and the exit states are (Tables A-11 through A-13) P1 = 120 kPa ü ïï h1 = h f + x1h fg = 22.47 + 0.3´ 214.52 = 86.83 kJ/kg ý ïïþ s1 = s f + x1 s fg = 0.09269 + 0.3(0.85520) = 0.3492 kJ/kg ×K x1 = 0.3

T2 = 120 kPaïü ïý h2 = hg @ 120 kPa = 236.99 kJ/kg sat. vapor ïïþ s2 = hg @ 120 kPa = 0.9479 kJ/kg ×K

Analysis (a) The mass flow rate of air is

(100 kPa )(6 m3 /min ) P3V3 mair = = = 6.968 kg/min RT3 (0.287 kPa ×m3 /kg ×K )(300 K ) We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as Mass balance ( for each fluid stream): min - mout = D msystem ¿ 0 (steady) = 0 ® min = mout ® m1 = m2 = mair and m3 = m4 = mR

6-310

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 + m3 h3 = m2 h2 + m4 h4

(since Q = W = D ke @D pe @ 0)

Combining the two, mR (h2 - h1 ) = mair (h3 - h4 ) = mair c p (T3 - T4 ) Solving for T4 , T4 = T3 -

mR (h2 - h1 ) mair c p

Substituting, T4 = 27°C -

(2 kg/min)(236.99 - 86.83) kJ/kg = - 15.9°C=257.1 K (6.968 kg/min)(1.005 kJ/kg ×K)

Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as

Sin - Sout

Rate of net entropy transfer by heat and mass

= D Ssystem ¿ 0 (steady)

Sgen Rate of entropy generation

Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 (since Q = 0)

mR s1 + mair s3 - mR s2 - mair s4 + Sgen = 0 or,

Sgen = mR (s2 - s1 )+ mair (s4 - s3 )

where 0

T P¿ T 257.1 K s4 - s3 = c p ln 4 - R ln 4 = c p ln 4 = (1.005 kJ/kg ×K) ln = - 0.1551 kJ/kg ×K T3 P3 T3 300 K

Substituting,

Sgen = (2 kg/min )(0.9479-0.3492 kJ/kg ×K)+ (6.968 kg/min)( - 0.1551 kJ/kg ×K) = 0.1175 kJ/min ×K = 0.00196 kW / K

(b) When there is a heat gain from the surroundings at a rate of 30 kJ / min , the steady-flow energy equation reduces to

Qin = mR (h2 - h1 )+ mair c p (T4 - T3 ) Solving for T4 , T4 = T3 +

Qin - mR (h2 - h1 ) mair c p

Substituting, T4 = 27°C +

(30 kJ/min) - (2 kg/min)(236.99 - 86.83) kJ/kg = - 11.6°C = 261.4 K (6.968 kg/min)(1.005 kJ/kg ×K)

The entropy generation in this case is determined by applying the entropy balance on an extended system that includes the evaporator and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surrounding air at all times. The entropy balance for the extended system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem ¿ 0 (steady) Rate of change of entropy

6-311

Qin + mR s1 + mair s3 - mR s2 - mair s4 + Sgen = 0 Tsurr

Sgen = mR (s2 - s1 )+ mair (s4 - s3 )-

where

s4 - s3 = c p ln

Qin T0

T4 PZ 261.4 K - R ln 4 = (1.005 kJ/kg ×K) ln = - 0.1384 kJ/kg ×K T3 P3 300 K

Substituting,

Sgen = (2 kg/min )(0.9479 - 0.3492) kJ/kg ×K + (6.968 kg/min )(- 0.1384 kJ/kg ×K )-

30 kJ/min 305 K

= 0.1348 kJ/min ×K = 0.00225 kW / K

Discussion Note that the rate of entropy generation in the second case is greater because of the irreversibility associated with heat transfer between the evaporator and the surrounding air.

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=27 [C] Vol_dot_1=6 [m^3/min] P3=120 [kPa] x3=0.30 m_dot_3=2 [kg/min] P4=P3 x4=1 T_surr=32 [C] Q_dot_in=30 [kJ/min] "Properties" h1=enthalpy(air, T=T1) s1=entropy(air, T=T1, P=P1) h3=enthalpy(R134a, P=P3, x=x3) s3=entropy(R134a, P=P3, x=x3) h4=enthalpy(R134a, P=P4, x=x4) s4=entropy(R134a, P=P4, x=x4) C_p=CP(air, T=25) C_v=CV(air, T=25) R=C_p-C_v "Analysis" "(a)" v1=(R*(T1+273))/P1 m_dot_1=Vol_dot_1/v1 m_dot_1=m_dot_2 "mass balance on air" m_dot_3=m_dot_4 "mass balance on R134a" m_dot_1*h1+m_dot_3*h3=m_dot_2*h2_a+m_dot_4*h4 "energy balance on heat exchanger" T2_a=temperature(air, h=h2_a) m_dot_1*s1+m_dot_3*s3-m_dot_2*s2_a-m_dot_4*s4+S_dot_gen_a=0 "entropy balance" s2_a=entropy(air, T=T2_a, P=P2) P2=P1 "(b)" Q_dot_in+m_dot_1*h1+m_dot_3*h3=m_dot_2*h2_b+m_dot_4*h4 "energy balance on heat exchanger" T2_b=temperature(air, h=h2_b) Q_dot_in/(T_surr+273)+m_dot_1*s1+m_dot_3*s3-m_dot_2*s2_b-m_dot_4*s4+S_dot_gen_b=0 "entropy balance" s2_b=entropy(air, T=T2_b, P=P2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-312

8-73 Air flows in an adiabatic nozzle. The isentropic efficiency, the exit velocity, and the entropy generation are to be determined. Properties The gas constant of air is R  0.287 kJ / kg.K (Table A-1). Assumptions 1. Steady operating conditions exist. 2. Potential energy changes are negligible. Analysis (a) (b) Using variable specific heats, the properties can be determined from air table as follows

T1 = 400 K

¾¾ ®

T2 = 350 K

¾¾ ®

h1 = 400.98 kJ/kg s10 = 1.99194 kJ/kg.K

h2 = 350.49 kJ/kg s20 = 1.85708 kJ/kg.K

s2o = s1o + R ln( P2 / P1 ) = 1.99194 kJ/kg ×K + (0.287 kJ/kg ×K)ln(300/500) = 1.84533 kJ/kg ×K ¾ ¾ ® h2 s = 346.44 kJ/kg Energy balances on the control volume for the actual and isentropic processes give V12 V2 = h2 + 2 2 2 ö ö V22 æ (30 m/s) 2 æ 1 kJ/kg ÷ çç çç 1 kJ/kg ÷ 400.98 kJ/kg + = 350.49 kJ/kg + ÷ 2 2÷ 2 2÷ ÷ ç ç è1000 m /s ø 2 2 è1000 m /s ø h1 +

V2 = 319.1 m / s

h1 + 400.98 kJ/kg +

V2 V12 = h2 s + 2s 2 2

ö ö V2s2 æ (30 m/s) 2 æ çç 1 kJ/kg ÷ çç 1 kJ/kg ÷ = 346.44 kJ/kg + ÷ 2 2÷ 2 2÷ ÷ ç ç è1000 m /s ø 2 2 è1000 m /s ø

V2s = 331.6 m/s S The isentropic efficiency is determined from its definition,

hN =

V22 (319.1 m/s)2 = = 0.926 2 V2s (331.6 m/s)2

(c) Since the nozzle is adiabatic, the entropy generation is equal to the entropy increase of the air as it flows in the nozzle sgen = D sair = s20 - s10 - R ln

P2 P1

= (1.85708 - 1.99194)kJ/kg.K - (0.287 kJ/kg.K)ln

300 kPa 500 kPa

6-313

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=500 [kPa] T_1=400 [K] Vel_1=30 [m/s] P_2=300 [kPa] T_2=350 [K] "ANALYSIS" Fluid$='Air' h_1=enthalpy(Fluid$, T=T_1) s_1=entropy(Fluid$, T=T_1, P=P_1) h_2=enthalpy(Fluid$, T=T_2) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) s_2=entropy(Fluid$, T=T_2, P=P_2) h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg)=h_2+Vel_2^2/2*Convert(m^2/s^2, kJ/kg) "energy balance for actual process" h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg)=h_2s+Vel_2s^2/2*Convert(m^2/s^2, kJ/kg) "energy balance for isentropic process" eta_N=Vel_2^2/Vel_2s^2 "isentropic efficiency of diffuser" s_gen=s_2-s_1 "entropy generaton"

8-74E Helium is accelerated by a 94% efficient nozzle from a low velocity to 1000 ft / s . The pressure and temperature at the nozzle inlet are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Helium is an ideal gas with constant specific heats. 3. Potential energy changes are negligible. 4. The device is adiabatic and thus heat transfer is negligible. Properties The specific heat ratio of helium is k = 1.667. The constant pressure specific heat of helium is 1.25 Btu / lbm  R (Table A-2E). Analysis We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m(h1 + V12 / 2) = m(h2 +V22 /2) (since Q @W @D pe @ 0)

0 = h2 - h1 +

V22 - V12 2

¾¾ ®

0 = c p ,avg (T2 - T1 )+

V22 - V12 2

6-314

T1 = T2 a +

V22s - V12 2c p

= 180°F +

æ 1 Btu/lbm ö÷ çç ÷= 196.0°F = 656 R 2 (1.25 Btu/lbm ×R )èç 25,037 ft 2 /s 2 ø÷

(1000 ft/s)

From the isentropic efficiency relation,

hN =

or,

h2 a - h1 c p (T2 a - T1 ) = h2 s - h1 c p (T2 s - T1 )

T2 s = T1 + (T2 a - T1 ) / h N = 656 + (640 - 656) / (0.94) = 639 R

æP ÷ ö( T From the isentropic relation, 2 s = ççç 2 ÷ ÷ T1 çè P1 ÷ ø

k - 1)/ k

æT1 ö÷ ( P1 = P2 ççç ÷ ÷ èçT ø÷

k / k - 1)

1.667/0.667

æ656 R ÷ ö = (14 psia )çç ÷ çè639 R ÷ ø

= 14.9 psia

EES (Engineering Equation Solver) SOLUTION "Given" eta_N=0.94 Vel1=0 P2=14 [psia] T2=180+460 "[R]" Vel2=1000 [ft/s] "Properties" C_p=CP(Helium, T=540, P=14.7) C_v=CV(Helium, T=540, P=14.7) k=C_p/C_v h1=C_p*T1 h2=C_p*T2 h2_s=C_p*T2_s "Analysis" h1+Vel1^2/2*1/Convert(Btu/lbm, ft^2/s^2)=h2+Vel2^2/2*1/Convert(Btu/lbm, ft^2/s^2) "energy balance" eta_N=(h2-h1)/(h2_s-h1) T2_s/T1=(P2/P1)^((k-1)/k)

8-75 Refrigerant-134a is expanded adiabatically in a capillary tube. The rate of entropy generation is to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. Analysis The rate of entropy generation within the expansion device during this process can be determined by applying the rate form of the entropy balance on the system. Noting that the system is adiabatic and thus there is no heat transfer, the entropy balance for this steady-flow system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem 0 (steady) Rate of change of entropy

6-315

m1s1 - m2 s2 + Sgen = 0 Sgen = m( s2 - s1 )

sgen = s2 - s1 It may be easily shown with an energy balance that the enthalpy remains constant during the throttling process. The properties of the refrigerant at the inlet and exit states are (Tables A-11 through A-13) T1 = 70°C ü ïï h1 = 156.15 kJ/kg ×K ý ïïþ s1 = 0.53763 kJ/kg ×K x1 = 0

h2 - h f 156.15 - 25.47 x = = = 0.6136 ïü ïý 2 h fg 212.96 h2 = h1 = 156.15 kJ/kg ×K ïïþ s2 = s f + x2 s fg = 0.10456 + (0.6136)(0.84119) = 0.62075 kJ/kg ×K

T2 = - 20°C

Substituting,

Sgen = m(s2 - s1 ) = (0.2 kg/s)(0.62075 - 0.53763) kJ/kg ×K = 0.0166 kW / K

EES (Engineering Equation Solver) SOLUTION "GIVEN" x1=0 T1=70 [C] T2=-20 [C] m_dot=0.2 [kg/s] "ANALYSIS" Fluid$='R134a' h1=enthalpy(Fluid$, T=T1, x=x1) s1=entropy(Fluid$, T=T1, x=x1) h2=h1 h2_f=enthalpy(Fluid$, T=T2, x=0) h2_g=enthalpy(Fluid$, T=T2, x=1) h2_fg=h2_g-h2_f x2=(h2-h2_f)/h2_fg s2_f=entropy(Fluid$, T=T2, x=0) s2_g=entropy(Fluid$, T=T2, x=1) s2_fg=s2_g-s2_f s2=s2_f+x2*s2_fg S_dot_gen=m_dot*(s2-s1)

8-76 The work input and the entropy generation are to be determined for the compression of saturated liquid water in a pump and that of saturated vapor in a compressor. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Heat transfer to or from the fluid is zero. Analysis Pump Analysis: (properties from steam tables or EES)

6-316

ïüï h1 = 417.51 kJ/kg ý x1 = 0 (sat. liq.)ïïþ s1 = 1.3028 kJ/kg.K

P1 = 100 kPa

P2 = 1 MPa ïüï ý h2 s = 418.45 kJ/kg ïïþ s2 = s1 h2 = h1 +

h2 s - h1 418.45 - 417.51 = 417.51 + = 418.61 kJ/kg hP 0.85

ïü ïý s = 1.3032 kJ/kg.K 2 h2 = 418.61 kJ/kgïïþ P2 = 1 MPa

wP = h2 - h1 = 418.61- 417.51 = 1.10 kJ / kg

sgen, P = s2 - s1 = 1.3032 - 1.3028 = 0.0004 kJ / kg.K Compressor Analysis:

üïï h1 = 2675.0 kJ/kg ý x1 = 1 (sat. vap.)ïïþ s1 = 7.3589 kJ/kg.K

P1 = 100 kPa

P2 = 1 MPa ïüï ý h2 s = 3193.6 kJ/kg ïïþ s2 = s1 h2 = h1 +

h2 s - h1 3193.6 - 2675.0 = 2675.0 + = 3285.1 kJ/kg hC 0.85

ïü ïý s = 7.4974 kJ/kg.K 2 h2 = 3285.1 kJ/kgïïþ P2 = 1 MPa

wC = h2 - h1 = 3285.1- 2675.0 = 610.1kJ / kg

sgen, C = s2 - s1 = 7.4974 - 7.3589 = 0.1384 kJ / kg.K

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] P_2=1000 [kPa] eta_s=0.85 "ANALYSIS" Fluid$='Steam_iapws' "Pump analysis" h_p_1=enthalpy(Fluid$, P=P_1, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-317

s_p_1=entropy(Fluid$, P=P_1, x=0) h_p_2s=enthalpy(Fluid$, P=P_2, s=s_p_1) h_p_2=h_p_1+(h_p_2s-h_p_1)/eta_s s_p_2=entropy(Fluid$, P=P_2, h=h_p_2) w_p=h_p_2-h_p_1 "pump work input" s_p_gen=s_p_2-s_p_1 "entropy generation" "Compressor analysis" h_c_1=enthalpy(Fluid$, P=P_1, x=1) s_c_1=entropy(Fluid$, P=P_1, x=1) h_c_2s=enthalpy(Fluid$, P=P_2, s=s_c_1) h_c_2=h_c_1+(h_c_2s-h_c_1)/eta_s s_c_2=entropy(Fluid$, P=P_2, h=h_c_2) w_c=h_c_2-h_c_1 "compressor work input" s_c_gen=s_c_2-s_c_1 "entropy generation" 8-77 Carbon dioxide is compressed in a reversible isothermal process using a steady-flow device. The work required and the heat transfer are to be determined.

Assumptions 1. Steady operating conditions exist. 2. Changes in the kinetic and potential energies are negligible. 3.

CO2 is an ideal gas with constant specific heats.

Properties The gas constant of CO2 is R  0.1889 kPa  m3 / kg  K (Table A-2a). Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + Win - Qout = mh2

Win - Qout = m(h2 - h1 ) = mc p (T2 - T1 ) Win = Qout

(since T2 = T1 )

The work input for this isothermal, reversible process is win = RT ln

P2 400 kPa = (0.1889 kJ/kg ×K)(293 K)ln = 76.7 kJ / kg P1 100 kPa

6-318

qout = win = 76.7 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T=(20+273) [K] P2=400 [kPa] "Analysis" R=0.1889 [kPa-m^3/kg-K] w_in=R*T*ln(P2/P1) q_out=w_in

8-78 Carbon dioxide is compressed in an isentropic process using a steady-flow device. The work required and the heat transfer are to be determined. Assumptions 1. Steady operating conditions exist. 2. Changes in the kinetic and potential energies are negligible. 3.

CO2 is an ideal gas with constant specific heats.

Properties The gas constant of CO2 is R  0.1889 kPa  m3 / kg  K . Other properties at room temperature are

c p  0.846 kJ / kg  K and k =1.289 (Table A-2a). Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + Win = mh2 Win = m(h2 - h1 ) = mc p (T2 - T1 ) The temperature at the compressor exit for the isentropic process of an ideal gas is ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P1 ÷ ø

0.289/1.289

æ400 kPa ö÷ = (293 K) çç ÷ èç100 kPa ø÷

= 399.8 K

Substituting,

win = c p (T2 - T1 ) = (0.846 kJ/kg ×K)(399.8 - 293)K = 90.4 kJ The work input increases from 76.7 kJ / kg to 90.4 kJ / kg when the process is executed isentropically instead of isothermally. Since the process is isentropic (i.e., reversible, adiabatic), the heat transfer is zero.

6-319

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(20+273) [K] P2=400 [kPa] "Analysis" c_p=0.846 [kJ/kg-K] k=1.289 T2=T1*(P2/P1)^((k-1)/k) w_in=c_p*(T2-T1)

8-79 Air is compressed in a two-stage ideal compressor with intercooling. For a specified mass flow rate of air, the power input to the compressor is to be determined, and it is to be compared to the power input to a single-stage compressor. Assumptions 1. The compressor operates steadily. 2. Kinetic and potential energies are negligible. 3. The compression process is reversible adiabatic, and thus isentropic. 4. Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R  0.287 kPa  m3 / kg.K (Table A-1). The specific heat ratio of air is k = 1.4 (Table A-2). Analysis The intermediate pressure between the two stages is

Px =

P1 P2 = (100 kPa )(625 kPa ) = 250 kPa

The compressor work across each stage is the same, thus total compressor work is twice the compression work for a single stage:

wcomp, in = (2)(wcomp, in, I ) = 2 = 2

kRT1 (k - 1)/ k - 1 (Px P1 ) k- 1

(

)

0.4/1.4 ö (1.4)(0.287 kJ/kg ×K )(300 K )æ çæ ç 250 kPa ÷

ççç ÷ ÷ çè èç100 kPa ø

1.4 - 1

ö ÷ - 1÷ ÷ ÷ ø÷

= 180.4 kJ/kg and Win = mwcomp, in = (0.15 kg/s)(180.4 kJ/kg ) = 27.1 kW

The work input to a single-stage compressor operating between the same pressure limits would be

wcomp, in =

0.4/1.4 ö ö (1.4)(0.287 kJ/kg ×K )(300 K )æ kRT1 (k - 1)/ k ÷ ççæ çç625 kPa ÷ ÷ - 1 = 1 = 207.4 kJ/kg (P2 P1 ) ÷ ÷ ççèç100 kPa ÷ ÷ ø ÷ k- 1 1.4 - 1 è ø

(

)

6-320

and Win = mwcomp, in = (0.15 kg/s)(207.4 kJ/kg ) = 31.1 kW

Discussion Note that the power consumption of the compressor decreases significantly by using 2-stage compression with intercooling.

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(27+273) [K] P2=625 [kPa] m_dot=0.15 [kg/s] "Properties" k=1.4 "Table A-2a" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" P_x=sqrt(P1*P2) W_dot_comp_in_TwoStage=2*m_dot*(k*R*T1)/(k-1)*((P_x/P1)^((k-1)/k)-1) W_dot_comp_in_SingleStage=m_dot*(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1) w_comp_in_twostage=W_dot_comp_in_TwoStage/m_dot w_comp_in_singlestage=W_dot_comp_in_SingleStage/m_dot

8-80 Helium gas is compressed in an adiabatic closed system with an isentropic efficiency of 80%. The work input and the final temperature are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Helium is an ideal gas. Properties The properties of helium are cv  3.1156 kJ / kg  K and k = 1.667 (Table A-2b). Analysis We take the helium as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Win = D U = m(u2 - u1 ) = mcv (T2 - T1 ) The isentropic exit temperature is ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ èç P ø÷ 1

0.667/1.667

æ900 kPa ÷ ö = (300 K) çç ÷ çè100 kPa ÷ ø

= 722.7 K

The work input during isentropic process would be

Ws, in = mcv (T2 s - T1 ) = (3 kg)(3.1156 kJ/kg ×K)(722.7 - 300)K = 3950 kJ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-321

The work input during the actual process is

Win =

Ws, in h

3950 kJ = 4938 kJ 0.80

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=3 [kg] P1=100 [kPa] T1=(27+273) [K] P2=900 [kPa] eta_C=0.80 "PROPERTIES" c_v=3.1156 [kJ/kg-K] "Table A-2a" k=1.667 "Table A-2a" "ANALYSIS" T2_s=T1*(P2/P1)^((k-1)/k) W_s_in=m*c_v*(T2_s-T1) W_in=W_s_in/eta_C

8-81 Steam expands in a two-stage adiabatic turbine from a specified state to specified pressure. Some steam is extracted at the end of the first stage. The power output of the turbine is to be determined for the cases of 100% and 88% isentropic efficiencies. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The turbine is adiabatic and thus heat transfer is negligible.

Properties From the steam tables (Tables A-4 through 6)

P1 = 6 MPa ïüï h1 = 3423.1 kJ/kg ý T1 = 500°C ïïþ s1 = 6.8826 kJ/kg ×K

P2 = 1.2 MPaïüï ý h2 = 2962.8 kJ / kg ïïþ s2 = s1 s3 - s f 6.8826 - 0.8320 x = = = 0.8552 P3 = 20 kPa ïü ïý 3 s fg 7.0752 ïïþ s3 = s1 h3 = h f + x3 h fg = 251.42 + (0.8552)(2357.5) = 2267.5 kJ/kg

6-322

m3 = 0.9m1 = (0.9)(15 kg/s) = 13.5 kg/s

We take the entire turbine, including the connection part between the two stages, as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters the turbine and two fluid streams leave, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 = (m1 - m3 )h2 + Wout + m3 h3 Wout = m1h1 - (m1 - m3 )h2 - m3 h3 = m1 (h1 - h2 ) + m3 (h2 - h3 ) Substituting, the power output of the turbine is Wout = (15 kg/s)(3423.1- 2962.8)kJ/kg + (13.5 kg)(2962.8 - 2267.5)kJ/kg = 16, 290 kW

(b) If the turbine has an isentropic efficiency of 88%, then the power output becomes Wa = hT Ws = (0.88)(16,290 kW ) = 14, 335 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=6000 [kPa] T1=500 [C] P2=1200 [kPa] P3=20 [kPa] m1_dot=15 [kg/s] m2_dot=0.10*m1_dot eta_T=0.88 "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2=s1 "isentropic process" h2=enthalpy(Fluid$, P=P2, s=s2) s3=s1 "isentropic process" h3=enthalpy(Fluid$, P=P3, s=s3) x3=quality(Fluid$, P=P3, s=s3) "Analysis" "(a)" m3_dot=0.90*m1_dot m1_dot*h1=W_dot_out_s+m2_dot*h2+m3_dot*h3 "energy balance" "(b)" W_dot_out_a=eta_T*W_dot_out_s

8-82 Steam expands in an adiabatic turbine. Steam leaves the turbine at two different pressures. The process is to be sketched on a T-s diagram and the work done by the steam per unit mass of the steam at the inlet are to be determined. Assumptions 1. The kinetic and potential energy changes are negligible. Analysis (b) From the steam tables (Tables A-4 through A-6), © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-323

T1 = 500°C üïï h1 = 3423.1 kJ/kg ý P1 = 6 MPaïïþ s1 = 6.8826 kJ/kg ×K

P2 = 1 MPa üïï ý h2 s = 2921.3 kJ/kg ïïþ s2 = s1 P3 = 10 kPa ïüï h3 s = 2179.6 kJ/kg ý ïïþ x3 s = 0.831 s3 = s1 A mass balance on the control volume gives

m1 = m2 + m3 where

m2 = 0.1m1 m3 = 0.9m1

We take the turbine as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein = Eout m1h1 = Ws ,out + m2 h2 + m3 h3 m1h1 = Ws ,out + 0.1m1h2 + 0.9m1h3 or

h1 = ws ,out + 0.1h2 + 0.9h3 ws ,out = h1 - 0.1h2 - 0.9h3 = 3423.1- (0.1)(2921.3) - (0.9)(2179.6) = 1169.3 kJ/kg The actual work output per unit mass of steam at the inlet is

wout = hT ws ,out = (0.85)(1169.3 kJ/kg) = 993.9 kJ / kg

6-324

T1=500 [C] P2=1000 [kPa] P3=10 [kPa] y=0.10 eta_T=0.85 "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2=s1 h_2s=enthalpy(Fluid$, P=P2, s=s2) s3=s1 h_3s=enthalpy(Fluid$, P=P3, s=s3) "Analysis" w_s_out=h1-y*h_2s-(1-y)*h_3s w_out=eta_T*w_s_out

8-83 Refrigerant-134a is compressed by a 1.3-kW adiabatic compressor from a specified state to another specified state. The isentropic efficiency, the volume flow rate at the inlet, and the maximum flow rate at the compressor inlet are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Properties From the R-134a tables (Table A-13)

v1 = 0.14605 m3 /kg P1 = 140 kPaïü ïý h = 246.37 kJ/kg 1 T1 = - 10°C ïïþ s1 = 0.9724 kJ/kg ×K

P2 = 700 kPa üïï ý h2 = 298.44 kJ/kg T2 = 60°C ïïþ P2 = 700 kPa üïï ý h2 s = 281.18 kJ/kg ïïþ s2 s = s1 Analysis (a) The isentropic efficiency is determined from its definition, hC =

h2 s - h1 281.18 - 246.37 = = 0.668 = 66.8% h2 a - h1 298.44 - 246.37

6-325

(b) There is only one inlet and one exit, and thus m1 = m2 = m. We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Wa, in + mh1 = mh2

(since Q @Δke @Δpe @ 0)

Wa, in = m(h2 - h1 ) Then the mass and volume flow rates of the refrigerant are determined to be

Wa, in h2 a - h1

1.3 kJ/s = 0.02497 kg/s (298.44 - 246.37) kJ/kg

V1 = mv1 = (0.02497 kg/s)(0.14605 m3 /kg) = 0.003647 m3 /s = 219 L / min

(c) The volume flow rate will be a maximum when the process is isentropic, and it is determined similarly from the steadyflow energy equation applied to the isentropic process. It gives

mmax =

Ws, in h2 s - h1

1.3 kJ/s = 0.03735 kg/ (281.18 - 246.37) kJ/kg

V1, max = mmaxv1 = (0.03735 kg/s)(0.14605 m3 /kg ) = 0.005455 m 3 /s = 327 L / min

Discussion Note that the raising the isentropic efficiency of the compressor to 100% would increase the volumetric flow rate by about 50%.

EES (Engineering Equation Solver) SOLUTION "Given" P1=140 [kPa] T1=-10 [C] P2=700 [kPa] T2=60 [C] W_dot_in=1.3 [kW] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, T=T2) s2_s=s1 "isentropic process" h2_s=enthalpy(Fluid$, P=P2, s=s2_s) "Analysis" "(a)" eta_C=(h2_s-h1)/(h2-h1) "(b)" W_dot_in=m_dot*(h2-h1) Vol1_dot=m_dot*v1*Convert(m^3/s, L/min) "(c)" W_dot_in=m_dot_max*(h2_s-h1) Vol1_dot_max=m_dot_max*v1*Convert(m^3/s, L/min)

6-326

8-84 Air is expanded by an adiabatic turbine with an isentropic efficiency of 85%. The outlet temperature, the work produced, and the entropy generation are to be determined.

Assumptions 1. This is a steady-flow process since there is no change with time. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at the anticipated average temperature of 400 K are c p  1.013 kJ / kg. C and k = 1.395 (Table A-2b). Also,

R  0.287 kJ / kg  K (Table A-2a). Analysis We take the turbine as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the turbine, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 = Wa, out + mh2

(since Q @Δke @Δpe @0)

Wa , out = m(h1 - h2 ) = mc p (T1 - T2 ) The isentropic exit temperature is ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 s ÷ ÷ çè P1 ø÷

0.395/1.395

æ100 kPa ÷ ö = (400 + 273 K) çç ÷ èç 2800 kPa ÷ ø

= 262.0 K

From the definition of the isentropic efficiency,

wa , out = hT ws , out = hT c p (T1 - T2 s ) = (0.90)(1.013 kJ/kg ×K)(673 - 262.0)K = 374.7 kJ / kg The actual exit temperature is then wa , out = c p (T1 - T2 a ) ¾ ¾ ® T2 a = T1 -

wa , out cp

= T1 -

wa , out cp

= 673 K -

374.7 kJ/kg = 303.1 K 1.013 kJ/kg ×K

The rate of entropy generation in the turbine is determined by applying the rate form of the entropy balance on the turbine:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem ¿ 0 (steady) Rate of change of entropy

6-327

Sgen = m( s2 - s1 )

sgen = s2 - s1 Then, from the entropy change relation of an ideal gas, T P sgen = s2 - s1 = c p ln 2 - R ln 2 T1 P1

= (1.013 kJ/kg ×K)ln

303.1 K 100 kPa - (0.287 kJ/kg ×K)ln 673 K 2800 kPa

= 0.1482 kJ / kg×K

EES (Engineering Equation Solver) SOLUTION "GIVEN" eta_T=0.90 P1=2800 [kPa] T1=(400+273) [K] P2=100 [kPa] "PROPERTIES" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.013 [kJ/kg-K] "at 400 K, Table A-2b" k=1.395 "at 400 K, Table A-2a" "ANALYSIS" T2_s=T1*(P2/P1)^((k-1)/k) w_s_out=c_p*(T1-T2_s) w_a_out=eta_T*w_s_out T2=T1-w_a_out/c_p s_gen=c_p*ln(T2/T1)-R*ln(P2/P1)

8-85 Steam is expanded in an adiabatic turbine. Six percent of the inlet steam is bled for feedwater heating. The isentropic efficiencies for two stages of the turbine are given. The power produced by the turbine and the overall efficiency of the turbine are to be determined. Assumptions 1. This is a steady-flow process since there is no change with time. 2. The turbine is well-insulated, and there is no heat transfer from the turbine. Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

6-328

Wout = m1h1 - m2 h2 - m3 h3 wout = h1 - 0.06h2 - 0.94h3 wout = (h1 - h2 ) + 0.94(h2 - h3 ) The isentropic and actual enthalpies at three states are determined using steam tables as follows:

P1 = 4 MPa ïü ïý T1 = 350°C ïïþ

h1 = 3093.3 kJ/kg s1 = 6.5843 kJ/kg ×K

ü ïï ý s2 = s1 = 6.5843 kJ/kg ×K ïïþ

P2 = 800 kPa

h1 - h2 h1 - h2 s

hT ,1 =

¾¾ ®

x2 s = 0.9832 h2 s = 2734.0 kJ/kg

h2 = h1 - hT ,1 (h1 - h2 s ) = 3093.3 - (0.97)(3093.3 - 2734.0) = 2744.8 kJ/kg

ü ïï x2 = 0.9885 ý h2 = 2744.8 kJ/kg ïïþ s2 = 6.6086 kJ/kg ×K P3 = 30 kPa ïü ïý x3s = 0.8302 s3 = s2 = 6.6086 kJ/kg ×K ïïþ h3s = 2227.9 kJ/kg

P2 = 800 kPa

hT ,2 =

h2 - h3 h2 - h3 s

¾¾ ®

h3 = h2 - hT ,2 (h2 - h3 s ) = 2744.8 - (0.95)(2744.8 - 2227.9) = 2253.7 kJ/kg

Substituting,

wout = (h1 - h2 ) + 0.94(h2 - h3 ) = (3093.3 - 2744.8) + 0.94(2744.8 - 2253.7) = 810.1 kJ / kg - inlet

The overall isentropic efficiency of the turbine is hT =

(h1 - h2 ) + 0.94( h2 - h3 ) (3093.3 - 2744.8) + 0.94(2744.8 - 2253.7) 810.1 = = = 0.958 = 95.8% (h1 - h2 s ) + 0.94(h2 - h3 s ) (3093.3 - 2734.0) + 0.94(2744.8 - 2227.9) 845.2

EES (Engineering Equation Solver) SOLUTION "Given" P1=4000 [kPa] T1=350 [C] P2=800 [kPa] P3=30 [kPa] y=0.06 eta_T1=0.97 eta_T2=0.95 "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2s=s1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-329

x2s=quality(Fluid$, P=P2, s=s2s) h2s=enthalpy(Fluid$, P=P2, s=s2s) h2=h1-eta_T1*(h1-h2s) x2=quality(Fluid$, P=P2, h=h2) s2=entropy(Fluid$, P=P2, h=h2) s3s=s2 x3s=quality(Fluid$, P=P3, s=s3s) h3s=enthalpy(Fluid$, P=P3, s=s3s) h3=h2-eta_T2*(h2-h3s) w_out=h1-h2+(1-y)*(h2-h3) eta_T=(h1-h2+(1-y)*(h2-h3))/(h1-h2s+(1-y)*(h2-h3s))

8-86E R-134a vapor from a storage tank is used to drive an isentropic turbine. The work produced by the turbine is to be determined. Assumptions 1. This is an unsteady-flow process since the conditions in the tank changes. 2. Kinetic and potential energy changes are negligible. 3. The device is adiabatic and thus heat transfer is negligible. Analysis The inlet and exit enthalpies for the turbine are (Tables A-11E through A-13E),

üïï ý x1 = 1 (sat. vap.)ïïþ

h1 = hg = 113.96 Btu/lbm

P2 s = 10 psia ïü ïý ïïþ s2 = s1

x2 =

T1 = 80°F

s1 = sg = 0.21976 Btu/lbm ×R s2 s - s f s fg

0.21976 - 0.00741 = 0.95619 0.22208

h2 = h f + x2 h fg = 3.153 + (0.95619)(95.536) = 94.50 Btu/lbm

The initial mass of R-134a in the tank is minitial =

V v f @80° F

10 ft 3 = 749.6 lbm 0.01334 ft 3 /lbm

The amount of mass which passes through the turbine during this expansion is then

minitial 749.6 lbm = = 374.8 lbm 2 2

The amount of work produced from the turbine is then

Wout = m(h1 - h2 ) = (374.8 lbm)(113.96 - 94.50)Btu/lbm = 7294 Btu

6-330

x1=1 P2=10 [psia] Vol=10 [ft^3] "Analysis" Fluid$='R134a' h1=enthalpy(Fluid$,T=T1,x=x1) s1=entropy(Fluid$,T=T1,x=x1) s2=s1 x2=quality(Fluid$,P=P2,s=s2) h2=enthalpy(Fluid$,P=P2,s=s2) v_f=volume(Fluid$,T=T1,x=0) m_initial=Vol/v_f m=m_initial/2 W_out=m*(h1-h2)

8-87E A system consisting of a compressor, a storage tank, and a turbine as shown in the figure is considered. The total work required to fill the tank and the total heat transferred from the air in the tank as it is being filled are to be determined. Assumptions 1. Changes in the kinetic and potential energies are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R  0.3704 psia  ft 3 / lbm  R , c p  0.240 Btu / lbm  R ,

cv  0.171 Btu / lbm  R , k = 1.4 (Table A-2Ea). Analysis The initial mass of air in the tank is

minitial =

PinitialV (14.696 psia)(1´ 106 ft 3 )/lbm ×R) = = 74,860 lbm RTinitial (0.3704 psia ×ft 3 /lbm ×R)(530 R)

and the final mass in the tank is

mfinal =

PfinalV (146.96 psia)(1´ 106 ft 3 )/lbm ×R) = = 748, 600 lbm RTfinal (0.3704 psia ×ft 3 /lbm ×R)(530 R)

Since the compressor operates as an isentropic device, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P ÷ ø 1

where P2 matches the pressure in the tank. The conservation of mass applied to the tank gives

dm = m2 dt while the first law gives d (mu ) dm - h2 dt dt Employing the ideal gas equation of state and using constant specific heats, expands this result to Q=

V cv dP V dP - c pT2 R dt RT dt Using the temperature relation across the compressor and multiplying by dt puts this result in the form Q=

( k - 1)/ k

æP ö Vc ÷ Qdt = v dP - c pT1 ççç ÷ ÷ çè P1 ÷ R ø

V dP RT

When this integrated, it yields (i and f stand for initial and final states) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-331

é P ö V cv k c pV ê æ ç f÷ ( Pf - Pi ) êPf ç ÷ ÷ ÷ R 2k - 1 R ê çèç Pi ø ë

( k - 1)/ k

ù - Pi ú ú ú û

ö (1´ 106 )(0.171) 1.4 (0.24)(1´ 106 ) éê æ 147 ÷ (147 - 14.7) 147 çç ÷ ê ÷ ç è14.7 ø 0.3704 2(1.4) - 1 0.3704 ëê

0.4/1.4

ù - 14.7ú ú ú û

= - 7.454×107 Btu

The negative result show that heat is transferred from the tank. Applying the first law to the tank and compressor gives

(Q + W )dt = d (mu ) - h1dm which integrates to

Q + W = (m f u f - mi ui ) - h1 (m f - mi ) Upon rearrangement,

W = - Q - (c p - cv )T (m f - mi ) = 7.454´ 107 - (0.240 - 0.171)(530)(748,600 - 74,860) = 4.990×107 Btu

EES (Engineering Equation Solver) SOLUTION "Given" Vol=1E6 [ft^3] c_v=0.171 [Btu/lbm-F] c_p=0.24 [Btu/lbm-F] R=0.3704 [Btu/lbm-F] k=1.4 P_i=14.7 [psia] P_f=147 [psia] T=530 [R] "Analysis" m_i=(P_i*Vol)/(R*T) m_f=(P_f*Vol)/(R*T) Q=Vol*c_v/R*(P_f-P_i)-k/(2*k-1)*c_p*Vol/R*(P_f*(P_f/P_i)^((k-1)/k)-P_i) W=-Q-(c_p-c_v)*T*(m_f-m_i)

8-88E A system consisting of a compressor, a storage tank, and a turbine as shown in the figure is considered. The total work produced by the turbine and the total heat added to the air in the tank during the discharge are to be determined. Assumptions 1. Changes in the kinetic and potential energies are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R  0.3704 psia  ft 3 / lbm  R , c p  0.240 Btu / lbm.R ,

cv  0.171 Btu / lbm  R , k = 1.4 (Table A-2Ea). Analysis The initial mass of air in the tank is

minitial =

PinitialV (146.96 psia)(1´ 106 ft 3 )/lbm ×R) = = 748, 600 lbm RTinitial (0.3704 psia ×ft 3 /lbm ×R)(530 R)

and the final mass in the tank is

PfinalV (14.696 psia)(1´ 106 ft 3 )/lbm ×R) = = 74,860 lbm RTfinal (0.3704 psia ×ft 3 /lbm ×R)(530 R) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC. mfinal =

6-332

At any time, mt =

PV RT

and

dmt V dP = dt RT dt Since the turbine operates as an isentropic device, ( k - 1)/ k

æP ö ÷ T4 = T3 ççç 4 ÷ ÷ çè P ÷ ø 3

where P4 is the constant atmospheric pressure and P3 matches the pressure in the tank. The conservation of mass applied to the tank gives

dm =- m dt while the first law applied to the turbine is dm c p (T3 - T4 ) dt Substitution of the preceding results expands this result to W = m(h3 - h4 ) = mc p (T3 - T4 ) = -

éæP ö( k - 1)/ k ù V dP ÷ W = c pT3 êêççç 4 ÷ - 1ú ú ÷ ÷ ç RT dt êëè P3 ø ú û 3

Noting that P3 is the pressure in the tank and that P4 is the constant atmospheric pressure, multiplying of this result by dt and integration from the initial to the final state gives ( k - 1)/ k ù ö c pV f éêæ údP çç P2 ÷ ÷ W= 1 ê ÷ ú ÷ R òi êçèç P1 ø ú ë û 1/ k ù í ü é ï c pV ïïï P1 ö ê1- æ ú- ( P - P )ïï ÷ ç ÷ = kP ç ì f ê ç ÷ ú f i ý ÷ ú ïï R ïï êë çè P2 ø û ï îï þ =

ü é æ147 ö1/1.4 ù ï (0.240)(1´ 106 ) ïíï ê1- ç ú- (14.7 - 147)ïý ÷ (1.4)(14.7) ÷ ì ç ê ú ÷ ç ïï ïï 0.3704 êë è14.7 ø ú û ï îï þ

= 2.999×107 Btu

Adoption of the first law to the tank yields

Q = (m f u f - mi ui ) + h(m f - mi ) = (m f - mi )T (c p + cv ) = (748,600 - 74,860)(530)(0.240 + 0.171) = - 1.468×108 Btu

The negative result shows that heat is transferred from the turbine.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=1E6 [ft^3] c_v=0.171 [Btu/lbm-F] c_p=0.24 [Btu/lbm-F] R=0.3704 [Btu/lbm-F] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-333

k=1.4 P_i=147 [psia] P_f=14.7 [psia] T=530 [R] "Analysis" m_i=(P_i*Vol)/(R*T) m_f=(P_f*Vol)/(R*T) W=c_p*Vol/R*(k*P_f*(1-(P_i/P_f)^(1/k))-(P_f-P_i)) Q=(m_f-m_i)*T*(c_p+c_v)

8-89 Consider a cryogenic energy storage (CES) system in which nitrogen is liquefied during off-peak hours using surplus electricity generated by wind turbines and stored in a 500  m 3 cryogenic tank at T1  200 C and

P1  0.12 MPa . During peak times, the entire liquid nitrogen is pumped to P2  15 MPa and is heated by ambient air to T3  20 C at that pressure before it is routed to a nitrogen turbine where it expands to a pressure of

P4  100 MPa

1   825.9 kg / m3 , h1  440.1 kJ / kg , h4 s  245.5 kJ / kg (enthalpy at turbine exit for the case of isentropic expansion). The

properties

nitrogen

are

h3  33.36 kJ / kg ,

(a) Determine the work consumed by the pump and the work produced by the turbine if the isentropic efficiency of the pump is 73 percent and that of the turbine is 82 percent. (b) What fraction of work consumed during the liquefaction process is recovered when the stored liquid nitrogen is used to generate work? Assume that it takes 0.8 kWh of electricity to liquefy a unit mass of nitrogen. 8-89 A cryogenic energy storage (CES) system is considered. The amounts of work consumed by the pump and produced by the turbine are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. The losses in the pump are negligible. 3. The devices are adiabatic.

Properties (a) From the problem statement:

h1  440.1 kJ / kg , 1   825.9 kg / m3 , h3  33.36 kJ / kg , h4 s  245.5 kJ / kg Analysis Using the density value at pump inlet, the mass of nitrogen in the storage tank is determined from

m  V  (825.9 kg/m3 )(500 m3 )  412,962 kg Neglecting elevation and velocity changes across the pump, the work input to the pump per unit mass for the case of isentropic compression is determined as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-334

P2  P1

wpump,in,isen 





(15,000  120) kPa  18.02 kJ/kg 825.9 kg/m3

The actual work input to the pump is

wpump,in 

wpump,in,isen

pump



18.02 kJ/kg  24.68 kJ/kg 0.73

(since Q  0, ke  0, pe  0)

Therefore, the enthalpy of nitrogen at the exit of the pump is

h2  h1  wpump,in  440.1  24.68  415.4 kJ/kg The total work input to the pump is

 1 kWh  Wpump,in  mwpump,in  (412,962 kg)(24.68 kJ/kg)    2831 kW h  3600 kJ  We take the turbine as the system, which is a control volume since mass crosses the boundary. Noting that one fluid stream enters and leaves the turbine, the energy balance for this control volume can be expressed per unit mass basis as

ein  eout h3  h4  wturb,out

(since Q  0, ke  0, pe  0)

Using the relation for the isentropic efficiency of the turbine,

 turb 

wturb,out wturb,out,isen



h3  h4 h3  h4s

Then, the work output from the turbine becomes

 1 kWh  Wturb,out  turb m(h3  h4 s )  (0.82)(412,962 kg)[(33.36)  (245.5)] kJ/kg    19,959 kW h  3600 kJ  (b) The net work output is

Wnet,out  Wturb,out  Wpump,in  19,959  2831  17,127 kWh The total amount of work consumed for the liquefaction of nitrogen is

Wliq,in  mwliq,in  (412,962 kg)(0.8 kWh/kg)  330,370 kWh The fraction of work consumed during the liquefaction process is

f 

Wnet,out Wliq,in



17,127 kWh  0.0518 330,370 kWh

That is, only about 5.2 percent of work consumed during liquefaction of nitrogen is recovered by using the cryogenic storage of liquid nitrogen.

EES (Engineering Equation Solver) SOLUTION T1=-200 [C] P1=120 [kPa] V=500 [m^3] P2=15000 [kPa] T3=20 [C] P3=P2 P4=100 [kPa] eta_pump=0.73 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-335

eta_turb=0.82 w_liq=0.8 [kWh/kg] rho=density(Nitrogen, P=P1, T=T1) m=rho*V h1=enthalpy(Nitrogen, P=P1, T=T1) w_pump=1/eta_pump*(P2-P1)/rho h2=h1+w_pump h3=enthalpy(Nitrogen, P=P3, T=T3) s3=entropy(Nitrogen, P=P3, T=T3) h4s=enthalpy(Nitrogen, P=P4, s=s3) w_out_permass=eta_turb*(h3-h4s)*convert(kJ,kWh) W_in=m*w_pump*convert(kJ,kWh) W_out=m*eta_turb*(h3-h4s)*convert(kJ,kWh) W_net_out=W_out-W_in W_in_liq=m*w_liq f=W_net_out/W_in_liq

8-90 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses heat to the surroundings. The final temperature in each tank and the entropy generated during this process are to be determined. Assumptions 1. Tank A is insulated, and thus heat transfer is negligible. 2. The water that remains in tank A undergoes a reversible adiabatic process. 3. The thermal energy stored in the tanks themselves is negligible. 4. The system is stationary and thus kinetic and potential energy changes are negligible. 5. There are no work interactions. Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s 2  s1 . From the steam tables (Tables A-4 through A-6), Tank A: v1, A = v f + x1v fg = 0.001084 + (0.6)(0.46242 - 0.001084) = 0.27788 m3 /kg P1 = 400 kPaïü ïý u = u + x u = 604.22 + (0.6)(1948.9) = 1773.6 kJ/kg f 1 fg ïïþ 1, A x1 = 0.6 s1, A = s f + x1 s fg = 1.7765 + (0.6)(5.1191) = 4.8479 kJ/kg ×K

T2, A = Tsat @ 200 kPa = 120.2°C s - sf 4.8479 - 1.5302 P2 = 200 kPa ïü ïï x2, A = 2, A = = 0.5928 s 5.59680 s2 = s1 ý fg ï (sat. mixture)ïïïþv 2, A = v f + x2, Av fg = 0.001061 + (0.5928)(0.8858 - 0.001061) = 0.52552 m 3 /kg u2, A = u f + x2, Au fg = 504.50 + (0.5928)(2024.6 kJ/kg ) = 1704.7 kJ/kg

6-336

v1, B = 1.1989 m3 /kg ü P1 = 200 kPa ïï ý u1, B = 2731.4 kJ/kg ïïþ T1 = 250°C s1, B = 7.7100 kJ/kg ×K The initial and the final masses in tank A are and m1, A =

VA 0.3 m3 = = 1.080 kg v1, A 0.27788 m3 /kg

m2, A =

VA 0.3 m3 = = 0.5709 kg v 2, A 0.52552 m3 /kg

Thus, 1.080  0.5709 = 0.5091 kg of mass flows into tank B. Then,

m2, B = m1, B + 0.5091 = 2 + 0.5091 = 2.509 kg The final specific volume of steam in tank B is determined from

(m1v1 )B (2 kg)(1.1989 m /kg ) VB = = = 0.9558 m3 /kg m2, B m2, B 2.509 kg 3

v 2, B =

We take the entire contents of both tanks as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = (D U ) A + (D U ) B

(since W = KE = PE = 0)

- Qout = (m2 u2 - m1u1 )A + (m2 u2 - m1u1 )B Substituting, - 300 = {(0.5709)(1704.7 )- (1.080)(1773.6 )}+ {(2.509 )u2, B - (2 )(2731.4 )} u2, B = 2433.3 kJ/kg

Thus,

v2, B = 0.9558 m3 /kg üïï T2, B = 116.1°C ý u2, B = 2433.3 kJ/kg ïïþ s2, B = 6.9156 kJ/kg ×K (b) The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D SA + D SB Tb,surr

Rearranging and substituting, the total entropy generated during this process is determined to be

Sgen = D S A + D S B +

Qout Q = (m2 s2 - m1s1 )A + (m2 s2 - m1s1 )B + out Tb, surr Tb,surr

= {(0.5709)(4.8479)- (1.080)(4.8479)}+ {(2.509)(6.9156)- (2)(7.7100)}+

300 kJ 290 K

= 0.498 kJ / K

6-337

EES (Engineering Equation Solver) SOLUTION "Given" Vol_A=0.3 [m^3] P1_A=400 [kPa] x1_A=0.60 m1_B=2 [kg] P1_B=200 [kPa] T1_B=250 [C] P2_A=200 [kPa] Q_out=300 [kJ] T_surr=17 [C] "Properties" Fluid$='steam_iapws' v1_A=volume(Fluid$, P=P1_A, x=x1_A) u1_A=intenergy(Fluid$, P=P1_A, x=x1_A) s1_A=entropy(Fluid$, P=P1_A, x=x1_A) v1_B=volume(Fluid$, P=P1_B, T=T1_B) u1_B=intenergy(Fluid$, P=P1_B, T=T1_B) s1_B=entropy(Fluid$, P=P1_B, T=T1_B) "Analysis" "(a)" s2_A=s1_A "reversible-adiabatic (i.e., isentropic) process" T2_A=temperature(Fluid$, P=P2_A, s=s2_A) u2_A=intenergy(Fluid$, P=P2_A, s=s2_A) v2_A=volume(Fluid$, P=P2_A, s=s2_A) m1_A=Vol_A/v1_A m2_A=Vol_A/v2_A m2_B=m1_B+(m1_A-m2_A) Vol_B=m1_B*v1_B v2_B=Vol_B/m2_B -Q_out=DELTAU_A+DELTAU_B "energy balance" DELTAU_A=m2_A*u2_A-m1_A*u1_A DELTAU_B=m2_B*u2_B-m1_B*u1_B T2_B=temperature(Fluid$, v=v2_B, u=u2_B) s2_B=entropy(Fluid$, v=v2_B, u=u2_B) "(b)" -Q_out/(T_surr+273)+S_gen=DELTAS_A+DELTAS_B "entropy balance" DELTAS_A=m2_A*s2_A-m1_A*s1_A DELTAS_B=m2_B*s2_B-m1_B*s1_B v_f=volume(Fluid$, P=P2_A, x=0) v_g=volume(Fluid$, P=P2_A, x=1) u_f=intenergy(Fluid$, P=P2_A, x=0) u_g=intenergy(Fluid$, P=P2_A, x=1) u_fg=u_g-u_f s_f=entropy(Fluid$, P=P2_A, x=0) s_g=entropy(Fluid$, P=P2_A, x=1) s_fg=s_g-s_f h_f=enthalpy(Fluid$, P=P2_A, x=0) h_g=enthalpy(Fluid$, P=P2_A, x=1) x_2A=(s2_A-s_f)/s_fg

8-91 An electric resistance heater is immersed in water. The time it will take for the electric heater to raise the water temperature to a specified temperature and the entropy generated during this process are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-338

1. 2. 3.

Water is an incompressible substance with constant specific heats. The energy stored in the container itself and the heater is negligible. Heat loss from the container is negligible.

Properties The specific heat of water at room temperature is c  4.18 kJ / kg  C (Table A-3). Analysis Taking the water in the container as the system, which is a closed system, the energy balance can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We, in = (D U ) water We, in D t = mc(T2 - T1 ) water Substituting,

(1200 J / s)t  (40 kg)(4180 J / kg  C)(50 20) C Solving for t gives t = 4180 s = 69.7 min = 1.16 h Again we take the water in the tank to be the system. Noting that no heat or mass crosses the boundaries of this system and the energy and entropy contents of the heater are negligible, the entropy balance for it can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0 + Sgen = D Swater Therefore, the entropy generated during this process is T 323 K Sgen = D S water = mc ln 2 = (40 kg )(4.18 kJ/kg ×K ) ln = 16.3 kJ / K T1 293 K

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_e_in=1.2 [kW] D=0.005 [m] m=40 [kg] T1=(20+273) [K] T2=(50+273) [K] "Properties" c_p=4.18 [kJ/kg-K] "Table A-3" "Analysis" W_dot_e_in*time=m*c_p*(T2-T1) "energy balance" DELTAS_water=m*c_p*ln(T2/T1) S_gen=DELTAS_water "entropy balance"

6-339

8-92E A steel container that is filled with hot water is allowed to cool to the ambient temperature. The total entropy generated during this process is to be determined. Assumptions 1. Both the water and the steel tank are incompressible substances with constant specific heats at room temperature. 2. The system is stationary and thus the kinetic and potential energy changes are zero. 3. Specific heat of iron can be used for steel. 4. There are no work interactions involved. Properties The specific heats of water and the iron at room temperature are c p , water  1.00 Btu / lbm. F and c p , iron  0.107 Btu / lbm. C . The density of water at room temperature is 62.1 lbm / ft 3 (Table A-3E).

Analysis The mass of the water is

mwater = r V = (62.1 lbm/ft 3 )(15 ft 3 ) = 931.5 lbm We take the steel container and the water in it as the system, which is a closed system. The energy balance on the system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = D U container + D U water = [mc(T2 - T1 )]container + [mc(T2 - T1 )]water Substituting, the heat loss to the surrounding air is determined to be

Qout = [mc(T1 - T2 )]container + [mc(T1 - T2 )]water

= (75 lbm)(0.107 Btu/lbm ×o F)(120 - 70)°F + (931.5 lbm)(1.00 Btu/lbm ×°F)(120 - 70)°F = 46,976 Btu

We again take the container and the water In it as the system. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the container and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surrounding air at all times. The entropy balance for the extended system can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Scontainer + D S water Tb, out

where D Scontainer = mcavg ln

T2 530 R = (75 lbm)(0.107 Btu/lbm ×R )ln = - 0.72 Btu/R T1 580 R

T 530 R D S water = mcavg ln 2 = (931.5 lbm)(1.00 Btu/lbm ×R )ln = - 83.98 Btu/R T1 580 R

Therefore, the total entropy generated during this process is

6-340

Sgen = D Scontainer + D Swater +

Qout 46,976 Btu = - 0.72 - 83.98 + = 3.93 Btu / R Tb, out 70 + 460 R

EES (Engineering Equation Solver) SOLUTION "Given" V_w=15 [ft^3] m_container=75 [lbm] T1=(120+460) [R] T2=(70+460) [R] "Properties" c_container=c_('Iron', T_ave) rho_w=density(water, T=T_ave, P=P) c_w=CP(water, T=T_ave, P=P) T_ave=1/2*(T1+T2) P=14.7 [psia] "assumed" "Analysis" m_w=rho_w*V_w -Q_out=m_w*c_w*(T2-T1)+m_container*c_container*(T2-T1) "energy balance" -Q_out/T2+S_gen=DELTAS_container+DELTAS_w "entropy balance" DELTAS_container=m_container*C_container*ln(T2/T1) DELTAS_w=m_w*C_w*ln(T2/T1)

8-93 A 1- ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibrium temperature in the tank and the entropy generation are to be determined. Assumptions 1. Thermal properties of the ice and water are constant. 2. Heat transfer to the water tank is negligible. 3. There is no stirring by hand or a mechanical device (it will add energy). Properties The specific heat of water at room temperature is c  4.18 kJ / kg  C , and the specific heat of ice at about 0C is

c  2.11 kJ / kg  C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and 333.7 kJ / kg . Analysis (a) We take the ice and the water as the system, and disregard any heat transfer between the system and the surroundings. Then the energy balance for this process can be written as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = DU

6-341

[mc(0°C - T1 )solid + mhif + mc(T2 - 0°C)liquid ]ice + [mc(T2 - T1 )]water = 0 Substituting,

(140 kg){(2.11 kJ/kg ×°C)[0 - (-5)]°C + 333.7 kJ/kg + (4.18 kJ/kg ×°C)(T2 - 0)°C} + (1000 kg)(4.18 kJ/kg ×°C)(T2 - 20)°C = 0 It gives

T2  7.4 C which is the final equilibrium temperature in the tank. (b) We take the ice and the water as our system, which is a closed system. Considering that the tank is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0 + Sgen = D Sice + D Swater where

æ T ö 280.4 K ÷ D S water = çççmc ln 2 ÷ = - 183.3 kJ/K ÷ = (1000 kg )(4.18 kJ/kg ×K ) ln çè T1 ÷ 293 K øwater D Sice = (D Ssolid + D Smelting + D Sliquid )

ice

ææ æ ö ö Tmelting ö mhif ÷ ç ÷ ççmc ln T2 ÷ ÷ ÷ ÷ = çççççmc ln + + ÷ ÷ ÷ ÷ ç ççèç ÷ ÷ ç T T T è ø ø ÷ 1 melting 1 è liquid ø solid ice

æ 273 K 333.7 kJ/kg 280.4 K ö ÷ = (140 kg )çç(2.11 kJ/kg ×K )ln + + (4.18 kJ/kg ×K ) ln ÷ ÷ çè 268 K 273 K 273 K ø = 192.3 kJ/K

Then,

Sgen = D Swater + D Sice = - 183.3 + 192.3 = 9.01 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" m_w=1000 [kg] T1_w=(20+273) [K] m_ice=140 [kg] T1_ice=(-5+273) [K] T_melting=(0+273) [K] h_if=333.7 [kJ/kg] "Properties" c_w=4.18 [kJ/kg-C] c_ice=2.11 [kJ/kg-C] "Analysis" "(a)" 0=m_ice*c_ice*(T_melting-T1_ice)+m_ice*h_if+m_ice*c_w*(T2-T_melting)+m_w*c_w*(T2-T1_w) "energy balance on ice and water" T2_C=T2-273 "(b)" S_gen=DELTAS_ice+DELTAS_w DELTAS_ice=m_ice*c_ice*ln(T_melting/T1_ice)+(m_ice*h_if)/T_melting+m_ice*c_w*ln(T2/T_melting) DELTAS_w=m_w*c_w*ln(T2/T1_w)

6-342

8-94 A well-insulated room is heated by a steam radiator, and the warm air is distributed by a fan. The average temperature in the room after 30 min, the entropy changes of steam and air, and the entropy generated during this process are to be determined. Assumptions 1. Air is an ideal gas with constant specific heats at room temperature. 2. The kinetic and potential energy changes are negligible. 3. The air pressure in the room remains constant and thus the air expands as it is heated, and some warm air escapes. Properties The gas constant of air is R  0.287 kPa  m3 / kg.K (Table A-1). Also, c p  1.005 kJ / kg  K for air at room temperature (Table A-2). Analysis We first take the radiator as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = m(u2 - u1 )

(since W = KE = PE = 0)

Qout = m(u1 - u2 ) Using data from the steam tables (Tables A-4 through A-6), some properties are determined to be

v = 1.0805 m3 /kg P1 = 200 kPaïüï 1 ý u = 2654.6 kJ/kg T1 = 200°C ïïþ 1 s1 = 7.5081 kJ/kg.K v f = 0.001043, v g = 1.6941 m 3 /kg P2 = 100 kPa ïü ïý u = 417.40, u = 2088.2 kJ/kg f fg (v 2 = v1 ) ïïþ s f = 1.3028, s fg = 6.0562 kJ/kg.K

x2 =

v2 - v f v fg

1.0805 - 0.001043 = 0.6376 1.6941- 0.001043

u2 = u f + x2u fg = 417.40 + 0.6376´ 2088.2 = 1748.7 kJ/kg s2 = s f + x2 s fg = 1.3028 + 0.6376´ 6.0562 = 5.1642 kJ/kg.K m=

V1 0.015 m3 = = 0.01388 kg v1 1.0805 m3 /kg

Substituting,

Qout  (0.01388 kg)(2654.6 1748.7)kJ / kg  12.6 kJ The volume and the mass of the air in the room are V  4  4  5  80 m3 and

6-343

mair =

(100 kPa )(80 m3 ) P1V1 = = 98.5 kg RT1 (0.2870 kPa ×m3 /kg ×K) (283 K )

The amount of fan work done in 30 min is

Wfan, in = Wfan, in D t = (0.120 kJ/s)(30´ 60 s) = 216 kJ We now take the air in the room as the system. The energy balance for this closed system is expressed as

Ein - Eout = D Esystem Qin + Wfan, in - Wb,out = D U Qin + Wfan, in = D H @ mc p (T2 - T1 ) since the boundary work and U combine into H for a constant pressure expansion or compression process. Substituting, (12.6 kJ)  (216 kJ)  (98.5 kg)(1.005 kJ / kg

C) T2  10  C

which yields

T2  12.3o C (b) The entropy change of the steam is

D Ssteam = m (s2 - s1 ) = (0.01388 kg )(5.1642 - 7.5081) kJ/kg ×K = - 0.0325 kJ / K (c) Noting that air expands at constant pressure, the entropy change of the air in the room is D Sair = mc p ln

T2 P - mR ln 2 T1 P1

¿0

= (98.5 kg )(1.005 kJ/kg ×K ) ln

285.3 K = 0.8013 kJ / K 283 K

(d) We take the air in the room (including the steam radiator) as our system, which is a closed system. Noting that no heat or mass crosses the boundaries of this system, the entropy balance for it can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0 + Sgen = D Ssteam + D Sair Substituting, the entropy generated during this process is determined to be

Sgen = D Ssteam + D Sair = - 0.0325 + 0.8013 = 0.7688 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol_room=4*4*5 [m^3] T1_air=10 [C] P_air=100 [kPa] Vol_rad=0.015 [m^3] P1=200 [kPa] T1=200 [C] P2=100 [kPa] time=30*60 [s] W_dot_fan_in=0.120 [kW] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, T=T1) u1=intenergy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) v2=v1 "fixed mass, constant volume" u2=intenergy(Fluid$, P=P2, v=v2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-344

s2=entropy(Fluid$, P=P2, v=v2) x2=quality(Fluid$, P=P2, v=v2) C_p_air=1.005 [kJ/kg-C] R=0.287 [kJ/kg-K] "Analysis" "(a)" m_s=Vol_rad/v1 -Q_out_steam=m_s*(u2-u1) "energy balance on steam" m_air=(P_air*Vol_room)/(R*(T1_air+273)) W_fan_in=W_dot_fan_in*time Q_in_air=Q_out_steam Q_in_air+W_fan_in=m_air*C_p_air*(T2_air-T1_air) "energy balance on room air, DELTAU+W_b = DELTAH for a constant pressure process" "(b)" DELTAS_steam=m_s*(s2-s1) "(c)" DELTAS_air=m_air*C_p_air*ln((T2_air+273)/(T1_air+273)) "(d)" DELTAS_total=DELTAS_steam+DELTAS_air

8-95 The heating of a passive solar house at night is to be assisted by solar heated water. The length of time that the electric heating system would run that night and the amount of entropy generated that night are to be determined. Assumptions 1. Water is an incompressible substance with constant specific heats. 2. The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3. The house is maintained at 22°C at all times. Properties The density and specific heat of water at room temperature are ρ  997 kg / m3 and c  4.18 kJ / kg  C (Table A-3). Analysis The total mass of water is

mw = r V = (0.997 kg/L)(50´ 20 L) = 997 kg Taking the contents of the house, including the water as our system, the energy balance relation can be written as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We, in - Qout = D U = (D U ) water + (D U )air = (D U ) water = mc(T2 - T1 )water or, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-345

We, in D t - Qout = [mc(T2 - T1 )]water Substituting,

(15 kJ / s)t

(50,000 kJ / h)(10 h)  (997 kg)(4.18 kJ / kg  C)(22 80) C

It gives t = 17,219 s = 4.78 h We take the house as the system, which is a closed system. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the house and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for the extended system can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Swater + D Sair ¿ 0 = D Swater Tb,out

since the state of air in the house remains unchanged. Then the entropy generated during the 10-h period that night is Sgen = D S water +

æ ö Qout Q T ÷ = çççmc ln 2 ÷ + out ÷ ÷ ç Tb, out è T1 øwater Tsurr

= (997 kg)(4.18 kJ/kg ×K )ln

295 K 500,000 kJ + 353 K 276 K

= - 748 + 1811 = 1063 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_out=50000[kJ/h]*Convert(kJ/h, kW) T_surr=(3+273) [K] T1=(80+273) [K] T2=(22+273) [K] time_heatloss=10*3600 [s] V=50*0.020 [m^3] W_dot_e_in=15 [kW] "Properties" rho=997 [kg/m^3] "Table A-3" c_p=4.18 [kJ/kg-K] "Table A-3" "Analysis" m=rho*V W_dot_e_in*time-Q_dot_out*time_heatloss=m*c_p*(T2-T1) "energy balance on house" time_h=time*1/Convert(h, s) -Q_out/T_surr+S_gen=DELTAS_water "entropy balance" Q_out=Q_dot_out*time_heatloss DELTAS_water=m*c_p*ln(T2/T1)

8-96 An insulated cylinder initially contains a saturated liquid-vapor mixture of water at a specified temperature. The entire vapor in the cylinder is to be condensed isothermally by adding ice inside the cylinder. The amount of ice added and the entropy generation are to be determined. Assumptions 1. Thermal properties of the ice are constant. 2. The cylinder is well-insulated and thus heat transfer is negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-346

There is no stirring by hand or a mechanical device (it will add energy).

Properties The specific heat of ice at about 0C is c  2.11 kJ / kg  C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and 333.7 kJ / kg . Analysis (a) We take the contents of the cylinder (ice and saturated water) as our system, which is a closed system. Noting that the temperature and thus the pressure remains constant during this phase change process and thus Wb  ΔU  ΔH , the energy balance for this system can be written as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wb, in = D U DH = 0

D Hice + D H water = 0 or

[mc(0°C - T1 )solid + mhif + mc(T2 - 0°C)liquid ]ice + [m(h2 - h1 )]water = 0 -4) 3

v f = 0.001043, v g = 1.6720 m /kg

h f = 419.17, h fg = 2256.4 kJ.kg s f = 1.3072

s fg = 6.0490 kJ/kg.K

v1 = v f + x1v fg = 0.001043 + (0.1)(1.6720 - 0.001043) = 0.16814 m3 /kg h1 = h f + x1h fg = 419.17 + (0.1)(2256.4) = 644.81 kJ/kg

s1 = s f + x1 s fg = 1.3072 + (0.1)(6.0470) = 1.9119 kJ/kg ×K

h2 = h f @100 o C = 419.17 kJ/kg s2 = s f @100 o C = 1.3072 kJ/kg ×K msteam =

V1 0.02 m3 = = 0.119 kg v1 0.16814 m3 /kg

Noting that T1, ice  18 C and T2  100 C and substituting gives

m{(2.11 kJ / kg.K)[0 ( 18)]  333.7 kJ / kg  (4.18 kJ / kg  C)(100  0) C}  (0.119 kg)(419.17 644.81) kJ / kg  0 m = 0.034 kg = 34.0 g ice (b) We take the ice and the steam as our system, which is a closed system. Considering that the tank is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

6-347

0 + Sgen = D Sice + D Ssteam D Ssteam = m (s2 - s1 ) = (0.119 kg )(1.3072 - 1.9119) kJ/kg ×K = - 0.0719 kJ/K ææ ö ö æ ö ÷ Tmelting ÷ mhif ç ççmc ln T2 ÷ ÷ ÷ ÷ D Sice = (D Ssolid + D Smelting + D Sliquid ) = çççççmc ln + + ÷ ÷ ÷ ÷ ice ççèç T1 ÷ T1 ÷ øliquid ÷ øsolid Tmelting çèç è øice

æ ö 273.15 K 333.7 kJ/kg 373.15 K ÷ ÷ = (0.034 kg )ççç(2.11 kJ/kg ×K )ln + + (4.18 kJ/kg ×K )ln ÷ 255.15 K 273.15 K 273.15 K ÷ è ø

= 0.0907 kJ/K Then,

Sgen = D Ssteam + D Sice = - 0.0719 + 0.0907 = 0.0188 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.02 [m^3] x1=0.1 T1=(100+273.15) [K] T_ice=(-18+273.15) [K] x2=0 T2=(100+273.15) [K] T_melting=(0+273.15) [K] h_if=333.7 [kJ/kg] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=x1) s1=entropy(Fluid$, T=T1, x=x1) v1=volume(Fluid$, T=T1, x=x1) h2=enthalpy(Fluid$, T=T2, x=x2) s2=entropy(Fluid$, T=T2, x=x2) c_w=4.18 [kJ/kg-C] "Table A-3" c_ice=2.11 [kJ/kg-C] "Table A-3" "Analysis" "(a)" m_w=Vol1/v1 0=m_ice*c_ice*(T_melting-T_ice)+m_ice*h_if+m_ice*c_w*(T2-T_melting)+m_w*(h2-h1) "energy balance on ice and water" "(b)" S_gen=DELTAS_ice+DELTAS_w "entropy balance" DELTAS_ice=m_ice*c_ice*ln(T_melting/T_ice)+(m_ice*h_if)/T_melting+m_ice*c_w*ln(T2/T_melting) DELTAS_w=m_w*(s2-s1)

8-97 Water is heated from 16°C to 43°C by an electric resistance heater placed in the water pipe as it flows through a showerhead steadily at a rate of 10 L / min . The electric power input to the heater and the rate of entropy generation are to be determined. The reduction in power input and entropy generation as a result of installing a 50% efficient regenerator are also to be determined. Assumptions 1. This is a steady-flow process since there is no change with time at any point within the system and thus D mCV = 0 and D ECV = 0 . 2.

Water is an incompressible substance with constant specific heats.

6-348

The kinetic and potential energy changes are negligible, D ke @D pe @ 0 .

Heat losses from the pipe are negligible.

Properties The density of water is given to be ρ  1 kg / L . The specific heat of water at room temperature is

c  4.18 kJ / kg  C (Table A-3). Analysis (a) We take the pipe as the system. This is a control volume since mass crosses the system boundary during the process. We observe that there is only one inlet and one exit and thus m1 = m2 = m. Then the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0 ® Ein = Eout

Rate of change in internal, kinetic, potential, etc. energies

We, in + mh1 = mh2 (since D ke @D pe @ 0) We, in = m(h2 - h1 ) = mc(T2 - T1 )

m = rV = (1 kg/L)(10 L/min ) = 10 kg/min

where

Substituting, We, in = (10/60 kg/s)(4.18 kJ/kg ×°C)(43 - 16)°C = 18.8 kW The rate of entropy generation in the heating section during this process is determined by applying the entropy balance on the heating section. Noting that this is a steady-flow process and heat transfer from the heating section is negligible,

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

ms1 - ms2 + Sgen = 0 ¾ ¾ ® Sgen = m(s2 - s1 ) Noting that water is an incompressible substance and substituting, T 316 K Sgen = mc ln 2 = (10/60 kg/s )(4.18 kJ/kg ×K )ln = 0.0622 kJ / K T1 289 K

(b) The energy recovered by the heat exchanger is

Qsaved = eQmax = emC (Tmax - Tmin ) = 0.5(10/60 kg/s)(4.18 kJ/kg ×°C)(39 - 16)°C = 8.0 kJ/s = 8.0 kW Therefore, 8.0 kW less energy is needed in this case, and the required electric power in this case reduces to

Win, new = Win, old - Qsaved = 18.8 - 8.0 = 10.8 kW Taking the cold water stream in the heat exchanger as our control volume (a steady-flow system), the temperature at which the cold water leaves the heat exchanger and enters the electric resistance heating section is determined from

Q = mc (Tc, out - Tc, in ) Substituting,

8 kJ/s = (10/60 kg/s)(4.18 kJ/kg ×o C)(Tc, out - 16 o C) It yields Tc, out = 27.5 C = 300.5 K The rate of entropy generation in the heating section in this case is determined similarly to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-349

T 316 K Sgen = mc ln 2 = (10/60 kg/s)(4.18 kJ/kg ×K ) ln = 0.0350 kJ/K T1 300.5 K

Thus the reduction in the rate of entropy generation within the heating section is

Sreduction = 0.0622 - 0.0350 = 0.0272 kW / K

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=10 [L/min] T1=16 [C] T2=43 [C] T_DrainedWater=39 [C] epsilon=0.50 rho=1 [kg/L] "Properties" c_p=4.18 [kJ/kg-K] "Table A-3" "Analysis" "(a)" m_dot=rho*V_dot*1/Convert(min, s) W_dot_e_in_a=m_dot*c_p*(T2-T1) S_dot_gen_a=m_dot*c_p*ln((T2+273)/(T1+273)) "(b)" Q_dot_max=m_dot*c_p*(T_DrainedWater-T1) Q_dot_saved=epsilon*Q_dot_max W_dot_in_b=W_dot_e_in_a-Q_dot_saved Q_dot_saved=m_dot*c_p*(T_c_out-T1) S_dot_gen_b=m_dot*c_p*ln((T2+273)/(T_c_out+273)) S_dot_reduction=S_dot_gen_a-S_dot_gen_b

8-98 Using appropriate software, the work input to a multistage compressor is to be determined for a given set of inlet and exit pressures for any number of stages. The pressure ratio across each stage is assumed to be identical and the compression process to be polytropic. The compressor work is to be tabulated and plotted against the number of stages for P1  100 kPa , T1  25 C , P2  1000 kPa , and n = 1.35 for air. Analysis The problem is solved using EES, and the results are tabulated and plotted below. GAS$ = 'Air' Nstage = 2 "number of stages of compression with intercooling, each having same pressure ratio" n=1.35 MM=MOLARMASS(GAS$) R_u = 8.314 [kJ/kmol-K] R=R_u/MM k=1.4 P1=100 [kPa] T1=25 [C] P2=1000 [kPa] R_p = (P2/P1)^(1/Nstage) w_comp= Nstage*n*R*(T1+273)/(n-1)*((R_p)^((n-1)/n) - 1) Nstage

Wcomp [kJ / kg]

6-350

1 2 3 4 5 6 7 8 9 10

269.4 229.5 217.9 212.4 209.2 207.1 205.6 204.5 203.6 202.9

8-99 The inner and outer glasses of a double pane window are at specified temperatures. The rates of entropy transfer through both sides of the window and the rate of entropy generation within the window are to be determined. Assumptions Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. Analysis The entropy flows associated with heat transfer through the left and right glasses are

Sleft =

Qleft 110 W = = 0.378 W / K Tleft 291 K

6-351

Sright =

Qright Tright

110 W = 0.394 W / K 279 K

We take the double pane window as the system, which is a closed system. In steady operation, the rate form of the entropy balance for this system can be expressed as

Sin - Sout

Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

Q Qin - out + Sgen, system = 0 Tb, in Tb, out 110 W 110 W + Sgen, system = 0 ® Sgen, system = 0.016 W / K 291 K 279 K

EES (Engineering Equation Solver) SOLUTION "Given" A=2*2 [m^2] T1=(18+273) [K] T2=(6+273) [K] Q_dot=110 [W] "Analysis" S_dot_left=Q_dot/T1 S_dot_right=Q_dot/T2 Q_dot/T1-Q_dot/T2+S_dot_gen=0 "entropy balance"

8-100 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate. The rate of entropy generation in the surrounding air due to this heat transfer are to be determined. Assumptions Steady operating conditions exist. Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closed system. The system extends from the outer surface of the pipe to a distance at which the temperature drops to the surroundings temperature. In steady operation, the rate form of the entropy balance for this system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem

= 0

Rate of change of entropy

Q Qin - out + Sgen, system = 0 Tb, in Tb, out 1600 W 1600 W + Sgen, system = 0 ® Sgen, system = 1.22 W / K 353 K 278 K

6-352

T1=(80+273) [K] T2=(5+273) [K] Q_dot=1600 [W] "Analysis" Q_dot_in/T1-Q_dot_out/T2+S_dot_gen=0 "entropy balance" Q_dot_in=Q_dot Q_dot_out=Q_dot

8-101 A cryogenic turbine in a natural gas liquefaction plant produces 115 kW of power. The efficiency of the turbine is to be determined. Assumptions 1. The turbine operates steadily. Properties The density of natural gas is given to be 423.8 kg / m3 .

Analysis The maximum possible power that can be obtained from this turbine for the given inlet and exit pressures can be determined from Wmax =

m 20 kg/s ( Pin - Pout ) = (3000 - 300)kPa = 127.4 kW r 423.8 kg/m3

Given the actual power, the efficiency of this cryogenic turbine becomes

W 115 kW = = 0.903 = 90.3% Wmax 127.4 kW

This efficiency is also known as hydraulic efficiency since the cryogenic turbine handles natural gas in liquid state as the hydraulic turbine handles liquid water.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=3000 [kPa] T_1=-160 [C] m_dot=20 [kg/s] P_2=300 [kPa] W_dot=115 [kW] rho=423.8 [kg/m^3] "ANALYSIS" Vol_dot=m_dot/rho W_dot_max=Vol_dot*(P_1-P_2) eta=W_dot/W_dot_max

6-353

8-102 The turbocharger of an internal combustion engine consisting of a turbine driven by hot exhaust gases and a compressor driven by the turbine is considered. The air temperature at the compressor exit and the isentropic efficiency of the compressor are to be determined.

Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Exhaust gases have air properties and air is an ideal gas with constant specific heats. Properties The specific heat of exhaust gases at the average temperature of 425C is c p  1.075 kJ / kg  K and properties of air at an anticipated average temperature of 100C are c p  1.011 kJ / kg  K and k =1.397 (Table A-2). Analysis (a) The turbine power output is determined from WT = mexh c p (T1 - T2 )

= (0.02 kg/s)(1.075 kJ/kg.°C)(450-400)°C = 1.075 kW For a mechanical efficiency of 95% between the turbine and the compressor,

WC = hmWT = (0.95)(1.075 kW) = 1.021 kW Then, the air temperature at the compressor exit becomes

WC = mair c p (T2 - T1 ) 1.021 kW = (0.018 kg/s)(1.011 kJ/kg.°C)(T2 -70)°C T2 = 126.1°C @126°C (b) The air temperature at the compressor exit for the case of isentropic process is ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P1 ø÷

(1.397- 1)/1.397

æ135 kPa ö÷ = (70 + 273 K) çç ÷ èç 95 kPa ø÷

= 379 K = 106°C

The isentropic efficiency of the compressor is determined to be hC =

T2 s - T1 106 - 70 = = 0.642 or 64.2 percent T2 - T1 126.1- 70

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_exh_in=450 [C] T_exh_out=400 [C] m_dot_exh=0.02 [kg/s] T_air_in=70 [C] P_air_in=95 [kPa] P_air_out=135 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-354

m_dot_air=0.018 [kg/s] eta_mec=0.95 "PROPERTIES" c_p_exh=1.075 [kJ/kg-K] "at (450+400)/2=425 C, Table A-2b" c_p_air=1.011 [kJ/kg-K] "at 100 C, Table A-2b" k=1.397 [kJ/kg-K] "at 100 C, Table A-2b" "ANALYSIS" W_dot_T=m_dot_exh*c_p_exh*(T_exh_in-T_exh_out) "turbine power output" W_dot_C=eta_mec*W_dot_T "compressor power input" W_dot_C=m_dot_air*c_p_air*(T_air_out-T_air_in) (T_2s_air+273)=(T_air_in+273)*(P_air_out/P_air_in)^((k-1)/k) eta_s_Comp=(T_2s_air-T_air_in)/(T_air_out-T_air_in) "compressor isentropic efficiency"

8-103 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established and the amount of entropy generated are to be determined. Assumptions 1. This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2. Air is an ideal gas. 3. Kinetic and potential energies are negligible. 4. There are no work interactions involved. 5. The direction of heat transfer is to the air in the bottle (will be verified). Properties The gas constant of air is 0.287 kPa  m3 / kg.K (Table A-1).

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 (since W @ Eout = Einitial = ke @ pe @ 0) Combining the two balances:

Qin = m2 (u2 - hi ) where m2 =

(95 kPa )(0.050 m3 ) P2V = = 0.05517 kg RT2 (0.287 kPa ×m3 /kg ×K )(300 K )

6-355 A-17 Ti = T2 = 300 K ¾ Table ¾¾ ¾®

hi = 300.19 kJ/kg u2 = 214.07 kJ/kg

Substituting,

Qin  (0.05517 kg)(214.07  300.19) kJ / kg  4.751 kJ  Qout  4.751 kJ Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reverse the direction. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the bottle and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

mi si -

Entropy generation

Change in entropy

Qout + Sgen = D S tank = m2 s2 - m1 s1¿ 0 = m2 s2 Tsurr

Therefore, the entropy generated during this process is Sgen = - mi si + m2 s2 +

Qout Q Q 4.751 kJ 0 = m2 (s2 - si ) + out = out = = 0.0158 kJ / K Tsurr Tsurr Tsurr 300 K

EES (Engineering Equation Solver) SOLUTION "Given" V=0.050 [m^3] P_i=95 [kPa] T_i=27 [C] T2=T_i P2=P_i "Properties" Fluid$='air' h_i=enthalpy(Fluid$, T=T_i) s_i=entropy(Fluid$, T=T_i, P=P_i) u2=intenergy(Fluid$, T=T2) s2=entropy(Fluid$, T=T2, P=P2) R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" m_i=m2 "mass balance" m2=(P2*V)/(R*(T2+273)) Q_in+m_i*h_i=m2*u2 "energy balance" m_i*s_i+Q_in/(T_i+273)+S_gen=m2*s2 "entropy balance"

8-104 Air is allowed to enter an insulated piston-cylinder device until the volume of the air increases by 50%. The final temperature in the cylinder, the amount of mass that has entered, the work done, and the entropy generation are to be determined. Assumptions 1. Kinetic and potential energy changes are negligible. 2. Air is an ideal gas with constant specific heats.

6-356

Properties The gas constant of air is R  0.287 kJ / kg  K and the specific heats of air at room temperature are

c p  1.005 kJ / kg  K , cv  0.718 kJ / kg  K (Table A-2). Analysis The initial pressure in the cylinder is

P1 =

m1 RT1 (1.3 kg)(0.287 kPa ×m3 /kg ×K)(30 + 273 K) = = 282.6 kPa V1 0.40 m3

m2 =

P2V2 (282.6 kPa)(1.5´ 0.40 m3 ) 590.8 = = RT2 T2 (0.287 kPa ×m3 /kg ×K)T2

A mass balance on the system gives the expression for the mass entering the cylinder mi = m2 - m1 =

590.8 - 1.3 T2

Wb, out = P1 (V2 - V1 ) = (282.6 kPa)(1.5´ 0.40 - 0.40)m 3 = 56.52 kJ (a) An energy balance on the system may be used to determine the final temperature

mi hi - Wb, out = m2u2 - m1u1 mi c pTi - Wb, out = m2 cv T2 - m1cv T1 æ590.8 ö÷ æ590.8 ö÷ çç ÷(1.005)(70 + 273) - 56.52 = çç ÷ ÷(0.718)T2 - (1.3)(0.718)(30 + 273) çèç T - 1.3ø÷ ÷ èçç T ø÷ 2

There is only one unknown, which is the final temperature. By a trial-error approach or using EES, we find

T2  315.3 K (b) The final mass and the amount of mass that has entered are

m2 =

590.8 = 1.874 kg 315.3

mi = m2 - m1 = 1.874 - 1.3 = 0.5742 kg (d) The rate of entropy generation is determined from

Sgen = m2 s2 - m1s1 - mi si = m2 s2 - m1s1 - (m2 - m1 )si = m2 (s2 - si ) - m1 (s1 - si ) æ æ T Pö T Pö ÷ ÷ = m2 çççc p ln 2 - R ln 2 ÷ - m1 çççc p ln 1 - R ln 1 ÷ ÷ ÷ Ti Pi ø÷ Ti Pi ø÷ èç èç é æ315.3 K ÷ ö æ282.6 kPa ÷ öù ú = (1.874 kg) ê(1.005 kJ/kg.K)ln çç - (0.287 kJ/kg.K)ln çç ÷ ÷ ÷ çè 343 K ÷ ç êë ø è 500 kPa øú û é ù æ303 K ö ö ÷- (0.287 kJ/kg.K)ln æ ÷ú çç 282.6 kPa ÷ - (1.3 kg) ê(1.005 kJ/kg.K)ln çç ÷ ÷ ÷ êë èç343 K ø èç 500 kPa øú û = 0.09714 kJ / K

6-357

EES (Engineering Equation Solver) SOLUTION "GIVEN" V_1=0.40 [m^3] m_1=1.3 [kg] T_1=(30+273) [K] P_i=500 [kPa] T_i=(70+273) [K] V_2=1.5*V_1 "PROPERTIES" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_v=0.718 [kJ/kg-K] "Table A-2a" "ANALYSIS" P_1=(m_1*R*T_1)/V_1 P_2=P_1 m_2=(P_2*V_2)/(R*T_2) m_i=m_2-m_1 W_b_out=P_1*(V_2-V_1) m_i*C_p*T_i-W_b_out=m_2*c_v*T_2-m_1*c_v*T_1 "energy balance" S_gen=m_2*(c_p*ln(T_2/T_i)-R*ln(P_2/P_i))-m_1*(c_p*ln(T_1/T_i)-R*ln(P_1/P_i)) "entropy generation"

8-105E A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and is charged until the tank contains saturated liquid at a specified pressure. The mass of R-134a that entered the tank, the heat transfer with the surroundings at 100F, and the entropy generated during this process are to be determined. Assumptions 1. This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2. Kinetic and potential energies are negligible. 3. There are no work interactions involved. 4. The direction of heat transfer is to the tank (will be verified). Properties The properties of R-134a are (Tables A-11 through A-13)

v1 = v g @ 60 psia = 0.79462 ft 3 /lbm P1 = 60 psiaïü ïý u = u 1 g @ 60 psia = 101.31 Btu/lbm sat. vapor ïïþ s1 = sg @ 60 psia = 0.22132 Btu/lbm ×R

P2 = 100 psiaü ïï ý ïïþ sat. liquid

v 2 = v f @ 100 psia = 0.01332 ft 3 /lbm u2 = u f @ 100 psia = 37.62 Btu/lbm s2 = s f @ 100 psia = 0.07879 Btu/lbm ×R

6-358

Pi = 140 psiaüïï hi @ h f @80° F = 38.17 Btu/lbm ý Ti = 80°F ïïþ si @ s f @80° F = 0.07934 Btu/lbm ×R Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as

min - mout = D msystem ® mi = m2 - m1

Mass balance: Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 - m1u1 (since W @ ke @ pe @0) The initial and the final masses in the tank are

m1 =

V 5 ft 3 = = 6.292 lbm v1 0.79462 ft 3 /lbm

m2 =

V 5 ft 3 = = 375.30 lbm v 2 0.01332 ft 3 /lbm

Then from the mass balance,

mi = m2 - m1 = 375.30 - 6.292 = 369.0 lbm (b) The heat transfer during this process is determined from the energy balance to be

Qin = - mi hi + m2u2 - m1u1 = - (369.0 lbm)(38.17 Btu/lbm)+ (375.30 lbm)(37.62 Btu/lbm)- (6.292 lbm)(101.31 Btu/lbm) = - 602.5 Btu

The negative sign indicates that heat is lost from the tank in the amount of 602.5 Btu. (c) The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be expressed as

Sin - Sout + Sgen = D Ssystem ¾ ¾ ® Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin + mi si + Sgen = D S tank = m2 s2 - m1s1 Tb, in

Therefore, the total entropy generated during this process is

Sgen = - mi si + (m2 s2 - m1s1 ) -

Qin Tb, in

= - (369.0)(0.07934)+ (375.30)(0.07879)- (6.292)(0.22132)-

( - 602.5 Btu) 530 R

= 0.0369 Btu / R

EES (Engineering Equation Solver) SOLUTION "Given" Vol=5 [ft^3] P1=60 [psia] x1=1 P_i=140 [psia] T_i=80 [F] P2=100 [psia] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-359

x2=0 T_surr=70 [F] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) h_i=enthalpy(Fluid$, T=T_i, x=0) s_i=entropy(Fluid$, T=T_i, x=0) "Analysis" "(a)" m_i=m2-m1 "mass balance" m1=Vol/v1 m2=Vol/v2 "(b)" Q_in+m_i*h_i=m2*u2-m1*u1 "energy balance" "(c)" Q_in/(T_surr+460)+m_i*s_i+S_gen=m2*s2-m1*s1 "entropy balance"

8-106 It is to be shown that the difference between the steady-flow and boundary works is the flow energy. Analysis The total differential of flow energy Pv can be expressed as d (Pv ) = P dv + v dP = d wb - d w flow = d (wb - w flow )

Therefore, the difference between the reversible steady-flow work and the reversible boundary work is the flow energy.

8-107 A three-stage compressor with two stages of intercooling is considered. The two intermediate pressures that will minimize the work input are to be determined in terms of the inlet and exit pressures. Analysis The work input to this three-stage compressor with intermediate pressures Px and Py and two intercoolers can be expressed as

wcomp = wcomp, I + wcomp, II + wcomp, III =

(n- 1)/ n nRT1 nRT1 nRT1 (n- 1)/ n (n- 1)/ n 1- (Px P1 ) + 1- (Py Px ) + 1- (Px P1 ) n- 1 n- 1 n- 1

(n- 1)/ n nRT1 (n- 1)/ n (n- 1)/ n 1- (Px P1 ) + 1- (Py Px ) + 1- (Px P1 ) n- 1

(n- 1)/ n nRT1 (n- 1)/ n (n- 1)/ n 3 - (Px P1 ) - (Py Px ) - (Px P1 ) n- 1

(

)

(

)

(

)

The Px and Py values that will minimize the work input are obtained by taking the partial differential of w with respect to

Px and Py , and setting them equal to zero: n- 1

¶w = 0 ¶ Px

¾¾ ®

öæPx ö÷ n n - 1æ çç 1 ÷ çç ÷ ÷ ÷ ç ÷çç P1 ø÷ ÷ n èç P1 øè

- 1

n - 1æ 1 öæ ÷çç Px ö÷ ÷ ççç ÷ + ÷ ÷ ÷ ÷çç Py ø÷ ÷ n èç Py øè

n- 1 - 1 n

= 0

6-360 n- 1

¶w = 0 ¶ Py

n öæPy ö n - 1æ çç 1 ÷ ÷çç ÷ ÷ ÷ ÷ ÷çç Px ø ÷ n ççè Px øè

¾¾ ®

- 1

öæPy ö n - 1æ çç 1 ÷ ÷ ÷çç ÷ + ÷ ÷ ç ÷çç P2 ø ÷ ç n è P2 øè

n- 1 - 1 n

= 0

Simplifying, 1

ö n Pö 1æ 1æ çç Px ÷ ÷ ÷ ççç x ÷ = ÷ ÷ ç ÷ ÷ ÷ P1 èç P1 ø Py çè Py ø -

n ö ö 1æ ÷ = 1æ ÷ çç Py ÷ çç Py ÷ ÷ ÷ ÷ ÷ Px ççè Px ø P2 ççè P2 ø

2 n- 1 n

1- 2 n

¾¾ ®

ö 1 æ Pö 1 æ ÷ çç P1 ÷ ÷ = n ççç x ÷ ÷ ÷ n ç ÷ ÷ Py èç Py ø ÷ P1 èç Px ø

¾¾ ®

Px (

= (P1 Py )

¾¾ ®

ö 1 æPy ö 1 æ çç Px ÷ ÷ ÷ = n ççç ÷ ÷ ÷ n ç ÷ ÷ Px çè Py ÷ ø P2 èç P2 ø

¾¾ ®

Py (

= (Px P2 )

2 1- n)

1- n

1- 2 n 2 1- n)

1- n

which yield 13

Px2 = P1 Px P2

¾¾ ®

Px = (P12 P2 )

Py2 = P2 P1 Py

¾¾ ®

Py = (P1 P22 )

8-108 It is to be shown that for an ideal gas with constant specific heats the compressor and turbine isentropic efficiencies (T4 / T3 ) - 1 ( P / P )( k - 1)/ k may be written as hC = 2 1 and hT = . ( k - 1)/ k ( P - 1 (T2 / T1 ) - 1 4 / P3 ) Analysis The isentropic efficiency of a compressor for an ideal gas with constant specific heats is given by hC =

h2 s - h1 c p (T2 s - T1 ) T2 s - T1 = = h2 - h1 c p (T2 - T1 ) T2 - T1

The temperature at the compressor exit fort he isentropic case is expressed as ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P ÷ ø 1

Substituting,

éæP ö( k - 1)/ k ù ( k - 1)/ k êç 2 ÷ ú æ P2 ö÷ ç ÷ T 1 ç ÷ 1ê ú ç ÷ - T1 - 1 ÷ êçè P1 ø÷ ú 1 û= èç P1 ø÷ = ë éæT ö ù æT2 ö÷ T2 - T1 - 1 ÷ ççç ÷ T1 êêçç 2 ÷ - 1ú ÷ ÷ ú ÷ ÷ ç T è T1 ø ú ëêè 1 ø û ( k - 1)/ k

hC =

T2 s - T1 = T2 - T1

æP ö ÷ T1 çç 2 ÷ ÷ çè P ø÷

The isentropic efficiency of a turbine for an ideal gas with constant specific heats is given by hT =

c p (T4 - T3 ) h4 - h3 T - T3 = = 4 h4 s - h3 c p (T4 s - T3 ) T4 s - T3

The temperature at the turbine exit fort he isentropic case is expressed as ( k - 1)/ k

æP ö ÷ T4 s = T3 ççç 4 ÷ ÷ çè P ÷ ø 3

Substituting,

éæT ö ù æT4 ÷ ö çç ÷- 1 ÷- 1ú T3 êêçç 4 ÷ ÷ ÷ ú ç ÷ ç ÷ T - T3 T4 - T3 èT3 ø êëèT3 ø ú û = hC = 4 = = ( k - 1)/ k ( k - 1)/ k ( k - 1)/ k é ù T4 s - T3 æP ö æP4 ö÷ æP ö ÷ T3 ççç 4 ÷ - T3 T3 êêçç 4 ÷ - 1 ÷ ççç ÷ - 1ú ÷ ú ÷ ç ÷ ÷ P3 ø÷ P è P3 ø÷ è è ø êë 3 ú û © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-361

8-109 A container is filled through a single opening from a source of working fluid with constant properties. The final specific entropy is to be compared to the initial specific entropy. Assumptions The state of the fluid entering the control volume does not change with time. Analysis The conservation of mass principle is dmcv = å m- å m dt in out

which reduces to dmcv dt while the general expression of the second law is m=

d (ms )cv dS¥ + + å ms - å ms ³ 0 dt dt out in

which reduces to dScv - sin m ³ 0 dt

Combining these two expressions produces

d (ms)cv dmcv - sin ³ 0 dt dt When this result is multiplied by dt and integrated, it becomes

m2 s2 - m1 s1 - sin (m2 - m1 ) ³ 0 s2 ³

m1 m - m1 s1 + sin 2 m2 m2

where the subscript 1 and 2 stand for initial and final states.

8-110 The polytropic efficiency of a turbine is defined. A relation regarding this turbine operation is to be obtained. Analysis The efficiency is defined as

h¥ ,T =

dh dhs

For an ideal gas,

dh = c p dT Form Gibbs second equation, dh = Tds + v dP

For the isentropic case,

dhs = v dP Substituting, we obtain h¥ ,T =

c p dT v dP

Then,

h¥ ,T v dP = c p dT RT dP = c p dT P Integrating between inlet (1) and exit (2) states, h¥ ,T

6-362 2

ò h¥ ,T 1

h¥ ,T

R dP dT = ò cp P T 1

T R P2 ln = ln 2 c p P1 T1

This becomes h

k- 1

¥ ,T ö ¥ ,T k T2 æ P ö cp æ çç P2 ÷ ÷ ÷ = ççç 2 ÷ = ÷ ÷ ÷ T1 èç P1 ø èçç P1 ø÷

which is the desired expression.

Fundamentals of Engineering (FE) Exam Problems 8-111 Heat is lost through a plane wall steadily at a rate of 1500 W. If the inner and outer surface temperatures of the wall are 20C and 5C, respectively, the rate of entropy generation within the wall is (a) 0.07 W / K (b) 0.15 W / K (c) 0.28 W / K (d) 1.42 W / K (e) 5.21 W / K Answer (c) 0.28 W / K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q=1500 [W] T1=(20+273) [K] T2=(5+273) [K] "Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives" Q/T1-Q/T2+S_gen=0 "Some Wrong Solutions with Common Mistakes:" Q/(T1+273)-Q/(T2+273)+W1_Sgen=0 "Using C instead of K" W2_Sgen=Q/((T1+T2)/2)"Using avegage temperature in K" W3_Sgen=Q/((T1+T2)/2-273)"Using avegage temperature in C" W4_Sgen=Q/(T1-T2+273) "Using temperature difference in K"

8-112 Air is compressed steadily and adiabatically from 17C and 90 kPa to 200C and 400 kPa. Assuming constant specific heats for air at room temperature, the isentropic efficiency of the compressor is (a) 0.76 (b) 0.94 (c) 0.86 (d) 0.84 (e) 1 Answer (d) 0.84 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-363

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=90 [kPa] T1=17 [C] P2=400 [kPa] T2=200 [C] cp=1.005 [kJ/kg-K] k=1.4 T2s=(T1+273)*(P2/P1)^((k-1)/k)-273 Eta_comp=(cp*(T2s-T1))/(cp*(T2-T1)) "Some Wrong Solutions with Common Mistakes:" T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K in finding T2s" W1_Eta_comp=(cp*(T2sW1-T1))/(cp*(T2-T1)) W2_Eta_comp=T2s/T2 "Using wrong definition for isentropic efficiency, and using C" W3_Eta_comp=(T2s+273)/(T2+273) "Using wrong definition for isentropic efficiency, with K"

8-113 Argon gas expands in an adiabatic turbine steadily from 600C and 800 kPa to 80 kPa at a rate of 2.5 kg / s . For an isentropic efficiency of 88%, the power produced by the turbine is (a) 240 kW (b) 361 kW (c) 414 kW (d) 602 kW (e) 777 kW Answer (d) 602 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=2.5 [kg/s] T1=600 [C] P1=800 [kPa] P2=80 [kPa] cp=0.5203 [kJ/kg-K] k=1.667 T2s=(T1+273)*(P2/P1)^((k-1)/k)-273 Eta_turb=0.88 Eta_turb=(cp*(T2-T1))/(cp*(T2s-T1)) W_out=m*cp*(T1-T2) "Some Wrong Solutions with Common Mistakes:" T2sW1=T1*(P2/P1)^((k-1)/k) "Using C instead of K to find T2s" Eta_turb=(cp*(T2W1-T1))/(cp*(T2sW1-T1)) W1_Wout=m*Cp*(T1-T2W1) Eta_turb=(cp*(T2s-T1))/(cp*(T2W2-T1)) "Using wrong definition for isentropic efficiency, and using C" W2_Wout=m*cp*(T1-T2W2) W3_Wout=cp*(T1-T2) "Not using mass flow rate" cv=0.3122 [kJ/kg-K] W4_Wout=m*cv*(T1-T2) "Using cv instead of cp"

8-114 Water enters a pump steadily at 100 kPa at a rate of 35 L / s and leaves at 800 kPa. The flow velocities at the inlet and the exit are the same, but the pump exit where the discharge pressure is measured is 6.1 m above the inlet section. The minimum power input to the pump is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-364

(a) 34 kW (b) 22 kW (c) 27 kW (d) 52 kW (e) 44 kW Answer (c) 27 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=0.035 [m^3/s] P1=100 [kPa] T1=20 [C] P2=800 [kPa] h=6.1 [m] g=9.81 [m/s^2] "Pump power input is minimum when compression is reversible and thus w=v(P2-P1)+Dpe" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) m=V/v1 W_min=m*v1*(P2-P1)+m*g*h/1000 "The effect of 6.1 m elevation difference turns out to be small" "Some Wrong Solutions with Common Mistakes:" W1_Win=m*v1*(P2-P1) "Disregarding potential energy" W2_Win=m*v1*(P2-P1)-m*g*h/1000 "Subtracting potential energy instead of adding" W3_Win=m*v1*(P2-P1)+m*g*h "Not using the conversion factor 1000 in PE term" W4_Win=m*v1*(P2+P1)+m*g*h/1000 "Adding pressures instead of subtracting"

8-115 Air is to be compressed steadily and isentropically from 1 atm to 16 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be (a) 3 atm (b) 4 atm (c) 8.5 atm (d) 9 atm (e) 12 atm Answer (b) 4 atm Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1 [atm] P2=16 [atm] P_mid=SQRT(P1*P2) "Some Wrong Solutions with Common Mistakes:" W1_P=(P1+P2)/2 "Using average pressure" W2_P=P1*P2/2 "Half of product"

8-116 Liquid water is to be compressed by a pump whose isentropic efficiency is 85 percent from 0.2 MPa to 5 MPa at a rate of 0.15 m3 / min . The required power input to this pump is (a) 8.5 kW (b) 10.2 kW © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-365

(c) 12.0 kW (d) 14.1 kW (e) 15.3 kW Answer (d) 14.1 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=200 [kPa] P2=5000 [kPa] V=(0.15/60) [m^3/s] Eta_pump=0.85 rho=1000 [kg/m^3] v1=1/rho m=rho*V "Reversible pump power input is w =mv(P2-P1) = V(P2-P1)" W_rev=m*v1*(P2-P1) W_pump=W_rev/Eta_pump "Some Wrong Solutions with Common Mistakes:" W1_Wpump=W_rev*Eta_pump "Multiplying by efficiency" W2_Wpump=W_rev "Disregarding efficiency" W3_Wpump=m*v1*(P2+P1)/Eta_pump "Adding pressures instead of subtracting"

8-117 Steam enters an adiabatic turbine at 8 MPa and 500C at a rate of 18 kg / s , and exits at 0.2 MPa and 300C. The rate of entropy generation in the turbine is (a) 0 kW / K (b) 7.2 kW / K (c) 21 kW / K (d) 15 kW / K (e) 17 kW / K Answer (c) 21 kW / K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=8000 [kPa] T1=500 [C] m=18 [kg/s] P2=200 [kPa] T2=300 [C] s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) s2=ENTROPY(Steam_IAPWS,T=T2,P=P2) S_gen=m*(s2-s1) "Some Wrong Solutions with Common Mistakes:" W1_Sgen=0 "Assuming isentropic expansion"

8-118 Helium gas is compressed steadily from 90 kPa and 25C to 800 kPa at a rate of 2 kg / min by an adiabatic compressor. If the compressor consumes 80 kW of power while operating, the isentropic efficiency of this compressor is (a) 54.0% © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-366

(b) 80.5% (c) 75.8% (d) 90.1% (e) 100% Answer (d) 90.1% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=2/60 [kg/s] T1=25 [C] P1=90 [kPa] P2=800 [kPa] W_comp=80 [kW] cp=5.1926 [kJ/kg-K] cv=3.1156 [kJ/kg-K] k=1.667 T2s=(T1+273)*(P2/P1)^((k-1)/k)-273 W_s=m*cp*(T2s-T1) Eta_comp=W_s/W_comp "Some Wrong Solutions with Common Mistakes:" T2sA=T1*(P2/P1)^((k-1)/k) "Using C instead of K" W1_Eta_comp=m*cp*(T2sA-T1)/W_comp W2_Eta_comp=m*cv*(T2s-T1)/W_comp "Using cv instead of cp"

8-119 … 8-123 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

6-367

Chapter 9 EXERGY

6-368

CYU - Check Your Understanding CYU 9-1 ■ Check Your Understanding CYU 9-1.1 Select the incorrect statement about exergy: (a) The exergy of large bodies of air and water such as the atmosphere and the oceans is practically zero. (b) Two systems with the same amounts of energy will also have the same amounts of exergy. (c) A system which is in equilibrium with its environment at the dead state has zero exergy. (d) Exergy of a system depends on the initial state of the system and the dead state. (e) The useful work potential of a given amount of energy at some specified state is its exergy contentf. Answer: (b) Two systems with the same amounts of energy will also have the same amounts of exergy. CYU 9-1.2 100 kJ of heat is lost from a hot body at 500 K to its surroundings at 300 K. What is the exergy loss of the system? (a) 40 kJ (b) 60 kJ (c) 80 kJ (d) 100 kJ (e) 0 Answer: (a) 40 kJ X heat = Q (1- T0 / Ts ) = (100 kJ)(1- 300 / 500) = 40 kJ CYU 9-1.3 What is the exergy of water at an elevation of 100 m? (a) 0 (b) 980 kJ / kg (c) 0.98 kJ / kg (d) 100 kJ / kg (e) 0.1 kJ / kg Answer: (c) 0.98 kJ / kg x = pe = gz = (9.8 m / s2)(100 m) = 980 m 2 / s 2 = 0.98 kJ / kg

since 1000 m2 / s 2 = 1kJ / kg

CYU 9-2 ■ Check Your Understanding CYU 9-2.1 Of the relatioins below for irreversibility (or exergy destroyed), select the correct ones: I I = Wrev,out - Wuseful,out II

I = Wuseful,in - Wrev,in

III

I = Wrev,out - Wrev,in

6-369

(d) I and III (e) II and III Answer: (c) I and II

CYU 9-2.2 Air flows through a wind turbine at a velocity of 10 m / s with a flow rate of 10 kg / s . If 0.15 kW power is generated by the wind turbine, what is the rate of exergy destruction? (a) 0 (b) 0.15 kW (c) 0.35 kW (d) 0.50 kW (e) 1 kW Answer: (c) 0.35 kW X. = KE. = 0.5 m.V 2 = 0.5(10 kg / s)(10 m / s) 2 = 500 kg ×m 2 / s3 = 500 W = 0.5 kW

since 1000 m2 / s 2 = 1kJ / kg

I. = X. - W.act = 0.5 - 0.15 = 0.35 kW CYU 9-2.3 Heat is supplied to a power plant from a source at 1000 K at a rate of 100 kW and the waste heat is rejected to the atmosphere at 300 K. If 45 kW of power is produced by this power plant, what is the exergy destroyed? (a) 0 (b) 25 kW (c) 45 kW (d) 70 kW (e) 100 kW Answer: (b) 25 kW W.max = X.heat = Q.(1- T0 / Ts ) = (100 kW)(1 - 300 / 1000) = 70 kW X.destroyed = I. = W.max - W.act = 70 - 45 = 25 kW

CYU 9-3 ■ Check Your Understanding CYU 9-3.1 An electric resistance heater consumes 2 kW of electricity to maintain a room at 25C when the outdoors are at 2C. What are the first-law (or energy conversion) efficiency and the second-law efficiency of this resistance heater, respectively? (a) 0, 1- 275 / 298 (b) 0, 1- 2 / 25 (c) 1, 1- 275 / 298 (d) 1, 1- 2 / 25 (e) 1, 1 Answer: (c) 0, 1- 275 / 298

CYU 9-3.2 Heat is supplied to a power plant, from a source at 600 K at a rate of 100 kW and waste heat is rejected to the environment at 300 K. If the net power is produced 20 kW, what is the second-law efficiency of this power plant? (a) 20% (b) 40% © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-370

(c) 50% (d) 60% (e) 100% Answer: (b) 40% eta th,rev = 1- T0 / TH = (100 kW)(1- 300 / 600) = 0.5 eta th = W×net / /Q×H = 20 /100 = 0.2 eta II = eta th / eta th, rev = 0.2 / 0.5 = 0.4

CYU 9-3.3 A totally reversible power plant receives heat from a source at 600 K at a rate of 100 kW and rejects the waste heat its surroundings at an unknown temperature. What is the second-law efficiency of this power plant? (a) 30% (b) 50% (c) 60% (d) 100% (e) Insufficient information Answer: (e) 100% CYU 9-3.4 A compressor consumes 7 kW of electric power and it is determined that 1 kW of exergy is destroyed during the process. What are the exergy expended and the second-law efficiency for this process? (a) 7 kW, 6 / 7 (b) 7 kW, 6 / 8 (c) 8kW, 6 / 7 (d) 8kW, 7 / 8 (e) 6kW, 7 / 8 Answer: (a) 7 kW, 6 / 7 X.expended = W.in = 7 kW W×rev = X×expended - X.destroyed = 7 - 1 = 6 kW

eta II = W.rev / W.in = (6 kW) / (7 kW)

CYU 9-4 ■ Check Your Understanding CYU 9-4.1 Select the incorrect statement below regarding the operation of a turbine or compressor. (a) The difference between the exergy values at the inlet and exit states of a turbine represents the maximum possible work output. (b) The difference between the exergy values at the exit and inlet of a compressor represents the required minimum work input for compression. (c) When steam expands in a turbine, its exergy decreases. (d) When air is compressed in a compressor, its exergy decreases. Answer: (d) When air is compressed in a compressor, its exergy decreases. CYU 9-4.2 The exergy change relation for a closed system with negligible kinetic and potential energies is (a) D xnonflow = D u - T0D s © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-371

(b) D xnonflow = D u - P0D v (c) D xnonflow = D u + P0D v - T0D s (d) D xnonflow = D h + P0D v - T0D s (e) D xnonflow = D h - T0D s Answer: (c) D xnonflow = D u + P0D v - T0D s

CYU 9-4.3 Steam expands in an adiabatic turbine, producing power. The difference between the reversible work and actual work of the turbine is (a) T0D s (b) P0D v (c) D h (d) D h - T0D s (e) D h + P0D v - T0D s Answer: (a) T0D s

CYU 9-5 ■ Check Your Understanding CYU 9-5.1 A vertical cylinder fitted with a weightless and frictionless piston is filled with a gas at atmospheric pressure Pf . Heat is now transferred to the cylinder and the volume of the gas increases from V1 to V2 . What is the exergy transferred by the piston during this process? (a) P0 (V1 - V0 ) (b) P0 (V2 - V0 ) (c) P0 (V2 - V1) (d) T0 (V2 - V1) (e) 0 CYU 9-5.2 5 kJ of heat is transferred from the environment at 300 K to liquid nitrogen at 75 K during transportation. What is the change in the exergy of the nitrogen during this process? (a) Decreases by 5 kJ (b) Increases by 5 kJ (c) Decreases by 15 kJ (d) Increases by 15 kJ (e) Increases by 3.75 kJ Answer: (c) Decreases by 15 kJ X heat = Q (1- T0 / T ) = (5 kJ)(1- 300 / 75) = - 15 kJ CYU 9-5.3 A hot iron block at 600 K loses 100 kJ of heat to its environment at 300 K. What is the change in the exergy of the iron block during this process? (a) Decreases by 50 kJ (b) Increases by 50 kJ (c) Decreases by 100 kJ (d) Increases by 100 kJ © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-372

(e) Decreases by 200 kJ Answer: (a) Decreases by 50 kJ X heat = Q (1- T0 / T ) = (100 kJ)(1- 300 / 600) = 50 kJ

CYU 9-6 ■ Check Your Understanding CYU 9-6.1 Select the correct statement(s) below: I Exergy change of a system during a process cannot be negative. II Exergy destroyed during a process cannot be negative. III Exergy destroyed during a reversible process is zero. (a) Only III (b) I and II (c) I and III (d) II and III (e) I, II and III Answer: (d) II and III CYU 9-6.2 Heat is transferred from a medium A to a medium B. If the exergy change of medium A is 0.5 kJ and that of medium B is 0.8 kJ, this process is (a) Impossible (b) Irreversible (c) Reversible Answer: (a) Impossible Xin - Xout - X destroyed = dXsys

0 - Xdestroyed = dX A + dX B - Xdestroyed = - 0.5 kJ + 0.8 kJ X destroyed = - 0.3 kJ

CYU 9-6.3 A system undergoes a process during which 0.3 kJ / K of entropy is generated. What is the exergy destroyed during this process if the environment is at 27C. (a) 0 (b) 0.3 kJ (c) 0.81 kJ (d) 30 kJ (e) 90 kJ Answer: (e) 90 kJ Xdestroyed = T0 Sgen = (300 K)(0.3 kJ / K) = 90 kJ

CYU 9-7 ■ Check Your Understanding CYU 9-7.1 Select the incorrect statement below on the exergy balance for closed systems (a) For a reversible process, the exergy change of the system is equal to the exergy transfer. (b) In the exergy balance relation, the work term becomes the reversible work when the exergy destruction is zero. (c) Exergy can be transferred to or from a closed system by heat transfer only. (d) A process that involves electric resistive heating is necessarily an irreversible process. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-373

Answer: (c) Exergy can be transferred to or from a closed system by heat transfer only. CYU 9-7.2 An insulated rigid tank containing m kg liquid water is heated by a resistance heater and the temperature of water rises from T1 to T2 in an environment at T0 . The exergy destroyed during this process can be expressed as (a) mc ln (T2 / T1 ) (b) T0 c ln (T2 / T1 ) (c) T0 mc ln (T2 / T1 ) (d) T0 mc (T2 / T1 ) (e) mc (T2 / T1 ) Answer: (c) T0 mc ln (T2 / T1 ) Sin - Sout + Sgen = dSsys 0 + Sgen = dSsys = mc ln (T2 / T1 )

X destroyed = T0 Sgen = T0 mc ln (T2 / T1 )

CYU 9-7.3 Heat is transferred from a thermal energy reservoir to a closed system in an environment at 27C. If the entropy change of the reservoir is - 0.4 kJ / K and that of the system is 0.9 kJ / K , the exergy destroyed during this process is (a) 0 (b) 0.5 kJ (c) 13.5 kJ (d) 150 kJ (e) 300 kJ Answer: (d) 150 kJ Sgen = dSsys + dSres = (0.9 kJ / K) + (- 0.4 kJ / K) = 0.5 kJ / K

Xdestroyed = T0 Sgen = (300 K)(0.5 kJ / K) = 150 kJ

CYU 9-7.4 An insulated rigid tank containing air is heated by an electric resistance heater. 3000 Wh of electrical energy is supplied to the air and 2600 Wh of exergy is destroyed during the process. What is the reversible work input for this process? (a) 400 Wh (b) 800 Wh (c) 2600 Wh (d) 3000 Wh (e) 5600 Wh Answer: (a) 400 Wh (Note: The same heating could be accomplished by only 0.4 kWh of electricity) Xin - Xout - X destroyed = dXsys

3000 - 2600 = dXsys Wrev = dXsys = 400Wh

or Wrev = Wuseful - X destroyed = 3000 - 2600 = 400 Wh

6-374

CYU 9-8.1 What is the second-law efficiency of an isentropic compressor? (a) 0 (b) 1 (c) Between 0 and 1 (d) Insufficient information Answer: (b) 1 CYU 9-8.2 Steam expands in an adiabatic turbine, producing 80 kW power. The exergy of the steam decreases at a rate of 90 kW during this process. The reversible power and the rate of exergy destruction for this process, respectively, are (a) 10 kW, 90 kW (b) 10 kW, 80 kW (c) 70 kW, 10 kW (d) 80 kW, 10 kW (e) 90 kW, 10 kW Answer: (e) 90 kW, 10 kW Wrev = dXsteam = 90 kW Xdest = Wrev - Wactual = 90 - 80 = 10 kW CYU 9-8.3 Heat is transferred from a hot fluid to a cold fluid in an insulated heat exchanger. The exergy change of the cold fluid is 3 kW. If 2 kW of exergy is destroyed in the heat exchange process, the exergy change of the hot fluid is (a) 1 kW (b) 1 kW (c) 2 kW (d) 5 kW (e) 5 kW Answer: (d) 5 kW Xin - X out - X dest = dXsys = 0

X hot,in + X cold,in - X hot,out - X cold,out - X dest = dX sys = 0 - Xdest = X hot,in - X hot,out + X cold,in - X cold,out

- Xdest = dXhot + dXcold - 2 kW = dXhot + 3 kW dXhot = - 5 kW CYU 9-8.4 Air is compressed by an adiabatic compressor consuming 7 kW power. If the rate of exergy change of air during compression is 5 kW, the rate of exergy destruction and the second-law efficiency of the compressor, respectively, are (a) 2 kW, 2 / 5 (b) 2 kW, 2 / 7 (c) 2 kW, 5 / 7 (d) 5 kW, 2 / 5 (e) 5 kW, 2 / 7 Answer: (c) 2 kW, 5 / 7 Wrev = dXsteam = 5 kW Xdest = Wactual - Wrev = 7 - 5 = 2 kW

eta II = Wrev / Wactual = 5 kW / 7 kW

6-375

PROBLEMS Exergy, Irreversibility, Reversible Work, and Second-Law Efficiency 9-1C The dead state.

9-2C Yes; exergy is a function of the state of the surroundings as well as the state of the system.

9-3C Reversible work and irreversibility are identical for processes that involve no actual useful work.

9-4C Useful work differs from the actual work by the surroundings work. They are identical for systems that involve no surroundings work such as steady-flow systems.

9-5C Reversible work differs from the useful work by irreversibilities. For reversible processes both are identical. Wu = Wrev - I .

9-6C Yes.

9-7C Yes.

9-8C A processes with Wrev = 0 is reversible if it involves no actual useful work. Otherwise it is irreversible.

9-9C They would be identical.

9-10C The difference between the exergy values at two different states is 'reversible work' for a reversible process between those two states; which is independent of the reversible process path followed. For an actual process between two specified states, the exergy destruction (and entropy generation) and thus the actual work is different, depending on the amount of irreversibility the process followed involves. But when entropy generation or exergy destruction is zero, which is a requirement for reversible process, the work is bound to be the same regardless of the process.

9-11C No, not necessarily. The well with the higher temperature will have a higher exergy.

9-12C The system that is at the temperature of the surroundings has zero exergy. But the system that is at a lower temperature than the surroundings has some exergy since we can run a heat engine between these two temperature levels.

6-376

9-13C The second-law efficiency is a measure of the performance of a device relative to its performance under reversible conditions. It differs from the first law efficiency in that it is not a conversion efficiency.

9-14C No. The power plant that has a lower thermal efficiency may have a higher second-law efficiency.

9-15C No. The refrigerator that has a lower COP may have a higher second-law efficiency.

9-16E Saturated steam is generated in a boiler by transferring heat from the combustion gases. The wasted work potential associated with this heat transfer process is to be determined. Also, the effect of increasing the temperature of combustion gases on the irreversibility is to be discussed. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Analysis The properties of water at the inlet and outlet of the boiler and at the dead state are (Tables A-4E through A-6E)

üïï h1 = h f = 355.46 Btu/lbm ý x1 = 0 (sat. liq.) ïïþ s1 = s f = 0.54379 Btu/lbm ×R

P1 = 200 psia

ïüï h2 = hg = 1198.8 Btu/lbm ý x2 = 1 (sat. vap.) ïïþ s2 = sg = 1.5460 Btu/lbm ×R

P2 = 200 psia

ïü ïý h0 @ h f @80° F = 48.07 Btu/lbm P0 = 14.7 psiaïïþ s0 @ s f @80° F = 0.09328 Btu/lbm ×R

T0 = 80°F

The heat transfer during the process is

qin = h2 - h1 = 1198.8 - 355.46 = 843.3 Btu/lbm The entropy generation associated with this process is

sgen = D sw + D sR = ( s2 - s1 ) -

qin TR

= (1.5460 - 0.54379)Btu/lbm ×R -

843.3 Btu/lbm (700 + 460)R

= 0.2752 Btu/lbm ×R The wasted work potential (exergy destruction is) xdest = T0 sgen = (80 + 460 R)(0.2752 Btu/lbm ×R) = 149 Btu / lbm

The work potential (exergy) of the steam stream is D xflow, w = h2 - h1 - T0 ( s2 - s1 )

= (1198.8 - 355.46)Btu/lbm - (540 R)(1.5460 - 0.54379)Btu/lbm ×R

= 302.1 Btu/lbm Increasing the temperature of combustion gases does not effect the work potential of steam stream since it is determined by the states at which water enters and leaves the boiler. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-377

Discussion This problem may also be solved as follows: Exergy transfer by heat transfer:

æ T ö æ 540 ö÷ ÷ xheat = q ççç1- 0 ÷ = (843.3) çç1÷= 450.7 Btu/lbm ÷ çè TR ÷ èç 1160 ø÷ ø Exergy increase of steam: D xflow, w = 302.1 Btu/lbm

The net exergy destruction: xdest = xheat - D xflow, w = 450.7 - 302.1 = 149 Btu / lbm

EES (Engineering Equation Solver) SOLUTION "Given" x1=0 x2=1 P=200 [psia] T_R=700 [F] T0=80 [F] P0=14.7 [psia] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P, x=x1) s1=entropy(Fluid$, P=P, x=x1) h2=enthalpy(Fluid$, P=P, x=x2) s2=entropy(Fluid$, P=P, x=x2) h0=enthalpy(Fluid$, T=T0, x=0) s0=entropy(Fluid$, T=T0, x=0) "Analysis" q=h2-h1 DELTAs_w=s2-s1 DELTAs_R=-q/(T_R+460) s_gen=DELTAs_w+DELTAs_R x_dest=(T0+460)*s_gen DELTAx_flow_w=h2-h1-(T0+460)*(s2-s1) "Alternative Solution" x_heat=q*(1-(T0+460)/(T_R+460)) x_dest_alt=x_heat-DELTAx_flow_w

9-17 Water is to be pumped to a high elevation lake at times of low electric demand for use in a hydroelectric turbine at times of high demand. For a specified energy storage capacity, the minimum amount of water that needs to be stored in the lake is to be determined. Assumptions The evaporation of water from the lake is negligible.

6-378

Analysis The exergy or work potential of the water is the potential energy it possesses,

Exergy = PE = mgh Thus,

2 2ö öæ PE 5´ 106 kWh æ ç1000 m /s ÷ ÷ çç3600 s ÷ç ÷= 2.45´ 1010 kg = ÷ øèç 1 kW ×s/kg ÷ gh (9.8 m/s 2 )(75 m) èç 1 h ÷ç ø

EES (Engineering Equation Solver) SOLUTION "Given" X=5E+6 [kWh] h=75 [m] "Analysis" PE=X m=PE/(g*h)*Convert(h, s)*Convert(kW-s/kg, m^2/s^2) g=9.8 [m/s^2]

9-18 The thermal efficiency of a heat engine operating between specified temperature limits is given. The second-law efficiency of a engine is to be determined.

Analysis The thermal efficiency of a reversible heat engine operating between the same temperature reservoirs is Thus,

hth,rev = 1-

hII =

T0 293 K = 1= 0.801 TH 1200 + 273 K

hth 0.40 = = 49.9% hth,rev 0.801

6-379

T_0=(20+273) [K] eta_th=0.40 "Analysis" eta_th_rev=1-T_0/T_H eta_II=eta_th/eta_th_rev

9-19 A heat reservoir at a specified temperature can supply heat at a specified rate. The exergy of this heat supplied is to be determined.

Analysis The exergy of the supplied heat, in the rate form, is the amount of power that would be produced by a reversible heat engine,

hth,max = hth,rev = 1-

T0 298 K = 1= 0.8013 TH 1500 K

Exergy = Wmax,out = Wrev,out = hth,rev Qin = (0.8013)(150,000 / 3600 kJ/s)

= 33.4 kW

EES (Engineering Equation Solver) SOLUTION "Given" T_H=1500 [K] Q_dot_in=150000 [kJ/h] T_0=(25+273) [K] "Analysis" eta_th_rev=1-T_0/T_H eta_th_rev=W_dot_max_out/Q_dot_in*Convert(kW, kJ/h)

9-20 A heat engine receives heat from a source at a specified temperature at a specified rate, and rejects the waste heat to a sink. For a given power output, the reversible power, the rate of irreversibility, and the 2 nd law efficiency are to be determined.

6-380

Analysis (a) The reversible power is the power produced by a reversible heat engine operating between the specified temperature limits, hth,max = hth,rev = 1-

TL 320 K = 1= 0.7091 TH 1100 K

Wrev,out = hth,rev Qin = (0.7091)(400 kJ/s) = 283.6 kW (b) The irreversibility rate is the difference between the reversible power and the actual power output: I = Wrev,out - Wu,out = 283.6 - 120 = 163.6 kW

Wu,out Wrev,out

120 kW = 0.423 = 42.3% 283.6 kW

EES (Engineering Equation Solver) SOLUTION "Input Data" T_H= 1100 [K] Q_dot_H= 400 [kJ/s] T_L=320 [K] W_dot_out = 120 [kW] T_Lsurr =25 [C] "The reversible work is the maximum work done by the Carnot Engine between T_H and T_L:" Eta_Carnot=1 - T_L/T_H W_dot_rev=Q_dot_H*Eta_Carnot "The irreversibility is given as:" I_dot = W_dot_rev-W_dot_out "The thermal efficiency is, in percent:" Eta_th = Eta_Carnot*Convert( , %) "The second law efficiency is, in percent:" Eta_II = W_dot_out/W_dot_rev*Convert( , %)

9-21 Problem 9-20 is reconsidered. The effect of reducing the temperature at which the waste heat is rejected on the reversible power, the rate of irreversibility, and the second law efficiency is to be studied and the results are to be plotted. Analysis The problem is solved using EES, and the solution is given below. T_H= 1100 [K] Q_dot_H= 400 [kJ/s] T_L=320 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-381

W_dot_out = 120 [kW] T_Lsurr =25 [C] Eta_Carnot=1 - T_L/T_H W_dot_rev=Q_dot_H*Eta_Carnot I_dot = W_dot_rev-W_dot_out Eta_th = Eta_Carnot*Convert(, %) Eta_II = W_dot_out/W_dot_rev*Convert(, %)

TL [k] 500 477.6 455.1 432.7 410.2 387.8 365.3 342.9 320.4 298

Wrev [kJ / s] 218.2 226.3 234.5 242.7 250.8 259 267.2 275.3 283.5 291.6

I [kJ / s] 98.18 106.3 114.5 122.7 130.8 139 147.2 155.3 163.5 171.6

ηII [%] 55 53.02 51.17 49.45 47.84 46.33 44.92 43.59 42.33 41.15

6-382

9-22E The thermal efficiency and the second-law efficiency of a heat engine are given. The source temperature is to be determined.

Analysis From the definition of the second law efficiency, Thus,

hI =

hth,rev = 1-

hth h 0.25 ¾¾ ® hth,rev = th = = 0.50 hth,rev hII 0.50

TL ¾¾ ® TH = TL / (1- hth,rev ) = (510 R)/0.50 = 1020 R TH

6-383

EES (Engineering Equation Solver) SOLUTION "Given" T_L=510 [R] eta_th=0.25 eta_II=0.50 "Analysis" eta_th_rev=eta_th/eta_II eta_th_rev=1-T_L/T_H

9-23 A geothermal power produces 5.1 MW power while the exergy destruction in the plant is 7.5 MW. The exergy of the geothermal water entering to the plant, the second-law efficiency of the plant, and the exergy of the heat rejected from the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Water properties are used for geothermal water. Analysis (a) The properties of geothermal water at the inlet of the plant and at the dead state are (Tables A-4 through A-6)

T1 = 150°C ïüï h1 = 632.18 kJ/kg ý ïïþ s1 = 1.8418 kJ/kg ×K x1 = 0 T0 = 25°Cü ïï h0 = 104.83 kJ/kg ý ïïþ s0 = 0.36723 kJ/kg ×K x0 = 0 The exergy of geothermal water entering the plant is X in = m [h1 - h0 - T0 ( s1 - s0 ] = (210 kg/s) [(632.18 - 104.83) kJ/kg - (25 + 273 K)(1.8418 - 0.36723)kJ/kg ×K ] = 18, 460 kW = 18.46 MW

(b) The second-law efficiency of the plant is the ratio of power produced to the exergy input to the plant hII =

Wout 5100 kW = = 0.276 = 27.6% 18, 460 kW X in

(c) The exergy of the heat rejected from the plant may be determined from an exergy balance on the plant X heat,out = X in - Wout - X dest = 18, 460 - 5100 - 7500 = 5864 kW = 5.86 MW

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_geo=150 [C] m_dot_geo=210 [kg/s] W_dot_net=5.1 [MW] T_0=25 [C] EX_dot_dest=7.5 [MW] "ANALYSIS" Fluid$='Steam_iapws' h_geo=enthalpy(Fluid$, T=T_geo, x=0) "assumed saturated liquid" s_geo=entropy(Fluid$, T=T_geo, x=0) h_0=enthalpy(Fluid$, T=T_0, x=0) "dead state enthalpy" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-384

s_0=entropy(Fluid$, T=T_0, x=0) ex_in=h_geo-h_0-(T_0+273)*(s_geo-s_0) EX_dot_in=m_dot_geo*ex_in*Convert(kW, MW) "exergy input" eta_II=W_dot_net/EX_dot_in "second-law efficiency" EX_dot_heat_out=EX_dot_in-W_dot_net-EX_dot_dest "exergy out by heat"

9-24 A house is maintained at a specified temperature by electric resistance heaters. The reversible work for this heating process and irreversibility are to be determined. Analysis We consider a reversible heat pump operation as the reversible counterpart of the irreversible process of heating the house by resistance heaters. Instead of using electricity input for resistance heaters, it is used to power a reversible heat pump. The reversible work is the minimum work required to accomplish this process, and the irreversibility is the difference between the reversible work and the actual electrical work consumed. The actual power input is Win = Qout = QH = 35, 000 kJ/h = 9.722 kW

The COP of a reversible heat pump operating between the specified temperature limits is COPHP,rev =

1 1 = = 14.20 1- TL / TH 1- 277.15 / 298.15

Thus, and

Wrev,in =

QH 9.722 kW = = 0.685 kW COPHP,rev 14.20

I = Wu,in - Wrev,in = 9.722 - 0.658 = 9.04 kW

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_H=35000 [kJ/h] T_L=(4+273.15) [K] T_H=(25+273.15) [K] "Analysis" COP_HP_rev=1/(1-(T_L/T_H)) W_dot_rev_in=Q_dot_H/COP_HP_rev*1/Convert(kW, kJ/h) I_dot=W_dot_u_in-W_dot_rev_in W_dot_u_in=Q_dot_H*1/Convert(kW, kJ/h)

6-385

9-25E A freezer is maintained at a specified temperature by removing heat from it at a specified rate. The power consumption of the freezer is given. The reversible power, irreversibility, and the second-law efficiency are to be determined.

Analysis (a) The reversible work is the minimum work required to accomplish this task, which is the work that a reversible refrigerator operating between the specified temperature limits would consume, COPR,rev =

Wrev,in =

1 1 = = 8.73 TH / TL - 1 535 / 480 - 1

ö QL 75 Btu/min æ 1 hp ÷ çç = = 0.20 hp ÷ ÷ çè 42.41 Btu/min ø COPR,rev 8.73

(b) The irreversibility is the difference between the reversible work and the actual electrical work consumed, I = Wu,in - Wrev,in = 0.70 - 0.20 = 0.50 hp

Wrev 0.20 hp = = 28.9% 0.7 hp Wu

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(20+460) [R] T_H=(75+460) [R] Q_dot_L=75 [Btu/min] W_dot_u_in=0.70 [hp] "Analysis" "(a)" COP_R_rev=1/((T_H/T_L)-1) W_dot_rev_in=Q_dot_L/COP_R_rev*Convert(Btu/min, hp) "(b)" I_dot=W_dot_u_in-W_dot_rev_in "(c)" eta_II=W_dot_rev_in/W_dot_u_in

9-26 Windmills are to be installed at a location with steady winds to generate power. The minimum number of windmills that need to be installed is to be determined. Assumptions Air is at standard conditions of 1 atm and 25C

6-386

Properties The gas constant of air is 0.287 kPa.m3 / kg.K (Table A-1). Analysis The exergy or work potential of the blowing air is the kinetic energy it possesses, Exergy = ke =

ö V2 (6 m/s) 2 æ çç 1 kJ/kg ÷ = = 0.0180 kJ/kg 2 2÷ ÷ ç è ø 2 2 1000 m /s

At standard atmospheric conditions (25C, 101 kPa), the density and the mass flow rate of air are r=

P 101 kPa = = 1.18 m3 /kg RT (0.287 kPa ×m3 /kg ×K)(298 K)

and Thus,

m = r AV1 = r

p D2 V1 = (1.18 kg/m3 )(p / 4)(40 m)2 (6 m/s) = 8904 kg/s 4

Available Power = m ke = (8904 kg/s)(0.0180 kJ/kg) = 160.3 kW The minimum number of windmills that needs to be installed is N=

Wtotal 1500 kW = = 9.4 @10 windmills 160.3 kW W

EES (Engineering Equation Solver) SOLUTION "Given" D=40 [m] Vel=6 [m/s] W_dot_total=1500 [kW] "Properties" R=0.287 [kJ/kg-K] "gas constant, Table A-2a" "Analysis" P=101 [kPa] "assumed" T=(25+273) [K] "assumed" KE=Vel^2/2*Convert(m^2/s^2, kJ/kg) rho=P/(R*T) A=pi*D^2/4 m_dot=rho*A*Vel X_dot=m_dot*KE N=W_dot_total/X_dot

9-27 It is to be shown that the power produced by a wind turbine is proportional to the cube of the wind velocity and the square of the blade span diameter. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-387

Analysis The power produced by a wind turbine is proportional to the kinetic energy of the wind, which is equal to the product of the kinetic energy of air per unit mass and the mass flow rate of air through the blade span area. Therefore, Wind power = (Efficiency)(Kinetic energy)(Mass flow rate of air)

= hwind

V2 V 2 p D2 (r AV ) = hwind r V 2 2 4

= hwind r

pV 3 D 2 = (Constant)V 3 D 2 8

which completes the proof that wind power is proportional to the cube of the wind velocity and to the square of the blade span diameter.

9-28 A heat engine operates between two constant-pressure cylinders filled with air at different temperatures. The maximum work that can be produced and the final temperatures of the cylinders are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is 0.287 kPa.m3 / kg.K (Table A-1). The constant pressure specific heat of air at room temperature is c p = 1.005 kJ / kg.K (Table A-2).

Analysis For maximum power production, the entropy generation must be zero. We take the two cylinders (the heat source and heat sink) and the heat engine as the system. Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as

Sin - Sout + Sgen Z 0 = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0 + Sgen Z 0 = D Scylinder,source + D Scylinder,sink + D S heat engineZ 0

D Scylinder,source + D Scylinder,sink = 0 ¿0ö 0ö æ æ ççmc ln T2 - mR ln P2 ÷ ççmc ln T2 - mR ln P2 ÷ ÷ ÷ = 0 + 0 + ÷ ÷ çç p T ç p T ÷ ÷ P1 ø P1 ø è èç 1 1 source sink

T2 T2 = 0¾¾ ® T22 = T1 AT1B T1 A T1B

where T1A and T1B are the initial temperatures of the source and the sink, respectively, and T2 is the common final temperature. Therefore, the final temperature of the tanks for maximum power production is

6-388

The energy balance Ein - Eout = D Esystem for the source and sink can be expressed as follows: Source: - Qsource,out + Wb,in = D U ® Qsource,out = D H = mc p (T1 A - T2 )

Qsource,out = mc p (T1A - T2 ) = (30 kg)(1.005 kJ/kg ×K)(900 - 519.6)K = 11,469 kJ Sink: Qsink,in - Wb,out = D U ® Qsink,in = D H = mc p (T2 - T1 A ) Qsink,in = mc p (T2 - T1B ) = (30 kg)(1.005 kJ/kg ×K)(519.6 - 300)K = 6621 kJ

Then the work produced becomes Wmax,out = QH - QL = Qsource,out - Qsink,in = 11, 469 - 6621 = 4847 kJ

Therefore, a maximum of 4847 kJ of work can be produced during this process.

EES (Engineering Equation Solver) SOLUTION "Given" T1_A=900 [K] T1_B=300 [K] m=30 [kg] "Properties" c_p_A=1.005 [kJ/kg-K] c_p_B=c_p_A "Analysis" S_gen=DELTAS_A+DELTAS_B S_gen=0 "for maximum power production" DELTAS_A=m*c_p_A*ln(T2/T1_A) "since P2_A=P1_A" DELTAS_B=m*c_p_B*ln(T2/T1_B) "since P2_B=P1_B" Q_source_out=m*c_p_A*(T1_A-T2) Q_sink_in=m*c_p_B*(T2-T1_B) W_max_out=Q_source_out-Q_sink_in

9-29 Natural gas is typically over 90 percent methane, and it is often liquefied for transportation by large tankers. The temperature of liquid natural gas (LNG) drops to – 162ºC when it is liquified at 1 atm pressure, and its volume reduces by a factor of about 600. Approximating natural gas as methane, determine the minimum amount of work needed to liquify 1 kg of natural gas at the environmental conditions of 1 atm and 15ºC to LNG at 1 atm and −162ºC during steady operation. Take the enthalpy and entropy of natural gas at the initial and final states to be h1 = 601.2 kJ / kg.K , s1 = 11.53 kJ / kg.K and h2 = - 286.5 kJ / kg and s2 = 4.933 kJ / kg.K , respectively. 9-29 Natural gas which is approximated as methane at environmental conditions is to be liquefied at atmospheric pressure. The minimum amount of work needed to liquify 1 kg of natural gas is to be determined. Assumptions 1 The composition of natural gas closely approximates methane. 2 Steady operating conditions exist. 3 The environmental conditions are 1 atm and 15ºC. Properties The enthalpy and entropy of natural gas at the initial and final states are given to be h1 = 601.2 kJ / kg.K ,

s1 = 11.53 kJ / kg.K and h2 = - 286.5 kJ / kg and s2 = 4.933 kJ / kg.K .

6-389

Analysis The minimum work input required to liquify 1 kg of natural gas at 1 atm, 15°C (state 1) to 1 atm, −162°C (state 2, saturated liquid) during steady operation is simply the reversible work associated with the liquefaction process between these two states, which is simply the exergy change during that process: wmin in = wrev = ex2 - ex1 = (h2 - h1 )- T0 (s2 - s1 ) where T0 is the temperature of the environment, which is given to be T0 = 15° C = 288 K . Using the given enthalpy and entropy values, the reversible work is determined to be w min in = wrev = (- 286.5) - 601.2 - 288(4.933 - 11.53) = 1012kJ / kg

Therefore, a minimum of 1012 kJ of work input (more correctly, exergy input) is required to liquify 1 kg of gaseous natural gas at the environment conditions of 1 atm and 15ºC.

9-30 Hydrogen is considered a potential energy storage and transportation medium for renewable energy. The surplus electric power from the renewable resources such as solar and wind energy is used to produce hydrogen usually by the electrolysis of water. Because gaseous hydrogen requires a large volume of storage, it is sometimes liquefied using a liquefaction system in order to store a large mass of hydrogen in a small volume. Consider a Carnot refrigeration system used to liquefy hydrogen which enters at 100 kPa and 20C and leaves at 253C as a liquid, as shown in Fig. P9-30. Determine the work input, in kWh / kg and the COP of this ideal liquefaction system. If the same liquefaction process is carried out by an actual liquefaction system with a COP of 0.055, determine the second-law efficiency and the actual work input, in kWh / kg , of that system. Various properties of hydrogen before and after the liquefaction process are h1 = 4128.2 kJ / kg , h2 = 270.67 kJ / kg , s1 = 70.228 kJ / kg.K , s2 = 17.066 kJ / kg.K . 9-30 A Carnot refrigerator is used to liquefy hydrogen. The minimum work input and the COP of this Carnot refrigerator and the second-law efficiency and the actual work input for an actual liquefaction system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis The heat removed from hydrogen gas during the liquefaction process is qL = h1 - h2 = (4128.2- 270.67)kJ/kg = 3858 kJ/kg

Taking the dead state temperature to be T0 = T1 = 20C = 293.15 K, the minimum or reversible work input is determined from wrev = h2 - h1 - T0 (s2 - s1 ) = (270.67 - 4128.2) kJ/kg - (293.15 K)(17.066 - 70.228) kJ/kg ×K

= 11,725 kJ/kg = 3.26 kWh / kg since 1 kWh = 3600 kJ. The reversible COP is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-390

CO Prev =

qL 3858 kJ/kg = = 0.3289 w rev 11,725 kJ/kg

The second-law efficiency can be defined as the actual COP divided by the reversible COP. Then,

hII =

CO Pact CO Prev

0.055 = 0.1672 or 16.7 percent 0.3289

The second-law efficiency can also be defined as the reversible work divided by the actual work. Using this definition, the actual work is determined to be

hII =

w rev wact

® 0.1672 =

3.26 kW h/kg ® wact = 19.5 kWh / kg wact

Discussion It takes $1.95 to liquefy a unit mass of hydrogen at a unit electricity cost of $0.10 / kWh .

EES (Engineering Equation Solver) SOLUTION T_0=293.15 [K] T_1=20 [C] P_1=100 [kPa] P_2=P_1 x_2=0 COP_act=0.055 Fluid$='hydrogen' h_1=enthalpy(Fluid$,T=T_1,P=P_1) h_2=enthalpy(Fluid$,P=P_2,x=x_2) s_1=entropy(Fluid$,T=T_1,P=P_1) s_2=entropy(Fluid$,P=P_2,x=x_2) T_2=temperature(Fluid$,P=P_2,x=x_2) w_min=((h_2-h_1)-T_0*(s_2-s_1))*convert(kJ/kg, kWh/kg) q_L=h_1-h_2 COP_rev=q_L*convert(kJ/kg, kWh/kg)/w_min eta_ex=COP_act/COP_rev eta_ex=w_min/w_act

Exergy Analysis of Closed Systems 9-31C Yes, it can. For example, the 1st law efficiency of a reversible heat engine operating between the temperature limits of 300 K and 1000 K is 70%. However, the second law efficiency of this engine, like all reversible devices, is 100%.

9-32 A fixed mass of helium undergoes a process from a specified state to another specified state. The increase in the useful energy potential of helium is to be determined.

Assumptions 1 At specified conditions, helium can be treated as an ideal gas. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-391

2 Helium has constant specific heats at room temperature. Properties The gas constant of helium is R = 2.0769 kJ / kg ×K (Table A-1). The constant volume specific heat of helium is cv = 3.1156 kJ / kg ×K (Table A-2). Analysis From the ideal-gas entropy change relation,

T v s2 - s1 = cv ,avg ln 2 + R ln 2 T1 v1

= (3.1156 kJ/kg ×K) ln

353 K 0.5 m3 /kg + (2.0769 kJ/kg ×K) ln 288 K 3 m3 /kg

= - 3.087 kJ/kg ×K The increase in the useful potential of helium during this process is simply the increase in exergy, D X = m( xnonflow,2 - xnonflow,1 ) = - m[(u1 - u2 ) - T0 ( s1 - s2 ) + P0 (v1 - v 2 )] = - (8 kg){(3.1156 kJ / kg ×K)(288 - 353)K - (298 K)(3.087 kJ / kg ×K) + (100 kPa)(3 - 0.5)m 3 /kg [kJ / kPa ×m 3 ]}

= 6980 kJ EES (Engineering Equation Solver) SOLUTION "Given" m=8 [kg] v1=3 [m^3/kg] T1=(15+273) [K] v2=0.5 [m^3/kg] T2=(80+273) [K] T0=(25+273) [K] "Properties" R=2.0769 [kJ/kg-K] "Table A-2a" c_v=3.1156 [kJ/kg-K] "Table A-2a" "Analysis" P0=100 [kPa] "assumed" DELTAs=c_v*ln(T2/T1)+R*ln(v2/v1) DELTAu=c_v*(T2-T1) DELTAv=v2-v1 DELTAX=m*(DELTAu-T0*DELTAs+P0*DELTAv)

9-33E Air and helium at specified states are considered. The gas with the higher exergy content is to be identified. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air and helium are ideal gases with constant specific heats.

6-392

Properties The properties of air at room temperature are c p = 0.240 Btu / lbm ×R , cv = 0.171 Btu / lbm× R, k = 1.4, and R = 0.06855 Btu / 1bm ×R = 0.3704 psia ×ft 3 / 1bm ×R . For helium, c p = 1.25 Btu / lbm ×R , cv = 0.753 Btu / lbm , × R k = 1.667, and R = 0.4961 Btu / 1bm ×R = 2.6809 psia ×ft 3 / 1bm ×R . (Table A-2E). Analysis The mass of air in the system is m=

PV (100 psia)(15 ft 3 ) = = 5.704 lbm RT (0.3704 psia ×ft 3 /lbm ×R)(710 R)

The entropy change of air between the given state and the dead state is s - s0 = c p ln

T P - R ln T0 P0

= (0.240 Btu/lbm ×R)ln

710 R 100 psia - (0.06855 Btu/lbm ×R)ln 537 R 14.7 psia

= - 0.06441 Btu/lbm ×R The air’s specific volumes at the given state and dead state are v=

RT (0.3704 psia ×ft 3 /lbm ×R)(710 R) = = 2.630 ft 3 /lbm P 100 psia

v0 =

RT0 (0.3704 psia ×ft 3 /lbm ×R)(537 R) = = 13.53 ft 3 /lbm P0 14.7 psia

The specific closed system of nonflow exergy of the air is then xnonflow = u - u0 + P0 (v - v 0 ) - T0 (s - s0 )

= cv (T - T0 ) + P0 (v - v 0 ) - T0 (s - s0 ) æ ö 1 Btu ÷ ÷ = (0.171 Btu/lbm ×R)(300 - 77)R + (14.7 psia)(2.630 - 13.53)ft 3 /lbm çç ÷ ÷ çè5.404 psia ×ft 3 ø - (537 R)( - 0.06441) Btu/lbm ×R

= 34.52 Btu/lbm The total exergy available in the air for the production of work is then X = mxnonflow = (5.704 lbm)(34.52 Btu/lbm) = 197 Btu

We now repeat the calculations for helium: m=

PV (60 psia)(20 ft 3 ) = = 0.6782 lbm RT (2.6809 psia ×ft 3 /lbm ×R)(660 R)

s - s0 = c p ln

T P - R ln T0 P0

= (1.25 Btu/lbm ×R)ln

660 R 60 psia - (0.4961 Btu/lbm ×R)ln 537 R 14.7 psia

= - 0.4400 Btu/lbm ×R v=

RT (2.6809 psia ×ft 3 /lbm ×R)(660 R) = = 29.49 ft 3 /lbm P 60 psia

6-393

v0 =

RT0 (2.6809 psia ×ft 3 /lbm ×R)(537 R) = = 97.93 ft 3 /lbm P0 14.7 psia

xnonflow = u - u0 + P0 (v - v 0 ) - T0 (s - s0 ) = cv (T - T0 ) + P0 (v - v 0 ) - T0 (s - s0 ) æ ö 1 Btu ÷ ÷ - (537 R)( - 0.4400) Btu/lbm ×R = (0.753 Btu/lbm ×R)(200 - 77)R + (14.7 psia)(29.49 - 97.93)ft 3 /lbm çç ÷ çè5.404 psia ×ft 3 ÷ ø

= 142.7 Btu/lbm X = mxnonflow = (0.6782 lbm)(142.7 Btu/lbm) = 96.8 Btu

Comparison of two results shows that the air system has a greater potential for the production of work.

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol_air=15 [ft^3] P_air=100 [psia] T_air=(250+460) [R] Vol_He=20 [ft^3] P_He=60 [psia] T_He=(200+460) [R] T0=(77+460) [R] P0=14.7 [psia] "Properties" c_p_air=0.240 [Btu/lbm-R]; c_v_air=0.171 [Btu/lbm-R]; k_air=1.4; R_air=0.06855 [Btu/lbm-R]; R1_air=0.3704 [psia-ft^3/lbm-R] c_p_He=1.25 [Btu/lbm-R]; c_v_He=0.753 [Btu/lbm-R]; k_He=1.667; R_He=0.4961 [Btu/lbm-R]; R1_He=2.6809 [psia-ft^3/lbm-R] "ANALYSIS" m_air=(P_air*Vol_air)/(R1_air*T_air) DELTAs_air=c_p_air*ln(T_air/T0)-R_air*ln(P_air/P0) v_air=R1_air*T_air/P_air v0_air=R1_air*T0/P0 x_air=c_v_air*(T_air-T0)+P0*(v_air-v0_air)*Convert(psia-ft^3, Btu)-T0*DELTAs_air X_tot_air=m_air*x_air m_He=(P_He*Vol_He)/(R1_He*T_He) DELTAs_He=c_p_He*ln(T_He/T0)-R_He*ln(P_He/P0) v_He=R1_He*T_He/P_He v0_He=R1_He*T0/P0 x_He=c_v_He*(T_He-T0)+P0*(v_He-v0_He)*Convert(psia-ft^3, Btu)-T0*DELTAs_He X_tot_He=m_He*x_He

9-34 Steam and R-134a at the same states are considered. The fluid with the higher exergy content is to be identified. Assumptions Kinetic and potential energy changes are negligible.

6-394

Analysis The properties of water at the given state and at the dead state are u = 2594.7 kJ/kg P = 800 kPa ü ïï 3 v ý = 0.24720 m /kg T = 180°C ïïþ s = 6.7155 kJ/kg ×K

(Table A-6)

u0 @ u f @ 25°C = 104.83 kJ/kg ïü 3 ïý v @v 0 f @ 25° C = 0.001003 m /kg P0 = 100 kPa ïïþ s0 @ s f @ 25°C = 0.3672 kJ/kg ×K

T0 = 25°C

(Table A-4)

The exergy of steam is X = mxnonflow = m [u - u0 + P0 (v - v 0 ) - T0 ( s - s0 ) ]

é æ 1 kJ ÷ öù ê(2594.7 - 104.83)kJ/kg + (100 kPa)(0.24720 - 0.001003)m3 /kg çç ú 3÷ ÷ ç ê è1 kPa ×m øú = (1 kg) ê ú ê- (298 K)(6.7155 - 0.3672)kJ/kg ×K ú ë û = 622.7 kJ

For R-134a;

u = 386.99 kJ/kg P = 800 kPa ü ïï 3 v ý = 0.044554 m /kg T = 180°C ïïþ s = 1.3327 kJ/kg ×K

(Table A-13)

u0 = 252.60 kJ/kg ü ïï 3 v ý 0 = 0.001003 m /kg P0 = 100 kPaïïþ s0 = 1.1061 kJ/kg ×K

(from EES)

T0 = 25°C

X = mxnonflow = m [u - u0 + P0 (v - v 0 ) - T0 ( s - s0 ) ]

é æ 1 kJ ÷ öù ê(386.99 - 252.60)kJ/kg + (100 kPa)(0.044554 - 0.001003)m3 /kg çç ú 3÷ ÷ ç ê è1 kPa ×m øú = (1 kg) ê ú ê- (298 K)(1.3327 - 1.1061)kJ/kg ×K ú ë û = 47.5 kJ The steam can therefore has more work potential than the R-134a.

EES (Engineering Equation Solver) SOLUTION "Given" m=1 [kg] P=800 [kPa] T=180 [C] P0=100 [kPa] T0=25 [C] "Analysis" Fluid1$='steam_iapws' u_s=intenergy(Fluid1$, T=T, P=P) v_s=volume(Fluid1$, T=T, P=P) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-395

s_s=entropy(Fluid1$, T=T, P=P) u0_s=intenergy(Fluid1$, T=T0, x=0) v0_s=volume(Fluid1$, T=T0, x=0) s0_s=entropy(Fluid1$, T=T0, x=0) X_s=m*(u_s-u0_s+P0*(v_s-v0_s)-(T0+273)*(s_s-s0_s)) Fluid2$='R134a' u_R=intenergy(Fluid2$, T=T, P=P) v_R=volume(Fluid2$, T=T, P=P) s_R=entropy(Fluid2$, T=T, P=P) u0_R=intenergy(Fluid2$, T=T0, P=P0) v0_R=volume(Fluid2$, T=T0, P=P0) s0_R=entropy(Fluid2$, T=T0, P=P0) X_R=m*(u_R-u0_R+P0*(v_R-v0_R)-(T0+273)*(s_R-s0_R))

9-35 The radiator of a steam heating system is initially filled with superheated steam. The valves are closed, and steam is allowed to cool until the pressure drops to a specified value by transferring heat to the room. The amount of heat transfer to the room and the maximum amount of heat that can be supplied to the room are to be determined. Assumptions Kinetic and potential energies are negligible.

Properties From the steam tables (Tables A-4 through A-6), v1 = 1.0805 m3 /kg P1 = 200 kPa ü ïï ý u1 = 2654.6 kJ/kg T1 = 200 °C ïïþ s1 = 7.5081 kJ/kg ×K

x2 =

v2 - v f

1.0805 - 0.001029 = 0.3171 3.4053 - 0.001029

v fg T2 = 80 °C ïüï ý u2 = u f + x2 u fg = 334.97 + 0.3171´ 2146.6 = 1015.6 kJ/kg (v 2 = v1 ) ïïþ s2 = s f + x2 s fg = 1.0756 + 0.3171´ 6.5355 = 3.1479 kJ/kg ×K Analysis (a) The mass of the steam is m=

V 0.020 m3 = = 0.01851 kg v1 1.0805 m3 /kg

The amount of heat transfer to the room is determined from an energy balance on the radiator expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = m(u2 - u1 )

(since W = KE = PE = 0)

Qout = m(u1 - u2 ) or

Qout = (0.01851 kg)(2654.6 - 1015.6) kJ/kg = 30.3 kJ

(b) The reversible work output, which represents the maximum work output Wrev,out in this case can be determined from the exergy balance by setting the exergy destruction equal to zero, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-396

X in - X out

- X destroyed ¿ 0 (reversible) = D X system

Net exergy transfer by heat, work, and mass

Exergy destruction

- Wrev,out = X 2 - X 1

Change in exergy

Wrev,out = X 1 - X 2 = m( xnonflow,1 - xnonflow,2 )

Substituting the closed system exergy relation, the reversible work during this process is determined to be

Wrev,out = m éëê(u1 - u2 ) - T0 (s1 - s2 ) + P0 (v1¿ 0 - v 2 )ùûú = m [(u1 - u2 ) - T0 ( s1 - s2 ) ] = (0.01851 kg) [(2654.6 - 1015.6) kJ/kg - (273 K)(7.5081 - 3.1479) kJ/kg ×K ]= 8.305 kJ

When this work is supplied to a reversible heat pump, it will supply the room heat in the amount of QH = COPHP,revWrev =

Wrev 8.305 kJ = = 116 kJ 1- TL / TH 1- 273/294

Discussion Note that the amount of heat supplied to the room can be increased by about 3 times by eliminating the irreversibility associated with the irreversible heat transfer process.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.020 [m^3] P1=200 [kPa] T1=200 [C] T2=80 [C] T_air=ConvertTemp(C, K, 21) T0=ConvertTemp(C, K, 0) "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, P=P1) u1=intenergy(Fluid$, T=T1, P=P1) s1=entropy(Fluid$, T=T1, P=P1) v2=v1 "fixed mass, constant volume" u2=intenergy(Fluid$, T=T2, v=v2) s2=entropy(Fluid$, T=T2, v=v2) x2=quality(Fluid$, T=T2, v=v2) "Analysis" "(a)" m=Vol/v1 Q_out=m*(u1-u2) "(b)" W_rev=m*(u1-u2-T0*(s1-s2)) Q_H=COP_HP_rev*W_rev COP_HP_rev=1/(1-(T0/T_air))

9-36 Problem 9-35 is reconsidered. The effect of the final steam temperature in the radiator on the amount of actual heat transfer and the maximum amount of heat that can be transferred is to be investigated. Analysis The problem is solved using EES, and the solution is given below. T_1=200 [C] P_1=200 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-397

V=20 [L] T_2=80 [C] T_o=0 [C] P_o=100 [kPa] T_H = 294 [K] T_L = 273 [K] E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=0 E_out= Q_out u_1 =intenergy(steam_iapws,P=P_1,T=T_1) v_1 =volume(steam_iapws,P=P_1,T=T_1) s_1 =entropy(steam_iapws,P=P_1,T=T_1) v_2 = v_1 u_2 = intenergy(steam_iapws, v=v_2,T=T_2) s_2 = entropy(steam_iapws, v=v_2,T=T_2) m=V*convert(L,m^3)/v_1 W_rev=-m*(u_2 - u_1 -(T_o+273.15)*(s_2-s_1)+P_o*(v_1-v_2)) Q_H = COP_HP*W_rev COP_HP = T_H/(T_H-T_L)

Q out

Wrev

[kJ] 155.4 153.9 151.2 146.9 140.3 130.4 116.1

[kJ] 46.66 45.42 43.72 41.55 38.74 35.09 30.34

[C ] 21 30 40 50 60 70 80

[kJ] 11.1 11 10.8 10.49 10.02 9.318 8.293

6-398

9-37E An insulated rigid tank contains saturated liquid-vapor mixture of water at a specified pressure. An electric heater inside is turned on and kept on until all the liquid is vaporized. The exergy destruction and the second-law efficiency are to be determined. Assumptions Kinetic and potential energies are negligible.

Properties From the steam tables (Tables A-4 through A-6)

v = v f + x1v fg = 0.01708 + 0.25´ (11.901- 0.01708) = 2.9880 ft 3 / lbm P1 = 35 psia ïüï 1 ý u1 = u f + x1u fg = 227.92 + 0.25´ 862.19 = 443.47 Btu / lbm x1 = 0.25 ïïþ s1 = s f + x1s fg = 0.38093 + 0.25´ 1.30632 = 0.70751 Btu / lbm ×R v 2 = v1 ü ïï u2 = u g @v g = 2.9880 ft3 / lbm = 1110.9 Btu / lbm ý sat. vaporïïþ s2 = sg @ = 1.5692 Btu / lbm 3 v g = 2.9880 ft / lbm

Analysis (a) The irreversibility can be determined from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on the tank, which is an insulated closed system, Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Sgen = D Ssystem = m(s2 - s1 ) Substituting,

X destroyed = T0 Sgen = mT0 (s2 - s1 ) = (6 lbm)(535 R)(1.5692-0.70751)Btu/lbm ×R = 2766 Btu (b) Noting that V = constant during this process, the W and Wu are identical and are determined from the energy balance on the closed system energy equation, Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in = D U = m(u2 - u1 ) or,

We,in = (6 lbm)(1110.9 - 443.47) Btu/lbm = 4005 Btu Then the reversible work during this process and the second-law efficiency become

Wrev,in = Wu,in - X destroyed = 4005 - 2766 = 1239 Btu Thus,

hII =

Wrev 1239 Btu = = 30.9% Wu 4005 Btu

6-399

m=6 [lbm] P1=35 [psia] x1=0.25 x2=1 T0=(75+460) [R] P0=14.7 [psia] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) v2=v1 "fixed mass, constant volume" u2=intenergy(Fluid$, v=v2, x=x2) s2=entropy(Fluid$, v=v2, x=x2) "Analysis" "(a)" X_destroyed=T0*S_gen S_gen=m*(s2-s1) "(b)" W_e_in=m*(u2-u1) W_rev_in=W_e_in-X_destroyed eta_II=W_rev_in/W_e_in

9-38 A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensed at constant pressure. The exergy of the refrigerant at the initial and final states, and the exergy destroyed during this process are to be determined. Assumptions The kinetic and potential energies are negligible.

Properties From the refrigerant tables (Tables A-11 through A-13), v1 = 0.034875 m3 /kg P1 = 0.7 MPa ü ïï ý u1 = 274.03 kJ/kg ïïþ T1 = 60 °C s1 = 1.0257 kJ/kg ×K

v @v f @ 20°C =0.0008160 m3 / kg P2 = 0.7 MPa üïï 2 ý u2 @ u f @ 20°C =78.85 kJ/kg ïïþ T2 = 20 °C s2 @ s f @ 20°C =0.30062 kJ/kg ×K v 0 = 0.23373 m3 /kg P0 = 0.1 MPa ü ïï ý u0 = 248.81 kJ/kg ïïþ T0 = 20 °C s0 = 1.0919 kJ/kg ×K

Analysis (a) From the closed system exergy relation, X 1 = mxnonflow,1 = m{(u1 - u0 ) - T0 (s1 - s0 ) + P0 (v1 - v 0 )}

6-400 æ 1 kJ ÷ ö = (8 kg){(274.03 - 248.81)kJ / kg - (293 K)(1.0257 - 1.0919)kJ / kg ×K + (100 kPa)(0.034875 - 0.23373)m 3 / kg çç ÷} = 197.8 kJ çè1 kPa ×m3 ÷ ø

and X 2 = mxnonflow,2 = m{(u2 - u0 ) - T0 ( s2 - s0 ) + P0 (v 2 - v 0 )} æ 1 kJ ö ÷} = 308.6 kJ = (8 kg){(78.85 - 248.81) kJ / kg - (293 K)(0.30062 - 1.0919) kJ / kg ×K + (100 kPa)(0.0008160 - 0.23373)m3 / kg çç ÷ ÷ çè1 kPa ×m3 ø

(b) The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero,

X in - X out - X destroyed ¿ 0 (reversible) = D X system Net exergy transfer by heat, work, and mass

Exergy destruction

Change in exergy

Wrev,in = X 2 - X 1 = 308.6 - 197.8 = 110.8 kJ

Noting that the process involves only boundary work, the useful work input during this process is simply the boundary work in excess of the work done by the surrounding air,

Wu,in = Win - Wsurr,in = Win - P0 (V1 - V2 ) = P(V1 - V2 ) - P0 m(v1 - v 2 ) = m( P - P0 )(v1 - v 2 ) æ 1 kJ ö ÷ = (8 kg)(700 - 100 kPa)(0.034875 - 0.0008160m3 / kg) çç ÷ ÷= 163.5 kJ çè1 kPa ×m3 ø

Knowing both the actual useful and reversible work inputs, the exergy destruction or irreversibility that is the difference between the two is determined from its definition to be

X destroyed = I = Wu,in - Wrev,in = 163.5 - 110.8 = 52.7 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m=8 [kg] P1=700 [kPa] T1=60 [C] P2=P1 T2=20 [C] P0=100 [kPa] T0=20 [C] "Properties" Fluid$='R134a' v0=volume(Fluid$, T=T0, P=P0) u0=intenergy(Fluid$, T=T0, P=P0) s0=entropy(Fluid$, T=T0, P=P0) v1=volume(Fluid$, T=T1, P=P1) u1=intenergy(Fluid$, T=T1, P=P1) s1=entropy(Fluid$, T=T1, P=P1) v2=volume(Fluid$, T=T2, x=0) u2=intenergy(Fluid$, T=T2, x=0) s2=entropy(Fluid$, T=T2, x=0) "Analysis" "(a)" X1=m*(u1-u0-(T0+273)*(s1-s0)+P0*(v1-v0)) X2=m*(u2-u0-(T0+273)*(s2-s0)+P0*(v2-v0)) "(b)" W_rev_in=X2-X1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-401

"(c)" W_u_in=W_in-W_surr_in W_in=m*P2*(v1-v2) W_surr_in=m*P0*(v1-v2) X_destroyed=W_u_in-W_rev_in

9-39 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant expands in a reversible manner until the pressure drops to a specified value. The change in the exergy of the refrigerant during this process and the reversible work are to be determined. Assumptions 1 2 3 4

The kinetic and potential energy changes are negligible. The cylinder is well-insulated and thus heat transfer is negligible. The thermal energy stored in the cylinder itself is negligible. The process is stated to be reversible.

Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1 . From the refrigerant tables (Tables A-11 through A-13),

v = v g @ 0.6 MPa = 0.03433 m3 / kg P1 = 0.6 MPa ïüï 1 ý u1 = u g @ 0.6 MPa = 241.86 kJ/kg ïïþ sat. vapor s1 = sg @ 0.6 MPa = 0.9220 kJ/kg ×K The mass of the refrigerant is

V 0.018 m3 = = 0.5243 kg v1 0.03433 m3 / kg x2 =

s2 - s f s fg

0.9220 - 0.12686 = 0.9754 0.81517

P2 = 0.16 MPa ü ïï 3 ý v 2 = v f + x2v fg = 0.0007435 + 0.9754(0.12355 - 0.0007435) = 0.12053 m /kg ïïþ s2 = s1 u2 = u f + x2 u fg = 31.06 + 0.9754´ 190.31 = 216.69 kJ/kg

The reversible work output, which represents the maximum work output Wrev,out can be determined from the exergy balance by setting the exergy destruction equal to zero,

X in - X out - X destroyed ¿ 0 (reversible) = D X system Net exergy transfer by heat, work, and mass

Exergy destruction

Change in exergy

6-402

Therefore, the change in exergy and the reversible work are identical in this case. Using the definition of the closed system exergy and substituting, the reversible work is determined to be 0 Wrev,out = m( xnonflow,1 - xnonflow,2 ) = m éê(u1 - u2 ) - T0 ( s1 - s2 ) + P0 (v1 - v 2 )ù ú ë û

= m [(u1 - u2 ) + P0 (v1 - v 2 ) ]

= (0.5243 kg)[(241.86 - 216.69) kJ/kg + (100 kPa)(0.03433 - 0.12053) m3 / kg[kJ/kPa ×m3 ]

= 8.68 kJ EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.018 [m^3] P1=600 [kPa] x1=1 P2=160 [kPa] T0=(25+273) [K] P0=100 [kPa] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) s2=s1 "reversible-adiabatic (i.e., isentropic) process" v2=volume(Fluid$, P=P2, s=s2) u2=intenergy(Fluid$, P=P2, s=s2) x2=quality(Fluid$, P=P2, s=s2) "Analysis" m=Vol1/v1 W_rev_out=DELTAX DELTAX=m*(u1-u2-T0*(s1-s2)+P0*(v1-v2))

9-40E A rigid tank is initially filled with saturated mixture of R-134a. Heat is transferred to the tank from a source until the pressure inside rises to a specified value. The amount of heat transfer to the tank from the source and the exergy destroyed are to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There is no heat transfer with the environment. Properties From the refrigerant tables (Tables A-11E through A-13E), u1 = u f + x1u fg = 16.929 + 0.55´ 79.799 = 60.82 Btu / lbm P1 = 30 psia ü ïï ý s1 = s f + x1 s fg = 0.03792 + 0.55´ 0.18595 = 0.1402 Btu / lbm ×R ïïþ x1 = 0.55 v1 = v f + x1v fg = 0.01209 + 0.55´ (1.5506 - 0.01209) = 0.85825 ft 3 / lbm x2 =

v2 - v f v fg

0.85825 - 0.01252 = 0.9030 0.79361- 0.01252

P2 = 50 psia ü ïï ý s2 = s f + x2 s fg = 0.05412 + 0.9030´ 0.16780 = 0.2056 Btu/lbm ×R ïïþ (v 2 = v1 ) u2 = u f + x2 u fg = 24.824 + 0.9030´ 75.228 = 92.76 Btu/lbm

6-403

Analysis (a) The mass of the refrigerant is

V 12 ft 3 = = 13.98 lbm v1 0.85825 ft 3 / lbm

We take the tank as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = D U = m(u2 - u1 )

Substituting,

Qin = m(u2 - u1 ) = (13.98 lbm)(92.76 - 60.82) Btu/lbm = 446.5 Btu (b) The exergy destruction (or irreversibility) can be determined from its definition X destroyed = T0 Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the tank and the region in its immediate surroundings so that the boundary temperature of the extended system where heat transfer occurs is the source temperature,

Sin - Sout + Sgen = D Ssystem , Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin + Sgen = D Ssystem = m(s2 - s1 ) Tb,in Sgen = m( s2 - s1 ) -

Qin Tsource

Substituting, é 446.5 Btu ù ú= 77.8 Btu X destroyed = T0 Sgen = (535 R) ê(13.98 lbm)(0.2056 - 0.1402) Btu/lbm ×R ú 580 R û ëê

EES (Engineering Equation Solver) SOLUTION "Given" Vol=12 [ft^3] P1=30 [psia] x1=0.55 P2=50 [psia] T_source=(120+460) [R] T0=(75+460) [R] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-404

s1=entropy(Fluid$, P=P1, x=x1) v2=v1 "fixed mass, constant volume" u2=intenergy(Fluid$, P=P2, v=v2) s2=entropy(Fluid$, P=P2, v=v2) x2=quality(Fluid$, P=P2, v=v2) "Analysis" "(a)" m=Vol/v1 Q_in=m*(u2-u1) "(b)" X_destroyed=T0*S_gen S_gen=m*(s2-s1)-Q_in/T_source

9-41E Oxygen gas is compressed from a specified initial state to a final specified state. The reversible work and the increase in the exergy of the oxygen during this process are to be determined. Assumptions At specified conditions, oxygen can be treated as an ideal gas with constant specific heats.

Properties The gas constant of oxygen is R = 0.06206 Btu /1bm ×R (Table A-1E). The constant-volume specific heat of oxygen at the average temperature is Tavg = (T1 + T2 ) / 2 = (75 + 525) / 2 = 300°F ¾ ¾ ® cv ,avg = 0.164 Btu/lbm ×R

Analysis The entropy change of oxygen is

æT ö æv ö ÷ ÷ s2 - s1 = cv ,avg ln ççç 2 ÷ + R ln ççç 2 ÷ ÷ ÷ èç T1 ÷ ø èç v1 ø÷ æ1.5 ft 3 /lbm ö æ985 R ÷ ö ÷ ÷ = (0.164 Btu/lbm ×R) ln çç + (0.06206 Btu/lbm ×R) ln ççç ÷ ÷ 3 ÷ èç535 R ø è 12 ft /lbm ÷ ø

= - 0.02894 Btu/lbm ×R The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero,

X in - X out - X destroyedZ 0 (reversible) = D X system Net exergy transfer by heat, work, and mass

Exergy destruction

Wrev,in = X 2 - X 1

Change in exergy

Therefore, the change in exergy and the reversible work are identical in this case. Substituting the closed system exergy relation, the reversible work input during this process is determined to be

wrev,in = xnonflow,2 - xnonflow,1 = - [(u1 - u2 ) - T0 (s1 - s2 ) + P0 (v1 - v 2 )] = - {(0.164 Btu/lbm ×R)(535 - 985) R - (535 R)(0.02894 Btu/lbm ×R)+ (14.7 psia)(12 - 1.5) ft 3 /lbm [Btu/5.4039 psia ×ft 3 ]} = 60.7 Btu / lbm Also, the increase in the exergy of oxygen is

6-405

EES (Engineering Equation Solver) SOLUTION "Given" v1=12 [ft^3/lbm] T1=(75+460) [R] v2=1.5 [ft^3/lbm] T2=(525+460) [R] P0=14.7 [psia] T0=(75+460) [R] "Properties" R=0.06206 [Btu/lbm-R] "Table A-2Ea" c_v=0.164 [Btu/lbm-R] "at (75+525)/2=300 F, Table A-2Eb" "Analysis" DELTAs=c_v*ln(T2/T1)+R*ln(v2/v1) DELTAu=c_v*(T2-T1) DELTAv=(v2-v1) DELTAx=DELTAu-T0*DELTAs+P0*DELTAv*Convert(psia-ft^3, Btu) w_rev_in=DELTAx

9-42 A cylinder initially contains air at atmospheric conditions. Air is compressed to a specified state and the useful work input is measured. The exergy of the air at the initial and final states, and the minimum work input to accomplish this compression process, and the second-law efficiency are to be determined Assumptions 1 Air is an ideal gas with constant specific heats. 2 The kinetic and potential energies are negligible.

Properties The gas constant of air is R = 0.287 kPa.m3 / kg ×K (Table A-1). The specific heats of air at the average temperature of (298 + 423) / 2 = 360 K are c p = 1.009 kJ / kg ×K and cv = 0.722 kJ / kg ×K (Table A-2). Analysis (a) We realize that X1 = F 1 = 0 since air initially is at the dead state. The mass of air is

P1V1 (100 kPa)(0.002 m3 ) = = 0.00234 kg RT1 (0.287 kPa ×m3 / kg ×K)(298 K)

Also,

P2V2 PV PT (100 kPa)(423 K) = 1 1 ¾¾ ® V2 = 1 2 V1 = (2 L)=0.473 L T2 T1 P2T1 (600 kPa)(298 K) and

6-406

= (1.009 kJ/kg ×K) ln

423 K 600 kPa - (0.287 kJ/kg ×K) ln 298 K 100 kPa

= - 0.1608 kJ/kg ×K Thus, the exergy of air at the final state is X 2 = mxnonflow,2 = m éêëcv ,avg (T2 - T0 ) - T0 (s2 - s0 )ù ú û+ P0 (V2 - V0 ) = (0.00234 kg) [(0.722 kJ/kg ×K)(423-298)K - (298 K)( - 0.1608 kJ/kg ×K) ]+ (100 kPa)(0.000473 - 0.002)m3 [kJ/m3 ×kPa]

= 0.171 kJ (b) The minimum work input is the reversible work input, which can be determined from the exergy balance by setting the exergy destruction equal to zero,

X in - X out - X destroyedZ 0 (reversible) = D X system Net exergy transfer by heat, work, and mass

Exergy destruction

Change in exergy

Wrev,in = X 2 - X 1

= 0.171- 0 = 0.171 kJ (c) The second-law efficiency of this process is hII =

Wrev,in Wu,in

0.171 kJ = 14.3% 1.2 kJ

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.002 [m^3] P1=100 [kPa] T1=(25+273) [K] P2=600 [kPa] T2=(150+273) [K] W_u_in=1.2 [kJ] P0=100 [kPa] T0=(25+273) [K] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.009 [kJ/kg-K] "at (298+423)/2=360 K, Table A-2b" c_v=0.722 [kJ/kg-K] "at (298+423)/2=360 K, Table A-2b" "Analysis" "(a)" m=(P1*V1)/(R*T1) V2=(P1/P2)*(T2/T1)*V1 V0=V1 DELTAs=c_p*ln(T2/T0)-R*ln(P2/P0) X2=m*(c_v*(T2-T0)-T0*DELTAs)+P0*(V2-V0) "(b)" X1=0 W_rev_in=X2-X1 "(c)" eta_II=W_rev_in/W_u_in

6-407

9-43 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tank stirs the gas, and the pressure and temperature of CO2 rises. The actual paddle-wheel work and the minimum paddle-wheel work by which this process can be accomplished are to be determined. Assumptions 1 At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at the average temperature. 2

The surroundings temperature is 298 K.

Properties The gas constant of CO2 is 0.1889 kJ / kg K (Table A-1) Analysis (a) The initial and final temperature of CO2 are

T1 =

P1V1 (100 kPa)(0.8 m3 ) = = 275.0 K mR (1.54 kg)(0.1889 kPa ×m3 / kg ×K)

T2 =

P2V2 (135 kPa)(0.8 m3 ) = = 371.3 K mR (1.54 kg)(0.1889 kPa ×m3 / kg ×K)

Tavg = (T1 + T2 ) / 2 = (275.0 + 371.3) / 2 = 323.1 K ¾ ¾ ® cv ,avg = 0.680 kJ/kg ×K (Table A-2b)

The actual paddle-wheel work done is determined from the energy balance on the CO2 gas in the tank, We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wpw,in = D U = mcv (T2 - T1 ) or

Wpw,in = (1.54 kg)(0.680 kJ/kg ×K)(371.3 - 275.0)K = 100.8 kJ (b) The minimum paddle-wheel work with which this process can be accomplished is the reversible work, which can be determined from the exergy balance by setting the exergy destruction equal to zero,

X in - X out -

X destroyedZ 0 (reversible) = D X system

Net exergy transfer by heat, work, and mass

Exergy destruction

Wrev,in = X 2 - X 1

Change in exergy

Substituting the closed system exergy relation, the reversible work input for this process is determined to be 0 Wrev,in = m éê(u2 - u1 ) - T0 ( s2 - s1 ) + P0 (v 2¿ - v1 )ùú ë û

= m éëêcv ,avg (T2 - T1 ) - T0 ( s2 - s1 )ù ú û = (1.54 kg) [(0.680 kJ/kg ×K)(371.3 - 275.0) K - (298 K)(0.2041 kJ/kg ×K) ]

= 7.18 kJ since s2 - s1 = cv ,avg ln

T2 v + R ln 2 T1 v1

¿0

æ371.3 K ÷ ö = (0.680 kJ/kg ×K) ln çç = 0.2041 kJ/kg ×K ÷ çè 275.0 K ÷ ø

6-408

EES (Engineering Equation Solver) SOLUTION "Given" V=0.8 [m^3] m=1.54 [kg] P1=100 [kPa] P2=135 [kPa] T0=298 [K] "Properties" R=0.1889 [kJ/kg-K] "Table A-2a" c_v=0.680 [kJ/kg-K] "at T_avg = (T1+T2)/2=328 K, Table A-2b" "Analysis" "(a)" T1=(P1*V)/(m*R) T2=(P2*V)/(m*R) T_avg=(T1+T2)/2 W_pw_in=m*c_v*(T2-T1) "(b)" DELTAs=c_v*ln(T2/T1) W_rev_in=m*(c_v*(T2-T1)-T0*DELTAs)

9-44 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinder is turned on, and air is heated for 10 min at constant pressure. The exergy destruction during this process is to be determined. Assumptions Air is an ideal gas with variable specific heats.

Properties The gas constant of air is R = 0.287 kJ / kg ×K (Table A-1). Analysis The mass of the air and the electrical work done during this process are

P1V1 (140 kPa)(0.020 m3 ) = = 0.03250 kg RT1 (0.287 kPa ×m3 /kg ×K)(300 K)

We = We D t = (0.100 kJ/s)(10´ 60 s) = 60 kJ Also,

T1 = 300 K ¾ ¾ ® h1 = 300.19 kJ/kg

and

s1o = 1.70202 kJ/kg ×K

The energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in - Wb,out = D U We,in = m(h2 - h1 ) since D U + Wb = D H during a constant pressure quasi-equilibrium process. Thus,

6-409

h2 = h1 +

We,in m

= 300.19 +

T = 1915 K 60 kJ A- 17 = 2146.3 kJ/kg ¾ Table ¾¾ ¾® o2 s2 = 3.7452 kJ/kg ×K 0.03250 kg

Also, ¿0

æP ö ÷ = s2o - s1o = 3.7452 - 1.70202 = 2.0432 kJ/kg ×K s2 - s1 = s - s - R ln ççç 2 ÷ ÷ ÷ çè P ø o 2

o 1

The exergy destruction (or irreversibility) associated with this process can be determined from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on the cylinder, which is an insulated closed system,

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Sgen = D Ssystem = m(s2 - s1 ) Substituting, X destroyed = T0 Sgen = mT0 (s2 - s1 ) = (0.03250 kg)(300 K)(2.0432 kJ/kg ×K) = 19.9 kJ

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.020 [m^3] P1=140 [kPa] T1=(27+273) [K] P2=P1 time=10*60 [s] W_dot_e_in=0.100 [kW] T0=(27+273) [K] P0=100 [kPa] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" m=(P1*V1)/(R*T1) W_e_in=W_dot_e_in*time W_e_in=m*(h2-h1) "since U2-U1+W_b = H2-H1 for a constant pressure process" s2=entropy(Fluid$, P=P2, h=h2) X_destroyed=T0*S_gen S_gen=m*(s2-s1)

9-45 A rigid tank is divided into two equal parts by a partition. One part is filled with compressed liquid water while the other side is evacuated. The partition is removed and water expands into the entire tank. The exergy destroyed during this process is to be determined. Assumptions Kinetic and potential energies are negligible.

6-410

Analysis The properties of the water are (Tables A-4 through A-6)

v @v f @80°C =0.001029 m3 / kg P1 = 200 kPa ïüï 1 ý u1 @ u f @80°C =334.97 kJ/kg ïïþ T1 = 80 °C s1 @ s f @80°C =1.0756 kJ/kg ×K Noting that v 2 = 2v1 = 2´ 0.001029 = 0.002058 m3 / kg

x2 = ïüï ý 3 v 2 = 0.002058 m / kg ïïþ P2 = 40 kPa

v2 - v f v fg

0.002058 - 0.001026 = 0.0002584 3.9933 - 0.001026

u2 = u f + x2 u fg = 317.58 + 0.0002584´ 2158.8 = 318.14 kJ/kg s2 = s f + x2 s fg = 1.0261 + 0.0002584´ 6.6430 = 1.0278 kJ/kg ×K

Taking the direction of heat transfer to be to the tank, the energy balance on this closed system becomes

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = D U = m(u2 - u1 )

Qin = (4 kg)(318.14 - 334.97)kJ/kg= - 67.30 kJ ® Qout = 67.30 kJ The irreversibility can be determined from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times,

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Ssystem = m( s2 - s1 ) Tb,out

Sgen = m( s2 - s1 ) +

Qout Tsurr

Substituting,

æ Q ö ÷ X destroyed = T0 Sgen = T0 çççm( s2 - s1 ) + out ÷ ÷ çè Tsurr ø÷ é 67.30 kJ ù ú = (298 K) ê(4 kg)(1.0278 - 1.0756)kJ/kg ×K+ êë 298 K ú û = 10.3 kJ

6-411

P1=200 [kPa] T1=80 [C] P2=40 [kPa] T0=(25+273) [K] P0=100 [kPa] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=0) u1=intenergy(Fluid$, T=T1, x=0) s1=entropy(Fluid$, T=T1, x=0) v2=2*v1 "equally divided tank" u2=intenergy(Fluid$, P=P2, v=v2) x2=quality(Fluid$, P=P2, v=v2) s2=entropy(Fluid$, P=P2, v=v2) "Analysis" Q_in=m*(u2-u1) "energy balance" Q_in/T0+S_gen=m*(s2-s1) "entropy balance" X_destroyed=T0*S_gen

9-46 Problem 9-45 is reconsidered. The effect of final pressure in the tank on the exergy destroyed during the process is to be investigated. Analysis The problem is solved using EES, and the solution is given below. 800 700

Qout [kJ]

600 500 400 300 200 100 0 5

P2 [kPa]

T_1=80 [C] P_1=200 [kPa] m=4 [kg] P_2=40 [kPa] T_o=25 [C] P_o=100 [kPa] T_surr = T_o E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=0 E_out= Q_out u_1 =intenergy(steam_iapws,P=P_1,T=T_1) v_1 =volume(steam_iapws,P=P_1,T=T_1) s_1 =entropy(steam_iapws,P=P_1,T=T_1) v_2 = 2*v_1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-412

u_2 = intenergy(steam_iapws, v=v_2,P=P_2) s_2 = entropy(steam_iapws, v=v_2,P=P_2) S_in -S_out+S_gen=DELTAS_sys S_in=0 [kJ/K] S_out=Q_out/(T_surr+273) DELTAS_sys=m*(s_2 - s_1) X_destroyed = (T_o+273)*S_gen

P2 [kPa]

X destroyed

Q out [kJ]

5 10 15 20 25 30 35 40 45

[kJ] 74.41 64.4 53.8 43.85 34.61 25.99 17.91 10.3 3.091

788.4 571.9 435.1 332.9 250.5 181 120.7 67.15 18.95

80 70

Xdestroyed [kJ]

60 50 40 30 20 10 0 5

P2 [kPa]

9-47 One side of a partitioned insulated rigid tank contains argon gas at a specified temperature and pressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. The exergy destroyed during this process is to be determined. Assumptions Argon is an ideal gas with constant specific heats, and thus ideal gas relations apply. Properties The gas constant of argon is R = 0.2081 kJ / kg ×K (Table A-1).

6-413

Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = D U = m(u2 - u1 )

u2 = u1

® T2 = T1

since u = u(T) for an ideal gas. The exergy destruction (or irreversibility) associated with this process can be determined from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on the entire tank, which is an insulated closed system, Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Sgen = D Ssystem = m(s2 - s1 ) where

æ T ¿0 V ö V D Ssystem = m( s2 - s1 ) = m çççcv ,avg ln 2 + R ln 2 ÷÷÷= mR ln 2 çè T1 V1 ÷ø V1 = (3 kg)(0.2081 kJ/kg ×K) ln (2) = 0.433 kJ/K Substituting,

X destroyed = T0 Sgen = mT0 ( s2 - s1 ) = (298 K)(0.433 kJ/K) = 129 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m=3 [kg] P1=300 [kPa] T1=(70+273) [K] T0=(25+273) [K] "Properties" R=0.2081 [kJ/kg-K] "Table A-2a" "Analysis" X_destroyed=T0*S_gen S_gen=m*R*ln(V2\V1) "since T2=T1" V2\V1=2

9-48 An iron block and a copper block are dropped into a large lake where they cool to lake temperature. The amount of work that could have been produced is to be determined. Assumptions 1 The iron and copper blocks and water are incompressible substances with constant specific heats at room temperature. 2 Kinetic and potential energies are negligible. Properties The specific heats of iron and copper at room temperature are c p , iron = 0.45 kJ / kg.°C and c p , copper = 0.386 kJ / kg.°C (Table A-3).

6-414

Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature (15C) when the thermal equilibrium is established. We take both the iron and the copper blocks as the system, which is a closed system. The energy balance for this system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = D U iron + D U copper or,

Qout = [mc(T1 - T2 )]iron + [mc(T1 - T2 )]copper Substituting, Qout = (50 kg )(0.45 kJ/kg ×K )(353 - 288)K + (20 kg )(0.386 kJ/kg ×K )(353 - 288)K = 1964 kJ

The work that could have been produced is equal to the wasted work potential. It is equivalent to the exergy destruction (or irreversibility), and it can be determined from its definition X destroyed = T0 Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the blocks and the water in their immediate surroundings so that the boundary temperature of the extended system is the temperature of the lake water at all times, Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Ssystem = D Siron + D Scopper Tb,out

Sgen = D Siron + D Scopper +

Qout Tlake

where

æT ö æ288 K ö÷ ÷ D Siron = mcavg ln ççç 2 ÷ = (50 kg )(0.45 kJ/kg ×K ) ln çç ÷ ÷= - 4.579 kJ/K çè353 K ø÷ èç T1 ø÷ æT ö æ288 K ÷ ö ÷ D Scopper = mcavg ln ççç 2 ÷ = (20 kg )(0.386 kJ/kg ×K ) ln çç = - 1.571 kJ/K ÷ ÷ ÷ ç è353 K ø èç T1 ÷ ø Substituting, æ 1964 kJ ö ÷ X destroyed = T0 Sgen = (293 K) çç- 4.579 - 1.571 + ÷ ÷kJ/K=196 kJ çè 288 K ø

EES (Engineering Equation Solver) SOLUTION "Given" m_iron=50 [kg] m_Cu=20 [kg] T1=(80+273) [K] T_lake=(15+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-415

T0=(20+273) [K] "Properties" c_iron=0.45 [kJ/kg-K] "Table A-3" c_Cu=0.386 [kJ/kg-K] "Table A-3" "Analysis" -Q_out=m_Cu*c_Cu*(T_lake-T1)+m_iron*c_iron*(T_lake-T1) DELTAS_Cu=m_Cu*c_Cu*ln(T_lake/T1) DELTAS_iron=m_iron*c_iron*ln(T_lake/T1) S_gen=DELTAS_Cu+DELTAS_iron+Q_out/T_lake X_destroyed=T0*S_gen

9-49 Carbon steel balls are to be annealed at a rate of 2500 / h by heating them first and then allowing them to cool slowly in ambient air at a specified rate. The total rate of heat transfer from the balls to the ambient air and the rate of exergy destruction due to this heat transfer are to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 There are no changes in kinetic and potential energies. 3 The balls are at a uniform temperature at the end of the process. Properties The density and specific heat of the balls are given to be ρ = 7833 kg / m3 and c p = 0.465 kJ / kg.°C . Analysis (a) We take a single ball as the system. The energy balance for this closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U ball = m(u2 - u1 )

Qout = mc p (T1 - T2 )

The amount of heat transfer from a single ball is m = rV = r

p D3 p (0.008 m)3 = (7833 kg/m3 ) = 0.00210 kg 6 6

Qout = mc p (T1 - T2 ) = (0.0021 kg)(0.465 kJ/kg.°C)(900 - 100)°C = 781 J = 0.781 kJ (per ball)

Then the total rate of heat transfer from the balls to the ambient air becomes Qout = nball Qout = (1200 balls/h) ´ (0.781 kJ/ball) = 936 kJ/h = 260 W

(b) The exergy destruction (or irreversibility) can be determined from its definition X destroyed = T0 Sgen . The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the ball and its immediate surroundings so that the boundary temperature of the extended system is at 35C at all times:

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

6-416

Qout + Sgen = D Ssystem Tb

Sgen =

Qout + D Ssystem Tb

where

T 100+273 D Ssystem = m( s2 - s1 ) = mc p ln 2 = (0.00210 kg)(0.465 kJ/kg.K)ln = - 0.00112 kJ/K T1 900+273 Substituting,

Sgen =

Qout 0.781 kJ + D Ssystem = - 0.00112 kJ/K = 0.00142 kJ/K (per ball) Tb 308 K

Then the rate of entropy generation becomes

Sgen = Sgen nball = (0.00142 kJ/K ×ball)(1200 balls/h)=1.704 kJ/h.K=0.000473 kW/K Finally,

X destroyed = T0 Sgen = (308 K)(0.000473 kW / K) = 0.146 kW=146 W

EES (Engineering Equation Solver) SOLUTION "Given" D=0.008 [m] n_dot_ball=1200 [1/h] T1=(900+273) [K] T2=(100+273) [K] T0=(35+273) [K] rho=7833 [kg/m^3] c_p=0.465 [kJ/kg-C] "Analysis" "(a)" V=pi*D^3/6 m=rho*V Q_out=m*c_p*(T1-T2) Q_dot_total=n_dot_ball*Q_out*1/Convert(W, kJ/h) "(b)" -Q_out/T0+S_gen=DELTAS_sys "entropy balance" DELTAS_sys=m*c_p*ln(T2/T1) S_dot_gen=S_gen*n_dot_ball*1/Convert(W, kJ/h) X_dot_destroyed=T0*S_dot_gen

9-50E A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the work potential wasted during this process are to be determined. Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is no heat transfer. Properties The density and specific heat of water at the anticipated average temperature of 90F are ρ = 62.1 lbm / ft 3 and c p = 1.00 Btu / lbm. F . The specific heat of copper at the anticipated average temperature of 100F is c p = 0.0925 Btu / lbm. F (Table A-3E).

6-417

Analysis (a) We take the entire contents of the tank, water + copper block, as the system, which is a closed system. The energy balance for this system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = DU or

D U Cu + D U water = 0 [mc(T2 - T1 )]Cu + [mc(T2 - T1 )]water = 0 where

mw = r V = (62.1 lbm/ft 3 )(1.2 ft 3 ) = 74.52 lbm Substituting, 0 = (70 lbm)(0.0925 Btu/lbm ×°F)(T2 - 220°F) + (74.52 lbm)(1.0 Btu/lbm ×°F)(T2 - 65°F) T2 = 77.4°F = 537.4 R

(b) The wasted work potential is equivalent to the exergy destruction (or irreversibility), and it can be determined from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on the system, which is an insulated closed system, Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Sgen = D Ssystem = D Swater + D Scopper where

æT ö æ537.4 R ÷ ö ÷ D Scopper = mcavg ln ççç 2 ÷ = (70 lbm)(0.0925 Btu/lbm ×R) ln çç = - 1.5240 Btu/R ÷ ÷ ÷ çè T1 ø÷ èç 680 R ø æT ö æ537.4 R ÷ ö ÷= (74.52 lbm)(1.0 Btu/lbm ×R) ln çç D S water = mcavg ln ççç 2 ÷ ÷= 1.7384 Btu/R ÷ çè T1 ø÷ èç 525 R ÷ ø Substituting,

X destroyed = (525 R)( - 1.5240 + 1.7384)Btu/R = 113 Btu

EES (Engineering Equation Solver) SOLUTION "Given" m_Cu=70 [lbm] T1_Cu=(220+460) [R] V_w=1.2 [ft^3] T1_w=(65+460) [R] T0=(65+460) [R] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-418

"Properties" rho_w=62.1 [lbm/ft^3] "at 90 F, Table A3E" c_w=1.0 [Btu/lbm-F] "Table A-3E" c_Cu=0.0925 [Btu/lbm-F] "at anticipated average temperature of 100 F, Table A-3E" "Analysis" "(a)" m_w=rho_w*V_w m_w*c_w*(T2-T1_w)+m_Cu*c_Cu*(T2-T1_Cu)=0 T2_F=T2-460 "(b)" DELTAS_w=m_w*c_w*ln(T2/T1_w) DELTAS_Cu=m_Cu*c_Cu*ln(T2/T1_Cu) S_gen=DELTAS_w+DELTAS_Cu X_destroyed=T0*S_gen

9-51 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cooked and the amount of exergy destruction associated with this heat transfer process are to be determined. Assumptions 1 The egg is spherical in shape with a radius of r0 = 2.75 cm . 2

The thermal properties of the egg are constant.

Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible.

There are no changes in kinetic and potential energies.

5 The temperature of the surrounding medium is 25C. Properties The density and specific heat of the egg are given to be ρ = 1020 kg / m3 and c p = 3.32 kJ / kg.°C . Analysis We take the egg as the system. This is a closed system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = D U egg = m(u2 - u1 ) = mc(T2 - T1 )

Then the mass of the egg and the amount of heat transfer become m = rV = r

p D3 p (0.055 m)3 = (1020 kg/m3 ) = 0.08886 kg 6 6

Qin = mc p (T2 - T1 ) = (0.08886 kg)(3.32 kJ/kg.°C)(85 - 8)°C = 22.72 kJ The exergy destruction can be determined from its definition X destroyed = T0 Sgen . The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the egg and its immediate surroundings so that the boundary temperature of the extended system is at 97C at all times:

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Qin + Sgen = D Ssystem Tb

Change in entropy

Sgen = -

Qin + D Ssystem Tb

where

T 85 + 273 D Ssystem = m( s2 - s1 ) = mcavg ln 2 = (0.08886 kg)(3.32 kJ/kg.K) ln = 0.07144 kJ/K T1 8 + 273 Substituting, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-419

Sgen = -

Qin 22.72 kJ + D Ssystem = + 0.07144 kJ/K = 0.01005 kJ/K (per egg) Tb 370 K

Finally,

X destroyed = T0 Sgen = (298 K)(0.01005 kJ / K) = 3.00 kJ

EES (Engineering Equation Solver) SOLUTION "Given" D=0.055 [m] T1=(8+273) "ConvertTemp(C, K, 8[C])" T2=(85+273) "ConvertTemp(C, K, 70[C])" T_w=(97+273) "ConvertTemp(C, K, 97[C])" T0=(25+273) "ConvertTemp(C, K, 25)" rho=1020 [kg/m^3] c_p=3.32 [kJ/kg-C] "Analysis" V=pi*D^3/6 m=rho*V Q_in=m*c_p*(T2-T1) Q_in/T_w+S_gen=DELTAS_sys "entropy balance" DELTAS_sys=m*c_p*ln(T2/T1) X_destroyed=T0*S_gen

9-52 Heat is transferred to a piston-cylinder device with a set of stops. The work done, the heat transfer, the exergy destroyed, and the second-law efficiency are to be determined. Assumptions 1 The device is stationary and kinetic and potential energy changes are zero. 2 There is no friction between the piston and the cylinder. 3

Heat is transferred to the refrigerant from a source at 150C.

Analysis (a) The properties of the refrigerant at the initial and final states are (Tables A-11 through A-13)

v = 0.23373 m3 /kg P1 = 100 kPa üïï 1 ý u = 248.81 kJ/kg T1 = 20°C ïïþ 1 s1 = 1.0919 kJ/kg.K v = 0.23669 m3 /kg P2 = 120 kPa üïï 2 ý u = 296.94 kJ/kg T2 = 80°C ïïþ 2 s2 = 1.2419 kJ/kg.K © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-420

Noting that pressure remains constant at 120 kPa as the piston moves, the boundary work is determined to be Wb,out = mP2 (v 2 - v1 ) = (1.4 kg)(120 kPa)(0.23669 - 0.23373)m 3 /kg = 0.497 kJ

(b) The heat transfer can be determined from an energy balance on the system Qin = m(u2 - u1 ) + Wb,out = (1.4 kg)(296.94 - 248.81)kJ/kg + 0.497 kJ = 67.9 kJ

(c) The exergy destruction associated with this process can be determined from its definition X destroyed = T0 Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the piston-cylinder device and the region in its immediate surroundings so that the boundary temperature of the extended system where heat transfer occurs is the source temperature,

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin Qin + Sgen = D Ssystem = m( s2 - s1 ) ¾ ¾ ® Sgen = m( s2 - s1 ) Tb,in Tsource Substituting, é 67.9 kJ ù ú= 14.8 kJ X destroyed = T0 Sgen = (298 K) ê(1.4 kg)(1.2419 - 1.0919) kJ/kg ×K êë 150 + 273 K ú û (d) Exergy expended is the work potential of the heat extracted from the source at 150˚C,

æ T ÷ ö æ 25 + 273 K ÷ ö X expended = X Q = hth,rev Q = ççç1- L ÷ Q = çç1(67.9 kJ) = 20.06 kJ ÷ ÷ ÷ ç çè TH ÷ è 150 + 273 K ø ø Then the 2 nd law efficiency becomes hII =

X destroyed X recovered 14.80 kJ = 1= 1= 0.262 or 26.2% X expended X expended 20.06 kJ

Discussion The second-law efficiency can also be determined as follows: The exergy increase of the refrigerant is the exergy difference between the initial and final states, D X = m [u2 - u1 - T0 ( s2 - s1 ) + P0 (v 2 - v1 ) ] = (1.4 kg) éêë(296.94 - 248.81)kJ/kg - (298 K)(1.2419 - 1.0919)kg.K + (100 kPa)(0.23669 - 0.23373)m 3 /kgù ú û

= 5.177 kJ The useful work output for the process is Wu,out = Wb,out - mP0 (v 2 - v1 ) = 0.497 kJ - (1.4 kg)(100 kPa)(0.23669 - 0.23373)m3 /kg = 0.0828 kJ

The exergy recovered is the sum of the exergy increase of the refrigerant and the useful work output, X recovered = D X + Wu,out = 5.177 + 0.0828 = 5.260 kJ

Then the second-law efficiency becomes

hII =

X recovered 5.260 kJ = = 0.262 or 26.2% X expended 20.06 kJ

EES (Engineering Equation Solver) SOLUTION "GIVEN" m=1.4 [kg] P_1=100 [kPa] T_1=20 [C] P_2=120 [kPa] T_2=80 [C] P_0=100 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-421

T_0=25 [C] T_source=150 [C] "PROPERTIES" Fluid$='R134a' u_1=intenergy(Fluid$, T=T_1, P=P_1) v_1=volume(Fluid$, T=T_1, P=P_1) s_1=entropy(Fluid$, T=T_1, P=P_1) u_2=intenergy(Fluid$, T=T_2, P=P_2) v_2=volume(Fluid$, T=T_2, P=P_2) s_2=entropy(Fluid$, T=T_2, P=P_2) u_0=intenergy(Fluid$, T=T_0, P=P_0) v_0=volume(Fluid$, T=T_0, P=P_0) s_0=entropy(Fluid$, T=T_0, P=P_0) "ANALYSIS" "(a)" W_b_out=P_2*m*(v_2-v_1) "boundary work" "(b)" Q_in-W_b_out=m*(u_2-u_1) "energy balance" "(c)" S_gen=m*(s_2-s_1)-Q_in/(T_source+273) X_dest=(T_0+273)*S_gen "(d)" eta_th_rev=1-(T_0+273)/(T_source+273) X_expended=eta_th_rev*Q_in eta_II=1-X_dest/X_expended "second-law efficiency" "Alternative calculation for second-law efficiency" DELTAX=m*(u_2-u_1-(T_0+273)*(s_2-s_1)+P_0*(v_2-v_1)) W_u_out=W_b_out-m*P_0*(v_2-v_1) X_recovered=DELTAX+W_u_out eta_II_alt=X_recovered/X_expended

Exergy Analysis of Control Volumes

9-53 Steam is throttled from a specified state to a specified pressure. The wasted work potential during this throttling process is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3

The temperature of the surroundings is given to be 25C. 4 Heat transfer is negligible.

Properties The properties of steam before and after the throttling process are (Tables A-4 through A-6)

6-422

P2 = 6 MPa üïï ý ïïþ h2 = h1

s2 = 6.6806 kJ/kg ×K

Analysis The wasted work potential is equivalent to the exergy destruction (or irreversibility). It can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on the device, which is an adiabatic steady-flow system,

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem 0 = 0

Rate of entropy generation

Rate of change of entropy

ms1 - ms2 + Sgen = 0 ®

Sgen = m(s2 - s1 ) or sgen = s2 - s1

Substituting,

xdestroyed = T0 sgen = T0 (s2 - s1 ) = (298 K)(6.6806 - 6.5579) kJ/kg ×K = 36.6 kJ / kg Discussion Note that 36.6 kJ / kg of work potential is wasted during this throttling process.

EES (Engineering Equation Solver) SOLUTION "Given" P1=8000 [kPa] T1=450 [C] P2=6000 [kPa] T0=ConvertTemp(C, K, 25) "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, P=P1) s1=entropy(Fluid$, T=T1, P=P1) h2=h1 "throttling process" s2=entropy(Fluid$, P=P2, h=h2) "Analysis" s_gen=s2-s1 x_destroyed=T0*s_gen

9-54 R-134a is is throttled from a specified state to a specified pressure. The temperature of R-134a at the outlet of the expansion valve, the entropy generation, and the exergy destruction are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat transfer is negligible. (a) The properties of refrigerant at the inlet and exit states of the throttling valve are (from R134a tables)

P1 = 1200 kPa üïï h1 = 117.79 kJ/kg ý ïïþ s1 = 0.42449 kJ/kg ×K x1 = 0 ïüï T2 = - 10.1°C ý h2 = h1 = 117.79 kJ/kg ïïþ s2 = 0.45623 kJ/kg ×K P2 = 200 kPa

(b) Noting that the throttling valve is adiabatic, the entropy generation is determined from sgen = s2 - s1 = (0.45623 - 0.42449)kJ/kg ×K = 0.03174 kJ / kg ×K

Then the irreversibility (i.e., exergy destruction) of the process becomes © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-423

xdest = T0 sgen = (298 K)(0.03174 kJ/kg ×K) = 9.458 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=1200 [kPa] x1=0 P2=200 [kPa] T0=(25+273) [K] "Analysis" "(a)" h1=enthalpy(R134a, P=P1, x=x1) s1=entropy(R134a, P=P1, x=x1) h2=h1 T2=temperature(R134a, P=P2, h=h2) s2=entropy(R134a, P=P2, h=h2) "(b)" s_gen=s2-s1 ex_dest=T0*s_gen

9-55 Air is accelerated in a nozzle while losing some heat to the surroundings. The exit temperature of air and the exergy destroyed during the process are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily. Properties The gas constant of air is R = 0.287 kJ / kg ×K (Table A-1). The properties of air at the nozzle inlet are (Table A17) T1 = 338 K ¾ ¾ ® h1 = 338.40 kJ/kg s1o = 1.8219 kJ/kg ×K

Analysis (a) We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = m(h2 +V22 /2)+Qout

6-424

0 = qout +h2 - h1 +

V22 - V12 2

Therefore, h2 = h1 - qout -

ö V22 - V12 (240 m/s) 2 - (35 m/s) 2 æ çç 1 kJ/kg ÷ = 338.40 - 3 = 307.21 kJ/kg 2 2÷ ÷ ç è ø 2 2 1000 m / s

At this h2 value we read, from Table A-17, T2 = 307.0 K = 34.0°C and s2o = 1.7251 kJ/kg ×K (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation S gen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all times. It gives

Sin - Sout

Rate of net entropy transfer by heat and mass

ms1 - ms2 -

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

Qout + Sgen = 0 Tb,surr

Sgen = m (s2 - s1 )+

Qout Tsurr

where D sair = s2o - s1o - R ln

P2 95 kPa = (1.7251- 1.8219) kJ/kg ×K - (0.287 kJ/kg ×K) ln = 0.1169 kJ/kg ×K P1 200 kPa

Substituting, the entropy generation and exergy destruction per unit mass of air are determined to be

xdestroyed = T0 sgen = Tsurr sgen æ æ ö q ö 3 kJ/kg ÷ ÷ = T0 çççs2 - s1 + surr ÷ = (290 K) çç0.1169 kJ/kg ×K + ÷ ÷ ÷ çè èç Tsurr ø÷ 290 K ø = 36.9 kJ / kg

Alternative solution The exergy destroyed during a process can be determined from an exergy balance applied on the extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is environment temperature T0 (or Tsurr ) at all times. Noting that exergy transfer with heat is zero when the temperature at the point of transfer is the environment temperature, the exergy balance for this steady-flow system can be expressed as

X in - X out Rate of net exergy transfer by heat, work, and mass

- X destroyed = D X system ¿ 0 (steady) = 0 ® X destroyed = X in - X out = mxflow,1 - mxflow,2 = m( xflow,1 - xflow,2 ) Rate of exergy destruction

Rate of change of exergy

= m[(h1 - h2 ) - T0 ( s1 - s2 ) - D ke - D pe ¿ 0 ] = m[T0 ( s2 - s1 ) - (h2 - h1 + D ke)] = m[T0 ( s2 - s1 ) + qout ] since, from energy balance, - qout = h2 - h1 + D ke æ Q ö ÷ = T0 çççm( s2 - s1 ) + out ÷ ÷= T0 Sgen ÷ çè T0 ø Therefore, the two approaches for the determination of exergy destruction are identical.

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [kPa] T1=65 [C] Vel1=35 [m/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-425

P2=95 [kPa] Vel2=240 [m/s] q_surr_in=3 [kJ/kg] T0=(17+273) [K] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) "Analysis" "(a)" h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg)=h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)+q_out q_out=q_surr_in T2=temperature(Fluid$, h=h2) "(b)" s2=entropy(Fluid$, P=P2, h=h2) DELTAs_air=s2-s1 DELTAs_surr=q_surr_in/T0 DELTAs_total=DELTAs_air+DELTAs_surr s_gen=DELTAs_total x_destroyed=T0*s_gen

9-56 Problem 9-55 is reconsidered. The effect of varying the nozzle exit velocity on the exit temperature and exergy destroyed is to be investigated. Analysis The problem is solved using EES, and the solution is given below. P[1] = 200 [kPa] T[1] =65 [C] P[2] = 95 [kPa] Vel[1] = 35 [m/s] Vel[2] = 240 [m/s] T_o = 17 [C] T_surr = T_o q_loss = 3 [kJ/kg] WorkFluid$ = 'Air' h[1]=enthalpy(WorkFluid$,T=T[1]) s[1]=entropy(WorkFluid$,P=P[1],T=T[1]) ke[1] = Vel[1]^2/2 ke[2]=Vel[2]^2/2 h[1]+ke[1]*convert(m^2/s^2,kJ/kg) = h[2] + ke[2]*convert(m^2/s^2,kJ/kg)+q_loss T[2]=temperature(WorkFluid$,h=h[2]) s[2]=entropy(WorkFluid$,P=P[2],h=h[2]) s[1] - s[2] - q_loss/(T_surr+273) + s_gen = 0 x_destroyed = (T_o+273)*s_gen

Vel2 [ m / s] 100 140 180

X destroyed

[C]

[kJ / kg] 58.56 54.32 48.56

57.66 52.89 46.53

6-426

220 260 300

38.58 29.02 17.87

41.2 32.12 21.16

9-57 Steam is accelerated in a nozzle. The second law efficiency of the nozzle is to be determined. Assumptions 1 The diffuser operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A-4 through A-6)

P1 = 500 kPa ïüï h1 = 2855.8 kJ/kg ý ïïþ s1 = 7.0610 kJ/kg ×K T1 = 200°C ïüï h2 = 2706.3 kJ/kg ý x2 = 1 (sat. vapor) ïïþ s2 = 7.1270 kJ/kg ×K

P2 = 200 kPa

6-427

Analysis We take the nozzle to be the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = m(h2 +V22 /2) V22 - V12 = h1 - h2 = D keactual 2

Substituting,

D keactual = h1 - h2 = 2855.8 - 2706.3 = 149.5 kJ/kg An exergy balance on the nozzle gives

- X destroyed 0 (reversible) = D X system 0 (steady) = 0

X in - X out Rate of net exergy transfer by heat, work, and mass

Rate of exergy destruction

Rate of change of exergy

X in = X out mxflow,1 = mxflow,2

h1 - h0 +

V12 V2 - T0 ( s1 - s0 ) = h2 - h0 + 2 - T0 (s2 - s0 ) 2 2 V22 - V12 = h1 - h2 - T0 ( s1 - s2 ) 2 D kerev = h1 - h2 - T0 (s1 - s2 )

mxflow,1 = mxflow,2

h1 - h0 +

V12 V2 - T0 ( s1 - s0 ) = h2 - h0 + 2 - T0 (s2 - s0 ) 2 2 V22 - V12 = h1 - h2 - T0 ( s1 - s2 ) 2 D kerev = h1 - h2 - T0 (s1 - s2 )

Substituting, D kerev = h1 - h2 - T0 (s1 - s2 )

= (2855.8 - 2706.3)kJ/kg - (298 K)(7.0610 - 7.1270) kJ/kg ×K = 169.2 kJ/kg The second law efficiency is then hII =

D keactual 149.5 kJ/kg = = 0.884 = 88.4% D ke rev 169.2 kJ/kg

6-428

EES (Engineering Equation Solver) SOLUTION "Given" P1=500 [kPa] T1=200 [C] Vel1=30 [m/s] P2=200 [kPa] x2=1 T0=(25+273) [K] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) "Analysis" DELTAke_actual=h1-h2 DELTAke_rev=h1-h2-T0*(s1-s2) eta_II=DELTAke_actual/DELTAke_rev

9-58 Steam is decelerated in a diffuser. The mass flow rate of steam and the wasted work potential during the process are to be determined. Assumptions 1 The diffuser operates steadily. 2 The changes in potential energies are negligible.

Properties The properties of steam at the inlet and the exit of the diffuser are (Tables A-4 through A-6)

P1 = 10 kPa ïüï h1 = 2611.2 kJ/kg ý T1 = 60°C ïïþ s1 = 8.2326 kJ/kg ×K h2 = 2591.3 kJ/kg T2 = 50°C ïü ïý s = 8.0748 kJ/kg ×K 2 sat.vapor ïïþ v 2 = 12.026 m3 /kg

Analysis (a) The mass flow rate of the steam is m=

1 1 A2V2 = (3 m 2 )(70 m/s) = 17.46 kg / s v2 12.026 m 3 / kg

(b) We take the diffuser to be the system, which is a control volume. Assuming the direction of heat transfer to be from the stem, the energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem ¿ 0 (steady)

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

m(h1 + V12 / 2) = m(h2 +V22 /2) + Qout

6-429

æ V 2 - V12 ö÷ ÷÷ Qout = - m çççh2 - h1 + 2 çè 2 ÷ø Substituting,

é öù (70 m/s) 2 - (375 m/s) 2 æ çç 1 kJ/kg ÷ Qout = - (17.46 kg/s) ê2591.3 - 2611.2 + ÷ú= 1532 kJ/s 2 2 ê èç1000 m / s ø÷úû 2 ë The wasted work potential is equivalent to exergy destruction. The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation S gen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all times. It gives

Sin - Sout Rate of net entropy transfer by heat and mass

ms1 - ms2 -

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

Qout Q + Sgen = 0 ® Sgen = m (s2 - s1 )+ out Tb,surr Tsurr

Substituting, the exergy destruction is determined to be

æ Q ö X destroyed = T0 Sgen = T0 çççm( s2 - s1 ) + out ÷÷÷ çè T0 ø÷ æ 1532 kW ö ÷ = (298 K) çç(17.46 kg/s)(8.0748 - 8.2326)kJ/kg ×K + ÷ ÷ çè 298 K ø

= 711.1 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=10 [kPa] T1=60 [C] Vel1=375 [m/s] T2=50 [C] x2=1 Vel2=70 [m/s] A2=3 [m^2] T0=(25+273) [K] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, T=T2, x=x2) s2=entropy(Fluid$, T=T2, x=x2) v2=volume(Fluid$, T=T2, x=x2) "Analysis" "(a)" m_dot=1/v2*A2*Vel2 "(b)" m_dot*(h1+Vel1^2/2*Convert(m^2/s^2, kJ/kg))-Q_dot_out=m_dot*(h2+Vel2^2/2*Convert(m^2/s^2, kJ/kg)) "energy balance" m_dot*s1-m_dot*s2-Q_dot_out/T0+S_dot_gen=0 "entropy balance on extended stystem" X_dot_destroyed=T0*S_dot_gen

6-430

9-59 Argon enters an adiabatic compressor at a specified state, and leaves at another specified state. The reversible power input and irreversibility are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with constant specific heats. Properties For argon, the gas constant is R = 0.2081 kJ / kg ×K ; the specific heat ratio is k = 1.667; the constant pressure specific heat is c p = 0.5203 kJ / kg ×K (Table A-2). Analysis The mass flow rate, the entropy change, and the kinetic energy change of argon during this process are

v1 =

RT1 (0.2081 kPa ×m3 / kg ×K)(303 K) = = 0.5255 m3 /kg P1 (120 kPa)

1 1 AV (0.0130 m 2 )(20 m/s) = 0.495 kg/s 1 1 = v1 0.5255 m3 / kg

s2 - s1 = c p ln

T2 P - R ln 2 T1 P1

= (0.5203 kJ/kg ×K)ln

803 K 1200 kPa - (0.2081 kJ/kg ×K)ln 303 K 120 kPa

= 0.02793 kJ/kg ×K and Δke =

ö V22 - V12 (80 m/s) 2 - (20 m/s) 2 æ çç 1 kJ/kg ÷ = = 3.0 kJ/kg 2 2÷ ÷ ç è1000 m / s ø 2 2

The reversible (or minimum) power input is determined from the rate form of the exergy balance applied on the compressor, and setting the exergy destruction term equal to zero,

X in - X out

- X destroyed Z 0 (reversible) = D X systemZ 0 (steady) = 0

Rate of net exergy transfer by heat, work, and mass

Rate of exergy destruction

Rate of change of exergy

X in = X out

mxflow,1 + Wrev,in = mxflow,2 Wrev,in = m( xflow,2 - xflow,1 ) = m[(h2 - h1 ) - T0 ( s2 - s1 ) + Δke + ΔpeZ 0 ]

Substituting,

6-431

= (0.495 kg/s) [(0.5203 kJ/kg ×K)(530 - 30) K - (298 K)(0.02793 kJ/kg ×K) + 3.0] = 126 kW

The exergy destruction (or irreversibility) can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation is determined from an entropy balance on the system, which is an adiabatic steady-flow device,

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

ms1 - ms2 + Sgen = 0 ® Sgen = m(s2 - s1 ) Substituting, X destroyed = T0 m( s2 - s1 ) = (298 K)(0.495 kg/s)(0.02793 kJ/kg ×K) = 4.12 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=120 [kPa] T1=ConvertTemp(C, K, 30[C]) Vel1=20 [m/s] P2=1200 [kPa] T2=ConvertTemp(C, K, 530[C]) Vel2=80 [m/s] A1=130E-4 [m^2] T0=ConvertTemp(C, K, 25) "Properties, from Table A-2 of the text" C_p=0.5203 [kJ/kg-K] k=1.667 R=0.2081 [kJ/kg-K] "Analysis" v1=(R*T1/P1) m_dot=1/v1*A1*Vel1 DELTAh=C_p*(T2-T1) DELTAs=C_p*ln(T2/T1)-R*ln(P2/P1) DELTAke=(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) W_dot_rev_in=m_dot*(DELTAh-T0*DELTAs+DELTAke) S_dot_gen=m_dot*DELTAs X_dot_destroyed=T0*S_dot_gen

9-60E Air is compressed in a compressor from a specified state to another specified state. The minimum work input is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. 4 The inlet state of air is taken as the dead state.

6-432

Properties The properties of air at room temperature are c p = 0.240 Btu / lbm ×R and R = 0.06855 Btu /1bm ×R (Table A2Ea). Analysis The entropy change of air is

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1 = (0.240 Btu/lbm ×R) ln

660 R 140 psia - (0.06855 Btu/lbm ×R) ln 537 R 14.7 psia

= - 0.1050 Btu/lbm ×R The minimum (reversible) work is the exergy difference between the inlet and exit states

wrev,in = c p (T2 - T1 ) - T0 ( s2 - s1 ) = (0.240 Btu/lbm ×R)(200 - 77)R - (537 R)( - 0.1050 Btu/lbm ×R) = 85.9 Btu / lbm That is, a minimum of 85.9 Btu/lbm of work input is needed to change the state of air from 14.7 psia and 77°F to 140 psia, 200°F since the exergy of the exit state is 85.9 Btu/lbm higher than the exergy of the inlet state. A compression process that involves minimum work input is necessarily reversible (does not involve any irreversibilities, and thus the entropy generation and exergy destruction are necessarily zero. This also means that if the air undergoes a change from the exit state back to the inlet state, it will produce a maximum of 85.9 Btu/lbm work. Again this will be realized when the process is reversible. The compressor does not need to be adiabatic for the work input to be a minimum. The only requirement is for the process to be reversible. A reversible process may involve heat transfer provided that it occurs through a reversible heat engine or refrigerator so that there is an infinitesimal temperature difference during heat transfer. In fact, the negative entropy change for this process indicates that there is heat loss during this process. Discussion The amount of heat loss from the air during this reversible steady-flow compression process can be determined from the energy balance as follows:

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout h1 + win = h2 + qout win = h2 - h1 + qout qout = (h1 – h2) + win = cp(T1 -T2) – wrev,in = (0.240 Btu/lbm.R)(77 – 200)R – (85.9 Btu/lbm) = 56.4 Btu/lbm That is, for the specified exit state, heat is being lost from the compressed air at a rate of 56.4 Btu/lbm. Considering that the entropy generation (and thus exergy destruction) must be zero for a reversible process, this heat transfer to the surroundings must be occurring through a reversible heat engine. In reality, this is not practical, and heat will be lost to the surroundings directly via a finite temperature difference, rendering the process irreversible. Considering an extended system that borders the surrounding air at the surrounding temperature (= inlet temperature) and assuming the inner surface temperature of the compressor to be the average temperature between the inlet and the exit, the exergy destruction associated with this irreversible heat transfer process is xdest,heat loss = Tsurrsgen,heat loss =Tsurr(qloss/Tsurr – qloss/Tair,avg) = qloss(1 – Tsurr /Tair,avg) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-433

= (56.4 Btu/lbm)(1 – 537/[(537+660)/2]) = 5.80 Btu/lbm That is, an internally reversible but externally irreversible compressor will waste 5.80 Btu/lbm work potential instead of converting it to work, and thus will require a total of 85.9 + 5.8 = 91.7 Btu/lbm of work input compared to a totally reversible compressor which requires only 85.9 Btu/lbm of work input for the specified inlet and exit states.

9-61 Air is compressed steadily by a compressor from a specified state to another specified state. The reversible power is to be determined.

Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. 1) Properties The gas constant of air is R = 0.287 kJ / kg ×K (Table A-1). From the air table (Table A-17) T1 = 300 K

¾¾ ®

h1 = 300.19 kJ/kg s1o = 1.702 kJ/kg ×K

T2 = 493 K

¾¾ ®

h2 = 495.82 kJ/kg s2o = 2.20499 kJ/kg ×K

Analysis The reversible (or minimum) power input is determined from the rate form of the exergy balance applied on the compressor and setting the exergy destruction term equal to zero,

X in - X out Rate of net exergy transfer by heat, work, and mass

- X destroyed 0 (reversible) = D X system 0 (steady) = 0 Rate of exergy destruction

Rate of change of exergy

X in = X out mxflow,1 + Wrev,in = mxflow,2 Wrev,in = m( xflow,2 - xflow,1 ) = m[(h2 - h1 ) - T0 ( s2 - s1 ) + Δke 0 + Δpe 0 ]

where s2 - s1 = s2o - s1o - R ln

P2 P1

= (2.205 - 1.702) kJ/kg ×K - (0.287 kJ/kg ×K)ln

400 kPa 101 kPa

6-434

Substituting, Wrev,in = (0.15 kg/s) [(495.82 - 300.19) kJ/kg - (298 K)(0.1080 kJ/kg ×K) ]= 24.5 kW

Discussion Note that a minimum of 24.5 kW of power input is needed for this compression process.

EES (Engineering Equation Solver) SOLUTION "Given" P1=101 [kPa] T1=27 [C] P2=400 [kPa] T2=220 [C] m_dot=0.15 [kg/s] T0=ConvertTemp(C, K, 25) "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) h2=enthalpy(Fluid$, T=T2) s2=entropy(Fluid$, T=T2, P=P2) "Analysis" W_dot_rev_in=m_dot*(h2-h1-T0*(s2-s1))

9-62 Problem 9-61 is reconsidered. The effect of compressor exit pressure on reversible power is to be investigated. Analysis The problem is solved using EES, and the solution is given below. T_1=27 [C] P_1=101 [kPa] m_dot = 0.15 [kg/s] P_2=400 [kPa] T_2=220 [C] T_o=25 [C] P_o=100 [kPa] m_dot_in=m_dot m_dot_in = m_dot_out h_1 =enthalpy(air,T=T_1) h_2 = enthalpy(air, T=T_2) W_dot_rev=m_dot_in*(h_2 - h_1 -(T_1+273.15)*(s_2-s_1)) s_1=entropy(air, T=T_1,P=P_1) s_2=entropy(air,T=T_2,P=P_2)

6-435 30 28

Wrev [kW]

26 24 22 20 18 16 200

250

300

350

400

450

500

550

600

P2 [kPa]

P2 [kPa] 200 250 300 350 400 450 500 550 600

Wrev [kW] 15.55 18.44 20.79 22.79 24.51 26.03 27.4 28.63 29.75

9-63 Refrigerant-134a is compressed by an adiabatic compressor from a specified state to another specified state. The minimum power required is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.

6-436

Properties From the refrigerant tables (Tables A-11 through A-13)

üïï ý x1 = 1 (sat. vap.) ïïþ

P1 = 160 kPa

P2 = 800 kPa ïüï ý ïïþ T2 = 50°C

h1 = 241.14 kJ/kg s1 = 0.94202 kJ/kg ×K h2 = 286.71 kJ/kg s2 = 0.9803 kJ/kg ×K

The minimum (reversible) power is Wrev,in = m [h2 - h1 - T0 ( s2 - s1 )] = (0.1 kg/s) [(286.71- 241.14)kJ/kg - (298 K)(0.9803 - 0.94202) kJ/kg ×K ]

= 3.42 kW

9-64 Refrigerant-134a is compressed by an adiabatic compressor from a specified state to anothe

r specified state. The isentropic efficiency and the second-law efficiency of the compressor are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.

Properties From the refrigerant tables (Tables A-11E through A-13E)

h = 246.37 kJ/kg P1 = 140 kPa üïï 1 ý s1 = 0.9724 kJ/kg ×K T1 = - 10°C ïïþ v1 = 0.14605 m3 /kg

P2 = 700 kPa ïüï h2 = 298.44 kJ/kg ý ïïþ s2 = 1.0257 kJ/kg ×K T2 = 60°C

P2 s = 700 kPa üïï ý h2 s = 281.20 kJ/kg ïïþ s2 s = s1 Analysis (a) The isentropic efficiency is hc =

h2 s - h1 281.20 - 246.37 = = 0.669 = 66.9% h2 a - h1 298.44 - 246.37

6-437

(b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

(steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Wa,in + mh1 = mh2

(since Q @D ke @D pe @0)

Wa ,in = m(h2 - h1 )

Then the mass flow rate of the refrigerant becomes m=

Wa,in h2 a - h1

0.5 kJ/s = 0.009602 kg/s (298.44 - 246.37) kJ/kg

The reversible (or minimum) power input is determined from the exergy balance applied on the compressor and setting the exergy destruction term equal to zero,

- X destroyed 0 (reversible) = D X system 0 (steady) = 0

X in - X out Rate of net exergy transfer by heat, work, and mass

Rate of exergy destruction

Rate of change of exergy

X in = X out Wrev,in + mxflow,1 = mxflow,2

Wrev,in = m( xflow,2 - xflow,1 ) = m[(h2 - h1 ) - T0 ( s2 - s1 ) + Δke 0 + Δpe 0 ]

Substituting, Wrev,in = (0.009602 kg/s) [(298.44 - 246.37) kJ/kg - (300 K)(1.0257 - 0.9724) kJ/kg ×K ]= 0.3465 kW

and

hII =

Wrev,in Wa,in

0.3465 kW = 0.693 = 69.3% 0.5 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=140 [kPa] T1=-10 [C] P2=700 [kPa] T2=60 [C] W_dot_in=0.5 [kW] T0=ConvertTemp(C, K, 27) "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, T=T1) h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, T=T2) s2=entropy(Fluid$, P=P2, T=T2) s2_s=s1 "isentropic process" h2_s=enthalpy(Fluid$, P=P2, s=s2_s) "Analysis" "(a)" eta_C=(h2_s-h1)/(h2-h1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-438

"(b)" W_dot_in=m_dot*(h2-h1) W_dot_rev_in=m_dot*(h2-h1-T0*(s2-s1)) eta_II=W_dot_rev_in/W_dot_in

9-65 Air is compressed in a compressor that is intentionally cooled. The actual and reversible power inputs, the second law efficiency, and the mass flow rate of cooling water are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Air is an ideal gas with constant specific heats. Properties The gas constant of air is R = 0.287 kJ / kg ×K and the specific heat of air at room is c p = 1.005 kJ / kg ×K . the specific heat of water at room temperature is cw = 4.18 kJ / kg ×K (Tables A-2, A-3). Analysis (a) The mass flow rate of air is

m = r V1 =

P1 100 kPa V1 = (6.2 m3 /s) = 7.373 kg/s RT1 (0.287 kJ/kg.K)(20 + 273 K)

The power input for a reversible-isothermal process is given by

Wrev = mRT1 ln

æ900 kPa ÷ ö P2 = (7.373 kg/s)(0.287 kJ/kg.K)(20 + 273 K)ln çç = 1362 kW ÷ ÷ ç è100 kPa ø P1

Given the isothermal efficiency, the actual power may be determined from Wactual =

Wrev 1362 kW = = 1946 kW hT 0.70

(b) The given isothermal efficiency is actually the second-law efficiency of the compressor

hII = hT = 0.70 (c) An energy balance on the compressor gives

é V 2 - V22 ùú Qout = m êc p (T1 - T2 ) + 1 + Wactual,in ê 2 úû ë é öù 0 - (80 m/s) 2 æ çç 1 kJ/kg ÷÷ú+ 1946 kW = (7.373 kg/s) ê(1.005 kJ/kg.°C)(20 - 60)°C + 2 2 ÷ ê èç1000 m /s øúû 2 ë = 1626 kW © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-439

The mass flow rate of the cooling water is

mw =

Qout 1626 kW = = 38.9 kg / s cwD T (4.18 kJ/kg ×°C)(10°C)

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=100 [kPa] T1=(20+273) [K] Vel1=0 [m/s] V_dot=6.2 [m^3/s] P2=900 [kPa] T2=(60+273) [K] Vel2=80 [m/s] DELTAT_w=10 [C] eta_t=0.70 "PROPERTIES" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" c_w=4.18 [kJ/kg-K] "Table A-3" "ANALYSIS" "(a)" rho=P1/(R*T1) m_dot=rho*V_dot W_dot_rev=m_dot*R*T1*ln(P2/P1) "reversible isothermal power" W_dot_actual=W_dot_rev/eta_t "actual power" "(b)" eta_II=W_dot_rev/W_dot_actual "second-law efficiency" "(c)" Q_dot_out=W_dot_actual+m_dot*(c_p*(T1-T2)+(Vel1^2-Vel2^2)/2*Convert(m^2/s^2, kJ/kg)) "energy balance" m_dot_w=Q_dot_out/(c_w*DELTAT_w)

9-66 Combustion gases expand in a turbine from a specified state to another specified state. The exergy of the gases at the inlet and the reversible work output of the turbine are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The temperature of the surroundings is given to be 25C. 4 The combustion gases are ideal gases with constant specific heats. Properties The constant pressure specific heat and the specific heat ratio are given to be c p = 1.15 kJ / kg ×K and k = 1.3. The gas constant R is determined from R = c p - cv = c p - c p / k = c p (1- 1/ k ) = (1.15 kJ/kg ×K)(1- 1/1.3) = 0.265 kJ/kg ×K

6-440

Analysis (a) The exergy of the gases at the turbine inlet is simply the flow exergy, xflow,1 = h1 - h0 - T0 ( s1 - s0 ) +

0 V12 + gz1¿ 2

where

T P s1 - s0 = c p ln 1 - R ln 1 T0 P0

= (1.15 kJ/kg ×K)ln

1173 K 800 kPa - (0.265 kJ/kg ×K)ln 298 K 100 kPa

= 1.025 kJ/kg ×K Thus,

xflow,1 = (1.15 kJ/kg.K)(900 - 25)°C - (298 K)(1.025 kJ/kg ×K) +

ö (100 m/s) 2 æ çç 1 kJ/kg ÷ = 705.8 kJ / kg 2 2÷ ÷ ç è1000 m / s ø 2

(b) The reversible (or maximum) work output is determined from an exergy balance applied on the turbine and setting the exergy destruction term equal to zero,

X in - X out Rate of net exergy transfer by heat, work, and mass

- X destroyed ¿ 0 (reversible) = D X system ¿ 0 (steady) = 0 Rate of exergy destruction

Rate of change of exergy

X in = X out mxflow,1 = Wrev,out + mxflow,2 Wrev,out = m( xflow,1 - xflow,2 ) = m[(h1 - h2 ) - T0 ( s1 - s2 ) - Δke - Δpe ¿ 0 ]

where

D ke =

ö V22 - V12 (220 m/s) 2 - (100 m/s) 2 æ çç 1 kJ/kg ÷ = = 19.2 kJ/kg 2 2÷ ÷ ç è1000 m / s ø 2 2

and

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1 = (1.15 kJ/kg ×K)ln

923 K 400 kPa - (0.265 kJ/kg ×K)ln 1173 K 800 kPa

= - 0.09196 kJ/kg ×K Then the reversible work output on a unit mass basis becomes wrev,out = h1 - h2 + T0 ( s2 - s1 ) - D ke = c p (T1 - T2 ) + T0 ( s2 - s1 ) - D ke

6-441

= 240.9 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=(900+273) [K] P1=800 [kPa] Vel1=100 [m/s] T2=(650+273) [K] P2=400 [kPa] Vel2=220 [m/s] T0=(25+273) [K] P0=100 [kPa] c_p=1.15 [kJ/kg-C] k=1.3 "Analysis" R=c_p*(1-1/k) "(a)" x1=c_p*(T1-T0)-T0*(c_p*ln(T1/T0)-R*ln(P1/P0))+Vel1^2/2*Convert(m^2/s^2, kJ/kg) "(b)" -w_rev_out=c_p*(T2-T1)-T0*(c_p*ln(T2/T1)-R*ln(P2/P1))+(Vel2^2/2-Vel1^2/2)*Convert(m^2/s^2, kJ/kg)

9-67 Steam expands in a turbine, which is not insulated. The reversible power, the exergy destroyed, the second-law efficiency, and the possible increase in the turbine power if the turbine is well insulated are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible.

Analysis (a) The properties of the steam at the inlet and exit of the turbine are (Tables A-4 through A-6)

P1 = 9 MPa ïü ïý h1 = 3634.1 kJ/kg T1 = 600°C ïïþ s1 = 6.9605 kJ/kg.K P2 = 20 kPa ïü ïý h2 = 2491.1 kJ/kg ïïþ s2 = 7.5535 kJ/kg.K x2 = 0.95 The enthalpy at the dead state is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-442

T0 = 25°C ïü ïý h = 104.83 kJ/kg ïïþ 0 x= 0

The mass flow rate of steam may be determined from an energy balance on the turbine æ æ V2ö V2ö ÷= m ççh2 + 2 ÷ ÷+ Qout + Wa m çççh1 + 1 ÷ ÷ ÷ ÷ ÷ çè 2ø 2ø èçç 2 é é ù öù æ ö (60 m/s) 2 æ ÷ çç 1 kJ/kg ÷ ú= m ê2491.1 kJ/kg + (130 m/s) çç 1 kJ/kg ú m ê3634.1 kJ/kg + ® m = 4.137 ÷ ÷ 2 2 2 2 ÷ú ÷ú+ 220 kW + 4500 kW ¾ ¾ ê ê èç1000 m /s ø èç1000 m /s ø 2 2 ë û ë û

The reversible power may be determined from é V 2 -V 2 ù Wrev = m êh1 - h2 - T0 ( s1 - s2 ) + 1 2 ú ê 2 ú ë û

é öù (60 m/s) 2 - (130 m/s) 2 æ çç 1 kJ/kg ÷ ú = (2.693) ê(3634.1- 2491.1) - (298)(6.9605-7.5535) + ÷ 2 2 ê èç1000 m /s ÷ øú 2 ë û = 5451 kW

(b) The exergy destroyed in the turbine is

X dest = Wrev - Wa = 5451- 4500 = 951 kW (c) The second-law efficiency is

hII =

Wa 4500 kW = = 0.826 = 82.6% Wrev 5451 kW

(d) The energy of the steam at the turbine inlet in the given dead state is Q = m(h1 - h0 ) = (4.137 kg/s)(3634.1-104.83)kJ/kg = 14, 602 kW

The fraction of energy at the turbine inlet that is converted to power is

f =

Wa 4500 kW = = 0.3082 Q 14, 602 kW

Assuming that the same fraction of heat loss from the turbine could have been converted to work, the possible increase in the power if the turbine is to be well-insulated becomes

Wincrease = fQout = (0.3082)(220 kW) = 67.8 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[1]=9000 [kPa] T[1]=600 [C] Vel[1]=60 [m/s] P[2]=20 [kPa] Vel[2]=90 [m/s] x_2=0.95 Q_dot_out=220 [kW] W_dot_actual=4500 [kW] T[0]=25 [C] "PROPERTIES" Fluid$='Steam_iapws' h[0]=enthalpy(Fluid$, T=T[0], x=0) h[1]=enthalpy(Fluid$, P=P[1], T=T[1]) s[1]=entropy(Fluid$, P=P[1], T=T[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-443

h[2]=enthalpy(Fluid$, P=P[2], x=x_2) s[2]=entropy(Fluid$, P=P[2], x=x_2) "Analysis" "(a)" m_dot*h[1]+m_dot*Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)=m_dot*h[2]+m_dot*Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)+Q_dot_out+W_dot_actual "energy balance" W_dot_rev=m_dot*(h[1]-h[2]-(T[0]+273)*(s[1]-s[2])+(Vel[1]^2-Vel[2]^2)/2*Convert(m^2/s^2, kJ/kg)) "reversible power" "(b9" EX_destroyed=W_dot_rev-W_dot_actual "exergy destroyed" "(c)" eta_II=W_dot_actual/W_dot_rev "second-law efficiency" "(d)" Q_dot_max=m_dot*(h[1]-h[0]) "energy of steam at the turbine inlet in the given environment" f=W_dot_actual/Q_dot_max "fraction of steam energy converted to power" W_dot_increase=f*Q_dot_out "estimated possible increase of turbine power" "Alternative Solution" S_dot_gen=m_dot*(s[2]-s[1])+Q_dot_out/(T[0]+273) EX_destroyed_alt=(T[0]+273)*S_dot_gen W_dot_rev_alt=W_dot_actual+EX_destroyed_alt

9-68 Refrigerant-134a is condensed in a refrigeration system by rejecting heat to ambient air. The rate of heat rejected, the COP of the refrigeration cycle, and the rate of exergy destruction are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of refrigerant at the inlet and exit states of the condenser are (from R134a tables)

P1 = 700 kPa ïüï h1 = 288.54 kJ/kg ý T1 = 50°C ïïþ s1 = 0.9955 kJ/kg ×K P2 = 700 kPa ïüï h2 = 88.82 kJ/kg ý ïïþ s2 = 0.3323 kJ/kg ×K x2 = 0 The rate of heat rejected in the condenser is QH = mR (h1 - h2 ) = (0.05 kg/s)(288.54 - 88.82)kJ/kg = 9.986 kW

(b) From the definition of COP for a refrigerator, COP =

QL QL 6 kW = = = 1.505 (9.986 - 6) kW Win QH - QL

QH TH

= (0.05 kg/s)(0.3323 - 0.9955) kJ/kg ×K +

9.986 kW 298 K

= 0.0003517 kW/K

6-444

EES (Engineering Equation Solver) SOLUTION "Given" P1=700 [kPa] T1=50 [C] P2=P1 x2=0 m_dot=0.05 [kg/s] Q_dot_L=6 [kW] T0=(25+273) [K] "Analysis" h1=enthalpy(R134a, P=P1, T=T1) s1=entropy(R134a, P=P1, T=T1) h2=enthalpy(R134a, P=P2, x=x2) s2=entropy(R134a, P=P2, x=x2) "(a)" Q_dot_H=m_dot*(h1-h2) "(b)" COP=Q_dot_L/(Q_dot_H-Q_dot_L) "(c)" T_H=T0 S_dot_gen=m_dot*(s2-s1)+Q_dot_H/T_H Ex_dot_dest=T0*S_dot_gen "Exergy efficiency calculation" DELTAEx_dot=m_dot*((h1-h2)-T0*(s1-s2)) eta_ex=1-Ex_dot_dest/DELTAEx_dot

9-69 Refrigerant-134a is vaporized by air in the evaporator of an air-conditioner. For specified flow rates, the exit temperature of air and the rate of exergy destruction are to be determined for the cases of insulated and uninsulated evaporator. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 Air is an ideal gas with constant specific heats at room temperature.

Properties The gas constant of air is 0.287 kPa.m3 / kg ×K (Table A-1). The constant pressure specific heat of air at room temperature is c p = 1.005 kJ / kg ×K (Table A-2). The properties of R-134a at the inlet and the exit states are (Tables A-11 through A-13)

P1 = 120 kPa ü ïï h1 = h f + x1h fg = 22.47 + 0.3´ 214.52 = 86.83 kJ/kg ý ïïþ s1 = s f + x1 s fg = 0.09269 + 0.3(0.85520) = 0.34925 kJ/kg ×K x1 = 0.3 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-445

T2 = 120 kPa ü ïï h2 = hg @120 kPa = 236.99 kJ/kg ý ïïþ s2 = sg @120 kPa = 0.94789 kJ/kg ×K sat. vapor Analysis Air at specified conditions can be treated as an ideal gas with specific heats at room temperature. The properties of the refrigerant are

(100 kPa )(6 m /min ) P3V3 = = 6.969 kg/min RT3 (0.287 kPa ×m3 /kg ×K )(300 K ) 3

mair =

(a) We take the entire heat exchanger as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as: Mass balance ( for each fluid stream):

min - mout = D msystemZ 0 (steady) = 0 ® min = mout ® m1 = m2 = mair

m3 = m4 = mR

and

Energy balance (for the entire heat exchanger):

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 + m3 h3 = m2 h2 + m4 h4 (since Q = W = D ke @D pe @ 0)

Combining the two, mR (h2 - h1 ) = mair (h3 - h4 ) = mair c p (T3 - T4 ) Solving for T4 , T4 = T3 -

mR (h2 - h1 ) mair c p

Substituting, T4 = 27°C -

(2 kg/min)(236.99 - 86.83) kJ/kg = - 15.9°C = 257.1 K (6.969 kg/min)(1.005 kJ/kg ×K)

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation S gen is determined from an entropy balance on the evaporator. Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as

Sin - Sout

Rate of net entropy transfer by heat and mass

= D Ssystem ¿ 0 (steady)

Sgen Rate of entropy generation

Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 (since Q = 0)

mR s1 + mair s3 - mR s2 - mair s4 + Sgen = 0 or

Sgen = mR (s2 - s1 )+ mair (s4 - s3 )

where 0

s4 - s3 = c p ln

T4 P¿ T 257.1 K - R ln 4 = c p ln 4 = (1.005 kJ/kg ×K) ln = - 0.1550 kJ/kg ×K T3 P3 T3 300 K

Substituting, the exergy destruction is determined to be

X destroyed = T0 Sgen = T0 [mR (s2 - s1 ) + mair (s4 - s3 )] = (305 K) éë(2 kg/min )(0.94789 - 0.34925)kJ/kg ×K + (6.969 kg/min )(- 0.1551 kJ/kg ×K )ùû © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-446

= 35.50 kJ/min = 0.592 kW (b) When there is a heat gain from the surroundings, the steady-flow energy equation reduces to

Qin = mR (h2 - h1 )+ mair c p (T4 - T3 ) Solving for T4 ,

T4 = T3 +

Qin - mR (h2 - h1 ) mair c p

Substituting, T4 = 27°C+

(30 kJ/min) - (2 kg/min)(236.99 - 86.83) kJ/kg = - 11.6°C = 261.4 K (6.969 kg/min)(1.005 kJ/kg ×K)

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation S gen is determined from an entropy balance on an extended system that includes the evaporator and its immediate surroundings. It gives

Sin - Sout

Rate of net entropy transfer by heat and mass

= D Ssystem ¿ 0 (steady)

S gen Rate of entropy generation

Rate of change of entropy

Qin + m1 s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 Tb,in

Qin + mR s1 + mair s3 - mR s2 - mair s4 + Sgen = 0 T0 or

Sgen = mR (s2 - s1 )+ mair (s4 - s3 )-

Qin T0

where 0

s4 - s3 = c p ln

T4 PZ 261.4 K - R ln 4 = (1.005 kJ/kg ×K) ln = - 0.1384 kJ/kg ×K T3 P3 300 K

Substituting, the exergy destruction is determined to be

é Q ù X destroyed = T0 Sgen = T0 êmR ( s2 - s1 ) + mair (s4 - s3 ) - in ú ê T0 úû ë é 30 kJ/min ù ú = (305 K) ê(2 kg/min )(0.94789 - 0.34925)kJ/kg ×K + (6.969 kg/min )(- 0.1384 kJ/kg ×K )êë 305 K ú û

= 40.99 kJ/min = 0.683 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=27 [C] Vol_dot_1=6 [m^3/min] P3=120 [kPa] x3=0.30 m_dot_3=2 [kg/min] P4=P3 x4=1 Q_dot_in=30 [kJ/min] T0=(32+273) [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-447

"Properties" Fluid1$='air' Fluid2$='R134a' h1=enthalpy(Fluid1$, T=T1) s1=entropy(Fluid1$, T=T1, P=P1) h3=enthalpy(Fluid2$, P=P3, x=x3) s3=entropy(Fluid2$, P=P3, x=x3) h4=enthalpy(Fluid2$, P=P4, x=x4) s4=entropy(Fluid2$, P=P4, x=x4) C_p=1.005 [kJ/kg-C] R=0.287 [kJ/kg-K] "Analysis" "(a)" v1=(R*(T1+273))/P1 m_dot_1=Vol_dot_1/v1 m_dot_1=m_dot_2 "mass balance on air" m_dot_3=m_dot_4 "mass balance on R134a" m_dot_1*h1+m_dot_3*h3=m_dot_2*h2_a+m_dot_4*h4 "energy balance on heat exchanger" T2_a=temperature(Fluid1$, h=h2_a) m_dot_1*s1+m_dot_3*s3-m_dot_2*s2_a-m_dot_4*s4+S_dot_gen_a=0 "entropy balance" s2_a=entropy(Fluid1$, T=T2_a, P=P2) P2=P1 X_dot_destroyed_a=T0*S_dot_gen_a*1/Convert(kW, kJ/min) "(b)" Q_dot_in+m_dot_1*h1+m_dot_3*h3=m_dot_2*h2_b+m_dot_4*h4 "energy balance on heat exchanger" T2_b=temperature(air, h=h2_b) Q_dot_in/T0+m_dot_1*s1+m_dot_3*s3-m_dot_2*s2_b-m_dot_4*s4+S_dot_gen_b=0 "entropy balance" s2_b=entropy(air, T=T2_b, P=P2) X_dot_destroyed_b=T0*S_dot_gen_b*1/Convert(kW, kJ/min)

9-70E Refrigerant-134a is evaporated in the evaporator of a refrigeration system. the rate of cooling provided, the rate of exergy destruction, and the second-law efficiency of the evaporator are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. Analysis (a) The rate of cooling provided is QL = m(h2 - h1 ) = (0.08 lbm/s)(172.1- 107.5)Btu/lbm = 5.162 Btu/s = 18,580 Btu / h

(b) The entropy generation and the exergy destruction are Sgen = m( s2 - s1 ) -

QL TL

= (0.08 lbm/s)(0.4225 - 0.2851) Btu/lbm ×R -

5.162 Btu/s (50 + 460) R

= 0.0008691 Btu/s ×R

X dest = T0 Sgen = (537 R)(0.0008691 Btu/s ×R) = 0.4667 Btu / s (c) The exergy supplied (or expended) during this cooling process is the exergy decrease of the refrigerant as it evaporates in the evaporator: X 1 - X 2 = m(h1 - h2 ) - mT0 ( s1 - s2 )

6-448

= 0.7400 Btu/s The exergy efficiency is then hII,Evap = 1-

X dest 0.4667 = 1= 0.3693 = 36.9% 0.7400 X1 - X 2

EES (Engineering Equation Solver) SOLUTION "Given" m_dot=0.08 [lbm/s] T1=10 [F] x1=0.3 x2=1 T0=(77+460) [R] T_L=(50+460) [R] "Properties" h1=enthalpy(R22, x=x1, T=T1) s1=entropy(R22, x=x1, T=T1) T2=T1 h2=enthalpy(R22, x=x2, T=T2) s2=entropy(R22, x=x2, T=T2) "Analysis" "(a)" Q_dot_L=m_dot*(h2-h1) Q_dot_L_Btu\h=Q_dot_L*Convert(Btu/s, Btu/h) "(b)" S_dot_gen=m_dot*(s2-s1)-Q_dot_L/T_L Ex_dot_dest=T0*S_dot_gen "(c)" eta_ex=1-Ex_dot_dest/DELTAEx_dot DELTAEx_dot=m_dot*((h1-h2)-T0*(s1-s2)) Ex_dot_QL=-Q_dot_L*(1-T0/T_L)

9-71 Steam expands in a turbine steadily at a specified rate from a specified state to another specified state. The power potential of the steam at the inlet conditions and the reversible power output are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3

The temperature of the surroundings is given to be 25C.

6-449

P1 = 7 MPa ïüï h1 = 3650.6 kJ/kg ý T1 = 600°C ïïþ s1 = 7.0910 kJ/kg ×K P2 = 50 kPa ïü ïý h2 = 2645.2 kJ/kg sat. vapor ïïþ s2 = 7.5931 kJ/kg ×K

P0 = 100 kPa ïüï h0 @ h f @ 25°C = 104.83 kJ/kg ý T0 = 25°C ïïþ s0 @ s f @ 25°C = 0.36723 kJ/kg ×K Analysis (a) The power potential of the steam at the inlet conditions is equivalent to its exergy at the inlet state, æ 0 ö V 2¿ 0 ÷ X 1 = mxflow,1 = m çççh1 - h0 - T0 ( s1 - s0 ) + 1 + gz1¿ ÷ = m (h1 - h0 - T0 ( s1 - s0 )) ÷ ÷ çè 2 ø

= (18, 000 / 3600 kg/s) [(3650.6 - 104.83) kJ/kg - (298 K)(7.0910 - 0.36723) kJ/kg ×K ] = 7710 kW (b) The power output of the turbine if there were no irreversibilities is the reversible power, is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero,

X in - X out

- X destroyed Z 0 (reversible) = D X systemZ 0 (steady) = 0

Rate of net exergy transfer by heat, work, and mass

Rate of exergy destruction

Rate of change of exergy

X in = X out m xflow,1 = Wrev,out + mxflow,2

Wrev,out = m( xflow,1 - xflow,2 ) = m[(h1 - h2 ) - T0 ( s1 - s2 ) - ΔkeZ 0 - ΔpeZ 0 ]

Substituting, Wrev,out = m[(h1 - h2 ) - T0 ( s1 - s2 )] = (15,000/3600 kg/s) [(3650.6 - 2645.2) kJ/kg - (298 K)(7.0910 - 7.5931) kJ/kg ×K ]

= 5775 kW

EES (Engineering Equation Solver) SOLUTION "Given" m_dot=18000[kg/h]*Convert(kg/h, kg/s) P1=7000 [kPa] T1=600 [C] P2=50 [kPa] x2=1 P0=100 [kPa] T0=25 [C] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) h0=enthalpy(Fluid$, T=T0, x=0) s0=entropy(Fluid$, T=T0, x=0) "Analysis" "(a)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-450

EX1=m_dot*(h1-h0-(T0+273)*(s1-s0)) "(b)" W_dot_rev_out=m_dot*(h1-h2-(T0+273)*(s1-s2))

9-72 Air expands in an adiabatic turbine from a specified state to another specified state. The actual and maximum work outputs are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The device is adiabatic and thus heat transfer is negligible. 3 Air is an ideal gas with constant specific heats. 4 Potential energy changes are negligible. Properties At the average temperature of (425 + 325) / 2 = 375 K , the constant pressure specific heat of air is c p = 1.011 kJ / kg ×K (Table A-2b). The gas constant of air is R = 0.287 kJ / kg ×K (Table A-1).

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem

0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = Wout + m(h2 + V22 / 2)

é V 2 - V22 ù ú Wout = m êh1 - h2 + 1 ê ú 2 ë û

wout = c p (T1 - T2 ) +

V12 - V22 2

Substituting,

wout = c p (T1 - T2 ) +

V12 - V22 2

= (1.011 kJ/kg ×K)(425 - 325)K +

ö (150 m/s) 2 - (50 m/s) 2 æ ÷ çç 1 kJ/kg ÷ 2 2÷ ç è1000 m /s ø 2

6-451

= 111.1 kJ / kg The entropy change of air is s2 - s1 = c p ln

T2 P - R ln 2 T1 P1

= (1.011 kJ/kg ×K) ln

325 K 110 kPa - (0.287 kJ/kg ×K) ln 425 K 550 kPa

= 0.1907 kJ/kg ×K The maximum (reversible) work is the exergy difference between the inlet and exit states wrev,out = c p (T1 - T2 ) +

V12 - V22 - T0 ( s1 - s2 ) 2

= wout - T0 (s1 - s2 ) = 111.1 kJ/kg - (298 K)( - 0.1907 kJ/kg ×K) = 167.9 kJ / kg

Irreversibilites occurring inside the turbine cause the actual work production to be less than the reversible (maximum) work. The difference between the reversible and actual works is the irreversibility.

9-73E Air is compressed steadily by a 400-hp compressor from a specified state to another specified state while being cooled by the ambient air. The mass flow rate of air and the part of input power that is used to just overcome the irreversibilities are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Potential energy changes are negligible. 3

The temperature of the surroundings is given to be 60F.

Properties The gas constant of air is R = 0.06855 Btu /1bm ×R (Table A-1E). From the air table (Table A-17E) h = 124.27 Btu/lbm T1 = 520 R } o1 s1 = 0.59173 Btu/lbm ×R

h = 260.97 Btu/lbm T2 = 1080 R } o2 s2 = 0.76964 Btu/lbm ×R Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-452

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout æ ö V 2 - V12 ÷ ÷ Wa ,in + m(h1 + V12 / 2) = m(h2 + V22 / 2) + Qout ® Wa ,in - Qout = m çççh2 - h1 + 2 ÷ çè 2 ÷ ø Substituting, the mass flow rate of the refrigerant becomes 2 æ ö æ0.7068 Btu/s ÷ ö çç260.97 - 124.27 + (350 ft/s) - 0 1 Btu/lbm ÷ ÷ ÷(400 hp) çç (1500 / 60 Btu/s) = m ÷ 2 2÷ ÷ çè ç 1 hp 2 25,037 ft / s ÷ ø è ø

It yields

m = 1.852 lbm / s (b) The portion of the power output that is used just to overcome the irreversibilities is equivalent to exergy destruction, which can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation S gen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings. It gives

Sin - Sout Rate of net entropy transfer by heat and mass

ms1 - ms2 -

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

Qout Q + Sgen = 0 ® Sgen = m (s2 - s1 )+ out Tb,surr T0

where s2 - s1 = s20 - s10 - R ln

P2 150 psia = (0.76964 - 0.59173) Btu/lbm - (0.06855 Btu/lbm.R) ln = 0.02007 Btu/lbm.R P1 15 psia

Substituting, the exergy destruction is determined to be

æ Q ö ÷ X destroyed = T0 Sgen = T0 çççm( s2 - s1 ) + out ÷ ÷ çè T0 ø÷ æ ö 1500 / 60 Btu/s öæ 1 hp ÷ ÷ çç = (520 R) çç(1.852 lbm/s)(0.02007 Btu/lbm ×R) + ÷ ÷ ÷ ÷ çè ç øè0.7068 Btu/s ø 520 R = 62.72 hp

EES (Engineering Equation Solver) SOLUTION "Given" P1=15 [psia] T1=60 [F] Vel1=0 P2=150 [psia] T2=620 [F] Vel2=350 [ft/s] T0=ConvertTemp(F, R, 60) Q_dot_out=1500[Btu/min]*Convert(Btu/min, Btu/s) W_dot_in=400 [hp] "Properties" Fluid$='air' h1=enthalpy(Fluid$, T=T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-453

s1=entropy(Fluid$, T=T1, P=P1) h2=enthalpy(Fluid$, T=T2) s2=entropy(Fluid$, T=T2, P=P2) "Analysis" "(a)" W_dot_in*Convert(hp,Btu/s)+m_dot*(h1+Vel1^2/2*Convert(ft^2/s^2,Btu/lbm))=Q_dot_out+m_dot*(h2+Vel2^2/2* Convert(ft^2/s^2, Btu/lbm)) "(b)" S_dot_gen=m_dot*(s2-s1)+Q_dot_out/T0 X_dot_destroyed=T0*S_dot_gen*Convert(Btu/s, hp)

9-74 Hot combustion gases are accelerated in an adiabatic nozzle. The exit velocity and the decrease in the exergy of the gases are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 The combustion gases are ideal gases with constant specific heats. Properties The constant pressure specific heat and the specific heat ratio are given to be c p = 1.15 kJ / kg ×K and k = 1.3. The gas constant R is determined from

R = c p - cv = c p - c p / k = c p (1- 1/ k ) = (1.15 kJ/kg ×K)(1- 1/1.3) = 0.2654 kJ/kg ×K

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m. We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as

Ein - Eout Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = m( h2 + V22 /2) (since W = Q @D pe @ 0)

V22 - V12 2 Then the exit velocity becomes h2 = h1 -

V2 =

2c p (T1 - T2 ) + V12

æ1000 m 2 /s 2 ÷ ö ÷ 2(1.15 kJ/kg ×K)(627 - 450) K çç + (60 m/s) 2 ÷ çè 1 kJ/kg ÷ ø

= 641 m / s (b) The decrease in exergy of combustion gases is simply the difference between the initial and final values of flow exergy, and is determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-454 0

xflow,1 - xflow,2 = wrev = h1 - h2 - Δke - ΔpeZ + T0 (s2 - s1 ) = c p (T1 - T2 ) + T0 (s2 - s1 ) - D ke

where

D ke =

ö V22 - V12 (641 m/s) 2 - (60 m/s) 2 æ ÷ çç 1 kJ/kg ÷ = = 203.6 kJ/kg 2 2 ÷ çè1000 m / s ø 2 2

and

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1 = (1.15 kJ/kg ×K) ln

723 K 70 kPa - (0.2654 kJ/kg ×K) ln 900 K 230 kPa

= 0.06386 kJ/kg ×K Substituting,

Decrease in exergy = xflow,1 - xflow,2 = (1.15 kJ/kg ×K)(627 - 450)°C+(293 K)(0.06386 kJ/kg ×K) - 203.6 kJ/kg

= 18.7 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=230 [kPa] T1=(627+273) [K] Vel1=60 [m/s] P2=70 [kPa] T2=(450+273) [K] T0=(20+273) [K] c_p=1.15 [kJ/kg-C] k=1.3 "Properties" R=c_p*(1-1/k) "Analysis" "(a)" DELTAh=c_p*(T2-T1) DELTAke=(Vel2^2/2-Vel1^2/2)*Convert(m^2/s^2, kJ/kg) -DELTAh=DELTAke "energy balance" "(b)" DELTAs=c_p*ln(T2/T1)-R*ln(P2/P1) x_decrease=-DELTAh+T0*DELTAs-DELTAke

9-75 Air is compressed in a steady-flow device isentropically. The work done, the exit exergy of compressed air, and the exergy of compressed air after it is cooled to ambient temperature are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 3 The environment temperature and pressure are given to be 300 K and 100 kPa. 4 The kinetic and potential energies are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-455

Properties The gas constant of air is R = 0.287 kJ / kg ×K (Table A-1). The constant pressure specific heat and specific heat ratio of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4 (Table A-2).

Analysis (a) From the constant specific heats ideal gas isentropic relations, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P ø÷ 1

0.4/1.4

æ800 kPa ö÷ = (300 K )çç ÷ èç100 kPa ø÷

= 543.4 K

For a steady-flow isentropic compression process, the work input is determined from wcomp,in = =

kRT1 (k - 1)/ k - 1 (P2 P1 ) k- 1

{

}

(1.4)(0.287 kJ/kg ×K )(300 K ) 1.4 - 1

{(800/100)0.4/1.4 - 1}

= 244.5 kJ / kg (b) The exergy of air at the compressor exit is simply the flow exergy at the exit state, xflow,2 = h2 - h0 - T0 ( s2 - s0 )

V2 + 2 2

Z0 0

+ gz2Z (since the proccess 0 - 2 is isentropic)

= c p (T2 - T0 )

= (1.005 kJ/kg.K)(543.4 - 300)K = 244.7 kJ / kg

which is practically the same as the compressor work input. This is not surprising since the compression process is reversible. (c) The exergy of compressed air at 0.8 MPa after it is cooled to 300 K is again the flow exergy at that state, V2 xflow,3 = h3 - h0 - T0 (s3 - s0 ) + 3 2 0

= c p (T3 - T0 )Z - T0 ( s3 - s0 )

+ gz3Z

(since T3 = T0 = 300 K)

= - T0 (s3 - s0 ) where

T s3 - s0 = c p ln 3 T0

- R ln

P3 P 800 kPa = - R ln 3 = - (0.287 kJ/kg ×K)ln = - 0.5968 kJ/kg.K P0 P0 100 kPa

Substituting,

xflow,3 = - (300 K)(- 0.5968 kJ/kg.K) = 179 kJ / kg Note that the exergy of compressed air decreases from 245 to 179 kJ / kg as it is cooled to ambient temperature.

6-456

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=300 [K] P2=800 [kPa] P3=800 [kPa] T3=300 [K] T0=T1 P0=P1 "Properties" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.005 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" "(a)" T2=T1*(P2/P1)^((k-1)/k) w_comp_in=(k*R*T1)/(k-1)*((P2/P1)^((k-1)/k)-1) "(b)" x2=c_p*(T2-T0) "(c)" x3=-T0*(c_p*ln(T3/T0)-R*ln(P3/P0))

9-76 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer, the reversible work, and the exergy destruction during this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

Properties The properties of water are (Tables A-4 through A-6)

v = v f @ 135°C = 0.001075 m3 /kg T1 = 135°C üïï 1 ý u1 = u f @ 135°C = 567.41 kJ/kg sat. liquid ïïþ s1 = s f @ 135°C = 1.6872 kJ/kg ×K Te = 135°C ïü ïý he = h f @ 135°C = 567.75 kJ/kg sat. liquid ïïþ se = s f @ 135°C = 1.6872 kJ/kg ×K Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-457

(a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® me = m1 - m2 Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = me he + m2u2 - m1u1 (since W @ ke @ pe @0)

The initial and the final masses in the tank are

m1 =

V 0.6 m3 = = 558.32 kg v1 0.001075 m3 /kg

1 1 m1 = (558.32 kg ) = 279.16 kg = me 2 2 Now we determine the final internal energy and entropy, m2 =

v2 =

x2 =

V 0.6 m3 = = 0.002149 m3 /kg m2 279.16 kg

v2 - v f v fg

0.002149 - 0.001075 = 0.001851 0.58179 - 0.001075

ïüï u2 = u f + x2 u fg = 567.41 + (0.001851)(1977.3) = 571.07 kJ/kg ý x2 = 0.001851 ïïþ s2 = s f + x2 s fg = 1.6872 + (0.001851)(5.2901) = 1.6970 kJ/kg ×K

T2 = 135°C

The heat transfer during this process is determined by substituting these values into the energy balance equation,

Qin = me he + m2u2 - m1u1 = (279.16 kg )(567.75 kJ/kg )+ (279.16 kg )(571.07 kJ/kg )- (558.32 kg )(567.41 kJ/kg )

= 1115 kJ (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . The entropy generation S gen in this case is determined from an entropy balance on an extended system that includes the tank and the region between the tank and the source so that the boundary temperature of the extended system at the location of heat transfer is the source temperature Tsource at all times. It gives

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin - me se + Sgen = D Stank = (m2 s2 - m1 s1 )tank Tb,in

Sgen = m2 s2 - m1 s1 + me se -

Qin Tsource

Substituting, the exergy destruction is determined to be é Qin ù ú X destroyed = T0 Sgen = T0 êm2 s2 - m1 s1 + me se ê ú T source û ë

= (298 K) [279.16´ 1.6970 - 558.32 ´ 1.6872 + 279.16 ´ 1.6872 - (1115 kJ)/(483 K) ]

= 126 kJ For processes that involve no actual work, the reversible work output and exergy destruction are identical. Therefore, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-458

X destroyed = Wrev,out - Wact,out ® Wrev,out = X destroyed = 126 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.6 [m^3] T1=135 [C] x1=0 T_e=T1 x_e=x1 T2=T1 m2=m1/2 T_source=(210+273) [K] T0=(25+273) [K] P0=100 [kPa] "Properties" Fluid$='steam_iapws' v1=volume(Fluid$, T=T1, x=x1) u1=intenergy(Fluid$, T=T1, x=x1) s1=entropy(Fluid$, T=T1, x=x1) h_e=enthalpy(Fluid$, T=T_e, x=x_e) s_e=entropy(Fluid$, T=T_e, x=x_e) "Analysis" "(a)" m_e=m1-m2 "mass balance" m1=Vol/v1 v2=Vol/m2 u2=intenergy(Fluid$, T=T2, v=v2) s2=entropy(Fluid$, T=T2, v=v2) x2=quality(Fluid$, T=T2, v=v2) Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance" "(b)" Q_in/T_source-m_e*s_e+S_gen=m2*s2-m1*s1 "entropy balance" X_destroyed=T0*S_gen W_rev_out=X_destroyed "since there is no actual work"

9-77 A rigid tank initially contains saturated liquid of refrigerant-134a. R-134a is released from the vessel until no liquid is left in the vessel. The exergy destruction associated with this process is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. It can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.

6-459

Properties The properties of R-134a are (Table A-11)

v1 = v f @30°C = 0.0008421 m3 / kg ü T1 = 30°C ïï ý u1 = u f @30°C = 92.93 kJ/kg sat. liquid ïïþ s1 = s f @30°C = 0.34792 kJ/kg ×K v 2 = v g @ 30°C = 0.026648 m3 / kg T2 = 30°C ïüï u2 = u g @ 30°C = 246.17 kJ/kg ý sat. vapor ïïþ s2 = se = sg @ 30°C = 0.91897 kJ/kg ×K he = hg @ 30°C = 266.71 kJ/kg Analysis The volume of the container is

V = m1v1 = (1 kg)(0.0008421 m3 /kg) = 0.0008421 m3 The mass in the container at the final state is m2 =

V 0.0008421 m3 = = 0.03160 kg v2 0.026648 m3 /kg

The amount of mass leaving the container is

me = m1 - m2 = 1- 0.03160 = 0.9684 kg The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . The entropy generation S gen in this case is determined from an entropy balance on the system: Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

me se + Sgen = D Stank = (m2 s2 - m1s1 ) tank

Sgen = m2 s2 - m1s1 + me se Substituting, X destroyed = T0 Sgen = T0 (m2 s2 - m1 s1 + me se )

= (303 K)(0.03160´ 0.91897 - 1´ 0.34792 + 0.9684´ 0.91897)

= 173 kJ

6-460

x1=0 x2=1 P0=100 [kPa] T0=(30+273) [K] "Analysis" Fluid$='R134a' v1=volume(Fluid$, T=T, x=x1) u1=intenergy(Fluid$, T=T, x=x1) s1=entropy(Fluid$, T=T, x=x1) v2=volume(Fluid$, T=T, x=x2) u2=intenergy(Fluid$, T=T, x=x2) s2=entropy(Fluid$, T=T, x=x2) h_e=enthalpy(Fluid$, T=T, x=x2) s_e=s2 Vol=m1*v1 m2=Vol/v2 m_e=m1-m2 S_gen=m2*s2-m1*s1+m_e*s_e X_dest=T0*S_gen

9-78 A cylinder initially contains helium gas at a specified pressure and temperature. A valve is opened, and helium is allowed to escape until its volume decreases by half. The work potential of the helium at the initial state and the exergy destroyed during the process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the helium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats. Properties The gas constant of helium is R = 2.0769 kPa.m3 / kg.K = 2.0769 kJ / kg.K . The specific heats of helium are c p = 5.1926 kJ / kg ×K and cv = 3.1156 kJ / kg ×K (Table A-2).

Analysis (a) From the ideal gas relation, the initial and the final masses in the cylinder are determined to be m1 =

P1V (200 kPa)(0.12 m3 ) = = 0.03944 kg RT1 (2.0769 kPa ×m3 /kg ×K)(293 K)

me = m2 = m1 / 2 = 0.03944 / 2 = 0.01972 kg The work potential of helium at the initial state is simply the initial exergy of helium, and is determined from the closed-system exergy relation, X 1 = m1 xnonflow,1 = m1 [(u1 - u0 ) - T0 ( s1 - s0 ) + P0 (v1 - v 0 ) ]

6-461

where

v1 =

RT1 (2.0769 kPa ×m3 /kg ×K)(293 K) = = 3.043 m3 /kg P1 200 kPa

v0 =

RT0 (2.0769 kPa ×m3 /kg ×K)(293 K) = = 6.406 m3 /kg P0 95 kPa

and s1 - s0 = c p ln

T1 P - R ln 1 T0 P0

= (5.1926 kJ/kg ×K) ln

293 K 200 kPa - (2.0769 kJ/kg ×K) ln 293 K 95 kPa

= - 1.546 kJ/kg ×K Thus,

X1 = (0.03944 kg){(3.1156 kJ/kg ×K)(20 - 20)°C - (293 K)( - 1.546 kJ/kg ×K)+ (95 kPa)(3.043 - 6.406) m3 /kg [kJ/kPa ×m3 ]} = 5.27 kJ (b) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem ® me = m1 - m2 Energy balance: Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he + Wb,in = m2u2 - m1u1 Combining the two relations gives

Qin = (m1 - m2 )he + m2u2 - m1u1 - Wb,in = (m1 - m2 )he + m2 h2 - m1h1

= (m1 - m2 + m2 - m1 )h1 = 0 since the boundary work and U combine into H for constant pressure expansion and compression processes. The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation S gen can be determined from an entropy balance on the cylinder. Noting that the pressure and temperature of helium in the cylinder are maintained constant during this process and heat transfer is zero, it gives

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

- me se + Sgen = D Scylinder = (m2 s2 - m1 s1 )cylinder Sgen = m2 s2 - m1s1 + me se

= m2 s2 - m1s1 + (m1 - m2 )se = (m2 - m1 + m1 - m2 )s1

6-462

since the initial, final, and the exit states are identical and thus Se = S2 = S1 . Therefore, this discharge process is reversible, and X destroyed = T0 Sgen = 0

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.12 [m^3] T1=(20+273) [K] P1=200 [kPa] Vol2=0.5*Vol1 T2=T1 P2=P1 T_e=T1 P_e=P1 T0=(20+273) [K] P0=95 [kPa] "Properties" R=2.0769 [kJ/kg-K] "Table A-2a" c_p=5.1926 [kJ/kg-K] "Table A-2a" c_v=3.1156 [kJ/kg-K] "Table A-2a" "Analysis" "(a)" m_e=m1-m2 "mass balance" m1=(P1*Vol1)/(R*T1) m2=(P2*Vol2)/(R*T2) v1=(R*T1)/P1 v0=(R*T0)/P0 DELTAs=c_p*ln(T1/T0)-R*ln(P1/P0) X1=m1*(c_v*(T1-T0)-T0*DELTAs+P0*(v1-v0)) "(b)" Q_in=0 S_gen=0 X_destroyed=T0*S_gen

9-79 An insulated cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The amount of steam that entered the cylinder and the exergy destroyed are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A-4 through A-6)

ïüï h1 = h f + x1h fg = 561.43 + 0.8667 ´ 2163.5 = 2436.5 kJ/kg ý x1 = 13 /15 = 0.8667 ïïþ s1 = s f + x1 s fg = 1.6716 + 0.8667 ´ 5.3200 = 6.2824 kJ/kg ×K

P1 = 300 kPa

P2 = 300 kPa ïü ïý h2 = hg @300 kPa = 2724.9 kJ/kg sat.vapor ïïþ s2 = sg @300 kPa = 6.9917 kJ/kg ×K

6-463

Pi = 2 MPa ïüï hi = 3248.4 kJ/kg ý Ti = 400°C ïïþ si = 7.1292 kJ/kg ×K

Analysis (a) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this unsteady-flow system can be expressed as

min - mout = D msystem ® mi = m2 - m1

Mass balance: Energy balance:

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

mi hi = Wb,out + m2u2 - m1u1 (since Q @ ke @ pe @ 0)

Combining the two relations gives 0 = Wb,out - (m2 - m1 )hi + m2u2 - m1u1 or,

0 = - (m2 - m1 )hi + m2 h2 - m1h1

since the boundary work and U combine into H for constant pressure expansion and compression processes. Solving for m 2 and substituting, m2 =

hi - h1 (3248.4 - 2436.5) kJ/kg m1 = (15 kg) = 23.27 kg hi - h2 (3248.4 - 2724.9) kJ/kg

Thus, mi = m2 - m1 = 23.27 - 15 = 8.27 kg

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation S gen is determined from an entropy balance on the insulated cylinder,

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

mi si + Sgen = D Ssystem = m2 s2 - m1s1 Sgen = m2 s2 - m1s1 - mi si

Substituting, the exergy destruction is determined to be X destroyed = T0 Sgen = T0 [m2 s2 - m1 s1 - mi si ]

= (298 K)(23.27´ 6.9917 - 15´ 6.2824 - 8.27´ 7.1292)

= 2832 kJ

6-464

m1_v=13 [kg] P1=300 [kPa] P_i=2000 [kPa] T_i=400 [C] x2=1 P2=P1 T0=(25+273) [K] P0=100 [kPa] "Properties" Fluid$='steam_iapws' x1=m1_v/m1 h1=enthalpy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) h2=enthalpy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) h_i=enthalpy(Fluid$, T=T_i, P=P_i) s_i=entropy(Fluid$, T=T_i, P=P_i) "Analysis" "(a)" m_i=m2-m1 "mass balance" m_i*h_i=m2*h2-m1*h1 "energy balance, since U2-U1+W_b = H2-H1 for a constant pressure process" "(b)" m_i*s_i+S_gen=m2*s2-m1*s1 "entropy balance" X_destroyed=T0*S_gen

9-80 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam and the rate of exergy destruction are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions.

Properties Noting that T < Tsat @ 200 kPa = 120.23° C , the cold water and the exit mixture streams exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. From Tables A-4 through A-6,

h1 @ h f @ 15 C = 62.98 kJ/kg P1 = 200 kPa ïü ïý ïïþ s1 @ s T1 = 15°C = 0.22447 kJ/kg ×K f @ 15 C

P2 = 200 kPa üïï h2 = 2870.4 kJ/kg ý ïïþ s2 = 7.5081 kJ/kg ×K T2 = 200°C

6-465

P3 = 200 kPa ïüï h3 @ h f @80°C = 335.02 kJ/kg ý ïïþ s3 @ s f @ 80°C = 1.0756 kJ/kg ×K T3 = 80°C Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as

min - mout = D msystem 0 (steady) = 0

Mass balance:

¾ ¾®

m1 + m2 = m3

Energy balance:

Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + m2 h2 = Qout + m3 h3

Combining the two relations gives

Qout = m1h1 + m2 h2 - (m1 + m2 )h3 = m1 (h1 - h3 )+ m2 (h2 - h3 ) Solving for m2 and substituting, the mass flow rate of the superheated steam is determined to be

m2 =

Qout - m1 (h1 - h3 ) h2 - h3

(600/60 kJ/s) - (4 kg/s)(62.98 - 335.02) kJ/kg

(2870.4 - 335.02) kJ/kg

= 0.429 kg / s

Also,

m3 = m1 + m2 = 4 + 0.429 = 4.429 kg/s (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen where the entropy generation S gen is determined from an entropy balance on an extended system that includes the mixing chamber and its immediate surroundings. It gives

Sin - Sout

Rate of net entropy transfer by heat and mass

m1 s1 + m2 s2 - m3 s3 -

Sgen

= D Ssystem 0 = 0

Rate of entropy generation

Rate of change of entropy

Qout + Sgen = 0 Tb,surr

Sgen = m3 s3 - m1s1 - m2 s2 +

Qout T0

Substituting, the exergy destruction is determined to be

æ ö Q ÷ X destroyed = T0 Sgen = T0 çççm3 s3 - m2 s2 - m1 s1 + out ÷ ÷ çè Tb, surr ÷ ø = (298 K)(4.429´ 1.0756 - 0.429´ 7.5081- 4´ 0.22447 + 10 / 298)kW/K = 202 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [kPa] T1=15 [C] P2=200 [kPa] T2=200 [C] P3=200 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-466

T3=80 [C] m1_dot=4 [kg/s] Q_dot_out=600[kJ/min]*Convert(kJ/min, kJ/s) T0=(25+273) [K] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, T=T1, x=0) s1=entropy(Fluid$, T=T1, x=0) h2=enthalpy(Fluid$, P=P2, T=T2) s2=entropy(Fluid$, P=P2, T=T2) h3=enthalpy(Fluid$, T=T3, x=0) s3=entropy(Fluid$, T=T3, x=0) "Analysis" "(a)" m1_dot+m2_dot=m3_dot "mass balance" m1_dot*h1+m2_dot*h2=m3_dot*h3 "energy balance" "(b)" m1_dot*s1+m2_dot*s2-m3_dot*s3-Q_dot_out/T0+S_dot_gen=0 "entropy balance on extended system" X_dot_destroyed=T0*S_dot_gen

9-81 Each member of a family of four takes a shower every day. The amount of exergy destroyed by this family per year is to be determined. Assumptions 1 Steady operating conditions exist. 2 The kinetic and potential energies are negligible. 3

Heat losses from the pipes, mixing section are negligible and thus Q @ 0.

4 5 6 7

Showers operate at maximum flow conditions during the entire shower. Each member of the household takes a shower every day. Water is an incompressible substance with constant properties at room temperature. The efficiency of the electric water heater is 100%.

Properties The density and specific heat of water are at room temperature are ρ = 997 kg / m3 and c = 4.18 kJ / kg.°C (Table A-3). Analysis The mass flow rate of water at the shower head is m = r V = (0.997 kg/L)(10 L/min) = 9.97 kg/min

The mass balance for the mixing chamber can be expressed in the rate form as min - mout = D msystemZ 0 (steady) = 0 ® min = mout ® m1 + m2 = m3

where the subscript 1 denotes the cold water stream, 2 the hot water stream, and 3 the mixture. The rate of entropy generation during this process can be determined by applying the rate form of the entropy balance on a system that includes the electric water heater and the mixing chamber (the T-elbow). Noting that there is no entropy transfer associated with work transfer (electricity) and there is no heat transfer, the entropy balance for this steadyflow system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem ¿ 0 (steady) Rate of change of entropy

m1s1 + m2 s2 - m3 s3 + Sgen = 0 (since Q = 0 and work is entropy free) Sgen = m3 s3 - m1s1 - m2 s2

6-467

Noting from mass balance that m1 + m2 = m3 and s2 = s1 since hot water enters the system at the same temperature as the cold water, the rate of entropy generation is determined to be T Sgen = m3 s3 - (m1 + m2 ) s1 = m3 ( s3 - s1 ) = m3 c p ln 3 T1

42 + 273 = 3.735 kJ/min.K 15 + 273 Noting that 4 people take a 6-min shower every day, the amount of entropy generated per year is = (9.97 kg/min)(4.18 kJ/kg.K) ln

Sgen = (Sgen )D t (No. of people)(No. of days) = (3.735 kJ/min.K)(6 min/person ×day)(4 persons)(365 days/year)

=32,715 kJ/K (per year) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (298 K)(32,715 kJ/K) = 9,749,000 kJ Discussion The value above represents the exergy destroyed within the water heater and the T-elbow in the absence of any heat losses. It does not include the exergy destroyed as the shower water at 42C is discarded or cooled to the outdoor temperature. Also, an entropy balance on the mixing chamber alone (hot water entering at 55C instead of 15C) will exclude the exergy destroyed within the water heater.

EES (Engineering Equation Solver) SOLUTION "Given" N_people=4 time=6 [min/day] V_dot=0.010 [m^3/min] T1=(15+273) [K] T2=(55+273) [K] T3=(42+273) [K] T0=(25+273) [K] "Properties" rho=997 [kg/m^3] "Table A-3" c_p=4.18 [kJ/kg-K] "Table A-3" "Analysis" m_dot=rho*V_dot S_dot_gen=m_dot*c_p*ln(T3/T1) "entropy balance" S_gen=S_dot_gen*time*N_people*day_year day_year=365 [day/year] X_destroyed=T0*S_gen

9-82 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the rate of exergy destruction in the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-468

Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ / kg.°C , respectively. The gas constant of air is R = 0.287 kJ / kg ×K (Table A-1). Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 = Qout + mh2 (since D ke @D pe @ 0)

Qout = mc p (T1 - T2 ) Then the rate of heat transfer from the exhaust gases becomes

Q = [mc p (Tin - Tout )]gas. = (0.85 kg/s)(1.1 kJ/kg.°C)(350°C - 260°C) = 84.15 kW The mass flow rate of air is PV (101 kPa)(0.5 m3 /s) m= = = 0.5807 kg/s RT (0.287 kPa.m3 /kg.K)´ 303 K Noting that heat loss by exhaust gases is equal to the heat gain by the air, the air exit temperature becomes Q = éëêmc p (Tout - Tin )ù ú ® Tout = Tin + û air

Q 84.15 kW = 30°C + = 174.2°C mc p (0.5807 kg/s)(1.005 kJ/kg ×°C)

The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout

Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 (steady)

Rate of entropy generation

Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m3 s4 + Sgen = 0 (since Q = 0)

mexhaust s1 + mair s3 - mexhaust s2 - mair s4 + Sgen = 0 Sgen = mexhaust (s2 - s1 ) + mair (s4 - s3 ) Noting that the pressure of each fluid remains constant in the heat exchanger, the rate of entropy generation is Sgen = mexhaust c p ln

T2 T + mair c p ln 4 T1 T3

= (0.85 kg/s)(1.1 kJ/kg.K)ln

260 + 273 174.2 + 273 + (0.5807 kg/s)(1.005 kJ/kg.K)ln 350 + 273 30 + 273

= 0.08129 kW/K

6-469

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (303 K)(0.08129 kW/K) = 24.6 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=101 [kPa] T1=(30+273) [K] V1_dot_air=0.5 [m^3/s] T3=(350+273) [K] T4=(260+273) [K] m_dot_gas=0.85 [kg/s] c_p_air=1.005 [kJ/kg-C] c_p_gas=1.10 [kJ/kg-C] T0=T1 "assumed" "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" Q_dot=m_dot_gas*c_p_gas*(T3-T4) m_dot_air=(P1*V1_dot_air)/(R*T1) Q_dot=m_dot_air*c_p_air*(T2-T1) S_dot_gen=m_dot_air*c_p_air*ln(T2/T1)+m_dot_gas*c_p_gas*ln(T4/T3) "entropy balance" X_dot_destroyed=T0*S_dot_gen

9-83 Water is heated by hot oil in a heat exchanger. The outlet temperature of the oil and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.

Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ / kg.°C , respectively. Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-470

(a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the rate of heat transfer to the cold water in this heat exchanger becomes

Q = [mc p (Tout - Tin )]water = (4.5 kg/s)(4.18 kJ/kg ×°C)(70°C - 20°C)=940.5 kW Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot water is determined from Q = [mc p (Tin - Tout )]oil ®

Tout = Tin -

Q 940.5 kW = 170°C = 129.1°C mc p (10 kg/s)(2.3 kJ/kg.°C)

(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem 0 (steady)

Rate of entropy generation

Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m3 s4 + Sgen = 0 (since Q = 0) mwater s1 + moil s3 - mwater s2 - moil s4 + Sgen = 0 Sgen = mwater (s2 - s1 ) + moil (s4 - s3 ) Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be

T T Sgen = mwater c p ln 2 + moil c p ln 4 T1 T3 = (4.5 kg/s)(4.18 kJ/kg.K) ln

70+273 129.1+273 + (10 kg/s)(2.3 kJ/kg.K) ln = 0.736 kW/K 20+273 170+273

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (298 K)(0.736 kW/K) = 219 kW

EES (Engineering Equation Solver) SOLUTION "Given" T1=(20+273) [K] T2=(70+273) [K] m_dot_w=4.5 [kg/s] T3=(170+273) [K] m_dot_oil=10 [kg/s] c_p_w=4.18 [kJ/kg-C] c_p_oil=2.30 [kJ/kg-C] T0=(25+273) [K] "Analysis" "(a)" Q_dot=m_dot_w*c_p_w*(T2-T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-471

Q_dot=m_dot_oil*c_p_oil*(T3-T4) T4_C=T4-273 "(b)" S_dot_gen=m_dot_w*c_p_w*ln(T2/T1)+m_dot_oil*c_p_oil*ln(T4/T3) "entropy balance" X_dot_gen=T0*S_dot_gen

9-84E Steam is condensed by cooling water in a condenser. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5

The temperature of the environment is 77F.

Properties The specific heat of water is 1.0 Btu / lbm. F (Table A-3E). The enthalpy and entropy of vaporization of water at 120F are 1025.2 Btu / lbm and sfg = 1.7686 Btu / lbm.R (Table A-4E). Analysis (a) We take the tube-side of the heat exchanger where cold water is flowing as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Qin + mh1 = mh2

(since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the rate of heat transfer to the cold water in this heat exchanger becomes

Q = [mc p (Tout - Tin )]water = (115.3 lbm/s)(1.0 Btu/lbm.°F)(73°F - 60°F) = 1499 Btu / s Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate of condensation of the steam in the heat exchanger is determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-472

Q = (mh fg )steam = ¾ ¾ ® msteam =

Q 1499 Btu/s = = 1.462 lbm/s h fg 1025.2 Btu/lbm

(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout Rate of net entropy transfer by heat and mass

= D Ssystem ¿ 0 (steady)

Sgen Rate of entropy generation

Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 (since Q = 0) mwater s1 + msteam s3 - mwater s2 - msteam s4 + Sgen = 0 Sgen = mwater (s2 - s1 ) + msteam (s4 - s3 ) Noting that water is an incompressible substance and steam changes from saturated vapor to saturated liquid, the rate of entropy generation is determined to be

T T Sgen = mwater c p ln 2 + msteam ( s f - sg ) = mwater c p ln 2 - msteam s fg T1 T1 73 + 460 - (1.462 lbm/s)(1.7686 Btu/lbm.R) = 0.2613 Btu/s.R 60 + 460 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition = (115.3 lbm/s)(1.0 Btu/lbm.R)ln

X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (537 R)(0.2613 Btu/s.R) = 140.3 Btu / s

EES (Engineering Equation Solver) SOLUTION "Given" T_s=120 [F] T1=60 [F] T2=73 [F] m_dot_w=115.3 [lbm/s] T0=(77+460) [R] "Properties" Fluid$='steam_iapws' h_f=enthalpy(Fluid$, T=T_s, x=0) h_g=enthalpy(Fluid$, T=T_s, x=1) h_fg=h_g-h_f s_f=entropy(Fluid$, T=T_s, x=0) s_g=entropy(Fluid$, T=T_s, x=1) s_fg=s_g-s_f c_p=1.0 [Btu/lbm-F] "Table A-3E" "Analysis" "(a)" Q_dot=m_dot_w*c_p*(T2-T1) m_dot_s=Q_dot/h_fg "(b)" S_dot_gen=m_dot_w*c_p*ln((T2+460)/(T1+460))-m_dot_s*s_fg "entropy balance" X_dot_gen=T0*S_dot_gen

6-473

9-85 A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The mass of the R-134a that entered the tank and the exergy destroyed during this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified).

Properties The properties of refrigerant are (Tables A-11 through A-13)

v = v g @1.2 MPa =0.016728 m3 / kg P1 = 1.2 MPa ïüï 1 ý u1 = ug @1.2 MPa =253.84 kJ/kg ïïþ sat. vapor s1 = sg @1.2 MPa =0.91320 kJ/kg ×K v 2 = v g @1.4 MPa =0.014119 m3 / kg T2 = 1.4 MPa ïüï ý u2 = ug @1.4 MPa =256.40 kJ/kg ïïþ sat. vapor s2 = sg @1.4 MPa =0.91067 kJ/kg ×K Pi = 1.6 MPa ïüï hi = 93.58 kJ/kg ý ïïþ si = 0.34792 kJ/kg ×K Ti = 30°C Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - mout = D msystem

mi = m2 - m1

Energy balance:

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 - m1u1 (since W @ ke @ pe @ 0) (a) The initial and the final masses in the tank are

m1 =

V1 0.1 m3 = = 5.978 kg v1 0.016728 m3 /kg

m2 =

V2 0.1 m3 = = 7.083 kg v 2 0.014119 m3 /kg

6-474

mi = m2 - m1 = 7.083 - 5.978 = 1.105 kg

The heat transfer during this process is determined from the energy balance to be Qin = - mi hi + m2 u2 - m1u1 = - (1.105 kg )(93.58 kJ/kg) + (7.083) (256.40 kJ/kg )- (5.978 kg )(253.84 kJ/kg ) = 195.2 kJ

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . The entropy generation S gen in this case is determined from an entropy balance on an extended system that includes the tank and part of the source so that the boundary temperature is the source temperature Ts at all times. It gives Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin + mi si + Sgen = D Stank =(m2 s2 - m1 s1 )tank Tb,in

Sgen = m2 s2 - m1 s1 - mi si -

Qin Ts

Substituting, the exergy destruction is determined to be

é Q ù X destroyed = T0 Sgen = T0 êm2 s2 - m1s1 - mi si - in ú ê Ts ú ë û = (288 K) [7.083´ 0.91067 - 5.978´ 0.91320 - 1.105´ 0.34792 - (195.2 kJ)/(473 K) ]

= 55.9 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.1 [m^3] P1=1200 [kPa] x1=1 P_i=1600 [kPa] T_i=30 [C] P2=1400 [kPa] x2=1 T_s=(200+273) [K] T0=(15+273) [K] P0=100 [kPa] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) h_i=enthalpy(Fluid$, T=T_i, P=P_i) s_i=entropy(Fluid$, T=T_i, P=P_i) "Analysis" "(a)" m_i=m2-m1 "mass balance" m1=Vol/v1 m2=Vol/v2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-475

Q_in+m_i*h_i=m2*u2-m1*u1 "energy balance" "(b)" Q_in/T_s+m_i*s_i+S_gen=m2*s2-m1*s1 "entropy balance" X_destroyed=T0*S_gen

9-86 A rigid tank initially contains saturated R-134a vapor at a specified pressure. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The amount of heat transfer with the surroundings and the exergy destruction are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is from the tank (will be verified).

Properties The properties of refrigerant are (Tables A-11 through A-13) u1 = u g @1 MPa =250.71 kJ/kg P1 = 1 MPa ïü ïý s = s g @1 MPa =0.91574 kJ/kg ×K ïïþ 1 sat.vapor v1 = v g @1 MPa =0.020329 m3 / kg

Pi = 1.4 MPa ïüï hi = 285.47 kJ/kg ý ïïþ si = 0.9389 kJ/kg ×K Ti = 60°C Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as

min - mout = D msystem

Mass balance:

mi = m2 - m1

Energy balance:

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

mi hi - Qout = m2u2 - m1u1 (since W @ ke @ pe @0)

The initial and the final masses in the tank are m1 =

V 0.2 m3 = = 9.838 kg v1 0.020329 m3 / kg

6-476

m2 = m f + mg =

Vf vf

Vg vg

0.1 m3 0.1 m3 + = 111.92 + 5.978 = 117.89 kg 0.0008935 m3 / kg 0.016728 m3 / kg

U 2 = m2u2 = m f u f + mg u g = 111.92´ 116.72 + 5.978´ 253.84 = 14,581 kJ S2 = m2 s2 = m f s f + mg sg = 111.92´ 0.42449 + 5.978´ 0.91320 = 52.968 kJ/K

Then from the mass and energy balances, mi = m2 - m1 = 117.89 - 9.838 = 108.05 kg

The heat transfer during this process is determined from the energy balance to be Qout = mi hi - m2u2 + m1u1 = 108.05´ 285.47 - 14,581 + 9.838´ 250.71 = 18, 731 kJ

(b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . The entropy generation S gen in this case is determined from an entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times. It gives Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + mi si + Sgen = D S tank =(m2 s2 - m1s1 ) tank Tb,out

Sgen = m2 s2 - m1 s1 - mi si +

Qout T0

Substituting, the exergy destruction is determined to be

é Q ù X destroyed = T0 Sgen = T0 êm2 s2 - m1 s1 - mi si + out ú ê T0 ú ë û = (298 K)[52.968 - 9.838´ 0.91574 - 108.05´ 0.9389 + 18,731/ 298]

= 1599 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.2 [m^3] P1=1000 [kPa] x1=1 P_i=1400 [kPa] T_i=60 [C] P2=1200 [kPa] Vol2_f=Vol/2 Vol2_g=Vol/2 T0=(25+273) [K] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) h_i=enthalpy(Fluid$, T=T_i, P=P_i) s_i=entropy(Fluid$, T=T_i, P=P_i) v2_f=volume(Fluid$, P=P2, x=0) v2_g=volume(Fluid$, P=P2, x=1) v2=volume(Fluid$, P=P2, x=x2) u2=intenergy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-477

"Analysis" "(a)" m1=Vol/v1 m2_f=Vol2_f/v2_f m2_g=Vol2_g/v2_g m2=m2_f+m2_g x2=m2_g/m2 m_i=m2-m1 "mass balance" m_i*h_i-Q_out=m2*u2-m1*u1 "energy balance" "(b)" -Q_out/T0+m_i*s_i+S_gen=m2*s2-m1*s1 "entropy balance" X_destroyed=T0*S_gen

9-87 An expression is to be derived for the work potential of the single-phase contents of a rigid adiabatic container when the initially empty container is filled through a single opening from a source of working fluid whose properties remain fixed. Analysis The conservation of mass principle for this system reduces to dmCV = mi dt

where the subscript i stands for the inlet state. When the entropy generation is set to zero (for calculating work potential) and the combined first and second law is reduced to fit this system, it becomes

d (U - T0 S ) + (h - T0 S )i mi dt When these are combined, the result is Wrev = -

Wrev = -

d (U - T0 S ) dmCV + (h - T0 S )i dt dt

Recognizing that there is no initial mass in the system, integration of the above equation produces

Wrev = (h - T0 s)i m2 - m2 (h2 - T0 s2 ) Wrev = (hi - h2 ) - T0 ( si - s2 ) m2 where the subscript 2 stands for the final state in the container.

Review Problems

9-88 Natural gas is liquefied at a rate of 1.75 kg / s in a liquefaction plant. Natural gas enters the plant at 3 bar and 40C and liquefied at 160C as a liquid. The actual power input during this liquefaction process is 95 MW. Determine the minimum power input, the reversible and actual COPs, and the second-law efficiency of the plant. Methane properties are used for natural gas. Various properties of methane before and after liquefaction process are given as follows: h1 = 30.97 kJ / kg, h2 = - 905.46 kJ / kg, s1 = - 0.4588 kJ / kg ×K, s2 = - 6.6324 kJ / kg ×K

9-88 A natural gas liquefaction plant is considered. The the minimum power input, the reversible and actual COPs, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-478

Analysis The rate of heat removed from the natural gas is

QL  m(h1  h2 )  (1.75 kg/s) 30.97  (905.46) kJ/kg  1639 kW Taking the dead state temperature to be T0 = T1 = 40C = 313.15 K, the minimum work input is determined from Wmin  Wrev  m  h2  h1  T0 ( s2  s1 )   (1.75 kg/s)  (905.46)  30.97  kJ/kg  (313.15 K)  (6.6324  (0.4588) kJ/kg  K 

 1744 kW The reversible COP is COPrev 

QL 1639 kW   0.9394 Wrev 1744 kW

The second-law efficiency is defined as the reversible power divided by the actual power: II 

Wrev 1744 kW   0.1836  18.4% Wact 9500 kW

The second-law efficiency can also be defined as the actual COP divided by the reversible COP. Using this definition, the actual COP becomes

II 

COPact COPact   0.1836    COPact  0.1725 COPrev 0.9394

EES (Engineering Equation Solver) SOLUTION T_1=40 [C] P_1=300 [kPa] P_2=300 [kPa] T_2=-160 [C] m_dot=1.75[kg/s] W_dot_act=9500 [kW] h_1=enthalpy(Methane,T=T_1,P=P_1) h_2=enthalpy(Methane,P=P_2,T=T_2) s_1=entropy(Methane,T=T_1,P=P_1) s_2=entropy(Methane,P=P_2,T=T_2) T_0=(T_1+273.15) W_dot_min=m_dot*((h_2-h_1)-T_0*(s_2-s_1)) Q_dot_L=m_dot*(h_1-h_2) COP_rev=Q_dot_L/W_dot_min eta_ex=W_dot_min/W_dot_act eta_ex=COP_act/COP_rev

9-89E The second-law efficiency of a refrigerator and the refrigeration rate are given. The power input to the refrigerator is to be determined.

6-479

Analysis From the definition of the second-law efficiency, the COP of the refrigerator is determined to be COPR,rev =

hII =

1 1 = = 7.462 TH / TL - 1 550 / 485 - 1

COPR ¾¾ ® COPR = hII COPR,rev = 0.28´ 7.462 = 2.089 COPR,rev

Thus the power input is

Win =

ö QL 800 Btu/min æ 1 hp ÷= 9.03 hp çç = ÷ ÷ COPR 2.089 çè 42.41 Btu/min ø

EES (Engineering Equation Solver) SOLUTION "Given" eta_II=0.28 Q_dot_L=800 [Btu/min] T_L=(25+460) [R] T_H=(90+460) [R] "Analysis" COP_R_rev=1/((T_H/T_L)-1) eta_II=COP_R/COP_R_rev W_dot_in=Q_dot_L/COP_R*Convert(Btu/min, hp)

9-90 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat loss and the amount of exergy destruction in 5 h are to be determined Assumptions Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. Analysis We take the glass to be the system, which is a closed system. The amount of heat loss is determined from Q = QD t = (4.4 kJ/s)(5´ 3600 s) = 79, 200 kJ

6-480

Under steady conditions, the rate form of the entropy balance for the glass simplifies to

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem 0 = 0

Rate of entropy generation

Rate of change of entropy

Q Qin - out + Sgen,glass = 0 Tb,in Tb,out

4400 W 4400 W + Sgen,wall = 0 283 K 276 K

Sgen,glass = 0.3943 W/K

Then the amount of entropy generation over a period of 5 h becomes

Sgen,glass = Sgen,glass D t = (0.3943 W / K)(5´ 3600 s) = 7098 J/K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen , X destroyed = T0 Sgen = (278 K)(7.098 kJ/K) = 1973 kJ

Discussion The total entropy generated during this process can be determined by applying the entropy balance on an extended system that includes the glass and its immediate surroundings on both sides so that the boundary temperature of the extended system is the room temperature on one side and the environment temperature on the other side at all times. Using this value of entropy generation will give the total exergy destroyed during the process, including the temperature gradient zones on both sides of the window.

EES (Engineering Equation Solver) SOLUTION "Given" L=0.005 [m] A=2*2 [m^2] T1=(10+273) [K] T2=(3+273) [K] Q_dot=4.4 [kJ/s] time=5*3600 [s] T0=(5+273) [K] "Analysis" Q=Q_dot*time Q_dot_in/T1-Q_dot_out/T2+S_dot_gen=0 "entropy balance" Q_dot_in=Q_dot Q_dot_out=Q_dot S_gen=S_dot_gen*time X_destroyed=T0*S_gen

9-91 Heat is transferred steadily to boiling water in the pan through its bottom. The inner and outer surface temperatures of the bottom of the pan are given. The rate of exergy destruction within the bottom plate is to be determined. Assumptions Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values.

6-481

Analysis We take the bottom of the pan to be the system, which is a closed system. Under steady conditions, the rate form of the entropy balance for this system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem 0 = 0

Rate of entropy generation

Rate of change of entropy

Q Qin - out + Sgen,system = 0 Tb,in Tb,out

1100 W 1100 W + Sgen,system = 0 378 K 377 K

Sgen,system = 0.007719 W/K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (298 K)(0.007719 W/K) = 2.30 W

EES (Engineering Equation Solver) SOLUTION "Given" D=0.30 [m] Q_dot=1100 [W] T1=(105+273) [K] T2=(104+273) [K] T0=(25+273) [K] "Analysis" Q_dot_in/T1-Q_dot_out/T2+S_dot_gen=0 "entropy balance" Q_dot_in=Q_dot Q_dot_out=Q_dot X_dot_destroyed=T0*S_dot_gen

9-92 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate. The rate at which the work potential is wasted during this process is to be determined. Assumptions Steady operating conditions exist.

6-482

Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closed system.. The system extends from the outer surface of the pipe to a distance at which the temperature drops to the surroundings temperature. In steady operation, the rate form of the entropy balance for this system can be expressed as

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem 0 = 0

Rate of entropy generation

Rate of change of entropy

Qout Qin + Sgen,system = 0 Tb,in Tb,out

1175 W 1175 W + Sgen,system = 0 353 K 278 K

Sgen,system = 0.8980 W/K

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (278 K)(0.8980 W/K) = 250 W

EES (Engineering Equation Solver) SOLUTION "Given" D=0.05 [m] L=10 [m] T1=(80+273) [K] T2=(5+273) [K] Q_dot=1175 [W] T0=T2 "Analysis" Q_dot_in/T1-Q_dot_out/T2+S_dot_gen=0 "entropy balance" Q_dot_in=Q_dot Q_dot_out=Q_dot X_dot_destroyed=T0*S_dot_gen

9-93 Steam is condensed in a closed system at a constant pressure from a saturated vapor to a saturated liquid by rejecting heat to a thermal energy reservoir. The second law efficiency is to be determined. Assumptions 1

Kinetic and potential energy changes are negligible.

6-483

Analysis We take the steam as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Wb,in - Qout = D U = m(u2 - u1 )

From the steam tables (Table A-5),

v = v g = 2.2172 m3 /kg P1 = 75 kPa ïüï 1 ý u1 = ug = 2496.1 kJ/kg Sat. vapor ïïþ s1 = sg = 7.4558 kJ/kg ×K

v = v f = 0.001037 m3 /kg P2 = 75 kPa ïüï 2 ý u2 = u f = 384.36 kJ/kg Sat. liquid ïïþ s2 = s f = 1.2132 kJ/kg ×K

The boundary work during this process is æ 1 kJ ÷ ö wb,in = P(v1 - v 2 ) = (75 kPa)(2.2172 - 0.001037) m3 /kg çç = 166.2 kJ/kg ÷ çè1 kPa ×m3 ÷ ø

The heat transfer is determined from the energy balance: qout = wb,in - (u2 - u1 ) = 166.2 kJ/kg - (384.36 - 2496.1)kJ/kg = 2278 kJ/kg

The exergy change between initial and final states is

æ T ö ÷ xnonflow,1 - xnonflow,2 = u1 - u2 + P0 (v1 - v 2 ) - T0 ( s1 - s2 ) - qout ççç1- 0 ÷ ÷ çè TR ÷ ø

æ 1 kJ ÷ö = (2496.1- 384.36)kJ/kg + (100 kPa)(2.2172 - 0.001037) m3 /kg çç - (298 K)(7.4558 - 1.2132)kJ/kg ×K - (2278 kJ/kg) çè1 kPa ×m3 ÷÷ø = 384.9 kJ/kg

The second law efficiency is then

hII =

wb,in D xnonflow

166.2 kJ/kg = 0.432 = 43.2% 384.9 kJ/kg

6-484

EES (Engineering Equation Solver) SOLUTION "GIVEN" P=75 [kPa] x1=1 x2=0 T_R=(37+273) [K] T0=(25+273) [K] P0=100 [kPa] "ANALYSIS" Fluid$='steam_iapws' v1=volume(Fluid$, P=P, x=x1) u1=intenergy(Fluid$, P=P, x=x1) s1=entropy(Fluid$, P=P, x=x1) v2=volume(Fluid$, P=P, x=x2) u2=intenergy(Fluid$, P=P, x=x2) s2=entropy(Fluid$, P=P, x=x2) w_b_in=P*(v1-v2) q_out=w_b_in-(u2-u1) DELTAx=u1-u2+P0*(v1-v2)-T0*(s1-s2)-q_out*(1-T0/T_R) eta_II=w_b_in/DELTAx

9-94 R-134a is vaporized in a closed system at a constant pressure from a saturated liquid to a saturated vapor by transferring heat from a reservoir at two pressures. The pressure that is more effective from a second-law point of view is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible.

Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - Wb,out = D U = m(u2 - u1 )

Qin = Wb,out + D U Qin = D H = m(h2 - h1 )

At 100 kPa:

6-485

From the R-134a tables (Table A-12), u fg @100 kPa = 198.01 kJ/kg h fg @100 kPa = 217.19 kJ/kg s fg @100 kPa = 0.88008 kJ/kg ×K v fg @100 kPa = v g - v f = 0.19255 - 0.0007258 = 0.19182 m3 /kg

The boundary work during this process is æ 1 kJ ÷ ö wb,out = P(v 2 - v1 ) = Pv fg = (100 kPa)(0.19182) m 3 /kg çç = 19.18 kJ/kg ÷ çè1 kPa ×m3 ÷ ø

The useful work is determined from

wu = wb,out - wsurr = P(v 2 - v1 ) - P0 (v 2 - v1 ) = 0 kJ/kg since P = P0 = 100 kPa . The heat transfer from the energy balance is

qin = h fg = 217.19 kJ/kg The exergy change between initial and final states is

æ T ö xnonflow,1 - xnonflow,2 = u1 - u2 + P0 (v1 - v 2 ) - T0 ( s1 - s2 ) + qin ççç1- 0 ÷÷÷ çè TR ø÷ æ T ö = - u fg - P0v fg + T0 s fg + qin ççç1- 0 ÷÷÷ çè TR ø÷ æ 1 kJ ö ÷ = - 198.01 kJ/kg - (100 kPa)(0.19182 m 3 /kg) çç ÷ ÷+ (298 K)(0.88008 kJ/kg ×K) + (217.19 kJ/kg) çè1 kPa ×m3 ø æ 298 K ö ÷ çç1÷ çè 279 K ÷ ø = 30.28 kJ/kg

The second law efficiency is then

hII =

wu 0 kJ/kg = = 0 D xnonflow 30.28 kJ/kg

At 180 kPa: u fg @180 kPa = 188.20 kJ/kg

h fg @180 kPa = 207.95 kJ/kg s fg @180 kPa = 0.79848 kJ/kg ×K v fg @180 kPa = v g - v f = 0.11049 - 0.0007485 = 0.10974 m3 /kg

æ 1 kJ ÷ ö wb,out = P(v 2 - v1 ) = Pv fg = (180 kPa)(0.10974) m 3 /kg çç = 19.75 kJ/kg ÷ çè1 kPa ×m3 ÷ ø

6-486

wu = wb,out - wsurr = P(v 2 - v1 ) - P0 (v 2 - v1 )

æ 1 kJ ÷ ö = ( P - P0 )v fg = (180 - 100) kPa(0.10974) m 3 /kg çç = 8.779 kJ/kg ÷ çè1 kPa ×m3 ÷ ø

qin = h fg = 207.95 kJ/kg

æ T ö xnonflow,1 - xnonflow,2 = u1 - u2 + P0 (v1 - v 2 ) - T0 ( s1 - s2 ) + qin ççç1- 0 ÷÷÷ çè TR ø÷ æ T ö = - u fg - P0v fg + T0 s fg + qin ççç1- 0 ÷÷÷ çè TR ø÷ æ 1 kJ ÷ ö = - 188.20 kJ/kg - (100 kPa)(0.10974 m 3 /kg) çç + (298 K)(0.79848 kJ/kg ×K) + (207.95 kJ/kg) ÷ 3 ÷ çè1 kPa ×m ø æ 298 K ö ÷ çç1÷ ÷ çè 279 K ø

= 24.61 kJ/kg hII =

wu 8.779 kJ/kg = = 0.3567 = 35.7% D xnonflow 24.61 kJ/kg

The process at 180 kPa is more effective from a work production standpoint.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P=100 [kPa] "or P =180 [kPa]" x1=0 x2=1 T_R=(6+273) [K] T0=(25+273) [K] P0=100 [kPa] "ANALYSIS" Fluid$='R134a' v1=volume(Fluid$, P=P, x=x1) u1=intenergy(Fluid$, P=P, x=x1) h1=enthalpy(Fluid$, P=P, x=x1) s1=entropy(Fluid$, P=P, x=x1) v2=volume(Fluid$, P=P, x=x2) u2=intenergy(Fluid$, P=P, x=x2) h2=enthalpy(Fluid$, P=P, x=x2) s2=entropy(Fluid$, P=P, x=x2) w_b_out=P*(v2-v1) w_surr=P0*(v2-v1) w_u=w_b_out-w_surr q_in=h2-h1 DELTAx=u1-u2+P0*(v1-v2)-T0*(s1-s2)+q_in*(1-T0/T_R) eta_II=w_u/DELTAx

9-95 Oil is to be cooled by water in a thin-walled heat exchanger. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-487

Properties The specific heats of water and oil are given to be 4.18 and 2.20 kJ / kg.°C , respectively. Analysis (a) We take the oil tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

D EsystemZ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 = Qout + mh2 (since D ke @D pe @ 0)

Qout = mc p (T1 - T2 ) Then the rate of heat transfer from the oil becomes

Qout = [mc p (Tin - Tout )]oil = (2 kg/s)(2.2 kJ/kg.°C)(150°C - 40°C) = 484 kW Noting that heat lost by the oil is gained by the water, the outlet temperature of water is determined from Q = [mc p (Tin - Tout )]water ¾ ¾ ® Tout = Tin +

Q 484 kW = 22°C = 99.2°C mc p (1.5 kg/s)(4.18 kJ/kg.°C)

(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 (steady)

Rate of entropy generation

Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m3 s4 + Sgen = 0 (since Q = 0) moil s1 + mwater s3 - moil s2 - mwater s4 + Sgen = 0 Sgen = moil (s2 - s1 ) + mwater (s4 - s3 ) Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be

T T Sgen = moil c p ln 2 + mwater c p ln 4 T1 T3

6-488

= (2 kg/s)(2.2 kJ/kg.K) ln

40 + 273 99.2 + 273 + (1.5 kg/s)(4.18 kJ/kg.K) ln = 0.132 kW/K 150 + 273 22 + 273

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (295 K)(0.132 kW/K) = 38.9 kW

EES (Engineering Equation Solver) SOLUTION "Given" T1=150[C]+273 [K] T2=40[C]+273 [K] m_dot_oil=2 [kg/s] T3=22[C]+273 [K] m_dot_w=1.5 [kg/s] C_p_w=4.18 [kJ/kg-C] C_p_oil=2.20 [kJ/kg-C] T0=T3 "assumed" "Analysis" "(a)" Q_dot=m_dot_oil*C_p_oil*(T1-T2) Q_dot=m_dot_w*C_p_w*(T4-T3) "(b)" S_dot_gen=m_dot_oil*C_p_oil*ln(T2/T1)+m_dot_w*C_p_w*ln(T4/T3) "entropy balance" X_dot_destroyed=T0*S_dot_gen

9-96 Water is heated in a heat exchanger by geothermal water. The rate of heat transfer to the water and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5

The environment temperature is 25C.

Properties The specific heats of water and geothermal fluid are given to be 4.18 and 4.31 kJ / kg.°C , respectively. Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-489

(a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout Qin + mh1 = mh2 (since D ke @D pe @ 0)

Qin = mc p (T2 - T1 ) Then the rate of heat transfer to the cold water in the heat exchanger becomes

Qin,water = [mc p (Tout - Tin )]water = (0.4 kg/s)(4.18 kJ/kg ×°C)(60°C - 25°C)=58.52 kW Noting that heat transfer to the cold water is equal to the heat loss from the geothermal water, the outlet temperature of the geothermal water is determined from Qout = [mc p (Tin - Tout )]geo ¾ ¾ ® Tout = Tin -

Qout 58.52 kW = 140°C = 94.7°C mc p (0.3 kg/s)(4.31 kJ/kg.°C)

(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger:

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen Rate of entropy generation

= D Ssystem 0 (steady) Rate of change of entropy

m1s1 + m3 s3 - m2 s2 - m4 s4 + Sgen = 0 (since Q = 0) mwater s1 + mgeo s3 - mwater s2 - mgeo s4 + Sgen = 0 Sgen = mwater (s2 - s1 ) + mgeo (s4 - s3 ) Noting that both fresh and geothermal water are incompressible substances, the rate of entropy generation is determined to be Sgen = mwater c p ln

T2 T + mgeo c p ln 4 T1 T3

= (0.4 kg/s)(4.18 kJ/kg.K)ln

60+273 94.7+273 + (0.3 kg/s)(4.31 kJ/kg.K)ln = 0.0356 kW/K 25+273 140+273

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (298 K)(0.0356 kW/K)=10.61 kW

EES (Engineering Equation Solver) SOLUTION "Given" T1=(25+273) [K] T2=(60+273) [K] m_dot_w=0.4 [kg/s] T3=(140+273) [K] m_dot_geo=0.3 [kg/s] c_p_w=4.18 [kJ/kg-C] c_p_geo=4.31 [kJ/kg-C] T0=T1 "assumed" "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-490

"(a)" Q_dot=m_dot_w*c_p_w*(T2-T1) Q_dot=m_dot_geo*c_p_geo*(T3-T4) "(b)" S_dot_gen=m_dot_w*c_p_w*ln(T2/T1)+m_dot_geo*c_p_geo*ln(T4/T3) "entropy balance" X_dot_destroyed=T0*S_dot_gen

9-97 Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an adiabatic heat exchanger. The rate at which the steam is obtained, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air properties are used for exhaust gases. 4 Pressure drops in the heat exchanger are negligible. Properties The gas constant of air is R = 0.287 kJ / kg ×K . The specific heat of air at the average temperature of exhaust gases (650 K) is c p = 1.063 kJ / kg ×K (Table A-2).

Analysis (a) We denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3) and (4). The properties of water are (Table A-4)

T3 = 20°Cüïï h3 = 83.91 kJ/kg ý x3 = 0 ïïþ s3 = 0.29649 kJ/kg.K T4 = 200°Cüïï h4 = 2792.0 kJ/kg ý ïïþ s4 = 6.4302 kJ/kg.K x4 = 1 An energy balance on the heat exchanger gives

ma h1 + mw h3 = ma h2 + mw h4 ma c p (T1 - T2 ) = mw (h4 - h3 )

(0.8 kg/s)(1.063 kJ/kg°C)(400 - 350)°C = mw (2792.0 - 83.91)kJ/kg mw = 0.01570 kg / s (b) The specific exergy changes of each stream as it flows in the heat exchanger is D sa = c p ln

T2 (350 + 273) K = (0.8 kg/s)(1.063 kJ/kg.K)ln = - 0.08206 kJ/kg.K T1 (400 + 273) K

D xflow,a = c p (T2 - T1 ) - T0D sa = (1.063 kJ/kg.°C)(350 - 400)°C - (20 + 273 K)( - 0.08206 kJ/kg.K)

= - 29.106 kJ/kg D xflow, w = h4 - h3 - T0 ( s4 - s3 )

6-491

= (2792.0 - 83.91)kJ/kg - (20 + 273 K)(6.4302 - 0.29649)kJ/kg.K = 910.913 kJ/kg

The exergy destruction is determined from an exergy balance on the heat exchanger to be Or - X dest = ma D xflow, a + mwD xflow, w = (0.8 kg/s)(-29.106 kJ/kg) + (0.01570 kg/s)(910.913) kJ/kg = - 8.98 kW X dest = 8.98 kW

(c) The second-law efficiency for a heat exchanger may be defined as the exergy increase of the cold fluid divided by the exergy decrease of the hot fluid. That is, hII =

mwD xflow, w - ma D xflow, a

(0.01570 kg/s)(910.913 kJ/kg) = 0.614 - (0.8 kg/s)(-29.106 kJ/kg)

EES (Engineering Equation Solver) SOLUTION "GIVEN" T[1]=400 [C] P[1]=150 [kPa] m_dot_gas=0.8 [kg/s] T[2]=350 [C] P[2]=P[1] T[3]=20 [C] x[3]=0 T[4]=200 [C] x[4]=1 T0=(20+273) [K] "PROPERTIES" Fluid1$='Air' Fluid2$='Steam_iapws' h[1]=enthalpy(Fluid1$, T=T[1]) s[1]=entropy(Fluid1$, T=T[1], P=P[1]) h[2]=enthalpy(Fluid1$, T=T[2]) s[2]=entropy(Fluid1$, T=T[2], P=P[2]) h[3]=enthalpy(Fluid2$, T=T[3], x=x[3]) s[3]=entropy(Fluid2$, T=T[3], x=x[3]) h[4]=enthalpy(Fluid2$, T=T[4], x=x[4]) s[4]=entropy(Fluid2$, T=T[4], x=x[4]) "ANALYSIS" m_dot_gas*(h[1]-h[2])=m_dot_steam*(h[4]-h[3]) "energy balance" DELTAx_gas=h[1]-h[2]-T0*(s[1]-s[2]) DELTAx_steam=h[4]-h[3]-T0*(s[4]-s[3]) EX_destroyed=m_dot_gas*DELTAx_gas-m_dot_steam*DELTAx_steam "exergy destroyed" eta_II=(m_dot_steam*DELTAx_steam)/(m_dot_gas*DELTAx_gas) "second-law efficiency"

9-98 Elevation, base area, and the depth of a crater lake are given. The maximum amount of electricity that can be generated by a hydroelectric power plant is to be determined. Assumptions The evaporation of water from the lake is negligible.

6-492

Analysis The exergy or work potential of the water is the potential energy it possesses relative to the ground level,

Exergy = PE = mgh Therefore, Exergy = PE = ò dPE = ò gz dm = ò gz (r Adz )

= r Ag ò

zdz = r Ag ( z22 - z12 ) / 2

æ 1h ÷ öæ 1 kJ/kg ö÷ çç = 0.5(1000 kg/m3 )(2´ 104 m 2 )(9.81 m/s 2 ) ((117 m) 2 - (105 m 2 ))çç ÷ ÷ ÷ èç3600 s øèç1000 m 2 / s 2 ø÷ = 72, 600 kWh

EES (Engineering Equation Solver) SOLUTION "Given" A=20000 [m^2] h1=105 [m] h2=12 [m] "Properties" g=9.81 [m/s^2] rho=1000 [kg/m^3] "assumed" "Analysis" PE=rho*A*g*(z2^2-z1^2)/2*Convert(kg-m^2/s^2, kWh) z1=h1 z2=h1+h2 X=PE

9-99 An electric resistance heater is immersed in water. The time it will take for the electric heater to raise the water temperature to a specified temperature, the minimum work input, and the exergy destroyed during this process are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the container itself and the heater is negligible. 3 Heat loss from the container is negligible. 4 The environment temperature is given to be T0 = 20° C . Properties The specific heat of water at room temperature is c = 4.18 kJ / kg ×°C (Table A-3).

6-493

Analysis Taking the water in the container as the system, which is a closed system, the energy balance can be expressed as Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in = (D U ) water We,in D t = mc(T2 - T1 ) water

Substituting,

(1500 J / s)D t = (70 kg) (4180 J / kg ×° C)(80 - 20) ° C Solving for t gives t = 11,704 s = 195 min = 3.25 h Again we take the water in the tank to be the system. Noting that no heat or mass crosses the boundaries of this system and the energy and entropy contents of the heater are negligible, the entropy balance for it can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0 + Sgen = D S water

Therefore, the entropy generated during this process is

T 353 K Sgen = D S water = mc ln 2 = (70 kg )(4.18 kJ/kg ×K ) ln = 54.51 kJ/K T1 293 K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (293 K)(54.51 kJ/K) = 15,971 kJ The actual work input for this process is Wact,in = Wact,in D t = (1.5 kJ/s)(11,704 s) = 17,556 kJ

Then the reversible (or minimum required) work input becomes

Wrev,in = Wact,in - X destroyed = 17,556 - 15,971 = 1585 kJ

EES (Engineering Equation Solver) SOLUTION "Given" L=0.30 [m] W_dot_e_in=1.500 [kW] D=0.012 [m] m=70 [kg] T1=(20+273) [K] T2=(80+273) [K] T0=(20+273) [K] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-494

c=4.18 [kJ/kg-K] "Table A-3" "Analysis" W_dot_e_in*time=m*C*(T2-T1) S_gen=m*C*ln(T2/T1) X_destroyed=T0*S_gen W_e_in=W_dot_e_in*time W_rev_e_in=W_e_in-X_destroyed

9-100 Heat is lost from the air flowing in a diffuser. The exit temperature, the rate of exergy destruction, and the second law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Nitrogen is an ideal gas with variable specific heats. Properties The gas constant of nitrogen is R = 0.2968 kJ / kg.K . Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. At the inlet of the diffuser and at the dead state, we have T1 = 110°C = 383 Kïü ïý h1 = 88.39 kJ/kg ïïþ s1 = 7.101 kJ/kg ×K P1 = 100 kPa

T1 = 300 K ïü ïý h0 = 1.93 kJ/kg P1 = 100 kPaïïþ s0 = 6.846 kJ/kg ×K

An energy balance on the diffuser gives

h1 + 88.39 kJ/kg +

V12 V2 = h2 + 2 + qout 2 2

ö ö (205 m/s) 2 æ (45 m/s) 2 æ çç 1 kJ/kg ÷ çç 1 kJ/kg ÷ = h + ® h2 = 105.9 kJ/kg ÷ 2 2 2÷ 2 2÷ ÷+ 2.5 kJ/kg ¾ ¾ ç ç è1000 m /s ø è1000 m /s ø 2 2

The corresponding properties at the exit of the diffuser are

h2 = 105.9 kJ/kgïü ïýT2 = 127°C = 400 K P1 = 110 kPa ïïþ s2 = 7.117 kJ/kg ×K (b) The mass flow rate of the nitrogen is determined to be m = r 2 A2V2 =

P2 110 kPa A2V2 = (0.04 m 2 )(45 m/s) = 1.669 kg/s RT2 (0.2968 kJ/kg.K)(400 K)

The exergy destruction in the nozzle is the exergy difference between the inlet and exit of the diffuser:

6-495 2 2 é æ öù ÷ ê(88.39 - 105.9)kJ/kg + (205 m/s) - (45 m/s) çç 1 kJ/kg ú ÷ çè1000 m 2 /s 2 ÷ = (1.669 kg/s) ê øú= 12.4 kW 2 ê ú ê- (300 K)(7.101- 7.117)kJ/kg.K ú ë û

é ù V2 X 1 = m êh1 - h0 + 1 - T0 ( s1 - s0 )ú ê ú 2 ë û é ù ö (205 m/s) 2 æ çç 1 kJ/kg ÷ = (1.669 kg/s) ê(88.39 - 1.93) kJ/kg + - (300 K)(7.101- 6.846)kJ/kg.K ú ÷ 2 2 ÷ ê ú èç1000 m /s ø 2 ë û

= 51.96 kW hII =

X X2 12.4 kW = 1- dest = 1= 0.761 = 76.1% 51.96 kW X1 X1

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] T_1=(110+273) [K] Vel_1=205 [m/s] P_2=110 [kPa] Vel_2=45 [m/s] q_out=2.5 [kJ/kg] A_2=0.04 [m^2] P_0=100 [kPa] T_0=(27+273) [K] "PROPERTIES" Fluid$='N2' R=0.2968 [kJ/kg-K] "Table A-2a" h_0=enthalpy(Fluid$, T=T_0) s_0=entropy(Fluid$, T=T_0, P=P_0) h_1=enthalpy(Fluid$, T=T_1) s_1=entropy(Fluid$, T=T_1, P=P_1) "ANALYSIS" "(a)" h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg)=h_2+Vel_2^2/2*Convert(m^2/s^2, kJ/kg)+q_out "energy balance" T_2=temperature(Fluid$, h=h_2) "exit temperature" T_2_C=T_2-273 s_2=entropy(Fluid$, h=h_2, P=P_2) v_2=(R*T_2)/P_2 "(b)" m_dot=1/v_2*A_2*Vel_2 x_1=h_1-h_0+Vel_1^2/2*Convert(m^2/s^2, kJ/kg)-T_0*(s_1-s_0) x_2=h_2-h_0+Vel_2^2/2*Convert(m^2/s^2, kJ/kg)-T_0*(s_2-s_0) X_dot_dest=m_dot*(x_1-x_2) "exergy destruction" "(c)" eta_II=x_2/x_1 "Alternative method" s_gen=s_2-s_1+q_out/T_0 X_dot_dest_b=m_dot*T_0*s_gen X_dot_1=m_dot*x_1 eta_II_b=1-X_dot_dest/X_dot_1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-496

9-101 Steam is accelerated in a nozzle. The actual and maximum outlet velocities are to be determined. Assumptions 1 The nozzle operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A-4 through A-6)

P1 = 300 kPa ïüï h1 = 2761.2 kJ/kg ý T1 = 150°C ïïþ s1 = 7.0792 kJ/kg ×K

ïüï h2 = 2693.1 kJ/kg ý x2 = 1 (sat. vapor) ïïþ s2 = 7.2231 kJ/kg ×K

P2 = 150 kPa

Analysis We take the nozzle to be the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

D Esystem ¿ 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout 2 1

m(h1 + V / 2) = m(h2 +V22 /2)

V22 - V12 = h1 - h2 = D keactual 2 Substituting,

D keactual = h1 - h2 = 2761.2 - 2693.1 = 68.1 kJ/kg The actual velocity at the exit is then V22 - V12 = D keactual 2

V2 =

V12 + 2D keactual =

æ1000 m 2 /s 2 ÷ ö ÷ (45 m/s) 2 + 2(68.1 kJ/kg) çç = 372 m / s ÷ çè 1 kJ/kg ÷ ø

The maximum kinetic energy change is determined from D kemax = h1 - h2 - T0 (s1 - s2 ) = 68.1- (298)(7.0792 - 7.2231) = 111.0 kJ/kg

The maximum velocity at the exit is then or

2 V2,max - V12

V2,max =

= D kemax

V12 + 2D kemax =

æ1000 m 2 /s 2 ÷ ö ÷ (45 m/s) 2 + 2(111.0 kJ/kg) çç = 473 m / s ÷ çè 1 kJ/kg ÷ ø

6-497

T1=150 [C] Vel1=45 [m/s] P2=150 [kPa] x2=1 T0=(25+273) [K] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, x=x2) s2=entropy(Fluid$, P=P2, x=x2) "Analysis" DELTAke_actual=h1-h2 DELTAke_actual=(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) DELTAke_max=h1-h2-T0*(s1-s2) DELTAke_max=(Vel2_max^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg)

9-102 Steam is accelerated in an adiabatic nozzle. The exit velocity, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy changes are negligible.

Analysis (a) The properties of the steam at the inlet and exit of the turbine and at the dead state are (Tables A -4 through A-6)

P1 = 3.5 MPa üïï h1 = 2978.4 kJ/kg ý T1 = 300°C ïïþ s1 = 6.4484 kJ/kg.K P2 = 1.6 MPa üïï h2 = 2919.9 kJ/kg ý T2 = 250°C ïïþ s2 = 6.6753 kJ/kg.K T0 = 18°Cïü ïý h0 = 75.54 kJ/kg ïïþ s0 = 0.2678 kJ/kg.K x= 0

The exit velocity is determined from an energy balance on the nozzle h1 +

2978.4 kJ/kg +

V12 V2 = h2 + 2 2 2

ö ö V22 æ (0 m/s) 2 æ çç 1 kJ/kg ÷ çç 1 kJ/kg ÷ = 2919.9 kJ/kg + ÷ 2 2 2 2÷ ÷ ç ç è ø è 2 2 1000 m /s ø÷ 1000 m /s V2 = 342.0 m / s

(b) The rate of exergy destruction is the exergy decrease of the steam in the nozzle

6-498 2 é ù æ ö ÷ ê(2919.9 - 2978.4)kJ/kg + (342 m/s) - 0 çç 1 kJ/kg ú ÷ 2 2 ÷ = - (0.4 kg/s) ê èç1000 m /s øú 2 ê ú ê- (291 K)(6.6753 - 6.4484)kJ/kg.K ú ë û

= 26.41kW (c) The exergy of the refrigerant at the inlet is

é ù V2 X 1 = m êh1 - h0 + 1 - T0 ( s1 - s0 )ú ê ú 2 ë û = (0.4 kg/s) [(2978.4 - 75.54) kJ/kg + 0 - (291 K)(6.4484 - 0.2678)kJ/kg.K ]

= 441.72 kW The second-law efficiency for a nozzle may be defined as the exergy recovered divided by the exergy expended:

ö (342 m/s) 2 - 0 æ çç 1 kJ/kg ÷ 2 2÷ ç xrecovered V /2 è ø 2 1000 m /s ÷ hII = = = 2 xexpended h1 - h2 - T0 ( s1 - s2 ) + V1 / 2 (2978.4 - 2929.9)kJ/kg - (291 K)(6.4484 - 6.6753)kJ/kg.K + 0 2 2

= 0.886 = 88.6% Discussion Alternatively, the second-law efficiency for this device may be defined as the exergy output divided by the exergy input:

hII =

X X2 26.41 kW = 1- dest = 1= 0.940 = 94.0% 441.72 kW X1 X1

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=3500 [kPa] T_1=300 [C] Vel_1=0 [m/s] P_2=1600 [kPa] T_2=250 [C] m_dot=0.4 [kg/s] T_0=18 [C] P_0=100 [kPa] "PROPERTIES" Fluid$='Steam_iapws' h_0=enthalpy(Fluid$, P=P_0, T=T_0) s_0=entropy(Fluid$, P=P_0, T=T_0) h_1=enthalpy(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) h_2=enthalpy(Fluid$, P=P_2, T=T_2) s_2=entropy(Fluid$, P=P_2, T=T_2) "ANALYSIS" h_1+Vel_1^2/2*Convert(m^2/s^2, kJ/kg)=h_2+Vel_2^2/2*Convert(m^2/s^2, kJ/kg) "energy balance" ex_1=h_1-h_0+(Vel_1^2/2*Convert(m^2/s^2, kJ/kg))-(T_0+273)*(s_1-s_0) ex_2=h_2-h_0+(Vel_2^2/2*Convert(m^2/s^2, kJ/kg))-(T_0+273)*(s_2-s_0) EX_dot_dest=m_dot*(ex_1-ex_2) "exergy destruction" eta_II=ex_2/ex_1 "Alternative method" s_gen=s_2-s_1 EX_dot_dest_b=m_dot*(T_0+273)*s_gen EX_dot_1=m_dot*ex_1 eta_II_b=1-EX_dot_dest/EX_dot_1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-499

"Alternative method for second-law efficiency" eta_II_alt=Vel_2^2/2*Convert(m^2/s^2, kJ/kg)/(ex_1-ex_2)

9-103 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses heat to the surroundings. The final temperature in each tank and the work potential wasted during this process are to be determined. Assumptions 1 Tank A is insulated and thus heat transfer is negligible. 2 The water that remains in tank A undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible. 4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no work interactions.

Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1 . From the steam tables (Tables A-4 through A-6), Tank A : v1, A = v f + x1v fg = 0.001084 + (0.8)(0.46242 - 0.001084) = 0.37015 m 3 /kg P1 = 400 kPa ü ïï ý u1, A = u f + x1u fg = 604.22 + (0.8)(1948.9) = 2163.3 kJ/kg ïïþ x1 = 0.8 s1, A = s f + x1 s fg = 1.7765 + (0.8)(5.1191) = 5.8717 kJ/kg ×K

T2, A = Tsat @ 300 kPa = 133.52°C P2 = 300 kPa üïï x = s2, A - s f = 5.8717 - 1.6717 = 0.7895 ï 2, A s fg 5.3200 s2 = s1 ý ïï sat. mixture ( ) ïþï v 2, A = v f + x2, Av fg = 0.001073 + (0.7895)(0.60582 - 0.001073) = 0.47850 m 3 /kg u2, A = u f + x2, Au fg = 561.11 + (0.7895)(1982.1 kJ/kg ) = 2125.9 kJ/kg Tank B : v1, B = 1.1989 m3 /kg P1 = 200 kPa ü ïï ý u1, B = 2731.4 kJ/kg T1 = 250°C ïïþ s1, B = 7.7100 kJ/kg ×K

The initial and the final masses in tank A are and

m1, A =

VA 0.2 m3 = = 0.5403 kg v1, A 0.37015 m3 /kg

m2, A =

VA 0.2 m3 = = 0.4180 kg v 2, A 0.479 m3 /kg

6-500

m2, B = m1, B - 0.122 = 3 + 0.122 = 3.122 kg

The final specific volume of steam in tank B is determined from

(m1v1 )B (3 kg)(1.1989 m /kg) VB = = = 1.152 m3 /kg m2, B m2, B 3.122 m3 3

v2, B =

We take the entire contents of both tanks as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U = (D U ) A + (D U ) B

(since W = KE = PE = 0)

- Qout = (m2 u2 - m1u1 )A + (m2 u2 - m1u1 )B

Substituting,

- 900 = {(0.418)(2125.9)- (0.5403)(2163.3)}+ {(3.122)u2, B - (3)(2731.4)} u2, B = 2425.9 kJ/kg

Thus,

ïï v 2, B = 1.152 m3 /kg ü T2, B = 110.1°C ý ï s u2, B = 2425.9 kJ/kg ïþ 2, B = 6.9772 kJ/kg ×K (b) The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D SA + D SB Tb,surr

Rearranging and substituting, the total entropy generated during this process is determined to be

Sgen = D S A + D S B +

Qout Q = (m2 s2 - m1s1 )A + (m2 s2 - m1s1 )B + out Tb,surr Tb,surr

= {(0.418)(5.8717)- (0.5403)(5.8717)}+ {(3.122)(6.9772)- (3)(7.7100)}+

900 kJ 273 K

= 1.234 kJ/K The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen , X destroyed = T0 Sgen = (273 K)(1.234 kJ/K) = 337 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol_A=0.2 [m^3] P1_A=400 [kPa] x1_A=0.80 m1_B=3 [kg] P1_B=200 [kPa] T1_B=250 [C] P2_A=300 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-501

Q_out=900 [kJ] T0=(0+273) [K] "Properties" Fluid$='steam_iapws' v1_A=volume(Fluid$, P=P1_A, x=x1_A) u1_A=intenergy(Fluid$, P=P1_A, x=x1_A) s1_A=entropy(Fluid$, P=P1_A, x=x1_A) v1_B=volume(Fluid$, P=P1_B, T=T1_B) u1_B=intenergy(Fluid$, P=P1_B, T=T1_B) s1_B=entropy(Fluid$, P=P1_B, T=T1_B) "Analysis" "(a)" s2_A=s1_A "reversible-adiabatic (i.e., isentropic) process" T2_A=temperature(Fluid$, P=P2_A, s=s2_A) u2_A=intenergy(Fluid$, P=P2_A, s=s2_A) v2_A=volume(Fluid$, P=P2_A, s=s2_A) x2_A=quality(Fluid$, P=P2_A, s=s2_A) m1_A=Vol_A/v1_A m2_A=Vol_A/v2_A m2_B=m1_B+(m1_A-m2_A) Vol_B=m1_B*v1_B v2_B=Vol_B/m2_B -Q_out=DELTAU_A+DELTAU_B "energy balance" DELTAU_A=m2_A*u2_A-m1_A*u1_A DELTAU_B=m2_B*u2_B-m1_B*u1_B T2_B=temperature(Fluid$, v=v2_B, u=u2_B) s2_B=entropy(Fluid$, v=v2_B, u=u2_B) "(b)" -Q_out/T0+S_gen=DELTAS_A+DELTAS_B "entropy balance" DELTAS_A=m2_A*s2_A-m1_A*s1_A DELTAS_B=m2_B*s2_B-m1_B*s1_B X_destroyed=T0*S_gen

9-104E A cylinder initially filled with helium gas at a specified state is compressed polytropically to a specified temperature and pressure. The actual work consumed and the minimum useful work input needed are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. 5 The environment temperature is 70F. Properties The gas constant of helium is R = 2.6805 psia.ft 3 / 1bm.R = 0.4961 Btu / 1bm ×R (Table A-1E). The specific heats of helium are cv = 0.753 and c p = 1.25 Btu / lbm ×R (Table A-2E).

6-502

Analysis (a) Helium at specified conditions can be treated as an ideal gas. The mass of helium is m=

P1V1 (40 psia)(8 ft 3 ) = = 0.2252 lbm RT1 (2.6805 psia ×ft 3 /lbm ×R)(530 R)

The exponent n and the boundary work for this polytropic process are determined to be

P1V1 P2V2 T P (780 R)(40 psia) = ¾¾ ® V2 = 2 1 V1 = (8 ft 3 ) = 3.364 ft 3 T1 T2 T1 P2 (530 R)(140 psia) n

n æP ö æV1 ö æ140 ÷ ö æ 8 ö çç ÷ ÷ ç ç ÷ ÷ P2V2n = P1V1n ¾ ¾ ® ççç 2 ÷ = ¾ ¾ ® = ® n = 1.446 ÷ ÷ ç ç ÷ ÷ ÷ ¾¾ çè 40 ÷ ÷ ççèV2 ø ÷ ø èç3.364 ø èç P1 ø

Then the boundary work for this polytropic process can be determined from 2

Wb,in = - ò P dV = 1

mR (T2 - T1 ) P2V2 - P1V1 =1- n 1- n

(0.2252 lbm)(0.4961 Btu/lbm ×R )(780 - 530)R 1- 1.446

= 62.62 Btu

Also, Thus,

æ ö 1 Btu ÷ ÷= 12.61 Btu Wsurr,in = - P0 (V2 - V1 ) = - (14.7 psia)(3.364 - 8) ft 3 çç 3÷ èç5.4039 psia ×ft ÷ ø Wu,in = Wb,in - Wsurr,in = 62.62 - 12.61 = 50.0 Btu

(b) We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heat transfer to be from the cylinder, the energy balance for this stationary closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout + Wb,in = D U = m(u2 - u1 ) - Qout = m(u2 - u1 ) - Wb,in Qout = Wb,in - mcv (T2 - T1 )

Substituting, Qout = 62.62 Btu - (0.2252 lbm)(0.753 Btu/lbm ×R )(780 - 530)R = 20.69 Btu

The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Ssys Tb,surr

where the entropy change of helium is

æ T Pö D Ssys = D Shelium = m çççc p , avg ln 2 - R ln 2 ÷÷÷ çè T1 P1 ø÷ é 780 R 140 psia ù ú = (0.2252 lbm) ê(1.25 Btu/lbm ×R)ln - (0.4961 Btu/lbm ×R)ln ê 530 R 40 psia ú ë û

6-503

Rearranging and substituting, the total entropy generated during this process is determined to be

Sgen = D S helium +

Qout 20.69 Btu = (- 0.03201 Btu/R) + = 0.007022 Btu/R T0 530 R

The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (530 R)(0.007022 Btu/R) = 3.722 Btu The minimum work with which this process could be accomplished is the reversible work input, Wrev,in which can be determined directly from

Wrev,in = Wu,in - X destroyed = 50.0 - 3.722 = 46.3 Btu Discussion The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction term equal to zero,

X in - X out - X destroyedZ 0 (reversible) = D X system ® Wrev,in = X 2 - X 1 Net exergy transfer by heat, work, and mass

Exergy destruction

Change in exergy

Substituting the closed system exergy relation, the reversible work input during this process is determined to be

Wrev = (U 2 - U1 ) - T0 (S2 - S1 ) + P0 (V2 - V1 ) = (0.2252 lbm)(0.753 Btu/lbm ×R)(320 - 70)°F - (530 R)( - 0.03201 Btu/R) + (14.7 psia)(3.364 - 8) ft 3 [Btu/5.4039 psia ×ft 3 ]

= 46.7 Btu The slight difference is due to round-off error.

EES (Engineering Equation Solver) SOLUTION "Given" V1=8 [ft^3] P1=40 [psia] T1=(70+460) [R] P2=140 [psia] T2=(320+460) [R] P0=14.7 [psia] T0=(70+460) [R] "Properties" c_p=1.25 [Btu/lbm-R] "Table A-2Ea" c_v=0.753 [Btu/lbm-R] "Table A-2Ea" R1=2.6805 [psia-ft^3/lbm-R] "Table A-1E" R2=0.4961 [Btu/lbm-R] "Table A-1E" "Analysis" "(a)" m=(P1*V1)/(R1*T1) (P1*V1)/T1=(P2*V2)/T2 "from ideal gas relation" P1*V1^n=P2*V2^n W_b_out=m*R2*(T2-T1)/(1-n) W_b_in=-W_b_out W_surr_in=-P0*(V2-V1)*Convert(psia-ft^3, Btu) W_u_in=W_b_in-W_surr_in "(b)" -Q_out+W_b_in=m*c_v*(T2-T1) "energy balance" Q_surr_in=Q_out DELTAS_sys=m*(c_p*ln(T2/T1)-R2*ln(P2/P1)) DELTAS_surr=Q_surr_in/T0 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-504

DELTAS_total=DELTAS_sys+DELTAS_surr S_gen=DELTAS_total X_destroyed=T0*S_gen W_rev_in=W_u_in-X_destroyed

9-105 Air expands in an adiabatic turbine from a specified state to another specified state. The second-law efficiency is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The device is adiabatic and thus heat transfer is negligible. 3 Air is an ideal gas with constant specific heats. 4 Kinetic and potential energy changes are negligible. Properties At the average temperature of (425 + 325) / 2 = 375 K , the constant pressure specific heat of air is c p = 1.011 kJ / kg ×K (Table A-2b). The gas constant of air is R = 0.287 kJ / kg.K (Table A-1).

Ein - Eout

D Esystem 0 (steady)

Rate of net energy transfer by heat, work, and mass

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 = Wout + mh2 Wout = m(h1 - h2 ) wout = c p (T1 - T2 )

Substituting,

wout = c p (T1 - T2 ) = (1.011 kJ/kg ×K)(425 - 325)K = 101.1 kJ/kg The entropy change of air is

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1

= (1.011 kJ/kg ×K) ln

325 K 110 kPa - (0.287 kJ/kg ×K) ln 425 K 550 kPa

= 0.1907 kJ/kg ×K The maximum (reversible) work is the exergy difference between the inlet and exit states © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-505

wrev,out = c p (T1 - T2 ) - T0 ( s1 - s2 )

= wout - T0 (s1 - s2 ) = 101.1 kJ/kg - (298 K)( - 0.1907 kJ/kg ×K)

= 157.9 kJ / kg The second law efficiency is then hII =

wout 101.1 kJ/kg = = 0.640 = 64.0% wrev,out 157.9 kJ/kg

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=550 [kPa] T1=425 [K] P2=110 [kPa] T2=325 [K] T0=(25+273) [K] "PROPERTIES" R=0.287 [kJ/kg-K] "Table A-2a" c_p=1.011 [kJ/kg-K] "at (425+325)/2= 375 K, Table A-2b" "ANALYSIS" w_out=c_p*(T1-T2) DELTAs=c_p*ln(T2/T1)-R*ln(P2/P1) -w_rev_out=c_p*(T2-T1)-T0*DELTAs eta_II=w_out/w_rev_out

9-106 Steam expands in a two-stage adiabatic turbine from a specified state to specified pressure. Some steam is extracted at the end of the first stage. The wasted power potential is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible. 4

The environment temperature is given to be T0 = 25° C .

Analysis The wasted power potential is equivalent to the rate of exergy destruction during a process, which can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen .

6-506

The total rate of entropy generation during this process is determined by taking the entire turbine, which is a control volume, as the system and applying the entropy balance. Noting that this is a steady-flow process and there is no heat transfer,

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

m1s1 - m2 s2 - m3 s3 + Sgen = 0

m1s1 - 0.1m1s2 - 0.9m1s3 + Sgen = 0 ® Sgen = m1[0.9s3 + 0.1s2 - s1 ] and

X destroyed = T0 Sgen = T0 m1[0.9s3 + 0.1s2 - s1 ]

From the steam tables (Tables A-4 through 6)

P1 = 7 MPa üïï h1 = 3159.2 kJ / kg ý T1 = 400°C ïïþ s1 = 6.4502 kJ / kg ×K P2 = 1.8 MPa ïüï ý h2 s = 2831.3 kJ / kg ïïþ s2 s = s1 and

hT =

h1 - h2 ¾¾ ® h2 = h1 - hT (h1 - h2 s ) h1 - h2 s

= 3159.2 - 0.88(3159.2 - 2831.3) = 2870.6 kJ/kg

ïüï ý s2 = 6.5290 kJ / kg ×K h2 = 2870.6 kJ/kg ïïþ P2 = 1.8 MPa

s3 s - s f 6.64502 - 0.6492 x = = = 0.7735 P3 = 10 kPa ïü ïý 3s s fg 7.4996 ïïþ s3 s = s1 h3s = h f + x3s h fg = 191.81 + 0.7735´ 2392.1 = 2042.1 kJ/kg

and

hT =

h1 - h3 ¾¾ ® h3 = h1 - hT (h1 - h3 s ) h1 - h3 s

= 3159.2 - 0.88(3159.2 - 2042.1) = 2176.1 kJ/kg h3 - h f 2176.1- 191.81 = = 0.8295 ü ïï x3 = h fg 2392.1 ý h3 = 2176.1 kJ/kg ïïþ s3 = s f + x3 s fg = 0.6492 + 0.8295´ 7.4996 = 6.8705 kJ/kg ×K P3 = 10 kPa

Substituting, the wasted work potential is determined to be

X destroyed = T0 Sgen = (298 K)(15 kg/s)(0.9´ 6.8705 + 0.1´ 6.5290 - 6.4502)kJ/kg = 1726 kW

EES (Engineering Equation Solver) SOLUTION "Given" P1=7000 [kPa] T1=400 [C] P2=1800 [kPa] P3=10 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-507

m1_dot=15 [kg/s] m2_dot=0.10*m1_dot eta_T=0.88 T0=(25+273) [K] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2_s=s1 "isentropic process" h2_s=enthalpy(Fluid$, P=P2, s=s2_s) s3_s=s1 "isentropic process" h3_s=enthalpy(Fluid$, P=P3, s=s3_s) x3_s=quality(Fluid$, P=P3, s=s3_s) "Analysis" m3_dot=0.90*m1_dot eta_T=(h1-h2)/(h1-h2_s) s2=entropy(Fluid$, P=P2, h=h2) eta_T=(h1-h3)/(h1-h3_s) s3=entropy(Fluid$, P=P3, h=h3) x3=quality(Fluid$, P=P3, h=h3) m1_dot*s1-m2_dot*s2-m3_dot*s3+S_dot_gen=0 "entropy balance" X_dot_destroyed=T0*S_dot_gen

9-107E Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at a specified pressure. The isentropic and second-law efficiencies of the turbine are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with constant specific heats. Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is c p = 0.1253 Btu / lbm ×R . The gas constant is R = 0.04971 Btu /1bm.R (Table A-2E).

Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the isentropic turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

6-508

mh1 = Ws ,out + mh2 s

(since Q @Δke @Δpe @0)

Ws ,out = m(h1 - h2 s )

From the isentropic relations, ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ èç P ø÷ 1

0.667/1.667

æ 20 psia ÷ ö ÷ = (1350 + 460 R) ççç ÷ è 200 psia ÷ ø

= 720.4 R

Then the power output of the isentropic turbine becomes æ ö 1 hp ÷= 128.8 hp Ws ,out = mc p (T1 - T2 s ) = (40 lbm/min)(0.1253 Btu/lbm ×R)(1810 - 720.4) R çç ÷ ÷ çè 42.41 Btu/min ø

Then the isentropic efficiency of the turbine is determined from

hT =

Wa ,out Ws ,out

105 hp = 0.8153 = 81.5% 128.8 hp

(b) Using the steady-flow energy balance relation Wa ,out = mc p (T1 - T2 ) above, the actual turbine exit temperature is determined to be

T2 = T1 -

Wa ,out mc p

= 1350°F -

æ42.41 Btu/min ö÷ 105 hp çç ÷ ÷= 461.5°F = 921.6 R (40 lbm/min)(0.1253 Btu/lbm ×R) çè 1 hp ø÷

The entropy generation during this process can be determined from an entropy balance on the turbine,

Sin - Sout

Sgen

= D Ssystem ¿ 0 = 0

Rate of entropy generation

Rate of change of entropy

Rate of net entropy transfer by heat and mass

ms1 - ms2 + Sgen = 0 Sgen = m( s2 - s1 ) where

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1 = (0.1253 Btu/lbm ×R) ln

921.6 R 20 psia - (0.04971 Btu/lbm ×R) ln 1810 R 200 psia

= 0.02988 Btu/lbm ×R The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = mT0 (s2 - s1 ) æ ö 1 hp ÷ = (40 lbm/min)(537 R)(0.02988 Btu/lbm ×R) çç ÷ çè 42.41 Btu/min ÷ ø = 15.14 hp Then the reversible power and second-law efficiency become Wrev,out = Wa ,out + X destroyed = 105 + 15.14 = 120.1 hp

and

hII =

Wa,out Wrev,out

105 hp = 0.874 = 87.4% 120.1 hp

6-509

EES (Engineering Equation Solver) SOLUTION "Given" T1=(1350+460) [R] P1=200 [psia] P2=20 [psia] m_dot=40 [lbm/min] W_dot_out=105 [hp] T0=(77+460) [R] "Properties, from Table A-2" R=0.04971 [Btu/lbm-R] c_p=0.1253 [Btu/lbm-R] k=1.667 "Analysis" "(a)" T2_s=T1*(P2/P1)^((k-1)/k) W_dot_s_out*Convert(hp, Btu/min)=m_dot*c_p*(T1-T2_s) eta_T=W_dot_out/W_dot_s_out "(b)" W_dot_out*Convert(hp, Btu/min)=m_dot*c_p*(T1-T2) DELTAS_Ar=c_p*ln(T2/T1)-R*ln(P2/P1) DELTAS_dot_Argon=m_dot*(c_p*ln(T2/T1)-R*ln(P2/P1)) S_dot_gen=DELTAS_dot_Argon X_dot_destroyed=T0*S_dot_gen*Convert(Btu/min, hp) W_dot_rev_out=W_dot_out+X_dot_destroyed eta_II=W_dot_out/W_dot_rev_out

9-108 Steam expands in a two-stage adiabatic turbine from a specified state to another specified state. Steam is reheated between the stages. For a given power output, the reversible power output and the rate of exergy destruction are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible. 4 The environment temperature is given to be T0 = 25° C . Properties From the steam tables (Tables A-4 through 6) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-510

P1 = 8 MPa üïï h1 = 3399.5 kJ / kg ý T1 = 500°C ïïþ s1 = 6.7266 kJ / kg ×K

P2 = 2 MPa ïüï h2 = 3137.7 kJ / kg ý T2 = 350°C ïïþ s2 = 6.9583 kJ / kg ×K P3 = 2 MPa ïüï h3 = 3468.3 kJ / kg ý T3 = 500°C ïïþ s3 = 7.4337 kJ / kg ×K P4 = 30 kPa üïï h4 = h f + x4 h fg = 289.27 + 0.97 ´ 2335.3 = 2554.5 kJ/kg ý x4 = 0.97 ïïþ s4 = s f + x4 s fg = 0.9441 + 0.97 ´ 6.8234 = 7.5628 kJ/kg ×K Analysis We take the entire turbine, excluding the reheat section, as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout mh1 + mh3 = mh2 + mh4 + Wout Wout = m[(h1 - h2 ) + (h3 - h4 )]

Substituting, the mass flow rate of the steam is determined from the steady-flow energy equation applied to the actual process, m=

Wout 5000 kJ/s = = 4.253 kg/s h1 - h2 + h3 - h4 (3399.5 - 3137.7 + 3468.3 - 2554.5)kJ/kg

The reversible (or maximum) power output is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero,

X in - X out Rate of net exergy transfer by heat, work, and mass

- X destroyed Z 0 (reversible) = D X systemZ 0 (steady) = 0 Rate of exergy destruction

Rate of change of exergy

X in = X out

mxflow,1 + mxflow,3 = mxflow,2 + mxflow,4 + Wrev,out Wrev,out = m( xflow,1 - xflow,2 ) + m( xflow,3 - xflow,4 ) = m[(h1 - h2 ) + T0 ( s2 - s1 ) - ΔkeZ 0 - ΔpeZ 0 ] + m[(h3 - h4 ) + T0 ( s4 - s3 ) - ΔkeZ 0 - ΔpeZ 0 ]

Then the reversible power becomes Wrev,out = m [h1 - h2 + h3 - h4 + T0 ( s2 - s1 + s4 - s3 )]

= (4.253 kg/s)[(3399.5 - 3137.7 + 3468.3 - 2554.5)kJ/kg + (298 K)(6.9583 - 6.7266 + 7.5628 - 7.4337)kJ/kg ×K] = 5457 kW Then the rate of exergy destruction is determined from its definition, X destroyed = Wrev,out - Wout = 5457 - 5000 = 457 kW

6-511

EES (Engineering Equation Solver) SOLUTION "Given" P1=8000 [kPa] T1=500 [C] P2=2000 [kPa] T2=350 [C] P3=P2 T3=500 [C] P4=30 [kPa] x4=0.97 W_dot_out=5000 [kW] T0=(25+273) [K] "Properties" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) h2=enthalpy(Fluid$, P=P2, T=T2) s2=entropy(Fluid$, P=P2, T=T2) h3=enthalpy(Fluid$, P=P3, T=T3) s3=entropy(Fluid$, P=P3, T=T3) h4=enthalpy(Fluid$, P=P4, x=x4) s4=entropy(Fluid$, P=P4, x=x4) "Analysis" m_dot*h1+m_dot*h3=W_dot_out+m_dot*h2+m_dot*h4 "energy balance" W_dot_rev_out=m_dot*(h1-h2+h3-h4-T0*(s1-s2+s3-s4)) X_dot_destroyed=W_dot_rev_out-W_dot_out

9-109 A throttle valve is placed in the steam line supplying the turbine inlet in order to control an isentropic steam turbine. The second-law efficiency of this system when the valve is partially open to when it is fully open is to be compared. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is well-insulated, and there is no heat transfer from the turbine. Analysis

Valve is fully open: The properties of steam at various states are P0 = 100 kPa ü ïï h0 @ h f @25°C = 104.83 kJ/kg ý T1 = 25°C ïïþ s0 @ s f @25°C = 0.3672 kJ/kg ×K P1 = P2 = 6 MPa ü ïï h1 = h2 = 3658.8 kJ/kg ý T1 = T2 = 600°C ïïþ s1 = s2 = 7.1693 kJ/kg ×K

6-512

P3 = 40 kPa ü ïï x3 = 0.9248 ý ïïþ h3 = 2461.7 kJ/kg s2 = s1 The stream exergy at the turbine inlet is xflow,1 = h1 - h0 - T0 ( s1 - s0 ) = 3658.8 - 104.83 - (298)(7.1693 - 0.3672) = 1527 kJ / kg

The second law efficiency of the entire system is then hII =

wout h1 - h3 h - h3 = = 1 = 1.0 wrev h1 - h3 - T0 ( s1 - s3 ) h1 - h3

since s1 = s3 for this system. Valve is partly open: ïü ïý s = 7.6674 kJ/kg ×K (from EES) h2 = h1 = 3658.8 kJ/kg ïïþ 2

P2 = 2 MPa

P3 = 40 kPa ïü ïý h = 2635.5 kJ/kg 3 s3 = s2 ïïþ

(from EES)

xflow,2 = h2 - h0 - T0 (s2 - s0 ) = 3658.8 - 104.83 - (298)(7.6674 - 0.3672) = 1378 kJ / kg

hII =

wout h2 - h3 3658.8 - 2635.5 = = = 1.0 wrev h2 - h3 - T0 ( s2 - s3 ) 3658.8 - 2635.5 - (298)(7.6674 - 7.6674)

EES (Engineering Equation Solver) SOLUTION "Given" P1=6000 [kPa] T1=600 [C] P3=40 [kPa] P2=2000 [kPa] T0=25 [C] "Analysis" Fluid$='steam_iapws' "Throttle valve is fully open" h0=enthalpy(Fluid$, T=T0, x=0) "assumed sat. liquid" s0=entropy(Fluid$, T=T0, x=0) "assumed sat. liquid" h1=enthalpy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s3_a=s1 h3_a=enthalpy(Fluid$, P=P3, s=s3_a) x3_a=quality(Fluid$, P=P3, s=s3_a) x_1=h1-h0-(T0+273)*(s1-s0) w_out_a=h1-h3_a w_rev_a=h1-h3_a-(T0+273)*(s1-s3_a) eta_II_a=w_out_a/w_rev_a "Throttle valve is partially closed" h2=h1 s2=entropy(Fluid$, P=P2, h=h2) s3_b=s2 h3_b=enthalpy(Fluid$, P=P3, s=s3_b) x3_b=quality(Fluid$, P=P3, s=s3_b) x_2=h2-h0-(T0+273)*(s2-s0) w_out_b=h2-h3_b w_rev_b=h2-h3_b-(T0+273)*(s2-s3_b) eta_II_b=w_out_b/w_rev_b © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-513

9-110 One ton of liquid water at 65°C is brought into a room. The final equilibrium temperature in the room and the entropy generated are to be determined. Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are constant at room temperature. 3 The system is stationary and thus the kinetic and potential energy changes are zero. 4 There are no work interactions involved. Properties The gas constant of air is R = 0.287 kPa.m3 / kg.K (Table A-1). The constant volume specific heat of water at room temperature is cv = 0.718 kJ / kg.°C (Table A-2). The specific heat of water at room temperature is c = 4.18 kJ / kg ×°C (Table A-3).

Analysis (a) The volume and the mass of the air in the room are V = 3´ 4´ 7 = 84 m3

mair =

P1V (100 kPa)(84 m3 ) = = 101.3 kg RT1 (0.2870 kPa ×m3 /kg ×K)(289 K)

Taking the contents of the room, including the water, as our system, the energy balance can be written as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

0 = D U = (D U )water + (D U )air

émc (T2 - T1 )ù + émcv (T2 - T1 )ù = 0 ë ûwater ë ûair Substituting,

(1000 kg)(4.18 kJ/kg ×°C)(Tf - 65)°C + (101.3 kg)(0.718 kJ/kg ×°C)(Tf - 16)°C = 0 It gives the final equilibrium temperature in the room to be T f = 64.2° C

(b) We again take the room and the water in it as the system, which is a closed system. Considering that the system is wellinsulated and no mass is entering and leaving, the entropy balance for this system can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

6-514

0 + Sgen = D Sair + D Swater

where 0

D Sair = mcv ln

T2 V + mR ln 2 T1 V1

D S water = mc ln

T2 337.2 K = (1000 kg )(4.18 kJ/kg ×K ) ln = - 10.37 kJ / K T1 338 K

= (101.3 kg )(0.718 kJ/kg ×K )ln

337.2 K = 11.21 kJ / K 289 K

Substituting, the entropy generation is determined to be Sgen = 11.21- 10.37 = 0.834 kJ / K

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (283 K)(0.834 kJ/K) = 236 kJ (c) The work potential (the maximum amount of work that can be produced) during a process is simply the reversible work output. Noting that the actual work for this process is zero, it becomes X destroyed = Wrev,out - Wact,out

Wrev,out = X destroyed = 236 kJ

EES (Engineering Equation Solver) SOLUTION "Given" m_w=1000 [kg] T1_w=(65+273) [K] Vol=3*4*7 [m^3] T1_air=(16+273) [K] P1_air=100 [kPa] T0=(10+273) [K] "Properties of air at room temperature" c_p_w=4.18 [kJ/kg-C] "Table A-3" c_v_air=0.718 [kJ/kg-C] "Table A-2a" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" "(a)" m_air=(P1_air*Vol)/(R*T1_air) m_w*c_p_w*(T2-T1_w)+m_air*c_v_air*(T2-T1_air)=0 "energy balance" T2_C=T2-273 "(b)" DELTAS_w=m_w*c_p_w*ln(T2/T1_w) DELTAS_air=m_air*c_v_air*ln(T2/T1_air) DELTAS_total=DELTAS_w+DELTAS_air S_gen=DELTAS_total X_destroyed=T0*S_gen "(c)" W_rev_out=X_destroyed "since there is no actual work"

9-111 The feedwater of a steam power plant is preheated using steam extracted from the turbine. The ratio of the mass flow rates of the extracted steam and the feedwater are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-515

3 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.

Properties The properties of steam and feedwater are (Tables A-4 through A-6)

P1 = 1.6 MPa üïï h1 = 2919.9 kJ/kg ý T1 = 250°C ïïþ s1 = 6.6753 kJ/kg ×K h2 = h f @1.6 MPa = 858.44 kJ/kg P2 = 1.6 MPa ü ïï s ý 2 = s f @ 1.6 MPa = 2.3435 kJ/kg ×K ïïþ sat. liquid T2 = 201.4°C

P3 = 4 MPa ïüï h3 @ h f @ 30 C = 129.37 kJ/kg ý T3 = 30°C ïïþ s3 @ s = 0.4355 kJ/kg ×K f @ 30 C

üïï h4 @ h f @ 191.4°C = 814.78 kJ/kg ý T4 = T2 - 10°C @191.4°C ïïþ s4 @ s f @ 191.4°C = 2.2446 kJ/kg ×K P4 = 4 MPa

Analysis (a) We take the heat exchanger as the system, which is a control volume. The mass and energy balances for this steadyflow system can be expressed in the rate form as follows: Mass balance (for each fluid stream): min - mout = D msystemZ 0 (steady) = 0 ® min = mout ® m1 = m2 = ms

and m3 = m4 = m fw

Energy balance (for the heat exchanger):

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0¾¾ ® Ein = Eout

Rate of change in internal, kinetic, potential, etc. energies

m1h1 + m3 h3 = m2 h2 + m4 h4

(since Q = W = D ke @D pe @ 0)

Combining the two, ms (h2 - h1 ) = m fw (h3 - h4 )

Dividing by m fw and substituting,

(129.37 - 814.78)kJ/kg ms h - h4 = 3 = = 0.3325 m fw h2 - h1 (858.44 - 2919.9)kJ/kg (b) The entropy generation during this process per unit mass of feedwater can be determined from an entropy balance on the feedwater heater expressed in the rate form as © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-516

Sin - Sout Rate of net entropy transfer by heat and mass

Sgen

= D SsystemZ 0 = 0

Rate of entropy generation

Rate of change of entropy

m1s1 - m2 s2 + m3 s3 - m4 s4 + Sgen = 0 ms (s1 - s2 ) + m fw (s3 - s4 ) + Sgen = 0 Sgen m fw

ms (s2 - s1 )+ (s4 - s3 ) = (0.3325)(2.3435 - 6.6753)+ (2.2446 - 0.4355) = 0.3688 kJ/K ×kg fw m fw

Noting that this process involves no actual work, the reversible work and exergy destruction become equivalent since X destroyed = Wrev,out - Wact,out ® Wrev,out = X destroyed . The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen , X destroyed = T0 Sgen = (298 K)(0.3688 kJ / K ×kg fw) = 109.9 kJ / kg feedwater

EES (Engineering Equation Solver) SOLUTION "Input Data" "Steam (let st=steam data):" Fluid$='steam_iapws' T_st[1]=250 [C] P_st[1]=1600 [kPa] P_st[2] = P_st[1] x_st[2]=0 "saturated liquid, quality = 0%" T_st[2]=temperature(Fluid$, P=P_st[2], x=x_st[2]) "Feedwater (let fw=feedwater data):" T_fw[1]=30 [C] P_fw[1]=4000 [kPa] P_fw[2]=P_fw[1] "assume no pressure drop for the feedwater" T_fw[2]=T_st[2]-10 "Surroundings:" T_o = 25 [C] P_o = 100 [kPa] "Assumed value for the surrroundings pressure" "Conservation of mass:" "There is one entrance, one exit for both the steam and feedwater." "Steam: m_dot_st[1] = m_dot_st[2]" "Feedwater: m_dot_fw[1] = m_dot_fw[2]" "Let m_ratio = m_dot_st/m_dot_fw" "Conservation of Energy:" "We write the conservation of energy for steady-flow control volume having two entrances and two exits with the above assumptions. Since neither of the flow rates is know or can be found, write the conservation of energy per unit mass of the feedwater." E_in - E_out =DELTAE_cv DELTAE_cv=0 "Steady-flow requirement" E_in = m_ratio*h_st[1] + h_fw[1] h_st[1]=enthalpy(Fluid$, T=T_st[1], P=P_st[1]) h_fw[1]=enthalpy(Fluid$,T=T_fw[1], P=P_fw[1]) E_out = m_ratio*h_st[2] + h_fw[2] h_fw[2]=enthalpy(Fluid$, T=T_fw[2], P=P_fw[2]) h_st[2]=enthalpy(Fluid$, x=x_st[2], P=P_st[2]) "The reversible work is found when the Exergy destroyed is set equal to zero and where the heat transfer is zero (the feedwater heater is adiabatic)." © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-517

W_rev = m_ratio*(Psi_st[1]-Psi_st[2]) +(Psi_fw[1]-Psi_fw[2]) Psi_st[1]=h_st[1]-h_st_o -(T_o + 273)*(s_st[1]-s_st_o) s_st[1]=entropy(Fluid$,T=T_st[1], P=P_st[1]) h_st_o=enthalpy(Fluid$, T=T_o, P=P_o) s_st_o=entropy(Fluid$, T=T_o, P=P_o) Psi_st[2]=h_st[2]-h_st_o -(T_o + 273)*(s_st[2]-s_st_o) s_st[2]=entropy(Fluid$,x=x_st[2], P=P_st[2]) Psi_fw[1]=h_fw[1]-h_fw_o -(T_o + 273)*(s_fw[1]-s_fw_o) h_fw_o=enthalpy(Fluid$, T=T_o, P=P_o) s_fw[1]=entropy(Fluid$,T=T_fw[1], P=P_fw[1]) s_fw_o=entropy(Fluid$, T=T_o, P=P_o) Psi_fw[2]=h_fw[2]-h_fw_o -(T_o + 273)*(s_fw[2]-s_fw_o) s_fw[2]=entropy(Fluid$,T=T_fw[2], P=P_fw[2])

9-112 Problem 9-111 is reconsidered. The effect of the state of the steam at the inlet of the feedwater heater on the ratio of mass flow rates and the reversible power is to be investigated. Analysis Using EES, the problem is solved as follows: Fluid$='Steam_IAPWS' T_st[1]=250 [C] P_st[1]=1600 [kPa] P_st[2] = P_st[1] x_st[2]=0 T_st[2]=temperature(steam, P=P_st[2], x=x_st[2]) "Surroundings:" T_o = 25 [C] P_o = 100 [kPa] "Feedwater (let fw=feedwater data):" T_fw[1]=30 [C] P_fw[1]=4000 [kPa] P_fw[2]=P_fw[1] T_fw[2]=T_st[2]-10 "Conservation of mass:" "Steam: m_dot_st[1] = m_dot_st[2]" "Feedwater: m_dot_fw[1] = m_dot_fw[2]" "Let m_ratio = m_dot_st/m_dot_fw" "Conservation of Energy:" E_in - E_out =DELTAE_cv DELTAE_cv=0 E_in = m_ratio*h_st[1] + h_fw[1] h_st[1]=enthalpy(Fluid$, T=T_st[1], P=P_st[1]) h_fw[1]=enthalpy(Fluid$,T=T_fw[1], P=P_fw[1]) E_out = m_ratio*h_st[2] + h_fw[2] h_fw[2]=enthalpy(Fluid$, T=T_fw[2], P=P_fw[2]) h_st[2]=enthalpy(Fluid$, x=x_st[2], P=P_st[2]) W_rev = m_ratio*(Psi_st[1]-Psi_st[2]) +(Psi_fw[1]-Psi_fw[2]) Psi_st[1]=h_st[1]-h_st_o -(T_o + 273)*(s_st[1]-s_st_o) s_st[1]=entropy(Fluid$,T=T_st[1], P=P_st[1]) h_st_o=enthalpy(Fluid$, T=T_o, P=P_o) s_st_o=entropy(Fluid$, T=T_o, P=P_o) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-518

Psi_st[2]=h_st[2]-h_st_o -(T_o + 273)*(s_st[2]-s_st_o) s_st[2]=entropy(Fluid$,x=x_st[2], P=P_st[2]) Psi_fw[1]=h_fw[1]-h_fw_o -(T_o + 273)*(s_fw[1]-s_fw_o) h_fw_o=enthalpy(Fluid$, T=T_o, P=P_o) s_fw[1]=entropy(Fluid$,T=T_fw[1], P=P_fw[1]) s_fw_o=entropy(Fluid$, T=T_o, P=P_o) Psi_fw[2]=h_fw[2]-h_fw_o -(T_o + 273)*(s_fw[2]-s_fw_o) s_fw[2]=entropy(Fluid$,T=T_fw[2], P=P_fw[2])

Pst, 1

mratio

Wrev

[kPa] 200 400 600 800 1000 1200 1400 1600 1800 2000

[kg / kg] 0.1361 0.1843 0.2186 0.2466 0.271 0.293 0.3134 0.3325 0.3508 0.3683

[kJ / kg] 42.07 59.8 72.21 82.06 90.35 97.58 104 109.9 115.3 120.3

6-519

9-113 The heating of a passive solar house at night is to be assisted by solar heated water. The amount of heating this water will provide to the house at night and the exergy destruction during this heat transfer process are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3 The house is maintained at 22°C at all times. 4

The outdoor temperature is given to be 5C.

Properties The density and specific heat of water at room temperature are ρ = 997 kg / m3 and c = 4.18 kJ / kg ×°C (Table A3).

Analysis The total mass of water is mw = r V = (997 kg/m3 )(0.270 m3 ) = 269.2 kg

The amount of heat this water storage system can provide is determined from an energy balance on the 350-L water storage system Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Qout = D U system = mc(T2 - T1 ) water

Substituting,

6-520

The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the water and its immediate surroundings so that the boundary temperature of the extended system is the environment temperature at all times. It gives

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Swater Tb,out

Substituting,

Sgen = D S water +

Qout æ Q T ö ÷ = çççmc ln 2 ÷ + out ÷ ÷ Tb,out çè T1 øwater Troom

= (269.2 kg)(4.18 kJ/kg ×K )ln

295 K 25,880 kJ + 318 K 295 K

= 3.252 kJ / K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (278 K)(3.252 kJ/K) = 904 kJ

EES (Engineering Equation Solver) SOLUTION "Given" T1=45[C]+273 [K] T2=22[C]+273 [K] V=0.270 [m^3] T0=5[C]+273 [K] "Properties, from Table A-3" rho=997 [kg/m^3] c_p=4.18 [kJ/kg-K] "Analysis" m=rho*V -Q_out=m*c_p*(T2-T1) "energy balance" -Q_out/T2+S_gen=DELTAS_water "entropy balance" DELTAS_water=m*c_p*ln(T2/T1) X_destroyed=T0*S_gen

9-114 The heating of a passive solar house at night is to be assisted by solar heated water. The length of time that the electric heating system would run that night, the exergy destruction, and the minimum work input required that night are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3 The house is maintained at 22°C at all times. 4

The environment temperature is given to be T0 = 5° C .

Properties The density and specific heat of water at room temperature are ρ = 1 kg / L and c = 4.18 kJ / kg ×°C (Table A3).

6-521

Analysis (a) The total mass of water is mw = r V = (1 kg/L)(50´ 20 L) = 1000 kg

Taking the contents of the house, including the water as our system, the energy balance relation can be written as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in - Qout = D U = (D U ) water + (D U )air Z 0

= (D U )water = mc(T2 - T1 )water

We,in D t - Qout = [mc(T2 - T1 )]water

Substituting,

(15 kJ / s)D t - (50,000 kJ / h)(10 h) = (1000 kg) (4.18 kJ / kg ×° C)(22 - 80)° C It gives t = 17,170 s = 4.77 h (b) We take the house as the system, which is a closed system. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the house and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for the extended system can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qout + Sgen = D Swater + D Sair ¿ 0 = D Swater Tb,out

since the state of air in the house remains unchanged. Then the entropy generated during the 10-h period that night is

Sgen = D S water +

ö Qout æ Q T ÷ = çççmc ln 2 ÷ + out ÷ ÷ ç Tb,out è T1 øwater T0

= (1000 kg )(4.18 kJ/kg ×K )ln

295 K 500,000 kJ + = 1048 kJ/K 353 K 278 K

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (278 K)(1048 kJ/K) = 291,400 kJ (c) The actual work input during this process is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-522

Wact,in = Wact,in D t = (15 kJ/s)(17,170 s) = 257,550 kJ

The minimum work with which this process could be accomplished is the reversible work input, Wrev, in . which can be determined directly from Wrev,in = Wact,in - X destroyed = 257,550 - 291, 400 = - 33,850 kJ Wrev,out = 33,850 kJ = 9.40 kWh

That is, 9.40 kWh of electricity could be generated while heating the house by the solar heated water (instead of consuming electricity) if the process was done reversibly.

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_out=50000[kJ/h]*Convert(kJ/h, kW) T0=3[C]+273 [K] T1=80[C]+273 [K] T2=22[C]+273 [K] time_heatloss=10*3600 [s] V=50*0.020 [m^3] W_dot_e_in=15 [kW] "Properties" rho=1000 [kg/m^-3] c_p=4.18 [kJ/kg-K] "Analysis" "(a)" m=rho*V W_dot_e_in*time-Q_dot_out*time_heatloss=m*C_p*(T2-T1) "energy balance on house" time_h=time*1/Convert(h, s) "(b)" Q_out=Q_dot_out*time_heatloss DELTAS_water=m*C_p*ln(T2/T1) -Q_out/T0+S_gen=DELTAS_water "entropy balance" X_destroyed=T0*S_gen "(c)" W_e_in=W_dot_e_in*time W_rev_in=(W_e_in-X_destroyed)*1/Convert(kWh, kJ)

9-115 A rigid tank containing nitrogen is considered. Nitrogen is allowed to escape until the mass of nitrogen becomes onehalf of its initial mass. The change in the nitrogen's work potential is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing duri ng the process. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Nitrogen is an ideal gas with constant specific heats.

6-523

Properties The properties of nitrogen at room temperature are c p = 1.039 kJ / kg ×K , cv = 0.743 kJ / kg ×K , k = 1.4, and R = 0.2968 kJ / kg.K (Table A-2a).

Analysis The initial and final masses in the tank are m1 =

PV (1000 kPa)(0.100 m3 ) = = 1.150 kg RT1 (0.2968 kPa ×m3 /kg ×K)(293 K)

m2 = me =

m1 1.150 kg = = 0.575 kg 2 2

We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - mout = D msystem ® me = m2

Energy balance: Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- me he = m2u2 - m1u1

Using the average of the initial and final temperatures for the exiting nitrogen, this energy balance equation becomes

- me he = m2u2 - m1u1 - me c pTe = m2 cv T2 - m1cv T1

- (0.575)(1.039)(0.5)(293 + T2 ) = (0.575)(0.743)T2 - (1.150)(0.743)(293) Solving for the final temperature, we get T2 = 224.3 K

The final pressure in the tank is P2 =

m2 RT2 (0.575 kg)(0.2968 kPa ×m3 /kg ×K)(224.3 K) = = 382.8 kPa V 0.100 m3

The average temperature and pressure for the exiting nitrogen is Te = 0.5(T1 + T2 ) = 0.5(293 + 224.3) = 258.7 K Pe = 0.5( P1 + P2 ) = 0.5(1000 + 382.8) = 691.4 kPa

The work potential associated with this process is equal to the exergy destroyed during the process. The exergy destruction during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . The entropy generation S gen in this case is determined from an entropy balance on the system:

6-524

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

- me se + Sgen = D S tank = m2 s2 - m1s1

Sgen = m2 s2 - m1s1 + me se

Rearranging and substituting gives Sgen = m2 s2 - m1s1 + me se = m2 (c p ln T2 - R ln P2 ) - m1 (c p ln T1 - R ln P1 ) + me (c p ln Te - R ln Pe )

= (0.575) [1.039 ln(224.3) - (0.2968) ln(382.8) ]- (1.15) [1.039 ln(293) - (0.2968) ln(1000) ]+ (0.575) [1.039 ln(258.7) - (0.2968) ln

= 2.2188 - 4.4292 + 2.2032 = - 0.007152 kJ/K Then, Wrev = X destroyed = T0 Sgen = (293 K)(- 0.007152 kJ/K) = - 2.10 kJ

The entropy generation cannot be negative for a thermodynamically possible process. This result is probably due to using average temperature and pressure values for the exiting gas and using constant specific heats for nitrogen. This sensitivity occurs because the entropy generation is very small in this process. Alternative More Accurate Solution This problem may also be solved by considering the variation of gas temperature and pressure at the outlet of the tank. The mass balance in this case is me = -

dm dt

which when combined with the reduced first law gives d (mu ) dm = h dt dt

Using the specific heats and the ideal gas equation of state reduces this to cv

V dP dm = c pT R dt dt

which upon rearrangement and an additional use of ideal gas equation of state becomes

1 dP c p 1 dm = P dt cv m dt When this is integrated, the result is k

1.4 æm ö æ1 ö ÷ ç ÷ P2 = P1 ççç 2 ÷ = 1000 = 378.9 kPa ÷ ç ÷ çè 2 ÷ ÷ ø èç m ø 1

The final temperature is then

T2 =

P2V (378.9 kPa)(0.100 m3 ) = = 222.0 K m2 R (0.575 kg)(0.2968 kPa ×m3 /kg ×K)

The process is then one of mk mk - 1 = const or = const P T The reduced combined first and second law becomes Wrev = -

d (U - T0 S ) dm + (h - T0 s) dt dt

6-525

when the mass balance is substituted and the entropy generation is set to zero (for maximum work production). Replacing the enthalpy term with the first law result and canceling the common dU /dt term reduces this to Wrev = T0

d (ms) dm - T0 s dt dt

Expanding the first derivative and canceling the common terms further reduces this to Wrev = T0 m

ds dt

Letting a = P1 / m1k and b = T1 / m1k - 1 , the pressure and temperature of the nitrogen in the system are related to the mass by P = am k and

T = bm k - 1

according to the first law. Then, k- 2 dP = akm k - 1dm and dT = b(k - 1)m dm

The entropy change relation then becomes

dT dP é dm - R = (k - 1)c p - Rk ùúû T P êë m Now, multiplying the combined first and second laws by dt and integrating the result gives ds = c p

Wrev = T0 ò mds = T0 ò mds éêë(k - 1)c p - Rk ùúûdm = T0 éêë(k - 1)c p - Rk ù ú û(m2 - m1 ) = (293) [(1.4 - 1)(1.039) - (0.2968)(1.4) ](0.575 - 1.15)

= - 0.0135 kJ Once again the entropy generation is negative, which cannot be the case for a thermodynamically possible process. This is probably due to using constant specific heats for nitrogen. This sensitivity occurs because the entropy generation is very small in this process.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.100 [m^3] P1=1000 [kPa] T1=(20+273) [K] m2=0.5*m1 T0=(20+273) [K] "assumed" "Properties of air at room temperature, Table A-2a" c_p=1.039 [kJ/kg-K] c_v=0.743 [kJ/kg-K] R=0.2968 [kJ/kg-K] k=1.4 "Analysis" m1=(P1*Vol)/(R*T1) m_e=m2 -m_e*c_p*T_e=m2*c_v*T2-m1*c_v*T1 T_e=0.5*(T1+T2) P2=(m2*R*T2)/Vol P_e=0.5*(P1+P2) S_gen=m2*(c_p*ln(T2)-R*ln(P2))-m1*(c_p*ln(T1)-R*ln(P1))+m_e*(c_p*ln(T_e)-R*ln(P_e)) W_rev=T0*S_gen "Alternative more accurate solution" P2_alt=P1*(m2/m1)^k © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-526

T2_alt=(P2_alt*Vol)/(m2*R) W_rev_alt=T0*((k-1)*c_p-R*k)*(m2-m1)

9-116 A pressure cooker is initially half-filled with liquid water. It is kept on the heater for 30 min. The amount water that remained in the cooker and the exergy destroyed are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of water vapor leaving the cooker remains constant. 2 Kinetic and potential energies are negligible. 3 Heat loss from the cooker is negligible.

Properties The properties of water are (Tables A-4 through A-6) P1 = 175 kPa ® v f = 0.001057 m3 /kg, v g = 1.0037 m 3 /kg u f = 486.82 kJ/kg, ug = 2524.5 kJ/kg

s f = 1.4850 kJ/kg.K, sg = 7.1716 kJ/kg.K Pe = 175 kPa ïü ïý he = hg @175 kPa = 2700.2 kJ/kg ïïþ se = sg @175 kPa = 7.1716 kJ/kg ×K sat. vapor

Analysis (a) We take the cooker as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min - mout = D msystem ® me = m1 - m2 Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

We,in - me he = m2u2 - m1u1 (since Q @ ke @ pe @ 0)

The initial mass, initial internal energy, initial entropy, and final mass in the tank are V f = Vg = 2 L = 0.002 m3

m1 = m f + mg =

Vf vf

Vg vg

0.002 m3 0.002 m 3 + = 1.893 + 0.002 = 1.8945 kg 3 0.001057 m /kg 1.0037 m3 /kg

U1 = m1u1 = m f u f + mg ug = 1.893´ 486.82 + 0.002´ 2524.5 = 926.6 kJ S1 = m1s1 = m f s f + mg sg = 1.892´ 1.4850 + 0.002´ 7.1716 = 2.8239 kJ/K

6-527

m2 =

V 0.004 m3 = v2 v2

The amount of electrical energy supplied during this process is We,in = We ,in D t = (0.750 kJ/s)(20´ 60 s) = 900 kJ

Then from the mass and energy balances,

me = m1 - m2 = 1.894 900 kJ = (1.894 -

0.004 v2

0.004 0.004 )(2700.2 kJ/kg) + ( )(u2 ) - 926.6 kJ v2 v2

Substituting u2 = u f + x2u fg and v 2 = v f + x2v fg , and solving for x2 yields

x2 = 0.001918 Thus, v 2 = v f + x2v fg = 0.001057 + 0.001918´ (1.0037 - 0.001057) = 0.002654 m3 / kg

s2 = s f + x2 s fg = 1.4850 + 0.001918´ 5.6865 = 1.5642 kJ / kg ×K

and m2 =

V 0.004 m3 = = 1.507 kg v2 0.002654 m3 / kg

(b) The entropy generated during this process is determined by applying the entropy balance on the cooker. Noting that there is no heat transfer and some mass leaves, the entropy balance can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

- me se + Sgen = D Ssys = m2 s2 - m1s1 Sgen = me se + m2 s2 - m1s1

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . Using the S gen relation obtained above and substituting, X destroyed = T0 Sgen = T0 (me se + m2 s2 - m1s1 ) = (298 K) [(1.894 - 1.507) ´ 7.1716 + 1.507 ´ 1.5642 - 2.8239]

= 689 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.004 [m^3] P1=175 [kPa] Vol1_f=0.50*Vol Vol1_g=0.50*Vol P_e=P1 x_e=1 W_dot_e_in=0.750 [kW] time=20*60 [s] T0=25[C]+273 [K] P0=100 [kPa] "Properties" Fluid$='steam_iapws' v1_f=volume(Fluid$, P=P1, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-528

v1_g=volume(Fluid$, P=P1, x=1) h_e=enthalpy(Fluid$, P=P_e, x=x_e) s_e=entropy(Fluid$, P=P_e, x=x_e) "Analysis" "(a)" m1_f=Vol1_f/v1_f m1_g=Vol1_g/v1_g m1=m1_f+m1_g x1=m1_g/m1 u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) m2=Vol/v2 m_e=m1-m2 "mass balance" W_e_in=W_dot_e_in*time W_e_in=m_e*h_e+m2*u2-m1*u1 "energy balance" v2=volume(Fluid$, u=u2, x=x2) u2=intenergy(Fluid$, v=v2, x=x2) s2=entropy(Fluid$, v=v2, x=x2) "(b)" -m_e*s_e+S_gen=m2*s2-m1*s1 "entropy balance" X_destroyed=T0*S_gen

9-117 A pressure cooker is initially half-filled with liquid water. Heat is transferred to the cooker for 30 min. The amount water that remained in the cooker and the exergy destroyed are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of water vapor leaving the cooker remains constant. 2 Kinetic and potential energies are negligible. 3 Heat loss from the cooker is negligible.

Properties The properties of water are (Tables A-4 through A-6) P1 = 175 kPa ® v f = 0.001057m3 /kg, v g = 1.0037 m3 /kg

u f = 486.82 kJ/kg, ug = 2524.5 kJ/kg s f = 1.4850 kJ/kg.K, sg = 7.1716 kJ/kg.K Pe = 175 kPa ü ïï he = hg @175 kPa = 2700.2 kJ/kg ý ïïþ se = sg @175 kPa = 7.1716 kJ/kg ×K sat. vapor

Analysis

6-529

(a) We take the cooker as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: Energy balance:

min - mout = D msystem ® me = m1 - m2 Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 (since W @ ke @ pe @0) The initial mass, initial internal energy, initial entropy, and final mass in the tank are V f = Vg = 2 L = 0.002 m3

m1 = m f + mg =

Vg vg

0.002 m3 0.002 m 3 + = 1.893 + 0.002 = 1.8945 kg 0.001057 m3 /kg 1.0037 m3 /kg

U1 = m1u1 = m f u f + mg ug = 1.893´ 486.82 + 0.002´ 2524.5 = 926.6 kJ

S1 = m1s1 = m f s f + mg sg = 1.892´ 1.4850 + 0.002´ 7.1716 = 2.8239 kJ/K m2 =

V 0.004 m3 = v2 v2

The amount of heat transfer during this process is Q = QD t = (0.750 kJ/s)(20´ 60 s) = 900 kJ

Then from the mass and energy balances,

me = m1 - m2 = 1.894 -

900 kJ = (1.894 Substituting u2 = u f + x2u fg

and

0.004 v2

0.004 0.004 ) (2700.2 kJ/kg) + ( ) (u2 ) - 926.6 kJ v2 v2 v 2 = v f + x2v fg , and solving for x2 yields

x2 = 0.001918 Thus, v 2 = v f + x2v fg = 0.001057 + 0.001918´ (1.0037 - 0.001057) = 0.002654 m3 / kg

s2 = s f + x2 s fg = 1.4850 + 0.001918´ 5.6865 = 1.5642 kJ / kg ×K

and

m2 =

V 0.004 m3 = = 1.507 kg v 2 0.002654 m3 / kg

(b) The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the cooker and its immediate surroundings so that the boundary temperature of the extended system at the location of heat transfer is the heat source temperature, Tsource = 180° C at all times. The entropy balance for it can be expressed as Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin - me se + Sgen = D Ssys = m2 s2 - m1s1 Tb,in Sgen = me se + m2 s2 - m1 s1 -

Qin Tsource

6-530

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . Using the S gen relation obtained above and substituting,

æ ö Qin ÷ ÷ X destroyed = T0 Sgen = T0 çççme se + m2 s2 - m1 s1 ÷ çè Tsource ÷ ø = (298 K) [(1.894 - 1.507) ´ 7.1716 + 1.507 ´ 1.5642 - 2.8239 - 900 / 453]

= 96.8 kJ Note that the exergy destroyed is much less when heat is supplied from a heat source rather than an electric resistance heater.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.004 [m^3] P1=175 [kPa] Vol1_f=0.50*Vol Vol1_g=0.50*Vol P_e=P1 x_e=1 time=20*60 [s] Q_dot_in=0.750 [kW] T_source=180[C]+273 [K] T0=25[C]+273 [K] P0=100 [kPa] "Properties" Fluid$='steam_iapws' v1_f=volume(Fluid$, P=P1, x=0) v1_g=volume(Fluid$, P=P1, x=1) h_e=enthalpy(Fluid$, P=P_e, x=x_e) s_e=entropy(Fluid$, P=P_e, x=x_e) "Analysis" "(a)" m1_f=Vol1_f/v1_f m1_g=Vol1_g/v1_g m1=m1_f+m1_g x1=m1_g/m1 u1=intenergy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) m2=Vol/v2 m_e=m1-m2 "mass balance" Q_in=Q_dot_in*time Q_in=m_e*h_e+m2*u2-m1*u1 "energy balance" v2=volume(Fluid$, u=u2, x=x2) u2=intenergy(Fluid$, v=v2, x=x2) s2=entropy(Fluid$, v=v2, x=x2) "(b)" Q_in/T_source-m_e*s_e+S_gen=m2*s2-m1*s1 "entropy balance" X_destroyed=T0*S_gen

9-118 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established and the amount of exergy destroyed are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-531

Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). Properties The gas constant of air is 0.287 kPa.m3 / kg.K (Table A-1).

Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances can be expressed as Mass balance: min - mout = D msystem ® mi = m2 (since mout = minitial = 0) Energy balance:

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin + mi hi = m2u2 (since W @ Eout = Einitial = ke @ pe @ 0) Combining the two balances: Qin = m2 (u2 - hi ) where

(100 kPa )(0.020 m ) P2V = = 0.02338 kg RT2 (0.287 kPa ×m 3 /kg ×K )(298 K ) 3

m2 =

A-17 Ti = T2 = 298 K ¾ Table ¾¾ ¾®

hi = 298.18 kJ/kg u2 = 212.64 kJ/kg

Substituting,

Qin = (0.02338 kg)(212.64 - 298.18) kJ / kg = - 2.0 kJ ® Qout = 2.0kJ Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the bottle and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be expressed as

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

mi si -

Entropy generation

Change in entropy

Qout + Sgen = D Stank = m2 s2 - m1s1¿ 0 = m2 s2 Tb,in

6-532

Therefore, the total entropy generated during this process is

Sgen = - mi si + m2 s2 +

Qout Q Q 2.0 kJ ¿0 = m2 (s2 - si ) + out = out = = 0.006711 kJ/K Tb,out Tb,out Tsurr 298 K

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (298 K)(0.006711 kJ/K) = 2.0 kJ

EES (Engineering Equation Solver) SOLUTION "Given" V=0.020 [m^3] P_i=100 [kPa] T_i=(25+273) [K] T2=T_i P2=P_i T0=T_i "Properties" Fluid$='air' h_i=enthalpy(Fluid$, T=T_i) s_i=entropy(Fluid$, T=T_i, P=P_i) u2=intenergy(Fluid$, T=T2) s2=entropy(Fluid$, T=T2, P=P2) R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" m_i=m2 "mass balance" m2=(P2*V)/(R*T2) Q_in+m_i*h_i=m2*u2 "energy balance" m_i*s_i+Q_in/T_i+S_gen=m2*s2 "entropy balance" X_destroyed=T0*S_gen

9-119 A rigid tank containing nitrogen is considered. Heat is now transferred to the nitrogen from a reservoir and nitrogen is allowed to escape until the mass of nitrogen becomes one-half of its initial mass. The change in the nitrogen's work potential is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Nitrogen is an ideal gas with constant specific heats.

6-533

Properties The properties of nitrogen at room temperature are c p = 1.039 kJ / kg ×K , cv = 0.743 kJ / kg ×K , and

R = 0.2968 kJ / kg.K (Table A-2a). Analysis The initial and final masses in the tank are

m1 =

PV (1200 kPa)(0.050 m3 ) = = 0.690 kg RT1 (0.2968 kPa ×m3 /kg ×K)(293 K)

m1 0.690 kg = = 0.345 kg 2 2 The final temperature in the tank is m2 = me =

T2 =

PV (1200 kPa)(0.050 m3 ) = = 586 K m2 R (0.345 kg)(0.2968 kPa ×m3 /kg ×K)

me = m2

Energy balance:

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - me he = m2u2 - m1u1 Qout = me he + m2u2 - m1u1 Using the average of the initial and final temperatures for the exiting nitrogen, Te = 0.5(T1 + T2 ) = 0.5(293 + 586) = 439.5 K this energy balance equation becomes Qout = me he + m2u2 - m1u1

= me c pTe + m2 cv T2 - m1cv T1

= (0.345)(1.039)(439.5) + (0.345)(0.743)(586) - (0.690)(0.743)(293)

= 157.5 kJ The work potential associated with this process is equal to the exergy destroyed during the process. The exergy destruction during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen . The entropy generation S gen in this case is determined from an entropy balance on the system:

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

6-534

Qin - me se + Sgen = D S tank =m2 s2 - m1 s1 TR

Sgen = m2 s2 - m1 s1 + me se -

Qin TR

Noting that pressures are same, rearranging and substituting gives

Sgen = m2 s2 - m1 s1 + me se -

Qin TR

= m2 c p ln T2 - m1c p ln T1 + me c p ln Te -

Qin TR

= (0.345)(1.039) ln(586) - (0.690)(1.039) ln(293) + (0.345)(1.039) ln(439.5) -

157.5 773

= 0.190 kJ/K Then,

Wrev = X destroyed = T0 Sgen = (293 K)(0.190 kJ/K) = 55.7 kJ Alternative More Accurate Solution This problem may also be solved by considering the variation of gas temperature at the outlet of the tank. The mass and energy balances are me = Q=

dm dt

d (mu ) dm cv d (mT ) dm - h = - c pT dt dt dt dt

Combining these expressions and replacing T in the last term gives Q = cv

d (mT ) c p PV dm dt Rm dt

Integrating this over the time required to release one-half the mass produces

Q = cv (m2T2 - m1T1 ) -

c p PV R

m2 m1

The reduced combined first and second law becomes

æ T ö d (U - T0 S ) dm ÷ Wrev = Q ççç1- 0 ÷ + (h - T0 s) ÷ çè TR ÷ dt dt ø when the mass balance is substituted and the entropy generation is set to zero (for maximum work production). Expanding the system time derivative gives æ T0 ö ÷- d (mu - T0 ms ) + (h - T s ) dm ÷ Wrev = Q ç ç 0 ç1- T ÷ ÷ ç dt dt è R ø

æ T ö d (mu ) ds dm dm ÷ = Q ççç1- 0 ÷ + T0 m + T0 s + (h - T0 s ) ÷ çè TR ÷ dt dt dt dt ø æ T ö d (mu ) T dh dm ÷ = Q ççç1- 0 ÷ +h +m 0 ÷ çè TR ø÷ dt dt T dt Substituting Q from the first law,

éd (mu ) T ö éd (mu ) T dh dm ùæ dm ù ÷ úççç1- 0 ÷ ú+ m 0 Wrev = ê - h - ê - h ÷ dt ûúçè TR ø÷ ëê dt dt ûú T dt ëê dt © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-535

T0 éd (mu ) dm dh ù ê - h - m ú ú TR ëê dt dt dt û

T0 é d (mT ) dm dT ù êcv ú - c pT - mc p TR êë dt dt dt ú û

At any time, T=

PV mR

which further reduces this result to

Wrev =

æc p dT R dP ö÷ T0 PV dm ÷ cp + T0 m ççç ÷ çè T dt P dt ø÷ TR mR dt

When this integrated over the time to complete the process, the result is

Wrev =

ö c p PV æ T0 c p PV m2 çç 1 - 1 ÷ ÷ ln + T0 ÷ ç ÷ ç TR R m1 R èT1 T2 ø

ö 293 (1.039)(1200)(0.050) 1 (1.039)(1200)(0.050) æ ÷ çç 1 - 1 ÷ ln + (293) ÷ çè293 586 ø 773 0.2968 2 0.2968

= 49.8 kJ

EES (Engineering Equation Solver) SOLUTION "Given" Vol=0.050 [m^3] P=1200 [kPa] P1=1200 [kPa] T1=(20+273) [K] T_R=(500+273) [K] T0=(20+273) [K] "Properties of air at room temperature, Table A-2a" c_p=1.039 [kJ/kg-K] c_v=0.743 [kJ/kg-K] R=0.2968 [kJ/kg-K] "Analysis" m1=(P1*Vol)/(R*T1) m2=m1/2 m_e=m2 T2=(P*Vol)/(m2*R) T_e=0.5*(T1+T2) Q_out=m_e*c_p*T_e+m2*c_v*T2-m1*c_v*T1 S_gen=m2*c_p*ln(T2)-m1*c_p*ln(T1)+m_e*c_p*ln(T_e)-Q_out/T_R W_rev=T0*S_gen "Alternative more accurate solution" W_rev_alt=T0/T_R*c_p*P*Vol/R*ln(m2/m1)+T0*c_p*P*Vol/R*(1/T1-1/T2)

9-120 Argon gas in a piston–cylinder device expands isothermally as a result of heat transfer from a furnace. The useful work output, the exergy destroyed, and the reversible work are to be determined. Assumptions 1 Argon at specified conditions can be treated as an ideal gas since it is well above its critical temperature of 151 K. 2 The kinetic and potential energies are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-536

Analysis We take the argon gas contained within the piston–cylinder device as the system. This is a closed system since no mass crosses the system boundary during the process. We note that heat is transferred to the system from a source at 1200 K, but there is no heat exchange with the environment at 300 K. Also, the temperature of the system remains constant during the expansion process, and its volume doubles, that is, T2 = T1 and V2 = 2V1 . (a) The only work interaction involved during this isothermal process is the quasi-equilibrium boundary work, which is determined from 2

W = Wb = ò PdV = P1V1 ln 1

V2 0.02 m3 = (350 kPa)(0.01 m3 )ln = 2.43 kPa ×m3 = 2.43 kJ V1 0.01 m3

This is the total boundary work done by the argon gas. Part of this work is done against the atmospheric pressure P0 to push the air out of the way, and it cannot be used for any useful purpose. It is determined from

Wsurr = P0 (V2 - V1 ) = (100 kPa)(0.02 - 0.01)m3 = 1 kPa ×m3 = 1 kJ The useful work is the difference between these two:

Wu = W - Wsurr = 2.43 - 1 = 1.43 kJ That is, 1.43 kJ of the work done is available for creating a useful effect such as rotating a shaft. Also, the heat transfer from the furnace to the system is determined from an energy balance on the system to be Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin - Qb,out = D U = mcv D T = 0 Qin = Qb,out = 2.43 kJ

(b) The exergy destroyed during a process can be determined from an exergy balance, or directly from X destroyed = T0 Sgen . We will use the second approach since it is usually easier. But first we determine the entropy generation by applying an entropy balance on an extended system (system + immediate surroundings), which includes the temperature gradient zone between the cylinder and the furnace so that the temperature at the boundary where heat transfer occurs is TR = 1200 K . This way, the entropy generation associated with the heat transfer is included. Also, the entropy change of the argon gas can be determined from Q / Tsys since its temperature remains constant.

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Q Q + Sgen = D Ssys = TR Tsys Therefore, Sgen =

Q Q 2.43 kJ 2.43 kJ = = 0.00405 kJ/K Tsys TR 400 K 1200 K

and X dest = T0 Sgen = (300 K)(0.00405 kJ/K) = 1.22 kJ

(c) The reversible work, which represents the maximum useful work that could be produced Wrev,out, can be determined from the exergy balance by setting the exergy destruction equal to zero, X in - X out Net exergy transfer by heat, work, and mass

- X destroyedZ 0 (reversible) = D X system Exergy destruction

Change in exergy

æ T0 ö÷ çç1- ÷Q - W rev,out = X 2 - X 1 ççè T ÷ ÷ ø b

= (U 2 - U1 ) + P0 (V2 - V1 ) - T0 (S2 - S1 )

6-537

= 0 + Wsurr - T0

Q Tsys

since KE = PE = 0 and U = 0 (the change in internal energy of an ideal gas is zero during an isothermal process), and D Ssys = Q / Tsys for isothermal processes in the absence of any irreversibilities. Then,

Wrev,out = T0

æ T ö Q ÷ - Wsurr + ççç1- 0 ÷ ÷Q ÷ çè TR ø Tsys

= (300 K)

æ 300 K ö÷ 2.43 kJ - (1 kJ) + çç1÷(2.43 kJ) çè 1200 K ø÷ 400 K

= 2.65 kJ Therefore, the useful work output would be 2.65 kJ instead of 1.43 kJ if the process were executed in a totally reversible manner. Alternative Approach The reversible work could also be determined by applying the basics only, without resorting to exergy balance. This is done by replacing the irreversible portions of the process by reversible ones that create the same effect on the system. The useful work output of this idealized process (between the actual end states) is the reversible work. The only irreversibility the actual process involves is the heat transfer between the system and the furnace through a finite temperature difference. This irreversibility can be eliminated by operating a reversible heat engine between the furnace at 1200 K and the surroundings at 300 K. When 2.43 kJ of heat is supplied to this heat engine, it produces a work output of

æ T ö æ 300 K ÷ ö ÷ WHE = hrev QH = ççç1- L ÷ QH = çç1(2.43 kJ) = 1.82 kJ ÷ ÷ ÷ ç è 1200 K ø èç TH ø÷ The 2.43 kJ of heat that was transferred to the system from the source is now extracted from the surrounding air at 300 K by a reversible heat pump that requires a work input of

WHP,in =

QH QH 2.43 kJ = = = 0.61 kJ COPHP TH / (TH - TL ) (400 K) / (400 - 300)K

Then the net work output of this reversible process (i.e., the reversible work) becomes

Wrev = Wu + WHE - WHP,in = 1.43 + 1.82 - 0.61 = 2.64 kJ which is practically identical to the result obtained before. Also, the exergy destroyed is the difference between the reversible work and the useful work, and is determined to be

X dest = Wrev,out - Wu,out = 2.65 - 1.43 = 1.22 kJ which is identical to the result obtained before.

EES (Engineering Equation Solver) SOLUTION "Given" Vol1=0.01 [m^3] T1=400 [K] P1=350 [kPa] T_R=1200 [K] T2=T1 Vol2=2*Vol1 T0=300 [K] P0=100 [kPa] "Analysis" "(a)" W_b_out=P1*Vol1*ln(Vol2/Vol1) W_surr=P0*(Vol2-Vol1) W_u=W_b_out-W_surr Q_in-W_b_out=DELTAU DELTAU=0 "Temperature is constant" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-538

"(b)" S_gen=Q_in/T1-Q_in/T_R X_dest=T0*S_gen "(c)" W_rev=T0*Q_in/T1-W_surr+(1-T0/T_R)*Q_in "Alternative Solution" eta_rev=1-T0/T_R W_HE=eta_rev*Q_in COP_HP_rev=T1/(T1-T0) W_HP=Q_in/COP_HP_rev W_rev_alt=W_u+W_HE-W_HP X_dest_alt=W_rev_alt-W_u

9-121 A heat engine operates between a nitrogen tank and an argon cylinder at different temperatures. The maximum work that can be produced and the final temperatures are to be determined. Assumptions Nitrogen and argon are ideal gases with constant specific heats at room temperature. Properties The constant volume specific heat of nitrogen at room temperature is cv = 0.743 kJ / kg ×K . The constant pressure specific heat of argon at room temperature is c p = 0.5203 kJ / kg ×K (Table A-2).

Analysis For maximum power production, the entropy generation must be zero. We take the tank, the cylinder (the heat source and the heat sink) and the heat engine as the system. Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as

Sin - Sout + Sgen Z 0 = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

0 + Sgen Z 0 = D S tank,source + D Scylinder,sink + D Sheat engineZ 0

(D S )source + (D S )sink = 0 ¿0ö ¿0ö æ æ ççmc ln T2 - mR ln V2 ÷ ççmc ln T2 - mR ln P2 ÷ ÷ ÷ + 0 + ÷ ÷ = 0 çç v T ç p T ÷ ÷ V1 ø P1 ø è èç 1 1 source sink

Substituting,

(30 kg)(0.743 kJ/kg ×K) ln

T2 T + (15 kg)(0.5203 kJ/kg ×K) ln 2 = 0 900 K 300 K

Solving for T2 yields

6-539

where T2 is the common final temperature of the tanks for maximum power production. The energy balance

Ein - Eout = D Esystem for the source and sink can be expressed as follows: Source: - Qsource,out = D U = mcv (T2 - T1 A ) ® Qsource,out = mcv (T1 A - T2 )

Qsource,out = mcv (T1 A - T2 ) = (30 kg)(0.743 kJ/kg ×K)(900 - 676.9)K = 4973 kJ Sink: Qsink,in - Wb,out = D U ® Qsink,in = D H = mc p (T2 - T1 A )

Qsink,in = mcv (T2 - T1A ) = (15 kg)(0.5203 kJ/kg ×K)(676.9 - 300)K = 2941 kJ Then the work produced becomes

Wmax,out = QH - QL = Qsource,out - Qsink,in = 4973 - 2941 = 2032 kJ Therefore, a maximum of 2032 kJ of work can be produced during this process.

EES (Engineering Equation Solver) SOLUTION "Given" m_N2=30 [kg] T1_N2=900 [K] m_Ar=15 [kg] T1_Ar=300 [K] "Properties at room temperature, Table A-2a" c_v_N2=0.743 [kJ/kg-K] c_p_Ar=0.5203 [kJ/kg-K] "Analysis" S_gen=DELTAS_N2+DELTAS_Ar S_gen=0 "for maximum power production" DELTAS_N2=m_N2*c_v_N2*ln(T2/T1_N2) "since V2=V1" DELTAS_Ar=m_Ar*c_p_Ar*ln(T2/T1_Ar) "since P2=P1" Q_source_out=m_N2*c_v_N2*(T1_N2-T2) Q_sink_in=m_Ar*c_p_Ar*(T2-T1_Ar) W_max_out=Q_source_out-Q_sink_in

9-122 A system consisting of a compressor, a storage tank, and a turbine as shown in the figure is considered. The change in the exergy of the air in the tank and the work required to compress the air as the tank was being filled are to be determined. Assumptions 1 Changes in the kinetic and potential energies are negligible. 2 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa.m3 / kg.K , c p = 1.005 kJ / kg ×K ,

cv = 0.718 kJ / kg ×K , k = 1.4 (Table A-2a). Analysis The initial mass of air in the tank is minitial =

PinitialV (100 kPa)(5´ 105 m3 ) = = 0.5946´ 106 kg RTinitial (0.287 kPa ×m3 /kg ×K)(293 K)

and the final mass in the tank is

mfinal =

PfinalV (600 kPa)(5´ 105 m3 ) = = 3.568´ 106 kg RTfinal (0.287 kPa ×m3 /kg ×K)(293 K)

6-540 ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

The conservation of mass applied to the tank gives dm = min dt

while the first law gives Q=

d (mu ) dm - h dt dt

Employing the ideal gas equation of state and using constant specific heats, expands this result to

V cv dP V dP - c pT2 R dt RT dt Using the temperature relation across the compressor and multiplying by dt puts this result in the form Q=

( k - 1)/ k

æP ÷ ö V cv Qdt = dP - c pT1 ççç ÷ ÷ çè P1 ÷ R ø

V dP RT

When this integrated, it yields (i and f stand for initial and final states) é P ö( k - 1)/ k ù V cv k c pV ê æ ç f÷ Q= ( Pf - Pi ) - Pi ú êPf ççç ÷ ú ÷ ÷ R 2k - 1 R ê è Pi ø ú ë û

ö (5´ 105 )(0.718) 1.4 (1.005)(5´ 105 ) éê æ 600 ÷ (600 - 100) 600 çç ÷ ê ÷ ç è100 ø 0.287 2(1.4) - 1 0.287 ëê

0.4/1.4

ù - 100ú ú ú û

= - 6.017 ´ 108 kJ The negative result show that heat is transferred from the tank. Applying the first law to the tank and compressor gives (Q - Wout )dt = d (mu ) - h1dm

which integrates to

Q - Wout = (m f u f - mi ui ) - h1 (m f - mi ) Upon rearrangement, Wout = Q + (c p - cv )T (m f - mi )

= - 6.017´ 108 + (1.005 - 0.718)(293)[(3.568 - 0.5946)´ 106 ]

= - 3.516´ 108 kJ The negative sign shows that work is done on the compressor. When the combined first and second laws is reduced to fit the compressor and tank system and the mass balance incorporated, the result is

6-541

æ T ö d (U - T0 S ) dm ÷Wrev = Q ççç1- 0 ÷ + (h - T0 s ) ÷ çè TR ÷ dt dt ø which when integrated over the process becomes

æ T ÷ ö Wrev = Q ççç1- 0 ÷ + mi [(ui - h1 ) - T0 ( si - s1 ) ]- m f éêë(u f - h1 ) - T0 ( s f - s1 )ùúû ÷ çè TR ÷ ø é æPf öù æ T ö ú ÷ ÷ = Q ççç1- 0 ÷ + mi éëêTi (cv - c p )ù - m f êêT f (cv - c p ) - T0 R ln ççç ÷ ÷ú ÷ ú û ÷ èç TR ø èç Pi ÷ øú êë û

æ 293ö÷ 600 ù 6 6 é = - 6.017 ´ 108 çç1÷+ 0.5946´ 10 [(0.718 - 1.005)293]- 3.568´ 10 êê(0.718 - 1.005)293 + 293(0.287) ln 100 úú çè 293 ø÷ ë û

= 7.875 ´ 108 kJ This is the exergy change of the air stored in the tank.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=500000 [m^3] P_i=100 [kPa] "i: initial state" T_i=(20+273) [K] P_f=600 [kPa] "f: initial state" T_f=(20+273) [K] T0=(20+273) [K] "Properties of air at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.4 "Analysis" m_i=(P_i*Vol)/(R*T_i) m_f=(P_f*Vol)/(R*T_f) Q=(Vol*c_v)/R*(P_f-P_i)-k/(2*k-1)*(c_p*Vol)/R*(P_f*(P_f/P_i)^((k-1)/k)-P_i) W_out=Q+(c_p-c_v)*T_i*(m_f-m_i) W_rev=Q*(1-T0/T0)+m_i*(T_i*(c_v-c_p))-m_f*(T_f*(c_v-c_p)-T0*R*ln(P_f/P_i))

9-123 The air stored in the tank of the system shown in the figure is released through the isentropic turbine. The work produced and the change in the exergy of the air in the tank are to be determined. Assumptions 1 2

Changes in the kinetic and potential energies are negligible. Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are R = 0.287 kPa.m3 / kg.K , c p = 1.005 kJ / kg ×K ,

cv = 0.718 kJ / kg ×K , k = 1.4 (Table A-2a).

6-542

Analysis The initial mass of air in the tank is

minitial =

PinitialV (600 kPa)(5´ 105 m3 ) = = 3.568´ 106 kg 3 RTinitial (0.287 kPa ×m /kg ×K)(293 K)

and the final mass in the tank is

mfinal =

PfinalV (100 kPa)(5´ 105 m3 ) = = 0.5946´ 106 kg RTfinal (0.287 kPa ×m3 /kg ×K)(293 K)

The conservation of mass is dm = min dt

while the first law gives d (mu ) dm - h dt dt

Employing the ideal gas equation of state and using constant specific heats, expands this result to Q=

V cv dP V dP - c pT R dt RT dt

cv - c p R

dP dt

1 When this is integrated over the process, the result is (i and f stand for initial and final states) Q = - V ( Pf - Pi ) = - 5´ 105 (100 - 600) = 2.5´ 108 kJ

Applying the first law to the tank and compressor gives (Q - Wout )dt = d (mu ) - hdm

which integrates to Q - Wout = (m f u f - mi ui ) + h(mi - m f ) - Wout = - Q + m f u f - mi ui + h(mi - m f ) Wout = Q - m f u f + mi ui - h(mi - m f ) = Q - m f cvT + mi c pT - c pT (mi - m f ) = 2.5´ 108 - (0.5946´ 106 )(0.718)(293) + (3.568´ 106 )(1.005)(293) - (1.005)(293)(3.568´ 106 - 0.5946´ 106 )

= 3.00´ 108 kJ This is the work output from the turbine. When the combined first and second laws is reduced to fit the turbine and tank system and the mass balance incorporated, the result is

6-543

æ T ö d (U - T0 S ) dm ÷ Wrev = Q ççç1- 0 ÷ + (h - T0 s) ÷ çè TR ÷ dt dt ø

æ T ö d (u - T0 s ) dm dm = Q ççç1- 0 ÷÷÷- (u - T0 s ) - m + (h - T0 s ) çè TR ÷ø dt dt dt æ T ö dm ds ÷+ (c p - cv )T = Q ççç1- 0 ÷ + mT0 ÷ çè TR ø÷ dt dt

æ T ö T dm ÷+ (c p - cv )T = Q ççç1- 0 ÷ + V 0 ( Pf - Pi ) ÷ çè TR ÷ dt T ø where the last step uses entropy change equation. When this is integrated over the process it becomes

æ T ö T ÷+ (c p - cv )T (m f - mi ) + V 0 ( Pf - Pi ) Wrev = Q ççç1- 0 ÷ ÷ çè TR ÷ T ø

æ 293 ÷ ö 293 = 3.00´ 108 çç1+ (1.005 - 0.718)(293)(0.5946 - 3.568) ´ 10 6 + 5´ 105 (100 - 600) ÷ çè 293 ÷ ø 293

= 0 - 2.500´ 108 - 2.5´ 108 = - 5.00´ 108 kJ This is the exergy change of the air in the storage tank.

EES (Engineering Equation Solver) SOLUTION "Given" Vol=500000 [m^3] P_i=600 [kPa] "i: initial state" T_i=(20+273) [K] P_f=100 [kPa] "f: initial state" T_f=(20+273) [K] T0=(20+273) [K] "Properties of air at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.4 "Analysis" m_i=(P_i*Vol)/(R*T_i) m_f=(P_f*Vol)/(R*T_f) Q=-Vol*(P_f-P_i) W_out=Q-m_f*c_v*T_f+m_i*c_p*T_i-c_p*T_i*(m_i-m_f) W_rev=Q*(1-T0/T0)+(c_p-c_v)*T_i*(m_f-m_i)+Vol*T0/T_i*(P_f-P_i)

9-124E Large brass plates are heated in an oven at a rate of 175 / min . The rate of heat transfer to the plates in the oven and the rate of exergy destruction associated with this heat transfer process are to be determined.

6-544

Assumptions 1 The thermal properties of the plates are constant. 2 The changes in kinetic and potential energies are negligible. 3

The environment temperature is 75F.

Properties The density and specific heat of the brass are given to be ρ = 532.5 lbm / ft 3 and c p = 0.091 Btu / lbm. F . Analysis We take the plate to be the system. The energy balance for this closed system can be expressed as Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = D U plate = m(u2 - u1 ) = mc(T2 - T1 )

The mass of each plate and the amount of heat transfer to each plate is

m = r V = r LA = (532.5 lbm/ft 3 )[(1.5 /12 ft)(1 ft)(3 ft)] = 199.7 lbm Qin = mc(T2 - T1 ) = (199.7 lbm/plate)(0.091 Btu/lbm.°F)(1000 - 75)°F = 16,809 Btu/plate

Then the total rate of heat transfer to the plates becomes

Qtotal = nplateQin, per plate = (175 plates/min)´ (16,809 Btu/plate) = 2,941,575 Btu/min = 49,025 Btu / s We again take a single plate as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the plate and its immediate surroundings so that the boundary temperature of the extended system is at 1300F at all times:

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Change in entropy

Qin Q + Sgen = D Ssystem ® Sgen = - in + D Ssystem Tb Tb

where D Ssystem = m( s2 - s1 ) = mcavg ln

T2 (1000 + 460) R = (199.7 lbm)(0.091 Btu/lbm.R) ln = 18.24 Btu/R T1 (75 + 460) R

Substituting, Sgen = -

Qin 16,809 Btu + D Ssystem = + 18.24 Btu/R = 8.692 Btu/R (per plate) Tb 1300 + 460 R

Then the rate of entropy generation becomes

Sgen = Sgen nball = (8.692 Btu/R ×plate)(175 plates/min) = 1521 Btu/min.R = 25.35 Btu/s.R © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-545

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (535 R)(25.35 Btu/s.R) = 13,560 Btu / s

EES (Engineering Equation Solver) SOLUTION "Given" L=1.5[in]*Convert(in, ft) A=1*3 [ft^2] n_dot_plate=175 [1/min] T1=(75+460) [R] T2=(1000+460) [R] T_oven=(1300+460) [R] rho=532.5 [lbm/ft^3] c_p=0.091 [Btu/lbm-F] T0=(75+460) [R] "assumed" "Analysis" V=L*A m=rho*V Q_in=m*c_p*(T2-T1) Q_dot_total=n_dot_plate*Q_in*1/Convert(min, s) Q_in/T_oven+S_gen=DELTAS_sys "entropy balance" DELTAS_sys=m*c_p*ln(T2/T1) S_dot_gen=S_gen*n_dot_plate*1/Convert(min, s) X_dot_destroyed=T0*S_dot_gen

9-125 Long cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in the oven and the rate of exergy destruction associated with this heat transfer process are to be determined. Assumptions 1 The thermal properties of the rods are constant. 2 The changes in kinetic and potential energies are negligible. 3 The environment temperature is 30C. Properties The density and specific heat of the steel rods are given to be ρ = 7833 kg / m3 and c p = 0.465 kJ / kg.°C . Analysis Noting that the rods enter the oven at a velocity of 3 m/min and exit at the same velocity, we can say that a 3-m long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is m = r V = r LA = r L(p D 2 / 4) = (7833 kg/m3 )(3 m)[p (0.1 m)2 / 4] = 184.6 kg

We take the 3-m section of the rod in the oven as the system. The energy balance for this closed system can be expressed as

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = D U rod = m(u2 - u1 ) = mc(T2 - T1 ) Substituting,

6-546

We again take the 3-m long section of the rod as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the rod and its immediate surroundings so that the boundary temperature of the extended system is at 900C at all times:

Sin - Sout + Sgen = D Ssystem Net entropy transfer by heat and mass

Entropy generation

Qin + Sgen = D Ssystem Tb

Change in entropy

Sgen = -

Qin + D Ssystem Tb

where

T 700+273 D Ssystem = m( s2 - s1 ) = mcavg ln 2 = (184.6 kg)(0.465 kJ/kg.K) ln = 100.1 kJ/K T1 30+273 Substituting, Sgen = -

Qin 57,512 kJ + D Ssystem = + 100.1 kJ/K = 51.1 kJ/K Tb (900+273) R

Noting that this much entropy is generated in 1 min, the rate of entropy generation becomes Sgen =

Sgen Dt

51.1 kJ/K = 51.1 kJ/min.K=0.852 kW/K 1 min

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

X destroyed = T0 Sgen = (298 K)(0.852 kW/K)=254 kW

EES (Engineering Equation Solver) SOLUTION "Given" D=0.10 [m] L=6 [m] Vel=3 [m/min] T_oven=(900+273) [K] T1=(30+273) [K] T2=(700+273) [K] T0=(25+273) [K] rho=7833 [kg/m^3] c_p=0.465 [kJ/kg-C] "Analysis" "(a)" A=pi*D^2/4 V_dot=Vel*A m_dot=rho*V_dot Q_dot_in=m_dot*c_p*(T2-T1)*1/Convert(kW, kJ/min) "(b)" Q_dot_in/T_oven+S_dot_gen=DELTAS_dot_sys "entropy balance" DELTAS_dot_sys=m_dot*c_p*ln(T2/T1)*1/Convert(kW, kJ/min) X_dot_destroyed=T0*S_dot_gen

9-126E Refrigerant-134a enters an adiabatic compressor with an isentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specified pressure. The actual power input and the second-law efficiency to the compressor are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-547

1 2 3

This is a steady-flow process since there is no change with time. Kinetic and potential energy changes are negligible. The device is adiabatic and thus heat transfer is negligible.

Properties From the refrigerant tables (Tables A-11E through A-13E) h1 = hg @ 30 psia =105.34 Btu / lbm P1 = 30 psia ü ïï ý s1 = sg @ 30 psia =0.22386 Btu/lbm ×R ïïþ sat. vapor v1 = v g @ 30 psia =1.5506 ft 3 /lbm

P2 = 70 psia üïï ý h2 s = 112.83 Btu/lbm ïïþ s2 s = s1 Analysis (a) From the isentropic efficiency relation, h - h1 hc = 2 s ¾¾ ® h2 a = h1 + (h2 s - h1 ) / hc h2 a - h1 = 105.34 + (112.83 - 105.34) / 0.80

= 114.70 Btu/lbm Then,

P2 = 70 psia ïüï ý s2 = 0.22739 Btu/lbm h2 a = 114.70 ïïþ

V1 20 / 60 ft 3 / s = = 0.2150 lbm/s v1 1.5506 ft 3 / lbm There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as Ein - Eout = D Esystem 0 (steady) = 0 Also,

Rate of net energy transfer by heat, work, and mass

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

Wa,in + mh1 = mh2

(since Q @D ke @D pe @0)

Wa ,in = m(h2 - h1 )

Substituting, the actual power input to the compressor becomes æ 1 hp ö ÷ Wa,in = (0.2150 lbm/s)(114.70 - 105.34) Btu/lbm çç = 2.847 hp ÷ çè0.7068 Btu/s ÷ ø (b) The reversible (or minimum) power input is determined from the exergy balance applied on the compressor and setting the exergy destruction term equal to zero, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-548

- X destroyed 0 (reversible) = D X system 0 (steady) = 0

X in - X out Rate of net exergy transfer by heat, work, and mass

Rate of exergy destruction

Rate of change of exergy

X in = X out

Wrev,in + mxflow,1 = mxflow,2 Wrev,in = m( xflow,2 - xflow,1 ) = m[(h2 - h1 ) - T0 ( s2 - s1 ) + Δke 0 + Δpe 0 ]

Substituting, Wrev,in = (0.2150 lbm/s) [(114.70 - 105.34) Btu/lbm - (535 R)(0.22739 - 0.22386) Btu/lbm ×R ]

= 1.606 Btu/s=2.273 hp

(since 1 hp=0.7068 Btu/s)

Thus,

hII =

Wrev,in Wact,in

2.273 hp = 0.798 = 79.8% 2.847hp

EES (Engineering Equation Solver) SOLUTION "Given" P1=30 [psia] x1=1 P2=70 [psia] Vol1_dot=20[ft^3/min]*Convert(ft^3/min, ft^3/s) eta_C=0.80 T0=(75+460) [R] "Properties" Fluid$='R134a' v1=volume(Fluid$, P=P1, x=x1) h1=enthalpy(Fluid$, P=P1, x=x1) s1=entropy(Fluid$, P=P1, x=x1) s2_s=s1 "isentropic process" h2_s=enthalpy(Fluid$, P=P2, s=s2_s) "Analysis" "(a)" eta_C=(h2_s-h1)/(h2-h1) s2=entropy(Fluid$, P=P2, h=h2) m_dot=Vol1_dot/v1 W_dot_in=m_dot*(h2-h1)*Convert(Btu/s, hp) "(b)" W_dot_rev_in=m_dot*(h2-h1-T0*(s2-s1))*Convert(Btu/s, hp) eta_II=W_dot_rev_in/W_dot_in

9-127 R-134a is expanded in an adiabatic process with an isentropic efficiency of 0.85. The second law efficiency is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 The device is adiabatic and thus heat transfer is negligible.

6-549

Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

Ein - Eout

Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

- Wout = D U = m(u2 - u1 )

From the R-134a tables (Tables A-11 through A-13), v1 = 0.014362 m3 /kg P1 = 1600 kPa ü ïï ý u1 = 282.11 kJ/kg ïïþ T1 = 80°C s1 = 0.9875 kJ/kg ×K

P2 = 100 kPa s2 s = s1

üïï ý u2 s = 223.20 kJ/kg ïïþ

The actual work input is

wa ,out = hT ws ,out = hT (u1 - u2 s ) = (0.85)(282.11- 223.20)kJ/kg = 50.07 kJ/kg The actual internal energy at the end of the expansion process is wa ,out = (u1 - u2 ) ¾ ¾ ® u2 = u1 - wa ,out = 282.11- 50.07 = 232.04 kJ/kg

Other actual properties at the final state are (Table A-13)

ïüï v 2 = 0.21392 m3 /lbm ý u2 = 232.04 kJ/kg ïïþ s2 = 1.0252 kJ/kg ×K P2 = 100 kPa

The useful work is determined from wu = wa ,out - wsurr = wa ,out - P0 (v 2 - v1 )

æ 1 kJ ÷ ö = 50.07 kJ/kg - (100 kPa)(0.21392 - 0.014362) m 3 /kg çç ÷ çè1 kPa ×m3 ÷ ø = 30.11 kJ/kg The exergy change between initial and final states is xnonflow,1 - xnonflow,2 = u1 - u2 + P0 (v1 - v 2 ) - T0 ( s1 - s2 )

æ 1 kJ ÷ ö = (282.11- 232.04)kJ/kg + (100 kPa)(0.014362 - 0.21392) m3 /kg çç ÷ çè1 kPa ×m3 ÷ ø (298 K)(0.9875 - 1.0252)kJ/kg ×K

= 41.35 kJ/kg The second law efficiency is then

hII =

wu 30.11 kJ/kg = = 0.728 = 72.8% D xnonflow 41.35 kJ/kg

6-550

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=1600 [kPa] T1=80 [C] P2=100 [kPa] eta=0.85 T0=(25+273) [K] P0=100 [kPa] "ANALYSIS" Fluid$='R134a' v1=volume(Fluid$, P=P1, T=T1) u1=intenergy(Fluid$, P=P1, T=T1) s1=entropy(Fluid$, P=P1, T=T1) s2_s=s1 u2_s=intenergy(Fluid$, P=P2, s=s2_s) w_s_out=u1-u2_s w_a_out=eta*w_s_out u2=u1-w_a_out v2=volume(Fluid$, P=P2, u=u2) s2=entropy(Fluid$, P=P2, u=u2) w_surr=P0*(v2-v1) w_u=w_a_out-w_surr DELTAx=u1-u2+P0*(v1-v2)-T0*(s1-s2) eta_II=w_u/DELTAx

9-128 Exhaust gases are expanded in a turbine, which is not well-insulated. Tha actual and reversible power outputs, the exergy destroyed, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Air is an ideal gas with constant specific heats.

Properties The gas constant of air is R = 0.287 kJ / kg.K and the specific heat of air at the average temperature of (627 + 527) / 2 = 577° C = 850 K is c p = 1.11 kJ / kg.°C (Table A-2). Analysis (a) The enthalpy and entropy changes of air across the turbine are D h = c p (T2 - T1 ) = (1.11 kJ/kg.°C)(527 - 627)°C = - 111 kJ/kg

6-551

D s = c p ln

T2 P - R ln 2 T1 P1

= (1.11 kJ/kg.K)ln

(527 + 273) K 500 kPa - (0.287 kJ/kg.K) ln (627 + 273) K 1200 kPa

= 0.1205 kJ/kg.K

The actual and reversible power outputs from the turbine are

- Wa,out = mD h + Qout = (2.5 kg/s)( - 111 kJ/kg) + 20 kW = - 257.5 kW - Wrev,out = m(D h - T0D s ) = (2.5 kg/s)(111 kJ/kg) - (25 + 273 K)(0.1205 kJ/kg.K) = - 367.3 kW

or Wa,out = 257.5 kW Wrev,out = 367.3 kW

(b) The exergy destroyed in the turbine is X dest = Wrev - Wa = 367.3 - 257.5 = 109.8 kW

Wa 257.5 kW = = 0.701 = 70.1% 367.3 kW Wrev

EES (Engineering Equation Solver) SOLUTION "Given" T_1=(627+273) [K] P_1=1200 [kPa] T_2=(527+273) [K] P_2=500 [kPa] m_dot=2.5 [kg/s] Q_dot_out=20 [kW] T_0=(25+273) [K] P_0=100 [kPa] "Properties" c_p=1.11 [kJ/kg-K] "at (627+527)/2=577 C = 850 K, Table A-2b" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" "(a)" DELTAh=c_p*(T_2-T_1) DELTAs=c_p*ln(T_2/T_1)-R*ln(P_2/P_1) -W_dot_actual-Q_dot_out=m_dot*DELTAh "actual power" -W_dot_rev=m_dot*(DELTAh-T_0*DELTAs) "reversible power" "(b)" EX_destroyed=W_dot_rev-W_dot_actual "exergy destroyed" "(c)" eta_II=W_dot_actual/W_dot_rev "second-law efficiency"

9-129 The isentropic efficiency of a water pump is specified. The actual power output, the rate of frictional heating, the exergy destruction, and the second-law efficiency are to be determined.

6-552

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Using saturated liquid properties at the given temperature for the inlet state (Table A-4)

h1 = 125.82 kJ/kg T1 = 30°Cüïï ý s1 = 0.4367 kJ/kg.K x1 = 0 ïïþ v1 = 0.001004 m3 /kg The power input if the process was isentropic is Ws = mv1 ( P2 - P1 ) = (1.35 kg/s)(0.001004 m3 /kg)(4000 - 100)kPa = 5.288 kW

Given the isentropic efficiency, the actual power may be determined to be

Wa =

Ws 5.288 kW = = 7.554 kW hs 0.70

(b) The difference between the actual and isentropic works is the frictional heating in the pump Qfrictional = Wa - Ws = 7.554 - 5.288 = 2.266 kW

Wa = m(h2 - h1 ) ¾ ¾ ® 7.554 kW = (1.35 kg/s)(h2 - 125.82)kJ/kg ¾ ¾ ® h2 = 131.42 kJ/kg The entropy at the exit is

üïï ý s2 = 0.4423 kJ/kg.K h2 = 131.42 kJ/kgïïþ The reversible power and the exergy destruction are P2 = 4 MPa

Wrev = m [h2 - h1 - T0 ( s2 - s1 )] = (1.35 kg/s) [(131.42 - 125.82)kJ/kg - (20 + 273 K)(0.4423 - 0.4367)kJ/kg.K ]= 5.362 kW

X dest = Wa - Wrev = 7.554 - 5.362 = 2.193 kW (d) The second-law efficiency is

hII =

Wrev 5.362 kW = = 0.710 = 71.0% 7.554 kW Wa

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] T_1=30 [C] P_2=4000 [kPa] m_dot=1.35 [kg/s] eta_P=0.70 T_0=(20+273) [K] "PROPERTIES" Fluid$='Steam_iapws' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-553

h_1=enthalpy(Fluid$, P=P_1, T=T_1) v_1=volume(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) "ANALYSIS" "(a)" W_dot_isentropic=m_dot*v_1*(P_2-P_1) "isentropic power" W_dot_actual=W_dot_isentropic/eta_P "actual power" "(b)" Q_dot_friction=W_dot_actual-W_dot_isentropic "frictional heating" "(c)" h_2=h_1+(v_1*(P_2-P_1))/eta_P s_2=entropy(Fluid$, P=P_2, h=h_2) W_dot_rev=m_dot*(h_2-h_1-T_0*(s_2-s_1)) "reversible power" EX_dot_dest=W_dot_actual-W_dot_rev "exergy destruction" "(d)" eta_II=W_dot_rev/W_dot_actual "second-law efficiency"

9-130 Argon gas is expanded adiabatically in an expansion valve. The exergy of argon at the inlet, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are zero. 3 Argon is an ideal gas with constant specific heats.

Properties The properties of argon gas are R = 0.2081 kJ / kg.K , c p = 0.5203 kJ / kg.°C (Table A-2). Analysis (a) The exergy of the argon at the inlet is xflow,1 = h1 - h0 - T0 (s1 - s0 )

é T Pù = c p (T1 - T0 ) - T0 êc p ln 1 - R ln 1 ú ê T0 P0 úû ë

é 373 K 3500 kPa ù ú = (0.5203 kJ/kg.K)(100 - 25)°C - (298 K) ê(0.5203 kJ/kg.K)ln - (0.2081 kJ/kg.K)ln êë 298 K 100 kPa úû = 224.7 kJ / kg (b) Noting that the temperature remains constant in a throttling process of an ideal gas, the exergy destruction is determined from xdest = T0 sgen

= T0 (s2 - s1 ) æ ö é æ 500 kPa ö÷ù P÷ = T0 ççç- R ln 1 ÷ = (298 K) ê- (0.2081 kJ/kg.K)ln çç ÷ú ÷ ê ÷ çè èç3500 kPa ø÷ú P0 ø ë û = 120.7 kJ / kg (c) The second-law efficiency can be determined from x 0 0 kJ/kg hII = recovered = = = 0 = 0% xexpended xflow,1 - xflow,2 (224.7 - 104.0) kJ/kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-554

where

xflow,2 = xflow,1 - xdest = [224.7 - (224.7 - 120.7)] kJ/kg = 104.0 kJ/kg Discussion An alternative (and different) definition for the second-law efficiency of an expansion valve can be expressed using an “exergy out/exergy in” approach as hII =

x xout 104.0 kJ/kg = flow,2 = = 0.463 = 46.3% xin xflow,1 224.7 kJ/kg

Note that an expansion valve is a very wasteful device and it destructs all of the exergy expended in the process. Therefore, we prefer the first approach for the calculation of second-law efficiency.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=3500 [kPa] T_1=(100+273) [K] P_2=500 [kPa] T_0=(25+273) [K] P_0=100 [kPa] "PROPERTIES" R=0.2081 [kJ/kg-K] "Table A-2a" c_p=0.5203 [kJ/kg-K] "Table A-2a" "ANALYSIS" "(a)" c_p*T_1=c_p*T_2 "energy balance" x_1=c_p*(T_1-T_0)-T_0*DELTAs_1 "exergy at the inlet" x_2=c_p*(T_2-T_0)-T_0*DELTAs_2 "exergy at the inlet" DELTAs_2=c_p*ln(T_2/T_0)-R*ln(P_2/P_0) "(b)" DELTAs_1=c_p*ln(T_1/T_0)-R*ln(P_1/P_0) s_gen=c_p*ln(T_2/T_1)-R*ln(P_2/P_1) x_dest=T_0*s_gen "exergy destruction" "(c)" eta_II=(x_1-x_dest)/x_1 "second-law efficiency"

9-131 A relation for the second-law efficiency of a heat engine operating between a heat source and a heat sink at specified temperatures is to be obtained.

6-555

Analysis The second-law efficiency is defined as the ratio of the exergy recovered to exergy supplied during a process. The work W produced is the exergy recovered. The decrease in the exergy of the heat supplied QH is the exergy supplied or invested. Therefore, hII =

W æ T0 ö æ T0 ö ÷ ÷Q çç1ç ÷ ÷QH - çç1- T ÷ ÷ ÷ ÷ L çè TH ø è Lø

Note that the first term in the denominator is the exergy of heat supplied to the heat engine whereas the second term is the exergy of the heat rejected by the heat engine. The difference between the two is the exergy expended during the process.

9-132 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a closed system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount

QR . Assumptions Kinetic and potential changes are negligible.

Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as Energy balance: Ein - Eout = D Esystem ® Q0 + QR - W = U 2 - U1 ¾ ¾ ® W = U1 - U 2 + Q0 + QR (1) Entropy balance: Sin - Sout + Sgen = D Ssystem ® Sgen = ( S 2 - S1 ) +

- QR - Q0 + TR T0

(2)

6-556

Solving for Q0 from (2) and substituting in (1) yields

æ T ö ÷- T0 Sgen W = (U1 - U 2 ) - T0 ( S1 - S 2 ) - QR ççç1- 0 ÷ ÷ çè TR ÷ ø The useful work relation for a closed system is obtained from

Wu = W - Wsurr æ T ö ÷- T0 Sgen - P0 (V2 - V1 ) = (U1 - U 2 ) - T0 ( S1 - S2 ) - QR ççç1- 0 ÷ ÷ çè TR ø÷ Then the reversible work relation is obtained by substituting Sgen = 0 ,

æ T ö ÷ Wrev = (U1 - U 2 ) - T0 ( S1 - S2 ) + P0 (V1 - V2 ) - QR ççç1- 0 ÷ ÷ çè TR ø÷ A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positive quantity for heat transfer to the system, and a negative quantity for heat transfer from the system.

9-133 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a steady-flow system that exchanges heat with surroundings at T0 at a rate of Q0 as well as a heat reservoir at temperature TR in the amount QR . Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as

Energy balance: Ein - Eout = D Esystem ® Ein = Eout Ve2 V2 + gze ) - å mi (hi + i + gzi ) 2 2 Vi 2 Ve2 or W = å mi (hi + + gzi ) - å me (he + + gze ) + Q0 + QR (1) 2 2 Entropy balance: Q0 + QR - W = å me (he +

Sin - Sout + Sgen = D Ssystem = 0 Sgen = Sout - Sin

Sgen = å me se - å mi si +

- QR - Q0 + TR T0

(2)

Solving for Q0 from (2) and substituting in (1) yields

W = å mi (hi +

æ T ö Vi 2 V2 ÷ + gzi - T0 si ) - å me (he + e + gze - T0 se ) - T0 S gen - QR ççç1- 0 ÷ ÷ çè TR ÷ 2 2 ø

6-557

Then the reversible work relation is obtained by substituting Sgen = 0 ,

Wrev = å mi (hi +

æ T ö Vi 2 V2 ÷ + gzi - T0 si ) - å me (he + e + gze - T0 se ) - QR ççç1- 0 ÷ ÷ çè TR ÷ 2 2 ø

A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positive quantity for heat transfer to the system, and a negative quantity for heat transfer from the system.

9-134 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a uniform-flow system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount

QR . Assumptions Kinetic and potential changes are negligible.

æ V2 ö æ Vi 2 ö ç Q0 + QR - W = å me çç he + e + gze ÷ m + gzi ÷ i ç hi + å ÷ ÷+ (U 2 - U1 ) cv 2 2 è ø è ø æ Vi 2 ö æ Ve2 ö ç ÷ + gzi ÷ or, W = å mi çç hi + ÷ - å me ç he + 2 + gze ÷ - (U 2 - U1 ) cv + Q0 + QR (1) 2 è ø è ø Entropy balance: Sin  Sout  Sgen  Ssystem Sgen = ( S 2 - S1 ) cv + å me se - å mi si +

- QR - Q0 + (2) TR T0

Solving for Q0 from (2) and substituting in (1) yields

æ V2 ö æ Ve2 ö æ T ö ç W = å mi çç hi + i + gzi - T0 si ÷ m + gze - T0 se ÷ + éë(U1 - U 2 ) - T0 ( S1 - S 2 ) ùûcv - T0 S gen - QR çç1 - 0 ÷ e ç he + å ÷ ÷ ÷ 2 2 è TR ø è ø è ø The useful work relation for a closed system is obtained from æ V2 ö æ Ve2 ö æ ö ç he + + gze - T0 se ÷+ é(U1 - U 2 ) - T0 ( S1 - S 2 ) ù - T0 S gen - QR ç1 - T0 ÷- P0 (V2 - V1 ) Wu = W - Wsurr = å mi çç hi + i + gzi - T0 si ÷ m e å ë û ç ÷ ÷ ç ÷ cv 2 2 è TR ø è ø è ø

Then the reversible work relation is obtained by substituting Sgen = 0 ,

æ V2 ö æ Ve2 ö æ ö ç he + ÷+ é(U1 - U 2 ) - T0 ( S1 - S 2 ) + P0 (V1 - V2 ) ù - QR ç1 - T0 ÷ Wrev = å mi çç hi + i + gzi - T0 si ÷ m + gz T s e e 0 e å ë û ç T ÷ ÷ ç ÷ cv 2 2 è R ø è ø è ø

6-558

Fundamentals of Engineering (FE) Exam Problems

9-135 Keeping the limitations imposed by the second-law of thermodynamics in mind, choose the wrong statement below: (a) A heat engine cannot have a thermal efficiency of 100%. (b) For all reversible processes, the second-law efficiency is 100%. (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. (d) The second-law efficiency of a process is 100% if no entropy is generated during that process. (e) The coefficient of performance of a refrigerator can be greater than 1. Answer (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency.

9-136 Heat is lost through a plane wall steadily at a rate of 800 W. If the inner and outer surface temperatures of the wall are 20C and 9C, respectively, and the environment temperature is 0C, the rate of exergy destruction within the wall is (a) 0 W (b) 11 W (c) 15 W (d) 29 W (e) 76 W Answer (d) 29 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q=800 [W] T1=20 [C] T2=9 [C] To=0 [C] "Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives" Q/(T1+273)-Q/(T2+273)+S_gen=0 X_dest=(To+273)*S_gen "Some Wrong Solutions with Common Mistakes:" Q/T1-Q/T2+Sgen1=0; W1_Xdest=(To+273)*Sgen1 "Using C instead of K in Sgen" Sgen2=Q/((T1+T2)/2); W2_Xdest=(To+273)*Sgen2 "Using avegage temperature in C for Sgen" Sgen3=Q/((T1+T2)/2+273); W3_Xdest=(To+273)*Sgen3 "Using avegage temperature in K" W4_Xdest=To*S_gen "Using C for To"

9-137 Liquid water enters an adiabatic piping system at 15C at a rate of 3 kg / s . It is observed that the water temperature rises by 0.3C in the pipe due to friction. If the environment temperature is also 15C, the rate of exergy destruction in the pipe is (a) 3.8 kW (b) 24 kW (c) 72 kW (d) 98 kW © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-559

(e) 124 kW Answer (a) 3.8 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=3 [kg/s] T1=15 [C] T2=15.3 [C] To=15 [C] cp=4.18 [kJ/kg-K] S_gen=m*cp*ln((T2+273)/(T1+273)) X_dest=(To+273)*S_gen "Some Wrong Solutions with Common Mistakes:" W1_Xdest=(To+273)*m*Cp*ln(T2/T1) "Using deg. C in Sgen" W2_Xdest=To*m*Cp*ln(T2/T1) "Using deg. C in Sgen and To" W3_Xdest=(To+273)*Cp*ln(T2/T1) "Not using mass flow rate with deg. C" W4_Xdest=(To+273)*Cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with K"

9-138 A water reservoir contains 100 tons of water at an average elevation of 60 m. The maximum amount of electric power that can be generated from this water is (a) 8 kWh (b) 16 kWh (c) 1630 kWh (d) 16,300 kWh (e) 58,800 kWh Answer (b) 16 kWh Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=100000 [kg] h=60 [m] g=9.81 [m/s^2] W_max=m*g*h/1000 "Maximum power is simply the potential energy change" W_max_kWh=W_max/3600 "Some Wrong Solutions with Common Mistakes:" W1_Wmax =m*g*h/3600 "Not using the conversion factor 1000" W2_Wmax =m*g*h/1000 "Obtaining the result in kJ instead of kWh" W3_Wmax =m*g*h*3.6/1000 "Using worng conversion factor" W4_Wmax =m*h/3600"Not using g and the factor 1000 in calculations"

9-139 A house is maintained at 21C in winter by electric resistance heaters. If the outdoor temperature is 3C, the second-law efficiency of the resistance heaters is (a) 0% (b) 4.1% (c) 6.1% (d) 8.6% (e) 16.3% Answer (c) 6.1%

6-560

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=(3+273) [K] TH=(21+273) [K] To=TL COP_rev=TH/(TH-TL) COP=1 Eta_II=COP/COP_rev "Some Wrong Solutions with Common Mistakes:" W1_Eta_II=COP/COP_rev1; COP_rev1=TL/(TH-TL) "Using wrong relation for COP_rev" W2_Eta_II=1-(TL-273)/(TH-273) "Taking second-law efficiency to be reversible thermal efficiency with C for temp" W3_Eta_II=COP_rev "Taking second-law efficiency to be reversible COP" W4_Eta_II=COP_rev2/COP; COP_rev2=(TL-273)/(TH-TL) "Using C in COP_rev relation instead of K, and reversing"

9-140 A furnace can supply heat steadily at a 1300 K at a rate of 500 kJ / s . The maximum amount of power that can be produced by using the heat supplied by this furnace in an environment at 300 K is (a) 115 kW (b) 192 kW (c) 385 kW (d) 500 kW (e) 650 kW Answer (c) 385 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_in=500 [kJ/s] TL=300 [K] TH=1300 [K] W_max=Q_in*(1-TL/TH) "Some Wrong Solutions with Common Mistakes:" W1_Wmax=W_max/2 "Taking half of Wmax" W2_Wmax=Q_in/(1-TL/TH) "Dividing by efficiency instead of multiplying by it" W3_Wmax =Q_in*TL/TH "Using wrong relation" W4_Wmax=Q_in "Assuming entire heat input is converted to work"

9-141 A heat engine receives heat from a source at 1500 K at a rate of 600 kJ/s and rejects the waste heat to a sink at 300 K. If the power output of the engine is 400 kW, the second-law efficiency of this heat engine is (a) 42% (b) 53% (c) 83% (d) 67% (e) 80% Answer (c) 83% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Qin=600 [kJ/s] W=400 [kW] TL=300 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-561

TH=1500 [K] Eta_rev=1-TL/TH Eta_th=W/Qin Eta_II=Eta_th/Eta_rev "Some Wrong Solutions with Common Mistakes:" W1_Eta_II=Eta_th1/Eta_rev; Eta_th1=1-W/Qin "Using wrong relation for thermal efficiency" W2_Eta_II=Eta_th "Taking second-law efficiency to be thermal efficiency" W3_Eta_II=Eta_rev "Taking second-law efficiency to be reversible efficiency" W4_Eta_II=Eta_th*Eta_rev "Multiplying thermal and reversible efficiencies instead of dividing"

9-142 Air is throttled from 50C and 800 kPa to a pressure of 200 kPa at a rate of 0.5 kg / s in an environment at 25C. The change in kinetic energy is negligible, and no heat transfer occurs during the process. The power potential wasted during this process is (a) 0 (b) 0.20 kW (c) 47 kW (d) 59 kW (e) 119 kW Answer (d) 59 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.5 [kg/s] T1=(50+273) [K] P1=800 [kPa] To=25 [C] P2=200 [kPa] R=0.287 [kJ/kg-K] cp=1.005 [kJ/kg-K] T2=T1 "Temperature of an ideal gas remains constant during throttling since h=const and h=h(T)" ds=cp*ln(T2/T1)-R*ln(P2/P1) X_dest=(To+273)*m*ds "Some Wrong Solutions with Common Mistakes:" W1_dest=0 "Assuming no loss" W2_dest=(To+273)*ds "Not using mass flow rate" W3_dest=To*m*ds "Using C for To instead of K" W4_dest=m*(P1-P2) "Using wrong relations"

9-143 Steam enters a turbine steadily at 4 MPa and 600C and exits at 0.2 MPa and 150C in an environment at 25C. The decrease in the exergy of the steam as it flows through the turbine is (a) 879 kJ / kg (b) 1123 kJ / kg (c) 1645 kJ / kg (d) 1910 kJ / kg (e) 4260 kJ / kg Answer (a) 879 kJ / kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF MCGRAW HILL LLC.

6-562

P1=4000 [kPa] T1=600 [C] P2=200 [kPa] T2=150 [C] To=25 [C] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) s2=ENTROPY(Steam_IAPWS,T=T2,P=P2) -Dx=h2-h1-(To+273)*(s2-s1) "Some Wrong Solutions with Common Mistakes:" -W1_Dx=0 "Assuming no exergy destruction" -W2_Dx=h2-h1 "Using enthalpy change" -W3_Dx=h2-h1-To*(s2-s1) "Using C for To instead of K" -W4_Dx=(h2+(T2+273)*s2)-(h1+(T1+273)*s1) "Using wrong relations for exergy"

9-144 A 12-kg solid whose specific heat is 2.8 kJ / kg.°C is at a uniform temperature of -10C. For an environment temperature of 20C, the exergy content of this solid is (a) Less than zero (b) 0 kJ (c) 4.6 kJ (c) 55 kJ (d) 1008 kJ Answer (d) 55 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=12 [kg] cp=2.8 [kJ/kg-K] T=(-10+273) [K] To=(20+273) [K] x=m*(cp*(T-To)-To*cp*ln(T/To)) "Exergy content of a fixed mass is x=u-uo-To*(s-so)+Po*(v-vo)" "Some Wrong Solutions with Common Mistakes:" W1_x=m*cp*(To-T) "Taking the energy content as the exergy content" W2_x=m*(cp*(T-To)+To*cp*ln(T/To)) "Using + for the second term instead of -" W3_x=cp*(T-To)-To*cp*ln(T/To) "Using exergy content per unit mass" W4_x=0 "Taking the exergy content to be zero"

9-145 … 9-150 Design and Essay Problems

10-1

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 10 GAS POWER CYCLES

10-2

CYU - Check Your Understanding CYU 10-1 ■ Check Your Understanding CYU 10-1.1 Select the correct statement(s) regarding ideal power cycles. I. Ideal power cycles are internally reversible, but they are not necessarily externally reversible. II. The Carnot cycle is internally reversible, but it is not necessarily externally reversible. III. Ideal power cycles are better representation of actual cycles compared to the Carnot cycle. (a) Only I (b) Only II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III CYU 10-1.2 Select the incorrect statement regarding property diagrams of power cycles. (a) On a T-s diagram, a heat-addition process proceeds in the direction of increasing entropy. (b) On a T-s diagram, the increase in entropy for a process is a measure of irreversibilities for that process. (c) On both the P-v and T-s diagrams, the area enclosed by the process curves of a cycle represents the net work produced. (d) On both the P-v and T-s diagrams, the area enclosed by the process curves of a cycle represents the net heat transfer. (e) The area under the heat rejection process on a T-s diagram is a measure of the total heat rejected during the cycle. Answer: (b) On a T-s diagram, the increase in entropy for a process is a measure of irreversibilities for that process.

CYU 10-2 ■ Check Your Understanding CYU 10-2.1 Select the incorrect statement regarding the Carnot cycle. (a) The thermal efficiency of the Carnot cycle is a function of the sink and source temperatures only. (b) The Carnot cycle is the most efficient cycle operating between specified source and sink temperatures. (c) Thermal efficiency increases with an increase in temperature during the heat addition process. (d) Thermal efficiency increases with an increase in temperature during the heat rejection process. Answer: (d) Thermal efficiency increases with an increase in temperature during the heat rejection process. CYU 10-2.2 Which equation(s) below can be used to determine the thermal efficiency of the Carnot cycle? I. II. III. (a) Only I (b) Only II (c) I and II (d) I and III (e) I, II, and III Answer: (e) I, II, and III © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-3

CYU 10-4 ■ Check Your Understanding CYU 10-4.1 Select the wrong statement regarding reciprocating engines. (a) The distance between the top dead center (TDC) and the bottom dead center (BDC) is called the stroke. (b) The ratio of the maximum pressure in the cylinder to the minimum pressure is called the compression ratio of the engine. (c) The volume displaced by the piston as it moves between TDC and BDC is called the displacement volume. (d) The minimum volume formed in the cylinder when the piston is at TDC is called the clearance volume (e) The diameter of the piston is called the bore. Answer: (b) The ratio of the maximum pressure in the cylinder to the minimum pressure is called the compression ratio of the engine.

CYU 10-5 ■ Check Your Understanding CYU 10-5.1 Select the correct statements regarding the ideal Otto cycle and actual spark-ignition engines. I. The thermal efficiency of an ideal Otto cycle depends only on the compression ratio of the engine. II. For a given compression ratio, the thermal efficiency of an actual spark-ignition engine is less than that of an ideal Otto cycle. III. In actual spark-ignition engine, ignition takes place shortly after the top dead center. (a) Only I (b) Only II (c) I and II (d) II and III (e) I, II, and III Answer: (b) Only II CYU 10-5.2 Which measures help avoid engine knock in spark-ignition engines? I. Higher compression ratio II. Higher Octane number gasoline III. Cooler intake air (a) Only I (b) Only II (c) I and II (d) II and III (e) I, II, and III Answer: (d) II and III CYU 10-5.3 Select the incorrect statement regarding the two-stroke engines in comparison to four-stroke engines. (a) Intake and exhaust strokes do not exist. (b) They produce less power for a given engine displacement. (c) They have high power-to-weight ratios. (d) They emit more pollutants. Answer: (b) They produce less power for a given engine displacement. CYU 10-5.4 For Otto cycle, which equation(s) below can be used to determine the heat input? I. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-4

II. III. IV. (a) Only I (b) Only II (c) I and II (d) III and IV (e) I, II, III, and IV Answer: (c) I and II CYU 10-5.54 If the work output of a four-stroke engine is is the power produced by the engine? (a) 5 kW (b) 10 kW (c) 20 kW (d) 24 kW (e) 60 kW

and the engine speed is 1200 rpm (

), what

Answer: (a) 5 kW There are 2 revolutions per cycle in four-stroke engines.

CYU 10-6 ■ Check Your Understanding CYU 10-6.1 Select the correct statement(s) below regarding power cycle efficiencies. I. The thermal efficiency of the dual cycle is greater than that of the Diesel cycle for the same compression ratio. II. The thermal efficiency of the Diesel cycle is greater than that of the Otto cycle for the same compression ratio. III. For a given compression, the thermal efficiency of the Diesel cycle increases as the cutoff ratio increases,. (a) Only I (b) Only II (c) I and II (d) II and III (e) I, II, and III Answer: (a) Only I CYU 10-6.2 Select the incorrect statement regarding Diesel cycles. (a) Diesel engines operate at higher compression ratios than the gasoline engines. (b) There are no spark plugs in diesel engines. (c) The dual cycle is a better representation of modern, high-speed diesel engines than the Diesel cycle. (d) Only air is compressed in the operation of diesel engines. (e) None of the above. Answer: (e) None of the above CYU 10-6.3 For Diesel cycle, which equation(s) below can be used to determine the heat rejection? I. II. III. IV. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-5

(a) I and II (b) II and III (c) I and IV (d) III and IV (e) I, II, and IV Answer: (a) I and II CYU 10-6.4 If the work output of a four-stroke engine is is the power produced by the engine? (a) 5 kW (b) 10 kW (c) 20 kW (d) 24 kW (e) 60 kW

and the engine speed is 2400 rpm (

), what

Answer: (b) 10 kW There are 2 revolutions per cycle in four-stroke engines.

CYU 10-8 ■ Check Your Understanding CYU 10-8.1 Select the correct statement(s) below about the ideal Brayton cycle. I. The thermal efficiency of the Brayton cycle is a function of pressure ratio only. II. In the Brayton cycle, there are two adiabatic and two isobaric processes. III. As the pressure ratio increases, the thermal efficiency of the Brayton cycle decreases. IV. The phrases compression ratio and pressure ratio are identical and can be used interchangibly. (a) Only I (b) Only II (c) I and II (d) II and IV (e) I, II, and III Answer: (b) Only II CYU 10-8.2 Which of the suggested measures below will not help increase the thermal efficiencies of gas-turbine engines? (a) Insulating the compressor (b) Insulating the turbine (c) Incorporating regeneration (d) Increasing the pressure ratio (e) Increasing turbine inlet temperature Answer: (a) Insulating the compressor CYU 10-8.3 Select the incorrect statement below regarding the back work ratio. (a) The back work ratios in gas-turbine power plants are higher than in steam power plants. (b) The greater the isentropic efficiencies of the compressor and the turbine, the smaller the back work ratio. (c) The greater the back work ratio, the greater the thermal efficiency of the Brayton cycle. (d) The back work ratio of a gas turbine is the ratio of the compressor work to the turbine work. Answer: (c) The greater the back work ratio, the greater the thermal efficiency of the Brayton cycle. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-6

CYU 10-8.4 Which equation(s) below can be used to determine the heat input for the Brayton cycle? I. II. III. IV. (a) I and II (b) I and III (c) III and IV (d) I, III and IV (e) I, II, III, and IV Answer: (b) I and III

CYU 10-9 ■ Check Your Understanding CYU 10-9.1 Select the correct statement regarding the Brayton cycle with regeneration. (a) The greater the effectiveness of the regenerator, the greater the thermal efficiency of the Brayton cycle. (b) The thermal efficiency of the Brayton cycle increases as a result of regeneration. (c) The use of a regenerator is only recommended when the turbine exhaust temperature is higher than the compressor exit temperature. (d) All of the above Answer: (d) All of the above CYU 10-9.2 Select the correct statement regarding the Brayton cycle with regeneration. (a) In an ideal regenerator, the air leaving the compressor is heated to the turbine exit temperature. (b) Regeneration is less effective at lower pressure ratios. (c) The thermal efficiency of an ideal Brayton cycle with regeneration depends only on the ratio of the minimum to maximum temperatures and the specific heat ratio. (d) None of the above Answer: (a) In an ideal regenerator, the air leaving the compressor is heated to the temperature at the turbine exit.

CYU 10-10 ■ Check Your Understanding CYU 10-10.1 Select the correct statement(s) below regarding the Brayton cycle with intercooling, reheating, and regeneration. I. Cooling the gas during compression decreases the compressor work input. II. Heating the gas during expansion decreases the turbine work output. III. The gas leaves the compressor at a lower temperature when intercooling is utilized. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-7

CYU 10-10.2 Select the incorrect statement below regarding the Brayton cycle with intercooling, reheating, and regeneration. I. For highest thermal efficiency, equal pressure ratios should be maintained across each stage of the compression and expansion. II. Intercooling and reheating always decrease the thermal efficiency if they are not accompanied by regeneration. III. The back work ratio of a gas-turbine cycle increases as a result of intercooling and reheating. (a) Only I (b) Only III (c) I and II (d) I and III (e) II and III Answer: (b) Only III

CYU 10-11 ■ Check Your Understanding CYU 10-11.1 Select the incorrect equation below regarding the analysis of the ideal jet-propulsion cycle. (a) ̇ (b) ̇ ̇ ̇ ̇ (c) ̇ (d) ̇ (e) None of them Answer: (e) None of them CYU 10-11.2 In a jet engine working on the ideal jet-propulsion cycle, the total heat supplied is 10,000 kW, the thermal energy rejected is 5000 kW, and the excess kinetic energy is 3000 kW. What is the propulsive efficiency of this engine? (a) 20% (b) 25% (c) 30% (d) 50% (e) 80% Answer: (a) 20%

CYU 10-12 ■ Check Your Understanding CYU 10-12.1 For which cycle pair below is the total exergy destruction zero? (a) Stirling and Jet-Propulsion cycles (b) Carnot and Brayton cycles (c) Brayton and Jet-Propulsion cycles (d) Ericsson and Carnot cycles (e) Brayton and Ericsson cycles Answer: (d) Ericsson and Carnot cycles

10-8

CYU 10-12.2 A cycle involves of heat supply from a source at 1200 K and environment at 300 K. What is the exergy destruction for this cycle? (a) (b) (c) (d) (e)

of heat rejection to the

Answer: (e)

CYU 10-12.3 An ideal Brayton cycle involves a heat input of whose exergy content is output of to the environment at 15ºC. What is the second-law efficiency of this cycle? (a) 20% (b) 25% (c) 30% (d) 40% (e) 50%

and a heat

Answer: (e) 50%

10-9

PROBLEMS Actual and Ideal Cycles, Carnot cycle, Air-Standard Assumptions, Reciprocating Engines 10-1C The air standard assumptions are: (1) the working fluid is air which behaves as an ideal gas, (2) all the processes are internally reversible, (3) the combustion process is replaced by the heat addition process, and (4) the exhaust process is replaced by the heat rejection process which returns the working fluid to its original state.

10-2C The cold air standard assumptions involves the additional assumption that air can be treated as an ideal gas with constant specific heats at room temperature.

10-3C The Carnot cycle is not suitable as an ideal cycle for all power producing devices because it cannot be approximated using the hardware of actual power producing devices.

10-4C It is less than the thermal efficiency of a Carnot cycle.

10-5C Under the air standard assumptions, the combustion process is modeled as a heat addition process, and the exhaust process as a heat rejection process.

10-6C It represents the net work on both diagrams.

10-7C It is the ratio of the maximum to minimum volumes in the cylinder.

10-8C The MEP is the fictitious pressure which, if acted on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle.

10-9C Assuming no accumulation of carbon deposits on the piston face, the compression ratio will remain the same (otherwise it will increase). The mean effective pressure, on the other hand, will decrease as a car gets older as a result of wear and tear.

10-10C Yes.

10-11C The SI and CI engines differ from each other in the way combustion is initiated; by a spark in SI engines, and by compressing the air above the self-ignition temperature of the fuel in CI engines.

10-10

10-12C Stroke is the distance between the TDC and the BDC, bore is the diameter of the cylinder, TDC is the position of the piston when it forms the smallest volume in the cylinder, and clearance volume is the minimum volume formed in the cylinder.

10-13C The clearance volume is the minimum volume formed in the cylinder whereas the displacement volume is the volume displaced by the piston as the piston moves between the top dead center and the bottom dead center.

10-14 The temperatures of the energy reservoirs of an ideal gas power cycle are given. It is to be determined if this cycle can have a thermal efficiency greater than 55 percent. Analysis The maximum efficiency any engine using the specified reservoirs can have is

Therefore, an efficiency of 55 percent is possible.

10-15 The three processes of an air-standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the back work ratio and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air are given as , , , and k = 1.4. Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures. (b) The temperature at state 2 is

During process 1-3, we have ∫

During process 2-3, we have

10-11

∫

The back work ratio is then

Heat input is determined from an energy balance on the cycle during process 1-3,

The net work output is (c) The thermal efficiency is then

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(27+273) [K] P2=850 [kPa] Vol3\Vol2=7 c_v=0.718 [kJ/kg-K] c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.4 "Analysis" "(b)" T2=T1*(P2/P1) T3=T2 w_31_in=-R*(T1-T3) w_23_out=R*T2*ln(Vol3\Vol2) r_bw=w_31_in/w_23_out q_13_in=c_v*(T3-T1)+w_23_out w_net=w_23_out-w_31_in "(c)" eta_th=w_net/q_13_in

10-12

10-16 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work per cycle and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis

(b) The properties of air at various states are → → → →

(c)

EES (Engineering Equation Solver) SOLUTION "Given" m=0.003 [kg] P1=95 [kPa] T1=17[C]+273 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-13

P2=380 [kPa] P3=95 [kPa] "Analysis" Fluid$='air' "(b)" u1=intenergy(Fluid$, T=T1) h1=enthalpy(Fluid$, T=T1) T2=(P2/P1)*T1 "from ideal gas relation" u2=intenergy(Fluid$, T=T2) s2=entropy(Fluid$, T=T2, P=P2) s3=s2 "isentropic process" h3=enthalpy(Fluid$, P=P3, s=s3) Q_in=m*(u2-u1) Q_out=m*(h3-h1) W_net_out=Q_in-Q_out "(c)" eta_th=W_net_out/Q_in

10-17 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work per cycle and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A2). Analysis

(b) From the isentropic relations and energy balance, → ( )

(

)

10-14

(c)

EES (Engineering Equation Solver) SOLUTION "Given" m=0.003 [kg] P1=95 [kPa] T1=17[C]+273 [K] P2=380 [kPa] P3=95 [kPa] "Properties" Fluid$='air' C_p=CP(Fluid$, T=T) C_v=CV(Fluid$, T=T) T=25[C]+273 [K] "room temperature" k=C_p/C_v "Analysis" "(b)" T2=(P2/P1)*T1 "from ideal gas relation" T3=T2*(P3/P2)^((k-1)/k) "isentropic process" Q_in=m*C_v*(T2-T1) Q_out=m*C_p*(T3-T1) W_net_out=Q_in-Q_out "(c)" eta_th=W_net_out/Q_in

10-18E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis

10-15

(b) The properties of air at various states are → → → → (

)

→ From energy balance,

EES (Engineering Equation Solver) SOLUTION "Given" P1=14.7 [psia] T1=(80+460) [R] q_in_12=300 [Btu/lbm] T3=3200 [R] P4=14.7 [psia] "Analysis" Fluid$='air' "(b)" u1=intenergy(Fluid$, T=T1) h1=enthalpy(Fluid$, T=T1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-16

q_in_12=u2-u1 T2=temperature(Fluid$, u=u2) h2=enthalpy(Fluid$, u=u2) P2=(T2/T1)*P1 "from ideal gas relation" P3=P2 h3=enthalpy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) s4=s3 "isentropic process" h4=enthalpy(Fluid$, P=P4, s=s4) q_in_23=h3-h2 q_in=q_in_12+q_in_23 "(c)" q_out=h4-h1 eta_th=1-q_out/q_in

10-19E The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the total heat input and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A-2E). Analysis

(b) The temperature at state 2 and the heat input are

→

Process 3-4 is isentropic: ( )

(

)

10-17

EES (Engineering Equation Solver) SOLUTION "Given" P1=14.7 [psia] T1=(80+460) [R] q_in_12=300 [Btu/lbm] T3=3200 [R] P4=14.7 [psia] "Properties at room temperature, Table A-2Ea" c_p=0.240 [Btu/lbm-R] c_v=0.171 [Btu/lbm-R] k=1.4 "Analysis" "(b)" q_in_12=c_v*(T2-T1) P2=(T2/T1)*P1 "from ideal gas relation" P3=P2 q_in_23=c_p*(T3-T2) T4=T3*(P4/P3)^((k-1)/k) "isentropic process" q_in=q_in_12+q_in_23 "(c)" q_out=c_p*(T4-T1) eta_th=1-q_out/q_in

10-20 An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to be sketcehed on the P-v and T-s diagrams; the heat and work interactions and the thermal efficiency of the cycle are to be determined; and an expression for thermal efficiency as functions of compression ratio and specific heat ratio is to be obtained. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air are given as and . Analysis (a) The P-v and T-s diagrams of the cycle are shown in the figures. (b) Noting that © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-18

Process 1-2: Isentropic compression ( )

From ideal gas relation, → Process 2-3: Constant pressure heat addition ∫

Process 3-1: Constant volume heat rejection

(d) The expression for the cycle thermal efficiency is obtained as follows:

10-19

(

)

(

)

(

)

(

) (

)

since

EES (Engineering Equation Solver) SOLUTION "Given" T1=(20+273) [K] r_c=5 c_v=0.7 [kJ/kg-K] R=0.3 [kJ/kg-K] "Analysis" "(b)" c_p=c_v+R k=c_p/c_v T2=T1*r_c^(k-1) w_12_in=c_v*(T2-T1) q_12=0 T3=T2*r_c w_23_out=R*(T3-T2) q_23_in=c_p*(T3-T2) q_31_out=c_v*(T3-T1) w_31=0 "(c)" w_net=w_23_out-w_12_in eta_th=w_net/q_23_in

10-21 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle, the heat transfer to the working fluid, and the mass of the working fluid are to be determined. Assumptions Air is an ideal gas with variable specific heats. Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process, which is state 1. From Table A-17:

10-20

→ → Using isentropic relation:

(

)

(b) The heat input is determined from

EES (Engineering Equation Solver) SOLUTION "Given" T1=1200 [K] T2=T1 T3=350 [K] T4=T3 P3=150 [kPa] P4=300 [kPa] W_net_out=0.5 [kJ] "Analysis" Fluid$='air' "(a)" s4=entropy(Fluid$, T=T4, P=P4) s1=s4 "isentropic process" P1=pressure(Fluid$, T=T1, s=s1) P_max=P1 "(b)" eta_th=1-T3/T1 Q_in=W_net_out/eta_th © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-21

"(c)" s3=entropy(Fluid$, T=T3, P=P3) DELTAs=s3-s4 w_net=DELTAs*(T1-T3) m=W_net_out/w_net

10-22 A Carnot cycle with specified temperature limits is considered. The maximum pressure in the cycle, the heat transfer to the working fluid, and the mass of the working fluid are to be determined. Assumptions Helium is an ideal gas with constant specific heats. Properties The properties of helium at room temperature are and k = 1.667 (Table A-2). Analysis (a) In a Carnot cycle, the maximum pressure occurs at the beginning of the expansion process, which is state 1. or,

( ) ( )

(

)

(b) The heat input is determined from

EES (Engineering Equation Solver) SOLUTION "Given" T1=1200 [K] T2=T1 T3=350 [K] T4=T3 P3=150 [kPa] P4=300 [kPa] W_net_out=0.5 [kJ] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-22

Fluid$='Helium' C_p=CP(Fluid$, T=T, P=P) C_v=CV(Fluid$, T=T, P=P) T=25[C]+273 [K] "room temperature" P=100 [kPa] "assumed" k=C_p/C_v R=C_p-C_v "Analysis" "(a)" P1=P4*(T1/T4)^(k/(k-1)) "isentropic process" P_max=P1 "(b)" eta_th=1-T3/T1 Q_in=W_net_out/eta_th "(c)" DELTAs=C_p*ln(T4/T3)-R*ln(P4/P3) w_net=-DELTAs*(T1-T3) m=W_net_out/w_net

10-23E The temperatures of the energy reservoirs of an ideal gas Carnot cycle are given. The heat supplied and the work produced per cycle are to be determined. Analysis According to the thermodynamic definition of temperature,

Applying the first law to the cycle gives

10-24 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2). Analysis The minimum pressure in the cycle is and the maximum pressure is . Then,

10-23

( ) ( )

(

)

The heat input is determined from Then,

EES (Engineering Equation Solver) SOLUTION "Given" m=0.6 [kg] T1=1100 [K] T2=T1 T3=300 [K] T4=T3 P1=3000 [kPa] P3=20 [kPa] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 R=0.287 [kJ/kg-K] "Analysis" P2=P3*(T2/T3)^(k/(k-1)) "isentropic process" DELTAs=c_p*ln(T2/T1)-R*ln(P2/P1) Q_in=m*T1*DELTAs eta_th=1-T3/T1 W_net_out=eta_th*Q_in

10-25 A Carnot cycle executed in a closed system with air as the working fluid is considered. The minimum pressure in the cycle, the heat rejection from the cycle, the thermal efficiency of the cycle, and the second-law efficiency of an actual cycle operating between the same temperature limits are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperatures are and k = 1.4 (Table A-2). Analysis (a) The minimum temperature is determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-24

→

The pressure at state 4 is determined from

( ) ( ) (

)

→

The minimum pressure in the cycle is determined from

→ (b) The heat rejection from the cycle is (c) The thermal efficiency is determined from

(d) The power output for the Carnot cycle is ̇

̇ Then, the second-law efficiency of the actual cycle becomes ̇ ̇

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=1300 [kPa] T_1=950 [K] DELTAs_12=0.25 [kJ/kg-K] w_net=110 [kJ/kg] m_dot=95 [kg/s] W_dot_net=5200 [kW] "PROPERTIES, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 R=0.287 [kJ/kg-K] "ANALYSIS" "(a)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-25

w_net=DELTAs_12*(T_1-T_4) "for T_4" P_1=P_4*(T_1/T_4)^k/(k-1) "for P_1" -DELTAs_12=-R*ln(P_4/P_3) "for P_3, minimum pressure" "(b)" q_out=T_4*DELTAs_12 "(c)" eta_th=1-T_4/T_1 "for eta_th" "(d)" eta_th=1-q_out/q_in "for q_in" W_dot_Carnot=m_dot*w_net eta_II=W_dot_net/W_dot_Carnot

Otto Cycle 10-26C The four processes that make up the Otto cycle are (1) isentropic compression, (2) v = constant heat addition, (3) isentropic expansion, and (4) v = constant heat rejection.

10-27C They are analyzed as closed system processes because no mass crosses the system boundaries during any of the processes.

10-28C The ideal Otto cycle involves external irreversibilities, and thus it has a lower thermal efficiency. 10-29C It increases with both of them.

10-30C Because high compression ratios cause engine knock.

10-31C The thermal efficiency will be the highest for argon because it has the highest specific heat ratio, k = 1.667.

10-32C For actual four-stroke engines, the rpm is twice the number of thermodynamic cycles; for two-stroke engines, it is equal to the number of thermodynamic cycles.

10-33C The fuel is injected into the cylinder in both engines, but it is ignited with a spark plug in gasoline engines.

10-34E The properties at various states of an ideal Otto cycle are given. The mean effective pressure is to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are (Table A-1E), , , and k = 1.4 (Table A-2Ea). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-26

Analysis At the end of the compression, the temperature is ( ) while the air temperature at the end of the expansion is ( )

( )

Application of the first law to the compression and expansion processes gives

At the beginning of the compression, the specific volume is

while the specific volume at the end of the compression is

The engine’s mean effective pressure is then .

10-35E The power produced by an ideal Otto cycle is given. The rate of heat addition and rejection are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are (Table A-1E), , , and k = 1.4 (Table A-2Ea).

10-27

According to the definition of the thermal efficiency, the rate of heat addition to this cycle is ̇ ̇ ( ) The rate of heat rejection is then ̇ ̇ ̇

10-36 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is . The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression. →

→ →

(

)

Process 2-3: v = constant heat addition. →

→ →

(

)

(b) Process 3-4: isentropic expansion. → Process 4-1: v = constant heat rejection.

10-28

(

)

EES (Engineering Equation Solver) SOLUTION "Given" r=8 "r=v1/v2=v4/v3" P1=95 [kPa] T1=27[C]+273 [K] q_in=750 [kJ/kg] "Properties" Fluid$='Air' C_p=CP(Fluid$, T=T) C_v=CV(Fluid$, T=T) T=25[C]+273 [K] "room temperature" R_air=C_p-C_v "Analysis" "(a)" "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" P2=r*(T2/T1)*P1 "from ideal gas relation" u2=intenergy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant volume heat addition" q_in=u3-u2 T3=temperature(Fluid$, u=u3) s3=entropy(Fluid$, u=u3, P=P3) P3=(T3/T2)*P2 "from ideal gas relation" "(b)" "Process 3-4: isentropic expansion" s4=s3 "isentropic process" P4=1/r*(T4/T3)*P3 "from ideal gas relation" u4=intenergy(Fluid$, P=P4, s=s4) T4=temperature(Fluid$, P=P4, s=s4) "Process 4-1: constant volume heat rejection" q_out=u4-u1 w_net_out=q_in-q_out "(c)" eta_th=w_net_out/q_in "(d)" v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2)

10-29

10-37 Problem 10-36 is reconsidered. The effect of the compression ratio on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: T[1]=300 [K] P[1]=95 [kPa] q_23 = 750 [kJ/kg] r_comp = 8 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] q_23 - w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] q_34 -w_34 = DELTAu_34 q_34 =0"isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 MEP = w_net/(V[1]-V[2])

MEP [kPa] 5 6 7 8 9 10

43.78 47.29 50.08 52.36 54.28 55.93

452.9 469.6 483.5 495.2 505.3 514.2

328.4 354.7 375.6 392.7 407.1 419.5

10-30

10-31

10-38 An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The pressure and temperature at the end of the heat addition process, the net work output, the thermal efficiency, and the mean effective pressure for the cycle are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2).

Analysis (a) Process 1-2: isentropic compression. ( )

10-32

→

(

)

Process 2-3: v = constant heat addition.

→

(

)

(b) Process 3-4: isentropic expansion. ( )

( )

Process 4-1: v = constant heat rejection.

(

)

EES (Engineering Equation Solver) SOLUTION "Given" r=8 "r=v1/v2=v4/v3" P1=95 [kPa] T1=27[C]+273 [K] q_in=750 [kJ/kg] "Properties" Fluid$='air' c_p=CP(Fluid$, T=T) c_v=CV(Fluid$, T=T) T=25[C]+273 [K] "room temperature" k=c_p/c_v R_air=c_p-c_v "Analysis" "(a)" "Process 1-2: isentropic compression" T2=T1*r^(k-1) P2=r*(T2/T1)*P1 "from ideal gas relation" "Process 2-3: constant volume heat addition" q_in=c_v*(T3-T2) P3=(T3/T2)*P2 "from ideal gas relation" "(b)" T4=T3*(1/r)^(k-1) "Process 3-4: isentropic expansion" q_out=c_v*(T4-T1) "Process 4-1: constant volume heat rejection" w_net_out=q_in-q_out "(c)" eta_th=w_net_out/q_in "(d)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-33

v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2)

10-39E An Otto cycle with non-isentropic compression and expansion processes is considered. The thermal efficiency, the heat addition, and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A-2Ea).

(Table A-1E),

Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then

( ) With the isentropic compression efficiency, the actual temperature at the end of the compression is → Similarly for the expansion, ( )

( )

(

)

→ The specific heat addition is that of process 2-3, The net work production is the difference between the work produced by the expansion and that used by the compression,

The thermal efficiency of this cycle is then

At the beginning of compression, the maximum specific volume of this cycle is

10-34

while the minimum specific volume of the cycle occurs at the end of the compression

The engine’s mean effective pressure is then .

EES (Engineering Equation Solver) SOLUTION "Given" r=10 "v1/v2=v4/v3=8" eta_comp=0.85 eta_exp=0.95 P1=13 [psia] T1=(60+460) [R] T3=(2300+460) [R] "Properties at room temperature, Table A-2Ea" c_p=0.240 [Btu/lbm-R] c_v=0.171 [Btu/lbm-R] k=1.4 R_a=0.3704 [psia-ft^3/lbm-R] "Analysis" T2_s=T1*r^(k-1) eta_comp=(T2_s-T1)/(T2-T1) "solve for T2" T4_s=T3*(1/r)^(k-1) eta_exp=(T3-T4)/(T3-T4_s) "solve for T4" q_in=c_v*(T3-T2) w_net=c_v*(T3-T4)-c_v*(T2-T1) eta_th=w_net/q_in v1=(R_a*T1)/P1 v2=v1/r MEP=w_net/(v1-v2)*Convert(Btu, psia-ft^3)

10-40 An ideal Otto cycle is considered. The heat rejection, the net work production, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a). Analysis The mass in this system is

10-35

The two unknown temperatures are ( ) ( )

( )

Application of the first law to four cycle processes gives

The net work is The thermal efficiency is then

The minimum volume of the cycle occurs at the end of the compression

The engine’s mean effective pressure is then .

10-41 The power produced by an ideal Otto cycle is given. The rate of heat addition is to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A-2a). Analysis The compression ratio is

10-36

and the thermal efficiency is

The rate at which heat must be added to this engine is then ̇ ̇ (

)

10-42E An ideal Otto cycle with air as the working fluid has a compression ratio of 8. The amount of heat transferred to the air during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions

1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Process 1-2: isentropic compression. → → Process 2-3: v = constant heat addition. →

(b) Process 3-4: isentropic expansion. → Process 4-1: v = constant heat rejection.

10-37

EES (Engineering Equation Solver) SOLUTION "Given" r=8 "r=v1/v2=v4/v3" T1=540 [R] T3=2400 [R] P1=14.7 [psia] "assumed" "Analysis" Fluid$='air' "(a)" "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 P2=r*(T2/T1)*P1 "from ideal gas relation" u2=intenergy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant volume heat addition" P3=(T3/T2)*P2 "from ideal gas relation" u3=intenergy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) q_in=u3-u2 "(b)" "Process 3-4: isentropic expansion" s4=s3 P4=1/r*(T4/T3)*P3 "from ideal gas relation" T4=temperature(Fluid$, P=P4, s=s4) u4=intenergy(Fluid$, P=P4, s=s4) q_out=u4-u1 eta_th=1-q_out/q_in "(c)" eta_th_C=1-T1/T3

10-43E An ideal Otto cycle with argon as the working fluid has a compression ratio of 8. The amount of heat transferred to the argon during the heat addition process, the thermal efficiency, and the thermal efficiency of a Carnot cycle operating between the same temperature limits are to be determined. Assumptions 1. The air-standard assumptions are applicable with argon as the working fluid. 2. Kinetic and potential energy changes are negligible. 3. Argon is an ideal gas with constant specific heats. Properties The properties of argon are , , and k = 1.667 (Table A-2E). Analysis (a) Process 1-2: isentropic compression.

10-38

( ) Process 2-3: v = constant heat addition.

(b) Process 3-4: isentropic expansion. ( )

( )

Process 4-1: v = constant heat rejection.

EES (Engineering Equation Solver) SOLUTION "Given" r=8 "r=v1/v2=v4/v3" T1=540 [R] T3=2400 [R] "Properties at room temperature, Table A-2Ea" c_p=0.1253 [Btu/lbm-R] c_v=0.0756 [Btu/lbm-R] k=1.667 "Analysis" "(a)" T2=T1*r^(k-1)"Process 1-2: isentropic compression" q_in=c_v*(T3-T2) "Process 2-3: constant volume heat addition" "(b)" T4=T3*(1/r)^(k-1) "Process 3-4: isentropic expansion" q_out=c_v*(T4-T1) eta_th=1-q_out/q_in "(c)" eta_th_C=1-T1/T3

10-44 The two isentropic processes in an Otto cycle are replaced with polytropic processes. The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a). Analysis The temperature at the end of the compression is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-39

( ) And the temperature at the end of the expansion is ( )

( )

The integral of the work expression for the polytropic compression gives 0( )

Similarly, the work produced during the expansion is 0( )

0( )

Application of the first law to each of the four processes gives

The head added and rejected from the cycle are

The thermal efficiency of this cycle is then

EES (Engineering Equation Solver) SOLUTION "Given" n=1.3 r=8 "r=v1/v2=v4/v3" P1=95 [kPa] T1=(15+273) [K] T3=(1200+273) [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 R_a=0.287 [kJ/kg-K] "Analysis" T2=T1*r^(n-1) T4=T3*(1/r)^(n-1) w_12=(R_a*T1)/(n-1)*(r^(n-1)-1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-40

w_34=-(R_a*T3)/(n-1)*((1/r)^(n-1)-1) q_12=w_12-c_v*(T2-T1) q_23=c_v*(T3-T2) q_34=w_34-c_v*(T3-T4) q_41=c_v*(T4-T1) q_in=q_23+q_34 q_out=q_12+q_41 eta_th=1-q_out/q_in

10-45 An ideal Otto cycle is considered. The heat added to and rejected from this cycle, and the cycle’s thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a). Analysis The temperature at the end of the compression is

( ) and the temperature at the end of the expansion is ( )

( )

Application of the first law to the heat addition process gives Similarly, the heat rejected is The thermal efficiency of this cycle is then

EES (Engineering Equation Solver) SOLUTION "Given" r=8 "r=v1/v2=v4/v3" P1=95 [kPa] T1=(15+273) [K] T3=(1200+273) [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-41

k=1.4 R_a=0.287 [kJ/kg-K] "Analysis" T2=T1*r^(k-1) T4=T3*(1/r)^(k-1) q_in=c_v*(T3-T2) q_out=c_v*(T4-T1) eta_th=1-q_out/q_in

10-46 The expressions for the maximum gas temperature and pressure of an ideal Otto cycle are to be determined when the compression ratio is doubled. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis The temperature at the end of the compression varies with the compression ratio as ( )

since is fixed. The temperature rise during the combustion remains constant since the amount of heat addition is fixed. Then, the maximum cycle temperature is given by The smallest gas specific volume during the cycle is

When this is combined with the maximum temperature, the maximum pressure is given by

10-47 It is to be determined if the polytropic exponent to be used in an Otto cycle model will be greater than or less than the isentropic exponent. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis During a polytropic process,

10-42

and for an isentropic process,

If heat is lost during the expansion of the gas, where when

is the temperature that would occur if the expansion were reversible and adiabatic (n=k). This can only occur

Diesel Cycle 10-48C A diesel engine differs from the gasoline engine in the way combustion is initiated. In diesel engines combustion is initiated by compressing the air above the self-ignition temperature of the fuel whereas it is initiated by a spark plug in a gasoline engine.

10-49C The Diesel cycle differs from the Otto cycle in the heat addition process only; it takes place at constant volume in the Otto cycle, but at constant pressure in the Diesel cycle.

10-50C Cutoff ratio is the ratio of the cylinder volumes after and before the combustion process. As the cutoff ratio decreases, the efficiency of the diesel cycle increases.

10-51C The gasoline engine.

10-52C Diesel engines operate at high compression ratios because the diesel engines do not have the engine knock problem.

10-53 An ideal diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-43

Analysis We begin by using the process types to fix the temperatures of the states.

( ) ( ) Combining the first law as applied to the various processes with the process equations gives

According to the definition of the thermal efficiency, ̇ ̇

(

)

10-54 A Diesel cycle with non-isentropic compression and expansion processes is considered. The maximum temperature of the air and the rate of heat addition are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a). Analysis We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then

( ) With the isentropic compression efficiency, the actual temperature at the end of the compression is → The maximum temperature is ( ) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-44

For the isentropic expansion process, ( )

( )

(

)

since } The actual temperature at the end of expansion process is then → The net work production is the difference between the work produced by the expansion and that used by the compression,

The heat addition occurs during process 2-3, The thermal efficiency of this cycle is then

According to the definition of the thermal efficiency, ̇ ̇

(

)

10-55 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The temperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is . The properties of air are given in Table A-17.

Analysis (a) Process 1-2: isentropic compression. → → Process 2-3: P = constant heat addition. →

→

10-45

(b) Process 3-4: isentropic expansion. → Process 4-1: v = constant heat rejection.

(c)

EES (Engineering Equation Solver) SOLUTION "Given" r=16 "r=v1/v2" r_c=2 "r_c=v3/v2" P1=95 [kPa] T1=27[C]+273 [K] "Properties" Fluid$='air' C_p=CP(Fluid$, T=T1) C_v=CV(Fluid$, T=T1) R_air=C_p-C_v "Analysis" "(a)" "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" P2=r*(T2/T1)*P1 "from ideal gas relation" h2=enthalpy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant pressure heat addition" T3=r_c*T2 h3=enthalpy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) P3=P2 "(b)" q_in=h3-h2 "Process 3-4: isentropic expansion" s4=s3 "isentropic process" P4=(T4/T1)*P1 "from ideal gas relation" T4=temperature(Fluid$, P=P4, s=s4) u4=intenergy(Fluid$, P=P4, s=s4) "Process 4-1: constant volume heat rejection" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-46

q_out=u4-u1 eta_th=1-q_out/q_in "(c)" w_net_out=q_in-q_out v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2)

10-56 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The temperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1. 2. 3.

The air-standard assumptions are applicable. Kinetic and potential energy changes are negligible. Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are , and k = 1.4 (Table A-2).

Analysis (a)

Process 1-2: isentropic compression.

( ) Process 2-3: P = constant heat addition. → (b) Process 3-4: isentropic expansion. ( )

(

)

(

)

Process 4-1: v = constant heat rejection.

(c)

10-47

EES (Engineering Equation Solver) SOLUTION "Given" r=16 "r=v1/v2" r_c=2 "r_c=v3/v2" P1=95 [kPa] T1=27[C]+273 [K] "Properties" Fluid$='air' c_p=CP(Fluid$, T=T) c_v=CV(Fluid$, T=T) T=25[C]+273 [K] "room temperature" k=c_p/c_v R_air=c_p-c_v "Analysis" "(a)" T2=T1*r^(k-1) "Process 1-2: isentropic compression" "Process 2-3: constant pressure heat addition" T3=r_c*T2 "(b)" q_in=c_p*(T3-T2) T4=T3*(r_c/r)^(k-1) "Process 3-4: isentropic expansion" q_out=c_v*(T4-T1) "Process 4-1: constant volume heat rejection" eta_th=1-q_out/q_in "(c)" w_net_out=q_in-q_out v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2)

10-57E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a)

Process 1-2: isentropic compression. →

10-48

→ (b) →

Process 3-4: isentropic expansion. (

)

→ Process 4-1: v = constant heat rejection: (c)

EES (Engineering Equation Solver) SOLUTION "Given" r=18.2 "r=v1/v2" P1=14.7 [psia] T1=(120+460) [R] T3=3200 [R] "Analysis" Fluid$='air' "(a)" "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" P2=r*(T2/T1)*P1 "from ideal gas relation" h2=enthalpy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant pressure heat addition" r_c=T3/T2 "r_c=v3/v2" "(b)" h3=enthalpy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) P3=P2 q_in=h3-h2 "Process 3-4: isentropic expansion" s4=s3 "isentropic process" P4=(T4/T1)*P1 "from ideal gas relation" T4=temperature(Fluid$, P=P4, s=s4) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-49

u4=intenergy(Fluid$, P=P4, s=s4) "Process 4-1: constant volume heat rejection" q_out=u4-u1 "(c)" eta_th=1-q_out/q_in

10-58E An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A-2E). Analysis (a)

Process 1-2: isentropic compression. ( )

Process 2-3: P = constant heat addition. → (b) Process 3-4: isentropic expansion. ( )

(

)

(

)

Process 4-1: v = constant heat rejection. (c)

EES (Engineering Equation Solver) SOLUTION "Given" r=18.2 "r=v1/v2" P1=14.7 [psia] T1=(120+460) [R] T3=3200 [R] "Properties at room temperature, Table A-2Ea" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-50

c_p=0.240 [Btu/lbm-R] c_v=0.171 [Btu/lbm-R] k=1.4 "Analysis" "(a)" T2=T1*r^(k-1) "Process 1-2: isentropic compression" "Process 2-3: constant pressure heat addition" r_c=T3/T2 "r_c=v3/v2" "(b)" q_in=c_p*(T3-T2) T4=T3*(r_c/r)^(k-1) "Process 3-4: isentropic expansion" q_out=c_v*(T4-T1) "Process 4-1: constant volume heat rejection" "(c)" eta_th=1-q_out/q_in

10-59 An ideal diesel cycle is considered. The power produced is to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , and k = 1.4 (Table A-2a). Analysis The specific volume of the air at the start of the compression is

The total air mass taken by all 8 cylinders when they are charged is

The rate at which air is processed by the engine is determined from ̇ ̇ since there are two revolutions per cycle in a four-stroke engine. The compression ratio is

At the end of the compression, the air temperature is Application of the first law and work integral to the constant pressure heat addition gives The cutoff ratio is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-51

while the thermal efficiency is

The power produced by this engine is then ̇ ̇

10-60 An ideal diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2).

Analysis (a) Process 1-2: isentropic compression. ( ) Process 2-3: P = constant heat addition. → Process 3-4: isentropic expansion. ( )

(b)

(

)

(

)

(

)

10-52

EES (Engineering Equation Solver) SOLUTION "Given" r=20 "r=v1/v2" P1=95 [kPa] T1=20[C]+273 [K] T3=2200 [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 R_air=0.287 [kJ/kg-K] "Analysis" "(a)" T2=T1*r^(k-1) "Process 1-2: isentropic compression" r_c=T3/T2 "r_c=v3/v2, Process 2-3: constant pressure heat addition" T4=T3*(r_c/r)^(k-1) "Process 3-4: isentropic expansion" q_in=c_p*(T3-T2) q_out=c_v*(T4-T1) w_net_out=q_in-q_out eta_th=w_net_out/q_in "(b)" v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2)

10-61 A diesel engine with air as the working fluid has a compression ratio of 20. The thermal efficiency and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The gas constant of air is (Table A-2). Analysis

(a)

Process 1-2: isentropic compression.

10-53

→ → Process 2-3: P = constant heat addition. → →

(Table A-17)

Process 3-4: polytropic expansion. ( )

(

→ An energy balance gives

)

(

)

(

)

(Table A-17)

→

Process 4-1: v = constant heat rejection. Total heat rejected is and

(b)

(

)

EES (Engineering Equation Solver) SOLUTION "Given" r=20 "r=v1/v2" P1=95 [kPa] T1=(20+273) [K] T3=2200 [K] n=1.35 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-54

R_air=0.287 [kJ/kg-K] "Analysis" "(a)" T2=T1*r^(k-1) "Process 1-2: isentropic compression" r_c=T3/T2 "r_c=v3/v2, Process 2-3: constant pressure heat addition" T4=T3*(r_c/r)^(n-1) "Process 3-4: polytropic expansion" q_in=c_p*(T3-T2) q_41_out=c_v*(T4-T1) w_34_out=(R_air*(T4-T3))/(1-n) -q_34_out-w_34_out=u4-u3 "using constant specific heat assumption gives heat input for expansion process, which is unrealistic. Therefore, we use u data for this process" Fluid$='air' u3=intenergy(Fluid$, T=T3) u4=intenergy(Fluid$, T=T4) q_out=q_34_out+q_41_out w_net_out=q_in-q_out eta_th=w_net_out/q_in "(b)" v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2)

10-62 Problem 10-61 is reconsidered. The effect of the compression ratio on the net work output, mean effective pressure, and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: Procedure QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) q_in_total = 0 q_out_total = 0 IF (q_12 > 0) THEN q_in_total = q_12 ELSE q_out_total = - q_12 If q_23 > 0 then q_in_total = q_in_total + q_23 else q_out_total = q_out_total - q_23 If q_34 > 0 then q_in_total = q_in_total + q_34 else q_out_total = q_out_total - q_34 If q_41 > 0 then q_in_total = q_in_total + q_41 else q_out_total = q_out_total - q_41 END "Input Data" T[1]=293 [K] P[1]=95 [kPa] T[3] = 2200 [K] n=1.35 r_comp = 20 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp q_12 - w_12 = DELTAu_12 q_12 =0"isentropic process" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-55

DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant pressure heat addition" P[3]=P[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] q_23 - w_23 = DELTAu_23 w_23 =P[2]*(V[3] - V[2])"constant pressure process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is polytropic expansion" P[3]/P[4] =(V[4]/V[3])^n s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] q_34 - w_34 = DELTAu_34 "q_34 is not 0 for the ploytropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) P[3]*V[3]^n = Const w_34=(P[4]*V[4]-P[3]*V[3])/(1-n) "Process 4-1 is constant volume heat rejection" V[4] = V[1] q_41 - w_41 = DELTAu_41 w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) Call QTotal(q_12,q_23,q_34,q_41: q_in_total,q_out_total) w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*100 MEP = w_net/(V[1]-V[2])

14 16 18 20 22 24

47.69 50.14 52.16 53.85 55.29 56.54

MEP [kPa] 970.8 985 992.6 995.4 994.9 992

797.9 817.4 829.8 837.0 840.6 841.5

10-56

10-57

10-63 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at a given rpm is to be determined. Assumptions 1. The cold air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2).

Analysis Process 1-2: isentropic compression. ( ) Process 2-3: P = constant heat addition. → Process 3-4: isentropic expansion. ( )

(

)

(

)

(

)

For the cycle:

10-58

̇ ̇

Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.0024 [m^3] r=22 "r=v1/v2" r_c=1.8 "r_c=v3/v2" P1=97 [kPa] T1=(70+273) [K] n_dot=4250 [1/min] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 R_air=0.287 [kJ/kg-K] "Analysis" T2=T1*r^(k-1) "Process 1-2: isentropic compression" T3=r_c*T2 "Process 2-3: constant pressure heat addition" T4=T3*(r_c/r)^(k-1) "Process 3-4: isentropic expansion" m=(P1*V1)/(R_air*T1) Q_in=m*c_p*(T3-T2) Q_out=m*c_v*(T4-T1) W_net_out=Q_in-Q_out W_dot_net_out=n_dot*W_net_out*1/Convert(min, s)

10-64 A four-cylinder ideal diesel engine with nitrogen as the working fluid has a compression ratio of 22 and a cutoff ratio of 1.8. The power the engine will deliver at a given rpm is to be determined. Assumptions 1. The air-standard assumptions are applicable with nitrogen as the working fluid. 2. Kinetic and potential energy changes are negligible. 3. Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at room temperature are , , , and k = 1.4 (Table A-2). Analysis Process 1-2: isentropic compression.

10-59

( ) Process 2-3: P = constant heat addition. → Process 3-4: isentropic expansion. ( )

(

)

(

)

(

)

For the cycle:

̇ ̇

Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions).

EES (Engineering Equation Solver) SOLUTION "Given" V1=0.0024 [m^3] r=22 "r=v1/v2" r_c=1.8 "r_c=v3/v2" P1=97 [kPa] T1=(70+273) [K] n_dot=4250 [1/min] "Properties of nirogen at room temperature, Table A-2a" c_p=1.039 [kJ/kg-K] c_v=0.743 [kJ/kg-K] k=1.4 R_N2=0.2968 [kJ/kg-K] "Analysis" T2=T1*r^(k-1) "Process 1-2: isentropic compression" T3=r_c*T2 "Process 2-3: constant pressure heat addition" T4=T3*(r_c/r)^(k-1) "Process 3-4: isentropic expansion" m=(P1*V1)/(R_N2*T1) Q_in=m*c_p*(T3-T2) Q_out=m*c_v*(T4-T1) W_net_out=Q_in-Q_out W_dot_net_out=n_dot*W_net_out*1/Convert(min, s)

10-65 An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heat transferred at constant volume and the thermal efficiency of the cycle are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-60

Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression.

→ → Process 2-x, x-3: heat addition, →

By trial and error, we get and

Thus, and

(b)

→

(

)

→

Process 4-1: v = constant heat rejection.

EES (Engineering Equation Solver) SOLUTION "Input Data" R=0.287 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-61

T[1]=300 [K] P[1]=100 [kPa] T[4]=2200 [K] q_in_total=1520 [kJ/kg] v[1]/v[2]=14 "Compression ratio" "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] "Conservation of energy for process 1 to 2" q_12 -w_12 = DELTAu_12 q_12 =0"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} P[3]*v[3]=R*T[3] v[3]=v[2] "Conservation of energy for process 2 to 3" q_23 -w_23 = DELTAu_23 w_23 =0 "constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is constant pressure heat addition" s[4]=entropy(air, T=T[4], P=P[4]) {P[4]*v[4]/T[4]=P[3]*v[3]/T[3]} P[4]*v[4]=R*T[4] P[4]=P[3] "Conservation of energy for process 3 to4" q_34 -w_34 = DELTAu_34 w_34 =P[3]*(v[4]-v[3]) "constant pressure process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) q_in_total=q_23+q_34 f_constantVol=q_23/q_in_total "Process 4-5 is isentropic expansion" s[5]=entropy(air,T=T[5],P=P[5]) s[5]=s[4] P[5]*v[5]/T[5]=P[4]*v[4]/T[4] {P[5]*v[5]=R*T[5]} "Conservation of energy for process 4 to 5" q_45 -w_45 = DELTAu_45 q_45 =0"isentropic process" DELTAu_45=intenergy(air,T=T[5])-intenergy(air,T=T[4]) "Process 5-1 is constant volume heat rejection" v[5]=v[1] "Conservation of energy for process 2 to 3" q_51 -w_51 = DELTAu_51 w_51 =0"constant volume process" DELTAu_51=intenergy(air,T=T[1])-intenergy(air,T=T[5]) w_net = w_12+w_23+w_34+w_45+w_51 Eta_th=w_net/q_in_total*Convert( ,%) "Thermal efficiency, in percent"

10-62

10-66 Problem 10-65 is reconsidered. The effect of the compression ratio on the net work output and thermal efficiency is to be investigated. Also, T-s and P-v diagrams for the cycle are to be plotted. Analysis Using EES, the problem is solved as follows: "Input Data" T[1]=300 [K] P[1]=100 [kPa] T[4]=2200 [K] q_in_total=1520 [kJ/kg] r_v = 14 v[1]/v[2]=r_v "Compression ratio" "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] s[2]=entropy(air, T=T[2], v=v[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] q_12 -w_12 = DELTAu_12 q_12 =0 [kJ/kg]"isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] {P[3]*v[3]/T[3]=P[2]*v[2]/T[2]} v[3]=v[2] q_23 -w_23 = DELTAu_23 w_23 =0"constant volume process" DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is constant pressure heat addition" s[4]=entropy(air, T=T[4], P=P[4]) P[4]*v[4]=R*T[4] {P[4]*v[4]/T[4]=P[3]*v[3]/T[3]} P[4]=P[3] q_34 -w_34 = DELTAu_34 w_34 =P[3]*(v[4]-v[3]) "constant pressure process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) q_in_total=q_23+q_34 "Process 4-5 is isentropic expansion" s[5]=entropy(air,T=T[5],P=P[5]) s[5]=s[4] P[5]*v[5]/T[5]=P[4]*v[4]/T[4] {P[5]*v[5]=0.287*T[5]} q_45 -w_45 = DELTAu_45 q_45 =0 [kJ/kg] "isentropic process" DELTAu_45=intenergy(air,T=T[5])-intenergy(air,T=T[4]) "Process 5-1 is constant volume heat rejection" v[5]=v[1] q_51 -w_51 = DELTAu_51 w_51 =0 [kJ/kg] "constant volume process" DELTAu_51=intenergy(air,T=T[1])-intenergy(air,T=T[5]) w_net = w_12+w_23+w_34+w_45+w_51 Eta_th=w_net/q_in_total*Convert(, %)

10-63

10 11 12 13 14 15 16 17 18

52.33 53.43 54.34 55.09 55.72 56.22 56.63 56.94 57.17

795.4 812.1 826 837.4 846.9 854.6 860.7 865.5 869

10-64

10-67 An ideal dual cycle with air as the working fluid has a compression ratio of 14. The fraction of heat transferred at constant volume and the thermal efficiency of the cycle are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K

, and k = 1.4 (Table A-2).

Analysis

(a)

Process 1-2: isentropic compression.

10-65

( ) Process 2-x, x-3: heat addition,

Solving for we get which is impossible. Therefore, constant specific heats at room temperature turned out to be an unreasonable assumption in this case because of the very high temperatures involved.

EES (Engineering Equation Solver) SOLUTION "Given" r=14 "r=v1/v2" P1=100 [kPa] T1=300 [K] T3=2200 [K] q_in=1520.4 [kJ/kg] "Properties at room temperature" c_p=1.005 [kJ/kg-K] c_v=0.718 [kJ/kg-K] k=1.4 R_air=0.287 [kJ/kg-K] "Analysis" T2=T1*r^(k-1) "Process 1-2: isentropic compression" q_in=c_v*(T_x-T2)+c_p*(T3-T_x) "Process 2-x-3: heat addition" "Solving above equations, we obtain T_x = 263.9 K, which is impossible. Therefore, constant specific heat assumption is not reasonable in this case."

10-68E An ideal dual cycle has a compression ratio of 20 and cutoff ratio of 1.3. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are

(Table A-1E),

10-66

( )

(

) ( ) ( )

( )

(

)

Applying the first law and work expression to the heat addition processes gives

The heat rejected is Then,

10-69E An ideal dual cycle has a compression ratio of 12 and cutoff ratio of 1.3. The thermal efficiency, amount of heat added, and the maximum gas pressure and temperature are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are (Table A-1E), , , and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are ( )

( )

(

) ( ) ( )

( )

(

)

10-67

Applying the first law and work expression to the heat addition processes gives

The heat rejected is Then,

10-70 An expression for cutoff ratio of an ideal diesel cycle is to be developed. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis Employing the isentropic process equations,

while the ideal gas law gives When the first law and the closed system work integral is applied to the constant pressure heat addition, the result is When this is solved for cutoff ratio, the result is

10-71 An expression for the thermal efficiency of a dual cycle is to be developed and the thermal efficiency for a given case is to be calculated. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-68

Analysis The thermal efficiency of a dual cycle may be expressed as

By applying the isentropic process relations for ideal gases with constant specific heats to the processes 1-2 and 3-4, as well as the ideal gas equation of state, the temperatures may be eliminated from the thermal efficiency expression. This yields the result

where and When

, we obtain

For the case r = 20 and

, .

10-72 An expression regarding the thermal efficiency of a dual cycle for a special case is to be obtained. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis The thermal efficiency of a dual cycle may be expressed as

Where

10-69

and When

, we obtain .

Rearrangement of this result gives

10-73 The five processes of the dual cycle is described. The P-v and T-s diagrams for this cycle is to be sketched. An expression for the cycle thermal efficiency is to be obtained and the limit of the efficiency is to be evaluated for certain cases. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis (a) The P-v and T-s diagrams for this cycle are as shown.

(b) Apply first law to the closed system for processes 2-3, 3-4, and 5-1 to show:

The cycle thermal efficiency is given by

10-70

( ) Process 2-3 is constant volume; therefore,

Process 3-4 is constant pressure; therefore,

Process 4-5 is isentropic; therefore, ( ) Process 5-1 is constant volume; however

( )

(

)

(

)

( )

is found from the following. ( )

The ratio

is found from the following.

The efficiency becomes ( (c) In the limit as

)

approaches unity, the cycle thermal efficiency becomes 2

( 0

(d) In the limit as

) 1

approaches unity, the cycle thermal efficiency becomes 2

(

)

(

3 )

Stirling and Ericsson Cycles 10-74C The Stirling cycle.

10-75C The two isentropic processes of the Carnot cycle are replaced by two constant pressure regeneration processes in the Ericsson cycle.

10-76C The efficiencies of the Carnot and the Stirling cycles would be the same, the efficiency of the Otto cycle would be less.

10-71

10-77C The efficiencies of the Carnot and the Ericsson cycles would be the same, the efficiency of the Diesel cycle would be less.

10-78E An ideal Ericsson engine with helium as the working fluid operates between the specified temperature and pressure limits. The thermal efficiency of the cycle, the heat transfer rate in the regenerator, and the power delivered are to be determined. Assumptions Helium is an ideal gas with constant specific heats. Properties The gas constant and the specific heat of helium at room temperature are and (Table A-2E). Analysis (a) The thermal efficiency of this totally reversible cycle is determined from

(b) The amount of heat transferred in the regenerator is ̇ ̇ ̇ ̇

̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" T1=3000 [R] T3=550 [R] P1=200 [psia] P2=25 [psia] m_dot=14 [lbm/s] "Properties" c_p=1.25 [Btu/lbm-R] R=0.4961 [Btu/lbm-R] "Analysis" "(a)" eta_th=1-T3/T1 "(b)" Q_dot_regen=m_dot*c_p*(T1-T4) T4=T3 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-72

"(c)" DELTAs=c_p*ln(T2/T1)-R*ln(P2/P1) T2=T1 Q_dot_in=m_dot*T1*DELTAs W_dot_net_out=eta_th*Q_dot_in

10-79 An ideal Stirling engine with helium as the working fluid operates between the specified temperature and pressure limits. The thermal efficiency of the cycle, the amount of heat transfer in the regenerator, and the work output per cycle are to be determined. Assumptions Helium is an ideal gas with constant specific heats. Properties The gas constant and the specific heat of helium at room temperature are , and (Table A-2). Analysis

(a) The thermal efficiency of this totally reversible cycle is determined from

(b) The amount of heat transferred in the regenerator is

EES (Engineering Equation Solver) SOLUTION "Given" T1=2000 [K] T3=300 [K] P1=3000 [kPa] P2=150 [kPa] m=0.12 [kg] "Properties" Fluid$='Helium' c_p=CP(Fluid$, T=T, P=P) c_v=CV(Fluid$, T=T, P=P) T=25[C]+273 [K] "room temperature" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-73

P=101.3 [kPa] "atmospheric pressure" k=c_p/c_v R=c_p-c_v "Analysis" "(a)" eta_th=1-T3/T1 "(b)" Q_regen=m*c_v*(T1-T4) T4=T3 "(c)" Ratio_v=(T3/T1)*(P1/P3) "Ratio_v = v3/v1" P3=P2 DELTAs=c_v*ln(T2/T1)+R*ln(Ratio_v) T2=T1 Q_in=m*T1*DELTAs W_net_out=eta_th*Q_in

10-80E An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat added to and rejected by this cycle, and the net work produced by the cycle are to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2Ea). Analysis Applying the ideal gas equation to the isothermal process 3-4 gives

Since process 4-1 is a constant volume process,, ( )

(

)

According to first law and work integral,

and (

)

The net work is then

10-74

10-81E An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat transfer in the regenerator is to be determined. Assumptions Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2Ea). Analysis Applying the ideal gas equation to the isothermal process 1-2 gives

(

)

Since process 2-3 is a constant-volume process, ( )

(

)

Application of the first law to process 2-3 gives

10-82 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximum pressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is

(Table A-1).

Analysis

(a) The entropy change during process 3-4 is

and

10-75

(b) For reversible cycles, → Thus, (c) The thermal efficiency of this totally reversible cycle is determined from

EES (Engineering Equation Solver) SOLUTION "Given" T3=(27+273) [K] P3=120 [kPa] q_out=150 [kJ/kg] T1=950 [K] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" "(a)" DELTAs=-q_out/T3 DELTAs=-R*ln(P4/P3) "(b)" q_out/q_in=T3/T1 w_net_out=q_in-q_out "(c)" eta_th=1-T3/T1

10-83E An ideal Stirling engine with hydrogen as the working fluid operates between the specified temperature limits. The amount of external heat addition, external heat rejection, and heat transfer between the working fluid and regenerator per cycle are to be determined. Assumptions Hydrogen is an ideal gas with constant specific heats. Properties The properties of hydrogen at room temperature are , , , and k = 1.404 (Table A-2Ea). Analysis The mass of the air contained in this engine is

At the end of the compression, the pressure will be

10-76

The entropy change is

(

)

Since the processes are reversible,

Applying the first law to the process where the gas passes through the regenerator gives

Ideal and Actual Gas-Turbine (Brayton) Cycles 10-84C They are (1) isentropic compression (in a compressor), (2) P = constant heat addition, (3) isentropic expansion (in a turbine), and (4) P = constant heat rejection.

10-85C For fixed maximum and minimum temperatures, (a) the thermal efficiency increases with pressure ratio, (b) the net work first increases with pressure ratio, reaches a maximum, and then decreases.

10-86C Back work ratio is the ratio of the compressor (or pump) work input to the turbine work output. It is usually between 0.40 and 0.6 for gas turbine engines.

10-87C In gas turbine engines a gas is compressed, and thus the compression work requirements are very large since the steady-flow work is proportional to the specific volume.

10-88C As a result of turbine and compressor inefficiencies, (a) the back work ratio increases, and (b) the thermal efficiency decreases.

10-89E A simple ideal Brayton cycle with air as the working fluid has a pressure ratio of 10. The air temperature at the compressor exit, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E. Analysis (a) Noting that process 1-2 is isentropic,

10-77

→ → (b)

Process 3-4 is isentropic, and thus → →

Then the back-work ratio becomes

(c)

EES (Engineering Equation Solver) SOLUTION "Given" r_p=10 r_p=P[2]/P[1] T[1]=520 [R] T[3]=2000 [R] P[1]=14.7 [psia] "assumed" "Analysis" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic compression" T[2]=temperature(Fluid$, P=P[2], s=s[2]) T2=T[2] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "(b)" h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic expansion" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-78

h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) P[4]=P[1] w_C_in=h[2]-h[1] w_T_out=h[3]-h[4] r_bw=w_C_in/w_T_out "(c)" q_in=h[3]-h[2] w_net_out=w_T_out-w_C_in eta_th=w_net_out/q_in

10-90 A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The power delivered by this plant is to be determined assuming constant and variable specific heats. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas. Analysis

(a)

Assuming constant specific heats, ( ) ( )

̇ ̇

(b)

( )

Assuming variable specific heats (Table A-17), → → → ( )

→

10-79

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=95 [kPa] T[1]=290 [K] P[3]=760 [kPa] T[3]=1100 [K] Q_dot_in=35000 [kJ/s] "Properties" Fluid$='air' C_p=CP(Fluid$, T=T) C_v=CV(Fluid$, T=T) T=25[C]+273 [K] "room temperature" k=C_p/C_v "Analysis" "(a)" T[2]=T[1]*(P[2]/P[1])^((k-1)/k) P[2]=P[3] T[4]=T[3]*(P[4]/P[3])^((k-1)/k) P[4]=P[1] eta_th_a=1-((T[4]-T[1])/(T[3]-T[2])) W_dot_net_out_a=eta_th_a*Q_dot_in "(b)" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic compression" h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) s[4]=s[3] "isentropic expansion" h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) eta_th_b=1-((h[4]-h[1])/(h[3]-h[2])) W_dot_net_out_b=eta_th_b*Q_dot_in

10-91 A simple Brayton cycle with air as the working fluid operates between the specified temperature and pressure limits. The effects of non-isentropic compressor and turbine on the back-work ratio is to be compared. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis For the compression process,

10-80

( ) →

For the expansion process, ( )

(

)

→

The isentropic and actual work of compressor and turbine are

The back work ratio for 90% efficient compressor and isentropic turbine case is

The back work ratio for 90% efficient turbine and isentropic compressor case is

The two results are identical.

EES (Engineering Equation Solver) SOLUTION "Given" r_p=12 "r_p=P2/P1=P3/P4" T3=(600+273) [K] P1=100 [kPa] T1=(15+273) [K] eta_C=0.80 eta_T=0.80 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-81

T2_s=T1*r_p^((k-1)/k) eta_C=(T2_s-T1)/(T2-T1) T4_s=T3*(1/r_p)^((k-1)/k) eta_T=(T3-T4)/(T3-T4_s) w_Comp_s=c_p*(T2_s-T1) w_Comp=c_p*(T2-T1) w_Turb_s=c_p*(T3-T4_s) w_Turb=c_p*(T3-T4) r_bw_1=w_Comp/w_Turb_s r_bw_2=w_Comp_s/w_Turb

10-92 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The mass flow rate of air through the cycle is to be determined. Assumptions

1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis Using variable specific heats, → → → →

and ̇

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net_out=32000 [kW] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-82

T[1]=310 [K] T[3]=900 [K] r_p=8 r_p=P[2]/P[1] eta_C=0.80 eta_T=0.86 P[1]=100 [kPa] "assumed" "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic process" h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic process" h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) P[4]=P[1] w_C_in=(h[2]-h[1])/eta_C w_T_out=eta_T*(h[3]-h[4]) w_net_out=w_T_out-w_C_in m_dot=W_dot_net_out/w_net_out

10-93 A 32-MW gas-turbine power plant operates on a simple Brayton cycle with air as the working fluid. The mass flow rate of air through the cycle is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are

and k = 1.4 (Table A-2).

Analysis Using constant specific heats,

( ) ( )

( )

10-83

and ̇

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net_out=32000 [kW] T[1]=310 [K] T[3]=900 [K] r_p=8 r_p=P[2]/P[1] eta_C=0.80 eta_T=0.86 P[1]=100 [kPa] "assumed" "Properties" Fluid$='air' c_p=CP(Fluid$, T=T) c_v=CV(Fluid$, T=T) T=25[C]+273 [K] "room temperature" k=c_p/c_v "Analysis" "(a)" T[2]=T[1]*r_p^((k-1)/k) "isentropic process" T[4]=T[3]*(1/r_p)^((k-1)/k) "isentropic process" w_C_in=C_p*(T[2]-T[1])/eta_C w_T_out=eta_T*C_p*(T[3]-T[4]) w_net_out=w_T_out-w_C_in m_dot=W_dot_net_out/w_net_out

10-94 A gas turbine power plant that operates on the simple Brayton cycle with air as the working fluid has a specified pressure ratio. The required mass flow rate of air is to be determined for two cases. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are and k = 1.4 (Table A-2).

10-84

(a) Using the isentropic relations, ( ) ( )

(

)

̇ ̇

(b) The net work output is determined to be

EES (Engineering Equation Solver) SOLUTION "Given" r_p=12 "r_p=P2/P1=P3/P4" T1=300 [K] T3=1000 [K] W_dot_net_out=70000 [kW] eta_C=0.85 eta_T=0.85 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "(a)" T2s=T1*r_p^((k-1)/k) T4s=T3*(1/r_p)^((k-1)/k) w_s_C_in=c_p*(T2s-T1) w_s_T_out=c_p*(T3-T4s) w_s_net_out=w_s_T_out-w_s_C_in m_dot_s=W_dot_net_out/w_s_net_out "(b)" w_a_C_in=w_s_C_in/eta_C w_a_T_out=eta_T*w_s_T_out w_a_net_out=w_a_T_out-w_a_C_in m_dot_a=W_dot_net_out/w_a_net_out

10-95 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-85

4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are

and k = 1.4 (Table A-2a).

Analysis For the isentropic compression process, The heat addition is

̇ ̇ Applying the first law to the heat addition process,

The temperature at the exit of the turbine is . /

(

)

Applying the first law to the adiabatic turbine and the compressor produce

The net power produced by the engine is then ̇ ̇ Finally the thermal efficiency is ̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" r_p=10 "r_p=P2/P1=P3/P4" Q_dot_in=500 [kW] m_dot=1 [kg/s] P1=70 [kPa] T1=(0+273) [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T2=T1*r_p^((k-1)/k) q_in=Q_dot_in/m_dot T3=T2+q_in/c_p T4=T3*(1/r_p)^((k-1)/k) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-86

w_T=c_p*(T3-T4) w_C=c_p*(T2-T1) w_net=w_T-w_C W_dot_net=m_dot*w_net eta_th=W_dot_net/Q_dot_in

10-96 An aircraft engine operates as a simple ideal Brayton cycle with air as the working fluid. The pressure ratio and the rate of heat input are given. The net power and the thermal efficiency are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis For the isentropic compression process, The heat addition is

̇ ̇ Applying the first law to the heat addition process,

The temperature at the exit of the turbine is . /

(

)

Applying the first law to the adiabatic turbine and the compressor produce

The net power produced by the engine is then ̇ ̇ Finally the thermal efficiency is ̇ ̇

10-87

EES (Engineering Equation Solver) SOLUTION "Given" r_p=15 "r_p=P2/P1=P3/P4" Q_dot_in=500 [kW] m_dot=1 [kg/s] P1=70 [kPa] T1=(0+273) [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T2=T1*r_p^((k-1)/k) q_in=Q_dot_in/m_dot T3=T2+q_in/c_p T4=T3*(1/r_p)^((k-1)/k) w_T=c_p*(T3-T4) w_C=c_p*(T2-T1) w_net=w_T-w_C W_dot_net=m_dot*w_net eta_th=W_dot_net/Q_dot_in

10-97 A gas-turbine plant operates on the simple Brayton cycle. The net power output, the back work ratio, and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is (Table A-1). Other properties of air are given as , , and k = 1.35.

Analysis (a) For the compression process,

( )

10-88

→ For the expansion process, ( )

(

)

→

The mass flow rate is determined from ̇ ̇ The net power output is ̇ ̇

̇ ̇ ̇

(b) The back work ratio is ̇ ̇ (c) The rate of heat input and the thermal efficiency are ̇ ̇ ̇ ̇

EES (Engineering Equation Solver) SOLUTION "GIVEN" P1=100 [kPa] P2=1600 [kPa] T1=40+273 [C] Vol_dot_1=850[m^3/min]*Convert(m^3/min, m^3/s) T4=650+273 [C] eta_C=0.85 eta_T=0.88 "Properties" c_p=1.108 [kJ/kg-K] c_v=0.821 R=0.287 k=1.35 "Analysis" r_p=P2/P1 T2_s=T1*r_p^((k-1)/k) eta_C=(T2_s-T1)/(T2-T1) T4_s=T3*(1/r_p)^((k-1)/k) eta_T=(T3-T4)/(T3-T4_s) m_dot=(P1*Vol_dot_1)/(R*T1) W_dot_C_in=m_dot*c_p*(T2-T1) W_dot_T_out=m_dot*c_p*(T3-T4) W_dot_net=W_dot_T_out-W_dot_C_in r_bw=W_dot_C_in/W_dot_T_out © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-89

Q_dot_in=m_dot*c_p*(T3-T2) eta_th=W_dot_net/Q_dot_in

10-98E A simple ideal Brayton cycle with argon as the working fluid operates between the specified temperature and pressure limits. The rate of heat addition, the power produced, and the thermal efficiency are to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are (Table A-1E), and k = 1.667 (Table A-2Ea). Analysis At the compressor inlet,

̇ According to the isentropic process expressions for an ideal gas, ( )

(

( )

(

) )

Applying the first law to the constant-pressure heat addition process 2-3 gives ̇ ̇ The net power output is ̇ ̇

The thermal efficiency of this cycle is then ̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=120 [psia] T[3]=2000 [R] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-90

P[4]=15 [psia] T[4]=1200 [R] Q_dot_out=6400 [Btu/s] m_dot=40 [lbm/s] eta_C=0.80 "Analysis" Fluid$='air' r_p=P[3]/P[4] h[3]=enthalpy(Fluid$, T=T[3]) h[4]=enthalpy(Fluid$, T=T[4]) Q_dot_out=m_dot*(h[4]-h[1]) s[1]=entropy(Fluid$, h=h[1], P=P[1]) P[1]=P[4] s[2]=s[1] "isentropic process" h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) P[2]=P[3] W_dot_C_in=m_dot*(h[2]-h[1])/eta_C W_dot_T_out=m_dot*(h[3]-h[4]) W_dot_net_out=W_dot_T_out-W_dot_C_in W_dot_net_out_kW=W_dot_net_out*Convert(Btu/s, kW)

10-99 A modified Brayton cycle with air as the working fluid operates at a specified pressure ratio. The T-s diagram is to be sketched and the temperature and pressure at the exit of the high-pressure turbine and the mass flow rate of air are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air are given as , , , k = 1.4. Analysis (b) For the compression process, ( ) The power input to the compressor is equal to the power output from the high-pressure turbine. Then,

̇ ̇

10-91

The pressure at this state is ( )

→

( )

(

)

(

)

The net power is that generated by the low-pressure turbine since the power output from the high-pressure turbine is equal to the power input to the compressor. Then, ̇ ̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" r_p=8 "r_p=P2/P1=P3/P4" T1=(0+273) [K] P1=100 [kPa] T3=1500 [K] c_v=0.718 [kJ/kg-K] c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.4 W_dot_net=200000 [kW] "Analysis" "(b)" T2=T1*r_p^((k-1)/k) T2-T1=T3-T4 "W_dot_comp=W_dot_turb" P4=r_p*P1*(T4/T3)^(k/(k-1)) "(c)" P5=P1 T5=T4*(P5/P4)^((k-1)/k) W_dot_LP_turb=W_dot_net m_dot=W_dot_LP_turb/(c_p*(T4-T5))

10-100 A simple Brayton cycle with air as the working fluid operates at a specified pressure ratio and between the specified temperature and pressure limits. The cycle is to be sketched on the T-s cycle and the volume flow rate of the air into the compressor is to be determined. Also, the effect of compressor inlet temperature on the mass flow rate and the net power output are to be investigated. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The properties of air are given as , , , k = 1.4. Analysis (b) For the compression process, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-92

( ) ̇ ̇ ̇ ̇ → For the expansion process,

( )

( ) ̇ ̇

̇ ̇

→ Given the net power, the mass flow rate is determined from ̇ ̇ ̇ ̇

̇ ̇

The specific volume and the volume flow rate at the inlet of the compressor are

̇ (c) For a fixed compressor inlet velocity and flow area, when the compressor inlet temperature increases, the specific ̇

. When specific volume increases, the mass flow rate decreases since ̇ . Note that volume flow rate is the same since inlet velocity and flow area are fixed ( ̇ ). When mass flow rate decreases, the net power decreases since ̇ ̇ . Therefore, when the inlet temperature increases, both mass flow rate and the net power decrease. volume increases since

EES (Engineering Equation Solver) SOLUTION "Given" r_p=7 "r_p=P2/P1=P3/P4" T1=(0+273) [K] P1=100 [kPa] T3=1500 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-93

eta_C=0.80 eta_T=0.90 W_dot_net=150000 [kW] c_v=0.718 [kJ/kg-K] c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.4 "Analysis" "(b)" T2_s=T1*r_p^((k-1)/k) eta_C=(T2_s-T1)/(T2-T1) T4_s=T3*(1/r_p)^((k-1)/k) eta_T=(T3-T4)/(T3-T4_s) m_dot=W_dot_net/(c_p*((T3-T4)-(T2-T1))) v1=(R*T1)/P1 Vol_dot_1=m_dot*v1

Brayton Cycle with Regeneration 10-101C Regeneration increases the thermal efficiency of a Brayton cycle by capturing some of the waste heat from the exhaust gases and preheating the air before it enters the combustion chamber.

10-102C The extent to which a regenerator approaches an ideal regenerator is called the effectiveness , and is defined as .

10-103C Yes. At very high compression ratios, the gas temperature at the turbine exit may be lower than the temperature at the compressor exit. Therefore, if these two streams are brought into thermal contact in a regenerator, heat will flow to the exhaust gases instead of from the exhaust gases. As a result, the thermal efficiency will decrease.

10-104C (b) turbine exit.

10-105C The steam injected increases the mass flow rate through the turbine and thus the power output. This, in turn, increases the thermal efficiency since and W increases while remains constant. Steam can be obtained by utilizing the hot exhaust gases.

10-106 An expression for the thermal efficiency of an ideal Brayton cycle with an ideal regenerator is to be developed. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Analysis The expressions for the isentropic compression and expansion processes are

10-94

. / For an ideal regenerator,

The thermal efficiency of the cycle is

10-107 A Brayton cycle with regeneration produces 115 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas,

. /

( )

10-95

while the regenerator temperature specification gives The simultaneous solution of these two results gives Application of the first law to the turbine and compressor gives

Then, ̇

Applying the first law to the combustion chamber produces ̇ ̇ Similarly, ̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(30+273) [K] r_p=8 "r_p=P2/P1" T4=(800+273) [K] DELTAT=10 [K] W_dot_net=115 [kW] eta_C=1 eta_T=1 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T2_s=T1*r_p^((k-1)/k) eta_C=(T2_s-T1)/(T2-T1) T5_s=T4*(1/r_p)^((k-1)/k) eta_T=(T4-T5)/(T4-T5_s) T3=T5-DELTAT T3-T2=T5-T6 "energy balance on heat exchanger" w_net=c_p*(T4-T5)-c_p*(T2-T1) m_dot=W_dot_net/w_net Q_dot_in=m_dot*c_p*(T4-T3) Q_dot_out=m_dot*c_p*(T6-T1)

10-108 A Brayton cycle with regeneration produces 115 kW power. The rates of heat addition and rejection are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-96

3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are

and k = 1.4 (Table A-2a).

Analysis For the compression and expansion processes we have

→

. /

( ) →

When the first law is applied to the heat exchanger, the result is while the regenerator temperature specification gives The simultaneous solution of these two results gives Application of the first law to the turbine and compressor gives

Then, ̇

Applying the first law to the combustion chamber produces ̇ ̇ Similarly, ̇ ̇

10-97

P1=100 [kPa] T1=(30+273) [K] r_p=8 "r_p=P2/P1" T4=(800+273) [K] DELTAT=10 [K] W_dot_net=115 [kW] eta_C=0.87 eta_T=0.93 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T2_s=T1*r_p^((k-1)/k) eta_C=(T2_s-T1)/(T2-T1) T5_s=T4*(1/r_p)^((k-1)/k) eta_T=(T4-T5)/(T4-T5_s) T3=T5-DELTAT T3-T2=T5-T6 "energy balance on heat exchanger" w_net=c_p*(T4-T5)-c_p*(T2-T1) m_dot=W_dot_net/w_net Q_dot_in=m_dot*c_p*(T4-T3) Q_dot_out=m_dot*c_p*(T6-T1)

10-109 A Brayton cycle with regeneration is considered. The thermal efficiencies of the cycle for parallel-flow and counterflow arrangements of the regenerator are to be compared. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis According to the isentropic process expressions for an ideal gas,

. /

( )

When the first law is applied to the heat exchanger as originally arranged, the result is while the regenerator temperature specification gives The simultaneous solution of these two results gives © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-98

The thermal efficiency of the cycle is then

For the rearranged version of this cycle,

An energy balance on the heat exchanger gives The solution of these two equations is

The thermal efficiency of the cycle is then

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=(20+273) [K] r_p=7 "r_p=P2/P1" T4=(727+273) [K] DELTAT=6 [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T2=T1*r_p^((k-1)/k) T5=T4*(1/r_p)^((k-1)/k) T3=T5-DELTAT T3-T2=T5-T6 "energy balance on heat exchanger" eta_th=1-(T6-T1)/(T4-T3) "For the rearranged version of the cycle" T3_a=T6_a-DELTAT T3_a-T2=T5-T6_a "energy balance on heat exchanger" eta_th_a=1-(T6_a-T1)/(T4-T3_a)

10-110E An ideal Brayton cycle with regeneration has a pressure ratio of 11. The thermal efficiency of the cycle is to be determined with and without regenerator cases. Assumptions 1. The air standard assumptions are applicable. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-99

2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are

and k = 1.4 (Table A-2Ea).

Analysis According to the isentropic process expressions for an ideal gas,

. /

(

)

(

)

The regenerator is ideal (i.e., the effectiveness is 100%) and thus,

The thermal efficiency of the cycle is then

The solution without a regenerator is as follows:

. /

EES (Engineering Equation Solver) SOLUTION "Given" P1=16 [psia] T1=(100+460) [R] r_P=11 T4=(1940+460) [R] "Properties at room temperature, Table A-2Ea" c_p=0.240 [Btu/lbm-R] k=1.4 "Analysis" T2=T1*r_P^((k-1)/k) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-100

T3=T4*(1/r_P)^((k-1)/k) T5=T3 T6=T2 eta_th=1-(T6-T1)/(T4-T3) eta_th_noreg=1-(T5-T1)/(T4-T2)

10-111 The thermal efficiency and power output of an actual gas turbine are given. The isentropic efficiency of the turbine and of the compressor, and the thermal efficiency of the gas turbine modified with a regenerator are to be determined. Assumptions 1. Air is an ideal gas with variable specific heats. 2. Kinetic and potential energy changes are negligible. 3. The mass flow rates of air and of the combustion gases are the same, and the properties of combustion gases are the same as those of air. Properties The properties of air are given in Table A-17.

Analysis The properties at various states are → → → → The net work output and the heat input per unit mass are ̇ ̇

(

)

Then the compressor and turbine efficiencies become

10-101

When a regenerator is added, the new heat input and the thermal efficiency become

Discussion Note a 65% efficient regenerator would increase the thermal efficiency of this gas turbine from 35.9% to 44.1%.

EES (Engineering Equation Solver) SOLUTION "Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 30 [C] P[1]= 100 [kPa] W_dot_net=159 [MW] m_dot = 1536000 [kg/h]*Convert(kg/h, kg/s) Eta_th_noreg=0.359 Eta_reg = 0.65 "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_compisen" "Conservation of energy for the compressor for the isentropic case: E_dot_in - E_dot_out = DELTAE_dot=0 for steady-flow" m_dot*h[1] + W_dot_comp_isen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" "SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0 E_dot_in - E_dot_out =DELTAE_dot_cv =0 for steady flow" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2] "process 2-3 is SSSF constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4])"T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turbisen > W_dot_turb" "SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0 E_dot_in -E_dot_out = DELTAE_dot_cv = 0 for steady-flow" m_dot*h[3] = W_dot_turb_isen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-102

h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" "Using the definition of the net cycle work and the convert function from MW to kW:" W_dot_net*convert(MW, kw)=W_dot_turb-W_dot_comp Eta_th_noreg=W_dot_net*convert(MW, kw)/Q_dot_in_noreg"Cycle thermal efficiency" Bwr=W_dot_comp/W_dot_turb"Back work ratio" "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*convert(MW, kw)/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]

10-112 Problem 10-111 is reconsidered. A solution that allows different isentropic efficiencies for the compressor and turbine is to be developed and the effect of the isentropic efficiencies on net work done and the heat supplied to the cycle is to be studied. Also, the T-s diagram for the cycle is to be plotted. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1288 [C] Pratio = 14.7 T[1] = 30 [C] P[1]= 100 [kPa] {T[4]=659 [C]} {W_dot_net=159 [MW] }"We omit the information about the cycle net work" m_dot = 1536000 [kg/h]*Convert(kg/h,kg/s) {Eta_th_noreg=0.359} "We omit the information about the cycle efficiency." Eta_reg = 0.65 Eta_c = 0.84 "Compressor isentropic efficiency" Eta_t = 0.95 "Turbien isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] P[2] = Pratio*P[1] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-103

s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) Eta_c = W_dot_compisen/W_dot_comp m_dot*h[1] + W_dot_compisen = m_dot*h_s[2] h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" m_dot*h[2] + Q_dot_in_noreg = m_dot*h[3] q_in_noreg=Q_dot_in_noreg/m_dot h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2] "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) Eta_t = W_dot_turb /W_dot_turbisen m_dot*h[3] = W_dot_turbisen + m_dot*h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" W_dot_net*1000=W_dot_turb-W_dot_comp Eta_th_noreg=W_dot_net*1000/Q_dot_in_noreg Bwr=W_dot_comp/W_dot_turb "With the regenerator the heat added in the external heat exchanger is" m_dot*h[5] + Q_dot_in_withreg = m_dot*h[3] q_in_withreg=Q_dot_in_withreg/m_dot h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] Eta_reg = (h[5]-h[2])/(h[4]-h[2]) m_dot*h[2] + m_dot*h[4]=m_dot*h[5] + m_dot*h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=W_dot_net*1000/Q_dot_in_withreg "The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-104

s_s[6]=s[6] T_s[6]=T[6]

0.7 0.75 0.8 0.85 0.9 0.95 1

0.84 0.84 0.84 0.84 0.84 0.84 0.84

0.2044 0.2491 0.2939 0.3386 0.3833 0.4281 0.4728

0.27 0.3169 0.3605 0.4011 0.4389 0.4744 0.5076

422152 422152 422152 422152 422152 422152 422152

319582 331856 344129 356403 368676 380950 393223

86.3 105.2 124.1 142.9 161.8 180.7 199.6

T-s Diagram for Gas Turbine with Regeneration 1700 1500

1470 kPa

1300

T [C]

1100 900

100 kPa

700

5 2

500

300 100 -100 4.5

4 4s

1 5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

0.9

0.95

0.9

0.95

s [kJ/kg-K] 200 180

Wnet [kW]

160 140 120 100 80 0.7

0.75

0.8

0.85

t 430000

Qin [kW]

410000

no regeneration

390000 370000

with regeneration

350000 330000 310000 0.7

0.75

0.8

0.85

10-105 0.55 0.5 0.45

with regeneration th

0.4 0.35

no regeneration

0.3 0.25 0.2 0.7

0.75

0.8

0.85

0.9

0.95

10-113E A car is powered by a gas turbine with a pressure ratio of 4. The thermal efficiency of the car and the mass flow rate of air for a net power output of 130 hp are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air is an ideal gas with variable specific heats. 3. The ambient air is 540 R and 14.5 psia. 4. The effectiveness of the regenerator is 0.9, and the isentropic efficiencies for both the compressor and the turbine are 80%. 5. The combustion gases can be treated as air. Properties The properties of air at the compressor and turbine inlet temperatures can be obtained from Table A-17E. Analysis The gas turbine cycle with regeneration can be analyzed as follows: → → → → and

Then the thermal efficiency of the gas turbine cycle becomes

10-106

Finally, the mass flow rate of air through the turbine becomes ̇ ̇

(

)

EES (Engineering Equation Solver) SOLUTION "Given" r_p=4 r_p=P[2]/P[1] T[1]=510 [R] P[1]=14.5 [psia] T[3]=(1700+460) [R] epsilon=0.90 eta_C=0.80 eta_T=0.80 W_dot_net_out=130 [hp] "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic process" h_s[2]=enthalpy(Fluid$, P=P[2], s=s[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic process" h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) P[4]=P[1] eta_C=(h_s[2]-h[1])/(h[2]-h[1]) eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) q_in=(h[3]-h[2])-q_regen w_C_in=h[2]-h[1] w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in eta_th=w_net_out/q_in m_dot=W_dot_net_out/w_net_out*Convert(hp, Btu/s)

10-114 A Brayton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions

10-107

1. The air standard assumptions are applicable. 2. Air is an ideal gas with variable specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis (a) The properties of air at various states are → → → → → → Thus, (b)

(c)

→

Then,

10-108

r_p=7 r_p=P[2]/P[1] T[1]=310 [K] T[3]=1150 [K] eta_C=0.75 eta_T=0.82 epsilon=0.65 P[1]=100 [kPa] "assumed" "Analysis" "(a)" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic process" h_s[2]=enthalpy(Fluid$, P=P[2], s=s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic process" h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) T[4]=temperature(Fluid$, h=h[4]) T4=T[4] "(b)" w_C_in=h[2]-h[1] w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in "(c)" epsilon=(h[5]-h[2])/(h[4]-h[2]) q_in=h[3]-h[5] eta_th=w_net_out/q_in

10-115 A stationary gas-turbine power plant operating on an ideal regenerative Brayton cycle with air as the working fluid is considered. The power delivered by this plant is to be determined for two cases. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas. 3. Kinetic and potential energy changes are negligible. Properties When assuming constant specific heats, the properties of air at room temperature are and k = 1.4 (Table A-2a). When assuming variable specific heats, the properties of air are obtained from Table A-17.

10-109

(a) Assuming constant specific heats, ( ) ( )

(

)

→

̇ ̇

(b) Assuming variable specific heats, → → → (

)

→

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=95 [kPa] T[1]=290 [K] P[3]=880 [kPa] T[3]=1100 [K] Q_dot_in=30000 [kJ/s] epsilon=1 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "(a)" T[2]=T[1]*(P[2]/P[1])^((k-1)/k) P[2]=P[3] T[4]=T[3]*(P[4]/P[3])^((k-1)/k) P[4]=P[1] q_regen_a=epsilon*c_p*(T[4]-T[2]) q_in_a=c_p*(T[3]-T[2])-q_regen_a q_out_a=c_p*(T[4]-T[1])-q_regen_a eta_th_a=1-q_out_a/q_in_a W_dot_net_out_a=eta_th_a*Q_dot_in "(b)" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-110

s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic compression" h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) s[4]=s[3] "isentropic expansion" h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) q_regen_b=epsilon*(h[4]-h[2]) q_in_b=(h[3]-h[2])-q_regen_b q_out_b=(h[4]-h[1])-q_regen_b eta_th_b=1-q_out_b/q_in_b W_dot_net_out_b=eta_th_b*Q_dot_in

10-116 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions

→ →

(b)

10-111

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=310 [K] P[1]=100 [kPa] P[2]=900 [kPa] T[2]=650 [K] T[3]=1400 [K] eta_T=0.90 epsilon=0.80 "Analysis" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) h[2]=enthalpy(Fluid$, T=T[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) "(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in

10-117 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a).

10-112

( )

( ) →

(b)

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=310 [K] P[1]=100 [kPa] P[2]=900 [kPa] T[2]=650 [K] T[3]=1400 [K] eta_T=0.90 epsilon=0.80 "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "(a)" r_P=P[2]/P[1] T_s[4]=T[3]*(1/r_P)^((k-1)/k) eta_T=(T[3]-T[4])/(T[3]-T_s[4]) q_regen=epsilon*c_p*(T[4]-T[2]) "(b)" w_T_out=c_p*(T[3]-T[4]) w_C_in=c_p*(T[2]-T[1]) w_net_out=w_T_out-w_C_in q_in=c_p*(T[3]-T[2])-q_regen eta_th=w_net_out/q_in

10-118 A regenerative gas-turbine engine using air as the working fluid is considered. The amount of heat transfer in the regenerator and the thermal efficiency are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with variable specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-113

Analysis

(a) The properties at various states are → → →

→ →

(b)

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=310 [K] P[1]=100 [kPa] P[2]=900 [kPa] T[2]=650 [K] T[3]=1400 [K] eta_T=0.90 epsilon=0.70 "Analysis" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) h[2]=enthalpy(Fluid$, T=T[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-114

P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) "(b)" w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in

Brayton Cycle with Intercooling, Reheating, and Regeneration 10-119C Because the steady-flow work is proportional to the specific volume of the gas. Intercooling decreases the average specific volume of the gas during compression, and thus the compressor work. Reheating increases the average specific volume of the gas, and thus the turbine work output.

10-120C (c) The Carnot (or Ericsson) cycle efficiency.

10-121C As the number of compression and expansion stages are increased and regeneration is employed, the ideal Brayton cycle will approach the Ericsson cycle.

10-122C The single-stage compression process of an ideal Brayton cycle without regeneration is replaced by a multistage compression process with intercooling between the same pressure limits. As a result of this modification, (a) decrease, (b) decrease, and (c) decrease.

10-123C The single-stage expansion process of an ideal Brayton cycle without regeneration is replaced by a multistage expansion process with reheating between the same pressure limits. As a result of this modification, (a) increase, (b) decrease, and (c) decrease.

10-124C A simple ideal Brayton cycle without regeneration is modified to incorporate multistage compression with intercooling and multistage expansion with reheating, without changing the pressure or temperature limits of the cycle. As a result of these two modifications, (a) increase, (b) decrease, (c) decrease, and (d) increase.

10-115

10-125C A simple ideal Brayton cycle is modified to incorporate multistage compression with intercooling, multistage expansion with reheating, and regeneration without changing the pressure limits of the cycle. As a result of these modifications, (a) increase, (b) decrease, (c) increase, and (d) decrease.

10-126 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with variable specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is Then the work inputs to each stage of compressor are identical, so are the √ work outputs of each stage of the turbine. → → → →

10-116

T[1]=300 [K] T[3]=300 [K] T[5]=1200 [K] T[7]=1200 [K] W_dot_net_out=110000 [kW] r_p=sqrt(9) r_p=P[2]/P[1] r_p=P[4]/P[3] r_p=P[5]/P[6] r_p=P[7]/P[8] P[1]=100 [kPa] "assumed" P[3]=P[2] P[5]=P[4] P[7]=P[6] "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic process" h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) h[5]=enthalpy(Fluid$, T=T[5]) s[5]=entropy(Fluid$, T=T[5], P=P[5]) s[6]=s[5] "isentropic process" h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) w_C_in=2*(h[2]-h[1]) w_T_out=2*(h[5]-h[6]) w_net_out=w_T_out-w_C_in m_dot=W_dot_net_out/w_net_out

10-127 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1. Argon is an ideal gas with constant specific heats. 2. Kinetic and potential energy changes are negligible. Properties The properties of argon at room temperature are and k = 1.667 (Table A-2a). Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is . Then the work inputs to each stage of compressor are identical, so are the √ work outputs of each stage of the turbine.

10-117

( ) ( )

( )

wC, in = 2 (h2 - h1 ) = 2c p (T2 - T1 ) = 2 (0.5203 kJ/kg ×K )(465.6 - 300)K = 172.3 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" r_p=sqrt(9) "r_p = P2/P1 = P4/P3 = P5/P6 = P7/P8" T[1]=300 [K] T[3]=300 [K] T[5]=1200 [K] T[7]=1200 [K] W_dot_net_out=110000 [kW] "Properties" Fluid$='Argon' c_p=CP(Fluid$, T=T, P=P) c_v=CV(Fluid$, T=T, P=P) T=25+273 "[K], room temperature" P=101.3 "[kPa], atmospheric pressure" k=c_p/c_v "Analysis" T[2]=T[1]*r_p^((k-1)/k) T[6]=T[5]*(1/r_p)^((k-1)/k) w_C_in=2*c_p*(T[2]-T[1]) w_T_out=2*c_p*(T[5]-T[6]) w_net_out=w_T_out-w_C_in m_dot=W_dot_net_out/w_net_out

10-128 An ideal gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with variable specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. Analysis

10-118

(a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine since this is an ideal cycle. Then, → → → →

Thus,

(b) When a regenerator is used,

remains the same. The thermal efficiency in this case becomes

EES (Engineering Equation Solver) SOLUTION "Given" r_p=3 "r_p = P2/P1 = P4/P3 = P5/P6 = P7/P8" T[1]=300 [K] T[3]=300 [K] T[5]=1200 [K] T[7]=1200 [K] epsilon=0.75 "Analysis" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) P[1]=100 [kPa] "assumed" s[1]=entropy(Fluid$, T=T[1], P=P[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-119

s[2]=s[1] "isentropic compression" P[2]=r_p*P[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) h[3]=h[1] h[4]=h[2] h[5]=enthalpy(Fluid$, T=T[5]) P[5]=r_p*P[2] s[5]=entropy(Fluid$, T=T[5], P=P[5]) s[6]=s[5] "isentropic expansion" P[6]=P[2] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) h[7]=h[5] h[8]=h[6] w_C_in=2*(h[2]-h[1]) w_T_out=2*(h[5]-h[6]) r_bw=w_C_in/w_T_out q_in_a=(h[5]-h[4]+h[7]-h[6]) w_net_out=w_T_out-w_C_in eta_th_a=w_net_out/q_in_a "(b)" "When a regenerator is used" q_regen=epsilon*(h[8]-h[4]) q_in_b=q_in_a-q_regen eta_th_b=w_net_out/q_in_b

10-129 A gas-turbine cycle with two stages of compression and two stages of expansion is considered. The back work ratio and the thermal efficiency of the cycle are to be determined for the cases of with and without a regenerator. Assumptions

1. 2. 3.

The air standard assumptions are applicable. Air is an ideal gas with variable specific heats. Kinetic and potential energy changes are negligible.

Properties The properties of air are given in Table A17. Analysis (a) The work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. Then, → → →

10-120

→ → →

Thus,

(b) When a regenerator is used,

remains the same. The thermal efficiency in this case becomes

EES (Engineering Equation Solver) SOLUTION "Given" r_p=3 "r_p = P2/P1 = P4/P3 = P5/P6 = P7/P8" T[1]=300 [K] T[3]=300 [K] T[5]=1200 [K] T[7]=1200 [K] epsilon=0.75 eta_C=0.86 eta_T=0.90 "Analysis" Fluid$='air' "(a)" h[1]=enthalpy(Fluid$, T=T[1]) P[1]=100 [kPa] "assumed" s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" P[2]=r_p*P[1] h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=h[1] h[4]=h[2] h[5]=enthalpy(Fluid$, T=T[5]) P[5]=r_p*P[2] s[5]=entropy(Fluid$, T=T[5], P=P[5]) s_s[6]=s[5] "isentropic expansion" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-121

P[6]=P[2] h_s[6]=enthalpy(Fluid$, P=P[6], s=s_s[6]) eta_T=(h[5]-h[6])/(h[5]-h_s[6]) h[7]=h[5] h[8]=h[6] w_C_in=2*(h[2]-h[1]) w_T_out=2*(h[5]-h[6]) r_bw=w_C_in/w_T_out q_in_a=(h[5]-h[4]+h[7]-h[6]) w_net_out=w_T_out-w_C_in eta_th_a=w_net_out/q_in_a "(b)" "When a regenerator is used" q_regen=epsilon*(h[8]-h[4]) q_in_b=q_in_a-q_regen eta_th_b=w_net_out/q_in_b

10-130E A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The mass flow rate of air and the rates of heat addition and rejection for a specified net power output are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2Ea).

Analysis According to the isentropic process expressions for an ideal gas,

. /

( )

The regenerator is ideal (i.e., the effectiveness is 100%) and thus,

The net work output is determined as follows

10-122

The mass flow rate is then ̇

(

Applying the first law to the heat addition processes gives ̇ ̇

)

Similarly, ̇ ̇

10-131E A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The mass flow rate of air and the rates of heat addition and rejection for a specified net power output are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2Ea).

Analysis For the compression and expansion processes, we have

→

. /

( ) →

The regenerator is ideal (i.e., the effectiveness is 100%) and thus,

10-123

The net work output is determined as follows

The mass flow rate is then ̇

(

)

The rate of heat addition is then ̇ ̇

Jet-Propulsion Cycles 10-132C The power developed from the thrust of the engine is called the propulsive power. It is equal to thrust times the aircraft velocity.

10-133C The ratio of the propulsive power developed and the rate of heat input is called the propulsive efficiency. It is determined by calculating these two quantities separately, and taking their ratio.

10-134C It reduces the exit velocity, and thus the thrust.

10-135 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with variable specific heats. 4. Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of Taking the entire engine as our control volume and writing the steady-flow energy balance yield

10-124

̇ ̇

̇ ̇ ̇ 0

̇ ̇

̇ .

/ (

It gives Thus, ̇

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=(7+273) [K] Vel1=220 [m/s] m_dot=16 [kg/s] T[2]=(427+273) [K] Q_dot_in=15000 [kJ/s] "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) h[2]=enthalpy(Fluid$, T=T[2]) Q_dot_in=m_dot*(h[2]-h[1]+(Vel2^2/2-Vel1^2/2)*Convert(m^2/s^2, kJ/kg)) "energy balance" F_p=m_dot*(Vel2-Vel1)

10-136E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with constant specific heats at room temperature. 4. Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5. The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are and k = 1.4 (Table A-2Ea). Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of . Ideally, the air will leave the diffuser with a negligible velocity ( ). Diffuser:

10-125

̇ ̇

( ( )

(

)

Compressor: ( ) ( ) Turbine: →

→

( )

(

)

( )

(

)

(b) Nozzle:

̇ ̇

or √

10-126

(

)

EES (Engineering Equation Solver) SOLUTION "Given" Vel1=900 [ft/s] H=20000 [ft] P[1]=7 [psia] T[1]=(10+460) [R] T[4]=2400 [R] r_p=13 "Properties" Fluid$='air' c_p=CP(Fluid$, T=T) c_v=CV(Fluid$, T=T) T=77+460 "[R], room temperature" k=c_p/c_v "Analysis" "(a)" "Diffuser" c_p*(T[2]-T[1])+(Vel2^2/2-Vel1^2/2)*1/Convert(Btu/lbm, ft^2/s^2)=0 "energy balance" Vel2=0 "ideal diffuser operation" P[2]=P[1]*(T[2]/T[1])^(k/(k-1)) "Compressor" P[3]=r_p*P[2] P[4]=P[3] T[3]=T[2]*(P[3]/P[2])^((k-1)/k) "Turbine" w_C_in=c_p*(T[3]-T[2]) w_T_out=c_p*(T[4]-T[5]) w_C_in=w_T_out P[5]=P[4]*(T[5]/T[4])^(k/(k-1)) P5=P[5] "(b)" "Nozzle" T[6]=T[5]*(P[6]/P[5])^((k-1)/k) P[6]=P[1] c_p*(T[6]-T[5])+(Vel6^2/2-Vel5^2/2)*1/Convert(Btu/lbm, ft^2/s^2)=0 "energy balance" Vel5=0 "ideal nozzle operation" "(c)" w_p=(Vel6-Vel1)*Vel1*1/Convert(Btu/lbm, ft^2/s^2) q_in=c_p*(T[4]-T[3]) eta_p=w_p/q_in

10-137E A turbojet engine operating on an ideal cycle is flying at an altitude of 20,000 ft. The pressure at the turbine exit, the velocity of the exhaust gases, and the propulsive efficiency are to be determined. Assumptions 1. Steady operating conditions exist. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-127

2. 3. 4.

The air standard assumptions are applicable. Air is an ideal gas with variable specific heats. Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air are given in Table A-17E. Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of . Ideally, the air will leave the diffuser with a negligible velocity ( ).

Diffuser: → ̇

̇ ̇

(

)

→

(

)

Compressor: ( ) (

)

→

Turbine: →

or →

(

)

10-128

(b) Nozzle: ( ̇

)

→

̇ ̇

or √

√

(

)

EES (Engineering Equation Solver) SOLUTION "Given" Vel1=900 [ft/s] H=20000 [ft] P[1]=7 [psia] T[1]=(10+460) [R] T[4]=2400 [R] r_p=13 "Analysis" Fluid$='air' "(a)" "Diffuser" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) h[2]-h[1]+(Vel2^2/2-Vel1^2/2)*1/Convert(Btu/lbm, ft^2/s^2)=0 "energy balance" Vel2=0 "ideal diffuser operation" s[2]=s[1] P[2]=pressure(Fluid$, h=h[2], s=s[2]) "Compressor" P[3]=r_p*P[2] P[4]=P[3] s[3]=s[2] h[3]=enthalpy(Fluid$, P=P[3], s=s[3]) "Turbine" h[4]=enthalpy(Fluid$, T=T[4]) s[4]=entropy(Fluid$, T=T[4], P=P[4]) w_C_in=h[3]-h[2] w_T_out=h[4]-h[5] w_C_in=w_T_out © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-129

s[5]=s[4] P[5]=pressure(Fluid$, h=h[5], s=s[5]) P5=P[5] "(b)" "Nozzle" s[6]=s[5] P[6]=P[1] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) h[6]-h[5]+(Vel6^2/2-Vel5^2/2)*1/Convert(Btu/lbm, ft^2/s^2)=0 "energy balance" Vel5=0 "ideal nozzle operation" "(c)" w_p=(Vel6-Vel1)*Vel1*1/Convert(Btu/lbm, ft^2/s^2) q_in=h[4]-h[3] eta_p=w_p/q_in

10-138 A pure jet engine operating on an ideal cycle is considered. The velocity at the nozzle exit and the thrust produced are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with constant specific heats at room temperature. 4. The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are , and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of the air will leave the diffuser with a negligible velocity ( ).

. Ideally,

Diffuser: ̇

→

( (

)

(

)

10-130

→

or Nozzle: ( )

( ̇

)

→

→ or

(

√

)

The mass flow rate through the engine is

̇ The thrust force generated is then ̇

(

)

EES (Engineering Equation Solver) SOLUTION "Given" Vel1=240 [m/s] P1=45 [kPa] T1=(-13+273) [K] D1=1.6 [m] r_p=13 T4=(557+273) [K] "Properties from Table A-2a" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" T2=T1+Vel1^2/(2*c_p*Convert(kJ/kg, m^2/s^2)) P2=P1*(T2/T1)^(k/(k-1)) P3=r_P*P2 P4=P3 P6=P1 T3=T2*(P3/P2)^((k-1)/k) T5=T4-T3+T2 T6=T4*(P6/P4)^((k-1)/k) Vel6^2=2*c_p*(T5-T6)*Convert(kJ/kg, m^2/s^2) v1=(R*T1)/P1 m_dot=pi*D1^2/4*Vel1/v1 F=m_dot*(Vel6-Vel1)

10-139 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-131

3. 4. 5.

Air is an ideal gas with constant specific heats at room temperature. Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. The turbine work output is equal to the compressor work input.

Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis

(a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of the air will leave the diffuser with a negligible velocity ( ).

. Ideally,

Diffuser: ̇

→

( ( )

(

)

Compressor: ( ) ( ) Turbine: →

→

or Nozzle: ( )

( ̇

)

→

(b) (c)

(

√

)

(

)

̇ ̇

10-132

EES (Engineering Equation Solver) SOLUTION "Given" Vel1=280 [m/s] H=9150 [m] P[1]=32 [kPa] T[1]=(-32+273) [K] T[4]=1100 [K] r_p=12 m_dot=50 [kg/s] HV=42700 [kJ/kg] "Properties, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "(a)" "Diffuser" c_p*(T[2]-T[1])+(Vel2^2/2-Vel1^2/2)*Convert(m^2/s^2, kJ/kg)=0 "energy balance" Vel2=0 "ideal diffuser operation" P[2]=P[1]*(T[2]/T[1])^(k/(k-1)) "Compressor" P[3]=r_p*P[2] P[4]=P[3] T[3]=T[2]*(P[3]/P[2])^((k-1)/k) "Turbine" w_C_in=c_p*(T[3]-T[2]) w_T_out=c_p*(T[4]-T[5]) w_C_in=w_T_out "Nozzle" T[6]=T[4]*(P[6]/P[4])^((k-1)/k) P[6]=P[1] c_p*(T[6]-T[5])+(Vel6^2/2-Vel5^2/2)*Convert(m^2/s^2, kJ/kg)=0 "energy balance" Vel5=0 "ideal nozzle operation" "(b)" W_dot_p=m_dot*(Vel6-Vel1)*Vel1*Convert(m^2/s^2, kJ/kg) "(c)" Q_dot_in=m_dot*c_p*(T[4]-T[3]) m_dot_fuel=Q_dot_in/HV

10-140 A turbojet aircraft is flying at an altitude of 9150 m. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1. 2. 3. 4.

Steady operating conditions exist. The air standard assumptions are applicable. Air is an ideal gas with constant specific heats at room temperature. Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit.

Properties The properties of air at room temperature are

and k = 1.4 (Table A-2a).

Analysis (a) For convenience, we assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of . Ideally, the air will leave the diffuser with a negligible velocity ( ). Diffuser: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-133

̇ ̇

( ( )

(

)

Compressor: ( ) ( )

Turbine: or,

→

(

)

(

)

Nozzle: ( ) ̇

( ̇

)

̇ ̇

10-134

or, √ ̇

(b)

̇ (

(c)

̇ ̇

)

̇ ̇

EES (Engineering Equation Solver) SOLUTION "Given" Vel1=280 [m/s] H=9150 [m] P[1]=32 [kPa] T[1]=(-32+273) [K] T[4]=1100 [K] r_p=12 m_dot=50 [kg/s] HV=42700 [kJ/kg] eta_C=0.80 eta_T=0.85 "Properties, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "(a)" "Diffuser" c_p*(T[2]-T[1])+(Vel2^2/2-Vel1^2/2)*Convert(m^2/s^2, kJ/kg)=0 "energy balance" Vel2=0 "ideal diffuser operation" P[2]=P[1]*(T[2]/T[1])^(k/(k-1)) "Compressor" P[3]=r_p*P[2] P[4]=P[3] T_s[3]=T[2]*(P[3]/P[2])^((k-1)/k) eta_C=(T_s[3]-T[2])/(T[3]-T[2]) "Turbine" w_C_in=c_p*(T[3]-T[2]) w_T_out=c_p*(T[4]-T[5]) w_C_in=w_T_out eta_T=(T[4]-T[5])/(T[4]-T_s[5]) P[5]=P[4]*(T_s[5]/T[4])^(k/(k-1)) "Nozzle" T[6]=T[5]*(P[6]/P[5])^((k-1)/k) P[6]=P[1] c_p*(T[6]-T[5])+(Vel6^2/2-Vel5^2/2)*Convert(m^2/s^2, kJ/kg)=0 "energy balance" Vel5=0 "ideal nozzle operation" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-135

"(b)" W_dot_p=m_dot*(Vel6-Vel1)*Vel1*Convert(m^2/s^2, kJ/kg) "(c)" Q_dot_in=m_dot*c_p*(T[4]-T[3]) m_dot_fuel=Q_dot_in/HV

10-141E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with constant specific heats at room temperature. 4. The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are (Table A-1E), and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields

. /

(

)

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, ̇

which when solved for the velocity at which the air leaves the propeller gives 0 0

̇ .

The mass flow rate through the propeller is

10-136

The thrust force generated by this propeller is then ̇

(

)

EES (Engineering Equation Solver) SOLUTION "Given" P1=8 [psia] T1=(-10+460) [R] Vel_inlet=600 [ft/s] r_p=10 T3=(940+460) [R] D=10 [ft] MassRatio=20 "MassRatio = m_dot_p/m_dot_e" "Properties at room temperature, Table A-2Ea" R=0.3704 [psia-ft^3/lbm-R] c_p=0.240 [Btu/lbm-R] k=1.4 "Analysis" T2=T1*r_p^((k-1)/k) T5=T3*(1/r_p)^((k-1)/k) T4=T3-(T2-T1) Vel_exit=sqrt(2*1/MassRatio*c_p*(T4-T5)*Convert(Btu/lbm, ft^2/s^2)+Vel_inlet^2) v1=(R*T1)/P1 A=pi*D^2/4 m_dot_p=(A*Vel_inlet)/v1 F=m_dot_p*(Vel_exit-Vel_inlet)*Convert(lbm-ft/s^2, lbf)

10-142E A turboprop engine operating on an ideal cycle is considered. The thrust force generated is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with constant specific heats at room temperature. 4. The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are (Table A-1E), and k = 1.4 (Table A-2Ea). Analysis Working across the two isentropic processes of the cycle yields

. /

(

)

Since the work produced by expansion 3-4 equals that used by compression 1-2, an energy balance gives

10-137

The mass flow rate through the propeller is

̇ According to the previous problem, ̇ ̇

The excess enthalpy generated by expansion 4-5 is used to increase the kinetic energy of the flow through the propeller, ̇

which when solved for the velocity at which the air leaves the propeller gives 0

The thrust force generated by this propeller is then ̇

(

)

EES (Engineering Equation Solver) SOLUTION "Given" P1=8 [psia] T1=(-10+460) [R] Vel_inlet=600 [ft/s] r_p=10 T3=(940+460) [R] D=8 [ft] m_dot_e=113.1 [lbm/s] "from the previous problem" "Properties at room temperature, Table A-2Ea" R=0.3704 [psia-ft^3/lbm-R] c_p=0.240 [Btu/lbm-R] k=1.4 "Analysis" T2=T1*r_p^((k-1)/k) T5=T3*(1/r_p)^((k-1)/k) T4=T3-(T2-T1) v1=(R*T1)/P1 A=pi*D^2/4 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-138

m_dot_p=(A*Vel_inlet)/v1 Vel_exit=sqrt(2*m_dot_e/m_dot_p*c_p*(T4-T5)*Convert(Btu/lbm, ft^2/s^2)+Vel_inlet^2) F=m_dot_p*(Vel_exit-Vel_inlet)*Convert(lbm-ft/s^2, lbf)

10-143 A turbojet aircraft that has a pressure rate of 9 is stationary on the ground. The force that must be applied on the brakes to hold the plane stationary is to be determined. Assumptions

1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with variable specific heats. 4. Kinetic and potential energies are negligible, except at the nozzle exit. Properties The properties of air are given in Table A-17. Analysis (a) Using variable specific heats for air, Compressor: → → ̇ ̇ ̇ ̇ → Turbine: or

→ →

Nozzle: → ̇

̇ ̇

10-139

√

Brake force = Thrust = ̇

. (

/ )

EES (Engineering Equation Solver) SOLUTION "Given" r_p=9 r_p=P[2]/P[1] T[1]=(7+273) [K] P[1]=95 [kPa] m_dot=20 [kg/s] m_dot_fuel=0.5 [kg/s] HV=42700 [kJ/kg] "Analysis" Fluid$='air' "Compressor" h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) Q_dot_in=m_dot_fuel*HV q_in=Q_dot_in/m_dot q_in=h[3]-h[2] "Turbine" w_C_in=h[2]-h[1] w_T_out=h[3]-h[4] w_C_in=w_T_out s[3]=entropy(Fluid$, h=h[3], P=P[3]) P[3]=P[2] "Nozzle" s[5]=s[3] h[5]=enthalpy(Fluid$, P=P[5], s=s[5]) P[5]=P[1] h[5]-h[4]+(Vel5^2/2-Vel4^2/2)*Convert(m^2/s^2, kJ/kg)=0 "energy balance" Vel4=0 "ideal nozzle operation" Thrust=m_dot*(Vel5-Vel1) Vel1=0 F_brake=Thrust

10-144 Problem 10-143 is reconsidered. The effect of compressor inlet temperature on the force that must be applied to the brakes to hold the plane stationary is to be investigated. Analysis Using EES, the problem is solved as follows: P_ratio =9 T_1 = 7 [C] T[1] = (T_1+273) P[1]= 95 [kPa] P[5]=P[1] Vel[1]=0 [m/s] V_dot[1] = 18.1 [m^3/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-140

HV_fuel = 42700 [kJ/kg] m_dot_fuel = 0.5 [kg/s] Eta_c = 1.0 Eta_t = 1.0 Eta_N = 1.0 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) v[1]=volume(Air,T=T[1],P=P[1]) m_dot = V_dot[1]/v[1] "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1] "Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2]"process 2-3 is constant pressure" h[3]=ENTHALPY(Air,T=T[3]) Q_dot_in = m_dot_fuel*HV_fuel m_dot*h[2] + Q_dot_in= m_dot*h[3] "First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" {P_ratio= P[3] /P[4]} T_s[4]=TEMPERATURE(Air,h=h_s[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" {h_s[4]=ENTHALPY(Air,T=T_s[4])} "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "First Law for the actual turbine, assuming: adiabatic, ke=pe=0" T[4]=TEMPERATURE(Air,h=h[4]) P[4]=pressure(Air,s=s_s[4],h=h_s[4]) "Cycle analysis" W_dot_net=W_dot_t-W_dot_c W_dot_net = 0 [kW] "Exit nozzle analysis:" s[4]=entropy('air',T=T[4],P=P[4]) s_s[5]=s[4] "For the ideal case the entropies are constant across the nozzle" T_s[5]=TEMPERATURE(Air,s=s_s[5], P=P[5]) "T_s[5] is the isentropic value of T[5] at nozzle exit" h_s[5]=ENTHALPY(Air,T=T_s[5]) Eta_N=(h[4]-h[5])/(h[4]-h_s[5]) m_dot*h[4] = m_dot*(h_s[5] + Vel_s[5]^2/2*convert(m^2/s^2,kJ/kg)) m_dot*h[4] = m_dot*(h[5] + Vel[5]^2/2*convert(m^2/s^2,kJ/kg)) T[5]=TEMPERATURE(Air,h=h[5]) s[5]=entropy('air',T=T[5],P=P[5]) "Brake Force to hold the aircraft:" Thrust = m_dot*(Vel[5] - Vel[1]) BrakeForce = Thrust "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) s[2]=entropy('air',T=T[2],P=P[2]) Brake Force [N] 21232

23.68

1284

−20

10-141

21007 20788 20576 20369 20168 19972 19782 19596 19415 19238

23.22 22.78 22.35 21.94 21.55 21.17 20.8 20.45 20.1 19.77

1307 1330 1352 1375 1398 1420 1443 1466 1488 1510

−15 −10 −5 0 5 10 15 20 25 30

10-145 A turbofan engine operating on an ideal cycle produces 50,000 N of thrust. The air temperature at the fan outlet needed to produce this thrust is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with constant specific heats at room temperature. 4. The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are , and k = 1.4 (Table A-2a). Analysis The total mass flow rate is

10-142

Now, ̇ ̇ The mass flow rate through the fan is ̇

In order to produce the specified thrust force, the velocity at the fan exit will be ̇ ̇

An energy balance on the stream passing through the fan gives

(

)

EES (Engineering Equation Solver) SOLUTION "Given" Vel_inlet=200 [m/s] P1=50 [kPa] T1=(-20+273) [K] F=50000 [N] D=2.5 [m] r_p=12 MassRatio=8 "m_dot/m_dot_e = 20" "Properties at room temperature, Table A-2Ea" R=0.287 [kPa-m^3/kg-K] c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" v1=(R*T1)/P1 A=pi*D^2/4 m_dot=(A*Vel_inlet)/v1 m_dot_e=m_dot/MassRatio m_dot_f=m_dot-m_dot_e Vel_exit=Vel_inlet+F/m_dot_f T4=T1 T5=T4-(Vel_exit^2-Vel_inlet^2)/(2*c_p)*Convert(m^2/s^2, kJ/kg) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-143

Second-Law Analysis of Gas Power Cycles 10-146 The total exergy destruction associated with an Otto cycle with air as the working fluid and the exergy at the end of the power stroke are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is . The properties of air are given in Table A-17. Analysis Process 1-2: isentropic compression.

→

→ → →

(

)

Process 2-3: v = constant heat addition. → → →

(

)

Process 3-4: isentropic expansion. → Process 4-1: v = constant heat rejection. The total exergy destruction associated with this Otto cycle is determined from (

)

(

)

Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from where

10-144

Thus,

EES (Engineering Equation Solver) SOLUTION "Given" r=8 "r=v1/v2=v4/v3" P1=95 [kPa] T1=(27+273) [K] q_in=750 [kJ/kg] T_H=2000 [K] T0=300 [K] "Properties" Fluid$='Air' R_air=R_u/MM MM=Molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] "Analysis" "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" P2=r*(T2/T1)*P1 "from ideal gas relation" u2=intenergy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant volume heat addition" q_in=u3-u2 T3=temperature(Fluid$, u=u3) s3=entropy(Fluid$, u=u3, P=P3) P3=(T3/T2)*P2 "from ideal gas relation" "Process 3-4: isentropic expansion" s4=s3 "isentropic process" P4=1/r*(T4/T3)*P3 "from ideal gas relation" u4=intenergy(Fluid$, P=P4, s=s4) T4=temperature(Fluid$, P=P4, s=s4) "Process 4-1: constant volume heat rejection" q_out=u4-u1 w_net_out=q_in-q_out eta_th=w_net_out/q_in v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2) x_destroyed=T0*(q_out/T0-q_in/T_H) x4=DELTAu-T0*DELTAs+P0*DELTAv DELTAu=q_out "since u0=u1" DELTAs=s4-s1 DELTAv=0 "since v4=v1" P0=P1

10-147 The total exergy destruction associated with a Diesel cycle and the exergy at the end of the compression stroke are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-145

Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is . The properties of air are given in Table A-17. Analysis

(a) Process 1-2: isentropic compression. →

→ Process 2-3: P = constant heat addition. →

→

Process 3-4: isentropic expansion.

→ Process 4-1: v = constant heat rejection.

The total exergy destruction associated with this Otto cycle is determined from (

)

(

)

Noting that state 1 is identical to the state of the surroundings, the exergy at the end of the compression stroke (state 2) is determined from

(

)

10-146

EES (Engineering Equation Solver) SOLUTION "Given" r=16 "r=v1/v2" r_c=2 "r_c=v3/v2" P1=95 [kPa] T1=(27+273) [K] T_H=2000 [K] T0=300 [K] "Properties" Fluid$='Air' R_air=R_u/MM MM=Molarmass(Fluid$) R_u=8.314 [kPa-m^3/kmol-K] “Analysis” "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" P2=r*(T2/T1)*P1 "from ideal gas relation" h2=enthalpy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant volume heat addition" T3=r_c*T2 h3=enthalpy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) P3=P2 q_in=h3-h2 "Process 3-4: isentropic expansion" s4=s3 "isentropic process" P4=(r_c/r)*(T4/T3)*P3 "from ideal gas relation" T4=temperature(Fluid$, P=P4, s=s4) u4=intenergy(Fluid$, P=P4, s=s4) "Process 4-1: constant volume heat rejection" q_out=u4-u1 eta_th=1-q_out/q_in w_net_out=q_in-q_out v1=(R_air*T1)/P1 v2=v1/r MEP=w_net_out/(v1-v2) x_destroyed=T0*(q_out/T0-q_in/T_H) x2=DELTAu-T0*DELTAs+P0*DELTAv DELTAu=u2-u1 u2=intenergy(Fluid$, P=P2, s=s2) DELTAs=s2-s1 DELTAv=v2-v1 P0=P1

10-148E The exergy destruction associated with the heat rejection process of a Diesel cycle and the exergy at the end of the expansion stroke are to be determined. Assumptions 1. 2.

The air-standard assumptions are applicable. Kinetic and potential energy changes are negligible.

10-147

3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17E.

Analysis Process 1-2: isentropic compression. →

→ Process 2-3: P = constant heat addition. → →

Process 3-4: isentropic expansion. (

)

→ Process 4-1: v = constant heat rejection:

The entropy change during process 4-1 is

Thus, (

)

(

)

Noting that state 4 is identical to the state of the surroundings, the exergy at the end of the power stroke (state 4) is determined from where

10-148

Discussion Note that the exergy at state 4 is identical to the exergy destruction for the process 4-1 since state 1 is identical to the dead state, and the entire exergy at state 4 is wasted during process 4-1.

EES (Engineering Equation Solver) SOLUTION "Given" r=18.2 "r=v1/v2" P1=14.7 [psia] T1=(120+460) [R] T3=3200 [R] T_H=3200 [R] T0=540 [R] "Analysis" Fluid$='air' "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" P2=r*(T2/T1)*P1 "from ideal gas relation" h2=enthalpy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant volume heat addition" r_c=T3/T2 "r_c=v3/v2" h3=enthalpy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) P3=P2 q_in=h3-h2 "Process 3-4: isentropic expansion" s4=s3 "isentropic process" P4=(r_c/r)*(T4/T3)*P3 "from ideal gas relation" T4=temperature(Fluid$, P=P4, s=s4) u4=intenergy(Fluid$, P=P4, s=s4) "Process 4-1: constant volume heat rejection" q_out=u4-u1 eta_th=1-q_out/q_in x_destroyed=T0*(s1-s4+q_out/T0) x4=DELTAu-T0*DELTAs+P0*DELTAv*Convert(psia-ft^3, Btu) DELTAu=q_out DELTAs=s4-s1 DELTAv=0 P0=P1

10-149E The exergy loss of an ideal dual cycle is to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , and k = 1.4 (Table A-2Ea). Analysis Working around the cycle, the germane properties at the various states are

(Table A-1E),

10-149

( ) ( )

( )

(

) ( ) ( )

( )

(

)

Applying the first law and work expression to the heat addition processes gives

The heat rejected is Also,

The exergy destruction during a process of the cycle is (

)

Application of this equation for each process of the cycle gives (isentropic process) -

(

)

(

)

(

)

(

)

(isentropic process)

10-150

(

)

(

)

The largest exergy destruction in the cycle occurs during the heat-rejection process. The total exergy destruction in the cycle is

10-150E The entropy generated by a Brayton cycle is to be determined. Assumptions 1. Steady operating conditions exist. 2. Kinetic and potential energy changes are negligible. 3. Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are and k = 1.667 (Table A-2Ea). Analysis At the compressor inlet,

(Table A-1E),

̇ According to the isentropic process expressions for an ideal gas, ( )

(

)

( )

(

)

Applying the first law to the constant-pressure heat addition process 2-3 gives ̇ ̇ The net power output and heat rejected are ̇

Note that No entropy is generated by the working fluid since it always returns to its original state. Then, ̇ ̇ ̇

10-151

10-151 A gas-turbine plant uses diesel fuel and operates on simple Brayton cycle. The isentropic efficiency of the compressor, the net power output, the back work ratio, the thermal efficiency, and the second-law efficiency are to be determined.

Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at 500ºC = 773 K are , , and k = 1.357 (Table A-2b). Analysis (a) The isentropic efficiency of the compressor may be determined if we first calculate the exit temperature for the isentropic case ( )

(

)

(b) The total mass flowing through the turbine and the rate of heat input are ̇ ̇ ̇ ̇ ̇ ̇ ̇

The temperature at the exit of combustion chamber is ̇ ̇ →

→

The temperature at the turbine exit is determined using isentropic efficiency relation ( )

( →

) →

The net power and the back work ratio are ̇ ̇ ̇ ̇ ̇

̇ ̇ ̇

10-152

̇ ̇ The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,

and

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] P_2=700 [kPa] T_1=(30+273) [K] T_2=(260+273) [K] m_dot_a=12.6 [kg/s] HV=42000 [kJ/kg] AF=60 eta_comb=0.97 eta_T=0.85 "PROPERTIES at 500 C = 773 K, Table A-2b" c_p=1.093 [kJ/kg-K] c_v=0.806 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.357 "ANALYSIS" "(a)" T_2s=T_1*(P_2/P_1)^((k-1)/k) eta_C=(T_2s-T_1)/(T_2-T_1) "isentropic efficiency" "(b)" m_dot_f=m_dot_a/AF m_dot=m_dot_a+m_dot_f Q_dot_in=m_dot_f*HV*eta_comb T_3=T_2+Q_dot_in/(m_dot*C_p) T_4s=T_3*(P_1/P_2)^((k-1)/k) T_4=T_3-eta_T*(T_3-T_4s) W_dot_C_in=m_dot_a*c_p*(T_2-T_1) W_dot_T_out=m_dot*c_p*(T_3-T_4) W_dot_net=W_dot_T_out-W_dot_C_in "net power" r_bw=W_dot_C_in/W_dot_T_out "back-work ratio" "(c)" eta_th=W_dot_net/Q_dot_in "thermal efficiency" eta_th_max=1-T_1/T_3 "Carnot efficiency" eta_II=eta_th/eta_th_max "second-law efficiency"

10-152 The exergy loss of each process for a regenerative Brayton cycle is to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-153

3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are

and k = 1.4 (Table A-2a).

Analysis For the compression and expansion processes we have

→

. /

( ) →

The heat input and heat output are

The exergy destruction during a process of a stream from an inlet state to exit state is given by (

)

Application of this equation for each process of the cycle gives ( (

)

[ )

] [

]

10-154

(

)

[

(

)

( )] [

] (

[

) ]

10-153 The total exergy destruction associated with a regenerative Brayton cycle and the exergy at the exhaust gases at the turbine exit are to be determined. Assumptions

1. The air standard assumptions are applicable. 2. Air is an ideal gas with variable specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17. The gas constant of air is (Table A-1). Analysis The properties of air at various states are → → → → hC =

h2 s - h1 ¾¾ ® h2 = h1 + (h2 s - h1 ) / hC = 310.24 + (541.35 - 310.24)/ (0.75) = 618.39 kJ/kg h2 - h1

→

Thus,

→

10-155

Then,

Also, → → → → → → Thus, (

)

(

) (

)

(

)

(

)

(

) (

)

( Noting that and the regenerator (state 6) is determined from

/ )

→

, the flow exergy at the exit of

where

Thus,

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=310 [K] P[1]=100 [kPa] Ratio_P=7 P[2]=Ratio_P*P[1] T[3]=1150 [K] eta_C=0.75 eta_T=0.82 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-156

epsilon=0.65 T_H=1500 [K] T0=290 [K] P0=100 [kPa] "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in "Analysis" q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid$, P=P[2], h=h[2]) s[4]=entropy(Fluid$, h=h[4], P=P[4]) s[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x_total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_destroyed_53+x_destroyed_61 x6=h[6]-h[0]-T0*(s[6]-s[0])

10-154 Prob. 10-153 is reconsidered. The effect of the cycle pressure on the total irreversibility for the cycle and the exergy of the exhaust gas leaving the regenerator is to be investigated. Analysis Using EES, the problem is solved as follows: T[1]=310 [K] P[1]=100 [kPa] Ratio_P=7 P[2]=Ratio_P*P[1] T[3]=1150 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-157

eta_C=0.75 eta_T=0.82 epsilon=0.65 T_H=1500 [K] T0=290 [K] P0=100 [kPa] Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] "isentropic compression" h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) eta_C=(h_s[2]-h[1])/(h[2]-h[1]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4])

q_regen=epsilon*(h[4]-h[2]) w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[2]=entropy(Fluid$, P=P[2], h=h[2]) s[4]=entropy(Fluid$, h=h[4], P=P[4]) s[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x_total=x_destroyed_12+x_destroyed_34+x_destroyed_regen+x_destroyed_53+x_destroyed_61 x6=h[6]-h[0]-T0*(s[6]-s[0]) "since state 0 and state 1 are identical" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-158

Ratio_P 6 7 8 9 10 11 12 13 14

270.1 280 289.9 299.5 308.8 317.8 326.6 335.1 343.3

137.2 143.5 149.6 155.5 161.1 166.6 171.9 177.1 182.1

10-155 The exergy destruction associated with each of the processes of the regenerative Brayton cycle and the exergy at the end of the exhaust gases at the exit of the regenerator are to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with variable specific heats. 3. Kinetic and potential energy changes are negligible. Properties The properties of air are given in Table A-17.

10-159

→

→ →

→ → → → → → Thus, (

)

(

) (

)

(

)

(

)

(

) (

)

( Noting that at

/ )

and

the stream exergy at the exit of

the regenerator (state 6) is determined from

where

10-160

Thus,

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=310 [K] P[1]=100 [kPa] P[2]=900 [kPa] T[2]=650 [K] T[3]=1400 [K] eta_T=0.90 epsilon=0.80 T_H=1260 [K] T0=300 [K] P0=100 [kPa] "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) h[2]=enthalpy(Fluid$, T=T[2]) h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s_s[4]=s[3] "isentropic expansion" h_s[4]=enthalpy(Fluid$, P=P[4], s=s_s[4]) P[4]=P[1] eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_regen=epsilon*(h[4]-h[2]) w_C_in=(h[2]-h[1]) w_T_out=h[3]-h[4] w_net_out=w_T_out-w_C_in q_in=(h[3]-h[2])-q_regen eta_th=w_net_out/q_in q_in=h[3]-h[5] q_out=h[6]-h[1] h[5]-h[2]=h[4]-h[6] s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=entropy(Fluid$, T=T[2], P=P[2]) s[4]=entropy(Fluid$, h=h[4], P=P[4]) s[5]=entropy(Fluid$, h=h[5], P=P[5]) P[5]=P[2] s[6]=entropy(Fluid$, h=h[6], P=P[6]) P[6]=P[1] h[0]=enthalpy(Fluid$, T=T0) s[0]=entropy(Fluid$, T=T0, P=P0) x_destroyed_12=T0*(s[2]-s[1]) x_destroyed_34=T0*(s[4]-s[3]) x_destroyed_regen=T0*(s[5]-s[2]+s[6]-s[4]) x_destroyed_53=T0*(s[3]-s[5]-q_in/T_H) x_destroyed_61=T0*(s[1]-s[6]+q_out/T0) x6=h[6]-h[0]-T0*(s[6]-s[0]) "since state 0 and state 1 are identical"

10-161

10-156E The exergy loss of each process for a reheat-regenerative Brayton cycle with intercooling is to be determined. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2Ea). Analysis For the compression and expansion processes, we have

→

. /

( ) →

The regenerator is ideal (i.e., the effectiveness is 100%) and thus,

Also,

The exergy destruction during a process of a stream from an inlet state to exit state is given by (

)

Application of this equation for each process of the cycle gives (

)

( (

[ )

)

] [

[

] ( )]

10-162

æ é q ö T P 520 52.27 ù ÷ ú= 8.61 Btu / lbm xdest, 10-1 = xdest, 2-3 = T0 çççc p ln 1 - R ln 1 + out ÷ = (520) ê(0.24)ln - 0+ ÷ ÷ êë çè T10 P10 Tsink ø 737.8 520 ú û

(

)

The greatest exergy destruction occurs during the heat rejection processes.

10-157 A gas-turbine plant operates on the regenerative Brayton cycle. The isentropic efficiency of the compressor, the effectiveness of the regenerator, the air-fuel ratio in the combustion chamber, the net power output, the back work ratio, the thermal efficiency, the second law efficiency, the exergy efficiencies of the compressor, the turbine, and the regenerator, and the rate of the exergy of the combustion gases at the regenerator exit are to be determined.

Assumptions 1. 2. 3.

The air-standard assumptions are applicable. Kinetic and potential energy changes are negligible. Air is an ideal gas with constant specific heats.

Properties The properties of air at 500C = 773 K are , , , and k = 1.357 (Table A-2b). Analysis (a) For the compressor and the turbine: ( )

( ) →

(

)

(

) →

(b) The effectiveness of the regenerator is

10-163

̇ ̇

→

̇ ̇ Also, ̇ ̇

(d) The net power and the back work ratio are ̇ ̇ ̇ ̇ ̇

̇ ̇ ̇

(e) The thermal efficiency is ̇ ̇ (f) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,

and

(g) The exergy efficiency for the compressor is defined as the ratio of stream exergy difference between the inlet and exit of the compressor to the actual power input: ̇ ̇

̇ { {

[ [

(

]} )

(

)]}

̇ ̇ The exergy efficiency for the turbine is defined as the ratio of actual turbine power to the stream exergy difference between the inlet and exit of the turbine: ̇

̇ {

[

{

]} [

(

)

(

)]}

̇ ̇ An energy balance on the regenerator gives ̇

̇ →

10-164

The exergy efficiency for the regenerator is defined as the ratio of the exergy increase of the cold fluid to the exergy decrease of the hot fluid: ̇

̇ {

[

]}

{

[

̇ {

[

(

)

]}

{

[

(

)

]}

̇ ̇ The exergy of the combustion gases at the regenerator exit: ̇

̇ { {

[

]} [

(

)

]}

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] P_2=700 [kPa] T_1=30+273 [K] T_2=260+273 [K] T_3=871+273 [K] T_5=400+273 [K] m_dot_a=12.6 [kg/s] HV=42000 [kJ/kg] eta_comb=0.97 eta_T=0.85 T_0=T_1 "dead state temperature" "PROPERTIES" Fluid$='Air' c_p=CP(Fluid$, T=500+273) c_v=CV(Fluid$, T=500+273) k=c_p/c_v R=c_p-c_v "ANALYSIS" T_2s=T_1*(P_2/P_1)^((k-1)/k) eta_C=(T_2s-T_1)/(T_2-T_1) "compressor isentropic efficiency" T_4s=T_3*(P_1/P_2)^((k-1)/k) T_4=T_3-eta_T*(T_3-T_4s) epsilon_regen=(T_5-T_2)/(T_4-T_2) m_dot=m_dot_a+m_dot_f Q_dot_in=m_dot*c_p*(T_3-T_5) Q_dot_in=m_dot_f*HV*eta_comb AF=m_dot_a/m_dot_f "air-fuel ratio" W_dot_C_in=m_dot_a*c_p*(T_2-T_1) W_dot_T_out=m_dot*c_p*(T_3-T_4) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-165

W_dot_net=W_dot_T_out-W_dot_C_in "net power" r_bw=W_dot_C_in/W_dot_T_out "back work ratio" eta_th=W_dot_net/Q_dot_in "thermal efficiency" "Exergy analysis" eta_th_max=1-T_1/T_3 "Carnot efficiency" eta_II=eta_th/eta_th_max "second-law efficiency of cycle" DELTAEX_dot_C=m_dot_a*(c_p*(T_2-T_1)-T_0*(c_p*ln(T_2/T_1)-R*ln(P_2/P_1))) DELTAEX_dot_T=m_dot*(c_p*(T_3-T_4)-T_0*(c_p*ln(T_3/T_4)-R*ln(P_2/P_1))) DELTAEX_dot_hot=m_dot*(c_p*(T_4-T_6)-T_0*(c_p*ln(T_4/T_6))) DELTAEX_dot_cold=m_dot_a*(c_p*(T_5-T_2)-T_0*(c_p*ln(T_5/T_2))) eta_ex_C=DELTAEX_dot_C/W_dot_C_in "exergy efficiency of compressor" eta_ex_T=W_dot_T_out/DELTAEX_dot_T "exergy efficiency of turbine" eta_ex_regen=DELTAEX_dot_cold/DELTAEX_dot_hot "exergy efficiency of regenerator" m_dot_a*c_p*(T_5-T_2)=m_dot*c_p*(T_4-T_6) "energy balance on regenerator" EX_dot_6=m_dot_a*(c_p*(T_6-T_0)-T_0*(c_p*ln(T_6/T_0))) "exergy at regenerator exit"

10-158 A modern compression ignition engine operates on the ideal dual cycle. The maximum temperature in the cycle, the net work output, the thermal efficiency, the mean effective pressure, the net power output, the second-law efficiency of the cycle, and the rate of exergy of the exhaust gases are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at 1000 K are , , , and k = 1.336 (Table A-2b). Analysis (a) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are →

→

Process 1-2: Isentropic compression ( ) ( ) Process 2-x and x-3: Constant-volume and constant pressure heat addition processes:

10-166

→ (b)

→

Process 3-4: isentropic expansion. ( )

( )

Process 4-1: constant voume heat rejection. The net work output and the thermal efficiency are

. (d) The power for engine speed of 3500 rpm is ̇ ̇

(

)

Note that there are two revolutions in one cycle in four-stroke engines. (e) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible

and

The rate of exergy of the exhaust gases is determined as follows

[

]

10-167

(

)

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol_d=0.0018 [m^3] "total engine displacement volume" r=16 " r = V_1 / V_2 " P_1=95 [kPa] T_1=(70+273) [K] N_dot=2200 [1/min] P_3=7500 [kPa] "PROPERTIES at 1000 K, Table A-2b" c_p=1.142 [kJ/kg-K] c_v=0.855 [kJ/kg-K] k=1.336 R_air=0.287 [kJ/kg-K] "ANALYSIS" "(a)" r=(Vol_c+Vol_d)/Vol_c "Vol_c is the total engine clearence volume" Vol_1=Vol_d+Vol_c Vol_2=Vol_1/r T_2=T_1*r^(k-1) "isentropic compression" P_2=P_1*r^k Vol_x=Vol_2 P_x=P_3 T_x=T_2*(P_x/P_2) q_2x=c_v*(T_x-T_2) q_x3=c_p*(T_3-T_x) q_2x=q_x3 "(b)" q_in=q_2x+q_x3 Vol_3=Vol_x*(T_3/T_x) Vol_4=Vol_1 T_4=T_3*(Vol_3/Vol_4)^(k-1) "isentropic expansion" P_4=P_3*(Vol_3/Vol_4)^k q_out=c_v*(T_4-T_1) "heat rejection during the constant volume heat rejection" w_net=q_in-q_out eta_th=w_net/q_in "(c)" m_t=(P_1*Vol_1)/(R_air*T_1) "total air+fuel mass in engine" MEP=(m_t*w_net)/(Vol_1-Vol_2) "(d)" W_dot_net=(m_t*w_net*N_dot*Convert(1/min, 1/s))/n_rev n_rev=2 "for four-stroke cycle, there are two revolutions per cycle" "(e)" "exergy analysis" eta_max=1-T_0/T_3 T_0=298 [K] "assumed ambient temperature" eta_II=eta_th/eta_max ex_4=DELTAu-T_0*DELTAs DELTAu=c_v*(T_4-T_0) DELTAs=c_p*ln(T_4/T_0)-R_air*ln(P_4/P_1) EX_dot_4=(m_t*ex_4*N_dot*Convert(1/min, 1/s))/n_rev

10-168

Review Problems 10-159 The three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions

1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (b) We treat air as an ideal gas with variable specific heats, →

→

(

)

(

)

→

(c)

10-169

P1=100 "[kPa]" T1=27+273 "[K]" P2=700 "[kPa]" P3=P2 "Analysis" Fluid$='air' "(b)" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 "isentropic process" h2=enthalpy(Fluid$, P=P2, s=s2) T3=(P3/P1)*T1 "from ideal gas relation" T_max=T3 "(c)" u3=intenergy(Fluid$, T=T3) h3=enthalpy(Fluid$, T=T3) q_in=h3-h2 q_out=u3-u1 eta_th=1-q_out/q_in

10-160 All three processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined. Assumptions

1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are 2). Analysis (b) We treat air as an ideal gas with constant specific heats. Process 1-2 is isentropic: ( )

( →

, and k = 1.4 (Table A-

) (

)

10-170

(c)

EES (Engineering Equation Solver) SOLUTION "Given" m=0.003 "[kg]" P1=100 "[kPa]" T1=27+273 "[K]" P2=700 "[kPa]" P3=P2 "Properties" Fluid$='air' C_p=CP(Fluid$, T=T) C_v=CV(Fluid$, T=T) T=298 "[K], room temperature" k=C_p/C_v "Analysis" "(b)" T2=T1*(P2/P1)^((k-1)/k) "isentropic process" T3=(P3/P1)*T1 "from ideal gas relation" T_max=T3 "(c)" q_in=C_p*(T3-T2) q_out=C_v*(T3-T1) eta_th=1-q_out/q_in

10-161 A Carnot cycle executed in a closed system uses air as the working fluid. The net work output per cycle is to be determined. Assumptions 1.

Air is an ideal gas with variable specific heats.

Properties The gas constant of air is

(Table A-1).

10-171

Analysis (a) The maximum temperature is determined from →

→

EES (Engineering Equation Solver) SOLUTION "Given" m=0.0025 [kg] eta_th=0.60 T3=300 [K] P2=700 [kPa] P1=1000 [kPa] "Analysis" Fluid$='air' eta_th=1-T3/T1 s1=entropy(Fluid$, T=T1, P=P1) s2=entropy(Fluid$, T=T2, P=P2) T2=T1 W_net=m*(s2-s1)*(T1-T3)

10-162E A turbocharged four-stroke V-16 diesel engine produces 4400 hp at 1500 rpm. The amount of work produced per cylinder per mechanical and per thermodynamic cycle is to be determined. Analysis Noting that there are 16 cylinders and each thermodynamic cycle corresponds to 2 mechanical cycles (revolutions), we have (a)

(

)

10-172

(

)

EES (Engineering Equation Solver) SOLUTION "Given" N_cyl=16 W_dot=4400 [hp] n_dot=1500 [1/min] "Analysis" "(a)" w_mechanical=W_dot/(N_cyl*n_dot)*Convert(hp, Btu/min) "(b)" w_thermodynamic=2*W_dot/(N_cyl*n_dot)*Convert(hp, Btu/min) "Note that 1 thermodynamic cycle corresponds to 2 mechanical cycles"

10-163 An Otto cycle with a compression ratio of 8 is considered. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a). Analysis (a) Constant specific heats:

(b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression.

→ → Process 2-3: v = constant heat addition. →

10-173

→ Process 4-1: v = constant heat rejection.

10-164 An ideal diesel engine with air as the working fluid has a compression ratio of 22. The thermal efficiency is to be determined using constant and variable specific heats. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2a).

Analysis (a) Constant specific heats: Process 1-2: isentropic compression. ( ) Process 2-3: P = constant heat addition. → Process 3-4: isentropic expansion. ( )

(

)

(

)

(

)

(b) Variable specific heats: (using air properties from Table A-17) Process 1-2: isentropic compression. → © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-174

→ Process 2-3: P = constant heat addition. → →

Process 3-4: isentropic expansion. → Process 4-1: v = constant heat rejection. Then

10-165 An engine operating on the ideal diesel cycle with air as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is (Table A-1). The properties of air are given in Table A-17. Analysis (a) The compression and the cutoff ratios are

Process 1-2: isentropic compression. →

10-175

→

Process 3-4: isentropic expansion.

→ Process 4-1: v = constant heat rejection. → (b)

(

)

(c)

(

)

EES (Engineering Equation Solver) SOLUTION "Given" V1=1200E-6 "[m^3]" V2=75E-6 "[m^3]" V3=150E-6 "[m^3]" T1=17+273 "[K]" P1=100 "[kPa]" "Properties" Fluid$='air' C_p=CP(Fluid$, T=T) C_v=CV(Fluid$, T=T) T=25+273 "[K], room temperature" R_air=C_p-C_v "Analysis" "(a)" r=V1/V2 r_c=V3/V2 "Process 1-2: isentropic compression" u1=intenergy(Fluid$, T=T1) s1=entropy(Fluid$, T=T1, P=P1) s2=s1 P2=r*(T2/T1)*P1 "from ideal gas relation" h2=enthalpy(Fluid$, P=P2, s=s2) T2=temperature(Fluid$, P=P2, s=s2) "Process 2-3: constant pressure heat addition" T3=r_c*T2 h3=enthalpy(Fluid$, T=T3) s3=entropy(Fluid$, T=T3, P=P3) P3=P2 "Process 3-4: isentropic expansion" s4=s3 "Process 4-1: constant volume heat rejection" P4=(T4/T1)*P1 "from ideal gas relation" T4=temperature(Fluid$, P=P4, s=s4) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-176

u4=intenergy(Fluid$, P=P4, s=s4) "(b)" m=(P1*V1)/(R_air*T1) Q_in=m*(h3-h2) Q_out=m*(u4-u1) W_net=Q_in-Q_out "(c)" MEP=W_net/(V1-V2)

10-166 An engine operating on the ideal diesel cycle with argon as the working fluid is considered. The pressure at the beginning of the heat-rejection process, the net work per cycle, and the mean effective pressure are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Argon is an ideal gas with constant specific heats. Properties The properties of argon at room temperature are , , and k = 1.667 (Table A-2). Analysis (a) The compression and the cutoff ratios are

Process 1-2: isentropic compression.

( ) Process 2-3: P = constant heat addition. → Process 3-4: isentropic expansion. ( )

(

)

( )

(

)

Process 4-1: v = constant heat rejection. → (b)

(

)

10-177

(c)

(

)

EES (Engineering Equation Solver) SOLUTION "Given" V1=1200E-6 "[m^3]" V2=75E-6 "[m^3]" V3=150E-6 "[m^3]" T1=17+273 "[K]" P1=100 "[kPa]" "Properties" Fluid$='Argon' C_p=CP(Fluid$, T=T, P=P) C_v=CV(Fluid$, T=T, P=P) T=25+273 "[K], room temperature" P=101.3 "[kPa], atmospheric pressure" k=C_p/C_v R_Ar=C_p-C_v "Analysis" "(a)" r=V1/V2 r_c=V3/V2 T2=T1*r^(k-1) "Process 1-2: isentropic compression" T3=r_c*T2 "Process 2-3: constant pressure heat addition" T4=T3*(r_c/r)^(k-1) "Process 3-4: isentropic expansion" P4=(T4/T1)*P1 "from ideal gas relation" "(b)" m=(P1*V1)/(R_Ar*T1) Q_in=m*C_p*(T3-T2) Q_out=m*C_v*(T4-T1) W_net=Q_in-Q_out "(c)" MEP=W_net/(V1-V2)

10-167 A spark-ignition engine operates on an ideal Otto cycle with a compression ratio of 11. The maximum temperature and pressure in the cycle, the net work per cycle and per cylinder, the thermal efficiency, the mean effective pressure, and the power output for a specified engine speed are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is (Table A-1). Other properties of air are given as , , and k = 1.35. Analysis

10-178

(a) First the mass in one cylinder is determined as follows

Process 1-2: isentropic compression. ( ) →

(

)

Process 2-3: v = constant heat addition.

→

(

)

(b) Process 3-4: isentropic expansion. ( )

(

)

Process 4-1: v = constant heat rejection. The net work output and the thermal efficiency are (per cycle per cylinder)

(

)

Note that there are two revolutions in one cycle in four-stroke engines.

10-179

n_cyl=4 "number of cylinders" r=11 " r = V_1 / V_2 = V_4 / V_3 " V_engine=0.0018 [m^3] "total displacement volume of engine" P1=90 [kPa] T1=50[C]+273 [K] Q_in=0.50 [kJ] "per cycle per cylinder" n_dot=3000 [1/min] "Properties" Fluid$='air' c_p=1.108 c_v=0.821 k=1.35 R_air=0.287 "Analysis" V_d=V_engine/n_cyl "displacement volume for one cylinder" r=(V_c+V_d)/V_c "V_c is the clearence volume" V1=V_d+V_c V2=V1/r m=(P1*V1)/(R_air*T1) "air mass in one cylinder" "Process 1-2: isentropic compression" T2=T1*r^(k-1) P2=r*(T2/T1)*P1 "from ideal gas relation" "Process 2-3: constant volume heat addition" Q_in=m*c_v*(T3-T2) P3=(T3/T2)*P2 "from ideal gas relation" T4=T3*(1/r)^(k-1) "Process 3-4: isentropic expansion" Q_out=m*c_v*(T4-T1) "Process 4-1: constant volume heat rejection" W_net=Q_in-Q_out eta_th=W_net/Q_in MEP=W_net/(V1-V2) W_dot_net=n_cyl*(W_net*n_dot*Convert(1/min, 1/s))/n_rev n_rev=2 "for four-stroke cycle, there are 2 revolutions per cycle"

10-168 The compression ratio required for an ideal Otto cycle to produce certain amount of work when consuming a given amount of fuel is to be determined. Assumptions 1. 2. 3. 4.

The air-standard assumptions are applicable. Kinetic and potential energy changes are negligible. Air is an ideal gas with constant specific heats. The combustion efficiency is 100 percent.

Properties The properties of air at room temperature are k = 1.4 (Table A-2). Analysis The heat input to the cycle for 0.039 grams of fuel consumption is

10-180

The thermal efficiency is then

From the definition of thermal efficiency, we obtain the required compression ratio to be →

EES (Engineering Equation Solver) SOLUTION "Given" q_HV=43000 [kJ/kg] m_fuel=0.039E-3 [kg] W_net=1 [kJ] "Properties" k=1.4 "Table A-2a" "Analysis" Q_in=m_fuel*q_HV eta_th=W_net/Q_in eta_th=1-1/r^(k-1)

10-169E An ideal dual cycle with air as the working fluid with a compression ratio of 12 is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1. 2. 3.

The air-standard assumptions are applicable. Kinetic and potential energy changes are negligible. Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are (Table A-2E).

, and k = 1.4

Analysis The mass of air is

10-181

Process 1-2: isentropic compression. ( ) Process 2-x: v = constant heat addition, → Process x-3: P = constant heat addition. → → Process 3-4: isentropic expansion. ( )

(

)

(

)

(

)

Process 4-1: v = constant heat rejection.

EES (Engineering Equation Solver) SOLUTION "Given" r=14 "r=v1/v2" P1=14.7 [psia] T1=(120+460) [R] V1=98*Convert(in^3, ft^3) Q_in_2x=0.6 [Btu] Q_in_x3=1.1 [Btu] "Properties" Fluid$='air' c_p=0.240 [Btu/lbm-R] c_v=0.171 [Btu/lbm-R] k=1.4 R_air=0.3704 [psia-ft^3/lbm-R] "Analysis" m=(P1*V1)/(R_air*T1) T2=T1*r^(k-1) "isentropic compression" Q_in_2x=m*c_v*(T_x-T2) "constant volume heat addition" Q_in_x3=m*c_p*(T3-T_x) "constant pressure heat addition" r_c=T3/T_x "r_c=V3/V_x" T4=T3*(r_c/r)^(k-1) "isentropic expansion" Q_out=m*c_v*(T4-T1) "constant volume heat rejection" Q_in=Q_in_2x+Q_in_x3 eta_th=1-Q_out/Q_in

10-170 A gasoline engine operates on an Otto cycle. The compression and expansion processes are modeled as polytropic. The temperature at the end of expansion process, the net work output, the thermal efficiency, the mean effective pressure, the engine speed for a given net power, and the specific fuel consumption are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-182

Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at 850 K are 1.349 (Table A-2b). Analysis (a) Process 1-2: polytropic compression

, and k =

( ) ( )

Process 2-3: constant volume heat addition ( )

(

)

Process 3-4: polytropic expansion. ( ) ( )

( (

)

Process 4-1: constant volume heat rejection. (b) The net work output and the thermal efficiency are

(d) The clearance volume and the total volume of the engine at the beginning of compression process (state 1) are © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-183

→

The total mass contained in the cylinder is

The engine speed for a net power output of 50 kW is ̇ ̇

(

)

Note that there are two revolutions in one cycle in four-stroke engines. (e) The mass of fuel burned during one cycle is →

→

Finally, the specific fuel consumption is (

)(

)

EES (Engineering Equation Solver) SOLUTION "GIVEN" Vol_d=0.0016 [m^3] "total engine displacement volume" r=11 " r = v_1 / v_2 = v_4 / v_3 " P_1=100 [kPa] T_1=(37+273) [K] P_3=8000 [kPa] n=1.3 "polytropic constant" W_dot_net=50 [kW] AF=16 " AF = m_air / m_fuel " "PROPERTIES" c_p=1.110 [kJ/kg-K] "at 850 K, Table A-2b" c_v=0.823 [kJ/kg-K] "at 850 K, Table A-2b" R_air=0.287 [kJ/kg-K] "Table A-2a" "ANALYSIS" T_2=T_1*r^(n-1) "Process 1-2: polytropic compression" P_2=P_1*r^n w_12=R_air*(T_1-T_2)/(1-n) T_3=T_2*(P_3/P_2) "Process 2-3: constant volume heat addition" q_in=C_v*(T_3-T_2) T_4=T_3*(1/r)^(n-1) "Process 3-4: polytropic expansion" P_4=P_3*(1/r)^n w_34=R_air*(T_4-T_3)/(1-n) q_out=C_v*(T_4-T_1) "Process 4-1: constant volume heat rejection" w_net=w_34-w_12 eta_th=w_net/q_in v_1=(R_air*T_1)/P_1 v_2=v_1/r MEP=w_net/(v_1-v_2) r=(Vol_c+Vol_d)/Vol_c "Vol_c is the total engine clearence volume" Vol_1=Vol_d+Vol_c m_t=(P_1*Vol_1)/(R_air*T_1) "total air+fuel mass in engine" n_dot=n_rev*W_dot_net/(m_t*w_net)*Convert(1/s, 1/min) n_rev=2 "for four-stroke cycle, there are two revolutions per cycle" m_t=m_air+m_fuel © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-184

AF=m_air/m_fuel sfc=m_fuel/(m_t*w_net)*Convert(kg/kJ, g/kWh) "specific fuel consumption"

10-171 An ideal Stirling cycle with air as the working fluid is considered. The maximum pressure in the cycle and the net work output are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2).

Analysis (a) The entropy change during process 1-2 is

and →

→

(

)

(b) The net work output is (

)

(

)

EES (Engineering Equation Solver) SOLUTION "Given" T3=400 [K] P3=200 [kPa] T1=1800 [K] q_in=900 [kJ/kg] "Properties" R=0.287 [kJ/kg-K] "Table A-2a" "Analysis" "(a)" DELTAs=q_in/T1 "DELTAs = s2-s1" DELTAs=R*ln(Ratio_v) "ratio_v = v2/v1" P1=P3*Ratio_v*(T1/T3) "since v3/v1 = v2/v1" "(b)" eta_th=1-T3/T1 w_net=eta_th*q_in © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-185

10-172 A simple ideal Brayton cycle operating between the specified temperature limits is considered. The pressure ratio for which the compressor and the turbine exit temperature of air are equal is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. 4. Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k =1.4 (Table A-2). Analysis We treat air as an ideal gas with constant specific heats. Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as

Setting

and solving for

( )

. /

gives ( )

(

)

Therefore, the compressor and turbine exit temperatures will be equal when the compression ratio is 12.2.

EES (Engineering Equation Solver) SOLUTION "Given" T1=300 [K] T3=1250 [K] "Properties" k=1.4 "Table A-2a" "Analysis" T2=T1*r_p^((k-1)/k) "r_p=P2/P1=P3/P4" T4=T3*(1/r_p)^((k-1)/k) T2=T4

10-173 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-186

Analysis The properties at various states are → → For

, → →

For

, → →

Thus, (a) (b)

EES (Engineering Equation Solver) SOLUTION "Given" r_p=P[2]/P[1] T[1]=300 [K] T[3]=1300 [K] P[1]=100 [kPa] "assumed" "Analysis" Fluid$='air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] "isentropic compression" h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-187

h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) P[3]=P[2] s[4]=s[3] "isentropic expansion" h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) P[4]=P[1] q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out eta_th=w_net/q_in "Press F3 or choose Solve Table from Calculate menu" "This problem is solved using Parametric Table, where it is found that for r_p = 6, w_net = 339.02 kJ/kg, and eta_th = 0.3792 for r_p = 12, w_net = 380.70 kJ/kg, and eta_th = 0.4852 Therefore, DELTAw_net = 41.68 kJ/kg and DELTAeta_th=0.106"

10-174 A simple ideal Brayton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are , , , and k = 1.4 (Table A-2). Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For ,

( ) ( )

For

( )

10-188

( ) ( )

(

)

Thus, (a) (b)

EES (Engineering Equation Solver) SOLUTION "Given" r_p=P[2]/P[1] T[1]=300 [K] T[3]=1300 [K] P[1]=100 [kPa] "assumed" "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T[2]=T[1]*r_p^((k-1)/k) T[4]=T[3]*(1/r_p)^((k-1)/k) q_in=c_p*(T[3]-T[2]) q_out=c_p*(T[4]-T[1]) w_net=q_in-q_out eta_th=w_net/q_in DELTAw_net=352.42-321.91 DELTAeta_th=0.5083-0.4007 "Press F3 or choose Solve Table from Calculate menu" "This problem is solved using Parametric Table, where it is found that for r_p = 6, w_net = 321.91 kJ/kg, and eta_th = 0.4007 for r_p = 12, w_net = 352.42 kJ/kg, and eta_th = 0.5083 Therefore, DELTAw_net = 30.51 kJ/kg and DELTAeta_th=0.1076"

10-175 A simple ideal Brayton cycle with air as the working fluid operates between the specified temperature limit. The net work is to be determined using constant and variable specific heats. Assumptions 1. Steady operating conditions exist. 2. The air-standard assumptions are applicable. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-189

(a) Constant specific heats:

. /

(

)

(b) Variable specific heats: (using air properties from Table A-17) → → → →

10-176 A regenerative Brayton cycle with helium as the working fluid is considered. The thermal efficiency and the required mass flow rate of helium are to be determined for 100 percent and 80 percent isentropic efficiencies for both the compressor and the turbine. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Helium is an ideal gas with constant specific heats. Properties The properties of helium are and k = 1.667 (Table A-2).

10-190

Analysis (a) Assuming

, ( ) ( )

( )

→

̇ ̇

(b) Assuming

, ( ) → ( )

( )

→ →

10-191

EES (Engineering Equation Solver) SOLUTION "Given" r_p=8 r_p=P[2]/P[1] T[1]=300 [K] T[3]=1800 [K] epsilon=0.75 W_dot_net=60000 [kW] eta_C=0.80 eta_T=0.80 P[1]=100 [kPa] "assumed" "Properties" Fluid$='Helium' c_p=CP(Fluid$, T=T, P=P) c_v=CV(Fluid$, T=T, P=P) T=(25+273) [K] "room temperature" P=101.3 [kPa] "atmospheric pressure" k=c_p/c_v "Analysis" "(b)" T_s[2]=T[1]*r_p^((k-1)/k) eta_C=(T_s[2]-T[1])/(T[2]-T[1]) T_s[4]=T[3]*(1/r_p)^((k-1)/k) eta_T=(T[3]-T[4])/(T[3]-T_s[4]) w_C_in=c_p*(T[2]-T[1]) w_T_out=c_p*(T[3]-T[4]) w_net=w_T_out-w_C_in m_dot=W_dot_net/w_net q_regen=epsilon*c_p*(T[4]-T[2]) q_in=c_p*(T[3]-T[2])-q_regen eta_th=w_net/q_in "(a) Solution for 100% efficient compressor and turbine case can be obtained by changing eta_C and eta_T values to 1 in the GIVEN section above."

10-177 An ideal gas-turbine cycle with one stage of compression and two stages of expansion and regeneration is considered. The thermal efficiency of the cycle as a function of the compressor pressure ratio and the high-pressure turbine to compressor inlet temperature ratio is to be determined, and to be compared with the efficiency of the standard regenerative cycle. Analysis The T-s diagram of the cycle is as shown in the figure. If the overall pressure ratio of the cycle is , which is the pressure ratio across the compressor, then the pressure ratio across each turbine stage in the ideal case becomes √ . Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as

10-192

( ) ( )

( )

√

( ) .

√

Then, (

)

(

)

and thus (

)

(

)

which simplifies to

The thermal efficiency of the single stage ideal regenerative cycle is given as

Therefore, the regenerative cycle with two stages of expansion has a higher thermal efficiency than the standard regenerative cycle with a single stage of expansion for any given value of the pressure ratio .

10-178 A gas-turbine plant operates on the regenerative Brayton cycle with reheating and intercooling. The back work ratio, the net work output, the thermal efficiency, the second-law efficiency, and the exergies at the exits of the combustion chamber and the regenerator are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Properties The gas constant of air is . Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-193

(a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

Optimum intercooling and reheating pressure is √

√

Process 1-2, 3-4: Compression → } } →

→ → } }

→

Process 6-7, 8-9: Expansion → } } →

→ → } }

→

Cycle analysis:

10-194

Regenerator analysis: →

→ } →

→

(b)

(c) The second-law efficieny of the cycle is defined as the ratio of actual thermal efficiency to the maximum possible thermal efficiency (Carnot efficiency). The maximum temperature for the cycle can be taken to be the turbine inlet temperature. That is,

and

(d) The exergies at the combustion chamber exit and the regenerator exit are

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[1]=100 [kPa]; P[4]=1200 [kPa] P[2]=sqrt(P[1]*P[4]) P[3]=P[2]; P[5]=P[4]; P[6]=P[4] P[7]=P[2]; P[8]=P[2]; P[9]=P[1]; P[10]=P[1] T[1]=300 [K]; T[3]=350 [K] T[6]=1400 [K]; T[8]=1300 [K] eta_C=0.80 eta_T=0.80 epsilon_regen=0.75 "PROPERTIES" Fluid$='Air' h[1]=enthalpy(Fluid$, T=T[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) h_s[2]=enthalpy(Fluid$, P=P[2], s=s[1]) h[2]=h[1]+(h_s[2]-h[1])/eta_C h[3]=enthalpy(Fluid$, T=T[3]) s[3]=entropy(Fluid$, T=T[3], P=P[3]) h_s[4]=enthalpy(Fluid$, P=P[4], s=s[3]) h[4]=h[3]+(h_s[4]-h[3])/eta_C h[6]=enthalpy(Fluid$, T=T[6]) s[6]=entropy(Fluid$, T=T[6], P=P[6]) h_s[7]=enthalpy(Fluid$, P=P[7], s=s[6]) h[7]=h[6]-eta_T*(h[6]-h_s[7]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-195

h[8]=enthalpy(Fluid$, T=T[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8]) h_s[9]=enthalpy(Fluid$, P=P[9], s=s[8]) h[9]=h[8]-eta_T*(h[8]-h_s[9]) "ANALYSIS" "(a)" w_C_in=h[2]-h[1]+h[4]-h[3] w_T_out=h[6]-h[7]+h[8]-h[9] r_bw=w_C_in/w_T_out w_net_out=w_T_out-w_C_in h[10]=h[9]-epsilon_regen*(h[9]-h[4]) q_regen=h[9]-h[10] h[5]=q_regen+h[4] "(b)" q_in=h[6]-h[5] eta_th=w_net_out/q_in "(c)" "Exergy analysis" eta_th_max=1-T[1]/T[6] eta_II=eta_th/eta_th_max "(d)" h[0]=h[1] "Subscript '0' stands for dead state properties" s[0]=s[1] T[0]=T[1] s[10]=entropy(Fluid$, h=h[10], P=P[10]) ex_6=h[6]-h[0]-T[0]*(s[6]-s[0]) ex_10=h[10]-h[0]-T[0]*(s[10]-s[0])

10-179 The thermal efficiency of a two-stage gas turbine with regeneration, reheating and intercooling to that of a threestage gas turbine is to be compared. Assumptions 1. The air standard assumptions are applicable. 2. Air is an ideal gas with constant specific heats at room temperature. 3. Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are and k = 1.4 (Table A-2a). Analysis Two Stages:

The pressure ratio across each stage is √ The temperatures at the end of compression and expansion are

10-196

. /

( )

The heat input and heat output are

The thermal efficiency of the cycle is then

Three Stages: The pressure ratio across each stage is The temperatures at the end of compression and expansion are

. /

(

)

The heat input and heat output are

The thermal efficiency of the cycle is then

EES (Engineering Equation Solver) SOLUTION "Given" r_p_total=16 P1=100 [kPa] T_min=(20+273) [K] T_max=(800+273) [K] "Properties at room temperature, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "(a)" r_p_a=sqrt(r_p_total) T_c_a=T_min*r_p_a^((k-1)/k) T_e_a=T_max*(1/r_p_a)^((k-1)/k) q_in_a=2*c_p*(T_max-T_e_a) q_out_a=2*c_p*(T_c_a-T_min) eta_th_a=1-q_out_a/q_in_a "(b)" r_p_b=r_p_total^(1/3) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-197

T_c_b=T_min*r_p_b^((k-1)/k) T_e_b=T_max*(1/r_p_b)^((k-1)/k) q_in_b=2*c_p*(T_max-T_e_b) q_out_b=2*c_p*(T_c_b-T_min) eta_th_b=1-q_out_b/q_in_b

10-180E A pure jet engine operating on an ideal cycle is considered. The thrust force produced per unit mass flow rate is to be determined. Assumptions 1. Steady operating conditions exist. 2. The air standard assumptions are applicable. 3. Air is an ideal gas with constant specific heats at room temperature. 4. The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are (Table A-1E), and k = 1.4 (Table A-2Ea). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of . Ideally, the air will leave the diffuser with a negligible velocity ( ). Diffuser:

→

( ( )

(

)

Compressor: ( ) ( ) Turbine: →

→

or Nozzle: ( )

(

)

10-198

→

→ √

or,

(

)

The specific impulse is then ̇

EES (Engineering Equation Solver) SOLUTION "Given" P1=10 [psia] T1=(30+460) [R] Vel1=1200 [ft/s] r_p=9 T4=(700+460) [R] "Properties at room temperature, Table A-2Ea" R=0.3704 [psia-ft^3/lbm-R] c_p=0.240 [Btu/lbm-R] k=1.4 "Analysis" T2=T1+Vel1^2/(2*c_p)*Convert(ft^2/s^2, Btu/lbm) P2=P1*(T2/T1)^(k/(k-1)) P3=r_p*P2 P4=P3 T3=T2*(P3/P2)^((k-1)/k) T5=T4-T3+T2 P6=P1 T6=T4*(P6/P4)^((k-1)/k) Vel6=sqrt(2*c_p*(T5-T6)*Convert(Btu/lbm, ft^2/s^2)) Vel_exit=Vel6 Vel_inlet=Vel1 F_specific=Vel_exit-Vel_inlet

10-181 The electricity and the process heat requirements of a manufacturing facility are to be met by a cogeneration plant consisting of a gas-turbine and a heat exchanger for steam production. The mass flow rate of the air in the cycle, the back work ratio, the thermal efficiency, the rate at which steam is produced in the heat exchanger, and the utilization efficiency of the cogeneration plant are to be determined. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure.

10-199

Process 1-2: Compression → } } →

→

Process 3-4: Expansion → → We cannot find the enthalpy at state 3 directly. However, using EES, we find ,

The solution by hand would require a trial-error approach. Also, → The inlet water is compressed liquid at 15C and at the saturation pressure of steam at 200C (1555 kPa). This is not available in the tables but we can obtain it in EES. The alternative is to use saturated liquid enthalpy at the given temperature. } } The net work output is

The mass flow rate of air is ̇

(b) The back work ratio is

10-200

̇ ̇

̇ ̇ (c) An energy balance on the heat exchanger gives ̇

̇ → ̇ (d) The heat supplied to the water in the heat exchanger (process heat) and the utilization efficiency are ̇ ̇ ̇ ̇ ̇

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=100 [kPa] P_2=1000 [kPa] T_1=20 [C] T_4=450 [C] T_hx_out=325 [C] W_dot_net=1500 [kW] T_w1=15 [C] T_w2=200 [C] eta_C=0.86 eta_T=0.88 "PROPERTIES" Fluid$='Air' h_1=enthalpy(Fluid$, T=T_1) s_1=entropy(Fluid$, T=T_1, P=P_1) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) h_2=h_1+(h_2s-h_1)/eta_C h_4=enthalpy(Fluid$, T=T_4) s_3=entropy(Fluid$, T=T_3, P=P_2) h_3=enthalpy(Fluid$, T=T_3) h_4s=enthalpy(Fluid$, P=P_1, s=s_3) h_3=h_4+eta_T*(h_3-h_4s) h_hx_out=enthalpy(Fluid$, T=T_hx_out) FluidW$='Steam_iapws' P_w=pressure(FluidW$, T=T_w2, x=1) h_w1=enthalpy(FluidW$, T=T_w1, P=P_w) h_w2=enthalpy(FluidW$, T=T_w2, x=1) "ANALYSIS" "(a)" w_C_in=h_2-h_1 w_T_out=h_3-h_4 w_net=w_T_out-w_C_in m_dot_air=W_dot_net/w_net "(b)" r_bw=w_C_in/w_T_out Q_dot_in=m_dot_air*(h_3-h_2) eta_th=W_dot_net/Q_dot_in "(c)" m_dot_steam*(h_w2-h_w1)=m_dot_air*(h_4-h_hx_out) "energy balance on heat exchanger" "(d)" Q_dot_hx=m_dot_steam*(h_w2-h_w1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-201

eta_utilization=(W_dot_net+Q_dot_hx)/Q_dot_in "utilization efficiency"

10-182 The three processes of an air standard cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the expressions for back work ratio and the thermal efficiency are to be obtained. Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis (a) The P-v and T-s diagrams for this cycle are as shown. (b) The work of compression is found by the first law for process 1-2:

The expansion work is found by ∫ The back work ratio is

Process 1-2 is isentropic; therefore, ( )

and

( )

Process 2-3 is constant pressure; therefore,

10-202

The back work ratio becomes (

)

The cycle thermal efficiency is given by

The efficiency becomes

(d) Determine the value of the back work ratio and efficiency as r goes to unity.

lim r® 1

wcomp wexp

í ü (k - 1)r k - 2 ïüï 1 ïïï 1 r k - 1 - 1ïïï 1 ïíï r k - 1 - 1 ïüï 1 ïíï = lim ì lim k - 1 ý= ì lim ý ì ý k - 1 ïï r - 1 ïï k - 1 ïïî r ® 1 r k - r k - 1 ïïþ k - 1 ïïî r ® 1 kr k - 1 - (k - 1)r k - 2 ïïþ r îï r ® 1 þï ,

2 {

3 }

{ }

These results show that if there is no compression (i.e. r = 1), there can be no expansion and no net work will be done even though heat may be added to the system.

10-183 An ideal regenerative Brayton cycle is considered. The pressure ratio that maximizes the thermal efficiency of the cycle is to be determined, and to be compared with the pressure ratio that maximizes the cycle net work. Analysis Using the isentropic relations, the temperatures at the compressor and turbine exit can be expressed as

( )

10-203

( )

. /

Then, (

)

(

)

(

)

To maximize the net work, we must have ( Solving for

)

gives ( )

Similarly, (

)

(

)

which simplifies to

When , the thermal efficiency becomes maximum when , and must decrease as cannot be less than 1, and the factor

, which is the Carnot efficiency. Therefore, the efficiency is a increases for the fixed values of and . Note that the compression ratio

is always greater than 1 for . Also note that the net work for . This being the case, the pressure ratio for maximum thermal efficiency, which is , is always less than the pressure ratio for maximum work.

10-184 An equation is to be developed for

in terms of k,

and

Assumptions 1. The air-standard assumptions are applicable. 2. Kinetic and potential energy changes are negligible. 3. Air is an ideal gas with constant specific heats. Analysis The temperatures at various points of the dual cycle are given by

10-204

( ) Application of the first law to the two heat addition processes gives

or upon rearrangement

10-185 The effect of variable specific heats on the thermal efficiency of the ideal Otto cycle using air as the working fluid is to be investigated. The percentage of error involved in using constant specific heat values at room temperature for different combinations of compression ratio and maximum cycle temperature is to be determined. Analysis Using EES, the problem is solved as follows: Procedure ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) c_v = 0.718 [kJ/kg-K] k = 1.4 T2 = T[1]*r_comp^(k-1) P2 = P[1]*r_comp^k q_in_23 = c_v*(T[3]-T2) T4 = T[3]*(1/r_comp)^(k-1) q_out_41 = c_v*(T4-T[1]) Eta_th_ConstProp = (1-q_out_41/q_in_23)*Convert(, %) "The Easy Way to calculate the Otto cycle efficiency for constant specific heats is:" Eta_th_easy = (1 - 1/r_comp^(k-1))*Convert(, %) END "Input Data" T[1]=300 [K] P[1]=100 [kPa] T[3] = 1000 [K] r_comp = 12 "Process 1-2 is isentropic compression" s[1]=entropy(air,T=T[1],P=P[1]) s[2]=s[1] T[2]=temperature(air, s=s[2], P=P[2]) P[2]*v[2]/T[2]=P[1]*v[1]/T[1] P[1]*v[1]=R*T[1] R=0.287 [kJ/kg-K] V[2] = V[1]/ r_comp q_12 - w_12 = DELTAu_12 "Conservation of energy for process 1 to 2" q_12 =0 "isentropic process" DELTAu_12=intenergy(air,T=T[2])-intenergy(air,T=T[1]) "Process 2-3 is constant volume heat addition" v[3]=v[2] s[3]=entropy(air, T=T[3], P=P[3]) P[3]*v[3]=R*T[3] q_23 - w_23 = DELTAu_23 "Conservation of energy for process 2 to 3" w_23 =0 "constant volume process" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-205

DELTAu_23=intenergy(air,T=T[3])-intenergy(air,T=T[2]) "Process 3-4 is isentropic expansion" s[4]=s[3] s[4]=entropy(air,T=T[4],P=P[4]) P[4]*v[4]=R*T[4] q_34 -w_34 = DELTAu_34 "Conservation of energy for process 3 to 4" q_34 =0 "isentropic process" DELTAu_34=intenergy(air,T=T[4])-intenergy(air,T=T[3]) "Process 4-1 is constant volume heat rejection" V[4] = V[1] q_41 - w_41 = DELTAu_41 "Conservation of energy for process 4 to 1" w_41 =0 "constant volume process" DELTAu_41=intenergy(air,T=T[1])-intenergy(air,T=T[4]) q_in_total=q_23 q_out_total = -q_41 w_net = w_12+w_23+w_34+w_41 Eta_th=w_net/q_in_total*Convert(, %) Call ConstPropResult(T[1],P[1],r_comp,T[3]:Eta_th_ConstProp,Eta_th_easy) PerCentError = ABS(Eta_th - Eta_th_ConstProp)/Eta_th*Convert(, %) PerCentError [%] 3.604 6.681 9.421 11.64

12 12 12 12

60.8 59.04 57.57 56.42

62.99 62.99 62.99 62.99

1000 1500 2000 2500

10-206

10-186 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine inefficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: "Given" r_p=5 "r_p=P2/P1=P3/P4" T1=300 [K] T3=800 [K] eta_C=0.80 eta_T=0.80 "Properties" Fluid$='air' c_p=CP(Fluid$, T=T) c_v=CV(Fluid$, T=T) T=300 [K] k=c_p/c_v "Analysis" T2s=T1*r_p^((k-1)/k) T4s=T3*(1/r_p)^((k-1)/k) eta_C=(T2s-T1)/(T2-T1) eta_T=(T3-T4)/(T3-T4s) q_in=c_p*(T3-T2) q_out=c_p*(T4-T1) w_net_out=q_in-q_out eta_th=w_net_out/q_in

800 900 1000 1100 1200 1300

17.07 46.7 76.32 105.9 135.6 165.2

0.06044 0.122 0.1579 0.1815 0.1981 0.2105

10-207

1400 1500 1600

194.8 224.4 254.1

0.2201 0.2277 0.2339

0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1 ,

17.07 28.36 39.39 50.19 60.76 71.13 81.31 91.3 101.1 110.8 120.3

0.06044 0.09855 0.1345 0.1686 0.2009 0.2318 0.2614 0.2897 0.3169 0.3432 0.3686

10-208

10-187 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-209

T[3] = 1160 [K] m_dot = 1 [kg/s] Eta_c = 75/100 Eta_t = 82/100 "Inlet conditions" h[1]=ENTHALPY(Air,T=T[1]) s[1]=ENTROPY(Air,T=T[1],P=P[1]) "Compressor anaysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1] "Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(Air,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=ENTHALPY(Air,T=T_s[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2] "process 2-3 is constant pressure" h[3]=ENTHALPY(Air,T=T[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3] "First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(Air,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=ENTHALPY(Air,T=T_s[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c Eta=W_dot_net/Q_dot_in Bwr=W_dot_c/W_dot_t "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) Bwr



0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864 0.9192 0.9491 0.9767

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131 0.1041 0.07272 0.03675

2 4 6 8 10 12 14 16 18 20

1818 4033 5543 6723 7705 8553 9304 9980 10596 11165

1659 2364 2333 2110 1822 1510 1192 877.2 567.9 266.1

3477 6396 7876 8833 9527 10063 10496 10857 11164 11431

16587 14373 12862 11682 10700 9852 9102 8426 7809 7241

10-210

10-188 The effects of pressure ratio, maximum cycle temperature, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a simple Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: Function hFunc(WorkFluid$,T,P) "The EES functions teat helium as a real gas; thus, T and P are needed for helium's enthalpy." IF WorkFluid$ = 'Air' then hFunc:=enthalpy(Air,T=T) ELSE hFunc: = enthalpy(Helium,T=T,P=P) endif END Procedure EtaCheck(Eta_th:EtaError$) If Eta_th < 0 then EtaError$ = 'Why are the net work done and efficiency < 0?' Else EtaError$ = '' END © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-211

P_ratio = 8 T[1] = 300 [K] P[1]= 100 [kPa] T[3] = 800 [K] m_dot = 1 [kg/s] Eta_c = 0.8 Eta_t = 0.8 WorkFluid$ = 'Helium' "Inlet conditions" h[1]=hFunc(WorkFluid$,T[1],P[1]) s[1]=ENTROPY(WorkFluid$,T=T[1],P=P[1]) "Compressor analysis" s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P_ratio=P[2]/P[1]"Definition of pressure ratio - to find P[2]" T_s[2]=TEMPERATURE(WorkFluid$,s=s_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" h_s[2]=hFunc(WorkFluid$,T_s[2],P[2]) Eta_c =(h_s[2]-h[1])/(h[2]-h[1]) "Compressor adiabatic efficiency; Eta_c = W_dot_c_ideal/W_dot_c_actual. " m_dot*h[1] +W_dot_c=m_dot*h[2] "First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "External heat exchanger analysis" P[3]=P[2] "process 2-3 is constant pressure" h[3]=hFunc(WorkFluid$,T[3],P[3]) m_dot*h[2] + Q_dot_in= m_dot*h[3] "First Law for the heat exchanger, assuming W=0, ke=pe=0" "Turbine analysis" s[3]=ENTROPY(WorkFluid$,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P_ratio= P[3] /P[4] T_s[4]=TEMPERATURE(WorkFluid$,s=s_s[4],P=P[4]) "Ts[4] is the isentropic value of T[4] at turbine exit" h_s[4]=hFunc(WorkFluid$,T_s[4],P[4]) "Eta_t = W_dot_t /Wts_dot turbine adiabatic efficiency, Wts_dot > W_dot_t" Eta_t=(h[3]-h[4])/(h[3]-h_s[4]) m_dot*h[3] = W_dot_t + m_dot*h[4] "First Law for the actual compressor, assuming: adiabatic, ke=pe=0" "Cycle analysis" W_dot_net=W_dot_t-W_dot_c Eta_th=W_dot_net/Q_dot_in Call EtaCheck(Eta_th:EtaError$) Bwr=W_dot_c/W_dot_t "The following state points are determined only to produce a T-s plot" T[2]=temperature('air',h=h[2]) T[4]=temperature('air',h=h[4]) s[2]=entropy('air',T=T[2],P=P[2]) s[4]=entropy('air',T=T[4],P=P[4]) Bwr



0.5229 0.6305 0.7038 0.7611 0.8088 0.85 0.8864

0.1 0.1644 0.1814 0.1806 0.1702 0.1533 0.131

2 4 6 8 10 12 14

1818 4033 5543 6723 7705 8553 9304

1659 2364 2333 2110 1822 1510 1192

3477 6396 7876 8833 9527 10063 10496

16587 14373 12862 11682 10700 9852 9102

10-212

0.9192 0.9491 0.9767

0.1041 0.07272 0.03675

16 18 20

9980 10596 11165

877.2 567.9 266.1

10857 11164 11431

8426 7809 7241

10-189 The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Constant specific heats at room temperature are to be used. Analysis Using EES, the problem is solved as follows: T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-213

Eta_c =0.8 Eta_t =0.9 c_p = 1.005 [kJ/kg-K] k = 1.4 "Compressor anaysis" T_s[2] = T[1]*Pratio^((k-1)/k) "T_s[2] is the isentropic value of T[2] at compressor exit" P[2] = Pratio*P[1] Eta_c = w_compisen/w_comp w_compisen = c_p*(T_s[2]-T[1]) "Conservation of energy for the compressor for the isentropic case" w_comp = c_p*(T[2]-T[1]) "Actual compressor work" "External heat exchanger analysis" q_in_noreg = c_p*(T[3] - T[2]) P[3]=P[2] "process 2-3 is constant pressure" "Turbine analysis" P[4] = P[3] /Pratio T_s[4] = T[3]*(1/Pratio)^((k-1)/k) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen w_turbisen = c_p*(T[3] - T_s[4]) w_turb = c_p*(T[3] - T[4]) "Cycle analysis" w_net=w_turb-w_comp Eta_th_noreg=w_net/q_in_noreg*Convert(, %) Bwr=w_comp/w_turb "With the regenerator the heat added in the external heat exchanger is" q_in_withreg = c_p*(T[3] - T[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (T[5]-T[2])/(T[4]-T[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" T[2] + T[4]=T[5] + T[6] P[6]=P[4] Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "Cycle thermal efficiency with regenerator"

0.6 0.65 0.7 0.75 0.8 0.85 0.9

0.9 0.9 0.9 0.9 0.9 0.9 0.9

14.76 20.35 24.59 27.91 30.59 32.79 34.64

13.92 20.54 26.22 31.14 35.44 39.24 42.61

510.9 546.8 577.5 604.2 627.5 648 666.3

541.6 541.6 541.6 541.6 541.6 541.6 541.6

75.4 111.3 142 168.6 192 212.5 230.8

10-190 The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with air as the working fluid is to be investigated. Variable specific heats are to be used. Analysis Using EES, the problem is solved as follows: "Input data" T[3] = 1200 [K] Pratio = 10 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-214

T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 "Compressor isentorpic efficiency" Eta_t =0.9 "Turbine isentropic efficiency" "Isentropic Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s_s[2]=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = Pratio*P[1] s_s[2]=ENTROPY(Air,T=T_s[2],P=P[2]) "T_s[2] is the isentropic value of T[2] at compressor exit" Eta_c = w_compisen/w_comp h[1] + w_compisen = h_s[2] "Conservation of energy for the compressor for the isentropic case" h[1]=ENTHALPY(Air,T=T[1]) h_s[2]=ENTHALPY(Air,T=T_s[2]) "Actual compressor analysis:" h[1] + w_comp = h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,T=T[2], P=P[2]) "External heat exchanger analysis" h[2] + q_in_noreg = h[3] h[3]=ENTHALPY(Air,T=T[3]) P[3]=P[2] "process 2-3 is constant pressure" "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s_s[4]=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[3] /Pratio s_s[4]=ENTROPY(Air,T=T_s[4],P=P[4]) "T_s[4] is the isentropic value of T[4] at turbine exit" Eta_t = w_turb /w_turbisen "turbine adiabatic efficiency, w_turbisen > w_turb" h[3] = w_turbisen + h_s[4] h_s[4]=ENTHALPY(Air,T=T_s[4]) "Actual Turbine analysis:" h[3] = w_turb + h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,T=T[4], P=P[4]) "Cycle analysis" w_net=w_turb-w_comp Eta_th_noreg=w_net/q_in_noreg*Convert(, %) Bwr=w_comp/w_turb "With the regenerator the heat added in the external heat exchanger is" h[5] + q_in_withreg = h[3] h[5]=ENTHALPY(Air, T=T[5]) s[5]=ENTROPY(Air,T=T[5], P=P[5]) P[5]=P[2] "The regenerator effectiveness gives h[5] and thus T[5] as:" Eta_reg = (h[5]-h[2])/(h[4]-h[2]) "Energy balance on regenerator gives h[6] and thus T[6] as:" h[2] + h[4]=h[5] + h[6] h[6]=ENTHALPY(Air, T=T[6]) s[6]=ENTROPY(Air,T=T[6], P=P[6]) P[6]=P[4] Eta_th_withreg=w_net/q_in_withreg*Convert(, %) "Cycle thermal efficiency with regenerator" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-215

"The following data is used to complete the Array Table for plotting purposes." s_s[1]=s[1] T_s[1]=T[1] s_s[3]=s[3] T_s[3]=T[3] s_s[5]=ENTROPY(Air,T=T[5],P=P[5]) T_s[5]=T[5] s_s[6]=s[6] T_s[6]=T[6]

0.6 0.65 0.7 0.75 0.8 0.85 0.9

0.9 0.9 0.9 0.9 0.9 0.9 0.9

14.76 20.35 24.59 27.91 30.59 32.79 34.64

13.92 20.54 26.22 31.14 35.44 39.24 42.61

510.9 546.8 577.5 604.2 627.5 648 666.3

541.6 541.6 541.6 541.6 541.6 541.6 541.6

75.4 111.3 142 168.6 192 212.5 230.8

10-216

10-217

10-191 The effects of pressure ratio, maximum cycle temperature, regenerator effectiveness, and compressor and turbine efficiencies on the net work output per unit mass and the thermal efficiency of a regenerative Brayton cycle with helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: c_p = 5.1926 [kJ/kg-K] k = 1.667 T[3] = 1200 [K] Pratio = 10 T[1] = 300 [K] P[1]= 100 [kPa] Eta_reg = 1.0 Eta_c =0.8 Eta_t =0.9 "Compressor anaysis" T_s[2] = T[1]*Pratio^((k-1)/k) P[2] = Pratio*P[1] Eta_c = w_compisen/w_comp w_compisen = c_p*(T_s[2]-T[1]) w_comp = c_p*(T[2]-T[1]) "External heat exchanger analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-218

q_in_noreg = c_p*(T[3] - T[2]) P[3]=P[2] "Turbine analysis" P[4] = P[3] /Pratio T_s[4] = T[3]*(1/Pratio)^((k-1)/k) Eta_t = w_turb /w_turbisen w_turbisen = c_p*(T[3] - T_s[4]) w_turb = c_p*(T[3] - T[4]) "Cycle analysis" w_net=w_turb-w_comp Eta_th_noreg=w_net/q_in_noreg*Convert(, %) Bwr=w_comp/w_turb "With the regenerator the heat added in the external heat exchanger is" q_in_withreg = c_p*(T[3] - T[5]) P[5]=P[2] Eta_reg = (T[5]-T[2])/(T[4]-T[2]) T[2] + T[4]=T[5] + T[6] P[6]=P[4] "Cycle thermal efficiency with regenerator" Eta_th_withreg=w_net/q_in_withreg*Convert(, %)

0.6 0.65 0.7 0.75 0.8 0.85 0.9

0.9 0.9 0.9 0.9 0.9 0.9 0.9

14.76 20.35 24.59 27.91 30.59 32.79 34.64

13.92 20.54 26.22 31.14 35.44 39.24 42.61

510.9 546.8 577.5 604.2 627.5 648 666.3

541.6 541.6 541.6 541.6 541.6 541.6 541.6

75.4 111.3 142 168.6 192 212.5 230.8

10-192 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and air as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: c_p = 1.005 [kJ/kg-K] k = 1.4 Nstages = 1 "Nstages is the number of compression and expansion stages" T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 Eta_c =1.0 Eta_t =1.0 R_p = Pratio^(1/Nstages) "Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 Eta_c = w_compisen/w_comp © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-219

w_compisen = c_p*(T_2s-T_1) w_comp = c_p*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp "The heat added in the external heat exchanger + the reheat between turbines is" q_in_total = c_p*(T_6 - T_5) +(Nstages - 1)*c_p*(T_8 - T_7) "Reheat is assumed to occur until:" T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p T_7s = T_6*(1/R_p)^((k-1)/k) Eta_t = w_turb /w_turbisen w_turbisen = c_p*(T_6 - T_7s) w_turb = c_p*(T_6 - T_7) w_turb_total = Nstages*w_turb "Cycle analysis" w_net=w_turb_total-w_comp_total Bwr=w_comp/w_turb P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 T_4 + T_9=T_5 + T_10 Eta_th_regenerative=w_net/q_in_total*Convert(, %) "Cycle thermal efficiency with regenerator" "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %)

Nstages 75 75 75 75 75 75 75 75

49.15 64.35 68.32 70.14 72.33 73.79 74.05 74.18

1 2 3 4 7 15 19 22

10-220

10-193 The effect of the number of compression and expansion stages on the thermal efficiency of an ideal regenerative Brayton cycle with multistage compression and expansion and helium as the working fluid is to be investigated. Analysis Using EES, the problem is solved as follows: c_p = 5.1926 [kJ/kg-K] k = 1.667 Nstages = 1 "Nstages is the number of compression and expansion stages" T_6 = 1200 [K] Pratio = 12 T_1 = 300 [K] P_1= 100 [kPa] Eta_reg = 1.0 "regenerator effectiveness" Eta_c =1.0 "Compressor isentorpic efficiency" Eta_t =1.0 "Turbine isentropic efficiency" R_p = Pratio^(1/Nstages) "Compressor anaysis" T_2s = T_1*R_p^((k-1)/k) P_2 = R_p*P_1 Eta_c = w_compisen/w_comp w_compisen = c_p*(T_2s-T_1) w_comp = c_p*(T_2 - T_1) "Since intercooling is assumed to occur such that T_3 = T_1 and the compressors have the same pressure ratio, the work input to each compressor is the same. The total compressor work is:" w_comp_total = Nstages*w_comp q_in_total = c_p*(T_6 - T_5) +(Nstages - 1)*c_p*(T_8 - T_7) T_8 = T_6 "Turbine analysis" P_7 = P_6 /R_p T_7s = T_6*(1/R_p)^((k-1)/k) Eta_t = w_turb /w_turbisen w_turbisen = c_p*(T_6 - T_7s) w_turb = c_p*(T_6 - T_7) w_turb_total = Nstages*w_turb © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-221

"Cycle analysis" w_net=w_turb_total-w_comp_total Bwr=w_comp/w_turb P_4=P_2 P_5=P_4 P_6=P_5 T_4 = T_2 Eta_reg = (T_5 - T_4)/(T_9 - T_4) T_9 = T_7 T_4 + T_9=T_5 + T_10 "Cycle thermal efficiency with regenerator" Eta_th_regenerative=w_net/q_in_total*Convert(, %) "The efficiency of the Ericsson cycle is the same as the Carnot cycle operating between the same max and min temperatures, T_6 and T_1 for this problem." Eta_th_Ericsson = (1 - T_1/T_6)*Convert(, %)

Nstages 75 75 75 75 75 75 75 75

32.43 58.9 65.18 67.95 71.18 73.29 73.66 73.84

1 2 3 4 7 15 19 22

Fundamentals of Engineering (FE) Exam Problems 10-194 For specified limits for the maximum and minimum temperatures, the ideal cycle with the lowest thermal efficiency is (a) Carnot © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-222

(b) Stirling (c) Ericsson (d) Otto (e) All are the same Answer (d) Otto

10-195 A Carnot cycle operates between the temperatures limits of 300 K and 2000 K, and produces 400 kW of net power. The rate of entropy change of the working fluid during the heat addition process is (a) 0 (b) (c) (d) (e) Answer (d) Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=300 [K] TH=2000 [K] Wnet=400 [kJ/s] Wnet= (TH-TL)*DS "Some Wrong Solutions with Common Mistakes:" W1_DS = Wnet/TH "Using TH instead of TH-TL" W2_DS = Wnet/TL "Using TL instead of TH-TL" W3_DS = Wnet/(TH+TL) "Using TH+TL instead of TH-TL"

10-196 An Otto cycle with air as the working fluid has a compression ratio of 10.4. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 10% (b) 39% (c) 61% (d) 79% (e) 82% Answer (c) 61% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). r=10.4 k=1.4 Eta_Otto=1-1/r^(k-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/r "Taking efficiency to be 1/r" W2_Eta = 1/r^(k-1) "Using incorrect relation" W3_Eta = 1-1/r^(k1-1); k1=1.667 "Using wrong k value" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-223

10-197 Air in an ideal Diesel cycle is compressed from 2 L to 0.13 L, and then it expands during the constant pressure heat addition process to 0.30 L. Under cold air standard conditions, the thermal efficiency of this cycle is (a) 41% (b) 59% (c) 66% (d) 70% (e) 78% Answer (b) 59% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=2 [L] V2= 0.13 [L] V3= 0.30 [L] k=1.4 r=V1/V2 rc=V3/V2 Eta_Diesel=1-(1/r^(k-1))*(rc^k-1)/k/(rc-1) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-(1/r1^(k-1))*(rc^k-1)/k/(rc-1); r1=V1/V3 "Wrong r value" W2_Eta = 1-Eta_Diesel "Using incorrect relation" W3_Eta = 1-(1/r^(k1-1))*(rc^k1-1)/k1/(rc-1); k1=1.667 "Using wrong k value" W4_Eta = 1-1/r^(k-1) "Using Otto cycle efficiency"

10-198 Helium gas in an ideal Otto cycle is compressed from 20C and 2.5 L to 0.25 L, and its temperature increases by an additional 700C during the heat addition process. The temperature of helium before the expansion process is (a) 1790C (b) 2060C (c) 1240C (d) 620C (e) 820C Answer (a) 1790C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V1=2.5 V2=0.25 T1=(20+273) [K] DELTAT=700 [K] k=1.667 r=V1/V2 T2=T1*r^(k-1) T3=T2+DELTAT-273 "Some Wrong Solutions with Common Mistakes:" W1_T3 =T22+700-273; T22=T1*r^(k1-1); k1=1.4 "Using wrong k value" W2_T3 = T3+273 "Using K instead of C" W3_T3 = T1+700-273 "Disregarding temp rise during compression" W4_T3 = T222+700-273; T222=(T1-273)*r^(k-1) "Using C for T1 instead of K" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-224

10-199 In an ideal Otto cycle, air is compressed from and 2.2 L to 0.26 L, and the net work output of the cycle is . The mean effective pressure (MEP) for this cycle is (a) 612 kPa (b) 599 kPa (c) 528 kPa (d) 416 kPa (e) 367 kPa Answer (b) 599 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho1=1.20 [kg/m^3] V1=2.2 [L] V2=0.26 [L] w_net=440 [kJ/kg] k=1.4 m=rho1*V1/1000 Wtotal=m*w_net MEP=Wtotal/((V1-V2)/1000) "Some Wrong Solutions with Common Mistakes:" W1_MEP = w_net/((V1-V2)/1000) "Disregarding mass" W2_MEP = Wtotal/(V1/1000) "Using V1 instead of V1-V2" W3_MEP = (rho1*V2/1000)*w_net/((V1-V2)/1000); "Finding mass using V2 instead of V1" W4_MEP = Wtotal/((V1+V2)/1000) "Adding V1 and V2 instead of subtracting"

10-200 Air enters a turbojet engine at thrust developed by the engine is (a) 2.5 kN (b) 5.0 kN (c) 7.5 kN (d) 10 kN (e) 12.5 kN

at a rate of

, and exits at

relative to the aircraft. The

Answer (c) 7.5 kN Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Vel1=320 [m/s] Vel2=570 [m/s] m= 30 [kg/s] Thrust=m*(Vel2-Vel1)/1000 "Some Wrong Solutions with Common Mistakes:" W1_thrust = (Vel2-Vel1)/1000 "Disregarding mass flow rate" W2_thrust = m*Vel2/1000 "Using incorrect relation"

10-201 In an ideal Brayton cycle, air is compressed from 95 kPa and 25C to 1400 kPa. Under cold air standard conditions, the thermal efficiency of this cycle is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-225

(a) 40% (b) 44% (c) 49% (d) 54% (e) 58% Answer (d) 54% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=95 [kPa] P2=1400 [kPa] T1=(25+273) [K] k=1.4 rp=P2/P1 Eta_Brayton=1-1/rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1/rp "Taking efficiency to be 1/rp" W2_Eta = 1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-1/rp^((k1-1)/k1); k1=1.667 "Using wrong k value"

10-202 In an ideal Brayton cycle, air is compressed from 100 kPa and 25C to 1 MPa, and then heated to 927C before entering the turbine. Under cold air standard conditions, the air temperature at the turbine exit is (a) 349C (b) 426C (c) 622C (d) 733C (e) 825C Answer (a) 349C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 [kPa] P2=1000 [kPa] T1=(25+273) [K] T3=(927+273) [K] k=1.4 rp=P2/P1 T4=T3*(1/rp)^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp "Using wrong relation" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T1+800-273 "Disregarding temp rise during compression"

10-203 Consider an ideal Brayton cycle executed between the pressure limits of 1200 kPa and 100 kPa and temperature limits of 20C and 1000C with argon as the working fluid. The net work output of the cycle is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-226

(a) (b) (c) (d) (e) Answer (c) Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 [kPa] P2=1200 [kPa] T1=(20+273) [K] T3=(1000+273) [K] k=1.667 cp=0.5203 [kJ/kg-K] cv=0.3122 [kJ/kg-K] rp=P2/P1 T2=T1*rp^((k-1)/k) q_in=cp*(T3-T2) Eta_Brayton=1-1/rp^((k-1)/k) w_net=Eta_Brayton*q_in "Some Wrong Solutions with Common Mistakes:" W1_wnet = (1-1/rp^((k-1)/k))*qin1; qin1=cv*(T3-T2) "Using cv instead of cp" W2_wnet = (1-1/rp^((k-1)/k))*qin2; qin2=1.005*(T3-T2) "Using cp of air instead of argon" W3_wnet = (1-1/rp^((k1-1)/k1))*cp*(T3-T22); T22=T1*rp^((k1-1)/k1); k1=1.4 "Using k of air instead of argon" W4_wnet = (1-1/rp^((k-1)/k))*cp*(T3-T222); T222=(T1-273)*rp^((k-1)/k) "Using C for T1 instead of K"

10-204 An ideal Brayton cycle has a net work output of and a backwork ratio of 0.4. If both the turbine and the compressor had an isentropic efficiency of 85%, the net work output of the cycle would be (a) (b) (c) (d) (e) Answer (b) Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). wcomp/wturb=0.4 wturb-wcomp=150 [kJ/kg] Eff=0.85 w_net=Eff*wturb-wcomp/Eff "Some Wrong Solutions with Common Mistakes:" W1_wnet = Eff*wturb-wcomp*Eff "Making a mistake in Wnet relation" W2_wnet = (wturb-wcomp)/Eff "Using a wrong relation" W3_wnet = wturb/eff-wcomp*Eff "Using a wrong relation"

10-227

10-205 In an ideal Brayton cycle with regeneration, argon gas is compressed from 100 kPa and 25C to 400 kPa, and then heated to 1200C before entering the turbine. The highest temperature that argon can be heated in the regenerator is (a) 246C (b) 846C (c) 689C (d) 368C (e) 573C Answer (e) 573C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=100 [kPa] P2=400 [kPa] T1=(25+273) [K] T3=(1200+273) [K] k=1.667 "The highest temperature that argon can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2=T1*rp^((k-1)/k) T4=T3/rp^((k-1)/k)-273 "Some Wrong Solutions with Common Mistakes:" W1_T4 = T3/rp "Using wrong relation" W2_T4 = (T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_T4 = T4+273 "Using K instead of C" W4_T4 = T2-273 "Taking compressor exit temp as the answer"

10-206 In an ideal Brayton cycle with regeneration, air is compressed from 80 kPa and 10C to 400 kPa and 175C, is heated to 450C in the regenerator, and then further heated to 1000C before entering the turbine. Under cold air standard conditions, the effectiveness of the regenerator is (a) 33% (b) 44% (c) 62% (d) 77% (e) 89% Answer (d) 77% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=80 [kPa] P2=400 [kPa] T1=(10+273) [K] T2=(175+273) [K] T3=(1000+273) [K] T5=(450+273) [K] k=1.4 "The highest temperature that the gas can be heated in the regenerator is the turbine exit temperature," rp=P2/P1 T2check=T1*rp^((k-1)/k) "Checking the given value of T2. It checks." © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-228

T4=T3/rp^((k-1)/k) Effective=(T5-T2)/(T4-T2) "Some Wrong Solutions with Common Mistakes:" W1_eff = (T5-T2)/(T3-T2) "Using wrong relation" W2_eff = (T5-T2)/(T44-T2); T44=(T3-273)/rp^((k-1)/k) "Using C instead of K for T3" W3_eff = (T5-T2)/(T444-T2); T444=T3/rp "Using wrong relation for T4"

10-207 Consider a gas turbine that has a pressure ratio of 6 and operates on the Brayton cycle with regeneration between the temperature limits of 20C and 900C. If the specific heat ratio of the working fluid is 1.3, the highest thermal efficiency this gas turbine can have is (a) 38% (b) 46% (c) 62% (d) 58% (e) 97% Answer (c) 62% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.3 rp=6 T1=(20+273) [K] T3=(900+273) [K] Eta_regen=1-(T1/T3)*rp^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" W1_Eta = 1-((T1-273)/(T3-273))*rp^((k-1)/k) "Using C for temperatures instead of K" W2_Eta = (T1/T3)*rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k1-1)/k1); k1=1.4 "Using wrong k value (the one for air)"

10-208 An ideal gas turbine cycle with many stages of compression and expansion and a regenerator of 100 percent effectiveness has an overall pressure ratio of 10. Air enters every stage of compressor at 290 K, and every stage of turbine at 1200 K. The thermal efficiency of this gas-turbine cycle is (a) 36% (b) 40% (c) 52% (d) 64% (e) 76% Answer (e) 76% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rp=10 T1=290 [K] T3=1200 [K] Eff=1-T1/T3 "Some Wrong Solutions with Common Mistakes:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-229

W1_Eta = 100 k=1.4 W2_Eta = 1-1/rp^((k-1)/k) "Using incorrect relation" W3_Eta = 1-(T1/T3)*rp^((k-1)/k) "Using wrong relation" W4_Eta = T1/T3 "Using wrong relation"

10-209 … 10-215 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

10-230

Chapter 11 VAPOR AND COMBINED POWER CYCLES

10-231

CYU - Check Your Understanding CYU 11-2 ■ Check Your Understanding CYU 11-2.1 Select the equation which cannot be used to determine the pump work input or turbine work output in the ideal Rankine cycle. (a) win = h2 - h1 (b) win = v1 ( P2 - P1 ) (c) wout = h3 - h4 (d) wout = v 3 ( P3 - P4 ) Answer: (d) w out = v3(P3 - P4 ) CYU 11-2.2 A Rankine cycle operates between a boiler exit temperature of 900 K and a sink temperature of 300 K. Heat is transferred to the boiler at a rate of 100 kW and rejected from the condenser at a rate of 60 kW. What is the thermal efficiency of this Rankine cycle? (a) 90% (b) 67% (c) 60% (d) 40% (e) 33% Answer: (d) 40% eta th  1  QL / QH  1  (60 kW) / (100 kW)  0.40 CYU 11-2.3 In a Rankine cycle, heat is supplied to the boiler at a rate of 100 kW and rejected from the condenser at a rate of 60 kW. If the pump power input is 5 kW, what is the turbine power output? (a) 35 kW (b) 40 kW (c) 45 kW (d) 60 kW (e) 70 kW Answer: (c) 45 kW Wnet  QH  QL  100  60  40 kW Wturb  Wnet  Wpump  40  5  45 kW

CYU 11-4 ■ Check Your Understanding CYU 11-4.1 Which of the following does not help increase the efficiency of the Rankine cycle? (a) Using steam at 3 MPa and 500C at the turbine inlet instead of 3 MPa and 400C. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-232

(b) Using steam at 2 MPa and 500C at the turbine inlet instead of 3 MPa and 500C. (c) Using a cool lake at 15C as the cooling fluid in the condenser instead of ambient air at 25C. (d) Steam exiting the turbine and entering the condenser at 10 kPa instead of 20 kPa. Answer: (b) Using steam at 2 MPa and 500C at the turbine inlet instead of steam at 3 MPa and 500C. CYU 11-4.2 Which modifications to the Rankine cycle can cause higher moisture content at the turbine exit? I. Increasing the turbine inlet temperature for a fixed boiler pressure II. Increasing the boiler pressure for a fixed turbine inlet temperature III. Decreasing the condenser pressure (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (d) II and III

CYU 11-5 ■ Check Your Understanding CYU 11-5.1 Increasing the boiler pressure in the Rankine cycle increases the thermal efficiency, but it can cause higher moisture content at the turbine exit. How can this problem be avoided? (a) Regeneration (b) Reheating (c) Intercooling (d) Cogeneration (e) Lowering the condenser pressure Answer: (b) Reheating CYU 11-5.2 What is the approximate optimum reheat pressure in an ideal steam reheat Rankine cycle if the boiler pressure is 8 MPa and the condenser pressure is 100 kPa? (a) 4 MPa (b) 8/3 MPa (c) 81/ 2 MPa (d) 2 MPa (e) 0.2 MPa Answer: (d) 2 MPa Preheat  0.25Pmax  (0.25)(8 MPa)  2 MPa

10-233

CYU 11-6 ■ Check Your Understanding CYU 11-6.1 Which modification to the Rankine cycle increases the average heat-addition temperature by incorporating feedwater heating? (a) Regeneration (b) Reheating (c) Intercooling (d) Cogeneration Answer: (a) Regeneration CYU 11-6.2 Which of the following is not among the benefits of regeneration in the Rankine cycle? (a) It increases the thermal efficiency of the cycle. (b) It decreases the percentage of moisture at the turbine exit. (c) It removes the air that leaks in at the condenser. (d) It helps control the large volume flow rate of the steam at the final stages of the turbine. Answer: (b) It decreases the percentage of moisture at the turbine exit. CYU 11-6.3 In an ideal regenerative Rankine cycle with an open feedwater heater, steam enters the turbine at state 5. Some steam extracted from the turbine at state 6 is used to heat the feedwater. The remaining steam continues to expand in the turbine to state 7. The fraction of steam extracted from the turbine for feedwater heating is y. Which equation can be used to determine the work output from the turbine per unit mass through the boiler? (a) w out = h5 - h6 + h6 - h7 (b) w out = (h5 - h6 )+ y(h6 - h7 ) (c) w out = (1- y)(h5 - h6 )+ y(h6 - h7 ) (d) w out = y(h5 - h6 )+ (1- y)(h6 - h7 ) (e) w out = (h5 - h6 )+ (1- y)(h6 - h7 ) Answer: (e) w out = (h5 - h6 )+ (1- y)(h6 - h7 ) CYU 11-6.4 In an ideal regenerative Rankine cycle with an open feedwater heater, steam enters the turbine at state 5. Some steam extracted from the turbine at state 6 is used to heat the feedwater. The remaining steam continues to expand in the turbine to state 7 and it is condensed to state 1. The fraction of steam extracted from the turbine for feedwater heating is y. Which equation below can be used to determine the heat rejection in the condenser per unit mass through the boiler? (a) qout = (1- y)(h7 - h1) (b) qout = y(h7 - h1) (c) qout = (1- y)(h7 - h1)+ y(h6 - h1) (d) qout = y(h7 - h1)+ (1- y)(h6 - h1) (e) qout = (h7 - h1)+ y(h6 - h1) Answer: (a) qout = (1- y)(h7 - h1)

10-234

CYU 11-7 ■ Check Your Understanding CYU 11-7.1 In a Rankine cycle, heat is supplied in the boiler at a rate of 400 kW whose exergy content is 200 kW. If waste heat is rejected from the condenser to the environment at,30ºC at a rate of 300 kW, what is the second-law efficiency of this cycle? (a) 75% (b) 50% (c) 30% (d) 25% (e) 20% Answer: (b) 50% Wnet  QH  QL  400  300  100 kW eta II  Wnet / Xexpended  100 / 200  0.5

CYU 11-7.2 The analysis of a heat engine operating on the Rankine cycle indicates a heat input of 100 kJ / kg with an exergy content of 50 kJ / kg . If the total exergy destruction is 30 kJ / kg , what is the net work output of this cycle? (a) 100 kJ / kg (b) 50 kJ / kg (c) 30 kJ / kg (d) 20 kJ / kg (e) 10 kJ / kg Answer: (d) 20 kJ / kg w net  x expended  x destroyed  50  30  20 kJ / kg

CYU 11-8 ■ Check Your Understanding CYU 11-8.1 A cogeneration plant with a heat input of 1000 kW in the boiler produces 300 kW power and 250 kW process. What is the utilization factor of this plant? (a) 100% (b) 55% (c) 30% (d) 25% (e) 15% Answer: (b) 55%

eps u   Wnet  Qprocess  / Qin  (300  250) / 1000  0.55

10-235

CYU 11-8.2 An ideal cogeneration plant produces 150 kW power together with 350 kW process heat What is the rate of heat input to the plant? (a) 500 kW (b) 350 kW (c) 200 kW (d) 150 kW (e) 100 kW Answer: (a) 500 kW epsu  1 for an ideal cogeneration plant eps u   Wnet  Qprocess  / Qin

1  (150  350) / Qin Qin  500 kW

CYU 11-9 ■ Check Your Understanding CYU 11-9.1 Fill in the blanks: The common configuration of the combined cycle uses the _________ cycle as the topping cycle and __________ cycle as the bottoming cycle. (a) Rankine, Otto (b) Brayton, Stirling (c) Carnot, Rankine (d) Rankine, Brayton (e) Brayton, Rankine Answer: (e) Brayton, Rankine CYU 11-9.2 Of all the heat engines (including thermal power plants) currently operating in the real word, the one with the highest thermal efficiency is most probably operating on the (a) Rankine cycle (b) Brayton cycle (c) Combined cycle (d) Otto cycle (e) Diesel cycle Answer: (c) Combined cycle

10-236

PROBLEMS Carnot Vapor Cycle 11-1C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases.

11-2C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%.

11-3 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH  250 C  523 K and TL  Tsat @ 20 kPa  60.06 C  333.1 K , the thermal efficiency becomes

hth, C = 1-

TL 333.1 K = 1= 0.3632 = 36.3% TH 523 K

(b) The heat supplied during this cycle is simply the enthalpy of vaporization, Thus,

qin = h fg @ 250 oC = 1715.3 kJ/kg qout = qL =

æ333.1 K ÷ ö TL qin = çç (1715.3 kJ/kg) = 1092.3 kJ/kg ÷ ÷ ç è 523 K ø TH

wnet = hth qin = (0.3632)(1715.3 kJ/kg) = 623.0 kJ/kg

10-237

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=250 [C] x[1]=0 x[2]=1 P[3]=20 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" T_H=T[1]+273 T[3]=temperature(Fluid$, P=P[3], x=1) T_L=T[3]+273 Eta_th_C=1-T_L/T_H "(b)" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) T[2]=T[1] h[2]=enthalpy(Fluid$, T=T[2], x=x[2]) q_in=h[2]-h[1] q_out=q_in*T_L/T_H "(c)" w_net=Eta_th_C*q_in

11-4 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH  250 C  523 K and TL  Tsat @ 10 kPa  45.81 C  318.8 K , the thermal efficiency becomes

hth, C = 1-

TL 318.8 K = 1= 39.04% TH 523 K

(b) The heat supplied during this cycle is simply the enthalpy of vaporization, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-238

Thus,

qin = h fg @ 250°C = 1715.3 kJ/kg

qout = qL =

æ318.8 K ÷ ö TL qin = çç (1715.3 kJ/kg ) = 1045.6 kJ/kg ÷ ÷ ç è ø TH 523 K

wnet = hth qin = (0.3904)(1715.3kJ/kg ) = 669.7 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=250 [C] x[1]=0 x[2]=1 P[3]=10 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" T_H=T[1]+273 T[3]=temperature(Fluid$, P=P[3], x=1) T_L=T[3]+273 Eta_th_C=1-T_L/T_H "(b)" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) T[2]=T[1] h[2]=enthalpy(Fluid$, T=T[2], x=x[2]) q_in=h[2]-h[1] q_out=q_in*T_L/T_H "(c)" w_net=Eta_th_C*q_in

11-5 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-239

1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The thermal efficiency is determined from

hth, C = 1-

TL 60 + 273 K = 1= 46.5% TH 350 + 273 K

(b) Note that s2  s3  s f  x3 s fg

 0.8313  0.891 7.0769  7.1368 kJ / kg  K

Thus, ü ïï ý P2 @1.40 MPa (Table A-6) s2 = 7.1368 kJ/kg ×Kïïþ

T2 = 350°C

s4 = s f + x4 s fg = 0.8313 + (0.1)(7.0769) = 1.5390 kJ/kg ×K wnet = Area = (TH - TL )(s3 - s4 ) = (350 - 60)(7.1368 - 1.5390) = 1623 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=350 [C] T[3]=60 [C] x[3]=0.891 x[4]=0.1 "Analysis" Fluid$='steam_iapws' "(a)" T_H=T[1]+273 T_L=T[3]+273 Eta_th_C=1-T_L/T_H "(b)" s[3]=entropy(Fluid$, T=T[3], x=x[3]) s[2]=s[3] T[2]=T[1] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-240

P[2]=pressure(Fluid$, T=T[2], s=s[2]) P_2=P[2] "(c)" T[4]=T[3] s[4]=entropy(Fluid$, T=T[4], x=x[4]) w_net=(T_H-T_L)*(s[3]-s[4])

11-6E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We note that and TH = Tsat @300 psia = 417.4°F = 877.4 R TL = Tsat @ 20 psia = 227.9°F = 687.9 R

hth, C = 1-

TL 687.9 R = 1= 0.2159 = 21.6% TH 877.4 R

(b) Noting that s4  s1  s f @ 300 psia  0.58818 Btu / lbm  R,

x4 =

s4 - s f s fg

0.58818 - 0.3358 = 0.181 1.3961

h2 = h f + x2 h fg = 393.94 + (0.95)(809.41) = 1162.9 Btu/lbm Thus, and qin = h2 - h1 = 1162.9 - 393.94 = 768.9 Btu/lbm

wnet = hth qin = (0.2159)(768.9 Btu/lbm) = 166 Btu / lbm

10-241

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=300 [psia] x[1]=0 x[2]=0.95 P[3]=20 [psia] "Analysis" Fluid$='steam_iapws' "(a)" T[1]=temperature(Fluid$, P=P[1], x=x[1]) T_H=T[1]+460 T[3]=temperature(Fluid$, P=P[3], x=1) T_L=T[3]+460 Eta_th_C=1-T_L/T_H "(b)" s[4]=entropy(Fluid$, P=P[1], x=x[1]) P[4]=P[3] x[4]=quality(Fluid$, P=P[4], s=s[4]) x_4=x[4] "(c)" h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) P[1]=P[2] h[2]=enthalpy(Fluid$, P=P[2], x=x[2]) q_in=h[2]-h[1] w_net=Eta_th_C*q_in s4_f=entropy(Fluid$, P=P[4], x=0) s4_g=entropy(Fluid$, P=P[4], x=1) s4_fg=s4_g-s4_f h2_f=enthalpy(Fluid$, P=P[2], x=0) h2_g=enthalpy(Fluid$, P=P[2], x=1) h2_fg=h2_g-h2_f

The Simple Rankine Cycle 11-7C The four processes that make up the simple ideal cycle are (1) Isentropic compression in a pump, (2) P = constant heat addition in a boiler, (3) Isentropic expansion in a turbine, and (4) P = constant heat rejection in a condenser.

11-8C A simple ideal Rankine cycle with fixed turbine inlet conditions is considered. The effect of lowering the condenser pressure on various quantities are: Heat rejected decreases; everything else increases.

11-9C A simple ideal Rankine cycle with fixed turbine inlet temperature and condenser pressure is considered. The effect of increasing the boiler pressure on various quantities are: Heat supplied and heat rejected decrease; everything else increases.

11-10C A simple ideal Rankine cycle with fixed boiler and condenser pressures is considered. The effect of superheating the steam to a higher temperature on various quantities are The pump work remains the same, the moisture content decreases, everything else increases. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-242

11-11C The actual vapor power cycles differ from the idealized ones in that the actual cycles involve friction and pressure drops in various components and the piping, and heat loss to the surrounding medium from these components and piping.

11-12C The boiler exit pressure will be (a) lower than the boiler inlet pressure in actual cycles, and (b) the same as the boiler inlet pressure in ideal cycles.

11-13C We would reject this proposal because wturb  h1  h2  qout , and any heat loss from the steam will adversely affect the turbine work output.

11-14C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than the temperature of the cooling water.

11-15 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The maximum thermal efficiency of the cycle for a given quality at the turbine exit is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis For maximum thermal efficiency, the quality at state 4 would be at its minimum of 85% (most closely approaches the Carnot cycle), and the properties at state 4 would be (Table A-5)

P4 = 30 kPa x4 = 0.85

ïüï h4 = h f + x4 h fg = 289.27 + (0.85)(2335.3) = 2274.3 kJ/kg ý ïïþ s4 = s f + x4 s fg = 0.9441 + (0.85)(6.8234) = 6.7440 kJ/kg ×K

Since the expansion in the turbine is isentropic,

ïüï ý h = 3115.5 kJ/kg s3 = s4 = 6.7440 kJ/kg ×K ïïþ 3

P3 = 3000 kPa

Other properties are obtained as follows (Tables A-4, A-5, and A-6), h1 = h f @ 30 kPa = 289.27 kJ/kg

v1 = v f @ 30 kPa = 0.001022 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001022 m3 /kg)(3000 - 30)kPa çç ÷ çè1 kPa ×m3 ÷ ø

= 3.04 kJ/kg h2 = h1 + wp, in = 289.27 + 3.04 = 292.31 kJ/kg

Thus,

qin = h3 - h2 = 3115.5 - 292.31 = 2823.2 kJ/kg qout = h4 - h1 = 2274.3 - 289.27 = 1985.0 kJ/kg and the thermal efficiency of the cycle is

hth = 1-

qout 1985.0 = 1= 0.297 = 29.7% qin 2823.2

10-243

11-16 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine and consumed by the pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ö ÷ = (0.001017 m3 /kg)(4000 - 20)kPa çç ÷ ÷ çè1 kPa ×m3 ø

= 4.05 kJ/kg h2 = h1 + wp, in = 251.42 + 4.05 = 255.47 kJ/kg

ïü ïý ïïþ

P3 = 4000 kPa T3 = 700°C P4 = 20 kPa s4 = s3

ïü ïý ïïþ

h3 = 3906.3 kJ/kg s3 = 7.6214 kJ/kg ×K x4 =

s4 - s f s fg

7.6214 - 0.8320 = 0.9596 7.0752

h4 = h f + x4 h fg = 251.42 + (0.9596)(2357.5) = 2513.7 kJ/kg

10-244

The power produced by the turbine and consumed by the pump are

WT,out = m(h3 - h4 ) = (50 kg/s)(3906.3 - 2513.7)kJ/kg = 69, 630 kW WP,in = mwP,in = (50 kg/s)(4.05 kJ/kg) = 203 kW

11-17 A simple ideal Rankine cycle with water as the working fluid is considered. The work output from the turbine, the heat addition in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), P1 = Psat @ 40°C = 7.385 kPa P2 = Psat @ 250°C = 3976 kPa h1 = h f @ 40°C = 167.53 kJ/kg

v1 = v f @ 40°C = 0.001008 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ö ÷ = (0.001008 m3 /kg)(3976 - 7.385)kPa çç ÷ ÷ çè1 kPa ×m3 ø

= 4.00 kJ/kg h2 = h1 + wp, in = 167.53 + 4.00 = 171.54 kJ/kg

T3 = 250°C ïü ïý h3 = 2801.0 kJ/kg ïïþ s3 = 6.0721 kJ/kg ×K x3 = 1 s4 - s f 6.0721- 0.5724 = = 0.7158 T4 = 40°C üïï x4 = s 7.6832 ý fg ïïþ s4 = s3 h4 = h f + x4 h fg = 167.53 + (0.7158)(2406.0) = 1889.8 kJ/kg

10-245

Thus,

wT, out = h3 - h4 = 2801.0 - 1889.8 = 911.2 kJ / kg qin = h3 - h2 = 2801.0 - 171.54 = 2629.4 kJ / kg qout = h4 - h1 = 1889.8 - 167.53 = 1722.2 kJ/kg The thermal efficiency of the cycle is

hth = 1-

qout 1722.2 = 1= 0.345 qin 2629.4

EES (Engineering Equation Solver) SOLUTION "Given" T[4]=40 [C] T[3]=250 [C] x[3]=1 "Analysis" Fluid$='steam_iapws' T[1]=T[4] x[1]=0 P[1]=pressure(Fluid$, T=T[1], x=x[1]) h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) v[1]=volume(Fluid$, T=T[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in P[3]=pressure(Fluid$, T=T[3], x=x[3]) P[2]=P[3] h[3]=enthalpy(Fluid$, T=T[3], x=x[3]) s[3]=entropy(Fluid$, T=T[3], x=x[3]) s[4]=s[3] s4_f=entropy(Fluid$, T=T[4], x=0) s4_g=entropy(Fluid$, T=T[4], x=1) s4_fg=s4_g-s4_f x[4]=(s[4]-s4_f)/s4_fg h4_f=enthalpy(Fluid$, T=T[4], x=0) h4_g=enthalpy(Fluid$, T=T[4], x=1) h4_fg=h4_g-h4_f h[4]=h4_f+x[4]*h4_fg w_t_out=h[3]-h[4] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-246

q_in=h[3]-h[2] q_out=h[4]-h[1] Eta_th=1-q_out/q_in

11-18E A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

h1  h f @ 3 psia  109.40 Btu/lbm v1  v f @ 3 psia  0.01630 ft 3 /lbm wp, in  v1 ( P2  P1 )

  1 Btu  (0.01630 ft 3 /lbm)(800  3)psia  3   5.404 psia  ft   2.40 Btu/lbm h2  h1  wp, in  109.40  2.40  111.81 Btu/lbm

P3  800 psia

  

T3  900F P4  3 psia s4  s3

  

h3  1456.0 Btu/lbm s3  1.6413 Btu/lbm  R x4 

s4  s f s fg



1.6413  0.2009  0.8549 1.6849

h4  h f  x4 h fg  109.40  (0.8549)(1012.8)  975.24 Btu/lbm

Knowing the power output from the turbine the mass flow rate of steam in the cycle is determined from W T, out 1750 kJ/s  0.94782 Btu  W T, out  m (h3  h4 )   m      3.450 lbm/s h3  h4 (1456.0  975 .24)Btu/lbm  1 kJ  The rates of heat addition and rejection are

Qin  m(h3  h2 )  (3.450 lbm/s)(1456.0  111.81)Btu/lbm  4637 Btu / s Qout  m(h4  h1 )  (3.450 lbm/s)(975.24  109.40)Btu/lbm  2987 Btu / s © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-247

and the thermal efficiency of the cycle is Q 2987  th  1  out  1   0.3559  35.6%  4637 Qin

EES (Engineering Equation Solver) SOLUTION "GIVEN" W_dot_T=1750 [kW] P[2]=800 [psia] P[4]=3 [psia] T[3]=900 [F] "ANALYSIS" Fluid$='Steam_iapws' P[1]=P[4] P[3]=P[2] h[1]=enthalpy(Fluid$, P=P[1], x=0) v[1]=volume(Fluid$, P=P[1], x=0) s[1]=entropy(Fluid$, P=P[1], x=0) w_pump=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_pump h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) h[4]=enthalpy(Fluid$, P=P[4], s=s[3]) x[4]=quality(Fluid$, P=P[4], s=s[3]) W_dot_T*Convert(kW, Btu/s)=m_dot*(h[3]-h[4]) Q_dot_in=m_dot*(h[3]-h[2]) Q_dot_out=m_dot*(h[4]-h[1]) eta_th=1-Q_dot_out/Q_dot_in s4_f=entropy(Fluid$, P=P[4], x=0) s4_g=entropy(Fluid$, P=P[4], x=1) s4_fg=s4_g-s4_f h4_f=enthalpy(Fluid$, P=P[4], x=0) h4_g=enthalpy(Fluid$, P=P[4], x=1) h4_fg=h4_g-h4_f

11-19 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13), h1 = h f @ 0.7 MPa = 88.82 kJ/kg

v1 = v f @ 0.7 MPa = 0.0008331 m3 /kg wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ ÷ = (0.0008331 m3 /kg )(1400 - 700 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

10-248

= 0.58 kJ/kg h2 = h1 + wp , in = 88.82 + 0.58 = 89.40 kJ/kg

P3 = 1.4 MPa üïï h3 = hg @ 1.4 MPa = 276.17 kJ/kg ý ïïþ s3 = sg @ 1.4 MPa = 0.91067 kJ/kg ×K sat. vapor s4 - s f P4 = 0.7 MPa ïüï 0.91067 - 0.33232 = = 0.9839 ý x4 = ïïþ s4 = s3 s fg 0.58780 h4 = h f + x4 h fg = 88.82 + (0.9839)(176.26) = 262.24 kJ/kg

Thus , qin = h3 - h2 = 276.17 - 89.40 = 186.77 kJ/kg qout = h4 - h1 = 262.24 - 88.82 = 173.42 kJ/kg

wnet = qin - qout = 186.77 - 173.42 = 13.35 kJ/kg

and hth =

wnet 13.35 kJ/kg = = 0.0715 = 7.15% qin 186.77 kJ/kg

(b) Wnet = mwnet = (3 kg/s)(13.35 kJ/kg ) = 40.05 kW

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=1400 [kPa] x[3]=1 P[4]=700 [kPa] m_dot=3 [kg/s] "Analysis" Fluid$='R134a' "(a)" P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-249

w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) s[3]=entropy(Fluid$, P=P[3], x=x[3]) s[4]=s[3] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in "(b)" W_dot_net=m_dot*w_net

11-20 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /k wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.00101 m3 /kg )(10, 000 - 10 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 10.09 kJ/kg h2 = h1 + wp , in = 191.81 + 10.09 = 201.90 kJ/kg

P3 = 10 MPa üïï h3 = 3375.1 kJ/kg ý T3 = 500 °C ïïþ s3 = 6.5995 kJ/kg ×K

10-250

s4 - s f P4 = 10 kPa üïï 6.5995 - 0.6492 = = 0.7934 ý x4 = ïïþ s4 = s3 s fg 7.4996 h4 = h f + x4 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg (b) qin = h3 - h2 = 3375.1- 201.90 = 3173.2 kJ/kg (c) qout = h4 - h1 = 2089.7 - 191.81 = 1897.9 kJ/kg (d)

wnet = qin - qout = 3173.2 - 1897.9 = 1275.4 kJ/kg

and hth =

wnet 1275.4 kJ/kg = = 40.2% qin 3173.2 kJ/kg

(e) m = Wnet = 210,000 kJ/s = 164.7 kg / s wnet

1275.4 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=210000 [kW] P[3]=10000 [kPa] T[3]=500 [C] P[4]=10 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) x_4=x[4] "(b)" q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in "(c)" m_dot=W_dot_net/w_net

11-21 A steam power plant that operates on a simple nonideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-251

2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg wp , in = v1 (P2 - P1 ) / h p æ 1 kJ ÷ ö = (0.00101 m3 /kg )(10,000 - 10 kPa )çç ÷/ (0.85) çè1 kPa ×m3 ÷ ø

= 11.87 kJ/kg h2 = h1 + wp , in = 191.81 + 11.87 = 203.68 kJ/kg

P3 = 10 MPa ü ïï h3 = 3375.1 kJ/kg ý T3 = 500°C ïïþ s3 = 6.5995 kJ/kg ×K

s4 s - s f 6.5995 - 0.6492 = = 0.7934 P4 s = 10 kPa ü ïï x4 s = s 7.4996 ý fg ïïþ s4 s = s3 h4 s = h f + x4 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg

hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s = 3375.1- (0.85)(3375.1- 2089.7) = 2282.5 kJ/kg

ü ïïý x = h4 - h f = 2282.5 - 191.81 = 0.874 4 h4 = 2282.5 kJ/kg ïïþ h fg 2392.1

P4 = 10 kPa

(b) qin = h3 - h2 = 3375.1- 203.68 = 3171.4 kJ/kg (c) qout = h4 - h1 = 2282.5 - 191.81 = 2090.7 kJ/kg (d) wnet = qin - qout = 3171.4 - 2090.7 = 1080.7 kJ/kg and

hth = (c) m =

wnet 1080.7 kJ/kg = = 34.1% qin 3171.5 kJ/kg

Wnet 210,000 kJ/s = = 194.3 kg / s wnet 1080.7 kJ/kg

10-252

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=210000 "[kW]" P[3]=10000 "[kPa]" T[3]=500 "[C]" P[4]=10 "[kPa]" Eta_T=0.85 Eta_P=0.85 "Analysis" Fluid$='steam_iapws' "(a)" P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_s[4]) x[4]=quality(Fluid$, P=P[4], h=h[4]) x_4=x[4] "(b)" q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in "(c)" m_dot=W_dot_net/w_net

11-22 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits. The power produced by the turbine, the heat added in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg v1 = v f @100 kPa = 0.001043 m3 /kg

wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001043 m3 /kg)(15, 000 - 100)kPa çç ÷ çè1 kPa ×m3 ÷ ø = 15.54 kJ/kg h2 = h1 + wp, in = 417.51 + 15.54 = 433.05 kJ/kg

P3 = 15, 000 kPa x3 = 1

ü ïï ý ïïþ

h3 = 2610.8 kJ/kg s3 = 5.3108 kJ/kg ×K

10-253

P4 = 100 kPa s4 = s3

ïüï ý ïïþ

x4 =

s4 - s f s fg

5.3108 - 1.3028 = 0.6618 6.0562

h4 = h f + x4 h fg = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg

Thus, wT,out = h3 - h4 = 2610.8 - 1911.5 = 699.3 kJ / kg

qin = h3 - h2 = 2610.8 - 433.05 = 2177.8 kJ / kg qout = h4 - h1 = 1911.5 - 417.51 = 1494.0 kJ/kg

The thermal efficiency of the cycle is hth = 1-

qout 1494.0 = 1= 0.314 = 31.4% qin 2177.8

11-23 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The isentropic efficiency of the turbine, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 100 kPa = 417.51 kJ/kg v1 = v f @ 100 kPa = 0.001043 m3 /kg

wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001043 m3 /kg)(15, 000 - 100)kPa çç ÷ çè1 kPa ×m3 ÷ ø = 15.54 kJ/kg h2 = h1 + wp, in = 417.51 + 15.54 = 433.05 kJ/kg

P3 = 15, 000 kPa x3 = 1

ïü ïý ïïþ

h3 = 2610.8 kJ/kg s3 = 5.3108 kJ/kg ×K

10-254

P4 = 100 kPa s4 = s3 P4 = 100 kPa x4 = 0.70

ü ïï ý ïïþ

x4 s =

s4 - s f s fg

5.3108 - 1.3028 = 0.6618 6.0562

h4 s = h f + x4 s h fg = 417.51 + (0.6618)(2257.5) = 1911.5 kJ/kg

ïü ïý h = h + x h = 417.51 + (0.70)(2257.5) = 1997.8 kJ/kg f 4 fg ïïþ 4

The isentropic efficiency of the turbine is hT =

h3 - h4 2610.8 - 1997.8 = = 0.877 = 87.7% h3 - h4 s 2610.8 - 1911.5

Thus,

qin = h3 - h2 = 2610.8 - 433.05 = 2177.8 kJ/kg qout = h4 - h1 = 1997.8 - 417.51 = 1580.3 kJ/kg

The thermal efficiency of the cycle is hth = 1-

qout 1580.3 = 1= 0.274 = 27.4% qin 2177.8

11-24 A simple Rankine cycle with water as the working fluid operates between the specified pressure limits. The rate of heat addition in the boiler, the power input to the pumps, the net power, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), ü ïï h1 @ h f @ 75°C = 314.03 kJ/kg ý T1 = Tsat @ 50 kPa - 6.3 = 81.3 - 6.3 = 75°Cïïþ v1 = v f @ 75° C = 0.001026 m 3 /kg P1 = 50 kPa

10-255

wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001026 m3 /kg)(6000 - 50)kPa çç ÷ çè1 kPa ×m3 ÷ ø = 6.10 kJ/kg h2 = h1 + wp, in = 314.03 + 6.10 = 320.13 kJ/kg

P3 = 6000 kPaïü ïý h3 = 3302.9 kJ/kg T3 = 450°C ïïþ s3 = 6.7219 kJ/kg ×K

s4 - s f 6.7219 - 1.0912 = = 0.8660 P4 = 50 kPa üïï x4 s = s 6.5019 ý fg ïïþ s4 = s3 h4 s = h f + x4 s h fg = 340.54 + (0.8660)(2304.7) = 2336.4 kJ/kg hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4s ) = 3302.9 - (0.94)(3302.9 - 2336.4) = 2394.4 kJ/kg h3 - h4 s

Thus, Qin = m(h3 - h2 ) = (20 kg/s)(3302.9 - 320.13)kJ/kg = 59,660 kW WT, out = m(h3 - h4 ) = (20 kg/s)(3302.9 - 2394.4)kJ/kg = 18,170 kW WP, in = mwP, in = (20 kg/s)(6.10 kJ/kg) = 122 kW

Wnet = WT, out - WP, in = 18,170 - 122 = 18,050 kW

and

hth =

Wnet 18,050 = = 0.3025 59,660 Qin

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=6000 [kPa] P[4]=50 [kPa] T[3]=450 [C] Eta_T=0.94 DELTAT_subcool=6.3 [C] m_dot=20 [kg/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-256

"Analysis" Fluid$='steam_iapws' P[1]=P[4] x[1]=0 "state 1 approximated as saturated liquid at the given temperature" T_sat1=temperature(Fluid$, P=P[1], x=x[1]) T[1]=T_sat1-DELTAT_subcool h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) v[1]=volume(Fluid$, T=T[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s[4]) x_4=quality(Fluid$, P=P[4], s=s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_4s) Q_dot_in=m_dot*(h[3]-h[2]) W_dot_T_out=m_dot*(h[3]-h[4]) W_dot_P_in=m_dot*w_p_in W_dot_net=W_dot_T_out-W_dot_P_in eta_th=W_dot_net/Q_dot_in

11-25 The change in the thermal efficiency of the cycle in Prob. 11-24 due to a pressure drop in the boiler is to be determined. Analysis We use the following EES routine to obtain the solution. "Given" P[2]=6000 [kPa] DELTAP=50 [kPa] "set this value to zero to calculate thermal efficiency for no pressure drop case" P[3]=6000-DELTAP [kPa] T[3]=450 [C] P[4]=50 [kPa] Eta_T=0.94 DELTAT_subcool=6.3 [C] T[1]=temperature(Fluid$, P=P[1], x=x[1])-DELTAT_subcool m_dot=20 [kg/s] "Analysis" Fluid$='steam_iapws' P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], T=T[1]) v[1]=volume(Fluid$, P=P[1], T=T[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h_s[4]=enthalpy(Fluid$, P=P[4], s=s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_s[4]) q_in=h[3]-h[2] q_out=h[4]-h[1] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-257

w_net=q_in-q_out Eta_th=1-q_out/q_in Solution DELTAP=50 [kPa] Eta_th=0.3022

DELTAT_subcool=6.3 [C] Eta_T=0.94 Fluid$='steam_iapws' h[1]  314.11[kJ / kg]

h[2]  320.21[kJ / kg] h[3]  3303.64 [kJ / kg] h[4]  2396.01[kJ / kg] P[1]=50 h_s[4]  2338.1[kJ / kg] m dot  20 [kg / s] P[2]=6000 P[3]=5950 P[4]=50 q_in  2983.4 [kJ / kg] q_out  2081.9 [kJ / kg] s[3]  6.7265 [kJ / kg  K] T[1]=75.02 [C] T[3]=450 [C] s[4]  6.7265 [kJ / kg  K]  w_net  901.5 [kJ / kg] w_p_in  6.014 [kJ / kg] v[1]  0.001026   m 3 / kg   x[1]=0 Discussion The thermal efficiency without a pressure drop was obtained to be 0.3025.

11-26 The net work outputs and the thermal efficiencies for a Carnot cycle and a simple ideal Rankine cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 50 kPa = 340.54 kJ/kg v1 = v f @ 50 kPa = 0.001030 m3 /kg

wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.001030 m3 /kg )(5000 - 50) kPa çç ÷ ÷ çè1 kPa ×m3 ø

= 5.10 kJ/kg h2 = h1 + wp , in = 340.54 + 5.10 = 345.64 kJ/kg

P3 = 5 MPa üïï h3 = 2794.2 kJ/kg ý ïïþ s3 = 5.9737 kJ/kg ×K x3 = 1 s4 - s f 5.9737 - 1.09120 P4 = 50 kPaüïï = = 0.7509 ý x4 = ï s4 = s3 s fg 6.5019 ïþ h4 = h f + x4 h fg = 340.54 + (0.7509)(2304.7)

= 2071.2 kJ/kg

10-258

qin = h3 - h2 = 2794.2 - 345.64 = 2448.6 kJ/kg qout = h4 - h1 = 2071.2 - 340.54 = 1730.7 kJ/kg wnet = qin - qout = 2448.6 - 1730.7 = 717.9 kJ / kg hth = 1-

qout 1730.7 = 1= 0.2932 = 29.3% qin 2448.6

(b) Carnot Cycle analysis: P3 = 5 MPa ïü ïý h3 = 2794.2 kJ/kg ïïþ T3 = 263.9°C x3 = 1

T2 = T3 = 263.9°Cüïï h2 = 1154.5 kJ/kg ý ïïþ s2 = 2.9207 kJ/kg ×K x2 = 0 s1 - s f 2.9207 - 1.0912 = = 0.2814 P1 = 50 kPa ïüï x1 = s fg 6.5019 ý ïïþ s1 = s2 h1 = h f + x1h fg = 340.54 + (0.2814)(2304.7) = 989.05 kJ/kg

qin = h3 - h2 = 2794.2 - 1154.5 = 1639.7 kJ/kg

qout = h4 - h1 = 2071.2 - 340.54 = 1082.2 kJ/kg wnet = qin - qout = 1639.7 - 1082.2 = 557.5 kJ / kg

10-259

hth = 1-

qout 1082.2 = 1= 0.3400 = 34.0% qin 1639.7

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[3]=5000 [kPa] x_3=1 P[4]=50 [kPa] "ANALYSIS" Fluid$='Steam_iapws' "Rankine cycle" P[1]=P[4] P[2]=P[3] h[1]=enthalpy(Fluid$, P=P[1], x=0) v[1]=volume(Fluid$, P=P[1], x=0) s[1]=entropy(Fluid$, P=P[1], x=0) w_pump=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pump h[3]=enthalpy(Fluid$, P=P[3], x=x_3) s[3]=entropy(Fluid$, P=P[3], x=x_3) h[4]=enthalpy(Fluid$, P=P[4], s=s[3]) x[4]=quality(Fluid$, P=P[4], s=s[3]) q_in_R=h[3]-h[2] q_out=h[4]-h[1] w_net_R=q_in_R-q_out eta_th_R=1-q_out/q_in_R s4_f=entropy(Fluid$, P=P[4], x=0) s4_g=entropy(Fluid$, P=P[4], x=1) s4_fg=s4_g-s4_f h4_f=enthalpy(Fluid$, P=P[4], x=0) h4_g=enthalpy(Fluid$, P=P[4], x=1) h4_fg=h4_g-h4_f "Carnot cycle" T_3=temperature(Fluid$, P=P[3], x=x_3) T_2=T_3 x_2=0 h_2_C=enthalpy(Fluid$, T=T_2, x=x_2) s_2_C=entropy(Fluid$, T=T_2, x=x_2) h_1_C=enthalpy(Fluid$, P=P[1], s=s_2_C) q_in_C=h[3]-h_2_C q_out_C=h[4]-h_1_C w_net_C=q_in_C-q_out_C eta_th_C=1-q_out_C/q_in_C

11-27 A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-260

Properties The specific heat of geothermal water is taken to be 4.258 kJ/kg.ºC. The properties are isobutane are given as h1  273.01 kJ / kg , v1  0.001842 m3 / kg , h3  761.54 kJ / kg , h4  689.74 kJ / kg , h4 s  670.40 kJ / kg . Analysis (a) An analysis on the pump gives the enthalpy of isobutane at state 2: wp , in = v1 (P2 - P1 ) / h P

æ 1 kJ ÷ö ÷/ 0.90 = (0.001842 m3 /kg )(3250 - 410 ) kPa ççç 3÷ è1 kPa ×m ÷ø = 5.81 kJ/kg h2 = h1 + wp , in = 273.01 + 5.81 = 278.82 kJ/kg

An energy balance on the heat exchanger gives mgeo cgeo (Tin - Tout ) = miso (h3 - h2 )

(556 kJ/kg)(4.258kJ/kg.°C)(160 - 90)°C = miso (761.54 - 278.82)kJ/kg

miso = 343.3 kJ/kg The isentropic efficiency of the turbine is

hT =

h3 - h4 761.54 - 689.74 = = 0.788 = 78.8% h3 - h4 s 761.54 - 670.40

(b) The net power is determined as follows: WT,out = miso (h3 - h4 ) = (343.3 kJ/kg)(761.54 - 689.74)kJ/kg = 24, 648 kW WP,in = miso (h2 - h1 ) = miso wp,in = (343.3 kJ/kg)(5.81 kJ/kg) = 1995 kW

Wnet = WT,out - WP,in = 24, 648 - 1995 = 22, 650 kW

Qin = mgeo cgeo (Tin - Tout ) = (556 kJ/kg)(4.1258 kJ/kg.°C)(160 - 90)°C = 165,710 kW (c) Then the thermal efficiency is

hth =

Wnet 22,650 = = 0.137 = 13.7% Qin 165,710

10-261

T_geo_out=90 [C] m_dot_geo=556 [kg/s] T[3]=147 [C] P[3]=3250 [kPa] T[4]=79.5 [C] P[4]=410 [kPa] eta_P=0.90 h[1]=273.01 [kJ/kg] v_1=0.001842 [m^3/kg] h[3]=761.54 [kJ/kg] h[4]=689.74 [kJ/kg] h_4s=670.40 [kJ/kg] "ANALYSIS" "Pump" P[1]=P[4] P[2]=P[3] h[2]=h[1]+v_1*(P[2]-P[1])/eta_P C_geo=CP(water, T=T_geo_ave, x=0) T_geo_ave=1/2*(T_geo_in+T_geo_out) m_dot_geo*C_geo*(T_geo_in-T_geo_out)=m_dot_iso*(h[3]-h[2]) "energy balance on heat exchanger" eta_T=(h[3]-h[4])/(h[3]-h_4s) "turbine isentropic efficiency" W_dot_T=m_dot_iso*(h[3]-h[4]) "turbine power output" W_dot_P=m_dot_iso*(h[2]-h[1]) "pump power input" W_dot_net=W_dot_T-W_dot_P "net power" Q_dot_in=m_dot_geo*C_geo*(T_geo_in-T_geo_out) "heat input" eta_th=W_dot_net/Q_dot_in "thermal efficiency"

11-28 A 175-MW coal-fired steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The overall plant efficiency and the required rate of the coal supply are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 15 kPa = 225.94 kJ/kg

v1 = v f @ 15 kPa = 0.0010140 m3 /kg wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.001014 m3 /kg )(7000 - 15 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 7.08 kJ/kg h2 = h1 + wp , in = 225.94 + 7.08 = 233.03 kJ/kg

P3 = 7 MPa ïüï h3 = 3531.6 kJ/kg ý T3 = 550°C ïïþ s3 = 6.9507 kJ/kg ×K s4 - s f 6.9507 - 0.7549 P4 = 15 kPa ïüï = = 0.8543 ý x4 = ïïþ s4 = s3 s fg 7.2522 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-262 h4 = h f + x4 h fg = 225.94 + (0.8543)(2372.4) = 2252.7 kJ/kg

The thermal efficiency is determined from qin = h3 - h2 = 3531.6 - 233.03 = 3298.6 kJ/kg qout = h4 - h1 = 2252.7 - 225.94 = 2026.8 kJ/kg

and

hth = 1-

Thus,

qout 2026.8 = 1= 0.3856 qin 3298.6

hoverall = hth ´ hcomb ´ hgen = (0.3856)(0.85)(0.96) = 0.3146 = 31.5% (b) Then the required rate of coal supply becomes and Qin = mcoal =

Wnet 175,000 kJ/s = = 556, 220 kJ/s hoverall 0.3146 Qin 556,220 kJ/s = = 18.98 kg/s = 68.3 tons / h HVcoal 29,300 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=175000 [kW] P[3]=7000 [kPa] T[3]=550 [C] P[4]=15 [kPa] HV_coal=29300 [kJ/kg] Eta_comb=0.85 Eta_gen=0.96 "Analysis" Fluid$='steam_iapws' "(a)" P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-263

s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) q_in=h[3]-h[2] q_out=h[4]-h[1] Eta_th=1-q_out/q_in Eta_overall=Eta_th*Eta_comb*Eta_gen "(b)" Q_dot_in=W_dot_net/Eta_overall m_dot_coal=(Q_dot_in/HV_coal) s4_f=entropy(Fluid$, P=P[4], x=0) s4_g=entropy(Fluid$, P=P[4], x=1) s4_fg=s4_g-s4_f h4_f=enthalpy(Fluid$, P=P[4], x=0) h4_g=enthalpy(Fluid$, P=P[4], x=1) h4_fg=h4_g-h4_f

The Reheat Rankine Cycle 11-29C The T-s diagram of the ideal Rankine cycle with 3 stages of reheat is shown on the side. The cycle efficiency will increase as the number of reheating stages increases.

11-30C The T-s diagram shows two reheat cases for the reheat Rankine cycle similar to the one shown in Figure 11-11. In the first case there is expansion through the high-pressure turbine from 6000 kPa to 4000 kPa between states 1 and 2 with reheat at 4000 kPa to state 3 and finally expansion in the low-pressure turbine to state 4. In the second case there is expansion through the high-pressure turbine from 6000 kPa to 500 kPa between states 1 and 5 with reheat at 500 kPa to state 6 and finally expansio in the low-pressure turbine to state 7. Increasing the pressure for reheating increases the average temperature for heat addition makes the energy of the steam more available for doing work, see the reheat process 2 to 3 versus the reheat process 5 to 6. Increasing the reheat pressure will increase the cycle efficiency. However, as the reheating pressure increases, the amount of condensation increases during the expansion process in the low-pressure turbine, state 4 versus state 7. An optimal pressure for reheating generally allows for the moisture content of the steam at the low-pressure turbine exit to be in the range of 10 to 15% and this corresponds to quality in the range of 85 to 90%.

10-264

11-31C When a simple ideal Rankine cycle is modified with reheating, the pump work remains the same, the moisture content decreases, everything else increases.

11-32C The thermal efficiency of the simple ideal Rankine cycle will probably be higher since the average temperature at which heat is added will be higher in this case.

11-33 An ideal reheat Rankine cycle with water as the working fluid is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 50 kPa = 340.54 kJ/kg

v1 = v f @ 50 kPa = 0.001030 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ö ÷ = (0.001030 m3 /kg)(17500 - 50)kPa çç ÷ ÷ çè1 kPa ×m3 ø = 17.97 kJ/kg h2 = h1 + wp, in = 340.54 + 17.97 = 358.51 kJ/kg

10-265

P3 = 17,500 kPa T3 = 550°C s4 = s3 P5 = 2000 kPa T5 = 300°C

s6 = s5

h3 = 3423.6 kJ/kg s3 = 6.4266 kJ/kg ×K

ü ïï ý h4 = 2841.5 kJ/kg ïïþ ü ïï h5 = 3024.2 kJ/kg ý ïïþ s5 = 6.7684 kJ/kg ×K

P4 = 2000 kPa

P6 = 50 kPa

ü ïï ý ïïþ

ïüï ý ïïþ

x6 =

s6 - s f s fg

6.7684 - 1.0912 = 0.8732 6.5019

h6 = h f + x6 h fg = 340.54 + (0.8732)(2304.7) = 2352.9 kJ/kg

Thus, qin = (h3 - h2 ) + (h5 - h4 ) = 3423.6 - 358.51 + 3024.2 - 2841.5 = 3247.8 kJ/kg qout = h6 - h1 = 2352.9 - 340.54 = 2012.4 kJ/kg

and hth = 1-

qout 2012.4 = 1= 0.380 = 38.0% qin 3247.8

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=17500 "[kPa]" T[3]=550 "[C]" P[5]=2000 "[kPa]" T[5]=300 "[C]" P[6]=50 "[kPa]" "Analysis" Fluid$='steam_iapws' "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-266

w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) T[4]=temperature(Fluid$, P=P[4], s=s[4]) "Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) T[6]=temperature(Fluid$, P=P[6], h=h[6]) x[6]=quality(Fluid$, P=P[6], h=h[6]) q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] w_net=q_in-q_out Eta_th=w_net/q_in

11-34 An ideal reheat Rankine cycle with water as the working fluid is considered. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 50 kPa = 340.54 kJ/kg

v1 = v f @ 50 kPa = 0.001030 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ö ÷ = (0.001030 m3 /kg)(17500 - 50)kPa çç ÷ ÷ çè1 kPa ×m3 ø

= 17.97 kJ/k h2 = h1 + wp, in = 340.54 + 17.97 = 358.52 kJ/kg

P3 = 17,500 kPa T3 = 550°C s4 = s3 P5 = 2000 kPa T5 = 550°C

s6 = s5

h3 = 3423.6 kJ/kg s3 = 6.4266 kJ/kg ×K

ü ïï ý h4 = 2841.5 kJ/kg ïïþ ü ïï h5 = 3579.0 kJ/kg ý ïïþ s5 = 7.5725 kJ/kg ×K

P4 = 2000 kPa

P6 = 50 kPa

ïü ïý ïïþ

ü ïï ý ïïþ

x6 =

s6 - s f s fg

7.5725 - 1.0912 = 0.9968 6.5019

h6 = h f + x6 h fg = 340.54 + (0.9968)(2304.7) = 2638.0 kJ/kg

10-267

Thus,

qin = (h3 - h2 ) + (h5 - h4 ) = 3423.6 - 358.52 + 3579.0 - 2841.5 = 3802.6 kJ/kg qout = h6 - h1 = 2638.0 - 340.54 = 2297.4 kJ/kg and

hth = 1-

qout 2297.4 = 1= 0.396 = 39.6% qin 3802.6

The thermal efficiency was determined to be 0.380 when the temperature at the inlet of low-pressure turbine was 300C. When this temperature is increased to 550C, the thermal efficiency becomes 0.396. This corresponding to a percentage increase of 4.2% in thermal efficiency.

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=17500 "[kPa]" T[3]=550 "[C]" P[5]=2000 "[kPa]" T[5]=550 "[C]" P[6]=50 "[kPa]" "Analysis" Fluid$='steam_iapws' "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) "Low pressure turbine" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-268

h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) T[6]=temperature(Fluid$, P=P[6], h=h[6]) x[6]=quality(Fluid$, P=P[6], h=h[6]) q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] w_net=q_in-q_out Eta_th=w_net/q_in

11-35 An ideal reheat steam Rankine cycle produces 2000 kW power. The mass flow rate of the steam, the rate of heat transfer in the reheater, the power used by the pumps, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 100 kPa = 417.51 kJ/kg

v1 = v f @ 100 kPa = 0.001043 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ö ÷ = (0.001043 m3 /kg)(15000 - 100)kPa çç ÷ ÷ çè1 kPa ×m3 ø

= 15.54 kJ/kg h2 = h1 + wp, in = 417.51 + 15.54 = 433.05 kJ/kg

P3 = 15, 000 kPa ïü ïý h3 = 3157.9 kJ/kg ïïþ s3 = 6.1434 kJ/kg ×K T3 = 450°C s4 - s f 6.1434 - 2.4467 = = 0.9497 P4 = 2000 kPa ü ïï x4 = s 3.8923 ý fg ïïþ s4 = s3 h4 = h f + x4 h fg = 908.47 + (0.9497)(1889.8) = 2703.3 kJ/kg

P5 = 2000 kPa ïü ïý h5 = 3358.2 kJ/kg T5 = 450°C ïïþ s5 = 7.2866 kJ/kg ×K © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-269

s6 - s f 7.2866 - 1.3028 = = 0.9880 P6 = 100 kPa üïï x6 = s 6.0562 ý fg ïïþ s6 = s5 h6 = h f + x6 h fg = 417.51 + (0.9880)(22257.5) = 2648.0 kJ/kg Thus,

qin = (h3 - h2 ) + (h5 - h4 ) = 3157.9 - 433.05 + 3358.2 - 2703.3 = 3379.8 kJ/kg qout = h6 - h1 = 2648.0 - 417.51 = 2230.5 kJ/kg wnet = qin - qout = 379.8 - 2230.5 = 1149.2 kJ/kg The power produced by the cycle is Wnet = mwnet = (1.74 kg/s)(1149.2 kJ/kg) = 2000 kW

The rate of heat transfer in the rehetaer is

Qreheater = m(h5 - h4 ) = (1.740 kg/s)(3358.2 - 2703.3) kJ/kg = 1140 kW WP, in = mwP, in = (1.740 kg/s)(15.54 kJ/kg) = 27 kW and the thermal efficiency of the cycle is

hth = 1-

qout 2230.5 = 1= 0.340 qin 3379.8

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=15000 [kPa] T[3]=450 [C] P[5]=2000 [kPa] T[5]=450 [C] P[6]=100 [kPa] m_dot=1.74 [kg/s] "Analysis" Fluid$='steam_iapws' "(a)" "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) "Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) T[6]=temperature(Fluid$, P=P[6], h=h[6]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-270

T_6=T[6] x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] "(b)" w_T_out=h[3]-h[4]+h[5]-h[6] q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] w_net=w_T_out-w_p_in Eta_th=w_net/q_in "(c)" W_dot_net=m_dot*w_net Q_dot_in=m_dot*q_in Q_dot_out=m_dot*q_out Q_dot_reheater=m_dot*(h[5]-h[4]) W_dot_p_in=m_dot*w_p_in

11-36E A steam power plant that operates on the ideal reheat Rankine cycle is considered. The pressure at which reheating takes place, the net power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = hsat @ 1 psia = 69.72 Btu/lbm

v1 = v sat @ 1 psia = 0.01614 ft 3 /lbm T1 = Tsat @ 1 psia = 101.69°F

wp , in = v1 (P2 - P1 ) æ ö 1 Btu ÷ ÷ = (0.01614 ft 3 /lbm)(800 - 1 psia )ççç 3÷ è5.4039 psia ×ft ÷ ø = 2.39 Btu/lbm h2 = h1 + wp , in = 69.72 + 2.39 = 72.11 Btu/lbm

10-271

P3 = 800 psia ü ïï h3 = 1456.0 Btu/lbm ý T3 = 900°F ïïþ s3 = 1.6413 Btu/lbm ×R

üï h4 = hg @ sg = s4 = 1178.5 Btu/lbm ïý (sat.vapor )ïïþ P4 = Psat @ sg = s4 = 62.23 psia (the reheat pressure) s4 = s3

P5 = 62.23 psia ïü ïý h5 = 1431.4 Btu/lbm ïïþ s5 = 1.8985 Btu/lbm ×R T5 = 800°F

s6 - s f 1.8985 - 0.13262 x = = = 0.9572 P6 = 1 psia ïü ïý 6 s fg 1.84495 ïïþ s6 = s5 h6 = h f + x6 h fg = 69.72 + (0.9572)(1035.7) = 1061.0 Btu/lbm

(b) qin = (h3 - h2 )+ (h5 - h4 ) = 1456.0 - 72.11 + 1431.4 - 1178.5 = 1636.8 Btu/lbm (c) qout = h6 - h1 = 1061.0 - 69.72 = 991.3 Btu/lbm Thus, hth = 1-

qout 991.3 Btu/lbm = 1= 39.4% qin 1636.8 Btu/lbm

(d) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the

Qout = Qin - Wnet = (1- hth )Qin = (1- 0.3943)(6´ 104 Btu/s) = 3.634´ 10 4 Btu/s

mcool =

Qout 3.634´ 104 Btu/s = = 641.0 lbm / s cD T (1.0 Btu/lbm ×°F)(101.69 - 45)°F

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=800 [psia] T[3]=900 [F] x[4]=1 T[5]=800 [F] P[6]=1 [psia] Q_dot_in=60000 [Btu/s] T_water_in=45 [F] "Analysis" Fluid$='steam_iapws' "(a)" "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) T[1]=temperature(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1])*convert(psia*ft^3/lbm, Btu/lbm) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-272

h[4]=enthalpy(Fluid$, x=x[4], s=s[4]) P[4]=pressure(Fluid$, x=x[4], s=s[4]) P_4=P[4] "Low pressure turbine" P[5]=P[4] h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) "(b)" q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] Eta_th=1-q_out/q_in "(c)" Q_dot_out=(1-Eta_th)*Q_dot_in C_water=1.0 [Btu/lbm-F] T_water_out=T[1] DELTAT_water=T_water_out-T_water_in m_dot_water=Q_dot_out/(C_water*DELTAT_water)

11-37 An ideal reheat steam Rankine cycle produces 5000 kW power. The rates of heat addition and rejection, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.001010 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001010 m3 /kg)(8000 - 10)kPa çç ÷ çè1 kPa ×m3 ÷ ø

= 8.07 kJ/kg h2 = h1 + wp, in = 191.81 + 8.07 = 199.88 kJ/kg c

10-273

ü ïï ý ïïþ

P3 = 8000 kPa T3 = 450°C

P4 = 500 kPaïü ïý s4 = s3 ïïþ

s4 - s j sk

6.5579 - 1.8604 = 0.9470 4.96013

h4 = h j + x4 hk = 640.09 + (0.9470)(2108.0) = 2636.4 kJ / kg ïü ïý ïïþ

T5 = 500°C

s6 = s5

s3 = 6.5579 kJ/kg ×K

x4 =

P5 = 500 kPa

P6 = 10 kPa

h3 = 3273.3 kJ/kg

ïüï ý ïïþ

h5 = 3484.5 kJ/kg s5 = 8.0893 kJ/kg ×K

x6 =

s6 - s f s fg

8.0893 - 0.6492 = 0.9921 7.4996

h6 = h f + x6 h fg = 191.81 + (0.9921)(2392.1) = 2564.9 kJ/kg

Thus, qin = (h3 - h2 ) + (h5 - h4 ) = 3273.3 - 199.88 + 3484.5 - 2636.4 = 3921.5 kJ/kg

qout = h6 - h1 = 2564.9 - 191.81 = 2373.1 kJ/kg wnet = qin - qout = 3921.5 - 2373.1 = 1548.5 kJ/kg The mass flow rate of steam in the cycle is determined from Wnet = m(h3 - h4 ) ¾ ¾ ® m=

Wnet 5000 kJ/s = = 3.229 kg / s wnet 1548.5 kJ/kg

and the thermal efficiency of the cycle is hth = 1-

qout 2373.1 = 1= 0.395 = 39.5% qin 3921.5

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=5000 [kW] P[3]=8000 [kPa] T[3]=450 [C] P[5]=500 [kPa] T[5]=500 [C] P[6]=10 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-274

"Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) T[6]=temperature(Fluid$, P=P[6], h=h[6]) T_6=T[6] x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] "(b)" w_T_out=h[3]-h[4]+h[5]-h[6] q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] w_net=w_T_out-w_p_in Eta_th=w_net/q_in "(c)" m_dot=W_dot_net/w_net Q_dot_in=m_dot*q_in Q_dot_out=m_dot*q_out

11-38 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = hsat @ 10 kPa = 191.81 kJ / kg v1 = v sat @ 10 kPa = 0.00101 m3 /kg

wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.00101 m3 /kg )(15, 000 - 10 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

10-275 h2 = h1 + wp , in = 191.81 + 15.14 = 206.95 kJ/kg

P3 = 15 MPa ïü ïý h3 = 3310.8 kJ/kg T3 = 500°C ïïþ s3 = 6.3480 kJ/kg ×K

P6 = 10 kPa ïüï h6 = h f + x6 h fg = 191.81 + (0.95)(2392.1) = 2464.3 kJ/kg ý ïïþ s6 = s f + x6 s fg = 0.6492 + (0.95)(7.4996) = 7.7739 kJ/kg ×K s6 = s5

T5 = 500°C ïüï P5 = 980 kPa (the reheat pressure) ý ïïþ h5 = 3479.3 kJ/kg s5 = s6 P4 = 980 kPa üïï ý h4 = 2666.0 kJ/kg ïïþ s4 = s3 (b) The rate of heat supply is

Qin = m éë(h3 - h2 )+ (h5 - h4 )ùû = (12 kg/s)(3310.8 - 206.95 + 3479.3 - 2666.0) kJ/kg

= 47,000 kW (c) The thermal efficiency is determined from Thus,

Qout = m(h6 - h1 ) = (12 kJ/s)(2464.3 - 191.81) kJ/kg = 27,270 kJ/s

hth = 1-

Qout 27,270 kJ/s = 1= 0.420 = 42.0% 47,000 kJ/s Qin

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=15000 [kPa] T[3]=500 [C] T[5]=500 [C] P[6]=10 [kPa] m_dot=12 [kg/s] x[6]=0.95 "Analysis" Fluid$='steam_iapws' "(a)" "pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) h[6]=enthalpy(Fluid$, P=P[6], x=x[6]) s[6]=entropy(Fluid$, P=P[6], x=x[6]) s[5]=s[6] h[5]=enthalpy(Fluid$, T=T[5], s=s[5]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-276

P[5]=pressure(Fluid$, T=T[5], s=s[5]) P_5=P[5] P[4]=P[5] s[4]=s[3] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) "(b)" q_in=h[3]-h[2]+h[5]-h[4] Q_dot_in=m_dot*q_in "(c)" Q_dot_out=m_dot*(h[6]-h[1]) Eta_th=1-Q_dot_out/Q_dot_in

11-39 A steam power plant that operates on a reheat Rankine cycle is considered. The condenser pressure, the net power output, and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

P3 = 12.5 MPaïüï h3 = 3476.5 kJ/kg ý T3 = 550°C ïïþ s3 = 6.6317 kJ/kg ×K P4 = 2 MPa ü ïï ý h4 s = 2948.1 kJ/kg ïïþ s4 s = s3

hT =

h3 - h4 h3 - h4 s

® h4 = h3 - hT (h3 - h4 s )

10-277

P5 = 2 MPa ïü ïý h5 = 3358.2 kJ/kg T5 = 450°C ïïþ s5 = 7.2815 kJ/kg ×K ïü ïý h = 6 x6 = 0.95ïïþ

(Eq. 1)

P6 = ? ü ïï ý h6 s = s6 = s5 ïïþ

(Eq. 2)

P6 = ?

hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) = 3358.2 - (0.85)(3358.2 - h6 s ) (Eq. 3) h5 - h6 s

The pressure at state 6 may be determined by a trial-error approach from the steam tables or by using EES from the above three equations: P6  9.73 kPa , h6  2463.3 kJ / kg , (b) Then, h1 = h f @ 9.73 kPa = 189.57 kJ/kg

v1 = v f @ 10 kPa = 0.001010 m3 /kg wp , in = v1 (P2 - P1 ) / h p æ 1 kJ ö ÷ = (0.00101 m3 /kg )(12,500 - 9.73 kPa )çç ÷ ÷/ (0.90) çè1 kPa ×m3 ø

= 14.02 kJ/kg h2 = h1 + wp , in = 189.57 + 14.02 = 203.59 kJ/kg

Cycle analysis: qin = (h3 - h2 )+ (h5 - h4 ) = 3476.5 - 203.59 + 3358.2 - 2463.3 = 3603.8 kJ/kg

qout = h6 - h1 = 2463.3 - 189.57 = 2273.7 kJ/kg Wnet = m(qin - qout ) = (7.7 kg/s)(3603.8-2273.7)kJ/kg = 10,242 kW

qout 2273.7 kJ/kg = 1= 0.369 = 36.9% qin 3603.8 kJ/kg

10-278

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[3]=12500 [kPa] T[3]=550 [C] m_dot=7.7 [kg/s] P[4]=2000 [kPa] P[5]=P[4] T[5]=450 [C] eta_T=0.85 eta_P=0.90 x[1]=0 x[6]=0.95 "PROPERTIES" Fluid$='Steam_iapws' P[1]=P[6] P[2]=P[3] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) h_4s=enthalpy(Fluid$, P=P[4], s=s[3]) h[4]=h[3]-eta_T*(h[3]-h_4s) h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) h_6s=enthalpy(Fluid$, P=P[6], s=s[5]) h[6]=h[5]-eta_T*(h[5]-h_6s) h[6]=enthalpy(Fluid$, P=P[6], x=x[6]) P_6=P[6] h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v_1=volume(Fluid$, P=P[1], x=x[1]) w_pump=v_1*(P[2]-P[1])/eta_P h[2]=h[1]+w_pump "ANALYSIS" q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] W_dot_net=m_dot*(q_in-q_out) eta_th=1-q_out/q_in

11-40 A steam power plant that operates on a reheat Rankine cycle is considered. The quality (or temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg v1 = v f @ 10 kPa = 0.001010 m3 /kg

wp , in = v1 (P2 - P1 ) / h p æ 1 kJ ÷ ö = (0.00101 m3 /kg )(10,000 - 10 kPa )çç / (0.95) ÷ çè1 kPa ×m3 ÷ ø

10-279

= 10.62 kJ/kg h2 = h1 + wp , in = 191.81 + 10.62 = 202.43 kJ/kg

P3 = 10 MPa ïü ïý h3 = 3375.1 kJ/kg T3 = 500°C ïïþ s3 = 6.5995 kJ/kg ×K P4 s = 1 MPa ïü ïý h = 2783.8 kJ/kg ïïþ 4 s s4 s = s3

hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s = 3375.1- (0.80)(3375.1 - 2783.7 ) = 2902.0 kJ/kg

P5 = 1 MPa ü ïï h5 = 3479.1 kJ/kg ý T5 = 500°C ïïþ s5 = 7.7642 kJ/kg ×K

s6 s - s f 7.7642 - 0.6492 = = 0.9487 (at turbine exit ) P6 s = 10 kPa üïï x6 s = s fg 7.4996 ý ïïþ s6 s = s5 h6 s = h f + x6 s h fg = 191.81 + (0.9487)(2392.1) = 2461.2 kJ/kg hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) h5 - h6 s

= 3479.1- (0.80)(3479.1- 2461.2)

= 2664.8 kJ/kg > hg (superheated vapor )

From steam tables at 10 kPa we read T6  88.1 C . (b) wT, out = (h3 - h4 )+ (h5 - h6 ) = 3375.1- 2902.0 + 3479.1- 2664.8 = 1287.4 kJ/kg (c) qin = (h3 - h2 )+ (h5 - h4 ) = 3375.1- 202.43 + 3479.1- 2902.0 = 3749.8 kJ/kg (d) wnet = wT, out - wp, in = 1287.4 - 10.62 = 1276.8 kJ/kg Thus the thermal efficiency is hth =

wnet 1276.8 kJ/kg = = 34.1% qin 3749.8 kJ/kg

(e) The mass flow rate of the steam is

Wnet 80,000 kJ/s = = 62.7 kg / s wnet 1276.9 kJ/kg

10-280

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=80000 [kW] P[3]=10000 [kPa] T[3]=500 [C] P[5]=1000 [kPa] T[5]=500 [C] P[6]=10 [kPa] Eta_T=0.80 Eta_P=0.95 "Analysis" Fluid$='steam_iapws' "(a)" "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_4s=s[3] P[4]=P[5] h_4s=enthalpy(Fluid$, P=P[4], s=s_4s) Eta_T=(h[3]-h[4])/(h[3]-h_4s) "Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s_6s=s[5] h_6s=enthalpy(Fluid$, P=P[6], s=s_6s) x_6s=quality(Fluid$, P=P[6], s=s_6s) Eta_T=(h[5]-h[6])/(h[5]-h_6s) x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] T[6]=temperature(Fluid$, P=P[6], h=h[6]) T_6=T[6] "(b)" w_T_out=h[3]-h[4]+h[5]-h[6] q_in=h[3]-h[2]+h[5]-h[4] w_net=w_T_out-w_p_in Eta_th=w_net/q_in "(c)" m_dot=W_dot_net/w_net

11-41 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The quality (or temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-281

2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 10 kPa = 191.81 kJ / kg v1 = v f @ 10 kPa = 0.00101 m3 /kg wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.00101 m3 /kg )(10, 000 - 10 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 10.09 kJ/kg h2 = h1 + wp , in = 191.81 + 10.09 = 201.90 kJ/kg

P3 = 10 MPa ü ïï h3 = 3375.1 kJ/kg ý T3 = 500°C ïïþ s3 = 6.5995 kJ/kg ×K P4 = 1 MPaü ïï ý h4 = 2783.8 kJ/kg ïïþ s4 = s3 P5 = 1 MPa ü ïï h5 = 3479.1 kJ/kg ý T5 = 500°C ïïþ s5 = 7.7642 kJ/kg ×K

s6 - s f 7.7642 - 0.6492 = = 0.9487 (at turbine exit ) P6 = 10 kPa ü ïï x6 = s 7.4996 ý fg ïïþ s6 = s5 h6 = h f + x6 h fg = 191.81 + (0.9487)(2392.1) = 2461.2 kJ/kg

(b) wT, out = (h3 - h4 )+ (h5 - h6 ) = 3375.1- 2783.7 + 3479.1- 2461.2 = 1609.3 kJ/kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-282

(c) qin = (h3 - h2 )+ (h5 - h4 ) = 3375.1- 201.90 + 3479.1- 2783.7 = 3868.5 kJ/kg (d) wnet = wT, out - wp , in = 1609.4 - 10.09 = 1599.3 kJ/kg Thus the thermal efficiency is hth =

wnet 1599.3 kJ/kg = = 41.3% qin 3868.5 kJ/kg

(e) The mass flow rate of the steam is

Wnet 80,000 kJ/s = = 50.0 kg / s wnet 1599.3 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=80000 [kW] P[3]=10000 [kPa] T[3]=500 [C] P[5]=1000 [kPa] T[5]=500 [C] P[6]=10 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) "Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) T[6]=temperature(Fluid$, P=P[6], h=h[6]) T_6=T[6] x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] "(b)" w_T_out=h[3]-h[4]+h[5]-h[6] q_in=h[3]-h[2]+h[5]-h[4] w_net=w_T_out-w_p_in Eta_th=w_net/q_in "(c)" m_dot=W_dot_net/w_net © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-283

Regenerative Rankine Cycle 11-42C To have the same thermal efficiency as the Carnot cycle, the cycle must receive and reject heat isothermally. Thus the liquid should be brought to the saturated liquid state at the boiler pressure isothermally, and the steam must be a saturated vapor at the turbine inlet. This will require an infinite number of heat exchangers (feedwater heaters), as shown on the T-s diagram.

11-43C This is a smart idea because we waste little work potential but we save a lot from the heat input. The extracted steam has little work potential left, and most of its energy would be part of the heat rejected anyway. Therefore, by regeneration, we utilize a considerable amount of heat by sacrificing little work output.

11-44C Both cycles would have the same efficiency.

11-45C In open feedwater heaters, the two fluids actually mix, but in closed feedwater heaters there is no mixing.

19-46C When the simple ideal Rankine cycle is modified with regeneration, moisture content remains the same, everything else decreases.

11-47E In a regenerative Rankine cycle, the closed feedwater heater with a pump as shown in the figure is arranged so that the water at state 5 is mixed with the water at state 2 to form a feedwater which is a saturated liquid. The mass flow rate of bleed steam required to operate this unit is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible. Properties From the steam tables (Tables A-4E through A-6E), © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-284

P1 = 200 psia T1 = 350°F

P3 = 160 psia T3 = 400°F

ü ïï ý h1 @ h f @ 350° F = 321.73 Btu/lbm ïïþ ü ïï ý h3 = 1218.0 Btu/lbm ïïþ ïü ïý h = h f @ 200 psia = 355.46 Btu/lbm ïïþ 6

P6 = 200 psia x6 = 0

Analysis We take the entire unit as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout m1h1 + m3 h3 + m3 wP,in = m6 h6 m1h1 + m3 h3 + m3 wP,in = (m1 + m3 )h6

Solving this for m3 , m3 = m1

h6 - h1 355.46 - 321.73 = (2 lbm/s) = 0.0782 lbm / s (h3 - h6 ) + wP,in 1218.0 - 355.46 + 0.1344

where

wP,in = v 4 ( P5 - P4 ) = v f @ 160 psia ( P5 - P4 ) æ ö 1 Btu ÷ ÷ = (0.01815 ft 3 /lbm)(200 - 160) psia ççç 3÷ ÷ è5.404 psia ×ft ø

= 0.1344 Btu/lbm

11-48 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-285

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wpI , in = v1 (P2 - P1 ) æ 1 kJ ÷ ö = (0.001017 m3 /kg )(400 - 20 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 0.39 kJ/kg h2 = h1 + wpI , in = 251.42 + 0.39 = 251.81 kJ/kg

P3 = 0.4 MPaïüï h3 = h f @ 0.4 MPa = 604.66 kJ/kg ý ïïþv 3 = v f @ 0.4 MPa = 0.001084 m3 /kg sat.liquid æ 1 kJ ÷ ö wpII , in = v 3 (P4 - P3 ) = (0.001084 m3 /kg )(6000 - 400 kPa )çç = 6.07 kJ/kg ÷ çè1 kPa ×m3 ÷ ø h4 = h3 + wpII ,in = 604.66 + 6.07 = 610.73 kJ/kg

P5 = 6 MPa ü ïï h5 = 3302.9 kJ/kg ý T5 = 450°C ïïþ s5 = 6.7219 kJ/kg ×K

s6 - s f 6.7219 - 1.7765 = = 0.9661 P6 = 0.4 MPa ü ïï x6 = s 5.1191 ý fg ïïþ s6 = s5 h6 = h f + x6 h fg = 604.66 + (0.9661)(2133.4) = 2665.7 kJ/kg s7 - s f 6.7219 - 0.8320 = = 0.8325 P7 = 20 kPaü ïï x7 = s fg 7.0752 ý ïïþ s7 = s5 h7 = h f + x7 h fg = 251.42 + (0.8325)(2357.5) = 2214.0 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q @W @D ke @D pe @ 0,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

6 6

2 2

3 3

10-286

where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y, y=

h3 - h2 604.66 - 251.81 = = 0.1462 h6 - h2 2665.7 - 251.81

Then, qin = h5 - h4 = 3302.9 - 610.73 = 2692.2 kJ/kg

qout = (1- y )(h7 - h1 ) = (1- 0.1462)(2214.0 - 251.42) = 1675.7 kJ/kg and wnet = qin - qout = 2692.2 - 1675.7 = 1016.5 kJ / kg (b) The thermal efficiency is determined from

hth = 1-

qout 1675.7 kJ/kg = 1= 37.8% qin 2692.2 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[5]=6000 [kPa] T[5]=450 [C] P[6]=400 [kPa] P[7]=20 [kPa] x[3]=0 "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-287

w_net=q_in-q_out "(b)" Eta_th=1-q_out/q_in

11-49 A steam power plant that operates on an ideal regenerative Rankine cycle with a closed feedwater heater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wpI , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.001017 m3 /kg )(6000 - 20 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 6.08 kJ/kg h2 = h1 + wpI , in = 251.42 + 6.08 = 257.50 kJ/kg

P3 = 0.4 MPaüïï h3 = h f @ 0.4 MPa = 604.66 kJ/kg ý ïïþv 3 = v f @ 0.4 MPa = 0.001084 m3 /kg sat. liquid

æ 1 kJ ÷ ö wpII , in = v 3 (P9 - P3 ) = (0.001084 m3 /kg )(6000 - 400 kPa )çç = 6.07 kJ/kg ÷ çè1 kPa ×m3 ÷ ø h9 = h3 + wpII , in = 604.66 + 6.07 = 610.73 kJ/kg h8 = h3 + v 3 (P8 - P3 ) = h9 = 610.73 kJ/kg

Also, h4  h9  h8  610.73 kJ / kg since the two fluid streams which are being mixed have the same enthalpy. P5 = 6 MPa ïü ïý h5 = 3302.9 kJ/kg T5 = 450°C ïïþ s5 = 6.7219 kJ/kg ×K

10-288

s6 - s f 6.7219 - 1.7765 = = 0.9661 P6 = 0.4 MPa ïüï x6 = s 5.1191 ý fg ïïþ s6 = s5 h6 = h f + x6 h fg = 604.66 + (0.9661)(2133.4) = 2665.7 kJ/kg s7 - s f 6.7219 - 0.8320 x = = = 0.8325 P7 = 20 kPaïü ïý 7 s fg 7.0752 ïïþ s7 = s5 h7 = h f + x7 h fg = 251.42 + (0.8325)(2357.5) = 2214.0 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q @W @ Δke @ Δpe @ 0, Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) ¾ ¾® (1- y )(h - h ) = y (h - h ) i

where y is the fraction of steam extracted from the turbine (= m6 / m5 ). Solving for y, y=

h8 - h2

(h6 - h3 )+ (h8 - h2 )

610.73 - 257.50 = 0.1463 2665.7 - 604.66 + 610.73 - 257.50

Then, qin = h5 - h4 = 3302.9 - 610.73 = 2692.2 kJ/kg qout = (1- y )(h7 - h1 ) = (1- 0.1463)(2214.0 - 251.42) = 1675.4 kJ/kg

and

wnet = qin - qout = 2692.2 - 1675.4 = 1016.8 kJ / kg (b) The thermal efficiency is determined from

hth = 1-

qout 1675.4 kJ/kg = 1= 37.8% qin 2692.2 kJ/kg

10-289

P[9]=P[5] w_pII_in=v[3]*(P[9]-P[3]) h[9]=h[3]+w_pII_in h[8]=h[9] h[4]=h[9] "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) "closed feedwater heater" (1-y)*(h[8]-h[2])=y*(h[6]-h[3]) "y=m_dot_6/m_dot_5" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out "(b)" Eta_th=1-q_out/q_in

11-50E An ideal regenerative Rankine cycle with an open feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat rejected in the condenser are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

h1  h f @ 5 psia  130.18 Btu/lbm v1  v f @ 5 psia  0.01641 ft 3 /lbm wPI, in  v1 ( P2  P1 )

  1 Btu  (0.01641 ft 3 /lbm)(40  5)psia  3  5.404 psia  ft    0.11 Btu/lbm h2  h1  wpI, in  130.18  0.11  130.29 Btu/lbm

10-290

h3  h f @ 40 psia  236.14 Btu/lbm v 3  v f @ 40 psia  0.01715 ft 3 /lbm wPII, in  v 3 ( P4  P3 )   1 Btu  (0.01715 ft 3 /lbm)(500  40)psia  3  5.404 psia  ft  

 1.46 Btu/lbm h4  h3  wpII, in  236.14  1.46  237.60 Btu/lbm

P5  500 psia

 h5  1298.6 Btu/lbm   s5  1.5590 Btu/lbm  R s6  s f 1.5590  0.39213   0.9085  x6  s fg 1.28448   h6  h f  x6 h fg  236.14  (0.9085)(933.69)  1084.4 Btu/lbm

T5  600F P6  40 psia s6  s5

P7  5 psia s7  s5

  

x7 

s4  s f s fg



1.5590  0.23488  0.8230 1.60894

h7  h f  x7 h fg  130.18  (0.8230)(1000.5)  953.63 Btu/lbm

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q  W  ke  pe  0 , Ein  Eout  Esystem 0 (steady)  0

Ein  Eout

m h  m h i i

e e

 

m6 h6  m2 h2  m3 h3

 

yh6  (1  y )h2  1h3

6 /m  3 ). Solving for y, where y is the fraction of steam extracted from the turbine (  m y

h3  h2 236 .14  130 .29   0.1109 h6  h2 1084 .4  130 .29

Then,

wT,out  h5  h6  (1  y)(h6  h7 )  1298.6  1084.4  (1  0.1109)(1084.4  953.63)  330.5 Btu / lbm wP,in  wPI,in  wPII,in  0.11  1.46  1.57 Btu / lbm

qout  (1  y)(h7  h1 )  (1  0.1109)(953.63  130.18)  732.1Btu / lbm © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-291

Also, qin  h5  h4  1298 .6  237.60  1061 Btu/lbm

 th  1 

q out 732 .1  1  0.3100 q in 1061

EES (Engineering Equation Solver) SOLUTION "Given" P[5]=500 [psia] T[5]=600 [F] P[6]=40 [psia] P[7]=5 [psia] x[3]=0 "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])*Convert(psia-ft^3, Btu) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out w_T_out=h[5]-h[6]+(1-y)*(h[6]-h[7]) w_P_in=w_pI_in+w_pII_in "(b)" Eta_th=1-q_out/q_in

10-292

11-51E An ideal regenerative Rankine cycle with an open feedwater heater is considered. The change in thermal efficiency when the steam supplied to the open feedwater heater is at 60 psia rather than 40 psia is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4E, A-5E, and A-6E),

h1  h f @ 5 psia  130.18 Btu/lbm v1  v f @ 5 psia  0.01641 ft 3 /lbm wPI, in  v1 ( P2  P1 )   1 Btu  (0.01641 ft 3 /lbm)(60  5)psia  3  5.404 psia  ft  

 0.17 Btu/lbm h2  h1  wp, in  130.18  0.17  130.35 Btu/lbm

h3  h f @ 60 psia  262.20 Btu/lbm v 3  v f @ 60 psia  0.01738 ft 3 /lbm wPII, in  v 3 ( P4  P3 )   1 Btu  (0.01738 ft 3 /lbm)(500  60)psia  3   5.404 psia  ft 

 1.42 Btu/lbm h4  h3  wp, in  262.20  1.42  263.62 Btu/lbm

P5  500 psia

 h5  1298.6 Btu/lbm   s5  1.5590 Btu/lbm  R

T5  600F P6  60 psia

  

s6  s5 P7  5 psia s7  s5

  

x6 

s4  s f s fg



1.5590  0.42728  0.9300 1.21697

h6  h f  x6 h fg  262.20  (0.9300)(915.61)  1113.7 Btu/lbm x7 

s4  s f s fg



1.5590  0.23488  0.8230 1.60894

h7  h f  x7 h fg  130.18  (0.8230)(1000.5)  953.63 Btu/lbm

10-293

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q  W  ke  pe  0 ,

Ein  Eout  Esystem 0 (steady)  0 Ein  Eout

m h  m h i i

e e

 

m6 h6  m2 h2  m3 h3

 

yh6  (1  y )h2  1h3

6 /m  3 ). Solving for y, where y is the fraction of steam extracted from the turbine (  m y

h3  h2 262 .20  130 .35   0.1341 h6  h2 1113 .7  130 .35

Then, qin  h5  h4  1298.6  263.62  1035 Btu/lbm qout  (1  y)(h7  h1 )  (1  0.1341)(953.63  130.18)  713.0 Btu/lbm

and  th  1 

q out 713 .0  1  0.3111 q in 1035

When the reheater pressure is increased from 40 psia to 60 psia, the thermal efficiency increases from 0.3100 to 0.3111, which is an increase of 0.35%.

EES (Engineering Equation Solver) SOLUTION "Given" P[5]=500 [psia] T[5]=600 [F] P[6]=60 [psia] "Change to 40 psia, calculate thermal efficiency, and compare" P[7]=5 [psia] x[3]=0 "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])*Convert(psia-ft^3, Btu) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-294

x[7]=quality(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out w_T_out=h[5]-h[6]+(1-y)*(h[6]-h[7]) w_P_in=w_pI_in+w_pII_in "(b)" Eta_th=1-q_out/q_in

11-52E The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by EES. Analysis The EES program used to solve this problem as well as the solutions are given below. P[5]=500 [psia] T[5]=600 [F] P[6]=60 [psia] P[7]=5 [psia] x[3]=0 "Analysis" Fluid$='steam_iapws' "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])*Convert(psia-ft^3, Btu) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-295

w_net=q_in-q_out Eta_th=1-q_out/q_in P6 [kPa] 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200

ηth 0.3011 0.3065 0.3088 0.3100 0.3107 0.3111 0.31126 0.31129 0.3112 0.3111 0.3109 0.3107 0.3104 0.3101 0.3098 0.3095 0.3091 0.3087 0.3084 0.3080

11-53 A steam power plant operates on an ideal regenerative Rankine cycle with two open feedwater heaters. The net power output of the power plant and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

10-296

(a) From the steam tables (Tables A-4, A-5, and A-6),

10-297

h1 = h f @ 15 kPa = 225.94 kJ / kg v1 = v f @ 15 kPa = 0.001014 m3 /kg

æ 1 kJ ö ÷= 0.188 kJ/kg wpI , in = v1 (P2 - P1 ) = (0.001014 m3 /kg )(200 - 15 kPa )çç ÷ çè1 kPa ×m3 ÷ ø h2 = h1 + wpI , in = 225.94 + 0.188 = 226.13 kJ/kg

P3 = 0.2 MPaüïï h3 = h f @ 0.2 MPa = 504.71 kJ/kg ý ïïþv 3 = v f @ 0.2 MPa = 0.001061 m3 /kg sat. liquid æ 1 kJ ö÷ wpII , in = v 3 (P4 - P3 ) = (0.001061 m3 /kg )(600 - 200 kPa )çç ÷ çè1 kPa ×m3 ø÷ = 0.42 kJ/kg h4 = h3 + wpII , in = 504.71 + 0.42 = 505.13 kJ/kg

P5 = 0.6 MPaüïï h5 = h f @ 0.6 MPa = 670.38 kJ/kg ý ïïþv 5 = v f @ 0.6 MPa = 0.001101 m3 /kg sat. liquid æ 1 kJ ÷ ö wpIII , in = v 5 (P6 - P5 ) = (0.001101 m3 /kg )(8000 - 600 kPa )çç ÷ çè1 kPa ×m3 ÷ ø = 8.144 kJ/kg h6 = h5 + wpIII , in = 670.38 + 8.144 = 678.52 kJ/kg

P7 = 8 MPa ü ïï h7 = 3521.8 kJ/kg ý T7 = 550°C ïïþ s7 = 6.8800 kJ/kg ×K P8 = 0.6 MPa ïü ïý h = 2809.6 kJ/kg ïïþ 8 s8 = s7

s9 - s f 6.8800 - 1.5302 x = = = 0.9559 P9 = 0.2 MPa ïü ïý 9 s fg 5.5968 ïïþ s9 = s7 h9 = h f + x9 h fg = 504.71 + (0.9559)(2201.6) = 2609.1 kJ/kg

s10 - s f 6.8800 - 0.7549 = = 0.8446 P10 = 15 kPa üïï x10 = s fg 7.2522 ý ïïþ s10 = s7 h10 = h f + x10 h fg = 225.94 + (0.8446)(2372.3) = 2229.6 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0, FWH-2: Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i

8 8

where y is the fraction of steam extracted from the turbine (= m8 / m5 ). Solving for y, y=

h5 - h4 670.38 - 505.13 = = 0.07171 h8 - h4 2809.6 - 505.13

FWH-1:

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® zh + (1- y - z )h = (1- y )h i i

e e

9 9

2 2

3 3

where z is the fraction of steam extracted from the turbine (= m9 / m5 ) at the second stage. Solving for z, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-298

h3 - h2 504.71- 226.13 (1- y ) = (1- 0.07171) = 0.1085 h9 - h2 2609.1- 226.13

Then, qin = h7 - h6 = 3521.8 - 678.52 = 2843.3 kJ/kg

qout = (1- y - z )(h10 - h1 ) = (1- 0.07171- 0.1085)(2229.6 - 225.94) = 1642.5 kJ/kg wnet = qin - qout = 2843.3 - 1642.5 = 1200.8 kJ/kg

and Wnet = mwnet = (24 kg/s)(1200.8 kJ/kg ) = 28,819 kW @28.8 MW

(b) The thermal efficiency is

hth = 1-

qout 1642.5 kJ/kg = 1= 0.4223 = 42.2% qin 2843.3 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[7]=8000 [kPa] T[7]=550 [C] P[8]=600 [kPa] P[9]=200 [kPa] P[10]=15 [kPa] x[3]=0 x[5]=0 m_dot=24 [kg/s] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[10] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[9] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[5]=P[8] P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "pump III" h[5]=enthalpy(Fluid$, P=P[5], x=x[5]) v[5]=volume(Fluid$, P=P[5], x=x[5]) P[6]=P[7] w_pIII_in=v[5]*(P[6]-P[5]) h[6]=h[5]+w_pIII_in "turbine" h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-299

s[8]=s[7] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) s[9]=s[7] h[9]=enthalpy(Fluid$, P=P[9], s=s[9]) x[9]=quality(Fluid$, P=P[9], s=s[9]) s[10]=s[7] h[10]=enthalpy(Fluid$, P=P[10], s=s[10]) x[10]=quality(Fluid$, P=P[10], s=s[10]) "open feedwater heaters" y*h[8]+(1-y)*h[4]=h[5] "y=m_8/m_5" z*h[9]+(1-y-z)*h[2]=(1-y)*h[3] "z=m_9/m_5" "cycle" q_in=h[7]-h[6] q_out=(1-y-z)*(h[10]-h[1]) w_net=q_in-q_out W_dot_net=m_dot*w_net "kW" "(b)" Eta_th=1-q_out/q_in

11-54 A steam power plant operates on an ideal regenerative Rankine cycle with two feedwater heaters, one closed and one open. The mass flow rate of steam through the boiler for a net power output of 400 MW and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 / kg wpI , in = v1 (P2 - P1 )

æ 1 kJ ÷ ö = (0.00101 m3 /kg )(600 - 10 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 0.60 kJ/kg h2 = h1 + wpI , in = 191.81 + 0.60 = 192.40 kJ/kg

10-300

P3 = 0.6 MPaïüï h3 = h f @ 0.3 MPa = 670.38 kJ/kg ý ïïþv 3 = v f @ 0.3 MPa = 0.001101 m3 /kg sat. liquid wpII , in = v 3 (P4 - P3 ) æ 1 kJ ö ÷ = (0.001101 m3 /kg )(10000 - 600 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 10.35 kJ/kg h4 = h3 + wpII , in = 670.38 + 10.35 = 680.73 kJ/kg

h = h7 = h f @1.2 MPa = 798.33 kJ/kg P6 = 1.2 MPa ïü ïý 6 ïïþ T6 = Tsat@1.2 MPa = 188.0°C sat. liquid

T5 = T5 ,

P5 = 10MPa ® h5 = 798.33 kJ / kg

10-301

10-302

P8 = 10 MPaïüï h8 = 3625.8 kJ/kg ý T8 = 600°C ïïþ s8 = 6.9045 kJ/kg ×K P9 = 1.2 MPa ü ïï ý h9 = 2974.5 kJ/kg ïïþ s9 = s8 P10 = 0.6 MPa ïü ïý h = 2820.9 kJ/kg ïïþ 10 s10 = s8

s11 - s f 6.9045 - 0.6492 x = = = 0.8341 P11 = 10 kPa ïü ïý 11 s fg 7.4996 ïïþ s11 = s8 h11 = h f + x11h fg = 191.81 + (0.8341)(2392.1) = 2187.0 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) ¾ ¾® y (h - h ) = (h - h ) i i

e e

where y is the fraction of steam extracted from the turbine (= m10 / m5 ) . Solving for y,

h5 - h4 798.33 - 680.73 = = 0.05404 h9 - h6 2974.5 - 798.33

For the open FWH,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h + m h + m h = m h ¾ ¾® yh + (1- y - z )h + zh = (1)h i i

e e

7 7

2 2

10 10

3 3

where z is the fraction of steam extracted from the turbine (= m9 / m5 ) at the second stage. Solving for z, z=

(h3 - h2 )- y (h7 - h2 ) h10 - h2

670.38 - 192.40 - (0.05404)(798.33 - 192.40) 2820.9 - 192.40

= 0.1694

Then, qin = h8 - h5 = 3625.8 - 798.33 = 2827 kJ/kg

qout = (1- y - z )(h11 - h1 ) = (1- 0.05404 - 0.1694)(2187.0 - 191.81) = 1549 kJ/kg wnet = qin - qout = 2827 - 1549 = 1278 kJ/kg and m=

Wnet 400,000 kJ/s = = 313.0 kg / s wnet 1278 kJ/kg

(b) hth = 1-

qout 1549 kJ/kg = 1= 0.452 = 45.2% qin 2827 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Input Data" P[8] = 10000 [kPa] T[8] = 600 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-303

P[9] = 1200 [kPa] P_cfwh=600 [kPa] P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=400[MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency" Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages" Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages" Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages" Eta_pump = 100/100 "Pump isentropic efficiency" Fluid$='Steam_IAPWS' "Condenser exit pump or Pump 1 analysis" P[1] = P[11] P[2]=P[10] h[1]=enthalpy(Fluid$,P=P[1],x=0) "Sat'd liquid" v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1]) "SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] "Steady-flow conservation of energy" h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[5]=P[8] P[4] = P[5] P[3]=P[10] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3]) "SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Closed Feedwater Heater analysis" P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] "Steady-flow conservation of energy" h[5]=enthalpy(Fluid$,P=P[6],x=0) "h[5] = h(T[5], P[5]) where T[5]=Tsat at P[9]" T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Condensate leaves heater as sat. liquid at P[6]" s[5]=entropy(Fluid$,P=P[6],h=h[5]) h[6]=enthalpy(Fluid$,P=P[6],x=0) T[6]=temperature(Fluid$,P=P[6],x=0) "Condensate leaves heater as sat. liquid at P[6]" s[6]=entropy(Fluid$,P=P[6],x=0) "Trap analysis" P[7] = P[10] y*h[6] = y*h[7] "Steady-flow conservation of energy for the trap operating as a throttle" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "Boiler analysis" q_in + h[5]=h[8]"SSSF conservation of energy for the Boiler" h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-304

"Turbine analysis" ss[9]=s[8] hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9]) Ts[9]=temperature(Fluid$,s=ss[9],P=P[9]) h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9]) "Definition of turbine efficiency for high pressure stages" T[9]=temperature(Fluid$,P=P[9],h=h[9]) s[9]=entropy(Fluid$,P=P[9],h=h[9]) x[9]=quality(Fluid$,P=P[9],h=h[9]) ss[10]=s[8] hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10]) Ts[10]=temperature(Fluid$,s=ss[10],P=P[10]) h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10]) "Definition of turbine efficiency for Intermediate pressure stages" T[10]=temperature(Fluid$,P=P[10],h=h[10]) s[10]=entropy(Fluid$,P=P[10],h=h[10]) x[10]=quality(Fluid$,P=P[10],h=h[10]) ss[11]=s[8] hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11]) Ts[11]=temperature(Fluid$,s=ss[11],P=P[11]) h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11]) "Definition of turbine efficiency for low pressure stages" T[11]=temperature(Fluid$,P=P[11],h=h[11]) s[11]=entropy(Fluid$,P=P[11],h=h[11]) x[11]=quality(Fluid$,P=P[11],h=h[11]) h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "SSSF conservation of energy for turbine" "Condenser analysis" (1-y-z)*h[11]=q_out+(1-y-z)*h[1] "SSSF First Law for the Condenser" "Cycle Statistics" w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) Eta_th=w_net/q_in W_dot_net = m_dot * w_net "s10_f=entropy(Fluid$, P=P[10], x=0); s10_g=entropy(Fluid$, P=P[10], x=1); s10_fg=s10_g-s10_f h10_f=enthalpy(Fluid$, P=P[10], x=0); h10_g=enthalpy(Fluid$, P=P[10], x=1); h10_fg=h10_g-h10_f s6_f=entropy(Fluid$, P=P[6], x=0); s6_g=entropy(Fluid$, P=P[6], x=1); s6_fg=s6_g-s6_f h6_f=enthalpy(Fluid$, P=P[6], x=0); h6_g=enthalpy(Fluid$, P=P[6], x=1); h6_fg=h6_g-h6_f"

11-55 Problem 11-54 is reconsidered. The effects of turbine and pump efficiencies on the mass flow rate and thermal efficiency are to be investigated. Also, the T-s diagram is to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[8] = 10000 [kPa] T[8] = 600 [C] P[9] = 1200 [kPa] P_cfwh=600 [kPa] P[10] = P_cfwh P_cond=10 [kPa] P[11] = P_cond W_dot_net=400 [MW]*Convert(MW, kW) Eta_turb= 100/100 "Turbine isentropic efficiency" Eta_turb_hp = Eta_turb "Turbine isentropic efficiency for high pressure stages" Eta_turb_ip = Eta_turb "Turbine isentropic efficiency for intermediate pressure stages" Eta_turb_lp = Eta_turb "Turbine isentropic efficiency for low pressure stages" Eta_pump = 100/100 "Pump isentropic efficiency" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-305

"Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[11] P[2]=P[10] h[1]=enthalpy(Fluid$,P=P[1],x=0) v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1]) w_pump1=w_pump1_s/Eta_pump h[1]+w_pump1= h[2] s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" z*h[10] + y*h[7] + (1-y-z)*h[2] = 1*h[3] h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[5]=P[8] P[4] = P[5] P[3]=P[10] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3]) w_pump2=w_pump2_s/Eta_pump h[3]+w_pump2= h[4] s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) "Closed Feedwater Heater analysis" P[6]=P[9] y*h[9] + 1*h[4] = 1*h[5] + y*h[6] h[5]=enthalpy(Fluid$,P=P[6],x=0) T[5]=temperature(Fluid$,P=P[5],h=h[5]) s[5]=entropy(Fluid$,P=P[6],h=h[5]) h[6]=enthalpy(Fluid$,P=P[6],x=0) T[6]=temperature(Fluid$,P=P[6],x=0) s[6]=entropy(Fluid$,P=P[6],x=0) "Trap analysis" P[7] = P[10] y*h[6] = y*h[7] T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "Boiler analysis" q_in + h[5]=h[8] h[8]=enthalpy(Fluid$, T=T[8], P=P[8]) s[8]=entropy(Fluid$, T=T[8], P=P[8]) "Turbine analysis" ss[9]=s[8] hs[9]=enthalpy(Fluid$,s=ss[9],P=P[9]) Ts[9]=temperature(Fluid$,s=ss[9],P=P[9]) h[9]=h[8]-Eta_turb_hp*(h[8]-hs[9]) T[9]=temperature(Fluid$,P=P[9],h=h[9]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-306

s[9]=entropy(Fluid$,P=P[9],h=h[9]) ss[10]=s[8] hs[10]=enthalpy(Fluid$,s=ss[10],P=P[10]) Ts[10]=temperature(Fluid$,s=ss[10],P=P[10]) h[10]=h[9]-Eta_turb_ip*(h[9]-hs[10]) T[10]=temperature(Fluid$,P=P[10],h=h[10]) s[10]=entropy(Fluid$,P=P[10],h=h[10]) ss[11]=s[8] hs[11]=enthalpy(Fluid$,s=ss[11],P=P[11]) Ts[11]=temperature(Fluid$,s=ss[11],P=P[11]) h[11]=h[10]-Eta_turb_lp*(h[10]-hs[11]) T[11]=temperature(Fluid$,P=P[11],h=h[11]) s[11]=entropy(Fluid$,P=P[11],h=h[11]) h[8] =y*h[9] + z*h[10] + (1-y-z)*h[11] + w_turb "Condenser analysis" (1-y-z)*h[11]=q_out+(1-y-z)*h[1] "Cycle Statistics" w_net=w_turb - ((1-y-z)*w_pump1+ w_pump2) ηth Eta_th=w_net/q_in W_dot_net = m_dot * w_net

turb

0.7 0.75 0.8 0.85 0.9 0.95 1

0.3834 0.397 0.4096 0.4212 0.4321 0.4423 0.452

m [kg / s] 369 356.3 345.4 335.8 327.4 319.8 313

0.7 0.75 0.8 0.85 0.9 0.95 1

0.4509 0.4511 0.4513 0.4515 0.4517 0.4519 0.452

m [kg / s] 313.8 313.6 313.4 313.3 313.2 313.1 313

Pcfwh [kPa] 100 200 300 400 500 600 700 800 900 1000

0.1702 0.1301 0.1041 0.08421 0.06794 0.05404 0.04182 0.03088 0.02094 0.01179

0.05289 0.09634 0.1226 0.1418 0.1569 0.1694 0.1801 0.1895 0.1979 0.2054

ηpump

10-307

10-308

11-56 An ideal regenerative Rankine cycle with a closed feedwater heater is considered. The work produced by the turbine, the work consumed by the pumps, and the heat added in the boiler are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-309

Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001017 m3 /kg)(3000 - 20)kPa çç ÷ çè1 kPa ×m3 ÷ ø

= 3.03 kJ/kg h2 = h1 + wp, in = 251.42 + 3.03 = 254.45 kJ/kg

P4 = 3000 kPa ïü ïý ïïþ T4 = 350°C

h4 = 3116.1 kJ/kg s4 = 6.7450 kJ/kg ×K

10-310

P5 = 1000 kPa ü ïï ý h5 = 2851.9 kJ/kg ïïþ s5 = s4 P6 = 20 kPa ïü ïý ïïþ s6 = s4

x6 =

s6 - s f s fg

6.7450 - 0.8320 = 0.8357 7.0752

h6 = h f + x6 h fg = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure. P7 = 1000 kPaü ïï h7 = 762.51 kJ/kg ý ïïþ T7 = 179.9°C x7 = 0 h8 = h7 = 762.51 kJ/kg

ïü ïý h = 763.53 kJ/kg T3 = T7 = 209.9°Cïïþ 3 P3 = 3000 kPa

An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine (= m5 / m4 ) for closed feedwater heater:

å mh = å m h i i

e e

m5 h5 + m2 h2 = m3 h3 + m7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging, h - h2 763.53 - 254.45 y= 3 = = 0.2437 h5 - h7 2851.9 - 762.51 Then, wT, out = h4 - h5 + (1- y )(h5 - h6 ) = 3116.1- 2851.9 + (1- 0.2437)(2851.9 - 2221.7) = 740.9 kJ / kg

wP, in = 3.03 kJ / kg q in = h4 - h3 = 3116.1- 763.53 = 2353 kJ / kg

Also, wnet = wT, out - wP, in = 740.9 - 3.03 = 737.8 kJ/kg

hth =

wnet 737.8 = = 0.3136 qin 2353

EES (Engineering Equation Solver) SOLUTION "Given" P[4]=3000 [kPa] T[4]=350 [C] P[5]=1000 [kPa] P[6]=20 [kPa] "Analysis" Fluid$='steam_iapws' P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "(a)" "pump I" x[1]=0 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-311

h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in

11-57 Problem 11-56 is reconsidered. The optimum bleed pressure for the open feedwater heater that maximizes the thermal efficiency of the cycle is to be determined by appropriate software. Analysis The EES program used to solve this problem and the solution is given below. P[4]=3000 [kPa] T[4]=350 [C] P[5]=600 [kPa] P[6]=20 [kPa] Fluid$='steam_iapws' P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-312

h[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x[6]=quality(Fluid$, P=P[6], s=s[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in

P6 [kPa] 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280 290

ηth 0.32380 0.32424 0.32460 0.32490 0.32514 0.32534 0.32550 0.32563 0.32573 0.32580 0.32585 0.32588 0.32590 0.32589 0.32588 0.32585 0.32581 0.32576 0.32570 0.32563

10-313

11-58 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-314

Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001017 m3 /kg)(3000 - 20)kPa çç ÷ çè1 kPa ×m3 ÷ ø

10-315

h2 = h1 + wp, in = 251.42 + 3.03 = 254.45 kJ/kg

P4 = 3000 kPaïü ïý h4 = 3116.1 kJ/kg T4 = 350°C ïïþ s4 = 6.7450 kJ/kg ×K P5 = 1000 kPaü ïï ý h5 s = 2851.9 kJ/kg ïïþ s5 s = s4 s6 s - s f 6.7450 - 0.8320 x = = = 0.8357 P6 = 20 kPa ïü ïý 6 s s fg 7.0752 ïïþ s6 s = s4 h6 s = h f + x6 s h fg = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg

hT =

h4 - h5 ¾¾ ® h5 = h4 - hT (h4 - h5s ) = 3116.1- (0.90)(3116.1- 2851.9) = 2878.3 kJ/kg h4 - h5 s

hT =

h4 - h6 ¾¾ ® h6 = h4 - hT (h4 - h6s ) = 3116.1- (0.90)(3116.1- 2221.7) = 2311.1 kJ/kg h4 - h6 s

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.

P7 = 1000 kPaïü ïý h7 = 762.51 kJ/kg ïïþ T7 = 179.9°C x7 = 0 ïü ïý h = 763.53 kJ/kg T3 = T7 = 209.9°Cïïþ 3 P3 = 3000 kPa

An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine (= m5 / m4 ) for closed feedwater heater:

å mh = å m h i i

e e

m5 h5 + m2 h2 = m3 h3 + m7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging,

h3 - h2 763.53 - 254.45 = = 0.2406 h5 - h7 2878.3 - 762.51

Then, wT, out = h4 - h5 + (1- y )(h5 - h6 ) = 3116.1- 2878.3 + (1- 0.2406)(2878.3 - 2311.1) = 668.5 kJ/kg wP, in = 3.03 kJ/kg q in = h4 - h3 = 3116.1- 763.53 = 2353 kJ/kg

Also,

wnet = wT, out - wP, in = 668.5 - 3.03 = 665.5 kJ/kg

hth =

wnet 665.5 = = 0.2829 = 28.3% qin 2353

EES (Engineering Equation Solver) SOLUTION "Given" P[4]=3000 [kPa] T[4]=350 [C] P[5]=1000 [kPa] P[6]=20 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-316

eta_T=0.90 "Analysis" Fluid$='steam_iapws' P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "(a)" "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x_s[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h_s[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x_s[6]=quality(Fluid$, P=P[6], s=s[6]) h[5]=h[4]-eta_T*(h[4]-h_s[5]) h[6]=h[4]-eta_T*(h[4]-h_s[6]) x[5]=quality(Fluid$, P=P[5], h=h[5]) x[6]=quality(Fluid$, P=P[6], h=h[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in

11-59 A regenerative Rankine cycle with a closed feedwater heater is considered. The thermal efficiency is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-317

Analysis From the steam tables (Tables A-4, A-5, and A-6 or EES), saturated liquid at the condenser pressure, the enthalpies become ïü ïý h1 @ h f@50°C = 209.34 kJ/kg T1 = Tsat@20 kPa - 10 = 60.06 - 10 @ 50°Cïïþv1 @v f@50°C = 0.001012 m 3 /kg P1 = 20 kPa

wp, in = v1 ( P2 - P1 )

æ 1 kJ ö ÷ = (0.001012 m3 /kg)(3000 - 20)kPa çç ÷ ÷ çè1 kPa ×m3 ø

= 3.02 kJ/kg h2 = h1 + wp, in = 209.34 + 3.02 = 212.36 kJ/kg

10-318

P4 = 3000 kPaü ïï h4 = 3116.1 kJ/kg ý T4 = 350°C ïïþ s4 = 6.7450 kJ/kg ×K

P5 = 1000 kPaïü ïý h = 2851.9 kJ/kg ïïþ 5 s s5 s = s4 s6 s - s f 6.7450 - 0.8320 = = 0.8357 P6 = 20 kPa ü ïï x6 s = s 7.0752 ý fg ïïþ s6 s = s4 h6 s = h f + x6 s h fg = 251.42 + (0.8357)(2357.5) = 2221.7 kJ/kg

hT =

h4 - h5 ¾¾ ® h5 = h4 - hT (h4 - h5s ) = 3116.1- (0.90)(3116.1- 2851.9) = 2878.3 kJ/kg h4 - h5 s

hT =

h4 - h6 ¾¾ ® h6 = h4 - hT (h4 - h6s ) = 3116.1- (0.90)(3116.1- 2221.7) = 2311.1 kJ/kg h4 - h6 s

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.

P7 = 1000 kPaü ïï h7 = 762.51 kJ/kg ý ïïþ T7 = 179.9°C x7 = 0 ïü ïý h = 763.53 kJ/kg T3 = T7 = 209.9°Cïïþ 3 P3 = 3000 kPa

An energy balance on the heat exchanger gives the fraction of steam extracted from the turbine ( = m5 /m4 ) for closed feedwater heater:

å mh = å m h i

m5 h5 + m2 h2 = m3 h3 + m7 h7 yh5 + 1h2 = 1h3 + yh7 Rearranging,

h3 - h2 763.53 - 212.36 = = 0.2605 h5 - h7 2878.3 - 762.51

Then, wT, out = h4 - h5 + (1- y )(h5 - h6 ) = 3116.1- 2878.3 + (1- 0.2605)(2878.3 - 2311.1) = 657.2 kJ/kg

wP, in = 3.03 kJ/kg

qin = h4 - h3 = 3116.1- 763.53 = 2353 kJ/kg Also, wnet = wT, out - wP, in = 657.2 - 3.03 = 654.2 kJ/kg

hth =

wnet 654.2 = = 0.2781 qin 2353

10-319

eta_T=0.90 DELTAT_subcool=10 [C] "Analysis" Fluid$='steam_iapws' P[3]=P[4] P[2]=P[3] P[7]=P[5] P[1]=P[6] "(a)" "pump I" T_sat=temperature(Fluid$, P=P[1], x=0) T[1]=T_sat-DELTAT_subcool h[1]=enthalpy(Fluid$, x=0, T=T[1]) v[1]=volume(Fluid$, x=0, T=T[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x_s[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h_s[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x_s[6]=quality(Fluid$, P=P[6], s=s[6]) h[5]=h[4]-eta_T*(h[4]-h_s[5]) h[6]=h[4]-eta_T*(h[4]-h_s[6]) x[5]=quality(Fluid$, P=P[5], h=h[5]) x[6]=quality(Fluid$, P=P[6], h=h[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in

11-60 The effect of pressure drop and non-isentropic turbine on the rate of heat input is to be determined for a given power plant. Analysis The EES program used to solve this problem as well as the solutions are given below. P[3]=3000 [kPa] DELTAP_boiler=10 [kPa] "Change to zero for no pressure drop" P[4]=P[3]-DELTAP_boiler T[4]=350 [C] P[5]=1000 [kPa] P[6]=20 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-320

eta_T=0.9 "Change to 1 for isentropic turbine " Fluid$='steam_iapws' P[2]=P[3] P[7]=P[5] P[1]=P[6] "pump I" x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "turbine" h[4]=enthalpy(Fluid$, P=P[4], T=T[4]) s[4]=entropy(Fluid$, P=P[4], T=T[4]) s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s[5]) T[5]=temperature(Fluid$, P=P[5], s=s[5]) x_s[5]=quality(Fluid$, P=P[5], s=s[5]) s[6]=s[4] h_s[6]=enthalpy(Fluid$, P=P[6], s=s[6]) x_s[6]=quality(Fluid$, P=P[6], s=s[6]) h[5]=h[4]-eta_T*(h[4]-h_s[5]) h[6]=h[4]-eta_T*(h[4]-h_s[6]) x[5]=quality(Fluid$, P=P[5], h=h[5]) x[6]=quality(Fluid$, P=P[6], h=h[6]) "closed feedwater heater" x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) T[7]=temperature(Fluid$, P=P[7], x=x[7]) T[3]=T[7] h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) y=(h[3]-h[2])/(h[5]-h[7]) "y=m_dot_5/m_dot_4" "cycle" q_in=h[4]-h[3] w_T_out=h[4]-h[5]+(1-y)*(h[5]-h[6]) w_net=w_T_out-w_p_in Eta_th=w_net/q_in Solution with 10 kPa pressure drop in the boiler: DELTAP_boiler=10 [kPa] eta_T=0.9 Eta_th=0.2827 Fluid$='steam_iapws' P[3]=3000 [kPa] P[4]=2990 [kPa] q_in = 2352.8 [kJ / kg] w_net  665.1[kJ / kg] w_p_in  3.031[m  3  kPa / kg] w_T_out  668.1[kJ / kg] y=0.2405 Solution without any pressure drop in the boiler: DELTAP_boiler=0 [kPa] eta_T=1 Eta_th=0.3136 Fluid$='steam_iapws' P[3]=3000 [kPa] P[4]=3000 [kPa] q_in = 2352.5 [kJ / kg] w_net  737.8 [kJ / kg] w_T_out  740.9 [kJ / kg] w_p_in  3.031[m  3  kPa / kg] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-321

y=0.2437

11-61 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwater heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg æ 1 kJ ö ÷ wpI , in = v1 (P2 - P1 ) = (0.00101 m3 /kg )(800 - 10 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 0.80 kJ/kg h2 = h1 + wpI , in = 191.81 + 0.80 = 192.61 kJ/kg

P3 = 0.8 MPaüïï h3 = h f @ 0.8 MPa = 720.87 kJ/kg ý ïïþv 3 = v f @ 0.8 MPa = 0.001115 m3 /kg sat.liquid © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-322

æ 1 kJ ö÷ wpII , in = v 3 (P4 - P3 )= (0.001115 m 3 /kg )(10,000 - 800 kPa )çç ÷ çè1 kPa ×m3 ø÷ = 10.26 kJ/kg h4 = h3 + wpII , in = 720.87 + 10.26 = 731.12 kJ/kg

P5 = 10 MPa ïü ïý h5 = 3502.0 kJ/kg T5 = 550°C ïïþ s5 = 6.7585 kJ/kg ×K P6 = 0.8 MPa ïü ïý h = 2812.1 kJ/kg ïïþ 6

s6 = s5

P7 = 0.8 MPa ü ïï h7 = 3481.3 kJ/kg ý T7 = 500°C ïïþ s7 = 7.8692 kJ/kg ×K

s8 - s f 7.8692 - 0.6492 = = 0.9627 P8 = 10 kPa ü ïï x8 = s fg 7.4996 ý ïïþ s8 = s7 h8 = h f + x8 h fg = 191.81 + (0.9627)(2392.1) = 2494.7 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0, Ein - Eout = D Esystem ¿ 0 (steady) = 0 ®

Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

6 6

2 2

3 3

where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y,

h3 - h2 720.87 - 192.61 = = 0.2017 h6 - h2 2812.1- 192.61

Then,

qin = (h5 - h4 )+ (1- y )(h7 - h6 ) = (3502.0 - 731.12)+ (1- 0.2017)(3481.3 - 2812.1) = 3305.1 kJ/kg qout = (1- y )(h8 - h1 ) = (1- 0.2017)(2494.7 - 191.81) = 1838.5 kJ/kg wnet = qin - qout = 3305.1- 1838.5 = 1466.6 kJ/kg and m =

Wnet 80,000 kJ/s = = 54.5 kg / s wnet 1466.1 kJ/kg

(b) hth =

wnet 1466.1 kJ/kg = = 44.4% qin 3305.1 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=80000 [kW] P[5]=10000 [kPa] T[5]=550 [C] P[6]=800 [kPa] T[7]=500 [C] P[8]=10 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-323

P[1]=P[8] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "high pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) "low pressure turbine" P[7]=P[6] h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) s[8]=s[7] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) x[8]=quality(Fluid$, P=P[8], s=s[8]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4]+(1-y)*(h[7]-h[6]) q_out=(1-y)*(h[8]-h[1]) w_net=q_in-q_out m_dot=W_dot_net/w_net "(b)" Eta_th=w_net/q_in

11-62 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with a closed feedwater heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-324

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg æ 1 kJ ö ÷ wpI , in = v1 (P2 - P1 ) = (0.00101 m3 /kg )(10,000 - 10 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 10.09 kJ/kg h2 = h1 + wpI , in = 191.81 + 10.09 = 201.90 kJ/kg

10-325

P3 = 0.8 MPaïüï h3 = h f @ 0.8 MPa = 720.87 kJ/kg ý ïïþv 3 = v f @ 0.8 MPa = 0.001115 m3 /kg sat.liquid æ 1 kJ ÷ ö wpII , in = v 3 (P4 - P3 ) = (0.001115 m3 /kg )(10,000 - 800 kPa )çç ÷ çè1 kPa ×m3 ÷ ø = 10.26 kJ/kg h4 = h3 + wpII , in = 720.87 + 10.26 = 731.13 kJ/kg

Also, h4  h9  h10  731.12 kJ / kg since the two fluid streams that are being mixed have the same enthalpy. P5 = 10 MPa ü ïï h5 = 3502.0 kJ/kg ý T5 = 550°C ïïþ s5 = 6.7585 kJ/kg ×K

P6 = 0.8 MPa ïü ïý h = 2812.7 kJ/kg ïïþ 6 s6 = s5 P7 = 0.8 MPa ü ïï h7 = 3481.3 kJ/kg ý T7 = 500°C ïïþ s7 = 7.8692 kJ/kg ×K

s8 - s f 7.8692 - 0.6492 x = = = 0.9627 P8 = 10 kPa ïü ïý 8 s fg 7.4996 ïïþ s8 = s7 h8 = h f + x8 h fg = 191.81 + (0.9627)(2392.1) = 2494.7 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @D ke @ Δpe @ 0,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) ¾ ¾® (1- y )(h - h ) = y (h - h ) i

where y is the fraction of steam extracted from the turbine (= m3 / m4 ). Solving for y,

h9 - h2 731.13 - 201.90 = = 0.2019 h h + h h 2812.7 720.87 + 731.13 - 201.90 ( 6 3) ( 9 2)

Then,

qin = (h5 - h4 )+ (1- y )(h7 - h6 ) = (3502.0 - 731.13)+ (1- 0.2019)(3481.3 - 2812.7) = 3304.5 kJ/kg qout = (1- y )(h8 - h1 ) = (1- 0.2019)(2494.7 - 191.81) = 1837.9 kJ/kg

wnet = qin - qout = 3304.5 - 1837.8 = 1466.6 kJ/kg and

Wnet 80,000 kJ/s = = 54.5 kg / s wnet 1467.1 kJ/kg

(b) hth = 1-

qout 1837.8 kJ/kg = 1= 44.4% qin 3304.5 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=80000 [kW] P[5]=10000 [kPa] T[5]=550 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-326

P[6]=800 [kPa] T[7]=500 [C] P[8]=10 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[8] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[5] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] P[3]=P[6] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in h[9]=h[4] h[10]=h[4] "high pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) "low pressure turbine" P[7]=P[6] h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) s[8]=s[7] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) "closed feedwater heater" (1-y)*(h[9]-h[2])=y*(h[6]-h[3]) "y=m_dot_3/m_dot_4" "cycle" q_in=h[5]-h[4]+(1-y)*(h[7]-h[6]) q_out=(1-y)*(h[8]-h[1]) w_net=q_in-q_out m_dot=W_dot_net/w_net "(b)" Eta_th=w_net/q_in

11-63E A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two open feedwater heaters. The mass flow rate of steam through the boiler, the net power output of the plant, and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-327

(a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 1 psia = 69.72 Btu/lbm

v1 = v f @ 1 psia = 0.01614 ft 3 /lbm wpI , in = v1 (P2 - P1 )

æ ö 1 Btu ÷ ÷ = (0.01614 ft 3 /lbm)(40 - 1 psia )çç 3÷ ÷ çè5.4039 psia ×ft ø © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-328

= 0.12 Btu/lbm h2 = h1 + wpI , in = 69.72 + 0.12 = 69.84 Btu/lbm

ïü ïý h3 = h f @ 40 psia = 236.14 Btu/lbm ïïþ v 3 = v f @ 40 psia = 0.01715 ft 3 /lbm

P3 = 40 psia sat. liquid

wpII , in = v 3 (P4 - P3 ) æ ö 1 Btu ÷ ÷ = (0.01715 ft 3 /lbm)(250 - 40 psia )çç ÷ çè5.4039 psia ×ft 3 ÷ ø

= 0.67 Btu/lbm h4 = h3 + wpII , in = 236.14 + 0.67 = 236.81 Btu/lbm

P5 = 250 psia ïü ïý h5 = h f @ 250 psia = 376.09 Btu/lbm ïïþ v 5 = v f @ 250 psia = 0.01865 ft 3 /lbm sat. liquid

wpIII , in = v 5 (P6 - P5 )

æ ö 1 Btu ÷ ÷ = (0.01865 ft 3 /lbm)(1500 - 250 psia )ççç 3÷ è5.4039 psia ×ft ÷ ø = 4.31 Btu/lbm

h6 = h5 + wpIII , in = 376.09 + 4.31 = 380.41 Btu/lbm

P7 = 1500 psia ü ïï h7 = 1550.5 Btu/lbm ý ïïþ s7 = 1.6402 Btu/lbm ×R T7 = 1100°F ü ïï ý h8 = 1308.5 Btu/lbm ïïþ

P8 = 250 psia s8 = s7

ü ïï ý h9 = 1248.8 Btu/lbm ïïþ

P9 = 140 psia s9 = s7 P10 = 140 psia T10 = 1000°F

ïü ïý h10 = 1531.3 Btu/lbm ïïþ s10 = 1.8832 Btu/lbm ×R

ïü ïý h = 1356.0 Btu/lbm ïïþ 11

P11 = 40 psia s11 = s10

x12 =

s12 - s f

1.8832 - 0.13262 = 0.9488 1.84495

s fg ïü ïý h = h + x h = 69.72 + (0.9488)(1035.7) f 12 fg ïïþ 12 = 1052.4 Btu/lbm

P12 = 1 psia s12 = s10

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0 , FWH-2:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout

å mh = å m h i i

e e

¾¾ ®

m8 h8 + m4 h4 = m5 h5

¾¾ ®

yh8 + (1- y )h4 = 1(h5 )

where y is the fraction of steam extracted from the turbine ( = m8 / m5 ). Solving for y, y=

h5 - h4 376.09 - 236.81 = = 0.1300 h8 - h4 1308.5 - 236.81

10-329 Ein - Eout = D Esystem

0 (steady)

= 0

Ein = Eout

å mh = å m h i i

e e

¾¾ ®

m11h11 + m2 h2 = m3 h3

¾¾ ®

zh11 + (1- y - z )h2 = (1- y )h3

where z is the fraction of steam extracted from the turbine ( = m9 / m5 ) at the second stage. Solving for z, z=

h3 - h2 236.14 - 69.84 (1- y ) = (1- 0.1300) = 0.1125 h11 - h2 1356.0 - 69.84

Then,

qin = h7 - h6 + (1- y )(h10 - h9 ) = 1550.5 - 380.41 + (1- 0.1300)(1531.3 - 1248.8) = 1415.8 Btu/lbm qout = (1- y - z )(h12 - h1 ) = (1- 0.1300 - 0.1125)(1052.4 - 69.72 ) = 744.4 Btu/lbm

wnet = qin - qout = 1415.8 - 744.4 = 671.4 Btu/lbm and

Qin 4´ 105 Btu/s = = 282.5 lbm / s qin 1415.8 Btu/lbm

æ1.055 kJ ÷ ö ÷ = 200.1 MW (b) Wnet = mwnet = (282.5 lbm/s)(671.4 Btu/lbm)çç ÷ çè 1 Btu ÷ ø (c) hth = 1-

qout 744.4 Btu/lbm = 1= 0.474 = 47.4% qin 1415.8 Btu/lbm

EES (Engineering Equation Solver) SOLUTION "Given" P[7]=1500 [psia] T[7]=1100 [F] P[8]=250 [psia] P[10]=140 [psia] T[10]=1000 [F] P[11]=40 [psia] P[12]=1 [psia] x[3]=0 x[5]=0 Q_dot_in=400000 [Btu/s] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[12] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[11] w_pI_in=v[1]*(P[2]-P[1])*Convert(psia*ft^3/lbm, Btu/lbm) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[3]=P[11] P[4]=P[8] w_pII_in=v[3]*(P[4]-P[3])*Convert(psia*ft^3/lbm, Btu/lbm) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-330

h[4]=h[3]+w_pII_in "pump III" h[5]=enthalpy(Fluid$, P=P[5], x=x[5]) v[5]=volume(Fluid$, P=P[5], x=x[5]) P[5]=P[8] P[6]=P[7] w_pIII_in=v[5]*(P[6]-P[5])*Convert(psia*ft^3/lbm, Btu/lbm) h[6]=h[5]+w_pIII_in "high pressure turbine" h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) s[8]=s[7] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) s[9]=s[7] P[9]=P[10] h[9]=enthalpy(Fluid$, P=P[9], s=s[9]) "low pressure turbine" h[10]=enthalpy(Fluid$, P=P[10], T=T[10]) s[10]=entropy(Fluid$, P=P[10], T=T[10]) s[11]=s[10] h[11]=enthalpy(Fluid$, P=P[11], s=s[11]) s[12]=s[10] h[12]=enthalpy(Fluid$, P=P[12], s=s[12]) x[12]=quality(Fluid$, P=P[12], s=s[12]) "open feedwater heaters" y*h[8]+(1-y)*h[4]=h[5] "y=m_dot_8/m_dot_5" z*h[11]+(1-y-z)*h[2]=(1-y)*h[3] "z=m_dot_9/m_dot_5" "cycle" q_in=h[7]-h[6]+(1-y)*(h[10]-h[9]) q_out=(1-y-z)*(h[12]-h[1]) w_net=q_in-q_out m_dot=Q_dot_in/q_in "(b)" W_dot_net=m_dot*w_net*Convert(Btu/s, MW) "(c)" Eta_th=1-q_out/q_in

Second-Law Analysis of Vapor Power Cycles 11-64 The exergy destructions associated with each of the processes of a Rankine cycle are to be determined for the specified source and sink temperatures. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wp, in = v1 ( P2 - P1 ) æ 1 kJ ö ÷ = (0.001017 m3 /kg)(4000 - 20)kPa çç ÷ ÷ çè1 kPa ×m3 ø

10-331

= 4.05 kJ/kg h2 = h1 + wp, in = 251.42 + 4.05 = 255.47 kJ/kg

ü ïï ý ïïþ

P3 = 4000 kPa T3 = 700°C P4 = 20 kPa s4 = s3

ü ïï ý ïïþ

h3 = 3906.3 kJ/kg s3 = 7.6214 kJ/kg ×K x4 =

s4 - s f s fg

7.6214 - 0.8320 = 0.9596 7.0752

h4 = h f + x4 h fg = 251.42 + (0.9596)(2357.5) = 2513.7 kJ/kg

Also, s1 = s2 = s f @ 20 kPa = 0.8320 kJ/kg ×K

s3 = s4 = 7.6214 kJ/kg ×K

qin = h3 - h2 = 3906.3 - 255.47 = 3650.8 kJ/kg qout = h4 - h1 = 2513.7 - 251.42 = 2262.3 kJ/kg The exergy destruction during a process of a stream from an inlet state to exit state is given by æ q ö qin ÷ xdest = T0 sgen = T0 çççse - si + out ÷ ÷ çè Tsource Tsink ÷ ø

Application of this equation for each process of the cycle gives æ æ ö qin ö 3650.8 kJ/kg ÷ ÷ ÷ xdestroyed, 23 = T0 çççs3 - s2 = (288 K) çç7.6214 - 0.8320 = 928 kJ / kg ÷ ÷ ÷ ç ÷ è Tsource ø 1023 K ø èç æ æ ö q ö 2262.3 kJ/kg ÷ ÷ xdestroyed, 41 = T0 çççs1 - s4 + out ÷ = (288 K) çç0.8320 - 7.6214 + = 307 kJ / kg ÷ ÷ ÷ ç ÷ è Tsink ø 288 K ø èç

Processes 1-2 and 3-4 are isentropic, and thus xdestroyed, 12 = 0 xdestroyed, 34 = 0

10-332

11-65 The exergy destructions associated with each of the processes of a Rankine cycle are to be determined for the specified source and sink temperatures. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg wp , in = v1 (P2 - P1 ) æ 1 kJ ö ÷ = (0.00101 m3 /kg )(10, 000 - 10 kPa )çç ÷ ÷ çè1 kPa ×m3 ø

= 10.09 kJ/kg h2 = h1 + wp , in = 191.81 + 10.09 = 201.90 kJ/kg

P3 = 10 MPa ïü ïý h3 = 3375.1 kJ/kg T3 = 500 °C ïïþ s3 = 6.5995 kJ/kg ×K

s4 - s f P4 = 10 kPa ïüï 6.5995 - 0.6492 = = 0.7934 ý x4 = ï s4 = s3 s fg 7.4996 ïþ h4 = h f + x4 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg

qin = h3 - h2 = 3375.1- 201.90 = 3173.2 kJ/kg qout = h4 - h1 = 2089.7 - 191.81 = 1897.9 kJ/kg

Also,

s1 = s2 = s f @ 10 kPa = 0.6492 kJ/kg ×K s3 = s4 = 6.5995 kJ/kg ×K

Processes 1-2 and 3-4 are isentropic. Thus, i12  0 and i34  0 . Also, æ ö q æ ö - 3173.2 kJ/kg ÷ ÷ xdestroyed, 23 = T0 çççs3 - s2 + R , 23 ÷ = (290 K )çç6.5995 - 0.6492 + = 1112 kJ / kg ÷ ÷ ÷ ç è ø TR ÷ 1500 K èç ø

10-333

æ ö q æ ö 1897.9 kJ/kg ÷ ÷ xdestroyed, 41 = T0 çççs1 - s4 + R , 41 ÷ = (290 K )çç0.6492 - 6.5995 + ÷= 172 kJ / kg ÷ ÷ èç ø TR ø 290 K ÷ èç

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=210000 [kW] P[3]=10000 [kPa] T[3]=500 [C] P[4]=10 [kPa] T_R=1500 [K] T_0=290 [K] "Analysis" Fluid$='steam_iapws' P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) x_4=x[4] q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=1-q_out/q_in m_dot=W_dot_net/w_net s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] i_12=T_0*(s[2]-s[1]) i_34=T_0*(s[4]-s[3]) i_23=T_0*(s[3]-s[2]-q_in/T_R) i_41=T_0*(s[1]-s[4]+q_out/T_0)

11-66 The component of an ideal reheat Rankine cycle with the largest exergy destruction is to be identified. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.001010 m3 /kg wp, in = v1 ( P2 - P1 )

10-334

æ 1 kJ ÷ ö = (0.001010 m3 /kg)(8000 - 10)kPa çç ÷ çè1 kPa ×m3 ÷ ø

= 8.07 kJ/kg h2 = h1 + wp, in = 191.81 + 8.07 = 199.88 kJ/kg

ïü ïý ïïþ

P3 = 8000 kPa T3 = 450°C ïü ïý ïïþ

P4 = 500 kPa s4 = s3

P6 = 10 kPa s6 = s5

ïü ïý ïïþ

h3 = 3273.3 kJ/kg s3 = 6.5579 kJ/kg ×K x4 =

s4 - s f s fg

6.5579 - 1.8604 = 0.9470 4.9603

h4 = h f + x4 h fg = 640.09 + (0.9470)(2108.0) = 2636.4 kJ/kg

x6 =

s6 - s f s fg

8.0893 - 0.6492 = 0.9921 7.4996

h6 = h f + x6 h fg = 191.81 + (0.9921)(2392.1) = 2564.9 kJ/kg

Thus,

qin = (h3 - h2 ) + (h5 - h4 ) = 3273.3 - 199.88 + 3484.5 - 2636.4 = 3921.5 kJ/kg qout = h6 - h1 = 2564.9 - 191.81 = 2373.1 kJ/kg Also, s1 = s2 = s f @ 10 kPa = 0.6492 kJ/kg ×K

s3 = s4 = 6.5579 kJ/kg ×K s5 = s6 = 8.0893 kJ/kg ×K qin, 2-3 = h3 - h2 = 3273.3 - 199.88 = 3073.4 kJ/kg qin, 4-5 = h5 - h4 = 3485.4 - 2636.4 = 848.1 kJ/kg

The exergy destruction during a process of a stream from an inlet state to exit state is given by æ q ö qin ÷ xdest = T0 sgen = T0 çççse - si + out ÷ ÷ çè Tsource Tsink ÷ ø

Application of this equation for each process of the cycle gives

10-335 æ ö q æ 3073.4 kJ/kg ö ÷ ÷= (283 K) ç xdestroyed, 23 = T0 ç s3 - s2 - in, 2-3 ÷ 6.5579 - 0.6492 = 687 kJ / kg ç ÷ ç ÷ ÷ ç ç ÷ ç è ø Tsource ø 883 K è

æ ö q æ 848.1 kJ/kg ö ÷ ÷ xdestroyed, 45 = T0 çççs5 - s4 - in, 4-5 ÷ = (283 K) çç8.0893 - 6.5579 ÷= 162 kJ / kg ÷ ÷ çè èç ø Tsource ø 883 K ÷

æ æ q ö 2373.1 kJ/kg ö ÷= 268 kJ / kg ÷ xdestroyed, 61 = T0 çççs1 - s6 + out ÷ ÷ ÷= (283 K) ççèç0.6492 - 8.0893 + ø Tsink ÷ 283 K ÷ èç ø

Processes 1-2, 3-4, and 5-6 are isentropic, and thus, xdestroyed, 12 = 0 xdestroyed, 34 = 0 xdestroyed, 56 = 0

The greatest exergy destruction occurs during heat addition process 2-3.

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=5000 [kW] P[3]=8000 [kPa] T[3]=450 [C] P[5]=500 [kPa] T[5]=500 [C] P[6]=10 [kPa] T_0=(10+273) [K] T_sink=T_0 T_source=(600+273) [K] "Analysis" Fluid$='steam_iapws' "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) "Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) T[6]=temperature(Fluid$, P=P[6], h=h[6]) T_6=T[6] x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] w_T_out=h[3]-h[4]+h[5]-h[6] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-336

q_in=h[3]-h[2]+h[5]-h[4] q_out=h[6]-h[1] w_net=w_T_out-w_p_in Eta_th=w_net/q_in m_dot=W_dot_net/w_net Q_dot_in=m_dot*q_in Q_dot_out=m_dot*q_out s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] q_in_23=h[3]-h[2] q_in_45=h[5]-h[4] x_dest_23=T_0*(s[3]-s[2]-q_in_23/T_source) x_dest_45=T_0*(s[5]-s[4]-q_in_45/T_source) x_dest_61=T_0*(s[1]-s[6]+q_out/T_sink) x_dest_12=0 "isentropic process" x_dest_34=0 "isentropic process" x_dest_56=0 "isentropic process"

11-67 The exergy destruction associated with the heat addition process and the expansion process in a reheat Rankine cycle are to be determined for the specified source and sink temperatures. The exergy of the steam at the boiler exit is also to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.001010 m3 /kg wp , in = v1 (P2 - P1 ) / h p

æ 1 kJ ÷ ö = (0.00101 m3 /kg )(10,000 - 10 kPa )çç / 0.95) ÷ 3 ÷ çè1 kPa ×m ø (

= 10.62 kJ/kg h2 = h1 + wp , in = 191.81 + 10.62 = 202.43 kJ/kg

10-337

P3 = 10 MPaïüï h3 = 3375.1 kJ/kg ý T3 = 500°C ïïþ s3 = 6.5995 kJ/kg ×K P4 s = 1 MPa üïï ý h4 s = 2783.8 kJ/kg ïïþ s4 s = s3 hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s = 3375.1- (0.80)(3375.1 - 2783.7 ) = 2902.0 kJ/kg

P5 = 1 MPaïüï h5 = 3479.1 kJ/kg ý T5 = 500°C ïïþ s5 = 7.7642 kJ/kg ×K

s6 s - s f 7.7642 - 0.6492 = = 0.9487 (at turbine exit ) P6 s = 10 kPa üïï x6 s = s 7.4996 ý fg ïïþ s6 s = s5 h6 s = h f + x6 s h fg = 191.81 + (0.9487)(2392.1) = 2461.2 kJ/kg hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) h5 - h6 s

= 3479.1- (0.80)(3479.1- 2461.2) = 2664.8 kJ/kg > hg (superheated vapor ) Also, s1 = s2 = s f @ 10 kPa = 0.6492 kJ/kg ×K

s3 = 6.5995 kJ/kg ×K s4 = 6.8464 kJ/kg ×K (P4 = 1 MPa, h4 = 2902.0 kJ/kg )

s5 = 7.7642 kJ/kg ×K s6 = 8.3870 kJ/kg ×K (P6 = 10 kPa, h6 = 2664.8 kJ/kg ) qin = (h3 - h2 )+ (h5 - h4 ) = 3375.1- 202.43 + 3479.1- 2902.0 = 3749.8 kJ/kg The exergy destruction associated with the combined pumping and the heat addition processes is

æ qR , 15 ÷ö ÷÷ xdestroyed = T0 çççs3 - s1 + s5 - s4 + çè TR ÷ø æ ö - 3749.8 kJ/kg ÷ = (285 K )çç6.5995 - 0.6492 + 7.7642 - 6.8464 + ÷ ÷ çè ø 1600 K

= 1289.5 kJ/kg The exergy destruction associated with the pumping process is Thus, xdestroyed,12 @ wp , a - wp , s = wp , a - vD P = 10.62 - 10.09 = 0.53 kJ/kg xdestroyed, heating = xdestroyed - xdestroyed,12 = 1289.5 - 0.5 = 1289 kJ / kg

The exergy destruction associated with the expansion process is ¿0ö æ qR ,36 ÷ çç ÷ xdestroyed, 34 = T0 ç(s4 - s3 )+ (s6 - s5 )+ ÷ çè TR ø÷ ÷

= (285 K )(6.8464 - 6.5995 + 8.3870 - 7.7642) kJ/kg ×K

10-338

The exergy of the steam at the boiler exit is determined from xflow,3 = (h3 - h0 )- T0 (s3 - s0 )+

V32 2

¿0

+ qz3¿ 0

= (h3 - h0 )- T0 (s3 - s0 ) where h0 = h @ (285 K, 100 kPa ) @ h f@285 K = 50.51 kJ/kg

s0 = s @ (285 K, 100 kPa) @ s f@285 K = 0.1806 kJ/kg ×K Thus, xflow,3 = (3375.1- 50.51)kJ/kg - (285 K )(6.5995 - 0.1806)kJ/kg ×K = 1495 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=80000 [kW] P[3]=10000 [kPa] T[3]=500 [C] P[5]=1000 [kPa] T[5]=500 [C] P[6]=10 [kPa] Eta_T=0.80 Eta_P=0.95 T_R=1600 [K] T_0=285 [K] P[0]=100 [kPa] "Analysis" Fluid$='steam_iapws' "Pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_p_in "High pressure turbine" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_4s=s[3] P[4]=P[5] h_4s=enthalpy(Fluid$, P=P[4], s=s_4s) Eta_T=(h[3]-h[4])/(h[3]-h_4s) "Low pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s_6s=s[5] h_6s=enthalpy(Fluid$, P=P[6], s=s_6s) Eta_T=(h[5]-h[6])/(h[5]-h_6s) x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] T[6]=temperature(Fluid$, P=P[6], h=h[6]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-339

T_6=T[6] "cycle" w_T_out=h[3]-h[4]+h[5]-h[6] q_in=h[3]-h[2]+h[5]-h[4] w_net=w_T_out-w_p_in Eta_th=w_net/q_in m_dot=W_dot_net/w_net s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[4]=entropy(Fluid$, P=P[4], h=h[4]) i_15=T_0*(s[3]-s[1]+s[5]-s[4]-q_in/T_R) w_p_in_s=v[1]*(P[2]-P[1]) i_12=w_p_in-w_p_in_s i_23=i_15-i_12 s[6]=entropy(Fluid$, P=P[6], h=h[6]) i_34=T_0*(s[4]-s[3]+s[6]-s[5]) T[0]=T_0-273 "[C]" h[0]=enthalpy(Fluid$, P=P[0], T=T[0]) s[0]=entropy(Fluid$, P=P[0], T=T[0]) ex_3=h[3]-h[0]-T_0*(s[3]-s[0])

11-68 The exergy destruction associated with a regenerative cycle is to be determined for the specified source and sink temperatures. Assumptions

1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wpI , in = v1 (P2 - P1 )

æ 1 kJ ÷ ö = (0.001017 m3 /kg )(400 - 20 kPa )çç ÷ 3 ÷ çè1 kPa ×m ø

= 0.39 kJ/kg h2 = h1 + wpI , in = 251.42 + 0.39 = 251.81 kJ/kg

10-340

h = h f @ 0.4 MPa = 604.66 kJ/kg P3 = 0.4 MPaïü ïý 3 ïïþv 3 = v f @ 0.4 MPa = 0.001084 m3 /kg sat.liquid

æ 1 kJ ÷ ö wpII , in = v 3 (P4 - P3 ) = (0.001084 m3 /kg )(6000 - 400 kPa )çç = 6.07 kJ/kg ÷ 3 ÷ çè1 kPa ×m ø h4 = h3 + wpII ,in = 604.66 + 6.07 = 610.73 kJ/kg

P5 = 6 MPa ïüï h5 = 3302.9 kJ/kg ý T5 = 450°C ïïþ s5 = 6.7219 kJ/kg ×K s6 - s f 6.7219 - 1.7765 x = = = 0.9661 P6 = 0.4 MPa ïü ïý 6 s fg 5.1191 ïïþ s6 = s5 h6 = h f + x6 h fg = 604.66 + (0.9661)(2133.4) = 2665.7 kJ/kg s7 - s f 6.7219 - 0.8320 = = 0.8325 P7 = 20 kPaü ïï x7 = s fg 7.0752 ý ïïþ s7 = s5 h7 = h f + x7 h fg = 251.42 + (0.8325)(2357.5) = 2214.0 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heater. Noting that Q @W @D ke @D pe @ 0,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

6 6

2 2

3 3

where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y, y=

Then,

h3 - h2 604.66 - 251.81 = = 0.1462 h6 - h2 2665.7 - 251.81

qin = h5 - h4 = 3302.9 - 610.73 = 2692.2 kJ/kg

qout = (1- y )(h7 - h1 ) = (1- 0.1462)(2214.0 - 251.42) = 1675.7 kJ/kg Then the exergy destruction associated with this regenerative cycle is æq æ1675.7 kJ/kg 2692.2 kJ/kg ö q ö ÷ ÷ xdestroyed, cycle = T0 ççç out - in ÷ = (290 K )çç ÷ ÷ ÷= 1097 kJ / kg çè 290 K TH ÷ 1350 K ø èç TL ø

EES (Engineering Equation Solver) SOLUTION "Given" P[5]=6000 [kPa] T[5]=450 [C] P[6]=400 [kPa] P[7]=20 [kPa] x[3]=0 T_R=1350 [K] T_0=290 [K] "Analysis" Fluid$='steam_iapws' "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-341

v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) s[7]=s[5] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out Eta_th=1-q_out/q_in i_cycle=T_0*(q_out/T_0-q_in/T_R)

11-69 The exergy destruction associated with a reheating and regeneration processes on an ideal reheat-regenerative Rankine cycle with an open feedwater heater are to be determined for the specified source and sink temperatures. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-342

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg

æ 1 kJ ö÷ wpI , in = v1 (P2 - P1 ) = (0.00101 m3 /kg )(800 - 10 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 0.80 kJ/kg h2 = h1 + wpI , in = 191.81 + 0.80 = 192.61 kJ/kg

æ 1 kJ ö ÷ wpII , in = v 3 (P4 - P3 )= (0.001115 m 3 /kg )(10,000 - 800 kPa )çç ÷ ÷ çè1 kPa ×m3 ø = 10.26 kJ/kg h4 = h3 + wpII , in = 720.87 + 10.26 = 731.12 kJ/kg

P5 = 10 MPa ïüï h5 = 3502.0 kJ/kg ý T5 = 550°C ïïþ s5 = 6.7585 kJ/kg ×K P6 = 0.8 MPa ïüï ý h6 = 2812.1 kJ/kg ïïþ s6 = s5

P7 = 0.8 MPa ïüï h7 = 3481.3 kJ/kg ý T7 = 500°C ïïþ s7 = 7.8692 kJ/kg ×K s8 - s f 7.8692 - 0.6492 = = 0.9627 P8 = 10 kPa ïüï x8 = s 7.4996 ý fg ïïþ s8 = s7 h8 = h f + x8 h fg = 191.81 + (0.9627)(2392.1) = 2494.7 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-343

Ein - Eout = D Esystem ¿ 0 (steady) = 0 ® Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

6 6

2 2

3 3

where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y, y=

h3 - h2 720.87 - 192.61 = = 0.2017 h6 - h2 2812.1- 192.61

Then,

qin = (h5 - h4 )+ (1- y )(h7 - h6 ) = (3502.0 - 731.12)+ (1- 0.2017)(3481.3 - 2812.1) = 3305.1 kJ/kg qout = (1- y )(h8 - h1 ) = (1- 0.2017)(2494.7 - 191.81) = 1838.5 kJ/kg

qreheat = h7 - h6 = 3481.3 - 2812.7 = 668.6 kJ/kg From the steam tables, s3 = s f @ 0.8 MPa = 2.0457 kJ/kg ×K

s5 = s6 = 6.7585 kJ/kg ×K

s7 = 7.8692 kJ/kg ×K s1 = s2 = s f @ 10 kPa = 0.6492 kJ/kg ×K

Then the exergy destruction associated with reheat and regeneration processes are

æ ö q ÷ xdestroyed, reheat = T0 çççs7 - s6 + R ,67 ÷ ÷ çè TR ø÷

æ - 668.6 kJ/kg ö÷ = (290 K )çç7.8692 - 6.7585 + ÷ çè 1800 K ø÷ = 214.3 kJ / kg n0 ö æ q ÷ xdestroyed, regen = T0 sgen = T0 çççå me se - å mi si + surr ÷ = T s - ys6 - (1- y )s2 ) ÷ ÷ 0( 3 çè T0 ø

= (290 K )éë2.0457 - (0.2016)(6.7585)- (1- 0.2016)(0.6492)ùû = 47.8 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=80000 [kW] P[5]=10000 [kPa] T[5]=550 [C] P[6]=800 [kPa] T[7]=500 [C] P[8]=10 [kPa] T_R=1800 [K] T_0=290 [K] "Analysis" Fluid$='steam_iapws' "pump I" P[1]=P[8] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-344

v[1]=volume(Fluid$, P=P[1], x=x[1]) P[3]=P[6] P[2]=P[3] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "high pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) "low pressure turbine" P[7]=P[6] h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) s[8]=s[7] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4]+(1-y)*(h[7]-h[6]) q_out=(1-y)*(h[8]-h[1]) w_net=q_in-q_out m_dot=W_dot_net/w_net Eta_th=w_net/q_in q_reheat=h[7]-h[6] i_reheat=T_0*(s[7]-s[6]-q_reheat/T_R) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] s[3]=entropy(Fluid$, P=P[3], x=x[3]) i_regen=T_0*(s[3]-y*s[6]-(1-y)*s[2])

11-70 A singlefrom the turbine, the thermal efficiency of the plant, the exergy of the geothermal liquid at the exit of the flash chamber, and the exergy destructions and exergy efficiencies for the flash chamber, the turbine, and the entire plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) We use properties of water for geothermal water (Tables A-4, A-5, and A-6)

T1 = 230°Cüïï h1 = 990.14 kJ/kg ý ïïþ s1 = 2.6100 kJ/kg.K x1 = 0 üïï x2 = 0.1661 ý h2 = h1 = 990.14 kJ/kgïïþ s2 = 2.6841 kJ/kg.K P2 = 500 kPa

10-345

m3 = x2 m1

= (0.1661)(230 kg/s)

= 38.19 kg/s P3 = 500 kPaïü ïý h3 = 2748.1 kJ/kg ïïþ s3 = 6.8207 kJ/kg ×K x3 = 1

P4 = 10 kPa ïüï h4 = 2464.3 kJ/kg ý x4 = 0.95 ïïþ s4 = 7.7739 kJ/kg ×K P6 = 500 kPa ïü ïý h6 = 640.09 kJ/kg ïïþ s6 = 1.8604 kJ/kg ×K x6 = 0

m6 = m1 - m3 = 230 - 38.19 = 191.81 kg/s

The power output from the turbine is WT = m3 (h3 - h4 ) = (38.19 kJ/kg)(2748.1- 2464.3)kJ/kg = 10, 842 kW

We use saturated liquid state at the standard temperature for dead state properties T0 = 25°Cü ïï h0 = 104.83 kJ/kg ý ïïþ s0 = 0.3672 kJ/kg x0 = 0 Ein = m1 (h1 - h0 ) = (230 kJ/kg)(990.14 - 104.83)kJ/kg = 203, 622 kW

hth =

WT, out Ein

10,842 = 0.0532 = 5.3% 203, 622

(b) The specific exergies at various states are xflow,1 = h1 - h0 - T0 ( s1 - s0 ) = (990.14 - 104.83)kJ/kg - (298 K)(2.6100 - 0.3672)kJ/kg.K = 216.53 kJ/kg xflow,2 = h2 - h0 - T0 (s2 - s0 ) = (990.14 - 104.83)kJ/kg - (298 K)(2.6841- 0.3672)kJ/kg.K = 194.44 kJ/kg xflow,3 = h3 - h0 - T0 (s3 - s0 ) = (2748.1- 104.83)kJ/kg - (298 K)(6.8207 - 0.3672)kJ/kg.K = 719.10 kJ/kg

10-346

xflow,4 = h4 - h0 - T0 (s4 - s0 ) = (2464.3 - 104.83)kJ/kg - (298 K)(7.7739 - 0.3672)kJ/kg.K = 151.05 kJ/kg

xflow,6 = h6 - h0 - T0 (s6 - s0 ) = (640.09 - 104.83)kJ/kg - (298 K)(1.8604 - 0.3672)kJ/kg.K = 89.97 kJ/kg

The exergy of geothermal water at state 6 is

X 6 = m6 xflow,6 = (191.81 kg/s)(89.97 kJ/kg) = 17,257 kW (c) Turbine:

X dest, T = m3 ( xflow,3 - xflow,4 ) - WT = (38.19 kg/s)(719.10 - 151.05)kJ/kg - 10,842 kW = 10,854 kW

hII, T =

WT 10,842 kW = = 0.500 = 50.0% m3 ( xflow,3 - xflow,4 ) (38.19 kg/s)(719.10 - 151.05)kJ/kg

(d) Plant:

X in, Plant = m1 xflow,1 = (230 kg/s)(216.53 kJ/kg) = 49,802 kW X dest, Plant = X in, Plant - WT = 49,802 - 10,842 = 38,960 kW

hII, Plant =

WT 10,842 kW = = 0.2177 = 21.8% 49,802 kW X in, Plant

EES (Engineering Equation Solver) SOLUTION "GIVEN" T[1]=230 [C] x[1]=0 m_dot_1=230 [kg/s] P[2]=500 [kPa] P[4]=10 [kPa] x[4]=0.95 T[0]=25 [C] P_0=101.325 [kPa] "PROPERTIES" Fluid$='Steam_iapws' h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) h[2]=h[1] "constant enthalpy throttling process" x[2]=quality(Fluid$, P=P[2], h=h[2]) s[2]=entropy(Fluid$, P=P[2], h=h[2]) m_dot_3=x[2]*m_dot_1 "Mass flow rate of steam through turbine" x[3]=1 P[3]=P[2] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) s[3]=entropy(Fluid$, P=P[3], x=x[3]) h[4]=enthalpy(Fluid$, P=P[4], x=x[4]) s[4]=entropy(Fluid$, P=P[4], x=x[4]) h[0]=enthalpy(Fluid$, T=T[0], P=P_0) s[0]=entropy(Fluid$, T=T[0], P=P_0) h[6]=enthalpy(Fluid$, P=P[2], x=0) s[6]=entropy(Fluid$, P=P[2], x=0) m_dot_6=m_dot_1-m_dot_3 "ANALYSIS" W_dot_T=m_dot_3*(h[3]-h[4]) "turbine power output" E_dot_in=m_dot_1*(h[1]-h[0]) "energy input to plant" eta_th=W_dot_T/E_dot_in "thermal efficiency of the plant" ex[1]=h[1]-h[0]-(T[0]+273.15)*(s[1]-s[0]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-347

ex[2]=h[2]-h[0]-(T[0]+273.15)*(s[2]-s[0]) ex[3]=h[3]-h[0]-(T[0]+273.15)*(s[3]-s[0]) ex[4]=h[4]-h[0]-(T[0]+273.15)*(s[4]-s[0]) ex[6]=h[6]-h[0]-(T[0]+273.15)*(s[6]-s[0]) EX_dot_6=m_dot_6*ex[6] "Exergy of liquid at the exit of separator" EX_dot_dest_FC=m_dot_1*(ex[1]-ex[2]) "Exergy destruction in flash chamber" eta_II_FC=ex[2]/ex[1] "Exergy efficiency of flash chamber" EX_dot_dest_Turb=m_dot_3*(ex[3]-ex[4])-W_dot_T "Exergy destruction in turbine" eta_II_Turb=W_dot_T/(m_dot_3*(ex[3]-ex[4])) "Exergy efficiency of turbine" EX_dot_in=m_dot_1*ex[1] "Exergy input to plant" EX_dot_dest_plant=EX_dot_in-W_dot_T "Exergy destruction in the plant" eta_II_plant=W_dot_T/EX_dot_in "Exergy efficiency of the plant"

Cogeneration 11-71C Cogeneration is the production of more than one useful form of energy from the same energy source. Regeneration is the transfer of heat from the working fluid at some stage to the working fluid at some other stage.

11-72C The utilization factor of a cogeneration plant is the ratio of the energy utilized for a useful purpose to the total energy supplied. It could be unity for a plant that does not produce any power.

11-73C No. A cogeneration plant may involve throttling, friction, and heat transfer through a finite temperature difference, and still have a utilization factor of unity.

11-74C Yes, if the cycle involves no irreversibilities such as throttling, friction, and heat transfer through a finite temperature difference.

11-75E A cogeneration plant is to generate power while meeting the process steam requirements for a certain industrial application. The net power produced, the rate of process heat supply, and the utilization factor of this plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-348

Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 @ h f @ 240° F = 208.49 Btu/lbm

h2 @ h1 P3 = 600 psia ïüï h3 = 1321.3 Btu/lbm ý T3 = 650°F ïïþ s3 = s5 = s7 = 1.5614 Btu/lbm ×R h3 = h4 = h5 = h6

10-349 Wnet = m5 (h5 - h7 )

= (2/3´ 32 lbm/s)(1321.3 - 1169.2)Btu/lbm = 3244 Btu/s = 3422 kW

(b) Qprocess = å mi hi - å me he

= m6 h6 + m7 h7 - m1h1

(c)

(d) = (1/ 3´ 32)(1321.3)+ (2 / 3´ 32)(1169.2)- (32)(208.49 ) (e) = 32, 365 Btu / s (f)

ηu  1 since all the energy is utilized.

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=600 [psia] T[3]=650 [F] m_dot=32 [lbm/s] y=1/3 "fraction of steam throttled" P[6]=120 [psia] T[1]=240 "Analysis" Fluid$='steam_iapws' "(a)" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) h[2]=h[1] "pump work neglected" h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) "state 3 is the boiler exit" s[3]=entropy(Fluid$, P=P[3], T=T[3]) h[4]=h[3] "state 4 is the throttle inlet" h[5]=h[3] "state 5 is the turbine inlet" h[6]=h[3] "state 6 is the throttle exit" s[7]=s[3] P[7]=P[6] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) "state 7 is the turbine exit" m_dot[5]=(1-y)*m_dot W_dot_net=m_dot[5]*(h[5]-h[7])*Convert(Btu/s, kW) "(b)" m_dot[1]=m_dot m_dot[6]=y*m_dot m_dot[7]=(1-y)*m_dot Q_dot_process=m_dot[6]*h[6]+m_dot[7]*h[7]-m_dot[1]*h[1] Q_dot_in=m_dot*(h[3]-h[2]) "(c)" Epsilon_u=(W_dot_net*Convert(kW, Btu/s)+Q_dot_process)/Q_dot_in

11-76E A large food-processing plant requires steam at a relatively high pressure, which is extracted from the turbine of a cogeneration plant. The rate of heat transfer to the boiler and the power output of the cogeneration plant are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-350

1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Analysis (a) From the steam tables (Tables A-4E, A-5E, and A-6E), h1 = h f @ 2 psia = 94.02 Btu/lbm

v1 = v f @ 2 psia = 0.01623 ft 3 /lbm wpI, in = v1 (P2 - P1 ) / h p =

æ ö 1 ÷ 1 Btu ÷= 0.46 Btu (0.01623 ft 3 /lbm)(140 - 2) psia çç ÷ ÷ ç 0.91 è5.4039 psia ×ft 3 ø

10-351

h3 = h f @ 140 psia = 324.92 Btu/lbm Mixing chamber:

Ein - Eout = D EsystemZ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h = m h + m h i i

e e

4 4

2 2

3 3

or, h4 =

m2 h2 + m3 h3 (8.5)(94.50)+ (1.5)(324.92) = = 129.07 Btu/lbm m4 10

v 4 @v f @ h f = 129.07 Btu/lbm = 0.01640 ft 3 /lbm wpII, in = v 4 (P5 - P4 ) / h p

æ ö 1 Btu ÷ ÷ = (0.01640 ft 3 /lbm)(800 - 140 psia )çç ÷/ (0.91) ÷ çè5.4039 psia ×ft 3 ø

= 2.20 Btu/lbm h5 = h4 + wpII , in = 129.07 + 2.20 = 131.24 Btu/lbm

P6 = 800 psiaüïï h6 = 1512.2 Btu/lbm ý T6 = 1000°F ïïþ s6 = 1.6812 Btu/lbm ×R P7 s = 140 psiaïü ïý h = 1287.5 Btu/lbm ïïþ 7 s s7 s = s6

s8 s - s f 1.6812 - 0.17499 = = 0.8634 P8 s = 2 psiaüïï x8 s = s fg 1.74444 ý ïïþ s8 s = s6 h8 s = h f + x8 s h fg = 94.02 + (0.8634)(1021.7) = 976.21 Btu/lbm Then,

Qin = m5 (h6 - h5 ) = (10 lbm/s)(1512.2 - 131.39) Btu/lbm = 13,810 Btu / s (b)

WT,out = hT WT , s = hT éëm6 (h6 - h7 s )+ m8 (h7 s - h8s )ù û = (0.91)éë(10 lbm/s)(1512.2 - 1287.5)Btu/lbm + (1.5 lbm/s)(1287.5 - 976.21)Btu/lbmù û

= 4453 Btu/s = 4698 kW

EES (Engineering Equation Solver) SOLUTION "Given" P[6]=800 [psia] T[6]=1000 [F] P[7]=140 [psia] P[8]=2 [psia] m_dot=10 [lbm/s] m_dot[7]=1.5 [lbm/s] x[3]=0 Eta_T=0.91 Eta_P=0.91 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-352

"Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[8] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[7] w_pI_in=v[1]*(P[2]-P[1])/Eta_P*Convert(psia*ft^3/lbm, Btu/lbm) h[2]=h[1]+w_pI_in P[3]=P[7] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) "mixing chamber" m_dot[2]=m_dot-m_dot[7] m_dot[3]=m_dot[7] m_dot[4]=m_dot m_dot[4]*h[4]=m_dot[2]*h[2]+m_dot[3]*h[3] x[4]=0 v[4]=volume(Fluid$, h=h[4], x=x[4]) "pump II" P[4]=P[7] P[5]=P[6] w_pII_in=v[4]*(P[5]-P[4])/Eta_P*Convert(psia*ft^3/lbm, Btu/lbm) h[5]=h[4]+w_pII_in "turbine" h[6]=enthalpy(Fluid$, P=P[6], T=T[6]) s[6]=entropy(Fluid$, P=P[6], T=T[6]) s_s[7]=s[6] h_s[7]=enthalpy(Fluid$, P=P[7], s=s_s[7]) s_s[8]=s[6] h_s[8]=enthalpy(Fluid$, P=P[8], s=s_s[8]) x_8s=quality(Fluid$, P=P[8], s=s_s[8]) "cycle" m_dot[5]=m_dot Q_dot_in=m_dot[5]*(h[6]-h[5]) m_dot[6]=m_dot m_dot[8]=m_dot-m_dot[7] "(b)" W_dot_T_out=Eta_T*(m_dot[6]*(h[6]-h_s[7])+m_dot[8]*(h_s[7]-h_s[8]))*Convert(Btu/s, kW) "s8_f=entropy(Fluid$, P=P[8], x=0) s8_g=entropy(Fluid$, P=P[8], x=1) s8_fg=s8_g-s8_f h8_f=enthalpy(Fluid$, P=P[8], x=0) h8_g=enthalpy(Fluid$, P=P[8], x=1) h8_fg=h8_g-h8_f"

11-77 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively high pressure is used for process heating. The mass flow rate of steam that must be supplied by the boiler, the net power produced, and the utilization factor of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-353

Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg

wpI, in = v1 (P2 - P1 )

æ 1 kJ ö÷ = (0.00101 m3 /kg )(600 - 10 kPa )çç ÷ çè1 kPa ×m3 ø÷ = 0.596 kJ/kg h2 = h1 + wpI, in = 191.81 + 0.596 = 192.40 kJ/kg

10-354

h3 = h f @ 0.6 MPa = 670.38 kJ/kg

Mixing chamber:

m3 h3 + m2 h2 = m4 h4 (0.25)(670.38 kJ/kg) + (0.75)(192.40 kJ/kg)) = (1)h4 ¾ ¾ ® h4 = 311.90 kJ/kg

v 4 @v f @ h f = 311.90 kJ/kg = 0.001026 m3 /kg wpII, in = v 4 (P5 - P4 )

æ 1 kJ ö÷ = (0.001026 m3 /kg )(7000 - 600 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 6.563 kJ/kg h5 = h4 + wpII, in = 311.90 + 6.563 = 318.47 kJ/kg

P6 = 7 MPa T6 = 500°C

üïï h6 = 3411.4 kJ/kg ý ïïþ s6 = 6.8000 kJ/kg ×K

P7 = 0.6 MPa s7 = s6

P8 = 10 kPa s8 = s6

üïï ý h7 = 2773.9 kJ/kg ïïþ

ïüï ý h8 = 2153.6 kJ/kg ïïþ

Qprocess = m7 (h7 - h3 ) 8600 kJ/s = m7 (2773.9 - 670.38) kJ/kg

m7 = 4.088 kg/s This is one-fourth of the mass flowing through the boiler. Thus, the mass flow rate of steam that must be supplied by the boiler becomes

m6 = 4m7 = 4(4.088 kg/s) = 16.35 kg / s (b) Cycle analysis: (c) = (4.088 kg/s)(3411.4 - 2773.9) kJ/kg + (16.35-4.088 kg/s)(3411.4 - 2153.6) kJ/kg (d) = 18, 033 kW (e) Wp, in = m1 wpI, in + m4 wpII, in (f) (g)

= (16.35-4.088 kg/s)(0.596 kJ/kg )+ (16.35 kg/s )(6.563 kJ/kg ) = 114.6 kW

Wnet = WT,out - Wp, in = 18, 033 - 115 = 17, 919 kW WT,out = m7 (h6 - h7 )+ m8 (h6 - h8 ) (h) Then,

and Qin = m5 (h6 - h5 ) = (16.35 kg/s)(3411.4 - 318.46) = 50,581 kW eu =

Wnet + Qprocess Qin

17,919 + 8600 = 0.524 = 52.4% 50,581

10-355

Q_dot_process=8600 [kW] m_dot_process=1/4*m_dot P[7]=600 [kPa]; P[3]=P[7]; P[2]=P[7]; P[4]=P[7] P[6]=7000 [kPa]; P[5]=P[6] T[6]=500 [C] P[8]=10 [kPa]; P[1]=P[8] eta_T=1; eta_P=1 "PROPERTIES" Fluid$='Steam_iapws' h[1]=enthalpy(Fluid$, P=P[1], x=0) v_1=volume(Fluid$, P=P[1], x=0) w_PI=v_1*(P[2]-P[1])/eta_P h[2]=h[1]+w_PI h[3]=enthalpy(Fluid$, P=P[3], x=0) m_dot_process*h[3]+(m_dot-m_dot_process)*h[2]=m_dot*h[4] "energy balance on the mixing chamber" v_4=volume(Fluid$, x=0, h=h[4]) w_PII=v_4*(P[5]-P[4])/eta_P h[5]=h[4]+w_PII h[6]=enthalpy(Fluid$, P=P[6], T=T[6]) s[6]=entropy(Fluid$, P=P[6], T=T[6]) h_7s=enthalpy(Fluid$, P=P[7], s=s[6]) h[7]=h[6]-eta_T*(h[6]-h_7s) h_8s=enthalpy(Fluid$, P=P[8], s=s[6]) h[8]=h[6]-eta_T*(h[6]-h_8s) "ANALYSIS" Q_dot_process=m_dot_process*(h[7]-h[3]) "process heat" W_dot_T=m_dot_process*(h[6]-h[7])+(m_dot-m_dot_process)*(h[6]-h[8]) W_dot_P=(m_dot-m_dot_process)*w_PI+m_dot*w_PII W_dot_net=W_dot_T-W_dot_P "net power" Q_dot_in=m_dot*(h[6]-h[5]) eta_utilization=(W_dot_net+Q_dot_process)/Q_dot_in "utilization efficiency"

11-78 A cogeneration plant has two modes of operation. In the first mode, all the steam leaving the turbine at a relatively high pressure is routed to the process heater. In the second mode, 60 percent of the steam is routed to the process heater and remaining is expanded to the condenser pressure. The power produced and the rate at which process heat is supplied in the first mode, and the power produced and the rate of process heat supplied in the second mode are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-356

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kgg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wpI, in = v1 (P2 - P1 ) æ 1 kJ ÷ ö = (0.001017 m3 /kg )(10,000 - 20 kPa )çç ÷ 3 ÷ çè1 kPa ×m ø

= 10.15 kJ/kg h2 = h1 + wpI, in = 251.42 + 10.15 = 261.57 kJ/kg

h3 = h f @ 0.5 MPa = 640.09 kJ/kg

10-357

v 3 = v f @ 0.5 MPa = 0.001093 m3 /kg

wpII, in = v 3 (P4 - P3 ) æ 1 kJ ÷ ö = (0.001093 m3 /kg )(10,000 - 500 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 10.38 kJ/kg h4 = h3 + wpII, in = 640.09 + 10.38 = 650.47 kJ/k

Mixing chamber: Ein - Eout = D EsystemZ 0 (steady) = 0 ® Ein = Eout

å m h = å m h ¾ ¾® m h = m h + m h i i

e e

5 5

2 2

4 4

or, h5 =

(2)(261.57)+ (3)(650.47) m2 h2 + m4 h4 = = 494.91 kJ/kg m5 5

P6 = 10 MPaïüï h6 = 3242.4 kJ/kg ý T6 = 450°C ïïþ s6 = 6.4219 kJ/kg ×K s7 - s f 6.4219 - 1.8604 = = 0.9196 P7 = 0.5 MPaü ïï x7 = s 4.9603 ý fg ïïþ s7 = s6 h7 = h f + x7 h fg = 640.09 + (0.9196)(2108.0) = 2578.6 kJ/kg

s8 - s f 6.4219 - 0.8320 = = 0.7901 P8 = 20 kPaïü ïý x8 = s 7.0752 fg ïïþ s8 = s6 h8 = h f + x8 h fg = 251.42 + (0.7901)(2357.5) = 2114.0 kJ/kg When the entire steam is routed to the process heater,

WT, out = m6 (h6 - h7 ) = (5 kg/s)(3242.4 - 2578.6) kJ/kg = 3319 kW

Qprocess = m7 (h7 - h3 ) = (5 kg/s)(2578.6 - 640.09) kJ/kg = 9693 kW (b) When only 60% of the steam is routed to the process heater,

WT, out = m6 (h6 - h7 )+ m8 (h7 - h8 ) = (5 kg/s)(3242.4 - 2578.6)kJ/kg + (2 kg/s )(2578.6 - 2114.0)kJ/kg = 4248 kW

Qprocess = m7 (h7 - h3 ) = (3 kg/s)(2578.6 - 640.09) kJ/kg = 5816 kW

EES (Engineering Equation Solver) SOLUTION "Given" P[6]=10000 [kPa] T[6]=450 [C] P[7]=500 [kPa] P[8]=20 [kPa] m_dot=5 [kg/s] x[3]=0 "Analysis" Fluid$='steam_iapws' "pump I" P[1]=P[8] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-358

x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[7] w_pI_in=v[1]*(P[6]-P[1]) h[2]=h[1]+w_pI_in P[3]=P[7] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) "pump II" P[4]=P[6] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "mixing chamber" m_dot[2]=0.4*m_dot m_dot[4]=0.6*m_dot m_dot[5]=m_dot m_dot[5]*h[5]=m_dot[2]*h[2]+m_dot[4]*h[4] "turbine" h[6]=enthalpy(Fluid$, P=P[6], T=T[6]) s[6]=entropy(Fluid$, P=P[6], T=T[6]) s[7]=s[6] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) s[8]=s[6] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) x[8]=quality(Fluid$, P=P[8], s=s[8]) "(a) Entire steam is routed to the process heater" m_dot[6]=m_dot W_dot_T_out_a=m_dot[6]*(h[6]-h[7]) m_dot_7a=m_dot Q_dot_process_a=m_dot_7a*(h[7]-h[3]) "(b) Only 60% of the steam is routed to the process heater" m_dot[8]=0.40*m_dot W_dot_T_out_b=m_dot[6]*(h[6]-h[7])+m_dot[8]*(h[7]-h[8]) m_dot_7b=0.60*m_dot Q_dot_process_b=m_dot_7b*(h[7]-h[3])

11-79 A cogeneration plant modified with regeneration is to generate power and process heat. The mass flow rate of steam through the boiler for a net power output of 25 MW is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-359

Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg wpI, in = v1 (P2 - P1 )

æ 1 kJ ÷ ö = (0.00101 m3 /kg )(1600 - 10 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 1.61 kJ/kg h2 = h1 + wpI, in = 191.81 + 1.61 = 193.41 kJ/kg h3 = h4 = h9 = h f @ 1.6 MPa = 858.44 kJ/kg

v 4 = v f @ 1.6 MPa = 0.001159 m3 /kg wpII, in = v 4 (P5 - P4 )

æ 1 kJ ö÷ = (0.001159 m3 /kg )(9000 - 400 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 8.57 kJ/kg h5 = h4 + wpII, in = 858.44 + 8.57 = 867.02 kJ/kg

10-360

P6 = 9 MPaïü ïý h6 = 3118.8 kJ/kg T6 = 400°C ïïþ s6 = 6.2876 kJ/kg ×K s7 - s f 6.2876 - 2.3435 = = 0.9675 P7 = 1.6 MPaü ïï x7 = s 4.0765 ý fg ïïþ s7 = s6 h7 = h f + x7 h fg = 858.44 + (0.9675)(1934.4) = 2730.0 kJ/kg s8 - s f 6.2876 - 0.6492 = = 0.7518 P8 = 10 kPaïü ïý x8 = s 7.4996 fg ï s8 = s6 ïþ h = h + x h = 191.81 + 0.7518 2392.1 = 1990.2 kJ/kg ( )( ) 8 f 8 fg Then, per kg of steam flowing through the boiler, we have

wT, out = (h6 - h7 )+ (1- y ) (h7 - h8 ) = (3118.8 - 2730.0) kJ/kg + (1- 0.35)(2730.0 - 1990.2) kJ/kg

= 869.7 kJ/kg wp, in = (1- y)wpI, in + wpII, in

= (1- 0.35)(1.61 kJ/kg )+ (8.57 kJ/kg )

= 9.62 kJ/kg wnet = wT, out - wp, in = 869.7 - 9.62 = 860.1 kJ/kg

Thus,

Wnet 25,000 kJ/s = = 29.1kg / s wnet 860.1 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[6]=9000 [kPa] T[6]=400 [C] P[7]=1600 [kPa] P[8]=10 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-361

y=0.35 x[3]=0 W_dot_net=25000 [kW] "Calculations" Fluid$='steam_iapws' "pump I" P[1]=P[8] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[7] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in P[3]=P[7] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] h[9]=h[3] "pump II" P[4]=P[7] x[4]=x[3] v[4]=volume(Fluid$, P=P[4], x=x[4]) P[5]=P[6] w_pII_in=v[4]*(P[5]-P[4]) h[5]=h[4]+w_pII_in "turbine" h[6]=enthalpy(Fluid$, P=P[6], T=T[6]) s[6]=entropy(Fluid$, P=P[6], T=T[6]) s[7]=s[6] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) x[7]=quality(Fluid$, P=P[7], s=s[7]) s[8]=s[6] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) x[8]=quality(Fluid$, P=P[8], s=s[8]) w_T_out=h[6]-h[7]+(1-y)*(h[7]-h[8]) w_P_in=(1-y)*w_PI_in+w_PII_in w_net=w_T_out-w_P_in m_dot=W_dot_net/w_net s7_f=entropy(Fluid$, P=P[7], x=0) s7_g=entropy(Fluid$, P=P[7], x=1) s7_fg=s7_g-s7_f h7_f=enthalpy(Fluid$, P=P[7], x=0) h7_g=enthalpy(Fluid$, P=P[7], x=1) h7_fg=h7_g-h7_f

11-80 Problem 11-79 is reconsidered. The effect of the extraction pressure for removing steam from the turbine to be used for the process heater and open feedwater heater on the required mass flow rate is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" y = 0.35 "fraction of steam extracted from turbine for feedwater heater and process heater" P[6] = 9000 [kPa] T[6] = 400 [C] P_extract=1600 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-362

P[7] = P_extract P_cond=10 [kPa] P[8] = P_cond W_dot_net=25 [MW]*Convert(MW, kW) Eta_turb= 100/100 Eta_pump = 100/100 P[1] = P[8] P[2]=P[7] P[3]=P[7] P[4] = P[7] P[5]=P[6] P[9] = P[7] " Pump 1 analysis" Fluid$='Steam_IAPWS' h[1]=enthalpy(Fluid$,P=P[1],x=0) v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1]) w_pump1=w_pump1_s/Eta_pump h[1]+w_pump1= h[2] s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis:" z*h[7] + (1- y)*h[2] = (1- y + z)*h[3] "fraction of steam extracted for the closed feedwater heater" h[3]=enthalpy(Fluid$,P=P[3],x=0) T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Process heater analysis:" (y - z)*h[7] = q_process + (y - z)*h[9] Q_dot_process = m_dot*(y - z)*q_process h[9]=enthalpy(Fluid$,P=P[9],x=0) T[9]=temperature(Fluid$,P=P[9],x=0) s[9]=entropy(Fluid$,P=P[9],x=0) "Mixing chamber at 3, 4, and 9:" (y-z)*h[9] + (1-y+z)*h[3] = 1*h[4] T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,P=P[4],h=h[4]) "Boiler condensate pump or Pump 2 analysis" v4=volume(Fluid$,P=P[4],x=0) w_pump2_s=v4*(P[5]-P[4]) w_pump2=w_pump2_s/Eta_pump h[4]+w_pump2= h[5] s[5]=entropy(Fluid$,P=P[5],h=h[5]) T[5]=temperature(Fluid$,P=P[5],h=h[5]) "Boiler analysis" q_in + h[5]=h[6] h[6]=enthalpy(Fluid$, T=T[6], P=P[6]) s[6]=entropy(Fluid$, T=T[6], P=P[6]) "Turbine analysis" ss[7]=s[6] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-363

hs[7]=enthalpy(Fluid$,s=ss[7],P=P[7]) Ts[7]=temperature(Fluid$,s=ss[7],P=P[7]) h[7]=h[6]-Eta_turb*(h[6]-hs[7]) T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) ss[8]=s[7] hs[8]=enthalpy(Fluid$,s=ss[8],P=P[8]) Ts[8]=temperature(Fluid$,s=ss[8],P=P[8]) h[8]=h[7]-Eta_turb*(h[7]-hs[8]) T[8]=temperature(Fluid$,P=P[8],h=h[8]) s[8]=entropy(Fluid$,P=P[8],h=h[8]) h[6] =y*h[7] + (1- y)*h[8] + w_turb "Condenser analysis" (1- y)*h[8]=q_out+(1- y)*h[1] "Cycle Statistics" w_net=w_turb - ((1- y)*w_pump1+ w_pump2) Eta_th=w_net/q_in W_dot_net = m_dot * w_net

Pextract

ηth

[kg / s]

[kPa] 200 400 600 800 1000 1200 1400 1600 1800 2000

0.3778 0.3776 0.3781 0.3787 0.3794 0.3802 0.3811 0.3819 0.3828 0.3837

25.4 26.43 27.11 27.63 28.07 28.44 28.77 29.07 29.34 29.59

Qprocess [kW] 2770 2137 1745 1459 1235 1053 900.7 770.9 659 561.8

10-364

Combined Gas-Vapor Power Cycles 11-81C The energy source of the steam is the waste energy of the exhausted combustion gases.

11-82C Because the combined gas-steam cycle takes advantage of the desirable characteristics of the gas cycle at high temperature, and those of steam cycle at low temperature, and combines them. The result is a cycle that is more efficient than either cycle executed operated alone.

11-83 A combined gas-steam power cycle is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a simple ideal Rankine cycle. The mass flow rate of the steam, the net power output, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-365

Properties The properties of air at room temperature are c p  1.005 kJ / kg.K and k = 1.4 (Table A-2). Analysis (a) The analysis of gas cycle yields

æP ö( ÷ T6 = T5 ççç 6 ÷ ÷ çè P5 ÷ ø

k - 1)/ k 0.4/1.4

= (300 K )(16)

= 662.5 K

Qin = mair (h7 - h6 ) = mair c p (T7 - T6 ) = (14 kg/s)(1.005 kJ/kg ×K )(1500 - 662.5)K = 11, 784 kW WC , gas = mair (h6 - h5 ) = mair c p (T6 - T5 ) = (14 kg/s)(1.005 kJ/kg ×K )(662.5 - 300)K = 5100 kW æP ö( ÷ T8 = T7 ççç 8 ÷ ÷ èç P7 ÷ ø

k - 1)/ k

0.4/1.4

æ1 ö ÷ = (1500 K )çç ÷ ÷ èç16 ø

= 679.3 K

WT , gas = mair (h7 - h8 ) = mair c p (T7 - T8 ) = (14 kg/s)(1.005 kJ/kg ×K )(1500 - 679.3) K = 11,547 kW Wnet, gas = WT , gas - WC , gas = 11,547 - 5,100 = 6447 kW From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 15 kPa = 225.94 kJ/kg

v1 = v f @ 15 kPa = 0.001014 m3 /kg

10-366

æ 1 kJ ÷ ö wpI, in = v1 (P2 - P1 ) = (0.001014 m3 /kg )(10,000 - 15 kPa )çç = 10.12 kJ/kg ÷ çè1 kPa ×m3 ÷ ø h2 = h1 + wpI, in = 225.94 + 10.13 = 236.06 kJ/kg

P3 = 10 MPaüï h3 = 3097.0 kJ/kg ý T3 = 400°C ïïþ s3 = 6.2141 kJ/kg ×K s4 - s f 6.2141- 0.7549 = = 0.7528 P4 = 15 kPaü ïï x4 = s 7.2522 ý fg ïïþ s4 = s3 h4 = h f + x4 h fg = 225.94 + (0.7528)(2372.3) = 2011.8 kJ/kg Noting that Q @W @ Δke @ Δpe @ 0 for the heat exchanger, the steady-flow energy balance equation yields

Ein - Eout = D Esystem ¿ 0 (steady) = 0 ¾ ¾ ® Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) i i

ms =

e e

air

c p (T8 - T9 ) (1.005 kJ/kg ×K )(679.3 - 420) K h8 - h9 mair = mair = (14 kg/s) = 1.275 kg / s h3 - h2 h3 - h2 (3097.0 - 236.06) kJ/kg

(b) WT, steam = ms (h3 - h4 ) = (1.275 kg/s)(3097.0 - 2011.5) kJ/kg = 1384 kW (c) Wp, steam = ms wp = (1.275 kg/s)(10.12 kJ/kg) = 12.9 kW (d) Wnet, steam = WT, steam - Wp, steam = 1384 - 12.9 = 1371 kW and

Wnet = Wnet, steam + Wnet, gas = 1371 + 6448 = 7819 kW (e) hth =

Wnet 7819 kW = = 66.4% Qin 11,784 kW

EES (Engineering Equation Solver) SOLUTION "Given" r_P=16 T[5]=300 [K] m_dot_air=14 [kg/s] T[7]=1500 [K] T[9]=420 [K] P[3]=10000 [kPa] T[3]=(400+273) [K] P[4]=15 [kPa] "Properties" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" "Gas Turbine cycle" T[6]=T[5]*r_P^((k-1)/k) Q_dot_in=m_dot_air*c_p*(T[7]-T[6]) W_dot_C_gas=m_dot_air*c_p*(T[6]-T[5]) T[8]=T[7]*(1/r_P)^((k-1)/k) W_dot_T_gas=m_dot_air*c_p*(T[7]-T[8]) W_dot_net_gas=W_dot_T_gas-W_dot_C_gas "Steam cycle" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-367

Fluid$='steam_iapws' P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) x[4]=quality(Fluid$, P=P[4], s=s[4]) "Combined cycle" "(a)" m_dot_steam*(h[3]-h[2])=m_dot_air*C_p*(T[8]-T[9]) "heat exchanger energy balance" "(b)" W_dot_T_steam=m_dot_steam*(h[3]-h[4]) W_dot_P_steam=m_dot_steam*w_p_in W_dot_net_steam=W_dot_T_steam-W_dot_P_steam W_dot_net=W_dot_net_steam+W_dot_net_gas "(c)" Eta_th=W_dot_net/Q_dot_in

11-84 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

10-368

Properties The properties of air at room temperature are c p  1.005 kJ / kg.K and k = 1.4 (Table A-2a). Analysis Working around the topping cycle gives the following results: ( k - 1)/ k

æP ö ÷ T6 s = T5 ççç 6 ÷ ÷ ÷ çè P ø

= (293 K)(8)0.4/1.4 = 530.8 K

hC =

h6 s - h5 c p (T6 s - T5 ) = h6 - h5 c p (T6 - T5 )

¾¾ ®

T6 = T5 +

T6 s - T5 hC

= 293 + ( k - 1)/ k

æP ö ÷ T8 s = T7 ççç 8 ÷ ÷ çè P ÷ ø 7

hT =

530.8 - 293 = 572.8 K 0.85 0.4/1.4

æ1 ö = (1373 K) çç ÷ ÷ çè8 ÷ ø

c p (T7 - T8 ) h7 - h8 = h7 - h8 s c p (T7 - T8 s )

= 758.0 K

¾¾ ®

T8 = T7 - hT (T7 - T8 s )

= 1373 - (0.90)(1373 - 758.0)¶ = 819.5 K

T9 = Tsat @ 6000 kPa = 275.6°C = 548.6 K

10-369

Fixing the states around the bottom steam cycle yields (Tables A-4, A-5, A-6): h1 = h f @ 20 kPa = 251.42 kJ/kg wp, in = v1 ( P2 - P1 )

æ 1 kJ ÷ ö = (0.001017 m3 /kg)(6000 - 20)kPa çç ÷ çè1 kPa ×m3 ÷ ø

= 6.08 kJ/kg h2 = h1 + wp, in = 251.42 + 6.08 = 257.5 kJ/kg

P3 = 6000 kPa T3 = 320°C

h3 = 2953.6 kJ/kg s3 = 6.1871 kJ/kg ×K

ïü ïý h = 2035.8 kJ/kg ïïþ 4 s

P4 = 20 kPa s4 = s3 hT =

ü ïï ý ïïþ

h3 - h4 h3 - h4 s

¾¾ ® h4 = h3 - hT (h3 - h4 s )

= 2953.6 - (0.90)(2953.6 - 2035.8)

= 2127.6 kJ/kg The net work outputs from each cycle are wnet, gas cycle = wT,out - wC,in = c p (T7 - T8 ) - c p (T6 - T5 )

= (1.005 kJ/kg ×K)(1373 - 819.5 - 572.7 + 293)K

= 275.2 kJ/kg wnet, steam cycle = wT,out - wP,in

= (h3 - h4 ) - wP,in

= (2953.6 - 2127.6) - 6.08 = 819.9 kJ/kg An energy balance on the heat exchanger gives

ma c p (T8 - T9 ) = mw (h3 - h2 ) ¾ ¾ ® mw =

c p (T8 - T9 ) h3 - h2

ma =

(1.005)(819.5 - 548.6) = 0.1010ma 2953.6 - 257.5

That is, 1 kg of exhaust gases can heat only 0.1010 kg of water. Then, the mass flow rate of air is

ma =

Wnet 100,000 kJ/s = = 279.3 kg / s wnet (1´ 275.2 + 0.1010´ 819.9) kJ/kg air

EES (Engineering Equation Solver) SOLUTION "Given" P5=101 [kPa] T5=(20+273) [K] T7=(1100+273) [K] r_p=8 eta_C=0.85 eta_T=0.90 P[2]=6000 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-370

T[3]=320 [C] P[4]=20 [kPa] W_dot_net=100000 [kW] "Properties" c_p=1.005 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" "Gas turbine cycle" T6_s=T5*r_p^((k-1)/k) T6=T5+(T6_s-T5)/eta_C T8_s=T7*(1/r_p)^((k-1)/k) T8=T7-eta_T*(T7-T8_s) T9=temperature(Fluid$, P=P[2], x=1)+273 "K" "Rankine cycle" Fluid$='steam_iapws' P[1]=P[4] P[3]=P[2] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s_s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_4s) w_net_gas=c_p*(T7-T8)-c_p*(T6-T5) w_net_steam=h[3]-h[4]-w_p_in m_a*c_p*(T8-T9)=m_w*(h[3]-h[2]) "energy balance on heat exchanger" m_a=1 "assumed" w_net=m_a*w_net_gas+m_w*w_net_steam m_dot_a=W_dot_net/w_net

11-85 A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The mass flow rate of air for a specified power output is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats.

10-371

Properties The properties of air at room temperature are c p  1.005 kJ / kg.K and k = 1.4 (Table A-2a). Analysis With an ideal regenerator, the temperature of the air at the compressor exit will be heated to the to the temperature at the turbine exit. Representing this state by “6a”

T6a = T8 = 819.5 K The rate of heat addition in the cycle is

Qin = ma c p (T7 - T6 a ) = (279.3 kg/s)(1.005 kJ/kg ×°C)(1373 - 819.5) K = 155,370 kW The thermal efficiency of the cycle is then hth =

Wnet 100, 000 kW = = 0.6436 155,370 kW Qin

Without the regenerator, the rate of heat addition and the thermal efficiency are

Qin = ma c p (T7 - T6 ) = (279.3 kg/s)(1.005 kJ/kg ×°C)(1373 - 572.7) K = 224,640 kW

hth =

Wnet 100,000 kW = = 0.4452 224,640 kW Qin

The change in the thermal efficiency due to using the ideal regenerator is D hth = 0.6436 - 0.4452 = 0.1984

10-372

EES (Engineering Equation Solver) SOLUTION "Given" P5=101 [kPa] T5=(20+273) [K] T7=(1100+273) [K] r_p=8 eta_C=0.85 eta_T=0.90 P[2]=6000 [kPa] T[3]=320 [C] P[4]=20 [kPa] W_dot_net=100000 [kW] "Properties" c_p=1.005 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" "Analysis" "Gas turbine cycle" T6_s=T5*r_p^((k-1)/k) T6=T5+(T6_s-T5)/eta_C T8_s=T7*(1/r_p)^((k-1)/k) T8=T7-eta_T*(T7-T8_s) T9=temperature(Fluid$, P=P[2], x=1)+273 "K" "Rankine cycle" Fluid$='steam_iapws' P[1]=P[4] P[3]=P[2] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s_s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_4s) w_net_gas=c_p*(T7-T8)-c_p*(T6-T5) w_net_steam=h[3]-h[4]-w_p_in m_a*c_p*(T8-T9)=m_w*(h[3]-h[2]) "energy balance on heat exchanger" m_a=1 "assumed" w_net=m_a*w_net_gas+m_w*w_net_steam m_dot_a=W_dot_net/w_net Q_dot_in=m_dot_a*c_p*(T7-T6) eta_th=W_dot_net/Q_dot_in "If an ideal regenrator is added" T6_a=T8 Q_dot_in_a=m_dot_a*c_p*(T7-T6_a) eta_th_a=W_dot_net/Q_dot_in_a DELTAeta_th=eta_th_a-eta_th

11-86 The component of the combined cycle with the largest exergy destruction of the component of the combined cycle in Prob. 11-84 is to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-373

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From Problem 11-84, Tsource, gas cycle = 1373 K

Tsource, steam cycle = T8 = 819.5 K s1 = s2 = s f @ 20 kPa = 0.8320 kJ/kg ×K

s3 = 6.1871 kJ/kg ×K

s4 = 6.4627 kJ/kg ×K qin,67 = c p (T7 - T6 ) = 804.3 kJ/kg qin,23 = h3 - h2 = 2696.1 kJ/kg

qout = h4 - h1 = 1876.2 kJ/kg mw = 0.1010ma = 0.1010(279.3) = 28.21 kg/s X destroyed, 12 = 0 (isentropic process) X destroyed, 34 = mwT0 (s4 - s3 ) = (28.21 kg/s)(293 K)(6.4627 - 6.1871) = 2278 kW

æ q ö ÷ X destroyed, 41 = mwT0 çççs1 - s4 + out ÷ ÷ çè Tsink ÷ ø

æ 1876.2 kJ/kg ö÷ = (28.21 kg/s)(293 K) çç0.8320 - 6.1871 + ÷= 8665 kW çè 293 K ø÷ æ T ö ÷ X destroyed,heat exchanger = maT0D s89 + mwT0D s23 = maT0 çççc p ln 9 ÷ ÷+ maT0 (s3 - s2 ) çè T8 ÷ ø

é 548.6 ù ú+ (28.21)(293)(6.1871- 0.8320) = (279.3)(293) ê(1.005) ln êë 819.5 úû = 11260 kW æ ù T Pö 572.7 ÷= (279.3)(293) é ê(1.005)ln X destroyed, 56 = maT0 ç c p ln 6 - R ln 6 ÷ - (0.287) ln(8) ú= 6280 kW ç ÷ ç ÷ ê ú ç T P 293 è ë û 5 5ø

æ é T qin ö 1373 804.3 ù ÷ ÷ ú= 23, 970 kW X destroyed, 67 = maT0 çççc p ln 7 = (279.3)(293) ê(1.005)ln ÷ ú çè T6 Tsource ÷ 572.7 1373 û ø ëê æ é æ1 öù T Pö 819.5 ú ÷= (279.3)(293) ê(1.005)ln X destroyed, 78 = maT0 çççc p ln 8 - R ln 8 ÷ - (0.287) ln çç ÷ ÷ ÷ ÷ú= 6396 kW çè8 ø ê ÷ çè T7 P7 ø 1373 ë û

The largest exergy destruction occurs during the heat addition process in the combustor of the gas cycle.

EES (Engineering Equation Solver) SOLUTION "Given" P5=101 [kPa] T5=(20+273) [K] T7=(1100+273) [K] r_p=8 eta_C=0.85 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-374

eta_T=0.90 P[2]=6000 [kPa] T[3]=320 [C] P[4]=20 [kPa] W_dot_net=100000 [kW] "Properties" c_p=1.005 [kJ/kg-K] "Table A-2a" k=1.4 "Table A-2a" R=0.287 [kJ/kg-K] "Table A-2a" T_source=T7 T0=T5 T_sink=T5 s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] s[4]=entropy(Fluid$, P=P[4], h=h[4]) q_in_67=c_p*(T7-T6) q_in_23=h[3]-h[2] q_out=h[4]-h[1] m_dot_w=m_w*m_dot_a X_dot_dest_12=0 "isentropic process" X_dot_dest_34=m_dot_w*T0*(s[4]-s[3]) X_dot_dest_41=m_dot_w*T0*(s[1]-s[4]+q_out/T_sink) X_dot_dest_HX=m_dot_a*T0*(c_p*ln(T9/T8))+m_dot_w*T0*(s[3]-s[2]) X_dot_dest_56=m_dot_a*T0*(c_p*ln(T6/T5)-R*ln(r_p)) X_dot_dest_67=m_dot_a*T0*(c_p*ln(T7/T6)-q_in_67/T_source) X_dot_dest_78=m_dot_a*T0*(c_p*ln(T8/T7)-R*ln(1/r_p)) "Gas turbine cycle" T6_s=T5*r_p^((k-1)/k) T6=T5+(T6_s-T5)/eta_C T8_s=T7*(1/r_p)^((k-1)/k) T8=T7-eta_T*(T7-T8_s) T9=temperature(Fluid$, P=P[2], x=1)+273 "K" "Rankine cycle" Fluid$='steam_iapws' P[1]=P[4] P[3]=P[2] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s_s[4]) Eta_T=(h[3]-h[4])/(h[3]-h_4s) w_net_gas=c_p*(T7-T8)-c_p*(T6-T5) w_net_steam=h[3]-h[4]-w_p_in m_a*c_p*(T8-T9)=m_w*(h[3]-h[2]) "energy balance on heat exchanger" m_a=1 "assumed" w_net=m_a*w_net_gas+m_w*w_net_steam m_dot_a=W_dot_net/w_net

10-375

11-87 A 280-MW combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal Rankine cycle with an open feedwater heater. The mass flow rate of air to steam, the required rate of heat input in the combustion chamber, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) Using the properties of air from Table A-17, the analysis of gas cycle yields

T8 = 300 K ¾ ¾ ® s9o = s8o + R ln hC =

h8 = 300.19 kJ/kg s8o = 1.70203 kJ/kg ×K

P9 = 1.70203 + (0.287 kJ/kg ×K) ln(11) = 2.39023 kJ/kg ×K ¾ ¾ ® h9 s = 595.83 kJ/kg P8

h9 s - h8 ¾¾ ® h9 = h8 + (h9 s - h8 ) / hC h9 - h8

= 300.19 + (595.83 - 300.19) / (0.82)

= 660.73 kJ/kg T10 = 1100 K ¾ ¾ ®

h10 = 1161.07 kJ/kg s10o = 3.07732 kJ/kg ×K

s11o = s10o + R ln

hT =

P11 = 3.07732 + (0.287 kJ/kg ×K) ln(1/ 11) = 2.38912 kJ/kg ×K ¾ ¾ ® h11s = 595.19 kJ/kg P10

h10 - h11 ¾¾ ® h11 = h10 - hT (h10 - h11s ) h10 - h11s

= 1161.07 - (0.86)(1161.07 - 595.19)

= 674.41 kJ/kg T12 = 420 K ¾ ¾ ® h12 = 421.26 kJ/kg

10-376

From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg wpI, in = v1 (P2 - P1 )

æ 1 kJ ö÷ = (0.00101 m3 /kg )(800 - 10 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 0.80 kJ/kg h2 = h1 + wpI, in = 191.81 + 0.80 = 192.60 kJ/kg h3 = h f @ 0.8 MPa = 720.87 kJ/kg

v 3 = v f @ 0.8 MPa = 0.001115 m3 /kg

wpII, in = v 3 (P4 - P3 )

æ 1 kJ ÷ ö = (0.001115 m3 /kg )(5000 - 800 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 4.68 kJ/kg h4 = h3 + wpI, in = 720.87 + 4.68 = 725.55 kJ/kg

P5 = 5 MPaïü h5 = 3069.3 kJ/kg ý T5 = 350°C ïïþ s5 = 6.4516 kJ/kg ×K s6 s - s f 6.4516 - 2.0457 = = 0.9545 P6 = 0.8 MPaïü ïý x6 s = s 4.6160 fg ïïþ s6 s = s5 h6 s = h f + x6 s h fg = 720.87 + (0.9545)(2085.8) = 2675.1 kJ/kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-377

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) = 3069.3 - (0.86)(3069.3 - 2675.1) = 2730.3 kJ/kg h5 - h6 s

hT =

s7 - s f 6.4516 - 0.6492 = = 0.7737 P7 = 10 kPaüïï x7 s = s fg 7.4996 ý ï s7 = s5 ïþ h = h + x h = 191.81 + (0.7737)(2392.1) = 2042.5 kJ/kg 7s f 7 fg h5 - h7 ¾¾ ® h7 = h5 - hT (h5 - h7 s ) = 3069.3 - (0.86)(3069.3 - 2042.5) = 2186.3 kJ/kg h5 - h7 s

hT =

Noting that Q @W @ Δke @ Δpe @ 0 for the heat exchanger, the steady-flow energy balance equation yields

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) i i

e e

air

mair h - h4 3069.3 - 725.55 = 5 = = 9.258 kg air / kg steam ms h11 - h12 674.41- 421.26

(b) Noting that Q @W @ Δke @ Δpe @ 0 for the open FWH, the steady-flow energy balance equation yields

Ein - Eout = D Esystem ¿ 0 (steady) = 0 ® Ein = Eo3ut

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = (1)h i i

e e

2 2

6 6

3 3

Thus, y=

h3 - h2 720.87 - 192.60 = = 0.2082 (the fraction of steam extracted ) h6 - h2 2730.3 - 192.60

wT = hT éëh5 - h6 + (1- y )(h6 - h7 )ùû = (0.86)éë3069.3 - 2730.3 + (1- 0.2082)(2730.3 - 2186.3)ù û= 769.8 kJ/kg wnet, steam = wT - wp, in = wT - (1- y )wp, I - wp, II

= 769.8 - (1- 0.2082)(0.80)- 4.68 = 764.5 kJ/kg wnet, gas = wT - wC , in = (h10 - h11 )- (h9 - h8 ) = 1161.07 - 674.41- (660.73 - 300.19) = 126.12 kJ/kg The net work output per unit mass of gas is

wnet = wnet, gas + mair =

1 1 wnet, steam = 126.12 + (764.5) = 208.70 kJ/kg 6.425 9.258

Wnet 280,000 kJ/s = = 1341.6 kg/s wnet 208.70 kJ/kg

and Qin = mair (h10 - h9 ) = (1341.6 kg/s)(1161.07 - 660.73) kJ/kg = 671, 300 kW (c) hth =

Wnet 280,000 kW = = 0.4171 = 41.7% 671,300 kW Qin

10-378

EES (Engineering Equation Solver) SOLUTION $Array on "Input data" T[8] = 300 [K] "Gas compressor inlet" P[8] = 14.7 [kPa] "Assumed air inlet pressure" Pratio = 11 "Pressure ratio for gas compressor" T[10] = 1100 [K] "Gas turbine inlet" T[12] = 420 [K] "Gas exit temperature from Gas-to-steam heat exchanger " P[12] = P[8] "Assumed air exit pressure" W_dot_net=280 [MW] Eta_comp = 0.82 Eta_gas_turb =0.86 Eta_pump = 1.0 Eta_steam_turb =0.86 P[5] = 5000 [kPa] "Steam turbine inlet" T[5] =(350+273.15) "[K]" "Steam turbine inlet" P[6] = 800 [kPa] "Extraction pressure for steam open feedwater heater" P[7] = 10 [kPa] "Steam condenser pressure" "GAS POWER CYCLE ANALYSIS" "Gas Compressor anaysis" s[8]=ENTROPY(Air,T=T[8],P=P[8]) ss9=s[8] "For the ideal case the entropies are constant across the compressor" P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9]) "Ts9 is the isentropic value of T[9] at compressor exit" Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp > w_comp_isen" h[8] + w_gas_comp_isen =hs9 "SSSF conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9] "SSSF conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10] "SSSF conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is SSSF constant pressure" Q_dot_in*Convert(MW, kW)=m_dot_gas*q_in "Gas Turbine analysis" s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11]) "Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb" h[10] = w_gas_turb_isen + hs11"SSSF conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11] "SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11]) "Gas-to-Steam Heat Exchanger" "SSSF conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-379

h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid$='steam_iapws' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid$,P=P[1],x=0) {Saturated liquid} v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1]) "SSSF isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3]) "SSSF isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s[5]=entropy(Fluid$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid$,s=ss6,P=P[6]) x_6s=quality(Fluid$,s=ss6,P=P[6]) Ts6=temperature(Fluid$,s=ss6,P=P[6]) h[6]=h[5]-Eta_steam_turb*(h[5]-hs6)"Definition of steam turbine efficiency" T[6]=temperature(Fluid$,P=P[6],h=h[6]) s[6]=entropy(Fluid$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid$,s=ss7,P=P[7]) x_7s=quality(Fluid$,s=ss7,P=P[7]) Ts7=temperature(Fluid$,s=ss7,P=P[7]) h[7]=h[5]-Eta_steam_turb*(h[5]-hs7)"Definition of steam turbine efficiency" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1]"SSSF conservation of energy for the Condenser per unit mass" Q_dot_out*Convert (MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-380

W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb w_steam_pumps) "definition of the net cycle work" Eta_th=W_dot_net/Q_dot_in*convert( ,%) "Cycle thermal efficiency, in percent" Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

11-88 Problem 11-87 is reconsidered. The effect of the gas cycle pressure ratio on the ratio of gas flow rate to steam flow rate and cycle thermal efficiency is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input data" T[8] = 300 [K] "Gas compressor inlet" P[8] = 100 [kPa] "Assumed air inlet pressure" Pratio = 11 "Pressure ratio for gas compressor" T[10] = 1100 [K] "Gas turbine inlet" T[12] = 420 [K] "Gas exit temperature from Gas-to-steam heat exchanger " P[12] = P[8] "Assumed air exit pressure" W_dot_net=280 [MW] Eta_comp = 0.82 Eta_gas_turb = 0.86 Eta_pump = 1.0 Eta_steam_turb = 0.86 P[5] = 5000 [kPa] "Steam turbine inlet" T[5] =(350+273.15) [K] "Steam turbine inlet" P[6] = 800 [kPa] "Extraction pressure for steam open feedwater heater" P[7] = 10 [kPa] "Steam condenser pressure" "GAS POWER CYCLE ANALYSIS" "Gas Compressor anaysis" s[8]=ENTROPY(Air,T=T[8],P=P[8]) ss9=s[8] "For the ideal case the entropies are constant across the compressor" P[9] = Pratio*P[8] Ts9=temperature(Air,s=ss9,P=P[9]) "Ts9 is the isentropic value of T[9] at compressor exit" Eta_comp = w_gas_comp_isen/w_gas_comp "compressor adiabatic efficiency, w_comp > w_comp_isen" h[8] + w_gas_comp_isen =hs9 "conservation of energy for the isentropic compressor, assuming: adiabatic, ke=pe=0 per unit gas mass flow rate in kg/s" h[8]=ENTHALPY(Air,T=T[8]) hs9=ENTHALPY(Air,T=Ts9) h[8] + w_gas_comp = h[9] "conservation of energy for the actual compressor, assuming: adiabatic, ke=pe=0" T[9]=temperature(Air,h=h[9]) s[9]=ENTROPY(Air,T=T[9],P=P[9]) "Gas Cycle External heat exchanger analysis" h[9] + q_in = h[10] "conservation of energy for the external heat exchanger, assuming W=0, ke=pe=0" h[10]=ENTHALPY(Air,T=T[10]) P[10]=P[9] "Assume process 9-10 is constant pressure" Q_dot_in*1000=m_dot_gas*q_in "Gas Turbine analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-381

s[10]=ENTROPY(Air,T=T[10],P=P[10]) ss11=s[10] "For the ideal case the entropies are constant across the turbine" P[11] = P[10] /Pratio Ts11=temperature(Air,s=ss11,P=P[11]) "Ts11 is the isentropic value of T[11] at gas turbine exit" Eta_gas_turb = w_gas_turb /w_gas_turb_isen "gas turbine adiabatic efficiency, w_gas_turb_isen > w_gas_turb" h[10] = w_gas_turb_isen + hs11 "conservation of energy for the isentropic gas turbine, assuming: adiabatic, ke=pe=0" hs11=ENTHALPY(Air,T=Ts11) h[10] = w_gas_turb + h[11] "SSSF conservation of energy for the actual gas turbine, assuming: adiabatic, ke=pe=0" T[11]=temperature(Air,h=h[11]) s[11]=ENTROPY(Air,T=T[11],P=P[11]) "Gas-to-Steam Heat Exchanger" m_dot_gas*h[11] + m_dot_steam*h[4] = m_dot_gas*h[12] + m_dot_steam*h[5] "conservation of energy for the gas-to-steam heat exchanger, assuming: adiabatic, W=0, ke=pe=0" h[12]=ENTHALPY(Air, T=T[12]) s[12]=ENTROPY(Air,T=T[12],P=P[12]) "STEAM CYCLE ANALYSIS" "Steam Condenser exit pump or Pump 1 analysis" Fluid$='Steam_IAPWS' P[1] = P[7] P[2]=P[6] h[1]=enthalpy(Fluid$,P=P[1],x=0) v1=volume(Fluid$,P=P[1],x=0) s[1]=entropy(Fluid$,P=P[1],x=0) T[1]=temperature(Fluid$,P=P[1],x=0) w_pump1_s=v1*(P[2]-P[1]) "isentropic pump work assuming constant specific volume" w_pump1=w_pump1_s/Eta_pump "Definition of pump efficiency" h[1]+w_pump1= h[2] "Steady-flow conservation of energy" s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Open Feedwater Heater analysis" y*h[6] + (1-y)*h[2] = 1*h[3] "Steady-flow conservation of energy" P[3]=P[6] h[3]=enthalpy(Fluid$,P=P[3],x=0) "Condensate leaves heater as sat. liquid at P[3]" T[3]=temperature(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) "Boiler condensate pump or Pump 2 analysis" P[4] = P[5] v3=volume(Fluid$,P=P[3],x=0) w_pump2_s=v3*(P[4]-P[3]) "isentropic pump work assuming constant specific volume" w_pump2=w_pump2_s/Eta_pump "Definition of pump efficiency" h[3]+w_pump2= h[4] "Steady-flow conservation of energy" s[4]=entropy(Fluid$,P=P[4],h=h[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) w_steam_pumps = (1-y)*w_pump1+ w_pump2 "Total steam pump work input/ mass steam" "Steam Turbine analysis" h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s[5]=entropy(Fluid$,P=P[5],T=T[5]) ss6=s[5] hs6=enthalpy(Fluid$,s=ss6,P=P[6]) Ts6=temperature(Fluid$,s=ss6,P=P[6]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-382

h[6]=h[5]-Eta_steam_turb*(h[5]-hs6) "Definition of steam turbine efficiency" T[6]=temperature(Fluid$,P=P[6],h=h[6]) s[6]=entropy(Fluid$,P=P[6],h=h[6]) ss7=s[5] hs7=enthalpy(Fluid$,s=ss7,P=P[7]) Ts7=temperature(Fluid$,s=ss7,P=P[7]) h[7]=h[5]-Eta_steam_turb*(h[5]-hs7) "Definition of steam turbine efficiency" T[7]=temperature(Fluid$,P=P[7],h=h[7]) s[7]=entropy(Fluid$,P=P[7],h=h[7]) h[5] = w_steam_turb + y*h[6] +(1-y)*h[7] "SSSF conservation of energy for the steam turbine: adiabatic, neglect ke and pe" "Steam Condenser analysis" (1-y)*h[7]=q_out+(1-y)*h[1] "conservation of energy for the Condenser per unit mass" Q_dot_out*Convert(MW, kW)=m_dot_steam*q_out "Cycle Statistics" MassRatio_gastosteam =m_dot_gas/m_dot_steam W_dot_net*Convert(MW, kW)=m_dot_gas*(w_gas_turb-w_gas_comp)+ m_dot_steam*(w_steam_turb w_steam_pumps) "definition of the net cycle work" Eta_th=W_dot_net/Q_dot_in*Convert(, %) "Cycle thermal efficiency, in percent" Bwr=(m_dot_gas*w_gas_comp + m_dot_steam*w_steam_pumps)/(m_dot_gas*w_gas_turb + m_dot_steam*w_steam_turb) "Back work ratio" W_dot_net_steam = m_dot_steam*(w_steam_turb - w_steam_pumps) W_dot_net_gas = m_dot_gas*(w_gas_turb - w_gas_comp) NetWorkRatio_gastosteam = W_dot_net_gas/W_dot_net_steam

Pratio

MassRatiogastosteam

η th

10 11 12 13 14 15 16 17 18 19 20

8.775 9.262 9.743 10.22 10.7 11.17 11.64 12.12 12.59 13.07 13.55

[%] 42.03 41.67 41.22 40.68 40.08 39.4 38.66 37.86 36.99 36.07 35.08

10-383

10-384

11-89 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The moisture percentage at the exit of the low-pressure turbine, the steam temperature at the inlet of the high-pressure turbine, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) We obtain the air properties from EES. The analysis of gas cycle is as follows T7 = 15°C ¾ ¾ ® h7 = 288.53 kJ/kg

ïüï ý s7 = 5.6649 kJ/kg P7 = 100 kPaïïþ

T7 = 15°C

P8 = 700 kPa ïüï ý h8 s = 503.70 kJ/kg ïïþ s8 = s7 hC =

h8 s - h7 ¾¾ ® h8 = h7 + (h8 s - h7 ) / hC h8 - h7

= 288.53 + (503.70 - 288.53) / (0.80)

= 557.49 kJ/kg T9 = 950°C ¾ ¾ ® h9 = 1305.2 kJ/kg

T9 = 950°C ïüï ý s9 = 6.6456 kJ/kg P9 = 700 kPa ïïþ

P10 = 100 kPa ïüï ý h10 s = 764.01 kJ/kg ïïþ s10 = s9 hT =

h9 - h10 ¾¾ ® h10 = h9 - hT (h9 - h10 s ) h9 - h10 s

= 1305.2 - (0.80)(1305.2 - 764.01)

= 872.25 kJ/kg T11 = 200°C ¾ ¾ ® h11 = 475.77 kJ/kg

10-385

From the steam tables (Tables A-4, A-5, and A-6 or from EES), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg wpI, in = v1 (P2 - P1 ) / h p æ 1 kJ ö ÷ = (0.00101 m3 /kg)(6000 - 10 kPa )çç ÷ ÷/ 0.80 çè1 kPa ×m3 ø

= 7.56 kJ/kg h2 = h1 + wpI, in = 191.81 + 7.65 = 199.37 kJ/kg

P5 = 1 MPaïü h5 = 3264.5 kJ/kg ý T5 = 400°Cïïþ s5 = 7.4670 kJ/kg ×K

s6 s - s f 7.4670 - 0.6492 = = 0.9091 P6 = 10 kPaüïï x6 s = s fg 7.4996 ý ïïþ s6 s = s5 h6 s = h f + x6 s h fg = 191.81 + (0.9091)(2392.1) = 2366.4 kJ/kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-386

hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) h5 - h6 s = 3264.5 - (0.80)(3264.5 - 2366.4) = 2546.0 kJ/kg

üï P6 = 10 kPa ý x = 0.9842 h6 = 2546.5 kJ/kgïïþ 6

Moisture Percentage = 1- x6 = 1- 0.9842 = 0.0158 = 1.58%

(b) Noting that Q @W @ Δke @ Δpe @ 0 for the heat exchanger, the steady-flow energy balance equation yields Ein = Eout

å mh = å m h i

ms (h3 - h2 )+ ms (h5 - h4 ) = mair (h10 - h11 )

(4.6) [(3350.3 - 199.37) + (3264.5 - h4 ) ]= (40)(872.25 - 475.77) ¾ ¾ ® h4 = 2967.8 kJ/kg Also, P3 = 6 MPa ü ïý h3 = T3 = ? ïï s3 = þ

hT =

P4 = 1 MPaü ï ý h4 s = s4 s = s3 ïï þ

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s

The temperature at the inlet of the high-pressure turbine may be obtained by a trial-error approach or using EES from the above relations. The answer is T3  469.6o C . Then, the enthalpy at state 3 becomes: h3  3350.3 kJ / kg (c) WT, gas = mair (h9 - h10 ) = (40 kg/s)(1305.2 - 872.25) kJ/kg = 17,318 kW

WC, gas = mair (h8 - h7 ) = (40 kg/s)(557.49 - 288.53) kJ/kg = 10, 758 kW © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-387

Wnet, gas = WT, gas - WC, gas = 17,318 - 10, 758 = 6560 kW WT, steam = ms (h3 - h4 + h5 - h6 ) = (4.6 kg/s)(3350.3 - 2967.8 + 3264.5 - 2546.0) kJ/kg = 5065 kW

WP, steam = ms wpump = (4.6 kg/s)(7.564) kJ/kg = 34.8 kW Wnet, steam = WT, steam - WP, steam = 5065 - 34.8 = 5030 kW Wnet, plant = Wnet, gas + Wnet, steam = 6560 + 5030 = 11,590 kW

(d) Qin = mair (h9 - h8 ) = (40 kg/s)(1305.2 - 557.49) kJ/kg = 29,908 kW

hth =

Wnet, plant Qin

11,590 kW = 0.3875 = 38.8% 29,908 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[7]=100 [kPa] "assumed" r=7; P[8]=r*P[7]; P[9]=P[8]; P[10]=P[7]; P[11]=P[7] T[7]=15 [C]; T[9]=950 [C]; T[11]=200 [C] m_dot_gas=40 [kg/s]; m_dot_steam=4.6 [kg/s] P[3]=6000 [kPa]; P[6]=10 [kPa]; P[2]=P[3]; P[1]=P[6] T[5]=400 [C]; P[5]=1000 [kPa]; P[4]=P[5] eta_C=0.80; eta_T=0.80; eta_P=0.80 "PROPERTIES, gas cycle" GasFluid$='Air' h[7]=enthalpy(GasFluid$, T=T[7]) s[7]=entropy(GasFluid$, T=T[7], P=P[7]) h_8s=enthalpy(GasFluid$, P=P[8], s=s[7]) h[8]=h[7]+(h_8s-h[7])/eta_C h[9]=enthalpy(GasFluid$, T=T[9]) s[9]=entropy(GasFluid$, P=P[9], T=T[9]) h_10s=enthalpy(GasFluid$, P=P[10], s=s[9]) h[10]=h[9]-eta_T*(h[9]-h_10s) T[10]=temperature(GasFluid$, h=h[10]) h[11]=enthalpy(GasFluid$, T=T[11]) "PROPERTIES, steam cycle" SteamFluid$='Steam_iapws' h[1]=enthalpy(SteamFluid$, P=P[1], x=0) v_1=volume(SteamFluid$, P=P[1], x=0) w_pump=v_1*(P[2]-P[1])/eta_P h[2]=h[1]+w_pump h[5]=enthalpy(SteamFluid$, P=P[5], T=T[5]) s[5]=entropy(SteamFluid$, P=P[5], T=T[5]) h_6s=enthalpy(SteamFluid$, P=P[6], s=s[5]) x_6s=quality(SteamFluid$, P=P[6], s=s[5]) h[6]=h[5]-eta_T*(h[5]-h_6s) h[3]=enthalpy(SteamFluid$, P=P[3], T=T[3]) s[3]=entropy(SteamFluid$, P=P[3], T=T[3]) h_4s=enthalpy(SteamFluid$, P=P[4], s=s[3]) h[4]=h[3]-eta_T*(h[3]-h_4s) "ANALYSIS" m_dot_gas*(h[10]-h[11])=m_dot_steam*(h[3]-h[2])+m_dot_steam*(h[5]-h[4]) "energy balance on heat exchanger" T_3=T[3] "temperature of steam at the turbine inlet" x_6=quality(SteamFluid$, P=P[6], h=h[6]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-388

MoisturePercentage=(1-x_6)*Convert(, %) "moisture percentage at the turbine exit" W_dot_T_gas=m_dot_gas*(h[9]-h[10]) W_dot_C_gas=m_dot_gas*(h[8]-h[7]) W_dot_net_gas=W_dot_T_gas-W_dot_C_gas W_dot_T_steam=m_dot_steam*(h[3]-h[4]+h[5]-h[6]) W_dot_P_steam=m_dot_steam*w_pump W_dot_net_steam=W_dot_T_steam-W_dot_P_steam W_dot_net_plant=W_dot_net_gas+W_dot_net_steam "net power from the plant" Q_dot_in=m_dot_gas*(h[9]-h[8]) Q_dot_out=m_dot_steam*(h[6]-h[1])+m_dot_gas*(h[11]-h[7]) eta_th=W_dot_net_plant/Q_dot_in "thermal efficiency" "Alternatively" W_dot_net_plant_alt=Q_dot_in-Q_dot_out SOLUTIONS eta_C=0.8 eta_P=0.8 eta_T=0.8 eta_th=0.3875 GasFluid$='Air' h_10s  764.03 [kJ / kg] h_4s  2872.26 [kJ / kg]

h_6s  2366.38 [kJ / kg] h_8s  503.74 [kJ / kg] MoisturePercentage=1.583 [%] m dot  gas  40 [kg / s] m dot  steam  4.6 [kg / s] Q_dot_in=29909 [kW] Q_dot_out=18319 [kW] r=7 SteamFluid$='Steam_iapws' T_3=469.6 [C] v_1  0.00101[m 3 / kg]

W_dot_C_gas=10760 [kW] W_dot_net_gas=6560 [kW] W_dot_net_plant=11590 [kW] W_dot_net_plant_alt=11590 [kW] W_dot_net_steam=5030 [kW] W_dot_P_steam=34.8 [kW] W_dot_T_gas=17320 [kW] W_dot_T_steam=5065 [kW] w_pump  7.564 [kJ / kg] x_6=0.9842 x_6s=0.9091

Review Problems 11-90 The closed feedwater heater of a regenerative Rankine cycle is to heat feedwater to a saturated liquid. The required mass flow rate of bleed steam is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-389

2 3 4

Kinetic and potential energy changes are negligible. There are no work interactions. Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.

Properties From the steam tables (Tables A-4 through A-6), P1 = 4000 kPa T1 = 200°C

P2 = 4000 kPa T2 = 245°C P3 = 3000 kPa x3 = 0.90

ü ïï ý h1 @ h f @ 200°C = 852.26 kJ/kg ïïþ

ïüï ý h2 @ h f @ 245°C = 1061.5 kJ/kg ïïþ ïü ïý h = h + x h f 3 fg ïïþ 3

= 1008.3 + (0.9)(1794.9) = 2623.7 kJ/kg P4 = 3000 kPa x4 = 0

ïü ïý h = h f @ 3000 kPa = 1008.3 kJ/kg ïïþ 4

min - mout = D msystem 0 (steady) = 0 ® min = mout ® m1 = m2 = m fw and

m3 = m4 = ms

Energy balance (for the heat exchanger):

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D Esystem 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

10-390

Combining the two, m fw (h2 - h1 ) = ms (h3 - h4 )

Solving for ms : ms =

h2 - h1 m fw h3 - h4

Substituting,

ms =

1061.5 - 852.26 (6 kg/s) = 0.777 kg / s 2623.7 - 1008.3

11-91 A steam power plant operates on the simple ideal Rankine cycle. The turbine inlet temperature, the net power output, the thermal efficiency, and the minimum mass flow rate of the cooling water required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 7.5kPa = 168.75 kJ / kg v1 = v f @ 7.5 kPa = 0.001008 m3 /kg T1 = Tsat @ 7.5 kPa = 40.29°C

wp, in = v1 (P2 - P1 )

æ 1 kJ ö÷ = (0.001008 m3 /kg)(6000 - 7.5 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 6.04 kJ/kg

h2 = h1 + wp, in = 168.75 + 6.04 = 174.79 kJ/kg

h4 = hg @ 7.5 kPa = 2574.0 kJ/kg s4 = sg @ 7.5 kPa = 8.2501 kJ/kg

P3 = 6 MPaïü h3 = 4852.2 kJ/kg ýT = 1089.2°C s3 = s4 ïïþ 3 (b) qin = h3 - h2 = 4852.2 - 174.79 = 4677.4 kJ/kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-391

wnet 2272.1 kJ/kg = = 48.6% qin 4677.4 kJ/kg

Thus,

Wnet = hth Qin = (0.4857)(40,000 kJ/s) = 19,428 kJ / s (e) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the

Qout = Qin - Wnet = 40,000 - 19, 428 = 20,572 kJ/s mcool =

Qout 20,572 kJ/s = = 194.6 kg / s cD T (4.18 kJ/kg ×°C)(40.29 - 15°C)

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=6000 [kPa] P[4]=7.5 [kPa] x[4]=1 Q_dot_in=40000 [kJ/s] T_water_in=15 [C] "Properties" c_water=4.18 [kJ/kg-C] "at room conditions" "Analysis" Fluid$='steam_iapws' "(a)" P[1]=P[4] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) T[1]=temperature(Fluid$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in h[4]=enthalpy(Fluid$, P=P[4], x=x[4]) s[4]=entropy(Fluid$, P=P[4], x=x[4]) s[3]=s[4] h[3]=enthalpy(Fluid$, P=P[3], s=s[3]) T[3]=temperature(Fluid$, P=P[3], s=s[3]) T_3=T[3] "(b)" q_in=h[3]-h[2] q_out=h[4]-h[1] w_net=q_in-q_out Eta_th=w_net/q_in W_dot_net=Eta_th*Q_dot_in "(c)" Q_dot_out=Q_dot_in-W_dot_net T_water_out=T[1] DELTAT=T_water_out-T_water_in © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-392

m_dot_water=Q_dot_out/(c_water*DELTAT)

11-92 A steam power plant operating on the ideal Rankine cycle with reheating is considered. The reheat pressures of the cycle are to be determined for the cases of single and double reheat. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) Single Reheat: From the steam tables (Tables A-4, A-5, and A-6 or EES),

P6 = 10 kPaïü ïý h6 = h f + x6 h fg = 191.81 + (0.95)(2392.1) = 2464.3 kJ/kg x6 = 0.95 ïïþ s6 = s f + x6 s fg = 0.6492 + (0.95)(7.4996) = 7.7739 kJ/kg ×K

10-393

P3 = 30 MPaüï ý s = 6.5599 kJ/kg ×K T3 = 700°C ïïþ 3

P4 = Px P = Px and 5 T5 = 700°C s4 = s3 Any pressure Px selected between the limits of 30 MPa and 2.908 MPa will satisfy the requirements, and can be used for the double reheat pressure.

EES (Engineering Equation Solver) SOLUTION "Given" P[3]=30000 [kPa] T[3]=700 [C] P[6]=10 [kPa] x[6]=0.95 "Analysis" Fluid$='steam_iapws' "Single reheat" h[6]=enthalpy(Fluid$, P=P[6], x=x[6]) s[6]=entropy(Fluid$, P=P[6], x=x[6]) s[5]=s[6] T[5]=T[3] P[5]=pressure(Fluid$, T=T[5], s=s[5]) P_x=P[5] "reheat pressure" "Double reheat" s[3]=entropy(Fluid$, P=P[3], T=T[3]) s[4]=s[3] "P_x1=P[4] first reheat pressure Any P_x1 selected between P[3] and P_x2 will satisfy the requirements" P_x2=pressure(Fluid$, T=T[3], s=s[6]) "second reheat pressure"

11-93 A steam power plant operating on an ideal Rankine cycle with two stages of reheat is considered. The thermal efficiency of the cycle and the mass flow rate of the steam are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 30 kPa = 289.18 kJ/kg

v1 = v f @ 30 kPa = 0.001022 m3 /kg wp, in = v1 (P2 - P1 )

æ 1 kJ ÷ ö = (0.001022 m3 /kg )(10,000 - 30 kPa )çç ÷ 3 ÷ çè1 kPa ×m ø

= 10.19 kJ/kg h2 = h1 + wp, in = 289.18 + 10.19 = 299.37 kJ/kg P3 = 10 MPa h3 = 3500.9 kJ/kg T3 = 550°C s3 = 6.7561 kJ/kg ×K

}

10-394

P4 = 4 MPaüï ý h4 = 3204.9 kJ/kg s4 = s3 ïïþ

P5 = 4 MPa h5 = 3559.7 kJ/kg T5 = 550°C s5 = 7.2335 kJ/kg ×K

}

P6 = 2 MPaïü ý h6 = 3321.1 kJ/kg s6 = s5 ïïþ P7 = 2 MPa h7 = 3578.4 kJ/kg T7 = 550°C s7 = 7.5706 kJ/kg ×K

}

s8 - s f 7.5706 - 0.9441 = = 0.9711 P8 = 30 kPaü ï x8 = s fg 6.8234 ý s8 = s7 ïïþ h8 = h f + x8 h fg = 289.27 + (0.9711)(2335.3) = 2557.1 kJ/kg

Then,

qin = (h3 - h2 )+ (h5 - h4 )+ (h7 - h6 )

= 3500.9 - 299.37 + 3559.7 - 3204.9 + 3578.4 - 3321.1 = 3813.7 kJ/kg qout = h8 - h1 = 2557.1- 289.18 = 2267.9 kJ/kg

wnet = qin - qout = 3813.7 - 2267.9 = 1545.8 kJ/kg Thus, hth =

wnet 1545.8 kJ/kg = = 0.4053 = 40.5% qin 3813.7 kJ/kg

(b) The mass flow rate of the steam is then

Wnet 75,000 kJ/s = = 48.5 kg / s wnet 1545.8 kJ/kg

10-395

W_dot_net=75000 [kW] P[3]=10000 [kPa] T[3]=550 [C] P[5]=4000 [kPa] T[5]=550 [C] P[7]=2000 [kPa] T[7]=550 [C] P[8]=30 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" "Pump" P[1]=P[8] x[1]=0 h[1]=enthalpy(steam, P=P[1], x=x[1]) v[1]=volume(steam, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "Turbine stage 1" h[3]=enthalpy(steam, P=P[3], T=T[3]) s[3]=entropy(steam, P=P[3], T=T[3]) s[4]=s[3] P[4]=P[5] h[4]=enthalpy(steam, P=P[4], s=s[4]) "Turbine stage 2" h[5]=enthalpy(steam, P=P[5], T=T[5]) s[5]=entropy(steam, P=P[5], T=T[5]) s[6]=s[5] P[6]=P[7] h[6]=enthalpy(steam, P=P[6], s=s[6]) "Turbine stage 3" h[7]=enthalpy(steam, P=P[7], T=T[7]) s[7]=entropy(steam, P=P[7], T=T[7]) s[8]=s[7] h[8]=enthalpy(steam, P=P[8], s=s[8]) x[8]=quality(steam, P=P[8], s=s[8]) "Cycle" q_in=h[3]-h[2]+h[5]-h[4]+h[7]-h[6] q_out=h[8]-h[1] w_net=q_in-q_out Eta_th=w_net/q_in "(b)" m_dot=W_dot_net/w_net s8_f=entropy(Fluid$, P=P[8], x=0) s8_g=entropy(Fluid$, P=P[8], x=1) s8_fg=s8_g-s8_f h8_f=enthalpy(Fluid$, P=P[8], x=0) h8_g=enthalpy(Fluid$, P=P[8], x=1) h8_fg=h8_g-h8_f

11-94 An 150-MW steam power plant operating on a regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-396

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 10 kPa = 191.81 kJ/kg v1 = v f @ 10 kPa = 0.00101 m3 /kg

wpI, in = v1 (P2 - P1 ) / h p

æ 1 kJ ö÷ = (0.00101 m3 /kg )(500 - 10 kPa )çç ÷/ (0.95) çè1 kPa ×m3 ø÷ = 0.52 kJ/kg

h2 = h1 + wpI, in = 191.81 + 0.52 = 192.33 kJ/kg

10-397

P3 = 0.5 MPa üï h3 = h f @ 0.5 MPa = 640.09 kJ/kg ýv = v 3 sat.liquid ïïþ 3 f @ 0.5 MPa = 0.001093 m /kg wpII, in = v 3 (P4 - P3 ) / h p

æ 1 kJ ö÷ = (0.001093 m3 /kg )(10,000 - 500 kPa )çç ÷/ (0.95) çè1 kPa ×m3 ø÷

= 10.93 kJ/kg h4 = h3 + wpII, in = 640.09 + 10.93 = 651.02 kJ/kg

P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500°C s5 = 6.5995 kJ/kg ×K

}

x6 s =

s6 s - s f s fg

6.5995 - 1.8604 = 0.9554 4.9603

P6 s = 0.5 MPaïü h6 s = h f + x6 s h fg = 640.09 + (0.9554)(2108.0) ý s6 s = s5 ïïþ = 2654.1 kJ/kg hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) h5 - h6 s

= 3375.1- (0.80)(3375.1 - 2654.1)

= 2798.3 kJ/kg s7 s - s f

6.5995 - 0.6492 = = 0.7934 s fg 7.4996 P7 s = 10 kPaü ï ý h7 s = h f + x7 s h fg = 191.81 + (0.7934)(2392.1) s7 s = s5 ïïþ = 2089.7 kJ/kg x7 s =

hT =

h5 - h7 ¾¾ ® h7 = h5 - hT (h5 - h7 s ) h5 - h7 s = 3375.1- (0.80)(3375.1- 2089.7) = 2346.8 kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @D ke @D pe @ 0,

Ein - Eout = D Esystem n 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

6 6

2 2

3 3

where y is the fraction of steam extracted from the turbine (= m6 / m3 ). Solving for y, h3 - h2 640.09 - 192.33 = = 0.1718 h6 - h2 2798.3 - 192.33

qin = h5 - h4 = 3375.1- 651.02 = 2724.1 kJ/kg

Then,

qout = (1- y )(h7 - h1 ) = (1- 0.1718)(2346.8 - 191.81) = 1784.7 kJ/kg

wnet = qin - qout = 2724.1- 1784.7 = 939.4 kJ/kg and

Wnet 150,000 kJ/s = = 159.7 kg / s wnet 939.4 kJ/kg

10-398

(b) The thermal efficiency is determined from hth = 1-

qout 1784.7 kJ/kg = 1= 34.5% qin 2724.1 kJ/kg

Also,

ïüï ý s6 = 6.9453 kJ/kg ×K h6 = 2798.3 kJ/kgïïþ

P6 = 0.5 MPa

s3 = s f @ 0.5 MPa = 1.8604 kJ/kg ×K

s2 = s1 = s f @10 kPa = 0.6492 kJ/kg ×K

Then the irreversibility (or exergy destruction) associated with this regeneration process is ¿0ö æ q ÷ iregen = T0 sgen = T0 çççå me se - å mi si + surr ÷ = T és - ys6 - (1- y )s2 ù ÷ û ÷ 0 ë3 çè TL ø

= (303 K )éë1.8604 - (0.1718)(6.9453)- (1- 0.1718)(0.6492)ù û = 39.25 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=150000 [kW] P[5]=10000 [kPa] T[5]=500 [C] P[6]=500 [kPa] P[7]=10 [kPa] x[3]=0 Eta_T=0.80 Eta_P=0.95 T_R=1300 [K] T_0=303 [K] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[6] w_pI_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_pI_in "pump II" P[3]=P[6] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])/Eta_P h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s_s[6]=s[5] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-399

h_s[6]=enthalpy(Fluid$, P=P[6], s=s_s[6]) x_6s=quality(Fluid$, P=P[6], s=s_s[6]) Eta_T=(h[5]-h[6])/(h[5]-h_s[6]) s_s[7]=s[5] h_s[7]=enthalpy(Fluid$, P=P[7], s=s_s[7]) x_7s=quality(Fluid$, P=P[7], s=s_s[7]) Eta_T=(h[5]-h[7])/(h[5]-h_s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out m_dot=W_dot_net/w_net "(b)" Eta_th=1-q_out/q_in s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] s[3]=entropy(Fluid$, P=P[3], x=x[3]) s[6]=entropy(Fluid$, P=P[6], h=h[6]) i_regen=T_0*(s[3]-y*s[6]-(1-y)*s[2])

11-95 A 150-MW steam power plant operating on an ideal regenerative Rankine cycle with an open feedwater heater is considered. The mass flow rate of steam through the boiler, the thermal efficiency of the cycle, and the irreversibility associated with the regeneration process are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

10-400

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @10 kPa = 191.81 kJ/kg v1 = v f @10 kPa = 0.00101 m3 /kg

wpI, in = v1 (P2 - P1 )

æ 1 kJ ö÷ = (0.00101 m3 /kg )(500 - 10 kPa )çç ÷= 0.50 kJ/kg çè1 kPa ×m3 ø÷ h2 = h1 + wpI, in = 191.81 + 0.50 = 192.30 kJ/kg

P3 = 0.5 MPa ïü h3 = h f @ 0.5 MPa = 640.09 kJ/kg ýv = v 3 sat.liquid ïïþ 3 f @ 0.5 MPa = 0.001093 m /kg

wpII, in = v 3 (P4 - P3 )

æ 1 kJ ö÷ = (0.001093 m3 /kg )(10,000 - 500 kPa )çç ÷= 10.38 kJ/kg çè1 kPa ×m3 ø÷ h4 = h3 + wpII, in = 640.09 + 10.38 = 650.47 kJ/kg

P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500°C s5 = 6.5995 kJ/kg ×K

}

s6 - s f 6.5995 - 1.8604 = = 0.9554 P6 = 0.5 MPa ü ï x6 = s 4.9603 ý fg s6 = s5 ïïþ h6 = h f + x6 h fg = 640.09 + (0.9554)(2108.0) = 2654.1 kJ/kg s7 - s f 6.5995 - 0.6492 = = 0.7934 P7 = 10 kPa ïü x7 = s fg 7.4996 ý s7 = s5 ïïþ h7 = h f + x7 h fg = 191.81 + (0.7934)(2392.1) = 2089.7 kJ/kg The fraction of steam extracted is determined from the steady-flow energy equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0 , Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout

10-401

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

6 6

2 2

3 3

where y is the fraction of steam extracted from the turbine ( = m6 / m3 ). Solving for y, h3 - h2 640.09 - 192.31 = = 0.1819 h6 - h2 2654.1- 192.31

qin = h5 - h4 = 3375.1- 650.47 = 2724.6 kJ/kg

Then,

qout = (1- y )(h7 - h1 ) = (1- 0.1819)(2089.7 - 191.81) = 1552.7 kJ/kg

wnet = qin - qout = 2724.6 - 1552.7 = 1172.0 kJ/kg and m =

Wnet 150,000 kJ/s = = 128.0 kg / s wnet 1171.9 kJ/kg

(b) The thermal efficiency is determined from hth = 1-

qout 1552.7 kJ/kg = 1= 43.0% qin 2724.7 kJ/kg

Also,

s6 = s5 = 6.5995 kJ/kg ×K s3 = s f @ 0.5 MPa = 1.8604 kJ/kg ×K s2 = s1 = s f @ 10 kPa = 0.6492 kJ/kg ×K

= (303 K)éë1.8604 - (0.1819)(6.5995)- (1- 0.1819)(0.6492)ù û = 39.0 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_net=150000 [kW] P[5]=10000 [kPa] T[5]=500 [C] P[6]=500 [kPa] P[7]=10 [kPa] x[3]=0 Eta_T=1 Eta_P=1 T_R=1300 [K] T_0=303 [K] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[7] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-402

P[2]=P[6] w_pI_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_pI_in "pump II" P[3]=P[6] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])/Eta_P h[4]=h[3]+w_pII_in "turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s_s[6]=s[5] h_s[6]=enthalpy(Fluid$, P=P[6], s=s_s[6]) Eta_T=(h[5]-h[6])/(h[5]-h_s[6]) s_s[7]=s[5] h_s[7]=enthalpy(Fluid$, P=P[7], s=s_s[7]) Eta_T=(h[5]-h[7])/(h[5]-h_s[7]) "open feedwater heater" y*h[6]+(1-y)*h[2]=h[3] "y=m_dot_6/m_dot_3" "cycle" q_in=h[5]-h[4] q_out=(1-y)*(h[7]-h[1]) w_net=q_in-q_out m_dot=W_dot_net/w_net "(b)" Eta_th=1-q_out/q_in s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] s[3]=entropy(Fluid$, P=P[3], x=x[3]) s[6]=entropy(Fluid$, P=P[6], h=h[6]) i_regen=T_0*(s[3]-y*s[6]-(1-y)*s[2])

11-96 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-403

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 15 kPa = 225.94 kJ/kg

v1 = v f @ 15 kPa = 0.001014 m3 /kg wpI, in = v1 (P2 - P1 ) æ 1 kJ ÷ ö = (0.001014 m3 /kg )(600 - 15 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 0.59 kJ/kg h2 = h1 + wpI, in = 225.94 + 0.59 = 226.53 kJ/kg

P3 = 0.6 MPa ïü h3 = h f @ 0.6 MPa = 670.38 kJ/kg ýv = v 3 sat. liquid ïïþ 3 f @ 0.6 MPa = 0.001101 m /kg

wpII, in = v 3 (P4 - P3 ) æ 1 kJ ÷ ö = (0.001101 m3 /kg )(10,000 - 600 kPa )çç ÷ 3 ÷ çè1 kPa ×m ø

= 10.35 kJ/kg h4 = h3 + wpII, in = 670.38 + 10.35 = 680.73 kJ/kg

10-404

P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500°C s5 = 6.5995 kJ/kg ×K

}

P6 = 1.0 MPaüï ý h6 = 2783.8 kJ/kg s6 = s5 ïïþ P7 = 1.0 MPa h7 = 3479.1 kJ/kg T7 = 500°C s7 = 7.7642 kJ/kg ×K

}

P8 = 0.6 MPaïü ý h8 = 3310.2 kJ/kg s8 = s7 ïïþ

s9 - s f 7.7642 - 0.7549 = = 0.9665 P9 = 15 kPaüï x9 = s 7.2522 ý fg s9 = s7 ïïþ h9 = h f + x9 h fg = 225.94 + (0.9665)(2372.3) = 2518.8 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0,

Ein - Eout = D Esystem n 0 (steady) = 0 ® Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i i

e e

8 8

2 2

3 3

where y is the fraction of steam extracted from the turbine (= m8 / m3 ). Solving for y, y=

h3 - h2 670.38 - 226.53 = = 0.144 h8 - h2 3310.2 - 226.53

(b) The thermal efficiency is determined from

qin = (h5 - h4 )+ (h7 - h6 ) = (3375.1- 680.73)+ (3479.1- 2783.8) = 3389.7 kJ/kg

qout = (1- y )(h9 - h1 ) = (1- 0.1440)(2518.8 - 225.94) = 1962.7 kJ/kg and hth = 1-

qout 1962.7 kJ/kg = 1= 42.1% qin 3389.7 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[5]=10000 [kPa] T[5]=500 [C] P[7]=1000 [kPa] P[8]=600 [kPa] T[7]=500 [C] P[9]=15 [kPa] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[9] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[8] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-405

x[3]=0 P[3]=P[8] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in "high pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s[6]=s[5] P[6]=P[7] h[6]=enthalpy(Fluid$, P=P[6], s=s[6]) "low pressure turbine" h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) s[8]=s[7] h[8]=enthalpy(Fluid$, P=P[8], s=s[8]) s[9]=s[7] h[9]=enthalpy(Fluid$, P=P[9], s=s[9]) x[9]=quality(Fluid$, P=P[9], s=s[9]) "open feedwater heater" y*h[8]+(1-y)*h[2]=h[3] "y=m_dot_8/m_dot_3" "(b)" q_in=h[5]-h[4]+(h[7]-h[6]) q_out=(1-y)*(h[9]-h[1]) Eta_th=1-q_out/q_in

11-97 A nonideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-406

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 15 kPa = 225.94 kJ/kg

v1 = v f @ 15 kPa = 0.001014 m3 /kg wpI , in = v1 (P2 - P1 ) / h p

æ 1 kJ ö÷ = (0.001014 m3 /kg)(600 - 15 kPa )çç ÷/ 0.89 = 0.67 kJ/kg çè1 kPa ×m3 ø÷ h2 = h1 + wpI ,in = 225.94 + 0.67 = 226.61 kJ/kg

P3 = 0.6 MPaïü h3 = h f @ 0.6 MPa = 670.38 kJ/kg ýv = v 3 sat. liquid ïïþ 3 f @ 0.6 MPa = 0.001101 m /kg wpII, in = v 3 (P4 - P3 ) / h p

æ 1 kJ ÷ ö = (0.001101 m3 /kg )(10,000 - 600 kPa )çç / 0.89 = 11.62 kJ/kg ÷ çè1 kPa ×m3 ÷ ø h4 = h3 + wpII, in = 670.38 + 11.62 = 682.00 kJ/kg

P5 = 10 MPa h5 = 3375.1 kJ/kg T5 = 500°C s5 = 6.5995 kJ/kg ×K

}

P6 s = 1.0 MPaïü ý h6 s = 2783.8 kJ/kg s6 s = s5 ïïþ hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) h5 - h6 s

= 3375.1- (0.84)(3375.1 - 2783.8)

= 2878.4 kJ/kg P7 = 1.0 MPa h7 = 3479.1 kJ/kg T7 = 500°C s7 = 7.7642 kJ/kg ×K

}

10-407

P8 s = 0.6 MPaüï ý h8 s = 3310.2 kJ/kg s8 s = s7 ïïþ h7 - h8 ¾¾ ® h8 = h7 - hT (h7 - h8 s ) = 3479.1- (0.84)(3479.1- 3310.2) h7 - h8 s = 3337.2 kJ/kg

hT =

s9 s - s f 7.7642 - 0.7549 = = 0.9665 P9 s = 15 kPaïü x9 s = s fg 7.2522 ý s9 s = s7 ïïþ h9 s = h f + x9 s h fg = 225.94 + (0.9665)(2372.3) = 2518.8 kJ/kg h7 - h9 ¾¾ ® h9 = h7 - hT (h7 - h9 s ) = 3479.1- (0.84)(3479.1- 2518.8) h7 - h9 s = 2672.5 kJ/kg

hT =

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h + m h = m h ¾ ¾® yh + (1- y )h = 1(h ) i

8 8

3 3

where y is the fraction of steam extracted from the turbine (= m8 / m3 ). Solving for y, y=

h3 - h2 670.38 - 226.61 = = 0.1427 h8 - h2 3335.3 - 226.61

(b) The thermal efficiency is determined from

qin = (h5 - h4 )+ (h7 - h6 ) = (3375.1- 682.00)+ (3479.1- 2878.4) = 3293.9 kJ/kg qout = (1- y )(h9 - h1 ) = (1- 0.1427)(2672.5 - 225.94) = 2097.2 kJ/kg and hth = 1-

qout 2097.2 kJ/kg = 1= 0.363 = 36.3% qin 3293.9 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[5]=10000 [kPa] T[5]=500 [C] P[7]=1000 [kPa] P[8]=600 [kPa] T[7]=500 [C] P[9]=15 [kPa] Eta_T=0.84 Eta_P=0.89 "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[9] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-408

v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[8] w_pI_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_pI_in "pump II" x[3]=0 P[3]=P[8] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[5] w_pII_in=v[3]*(P[4]-P[3])/Eta_P h[4]=h[3]+w_pII_in "high pressure turbine" h[5]=enthalpy(Fluid$, P=P[5], T=T[5]) s[5]=entropy(Fluid$, P=P[5], T=T[5]) s_s[6]=s[5] P[6]=P[7] h_s[6]=enthalpy(Fluid$, P=P[6], s=s_s[6]) Eta_T=(h[5]-h[6])/(h[5]-h_s[6]) "low pressure turbine" h[7]=enthalpy(Fluid$, P=P[7], T=T[7]) s[7]=entropy(Fluid$, P=P[7], T=T[7]) s_s[8]=s[7] h_s[8]=enthalpy(Fluid$, P=P[8], s=s_s[8]) Eta_T=(h[7]-h[8])/(h[7]-h_s[8]) s_s[9]=s[7] h_s[9]=enthalpy(Fluid$, P=P[9], s=s_s[9]) x_9s=quality(Fluid$, P=P[9], s=s_s[9]) Eta_T=(h[7]-h[9])/(h[7]-h_s[9]) "open feedwater heater" y*h[8]+(1-y)*h[2]=h[3] "y=m_dot_8/m_dot_3" "(b)" q_in=h[5]-h[4]+(h[7]-h[6]) q_out=(1-y)*(h[9]-h[1]) Eta_th=1-q_out/q_in

11-98 A cogeneration plant is to generate power and process heat. Part of the steam extracted from the turbine at a relatively high pressure is used for process heating. The rate of process heat, the net power produced, and the utilization factor of the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-409

Analysis From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 20 kPa = 251.42 kJ/kg

v1 = v f @ 20 kPa = 0.001017 m3 /kg wpI, in = v1 (P2 - P1 ) / h p æ 1 kJ ö ÷ = (0.001017 m3 /kg )(2000 - 20 kPa )çç ÷ ÷/ 0.88 çè1 kPa ×m3 ø

= 2.29 kJ/kg h2 = h1 + wpI, in = 251.42 + 2.29 = 253.71 kJ/kg h3 = h f @ 2 MPa = 908.47 kJ/kg

Mixing chamber: m3 h3 + m2 h2 = m4 h4 (4 kg/s)(908.47 kJ/kg) + (11- 4 kg/s)(253.71 kJ/kg) = (11 kg/s)h4 ¾ ¾ ® h4 = 491.81 kJ/kg

10-410 v 4 @v f @ h f = 491.81 kJ/kg = 0.001058 m3 /kg

wpII, in = v 4 (P5 - P4 ) / h p

æ 1 kJ ÷ ö = (0.001058 m3 /kg )(8000 - 2000 kPa )çç / 0.88 ÷ çè1 kPa ×m3 ÷ ø

= 7.21 kJ/kg h5 = h4 + wpII, in = 491.81 + 7.21 = 499.02 kJ/kg

P6 = 8 MPaüïï h6 = 3399.5 kJ/kg ý T6 = 500°C ïïþ s6 = 6.7266 kJ/kg ×K P7 = 2 MPaïüï ý h7 s = 3000.4 kJ/kg ïïþ s7 = s6 hT =

h6 - h7 ¾¾ ® h7 = h6 - hT (h6 - h7 s ) = 3399.5 - (0.88)(3399.5 - 3000.4) = 3048.3 kJ/kg h6 - h7 s

P8 = 20 kPaïüï ý h8 s = 2215.5 kJ/kg ïïþ s8 = s6 hT =

h6 - h8 ¾¾ ® h8 = h6 - hT (h6 - h8 s ) = 3399.5 - (0.88)(3399.5 - 2215.5) = 2357.6 kJ/kg h6 - h8 s

Then,

Qprocess = m7 (h7 - h3 ) = (4 kg/s)(3048.3 - 908.47) kJ/kg = 8559 kW (b) Cycle analysis:

WT, out = m7 (h6 - h7 )+ m8 (h6 - h8 ) = (4 kg/s)(3399.5 - 3048.3)kJ/kg + (7 kg/s )(3399.5 - 2357.6)kJ/kg = 8698kW

Wp, in = m1 wpI, in + m4 wpII, in = (7 kg/s)(2.29 kJ/kg )+ (11 kg/s)(7.21 kJ/kg ) = 95 kW Wnet = WT, out - Wp, in = 8698 - 95 = 8603 kW (c) Then, and Qin = m5 (h6 - h5 ) = (11 kg/s)(3399.5 - 499.02) = 31,905 kW

eu =

Wnet + Qprocess Qin

8603 + 8559 = 0.538 = 53.8% 31,905

EES (Engineering Equation Solver) SOLUTION "GIVEN" m_dot_process=4 [kg/s] P[7]=2000 [kPa]; P[3]=P[7]; P[2]=P[7]; P[4]=P[7] P[6]=8000 [kPa]; P[5]=P[6] T[6]=500 [C] m_dot=11 [kg/s] P[8]=20 [kPa]; P[1]=P[8] eta_T=0.88; eta_P=0.88 "PROPERTIES" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-411

Fluid$='Steam_iapws' h[1]=enthalpy(Fluid$, P=P[1], x=0) v_1=volume(Fluid$, P=P[1], x=0) w_PI=v_1*(P[2]-P[1])/eta_P h[2]=h[1]+w_PI h[3]=enthalpy(Fluid$, P=P[3], x=0) m_dot_process*h[3]+(m_dot-m_dot_process)*h[2]=m_dot*h[4] "energy balance on the mixing chamber" v_4=volume(Fluid$, x=0, h=h[4]) w_PII=v_4*(P[5]-P[4])/eta_P h[5]=h[4]+w_PII h[6]=enthalpy(Fluid$, P=P[6], T=T[6]) s[6]=entropy(Fluid$, P=P[6], T=T[6]) h_7s=enthalpy(Fluid$, P=P[7], s=s[6]) h[7]=h[6]-eta_T*(h[6]-h_7s) h_8s=enthalpy(Fluid$, P=P[8], s=s[6]) h[8]=h[6]-eta_T*(h[6]-h_8s) "ANALYSIS" Q_dot_process=m_dot_process*(h[7]-h[3]) "process heat" W_dot_T=m_dot_process*(h[6]-h[7])+(m_dot-m_dot_process)*(h[6]-h[8]) W_dot_P=(m_dot-m_dot_process)*w_PI+m_dot*w_PII W_dot_net=W_dot_T-W_dot_P "net power" Q_dot_in=m_dot*(h[6]-h[5]) eta_utilization=(W_dot_net+Q_dot_process)/Q_dot_in "utilization efficiency"

11-99 A cogeneration power plant is modified with reheat and that produces 3 MW of power and supplies 7 MW of process heat. The rate of heat input in the boiler and the fraction of steam extracted for process heating are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @15 kPa = 225.94 kJ/kg

h2 @ h1 h3 = h f @120 °C = 503.81 kJ/kg

P6 = 8 MPa h6 = 3399.5 kJ/kg T6 = 500°C s6 = 6.7266 kJ/kg ×K

}

P7 = 1 MPaïü ý h7 = 2843.7 kJ/kg s7 = s6 ïïþ P8 = 1 MPa h8 = 3479.1 kJ/kg T8 = 500°C s8 = 7.7642 kJ/kg ×K

}

s9 - s f

7.7642 - 0.7549 = 0.9665 s fg 7.2522 P9 = 15 kPaïü ý h9 = h f + x9 h fg = 225.94 + (0.9665)(2372.3) s9 = s8 ïïþ = 2518.8 kJ/kg x9 =

The mass flow rate through the process heater is

10-412

m3 =

Qprocess h7 - h3

7,000 kJ/s

(2843.7 - 503.81) kJ/kg

= 2.992 kg/s

Also, WT = m6 (h6 - h7 )+ m9 (h8 - h9 ) = m6 (h6 - h7 )+ (m6 - 2.993)(h8 - h9 )

3000 kJ/s = m6 (3399.5 - 2843.7)+ (m6 - 2.992)(3479.1 - 2518.8) It yields and

m6 = 3.873 kg/s m9 = m6 - m3 = 3.873 - 2.992 = 0.881 kg/s

Mixing chamber:

Ein - Eout = D EsystemZ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m h = m h + m h i i

e e

4 4

2 2

3 3

10-413

or, h4 @ h5 =

m2 h2 + m3 h3 (0.881)(225.94)+ (2.992)(503.81) = = 440.60 kJ/kg m4 3.873

Then, Qin = m6 (h6 - h5 )+ m8 (h8 - h7 )

= (3.873 kg/s)(3399.5 - 440.60 kJ/kg )+ (0.881 kg/s )(3479.1- 2843.7 kJ/kg )

= 12, 020 kW (b) The fraction of steam extracted for process heating is y=

m3 2.992 kg/s = = 77.3% mtotal 3.873 kg/s

EES (Engineering Equation Solver) SOLUTION "Given" W_dot_T=3000 [kW] Q_dot_process=7000 [kW] P[6]=8000 [kPa] T[6]=500 [C] P[7]=1000 [kPa] T[8]=500 [C] P[9]=15 [kPa] T[3]=120 [C] "Analysis" Fluid$='steam_iapws' "(a)" P[1]=P[9] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) h[2]=h[1] x[3]=0 h[3]=enthalpy(Fluid$, T=T[3], x=x[3]) "high pressure turbine" h[6]=enthalpy(Fluid$, P=P[6], T=T[6]) s[6]=entropy(Fluid$, P=P[6], T=T[6]) s[7]=s[6] h[7]=enthalpy(Fluid$, P=P[7], s=s[7]) "low pressure turbine" P[8]=P[7] h[8]=enthalpy(Fluid$, P=P[8], T=T[8]) s[8]=entropy(Fluid$, P=P[8], T=T[8]) s[9]=s[8] h[9]=enthalpy(Fluid$, P=P[9], s=s[9]) x[9]=quality(Fluid$, P=P[9], s=s[9]) m_dot[3]=Q_dot_process/(h[7]-h[3]) m_dot[9]=m_dot[6]-m_dot[3] W_dot_T=m_dot[6]*(h[6]-h[7])+m_dot[9]*(h[8]-h[9]) "mixing chamber" m_dot[4]=m_dot[6] m_dot[2]=m_dot[9] m_dot[4]*h[4]=m_dot[2]*h[2]+m_dot[3]*h[3] h[5]=h[4] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-414

m_dot[8]=m_dot[9] Q_dot_in=m_dot[6]*(h[6]-h[5])+m_dot[8]*(h[8]-h[7]) "(b)" y=m_dot[3]/m_dot[6]

11-100 A cogeneration plant is to produce power and process heat. There are two turbines in the cycle: a high-pressure turbine and a low-pressure turbine. The temperature, pressure, and mass flow rate of steam at the inlet of high-pressure turbine are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From the steam tables (Tables A-4, A-5, and A-6),

P4 = 1.4 MPaïü h4 = hg @1.4 MPa = 2788.9 kJ/kg ý sat. vapor ïïþ s4 = sg @1.4 MPa = 6.4675 kJ/kg ×K s4 s - s f 6.4675 - 0.6492 = = 0.7758 P5 = 10 kPaü ï x5 s = s 7.4996 ý fg s5 s = s4 ïïþ h5 s = h f + x5 s h fg = 191.81 + (0.7758)(2392.1)

= 2047.6 kJ/kg hT =

h4 - h5 ¾¾ ® h5 = h4 - hT (h4 - h5 s ) h4 - h5 s

= 2788.9 - (0.80)(2788.9 - 2047.6)

= 2195.8 kJ/kg and wturb, low = h4 - h5 = 2788.9 - 2195.8 = 593.0 kJ/kg

mlowturb =

Wturb, II wturb, low

1500 kJ/s = 2.529 kg/s = 151.7 kg/min 593.0 kJ/kg

10-415

Therefore,

mtotal = 1000 + 151.7 = 1152 kg/min = 19.20 kg / s

wturb, high =

Wturb, I mhigh, turb

5500 kJ/s = 286.5 kJ/kg = h3 - h4 19.20 kg/s

h3 = wturb, high + h4 = 286.5 + 2788.9 = 3075.4 kJ/kg

hT =

h3 - h4 ¾¾ ® h4 s = h3 - (h3 - h4 ) / hT h3 - h4 s

= 3075.4 - (3075.4 - 2788.9) / (0.85)

= 2738.3 kJ/kg h4 s - h f 2738.3 - 829.96 = = 0.9742 P4 s = 1.4 MPaïü x4 s = h fg 1958.9 ý s4 s = s3 ïïþ s4 s = s f + x4 s s fg = 2.2835 + (0.9742)(4.1840) = 6.3595 kJ/kg ×K Then from the tables or the software, the turbine inlet temperature and pressure becomes

h3 = 3075.4 kJ/kg üïï P3 = 6.40 MPa ý s3 = 6.3595 kJ/kg ×KïïþT3 = 365°C

EES (Engineering Equation Solver) SOLUTION "Given" Eta_T_h=0.85 P[4]=1400 [kPa] x[4]=1 W_dot_T_h=5500 [kW] m_dot_4=1000 [kg/min] m_dot_cg=m_dot_4*Convert(kg/min, kg/s) Eta_T_l=0.80 P[5]=10 [kPa] W_dot_T_l=1500 [kW] "Analysis" Fluid$='steam_iapws' "low pressure turbine" h[4]=enthalpy(Fluid$, P=P[4], x=x[4]) s[4]=entropy(Fluid$, P=P[4], x=x[4]) s_s[5]=s[4] h_s[5]=enthalpy(Fluid$, P=P[5], s=s_s[5]) x_5s=quality(Fluid$, P=P[5], s=s_s[5]) Eta_T_l=(h[4]-h[5])/(h[4]-h_s[5]) w_T_l=h[4]-h[5] m_dot_T_l=W_dot_T_l/w_T_l m_dot_total=m_dot_T_l+m_dot_cg "high pressure turbine" w_T_h=W_dot_T_h/m_dot_total w_T_h=h[3]-h[4] Eta_T_h=(h[3]-h[4])/(h[3]-h_s[4]) s_s[4]=entropy(Fluid$, P=P[4], h=h_s[4]) x_4s=quality(Fluid$, P=P[4], h=h_s[4]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-416

s[3]=s_s[4] P[3]=pressure(Fluid$, h=h[3], s=s[3]) T[3]=temperature(Fluid$, h=h[3], s=s[3]) P_3=P[3] T_3=T[3]

11-101E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable for Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p  0.240 Btu / lbm.R and k = 1.4 (Table A-2Ea).

Analysis Working around the topping cycle gives the following results: ( k - 1)/ k

æP ö ÷ T6 s = T5 ççç 6 ÷ ÷ ÷ çè P ø

= (540 R)(10)0.4/1.4 = 1043 R

hC =

h6 s - h5 c p (T6 s - T5 ) = h6 - h5 c p (T6 - T5 )

¾¾ ® T6 = T5 +

T6 s - T5 hC

10-417

1043 - 540 = 1099 R 0.90

= 540 +

( k - 1)/ k

æP ö ÷ T8 s = T7 ççç 8 ÷ ÷ ÷ èç P ø 7

hT =

0.4/1.4

æ1 ö = (2560 R) çç ÷ ÷ ÷ çè10 ø

= 1326 R

c p (T7 - T8 ) h7 - h8 = ¾¾ ® T8 = T7 - hT (T7 - T8 s ) h7 - h8 s c p (T7 - T8 s )

= 2560 - (0.90)(2560 - 1326) = 1449 R

T9 = Tsat @ 800 psia + 50 = 978.3 R + 50 = 1028 R

Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E): h1 = h f @ 5 psia = 130.18 Btu/lbm

v1 = v f @ 5 psia = 0.01641 ft 3 /lbm wp, in = v1 ( P2 - P1 )

æ 1 Btu ö ÷ ÷ = (0.01641 ft 3 /lbm)(800 - 5)psia çç ÷ çè5.404 psia ×ft 3 ÷ ø = 2.41 Btu/lbm h2 = h1 + wp, in = 130.18 + 2.41 = 132.59 Btu/lbm

P3 = 800 psiaü ïï h3 = 1270.9 Btu/lbm ý T3 = 600°F ïïþ s3 = 1.4866 Btu/lbm ×R

P4 = 5 psiaü ïï ý h4 s = 908.6 Btu/lbm ïïþ s4 = s3 hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s

= 1270.9 - (0.95)(1270.9 - 908.6) = 926.7 Btu/lbm

The net work outputs from each cycle are

wnet, gas cycle = wT, out - wC, in = c p (T7 - T8 ) - c p (T6 - T5 )

= (0.240 Btu/lbm ×R)(2560 - 1449 - 1099 + 540)R = 132.5 Btu/lbm

wnet, steam cycle = wT, out - wP, in = (h3 - h4 ) - wP, in

= (1270.9 - 926.7) - 2.41 = 341.8 Btu/lbm

An energy balance on the heat exchanger gives

ma c p (T8 - T9 ) = mw (h3 - h2 ) ¾ ¾ ® mw =

c p (T8 - T9 ) h3 - h2

ma =

(0.240)(1449 - 1028) = 0.08876ma 1270.9 - 132.59

10-418

That is, 1 lbm of exhaust gases can heat only 0.08876 lbm of water. Then the heat input, the heat output and the thermal efficiency are qin =

ma c p (T7 - T6 ) = (0.240 Btu/lbm ×R)(2560 - 1099)R = 350.6 Btu/lbm ma

qout =

ma m c p (T9 - T5 ) + w (h4 - h1 ) ma ma

= 1´ (0.240 Btu/lbm ×R)(1028 - 540)R + 0.08876´ (926.7 - 130.18) Btu/lbm

= 187.8 Btu/lbm

hth = 1-

qout 187.8 = 1= 0.4643 qin 350.6

EES (Engineering Equation Solver) SOLUTION "Given" P5=14.7 [psia] T5=(80+460) [R] r_p=10 T7=(2100+460) [R] eta_C=0.90 eta_T_gas=0.90 DELTAT=50 [F] P[2]=800 [psia] T[3]=600 [F] P[4]=5 [psia] eta_T_steam=0.95 "Properties" c_p=0.240 [Btu/lbm-R] "Table A-2Ea" k=1.4 "Table A-2Ea" "Analysis" "Gas turbine cycle" T6_s=T5*r_p^((k-1)/k) T6=T5+(T6_s-T5)/eta_C T8_s=T7*(1/r_p)^((k-1)/k) T8=T7-eta_T_gas*(T7-T8_s) T9=T_sat+DELTAT+460 "R" T_sat=temperature(Fluid$, P=P[2], x=1) "Rankine cycle" Fluid$='steam_iapws' P[1]=P[4] P[3]=P[2] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s_s[4]) Eta_T_steam=(h[3]-h[4])/(h[3]-h_4s) w_net_gas=c_p*(T7-T8)-c_p*(T6-T5) w_net_steam=h[3]-h[4]-w_p_in © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-419

m_a*c_p*(T8-T9)=m_w*(h[3]-h[2]) "energy balance on heat exchanger" m_a=1 "assumed" q_in=c_p*(T7-T6) q_out=c_p*(T9-T5)+m_w/m_a*(h[4]-h[1]) eta_th=1-q_out/q_in

11-102E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The thermal efficiency of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable fo Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p  0.240 Btu / lbm.R and k = 1.4 (Table A-2Ea).

Analysis Working around the topping cycle gives the following results: ( k - 1)/ k

æP ö ÷ T6 s = T5 ççç 6 ÷ ÷ ÷ çè P5 ø

hC =

= (540 R)(10)0.4/1.4 = 1043 R

h6 s - h5 c p (T6 s - T5 ) = h6 - h5 c p (T6 - T5 )

¾¾ ® T6 = T5 +

T6 s - T5 hC

10-420

= 540 +

1043 - 540 = 1099 R 0.90 ( k - 1)/ k

æP ö ÷ T8 s = T7 ççç 8 ÷ ÷ èç P7 ø÷ hT =

0.4/1.4

æ1 ö = (2560 R) çç ÷ ÷ çè10 ø÷

= 1326 R

c p (T7 - T8 ) h7 - h8 = ¾¾ ® T8 = T7 - hT (T7 - T8 s ) h7 - h8 s c p (T7 - T8 s )

= 2560 - (0.90)(2560 - 1326) = 1449 R

T9 = Tsat @ 800 psia + 50 = 978.3 R + 50 = 1028 R

Fixing the states around the bottom steam cycle yields (Tables A-4E, A-5E, A-6E): h1 = h f @ 10 psia = 161.25 Btu/lbm

v1 = v f @ 10 psia = 0.01659 ft 3 /lbm

wp, in = v1 ( P2 - P1 )

æ 1 Btu ö ÷ ÷ = (0.01659 ft 3 /lbm)(800 - 10)psia çç ÷ ÷ çè5.404 psia ×ft 3 ø = 2.43 Btu/lbm h2 = h1 + wp, in = 161.25 + 2.43 = 163.7 Btu/lbm

P3 = 800 psiaü ïï h3 = 1270.9 Btu/lbm ý T3 = 600°F ïïþ s3 = 1.4866 Btu/lbm ×R P4 = 10 psiaïü ïý h = 946.6 Btu/lbm ïïþ 4 s s4 = s3

hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s

= 1270.9 - (0.95)(1270.9 - 946.6) = 962.8 Btu/lbm

The net work outputs from each cycle are

wnet, gas cycle = wT, out - wC, in = c p (T7 - T8 ) - c p (T6 - T5 )

= (0.240 Btu/lbm ×R)(2560 - 1449 - 1099 + 540)R = 132.5 Btu/lbm

wnet, steam cycle = wT, out - wP, in = (h3 - h4 ) - wP, in

= (1270.9 - 962.8) - 2.43 = 305.7 Btu/lbm

An energy balance on the heat exchanger gives

10-421

ma c p (T8 - T9 ) = mw (h3 - h2 ) ¾ ¾ ® mw =

c p (T8 - T9 ) h3 - h2

ma =

(0.240)(1449 - 1028) = 0.09126ma 1270.9 - 163.7

That is, 1 lbm of exhaust gases can heat only 0.09126 lbm of water. Then the heat input, the heat output and the thermal efficiency are qin =

ma c p (T7 - T6 ) = (0.240 Btu/lbm ×R)(2560 - 1099)R = 350.6 Btu/lbm ma

qout =

ma m c p (T9 - T5 ) + w (h4 - h1 ) ma ma

= 1´ (0.240 Btu/lbm ×R)(1028 - 540)R + 0.09126´ (962.8 - 161.25) Btu/lbm = 190.3 Btu/lbm

hth = 1-

qout 190.3 = 1= 0.4573 qin 350.6

When the condenser pressure is increased from 5 psia to 10 psia, the thermal efficiency is decreased from 0.4643 to 0.4573.

EES (Engineering Equation Solver) SOLUTION "Given" P5=14.7 [psia] T5=(80+460) [R] r_p=10 T7=(2100+460) [R] eta_C=0.90 eta_T_gas=0.90 DELTAT=50 [F] P[2]=800 [psia] T[3]=600 [F] P[4]=10 [psia] eta_T_steam=0.95 "Properties" c_p=0.240 [Btu/lbm-R] "Table A-2Ea" k=1.4 "Table A-2Ea" "Analysis" "Gas turbine cycle" T6_s=T5*r_p^((k-1)/k) T6=T5+(T6_s-T5)/eta_C T8_s=T7*(1/r_p)^((k-1)/k) T8=T7-eta_T_gas*(T7-T8_s) T9=T_sat+DELTAT+460 "R" T_sat=temperature(Fluid$, P=P[2], x=1) "Rankine cycle" Fluid$='steam_iapws' P[1]=P[4] P[3]=P[2] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-422

s_s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s_s[4]) Eta_T_steam=(h[3]-h[4])/(h[3]-h_4s) w_net_gas=c_p*(T7-T8)-c_p*(T6-T5) w_net_steam=h[3]-h[4]-w_p_in m_a*c_p*(T8-T9)=m_w*(h[3]-h[2]) "energy balance on heat exchanger" m_a=1 "assumed" q_in=c_p*(T7-T6) q_out=c_p*(T9-T5)+m_w/m_a*(h[4]-h[1]) eta_th=1-q_out/q_in

11-103E A combined gas-steam power cycle uses a simple gas turbine for the topping cycle and simple Rankine cycle for the bottoming cycle. The cycle supplies a specified rate of heat to the buildings during winter. The mass flow rate of air and the net power output from the cycle are to be determined. Assumptions 1 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable to Brayton cycle. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p  0.240 Btu / lbm.R and k = 1.4 (Table A-2Ea).

Analysis The mass flow rate of water is

mw =

Qbuildings h4 - h1

2´ 106 Btu/h = 2495 lbm/h (962.8-161.25) Btu/lbm

10-423

ma =

mw 2495 = = 27,340 lbm / h 0.09126 0.09126

The power outputs from each cycle are

Wnet, gas cycle = ma ( wT, out - wC, in ) = ma c p (T7 - T8 ) - ma c p (T6 - T5 )

æ 1 kW ö÷ = (27,340 lbm/h)(0.240 Btu/lbm ×R)(2560 - 1449 - 1099 + 540)R çç ÷ çè3412.14 Btu/h ø÷ = 1062 kW

Wnet, steam cycle = ma ( wT, out - wP, in ) = ma (h3 - h4 - wP, in )

æ 1 kW ö ÷ = (2495 lbm/h)(1270.9 - 962.8 - 2.43) çç ÷ çè3412.14 Btu/h ÷ ø = 224 kW The net electricity production by this cycle is then Wnet = 1062 + 224 = 1286 kW

EES (Engineering Equation Solver) SOLUTION "Given" P5=14.7 [psia] T5=(80+460) [R] r_p=10 T7=(2100+460) [R] eta_C=0.90 eta_T_gas=0.90 DELTAT=50 [F] P[2]=800 [psia] T[3]=600 [F] P[4]=10 [psia] eta_T_steam=0.95 Q_dot_buildings=2E6 [Btu/h] "Properties" c_p=0.240 [Btu/lbm-R] "Table A-2Ea" k=1.4 "Table A-2Ea" "Analysis" "Gas turbine cycle" T6_s=T5*r_p^((k-1)/k) T6=T5+(T6_s-T5)/eta_C T8_s=T7*(1/r_p)^((k-1)/k) T8=T7-eta_T_gas*(T7-T8_s) T9=T_sat+DELTAT+460 "R" T_sat=temperature(Fluid$, P=P[2], x=1) "Rankine cycle" Fluid$='steam_iapws' P[1]=P[4] P[3]=P[2] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-424

v[1]=volume(Fluid$, P=P[1], x=x[1]) w_p_in=v[1]*(P[2]-P[1])*Convert(psia-ft^3, Btu) h[2]=h[1]+w_p_in h[3]=enthalpy(Fluid$, P=P[3], T=T[3]) s[3]=entropy(Fluid$, P=P[3], T=T[3]) s_s[4]=s[3] h_4s=enthalpy(Fluid$, P=P[4], s=s_s[4]) Eta_T_steam=(h[3]-h[4])/(h[3]-h_4s) m_a*c_p*(T8-T9)=m_w*(h[3]-h[2]) "energy balance on heat exchanger" m_a=1 "assumed" m_dot_w=Q_dot_buildings/(h[4]-h[1]) m_dot_a=m_dot_w/m_w W_dot_net_gas=(m_dot_a*c_p*(T7-T8)-m_dot_a*c_p*(T6-T5))*Convert(Btu/h, kW) W_dot_net_steam=m_dot_w*(h[3]-h[4]-w_p_in)*Convert(Btu/h, kW) W_dot_net=W_dot_net_gas+W_dot_net_steam

11-104 A combined gas-steam power plant is considered. The topping cycle is an ideal gas-turbine cycle and the bottoming cycle is an ideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

Analysis (a) The analysis of gas cycle yields

T7 = 310 K ¾ ¾ ®

h7 = 310.24 kJ/kg s7o = 1.73498 kJ/kg ×K

10-425

s8o = s7o + R ln

P8 = 1.73498 + (0.287 kJ/kg ×K) ln(12) = 2.44815 kJ/kg ×K P7 ¾¾ ® h8 = 630.93 kJ/kg

T9 = 1400 K ¾ ¾ ®

h9 = 1515.42 kJ/kg s9o = 3.36200 kJ/kg ×K

P s10o = s9o + R ln 10 = 3.36200 + (0.287 kJ/kg ×K) ln(1/12) = 2.64883 kJ/kg ×K P9 ¾¾ ® h10 = 768.40 kJ/kg T11 = 520 K ¾ ¾ ® h11 = 523.63 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg

æ 1 kJ ö÷ wpI,in = v1 (P2 - P1 ) = (0.00101 m3 /kg )(12,500 - 10 kPa )çç ÷= 12.62 kJ/kg çè1 kPa ×m3 ø÷ h2 = h1 + wpI, in = 191.81 + 12.62 = 204.42 kJ/kg

P3 = 12.5 MPaïü h3 = 3343.6 kJ/kg ý T3 = 500°C ïïþ s3 = 6.4651 kJ/kg ×K

P4 = 2.5 MPaïü ý h4 = 2909.6 kJ/kg s4 = s3 ïïþ P5 = 2.5 MPaüï h5 = 3574.4 kJ/kg ý T5 = 550°C ïïþ s5 = 7.4653 kJ/kg ×K s6 - s f 7.4653 - 0.6492 = = 0.9089 P6 = 10 kPaü ïï x6 = s 7.4996 ý fg ïïþ s6 = s5 h6 = h f + x6 h fg = 191.81 + (0.9089)(2392.1) = 2365.8 kJ/kg Noting that Q @W @ Δke @ Δpe @ 0 for the heat exchanger, the steady-flow energy balance equation yields

Ein = Eout ¾ ¾ ® å mi hi = å me he ¾ ¾ ® ms (h3 - h2 ) = mair (h10 - h11 ) mair =

h3 - h2 3343.6 - 204.42 ms = (12 kg/s) = 153.9 kg / s h10 - h11 768.40 - 523.63

(b) The rate of total heat input is

Qin = Qair + Qreheat = mair (h9 - h8 )+ mreheat (h5 - h4 ) = (153.9 kg/s)(1515.42 - 630.93) kJ/kg + (12 kg/s)(3574.4 - 2909.6)kJ/kg

= 144,100 kW @1.44×105 kW (c) The rate of heat rejection and the thermal efficiency are then

Qout = Qout, air + Qout, steam = mair (h11 - h7 )+ ms (h6 - h1 ) = (153.9 kg/s)(523.63 - 310.24)kJ/kg + (12 kg/s )(2365.8 - 191.81)kJ/kg

10-426

= 58,930 kW hth = 1-

Qout 58,930 kW = 1= 0.591 = 59.1% 144,100 kW Qin

EES (Engineering Equation Solver) SOLUTION "Given" r_p=12 T[7]=310 [K] P[7]=100 [kPa] "Assumed air inlet pressure to compressor" T[9]=1400 [K] T_11=247 [C] T[11]=T_11+273 "[K]" P[3]=12500 [kPa] T_3=500 [C] T[3]=T_3+273.15 "[K]" P[5]=2500 [kPa] T_5=550 [C] T[5]=T_5+273.15 "[K]" P[6]=10 [kPa] m_dot_steam=12 [kg/s] "Analysis" Fluid1$='air' Fluid2$='steam_iapws' "(a)" "Gas Turbine cycle" "compressor" h[7]=enthalpy(Fluid1$, T=T[7]) s[7]=entropy(Fluid1$, T=T[7], P=P[7]) P[8]=r_p*P[7] s[8]=s[7] T[8]=temperature(Fluid1$, P=P[8], s=s[8]) h[8]=enthalpy(Fluid1$, T=T[8]) "turbine" h[9]=enthalpy(Fluid1$, T=T[9]) P[9]=P[8] s[9]=entropy(Fluid1$, T=T[9], P=P[9]) P[10]=P[9]/r_p s[10]=s[9] T[10]=temperature(Fluid1$, P=P[10], s=s[10]) h[10]=enthalpy(Fluid1$, T=T[10]) h[11]=enthalpy(Fluid1$, T=T[11]) "Steam cycle" "pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid2$, P=P[1], x=x[1]) v[1]=volume(Fluid2$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_p_in "high pressure turbine" h[3]=enthalpy(Fluid2$, P=P[3], T=T[3]) s[3]=entropy(Fluid2$, P=P[3], T=T[3]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-427

s[4]=s[3] P[4]=P[5] h[4]=enthalpy(Fluid2$, P=P[4], s=s[4]) x[4]=quality(Fluid2$, P=P[4], s=s[4]) "low pressure turbine" h[5]=enthalpy(Fluid2$, P=P[5], T=T[5]) s[5]=entropy(Fluid2$, P=P[5], T=T[5]) s[6]=s[5] h[6]=enthalpy(Fluid2$, P=P[6], s=s[6]) x[6]=quality(Fluid2$, P=P[6], s=s[6]) "Combined cycle" m_dot_steam*(h[3]-h[2])=m_dot_air*(h[10]-h[11]) "heat exchanger energy balance" "(b)" Q_dot_air=m_dot_air*(h[9]-h[8]) Q_dot_reheat=m_dot_steam*(h[5]-h[4]) Q_dot_in=Q_dot_air+Q_dot_reheat "(c)" Q_dot_out_air=m_dot_air*(h[11]-h[7]) Q_dot_out_steam=m_dot_steam*(h[6]-h[1]) Q_dot_out=Q_dot_out_air+Q_dot_out_steam Eta_th=1-Q_dot_out/Q_dot_in

11-105 A combined gas-steam power plant is considered. The topping cycle is a gas-turbine cycle and the bottoming cycle is a nonideal reheat Rankine cycle. The mass flow rate of air in the gas-turbine cycle, the rate of total heat input, and the thermal efficiency of the combined cycle are to be determined.

10-428

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) The analysis of gas cycle yields (Table A-17) h7 = 310.24 kJ/kg T7 = 310 K ¾ ¾ ® o s7 = 1.73498 kJ/kg ×K

s8o = s7o + R ln

P8 = 1.73498 + (0.287 kJ/kg ×K) ln(12) = 2.44815 kJ/kg ×K P7 ¾¾ ® h8 s = 630.93 kJ/kg

hC =

h8 s - h7 ¾¾ ® h8 = h7 + (h8 s - h7 ) / hC h8 - h7

= 310.24 + (630.93 - 310.24) / (0.85)

= 687.52 kJ/kg T9 = 1400 K ¾ ¾ ®

h9 = 1515.42 kJ/kg s9o = 3.36200 kJ/kg ×K

P s10o = s9o + R ln 10 = 3.36200 + (0.287 kJ/kg ×K) ln(1/12) = 2.64883 kJ/kg ×K P9 ¾¾ ® h10 s = 768.40 kJ/kg hT =

h9 - h10 ¾¾ ® h10 = h9 - hT (h9 - h10 s ) h9 - h10 s

= 1515.42 - (0.90)(1515.42 - 768.40)

= 843.10 kJ/kg T11 = 520 K ¾ ¾ ® h11 = 523.63 kJ/kg

From the steam tables (Tables A-4, A-5, and A-6), h1 = h f @ 10 kPa = 191.81 kJ/kg

v1 = v f @ 10 kPa = 0.00101 m3 /kg wpI, in = v1 (P2 - P1 ) æ 1 kJ ÷ ö = (0.00101 m3 /kg )(15,000 - 10 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 15.14 kJ/kg h2 = h1 + wpI, in = 191.81 + 15.14 = 206.95 kJ/kg

P3 = 15 MPaïü h3 = 3157.9 kJ/kg ý T3 = 450°C ïïþ s3 = 6.1428 kJ/kg ×K s4 s - s f 6.1434 - 2.6454 = = 0.9880 P4 = 3 MPaü ï x4 s = s fg 3.5402 ý s4 s = s3 ïïþ h4 s = h f + x4 s h fg = 1008.3 + (0.9879)(1794.9) = 2781.7 kJ/kg

10-429

hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s = 3157.9 - (0.90)(3157.9 - 2781.7) = 2819.4 kJ/kg

P5 = 3 MPaïü h5 = 3457.2 kJ/kg ý T5 = 500°C ïïþ s5 = 7.2359 kJ/kg ×K s6 s - s f 7.2359 - 0.6492 = = 0.8783 P6 = 10 kPaïü ïý x6 s = s 7.4996 fg ïïþ s6 s = s5 h6 s = h f + x6 s h fg = 191.81 + (0.8782)(2392.1) = 2292.7 kJ/kg hT =

h5 - h6 ¾¾ ® h6 = h5 - hT (h5 - h6 s ) h5 - h6 s

= 3457.2 - (0.90)(3457.2 - 2292.7)

= 2409.1 kJ/kg Noting that Q @W @ Δke @ Δpe @ 0 for the heat exchanger, the steady-flow energy balance equation yields

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) i i

mair =

e e

air

h3 - h2 3157.9 - 206.95 ms = (12 kg/s) = 110.8 kg / s h10 - h11 843.10 - 523.63

(b) Qin = Qair + Qreheat = mair (h9 - h8 )+ mreheat (h5 - h4 )

= (110.8 kg/s)(1515.42 - 687.52) kJ/kg + (12 kg/s )(3457.2 - 2819.4) kJ/kg = 99, 385 kW

= (110.8 kg/s)(523.63 - 310.24) kJ/kg + (12 kg/s)(2409.1- 191.81) kJ/kg

= 50,250 kW hth = 1-

Qout 50,250 kW = 1= 0.494 = 49.4% 99,385 kW Qin

EES (Engineering Equation Solver) SOLUTION "Given" r_p=8 T[7]=290 [K] P[7]=100 [kPa] "Assumed air inlet pressure to compressor" T[9]=1400 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-430

T_11=247 [C] T[11]=T_11+273.15 "[K]" P[3]=15000 [kPa] T_3=450 [C] T[3]=T_3+273.15 "[K]" P[5]=3000 [kPa] T_5=500 [C] T[5]=T_5+273.15 "[K]" P[6]=10 [kPa] m_dot_steam=30 [kg/s] Eta_P=1 Eta_C=0.85 Eta_T_G=0.90 Eta_T_S=0.90 "Analysis" Fluid1$='air' Fluid2$='steam_iapws' "(a)" "Gas Turbine cycle" "compressor" h[7]=enthalpy(Fluid1$, T=T[7]) s[7]=entropy(Fluid1$, T=T[7], P=P[7]) P[8]=r_p*P[7] s_s[8]=s[7] T_s[8]=temperature(Fluid1$, P=P[8], s=s_s[8]) h_s[8]=enthalpy(Fluid1$, T=T_s[8]) Eta_C=(h_s[8]-h[7])/(h[8]-h[7]) "turbine" h[9]=enthalpy(Fluid1$, T=T[9]) P[9]=P[8] s[9]=entropy(Fluid1$, T=T[9], P=P[9]) P[10]=P[9]/r_p s_s[10]=s[9] T_s[10]=temperature(Fluid1$, P=P[10], s=s_s[10]) h_s[10]=enthalpy(Fluid1$, T=T_s[10]) Eta_T_G=(h[9]-h[10])/(h[9]-h_s[10]) h[11]=enthalpy(Fluid1$, T=T[11]) "Steam cycle" "pump" P[1]=P[6] x[1]=0 h[1]=enthalpy(Fluid2$, P=P[1], x=x[1]) v[1]=volume(Fluid2$, P=P[1], x=x[1]) P[2]=P[3] w_p_in=v[1]*(P[2]-P[1])/Eta_P h[2]=h[1]+w_p_in "high pressure turbine" h[3]=enthalpy(Fluid2$, P=P[3], T=T[3]) s[3]=entropy(Fluid2$, P=P[3], T=T[3]) s_s[4]=s[3] P[4]=P[5] h_s[4]=enthalpy(Fluid2$, P=P[4], s=s_s[4]) x_4s=quality(Fluid2$, P=P[4], s=s_s[4]) Eta_T_S=(h[3]-h[4])/(h[3]-h_s[4]) "low pressure turbine" h[5]=enthalpy(Fluid2$, P=P[5], T=T[5]) s[5]=entropy(Fluid2$, P=P[5], T=T[5]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-431

s_s[6]=s[5] h_s[6]=enthalpy(Fluid2$, P=P[6], s=s_s[6]) x_6s=quality(Fluid2$, P=P[6], s=s_s[6]) Eta_T_S=(h[5]-h[6])/(h[5]-h_s[6]) "Combined cycle" m_dot_steam*(h[3]-h[2])=m_dot_air*(h[10]-h[11]) "heat exchanger energy balance" "(b)" Q_dot_air=m_dot_air*(h[9]-h[8]) Q_dot_reheat=m_dot_steam*(h[5]-h[4]) Q_dot_in=Q_dot_air+Q_dot_reheat "(c)" Q_dot_out_air=m_dot_air*(h[11]-h[7]) Q_dot_out_steam=m_dot_steam*(h[6]-h[1]) Q_dot_out=Q_dot_out_air+Q_dot_out_steam Eta_th=1-Q_dot_out/Q_dot_in

11-106 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with one reheater and two feedwater heaters, one open and one closed. The fraction of steam extracted from the turbine for the open feedwater heater, the thermal efficiency of the cycle, and the net power output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-432

Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

h1 = h f @ 5 kPa = 137.75 kJ/kg v1 = v f @ 5 kPa = 0.001005 m3 /kg wpI, in = v1 (P2 - P1 )

æ 1 kJ ÷ ö = (0.001005 m3 /kg )(200 - 5 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 0.20 kJ/kg h2 = h1 + wpI, in = 137.75 + 0.20 = 137.95 kJ/kg

P3 = 0.2 MPaïü ïý h3 = h f @ 0.2 MPa = 504.71 kJ/kg ïïþv 3 = v f @ 0.2 MPa = 0.001061 m3 /kg sat.liquid wpII,in = v 3 (P4 - P3 )

æ 1 kJ ö÷ = (0.001061 m3 /kg )(15,000 - 200 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 15.70 kJ/kg h4 = h3 + wpII, in = 504.71 + 15.70 = 520.41 kJ/kg

P6 = 0.6 MPaü ïï h6 = h7 = h f @ 0.6 MPa = 670.38 kJ/kg ý ïïþv 6 = v f @ 0.6 MPa = 0.001101 m3 /kg sat.liquid T6 = T5 ¾ ¾ ® h5 = h6 + v6 (P5 - P6 )

æ 1 kJ ÷ ö = 670.38 + (0.001101 m3 /kg )(15,000 - 600 kPa )çç ÷ çè1 kPa ×m3 ÷ ø

= 686.23 kJ/kg P8 = 15 MPa h8 = 3583.1 kJ/kg T8 = 600°C s8 = 6.6796 kJ/kg ×K

}

10-433

P9 = 1.0 MPa üïï ý h9 = 2820.8 kJ/kg ïïþ s9 = s8 P10 = 1.0 MPa ïü h10 = 3479.1 kJ/kg ý T10 = 500°C ïïþ s10 = 7.7642 kJ/kg ×K P11 = 0.6 MPaïü ïý h = 3310.2 kJ/kg 11 s11 = s10 ïïþ P12 = 0.2 MPaü ïï ý h12 = 3000.9 kJ/kg s12 = s10 ïïþ

x13 = P13 = 5 kPa ïüï ý s13 = s10 ïïþ

s13 - s f

s fg

7.7642 - 0.4762 7.9176

= 0.9205 h13 = h f + x13 h fg

= 137.75 + (0.9205)(2423.0)

= 2368.1 kJ/kg The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the feedwater heaters. Noting that Q @W @ Δke @ Δpe @ 0,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) ¾ ¾® y (h - h ) = (h - h ) i i

e e

where y is the fraction of steam extracted from the turbine (= m11 / m5 ). Solving for y, y=

h5 - h4 686.23 - 520.41 = = 0.06287 h11 - h6 3310.2 - 670.38

For the open FWH,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

å mh = å m h i i

e e

m7 h7 + m2 h2 + m12 h12 = m3h3 yh7 + (1- y - z )h2 + zh12 = (1)h3 where z is the fraction of steam extracted from the turbine (= m12 / m5 ) at the second stage. Solving for z,

(h3 - h2 )- y (h7 - h2 ) 504.71- 137.95 - (0.06287)(670.38 - 137.95) h12 - h2

3000.0 - 137.95

= 0.1165

(b) qin = (h8 - h5 )+ (h10 - h9 ) (c) = (3583.1- 686.23)+ (3479.1- 2820.8) = 3555.3 kJ/kg (d) qout = (1- y - z )(h13 - h1 ) = (1- 0.06287 - 0.1165)(2368.0 - 137.75) = 1830.4 kJ/kg (e) wnet = qin - qout = 3555.3 - 1830.4 = 1724.9 kJ/kg and

10-434

hth = 1-

qout 1830.4 kJ/kg = 1= 48.5% qin 3555.3 kJ/kg

(f) Wnet = m wnet = (42 kg/s)(1724.9 kJ/kg ) = 72, 447 kW

EES (Engineering Equation Solver) SOLUTION "Given" P[8]=15000 [kPa] T[8]=600 [C] P[10]=1000 [kPa] T[10]=500 [C] P[11]=600 [kPa] P[12]=200 [kPa] P[13]=5 [kPa] x[6]=0 m_dot=42 [kg/s] "Analysis" Fluid$='steam_iapws' "(a)" "pump I" P[1]=P[13] x[1]=0 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) P[2]=P[12] w_pI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pI_in "pump II" x[3]=0 P[3]=P[12] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) v[3]=volume(Fluid$, P=P[3], x=x[3]) P[4]=P[8] w_pII_in=v[3]*(P[4]-P[3]) h[4]=h[3]+w_pII_in P[6]=P[11] h[6]=enthalpy(Fluid$, P=P[6], x=x[3]) v[6]=volume(Fluid$, P=P[6], x=x[3]) h[7]=h[6] P[5]=P[8] h[5]=h[6]+v[6]*(P[5]-P[6]) "high pressure turbine" h[8]=enthalpy(Fluid$, P=P[8], T=T[8]) s[8]=entropy(Fluid$, P=P[8], T=T[8]) s[9]=s[8] P[9]=P[10] h[9]=enthalpy(Fluid$, P=P[9], s=s[9]) "low pressure turbine" h[10]=enthalpy(Fluid$, P=P[10], T=T[10]) s[10]=entropy(Fluid$, P=P[10], T=T[10]) s[11]=s[10] h[11]=enthalpy(Fluid$, P=P[11], s=s[11]) s[12]=s[10] h[12]=enthalpy(Fluid$, P=P[12], s=s[12]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-435

s[13]=s[10] h[13]=enthalpy(Fluid$, P=P[13], s=s[13]) x[13]=quality(Fluid$, P=P[13], s=s[13]) "closed feedwater heater" y*(h[11]-h[6])=1*(h[5]-h[4]) "y=m_dot_11/m_dot_5" "open feedwater heater" y*h[7]+(1-y-z)*h[2]+z*h[12]=h[3] "z=m_dot_12/m_dot_5" "(b)" q_in=h[8]-h[5]+(h[10]-h[9]) q_out=(1-y-z)*(h[13]-h[1]) w_net=q_in-q_out Eta_th=1-q_out/q_in "(c)" W_dot_net=m_dot*w_net

11-107 A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater is considered. The T-s diagram for the ideal cycle is to be sketched. The net power output of the cycle and the minimum flow rate of the cooling water required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (b) Using the data from the problem statement, the enthalpies at various states are h1 = h f @ 20 kPa = 251.4 kJ/kg 1

= v f @ 20 kPa = 0.00102 m3 /kg

10-436

æ 1 kJ ö÷ = (0.00102 m3 /kg )(1400 - 20 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 1.41 kJ/kg h2 = h1 + wpI, in = 251.4 + 1.41 = 252.8 kJ/kg h4 = h f @ 1400 kPa = 830 kJ/kg

v 4 = v f @ 1400 kPa = 0.00115 m3 /kg wpII, in = v1 (P5 - P4 )

æ 1 kJ ö÷ = (0.00115 m3 /kg )(5000 - 1400 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 4.14 kJ/kg h5 = h4 + wpII, in = 830 + 4.14 = 834.1 kJ/kg

Also,

h3 = h12 = h f @ 245 kPa = 532 kJ/kg h13 = h12 (throttle valve operation) An energy balance on the open feedwater heater gives

yh7 + (1- y)h3 = 1(h4 ) where y is the fraction of steam extracted from the high-pressure turbine. Solving for y, y=

h4 - h3 830 - 532 = = 0.1039 h7 - h3 3400 - 532

An energy balance on the closed feedwater heater gives

zh10 + (1- y)h2 = (1- y)h3 + zh12 where z is the fraction of steam extracted from the low-pressure turbine. Solving for z, z=

(1- y )( h3 - h2 ) (1- 0.1039)(532 - 252.8) = = 0.09542 h10 - h12 3154 - 532

The heat input in the boiler is

qin = (h6 - h5 ) + (1- y)(h9 - h8 ) = (3894 - 834.1) + (1- 0.1039)(3692 - 3349) = 3367 kJ/kg The work output from the turbines is

wT,out = h6 - yh7 - (1- y)h8 + (1- y)h9 - zh10 - (1- y - z )h11 = 3894 - (0.1039)(3400) - (1- 0.1039)(3349) + (1- 0.1039)(3692) - (0.09542)(3154) - (1- 0.1039 - 0.09542)(2620)

= 1449 kJ/kg The net work output from the cycle is

wnet = wT,out - (1- y)wPI,in - wPII,in = 1449 - (1- 0.1039)(1.41) - 4.14

= 1444 kJ/kg The net power output is Wnet = mwnet = (100 kg/s)(1444 kJ/kg) = 144,400 kW = 144.4 MW

10-437

qout = (1- y - z)h11 + zh13 - (1- y)h1 = (1- 0.1039 - 0.09542)(2620) + (0.09542)(532) - (1- 0.1039)(251.4)

= 1923 kJ/kg The mass flow rate of cooling water will be minimum when the cooling water exit temperature is a maximum. That is, Tw,2 = T1 = Tsat @ 20 kPa = 60.1°C

Then an energy balance on the condenser gives

mqout = mw cp,w (Tw,2 - Tw,1 ) mw =

mqout (100 kg/s)(1923 kJ/kg) = = 1311 kg / s cp,w (Tw,2 - Tw,1 ) (4.18 kJ/kg ×K)(60.1- 25) K

11-108 A Rankine steam cycle modified for reheat and three closed feedwater heaters is considered. The T-s diagram for the ideal cycle is to be sketched. The net power output of the cycle and the flow rate of the cooling water required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (b) Using the data from the problem statement, the enthalpies at various states are h1 = h f @ 10 kPa = 191.8 kJ/kg

10-438

wpI, in = v1 (P2 - P1 )

æ 1 kJ ö÷ = (0.001010 m3 /kg )(5000 - 10 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 5.04 kJ/kg h2 = h1 + wpI, in = 191.8 + 5.04 = 196.8 kJ/kg

Also, h16 = h f @ 300 kPa = 561.4 kJ/kg

v16 = v f @ 300 kPa = 0.001073 m3 /kg wpII, in = v16 (P17 - P16 )

æ 1 kJ ö÷ = (0.001073 m3 /kg )(5000 - 300 kPa )çç ÷ çè1 kPa ×m3 ø÷

= 5.04 kJ/kg h17 = h16 + wpII, in = 561.4 + 5.04 = 566.4 kJ/kg

h3 = h18 = h f @ 75 kPa = 384.4 kJ/kg h4 = h16 = h f @ 300 kPa = 561.4 kJ/kg

h5 = h14 = h f @ 925 kPa = 747.7 kJ/kg

h15 = h14 (throttle valve operation) h19 = h18 (throttle valve operation) Energy balances on three closed feedwater heaters give

yh10 + (1- y - z )h4 = (1- y - z )h5 + yh15 zh11 + (1- y - z)h3 + yh15 = (1- y - z )h4 + ( y + z )h16

wh12 + (1- y - z )h2 = (1- y - z)h3 + wh18 The enthalpies are known, and thus there are three unknowns (y, z, w) and three equations. Solving these equations using an equation solver such as EES, we obtain y = 0.06335 z = 0.05863 w = 0.07063

The enthalpy at state 6 may be determined from an energy balance on mixing chamber:

h6 = (1- y - z)h5 + ( y + z)h17 = (1- 0.06335 - 0.05863)(747.7) + (0.06335 + 0.05863)(566.4) = 725.6 kJ/kg The heat input in the boiler is

qin = (h7 - h6 ) + (h9 - h8 ) = (3900 - 725.6) + (3687 - 3615) = 3246 kJ/kg The work output from the turbines is

wT,out = h7 - h8 + h9 - yh10 - zh11 - wh12 - (1- y - z - w)h13 = 3900 - 3615 + 3687 - (0.06335)(3330) - (0.05863)(3011)

- (0.07063)(2716) - (1- 0.06335 - 0.05863 - 0.07063)(2408)

10-439

The net work output from the cycle is

wnet = wT,out - (1- y - z ) wPI,in - ( y + z )wPII,in = 1448 - (1- 0.06335 - 0.05863)(5.04) - (0.06335 + 0.05863)(5.04) = 1443 kJ/kg

The net power output is Wnet = mwnet = (100 kg/s)(1443 kJ/kg) = 144,300 kW = 144.3 MW

qout = (1- y - z - w)h13 + wh19 - (1- y - z )h1 = (1- 0.06335 - 0.05863 - 0.07063)(2408) + (0.07063)(384.4) - (1- 0.06335 - 0.05863)(191.8) = 1803 kJ/kg Then an energy balance on the condenser gives

mqout = mw cp,w (Tw,2 - Tw,1 ) mw =

mqout (100 kg/s)(1803 kJ/kg) = = 4313 kg / s cp,w (Tw,2 - Tw,1 ) (4.18 kJ/kg ×K)(10 K)

11-109 The effect of the boiler pressure on the performance a simple ideal Rankine cycle is to be investigated. Analysis The problem is solved using EES, and the solution is given below. function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$='' if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)' end P[3] = 20000 [kPa] T[3] = 500 [C] P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[4] P[2]=P[3] x[1]=0 h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1]) W_p=W_p_s/Eta_p h[2]=h[1]+W_p s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Turbine analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-440

h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) x[4]=quality(Fluid$,h=h[4],P=P[4]) h[3] =W_t+h[4] x4s$=x4$(x[4]) "Boiler analysis" Q_in + h[2]=h[3] "Condenser analysis" h[4]=Q_out+h[1] "Cycle Statistics" W_net=W_t-W_0070 Eta_th=W_net/Q_in th

0.2792 0.3512 0.3752 0.3893 0.399 0.406 0.4114 0.4156 0.4188 0.4214

Wnet [kJ / kg] 919.1 1147 1216 1251 1270 1281 1286 1286 1283 1276

x4 0.9921 0.886 0.8462 0.8201 0.8001 0.7835 0.769 0.756 0.744 0.7328

Qin

Qout

[kPa]

[kJ / kg] 3292 3266 3240 3213 3184 3155 3125 3094 3062 3029

[kJ / kg] 2373 2119 2024 1962 1914 1874 1840 1808 1780 1753

[kJ / kg] 0.495 2.684 4.873 7.062 9.251 11.44 13.63 15.82 18.01 20.2

[kJ / kg] 919.6 1150 1221 1258 1280 1292 1299 1302 1301 1297

500 2667 4833 7000 9167 11333 13500 15667 17833 20000

10-441

11-110 The effect of the condenser pressure on the performance a simple ideal Rankine cycle is to be investigated. Analysis The problem is solved using EES, and the solution is given below. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-442

function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$='' if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)' end P[3] = 10000 [kPa] T[3] = 550 [C] P[4] = 5 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[4] P[2]=P[3] x[1]=0 h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1]) W_p=W_p_s/Eta_p h[2]=h[1]+W_p s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) x[4]=quality(Fluid$,h=h[4],P=P[4]) h[3] =W_t+h[4] x4s$=x4$(x[4]) "Boiler analysis" Q_in + h[2]=h[3] "Condenser analysis" h[4]=Q_out+h[1] "Cycle Statistics" W_net=W_t-W_p Eta_th=W_net/Q_in

P4 [kPa]

Wnet

[kJ / kg] 5 0.4268 1432 15 0.3987 1302 25 0.3841 1237 35 0.3739 1192 45 0.3659 1157 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-443

55 65 75 85 100

0.3594 0.3537 0.3488 0.3443 0.3385

1129 1105 1084 1065 1040

10-444

11-111 The effect of superheating the steam on the performance a simple ideal Rankine cycle is to be investigated. Analysis The problem is solved using EES, and the solution is given below. function x4$(x4) "this function returns a string to indicate the state of steam at point 4" x4$='' if (x4>1) then x4$='(superheated)' if (x4<0) then x4$='(compressed)' end P[3] = 3000 [kPa] T[3] = 600 [C] P[4] = 10 [kPa] Eta_t = 1.0 "Turbine isentropic efficiency" Eta_p = 1.0 "Pump isentropic efficiency" "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[4] P[2]=P[3] x[1]=0 h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1]) W_p=W_p_s/Eta_p h[2]=h[1]+W_p s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-445

Eta_t=(h[3]-h[4])/(h[3]-hs[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) x[4]=quality(Fluid$,h=h[4],P=P[4]) h[3] =W_t+h[4] x4s$=x4$(x[4]) "Boiler analysis" Q_in + h[2]=h[3] "Condenser analysis" h[4]=Q_out+h[1] "Cycle Statistics" W_net=W_t-W_p Eta_th=W_net/Q_in T3

ηth

Wnet

0.3241 0.3338 0.3466 0.3614 0.3774 0.3939 0.4106 0.4272 0.4424 0.456

[kJ / kg] 862.8 970.6 1083 1206 1340 1485 1639 1803 1970 2139

[C]

250 344.4 438.9 533.3 627.8 722.2 816.7 911.1 1006 1100

x4 0.752 0.81 0.8536 0.8909 0.9244 0.955 0.9835 100 100 100

10-446

11-112 The effect of reheat pressure on the performance an ideal Rankine cycle is to be investigated. Analysis The problem is solved using EES, and the solution is given below. function x6$(x6) "this function returns a string to indicate the state of steam at point 6" x6$='' if (x6>1) then x6$='(superheated)' if (x6<0) then x6$='(subcooled)' end P[6] = 10 [kPa] P[3] = 15000 [kPa] T[3] = 500 [C] P[4] = 3000 [kPa] T[5] = 500 [C] Eta_t = 100/100 "Turbine isentropic efficiency" Eta_p = 100/100 "Pump isentropic efficiency" "Pump analysis" Fluid$='Steam_IAPWS' P[1] = P[6] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-447

P[2]=P[3] x[1]=0 h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) v[1]=volume(Fluid$,P=P[1],x=x[1]) s[1]=entropy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,P=P[1],x=x[1]) W_p_s=v[1]*(P[2]-P[1]) W_p=W_p_s/Eta_p h[2]=h[1]+W_p v[2]=volume(Fluid$,P=P[2],h=h[2]) s[2]=entropy(Fluid$,P=P[2],h=h[2]) T[2]=temperature(Fluid$,P=P[2],h=h[2]) "High Pressure Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3]) s[3]=entropy(Fluid$,T=T[3],P=P[3]) v[3]=volume(Fluid$,T=T[3],P=P[3]) s_s[4]=s[3] hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4]) Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4]) Eta_t=(h[3]-h[4])/(h[3]-hs[4]) T[4]=temperature(Fluid$,P=P[4],h=h[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) v[4]=volume(Fluid$,s=s[4],P=P[4]) h[3] =W_t_hp+h[4] "Low Pressure Turbine analysis" P[5]=P[4] s[5]=entropy(Fluid$,T=T[5],P=P[5]) h[5]=enthalpy(Fluid$,T=T[5],P=P[5]) s_s[6]=s[5] hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6]) Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6]) vs[6]=volume(Fluid$,s=s_s[6],P=P[6]) Eta_t=(h[5]-h[6])/(h[5]-hs[6]) h[5]=W_t_lp+h[6] x[6]=quality(Fluid$,h=h[6],P=P[6]) "Boiler analysis" Q_in + h[2]+h[4]=h[3]+h[5] "Condenser analysis" h[6]=Q_out+h[1] T[6]=temperature(Fluid$,h=h[6],P=P[6]) s[6]=entropy(Fluid$,h=h[6],P=P[6]) x6s$=x6$(x[6]) "Cycle Statistics" W_net=W_t_hp+W_t_lp-W_p Eta_th=W_net/Q_in

ηth

Wnet

0.4128 0.4253 0.4283

[kJ / kg] 1668 1611 1567

[kPa] 500 1833 3167

x6 0.9921 0.9102 0.8747

10-448

4500 5833 7167 8500 9833 11167 12500

0.4287 0.428 0.4268 0.4252 0.4233 0.4212 0.4189

1528 1492 1458 1426 1395 1366 1337

0.8511 0.8332 0.8184 0.8058 0.7947 0.7847 0.7755

10-449

11-113 It is to be demonstrated that the thermal efficiency of a combined gas-steam power plant ηcc can be expressed as hcc = hg + hs - hg hs where hg = Wg / Qin and hs = Ws / Qg, out are the thermal efficiencies of the gas and steam cycles, respectively, and the efficiency of a combined cycle is to be obtained. Analysis The thermal efficiencies of gas, steam, and combined cycles can be expressed as hcc =

hg =

hs =

Wtotal Q = 1- out Qin Qin

Wg Qin

= 1-

Qg, out Qin

Ws Qout = 1Qg,out Qg, out

where Qin is the heat supplied to the gas cycle, where Qout is the heat rejected by the steam cycle, and where Qg, out is the heat rejected from the gas cycle and supplied to the steam cycle. Using the relations above, the expression hg + hs - hg hs can be expressed as æ Q ÷ ö æ Qg, out öæ Q ö æ Qg, out ö çç1÷ ÷ out ÷ ç out ÷ ÷ ÷ ÷ hg + hs - hg hs = ççç1+ ççç1- çç1÷ ÷ ÷ çç Q ÷ ÷ ÷ ÷ çè Qg, out ÷ ÷ çè ç ÷ Qin ø Q ø è in øè g,out ø

10-450

= 1= 1-

Qg, out Qin

+ 1-

Qg, out Qout Q Q - 1+ + out - out Qg,out Qin Qg, out Qin

Qout Qin

= hcc Therefore, the proof is complete. Using the relation above, the thermal efficiency of the given combined cycle is determined to be hcc = hg + hs - hg hs = 0.4 + 0.30 - 0.40´ 0.30 = 0.58

11-114 The thermal efficiency of a combined gas-steam power plant

can be expressed in terms of the thermal

efficiencies of the gas and the steam turbine cycles as hcc = hg + hs - hg hs . It is to be shown that the value of hcc is greater than either of hg or hs . Analysis By factoring out terms, the relation hcc = hg + hs - hg hs can be expressed as

hcc = hg + hs - hg hs = hg + hs (1- hg ) > hg Positive since hg <1

hcc = hg + hs - hg hs = hs + hg (1- hs ) > hs Positive since hs <1

Thus we conclude that the combined cycle is more efficient than either of the gas turbine or steam turbine cycles alone.

11-115 Thermal collection efficiency of a solar collector system is given. The operating temperature of the solar collector for maximum system efficiency is to be determined and an expression for maximum system efficiency is to be developed. Analysis (a) The overall system efficiency is a product of the solar collector and power plant efficiencies, or é æ T öù æ ö ú= F çç A - BT - A TL + BT ÷ ÷ ÷ (1) h = hsc hth = (A - BTH ) êêF ççç1- L ÷ H L÷ ÷ ú ç ÷ú ÷ ç TH èç ø ëê è TH øû The maximum temperature is found when

dh = 0, for which the optimal TH is dTH

T dh = - B + A L2 = 0 ¾ ¾ ® TH , optimal = dTH TH

T A L B

(2)

(b) Substituting Eq. (2) into Eq. (1) gives the maximum overall system efficiency: æ ö ÷ çç ÷ ÷ çç 2 ÷ T T hmax = F çç A - B A L - A L + BTL ÷ = F A - BTL ÷ ÷ ÷ çç B T ÷ ÷ A L çç ÷ è ø B

(

)

Discussion A higher value of TH increases the power plant efficiency but decreases the heat gained from the solar collector, and the converse. This problem illustrates how components integrated to form a system must be mutually adjusted to optimize system performance. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-451

11-116 It is to be shown that the exergy destruction associated with a simple ideal Rankine cycle can be expressed as xdestroyed = qin (hth, Carnot - hth ) , where th is efficiency of the Rankine cycle and th, Carnot is the efficiency of the Carnot cycle operating between the same temperature limits. Analysis The exergy destruction associated with a cycle is given on a unit mass basis as q xdestroyed = T0 å R TR where the direction of qin is determined with respect to the reservoir (positive if to the reservoir and negative if from the reservoir). For a cycle that involves heat transfer only with a source at TH and a sink at T0 , the irreversibility becomes

æq æqout T0 ö÷ T0 q ö ç ÷ ÷ xdestroyed = T0 ççç out - in ÷ = q q = q in in ç ÷ out ÷ çèç q TH TH ÷ ø èç T0 TH ÷ ø in = qin éê(1- hth )- (1- hth , C )ù = q h - hth ) ú ë û in ( th , C

Fundamentals of Engineering (FE) Exam Problems 11-117 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures. If the steam is superheated to a higher temperature, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease. Answer (d) the moisture content at turbine exit will decrease.

11-118 Consider a simple ideal Rankine cycle. If the condenser pressure is lowered while keeping turbine inlet state the same, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the cycle efficiency will decrease. (d) the moisture content at turbine exit will decrease. (e) the pump work input will decrease. Answer (b) the amount of heat rejected will decrease.

11-119 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with reheating, (a) the turbine work output will decrease. (b) the amount of heat rejected will decrease. (c) the pump work input will decrease. (d) the moisture content at turbine exit will decrease. (e) the amount of heat input will decrease. Answer (d) the moisture content at turbine exit will decrease. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-452

11-120 Consider a simple ideal Rankine cycle with fixed boiler and condenser pressures . If the cycle is modified with regeneration that involves one open feed water heater, (select the correct statement per unit mass of steam flowing through the boiler) (a) the turbine work output will decrease. (b) the amount of heat rejected will increase. (c) the cycle thermal efficiency will decrease. (d) the quality of steam at turbine exit will decrease. (e) the amount of heat input will increase. Answer (a) the turbine work output will decrease.

11-121 Consider a steady-flow Carnot cycle with water as the working fluid executed under the saturation dome between the pressure limits of 1 MPa and 10 kPa. Water changes from saturated liquid to saturated vapor during the heat addition process. The net work output of this cycle is (a) 596 kJ / kg (b) 666 kJ / kg (c) 708 kJ / kg (d) 822 kJ / kg (e) 1500 kJ / kg Answer (a) 596 kJ / kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=1000 [kPa] P2=10 [kPa] h_fg=ENTHALPY(Steam_IAPWS,x=1,P=P1)-ENTHALPY(Steam_IAPWS,x=0,P=P1) T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1)+273 T2=TEMPERATURE(Steam_IAPWS,x=0,P=P2)+273 q_in=h_fg Eta_Carnot=1-T2/T1 w_net=Eta_Carnot*q_in "Some Wrong Solutions with Common Mistakes:" W1_work = Eta1*q_in; Eta1=T2/T1 "Taking Carnot efficiency to be T2/T1" W2_work = Eta2*q_in; Eta2=1-(T2-273)/(T1-273) "Using C instead of K" W3_work = Eta_Carnot*ENTHALPY(Steam_IAPWS,x=1,P=P1) "Using h_g instead of h_fg" W4_work = Eta_Carnot*q2; q2=ENTHALPY(Steam_IAPWS,x=1,P=P2)-ENTHALPY(Steam_IAPWS,x=0,P=P2) "Using h_fg at P2"

11-122 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 5 MPa, with a turbine inlet temperature of 600C. The mass fraction of steam that condenses at the turbine exit is (a) 6% (b) 9% (c) 12% (d) 15% (e) 18% © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-453

Answer (c) 12% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 [kPa] P2=5000 [kPa] T3=600 [C] P3=P2 P4=P1 s4=s3 h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) x4=QUALITY(Steam_IAPWS,s=s4,P=P4) moisture=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_moisture = x4 "Taking quality as moisture" W2_moisture = 0 "Assuming superheated vapor"

11-123 A steam power plant operates on the simple ideal Rankine cycle between the pressure limits of 10 kPa and 5 MPa, with a turbine inlet temperature of 600C. The rate of heat transfer in the boiler is 450 kJ / s . Disregarding the pump work, the power output of this plant is (a) 118 kW (b) 140 kW (c) 177 kW (d) 286 kW (e) 450 kW Answer (c) 177 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 [kPa] P2=5000 [kPa] T3=600 [C] Q_rate=450 [kJ/s] P3=P2 P4=P1 s4=s3 m=Q_rate/q_in h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 "pump work is neglected" h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) q_in=h3-h2 W_turb=m*(h3-h4) "Some Wrong Solutions with Common Mistakes:" W1_power = Q_rate "Assuming all heat is converted to power" W3_power = Q_rate*Carnot; Carnot = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_power = m*(h3-h44); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Taking h4=h_g" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-454

11-124 A simple ideal Rankine cycle operates between the pressure limits of 10 kPa and 3 MPa, with a turbine inlet temperature of 600C. Disregarding the pump work, the cycle efficiency is (a) 24% (b) 37% (c) 52% (d) 63% (e) 71% Answer (b) 37% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 [kPa] P2=3000 [kPa] T3=600 [C] P3=P2 P4=P1 s4=s3 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) q_in=h3-h2 q_out=h4-h1 Eta_th=1-q_out/q_in "Some Wrong Solutions with Common Mistakes:" W1_Eff = q_out/q_in "Using wrong relation" W2_Eff = 1-(h44-h1)/(h3-h2); h44 = ENTHALPY(Steam_IAPWS,x=1,P=P4) "Using h_g for h4" W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_Eff = (h3-h4)/q_in "Disregarding pump work"

11-125 An ideal reheat Rankine cycle operates between the pressure limits of 10 kPa and 8 MPa, with reheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is 500C, and the enthalpy of steam is 3185 kJ / kg at the exit of the high-pressure turbine, and 2247 kJ / kg at the exit of the low-pressure turbine. Disregarding the pump work, the cycle efficiency is (a) 29% (b) 32% (c) 36% (d) 41% (e) 49% Answer (d) 41% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=10 [kPa] P2=8000 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-455

P4=4000 [kPa] T3=500 [C] T5=500 [C] P3=P2 P5=P4 P6=P1 s4=s3 s6=s5 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) h2=h1 h44=3185 [kJ/kg] "for checking given data" h66=2247 [kJ/kg] "for checking given data" h3=ENTHALPY(Steam_IAPWS,T=T3,P=P3) s3=ENTROPY(Steam_IAPWS,T=T3,P=P3) h4=ENTHALPY(Steam_IAPWS,s=s4,P=P4) h5=ENTHALPY(Steam_IAPWS,T=T5,P=P5) s5=ENTROPY(Steam_IAPWS,T=T5,P=P5) h6=ENTHALPY(Steam_IAPWS,s=s6,P=P6) q_in=(h3-h2)+(h5-h4) q_out=h6-h1 Eta_th=1-q_out/q_in "Some Wrong Solutions with Common Mistakes:" W1_Eff = q_out/q_in "Using wrong relation" W2_Eff = 1-q_out/(h3-h2) "Disregarding heat input during reheat" W3_Eff = 1-(T1+273)/(T3+273); T1=TEMPERATURE(Steam_IAPWS,x=0,P=P1) "Using Carnot efficiency" W4_Eff = 1-q_out/(h5-h2) "Using wrong relation for q_in"

11-126 Pressurized feedwater in a steam power plant is to be heated in an ideal open feedwater heater that operates at a pressure of 2 MPa with steam extracted from the turbine. If the enthalpy of feedwater is 252 kJ / kg and the enthalpy of extracted steam is 2810 kJ / kg , the mass fraction of steam extracted from the turbine is (a) 10% (b) 14% (c) 26% (d) 36% (e) 50% Answer (c) 26% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). h_feed=252 [kJ/kg] h_extracted=2810 [kJ/kg] P3=2000 [kPa] h3=ENTHALPY(Steam_IAPWS,x=0,P=P3) h3=x_ext*h_extracted+(1-x_ext)*h_feed "Energy balance on the FWH" "Some Wrong Solutions with Common Mistakes:" W1_ext = h_feed/h_extracted "Using wrong relation" W2_ext = h3/(h_extracted-h_feed) "Using wrong relation" W3_ext = h_feed/(h_extracted-h_feed) "Using wrong relation"

10-456

11-127 Consider a steam power plant that operates on the regenerative Rankine cycle with one open feedwater heater. The enthalpy of the steam is 3374 kJ / kg at the turbine inlet, 2797 kJ / kg at the location of bleeding, and 2346 kJ / kg at the turbine exit. The net power output of the plant is 120 MW, and the fraction of steam bled off the turbine for regeneration is 0.172. If the pump work is negligible, the mass flow rate of steam at the turbine inlet is (a) 117 kg / s (b) 126 kg / s (c) 219 kg / s (d) 288 kg / s (e) 679 kg / s Answer (b) 126 kg / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). h_in=3374 [kJ/kg] h_out=2346 [kJ/kg] h_extracted=2797 [kJ/kg] Wnet_out=120000 [kW] x_bleed=0.172 w_turb=(h_in-h_extracted)+(1-x_bleed)*(h_extracted-h_out) m=Wnet_out/w_turb "Some Wrong Solutions with Common Mistakes:" W1_mass = Wnet_out/(h_in-h_out) "Disregarding extraction of steam" W2_mass = Wnet_out/(x_bleed*(h_in-h_out)) "Assuming steam is extracted at trubine inlet" W3_mass = Wnet_out/(h_in-h_out-x_bleed*h_extracted) "Using wrong relation"

11-128 Consider a combined gas-steam power plant. Water for the steam cycle is heated in a well-insulated heat exchanger by the exhaust gases that enter at 800 K at a rate of 60 kg / s and leave at 400 K. Water enters the heat exchanger at 200C and 8 MPa and leaves at 350C and 8 MPa. If the exhaust gases are treated as air with constant specific heats at room temperature, the mass flow rate of water through the heat exchanger becomes (a) 11 kg / s (b) 24 kg / s (c) 46 kg / s (d) 53 kg / s (e) 60 kg / s Answer (a) 11 kg / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m_gas=60 [kg/s] T3=800 [K] T4=400 [K] P1=8000 [kPa] T1=200 [C] P2=8000 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-457

T2=350 [C] cp=1.005 [kJ/kg-K] Q_gas=m_gas*cp*(T3-T4) h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) Q_steam=m_steam*(h2-h1) Q_gas=Q_steam "Some Wrong Solutions with Common Mistakes:" m_gas*cp*(T3 -T4)=W1_msteam*4.18*(T2-T1) "Assuming no evaporation of liquid water" m_gas*cv*(T3 -T4)=W2_msteam*(h2-h1); cv=0.718 [kJ/kg-K] "Using cv for air instead of cp" W3_msteam = m_gas "Taking the mass flow rates of two fluids to be equal" m_gas*cp*(T3 -T4)=W4_msteam*(h2-h11); h11=ENTHALPY(Steam_IAPWS,x=0,P=P1) "Taking h1=hf@P1"

11-129 Consider a cogeneration power plant modified with regeneration. Steam enters the turbine at 6 MPa and 450C at a rate of 20 kg / s and expands to a pressure of 0.4 MPa. At this pressure, 60% of the steam is extracted from the turbine, and the remainder expands to a pressure of 10 kPa. Part of the extracted steam is used to heat feedwater in an open feedwater heater. The rest of the extracted steam is used for process heating and leaves the process heater as a saturated liquid at 0.4 MPa. It is subsequently mixed with the feedwater leaving the feedwater heater, and the mixture is pumped to the boiler pressure. The steam in the condenser is cooled and condensed by the cooling water from a nearby river, which enters the adiabatic condenser at a rate of 463 kg / s .

1. The total power output of the turbine is (a) 17.0 MW (b) 8.4 MW (c) 12.2 MW (d) 20.0 MW (e) 3.4 MW Answer (a) 17.0 MW

10-458

2. The temperature rise of the cooling water from the river in the condenser is (a) 8.0C (b) 5.2C (c) 9.6C (d) 12.9C (e) 16.2C Answer (a) 8.0C

3. The mass flow rate of steam through the process heater is (a) 1.6 kg / s (b) 3.8 kg / s (c) 5.2 kg / s (d) 7.6 kg / s (e) 10.4 kg / s Answer (e) 10.4 kg / s

4. The rate of heat supply from the process heater per unit mass of steam passing through it is (a) 246 kJ / kg (b) 893 kJ / kg (c) 1344 kJ / kg (d) 1891 kJ / kg (e) 2060 kJ / kg Answer (e) 2060 kJ / kg

5. The rate of heat transfer to the steam in the boiler is (a) 26.0 MJ / s (b) 53.8 MJ / s (c) 39.5 MJ / s (d) 62.8 MJ / s (e) 125.4 MJ / s Answer (b) 53.8 MJ / s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Note: The solution given below also evaluates all enthalpies given on the figure. P1=10 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-459

P2=400 [kPa] P5=6000 [kPa] T6=450 [C] m_total=20 [kg/s] m_cooling=463 [kg/s] y=0.6 c=4.18 [kJ/kg-C] P11=P1 P3=P2; P4=P2; P7=P2; P8=P2; P9=P2; P10=P2 P6=P5 m7=y*m_total m_cond=(1-y)*m_total s7=s6 s11=s6 h1=ENTHALPY(Steam_IAPWS,x=0,P=P1) v1=VOLUME(Steam_IAPWS,x=0,P=P1) w_pump=v1*(P2-P1) h2=h1+w_pump h3=ENTHALPY(Steam_IAPWS,x=0,P=P3) h4=h3 h9=h3 v4=VOLUME(Steam_IAPWS,x=0,P=P4) w_pump2=v4*(P5-P4) h5=h4+w_pump2 h6=ENTHALPY(Steam_IAPWS,T=T6,P=P6) s6=ENTROPY(Steam_IAPWS,T=T6,P=P6) h7=ENTHALPY(Steam_IAPWS,s=s7,P=P7) h8=h7 h10=h7 h11=ENTHALPY(Steam_IAPWS,s=s11,P=P11) W_turb=m_total*(h6-h7)+m_cond*(h7-h11) m_cooling*c*T_rise=m_cond*(h11-h1) m_cond*h2+m_feed*h10=(m_cond+m_feed)*h3 m_process=m7-m_feed q_process=h8-h9 Q_in=m_total*(h6-h5)

11-130 … 11-139 Design and Essay Problems

10-460

Chapter 12 REFRIGERATION CYCLES

10-461

CYU - Check Your Understanding CYU 12-1 ■ Check Your Understanding CYU 12-1.1 Select the incorrect statement below about refrigerators and heat pumps: (a) The COP of a refrigerator is less than that of a heat pump for fixed values of QL and QH . (b) The minimum COP value of a heat pump is one. (c) The COP of a refrigerator cannot be less than one. (d) COPHP = COPR + 1 for fixed values of QL and QH . Answer: (c) The COP of a refrigerator cannot be less than one. CYU 12-1.2 A vapor-compression refrigeration cycle absorbs heat from a cool space at a rate of 4 kW and transfers 6 kW of heat to a warm space. What are the COP values corresponding to the mode of operation as a refrigerator and as a heat pump? (a) COPR = 0.5, COPHP = 1.5 (b) COPR = 1.5, COPHP = 0.33 (c) COPR = 2, COPHP = 3 (d) COPR = 1.5, COPHP = 1.5 (e) COPR = 3, COPHP = 5 Answer: (c) (c) COPR = 2, COPHP = 3 Win = QH - QL = 6 - 4 = 2 kW COPR = QL / Win = 4 / 2 = 2 COPHP = QH / Win = 6 / 2 = 3

CYU 12-1.3 A heat pump absorbs heat from the cold outdoors at a rate of 2 kW and transfers 4 kW of heat to the warm indoors. What are the COP and the power input values for this heat pump? (a) COPHP = 1, Win = 6 kW (b) COPHP = 4, Win = 0.5 kW (c) COPHP = 1, Win = 4 kW (d) COPHP = 0.5, Win = 2 kW (e) COPHP = 2, Win = 2 kW Answer: (e) COPHP = 2, Win = 2 kW

Win = QH - QL = 4 - 2 = 2 kW COPHP = QH / Win = 4 / 2 = 2

CYU 12-3 ■ Check Your Understanding

10-462

CYU 12-3.1 The state of the refrigerant at the inlet or exit of the components of a refrigerator operating on the ideal vaporcompression refrigeration cycle are specified below. Select the item with the incorrect specification. (a) Saturated vapor at the compressor inlet (b) Saturated vapor at the expansion valve inlet (c) Saturated vapor evaporator exit (d) Superheated vapor at the condenser inlet (e) Superheated vapor at the compressor exit Answer: (b) Expansion valve inlet - Saturated vapor CYU 12-3.1 Select the correct statement below regarding the ideal vapor-compression refrigeration cycle. (a) It is an internally reversible cycle. (b) Replacing the expansion valve by a turbine increases the net work input to the cycle. (c) The COP of the cycle increases when the condensing temperature is raised. (d) The COP of the cycle increases when the evaporating temperature is raised. Answer: (c) The COP of the cycle increases when the condensing temperature is raised. CYU 12-3.3 Which equations can be used to determine the COPs of a household refrigerator, an air conditioner and a heat pump in the analysis of the ideal vapor-compression refrigeration cycle? 1: Compressor inlet, 2: Compressor exit, 3: Expansion valve inlet, 4: Expansion valve exit h1 - h4 h2 - h1

CO PR =

CO PA C =

h1 - h4 h2 - h1

III CO PH P =

h2 - h3 h2 - h1

(a) Only III (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (e) I, II, and III CYU 12-3.4 Consider a household refrigerator, an air-source air conditioner and air-source heat pump. Which system is most likely to have the highest COP value? (a) A household refrigerator operating in a kitchen at 30C. (b) A heat pump keeping a room at 22C when the outdoors is at 2C (c) A heat pump keeping a room at 22C when the outdoors is at 10C (d) An air-conditioner keeping a room at 22C when the outdoors is at 45C (e) An air-conditioner keeping a room at 22C when the outdoors is at 36C Answer: (c) A heat pump keeping a room at 22C when the outdoors is at 10C

10-463

CYU 12-4.1 Which feature(s) of actual vapor-compression refrigeration cycle are undesirable? I The compressor inlet state is superheated vapor instead of saturated vapor. II The condenser exit state is subcooled liquid instead of saturated liquid. III The pressure drops in the evaporator and the line connecting the evaporator to the compressor. IV Heat losses from the refrigerant during the compression process. (a) Only I (b) Only II (c) I and II (d) I and III (e) I, III, and IV Answer: (d) I and III

CYU 12-5 ■ Check Your Understanding CYU 12-5.1 Which equation(s) below can be used to determine the exergy destructions of the components in the vaporcompression refrigeration cycle? 1: Compressor inlet, 2: Compressor exit, 3: Expansion valve inlet, 4: Expansion valve exit I

é Q ù Xdest,Evap = T0 êêm (s1 - s4 )+ L úú TL úû êë

Xdest,ExpValve = mT0 (s4 - s3)

III

Xdest,Cond = T0m (s3 - s2 )

(a) Only I (b) Only II (c) I and II (d) II and III (e) I, II, and III Answer: (b) Only II CYU 12-5.2 Which equation(s) can be used to determine the second-law efficiency of a heat pump operating on the vaporcompression refrigeration cycle? I

hII,cycle = XQ /W in

hII,cycle = W in /W rev

III

hII,cycle = COPHP /COPHP,rev

hII,cycle = 1- (Xdest,total /W in )

(a) I and II (b) II and III (c) III and IV (d) I, II, and III (e) I, II, III, and IV Answer: (c) III and IV

10-464

CYU 12-6 ■ Check Your Understanding CYU 12-6.1 Select the incorrect statement below regarding the refrigerants. (a) Hydrochlorofluorocarbons (HCFCs) such as R-22 do not contribute to ozone depletion. (b) Hydrofluorocarbons (HFCs) such as R-134a do not contribute to ozone depletion. (c) Chlorofluorocarbons (CFCs) such as R-12 do the most damage to the ozone layer. (d) Ammonia does not contribute to ozone depletion. Answer: (a) Hydrochlorofluorocarbons (HCFCs) such as R-22 do not contribute to ozone depletion. CYU 12-6.2 Which two refrigerants are HFCs and are among the viable alternatives to R-22 in residential and commercial air-conditioning and refrigeration applications? (a) R-134a and R-22 (b) R-502 and R-22 (c) R-12 and R-22 (d) Ammonia and R-115 (e) R-410A and R-32 Answer: (e) R-410A and R-32

CYU 12-7 ■ Check Your Understanding CYU 12-7.1 Select the correct statements regarding heat pump systems. I The initial cost of ground-source heat pumps are higher than that of air-source heat pumps. II Ground-source heat pump systems have higher COPs than air-source heat pumps. III Air-source heat pump systems have higher COPs than water-source heat pumps. (a) Only III (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (b) I and II CYU 12-7.2 Heat pumps make most economic sense as thermal comfort systems in areas that have a _______ cooling load in summer and a _______ heating load in winter. (a) large, large (b) small, small (c) small, large (d) large, small Answer: (d) large, small CYU 12-7.3 Which system is the most likely to have the highest COP value? (a) A household refrigerator operating in a kitchen at 25C (b) An air-source air conditioner keeping a room at 25C when the outside air is at 40C (c) A ground-source air conditioner keeping a room at 25C. (d) An air-source heat pump keeping a room at 25C when the outside air is at 2C © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-465

(e) A ground-source heat pump keeping a room at 25C when the outside air is at 2C Answer: (e) A ground-source heat pump keeping a room at 25C when the outside air is at 2C

CYU 12-8 ■ Check Your Understanding CYU 12-8.1 Select the correct statement(s) regarding cascade refrigeration systems? I Cascading improves the COP of a refrigeration system. II The compressor work decreases as a result of cascading. III The amount of heat absorbed from the refrigerated space decreases as a result of cascading. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (b) I and II CYU 12-8.2 A multistage compression refrigeration system is preferred over a cascade refrigeration system when I The fluid used throughout the cascade refrigeration system is the same. II The pressure difference between the condenser and evaporator is too large. III The application requires very low temperatures. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (a) Only I CYU 12-8.3 Which refrigeration system is the most likely to have the lowest COP value? (a) A gas liquefaction system producing liquefied natural gas (LNG) at 160C (b) A household refrigerator operating in a kitchen at 25C (c) An air-conditioner keeping a room at 21C when the outside is at 40C (d) An industrial food freezing unit producing ice cream at 35C (e) A food refrigeration unit cooling apples to 3C Answer: (a) A gas liquefaction system producing liquefied natural gas (LNG) at 160C

CYU 12-9 ■ Check Your Understanding CYU 12-9.1 Select the incorrect statement regarding the gas refrigeration cycles. (a) If the heat absorption and rejection processes of the gas refrigeration cycle are replaced by the isothermal processes, the result would be the reversed Carnot cycle. (b) Very low temperatures cannot be obtained in gas refrigeration cycles. (c) The cycle involves simple, lighter components, and this makes them suitable for aircraft cooling. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-466

(d) Gas refrigeration cycles generally have lower COPs than vapor-compression refrigeration cycles. Answer: (b) Very low temperatures cannot be obtained in gas refrigeration cycles. CYU 12-9.2 A gas refrigeration cycle uses air as the working medium. The turbine produces 20 kJ / kg work and compressor consumes 25 kJ / kg work. The cycle removes 10 kJ / kg heat from the cooled space. What is the COP of this cycle and the heat rejected to the warm medium? (a) 0.4, 35 kJ / kg (b) 0.5, 10 kJ / kg (c) 1.0, 30 kJ / kg (d) 2.0, 15 kJ / kg (e) 2.5, 45 kJ / kg Answer: (d) 2.0, 15 kJ / kg COP = q_L / (w_comp - w_turb ) = 10 / (25 - 20) = 2.0

CYU 12-10 ■ Check Your Understanding CYU 12-10.1 Select the correct statement(s) regarding absorption refrigeration systems? I They are heat-driven systems. II The pump work is usually negligible. III They involve a pump, but no compressor. (a) Only I (b) Only II (c) I and III (d) II and III (e) I, II, and III Answer: (e) I, II, and III CYU 12-10.2 Select the correct relation(s) which can be used to determine the overall COP of a reversible absorption refrigeration system. hth,rev : Thermal efficiency of reversible heat engine, qL : Heat removal from the cooed space, qgen : Heat input in the generator, TL : Temperature of cooled space, T0 : Temperature of ambient air I

COPabs,rev = hth,revCOPR,rev

COPabs,rev = qL /qgen

III

COPabs,rev = TL /(T0 - TL )

(a) I and II (b) I and III (c) II and III (d) I, II and III (e) Only I Answer: (a) I and II

10-467

PROBLEMS The Reversed Carnot Cycle 12-1C The reversed Carnot cycle serves as a standard against which actual refrigeration cycles can be compared. Also, the COP of the reversed Carnot cycle provides the upper limit for the COP of a refrigeration cycle operating between the specified temperature limits.

12-2C Because the compression process involves the compression of a liquid-vapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of high-moisture content refrigerant.

12-3 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Analysis (a) Noting that TH = 60°C = 333 K and TL = Tsat @180 kPa = - 12.73°C = 260.3 K , the COP of this Carnot refrigerator is determined from COPR, C =

1 1 = = 3.58 TH / TL - 1 (333 K ) / (260.3 K )- 1

(b) From the refrigerant tables (Table A-11), h3 = hg @ 60°C = 278.47 kJ/kg

h4 = h f @ 60°C = 139.38 kJ/kg Thus, æ260.3 K ÷ ö qH T T = H ¾¾ ® qL = L qH = çç ÷ ÷(139.1 kJ/kg ) = 108.7 kJ / kg çè 333 K ø qL TL TH

10-468

wnet = qH - qL = 139.1- 108.7 = 30.4 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" x[3]=1 x[4]=0 T[3]=60 [C] P[2]=180 [kPa] "Analysis" Fluid$='R134a' "(a)" T_H=T[3]+273 "[K]" T[2]=temperature(Fluid$, P=P[2], x=1) T_L=T[2]+273 "[K]" COP_R_C=1/((T_H/T_L)-1) "(b)" h[3]=enthalpy(Fluid$, T=T[3], x=x[3]) T[4]=T[3] h[4]=enthalpy(Fluid$, T=T[4], x=x[4]) q_H=h[3]-h[4] q_L=q_H*(T_L/T_H) "(c)" w_net=q_H-q_L

12-4E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net work input are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78°F = 532.8 R and TL = Tsat @ 30 psia = 15.37°F = 475.4 R . COPR, C =

1 1 = = 8.28 TH / TL - 1 (532.8 R )/ (475.4 R )- 1

(b) Process 4-1 is isentropic, and thus

s1 = s4 = (s f + x4 s fg )

@ 90 psia

= 0.07481 + (0.05)(0.14529)

10-469

= 0.08208 Btu/lbm ×R æs1 - s f ö ÷ ÷ x1 = ççç ÷ ÷ ÷ èç s fg ø

@ 30 psia

0.08208 - 0.03792 = 0.237 0.18595

(c) Remembering that on a T-s diagram the area enclosed represents the net work, and s3 = sg @90 psia = 0.22011Btu / lbm ×R ,

wnet, in = (TH - TL )(s3 - s4 ) = (72.78 - 15.37) (0.22011- 0.08208) Btu/lbm ×R = 7.92 Btu / lbm

EES (Engineering Equation Solver) SOLUTION "Given" x[3]=1 x[4]=0.05 P[3]=90 [psia] P[2]=30 [psia] "Analysis" Fluid$='R134a' "(a)" T[3]=temperature(Fluid$, P=P[3], x=x[3]) T_H=T[3]+460 "[R]" T[2]=temperature(Fluid$, P=P[2], x=1) T_L=T[2]+460 "[R]" COP_R_C=1/((T_H/T_L)-1) "(b)" T[4]=T[3] s[4]=entropy(Fluid$, T=T[4], x=x[4]) s[1]=s[4] P[1]=P[2] x[1]=quality(Fluid$, P=P[1], s=s[1]) x_1=x[1] "(c)" s[3]=entropy(Fluid$, P=P[3], x=x[3]) w_net_in=(T_H-T_L)*(s[3]-s[4])

Ideal and Actual Vapor-Compression Refrigeration Cycles 12-5C Yes; the throttling process is an internally irreversible process.

12-6C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.

12-7C Allowing a temperature difference of 10C for effective heat transfer, the condensation temperature of the refrigerant should be 25C. The saturation pressure corresponding to 25C is 0.67 MPa. Therefore, the recommended pressure would be 0.7 MPa.

12-8C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for the reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-470

12-9C The cycle that involves saturated liquid at 30C will have a higher COP because, judging from the T-s diagram, it will require a smaller work input for the same refrigeration capacity.

12-10C No. Assuming the water is maintained at 10C in the evaporator, the evaporator pressure will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design refrigeration or air-conditioning devices that involve such extremely low pressures.

12-11C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.

12-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 400 kPa sat. vapor

ïü ïý ïïþ

h1 = hg @ 400 kPa = 255.61 kJ/kg s1 = sg @ 400 kPa = 0.92711 kJ/kg ×K

P2 = 800 kPa s2 = s1

ïü ïý ïïþ

h2 = 269.99 kJ/kg

P3 = 800 kPa sat. liquid

ïü ïý ïïþ

h3 = h f @ 800 kPa = 95.48 kJ/kg

h4 @ h3 = 95.48 kJ/kg

(throttling)

The mass flow rate of the refrigerant is determined from QL = m(h1 - h4 ) ¾ ¾ ® m=

QL 10 kJ/s = = 0.06245 kg / s h1 - h4 (255.61- 95.48) kJ/kg

The power requirement is

10-471

12-13 An air conditioner operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The evaporating temperature will be 22-2=20C. From the refrigerant tables (Tables A-11, A-12, and A-13), ïü ïý ïïþ

T1 = 20°C sat. vapor

h1 = hg @ 20°C = 261.64 kJ/kg s1 = sg @ 20°C = 0.92254 kJ/kg ×K

P2 = 1 MPa s2 = s1

ïü ïý ïïþ

h2 = 273.18 kJ/kg

P3 = 1 MPa sat. liquid

ïü ïý ïïþ

h3 = h f @ 1 MPa = 107.34 kJ/kg

h4 @ h3 = 107.34 kJ/kg (throttling) The COP of the air conditioner is determined from its definition, COPAC =

qL h -h 261.64 - 107.34 = 1 4= = 13.4 win h2 - h1 273.18 - 261.64

12-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system and the cooling load are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-472

ïü ïý ïïþ

h1 = hg @ - 10°C = 244.55 kJ/kg s1 = sg @ - 10°C = 0.93782 kJ/kg ×K

P2 = 600 kPa s2 = s1

ïü ïý ïïþ

h2 = 267.19 kJ/kg

P3 = 600 kPa sat. liquid

ïü ïý ïïþ

h3 = h f @ 600 kPa = 81.50 kJ/kg

h4 @ h3 = 81.50 kJ/kg (throttling) The COP of the air conditioner is determined from its definition, COPAC =

qL h -h 244.55 - 81.50 = 1 4= = 7.20 win h2 - h1 267.19 - 244.55

The cooling load is

QL = m(h1 - h4 ) = (7 kg/s)(244.55 - 81.50) kJ/kg = 1141 kW

12-15E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The increase in the COP if the throttling process were replaced by an isentropic expansion is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-473

P2 = 300 psiaü ïï ý h2 = 125.70 Btu/lbm ïïþ s2 = s1 P3 = 300 psiaïü ïý h3 = h f @300 psia = 66.347 Btu/lbm ïïþ s3 = s f @300 psia = 0.12717 Btu/lbm ×R sat. liquid h4 @ h3 = 66.347 Btu/lbm (throttling)

T4 = 20°Fïü ïý h4 s = 59.81 Btu/lbm (isentropic expansion) s4 = s3 ïïþ x4 s = 0.4724 The COP of the refrigerator for the throttling case is COPR =

qL h - h4 106.00 - 66.347 = 1 = = 2.013 win h2 - h1 125.70 - 106.00

The COP of the refrigerator for the isentropic expansion case is COPR =

h - h4 s qL 106.00 - 59.81 = 1 = = 2.344 win h2 - h1 125.70 - 106.00

The increase in the COP by isentropic expansion is

Percent Increase in COP =

2.344 - 3.013 = 0.165 = 16.5% 2.013

EES (Engineering Equation Solver) SOLUTION "Given" P[2]=300 [psia] T[1]=20 [F] "Analysis" Fluid$='R134a' "compressor" x[1]=1 h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" P[3]=P[2] x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "if expansion valve is replaced by an isentropic turbine" s[3]=entropy(Fluid$, P=P[3], x=x[3]) s_s[4]=s[3] T[4]=T[1] h_s[4]=enthalpy(Fluid$, T=T[1], s=s_s[4]) x_4s=quality(Fluid$, T=T[4], s=s_s[4]) "cycle" q_L_v=h[1]-h[4] "in case of valve" q_L_t=h[1]-h_s[4] "in case of turbine" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-474

w_in=h[2]-h[1] COP_R_v=q_L_v/W_in "in case of valve" COP_R_t=q_L_t/w_in "in case of turbine" "Percent Increase" COP_R_increase=(COP_R_t-COP_R_v)/COP_R_v*Convert(,%)

12-16 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13) P4 = 120 kPa ïü ïý h = 95.41 kJ/kg ïïþ 4 x4 = 0.34

h3 = h4 h3 = 95.41 kJ/kg ü ïï ý P3 = 799 kPa @ 800 kPa x3 = 0 (sat. liq.) ïïþ

P2 = P3 P2 = 799 kPa ü ïï ý h2 = 306.92 kJ/kg ïïþ T2 = 70°C

P1 = P4 = 120 kPa ü ïï ý h1 = 236.99 kJ/kg x1 = 1 (sat. vap.) ïïþ

10-475

The mass flow rate of the refrigerant is determined from m=

Win 0.45 kW = = 0.006435 kg / s h2 - h1 (306.92 - 236.99)kJ/kg

= (0.06435 kg/s)(236.99 - 95.41)kJ/kg

= 0.9111 kW COP =

QL 0.9111 kW = = 2.03 0.45 kW Win

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_4=120 [kPa] x_4=0.34 T_2=70 [C] W_dot_in=0.45 [kW] "PROPERTIES" Fluid$='R134a' h_4=enthalpy(Fluid$, P=P_4, x=x_4) x_3=0 "condenser exit is saturated liquid" h_3=h_4 P_3=pressure(Fluid$, h=h_3, x=x_3) P_2=P_3 h_2=enthalpy(Fluid$, P=P_2, T=T_2) x_1=1 "compressor inlet is saturated vapor" P_1=P_4 h_1=enthalpy(Fluid$, P=P_1, x=x_1) "ANALYSIS" m_dot=W_dot_in/(h_2-h_1) Q_dot_L=m_dot*(h_1-h_4) COP=Q_dot_L/W_dot_in 12-17E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-476

T1 = 5°F ü ïï ý sat. vapor ïïþ

h1 = hg @ 5°F = 103.83 Btu/lbm s1 = sg @ 5°F = 0.22488 Btu/lbm ×R

P2 = 180 psia s2 = s1

ïü ïý ïïþ

h2 = 122.03 Btu/lbm

P3 = 180 psia sat. liquid

ü ïï ý ïïþ

h3 = h f @ 180 psia = 51.507 Btu/lbm

@ h3 = 51.507 Btu/lbm (throttling)

The mass flow rate of the refrigerant is QL = m(h1 - h4 ) ¾ ¾ ® m=

QL 45, 000 Btu/h = = 860.0 lbm / h h1 - h4 (103.83 - 51.507) Btu/lbm

The power requirement is

æ 1 kW ö÷ Win = m(h2 - h1 ) = (860.0 lbm/h)(122.03 - 103.83) Btu/lbm çç ÷= 4.59 kW çè3412.14 Btu/h ø÷

EES (Engineering Equation Solver) SOLUTION "Given" x[1]=1 T[1]=5 [F] x[3]=0 P[3]=180 [psia] Q_dot_L=45000 [Btu/h] "Analysis" Fluid$='R134a' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-477

"cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) COP_R=Q_dot_L/(W_dot_in*Convert(kW, Btu/h))

12-1 8E Problem 12-17E is to be repeated if ammonia is used as the refrigerant. Analysis The problem is solved using EES, and the solution is given below. "Given" x[1]=1 T[1]=5 [F] x[3]=0 P[3]=180 [psia] Q_dot_L=45000 [Btu/h] "Analysis" Fluid$='ammonia' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) Solution for ammonia COP_R=4.515 Fluid$='ammonia' m_dot_R=95.8 [lbm / h]

Q _dot_ L = 45000 [Btu / h] W_dot_in = 2.921[kW] Solution for R-134a COP_R=2.878 Fluid$='R134a' m_dot_R=860.1[lbm / h]

Q _dot_ L = 45000 [Btu / h] W_dot_in = 2.921[kW]

12-19 A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of cooling, the power input, and the COP are to be determined. Also, the same parameters are to be determined if the cycle operated on the ideal vapor-compression refrigeration cycle between the same pressure limits. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-478

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13)

Tsat@200 kPa = - 10.1°C ïü ïý h1 = 253.07 kJ/kg T1 = - 10.1 + 10.1 = 0°Cïïþ s1 = 0.9699 kJ/kg ×K P1 = 200 kPa

P2 = 1400 kPa ïü ïý h = 295.95 kJ/kg ïïþ 2 s s1 = s1

Tsat@1400 kPa = 52.4°C ü ïï ý h3 @ h f@48°C = 120.41 kJ/kg T3 = 52.4 - 4.4 = 48°Cïïþ P3 = 1400 kPa

h4 = h3 = 120.41 kJ/kg

hC = 0.88 =

h2 s - h1 h2 - h1 295.95 - 253.07 ¾¾ ® h2 = 301.80 kJ/kg h2 - 253.07

QL = m(h1 - h4 ) = (0.025 kg/s)(253.07 - 120.41) = 3.317 kW QH = m(h2 - h3 ) = (0.025 kg/s)(301.80 - 120.41) = 4.535 kW

Win = m(h2 - h1 ) = (0.025 kg/s)(301.80 - 253.07) = 1.218 kW

COP =

QL 3.317 kW = = 2.723 Win 1.218 kW

(b) Ideal vapor-compression refrigeration cycle solution

10-479

From the refrigerant-134a tables (Tables A-11 through A-13)

P1 = 200 kPa üïï h1 = 244.50 kJ/kg ý ïïþ s1 = 0.93788 kJ/kg ×K x1 = 1 P2 = 1400 kPa üïï ý h2 = 285.13 kJ/kg ïïþ s1 = s1 P3 = 1400 kPa ïü ïý h = 127.25 kJ/kg ïïþ 3 x3 = 0

h4 = h3 = 127.25 kJ/kg QL = m(h1 - h4 ) = (0.025 kg/s)(244.50 - 127.25) = 2.931 kW QH = m(h2 - h3 ) = (0.025 kg/s)(285.13 - 127.25) = 3.947 kW

Win = m(h2 - h1 ) = (0.025 kg/s)(285.13 - 244.50) = 1.016 kW

COP =

QL 2.931 kW = = 2.885 Win 1.016 kW

Discussion The cooling load decreases by 11.6% while the COP increases by 5.9% when the cycle operates on the ideal vapor-compression cycle.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=200 [kPa] P_2=1400 [kPa] DELTAT_superheat=10.1 [C] DELTAT_subcool=4.4 [C] eta_C=0.88 m_dot=0.025 [kg/s] "PROPERTIES" Fluid$='R134a' T_sat_1=temperature(Fluid$, P=P_1, x=1) "T_1=T_sat_1+DELTAT_superheat" T_1=0 h_1=enthalpy(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) eta_C=(h_2s-h_1)/(h_2-h_1) T_sat_3=temperature(Fluid$, P=P_2, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-480

T_3=T_sat_3-DELTAT_subcool h_3=enthalpy(Fluid$, T=T_3, x=0) "subcooled state is approximated as saturated liquid at T_3" h_4=h_3 Q_dot_L=m_dot*(h_1-h_4) Q_dot_H=m_dot*(h_2-h_3) W_dot_in=m_dot*(h_2-h_1) COP=Q_dot_L/W_dot_in "Ideal vapor-compression refrigeration cycle solution" x_1=1 h_1_ideal=enthalpy(Fluid$, P=P_1, x=x_1) s_1_ideal=entropy(Fluid$, P=P_1, x=x_1) h_2_ideal=enthalpy(Fluid$, P=P_2, s=s_1_ideal) x_3=0 h_3_ideal=enthalpy(Fluid$, P=P_2, x=x_3) h_4_ideal=h_3_ideal Q_dot_L_ideal=m_dot*(h_1_ideal-h_4_ideal) Q_dot_H_ideal=m_dot*(h_2_ideal-h_3_ideal) W_dot_in_ideal=m_dot*(h_2_ideal-h_1_ideal) COP_ideal=Q_dot_L_ideal/W_dot_in_ideal

12-20 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-481

(a) From refrigerant-134a tables (Tables A-11 through A-13)

P1 = 60 kPa üïï ý h1 = 230.04 kJ/kg T1 = - 34°C ïïþ P2 = 1200 kPa üïï ý h2 = 295.18 kJ/kg ïïþ T2 = 65°C

P3 = 1200 kPa üïï ý h3 = 111.25 kJ/kg ïïþ T3 = 42°C h4 = h3 = 111.25 kJ/kg ïü ïý x = 0.4796 4 h4 = 111.25 kJ/kg ïïþ P4 = 60 kPa

Using saturated liquid enthalpy at the given temperature, for water we have (Table A-4) hw1 = h f @18°C = 75.47 kJ/kg

hw2 = h f @ 26°C = 108.94 kJ/kg (b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor

mR (h2 - h3 ) = mw (hw2 - hw1 ) mR (295.18 - 111.25)kJ/kg = (0.25 kg/s)(108.94 - 75.47)kJ/kg

¾¾ ® mR = 0.04549 kg/s The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are

QH = mR (h2 - h3 ) = (0.04549 kg/s)(295.18 - 111.25)kJ/kg = 8.367 kW Win = mR (h2 - h1 ) - Qin = (0.04549 kg/s)(295.18 - 230.04)kJ/kg - 0.450 kW = 2.513 kW QL = QH - Win - Qin = 8.367 - 2.513 - 0.450 = 5.404 kW (c) The COP of the refrigerator is determined from its definition COP =

QL 5.404 = = 2.15 2.513 Win

(d) The reversible COP of the refrigerator for the same temperature limits is

10-482

COPmax =

1 1 = = 5.063 TH / TL - 1 (18 + 273) / (- 30 + 273) - 1

Then, the maximum refrigeration load becomes

QL, max = COPmaxWin = (5.063)(2.513 kW) = 12.72 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=-30 [C] T_H=18 [C] T_w1=18 [C] m_dot_w=0.25 [kg/s] T_w2=26 [C] P_2=1200 [kPa] T_2=65 [C] P_3=P_2 T_3=42 [C] P_1=60 [kPa] T_1=-34 [C] Q_dot_in=0.45 [kW] "PROPERTIES" Fluid$='R134a' h_1=enthalpy(Fluid$, P=P_1, T=T_1) h_2=enthalpy(Fluid$, P=P_2, T=T_2) h_3=enthalpy(Fluid$, P=P_3, T=T_3) h_4=h_3 x_4=quality(Fluid$, P=P_1, h=h_4) h_w1=enthalpy(Water, T=T_w1, x=0) h_w2=enthalpy(Water, T=T_w2, x=0) "ANALYSIS" Q_dot_H=m_dot_w*(h_w2-h_w1) m_dot_R=Q_dot_H/(h_2-h_3) "mass flow rate of refrigerant" W_dot_in=m_dot_R*(h_2-h_1)-Q_dot_in "energy balance on the compressor" Q_dot_L=Q_dot_H-W_dot_in-Q_dot_in "refrigeration load" Q_dot_L_alt=m_dot_R*(h_1-h_4) COP=Q_dot_L/W_dot_in "actual COP" COP_max=1/((T_H+273)/(T_L+273)-1) "maximum COP" Q_dot_L_max=COP_max*W_dot_in "maximum refrigeration load"

12-21 A vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The hardware and the T-s diagram for this air conditioner are to be sketched. The heat absorbed by the refrigerant, the work input to the compressor and the heat rejected in the condenser are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In this normal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure.

10-483

(b) The properties as given in the problem statement are

h4 = h3 = h f @ 45 C = 101 kJ / kg h1 = hg @- 5 C = 248.1kJ / kg g. The heat absorbed by the refrigerant in the evaporator is

qL = h1 - h4 = 248.1- 101 = 147.1kJ / kg (c) The COP of the air conditioner is

æ 1W ö÷ æ Btu/h öæ ö÷ ÷çç 1 W COPR = SEER çç = çç16 ÷ ÷ ÷= 4.689 ÷ ÷ èç3.412 Btu/h ø èç W øèç3.412 Btu/h ø÷ The work input to the compressor is COPR =

qL qL 147.1 kJ/kg ¾¾ ® win = = = 31.4 kJ / kg win COPR 4.689

The enthalpy at the compressor exit is

win = h2 - h1 ¾ ¾® h2 = h1 + win = 248.1 kJ/kg + 31.4 kJ/kg = 279.5 kJ/kg The heat rejected from the refrigerant in the condenser is then qH = h2 - h3 = 279.5 - 101 = 178.5 kJ / kg

EES (Engineering Equation Solver) SOLUTION "GIVEN" SEER=16 [Btu/h-W] T_1=-5 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-484

T_3=45 [C] "ANALYSIS" Fluid$='R134a' "(b)" h_1=248.1 [kJ/kg] h_3=101 [kJ/kg] h_4=h_3 q_L=h_1-h_4 "(c)" COP_R=SEER*Convert(Btu/h-W, W/W) w_in=q_L/COP_R h_2=h_1+w_in q_H=h_2-h_3

12-22 A vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The amount of cooling, the work input, and the COP are to be determined. Also, the same parameters are to be determined if the cycle operated on the ideal vapor-compression refrigeration cycle between the same temperature limits. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The expansion process through the expansion valve is isenthalpic: h4 = h3 . Then,

qL = h1 - h4 = 402.49 - 243.19 = 159.3 kJ / kg qH = h2 - h3 = 454.00 - 243.19 = 210.8 kJ/kg win = h2 - h1 = 454.00 - 402.49 = 51.51kJ / kg COP =

qL 159.3 kJ/kg = = 3.093 win 51.51 kJ/kg

(b) Ideal vapor-compression refrigeration cycle solution: qL = h1 - h4 = 399.04 - 249.80 = 149.2 kJ / kg qH = h2 - h3 = 440.71- 249.80 = 190.9 kJ/kg win = h2 - h1 = 440.71- 399.04 = 41.67 kJ / kg COP =

qL 149.2 kJ/kg = = 3.582 win 41.67 kJ/kg

10-485

Discussion In the ideal operation, the refrigeration load decreases by 6.3% and the work input by 19.1% while the COP increases by 15.8%.

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_evap=-15 [C] T_cond=40 [C] DELTAT_superheat=5 [C] DELTAT_subcool=5 [C] eta_C=0.83 "ANALYSIS" Fluid$='R22' "(a)" T_1=T_evap+DELTAT_superheat P_1=pressure(Fluid$, T=T_evap, x=1) h_1=enthalpy(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) P_2=pressure(Fluid$, T=T_cond, x=1) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) eta_C=(h_2s-h_1)/(h_2-h_1) T_sat_3=temperature(Fluid$, P=P_2, x=0) T_3=T_sat_3-DELTAT_subcool h_3=enthalpy(Fluid$, T=T_3, P=P_2) h_4=h_3 q_L=h_1-h_4 q_H=h_2-h_3 w_in=h_2-h_1 COP=q_L/w_in "(b)" "Ideal vapor-compression refrigeration cycle solution" x_1=1 h_1_ideal=enthalpy(Fluid$, T=T_evap, x=x_1) s_1_ideal=entropy(Fluid$, T=T_evap, x=x_1) h_2_ideal=enthalpy(Fluid$, P=P_2, s=s_1_ideal) x_3=0 h_3_ideal=enthalpy(Fluid$, P=P_2, x=x_3) h_4_ideal=h_3_ideal q_L_ideal=h_1_ideal-h_4_ideal q_H_ideal=h_2_ideal-h_3_ideal w_in_ideal=h_2_ideal-h_1_ideal COP_ideal=q_L_ideal/w_in_ideal

10-486

Second-Law Analysis of Vapor-Compression Refrigeration Cycles 12-23C The second-law efficiency of a refrigerator operating on the vapor-compression refrigeration cycle is defined as hII, R =

X QL W

X Wmin = 1- dest, total W W

where X QL is the exergy of the heat transferred from the low-temperature medium and it is expressed as æ T ö ÷. X QL = - QL ççç1- 0 ÷ ÷ çè TL ÷ ø

X dest, total is the total exergy destruction in the cycle and W is the actual power input to the cycle. The second-law efficiency can also be expressed as the ratio of the actual COP to the Carnot COP: hII, R =

COPR COPCarnot

12-24C The second-law efficiency of a heat pump operating on the a vapor-compression refrigeration cycle is defined as hII, HP =

X QH W

X Wmin = 1- dest, total W W

Substituting W=

QH COPHP

and

æ T ö ÷ X QH = QH çç1- 0 ÷ ÷ ÷ çè TH ø

into the second-law efficiency equation

æ T ö ÷ QH ççç1- 0 ÷ ÷ æ T öCOPHP X QH ÷ COPHP COPHP è TH ø ÷ hII, HP = = = QH çç1- 0 ÷ = = ÷ ÷ T ç W QH COPCarnot è TH ø QH H TH - TL COPHP since T0 = TL for a heat pump.

12-25C In an isentropic compressor, s2 = s1 and h2 s = h2 . Applying these to the two the efficiency definitions, we obtain hs, Comp =

wisen h - h1 h - h1 = 2s = 2 = 1 = 100% w h2 - h1 h2 - h1

hII, Comp =

wrev h - h1 - T0 ( s2 - s1 ) h2 - h1 = 2 = = 1 = 100% w h2 - h1 h2 - h1

Thus, the isentropic efficiency and the second-law efficiency of an isentropic compressor are both 100%. The second-law efficiency of a compressor is not necessarily equal to its isentropic efficiency. The two definitions are different as shown in the above equations. In the calculation of isentropic efficiency, the exit enthalpy is found at the hypothetical exit state (at the exit pressure and the inlet entropy) while the second-law efficiency involves the actual exit state. The two efficiencies are usually close but different. In the special case of an isentropic compressor, the two efficiencies become equal to each other as proven above.

10-487

12-26 A vapor-compression refrigeration system is used to keep a space at a low temperature. The power input, the COP and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The power input is æ 1 kW ö ÷ Win = QH - QL = 5500 - 3500 = 2000 kJ/h = (2000 kJ/h) çç ÷ ÷= 0.5556 kW çè3600 kJ/h ø

The COP is COPR =

QL (3500 / 3600) kW = = 1.75 0.5556 kW Win

The COP of the Carnot cycle operating between the space and the ambient is COPCarnot =

TL 258 K = = 6.45 TH - TL (298 - 258) K

The second-law efficiency is then hII =

COPR 1.75 = = 0.271 = 27.1% COPCarnot 6.45

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=(-15+273) [K] T_H=(25+273) [K] Q_dot_L=3500 [kJ/h] Q_dot_H=5500 [kJ/h] "ANALYSIS" W_dot_in=(Q_dot_H-Q_dot_L)*Convert(kJ/h, kW) COP=Q_dot_L*Convert(kJ/h, kW)/W_dot_in COP_rev=T_L/(T_H-T_L) eta_II=COP/COP_rev

12-27 A refrigerator is used to cool bananas at a specified rate. The rate of heat absorbed from the bananas, the COP, The minimum power input, the second-law efficiency and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The rate of heat absorbed from the bananas is

QL = mc p (T1 - T2 ) = (1330 kg/h)(3.35 kJ/kg ×°C)(28 - 12)°C = 71,300 kJ / h The COP is COP =

QL (71,300 / 3600) kW 19.80 kW = = = 2.30 8.6 kW 8.6 kW Win

(b) The minimum power input is equal to the exergy of the heat transferred from the low-temperature medium:

10-488 æ T0 ÷ ö æ 28 + 273 ÷ ö ÷= - (19.80 kW) ç X QL = - QL ç 1= 0.541 kW ç1÷ ç ÷ ÷ ç ÷ ç è 20 + 273 ø è TL ø

where the dead state temperature is taken as the inlet temperature of the eggplants ( T0 = 28°C ) and the temperature of the low-temperature medium is taken as the average temperature of bananas T = (12 + 28) / 2 = 20 C . (c) The second-law efficiency of the cycle is hII =

X QL Win

0.541 = 0.0629 = 6.29% 8.6

The exergy destruction is the difference between the exergy expended (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium):

X dest = Win - X QL = 8.6 - 0.541 = 8.06 kW EES (Engineering Equation Solver) SOLUTION "GIVEN" T1=(28+273) [K] T2=(12+273) [K] m_dot=1330 [kg/h]*Convert(kg/h,kg/s) c_p=3.35 [kJ/kg-K] W_dot_in=8.6 [kW] "ANALYSIS" "(a)" Q_dot_L=m_dot*c_p*(T1-T2) COP=Q_dot_L/W_dot_in T_0=T1 "(b)" Ex_QL=-Q_dot_L*(1-T_0/T) "Rate of exergy of heat removal Q_L" T=0.5*(T1+T2) "Effective T_L" "(c)" eta_II=Ex_QL/W_dot_in Ex_dot_dest=W_dot_in-Ex_QL

12-28 A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The power input, the mass flow rate of water in the condenser, the second-law efficiency, and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The power input is

æ 1 kW ÷ ö (24, 000 Btu/h) çç ÷ ÷ ç QL è3412 Btu/h ø 7.034 kW Win = = = = 3.431 kW COP 2.05 2.05 (b) From an energy balance on the cycle, QH = QL + Win = 7.034 + 3.431 = 10.46 kW

The mass flow rate of the water is then determined from

10-489 QH = mc pwD Tw ¾ ¾ ® m=

QH 10.46 kW = = 0.2086 kg / s c pwD Tw (4.18 kJ/kg ×°C)(12°C)

(c) The exergy of the heat transferred from the low-temperature medium is æ T0 ÷ ö æ 20 + 273 ö ç1÷ ÷ X QL = - QL ç = - (7.034 kW) ç ÷= 0.5153 kW ç1÷ ç ç è ø TL ÷ 0 + 273 ÷ è ø

The second-law efficiency of the cycle is hII =

ExQL Win

0.5153 = 0.1502 = 15.0% 3.431

The exergy destruction is the difference between the exergy supplied (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium):

X dest = Win - X QL = 3.431- 0.5153 = 2.916 kW Alternative Solution The exergy efficiency can also be determined as follows: COPR, Carnot = hII =

TL 0 + 273 = = 13.65 TH - TL 20 - 0

COP 2.05 = = 0.1502 = 15.0% COPR, Carnot 13.65

The result is the same, as expected.

EES (Engineering Equation Solver) SOLUTION "GIVEN" Q_dot_L=24000 [Btu/h]*Convert(Btu/h, kW) T_0=(20+273) [K] DELTAT_w=12 [C] COP=2.05 c_p=4.18 [kJ/kg-C] T_L=(0+273) [K] "ANALYSIS" W_dot_in=Q_dot_L/COP Q_dot_H=Q_dot_L+W_dot_in Q_dot_H=m_dot*c_p*DELTAT_w Ex_dot_QL=-Q_dot_L*(1-T_0/T_L) "Rate of exergy of heat removal Q_L" eta_II=Ex_dot_QL/W_dot_in Ex_dot_dest=W_dot_in-Ex_dot_QL "Alternative Solution" COP_rev=T_L/(T_0-T_L) eta_II_alt=COP/COP_rev 12-29 A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The mass flow rate of R-134a, the COP, The exergy destruction in each component and the exergy efficiency of the compressor, the second-law efficiency, and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-490

Analysis (a) The properties of R-134a are (Tables A-11 through A-13)

P2 = 1.2 MPa ïüï h2 = 278.28 kJ/kg ý T2 = 50°C ïïþ s2 = 0.9268 kJ/kg ×K P3 = 1.2 MPa ïü ïý h3 = 117.79 kJ/kg ïïþ s3 = 0.4245 kJ/kg ×K x3 = 0

The rate of heat transferred to the water is the energy change of the water from inlet to exit

QH = mwc p (Tw,2 - Tw,1 ) = (0.13 kg/s)(4.18 kJ/kg ×°C)(28 - 20)°C = 4.347 kW The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser. That is, QH = mR (h2 - h3 ) ¾ ¾ ® mR =

QH 4.347 kW = = 0.02709 kg/s h2 - h3 (278.28 - 117.79) kJ/kg

The refrigeration load is æ3412 Btu/h ÷ ö QL = QH - Win = 4.347 - 1.9 = 2.447 kW = (2.447 kW) çç = 8350 Btu / h ÷ çè 1 kW ÷ ø

The COP of the refrigerator is determined from its definition, COP =

QL 2.447 kW = = 1.29 1.9 kW Win

(b) The COP of a reversible refrigerator operating between the same temperature limits is COPCarnot =

TL - 5 + 273 = = 10.72 TH - TL (20 + 273) - (- 5 + 273)

The minimum power input to the compressor for the same refrigeration load would be Win, min =

QL 2.447 kW = = 0.2283 kW COPCarnot 10.72

The second-law efficiency of the cycle is

hII =

Win, min Win

0.2283 = 0.120 = 12.0% 1.9

The total exergy destruction in the cycle is the difference between the actual and the minimum power inputs:

X dest, total = Win - Win, min = 1.9 - 0.2283 = 1.67 kW (c) The entropy generation in the condenser is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-491

æT ÷ ö Sgen, cond = mw c p ln ççç w,2 ÷ + mR ( s3 - s2 ) ÷ çèTw,1 ÷ ø æ28 + 273 ÷ ö = (0.13 kg/s)(4.18 kJ/kg ×°C)ln çç + (0.02709 kg/s)(0.4245 - 0.9268) kJ/kg ×K ÷ çè 20 + 273 ÷ ø

= 0.001032 kW/K The exergy destruction in the condenser is

X dest, cond = T0 Sgen, cond = (293 K)(0.001032 kW/K) = 0.303 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=(-5+273) [K] T_w1=(20+273) [K] m_dot_w=0.13 [kg/s] T_w2=(28+273) [K] P_1=1200 [kPa] T_1=50 [C] P_2=1200 [kPa] W_dot_in=1.9 [kW] T_0=(20+273) [K] T_H=T_0 "PROPERTIES" Fluid1$='R134a' Fluid2$='Steam_iapws' h_1=enthalpy(Fluid1$, P=P_1, T=T_1) s_1=entropy(Fluid1$, P=P_1, T=T_1) T_sat=temperature(Fluid1$, P=P_2, x=0) h_2=enthalpy(Fluid1$, P=P_2, x=0) s_2=entropy(Fluid1$, P=P_2, x=0) "ANALYSIS" c_p=4.18 [kJ/kg-C] "Table A-3" Q_dot_H=m_dot_w*c_p*(T_w2-T_w1) m_dot_R=Q_dot_H/(h_1-h_2) "mass flow rate of refrigerant" Q_dot_L=Q_dot_H-W_dot_in "refrigeration load" Q_dot_L_Btuh=Q_dot_L*Convert(kW, Btu/h) COP=Q_dot_L/W_dot_in "actual COP" COP_max=T_L/(T_H-T_L) eta_II=COP/COP_max eta_II=W_dot_in_min/W_dot_in Ex_dot_dest_total=W_dot_in-W_dot_in_min S_gen_condenser=DELTAS_w+DELTAS_R DELTAS_w=m_dot_w*c_p*ln(T_w2/T_w1) DELTAS_R=m_dot_R*(s_2-s_1) Ex_dot_dest_condenser=T_0*S_gen_condenser

12-30 An ideal vapor-compression refrigeration cycle is used to keep a space at a low temperature. The cooling load, the COP, the exergy destruction in each component, the total exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-492

Kinetic and potential energy changes are negligible.

Analysis (a) The properties of R-134a are (Tables A-11 through A-13) T1 = - 10°C ïü ïý h1 = 244.55 kJ/kg ïïþ s1 = 0.9378 kJ/kg ×K x1 = 1 P2 = Psat@57.9°C = 1600 kPa ïü ïý h = 287.89 kJ/kg ïï 2 s2 = s1 þ P3 = 1600 kPa ïü ïý h3 = 135.96 kJ/kg ïïþ s3 = 0.4792 kJ/kg ×K x3 = 0

h4 = h3 = 135.96 kJ/kg ü ïï ý s4 = 0.5252 kJ/kg ×K h4 = 135.96 kJ/kgïïþ

T4 = - 10°C

The energy interactions in the components and the COP are qL = h1 - h4 = 244.55 - 135.96 = 108.6 kJ / kg

qH = h2 - h3 = 287.89 - 135.96 = 151.9 kJ/kg win = h2 - h1 = 287.89 - 244.55 = 43.34 kJ/kg

COP =

qL 108.6 kJ/kg = = 2.506 win 43.34 kJ/kg

(b) The exergy destruction in each component of the cycle is determined as follows Compressor:

sgen, 1- 2 = s2 - s1 = 0 xdest, 1- 2 = T0 sgen, 1- 2 = 0 Condenser: sgen, 2- 3 = s3 - s2 +

qH 151.9 kJ/kg = (0.4792 - 0.9378) kJ/kg ×K + = 0.05125 kJ/kg ×K TH 298 K

xdest, 2- 3 = T0 sgen, 2- 3 = (298 K)(0.05125 kJ/kg ×K) = 15.27 kJ / kg Expansion valve:

10-493

xdest, 3- 4 = T0 sgen, 3- 4 = (298 K)(0.04595 kJ/kg ×K) = 13.69 kJ / kg Evaporator: sgen, 4- 1 = s1 - s4 -

qL 108.6 kJ/kg = (0.9378 - 0.5252) kJ/kg ×K = 0.02202 kJ/kg ×K TL 278 K

xdest, 4- 1 = T0 sgen, 4- 1 = (298 K)(0.02202 kJ/kg ×K) = 6.56 kJ / kg The total exergy destruction can be determined by adding exergy destructions in each component:

xdest, total = xdest, 1- 2 + xdest, 2- 3 + xdest, 3- 4 + xdest, 4- 1

= 0 + 15.27 + 13.69 + 6.56 = 35.52 kJ / kg (c) The exergy of the heat transferred from the low-temperature medium is æ T0 ÷ ö æ 298 ö ÷ ÷= - (108.6 kJ/kg) ç xqL = - qL ç 1ç1÷= 7.813 kJ/kg ç ÷ ç ÷ ç è ø T 278 ÷ è Lø

The second-law efficiency of the cycle is hII =

xqL win

7.813 = 0.180 = 18.0% 43.34

The total exergy destruction in the cycle can also be determined from

xdest, total = win - xqL = 43.34 - 7.813 = 35.53 kJ/kg The result is practically identical as expected. The second-law efficiency of the compressor is determined from hII, Comp =

m [h2 - h1 - T0 ( s2 - s1 )] X recovered Wrev = = m(h2 - h1 ) X expended Wact, in

since the compression through the compressor is isentropic ( s2 = s1 ), the second-law efficiency is

hII, Comp = 1 = 100% The second-law efficiency of the evaporator is determined from hII, Evap =

X dest,4-1 X recovered QL (T0 - TL ) / TL = = 1m [h4 - h1 - T0 ( s4 - s1 ) ] X expended X 4 - X1

where

xflow,4 - xflow,1 = h4 - h1 - T0 (s4 - s1 ) = (135.96 - 244.55) kJ/kg - (298 K)(0.5252 - 0.9378) kJ/kg ×K

= 14.37 kJ/kg Substituting,

hII, Evap = 1-

xdest, 4-1 xflow,4 - xflow,1

= 1-

6.56 kJ/kg = 0.544 = 54.4% 14.37 kJ/kg

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_1=-10 [C] T_3=57.9 [C] T_L=(5+273) [K] T_H=(25+273) [K] T_0=T_H © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-494

"PROPERTIES" Fluid$='R134a' x_1=1 x_3=0 h_1=enthalpy(Fluid$, T=T_1, x=x_1) s_1=entropy(Fluid$, T=T_1, x=x_1) "P_2=pressure(Fluid$, T=T_3, x=x_1)" P_2=1600 h_2=enthalpy(Fluid$, P=P_2, s=s_1) s_2=s_1 h_3=enthalpy(Fluid$, P=P_2, x=x_3) s_3=entropy(Fluid$, P=P_2, x=x_3) h_4=h_3 s_4=entropy(Fluid$, T=T_1, h=h_4) q_L=h_1-h_4 q_H=h_2-h_3 w_in=h_2-h_1 COP=q_L/w_in s_gen_12=s_2-s_1 Ex_dest_12=T_0*s_gen_12 s_gen_23=s_3-s_2+q_H/T_H Ex_dest_23=T_0*s_gen_23 s_gen_34=s_4-s_3 Ex_dest_34=T_0*s_gen_34 s_gen_41=s_1-s_4-q_L/T_L Ex_dest_41=T_0*s_gen_41 Ex_qL=-q_L*(1-T_0/T_L) eta_II=Ex_qL/w_in Ex_dest=w_in-Ex_qL DELTAex_12=h_2-h_1-T_0*(s_2-s_1) eta_II_comp=DELTAex_12/w_in DELTAex_41=h_4-h_1-T_0*(s_4-s_1) eta_II_evap=1-Ex_dest_41/DELTAex_41 eta_II_evap_alt=(q_L*(T_0-T_L)/T_L)/DELTAex_41

12-31E A vapor-compression refrigeration cycle is used to keep a space at a low temperature. The mass flow rate of R-134a, the COP, the exergy destruction in each component, the second-law efficiency of the compressor, and the second-law efficiency and exergy destruction of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-495

Analysis (a) The properties of R-134a are (Tables A-11E through A-13E) P1 = 20 psia ïü ïý h1 = 102.74 Btu/lbm ïïþ s1 = 0.2257 Btu/lbm ×R x1 = 1 P2 = 140 psia ü ïï h2 = 131.37 Btu/lbm ý T2 = 160 °F ïïþ s2 = 0.2444 Btu/lbm ×R P3 = 140 psia ü ïï h3 = 45.31 Btu/lbm ý ïïþ s3 = 0.09215 Btu/lbm ×R x3 = 0

h4 = h3 = 45.31 Btu/lbm ü ïï ý s4 = 0.1001 Btu/lbm ×R h4 = 45.31 Btu/lbm ïïþ P4 = 20 psia

The energy interactions in each component and the mass flow rate of R-134a are win = h2 - h1 = 131.37 - 102.74 = 28.63 Btu/lbm

qH = h2 - h3 = 131.37 - 45.31 = 86.06 Btu/lbm qL = qH - win = 86.06 - 28.63 = 57.43 Btu/lbm

QL (45, 000 / 3600) Btu/s = = 0.2177 lbm / s qL 57.43 Btu/lbm

The COP is COP =

qL 57.43 Btu/lbm = = 2.006 win 28.63 Btu/lbm

(b) The exergy destruction in each component of the cycle is determined as follows: Compressor: sgen, 1- 2 = s2 - s1 = 0.2444 - 0.2257 = 0.01874 Btu/lbm ×R X dest, 1- 2 = mT0 sgen, 1- 2 = (0.2177 lbm/s)(540 R)(0.01874 Btu/lbm ×R) = 2.203 Btu / s

Condenser: sgen, 2- 3 = s3 - s2 +

qH 86.06 Btu/lbm = (0.09215 - 0.2444) Btu/lbm ×R + = 0.007073 Btu/lbm ×R TH 540 R

X dest, 2- 3 = mT0 sgen, 2- 3 = (0.2177 lbm/s)(540 R)(0.007073 Btu/lbm ×R) = 0.8313 Btu / s © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-496

Expansion valve: sgen, 3- 4 = s4 - s3 = 0.1001- 0.09215 = 0.007961 Btu/lbm ×R

X dest, 3- 4 = mT0 sgen, 3- 4 = (0.2177 lbm/s)(540 R)(0.007961 Btu/lbm ×R) = 0.9358 Btu / s Evaporator: sgen, 4- 1 = s1 - s4 -

qL 57.43 Btu/lbm = (0.2257 - 0.1001) Btu/lbm ×R = 0.003400 Btu/lbm ×R TL 470 R

X dest, 4- 1 = mT0 sgen, 4- 1 = (0.2177 lbm/s)(540 R)(0.003400 Btu/lbm ×R) = 0.3996 Btu / s The power input and the second-law efficiency of the compressor is determined from

Win = mwin = (0.2177 lbm/s)(28.63 Btu/lbm) = 6.232 Btu/s hII = 1-

X dest, 1- 2 Win

= 1-

2.203 Btu/s = 0.6465 = 64.7% 6.232 Btu/s

(c) The exergy of the heat transferred from the low-temperature medium is æ T0 ÷ ö æ 540 ö ÷ ÷= - (45000 / 3600 Btu/s) ç X QL = - QL ç 1ç1÷= 1.862 Btu/s ç ç ÷ ç TL ÷ è ø 470 ÷ è ø

The second-law efficiency of the cycle is hII =

X QL Win

1.862 Btu/s = 0.2988 = 29.9% 6.232 Btu/s

The total exergy destruction in the cycle is the difference between the exergy supplied (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium):

X dest, total = Win - X QL = 6.232 - 1.862 = 4.370 Btu / s The total exergy destruction can also be determined by adding exergy destructions in each component: X dest, total = X dest, 1- 2 + X dest, 2- 3 + X dest, 3- 4 + X dest, 4- 1

= 2.203 + 0.8313 + 0.9358 + 0.3996 = 4.370 Btu/s The result is the same as expected.

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=(10+460) [R] T_H=(80+460) [R] P_1=20 [psia] x_1=1 P_2=140 [psia] T_2=160 [F] x_3=0 Q_dot_L=45000 [Btu/h]*Convert(Btu/h, Btu/s) T_0=T_H "ANALYSIS" Fluid$='R134a' "(a)" h_1=enthalpy(Fluid$, P=P_1, x=x_1) s_1=entropy(Fluid$, P=P_1, x=x_1) h_2=enthalpy(Fluid$, P=P_2, T=T_2) s_2=entropy(Fluid$, P=P_2, T=T_2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-497

h_3=enthalpy(Fluid$, P=P_2, x=x_3) s_3=entropy(Fluid$, P=P_2, x=x_3) h_4=h_3 s_4=entropy(Fluid$, P=P_1, h=h_4) w_in=h_2-h_1 q_H=h_2-h_3 q_L=q_H-w_in m_dot=Q_dot_L/q_L COP=q_L/w_in W_dot_in=m_dot*(h_2-h_1) "(b)" s_gen_12=s_2-s_1 Ex_dot_dest_12=m_dot*T_0*s_gen_12 s_gen_23=s_3-s_2+q_H/T_H Ex_dot_dest_23=m_dot*T_0*s_gen_23 s_gen_34=s_4-s_3 Ex_dot_dest_34=m_dot*T_0*s_gen_34 s_gen_41=s_1-s_4-q_L/T_L Ex_dot_dest_41=m_dot*T_0*s_gen_41 eta_ex_Comp=1-Ex_dot_dest_12/W_dot_in "(c)" Ex_dot_QL=-Q_dot_L*(1-T_0/T_L) eta_II=Ex_dot_QL/W_dot_in Ex_dot_dest=W_dot_in-Ex_dot_QL

Selecting the Right Refrigerant 12-32C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above the atmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above the temperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic, noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes the mass flow rate) and, of course, being available at low cost.

12-33C Allowing a temperature difference of 10C for effective heat transfer, the evaporation temperature of the refrigerant should be −20C. The saturation pressure corresponding to −20C is 0.133 MPa. Therefore, the recommended pressure would be 0.12 MPa.

12-34C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure of the refrigerant at 30C, which is 0.771 MPa. At lower pressures, the refrigerant will have to condense at temperatures lower than the temperature of the surroundings, which cannot happen.

12-35 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Allowing a temperature difference of 10C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be −20C and 35C, respectively. The saturation pressures corresponding to these © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-498

temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommended evaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively.

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_space=-10 [C] T_env=25 [C] "Analysis" Fluid$='R134a' DELTAT=10 [C] "assumed" T_evap=T_space-DELTAT T_cond=T_env+DELTAT P_evap=pressure(Fluid$, T=T_evap, x=0) P_cond=pressure(Fluid$, T=T_cond, x=0)

12-36 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. Reasonable pressures for the evaporator and the condenser are to be selected. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis Allowing a temperature difference of 10C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be 4C and 36C, respectively. The saturation pressures corresponding to these temperatures are 338 kPa and 912 kPa. Therefore, the recommended evaporator and condenser pressures are 338 kPa and 912 kPa, respectively.

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_house=26 [C] T_geo=14 [C] "Analysis" Fluid$='R134a' DELTAT=10 [C] "assumed" T_evap=T_geo-DELTAT T_cond=T_house+DELTAT P_evap=pressure(Fluid$, T=T_evap, x=0) P_cond=pressure(Fluid$, T=T_cond, x=0)

Heat Pump Systems 12-37C A heat pump system is more cost effective in Miami because of the low heating loads and high cooling loads at that location.

12-38C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps have higher COPs than the air-source systems because the temperature of water is higher than the temperature of air in winter. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-499

12-39 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

ïü ïý ïïþ

h1 = hg @ 200 kPa = 244.50 kJ/kg s1 = sg @ 200 kPa = 0.93788 kJ/kg ×K

P2 = 1000 kPa s2 = s1

ü ïï ý ïïþ

h2 = 278.07 kJ/kg

P3 = 1000 kPa sat. liquid

ïü ïý ïïþ

h3 = h f @ 1000 kPa = 107.34 kJ/kg

h4 @ h3 = 107.34 kJ/kg (throttling) The mass flow rate of the refrigerant is determined from Win = m(h2 - h1 ) ¾ ¾ ® m=

Win 6 kJ/s = = 0.1787 kg/s h2 - h1 (278.07 - 244.50) kJ/kg

Then the rate of heat supplied to the evaporator is

QL = m(h1 - h4 ) = (0.1787 kg/s)(244.50 - 107.34) kJ/kg = 24.5 kW The COP of the heat pump is determined from its definition, COPHP =

h - h3 qH 278.07 - 107.34 = 2 = = 5.09 win h2 - h1 278.07 - 244.50

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=200 [kPa] P[2]=1000 [kPa] W_dot_in=6 [kW] "Analysis" Fluid$='R134a' "compressor" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-500

x[1]=1 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" P[3]=P[2] x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" W_dot_in=m_dot*(h[2]-h[1]) Q_dot_H=m_dot*(h[2]-h[3]) Q_dot_L=m_dot*(h[1]-h[4]) COP_HP=Q_dot_H/W_dot_in

12-40 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the same pressure limits are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis

(a) The properties of refrigerant-134a are (Tables A-11 through A-13) P2 = 800 kPa ïü ïý h = 286.71 kJ/kg ïïþ 2 T2 = 50°C

T3 = Tsat@750 kPa = 29.06°C ü ïï ý h3 = 87.93 kJ/kg T3 = (29.06 - 3)°Cïïþ P3 = 750 kPa

h4 = h3 = 87.93 kJ/kg

Tsat@200 kPa = - 10.09°C ü ïï h1 = 247.88 kJ/kg ý T1 = (- 10.09 + 4)°Cïïþ s1 = 0.9507 kJ/kg ×K P1 = 200 kPa

10-501

P2 = 800 kPa ü ïï ý h2 s = 277.28 ïïþ s2 = s1

The isentropic efficiency of the compressor is hC =

h2 s - h1 277.28 - 247.88 = = 0.757 h2 - h1 286.71- 247.88

(b) The rate of heat supplied to the room is

QH = m(h2 - h3 ) = (0.022 kg/s)(286.71- 87.93)kJ/kg = 4.373 kW

COP =

QH 4.373 = = 5.12 0.8542 Win

(d) The ideal vapor-compression cycle analysis of the cycle is as follows:

h1 = hg @ 200 kPa = 244.50 kJ/kg

s1 = sg @ 200 kPa = 0.9379 kJ/kg.K P2 = 800 kPa ü ïï ý h2 = 273.29 kJ/kg ïïþ s2 = s1 h3 = h f @ 800 kPa = 95.48 kJ/kg

h4 = h3 COP =

h2 - h3 273.29 - 95.48 = = 6.18 h2 - h1 273.29 - 244.50

10-502

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_2=800 [kPa] T_2=50 [C] m_dot=0.022 [kg/s] P_3=750 [kPa] DELTAT_subcool=3 [C] P_1=200 [kPa] DELTAT_superheat=4 [C] "PROPERTIES" Fluid$='R134a' h_2=enthalpy(Fluid$, P=P_2, T=T_2) T_sat_3=temperature(Fluid$, P=P_3, x=0) T_3=T_sat_3-DELTAT_subcool h_3=enthalpy(Fluid$, P=P_3, T=T_3) h_4=h_3 "isenthalpic expansion is assumed" T_sat_1=temperature(Fluid$, P=P_1, x=1) T_1=T_sat_1+DELTAT_superheat h_1=enthalpy(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) "ANALYSIS" eta_C=(h_2s-h_1)/(h_2-h_1) "compressor isentropic efficiency" Q_dot_H=m_dot*(h_2-h_3) "heat supplied to the room" W_dot_in=m_dot*(h_2-h_1) "compressor power" COP=Q_dot_H/W_dot_in "Ideal vapor-compression cycle analysis" h_1_ideal=enthalpy(Fluid$, P=P_1, x=1) s_1_ideal=entropy(Fluid$, P=P_1, x=1) h_2_ideal=enthalpy(Fluid$, P=P_2, s=s_1_ideal) h_3_ideal=enthalpy(Fluid$, P=P_2, x=0) h_4_ideal=h_3_ideal COP_ideal=(h_2_ideal-h_3_ideal)/(h_2_ideal-h_1_ideal) Q_dot_H_ideal=m_dot*(h_2_ideal-h_3_ideal)

12-41E A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered. The power input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-503

P2 = 120 psia ïü ïý h = 116.64 Btu/lbm ïïþ 2 s2 = s1 P3 = 120 psia ïü ïý h = h f @120 psia = 41.79 Btu/lbm ïïþ 3 sat. liquid h4 @ h3 = 41.79 Btu/lbm (throttling )

The mass flow rate of the refrigerant and the power input to the compressor are determined from and

QH QH 80,000/3600 Btu/s = = = 0.2970 lbm/s qH h2 - h3 (116.64 - 41.79) Btu/lbm

Win = m (h2 - h1 ) = (0.2970 kg/s)(116.64 - 108.83) Btu/lbm

= 2.320 Btu/s = 3.28 hp since 1 hp = 0.7068 Btu/s

The electrical power required without the heat pump is Thus,

æ 1 hp ö ÷ We = QH = (80,000/3600 Btu/s)çç = 31.44 hp ÷ çè0.7068 Btu/s ÷ ø

Wsaved = We - Win = 31.44 - 3.28 = 28.2 hp = 21.0 kW since 1 hp = 0.7457 kW

EES (Engineering Equation Solver) SOLUTION "Given" T_house=75 [F] T_water=50 [F] Q_dot_H=80000 [Btu/h] P[1]=50 [psia] P[2]=120 [psia] "Analysis" Fluid$='R134a' "compressor" x[1]=1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-504

h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" x[3]=0 P[3]=P[2] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot=Q_dot_H/(h[2]-h[3])*Convert(lbm/h, lbm/s) W_dot_in=m_dot*(h[2]-h[1])*Convert(Btu/s, hp) W_dot_e=Q_dot_H*Convert(Btu/h, Btu/s)*Convert(Btu/s, hp) W_dot_saved=W_dot_e-W_dot_in

12-42E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of subcooling at the exit of the condenser on the power requirement is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In this cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as subcooled liquid. From the refrigerant tables (Tables A-11E, A-12E, and A-13E),

P1 = 50 psia sat. vapor P2 = 160 psia s2 = s1

ü ïï ý ïïþ

h1 = hg @ 50 psia = 108.83 Btu/lbm s1 = sg @ 50 psia = 0.22192 Btu/lbm ×R ïü ïý ïïþ

h2 = 119.22 Btu/lbm

T3 = Tsat @ 160 psia - 9.5 = 109.5 - 9.5 = 100°F P3 = 160 psia T3 = 100°F

ïü ïý ïïþ

h3 @ h f @ 100° F = 45.130 Btu/lbm

h4 @ h3 = 45.130 Btu/lbm

(throttling)

The states at the inlet and exit of the expansion valve when the refrigerant is saturated liquid at the condenser exit are © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-505

P3 = 160 psia sat. liquid

ü ïï ý ïïþ

h3 = h f @ 160 psia = 48.527 Btu/lbm

h4 @ h3 = 48.527 Btu/lbm (throttling) The mass flow rate of the refrigerant in the ideal case is QH = m(h2 - h3 ) ¾ ¾ ® m=

QH 100, 000 Btu/h = = 1414.6 lbm/h h2 - h3 (119.22 - 48.527) Btu/lbm

The power requirement in the ideal case is

æ 1 kW ö ÷ Win = m(h2 - h1 ) = (1414.6 lbm/h)(119.22 - 108.83) Btu/lbm çç = 4.307 kW ÷ ÷ çè3412.14 Btu/h ø With subcooling, the mass flow rate and the power input are QH = m(h2 - h3 ) ¾ ¾ ® m=

QH 100, 000 Btu/h = = 1349.7 lbm/h h2 - h3 (119.22 - 45.130) Btu/lbm

æ 1 kW ö ÷ Win = m(h2 - h1 ) = (1349.7 lbm/h)(119.22 - 108.83) Btu/lbm çç = 4.110 kW ÷ çè3412.14 Btu/h ÷ ø Subcooling decreases the power requirement by 4.6%.

12-43E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the power requirement is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 10F. The saturation temperature of the refrigerant at 50 psia is 40.23F. From the refrigerant tables (Tables A-11E, A-12E, and A-13E),

P1 = 50 psia T1 = 40.23 + 10 = 50.2°F P2 = 160 psia s2 = s1

ïü ïý ïïþ

ü ïï ý ïïþ

h1 = 111.02 Btu/lbm s1 = 0.22621 Btu/lbm ×R

h2 = 121.73 Btu/lbm

10-506

P3 = 160 psia sat. liquid

ü ïï ý ïïþ

h3 = h f @ 160 psia = 48.527 Btu/lbm

h4 @ h3 = 48.527 Btu/lbm (throttling) The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are P1 = 50 psia sat. vapor

ïü ïý ïïþ

h1 = hg @ 50 psia = 108.83 Btu/lbm s1 = sg @ 50 psia = 0.22192 kJ/kg ×K

P2 = 160 psia s2 = s1

ü ïï ý ïïþ

h2 = 119.22 Btu/lbm

The mass flow rate of the refrigerant in the ideal case is QH = m(h2 - h3 ) ¾ ¾ ® m=

QH 100, 000 Btu/h = = 1414.6 lbm/h h2 - h3 (119.22 - 48.527) Btu/lbm

The power requirement in the ideal case is

æ 1 kW ö ÷ Win = m(h2 - h1 ) = (1414.6 lbm/h)(119.22 - 108.83) Btu/lbm çç = 4.307 kW ÷ ÷ çè3412.14 Btu/h ø With superheating, the mass flow rate and the power input are QL = m(h1 - h4 ) ¾ ¾ ® m=

QH 100, 000 Btu/h = = 1366.1 lbm/h h2 - h3 (121.73 - 48.527) Btu/lbm

æ 1 kW ö÷ Win = m(h2 - h1 ) = (1366.1 lbm/h)(121.73 - 111.02) Btu/lbm çç = 4.288 kW ÷ çè3412.14 Btu/h ÷ ø Superheating decreases the power requirement slightly.

12-44 A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the heat source. The power input to the heat pump, the rate of heat absorption from the water, and the increase in electric power input if an electric resistance heater is used instead of a heat pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-507

Analysis (a) From the refrigerant tables (Tables A-12 and A-13),

}h = 250.85 kJ/kg P = 1.0 MPa }h = 293.40 kJ/kg T = 60°C P = 1.0 MPa }h @ h = 93.58 kJ/kg T = 30°C P1 = 280 kPa T1 = 0°C

f @ 30° C

h4 @ h3 = 93.58 kJ/kg (throttling )

The mass flow rate of the refrigerant is

mR =

QH QH 60,000/3,600 kJ/s = = = 0.08341 kg/s qH h2 - h3 (293.40 - 93.58) kJ/kg

Then the power input to the compressor becomes

Win = m (h2 - h1 ) = (0.08341 kg/s)(293.40 - 250.85) kJ/kg = 3.55 kW (b) The rate of hat absorption from the water is

QL = m(h1 - h4 ) = (0.08341 kg/s)(250.85 - 93.58) kJ/kg = 13.12 kW (c) The electrical power required without the heat pump is Thus, We = QH = 60, 000 / 3600 kJ/s = 16.67 kW

Wincrease = We - Win = 16.67 - 3.55 = 13.12 kW

EES (Engineering Equation Solver) SOLUTION "Given" T_water=8 [C] Q_dot_H=60000 [kJ/h] P[1]=280 [kPa] T[1]=0 [C] P[2]=1000 [kPa] T[2]=60 [C] T[3]=30 [C] "Analysis" Fluid$='R134a' "(a)" "compressor" h[1]=enthalpy(Fluid$, P=P[1], T=T[1]) h[2]=enthalpy(Fluid$, P=P[2], T=T[2]) "expansion valve" x[3]=0 "approximated as saturated liquid" h[3]=enthalpy(Fluid$, T=T[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_H/(h[2]-h[3])*Convert(kg/h, kg/s) W_dot_in=m_dot_R*(h[2]-h[1]) "(b)" Q_dot_L=m_dot_R*(h[1]-h[4]) "(c)" W_dot_e=Q_dot_H*Convert(kJ/h, kW) W_dot_increase=W_dot_e-W_dot_in © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-508

12-45 Problem 12-44 is reconsidered. The effect of the compressor isentropic efficiency on the power input to the compressor and the electric power saved by using a heat pump rather than electric resistance heating is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=280 [kPa] T[1] = 0 [C] P[2] = 1000 [kPa] T[3] = 30 [C] Q_dot_out = 60000 [kJ/h] Eta_c=1.0 "T[2] is not given. It changes with compressor isentropic efficiency. Use ETA_c = 0.623 to obtain T[2] = 60C" Fluid$='R134a' "Compressor" h[1]=enthalpy(Fluid$,P=P[1],T=T[1]) s[1]=entropy(Fluid$,P=P[1],T=T[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c "definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on actual compressor" s[2]=entropy(Fluid$,h=h[2],P=P[2]) T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],T=T[3]) s[3]=entropy(Fluid$,P=P[3],T=T[3]) h[2]=Qout+h[3] "energy balance on condenser" Q_dot_out*convert(kJ/h,kJ/s)=m_dot*Qout "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" Q_dot_in=m_dot*Q_in COP=Q_dot_out*convert(kJ/h,kJ/s)/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c E_dot_saved = Q_dot_out*convert(kJ/h,kJ/s) - W_dot_ct

10-509

2.641 2.415

0.9 1

14.03 14.25

12-46 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of compressor irreversibilities on the COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-510

Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. The saturation pressure of refrigerant at 1.25C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13),

P1 = 280 kPa sat. vapor

ïü ïý ïïþ

h1 = hg @ 280 kPa = 249.77 kJ/kg s1 = sg @ 280 kPa = 0.93228 kJ/kg ×K

P2 = 800 kPa s2 = s1

ïü ïý ïïþ

h2 s = 271.59 kJ/kg

P3 = 800 kPa sat. liquid

ü ïï ý ïïþ

h3 = h f @ 800 kPa = 95.48 kJ/kg

h4 @ h3 = 95.48 kJ/kg (throttling) The actual enthalpy at the compressor exit is determined by using the compressor efficiency: hC =

h2 s - h1 h - h1 271.59 - 249.77 ¾¾ ® h2 = h1 + 2 s = 249.77 + = 275.44 kJ/kg h2 - h1 hC 0.85

The COPs of the heat pump for isentropic and irreversible compression cases are COPHP, ideal =

h - h3 qH 271.59 - 95.48 = 2s = = 8.071 win h2 s - h1 271.59 - 249.77

COPHP, actual =

h - h3 qH 275.44 - 95.48 = 2 = = 7.011 win h2 - h1 275.44 - 249.77

The irreversible compressor decreases the COP by 13.1%.

12-47 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-511

Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 2C. The saturation pressure of refrigerant at 1.25C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13),

P1 = 280 kPa T1 = - 1.25 + 2 = 1.25°C

ü ïï ý ïïþ

h1 = 251.96 kJ/kg s1 = 0.9402 kJ/kg ×K

P2 = 800 kPa s2 = s1

ïü ïý ïïþ

h2 = 274.03 kJ/kg

P3 = 800 kPa sat. liquid

ü ïï ý ïïþ

h3 = h f @ 800 kPa = 95.48 kJ/kg

h4 @ h3 = 95.48 kJ/kg

(throttling)

The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are P1 = 280 kPa sat. vapor

ü ïï ý ïïþ

h1 = hg @ 280 kPa = 249.77 kJ/kg s1 = sg @ 280 kPa = 0.93228 kJ/kg ×K

P2 = 800 kPa s2 = s1

ü ïï ý ïïþ

h2 s = 271.59 kJ/kg

The COPs of the heat pump for the two cases are COPHP, ideal =

h - h3 qH 271.59 - 95.48 = 2s = = 8.071 win h2 s - h1 271.59 - 249.77

COPHP, actual =

h - h3 qH 274.03 - 95.48 = 2 = = 8.090 win h2 - h1 274.03 - 251.96

The effect of superheating on the COP is negligible.

Innovative Refrigeration Systems 12-48C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, two or more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, but they have higher COP's, they can incorporate two or more different refrigerants, and they can achieve much lower temperatures.

12-49C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits.

10-512

12-50C We would favor the two-stage compression refrigeration system with a flash chamber since it is simpler, cheaper, and has better heat transfer characteristics.

12-51C Yes, by expanding the refrigerant in stages in several throttling devices.

12-52C To take advantage of the cooling effect by throttling from high pressures to low pressures.

12-53C The saturation pressure of refrigerant-134a at −32C is 77 kPa, which is below the atmospheric pressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration system with a different refrigerant at the bottoming cycle is recommended in this case.

12-54 A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flash chamber is adiabatic.

Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A13) to be h1 = 234.46 kJ/kg,

h2 = 262.70 kJ/kg

h3 = 255.61 kJ/kg, h5 = 127.25 kJ/kg,

h6 = 127.25 kJ/kg

h7 = 63.92 kJ/kg,

h8 = 63.92 kJ/kg

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, x6 =

h6 - h f h fg

127.25 - 63.92 = 0.3304 191.68

(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-513

Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout

å m h = å mh e e

i i

(1)h9 = x6 h3 + (1- x6 )h2 h9 = (0.3304)(255.61)+ (1- 0.3304)(262.70 ) = 260.35 kJ/kg ïü ïý s = 0.9438 kJ/kg ×K 9 h9 = 260.35 kJ/kgïïþ P9 = 0.4 MPa

Also, ü ïï ý h4 = 287.10 kJ/kg s4 = s9 = 0.9438 kJ/kg ×K ïïþ P4 = 1.4 MPa

Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are mB = (1- x6 )mA = (1- 0.3304)(0.25 kg/s) = 0.1674 kg/s

QL = mB (h1 - h8 ) = (0.1674 kg/s)(234.46 - 63.92) kJ/kg = 28.55 kW Win = WcompI, in + WcompII, in = mA (h4 - h9 )+ mB (h2 - h1 ) = (0.25 kg/s)(287.10 - 260.35) kJ/kg + (0.1674 kg/s )(262.70 - 234.46) kJ/kg = 11.41 kW

COPR =

QL 28.55 kW = = 2.50 Wnet, in 11.41 kW

EES (Engineering Equation Solver) SOLUTION "Input Data" P[1]=100 [kPa] P[4] = 1400 [kPa] P[6]=400 [kPa] m_dot_A=0.25 [kg/s] Eta_compB =1.0 "Low Pressure Compressor B isentropic efficiency" Eta_compA =1.0 "High Pressure Compressor A isentropic efficiency" Fluid$='R134a' "High Pressure Compressor A" P[9]=P[6] h4s=enthalpy(Fluid$,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4" h[9]+w_compAs=h4s "energy balance on isentropic compressor" w_compA=w_compAs/Eta_compA"definition of compressor isentropic efficiency" h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic" s[4]=entropy(Fluid$,h=h[4],P=P[4]) "properties for state 4" T[4]=temperature(Fluid$,h=h[4],P=P[4]) W_dot_compA=m_dot_A*w_compA "Condenser" P[5]=P[4] "neglect pressure drops across condenser" T[5]=temperature(Fluid$,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit" h[5]=enthalpy(Fluid$,T=T[5],x=0) "properties for state 5" s[5]=entropy(Fluid$,T=T[5],x=0) h[4]=q_out+h[5] "energy balance on condenser" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-514

Q_dot_out = m_dot_A*q_out "Throttle Valve A" h[6]=h[5] "energy balance on throttle - isenthalpic" x6=quality(Fluid$,h=h[6],P=P[6]) "properties for state 6" s[6]=entropy(Fluid$,h=h[6],P=P[6]) T[6]=temperature(Fluid$,h=h[6],P=P[6]) "Flash Chamber" m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(Fluid$, P=P[7], x=0) s[7]=entropy(Fluid$,h=h[7],P=P[7]) T[7]=temperature(Fluid$,h=h[7],P=P[7]) "Mixing Chamber" x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6] h[3]=enthalpy(Fluid$, P=P[3], x=1) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=1) T[3]=temperature(Fluid$,P=P[3],x=x1) s[9]=entropy(Fluid$,h=h[9],P=P[9]) "properties for state 9" T[9]=temperature(Fluid$,h=h[9],P=P[9]) "Low Pressure Compressor B" x1=1 "assume flow to compressor inlet to be saturated vapor" h[1]=enthalpy(Fluid$,P=P[1],x=x1) "properties for state 1" T[1]=temperature(Fluid$,P=P[1], x=x1) s[1]=entropy(Fluid$,P=P[1],x=x1) P[2]=P[6] h2s=enthalpy(Fluid$,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit" h[1]+w_compBs=h2s "energy balance on isentropic compressor" w_compB=w_compBs/Eta_compB"definition of compressor isentropic efficiency" h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_compB=m_dot_B*w_compB "Throttle Valve B" h[8]=h[7] "energy balance on throttle - isenthalpic" x8=quality(Fluid$,h=h[8],P=P[8]) "properties for state 8" s[8]=entropy(Fluid$,h=h[8],P=P[8]) T[8]=temperature(Fluid$,h=h[8],P=P[8]) "Evaporator" P[8]=P[1] "neglect pressure drop across evaporator" q_in + h[8]=h[1] "energy balance on evaporator" Q_dot_in=m_dot_B*q_in "Cycle Statistics" W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_in/W_dot_in_total "definition of COP" h6_f=enthalpy(Fluid$, P=P[6], x=0) h6_g=enthalpy(Fluid$, P=P[6], x=1) h6_fg=h6_g-h6_f

12-55 A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-515

1 2 3

Steady operating conditions exist. Kinetic and potential energy changes are negligible. The flash chamber is adiabatic.

Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be h1 = 234.46 kJ/kg,

h2 = 271.42 kJ/kg

h3 = 262.46 kJ/kg, h5 = 127.25 kJ/kg,

h6 = 127.25 kJ/kg

h7 = 81.50 kJ/kg,

h8 = 81.50 kJ/kg

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6, h6 - h f 127.25 - 81.50 x6 = = = 0.2528 h fg 180.95 (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout

å m h = å mh e e

i i

(1)h9 = x6 h3 + (1- x6 )h2 h9 = (0.2528)(262.46)+ (1- 0.2528)(271.42) = 269.15 kJ/kg ü P9 = 0.6 MPa ïï ý s9 = 0.9444 kJ/kg ×K h9 = 269.15 kJ/kgïïþ

Also, ü P4 = 1.4 MPa ïï ý h4 = 287.31 kJ/kg s4 = s9 = 0.9444 kJ/kg ×K ïïþ

Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are mB = (1- x6 )mA = (1- 0.2528)(0.25 kg/s) = 0.1868 kg/s

QL = mB (h1 - h8 ) = (0.1868 kg/s)(234.46 - 81.50) kJ/kg = 28.57 kW Win = WcompI, in + WcompII, in = mA (h4 - h9 )+ mB (h2 - h1 )

10-516

= (0.25 kg/s)(287.31- 269.15)kJ/kg + (0.1868 kg/s )(271.42 - 234.46)kJ/kg = 11.44 kW

COPR =

QL 28.57 kW = = 2.50 Wnet, in 11.44 kW

EES (Engineering Equation Solver) SOLUTION "Input Data" P[1]=100 [kPa] P[4] = 1400 [kPa] P[6]=600 [kPa] m_dot_A=0.25 [kg/s] Eta_compB =1.0 "Low Pressure Compressor B isentropic efficiency" Eta_compA =1.0 "High Pressure Compressor A isentropic efficiency" Fluid$='R134a' "High Pressure Compressor A" P[9]=P[6] h4s=enthalpy(Fluid$,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4" h[9]+w_compAs=h4s "energy balance on isentropic compressor" w_compA=w_compAs/Eta_compA"definition of compressor isentropic efficiency" h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic" s[4]=entropy(Fluid$,h=h[4],P=P[4]) "properties for state 4" T[4]=temperature(Fluid$,h=h[4],P=P[4]) W_dot_compA=m_dot_A*w_compA "Condenser" P[5]=P[4] "neglect pressure drops across condenser" T[5]=temperature(Fluid$,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit" h[5]=enthalpy(Fluid$,T=T[5],x=0) "properties for state 5" s[5]=entropy(Fluid$,T=T[5],x=0) h[4]=q_out+h[5] "energy balance on condenser" Q_dot_out = m_dot_A*q_out "Throttle Valve A" h[6]=h[5] "energy balance on throttle - isenthalpic" x6=quality(Fluid$,h=h[6],P=P[6]) "properties for state 6" s[6]=entropy(Fluid$,h=h[6],P=P[6]) T[6]=temperature(Fluid$,h=h[6],P=P[6]) "Flash Chamber" m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(Fluid$, P=P[7], x=0) s[7]=entropy(Fluid$,h=h[7],P=P[7]) T[7]=temperature(Fluid$,h=h[7],P=P[7]) "Mixing Chamber" x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6] h[3]=enthalpy(Fluid$, P=P[3], x=1) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=1) T[3]=temperature(Fluid$,P=P[3],x=x1) s[9]=entropy(Fluid$,h=h[9],P=P[9]) "properties for state 9" T[9]=temperature(Fluid$,h=h[9],P=P[9]) "Low Pressure Compressor B" x1=1 "assume flow to compressor inlet to be saturated vapor" h[1]=enthalpy(Fluid$,P=P[1],x=x1) "properties for state 1" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-517

T[1]=temperature(Fluid$,P=P[1], x=x1) s[1]=entropy(Fluid$,P=P[1],x=x1) P[2]=P[6] h2s=enthalpy(Fluid$,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit" h[1]+w_compBs=h2s "energy balance on isentropic compressor" w_compB=w_compBs/Eta_compB"definition of compressor isentropic efficiency" h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_compB=m_dot_B*w_compB "Throttle Valve B" h[8]=h[7] "energy balance on throttle - isenthalpic" x8=quality(Fluid$,h=h[8],P=P[8]) "properties for state 8" s[8]=entropy(Fluid$,h=h[8],P=P[8]) T[8]=temperature(Fluid$,h=h[8],P=P[8]) "Evaporator" P[8]=P[1] "neglect pressure drop across evaporator" q_in + h[8]=h[1] "energy balance on evaporator" Q_dot_in=m_dot_B*q_in "Cycle Statistics" W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_in/W_dot_in_total "definition of COP" h6_f=enthalpy(Fluid$, P=P[6], x=0) h6_g=enthalpy(Fluid$, P=P[6], x=1) h6_fg=h6_g-h6_f

12-56 Problem 12-54 is reconsidered. The effects of the various refrigerants for compressor efficiencies of 80, 90, and 100 percent is to be investigated. Analysis The problem is solved using EES, and the results are tabulated and plotted below. Fluid$='R134a' "Input Data" P[1]=100 [kPa] P[4] = 1400 [kPa] P[6]=400 [kPa] Eta_comp =1.0 m_dot_A=0.25 [kg/s] "High Pressure Compressor A" P[9]=P[6] h4s=enthalpy(Fluid$,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4" h[9]+w_compAs=h4s "energy balance on isentropic compressor" w_compA=w_compAs/Eta_comp "definition of compressor isentropic efficiency" h[9]+w_compA=h[4] "energy balance on actual compressor" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) W_dot_compA=m_dot_A*w_compA "Condenser" P[5]=P[4] "neglect pressure drops across condenser" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-518

T[5]=temperature(Fluid$,P=P[5],x=0) h[5]=enthalpy(Fluid$,T=T[5],x=0) s[5]=entropy(Fluid$,T=T[5],x=0) h[4]=q_H+h[5] "energy balance on condenser" Q_dot_H = m_dot_A*q_H "Throttle Valve A" h[6]=h[5] "energy balance on throttle - isenthalpic" x6=quality(Fluid$,h=h[6],P=P[6]) s[6]=entropy(Fluid$,h=h[6],P=P[6]) T[6]=temperature(Fluid$,h=h[6],P=P[6]) "Flash Chamber" m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(Fluid$, P=P[7], x=0) s[7]=entropy(Fluid$,h=h[7],P=P[7]) T[7]=temperature(Fluid$,h=h[7],P=P[7]) "Mixing Chamber" x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6] h[3]=enthalpy(Fluid$, P=P[3], x=1) s[3]=entropy(Fluid$,P=P[3],x=1) T[3]=temperature(Fluid$,P=P[3],x=x1) s[9]=entropy(Fluid$,h=h[9],P=P[9]) T[9]=temperature(Fluid$,h=h[9],P=P[9]) "Low Pressure Compressor B" x1=1 "assume flow to compressor inlet to be saturated vapor" h[1]=enthalpy(Fluid$,P=P[1],x=x1) T[1]=temperature(Fluid$,P=P[1], x=x1) s[1]=entropy(Fluid$,P=P[1],x=x1) P[2]=P[6] h2s=enthalpy(Fluid$,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit" h[1]+w_compBs=h2s "energy balance on isentropic compressor" w_compB=w_compBs/Eta_comp "definition of compressor isentropic efficiency" h[1]+w_compB=h[2] "energy balance on actual compressor" s[2]=entropy(Fluid$,h=h[2],P=P[2]) T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_compB=m_dot_B*w_compB "Throttle Valve B" h[8]=h[7] "energy balance on throttle - isenthalpic" x8=quality(Fluid$,h=h[8],P=P[8]) s[8]=entropy(Fluid$,h=h[8],P=P[8]) T[8]=temperature(Fluid$,h=h[8],P=P[8]) "Evaporator" P[8]=P[1] "neglect pressure drop across evaporator" q_L + h[8]=h[1] "energy balance on evaporator" Q_dot_L=m_dot_B*q_L

10-519

"Cycle Statistics" W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_L/W_dot_in_total

ηcomp

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

QL [kW] 28.55 28.55 28.55 28.55 28.55 28.55 28.55 28.55 28.55

COP 1.438 1.57 1.702 1.835 1.968 2.101 2.235 2.368 2.502

10-520

12-57 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through the upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13): h1 = hg @ 160 kPa = 241.14 kJ/kg s1 = sg @ 160 kPa = 0.9420 kJ/kg.K

P2 = 500 kPa ïü ïý h = 264.55 kJ/kg ïïþ 2 s s2 = s1 hC =

h2 s - h1 h2 - h1

10-521

0.80 =

264.55 - 241.14 ¾¾ ® h2 = 270.41 kJ/kg h2 - 241.14

h3 = h f @500 kPa = 73.32 kJ/kg

h4 = h3 = 73.32 kJ/kg h5 = hg @ 400 kPa = 255.61 kJ/kg s5 = sg @ 400 kPa = 0.9271 kJ/kg.K

P6 = 1400 kPa ïü ïý h = 281.56 kJ/kg ïïþ 6 s

s6 = s5

hC =

h6 s - h5 h6 - h5

0.80 =

281.56 - 255.61 ¾¾ ® h6 = 288.04 kJ/kg h6 - 255.61

h7 = h f @1400 kPa = 127.25 kJ/kg

h8 = h7 = 127.25 kJ/kg

The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger

mA (h5 - h8 ) = mB (h2 - h3 )

mA (255.61- 127.25)kJ/kg = (0.25 kg/s)(270.41- 73.32)kJ/kg ¾ ¾ ® mA = 0.3839 kg / s (b) The rate of heat removal from the refrigerated space is

QL = mB (h1 - h4 ) = (0.25 kg/s)(241.14 - 73.32)kJ/kg = 41.96 kW (c) The power input and the COP are

10-522

= (0.3839 kg/s)(288.04 - 255.61)kJ/kg + (0.25 kg/s)(270.41- 241.14)kJ/kg = 19.77 kW

COP =

QL 41.96 = = 2.12 Win 19.77

EES (Engineering Equation Solver) SOLUTION "GIVEN" P[1]=160 [kPa] P[2]=500 [kPa] P[5]=400 [kPa] P[6]=1400 [kPa] x[1]=1 x[3]=0 x[5]=1 x[7]=0 eta_C=0.8 m_dot_B=0.25 [kg/s] "PROPERTIES" Fluid$='R134a' h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) h_s[2]=enthalpy(Fluid$, P=P[2], s=s[1]) h[3]=enthalpy(Fluid$, P=P[2], x=x[3]) h[4]=h[3] h[5]=enthalpy(Fluid$, P=P[5], x=x[5]) s[5]=entropy(Fluid$, P=P[5], x=x[5]) h_s[6]=enthalpy(Fluid$, P=P[6], s=s[5]) h[7]=enthalpy(Fluid$, P=P[6], x=x[7]) h[8]=h[7] "ANALYSIS" eta_C=(h_s[2]-h[1])/(h[2]-h[1]) eta_C=(h_s[6]-h[5])/(h[6]-h[5]) m_dot_A*(h[5]-h[8])=m_dot_B*(h[2]-h[3]) Q_dot_L=m_dot_B*(h[1]-h[4]) W_dot_in=m_dot_A*(h[6]-h[5])+m_dot_B*(h[2]-h[1]) COP=Q_dot_L/W_dot_in

12-58 A two-stage compression refrigeration system with a separation unit is considered. The mass flow rate through the two compressors, the power used by the compressors, and the system’s COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-523

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

T1 = - 10.1°C sat. vapor

ü ïï ý ïïþ

h1 = hg @ - 10.1°C = 244.50 kJ/kg s1 = sg @ - 10.1°C = 0.93788 kJ/kg ×K

P2 = 800 kPa s2 = s1

ïü ïý ïïþ

h2 = 273.31 kJ/kg

P3 = 800 kPa sat. liquid

ü ïï ý ïïþ

h3 = h f @ 800 kPa = 95.48 kJ/kg

h4 @ h3 = 95.48 kJ/kg (throttling)

ïü ïý ïïþ

T5 = - 10.1°C sat. liquid

h5 = h f @ - 10.1°C = 38.41 kJ/kg

h6 @ h5 = 38.41 kJ/kg (throttling)

T7 = - 40°C sat. vapor

ïü ïý ïïþ

h7 = hg @ - 40°C = 225.86 kJ/kg s7 = sg @ - 40°C = 0.96869 kJ/kg ×K

P8 = Psat @ - 10.1°C = 200 kPa ïü ïý ïïþ s8 = s7

h8 = 252.75 kJ/kg

The mass flow rate through the evaporator is determined from QL = m6 (h7 - h6 )

m6 =

QL 30 kJ/s = = 0.1600 kg / s h7 - h6 (225.86 - 38.41) kJ/kg

10-524

An energy balance on the separator gives m6 (h8 - h5 ) = m2 (h1 - h4 ) ¾ ¾ ® m2 = m6

h8 - h5 252.75 - 38.41 = (0.1600) = 0.2301 kg / s h1 - h4 244.50 - 95.48

The total power input to the compressors is

Win = m6 (h8 - h7 ) + m2 (h2 - h1 ) = (0.1600 kg/s)(252.75 - 225.86) kJ/kg + (0.2301 kg/s)(273.31- 244.50) kJ/kg = 10.93 kW

The COP of this refrigeration system is determined from its definition, COPR =

QL 30 kW = = 2.74 Win 10.93 kW

12-59E A two-stage compression refrigeration system with a separation unit is considered. The cooling load and the system’s COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),

P1 = 120 psia sat. vapor

ü ïï ý ïïþ

h1 = hg @ 120 psia = 115.18 Btu/lbm s1 = sg @ 120 psia = 0.21928 Btu/lbm ×R

10-525

P2 = 300 psia s2 = s1

ü ïï ý ïïþ

h2 = 123.10 Btu/lbm

P3 = 300 psia sat. liquid

ïü ïý ïïþ

h3 = h f @ 300 psia = 66.347 Btu/lbm

h4 @ h3 = 66.347 Btu/lbm (throttling)

P5 = 120 psia sat. liquid

ïü ïý ïïþ

h5 = h f @ 120 psia = 41.791 Btu/lbm

h6 @ h5 = 41.791 kJ/kg (throttling)

P7 = 60 psia sat. vapor P8 = 120 psia s8 = s7

ïü ïý ïïþ

h7 = hg @ 60 psia = 110.13 Btu/lbm s7 = sg @ 60 psia = 0.22132 Btu/lbm ×R ïü ïý ïïþ

h8 = 116.31 Btu/lbm

An energy balance on the separator gives

m6 (h8 - h5 ) = m2 (h1 - h4 ) m2 = m6

h8 - h5 116.31- 41.791 = m6 = 1.5260m6 h1 - h4 115.18 - 66.347

The total power input to the compressors is given by Win = m6 (h8 - h7 ) + m2 (h2 - h1 )

= m6 (h8 - h7 ) + 1.5260m6 (h2 - h1 ) Solving for m6 , m6 =

Win (25´ 3412.14) Btu/h = = 4670 lbm/h (h8 - h7 ) + 1.5258(h2 - h1 ) (116.31- 110.13) + 1.5260(123.10 - 115.18) Btu/lbm

The cooling effect produced by this system is then QL = m6 (h7 - h6 ) = (4670 lbm/h)(110.13 - 41.791) Btu/lbm = 319,100 Btu / h

The COP of this refrigeration system is determined from its definition, COPR =

QL (319,100/3412.14) kW = = 3.74 25 kW Win

12-60 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling rate of the high-temperature evaporator, the power required by the compressor, and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-526

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

P3 = 800 kPa ïü ïý h = h f @800 kPa = 95.48 kJ/kg ïïþ 3 sat. liquid h4 = h6 @ h3 = 95.48 kJ/kg (throttling)

T5 = 0°C ü ïï ý h = hg @ 0°C = 250.50 kJ/kg sat. vapor ïïþ 5 T7 = - 26.4°C ü ïï ý h7 = hg @- 26.4°C = 234.44 kJ/kg ïïþ sat. vapor

The mass flow rate through the low-temperature evaporator is found by QL = m2 (h7 - h6 ) ¾ ¾ ® m2 =

QL 8 kJ/s = = 0.05757 kg/s h7 - h6 (234.44 - 95.48) kJ/kg

The mass flow rate through the warmer evaporator is then

m1 = m - m2 = 0.1- 0.05757 = 0.04243 kg/s Applying an energy balance to the point in the system where the two evaporator streams are recombined gives

m1h5 + m2 h7 = mh1 ¾ ¾ ® h1 =

m1h5 + m2 h7 (0.04243)(250.50) + (0.05757)(234.44) = = 241.26 kJ/kg m 0.1

Then,

10-527

P1 = Psat@- 26.4°C @100 kPa ü ïï ý s1 = 0.9791 kJ/kg ×K ïïþ h1 = 241.26 kJ/kg P2 = 800 kPa ïü ïý h = 286.32 kJ/kg ï 2 s2 = s1 ïþ The cooling rate of the high-temperature evaporator is

QL = m1 (h5 - h4 ) = (0.04243 kg/s)(250.50 - 95.48) kJ/kg = 6.578 kW The power input to the compressor is

Win = m(h2 - h1 ) = (0.1 kg/s)(286.32 - 241.26) kJ/kg = 4.506 kW The COP of this refrigeration system is determined from its definition, COPR =

QL (8 + 6.578) kW = = 3.24 4.506 kW Win

EES (Engineering Equation Solver) SOLUTION "GIVEN" T[5]=0 [C] T[7]=-26.4 [C] P[3]=800 [kPa] m_dot=0.1 [kg/s] Q_dot_L2=8 [kW] "PROPERTIES" Fluid$='R134a' h[3]=enthalpy(Fluid$, P=P[3], x=0) h[4]=h[3] h[6]=h[3] h[5]=enthalpy(Fluid$, T=T[5], x=1) h[7]=enthalpy(Fluid$, T=T[7], x=1) "ANALYSIS" m_dot_2=Q_dot_L2/(h[7]-h[6]) m_dot_1=m_dot-m_dot_2 m_dot_1*h[5]+m_dot_2*h[7]=m_dot*h[1] P[1]=pressure(Fluid$, T=T[7], x=0) s[1]=entropy(Fluid$, P=P[1], h=h[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) Q_dot_L1=m_dot_1*(h[5]-h[4]) W_dot_in=m_dot*(h[2]-h[1]) COP=(Q_dot_L1+Q_dot_L2)/W_dot_in

12-61 A two-stage cascade refrigeration cycle with a flash chamber with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant through the high-pressure compressor, the rate of refrigeration, the COP are to be determined. Also, the rate of refrigeration and the COP are to be determined if this refrigerator operated on a single-stage vapor-compression cycle under similar conditions. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-528

Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13) h1 = hg @- 10 °C = 244.55 kJ/kg

s1 = sg @- 10 °C = 0.9382 kJ/kg.K P2 = 450 kPa ïü ïý h = 261.10 kJ/kg ïïþ 2 s s2 = s1

hC =

h2 s - h1 h2 - h1

0.86 =

261.10 - 244.55 ¾¾ ® h2 = 263.79 kJ/kg h2 - 244.55

h3 = hg @ 450 kPa = 257.58 kJ/kg

h5 = h f @ 1600 kPa = 135.96 kJ/kg h6 = h5 = 135.96 kJ/kg

h7 = h f @ 450 kPa = 68.80 kJ/kg h8 = h7 = 68.80 kJ/kg h6 = 135.96 kJ/kgü ïï ý x6 = 0.35576 ïïþ P6 = 450 kPa

The mass flow rate of the refrigerant through the high pressure compressor is determined from a mass balance on the flash chamber m=

m7 0.11 kg/s = = 0.17074 kg / s 1- x6 1- 0.35576

Also,

m3 = m - m7 = 0.17074 - 0.11 = 0.06074 kg/s (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:

mh9 = m7 h2 + m3 h3

(0.17074 kg/s)h9 = (0.11 kg/s)(263.79 kJ/kg) + (0.06074 kg/s)(257.58 kJ/kg) ¾ ¾ ® h9 = 261.58 kJ/kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-529

Then, ïü ïý s = 0.9399 kJ/kg 9 h9 = 261.58 kJ/kgïïþ

P9 = 450 kPa

P4 = 1600 kPa ïü ïý h = 288.64 kJ/kg ïïþ 4 s

s4 = s9

hC =

h4 s - h9 h4 - h9

0.86 =

288.64 - 261.58 ¾¾ ® h4 = 293.05 kJ/kg h4 - 261.58

The rate of heat removal from the refrigerated space is QL = m7 (h1 - h8 ) = (0.11 kg/s)(244.55 - 68.80)kJ/kg = 19.33 kW

= (0.11 kg/s)(263.79 - 244.55)kJ/kg + (0.17074 kg/s)(293.05 - 261.58)kJ/kg = 7.490 kW COP =

QL 19.33 = = 2.58 7.490 Win

(d) If this refrigerator operated on a single-stage cycle between the same pressure limits, we would have h1 = hg @- 10 °C = 244.55 kJ/kg

s1 = sg @- 10 °C = 0.93782 kJ/kg.K P2 = 1600 kPa ïü ïý h = 287.94 kJ/kg ïïþ 2 s s2 = s1

hC =

h2 s - h1 h2 - h1

0.86 =

287.94 - 244.55 ¾¾ ® h2 = 295.00 kJ/kg h2 - 244.55

h3 = h f @ 1600 kPa = 135.96 kJ/kg h4 = h3 = 135.96 kJ/kg QL = m(h1 - h4 ) = (0.17074 kg/s)(244.55 - 135.96)kJ/kg = 18.54 kW

Win = m(h2 - h1 ) = (0.17074 kg/s)(295.00 - 244.55)kJ/kg = 8.614 kW

COP =

QL 18.54 = = 2.15 Win 8.614

10-530

Discussion The cooling load decreases by 4.1% while the COP decreases by 16.7% when the cycle operates on the singlestage vapor-compression cycle.

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_evap=-10 [C] P_medium=450 [kPa] P_high=1600 [kPa] x[1]=1 x[5]=0 x[7]=0 eta_C=0.86 m_dot_7=0.11 [kg/s] "PROPERTIES" Fluid$='R134a' h[1]=enthalpy(Fluid$, T=T_evap, x=x[1]) s[1]=entropy(Fluid$, T=T_evap, x=x[1]) h_s[2]=enthalpy(Fluid$, P=P_medium, s=s[1]) h[3]=enthalpy(Fluid$, P=P_medium, x=1) h[5]=enthalpy(Fluid$, P=P_high, x=x[5]) h[6]=h[5] h[7]=enthalpy(Fluid$, P=P_medium, x=0) h[8]=h[7] "ANALYSIS" eta_C=(h_s[2]-h[1])/(h[2]-h[1]) x[6]=quality(Fluid$, P=P_medium, h=h[6]) m_dot=m_dot_7/(1-x[6]) m_dot_3=m_dot-m_dot_7 m_dot*h[9]=m_dot_7*h[2]+m_dot_3*h[3] s[9]=entropy(Fluid$, P=P_medium, h=h[9]) h_s[4]=enthalpy(Fluid$, P=P_high, s=s[9]) eta_C=(h_s[4]-h[9])/(h[4]-h[9]) Q_dot_L=m_dot_7*(h[1]-h[8]) W_dot_in=m_dot_7*(h[2]-h[1])+m_dot*(h[4]-h[9]) COP=Q_dot_L/W_dot_in "Single-stage cycle analysis" h_1=h[1] s_1=s[1] h_s2=enthalpy(Fluid$, P=P_high, s=s_1) eta_C=(h_s2-h_1)/(h_2-h_1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-531

h_3=enthalpy(Fluid$, P=P_high, x=0) h_4=h_3 Q_dot_L_single=m_dot*(h_1-h_4) W_dot_in_single=m_dot*(h_2-h_1) COP_single=Q_dot_L_single/W_dot_in_single

12-62 An ideal Linde-Hampson cycle is used to liquefy nitrogen gas. Nitrogen is compressed isothermally from 100 kPa and 15C to 17.5 MPa. Various properties of nitrogen in the cycle are given as follows: h1  h7  h8  10.46 kJ/kg, h2  43.05 kJ/kg, h6  431.76 kJ/kg s1  0.03168 kJ/kg  K, s2  1.6746 kJ/kg  K, s6  4.0055 kJ/kg  K

(a) Show that the fraction of the gas that is liquefied in the cycle can be expressed by y = (h1 - h2 )(h1 - h6 ) (b) Determine the fraction of the gas liquefied, and the refrigeration loads per unit mass of the gas flowing through the compressor and per unit of the gas liquefied. (c) Determine the COP of the cycle. (d) Determine the second-law efficiency of the cycle. 12-62 A simple Linde-Hampson cycle is used to liquefy nitrogen. The simple Linde-Hampson cycle is considered. A relation for the fraction of the liquefied gas is to be obtained and various performance parameters are to be determined. Assumptions 1 2 3 4

Steady operating conditions exist. Nitrogen is an ideal gas during the compression process. Kinetic and potential energy changes are negligible. Heat exchanger is adiabatic.

Properties The gas constant of nitrogen is R = 0.2968 kJ / kg ×K (Table A-2a).

10-532

(a) Denoting the fraction of the liquefied gas y, and noting that h1 = h7 = h8 an energy balance on the heat exchanger can be written as (1) (1  y)(h1  h5 )  h2  h3 Throttling is a constant-enthalpy process. Therefore, h4  h3

(2)

and h3  yh6  (1  y)h5

(3)

Substituting Eq. 3 into Eq. 1,

(1  y)(h1  h5 )  h2  yh6  (1  y)h5

(1  y)h1  h2  yh6 h1  yh1  h2  yh6

h1  h2  y(h1  h6 ) That is, y

h1  h2 h1  h6

which is the desired expression. (b) The fraction of the gas in the cycle that is liquefied is determined from

(10.46)  (43.05) kJ/kg  0.07736 h h y 1 2  h1  h6  (10.46)  (431.76)  kJ/kg The refrigeration effect per unit mass of the gas in the cycle is qL  h1  h2   (10.46)  (43.05)  kJ/kg  32.59 kJ / kg gas

The refrigeration effect per unit mass of liquefaction is qL 

32.59 kJ/kg 32.59 kJ/kg   421.3 kJ /kg liq y 0.07736

Liquefaction is a heat removal process from the gas until the gas turns into a liquid. Therefore, it can also be determined from qL  h1  h6   (10.46)  ( 431.76)  kJ/kg  421.3 kJ/kg liq

(c) Nitrogen can be considered to be an ideal gas during the compression process. Then the work input per unit gas mass during the reversible isothermal compression is determined from wcomp  RT0 ln( P2 / P1 )  (0.2968 kJ/kg  K)(288.15) ln(17, 500 / 100)  441.7 kJ/kg gas

Taking the dead state temperature to be 15C = 288.15 K, the work input for this reversible isothermal compressor can also be determined from

wcomp  h2  h1  T1 ( s2  s1 )   (43.05)  (10.46)  kJ/kg  (288.15 K) ( 1.6746  ( 0.03168) kJ/kg  K 

 440.8 kJ/kg gas

The results are practically the same. The work input per unit mass of liquefaction is

10-533

wcomp 

440.8 kJ/kg 440.8 kJ/kg   5698 kJ/kg liq y 0.07736

The coefficient of performance (COP) of this cycle is COP 

qL 32.59 kJ/kg gas   0.07393 wcomp 440.8 kJ/kg gas

(d) The reversible or minimum work per unit mass of liquefaction is wrev  h6  h1  T0 (s6  s1 )   (431.76)  ( 10.46)  kJ/kg  (288.15 K) ( 4.0055)  ( 0.03168) kJ/kg  K 

 723.7 kJ/kg liq The reversible COP is COPrev 

qL 421.3 kJ/kg liq   0.5821 wrev 723.7 kJ/kg liq

The second-law efficiency is the reversible work input divided by the actual work input, both per unit mass of the liquefaction: hII =

w rev w com p,liq

723.7 kJ/kg liq = 0.127 or 12.7% 5698 kJ/kg liq

The second-law efficiency can also be determined from

II 

COP 0.07393   0.127 or 12.7% COPrev 0.5821

The results are the same, as expected.

EES (Engineering Equation Solver) SOLUTION "Comparison of liquefaction systems using air as working fluid" Fluid$='nitrogen' "Ideal reversible system" w_rev=-(h1-hf-T0*(s1-sf)) "Reversible work per unit mass liquefied" T0=288.15 [K] "Simple Linde-Hampson system (isothermal compressor, 100% effectiveness heat exc)" T1=15 [C] P1=100 [kPa] P2=17500 [kPa] P3=P2 T2=T1 R=0.2968 [kJ/kg-K] h1=enthalpy(Fluid$, T=T1, P=P1) h2=enthalpy(Fluid$, T=T2, P=P2) s1=entropy(Fluid$, T=T1, P=P1) s2=entropy(Fluid$, T=T2, P=P2) hf=enthalpy(Fluid$, x=0, P=P1) sf=entropy(Fluid$, x=0, P=P1) Tf=temperature(Fluid$, x=0, P=P1) hg=enthalpy(Fluid$, x=1, P=P1) h5=hg "h2-h3=x*(h1-h5) x=quality(Fluid$, h=h3, P=P1)" y=(h1-h2)/(h1-hf) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-534

w_c2=R*T0*ln(P2/P1) "work input for isothermal compressor per unit mass of gas" w_c=h2-h1-T0*(s2-s1) "work input for isothermal compressor per unit mass of gas" w_c_m=w_c/y "work input per unit mass liquefied" q_L=h1-h2 "refrigeration effect per unit mass of gas" q_L2=y*(h1-hf) "refrigeration effect per unit mass of gas" q_Lm=(h1-h2)/y "refrigeration effect per unit mass of gas" COP=q_L/w_c eta_ex=w_rev/w_c_m COP_rev=q_Lm/w_rev "w_rev=-q_Lm*(1-(T1+273)/T1a) COP_rev=1/(((T1+273)/T2a)-1)"

Gas Refrigeration Cycles 12-63C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature.

12-64C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reversed direction.

12-65C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reversed direction. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle is also totally reversible, and thus its COP is COPR, Stirling =

1 TH / TL - 1

12-66C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin.

12-67C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling ( h1 = h2 ) process.

12-68C By regeneration.

12-69 An ideal gas refrigeration cycle with air as the working fluid provides 15 kW of cooling. The mass flow rate of air and the rates of heat addition and rejection are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible.

10-535

Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4 (Table A-2a). Analysis From the isentropic relations, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ ÷ çè P ø 1

( k - 1)/ k

æP ö ÷ T4 = T3 ççç 4 ÷ ÷ ÷ èç P3 ø

0.4/1.4

æ500 kPa ÷ ö = (293 K) çç ÷ ÷ çè100 kPa ø

= 464.1 K

0.4/1.4

æ100 kPa ÷ ö = (303 K) çç ÷ ÷ èç500 kPa ø

= 191.3 K

The mass flow rate of the air is determined from QRefrig = mc p (T1 - T4 ) ¾ ¾ ® m=

QRefrig c p (T1 - T4 )

15 kJ/s = 0.1468 kg / s (1.005 kJ/kg ×K)(293 - 191.3) K

The rate of heat addition to the cycle is the same as the rate of cooling,

Qin = QRefrig = 15 kW The rate of heat rejection from the cycle is

QH = mc p (T2 - T3 ) = (0.1468 kg/s)(1.005 kJ/kg ×K)(464.1- 303) K = 23.8 kW

12-70 An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 280 K ¾ ¾ ® s2o = s1o + R ln

h1 = 280.13 kJ/kg s1o = 1.63279 kJ/kg ×K

T = 431.6 K æ160 kPa ÷ ö P2 = 1.63279 + (0.287 kJ/kg ×K) ln çç = 2.06898 kJ/kg ×K ¾ ¾ ® 2 ÷ ÷ ç è 35 kPa ø h2 = 433.02 kJ/kg P1

10-536

T3 = 310 K ¾ ¾ ® s4o = s3o + R ln

h3 = 310.24 kJ/kg s3o = 1.73498 kJ/kg ×K

T = 200.7 K æ35 kPa ÷ ö P4 = 1.73498 + (0.287 kJ/kg ×K) ln çç = 1.29879 kJ/kg ×K ¾ ¾ ® 4 ÷ ÷ ç è ø h4 = 200.63 kJ/kg P3 160 kPa

Then the rate of refrigeration is

QL = m(qL ) = m (h1 - h4 ) = (0.2 kg/s)(280.13 - 200.63) kJ/kg = 15.90 kW

(b) The net power input is determined from

Wnet, in = Wcomp, in - Wturb, out where Wcomp, in = m (h2 - h1 ) = (0.2 kg/s)(433.02 - 280.13) kJ/kg = 30.58 kW

Wturb, out = m (h3 - h4 ) = (0.2 kg/s)(310.24 - 200.63) kJ/kg = 21.92 kW

Thus,

Wnet, in = 30.58 - 21.92 = 8.66 kW (c) The COP of this ideal gas refrigeration cycle is determined from

COPR =

QL 15.90 kW = = 1.84 8.66 kW Wnet, in

EES (Engineering Equation Solver) SOLUTION "Input data" $Array on T[1] = 7 [C] P[1]= 35 [kPa] T[3] = 37 [C] P[3]=160 [kPa] m_dot=0.2 [kg/s] Eta_comp = 1 Eta_turb = 1 "Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s2s=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = P[3] s2s=ENTROPY(Air,T=Ts2,P=P[2])"Ts2 is the isentropic value of T[2] at compressor exit" Eta_comp = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-537

W_dot_comp > W_dot_comp_isen" m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor, assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s" h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3, assumed SSSF constant pressure process" m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit" Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen > W_dot_turb" m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0" hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect:" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor" COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s

12-71 An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 280 K ¾ ¾ ® s2o = s1o + R ln

h1 = 280.13 kJ/kg s1o = 1.63279 kJ/kg ×K

T2 = 431.6 K æ160 kPa ö P2 ÷ = 1.63279 + (0.287 kJ/kg ×K) ln çç = 2.06898 kJ/kg ×K ¾ ¾ ® ÷ ÷ ç è 35 kPa ø h2 s = 433.02 kJ/kg P1

10-538

T3 = 310 K ¾ ¾ ® s4o = s3o + R ln

h3 = 310.24 kJ/kg s3o = 1.73498 kJ/kg ×K

T4 = 200.7 K æ35 kPa ö P4 ÷= 1.29879 kJ/kg ×K ¾ ¾ = 1.73498 + (0.287 kJ/kg ×K) ln çç ® ÷ ÷ çè160 kPa ø h4 s = 200.63 kJ/kg P3

Also, hT =

h3 - h4 ¾¾ ® h4 = h3 - hT (h3 - h4 s ) h3 - h4 s

= 310.24 - (0.85)(310.24 - 200.63)

= 217.07 kJ/kg Then the rate of refrigeration is

QL = m(qL ) = m(h1 - h4 ) = (0.2 kg/s)(280.13 - 217.07) kJ/kg = 12.61 kW

(b) The net power input is determined from

Wnet, in = Wcomp, in - Wturb, out where Wcomp, in = m (h2 - h1 ) = m (h2 s - h1 ) / hC

= (0.2 kg/s)éë(433.02 - 280.13) kJ/kgù û/ (0.80) = 38.22 kW Wturb, out = m (h3 - h4 ) = (0.2 kg/s)(310.24 - 217.07) kJ/kg = 18.63 kW

Thus,

Wnet, in = 38.22 - 18.63 = 19.59 kW (c) The COP of this ideal gas refrigeration cycle is determined from

COPR =

QL 12.61 kW = = 0.644 Wnet, in 19.59 kW

EES (Engineering Equation Solver) SOLUTION "Input data" $Array on T[1] = 7 [C] P[1]= 35 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-539

T[3] = 37 [C] P[3]=160 [kPa] m_dot=0.2 [kg/s] Eta_comp =0.80 Eta_turb = 0.85 "Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s2s=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = P[3] s2s=ENTROPY(Air,T=Ts2,P=P[2])"Ts2 is the isentropic value of T[2] at compressor exit" Eta_comp = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_comp_isen" m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor, assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s" h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3, assumed SSSF constant pressure process" m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit" Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen > W_dot_turb" m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0" hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect:" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor" COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s

12-72 Problem 12-71 is reconsidered. The effects of compressor and turbine isentropic efficiencies on the rate of refrigeration, the net power input, and the COP are to be investigated. Analysis The problem is solved using EES, and the solution is given below.

10-540

"Input data" T[1] = 7 [C] P[1]= 35 [kPa] T[3] = 37 [C] P[3]=160 [kPa] m_dot=0.2 [kg/s] Eta_comp = 0.80 Eta_turb = 0.85 "Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s2s=s[1] P[2] = P[3] s2s=ENTROPY(Air,T=Ts2,P=P[2]) Eta_comp = W_dot_comp_isen/W_dot_comp m_dot*h[1] + W_dot_comp_isen = m_dot*hs2 h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2] h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3" m_dot*h[2] + Q_dot_out = m_dot*h[3] h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4]) Eta_turb = W_dot_turb /W_dot_turb_isen m_dot*h[3] = W_dot_turb_isen + m_dot*hs4 hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4] h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s

10-541

ηcomp

0.7 0.75 0.8 0.85 0.9 0.95 1

COP

QRefrig

Winnet [kW]

0.3291 0.3668 0.4077 0.4521 0.5006 0.5538 0.6123

[kW] 9.334 9.334 9.334 9.334 9.334 9.334 9.334

28.36 25.45 22.9 20.65 18.65 16.86 15.24

10-542

12-73E An ideal gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 0.240 Btu / lbm ×R and k = 1.4 (Table A-2Ea).

10-543

Analysis From the isentropic relations, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ ÷ çè P1 ø

( k - 1)/ k

æP ö ÷ T4 = T3 ççç 4 ÷ ÷ çè P ÷ ø 3

= (450 R)(4)0.4/1.4 = 668.7 R 0.4/1.4

æ1 ö = (560 R) çç ÷ ÷ çè 4 ÷ ø

= 376.8 R

The COP of this ideal gas refrigeration cycle is determined from

COPR =

qL qL = wnet, in wcomp, in - wturb, out

h1 - h4 (h2 - h1 ) - (h3 - h4 )

T1 - T4 (T2 - T1 ) - (T3 - T4 )

450 - 376.8 = 2.06 (668.7 - 450) - (560 - 376.8)

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=5 [psia] T[1]=(-10+460) [R] r_p=4 T_3=100 [F] T[3]=(100+460) [R] "Properties, Table A-2Ea" c_p=0.240 [Btu/lbm-R] k=1.4 "Analysis" T[2]=T[1]*r_P^((k-1)/k) T[4]=T[3]*(1/r_P)^((k-1)/k) COP_R=(T[1]-T[4])/((T[2]-T[1])-(T[3]-T[4]))

10-544

12-74E An gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 0.240 Btu / lbm ×R and k = 1.4 (Table A-2Ea). Analysis From the isentropic relations, ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ ÷ çè P1 ø

( k - 1)/ k

æP ö ÷ T4 s = T3 ççç 4 ÷ ÷ çè P3 ÷ ø

= (450 R)(4)0.4/1.4 = 668.7 R 0.4/1.4

æ6 psia ÷ ö ÷ = (560 R) çç ÷ ÷ çè19 psia ø

= 402.9 R

and

hT =

h3 - h4 T - T4 = 3 h3 - h4 s T3 - T4 s

¾¾ ®

T4 = T3 - hT (T3 - T4 s ) = 560 - (0.94)(560 - 402.9) = 412.3 R

hC =

h2 s - h1 T2 s - T1 = h2 - h1 T2 - T1

¾¾ ®

T2 = T1 + (T2 s - T1 ) / hC = 450 + (668.7 - 450) / (0.87) = 701.4 R

The COP of this gas refrigeration cycle is determined from COPR =

qL qL = wnet, in wcomp, in - wturb, out

h1 - h4 (h2 - h1 ) - (h3 - h4 )

T1 - T4 (T2 - T1 ) - (T3 - T4 )

450 - 412.3 = 0.364 (701.4 - 450) - (560 - 412.3)

10-545

P[1]=5 [psia] T[1]=(-10+460) [R] r_p=4 T_3=100 [F] T[3]=(100+460) [R] eta_C=0.87 eta_T=0.94 DELTAP=1 [psia] "Properties, Table A-2Ea" c_p=0.240 [Btu/lbm-R] k=1.4 "Analysis" T_s[2]=T[1]*r_P^((k-1)/k) P[4]=P[1]+DELTAP P[3]=r_p*P[1]-DELTAP T_s[4]=T[3]*(P[4]/P[3])^((k-1)/k) Eta_T=(T[3]-T[4])/(T[3]-T_s[4]) Eta_C=(T_s[2]-T[1])/(T[2]-T[1]) COP_R=(T[1]-T[4])/((T[2]-T[1])-(T[3]-T[4])) 12-75 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are c p = 5.1926 kJ / kg ×K and k = 1.667 (Table A-2).

Analysis (a) From the isentropic relations, (k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

0.667/1.667

= (267 K )(4)

æP ö( ÷ T4 s = T3 ççç 4 ÷ ÷ çè P3 ÷ ø

k - 1)/ k

= 465.0 K

0.667/1.667

æ1 ö = (323 K )çç ÷ ÷ ÷ èç 4 ø

= 185.5 K

and hT =

h3 - h4 T - T4 = 3 ¾¾ ® T4 = T3 - hT (T3 - T4 s ) = 323 - (0.88)(323 - 185.5) h3 - h4 s T3 - T4 s

10-546

= 202.0 K = Tmin hC =

h2 s - h1 T2 s - T1 = ¾¾ ® T2 = T1 + (T2 s - T1 ) / hC = 267 + (465.0 - 267 ) / (0.88) h2 - h1 T2 - T1

= 491.9 K (b) The COP of this gas refrigeration cycle is determined from COPR =

qL qL = wnet, in wcomp, in - wturb, out

h1 - h4 (h2 - h1 )- (h3 - h4 )

T1 - T4 (T2 - T1 )- (T3 - T4 )

267 - 202.0

(491.9 - 267)- (323 - 202.0)

= 0.626 (c) The mass flow rate of helium is determined from m=

Qrefrig qL

Qrefrig h1 - h4

Qrefrig c p (T1 - T4 )

25 kJ/s = 0.0741 kg / s (5.1926 kJ/kg ×K )(267 - 202.0) K

EES (Engineering Equation Solver) SOLUTION "Given" r_p=4 T_1=-6 [C] T[1]=T_1+273 "[K]" T_3=50 [C] T[3]=T_3+273 "[K]" Eta_C=0.88 Eta_T=0.88 Q_dot_L=25 [kW] "Properties, Table A-2a" c_p=5.1926 [kJ/kg-K] k=1.667 "Analysis" "(a)" T_s[2]=T[1]*r_P^((k-1)/k) T_s[4]=T[3]*(1/r_P)^((k-1)/k) Eta_T=(T[3]-T[4])/(T[3]-T_s[4]) T_4=T[4] Eta_C=(T_s[2]-T[1])/(T[2]-T[1]) T_2=T[2] "(b)" COP_R=(T[1]-T[4])/((T[2]-T[1])-(T[3]-T[4])) "(c)" q_L=c_p*(T[1]-T[4]) m_dot=Q_dot_L/q_L

10-547

12-76 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4 (Table A-2). Analysis (a) From the isentropic relations,

æP ö( ÷ T2 s = T1 ççç 2 ÷ ÷ çè P1 ø÷

k - 1)/ k 0.4/1.4

= (273.2 K )(5)

= 432.4 K

hC =

h2 s - h1 T2 s - T1 = h2 - h1 T2 - T1

0.80 =

432.4 - 273.2 ¾¾ ® T2 = 472.5 K T2 - 273.2

The temperature at state 4 can be determined by solving the following two equations simultaneously:

æP5 ÷ ö( ç ÷ T5 s = T4 çç ÷ çè P ÷ ø

k - 1)/ k

hT =

0.4/1.4

æ1 ö = T4 çç ÷ ÷ çè5 ÷ ø

h4 - h5 T - 193.2 ® 0.85 = 4 h4 - h5 s T4 - T5 s

Using EES, we obtain

T4 = 281.3 K .

10-548

An energy balance on the regenerator may be written as or,

mc p (T3 - T4 ) = mc p (T1 - T6 ) ¾ ¾ ® T3 - T4 = T1 - T6

T6 = T1 - T3 + T4 = 273.2 - 308.2 + 281.3 = 246.3 K The effectiveness of the regenerator is eregen =

h3 - h4 T3 - T4 308.2 - 281.3 = = = 0.434 h3 - h6 T3 - T6 308.2 - 246.3

(b) The refrigeration load is

QL = mc p (T6 - T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(246.3 - 193.2)K = 21.36 kW (c) The turbine and compressor powers and the COP of the cycle are

WC, in = mc p (T2 - T1 ) = (0.4 kg/s)(1.005 kJ/kg.K)(472.5 - 273.2)K = 80.13 kW WT, out = mc p (T4 - T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(281.3 - 193.2)kJ/kg = 35.43 kW COP =

QL QL 21.36 = = = 0.478 Wnet, in WC, in - WT, out 80.13 - 35.43

(d) The simple gas refrigeration cycle analysis is as follows:

( æ1 ö ÷ T4 s = T3 ç ÷ ç ÷ çr ø è

k - 1)/ k

0.4/1.4

æ1 ö ÷ = (308.2 K )ç ÷ ç ÷ ç5 ø è

= 194.6 K

10-549

hT =

T3 - T4 308.2 - T4 ¾¾ ® 0.85 = ¾¾ ® T4 = 211.6 K T3 - T4 s 308.2 - 194.6

QL = mc p (T1 - T4 ) = (0.4 kg/s)(1.005 kJ/kg.K)(273.2 - 211.6)kJ/kg

= 24.74 kW Wnet, in = mc p (T2 - T1 ) - mc p (T3 - T4 ) = (0.4 kg/s)(1.005 kJ/kg.K) [(472.5 - 273.2) - (308.2 - 211.6)kJ/kg ]

= 41.32 kW COP =

QL 24.74 = = 0.599 41.32 Wnet, in

EES (Engineering Equation Solver) SOLUTION "GIVEN" r=5 T_1=(0+273.15) [K] T_3=(35+273.15) [K] T_5=(-80+273.15) [K] m_dot=0.4 [kg/s] eta_C=0.80 eta_T=0.85 "PROPERTIES, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "ANALYSIS" T_2s=T_1*r^((k-1)/k) T_2=T_1+(T_2s-T_1)/eta_C T_4=T_5s/((1/r)^((k-1)/k)) T_4=T_5+eta_T*(T_4-T_5s) T_6=T_1-T_3+T_4 "energy balance on the regenerator" epsilon_regen=(T_3-T_4)/(T_3-T_6) Q_dot_L=m_dot*c_p*(T_6-T_5) W_dot_net_in=W_dot_C_in-W_dot_T_out W_dot_C_in=m_dot*c_p*(T_2-T_1) W_dot_T_out=m_dot*c_p*(T_4-T_5) COP=Q_dot_L/W_dot_net_in "Simple gas refrigeration cycle analysis, the subscript 'a' is used" T_1_a=T_1 T_2_a=T_2 T_3_a=T_3 T_4s_a=T_3_a*(1/r)^((k-1)/k) T_4_a=T_3_a-eta_T*(T_3_a-T_4s_a) Q_dot_L_a=m_dot*c_p*(T_1_a-T_4_a) W_dot_in_net_a=m_dot*c_p*(T_2_a-T_1_a)-m_dot*c_p*(T_3_a-T_4_a) COP_a=Q_dot_L_a/W_dot_in_net_a

12-77 An ideal gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system and the mass flow rate of air are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-550

1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4 (Table A-2a).

Analysis From the isentropic relations, ( k - 1)/ k

æP ö÷ T2 = T1 ççç 2 ÷ ÷ çè P ÷ ø

= (249 K)(3)0.4/1.4 = 340.8 K

( k - 1)/ k

æP ö ÷ T4 = T3 ççç 4 ÷ ÷ çè P3 ÷ ø

( k - 1)/ k

æP ö ÷ T6 = T5 ççç 6 ÷ ÷ èç P ÷ ø 5

= (278 K)(3)0.4/1.4 = 380.5 K 0.4/1.4

æ1 ö = (278 K) çç ÷ ÷ çè9 ÷ ø

= 148.4 K

The COP of this ideal gas refrigeration cycle is determined from COPR =

qL qL = wnet, in wcomp, in - wturb, out

10-551

h1 - h6 (h2 - h1 ) + (h4 - h3 ) - ( h5 - h6 )

T1 - T6 (T2 - T1 ) + (T4 - T3 ) - (T5 - T6 )

249 - 148.4 (340.8 - 249) + (380.5 - 278) - (278 - 148.4)

= 1.56 The mass flow rate of the air is determined from QRefrig = mc p (T1 - T6 ) ¾ ¾ ® m=

QRefrig c p (T1 - T6 )

(45, 000 / 3600) kJ/s = 0.124 kg / s (1.005 kJ/kg ×K)(249 - 148.4) K

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=90 [kPa] T[1]=(-24+273) [K] r_p=3 T[3]=(5+273) [K] T[5]=T[3] Q_dot_L=45000 [kJ/h] "Properties, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T[2]=T[1]*r_p^((k-1)/k) T[4]=T[3]*(r_p)^((k-1)/k) T[6]=T[5]*(1/r_p^2)^((k-1)/k) COP_R=(T[1]-T[6])/((T[2]-T[1])+(T[4]-T[3])-(T[5]-T[6])) m_dot=Q_dot_L*Convert(kJ/h, kW)/(c_p*(T[1]-T[6]))

12-78 A gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4 (Table A-2a).

10-552

Analysis From the isentropic relations, ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ çè P ÷ ø

= (249 K)(3)0.4/1.4 = 340.8 K

( k - 1)/ k

æP ö ÷ T4 s = T3 ççç 4 ÷ ÷ çè P ÷ ø

= (278 K)(3)0.4/1.4 = 380.5 K

( k - 1)/ k

æP ö ÷ T6 s = T5 ççç 6 ÷ ÷ çè P5 ÷ ø

0.4/1.4

æ1 ÷ ö = (278 K) çç ÷ ÷ èç9 ø

= 148.4 K

and hC =

h2 s - h1 T2 s - T1 = ¾¾ ® T2 = T1 + (T2 s - T1 ) / hC = 249 + (340.8 - 249) / 0.85 = 357.0 K h2 - h1 T2 - T1

hC =

h4 s - h3 T4 s - T3 = ¾¾ ® T4 = T3 + (T4 s - T3 ) / hC = 278 + (380.5 - 278) / 0.85 = 398.6 K h4 - h3 T4 - T3

hT =

h5 - h6 T - T6 = 5 ¾¾ ® T6 = T5 - hT (T5 - T6 s ) = 278 - (0.95)(278 - 148.4) = 154.9 K h5 - h6 s T5 - T6 s

10-553

COPR =

qL qL = wnet, in wcomp, in - wturb, out

h1 - h6 (h2 - h1 ) + (h4 - h3 ) - ( h5 - h6 )

T1 - T6 249 - 154.9 = = 0.892 (T2 - T1 ) + (T4 - T3 ) - (T5 - T6 ) (357.0 - 249) + (398.6 - 278) - (278 - 154.9)

The mass flow rate of the air is determined from QRefrig = mc p (T1 - T6 ) ¾ ¾ ® m=

QRefrig c p (T1 - T6 )

(45, 000 / 3600) kJ/s = 0.132 kg / s (1.005 kJ/kg ×K)(249 - 154.9) K

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=90 [kPa] T[1]=(-24+273) [K] r_p=3 T[3]=(5+273) [K] T[5]=T[3] Q_dot_L=45000 [kJ/h] eta_C=0.85 eta_T=0.95 "Properties, Table A-2a" c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T_s[2]=T[1]*r_p^((k-1)/k) Eta_C=(T_s[2]-T[1])/(T[2]-T[1]) T_s[4]=T[3]*(r_p)^((k-1)/k) Eta_C=(T_s[4]-T[3])/(T[4]-T[3]) T_s[6]=T[5]*(1/r_p^2)^((k-1)/k) Eta_T=(T[5]-T[6])/(T[5]-T_s[6]) COP_R=(T[1]-T[6])/((T[2]-T[1])+(T[4]-T[3])-(T[5]-T[6])) m_dot=Q_dot_L*Convert(kJ/h, kW)/(c_p*(T[1]-T[6]))

12-79 Ambient air at enters the compressor of a simple gas refrigeration cycle at 98 kPa and 17°C and the turbine at 70°C. The pressure ratio in the cycle is 5 and the cooled space temperature is 3°C. Assuming both the compressor and the turbine to be isentropic, determine (a) the COP of the cycle, (b) the exergy destruction in each component and the total exergy destruction, and (c) the second-law efficiency of the cycle. Take c p = 1.005 kJ / kg ×K and k = 1.4 for air.

12-79 A simple gas refrigeration cycle is considered. Various performance parameters are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Properties The air properties are given as c p = 1.005 kJ / kg ×K and k = 1.4.

10-554

Analysis (a) From the isentropic relations,

P  T2  T1  2   P1 

( k 1)/ k

P  T4  T3  4   P3 

 (290 K)(5)0.4/1.4  459.3 K ( k 1)/ k

1  (343 K)   5

0.4/1.4

 216.6 K

The heat removal per unit mass is

qL  c p (T1  T4 )  (1.005 kJ/kg  K)(290  216.6) K  73.80 kJ/kg The heat rejected is

qH  c p (T2  T3 )  (1.005 kJ/kg  K)(459.3  343) K  116.9 kJ/kg Net work input and the COP are

win  c p (T2  T1 )  (1.005 kJ/kg  K)(459.3  290) K  170.2 kJ/kg wout  c p (T3  T4 )  (1.005 kJ/kg  K)(343  216.6) K  127.1 kJ/kg wnet  win  wout  170.2  127.1  43.1 kJ/kg

COP 

qL 73.80   1.713 wnet 43.1

(b) The exergy destruction in each component of the cycle is determined as follows Compressor:

s gen,1 2  s 2  s1  0 (isentropic process) ex dest,1-2  T0 s gen,1 -2  0

Heat exchanger (Process 2-3): sgen,23  s3  s2 

T  P  q qH  c p ln  3   R ln  3   H TH  T2   P2  TH

116.9 kJ/kg  343 K   (1.005 kJ/kg  K) ln  0 459.3 K 290 K    0.1096 kJ/kg  K

10-555

Turbine: s gen,3 4  s 4  s 3  0 (isentropic process) ex dest,3-4  T0 s gen,3 -4  0

Heat exchanger (Process 4-1): sgen,41  s1  s4 

T  P  q qL  c p ln  1   R ln  1   L TL  T4   P4  TL

73.8 kJ/kg  290 K   (1.005 kJ/kg  K) ln    0)  216.6 K 276 K    0.02605 kJ/kg  K

exdest,4-1  T0 sgen,4-1  (290 K)(0.02605 kJ/kg  K)  7.56 kJ / kg The total exergy destruction can be determined by adding exergy destructions in each component: exdest,total  exdest,1-2  exdest,2-3  exdest,3-4  exdest,4-1  0  31.8  0  7.56  39.3 kJ / kg

(c) The exergy of the heat transferred from the low-temperature medium is  T   290  exqL  qL 1  0   (73.8 kJ/kg) 1    3.74 kJ/kg T  276   L 

The second-law efficiency of the cycle is

II 

exqL wnet



3.74  0.0869  8.7% 43.1

The total exergy destruction in the cycle can also be determined from

exdest,total  wnet  exqL  43.1  3.74  39.4 kJ/kg The result is practically the same, as expected.

EES (Engineering Equation Solver) SOLUTION "Given" r_p=5 T_1=17+273 [K] P_1=98 [kPa] T_3=70+273 [K] c_p=1.005 [kJ/kg-K] k=1.4 R=0.287 [kJ/kg-K] T_L=3+273 T_H=T_1 T_0=T_H "Analysis" P_2=r_P*P_1; P_3=P_2; P_4=P_1 T_2=T_1*r_P^((k-1)/k) T_4=T_3*(1/r_P)^((k-1)/k) q_L=c_p*(T_1-T_4) q_H=c_p*(T_2-T_3) w_in=c_p*(T_2-T_1) w_out=c_p*(T_3-T_4) w_net=w_in-w_out COP=q_L/w_net s_gen_12=DELTAs_12 DELTAs_12=c_p*ln(T_2/T_1)-R*ln(P_2/P_1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-556

ex_dest_12=T_0*s_gen_12 DELTAs_23=c_p*ln(T_3/T_2)-R*ln(P_3/P_2) s_gen_23=DELTAs_23+q_H/T_H ex_dest_23=T_0*s_gen_23 DELTAs_34=c_p*ln(T_4/T_3)-R*ln(P_4/P_3) s_gen_34=DELTAs_34 ex_dest_34=T_0*s_gen_34 DELTAs_41=c_p*ln(T_1/T_4)-R*ln(P_1/P_4) s_gen_41=DELTAs_41-q_L/T_L ex_dest_41=T_0*s_gen_41 Ex_qL=-q_L*(1-T_0/T_L) eta_II=Ex_qL/w_net ex_dest=w_net-ex_qL

Absorption Refrigeration Systems 12-80C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquid phase instead of in the vapor form.

12-81C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source. Its disadvantages include being expensive, complex, and requiring an external heat source.

12-82C In absorption refrigeration, water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point.

12-83C The rectifier separates the water from NH3 and returns it to the generator. The regenerator transfers some heat from the water-rich solution leaving the generator to the NH3 -rich solution leaving the pump.

12-84C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid in the generator is heated to maximize the refrigerant content of the vapor.

12-85C The coefficient of performance of absorption refrigeration systems is defined as

COPR =

QL Q desired output = @ L required input Qgen + Wpump, in Qgen

12-86E The conditions at which an absorption refrigeration system operates are specified. The COP is also given. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-557

QL = COPR Qgen = (0.55)(105 Btu/h ) = 0.55×105 Btu / h

EES (Engineering Equation Solver) SOLUTION "Given" T_s=250+460 "[R]" T_L=0+460 "[R]" T_0=80+460 "[R]" Q_dot_gen=1E5 "[Btu/h]" COP_R=0.55 "Analysis" Q_dot_L_max=COP_R*Q_dot_gen

12-87 The COP of an absorption refrigeration system that operates at specified conditions is given. It is to be determined whether the given COP value is possible. Analysis The maximum COP that this refrigeration system can have is æ T öæ TL ö æ 292 K öæ ö ÷ ÷ ç 273 ÷ ÷çç ÷= çç1COPR, max = ççç1- 0 ÷ ÷ ÷ ÷ ÷ ÷çç 292 - 273 ø ÷= 2.97 ÷ççT0 - TL ø ÷ çè 368 K øè çè Ts øè

which is smaller than 3.1. Thus the claim is not possible.

EES (Engineering Equation Solver) SOLUTION "Given" T_s=(95+273) [K] T_L=(0+273) [K] T_0=(19+273) [K] COP=3.1 "Analysis" COP_R_max=(1-T_0/T_s)*(T_L/(T_0-T_L))

12-88 The conditions at which an absorption refrigeration system operates are specified. The maximum COP this absorption refrigeration system can have is to be determined. Analysis The maximum COP that this refrigeration system can have is æ T öæ TL ö æ 298 K öæ ö ÷ çç ÷ çç 277 ÷ ÷ ÷ COPR, max = ççç1- 0 ÷ = çç1÷ ÷ ÷ ÷ ÷ ÷= 3.19 ç ç ç ÷çT0 - TL ø ÷ è 393 K øè 298 - 277 ø èç Ts øè

EES (Engineering Equation Solver) SOLUTION "Given" T_s=(120+273) [K] T_L=(4+273) [K] T_0=(25+273) [K] "Analysis" COP_R_max=(1-T_0/T_s)*(T_L/(T_0-T_L))

10-558

12-89 The conditions at which an absorption refrigeration system operates are specified. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis The maximum COP that this refrigeration system can have is æ T öæ TL ÷ ö æ 298 K öæ 255 ö çç ÷çç ÷= 1.316 ÷ ÷ COPR, max = ççç1- 0 ÷ = çç1÷ ÷ ÷ ÷ ÷ç 298 - 255 ø ÷ ç ÷çT0 - TL ÷ èç Ts øè ø çè 383 K øè

Thus,

QL, max = COPR, max Qgen = (1.316)(5´ 105 kJ/h ) = 6.58´ 105 kJ / h

EES (Engineering Equation Solver) SOLUTION "Given" T_s=(110+273) [K] T_L=(-18+273) [K] T_0=(25+273) [K] Q_dot_gen=5E5 [kJ/h] "Analysis" COP_R_max=(1-T_0/T_s)*(T_L/(T_0-T_L)) Q_dot_L_max=COP_R_max*Q_dot_gen

12-90 Consider an ammonia-water absorption refrigeration system. Ammonia enters the condenser at 400 psia and 150F at a rate of 0.86 lbm / s , and leaves the condenser as a saturated liquid. Ammonia is throttled to a pressure of 25 psia and leaves the evaporator as a saturated vapor. Heat is supplied to the generator by geothermal liquid water that enters at 260F at a rate of 550 lbm / min and leaves at 220F. Determine the rate of cooling provided by the system in tons of refrigeration and the mass flow rate of geothermal water if the COP of the system is 0.9. The enthalpies of ammonia at various states of the system are hcond,in = 646.0 Btu / lbm , hcond,out = 216.5 Btu / lbm , hevap,out = 616.8 Btu / lbm . Also, take the specific heat of geothermal water to be 1.0 Btu / lbm ×° F .

12-90 An absorption refrigeration system using ammonia-water solution is considered. The rate of cooling provided by the system and the mass flow rate of geothermal water are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Properties The enthalpies of ammonia at various states of the system are given as hcond,in = 646.0 Btu / lbm , hcond,out = 216.5 Btu / lbm , hevap,out = 616.8 Btu / lbm . The specific heat of geothermal water is given as 1.0 Btu / lbm ×°F .

Analysis Noting that the throttling process is isenthalpic,

hcond,out  hevap,in The rate of cooling provided by the system is

QL  mR (hevap,out  hevap,in )  (0.86 lbm/s)(616.8  216.5) Btu/lbm  344.3 Btu/s

 103.3 tons of refrigeration since 1 ton of refrigeration = 200 Btu / min . From the COP relation © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-559

COP 

QL 344.3 Btu/s   0.9    Qgen  382.5 Btu/s Qgen Qgen

Then the mass flow rate of geothermal water is

Qgen  mgeo c p (Tgeo;in  Tgeo;out ) 382.5 Btu/s  mgeo (1 Btu/lbm F)(260  240)F mgeo  9.56 lbm / s

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_3=400 [psia] T_3=150 [F] T_geo1=260 [F] T_geo2=220 [F] m_dot_R=0.86 [lbm/s] P_5=25 [psia] COP=0.9 "PROPERTIES" Fluid$='ammonia' h_3=enthalpy(Fluid$, P=P_3, T=T_3) h_4=enthalpy(Fluid$, P=P_3, x=0) h_5=h_4 h_6=enthalpy(Fluid$, P=P_5, x=1) Q_dot_gen=m_dot_geo*c_p*(T_geo1-T_geo2) c_p=1.0 Q_dot_L=m_dot_R*(h_6-h_5) Q_dot_H=m_dot_R*(h_3-h_4) COP=Q_dot_L/Q_dot_gen

12-91 Pure ammonia enters the condenser of an ammonia-water absorption refrigeration system at 3200 kPa and 70C and leaves the condenser as a saturated liquid. It is then throttled to a pressure of 220 kPa. Ammonia leaves the evaporator as a saturated vapor. Heat is supplied to the solution in the generator by solar energy, which is incident on the collector at a rate of 620 W / m2 . The total surface area of the collectors is 30 m2 and the efficiency of the collectors is 70 percent (i.e., 70 percent of the solar energy input is transferred to the ammonia-water solution). If the mass flow rate of ammonia through the evaporator is 0.012 kg / s , determine the COP of the system. The enthalpies of ammonia at various states of the system are given as hcond,in = 1491.9 kJ / kg , hcond,out = 537.0 kJ / kg , hevap,out = 1442.0 kJ / kg . 12-91 An absorption refrigeration system using ammonia-water solution is considered. The COP of the system is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Properties The enthalpies of ammonia at various states of the system are given as hcond,in = 1491.9 kJ / kg , hcond,out = 537.0 kJ / kg , hevap,out = 1442.0 kJ / kg .

10-560

Qsolar  qA  (0.62 kW/m 2 )(30 m 2 )  18.6 kW The rate of heat transferred to the solution is

Qgen  collector Qsolar  (0.70)(18.6 kW)/m2 )  13.02 kW Noting that the throttling process is isenthalpic,

hcond,out  hevap,in The rate of cooling provided by the system is

QL  mR (hevap,out  hevap,in )  (0.012 kg/s)(1442  537.0) kJ/kg  10.86 kW Then the COP becomes COP 

QL 10.86 kW   0.834 Qgen 13.02 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_3=3200 [kPa] T_3=70 [C] P_5=220 [kPa] q=0.62 [kW/m^2] A=30 [m^2] eff=0.70 m_dot_R=0.012 Fluid$='ammonia' h_3=enthalpy(Fluid$, P=P_3, T=T_3) h_4=enthalpy(Fluid$, P=P_3, x=0) h_5=h_4 h_6=enthalpy(Fluid$, P=P_5, x=1) Q_tot=q*A Q_dot_gen=eff*Q_tot Q_dot_L=m_dot_R*(h_6-h_5) COP=Q_dot_L/Q_dot_gen

12-92 An ammonia-water absorption refrigeration system is used to remove heat from a cooled space at 2C at a rate of 60 kW. The system operates in an environment at 20C. Heat is supplied to the cycle by condensing saturated steam at 200C at a rate of 0.038 kg / s . Determine the COP of the system and the COP of a reversible absorption refrigeration system operating between the same temperatures. 12-92 An absorption refrigeration system using ammonia-water solution is considered. The COP of the system and the reversible COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Properties The enthalpy of vaporization of water at 200C is h fg = 1940.34 kJ / kg . Analysis The rate of heat input in the generator is

Qgen  ms h fg  (0.038 kg/s)(1940.34 kJ/kg)  73.73 kW The COP of the cycle is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-561

COPabs 

QL 60 kW   0.814 Qgen 73.73 kW

The COP of the reversible absorption refrigerator is  T  TL   293  275    1  COPabs, rev  1  0     5.81  Ts  T0  TL   473  293  275  

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=2+273 T_0=20+273 [K] T_s=200+273 [K] Q_dot_L=66 [kW] m_dot_s=0.04 [kg/s] "PROPERTIES" Fluid$='Steam_iapws' h_f=enthalpy(Fluid$, T=T_s, x=0) h_g=enthalpy(Fluid$, T=T_s, x=1) h_fg=h_g-h_f "ANALYSIS" Q_dot_gen=m_dot_s*h_fg COP=Q_dot_L/Q_dot_gen COP_absorp_rev=(1-T_0/T_s)*(T_L/(T_0-T_L))

Special Topic: Thermoelectric Power Generation and Refrigeration Systems 12-93C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit.

12-94C When two wires made from different metals joined at both ends (junctions) forming a closed circuit and one of the joints is heated, a current flows continuously in the circuit. This is called the Seebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction is cooled. This is called the Peltier effect.

12-95C No.

12-96C No.

12-97C Yes.

12-98C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference, and the temperature can be measured by simply measuring voltages.

10-562

12-99C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals.

12-100C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because a thermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid.

12-101E A thermoelectric generator that operates at specified conditions is considered. The maximum thermal efficiency this thermoelectric generator can have is to be determined. Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency, hth, max = hth, Carnot = 1-

TL 550 R = 1= 31.3% TH 800 R

EES (Engineering Equation Solver) SOLUTION "Given" T_H=(340+460) [R] T_L=(90+460) [R] "Analysis" Eta_th_max=1-T_L/T_H

12-102 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined. Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits, Thus,

COPmax = COPR, Carnot = Win, min =

1 1 = = 10.72 T / T 1 293 K / ( H L) ( ) (268 K )- 1

QL 130 W = = 12.1 W COPmax 10.72

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(-5+273) [K] T_H=(20+273) [K] Q_dot_L=130 [W] "Analysis" COP_max=1/((T_H/T_L)-1) W_dot_in_min=Q_dot_L/COP_max

12-103 A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR , COPR =

QL QL 180 W ¾¾ ® Win = = = 1200 W COPR 0.15 Win

10-563

EES (Engineering Equation Solver) SOLUTION "Given" COP=0.15 Q_dot_L=180 [W] "Analysis" W_dot_in=Q_dot_L/COP

12-104 The maximum power a thermoelectric generator can produce is to be determined. Analysis The maximum thermal efficiency this thermoelectric generator can have is hth, max = 1-

TL 295 K = 1= 0.1873 TH 363 K

Thus,

Wout, max = hth, max Qin = (0.1873)(7´ 106 kJ/h) = 1.31´ 106 kJ/h = 364 kW

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_in=7E6 [kJ/h] T_H=(90+273) [K] T_L=(22+273) [K] "Analysis" Eta_th_max=1-T_L/T_H W_dot_out_max=Eta_th_max*Q_dot_in*Convert(kJ/h, kW)

12-105 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The average COP of this refrigerator is to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, r = 1 kg / L and cp = 4.18 kJ / kg ×°C (Table A-3). Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks,

m = rV = 9´ (1 kg/L)(0.350 L) = 3.15 kg Qcooling = mcD T = (3.15 kg)(4.18 kJ/kg ×°C)(25 - 3)°C = 290 kJ

Qcooling

290 kJ = = 0.00671 kW = 6.71 W Dt 12´ 3600 s The electric power consumed by the refrigerator is Qcooling =

Win = VI = (12 V)(3 A) = 36 W Then the COP of the refrigerator becomes

COP =

Qcooling Win

6.71 W = 0.186 » 0.20 36 W

10-564

EES (Engineering Equation Solver) SOLUTION "Given" V=12 [volt] I=3 [ampere] n_can=9 V_can=0.350 [liter] T_1=25 [C] T_2=3 [C] DELTAt=12 [h] "Properties, Table A-3" rho=1000 [kg/m^3] c=4.18 [kJ/kg-C] "Analysis" m=n_can*rho*V_can*Convert(liter, m^3) Q_cooling=m*c*(T_1-T_2) Q_dot_cooling=Q_cooling/DELTAt*Convert(kJ/h, W) W_dot_in=V*I COP=Q_dot_cooling/W_dot_in

Review Problems 12-106 A room is cooled adequately by a 5000 Btu / h window air-conditioning unit. The rate of heat gain of the room when the air-conditioner is running continuously is to be determined. Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist. Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the air-conditioning system,

Qheat gain = Qcooling = 5000 Btu / h

EES (Engineering Equation Solver) SOLUTION "Given" A_room=15 [m^2] Q_dot_cooling=5000 [Btu/h] COP=3.5 "Analysis" Q_dot_heatgain=Q_dot_cooling

12-107 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP, the condenser and evaporator pressures, and the net work input are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

10-565

Analysis (a) The COP of this refrigeration cycle is determined from COPR, C =

1 1 = = 5.060 (TH / TL )- 1 (303 K)/ (253 K)- 1

(b) The condenser and evaporative pressures are (Table A-11) Pevap = Psat @ - 20°C = 132.82 kPa

Pcond = Psat @ 30°C = 770.64 kPa (c) The net work input is determined from

h1 = (h f + x1h fg )

@ - 20° C

h2 = (h f + x2 h fg )

@ - 20° C

= 25.47 + (0.15)(212.96) = 57.42 kJ/kg

= 25.47 + (0.80)(212.96) = 195.84 kJ/kg

qL = h2 - h1 = 195.84 - 57.42 = 138.4 kJ/kg wnet, in =

qL 138.4 kJ/kg = = 27.36 kJ / kg COPR 5.060

EES (Engineering Equation Solver) SOLUTION "Given" T[3]=30 [C] T_H=T[3]+273 "[K]" T[1]=-20 [C] T_L=T[1]+273 "[K]" x[1]=0.15 x[2]=0.8 "Analysis" Fluid$='R134a' "(a)" COP_R_C=1/((T_H/T_L)-1) "(b)" P_evap=pressure(Fluid$, T=T[1], x=1) P_cond=pressure(Fluid$, T=T[3], x=1) "(c)" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) T[2]=T[1] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-566

h[2]=enthalpy(Fluid$, T=T[2], x=x[2]) q_L=h[2]-h[1] w_net_in=q_L/COP_R_C

12-108 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. Cooling water flows through the water jacket surrounding the condenser. To produce ice, potable water is supplied to the chiller section of the refrigeration cycle. The hardware and the T-s diagram for this refrigerant-ice making system are to be sketched. The mass flow rates of the refrigerant and the potable water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

T1 = 140 kPa sat. vapor

ïü ïý ïïþ

h1 = hg @ 140 kPa = 239.19 kJ/kg s1 = sg @ 140 kPa = 0.94467 kJ/kg ×K

P2 = 1200 kPa s2 = s1

ü ïï ý ïïþ

h2 = 284.14 kJ/kg

P3 = 1200 kPa sat. liquid

ü ïï ý ïïþ

h3 = h f @ 1200 kPa = 117.79 kJ/kg

h4 @ h3 = 117.79 kJ/kg

(throttling)

(b) An energy balance on the condenser gives

10-567

Solving for the mass flow rate of the refrigerant mR =

mw c p D T h2 - h3

(200 kg/s)(4.18 kJ/kg ×K)(10 K) = 50.26 kg / s (284.14 - 117.79)kJ/kg

QL = mR (h1 - h4 ) = mw hif Solving for the mass flow rate of the potable water mw =

mR (h1 - h4 ) (50.26 kg/s)(239.19 - 117.79)kJ/kg = = 18.32 kg / s hif 333 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=140 [kPa] P[2]=1200 [kPa] m_dot_w_cond=200 [kg/s] DELTAT_w=10 [C] h_if=333 [kJ/kg] "Analysis" Fluid$='R134a' "(b)" x[1]=1 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) x[3]=0 P[3]=P[2] h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] Q_dot_H=m_dot_R*(h[2]-h[3]) Q_dot_H=m_dot_w_cond*c_p*DELTAT_w c_p=4.18 [kJ/kg-C] "Table A-3" Q_dot_L=m_dot_R*(h[1]-h[4]) Q_dot_L=m_dot_w_evap*h_if

12-109 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the COP of this heat pump are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

10-568

h = hg@320 kPa = 251.93 kJ/kg P1 = 320 kPa ïüï 1 ý s1 = sg@320 kPa = 0.93026 kJ/kg ×K ïïþ sat. vapor v1 = v g@320 kPa = 0.063681 m3 /kg

P2 = 1.4 MPa ü ïï ý h2 = 282.60 kJ/kg ïïþ s2 = s1 P3 = 1.4 MPa ü ïï ý h3 = h f@.4 MPa = 127.25 kJ/kg ïïþ sat. liquid

h4 @ h3 = 127.25 kJ/kg (throttling )

The rate of heat supply to the house is determined from

QH = m(h2 - h3 ) = (0.25 kg/s)(282.60 - 127.25) kJ/kg = 38.8 kW (b) The volume flow rate of the refrigerant at the compressor inlet is

V1 = mv1 = (0.25 kg/s)(0.063681 m3 /kg ) = 0.0159 m 3 / s (c) The COP of this heat pump is determined from COPR =

h - h3 qL 282.60 - 127.25 = 2 = = 5.07 win h2 - h1 282.60 - 251.93

EES (Engineering Equation Solver) SOLUTION "Given" m_dot=0.25 [kg/s] P[1]=320 [kPa] P[2]=1400 [kPa] "Analysis" Fluid$='R134a' "(a)" "compressor" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-569

x[1]=1 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) v[1]=volume(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" P[3]=P[2] x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] Q_dot_H=m_dot*(h[2]-h[3]) "(b)" V_dot_1=m_dot*v[1] "(c)" q_L=h[2]-h[3] w_in=h[2]-h[1] COP_R=q_L/w_in

12-110 An ideal vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The evaporator is located inside the air handler of building. The hardware and the T-s diagram for this heat pump application are to be sketched. The COP of the unit and the ratio of volume flow rate of air entering the air handler to mass flow rate of R22 through the air handler are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-570

T1 = - 5°C sat. vapor

ü ïï ý ïïþ

h1 = hg @ - 5°C = 248.1 kJ/kg s1 = sg @ - 5°C = 0.9344 kJ/kg ×K

P2 = 1728 kPa s2 = s1

ïü ïý ïïþ

h2 = 283.7 kJ/kg

P3 = 1728 kPa sat. liquid

ïü ïý ïïþ

h3 = h f @ 1728 kPa = 101 kJ/kg

h4 @ h3 = 101 kJ/kg (throttling) (b) The COP of the heat pump is determined from its definition, COPHP =

h - h3 qH 283.7 - 101 = 2 = = 5.13 win h2 - h1 283.7 - 248.1

QH = mR (h2 - h3 ) = ma c p D T =

Va cpD T va

Rearranging, we obtain the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler

Va h2 - h3 (283.7 - 101) kJ/kg = = mR (1/ v )c p D T (1/ 0.847 m3 /kg)(1.005 kJ/kg ×K))(20 K)

= 7.699 (m3 air/s)/(kg R22/s)

= 462 (m3 air / min) / (kg R22 / s) Note that the specific volume of air is obtained from ideal gas equation taking the pressure of air to be 101 kPa and using the room temperature of air (25C = 298 K) to be 0.847m3 / kg .

10-571

12-111 A large refrigeration plant that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant, the power input to the compressor, and the mass flow rate of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa ü ïï h1 = hg @120 kPa = 236.99 kJ/kg ý ïïþ s1 = sg@120 kPa = 0.94789 kJ/kg ×K sat. vapor P2 = 0.7 MPa ïü ïý h = 273.52 kJ/kg (T = 34.95°C) 2 ïïþ 2 s2 = s1 P3 = 0.7 MPa ïü ïý h = h f@0.7 MPa = 88.82 kJ/kg ïïþ 3 sat. liquid h4 @ h3 = 88.82 kJ/kg (throttling )

The mass flow rate of the refrigerant is determined from

QL 100 kJ/s = = 0.6749 kg / s h1 - h4 (236.99 - 88.82) kJ/kg

(b) The power input to the compressor is

Win = m (h2 - h1 ) = (0.6749 kg/s)(273.52 - 236.99)kJ/kg = 24.7 kW (c) The mass flow rate of the cooling water is determined from

QH = m(h2 - h3 ) = (0.6749 kg/s)(273.52 - 88.82)kJ/kg = 124.7 kW mcooling =

QH 124.7 kJ/s = = 3.73 kg / s (c p D T ) water (4.18 kJ/kg ×°C)(8°C)

10-572

T_plant=-15 [C] Q_dot_L=100 [kW] DELTAT_water=8 [C] P[1]=120 [kPa] P[2]=700 [kPa] "Properties" c_water=4.18 [kJ/kg-K] "at room conditions" "Analysis" Fluid$='R134a' "(a)" "compressor" x[1]=1 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) T[2]=temperature(Fluid$, P=P[2], s=s[2]) "expansion valve" P[3]=P[2] x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] m_dot_R=Q_dot_L/(h[1]-h[4]) "(b)" W_dot_in=m_dot_R*(h[2]-h[1]) "(c)" Q_dot_H=m_dot_R*(h[2]-h[3]) m_dot_water=Q_dot_H/(c_water*DELTAT_water)

12-112 Problem 12-111 is reconsidered. The effect of evaporator pressure on the COP and the power input is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=120 [kPa] P[2] = 700 [kPa] Q_dot_in= 100 [kW] DELTAT_cw = 8 [C] C_P_cw = 4.18 [kJ/kg-K] Eta_c=1.0 Fluid$='R134a' "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) h[1]+Wcs=h2s Wc=Wcs/Eta_c h[1]+Wc=h[2] s[2]=entropy(Fluid$,h=h[2],P=P[2]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-573

T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] Q_dot_out=m_dot*Qout "Throttle Valve" h[4]=h[3] x[4]=quality(Fluid$,h=h[4],P=P[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] Q_dot_in=m_dot*Q_in COP=Q_dot_in/W_dot_c COP_plot = COP W_dot_in = W_dot_c m_dot_cw*C_P_cw*DELTAT_cw = Q_dot_out

P1 [kPa] 120 150 180 210 240 270 300 330 360 390

COP 4.056 4.743 5.475 6.271 7.147 8.127 9.236 10.51 11.99 13.75

WC [kW] 24.65 21.08 18.26 15.95 13.99 12.31 10.83 9.516 8.339 7.274

10-574

12-113 A large refrigeration plant operates on the vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of the refrigerant, the power input to the compressor, the mass flow rate of the cooling water, and the rate of exergy destruction associated with the compression process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa ü ïï h1 = hg @120 kPa = 236.99 kJ/kg ý ïïþ s1 = sg@120 kPa = 0.94789 kJ/kg ×K sat. vapor

P2 = 0.7 MPa ü ïï ý h2 s = 273.52 kJ/kg (T2 = 34.95°C) ïïþ s2 = s1 P3 = 0.7 MPa ïü ïý h = h f@0.7 MPa = 88.82 kJ/kg ïïþ 3 sat. liquid

h4 @ h3 = 88.82 kJ/kg (throttling )

The mass flow rate of the refrigerant is determined from

QL 100 kJ/s = = 0.6749 kg / s h1 - h4 (236.99 - 88.82) kJ/kg

(b) The actual enthalpy at the compressor exit is

hC =

h2 s - h1 ¾¾ ® h2 = h1 + (h2 s - h1 ) / hC = 236.99 + (273.52 - 236.99) / (0.75) h2 - h1 = 285.70 kJ/kg

Thus,

Win = m (h2 - h1 ) = (0.6749 kg/s)(285.70 - 236.99) kJ/kg = 32.9 kW (c) The mass flow rate of the cooling water is determined from and

QH = m(h2 - h3 ) = (0.6749 kg/s)(285.70 - 88.82) kJ/kg = 132.9 kW mcooling =

QH 132.9 kJ/s = = 3.97 kg / s (c p D T ) water (4.18 kJ/kg ×°C)(8°C)

The exergy destruction associated with this adiabatic compression process is determined from

10-575

where ïü ïý s = 0.98665 kJ/kg ×K 2 h2 = 285.70 kJ/kg ïïþ P2 = 0.7 MPa

Thus,

X destroyed = (298 K )(0.6749 kg/s)(0.98665 - 0.94789) kJ/kg ×K = 7.80 kW

EES (Engineering Equation Solver) SOLUTION "Given" T_plant=-15 [C] Q_dot_L=100 [kW] DELTAT_water=8 [C] P[1]=120 [kPa] P[2]=700 [kPa] Eta_C=0.75 T_0=25+273 "[K]" "Properties" c_water=4.18 [kJ/kg-C] "at room conditions" "Analysis" Fluid$='R134a' "(a)" "compressor" x[1]=1 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s_s[2]=s[1] h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) T_s2=temperature(Fluid$, P=P[2], s=s_s[2]) "expansion valve" P[3]=P[2] x[3]=0 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] m_dot_R=Q_dot_L/(h[1]-h[4]) "(b)" Eta_C=(h_s[2]-h[1])/(h[2]-h[1]) W_dot_in=m_dot_R*(h[2]-h[1]) "(c)" Q_dot_H=m_dot_R*(h[2]-h[3]) m_dot_water=Q_dot_H/(c_water*DELTAT_water) s[2]=entropy(Fluid$, P=P[2], h=h[2]) X_dot_destroyed=T_0*m_dot_R*(s[2]-s[1])

12-114 An air-conditioner with refrigerant-134a as the refrigerant is considered. The temperature of the refrigerant at the compressor exit, the rate of heat generated by the people in the room, the COP of the air-conditioner, and the minimum volume flow rate of the refrigerant at the compressor inlet are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-576

Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13)

h1 = 259.36 kJ/kg P1 = 500 kPa ïü ïýv = 0.04117 kJ/kg ïïþ 1 x1 = 1 s1 = 0.9242 kJ/kg

P2 = 1200 kPa ü ïï ý h2 s = 277.45 ïïþ s2 = s1

h3 = h f@1200 kPa = 117.79 kJ/kg h4 = h3 = 117.79 kJ/kg

hC =

h2 s - h1 h2 - h1

0.75 =

277.45 - 259.36 ¾¾ ® h2 = 283.48 kJ/kg h2 - 259.36

ü ïï ýT2 = 54.5°C h2 = 283.48 kJ/kgïïþ P2 = 1200 kPa

(b) The mass flow rate of the refrigerant is

V1 = v1

æ 1 m3 ÷ öæ1 min ö ÷ ÷çç (100 L/min) çç ÷ ÷ ÷ çè1000 L ÷ øèç 60 s ø 0.04117 m3 /kg

= 0.04048 kg/s

10-577

The refrigeration load is

QL = m(h1 - h4 ) = (0.04048 kg/s)(259.36 - 117.79)kJ/kg = 5.731 kW which is the total heat removed from the room. Then, the rate of heat generated by the people in the room is determined from

Qpeople = QL - Qheat - Qequip = (5.731- 250 / 60 - 0.9) kW = 0.665 kW (c) The power input and the COP are

Win = m(h2 - h1 ) = (0.04048 kg/s)(283.48 - 259.36)kJ/kg = 0.9764 kW COP =

QL 5.731 = = 5.87 0.9764 Win

(d) The reversible COP of the cycle is COPrev =

1 1 = = 37.38 TH /TL - 1 (34 + 273) / (26 + 273) - 1

The corresponding minimum power input is Win, min =

QL 5.731 kW = = 0.1533 kW COPrev 37.38

The minimum mass and volume flow rates are

mmin =

Win, min h2 - h1

0.1533 kW = 0.006358 kg/s (283.48 - 259.36)kJ/kg

V1, min = mminv1 = (0.006358 kg/s)(0.04117 m3 /kg) = (0.0002618 m3 /s) = 15.7 L / min

EES (Engineering Equation Solver) SOLUTION "GIVEN" T_L=26[C]+273 [K] T_H=34[C]+273 [K] Q_dot_heat=250[kJ/min]*Convert(kJ/min, kW) Q_dot_equipment=0.9 [kW] P_1=500 [kPa] P_2=1200 [kPa] P_4=P_1 P_3=P_2 x_1=1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-578

x_3=0 V_dot_1=100[L/min]*Convert(L/min, m^3/s) eta_C=0.75 "PROPERTIES" Fluid$='R134a' h_1=enthalpy(Fluid$, P=P_1, x=x_1) v_1=volume(Fluid$, P=P_1, x=x_1) s_1=entropy(Fluid$, P=P_1, x=x_1) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) h_3=enthalpy(Fluid$, P=P_3, x=x_3) h_4=h_3 "isenthalpic expansion" "ANALYSIS" eta_C=(h_2s-h_1)/(h_2-h_1) T_2=temperature(Fluid$, P=P_2, h=h_2) m_dot=V_dot_1/v_1 Q_dot_L=m_dot*(h_1-h_4) Q_dot_people=Q_dot_L-Q_dot_heat-Q_dot_equipment W_dot_in=m_dot*(h_2-h_1) COP=Q_dot_L/W_dot_in "actual COP" COP_max=1/(T_H/T_L-1) "maximum COP" W_dot_in_min=Q_dot_L/COP_max "minimum power input" m_dot_min=W_dot_in_min/(h_2-h_1) "minimum mass flow rate" V_dot_1_min=m_dot_min*v_1*Convert(m^3/s, L/min) "minimum volume flow rate"

12-115 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = - 10°C sat. vapor

ïü ïý ïïþ

h1 = hg @ - 10°C = 244.55 kJ/kg s1 = sg @ - 10°C = 0.93782 kJ/kg ×K

10-579

P2 = 700 kPa s2 = s1

ïü ïý ïïþ

h2 s = 270.45 kJ/kg

P3 = 700 kPa sat. liquid

ïü ïý ïïþ

h3 = h f @ 700 kPa = 88.82 kJ/kg s3 = s f @ 700 kPa = 0.33232 kJ/kg ×K

h4 @ h3 = 88.82 kJ/kg

(throttling)

ü T4 = - 10°C ïï ý s = 0.34606 kJ/kg ×K h4 = 88.82 kJ/kgïïþ 4 The actual enthalpy at the compressor exit is determined by using the compressor efficiency: hC =

h2 s - h1 h - h1 270.45 - 244.55 ¾¾ ® h2 = h1 + 2 s = 244.55 + = 275.02 kJ/kg h2 - h1 hC 0.85

and ü P2 = 700 kPa ïï ý h2 = 275.02 kJ/kgïïþ

s2 = 0.95263 kJ/kg

The heat added in the evaporator and that rejected in the condenser are

qL = h1 - h4 = (244.55 - 88.82) kJ/kg = 155.73 kJ/kg qH = h2 - h3 = (275.02 - 88.82) kJ/kg = 186.20 kJ/kg The exergy destruction during a process of a stream from an inlet state to exit state is given by æ q ö qin ÷ xdest = T0 sgen = T0 çççse - si + out ÷ ÷ çè Tsource Tsink ÷ ø

Application of this equation for each process of the cycle gives xdestroyed, 12 = T0 ( s2 - s1 ) = (295 K)(0.95263 - 0.93782) kJ/kg ×K = 4.369 kJ/kg

æ æ q ö 186.20 kJ/kg ÷ö xdestroyed, 23 = T0 çççs3 - s2 + H ÷÷÷= (295 K) çç0.33232 - 0.95263 + ÷= 3.209 kJ/kg çè èç TH ÷ø 295 K ÷ø xdestroyed, 34 = T0 ( s4 - s3 ) = (295 K)(0.34606 - 0.33232) kJ/kg ×K = 4.053 kJ/kg

æ æ q ö 155.73 kJ/kg ö÷ ÷ xdestroyed, 41 = T0 çççs1 - s4 - L ÷ = (295 K) çç0.93782 - 0.34606 ÷= 6.290 kJ / kg ÷ ç è ø TL ø÷ 273 K ÷ èç The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0C (273 K) and rejected to the ambient air at 22C (295 K), which is also taken as the dead state temperature.

12-116 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-580

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = - 10°C sat. vapor

P2 = 700 kPa s2 = s1

ïü ïý ïïþ

h1 = hg @ - 10°C = 244.55 kJ/kg s1 = sg @ - 10°C = 0.93782 kJ/kg ×K

ïü ïý ïïþ

h2 s = 270.45 kJ/kg

P3 = 700 kPa T3 = Tsat @ 700 kPa - 2.7 = 26.7 - 2.7 = 24°C

ü ïï ï ý ïï ïïþ

h3 @ h f @ 24°C = 84.98 kJ/kg s3 @ s f @ 24°C = 0.31959 kJ/kg ×K

h4 @ h3 = 84.98 kJ/kg (throttling) ü T4 = - 10°C ïï ý s = 0.33147 kJ/kg ×K h4 = 84.98 kJ/kgïïþ 4

The actual enthalpy at the compressor exit is determined by using the compressor efficiency: hC =

and

h2 s - h1 h - h1 270.45 - 244.55 ¾¾ ® h2 = h1 + 2 s = 244.55 + = 275.02 kJ/kg h2 - h1 hC 0.85

ü P2 = 700 kPa ïï ý h2 = 275.02 kJ/kgïïþ

s2 = 0.95263 kJ/kg

The heat added in the evaporator and that rejected in the condenser are

qL = h1 - h4 = (244.55 - 84.98) kJ/kg = 159.57 kJ/kg qH = h2 - h3 = (275.02 - 84.98) kJ/kg = 190.04 kJ/kg The exergy destruction during a process of a stream from an inlet state to exit state is given by æ q ö qin ÷ xdest = T0 sgen = T0 çççse - si + out ÷ ÷ çè Tsource Tsink ÷ ø

Application of this equation for each process of the cycle gives xdestroyed, 12 = T0 ( s2 - s1 ) = (295 K)(0.95263 - 0.93782) kJ/kg ×K = 4.369 kJ/kg

æ æ q ö 190.04 kJ/kg ÷ö xdestroyed, 23 = T0 çççs3 - s2 + H ÷÷÷= (295 K) çç0.31959 - 0.95263 + ÷= 3.293 kJ/kg ç ÷ çè è TH ø 295 K ÷ø xdestroyed, 34 = T0 ( s4 - s3 ) = (295 K)(0.33147 - 0.31959) kJ/kg ×K = 3.505 kJ/kg

10-581

æ æ q ö 159.57 kJ/kg ö÷ ÷ xdestroyed, 41 = T0 çççs1 - s4 - L ÷ = (295 K) çç0.93782 - 0.33147 ÷= 6.444 kJ / kg ÷ ÷ çè èç TL ø 273 K ø÷ The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0C (273 K) and rejected to the ambient air at 22C (295 K), which is also taken as the dead state temperature.

12-117 An air conditioner operates on the vapor-compression refrigeration cycle. The rate of cooling provided to the space, the COP, the isentropic efficiency and the exergetic efficiency of the compressor, the exergy destruction in each component of the cycle, the total exergy destruction, the minimum power input, and the second-law efficiency of the cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) The properties of R-134a are (Tables A-11 through A-13) Tsat@180 kPa = - 12.7°C

üïï h1 = 245.15 kJ/kg ý T1 = - 12.7 + 2.7 = 10°Cïïþ s1 = 0.9484 kJ/kg ×K P1 = 180 kPa

P2 = 1200 kPa ïü ïý h = 285.34 kJ/kg ïïþ 2 s s1 = s1

P2 = 1200 kPaïüï h2 = 289.66 kJ/kg ý ïïþ s2 = 0.9615 kJ/kg ×K T2 = 60°C Tsat@1200 kPa = 46.3°C

üïï h3 @ h f@40°C = 108.27 kJ/kg ý T3 = 46.3 - 6.3 = 40°Cïïþ s3 @ s f@40°C = 0.3949 kJ/kg ×K P3 = 1200 kPa

h4 = h3 = 108.27 kJ/kg

ïüï ý s4 = 0.4229 kJ/kg ×K h4 = 108.26 kJ/kgïïþ P4 = 180 kPa

10-582

The cooling load and the COP are QL = m(h1 - h4 ) = (0.06 kg/s)(245.15 - 108.27)kJ/kg = 8.213 kW

æ3412 Btu/h ÷ ö = (8.213 kW) çç = 28,020 Btu / h ÷ çè 1 kW ÷ ø

QH = m(h2 - h3 ) = (0.06 kg/s)(289.66 - 108.27)kJ/kg = 10.88 kW

Win = m(h2 - h1 ) = (0.06 kg/s)(289.66 - 245.15)kJ/kg = 2.670 kW COP =

QL 8.213 kW = = 3.076 2.670 kW Win

(b) The isentropic efficiency of the compressor is hC =

h2 s - h1 285.34 - 245.15 = = 0.903 = 90.3% h2 - h1 289.66 - 245.15

The reversible power and the exergy efficiency for the compressor are Wrev = m [(h2 - h1 ) - T0 ( s2 - s1 ) ]

= (0.06 kg/s) [(289.66 - 245.15)kJ/kg - (310 K)(0.9615 - 0.9484)kJ/kg ×K ] = 2.428 kW

hex , C =

Wrev 2.428 kW = = 0.909 = 90.9% 2.670 kW Win

Sgen, 1- 2 = m(s2 - s1 ) = (0.06 kg/s)(0.9615 - 0.9484) kJ/kg ×K = 0.0007827 kW/K Exdest, 1- 2 = T0 Sgen, 1- 2 = (310 K)(0.0007827 kW/K) = 0.2426 kW Condenser: Sgen, 2- 3 = m( s3 - s2 ) +

QH 10.88 kW = (0.06 kg/s)(0.3949 - 0.9615) kJ/kg ×K + = 0.001114 kW/K TH 310 K

Exdest, 2- 3 = T0 Sgen, 2- 3 = (310 K)(0.001114 kJ/kg ×K) = 0.3452 kW Expansion valve:

Sgen, 3- 4 = m(s4 - s3 ) = (0.06 kg/s)(0.4229 - 00.3949) kJ/kg ×K = 0.001678 kW/K Exdest, 3- 4 = T0 Sgen, 3- 4 = (310 K)(0.001678 kJ/kg ×K) = 0.5202 kJ / kg Evaporator: Sgen, 4- 1 = m( s1 - s4 ) -

QL 8.213 kW = (0.06 kg/s)(0.9484 - 0.4229) kJ/kg ×K = 0.003597 kW/K TL 294 K

Exdest, 4- 1 = T0 Sgen, 4- 1 = (310 K)(0.003597 kJ/kg ×K) = 1.115 kW The total exergy destruction can be determined by adding exergy destructions in each component: Exdest, total = Exdest, 1- 2 + Exdest, 2- 3 + Exdest, 3- 4 + Exdest, 4- 1

= 0.2426 + 0.3452 + 0.5202 + 1.115 = 2.223 kW

(d) The exergy of the heat transferred from the low-temperature medium is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-583

æ T ö æ 310 ö ÷= 0.4470 kW ÷ ExQL = - QL ççç1- 0 ÷ ÷ ÷= - (8.213 kW) ççèç1ø 294 ÷ èç TL ÷ ø

This is the minimum power input to the cycle:

Win, min = ExQL = 0.4470 kW The second-law efficiency of the cycle is

hII =

Win, min Win

0.4470 = 0.167 = 16.7% 2.670

The total exergy destruction in the cycle can also be determined from

Exdest, total = Win - ExQL = 2.670 - 0.4470 = 2.223 kW The result is the same as expected.

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_1=180 [kPa] DELTAT_superheat=2.7 [C] m_dot=0.06 [kg/s] P_2=1200 [kPa] T_2=60 [C] DELTAT_subcool=6.3 [C] T_L=(21+273) [K] T_H=(37+273) [K] T_0=T_H "PROPERTIES" Fluid$='R134a' T_sat_1=temperature(Fluid$, P=P_1, x=1) T_1=T_sat_1+DELTAT_superheat h_1=enthalpy(Fluid$, P=P_1, T=T_1) s_1=entropy(Fluid$, P=P_1, T=T_1) h_2=enthalpy(Fluid$, P=P_2, T=T_2) s_2=entropy(Fluid$, P=P_2, T=T_2) h_2s=enthalpy(Fluid$, P=P_2, s=s_1) eta_C=(h_2s-h_1)/(h_2-h_1) T_sat_3=temperature(Fluid$, P=P_2, x=0) T_3=T_sat_3-DELTAT_subcool h_3=enthalpy(Fluid$, T=T_3, x=0) "subcooled state is approximated as saturated liquid at T_3" s_3=entropy(Fluid$, T=T_3, x=0) h_4=h_3 s_4=entropy(Fluid$, P=P_1, h=h_4) Q_dot_L=m_dot*(h_1-h_4) Q_dot_L_Btuh=Q_dot_L*Convert(kW, Btu/h) Q_dot_H=m_dot*(h_2-h_3) W_dot_in=m_dot*(h_2-h_1) COP=Q_dot_L/W_dot_in W_dot_rev=m_dot*(h_2-h_1-T_0*(s_2-s_1)) eta_ex_Comp=W_dot_rev/W_dot_in S_dot_gen_12=m_dot*(s_2-s_1) Ex_dot_dest_12=T_0*S_dot_gen_12 S_dot_gen_23=m_dot*(s_3-s_2)+Q_dot_H/T_H Ex_dot_dest_23=T_0*S_dot_gen_23 S_dot_gen_34=m_dot*(s_4-s_3) Ex_dot_dest_34=T_0*S_dot_gen_34 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-584

S_dot_gen_41=m_dot*(s_1-s_4)-Q_dot_L/T_L Ex_dot_dest_41=T_0*S_dot_gen_41 Ex_dot_QL=-Q_dot_L*(1-T_0/T_L) W_dot_min=Ex_dot_QL eta_II=Ex_dot_QL/W_dot_in Ex_dot_dest=W_dot_in-Ex_dot_QL

12-118 A two-stage compression refrigeration system using refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space, the compressor work, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flashing chamber is adiabatic.

Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13)

h1 = 236.99 kJ/kg,

h2 = 266.29 kJ/kg

h3 = 259.36 kJ/kg, h5 = 127.25 kJ/kg,

h6 = 127.25 kJ/kg

h7 = 73.32 kJ/kg,

h8 = 73.32 kJ/kg

The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6,

x6 =

h6 - h f h fg

127.25 - 73.32 = 0.2899 186.04

(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: Ein - Eout = D Esystem ¿ 0 (steady) = 0 ® Ein = Eout

å m h = å mh e e

i i

(1)h9 = x6 h3 + (1- x6 )h2 h9 = (0.2899)(259.36)+ (1- 0.2899)(266.29) = 264.28 kJ/kg

10-585

ü ïï ý s9 = 0.94108 kJ/kg ×K h9 = 264.28 kJ/kg ïïþ P9 = 0.5 MPa

Also, ü P4 = 1.4 MPa ïï ý h4 = 286.19 kJ/kg s4 = s9 = 0.94108 kJ/kg ×K ïïþ

Then the amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are qL = (1- x6 )(h1 - h8 ) = (1- 0.2899)(236.99 - 73.32) kJ/kg = 116.2 kJ / kg

win = wcompI, in + wcompII, in = (1- x6 )(h2 - h1 )+ (1)(h4 - h9 ) = (1- 0.2899)(266.29 - 236.99) kJ/kg + (1) (286.19 - 264.28) kJ/kg

= 42.72 kJ / kg (c) The coefficient of performance is determined from COPR =

qL 116.2 kJ/kg = = 2.72 win 42.72 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=120 [kPa] P[2]=500 [kPa] P[4]=1400 [kPa] "Analysis" Fluid$='R134a' "(a)" "compressor B" "A: upper cycle, B: lower cycle" x[1]=1 h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) P[3]=P[2] x[3]=1 h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) "expansion valve A" P[5]=P[4] x[5]=0 h[5]=enthalpy(Fluid$, P=P[5], x=x[5]) h[6]=h[5] "expansion valve B" P[7]=P[2] x[7]=0 h[7]=enthalpy(Fluid$, P=P[7], x=x[7]) h[8]=h[7] P[6]=P[2] x[6]=quality(Fluid$, P=P[6], h=h[6]) x_6=x[6] "(b)" h[9]=x[6]*h[3]+(1-x[6])*h[2] "mixing chamber energy balance" P[9]=P[2] s[9]=entropy(Fluid$, P=P[9], h=h[9]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-586

s[4]=s[9] h[4]=enthalpy(Fluid$, P=P[4], s=s[4]) q_L=(1-x[6])*(h[1]-h[8]) w_in=(1-x[6])*(h[2]-h[1])+(h[4]-h[9]) "(c)" COP_R=q_L/w_in h6_f=enthalpy(Fluid$, P=P[6], x=0) h6_g=enthalpy(Fluid$, P=P[6], x=1) h6_fg=h6_g-h6_f

12-119E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling load of both evaporators per unit of flow through the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),

P3 = 160 psia ïü ïý h = h f @160 psia = 48.527 Btu/lbm ïïþ 3 sat. liquid h4 = h6 @ h3 = 48.527 Btu/lbm (throttling)

T5 = 30°F ïü ïý h = h g @30° F = 107.42 Btu/lbm sat. vapor ïïþ 5 T7 = - 29.5°F ïü ïý h = h g @- 29.5° F = 98.69 Btu/lbm ïïþ 7 sat. vapor

10-587

For a unit mass flowing through the compressor, the fraction of mass flowing through Evaporator II is denoted by x and that through Evaporator I is y (y = 1-x). From the cooling loads specification, QL, evap 1 = 2QL, evap 2

x(h5 - h4 ) = 2 y(h7 - h6 ) where

x = 1- y Combining these results and solving for y gives y=

h5 - h4 107.42 - 48.527 = = 0.3699 2(h7 - h6 ) + (h5 - h4 ) 2(98.69 - 48.527) + (107.42 - 48.527)

Then,

x = 1- y = 1- 0.3699 = 0.6301 Applying an energy balance to the point in the system where the two evaporator streams are recombined gives

xh5 + yh7 = h1 ¾ ¾ ® h1 =

xh5 + yh7 (0.6301)(107.42) + (0.3699)(98.69) = = 104.19 Btu/lbm 1 1

Then,

P1 = Psat@- 29.5° F @10 psia ïü ïý s = 0.2418 Btu/lbm ×R ïïþ 1 h1 = 104.19 Btu/lbm P2 = 160 psia ü ïï ý h2 = 131.15 Btu/lbm ïïþ s2 = s1 The cooling load of both evaporators per unit mass through the compressor is

qL = x(h5 - h4 ) + y(h7 - h6 ) = (0.6301)(107.42 - 48.527) Btu/lbm + (0.3699)(98.69 - 48.527) Btu/lbm = 55.66 Btu / lbm

The work input to the compressor is

win = h2 - h1 = (131.15 - 104.19) Btu/lbm = 26.96 Btu/lbm The COP of this refrigeration system is determined from its definition, COPR =

qL 55.66 Btu/lbm = = 2.07 win 26.96 Btu/lbm

EES (Engineering Equation Solver) SOLUTION "GIVEN" T[5]=30 [F]; T[4]=T[5] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-588

T[7]=-29.5 [F]; T[6]=T[7] P[2]=160 [psia]; P[3]=P[2] "Q_dot_L1=2*Q_dot_L2" "PROPERTIES" Fluid$='R134a' h[3]=enthalpy(Fluid$, P=P[3], x=0) h[4]=h[3] h[6]=h[3] h[5]=enthalpy(Fluid$, T=T[5], x=1) h[7]=enthalpy(Fluid$, T=T[7], x=1) "ANALYSIS" x=1-y "x: fraction of mass flowing through evap 1, y: fraction of mass flowing through evap 2" x*(h[5]-h[4])=2*y*(h[7]-h[6]) "Q_dot_L1=2*Q_dot_L2" x*h[5]+y*h[7]=h[1] P[1]=pressure(Fluid$, T=T[7], x=0) s[1]=entropy(Fluid$, P=P[1], h=h[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) q_L=x*(h[5]-h[4])+y*(h[7]-h[6]) w_in=h[2]-h[1] COP=q_L/w_in

12-120E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From Prob. 12-119E and the refrigerant tables (Tables A-11E, A-12E, and A-13E),

s1 = s2 = 0.2418 Btu/lbm ×R

s3 = 0.09776 Btu/lbm ×R s4 = 0.1024 Btu/lbm ×R s5 = 0.2226 Btu/lbm ×R s6 = 0.1129 Btu/lbm ×R s7 = 0.2295 Btu/lbm ×R

10-589

y = 1- x = 0.3699

qL, 45 = h5 - h4 = 58.89 Btu/lbm

qL, 67 = h7 - h6 = 50.16 Btu/lbm qH = 82.62 Btu/lbm The exergy destruction during a process of a stream from an inlet state to exit state is given by æ q ö qin ÷ xdest = T0 sgen = T0 çççse - si + out ÷ ÷ çè Tsource Tsink ÷ ø

Application of this equation for each process of the cycle gives the exergy destructions per unit mass flowing through the compressor: æ æ ö q ö 82.62 Btu/lbm ÷ ÷ xdestroyed, 23 = T0 ç s3 - s2 + H ÷ = (540 R) ç = 4.82 Btu/lbm ç ÷ ç0.09776 - 0.2418 + ÷ ÷ ç ç ÷ ç è ø TH ø 540 R è

xdestroyed, 346 = T0 ( xs4 + ys6 - s3 )

= (540 R)(0.6301´ 0.1024 + 0.3699´ 0.1129 - 0.09776) Btu/lbm ×R = 4.60 Btu/lbm

æ q ö xdestroyed, 45 = xT0 çççs5 - s4 - L , 45 ÷÷÷ çè TL ø÷ æ 58.89 Btu/lbm ö ÷ = (0.6301)(540 R) çç0.2226 - 0.1024 ÷ ÷= 0.84 Btu/lbm çè ø 500 R

æ ö q ÷ xdestroyed, 67 = yT0 çççs7 - s6 - L , 67 ÷ ÷ çè TL ÷ ø

æ 50.16 Btu/lbm ö÷ = (0.3699)(540 R) çç0.2295 - 0.1129 ÷= 0.77 Btu/lbm çè ø÷ 445 R X destroyed, mixing = T0 ( s1 - xs5 - ys7 ) = (540 R) [0.2418 - (0.6301)(0.2226) - (0.3699)(0.2295)]= 9.00 Btu / lbm

For isentropic processes, the exergy destruction is zero:

X destroyed, 12 = 0 The greatest exergy destruction occurs during the mixing process. Note that heat is absorbed in evaporator 2 from a reservoir at -15F (445 R), in evaporator 1 from a reservoir at 40F (500 R), and rejected to a reservoir at 80F (540 R), which is also taken as the dead state temperature.

EES (Engineering Equation Solver) SOLUTION "GIVEN" T[5]=30 [F]; T[4]=T[5] T[7]=-29.5 [F]; T[6]=T[7] P[2]=160 [psia]; P[3]=P[2] "Q_dot_L1=2*Q_dot_L2" T0=(80+460) [R] T_H=T0 T_L_45=(40+460) [R] T_L_67=(-15+460) [R] "PROPERTIES" Fluid$='R134a' h[3]=enthalpy(Fluid$, P=P[3], x=0) h[4]=h[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-590

h[6]=h[3] h[5]=enthalpy(Fluid$, T=T[5], x=1) h[7]=enthalpy(Fluid$, T=T[7], x=1) "ANALYSIS" x=1-y "x: fraction of mass flowing through evap 1, y: fraction of mass flowing through evap 2" x*(h[5]-h[4])=2*y*(h[7]-h[6]) "Q_dot_L1=2*Q_dot_L2" x*h[5]+y*h[7]=h[1] P[1]=pressure(Fluid$, T=T[7], x=0) s[1]=entropy(Fluid$, P=P[1], h=h[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) q_L=x*(h[5]-h[4])+y*(h[7]-h[6]) w_in=h[2]-h[1] COP=q_L/w_in s[3]=entropy(Fluid$, P=P[3], x=0) s[4]=entropy(Fluid$, T=T[4], h=h[4]) s[5]=entropy(Fluid$, T=T[5], x=1) s[6]=entropy(Fluid$, T=T[6], h=h[6]) s[7]=entropy(Fluid$, T=T[7], x=1) q_L_45=h[5]-h[4] q_L_67=h[7]-h[6] q_H=h[2]-h[3] x_dest_12=T0*(s[2]-s[1]) x_dest_23=T0*(s[3]-s[2]+q_H/T_H) x_dest_346=T0*(x*s[4]+y*s[6]-s[3]) x_dest_45=x*T0*(s[5]-s[4]-q_L_45/T_L_45) x_dest_67=y*T0*(s[7]-s[6]-q_L_67/T_L_67) x_dest_157_mixing=T0*(s[1]-x*s[5]-y*s[7])

12-121 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

10-591

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 900 kPa ïü ïý h = h f @ 900 kPa = 101.62 kJ/kg ïïþ 3 sat. liquid

ïü ïï = 35.51- 5.51 = 30°C ïý h4 @ h f @ 30° C = 93.58 kJ/kg ïï P4 = 900 kPa ïï þ

T4 = Tsat @ 900 kPa - 11.3

h5 @ h4 = 93.58 kJ/kg (throttling)

T6 = - 10.09°C ïü ïý h6 = hg @ - 10.09°C = 244.50 kJ/kg ïïþ P6 = Psat @ - 10.09°C = 200 kPa sat. vapor An energy balance on the heat exchanger gives

m(h1 - h6 ) = m(h3 - h4 ) ¾ ¾® h1 = h3 - h4 + h6 = 101.62 - 93.58 + 244.50 = 252.54 kJ/kg Then, ü P1 = 200 kPa ïï ý s = 0.9679 kJ/kg ×K h1 = 252.54 kJ/kg ïïþ 1

P2 = 900 kPa ü ïï ý h2 = 285.37 kJ/kg ïïþ s2 = s1 The COP of this refrigeration system is determined from its definition, COPR =

h - h5 qL 244.50 - 93.58 = 6 = = 4.60 win h2 - h1 285.37 - 252.54

EES (Engineering Equation Solver) SOLUTION "GIVEN" T6=-10.09 [C] P2=900 [kPa] DELTAT_subcool=5.51 [C] "ANALYSIS" Fluid$='R134a' P3=P2 h3=enthalpy(Fluid$, P=P3, x=0) P4=P2 T_sat=temperature(Fluid$, P=P4, x=0) T4=T_sat-DELTAT_subcool h4=enthalpy(Fluid$, T=T4, x=0) "approximated as sat. liq." © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-592

h5=h4 h6=enthalpy(Fluid$, T=T6, x=1) P6=pressure(Fluid$, T=T6, x=1) h1-h6=h3-h4 "energy balance on heat exchanger" P1=P6 s1=entropy(Fluid$, P=P1, h=h1) s2=s1 h2=enthalpy(Fluid$, P=P2, s=s2) COP_R=(h6-h5)/(h2-h1)

12-122 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 900 kPa ü ïï ý h3 = h f @ 900 kPa = 101.62 kJ/kg ïïþ sat. liquid

10-593

ü ïï ï = 35.51- 9.51 = 26°C ïý h4 @ h f @ 26°C = 87.82 kJ/kg ïï P4 = 900 kPa ïï þ

T4 = Tsat@900 kPa - 9.51

h5 @ h4 = 87.82 kJ/kg (throttling) T6 = - 10.09°C ïü ïý h6 = hg @- 10.09°C = 244.50 kJ/kg ïïþ P6 = Psat@- 10.09°C = 200 kPa sat. vapor

An energy balance on the heat exchanger gives

m(h1 - h6 ) = m(h3 - h4 ) ¾ ¾® h1 = h3 - h4 + h6 = 101.62 - 87.82 + 244.50 = 258.29 kJ/kg Then, ü P1 = 200 kPa ïï ý s = 0.9888 kJ/kg ×K h1 = 258.29 kJ/kg ïïþ 1

P2 = 900 kPa ïü ïý h = 292.19 kJ/kg ïïþ 2 s2 = s1 The COP of this refrigeration system is determined from its definition, COPR =

h - h5 qL 244.50 - 87.82 = 6 = = 4.62 win h2 - h1 292.19 - 258.29

EES (Engineering Equation Solver) SOLUTION "GIVEN" T6=-10.09 [C] P2=900 [kPa] DELTAT_subcool=9.51 [C] "ANALYSIS" Fluid$='R134a' P3=P2 h3=enthalpy(Fluid$, P=P3, x=0) P4=P2 T_sat=temperature(Fluid$, P=P4, x=0) T4=T_sat-DELTAT_subcool h4=enthalpy(Fluid$, T=T4, x=0) "approximated as sat. liq." h5=h4 h6=enthalpy(Fluid$, T=T6, x=1) P6=pressure(Fluid$, T=T6, x=1) h1-h6=h3-h4 "energy balance on heat exchanger" P1=P6 s1=entropy(Fluid$, P=P1, h=h1) s2=s1 h2=enthalpy(Fluid$, P=P2, s=s2) COP_R=(h6-h5)/(h2-h1)

12-123 An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle. The temperature of the air leaving the turbine is to be determined. Assumptions 1 Steady operating conditions exist. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-594

2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2). Analysis Assuming the turbine to be isentropic, the air temperature at the turbine exit is determined from ( æP4 ö ÷ ç ÷ T4 = T3 çç ÷ çè P ÷ ø

k - 1)/ k

0.4/1.4

æ100 kPa ö ÷ = (358 K )çç ÷ ÷ èç 250 kPa ø

= 275.5 K = 2.5°C

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=30 [C] P[1]=100 [kPa] P[2]=250 [kPa] T[3]=85 [C] "Properties" c_p=CP(air, T=25) "at room temperature" c_v=CV(air, T=25) k=c_p/c_v "Analysis" P[4]=P[1] P[3]=P[2] T_4=(T[3]+273)*(P[4]/P[3])^((k-1)/k)

12-124 A regenerative gas refrigeration cycle with helium as the working fluid is considered. The temperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are c p = 5.1926 kJ / kg ×K and k = 1.667 (Table A-2).

Analysis (a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator, Ein - Eout = D Esystem ¿ 0 (steady) = 0

10-595

Ein = Eout

å m h = å m h ¾ ¾® m (h - h ) = m (h - h ) e e

i i

or, Thus,

mc p (T3 - T4 ) = mc p (T1 - T6 ) ¾ ¾ ® T3 - T4 = T1 - T6 T4 = T3 - T1 + T6 = 20°C - (- 10°C)+ (- 25°C) = 5°C = 278 K

(b) From the isentropic relations, (k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ ÷ çè P1 ø

0.667/1.667

= (263 K )(3)

æP ö( ÷ T5 = T4 ççç 5 ÷ ÷ çè P4 ÷ ø

k - 1)/ k

= 408.2 K = 135.2°C

0.667/1.667

æ1 ö = (278 K )çç ÷ ÷ ÷ èç3ø

= 179.1 K = - 93.9°C

Then the COP of this ideal gas refrigeration cycle is determined from

COPR = =

h6 - h5 qL qL = = wnet, in wcomp, in - wturb, out (h2 - h1 )- (h4 - h5 ) - 25°C - (- 93.9°C) T6 - T5 = = 1.49 (T2 - T1 )- (T4 - T5 ) éë135.2 - (- 10)ùû°C - éë5 - (- 93.9)ùû°C

= mc p éë(T2 - T1 )- (T4 - T5 )ù û é ù = (0.45 kg/s)(5.1926 kJ/kg ×°C)(éë135.2 - (- 10)ù û- ë5 - (- 93.9)û)

= 108.2 kW

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=100 [kPa] T[1]=(-10+273) [K] P[2]=300 [kPa] T[3]=(20+273) [K] T_6=-25 [C] T[6]=(-25+273) [K] m_dot=0.45 [kg/s] "Properties, Table A-2a" c_p=5.1926 [kJ/kg-K] k=1.667 "Analysis" "(a)" T[3]-T[4]=T[1]-T[6] "regenerator energy balance" T_4=T[4]-273 "[C]" "(b)" T[2]=T[1]*(P[2]/P[1])^((k-1)/k) T[5]=T[4]*(P[1]/P[2])^((k-1)/k) COP_R=(T[6]-T[5])/((T[2]-T[1])-(T[4]-T[5])) "(c)" W_dot_comp_in=m_dot*c_p*(T[2]-T[1]) W_dot_turb_out=m_dot*c_p*(T[4]-T[5]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-596

W_dot_net_in=W_dot_comp_in-W_dot_turb_out

12-125 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

Analysis (a) For this problem, we use the properties of air from EES. Note that for an ideal gas enthalpy is a function of temperature only while entropy is functions of both temperature and pressure.

T1 = 0°C ¾ ¾ ® h1 = 273.40 kJ/kg P1 = 100 kPa ü ïï ý s1 = 5.6110 kJ/kg.K ïïþ T1 = 0°C

P2 = 500 kPa üïï ý h2 s = 433.50 kJ/kg ïïþ s2 = s1 hC =

h2 s - h1 h2 - h1

0.80 =

433.50 - 273.40 h2 - 273.40

h2 = 473.52 kJ/kg

T3 = 35°C ¾ ¾ ® h3 = 308.63 kJ/kg

10-597

For the turbine inlet and exit we have

T5 = - 80°C ¾ ¾® h5 = 193.45 kJ/kg T4 = ? ¾ ¾ ® h4 =

hT =

h4 - h5 h4 - h5 s

P1 = 100 kPa ïüï ý s1 = 5.6110 kJ/kg.K ïïþ T1 = 0°C P4 = 500 kPa ïüï ý s4 = ïïþ T4 = ? P5 = 500 kPa ïüï ý h5 s = ïïþ s5 = s4 We can determine the temperature at the turbine inlet from EES using the above relations. A hand solution would require a trial-error approach.

T4 = 281.3 K , h4 = 282.08 kJ / kg An energy balance on the regenerator gives

h6 = h1 - h3 + h4 = 273.40 - 308.63 + 282.08 = 246.85 kJ/kg The effectiveness of the regenerator is determined from eregen =

h3 - h4 308.63 - 282.08 = = 0.430 h3 - h6 308.63 - 246.85

(b) The refrigeration load is

QL = m(h6 - h5 ) = (0.4 kg/s)(246.85 - 193.45)kJ/kg = 21.36 kW (c) The turbine and compressor powers and the COP of the cycle are

WC, in = m(h2 - h1 ) = (0.4 kg/s)(473.52 - 273.40)kJ/kg = 80.05 kW WT, out = m(h4 - h5 ) = (0.4 kg/s)(282.08 - 193.45)kJ/kg = 35.45 kW COP =

QL QL 21.36 = = = 0.479 Wnet, in WC, in - WT, out 80.05 - 35.45

10-598

h1 = 273.40 kJ/kg h2 = 473.52 kJ/kg h3 = 308.63 kJ/kg P3 = 500 kPa ïü ïý s = 5.2704 kJ/kg ïïþ 3 T3 = 35°C

P1 = 100 kPa ü ïï ý h4 s = 194.52 kJ/kg.K ïïþ s4 = s3

hT =

h3 - h4 308.63 - h4 ¾¾ ® 0.85 = ¾¾ ® h4 = 211.64 kJ/kg h3 - h4 s 308.63 - 194.52

QL = m(h1 - h4 ) = (0.4 kg/s)(273.40 - 211.64)kJ/kg = 24.70 kW Wnet, in = m(h2 - h1 ) - m(h3 - h4 ) = (0.4 kg/s) [(473.52 - 273.40) - (308.63 - 211.64)kJ/kg ]= 41.25 kW COP =

QL 24.70 = = 0.599 41.25 Wnet, in

EES (Engineering Equation Solver) SOLUTION "GIVEN" r=5 T[1]=(0+273.15) [K] T[3]=(35+273.15) [K] T[5]=(-80+273.15) [K] m_dot=0.4 [kg/s] eta_C=0.80 eta_T=0.85 P_1=100 [kPa] "assumed" P_2=P_1*r "PROPERTIES" Fluid$='Air' h[1]=enthalpy(Fluid$, T=T[1]) s_1=entropy(Fluid$, T=T[1], P=P_1) h[3]=enthalpy(Fluid$, T=T[3]) h[5]=enthalpy(Fluid$, T=T[5]) "ANALYSIS" h_2s=enthalpy(Fluid$, P=P_2, s=s_1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-599

h[2]=h[1]+(h_2s-h[1])/eta_C h[4]=enthalpy(Fluid$, T=T[4]) h[4]=h[5]+eta_T*(h[4]-h_5s) h_5s=enthalpy(Fluid$, P=P_1, s=s_4) s_4=entropy(Fluid$, T=T[4], P=P_2) h[6]=h[1]-h[3]+h[4] "energy balance on the regenerator" epsilon_regen=(h[3]-h[4])/(h[3]-h[6]) "regenerator effectiveness" Q_dot_L=m_dot*(h[6]-h[5]) "refrigeration load" W_dot_net_in=W_dot_C_in-W_dot_T_out W_dot_C_in=m_dot*(h[2]-h[1]) W_dot_T_out=m_dot*(h[4]-h[5]) COP=Q_dot_L/W_dot_net_in "COP of cycle" "Simple gas refrigeration cycle analysis, the subscript 'a' is used" h_a[1]=h[1] h_a[2]=h[2] h_a[3]=h[3] s_3_a=entropy(Fluid$, h=h_a[3], P=P_2) h_4s_a=enthalpy(Fluid$, P=P_1, s=s_3_a) h_a[4]=h_a[3]-eta_T*(h_a[3]-h_4s_a) Q_dot_L_a=m_dot*(h_a[1]-h_a[4]) W_dot_in_net_a=m_dot*(h_a[2]-h_a[1])-m_dot*(h_a[3]-h_a[4]) COP_a=Q_dot_L_a/W_dot_in_net_a

12-126 An ideal gas refrigeration cycle with with three stages of compression with intercooling using air as the working fluid is considered. The COP of this system is to be determined. Assumptions 1 2 3

Steady operating conditions exist. Air is an ideal gas with constant specific heats. Kinetic and potential energy changes are negligible.

Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4 (Table A-2a).

Analysis From the isentropic relations, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

= (243 K)(7)0.4/1.4 = 423.7 K

10-600 ( k - 1)/ k

æP ö ÷ T4 = T6 = T3 ççç 4 ÷ ÷ ÷ çè P3 ø ( k - 1)/ k

æP ö ÷ T8 = T7 ççç 8 ÷ ÷ çè P ÷ ø 7

= (288 K)(7)0.4/1.4 = 502.2 K 0.4/1.4

æ 1 ÷ ö = (288 K) çç ÷ èç7´ 7´ 7 ÷ ø

= 54.3 K

The COP of this ideal gas refrigeration cycle is determined from

COPR =

qL qL = wnet, in wcomp, in - wturb, out

h1 - h8 (h2 - h1 ) + (h4 - h3 ) + (h6 - h5 ) - (h7 - h8 )

T1 - T8 (T2 - T1 ) + 2(T4 - T3 ) - (T7 - T8 )

243 - 54.3 (423.7 - 243) + 2(502.2 - 288) - (288 - 54.3)

= 0.503

EES (Engineering Equation Solver) SOLUTION "Given" P1=50 [kPa] T1=(-30+273) [K] r_P=7 T3=(15+273) [K] T5=T3 T7=T3 "Analysis" c_p=1.005 [kJ/kg-K] k=1.4 T2=T1*r_P^((k-1)/k) T4=T3*r_P^((k-1)/k) T6=T4 T8=T7*(1/r_P^3)^((k-1)/k) COP=q_L/w_net_in q_L=c_p*(T1-T8) w_net_in=w_comp_in-w_turb_out w_comp_in=c_p*(T2-T1)+c_p*(T4-T3)+c_p*(T6-T5) w_turb_out=c_p*(T7-T8) COP_alt=(T1-T8)/((T2-T1)+2*(T4-T3)-(T7-T8))

12-127 A natural gas liquefaction plant produces liquid natural gas at 1 atm and 161.5C when natural gas enters at 1 atm and 20C. Treating the natural gas as methane, determine the minimum work input and maximum (or reversible) COP. If the liquefaction is done by a Carnot refrigerator between temperature limits of TH = 20°C and TL with the same reversible COP, determine the temperature TL . Various properties of methane before and after liquefaction process are given as follows: h1 = - 12.12 kJ / kg , h2 = - 910.92 kJ / kg , s1 = - 0.04007 kJ / kg ×K , s2 = - 6.6767 kJ / kg ×K

10-601

12-127 A natural gas liquefaction plant is considered. The minimum work input and the reversible COP of the plant are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Analysis The heat rejection from the natural gas is

qL  h1  h2   (12.12)  (910.92)  kJ/kg  898.8 kJ/kg Taking the dead state temperature to be T0 = T1 = 20C = 293.15 K, the minimum work input is determined from

wmin  wrev  h2  h1  T0 (s2  s1 )   (910.92)  (12.12)  kJ/kg  (293.15 K) ( 6.6767  ( 0.04007) kJ/kg  K 

 1047 kJ / kg The maximum or reversible COP is COPrev 

qL 898.8 kJ/kg   0.8587 wrev 1047 kJ/kg

The temperature TL is determined from the definition of the reversible COP: COPR, rev 

1 1   0.8587    TL  135.4 K TH / TL  1 (293.15 K) / TL  1

It may also be determined from the definition of exergy of heat transfer:  T   293.15 K  xqL  wmin  qL 1  0   1047 kJ/kg  (898.8 kJ/kg) 1   TL  135.4 K   T TL  L   

EES (Engineering Equation Solver) SOLUTION T_0=293.15 [K] T_1=20 [C] P_1=101.325 [kPa] P_2=P_1 x_2=0 h_1=enthalpy(Methane,T=T_1,P=P_1) h_2=enthalpy(Methane,P=P_2,x=x_2) s_1=entropy(Methane,T=T_1,P=P_1) s_2=entropy(Methane,P=P_2,x=x_2) T_2=temperature(Methane,P=P_1,x=x_2) w_min=(h_2-h_1)-T_0*(s_2-s_1) q=h_1-h_2 COP_rev=q/w_min w_min=-q*(1-T_0/T1a) COP_rev=1/((T_0/T2a)-1)

12-128 Consider a gas refrigeration system with a pressure ratio of 4 and a regenerator. Air leaves the regenerator and enters the compressor at 20°C at a rate of 75 kg / min . Air leaving the compressor is cooled to 16°C by rejecting heat to the ambient at 7°C, and is further cooled to 30°C in the regenerator before entering the turbine. The isentropic efficiency of the compressor is 82 percent and that of the turbine is 84 percent. The temperature of the cooled space is 37°C. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-602

Determine the COP of the cycle, the minimum power input, the second-law efficiency of the cycle, and the total exergy destruction in the cycle. Use constant specific heat of c p = 1.005 kJ / kg ×K and k = 1.4 for air.

12-128 A regenerative gas refrigeration cycle with air as the working fluid is considered. Various performance parameters are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Properties Air properties are c p = 1.005 kJ / kg ×K and k = 1.4.

Analysis From the isentropic relations,

P  T2s  T1  2   P1 

 k 1 / k

P  T5s  T4  5   P4 

 k 1 / k

  253 K  4 

0.4/1.4

1   243 K    4

 376.0 K

0.4/1.4

 163.5 K

T 

h 4  h5 T  T5  4 h 4  h 5s T4  T5s

 

T5  T4  T  T4  T5s   243   0.84  243  163.5   176.2 K

C 

h 2s  h1 T2s  T1  h 2  h1 T2  T1

 

T2  T1   T2s  T1  / C  253   376.0  253  / 0.82  402.9 K

From an energy balance on the regenerator, or

mcp  T3  T4   mc p  T1  T6 

 

T3  T4  T1  T6

T6  T1  T3  T4  253  289  243  207 K

QL  mc p (T6  T5 )  (75 / 60 kg/s)(1.005 kJ/kg  K)(207  176.2) K  38.64 kW Wnet  mc p [(T2  T1 )  (T4  T5 )]  (75 / 60 kg/s)(1.005 kJ/kg  K)[(402.9  253)  (243  176.2)] K  104.5 kW

COP 

QL 38.64   0.370 Wnet 104.5

10-603

 T   280  ExQ  QL 1  0   (38.64 kW) 1    7.20 kW L T  236   L 

This is the minimum power input:

Wmin  ExQ  7.20 kW L

The second-law efficiency of the cycle is

II 

ExQ

Wnet



7.20  0.0689  6.89% 104.5

The total exergy destruction in the cycle can be determined from

Exdest,total  Wnet  ExQ  104.5  7.20  97.3 kW L

EES (Engineering Equation Solver) SOLUTION "Given" r_p=4 T_1=-20+273 [K] T_3=16+273 [K] T_4=-30+273 [K] m_dot=75/60 [kg/s] eta_C=0.82; eta_T=0.84 T_L=-37+273 [K]; T_H=7+273 [K]; T_0=T_H c_p=1.005 [kJ/kg-K]; R=0.287 [kJ/kg-K]; k=1.4 "Analysis" T_2s=T_1*r_P^((k-1)/k) eta_C=(T_2s-T_1)/(T_2-T_1) T_5s=T_4*(1/r_P)^((k-1)/k) eta_T=(T_4-T_5)/(T_4-T_5s) T_3-T_4=T_1-T_6 "regenerator energy balance" Q_dot_L=m_dot*c_p*(T_6-T_5) W_dot_in=m_dot*c_p*(T_2-T_1) W_dot_out=m_dot*c_p*(T_4-T_5) W_dot_net=W_dot_in-W_dot_out COP=Q_dot_L/W_dot_net Ex_dot_QL=-Q_dot_L*(1-T_0/T_L) eta_II=Ex_dot_QL/W_dot_net Ex_dot_dest=W_dot_net-Ex_dot_QL "Simple gas refrigeration cycle analysis, the subscript 'a' is used" T_1_a=T_1 T_2_a=T_2 T_3_a=T_3 T_4s_a=T_3_a*(1/r_P)^((k-1)/k) T_4_a=T_3_a-eta_T*(T_3_a-T_4s_a) Q_dot_L_a=m_dot*C_p*(T_1_a-T_4_a) W_dot_in_net_a=m_dot*C_p*(T_2_a-T_1_a)-m_dot*C_p*(T_3_a-T_4_a) COP_a=Q_dot_L_a/W_dot_in_net_a

12-129 Pure ammonia enters the condenser of an ammonia-water absorption refrigeration system at 2500 kPa and 60C at a rate of 1.15 kg / min . Ammonia leaves the condenser as saturated liquid and is throttled to a pressure of 150 kPa. Ammonia © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-604

leaves the evaporator as a saturated vapor. Heat is supplied to the generator by geothermal liquid water that enters at 134C at a rate of 0.35 kg / s and leaves at 117C. Determine the COP of the system. The enthalpies of ammonia at various states of the system are hcond,in = 1497.4 kJ / kg , hcond,out = 482.5 kJ / kg , hevap,out = 1430.0 kJ / kg . Also, take the specific heat of geothermal water to be 4.2 kJ / kg ×°C .

12-129 An absorption refrigeration system using ammonia-water solution is considered. The COP and the second-law efficiency of the system are to be determined. Assumptions 1 2

Steady operating conditions exist. Kinetic and potential energy changes are negligible.

Properties The enthalpies of ammonia at various states of the system are given as hcond,in = 1497.4 kJ / kg , hcond,out = 482.5 kJ / kg , hevap,out = 1430.0 kJ / kg . The specific heat of water is given as 4.2 kJ / kg ×°C .

Analysis Noting that the throttling process is isenthalpic,

hcond,out  hevap,in The rate of cooling provided by the system is

QL  mR (hevap,out  hevap,in )  (1.15 / 60 kg/s)(1430.0  482.5) kJ/kg  18.17 kW The rate of heat input to the generator is

Qgen  mgeo c p (Tgeo;in  Tgeo;out )  (0.35 kg/s)(4.2 kJ/kg C)(134  117) C  24.99 kW Then the COP becomes COP 

QL 18.17 kW   0.727 Qgen 24.99 kW

EES (Engineering Equation Solver) SOLUTION "GIVEN" P_3=2500 [kPa] T_3=60 [C] T_geo1=134 [C] T_geo2=117 [C] m_dot_geo=0.35 [kg/s] m_dot_R=1.15/60 [kg/s] P_5=150 [kPa] "PROPERTIES" Fluid$='ammonia' h_3=enthalpy(Fluid$, P=P_3, T=T_3) h_4=enthalpy(Fluid$, P=P_3, x=0) h_5=h_4 h_6=enthalpy(Fluid$, P=P_5, x=1) Q_dot_gen=m_dot_geo*c_p*(T_geo1-T_geo2) c_p=4.2 Q_dot_L=m_dot_R*(h_6-h_5) Q_dot_H=m_dot_R*(h_3-h_4) COP=Q_dot_L/Q_dot_gen © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-605

12-130 The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=100 [kPa] P[2] = 1400 [kPa] Fluid$='R134a' Eta_c=1.0 "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) h[1]+Wcs=h2s W_c=Wcs/Eta_c h[1]+W_c=h[2] s[2]=entropy(Fluid$,h=h[2],P=P[2]) T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "Throttle Valve" h[4]=h[3] x[4]=quality(Fluid$,h=h[4],P=P[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "Coefficient of Performance" COP=Q_in/W_c

10-606

COP

ηc

[kPa] 100 150 200 250 300 350 400 450 500

1.937 2.416 2.886 3.363 3.859 4.384 4.946 5.555 6.22

1 1 1 1 1 1 1 1 1

12-131 The effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below.

"Input Data" P[1]=150 [kPa] P[2] = 400 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) h[1]+Wcs=h2s W_c=Wcs/Eta_c h[1]+W_c=h[2] s[2]=entropy(Fluid$,h=h[2],P=P[2]) T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-607

"Throttle Valve" h[4]=h[3] x[4]=quality(Fluid$,h=h[4],P=P[4]) s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "Coefficient of Performance:" COP=Q_in/W_c

COP

ηc

[kPa] 400 500 600 700 800 900 1000 1100 1200 1300 1400

6.163 4.722 3.882 3.32 2.913 2.6 2.351 2.145 1.971 1.822 1.692

0.7 0.7 0.7 0.7 0.7 0.7 0.7 0.7 0.7 0.7 0.7

12-132 An absorption refrigeration system operating at specified conditions is considered. The minimum rate of heat supply required is to be determined. Analysis The maximum COP that this refrigeration system can have is æ T öæ TL ö æ 298 K ÷ öæ 275 ÷ ö ÷ çç ÷ ÷ COPR, max = ççç1- 0 ÷ = ç1÷ççç ÷= 2.274 ÷ ÷ ÷ççT0 - TL ø ÷ çèç 368 K ÷ çè Ts øè øè 298 - 275 ÷ ø

Thus,

10-608

Qgen, min =

QL 28 kW = = 12.3 kW COPR, max 2.274

EES (Engineering Equation Solver) SOLUTION "Given" T_L=(2+273) [K] Q_dot_L=28 [kW] T_0=(25+273) [K] T_s=(95+273) [K] "Analysis" COP_R_max=(1-T_0/T_s)*(T_L/(T_0-T_L)) Q_dot_gen_min=Q_dot_L/COP_R_max

12-133 Problem 12-132 is reconsidered. The effect of the source temperature on the minimum rate of heat supply is to be investigated. Analysis The problem is solved using EES, and the solution is given below.

T_L = 2 [C] T_0 = 25 [C] T_s = 95 [C] Q_dot_L = 28 [kW] COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L)) Q_dot_gen_min = Q_dot_L/COP_R_max

Qgenmin

[C]

[kW] 30.26 16.3 11.65 9.32 7.925 6.994 6.33 5.831 5.443

50 75 100 125 150 175 200 225 250

12-134 A relation for the COP of the two-stage refrigeration system with a flash chamber shown in Fig. 12-14 is to be derived. Analysis The coefficient of performance is determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-609

COPR =

qL win

where

qL = (1- x6 )(h1 - h8 ) with x6 =

h6 - h f h fg

win = wcompI, in + wcompII, in = (1- x6 )(h2 - h1 )+ (1)(h4 - h9 )

Fundamentals of Engineering (FE) Exam Problems

12-135 A refrigerator removes heat from a refrigerated space at 0°C at a rate of 1.5 kJ / s and rejects it to an environment at 20°C. The minimum required power input is (a) (a) 102 W (b) 110 W (c) 140 W (d) 150 W (e) 1500 W Answer (a) 102 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TH=(20+273) [K] TL=(0+273) [K] Q_L=1.5 [kJ/s] COP_max=TL/(TH-TL) w_min=Q_L/COP_max "Some Wrong Solutions with Common Mistakes:" W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump" W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = Q_L*TL/TH "Using the wrong relation" W4_work = Q_L "Taking the rate of refrigeration as power input"

12-136 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPa and 50C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of this refrigerator is (a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4 Answer (d) 3.2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=160 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-610

P2=800 [kPa] T2=50 [C] P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_R=qL/Win Win=h2-h1 qL=h1-h4 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h2-h3)/(h2-h1) "COP of heat pump" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropic compression"

12-137 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from the refrigerated space is 32 kJ / s , the mass flow rate of the refrigerant is (a) 0.19 kg / s (b) 0.15 kg / s (c) 0.23 kg / s (d) 0.28 kg / s (e) 0.81 kg / s Answer (c) 0.23 kg / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 [kPa] P2=800 [kPa] Q_refrig=32 [kJ/s] P3=P2 P4=P1 s2=s1 m=Q_refrig/(h1-h4) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in" W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H" W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (at P4)" W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2"

10-611

12-138 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) 28 kJ / kg (b) 34 kJ / kg (c) 49 kJ / kg (d) 144 kJ / kg (e) 275 kJ / kg Answer (a) 28 kJ / kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=800 [kPa] P2=140 [kPa] h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1) TH=TEMPERATURE(R134a,x=0,P=P1)+273 TL=TEMPERATURE(R134a,x=0,P=P2)+273 q_H=h_fg COP=TH/(TH-TL) w_net=q_H/COP "Some Wrong Solutions with Common Mistakes:" W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator" W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" W4_work = q_H*TL/TH "Using the wrong relation"

12-139 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerant is 0.193 kg / s , the rate of heat supply by the heat pump to the heated space is (a) 3.3 kW (b) 23 kW (c) 26 kW (d) 31 kW (e) 45 kW Answer (d) 31 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 [kPa] P2=1200 [kPa] m=0.193 [kg/s] P3=P2 P4=P1 s2=s1 Q_supply=m*(h2-h3) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-612

h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in" W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L" W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Using h_fg at P1"

12-140 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 120 kPa and 700 kPa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is (a) 0.69 (b) 0.63 (c) 0.58 (d) 0.43 (e) 0.35 Answer (a) 0.69 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 [kPa] P2=700 [kPa] P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 x4=QUALITY(R134a,h=h4,P=P4) liquid=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_liquid = x4 "Taking quality as liquid content" W2_liquid = 0 "Assuming superheated vapor" W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion" s3=ENTROPY(R134a,x=0,P=P3)

12-141 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.24 MPa and 1.2 MPa. The coefficient of performance of this heat pump is (a) 5.9 (b) 5.3 (c) 4.9 (d) 4.2 (e) 3.8 Answer (c) 4.9 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=240 [kPa] P2=1200 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-613

P3=P2 P4=P1 s2=s1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_HP=qH/Win Win=h2-h1 qH=h2-h3 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)"

12-142 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa. Air is cooled to 35C before entering the turbine. The lowest temperature of this cycle is (a) –58C (b) –26C (c) 0C (d) 11C (e) 24C Answer (a) –58C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1= 80 [kPa] P2=280 [kPa] T3=(35+273) [K] k=1.4 T4_K=T3*(P1/P2)^((k-1)/k) "Mimimum temperature is the turbine exit temperature" T4_C=T4_K-273 "Some Wrong Solutions with Common Mistakes:" W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent" W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent"

12-143 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kPa and 17°C and is compressed to 400 kPa. Helium is then cooled to 20°C before it enters the turbine. For a mass flow rate of 0.2 kg / s , the net power input required is (a) 28.3 kW (b) 40.5 kW (c) 64.7 kW (d) 93.7 kW (e) 113 kW Answer (d) 93.7 kW © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

10-614

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1= 100 [kPa] T1=(17+273) [K] P2=400 [kPa] T3=(20+273) [K] m=0.2 [kg/s] k=1.667 cp=5.1926 [kJ/kg-K] T2=T1*(P2/P1)^((k-1)/k) T4=T3*(P1/P2)^((k-1)/k) "Mimimum temperature is the turbine exit temperature" W_netin=m*cp*((T2-T1)-(T3-T4)) "Some Wrong Solutions with Common Mistakes:" W1_Win = m*cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relations for temps" W2_Win = m*cp*(T2-T1) "Ignoring turbine work" W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB-1)/kB); kB=1.4 "Using air properties" W4_Win=m*cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K"

12-144 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at a rate of 90 kJ / s minimum rate of heat supply required is (a) 13 kJ / s (b) 18 kJ / s (c) 30 kJ / s (d) 37 kJ / s (e) 90 kJ / s Answer (b) 18 kJ / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=(20+273) [K] Q_refrig=90 [kJ/s] To=35+273 [K] Ts=140+273 [K] COP_max=(1-To/Ts)*(TL/(To-TL)) Q_in=Q_refrig/COP_max "Some Wrong Solutions with Common Mistakes:" W1_Qin = Q_refrig "Taking COP = 1" W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression" W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP" W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"

12-145 … 12-153 Design and Essay Problems

13-1

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 13 THERMODYNAMIC PROPERTY RELATIONS

13-2

CYU - Check Your Understanding CYU 13-2 ■ Check Your Understanding CYU 13-2.1 Select the incorrect statement regarding the Gibbs relations and the Maxwell relations. (a) The Gibbs function is g = u  Ts (b) One of the Gibbs relation is du = T ds  P dv (c) One of the Gibbs relation is dh = T ds + v dP æ¶ T ö ö ÷ =- æ ÷ çç ¶ P ÷ (d) One of the Maxwell relations is çç ÷ ÷ ÷ ÷ ÷ èç ¶ v ø èç ¶ sø s

æ¶ T ö æ¶ v ö ÷ (e) One of the Maxwell relations is çç ÷ ÷ =ç ÷ ÷ ÷ çè ¶ P ø ÷ èçç ¶ sø ÷ s

Answer: (a) The Gibbs function is g = u  Ts

CYU 13-3 ■ Check Your Understanding CYU 13-3.1 Select the incorrect statement regarding the Clapeyron equation. (a) The Clapeyron equation enables us to determine the enthalpy of vaporization at a given temperature. (b) The Clapeyron equation enables us to determine the entropy change during a phase change process. (c) The Clapeyron equation is applicable to any phase-change process that occurs at constant temperature and pressure.

h fg ædP ö÷ = ÷ è dT ø÷sat T v fg

(d) The Clapeyron equation can be expressed as ççç (e) None of these

Answer: (b) The Clapeyron equation enables us to determine the entropy change during a phase change process.

CYU 13-4 ■ Check Your Understanding CYU 13-4.1 Select the incorrect statement regarding specific heats (a) The constant-pressure specific heat and the constant-volume specific heat are identical for ideal gases. (b) The specific heats of an ideal gas depend on temperature only. (c) The difference between the constant-pressure specific heat and the constant-volume specific heat approaches zero as the thermodynamic temperature approaches zero. (d) The constant-pressure specific heat is always greater than or equal to the constant-volume specific heat. (e) At low pressures, gases behave as ideal gases, and their specific heats essentially depend on temperature only. Answer: (a) The constant-pressure specific heat and the constant-volume specific heat are identical for ideal gases. CYU 13-4.2 Which relation for specific heats is wrong? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-3

(a)

æ¶ v ö æ¶ P ö ç ÷ c p - cv = T çç ÷ ÷ ÷ ÷ èçç¶ T ø ÷ çè¶ T ø P v

(b)

æ¶ v ö ö ÷ æ ç¶ P ÷ cp - cv = - T çç ÷ ÷ èçç ¶ v ÷ ÷ çè¶ T ø ÷ ø÷ P T

(cp - cp0 )T = - T (c)

P æ¶ 2v ö

ò ççèç¶ T ÷÷÷ø dP 0

æ¶ cp ÷ ö æ¶ 2v ö ÷ çç çç ÷ ÷ = T ÷ ÷ çç ¶ P ÷ ç 2 ÷ ç¶ T ÷ è øP è ø T (d) (e) None of these Answer: (e) None of these

CYU 13-5 ■ Check Your Understanding CYU 13-5.1 Select the incorrect statement regarding the Joule-Thomson coefficient: (a) The Joule-Thomson coefficient represents the slope of h = constant lines on a T-P diagram. (b) If the Joule-Thomson coefficient is negative, the temperature of the fluid decreases during a constant-enthalpy process. (c) The Joule-Thomson coefficient is a measure of the change in temperature with pressure during a constant-enthalpy process. (d) None of these Answer: (b) If the Joule-Thomson coefficient is negative, the temperature of the fluid decreases during a constant-enthalpy process. CYU 13-5.2 Select the incorrect statement regarding the Joule-Thomson coefficient. (a) The Joule-Thomson coefficient of an ideal gas is zero. (b) The temperature at a point where a constant-enthalpy line intersects the inversion line is called the inversion temperature. (c) The Joule-Thomson coefficient can be determined from a knowledge of the constant-pressure specific heat and the P-v -T behavior of the substance. (d) The temperature of a fluid decreases during a throttling process that takes place on the right-hand side of the inversion line. (e) A cooling effect cannot be achieved by throttling unless the fluid is below its maximum inversion temperature. Answer: (d) The temperature of a fluid decreases during a throttling process that takes place on the right-hand side of the inversion line.

CYU 13-6 ■ Check Your Understanding CYU 13-6.1 Select the incorrect statement regarding the property changes of real gases. (a) The difference between h and h* is called the enthalpy departure. (b) The enthalpy departure represents the variation of the enthalpy of a gas with temperature at a fixed pressure. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-4

(c) The generalized enthalpy departure chart is used to determine the deviation of the enthalpy of a gas at a given P and T from the enthalpy of an ideal gas at the same T. (d) The generalized entropy departure chart is used to determine the deviation of the entropy of a gas at a given P and T from the entropy of an ideal gas at the same P and T. (e) None of these Answer: (b) The enthalpy departure represents the variation of the enthalpy of a gas with temperature at a fixed pressure. CYU 13-6.2 Select the incorrect relation for the enthalpy or internal energy changes of a real gas. (a)

u2 - u1 = (h2 - h1)- Ru (Z2T2 - Z1T1)

(b) h2 - h1 = (h2 - h1)ideal - RTcr(Zh2 - Zh1) (c) u2 - u1 = (h2 - h1)ideal - R(Z2T2 - Z1T1) (d) None of these Answer: (c) u2 - u1 = (h2 - h1)ideal- R(Z2T2 - Z1T1) CYU 13-6.3 Select the correct relation(s) for the enthalpy or entropy changes of real gases. (a) s2 - s1 = (s2 - s1)ideal - Ru (Zs2 - Zs1) (b)

h2 - h1 = (h2 - h1)ideal- RuTcr(Zh2 - Zh1)

13-5

PROBLEMS Partial Derivatives and Associated Relations 13-1C For functions that depend on one variable, they are identical. For functions that depend on two or more variable, the partial differential represents the change in the function with one of the variables as the other variables are held constant. The ordinary differential for such functions represents the total change as a result of differential changes in all variables.

13-2C

13-3C Only when (z / x) y  0 . That is, when z does not depend on y and thus z = z(x).

13-4C It indicates that z does not depend on y. That is, z = z(x).

13-6

13-5C Yes.

13-6C Yes.

13-7C (a) (x) y  dx ; (b)

; and S (c) dz  (z ) x  (z ) y

13-8 Air at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Air is an ideal gas. Properties The gas constant of air is R  0.287kPa·m3 / kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = R T / v . Noting that R is a constant and P = P(T, v),

æ¶ P ö æ¶ P ö RdT RT d v ç ÷ dP = çç ÷ ÷ ÷ ÷ dT + èçç ¶v ø ÷ dv = v - v 2 çè¶T ø v T (a) The change in T can be expressed as dT  T = 350 × 0.01 = 3.5 K. At v = constant,

(dP)v = -

RdT (0.287 kPa ×m3 /kg ×K)(3.5 K) = = 1.339 kPa v 0.75 m3 /kg

(b) The change in v can be expressed as dv  v  0.75  0.01  0.0075 m3 / kg . At T = constant,

(dP)T = -

RT d v (0.287 kPa ×m3 /kg ×K)(350 K)(0.0075 m3 /kg) == - 1.339 kPa 2 v (0.75 m3 /kg)2

dP = (dP)v + (dP)T = 1.339 + (- 1.339) = 0 Thus the changes in T and v balance each other.

EES (Engineering Equation Solver) SOLUTION "Given" T=350 [K] v=0.75 [m^3/kg] y=0.01 "y*100: percentage change" "Properties" R=0.287 [kPa-m^3/kg-K] "Analysis" "Eq. 3 yields: dP=R*dT/v-R*T*dv/v^2" "(a)" dT=y*T dP_v=R*dT/v "(b)" dv=y*v dP_T=-R*T*dv/v^2 "(c)" dP=dP_v+dP_T

13-7

13-9 Helium at a specified temperature and specific volume is considered. The changes in pressure corresponding to a certain increase of different properties are to be determined. Assumptions Helium is an ideal gas Properties The gas constant of helium is R  2.0769kPa·m3 / kg·K (Table A-1). Analysis An ideal gas equation can be expressed as P = R T / v . Noting that R is a constant and P = P(T, v ),

æ¶ P ö æ¶ P ö R dT RT d v ç ÷ dP = çç ÷ ÷ ÷ ÷ dT + èçç ¶v ø ÷ dv = v - v 2 çè¶T ø v T (a) The change in T can be expressed as dT  T = 350 × 0.01 = 3.5 K. At v = constant,

(dP)v =

R dT (2.0769 kPa ×m3 /kg ×K)(3.5 K) = = 9.692 kPa v 0.75 m3 /kg

(b) The change in v can be expressed as dv  v  0.75  0.01  0.0075 m3 / kg . At T = constant,

(dP)T = -

RT d v (2.0769 kPa ×m3 /kg ×K)(350 K)(0.0075 m3 ) = = - 9.692 kPa v2 (0.75 m3 /kg)2

(c) When both v and T increases by 1%, the change in P becomes dP = (dP)v + (dP)T = 9.692 + (- 9.692) = 0 Thus the changes in T and v balance each other.

EES (Engineering Equation Solver) SOLUTION "Given" T=350 [K] v=0.75 [m^3/kg] y=0.01 "y*100: percentage change" "Properties" R=2.0769 [kPa-m^3/kg-K] "Analysis" "Eq. 3 yields: dP=R*dT/v-R*T*dV/v^2" "(a)" dT=y*T dP_v=R*dT/v "(b)" dv=y*v dP_T=-R*T*dv/v^2 "(c)" dP=dP_v+dP_T

13-10E Nitrogen gas at a specified state is considered. The c p and cv of the nitrogen are to be determined using Table A18E, and to be compared to the values listed in Table A-2Eb. Analysis The c p and cv of ideal gases depends on temperature only, and are expressed as c p (T )  dh(T ) / dT and

cv (T )  du(T ) / dT . Approximating the differentials as differences about 600 R, the c p and cv values are determined to be ædh(T ) ö æD h(T ) ö ÷ ÷ c p (800 R) = çç @çç ÷ ÷ ÷ ÷ èç dT ø èç D T ø T = 800 R T @ 800 R =

h (820 R )- h (780 R )

(820 - 780) R

13-8

(5704.7 - 5424.2)/28.013 Btu/lbm = 0.250 Btu / lbm ×R (820 - 780) R

(Compare: Table A-2Eb at 800 R  340F  c p  0.250Btu / lbm·R )

ædu (T ) ö æD u (T ) ö ÷ ÷ cv (800 R) = çç @çç ÷ ÷ ÷ ÷ çè dT ø ç è ø D T T = 800 R T @ 800 R =

u (820 R )- u (780 R )

(820 - 780) R (4076.3 - 3875.2)/28.013 Btu/lbm = 0.179 Btu / lbm ×R (820 - 780) R

(Compare: Table A-2Eb at 800 R  340F  cv  0.179Btu / lbm·R )

EES (Engineering Equation Solver) SOLUTION "Given" T=800 [R] P=50 [psia] "Analysis" Fluid$='N2' DELTAT=20 [R] T[1]=T-DELTAT T[2]=T+DELTAT h_T[1]=enthalpy(Fluid$, T=T[1]) h_T[2]=enthalpy(Fluid$, T=T[2]) c_p_T=(h_T[2]-h_T[1])/(T[2]-T[1]) u_T[1]=intenergy(Fluid$, T=T[1]) u_T[2]=intenergy(Fluid$, T=T[2]) c_v_T=(u_T[2]-u_T[1])/(T[2]-T[1]) "Tabulated (ees calculated) values" c_p=CP(Fluid$, T=T) c_v=CV(Fluid$, T=T)

13-11 The state of an ideal gas is altered slightly. The change in the specific volume of the gas is to be determined using differential relations and the ideal-gas relation at each state. Assumptions The gas is air and air is an ideal gas. Properties The gas constant of air is R  0.287kPa·m3 / kg·K (Table A-1). Analysis (a) The changes in T and P can be expressed as dT @D T = (404 - 400) K = 4 K

dP @D P = (96 - 100) kPa = - 4 kPa The ideal gas relation Pv = RT can be expressed as v = R T / P . Note that R is a constant and v = v (T, P). Applying the total differential relation and using average values for T and P,

æ¶ v ö æ¶ v ö÷ RdT RT dP ç d v = çç ÷ ÷ ÷ dT + èçç¶ P ø÷ ÷ dP = P - P 2 èç¶ T ø P T

æ 4K ö (402 K)( - 4 kPa) ÷ ÷ = (0.287 kPa ×m3 /kg ×K) ççç ÷ 2 ÷ (98 kPa) è98 kPa ø

13-9

= (0.01171 m3 /kg) + (0.04805 m3 /kg) = 0.0598 m3 / kg

(b) Using the ideal gas relation at each state,

v1 =

RT1 (0.287 kPa ×m3 /kg ×K)(400 K) = = 1.1480 m3 /kg P1 100 kPa

v2 =

RT2 (0.287 kPa ×m3 /kg ×K)(404 K) = = 1.2078 m3 /kg P2 96 kPa

Thus, D v = v 2 - v1 = 1.2078 - 1.1480 = 0.0598 m3 / kg

The two results are identical.

EES (Engineering Equation Solver) SOLUTION "Given" T[1]=400 [K] P[1]=100 [kPa] T[2]=404 [K] P[2]=96 [kPa] "Properties" R=0.287 [kPa-m^3/kg-K] "Analysis" "(a)" dT=T[2]-T[1] dP=P[2]-P[1] T=(T[1]+T[2])/2 P=(P[1]+P[2])/2 dv=R*dT/P-R*T*dP/P^2 "from Eq. 11-3" "(b)" v[1]=R*T[1]/P[1] v[2]=R*T[2]/P[2] DELTAv=v[2]-v[1]

13-12 Using the equation of state P(v  a) = RT, the cyclic relation, and the reciprocity relation at constant v are to be verified. Analysis (a) This equation of state involves three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. Replacing x, y, and z by P, v, and T, the cyclic relation can be expressed as

æ¶ P ö æ¶v ö æ¶ T ö çç ÷ çç ÷ çç ÷ ÷ ÷ ÷ ÷ ÷ ÷ =- 1 èç ¶ v øT èç¶T øP èç¶ P ø v where P=

æ¶ P ö RT - RT P ¾¾ ® çç ÷ = =÷ 2 ÷ ç è ¶v øT (v - a ) v- a v- a

æ¶v ö RT R + a ¾¾ ® çç ÷ = ÷ ÷ ç è¶ T øP P P

13-10

æ¶T ö P(v - a ) v- a ¾¾ ® çç ÷ = ÷ ÷ ç è¶ P øv R R

Substituting,

æ¶ P ö æ¶ v ö æ¶ T ö÷ æ P öæ öæv - a ö÷ ÷ çç ÷ ç ÷ ç ç çR÷ ç ÷ ÷ ÷ ÷ ÷ èçç¶ T ø ÷ èçç¶ P ø÷ ÷ = èçç- v - a øè ÷çç P øè ÷çç R ø÷ ÷= - 1 çè ¶ v ø T P v which is the desired result. (b) The reciprocity rule for this gas at v = constant can be expressed as

æ¶ P ÷ ö 1 çç ÷ = çè ¶ T ÷ øv (¶ T / ¶ P )v T=

æ¶T ö P(v - a ) v- a ¾¾ ® çç ÷ = ÷ ÷ ç è¶ P øv R R

æ¶ P ö RT R ¾¾ ® çç ÷ = ÷ ÷ ç è ¶ T øv v - a v- a

We observe that the first differential is the inverse of the second one. Thus the proof is complete.

13-13 It is to be proven for an ideal gas that the P = constant lines on a T-v diagram are straight lines and that the high pressure lines are steeper than the low-pressure lines. Analysis (a) For an ideal gas Pv = RT or T = P v / R . Taking the partial derivative of T with respect to v holding P constant yields

æ¶ T ö P çç ÷ = ÷ çè ¶ v ÷ øP R which remains constant at P = constant. Thus the derivative (T / v) P , which represents the slope of the P = const. lines on a T-v diagram, remains constant. That is, the P = const. lines are straight lines on a Tv diagram. (b) The slope of the P = const. lines on a T-v diagram is equal to P / R , which is proportional to P. Therefore, the high pressure lines are steeper than low pressure lines on the T-v diagram.

13-14 A relation is to be derived for the slope of the v = constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state. Analysis The van der Waals equation of state can be expressed as

13-11

T

1 a   P  2 v  b  R v 

Taking the derivative of T with respect to P holding v constant,

1 v b  T     1  0v  b    P R R  v which is the slope of the v = constant lines on a T-P diagram.

The Maxwell Relations 13-15E The validity of the last Maxwell relation for steam at a specified state is to be verified. Analysis We do not have exact analytical property relations for steam, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state,

æ¶ s ö÷ ? æ¶v ö÷ çç ÷ = - çç ÷ èç¶ P ø÷T èç¶ T ø÷P ? æD s ö æD v ö ÷ ÷ çç @- çç ÷ ÷ ÷ ÷ çèD P ø èçD T ø T = 800° F P = 400 psia

æ s450 psia - s350 psia ö÷ æv ? - v 750° F ö÷ çç ÷ ÷ @- ççç 850° F ÷ çç(450 - 350) psia ÷ è ø÷T = 800° F èç(850 - 750)°F ø÷P= 400 psia (1.6706 - 1.7009) Btu/lbm ×R ? (1.8976 - 1.7344) ft 3 /lbm @(450 - 350) psia (850 - 750)°F

- 1.639´ 10- 3 ft 3 /lbm ×R @- 1.632´ 10- 3 ft 3 /lbm ×R since 1Btu  5.4039 psia·ft 3 , and R = F for temperature differences. Thus the fourth Maxwell relation is satisfied.

EES (Engineering Equation Solver) SOLUTION "Given" T=800 [F] P=400 [psia] "Analysis" Fluid$='steam_iapws' DELTAT=50 [F] DELTAP=50 [psia] T[1]=T-DELTAT T[2]=T+DELTAT P[1]=P-DELTAP P[2]=P+DELTAP v[1]=volume(Fluid$, T=T[1], P=P) v[2]=volume(Fluid$, T=T[2], P=P) s[1]=entropy(Fluid$, P=P[1], T=T) s[2]=entropy(Fluid$, P=P[2], T=T) A=(s[2]-s[1])/(P[2]-P[1])*Convert(Btu, psia-ft^3) "A=(ds/dP)_T" B=-(v[2]-v[1])/(T[2]-T[1]) "B=(dv/dT)_P"

13-16 The validity of the last Maxwell relation for refrigerant-134a at a specified state is to be verified. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-12

Analysis We do not have exact analytical property relations for refrigerant-134a, and thus we need to replace the differential quantities in the last Maxwell relation with the corresponding finite quantities. Using property values from the tables about the specified state,

æ¶ s ÷ ö ? æ ö çç ÷ = - çç ¶ v ÷ ÷ èç¶ P ÷ ø èç¶ T ø÷ T

? æD s ö÷ æD v ö÷ çç @- çç ÷ ÷ çèD P ø÷ èçD T ø÷P= 700 kPa T = 50° C

æs ö æv ? - v 30°C ö ÷ çç 900 kPa - s500 kPa ÷ ÷ ÷ @- ççç 70°C ÷ çç(900 - 500) kPa ÷ ÷ ÷ ç 70 30 ° C ( ) è øT = 50°C è ø P = 700 kPa

(0.9661- 1.0309) kJ/kg ×K ? (0.036373 - 0.029966) m3 /kg @(900 - 500) kPa (70 - 30)°C - 1.621´ 10- 4 m3 /kg ×K @- 1.602´ 10- 4 m3 /kg ×K

since kJ  kPa·m3 , and K = C for temperature differences. Thus the last Maxwell relation is satisfied.

EES (Engineering Equation Solver) SOLUTION "Given" T=50 [C] P=700 [kPa] "Analysis" Fluid$='R134a' DELTAT=20 [C] T[1]=T-DELTAT T[2]=T+DELTAT DELTAP=200 [kPa] P[1]=P-DELTAP P[2]=P+DELTAP v[1]=volume(Fluid$, T=T[1], P=P) v[2]=volume(Fluid$, T=T[2], P=P) s[1]=entropy(Fluid$, P=P[1], T=T) s[2]=entropy(Fluid$, P=P[2], T=T) A=(s[2]-s[1])/(P[2]-P[1]) "A=(ds/dP)_T" B=-(v[2]-v[1])/(T[2]-T[1]) "B=(dv/dT)_P"

13-17 Problem 13-16 is reconsidered. The validity of the last Maxwell relation for refrigerant 134a at the specified state is to be verified. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T=50 [C] P=700 [kPa] P_increment=200 [kPa] T_increment=20 [C] P[2]=P+P_increment P[1]=P-P_increment T[2]=T+T_increment T[1]=T-T_increment

13-13

DELTAP=P[2]-P[1] DELTAT=T[2]-T[1] v[1]=volume(R134a,T=T[1],P=P) v[2]=volume(R134a,T=T[2],P=P) s[1]=entropy(R134a,T=T,P=P[1]) s[2]=entropy(R134a,T=T,P=P[2]) DELTAs=s[2] - s[1] DELTAv=v[2] - v[1] "The partial derivatives in the last Maxwell relation is associated with the Gibbs function and are approximated by the ratio of ordinary differentials:" LeftSide =DELTAs/DELTAP*Convert(kJ,m^3-kPa) "[m^3/kg-K] at T = const." RightSide=-DELTAv/DELTAT "[m^3/kg-K] at P = const." SOLUTION DELTAP=400 [kPa] DELTAs=-0.06484 [kJ/kg-K] DELTAT=40 [C] DELTAv=0.006407 [m^3/kg] LeftSide=-0.0001621 [m^3/kg-K] P=700 [kPa] P[1]=500 [kPa] P[2]=900 [kPa] P_increment=200 [kPa] RightSide=-0.0001602 [m^3/kg-K] s[1]=1.0309 [kJ/kg-K] s[2]=0.9661 [kJ/kg-K] T=50 [C] T[1]=30 [C] T[2]=70 [C] T_increment=20 [C] v[1]=0.02997 [m^3/kg] v[2]=0.03637 [m^3/kg]

13-18 It is to be shown how T, v, u, a, and g could be evaluated from the thermodynamic function h = h(s, P). Analysis Forming the differential of the given expression for h produces

æ¶ h ö æ¶ h ö dh = çç ÷ ds + çç ÷ ÷ dP ÷ çè ¶ s ÷ ç øP è¶ P ø÷s Solving the dh Gibbs equation gives

dh = Tds + v dP Comparing the coefficient of these two expressions æ¶ h ö T = çç ÷ ÷ çè ¶ s ÷ øP æ¶ h ö v = çç ÷ ÷ çè¶ P ø÷ s both of which can be evaluated for a given P and s. From the definition of the enthalpy,

13-14

æ¶ h ö u = h - Pv = h - P çç ÷ ÷ çè¶ P ÷ øs Similarly, the definition of the Helmholtz function,

æ¶ h ö æ¶ h ö÷ ç a = u - Ts = h - P çç ÷ ÷ ÷ - s èçç ¶ s ø÷ ÷ çè¶ P ø s P while the definition of the Gibbs function gives

æ¶ h ö q = h - Ts = h - s çç ÷ ÷ çè ¶ s ÷ øP All of these can be evaluated for a given P and s and the fundamental h(s,P) equation.

13-19 Using the Maxwell relations, a relation for (s / P)T for a gas whose equation of state is P(v-b) = RT is to be obtained. Analysis This equation of state can be expressed as v =

RT + b. Then, P

æ¶ v ö R çç ÷ = ÷ çè¶ T ÷ øP P From the fourth Maxwell relation,

æ¶ s ö÷ æ ö R çç ÷ = - çç ¶ v ÷ ÷ =- P çè¶ P ø÷T çè¶T ø÷ P

13-20 Using the Maxwell relations, a relation for (s / v)T for a gas whose equation of state is ( P  a / v 2 )(v  b)  RT is to be obtained. Analysis This equation of state can be expressed as P =

RT a + . Then, v - b v2

æ¶ P ö÷ çç ÷ = R çè ¶ T ÷ øv v - b From the third Maxwell relation,

æ¶ s ö÷ æ¶ P ö÷ çç ÷ = çç ÷ = R çè¶ v ø÷T èç ¶ T ø÷v v - b

13-21 Using the Maxwell relations and the ideal-gas equation of state, a relation for (s / v)T for an ideal gas is to be obtained. Analysis The ideal gas equation of state can be expressed as P =

RT . Then, v

æ¶ P ö R çç ÷ = ÷ çè ¶ T ÷ øv v From the third Maxwell relation,

13-15

æ¶ s ö÷ æ¶ P ö÷ çç ÷ = çç ÷ = R çè¶ v ø÷T èç ¶ T ø÷v v

æ¶ P ö ö k çæ¶ P ÷ = 13-22 It is to be proven that çç ÷ çç ÷ ÷ ÷ çè ¶ T ÷ øs k - 1 è ¶ T øv Analysis Using the definition of cv ,

æ¶ s ö æ¶ s ö÷ æ¶ P ö÷ ç ç cv = T çç ÷ ÷ = T èçç¶ P ø÷ ÷ èçç ¶ T ø÷ ÷ çè¶ T ø÷ v v v æ¶ s ö ö ÷ =- æ ÷, çç ¶ v ÷ Substituting the first Maxwell relation çç ÷ ÷ ÷ èç¶ P øv èç¶ T ø s

æ¶ v ö æ¶ P ö÷ ç cv = - T çç ÷ ÷ èçç ¶ T ø÷ ÷ çè¶ T ø÷ s v Using the definition of c p ,

æ¶ s ö æ¶ s ö÷ æ¶ v ö÷ ç ç c p = T çç ÷ ÷ ÷ = T èçç¶ v ø÷ ÷ èçç¶ T ø÷ ÷ çè¶ T ø P P P æ¶ s ö ö÷ ÷ =æ çç¶ P ÷ , Substituting the second Maxwell relation çç ÷ ÷ èç¶ v øP èç ¶ T ø÷s æ¶ P ö æ¶ v ö÷ ç c p = T çç ÷ ÷ èçç¶ T ø÷ ÷ çè ¶ T ø÷ s P From Eq. 13-46, 2

æ¶v ö ö ÷æ ÷ ç¶ P ÷ c p - cv = - T çç ÷ ÷ èçç ¶v ø ÷ èç¶T ø P T

Also, cp k = k - 1 c p - cv

Then, æ¶ P ö æ¶ v ö çç ÷ ç ÷ ÷ ÷ ÷ èçç¶ T ø ÷ æ¶ P ö æ¶ T ö æ¶ v ö èç ¶ T ø k s P çç ÷ çç ÷ == - çç ÷ ÷ ÷ ÷ 2 ÷ ÷ ÷ ç ç è ¶T øs è ¶ v øP èç¶ P ø k- 1 æ¶ v ö æ ö ¶ P T ÷ çç ÷ ç ÷ ÷ ç ÷ èç ¶ v ø ÷ çè¶ T ø P T

Substituting this into the original equation in the problem statement produces æ¶ P ö÷ æ ö æ¶ T ö÷ æ¶ v ö÷ æ¶ P ö÷ çç ÷ = - çç¶ P ÷ ÷ çç ÷ çç ÷ çç ÷ èç ¶ T ø÷ èç ¶ T ø÷ èç ¶ v ø÷ èç¶ P ø÷ èç ¶ T ø÷ s

But, according to the cyclic relation, the last three terms are equal to 1. Then,

æ¶ P ö÷ æ¶ P ö÷ çç ÷ = çç ÷ çè ¶ T ø÷s èç ¶ T ø÷s

13-16

The Clapeyron Equation 13-23C It enables us to determine the enthalpy of vaporization from h fg at a given temperature from the P, v, T data alone.

13-24C It is assumed that vfg  vg  RT / P , and h fg  constant for small temperature intervals .

13-25 Using the Clapeyron equation, the enthalpy of vaporization of refrigerant-134a at a specified temperature is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation,

ædP ö h fg = T v fg çç ÷ ÷ ÷ çè dT ø sat æD P ÷ ö @T (v g - v f )@ 40°C çç ÷ çèD T ÷ øsat, 40°C

æPsat @ 42°C - Psat @ 38°C ÷ ö ÷ = T (v g - v f )@ 40°C ççç ÷ ÷ çè 42°C - 38°C ø

æ(1072.8 - 963.68) kPa ö ÷ = (40 + 273.15 K)(0.019968 - 0.0008720 m3 /kg) çç ÷ ÷ çè ø 4K

= 163.13 kJ / kg The tabulated value of h fg at 40°C is 163.03 kJ / kg .

EES (Engineering Equation Solver) SOLUTION "Given" T=40 [C] "Analysis" Fluid$='R134a' v_f=volume(Fluid$, T=T, x=0) v_g=volume(Fluid$, T=T, x=1) T1=T-2 T2=T+2 P_sat_T1=pressure(Fluid$, T=T1, x=0) P_sat_T2=pressure(Fluid$, T=T2, x=0) DELTAP=P_sat_T2-P_sat_T1 DELTAT=T2-T1 v_fg=v_g-v_f h_fg=(T+273.15)*v_fg*(DELTAP/DELTAT) "Tabulated (calculated by ees) values" h_f=enthalpy(Fluid$, T=T, x=0) h_g=enthalpy(Fluid$, T=T, x=1) h_fg_tabulated=h_g-h_f

13-26 Problem 13-25 is reconsidered. The enthalpy of vaporization of refrigerant 134-a as a function of temperature over the temperature range 20 to 80°C by using the Clapeyron equation and the refrigerant 134-a data in EES is to be plotted. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-17

Analysis The problem is solved using EES, and the results are tabulated and plotted below. T=40 [C] T_increment=5 [C] T[2]=T+T_increment T[1]=T–T_increment P[1]=pressure(R134a,T=T[1],x=0) P[2]=pressure(R134a,T=T[2],x=0) DELTAP=P[2]-P[1] DELTAT=T[2]-T[1] v_f=volume(R134a,T=T,x=0) v_g=volume(R134a,T=T,x=1) h_f=enthalpy(R134a,T=T,x=0) h_g=enthalpy(R134a,T=T,x=1) h_fg=h_g – h_f v_fg=v_g – v_f "The Clapeyron equation provides a means to calculate the enthalpy of vaporization, h_fg at a given temperature by determining the slope of the saturation curve on a P-T diagram and the specific volume of the saturated liquid and satruated vapor at the temperature." h_fg_Clapeyron=(T+273.2)*v_fg*DELTAP/DELTAT*Convert(m^3-kPa,kJ) PercentError=ABS(h_fg_Clapeyron-h_fg)/h_fg*Convert(, %)

T[C]

h fg [kJ / kg]

h fg, Clapeyron [kJ / kg]

−20 −10 0 10 20 30 40 50 60 70 80

212.96 206.02 198.67 190.80 182.33 173.13 163.03 151.82 139.09 124.37 106.35

213.76 206.73 199.28 191.32 182.74 173.46 163.31 152.08 139.26 124.61 106.51

PercentError [%] 0.3752 0.3454 0.3118 0.2711 0.228 0.1917 0.1724 0.1748 0.1222 0.1931 0.1437

13-18

13-27 Using the Clapeyron equation, the enthalpy of vaporization of steam at a specified pressure is to be estimated and to be compared to the tabulated data. Analysis From the Clapeyron equation,

ædP ö h fg = T v fg çç ÷ ÷ ÷ çè dT ø sat æD P ÷ ö @T (v g - v f )@300 kPa çç ÷ çèD T ÷ øsat, 300 kPa æ (325 - 275) kPa ö ÷ ÷ = Tsat@300 kPa (v g - v f )@300 kPa ççç ÷ ÷ çèTsat@325 kPa - Tsat@275 kPa ø ÷

æ ö 50 kPa ÷ ÷ = (133.52 + 273.15 K)(0.60582 - 0.001073 m3 /kg) çç ÷ ÷ çè(136.27 - 130.58)°C ø

= 2159.9 kJ / kg The tabulated value of h fg at 300 kPa is 2163.5 kJ / kg .

EES (Engineering Equation Solver) SOLUTION "Given" P=300 [kPa] "Analysis" Fluid$='steam_iapws' v_f=volume(Fluid$, P=P, x=0) v_g=volume(Fluid$, P=P, x=1) dP=25 [kPa] P1=P-dP P2=P+dP T_sat_P1=temperature(Fluid$, P=P1, x=0) T_sat_P2=temperature(Fluid$, P=P2, x=0) T=temperature(Fluid$, P=P, x=0) DELTAT=T_sat_P2-T_sat_P1 DELTAP=P2-P1 v_fg=v_g-v_f h_fg=(T+273.15)*v_fg*(DELTAP/DELTAT) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-19

"Tabulated (calculated by ees) values" h_f=enthalpy(Fluid$, P=P, x=0) h_g=enthalpy(Fluid$, P=P, x=1) h_fg_tabulated=h_g-h_f

13-28E The h fg of refrigerant-134a at a specified temperature is to be calculated using the Clapeyron equation and Clapeyron-Clausius equation and to be compared to the tabulated data. Analysis (a) From the Clapeyron equation,

ædP ö h fg = T v fg çç ÷ ÷ ÷ çè dT ø sat

æD P ÷ ö @T (v g - v f )@ 10°F çç ÷ çèD T ÷ øsat, 10° F

æPsat @ 15° F - Psat @ 5° F ÷ ö ÷ = T (v g - v f )@ 10° F ççç ÷ ÷ èç 15°F - 5°F ø

æ(29.759 - 23.793) psia ÷ ö = (10 + 459.67 R)(1.7358 - 0.01200 ft 3 /lbm) çç ÷ ÷ çè ø 10 R = 483.0 psia ×ft 3 /lbm = 89.38 Btu / lbm (0.1% error)

since 1 Btu  5.4039 psia·ft 3 . (b) From the Clapeyron-Clausius equation,

æP ö h fg æ1 1 ö÷ ÷ ÷ ln ççç 2 ÷ @ ççç ÷ ÷ çè P1 ø÷ R èçT1 T2 ø÷sat sat æ23.793 psia ö h fg æ ö 1 1 ÷ ÷ çç ÷@ n ççç ÷ ÷ ÷ ç ÷ 0.01946 Btu/lbm ×R è15 + 459.67 R 5 + 459.67 R ø è 29.759 psia ø h fg = 96.04 Btu / lbm (7.6% error)

The tabulated value of h fg at 10F is 89.25Btu / lbm .

EES (Engineering Equation Solver) SOLUTION Fluid$='R134a' R=R_u/M R_u=1.9858 [Btu/lbmol-R] M=MolarMass(Fluid$) T=10 [F] "EES solution" h_fg_EES=enthalpy(Fluid$, T=T, x=1)-enthalpy(Fluid$, T=T, x=0) "Clapeyron-Clausius solution" ln(P_2/P_1)=(h_fg_CC/R)*(1/ConvertTemp(F,R,T_1)-1/ConvertTemp(F,R,T_2)) "Watch temperature units" dT=5 [F] T_1=T+dT T_2=T-dT P_1=pressure(fluid$, T=T_1, x=0) P_2=pressure(fluid$, T=T_2, x=0) "Could use either x=0 or x=1" "Clapeyron solution - note crazy units conversion." (DELTAP/DELTAT)*Convert(lbf/in^2, Btu/ft^3)=h_fg_C/(ConvertTemp(F,R,T)*v_fg) v_fg=volume(Fluid$, T=T, x=1)-volume(Fluid$, T=T, x=0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-20

DELTAT=T_1-T_2 DELTAP=P_1-P_2

13-29 The sublimation pressure of water at 30C is to be determined using Clapeyron-Clasius equation and the triple point data of water. Analysis The sublimation pressure may be determined using Clapeyron-Clasius equation from

æP ö hig æ1 1 ö÷ çç ÷ ÷ ln ççç sub, CC ÷ ÷= ÷ èç P1 ø÷ R èççT1 T2 ø÷ where the triple point properties of water are P1  0.6117kPa and T1  0.01C  273.16 K (first line in Table A-4). Also, the enthalpy of sublimation of water at 30C is determined from Table A-8 to be 2838.4 kJ / kg . Substituting,

æP ö hig æ1 1 ö÷ çç ÷ ÷ ln ççç sub, CC ÷ ÷= ÷ èç P1 ø÷ R èççT1 T2 ø÷ æ Psub, CC ÷ ö ö 2838.4 kJ/kg æ 1 1 ÷ çç ÷= ln ççç ÷ ÷ ÷ ç è0.6117 kPa ÷ ø 0.4615 kJ/kg.K è273.16 K - 30 + 273.15 K ø Psub, CC = 0.03799 kPa

The sublimation pressure of water at 30C is given in Table A-8 to be 0.03802 kPa. Then, the error involved in using Clapeyron-Clasius equation becomes

PercentError =

0.03802 - 0.03799 ´ 100 = 0.08% 0.03802

EES (Engineering Equation Solver) SOLUTION "Input Data:" Psub_TABLE=0.03802 [kPa] P_1=0.6117 [kPa] T_1=(0.01+273.15) [K] T_2=(-30+273.15) [K] h_ig=2838.4 [kJ/kg] MM=molarmass(steam) R_u=8.314 [kJ/kmol-K] R=R_u/MM "[kJ/kg-K]" "The Clapeyron-Clausius equation (Eq. 11-24) provides a means to calculate the sublimation pressure at a temperature when the enthalpy of sublimatin h_ig is known at a given temperature and pressure. 23803" ln(Psub_CC/P_1)=h_ig/R*(1/T_1-1/T_2) PercentError=ABS(Psub_CC-Psub_TABLE)/Psub_TABLE*100"[%]"

13-30 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature. The saturation pressure of this substance at a different temperature is to be estimated. Analysis From the Clapeyron equation, h fg ædP ö (5 kPa ×m3 )/(0.002 kg) çç ÷ = = = 14.16 kPa/K ÷ ÷ çè dT ø T v fg (353 K)(1´ 10- 3 m3 )/(0.002 kg) sat

Using the finite difference approximation,

13-21

æ ö ædP ÷ ö çç ÷ » çç P2 - P1 ÷ ÷ ÷ èç dT ø÷sat ççèT2 - T1 ø÷sat Solving for P2 ,

P2 = P1 +

dP (T2 - T1 ) = 200 kPa + (14.16 kPa/K)(373 - 353)K = 483.2 kPa dT

13-31 A substance is heated in a piston-cylinder device until it turns from saturated liquid to saturated vapor at a constant pressure and temperature. The s fg of this substance at a different temperature is to be estimated. Analysis From the Clapeyron equation, h fg s fg ædP ö çç ÷ = = ÷ çè dT ÷ øsat T v fg v fg

Solving for s fg , s fg =

h fg T

(5 kJ)/(0.002 kg) = 7.082 kJ / kg ×K 353 K

Alternatively,

ædP ö 1´ 10-3 m3 s fg = çç ÷ v = (14.16 kPa/K) = 7.08 kPa ×m3 /kg ×K = 7.08 kJ/kg ×K ÷ çè dT ÷ øsat fg 0.002 kg

13-22

æ¶ (h fg / T ) ö÷ æ¶ P ö ÷ + v fg çç ÷ 13-32 It is to be shown that c p , g - c p , f = T ççç ÷ . ÷ çè ¶ T ø÷ èç¶ T ø÷sat P Analysis The definition of specific heat and Clapeyron equation are

æ¶ h ö c p = çç ÷ ÷ çè¶ T ø÷ P h ædP ÷ ö çç ÷ = fg çè dT ÷ øsat T v fg

According to the definition of the enthalpy of vaporization, h fg

hg h f = T T T Differentiating this expression gives

æ¶ h fg /T ö÷ æ¶ hg /T ö æ¶ h f /T ö÷ ÷ çç ÷ = çç ÷ ÷ - ççç ÷ çèç ¶ T ø÷ ç ÷ çè ¶ T ø ÷ èç ¶ T ø÷ ÷ P

ö hg 1 æ¶ h f ö÷ hf 1æ çç¶ hg ÷ ÷ ÷ - 2 - ççç + 2 ÷ ÷ ç T èç ¶ T ø÷P T T èç ¶ T ø÷P T c p, g

c p, f

hg - h f

T T T2 Using Clasius-Clapeyron to replace the last term of this expression and solving for the specific heat difference gives

æ¶ (h fg / T ) ö÷ æ¶ P ö ÷ + v fg çç ÷ c p , g - c p , f = T ççç ÷ ÷ çè ¶ T ø÷ èç¶ T ø÷sat P

13-33 Saturation properties for R-134a at a specified temperature are given. The saturation pressure is to be estimated at two different temperatures. Analysis From the Clapeyron equation, h fg ædP ö 225.86 kPa ×m3 /kg çç ÷ = = = 2.693 kPa/K ÷ ÷ çè dT ø T v fg (233 K)(0.35993 m3 /kg) sat

Using the finite difference approximation,

æ ö ædP ÷ ö çç ÷ » çç P2 - P1 ÷ ÷ çèT - T ø÷ çè dT ø÷ ÷ ç sat 2 1 sat Solving for P2 at 50C

P2 = P1 +

dP (T2 - T1 ) = 51.25 kPa + (2.693 kPa/K)(223 - 233)K = 24.32 kPa dT

Solving for P2 at 30C P2 = P1 +

dP (T2 - T1 ) = 51.25 kPa + (2.693 kPa/K)(243 - 233)K = 78.18 kPa dT

13-34E Saturated liquid in a piston-cylinder device is cooled at a constant pressure and temperature until it becomes saturated liquid. The boiling temperature of this substance at a different pressure is to be estimated. Analysis From the Clapeyron equation,

13-23

 5.404 psia  ft 3   /(0.5 lbm) (250 Btu)    1 Btu h fg  dP      1.896 psia/R    (475 R)(1.5 ft 3 )/(0.5 lbm)  dT  sat Tv fg

Using the finite difference approximation,

P P   dP      2 1   dT  sat  T2  T1  sat Solving for T2 ,

T2  T1 

P2  P1 (60  50)psia  475 R   480.3 R dP / dT 1.896 psia/R

EES (Engineering Equation Solver) SOLUTION "Given" m=0.5 [lbm] P1=50 [psia] DELTAVol=1.5 [ft^3] Q=250 [Btu] T1=(15+460) [R] P2=60 [psia] "Analysis" h_fg=Q/m v_fg=DELTAVol/m dP\dT_sat=h_fg*Convert(Btu, psia-ft^3)/(T1*v_fg) T2=T1+(P2-P1)/dP\dT_sat

General Relations for du, dh, ds, cv , and c p 13-35C Yes, through the relation

æ¶ c p ÷ ö æ¶ 2v ÷ ö çç çç ÷ ÷ = T ÷ ÷ 2 ÷ çè¶T ÷ ççè ¶ P ø ø T

13-36 The internal energy change of air between two specified states is to be compared for two equations of states. Assumptions Constant specific heats for air can be used. Properties For air at the average temperature (20  300) / 2  160C  433 K , cv  0.731kJ / kg·K (Table A-2b). Analysis Solving the equation of state for P gives © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-24

RT v- a

Then,

æ¶ P ö÷ çç ÷ = R çè¶ T ÷ øv v - a Using equation 13-29,

é æ¶ P ö ù ú du = cv dT + êT çç ÷ ÷ - P úd v ê çè¶ T ø÷ v ë û Substituting, æ RT RT ö ÷ du = cv dT + çç ÷ ÷d v = cv dT çèv - a v - a ø

Integrating this result between the two states with constant specific heats gives

u2 - u1 = cv (T2 - T1 ) = (0.731 kJ/kg ×K)(300 - 20)K = 205 kJ / kg The ideal gas model for the air gives

du = cv dT which gives the same answer.

13-37 The enthalpy change of air between two specified states is to be compared for two equations of states. Assumptions Constant specific heats for air can be used. Properties For air at the average temperature (20  300) / 2  160C  433 K , c p  1.018 kJ / kg·K (Table A-2b). Analysis Solving the equation of state for v gives v=

RT +a P

Then,

æ¶ v ÷ ö çç ÷ = R çè¶ T ø÷P P Using equation 13-35,

é æ¶ v ö ù dh = c p dT + êv - T çç ÷ ÷ údP ê èç¶ T ø÷P úû ë Substituting,

æRT RT ö ÷ dh = c p dT + çç + a÷dP çè P ø P ÷ = c p dT + adP Integrating this result between the two states with constant specific heats gives h2 - h1 = c p (T2 - T1 ) + a( P2 - P1 ) = (1.018 kJ/kg ×K)(300 - 20)K + (0.10 m3 /kg)(600 - 100)kPa

= 335.0 kJ / kg For an ideal gas,

dh = c p dT which when integrated gives

13-25

h2 - h1 = c p (T2 - T1 ) = (1.018 kJ/kg ×K)(300 - 20)K = 285.0 kJ / kg

13-38 The entropy change of air between two specified states is to be compared for two equations of states. Assumptions Constant specific heats for air can be used. Properties For air at the average temperature (20  300) / 2  160C  433 K , c p  1.018 kJ / kg·K (Table A-2b) and

R  0.287 kJ / kg  K (Table A-1). Analysis Solving the equation of state for v gives RT +a P

Then,

æ¶ v ÷ ö çç ÷ = R çè¶ T ø÷P P The entropy differential is

ds = c p

dT æ ¶v ö - çç ÷ ÷ ÷ dP ç T è¶ T ø P

dT dP - R T P which is the same as that of an ideal gas. Integrating this result between the two states with constant specific heats gives = cp

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1 = (1.018 kJ/kg ×K)ln

573 K 600 kPa - (0.287 kJ/kg ×K)ln 293 K 100 kPa

= 0.1686 kJ / kg ×K

13-39 The internal energy change of helium between two specified states is to be compared for two equations of states. Properties For helium, cv  3.1156 kJ / kg·K (Table A-2a). Analysis Solving the equation of state for P gives P=

RT v- a

Then,

æ¶ P ö÷ çç ÷ = R çè¶ T ÷ øv v - a Using equation 13-29,

é æ¶ P ö ù úd v du = cv dT + êT çç ÷ P ÷ ê èç¶ T ø÷v ú ë û Substituting,

æ RT RT ö÷ du = cv dT + çç dv ÷ çèv - a v - a ÷ ø

13-26

Integrating this result between the two states gives

u2 - u1 = cv (T2 - T1 ) = (3.1156 kJ/kg ×K)(300 - 20)K = 872.4 kJ / kg The ideal gas model for the helium gives

du = cv dT which gives the same answer.

13-40 The enthalpy change of helium between two specified states is to be compared for two equations of states. Properties For helium, c p  5.1926 kJ / kg·K (Table A-2a). Analysis Solving the equation of state for v gives v=

RT +a P

Then,

æ¶ v ÷ ö çç ÷ = R çè¶ T ø÷P P Using equation 13-35,

é æ¶ v ö ù ú dh = c p dT + êv - T çç ÷ ÷ údP çè¶ T ø÷ ê Pû ë Substituting,

æRT ö RT ÷ dh = c p dT + çç + adP ÷ ÷ çè P P ø = c p dT + adP

Integrating this result between the two states gives h2 - h1 = c p (T2 - T1 ) + a( P2 - P1 ) = (5.1926 kJ/kg ×K)(300 - 20)K + (0.10 m3 /kg)(600 - 100)kPa

= 1504 kJ / kg For an ideal gas,

dh = c p dT which when integrated gives

h2 - h1 = c p (T2 - T1 ) = (5.1926 kJ/kg ×K)(300 - 20)K = 1454 kJ / kg

13-41 The entropy change of helium between two specified states is to be compared for two equations of states. Properties For helium, c p  5.1926 kJ / kg·K and R  2.0769 kJ / kg·K (Table A-2a). Analysis Solving the equation of state for v gives v=

RT +a P

Then,

æ¶ v ÷ ö çç ÷ = R çè¶ T ø÷P P

13-27

The entropy differential is

ds = c p

dT æ ¶v ö - çç ÷ ÷ ÷ dP ç è T ¶T ø P

dT dP - R T P which is the same as that of an ideal gas. Integrating this result between the two states gives = cp

T P s2 - s1 = c p ln 2 - R ln 2 T1 P1 = (5.1926 kJ/kg ×K)ln

573 K 600 kPa - (2.0769 kJ/kg ×K)ln 293 K 100 kPa

= - 0.2386 kJ / kg ×K

13-42 The volume expansivity  and the isothermal compressibility  of refrigerantestimated. Analysis The volume expansivity and isothermal compressibility are expressed as

ö 1æ 1 æ¶ v ö çç ¶ v ÷ and a = - çç ÷ ÷ ÷ ÷ ç v è¶ T øP v èç¶ P ø÷T

Approximating differentials by differences about the specified state, ö - v 20°C ÷ 1 æD v ö 1 æv ÷ ÷ b @ çç = ççç 40°C ÷ ÷ ÷ ç v èD T øP= 200 kPa v è(40 - 20)°C ÷ øP= 200 kPa

æ(0.12322 - 0.11418) m3 /kg ö 1 ÷ çç ÷ ÷ ÷ 20 K 0.11874 m3 /kg çè ø

= 0.00381 K - 1 and a @-

ö ö - v180 kPa ÷ 1æ 1 æv ççD v ÷ ÷ = - çç 240 kPa ÷ ÷ ÷ ç ÷ ç v èD P øT = 30°C v è(240 - 180) kPa øT = 30°C

æ(0.09812 - 0.13248) m3 /kg ö÷ 1 çç ÷ ÷ ÷ 60 kPa 0.11874 m3 /kg çè ø

= 0.00482 kPa- 1

EES (Engineering Equation Solver) SOLUTION "Given" P=200 [kPa] T=30 [C] "Analysis" Fluid$='R134a' dT=10 [C] T[1]=T-dT T[2]=T+dT v_T[1]=volume(Fluid$, T=T[1], P=P) v_T[2]=volume(Fluid$, T=T[2], P=P) DELTAT=T[2]-T[1] DELTAv_P=v_T[2]-v_T[1]

13-28

v=volume(Fluid$, T=T, P=P) Beta=(1/v)*(DELTAv_P/DELTAT) P[1]=P-20 [kPa] P[2]=P+40 [kPa] v_P[1]=volume(Fluid$, P=P[1], T=T) v_P[2]=volume(Fluid$, P=P[2], T=T) DELTAP=P[2]-P[1] DELTAv_T=v_P[2]-v_P[1] Alpha=(-1/v)*(DELTAv_T/DELTAP)

13-43E The specific heat difference c p  cv for liquid water at 1000 psia and 300F is to be estimated. Analysis The specific heat difference c p  cv is given as 2

æ¶v ö æ¶ P ö c p - cv = - T çç ÷ ÷ ççç ÷ ÷ ÷ ÷ èç¶T ø è ø P ¶v T

Approximating differentials by differences about the specified state, 2

æD v ö÷ æD P ö÷ çç c p - cv @- T çç ÷ ÷ çèD T ø÷ èç ø÷ P = 1000 psia D v T = 300° F 2 æ(1500 - 500) psia ÷ ö æv 325 ° F - v 275 ° F ÷ ö çç ÷ ÷ = - (300 + 459.67 R) ççç ÷ ÷ ç ÷ ÷ çv - v 500 psia ÷ èç(325 - 275) °F ø øT = 300 ° F P = 1000 psia è 1500 psia 2

æ(0.017633 - 0.017151) ft 3 /lbm ö æ ö 1000 psia ÷ ÷ çç ÷ ÷ = - (609.67 R) çç ÷ ÷ 3 ÷ ÷ çè(0.017345 - 0.017417) ft /lbm ø çè 50 R ø = 0.986 psia ×ft 3 /lbm ×R = 0.183 Btu / lbm ×R

(1 Btu = 5.4039 psia ×ft 3 )

Properties are obtained from Table A-7E.

EES (Engineering Equation Solver) SOLUTION "Given" P=1000 [psia] T=300 [F] "Analysis" Fluid$='steam_iapws' dT=25 [F] T[1]=T-dT T[2]=T+dT dP=500 [psia] P[1]=P-dP P[2]=P+dP v_T[1]=volume(Fluid$, T=T[1], P=P) v_T[2]=volume(Fluid$, T=T[2], P=P) v_P[1]=volume(Fluid$, P=P[1], T=T) v_P[2]=volume(Fluid$, P=P[2], T=T) DELTAT=T[2]-T[1] DELTAP=P[2]-P[1] DELTAv_P=v_T[2]-v_T[1] DELTAv_T=v_P[2]-v_P[1]

13-29

R=-(T+459.67)*(DELTAv_P/DELTAT)^2*(DELTAP/DELTAv_T)*Convert(psia-ft^3/lbm-R, Btu/lbm-R) "R=C_pC_v"

13-44 General expressions for u, h, and s for a gas whose equation of state is P(v-a) = RT for an isothermal process are to be derived. Analysis (a) A relation for u is obtained from the general relation

ö T2 v 2 æ æ¶ P ö ÷d v ÷ D u = u2 - u1 = ò cv dT + ò ççT çç ÷ P ÷ ÷ ÷ T1 v1 ç øv è èç ¶ T ÷ ø The equation of state for the specified gas can be expressed as Thus, P =

æ¶ P ö RT R ¾¾ ® çç ÷ = ÷ ÷ ç è ¶ T øv v - a v- a

æ¶ P ö RT T çç ÷ ÷ ÷ - P = v - a - P = P- P = 0 çè ¶ T ø v Substituting, T2

D u = ò cv dT T1

(b) A relation for h is obtained from the general relation T2 P2 æ æ¶ v ö ö÷ ÷ D h = h2 - h1 = ò cP dT + ò çççv - T çç ÷ dP ÷ ÷ çè¶ T ÷ ÷ T1 P1 è øP ø

The equation of state for the specified gas can be expressed as Thus, v =

æ¶v ö RT ÷ = R + a ¾¾ ® çç ÷ ÷ çè¶ T ø P P P

æ¶ P ö RT T çç ÷ ÷ ÷ - P = v - a - P = P- P = 0 çè ¶ T ø v Substituting, T2

D h = ò c p dT + ò adP = ò c p dT + a (P2 - P1 )

D s = s2 - s1 = ò

P2 æ¶ v ö dT - ò çç ÷ ÷ dP P1 ç è¶ T ÷ øP T

Substituting (v / T ) P  R / T ,

Ds = ò

P2 æR ö T2 c P p dT - ò çç ÷ dP = dT - R ln 2 ÷ ò P1 ç T1 T èP÷ øP T P1

For an isothermal process dT = 0 and these relations reduce to

D u = 0, D h = a (P2 - P1 ), and D s = - R ln

P2 P1

13-45 Expressions for the specific heat difference c p  cv for three substances are to be derived. Analysis The general relation for the specific heat difference c p  cv is

13-30

æ¶v ö æ¶ P ö c p - cv = - T çç ÷ ÷ ççç ÷ ÷ ÷ ÷ èç¶T ø è ø P ¶v T

(a) For an ideal gas Pv  RT . Then,

æ¶ v ö RT R ¾¾ ® çç ÷ ÷ ÷ = P çè¶ T ø P P

æ¶ P ö RT RT P ¾¾ ® çç ÷ ÷ =- 2 =÷ çè ¶ v ø v v v T

Substituting, 2

æ P ö æR ö TR ç ÷ c p - cv = - T çç- ÷ ÷ ÷ ÷ èçç P ø ÷= Pv R = R çè v ø æ aö (b) For a van der Waals gas ççP + 2 ÷ ÷ ÷(v - b) = RT . Then, çè v ø

ö æ¶ T ö 1æ 1 æ 2a ö 1æ aö ççP + a ÷ (v - b) ¾ ¾® ççç ÷÷÷ = ççç- 3 ÷÷÷(v - b)+ çççP + 2 ÷÷÷ 2÷ ÷ ç è ø è ø è ø è R ¶v P R v R v v ø

2a (b - v ) Rv 3

T v- b

Inverting,

æ¶v ö 1 çç ÷ ÷ ÷ = 2a (b - v ) çè¶T ø P Rv 3

T v- b

Also, P=

æ¶ P ö RT a RT 2a ¾¾ ® çç ÷ + 3 ÷ =2 ÷ çè ¶v ø v - b v2 v (v - b) T

Substituting, 2

æ ö÷ çç ÷æ ö ÷ çç 1 2a ÷ çç RT ÷ ÷ ÷ c p - cv = T çç ÷ ÷ ç çç(v - b)2 v 3 ÷ T ÷ ÷ çç 2a (b - v ) ÷ ø ÷ ÷è ççè Rv 3 + v - b ÷ ø (c) For an incompressible substance v = constant and thus (v / T ) P  0 . Therefore,

c p - cv = 0

13-46 An expression for the specific heat difference of a substance whose equation of state is P =

RT a is v - b v (v + b)T 1/ 2

to be derived. Analysis The specific heat difference is expressed by 2

æ¶ v ö æ¶ P ö ç ÷ c p - cv = - T çç ÷ ÷ ÷ ÷ èçç¶ v ø ÷ èç¶ T ø P T

According to the cyclic relation,

æ¶ v ö ö æ¶ T ÷ ö ÷æ ÷ çç ÷ çç¶ P ÷ çç ÷ = - 1 ÷ ÷ ÷ èç¶ T øP èç¶ v øT çè ¶ P øv

13-31

which on rearrangement becomes - 1

æ¶ v ö æ¶ P ö æ¶ P ö çç ÷ ç ÷ ç ÷ ÷ ÷ ÷ ÷ = - èçç¶ T ø ÷ ççè¶ v ø ÷ çè¶ T ø P v T

Substituting this result into the expression for the specific heat difference gives 2

- 1

æ¶ P ö æ¶ P ö c p - cv = - T çç ÷ ÷ ççç ÷ ÷ ÷ ÷ èç¶ T ø è ø v ¶v T

The appropriate partial derivatives of the equation of state are

æ¶ P ÷ ö a/2 çç ÷ = R + çè¶ T ÷ øv v - b v (v + b)T 3/ 2

æ¶ P ö RT a éê 1 1 ù çç ÷ ú ÷ ÷ = - (v - b)2 + bT 1/ 2 êv 2 - (v + b)2 ú çè¶ v ø T ë û RT a 2v + b + 1/ 2 2 2 (v - b) T v (v + b) 2

The difference in the specific heats is then 2

c p - cv =

é R ù a/2 ú - Tê + êëv - b v (v + b)T 3/ 2 ú û -

RT a 2v + b + (v - b) 2 T 1/ 2 v 2 (v + b) 2

13-47 An expression for the isothermal compressibility of a substance whose equation of state is P =

RT a - 2 is to be v- b v T

derived. Analysis The definition for the isothermal compressibility is

a=-

ö 1æ çç ¶ v ÷ ÷ ÷ ç v è¶ P øT

Now,

æ¶ P ÷ ö çç ÷ = - RT + 2a çè ¶ v ÷ øT (v - b) 2 v 3T Using the partial derivative properties,

a=-

1 1 =æ¶ P ÷ ö - RT v 2a + 2 v çç ÷ 2 ÷ (v - b) v T èç¶ v øT

13-48 An expression for the volume expansivity of a substance whose equation of state is P(v  a)  RT is to be derived. Analysis Solving the equation of state for v gives RT a P The specific volume derivative is then

v

13-32

R  v      T  P P The definition for volume expansivity is



1  v    v  T  P

Combining these two equations gives 

R RT  aP

æ¶ P ÷ ö æ¶ v ÷ ö çç ÷ . 13-49 It is to be shown that c p - cv = T çç ÷ ÷ ÷ èç¶ T øv èç¶ T øP Analysis We begin by taking the entropy to be a function of specific volume and temperature. The differential of the entropy is then

æ¶ s ö æ¶ s ö ç ÷ ds = çç ÷ ÷ ÷ ÷ dT + èçç¶ v ø ÷ dv çè¶ T ø v T

æ¶ s ö c Substituting çç ÷ ÷ = v from Eq. 13-28 and the third Maxwell equation changes this to çè¶ T ø÷ T v

ds =

æ¶ P ö cv dT + çç ÷ ÷ dv çè¶ T ø÷ T v

Taking the entropy to be a function of pressure and temperature,

æ¶ s ö æ¶ s ö ds = çç ÷ ÷ dT + ççç ÷ ÷ ÷ dP èç¶ T ÷ øP è¶ P ø T

cp æ¶ s ö Combining this result with çç ÷ ÷ ÷ = T from Eq. 13-34 and the fourth Maxwell equation produces çè¶ T ø P ds =

cp T

æ¶ v ö dT - çç ÷ ÷ ÷ dP çè¶ T ø P

Equating the two previous ds expressions and solving the result for the specific heat difference,

æ¶ v ö÷ æ¶ P ö÷ (c p - cv )dT = T çç ÷ dP + çç ÷ dv ÷ èç¶ T øP èç¶ T ø÷v Taking the pressure to be a function of temperature and volume,

æ¶ P ö æ¶ P ö dP = çç ÷ dT + çç ÷ ÷ ÷ ÷ ÷ dv èç¶ T øv èç¶ v ø T When this is substituted into the previous expression, the result is

éæ ö æ¶ P ö æ¶ P ö ù æ¶ v ö æ¶ P ÷ ö çç ÷ dT + T êçç ¶ v ÷ çç ÷ (c p - cv )dT = T çç ÷ dv ÷ ÷ ççç ÷ ÷ú ÷ çè¶ T ÷ ÷ êçè¶ T ÷ èç¶ T ø øv øP çè¶ v ÷ øT è¶ T ÷ øv ú P ë û According to the cyclic relation, the term in the bracket is zero. Then, canceling the common dT term, æ¶ P ö æ¶ v ÷ ö çç ÷ c p - cv = T çç ÷ ÷ ÷ çè¶ T ÷ ç øv è¶ T øP

13-33

13-50 It is to be shown that the enthalpy of an ideal gas is a function of temperature only and that for an incompressible substance it also depends on pressure. Analysis The change in enthalpy is expressed as

æ æ¶ v ö ö÷ ÷dP dh = cP dT + ççv - T çç ÷ ÷÷ çè èç¶ T ÷ øP ø÷ For an ideal gas v = R T / P . Then, Thus,

æ¶v ö ÷ = v - T ççæR ö ÷= v - v = 0 v - T çç ÷ ÷ ÷ ÷ çè P ø èç¶ T ø P dh = c p dT

To complete the proof we need to show that c p is not a function of P either. This is done with the help of the relation

æ¶ c p ÷ ö æ¶ 2v ö çç ÷ = - T çç 2 ÷ ÷ ÷ ÷ ÷ çè¶T ÷ ççè ¶ P ø ø T

For an ideal gas, Thus,

æ 2 ö æ¶ ( R / P) ö æ¶ v ö÷ ÷ = 0 çç ÷ = R and çç ¶ v ÷ ÷ =ç ÷ ÷ ÷ ÷ èçç ¶ T ø÷P çè¶ T 2 ø èç¶ T øP P P æ¶ cP ö ÷ çç ÷ ÷ = 0 èç ¶ P ø T

Therefore we conclude that the enthalpy of an ideal gas is a function of temperature only. For an incompressible substance v = constant and thus v / T  0 . Then,

dh = c p dT + v dP Therefore we conclude that the enthalpy of an incompressible substance is a function of temperature and pressure.

13-51 It is to be proven that the definition for temperature T = (¶ u / ¶ s)v reduces the net entropy change of two constantvolume systems filled with simple compressible substances to zero as the two systems approach thermal equilibrium. Analysis The two constant-volume systems form an isolated system shown here For the isolated system

dStot = dSA + dSB ³ 0 Assume

S = S (u, v ) Then,

æ¶ s ö æ¶ s ö÷ ç ds = çç ÷ ÷ du + èçç¶ v ø÷ ÷ dv çè¶ u ø÷ v u

13-34

Since v = const. and

d v = 0,

æ¶ s ö÷ ds = çç ÷ du çè¶ u ø÷v and from the definition of temperature from the problem statement,

du du = (¶ u / ¶ s )v T Then,

dS tot = mA

du A du + mB B TA TB

The first law applied to the isolated system yields

Ein - Eout = dU 0 = dU ¾ ¾ ® mA du A + mB duB = 0 ¾ ¾ ® mB duB = - mA du A

Now, the entropy change may be expressed as

æ1 æT - TA ö÷ 1 ö÷ ÷ ÷ dS tot = mA du A ççç = mA du A ççç B ÷ ÷ çèTA TB ø÷ èç TATB ø÷ As the two systems approach thermal equilibrium,

lim dStot = 0 TA ® TB

æ¶ P ö 13-52 It is to be shown that b = a çç ÷ . ÷ çè¶ T ÷ øv Analysis The definition for the volume expansivity is

ö 1æ çç ¶ v ÷ ÷ ç v è¶ T ø÷P

The definition for the isothermal compressibility is

a=-

ö 1æ çç ¶ v ÷ ÷ ÷ ç v è¶ P øT

According to the cyclic relation,

æ¶ v ö æ¶ P ÷ ö æ¶ T ÷ ö çç ÷ ç çç ÷ = - 1 ÷ ÷ ççè¶ v ÷ ÷ ÷ çè¶ T ø ç øT è ¶ P øv P

13-35

which on rearrangement becomes

æ¶ v ö÷ æ ö æ¶ P ö÷ çç ÷ = - çç¶ v ÷ ÷ çç ÷ èç¶ T ø÷ èç¶ P ø÷ èç¶ T ø÷ P

When this is substituted into the definition of the volume expansivity, the result is

b=-

ö æ¶ P ö÷ 1æ çç ¶ v ÷ ÷ çç ÷ v èç¶ P ø÷T èç¶ T ø÷v

æ¶ P ö = - a çç ÷ ÷ ÷ çè¶ T ø v

13-53 General expressions for (u / P)T and (h / v)T in terms of P, v, and T only are to be derived. Analysis The general relation for du is

æ æ¶ P ö ö ÷ du = cv dT + çççT çç ÷ - P÷ dv ÷ ÷ ÷ ÷ è èç ¶ T øv ø Differentiating each term in this equation with respect to P at T = constant yields

æ æ ö öæ¶ v ö æ¶ u ö÷ æ ö æ¶ v ö÷ æ ö ÷ çç ÷ = 0 + ççT çç¶ P ÷ ç ÷ = T çç¶ P ÷ çç ÷ - P çç ¶ v ÷ ÷ ÷ ÷èçç¶ P ø÷ ÷ - Pø÷ ÷ ÷ ÷ çè èç ¶ T ø÷ çè¶ P ø÷T ç ç ÷ è ¶ T øv è¶ P øT èç¶ P ø÷T v T Using the properties P, T, v, the cyclic relation can be expressed as

æ¶ P ö æ ö æ ö çç ÷ ç¶ T ÷ çç ¶ v ÷ ÷ ÷ ççè ¶ v ø÷ ÷ çè¶ P ø÷ ÷ =- 1 çè ¶ T ø v P T

æ ö æ¶ v ÷ ö æ¶ v ö çç¶ P ÷ ç = - çç ÷ ÷ ÷ ÷ ççè¶ P ÷ ÷ çè ¶ T ø ç øT è¶ T ÷ øP v

¾¾ ®

Substituting, we get

æ¶ u ö æ¶ v ö÷ æ¶ v ö÷ çç ÷ ç ç ÷ ÷ = - T èçç¶ T ø÷ ÷ - P èçç¶ P ø÷ ÷ çè¶ P ø T P T The general relation for dh is

æ æ¶ v ö ö÷ dh = c p dT + ççv - T çç ÷ ÷ ÷dP ÷ èç¶ T ø÷P ø÷ èç Differentiating each term in this equation with respect to v at T = constant yields

æ æ¶ h ö÷ æ ö ö÷æ¶ P ö÷ æ ö æ¶ v ö æ¶ P ö÷ çç ÷ = 0 + ççv - T çç ¶ v ÷ çç ÷ = v çç¶ P ÷ - T çç ÷ ÷ ÷ çç ÷ ÷ ÷÷ ÷ ÷ çè çè¶ v ø÷T çè¶ T ø÷ ç ç ÷è ø è ¶ v øT èç¶ T ø÷P èç ¶ v ø÷T P ø ¶v T Using the properties v, T, P, the cyclic relation can be expressed as

æ¶ v ö æ¶ T ö÷ æ¶ P ö çç ÷ ç ç ÷ ÷ ÷ ÷ èçç¶ P ø÷ ÷ ççè ¶ v ø ÷ =- 1 çè¶ T ø P v T

æ ö æ¶ P ö÷ æ ö çç ¶ v ÷ ç ç¶T ÷ ÷ ççè ¶ v ø÷ ÷ = - ççè¶ P ø÷ ÷ çè¶ T ø÷ P T v

¾¾ ®

Substituting, we get

æ¶ h ö æ¶ P ö æ¶ T ö çç ÷ ç ÷ ç ÷ ÷ ÷ ÷ ÷ = v èçç ¶ v ø ÷ + T èçç¶ P ø ÷ çè¶ v ø T T v

13-54 It is to be demonstrated that k =

cp cv

va . (¶ v / ¶ P)s

13-36

ds = cv

dT æ ¶Pö + çç ÷ ÷ ÷ dv ç T è¶ T ø v

ds = c p

ö dT æ ¶v ÷ - çç ÷ dP ÷ T çè¶ T øP

For fixed s, these basic equations reduce to

æ¶ P ö dT = - çç ÷ ÷ dv çè¶ T ø÷ T v

dT æ ¶v ö = çç ÷ ÷ dP çè¶ T ø÷ T P

Also, when s is fixed,

¶v æ ¶v ö ÷ = çç ÷ øs ¶ P çè¶ P ÷ Forming the specific heat ratio from these expressions gives

æ¶ v ö æ¶ T ö÷ çç ÷ ç ÷ ÷ èçç ¶ P ø÷ ÷ çè¶ T ø P v k= æ¶ v ö ÷ çç ÷ çè¶ P ø÷s The cyclic relation is

æ¶ v ö æ¶ P ÷ ö æ¶ T ÷ ö çç ÷ ç çç ÷ = - 1 ÷ ÷ ççè¶ v ÷ ÷ ÷ çè¶ T ø ç øT è ¶ P øv P Solving this for the numerator of the specific heat ratio expression and substituting the result into this numerator produces

æ¶ v ÷ ö çç ÷ ÷ çè¶ P ø va T k= =æ¶ v ö æ ö ¶ çç ÷ çç v ÷ ÷ ÷ ÷ ÷ ç èç¶ P ø è ¶ Pø s s

13-55 The Helmholtz function of a substance has the form a = - RT ln

æ T T T ö÷ v - cT0 ççç1+ ln ÷. It is to be shown how to ÷ çè T0 T0 T0 ø÷ v0

obtain P, h, s, cv , and c p from this expression. Analysis Taking the Helmholtz function to be a function of temperature and specific volume yields

æ¶ a ö æ ö ç ¶ a ÷ dv da = çç ÷ ÷ ÷ dT + ççè¶ v ø÷ ÷ çè¶ T ø v T while the applicable Helmholtz equation is

da = - Pdv - sdT Equating the coefficients of the two results produces æ¶ a ö P = - çç ÷ ÷ ÷ çè¶ v ø T æ¶ a ö s = - çç ÷ ÷ çè¶ T ø÷ v Taking the indicated partial derivatives of the Helmholtz function given in the problem statement reduces these expressions to

13-37

RT v

s = R ln

v T + c ln v0 T0

The definition of the enthalpy (h = u + Pv) and Helmholtz function (a = uTs) may be combined to give

h = u + Pv = a + Ts + Pv æ¶ a ö æ¶ a ö = a - T çç ÷ ÷ - v ççç ÷ ÷ èç¶ T ø÷v è¶ v ø÷T

= - RT ln

æ T T Tö v v T ÷+ RT ln - cT ln + RT - cT0 ççç1+ ln ÷ ÷ çè T0 T0 T0 ÷ v0 v0 T0 ø

= cT0 + cT + RT æ¶ s ö cv According to çç ÷ ÷ = T given in the text (Eq. 13-28), çè¶ T ø÷ v æ¶ s ö c cv = T çç ÷ =T = c ÷ çè¶ T ÷ øv T The preceding expression for the temperature indicates that the equation of state for the substance is the same as that of an ideal gas. Then,

c p = R + cv = R + c

The Joule-Thomson Coefficient 13-56C It represents the variation of temperature with pressure during a throttling process.

13-57C The line that passes through the peak points of the constant enthalpy lines on a T-P diagram is called the inversion line. The maximum inversion temperature is the highest temperature a fluid can be cooled by throttling.

13-58C Yes.

13-59C No. The temperature may even increase as a result of throttling.

13-60C No. Helium is an ideal gas and h = h(T) for ideal gases. Therefore, the temperature of an ideal gas remains constant during a throttling (h = constant) process.

13-61 The Joule-Thomson coefficient of refrigerant-134a at a given state is to be estimated. Analysis The Joule-Thomson coefficient is defined as

æ¶ T ö÷ ÷ çè ¶ P ø÷ h

 = çç

13-38

T - T1 (at constant enthalpy) P2 - P1

@ 2

At the given state (we call it state 1), the enthalpy of R-134a is

P1 = 200 kPa ïü ïý h = 270.20 kJ/kg (Table A-13) ïïþ 1 T1 = 20°C The second state will be selected for a pressure of 180 kPa. At this pressure and the same enthalpy, we have

P2 = 180 kPa h2 = h1 = 270.20 kJ/kg

ü ïï ý T2 = 19.53°C (Table A-13) ïïþ

Substituting,

T - T1 (19.53 - 20)K = = 0.0235 K / kPa P2 - P1 (180 - 200)kPa

@ 2

13-62E The Joule-Thomson coefficient of refrigerant-134a at a given state is to be estimated. Analysis The Joule-Thomson coefficient is defined as

æ¶ T ö÷ ÷ çè ¶ P ø÷ h

 = çç

We use a finite difference approximation as

T - T1 (at constant enthalpy) P2 - P1

@ 2

At the given state (we call it state 1), the enthalpy of R-134a is

P1 = 30 psia ïü ïý h = 106.27 Btu/lbm (Table A-13E) T1 = 20°F ïïþ 1 The second state will be selected for a pressure of 20 psia. At this pressure and the same enthalpy, we have

P2 = 20 psia h2 = h1 = 106.27 Btu/lbm

ü ïï ý T2 = 15.65°F (Table A-13E) ïïþ

Substituting,

T - T1 (15.65 - 20)R = = 0.435 R / psia P2 - P1 (20 - 30)psia

@ 2

13-63 The Joule-Thompson coefficient of steam at two states is to be estimated. Analysis (a) The enthalpy of steam at 3 MPa and 300C is h  2994.3 kJ / kg . Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as  T   T       P  h  P  h  2994.3 kJ/kg

Considering a throttling process from 3.5 MPa to 2.5 MPa at h  2994.3 kJ / kg , the Joule-Thomson coefficient is determined to be  T3.5 MPa  T2.5 MPa  (306 .3  294 )C       12.3 C/MPa  (3.5  2 . 5 ) MPa   h  2994.3 kJ/kg (3.5  2.5) MPa

13-39

(b) The enthalpy of steam at 6 MPa and 500C is h  3423.1kJ / kg . Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as  T   T       P   h  P  h 3423.1 kJ/kg

Considering a throttling process from 7.0 MPa to 5.0 MPa at h  3423.1kJ / kg , the Joule-Thomson coefficient is determined to be  T7.0 MPa  T5.0 MPa  (504 .8  495 .1)C    4.9 C/MPa  (7.0  5.0) MPa  (7.0  5.0) MPa  h 3423.1 kJ/kg

  

EES (Engineering Equation Solver) SOLUTION "Given" P=3000 [kPa] T=300 [C] "Change the given values above to P=6000 kPa and T=500 C to get the solution for part (b)" "Analysis" Fluid$='steam_iapws' P[1]=P-1000 P[2]=P+1000 h=enthalpy(Fluid$, P=P, T=T) T[1]=temperature(Fluid$, P=P[1], h=h) T[2]=temperature(Fluid$, P=P[2], h=h) DELTAP=P[2]-P[1] DELTAT=T[2]-T[1] mu=DELTAT/DELTAP

13-64 Steam is throttled slightly from 1 MPa and 300C. It is to be determined if the temperature of the steam will increase, decrease, or remain the same during this process. Analysis The enthalpy of steam at 1 MPa and T = 300C is h  3051.6 kJ / kg . Now consider a throttling process from this state to 0.8 MPa, which is the next lowest pressure listed in the tables. The temperature of the steam at the end of this throttling process will be

ü P = 0.8 MPa ïï ý T2 = 297.5°C h = 3051.6 kJ/kg ïïþ Therefore, the temperature will decrease.

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [kPa] T1=300 [C] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) h2=h1 dP=200 [kPa] "assumed" P2=P1-dP T2=temperature(Fluid$, P=P2, h=h2)

13-40

13-65 The most general equation of state for which the Joule-Thomson coefficient is always zero is to be determined. Analysis From Eq. 13-52 of the text,

ù 1 éê æ ¶v ö T çç ÷ - vú ÷ ÷ ç ê ú m ë è¶ T øP û

cp =

When  = 0, this equation becomes

æ¶ v ÷ ö çç ÷ = v çè¶ T ø÷P T This can only be satisfied by an equation of state of the form v = f ( P) T where f(P) is an arbitrary function of the pressure.

13-66 It is to be demonstrated that the Joule-Thomson coefficient is given by m =

ö T2 æ çç¶ (v / T )÷ ÷ . ÷ ç ÷ c p èç ¶ T øP

Analysis From Eq. 13-52 of the text,

ù 1 éê æ ¶v ö T çç ÷ - vú ÷ ÷ ç ê ú m ë è¶ T øP û

cp =

Expanding the partial derivative of v / T produces

æ¶ v / T ö÷ 1 æ¶ v ö v çç ÷ = çç ÷ ÷ èç ¶ T ø÷P T èç¶ T ø÷P T 2 When this is multiplied by T 2 , the right-hand side becomes the same as the bracketed quantity above. Then, m=

ö T2 æ çç¶ (v / T )÷ ÷ ÷ ç c p èç ¶ T ÷ øP

13-67 The equation of state of a gas is given to be P(v-a) = RT. It is to be determined if it is possible to cool this gas by throttling. Analysis The equation of state of this gas can be expressed as

æ¶ v ö RT R + a ¾¾ ® çç ÷ = ÷ ÷ ç è ø P ¶T P P

Substituting into the Joule-Thomson coefficient relation,

m= -

ö ö÷ ö 1æ 1æ 1 a ççv - T æ çç ¶ v ÷ ççv - T R ÷ ÷ ==v - v + a) = <0 ( ÷ ÷ ÷ ÷ ÷ ç ç ç è¶T øP ÷ cp è cp è Pø cp cp ø

Therefore, this gas cannot be cooled by throttling since  is always a negative quantity.

13-68 The equation of state of a gas is given by v =

RT a + b . An equation for the Joule-Thomson coefficient inversion P T

line using this equation is to be derived.

13-41

Analysis From Eq. 13-52 of the text,

cp =

ù 1 éê æ ¶v ö T çç ÷ - vú ÷ ÷ ç ú m êë è¶ T øP û

When  = 0 as it does on the inversion line, this equation becomes

æ¶ v ö T çç ÷ ÷ =v çè¶ T ø÷ P Using the equation of state to evaluate the partial derivative,

æ¶ v ÷ ö çç ÷ = R + a çè¶ T ÷ øP P T 2 Substituting this result into the previous expression produces æR a ö RT a T çç + 2 ÷ ÷ ÷= P - T + b çè P T ø

Solving this for the temperature gives 2a b

as the condition along the inversion line.

The dh, du, and ds of Real Gases 13-69C It is the variation of enthalpy with pressure at a fixed temperature.

13-70C As PR approaches zero, the gas approaches ideal gas behavior. As a result, the deviation from ideal gas behavior diminishes.

13-71C So that a single chart can be used for all gases instead of a single particular gas.

13-72 The errors involved in the enthalpy and internal energy of CO2 at 350 K and 10 MPa if it is assumed to be an ideal gas are to be determined. Analysis (a) The enthalpy departure of CO2 at the specified state is determined from the generalized chart to be (Fig. A-29)

TR = and

T 350 = = 1.151 Tcr 304.2

P 10 PR = = = 1.353 Pcr 7.39

ü ïï ïï ï ý ïï ïï ïþ

¾¾ ®

Zh =

(hideal - h )T , P Ru Tcr

= 1.5

Thus, and, h = hideal - Z h RuTcr = 11,351- éë(1.5)(8.314)(304.2)ù û= 7,557 kJ/kmol

Error =

(hideal - h )T , P h

11,351- 7,557 = 50.2% 7,557

13-42

(b) At the calculated TR and PR the compressibility factor is determined from the compressibility chart to be Z = 0.65. Then using the definition of enthalpy, the internal energy is determined to be and, u = h - Pv = h - ZRuT = 7557 - éë(0.65)(8.314)(350)ù û= 5, 666 kJ/kmol

Error =

uideal - u 8, 439 - 5, 666 = = 48.9% u 5, 666

EES (Engineering Equation Solver) SOLUTION "Given" T=350 [K] P=10000 [kPa] "Properties" Fluid$='CO2' R_u=8.314 [kJ/kmol-K] T_cr=304.2 [K] "or from T_cr = T_CRIT(Fluid$)" P_cr=7390 [kPa] "P_cr = P_CRIT(Fluid$)" "Analysis" "(a)" T_R=T/T_cr P_R=P/P_cr Z_h=ENTHDEP(T_R, P_R) "the function that returns enthalpy departure factor at T_R and P_R" Z=COMPRESS(T_R, P_R) "the function that returns compressibility factor at T_R and P_R" h_o=enthalpy(Fluid$, T=0) "enthalpy at zero Kelvin" h_ideal_EES=enthalpy(Fluid$, T=T)-h_o h_ideal=11351 [kJ/kmol] "from Table A-20" h=h_ideal-Z_h*R_u*T_cr Error_h=(h_ideal-h)/h*Convert(,%) "(b)" u_o=intenergy(Fluid$, T=0) "internal energy at zero Kelvin" u_ideal_EES=intenergy(Fluid$, T=T)-u_o u_ideal=8439 [kJ/kmol] "from Table A-20" u=h-Z*R_u*T Error_u=(u_ideal-u)/u*Convert(,%)

13-73E The enthalpy of nitrogen at 400 R and 2000 psia is to be determined using data from the ideal-gas nitrogen table and the generalized enthalpy departure chart. Analysis (a) From the ideal gas table of nitrogen (Table A-18E) we read h = 2777.0 Btu/lbmol = 99.18 Btu / lbm ( M N2 = 28 lbm/lbmol)

at the specified temperature. This value involves 44.2% error.

13-43

(b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be (Fig. A-29)

T 400 ïü = = 1.761ïï ïï Tcr 227.1 (h - h )T , P ® Z h = ideal = 1.18 ý¾¾ ïï P 2000 RuTcr PR = = = 4.065ï ïï Pcr 492 þ Thus, TR =

or,

h = hideal - Z h RuTcr = 2777.0 - éë(1.18)(1.986)(227.1)ù û= 2244.8 Btu/lbmol

h 2244.8 Btu/lbmol = = 80.17 Btu / lbm M 28 lbm/lbmol

(54.9% error)

EES (Engineering Equation Solver) SOLUTION "Given" T=400 [R] P=2000 [psia] h_actual=enthalpy(Nitrogen, T=T, P=P) "Properties" Fluid$='N2' R_u=1.986 [Btu/lbmol-R] T_cr=T_CRIT(Fluid$) P_cr=P_CRIT(Fluid$) MM=molarmass(Fluid$) "Analysis" "(a)" h_ideal=enthalpy(Fluid$, T=T) Error_ideal=(h_ideal-h_actual)/h_actual*Convert(,%) "(b)" T_R=T/T_cr P_R=P/P_cr Z_h=ENTHDEP(T_R, P_R) "the function that returns enthalpy departure at T_R and P_R" h_chart=h_ideal-(Z_h*R_u*T_cr)/MM Error_chart=(h_chart-h_actual)/h_actual*Convert(,%)

13-74 The enthalpy and entropy changes of water vapor during a change of state are to be determined using the departure charts and the property tables. Properties The properties of water are (Table A-1)

M = 18.015 kg/kmol, Tcr = 647.1 K, Pcr = 22.06 MPa Analysis (a) The pressure of water vapor during this process is

P1 = P2 = Psat @ 300°C = 8588 kPa Using data from the ideal gas property table of water vapor (Table A-23), and (h2 - h1 )ideal = h2, ideal - h1, ideal = 34, 775 - 19, 426 = 15,349 kJ/kmol © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-44

( s2 - s1 )ideal = s2 - s1 - Ru ln

P2 = 231.473 - 211.263 - 0 = 20.210 kJ/kmol ×K P1

The enthalpy and entropy departures of water vapor at the specified states are determined from the generalized charts to be (Figs. A-29, A-30 or from EES)

ü ïï ïï ï ý ï P1 8.588 PR1 = = = 0.389 ïïï Pcr 22.06 ïþ

TR1 =

T1 573 = = 0.885 Tcr 647.1

¾¾ ®

Z h1 = 0.609

and

Z s1 = 0.481

and

Z s 2 = 0.105

and

T2 973 ïü = = 1.504 ïï ïï Tcr 647.1 ® Z h 2 = 0.204 ý ¾¾ ïï P2 8.588 PR 2 = = = 0.389 ïï Pcr 22.06 ïþ The enthalpy and entropy changes per mole basis are TR 2 =

h2 - h1 = (h2 - h1 )ideal - RuTcr ( Z h 2 - Z h1 )

= 15,349 - (8.314)(647.1)(0.204 - 0.609) = 17,528 kJ/kmol s2 - s1 = (s2 - s1 )ideal - Ru (Z s 2 - Z s1 ) = 20.210 - (8.314)(0.105 - 0.481) = 23.336 kJ/kmol ×K The enthalpy and entropy changes per mass basis are

h2 - h1 =

h2 - h1 17,528 kJ/kmol = = 973.0 kJ / kg M 18.015 kg/kmol

s2 - s1 =

s2 - s1 23.336 kJ/kmol ×K = = 1.2954 kJ / kg ×K M 18.015 kg/kmol

(b) The inlet and exit state properties of water vapor from Table A-6 are

T1 = 300°Cü ïï h1 = 2749.6 kJ/kg ý ïïþ s1 = 5.7059 kJ/kg ×K x1 = 1 P2 = 8588 kPa ïü ïý h2 = 3878.6 kJ/kg ïïþ s2 = 7.2465 kJ/kg ×K T2 = 700°C

(from EES)

The enthalpy and entropy changes are

h2 - h1 = 3878.6 - 2749.6 = 1129 kJ / kg s2 - s1 = 7.2465 - 5.7059 = 1.5406 kJ / kg ×K

13-75 Oxygen undergoes a process between two specified states. The enthalpy and entropy changes are to be determined by assuming ideal-gas behavior and by accounting for the deviation from ideal-gas behavior. Analysis The critical temperature and pressure of oxygen are Tcr  154.8 K and Pcr  5.08MPa (Table A–1), respectively. The oxygen remains above its critical temperature; therefore, it is in the gas phase, but its pressure is quite high. Therefore, the oxygen will deviate from ideal-gas behavior and should be treated as a real gas. (a) If the O2 is assumed to behave as an ideal gas, its enthalpy will depend on temperature only, and the enthalpy values at the initial and the final temperatures can be determined from the ideal-gas table of O2 (Table A–19) at the specified temperatures: (h2 - h1 )ideal = h2, ideal - h1, ideal = 8736 - 6404 = 2332 kJ / kmol

13-45

The entropy depends on both temperature and pressure even for ideal gases. Under the ideal-gas assumption, the entropy change of oxygen is determined from

( s2 - s1 )ideal = s2 - s1 - Ru ln

P2 10 MPa = 205.213 - 196.171 - (8.314 kJ/kmol ×K) ln = 3.28 kJ / kmol ×K P1 5 MPa

(b) The deviation from the ideal-gas behavior can be accounted for by determining the enthalpy and entropy departures from the generalized charts at each state:

ü ïï ïï ï ý ï P1 5 MPa PR1 = = = 0.98ïïï Pcr 5.08 MPa ïþ

TR1 =

T1 220 K = = 1.42 Tcr 154.8 K

¾¾ ®

Z h1 = 0.53

and

Z s1 = 0.25

and

ü ïï ïï ï ® Z h 2 = 0.48 and Z s 2 = 0.20 ý ¾¾ ï P2 10 MPa PR 2 = = = 1.97ïïï Pcr 5.08 MPa ïþ Then the enthalpy and entropy changes of oxygen during this process are determined by substituting the values above into Eqs. 12–58 and 12–63, TR 2 =

T2 300 K = = 1.94 Tcr 154.8 K

h2 - h1 = (h2 - h1 )ideal - RuTcr ( Z h 2 - Z h1 )

= 2332 kJ/kmol - (8.314 kJ/kmol ×K)(154.8 K)(0.48 - 0.53) = 2396 kJ / kmol s2 - s1 = (s2 - s1 )ideal - Ru (Z s 2 - Z s1 ) = 3.28 kJ/kmol ×K - (8.314 kJ/kmol ×K)(0.20 - 0.25) = 3.70 kJ / kmol ×K Discussion Note that the ideal-gas assumption would underestimate the enthalpy change of the oxygen by 2.7 percent and the entropy change by 11.4 percent.

13-76 Methane is compressed adiabatically by a steady-flow compressor. The required power input to the compressor is to be determined using the generalized charts. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The steady-flow energy balance equation for this compressor can be expressed as Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout WC,in + mh1 = mh2 WC,in = m (h2 - h1 )

The enthalpy departures of CH4 at the specified states are determined from the generalized charts to be (Fig. A-29)

13-46

T1 263 ïü = = 1.376ïï ïï Tcr 191.1 and ® Z h1 = 0.075 ý¾¾ ï P 0.8 PR1 = 1 = = 0.172 ïïï Pcr 4.64 ïþ TR1 =

T2 448 ïü = = 2.34ïï ïï Tcr 191.1 ® Z h 2 = 0.25 ý¾¾ ï P 6 PR 2 = 2 = = 1.29 ïïï Pcr 4.64 ïþ

TR 2 =

Thus,

h2 - h1 = RTcr (Z h1 - Z h 2 ) + (h2 - h1 )ideal = (0.5182)(191.1)(0.075 - 0.25)+ 2.2537 (175 - (- 10))

= 399.6 kJ/kg Substituting, WC,in = (0.2 kg/s)(399.6 kJ/kg ) = 79.9 kW

EES (Engineering Equation Solver) SOLUTION "Given" T1=(-10+273) [K] P1=800 [kPa] T2=(175+273) [K] P2=6000 [kPa] m_dot=0.2 [kg/s] "Properties" Fluid$='CH4' R_u=8.314 [kJ/kmol-K] T_cr=191.1 [K] P_cr=4640 [kPa] c_p=2.2537 [kJ/kg-K] R=0.5182 [kJ/kg-K] "Analysis" DELTAh_ideal=c_p*(T2-T1) T_R1=T1/T_cr P_R1=P1/P_cr T_R2=T2/T_cr P_R2=P2/P_cr © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-47

"Z_h1=ENTHDEP(T_R1, P_R1)" "the function that returns enthalpy departure factor at T_R1 and P_R1" "Z_h2=ENTHDEP(T_R2, P_R2)" "the function that returns enthalpy departure factor at T_R2 and P_R2" Z_h1=0.075 "from Fig. A-29" Z_h2=0.25 "from Fig. A-29" DELTAh=R*T_cr*(Z_h1-Z_h2)+DELTAh_ideal W_dot_C_in=m_dot*DELTAh

13-77 Carbon dioxide passes through an adiabatic nozzle. The exit velocity is to be determined using the generalized enthalpy departure chart. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The nozzle is adiabatic and thus heat transfer is negligible Properties The gas constant of CO2 is 0.1889 kJ / kg.K (Table A-1). Analysis The steady-flow energy balance equation for this nozzle can be expressed as Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout h1 + (V12 /2)¿ 0 = h2 + (V22 /2) V2 =

2(h1 - h2 )

The enthalpy departures of CO2 at the specified states are determined from the generalized enthalpy departure chart to be T1 450 ïü = = 1.48ïï ïï Tcr 304.2 and ® Z h1 = 0.55 ý¾¾ ïï P1 8 PR1 = = = 1.08 ïï Pcr 7.39 ïþ TR1 =

T2 350 ïü = = 1.15ïï ïï Tcr 304.2 ® Z h 2 = 0.20 ý¾¾ ïï P2 2 PR 2 = = = 0.27 ïï Pcr 7.39 ïþ Thus, TR 2 =

h2 - h1 = RTcr (Z h1 - Z h 2 ) + (h2 - h1 )ideal = (0.1889)(304.2)(0.55 - 0.2)+ (11,351 - 15, 483)/ 44 = - 73.8 kJ/kg

Substituting,

13-48

V2 =

æ1000 m2 /s 2 ÷ ö ÷ 2 (73.8 kJ/kg )ççç = 384 m / s ÷ è 1 kJ/kg ø÷

EES (Engineering Equation Solver) SOLUTION "Given" T1=450 [K] P1=8000 [kPa] T2=350 [K] P2=2000 [kPa] "Properties" Fluid$='CO2' R_u=8.314 [kJ/kmol-K] T_cr=T_CRIT(Fluid$) P_cr=P_CRIT(Fluid$) MM=molarmass(Fluid$) "Analysis" h1_ideal=enthalpy(CO2, T=T1) h2_ideal=enthalpy(CO2, T=T2) DELTAh_ideal=(h2_ideal-h1_ideal) T_R1=T1/T_cr P_R1=P1/P_cr T_R2=T2/T_cr P_R2=P2/P_cr Z_h1=ENTHDEP(T_R1, P_R1) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor at T_R2 and P_R2" DELTAh=R_u/MM*T_cr*(Z_h1-Z_h2)+DELTAh_ideal Vel_2=sqrt(2*(-DELTAh)*Convert(kJ/kg, m^2/s^2))

13-78 Problem 13-77 is reconsidered. the exit velocity to the nozzle assuming ideal gas behavior, the generalized chart data, and software data for carbon dioxide are to be compared. Analysis The problem is solved using EES, and the results are given below. Procedure INFO(Name$: Fluid$, T_critical, p_critical) If Name$='CarbonDioxide' then T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2' endif END T[1]=450 [K] P[1]=8000 [kPa] P[2]=2000 [kPa] T[2]=350 [K] Name$='CarbonDioxide' Call INFO(Name$: Fluid$, T_critical, P_critical) R_u=8.314 M=molarmass(Fluid$) R=R_u/M "IDEAL GAS SOLUTION" "State 1 nd 2" h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas"

13-49

h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas" V_2_ideal=SQRT(2*(h_ideal[1]-h_ideal[2])*convert(kJ/kg,m^2/s^2)) "Exit velocity" "COMPRESSIBILITY CHART SOLUTION" "State 1" Tr[1]=T[1]/T_critical Pr[1]=P[1]/P_critical DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure" h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts" "State 2" Tr[2]=T[2]/T_critical Pr[2]=P[2]/P_critical DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure" h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts" V_2_EnthDep=SQRT(2*(h[1]-h[2])*convert(kJ/kg,m^2/s^2)) "Exit velocity" "SOLUTION USING EES BUILT-IN PROPERTY DATA" "At state 1 and 2" h_ees[1]=enthalpy(Name$,T=T[1],P=P[1]) h_ees[2]=enthalpy(Name$,T=T[2],P=P[2]) V_2_ees=SQRT(2*(h_ees[1]-h_ees[2])*convert(kJ/kg,m^2/s^2)) "Exit velocity"

SOLUTION DELTAh[1]=34.19 [kJ/kg] DELTAh[2]=13.51 [kJ/kg] Fluid$='CO2' h[1]= – 8837 [kJ/kg] h[2]= – 8910 [kJ/kg] h_ees[1]=106.4 [kJ/kg] h_ees[2]=31.38 [kJ/kg] h_ideal[1]= – 8803 [kJ/kg] h_ideal[2]= – 8897 [kJ/kg] M=44.01 [kg/kmol] Name$='CarbonDioxide' P[1]=8000 [kPa] P[2]=2000 [kPa]

Pr[1]=1.083 Pr[2]=0.2706 P_critical=7390 [kPa] R=0.1889 [kJ/kg-K] R_u=8.314 [kJ/kmol-K] T[1]=450 [K] T[2]=350 [K] Tr[1]=1.479 Tr[2]=1.151 T_critical=304.2 [K] V_2_ees=387.4 [m/s] V_2_EnthDep=382.3 [m/s] V_2_ideal=433 [m/s]

13-79E Oxygen is to be adiabatically and reversibly expanded in a nozzle. The exit velocity is to be determined using the departure charts and treating the oxygen as an ideal gas with temperature variable specific heats. Properties The properties of oxygen are (Table A-1) M = 31.999 lbm/lbmol, R = 0.06206 Btu/lbm ×R, Tcr = 278.6 R, Pcr = 736 psia Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero. From the entropy change equation for an ideal gas with variable specific heats: ( s2 - s1 )ideal = 0

s2o - s1o = Ru ln

P2 70 = (1.9858) ln = - 2.085 Btu/lbmol ×R P1 200

13-50

Then from Table A-19E, T1 = 1060 R ¾ ¾ ® h1,ideal = 7543.6 Btu/lbmol, s1o = 53.921 Btu/lbmol ×R

s2o = s1o - 2.085 = 53.921- 2.085 = 51.836 Btu/lbmol ×R s2o = 51.836 Btu/lbmol ×R ¾ ¾ ® T2 = 802 R, h2,ideal = 5614.1 Btu/lbmol s

The enthalpy change per mole basis is (h2 - h1 )ideal = h2,ideal - h1,ideal = 5614.1- 7543.6 = - 1929.5 Btu/lbmol

The enthalpy change per mass basis is (h2 - h1 )ideal - 1929.5 Btu/lbmol = = - 60.30 Btu/lbm M 31.999 lbm/lbmol An energy balance on the nozzle gives (h2 - h1 )ideal =

Ein = Eout 2 1

m(h1 + V /2) = m(h2 +V22 /2)

h1 + V12 /2 = h2 +V22 /2

Solving for the exit velocity, 0.5

é æ25,037 ft 2 /s 2 ÷ öù 0.5 ú = 1738 ft / s ÷ V2 = éëêV12 + 2(h1 - h2 )ùûú = êê(0 ft/s) 2 + 2(60.30 Btu/lbm) ççç ÷ ú ÷ 1 Btu/lbm è ø ëê ûú The enthalpy departures of oxygen at the specified states are determined from the generalized charts to be (Fig. A-29 or from EES) T T 1060 802 ïü ïü TR1 = 1 = = 3.805ïï TR 2 = 2 = = 2.879ïï ï ïï Tcr 278.6 Tcr 278.6 ï Z = 0.000759 ý h1 ý Z h 2 = 0.00894 ï ï P P 200 70 PR1 = 1 = = 0.272 ïïï PR 2 = 2 = = 0.0951 ïïï Pcr 736 Pcr 736 ïþ ïþ The enthalpy change is

h2 - h1 = (h2 - h1 )ideal - RTcr (Z h 2 - Z h1 ) = - 60.30 Btu/lbm - (0.06206 Btu/lbm ×R)(278.6 R)(0.00894 - 0.000759)

= - 60.44 Btu/lbm The exit velocity is 0.5

é æ25,037 ft 2 /s 2 ÷ öù 0.5 ú = 1740 ft / s ÷ V2 = éëêV12 + 2(h1 - h2 )ùûú = êê(0 ft/s) 2 + 2(60.44 Btu/lbm) ççç ÷ ú ÷ 1 Btu/lbm è ø ëê ûú

EES (Engineering Equation Solver) SOLUTION "Given" P1=200 [psia] T1=(600+460) [R] P2=70 [psia] Vel1=0 [ft/s] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-51

T_cr=278.6 [R] P_cr=736 [psia] R=0.06206 [Btu/lbm-R] M=31.999 [lbm/lbmol] "Analysis" Fluid$='O2' h1_ideal=enthalpy(Fluid$, T=T1) s1_ideal=entropy(Fluid$, T=T1, P=P1) s2_ideal=s1_ideal "isentropic process" h2_ideal=enthalpy(Fluid$, P=P2, s=s2_ideal) T2=temperature(Fluid$, P=P2, s=s2_ideal) DELTAh_ideal=(h2_ideal-h1_ideal)/M Vel2_ideal^2=Vel1^2+2*(-DELTAh_ideal)*Convert(Btu/lbm, ft^2/s^2) T_R1=T1/T_cr P_R1=P1/P_cr T_R2=T2/T_cr P_R2=P2/P_cr Z_h1=ENTHDEP(T_R1, P_R1) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor at T_R2 and P_R2" DELTAh=DELTAh_ideal-R*T_cr*(Z_h2-Z_h1) Vel2_chart^2=Vel1^2+2*(-DELTAh_ideal)*Convert(Btu/lbm, ft^2/s^2)

13-80 A paddle-wheel placed in a well-insulated rigid tank containing oxygen is turned on. The final pressure in the tank and the paddle-wheel work done during this process are to be determined. Assumptions 1 The tank is well-insulated and thus heat transfer is negligible. 2 Kinetic and potential energy changes are negligible. Properties The gas constant of O2 is R  0.2598 kJ / kg·K (Table A-1). Analysis (a) For this problem, we use critical properties, compressibility factor, and enthalpy departure factors in EES. The compressibility factor of oxygen at the initial state is determined from EES to be

T1 175 ïü = = 1.13 ïï ïï Tcr 154.6 ® Z1 = 0.682 and Z h1 = 1.33 ý¾¾ P1 ïï 6 PR1 = = = 1.19ï ïï Pcr 5.043 þ

TR1 =

Then,

Pv = ZRT ¾ ¾ ® v1 = m=

(0.682)(0.2598 kPa ×m3 /kg ×K)(175 K) = 0.00516 m3 /kg 6000 kPa

V 0.05 m3 = = 9.68 kg v1 0.00516 m3 /kg

The specific volume of oxygen remains constant during this process, v2  v1 . Thus, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-52

ü T2 225 ïï Z = 0.853 = = 1.46 ïï 2 Tcr 154.8 3 ý Z h 2 = 1.09 v2 0.00516 m /kg ïï v R2 = = = 0.649 ïï PR 2 = 1.91 RTcr / Pcr (0.2598 kPa ×m3 /kg ×K)(154.6 K) / (5043 kPa) ïþ TR 2 =

P2 = PR 2 Pcr = (1.91)(5043) = 9652 kPa

(b) The energy balance relation for this closed system can be expressed as Ein - Eout = D Esystem

Win = D U = m(u2 - u1 ) Win = m [h2 - h1 - ( P2v 2 - P1v1 )]= m [h2 - h1 - R( Z 2T2 - Z1T1 ) ]

where

h2 - h1 = RTcr (Z h1 - Z h 2 ) + (h2 - h1 )ideal = (0.2598)(154.6)(1.33 - 1.09) + 52.96

= 62.51 kJ/kg Substituting,

Win = (9.68 kg) éëê62.51- (0.2598 kJ/kg ×K) {(0.853)(225)- (0.682)(175)}K ù ú= 423 kJ û Discussion The following routine in EES is used to get the solution above. Reading values from Fig. A-15 and A-29 together with properties in the book could yield different results.

EES (Engineering Equation Solver) SOLUTION "Given" V=0.05 [m^3] T1=175 [K] P1=6000 [kPa] T2=225 [K] "Properties" Fluid$='O2' R_u=8.314 [kJ/kmol-K] T_cr=T_CRIT(Fluid$) P_cr=P_CRIT(Fluid$) MM=molarmass(Fluid$) R=R_u/MM "Analysis" "(a)" T_R1=T1/T_cr P_R1=P1/P_cr Z_h1=ENTHDEP(T_R1, P_R1) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_1=COMPRESS(T_R1, P_R1) "the function that returns compressibility factor at T_R1 and P_R1" T_R2=T2/T_cr v_R2=(v2*P_cr)/(R*T_cr) v1=Z_1*R*T1/P1 m=V/v1 v2=v1 Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor at T_R2 and P_R2" Z_2=COMPRESS(T_R2, P_R2) "the function that returns compressibility factor at T_R2 and P_R2" P2=Z_2*R*T2/v2 P2=P_R2*P_cr "(b)" h1_ideal=enthalpy(Fluid$, T=T1) h2_ideal=enthalpy(Fluid$, T=T2) DELTAh_ideal=(h2_ideal-h1_ideal) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-53

DELTAh=R*T_cr*(Z_h1-Z_h2)+DELTAh_ideal DELTAu=DELTAh-R*(Z_2*T2-Z_1*T1) W_in=m*DELTAu Solution DELTAh=62.51 [kJ/kg] DELTAh_ideal=52.96 [kJ/kg] DELTAu=43.65 [kJ/kg] Fluid$='O2' h1_ideal=-121.8 [kJ/kg] h2_ideal=-68.8 [kJ/kg] m=9.682 [kg] MM=32 [kg/kmol] P1=6000 [kPa] P2=9652 [kPa] P_cr=5043 [kPa] P_R1=1.19 P_R2=1.914 R=0.2598 [kJ/kg-K] R_u=8.314 [kJ/kmol-K]

T1=175 [K] T2=225 [K] T_cr=154.6 [K] T_R1=1.132 T_R2=1.456 V=0.05 [m^3] v1=0.005164 [m^3/kg] v2=0.005164 [m^3/kg] v_R2=0.6485 W_in=422.6 [kJ] Z_1=0.6815 Z_2=0.8527 Z_h1=1.331 Z_h2=1.094

13-81 Propane is compressed isothermally by a piston-cylinder device. The work done and the heat transfer are to be determined using the generalized charts. Assumptions 1 The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible. Analysis (a) The enthalpy departure and the compressibility factors of propane at the initial and the final states are determined from the generalized charts to be (Figs. A-29, A-15) ü T1 373 ï = = 1.008 ïï ïï Tcr 370 and ® Z h1 = 0.28 and Z1 = 0.92 ý¾¾ ïï P1 1 PR1 = = = 0.235ïï Pcr 4.26 ïþ TR1 =

ü T2 373 ï = = 1.008 ïï ïï Tcr 370 ® Z h 2 = 1.8 and Z 2 = 0.50 ý¾¾ ï P 4 PR 2 = 2 = = 0.939ïïï Pcr 4.26 ïþ

TR 2 =

Treating propane as a real gas with Z avg   Z1  Z 2  / 2  (0.92  0.50) / 2  0.71 ,

13-54

Then the boundary work becomes 2

wb ,in = - ò P d v = - ò 1

v Z RT / P2 Z P C d v = - C ln 2 = Z avg RT ln 2 = - Z ave RT ln 2 1 v v1 Z1 RT / P1 Z1 P2

= - (0.71)(0.1885 kJ/kg ×K )(373 K )ln

(0.50)(1) (0.92)(4)

= 99.6 kJ / kg s Also, h2 - h1 = RTcr ( Z h1 - Z h 2 ) + (h2 - h1 )ideal = (0.1885)(370)(0.28 - 1.8)+ 0 = - 106 kJ/kg

u2 - u1 = (h2 - h1 ) - R(Z 2T2 - Z1T1 ) = - 106 - (0.1885)éë(0.5)(373)- (0.92)(373)ù û= - 76.5 kJ/kg Then the heat transfer for this process is determined from the closed system energy balance to be Ein - Eout = D Esystem qin + wb,in = D u = u2 - u1 qin = (u2 - u1 )- wb,in = - 76.5 - 99.6 = - 176.1 kJ/kg ® qout = 176.1 kJ / kg

EES (Engineering Equation Solver) SOLUTION Procedure IdealGasName(Name$ : Fluid$) "This procedure is only needed when the diagram window is being used for input." If Name$='Propane' then Fluid$='C3H8'; goto 10 endif If Name$='Methane' then Fluid$='CH4'; goto 10 endif If Name$='Nitrogen' then Fluid$='N2'; goto 10 endif If Name$='Oxygen' then Fluid$='O2'; goto 10 endif If Name$='CarbonDioxide' then Fluid$='CO2' ; goto 10 endif 10: end "Data from the Diagram Window. If you wish to use the Diagram Window for input then comment out the following 4 equations and bring the Diagram Window to the front." T[1]=(100+273.15) [K] p[1]=1000 [kPa] p[2]=4000 [kPa] Name$='Propane' T_critical=T_crit(Name$) P_critical=P_crit(Name$) Call IdealGasName(Name$ : Fluid$) R_u=8.314 [kJ/kmol-K] M=molarmass(Fluid$) R=R_u/M

13-55

"!****** IDEAL GAS SOLUTION ******" "State 1" h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas" s_ideal[1]=entropy(Fluid$, T=T[1], p=p[1]) "Entropy of ideal gas" u_ideal[1]=h_ideal[1]-R*T[1] "Internal energy of ideal gas" "State 2" h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas" s_ideal[2]=entropy(Fluid$, T=T[2], p=p[2]) "Entropy of ideal gas" u_ideal[2]=h_ideal[2]-R*T[2] "Internal energy of ideal gas" "Work is the integral of p dv, which can be done analytically." w_ideal=R*T[1]*Ln(p[1]/p[2]) "First Law - note that u_ideal[2] is equal to u_ideal[1]" q_ideal-w_ideal=u_ideal[2]-u_ideal[1] "Entropy change" DELTAs_ideal=s_ideal[2]-s_ideal[1] "!***** COMPRESSABILITY CHART SOLUTION ******" "State 1" Tr[1]=T[1]/T_critical pr[1]=p[1]/p_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure" h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts" u[1]=h[1]-Z[1]*R*T[1] "Internal energy of gas using charts" DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" s[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts" "State 2" T[2]=T[1] Tr[2]=Tr[1] pr[2]=p[2]/p_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure" DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts" s[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts" u[2]=h[2]-Z[2]*R*T[2] "Internal energy of gas using charts" "Work using charts - note use of EES integral function to evaluate the integral of p dv." w_chart=Integral(p,v,v[1],v[2]) "We need an equation to relate p and v in the above INTEGRAL function. " p*v=COMPRESS(Tr[2],p/p_critical)*R*T[1] "To specify relationship between p and v" "Find the limits of integration" p[1]*v[1]=Z[1]*R*T[1] "to get v[1], the lower bound" p[2]*v[2]=Z[2]*R*T[2] "to get v[2], the upper bound" "First Law - note that u[2] is not equal to u[1]" q_chart-w_chart=u[2]-u[1] "Entropy Change" DELTAs_chart=s[2]-s[1] "!***** SOLUTION USING EES BUILT-IN PROPERTY DATA *****" "At state 1" u_ees[1]=intEnergy(Name$,T=T[1],p=p[1]) s_ees[1]=entropy(Name$,T=T[1],p=p[1]) "At state 2" u_ees[2]=IntEnergy(Name$,T=T[2],p=p[2]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-56

s_ees[2]=entropy(Name$,T=T[2],p=p[2]) "Work using EES built-in properties- note use of EES Integral funcion to evaluate the integral of pdv." w_ees=integral(p_ees, v_ees, v_ees[1],v_ees[2], .002) "The following equation relates p and v in the above INTEGRAL" p_ees=pressure(Name$,T=T[1], v=v_ees) "To specify relationship between p and v" "Find the limits of integration" v_ees[1]=volume(Name$, T=T[1],p=p[1]) "to get lower bound" v_ees[2]=volume(Name$, T=T[2],p=p[2]) "to get upper bound" "First law - note that u_ees[2] is not equal to u_ees[1]" q_ees-w_ees=u_ees[2]-u_ees[1] "Entropy change" DELTAs_ees=s_ees[2]-s_ees[1] "Note: In all three solutions to this problem we could have calculated the heat transfer by q/T=DELTA_s since T is constant. Then the first law could have been used to find the work. The use of integral of p dv to find the work is a more fundamental approach and can be used if T is not constant."

13-82 Problem 13-81 is reconsidered. This problem is to be extended to compare the solutions based on the ideal gas assumption, generalized chart data and real fluid data. Also, the solution is to be extended to carbon dioxide, nitrogen and methane. Analysis The problem is solved using EES, and the solution is given below. Procedure INFO(Name$, T[1] : Fluid$, T_critical, p_critical) If Name$='Propane' then T_critical=370 ; p_critical=4620 ; Fluid$='C3H8'; goto 10 endif If Name$='Methane' then T_critical=191.1 ; p_critical=4640 ; Fluid$='CH4'; goto 10 endif If Name$='Nitrogen' then T_critical=126.2 ; p_critical=3390 ; Fluid$='N2'; goto 10 endif If Name$='Oxygen' then T_critical=154.8 ; p_critical=5080 ; Fluid$='O2'; goto 10 endif If Name$='CarbonDioxide' then T_critical=304.2 ; p_critical=7390 ; Fluid$='CO2' ; goto 10 endif If Name$='n-Butane' then T_critical=425.2 ; p_critical=3800 ; Fluid$='C4H10' ; goto 10 endif 10: If T[1]<=T_critical then CALL ERROR('The supplied temperature must be greater than the critical temperature for the fluid. A value of XXXF1 K was supplied',T[1]) endif end T[1]=(100+273.15) [K]

13-57

p[1]=1000 [kPa] p[2]=4000 [kPa] Name$='Propane' Fluid$='C3H8' Call INFO(Name$, T[1] : Fluid$, T_critical, p_critical) R_u=8.314 M=molarmass(Fluid$) R=R_u/M "IDEAL GAS SOLUTION" "State 1" h_ideal[1]=enthalpy(Fluid$, T=T[1]) "Enthalpy of ideal gas" s_ideal[1]=entropy(Fluid$, T=T[1], p=p[1]) "Entropy of ideal gas" u_ideal[1]=h_ideal[1]-R*T[1] "Internal energy of ideal gas" "State 2" h_ideal[2]=enthalpy(Fluid$, T=T[2]) "Enthalpy of ideal gas" s_ideal[2]=entropy(Fluid$, T=T[2], p=p[2]) "Entropy of ideal gas" u_ideal[2]=h_ideal[2]-R*T[2] "Internal energy of ideal gas" w_ideal=R*T[1]*Ln(p[1]/p[2]) "Work is the integral of p dv, which can be done analytically." q_ideal-w_ideal=u_ideal[2]-u_ideal[1] "First Law - note that u_ideal[2] is equal to u_ideal[1]" DELTAs_ideal=s_ideal[2]-s_ideal[1] "Entropy change" "COMPRESSABILITY CHART SOLUTION" "State 1" Tr[1]=T[1]/T_critical pr[1]=p[1]/p_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure" h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts" u[1]=h[1]-Z[1]*R*T[1] "Internal energy of gas using charts" DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" s[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts" "State 2" T[2]=T[1] Tr[2]=Tr[1] pr[2]=p[2]/p_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure" DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts" s[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts" u[2]=h[2]-Z[2]*R*T[2] "Internal energy of gas using charts" "Work using charts - note use of EES integral function to evaluate the integral of p dv." w_chart=Integral(p,v,v[1],v[2]) "We need an equation to relate p and v in the above INTEGRAL function." p*v=COMPRESS(Tr[2],p/p_critical)*R*T[1] "To specify relationship between p and v" "Find the limits of integration" p[1]*v[1]=Z[1]*R*T[1] "to get v[1], the lower bound" p[2]*v[2]=Z[2]*R*T[2] "to get v[2], the upper bound" q_chart-w_chart=u[2]-u[1] "First Law - note that u[2] is not equal to u[1]" DELTAs_chart=s[2]-s[1] "Entropy Change" "SOLUTION USING EES BUILT-IN PROPERTY DATA" "At state 1" u_ees[1]=intEnergy(Name$,T=T[1],p=p[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-58

s_ees[1]=entropy(Name$,T=T[1],p=p[1]) "At state 2" u_ees[2]=IntEnergy(Name$,T=T[2],p=p[2]) s_ees[2]=entropy(Name$,T=T[2],p=p[2]) "Work using EES built-in properties- note use of EES Integral funcion to evaluate the integral of pdv." w_ees=integral(p_ees, v_ees, v_ees[1],v_ees[2]) "The following equation relates p and v in the above INTEGRAL" p_ees=pressure(Name$,T=T[1], v=v_ees) "To specify relationship between p and v" "Find the limits of integration" v_ees[1]=volume(Name$, T=T[1],p=p[1]) "to get lower bound" v_ees[2]=volume(Name$, T=T[2],p=p[2]) "to get upper bound" q_ees-w_ees=u_ees[2]-u_ees[1] "First law - note that u_ees[2] is not equal to u_ees[1]" DELTAs_ees=s_ees[2]-s_ees[1] "Entropy change" "Note: In all three solutions to this problem we could have calculated the heat transfer by q/T=DELTA_s since T is constant. Then the first law could have been used to find the work. The use of integral of p dv to find the work is a more fundamental approach and can be used if T is not constant." SOLUTION DELTAh[1]=16.48 [kJ/kg] DELTAh[2]=91.96 [kJ/kg] DELTAs[1]=0.03029 [kJ/kg-K] DELTAs[2]=0.1851 [kJ/kg-K] DELTAs_chart=−0.4162 [kJ/kg-K] DELTAs_ees=-0.4711 [kJ/kg-K] DELTAs_ideal=−0.2614 [kJ/kg-K] Fluid$='C3H8' h[1]=−2232 [kJ/kg] h[2]= −2308 [kJ/kg] h_ideal[1]= −2216 [kJ/kg] h_ideal[2]= −2216 [kJ/kg] M=44.1 Name$='Propane' p=4000 p[1]=1000 [kPa] p[2]=4000 [kPa] pr[1]=0.2165 pr[2]=0.8658 p_critical=4620 [kPa] p_ees=4000 q_chart=-155.3 [kJ/kg] q_ees=-175.8 [kJ/kg] q_ideal=-97.54 [kJ/kg] R=0.1885 [kJ/kg-K] R_u=8.314 [kJ/mole-K] s[1]=6.073 [kJ/kg-K]

s[2]=5.657 [kJ/kg-K] s_ees[1]=2.797 [kJ/kg-K] s_ees[2]=2.326 [kJ/kg-K] s_ideal[1]=6.103 [kJ/kg-K] s_ideal[2]=5.842 [kJ/kg-K] T[1]=373.2 [K] T[2]=373.2 [K] Tr[1]=1.009 Tr[2]=1.009 T_critical=370 [K] u[1]=-2298 [kJ/kg] u[2]=-2351 [kJ/kg] u_ees[1]=688.4 [kJ/kg] u_ees[2]=617.1 [kJ/kg] u_ideal[1]=-2286 [kJ/kg] u_ideal[2]=-2286 [kJ/kg] v=0.01074 v[1]=0.06506 [m^3/kg] v[2]=0.01074 [m^3/kg] v_ees=0.009426 v_ees[1]=0.0646 [m^3/kg] v_ees[2]=0.009426 [m^3/kg] w_chart=-101.9 [kJ/kg] w_ees=-104.5 [kJ/kg] w_ideal=-97.54 [kJ/kg] Z[1]=0.9246 Z[2]=0.6104

13-83 Propane is compressed isothermally by a piston-cylinder device. The exergy destruction associated with this process is to be determined. Assumptions 1 The compression process is quasi-equilibrium. 2 Kinetic and potential energy changes are negligible. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-59

Properties The gas constant of propane is R  0.1885 kJ / kg·K (Table A-1). Analysis The exergy destruction is determined from its definition xdestroyed = T0 sgen where the entropy generation is determined from an entropy balance on the contents of the cylinder. It gives Sin - Sout + Sgen = D Ssystem

Qout q + Sgen = m( s2 - s1 ) ® sgen = ( s2 - s1 ) + out Tb,surr Tsurr

where D ssys = s2 - s1 = R(Z s1 - Z s 2 ) + (s2 - s1 )ideal

T ( s2 - s1 )ideal = c p ln 2 T1

Z 0

- R ln

P2 4 = 0 - 0.1885ln = - 0.261 kJ/kg ×K P1 1

ü T1 373 ï = = 1.008 ïï ïï Tcr 370 and ® Z s1 = 0.21 ý¾¾ ï P 1 PR1 = 1 = = 0.235ïïï Pcr 4.26 ïþ TR1 =

ü T2 373 ï = = 1.008 ïï ïï Tcr 370 ® Z s 2 = 1.5 ý¾¾ ïï P2 4 PR 2 = = = 0.939ïï Pcr 4.26 ïþ

TR 2 =

Thus, D ssys = s2 - s1 = R( Z s1 - Z s 2 ) + ( s2 - s1 )ideal = (0.1885)(0.21- 1.5)- 0.261 = - 0.504 kJ/kg ×K

and

æ q ö ÷ xdestroyed = T0 sgen = T0 ççç( s2 - s1 ) + out ÷ ÷ çè Tsurr ÷ ø

æ ö 176.1 kJ/kg ÷ = (303 K )çç- 0.504 + ÷kJ/kg ×K çè 298 K ø÷

= 26.3 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T1=(100+273) [K] T2=T1 P1=1000 [kPa] P2=4000 [kPa] T0=(30+273) [K] "Properties" Fluid$='C3H8' R_u=8.314 [kJ/kmol-K] T_cr=T_CRIT(Fluid$) P_cr=P_CRIT(Fluid$) MM=molarmass(Fluid$) "Analysis" h1_ideal=enthalpy(Fluid$, T=T1) h2_ideal=enthalpy(Fluid$, T=T2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-60

DELTAh_ideal=h2_ideal-h1_ideal T_R1=T1/T_cr P_R1=P1/P_cr T_R2=T2/T_cr P_R2=P2/P_cr Z_h1=ENTHDEP(T_R1, P_R1) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_1=COMPRESS(T_R1, P_R1) "the function that returns compressibility factor at T_R1 and P_R1" Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor at T_R2 and P_R2" Z_2=COMPRESS(T_R2, P_R2) "the function that returns compressibility factor at T_R2 and P_R2" Z_ave=(Z_1+Z_2)/2 w_b_in=-Integral(P, v, v1, v2) "Ees integral function to evaluate boundary work" P*v=Z_ave*R_u/MM*T1 "The relationship between P and v for the above integral function" P1*v1=Z_1*R_u/MM*T1 "lower limit v1 in the above integral function" P2*v2=Z_2*R_u/MM*T2 "upper limit v2 in the above integral function" DELTAh=(R_u*T_cr)/MM*(Z_h1-Z_h2)+DELTAh_ideal DELTAu=DELTAh-R_u/MM*(Z_2*T2-Z_1*T1) q_in=DELTAu-w_b_in s1_ideal=entropy(Fluid$, T=T1, P=P1) s2_ideal=entropy(Fluid$, T=T2, P=P2) DELTAs_ideal=s2_ideal-s1_ideal Z_s1=ENTRDEP(T_R1, P_R1) "the function that returns entropy departure factor at T_R1 and P_R1" Z_s2=ENTRDEP(T_R2, P_R2) "the function that returns entropy departure factor at T_R2 and P_R2" DELTAs=R_u/MM*(Z_s1-Z_s2)+DELTAs_ideal s_gen=DELTAs-q_in/T0 x_destroyed=T0*s_gen

Review Problems æ b /T ö ÷ 13-84 Expressions for h, u s o , Pr , and vr for an ideal gas whose cp o is given by c op = å aiT i- n + a0 eb / T çç b / T ÷ ÷ are çèe - 1ø to be developed. Analysis The enthalpy of this substance relative to a reference state is given by T

h = ò c p dT = å Tref

é æ b ö÷ù æ bö a1 b /T çç1÷ú T i- n+ 1 - Trefi- n+ 1 )+ a0 êêeb / T - e ref - eE1 çç1- ÷ eE ÷ ( 1 çè T ø÷ ÷úú ççè Tref ø÷ i- n+ 1 êë û

where E1 ( x) is the exponential integral function of order 1. Similarly, s 0 is given by T

so = ò Tref

cp T

dT = å

é æ b ÷ öù æ bö a1 ú ÷ T i- n - Trefi- n )- a0 êêeb / T - eb / Tref - eE1 çç1- ÷ - eE1 ççç1÷ ( ÷ ÷ èç T ø i- n èç Tref ÷ øúûú ëê

With these two results,

u = h - Pv o

Pr = e s / R

According to the du form of Gibbs equations, du dv =- R T v

Noting that for ideal gases, cv = c p - R and du = cv dT , this expression reduces to dT dv =- R T v When this is integrated between the reference and actual states, the result is (c p - R )

13-61

ò c T - R ln T p

= - R ln

ref

v v ref

Solving this for the specific volume ratio gives

v =v ref

æ so ö ççe - e R T ÷ ÷ ÷ çè T ÷ ø ref

e + eR

T o Tref es T T = + = - exp( s o - R) + Tref eR e R Tref

The ratio of the specific volumes at two states which have the same entropy is then

v 2 v r,2 T = = - exp( s2o - s1o - R) + 2 v1 v r,1 T1 Inspection of this result gives

v r = - exp( s o - R) + T

13-85 It is to be shown that the slope of a constant-pressure line on an h-s diagram is constant in the saturation region and increases with temperature in the superheated region. Analysis For P = constant, dP = 0 and the given relation reduces to dh = Tds, which can also be expressed as

æ¶ h ö÷ çç ÷ = T çè ¶ s ø÷ P

Thus the slope of the P = constant lines on an h-s diagram is equal to the temperature. (a) In the saturation region, T = constant for P = constant lines, and the slope remains constant. (b) In the superheat region, the slope increases with increasing temperature since the slope is equal temperature.

13-86 Using the cyclic relation and the first Maxwell relation, the other three Maxwell relations are to be obtained. Analysis (1) Using the properties P, s, v, the cyclic relation can be expressed as

æ¶ P ö ÷ ççæ¶ s ö ÷ ççæ¶ v ö ÷ =- 1 çç ÷ ÷ ÷ ÷ ÷ çè¶ P ø ÷ èç ¶ s ø çè¶ v ø v

æ¶ T ö æ¶ P ö Substituting the first Maxwell relation, çç ÷ ÷ = - ççç ÷ ÷ èç ¶ v ÷ øs è¶s ÷ øv æ¶T ö æ¶ s ÷ ö æ¶ v ö ç çç ÷ - çç ÷ ÷ ÷ ÷ èçç¶v ÷ ÷ ÷=- 1 çè ¶v ø øP èç¶ P ø s s

¾¾ ®

æ¶T ö æ¶ s ÷ ö çç ÷ ç =1 ÷ ÷ èçç¶ v ÷ ÷ çè¶ P ø øP s

¾¾ ®

æ¶ T ÷ ö æ ö çç ÷ = çç¶ v ÷ ÷ çè¶ P ÷ øs èç ¶ s ÷ øP

13-62

(2) Using the properties T, v, s, the cyclic relation can be expressed as

æ¶T ö÷ æ¶ v ö÷ æ¶ s ö÷ çç ÷ çç ÷ çç ÷ = - 1 èç ¶v ø÷s èç ¶ s ø÷T èç¶ T ø÷v æ¶ T ö æ¶ P ö = - çç ÷ Substituting the first Maxwell relation, çç ÷ ÷ ÷ çè ¶ v ÷ øs èç ¶ s ÷ øv æ¶ P ö æ¶ v ö÷ æ¶ s ö÷ ç ç - çç ÷ ÷ èçç ¶ s ø÷ ÷ èçç¶T ø÷ ÷ =- 1 çè ¶ s ø÷ v T v

¾¾ ®

æ¶ P ö÷ æ¶ v ö÷ çç ÷ çç ÷ = 1 èç ¶T ø÷v èç ¶ s ø÷T

¾¾ ®

æ¶ s ö÷ æ¶ P ö÷ çç ÷ = çç ÷ èç¶ v ø÷T èç ¶T ø÷v

(3) Using the properties P, T, v, the cyclic relation can be expressed as

æ¶ P ö æ¶T ö÷ æ¶ v ö÷ çç ÷ ç ç ÷ ÷ èçç ¶ v ø÷ ÷ èçç¶ P ø÷ ÷ =- 1 çè ¶ T ø v P T æ¶ s ö ö÷ ÷ =æ çç¶ P ÷ Substituting the third Maxwell relation, çç ÷ ÷ èç¶ v øT èç ¶ T ø÷v æ¶ s ö æ¶ T ö æ¶ v ö÷ çç ÷ ç ÷ ç ÷ ÷ ÷ èçç ¶v ø ÷ èçç¶ P ø÷ ÷ =- 1 èç¶v ø T

æ¶ s ö÷ æ¶ T ö÷ çç ÷ çç ÷ = - 1 èç¶ P ø÷ èç ¶ v ø÷

¾¾ ®

æ¶ s ö÷ æ ö çç ÷ = - çç ¶ v ÷ ÷ èç¶ P ø÷ èç¶ T ø÷ T

æ¶ u ö÷ æ¶ u ö ÷ that for ideal 13-87 It is to be proven by using the definitions of pressure and temperature, T = çç ÷ and P = - çç ÷ ÷ çè ¶ s øv çè¶ v ÷ øs æ¶ u ö gases, the development of the constant-volume specific heat yields çç ÷ ÷ = 0. çè¶ v ø÷ T Analysis Assume u = u (s, v ) Then,

æ¶ u ö ö÷ ÷ ds + æ çç ¶ u ÷ du = çç ÷ dv ÷ èç ¶ s øv èç¶ v ø÷s

æ¶ u ö÷ æ¶ u ö÷ æ¶ s ö÷ æ¶ u ö÷ æ¶ v ö÷ çç ÷ = çç ÷ çç ÷ + çç ÷ çç ÷ èç¶ v ø÷ èç ¶ s ø÷ èç¶ v ø÷ èç¶ v ø÷ èç¶ v ø÷ v

æ¶ s ö = T çç ÷ +P ÷ çè¶ v ÷ øT From Maxwell equation,

æ¶ s ö æ ö ç¶ P ÷ T çç ÷ ÷ ÷ ÷ + P = T ççè¶ T ø ÷- P èç¶ v ø T v For ideal gases

RT v

æ¶ P ö R and çç ÷ ÷ ÷ = v çè¶ T ø v

Then,

æ¶ u ö R çç ÷ ÷ = T - P = P- P = 0 ÷ çè¶ v ø v T

13-88 It is to be shown that

13-63

æ¶ P ö÷ æ¶ v ö÷ æ¶ v ö æ¶ P ö÷ ç ç ç cv = - T çç ÷ ÷ èçç ¶ T ø÷ ÷ and c p = T ççè ¶ T ø÷ ÷ èçç¶ T ø÷ ÷ çè¶ T ø÷ s P s v Analysis Using the definition of cv ,

æ¶ s ö æ¶ s ö÷ æ¶ P ö÷ ç ç cv = T çç ÷ ÷ = T èçç¶ P ø÷ ÷ èçç ¶ T ø÷ ÷ çè¶ T ø÷ v v v æ¶ s ö æ¶ v ö ç ÷ Substituting the first Maxwell relation çç ÷ ÷ ÷ ÷ = - èçç¶ T ø ÷, çè¶ P ø v s

æ¶ v ö÷ æ¶ P ö÷ çç ÷ cv = - T çç ÷ èç¶ T ø÷s èç ¶ T ø÷v Using the definition of c p ,

æ¶ s ö æ¶ s ö÷ æ¶ v ö÷ ç ç c p = T çç ÷ ÷ ÷ = T èçç¶ v ø÷ ÷ èçç¶ T ø÷ ÷ èç¶ T ø P P P æ¶ s ö æ¶ P ö ç ÷ Substituting the second Maxwell relation çç ÷ ÷ ÷ ÷ = èçç ¶ T ø ÷, çè¶ v ø P s æ¶ P ö æ¶ v ö÷ ç c p = T çç ÷ ÷ èçç¶ T ø÷ ÷ çè ¶ T ø÷ s P

æ¶ s ö P = . 13-89 It is to be proven that for a simple compressible substance çç ÷ ÷ çè¶ v ÷ øu T Analysis The proof is simply obtained as æ¶ u ö - çç ÷ ÷ ç ÷ æ¶ s ö÷ çç ÷ = è¶ v øs = - - P = P ÷ æ¶ u ö÷ èç¶ v øu T T çç ÷ çè ¶ s ÷ øv

dv = b dT - a dP . Also, a relation is to be obtained for the ratio of specific volumes v2 / v1 as a v homogeneous system undergoes a process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is

13-90 It is to be shown that

æ¶ v ö æ¶ v ö d v = çç ÷ ÷ dT + ççç ÷ ÷ dP ÷ ÷ èç¶ T ø è¶ P ø P T Dividing by v, dv 1 æ ¶v ö 1 æ¶ v ö = çç ÷ dT + çç ÷ ÷ ÷ dP ÷ v v èç¶ T øP v çè¶ P ø÷T Using the definitions of  and , dv = b dT - a dP v

Taking  and  to be constants, integration from 1 to 2 yields

v2 = b (T2 - T1 )- a (P2 - P1 ) v1

13-64

which is the desired relation.

dv = b dT - a dP. Also, a relation is to be obtained for the ratio of specific volumes v2 / v1 as a v homogeneous system undergoes an isobaric process from state 1 to state 2. Analysis We take v = v (P, T). Its total differential is

13-91 It is to be shown that

æ¶ v ö ö ÷ dT + æ ÷ dP çç ¶ v ÷ d v = çç ÷ ÷ ÷ èç¶ T øP èç¶ P ø T which, for a constant pressure process, reduces to

æ¶ v ö d v = çç ÷ dT ÷ çè¶ T ÷ øP Dividing by v,

dv 1 æ ¶v ö = çç ÷ ÷ dT ç v v è¶ T ø÷P Using the definition of , dv = b dT v

Taking  to be a constant, integration from 1 to 2 yields or

v2 = b (T2 - T1 ) v1

v2 = exp éëb (T2 - T1 )ù û v1 which is the desired relation.

13-92 It is to be shown that the position of the Joule-Thompson coefficient inversion curve on the T-P plane is given by (Z / T ) P  0 . Analysis The inversion curve is the locus of the points at which the Joule-Thompson coefficient  is zero,

ö ö 1æ ççT æ çç ¶ v ÷ ÷ - v÷ = 0 ÷ ÷ ÷ ç ç ÷ c p è è¶ T øP ø

which can also be written as æ¶ v ö ZRT T çç ÷ ÷ ÷ - P = 0 çè¶ T ø P

(a)

since it is given that ZRT (b) P Taking the derivative of (b) with respect to T holding P constant gives æ¶ (ZRT / P )ö ö æ¶ v ö R æ æ¶ Z ö ÷ ç çç ÷ ÷ ÷ = çççT çç ÷ + Z÷ ÷ ÷ ÷ ÷ ÷ = ççèç ÷ çè¶ T ø ç ÷ ÷ ¶T P è è¶ T øP ø ø P v=

Substituting in (a),

13-65

æ¶ Z ö T çç ÷ + Z- Z = 0 ÷ çè ¶ T ÷ øP æ¶ Z ö÷ çç ÷ = 0 çè ¶ T ø÷ P which is the desired relation.

13-93 It is to be shown that for an isentropic expansion or compression process Pv k  constant . It is also to be shown that the isentropic expansion exponent k reduces to the specific heat ratio c p / cv for an ideal gas. Analysis We note that ds = 0 for an isentropic process. Taking s = s(P, v), the total differential ds can be expressed as æ¶ s ö æ¶ s ö÷ ç ds = çç ÷ (a) ÷ dP + èçç¶ v ø÷ ÷ dv = 0 çè¶ P ø÷ v P We now substitute the Maxwell relations below into (a) æ¶ s ö æ¶ v ö÷ æ¶ s ö÷ æ¶ P ö÷ ç ç ç to get çç ÷ ÷ = - èçç¶ T ø÷ ÷ and èçç¶ v ø÷ ÷ = èçç ¶ T ø÷ ÷ çè¶ P ø÷ v s P s

æ¶v ö æ¶ P ö÷ ç - çç ÷ ÷ ÷ dP + èçç ¶T ø÷ ÷ dv = 0 çè¶T ø s s Rearranging, æ¶ T ö æ¶ P ö÷ æ¶ P ö ç dP - çç ÷ ® dP - çç ÷ ÷ ÷ dv = 0 ÷ èçç ¶ T ø÷ ÷ dv = 0 ¾ ¾ èç ¶ v ø èç ¶ v ø÷ s

Dividing by P, dP 1 æ ¶Pö - çç ÷ ÷ dv = 0 ç øs P P è ¶v ÷

(b)

We now define isentropic expansion exponent k as v æ¶ P ö k = - çç ÷ ÷ ÷ P çè ¶ v ø s Substituting in (b), dP dv +k = 0 P v Taking k to be a constant and integrating,

Thus,

ln P + k ln v = constant ¾ ¾ ® ln Pv k = constant

Pv k = constant To show that k  c p / cv for an ideal gas, we write the cyclic relations for the following two groups of variables:

æ¶ s ÷ ö æ¶ v ö æ¶ T ÷ ö c æ¶ v ö æ ¶T ö = - 1¾¾ ® v çç ÷ ÷ ççç ÷ ÷ ççç ÷ ÷ ççç ÷ ÷ = - 1 (c) è¶ T ÷ øv è ¶ s ÷ øT è ¶ v ÷ øs øT è ¶ v ÷ øs T èç ¶ s ÷

(s, T , v ) ¾ ¾® ççç

c p æ¶ P ö æ¶T ö æ¶ s ö æ¶ P ö æ¶T ÷ ö ÷ çç ÷ çç ÷ = - 1 ¾ ¾ çç ÷ ® ÷ ÷ ÷ ççç ÷ ÷ = - 1 (d ) ÷ ÷ ÷ ç ç è¶T øP è ¶ s øT è¶ P øs øT è¶ P ÷ øs T çè ¶ s ÷

(s, T , P) ¾ ¾® ççç

where we used the relations æ¶ s ö æ¶ s ö cv = T çç ÷ and c p = T çç ÷ ÷ ÷ çè¶T ÷ øv èç¶T ø÷P Setting Eqs. (c) and (d) equal to each other, c p æ¶ P ö æ¶T ö ö c æ¶v ö æ çç ÷ çç ÷ çç¶T ÷ = v çç ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ç ç ç ç T è ¶ s ø è¶ P ø T è ¶ s ø è ¶v ø T

13-66

or,

æ¶ s ö æ¶ P ö æ¶v ö æ¶T ö æ¶ s ¶v ö æ¶ P ¶T ö æ¶v ö æ¶ P ö÷ ÷ ÷ çç ÷ çç ÷ çç ÷ çç = çç ÷ = çç = çç ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ çç ÷ ÷ ÷ ÷ ÷ ÷ ÷ cv èç¶ P øT èç¶T øs èç ¶ s øT èç ¶v øs èç¶ P ¶ s øT èç¶T ¶v øs èç¶ P ø÷T èç ¶v ø÷s

æ¶ (RT / P )ö æ¶ v ö v ÷ ç ÷ but çç ÷ = =÷ ÷ ÷ çççè ¶ P ÷ èç¶ P ø P øT T

Substituting, cp v æ¶ P ö = - çç ÷ ÷= k øs cv P çè ¶v ÷ which is the desired relation.

13-94 The c p of nitrogen at 300 kPa and 400 K is to be estimated using the relation given and its definition, and the results are to be compared to the value listed in Table A-2b. Analysis (a) We treat nitrogen as an ideal gas with R  0.297 kJ / kg·K and k = 1.397. Note that PT  k / ( k 1)  C = constant for the isentropic processes of ideal gases. The c p relation is given as

æ¶ P ö÷ æ¶ v ö÷ çç ÷ c p = T çç ÷ èç ¶ T ø÷s èç¶ T ø÷P v=

æ¶ v ö RT R ¾¾ ® çç ÷ ÷ ÷ = P çè¶ T ø P P

æ¶ P ö k k kP P = CT k / ( k - 1) ¾ ¾ ® çç ÷ = CT k / ( k - 1)- 1 = PT - k / ( k - 1) ) T k / ( k - 1)- 1 = ( ÷ çè ¶ T ÷ øs k - 1 k- 1 T (k - 1) Substituting,

æ kP ÷ öæR ö kR 1.397(0.297 kJ/kg ×K) çç ÷ ÷ c p = T ççç = = = 1.045 kJ/kg ×K ÷ ÷ ÷ ç ÷ 1.397 - 1 èT (k - 1) øè P ø k - 1

æ¶ h ö (b) The c p is defined as c p = çç ÷ ÷ .Replacing the differentials by differences, çè¶ T ø÷ P h (410 K )- h (390 K ) (11,932 - 11,347)/28.0 kJ/kg æD h ö ÷ c p @çç = = = 1.045 kJ / kg ×K ÷ çèD T ÷ øP= 300 kPa (410 - 390)K (410 - 390)K

(Compare: Table A-2b at 400 K  c p  1.044 kJ / kg·K )

EES (Engineering Equation Solver) SOLUTION "Given" T=400 [K] P=300 [kPa] "Properties, Table A-2b" R=0.297 [kJ/kg-K] k=1.397 Fluid$='N2' "Analysis" "(a)" c_p_a=k*R/(k-1) "(b)" dT=10 [K] T[1]=T-dT T[2]=T+dT

13-67

h_T[1]=enthalpy(Fluid$, T=T[1]) h_T[2]=enthalpy(Fluid$, T=T[2]) c_p_b=(h_T[2]-h_T[1])/(T[2]-T[1])

13-95 The volume expansivity of water is given. The change in volume of water when it is heated at constant pressure is to be determined. Properties The volume expansivity of water is given to be 0.207 106 K 1 at 20C. Analysis We take v = v (P, T). Its total differential is

æ¶ v ö æ¶ v ö ç ÷ d v = çç ÷ ÷ ÷ ÷ dT + èçç¶ P ø ÷ dP çè¶ T ø P T which, for a constant pressure process, reduces to

æ¶ v ö d v = çç ÷ dT ÷ çè¶ T ÷ øP Dividing by v and using the definition of ,

dv 1 æ ¶v ö = çç ÷ ÷ ÷ dT = b dT ç v v è¶ T ø P Taking  to be a constant, integration from 1 to 2 yields or

v2 = b (T2 - T1 ) v1

v2 = exp éëb (T2 - T1 )ù û v1 Substituting the given values and noting that for a fixed mass V2 / V1  v2 / v1 , 3 - 6 - 1 é ù= 0.50000207 m 3 V2 = V1 exp éëb (T2 - T1 )ù ú û= (0.5 m )exp ëê(0.207 ´ 10 K )(30 - 10)°Cû

Therefore,

D V =V2 - V1 = 0.50000207 - 0.5 = 0.00000207 m3 = 2.07 cm 3

EES (Engineering Equation Solver) SOLUTION "Given" beta=0.207E-6 [1/K] "at 20 C" V1=0.5 [m^3] T1=10 [C] T2=30 [C] "Analysis" "Using the equations dv=(deltav/deltaT)_P*dT and using the definition of beta, following equation is obtained" V2=V1*exp(beta*(T2-T1)) DELTAV=(V2-V1)*Convert(m^3, cm^3)

13-96 The temperature change of steam and the average Joule-Thompson coefficient during a throttling process are to be estimated. Analysis The enthalpy of steam at 4.5 MPa and T = 300C is h  2944.2 kJ / kg . Now consider a throttling process from this state to 2.5 MPa. The temperature of the steam at the end of this throttling process will be

13-68

ü P = 2.5 MPa ïï ýT2 = 273.7°C h = 2944.2 kJ/kgïïþ Thus the temperature drop during this throttling process is

D T = T2 - T1 = 273.7 - 300 = - 26.3°C The average Joule-Thomson coefficient for this process is determined from æ¶T ö æD T ö (273.7 - 300)°C ÷ m = çç ÷ = = 13.1°C / MPa ÷ @ççç ÷ ÷ ÷ èç¶ P ø è ø D P (2.5 - 4.5) MPa h h= 2944.2 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=4500 [kPa] T1=300 [C] P2=2500 [kPa] "Analysis" Fluid$='steam_iapws' h1=enthalpy(Fluid$, P=P1, T=T1) h2=h1 T2=temperature(Fluid$, P=P2, h=h2) DELTAT=T2-T1 mu=(T2-T1)/(P2-P1)

13-97E Argon gas enters a turbine at a specified state and leaves at another specified state. The power output of the turbine and the exergy destruction associated with the process are to be determined using the generalized charts. Properties The gas constant and critical properties of argon are R  0.04971Btu / lbm.R , Tcr  272R , and Pcr  705 psia (Table A-1E). Analysis (a) The enthalpy and entropy departures of argon at the specified states are determined from the generalized charts to be

ü ïï ïï ï Z @ 0 and Z @ 0 ý h1 s1 ï P1 1000 PR1 = = = 1.418 ïïï Pcr 705 ïþ Thus argon behaves as an ideal gas at turbine inlet. Also, TR1 =

T1 1000 = = 3.68 Tcr 272

ü ïï ïï ï Z = 0.04 ý h2 ï P2 150 PR2 = = = 0.213 ïïï Pcr 705 ïþ

TR2 =

T2 500 = = 1.838 Tcr 272

and

Z s2 = 0.02

13-69

Thus,

h2 - h1 = RTcr (Z h1 - Z h2 )+ (h2 - h1 )ideal = (0.04971)(272)(0 - 0.04)+ 0.1253(500 - 1000) = - 63.2 Btu/lbm

The power output of the turbine is to be determined from the energy balance equation, Ein - Eout = D Esystem = 0 (steady)

Ein = Eout

é V 2 - V12 ù ú- Qout m(h1 + V12 /2) = m(h2 + V22 /2) + Qout + Wout ¾ ¾ ® Wout = - m ê(h2 - h1 ) + 2 ê ú 2 ë û æ ö ö (450 ft/s) 2 - (300 ft/s) 2 æ ÷÷ çç 1 Btu/lbm ÷ ÷- 80 Btu/s Wout = - (12 lbm/s) ççç- 63.2 + 2 2 ÷÷ ç çè 2 è 25,037 ft /s ø÷ ø = 651.4 Btu / s = 922 hp (b) Under steady conditions, the rate form of the entropy balance for the turbine simplifies to Sin - Sout + Sgen = D SsystemZ 0 = 0

ms1 - ms2 -

Qout Q + Sgen = 0 ® Sgen = m( s2 - s2 ) + out Tb,out T0

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 Sgen ,

æ Q ö ÷ X destroyed = T0 Sgen = T0 çççm( s2 - s2 ) + out ÷ ÷ çè T0 ÷ ø where

s2 - s1 = R (Z s1 - Z s2 )+ (s2 - s1 )ideal

and

T T1

P P1

(s2 - s1 )ideal = c p ln 2 - R ln 2 = 0.1253ln

500 150 - 0.04971 ln = 0.00745 Btu/lbm ×R 1000 1000

Thus

s2 - s1 = R (Z s1 - Z s2 )+ (s2 - s1 )ideal = (0.04971) [0 - (0.02) ]+ 0.00745 = 0.00646 Btu/lbm ×R Substituting, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-70

æ 80 Btu/s ö ÷ X destroyed = (535 R )çç(12 lbm/s)(0.00646 Btu/lbm ×R )+ ÷ ÷= 121.5 Btu/s @122 Btu / s çè 535 R ø

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=1000 [psia] T[1]=1000 [R] Vel[1]=300 [ft/s] P[2]=150 [psia] T[2]=500 [R] Vel[2]=450 [ft/s] m_dot=12 [lbm/s] Q_dot_out=80 [Btu/s] T0=75+460 [R] "Properties" Fluid$='Argon' R_u=1.986 [Btu/lbmol-R] T_cr=T_CRIT(Fluid$) P_cr=P_CRIT(Fluid$) MM=molarmass(Fluid$) R=R_u/MM C_p=CP(Fluid$, T=537, P=14.7) "Analysis" "(a)" T_R[1]=T[1]/T_cr P_R[1]=P[1]/P_cr T_R[2]=T[2]/T_cr P_R[2]=P[2]/P_cr Z_h[1]=ENTHDEP(T_R[1], P_R[1]) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_s[1]=ENTRDEP(T_R[1], P_R[1]) "the function that returns entropy departure factor at T_R1 and P_R1" Z_h[2]=ENTHDEP(T_R[2], P_R[2]) "the function that returns enthalpy departure factor at T_R2 and P_R2" Z_s[2]=ENTRDEP(T_R[2], P_R[2]) "the function that returns entropy departure factor at T_R2 and P_R2" DELTAh_ideal=C_p*(T[2]-T[1]) DELTAh=R*T_cr*(Z_h[1]-Z_h[2])+DELTAh_ideal W_dot_out=-m_dot*(DELTAh+(Vel[2]^2-Vel[1]^2)/2*Convert(ft^2/s^2, Btu/lbm))-Q_dot_out W_dot_out_hp=W_dot_out*Convert(Btu/s, hp) "(b)" DELTAs_ideal=C_p*ln(T[2]/T[1])-R*ln(P[2]/P[1]) DELTAs=R*(Z_s[1]-Z_s[2])+DELTAs_ideal S_dot_gen=m_dot*DELTAs+Q_dot_out/T0 X_dot_destroyed=T0*S_dot_gen

13-98E Methane is to be adiabatically and reversibly compressed in a steady-flow device. The specific work required for this compression is to be determined using the departure charts and treating the methane as an ideal gas with temperature variable specific heats. Properties The properties of methane are (Table A-1E)

M = 16.043 lbm/lbmol, R = 0.1238 Btu/lbm ×R, Tcr = 343.9 R, Pcr = 673 psia Analysis The temperature at the exit state may be determined by the fact that the process is isentropic and the molar entropy change between the inlet and exit is zero. When the expression of Table A-2Ec is substituted for c p and the integration performed, we obtain

13-71

( s2 - s1 )ideal = ò 1

cp T

dT - Ru ln

2 æa ö P2 P = ò çç + b + cT + dT 2 ÷ dT - Ru ln 2 ÷ ÷ ç èT ø P1 P1 1

T P c d = a ln 2 + b(T2 - T1 ) + (T22 - T12 ) + (T23 - T13 ) - Ru ln 2 T1 2 3 P1 Substituting,

0 = 4.75ln

T2 0.09352´ 10- 5 2 + 0.006666 (T2 - 560)+ (T2 - 5602 ) 560 2

0.4510´ 10- 9 3 500 (T2 - 5603 ) - (1.9858) ln 3 50 Solving this equation by EES or an iterative solution gives -

T2 = 892 R When en energy balance is applied to the compressor, it becomes 2

win = (h2 - h1 )ideal = ò c p dT = ò (a + bT + cT 2 + dT 3 )dT 1

= a(T2 - T1 ) +

b 2 c d (T2 - T12 ) + (T23 - T13 ) + (T24 - T14 ) 2 3 4

= 4.75(892 - 560) + -

0.006666 0.09352´ 10- 5 (8922 - 5602 ) + (8923 - 5603 ) 2 3

0.4510´ 10- 9 (8924 - 5604 ) 4

= 3290 Btu/lbmol The work input per unit mass basis is win =

win 3290 Btu/lbmol = = 205.1Btu / lbm M 16.043 lbm/lbmol

The enthalpy departures of propane at the specified states are determined from the generalized charts to be (Fig. A-29 or from EES)

ü T1 560 ï = = 1.628ïï ïï Tcr 343.9 ® Z h1 = 0.0332 ý¾¾ ïï P1 50 PR1 = = = 0.0743 ïï Pcr 673 ïþ

TR1 =

and

13-72

ü T2 892 ï = = 2.594ïï ïï Tcr 343.9 ® Z h 2 = 0.0990 ý¾¾ ïï P2 500 PR 2 = = = 0.743 ïï Pcr 673 ïþ The work input is determined to be TR 2 =

win = h2 - h1 = (h2 - h1 )ideal - RTcr (Z h 2 - Z h1 ) = 205.1 Btu/lbm - (0.1238 Btu/lbm ×R)(343.9 R)(0.0990 - 0.0332)

= 202.3 Btu / lbm

EES (Engineering Equation Solver) SOLUTION "Given" P1=50 [psia] T1=(100+460) [R] P2=500 [psia] "Properties" T_cr=343.9 [R] P_cr=673 [psia] R=0.1238 [Btu/lbm-R] R_u=1.9858 [Btu/lbmol-R] M=16.043 [lbm/lbmol] "Analysis" DELTAs_ideal=0 DELTAs_ideal=a*ln(T2/T1)+b*(T2-T1)+c/2*(T2^2-T1^2)+d/3*(T2^3-T1^3)-R_u*ln(P2/P1) a=4.75 b=0.006666 c=0.09352E-5 d=-0.4510E-9 w_in_bar=a*(T2-T1)+b/2*(T2^2-T1^2)+c/3*(T2^3-T1^3)+d/4*(T2^4-T1^4) w_in=w_in_bar/M T_R1=T1/T_cr P_R1=P1/P_cr T_R2=T2/T_cr P_R2=P2/P_cr Z_h1=ENTHDEP(T_R1, P_R1) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor at T_R2 and P_R2" DELTAh_ideal=w_in w_in_real=DELTAh_ideal-R*T_cr*(Z_h2-Z_h1)

13-99 Propane is compressed in a steady-flow device. The entropy change and the specific work required for this compression are to be determined using the departure charts and treating the propane as an ideal gas with temperature variable specific heats. Properties The properties of propane are (Table A-1, A-2a)

M = 44.097 kg/kmol, R = 0.1885 kJ/kg ×K, Tcr = 370 K, Pcr = 4.26 MPa, c p = 1.6794 kJ/kg ×K

13-73

Analysis (a) Using empirical correlation for the c p of propane as given in Table A-2c gives 2

win = (h2 - h1 )ideal = ò c p dT = ò (a + bT + cT 2 + dT 3 )dT 1

= a(T2 - T1 ) +

b 2 c d (T2 - T12 ) + (T23 - T13 ) + (T24 - T14 ) 2 3 4

= - 4.04(773 - 373) +

0.3048 15.72´ 10- 5 31.74´ 10- 9 (7732 - 3732 ) (7733 - 3733 ) + (7734 - 3734 ) 2 3 4

= 49, 440 kJ/kmol The work input per unit mass is

win =

win 49, 440 kJ/kmol = = 1121 kJ / kg M 44.097 kg/kmol

Similarly, the entropy change is given by 2

( s2 - s1 )ideal = ò 1

cp T

dT - Ru ln

2 æa ö P2 P = ò çç + b + cT + dT 2 ÷ dT - Ru ln 2 ÷ ÷ ç è ø P1 T P1 1

T P c d = a ln 2 + b(T2 - T1 ) + (T22 - T12 ) + (T23 - T13 ) - Ru ln 2 T1 2 3 P1

= - 4.04 ln

773 15.72´ 10- 5 31.74´ 10- 9 + 0.3048(773 - 373) (7732 - 3732 ) + (7733 - 3733 ) 373 2 3

- (8.314) ln

4000 500

= 69.995 kJ/kmol ×K The entropy change per unit mass is ( s2 - s1 )ideal =

( s2 - s1 )ideal 69.995 kJ/kmol ×K = = 1.587 kJ / kg ×K M 44.097 kg/kmol

(b) The enthalpy and entropy departures of propane at the specified states are determined from the generalized charts to be (Fig. A-29, A-30 or from EES)

13-74

ü ïï ïï ï Z h1 = 0.124 ý ï Z s1 = 0.0837 P1 0.5 PR1 = = = 0.117 ïïï Pcr 4.26 ïþ

TR1 =

T1 373 = = 1.008 Tcr 370

ü T2 773 ï = = 2.089 ïï ïï Tcr 370 Z h 2 = 0.233 ý ï Z s 2 = 0.105 P 4 PR 2 = 2 = = 0.939 ïïï Pcr 4.26 ïþ The work input is determined from TR 2 =

win = h2 - h1 = (h2 - h1 )ideal - RTcr (Z h 2 - Z h1 ) = 1121- (0.1885)(370)(0.233 - 0.124) = 1113 kJ / kg and the entropy change is determined to be

s2 - s1 = (s2 - s1 )ideal - R(Z s 2 - Z s1 ) = 1.587 - (0.1885)(0.105 - 0.0837) = 1.583 kJ / kg ×K Discussion Let us see what happens when constant specific heats for propane at the room temperature are used when calculating enthalpy and entropy changes under ideal gas assumption. The entropy and enthalpy changes are determined from (h2 - h1 )ideal = c p (T2 - T1 ) = (1.6794 kJ/kg ×K)(500 - 100)K = 671.8 kJ/kg

T P ( s2 - s1 )ideal = c p ln 2 - R ln 2 T1 P1 = (1.6794) ln

773 4000 - (0.1885) ln = 0.8318 kJ/kg ×K 373 500

With departure factors,

win = h2 - h1 = (h2 - h1 )ideal - RTcr (Z h 2 - Z h1 ) = 671.8 - (0.1885)(370)(0.233 - 0.124) = 664.2 kJ/kg s2 - s1 = (s2 - s1 )ideal - R(Z s 2 - Z s1 ) = 0.8318 - (0.1885)(0.105 - 0.0837) = 0.8278 kJ / kg×K These are not any close to the results obtained using variable specific heats. This shows that using constant specific heats may yield unacceptable results.

13-100 Propane is compressed in a steady-flow device. The second-law efficiency of the compression process is to be determined. Properties The properties of propane are (Table A-1, A-2a)

M = 44.097 kg/kmol, R = 0.1885 kJ/kg ×K, Tcr = 370 K, Pcr = 4.26 MPa, c p = 1.6794 kJ/kg ×K Analysis Using the variable specific heat results of the previous problem, the actual and reversible works are

win = h2 - h1 = 1113 kJ/kg wrev,in = h2 - h1 - T0 (s2 - s1 ) = 1113 kJ/kg - (298 K)(1.583 kJ/kg ×K) = 641.3 kJ/kg

The second-law efficiency is then hII =

wrev,in wrev,in

641.3 = 0.576 1113

13-75

13-101 The initial state and the final temperature of argon contained in a rigid tank are given. The mass of the argon in the tank, the final pressure, and the heat transfer are to be determined using the generalized charts. Analysis (a) The compressibility factor of argon at the initial state is determined from the generalized chart to be

T1 173 ïü = = 1.146ïï ïï Tcr 151.0 ý Z1 = 0.95 and Z h1 = 0.18 ï P 1 PR1 = 1 = = 0.206 ïïï Pcr 4.86 ïþ Then, TR1 =

Pv = ZRT ¾ ¾ ® v=

ZRT (0.95)(0.2081 kPa ×m3 /kg ×K)(173 K) = = 0.0342 m3 /kg P 1000 kPa

V 1.2 m3 = = 35.1 kg v 0.0342 m3 /kg

(b) The specific volume of argon remains constant during this process, v2  v1 . Thus, ü ïï ïï PR2 = 0.315 ï ý Z 2 = 0.99 ï v2 0.0342 m 3 /kg v R2 = = = 5.29ïïï Z h2 @ 0 3 RTcr / Pcr (0.2081 kPa ×m /kg ×K)(151 K)(4860 kPa) ïïþ TR2 =

T2 273 = = 1.808 Tcr 151.0

P2 = PR2 Pcr = (0.315)(4860) = 1531 kPa

Qin = D U = m(u2 - u1 ) Qin = m [h2 - h1 - ( P2v 2 - P1v1 )]= m [h2 - h1 - R( Z 2T2 - Z1T1 ) ]

where

h2 - h1 = RTcr (Z h1 - Z h2 )+ (h2 - h1 )ideal = (0.2081)(151)(0.18 - 0)+ 0.5203(0 - (- 100)) = 57.69 kJ/kg Thus,

Qin = (35.1 kg)éë57.69 - (0.2081 kJ/kg ×K) [(0.99)(273) - (0.95)(173)]K ù û= 1251 kJ

EES (Engineering Equation Solver) SOLUTION "Given" V=1.2 [m^3] T1=(-100+273) [K] P1=1000 [kPa] T2=(0+273) [K] "Properties" Fluid$='Argon' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-76

R_u=8.314 [kJ/kmol-K] T_cr=T_CRIT(Fluid$) P_cr=P_CRIT(Fluid$) MM=molarmass(Fluid$) R=R_u/MM C_p=CP(Fluid$, T=298, P=100) "Analysis" "(a)" T_R1=T1/T_cr P_R1=P1/P_cr Z_h1=ENTHDEP(T_R1, P_R1) "the function that returns enthalpy departure factor at T_R1 and P_R1" Z_1=COMPRESS(T_R1, P_R1) "the function that returns compressibility factor at T_R1 and P_R1" T_R2=T2/T_cr v1=Z_1*R*T1/P1 m=V/v1 "(b)" v2=v1 v_R2=(v2*P_cr)/(R*T_cr) Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor at T_R2 and P_R2" Z_2=COMPRESS(T_R2, P_R2) "the function that returns compressibility factor at T_R2 and P_R2" P2=Z_2*R*T2/v2 P2=P_R2*P_cr "(c)" DELTAh_ideal=C_p*(T2-T1) DELTAh=R*T_cr*(Z_h1-Z_h2)+DELTAh_ideal DELTAu=DELTAh-R*(Z_2*T2-Z_1*T1) Q_in=m*DELTAu

13-102 The heat transfer, work, and entropy changes of methane during a process in a piston-cylinder device are to be determined assuming ideal gas behavior, using generalized charts, and real fluid data. Analysis The ideal gas solution: (Properties are obtained from EES) State 1: T1 = 100°C ¾ ¾ ® h1 = - 4473.3 kJ/kg T1 = 100°C, P1 = 5 MPa ¾ ¾ ® s1 = 10.116 kJ/kg.K

u1 = h1 - RT1 = (- 4473.3) - (0.5182)(100 + 273.15) = - 4666.7 kJ/kg æ100 + 273.15 K ö T 3 ÷ v1 = R 1 = (0.5182 kJ/kg.K) çç ÷= 0.03868 m /kg ÷ çè 5000 kPa ø P1 State 2: T2 = 250°C ¾ ¾ ® h2 = - 4063.0 kJ/kg T2 = 250°C, P2 = 5 MPa ¾ ¾ ® s2 = 11.035 kJ/kg.K

u2 = h2 - RT2 = (- 4063.0) - (0.5182)(250 + 273.15) = - 4334.1 kJ/kg

wideal = P(v 2 - v1 ) = (5000 kPa)(0.05422 - 0.03868)m3 /kg = 77.73 kJ / kg qideal = wideal + (u2 - u1 ) = 77.73 + [(- 4334.1) - (- 4666.7) ]= 410.3 kJ / kg

D sideal = s2 - s1 = 11.035 - 10.116 = 0.919 kJ / kg For the generalized chart solution we first determine the following factors using EES as

13-77

ü T1 373.2 ï = = 1.227 ïï ïï Tcr 304.2 ® Z1 = 0.8773, Z h1 = 0.5493 and Z s1 = 0.3269 ý¾¾ ïï P1 5 PR1 = = = 0.6766ïï Pcr 7.39 ïþ

TR1 =

ü T2 523.2 ï = = 1.720 ïï ïï Tcr 304.2 ® Z 2 = 0.9739, Z h 2 = 0.2676 and Z s 2 = 0.1281 ý¾¾ ïï P2 5 PR 2 = = = 0.6766ïï Pcr 7.39 ïþ

TR 2 =

State 1:

D h1 = Zh1 RTcr = (0.5493)(0.5182 kJ/kg.K)(304.2 K) = 86.60 kJ/kg h1 = h1,ideal - D h1 = (- 4473.3) - 86.60 = - 4559.9 kJ/kg

u1 = h1 - Z1 RT1 = (- 4559.9) - (0.8773)(0.5182)(373.15) = - 4729.6 kJ/kg T 373.15 v1 = Z1 R 1 = (0.8773)(0.5182) = 0.03393 m 3 /kg P1 5000

D s1 = Z s1 R = (0.3269)(0.5182 kJ/kg.K) = 0.1694 kJ/kg.K s1 = s1,ideal - D s1 = 10.116 - 0.1694 = 9.946 kJ/kg.K

State 2:

D h2 = Z h 2 RTcr = (0.2676)(0.5182 kJ/kg.K)(304.2 K) = 42.19 kJ/kg h2 = h2,ideal - D h2 = (- 4063.0) - 42.19 = - 4105.2 kJ/kg

u2 = h2 - Z 2 RT2 = (- 4105.2) - (0.9739)(0.5182)(523.15) = - 4369.2 kJ/kg T 523.15 v 2 = Z 2 R 2 = (0.9739)(0.5182) = 0.05281 m3 /kg P2 5000

D s2 = Z s 2 R = (0.1281)(0.5182 kJ/kg.K) = 0.06639 kJ/kg.K s2 = s2,ideal - D s2 = 11.035 - 0.06639 = 10.968 kJ/kg.K

Then,

wchart = P(v 2 - v1 ) = (5000 kPa)(0.05281- 0.03393)m3 /kg = 94.38 kJ / kg qchart = wchart + (u2 - u1 ) = 94.38 + [(- 4369.2) - (- 4729.6)]= 454.7 kJ / kg

D schart = s2 - s1 = 10.968 - 9.946 = 1.02 kJ / kg The solution using EES built-in property data is as follows:

v1 = 0.03753 m3 /kg T1 = 100°C ü ïï ý u1 = - 45.10 kJ/kg P1 = 5 MPa ïïþ s1 = - 1.569 kJ/kg.K

v = 0.05443 m3 /kg T2 = 250°C üïï 2 ý u2 = 293.22 kJ/kg P2 = 5 MPa ïïþ s2 = - 0.6204 kJ/kg.K

wEES = P(v 2 - v1 ) = (5000 kPa)(0.05443 - 0.03753)m3 /kg = 84.47 kJ / kg qEES = wEES + (u2 - u1 ) = 84.47 + [293.22 - (- 45.10) ]= 422.8 kJ / kg

D sEES = s2 - s1 = - 0.6204 - (- 1.569) = 0.948 kJ / kg

13-78

EES (Engineering Equation Solver) SOLUTION "Input data" T_critical=304.2 [K] P_critical=7390 [kPa] T[1]=100+273.15 [K] T[2]=250+273.15 [K] P[1]=5000 [kPa] P[2]=5000 [kPa] R_u=8.314 [kJ/kmol-K] MM=molarmass(CH4) R=R_u/MM "IDEAL GAS SOLUTION" "State 1" h_ideal[1]=enthalpy(CH4, T=T[1]) "Enthalpy of ideal gas" s_ideal[1]=entropy(CH4, T=T[1], P=P[1]) "Entropy of ideal gas" u_ideal[1]=h_ideal[1]-R*T[1] "Internal energy of ideal gas" v_ideal[1]=R*T[1]/P[1] "specific volume" "State 2" h_ideal[2]=enthalpy(CH4, T=T[2]) "Enthalpy of ideal gas" s_ideal[2]=entropy(CH4, T=T[2], P=P[2]) "Entropy of ideal gas" u_ideal[2]=h_ideal[2]-R*T[2] "Internal energy of ideal gas" v_ideal[2]=R*T[2]/P[2] "specific volume" w_ideal = P[1]*(v_ideal[2]-v_ideal[1]) "For a constant volume process the work is zero." q_ideal-w_ideal=u_ideal[2]-u_ideal[1] "Conservation of energy on a mass basis:" DELTAs_ideal=s_ideal[2]-s_ideal[1] "Entropy change" "COMPRESSABILITY CHART SOLUTION" "State 1" Tr[1]=T[1]/T_critical pr[1]=p[1]/p_critical Z[1]=COMPRESS(Tr[1], Pr[1]) DELTAh[1]=ENTHDEP(Tr[1], Pr[1])*R*T_critical "Enthalpy departure" Z_h1=ENTHDEP(Tr[1], Pr[1]) h[1]=h_ideal[1]-DELTAh[1] "Enthalpy of real gas using charts" u[1]=h[1]-Z[1]*R*T[1] "Internal energy of gas using charts" v_chart[1]=Z[1]*R*T[1]/P[1] DELTAs[1]=ENTRDEP(Tr[1], Pr[1])*R "Entropy departure" Z_s1=ENTRDEP(Tr[1], Pr[1]) s[1]=s_ideal[1]-DELTAs[1] "Entropy of real gas using charts" "State 2" Tr[2]=T[2]/T_critical Pr[2]=p[2]/P_critical Z[2]=COMPRESS(Tr[2], Pr[2]) DELTAh[2]=ENTHDEP(Tr[2], Pr[2])*R*T_critical "Enthalpy departure" Z_h2=ENTHDEP(Tr[2], Pr[2]) DELTAs[2]=ENTRDEP(Tr[2], Pr[2])*R "Entropy departure" Z_s2=ENTRDEP(Tr[2], Pr[2]) h[2]=h_ideal[2]-DELTAh[2] "Enthalpy of real gas using charts" s[2]=s_ideal[2]-DELTAs[2] "Entropy of real gas using charts" u[2]=h[2]-Z[2]*R*T[2] "Internal energy of gas using charts" v_chart[2]=Z[2]*R*T[2]/P[2] w_chart = P[1]*(v_chart[2]-v_chart[1]) "Constant volume process" q_chart-w_chart=u[2]-u[1] "Conservation of energy" DELTAs_chart=s[2]-s[1] "Entropy Change" "SOLUTION USING EES BUILT-IN PROPERTY DATA" "At state 1" v_EES[1]=volume(Methane, T=T[1], P=P[1]) v_EES[2]=volume(Methane, T=T[2], P=P[2]) u_EES[1]=intEnergy(Methane,T=T[1],P=P[1]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-79

s_EES[1]=entropy(Methane,T=T[1],P=P[1]) "At state 2" u_EES[2]=IntEnergy(Methane,T=T[2],P=P[2]) s_EES[2]=entropy(Methane,T=T[2],P=P[2]) w_EES=P[1]*(v_EES[2]-v_EES[1]) "constant volume process" q_EES-w_EES=u_EES[2]-u_EES[1] "Conservation of energy" DELTAs_EES=s_EES[2]-s_EES[1] "Entropy change" SOLUTION DELTAs_chart=1.022 [kJ/kg-K] DELTAs_EES=0.9481 [kJ/kg-K] DELTAs_ideal=0.9192 [kJ/kg-K] MM=16.04 [kg/kmol] P_critical=7390 [kPa] q_chart=454.72 [kJ/kg] q_EES=422.79 [kJ/kg] q_ideal=410.31 [kJ/kg] R=0.5182 [kJ/kg-K] R_u=8.314 [kJ/kmol-K] T_critical=304.2 [K] w_chart=94.38 [kJ/kg] w_EES=84.47 [kJ/kg] w_ideal=77.73 [kJ/kg] Z_h1=0.5493 Z_h2=0.2676 Z_s1=0.3269 Z_s2=0.1281

13-103 An adiabatic storage tank that is initially evacuated is connected to a supply line that carries nitrogen. A valve is opened, and nitrogen flows into the tank. The final temperature in the tank is to be determined by treating nitrogen as an ideal gas and using the generalized charts, and the results are to be compared to the given actual value. Assumptions 1 Uniform flow conditions exist. 2 Kinetic and potential energies are negligible. Analysis We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance:

min - mout = D msystem

Energy balance: Ein - Eout = D Esystem Combining the two balances:

mi = m2

(since mout = minitial = 0)

® 0 + mi hi = m2u2

u2  hi

(a) From the ideal gas property table of nitrogen, at 225 K we read u2 = hi = h@225 K = 6,537 kJ/kmol

The temperature that corresponds to this u2 value is

T2 = 314.8 K

(7.4% error)

(b) Using the generalized enthalpy departure chart, hi is determined to be

13-80

ü Ti 225 ï = = 1.78ïï ïï Tcr 126.2 hi ,ideal - hi = 0.9 ý Z h ,i = ï P RuTcr 10 PR ,i = i = = 2.95 ïïï Pcr 3.39 ïïþ Thus, TR ,i =

(Fig. A-29)

and hi = hi ,ideal - 0.9RuTcr = 6,537 - (0.9)(8.314)(126.2) = 5,593 kJ/kmol u2 = hi = 5,593 kJ/kmol

Try T2  280 K . Then at PR 2  2.95 and TR 2  2.22 we read Z 2  0.98 and (h2, ideal - h2 ) / RuTcr = 0.55 Thus, h2 = h2,ideal - 0.55RuTcr = 8,141- (0.55)(8.314)(126.2) = 7,564 kJ/kmol u2 = h2 - ZRuT2 = 7,564 - (0.98)(8.314)(280) = 5, 283 kJ/kmol

Try T2  300 K . Then at PR 2  2.95 and TR 2  2.38 we read Z 2  1.0 and (h2, ideal - h2 ) / RuTcr = 0.50 Thus, h2 = h2,ideal - 0.50RuTcr = 8, 723 - (0.50)(8.314)(126.2) = 8,198 kJ/kmol u2 = h2 - ZRuT2 = 8,198 - (1.0)(8.314)(300) = 5, 704 kJ/kmol

By linear interpolation,

T2 = 294.7 K

(0.6% error)

EES (Engineering Equation Solver) SOLUTION "Given" V=0.2 [m^3] T_i=225 [K] P_i=10000 [kPa] P2=10000 [kPa] "Properties" Fluid$='N2' R_u=8.314 [kJ/kmol-K] T_cr=126.2 [K] "Table A-1" P_cr=3390 [kPa] "Table A-1" R=0.2598 [kJ/kg-K] "Analysis" "(a)" h_i_ideal=enthalpy(Fluid$, T=T_i) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-81

u2_ideal=h_i_ideal T2_ideal=temperature(Fluid$, u=u2_ideal) "(b)" T_R_i=T_i/T_cr P_R_i=P_i/P_cr Z_h_i=ENTHDEP(T_R_i, P_R_i) "the function that returns enthalpy departure factor" h_i=h_i_ideal--Z_h_i*R*T_cr u2=h_i u2=h2-Z_2*R*T2 h2=h2_ideal-Z_h2*R*T_cr h2_ideal=enthalpy(Fluid$, T=T2) Z_2=COMPRESS(T_R2, P_R2) "The function that returns compressibility factor" Z_h2=ENTHDEP(T_R2, P_R2) "the function that returns enthalpy departure factor" T_R2=T2/T_cr P_R2=P2/P_cr

Fundamentals of Engineering (FE) Exam Problems 13-104 A substance whose Joule-Thomson coefficient is negative is throttled to a lower pressure. During this process, (select the correct statement) (a) the temperature of the substance will increase. (b) the temperature of the substance will decrease. (c) the entropy of the substance will remain constant. (d) the entropy of the substance will decrease. (e) the enthalpy of the substance will decrease. Answer (a) the temperature of the substance will increase.

13-105 Consider the liquid-vapor saturation curve of a pure substance on the P-T diagram. The magnitude of the slope of the tangent line to this curve at a temperature T (in Kelvin) is (a) proportional to the enthalpy of vaporization h fg at that temperature, (b) proportional to the temperature T, (c) proportional to the square of the temperature T, (d) proportional to the volume change v fg at that temperature, (e) inversely proportional to the entropy change s fg at that temperature, Answer (a) proportional to the enthalpy of vaporization h fg at that temperature,

13-106 For a gas whose equation of state is P(v – b) = RT, the specific heat difference c p  cv is equal to (a) R (b) R – b (c) R + b (d) 0 (e) R(1  v / b) Answer (a) R © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-82

Solution The general relation for the specific heat difference c p  cv is 2

æ¶v ö æ¶ P ö c p - cv = - T çç ÷ ÷ ççç ÷ ÷ ÷ ÷ èç¶T ø è ø P ¶v T

For the given gas, P(v - b) = RT. Then,

æ¶ v ö RT R + b ¾¾ ® çç ÷ ÷ ÷ = P çè¶ T ø P P

æ¶ P ö RT RT P ¾¾ ® çç ÷ =÷ =çè ¶ v ÷ øT v- b v- b (v - b) 2

Substituting, 2

æR ö æ P ö TR 2 ÷ ççc p - cv = - T çç ÷ = ÷ ÷ ÷ P(v - b) = R çè P ÷ ø èç v - b ø

13-107 Based on the generalized charts, the error involved in the enthalpy of CO2 at 300 K and 5 MPa if it is assumed to be an ideal gas is (a) 0 (b) 9% (c) 16% (d) 22% (e) 27% Answer (e) 27% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=300 [K] P=5000 [kPa] Pcr=P_CRIT(CarbonDioxide) Tcr=T_CRIT(CarbonDioxide) Tr=T/Tcr Pr=P/Pcr hR=ENTHDEP(Tr, Pr) h_ideal=11351/Molarmass(CO2) "Table A-20 of the text" h_chart=h_ideal-R*Tcr*hR R=0.1889 Error=(h_chart-h_ideal)/h_chart*Convert(, %)

13-108 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134a at 0.8 MPa and 60C is approximately (a) 0 (b) 5C / MPa (c) 11C / MPa (d) 16C / MPa (e) 25C / MPa

13-83

Answer (d) 16C / MPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=60 [C] P1=800 [kPa] h1=ENTHALPY(R134a,T=T1,P=P1) dP=100 [kPa] Tlow=TEMPERATURE(R134a,h=h1,P=P1+dP) Thigh=TEMPERATURE(R134a,h=h1,P=P1-dP) JT=(Tlow-Thigh)/(2*dP)

13-109 … 13-111 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 14 GAS MIXTURES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw Hill LLC and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw Hill LLC: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw Hill LLC.

13-84

CYU - Check Your Understanding CYU 14-1 ■ Check Your Understanding CYU 14-1.1 A gas mixture consists of 3 kg of O2 , 7 kg of N 2 , and 2 kg of CO2 . What is the mass fraction of O2 ? (a) 0.2 (b) 0.25 (c) 0.40 (d) 0.50 (e) 0.75 Answer: (b) 0.25 mfO2  mO2 / mm  3 / (3  7  2)  0.25 CYU 14-1.2 A gas mixture consists of 2 kg of H 2 and 32 kg of O2 . What are the mole fractions of H 2 and O2 , respectively? (a) 2 / 34, 32 / 34 (b) 0.2, 0.8 (c) 0.4, 0.6 (d) 0.5, 0.5 (e) 0.1, 0.9 Answer: (d) 0.5, 0.5 yO2  1 kmol

y N2  1 kmol ym  2 kmol yO2  yO2 / ym  (1 kmol) / (2 kmol)  0.5 yN2  yN2 / ym  (1 kmol) / (2 kmol)  0.5 CYU 14-1.3 A gas mixture consists of 16 kmol of H 2 and 2 kmol of O2 . What are the mass and mole fractions of H 2 , respectively? (a) 1/ 3, 2 / 3 (b) 2 /18, 16 /18 (c) 16 /18, 2 /18 (d) 2 / 3, 2 /18 (e) 1/ 3, 16 /18 Answer: (e) 1/ 3, 16 /18

mO2  64 kg mH2  32 kg mm  96 kg mf H2  mH2 / mm  (32 kg) / (96 kg)  1/ 3 yH2  yH2 / ym  (16 kmol) / (18 kmol)  16 /18 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-85

CYU 14-1.4 A gas mixture consists of 1 kmol of O2 and 1 kmol of N 2 . What is the molar mass of the mixture? (a) 30 kg / kmol (b) 60 kg / kmol (c) 0.5 kg / kmol (d) 2 kg / kmol (e) 4 kg / kmol Answer: (a) 30 kg / kmol

Mm  yO2 MO2  yN2 MN2  (0.5)(32)  (0.5)(28)  30 kg / kmol

CYU 14-2 ■ Check Your Understanding CYU 14-2.1 Select the incorrect statement regarding the behavior of ideal gas mixtures. (a) The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if it existed alone at the mixture temperature and volume (b) The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if it existed alone at the mixture temperature and pressure (c) The mass of a gas mixture is equal to the sum of the masses of the constituents of the mixture. (d) In a vessel that holds a gas mixture, the volume of each gas is equal to the volume of the vessel. (e) None of the above. Answer: (e) None of the above CYU 14-2.2 0.4 m3 of oxygen gas at 100 kPa and 300 K is mixed with 0.6 m3 of nitrogen gas at 100 kPa and 300 K. If the mixture temperature and pressure are also at 100 kPa and 300 K, what is the volume of this mixture? (a) 0.4 m3 (b) 0.6 m3 (c) 1 m3 (d) 1.4 m3 (e) 1.6 m3 Answer: (c) 1 m3 From Amagad's law of additive volumes: Vm  VO2  VN2  0.4  0.6  1 m3 CYU 14-2.3 0.5 m3 of oxygen gas at 100 kPa and 300 K is mixed with 0.5 m3 of nitrogen gas at 100 kPa and 300 K. If the mixture volume and temperature are 1 m3 and 300 K, what is the pressure of the mixture? (a) 10 kPa (b) 25 kPa (c) 50 kPa (d) 100 kPa © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-86

(e) 200 kPa Answer: (d) 100 kPa From Amagad's law of additive volumes: Vm  VO2  VN2  0.4  0.6  1 m3 Pm  PO2  PN2  100 kPa CYU 14-2.4 A 2  m3 tank contains a gas mixture that consist of 1 kmol of O2 and 3 kmol of N 2 at 100 kPa and 500 K. What would be the pressure of O2 if it existed alone in this tank at 500 K? (a) 25 kPa (b) 50 kPa (c) 75 kPa (d) 100 kPa (e) 200 kPa Answer: (a) 25 kPa From Dalton's law of additive pressures: yO2  NO2 / Nm  (1 kmol) / (4 kmol)  0.25 PO2  yO2 Pm  (0.25)(100 kPa)  25 kPa CYU 14-2.5 A 2  m3 tank contains an gas mixture that consist of 1 kmol of O2 and 3 kmol of N 2 at 100 kPa and 500 K. What would be the volume of N 2 if it existed alone in a tank at 100 kPa and 500 K? (a) 0.25 m3 (b) 0.5 m3 (c) 1 m3 (d) 1.5 m3 (e) 2 m3 Answer: (d) 1.5 m3 From Amagad's law of additive volumes: yN2  NN2 / Nm  (3 kmol) / (4 kmol)  0.75 VO2  yO2 Vm  (0.75) / (2 m3 )  1.5 m3

13-87

PROBLEMS Composition of Gas Mixtures 14-1C The ratio of the mass of a component to the mass of the mixture is called the mass fraction (mf), and the ratio of the mole number of a component to the mole number of the mixture is called the mole fraction (y).

14-2C The mass fractions will be identical, but the mole fractions will not.

14-3C Yes.

14-4C Yes, because both CO2 and N2 O has the same molar mass, M  44 kg / kmol .

14-5C No. We can do this only when each gas has the same mole fraction.

14-6C It is the average or the equivalent molar mass of the gas mixture. No.

14-7C It is the average or the equivalent gas constant of the gas mixture. No.

14-8 The molar fractions of the constituents of moist air are given. The mass fractions of the constituents are to be determined. Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N 2 and O2 only. Properties The molar masses of N 2 , O2 , and H 2 O are 28.0, 32.0, and 18.0 kg / kmol , respectively (Table A-1). Analysis The molar mass of moist air is

M = å yi M i = 0.78´ 28.0 + 0.20´ 32.0 + 0.02´ 18 = 28.6 kg/kmol Then the mass fractions of constituent gases are determined to be N 2 : mf N2 = yN2 O 2 : mf O2 = yO2

M N2 M M O2 M

H 2 O: mf H2 O = yH2 O

= (0.78)

28.0 = 0.764 28.6

= (0.20)

32.0 = 0.224 28.6

M H2 O M

= (0.02)

18.0 = 0.013 28.6

13-88

Therefore, the mass fractions of N 2 , O2 , and H 2 O in the air are 76.4%, 22.4%, and 1.3%, respectively.

EES (Engineering Equation Solver) SOLUTION "Given" y_N2=0.78 y_O2=0.20 y_H2O=0.02 "Properties" M_N2=molarmass(N2) M_O2=molarmass(O2) M_H2O=molarmass(H2O) "Analysis" M=y_N2*M_N2+y_O2*M_O2+y_H2O*M_H2O w_N2=y_N2*M_N2/M w_O2=y_O2*M_O2/M w_H2O=y_H2O*M_H2O/M

14-9 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of the mixture, its molar mass, and gas constant are to be determined. Properties The molar masses of N 2 , and CO2 are 28.0 and 44.0 kg / kmol , respectively (Table A-1) Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are

N N2 = 60 kmol ¾ ¾ ® mN2 = N N2 M N2 = (60 kmol)(28 kg/kmol) = 1680 kg NCO2 = 40 kmol ¾ ¾ ® mCO2 = NCO2 M CO2 = (40 kmol)(44 kg/kmol) = 1760 kg

mm = mN2 + mCO2 = 1680 kg + 1760 kg = 3440 kg Then the mass fraction of each component (gravimetric analysis) becomes

mf N2 = mf CO2 =

mN2 mm

mCO2 mm

1680 kg = 0.488 or 48.8% 3440 kg =

1760 kg = 0.512 or 51.2% 3440 kg

13-89

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

mm 3,440 kg = = 34.40 kg / kmol N m 100 kmol

Rm =

Ru 8.314 kJ/kmol ×K = = 0.242 kJ / kg ×K Mm 34.4 kg/kmol

and

EES (Engineering Equation Solver) SOLUTION "Given" y_N2=0.60 y_CO2=0.40 "Properties" R_u=8.314 [kJ/kmol-K] M_N2=molarmass(N2) M_CO2=molarmass(CO2) "Analysis" N_m=100 "[kmol], Consider 100 kmol of mixture" N_N2=N_m*y_N2 N_CO2=N_m*y_CO2 mass_N2=N_N2*M_N2 mass_CO2=N_CO2*M_CO2 mass_m=mass_N2+mass_CO2 mf_N2=mass_N2/mass_m mf_CO2=mass_CO2/mass_m M_m=mass_m/N_m R_m=R_u/M_m

14-10 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of the mixture, its molar mass, and gas constant are to be determined. Properties The molar masses of O2 and CO2 are 32.0 and 44.0 kg / kmol , respectively (Table A-1) Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are

NO2 = 60 kmol ¾ ¾ ® mO2 = NO2 M O2 = (60 kmol)(32 kg/kmol) = 1920 kg NCO2 = 40 kmol ¾ ¾ ® mCO2 = NCO2 M CO2 = (40 kmol)(44 kg/kmol) = 1760 kg

mm = mO2 + mCO2 = 1920 kg + 1760 kg = 3680 kg Then the mass fraction of each component (gravimetric analysis) becomes

mf O2 =

mf CO2 =

mO2 mm

mCO2 mm

1920 kg = 0.522 or 52.2% 3680 kg

1760 kg = 0.478 or 47.8% 3680 kg

13-90

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

mm 3680 kg = = 36.80 kg / kmol N m 100 kmol

Rm =

Ru 8.314 kJ/kmol ×K = = 0.226 kJ / kg ×K Mm 36.8 kg/kmol

and

EES (Engineering Equation Solver) SOLUTION "Given" y_O2=0.60 y_CO2=0.40 "Properties" R_u=8.314 [kJ/kmol-K] M_O2=molarmass(O2) M_CO2=molarmass(CO2) "Analysis" N_m=100 "[kmol], Consider 100 kmol of mixture" N_O2=N_m*y_O2 N_CO2=N_m*y_CO2 mass_O2=N_O2*M_O2 mass_CO2=N_CO2*M_CO2 mass_m=mass_O2+mass_CO2 mf_O2=mass_O2/mass_m mf_CO2=mass_CO2/mass_m M_m=mass_m/N_m R_m=R_u/M_m

14-11 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of the mixture and the apparent gas constant are to be determined. Properties The molar masses of O2 , N 2 and CO2 are 32.0, 28, and 44.0 kg / kmol , respectively (Table A-1) Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component and the total number of moles are mO2

mO2 = 20 kg

¾¾ ®

N O2 =

mN2 = 20 kg

¾¾ ®

N N2 =

mCO2 = 50 kg

¾¾ ®

N CO2 =

M O2

mN2 M N2

20 kg = 0.625 kmol 32 kg/kmol

30 kg = 1.071 kmol 28 kg/kmol

mCO2 M CO2

50 kg = 1.136 kmol 44 kg/kmol

13-91

Noting that the volume fractions are same as the mole fractions, the volume fraction of each component becomes

yO2 = yN 2 = yCO2 =

N O2 Nm

0.625 kmol = 0.221 2.832 kmol

22.1%

1.071 kmol = 0.378 2.832 kmol

37.8%

40.1%

N N2 Nm

NCO2 Nm

1.136 kmol = 0.401 2.832 kmol

The molar mass and the gas constant of the mixture are determined from their definitions, Mm =

mm 100 kg = = 35.31 kg / kmol N m 2.832 kmol

and Rm =

Ru 8.314 kJ/kmol ×K = = 0.235 kJ / kg ×K Mm 35.31 kg/kmol

EES (Engineering Equation Solver) SOLUTION "Given" mf_O2=0.20 mf_N2=0.30 mf_CO2=0.50 "Properties" M_O2=32 [kg/kmol] M_N2=28 [kg/kmol] M_CO2=44 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" mass_m=100 mass_O2=mass_m*mf_O2 mass_N2=mass_m*mf_N2 mass_CO2=mass_m*mf_CO2 N_O2=mass_O2/M_O2 N_N2=mass_N2/M_N2 N_CO2=mass_CO2/M_CO2 N_m=N_O2+N_N2+N_CO2 y_O2=N_O2/N_m y_N2=N_N2/N_m y_CO2=N_CO2/N_m M_m=mass_m/N_m R_m=R_u/M_m

13-92

14-12 The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the apparent gas constant are to be determined. Properties The molar masses of H 2 , and N 2 are 2.0 and 28.0 kg / kmol , respectively (Table A-1) Analysis The mass of each component is determined from

NH2  6 kmol

 

mH2  N H2 MH2   6 kmol  2.0 kg/kmol   12 kg

N N2  2 kmol

 

mN2  N N2 M N2   2 kmol  28 kg/kmol   56 kg

The total mass and the total number of moles are

m m  m H2  m N2  12 kg  56 kg  68 kg N m  N H2  N N2  6 kmol  2 kmol  8 kmol

The molar mass and the gas constant of the mixture are determined from their definitions, Mm 

mm 68 kg   8.5 kg / kmol N m 8 kmol

Rm 

R u 8.314 kJ/kmol  K   0.978 kJ / kg  K Mm 8.5 kg/kmol

and

EES (Engineering Equation Solver) SOLUTION "Given" N_H2=6 [kmol] N_N2=2 [kmol] "Properties" M_H2=2 [kg/kmol] M_N2=28 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" mass_H2=N_H2*M_H2 mass_N2=N_N2*M_N2 mass_m=mass_H2+mass_N2 N_m=N_H2+N_N2 M_m=mass_m/N_m R_m=R_u/M_m

14-13 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, the average molar mass, and gas constant are to be determined. Properties The molar masses of O2 , N 2 , and CO2 are 32.0, 28.0 and 44.0 kg / kmol , respectively (Table A-1) Analysis (a) The total mass of the mixture is

13-93

mm = mO2 + mN2 + mCO2 = 4 kg + 5 kg + 7 kg = 16 kg Then the mass fraction of each component becomes

mO2

mf O2 =

4 kg = 0.25 16 kg

5 kg = 0.3125 16 kg

mN2

mf N2 =

mf CO2 =

mCO2 mm

7 kg = 0.4375 16 kg

(b) To find the mole fractions, we need to determine the mole numbers of each component first, mO2

N O2 =

M O2

mN2

N N2 =

M N2

N CO2 =

4 kg = 0.125 kmol 32 kg/kmol

5 kg = 0.1786 kmol 28 kg/kmol

mCO2 M CO2

7 kg = 0.1591 kmol 44 kg/kmol

Thus,

N m = N O2 + N N2 + N CO2 = 0.125 kmol + 0.1786 kmol + 0.1591 kmol = 0.4627 kmol and

N O2

yO2 =

0.125 kmol = 0.2702 0.4627 kmol

0.1786 kmol = 0.386 0.4627 kmol

N N2

yN2 =

yCO2 =

NCO2 Nm

0.1591 kmol = 0.3439 0.4627 kmol

mm 16 kg = = 34.58 kg / kmol N m 0.4627 kmol

Rm =

Ru 8.314 kJ/kmol ×K = = 0.2404 kJ / kg ×K Mm 34.58 kg/kmol

and

13-94

mass_N2=5 [kg] mass_CO2=7 [kg] "Properties" M_O2=32 [kg/kmol] M_N2=28 [kg/kmol] M_CO2=44 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" mass_m=mass_O2+mass_N2+mass_CO2 mf_O2=mass_O2/mass_m mf_N2=mass_N2/mass_m mf_CO2=mass_CO2/mass_m N_O2=mass_O2/M_O2 N_N2=mass_N2/M_N2 N_CO2=mass_CO2/M_CO2 N_m=N_O2+N_N2+N_CO2 y_O2=N_O2/N_m y_N2=N_N2/N_m y_CO2=N_CO2/N_m M_m=mass_m/N_m R_m=R_u/M_m

14-14 From the definition of mass fraction, mf i =

æM ÷ ö mi Ni M i = = yi ççç i ÷ ÷ çè M m ÷ mm Nm M m ø

14-15 A mixture consists of two gases. Relations for mole fractions when mass fractions are known are to be obtained. Analysis The mass fractions of A and B are expressed as mf A =

mA N M M = A A = yA A mm Nm M m Mm

and mf B = yB

MB Mm

Where m is mass, M is the molar mass, N is the number of moles, and y is the mole fraction. The apparent molar mass of the mixture is Mm =

mm N M + NB M B = A A = y A M A + yB M B Nm Nm

Combining the two equation above and noting that y A + yB = 1 gives the following convenient relations for converting mass fractions to mole fractions,

yA =

MB and yB = 1- yA M A (1/ mf A - 1) + M B

which are the desired relations.

P-v-T Behavior of Gas Mixtures 14-16C Normally yes. Air, for example, behaves as an ideal gas in the range of temperatures and pressures at which oxygen and nitrogen behave as ideal gases.

13-95

14-17C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existed alone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but only approximately for real gas mixtures.

14-18C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure. This law holds exactly for ideal gas mixtures, but only approximately for real gas mixtures.

14-19C With Kay's rule, a real-gas mixture is treated as a pure substance whose critical pressure and temperature are defined in terms of the critical pressures and temperatures of the mixture components as

Pcr,¢ m = å yi Pcr, i

and

Tcr,¢ m = å yiTcr, i

The compressibility factor of the mixture (Z m ) is then easily determined using these pseudo-critical point values.

14-20C The P-v-T behavior of a component in an ideal gas mixture is expressed by the ideal gas equation of state using the properties of the individual component instead of the mixture, Pv i i  Ri Ti . The P-v-T behavior of a component in a real gas mixture is expressed by more complex equations of state, or by Pv i i  Zi Ri Ti , where Z i is the compressibility factor.

14-21C Component pressure is the pressure a component would exert if existed alone at the mixture temperature and volume. Partial pressure is the quantity yi Pm , where yi is the mole fraction of component i. These two are identical for ideal gases.

14-22C Component volume is the volume a component would occupy if existed alone at the mixture temperature and pressure. Partial volume is the quantity yiVm , where yi is the mole fraction of component i. These two are identical for ideal gases.

14-23C The one with the higher mole number.

14-24C The partial pressures will decrease but the pressure fractions will remain the same.

14-25C The partial pressures will increase but the pressure fractions will remain the same.

14-26C No. The correct expression is “the temperature of a gas mixture is equal to the temperature of the individual gas components.”

14-27C No. The correct expression is “the volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed alone at the mixture temperature and pressure.”

13-96

14-29 The masses of the constituents of a gas mixture at a specified pressure and temperature are given. The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined. Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg / kmol , respectively (Table A-1) Analysis The mole numbers of the constituents are mCO2 = 1 kg ® N CO2 =

mCO2 1 kg = = 0.0227 kmol M CO2 44 kg/kmol

mCH4 = 3 kg ® N CH4 =

mCH4 3 kg = = 0.1875 kmol M CH4 16 kg/kmol

N m = N CO2 + N CH4 = 0.0227 kmol + 0.1875 kmol = 0.2102 kmol yCO2 = yCH4 =

N CO2 Nm N CH4 Nm

0.0227 kmol = 0.108 0.2102 kmol

0.1875 kmol = 0.892 0.2102 kmol

Then the partial pressures become

PCO2 = yCO2 Pm = (0.108)(200 kPa ) = 21.6 kPa PCH4 = yCH4 Pm = (0.892)(200 kPa ) = 178.4 kPa The apparent molar mass of the mixture is Mm =

mm 4 kg = = 19.03 kg / kmol N m 0.2102 kmol

EES (Engineering Equation Solver) SOLUTION "Given" T_m=300 "[K]" P_m=200 "[kPa]" mass_CO2=1 "[kg]" mass_CH4=3 "[kg]" "Properties" M_CO2=molarmass(CO2) M_CH4=molarmass(CH4) "Analysis" N_CO2=mass_CO2/M_CO2 N_CH4=mass_CH4/M_CH4 N_m=N_CO2+N_CH4 y_CO2=N_CO2/N_m y_CH4=N_CH4/N_m © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-97

P_CO2=y_CO2*P_m P_CH4=y_CH4*P_m mass_m=mass_CO2+mass_CH4 M_m=mass_m/N_m

14-30 The masses of the constituents of a gas mixture at a specified temperature are given. The partial pressure of each gas and the total pressure of the mixture are to be determined. Assumptions Under specified conditions both N 2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The partial pressures of constituent gases are

æmRT ÷ ö (0.6 kg)(0.2968 kPa ×m3 /kg ×K)(300 K) PN2 = çç = = 178.1 kPa ÷ çè V ÷ øN2 0.3 m3 æmRT ÷ ö (0.4 kg)(0.2598 kPa ×m3 /kg ×K)(300 K) PO2 = çç = = 103.9 kPa ÷ ÷ çè V øO 0.3 m3 2

and

Pm = PN2 + PO2 = 178.1 kPa + 103.9 kPa = 282.0 kPa

EES (Engineering Equation Solver) SOLUTION "Given" V=0.3 "[m^3]" mass_N2=0.6 "[kg]" mass_O2=0.4 "[kg]" T=300 "[K]" "Properties" R_u=8.314 "[kJ/kmol-K]" M_N2=molarmass(N2) M_O2=molarmass(O2) R_N2=R_u/M_N2 R_O2=R_u/M_O2 "Analysis" P_N2=mass_N2*R_N2*T/V P_O2=mass_O2*R_O2*T/V P_m=P_N2+P_O2

14-31 The volumetric fractions of components of a gas mixture before and after a separation unit are given. The changes in partial pressures of the components in the mixture before and after the separation unit are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-98

Analysis The partial pressures before the separation unit are

PCH4 = yCH4 Pm = (0.60)(100 kPa) = 60 kPa PC2H6 = yC2H6 Pm = (0.30)(100 kPa) = 30 kPa PC3H8 = yC3H8 Pm = (0.10)(100 kPa) = 10 kPa The mole fraction of propane is 0.01 after the separation unit. The corresponding mole fractions of methane and ethane are determined as follows:

x = 0.01 ¾ ¾ ® x = 0.00909 0.6 + 0.3 + x yCH4 =

0.60 = 0.6600 0.6 + 0.3 + 0.00909

yC2H6 =

0.30 = 0.3708 0.6 + 0.2 + 0.00909

yC3H8 =

0.00909 = 0.01 0.6 + 0.3 + 0.00909

The partial pressures after the separation unit are

PCH4 = yCH4 Pm = (0.6600)(100 kPa) = 66.00 kPa PC2H6 = yC2H6 Pm = (0.3708)(100 kPa) = 37.08 kPa PC3H8 = yC3H8 Pm = (0.01)(100 kPa) = 1 kPa The changes in partial pressures are then

D PCH4 = 66.00 - 60 = 6.00 kPa D PC2H6 = 37.08 - 30 = 7.08 kPa D PC3H8 = 1- 10 = - 9 kPa

14-32 The volume fractions of components of a gas mixture are given. The mass fractions and apparent molecular weight of the mixture are to be determined. Properties The molar masses of H 2 , He, and N 2 are 2.0, 4.0, and 28.0 kg / kmol , respectively (Table A-1). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are

mH2 = NH2 M H2 = (30 kmol)(2 kg/kmol) = 60 kg mHe = NHe M He = (40 kmol)(4 kg/kmol) = 160 kg mN2 = N N2 M N2 = (30 kmol)(28 kg/kmol) = 840 kg The total mass is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-99

mm = mH2 + mHe + N N2 = 60 + 160 + 840 = 1060 kg

Then the mass fractions are mf H2 =

mH2 60 kg = = 0.05660 mm 1060 kg

mf He =

mHe 160 kg = = 0.1509 mm 1060 kg

mf N2 =

mN2 840 kg = = 0.7925 mm 1060 kg

The apparent molecular weight of the mixture is Mm =

mm 1060 kg = = 10.60 kg / kmol N m 100 kmol

EES (Engineering Equation Solver) SOLUTION "Given" y_H2=0.30 y_He=0.40 y_N2=0.30 "Properties" M_H2=2 [kg/kmol] M_He=4 [kg/kmol] M_N2=28 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" "We consider 100 kmol of this mixture:" N_m=100 [kmol] N_H2=y_H2*N_m N_He=y_He*N_m N_N2=y_N2*N_m mass_H2=N_H2*M_H2 mass_He=N_He*M_He mass_N2=N_N2*M_N2 mass_m=mass_H2+mass_He+mass_N2 mf_H2=mass_H2/mass_m mf_He=mass_He/mass_m mf_N2=mass_N2/mass_m M_m=mass_m/N_m

14-33 The mass fractions of components of a gas mixture are given. The mole fractions of each constituent, the mixture’s apparent molecular weight, the partial pressure of each constituent, and the apparent specific heats of the mixture are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-100

Properties The molar masses of N 2 , He, CH4 , and C2 H6 are 28.0, 4.0, 16.0, and 30.0 kg / kmol , respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 1.039, 5.1926, 2.2537, and 1.7662 kJ / kg  K , respectively (Table A-2a). Analysis We consider 100 kg of this mixture. The mole numbers of each component are m 15 kg N N2 = N2 = = 0.5357 kmol M N2 28 kg/kmol N He =

mHe 5 kg = = 1.25 kmol M He 4 kg/kmol

N CH4 =

mCH4 60 kg = = 3.75 kmol M CH4 16 kg/kmol

N C2H6 =

mC2H6 20 kg = = 0.6667 kmol M C2H6 30 kg/kmol

The mole number of the mixture is Nm = N N2 + NHe + NCH4 + NC2H6 = 0.5357 + 1.25 + 3.75 + 0.6667 = 6.2024 kmol and the mole fractions are N 0.5357 kmol yN2 = N2 = = 0.08637 Nm 6.2024 kmol yHe =

N He 1.25 kmol = = 0.2015 Nm 6.2024 kmol

yCH4 =

N CH4 3.75 kmol = = 0.6046 Nm 6.2024 kmol

yC2H6 =

N C2H6 0.6667 kmol = = 0.1075 Nm 6.2024 kmol

The apparent molecular weight of the mixture is m 100 kg Mm = m = = 16.12 kg / kmol N m 6.2024 kmol The partial pressure of each constituent for a mixture pressure of 1200 kPa are PN2 = yN2 Pm = (0.08637)(1200 kPa) = 103.6 kPa

PHe = yHe Pm = (0.2015)(1200 kPa) = 241.8 kPa PCH4 = yCH4 Pm = (0.6046)(1200 kPa) = 725.5 kPa PC2H6 = yC2H6 Pm = (0.1075)(1200 kPa) = 129.0 kPa The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p ,N2 + mf He c p ,He + mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6 = 0.15´ 1.039 + 0.05´ 5.1926 + 0.60´ 2.2537 + 0.20´ 1.7662 = 2.121 kJ / kg ×K The apparent gas constant of the mixture is

13-101

Ru 8.314 kJ/kmol ×K = = 0.5158 kJ/kg ×K Mm 16.12 kg/kmol

Then the constant-volume specific heat is cv = c p - R = 2.121- 0.5158 = 1.605 kJ / kg ×K

14-34 The volume fractions of components of a gas mixture are given. The mixture’s apparent molecular weight and the apparent specific heats of the mixture are to be determined. Properties The molar masses of O2 , N 2 , CO2 , and CH4 are 32.0, 28.0, 44.0, and 16.0 kg / kmol , respectively (Table A1). The constant-pressure specific heats of these gases at room temperature are 0.918, 1.039, 0.846, and 2.2537 kJ / kg  K , respectively (Table A-2). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are

mO2 = NO2 M O2 = (30 kmol)(32 kg/kmol) = 960 kg mN2 = N N2 M N2 = (40 kmol)(28 kg/kmol) = 1120 kg mCO2 = NCO2 M CO2 = (10 kmol)(44 kg/kmol) = 440 kg mCH4 = NCH4 M CH4 = (20 kmol)(16 kg/kmol) = 320 kg

The total mass is

mm = mO2 + mN2 + mCO2 + mCH4 = 960 + 1120 + 440 + 320 = 2840 kg Then the mass fractions are mf O2 =

mO2 960 kg = = 0.3380 mm 2840 kg

mf N2 =

mN2 1120 kg = = 0.3944 mm 2840 kg

mf CO2 =

mCO2 440 kg = = 0.1549 mm 2840 kg

mf CH4 =

mCH4 320 kg = = 0.1127 mm 2840 kg

The apparent molecular weight of the mixture is Mm =

mm 2840 kg = = 28.40 kg / kmol N m 100 kmol

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf N2 c p ,N2 + mf CO2 c p ,CO2 + mf CH4 c p ,CH4

13-102

= 0.3380´ 0.918 + 0.3944´ 1.039 + 0.1549´ 0.846 + 0.1127´ 2.2537 = 1.1051 kJ / kg ×K The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ×K = = 0.2927 kJ/kg ×K Mm 28.40 kg/kmol

Then the constant-volume specific heat is

cv = c p - R = 1.1051- 0.2927 = 0.8124 kJ / kg ×K

14-35 An additional 5% of oxygen is mixed with standard atmospheric air. The molecular weight of this mixture is to be determined. Properties The molar masses of N 2 and O2 are 28.0 and 32.0 kg / kmol , respectively (Table A-1). Analysis Standard air is taken as 79% nitrogen and 21% oxygen by mole. That is,

yO2 = 0.21 yN2 = 0.79 Adding another 0.05 moles of O2 to 1 kmol of standard air gives

0.26 = 0.2476 1.05 0.79 yN2 = = 0.7524 1.05 yO2 =

Then,

M m = yO2 M O2 + yN2 M N2 = 0.2476´ 32 + 0.7524´ 28 = 28.99 kg / kmol

EES (Engineering Equation Solver) SOLUTION "Given" N_O2_add=0.05 [kmol] "Properties, Tables A-1" M_O2=32 [kg/kmol] M_N2=28 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" "Standard air, for 1 kmol air" N_O2_s=0.21 [kmol] N_N2_s=0.79 [kmol] "If 5% O2 is added:" N_O2=N_O2_s+N_O2_add N_N2=N_N2_s N_total=N_O2+N_N2 y_O2=N_O2/N_total y_N2=N_N2/N_total M_m=y_O2*M_O2+y_N2*M_N2

14-36 A tank contains a mixture of two gases of known masses at a specified pressure and temperature. The mixture is now heated to a specified temperature. The volume of the tank and the final pressure of the mixture are to be determined.

13-103

Assumptions Under specified conditions both Ar and N 2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Analysis The total number of moles is and N m = N Ar + N N2 = 0.5 kmol + 2 kmol = 2.5 kmol

Vm =

N m RuTm (2.5 kmol)(8.314 kPa ×m3 /kmol ×K)(280 K) = = 23.3 m3 Pm 250 kPa

Also, P2V2 PV T 400 K  1 1  P2  2 P1  (250kPa)  357.1kPa T2 T1 T1 280 K

EES (Engineering Equation Solver) SOLUTION "Given" N_Ar=0.5 [kmol] N_N2=2 [kmol] T1=280 [K] P1=250 [kPa] T2=400 [K] "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" N_m=N_Ar+N_N2 V_m=N_m*R_u*T1/P1 P2=P1*T2/T1

14-37 The masses of components of a gas mixture are given. The apparent molecular weight of this mixture, the volume it occupies, the partial volume of the oxygen, and the partial pressure of the helium are to be determined. Properties The molar masses of O2 , CO2 , and He are 32.0, 44.0, and 4.0 kg / kmol , respectively (Table A-1). Analysis The total mass of the mixture is

mm = mO2 + mCO2 + mHe = 0.9 + 0.7 + 0.2 = 1.8 kg The mole numbers of each component are N O2 =

mO2 0.9 kg = = 0.02813 kmol M O2 32 kg/kmol

N CO2 =

mCO2 0.7 kg = = 0.01591 kmol M CO2 44 kg/kmol

N He =

mHe 0.2 kg = = 0.05 kmol M He 4 kg/kmol

13-104

The mole number of the mixture is

Nm = NO2 + NCO2 + NHe = 0.02813 + 0.01591 + 0.05 = 0.09403 kmol Then the apparent molecular weight of the mixture becomes Mm =

mm 1.8 kg = = 19.14 kg / kmol N m 0.09403 kmol

The volume of this ideal gas mixture is Vm =

N m RuT (0.09403 kmol)(8.314 kPa ×m3 /kmol ×K)(300 K) = = 2.345 m3 P 100 kPa

The partial volume of oxygen in the mixture is VO2 = yO2Vm =

N O2 0.02813 kmol Vm = (2.345 m3 ) = 0.7015 m3 Nm 0.09403 kmol

The partial pressure of helium in the mixture is PHe = yHe Pm =

N He 0.05 kmol Pm = (100 kPa) = 53.17 kPa Nm 0.09403 kmol

EES (Engineering Equation Solver) SOLUTION "Given" mass_O2=0.9 [kg] mass_CO2=0.7 [kg] mass_He=0.2 [kg] P=100 [kPa] T=(27+273) [K] "Properties" R_u=8.314 [kJ/kmol-K] M_O2=32 [kg/kmol] M_CO2=44 [kg/kmol] M_He=4 [kg/kmol] "Analysis" mass_m=mass_O2+mass_CO2+mass_He N_O2=mass_O2/M_O2 N_CO2=mass_CO2/M_CO2 N_He=mass_He/M_He N_m=N_O2+N_CO2+N_He M_m=mass_m/N_m V_m=(N_m*R_u*T)/P y_O2=N_O2/N_m y_He=N_He/N_m V_O2=y_O2*V_m P_He=y_He*P

14-38 A container contains a mixture of two fluids whose volumes are given. The density of the mixture is to be determined. Assumptions The volume of the mixture is the sum of the volumes of the two constituents. Properties The specific volumes of the two fluids are given to be 0.0003 m3 / kg and 0.00023 m3 / kg . Analysis The mass of the two fluids are

mA =

VA 0.001 m3 = = 3.333 kg v A 0.0003 m3 /kg

13-105

mB =

VB 0.002 m3 = = 8.696 kg v B 0.00023 m3 /kg

The density of the mixture is then mA + mB (3.333 + 8.696) lbf = = 4010 kg / m 3 VA + VB (0.001 + 0.002) ft 3

14-39E A mixture is obtained by mixing two gases at constant pressure and temperature. The volume and specific volume of the mixture are to be determined. Properties The densities of two gases are given in the problem statement. Analysis The volume of constituent gas A is VA =

mA 1 lbm = = 1000 ft 3 rA 0.001 lbm/ft 3

and the volume of constituent gas B is VB =

mB 2 lbm = = 1000 ft 3 rB 0.002 lbm/ft 3

Hence, the volume of the mixture is

V = VA + VB = 1000 + 1000 = 2000 ft 3 The specific volume of the mixture will then be

V 2000 ft 3 = = 666.7 ft 3 / lbm m (1 + 2) lbm

14-40 The mass fractions of components of a gas mixture are given. The partial pressure of ethane is to be determined. Properties The molar masses of CH4 and C2 H6 are 16.0 and 30.0 kg / kmol , respectively (Table A-1). Analysis We consider 100 kg of this mixture. The mole numbers of each component are N CH4 =

mCH4 70 kg = = 4.375 kmol M CH4 16 kg/kmol

13-106

N C2H6 =

mC2H6 30 kg = = 1.0 kmol M C2H6 30 kg/kmol

The mole number of the mixture is

Nm = NCH4 + NC2H6 = 4.375 + 1.0 = 5.375 kmol The mole fractions are yCH4 =

N CH4 4.375 kmol = = 0.8139 Nm 5.375 kmol

yC2H6 =

N C2H6 1.0 kmol = = 0.1861 Nm 5.375 kmol

The final pressure of ethane in the final mixture is

PC2H6 = yC2H6 Pm = (0.1861)(130 kPa) = 24.19 kPa

EES (Engineering Equation Solver) SOLUTION "Given" mf_C2H6=0.30 mf_CH4=0.70 V=100 [m^3] P=130 [kPa] T=(25+273) [K] "Properties" R_u=8.314 [kJ/kmol-K] M_CH4=16 [kg/kmol] M_C2H6=30 [kg/kmol] "Analysis" mass_m=100 [kg] "100 kg mass is considered" mass_CH4=mf_CH4*mass_m mass_C2H6=mf_C2H6*mass_m N_CH4=mass_CH4/M_CH4 N_C2H6=mass_C2H6/M_C2H6 N_m=N_CH4+N_C2H6 y_CH4=N_CH4/N_m y_C2H6=N_C2H6/N_m P_C2H6=y_C2H6*P

14-41E The Orsat analysis (molar fractions) of components of a gas mixture are given. The mass flow rate of the mixture is to be determined. Properties The molar masses of CO2 , O2 , N 2 , and CO are 44.0, 32.0, 28.0, and 28.0 lbm / lbmol , respectively (Table A-1E).

13-107

Analysis The molar fraction of N 2 is

yN2 = 1- yCO2 - yO2 - yCO = 1- 0.15 - 0.15 - 0.01 = 0.69 The molar mass of the mixture is determined from

M m = yCO2 M CO2 + yO2 M O2 + yCO M CO + yN2 M N2 = 0.15´ 44 + 0.15´ 32 + 0.01´ 28 + 0.69´ 28 = 31.00 lbm/lbmol

The specific volume of the mixture is

RuT (10.73 psia ×ft 3 /lbmol ×R)(660 R) = = 15.54 ft 3 /lbm MmP (31.00 lbm/lbmol)(14.7 psia)

The mass flow rate of these gases is then m=

AV (10 ft 2 )(20 ft/s) = = 12.87 lbm / s v 15.54 ft 3 /lbm

14-42 The mole fractions of components of a gas mixture are given. The mass flow rate of the mixture is to be determined. Properties The molar masses of air and CH4 are 28.97 and 16.0 kg / kmol , respectively (Table A-1). Analysis The molar fraction of air is

yair = 1- yCH4 = 1- 0.15 = 0.85 The molar mass of the mixture is determined from

M m = yCH4 M CH4 + yair M air = 0.15´ 16 + 0.85´ 28.97

= 27.02 kg/kmol

13-108

Given the engine displacement and speed and assuming that this is a 4-stroke engine (2 revolutions per cycle), the volume flow rate is determined from

nVd (3000 rev/min)(0.005 m3 ) = = 7.5 m3 /min 2 2 rev/cycle

The specific volume of the mixture is

RuT (8.314 kPa ×m3 /kmol ×K)(293 K) = = 1.127 m3 /kg MmP (27.02 kg/kmol)(80 kPa)

Hence the mass flow rate is

V 7.5 m3 /min = = 6.65 kg / min v 1.127 m3 /kg

EES (Engineering Equation Solver) SOLUTION "Given" y_CH4=0.15 n_dot=3000 [1/min] Vol_d=0.005 [m^3] P=80 [kPa] T=(20+273) [K] "Properties" R_u=8.314 [kJ/kmol-K] M_CH4=16 [kg/kmol] M_air=28.97 [kg/kmol] "Analysis" y_air=1-y_CH4 M_m=y_CH4*M_CH4+y_air*M_air Vol_dot=(n_dot*Vol_d)/2 v=(R_u*T)/(M_m*P) m_dot=Vol_dot/v

14-43 The masses, temperatures, and pressures of two gases contained in two tanks connected to each other are given. The valve connecting the tanks is opened and the final temperature is measured. The volume of each tank and the final pressure are to be determined. Assumptions Under specified conditions both N 2 and O2 can be treated as ideal gases, and the mixture as an ideal gas mixture Properties The molar masses of N 2 and O2 are 28.0 and 32.0 kg / kmol , respectively. The gas constants of N 2 and O2 are 0.2968 and 0.2598 kPa  m3 / kg  K , respectively (Table A-1). Analysis The volumes of the tanks are

13-109

3 æmRT ö ÷ = (2 kg)(0.2968 kPa ×m /kg ×K)(298 K) = 0.322 m 3 VN2 = çç ÷ ÷ çè P øN 550 kPa 2

æmRT ö (4 kg)(0.2598 kPa ×m3 /kg ×K)(298 K) ÷ VO2 = çç = = 2.065 m 3 ÷ çè P ÷ øO2 150 kPa

Vtotal = VN2 + VO2 = 0.322 m3 + 2.065 m3 = 2.386 m3

Also,

N N2 = N O2 =

mN2 M N2 mO2 M O2

2 kg = 0.07143 kmol 28 kg/kmol

4 kg = 0.125 kmol 32 kg/kmol

N m = N N2 + N O2 = 0.07143 kmol + 0.125 kmol = 0.1964 kmol Thus,

æNR T ÷ ö (0.1964 kmol)(8.314 kPa ×m3 /kmol ×K)(298 K) Pm = çç u ÷ = = 204 kPa ÷ çè V øm 2.386 m3

EES (Engineering Equation Solver) SOLUTION "Given" mass_N2=2 [kg] T_N2=(25+273) [K] P_N2=550 [kPa] mass_O2=4 [kg] T_O2=(25+273) [K] P_O2=150 [kPa] T_m=(25+273) [K] "Properties" R_u=8.314 [kJ/kmol-K] M_N2=28 [kg/kmol] M_O2=32 [kg/kmol] R_N2=0.2968 [kJ/kg-K] R_O2=0.2598 [kJ/kg-K] "Analysis" V_N2=mass_N2*R_N2*T_N2/P_N2 V_O2=mass_O2*R_O2*T_O2/P_O2 V_m=V_N2+V_O2 N_N2=mass_N2/M_N2 N_O2=mass_O2/M_O2 N_m=N_N2+N_O2 P_m=N_m*R_u*T_m/V_m

13-110

14-44 The volumes, temperatures, and pressures of two gases forming a mixture are given. The volume of the mixture is to be determined using three methods. Analysis (a) Under specified conditions both O2 and N 2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas, æPV ÷ ö (8000 kPa)(0.3 m3 ) ÷ N O2 = ççç = = 1.443 kmol ÷ (8.314 kPa ×m3 /kmol ×K)(200 K) èç Ru T ÷ øO 2

æPV ö (8000 kPa)(0.5 m3 ) ÷ ÷ N N2 = ççç = = 2.406 kmol ÷ çè RuT ÷ (8.314 kPa ×m3 /kmol ×K)(200 K) øN 2

N m = N O2 + N N2 = 1.443 kmol + 2.406 kmol

= 3.849 kmol Vm =

N m RuTm (3.849 kmol)(8.314 kPa ×m3 /kmol ×K)(200 K) = = 0.8 m 3 Pm 8000 kPa

(b) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of O2 and N 2 from Table A-1. But we first need to determine the Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15), T 200 K ïüï TR , O2 = m = = 1.292 ïï Tcr, O2 154.8 K ïï O2 : ý Z O2 = 0.77 ï Pm 8 MPa PR , O2 = = = 1.575 ïïï Pcr, O2 5.08 MPa ïïþ

üï ïï ïï ïý Z = 0.863 N2 : N2 ï Pm 8 MPa PR , N2 = = = 2.360 ïïï Pcr, N2 3.39 MPa ïïþ TR , N2 =

Tm 200 K = = 1.585 Tcr, N2 126.2 K

æ PV ö÷ (8000 kPa)(0.3 m3 ) ÷ N O2 = ççç = = 1.874 kmol ÷ (0.77)(8.314 kPa ×m3 /kmol ×K)(200 K) èç ZRuT ø÷O 2

æ PV ÷ ö (8000 kPa)(0.5 m3 ) ÷ N N2 = ççç = = 2.787 kmol ÷ çè ZRuT ÷ (0.863)(8.314 kPa ×m3 /kmol ×K)(200 K) øN 2

N m = N O2 + N N2 = 1.874 kmol + 2.787 kmol = 4.661 kmol The mole fractions are

13-111

N O2

yO2 =

yN2 =

N N2 Nm

1.874 kmol = 0.402 4.661 kmol

2.787 kmol = 0.598 4.661 kmol

Tcr,¢ m = å yiTcr, i = yO2 Tcr, O2 + yN2 Tcr, N2 = (0.402)(154.8 K) + (0.598)(126.2 K) = 137.7 K

Pcr,¢ m = å yi Pcr, i = yO2 Pcr, O2 + yN2 Pcr, N2 = (0.402)(5.08 MPa) + (0.598)(3.39 MPa) = 4.07 MPa Then,

üï ïï ïï ïý Z = 0.82 (Fig. A-15) m ï Pm 8 MPa PR = ' = = 1.966 ïïï 4.07 MPa Pcr, O2 ïïþ Thus, TR =

Tm 200 K = = 1.452 ' Tcr, O2 137.7 K

Vm =

Z m N m RuTm (0.82)(4.661 kmol)(8.314 kPa ×m3 /kmol ×K)(200 K) = = 0.79 m 3 Pm 8000 kPa

(c) To use the Amagat’s law for this real gas mixture, we first need the Z of each component at the mixture temperature and pressure, which are determined in part (b). Then,

Z m = å yi Zi = yO2 ZO2 + yN2 Z N2 = (0.402)(0.77)+ (0.598)(0.863) = 0.83 Thus,

Vm =

Z m N m RuTm (0.83)(4.661 kmol)(8.314 kPa ×m3 /kmol ×K)(200 K) = = 0.80 m 3 Pm 8000 kPa

EES (Engineering Equation Solver) SOLUTION "Given" V_O2=0.3 [m^3] T_O2=200 [K] P_O2=8000 [kPa] V_N2=0.5 [m^3] T_N2=200 [K] P_N2=8000 [kPa] T_m=200 [K] P_m=8000 [kPa] "Properties" R_u=8.314 [kJ/kmol-K] T_cr_O2=T_CRIT(O2) P_cr_O2=P_CRIT(O2) T_cr_N2=T_CRIT(N2) P_cr_N2=P_CRIT(N2) "Analysis" "(a)" N_O2_a=(P_O2*V_O2)/(R_u*T_O2) N_N2_a=(P_N2*V_N2)/(R_u*T_N2) N_m_a=N_O2_a+N_N2_a V_m_a=N_m_a*R_u*T_m/P_m "(b)" T_R_O2=T_m/T_cr_O2

13-112

P_R_O2=P_m/P_cr_O2 Z_O2=COMPRESS(T_R_O2, P_R_O2) "the function that returns compressibility factor" T_R_N2=T_m/T_cr_N2 P_R_N2=P_m/P_cr_N2 Z_N2=COMPRESS(T_R_N2, P_R_N2) "the function that returns compressibility factor" N_O2_b=(P_O2*V_O2)/(Z_O2*R_u*T_O2) N_N2_b=(P_N2*V_N2)/(Z_N2*R_u*T_N2) N_m_b=N_O2_b+N_N2_b y_O2=N_O2_b/N_m_b y_N2=N_N2_b/N_m_b T_cr_m=y_O2*T_cr_O2+y_N2*T_cr_N2 P_cr_m=y_O2*P_cr_O2+y_N2*P_cr_N2 T_R_m=T_m/T_cr_m P_R_m=P_m/P_cr_m Z_m=COMPRESS(T_R_m, P_R_m) "the function that returns compressibility factor" V_m_b=Z_m*N_m_b*R_u*T_m/P_m "(c)" Z_m_c=y_O2*Z_O2+y_N2*Z_N2 V_m_c=Z_m_c*N_m_b*R_u*T_m/P_m

14-45E The volumetric analysis of a mixture of gases is given. The volumetric and mass flow rates are to be determined using three methods. Properties The molar masses of O2 , N 2 , CO2 , and CH4 are 32.0, 28.0, 44.0, and 16.0 lbm / lbmol , respectively (Table A1E). Analysis (a) We consider 100 lbmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are mO2 = NO2 M O2 = (30 lbmol)(32 lbm/lbmol) = 960 lbm

mN2 = N N2 M N2 = (40 lbmol)(28 lbm/lbmol) = 1120 lbm mCO2 = NCO2 M CO2 = (10 lbmol)(44 lbm/lbmol) = 440 lbm mCH4 = NCH4 M CH4 = (20 lbmol)(16 lbm/lbmol) = 320 lbm

The total mass is mm = mO2 + mN2 + mCO2 + mCH4 = 960 + 1120 + 440 + 320 = 2840 lbm

The apparent molecular weight of the mixture is Mm =

mm 2840 lbm = = 28.40 lbm/lbmol N m 100 lbmol

13-113

Ru 10.73 psia ×ft 3 /lbmol ×R = = 0.3778 psia ×ft 3 /lbm ×R Mm 28.40 lbm/lbmol

The specific volume of the mixture is

RT (0.3778 psia ×ft 3 /lbm ×R)(530 R) = = 0.1335 ft 3 /lbm P 1500 psia

The volume flow rate is

p D2 p (1/12 ft)2 V= (10 ft/s) = 0.05454 ft 3 / s 4 4 and the mass flow rate is V = AV =

V 0.05454 ft 3 /s = = 0.4085 lbm / s v 0.1335 ft 3 /lbm (b) To use the Amagat’s law for this real gas mixture, we first need the mole fractions and the Z of each component at the mixture temperature and pressure. The compressibility factors are obtained using Fig. A-15 to be T 530 R ïü ïï TR , O2 = m = = 1.902 ïï Tcr, O2 278.6 R ïý Z = 0.94 O2 ï Pm 1500 psia PR , O2 = = = 2.038 ïïï Pcr, O2 736 psia ïïþ ïü 530 R ïï TR , N2 = = 2.334 ïï 227.1 R ïï ý Z N2 = 0.99 ïï ï 1500 psia PR , CN = = 3.049 ïï ïïþ 492 psia m=

TR , CO2 =

530 R = 0.968 547.5 R

1500 psia PR , CO2 = = 1.401 1071 psia TR , CH4 =

PR , CH4 =

530 R = 1.541 343.9 R 1500 psia = 2.229 673 psia

ü ïï ïï ïï ïý ïï ïï ïï ïþ

Z CO2 = 0.21

ü ïï ïï ïï ïý ïï ïï ïï ïþ

Z CO2 = 0.85

and Z m = å yi Z i = yO2 Z O2 + yO2 Z O2 + yCO2 Z CO2 + yCH4 Z CH4

= (0.30)(0.94) + (0.40)(0.99) + (0.10)(0.21) + (0.20)(0.85) = 0.869 Then,

13-114

Z m RT (0.869)(0.3778 psia ×ft 3 /lbm ×R)(530 R) = = 0.1160 ft 3 /lbm P 1500 psia

V = 0.05454 ft 3 / s V 0.05454 ft 3 /s = = 0.4702 lbm / s v 0.1160 ft 3 /lbm (c) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of mixture gases. Tcr,¢ m = å yiTcr, i = yO2Tcr, O2 + yN2Tcr, N2 + yCO2Tcr, CO2 + yCH4Tcr, CH4 m=

= (0.30)(278.6 R) + (0.40)(227.1 R) + (0.10)(547.5 R) + (0.20)(343.9 R) = 298.0 R Pcr,¢ m = å yi Pcr, i = yO2 Pcr, O2 + + yN2 Pcr, N2 + yCO2 Pcr, CO2 + yCH4 Pcr, CH4

= (0.30)(736 psia) + (0.40)(492 psia) + (0.10)(1071 psia) + (0.20)(673 psia) = 659.3 psia and ü ïï ïï ïï ý Z m = 0.915 ï Pm 1500 psia PR = ' = = 2.275 ïïï 659.3 psia Pcr, m ïïþ Then, TR =

Tm 530 R = = 1.779 298.0 R Tcr,' m

(Fig. A-15)

Z m RT (0.915)(0.3778 psia ×ft 3 /lbm ×R)(530 R) = = 0.1221 ft 3 /lbm P 1500 psia

V = 0.05454 ft 3 / s m=

V 0.05454 ft 3 /s = = 0.4467 lbm / s v 0.1221 ft 3 /lbm

14-46 The mole numbers, temperatures, and pressures of two gases forming a mixture are given. The final temperature is also given. The pressure of the mixture is to be determined using two methods. Analysis (a) Under specified conditions both Ar and N 2 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas,

Initial state : P1V1  N1 Ru T1  N 2 T2 (6)(230 K) P1  (3 MPa )  7.39 MPa  P2  Final state : P2V 2  N 2 Ru T2  N1T1 (2)(280 K) (b) Initially,

  Tcr, Ar   Z Ar  0.985 (Fig. A-15 or EES) P1 3 MPa PR    0.6173   Pcr, Ar 4.86 MPa 

TR 



280 K  1.854 151.0 K

13-115

Then the volume of the tank is

V 

ZN Ar Ru T (0.985)(2 kmol)(8.314 kPa  m 3 /kmol  K)(280 K)   1.529 m 3 P 3000 kPa

After mixing,     V m / N Ar v Ar  Ar: v R ,Ar    PR  0.496 Ru Tcr, Ar / Pcr, Ar Ru Tcr, Ar / Pcr, Ar   3 (1.529 m )/(2 kmol)    2 . 96  (8.314 kPa  m 3 /kmol  K)(151.0 K)/(4860 kPa)  TR ,Ar 

Tm 230 K   1.523 Tcr, Ar 151.0 K

(Fig. A-15 or EES)

    v N2 V m / N N2  N 2 : V R,N 2    PR  1.43 (Fig. A-15 or EES) Ru Tcr, N 2 / Pcr, N 2 Ru Tcr, N 2 / Pcr, N 2   3 (1.529 m )/(4 kmol)    1 . 235  (8.314 kPa  m 3 /kmol  K)(126.2 K)/(3390 kPa)  TR , N 2 

Tm 230 K   1.823 Tcr, N 2 126.2 K

Thus,

PAr  (PR Pcr )Ar  (0.496)(4.86 MPa)  2.41 MPa

PN2  (PR Pcr ) N2  (1.43)(3.39 MPa)  4.85 MPa and

Pm  PAr  PN 2  2.41 MPa  4.85 MPa  7.26 MPa

EES (Engineering Equation Solver) SOLUTION "Input Data" R_u = 8.314 [kJ/kmol-K] "universal Gas Constant" T_Ar = 280 [K] P_Ar = 3000 [kPa] "Pressure for only Argon in the tank initially." N_Ar = 2 [kmol] N_N2 = 4 [kmol] T_mix = 230 [K] T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text" P_cr_Ar=4860 [kPa] T_cr_N2=126.2 [K] P_cr_N2=3390 [kPa] "Ideal-gas Solution:" P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank." P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure"

13-116

N_mix=N_Ar + N_N2 "Moles of mixture" "Real Gas Solution:" P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank" T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar" P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar" Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar" P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture" T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture" P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture" Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture" P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture" T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture" P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture" Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture" P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law"

14-47 Problem 14-46 is reconsidered. The effect of the moles of nitrogen supplied to the tank on the final pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibility chart with Dalton's law. Analysis The problem is solved using EES, and the solution is given below. "Input Data" R_u = 8.314 [kJ/kmol-K] "universal Gas Constant" T_Ar = 280 [K] P_Ar = 3000 [kPa] "Pressure for only Argon in the tank initially." N_Ar = 2 [kmol] {N_N2 = 4 [kmol]} T_mix = 230 [K] T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text" P_cr_Ar=4860 [kPa] T_cr_N2=126.2 [K] P_cr_N2=3390 [kPa] "Ideal-gas Solution:" P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank." P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure" N_mix=N_Ar + N_N2 "Moles of mixture" "Real Gas Solution:" P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank" T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar" P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar" Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar" P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture" T_R_Ar_mix=T_mix/T_cr_Ar "Reduced Temp. of Ar in mixture" P_R_Ar_mix=P_Ar_mix/P_cr_Ar "Reduced Press. of Ar in mixture" Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture" P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture" T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture" P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture" Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture" P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law. 23800"

13-117

N N2

Pmix

Pmix,IG

[kmol]

[ kPa]

1 2 3 4 5 6 7 8 9 10

3647 4863 6063 7253 8438 9626 10822 12032 13263 14521

[kPa] 3696 4929 6161 7393 8625 9857 11089 12321 13554 14786

Properties of Gas Mixtures 14-48C Yes. Yes (extensive property).

14-49C No (intensive property).

14-50C The answers are the same for entropy.

14-51C We have to use the partial pressure.

14-52C No, this is an approximate approach. It assumes a component behaves as if it existed alone at the mixture temperature and pressure (i.e., it disregards the influence of dissimilar molecules on each other.)

13-118

14-53 Volumetric fractions of the constituents of a mixture are given. The mixture undergoes an adiabatic compression process. The makeup of the mixture on a mass basis and the internal energy change per unit mass of mixture are to be determined. Assumptions Under specified conditions all CO2 , CO, O2 , and N 2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties 1

The molar masses of CO2 , CO, O2 , and N 2 are 44.0, 28.0, 32.0, and 28.0 kg / kmol , respectively (Table A-1).

2 The process is reversible. Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be

M m = yCO2 M CO2 + yCO M CO + yO2 M O2 + yN2 M N2 = (0.15)(44) + (0.05)(28) + (0.10)(32) + (0.70)(28) = 30.80 kg/kmol

The mass fractions are

mf CO2 = yCO2 mf CO = yCO

mf O2 = yO2 mf N2 = yN2

M CO2 Mm

= (0.15)

44 kg/kmol = 0.2143 30.80 kg/kmol

M CO 28 kg/kmol = (0.05) = 0.0454 Mm 30.80 kg/kmol

M O2 Mm M N2 Mm

= (0.10)

32 kg/kmol = 0.1039 30.80 kg/kmol

= (0.70)

28 kg/kmol = 0.6364 30.80 kg/kmol

The final pressure of mixture is expressed from ideal gas relation to be P2 = Pr 1

T2 T = (100 kPa)(8) 2 = 2.667T2 T1 300 K

(Eq. 1)

since the final temperature is not known. We assume that the process is reversible as well being adiabatic (i.e. isentropic). Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be:

T = 300 K, P = (0.2143´ 100) = 21.43 kPa ¾ ¾ ® sCO2 ,1 = 5.2190 kJ/kg ×K

13-119

T = 300 K, P = (0.1039´ 100) = 10.39 kPa ¾ ¾ ® sN2 ,1 = 6.9485 kJ/kg ×K T = 300 K, P = (0.6364´ 100) = 63.64 kPa ¾ ¾ ® sO2 ,1 = 7.0115 kJ/kg ×K The final state entropies cannot be determined at this point since the final pressure and temperature are not known. However, for an isentropic process, the entropy change is zero and the final temperature and the final pressure may be determined from

D stotal = mf CO2 D sCO2 + mf CO D sCO + mf O2 D sO2 + mf N2 D sN2 = 0 and using Eq. (1). The solution may be obtained using EES to be T2  631.4 K , P2  1684 kPa The initial and final internal energies are (from EES)

uCO2 ,1 = - 8997 kJ/kg T1 = 300 K ¾ ¾ ®

uCC,1 = - 4033 kJ/kg uO2 ,1 = - 76.24 kJ/kg uN2 ,1 = - 87.11 kJ/kg uCO2 ,2 = - 8734 kJ/kg

T2 = 631.4 K ¾ ¾ ®

uCO,2 = - 3780 kJ/kg uO2 ,2 = 156.8 kJ/kg uN2 ,2 = 163.9 kJ/kg

The internal energy change per unit mass of mixture is determined from

D umixture = mf CO2 (uCO2 ,2 - uCO2 ,1 ) + mf CO (uCO,2 - uCO,1 ) + mfO2 (uO2 ,2 - uO2 ,1 ) + mf N 2 (uN 2 ,2 - uN 2 ,1 ) = 0.2143[(- 8734) - (- 8997) ]+ 0.0454[(- 3780) - (- 4033)]

+ 0.1039[156.8 - (- 76.24) ]6 + 0.6364 [163.9 - (- 87.11) ] = 251.8 kJ / kg

EES (Engineering Equation Solver) SOLUTION T_1 = 300 [K] P_1=100 [kPa] r_v=8 "The compression ratio is r_v=V_1/V_2" R_u=8.314 [kJ/kmol-K] "Assume the mixture is an ideal gas mixture. Note that volume fractions equal mole fractions in ideal gas mixtures." "The mole fractions are:" y_CO2 = 0.15 y_CO =0.05 y_O2 = 0.1 y_N2 = 0.7 "The molar mass of the mixture is:" MM_Mix = (y_CO2 *MOLARMASS(CO2)+y_CO *MOLARMASS(CO) + y_O2 *MOLARMASS(O2)+y_N2*MOLARMASS(N2))"[kg/kmol]" R_Mix=R_u/MM_Mix "The mass fractions are:" mf_CO2 = y_CO2 *MOLARMASS(CO2)/MM_Mix"[kg_CO2/kg_Mix]" mf_CO = y_CO*MOLARMASS(CO)/MM_Mix"[kg_CO/kg_Mix]" mf_O2 = y_O2 *MOLARMASS(O2)/MM_Mix"[kg_O2/kg_Mix]" mf_N2 = y_N2 *MOLARMASS(N2)/MM_Mix"[kg_N2/kg_Mix]" "Use the combined ideal gas law to determine mixture final pressure P_2." P_2=P_1*r_v*T_2/T_1 "Assume the process is reversible as well as adiabatic, therefore, isentorpic:" "Daltons Law is used to determine the partial pressures." © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-120

"The changes in entropy are:" DELTAs_CO2 = entropy(CO2,T=T_2, P=y_CO2*P_2)-entropy(CO2,T=T_1, P=y_CO2*P_1)"[kJ/kg_CO2 - K]" s_CO2_1=entropy(CO2,T=T_1, P=y_CO2*P_1) DELTAs_CO = entropy(CO,T=T_2, P=y_CO*P_2)-entropy(CO,T=T_1, P=y_CO*P_1)"[kJ/kg_CO - K]" s_CO_1=entropy(CO,T=T_1, P=y_CO*P_1) DELTAs_O2 = entropy(O2,T=T_2, P=y_O2*P_2)-entropy(O2,T=T_1, P=y_O2*P_1)"[kJ/kg_O2 - K]" s_O2_1=entropy(O2,T=T_1, P=y_O2*P_1) DELTAs_N2 = entropy(N2,T=T_2, P=y_N2*P_2)-entropy(N2,T=T_1, P=y_N2*P_1)"[kJ/kg_N2 - K]" s_N2_1=entropy(N2,T=T_1, P=y_N2*P_1) "For the isnetropic pocess DELTAs_Mix=0. This allows the determination of T_2." DELTAs_Mix= mf_CO2*DELTAs_CO2+mf_CO*DELTAs_CO +mf_O2*DELTAs_O2 +mf_N2*DELTAs_N2 "[kJ/kg_Mix - K]" DELTAs_Mix=0 "The changes in internal energy are:" DELTAu_CO2 = intenergy(CO2,T=T_2)-intenergy(CO2,T=T_1)"[kJ/kg_CO2]" DELTAu_CO = intenergy(CO,T=T_2)-intenergy(CO,T=T_1)"[kJ/kg_CO]" DELTAu_O2 = intenergy(O2,T=T_2)-intenergy(O2,T=T_1)"[kJ/kg_O2]" DELTAu_N2 = intenergy(N2,T=T_2)-intenergy(N2,T=T_1)"[kJ/kg_N2]" u_CO2_1= intenergy(CO2,T=T_1) u_CO2_2= intenergy(CO2,T=T_2) u_CO_1= intenergy(CO,T=T_1) u_CO_2= intenergy(CO,T=T_2) u_O2_1= intenergy(O2,T=T_1) u_O2_2= intenergy(O2,T=T_2) u_N2_1= intenergy(N2,T=T_1) u_N2_2= intenergy(N2,T=T_2) DELTAu_Mix = mf_CO2*DELTAu_CO2+mf_CO*DELTAu_CO +mf_O2*DELTAu_O2 +mf_N2*DELTAu_N2 "[kJ/kg_Mix]"

14-54 The volume fractions of components of a gas mixture are given. This mixture is heated while flowing through a tube at constant pressure. The heat transfer to the mixture per unit mass of the mixture is to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of O2 , N 2 , CO2 , and CH4 are 32.0, 28.0, 44.0, and 16.0 kg / kmol , respectively (Table A1). The constant-pressure specific heats of these gases at room temperature are 0.918, 1.039, 0.846, and 2.2537 kJ / kg  K , respectively (Table A-2a). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are

mO2 = NO2 M O2 = (30 kmol)(32 kg/kmol) = 960 kg mN2 = N N2 M N2 = (40 kmol)(28 kg/kmol) = 1120 kg mCO2 = NCO2 M CO2 = (10 kmol)(44 kg/kmol) = 440 kg mCH4 = NCH4 M CH4 = (20 kmol)(16 kg/kmol) = 320 kg The total mass is

mm = mO2 + mN2 + mCO2 + mCH4 = 960 + 1120 + 440 + 320

= 2840 kg

13-121

Then the mass fractions are mf O2 =

mO2 960 kg = = 0.3380 mm 2840 kg

mf N2 =

mN2 1120 kg = = 0.3944 mm 2840 kg

mf CO2 =

mCO2 440 kg = = 0.1549 mm 2840 kg

mf CH4 =

mCH4 320 kg = = 0.1127 mm 2840 kg

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf N2 c p ,N2 + mf CO2 c p ,CO2 + mf CH4 c p ,CH4

= 0.3380´ 0.918 + 0.3944´ 1.039 + 0.1549´ 0.846 + 0.1127´ 2.2537

= 1.1051 kJ/kg ×K An energy balance on the tube gives

qin = c p (T2 - T1 ) = (1.1051 kJ/kg ×K)(200 - 20) K = 199 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" y_O2=0.30 y_N2=0.40 y_CO2=0.10 y_CH4=0.20 T1=(20+273) [K] T2=(200+273) [K] P=150 [kPa] "Properties" R_u=8.314 [kJ/kmol-K] M_O2=32 [kg/kmol] M_N2=28 [kg/kmol] M_CO2=44 [kg/kmol] M_CH4=16 [kg/kmol] c_p_O2=0.918 [kJ/kg-K] c_p_N2=1.039 [kJ/kg-K] c_p_CO2=0.846 [kJ/kg-K] c_p_CH4=2.2537 [kJ/kg-K] "Analysis" N_m=100 [kmol] "We consider 100 kmol of mixture" N_O2=y_O2*N_m N_N2=y_N2*N_m N_CO2=y_CO2*N_m N_CH4=y_CH4*N_m © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-122

mass_O2=N_O2*M_O2 mass_N2=N_N2*M_N2 mass_CO2=N_CO2*M_CO2 mass_CH4=N_CH4*M_CH4 mass_m=mass_O2+mass_N2+mass_CO2+mass_CH4 mf_O2=mass_O2/mass_m mf_N2=mass_N2/mass_m mf_CO2=mass_CO2/mass_m mf_CH4=mass_CH4/mass_m c_p=mf_O2*c_p_O2+mf_N2*c_p_N2+mf_CO2*c_p_CO2+mf_CH4*c_p_CH4 q_in=c_p*(T2−T1)

14-55 A mixture of nitrogen and carbon dioxide is heated at constant pressure in a closed system. The work produced is to be determined. Assumptions 1 Nitrogen and carbon dioxide are ideal gases. 2 The process is reversible. Properties The mole numbers of nitrogen and carbon dioxide are 28.0 and 44.0 kg/kmol, respectively (Table A-1). Analysis One kg of this mixture consists of 0.5 kg of nitrogen and 0.5 kg of carbon dioxide or 0.5 kg/28.0 kg/kmol=0.01786 kmol of nitrogen and 0.5 kg / 44.0 kg / kmol  0.01136 kmol of carbon dioxide. The constituent mole fraction are then yN2 =

N N2 0.01786 kmol = = 0.6112 N total 0.02922 kmol

yCO2 =

N CO2 0.01136 kmol = = 0.3888 N total 0.02922 kmol

The effective molecular weight of this mixture is

M = yN2 M N2 + yCO2 M CO2 = (0.6112)(28) + (0.3888)(44) = 34.22 kg/kmol The work done is determined from 2

w = ò PdV = P2v 2 - P1v1 = R (T2 - T1 ) 1

Ru 8.314 kJ/kmol ×K (T2 - T1 ) = (200 - 30)K = 41.3 kJ / kg M 34.22 kg/kmol

13-123

14-56E The mass fractions of components of a gas mixture are given. This mixture is compressed in an isentropic process. The final mixture temperature and the work required per unit mass of the mixture are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N 2 , He, CH4 , and C2 H6 are 28.0, 4.0, 16.0, and 30.0 lbm / lbmol , respectively (Table A1E). The constant-pressure specific heats of these gases at room temperature are 0.248, 1.25, 0.532, and 0.427 Btu / lbm  R , respectively (Table A-2Ea). Analysis We consider 100 lbm of this mixture. The mole numbers of each component are N N2 =

mN2 15 lbm = = 0.5357 lbmol M N2 28 lbm/lbmol

N He =

mHe 5 lbm = = 1.25 lbmol M He 4 lbm/lbmol

N CH4 =

mCH4 60 lbm = = 3.75 lbmol M CH4 16 lbm/lbmol

N C2H6 =

mC2H6 20 lbm = = 0.6667 lbmol M C2H6 30 lbm/lbmol

The mole number of the mixture is

Nm = N N2 + NHe + NCH4 + NC2H6 = 0.5357 + 1.25 + 3.75 + 0.6667 = 6.2024 lbmol The apparent molecular weight of the mixture is Mm =

mm 100 lbm = = 16.12 lbm/lbmol N m 6.2024 lbmol

The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p ,N2 + mf He c p ,He + mf CH4 c p ,CH4 + mf C2H6 c p ,C2H6

= 0.15´ 0.248 + 0.05´ 1.25 + 0.60´ 0.532 + 0.20´ 0.427 = 0.5043 Btu/lbm ×R

The apparent gas constant of the mixture is R=

Ru 1.9858 Btu/lbmol ×R = = 0.1232 Btu/lbm ×R Mm 16.12 lbm/lbmol

Then the constant-volume specific heat is

cv = c p - R = 0.5043 - 0.1232 = 0.3811 Btu/lbm ×R The specific heat ratio is

cp cv

0.5043 = 1.323 0.3811

13-124

( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P1 ÷ ø

0.323/1.323

æ200 psia ÷ ö ÷ = (560 R) ççç ÷ è 20 psia ÷ ø

= 982 R

An energy balance on the adiabatic compression process gives

win = c p (T2 - T1 ) = (0.5043 Btu/lbm ×R)(982 - 560) R = 213 Btu / lbm

14-57 The masses of components of a gas mixture are given. This mixture is heated at constant pressure. The change in the volume of the mixture and the total heat transferred to the mixture are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of O2 , CO2 , and He are 32.0, 44.0, and 4.0 kg / kmol , respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 0.846, and 5.1926 kJ / kg  K , respectively (Table A-2a). Analysis The total mass of the mixture is

mm = mO2 + mCO2 + mHe = 0.1+ 1+ 0.5 = 1.6 kg

The mole numbers of each component are N O2 =

mO2 0.1 kg = = 0.003125 kmol M O2 32 kg/kmol

N CO2 =

mCO2 1 kg = = 0.02273 kmol M CO2 44 kg/kmol

N He =

mHe 0.5 kg = = 0.125 kmol M He 4 kg/kmol

The mole number of the mixture is Nm = NO2 + NCO2 + NHe = 0.003125 + 0.02273 + 0.125 = 0.15086 kmol The apparent molecular weight of the mixture is Mm =

mm 1.6 kg = = 10.61 kg/kmol N m 0.15086 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ×K = = 0.7836 kJ/kg ×K Mm 10.61 kg/kmol

The mass fractions are mf O2 =

mO2 0.1 kg = = 0.0625 mm 1.6 kg

13-125

mf CO2 =

mCO2 1 kg = = 0.625 mm 1.6 kg

mf He =

mHe 0.5 kg = = 0.3125 mm 1.6 kg

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf CO2 c p ,CO2 + mf He c p ,He

= 0.0625´ 0.918 + 0.625´ 0.846 + 0.3125´ 5.1926 = 2.209 kJ/kg ×K The change in the volume of this ideal gas mixture is mm RD T (1.6 kg)(0.7836 kPa ×m3 /kg ×K)(260 - 10) K = = 0.8955 m 3 P 350 kPa The heat transfer is determined to be D Vm =

qin = c p (T2 - T1 ) = (2.209 kJ/kg ×K)(260 - 10) K = 552 kJ / kg

14-58 The states of two gases contained in two tanks are given. The gases are allowed to mix to form a homogeneous mixture. The final pressure, the heat transfer, and the entropy generated are to be determined. Assumptions 1

Under specified conditions both O2 and N 2 can be treated as ideal gases, and the mixture as an ideal gas mixture.

2 3

The tank containing oxygen is insulated. There are no other forms of work involved.

Properties The constant volume specific heats of O2 and N 2 are 0.658 kJ / kg  C and 0.743 kJ / kg  C , respectively. (Table A-2). Analysis (a) The volume of the O2 tank and mass of the nitrogen are 3 æmRT1 ö ÷ = (1 kg)(0.2598 kPa ×m /kg ×K)(288 K) = 0.25 m3 ÷ V1, O2 = ççç ÷ ÷ çè P1 ø 300 kPa O2

æPV ö (500 kPa)(2 m3 ) ÷ mN2 = ççç 1 1 ÷ = = 10.43 kg ÷ (0.2968 kPa ×m3 /kg ×K)(323 K) èç RT1 ÷ øN 2

Vtotal = V1, O2 + V1, N2 = 0.25 m3 + 2.0 m3 = 2.25 m3

Also,

mO2 = 1 kg ¾ ¾ ® N O2 =

mO2 M O2

1 kg = 0.03125 kmol 32 kg/kmol

13-126

mN2 = 10.43 kg ¾ ¾ ® N N2 =

mN2 M N2

10.43 kg = 0.3725 kmol 28 kg/kmol

N m = N N2 + N O2 = 0.3725 kmol + 0.03125 kmol = 0.40375 kmol Thus,

æNR T ö (0.40375 kmol)(8.314 kPa ×m3 /kmol ×K)(298 K) Pm = çç u ÷ = = 444.6 kPa ÷ çè V ÷ øm 2.25 m3 (b) We take both gases as the system. No work or mass crosses the system boundary, and thus this is a closed system with W = 0. Taking the direction of heat transfer to be from the system (will be verified), the energy balance for this closed system reduces to Ein - Eout = D Esystem

é ù - Qout = D U = D U O2 + D U N2 ¾ ¾ ® Qout = éëmcv (T1 - Tm )ù ûO + ëmcv (T1 - Tm )ûN 2

Using cv values at room temperature (Table A-2), the heat transfer is determined to be Qout = (1 kg )(0.658 kJ/kg ×°C)(15 - 25)°C + (10.43 kg )(0.743 kJ/kg ×°C)(50 - 25)°C

= 187.2 kJ

(from the system)

(c) For and extended system that involves the tanks and their immediate surroundings such that the boundary temperature is the surroundings temperature, the entropy balance can be expressed as Sin - Sout + Sgen = D Ssystem

Qout + Sgen = m(s2 - s1 ) Tb,surr Sgen = m(s2 - s1 ) +

Qout Tsurr

The mole fraction of each gas is N O2

0.03125 = 0.077 Nm 0.40375 N N2 0.3725 yN2 = = = 0.923 Nm 0.40375 yO2 =

Thus,

æ y Pm, 2 ö T ÷ ÷ (s2 - s1 )O2 = çççc p ln 2 - Rln ÷ ÷ çè T1 P1 ø O2 = (0.918 kJ/kg ×K) ln

298 K (0.077)(444.6 kPa) - (0.2598 kJ/kg ×K) ln 288 K 300 kPa

= 0.5952 kJ/kg ×K æ y Pm, 2 ö T ÷ ÷ ( s2 - s1 ) N2 = çççc p ln 2 - Rln ÷ ÷ çè T1 P1 ø N2 = (1.039 kJ/kg ×K) ln

298 K (0.923)(444.6 kPa) - (0.2968 kJ/kg ×K) ln 323 K 500 kPa

= - 0.0251 kJ/kg ×K Substituting,

Sgen = (1 kg)(0.5952 kJ/kg ×K )+ (10.43 kg )(- 0.0251 kJ/kg ×K )+

187.2 kJ = 0.962 kJ / K 298 K

13-127

EES (Engineering Equation Solver) SOLUTION "Given" mass_O2=1 [kg] T_O2=(15+273) [K] P_O2=300 [kPa] V_N2=2 [m^3] T_N2=(50+273) [K] P_N2=500 [kPa] T_m=(25+273) [K] T0=(25+273) [K] "Properties" R_u=8.314 [kPa-m^3/kmol-K] M_O2=molarmass(O2) M_N2=molarmass(N2) R_O2=R_u/M_O2 R_N2=R_u/M_N2 C_v_O2=CV(O2, T=T_ave_O2) "at the average temperature" C_p_O2=CP(O2, T=T_ave_O2) "at the average temperature" T_ave_O2=1/2*(T_O2+T_m) C_v_N2=CV(N2, T=T_ave_N2) "at the average temperature" C_p_N2=CP(N2, T=T_ave_N2) "at the average temperature" T_ave_N2=1/2*(T_N2+T_m) "Analysis" "(a)" V_O2=mass_O2*R_O2*T_O2/P_O2 mass_N2=(P_N2*V_N2)/(R_N2*T_N2) V_m=V_O2+V_N2 N_O2=mass_O2/M_O2 N_N2=mass_N2/M_N2 N_m=N_O2+N_N2 P_m=N_m*R_u*T_m/V_m "(b)" Q_out=mass_O2*C_v_O2*(T_O2−T_m)+mass_N2*C_v_N2*(T_N2−T_m) "(c)" y_O2=N_O2/N_m y_N2=N_N2/N_m DELTAs_O2=C_p_O2*ln(T_m/T_O2)-R_O2*ln(y_O2*P_m/P_O2) DELTAs_N2=C_p_N2*ln(T_m/T_N2)-R_N2*ln(y_N2*P_m/P_N2) S_gen=mass_O2*DELTAs_O2+mass_N2*DELTAs_N2+Q_out/T0

14-59 Problem 14-58 is reconsidered. The results obtained assuming ideal gas behavior with constant specific heats at the average temperature, and using real gas data obtained from appropriate software by assuming variable specific heats over the temperature range are to be compared. Analysis The problem is solved using EES, and the solution is given below. T_O2[1] =15 [C] T_N2[1] =50 [C] T[2] =25 [C] T_o = 25 [C] m_O2 = 1 [kg] P_O2[1]=300 [kPa] V_N2[1]=2 [m^3] P_N2[1]=500 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-128

R_u=8.314 [kJ/kmol-K] MM_O2=molarmass(O2) MM_N2=molarmass(N2) P_O2[1]*V_O2[1]=m_O2*R_u/MM_O2*(T_O2[1]+273) P_N2[1]*V_N2[1]=m_N2*R_u/MM_N2*(T_N2[1]+273) V_total=V_O2[1]+V_N2[1] N_O2=m_O2/MM_O2 N_N2=m_N2/MM_N2 N_total=N_O2+N_N2 P[2]*V_total=N_total*R_u*(T[2]+273) P_Final =P[2] E_in - E_out = DELTAE_sys "Conservation of energy for the combined system" E_in = 0 [kJ] E_out = Q DELTAE_sys=m_O2*(intenergy(O2,T=T[2]) - intenergy(O2,T=T_O2[1])) + m_N2*(intenergy(N2,T=T[2]) intenergy(N2,T=T_N2[1])) P_O2[2]=P[2]*N_O2/N_total P_N2[2]=P[2]*N_N2/N_total - Q/(T_o+273) + S_gen = DELTAS_O2 + DELTAS_N2 "Entropy generation" DELTAS_O2 = m_O2*(entropy(O2,T=T[2],P=P_O2[2]) - entropy(O2,T=T_O2[1],P=P_O2[1])) DELTAS_N2 = m_N2*(entropy(N2,T=T[2],P=P_N2[2]) - entropy(N2,T=T_N2[1],P=P_N2[1])) "Constant Property Solution" -Q_ConstP=m_O2*cv_O2*(T[2]-T_O2[1])+m_N2*cv_N2*(T[2]-T_N2[1]) Tav_O2 =(T[2]+T_O2[1])/2 cv_O2 = SPECHEAT(O2,T=Tav_O2) - R_u/MM_O2 Tav_N2 =(T[2]+T_N2[1])/2 cv_N2 = SPECHEAT(N2,T=Tav_N2) - R_u/MM_N2 - Q_ConstP/(T_o+273) + S_gen_ConstP = DELTAS_O2_ConstP + DELTAS_N2_ConstP DELTAS_O2_ConstP = m_O2*( SPECHEAT(O2,T=Tav_O2)*LN((T[2]+273)/(T_O2[1]+273))R_u/MM_O2*LN(P_O2[2]/P_O2[1])) DELTAS_N2_ConstP = m_N2*( SPECHEAT(N2,T=Tav_N2)*LN((T[2]+273)/(T_N2[1]+273))R_u/MM_N2*LN(P_N2[2]/P_N2[1])) SOLUTION cv_N2=0.741 [kJ/kg-K] cv_O2=0.6528 [kJ/kg-K] DELTAE_sys=-186.7 [kJ] DELTAS_N2=-0.2581 [kJ/K] DELTAS_N2_ConstP=-0.2589 [kJ/K] DELTAS_O2=0.5937 [kJ/K] DELTAS_O2_ConstP=0.5937 [kJ/K] E_in=0 [kJ] E_out=186.7 [kJ] MM_N2=28.01 [kg/kmol] MM_O2=32 [kg/kmol] m_N2=10.43 [kg] m_O2=1 [kg] N_N2=0.3724 [kmol] N_O2=0.03125 [kmol] N_total=0.4036 [kmol] P_Final=444.6 [kPa] Q=186.7 [kJ] Q_ConstP=186.7 [kJ] R_u=8.314 [kJ/kmol-K] S_gen=0.9622 [kJ] S_gen_ConstP=0.9614 [kJ] Tav_N2=37.5 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-129

Tav_O2=20 [C] T_o=25 [C] V_total=2.249 [m^3]

14-60 A rigid tank contains two gases separated by a partition. The pressure and temperature of the mixture are to be determined after the partition is removed. Assumptions 1 We assume both gases to be ideal gases, and their mixture to be an ideal-gas mixture. This assumption is reasonable since both the oxygen and nitrogen are well above their critical temperatures and well below their critical pressures. 2 The tank is insulated and thus there is no heat transfer. 3 There are no other forms of work involved. Properties The constant volume specific heats of O2 and N 2 at room temperature are 0.658 kJ / kg  C and 0.743 kJ / kg  C , respectively (Table A-2a).

Analysis We take the entire contents of the tank (both compartments) as the system. This is a closed system since no mass crosses the boundary during the process. We note that the volume of a rigid tank is constant and thus there is no boundary work done. (a) Noting that there is no energy transfer to or from the tank, the energy balance for the system can be expressed as

Ein - Eout = D Esystem 0 = D U = D U O2 + D U N 2 é ù 0 = éëmcv (Tm - T1 )ù ûO + ëmcv (Tm - T1 )ûN 2

Using cv values at room temperature, the final temperature of the mixture is determined to be

(7 kg)(0.658 kJ/kg ×°C)(Tm - 40)°C + (4 kg )(0.743 kJ/kg ×°C)(Tm - 20)°C=0 Tm = 32.2°C (b) The final pressure of the mixture is determined from the ideal-gas relation

PmVm = Nm RuTm where

N O2 =

N N2 =

mO2 M O2

mN2 M N2

7 kg = 0.219 kmol 32 kg/kmol

4 kg = 0.143 kmol 28 kg/kmol

13-130

æNR T ö (0.219 kmol)(8.314 kPa ×m3 /kmol ×K)(313 K) ÷ VO2 = ççç u 1 ÷ = = 5.70 m3 ÷ ÷ çè P1 ø 100 kPa O2 æNR T ö (0.143 kmol)(8.314 kPa ×m3 /kmol ×K)(293 K) ÷ VN2 = ççç u 1 ÷ = = 2.32 m3 ÷ ÷ çè P1 ø 150 kPa N 2

Vm = VO2 + VN2 = 5.70 m3 + 2.32 m3 = 8.02 m3 Thus,

Pm =

N m RuTm (0.362 kmol)(8.314 kPa ×m3 /kmol ×K)(305.2 K) = = 114.5 kPa Vm 8.02 m3

Discussion We could also determine the mixture pressure by using PmVm = Nm RuTm where Rm is the apparent gas constant of the mixture. This would require a knowledge of mixture composition in terms of mass or mole fractions.

14-61 The mass fractions of components of a gas mixture are given. This mixture is compressed in a reversible, isothermal, steady-flow compressor. The work and heat transfer for this compression per unit mass of the mixture are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of CH4 , C3 H8 , and C4 H10 are 16.0, 44.0, and 58.0 kg / kmol , respectively (Table A-1). Analysis The mole numbers of each component are N CH4 =

mCH4 60 kg = = 3.75 kmol M CH4 16 kg/kmol

N C3H8 =

mC3H8 25 kg = = 0.5682 kmol M C3H8 44 kg/kmol

N C4H10 =

mC4H10 15 kg = = 0.2586 kmol M C4H10 58 kg/kmol

The mole number of the mixture is Nm = NCH4 + NC3H8 + NC4H10 = 3.75 + 0.5682 + 0.2586 = 4.5768 kmol

The apparent molecular weight of the mixture is Mm =

mm 100 kg = = 21.85 kg/kmol N m 4.5768 kmol

The apparent gas constant of the mixture is

13-131

Ru 8.314 kJ/kmol ×K = = 0.3805 kJ/kg ×K Mm 21.85 kg/kmol

For a reversible, isothermal process, the work input is æP ö æ800 kPa ö ÷= 232 kJ / kg ÷ win = RT ln ççç 2 ÷ = (0.3805 kJ/kg ×K)(293 K)ln çç ÷ ÷ ÷ èç100 kPa ø èç P1 ÷ ø

An energy balance on the control volume gives

Ein - Eout

Rate of net energy transfer by heat, work, and mass

D EsystemZ 0 (steady)

= 0

Rate of change in internal, kinetic, potential, etc. energies

Ein = Eout

mh1 + Win = mh2 + Qout Win - Qout = m(h2 - h1 ) win - qout = c p (T2 - T1 ) = 0

since T2 = T1

win = qout That is,

qout = win = 232 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" mf_CH4=0.60 mf_C3H8=0.25 mf_C4H10=0.15 P1=100 [kPa] T=(20+273) [K] P2=800 [kPa] "Properties" R_u=8.314 [kJ/kmol-K] M_CH4=16 [kg/kmol] M_C3H8=44 [kg/kmol] M_C4H10=58 [kg/kmol] "Analysis" mass_m=100 [kg] "100 kg mass considered" mass_CH4=mf_CH4*mass_m mass_C3H8=mf_C3H8*mass_m mass_C4H10=mf_C4H10*mass_m N_CH4=mass_CH4/M_CH4 N_C3H8=mass_C3H8/M_C3H8 N_C4H10=mass_C4H10/M_C4H10 N_m=N_CH4+N_C3H8+N_C4H10 M_m=mass_m/N_m R=R_u/M_m w_in=R*T*ln(P2/P1) q_out=w_in "isothermal process"

14-62E A gas mixture with known mass fractions is accelerated through a nozzle from a specified state to a specified pressure. For a specified isentropic efficiency, the exit temperature and the exit velocity of the mixture are to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-132

Under specified conditions both N 2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture.

2 3 4

The nozzle is adiabatic and thus heat transfer is negligible. This is a steady-flow process. Potential energy changes are negligible.

Properties The specific heats of N 2 and CO2 are c p ,N2  0.248 Btu / lbm  R , cv ,N2  0.177 Btu / lbm  R , c p ,CO2  0.203 Btu / lbm  R , and cv ,CO2  0.158 Btu / lbm  R . (Table A-2E).

Analysis (a) Under specified conditions both N 2 and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. The c p , cv , and k values of this mixture are determined from

c p, m = å mfi c p , i = mf N2 c p , N2 + mfCO2 c p, CO2

= (0.65)(0.248)+ (0.35)(0.203) = 0.2323 Btu/lbm ×R

cv , m = å mfi cv , i = mf N2 cv , N2 + mf CO2 cv , CO2 = (0.65)(0.177)+ (0.35)(0.158) = 0.1704 Btu/lbm ×R

km =

c p, m cv , m

0.2323 Btu/lbm ×R = 1.363 0.1704 Btu/lbm ×R

Therefore, the N2  CO2 mixture can be treated as a single ideal gas with above properties. Then the isentropic exit temperature can be determined from

æP ö( ÷ T2 s = T1 ççç 2 ÷ ÷ çè P1 ø÷

k - 1)/ k

0.363/1.363

æ12 psia ÷ ö ÷ = (1400 R )çç ÷ çè60 psia ÷ ø

= 911.7 R

From the definition of isentropic efficiency, hN =

h1 - h 2 h1 - h 2 s

c p (T1 - T2 ) c p (T1 - T2 s )

¾¾ ® 0.88 =

1400 - T2 ¾¾ ® T2 = 970.3 R 1400 - 911.7

(b) Noting that, q = w = 0, from the steady-flow energy balance relation,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout 2 1

h1 + V / 2 = h2 + V22 / 2

0 = c p (T2 - T1 )+

V22 - V12 ¿ 0 2

13-133

V2 =

2c p (T1 - T2 ) =

æ25,037 ft 2 /s 2 ö ÷ ÷ 2 (0.2323 Btu/lbm ×R )(1400 - 970.3) R ççç = 2236 ft / s ÷ ÷ è 1 Btu/lbm ø

EES (Engineering Equation Solver) SOLUTION "Given" mf_N2=0.65 mf_CO2=0.35 P1=60 [psia] T1=1400 [R] Vel1=0 [ft/s] P2=12 [psia] eta_N=0.88 "Properties" c_p_N2=0.248 [Btu/lbm-R] c_v_N2=0.177 [Btu/lbm-R] c_p_CO2=0.203 [Btu/lbm-R] c_v_CO2=0.158 [Btu/lbm-R] "Analysis" c_p_m=mf_N2*c_p_N2+mf_CO2*c_p_CO2 c_v_m=mf_N2*c_v_N2+mf_CO2*c_v_CO2 k_m=c_p_m/c_v_m T2_s=T1*(P2/P1)^((k_m-1)/k_m) eta_N=(T1−T2)/(T1−T2_s) 0=c_p_m*(T2−T1)+(Vel2^2−Vel1^2)/2*Convert(ft^2/s^2, Btu/lbm)

14-63E Problem 14-62E is reconsidered. The problem is first to be solved and then, for all other conditions being the same, the problem is to be resolved to determine the composition of the nitrogen and carbon dioxide that is required to have an exit velocity of 2200 ft / s at the nozzle exit. Analysis The problem is solved using EES, and the solution is given below. "Given" mf_N2=0.65 mf_CO2=1−mf_N2 P1=60 [psia] T1=1400 [R] Vel1=0 [ft/s] P2=12 [psia] eta_N=0.88 "Vel2=2200 [ft/s]" "Properties" c_p_N2=0.248 [Btu/lbm–R] c_v_N2=0.177 [Btu/lbm–R] c_p_CO2=0.203 [Btu/lbm–R] c_v_CO2=0.158 [Btu/lbm–R] MM_N2=28 [lbm/lbmol] MM_CO2=44 [lbm/lbmol] "Analysis" c_p_m=mf_N2*c_p_N2+mf_CO2*c_p_CO2 c_v_m=mf_N2*c_v_N2+mf_CO2*c_v_CO2 k_m=c_p_m/c_v_m T2_s=T1*(P2/P1)^((k_m−1)/k_m) eta_N=(T1−T2)/(T1−T2_s) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-134

0=c_p_m*(T2−T1)+(Vel2^2−Vel1^2)/2*Convert(ft^2/s^2, Btu/lbm) N_N2=mf_N2/MM_N2 N_CO2=mf_CO2/MM_CO2 N_total=N_N2+N_CO2 y_N2=N_N2/N_total y_CO2=N_CO2/N_total SOLUTION of the stated problem c_p_CO2=0.203 [Btu/lbm–R] c_p_m=0.2323 [Btu/lbm–R] c_p_N2=0.248 [Btu/lbm–R] c_v_CO2=0.158 [Btu/lbm–R] c_v_m=0.1704 [Btu/lbm–R] c_v_N2=0.177 [Btu/lbm–R] eta_N=0.88 k_m=1.363 mf_CO2=0.35 mf_N2=0.65 MM_CO2=44 [lbm/lbmol] MM_N2=28 [lbm/lbmol] N_CO2=0.007955 N_N2=0.02321 N_total=0.03117 P1=60 [psia] P2=12 [psia] T1=1400 [R] T2=970.3 [R] T2_s=911.7 [R] Vel1=0 [ft/s] Vel2=2236 [ft/s] y_CO2=0.2552 y_N2=0.7448 SOLUTION of the problem with exit velocity of 2200 ft/s c_p_CO2=0.203 [Btu/lbm–R] c_p_m=0.2285 [Btu/lbm–R] c_p_N2=0.248 [Btu/lbm–R] c_v_CO2=0.158 [Btu/lbm–R] c_v_m=0.1688 [Btu/lbm–R] c_v_N2=0.177 [Btu/lbm–R] eta_N=0.88 k_m=1.354 mf_CO2=0.434 mf_N2=0.566 MM_CO2=44 [lbm/lbmol] MM_N2=28 [lbm/lbmol] N_CO2=0.009863 N_N2=0.02022 N_total=0.03008 P1=60 [psia] P2=12 [psia] T1=1400 [R] T2=976.9 [R] T2_s=919.3 [R] Vel1=0 [ft/s] Vel2=2200 [ft/s] y_CO2=0.3279 y_N2=0.6721

14-64 An equimolar mixture of helium and argon gases expands in a turbine. The isentropic work output of the turbine is to be determined. Assumptions 1 Under specified conditions both He and Ar can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 The turbine is insulated and thus there is no heat transfer. 3 This is a steady-flow process. 4 The kinetic and potential energy changes are negligible. Properties The molar masses and specific heats of He and Ar are 4.0 kg / kmol , 40.0 kg / kmol, 5.1926 kJ / kg C , and 0.5203 kJ / kg  C , respectively. (Table A-1 and Table A-2).

Analysis The C p and k values of this equimolar mixture are determined from

13-135

= å y M = y M + y M = 0.5´ 4 + 0.5´ 40 = 22 kg/kmol i i He He Ar Ar m NM yM i i = i i mf = i = i m N M M m m m m M

= å mf c = p, m i p, i =

y M y M He He c + Ar Ar c p, He p, Ar M M m m

0.5´ 4 kg/kmol 0.5´ 40 kg/kmol (5.1926 kJ/kg ×K )+ (0.5203 kJ/kg ×K ) 22 kg/kmol 22 kg/kmol

= 0.945 kJ/kg ×K and

km  1.667 since k = 1.667 for both gases. Therefore, the He-Ar mixture can be treated as a single ideal gas with the properties above. For isentropic processes,

æP ö( ÷ T2 = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

k - 1)/ k

0.667/1.667

æ 200 kPa ÷ ö = (1300 K )çç ÷ èç 2500 kPa ÷ ø

= 473.2 K

From an energy balance on the turbine,

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout

h1 = h2 + wout wout = h1 - h2 wout = c p (T1 - T2 ) = (0.945 kJ/kg ×K )(1300 - 473.2)K = 781.3 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P1=2500 [kPa] T1=1300 [K] P2=200 [kPa] "Properties" M_He=4 M_Ar=40 c_p_He=5.1926 k_He=1.667 c_p_Ar=0.5203 k_Ar=1.667 "Analysis" y_He=0.5 y_Ar=0.5 M_m=y_He*M_He+y_Ar*M_Ar mf_He=(y_He*M_He)/M_m mf_Ar=(y_Ar*M_Ar)/M_m c_p_m=mf_He*c_p_He+mf_Ar*c_p_Ar k_m=mf_He*k_He+mf_Ar*k_Ar T2=T1*(P2/P1)^((k_m−1)/k_m) w_out=c_p_m*(T1−T2)

13-136

14-65 The volumetric fractions of the constituents of a mixture of products of combustion are given. The average molar mass of the mixture, the average specific heat, and the partial pressure of the water vapor in the mixture are to be determined. Assumptions Under specified conditions all N 2 , O2 , H 2 O , and CO2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 , H 2 O , O2 , and N 2 are 44.0, 18.0, 32.0, and 28.0 kg / kmol , respectively (Table A1). The specific heats of CO2 , H 2 O , O2 , and N 2 at 600 K are 1.075, 2.015, 1.003, and 1.075 kJ / kg  K , respectively (Table A-2b). The specific heat of water vapor at 600 K is obtained from EES. Analysis For convenience, consider 100 kmol of mixture. Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the average molar mass of the mixture is determined to be Mm =

N CO2 M CO2 + N H2 O M H2 O + N O2 M O2 + N N2 M N2 N CO2 + N H2 O + N O2 + N N2

(4.89 kmol)(44 kg/kmol) + (6.50)(18) + (12.20)(32) + (76.41)(28) (4.89 + 6.50 + 12.20 + 76.41) kmol

= 28.62 kg / kmol

The average specific heat is determined from

N CO2 c p ,CO2 M CO2 + N H2O c p ,H2 O M H2 O + N O2 c p ,O2 M O2 + N N2 c p ,N2 M N2

c p,m =

N CO2 + N H2 O + N O2 + N N2

(4.89 kmol)(1.075 kJ/kg.K)(44 kg/kmol) + (6.50)(2.015)(18) + (12.20)(1.003)(32) + (76.41)(1.075)(28) (4.89 + 6.50 + 12.20 + 76.41) kmol

= 31.59 kJ / kmol ×K The partial pressure of the water in the mixture is yv =

N H2 O N CO2 + N H2 O + N O2 + N N2

6.50 kmol = 0.0650 (4.89 + 6.50 + 12.20 + 76.41) kmol

Pv = yv Pm = (0.0650)(200 kPa) = 13.0 kPa

EES (Engineering Equation Solver) SOLUTION T_Mix = 600"[K]" P_Mix=200"[kPa]" "Assume the mixture is an ideal gas mixture and there are 100 kmol of mixture. Note that volume fractions equal mole fractions in ideal gas mixtures." "The molar mass of the mixture is:" MM_Mix = (4.89*MOLARMASS(CO2) + 6.50*MOLARMASS(H2O) +12.20*MOLARMASS(O2)+ 76.41*MOLARMASS(N2))/(4.89+6.50+12.20+76.41) "[kg/kmol]" "The average specific heat at constant pressure of the mixture is:"

13-137

Cp_Mix = (4.89*SPECHEAT(CO2, T=T_Mix) + 6.50*SPECHEAT(H2O, T=T_Mix) + 12.20*SPECHEAT(O2, T=T_Mix)+ 76.41*SPECHEAT(N2, T=T_Mix))/(4.89+6.50+12.20+76.41) "[kJ/kmol-K]" "The partial pressure of the water in the mixture is:" y_v= 6.50/(4.89+6.50+12.20+76.41) P_v=y_v *P_Mix"[kPa]"

14-66 In a liquid-oxygen plant, it is proposed that the pressure and temperature of air be adiabatically reduced. It is to be determined whether this process is possible and the work produced is to be determined using Kay's rule and the departure charts. Assumptions Air is a gas mixture with 21% O2 and 79% N 2 , by mole. Properties The molar masses of O2 and N 2 are 32.0 and 28.0 kg / kmol , respectively. The critical properties are 154.8 K, 5.08 MPa for O2 and 126.2 K and 3.39 MPa for N 2 (Table A-1). Analysis To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of gases.

Tcr,¢ m = å yiTcr, i = yO2Tcr, O2 + yN2Tcr, N2 = (0.21)(154.8 K) + (0.79)(126.2 K) = 132.2 K

Pcr,¢ m = å yi Pcr, i = yO2 Pcr, O2 + yN2 Pcr, N2 = (0.21)(5.08 MPa) + (0.79)(3.39 MPa) = 3.745 MPa

The enthalpy and entropy departure factors at the initial and final states are (from EES) T 283 K ïü ïï TR1 = 'm1 = = 2.141 ïï Z = 0.513 Tcr, m 132.2 K ïý h1 ï Z s1 = 0.235 Pm1 9 MPa PR1 = ' = = 2.403 ïïï Pcr, m 3.745 MPa ïïþ ü T 200 K ïï TR 2 = m' 2 = = 1.513 ïï Tcr, m 132.2 K ïïý Z h 2 = 0.0069 ï Z s 2 = 0.0035 Pm 2 0.050 MPa PR 2 = ' = = 0.0134 ïïï Pcr, m 3.745 MPa ïïþ The enthalpy and entropy changes of the air under the ideal gas assumption is (Properties are from Table A-17)

(h2 - h1 )ideal = 199.97 - 283.14 = - 83.2 kJ/kg ( s2 - s1 )ideal = s2o - s1o - R ln

P2 50 = 1.29559 - 1.64345 - (0.287) ln = 1.1425 kJ/kg ×K P1 9000

With departure factors, the enthalpy change (i.e., the work output) and the entropy change are

wout = h1 - h2 = (h1 - h2 )ideal - RTcr' (Z h1 - Z h 2 )

13-138

= 83.2 - (0.287)(132.2)(0.513 - 0.0069) = 64.0 kJ / kg

s2 - s1 = (s2 - s1 )ideal - R(Z s 2 - Z s1 ) = 1.1425 - (0.287)(0.0035 - 0.235) = 1.209 kJ / kg ×K The entropy change in this case is equal to the entropy generation during the process since the process is adiabatic. The positive value of entropy generation shows that this process is possible.

14-67E The mass fractions of a natural gas mixture at a specified pressure and temperature trapped in a geological location are given. This natural gas is pumped to the surface. The work required is to be determined using Kay's rule and the enthalpy-departure method. Properties The molar masses of CH4 and C2 H6 are 16.0 and 30.0 lbm / lbmol , respectively. The critical properties are 343.9 R, 673 psia for CH4 and 549.8 R and 708 psia for C2 H6 (Table A-1E). The constant-pressure specific heats of these gases at room temperature are 0.532 and 0.427 Btu / lbm  R , respectively (Table A-2Ea). Analysis We consider 100 lbm of this mixture. Then the mole numbers of each component are N CH4 =

mCH4 75 lbm = = 4.6875 lbmol M CH4 16 lbm/lbmol

N C2H6 =

mC2H6 25 lbm = = 0.8333 lbmol M C2H6 30 lbm/lbmol

The mole number of the mixture and the mole fractions are

Nm = 4.6875 + 0.8333 = 5.5208 lbmol yCH4 =

N CH4 4.6875 lbmol = = 0.8491 Nm 5.5208 lbmol

yC2H6 =

N C2H6 0.8333 lbmol = = 0.1509 Nm 5.5208 lbmol

Then the apparent molecular weight of the mixture becomes Mm =

mm 100 lbm = = 18.11 lbm/lbmol N m 5.5208 lbmol

The apparent gas constant of the mixture is R=

Ru 1.9858 Btu/lbmol ×R = = 0.1097 Btu/lbm ×R Mm 18.11 lbm/lbmol

The constant-pressure specific heat of the mixture is determined from

c p = mfCH4 c p,CH4 + mfC2H6 c p ,C2H6 = 0.75´ 0.532 + 0.25´ 0.427 = 0.506 Btu/lbm ×R To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of gases.

13-139

= (0.8491)(343.9 R) + (0.1509)(549.8 R) = 375.0 R

Pcr,¢ m = å yi Pcr, i = yCh4 Pcr, Ch4 + yC2H6 Pcr, C2H6 = (0.8491)(673 psia) + (0.1509)(708 psia) = 678.3 psia The compressibility factor of the gas mixture in the reservoir and the mass of this gas are T 760 R ïü TR = ' m = = 2.027 ïï ïï Tcr, m 375.0 R ïý Z = 0.963 (Fig. A-15 or EES, we used EES.) ï m Pm 1300 psia PR = ' = = 1.917 ïïï Pcr, m 678.3 psia ïïþ

PV (1300 psia)(2´ 106 ft 3 ) = = 5.996´ 106 lbm 3 Z m RT (0.963)(0.5925 psia ×ft /lbm ×R)(760 R)

The enthalpy departure factors in the reservoir and the surface are (from EES or Fig. A-29, we used EES.) T 760 R ïü TR1 = ' m = = 2.027 ïï ïï Tcr, m 375.0 R ïý Z = 0.487 ï h1 Pm 1300 psia PR1 = ' = = 1.917 ïïï Pcr, m 678.3 psia ïïþ T 660 R ïü ïï TR 2 = ' m = = 1.76 ïï Tcr, m 375.0 R ïý Z = 0.0112 ï h2 Pm 20 psia PR 2 = ' = = 0.0295 ïïï Pcr, m 678.3 psia ïïþ The enthalpy change for the ideal gas mixture is

(h1 - h2 )ideal = c p (T1 - T2 ) = (0.506 Btu/lbm ×R)(760 - 660)R = 50.6 Btu/lbm The enthalpy change with departure factors is

h1 - h2 = (h1 - h2 )ideal - RTcr,¢ m ( Z h1 - Z h 2 ) = 50.6 - (0.1096)(375)(0.487 - 0.0112) = 31.02 Btu/lbm

The work input is then

Win = m(h1 - h2 ) = (5.996´ 106 lbm)(31.02 Btu/lbm) = 1.86´ 108 Btu

EES (Engineering Equation Solver) SOLUTION "Given" mf_CH4=0.75 mf_C2H6=0.25 Vol=2E6 [ft^3] P1=1300 [psia] T1=(300+460) [R] z=6000 [ft] P2=20 [psia] T2=(200+460) [R] "Properties" R_u=1.9858 [Btu/lbmol-R] M_CH4=16 [lbm/lbmol] M_C2H6=30 [lbm/lbmol] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-140

T_cr_CH4=343.9 [R] P_cr_CH4=673 [psia] T_cr_C2H6=549.8 [R] P_cr_C2H6=708 [psia] c_p_CH4=0.532 [Btu/lbm-R] c_p_C2H6=0.427 [Btu/lbm-R] "Analysis" mass_m=100 [lbm] "We consdier 100 lbm of mixture" mass_CH4=mf_CH4*mass_m mass_C2H6=mf_C2H6*mass_m N_CH4=mass_CH4/M_CH4 N_C2H6=mass_C2H6/M_C2H6 N_m=N_CH4+N_C2H6 y_CH4=N_CH4/N_m y_C2H6=N_C2H6/N_m M_m=mass_m/N_m R=R_u/M_m c_p=mf_CH4*c_p_CH4+mf_C2H6*c_p_C2H6 T_cr_m=y_CH4*T_cr_CH4+y_C2H6*T_cr_C2H6 P_cr_m=y_CH4*P_cr_CH4+y_C2H6*P_cr_C2H6 T_R1=T1/T_cr_m P_R1=P1/P_cr_m Z_m=0.963 "COMPRESS(T_R1, P_R1)" mass=(P1*Vol)/(Z_m*R*Convert(Btu, psia-ft^3)*T1) Z_h1=ENTHDEP(T_R1, P_R1) T_R2=T2/T_cr_m P_R2=P2/P_cr_m Z_h2=ENTHDEP(T_R2, P_R2) DELTAh_ideal=c_p*(T1-T2) DELTAh=DELTAh_ideal-R*T_cr_m*(Z_h1−Z_h2) W_in=mass*DELTAh

14-68 A mixture of hydrogen and oxygen is considered. The entropy change of this mixture between the two specified states is to be determined. Assumptions Hydrogen and oxygen are ideal gases. Properties The gas constants of hydrogen and oxygen are 4.124 and 0.2598 kJ / kg  K , respectively (Table A-1). Analysis The effective gas constant of this mixture is

R = mf H2 RH2 + mfO2 RO2 = (0.33)(4.1240) + (0.67)(0.2598) = 1.5350 kJ/kg ×K Since the temperature of the two states is the same, the entropy change is determined from s2 - s1 = - R ln

P2 150 kPa = - (1.5350 kJ/kg ×K) ln = 2.470 kJ / kg ×K P1 750 kPa

EES (Engineering Equation Solver) SOLUTION "Given" mf_H2=0.33 P1=750 [kPa] T1=(150+273) [K] P2=150 [kPa] T2=(150+273) [K] "Properties" R_H2=4.124 [kJ/kg-K] R_O2=0.2598 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-141

"Analysis" mf_O2=1-mf_H2 R=mf_H2*R_H2+mf_O2*R_O2 DELTAs=-R*ln(P2/P1)

14-69 A piston-cylinder device contains a gas mixture at a given state. Heat is transferred to the mixture. The amount of heat transfer and the entropy change of the mixture are to be determined. Assumptions 1 Under specified conditions both H 2 and N 2 can be treated as ideal gases, and the mixture as an ideal gas mixture. 2 Kinetic and potential energy changes are negligible. Properties The constant pressure specific heats of H 2 and N 2 at 450 K are 14.501 kJ / kg  K and 1.049 kJ / kg  K , respectively. (Table A-2b).

Analysis (a) Noting that P2  P1 and V2  2V1 , P2V2 PV 2V = 1 1 ¾¾ ® T2 = 1 T1 = 2T1 = (2)(300 K ) = 600 K T2 T1 V1 From the closed system energy balance relation, Ein - Eout = D Esystem Qin - Wb,out = D U ® Qin = D H

since Wb and U combine into H for quasi-equilibrium constant pressure processes. é ù Qin = D H = D H H2 + D H N2 = éêëmc p , avg (T2 - T1 )ù ú ûH2 + êëmc p , avg (T2 - T1 )ú ûN2

= (0.5 kg )(14.501 kJ/kg ×K )(600 - 300)K + (1.2 kg )(1.049 kJ/kg ×K )(600 - 300)K = 2553 kJ (b) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of the mixture during this process is æ T P D S H2 = éëm (s2 - s1 )ùûH = mH2 çççc p ln 2 - R ln 2 2 çè T1 P1 = (0.5 kg)(14.501 kJ/kg ×K )ln

ö÷ æ T ö ÷ ÷ = mH2 çççc p ln 2 ÷ ÷ ÷ ç ÷ T1 ø÷H è ø÷ H2 2

600 K = 5.026 kJ/K 300 K

æ T2 P çç D S N2 = éëm (s2 - s1 )ù = m - R ln 2 N 2 çc p ln ûN2 T1 P1 èç

= (1.2 kg)(1.049 kJ/kg ×K )ln

ö æ T ö ÷ ÷ ÷ = mN2 çççc p ln 2 ÷ ÷ ÷ ÷ ÷ ÷ T1 ø èç øN2 N2

600 K = 0.8725 kJ/K 300 K

13-142

D S total = D SH2 + D S N2 = 5.026 kJ/K + 0.8725 kJ/K = 5.90 kJ / K

EES (Engineering Equation Solver) SOLUTION "Given" mass_H2=0.5 [kg] mass_N2=1.2 [kg] P1=100 [kPa] T1=300 [K] P2=P1 V2=2*V1 V1=1 [m^3] "assumed" "Properties, at 450 K" c_p_H2=14.501 [kJ/kg-K] c_p_N2=1.049 [kJ/kg-K] "Analysis" "(a)" (P1*V1)/T1=(P2*V2)/T2 "from ideal gas relation" Q_in=mass_H2*c_p_H2*(T2−T1)+mass_N2*c_p_N2*(T2−T1) "energy balance" "(b)" DELTAS_H2=mass_H2*c_p_H2*ln(T2/T1) DELTAS_N2=mass_N2*c_p_N2*ln(T2/T1) DELTAS_total=DELTAS_H2+DELTAS_N2

14-70E The volume fractions of components of a gas mixture during the expansion process of the ideal Otto cycle are given. The thermal efficiency of this cycle is to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N 2 , O2 , H 2 O , and CO2 are 28.0, 32.0, 18.0, and 44.0 lbm / lbmol , respectively (Table A-1E). The constant-pressure specific heats of these gases at room temperature are 0.248, 0.219, 0.445, and 0.203 Btu/lbmR, respectively. The air properties at room temperature are c p  0.240 Btu / lbm  R , cv  0.171 Btu / lbm  R , k = 1.4 (Table A-2Ea). Analysis We consider 100 lbmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are

mN2 = N N2 M N2 = (25 lbmol)(28 lbm/lbmol) = 700 lbm mO2 = NO2 M O2 = (7 lbmol)(32 lbm/lbmol) = 224 lbm mH2O = NH2O M H2O = (28 lbmol)(18 lbm/lbmol) = 504 lbm mCO2 = NCO2 M CO2 = (40 lbmol)(44 lbm/lbmol) = 1760 lbm

The total mass is

13-143

mm = mN2 + mO2 + mH2O + mCO2 = 700 + 224 + 504 + 1760 = 3188 lbm

Then the mass fractions are mf N2 =

mN2 700 lbm = = 0.2196 mm 3188 lbm

mf O2 =

mO2 224 lbm = = 0.07026 mm 3188 lbm

mf H2O =

mH2O 504 lbm = = 0.1581 mm 3188 lbm

mf CO2 =

mCO2 1760 lbm = = 0.5521 mm 3188 lbm

The constant-pressure specific heat of the mixture is determined from c p = mf N2 c p ,N2 + mf O2 c p,O2 + mf H2O c p,H2O + mf CO2 c p,CO2

= 0.2196´ 0.248 + 0.07026´ 0.219 + 0.1581´ 0.445 + 0.5521´ 0.203 = 0.2523 Btu/lbm ×R

The apparent molecular weight of the mixture is Mm =

mm 3188 lbm = = 31.88 lbm/lbmol N m 100 lbmol

The apparent gas constant of the mixture is R=

Ru 1.9858 Btu/lbmol ×R = = 0.06229 Btu/lbm ×R Mm 31.88 lbm/lbmol

Then the constant-volume specific heat is

cv = c p - R = 0.2523 - 0.06229 = 0.1900 Btu/lbm ×R The specific heat ratio is

cp cv

0.2523 = 1.328 0.1900

The average of the air properties at room temperature and combustion gas properties are

c p ,avg = 0.5(0.2523 + 0.240) = 0.2462 Btu/lbm ×R cv ,avg = 0.5(0.1900 + 0.171) = 0.1805 Btu/lbm ×R kavg = 0.5(1.328 + 1.4) = 1.364

These average properties will be used for heat addition and rejection processes. For compression, the air properties at room temperature and during expansion, the mixture properties will be used. During the compression process,

13-144

During the heat addition process,

qin = cv ,avg (T3 - T2 ) = (0.1805 Btu/lbm ×R)(2060 - 1122) R = 169.3 Btu/lbm During the expansion process, k- 1

æ1 ö T4 = T3 çç ÷ ÷ çè r ø÷

0.364

æ1 ö = (2060 R) çç ÷ ÷ èç7 ø÷

= 1014 R

During the heat rejection process,

qout = cv ,avg (T4 - T1 ) = (0.1805 Btu/lbm ×R)(1014 - 515) R = 90.1 Btu/lbm The thermal efficiency of the cycle is then hth = 1-

qout 90.1 Btu/lbm = 1= 0.468 qin 169.3 Btu/lbm

14-71E The thermal efficiency of the cycle in the previous problem is to be compared to that predicted by air standard analysis? Assumptions Air-standard assumptions are applicable.

Properties The air properties at room temperature are c p  0.240 Btu / lbm  R , cv  0.171 Btu / lbm  R , k = 1.4 (Table A-2Ea). Analysis In the previous problem, the thermal efficiency of the cycle was determined to be 0.468 (46.8%). The thermal efficiency with air-standard model is determined from

1 = 0.541 r 70.4 which is significantly greater than that calculated with gas mixture analysis in the previous problem. hth = 1-

k- 1

= 1-

14-72 The volume fractions of components of a gas mixture passing through the turbine of a simple ideal Brayton cycle are given. The thermal efficiency of this cycle is to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of N 2 , O2 , H 2 O , and CO2 are 28.0, 32.0, 18.0, and 44.0 kg / kmol , respectively (Table A1). The constant-pressure specific heats of these gases at room temperature are 1.039, 0.918, 1.8723, and 0.846 kJ/kgK, respectively. The air properties at room temperature are c p  1.005 kJ / kg  K , cv  0.718 kJ / kg  K , k = 1.4 (Table A-2a). Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each component are

mN2 = N N2 M N2 = (30 kmol)(28 kg/kmol) = 840 kg

13-145

mO2 = NO2 M O2 = (10 kmol)(32 kg/kmol) = 320 kg mH2O = NH2O M H2O = (20 kmol)(18 kg/kmol) = 360 kg mCO2 = NCO2 M CO2 = (40 kmol)(44 kg/kmol) = 1760 kg

The total mass is

mm = mN2 + mO2 + mH2O + mCO2 ] = 840 + 320 + 360 + 1760

= 3280 kg Then the mass fractions are mf N2 =

mN2 840 kg = = 0.2561 mm 3280 kg

mf O2 =

mO2 320 kg = = 0.09756 mm 3280 kg

mf H2O =

mH2O 360 kg = = 0.1098 mm 3280 kg

mf CO2 =

mCO2 1760 kg = = 0.5366 mm 3280 kg

The constant-pressure specific heat of the mixture is determined from

c p = mf N2 c p ,N2 + mf O2 c p,O2 + mf H2O c p,H2O + mf CO2 c p,CO2 = 0.2561´ 1.039 + 0.09756´ 0.918 + 0.1098´ 1.8723 + 0.5366´ 0.846 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-146

= 1.015 kJ/kg ×K The apparent molecular weight of the mixture is Mm =

mm 3280 kg = = 32.80 kg/kmol N m 100 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ×K = = 0.2535 kJ/kg ×K Mm 32.80 kg/kmol

Then the constant-volume specific heat is

cv = c p - R = 1.015 - 0.2535 = 0.762 kJ/kg ×K The specific heat ratio is

cp cv

1.015 = 1.332 0.762

The average of the air properties at room temperature and combustion gas properties are

c p ,avg = 0.5(1.015 + 1.005) = 1.010 kJ/kg ×K cv ,avg = 0.5(0.762 + 0.718) = 0.740 kJ/kg ×K kavg = 0.5(1.332 + 1.4) = 1.366

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P ÷ ø

= (293 K)(8)0.4/1.4 = 531 K

During the heat addition process,

qin = c p ,avg (T3 - T2 ) = (1.010 kJ/kg ×K)(1273 - 531) K = 749.4 kJ/kg During the expansion process, ( k - 1)/ k

æP ö ÷ T4 = T3 ççç 4 ÷ ÷ èç P ø÷ 3

0.332/1.332

æ1 ö = (1273 K) çç ÷ ÷ çè8 ø÷

= 758 K

During the heat rejection process,

qout = cp ,avg (T4 - T1 ) = (1.010 kJ/kg ×K)(758 - 293) K = 469.7 kJ/kg The thermal efficiency of the cycle is then hth = 1-

qout 469.7 kJ/kg = 1= 0.373 qin 749.4 kJ/kg

14-73 The thermal efficiency of the cycle in the previous problem is to be compared to that predicted by air standard analysis? Assumptions Air-standard assumptions are applicable.

13-147

Properties The air properties at room temperature are c p  1.005 kJ / kg  K , cv  1.4 kJ / kg  K , k = 1.4 (Table A-2a). Analysis In the previous problem, the thermal efficiency of the cycle was determined to be 0.373 (37.3%). The thermal efficiency with air-standard model is determined from

hth = 1-

1 ( k - 1)/ k p

= 1-

1 0.4/1.4

= 0.448

which is significantly greater than that calculated with gas mixture analysis in the previous problem.

14-74 Heat is transferred to a gas mixture contained in a piston cylinder device. The initial state and the final temperature are given. The heat transfer is to be determined for the ideal gas and non-ideal gas cases. Properties The molar masses of H 2 and N 2 are 2.0, and 28.0 kg / kmol . (Table A-1). Analysis From the energy balance relation, Ein - Eout = D E Qin - Wb,out = D U Qin = D H = D H H 2 + D H N 2 = N H 2 (h2 - h1 ) + N N 2 (h2 - h1 ) H2

since Wb and U combine into H for quasi-equilibrium constant pressure processes

N H2 =

N N2 =

mH2 M H2

mN2 M N2

6 kg = 3 kmol 2 kg/kmol

21 kg = 0.75 kmol 28 kg/kmol

(a) Assuming ideal gas behavior, the inlet and exit enthalpies of H 2 and N 2 are determined from the ideal gas tables to be

H 2 : h1 = h@ 160 K = 4,535.4 kJ/kmol,

h2 = h@ 200 K = 5,669.2 kJ/kmol

13-148

N2 : h1 = h@ 160 K = 4,648 kJ/kmol, Thus,

h2 = h@ 200 K = 5,810 kJ/kmol

Qideal = 3´ (5, 669.2 - 4,535.4)+ 0.75´ (5,810 - 4, 648) = 4273 kJ

(b) Using Amagat's law and the generalized enthalpy departure chart, the enthalpy change of each gas is determined to be ü ïï Tm , 1 160 ïï TR1 , H2 = = = 4.805 ïï Tcr, H2 33.3 ïï Z h1 @ 0 ï Pm 5 (Fig. A-29) H 2 : PR1 , H2 = PR2 , H2 = = = 3.846 ïý ïï Pcr, H2 1.30 ïï Z h2 @ 0 ïï Tm , 2 200 ïï TR2 , H2 = = = 6.006 ïï Tcr, H2 33.3 þ Thus H 2 can be treated as an ideal gas during this process. ü ïï ïï ïï Tcr, N2 ïï Z h1 = 1.3 ï Pm 5 (Fig. A-29) N 2 : PR1 , N2 = PR2 , N2 = = = 1.47 ïý ïï Pcr, N2 3.39 ïï Z h2 = 0.7 ïï Tm , 2 200 ïï TR2 , N2 = = = 1.58 ïï Tcr, N2 126.2 þ Therefore, = 5, 669.2 - 4,535.4 = 1,133.8 kJ/kmol (h2 - h1 ) = (h2 - h1 ) TR1 , N2 =

Tm , 1

160 = 1.27 126.2

H 2 , ideal

(h - h ) = R T (Z - Z )+ (h - h ) 2

1 N 2

u cr

1 ideal

= (8.314 kPa ×m /kmol ×K)(126.2 K)(1.3 - 0.7) + (5,810 - 4,648) kJ/kmol = 1,791.5 kJ/kmol Substituting, Qin = (3 kmol)(1,133.8 kJ/kmol)+ (0.75 kmol)(1,791.5 kJ/kmol) = 4745 kJ 3

EES (Engineering Equation Solver) SOLUTION "Given" mass_H2=6 [kg] mass_N2=21 [kg] T1=160 [K] P1=5000 [kPa] T2=200 [K] P2=P1 "Properties" R_u=8.314 [kPa-m^3/kmol-K] M_H2=2 [kg/kmol] M_N2=28 [kg/kmol] T_cr_H2=33.3 [K] T_cr_N2=126.2 [K] P_cr_H2=1300 [kPa] P_cr_N2=3390 [kPa] "Analysis" "(a)" N_H2=mass_H2/M_H2 N_N2=mass_N2/M_N2 h1_H2=enthalpy(H2, T=T1) h2_H2=enthalpy(H2, T=T2) h1_N2=enthalpy(N2, T=T1) h2_N2=enthalpy(N2, T=T2)

13-149

Q_in_ideal=N_H2*(h2_H2-h1_H2)+N_N2*(h2_N2-h1_N2) "(b)" T_R1_H2=T1/T_cr_H2 P_R1_H2=P1/P_cr_H2 Z_h1_H2=0 "ENTHDEP(T_R1_H2, P_R1_H2)" "the function that returns enthalpy departure factor" T_R2_H2=T2/T_cr_H2 P_R2_H2=P2/P_cr_H2 Z_h2_H2=0 "ENTHDEP(T_R2_H2, P_R2_H2)" "the function that returns enthalpy departure factor" T_R1_N2=T1/T_cr_N2 P_R1_N2=P1/P_cr_N2 Z_h1_N2=ENTHDEP(T_R1_N2, P_R1_N2) "the function that returns enthalpy departure factor" T_R2_N2=T2/T_cr_N2 P_R2_N2=P2/P_cr_N2 Z_h2_N2=ENTHDEP(T_R2_N2, P_R2_N2) "the function that returns enthalpy departure factor" DELTAh_H2_ideal=h2_H2-h1_H2 DELTAh_N2_ideal=h2_N2-h1_N2 DELTAh_H2=DELTAh_H2_ideal-R_u*T_cr_H2*(Z_h2_H2−Z_h1_H2) DELTAh_N2=DELTAh_N2_ideal-R_u*T_cr_N2*(Z_h2_N2−Z_h1_N2) Q_in=N_H2*DELTAh_H2+N_N2*DELTAh_N2

14-75 Heat is transferred to a gas mixture contained in a piston cylinder device discussed in previous problem. The total entropy change and the exergy destruction are to be determined for two cases. Analysis The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the piston-cylinder device and its immediate surroundings so that the boundary temperature of the extended system is the environment temperature at all times. It gives Sin - Sout + Sgen = D Ssystem

Qin Tboundary

+ Sgen = D Swater

Sgen = m( s2 - s1 ) -

Qin Tsurr

Then the exergy destroyed during a process can be determined from its definition X destroyed = T0 Sgen . (a) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the entropy change of a component in the mixture during this process is ¿0 æ T P ö÷ T2 D Si = mi çççc p ln 2 - R ln 2 ÷ ÷ = mi c p , i ln çè T1 P1 ø÷i T1 Assuming ideal gas behavior and using c p values at the average temperature, the S of H 2 and N 2 are determined from D SH2 , ideal = (6 kg)(13.60 kJ/kg ×K ) ln

200 K = 18.21 kJ/K 160 K

D S N2 , ideal = (21 kg)(1.039 kJ/kg ×K ) ln

200 K = 4.87 kJ/K 160 K

and

Sgen = 18.21 kJ/K + 4.87 kJ/K -

4273 kJ = 8.496 kJ / K 293 K

X destroyed = T0 Sgen = (293 K )(8.496 kJ/K ) = 2489 kJ (b) Using Amagat's law and the generalized entropy departure chart, the entropy change of each gas is determined to be

13-150

ü ïï ïï ïï Tcr, H2 ïï Z s1 @1 ï Pm 5 H 2 : PR1 , H2 = PR2 , H2 = = = 3.846 ïý ïï Pcr, H2 1.30 ïï Z s2 @1 ïï Tm , 2 200 ïï TR2 , H2 = = = 6.006 ïï Tcr, H2 33.3 þ Thus H 2 can be treated as an ideal gas during this process. TR1 , H2 =

Tm , 1

160 = 4.805 33.3

ü ïï ïï ïï Tcr, N2 ïï Z s1 = 0.8 ï Pm 5 N 2 : PR1 , N2 = PR2 , N2 = = = 1.475 ïý ïï Pcr, N2 3.39 ïï Z s2 = 0.4 ïï Tm , 2 200 ïï TR2 , N2 = = = 1.585 ïï Tcr, N2 126.2 þ Therefore, TR1 , N2 =

Tm , 1

(Table A-30)

160 = 1.268 126.2

(Table A-30)

D SH2 = D SH2 , ideal = 18.21 kJ/K

D S N2 = N N2 Ru (Z s1 - Z s2 )+ D S N2 , ideal = (0.75 kmol)(8.314 kPa ×m3 /kmol ×K)(0.8 - 0.4) + (4.87 kJ/K) = 7.37 kJ/K

D Ssurr =

Qsurr - 4745 kJ = = - 16.19 kJ/K T0 293 K

and Sgen = 18.21 kJ/K + 7.37 kJ/K - 16.19 = 9.390 kJ / K

X destroyed = T0 Sgen = (293 K )(9.385 kJ/K ) = 2751 kJ

EES (Engineering Equation Solver) SOLUTION "Given" mass_H2=6 [kg] mass_N2=21 [kg] T1=160 [K] P1=5000 [kPa] T2=200 [K] P2=P1 T0=(20+273) [K] "Properties" R_u=8.314 [kPa-m^3/kmol-K] M_H2=2 [kg/kmol] M_N2=28 [kg/kmol] T_cr_H2=33.3 [K] T_cr_N2=126.2 [K] P_cr_H2=1300 [kPa] P_cr_N2=3390 [kPa] c_p_H2=13.60 [kJ/kg-K]*M_H2 "kJ/kmol-K" c_p_N2=1.039 [kJ/kg-K]*M_N2 "kJ/kmol-K" "Analysis" N_H2=mass_H2/M_H2 N_N2=mass_N2/M_N2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-151

h1_H2=enthalpy(H2, T=T1) h2_H2=enthalpy(H2, T=T2) h1_N2=enthalpy(N2, T=T1) h2_N2=enthalpy(N2, T=T2) Q_in_ideal=N_H2*(h2_H2-h1_H2)+N_N2*(h2_N2-h1_N2) T_R1_H2=T1/T_cr_H2 P_R1_H2=P1/P_cr_H2 Z_h1_H2=0 "ENTHDEP(T_R1_H2, P_R1_H2)" "the function that returns enthalpy departure factor" T_R2_H2=T2/T_cr_H2 P_R2_H2=P2/P_cr_H2 Z_h2_H2=0 "ENTHDEP(T_R2_H2, P_R2_H2)" "the function that returns enthalpy departure factor" T_R1_N2=T1/T_cr_N2 P_R1_N2=P1/P_cr_N2 Z_h1_N2=ENTHDEP(T_R1_N2, P_R1_N2) "the function that returns enthalpy departure factor" T_R2_N2=T2/T_cr_N2 P_R2_N2=P2/P_cr_N2 Z_h2_N2=ENTHDEP(T_R2_N2, P_R2_N2) "the function that returns enthalpy departure factor" DELTAh_H2_ideal=h2_H2−h1_H2 DELTAh_N2_ideal=h2_N2−h1_N2 DELTAh_H2=DELTAh_H2_ideal-R_u*T_cr_H2*(Z_h2_H2−Z_h1_H2) DELTAh_N2=DELTAh_N2_ideal-R_u*T_cr_N2*(Z_h2_N2−Z_h1_N2) Q_in=N_H2*DELTAh_H2+N_N2*DELTAh_N2 "(a)" DELTAS_H2_ideal=N_H2*c_p_H2*ln(T2/T1) DELTAS_N2_ideal=N_N2*c_p_N2*ln(T2/T1) S_gen_ideal=DELTAS_H2_ideal+DELTAS_N2_ideal-Q_in_ideal/T0 X_destroyed_ideal=T0*S_gen_ideal "(b)" Z_s1_H2=ENTRDEP(T_R1_H2, P_R1_H2) "the function that returns entropy departure factor" Z_s2_H2=ENTRDEP(T_R2_H2, P_R2_H2) "the function that returns entropy departure factor" Z_s1_N2=ENTRDEP(T_R1_N2, P_R1_N2) "the function that returns entropy departure factor" Z_s2_N2=ENTRDEP(T_R2_N2, P_R2_N2) "the function that returns entropy departure factor" DELTAS_H2=DELTAS_H2_ideal-N_H2*R_u*(Z_s2_H2−Z_s1_H2) DELTAS_N2=DELTAS_N2_ideal-N_N2*R_u*(Z_s2_N2−Z_s1_N2) DELTAS_surr=-Q_in/T0 S_gen=DELTAS_H2+DELTAS_N2+DELTAS_surr X_destroyed=T0*S_gen

Special Topic: Chemical Potential and the Separation Work of Mixtures 14-76C Mixtures or solutions in which the effects of molecules of different components on each other are negligible are called ideal solutions (or ideal mixtures). The ideal-gas mixture is just one category of ideal solutions. For ideal solutions, the enthalpy change and the volume change due to mixing are zero, but the entropy change is not. The chemical potential of a component of an ideal mixture is independent of the identity of the other constituents of the mixture. The chemical potential of a component in an ideal mixture is equal to the Gibbs function of the pure component.

14-77C No, a process that separates a mixture into its components without requiring any work (exergy) input is impossible since such a process would violate the 2nd law of thermodynamics.

14-78C Yes, the volume of the mixture can be more or less than the sum of the initial volumes of the mixing liquids because of the attractive or repulsive forces acting between dissimilar molecules.

13-152

14-79C The person who claims that the temperature of the mixture can be higher than the temperatures of the components is right since the total enthalpy of the mixture of two components at the same pressure and temperature, in general, is not equal to the sum of the total enthalpies of the individual components before mixing, the difference being the enthalpy (or heat) of mixing, which is the heat released or absorbed as two or more components are mixed isothermally.

14-80 Brackish water is used to produce fresh water. The minimum power input and the minimum height the brackish water must be raised by a pump for reverse osmosis are to be determined. Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 18C. Properties The molar masses of water and salt are M w  18.0 kg / kmol and M s  58.44 kg / kmol . The gas constant of pure water is Rw  0.4615 kJ / kg  K (Table A-1). The density of fresh water is 1000 kg / m3 . Analysis First we determine the mole fraction of pure water in brackish water using Eqs. 14-4 and 14-5. Noting that mf s  0.00078 and mf w  1  mf s  0.99922 ,

Mm =

1 1 1 = = = 18.01 kg/kmol mf i mf s mf w 0.00078 0.99922 + å M M +M 58.44 18.0 i s w

yi = mf i

Mm M 18.01 kg/kmol ® yw = mf w m = (0.99922) = 0.99976 Mi Mw 18.0 kg/kmol

The minimum work input required to produce 1 kg of freshwater from brackish water is

wmin, in = RwT0 ln(1/ yw ) = (0.4615 kJ/kg ×K)(291.15 K) ln(1/0.99976) = 0.03230 kJ/kg fresh water Therefore, 0.03159 kJ of work is needed to produce 1 kg of fresh water is mixed with seawater reversibly. Therefore, the required power input to produce fresh water at the specified rate is æ1 kW ÷ ö Wmin, in = r V wmin, in = (1000 kg/m3 )(0.175 m3 /s)(0.03230 kJ/kg) çç = 5.65 kW ÷ çè1 kJ/s ÷ ø

The minimum height to which the brackish water must be pumped is D zmin =

wmin, in g

æ1 kg.m/s 2 ö æ0.03230 kJ/kg ö æ1000 N.m ö ÷ ç ÷ ÷= 3.29 m ÷çç = çç ÷ç ÷ 2 ÷ ÷ ÷ èç 9.81 m/s øèç 1 N ÷ øèç 1 kJ ø

EES (Engineering Equation Solver) SOLUTION "Given" T=(18+273.15) [K] mass_s=0.078 [kg] V_dot=0.175 [m^3/s] "Properties" R_u=8.314 [kPa-m^3/kmol-K] M_w=18 [kg/kmol] M_s=58.44 [kg/kmol] "molar mass of salt" R_w=0.4615 [kJ/kg-K] rho=1000 [kg/m^3] g=9.81 [m/s^2] "Analysis" mass_m=100 [kg] "consider 100 kg of mixture" mass_w=mass_m-mass_s N_s=mass_s/M_s N_w=mass_w/M_w N_m=N_s+N_w © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-153

y_w=N_w/N_m m_dot=rho*V_dot W_dot_min_in=m_dot*R_w*T*ln(1/y_w) w_min_in=W_dot_min_in/m_dot DELTAz_min=w_min_in/g*Convert(kJ, N-m)

14-81 A river is discharging into the ocean at a specified rate. The amount of power that can be generated is to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is also 15C. Properties The molar masses of water and salt are M w  18.0 kg / kmol and M s  58.44 kg / kmol . The gas constant of pure water is Rw  0.4615 kJ / kg  K (Table A-1). The density of river water is 1000 kg / m3 . Analysis First we determine the mole fraction of pure water in ocean water using Eqs. 14-4 and 14-5. Noting that mf s  0.025 and mf w  1  mf s  0.975 ,

Mm =

1 1 1 = = = 18.32 kg/kmol mf i mf s mf w 0.025 0.975 + + å M M M 58.44 18.0 i s w

yi = mf i

Mm M 18.32 kg/kmol ® yw = mf w m = (0.975) = 0.9922 Mi Mw 18.0 kg/kmol

The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is

wmax, out = RwT0 ln(1/ yw ) = (0.4615 kJ/kg ×K)(288.15 K)ln(1/0.9922) = 1.046 kJ/kg fresh water Therefore, 1.046 kJ of work can be produced as 1 kg of fresh water is mixed with seawater reversibly. Therefore, the power that can be generated as a river with a flow rate of 400, 000 m3 / s mixes reversibly with seawater is æ1 kW ö 6 ÷ Wmax out = r V wmax out = (1000 kg/m3 )(1.5´ 105 m3 /s)(1.046 kJ/kg) çç ÷ ÷= 157 ´ 10 kW çè1 kJ/s ø

Discussion This is more power than produced by all nuclear power plants in the U.S., which shows the tremendous amount of power potential wasted as the rivers discharge into the seas.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=150000 [m^3/s] salinity=2.5 T=(15+273.15) [K] "Properties" M_w=18 [kg/kmol] "molarmass(H2O)" M_s=58.44 [kg/kmol] "molar mass of salt" R_w=0.4615 [kJ/kg-K] "gas constant of water" rho=1000 [kg/m^3] "Analysis" mass_w=100-salinity mf_s=salinity/100 mf_w=mass_w/100 M_m=1/(mf_s/M_s+mf_w/M_w) y_w=mf_w*M_m/M_w © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-154

w_max_out=R_w*T*ln(1/y_w) W_dot_max_out=rho*V_dot*w_max_out

14-82 Problem 14-81 is reconsidered. The effect of the salinity of the ocean on the maximum power generated is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Given" V_dot = 150000 [m^3/s] salinity = 2.5 T = (15+273.15) [K] "Properties" M_w=18 [kg/kmol] "molarmass(H2O)" M_s=58.44 [kg/kmol] "molar mass of salt" R_w=0.4615 [kJ/kg–K] "gas constant of water" rho=1000 [kg/m^3] "Analysis" mass_w=100–salinity mf_s=salinity/100 mf_w=mass_w/100 M_m=1/(mf_s/M_s+mf_w/M_w) y_w=mf_w*M_m/M_w w_max_out=R_w*T*ln(1/y_w) W_dot_max_out=rho*V_dot*w_max_out Salinity [%] 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Wmax, out [kW] 0 3.085E+07 6.196E+07 9.334E+07 1.249E+08 1.569E+08 1.891E+08 2.216E+08 2.544E+08 2.874E+08 3.208E+08

13-155

14-83E Brackish water is used to produce fresh water. The mole fractions, the minimum work inputs required to separate 1 lbm of brackish water and to obtain 1 lbm of fresh water are to be determined. Assumptions 1 The brackish water is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is equal to the water temperature. Properties The molar masses of water and salt are M w  18.0 lbm / lbmol and M s  58.44 lbm / lbmol . The gas constant of pure water is Rw  0.1102 Btu / lbm  R (Table A-1E). Analysis (a) First we determine the mole fraction of pure water in brackish water using Eqs. 14-4 and 14-5. Noting that mfs  0.0012 and mf w  1  mfs  0.9988 ,

Mm =

1 1 1 = = = 18.015 lbm/lbmol mf i mf s mf w 0.0012 0.9988 + å M M +M 58.44 18.0 i s w

yi = mf i

Mm M 18.015 lbm/lbmol ® yw = mf w m = (0.9988) = 0.99963 Mi Mw 18.0 lbm/lbmol

ys = 1- yw = 1- 0.99963 = 0.00037 (b) The minimum work input required to separate 1 lbmol of brackish water is wmin,in = - RwT0 ( yw ln yw + ys ln ys )

= - (0.1102 Btu/lbmol.R)(525 R)[0.99963ln(0.99963) + 0.00037ln(0.00037)] = - 0.191Btu / lbm brackish water

wmin, in = RwT0 ln(1/ yw ) = (0.1102 Btu/lbm ×R)(525 R)ln(1/0.99963) = 0.0214 Btu / lbm fresh water

13-156

Discussion Note that it takes about 9 times work to separate 1 lbm of brackish water into pure water and salt compared to producing 1 lbm of fresh water from a large body of brackish water.

EES (Engineering Equation Solver) SOLUTION "Given" T=(65+460) [R] mass_s=0.12 [lbm] "Properties" R_u=1.9858 [Btu/lbmol-R] M_w=18 [lbm/lbmol] M_s=58.44 [lbm/lbmol] "molar mass of salt" R_w=0.1102 [Btu/lbm-R] "gas constant of water" "Analysis" "(a)" mass_m=100 [lbm] "consider 100 lbm of mixture" mass_w=mass_m-mass_s N_s=mass_s/M_s N_w=mass_w/M_w N_m=N_s+N_w y_w=N_w/N_m y_s=1-y_w "(b)" w_min_in_b=-R_w*T*(y_w*ln(y_w)+y_s*ln(y_s)) "(c)" w_min_in_c=R_w*T*ln(1/y_w)

14-84 The power consumption and the second law efficiency of a desalination plant are given. The power that can be produced if the fresh water produced is mixed with the seawater reversibly is to be determined. Assumptions 1 This is a steady-flow process. 2 The kinetic and potential energy changes are negligible. Analysis From the definition of the second law efficiency

hII =

Wrev Wrev ® 0.20 = ® Wrev = 2300 kW 11,500 kW Wactual

which is the maximum power that can be generated.

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=1.5 [m^3/s] W_dot_actual=11500 [kW] Eta_II=0.20 "Analysis" W_dot_rev=Eta_II*W_dot_actual

14-85E It is to be determined if it is it possible for an adiabatic liquid-vapor separator to separate wet steam at 100 psia and 90 percent quality, so that the pressure of the outlet streams is greater than 100 psia. Analysis Because the separator divides the inlet stream into the liquid and vapor portions,

13-157

According to the water property tables at 100 psia (Table A-5E),

s1 = s f + xs fg = 0.47427 + 0.9´ 1.12888 = 1.4903 Btu/lbm ×R When the increase in entropy principle is adapted to this system, it becomes

m2 s2 + m3 s3 ³ m1s1 xm1s2 + (1- x)m1s3 ³ m1s1 0.9s2 + 0.1s3 ³ s1

³ 1.4903 Btu/lbm ×R To test this hypothesis, let’s assume the outlet pressures are 110 psia. Then, s2 = sg = 1.5954 Btu/lbm ×R s3 = s f = 0.48341 Btu/lbm ×R

The left-hand side of the above equation is

0.9s2 + 0.1s3 = 0.9´ 1.5954 + 0.1´ 0.48341 = 1.4842 Btu/lbm ×R which is less than the minimum possible specific entropy. Hence, the outlet pressure cannot be 110 psia. Inspection of the water table in light of above equation proves that the pressure at the separator outlet cannot be greater than that at the inlet.

EES (Engineering Equation Solver) SOLUTION "Given" P=100 [psia] x=0.90 "Properties" R_u=1.9858 [Btu/lbmol-R] M_w=18 [lbm/lbmol] M_s=58.44 [lbm/lbmol] "molar mass of salt" R_w=0.1102 [Btu/lbm-R] "gas constant of water" "Analysis" Fluid$='steam_iapws' s_f=entropy(Fluid$, P=P, x=0) s_g=entropy(Fluid$, P=P, x=1) s_fg=s_g-s_f s1=s_f+x*s_fg "Let's assume the outlet pressures are 110 psia:" P_out=110 [psia] s2=entropy(Fluid$, P=P_out, x=1) s3=entropy(Fluid$, P=P_out, x=0) s_new=x*s2+(1-x)*s3 "which is less than the minimum possible specific entropy. Hence, the outlet pressure cannot be 110 psia. Inspection of the water table in light of above equation proves that the pressure at the separator outlet cannot be greater than that at the inlet."

13-158

14-86 A desalination plant produces fresh water from seawater. The second law efficiency of the plant is to be determined. Assumptions 1 The seawater is an ideal solution since it is dilute. 2 The total dissolved solids in water can be treated as table salt (NaCl). 3 The environment temperature is equal to the seawater temperature. Properties The molar masses of water and salt are M w  18.0 kg / kmol and M s  58.44 kg / kmol . The gas constant of pure water is Rw  0.4615 kJ / kg  K (Table A-1). The density of river water is 1000 kg / m3 . Analysis First we determine the mole fraction of pure water in seawater using Eqs. 14-4 and 14-5. Noting that mfs  0.032 and mf w  1  mfs  0.968 ,

Mm =

1 1 1 = = = 18.4076 kg/kmol mf i mf s mf w 0.032 0.968 å M M + M 58.44 + 18.0 i s w

yi = mf i

Mm M 18.4076 kg/kmol ® yw = mf w m = (0.968) = 0.98992 Mi Mw 18.0 kg/kmol

The maximum work output associated with mixing 1 kg of seawater (or the minimum work input required to produce 1 kg of freshwater from seawater) is

wmax, out = RwT0 ln(1/ yw ) = (0.4615 kJ/kg ×K)(283.15 K)ln(1/0.98992) = 1.3238 kJ/kg fresh water The power that can be generated as 1.4 m3 / s fresh water mixes reversibly with seawater is æ1 kW ÷ ö Wmax out = r V wmax out = (1000 kg/m3 )(1.2 m3 /s)(1.3238 kJ/kg) çç = 1589 kW ÷ çè1 kJ/s ÷ ø

Then the second law efficiency of the plant becomes

hII =

Wmin,in Win

1.589 MW = 0.187 = 18.7% 8.5 MW

EES (Engineering Equation Solver) SOLUTION "Given" T=(10+273.15) [K] mass_s=3.2 [kg] V_dot=1.2 [m^3/s] W_dot_in=8500 [kW] "Properties" R_u=8.314 [kPa-m^3/kmol-K] M_w=18 [kg/kmol] M_s=58.44 [kg/kmol] "molar mass of salt" R_w=0.4615 [kJ/kg-K] rho=1000 [kg/m^3] "Analysis" mass_m=100 [kg] "consider 100 kg of mixture" mass_w=mass_m-mass_s N_s=mass_s/M_s N_w=mass_w/M_w N_m=N_s+N_w y_w=N_w/N_m m_dot=rho*V_dot W_dot_min_in=m_dot*R_w*T*ln(1/y_w) Eta_II=W_dot_min_in/W_dot_in

13-159

Review Problems 14-87 The molar fractions of constituents of air are given. The gravimetric analysis of air and its molar mass are to be determined. Assumptions All the constituent gases and their mixture are ideal gases. Properties The molar masses of O2 , N 2 , and Ar are 32.0, 28.0, and 40.0 kg / kmol . (Table A-1). Analysis For convenience, consider 100 kmol of air. Then the mass of each component and the total mass are

NO2 = 21 kmol

¾¾ ®

mO2 = NO2 M O2 = (21 kmol)(32 kg/kmol) = 672 kg

N N2 = 78 kmol

¾¾ ®

mN2 = N N2 M N2 = (78 kmol)(28 kg/kmol) = 2184 kg

N Ar = 1 kmol

¾¾ ®

mAr = N Ar M Ar = (1 kmol)(40 kg/kmol) = 40 kg

mm = mO2 + mN2 + mAr = 672 kg + 2184 kg + 40 kg = 2896 kg

Then the mass fraction of each component (gravimetric analysis) becomes

mf O2 = mf N2 = mf Ar =

mO2

672 kg = 0.232 2896 kg

23.2%

2184 kg = 0.754 2896 kg

75.4%

mAr 40 kg = = 0.014 mm 2896 kg

1.4%

mm mN2 mm

The molar mass of the mixture is determined from its definitions, Mm =

mm 2896 kg = = 28.96 kg / kmol N m 100 kmol

EES (Engineering Equation Solver) SOLUTION "Given" y_O2=0.21 y_N2=0.78 y_Ar=0.01 "Properties" M_O2=molarmass(O2) M_N2=molarmass(N2) M_Ar=molarmass(Argon) "Analysis" N_m=100 "[kmol], consider 100 kmol of mixture" N_O2=N_m*y_O2 N_N2=N_m*y_N2 N_Ar=N_m*y_Ar mass_O2=N_O2*M_O2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-160

mass_N2=N_N2*M_N2 mass_Ar=N_Ar*M_Ar mass_m=mass_O2+mass_N2+mass_Ar mf_O2=mass_O2/mass_m mf_N2=mass_N2/mass_m mf_Ar=mass_Ar/mass_m M_m=mass_m/N_m

14-88 The mole numbers of combustion gases are given. The partial pressure of water vapor and the condensation temperature of water vapor are to be determined. Properties The molar masses of CO2 , H 2 O , O2 and N 2 are 44.0, 18.0, 32.0, and 28.0 kg / kmol , respectively (Table A1). Analysis The total mole of the mixture and the mole fraction of water vapor are

Ntotal = 8 + 9 + 4 + 94 = 115 kmol yH2O =

N H2O 9 = = 0.07826 N total 115

Noting that molar fraction is equal to pressure fraction, the partial pressure of water vapor is

PH2O = yH2O Ptotal = (0.07826)(101 kPa) = 7.90 kPa The temperature at which the condensation starts is the saturation temperature of water at this pressure. This is called the dew-point temperature. Then,

Tcond = Tsat@7.90 kPa = 41.3°C

(Table A-5)

Water vapor in the combustion gases will start to condense when the temperature of the combustion gases drop to 41.3C.

EES (Engineering Equation Solver) SOLUTION "Given" N_CO2=8 [kmol] N_H2O=9 [kmol] N_O2=4 [kmol] N_N2=94 [kmol] P=101 [kPa] "Properties, Tables A-1, A-2" M_CO2=44 [kg/kmol] M_H2O=18 [kg/kmol] M_N2=28 [kg/kmol] "Analysis" N_total=N_CO2+N_H2O+N_O2+N_N2 y_H2O=N_H2O/N_total P_H2O=y_H2O*P T_cond=temperature(steam_iapws, P=P_H2O, x=1)

14-89 A mixture of gases is assembled by filling an evacuated tank with neon, oxygen, and nitrogen added one after another. The mass of each constituent in the resulting mixture, the apparent molecular weight of the mixture, and the fraction of the tank volume occupied by nitrogen are to be determined. Properties The molar masses of Ne, O2 , and N 2 are 20.18, 32.0, 28.0 kg / kmol , respectively and the gas constants are 0.4119, 0.2598, and 0.2968 kJ / kg  K , respectively (Table A-1). Analysis The mass of each constituent is calculated by © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-161

mNe =

P NeVm (35 kPa)(0.15 m3 ) = = 0.03828 kg R NeT (0.4119 kPa ×m3 /kg ×K)(333 K)

mO2 =

P O2Vm (70 kPa)(0.15 m3 ) = = 0.1214 kg R O2T (0.2598 kPa ×m3 /kg ×K)(333 K)

mN2 =

P N2Vm (35 kPa)(0.15 m3 ) = = 0.05312 kg R N2T (0.2968 kPa ×m3 /kg ×K)(333 K)

The mole number of each constituent is N Ne =

mNe 0.03828 kg = = 0.001896 kmol M Ne 20.18 kg/kmol

N O2 =

mO2 0.1214 kg = = 0.003794 kmol M O2 32.0 kg/kmol

N N2 =

mN2 0.05312 kg = = 0.001897 kmol M N2 28.0 kg/kmol

The apparent molecular weight of the mixture is Mm =

mm (0.03828 + 0.1214 + 0.05312) kg 0.2128 kg = = = 28.05 kg / kmol N m (0.001896 + 0.003794 + 0.001897) kmol 0.007586 kmol

The mole fraction of nitrogen is yN2 =

PN2 35 kPa = = 0.25 Pm 140 kPa

The partial volume occupied by nitrogen is then

VN2 = y N2Vm = (0.25)(0.15 m3 ) = 0.0375 m3

14-90 A mixture of carbon dioxide and nitrogen flows through a converging nozzle. The required make up of the mixture on a mass basis is to be determined. Assumptions Under specified conditions CO2 and N 2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 and N 2 are 44.0 and 28.0 kg / kmol , respectively (Table A-1). The specific heat ratios of CO2 and N 2 at 500 K are kCO2  1.229 and kN2  1.391 (Table A-2). Analysis The molar mass of the mixture is determined from

M m = yCO2 M CO2 + yN2 M N2

13-162

The molar fractions are related to each other by

yCO2 + yN2 = 1 The gas constant of the mixture is given by Rm =

Ru Mm

The specific heat ratio of the mixture is expressed as

k = mf CO2 kCO2 + mf N2 kN2 The mass fractions are

mf CO2 = yCO2 mf N2 = yN2

M CO2 Mm

M N2 Mm

The exit velocity equals the speed of sound at 500 K

Vexit =

æ1000 m 2 /s 2 ö÷ ÷ kRmT çç ÷ çè 1 kJ/kg ø÷

Substituting the given values and known properties and solving the above equations simultaneously using EES, we find mf CO2 = 0.838

mf N2 = 0.162

EES (Engineering Equation Solver) SOLUTION T = 500 [K] V_exit=360 [m/s] R_u=8.314 [kJ/kmol-K] "Assume the mixture is an ideal gas mixture. " "The mole fractions are y_CO2 and y_N2 " "The molar mass of the mixture is:" MM_Mix = y_CO2 *MOLARMASS(CO2)+y_N2*MOLARMASS(N2) "The gas constants are:" R_Mix=R_u/MM_Mix R_CO2=R_u/MOLARMASS(CO2) R_N2=R_u/MOLARMASS(N2) "The mass fractions are:" mf_CO2 = y_CO2 *MOLARMASS(CO2)/MM_Mix mf_N2 = y_N2 *MOLARMASS(N2)/MM_Mix "The gas constant of the mixture is:" k=(mf_CO2*CP(CO2,T=T)+mf_N2*CP(CO2,T=T))/(mf_CO2*(CP(CO2,T=T)-R_CO2)+mf_N2*(CP(CO2,T=T)R_N2)) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-163

"The exit velocity equals the speed of sound at 500K." V_exit=sqrt(k*R_Mix*T*convert(kJ/kg,m^2/s^2)) "The mole fractions are related by:" y_CO2 =1- y_N2

14-91 The volumetric fractions of the constituents of combustion gases are given. The mixture undergoes a reversible adiabatic expansion process in a piston-cylinder device. The work done is to be determined. Assumptions Under specified conditions all CO2 , H 2 O , O2 , and N 2 can be treated as ideal gases, and the mixture as an ideal gas mixture. Properties The molar masses of CO2 , H 2 O , O2 , and N 2 are 44.0, 18.0, 32.0, and 28.0 kg / kmol , respectively (Table A1). Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of the mixture is determined to be

M m = yCO2 M CO2 + yH2 O M H2 O + yO2 M O2 + yN2 M N2 = (0.0489)(44) + (0.0650)(18) + (0.1220)(32) + (0.7641)(28) = 28.63 kg/kmol

The mass fractions are

mf CO2 = yCO2

mf H2 O = yH2 O mf O2 = yO2 mf N2 = yN2

M CO2 Mm

M H2 O Mm

M O2 Mm M N2 Mm

= (0.0489)

44 kg/kmol = 0.07516 28.63 kg/kmol

= (0.0650)

18 kg/kmol = 0.0409 28.63 kg/kmol

= (0.1220)

32 kg/kmol = 0.1363 28.63 kg/kmol

= (0.7641)

28 kg/kmol = 0.7476 28.63 kg/kmol

Using Dalton’s law to find partial pressures, the entropies at the initial state are determined from EES to be:

T = 1800 K, P = (0.0489´ 1000) = 48.9 kPa ¾ ¾ ® sCO2 ,1 = 7.0148 kJ/kg.K T = 1800 K, P = (0.0650´ 1000) = 65 kPa ¾ ¾ ® sH2O,1 = 14.590 kJ/kg.K T = 1800 K, P = (0.1220´ 1000) = 122 kPa ¾ ¾ ® sN2 ,1 = 8.2570 kJ/kg.K © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-164

T = 1800 K, P = (0.7641´ 1000) = 764.1 kPa ¾ ¾ ® sO2 ,1 = 8.2199 kJ/kg.K The final state entropies cannot be determined at this point since the final temperature is not known. However, for an isentropic process, the entropy change is zero and the final temperature may be determined from

D stotal = mf CO2 D sCO2 + mf H2 O D sH2 C + mf O2 D sO2 + mf N2 D sN2 = 0 The solution may be obtained using EES to be

T2  1253 K The initial and final internal energies are (from EES)

uCO2 ,1 = - 7478 kJ/kg T1 = 1800 K ¾ ¾ ®

uH2 C,1 = - 10, 779 kJ/kg uO2 ,1 = 1147 kJ/kg uN2 ,1 = 1214 kJ/kg, uCO2 ,2 = - 8102 kJ/kg

T2 = 1253 K ¾ ¾ ®

uH2 C,2 = - 11,955 kJ/kg uO2 ,2 = 662.8 kJ/kg uN2 ,2 = 696.5 kJ/kg

Noting that the heat transfer is zero, an energy balance on the system gives

qin - wout = D um ¾ ¾ ® wout = - D um where

D um = mf CO2 (uCO2 ,2 - uCO2 ,1 ) + mf CO (uH2O,2 - uH2O,1 ) + mfO2 (uO2 ,2 - uO2 ,1 ) + mf N 2 (uN 2 ,2 - uN 2 ,1 ) Substituting,

wout = - D um = - 0.07516[(- 8102) - (- 7478) ]- 0.0409 [(- 11,955) - (- 10, 779)]- 0.1363[662.8 - 1147 ]- 0.7476 [696.5 - 1214 ]

= 547.8 kJ / kg

EES (Engineering Equation Solver) SOLUTION T_1 = 1800 [K] P_1=1000 [kPa] P_2=200 [kPa] R_u=8.314 [kJ/kmol-K] "Assume the mixture is an ideal gas mixture. Note that volume fractions equal mole fractions in ideal gas mixtures." "The mole fractions are:" y_CO2 = 0.0489 y_H2O =0.065 y_O2 = 0.122 y_N2 = 0.7641 "The molar mass of the mixture is:" MM_Mix = y_CO2 *MOLARMASS(CO2)+y_H2O *MOLARMASS(H2O) + y_O2 *MOLARMASS(O2)+y_N2*MOLARMASS(N2)"[kg/kmol]" R_Mix=R_u/MM_Mix "The mass fractions are:" mf_CO2 = y_CO2 *MOLARMASS(CO2)/MM_Mix"[kg_CO2/kg_Mix]" mf_H2O = y_H2O*MOLARMASS(H2O)/MM_Mix"[kg_H2O/kg_Mix]" mf_O2 = y_O2 *MOLARMASS(O2)/MM_Mix"[kg_O2/kg_Mix]" mf_N2 = y_N2 *MOLARMASS(N2)/MM_Mix"[kg_N2/kg_Mix]" "Assume the process is reversible and adiabatic, therefore, isentorpic:" "Daltons Law is used to determine the partial pressures." © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-165

"The changes in entropy are:" DELTAs_CO2 = entropy(CO2,T=T_2, P=y_CO2*P_2)-entropy(CO2,T=T_1, P=y_CO2*P_1)"[kJ/kg_CO2 - K]" DELTAs_H2O = entropy(H2O,T=T_2, P=y_H2O*P_2)-entropy(H2O,T=T_1, P=y_H2O*P_1)"[kJ/kg_H2O - K]" DELTAs_O2 = entropy(O2,T=T_2, P=y_O2*P_2)-entropy(O2,T=T_1, P=y_O2*P_1)"[kJ/kg_O2 - K]" DELTAs_N2 = entropy(N2,T=T_2, P=y_N2*P_2)-entropy(N2,T=T_1, P=y_N2*P_1)"[kJ/kg_N2 - K]" s_CO2_1=entropy(CO2,T=T_1, P=y_CO2*P_1) s_CO2_2=entropy(CO2,T=T_2, P=y_CO2*P_2) s_H2O_1=entropy(H2O,T=T_1, P=y_H2O*P_1) s_H2O_2=entropy(H2O,T=T_2, P=y_H2O*P_2) s_O2_1=entropy(O2,T=T_1, P=y_O2*P_1) s_O2_2=entropy(O2,T=T_2, P=y_O2*P_2) s_N2_1=entropy(N2,T=T_1, P=y_N2*P_1) s_N2_2=entropy(N2,T=T_2, P=y_N2*P_2) "For the isentropic pocess DELTAs_Mix=0. This allows the determination of T_2." DELTAs_Mix= mf_CO2*DELTAs_CO2+mf_H2O*DELTAs_H2O +mf_O2*DELTAs_O2 +mf_N2*DELTAs_N2 "[kJ/kg_Mix - K]" DELTAs_Mix=0 "The changes in internal energy are:" DELTAu_CO2 = intenergy(CO2,T=T_2)-intenergy(CO2,T=T_1)"[kJ/kg_CO2]" DELTAu_H2O = intenergy(H2O,T=T_2)-intenergy(H2O,T=T_1)"[kJ/kg_H2O]" DELTAu_O2 = intenergy(O2,T=T_2)-intenergy(O2,T=T_1)"[kJ/kg_O2]" DELTAu_N2 = intenergy(N2,T=T_2)-intenergy(N2,T=T_1)"[kJ/kg_N2]" u_CO2_1=intenergy(CO2,T=T_1) u_CO2_2=intenergy(CO2,T=T_2) u_H2O_1=intenergy(H2O,T=T_1) u_H2O_2=intenergy(H2O,T=T_2) u_O2_1=intenergy(O2,T=T_1) u_O2_2=intenergy(O2,T=T_2) u_N2_1=intenergy(N2,T=T_1) u_N2_2=intenergy(N2,T=T_2) DELTAu_Mix = mf_CO2*DELTAu_CO2+mf_H2O*DELTAu_H2O +mf_O2*DELTAu_O2 +mf_N2*DELTAu_N2 "[kJ/kg_Mix]" "Conservation of energy:" "The heat transfer is zero, isentropic pocess:" q=0"[kJ/kg]" q - w=DELTAu_Mix

14-92 A mixture of carbon dioxide, nitrogen, and oxygen is compressed isothermally. The required work is to be determined. Assumptions 1 Nitrogen, oxygen, and carbon dioxide are ideal gases. 2 The process is reversible. Properties The mole numbers of nitrogen, oxygen, and carbon dioxide are 28.0, 32.0, and 44.0 kg / kmol , respectively (Table A-1). Analysis The mole fractions are yCO2 =

N CO2 1 kmol = = 0.4348 N total 2.3 kmol

yN2 =

N N2 1 kmol = = 0.4348 N total 2.3 kmol

yO2 =

N O2 0.3 kmol = = 0.1304 N total 2.3 kmol

13-166

The gas constant for this mixture is then R= =

Ru yCO2 M CO2 + yN2 M N2 + yO2 M O2 8.314 kJ/kmol ×K = 0.2343 kJ/kg ×K (0.4348´ 44 + 0.4348´ 28 + 0.1304´ 32)kg/kmol

The mass of this mixture of gases is

m = NCO2 M CO2 + N N2 M N2 + NO2 M O2 = 1´ 44 + 1´ 28 + 0.3´ 32 = 81.6 kg Noting that Pv = RT for an ideal gas, the work done for this process is then 2

v P dv = mRT ln 2 = mRT ln 1 v v1 P2 1

Wout = m ò Pd v = mRT ò 1

= (81.6 kg)(0.2343 kJ/kg ×K)(300 K) ln

10 kPa 100 kPa

= - 13, 200 kJ

The negative sign shows that the work is done on the system.

14-93 The mole numbers, pressure, and temperature of the constituents of a gas mixture are given. The volume of the tank containing this gas mixture is to be determined using three methods. Analysis (a) Under specified conditions both N 2 and CH4 will considerably deviate from the ideal gas behavior. Treating the mixture as an ideal gas gives N m = N N2 + N CH4 = 2 kmol + 6 kmol = 8 kmol and

N m RuTm (8 kmol)(8.314 kPa ×m3 /kmol ×K)(200 K) = = 1.11 m 3 Pm 12,000 kPa

Vm =

(b) To use Kay's rule, we first need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture using the critical point properties of N 2 and CH4 from Table A-1,

N N2

yN 2 =

yCH4 =

NCH4 Nm

2 kmol = 0.25 8 kmol =

6 kmol = 0.75 8 kmol

Tcr,¢ m = å yiTcr, i = yN2 Tcr, N2 + yCH4 Tcr, CH4 = (0.25)(126.2 K) + (0.75)(191.1 K) = 174.9 K

13-167

= (0.25)(3.39 MPa) + (0.75)(4.64 MPa) = 4.33 MPa

Then, Tm 200 = = 1.144 Tcr,' m 174.9

TR =

P 12 PR = ' m = = 2.77 4.33 Pcr, m

ü ïï ïï ïï ý Z m = 0.47 ïï ïï ïïþ

(Fig. A-15)

Thus, Vm =

Z m N m Ru Tm = Z mVideal = (0.47)(1.11 m 3 ) = 0.522 m 3 Pm

(c) To use the Amagat's law for this real gas mixture, we first need to determine the Z of each component at the mixture temperature and pressure, ü T 200 ï TR , N2 = m = = 1.585 ïï ïï Tcr, N2 126.2 (Fig. A-15) N2 : ý Z N2 = 0.85 Pm ïï 12 PR , N2 = = = 3.54 ïï Pcr, N2 3.39 ïï þ Tm 200 ïü TR , CH4 = = = 1.047 ïï ïï Tcr, CH4 191.1 (Fig. A-15) CH4 : ý Z CH4 = 0.37 Pm ï 12 PR , CH4 = = = 2.586 ïï ïï Pcr, CH4 4.64 ïþ Mixture:

Z m = å yi Zi = yN2 Z N2 + yCH4 ZCH4 = (0.25)(0.85)+ (0.75)(0.37) = 0.49 Thus, Vm =

Z m N m RuTm = Z mVideal = (0.49)(1.11 m3 ) = 0.544 m 3 Pm

EES (Engineering Equation Solver) SOLUTION "Given" N_N2=2 [kmol] N_CH4=6 [kmol] T=200 [K] P=12000 [kPa] "Properties" R_u=8.314 [kJ/kmol-K] T_cr_N2=T_CRIT(N2) P_cr_N2=P_CRIT(N2) T_cr_CH4=T_CRIT(CH4) P_cr_CH4=P_CRIT(CH4) "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-168

"(a)" N_m=N_N2+N_CH4 V_m_ideal=N_m*R_u*T/P "(b)" y_N2=N_N2/N_m y_CH4=N_CH4/N_m T_cr_m=y_N2*T_cr_N2+y_CH4*T_cr_CH4 P_cr_m=y_N2*P_cr_N2+y_CH4*P_cr_CH4 T_R_m=T/T_cr_m P_R_m=P/P_cr_m Z_m_Kay=COMPRESS(T_R_m, P_R_m) "compressibility factor function" V_m_Kay=Z_m_Kay*V_m_ideal "(c)" T_R_N2=T/T_cr_N2 P_R_N2=P/P_cr_N2 Z_N2=COMPRESS(T_R_N2, P_R_N2) "compressibility factor function" T_R_CH4=T/T_cr_CH4 P_R_CH4=P/P_cr_CH4 Z_CH4=COMPRESS(T_R_CH4, P_R_CH4) "compressibility factor function" Z_m_Amagat=y_N2*Z_N2+y_CH4*Z_CH4 V_m_Amagat=Z_m_Amagat*V_m_ideal

14-94 The specific heat ratio and an apparent molecular weight of a mixture of ideal gases are given. The work required to compress this mixture isentropically in a closed system is to be determined. Analysis For an isentropic process of an ideal gas with constant specific heats, the work is expressed as 2

wout = ò Pd v = P1v1k ò v - k d v 1

1- k ù P1v1k 1- k P1v1k éêæ v2 ö 1- k ÷ ç ÷ = (v 2 - v1 ) = - 1ú êççç ÷ ú ÷ 1- k 1- k êè v1 ø ú ë û

since P1v1k = Pv k for an isentropic process. Also,

P1v1 = RT1 (v 2 / v1 )k = P1 / P2

Substituting, we obtain ( k - 1)/ k ù ö Ru T1 éêæ P2 ÷ ç ÷ wout = - 1úú êççç ÷ M (1- k ) êè P1 ø÷ úû ë

13-169

(1.35- 1)/1.35 ù ö (8.314 kJ/kmol ×K)(308 K) éêæ ú çç700 kPa ÷ 1 ÷ ú ÷ (32 kg/kmol)(1- 1.35) êëêçè100 kPa ø ú û

= - 150 kJ / kg The negative sign shows that the work is done on the system.

EES (Engineering Equation Solver) SOLUTION "Given" k=1.35 M=32 [kg/kmol] P1=100 [kPa] T1=(35+273) [K] P2=700 [kPa] "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" w_out=(R_u*T1)/(M*(1−k))*((P2/P1)^((k−1)/k) −1)

14-95 The masses, pressures, and temperatures of the constituents of a gas mixture in a tank are given. Heat is transferred to the tank. The final pressure of the mixture and the heat transfer are to be determined. Assumptions He is an ideal gas and O2 is a nonideal gas. Properties The molar masses of He and O2 are 4.0 and 32.0 kg / kmol . (Table A-1) Analysis (a) The number of moles of each gas is N He =

N O2 =

mHe 4 kg = = 1 kmol M He 4.0 kg/kmol

mO2 M O2

8 kg = 0.25 kmol 32 kg/kmol

N m = N He + N O2 = 1 kmol + 0.25 kmol = 1.25 kmol

Then the partial volume of each gas and the volume of the tank are He:

VHe =

N He RuT1 (1 kmol)(8.314 kPa ×m3 /kmol ×K)(170 K) = = 0.202 m3 Pm, 1 7000 kPa

O2 :

13-170

ü ï 7 = 1.38 ïï ïï Pcr, O2 5.08 ý Z1 = 0.53 (Fig. A-15) ï T1 170 TR1 = = = 1.10 ïï ïï Tcr, O2 154.8 ïþ Pm , 1

PR1 =

VO2 =

ZN O2 RuT1 Pm, 1

(0.53)(0.25 kmol)(8.314 kPa ×m3 /kg ×K)(170 K) = 0.027 m3 7000 kPa

Vtank = VHe + VO2 = 0.202 m3 + 0.027 m3 = 0.229 m3 The partial pressure of each gas and the total final pressure is He:

N He RuT2 (1 kmol)(8.314 kPa ×m3 /kmol ×K)(220 K) = = 7987 kPa Vtank 0.229 m3

PHe, 2 = O2 :

ü ïï ïï ïï ïï ïý P = 0.39 ïï R ïï ïï (0.229 m3 )/(0.25 kmol) = = 3.616 ïï (8.314 kPa ×m3 /kmol ×K)(154.8 K)/(5080 kPa) ïþ T2

220 = 1.42 Tcr, O2 154.8 v O2 Vm / N O2 v R , O2 = = Ru Tcr, O2 / Pcr, O2 RuTcr, O2 / Pcr, O2 TR2 =

(Fig. A-15)

PO2 = (PR Pcr )O = (0.39)(5080 kPa ) = 1981 kPa = 1.981 MPa 2

Pm, 2 = PHe + PO2 = 7.987 MPa + 1.981 MPa = 9.97 MPa (b) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a closed system with no work interactions. Then the energy balance for this closed system reduces to Ein - Eout = D Esystem

Qin = D U = D U He + D U O2 He:

D U He = mcv (Tm - T1 ) = (4 kg )(3.1156 kJ/kg ×K )(220 - 170)K = 623.1 kJ

O2 : TR1 = 1.10 ü ïï ý Z = 2.2 PR1 = 1.38 ïïþ h1 ü TR2 = 1.42 ïï ï Z = 1.2 ý h 9.97 PR2 = = 1.963 ïï 2 ïþ 5.08

(Fig. A-29)

h2 - h1 = RuTcr (Z h1 - Z h2 ) + (h2 - h1 )ideal

= (8.314 kJ/kmol ×K)(154.8 K)(2.2 - 1.2) + (6404 - 4949) kJ/kmol = 2742 kJ/kmol

Also,

PHe, 1 =

N He RuT1 (1 kmol)(8.314 kPa ×m3 /kg ×K)(170 K) = = 6,172 kPa Vtank 0.229 m3

13-171

Thus, D U O2 = N O2 (h2 - h1 ) - ( P2V2 - P1V1 ) = NO2 (h2 - h1 ) - ( PO2 , 2 - PO2 , 1 )Vtank = (0.25 kmol)(2742 kJ/kmol) - (1981- 828)(0.229) kPa ×m3 = 421.5 kJ

Substituting,

Qin = 623.1 kJ + 421.5 kJ = 1045 kJ

EES (Engineering Equation Solver) SOLUTION "Given" mass_He=4 [kg] mass_O2=8 [kg] T1=170 [K] P1=7000 [kPa] T2=220 [K] "Properties" R_u=8.314 [kJ/kmol-K] T_cr_O2=154.8 [K] P_cr_O2=5080 [kPa] M_O2=32 [kg/kmol] M_He=4 [kg/kmol] c_v_He=3.1156 [kJ/kg-K]*M_He "Analysis" "(a)" N_He=mass_HE/M_He N_O2=mass_O2/M_O2 N_m=N_He+N_O2 V_He=N_He*R_u*T1/P1 T_R1_O2=T1/T_cr_O2 P_R1_O2=P1/P_cr_O2 Z_1=COMPRESS(T_R1_O2, P_R1_O2) "the function that returns compressibility factor" V_O2=Z_1*N_O2*R_u*T1/P1 V_tank=V_He+V_O2 P2_He=N_He*R_u*T2/V_tank T_R2_O2=T2/T_cr_O2 P_R2_O2=P2_O2/P_cr_O2 Z_2=COMPRESS(T_R2_O2, P_R2_O2) "the function that returns compressibility factor" P2_O2=(Z_2*N_O2*R_u*T2)/V_tank P2_m=P2_He+P2_O2 "(b)" DELTAU_He=N_He*C_v_He*(T2-T1) Z_h1=ENTHDEP(T_R1_O2, P_R1_O2) "the function that returns enthalpy departure factor" P_R2_O2_m=P2_m/P_cr_O2 Z_h2=ENTHDEP(T_R2_O2, P_R2_O2_m) "the function that returns enthalpy departure factor" h1_O2=enthalpy(O2, T=T1) h2_O2=enthalpy(O2, T=T2) DELTAh_ideal_O2=h2_O2-h1_O2 DELTAh_O2=DELTAh_ideal_O2-R_u*T_cr_O2*(Z_h2-Z_h1) P1_He=N_He*R_u*T1/V_tank P1_O2=P1-P1_He DELTAU_O2=N_O2*DELTAh_O2-(P2_O2-P1_O2)*V_tank Q_in=DELTAU_He+DELTAU_O2

13-172

14-96 A mixture of gases is placed in a spring-loaded piston-cylinder device. The device is now heated until the pressure rises to a specified value. The total work and heat transfer for this process are to be determined. Properties The molar masses of Ne, O2 , and N 2 are 20.18, 32.0, 28.0 kg / kmol , respectively and the gas constants are 0.4119, 0.2598, and 0.2968 kJ / kg  K , respectively (Table A-1). The constant-volume specific heats are 0.6179, 0.658, and

0.743 kJ / kg  K , respectively (Table A-2a). Analysis The total pressure is 200 kPa and the partial pressures are

PNe = yNe Pm = (0.25)(200 kPa) = 50 kPa PO2 = yO2 Pm = (0.50)(200 kPa) = 100 kPa PN2 = yN2 Pm = (0.25)(200 kPa) = 50 kPa

The mass of each constituent for a volume of 0.1 m3 and a temperature of 10C are

mNe =

P NeVm (50 kPa)(0.1 m3 ) = = 0.04289 kg R NeT (0.4119 kPa ×m3 /kg ×K)(283 K)

mO2 =

P O2Vm (100 kPa)(0.1 m3 ) = = 0.1360 kg R O2T (0.2598 kPa ×m3 /kg ×K)(283 K)

mN2 =

P N2Vm (50 kPa)(0.1 m3 ) = = 0.05953 kg R N2T (0.2968 kPa ×m3 /kg ×K)(283 K)

mtotal = 0.04289 + 0.1360 + 0.05953 = 0.2384 kg

The mass fractions are

13-173

mf Ne =

mNe 0.04289 kg = = 0.1799 mm 0.2384 kg

mf O2 =

mO2 0.1360 kg = = 0.5705 mm 0.2384 kg

mf N2 =

mN2 0.05953 kg = = 0.2497 mm 0.2384 kg

The constant-volume specific heat of the mixture is determined from

cv = mf Ne cv ,Ne + mfO2 cv ,O2 + mf N2 cv ,N2 = 0.1799´ 0.6179 + 0.5705´ 0.658 + 0.2497´ 0.743

= 0.672 kJ/kg ×K The moles are N Ne =

mNe 0.04289 kg = = 0.002126 kmol M Ne 20.18 kg/kmol

N O2 =

mO2 0.1360 kg = = 0.00425 kmol M O2 32 kg/kmol

N N2 =

mN2 0.05953 kg = = 0.002126 kmol M N2 28 kg/kmol

Nm = N Ne + NO2 + N N2 = 0.008502 kmol Then the apparent molecular weight of the mixture becomes Mm =

mm 0.2384 kg = = 28.04 kg/kmol N m 0.008502 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ×K = = 0.2964 kJ/kg ×K Mm 28.05 kg/kmol

The mass contained in the system is

P 1V1 (200 kPa)(0.1 m3 ) = = 0.2384 kg RT1 (0.2964 kPa ×m3 /kg ×K)(283 K)

Noting that the pressure changes linearly with volume, the final volume is determined by linear interpolation to be

500 - 200 V2 - 0.1 = ¾¾ ® V2 = 0.4375 m3 1000 - 200 1.0 - 0.1 The final temperature is

T2 =

P 2V2 (500 kPa)(0.4375 m3 ) = = 3096 K mR (0.2384 kg)(0.2964 kPa ×m3 /kg ×K)

The work done during this process is

Wout =

P 1+ P 2 (500 + 200) kPa (V2 - V1 ) = (0.4375 - 0.1) m3 = 118 kJ 2 2

An energy balance on the system gives

Qin = Wout + mcv (T2 - T1 ) = 118 + (0.2384 kg)(0.672 kJ/kg ×K)(3096 - 283) K = 569 kJ

13-174

y_Ne=0.25 y_O2=0.50 y_N2=0.25 Vol1a=0.1 [m^3] P1a=200 [kPa] Vol2a=1.0 [m^3] P2a=1000 [kPa] Vol1=0.1 [m^3] P1=200 [kPa] T1=(10+273) [K] P2=500 [kPa] "Properties, Tables A-1, A-2" R_u=8.314 [kJ/kmol-K] M_Ne=20.18 [kg/kmol] M_O2=32 [kg/kmol] M_N2=28 [kg/kmol] R_Ne=0.4119 [kJ/kg-K] R_O2=0.2598 [kJ/kg-K] R_N2=0.2968 [kJ/kg-K] c_v_Ne=0.6179 [kJ/kg-K] c_v_O2=0.658 [kJ/kg-K] c_v_N2=0.743 [kJ/kg-K] "Analysis" P_Ne=y_Ne*P1 P_O2=y_O2*P1 P_N2=y_N2*P1 mass_Ne=(P_NE*Vol1)/(R_Ne*T1) mass_O2=(P_O2*Vol1)/(R_O2*T1) mass_N2=(P_N2*Vol1)/(R_N2*T1) mass_m=mass_Ne+mass_O2+mass_N2 mf_Ne=mass_Ne/mass_m mf_O2=mass_O2/mass_m mf_N2=mass_N2/mass_m N_Ne=mass_Ne/M_Ne N_O2=mass_O2/M_O2 N_N2=mass_N2/M_N2 N_m=N_Ne+N_O2+N_N2 M_m=mass_m/N_m c_v=mf_Ne*c_v_Ne+mf_O2*c_v_O2+mf_N2*c_v_N2 R=R_u/M_m mass=(P1*Vol1)/(R*T1) (P2-P1)/(P2a-P1a)=(Vol2-Vol1)/(Vol2a-Vol1a) T2=(P2*Vol2)/(mass*R) W_out=(P1+P2)/2*(Vol2-Vol1) Q_in=W_out+mass*c_v*(T2-T1)

14-97 A spring-loaded piston-cylinder device is filled with a mixture of nitrogen and carbon dioxide whose mass fractions are given. The gas is heated until the volume has doubled. The total work and heat transfer for this process are to be determined. Properties The molar masses of N 2 and CO2 are 28.0 and 44.0 kg / kmol , respectively (Table A-1). The constant-volume specific heats of these gases at room temperature are 0.743 and 0.657 kJ / kg  K , respectively (Table A-2a). Analysis We consider 100 kg of this mixture. The mole numbers of each component are

13-175

N N2 =

mN2 55 kg = = 1.964 kmol M N2 28 kg/kmol

N CO2 =

mCO2 45 kg = = 1.023 kmol M CO2 44 kg/kmol

The mole number of the mixture is Nm = N N2 + NCO2 = 1.964 + 1.023 = 2.987 kmol The apparent molecular weight of the mixture is m 100 kg Mm = m = = 33.48 kg/kmol Nm 2.987 kmol The constant-volume specific heat of the mixture is determined from cv = mf N2 cv ,N2 + mf CO2 cv ,CO2

= 0.55´ 0.743 + 0.45´ 0.657 = 0.7043 kJ/kg ×K

V - 0.1 200 - 200 = 1 ¾¾ ® V1 = 0.1 m3 1000 - 200 1.0 - 0.1 The final volume is

13-176

The final pressure is similarly determined by linear interpolation using the data of the previous problem to be

P2 - 200 0.2 - 0.1 = ¾¾ ® P2 = 288.9 kPa 1000 - 200 1.0 - 0.1 The mass contained in the system is m=

P 1V1 (200 kPa)(0.1 m3 ) = = 0.2533 kg RT1 (0.2483 kPa ×m3 /kg ×K)(318 K)

The final temperature is

T2 =

P 2V2 (288.9 kPa)(0.2 m3 ) = = 918.7 K mR (0.2533 kg)(0.2483 kPa ×m3 /kg ×K)

The work done during this process is P + P2 (200 + 288.9) kPa Wout = 1 (V2 - V1 ) = (0.2 - 0.1) m 3 = 24.4 kJ 2 2 An energy balance on the system gives Qin = Wout + mcv (T2 - T1 ) = 24.4 + (0.2533 kg)(0.7043 kJ/kg ×K)(918.7 - 318) K = 132 kJ

EES (Engineering Equation Solver) SOLUTION "Given" mf_N2=0.55 mf_CO2=0.45 P1=200 [kPa] T1=(45+273) [K] V2=2*V1 "Properties" R_u=8.314 [kJ/kmol-K] MM_N2=28 [kg/kmol] MM_CO2=44 [kg/kmol] c_v_N2=0.743 [kJ/kg-K] c_v_CO2=0.657 [kJ/kg-K] "Analysis" m_tot=100 [kg] "we consider 100 kg mixture" m_N2=mf_N2*m_tot m_CO2=mf_CO2*m_tot m_m=m_N2+m_CO2 N_N2=m_N2/MM_N2 N_CO2=m_CO2/MM_CO2 N_m=N_N2+N_CO2 MM_m=m_m/N_m c_v=mf_N2*c_v_N2+mf_CO2*c_v_CO2 R=R_u/MM_m V1a=0.1 [m^3]; P1a=200 [kPa]; P2a=1000 [kPa]; V2a=1 [m^3] "From previous problem" (P1-P1a)/(P2a-P1a)=(V1-V1a)/(V2a-V1a) (P2-P1a)/(P2a-P1a)=(V2-V1a)/(V2a-V1a) m=(P1*V1)/(R*T1) T2=(P2*V2)/(m*R) W_out=(P1+P2)/2*(V2-V1) Q_in=W_out+m*c_v*(T2-T1)

14-98 A spring-loaded piston-cylinder device is filled with a mixture of nitrogen and carbon dioxide whose mass fractions are given. The gas is heated until the pressure has tripled. The total work and heat transfer for this process are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-177

Properties The molar masses of N 2 and CO2 are 28.0 and 44.0 kg / kmol , respectively (Table A-1). The constant-volume specific heats of these gases at room temperature are 0.743 and 0.657 kJ / kg  K , respectively (Table A-2a). Analysis We consider 100 kg of this mixture. The mole numbers of each component are m 55 kg N N2 = N2 = = 1.964 kmol M N2 28 kg/kmol N CO2 =

mCO2 45 kg = = 1.023 kmol M CO2 44 kg/kmol

= 0.55´ 0.743 + 0.45´ 0.657 = 0.7043 kJ/kg ×K

The apparent gas constant of the mixture is R 8.134 kJ/kmol ×K R= u = = 0.2483 kJ/kg ×K Mm 33.48 kg/kmol Noting that the pressure changes linearly with volume, the initial volume is determined by linear interpolation using the data of the previous problem to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-178

V - 0.1 200 - 200 = 1 ¾¾ ® V1 = 0.1 m3 1000 - 200 1.0 - 0.1 The final pressure is P2 = 3P1 = 3(200 kPa) = 600 kPa The final volume is similarly determined by linear interpolation using the data of the previous problem to be

600 - 200 V2 - 0.1 = ¾¾ ® V2 = 0.55 m3 1000 - 200 1.0 - 0.1 The mass contained in the system is m=

P 1V1 (200 kPa)(0.1 m3 ) = = 0.2533 kg RT1 (0.2483 kPa ×m3 /kg ×K)(318 K)

The final temperature is

T2 =

P 2V2 (600 kPa)(0.55 m3 ) = = 5247 K mR (0.2533 kg)(0.2483 kPa ×m3 /kg ×K)

The work done during this process is P + P2 (200 + 600) kPa Wout = 1 (V2 - V1 ) = (0.55 - 0.1) m 3 = 180 kJ 2 2 An energy balance on the system gives Qin = Wout + mcv (T2 - T1 ) = 180 + (0.2533 kg)(0.7043 kJ/kg ×K)(5247 - 318) K = 1059 kJ

EES (Engineering Equation Solver) SOLUTION "Given" mf_N2=0.55 mf_CO2=0.45 P1=200 [kPa] T1=(45+273) [K] P2=3*P1 "Properties" R_u=8.314 [kJ/kmol-K] MM_N2=28 [kg/kmol] MM_CO2=44 [kg/kmol] c_v_N2=0.743 [kJ/kg-K] c_v_CO2=0.657 [kJ/kg-K] "Analysis" m_tot=100 [kg] "we consider 100 kg mixture" m_N2=mf_N2*m_tot m_CO2=mf_CO2*m_tot m_m=m_N2+m_CO2 N_N2=m_N2/MM_N2 N_CO2=m_CO2/MM_CO2 N_m=N_N2+N_CO2 MM_m=m_m/N_m c_v=mf_N2*c_v_N2+mf_CO2*c_v_CO2 R=R_u/MM_m V1a=0.1 [m^3]; P1a=200 [kPa]; P2a=1000 [kPa]; V2a=1 [m^3] "From previous problem" (P1-P1a)/(P2a-P1a)=(V1-V1a)/(V2a-V1a) (P2-P1a)/(P2a-P1a)=(V2-V1a)/(V2a-V1a) m=(P1*V1)/(R*T1) T2=(P2*V2)/(m*R) W_out=(P1+P2)/2*(V2-V1) Q_in=W_out+m*c_v*(T2-T1)

13-179

14-99 The masses of components of a gas mixture are given. This mixture is expanded in an adiabatic, steady-flow turbine of specified isentropic efficiency. The second law efficiency and the exergy destruction during this expansion process are to be determined. Assumptions All gases will be modeled as ideal gases with constant specific heats. Properties The molar masses of O2 , CO2 , and He are 32.0, 44.0, and 4.0 kg / kmol , respectively (Table A-1). The constant-pressure specific heats of these gases at room temperature are 0.918, 0.846, and 5.1926 kJ / kg  K , respectively (Table A-2a). Analysis The total mass of the mixture is

mm = mO2 + mCO2 + mHe = 0.1+ 1+ 0.5 = 1.6 kg

The mole numbers of each component are N O2 =

mO2 0.1 kg = = 0.003125 kmol M O2 32 kg/kmol

N CO2 =

mCO2 1 kg = = 0.02273 kmol M CO2 44 kg/kmol

N He =

mHe 0.5 kg = = 0.125 kmol M He 4 kg/kmol

The mole number of the mixture is

Nm = NO2 + NCO2 + NHe = 0.003125 + 0.02273 + 0.125 = 0.15086 kmol The apparent molecular weight of the mixture is Mm =

mm 1.6 kg = = 10.61 kg/kmol N m 0.15086 kmol

The apparent gas constant of the mixture is R=

Ru 8.314 kJ/kmol ×K = = 0.7836 kJ/kg ×K Mm 10.61 kg/kmol

The mass fractions are mf O2 =

mO2 0.1 kg = = 0.0625 mm 1.6 kg

mf CO2 =

mCO2 1 kg = = 0.625 mm 1.6 kg

mf He =

mHe 0.5 kg = = 0.3125 mm 1.6 kg

13-180

The constant-pressure specific heat of the mixture is determined from c p = mf O2 c p ,O2 + mf CO2 c p ,CO2 + mf He c p ,He

= 0.0625´ 0.918 + 0.625´ 0.846 + 0.3125´ 5.1926

= 2.209 kJ/kg ×K Then the constant-volume specific heat is

cv = c p - R = 2.209 - 0.7836 = 1.425 kJ/kg ×K The specific heat ratio is

cp cv

2.209 = 1.550 1.425

The temperature at the end of the expansion for the isentropic process is ( k - 1)/ k

æP ö ÷ T2 s = T1 ççç 2 ÷ ÷ èç P1 ÷ ø

0.55/1.55

æ100 kPa ÷ ö = (600 K) çç ÷ çè1000 kPa ÷ ø

= 265 K

Using the definition of turbine isentropic efficiency, the actual outlet temperature is

T2 = T1 - hturb (T1 - T2 s ) = (600 K) - (0.90)(600 - 265) = 299 K The entropy change of the gas mixture is T P 299 100 s2 - s1 = c p ln 2 - R ln 2 = (2.209) ln - (0.7836) ln = 0.2658 kJ/kg ×K T1 P1 600 1000

The actual work produced is

wout = h1 - h2 = c p (T1 - T2 ) = (2.209 kJ/kg ×K)(600 - 299) K = 665 kJ/kg The reversible work output is

wrev,out = h1 - h2 - T0 (s1 - s2 ) = 665 kJ/kg - (298 K)(- 0.2658 kJ/kg ×K) = 744 kJ/kg The second-law efficiency and the exergy destruction are then

hII =

wout 665 = = 0.894 wrev,out 744

xdest = wrev,out - wout = 744 - 665 = 79 kJ / kg

14-100 A program is to be written to determine the mole fractions of the components of a mixture of three gases with known molar masses when the mass fractions are given, and to determine the mass fractions of the components when the mole fractions are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. Procedure Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C) {If Type$ <> ('mass fraction' OR 'mole fraction' ) then Call ERROR('Type$ must be set equal to "mass fraction" or "mole fraction".') GOTO 10 endif} Sum = A+B+C If ABS(Sum - 1) > 0 then goto 20 MM_A = molarmass(A$) MM_B = molarmass(B$) MM_C = molarmass(C$) If Type$ = 'mass fraction' then © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-181

mf_A = A mf_B = B mf_C = C sumM_mix = mf_A/MM_A+ mf_B/MM_B+ mf_C/MM_C y_A = mf_A/MM_A/sumM_mix y_B = mf_B/MM_B/sumM_mix y_C = mf_C/MM_C/sumM_mix GOTO 10 endif if Type$ = 'mole fraction' then y_A = A y_B = B y_C = C MM_mix = y_A*MM_A+ y_B*MM_B+ y_C*MM_C mf_A = y_A*MM_A/MM_mix mf_B = y_B*MM_B/MM_mix mf_C = y_C*MM_C/MM_mix GOTO 10 Endif Call ERROR('Type$ must be either mass fraction or mole fraction.') GOTO 10 20: Call ERROR('The sum of the mass or mole fractions must be 1') 10: END "Either the mole fraction y_i or the mass fraction mf_i may be given by setting the parameter Type$='mole fraction' when the mole fractions are given or Type$='mass fraction' is given" Type$='mole fraction' A$ = 'N2' B$ = 'O2' C$ = 'Argon' A = 0.71 "When Type$='mole fraction' A, B, C are the mole fractions" B = 0.28 "When Type$='mass fraction' A, B, C are the mass fractions" C = 0.01 Call Fractions(Type$,A$,B$,C$,A,B,C:mf_A,mf_B,mf_C,y_A,y_B,y_C) SOLUTION A=0.71 A$='N2' B=0.28 B$='O2' C=0.01 C$='Argon' mf_A=0.680 mf_B=0.306 mf_C=0.014 Type$='mole fraction' y_A=0.710 y_B=0.280 y_C=0.010

13-182

14-101 A program is to be written to determine the apparent gas constant, constant volume specific heat, and internal energy of a mixture of 3 ideal gases when the mass fractions and other properties of the constituent gases are given. Also, the program is to be run for a sample case. Analysis The problem is solved using EES, and the solution is given below. T=300 [K] A$='N2' B$='O2' C$='CO2' mf_A=0.71 mf_B=0.28 mf_C=0.01 R_u=8.314 [kJ/kmol–K] MM_A=molarmass(A$) MM_B=molarmass(B$) MM_C=molarmass(C$) SumM_mix=mf_A/MM_A+ mf_B/MM_B+mf_C/MM_C y_A=mf_A/MM_A/SumM_mix y_B=mf_B/MM_B/SumM_mix y_C=mf_C/MM_C/SumM_mix MM_mix=y_A*MM_A+y_B*MM_B+y_C*MM_C R_mix=R_u/MM_mix c_p_mix=mf_A*specheat(A$,T=T)+mf_B*specheat(B$,T=T)+mf_C*specheat(C$,T=T) c_v_mix=c_p_mix – R_mix u_mix=c_v_mix*T h_mix=c_p_mix*T SOLUTION A$='N2' B$='O2' C$='CO2' c_p_mix=1.006 [kJ/kg–K] c_v_mix=0.7206 [kJ/kg–K] h_mix=301.8 [kJ/kg] mf_A=0.71 mf_B=0.28 mf_C=0.01 MM_A=28.01 [kg/kmol] MM_B=32 [kg/kmol] MM_C=44.01 [kg/kmol] MM_mix=29.14 [kg/kmol] R_mix=0.2854 [kJ/kg–K] R_u=8.314 [kJ/kmol–K] SumM_mix=0.03432 T=300 [K] u_mix=216.2 [kJ/kg] y_A=0.7384 y_B=0.2549 y_C=0.00662

13-183

14-102 Using Dalton’s law, it is to be shown that Z m = å yi Z i for a real-gas mixture. i= 1

Analysis Using the compressibility factor, the pressure of a component of a real-gas mixture and of the pressure of the gas mixture can be expressed as Pi =

Z i N i RuTm Vm

and

Pm =

Z m N m RuTm Vm

Dalton's law can be expressed as Pm = å Pi (Tm ,Vm ) . Substituting, Z m N m Ru Tm ZNRT =å i i u m Vm Vm

Simplifying,

Z m N m = å Zi Ni Dividing by N m ,

Z m = å yi Z i where Z i is determined at the mixture temperature and volume.

14-103 Two mass streams of two different ideal gases are mixed in a steady-flow chamber while receiving energy by heat transfer from the surroundings. Expressions for the final temperature and the exit volume flow rate are to be obtained and two special cases are to be evaluated. Assumptions Kinetic and potential energy changes are negligible. Analysis (a) Mass and energy balances for the mixing process:

m1 + m2 = m3 m1h1 + m2 h2 + Qin = m3 h3 h = c pT

m1c p, 1T1 + m2 c p, 2T2 + Qin = m3c p, mT3 c p, m =

T3 =

m1 m c p, 1 + 2 c p, 2 m3 m3

m1c p , 1 m3c p , m

T1 +

m2 c p , 2 m3c p , m

T2 +

Qin m3c p , m

13-184

V3 = m3v 3 = m3

V3 =

R3T3 P3

m2 c p , 2 m3 R3 éêm1c p , 1 Qin ù ú T + T + 1 2 P3 êëêm3c p , m m3c p , m m3c p , m ú ú û

c p , 1 R3 m1 R1T1 c p , m R1

c p , 2 R3 m2 R2T2 c p , m R2

R3Qin P3c p , m

P3 = P1 = P2 V3 = R=

V3 =

c p , 1 R3 c p , m R1

V1 +

c p , 2 R3 c p , m R2

V2 +

R3Qin P3 c p , m

Ru R3 R M M R M , = u 1= 1, 3 = 2 M R1 M 3 Ru M 3 R2 M3

c p , 1 M1 c p, m M 3

V1 +

c p, 2 M 2 c p, m M 3

V2 +

Ru Qin P3 M 3c p , m

The mixture molar mass M3 is found as follows:

M 3 = å yi M i , yi =

mfi / M i mi , mfi = å mfi / M i å mi

V3 =

c p , 1M1 c p, m M 3

V1 +

c p, 2 M 2 c p, m M 3

(d) When adiabatically mixing the same two ideal gases, the mixture volume flow rate becomes M 3 = M1 = M 2 c p ,3 = c p ,1 = c p , 2

V3 = V1 + V2

Fundamentals of Engineering (FE) Exam Problems 14-104 An ideal gas mixture consists of 3 kmol of N 2 and 6 kmol of CO2 . The mass fraction of CO2 in the mixture is (a) 0.241 (b) 0.333 (c) 0.500 (d) 0.667 (e) 0.759 Answer (e) 0.759 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N1=3 [kmol] N2=6 [kmol] N_mix=N1+N2 MM1=28 [kg/kmol] MM2=44 [kg/kmol] m_mix=N1*MM1+N2*MM2 mf2=N2*MM2/m_mix © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-185

"Some Wrong Solutions with Common Mistakes:" W1_mf = N2/N_mix "Using mole fraction" W2_mf = 1-mf2 "The wrong mass fraction"

14-105 An ideal gas mixture whose apparent molar mass is 20 kg / kmol consists of nitrogen N 2 and three other gases. If the mole fraction of nitrogen is 0.55, its mass fraction is (a) 0.15 (b) 0.23 (c) 0.39 (d) 0.55 (e) 0.77 Answer (e) 0.77 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). M_mix=20 [kg/kmol] y_N2=0.55 M_N2=28 [kg/kmol] mf_N2=(M_N2/M_mix)*y_N2 "Some Wrong Solutions with Common Mistakes:" W1_mf = y_N2 "Taking mass fraction to be equal to mole fraction" W2_mf= y_N2*(M_mix/M_N2) "Using the molar mass ratio backwords" W3_mf= 1-mf_N2 "Taking the complement of the mass fraction"

14-106 An ideal gas mixture consists of 2 kmol of N 2 and 4 kmol of CO2 . The apparent gas constant of the mixture is (a) 0.215 kJ / kg  K (b) 0.225 kJ / kg  K (c) 0.243 kJ / kg  K (d) 0.875 kJ / kg  K (e) 1.24 kJ / kg  K Answer (a) 0.215 kJ / kg  K

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N1=2 [kmol] N2=4 [kmol] Ru=8.314 [kJ/kmol-K] MM1=28 [kg/kmol] MM2=44 [kg/kmol] R1=Ru/MM1 R2=Ru/MM2 N_mix=N1+N2 y1=N1/N_mix y2=N2/N_mix © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-186

MM_mix=y1*MM1+y2*MM2 R_mix=Ru/MM_mix "Some Wrong Solutions with Common Mistakes:" W1_Rmix =(R1+R2)/2 "Taking the arithmetic average of gas constants" W2_Rmix= y1*R1+y2*R2 "Using wrong relation for Rmixture"

14-107 A rigid tank is divided into two compartments by a partition. One compartment contains 3 kmol of N 2 at 400 kPa pressure and the other compartment contains 7 kmol of CO2 at 200 kPa. Now the partition is removed, and the two gases form a homogeneous mixture at 250 kPa. The partial pressure of N 2 in the mixture is (a) 75 kPa (b) 90 kPa (c) 125 kPa (d) 175 kPa (e) 250 kPa Answer (a) 75 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1 = 400 [kPa] P2 = 200 [kPa] P_mix=250 [kPa] N1=3 [kmol] N2=7 [kmol] MM1=28 [kg/kmol] MM2=44 [kg/kmol] N_mix=N1+N2 y1=N1/N_mix y2=N2/N_mix P_N2=y1*P_mix "Some Wrong Solutions with Common Mistakes:" W1_P1= P_mix/2 "Assuming equal partial pressures" W2_P1= mf1*P_mix; mf1=N1*MM1/(N1*MM1+N2*MM2) "Using mass fractions" W3_P1 = P_mix*N1*P1/(N1*P1+N2*P2) "Using some kind of weighed averaging"

14-108 A 60-L rigid tank contains an ideal gas mixture of 5 g of N 2 and 5 g of CO2 at a specified pressure and temperature. If N 2 were separated from the mixture and stored at mixture temperature and pressure, its volume would be (a) 30 L (b) 37 L (c) 42 L (d) 49 L (e) 60 L Answer (b) 37 L Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-187

V_mix=60 [L] m1=5 [g] m2=5 [g] MM1=28 [kg/kmol] MM2=44 [kg/kmol] N1=m1/MM1 N2=m2/MM2 N_mix=N1+N2 y1=N1/N_mix V1=y1*V_mix "Some Wrong Solutions with Common Mistakes:" W1_V1=V_mix*m1/(m1+m2) "Using mass fractions" W2_V1= V_mix "Assuming the volume to be the mixture volume"

14-109 An ideal gas mixture consists of 3 kg of Ar and 6 kg of CO2 gases. The mixture is now heated at constant volume from 250 K to 350 K. The amount of heat transfer is (a) 374 kJ (b) 436 kJ (c) 488 kJ (d) 525 kJ (e) 664 kJ Answer (c) 488 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m1=3 [kg] m2=6 [kg] T1=250 [K] T2=350 [K] cv1=0.3122 [kJ/kg-K] cp1=0.5203 [kJ/kg-K] cv2=0.657 [kJ/kg-K] cp2=0.846 [kJ/kg-K] MM1=39.95 [kg/kmol] MM2=44 [kg/kmol] "Applying Energy balance gives Q=DeltaU=DeltaU_Ar+DeltaU_CO2" Q=(m1*cv1+m2*cv2)*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Q = (m1+m2)*(cv1+cv2)/2*(T2-T1) "Using arithmetic average of properties" W2_Q = (m1*cp1+m2*cp2)*(T2-T1) "Using cp instead of cv" W3_Q = (m1*cv1+m2*cv2)*T2 "Using T2 instead of T2-T1"

14-110 One compartment of an insulated rigid tank contains 2 kmol of CO2 at 20C and 150 kPa while the other compartment contains 5 kmol of H 2 gas at 35C and 300 kPa. Now the partition between the two gases is removed, and the two gases form a homogeneous ideal gas mixture. The temperature of the mixture is (a) 25C (b) 30C (c) 22C © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-188

(d) 32C (e) 34C Answer (b) 30C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N_H2=5 [kmol] T1_H2=35 [C] P1_H2=300 [kPa] N_CO2=2 [kmol] T1_CO2=20 [C] P1_CO2=150 [kPa] cv_H2=10.183 [kJ/kg-K] cp_H2=14.307 [kJ/kg-K] cv_CO2=0.657 [kJ/kg-K] cp_CO2=0.846 [kJ/kg-K] MM_H2=2 [kg/kmol] MM_CO2=44 [kg/kmol] m_H2=N_H2*MM_H2 m_CO2=N_CO2*MM_CO2 "Applying Energy balance gives 0=DeltaU=DeltaU_H2+DeltaU_CO2" 0=m_H2*cv_H2*(T2-T1_H2)+m_CO2*cv_CO2*(T2-T1_CO2) "Some Wrong Solutions with Common Mistakes:" 0=m_H2*cp_H2*(W1_T2-T1_H2)+m_CO2*cp_CO2*(W1_T2-T1_CO2) "Using cp instead of cv" 0=N_H2*cv_H2*(W2_T2-T1_H2)+N_CO2*cv_CO2*(W2_T2-T1_CO2) "Using N instead of mass" W3_T2 = (T1_H2+T1_CO2)/2 "Assuming averate temperature"

14-111 A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas at 70C and 400 kPa. Now the gas expands at constant pressure until its volume doubles. The amount of heat transfer to the gas mixture is (a) 286 MJ (b) 71 MJ (c) 30 MJ (d) 15 MJ (e) 6.6 MJ Answer (b) 71 MJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). N_He=3 [kmol] N_Ar=7 [kmol] T1=(70+273) [K] P1=400 [kPa] P2=P1 T2=2*T1 "T2=2T1 since PV/T=const for ideal gases and it is given that P=constant" MM_He=4 [kg/kmol] MM_Ar=39.95 [kg/kmol] m_He=N_He*MM_He m_Ar=N_Ar*MM_Ar cp_Ar=0.5203 [kJ/kg-K] cv_Ar = 3122 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-189

cp_He=5.1926 [kJ/kg-K] cv_He = 3.1156 [kJ/kg-K] Q=m_Ar*cp_Ar*(T2-T1)+m_He*cp_He*(T2-T1) "For a P=const process, Q=DeltaH since DeltaU+Wb is DeltaH" "Some Wrong Solutions with Common Mistakes:" W1_Q =m_Ar*cv_Ar*(T2-T1)+m_He*cv_He*(T2-T1) "Using cv instead of cp" W2_Q=N_Ar*cp_Ar*(T2-T1)+N_He*cp_He*(T2-T1) "Using N instead of mass" W3_Q=m_Ar*cp_Ar*(T22-T1)+m_He*cp_He*(T22-T1); T22=2*(T1-273)+273 "Using C for T1" W4_Q=(m_Ar+m_He)*0.5*(cp_Ar+cp_He)*(T2-T1) "Using arithmetic averate of cp"

14-112 An ideal gas mixture of helium and argon gases with identical mass fractions enters a turbine at 1500 K and 1 MPa at a rate of 0.12 kg / s , and expands isentropically to 100 kPa. The power output of the turbine is (a) 253 kW (b) 310 kW (c) 341 kW (d) 463 kW (e) 550 kW Answer (b) 310 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=0.12 [kg/s] T1=1500 [K] P1=1000 [kPa] P2=100 [kPa] mf_He=0.5 mf_Ar=0.5 k_He=1.667 k_Ar=1.667 cp_Ar=0.5203 [kJ/kg-K] cp_He=5.1926 [kJ/kg-K] cp_mix=mf_He*cp_He+mf_Ar*cp_Ar k_mix=1.667 "The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases" T2=T1*(P2/P1)^((k_mix-1)/k_mix) -W_out=m*cp_mix*(T2-T1) "Some Wrong Solutions with Common Mistakes:" W1_Wout= - m*cp_mix*(T22-T1); T22 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix)+273 "Using C for T1 instead of K" W2_Wout= - m*cp_mix*(T222-T1); T222 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air" W3_Wout= - m*cp_mix*(T2222-T1); T2222 = T1*P2/P1 "Assuming T to be proportional to P" W4_Wout= - m*0.5*(cp_Ar+cp_He)*(T2-T1) "Using arithmetic average for cp"

14-113 An ideal gas mixture consists of 60% helium and 40% argon gases by mass. The mixture is now expanded isentropically in a turbine from 400C and 1.2 MPa to a pressure of 200 kPa. The mixture temperature at turbine exit is (a) 56C (b) 195C (c) 130C (d) 112C (e) 400C

13-190

Answer (a) 56C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). mf_He=0.6 mf_Ar=0.4 T1=(400+273) [K] P1=1200 [kPa] P2=200 [kPa] k1=1.667 k2=1.667 k_mix=1.667 "The specific heat ratio k of the mixture is also 1.667 since k=1.667 for all componet gases" T2=T1*(P2/P1)^((k_mix-1)/k_mix)-273 "Some Wrong Solutions with Common Mistakes:" W1_T2 = (T1-273)*(P2/P1)^((k_mix-1)/k_mix) "Using C for T1 instead of K" W2_T2 = T1*(P2/P1)^((k_air-1)/k_air)-273; k_air=1.4 "Using k value for air" W3_T2 = T1*P2/P1 "Assuming T to be proportional to P"

14-114 … 14-117 Design and Essay Problem

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

13-191

Chapter 15 GAS-VAPOR MIXTURES AND AIR-CONDITIONING

13-192

CYU - Check Your Understanding CYU 15-1 ■ Check Your Understanding CYU 15-1.1 Select the incorrect statement regarding dry and atmospheric air: (a) The atmospheric air can be treated as an ideal-gas mixture regardless of the amount of water vapor it contains. (b) Water vapor in the air behaves as an ideal gas despite being close to a saturation state. (c) The enthalpy of water vapor in air can be taken to be a function of temperature only as a good approximation. (d) The pressure of atmospheric air can be taken to be equal to the partial pressure of dry air. (e) The enthalpy of water vapor in air can be taken to be equal to the enthalpy of saturated vapor at the air temperature with good accuracy. Answer: (d) The pressure of atmospheric air can be taken to be equal to the partial pressure of dry air. CYU 15-1.2 The pressure of atmospheric air in a location is measured to be 95 kPa. If the partial pressure of water vapor in the air is 2 kPa, what is the partial pressure of dry air? (a) 0 kPa (b) 2 kPa (c) 93 kPa (d) 95 kPa (e) 101.3 kPa Answer: (c) 93 kPa

Pa = P - Pv = 95 - 2 = 93 kPa

CYU 15-2 ■ Check Your Understanding CYU 15-2.1 Select the incorrect statement regarding specific humidity and relative humidity. (a) The relative humidity of dry air is zero. (b) The relative humidity of saturated air is one. (c) The specific humidity of air increases when it is cooled. (d) The relative humidity of air decreases when it is heated. (e) The air which contains the highest possible amount of water vapor is called saturated air. Answer: (c) The specific humidity of air increases when it is cooled. CYU 15-2.2 Select the incorrect relation below for specific humidity. (a) w =

(b) w =

mv ma + mv

0.622f Pg P - f Pg P Pa

13-193

(d) w =

0.622Pv P - Pa

(e) None of these

Answer: (a) w =

mv ma + mv

CYU 15-2.3 Select the incorrect relation for relative humidity. (a) f =

mv mg

(b) f =

Pv Pg

P - Pa Pg

(d) f =

wP (0.622 + w)Pg

(e) None of these Answer: (e) None of these

CYU 15-3 ■ Check Your Understanding CYU 15-3.1 Select the incorrect statement regarding the dew-point temperature of air. (a) The dew-point temperature is defined as the temperature at which condensation begins when the air is cooled at constant pressure. (b) The dew-point temperature is the saturation temperature of water corresponding to the vapor pressure. (c) The dry-bulb temperature and the dew-point temperature of saturated air are identical. (d) The dry-bulb temperature cannot be above the dew-point temperature. Answer: (d) The dry-bulb temperature cannot be above the dew-point temperature. CYU 15-3.2 Select the correct statements regarding the dew-point temperature. I The dew-point temperature of air with a relative humidity of 100 percent is equal to the dry-bulb temperature. II The dew-point temperature of air at 30C with a relative humidity of 70 percent is less than that of air at 30C with a relative humidity of 50 percent. III The dew-point temperature of air at 30C with a relative humidity of 60 percent is less than that of air at 25C with a relative humidity of 60 percent. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II and III Answer: (a) Only I

13-194

CYU 15-4 ■ Check Your Understanding CYU 15-4.1 Select the incorrect statement regarding adiabatic saturation. (a) During an adiabatic saturation process, the partial pressure of dry air remains constant. (b) For air-water vapor mixtures at atmospheric pressure, the wet-bulb temperature is approximately equal to the adiabatic saturation temperature. (c) The adiabatic saturation temperature of a given moist air is generally higher than its dew-point temperature. (d) None of these Answer: (a) During an adiabatic saturation process, the partial pressure of dry air remains constant. CYU 15-4.2 Which quantity remains constant during an adiabatic saturation process? (a) Specific humidity (b) Relative humidity (c) The enthalpy of the moist air (d) Mass flow rate of dry air Answer: (d) Mass flow rate of dry air

CYU 15-5 ■ Check Your Understanding CYU 15-5.1 Select the incorrect statement regarding the psychrometric chart. (a) A horizontal process line indicates that the relative humidity remains constant. (b) A horizontal process line indicates that there is no humidification or dehumidification. (c) A downward vertical process line indicates moisture removal from at constant temperature. (d) Relative humidity is 100 percent at any point of the saturation line. (e) Lines of constant enthalpy lie very nearly parallel to the lines of constant wet-bulb temperature. Answer: (a) A horizontal process line indicates that the relative humidity remains constant. CYU 15-5.2 Which quantity remains constant for a horizontal process line in the psychrometric chart? (a) Dry-bulb temperature (b) Relative humidity (c) Specific humidity (d) Wet-bulb temperature (e) Specific volume Answer: (c) Specific humidity

CYU 15-6 ■ Check Your Understanding CYU 15-6.1 Which parameter has the most effect on the human thermal comfort? (a) Temperature (b) Altitude (c) Air motion (d) Air cleanliness (e) Relative humidity © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-195

Answer: (a) Temperature CYU 15-6.2 The wind-chill effect is related to heat transfer from the body by __________. (a) Radiation (b) Conduction (c) Convection (d) Evaporation (e) Perspiration Answer: (c) Convection CYU 15-6.3 The relative humidity affects the amount of heat a body can dissipate per unit time by __________. (a) Radiation (b) Conduction (c) Convection (d) Evaporation Answer: (d) Evaporation

CYU 15-7 ■ Check Your Understanding CYU 15-7.1 Which air-conditioning process involves lowering air temperature and removing moisture from the air? (a) Simple heating (b) Humidification (c) Simple cooling (d) Heating with humidification (e) Cooling with dehumidification Answer: (e) Cooling with dehumidification CYU 15-7.2 Which air-conditioning processes appear as horizontal lines on the psychrometric chart? (a) Simple heating and cooling (b) Simple heating and humidification (c) Simple cooling and humidification (d) Simple cooling and dehumidification (e) Humidification and dehumidification Answer: (a) Simple heating and cooling CYU 15-7.3 What happens during a simple heating process? (a) Both temperature and relative humidity increase. (b) Both temperature and specific humidity increase. (c) Temperature increases and relative humidity decreases. (d) Both specific humidity and relative humidity increase. (e) Relative humidity decreases and specific humidity increases. Answer: (c) Temperature increases and relative humidity decreases. CYU 15-7.4 What happens during a simple cooling process? © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-196

(a) Both temperature and relative humidity decrease. (b) Relative humidity remains constant and specific humidity decreases. (c) Specific humidity decreases and relative humidity increases. (d) Relative humidity increases and specific humidity remains constant. (e) Relative humidity decreases and specific humidity increases. Answer: (d) Relative humidity increases and specific humidity remains constant. CYU 15-7.5 Which two parameters remain nearly constant during an evaporative cooling process? (a) Enthalpy and wet-bulb temperature (b) Enthalpy and dry-bulb temperature (c) Specific humidity and wet-bulb temperature (d) Wet-bulb temperature and relative humidity (e) Specific humidity and dry-bulb temperature Answer: (a) Enthalpy and wet-bulb temperature CYU 15-7.6 Which relation is applicable on a cooling and dehumidification process? 1: inlet, 2: exit (a) Q out = m w (h1 - h2 ) (b) Q out = m a (h1 - h2 ) (c) Q out = m a (h1 - h2 )- m w hw (d) Q out = m a (h2 - h1)+ m w hw (e) Q out = m a (h2 - h1)- m w hw Answer: (c) Q out = m a (h1 - h2 )- m w hw

13-197

PROBLEMS Dry and Atmospheric Air: Specific and Relative Humidity 15-1C Dry air does not contain any water vapor, but atmospheric air does.

15-2C The partial pressure of the water vapor in atmospheric air is called vapor pressure.

15-3C Specific humidity is the amount of water vapor present in a unit mass of dry air. Relative humidity is the ratio of the actual amount of vapor in the air at a given temperature to the maximum amount of vapor air can hold at that temperature.

15-4C Yes, the water vapor in the air can be treated as an ideal gas because of its very low partial pressure.

15-5C When the temperature, total pressure, and the relative humidity are given, the vapor pressure can be determined from the psychrometric chart or the relation Pv = f Psat where Psat is the saturation (or boiling) pressure of water at the specified temperature and  is the relative humidity.

15-6C Yes.

15-7C Specific humidity will decrease but relative humidity will increase.

15-8C The specific humidity will remain constant, but the relative humidity will decrease as the temperature rises in a wellsealed room.

15-9C The specific humidity will remain constant, but the relative humidity will increase as the temperature drops in a wellsealed room.

15-10C A tank that contains moist air at 3 atm is located in moist air that is at 1 atm. The driving force for moisture transfer is the vapor pressure difference, and thus it is possible for the water vapor to flow into the tank from surroundings if the vapor pressure in the surroundings is greater than the vapor pressure in the tank.

15-11C Yes; by cooling the air at constant pressure.

13-198

15-12C Insulations on chilled water lines are always wrapped with vapor barrier jackets to eliminate the possibility of vapor entering the insulation. This is because moisture that migrates through the insulation to the cold surface will condense and remain there indefinitely with no possibility of vaporizing and moving back to the outside.

15-13C The same. This is because water vapor behaves as an ideal gas at low pressures, and the enthalpy of an ideal gas depends on temperature only.

15-14 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined.

Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition, w=

mv 0.17 kg = = 0.01133 kg H 2O / kg dry air ma 15 kg

(b) The saturation pressure of water at 30C is

Pg = Psat @ 30°C = 4.2469 kPa Then the relative humidity can be determined from

f =

wP (0.01133)(100 kPa) = = 0.4214 = 42.1% (0.622 + w) Pg (0.622 + 0.01133)(4.2469 kPa)

(c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = f Pg = (0.4214)(4.2469 kPa) = 1.789 kPa

Pa = P - Pv = 100 - 1.789 = 98.211 kPa V=

ma RaT (15 kg)(0.287 kJ/kg ×K)(303 K) = = 13.3 m 3 Pa 98.211 kPa

EES (Engineering Equation Solver) SOLUTION "Given" m_a=15 [kg] m_v=0.17 [kg] T=30 [C] P=100 [kPa] "Properties" R_a=R_u/MM MM=Molarmass(air) R_u=8.314 [kJ/kmol-K] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-199

"(a)" w=m_v/m_a "kg H2O/kg dry air" "(b)" P_g=pressure(steam_iapws, T=T, x=1) phi=(w*P)/((0.622+w)*P_g) "(c)" P_v=phi*P_g P_a=P-P_v V=(m_a*R_a*(T+273))/P_a

15-15 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity, and the volume of the tank are to be determined.

Assumptions The air and the water vapor are ideal gases. Analysis (a) The specific humidity can be determined form its definition, w=

mv 0.17 kg = = 0.01133 kg H 2 O / kg dry air ma 15 kg

(b) The saturation pressure of water at 20C is

Pg = Psat @40°C = 7.385 kPa Then the relative humidity can be determined from

f =

wP (0.01133)(100 kPa) = = 0.2423 = 24.3% (0.622 + w) Pg (0.622 + 0.01133)(7.385 kPa)

(c) The volume of the tank can be determined from the ideal gas relation for the dry air, Pv = f Pg = (0.2423)(7.385 kPa) = 1.789 kPa

Pa = P - Pv = 100 - 1.789 = 98.211 kPa V=

ma RaT (15 kg)(0.287 kJ/kg ×K)(313 K) = = 13.7 m 3 Pa 98.211 kPa

EES (Engineering Equation Solver) SOLUTION "Given" m_a=15 [kg] m_v=0.17 [kg] T=40 [C] P=100 [kPa] "Properties" R_a=R_u/MM MM=Molarmass(air) R_u=8.314 [kJ/kmol-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-200

"Analysis" "(a)" w=m_v/m_a "kg H2O/kg dry air" "(b)" P_g=pressure(steam_iapws, T=T, x=1) phi=(w*P)/((0.622+w)*P_g) "(c)" P_v=phi*P_g P_a=P-P_v V=(m_a*R_a*(T+273))/P_a

15-16 A tank contains saturated air at a specified temperature and pressure. The mass of dry air, the specific humidity, and the enthalpy of the air are to be determined. Assumptions The air and the water vapor are ideal gases.

Analysis (a) The mass of dry air can be determined from the ideal gas relation for the dry air, ma =

[(105 - 4.2469) kPa ](8 m3 ) PaV = = 9.264 kg RaT (0.287 kJ/kg.K)(30 + 273.15 K)

(b) The relative humidity of air is 100 percent since the air saturated. The vapor pressure is equal to the saturation pressure of water at 30C

Pv = Pg = Psat @ 30°C = 4.2469 kPa The specific humidity can be determined from w=

0.622 Pv (0.622)(4.2469 kPa) = = 0.0262 kg H 2 O / kg dry air P - Pv (105 - 4.2469) kPa

= (1.005 kJ/kg ×°C)(30°C) + (0.0262)(2555.6 kJ/kg) = 97.1 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION P[1] =105 [kPa] T[1] = 30 [C] RH[1] = 100/100 " relative humidity for saturated air" "Mass of dry air m_dot_a is:" Vol= 8 [m^3] v=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) "[m^3/kg_a]" m_a = Vol/v "[kg_a]" "The specific humidity is:" w=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1])"[kg_v/kg_a]" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-201

"The enthalpy per unit mass dry air is:" h=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w)"[kJ/kg_a]"

15-17 The masses of dry air and the water vapor contained in a room at specified conditions and relative humidity are to be determined. Assumptions The air and the water vapor are ideal gases.

Analysis The partial pressure of water vapor and dry air are determined to be

Pv = f Pg = f Psat@15°C = (0.50)(1.706 kPa) = 0.8528 kPa

Pa = P - Pv = 93 - 0.8528 = 92.15 kPa The masses are determined to be

ma =

PaV (92.15 kPa)(90 m3 ) = = 100.3 kg RaT (0.287 kPa ×m3 /kg ×K)(288 K)

mv =

PvV (0.8528 kPa)(90 m3 ) = = 0.578 kg RvT (0.4615 kPa ×m3 /kg ×K)(288 K)

EES (Engineering Equation Solver) SOLUTION "Given" V=90 [m^3] P=93 [kPa] T=15 [C] phi=0.50 "Properties" Fluid$='steam_iapws' R_a=R_u/MM_a R_v=R_u/MM_v MM_a=Molarmass(air) MM_v=Molarmass(Fluid$) R_u=8.314 [kJ/kmol-K] "Analysis" P_g=pressure(Fluid$, T=T, x=1) P_v=phi*P_g P_a=P-P_v m_a=(P_a*V)/(R_a*(T+273)) m_v=(P_v*V)/(R_v*(T+273))

15-18 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-202

Analysis (a) The partial pressure of dry air can be determined from Pv   Pg   Psat @ 20C  (0.85)(2.3392 kPa)=1.988 kPa

Pa  P  Pv  98  1.988  96.01 kPa (b) The specific humidity of air is determined from



0.622 Pv (0.622 )(1.988 kPa)   0.0129 kg H 2 O/kg dry air P  Pv (98  1.988 ) kPa

 (1.005 kJ/kg C)(20C)+(0.0129)(2537.4 kJ/kg) =52.78 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" T=20 [C] P=98 [kPa] phi=0.85 "Properties" c_p=1.005 [kJ/kg-C] "Table A-2a" "Analysis" "(a)" Fluid$='steam_iapws' P_g=pressure(Fluid$, T=T, x=1) P_v=phi*P_g P_a=P-P_v "(b)" w=(0.622*P_v)/(P-P_v) "kg H2O/kg dry air" "(c)" h_g=enthalpy(Fluid$, T=T, x=1) h=c_p*T+w*h_g

15-19 A room contains air at specified conditions and relative humidity. The partial pressure of air, the specific humidity, and the enthalpy per unit mass of dry air are to be determined. Assumptions The air and the water vapor are ideal gases.

13-203

Analysis (a) The partial pressure of dry air can be determined from Pv   Pg   Psat @ 20C  (0.85)(2.3392 kPa)=1.988 kPa

Pa  P  Pv  85  1.988  83.01 kPa (b) The specific humidity of air is determined from



0.622 Pv (0.622 )(1.988 kPa)   0.0149 kg H 2 O/kg dry air P  Pv (85  1.988 ) kPa

 (1.005 kJ/kg C)(20C)+(0.0149)(2537.4 kJ/kg) =57.90 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" T=20 [C] P=85 [kPa] phi=0.85 "Properties" c_p=1.005 [kJ/kg-C] "Table A-2a" "Analysis" Fluid$='steam_iapws' "(a)" P_g=pressure(Fluid$, T=T, x=1) P_v=phi*P_g P_a=P-P_v "(b)" w=(0.622*P_v)/(P-P_v) "kg H2O/kg dry air" "(c)" h_g=enthalpy(Fluid$, T=T, x=1) h=c_p*T+w*h_g

15-20 Humid air is compressed in an isentropic compressor. The relative humidity of the air at the compressor outlet is to be determined. Assumptions The air and the water vapor are ideal gases. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The saturation properties of water are to be obtained from water tables.

13-204

Analysis At the inlet,

Pv , 1 = f 1 Pg , 1 = f 1 Psat @ 20°C = (0.90)(2.3392 kPa) = 2.105 kPa 0.622 Pv ,1

w2 = w1 =

P - Pv ,1

(0.622)(2.105 kPa) = 0.0134 kg H 2 O/kg dry air (100 - 2.105) kPa

Since the mole fraction of the water vapor in this mixture is very small, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P ÷ ø 1

0.4/1.4

æ800 kPa ö ÷ = (293 K) çç ÷ ÷ çè100 kPa ø

= 531 K

The saturation pressure at this temperature is

Pg , 2 = Psat @ 258°C = 4542 kPa (from EES) The vapor pressure at the exit is Pv , 2 =

w2 P2 (0.0134)(800) = = 16.87 kPa w2 + 0.622 0.0134 + 0.622

The relative humidity at the exit is then f2=

Pv , 2 Pg , 2

16.87 = 0.0037 = 0.37% 4542

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [kPa] T1=20 [C] phi_1=0.90 P2=800 [kPa] "Properties" k=1.4 "Analysis" P_g_1=pressure(steam_iapws, T=T1, x=1) P_v_1=phi_1*P_g_1 w1=(0.622*P_v_1)/(P1-P_v_1) w2=w1 T2_K=(T1+273)*(P2/P1)^((k-1)/k) T2=T2_K-273 P_g_2=pressure(steam_iapws, T=T2, x=1) P_v_2=(w2*P2)/(w2+0.622) phi_2=P_v_2/P_g_2

13-205

15-21E Humid air is expanded in an isentropic nozzle. The amount of water vapor that has condensed during the process is to be determined. Assumptions The air and the water vapor are ideal gases. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2a). The saturation properties of water are to be obtained from water tables.

Analysis Since the mole fraction of the water vapor in this mixture is very small, ( k - 1)/ k

æP ÷ ö T2 = T1 ççç 2 ÷ ÷ çè P ÷ ø 1

0.4/1.4

æ15 psia ÷ ö ÷ = (860 R) çç ÷ çè100 psia ÷ ø

= 500 R

We will assume that the air leaves the nozzle at a relative humidity of 100% (will be verified later). The vapor pressure and specific humidity at the outlet are then

Pv ,2 = f 2 Pg ,2 = f 2 Psat @ 40°F = (1.0)(0.12173 psia)=0.1217 psia w2 =

0.622 Pv ,2 P - Pv ,2

(0.622)(0.1217 psia) = 0.00509 lbm H 2 O/lbm dry air (15 - 0.1217) psia

This is less than the inlet specific humidity ( 0.025 lbm / lbm dry air), the relative humidity at the outlet must be 100% as originally assumed. The amount of liquid formation is then

D w = w1 - w2 = 0.025 - 0.00509 = 0.0199 lbm H2O / lbm dry air

EES (Engineering Equation Solver) SOLUTION "Given" P1=100 [psia] T1=400 [F] w1=0.025 P2=15 [psia] "Properties" k=1.4 "Analysis" T2_R=(T1+460)*(P2/P1)^((k-1)/k) T2=T2_R-460 P_g_2=pressure(steam_iapws, T=T2, x=1) phi_2=1 "assumed" P_v_2=phi_2*P_g_2 w2=(0.622*P_v_2)/(P2-P_v_2) DELTAw=w1-w2

Dew-Point, Adiabatic Saturation, and Wet-bulb Temperatures 15-22C Dew-point temperature is the temperature at which condensation begins when air is cooled at constant pressure.

13-206

15-23C The outer surface temperature of the glass may drop below the dew-point temperature of the surrounding air, causing the moisture in the vicinity of the glass to condense. After a while, the condensate may start dripping down because of gravity.

15-24C When the temperature falls below the dew-point temperature, dew forms on the outer surfaces of the car. If the temperature is below 0C, the dew will freeze. At very low temperatures, the moisture in the air will freeze directly on the car windows.

15-25C Addison’s. The temperature of his glasses may be below the dew-point temperature of the room, causing condensation on the surface of the glasses.

15-26C When the air is saturated (100% relative humidity).

15-27C These two are approximately equal at atmospheric temperatures and pressure.

15-28 A house contains air at a specified temperature and relative humidity. It is to be determined whether any moisture will condense on the inner surfaces of the windows when the temperature of the window drops to a specified value. Assumptions The air and the water vapor are ideal gases.

Analysis The vapor pressure Pv is uniform throughout the house, and its value can be determined from

Pv = f Pg @ 25°C = (0.65)(3.1698 kPa) = 2.06 kPa The dew-point temperature of the air in the house is Tdp = Tsat @ Pv = Tsat @ 2.06 kPa = 18.0°C

That is, the moisture in the house air will start condensing when the temperature drops below 18.0C. Since the windows are at a lower temperature than the dew-point temperature, some moisture will condense on the window surfaces.

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] phi=0.65 T_window=10 [C] "Analysis" Fluid$='steam_iapws' P_g=pressure(Fluid$, T=T, x=1) P_v=phi*P_g T_dp=temperature(Fluid$, P=P_v, x=1) "If the window temperature (=10 C) is smaller than T_dp, the moisture condenses" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-207

15-29E A person drinks a cool canned soda in a room at a specified temperature and relative humidity. It is to be determined whether the can will sweat. Assumptions The air and the water vapor are ideal gases.

Analysis The vapor pressure Pv of the air in the house is uniform throughout, and its value can be determined from

Pv = f Pg @ 70°F = (0.38)(0.3633 psia) = 0.1381 psia The dew-point temperature of the air in the house is Tdp = Tsat @ Pv = Tsat @ 0.1381 psia = 43.3°F (from EES) That is, the moisture in the house air will start condensing when the air temperature drops below 43.3C. Since the canned drink is at a lower temperature than the dew-point temperature, some moisture will condense on the can, and thus it will sweat.

EES (Engineering Equation Solver) SOLUTION "Given" T=70 [F] phi=0.38 T_drink=40 [F] "Analysis" Fluid$='steam_iapws' P_g=pressure(Fluid$, T=T, x=1) P_v=phi*P_g T_dp=temperature(Fluid$, P=P_v, x=1) "If the temperature of the canned drink (=40 F) is smaller than T_dp, the can will sweat."

15-30 The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases.

Analysis (a) We obtain the properties of water vapor from EES. The specific humidity w1 is determined from w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

Where T2 is the wet-bulb temperature, and w2 is determined from

13-208

0.622 Pg 2

w2 =

P2 - Pg 2

(0.622)(2.488 kPa) = 0.01587 kg H 2 O/kg dry air (100 - 2.488) kPa

Thus, w1 =

(1.005 kJ/kg ×°C)(21- 26)°C + (0.01587)(2451.2 kJ/kg) = 0.01377 kg H 2 O / kg dry air (2548.3 - 88.10) kJ/kg

(b) The relative humidity f 1 is determined from

f1=

w1 P1 (0.01377)(100 kPa) = = 0.644 or 64.4% (0.622 + w1 ) Pg1 (0.622 + 0.01377)(3.3638 kPa)

Pv1 = f 1 Pg1 = f 1 Psat @ 26°C = (0.644)(3.3638 kPa) = 2.166 kPa Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 2.166 kPa = 18.8°C

EES (Engineering Equation Solver) SOLUTION "Given" T_db=26 [C] T_wb=21 [C] P=100 [kPa] "Properties" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_db, x=1) P_g2=pressure(Fluid$, T=T_wb, x=1) h_g1=enthalpy(Fluid$, T=T_db, x=1) h_g2=enthalpy(Fluid$, T=T_wb, x=1) h_f2=enthalpy(Fluid$, T=T_wb, x=0) h_fg2=h_g2-h_f2 c_p=1.005 [kJ/kg-C] "Table A-2a" "Analysis" "(a)" w_2=(0.622*P_g2)/(P-P_g2) "kg H2O/kg dry air" w_1=(c_p*(T_wb-T_db)+w_2*h_fg2)/(h_g1-h_f2) "(b)" phi_1=(w_1*P)/((0.622+w_1)*P_g1) "(c)" P_v1=phi_1*P_g1 T_dp=temperature(Fluid$, P=P_v1, x=1)

15-31 Problem 15-30 is reconsidered. The required properties are to be determined using appropriate software at 100 and 300 kPa pressures. Analysis The problem is solved using EES, and the solution is given below. Tdb=26 [C] Twb=21 [C] P1=100 [kPa] P2=300 [kPa]

13-209

h1=enthalpy(AirH2O,T=Tdb,P=P1,B=Twb) v1=volume(AirH2O,T=Tdb,P=P1,B=Twb) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,B=Twb) w1=humrat(AirH2O,T=Tdb,P=P1,B=Twb) Rh1=relhum(AirH2O,T=Tdb,P=P1,B=Twb) h2=enthalpy(AirH2O,T=Tdb,P=P2,B=Twb) v2=volume(AirH2O,T=Tdb,P=P2,B=Twb) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,B=Twb) w2=humrat(AirH2O,T=Tdb,P=P2,B=Twb) Rh2=relhum(AirH2O,T=Tdb,P=P2,B=Twb) SOLUTION h1=61.25 [kJ/kg] h2=34.16 [kJ/kg] P1=100 [kPa] P2=300 [kPa] Rh1=0.6437 Rh2=0.4475 Tdb=26 [C] Tdp1=18.76 Tdp2=13.07 Twb=21 [C] v1=0.8777 [m^3/kg] v2=0.2877 [m^3/kg] w1=0.01376 [kg/kg] w2=0.003136 [kg/kg] Alternative Solution The following EES routine can also be used to solve this problem. The above EES routine uses built-in psychrometric functions whereas the one below uses analytical expressions together with steam properties. "Given" T_db=26 [C] T_wb=21 [C] P=100 [kPa] "Properties" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_db, x=1) P_g2=pressure(Fluid$, T=T_wb, x=1) h_g1=enthalpy(Fluid$, T=T_db, x=1) h_g2=enthalpy(Fluid$, T=T_wb, x=1) h_f2=enthalpy(Fluid$, T=T_wb, x=0) h_fg2=h_g2-h_f2 c_p=1.005 [kJ/kg-C] "for air" "Analysis" "(a)" w_2=(0.622*P_g2)/(P-P_g2) "kg H2O/kg dry air" w_1=(c_p*(T_wb-T_db)+w_2*h_fg2)/(h_g1-h_f2) "(b)" phi_1=(w_1*P)/((0.622+w_1)*P_g1) "(c)" P_v1=phi_1*P_g1 T_dp=temperature(Fluid$, P=P_v1, x=1) SOLUTION c_p=1.005 [kJ/kg-C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-210

Fluid$='steam_iapws' h_f2=88.1 [kJ/kg] h_fg2=2451.2 [kJ/kg] h_g1=2548.3 [kJ/kg] h_g2=2539.3 [kJ/kg] P=100 [kPa] phi_1=0.6439 P_g1=3.3638 [kPa] P_g2=2.488 [kPa] P_v1=2.166 [kPa] T_db=26 [C] T_dp=18.8 [C] T_wb=21 [C] w_1=0.01377 w_2=0.01587

15-32E The dry- and wet-bulb temperatures of air in room at a specified pressure are given. The specific humidity, the relative humidity, and the dew-point temperature are to be determined. Assumptions The air and the water vapor are ideal gases.

Analysis (a) The specific humidity w1 is determined from w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

where T2 is the wet-bulb temperature, and w2 is determined from w2 =

0.622 Pg 2 P2 - Pg 2

(0.622)(0.25638 psia) = 0.01136 lbm H 2 O/lbm dry air (14.3 - 0.25638) psia

where Pg 2 = Psat @60° F = 0.25638 psia. Thus, w1 =

(0.24 Btu/lbm ×°F)(60 - 75)°F + (0.01136)(1059.4 Btu/lbm) = 0.007908 lbm H 2 O / lbm dry air (1093.9 - 28.08) Btu/lbm

where h fg 2 = h fg @ 60 F = 1059.4Btu / lbm , hg1 = hg@75o F = 1093.9Btu / lbm and h f 2 = h f @60 F = 28.08Btu / lbm (b) The relative humidity f 1 is determined from

f1=

w1 P1 (0.007908)(14.3 psia) = = 0.4174 or 41.7% (0.622 + w1 ) Pg1 (0.622 + 0.007908)(0.4302 psia)

Where Pg1 = Psat @75° F = 0.4302 psia. (c) The vapor pressure at the inlet conditions is

Pv1 = f 1 Pg1 = f 1 Psat @ 75°F = (0.4174)(0.4302 psia) = 0.1795 psia

13-211

Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 0.1795 psia = 50.2°F

(from EES)

EES (Engineering Equation Solver) SOLUTION "Given" T_db=75 [F] T_wb=60 [F] P=14.3 [psia] "Properties" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_db, x=1) P_g2=pressure(Fluid$, T=T_wb, x=1) h_g1=enthalpy(Fluid$, T=T_db, x=1) h_g2=enthalpy(Fluid$, T=T_wb, x=1) h_f2=enthalpy(Fluid$, T=T_wb, x=0) h_fg2=h_g2-h_f2 c_p=0.24 [Btu/lbm-F] "for air, Table A-2Ea" "Analysis" "(a)" w_2=(0.622*P_g2)/(P-P_g2) "lbm H2O/lbm dry air" w_1=(c_p*(T_wb-T_db)+w_2*h_fg2)/(h_g1-h_f2) "(b)" phi_1=(w_1*P)/((0.622+w_1)*P_g1) "(c)" P_v1=phi_1*P_g1 T_dp=temperature(Fluid$, P=P_v1, x=1) 15-33 The dry- and wet-bulb temperatures of atmospheric air at a specified pressure are given. The specific humidity, the relative humidity, and the enthalpy of air are to be determined. Assumptions The air and the water vapor are ideal gases.

Analysis (a) We obtain the properties of water vapor from EES. The specific humidity w1 is determined from w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

where T2 is the wet-bulb temperature, and w2 is determined from w2 =

0.622 Pg 2 P2 - Pg 2

(0.622)(1.938 kPa) = 0.01295 kg H 2 O/kg dry air (95 - 1.938) kPa

Thus, w1 =

(1.005 kJ/kg ×°C)(17 - 25)°C + (0.01295)(2460.6 kJ/kg) = 0.00963 kg H 2 O / kg dry air (2546.5 - 71.36) kJ/kg

(b) The relative humidity f 1 is determined from

13-212

f1=

w1 P1 (0.00963)(95 kPa) = = 0.457 or 45.7% (0.622 + w1 ) Pg1 (0.622 + 0.00963)(3.1698 kPa)

h1 = ha1 + w1hv1 @ c pT1 + w1hg1

= (1.005 kJ/kg ×°C)(25°C) + (0.00963)(2546.5 kJ/kg) = 49.65 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" T_db=25 [C] T_wb=17 [C] P=95 [kPa] "Properties" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_db, x=1) P_g2=pressure(Fluid$, T=T_wb, x=1) h_g1=enthalpy(Fluid$, T=T_db, x=1) h_g2=enthalpy(Fluid$, T=T_wb, x=1) h_f2=enthalpy(Fluid$, T=T_wb, x=0) h_fg2=h_g2-h_f2 c_p=1.005 [kJ/kg-C] "Table A-2a" "Analysis" "(a)" w_2=(0.622*P_g2)/(P-P_g2) "kg H2O/kg dry air" w_1=(c_p*(T_wb-T_db)+w_2*h_fg2)/(h_g1-h_f2) "(b)" phi_1=(w_1*P)/((0.622+w_1)*P_g1) "(c)" h_1=c_p*T_db+w_1*h_g1

15-34 Atmospheric air flows steadily into an adiabatic saturation device and leaves as a saturated vapor. The relative humidity and specific humidity of air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-213

Analysis The exit state of the air is completely specified, and the total pressure is 98 kPa. The properties of the moist air at the exit state may be determined from EES to be

h2 = 78.11 kJ/kg dry air w2 = 0.02079 kg H2 O/kg dry air The enthalpy of makeup water is hw2 = h f@ 25°C = 104.83 kJ/kg

(Table A-4)

An energy balance on the control volume gives

h1 + (w2 - w1 )hw = h2 h1 + (0.02079 - w1 )(104.83 kJ/kg) = 78.11 kJ/kg Pressure and temperature are known for inlet air. Other properties may be determined from this equation using EES. A hand solution would require a trial-error approach. The results are

h1 = 77.66 kJ/kg dry air w1 = 0.01654 kg H2O / kg dry air

f 1 = 0.4511

EES (Engineering Equation Solver) SOLUTION T_1 = 35 [C] P_1 =98 [kPa] T_2 = 25 [C] P_2 =98 [kPa] "Assume the process is constant pressure" RH_2 = 100/100 "relative humidity at state 2" T_w = 25 [C] "supply water temperature" "Energy balance for the steady flow process:" e_in - e_out = DELTAe_cv e_in = h_1 + (w_2-w_1)*h_w e_out = h_2 DELTAe_cv=0 "The specific humidity at entrance (state 1) and exit (state 2):" w_1=HUMRAT(AirH2O,T=T_1,P=P_1,R=RH_1) w_2=HUMRAT(AirH2O,T=T_2,P=P_2,R=RH_2) "The enthalpies per unit mass dry air flowing through the device are:" h_1=ENTHALPY(AirH2O,T=T_1,P=P_1,w=w_1) h_2=ENTHALPY(AirH2O,T=T_2,P=P_2,R=RH_2) "Enthalpy of water entering the process from the water supply:"

13-214

h_w=enthalpy(steam_iapws,T=T_w,x=0) "saturated liquid value at T_w"

Psychrometric Chart 15-35C They are very nearly parallel to each other.

15-36C The saturation states (located on the saturation curve).

15-37C By drawing a horizontal line until it intersects with the saturation curve. The corresponding temperature is the dewpoint temperature.

15-38C No, they cannot. The enthalpy of moist air depends on , which depends on the total pressure.

15-39 The pressure, temperature, and relative humidity of air are specified. Using the psychrometric chart, the wet-bulb temperature, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined.

Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain (a) Twb = 27.1°C (b) w = 0.0217 kg H2O / kg dry air (c) h = 85.5 kJ/kg dry air (d) Tdp = 26.2°C (e) Pv = f Pg = f Psat @ 30°C = (0.80)(4.2469 kPa)=3.40 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 "[kPa]" T=30 "[C]" phi=0.80 "Analysis" Fluid$='AirH2O' w=humrat(Fluid$, T=T, P=P, R=phi) h=enthalpy(Fluid$, T=T, P=P, R=phi) T_wb=wetbulb(Fluid$, T=T, P=P, R=phi) T_dp=dewpoint(Fluid$, T=T, P=P, R=phi) v=volume(Fluid$, T=T, P=P, R=phi)

13-215

15-40E The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined.

Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain (a) f = 0.816 = 81.6% (b) w = 0.0252 lbm H2 O / lbm dry air (c) h = 49.4 Btu/lbm dry air (d) Tdp = 83.7°F (e) Pv = f Pg = f Psat @ 90° F = (0.816)(0.69904 psia)=0.570 psia

EES (Engineering Equation Solver) SOLUTION "Given" P=14.696 [psia] T=90 [F] T_wb=85 [F] "Analysis" Fluid$='AirH2O' phi=relhum(Fluid$, T=T, P=P, B=T_wb) w=humrat(Fluid$, T=T, P=P, B=T_wb) h=enthalpy(Fluid$, T=T, P=P, B=T_wb) T_dp=dewpoint(Fluid$, T=T, P=P, B=T_wb) Pv=phi*P_s P_s=pressure(steam_iapws, T=T, x=1) h_g=enthalpy(steam_iapws, T=T, x=1)

15-41E The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature is to be determined. Analysis For an adiabatic saturation process, we obtained Eq. 15-14 in the text, w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

This requires a trial-error solution for the adiabatic saturation temperature, T2 . The inlet state properties are © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-216

w1 = 0.0252 lbm H2O / lbm dry air hg1 = hg @ 90°F = 1100.4 Btu/lbm As a first estimate, let us take T2 = 85° F (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is 100% (  = 1) and the pressure is 1 atm. Other properties at the exit state are

w2 = 0.0264 lbm H2 O / lbm dry air h f 2 = h f @ 85° F = 53.06 Btu/lbm (Table A-4E)

Substituting, w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

(0.240)(85 - 90) + (0.0264)(1045.2) = 0.0252 lbm H 2 O / lbm dry air 1100.4 - 53.06

which is equal to the inlet specific humidity. Therefore, the adiabatic saturation temperature is T2 = 85° F

Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately equal to each other for air-water mixtures at atmospheric pressure.

15-42 The pressure and the dry- and wet-bulb temperatures of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the relative humidity, the dew-point temperature, and the specific volume of the air are to be determined. Analysis The air is at 1 atm, a dry-bulb temperature of 24°C, and a wet-bulb temperature of 17°C. From the psychrometric chart (Fig. A-31) we read (a) w = 0.0092 kg H2 O/kg dry air (b) h = 47.6 kJ/kg dry air (c) f = 49.6% (d) Tdp = 12.8°C (e) v = 0.855 m3 /kg dry air

13-217

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_db=24 [C] T_wb=17 [C] "Analysis" Fluid$='AirH2O' w=humrat(Fluid$, T=T_db, B=T_wb, P=P) h=enthalpy(Fluid$, T=T_db, B=T_wb, P=P) phi=relhum(Fluid$, T=T_db, B=T_wb, P=P) T_dp=dewpoint(Fluid$, T=T_db, B=T_wb, P=P) v=volume(Fluid$, T=T_db, B=T_wb, P=P)

15-43 Problem 15-42 is reconsidered. The required properties are to be determined using EES. Also, the properties are to be obtained at an altitude of 3000 m. Analysis The problem is solved using EES, and the solution is given below. Tdb=24 [C] Twb=17 [C] P1=101.325 [kPa] Z = 3000 [m] P2=101.325*(1-0.02256*Z*convert(m,km))^5.256 "Relation giving P as function of altitude" h1=enthalpy(AirH2O,T=Tdb,P=P1,B=Twb) v1=volume(AirH2O,T=Tdb,P=P1,B=Twb) Tdp1=dewpoint(AirH2O,T=Tdb,P=P1,B=Twb) w1=humrat(AirH2O,T=Tdb,P=P1,B=Twb) Rh1=relhum(AirH2O,T=Tdb,P=P1,B=Twb) h2=enthalpy(AirH2O,T=Tdb,P=P2,B=Twb) v2=volume(AirH2O,T=Tdb,P=P2,B=Twb) Tdp2=dewpoint(AirH2O,T=Tdb,P=P2,B=Twb) w2=humrat(AirH2O,T=Tdb,P=P2,B=Twb) Rh2=relhum(AirH2O,T=Tdb,P=P2,B=Twb) SOLUTION h1=47.61 [kJ/kg] h2=61.68 [kJ/kg] P1=101.3 [kPa] P2=70.11 [kPa] Rh1=0.4956 Rh2=0.5438 Tdb=24 [C] Tdp1=12.81 [C] Tdp2=14.24 [C] Twb=17 [C] v1=0.8542 [m^3/kg] v2=1.245 [m^3/kg] w1=0.009219 [kg/kg] w2=0.01475 [kg/kg] Z=3000 [m] Discussion The atmospheric pressure for a given elevation can also be obtained from Table A-16 of the book.

13-218

15-44 The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be determined.

Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain (a) j = 0.618 = 61.8% (b) w = 0.0148 kg H2O / kg dry air (c) h = 65.8 kJ/kg dry air (d) Twb = 22.4°C (e) Pv = f Pg = f Psat @ 28°C = (0.618)(3.780 kPa)=2.34 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T=28 [C] T_dp=20 [C] "Analysis" Fluid$='AirH2O' phi=relhum(Fluid$, T=T, P=P, D=T_dp) w=humrat(Fluid$, T=T, P=P, D=T_dp) h=enthalpy(Fluid$, T=T, P=P, D=T_dp) T_wb=wetbulb(Fluid$, T=T, P=P, D=T_dp) Pv=phi*P_s P_s=pressure(steam_iapws, T=T, x=1) h_g=enthalpy(steam_iapws, T=T, x=1)

15-45 The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature is to be determined. Analysis For an adiabatic saturation process, we obtained Eq. 15-14 in the text, w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

13-219

w1 = 0.0148 kg H2 O / kg dry air

(Fig. A-31)

hg1 = hg @ 28°C = 2551.9 kJ/kg (Table A-4) As a first estimate, let us take T2 = 22 C (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is 100% (

f 2 = 1 ) and the pressure is 1 atm. Other properties at the exit state are w2 = 0.0167 kg H2 O / kg dry air h f 2 = h f @ 22°C = 92.28 kJ/kg (Table A-4)

Substituting, w1 =

c p (T2 - T1 ) + w2 h fg 2 hg1 - h f 2

(1.005)(22 - 28) + (0.0167)(2448.8) = 0.0142 kg H 2 O / kg dry air 2551.9 - 92.28

which is sufficiently close to the inlet specific humidity (0.0148). Therefore, the adiabatic saturation temperature is T2 @ 22o C

Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately equal to each other for air-water mixtures at atmospheric pressure.

Human Comfort and Air-Conditioning 15-46C It humidifies, dehumidifies, cleans and even deodorizes the air.

15-47C (a) Perspires more, (b) cuts the blood circulation near the skin, and (c) sweats excessively.

15-48C It affects by removing the warm, moist air that builds up around the body and replacing it with fresh air.

13-220

15-49C The spectators. Because they have a lower level of activity, and thus a lower level of heat generation within their bodies.

15-50C Because they have a large skin area to volume ratio. That is, they have a smaller volume to generate heat but a larger area to lose it from.

15-51C It is the direct heat exchange between the body and the surrounding surfaces. It can make a person feel chilly in winter, and hot in summer.

15-52C It affects a body’s ability to perspire, and thus the amount of heat a body can dissipate through evaporation.

15-53C Humidification is to add moisture into an environment, dehumidification is to remove it.

15-54C The metabolism refers to the burning of foods such as carbohydrates, fat, and protein in order to perform the necessary bodily functions. The metabolic rate for an average male ranges from 108 W while reading, writing, typing, or listening to a lecture in a classroom in a seated position to 1250 W at age 20 (730 at age 70) during strenuous exercise. The corresponding rates for female are about 30 percent lower. Maximum metabolic rates of trained athletes can exceed 2000 W. We are interested in metabolic rate of the occupants of a building when we deal with heating and air conditioning because the metabolic rate represents the rate at which a body generates heat and dissipates it to the room. This body heat contributes to the heating in winter, but it adds to the cooling load of the building in summer.

15-55C The metabolic rate is proportional to the size of the body, and the metabolic rate of females, in general, is lower than that of males because of their smaller size. Clothing serves as insulation, and the thicker the clothing, the lower the environmental temperature that feels comfortable.

15-56C Sensible heat is the energy associated with a temperature change. The sensible heat loss from a human body increases as (a) the skin temperature increases, (b) the environment temperature decreases, and (c) the air motion (and thus the convection heat transfer coefficient) increases.

15-57C Latent heat is the energy released as water vapor condenses on cold surfaces, or the energy absorbed from a warm surface as liquid water evaporates. The latent heat loss from a human body increases as (a) the skin wetness increases and (b) the relative humidity of the environment decreases. The rate of evaporation from the body is related to the rate of latent heat loss by Qlatent = mvapor h fg where h fg is the latent heat of vaporization of water at the skin temperature.

15-58 A department store expects to have a specified number of people at peak times in summer. The contribution of people to the sensible, latent, and total cooling load of the store is to be determined. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people doing light work is 115 W, and 70% of is in sensible form (see Sec. 15-6). Analysis The contribution of people to the sensible, latent, and total cooling load of the store are

Qpeople, total = (No. of people)´ Qperson, total = 245´ (115 W) = 28,180 W © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-221

Qpeople, sensible = (No. of people)´ Qperson, sensible = 245´ (0.7´ 115 W) = 19,720 W Qpeople, latent = (No. of people)´ Qperson, latent = 245´ (0.3´ 115 W) = 8450 W

EES (Engineering Equation Solver) SOLUTION "Given" NumberofCustomers=225 NumberofEmployees=20 "Analysis" "The average rate of heat generation from a person doing light work is 115 W, 70% of it is in sensible form and 30% in latent form. See Sec. 15-6 of the text." Q_dot_person=115 [W] Q_dot_sensible=0.70*Q_dot_person Q_dot_latent=0.30*Q_dot_person NumberofPeople=NumberofCustomers+NumberofEmployees Q_dot_people_total=NumberofPeople*Q_dot_person Q_dot_people_sensible=NumberofPeople*Q_dot_sensible Q_dot_people_latent=NumberofPeople*Q_dot_latent

15-59E There are a specified number of people in a movie theater in winter. It is to be determined if the theater needs to be heated or cooled. Assumptions There is a mix of men, women, and children in the classroom. Properties The average rate of heat generation from people in a movie theater in sensible form is 80 W per person. Analysis Noting that only the sensible heat from a person contributes to the heating load of a building, the contribution of people to the heating of the building is

Qpeople, sensible = (No. of people)´ Qperson, sensible = 500´ (80 W) = 40,000 W = 136,500 Btu / h since 1 W = 3.412 Btu / h. The building needs to be cooled since the heat gain from people is more than the rate of heat loss of 130,000 Btu / h from the building.

EES (Engineering Equation Solver) SOLUTION "Given" NumberofPeople=500 Q_dot_sensible=80 [W] Q_dot_loss=130000 [Btu/h] "Analysis" Q_dot_people_sensible=NumberofPeople*Q_dot_sensible*Convert(W, Btu/h)

15-60 The infiltration rate of a building is estimated to be 1.2 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3

Excess moisture condenses at room temperature of 24C.

13-222

The effect of water vapor on air density is negligible.

Properties The gas constant and the specific heat of air are R = 0.287 kPa ×m3 / kg.K and c p = 1.005 kJ / kg. C (Table A2). The heat of vaporization of water at 24C is h fg = h fg @ 24°C = 2444.1 kJ/kg (Table A-4). The properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31 or EES) to be Tambient = 32°Cïü ïý w ambient = 0.01039 kg/kg dry air f ambient = 35% ïïþ

Troom = 24°Cü ïï ý wroom = 0.01024 kg/kg dry air f room = 55% ïïþ

Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 0.9 times every hour, the air will enter the room at a mass flow rate of r ambient =

P0 101.325 kPa = = 1.158 kg/m3 RT0 (0.287 kPa.m3 /kg.K)(32 + 273 K)

mair = r ambientVroom ACH = (1.158 kg/m3 )(20´ 13´ 3 m3 )(1.2 h -1 ) = 1084 kg/h = 0.3010 kg/s Then the sensible, latent, and total infiltration heat loads of the room are determined to be

Qinfiltration, sensible = mair c p (Tambient - Troom ) = (0.3010 kg/s)(1.005 kJ/kg.°C)(32 - 24)°C = 2.42 kW Qinfiltration, latent = mair (wambient - wroom )h fg @ 24°C = (0.3010 kg/s)(0.01039 - 0.01024)(2444.1 kJ/kg) = 0.110 kW Qinfiltration, total = Qinfiltration, sensible + Qinfiltration, latent = 2.42 + 0.110 = 2.53 kW Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions.

EES (Engineering Equation Solver) SOLUTION "Given" ACH=1.2 [1/h] L=20 [m] W=13 [m] H=3 [m] T_ambient=32 [C] phi_ambient=0.35 T_room=24 [C] phi_room=0.55 "Properties" c_p=1.005 [kJ/kg-C] R=0.287 [kJ/kg-C] h_f=enthalpy(steam_iapws, T=T_room, x=0) h_g=enthalpy(steam_iapws, T=T_room, x=1) h_fg=h_g-h_f P=101.325 [kPa] w_ambient=humrat(AirH2O, T=T_ambient, P=P, R=phi_ambient) w_room=humrat(AirH2O, T=T_room, P=P, R=phi_room) "Analysis" rho_ambient=P/(R*(T_ambient+273)) V_room=L*W*H m_dot_air=rho_ambient*V_room*ACH*1/Convert(h, s) Q_dot_infiltration_sensible=m_dot_air*C_p*(T_ambient-T_room) Q_dot_infiltration_latent=m_dot_air*(w_ambient-w_room)*h_fg Q_dot_infiltration_total=Q_dot_infiltration_sensible+Q_dot_infiltration_latent

13-223

15-61 The infiltration rate of a building is estimated to be 1.8 ACH. The sensible, latent, and total infiltration heat loads of the building at sea level are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air infiltrates at the outdoor conditions, and exfiltrates at the indoor conditions. 3 4

Excess moisture condenses at room temperature of 24C. The effect of water vapor on air density is negligible.

Properties The gas constant and the specific heat of air are R = 0.287 kPa ×m3 / kg.K and c p = 1.005 kJ / kg. C (Table A2). The heat of vaporization of water at 24C is h fg = h fg @ 24°C = 2444.1 kJ/kg (Table A-4). The properties of the ambient and room air are determined from the psychrometric chart (Fig. A-31) to be Tambient = 32°Cïü ïý w ambient = 0.01039 kg/kg dry air f ambient = 35% ïïþ

Troom = 24°Cïü ïý w room = 0.01024 kg/kg dry air f room = 55% ïïþ

Analysis Noting that the infiltration of ambient air will cause the air in the cold storage room to be changed 1.8 times every hour, the air will enter the room at a mass flow rate of r ambient =

P0 101.325 kPa = = 1.158 kg/m3 RT0 (0.287 kPa.m3 /kg.K)(32 + 273 K)

mair = r ambientVroom ACH = (1.158 kg/m3 )(20´ 13´ 3 m3 )(1.8 h -1 ) = 1625 kg/h = 0.4514 kg/s Then the sensible, latent, and total infiltration heat loads of the room are determined to be

Qinfiltration, sensible = mair c p (Tambient - Troom ) = (0.4514 kg/s)(1.005 kJ/kg.°C)(32 - 24)°C = 3.63 kW Qinfiltration, latent = mair (wambient - wroom )h fg = (0.4514 kg/s)(0.01039 - 0.01024)(2444.1 kJ/kg) = 0.166 kW Qinfiltration, total = Qinfiltration, sensible + Qinfiltration, latent = 3.63 + 0.166 = 3.80 kW Discussion The specific volume of the dry air at the ambient conditions could also be determined from the psychrometric chart at ambient conditions.

EES (Engineering Equation Solver) SOLUTION "Given" ACH=1.8 [1/h] L=20 [m] W=13 [m] H=3 [m] T_ambient=32 [C] phi_ambient=0.35 T_room=24 [C] phi_room=0.55 "Properties" c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] h_f=enthalpy(steam_iapws, T=T_room, x=0) h_g=enthalpy(steam_iapws, T=T_room, x=1) h_fg=h_g-h_f P=101.325 [kPa] w_ambient=humrat(AirH2O, T=T_ambient, P=P, R=phi_ambient) w_room=humrat(AirH2O, T=T_room, P=P, R=phi_room) "Analysis" rho_ambient=P/(R*(T_ambient+273)) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-224

V_room=L*W*H m_dot_air=rho_ambient*V_room*ACH*1/Convert(h, s) Q_dot_infiltration_sensible=m_dot_air*c_p*(T_ambient-T_room) Q_dot_infiltration_latent=m_dot_air*(w_ambient-w_room)*h_fg Q_dot_infiltration_total=Q_dot_infiltration_sensible+Q_dot_infiltration_latent

15-62 There are 100 chickens in a breeding room. The rate of total heat generation and the rate of moisture production in the room are to be determined. Assumptions All the moisture from the chickens is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2430 kJ / kg . The average metabolic rate of chicken during normal activity is 10.2 W (3.78 W sensible and 6.42 W latent). Analysis The total rate of heat generation of the chickens in the breeding room is

Qgen, total = qgen, total (No. of chickens) = (10.2 W/chicken)(100 chickens) = 1020 W The latent heat generated by the chicken and the rate of moisture production are Qgen, latent = qgen, latent (No. of chickens)

= (6.42 W/chicken)(100 chickens) = 642 W

= 0.642 kW

mmoisture =

Qgen, latent h fg

0.642 kJ/s = 0.000264 kg/s = 0.264 g / s 2430 kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" m_chicken=1.82 [kg] metabolic_basal=5.47 [W] metabolic_total=10.2 [W] metabolic_sensible=3.78 [W] metabolic_latent=6.42 [W] numberofchickens=100 h_fg=2430 [kJ/kg] "Analysis" Q_dot_gen_total=metabolic_total*numberofchickens Q_dot_gen_latent=metabolic_latent*numberofchickens*Convert(W, kW) m_dot_moisture=Q_dot_gen_latent/h_fg*Convert(kg, g)

15-63 An average person produces 0.25 kg of moisture while taking a shower. The contribution of showers of a family of four to the latent heat load of the air-conditioner per day is to be determined. Assumptions All the water vapor from the shower is condensed by the air-conditioning system. Properties The latent heat of vaporization of water is given to be 2430 kJ / kg . Analysis The amount of moisture produced per day is

mvapor = (Moisture produced per person)(No. of persons)

13-225

= (0.25 kg/person)(4 persons/day) = 1 kg/day Then the latent heat load due to showers becomes

Qlatent = mvapor h fg = (1 kg/day)(2450 kJ/kg) = 2450 kJ / day

EES (Engineering Equation Solver) SOLUTION "Given" moistureperperson=0.25 [kg] numberofpersons=4 h_fg=2450 [kJ/kg] "Analysis" m_vapor=moistureperperson*numberofpersons Q_latent=m_vapor*h_fg

Simple Heating and Cooling 15-64C Relative humidity decreases during a simple heating process and increases during a simple cooling process. Specific humidity, on the other hand, remains constant in both cases.

15-65C Because a horizontal line on the psychrometric chart represents a  = constant process, and the moisture content  of air remains constant during these processes.

15-66 Humid air at a specified state is cooled at constant pressure to the dew-point temperature. The cooling required for this process is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Analysis The amount of moisture in the air remains constant (s) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 150 kPa. The properties of the air at the inlet and exit states are determined to be Pv1 = f 1 Pg1 = f 1 Psat @ 40°C = (0.70)(7.3851 kPa) = 5.1696 kPa hg1 = hg @ 40°C = 2573.5 kJ/kg

w1 =

0.622 Pv1 0.622(5.1696 kPa) = = 0.02220 kg H 2 O/kg dry air P1 - Pv1 (150 - 5.1696) kPa

13-226

h1 = c pT1 + w1hg1

= (1.005 kJ/kg ×°C)(40°C)+(0.02220)(2573.5 kJ/kg)

= 97.33 kJ/kg dry air

Pv 2 = Pv1 = 5.1696 kPa Pg 2 =

Pv 2 5.1696 kPa = = 5.1696 kPa f2 1

T2 = Tsat @5.1695 kPa = 33.5°C

hg 2 = hg @33.5°C = 2561.9 kJ/kg

w2 = w1 h2 = c pT2 + w2 hg 2

= (1.005 kJ/kg ×°C)(33.5°C)+(0.02220)(2561.9 kJ/kg)

= 90.55 kJ/kg dry air From the energy balance on air in the cooling section,

qout = h1 - h2 = 97.33 - 90.55 = 6.78 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 "[kPa]" T=20 "[C]" phi=1.0 "Analysis" Fluid$='AirH2O' w=humrat(Fluid$, T=T, P=P, R=phi) h=enthalpy(Fluid$, T=T, P=P, R=phi) T_wb=wetbulb(Fluid$, T=T, P=P, R=phi) T_dp=dewpoint(Fluid$, T=T, P=P, R=phi) v=volume(Fluid$, T=T, P=P, R=phi) h2=enthalpy(Fluid$, R=1, P=P, T=T_dp) T_w=temperature(steam_iapws, P=5.1696, x=1) h_g2=enthalpy(steam_iapws, T=33.5, x=1) h_f=enthalpy(steam_iapws, T=22, x=0)

15-67E Humid air at a specified state is heated at constant pressure to a specified temperature. The relative humidity at the exit and the amount of heat required are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-227

Analysis The amount of moisture in the air remains constant ( w1 = w2 ) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 40 psia. The properties of the air at the inlet and exit states are determined to be Pv1 = f 1 Pg1 = f 1 Psat @50°F = (0.90)(0.17812 psia) = 0.16031 psia

hg1 = hg @50°F = 1083.1 Btu/lbm

w1 =

0.622 Pv1 0.622(0.16031 psia) = P1 - Pv1 (40 - 0.16031) psia

= 0.002503 lbm H2 O/lbm dry air

h1 = c pT1 + w1hg1 = (0.240 Btu/lbm ×°F)(50°F) + (0.002503)(1083.1 Btu/lbm)

= 14.71 Btu/lbm dry air Pv 2 = Pv1 = 0.16031 psia Pg 2 = Psat @120°F = 1.6951 psia

f2=

Pv 2 0.16031 psia = = 0.0946 = 9.46% Pg 2 1.6951 psia

hg 2 = hg @120° F = 1113.2 Btu/lbm

w2 = w1 h2 = c pT2 + w2 hg 2

= (0.240 Btu/lbm ×°F)(120°F) + (0.002503)(1113.2 Btu/lbm)

= 31.59 Btu/lbm dry air From the energy balance on air in the heating section,

qin = h2 - h1 = 31.59 - 14.71 = 16.9 Btu / lbm dry air

EES (Engineering Equation Solver) SOLUTION "Given" P=40 [psia] T1=50 [F] phi_1=0.90 T2=120 [F] "Analysis" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T1, x=1) P_v1=phi_1*P_g1 h_g1=enthalpy(Fluid$, T=T1, x=1) w1=(0.622*P_v1)/(P-P_v1) c_p=0.24 [Btu/lbm-F] "for air, Table A-2Ea" h1=c_p*T1+w1*h_g1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-228

P_v2=P_v1 P_g2=pressure(Fluid$, T=T2, x=1) phi_2=P_v2/P_g2 h_g2=enthalpy(Fluid$, T=T2, x=1) w2=w1 h2=c_p*T2+w2*h_g2 q_in=h2-h1

15-68 Air enters a heating section at a specified state and relative humidity. The rate of heat transfer in the heating section and the relative humidity of the air at the exit are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible.

Analysis (a) The amount of moisture in the air remains constant ( w1 = w2 ) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 95 kPa. The properties of the air are determined to be Pv1 = f 1 Pg1 = f 1 Psat@10°C = (0.3)(1.403 kPa) = 0.3684 kPa

Pa1 = P1 - Pv1 = 95 - 0.3684 = 94.63 kPa v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(283 K) = Pa1 94.63 kPa

= 0.8583 m3 /kg dry air

w1 =

0.622 Pv1 0.622(0.3684 kPa) = P1 - Pv1 (95 - 0.3684) kPa

= 0.002422 kg H2 O/kg dry air ( = w2 ) h1 = c pT1 + w1hg1

= (1.005 kJ/kg ×°C)(10°C) + (0.002422)(2519.2 kJ/kg)

= 16.15 kJ/kg dry air and h2 = c pT2 + w2 hg 2

= (1.005 kJ/kg ×°C)(25°C) + (0.002422)(2546.5 kJ/kg)

= 31.29 kJ/kg dry air Also,

ma1 =

V1 6 m3 /min = = 6.991 kg/min v1 0.8583 m3 /kg dry air

13-229

Then the rate of heat transfer to the air in the heating section is determined from an energy balance on air in the heating section to be Qin = ma (h2 - h1 ) = (6.991 kg/min)(31.29 - 16.15) kJ/kg = 106 kJ / min

(b) Noting that the vapor pressure of air remains constant ( Pv 2 = Pv1 ) during a simple heating process, the relative humidity of the air at leaving the heating section becomes

f2=

Pv 2 Pv 2 0.3684 kPa = = = 0.1162 or 11.6% Pg 2 Psat@25°C 3.1698 kPa

EES (Engineering Equation Solver) SOLUTION "Given" P_1=95 [kPa] T_1=10 [C] phi_1=0.30 V_dot_1=6 [m^3/min] T_2=25 [C] "Properties" Fluid$='steam_iapws' c_p=1.005 [kJ/kg-K] R_a=0.287 [kJ/kg-K] P_g1=pressure(Fluid$, T=T_1, x=1) P_g2=pressure(Fluid$, T=T_2, x=1) h_g1=enthalpy(Fluid$, T=T_1, x=1) h_g2=enthalpy(Fluid$, T=T_2, x=1) "Analysis" "(a)" P_v1=phi_1*P_g1 P_a1=P_1-P_v1 v_1=(R_a*(T_1+273))/P_a1 w_1=(0.622*P_v1)/(P_1-P_v1) h_1=C_p*T_1+w_1*h_g1 w_2=w_1 h_2=c_p*T_2+w_2*h_g2 m_dot_a1=V_dot_1/v_1 Q_dot_in=m_dot_a1*(h_2-h_1) "(b)" P_v2=P_v1 phi_2=P_v2/P_g2

15-69 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1

This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-230

Analysis (a) The amount of moisture in the air remains constant ( w1 = w2 ) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31 or EES) to be

h1 = 76.14 kJ/kg dry air

w1 = 0.01594 kg H2 O/kg dry air ( = w2 ) v1 = 0.8953 m3 /kg dry air

The mass flow rate of dry air through the cooling section is ma = =

1 V1 A1 v1 1 (18 m/s)(p ´ 0.32 /4 m 2 ) (0.8953 m3 /kg)

= 1.421 kg/s From the energy balance on air in the cooling section, - Qout = ma (h2 - h1 )

- (750 / 60) kJ/s = (1.421 kg/s)(h2 - 76.14) kJ/kg h2 = 67.35 kJ/kg dry air (b) The exit state of the air is fixed now since we know both h2 and w2 . From the psychrometric chart at this state we read

T2 = 26.5°C f 2 = 73.1% v 2 = 0.8706 m3 /kg dry air

ma1 = ma 2 ¾ ¾ ® V2 =

V1 V2 VA V A = ¾¾ ® 1 = 2 v1 v 2 v1 v2

v2 0.8706 V1 = (18 m/s) = 17.5 m / s v1 0.8953

EES (Engineering Equation Solver) SOLUTION "Given" D=0.30 [m] Q_dot_out=750 [kJ/min] P_1=101.325 [kPa] T_1=35 [C] phi_1=0.45 Vel_1=18 [m/s] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-231

Fluid$='AirH2O' "Inlet Properties" h_1=enthalpy(Fluid$, T=T_1, P=P_1, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P_1, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P_1, R=phi_1) A_1=pi*D^2/4 m_dot_a=(1/v_1)*Vel_1*A_1 -Q_dot_out*1/Convert(min, s)=m_dot_a*(h_2-h_1) "Exit Properties" w_2=w_1 P_2=P_1 T_2=temperature(Fluid$, h=h_2, P=P_2, w=w_2) phi_2=relhum(Fluid$, h=h_2, P=P_2, w=w_2) v_2=volume(Fluid$, T=T_2, P=P_2, w=w_2) Vel_2=(v_2/v_1)*Vel_1

15-70 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air remains constant ( w1 = w2 ) as it flows through the cooling section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31 or EES) to be

h1 = 76.14 kJ/kg dry air w1 = 0.01594 kg H2 O/kg dry air ( = w2 ) v1 = 0.8953 m3 /kg dry air

The mass flow rate of dry air through the cooling section is ma = =

1 V1 A1 v1 1 (18 m/s)(p ´ 0.32 /4 m 2 ) 3 (0.8953 m /kg)

= 1.421 kg/s From the energy balance on air in the cooling section, - Qout = ma (h2 - h1 )

13-232

h2 = 65.00 kJ/kg dry air (b) The exit state of the air is fixed now since we know both h2 and w2 . From the psychrometric chart (or EES) at this state we read

T2 = 24.3°C f 2 = 83.6% v 2 = 0.8641 m3 /kg dry air

ma1 = ma 2 ¾ ¾ ® V2 =

V1 V2 VA V A = ¾¾ ® 1 = 2 v1 v 2 v1 v2

v2 0.8641 V1 = (18 m/s) = 17.4 m / s v1 0.8953

EES (Engineering Equation Solver) SOLUTION "Given" D=0.30 [m] Q_dot_out=950 [kJ/min] P_1=101.325 [kPa] T_1=35 [C] phi_1=0.45 Vel_1=18 [m/s] "Analysis" Fluid$='AirH2O' "Inlet Properties" h_1=enthalpy(Fluid$, T=T_1, P=P_1, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P_1, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P_1, R=phi_1) A_1=pi*D^2/4 m_dot_a=(1/v_1)*Vel_1*A_1 -Q_dot_out*1/Convert(min, s)=m_dot_a*(h_2-h_1) "Exit Properties" w_2=w_1 P_2=P_1 T_2=temperature(Fluid$, h=h_2, P=P_2, w=w_2) phi_2=relhum(Fluid$, h=h_2, P=P_2, w=w_2) v_2=volume(Fluid$, T=T_2, P=P_2, w=w_2) Vel_2=(v_2/v_1)*Vel_1

15-71E Air enters a heating section at a specified pressure, temperature, velocity, and relative humidity. The exit temperature of air, the exit relative humidity, and the exit velocity are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-233

Analysis (a) The amount of moisture in the air remains constant ( w1 = w2 ) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31E) to be

h1 = 15.3 Btu/lbm dry air w1 = 0.0030 lbm H2 O/lbm dry air ( = w2 ) v1 = 12.9 ft 3 / lbm dry air

The mass flow rate of dry air through the heating section is ma =

1 V1 A1 v1

1 (25 ft/s)(p ´ (15/12) 2 /4 ft 2 ) = 2.38 lbm/s (12.9 ft 3 / lbm)

From the energy balance on air in the heating section, Qin = ma (h2 - h1 )

æ0.9478 Btu/s ö ÷ 4 kW çç ÷ ÷= (2.38 lbm/s)(h2 - 15.3) Btu/lbm çè 1 kW ø

h2 = 16.9 Btu/lbm dry air The exit state of the air is fixed now since we know both h2 and w2 . From the psychrometric chart at this state we read

T2 = 56.6°F (b) f 2 = 31.4% (c) v 2 = 13.1 ft 3 /lbm dry air (d) The exit velocity is determined from the conservation of mass of dry air,

ma1 = ma 2 ¾ ¾ ®

V1 V2 VA V A = ¾¾ ® 1 = 2 v1 v 2 v1 v2

Thus, V2 =

v2 13.1 V1 = (25 ft/s) = 25.4 ft / s v1 12.9

EES (Engineering Equation Solver) SOLUTION "Given" D=15 [in] Q_dot_in=4 [kW] P_1=14.7 [psia] T_1=50 [F] phi_1=0.40 Vel_1=25 [ft/s] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-234

Fluid$='AirH2O' "Inlet Properties" h_1=enthalpy(Fluid$, T=T_1, P=P_1, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P_1, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P_1, R=phi_1) A_1=pi*D^2/4*Convert(in^2, ft^2) m_dot_a=(1/v_1)*Vel_1*A_1 Q_dot_in*Convert(kW, Btu/s)=m_dot_a*(h_2-h_1) "Exit Properties" w_2=w_1 P_2=P_1 T_2=temperature(Fluid$, h=h_2, P=P_2, w=w_2) phi_2=relhum(Fluid$, h=h_2, P=P_2, w=w_2) v_2=volume(Fluid$, T=T_2, P=P_2, w=w_2) Vel_2=(v_2/v_1)*Vel_1

Heating with Humidification 15-72C To achieve a higher level of comfort. Very dry air can cause dry skin, respiratory difficulties, and increased static electricity.

15-73 Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

h1 = 31.1 kJ/kg dry air

w1 = 0.0064 kg H2O/kg dry air ( = w2 ) h2 = 36.2 kJ/kg dry air h3 = 58.1 kJ/kg dry air w3 = 0.0129 kg H2O/kg dry air Analysis

13-235

(a) The amount of moisture in the air remains constant it flows through the heating section ( w1 = w2 ), but increases in the humidifying section w3 > w2 ). The amount of steam added to the air in the heating section is

D w = w3 - w2 = 0.0129 - 0.0064 = 0.0065 kg H2O / kg dry air (b) The heat transfer to the air in the heating section per unit mass of air is

qin = h2 - h1 = 36.2 - 31.1 = 5.1 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" P_1=101.325 [kPa] T_1=15[C] phi_1=0.60 T_2=20 [C] T_3=25 [C] phi_3=0.65 "Properties" Fluid$='AirH2O' h_1=enthalpy(Fluid$, T=T_1, P=P_1, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P_1, R=phi_1) P_2=P_1 w_2=w_1 h_2=enthalpy(Fluid$, T=T_2, P=P_2, w=w_2) P_3=P_1 h_3=enthalpy(Fluid$, T=T_3, P=P_3, R=phi_3) w_3=humrat(Fluid$, T=T_3, P=P_3, R=phi_3) "Analysis" "(a)" DELTAw=w_3-w_2 "(b)" q_in=h_2-h_1

15-74E Air is first heated and then humidified by water vapor. The amount of steam added to the air and the amount of heat transfer to the air are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Figure A-31E or EES) to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-236

h1 = 10.67 Btu/lbm dry air w1 = 0.002122 lbm H2 O/lbm dry air w2 = w1 = 0.002122 lbm H2 O/lbm dry air

h2 = 17.89 Btu/lbm dry air h3 = 35.98 Btu/lbm dry air

w3 = 0.01420 lbm H2 O/lbm dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section ( w1 = w2 ), but increases in the humidifying section ( w3 > w2 ). The amount of steam added to the air in the heating section is

D w = w3 - w2 = 0.01420 - 0.002122 = 0.0121lbm H2O / lbm dry air (b) The heat transfer to the air in the heating section per unit mass of air is

qin = h2 - h1 = 17.89 - 10.67 = 7.22 Btu / lbm dry air

EES (Engineering Equation Solver) SOLUTION "Given" P_1=14.7 [psia] T_1=35 [F] phi_1=0.50 T_2=65 [F] T_3=85 [F] phi_3=0.55 "Properties" Fluid$='AirH2O' h_1=enthalpy(Fluid$, T=T_1, P=P_1, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P_1, R=phi_1) P_2=P_1 w_2=w_1 h_2=enthalpy(Fluid$, T=T_2, P=P_2, w=w_2) P_3=P_1 h_3=enthalpy(Fluid$, T=T_3, P=P_3, R=phi_3) w_3=humrat(Fluid$, T=T_3, P=P_3, R=phi_3) "Analysis" "(a)" DELTAw=w_3-w_2 "(b)" q_in=h_2-h_1

15-75 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined. Assumptions 1

This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-237

h1 = 23.5 kJ/kg dry air w1 = 0.0053 kg H2 O/kg dry air ( = w2 ) v1 = 0.809 m3 /kg dry air

h3 = 42.3 kJ/kg dry air w3 = 0.0087 kg H2O/kg dry air Analysis (a) The amount of moisture in the air remains constant it flows through the heating section ( w1 = w2 ), but increases in the humidifying section ( w3 > w2 ). The mass flow rate of dry air is

ma =

V1 35 m3 /min = = 43.3 kg/min v1 0.809 m3 /kg

Noting that Q = W =0, the energy balance on the humidifying section can be expressed as Ein - Eout = D EsystemZ 0 (steady) = 0

Ein = Eout

å mi hi = å me he

¾ ¾®

mw hw + ma 2 h2 = ma h3 (w3 - w2 )hw + h2 = h3

Solving for h2 ,

h2 = h3 - (w3 - w2 )hg @100°C = 42.3 - (0.0087 - 0.0053)(2675.6) = 33.2 kJ/kg dry air Thus at the exit of the heating section we have w2 = 0.0053 kg H2 O dry air and h2 = 33.2 kJ / kg dry air, which completely fixes the state. Then from the psychrometric chart we read

T2 = 19.5°C f 2 = 37.8% (b) The rate of heat transfer to the air in the heating section is Qin = ma (h2 - h1 ) = (43.3 kg/min)(33.2 - 23.5) kJ/kg = 420 kJ / min

(c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section,

mw = ma (w3 - w2 ) = (43.3 kg/min)(0.0087 - 0.0053) = 0.15 kg / min

13-238

P=101.325 [kPa] T_steam=100 [C] T_1=10 [C] phi_1=0.70 V_dot_1=35 [m^3/min] T_3=20 [C] phi_3=0.60 "Properties" Fluid$='AirH2O' h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) w_2=w_1 h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3) w_3=humrat(Fluid$, T=T_3, P=P, R=phi_3) h_g=enthalpy(steam_iapws, T=T_steam, x=1) "Analysis" "(a)" m_dot_a=V_dot_1/v_1 (w_3-w_2)*h_g+h_2=h_3 "energy balance on the humidifying section" T_2=temperature(Fluid$, h=h_2, P=P, w=w_2) phi_2=relhum(Fluid$, h=h_2, P=P, w=w_2) "(b)" Q_dot_in=m_dot_a*(h_2-h_1) "(c)" m_dot_w=m_dot_a*(w_3-w_2)

15-76 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be determined.

Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The amount of moisture in the air also remains constant it flows through the heating section ( w1 = w2 ), but increases in the humidifying section ( w3 > w2 ). The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. The properties of the air at various states are determined to be Pv1 = f 1 Pg1 = f 1 Psat @10°C = (0.70)(1.2281 kPa) = 0.860 kPa ( = Pv 2 )

Pa1 = P1 - Pv1 = 95 - 0.860 = 94.14 kPa

13-239

v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(283 K) = = 0.863 m3 /kg dry air Pa1 94.14 kPa

w1 =

0.622 Pv1 0.622(0.86 kPa) = = 0.00568 kg H 2 O/kg dry air ( = w2 ) P1 - Pv1 (95 - 0.86) kPa

h1 = c pT1 + w1hg1 = (1.005 kJ/kg ×°C)(10°C) + (0.00568)(2519.2 kJ/kg) = 24.36 kJ/kg dry air Pv 3 = f 3 Pg 3 = f 3 Psat @ 20°C = (0.60)(2.3392 kPa) = 1.40 kPa

w3 =

0.622 Pv 3 0.622(1.40 kPa) = = 0.00930 kg H 2 O/kg dry air P3 - Pv 3 (95 - 1.40) kPa

h3 = c pT3 + w3 hg 3 = (1.005 kJ/kg ×°C)(20°C) + (0.0093)(2537.4 kJ/kg) = 43.70 kJ/kg dry air

Also,

ma =

V1 35 m3 / min = = 40.6 kg/min v1 0.863 m3 / kg

Noting that Q = W = 0, the energy balance on the humidifying section gives Ein - Eout = D EsystemZ 0 (steady) = 0 ¾ ¾ ® Ein = Eout

å me he = å mi hi ¾ ¾ ® mw hw + ma 2 h2 = ma h3 ¾ ¾ ® (w3 - w2 )hw + h2 = h3

h2 = h3 - (w3 - w2 )hg @100°C = 43.7 - (0.0093 - 0.00568)´ 2675.6 = 34.0 kJ/kg dry air Thus at the exit of the heating section we have w = 0.00568 kg H2 O dry air and h2 = 34.0 kJ / kg dry air, which completely fixes the state. The temperature of air at the exit of the heating section is determined from the definition of enthalpy,

h2 = c pT2 + w2 hg 2 @ c pT2 + w2 (2500.9 + 1.82T2 )

34.0 = (1.005)T2 + (0.00568)(2500.9 + 1.82T2 ) Solving for h2 , yields

T2 = 19.5°C The relative humidity at this state is

f2=

Pv 2 Pv 2 0.860 kPa = = = 0.377 or 37.7% Pg 2 Psat @19.5°C 2.2759 kPa

(b) The rate of heat transfer to the air in the heating section becomes Qin = ma (h2 - h1 ) = (40.6 kg/min)(34.0 - 24.36) kJ/kg = 391 kJ / min

(c) The amount of water added to the air in the humidifying section is determined from the conservation of mass equation of water in the humidifying section, mw = ma (w3 - w2 ) = (40.6 kg/min)(0.0093 - 0.00568) = 0.147 kg / min

EES (Engineering Equation Solver) SOLUTION "Given" P=95 [kPa] T_steam=100 [C] T1=10 [C] phi_1=0.70 V_dot_1=35 [m^3/min] T3=20 [C] phi_3=0.60 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-240

"Properties" R_a=0.287 [kJ/kg-C] "for air, Table A-2a" c_p=1.005 [kJ/kg-C] "for air, Table A-2a" "Analysis" "(a)" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T1, x=1) P_v1=phi_1*P_g1 P_a1=P-P_v1 v1=(R_a*(T1+273))/P_a1 w1=(0.622*P_v1)/(P-P_v1) h_g1=enthalpy(Fluid$, T=T1, x=1) h1=c_p*T1+w1*h_g1 P_g3=pressure(Fluid$, T=T3, x=1) P_v3=phi_3*P_g3 w3=(0.622*P_v3)/(P-P_v3) h_g3=enthalpy(Fluid$, T=T3, x=1) h3=c_p*T3+w3*h_g3 m_dot_a=V_dot_1/v1 h_s=enthalpy(Fluid$, T=T_steam, x=1) w2=w1 h2=h3-(w3-w2)*h_s h_g2=2500.9+1.82*T2 h2=c_p*T2+w2*h_g2 P_v2=P_v1 P_g2=pressure(Fluid$, T=T2, x=1) phi_2=P_v2/P_g2 "(b)" Q_dot_in=m_dot_a*(h2-h1) "(c)" m_dot_w=m_dot_a*(w3-w2)

Cooling with Dehumidification 15-77C To drop its relative humidity to more desirable levels.

15-78 Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the cooling requirement are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-241

h1 = 85.5 kJ/kg dry air w1 = 0.0217 kg H2 O/kg dry air and

f 2 = 1.0 h2 = 57.6 kJ/kg dry air

w2 = 0.0148 kg H2 O/kg dry air Also,

hw @ h f @ 22°C = 92.28 kJ/kg

(Table A-4)

Analysis The amount of moisture in the air decreases due to dehumidification ( w2 < w1 ). Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

D w = w1 - w2 = 0.0217 - 0.0148 = 0.0069 kg H2O / kg dry air Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = Qout + å me he Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw

qout = h1 - h2 - (w1 - w2 )hw

= (85.5 - 57.6)kJ/kg - (0.0069)(92.28) = 27.3 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 "[kPa]" T=20 "[C]" phi=1.0 "Analysis" Fluid$='AirH2O' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-242

w=humrat(Fluid$, T=T, P=P, R=phi) h=enthalpy(Fluid$, T=T, P=P, R=phi) T_wb=wetbulb(Fluid$, T=T, P=P, R=phi) T_dp=dewpoint(Fluid$, T=T, P=P, R=phi) v=volume(Fluid$, T=T, P=P, R=phi) h2=enthalpy(Fluid$, R=1, P=P, T=T_dp) T_w=temperature(steam_iapws, P=5.1696, x=1) h_g2=enthalpy(steam_iapws, T=33.5, x=1) h_f=enthalpy(steam_iapws, T=22, x=0)

15-79E Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

h1 = 37.8 Btu/lbm dry air w1 = 0.0158 lbm H2 O/lbm dry air

v1 = 14.08 ft 3 /lbm dry air

and

f 2 = 1.0 h2 = 26.5 Btu/lbm dry air

w2 = 0.0111 lbm H2 O/lbm dry air Also,

hw @ h f @ 65°F = 33.08 Btu/lbm (Table A-4E) Analysis The amount of moisture in the air decreases due to dehumidification ( w2 < w1 ). The mass flow rate of air is

ma1 =

V1 (10, 000 / 3600) ft 3 / s = = 0.1973 lbm/s v1 14.08 ft 3 / lbm dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

13-243

mw = ma (w1 - w2 ) = (0.1973 lbm/s)(0.0158 - 0.0111) = 0.000927 lbm / s Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = Qout + å me he Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw Qout = (0.1973 lbm/s)(37.8 - 26.5)Btu/lbm - (0.000927 lbm/s)(33.08 Btu/lbm)

= 2.20 Btu / s

EES (Engineering Equation Solver) SOLUTION "Given" P=14.696 [psia] T1=85 [F] T_dp1=70 [F] T2=60 [F] phi_2=1 "Analysis" Fluid$='AirH2O' w1=humrat(Fluid$, T=T1, P=P, D=T_dp1) h1=enthalpy(Fluid$, T=T1, P=P, D=T_dp1) T_wb1=wetbulb(Fluid$, T=T1, P=P, D=T_dp1) v1=volume(Fluid$, T=T1, P=P, D=T_dp1) h2=enthalpy(Fluid$, T=T2, P=P, R=phi_2) w2=humrat(Fluid$, T=T2, P=P, R=phi_2) h_w=enthalpy(steam_iapws, T=65, x=0)

15-80 Air is cooled by passing it over a cooling coil through which chilled water flows. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The saturation pressure of water at 32C is 4.76 kPa (Table A-4). Then the dew point temperature of the incoming air stream at 32C becomes Tdp = Tsat @ Pv = Tsat @ 0.7´ 4.76 kPa = 25.8°C (Table A-5)

13-244

since air is cooled to 20C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification (w2 < w1 ). The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. Then the properties of the air at both states are determined from the psychrometric chart (Fig. A-31 or EES) to be

h1 = 86.35 kJ/kg dry air w1 = 0.02114 kg H2 O/kg dry air v1 = 0.8939 m3 /kg dry air

and

h2 = 57.43 kJ/kg dry air w2 = 0.0147 kg H2O/kg dry air v 2 = 0.8501 m3 /kg dry air

Also,

hw @ h f @ 20°C = 83.91 kJ/kg

(Table A-4)

Then, V1 = V1 A1 = V1

ma1 =

æp (0.4 m) 2 ÷ ö p D2 ÷= 15.08 m3 /min = (120 m/min) çç ÷ ÷ ç 4 4 è ø

V1 15.08 m3 /min = = 16.87 kg/min v1 0.8939 m3 /kg dry air

Applying the water mass balance and the energy balance equations to the combined cooling and dehumidification section (excluding the water), Water Mass Balance: å mw, i = å mw, e

¾¾ ®

ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) = (16.87 kg/min)(0.02114 - 0.0147) = 0.1086 kg/min Energy Balance: å mi hi = å me he + Qout

¾¾ ®

Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw

Qout = (16.87 kg/min)(86.35 - 57.43) kJ/kg - (0.1086 kg/min)(83.91 kJ/kg) = 478.7 kJ / min

(b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from

Qcooling water = mcooling water D h = mcooling water c p D T mcooling water =

Qw 478.7 kJ/min = = 19.09 kg / min cpD T (4.18 kJ/kg ×°C)(6°C)

ma1 = ma 2 V2 =

¾¾ ®

V1 V2 = v1 v 2

¾¾ ®

V1 A V2 A = v1 v2

v2 0.8501 V1 = (120 m/min) = 114 m / min v1 0.8939

EES (Engineering Equation Solver) SOLUTION D=0.4 P[1] =101.32 [kPa] T[1] = 32 [C] RH[1] = 70/100 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-245

Vel[1] = 120[m/min]*Convert(m/min, m/s) DELTAT_cw =6 [C] P[2] = 101.32 [kPa] T[2] = 20 [C] RH[2] = 100/100 $Array Off P_sat=pressure(steam_iapws,T=T[1],x=1) T_dp=temperature(steam_iapws,P=0.7*P_sat, x=1) "Dry air flow rate, m_dot_a, is constant" Vol_dot[1]= (pi * D^2)/4*Vel[1] v[1]=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_a = Vol_dot[1]/v[1] "Exit vleocity" Vol_dot[2]= (pi * D^2)/4*Vel[2] v[2]=VOLUME(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_a = Vol_dot[2]/v[2] "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2] "SSSF conservation of energy for the air" m_dot_a *(h[1] + (1+w[1])*Vel[1]^2/(2*Convert(kJ/kg, m^2/s^2))) + Q_dot = m_dot_a*(h[2] +(1+w[2])*Vel[2]^2/(2*Convert(kJ/kg, m^2/s^2))) +m_dot_w*h_liq_2 h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) h_liq_2=ENTHALPY(steam_iapws,T=T[2],x=0) "SSSF conservation of energy for the cooling water" -Q_dot =m_dot_cw*Cp_cw*DELTAT_cw "Note: Q_netwater=-Q_netair" Cp_cw = 4.18 [kJ/kg-K]

15-81 Problem 15-80 is reconsidered. A general solution of the problem in which the input variables may be supplied and parametric studies performed is to be developed and the process is to be shown in the psychrometric chart for each set of input variables. Analysis The problem is solved using EES, and the solution is given below. D=0.4 P[1] =101.32 [kPa] T[1] = 32 [C] RH[1] = 70/100 Vel[1] = (120/60) [m/s] DELTAT_cw =6 [C] P[2] = 101.32 [kPa] T[2] = 20 [C] RH[2] = 100/100 "Dry air flow rate, m_dot_a, is constant" Vol_dot[1]= (pi * D^2)/4*Vel[1] v[1]=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_a = Vol_dot[1]/v[1] "Exit vleocity" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-246

Vol_dot[2]= (pi * D^2)/4*Vel[2] v[2]=VOLUME(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_a = Vol_dot[2]/v[2] "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2] "Conservation of energy for the air" m_dot_a *(h[1] + (1+w[1])*Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) + Q_dot = m_dot_a*(h[2] +(1+w[2])*Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)) +m_dot_w*h_liq_2 h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) h_liq_2=ENTHALPY(Water,T=T[2],P=P[2]) "Conservation of energy for the cooling water" -Q_dot =m_dot_cw*cp_cw*DELTAT_cw "Note: Q_netwater=-Q_netair" cp_cw = SpecHeat(water,T=10,P=P[2]) Q_dot_out=-Q_dot

RH1 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9

mcw

Qout

Vel 2

[kg / s] 0.1475 0.19 0.2325 0.275 0.3175 0.36 0.4025 0.445 0.4875

[kW] 3.706 4.774 5.842 6.91 7.978 9.046 10.11 11.18 12.25

[m / s] 1.921 1.916 1.911 1.907 1.902 1.897 1.893 1.888 1.884

13-247

15-82 Air is cooled by passing it over a cooling coil. The rate of heat transfer, the mass flow rate of water, and the exit velocity of airstream are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible.

Analysis (a) The dew point temperature of the incoming air stream at 35C is Pv1 = f 1 Pg1 = f 1 Psat @32°C

= (0.7)(4.76 kPa) = 3.332 kPa Tdp = Tsat @ Pv = Tsat @ 3.332 kPa = 25.8°C

Since air is cooled to 20C, which is below its dew point temperature, some of the moisture in the air will condense. The amount of moisture in the air decreases due to dehumidification (w2 < w1 ) . The inlet and the exit states of the air are completely specified, and the total pressure is 95 kPa. Then the properties of the air at both states are determined to be

Pa1 = P1 - Pv1 = 88 - 3.332 = 84.67 kPa v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(305 K) = = 1.034 m3 /kg dry air Pa1 84.67 kPa

w1 =

0.622 Pv1 0.622(3.332 kPa) = = 0.02448 kg H 2 O/kg dry air P1 - Pv1 (88 - 3.332) kPa

h1 = c pT1 + w1hg1 = (1.005 kJ/kg ×°C)(32°C) + (0.02448)(2559.2 kJ/kg) = 94.80 kJ/kg dry air

13-248

Pv 2 = f 2 Pg 2 = (1.00) Psat @ 20°C = 2.339 kPa

v2 =

RaT2 (0.287 kPa ×m3 /kg ×K)(293 K) = = 0.9817 m3 /kg dry air Pa 2 (88 - 2.339) kPa

w2 =

0.622 Pv 2 0.622(2.339 kPa) = = 0.01699 kg H 2 O/kg dry air P2 - Pv 2 (88 - 2.339) kPa

h2 = c pT2 + w2 hg 2 = (1.005 kJ/kg ×°C)(20°C) + (0.01699)(2537.4 kJ/kg) = 63.20 kJ/kg dry air

Also,

hw @ h f @ 20°C = 83.915 kJ/kg (Table A-4) Then, V1 = V1 A1 = V1

ma1 =

æp (0.4 m) 2 ÷ ö p D2 ÷ = (120 m/min) çç = 15.08 m3 /min ÷ ÷ çè 4 4 ø

V1 15.08 m3 /min = = 14.59 kg/min v1 1.034 m3 /kg dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section (excluding the water), Water Mass Balance: å mw, i = å mw, e

¾¾ ®

ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) = (14.59 kg/min)(0.02448 - 0.01699) = 0.1093 kg/min Energy Balance: Ein - Eout = D EsystemZ 0 (steady) = 0

Ein = Eout å mi hi = å me he + Qout ®

Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw

Qout = (14.59 kg/min)(94.80 - 63ç.20) kJ/kg - (0.1093 kg/min)(83.915 kJ/kg) = 451.7 kJ / min

(b) Noting that the heat lost by the air is gained by the cooling water, the mass flow rate of the cooling water is determined from Qcooling water = mcooling water D h = mcooling water c p D T

mcooling water =

Qw 451.7 kJ/min = = 18.01 kg / min c p D T (4.18 kJ/kg ×°C)(6°C)

ma1 = ma 2 ¾ ¾ ® V2 =

V1 V2 VA V A = ¾¾ ® 1 = 2 v1 v 2 v1 v2

v2 0.9817 V1 = (120 m/min) = 113.9 m / min v1 1.034

EES (Engineering Equation Solver) SOLUTION "Given" D=0.4 [m] P=88 [kPa] T1=32 [C] phi_1=0.70 Vel1=120 [m/min] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-249

DELTAT_water=6 [C] T2=20 [C] phi_2=1 "Properties" R_a=0.287 [kJ/kg-K] c_p_a=1.005 [kJ/kg-C] c_p_w=4.18 [kJ/kg-C] Fluid1$='steam_iapws' Fluid2$='AirH2O' P_g1=pressure(Fluid1$, T=T1, x=1) P_v1=phi_1*P_g1 T_dp=temperature(Fluid1$, P=P_v1, x=1) P_a1=P-P_v1 v1=(R_a*(T1+273))/P_a1 w1=(0.622*P_v1)/(P-P_v1) h1=c_p_a*T1+w1*h_g1 h_g1=enthalpy(Fluid1$, T=T1, x=1) P_g2=pressure(Fluid1$, T=T2, x=1) P_v2=phi_2*P_g2 P_a2=P-P_v2 v2=(R_a*(T2+273))/P_a2 w2=(0.622*P_v2)/(P-P_v2) h2=c_p_a*T2+w2*h_g2 h_g2=enthalpy(Fluid1$, T=T2, x=1) h_w=enthalpy(Fluid1$, T=T2, x=0) "Analysis" "(a)" A1=pi*D^2/4 V_dot_1=Vel1*A1 m_dot_a=V_dot_1/v1 m_dot_w=m_dot_a*(w1-w2) Q_dot_out=m_dot_a*(h1-h2)-m_dot_w*h_w "(b)" m_dot_coolingwater=Q_dot_out/(c_p_w*DELTAT_water) "(c)" Vel2=(v2/v1)*Vel1

15-83 Air is cooled and dehumidified at constant pressure. The amount of water removed from the air and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-250

h1 = 106.8 kJ/kg dry air

w1 = 0.0292 kg H2O/kg dry air v1 = 0.905 m3 /kg dry air

and

h2 = 52.7 kJ/kg dry air w2 = 0.0112 kg H2 O/kg dry air We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then,

hw @ h f @ 28°C = 117.4 kJ/kg

(Table A-4)

Analysis The amount of moisture in the air decreases due to dehumidification ( w2 < w1 ). The mass of air is

ma =

V1 1000 m3 = = 1105 kg v1 0.905 m3 / kg dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw, i = å mw, e

¾¾ ®

ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) = (1105 kg)(0.0292 - 0.0112) = 19.89 kg Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = Qout + å me he Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw Qout = ma (h1 - h2 ) - mw hw Qout = (1105 kg)(106.8 - 52.7)kJ/kg - (19.89 kg)(117.4 kJ/kg) = 57,450 kJ

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 "[kPa]" T=24 "[C]" phi=0.60 "Analysis" Fluid$='AirH2O' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-251

15-84 The humid air of the previous problem is reconsidered. The exit temperature of the air to produce the desired dehumidification is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

h1 = 106.8 kJ/kg dry air w1 = 0.0292 kg H2 O/kg dry air v1 = 0.905 m3 /kg dry air

and

h2 = 52.7 kJ/kg dry air w2 = 0.0112 kg H2 O/kg dry air

Analysis For the desired dehumidification, the air at the exit should be saturated with a specific humidity of 0.0112 kg water/kg dry air. That is,

f 2 = 1.0

w2 = 0.0112 kg H2 O/kg dry air The temperature of the air at this state is

T2 = 15.8°C

13-252

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 "[kPa]" T=24 "[C]" phi=0.60 "Analysis" Fluid$='AirH2O' w=humrat(Fluid$, T=T, P=P, R=phi) h=enthalpy(Fluid$, T=T, P=P, R=phi) T_wb=wetbulb(Fluid$, T=T, P=P, R=phi) T_dp=dewpoint(Fluid$, T=T, P=P, R=phi) v=volume(Fluid$, T=T, P=P, R=phi) T2=temperature(Fluid$, w=0.0112, P=P, R=1) h2=enthalpy(Fluid$, R=1, P=P, T=T_dp) T_w=temperature(steam_iapws, P=5.1696, x=1) h_g2=enthalpy(steam_iapws, T=33.5, x=1) h_f=enthalpy(steam_iapws, T=28, x=0)

15-85 Air is cooled and dehumidified at constant pressure. The syatem hardware and the psychrometric diagram are to be sketched and the inlet volume flow rate is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties (a) The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below. The psychrometric chart is obtained from the Property Plot feature of EES. AirH2O

0.050 Pressu re = 101.3 [kPa]

0.045

Humidity Ratio

0.040 0.035

0.8

0.030 0.6

0.025 0.020

1 0.4

0.015

0.010

0.2

0.005 0.000 0

T [°C] (b) The inlet and the exit states of the air are completely specified, and the total pressure is 101.3 kPa (1 atm). The properties of the air at the inlet and exit states are determined from the psychrometric chart (Figure A-31 or EES) to be

13-253

Tdp = 26.7°C

h1 = 96.5 kJ/kg dry air w1 = 0.0222 kg H2 O/kg dry air v1 = 0.916 m3 /kg dry air

and

T2 = T2 - 10 = 26.7 - 10 = 16.7°C f 2 = 1.0

h2 = 46.9 kJ/kg dry air w2 = 0.0119 kg H2 O/kg dry air Also,

hw @ h f @16.7°C = 70.10 kJ/kg

(Table A-4)

Analysis Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout

å mi hi = Qout + å me he

Qout = ma1h1 - (ma 2 h2 + mw hw )

= ma (h1 - h2 ) - mw hw

= ma (h1 - h2 ) - ma (w1 - w2 )hw Solving for the mass flow rate of dry air,

ma =

Qout 1340 kW = = 27.4 kg/s (h1 - h2 ) - (w1 - w2 )hw (96.5 - 46.9) kJ/kg - (0.0222 - 0.0119)(70.1 kJ/kg)

The volume flow rate is then determined from V1 = mav1 = (27.4 kg/s)(0.916 m3 /kg) = 25.1 m 3 / s

13-254

EES (Engineering Equation Solver) SOLUTION "Given" P=101.3 [kPa] T1=39 [C] phi_1=0.50 DELTAT=10 [C] T2=T_dp1-DELTAT Q_dot_out=1340 [kW] "Analysis" Fluid$='AirH2O' h1=enthalpy(Fluid$, T=T1, P=P, R=phi_1) w1=humrat(Fluid$, T=T1, P=P, R=phi_1) v1=volume(Fluid$, T=T1, P=P, R=phi_1) T_dp1=Dewpoint(Fluid$, T=T1, P=P, R=phi_1) phi_2=1 h2=enthalpy(Fluid$, T=T2, P=P, R=phi_2) w2=humrat(Fluid$, T=T2, P=P, R=phi_2) T_w=T2 "assumed" h_w=enthalpy(steam_iapws,T=T_w, x=0) m_dot_a=Q_dot_out/((h1-h2)-(w1-w2)*h_w) V_dot_1=m_dot_a*v1

15-86E Saturated humid air at a specified state is cooled at constant pressure to a specified temperature. The rate at which liquid water is formed and the rate of cooling are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Analysis The amount of moisture in the air remains constant ( w1 = w2 ) as it flows through the heating section since the process involves no humidification or dehumidification. The inlet state of the air is completely specified, and the total pressure is 70 psia. The properties of the air at the inlet and exit states are determined to be Pv1 = f 1 Pg1 = f 1 Psat @ 200° F = (1.0)(11.538 psia) = 11.538 psia

hg1 = hg @ 200° F = 1145.7 Btu/lbm

w1 =

0.622 Pv1 0.622(11.538 psia) = = 0.1228 lbm H 2 O/lbm dry air P1 - Pv1 (70 - 11.538) psia

h1 = c pT1 + w1hg1 = (0.240 Btu/lbm ×°F)(200°F)+(0.1228)(1145.7 Btu/lbm) = 188.7 Btu/lbm dry air

v1 =

RaT1 (0.3704 psia ×ft 3 / lbm ×R)(660 K) = = 4.182 ft 3 / lbm dry air Pa1 (70 - 11.538) psia

13-255

and

Pv 2 = f 2 Pg 2 = f 2 Psat @100°F = (1.0)(0.95052 psia) = 0.95052 psia

hg 2 = hg @100°F = 1104.7 Btu/lbm w2 =

0.622 Pv 2 0.622(0.95052 psia) = = 0.00856 lbm H 2 O/lbm dry air P2 - Pv 2 (70 - 0.95052) psia

h2 = c pT2 + w2 hg 2 = (0.240 Btu/lbm ×°F)(100°F)+(0.00856)(1104.7 Btu/lbm) = 33.46 Btu/lbm dry air

We assume that the condensate leaves this system at the average temperature of the air inlet and exit:

hw @ h f @150°F = 117.99 Btu/lbm

(Table A-4E)

The mass flow rate of the dry air is V1 = V1 A1 = V1

ma =

æp (3/12 ft) 2 ÷ ö p D2 ÷ = (50 ft/s) ççç = 2.454 ft 3 /s ÷ ÷ 4 4 è ø

V1 2.454 ft 3 / s = = 0.5868 lbm/s v1 4.182 ft 3 / lbm dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) = (0.5868 lbm/s)(0.1228 - 0.00856) = 0.0670 lbm / s Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = Qout + å me he

Qout = ma1h1 - (ma 2 h2 + mw hw )

= ma (h1 - h2 ) - mw hw

= (0.5868 lbm)(188.7 - 33.46)Btu/lbm - (0.06704 lbm)(117.99 Btu/lbm) = 83.2 Btu / s

EES (Engineering Equation Solver) SOLUTION "Given" phi_1=1 P=70 [psia] T1=200 [F] T2=100 [F] D=(3/12) [ft] Vel_1=50 [ft/s] "Properties" R_a=0.3704 [psia-ft^3/lbm-F] "for air, Table A-2Ea" c_p=0.240 [Btu/lbm-F] "for air, Table A-2Ea" "Analysis" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T1, x=1) P_v1=phi_1*P_g1 h_g1=enthalpy(Fluid$, T=T1, x=1) P_a1=P-P_v1 v1=(R_a*(T1+460))/P_a1 w1=(0.622*P_v1)/(P-P_v1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-256

h1=c_p*T1+w1*h_g1 P_g2=pressure(Fluid$, T=T2, x=1) P_v2=phi_2*P_g2 phi_2=P_v2/P_g2 h_g2=enthalpy(Fluid$, T=T2, x=1) w2=(0.622*P_v2)/(P-P_v2) h2=c_p*T2+w2*h_g2 T_w=0.5*(T1+T2) "assumed" h_w=enthalpy(Fluid$, T=T_w, x=0) A1=pi*D^2/4 V_dot_1=Vel_1*A1 m_dot_a=V_dot_1/v1 m_dot_w=m_dot_a*(w1-w2) Q_dot_out=m_dot_a*(h1-h2)-m_dot_w*h_w

15-87 Air is cooled, dehumidified and heated at constant pressure. The psychrometric diagram is to be sketched, the dew point temperatures, and the heat transfers are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below. The psychrometric chart is obtained from the Property Plot feature of EES.

13-257

AirH2O

0.050 Pressure = 101.3 [kPa]

0.045 0.040

Humidity Ratio

35°C

0.035

0.8

0.030 30°C 0.6

0.025 25°C

0.020

0.4

0.015

20°C

15°C

0.010 0.005 0.000 0

0.2

10°C

T [°C] (b) The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at three states are determined from the psychrometric chart (Figure A-31 or EES) to be

T1 = 39°C

Tdp1 = 26.7°C 

f 1 = 50%

h1 = 96.43 kJ/kg dry air

P = 101.325 kPa

w1 = 0.02225 kg H 2 O/kg dry air Tdp3 = 5.3°C

T3 = 17°C Twb3 = 10.8°C



P = 101.325 kPa

f 3 = 46.0% h3 = 31.03 kJ/kg dry air w3 = 0.005514 kg H 2 O/kg dry air

Tdp2 = 5.3°C

w2 = w3 = 0.005514 kg H 2 O/kg dry air 

f2=1 P = 101.325 kPa

T2 = 5.3°C

*******

h2 = 19.16 kJ/kg dry air

Also,

hw @ h f @5.3°C = 22.25 kJ/kg

(Table A-4)

The enthalpy of water is evaluated at 5.3C, which is the temperature of air at the end of the dehumidification process. (c) Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout

å mi hi = Qout + å me he Qout = ma1h1 - (ma 2 h2 + mw hw )

= ma (h1 - h2 ) - mw hw

13-258

= ma (h1 - h2 ) - ma (w1 - w2 )hw Substituting,

q12 = (h1 - h2 ) - (w1 - w2 )hw = (96.43 - 19.16) kJ/kg - (0.02225 - 0.005514)(22.25 kJ/kg) = 76.9 kJ / kg Applying the energy balance to the heating section,

q23 = h3 - h2 = (31.03 - 19.16) kJ/kg = 11.9 kJ / kg The net heat transfer is

q23 = q12 - q23 = (76.9 - 11.9) kJ/kg = 65.0 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 T_1=39 phi_1=0.50 T_3=17 T_wb3=10.8 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) T_wb1=wetbulb(Fluid1$, T=T_1, R=phi_1, P=P) T_dp1=dewpoint(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_3=enthalpy(Fluid1$, T=T_3, B=T_wb3, P=P) w_3=humrat(Fluid1$, T=T_3, B=T_wb3, P=P) phi_3=relhum(Fluid1$, T=T_3, B=T_wb3, P=P) T_dp3=dewpoint(Fluid1$, T=T_3, B=T_wb3, P=P) w_2=w_3 T_2=temperature(Fluid1$, R=1, w=w_2, P=P) h_2=enthalpy(Fluid1$, R=1, w=w_2, P=P) T_dp2=dewpoint(Fluid1$, R=1, w=w_2, P=P) T_w=T_2 h_w=enthalpy(Fluid2$, T=T_w, x=0) DELTAw=w_2-w_1 q_12=h_1-h_2-(w_1-w_2)*h_w q_23=h_3-h_2 q_net=q_12-q_23

15-88 Air is first cooled, then dehumidified, and finally heated. The temperature of air before it enters the heating section, the amount of heat removed in the cooling section, and the amount of heat supplied in the heating section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process  mal  ma 2  ma  2

Dry air and water vapor are ideal gases.

13-259

The kinetic and potential energy changes are negligible.

Analysis (a) The amount of moisture in the air decreases due to dehumidification ( w3 < w1 ), and remains constant during heating (

w3 = w2 ). The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The intermediate state (state 2) is also known since f 2 = 100% and w2 = w3 . Therefore, we can determine the properties of the air at all three states from the psychrometric chart (Fig. A-31) to be

h1  95.2 kJ / kg dry air

1  0.0238 kgH2 O / kg dry air and

h3  43.1kJ / kg dry air 3  0.0082 kgH 2 O / kg dry air   2  Also,

hw  h f @10 C  42.02 kJ / kg (Table A - 4)

h2  31.8 kJ / kg dry air T2  11.1 C

(b) The amount of heat removed in the cooling section is determined from the energy balance equation applied to the cooling section,

Ein  Eout  ΔEsystem 0(steady)  0 Ein  Eout

m h  m h  Q i i

e e

out cooling

Qout,cooling  ma1h1   ma 2 h2  mw hw   ma  h1  h2   mw hw or, per unit mass of dry air, qout,cooling  (h1  h2 )  (1  2 )hw

 (95.2  31.8)  (0.0238  0.0082)42.02  62.7 kJ / kg dry air (c) The amount of heat supplied in the heating section per unit mass of dry air is qinheating  h3  h2  43.1  31.8  11.3kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=34 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-260

phi_1=0.70 T_3=22 [C] phi_3=0.50 T_water=10 [C] "Properties" Fluid$='AirH2O' h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3) w_3=humrat(Fluid$, T=T_3, P=P, R=phi_3) h_w=enthalpy(steam_iapws, T=T_water, x=0) w_2=w_3 phi_2=1 h_2=enthalpy(Fluid$, R=phi_2, P=P, w=w_2) "Analysis" "(a)" T_2=temperature(Fluid$, R=phi_2, P=P, w=w_2) "(b)" q_out_cooling=(h_1-h_2)-(w_1-w_2)*h_w "(c)" q_in_heating=h_3-h_2

15-89 Air is cooled and dehumidified at constant pressure. The cooling required is provided by a simple ideal vaporcompression refrigeration system using refrigerant-134a as the working fluid. The exergy destruction in the total system per 1000 m3 of dry air is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

h1 = 106.8 kJ/kg dry air

13-261

v1 = 0.905 m3 /kg dry air

and

h2 = 52.7 kJ/kg dry air w2 = 0.0112 kg H2 O/kg dry air We assume that the condensate leaves this system at the average temperature of the air inlet and exit. Then, from Table A-4,

hw @ h f @ 28°C = 117.4 kJ/kg Analysis The amount of moisture in the air decreases due to dehumidification ( w2 < w1 ). The mass of air is

ma =

V1 1000 m3 = = 1105 kg v1 0.905 m3 / kg dry air

Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw, i = å mw, e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) = (1105 kg)(0.0292 - 0.0112) = 19.89 kg Energy Balance:

Ein - Eout = D Esystem ¿ 0 (steady) = 0 Ein = Eout å mi hi = Qout + å me he Qout = ma1h1 - (ma 2 h2 + mw hw ) = ma (h1 - h2 ) - mw hw

Qout = ma (h1 - h2 ) - mw hw Qout = (1105 kg)(106.8 - 52.7) kJ/kg - (19.89 kg)(117.4 kJ/kg) = 57,450 kJ We obtain the properties for the vapor-compression refrigeration cycle as follows (Tables A-11,through A-13):

T1 = 4°C ü ïï h1 = hg @ 4°C = 252.77 kJ/kg ý sat. vapor ïïþ s1 = sg @ 4°C = 0.92927 kJ/kg ×K P2 = Psat@39.4°C = 1 MPa ïü ïý h = 275.29 kJ/kg ïïþ 2 s2 = s1

P3 = 1 MPa ü ïï h3 = h f @1 MPa = 107.32 kJ/kg ý sat. liquid ïïþ s3 = s f @1 MPa = 0.39189 kJ/kg ×K

h4 @ h3 = 107.32 kJ/kg (throttling) ü T4 = 4°C ïï x4 = 0.2561 ý h4 = 107.32 kJ/kg ïïþ s4 = 0.4045 kJ/kg ×K

13-262

The mass flow rate of refrigerant-134a is mR =

QL 57, 450 kJ = = 395.0 kg h1 - h4 (252.77 - 107.32) kJ/kg

The amount of heat rejected from the condenser is

QH = mR (h2 - h3 ) = (395.0 kg)(275.29 - 107.32) kJ/kg = 66,350 kg Next, we calculate the exergy destruction in the components of the refrigeration cycle: X destroyed, 12 = mRT0 (s2 - s1 ) = 0 (since the process is isentropic)

æ Q ö ÷ X destroyed, 23 = T0 çççmR ( s3 - s2 ) + H ÷ ÷ çè TH ÷ ø æ 66,350 kJ ö ÷ = (305 K) çç(395 kg)(0.39189 - 0.92927) kJ/kg ×K + ÷= 1609 kJ ÷ çè 305 K ø X destroyed, 34 = mRT0 (s4 - s3 ) = (395 kg)(305 K)(0.4045 - 0.39189) kJ/kg ×K = 1519 kJ

The entropies of water vapor in the air stream are sg1 = sg @32°C = 8.4114 kJ/kg ×K sg 2 = sg @ 24°C = 8.5782 kJ/kg ×K

The entropy change of water vapor in the air stream is

D Svapor = ma (w2 sg 2 - w1 sg1 ) = (1105)(0.0112´ 8.5782 - 0.0292´ 8.4114) = - 165.2 kJ/K The entropy of water leaving the cooling section is

Sw = mw s f @ 28°C = (19.89 kg)(0.4091 kJ/kg ×K) = 8.14 kJ/K The partial pressures of water vapor and dry air for air streams are Pv1 = f 1 Pg1 = f 1 Psat @32°C = (0.95)(4.760 kPa) = 4.522 kPa Pa1 = P1 - Pv1 = 101.325 - 4.522 = 96.80 kPa

Pv 2 = f 2 Pg 2 = f 2 Psat @ 24°C = (0.60)(2.986 kPa) = 1.792 kPa Pa 2 = P2 - Pv 2 = 101.325 - 1.792 = 99.53 kPa

The entropy change of dry air is æ P ö T ÷ D S a = ma ( s2 - s1 ) = ma çççc p ln 2 - R ln a 2 ÷ ÷ ÷ çè T1 Pa1 ø

é 297 99.53 ù ú= - 38.34 kJ/kg dry air = (1105) ê(1.005) ln - (0.287) ln ú 305 96.80 û ëê The entropy change of R-134a in the evaporator is

13-263

An entropy balance on the evaporator gives

Sgen, evaporator = D SR,41 + D Svapor + D Sa + Sw = 207.3 + (- 165.2) + (- 38.34) + 8.14 = 11.90 kJ/K Then, the exergy destruction in the evaporator is

X dest = T0 Sgen, evaporator = (305 K)(11.90 kJ/K) = 3630 kJ Finally the total exergy destruction is X dest, total = X dest, compressor + X dest, condenser + X dest, throttle + X dest, evaporator

= 0 + 1609 + 1519 + 3630 = 6758 kJ

The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from humid air and rejected to the ambient air at 32C (305 K), which is also taken as the dead state temperature.

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T1=32 [C] phi_1=0.95 T2=24 [C] phi_2=0.60 T_evap=4 [C] T_cond=39.4 [C] Vol1=1000 [m^3] T0=(T1+273) "assumed" "Analysis" Fluid$='AirH2O' h1=enthalpy(Fluid$, T=T1, P=P, R=phi_1) w1=humrat(Fluid$, T=T1, P=P, R=phi_1) v1=volume(Fluid$, T=T1, P=P, R=phi_1) h2=enthalpy(Fluid$, T=T2, P=P, R=phi_2) w2=humrat(Fluid$, T=T2, P=P, R=phi_2) T_w=0.5*(T1+T2) "assumed" h_w=enthalpy(steam_iapws,T=T_w, x=0) m_a=Vol1/v1 m_w=m_a*(w1-w2) Q_out=m_a*(h1-h2)-m_w*h_w "Refrigeration cycle analysis" Ref$='R134a' h_R1=enthalpy(Ref$, T=T_evap, x=1) s_R1=entropy(Ref$, T=T_evap, x=1) P_R2=pressure(Ref$, T=T_cond, x=1) s_R2=s_R1 h_R2=enthalpy(Ref$, P=P_R2, s=s_R2) P_R3=P_R2 h_R3=enthalpy(Ref$, P=P_R3, x=0) s_R3=entropy(Ref$, P=P_R3, x=0) h_R4=h_R3 s_R4=entropy(Ref$, T=T_evap, h=h_R4) x_R4=quality(Ref$, T=T_evap, h=h_R4) Q_L=Q_out m_R=Q_L/(h_R1-h_R4) Q_H=m_R*(h_R2-h_R3) "Exergy calculations" X_dest_12=m_R*T0*(s_R2-s_R1) X_dest_23=T0*(m_R*(s_R3-s_R2)+Q_H/T_H) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-264

T_H=T0 X_dest_34=m_R*T0*(s_R4-s_R3) s_g1=entropy(steam_iapws, T=T1, x=1) s_g2=entropy(steam_iapws, T=T2, x=1) s_w=entropy(steam_iapws, T=T_w, x=0) DELTAS_vapor=m_a*(w2*s_g2-w1*s_g1) S_w_tot=m_w*s_w P_g1=pressure(steam_iapws, T=T1, x=1) P_v1=phi_1*P_g1 P_a1=P-P_v1 P_g2=pressure(steam_iapws, T=T2, x=1) P_v2=phi_2*P_g2 P_a2=P-P_v2 c_p=1.005 [kJ/kg-C] R_a=0.287[kJ/kg-C] DELTAS_air=m_a*(c_p*ln((T2+273)/(T1+273))-R_a*ln(P_a2/P_a1)) DELTAS_R_41=m_R*(s_R1-s_R4) S_gen_evap=DELTAS_R_41+DELTAS_vapor+DELTAS_air+S_w_tot X_dest_evap=T0*S_gen_evap X_dest_total=X_dest_12+X_dest_23+X_dest_34+X_dest_evap

Evaporative Cooling 15-90C Evaporative cooling is the cooling achieved when water evaporates in dry air. It will not work on humid climates.

15-91C During evaporation from a water body to air, the latent heat of vaporization will be equal to convection heat transfer from the air when conduction from the lower parts of the water body to the surface is negligible, and temperature of the surrounding surfaces is at about the temperature of the water surface so that the radiation heat transfer is negligible.

15-92C In steady operation, the mass transfer process does not have to involve heat transfer. However, a mass transfer process that involves phase change (evaporation, sublimation, condensation, melting etc.) must involve heat transfer. For example, the evaporation of water from a lake into air (mass transfer) requires the transfer of latent heat of water at a specified temperature to the liquid water at the surface (heat transfer).

15-93 Desert travelers often wrap their heads with a water-soaked porous cloth. The temperature of this cloth on a desert with specified temperature and relative humidity is to be determined.

Analysis Since the cloth behaves as the wick on a wet bulb thermometer, the temperature of the cloth will become the wetbulb temperature. According to the pshchrometric chart, this temperature is T2 = Twb1 = 23.3°C

This process can be represented by an evaporative cooling process as shown in the figure. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-265

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T1=45 [C] phi_1=0.15 "Analysis" Fluid$='AirH2O' T_wb1=wetbulb(Fluid$, T=T1, P=P, R=phi_1) T2=T_wb1

15-94E Air is cooled by an evaporative cooler. The exit temperature of the air is to be determined. Analysis The enthalpy of air at the inlet is determined from Pv1 = f 1 Pg1 = f 1 Psat @93°F = (0.30)(0.7676 psia) = 0.230 psia

w1 =

0.622 Pv1 0.622(0.230 psia) = P1 - Pv1 (14.5 - 0.230) psia

= 0.01004 lbm H2O/lbm dry air h1 = c pT1 + w1hg1

= (0.24 Btu/lbm ×°F)(93°F) + (0.01004)(1101.7 Btu/lbm) = 33.38 Btu/lbm dry air

Assuming the liquid water is supplied at a temperature not much different than the exit temperature of the air stream, the evaporative cooling process follows a line of constant wet-bulb temperature, which is almost parallel to the constant enthalpy lines. That is,

h2 @ h1 = 33.38 Btu/lbm dry air Also, w2 =

0.622 Pg 2 0.622 Pv 2 = P2 - Pv 2 14.5 - Pg 2

since air leaves the evaporative cooler saturated. Substituting this into the definition of enthalpy, we obtain h2 = c pT2 + w2 hg 2 @ c pT2 + w2 (1060.9 + 0.435T2 )

33.38 Btu/lbm = (0.24 Btu/lbm ×°F) T2 +

0.622 Pg 2 14.5 - Pg 2

(1060.9 + 0.435 T2 )

By trial and error, the exit temperature is determined to be

T2 = 69.0°F.

13-266

EES (Engineering Equation Solver) SOLUTION "Given" P=14.5 [psia] T_1=93 [F] phi_1=0.30 phi_2=1 "Analysis" Fluid$='steam_iapws' c_p=0.24 [Btu/lbm-F] P_g1=pressure(Fluid$, T=T_1, x=1) h_g1=enthalpy(Fluid$, T=T_1, x=1) P_v1=phi_1*P_g1 w_1=(0.622*P_v1)/(P-P_v1) h_1=c_p*T_1+w_1*h_g1 h_2=h_1 P_g2=pressure(Fluid$, T=T_2, x=1) P_v2=P_g2 h_g2=1060.9+0.435*T_2 w_2=(0.622*P_v2)/(P-P_v2) h_2=c_p*T_2+w_2*h_g2

15-95 Air is cooled steadily by an evaporative cooler. The temperature of discharged air and the lowest temperature to which the air can be cooled are to be determined.

Analysis The schematic of the evaporative cooler and the psychrometric chart of the process are shown in the figure. (a) If we assume the liquid water is supplied at a temperature not much different from the exit temperature of the airstream, the evaporative cooling process follows a line of constant wet-bulb temperature on the psychrometric chart. That is,

Twb @constant The wet-bulb temperature at 95°F and 20 percent relative humidity is determined from the psychrometric chart to be 66.0°F. The intersection point of the Twb = 66.0° F and the  = 80 percent lines is the exit state of the air. The temperature at this point is the exit temperature of the air, and it is determined from the psychrometric chart to be

T2 = 70.4°F (b) In the limiting case, air leaves the evaporative cooler saturated ( = 100 percent), and the exit state of the air in this case is the state where the Twb = 66.0° F line intersects the saturation line. For saturated air, the dry- and the wet-bulb

13-267

temperatures are identical. Therefore, the lowest temperature to which air can be cooled is the wet-bulb temperature, which is

Tmin = T2¢ = 66.0°F Discussion Note that the temperature of air drops by as much as 30°F in this case by evaporative cooling.

15-96 Air is cooled by an evaporative cooler. The exit temperature of the air and the required rate of water supply are to be determined. Analysis (a) From the psychrometric chart (Fig. A-31 or EES) at 40C and 20% relative humidity we read

Twb1 = 22.04°C w1 = 0.009199 kg H2 O/kg dry air v1 = 0.9002 m3 /kg dry air

Twb2 @Twb1 = 22.04°C At this wet-bulb temperature and 90% relative humidity we read

T2 = 23.3°C w2 = 0.01619 kg H2 O/kg dry air Thus air will be cooled to 23.3C in this evaporative cooler. (b) The mass flow rate of dry air is

ma =

V1 7 m3 / min = = 7.776 kg/min v1 0.9002 m3 / kg dry air

Then the required rate of water supply to the evaporative cooler is determined from msupply = mw2 - mw1

= ma (w2 - w1 )

= (7.776 kg/min)(0.01619 - 0.009199) = 0.0543 kg / min

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=40 [C] phi_1=0.20 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-268

V_dot_1=7 [m^3/min] phi_2=0.90 "Analysis" Fluid$='AirH2O' T_wb1=wetbulb(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) T_wb2=T_wb1 T_2=temperature(Fluid$, R=phi_2, P=P, B=T_wb2) w_2=humrat(Fluid$, T=T_2, P=P, B=T_wb2) m_dot_a=V_dot_1/v_1 m_dot_supply=m_dot_a*(w_2-w_1)

15-97 Air is first heated in a heating section and then passed through an evaporative cooler. The exit relative humidity and the amount of water added are to be determined.

Analysis (a) From the psychrometric chart (Fig. A-31 or EES) at 20C and 70% relative humidity we read

w1 = 0.01021 kg H2 O/kg dry air The specific humidity  remains constant during the heating process. Therefore,

w2 = w1 = 0.01021 kg H2 O/kg dry air At this  value and 35C we read Twb2 = 21.3° C

At this Twb value and 25C we read

f 3 = 0.7225 = 72.3% w3 = 0.01438 kg H2 O/kg dry air (b) The amount of water added to the air per unit mass of air is

D w23 = w3 - w2 = 0.01438 - 0.01021 = 0.00417 kg H2O / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=20 [C] phi_1=0.70 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-269

T_2=35 [C] T_3=25 [C] "Analysis" Fluid$='AirH2O' w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) w_2=w_1 T_wb2=wetbulb(Fluid$, T=T_2, P=P, w=w_2) T_wb3=T_wb2 phi_3=relhum(Fluid$, T=T_3, P=P, B=T_wb3) w_3=humrat(Fluid$, T=T_3, P=P, B=T_wb3) DELTAw_23=w_3-w_2

Adiabatic Mixing of Airstreams 15-98C This will occur when the straight line connecting the states of the two streams on the psychrometric chart crosses the saturation line.

15-99C Yes.

15-100 Two airstreams are mixed steadily. The mass flow ratio of the two streams for a specified mixture relative humidity and the temperature of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic.

Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or from EES) to be

h1 = 29.4 kJ/kg dry air w1 = 0.0077 kg H2 O/kg dry air

and

h2 = 94.6 kJ/kg dry air

13-270

w2 = 0.0244 kg H2O/kg dry air Analysis An application of Eq. 15-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams gives ma1 w2 - w3 h2 - h3 = = ma 2 w3 - w1 h3 - h1 ma1 0.0244 - w3 94.6 - h3 = = ma 2 w3 - 0.0077 h3 - 29.4

This equation cannot be solved directly. An iterative solution is needed. A mixture temperature T3 is selected. At this temperature and given relative humidity (70%), specific humidity and enthalpy are read from the psychrometric chart. These values are substituted into the above equation. If the equation is not satisfied, a new value of T3 is selected. This procedure is repeated until the equation is satisfied. Alternatively, EES software can be used. We used the following EES program to get the results: "Given" P=101.325 [kPa] T_1=10 [C] phi_1=1.0 T_2=32 [C] phi_2=0.80 phi_3=0.70 "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) (w_2-w_3)/(w_3-w_1)=(h_2-h_3)/(h_3-h_1) Ratio=(w_2-w_3)/(w_3-w_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, R=phi_3) h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3) The solution of this EES program is

T3 = 24.0°C,

w3 = 0.0149 kg H2O/kg dry air

h3 = 57.6 kJ/kg dry air,

ma1 = 1.31 ma 2

15-101 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic.

13-271

Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or EES) to be

h1 = 62.28 kJ/kg dry air w1 = 0.01054 kg H2 O/kg dry air v1 = 0.8877 m3 /kg dry air

and

h2 = 31.88 kJ/kg dry air

w2 = 0.007847 kg H2O/kg dry air v 2 = 0.8180 m3 /kg dry air

Analysis The mass flow rate of dry air in each stream is

ma1 =

V1 15 m3 /min = = 16.90 kg/min v1 0.8877 m3 /kg dry air

ma 2 =

V2 25 m3 /min = = 30.56 kg/min v 2 0.8180 m3 /kg dry air

From the conservation of mass,

ma3 = ma1 + ma 2 = 16.90 + 30.56 = 47.46 kg/min The specific humidity and the enthalpy of the mixture can be determined from Eqs. 15-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: ma1 w2 - w3 h2 - h3 = = ma 2 w3 - w1 h3 - h1 31.88 - h3 16.90 0.007847 - w3 = = 30.56 w3 - 0.01054 h3 - 62.28

which yields

h3 = 42.70 kJ/kg dry air w3 = 0.0088 kg H2O / kg dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart:

T3 = 20.2°C

f 3 = 59.7% v 3 = 0.8428 m3 /kg dry air

Finally, the volume flow rate of the mixture is determined from

13-272

V3 = ma 3v 3 = (47.46 kg/min)(0.8428 m 3 /kg dry air) = 40.0 m3 / min

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=35 [C] phi_1=0.30 V_dot_1=15 [m^3/min] T_2=12 [C] phi_2=0.90 V_dot_2=25 [m^3/min] "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) v_2=volume(Fluid$, T=T_2, P=P, R=phi_2) m_dot_a1=V_dot_1/v_1 m_dot_a2=V_dot_2/v_2 m_dot_a3=m_dot_a1+m_dot_a2 m_dot_a1/m_dot_a2=(w_2-w_3)/(w_3-w_1) m_dot_a1/m_dot_a2=(h_2-h_3)/(h_3-h_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, w=w_3) phi_3=relhum(Fluid$, h=h_3, P=P, w=w_3) v_3=volume(Fluid$, T=T_3, P=P, w=w_3) V_dot_3=m_dot_a3*v_3

15-102 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb temperature, and the volume flow rate of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic.

13-273

Analysis The properties of each inlet stream are determined to be Pv1 = f 1 Pg1 = f 1 Psat @35°C = (0.30)(5.629 kPa) = 1.689 kPa

Pa1 = P1 - Pv1 = 90 - 1.689 = 88.31 kPa v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(308 K) = Pa1 88.31 kPa

= 1.001 m3 /kg dry air

w1 =

0.622 Pv1 0.622(1.689 kPa) = P1 - Pv1 (90 - 1.689) kPa

= 0.01189 kg H2 O/kg dry air h1 = c pT1 + w1hg1

= (1.005 kJ/kg ×°C)(35°C) + (0.01189)(2564.6 kJ/kg)

= 65.68 kJ/kg dry air and Pv 2 = f 2 Pg 2 = f 2 Psat@12°C = (0.90)(1.403 kPa) = 1.262 kPa

Pa 2 = P2 - Pv 2 = 90 - 1.262 = 88.74 kPa v2 =

RaT2 (0.287 kPa ×m3 /kg ×K)(285 K) = = 0.9218 m3 /kg dry air Pa 2 88.74 kPa

w2 =

0.622 Pv 2 0.622(1.262 kPa) = = 0.008849 kg H 2 O/kg dry air P2 - Pv 2 (90 - 1.262) kPa

h2 = c pT2 + w2 hg 2 = (1.005 kJ/kg ×°C)(12°C) + (0.008849)(2522.9 kJ/kg) = 34.39 kJ/kg dry air

Then the mass flow rate of dry air in each stream is

ma1 =

V1 150 m3 /min = = 14.99 kg/min v1 1.001 m3 / kg dry air

ma 2 =

V2 25 m3 /min = = 27.12 kg/min v2 0.9218 m3 /kg dry air

From the conservation of mass,

ma3 = ma1 + ma 2 = 14.99 + 27.12 = 42.11 kg/min The specific humidity and the enthalpy of the mixture can be determined from Eqs. 15-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: ma1 w2 - w3 h2 - h3 14.99 0.008849 - w3 34.39 - h3 = = ¾¾ ® = = ma 2 w3 - w1 h3 - h1 27.12 w3 - 0.01189 h3 - 65.68

13-274

which yields

h3 = 45.52 kJ/kg dry air

w3 = 0.009933 kg H2 O / kg dry air These two properties fix the state of the mixture. Other properties are determined from h3 = c pT3 + w3 hg 3 @ c pT3 + w3 (2501.3 + 1.82T3 ) 45.52 kJ/kg = (1.005 kJ/kg ×°C)T3 + (0.009933)(2500.9 + 1.82 T3 ) kJ/kg ¾ ¾ ® T3 = 20.2°C

w3 = 0.009933 = f3=

0.622 Pv 3 P3 - Pv 3 0.622 Pv 3 ¾¾ ® Pv 3 = 1.415 kPa 90 - Pv 3 Pv 3 Pv 3 1.415 kPa = = = 0.597 or 59.7% Pg 3 Psat @ T3 2.370 kPa

Finally,

Pa3 = P3 - Pv3 = 90 - 1.415 = 88.59 kPa v3 =

RaT3 (0.287 kPa ×m3 /kg ×K)(293.2 K) = = 0.9500 m3 /kg dry air Pa 3 88.59 kPa

V3 = ma 3v 3 = (42.11 kg/min)(0.9500 m3 /kg) = 40.0 m 3 / min

EES (Engineering Equation Solver) SOLUTION "Given" P=90 [kPa] T1=35 [C] phi_1=0.30 V_dot_1=15 [m^3/min] T2=12 [C] phi_2=0.90 V_dot_2=25 [m^3/min] "Properties" R_a=0.287 [kJ/kg-C] "for air, Table A-2a" c_p=1.005 [kJ/kg-C] "for air, Table A-2a" "Analysis" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T1, x=1) P_v1=phi_1*P_g1 P_a1=P-P_v1 v1=(R_a*(T1+273))/P_a1 w1=(0.622*P_v1)/(P-P_v1) h_g1=enthalpy(Fluid$, T=T1, x=1) h1=c_p*T1+w1*h_g1 P_g2=pressure(Fluid$, T=T2, x=1) P_v2=phi_2*P_g2 P_a2=P-P_v2 v2=(R_a*(T2+273))/P_a2 w2=(0.622*P_v2)/(P-P_v2) h_g2=enthalpy(Fluid$, T=T2, x=1) h2=c_p*T2+w2*h_g2 m_dot_a1=V_dot_1/v1 m_dot_a2=V_dot_2/v2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-275

m_dot_a3=m_dot_a1+m_dot_a2 m_dot_a1/m_dot_a2=(w2-w3)/(w3-w1) m_dot_a1/m_dot_a2=(h2-h3)/(h3-h1) "mixture properties" h3=c_p*T3+w3*h_g3 h_g3=2500.9+1.82*T3 w3=(0.622*P_v3)/(P-P_v3) P_g3=pressure(Fluid$, T=T3, x=1) phi_3=P_v3/P_g3 P_a3=P-P_v3 v3=(R_a*(T3+273))/P_a3 V_dot_3=m_dot_a3*v3

15-103 A stream of warm air is mixed with a stream of saturated cool air. The temperature, the specific humidity, and the relative humidity of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic.

Properties The properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or EES) to be

h1 = 99.4 kJ/kg dry air w1 = 0.0246 kg H2 O/kg dry air and

h2 = 34.1 kJ/kg dry air w2 = 0.00873 kg H2 O/kg dry air Analysis (a) The specific humidity and the enthalpy of the mixture can be determined from Eqs. 15-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: ma1 w2 - w3 h2 - h3 = = ma 2 w3 - w1 h3 - h1 0.00873 - w3 34.1 - h3 8 = = 10 w3 - 0.0246 h3 - 99.4

which yields,

13-276

(b) w3 = 0.0158 kg H2O / kg dry air

h3 = 63.1 kJ/kg dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart: (a) T3 = 22.8°C (b) f 3 = 90.1%

EES (Engineering Equation Solver) SOLUTION "Given" P=1[atm]*Convert(atm, kPa) T_1=36 [C] T_wb1=30 [C] m_dot_a1=8 [kg/s] T_2=12 [C] phi_2=1 m_dot_a2=10 [kg/s] "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, B=T_wb1, P=P) w_1=humrat(Fluid$, T=T_1, B=T_wb1, P=P) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) m_dot_a1/m_dot_a2=(w_2-w_3)/(w_3-w_1) m_dot_a1/m_dot_a2=(h_2-h_3)/(h_3-h_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, w=w_3) phi_3=relhum(Fluid$, h=h_3, P=P, w=w_3)

15-104 Problem 15-103 is reconsidered. The effect of the mass flow rate of saturated cool air stream on the mixture temperature, specific humidity, and relative humidity is to be investigated. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =36 [C] Twb[1] =30 [C] m_dot[1] = 8 [kg/s] Tdb[2] =12 [C] Rh[2] = 1.0 m_dot[2] = 10 [kg/s] P[1]=P P[2]=P[1] P[3]=P[1] E_dot_in - E_dot_out = DELTAE_dot_sys "Energy balance for the steady-flow mixing process" DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-277

m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of dry air during mixing" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] "Conservation of mass of water vapor during mixing" m_dot[1]=V_dot[1]/v[1]*convert(1/min,1/s) m_dot[2]=V_dot[2]/v[2]*convert(1/min,1/s) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) Rh[1]=RELHUM(AirH2O,T=Tdb[1],P=P[1],B=Twb[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],w=w[3]) Rh[3]=RELHUM(AirH2O,T=Tdb[3],P=P[3],w=w[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3]) Twb[2]=WETBULB(AirH2O,T=Tdb[2],P=P[2],R=RH[2]) Twb[3]=WETBULB(AirH2O,T=Tdb[3],P=P[3],R=RH[3]) m_dot[3]=V_dot[3]/v[3]*convert(1/min,1/s)

Tdb3

[kga/s] 0 2 4 6 8 10 12 14 16

[C] 36 31.31 28.15 25.88 24.17 22.84 21.77 20.89 20.15

Rh 3

0.6484 0.7376 0.7997 0.8442 0.8768 0.9013 0.92 0.9346 0.9461

[kgw/kga] 0.02461 0.02143 0.01931 0.0178 0.01667 0.01579 0.01508 0.0145 0.01402

13-278

15-105E Two airstreams are mixed steadily. The temperature and the relative humidity of the mixture are to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31E or from EES) to be

h1 = 66.7 Btu/lbm dry air w1 = 0.0386 lbm H2 O/lbm dry air

v1 = 14.98 ft 3 /lbm dry air

13-279

and

h2 = 14.5 Btu/lbm dry air w2 = 0.0023 lbm H2 O/lbm dry air v 2 = 12.90 ft 3 /lbm dry air

Analysis The mass flow rate of dry air in each stream is

ma1 =

V1 3 ft 3 / s = = 0.2002 lbm/s v1 14.98 ft 3 / lbmdry air

ma 2 =

V2 1 ft 3 / s = = 0.07755 lbm/s v 2 12.90 ft 3 / lbm dry air

From the conservation of mass,

ma3 = ma1 + ma 2 = (0.2002 + 0.07755) lbm/s = 0.2778 lbm/s The specific humidity and the enthalpy of the mixture can be determined from Eqs. 15-24, which are obtained by combining the conservation of mass and energy equations for the adiabatic mixing of two streams: ma1 w2 - w3 h2 - h3 = = ma 2 w3 - w1 h3 - h1 0.0023 - w3 14.5 - h3 0.2002 = = 0.07755 w3 - 0.0386 h3 - 66.7

which yields

w3 = 0.0284 lbm H2 O/lbm dry air h3 = 52.1 Btu/lbm dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart:

T3 = 86.7°F f 3 = 1.0 = 100%

EES (Engineering Equation Solver) SOLUTION "Given" P=14.696 [psia] T_1=100 [F] phi_1=0.90 V_dot_1=3 [ft^3/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-280

T_2=50 [F] phi_2=0.30 V_dot_2=1 [ft^3/s] "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) v_2=volume(Fluid$, T=T_2, P=P, R=phi_2) m_dot_a1=V_dot_1/v_1 m_dot_a2=V_dot_2/v_2 m_dot_a3=m_dot_a1+m_dot_a2 m_dot_a1/m_dot_a2=(w_2-w_3)/(w_3-w_1) m_dot_a1/m_dot_a2=(h_2-h_3)/(h_3-h_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, w=w_3) phi_3=relhum(Fluid$, h=h_3, P=P, w=w_3) v_3=volume(Fluid$, T=T_3, P=P, w=w_3) V_dot_3=m_dot_a3*v_3

15-106E Two airstreams are mixed steadily. The rate of entropy generation is to be determined. Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing section is adiabatic. Properties Properties of each inlet stream are determined from the psychrometric chart (Fig. A-31 or from EES) to be

h1 = 66.7 Btu/lbm dry air w1 = 0.0386 lbm H2O/lbm dry air v1 = 14.98 ft 3 /lbm dry air

and

h2 = 14.5 Btu/lbm dry air w2 = 0.0023 lbm H2 O/lbm dry air v 2 = 12.90 ft 3 /lbm dry air

13-281

The entropies of water vapor in the air streams are sg1 = sg @100° F = 1.9819 Btu/lbm ×R sg 2 = sg @50° F = 2.1256 Btu/lbm ×R

Analysis The mass flow rate of dry air in each stream is

ma1 =

V1 3 ft 3 / s = = 0.2002 lbm/s v1 14.98 ft 3 / lbmdry air

ma 2 =

V2 1 ft 3 / s = = 0.07755 lbm/s v 2 12.90 ft 3 / lbm dry air

From the conservation of mass,

which yields

w3 = 0.0284 lbm H2 O/lbm dry air h3 = 52.1 Btu/lbm dry air These two properties fix the state of the mixture. Other properties of the mixture are determined from the psychrometric chart:

T3 = 86.7°F f 3 = 1.0 = 100% The entropy of water vapor in the mixture is

sg 3 = sg @86.7°F = 2.0168 Btu/lbm ×R An entropy balance on the mixing chamber for the water gives D S w = ma 3 w3 s3 - ma1w1 s1 - ma 2 w2 s2

= 0.2778´ 0.0284´ 2.0168 - 0.2002´ 0.0386´ 1.9819 - 0.07755´ 0.0023´ 2.1256

= 2.169´ 10- 4 Btu/s ×R The partial pressures of water vapor and dry air for all three air streams are Pv1 = f 1 Pg1 = f 1 Psat @100°F = (0.90)(0.95052 psia) = 0.8555 psia

13-282

Pv 2 = f 2 Pg 2 = f 2 Psat @50°F = (0.30)(0.17812 psia) = 0.0534 psia

Pa 2 = P2 - Pv 2 = 14.696 - 0.0534 = 14.64 psia Pv 3 = f 3 Pg 3 = f 3 Psat @86.7°C = (1.0)(0.6298 psia) = 0.6298 psia

Pa3 = P3 - Pv3 = 14.696 - 0.6298 = 14.07 psia An entropy balance on the mixing chamber for the dry air gives D Sa = ma1 ( s3 - s1 ) + ma 2 ( s3 - s2 )

æ æ T P ö T P ö ÷ ÷ = ma1 çççc p ln 3 - R ln a 3 ÷ + ma 2 çççc p ln 3 - R ln a 3 ÷ ÷ ÷ çè T1 Pa1 ø÷ T2 Pa 2 ø÷ èç é é 546.7 14.07 ù 546.7 14.07 ù ú+ 0.07755 ê(0.240) ln ú = 0.2002 ê(0.240) ln - (0.06855) ln - (0.06855) ln ú ú 560 13.84 û 510 14.64 û ëê ëê = (0.2002)(- 0.006899) + (0.07755)(0.01940) = 1.233´ 10- 4 Btu/s ×R

The rate of entropy generation is then

Sgen = D Sa + D Sw = 1.233´ 10- 4 + 2.169´ 10- 4 = 3.402×10-4 Btu / s ×R

EES (Engineering Equation Solver) SOLUTION "Given" P=14.696 [psia] T_1=100 [F] phi_1=0.90 V_dot_1=3 [ft^3/s] T_2=50 [F] phi_2=0.30 V_dot_2=1 [ft^3/s] "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) v_2=volume(Fluid$, T=T_2, P=P, R=phi_2) m_dot_a1=V_dot_1/v_1 m_dot_a2=V_dot_2/v_2 m_dot_a3=m_dot_a1+m_dot_a2 m_dot_a1/m_dot_a2=(w_2-w_3)/(w_3-w_1) m_dot_a1/m_dot_a2=(h_2-h_3)/(h_3-h_1) "mixture properties" T_3=temperature(Fluid$, h=h_3, P=P, w=w_3) phi_3=relhum(Fluid$, h=h_3, P=P, w=w_3) v_3=volume(Fluid$, T=T_3, P=P, w=w_3) V_dot_3=m_dot_a3*v_3 s_g3=entropy(steam_iapws, T=86.7, x=1) P_g3=pressure(steam_iapws, T=86.7, x=1)

13-283

Wet Cooling Towers 15-107C The working principle of a natural draft cooling tower is based on buoyancy. The air in the tower has a high moisture content, and thus is lighter than the outside air. This light moist air rises under the influence of buoyancy, inducing flow through the tower.

15-108C A spray pond cools the warm water by spraying it into the open atmosphere. They require 25 to 50 times the area of a wet cooling tower for the same cooling load.

15-109 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: å ma , i = å ma , e

¾¾ ®

ma1 = ma 2 = ma

¾¾ ®

m3 + ma1w1 = m4 + ma 2 w2

Water Mass Balance: å mw, i = å mw, e

m3 - m4 = ma (w2 - w1 ) = mmakeup

Energy Balance: Ein - Eout = D EsystemZ 0 (steady) = 0

Ein = Eout å mi hi = å me he

since Q = W = 0

0 = å me he - å mi hi 0 = ma 2 h2 + m4 h4 - ma1h1 - m3 h3 0 = ma (h2 - h1 ) + (m3 - mmakeup ) h4 - m3 h3

13-284

Solving for ma , ma =

m3 (h3 - h4 ) (h2 - h1 ) - (w2 - w1 ) h4

From the psychrometric chart (Fig. A-31),

h1 = 49.9 kJ/kg dry air w1 = 0.0105 kg H2 O/kg dry air

v1 = 0.853 m3 /kg dry air

and

h2 = 110.7 kJ/kg dry air

w2 = 0.0307 kg H2O/kg dry air From Table A-4, h3 @ h f @ 40°C = 167.53 kJ/kg H 2 O h4 @ h f @ 25°C = 104.83 kJ/kg H 2 O

Substituting, ma =

(90 kg/s)(167.53 - 104.83)kJ/kg = 96.2 kg/s (110.7 - 49.9) kJ/kg - (0.0307 - 0.0105)(104.83) kJ/kg

Then the volume flow rate of air into the cooling tower becomes V1 = mav1 = (96.2 kg/s)(0.854 m3 /kg) = 82.2 m 3 / s

(b) The mass flow rate of the required makeup water is determined from

mmakeup = ma (w2 - w1 ) = (96.2 kg/s)(0.0307 - 0.0105) = 1.94 kg / s

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_water_in=40 [C] m_dot_water=90 [kg/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-285

T_water_out=25 [C] T_1=23 [C] phi_1=0.60 T_2=32 [C] phi_2=1 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_3=enthalpy(Fluid2$, T=T_water_in, x=0) h_4=enthalpy(Fluid2$, T=T_water_out, x=0) 0=m_dot_a*(h_2-h_1)+(m_dot_water-m_dot_makeup)*h_4-m_dot_water*h_3 "energy balance" V_dot_1=m_dot_a*v_1 m_dot_makeup=m_dot_a*(w_2-w_1) "mass balance"

15-110 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: å ma , i = å ma , e ¾ ¾ ® ma1 = ma 2 = ma

Water Mass Balance: å mw, i = å mw, e ® m3 + ma1w1 = m4 + ma 2 w2

m3 - m4 = ma (w2 - w1 ) = mmakeup

Energy Balance: Ein - Eout = D EsystemZ 0 (steady) = 0

Ein = Eout å mi hi = å me he

(since Q = W = 0)

0 = å me he - å mi hi 0 = ma 2 h2 + m4 h4 - ma1h1 - m3 h3 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3 h3

13-286

Solving for ma , ma =

m3 (h3 - h4 ) (h2 - h1 ) - (w2 - w1 ) h4

From the psychrometric chart (Fig. A-31 or EES),

h1 = 44.70 kJ/kg dry air w1 = 0.008875 kg H2 O/kg dry air v1 = 0.8480 m3 /kg dry air

and h2 = 96.13 kJ/kg dry air

w2 = 0.02579 kg H2 O/kg dry air From Table A-4, h3 @ h f @ 40°C = 167.53 kJ/kg H 2 O

h4 @ h f @33°C = 138.28 kJ/kg H 2 O s

Substituting, ma =

(60 kg/s)(167.53 - 138.28)kJ/kg = 35.76 kg/s (96.13 - 44.70) kJ/kg - (0.02579 - 0.008875)(138.28) kJ/kg

Then the volume flow rate of air into the cooling tower becomes V1 = mav1 = (35.76 kg/s)(0.8480 m3 /kg) = 30.3 m 3 / s

(b) The mass flow rate of the required makeup water is determined from

mmakeup = ma (w2 - w1 ) = (35.76 kg/s)(0.02579 - 0.008875) = 0.605 kg / s

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_water=60 [kg/s] T_water_in=40 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-287

T_water_out=33 [C] P=101.325 [kPa] T_1=22 [C] T_wb1=16 [C] T_2=30 [C] phi_2=0.95 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, B=T_wb1, P=P) w_1=humrat(Fluid1$, T=T_1, B=T_wb1, P=P) v_1=volume(Fluid1$, T=T_1, B=T_wb1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_3=enthalpy(Fluid2$, T=T_water_in, x=0) h_4=enthalpy(Fluid2$, T=T_water_out, x=0) 0=m_dot_a*(h_2-h_1)+(m_dot_water-m_dot_makeup)*h_4-m_dot_water*h_3 "energy balance" V_dot_1=m_dot_a*v_1 m_dot_makeup=m_dot_a*(w_2-w_1) "mass balance"

15-111E Water is cooled by air in a cooling tower. The relative humidity of the air at the exit and the water’s exit temperature are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: å ma ,i = å ma , e

¾¾ ®

ma1 = ma 2 = ma

Water Mass Balance: å mw,i = å mw,e

m3 + ma1w1 = m4 + ma 2 w2

m3 - m4 = ma (w2 - w1 ) = mmakeup

Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = å me he

(since Q =W =0)

0 = å me he - å mi hi 0 = ma 2 h2 + m4 h4 - ma1h1 - m3 h3 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3 h3

13-288

Solving for h4 ,

h4 =

m3 h3 - ma (h2 - h1 ) m3 - mmakeup

From the psychrometric chart (Fig. A-31E),

h1 = 16.8 Btu/lbm dry air w1 = 0.00219 kg H2 O/kg dry air v1 = 13.15 ft 3 /lbm dry air

and

h2 = 37.7 Btu/lbm dry air f 2 = 0.957 = 95.7%

From Table A-4,

h3 @ h f @100° F = 68.03 Btu/lbm H 2 O Also,

mmakeup = ma (w2 - w1 ) = (7000 / 3600 lbm/s)(0.018 - 0.00219) = 0.03075 lbm/s Substituting,

h4 =

m3 h3 - ma (h2 - h1 ) (10, 000 / 3600)(68.03) - (7000 / 3600)(37.7 - 16.8) = = 53.99 Btu/lbm m3 - mmakeup (10, 000 / 3600) - 0.03075

The exit temperature of the water is then (Table A-4E) T4 = Tsat @ h f = 53.99 Btu/lbm = 85.9°F

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_water=10000/3600 "[lbm/s]" T_water_in=100 "[F]" P=14.696 "[psia]" T_1=60 "[F]" phi_1=0.20 T_2=75 "[F]" w_2=0.018 m_dot_a=7000/3600 [lbm/s] "Analysis" Fluid1$='AirH2O' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-289

Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, w=w_2, P=P) phi_2=relhum(Fluid1$, T=T_2, w=w_2, P=P) h_3=enthalpy(Fluid2$, T=T_water_in, x=0) m_dot_makeup=m_dot_a*(w_2-w_1) "mass balance" 0=m_dot_a*(h_2-h_1)+(m_dot_water-m_dot_makeup)*h_4-m_dot_water*h_3 "energy balance" T_4=temperature(Fluid2$, h=h_4, x=0)

15-112 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the required makeup water are to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis (a) The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: å ma , i = å ma , e ¾ ¾ ®

ma1 = ma 2 = ma

Water Mass Balance: å mw, i = å mw, e

m3 + ma1w1 = m4 + ma 2 w2 m3 - m4 = ma (w2 - w1 ) = mmakeup

Energy Balance:

13-290

Ein - Eout = D EsystemZ 0 (steady) = 0 ¾ ¾ ® Ein = Eout

å mi hi = å me he

(since Q = W = 0)

0 = å me he - å mi hi

0 = ma 2 h2 + m4 h4 - ma1h1 - m3 h3 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3 h3

ma =

m3 (h3 - h4 ) (h2 - h1 ) - (w2 - w1 ) h4

The properties of air at the inlet and the exit are Pv1 = f 1 Pg1 = f 1 Psat @ 20°C = (0.70)(2.3392 kPa) = 1.637 kPa

Pa1 = P1 - Pv1 = 96 - 1.637 = 94.363 kPa v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(293 K) = = 0.8911 m3 /kg dry air Pa1 94.363 kPa

w1 =

0.622 Pv1 0.622(1.637 kPa) = = 0.01079 kg H 2 O/kg dry air P1 - Pv1 (96 - 1.637) kPa

h1 = c pT1 + w1hg1 = (1.005 kJ/kg ×°C)(20°C) + (0.01079)(2537.4 kJ/kg) = 47.5 kJ/kg dry air

and Pv 2 = f 2 Pg 2 = f 2 Psat @35°C = (1.00)(5.6291 kPa) = 5.6291 kPa w2 =

0.622 Pv 2 0.622(5.6291 kPa) = = 0.03874 kg H 2 O/kg dry air P2 - Pv 2 (96 - 5.6291) kPa

h2 = c pT2 + w2 hg 2 = (1.005 kJ/kg ×°C)(35°C) + (0.03874)(2564.6 kJ/kg) = 134.5 kJ/kg dry air

From Table A-4, h3 @ h f @ 40°C = 167.53 kJ/kg H 2 O

h4 @ h f @30°C = 125.74 kJ/kg H 2 O

Substituting, ma =

(17 kg/s)(167.53 - 125.74)kJ/kg = 8.507 kg/s (134.5 - 47.5) kJ/kg - (0.03874 - 0.01079)(125.74) kJ/kg

Then the volume flow rate of air into the cooling tower becomes V1 = mav1 = (8.507 kg/s)(0.8911 m3 /kg) = 7.58 m 3 / s

(b) The mass flow rate of the required makeup water is determined from

mmakeup = ma (w2 - w1 ) = (8.507 kg/s)(0.03874 - 0.01079) = 0.238 kg / s

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_water=17 [kg/s] T_water_in=40 [C] T_water_out=30 [C] P=96 [kPa] T1=20 [C] phi_1=0.70 T2=35 [C] phi_2=1 "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-291

R_a=0.287 [kJ/kg-C] "for air, Table A-2a" c_p=1.005 [kJ/kg-C] "for air, Table A-2a" "Analysis" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T1, x=1) P_v1=phi_1*P_g1 P_a1=P-P_v1 v1=(R_a*(T1+273))/P_a1 w1=(0.622*P_v1)/(P-P_v1) h_g1=enthalpy(Fluid$, T=T1, x=1) h1=c_p*T1+w1*h_g1 P_g2=pressure(Fluid$, T=T2, x=1) P_v2=phi_2*P_g2 P_a2=P-P_v2 w2=(0.622*P_v2)/(P-P_v2) h_g2=enthalpy(Fluid$, T=T2, x=1) h2=c_p*T2+w2*h_g2 h3=enthalpy(Fluid$, T=T_water_in, x=0) h4=enthalpy(Fluid$, T=T_water_out, x=0) 0=m_dot_a*(h2-h1)+(m_dot_water-m_dot_makeup)*h4-m_dot_water*h3 "energy balance" V_dot_1=m_dot_a*v1 m_dot_makeup=m_dot_a*(w2-w1) "mass balance"

15-113 Water is cooled by air in a cooling tower. The mass flow rate of dry air is to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic. Analysis The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: å ma ,i = å ma , e

¾¾ ®

ma1 = ma 2 = ma

Water Mass Balance: å mw,i = å mw,e

m3 + ma1w1 = m4 + ma 2 w2

m3 - m4 = ma (w2 - w1 ) = mmakeup

Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = å me he

(since Q =W =0)

0 = å me he - å mi hi 0 = ma 2 h2 + m4 h4 - ma1h1 - m3 h3 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3 h3

13-292

Solving for ma , ma =

m3 (h3 - h4 ) (h2 - h1 ) - (w2 - w1 ) h4

From the psychrometric chart (Fig. A-31),

h1 = 21.8 kJ/kg dry air w1 = 0.00264 kg H2 O/kg dry air

v1 = 0.820 m3 /kg dry air

and h2 = 49.3 kJ/kg dry air

w2 = 0.0123 kg H2O/kg dry air From Table A-4, h3 @ h f @30°C = 125.74 kJ/kg H 2 O

h4 @ h f @ 22°C = 92.28 kJ/kg H 2 O

Substituting, ma =

(5 kg/s)(125.74 - 92.28)kJ/kg = 6.29 kg / s (49.3 - 21.8) kJ/kg - (0.0123 - 0.00264)(92.28) kJ/kg

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_water=5 "[kg/s]" T_water_in=30 "[C]" T_water_out=22 "[C]" P=101.325 "[kPa]" T_1=15 "[C]" phi_1=0.25 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-293

T_2=18 "[C]" phi_2=0.95 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_3=enthalpy(Fluid2$, T=T_water_in, x=0) h_4=enthalpy(Fluid2$, T=T_water_out, x=0) 0=m_dot_a*(h_2-h_1)+(m_dot_water-m_dot_makeup)*h_4-m_dot_water*h_3 "energy balance" V_dot_1=m_dot_a*v_1 m_dot_makeup=m_dot_a*(w_2-w_1) "mass balance"

15-114 Water is cooled by air in a cooling tower. The exergy lost in the cooling tower is to be determined. Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic.

Analysis The mass flow rate of dry air through the tower remains constant (ma1 = ma 2 = ma ) , but the mass flow rate of liquid water decreases by an amount equal to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields Dry Air Mass Balance: å ma ,i = å ma , e

¾¾ ®

ma1 = ma 2 = ma

13-294

Water Mass Balance: å mw,i = å mw,e

m3 + ma1w1 = m4 + ma 2 w2

m3 - m4 = ma (w2 - w1 ) = mmakeup

Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = å me he

(since Q =W =0)

0 = å me he - å mi hi

0 = ma 2 h2 + m4 h4 - ma1h1 - m3 h3 0 = ma (h2 - h1 ) + (m3 - mmakeup )h4 - m3 h3

Solving for ma , ma =

m3 (h3 - h4 ) (h2 - h1 ) - (w2 - w1 ) h4

From the psychrometric chart (Fig. A-31),

h1 = 21.8 kJ/kg dry air w1 = 0.00264 kg H2 O/kg dry air and h2 = 49.3 kJ/kg dry air

w2 = 0.0123 kg H2O/kg dry air From Table A-4, h3 @ h f @30°C = 125.74 kJ/kg H 2 O

h4 @ h f @ 22°C = 92.28 kJ/kg H 2 O

Substituting, ma =

(5 kg/s)(125.74 - 92.28)kJ/kg = 6.29 kg/s (49.3 - 21.8) kJ/kg - (0.0123 - 0.00264)(92.28) kJ/kg

The mass flow rate of water stream at state 3 per unit mass of dry air is m3 =

m3 5 kg water/s = = 0.7949 kg water/kg dry air ma 6.29 kg dry air/s

The mass flow rate of water stream at state 4 per unit mass of dry air is

m4 = m3 - (w2 - w1 ) = 0.7949 - (0.0123 - 0.00264) = 0.7852 kg water/kg dry air The entropies of water streams are s3 = s f @30°C = 0.4368 kJ/kg ×K

s4 = s f @ 22°C = 0.3249 kJ/kg ×K

The entropy change of water stream is

D swater = m4 s4 - m3 s3 = 0.7852´ 0.3249 - 0.7949´ 0.4368 = - 0.09210 kJ/K ×kg dry air The entropies of water vapor in the air stream are sg1 = sg @15°C = 8.7803 kJ/kg ×K sg 2 = sg @18°C = 8.7112 kJ/kg ×K

The entropy change of water vapor in the air stream is

13-295

D svapor = w2 sg 2 - w1 sg1 = 0.0123´ 8.7112 - 0.00264´ 8.7803 = 0.08397 kJ/K ×kg dry air The partial pressures of water vapor and dry air for air streams are Pv1 = f 1 Pg1 = f 1 Psat @15°C = (0.25)(1.7057 kPa) = 0.4264 kPa

Pa1 = P1 - Pv1 = 101.325 - 0.4264 = 100.90 kPa Pv 2 = f 2 Pg 2 = f 2 Psat @18°C = (0.95)(2.065 kPa) = 1.962 kPa

Pa 2 = P2 - Pv 2 = 101.325 - 1.962 = 99.36 kPa The entropy change of dry air is P T D sa = s2 - s1 = c p ln 2 - R ln a 2 T1 Pa1

291 99.36 - (0.287) ln = 0.01483 kJ/kg dry air 288 100.90 The entropy generation in the cooling tower is the total entropy change: = (1.005) ln

sgen = D swater + D svapor + D sa = - 0.09210 + 0.08397 + 0.01483 = 0.00670 kJ/K ×kg dry air Finally, the exergy destruction per unit mass of dry air is

xdest = T0 sgen = (288 K)(0.00670 kJ/K ×kg dry air) = 1.93 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" m_dot_water=5 "[kg/s]" T_water_in=30 "[C]" T_water_out=22 "[C]" P=101.325 "[kPa]" T_1=15 "[C]" phi_1=0.25 T_2=18 "[C]" phi_2=0.95 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_3=enthalpy(Fluid2$, T=T_water_in, x=0) h_4=enthalpy(Fluid2$, T=T_water_out, x=0) 0=m_dot_a*(h_2-h_1)+(m_dot_water-m_dot_makeup)*h_4-m_dot_water*h_3 "energy balance" V_dot_1=m_dot_a*v_1 m_dot_makeup=m_dot_a*(w_2-w_1) "mass balance" s_g2=entropy(Fluid2$, T=18, x=1) s_4=entropy(Fluid2$, T=22, x=0) P_sat=pressure(Fluid2$, T=18, x=1)

13-296

Review Problems 15-115 The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart, the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of the air are to be determined. Analysis The air in a room is at 1 atm, 32°C, and 60 percent relative humidity. From the psychrometric chart (Fig. A-31) we read (a) w = 0.0181 kg H2 O / kg dry air (b) h = 78.4 kJ/kg dry air (c) Twb = 25.5°C (d) Tdp = 23.3°C (e) v = 0.890 m3 / kg dry air

15-116 The mole fraction of the water vapor at the surface of a lake at a specified temperature is to be determined. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air at the lake surface is saturated. Properties The saturation pressure of water at 18C is 2.065 kPa (Table A-4).

Analysis The air at the water surface will be saturated. Therefore, the partial pressure of water vapor in the air at the lake surface will simply be the saturation pressure of water at 18C,

Pvapor = Psat @18°C = 2.065 kPa Assuming both the air and vapor to be ideal gases, the partial pressure and mole fraction of dry air in the air at the surface of the lake are determined to be

Pdry air = P - Pvapor = 100 - 2.065 = 97.94 kPa ydry air =

Pdry air P

97.94 kPa = 0.979 (or 97.9%) 100 kPa

Therefore, the mole fraction of dry air is 97.9 percent just above the air-water interface.

EES (Engineering Equation Solver) SOLUTION "Given" T=18 "[C]" P=100 "[kPa]" "Analysis" P_vapor=pressure(steam_iapws, T=T, x=1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-297

P_dryair=P-P_vapor y_dryair=P_dryair/P

15-117 Dry air flows over a water body at constant pressure and temperature until it is saturated. The molar analysis of the saturated air and the density of air before and after the process are to be determined. Assumptions The air and the water vapor are ideal gases. Properties The molar masses of N 2 , O2 , Ar, and H 2 O are 28.0, 32.0, 39.9 and 18 kg / kmol , respectively (Table A-1). The molar analysis of dry air is given to be 78.1 percent N 2 , 20.9 percent O2 , and 1 percent Ar. The saturation pressure of water at 25C is 3.1698 kPa (Table A-4). Also, 1 atm = 101.325 kPa. Analysis (a) Noting that the total pressure remains constant at 101.32 kPa during this process, the partial pressure of air becomes

P = Pair + Pvapor ® Pair = P - Pvapor = 101.325 - 3.1698 = 98.155 kPa

Then the molar analysis of the saturated air becomes

PH2 O

yH 2 O =

yN 2 = yO2 =

yAr =

PN2 P PO2 P

= =

3.1698 = 0.0313 101.325

yN2 , dry Pdry air P yO2 , dry Pdry air P

0.781(98.155 kPa) = 0.7566 101.325

0.209(98.155 kPa) = 0.2025 101.325

yAr, dry Pdry air 0.01(98.155 kPa) PAr = = = 0.0097 P P 101.325

(b) The molar masses of dry and saturated air are M dry air = å yi M i = 0.781´ 28.0 + 0.209´ 32.0 + 0.01´ 39.9 = 29.0 kg/kmol M sat. air = å yi M i = 0.7566´ 28.0 + 0.2025´ 32.0 + 0.0097 ´ 39.9 + 0.0313´ 18 = 28.62 kg/kmol

Then the densities of dry and saturated air are determined from the ideal gas relation to be r dry air =

P 101.325 kPa = = 1.186 kg / m 3 é(8.314 kPa ×m/kmol ×K ) / 29.0 kg/kmolù(25 + 273)K ( Ru / M dry air )T ë û

r sat. air =

P 101.325 kPa = = 1.170 kg / m 3 é(8.314 kPa ×m/kmol ×K ) / 28.62 kg/kmolù(25 + 273)K ( Ru / M sat air )T ë û

Discussion We conclude that the density of saturated air is less than that of the dry air, as expected. This is due to the molar mass of water being less than that of dry air.

13-298

EES (Engineering Equation Solver) SOLUTION "Given" y_N2_dry=0.781 y_O2_dry=0.209 y_Ar_dry=0.01 P=101.325 [kPa] T=25 [C] phi=1 "Properties" R_u=8.314 [kJ/kmol-K] M_N2=molarmass(N2) M_O2=molarmass(O2) M_Ar=molarmass(Argon) M_H2O=molarmass(H2O) P_v=pressure(steam_iapws, T=T, x=1) "Analysis" P_dryair=P-P_v P_H2O=P_v P_N2=y_N2_dry*P_dryair P_O2=y_O2_dry*P_dryair P_Ar=y_Ar_dry*P_dryair y_H2O=P_H2O/P y_N2=P_N2/P y_O2=P_O2/P y_Ar=P_Ar/P M_dryair=y_N2_dry*M_N2+y_O2_dry*M_O2+y_Ar_dry*M_Ar M_satair=y_N2*M_N2+y_O2*M_O2+y_Ar*M_Ar+y_H2O*M_H2O rho_dryair=P/((R_u/M_dryair)*(T+273)) rho_satair=P/((R_u/M_satair)*(T+273))

15-118 Air is compressed by a compressor and then cooled to the ambient temperature at high pressure. It is to be determined if there will be any condensation in the compressed air lines. Assumptions The air and the water vapor are ideal gases. Properties The saturation pressure of water at 20C is 2.3392 kPa (Table A-4). Analysis The vapor pressure of air before compression is

Pv1 = f 1 Pg = f 1 Psat @ 25°C = (0.50)(2.3392 kPa) = 1.17 kPa The pressure ratio during the compression process is (800 kPa) / (92 kPa) = 8.70 . That is, the pressure of air and any of its components increases by 8.70 times. Then the vapor pressure of air after compression becomes

Pv 2 = Pv1 ´ (Pressure ratio) = (1.17 kPa)(8.70) = 10.2 kPa The dew-point temperature of the air at this vapor pressure is Tdp = Tsat @ Pv 2 = Tsat @10.2 kPa = 46.1°C

which is greater than 20C. Therefore, part of the moisture in the compressed air will condense when air is cooled to 20C.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=92 [kPa] P_2=800 [kPa] T_1=20 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-299

phi_1=0.50 "Analysis" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_1, x=1) P_v1=phi_1*P_g1 r_P=P_2/P_1 P_v2=P_v1*r_P T_dp=temperature(Fluid$, P=P_v2, x=1) "If T_dp is greater than T_1(=25 C), some moisture will condense"

15-119E A room is cooled adequately by a 7500 Btu/h air-conditioning unit. If the room is to be cooled by an evaporative cooler, the amount of water that needs to be supplied to the cooler is to be determined. Assumptions 1 The evaporative cooler removes heat at the same rate as the air conditioning unit. 2

Water evaporates at an average temperature of 70F.

Properties The enthalpy of vaporization of water at 70F is 1053.7 Btu/h (Table A-4E). Analysis Noting that 1 lbm of water removes 1053.7 Btu of heat as it evaporates, the amount of water that needs to evaporate to remove heat at a rate of 7500 Btu/h is determined from Q = mwater h fg to be mwater =

Q 7500 Btu/h = = 7.12 lbm / h h fg 1053.7 Btu/lbm

EES (Engineering Equation Solver) SOLUTION "Given" Q_dot_cooling=7500 [Btu/h] "Analysis" Fluid$='steam_iapws' T=70 [F] "water evaporates at an assumed average temperature of 70 F" h_f=enthalpy(Fluid$, T=T, x=0) h_g=enthalpy(Fluid$, T=T, x=1) h_fg=h_g-h_f m_dot_water=Q_dot_cooling/h_fg

15-120 Shading the condenser can reduce the air-conditioning costs by up to 10 percent. The amount of money shading can save a homeowner per year during its lifetime is to be determined. Assumptions It is given that the annual air-conditioning cost is $500 a year, and the life of the air-conditioning system is 20 years. Analysis The amount of money that would be saved per year is determined directly from

($500 / year)(20 years)(0.10) = $1000 Therefore, the proposed measure will save about $1000 during the lifetime of the system.

13-300

Cost=500 [$/year] Life=20 [year] "Analysis" Saving=Cost*Life*f

15-121E It is estimated that 190,000 barrels of oil would be saved per day if the thermostat setting in residences in summer were raised by 6F (3.3C). The amount of money that would be saved per year is to be determined. Assumptions The average cooling season is given to be 120 days, and the cost of oil to be $20 / barrel. Analysis The amount of money that would be saved per year is determined directly from

(190,000 barrel/day)(120 days/year)($70/barrel) = $1,596,000,000 Therefore, the proposed measure will save more than one and half billion dollars a year.

EES (Engineering Equation Solver) SOLUTION "Given" OilSaved=190000 [1/day] CoolingSeason=120 [day/year] Cost=70 [$] "Analysis" Moneysaved=OilSaved*CoolingSeason*Cost

15-122 Air is cooled, dehumidified and heated at constant pressure. The system hardware and the psychrometric diagram are to be sketched and the heat transfer is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis (a) The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below. The psychrometric chart is obtained from the Property Plot feature of EES.

13-301

AirH2O 0.050 Pressure = 101.3 [kPa]

0.045 0.040

35°C

Humidity Ratio

0.035 0.8

0.030

30°C

0.025

0.6 25°C

0.020

0.4

20°C

0.015

15°C

0.010

10°C

0.2

0.005

0.000 0

T [°C]

(b) The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states are determined from the psychrometric chart (Figure A-31 or EES) to be

h1 = 71.2 kJ/kg dry air w1 = 0.0160 kg H2 O/kg dry air v1 = 0.881 m3 /kg dry air

and

f 2 = 0.376 h2 = 33.9 kJ/kg dry air w2 = 0.00544 kg H2 O/kg dry air v 2 = 0.838 m3 /kg dry air

Also,

hw @ h f @ 20°C = 83.9 kJ/kg

(Table A-4)

The mass flow rate of air in the room to be replaced is

ma =

V2 (700 / 60 m3 /s) = = 13.93 kg/s v2 0.838 m3 /kg

The mass flow rate of outside air is

mout = mout (1+ w1 ) = (13.93 kg/s)(1+ 0.0160) = 14.15 kg/s = 50,940 kg / h (c) Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) Substituting,

mw = ma (w1 - w2 ) = (13.93 kg/s)(0.0160 - 0.00544) = 0.147 kg/s = 8.83 kg / min (d) Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout

13-302

å mi hi = Qout + å me he

Qout = ma1h1 - (ma 2 h2 + mw hw )

= ma (h1 - h2 ) - mw hw Noting that the cooling water absorbs this heat, the mass flow rate of cooling water is determined from ma (h1 - h2 ) - mw hw = mcw c pwD Tcw

mcw = =

ma (h1 - h2 ) - mw hw c pwD Tcw (13.93 kg/s)(71.2 - 33.9) kJ/kg - (0.147 kg/s)(83.9 kJ/kg) = (4.18 kJ/kg ×°C)(15°C)

= 8.08 kg/s = 485 kg / min

EES (Engineering Equation Solver) SOLUTION "Given" V_dot=(700/60) [m^3/s] P=101.325 [kPa] T_1=30 [C] phi_1=0.60 T_2=20 [C] T_wb2=12 [C] DELTAT_w=15 [C] "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) T_wb1=wetbulb(Fluid1$, T=T_1, R=phi_1, P=P) T_dp1=dewpoint(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, B=T_wb2, P=P) w_2=humrat(Fluid1$, T=T_2, B=T_wb2, P=P) v_2=volume(Fluid1$, T=T_2, B=T_wb2, P=P) phi_2=relhum(Fluid1$, T=T_2, B=T_wb2, P=P) T_dp2=dewpoint(Fluid1$, T=T_2, B=T_wb2, P=P) h_w=enthalpy(Fluid2$, T=T_w, x=0) T_w=T_2 m_dot_a=V_dot/v_2 m_dot_outside=m_dot_a*(1+w_1) DELTAw=w_1-w_2 m_dot_w=m_dot_a*DELTAw Q_dot_out=m_dot_a*(h_1-h_2)-m_dot_w*h_w Q_dot_out=m_dot_cw*c_pw*DELTAT_w c_pw=4.18 [kJ/kg-C]

15-123 A tank contains saturated air at a specified state. The mass of the dry air, the specific humidity, and the enthalpy of the air are to be determined. Assumptions The air and the water vapor are ideal gases. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-303

Analysis (a) The air is saturated, thus the partial pressure of water vapor is equal to the saturation pressure at the given temperature,

Pv = Pg = Psat @ 20°C = 2.339 kPa

Pa = P - Pv = 90 - 2.339 = 87.66 kPa

Treating air as an ideal gas,

ma =

PaV (87.66 kPa)(1.8 m3 ) = = 1.88 kg RaT (0.287 kPa ×m3 /kg ×K)(293 K)

(b) The specific humidity of air is determined from w=

0.622 Pv (0.622)(2.339 kPa) = = 0.0166 kg H 2 O / kg dry air P - Pv (90 - 2.339) kPa

= (1.005 kJ/kg ×°C)(20°C) + (0.0166)(2537.4 kJ/kg) = 62.2 kJ / kg dry air

EES (Engineering Equation Solver) SOLUTION "Given" V=1.8 [m^3] T=20 [C] P=90 [kPa] "Properties" c_p=1.005 [kJ/kg-C] R_a=0.287 [kJ/kg-K] P_g=pressure(steam_iapws, T=T, x=1) h_g=enthalpy(steam_iapws, T=T, x=1) "Analysis" "(a)" P_v=P_g P_a=P-P_v m_a=(P_a*V)/(R_a*(T+273)) "(b)" w=(0.622*P_v)/(P-P_v) "(c)" h=c_p*T+w*h_g

15-124 Problem 15-123 is reconsidered. The properties of the air at the initial state are to be determined and the effects of heating the air at constant volume until the pressure is 110 kPa is to be studied. Analysis The problem is solved using EES, and the solution is given below.

13-304

Tdb[1] = 20 [C] P[1]=90 [kPa] Rh[1]=1.0 P[2]=110 [kPa] Vol = 1.8 [m^3] w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) m_a=Vol/v[1] h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) E_in - E_out = DELTAE_tank "Energy Balance for the constant volume tank" DELTAE_tank=m_a*(u[2] -u[1]) E_in = Q_in E_out = 0 [kJ] u[1]=INTENERGY(AirH2O,T=Tdb[1],P=P[1],w=w[1]) u[2]=INTENERGY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) "The ideal gas mixture assumption applied to the constant volume process yields:" P[1]/(Tdb[1]+273)=P[2]/(Tdb[2]+273) w[2]=w[1] "The mass of the water vapor and dry air are constant" Rh[2]=RELHUM(AirH2O,T=Tdb[2],P=P[2],w=w[2]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],w=w[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) PROPERTIES AT THE INITIAL STATE h[1]=62.15 [kJ/kga] m_a=1.875 [kga] v[1]=0.9599 [m^3/kga] w[1]=0.01658 [kgw/kga]

13-305

Qin

[kPa] 90 92 94 96 98 100 102 104 106 108 110

[k J] 0 9.071 18.14 27.22 36.29 45.37 54.44 63.52 72.61 81.69 90.78

15-125E Air at a specified state and relative humidity flows through a circular duct. The dew-point temperature, the volume flow rate of air, and the mass flow rate of dry air are to be determined. Assumptions The air and the water vapor are ideal gases.

Analysis (a) The vapor pressure of air is

Pv = f Pg = f Psat @ 60°F = (0.70)(0.2564 psia) = 0.1795 psia Thus the dew-point temperature of the air is Tdp = Tsat @ Pv = Tsat @ 0.1795 psia = 50.2°F (from EES)

(b) The volume flow rate is determined from V = VA = V

æp ´ (6 / 12 ft) 2 ö p D2 3 ÷ ÷ = (27 ft/s) çç ÷= 5.301 ft / s ÷ çè 4 4 ø

13-306

Pa = P - Pv = 15 - 0.1795 = 14.82 psia s v1 =

RaT1 (0.3704 psia ×ft 3 /lbm ×R)(520 R) = = 13.00 ft 3 /lbm dry air Pa1 14.82 psia

Thus,

ma1 =

V1 5.301 ft 3 /s = = 0.408 lbm / s v1 13.00 ft 3 /lbm dry air

EES (Engineering Equation Solver) SOLUTION "Given" P=15 [psia] T=60 [F] phi=0.70 D=6 [in] Vel=27 [ft/s] "Properties" R_a=R_u/MM MM=Molarmass(air) R_u=10.73 [psia-ft^3/lbmol-R] "Analysis" Fluid$='steam_iapws' "(a)" P_g=pressure(Fluid$, T=T, x=1) P_v=phi*P_g T_dp=temperature(Fluid$, P=P_v, x=1) "(b)" A=pi*D^2/4*Convert(in^2, ft^2) V_dot=Vel*A "(c)" P_a=P-P_v v=(R_a*(T+460))/P_a m_dot_a=V_dot/v

15-126 Air flows steadily through an isentropic nozzle. The pressure, temperature, and velocity of the air at the nozzle exit are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-307

Analysis The inlet state of the air is completely specified, and the total pressure is 200 kPa. The properties of the air at the inlet state may be determined from the psychrometric chart (Figure A-31) or using EES psychrometric functions to be (we used EES)

h1 = 57.65 kJ/kg dry air

w1 = w2 = 0.008803 kg H2O/kg dry air (no condensation) s1 = s2 = 5.613 kJ/kg.K dry air

(isentropic process)

We assume that the relative humidity at the nozzle exit is 100 percent since there is no condensation in the nozzle. Other exit state properties can be determined using EES built-in functions for moist air. The results are h2 = 42.53 kJ/kg dry air

P2 = 168.2 kPa T2 = 20°C An energy balance on the control volume gives the velocity at the exit h1 = h2 + (1 + w2 )

V22 2

57.65 kJ/kg = 42.53 kJ/kg + (1 + 0.008803)

ö V22 æ çç 1 kJ/kg ÷ 2 2÷ ÷ ç 2 è1000 m /s ø

V2 = 173.2 m / s

EES (Engineering Equation Solver) SOLUTION T_1 = (35+273) [K] P_1 =200 [kPa] RH_1 = 50/100 " relative humidity at state 1" {T_2 = ? "[k]" "Determine the exit temperature and pressure." P_2 =? "[kPa]" } RH_2 = 100/100 " Assume relative humidity at state 2 is 100% for no condensation" "Energy balance for the steady flow process through the isentropic nozzle, Q=0, W=0:" E_in - E_out = DELTAE_cv E_in = h_1"[kJ/kg_a]" E_out = h_2 +(1+w_2)*Vel_2^2/2*convert(m^2/s^2,kJ/kg)"[kJ/kg_a]" DELTAE_cv=0"[kJ/kg_a]" "For isentropic flow DELTAs =0:" s_2=s_1 s_1 = ENTROPY(AirH2O,T=T_1,P=P_1,R=RH_1) s_2 = ENTROPY(AirH2O,T=T_2,P=P_2,R=RH_2) "The specific humidity at entrance (state 1) and exit (state 2):" w_1=HUMRAT(AirH2O,T=T_1,P=P_1,R=RH_1)"[kg_v/kg_a]" w_2=HUMRAT(AirH2O,T=T_2,P=P_2,R=RH_2)"[kg_v/kg_a]" "For no condensation w_1=w_2" w_1=w_2 "The enthalpies per unit mass dry air flowing through the nozzle are:" h_1=ENTHALPY(AirH2O,T=T_1,P=P_1,R=RH_1)"[kJ/kg_a]" h_2=ENTHALPY(AirH2O,T=T_2,P=P_2,w=w_2)"[kJ/kg_a]"

15-127 Air is cooled by evaporating water into this air. The amount of water required and the cooling produced are to be determined. Assumptions

13-308

This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

h1 = 64.0 kJ/kg dry air

w1 = 0.0092 kg H2O/kg dry air and

h2 = 66.0 kJ/kg dry air w2 = 0.0160 kg H2 O/kg dry air Also,

hw @ h f @ 20°C = 83.92 kJ/kg

(Table A-4)

Analysis The amount of moisture in the air increases due to humidification (s). Applying the water mass balance and energy balance equations to the combined cooling and humidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

D w = w2 - w1 = 0.0160 - 0.0092 = 0.0068 kg H2O / kg dry air Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi = Qout + å me he

Qout = ma1h1 + mw hw - ma 2 h2 = ma (h1 - h2 ) + mw hw

qout = h1 - h2 + (w2 - w1 )hw

= (64.0 - 66.0)kJ/kg + (0.0068)(83.92) = - 1.43 kJ / kg dry air The negative sign shows that the heat is actually transferred to the system.

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 "[kPa]" T_1=40 "[C]" phi_1=0.20 T_2=25 "[C]" phi_2=0.80 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-309

"Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_w=enthalpy(Fluid2$, T=20, x=0) q_out=h_1-h_2+(w_2-w_1)*h_w

15-128 Air is humidified adiabatically by evaporating water into this air. The temperature of the air at the exit is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties The inlet state of the air is completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric chart (Figure A-31) to be

h1 = 64.0 kJ/kg dry air

w1 = 0.0092 kg H2O/kg dry air and hw @ h f @ 20°C = 83.92 kJ/kg

(Table A-4)

Analysis The amount of moisture in the air increases due to humidification ( w2 > w1 ). Applying the water mass balance and energy balance equations to the combined cooling and humidification section, Water Mass Balance: å mw,i = å mw,e ¾ ¾ ® ma1w1 = ma 2 w2 + mw mw = ma (w2 - w1 )

Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eo1ut å mi hi = å me he

ma1h1 + mw hw = ma 2 h2

mw hw = ma (h2 - h1 ) (w2 - w1 )hw = h2 - h Substituting,

13-310

(w2 - 0.0092)(83.92) = h2 - 64.0 The solution of this equation requires a trial-error method. An air exit temperature is assumed. At this temperature and given relative humidity, the enthalpy and specific humidity values are obtained from psychrometric chart and substituted into this equation. If the equation is not satisfied, a new value of exit temperature is assumed and this continues until the equation is satisfied. Alternatively, an equation solver such as EES may be used for the direct results. We used the following EES program.

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=40 [C] phi_1=0.20 phi_2=0.80 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_w=enthalpy(Fluid2$, T=20, x=0) q=0 q=h_1-h_2+(w_2-w_1)*h_w The results of these equations are

T2 = 24.6°C h2 = 64.56 kJ/kg dry air w2 = 0.01564 kg H2 O/kg dry air

15-129E Air at a specified state is heated to a specified temperature. The relative humidity after the heating is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Analysis There is no correspondence of inlet state from the psychrometric chart. Therefore, we have to use EES psychrometric functions to obtain the specific humidity: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-311

w1 = 0.0023 lbm H2 O/lbm dry air

As the outside air infiltrates into the dacha, it does not gain or lose any water. Therefore the humidity ratio inside the dacha is the same as that outside,

w2 = w1 = 0.0023 lbm H2O/lbm dry air From EES or Fig. A-31E, at this humidity ratio and the temperature inside the dacha gives

f 2 = 0.146 = 14.6%

EES (Engineering Equation Solver) SOLUTION "Given" P=14.696 "[psia]" T_1=32 "[F]" phi_1=0.60 "Analysis" Fluid1$='AirH2O' w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) w_2=w_1 T_2=70 phi_2=relhum(Fluid1$,T=T_2, P=P, w=w_2)

15-130E Air is humidified by evaporating water into this air. The amount of heating is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

h1 = 19.3 Btu/lbm dry air w1 = 0.0023 lbm H2 O/lbm dry air v1 = 13.40 ft 3 /lbm dry air

and

h2 = 27.1 Btu/lbm dry air

w2 = 0.0094 lbm H2O/lbm dry air Also,

hw @ h f @ 60°F = 28.08 Btu/lbm

(Table A-4E)

13-312

Energy Balance:

Ein - Eout = D Esystem 0 (steady) = 0 Ein = Eout å mi hi + Qin = å me he Qin = ma 2 h2 - ma1h1 - mw hw = ma (h2 - h1 ) - mw hw

= (27.1- 19.3)Btu/lbm - (0.0094 - 0.0023)(28.08)

= 7.59 Btu/lbm dry air The mass of air that has to be humidified is

ma =

V 16,000 ft 3 = = 1194 lbm dry air v1 13.40 ft 3 /lbm dry air

The total heat requirement is then

Qin = ma qin = (1194 lbm dry air)(7.59 Btu/lbm dry air) = 9062 Btu

EES (Engineering Equation Solver) SOLUTION "Given" P=14.696 "[psia]" T_1=70 "[F]" phi_1=0.146 T_2=70 "[F]" phi_2=0.60 "Analysis" Fluid1$='AirH2O' Fluid2$='steam_iapws' h_1=enthalpy(Fluid1$, T=T_1, R=phi_1, P=P) w_1=humrat(Fluid1$, T=T_1, R=phi_1, P=P) v_1=volume(Fluid1$, T=T_1, R=phi_1, P=P) h_2=enthalpy(Fluid1$, T=T_2, R=phi_2, P=P) w_2=humrat(Fluid1$, T=T_2, R=phi_2, P=P) h_w=enthalpy(Fluid2$, T=60, x=0) q_in=h_2-h_1-(w_2-w_1)*h_w

15-131 Air enters a cooling section at a specified pressure, temperature, and relative humidity. The temperature of the air at the exit and the rate of heat transfer are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-313

Analysis (a) The amount of moisture in the air also remains constant (w1 = w2 ) as it flows through the cooling section since the process involves no humidification or dehumidification. The total pressure is 97 kPa. The properties of the air at the inlet state are Pv1 = f 1 Pg1 = f 1 Psat @35°C = (0.3)(5.629 kPa) = 1.69 kPa

Pa1 = P1 - Pv1 = 97 - 1.69 = 95.31 kPa v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(308 K) = Pa1 95.31 kPa

= 0.927 m3 /kg dry air

w1 =

0.622 Pv1 0.622(1.69 kPa) = = 0.0110 kg H 2 O/kg dry air ( = w2 ) P1 - Pv1 (97 - 1.69) kPa

h1 = c pT1 + w1hg1 = (1.005 kJ/kg°C)(35°C) + (0.0110)(2564.6 kJ/kg) = 63.44 kJ/kg dry air

The air at the final state is saturated and the vapor pressure during this process remains constant. Therefore, the exit temperature of the air must be the dew-point temperature, Tdp = Tsat @ Pv = Tsat @1.69 kPa = 14.8°C

(b) The enthalpy of the air at the exit is

h2 = c pT2 + w2 hg 2 = (1.005 kJ/kg ×°C)(14.8°C) + (0.0110)(2528.1 kJ/kg) = 42.78 kJ/kg dry air Also

ma =

V1 6 m3 /s = = 6.47 kg/min v1 0.927 m3 /kg dry air

Then the rate of heat transfer from the air in the cooling section becomes Qout = ma (h1 - h2 ) = (6.47 kg/min)(63.44 - 42.78)kJ/kg = 134 kJ / min

EES (Engineering Equation Solver) SOLUTION "Given" P_1=97 [kPa] T_1=35 [C] phi_1=0.30 V_dot_1=6 [m^3/min] phi_2=1 [C] "Properties" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_1, x=1) h_g1=enthalpy(Fluid$, T=T_1, x=1) h_g2=enthalpy(Fluid$, T=T_2, x=1) c_p=1.005 [kJ/kg-C] R_a=0.287 [kJ/kg-K] "Analysis" "(a)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-314

P_v1=phi_1*P_g1 P_a1=P_1-P_v1 v_1=(R_a*(T_1+273))/P_a1 w_1=(0.622*P_v1)/(P_1-P_v1) h_1=c_p*T_1+w_1*h_g1 T_dp=temperature(Fluid$, P=P_v1, x=1) T_2=T_dp "(b)" w_2=w_1 h_2=c_p*T_2+w_2*h_g2 m_dot_a=V_dot_1/v_1 Q_dot_out=m_dot_a*(h_1-h_2)

15-132 The outdoor air is first heated and then humidified by hot steam in an air-conditioning system. The rate of heat supply in the heating section and the mass flow rate of the steam required in the humidifying section are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Properties The amount of moisture in the air also remains constants it flows through the heating section (w1 = w2 ), but increases in the humidifying section (w3 > w2 ). The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at various states are determined from the psychrometric chart (Fig. A-31 or EES) to be

h1 = 23.45 kJ/kg dry air

w1 = 0.005322 kg H2 O/kg dry air (= w2 ) h2 = 31.56 kJ/kg dry air

w2 = w1 = 0.005322 kg H2 O/kg dry air h3 = 47.72 kJ/kg dry air

w3 = 0.008879 kg H2O/kg dry air Analysis (a) The mass flow rate of dry air is

ma =

V1 26 m3 /min = = 32.14 kg/min v1 0.8090 m3 /kg

Then the rate of heat transfer to the air in the heating section becomes Qin = ma (h2 - h1 ) = (32.14 kg/min)(31.56 - 23.45)kJ/kg = 261 kJ / min

13-315

(b) The conservation of mass equation for water in the humidifying section can be expressed as

ma 2 w2 + mw = ma3w3 Thus,

or mw = ma (w3 - w2 )

mw = (32.14 kg/min)(0.008879 - 0.005322) = 0.114 kg / min

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=10 [C] phi_1=0.70 V_dot_1=26 [m^3/min] T_2=18 [C] T_3=25 [C] phi_3=0.45 "Properties" Fluid$='airH2O' h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) h_2=enthalpy(Fluid$, T=T_2, P=P, w=w_2) w_2=w_1 h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3) w_3=humrat(Fluid$, T=T_3, P=P, R=phi_3) "Analysis" "(a)" m_dot_a=V_dot_1/v_1 Q_dot_in=m_dot_a*(h_2-h_1) "(b)" m_dot_w=m_dot_a*(w_3-w_2)

15-133 Atmospheric air enters an air-conditioning system at a specified pressure, temperature, and relative humidity. The heat transfer, the rate of condensation of water, and the mass flow rate of the refrigerant are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

13-316

Analysis The inlet and exit states of the air are completely specified, and the total pressure is 1 atm. The properties of the air at the inlet and exit states may be determined from the psychrometric chart (Figure A-31) or using EES psychrometric functions to be (we used EES)

h1 = 78.24 kJ/kg dry air

w1 = 0.01880 kg H2 O/kg dry air v1 = 0.8847 m3 / kg dry air h2 = 27.45 kJ/kg dry air

The mass flow rate of dry air is

ma =

V1 4 m3 /min = = 4.521 kg/min v1 0.8847 m3

The mass flow rates of vapor at the inlet and exit are

mv1 = w1ma = (0.01880)(4.521 kg/min) = 0.0850 kg/min mv 2 = w2 ma = (0.002885)(4.521 kg/min) = 0.01304 kg/min An energy balance on the control volume gives ma h1 = Qout + ma h2 + mw hw 2

where the the enthalpy of condensate water is

hw2 = h f@ 20°C = 83.91 kJ/kg

(Table A-4)

and the rate of condensation of water vapor is

mw = mv1 - mv 2 = 0.0850 - 0.01304 = 0.07196 kg / min Substituting, ma h1 = Qout + ma h2 + mw hw 2 (4.521 kg/min)(78.24 kJ/kg) = Qout + (4.521 kg/min)(27.45 kJ/kg) + (0.07196 kg/min)(83.91 kJ/kg) Qout = 223.6 kJ/min = 3.727 kW

The properties of the R-134a at the inlet and exit of the cooling section are

PR1 = 350 kPaïüï ý hR1 = 97.52 kJ/kg xR1 = 0.20 ïïþ PR 2 = 350 kPa üïï ý hR 2 = 253.38 kJ/kg ïïþ xR 2 = 1.0 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-317

Noting that the rate of heat lost from the air is received by the refrigerant, the mass flow rate of the refrigerant is determined from mR hR1 + Qin = mR hR 2

mR =

Qin 223.6 kJ/min = = 1.435 kg / min hR 2 - hR1 (253.38 - 97.52) kJ/kg AirH2O

0,050 Pressure = 101.3 [kPa] 0,045 0,040

35°C 0.8

Humidity Ratio

0,035 0,030 30°C

0.6

0,025 25°C

0,020

0.4

20°C

0,015 15°C

0,010

0.2

10°C

0,005 0,000 0

2 5

T [°C]

EES (Engineering Equation Solver) SOLUTION P[1] =1 "[atm]"*convert(atm,kPa) "[kPa]" T[1] = 30"[C]" RH[1] = 70/100 " relative humidity" P[2] = P[1]"[kPa]" T[2] = 20"[C]" RH[2] = 20/100 P_R134a=350"[kPa]" x_R134a=.2 "Dry air flow rate, m_dot_a, is constant" Vol_dot[1]= 4 "[m^3/min]" v[1]=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) "[m^3/kg_a]" m_dot_a = Vol_dot[1]/v[1] "[kg_a/min]" "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2] "Steady flow conservation of energy for the air" m_dot_a *h[1] = Q_dot_out *convert(kW,kJ/min)+ m_dot_a*h[2] +m_dot_w*h_liq_2 h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-318

h_liq_2=ENTHALPY(steam_iapws,T=T[2],x=0)"[kJ/kg_v]" "Steady flow conservation of energy for R 134a" m_dot_R*h_R[1] + Q_dot_out *convert(kW,kJ/min)= m_dot_R*h_R[2] h_R[1]=ENTHALPY(R134a,x=x_R134a,P=P_R134a) h_R[2]=ENTHALPY(R134a,x=1,P=P_R134a)

15-134 Air is cooled and dehumidified at constant pressure. The system hardware and the psychrometric diagram are to be sketched and the heat transfer is to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Properties (a) The schematic of the cooling and dehumidifaction process and the the process on the psychrometric chart are given below. The psychrometric chart is obtained from the Property Plot feature of EES.

(b) The inlet and the exit states of the air are completely specified, and the total pressure is 101.3 kPa (1 atm). The properties of the air at the inlet and exit states are determined from the psychrometric chart (Figure A-31 or EES, we used EES) to be Tdp = 28.37°C

h1 = 99.60 kJ/kg dry air w1 = 0.02467 kg H2 O/kg dry air and T2 = Tdp - 10 = 28.37 - 10 = 18.37°C

13-319

h2 = 52.09 kJ/kg dry air w2 = 0.01324kg H2 O/kg dry air s

Also,

hw @ h f @18.37°C = 77.11 kJ/kg (Table A-4) Analysis Applying the water mass balance and energy balance equations to the combined cooling and dehumidification section, Water Mass Balance: å mw, i = å mw, e ¾ ¾ ® ma1w1 = ma 2 w2 + mw

mw = ma (w1 - w2 ) Energy Balance: Ein - Eout = D Esystem ¿ 0 (steady) = 0

Ein = Eout å mi hi = Qout + å me he Qout = ma1h1 - (ma 2 h2 + mw hw )

= ma (h1 - h2 ) - mw hw

= ma (h1 - h2 ) - ma (w1 - w2 )hw Dividing by ma

qout = (h1 - h2 ) - (w1 - w2 )hw Substituting,

qout = (h1 - h2 ) - (w1 - w2 )hw = (99.60 - 52.09) kJ/kg - (0.02467 - 0.01324)(77.11 kJ/kg) = 46.6 kJ / kg

EES (Engineering Equation Solver) SOLUTION "Given" T_1=36 RH_1=0.65 P=101.3 "Analysis" Fluid$='AirH2O' T_dp_1=dewpoint(Fluid$, T=T_1, P=P, R=RH_1) h_1=enthalpy(Fluid$, T=T_1, P=P, R=RH_1) w_1=humrat(Fluid$, T=T_1, P=P, R=RH_1) T_2=T_dp_1-10 RH_2=1 h_2=enthalpy(Fluid$, T=T_2, P=P, R=RH_2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-320

w_2=humrat(Fluid$, T=T_2, P=P, R=RH_2) h_w=enthalpy(steam_iapws,T=T_2, x=0) q_out=(h_1-h_2)-(w_1-w_2)*h_w

15-135 An automobile air conditioner using refrigerant 134a as the cooling fluid is considered. The inlet and exit states of moist air in the evaporator are specified. The volume flow rate of the air entering the evaporator of the air conditioner is to be determined. Assumptions 1 All processes are steady flow and the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. Analysis We assume that the total pressure of moist air is 100 kPa. Then, the inlet and exit states of the moist air for the evaporator are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES)

h1 = 55.88 kJ/kg dry air w1 = 0.01206 kg H2 O/kg dry air v1 = 0.8724 m3 /kg dry air

h2 = 23.31 kJ/kg dry air

w2 = 0.006064 kg H2O/kg dry air The mass flow rate of dry air is given by

ma =

V1 V1 = v1 0.8724 m3 /kg

The mass flow rate of condensate water is expressed as mw = ma (w1 - w2 ) =

V1 (0.01206-0.006064) = 0.006873V1 0.8724

The enthalpy of condensate water is

hw2 = h f@ 8°C = 33.63 kJ/kg

(Table A-4)

An energy balance on the control volume gives ma h1 = Qout + ma h2 + mw hw 2

V1 V1 (55.88) = Qout + (23.11) + 0.006873V1 (33.63) 0.8724 0.8724

(1)

13-321

The properties of the R-134a at the inlet of the compressor and the enthalpy at the exit for the isentropic process are (R-134a tables)

PR1 = 200 kPa ïüï hR1 = 244.50 kJ/kg ý ïïþ sR1 = 0.93788 kJ/kg.K xR1 = 1 PR 2 = 1600 kPaüïï ý hR 2, s = 287.96 kJ/kg ïïþ sR 2 = sR1 The enthalpies of R-134a at the condenser exit and the throttle exit are

hR3 = h f@ 1600 kPa = 135.96 kJ/kg

hR 4 = hR3 = 135.96 kJ/kg The mass flow rate of the refrigerant can be determined from the expression for the compressor power:

WC = mR

hR 2, s - hR1

6 kW = mR

hC (287.96 - 244.50) kJ/kg ¾¾ ® mR = 0.1173 kg/s = 7.041 kg/min 0.85

The rate of heat absorbed by the R-134a in the evaporator is

QR,in = mR (hR1 - hR 4 ) = (7.041 kg/min)(244.50 - 135.96) kJ/kg = 764.2 kJ/min The rate of heat lost from the air in the evaporator is absorbed by the refrigerant-134a. That is, QR,in = Qout . Then, the volume flow rate of the air at the inlet of the evaporator can be determined from Eq. (1) to be V1 V1 (55.88) = 764.2 + (23.11) + 0.006873V1 (33.63) ¾ ¾ ® V1 = 20.5 m 3 / min 0.8724 0.8724

EES code of this equation: V_dot_1/0.8724*55.88=764.2+V_dot_1/0.8724*23.11+0.006873*V_dot_1*33.63

EES (Engineering Equation Solver) SOLUTION "Atmospheric air input data:" P[1] =100 [kPa] "Assume atmospheric pressure = 100 kPa." T[1] = 25 [C] "dry bulb temperature" RH[1] = 60/100 "relative humidity" P[2] =100 [kPa] T[2] = 8 [C] "dry bulb temperature" RH[2]=90/100 "Refrigerant input data:" Pevap=(100+100) [kPa] Pcond=(1500+100) [kPa] W_dot_comp=6 [kW] eta_comp=0.85 "Dry air flow rate, m_dot_a, is constant and calculated by the energy balance on the evaporator." v_1=VOLUME(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_a = Vol_dot_air/v_1 "Mass flow rate of the condensed water" m_dot_v[1]=m_dot_v[2]+m_dot_w w[1]=HUMRAT(AirH2O,T=T[1],P=P[1],R=RH[1]) m_dot_v[1] = m_dot_a*w[1] w[2]=HUMRAT(AirH2O,T=T[2],P=P[2],R=RH[2]) m_dot_v[2] = m_dot_a*w[2] "Steady flow conservation of energy for the atmospheric air flowing across the evaporator." m_dot_a *h[1] = Qair_dot_out *convert(kW,kJ/min)+ m_dot_a*h[2] +m_dot_w*h_liq_2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-322

h[1]=ENTHALPY(AirH2O,T=T[1],P=P[1],w=w[1]) h[2]=ENTHALPY(AirH2O,T=T[2],P=P[2],w=w[2]) h_liq_2=ENTHALPY(Steam_iapws,T=T[2],x=0) "Steady flow analysis of the refrigerant:" hr[1]=enthalpy(r134a, P=Pevap, x=1) "evaporator exit" sr1=entropy(r134a, P=Pevap, x=1) "evaporator exit" sr2=sr1 "isentopic compression" hr_s[2]=enthalpy(r134a, P=Pcond, s=sr2) "isentropic state at exit to compressor" hr[3]=enthalpy(r134a, P=Pcond, x=0) "condenser exit" hr[4]=hr[3] "throttle exit" W_dot_comp=m_dot_R134a*(hr_s[2]-hr[1])/eta_comp*convert(kJ/min,kW) Qr_dot_in=m_dot_R134a*(hr[1]-hr[4])*convert(kJ/min,kW) "heat transfer rate to the refrigerant from the air" "Qr_dot_in = heat supplied to the refrigerant = Qair_dot_out from the atmospheric air:" Qr_dot_in = Qair_dot_out

15-136 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate of water added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined from the psychrometric chart (Fig. A-31 or EES) to be

h1 = 29.8 kJ/kg dry air T1 = 15°C ü ïï w ý 1 = 0.00581 kg/ H 2 O/kg dry air f 1 = 55% ïïþ v1 = 0.824 m3 /kg T2 = 32.5°C w2 = w1 ü ïï f ý 2 = 19.2% Twb2 = Twb3 ïïþ h2 @ h3 = 47.8 kJ / kg dry air h3 = 47.8 kJ/kg dry air T3 = 25°C ü ïï w ý 3 = 0.00888 kg/ H 2 O/kg dry air f 3 = 45% ïïþ Twb3 = 17.1°C (b) The mass flow rate of dry air is

13-323

ma =

V1 30 m3 /min = = 36.4 kg/min v1 0.824 m3 /kg dry air

Then the rate of heat transfer to air in the heating section becomes Qin = ma (h2 - h1 ) = (36.4 kg/min)(47.8 - 29.8)kJ/kg = 655 kJ / min

mw, added = mw3 - mw2 = ma (w3 - w2 ) = (36.4 kg/min)(0.00888 - 0.00581) = 0.112 kg / min

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=15 [C] phi_1=0.55 V_dot_1=30 [m^3/min] T_3=25 [C] phi_3=0.45 "Properties" Fluid$='AirH2O' h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) v_1=volume(Fluid$, T=T_1, P=P, R=phi_1) h_3=enthalpy(Fluid$, T=T_3, P=P, R=phi_3) w_3=humrat(Fluid$, T=T_3, P=P, R=phi_3) T_wb3=wetbulb(Fluid$, T=T_3, P=P, R=phi_3) w_2=w_1 T_wb2=T_wb3 h_2=h_3 "Analysis" "(a)" T_2=temperature(Fluid$, w=w_2, P=P, B=T_wb2) phi_2=relhum(Fluid$, w=w_2, P=P, B=T_wb2) "(b)" m_dot_a=V_dot_1/v_1 Q_dot_in=m_dot_a*(h_2-h_1) "(c)" m_dot_w=m_dot_a*(w_3-w_2)

15-137 Problem 15-136 is reconsidered. The effect of total pressure in the range 94 to 100 kPa on the results required in the problem is to be studied. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] Tdb[1] =15 [C] Rh[1] = 0.55 Vol_dot[1]= 30 [m^3/min] Tdb[3] = 25 [C] Rh[3] = 0.45 P[1]=P P[2]=P[1] P[3]=P[1] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-324

E_dot_in - E_dot_out = DELTAE_dot_sys "Energy balance for the steady-flow heating process 1 to 2" DELTAE_dot_sys = 0 [kJ/min] E_dot_in = m_dot_a*h[1]+Q_dot_in E_dot_out = m_dot_a*h[2] m_dot_a = Vol_dot[1]/v[1] "Conservation of mass of dry air during mixing: m_dot_a = constant" m_dot_a*w[1] = m_dot_a*w[2] "Conservation of mass of water vapor during the heating process" m_dot_a*w[2]+m_dot_w = m_dot_a*w[3] "Conservation of mass of water vapor during the evaporative cooler process" "During the evaporative cooler process:" Twb[2] = Twb[3] Twb[3] =WETBULB(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) {h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],B=Twb[2])} h[2]=h[3] Tdb[2]=TEMPERATURE(AirH2O,h=h[2],P=P[2],w=w[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) h[3]=ENTHALPY(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) w[3]=HUMRAT(AirH2O,T=Tdb[3],P=P[3],R=Rh[3])

P [kPa] 94 95 96 97 98 99 100

Qin

[kg / min] 0.1118 0.1118 0.1118 0.1118 0.1118 0.1117 0.1117

[k J/min] 628.6 632.2 635.8 639.4 643 646.6 650.3

Rh 2

Tdb 2

0.1833 0.1843 0.1852 0.186 0.1869 0.1878 0.1886

[C] 33.29 33.2 33.11 33.03 32.94 32.86 32.78

13-325

15-138 Air is heated and dehumidified in an air-conditioning system consisting of a heating section and an evaporative cooler. The temperature and relative humidity of the air when it leaves the heating section, the rate of heat transfer in the heating section, and the rate at which water is added to the air in the evaporative cooler are to be determined. Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the entire process

(ma1 = ma 2 = ma ). 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Analysis (a) Assuming the wet-bulb temperature of the air remains constant during the evaporative cooling process, the properties of air at various states are determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-326

Pv1 = f 1 Pg1 = f 1 Psat @15°C = (0.55)(1.7057 kPa) = 0.938 kPa

Pa1 = P1 - Pv1 = 96 - 0.938 = 95.06 kPa v1 =

RaT1 (0.287 kPa ×m3 /kg ×K)(288 K) = Pa1 95.06 kPa

= 0.8695 m3 /kg dry air

w1 =

0.622 Pv1 0.622(0.8695 kPa) = = 0.006138 kg H 2 O/kg dry air P1 - Pv1 (96 - 0.8695) kPa

h1 = c pT1 + w1hg1 = (1.005 kJ/kg ×°C)(15°C) + (0.006138)(2528.3 kJ/kg) = 30.59 kJ/kg dry air

and Pv 3 = f 3 Pg 3 = f 3 Psat @ 25°C = (0.45)(3.17 kPa) = 1.426 kPa

Pa3 = P3 - Pv3 = 96 - 1.426 = 94.57 kPa w3 =

0.622 Pv 3 0.622(1.426 kPa) = = 0.009381 kg H 2 O/kg dry air P3 - Pv 3 (96 - 1.426) kPa

h3 = c pT3 + w3 hg 3 = (1.005 kJ/kg ×°C)(25°C) + (0.009381)(2546.5 kJ/kg)

= 49.01 kJ/kg dry air Also,

h2 @ h3 = 49.01 kJ/kg w2 = w1 = 0.006138 kg H2 O/kg dry air Thus,

h2 = c pT2 + w2 hg 2 @ c pT2 + w2 (2500.9 + 1.82T2 ) = (1.005 kJ/kg ×°C)T2 + (0.006138)(2500.9 + 1.82T2 ) Solving for T2 , T2 = 33.1°C ¾ ¾ ® Pg 2 = Psat@33.1°C = 5.072 kPa

Thus,

f2=

w2 P2 (0.006138)(96) = = 0.185 or 18.5% (0.622 + w2 ) Pg 2 (0.622 + 0.006138)(5.072)

(b) The mass flow rate of dry air is

ma =

V1 30 m3 /min = = 34.5 kg/min v1 0.8695 m3 /kg dry air

Then the rate of heat transfer to air in the heating section becomes Qin = ma (h2 - h1 ) = (34.5 kg/min)(49.01 - 30.59)kJ/kg = 636 kJ / min

mw, added = mw3 - mw2 = ma (w3 - w2 ) = (34.5 kg/min)(0.009381- 0.006138)=0.112 kg / min

EES (Engineering Equation Solver) SOLUTION "Given" P=96 [kPa] T_1=15 [C] phi_1=0.55 V_dot_1=30 [m^3/min] T_3=25 [C] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-327

phi_3=0.45 "Properties" Fluid$='steam_iapws' P_g1=pressure(Fluid$, T=T_1, x=1) P_g3=pressure(Fluid$, T=T_3, x=1) h_g1=enthalpy(Fluid$, T=T_1, x=1) h_g3=enthalpy(Fluid$, T=T_3, x=1) c_p=1.005 [kJ/kg-C] R_a=0.287 [kJ/kg-K] "Analysis" "(a)" P_v1=phi_1*P_g1 P_a1=P-P_v1 v_1=(R_a*(T_1+273))/P_a1 w_1=(0.622*P_v1)/(P-P_v1) h_1=c_p*T_1+w_1*h_g1 P_v3=phi_3*P_g3 P_a3=P-P_v3 w_3=(0.622*P_v3)/(P-P_v3) h_3=c_p*T_3+w_3*h_g3 h_2=h_3 w_2=w_1 h_2=c_p*T_2+w_2*h_g2 h_g2=enthalpy(Fluid$, T=T_2, x=1) phi_2=(w_2*P)/((0.622+w_2)*P_g2) P_g2=pressure(Fluid$, T=T_2, x=1) "(b)" m_dot_a=V_dot_1/v_1 Q_dot_in=m_dot_a*(h_2-h_1) "(c)" m_dot_w=m_dot_a*(w_3-w_2)

15-139 Conditioned air is to be mixed with outside air. The ratio of the dry air mass flow rates of the conditioned- to-outside air, and the temperature of the mixture are to be determined. Assumptions 1 Steady operating conditions exist. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The mixing chamber is adiabatic.

Properties The properties of each inlet stream are determined from the psychrometric chart (Fig. A-31) to be

13-328

w1 = 0.0084 kg H2 O/kg dry air and

h2 = 68.5 kJ/kg dry air w2 = 0.0134 kg H2O/kg dry air Analysis (a) The ratio of the dry air mass flow rates of the Conditioned air to the outside air can be determined from ma1 w2 - w3 h2 - h3 = = ma 2 w3 - w1 h3 - h1

But state 3 is not completely specified. However, we know that state 3 is on the straight line connecting states 1 and 2 on the psychrometric chart. At the intersection point of this line and  = 60% line we read (b) T3 = 23.5°C

w3 = 0.0109 kg H2O/kg dry air h3 = 51.3 kJ/kg dry air Therefore, the mixture will leave at 23.5 C. The ma1 / ma 2 ratio is determined by substituting the specific humidity (or enthalpy) values into the above relation, (c)

ma1 0.0134 - 0.0109 = = 1.00 ma 2 0.0109 - 0.0084

Therefore, the mass flow rate of each stream must be the same.

EES (Engineering Equation Solver) SOLUTION "Given" P=101.325 [kPa] T_1=13 [C] phi_1=0.90 T_2=34 [C] phi_2=0.40 phi_3=0.60 "Analysis" Fluid$='AirH2O' "1st stream properties" h_1=enthalpy(Fluid$, T=T_1, P=P, R=phi_1) w_1=humrat(Fluid$, T=T_1, P=P, R=phi_1) "2nd stream properties" h_2=enthalpy(Fluid$, T=T_2, P=P, R=phi_2) w_2=humrat(Fluid$, T=T_2, P=P, R=phi_2) "mixture properties" w_3=humrat(Fluid$, T=T_3, P=P, R=phi_3) T_3=temperature(Fluid$, w=w_3, P=P, R=phi_3) Ratio=(w_2-w_3)/(w_3-w_1)

15-140 Problem 15-139 is reconsidered. The desired quantities are to be determined using appropriate software at 1 atm and 80 kPa pressures. Analysis The problem is solved using EES, and the solution is given below. P=101.325 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-329

Tdb[1] =13 [C] "State 1 is the conditioned air" Rh[1] = 0.90 Tdb[2] =34 [C] "State 2 is the outside air" Rh[2] = 0.40 Rh[3] = 0.60 P[1]=P P[2]=P[1] P[3]=P[1] m_dot[2] = 1 [kg/s] "Without loss of generality assume the mass flow rate of the outside air is 1 kg/s" MassRatio = m_dot[1]/m_dot[2] E_dot_in - E_dot_out = DELTAE_dot_sys "Energy balance for the steady-flow mixing process" DELTAE_dot_sys = 0 [kW] E_dot_in = m_dot[1]*h[1]+m_dot[2]*h[2] E_dot_out = m_dot[3]*h[3] m_dot[1]+m_dot[2] = m_dot[3] "Conservation of mass of dry air during mixing" m_dot[1]*w[1]+m_dot[2]*w[2] = m_dot[3]*w[3] "Conservation of mass of water vapor during mixing" h[1]=ENTHALPY(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) v[1]=VOLUME(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) w[1]=HUMRAT(AirH2O,T=Tdb[1],P=P[1],R=Rh[1]) h[2]=ENTHALPY(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) v[2]=VOLUME(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) w[2]=HUMRAT(AirH2O,T=Tdb[2],P=P[2],R=Rh[2]) Tdb[3]=TEMPERATURE(AirH2O,h=h[3],P=P[3],R=Rh[3]) w[3]=HUMRAT(AirH2O,T=Tdb[3],P=P[3],R=Rh[3]) v[3]=VOLUME(AirH2O,T=Tdb[3],P=P[3],w=w[3])

SOLUTION for P=1 atm (101.325 kPa) DELTAE_dot_sys=0 [kW] E_dot_out=102.9 [kW] h[2]=68.45 [kJ/kg] MassRatio=1.007 m_dot[2]=1 [kg/s]

E_dot_in=102.9 [kW] h[1]=34.26 [kJ/kg] h[3]=51.3 [kJ/kg] m_dot[1]=1.007 [kg/s] m_dot[3]=2.007 [kg/s]

13-330

P=101.3 [kPa] P[2]=101.3 [kPa] Rh[1]=0.9 Rh[3]=0.6 Tdb[2]=34 [C] v[1]=0.8215 [m^3/kg] v[3]=0.855 [m^3/kg] w[2]=0.01336

P[1]=101.3 [kPa] P[3]=101.3 [kPa] Rh[2]=0.4 Tdb[1]=13 [C] Tdb[3]=23.51 [C] v[2]=0.8888 [m^3/kg] w[1]=0.008387 w[3]=0.01086

SOLUTION for P=80 kPa DELTAE_dot_sys=0 E_dot_out=118.2 [kW] h[2]=77.82 [kJ/kga] MassRatio=1.009 m_dot[2]=1 [kga/s] P=80 [kPa] P[2]=80 [kPa] Rh[1]=0.9 Rh[3]=0.6 Tdb[2]=34 [C] v[1]=1.044 [m^3/kga] v[3]=1.088 [m^3/kga] w[2]=0.01701 [kgw/kga]

E_dot_in=118.2 [kW] h[1]=40 [kJ/kga] h[3]=58.82 [kJ/kga] m_dot[1]=1.009 [kga/s] m_dot[3]=2.009 [kga/s] P[1]=80 [kPa] P[3]=80 [kPa] Rh[2]=0.4 Tdb[1]=13 [C] Tdb[3]=23.51 [C] v[2]=1.132 [m^3/kga] w[1]=0.01066 [kgw/kga] w[3]=0.01382 [kgw/kga]

15-141 A natural-draft cooling tower is used to remove waste heat from the cooling water flowing through the condenser of a steam power plant. The mass flow rate of the cooling water, the volume flow rate of air into the cooling tower, and the mass flow rate of the required makeup water are to be determined. Assumptions 1 All processes are steady-flow and the mass flow rate of dry air remains constant during the entire process (ma1 = ma 2 = ma ) . 2 3

Dry air and water vapor are ideal gases. The kinetic and potential energy changes are negligible.

Analysis The inlet and exit states of the moist air for the tower are completely specified. The properties may be determined from the psychrometric chart (Fig. A-31) or using EES psychrometric functions to be (we used EES)

h1 = 50.74 kJ/kg dry air w1 = 0.01085 kg H2 O/kg dry air v1 = 0.8536 m3 /kg dry air

13-331

h2 = 142.83 kJ/kg dry air w2 = 0.04112 kg H2 O/kg dry air The enthalpies of cooling water at the inlet and exit of the condenser are (Table A-4) hw3 = h f@ 40°C = 167.53 kJ/kg hw4 = h f@ 26°C = 109.01 kJ/kg

The steam properties for the condenser are (Steam tables)

Ps1 = 200 kPa üïï ý hs1 = 504.71 kJ/kg ïïþ xs1 = 0 üïï ý hs 2 = 2524.3 kJ/kg ss 2 = 7.962 kJ/kg.Kïïþ Ps 2 = 10 kPa

Ps 3 = 10 kPa ïüï ý hs 3 = 191.81 kJ/kg ïïþ xs1 = 0 The mass flow rate of dry air is given by

ma =

V1 V1 = v1 0.8536 m3 /kg

The mass flow rates of vapor at the inlet and exit of the cooling tower are

mv1 = w1ma = (0.01085)

V1 = 0.01271V1 0.8536

mv 2 = w2 ma = (0.04112)

V1 = 0.04817V1 0.8536

Mass and energy balances on the cooling tower give

mv1 + mcw3 = mv 2 + mcw4 ma h1 + mcw3 hw3 = ma h2 + mcw4 hw4 The mass flow rate of the makeup water is determined from

mmakeup = mv 2 - mv1 = mcw3 - mcw4 An energy balance on the condenser gives

0.18ms hs1 + 0.82ms hs 2 + mcw4 hw4 + mmakeup hw4 = ms hs 3 + mcw3 hw3 Solving all the above equations simultaneously with known and determined values using EES, we obtain

mcw3 = 1413 kg / s V1 = 47, 700 m 3 / min

mmakeup = 28.19 kg / s

EES (Engineering Equation Solver) SOLUTION "Input Data for atmospheric air" P_atm =101.32 [kPa] T_db_1 = 23 [C] T_wb_1 = 18 [C] T_db_2 = 37 [C] RH_2 = 100/100 "Input Data for the steam cycle" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-332

m_dot_steam = 42 [kg/s] Psteam_lastFWH=200 [kPa] Psteam_Turbexit=10 [kPa] ssteam_turbexit=7.962 [kJ/kg-K] T_cw_3 = 40 [C] T_cw_4 = 26 [C] "Dry air mass flow rates:" v_1=VOLUME(AirH2O,T=T_db_1,P=P_atm,R=RH_1) RH_1=RELHUM(AirH2O,T=T_db_1,P=P_atm,B=T_wb_1) m_dot_a_1 = Vol_dot_1*convert(m^3/min,m^3/s)/v_1 "Conservaton of mass for the dry air (ma) in the mixing device:" m_dot_a_in - m_dot_a_out = DELTAm_dot_a_cv m_dot_a_in = m_dot_a_1 m_dot_a_out = m_dot_a_2 DELTAm_dot_a_cv = 0 "Steady flow requirement" "Conservation of mass for the water vapor (mv) and cooling water for the process:" m_dot_w_in - m_dot_w_out = DELTAm_dot_w_cv m_dot_w_in = m_dot_v_1 + m_dot_cw_3 m_dot_w_out = m_dot_v_2+m_dot_cw_4 DELTAm_dot_w_cv = 0 "Steady flow requirement" w_1=HUMRAT(AirH2O,T=T_db_1,P=P_atm,R=RH_1) m_dot_v_1 = m_dot_a_1*w_1 w_2=HUMRAT(AirH2O,T=T_db_2,P=P_atm,R=RH_2) m_dot_v_2 = m_dot_a_2*w_2 "Conservation of energy for the SSSF cooling tower process:" E_dot_in_tower - E_dot_out_tower = DELTAE_dot_tower_cv E_dot_in_tower= m_dot_a_1 *h[1] + m_dot_cw_3*h_w[3] E_dot_out_tower = m_dot_a_2*h[2] + m_dot_cw_4*h_w[4] DELTAE_dot_tower_cv = 0 "Steady flow requirement" h[1]=ENTHALPY(AirH2O,T=T_db_1,P=P_atm,w=w_1) h[2]=ENTHALPY(AirH2O,T=T_db_2,P=P_atm,w=w_2) h_w[3]=ENTHALPY(steam_iapws,T=T_cw_3,x=0) h_w[4]=ENTHALPY(steam_iapws,T=T_cw_4,x=0) "Energy balance on the condenser determines the cooling water flow rate:" "Assume the makeup water is supplied at a temperature equal to T_cw_4." "Steam enters condenser at 200 kPa from last feedwater heater and turbine at 10 kPa" "Steam leaves the condenser as satruated liquid at 10 kPa" E_dot_in_cond - E_dot_out_cond = DELTAE_dot_cond_cv E_dot_in_cond= 0.18*m_dot_steam*h_s1+0.82*m_dot_steam*h_s2+ m_dot_cw_4*h_w[4]+ m_dot_makeup * h_w[4] E_dot_out_cond = m_dot_steam*h_s3+m_dot_cw_3 *h_w[3] DELTAE_dot_cond_cv = 0 "Steady flow requirement" h_s1=enthalpy(steam_iapws,P=Psteam_lastFWH,x=0) h_s2=enthalpy(steam_iapws,P=Psteam_turbexit,s=ssteam_turbexit) h_s3=enthalpy(steam_iapws,P=10,x=0) Q_dot_Heatload=0.18*m_dot_steam*h_s1+0.82*m_dot_steam*h_s2-m_dot_steam*h_s3 "Conservation of mass on the condenser for the cooling water gives the makeup water flow rate." "Note: The makeup water flow rate equals the amount of water vaporized in the cooling tower." m_dot_cw_in - m_dot_cw_out = DELTAm_dot_cw_cv m_dot_cw_in = m_dot_cw_4 + m_dot_makeup m_dot_cw_out = m_dot_cw_3 DELTAm_dot_cw_cv = 0 "Steady flow requirement" EES Solution DELTAE_dot_cond_cv=0 [kW] DELTAE_dot_tower_cv=0 [kW] DELTAm_dot_a_cv=0 [kg/s] DELTAm_dot_cw_cv=0 [kg_cw/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-333

DELTAm_dot_w_cv=0 [kg_w/s] E_dot_in_cond=244790 [kW] E_dot_in_tower=284005 [kW] E_dot_out_cond=244790 [kW] E_dot_out_tower=284005 [kW] h_s1=504.71 h_s2=2524.27 [kJ/kg] h_s3=191.81 m_dot_a_1=932.4 [kg/s] m_dot_a_2=932.4 [kg_a/s] m_dot_a_in=932.4 [kg/s] m_dot_a_out=932.4 [kg/s] m_dot_cw_3=1413 [kg_cw/s] m_dot_cw_4=1385 [kg_cw/s] m_dot_cw_in=1413 [kg_cw/s] m_dot_cw_out=1413 [kg_cw/s] m_dot_makeup=28.17 [kg_cw/s] m_dot_steam=42 [kg/s] m_dot_v_1=10.12 [kg_w/s] m_dot_v_2=38.29 [kg_w/s] m_dot_w_in=1423 [kg_w/s] m_dot_w_out=1423 [kg_w/s] Psteam_lastFWH=200 [kPa] Psteam_Turbexit=10 [kPa] P_atm=101.3 [kPa] Q_dot_Heatload=82696 [kW] RH_1=0.6182 RH_2=1 ssteam_turbexit=7.962 [kJ/kg-K] T_cw_3=40 [C] T_cw_4=26 [C] T_db_1=23 [C] T_db_2=37 [C] T_wb_1=18 [C] Vol_dot_1=47755 [m^3/min] v_1=0.8536 [m^3/kg_a] w_1=0.01085 [kg_w/kg_a] w_2=0.04106 [kg_w/kg_a]

15-142E The required size of an evaporative cooler in cfm (ft 3 /min) for an 8-ft high house is determined by multiplying the floor area of the house by 4. An equivalent rule is to be obtained in SI units. Analysis Noting that 1 ft = 0.3048 m and thus 1 ft 2 = 0.0929 m 2 and 1 ft 3 = 0.0283 m3 , and noting that a flow rate of 4 ft 3 /min is required per ft 2 of floor area, the required flow rate in SI units per m2 of floor area is determined to

1 ft 2  4 ft 3 / min

0.0929 m 2  4  0.0283 m3 / min 1 m 2  1.22 m3 / min Therefore, a flow rate of 1.22 m 3 / min is required per m 2 of floor area.

13-334

Multiplyby=4*Convert(ft, m)

Fundamentals of Engineering (FE) Exam Problems 15-143 A room contains 65 kg of dry air and 0.43 kg of water vapor at 25°C and 90 kPa total pressure. The relative humidity of air in the room is (a) 29.9% (b) 35.2% (c) 41.5% (d) 60.0% (e) 66.2% Answer (a) 29.9% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 [C] P=90 [kPa] m_air=65 [kg] m_v=0.43 [kg] w=0.622*P_v/(P-P_v) w=m_v/m_air P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g "Some Wrong Solutions with Common Mistakes:" W1_RH=m_v/(m_air+m_v) "Using wrong relation" W2_RH=P_g/P "Using wrong relation"

15-144 A 40-m 3 room contains air at 30C and a total pressure of 90 kPa with a relative humidity of 75 percent. The mass of dry air in the room is (a) 24.7 kg (b) 29.9 kg (c) 39.9 kg (d) 41.4 kg (e) 52.3 kg Answer (c) 39.9 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). V=40 [m^3] T1=30 [C] P=90 [kPa] RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v R_air=0.287 [kJ/kg-K] m_air=P_air*V/(R_air*(T1+273)) "Some Wrong Solutions with Common Mistakes:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-335

W1_mass=P_air*V/(R_air*T1) "Using C instead of K" W2_mass=P*V/(R_air*(T1+273)) "Using P instead of P_air" W3_mass=m_air*RH "Using wrong relation"

15-145 A room is filled with saturated moist air at 25C and a total pressure of 100 kPa. If the mass of dry air in the room is 100 kg, the mass of water vapor is (a) 0.52 kg (b) 1.97 kg (c) 2.96 kg (d) 2.04 kg (e) 3.17 kg Answer (d) 2.04 kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 [C] P=100 [kPa] m_air=100 [kg] RH=1 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v w=0.622*P_v/(P-P_v) w=m_v/m_air "Some Wrong Solutions with Common Mistakes:" W1_vmass=m_air*w1; w1=0.622*P_v/P "Using P instead of P-Pv in w relation" W2_vmass=m_air "Taking m_vapor = m_air" W3_vmass=P_v/P*m_air "Using wrong relation"

15-146 A room contains air at 30C and a total pressure of 96.0 kPa with a relative humidity of 75 percent. The partial pressure of dry air is (a) 82.0 kPa (b) 85.8 kPa (c) 92.8 kPa (d) 90.6 kPa (e) 72.0 kPa Answer (c) 92.8 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=30 [C] P=96 [kPa] RH=0.75 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH=P_v/P_g P_air=P-P_v "Some Wrong Solutions with Common Mistakes:" W1_Pair=P_v "Using Pv as P_air" W2_Pair=P-P_g "Using wrong relation" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-336

W3_Pair=RH*P "Using wrong relation"

15-147 The air in a house is at 25C and 65 percent relative humidity. Now the air is cooled at constant pressure. The temperature at which the moisture in the air will start condensing is (a) 7.4C (b) 16.3C (c) 18.0C (d) 11.3C (e) 20.2C Answer (c) 18.0C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 [C] RH1=0.65 P_g=PRESSURE(Steam_IAPWS,T=T1,x=0) RH1=P_v/P_g T_dp=TEMPERATURE(Steam_IAPWS,x=0,P=P_v) "Some Wrong Solutions with Common Mistakes:" W1_Tdp=T1*RH1 "Using wrong relation" W2_Tdp=(T1+273)*RH1-273 "Using wrong relation" W3_Tdp=WETBULB(AirH2O,T=T1,P=P1,R=RH1); P1=100 "Using wet-bulb temperature"

15-148 Air is cooled and dehumidified as it flows over the coils of a refrigeration system at 100 kPa from 30C and a humidity ratio of 0.023 kg/kg dry air to 15C and a humidity ratio of 0.015 kg/kg dry air. If the mass flow rate of dry air is

0.4 kg/s , the rate of heat removal from the air is (a) 6 kJ/s (b) 8 kJ/s (c) 11 kJ/s (d) 14 kJ/s (e) 16 kJ/s Answer (d) 14 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=100 [kPa] T1=30 [C] w1=0.023 T2=15 [C] w2=0.015 m_air=0.4 [kg/s] m_water=m_air*(w1-w2) h1=ENTHALPY(AirH2O,T=T1,P=P,w=w1) h2=ENTHALPY(AirH2O,T=T2,P=P,w=w2) h_w=ENTHALPY(Steam_IAPWS,T=T2,x=0) Q=m_air*(h1-h2)-m_water*h_w "Some Wrong Solutions with Common Mistakes:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-337

W1_Q=m_air*(h1-h2) "Ignoring condensed water" W2_Q=m_air*cp_air*(T1-T2)-m_water*h_w; cp_air = 1.005 [kJ/kg-C] "Using dry air enthalpies" W3_Q=m_air*(h1-h2)+m_water*h_w "Using wrong sign"

15-149 Air at a total pressure of 90 kPa, 15°C, and 75 percent relative humidity is heated and humidified to 25°C and 75 percent relative humidity by introducing water vapor. If the mass flow rate of dry air is 4 kg/s , the rate at which steam is added to the air is (a) 0.032 kg/s (b) 0.013 kg/s (c) 0.019 kg/s (d) 0.0079 kg/s (e) 0 kg/s Answer (a) 0.032 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P=90 [kPa] T1=15 [C] RH1=0.75 T2=25 [C] RH2=0.75 m_air=4 [kg/s] w1=HUMRAT(AirH2O,T=T1,P=P,R=RH1) w2=HUMRAT(AirH2O,T=T2,P=P,R=RH2) m_water=m_air*(w2-w1) "Some Wrong Solutions with Common Mistakes:" W1_mv=0 "sine RH = constant" W2_mv=w2-w1 "Ignoring mass flow rate of air" W3_mv=RH1*m_air "Using wrong relation"

15-150 On the psychrometric chart, a cooling and dehumidification process appears as a line that is (a) horizontal to the left, (b) vertical downward, (c) diagonal upwards to the right (NE direction) (d) diagonal upwards to the left (NW direction) (e) diagonal downwards to the left (SW direction) Answer (e) diagonal downwards to the left (SW direction)

15-151 On the psychrometric chart, a heating and humidification process appears as a line that is (a) horizontal to the right, (b) vertical upward, (c) diagonal upwards to the right (NE direction) (d) diagonal upwards to the left (NW direction) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-338

(e) diagonal downwards to the right (SE direction) Answer (c) diagonal upwards to the right (NE direction)

15-152 An air stream at a specified temperature and relative humidity undergoes evaporative cooling by spraying water into it at about the same temperature. The lowest temperature the air stream can be cooled to is (a) the dry bulb temperature at the given state (b) the wet bulb temperature at the given state (c) the dew point temperature at the given state (d) the saturation temperature corresponding to the humidity ratio at the given state (e) the triple point temperature of water Answer (a) the dry bulb temperature at the given state

15-153 … 15-159 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

13-339

Chapter 16 CHEMICAL REACTIONS

13-340

CYU - Check Your Understanding CYU 16-1 ■ Check Your Understanding CYU 16-1.1 Which of the following can be considered a fuel? (a) Nitrogen (b) Carbon monoxide (c) Carbon dioxide (d) Oxygen (e) Water Answer: (b) Carbon monoxide CYU 16-1.2 1 kmol of carbon is burned with 1 kmol of oxygen, forming carbon dioxide. What are the amounts of mole number and mass of carbon dioxide? (a) 1 kmol, 1 kg (b) 1 kmol, 44 kg (c) 44 kmol, 44 kg (d) 2 kmol, 44 kg (e) 2 kmol, 88 kg Answer: (b) 1 kmol, 44 kg

C + O2 = CO2 NCO2 = 1 kmol mCO2 = 12 + (16)(2) = 44kg CYU 16-1.3 1 kmol of hydrogen ( H 2 ) is burned with 30 kg of air. What is the air-fuel ratio for this combustion process? (a) 30 (b) 16 (c) 15 (d) 2 (e) 1 Answer: (c) 15

AF = mair / mfuel = (30 kg) / (2 kg) = 15

CYU 16-2 ■ Check Your Understanding CYU 16-2.1 Propane ( C3 H8 ) is burned completely with excess air. Which of the components below cannot be found in the products? (a) Water (b) Nitrogen (c) Carbon dioxide (d) Carbon monoxide (e) Free oxygen Answer: (d) Carbon monoxide © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-341

CYU 16-2.2 Octane ( C8 H18 ) is burned completely with stoichiometric air. Which of the components below cannot be found in the products? (a) Free oxygen (b) Nitrogen (c) Carbon dioxide (d) Water (e) Octane Answer: (a) Free oxygen CYU 16-2.3 Fuel A is burned with 160 percent excess air while Fuel B is burned with 90 percent theoretical air. An alternative equivalent way of expressing the amounts of combustion air is (a) Fuel A: 160 percent theoretical air, Fuel B: 90 percent stoichiometric air (b) Fuel A: 260 percent theoretical air, Fuel B: 10 percent deficiency of air (c) Fuel A: 260 percent theoretical air, Fuel B: 90 percent excess air (d) Fuel A: 260 percent stoichiometric air, Fuel B: 10 percent excess air (e) Fuel A: 160 percent theoretical air, Fuel B: 10 percent deficiency of air Answer: (b) Fuel A: 260 percent theoretical air, Fuel B: 10 percent deficiency of air CYU 16-2.4 1 kmol of methane ( CH4 ) is burned with stoichiometric amount of air completely. What are the amounts of carbon dioxide ( CO2 ) and water vapor ( H 2 O ) in the products, respectively? (a) 22 kg, 18 kg (b) 44 kg, 18 kg (c) 44 kg, 36 kg (d) 88 kg, 36 kg (e) 1 kg, 2 kg Answer: (c) 44 kg, 36 kg

CH 4 + 2 (O 2 + 3.76N 2 ) = CO 2 + 2H 2 O + (2)(3.76)N 2 mCO2 = 44kg mH2O = (2)(18) = 36kg

CYU 16-3 ■ Check Your Understanding CYU 16-3.1 Select the correct statement(s) below regarding the enthalpy of formation. I. The enthalpy of formation of all stable elements such as O2 and N 2 is zero. II. A negative enthalpy of formation for a compound indicates that heat is released during the formation of that compound from its stable elements. III. Two enthalpy of formation values exist for each compound. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (b) I and II CYU 16-3.2 The relation between the higher and lower heating values (LHV and HHV) of a fuel is (a) LHV + mwater h fg , water = HHV © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-342

(b) LHV + mfuel h fg , water = HHV (c) LHV + mfuel h fg , fuel = HHV (d) LHV + mwater hc = HHV (e) LHV + mfuel hc = HHV Answer: (a) LHV + mwater h fg , water = HHV

CYU 16-3.3 When 1 kmol (16 kg) of methane ( CH4 ) is burned with stoichiometric amount of air completely, 2 kmol (36 kg) of water is produced. What is the difference between the higher and lower heating values of methane, in kJ / kg fuel? (a) 0.5h fg , water (b) h fg , water (c) 2h fg , water (d) (16 / 36)h fg , water (e) (36 / 16)h fg , water Answer: (e) (36 / 16)h fg , water HHV - LHV = mwater h fg / mfuel

HHV - LHV = (36 kg water ) (h fg kJ / kg water ) / (16 kg fuel ) = (36 /16)h fg , water kJ / kg fuel

CYU 16-4 ■ Check Your Understanding CYU 16-4.1 The internal energy of a component in a reaction chamber at the standard reference state (1 atm, 25C) can be expressed in its simplest form as o

(a) hf

(b) h - ho (c) hf + h - h

(d) hf + h - h - Pv (e)

hfo - Pv o

Answer: (e) hf - Pv

CYU 16-4.2 Select the correct energy balance relation(s) for a steady-flow combustion process. (a)

å N (h + h - h ) = Q + å N (h + h - h ) r

o f

out

o f

å N (h - h ) - å N (h - h ) (c) å n (h + h - h ) = Q + å n (h + h - h ) o

(b) Q out = - hC + r

o f

out

o f

(d) All of these Answer: (d) All of the above.

13-343

CYU 16-4.3 Select the correct energy balance relation(s) for a closed-system combustion process.

å N (h + h - h - Pv ) = Q + å N (h + h - h - Pv ) II. Q = - h - Pv + å N (h - h - Pv ) - å N (h - h - Pv ) I.

o f

o C

out

III.

out

o f

å n (h + h - h - Pv ) = Q + å n (h + h - h - Pv ) r

o f

out

o f

(a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III

CYU 16-5 ■ Check Your Understanding CYU 16-5.1 Select the correct statement regarding combustion chamber temperature. (a) The smaller the heat loss from the combustion chamber, the larger the temperature of the products. (b) The greater the amount of air used in the combustion, the larger the temperature of the products. (c) The adiabatic flame temperature does not depend on the temperature of combustion air. (d) All of the above Answer: (a) The smaller the heat loss from the combustion chamber, the larger the temperature of the products. CYU 16-5.2 Select the condition(s) to achieve maximum temperature in a combustion chamber: (a) No dissociation (b) No excess air (c) Complete combustion (d) No heat loss from the combustion chamber (e) All of the above Answer: (e) All of the above

CYU 16-6 ■ Check Your Understanding CYU 16-6.1 Select the incorrect statement regarding the entropy change of reacting systems. (a) Entropy is a function of both temperature and pressure. (b) The gaseous components of the reactants and the products are approximated as ideal gases. (c) When evaluating the entropy of a component of an ideal-gas mixture, we should use the temperature and the partial pressure of the component. (d) The temperature of a component is the same as the mixture temperature. (e) The partial pressure of a component is the same as the mixture pressure. Answer: (e) The partial pressure of a component is the same as the mixture pressure.

13-344

CYU 16-7 ■ Check Your Understanding CYU 16-7.1 Select the incorrect statement below regarding second-law analysis of reacting systems. (a) The difference between the exergy of the reactants and of the products during a chemical reaction is the reversible work associated with that reaction. (b) An adiabatic combustion process involves no exergy destruction. (c) The Gibbs function of formation at 25C and 1 atm for stable elements is zero. (d) During a combustion process with no actual work, the reversible work and exergy destroyed are identical. (e) The exergy content of a fuel decreases considerably as a result of the irreversible combustion process. Answer: (b) An adiabatic combustion process involves no exergy destruction. CYU 16-7.2 Consider the complete combustion reaction C + O2 ® CO2 with each of the reactants and the products at the state of the environment as a result of losing all the heat produced during combustion to the surrounding air. By considering the chemical exergy as well as the thermo-mechanical exergy, select the incorrect statement below. (a) The exergy of the reactants is zero. (b) The exergy of the products is zero. (c) The work potential of carbon is equal to the exergy destruction for this process. (d) The second-law efficiency of this process is zero. (e) The heat lost from the combustion chamber is equal to the heating value of carbon. Answer: (a) The exergy of the reactants is zero.

13-345

PROBLEMS Fuels and Combustion 16-1C Gasoline is C8 H18 , diesel fuel is C12 H26 , and natural gas is CH4 .

16-2C Nitrogen, in general, does not react with other chemical species during a combustion process but its presence affects the outcome of the process because nitrogen absorbs a large proportion of the heat released during the chemical process.

16-3C The number of atoms are preserved during a chemical reaction, but the total mole numbers are not.

16-4C Air-fuel ratio is the ratio of the mass of air to the mass of fuel during a combustion process. Fuel-air ratio is the inverse of the air-fuel ratio.

16-5C No. Because the molar mass of the fuel and the molar mass of the air, in general, are different.

16-6C Moisture, in general, does not react chemically with any of the species present in the combustion chamber, but it absorbs some of the energy released during combustion, and it raises the dew point temperature of the combustion gases.

16-7C The dew-point temperature of the product gases is the temperature at which the water vapor in the product gases starts to condense as the gases are cooled at constant pressure. It is the saturation temperature corresponding to the vapor pressure of the product gases.

16-8 Sulfur is burned with oxygen to form sulfur dioxide. The minimum mass of oxygen required and the mass of sulfur dioxide in the products are to be determined when 1 kg of sulfur is burned. Properties The molar masses of sulfur and oxygen are 32.06 kg / kmol and 32.00 kg / kmol , respectively (Table A-1). Analysis The chemical reaction is given by

S + O2 ¾ ¾® SO2

Hence, 1 kmol of oxygen is required to burn 1 kmol of sulfur which produces 1 kmol of sulfur dioxide whose molecular weight is

13-346

mO2 N M (1 kmol)(32 kg/kmol) = O2 O2 = = 0.998 kg O2 / kg S mS NS M S (1 kmol)(32.06 kg/kmol)

and mSO2 N M (1 kmol)(64.06 kg/kmol) = SO2 SO2 = = 1.998 kg SO2 / kg S mS NS M S (1 kmol)(32.06 kg/kmol)

EES (Engineering Equation Solver) SOLUTION "Given" m_S=1 [kg] "Analysis" "Chemical reaction: S+O2 = SO2" N_O2=1 [kmol] N_S=1 [kmol] N_SO2=1 [kmol] MM_S=32.06 [kg/kmol] MM_O2=32 [kg/kmol] MM_SO2=MM_S+MM_O2 m_O2\m_S=(N_O2*MM_O2)/(N_S*MM_S) m_SO2\m_S=(N_SO2*MM_SO2)/(N_S*MM_S)

16-9E Methane is burned with diatomic oxygen. The mass of water vapor in the products is to be determined when 1 lbm of methane is burned. Properties The molar masses of CH4 , O2 , CO2 , and H 2 O are 16, 32, 44, and 18 lbm / lbmol , respectively (Table A-1E).

Analysis The chemical reaction is given by

CH4 + 2O2 ¾ ¾® CO2 + 2H 2 O Hence, for each lbmol of methane burned, 2 lbmol of water vapor are formed. Then, mH2O N M (2 lbmol)(18 lbm/lbmol) = H2O H2O = = 2.25 lbm H2 O / lbm CH4 mCH4 N CH4 M CH4 (1 lbmol)(16 lbm/lbmol)

EES (Engineering Equation Solver) SOLUTION "Given" m_CH4=1 [lbm] "Properties" MM_CH4=16 [lbm/lbmol] MM_O2=32 [lbm/lbmol] MM_CO2=44 [lbm/lbmol] MM_H2O=18 [lbm/lbmol] "Analysis" "Chemical reaction: CH4 + 2O2 = CO2 + 2H2O" N_H2O=2 [kmol] N_CH4=1 [kmol] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-347

m_H2O\m_CH4=(N_H2O*MM_H2O)/(N_CH4*MM_CH4)

Theoretical and Actual Combustion Processes 16-10C No. The theoretical combustion is also complete, but the products of theoretical combustion does not contain any uncombined oxygen.

16-11C It represent the amount of air that contains the exact amount of oxygen needed for complete combustion. 16-12C Case (b).

16-13C The causes of incomplete combustion are insufficient time, insufficient oxygen, insufficient mixing, and dissociation.

16-14C CO. Because oxygen is more strongly attracted to hydrogen than it is to carbon, and hydrogen is usually burned to completion even when there is a deficiency of oxygen.

16-15 Methane is burned with the stoichiometric amount of air during a combustion process. The AF and FA ratios are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 only.

Properties The molar masses of C, N 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A1). Analysis This is a theoretical combustion process since methane is burned completely with stoichiometric amount of air. The stoichiometric combustion equation of CH4 is

CH 4 + ath [O2 + 3.76N 2 ] ¾ ¾® CO 2 + 2H 2 O + 3.76ath N 2

O2 balance:

ath = 1 + 1 ¾ ¾® ath = 2

Substituting,

CH 4 + 2[O2 + 3.76N 2 ] ¾ ¾® CO2 + 2H 2 O + 7.52N 2 The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

(2´ 4.76 kmol)(29 kg/kmol) mair = = 17.3 kg air / kg fuel mfuel (1 kmol)(12 kg/kmol)+ (2 kmol)(2 kg/kmol)

13-348

FA =

1 1 = = 0.0578 kg fuel / kg air AF 17.3 kg air/kg fuel

EES (Engineering Equation Solver) SOLUTION "Given" "Methane (CH4) is burned with stoichiometric amount of air Stoichiometric reaction is: CH4 + a_th (O2+3.76 N2) = CO2 + 2 H2O + 3.76*a_th N2" a_th=2 "[kmol], O2 balance" "Properties" MM_air=molarmass(air) MM_CH4=molarmass(CH4) "Analysis" m_air=N_air*MM_air m_fuel=N_fuel*MM_CH4 N_air=4.76*a_th N_fuel=1 [kmol] AF=m_air/m_fuel FA=1/AF

16-16 Acetylene is burned with the stoichiometric amount of air during a combustion process. The AF ratio is to be determined on a mass and on a mole basis. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 only.

Properties The molar masses of C, H 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A1). Analysis This is a theoretical combustion process since C2 H 2 is burned completely with stoichiometric amount of air. The stoichiometric combustion equation of C2 H 2 is

C2 H 2 + ath [O2 + 3.76N 2 ] ¾ ¾® 2CO 2 + H 2 O + 3.76ath N 2

ath = 2 + 0.5 ® ath = 2.5

O2 balance:

Substituting,

C2 H 2 + 2.5[O2 + 3.76N 2 ] ¾ ¾® 2CO2 + H 2 O + 9.4N 2 The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

(2.5´ 4.76 kmol)(29 kg/kmol) mair = = 13.3 kg air / kg fuel mfuel (2 kmol)(12 kg/kmol)+ (1 kmol)(2 kg/kmol)

On a mole basis, the air-fuel ratio is expressed as the ratio of the mole numbers of the air to the mole numbers of the fuel, AFmole basis =

N air (2.5´ 4.76) kmol = = 11.9 kmol air / kmol fuel N fuel 1 kmol fuel

13-349

EES (Engineering Equation Solver) SOLUTION "Given" "Acetylene (C2H2) is burned with stoichiometric air The stoichiometric combustion equation is: C2H2 + a_th (O2+3.76N2) = 2 CO2 + H2O + 3.76* a_th N2" a_th=2+0.5 "O2 balance" "Properties" MM_air=molarmass(air) MM_C2H2=molarmass(C2H2) "Analysis" m_air=N_air*MM_air m_fuel=N_fuel*MM_C2H2 N_air=4.76*a_th N_fuel=1 "[kmol]" AF=m_air/m_fuel AF_bar_molar=N_air/N_fuel

16-17 n-butane is burned with stoichiometric amount of air. The mass fraction of each product, the mass of CO2 and air per unit mass of fuel burned are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The reaction equation for 100% theoretical air is

C4 H10 + ath [O 2 + 3.76N 2 ] ¾ ¾®

B CO 2 + D H 2 O + E N 2

where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance: B=4 Hydrogen balance: 2 D = 10 ¾ ¾ ® D= 5 Oxygen balance:

2ath = 2B + D ¾ ¾® ath = 0.5(2´ 4 + 5) = 6.5

Nitrogen balance:

ath ´ 3.76 = E ¾ ¾® E = 6.5´ 3.76 = 24.44

Substituting, the balanced reaction equation is

C4 H10 + 6.5[O 2 + 3.76N 2 ] ¾ ¾® 4 CO 2 + 5 H 2 O + 24.44 N 2 The mass of each product and the total mass are

mCO2 = NCO2 M CO2 = (4 kmol)(44 kg/kmol) = 176 kg mH2O = NH2O M H2O = (5 kmol)(18 kg/kmol) = 90 kg © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-350

mN2 = N N2 M N2 = (24.44 kmol)(28 kg/kmol) = 684 kg mtotal = mCO2 + mH2O + mN2 = 176 + 90 + 684 = 950 kg Then the mass fractions are mf CO2 =

mCO2 176 kg = = 0.1853 mtotal 950 kg

mf H2O =

mH2O 90 kg = = 0.0947 mtotal 950 kg

mf O2 =

mO2 684 kg = = 0.720 B mtotal 950 kg

The mass of carbon dioxide per unit mass of fuel burned is mCO2 (4´ 44) kg = = 3.034 kg CO 2 / kg C4 H10 mC4H10 (1´ 58) kg

The mass of air required per unit mass of fuel burned is mair (6.5´ 4.76´ 29) kg = = 15.47 kg air / kg C4 H10 mC4H10 (1´ 58) kg

16-18 n-Octane is burned with stoichiometric amount of oxygen. The mass fractions of each of the products and the mass of water in the products per unit mass of fuel burned are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 and H 2 O .

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , and O2 are 12 kg / kmol , 2 kg / kmol , and 32 kg / kmol , respectively (Table A1). Analysis The combustion equation in this case is

C8 H18 + 12.5O2 ¾ ¾® 8CO2 + 9H 2 O

The mass of each product and the total mass are

mCO2 = NCO2 M CO2 = (8 kmol)(44 kg/kmol) = 352 kg mH2O = NH2O M H2O = (9 kmol)(18 kg/kmol) = 162 kg

mtotal = mCO2 + mH2O = 352 + 162 = 514 kg Then the mass fractions are mf CO2 =

mCO2 352 kg = = 0.6848 mtotal 514 kg

13-351

mf H2O =

mH2O 162 kg = = 0.3152 mtotal 514 kg

The mass of water in the products per unit mass of fuel burned is determined from mH2O (9´ 18) kg = = 1.42 kg H 2 O / kg C8 H18 mC8H18 (1´ 114) kg

16-19 Propane is burned with 75 percent excess air during a combustion process. The AF ratio is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A1). Analysis The combustion equation in this case can be written as

C3 H8 + 1.75ath [O2 + 3.76N 2 ] ¾ ¾® 3CO 2 + 4H 2 O + 0.75ath O 2 + (1.75´ 3.76) ath N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 75% excess air by using the factor 1.75ath instead of ath for air. The stoichiometric amount of oxygen ( ath O2 ) will be used to oxidize the fuel, and the remaining excess amount ( 0.75 ath O2 ) will appear in the products as free oxygen. The coefficient ath is determined from the

O2 balance,

O2 balance:

1.75ath = 3 + 2 + 0.75ath ® ath = 5

Substituting,

C3 H8 + 8.75[O 2 + 3.76N 2 ]¾ ¾® 3CO2 + 4H 2 O + 3.75O 2 + 32.9N 2 The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

(8.75´ 4.76 kmol)(29 kg/kmol) mair = = 27.5 kg air / kg fuel mfuel (3 kmol)(12 kg/kmol)+ (4 kmol)(2 kg/kmol)

EES (Engineering Equation Solver) SOLUTION "Given" "Propane (C3H8) is burned with 75% excess air" ex=1.75 "The combustion equation is: C3H8 + ex*a_th (O2+3.76N2) = 3 CO2 + 4 H2O + (ex-1)* a_th O2 + ex*3.76* a_th N2" ex*a_th=3+2+(ex-1)*a_th "O2 balance" "Properties" MM_air=molarmass(air) MM_C3H8=molarmass(C3H8) "Analysis"

13-352

AF=m_air/m_fuel m_air=N_air*MM_air m_fuel=N_fuel*MM_C3H8 N_air=4.76*ex*a_th N_fuel=1 "[kmol]"

16-20 Ethane is burned with air steadily. The mass flow rates of ethane and air are given. The percentage of excess air used is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 only.

Properties The molar masses of C, H 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A1). Analysis The theoretical combustion equation in this case can be written as

C2 H 6 + ath [O2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3H 2 O + 3.76ath N 2 where ath is the stoichiometric coefficient for air. It is determined from

O2 balance: ath = 2 + 1.5 ¾ ¾® ath = 3.5 The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the mass of the fuel for, AFth =

mair,th mfuel

(3.5´ 4.76 kmol)(29 kg/kmol) = 16.1 kg air/kg fuel (2 kmol)(12 kg/kmol)+ (3 kmol)(2 kg/kmol)

The actual air-fuel ratio used is AFact =

mair 176 kg/h = = 22 kg air/kg fuel mfuel 8 kg/h

Then the percent theoretical air used can be determined from Percent theoretical air =

AFact 22 kg air/kg fuel = = 137% AFth 16.1 kg air/kg fuel

Thus the excess air used during this process is 37%.

EES (Engineering Equation Solver) SOLUTION "Given" "Ehane (C2H6) is burned with air The combustion equation is: C2H6 + a_th (O2+3.76N2) = 2 CO2 + 3 H2O + 3.76*a_th N2" a_th=2+1.5 "O2 balance" m_dot_fuel=8 [kg/h] m_dot_air=176 [kg/h] "Properties" MM_air=molarmass(air) MM_C2H6=molarmass(C2H6) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-353

"Analysis" m_air=N_air*MM_air m_fuel=N_fuel*MM_C2H6 N_air=4.76*a_th N_fuel=1 [kmol] AF_th=m_air/m_fuel AF_act=m_dot_air/m_dot_fuel PercentTheorAir=AF_act/AF_th*Convert(, %) PecentExcessAir=PercentTheorAir-100

16-21 Methyl alcohol is burned with stoichiometric amount of air. The mole fraction of each product, the apparent molar mass of the product gas, and the mass of water in the products per unit mass of fuel burned are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 only.

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The balanced reaction equation for stoichiometric air is

CH3OH + 1.5[O 2 + 3.76N 2 ] ¾ ¾® CO 2 + 2 H 2 O + 5.64 N 2 The mole fractions of the products are

Nm = 1+ 2 + 5.64 = 8.64 kmol yCO2 =

N CO2 1 kmol = = 0.1157 Nm 8.64 kmol

yH2O =

N H2O 2 kmol = = 0.2315 Nm 8.64 kmol

yN2 =

N N2 5.64 kmol = = 0.6528 Nm 8.64 kmol

The apparent molecular weight of the product gas is Mm =

mm (1´ 44 + 2´ 18 + 5.64´ 28) kg = = 27.54 kg / kmol Nm 8.64 kmol

The mass of water in the products per unit mass of fuel burned is mH2O (2´ 18) kg = = 1.125 kg H 2O / kg CH 3OH mCH3OH (1´ 32) kg

16-22E Ethylene is burned with 175 percent theoretical air during a combustion process. The AF ratio and the dew-point temperature of the products are to be determined. Assumptions 1. Combustion is complete. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-354

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , and air are 12 lbm / lbmol , 2 lbm / lbmol , and 29 lbm / lbmol , respectively (Table A-1E). Analysis (a) The combustion equation in this case can be written as

C2 H 4 + 1.75ath [O 2 + 3.76N 2 ] ¾ ¾® 2CO 2 + 2H 2O + 0.75ath O 2 + (1.75´ 3.76)ath N 2 where ath is the stoichiometric coefficient for air. It is determined from

O2 balance: 1.75ath = 2 + 1 + 0.75ath ¾ ¾® ath = 3 Substituting,

C2 H 4 + 5.25[O 2 + 3.76N 2 ] ¾ ¾® 2CO 2 + 2H 2 O + 2.25O 2 + 19.74N 2 The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

(5.25´ 4.76 lbmol)(29 lbm/lbmol) mair = = 25.9 lbm air / lbm fuel mfuel (2 lbmol)(12 lbm/lbmol)+ (2 lbmol)(2 lbm/lbmol)

(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is,

æN ÷ ö æ 2 lbmol ö÷ Pv = ççç v ÷ Pprod = çç ÷ ÷(14.5 psia ) = 1.116 psia ÷ çè 25.99 lbmol ø÷ èç N prod ø÷ Thus, Tdp = Tsat@1.116 psia = 105.4°F

EES (Engineering Equation Solver) SOLUTION "Given" "Ethylene (C2H4) is burned with 175% theoretical air" ex=1.75 "The combustion equation is: C2H4 + ex*a_th (O2+3.76N2) = 2 CO2 + 2 H2O + (ex-1)* a_th O2 + ex*3.76* a_th N2" ex*a_th=2+1+(ex-1)*a_th "O2 balance" P_prod=14.5 [psia] "Properties" MM_air=29 [lbm/lbmol] MM_C2H4=28 [lbm/lbmol] "Analysis" "(a)" AF=m_air/m_fuel m_air=N_air*MM_air m_fuel=N_fuel*MM_C2H4 N_air=4.76*ex*a_th N_fuel=1 [lbmol] "(b)" P_v=(N_v/N_prod)*P_prod © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-355

N_v=2 [lbmol] "number of H2O in the products" N_prod=2+2+(ex-1)*a_th+ex*3.76*a_th T_dp=temperature(steam_iapws, P=P_v, x=1)

16-23 The fuel is burned completely with excess air. The AF and the dew point of the products are to be determined. Assumptions 1. Combustion is complete. 2. Combustion gases are ideal gases. Properties The molar masses of C, H 2 , and air are 12 lbm / lbmol , 2 lbm / lbmol , and 29 lbm / lbmol , respectively (Table A-1E). Analysis The combustion products contain CO2 , H 2 O , N 2 , and some excess O2 only. Then the combustion equation can be written as

C2 H 6 + 1.2ath [O 2 + 3.76N 2 ] ¾ ¾® 2CO 2 + 3H 2 O + 0.2ath O 2 + (1.2´ 3.76)ath N 2

where ath is the stoichiometric coefficient for air. We have automatically accounted for the 20 percent excess air by using the factor 1.2 ath instead of ath for air. The stoichiometric amount of oxygen ( ath O2 ) is used to oxidize the fuel, and the remaining excess amount ( 0.2 ath O2 ) appears in the products as unused oxygen. Notice that the coefficient of N 2 is the same on both sides of the equation, and that we wrote the C and H balances directly since they are so obvious. The coefficient ath is determined from the O2 balance to be

O2 balance: 1.2ath = 2 + 1.5 + 0.2ath ¾ ¾® ath = 3.5 Substituting,

C2 H 6 + 4.2[O 2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3H 2 O + 0.7O 2 + 15.79N 2 (a) The air–fuel ratio is determined from Eq. 15–3 by taking the ratio of the mass of the air to the mass of the fuel, AF =

(4.2´ 4.76 kmol)(29 kg/kmol) mair = = 19.3 kg air / kg fuel mfuel (2 kmol)(12 kg/kmol)+ (3 kmol)(2 kg/kmol)

That is, 19.3 kg of air is supplied for each kilogram of fuel during this combustion process. (b) The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled at constant pressure. Recall from Chap. 14 that the dew-point temperature of a gas– vapor mixture is the saturation temperature of the water vapor corresponding to its partial pressure. Therefore, we need to determine the partial pressure of the water vapor Pv in the products first. Assuming ideal-gas behavior for the combustion gases, we have

æN ÷ ö æ 3 kmol ö ÷ Pv = ççç v ÷ Pprod = çç ÷ ÷ ÷(100 kPa ) = 13.96 kPa çè 21.49 kmol ø çè N prod ÷ ÷ ø Thus,

Tdp = Tsat@13.96 kPa = 52.3°C

(Table A-5)

13-356

16-24 Octane is burned with 250 percent theoretical air during a combustion process. The AF ratio and the dew-pint temperature of the products are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A1). Analysis (a) The combustion equation in this case can be written as

C8 H18 + 2.5ath [O 2 + 3.76N 2 ]¾ ¾® 8CO2 + 9H 2 O + 1.5ath O 2 + (2.5´ 3.76)ath N 2 where ath is the stoichiometric coefficient for air. It is determined from

O2 balance: 2.5ath = 8 + 4.5 + 1.5ath

¾ ¾®

ath = 12.5

Substituting,

C8 H18 + 31.25[O 2 + 3.76N 2 ] ® 8CO2 + 9H 2 O + 18.75O 2 + 117.5N 2 Thus, AF =

(31.25´ 4.76 kmol)(29 kg/kmol) mair = = 37.8 kg air / kg fuel mfuel (8 kmol)(12 kg/kmol)+ (9 kmol)(2 kg/kmol)

(b) The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is,

æ N ö÷ æ 9 kmol ö÷ Pv = ççç v ÷ Pprod = çç (101.325 kPa ) = 5.951 kPa ÷ ÷ ÷ çè153.25 kmol ÷ ø çè N prod ø÷ Thus, Tdp = Tsat @ 5.951 kPa = 36.0°C

EES (Engineering Equation Solver) SOLUTION "Given" "Octane (C8H18) is burned with 250% theoretical air" ex=2.5 "The combustion equation is: C8H18 + ex*a_th (O2+3.76N2) = 8 CO2 + 9 H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=8+4.5+(ex-1)*a_th "O2 balance" P_prod=101.325 "[kPa]" "Properties" MM_air=molarmass(air) MM_C8H18=molarmass(C8H18) "Analysis" "(a)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-357

AF=m_air/m_fuel m_air=N_air*MM_air m_fuel=N_fuel*MM_C8H18 N_air=4.76*ex*a_th N_fuel=1 "[kmol]" "(b)" P_v=(N_v/N_prod)*P_prod N_v=9 "[kmol], number of H2O in the products" N_prod=8+9+(ex-1)*a_th+ex*3.76*a_th T_dp=temperature(steam_iapws, P=P_v, x=1)

16-25 Butane C4 H10 is burned with 200 percent theoretical air. The kmol of water that needs to be sprayed into the combustion chamber per kmol of fuel is to be determined. Assumptions 1.

Combustion is complete.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The reaction equation for 200% theoretical air without the additional water is

C4 H10 + 2ath [O 2 + 3.76N 2 ] ¾ ¾® B CO 2 + D H 2 O + E O 2 + F N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 100% excess air by using the factor 2 ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances

Carbon balance:

B=4

Hydrogen balance:

2 D = 10 ¾ ¾ ® D= 5

Oxygen balance:

2´ 2ath = 2B + D + 2E

ath = E Nitrogen balance:

2ath ´ 3.76 = F

Solving the above equations, we find the coefficients (E = 6.5, F = 48.88, and ath = 6.5 ) and write the balanced reaction equation as

C4 H10 + 13[O 2 + 3.76N 2 ] ¾ ¾® 4 CO 2 + 5 H 2 O + 6.5 O 2 + 48.88 N 2 With the additional water sprayed into the combustion chamber, the balanced reaction equation is

C4 H10 + 13[O 2 + 3.76N 2 ]+ N v H 2 O ¾ ¾® 4 CO 2 + (5 + N v ) H 2 O + 6.5 O 2 + 48.88 N 2 The partial pressure of water in the saturated product mixture at the dew point is

13-358

yv =

Pv ,prod Pprod

19.95 kPa = 0.1995 100 kPa

The amount of water that needs to be sprayed into the combustion chamber can be determined from yv =

N water 5 + Nv ¾¾ ® 0.1995 = ¾¾ ® Nv = 9.80 kmol N total,product 4 + 5 + Nv + 6.5 + 48.88

EES (Engineering Equation Solver) SOLUTION "The reaction equation for 200% theroetical air without the additional water is:" "C4H10 + 2*A_th (O2 +3.76N2)=B CO2 + D H2O + E O2 + F N2" "Carbon Balance:" 4=B "Hydrogen Balance:" 10=2*D "Oxygen Balance:" 2*A_th*2=B*2 + D + E*2 1*A_th = E "Nitrogen Balance:" 2*A_th*3.76 = F "With additional water (N_v) sprayed into the combustion chamber, the balanced reaction equation is:" "C4H10 + 2*A_th (O2 +3.76N2) + N_v H2O =B CO2 + (D+N_v) H2O + E O2 + F N2" "The product dew point temperature is:" T_DP = 60 [C] "The product pressure is:" P_prod = 100 [kPa] "The partial pressure of the water in the saturated product mixture at the dew point is:" P_v_prod = pressure(steam_iapws, T=T_DP, x=1) P_v_prod = y_v * P_prod "The additional moles of water required are determined from the vapor mole fraction in the products:" y_v = (D+N_v)/(B + D+N_v+ E + F)

16-26 A fuel mixture of 60% by mass methane, CH4 , and 40% by mass ethanol, C2 H6 O , is burned completely with theoretical air. The required flow rate of air is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H2 O , and N 2 only.

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1).

Analysis For 100 kg of fuel mixture, the mole numbers are N CH4 =

mf CH4 60 kg = = 3.75 kmol M CH4 16 kg/kmol

13-359

N C2H6O =

mf C2H6O 40 kg = = 0.8696 kmol M C2H6O 46 kg/kmol

The mole fraction of methane and ethanol in the fuel mixture are x=

N CH4 3.75 kmol = = 0.8118 N CH4 + N C2H6O (3.75 + 0.8696) kmol

N C2H6O 0.8696 kmol = = 0.1882 N CH4 + N C2H6O (3.75 + 0.8696) kmol

The combustion equation in this case can be written as

x CH 4 + y C2 H 6 O + ath [O 2 + 3.76N 2 ] ¾ ¾® B CO 2 + D H 2 O + F N 2 where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance: x + 2y = B Hydrogen balance:

4 x + 6 y = 2D

Oxygen balance:

2ath + y = 2B + D

Nitrogen balance:

3.76ath = F

Substituting x and y values into the equations and solving, we find the coefficients as

x = 0.8118

B = 1.188

y = 0.1882

D = 2.188

ath = 2.188

F = 8.228

Then, we write the balanced reaction equation as

0.8118 CH 4 + 0.1882 C 2 H 6O + 2.188 [O 2 + 3.76N 2 ] ¾ ¾® 1.188 CO 2 + 2.188 H 2 O + 8.228 N 2 The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

mair (2.188´ 4.76 kmol)(29 kg/kmol) = mfuel (0.8118 kmol)(12 + 4 ´ 1)kg/kmol + (0.1882 kmol)(2´ 12 + 6´ 1 + 16)kg/kmol

= 13.94 kg air/kg fuel Then, the required flow rate of air becomes

mair = AFmfuel = (13.94)(10 kg/s) = 139.4 kg / s

EES (Engineering Equation Solver) SOLUTION mf_CH4=0.6 mf_C2H6O=0.4 m_dot_fuel = 10 [kg/s] "For 100 kg of fuel, the moles of fuel are:" MM_CH4=1*12+4*1 N_CH4 = mf_CH4*100/MM_CH4 MM_C2H6O=2*12+6*1+16 N_C2H6O = mf_C2H6O*100/ MM_C2H6O "x is the mole fraction of the methane in the fuel mixture." "y is the mole fraction of the ethanol in the fuel mixture." x=N_CH4/(N_CH4+N_C2H6O) y=N_C2H6O/(N_CH4+N_C2H6O) "The reaction equation for theoretical air is:" "x CH4 + y C2H6O+ A_th (O2 +3.76N2)=B CO2 + D H2O + F N2" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-360

"Carbon Balance:" x + 2*y =B "Hydrogen Balance:" x*4 + y*6 =2*D "Oxygen Balance:" A_th*2+y=B*2 + D "Nitrogen Balance:" A_th*3.76 = F "Air-fuel ratio:" MM_air=molarmass(air) MM_fuel=x*MM_CH4+y*MM_C2H6O AF = A_th*4.76*MM_air/(MM_fuel) "[kg-air/kg-fuel]" "The mass flow rate of fuel is:" m_dot_air= AF*m_dot_fuel

16-27 Ethane is burned with an unknown amount of air during a combustion process. The AF ratio and the percentage of theoretical air used are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis (a) The combustion equation in this case can be written as

C2 H 6 + a [O2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3H 2 O + 3O 2 + 3.76aN 2 O2 balance:

a = 2 + 1.5 + 3 ¾ ¾ ® a = 6.5

Substituting,

C2 H 6 + 6.5[O 2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3H 2 O + 3O2 + 24.44N 2 The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

(6.5´ 4.76 kmol)(29 kg/kmol) mair = = 29.9 kg air / kg fuel mfuel (2 kmol)(12 kg/kmol)+ (3 kmol)(2 kg/kmol)

(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of C2 H6 ,

C2 H 6 + ath [O2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3H 2 O + 3.76ath N 2

O2 balance: ath = 2 + 1.5 ¾ ¾® ath = 3.5 Then, Percent theoretical air =

mair,act mair,th

N air,act N air,th

a 6.5 = = 186% ath 3.5

13-361

EES (Engineering Equation Solver) SOLUTION "Given" "One kmol of ethane (C2H6) is burned with an unknown amount of air" "The combustion equation is: C2H6 + a (O2+3.76N2) = 2 CO2 + 3 H2O + 2 O2 + 3.76*a N2" a=2+1.5+3 "O2 balance" "Properties" MM_air=molarmass(air) MM_C2H6=molarmass(C2H6) "Analysis" "(a)" AF=m_air/m_fuel m_air=N_air*MM_air m_fuel=N_fuel*MM_C2H6 N_air=4.76*a N_fuel=1 "[kmol]" "(b)" "The stoichiometric combustion equation is: C2H6 + a_th (O2+3.76N2) = 2 CO2 + 3 H2O + 3.76*a_th N2" a_th=2+1.5 "O2 balance" PercentTheorAir=(a/a_th)*Convert(, %)

16-28 The volumetric fractions of the constituents of a certain natural gas are given. The AF ratio is to be determined if this gas is burned with the stoichiometric amount of dry air. Assumptions 1.

Combustion is complete.

The combustion products contain CO2 , H 2 O , and N 2 only.

Properties The molar masses of C, H 2 , N 2 , O2 , and air are 12 kg / kmol , 2 kg / kmol , 28 kg / kmol , 32 kg / kmol , and

29 kg / kmol , respectively (Table A-1). Analysis Considering 1 kmol of fuel, the combustion equation can be written as

(0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O2 + 0.06CO 2 ) + ath (O 2 + 3.76N 2 ) ¾ ¾® xCO 2 + yH 2 O + zN2 The unknown coefficients in the above equation are determined from mass balances,

C : 0.65 + 0.06 = x ¾ ¾ ® x = 0.71

H : 0.65´ 4 + 0.08´ 2 = 2 y ¾ ¾® y = 1.38 O2 : 0.03 + 0.06 + ath = x + y / 2 ¾ ¾® ath = 1.31 N 2 : 0.18 + 3.76ath = z ¾ ¾ ® z = 5.106

Thus,

(0.65CH4 + 0.08H 2 + 0.18N 2 + 0.03O2 + 0.06CO2 ) + 1.31(O2 + 3.76N 2 ) ¾ ¾® 0.71CO2 + 1.38H2 O + 5.106N2 The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, mair = (1.31´ 4.76 kmol)(29 kg/kmol) = 180.8 kg

13-362

mfuel = (0.65´ 16 + 0.08´ 2 + 0.18´ 28 + 0.03´ 32 + 0.06´ 44)kg = 19.2 kg and AFth =

mair,th mfuel

180.8 kg = 9.42 kg air / kg fuel 19.2 kg

EES (Engineering Equation Solver) SOLUTION "Given" "Volumetric analysis of a certain natural gas is given." "Properties" MM_air=29 [kg/kmol] MM_CH4=16 [kg/kmol] MM_H2=2 [kg/kmol] MM_N2=28 [kg/kmol] MM_O2=32 [kg/kmol] MM_CO2=44 [kg/kmol] "Analysis" "The combustion equation is: 0.65 CH4 + 0.08 H2 + 0.18 N2 + 0.03 O2 +0.06 CO2 +a_th (O2+3.76N2) = x CO2 + y H2O + z N2" 0.65+0.06=x "C balance" 0.65*4+0.08*2=2*y "H balance" 0.03+0.06+a_th=x+y/2 "O2 balance" 0.18+3.76*a_th=z "N2 balance" m_air=a_th*4.76*MM_air m_fuel=0.65*MM_CH4+0.08*MM_H2+0.18*MM_N2+0.03*MM_O2+0.06*MM_CO2 AF=m_air/m_fuel

16-29 The composition of a certain natural gas is given. The gas is burned with stoichiometric amount of moist air. The AF ratio is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 only.

Properties The molar masses of C, H 2 , N 2 , O2 , and air are 12 kg / kmol , 2 kg / kmol , 28 kg / kmol , 32 kg / kmol , and

29 kg / kmol , respectively (Table A-1). Analysis The fuel is burned completely with the stoichiometric amount of air, and thus the products will contain only H 2 O ,

CO2 and N 2 , but no free O2 . The moisture in the air does not react with anything; it simply shows up as additional H2 O in the products. Therefore, we can simply balance the combustion equation using dry air, and then add the moisture to both sides of the equation. Considering 1 kmol of fuel, the combustion equation can be written as (0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O2 + 0.06CO2 ) + ath (O2 + 3.76N 2 ) ¾ ¾® xCO 2 + yH 2 O + zN2 The unknown coefficients in the above equation are determined from mass balances,

13-363

H : 0.65´ 4 + 0.08´ 2 = 2 y ¾ ¾ ® y = 1.38

O2 : 0.03 + 0.06 + ath = x + y / 2 ¾ ¾® ath = 1.31 N2 : 0.18 + 3.76ath = z ¾ ¾® z = 5.106 Thus,

(0.65CH4 + 0.08H2 + 0.18N2 + 0.03O2 + 0.06CO2 ) + 1.31(O2 + 3.76N2 ) ¾ ¾® 0.71CO2 + 1.38H2 O + 51.06N2 Next we determine the amount of moisture that accompanies 4.76ath = (4.76)(1.31) = 6.24 kmol of dry air. The partial pressure of the moisture in the air is

Pv,in = f air Psat @ 25°C = (0.85)(3.1698 kPa) = 2.694 kPa Assuming ideal gas behavior, the number of moles of the moisture in the air ( N v, in ) is determined to be æP ö æ 2.694 kPa ÷ ö ÷ N v ,in = ççç v ,in ÷ ÷ (6.24 + N v,in ) ¾ ¾® N v,air = 0.1703 kmol ÷N total = çççè ÷ çè Ptotal ÷ 101.325 kPa ø ø

The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.1703 kmol of H 2 O to both sides of the equation, (0.65CH 4 + 0.08H 2 + 0.18N 2 + 0.03O2 + 0.06CO2 ) + 1.31(O2 + 3.76N 2 ) + 0.1703H2 O) ¾ ¾ ® 0.71CO2 + 1.550H2 O + 51.06N 2

The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, mair = (1.31´ 4.76 kmol)(29 kg/kmol)+ (0.1703 kmol ´ 18 kg/kmol) = 183.9 kg

mfuel = (0.65´ 16 + 0.08´ 2 + 0.18´ 28 + 0.03´ 32 + 0.06´ 44)kg = 19.20 kg and AFth =

mair,th mfuel

183.9 kg = 9.58 kg air / kg fuel 19.20 kg

EES (Engineering Equation Solver) SOLUTION "Given" "Volumetric analysis of a certain natural gas is given." T_air=25 [C] P_air=101.325 [kPa] phi_air=0.85 "Properties" MM_air=29 [kg/kmol] MM_CH4=16 [kg/kmol] MM_H2=2 [kg/kmol] MM_N2=28 [kg/kmol] MM_O2=32 [kg/kmol] MM_CO2=44 [kg/kmol] MM_H2O=18 [kg/kmol] "Analysis" "The combustion equation is: 0.65 CH4 + 0.08 H2 + 0.18 N2 + 0.03 O2 +0.06 CO2 +a_th (O2+3.76N2) = x CO2 + y H2O + z N2" 0.65+0.06=x "C balance" 0.65*4+0.08*2=2*y "H balance" 0.03+0.06+a_th=x+y/2 "O2 balance" 0.18+3.76*a_th=z "N2 balance" P_sat=pressure(steam_iapws, T=T_air, x=1) P_v_in=phi_air*P_sat N_v_in=(P_v_in/P_air)*N_total © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-364

N_total=4.76*a_th+N_v_in "Then the balanced combustion equation is: : 0.65 CH4 + 0.08 H2 + 0.18 N2 + 0.03 O2 +0.06 CO2 + a_th (O2+3.76N2) + N_v_in H2O = x CO2 + (y+N_v_in) H2O + z N2" m_air=a_th*4.76*MM_air+N_v_in*MM_H2O m_fuel=0.65*MM_CH4+0.08*MM_H2+0.18*MM_N2+0.03*MM_O2+0.06*MM_CO2 AF=m_air/m_fuel

16-30 The composition of a gaseous fuel is given. It is burned with 130 percent theoretical air. The AF ratio and the fraction of water vapor that would condense if the product gases were cooled are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , N 2 , and air are 12 kg / kmol , 2 kg / kmol , 28 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis (a) The fuel is burned completely with excess air, and thus the products will contain H 2 O , CO2 , N 2 , and some free O2 . Considering 1 kmol of fuel, the combustion equation can be written as (0.45CH 4 + 0.35H 2 + 0.20N 2 ) + 1.3ath (O2 + 3.76N2 ) ¾ ¾ ® xCO2 + yH2 O + 0.3ath O2 + zN2

The unknown coefficients in the above equation are determined from mass balances,

C : 0.45 = x ¾ ¾ ® x = 0.45 H : 0.45´ 4 + 0.35´ 2 = 2 y ¾ ¾ ® y = 1.2 O2 :1.3ath = x + y / 2 + 0.3ath ¾ ¾ ® ath = 1.05 N 2 : 0.20 + 3.76´ 1.3ath = z ¾ ¾ ® z = 5.332

Thus, (0.45CH 4 + 0.35H 2 + 0.20N 2 ) + 1.365(O2 + 3.76N 2 ) ¾ ¾ ® 0.45CO2 + 1.2H2 O + 0.315O2 + 5.332N 2

The air-fuel ratio for the this reaction is determined by taking the ratio of the mass of the air to the mass of the fuel, mair = (1.365´ 4.76 kmol)(29 kg/kmol) = 188.4 kg

mfuel = (0.45´ 16 + 0.35´ 2 + 0.2´ 28)kg = 13.5 kg and AF =

mair 188.4 kg = = 14.0 kg air / kg fuel mfuel 13.5 kg

(b) For each kmol of fuel burned, 0.45 + 1.2 + 0.315 + 5.332 = 7.297 kmol of products are formed, including 1.2 kmol of H 2 O . Assuming that the dew-point temperature of the products is above 25C, some of the water vapor will condense as the products are cooled to 25C. If N w kmol of H 2 O condenses, there will be 1.2 - Nw kmol of water vapor left in the products. The mole number of the products in the gas phase will also decrease to 7.297 - N w as a result. Treating the product gases (including the remaining water vapor) as ideal gases, N w is determined by equating the mole fraction of the water vapor to its pressure fraction,

13-365

Nv N prod,gas

Pv 1.2 - N w 3.1698 kPa ¾¾ ® = ¾¾ ® N w = 1.003 kmol Pprod 7.297 - N w 101.325 kPa

since Pv = Psat @ 25°C = 3.1698 kPa . Thus the fraction of water vapor that condenses is

1.003 /1.2 = 0.836 or 83.6% .

EES (Engineering Equation Solver) SOLUTION "Given" "Volumetric analysis of a gaseous fuel is given." T_prod=25 [C] P_prod=101.325 [kPa] VR_CH4=0.45 VR_H2=0.35 VR_N2=0.2 "Properties" MM_air=29 [kg/kmol] MM_CH4=16 [kg/kmol] MM_H2=2 [kg/kmol] MM_N2=28 [kg/kmol] "Analysis" "(a)" "The gas is burned with 130% theoretical air" ex=1.3 "The combustion equation is: 0.60 CH4 + 0.30 H2 + 0.10 N2 + ex*a_th (O2+3.76N2) = x CO2 + y H2O + (ex-1)*a_th O2 + z N2" VR_CH4=x "C balance" VR_CH4*4+(ex-1)*2=2*y "H balance" ex*a_th=x+y/2+(ex-1)*a_th "O2 balance" VR_N2+3.76*ex*a_th=z "N2 balance" m_air=ex*a_th*4.76*MM_air m_fuel=VR_CH4*MM_CH4+VR_H2*MM_H2+VR_N2*MM_N2 AF=m_air/m_fuel "(b)" P_sat=pressure(steam_iapws, T=T_prod, x=1) P_v=P_sat N_v=y-N_w N_prod_gas=x+y+(ex-1)*a_th+z-N_w N_v/N_prod_gas=P_v/P_prod f_condensed=N_w/y

16-31 Problem 16-30 is reconsidered. The effects of varying the percentages of CH4 , H 2 and N 2 making up the fuel and the product gas temperature are to be studied. Analysis The problem is solved using EES, and the solution is given below. Let's modify this problem to include the fuels butane, ethane, methane, and propane in pull down menu. Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: aCxHy+bH2+cN2 + (a*y/4 + a*x+b/2) (Theo_air/100) (O2 + 3.76 N2) <--> a*xCO2 + ((a*y/2)+b) H2O + (c+3.76 (a*y/4 + a*x+b/2) (Theo_air/100)) N2 + (a*y/4 + a*x+b/2) (Theo_air/100 - 1) O2 T_prod is the product gas temperature.

13-366

Theo_air is the % theoretical air. " Procedure H20Cond(P_prod,T_prod,Moles_H2O,M_other:T_DewPoint,Moles_H2O_vap,Moles_H2O_liq,Result$) P_v=Moles_H2O/(M_other+Moles_H2O)*P_prod T_DewPoint=temperature(steam,P=P_v,x=0) IF T_DewPoint <= T_prod then Moles_H2O_vap=Moles_H2O Moles_H2O_liq=0 Result$='No condensation occurred' ELSE Pv_new=pressure(steam,T=T_prod,x=0) Moles_H2O_vap=Pv_new/P_prod*M_other/(1-Pv_new/P_prod) Moles_H2O_liq=Moles_H2O - Moles_H2O_vap Result$='There is condensation' ENDIF END "Input data" P_prod = 101.325 [kPa] Theo_air = 130 [%] a=0.45 b=0.35 c=0.20 T_prod = 25 [C] Fuel$='CH4' x=1 y=4 "Composition of Product gases:" A_th=a*y/4 +a* x+b/2 AF_ratio=4.76*A_th*Theo_air/100*molarmass(Air)/(a*16+b*2+c*28) Moles_O2=(a*y/4 +a* x+b/2) *(Theo_air/100 - 1) Moles_N2=c+(3.76*(a*y/4 + a*x+b/2))* (Theo_air/100) Moles_CO2=a*x Moles_H2O=a*y/2+b M_other=Moles_O2+Moles_N2+Moles_CO2 Call H20Cond(P_prod,T_prod,Moles_H2O,M_other:T_DewPoint,Moles_H2O_vap,Moles_H2O_liq,Result$) Frac_cond=Moles_H2O_liq/Moles_H2O*Convert(, %) "Reaction: aCxHy+bH2+cN2 + A_th Theo_air/100 (O2 + 3.76 N2) <--> a*xCO2 + (a*y/2+b) H2O + (c+3.76 A_th Theo_air/100) N2 + A_th (Theo_air/100 - 1) O2"

AFratio

Fraccond

[kg air / kg fuel ]

[%]

14.27 14.27 14.27 14.27 14.27 14.27 14.27 14.27

95.67 93.16 89.42 83.92 75.94 64.44 47.92 24.06

Moles H2O,liq

Moles H2O,vap

Tprod [C]

1.196 1.165 1.118 1.049 0.9492 0.8055 0.599 0.3008

0.05409 0.08549 0.1323 0.201 0.3008 0.4445 0.651 0.9492

5 11.67 18.33 25 31.67 38.33 45 51.67

13-367

14.27 14.27 14.27 14.27 14.27

0 0 0 0 0

1.25 1.25 1.25 1.25 1.25

58.33 65 71.67 78.33 85

16-32 Methane is burned with dry air. The volumetric analysis of the products is given. The AF ratio and the percentage of theoretical air used are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , and air are 12 kg / kmol , 2 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis Considering 100 kmol of dry products, the combustion equation can be written as

xCH 4 + a [O2 + 3.76N 2 ] ¾ ¾® 5.20CO2 + 0.33CO + 11.24O 2 + 83.23N 2 + bH 2 O The unknown coefficients x, a, and b are determined from mass balances,

N 2 : 3.76a = 83.23 ¾ ¾ ® a = 22.14 C : x = 5.20 + 0.33 ¾ ¾ ® x = 5.53 H : 4 x = 2b ¾ ¾ ® b = 11.06 (Check O2 : a = 5.20 + 0.165 + 11.24 + b / 2 ¾ ¾ ® 22.14 = 22.14)

Thus,

5.53CH 4 + 22.14[O2 + 3.76N 2 ] ¾ ¾® 5.20CO2 + 0.33CO + 11.24O2 + 83.23N 2 + 11.06H 2 O The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 5.53,

CH 4 + 4.0[O2 + 3.76N 2 ] ¾ ¾® 0.94CO2 + 0.06CO + 2.03O2 + 15.05N 2 + 2H 2 O © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-368

(a) The air-fuel ratio is determined from its definition, AF =

(4.0´ 4.76 kmol)(29 kg/kmol) mair = = 34.5 kg air / kg fuel mfuel (1 kmol)(12 kg/kmol)+ (2 kmol)(2 kg/kmol)

(b) To find the percent theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of the fuel, CH 4 + ath [O 2 + 3.76N 2 ] ¾ ¾ ® CO 2 + 2H 2 O + 3.76ath N 2

O2 :

ath = 1 + 1 ¾ ¾ ® ath = 2.0

Then, Percent theoretical air =

mair, act mair, th

N air, act N air, th

(4.0)(4.76) kmol = 200% (2.0)(4.76) kmol

EES (Engineering Equation Solver) SOLUTION "Given" "Methane is burned with dry air. The volumetric analysis of the products is given." "Properties" MM_air=29 [kg/kmol] MM_CH4=16 [kg/kmol] "Analysis" "Considering 100 kmol of dry products, the combustion equation is:" "x CH4 + a (O2+3.76N2) = 5.20 CO2 + 0.33 CO + 11.24 O2 + 83.23 N2 + b H2O" 3.76*a=83.23 "N2 balance" x=5.20+0.33 "C balance" 4*x=2*b "H balance" "Dividing the combustion equation by x, we obtain the combustion equation for 1 kmol of fuel" "(x/x) CH4 + (a/x) (O2+3.76N2) = (5.20/x) CO2 + (0.33/x) CO + (11.24/x) O2 + (83.23/x) N2 + (b/x) H2O" "(a)" m_air=N_air_act*MM_air m_fuel=N_fuel*MM_CH4 N_air_act=4.76*a/x N_fuel=1 [kmol] AF=m_air/m_fuel "(b)" "Theoretical combustion equation is: CH4 + a_th (O2+3.76N2) = CO2 + 2 H2O + 3.76a_th N2" a_th=1+1 "O2 balance" PercentTheorAir=N_air_act/N_air_th*Convert(, %) N_air_th=a_th*4.76

16-33 Methane is burned with air. The mass flow rates at the two inlets are to be determined. Properties The molar masses of CH4 , O2 , N 2 , CO2 , and H 2 O are 16, 32, 28, 44, and 18 kg / kmol , respectively (Table A-1). Analysis The stoichiometric combustion equation of CH4 is

CH 4 + ath [O 2 + 3.76N 2 ] ¾ ¾® CO2 + 2H 2 O + 3.76ath N 2

13-369

O2 balance: ath = 1 + 1

¾¾ ®

ath = 2

® Substituting, CH 4 + 2[O 2 + 3.76N 2 ] ¾ ¾

CO2 + 2H 2 O + 7.52N 2

The masses of the reactants are

mCH4 = NCH4 M CH4 = (1 kmol)(16 kg/kmol) = 16 kg mO2 = NO2 M O2 = (2 kmol)(32 kg/kmol) = 64 kg mN2 = N N2 M N2 = (2´ 3.76 kmol)(28 kg/kmol) = 211 kg The total mass is

mtotal = mCH4 + mO2 + N N2 = 16 + 64 + 211 = 291 kg Then the mass fractions are mf CH4 =

mCH4 16 kg = = 0.05498 mtotal 291 kg

mf O2 =

mO2 64 kg = = 0.2199 mtotal 291 kg

mf N2 =

mN2 211 kg = = 0.7251 mtotal 291 kg

For a mixture flow of 0.5 kg / s , the mass flow rates of the reactants are

mCH4 = mfCH4 m = (0.05498)(0.5 kg/s) = 0.02749 kg / s mair = m - mCH4 = 0.5 - 0.02749 = 0.4725 kg / s EES (Engineering Equation Solver) SOLUTION "Given" "Methane is burned with theoretical air." m_dot=0.5 [kg/s] "Properties" MM_CH4=16 [kg/kmol] MM_O2=32 [kg/kmol] MM_N2=28 [kg/kmol] "Analysis" "The combustion equation is: CH4 + a_th (O2 + 3.76 N2) = CO2 + 2 H2O + 3.76*a_th N2" a_th=1+2/2 "O2 balance" N_CH4=1 [kmol] N_O2=a_th N_N2=3.76*a_th m_CH4=N_CH4*MM_CH4 m_O2=N_O2*MM_O2 m_N2=N_N2*MM_N2 m_total=m_CH4+m_O2+m_N2 mf_CH4=m_CH4/m_total mf_O2=m_O2/m_total mf_N2=m_N2/m_total m_dot_CH4=mf_CH4*m_dot m_dot_air=m_dot-m_dot_CH4

16-34 n-Octane is burned with 60% excess air. The combustion is incomplete. The mole fractions of products and the dewpoint temperature of the water vapor in the products are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-370

Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 , N 2 and air are

12 kg / kmol , 2 kg / kmol , 32 kg / kmol , 28 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The combustion reaction for stoichiometric air is

C8 H18 + 12.5[O 2 + 3.76N 2 ] ¾ ¾® 8CO 2 + 9H 2 O + (12.5´ 3.76)N 2 The combustion equation with 60% excess air and incomplete combustion is

C8 H18 + 1.6´ 12.5[O 2 + 3.76N 2 ] ¾ ¾® (0.85´ 8) CO 2 + (0.15´ 8) CO + 9H 2O + x O 2 + (1.6´ 12.5´ 3.76) N 2 The coefficient for CO is determined from a mass balance,

O2 balance: 1.6´ 12.5 = 0.85´ 8 + 0.5´ 0.15´ 8 + 0.5´ 9 + x ¾ ¾® x = 8.100 Substituting,

C8 H18 + 20[O 2 + 3.76N 2 ] ¾ ¾® 6.8 CO 2 + 1.2 CO + 9 H 2 O + 8.1 O 2 + 75.2 N 2 The mole fractions of the products are

N prod = 6.8 + 1.2 + 9 + 8.1 + 75.2 = 100.3 kmol yCO2 =

NCO2 6.8 kmol = = 0.0678 N prod 100.3 kmol

yCO =

NCO 1.2 kmol = = 0.0120 N prod 100.3 kmol

yH2O =

N H2O 9 kmol = = 0.0897 N prod 100.3 kmol

yO2 =

N O2 8.1 kmol = = 0.0808 N prod 100.3 kmol

yN2 =

N N2 75.2 kmol = = 0.7498 N prod 100.3 kmol

The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is,

æ N ö÷ æ 9 kmol ö÷ Pv = ççç v ÷ Pprod = çç ÷(101.325 kPa) = 9.092 kPa ÷ ÷ çè100.3 kmol ø÷ çè N prod ø÷ Thus,

Tdp = Tsat @ 9.092 kPa = 44.0°C

(Table A-5 or EES)

13-371

"n-Octane (C8H18) is burned with 60 percent excess air with 15 percent of the carbon in the fuel forming carbon monoxide." P_prod=101.325 [kPa] "Analysis" "The combustion equation is: C8H18 + 1.6a_th (O2 + 3.76N2) = 0.85x8 CO2 + 0.15x8 CO + 9 H2O + x O2 + 1.6x3.76a_th N2" a_th=8+9/2 "from stoichiometric reaction" 1.6*a_th=0.85*8+0.5*0.15*8+9/2+x "O2 balance" N_CO2=0.85*8 [kmol] N_CO=0.15*8 [kmol] N_H2O=9 [kmol] N_O2=x N_N2=1.6*3.76*a_th N_prod=N_CO2+N_CO+N_H2O+N_O2+N_N2 y_CO2=N_CO2/N_prod y_CO=N_CO/N_prod y_H2O=N_H2O/N_prod y_O2=N_O2/N_prod y_N2=N_N2/N_prod P_v=(N_H2O/N_prod)*P_prod T_dp=temperature(steam_iapws, P=P_v, x=1)

16-35 Methyl alcohol is burned with 50% excess air. The combustion is incomplete. The mole fraction of carbon monoxide and the apparent molar mass of the products are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 , N 2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , 28 kg / kmol , and

29 kg / kmol , respectively (Table A-1).

Analysis The balanced reaction equation for stoichiometric air is

CH3OH + 1.5[O 2 + 3.76N 2 ] ¾ ¾® CO 2 + 2 H 2 O + 5.64 N 2 The reaction with 50% excess air and incomplete combustion can be written as

CH3OH + 1.5´ 1.5[O 2 + 3.76N 2 ] ¾ ¾® 0.90 CO 2 + 0.10 CO + 2 H 2O + x O 2 + 1.5´ 1.5´ 3.76 N 2 The coefficient for CO is determined from a mass balance,

O2 balance:

0.5 + 1.5´ 1.5 = 0.9 + 0.05 + 1 + x ¾ ¾ ® x = 0.8

Substituting,

CH3OH + 2.25[O 2 + 3.76N 2 ] ¾ ¾® 0.90 CO 2 + 0.10 CO + 2 H 2O + 0.8 O 2 + 8.46 N 2 The total moles of the products is

Nm = 0.9 + 0.10 + 2 + 0.8 + 8.46 = 12.26 kmol The mole fraction of carbon monoxide in the products is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-372

N CO 0.8 kmol = = 0.0653 Nm 12.26 kmol

yCO =

The apparent molecular weight of the product gas is mm (0.9´ 44 + 0.10´ 28 + 2´ 18 + 0.8´ 32 + 8.46´ 28) kg = = 27.80 kg / kmol Nm 12.26 kmol

Mm =

16-36 Coal whose mass percentages are specified is burned with 50 percent excess air. The fuel-air ratio is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , SO2 , N 2 , and O2 .

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , O2 , S, and air are 12, 2, 32, 32, and 29 kg / kmol , respectively (Table A-1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are mf C =

mC 79.61 kg 79.61 kg = = = 0.8712 mtotal (100-8.62) kg 91.38 kg

mf H2 =

mH2 4.66 kg = = 0.05100 mtotal 91.38 kg

mf O2 =

mO2 4.76 kg = = 0.05209 mtotal 91.38 kg

mf N2 =

mN2 1.83 kg = = 0.02003 mtotal 91.38 kg

mfS =

mS 0.52 kg = = 0.00569 mtotal 91.38 kg

We now consider 100 kg of this mixture. Then the mole numbers of each component are NC =

mC 87.12 kg = = 7.26 kmol M C 12 kg/kmol

N H2 =

mH2 5.10 kg = = 2.55 kmol M H2 2 kg/kmol

N O2 =

mO2 5.209 kg = = 0.1628 kmol M O2 32 kg/kmol

13-373

N N2 =

mN2 2.003 kg = = 0.07154 kmol M N2 28 kg/kmol

NS =

mS 0.569 kg = = 0.01778 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are

Nm = 7.26 + 2.55 + 0.1628 + 0.07154 + 0.01778 = 10.06 kmol yC =

NC 7.26 kmol = = 0.7215 N m 10.06 kmol

yH2 =

N H2 2.55 kmol = = 0.2535 Nm 10.06 kmol

yO2 =

N O2 0.1628 kmol = = 0.01618 Nm 10.06 kmol

yN2 =

N N2 0.07154 kmol = = 0.00711 Nm 10.06 kmol

yS =

NS 0.01778 kmol = = 0.00177 Nm 10.06 kmol

Then, the combustion equation in this case may be written as 0.7215C + 0.2535H 2 + 0.01618O2 + 0.00711N 2 + 0.00177S + 1.5ath (O2 + 3.76N 2 ) ¾ ¾ ® xCO2 + yH 2 O + zSO2 + kN 2 + 0.5ath O2

According to the species balances, C balance :

x = 0.7215

H2 balance : y = 0.2535 S balance :

z = 0.00177

O2 balance : 0.01618 + 1.5ath = x + 0.5 y + z + 0.5ath 1.5ath - 0.5ath = 0.7215 + 0.5(0.2535) + 0.00177 - 0.01617 ¾ ¾ ® ath = 0.8339 N 2 balance :

0.00711 + 1.5´ 3.76ath = k ¾ ¾ ® k = 0.00711 + 1.5´ 3.76´ 0.8339 = 4.710

Substituting, 0.7215C + 0.2535H 2 + 0.01617O 2 + 0.00711N 2 + 0.00177S + 1.2509(O2 + 3.76N 2 ) ¾ ¾ ® 0.7215CO 2 + 0.2535H 2O + 0.0018SO 2 + 4.71N 2 + 0.4170O2

The fuel-air mass ratio is then FA =

mfuel (0.7215´ 12 + 0.2535´ 2 + 0.01617 ´ 32 + 0.00711´ 28 + 0.00177´ 32) kg = x mair (1.2509´ 4.76´ 29) kg 9.938 kg = 0.0576 kg fuel / kg air 172.5 kg

13-374

16-37 Coal whose mass percentages are specified is burned with stoichiometric amount of air. The combustion is incomplete. The mass fractions of the products, the apparent molecular weight of the product gas, and the air-fuel ratio are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , SO2 , and N 2 .

Combustion gases are ideal gases.

mC 61.40 kg 61.40 kg = = = 0.6463 mtotal (100 - 5.00) kg 95.00 kg

mf H2 =

mH2 5.79 kg = = 0.06095 mtotal 95.00 kg

mf O2 =

mO2 25.31 kg = = 0.2664 mtotal 95.00 kg

mf N2 =

mN2 1.09 kg = = 0.01147 mtotal 95.00 kg

mfS =

mS 1.41 kg = = 0.01484 mtotal 95.00 kg

We now consider 100 kg of this mixture. Then the mole numbers of each component are NC =

mC 64.63 kg = = 5.386 kmol M C 12 kg/kmol

N H2 =

mH2 6.095 kg = = 3.048 kmol M H2 2 kg/kmol

N O2 =

mO2 26.64 kg = = 0.8325 kmol M O2 32 kg/kmol

N N2 =

mN2 1.147 kg = = 0.04096 kmol M N2 28 kg/kmol

NS =

mS 1.484 kg = = 0.04638 kmol M S 32 kg/kmol

13-375

The mole number of the mixture and the mole fractions are

Nm = 5.386 + 3.048 + 0.8325 + 0.04096 + 0.04638 = 9.354 kmol yC =

N C 5.386 kmol = = 0.5758 N m 9.354 kmol

yH2 =

N H2 3.048 kmol = = 0.3258 Nm 9.354 kmol

yO2 =

N O2 0.8325 kmol = = 0.0890 Nm 9.354 kmol

yN2 =

N N2 0.04096 kmol = = 0.00438 Nm 9.354 kmol

yS =

NS 0.04638 kmol = = 0.00496 Nm 9.354 kmol

Then, the combustion equation in this case may be written as

0.5758C + 0.3258H 2 + 0.0890O2 + 0.00438N 2 + 0.00496S + ath (O2 + 3.76N) ¾ ¾® x(0.95CO2 + 0.05CO) + yH2 O + zSO2 + kN2 According to the species balances,

C balance : x = 0.5758 H2 balance : y = 0.3258

S balance : z = 0.00496 O2 balance : 0.0890 + ath = 0.95x + 0.5´ 0.05x + 0.5 y + z ath = 0.95´ 0.5758 + 0.5´ 0.05´ 0.5758 + 0.5´ 0.3258 + 0.00496 - 0.0890 = 0.6403 N2 balance : k = 0.00438 + 3.76ath = 0.00438 + 3.76´ 0.6403 = 2.412 Substituting, 0.5758C + 0.3258H 2 + 0.0890O 2 + 0.00438N 2 + 0.00496S + 0.6403(O2 + 3.76N 2 ) ¾ ¾ ® 0.5470CO2 + 0.0288CO + 0.3258H 2O + 0.00496SO2 + 2.412N 2

The mass fractions of the products are

mtotal = 0.5470´ 44 + 0.0288´ 28 + 0.3258´ 18 + 0.00496´ 64 + 2.412´ 28 = 98.6 kg mf CO2 =

mCO2 (0.5470´ 44) kg = = 0.2441 mtotal 98.6 kg

mf CO =

mCO (0.0288´ 28) kg = = 0.0082 mtotal 98.6 kg

mf H2O =

mH2O (0.3258´ 18) kg = = 0.0595 mtotal 98.6 kg

mfSO2 =

mSO2 (0.00496´ 64) kg = = 0.0032 mtotal 98.6 kg

13-376

mf N2 =

mN2 (2.412´ 28) kg = = 0.6849 mtotal 98.6 kg

The total mole number of the products is

Nm = 0.5470 + 0.0288 + 0.3258 + 0.00496 + 2.712 = 3.319 kmol The apparent molecular weight of the product gas is Mm =

mm 98.6 kg = = 29.71 kg / kmol N m 3.319 kmol

The air-fuel mass ratio is then AF = =

mair (0.6403´ 4.76´ 29) kg = mfuel (0.5758´ 12 + 0.3258´ 2 + 0.0890´ 32 + 0.00438´ 28 + 0.00496´ 32) kg 88.39 kg = 8.27 kg air / kg fuel 10.69 kg

16-38 Methyl alcohol is burned with 100% excess air. The combustion is incomplete. The balanced chemical reaction is to be written and the air-fuel ratio is to be determined. Assumptions 1. Combustion is incomplete. 2.

The combustion products contain CO2 , CO, H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 , N 2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , 28 kg / kmol , and

29 kg / kmol , respectively (Table A-1). Analysis The balanced reaction equation for stoichiometric air is

CH3OH + ath [O2 + 3.76N 2 ]¾ ¾® CO 2 + 2 H 2 O + ath ´ 3.76 N 2 The stoicihiometric coefficient ath is determined from an O2 balance:

0.5 + ath = 1 + 1 ¾ ¾ ® ath = 1.5

Substituting,

CH3OH + 1.5[O 2 + 3.76N 2 ]¾ ¾® CO 2 + 2 H 2 O + 1.5´ 3.76 N 2 The reaction with 100% excess air and incomplete combustion can be written as

CH3OH + 2´ 1.5[O 2 + 3.76N 2 ]¾ ¾® 0.60 CO 2 + 0.40 CO + 2 H 2 O + x O 2 + 2´ 1.5´ 3.76 N 2 The coefficient for O2 is determined from a mass balance, ® x = 1.7 O2 balance: 0.5 + 2´ 1.5 = 0.6 + 0.2 + 1 + x ¾ ¾

Substituting,

CH3OH + 3[O 2 + 3.76N 2 ]¾ ¾® 0.6 CO 2 + 0.4 CO + 2 H 2 O + 1.7 O 2 + 11.28 N 2 The air-fuel mass ratio is AF =

mair (3´ 4.76´ 29) kg 414.1 kg = = = 12.94 kg air / kg fuel mfuel (1´ 32) kg 32 kg

13-377

EES (Engineering Equation Solver) SOLUTION "Given" "Methyl alcohol (CH3OH) is burned with 100 percent excess air. During the combustion process, 60 percent of the carbon in the fuel is converted to CO2 and 40 percent is converted to CO." "Properties" MM_CH3OH=32 [kg/kmol] MM_air=29 [kg/kmol] "Analysis" "The combustion equation is: CH3OH + 2a_th (O2 + 3.76N2) = 0.6 CO2 + 0.4 CO + 2 H2O + x O2 + 2x3.76a_th N2" 0.5+a_th=1+2/2 "O2 balance in stoichiometric reaction" 0.5+2*a_th=0.6+0.5*0.4+2/2+x "O2 balance" N_CH3OH=1 [kmol] N_air=2*a_th*4.76 m_air=N_air*MM_air m_CH3OH=N_CH3OH*MM_CH3OH AF=m_air/m_CH3OH

Enthalpy of Formation and Enthalpy of Combustion 16-39C Enthalpy of formation is the enthalpy of a substance due to its chemical composition. The enthalpy of formation is related to elements or compounds whereas the enthalpy of combustion is related to a particular fuel.

16-40C For combustion processes the enthalpy of reaction is referred to as the enthalpy of combustion, which represents the amount of heat released during a steady-flow combustion process.

16-41C If the combustion of a fuel results in a single compound, the enthalpy of formation of that compound is identical to the enthalpy of combustion of that fuel.

16-42C Yes.

16-43C The heating value is called the higher heating value when the H 2 O in the products is in the liquid form, and it is called the lower heating value when the H2 O in the products is in the vapor form. The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of that fuel.

16-44C No. The enthalpy of formation of N 2 is simply assigned a value of zero at the standard reference state for convenience.

16-45C 1 kmol of H 2 . This is evident from the observation that when chemical bonds of H 2 are destroyed to form H 2 O a large amount of energy is released.

13-378

16-46 The higher and lower heating values of coal from Utah are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , SO2 , and N 2 .

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , O2 , N 2 , S, and air are 12, 2, 32, 28, 32, and 29 kg / kmol , respectively (Table A1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are

mf C =

mC 61.40 kg 61.40 kg = = = 0.6463 mtotal (100 - 5.00) kg 95.00 kg

mf H2 =

mH2 5.79 kg = = 0.06095 mtotal 95.00 kg

mf O2 =

mO2 25.31 kg = = 0.2664 mtotal 95.00 kg

mf N2 =

mN2 1.09 kg = = 0.01147 mtotal 95.00 kg

mfS =

mS 1.41 kg = = 0.01484 mtotal 95.00 kg

We now consider 100 kg of this mixture. Then the mole numbers of each component are

NC =

mC 64.63 kg = = 5.386 kmol M C 12 kg/kmol

N H2 =

mH2 6.095 kg = = 3.048 kmol M H2 2 kg/kmol

N O2 =

mO2 26.64 kg = = 0.8325 kmol M O2 32 kg/kmol

N N2 =

mN2 1.147 kg = = 0.04096 kmol M N2 28 kg/kmol

NS =

mS 1.484 kg = = 0.04638 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are

13-379

Nm = 5.386 + 3.048 + 0.8325 + 0.04096 + 0.04638 = 9.354 kmol yC =

N C 5.386 kmol = = 0.5758 N m 9.354 kmol

yH2 =

N H2 3.048 kmol = = 0.3258 Nm 9.354 kmol

yO2 =

N O2 0.8325 kmol = = 0.0890 Nm 9.354 kmol

yN2 =

N N2 0.04096 kmol = = 0.00438 Nm 9.354 kmol

yS =

NS 0.04638 kmol = = 0.00496 Nm 9.354 kmol

Then, the combustion equation in this case may be written as

0.5758C + 0.3258H 2 + 0.0890O2 + 0.00438N 2 + 0.00496S + ath (O 2 + 3.76N2 ) ¾ ¾® 0.5758CO2 + 0.3258H 2O + 0.00496SO2 + kN 2 According to the species balances,

O2 balance : 0.0890 + ath = 0.5758 + 0.5´ 0.3258 + 0.00496 ¾ ¾® ath = 0.6547 N2 balance : k = 0.00438 + 3.76´ 0.6547 = 2.466 Substituting,

0.5758C + 0.3258H 2 + 0.0890O2 + 0.00438N 2 + 0.00496S + 0.6547(O2 + 3.76N) ¾ ¾® 0.5758CO2 + 0.3258H 2O + 0.00496SO2 + 2.466N2 Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values. The heat transfer for this process is equal to enthalpy of combustion. Note that C, S, H 2 , N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then,

q = hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2O

+ (Nh f )

SO2

For the HHV, the water in the products is taken to be liquid. Then,

hC = (0.5758 kmol)(- 393,520 kJ/kmol) + (0.3258 kmol)(- 285,830 kJ/kmol) + (0.00496 kmol)(- 297,100 kJ/kmol) = - 321,200 kJ/kmol coal

The apparent molecular weight of the coal is Mm =

mm (0.5758´ 12 + 0.3258´ 2 + 0.0890´ 32 + 0.00438´ 28 + 0.00496´ 32) kg = Nm (0.5758 + 0.3258 + 0.0890 + 0.00438 + 0.00496) kmol

10.69 kg = 10.69 kg/kmol coal 1.000 kmol The HHV of the coal is then =

HHV =

- hC 321,200 kJ/kmol coal = = 30, 000 kJ / kg coal Mm 10.69 kg/kmol coal

For the LHV, the water in the products is taken to be vapor. Then,

hC = (0.5758 kmol)(- 393,520 kJ/kmol) + (0.3258 kmol)(- 241,820 kJ/kmol) + (0.00496 kmol)(- 297,100 kJ/kmol) = - 306,850 kJ/kmol coal

13-380

LHV =

- hC 306,850 kJ/kmol coal = = 28, 700 kJ / kg coal Mm 10.69 kg/kmol coal

16-47 The enthalpy of combustion of methane at a 25C and 1 atm is to be determined using the data from Table A-26 and to be compared to the value listed in Table A-27. Assumptions The water in the products is in the liquid phase. Analysis The stoichiometric equation for this reaction is

CH 4 + 2[O2 + 3.76N 2 ] ¾ ¾® CO2 + 2H 2 O ( )+ 7.52N 2 Both the reactants and the products are at the standard reference state of 25C and 1 atm. Also, N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of CH4 becomes

hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2 O

- (Nh f )

CH 4

Using h f values from Table A-26,

hC = (1 kmol)(- 393,520 kJ/kmol)+ (2 kmol)(- 285,830 kJ/kmol)- (1 kmol)(- 74,850 kJ/kmol) = - 890, 330 kJ (per kmol CH 4 ) The listed value in Table A-27 is -890, 868 kJ / kmol , which is almost identical to the calculated value. Since the water in the products is assumed to be in the liquid phase, this hc value corresponds to the higher heating value of CH4 .

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] P=101.325 [kPa] "Analysis" "Stoichiometric combustion equation is: CH4 + a_th (O2+3.76N2) = CO2 + 2 H2O (l) + 3.76a_th N2" a_th=1+1 "O2 balance" N_CO2=1 [kmol] N_H2O=2 [kmol] N_CH4=1 [kmol] "Enthalpy of formation data from Table A-26" h_f_CO2=-393520 [kJ/kmol] h_f_H2O=-285830 [kJ/kmol] h_f_CH4=-74850 [kJ/kmol] H_P=N_CO2*h_f_CO2+N_H2O*h_f_H2O H_R=N_CH4*h_f_CH4 h_C=H_P-H_R

16-48 Problem 16-47 is reconsidered. The effect of temperature on the enthalpy of combustion is to be studied. Analysis The problem is solved using EES, and the solution is given below. Fuel$ = 'Methane (CH4)' T_comb =25 [C] T_fuel = T_comb +273 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-381

T_air1 = T_comb +273 T_prod =T_comb +273 h_bar_comb_TableA27 = -890360 [kJ/kmol] "For theoretical dry air, the complete combustion equation is" "CH4 + A_th(O2+3.76 N2)=1 CO2+2 H2O + A_th (3.76) N2 " A_th*2=1*2+2*1 "theoretical O balance" h_fuel_EES=enthalpy(CH4,T=298) h_fuel_TableA26=-74850 h_bar_fg_H2O=enthalpy(Steam_iapws,T=298,x=1)-enthalpy(Steam_iapws,T=298,x=0) HR=h_fuel_EES+ A_th*enthalpy(O2,T=T_air1)+A_th*3.76 *enthalpy(N2,T=T_air1) HP=1*enthalpy(CO2,T=T_prod)+2*(enthalpy(H2O,T=T_prod)-h_bar_fg_H2O)+A_th*3.76* enthalpy(N2,T=T_prod) h_bar_Comb_EES=(HP-HR) PercentError=ABS(h_bar_Comb_EES-h_bar_comb_TableA27)/ABS(h_bar_comb_TableA27)*Convert(, %)

h CombEES

TComb

[kJ / kmol]

[C]

–890335 –887336 –884186 –880908 –877508 –873985 –870339 –866568 –862675 –858661

25 88.89 152.8 216.7 280.6 344.4 408.3 472.2 536.1 600

16-49 The enthalpy of combustion of gaseous ethane at a 25C and 1 atm is to be determined using the data from Table A26 and to be compared to the value listed in Table A-27. Assumptions The water in the products is in the liquid phase. Analysis The stoichiometric equation for this reaction is

C2 H 6 + 3.5[O 2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3H 2 O ( )+ 13.16N 2 Both the reactants and the products are at the standard reference state of 25C and 1 atm. Also, N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C2 H6 becomes

hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO 2

+ (Nh f )

H2O

- (Nh f )

C2 H 6

Using h f values from Table A-26,

hC = (2 kmol)(- 393,520 kJ/kmol)+ (3 kmol)(- 285,830 kJ/kmol)- (1 kmol)(- 84,680 kJ/kmol) = - 1,559,850 kJ (per kmol C2 H 6 )

13-382

The listed value in Table A-27 is -1,560, 633 kJ / kmol , which is almost identical to the calculated value. Since the water in the products is assumed to be in the liquid phase, this hc value corresponds to the higher heating value of C2 H6 .

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] P=101.325 [kPa] "Analysis" "Stoichiometric combustion equation is: C2H6 + a_th (O2+3.76N2) = 2 CO2 + 3 H2O (l) + 3.76a_th N2" a_th=2+1.5 "O2 balance" N_CO2=2 [kmol] N_H2O=3 [kmol] N_C2H6=1 [kmol] "Enthalpy of formation data from Table A-26" h_f_CO2=-393520 [kJ/kmol] h_f_H2O=-285830 [kJ/kmol] h_f_C2H6=-84680 [kJ/kmol] H_P=N_CO2*h_f_CO2+N_H2O*h_f_H2O H_R=N_C2H6*h_f_C2H6 h_C=H_P-H_R

16-50 The enthalpy of combustion of liquid octane at a 25C and 1 atm is to be determined using the data from Table A-26 and to be compared to the value listed in Table A-27. Assumptions The water in the products is in the liquid phase. Analysis The stoichiometric equation for this reaction is

C8 H18 + 12.5[O2 + 3.76N 2 ] ¾ ¾® 8CO2 + 9H 2 O ( )+ 47N 2 Both the reactants and the products are at the standard reference state of 25C and 1 atm. Also, N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C8 H18 becomes

hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2 O

- (Nh f )

C8 H18

Using h f values from Table A-26, hC = (8 kmol)(- 393,520 kJ/kmol)+ (9 kmol)(- 285,830 kJ/kmol)- (1 kmol)(- 249,950 kJ/kmol)

= - 5, 470, 680 kJ

The listed value in Table A-27 is -5,470, 523 kJ / kmol , which is almost identical to the calculated value. Since the water in the products is assumed to be in the liquid phase, this hc value corresponds to the higher heating value of C8 H18 .

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] P=101.325 [kPa] "Analysis" "Stoichiometric combustion equation is: C8H18 + a_th (O2+3.76N2) = 8 CO2 + 9 H2O (l) + 3.76a_th N2" a_th=8+4.5 "O2 balance" N_CO2=8 [kmol] N_H2O=9 [kmol] N_C8H18=1 [kmol] "Enthalpy of formation data from Table A-26" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-383

h_f_CO2=-393520 [kJ/kmol] h_f_H2O=-285830 [kJ/kmol] h_f_C8H18=-249950 [kJ/kmol] H_P=N_CO2*h_f_CO2+N_H2O*h_f_H2O H_R=N_C8H18*h_f_C8H18 h_C=H_P-H_R

16-51 Ethane is burned with stoichiometric amount of air. The heat transfer is to be determined if both the reactants and products are at 25C. Assumptions The water in the products is in the vapor phase. Analysis The stoichiometric equation for this reaction is

C2 H 6 + 3.5[O2 + 3.76N 2 ]¾ ¾® 2CO2 + 3H 2 O + 13.16N 2 Since both the reactants and the products are at the standard reference state of 25C and 1 atm, the heat transfer for this process is equal to enthalpy of combustion. Note that N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then,

Q = hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2O

- (Nh f )

C2H6

Using h f° values from Table A-26,

Q = hC = (2 kmol)(- 393,520 kJ/kmol) + (3 kmol)(- 241,820 kJ/kmol) - (1 kmol)(- 84,680 kJ/kmol) = - 1,427,820 kJ / kmol C2H6

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] "Analysis" "Stoichiometric combustion equation is: C2H6 + a_th (O2+3.76N2) = 2 CO2 + 3 H2O (l) + 3.76a_th N2" a_th=2+3/2 "O2 balance" N_CO2=2 [kmol] N_H2O=3 [kmol] N_N2=3.76*a_th N_C2H6=1 [kmol] "Enthalpy of formation data from Table A-26" h_f_CO2=-393520 [kJ/kmol] h_f_H2O=-241820 [kJ/kmol] "water is vapor in products" h_f_C2H6=-84680 [kJ/kmol] H_P=N_CO2*h_f_CO2+N_H2O*h_f_H2O H_R=N_C2H6*h_f_C2H6 h_C=H_P-H_R

13-384

16-52 Ethane is burned with stoichiometric amount of air at 1 atm and 25C. The minimum pressure of the products which will assure that the water in the products will be in vapor form is to be determined. Assumptions The water in the products is in the vapor phase. Analysis The stoichiometric equation for this reaction is

C2 H 6 + 3.5[O2 + 3.76N 2 ]¾ ¾® 2CO2 + 3H 2 O + 13.16N 2 At the minimum pressure, the product mixture will be saturated with water vapor and Pv = Psat @ 25°C = 3.1698 kPa

The mole fraction of water in the products is yv =

N H2O 3 kmol = = 0.1652 N prod (2 + 3 + 13.16) kmol

The minimum pressure of the products is then Pmin =

Pv 3.1698 kPa = = 19.2 kPa yv 0.1652

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] "Analysis" "Stoichiometric combustion equation is: C2H6 + a_th (O2+3.76N2) = 2 CO2 + 3 H2O (l) + 3.76a_th N2" a_th=2+3/2 "O2 balance" N_CO2=2 [kmol] N_H2O=3 [kmol] N_N2=3.76*a_th N_prod=N_CO2+N_H2O+N_N2 y_v=N_H2O/N_prod P_v=pressure(steam_iapws, T=T, x=1) P_min=P_v/y_v

16-53 The higher and lower heating values of gaseous octane are to be determined and compared to the listed values. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 .

Combustion gases are ideal gases.

Properties The molar masses of C, O2 , H 2 , and air are 12, 32, 2, and 29 kg / kmol , respectively (Table A-1). Analysis The combustion reaction with stoichiometric air is

C8 H18 + 12.5(O 2 + 3.76N 2 ) ¾ ¾® 8 CO2 + 9H 2 O + 47N 2

Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-385

The heat transfer for this process is equal to enthalpy of combustion. Note that N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then,

q = hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2O

- (Nh f )

C8H18

For the HHV, the water in the products is taken to be liquid. Then,

hC = (8 kmol)(- 393,520 kJ/kmol) + (9 kmol)(- 285,830 kJ/kmol) - (1 kmol)(- 208,450 kJ/kmol) = - 5,512,180 kJ/kmol octane

The HHV of the gaseous octane is HHV =

- hC 5,512,180 kJ/kmol C8 H18 = = 48, 250 kJ / kg C8 H18 Mm 114.231 kg/kmol C8 H18

The listed value for liquid octane from Table A-27 is 47,890 kJ / kg . Adding the enthalpy of vaporization of octane to this value (47,890+363=48,253), the higher heating value of gaseous octane becomes 48, 253 kJ / kg octane. This value is practically identical to the calculated value. For the LHV, the water in the products is taken to be vapor. Then,

hC = (8 kmol)(- 393,520 kJ/kmol) + (9 kmol)(- 241,820 kJ/kmol) - (1 kmol)(- 208,450 kJ/kmol) = - 5,116,090 kJ/kmol octane

The LHV of the gaseous octane is then HHV =

- hC 5,116,090 kJ/kmol C8 H18 = = 44, 790 kJ / kg C8 H18 Mm 114.231 kg/kmol C8 H18

The listed value for liquid octane from Table A-27 is 44,430 kJ / kg . Adding the enthalpy of vaporization of octane to this value (44,430+363=44,793), the lower heating value of gaseous octane becomes 44, 793 kJ / kg octane. This value is practically identical to the calculated value.

EES (Engineering Equation Solver) SOLUTION "Given" "Calculate the HHV and LHV of gaseous n-octane fuel (C8H18)." "Properties" MM_C8H18=114.231 [kg/kmol] "Analysis" "Stoichiometric combustion equation is: C8H18 (g) + a_th (O2+3.76N2) = 8 CO2 + 9 H2O (l) + 3.76a_th N2" a_th=8+9/2 "O2 balance" N_CO2=8 [kmol] N_H2O=9 [kmol] N_N2=3.76*a_th N_C8H18=1 [kmol] "Enthalpy of formation data from Table A-26" h_f_CO2=-393520 [kJ/kmol] h_f_H2O_l=-285830 [kJ/kmol] "water is liquid in products" h_f_H2O_v=-241820 [kJ/kmol] "water is vapor in products" h_f_C8H18=-208450 [kJ/kmol] h_C_a=N_CO2*h_f_CO2+N_H2O*h_f_H2O_l-N_C8H18*h_f_C8H18 HHV=-h_C_a/MM_C8H18 h_C_b=N_CO2*h_f_CO2+N_H2O*h_f_H2O_v-N_C8H18*h_f_C8H18 LHV=-h_C_b/MM_C8H18 "Listed values in Table A-27" HHV_table=(47890+363) [kJ/kg] LHV_table=(44430+363) [kJ/kg]

13-386

First Law Analysis of Reacting Systems 16-54C The heat transfer will be the same for all cases. The excess oxygen and nitrogen enters and leaves the combustion chamber at the same state, and thus has no effect on the energy balance.

16-55C For case (b), which contains the maximum amount of nonreacting gases. This is because part of the chemical energy released in the combustion chamber is absorbed and transported out by the nonreacting gases.

16-56C In this case D U + Wb = D H , and the conservation of energy relation reduces to the form of the steady-flow energy relation.

16-57 Liquid propane is burned with 150 percent excess air during a steady-flow combustion process. The mass flow rate of air and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. Combustion is complete.

Properties The molar masses of propane and air are 44 kg / kmol and

29 kg / kmol , respectively (Table A-1). Analysis The fuel is burned completely with excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol of C3 H8 , the combustion equation can be written as

C3 H8 ( )+ 2.5ath (O2 + 3.76N 2 ) ¾ ¾® 3CO2 + 4H 2 O + 1.5ath O 2 + (2.5)(3.76ath )N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance, 2.5ath = 3 + 2 + 1.5ath ¾ ¾ ® ath = 5

Thus,

C3 H8 ( )+ 12.5(O 2 + 3.76N 2 ) ¾ ¾® 3CO2 + 4H 2 O + 7.5O 2 + 47N 2 (a) The air-fuel ratio for this combustion process is AF =

Thus,

(12.5´ 4.76 kmol)(29 kg/kmol) mair = = 39.22 kg air/kg fuel mfuel (3 kmol)(12 kg/kmol)+ (4 kmol)(2 kg/kmol)

mair = (AF)(mfuel ) = (39.22 kg air/kg fuel)(1.2 kg fuel/min ) = 47.1 kg air / min

13-387

(b) The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

h298

h1200

h 285

Substance

hf kJ / kmol

kJ / kmol

C3 H8 ( )

−118,910

---

8296.5

8682

38,447

8286.5

8669

36,777

H2 O ( g )

−241,820

---

9904

44,380

CO2

−393,520

---

9364

53,848

The h f of liquid propane is obtained by adding h fg of propane at 25C to h f of gas propane. Substituting, - Qout = (3)(- 393,520 + 53,848 - 9364)+ (4)(- 241,820 + 44,380 - 9904)+ (7.5)(0 + 38, 447 - 8682) + (47)(0 + 36, 777 - 8669)- (1)(- 118,910 + h298 - h298 )- (12.5)(0 + 8296.5 - 8682)- (47 )(0 + 8286.5 - 8669)

= - 190, 464 kJ / kmol C3H8 or

Qout = 190, 464 kJ / kmol C3H8

Then the rate of heat transfer for a mass flow rate of 1.2 kg / min for the propane becomes æ1.2 kg/min ÷ ö æm ö ÷(190,464 kJ/kmol) = 5194 kJ / min Qout = NQout = çç ÷ ÷Qout = çç ÷ çè 44 kg/kmol ÷ èç N ÷ ø ø

EES (Engineering Equation Solver) SOLUTION "Given" T1_fuel=25+273 "[K]" m_dot_fuel=1.2 "[kg/min]" T1_air=12+273 "[K]" T2=1200 "[K]" ex=2.5 "250% theoretical air" T0=25+273 "[K]" "Properties" MM_air=molarmass(air) MM_C3H8=molarmass(C3H8) "Analysis" "The combustion equation is: C3H8 + ex*a_th (O2+3.76N2) = 3 CO2 + 4 H2O + (ex-1)* a_th O2 + ex*3.76* a_th N2" ex*a_th=3+2+(ex-1)*a_th "O2 balance" "(a)" AF=m_air/m_fuel m_air=ex*a_th*4.76*MM_air m_fuel=1*MM_C3H8 m_dot_air=AF*m_dot_fuel "(b)" "Mol numbers of reactants and products in kmol"

13-388

N_C3H8=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_H2O=4 N_CO2=3 "Enthalpy of formation for C3H8" h_f_C3H8_gas=enthalpy(C3H8, T=T0) h_f_C3H8=h_f_C3H8_gas-h_fg h_fg=h_g-h_f h_f=enthalpy(propane, T=T0, x=0) h_g=enthalpy(propane, T=T0, x=1) "Ideal gas enthalpies in kJ/kmol" h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C3H8*h_f_C3H8+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R Q_dot_out=m_dot_fuel/MM_C3H8*Q_out

16-58E Liquid octane is burned with 180 percent theoretical air during a steady-flow combustion process. The AF ratio and the heat transfer from the combustion chamber are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. Combustion is complete. Properties The molar masses of C3 H18 and air are 54 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol of C8 H18 , the combustion equation can be written as

C8 H18 ( )+ 1.8ath (O2 + 3.76N 2 ) ¾ ¾® 8CO2 + 9H 2 O + 0.8ath O 2 + (1.8)(3.76ath )N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.8ath = 8 + 4.5 + 0.8ath ® ath = 12.5 Thus,

13-389

C8 H18 ( )+ 22.5(O 2 + 3.76N 2 ) ¾ ¾® 8CO2 + 9H 2 O + 10O 2 + 84.6N 2 (a) AF =

(22.5´ 4.76 lbmol)(29 lbm/lbmol) mair = = 27.2 lbm air / lbm fuel mfuel (8 lbmol)(12 lbm/lbmol)+ (9 lbmol)(2 lbm/lbmol)

(b) The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) = å N P (h f + h - h ) - å N R h f , R P

since all of the reactants are at 77F. Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

h537

Substance

hf Btu / Ibmol

Btu / Ibmol

C8 H18 ( )

−107,530

---

h 2500 Btu / Ibmol ---

3725.1

19,443

3729.5

18,590

CO2

−169,300

4027.5

27,801

H2 O ( g )

−104,040

4258.0

22,735

Thus, - Qout = (8)(- 169,300 + 27,801- 4027.5)+ (9)(- 104, 040 + 22, 735 - 4258)+ (10)(0 + 19, 443 - 3725.1) + (84.6)(0 + 18,590 - 3729.5)- (1)(- 107,530)- 0 - 0

= - 412,372 Btu/lbmol C8 H18 or

Qout = 412,372 Btu / lbmol C8 H18

EES (Engineering Equation Solver) SOLUTION "Given" T1=77+460 "[R]" T2=2500 "[R]" ex=1.8 "180% theoretical air" T0=77+460 "[R]" "Properties" MM_air=molarmass(air) MM_C8H18=molarmass(C8H18) "Analysis" "The combustion equation is: C8H18 + ex*a_th (O2+3.76N2) = 8 CO2 + 9 H2O + (ex-1)* a_th O2 + ex*3.76* a_th N2" ex*a_th=8+4.5+(ex-1)*a_th "O2 balance" "(a)" AF=m_air/m_fuel m_air=ex*a_th*4.76*MM_air m_fuel=1*MM_C8H18 "(b)" "Mol numbers of reactants and products in lbmol" N_C8H18=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-390

N_H2O=9 N_CO2=8 "Enthalpy of formation data from Table A-26E in Btu/lbmol" h_f_C8H18=-107530 "Ideal gas enthalpies in Btu/lbmol" h_O2_R=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C8H18*h_f_C8H18+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R

16-59 Propane is burned with 50% excess air. The heat transfer per kmol of fuel is to be determined for given reactants and products temperatures. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. Combustion is complete. 5. 6.

The reactants are at 17C and 1 atm. The fuel is in vapor phase.

Properties The molar masses of propane and air are 44 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis The combustion equation can be written as

C3 H8 + 1.5ath (O 2 + 3.76N 2 ) ¾ ¾® 3CO 2 + 4H 2 O + 0.5ath O 2 + 1.5ath ´ 3.76N 2 The stoichiometric coefficient is obtained from O2 balance: 1.5ath = 3 + 2 + 0.5ath ¾ ¾ ® ath = 5

Substituting,

C3 H8 + 7.5(O 2 + 3.76N 2 ) ¾ ¾® 3CO 2 + 4H 2 O + 2.5O 2 + 28.2N 2 The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

h290 K

h298 K

h370 K

kJ / kmol

13-391

C3 H8

−103,850

---

8443

8682

10,809

8432

8669

10,763

H2 O ( g )

−241,820

9904

12,331

CO2

−393,520

9364

12,148

Substituting,

- Qout = (3)(- 393,520 + 12,148 - 9364)+ (4)(- 241,820 + 12,331 - 9904 )+ (2.5)(0 + 10,809 - 8682) + (28.2)(0 + 10,763 - 8669)- (1)(- 103,850)- (7.5)(0 + 8443 - 8682 )- (28.2 )(0 + 8432 - 8669) = - 1,953,000 kJ/kmol C3 H8 S or

Qout = 1,953,000 kJ / kmol C3 H8

Discussion We neglected the enthalpy difference of propane between the standard temperature 25C and the given temperature 17C since the data is not available. It may be shown that the contribution of this would be small.

16-60 Propane is burned with an air-fuel ratio of 25. The heat transfer per kilogram of fuel burned when the temperature of the products is such that liquid water just begins to form in the products is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. Combustion is complete. 5. 6.

The reactants are at 25C and 1 atm. The fuel is in vapor phase.

Properties The molar masses of propane and air are 44 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis The mass of air per kmol of fuel is

mair = (AF)mfuel = (25 kg air/kg fuel)(1´ 44 kg/kmol fuel) = 1100 kg air/kmol fuel The mole number of air per kmol of fuel is then m 1100 kg air/kmol fuel N air = air = = 37.93 kmol air/kmol fuel M air 29 kg air/kmol air The combustion equation can be written as

13-392

C3 H8 + (37.93 / 4.76) (O2 + 3.76N 2 ) ¾ ¾® 3CO2 + 4H 2 O + xO 2 + (37.93 / 4.76)´ 3.76N 2 The coefficient for O2 is obtained from O2 balance:

(37.93/4.76) = 3 + 2 + x ¾ ¾® x = 2.968 Substituting,

C3 H8 + 7.968(O 2 + 3.76N 2 ) ¾ ¾® 3CO 2 + 4H 2 O + 2.968O 2 + 29.96N 2 The mole fraction of water in the products is N 4 kmol 4 kmol yv = H2O = = = 0.1002 N prod (3 + 4 + 2.968 + 29.96) kmol 39.93 kmol The partial pressure of water vapor at 1 atm total pressure is

Pv = yv P = (0.1002)(101.325 kPa) = 10.15 kPa When this mixture is at the dew-point temperature, the water vapor pressure is the same as the saturation pressure. Then, Tdp = Tsat @ 10.15 kPa = 46.1°C = 319.1 K @320 K We obtain properties at 320 K (instead of 319.1 K) to avoid iterations in the ideal gas tables. The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

h f , kJ / kmol

C3 H8 O2 N2 H2 O ( g ) CO2

–103,850

h298 K ,

h320 K ,

kJ / kmol

---

8682

9325

8669

9306

–241,820

9904

10,639

–393,520

9364

10,186

Substituting,

- Qout = (3)(- 393,520 + 10,186 - 9364)+ (4)(- 241,820 + 10,639 - 9904)+ (2.968)(0 + 9325 - 8682)+ (29.96)(0 + 9306 - 8669)- (1)(- 103,850)- 0 = - 2,017,590 kJ/kmol C3H8 or

Qout = 2, 017,590 kJkmol C3 H8

Then the heat transfer per kg of fuel is Qout =

Qout 2, 017,590 kJ/kmol fuel = = 45,850 kJ / kg C3H8 M fuel 44 kg/kmol

EES (Engineering Equation Solver) SOLUTION "Given" AF=25 "Properties" MM_air=29 [kg/kmol] MM_C3H8=44 [kg/kmol] "Analysis" m_fuel=N_C3H8*MM_C3H8 m_air=AF*m_fuel N_air=m_air/MM_air © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-393

"Combustion equation is: C3H8 + N_air/4.76 (O2+3.76N2) = 3 CO2 + 4 H2O + x O2 + N_air/4.76x3.76 N2" N_air/4.76=3+4/2+x "O2 balance" N_C3H8=1 [kmol] N_CO2=3 [kmol] N_H2O=4 [kmol] N_O2=x N_N2=N_air/4.76*3.76 N_prod=N_CO2+N_H2O+N_O2+N_N2 y_v=N_H2O/N_prod P_v=y_v*P0 P0=101.325 [kPa] T_dp=temperature(steam_iapws,P=P_v, x=1) h_CO2=enthalpy(CO2, T=T_dp) h_H2O=enthalpy(H2O, T=T_dp) h_O2=enthalpy(O2, T=T_dp) h_N2=enthalpy(N2, T=T_dp) h_C3H8=enthalpy(C3H8, T=T0) T0=25 [C] -Q_bar_out=N_CO2*h_CO2+N_H2O*h_H2O+N_O2*h_O2+N_N2*h_N2-N_C3H8*h_C3H8 Q_out=Q_bar_out/MM_C3H8

16-61 Benzene gas is burned with 95 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products and the heat transfer from the combustion chamber are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. Analysis

(a) The fuel is burned with insufficient amount of air, and thus the products will contain some CO as well as CO2 , H 2 O , and N 2 . The theoretical combustion equation of C6 H6 is

C6 H 6 + ath (O 2 + 3.76N 2 ) ¾ ¾® 6CO2 + 3H 2 O + 3.76ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

ath = 6 + 1.5 = 7.5 Then the actual combustion equation can be written as

C6 H 6 + 0.95´ 7.5(O 2 + 3.76N 2 ) ¾ ¾®

xCO 2 + (6 - x)CO + 3H 2 O + 26.79N 2

O2 balance:

0.95´ 7.5 = x + (6 - x) / 2 + 1.5

¾ ¾®

x = 5.25

Thus,

C6 H 6 + 7.125(O2 + 3.76N 2 ) ¾ ¾® 5.25CO2 + 0.75CO + 3H 2 O + 26.79N 2

13-394

yCO =

N CO 0.75 = = 0.021 or 2.1% N total 5.25 + 0.75 + 3 + 26.79

(b) The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) = å N P (h f + h - h ) - å N R h f , R P

since all of the reactants are at 25C. Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

h1000

h 298

Substance

kJ / kmol

C6 H6 ( g )

82,930

---

8682

31,389

8669

30,129

H2 O ( g )

−241,820

9904

35,882

−110,530

8669

30,355

CO2

−393,520

9364

42,769

Thus,

- Qout = (5.25)(- 393,520 + 42,769 - 9364)+ (0.75)(- 110,530 + 30,355 - 8669) + (3)(- 241,820 + 35,882 - 9904)+ (26.79)(0 + 30,129 - 8669 )- (1)(82,930)- 0 - 0 = - 2,112, 779 kJ / kmol C6 H6 or

Qout = 2,112,800 kJ/kmol C6 H 6 @ 2113 MJ / kmol C6 H 6

EES (Engineering Equation Solver) SOLUTION "Given" ex=0.95 "95% theoretical air" T1=25+273 "[K]" T2=1000 "[K]" T0=25+273 "[K]" "Analysis" "The stoichiometric combustion equation is: C6H6 + a_th (O2+3.76N2) = 6 CO2 + 3 H2O + 3.76*a_th N2" a_th=6+1.5 "O2 balance" "The actual combustion equation is: C6H6 + ex*a_th (O2+3.76N2) = x CO2 + (6-x) CO + 3 H2O + ex*a_th*3.76 N2" ex*a_th=x+(6-x)/2+1.5 "O2 balance" "(a)" y_CO=100*N_CO/N_tot N_CO=6-x N_tot=x+6-x+3+ex*a_th*3.76 "(b)" "Mol numbers of reactants and products in kmol" N_C6H6=1 N_O2=ex*a_th N_N2=ex*a_th*3.76 N_CO2=x

13-395

N_H2O=3 "Enthalpy of formation data from Table A-26 in kJ/kmol" h_f_C6H6=82930 "Ideal gas enthalpies in kJ/kmol" h_O2=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_N2_P=enthalpy(N2, T=T2) h_CO2=enthalpy(CO2, T=T2) h_CO=enthalpy(CO, T=T2) h_H2O=enthalpy(H2O, T=T2) H_P=N_CO2*h_CO2+N_CO*h_CO+N_H2O*h_H2O+N_N2*h_N2_P H_R=N_C6H6*h_f_C6H6+N_O2*h_O2+N_N2*h_N2_R -Q_out=H_P-H_R

16-62 Ethane gas is burned with stoichiometric amount of air during a steady-flow combustion process. The rate of heat transfer from the combustion chamber is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. Combustion is complete. Properties The molar mass of C2 H6 is 30 kg / kmol (Table A-1). Analysis The theoretical combustion equation of C2 H6 is

C2 H 6 + ath (O2 + 3.76N 2 ) ¾ ¾® 2CO 2 + 3H 2 O + 3.76ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

ath = 2 + 1.5 = 3.5 Then the actual combustion equation can be written as

C2 H 6 + 3.5(O 2 + 3.76N 2 ) ¾ ¾® 1.9CO 2 + 0.1CO + 3H 2 O + 0.05O 2 + 13.16N 2 The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

h 500

h 298

Substance

hf kJ / kmol

kJ / kmol

h 800 kJ / kmol

C2 H6 ( g )

−84,680

---

14,770

8682

24,523

13-396

14,581

8669

23,714

H2 O ( g )

−241,820

---

9904

27,896

−110,530

---

8669

23,844

CO2

−393,520

---

9364

32,179

Thus, - Qout = (1.9)(- 393,520 + 32,179 - 9364)+ (0.1)(- 110,530 + 23,844 - 8669 ) + (3)(- 241,820 + 27,896 - 9904)+ (0.05)(0 + 24,523 - 8682)+ (13.16 )(0 + 23, 714 - 8669) - (1)(- 84, 680 + h298 - h298 )- (3.5)(0 + 14, 770 - 8682)- (13.16)(0 + 14,581- 8669)

= - 1,201,005 kJ / kmol C2 H6 or

Qout = 1,201,005 kJ / kmol C2 H6

Then the rate of heat transfer for a mass flow rate of 3 kg / h for the ethane becomes æ 5 kg/h ÷ ö æm ö çç ÷ Qout = NQout = çç ÷ Q = (1,201,005 kJ/kmol) = 200,170 kJ/h @ 200 MJ / h ÷ out ÷ çè30 kg/kmol ÷ èç M ÷ ø ø

EES (Engineering Equation Solver) SOLUTION "Given" T1_fuel=25+273 "[K]" m_dot_fuel=5 "[kg/h]" T1_air=500 "[K]" T2=800 "[K]" T0=25+273 "[K]" "Properties" MM_C2H6=molarmass(C2H6) "Analysis" "The combustion equation is: C2H6 + a_th (O2+3.76N2) = 1.9 CO2 + 0.1 CO + 3 H2O + 0.05 O2 + 3.76*a_th N2" a_th=2+1.5 "O2 balance" "Mol numbers of reactants and products in kmol" N_C2H6=1 N_O2_R=a_th N_O2_P=0.05 N_N2=a_th*3.76 N_H2O=3 N_CO2=1.9 N_CO=0.1 "Ideal gas enthalpies in kJ/kmol" h_f_C2H6=enthalpy(C2H6, T=T0) h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) h_CO=enthalpy(CO, T=T2) H_P=N_CO2*h_CO2+N_CO*h_CO+N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C2H6*h_f_C2H6+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R Q_dot_out=m_dot_fuel/MM_C2H6*Q_out

13-397

16-63 A certain coal is burned steadily with 40% excess air. The heat transfer for a given product temperature is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , SO2 , and N 2 .

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , N 2 , O2 , S, and air are 12, 2, 28, 32, 32, and 29 kg / kmol , respectively (Table A1). Analysis We consider 100 kg of coal for simplicity. Noting that the mass percentages in this case correspond to the masses of the constituents, the mole numbers of the constituent of the coal are determined to be

NC =

mC 39.25 kg = = 3.271 kmol M C 12 kg/kmol

N H2 =

mH2 6.93 kg = = 3.465 kmol M H2 2 kg/kmol

N O2 =

mO2 41.11 kg = = 1.285 kmol M O2 32 kg/kmol

N N2 =

mN2 0.72 kg = = 0.0257 kmol M N2 28 kg/kmol

NS =

mS 0.79 kg = = 0.0247 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are

Nm = 3.271+ 3.465 + 1.285 + 0.0257 + 0.0247 = 8.071 kmol yC =

N C 3.271 kmol = = 0.4052 N m 8.071 kmol

yH2 =

N H2 3.465 kmol = = 0.4293 Nm 8.071 kmol

yO2 =

N O2 1.285 kmol = = 0.1592 Nm 8.071 kmol

yN2 =

N N2 0.0257 kmol = = 0.00319 Nm 8.071 kmol

13-398

Ash consists of the non-combustible matter in coal. Therefore, the mass of ash content that enters the combustion chamber is equal to the mass content that leaves. Disregarding this non-reacting component for simplicity, the combustion equation may be written as 0.4052C + 0.4293H 2 + 0.1592O2 + 0.00319N 2 + 0.00306S + 1.4ath (O2 + 3.76N2 ) ¾ ¾® 0.4052CO2 + 0.4293H 2 O + 0.4ath O2 + 0.00306SO2 + 1.4ath ´ 3.76N2

According to the O2 mass balance, 0.1592 + 1.4ath = 0.4052 + 0.5´ 0.4293 + 0.4ath + 0.00306 ¾ ¾ ® ath = 0.4637

Substituting, 0.4052C + 0.4293H 2 + 0.1592O2 + 0.00319N 2 + 0.00306S + 0.6492(O2 + 3.76N2 ) ¾ ¾® 0.4052CO2 + 0.4293H 2O + 0.1855O2 + 0.00306SO2 + 2.441N2

The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

h298 K

h400 K

Substance

kJ / kmol

8682

11,711

8669

11,640

H2 O ( g )

–241,820

9904

13,356

CO2

–393,520

9364

13,372

SO2

–297,100

–

The enthalpy change of sulfur dioxide between the standard temperature and the product temperature using constant specific heat assumption is

D hSO2 = c p D T = (41.7 kJ/kmol ×K)(127 - 25)K = 4253 kJ/kmol Substituting into the energy balance relation,

- Qout = (0.4052)(- 393,520 + 13,372 - 9364)+ (0.4293)(- 241,820 + 13,356 - 9904) + (0.1855)(0 + 11, 711- 8682)+ (2.441)(0 + 11, 640 - 8669)+ (0.00306)(- 297,100 + 4253)- 0

= - 253, 244 kJ/kmol C8 H18 or

Qout = 253, 244 kJ/kmol fuel

Then the heat transfer per kg of fuel is Qout =

Qout 253, 244 kJ/kmol fuel = M fuel (0.4052´ 12 + 0.4293´ 2 + 0.1592´ 32 + 0.00319´ 28 + 0.00306´ 32) kg/kmol

253, 244 kJ/kmol fuel 11.00 kg/kmol

= 23,020 kJ / kg coal

EES (Engineering Equation Solver) SOLUTION "Given" m_C=39.25 [kg] m_H2=6.93 [kg] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-399

m_O2=41.11 [kg] m_N2=0.72 [kg] m_S=0.79 [kg] m_ash=11.20 [kg] excessair=0.4 T0=25 [C] T_prod=127 [C] "h_SO2=-297100 [kJ/kmol] c_p_SO2=41.7 [kJ/kmol-K]" "Properties" MM_C=12 [kg/kmol] MM_H2=2 [kg/kmol] MM_O2=32 [kg/kmol] MM_N2=28 [kg/kmol] MM_S=32 [kg/kmol] MM_CO2=44 [kg/kmol] MM_H2O=18 [kg/kmol] MM_SO2=64 [kg/kmol] MM_air=29 [kg/kmol] "Analysis" N_C=m_C/MM_C N_H2=m_H2/MM_H2 N_O2=m_O2/MM_O2 N_N2=m_N2/MM_N2 N_S=m_S/MM_S N_m=N_C+N_H2+N_O2+N_N2+N_S y_C=N_C/N_m y_H2=N_H2/N_m y_O2=N_O2/N_m y_N2=N_N2/N_m y_S=N_S/N_m N_C_reac=y_C N_H2_reac=y_H2 N_O2_reac=y_O2 N_N2_reac=y_N2 N_S_reac=y_S x=y_C "C balance" y=y_H2 "H2 balance" z=y_S "S balance" y_O2+1.4*a_th=x+0.5*y+0.4*a_th+z "O2 balance" k=y_N2+(1+excessair)*3.76*a_th "N2 balance" m=excessair*a_th N_CO2_prod=x N_H2O_prod=y N_SO2_prod=z N_N2_prod=k N_O2_prod=m h_CO2=enthalpy(CO2, T=T_prod) h_H2O=enthalpy(H2O, T=T_prod) h_N2=enthalpy(N2, T=T_prod) h_O2=enthalpy(O2, T=T_prod) h_SO2=enthalpy(SO2, T=T_prod) H_P=N_CO2_prod*h_CO2+N_H2O_prod*h_H2O+N_O2_prod*h_O2+N_N2_prod*h_N2+N_SO2_prod*h_SO2 H_R=0 -Q_bar_out=H_P-H_R Q_out=Q_bar_out/MM_fuel MM_fuel=N_C_reac*MM_C+N_H2_reac*MM_H2+N_O2_reac*MM_O2+N_N2_reac*MM_N2+N_S_reac*MM_S

13-400

16-64 Octane gas is burned with 30 percent excess air during a steady-flow combustion process. The heat transfer per unit mass of octane is to be determined. Assumptions 1. 2. 3. 4.

Steady operating conditions exist. Air and combustion gases are ideal gases. Kinetic and potential energies are negligible. Combustion is complete.

Properties The molar mass of C8 H18 is 114 kg / kmol (Table A-1). Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . The moisture in the air does not react with anything; it simply shows up as additional H 2 O in the products. Therefore, for simplicity, we will balance the combustion equation using dry air, and then add the moisture to both sides of the equation. Considering 1 kmol of C8 H18 , the combustion equation can be written as

C8 H18 (g )+ 1.8ath (O 2 + 3.76N 2 ) ¾ ¾® 8CO 2 + 9H 2 O + 0.8ath O 2 + (1.8)(3.76ath )N 2 where ath is the stoichiometric coefficient for air. It is determined from O2 balance:1.8ath = 8 + 4.5 + 0.8ath ¾ ¾ ® ath = 12.5

Thus,

C8 H18 (g )+ 22.5(O 2 + 3.76N 2 ) ¾ ¾® 8CO 2 + 9H 2 O + 10O 2 + 84.6N 2 Therefore, 22.5  4.76 = 107.1 kmol of dry air will be used per kmol of the fuel. The partial pressure of the water vapor present in the incoming air is

Pv , in = f air Psat @ 25°C = (0.40)(3.1698 kPa ) = 1.268 kPa Assuming ideal gas behavior, the number of moles of the moisture that accompanies 107.1 kmol of incoming dry air is determined to be æPv , in ö æ 1.268 kPa ö ÷ ÷ ÷ N v , in = ççç N total = çç ® N v , in = 1.36 kmol ÷ ÷ ÷(107.1 + N v , in ) ¾ ¾ çè101.325 kPa ø èç Ptotal ÷ ø

The balanced combustion equation is obtained by adding 1.36 kmol of H 2 O to both sides of the equation,

C8 H18 (g )+ 22.5(O 2 + 3.76N 2 )+ 1.36H 2 O ¾ ¾® 8CO 2 + 10.36H 2O + 10O 2 + 84.6N 2 The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) = å N P (h f + h - h ) - å N R h f , R P

since all of the reactants are at 25C. Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

h f , kJ / kmol

h298 K , kJ / kmol

h1000 K , kJ / kmol

C8 H18 ( g )

–208,450

---

13-401

8682

31,389

8669

30,129

H2 O ( g )

–241,820

9904

35,882

CO2

–393,520

9364

42,769

Substituting, - Qout = (8)(- 393,520 + 42, 769 - 9364)+ (10.36)(- 241,820 + 35,882 - 9904 ) + (10)(0 + 31,389 - 8682)+ (84.6)(0 + 30,129 - 8669 ) - (1)(- 208, 450)- (1.36)(- 241,820)- 0 - 0

= - 2,537,130 kJ/kmol C8 H18 Thus 2,537,130 kJ of heat is transferred from the combustion chamber for each kmol (114 kg) of C8 H18 . Then the heat transfer per kg of C8 H18 becomes q=

Qout 2,537,130 kJ = = 22,260 kJ / kg C8H18 M 114 kg

EES (Engineering Equation Solver) SOLUTION Fuel$ = 'Octane (C8H18)' T_fuel = (25+273) [K] Ex = 0.8 "Excess air/100" P_air1 = 101.325 [kPa] T_air1 = (25+273) [K] RH_1 = 40/100 "inlet relative humidity" T_prod = 1000 [K] M_air = 28.97 [kg/kmol] M_water = 18 [kg/kmol] M_C8H18=(8*12+18*1) [kg/kmol] "For theoretical dry air, the complete combustion equation is" "C8H18 + A_th(O2+3.76 N2)=8 CO2+9 H2O + A_th (3.76) N2 " A_th*2=8*2+9*1 "theoretical O balance" "now to find the amount of water vapor associated with the dry air" w_1=HUMRAT(AirH2O,T=T_air1,P=P_air1,R=RH_1) "Humidity ratio, lbmv/lbma" N_w=w_1*(A_th*4.76*M_air)/M_water "Moles of water in the atmoshperic air, kmol/kmol_fuel" "The balanced combustion equation with Ex% excess moist air is:" "C8H18 + (1+EX)[A_th(O2+3.76 N2)+N_w H2O]=8 CO2+(9+(1+Ex)*N_w) H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " "Apply First Law SSSF" H_fuel = -208450 "[kJ/kmol] from the text" HR=H_fuel+ (1+Ex)*A_th*enthalpy(O2,T=T_air1)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air1)+(1+Ex)*N_w*enthalpy(H2O,T=T_air1) HP=8*enthalpy(CO2,T=T_prod)+(9+(1+Ex)*N_w)*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) Q_net=(HP-HR)"kJ/kmol"/(M_C8H18 "kg/kmol") "kJ/kg_C8H18" "This solution used the EES built-in function for humidity ratio to determine the moles of water vapor in atomspheric air. One could calculate the moles of water contained in the atmospheric air by the method shown in Chapter 15 which uses the relative humidity to find the partial pressure of the water vapor and, thus, the moles of water vapor. Explore what happens to the results as you vary the percent excess air, relative humidity, and product temperature." N_v=1.268/101.325*(107.1+N_v)

13-402

16-65 Problem 16-64 is reconsidered. The effect of the amount of excess air on the heat transfer for the combustion process is to be investigated. Analysis The problem is solved using EES, and the solution is given below. Fuel$ = 'Octane (C8H18)' T_fuel = (25+273) [K] PercentEX = 80 Ex = PercentEX/100 P_air1 = 101.3 [kPa] T_air1 = (25+273) [K] RH_1 = 40/100 T_prod = 1000 [K] M_air = 28.97 [kg/kmol] M_water = 18 [kg/kmol] M_C8H18=(8*12+18*1) [kg/kmol] "For theoretical dry air, the complete combustion equation is" "C8H18 + A_th(O2+3.76 N2)=8 CO2+9 H2O + A_th (3.76) N2" A_th*2=8*2+9*1 "theoretical O balance" "Now to find the amount of water vapor associated with the dry air" w_1=HUMRAT(AirH2O,T=T_air1,P=P_air1,R=RH_1) N_w=w_1*(A_th*4.76*M_air)/M_water "The balanced combustion equation with Ex% excess moist air is" "C8H18 + (1+EX)[A_th(O2+3.76 N2)+N_w H2O]=8 CO2+(9+(1+Ex)*N_w) H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " H_fuel = -208450 [kJ/kmol] "from Table A-26" HR=H_fuel+ (1+Ex)*A_th*enthalpy(O2,T=T_air1)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air1)+(1+Ex)*N_w*enthalpy(H2O,T=T_air1) HP=8*enthalpy(CO2,T=T_prod)+(9+(1+Ex)*N_w)*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) Q_net=(HP-HR)/(M_C8H18) Q_out = -Q_net "This solution used the humidity ratio form psychrometric data to determine the moles of water vapor in atomspheric air. One should calculate the moles of water contained in the atmospheric air by the method shown in Chapter 14 which uses the relative humidity to find the partial pressure of the water vapor and, thus, the moles of water vapor. Explore what happens to the results as you vary the percent excess air, relative humidity, and product temperature." PercentEX [%] 0 20 40 60 80 100 120 140 160 180 200

Qout [kJ / kgC8H18] 31444 29139 26834 24529 22224 19919 17614 15309 13003 10698 8393

13-403

16-66 Liquid ethyl alcohol, C2 H5OH (liq) , is burned in a steady-flow combustion chamber with 40 percent excess air. The required volume flow rate of the liquid ethyl alcohol is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The reaction equation for 40% excess air is

C2 H5 OH + 1.4ath [O2 + 3.76N 2 ] ¾ ¾® B CO2 + D H 2 O + E O2 + F N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 40% excess air by using the factor 1.4ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances

Carbon balance:

B=2

Hydrogen balance:

2D = 6 ¾ ¾ ® D= 3

Oxygen balance:

1+ 2´ 1.4ath = 2B + D + 2E 0.4ath = E

Nitrogen balance:

1.4ath ´ 3.76 = F

Solving the above equations, we find the coefficients (E = 1.2, F = 15.79, and ath = 3 ) and write the balanced reaction equation as

C2 H5 OH + 4.2[O 2 + 3.76N 2 ] ¾ ¾® 2 CO 2 + 3 H 2 O + 1.2 O 2 + 15.79 N 2 The steady-flow energy balance is expressed as

N fuel H R = Qout + N fuel H P where

H R = (h f - h fg )fuel + 4.2hO2 @ 298.15 K + (4.2´ 3.76)hN2 @ 298.15 K

= - 235,310 kJ/kmol - 42,340 kJ/kmol + 4.2( - 4.425 kJ/kmol) + (4.2´ 3.76)(- 4.376 kJ/kmol) = - 277, 650 kJ/kmol

H P = 2hCO2 @ 600 K + 3hH2O @ 600 K + 1.2hO2 @ 600 K + 15.79hN2 @ 600 K

= 2(- 380,623 kJ/kmol) + 3( - 231,333 kJ/kmol) + 1.2(9251 kJ/kmol) + 15.79(8889 kJ/kmol) = - 1.304´ 106 kJ/kmol

The enthalpies are obtained from EES except for the enthalpy of formation of the fuel, which is obtained in Table A-27 of the book. Substituting into the energy balance equation, N fuel H R = Qout + N fuel H P Nfuel (- 277, 650 kJ/kmol) = 2000 kJ/s + Nfuel (- 1.304 ´ 10 6 kJ/kmol) ¾ ¾ ® Nfuel = 0.001949 kmol/s

The fuel mass flow rate is

mfuel = Nfuel M fuel = (0.001949 kmol/s)(2´ 12 + 6´ 1 + 16) kg/kmol = 0.08966 kg/s Then, the volume flow rate of the fuel is determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-404

Vfuel =

ö mfuel 0.08966 kg/s æ ç 6000 L/min ÷ = = 6.81 L / min ÷ 3 ç 3 ÷ ç r fuel 790 kg/m è 1 m /s ø

EES (Engineering Equation Solver) SOLUTION "The reaction equation for 40% excess air is:" "C2H5OH(liq)+ 1.4*A_th (O2 +3.76N2)=B CO2 + D H2O + E O2 + F N2" "Carbon Balance:" 2=B "Hydrogen Balance:" 6=2*D "Oxygen Balance:" 1 + 1.4*A_th*2=B*2 + D + E*2 "Excess oxygen in the products is:" 0.4*A_th = E "Nitrogen Balance:" 1.4*A_th*3.76 = F "The product temperature is:" T_P = 600 "[K]" "The reactant temperature is:" T_R= 25+273.15 "[K]" "The required heat transfer rate is" Q_dot_out = 2000 "[kW]" "The steady-flow energy balance is:" N_dot_fuel*H_R = Q_dot_out+N_dot_fuel*H_P h_bar_fg=42340 "[kJ/kmol]" h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]" H_R=1*(h_bar_f_C2H5OHgas - h_bar_fg) +1.4*A_th*ENTHALPY(O2,T=T_R)+1.4*A_th*3.76*ENTHALPY(N2,T=T_R) "[kJ/kmol]" "h_O2_R=ENTHALPY(O2,T=T_R) h_N2_R=ENTHALPY(N2,T=T_R)" H_P=B*ENTHALPY(CO2,T=T_P)+D*ENTHALPY(H2O,T=T_P) +E*ENTHALPY(O2,T=T_P)+F*ENTHALPY(N2,T=T_P) "[kJ/kmol]" "h_CO2_P=ENTHALPY(CO2,T=T_P) h_H2O_P=ENTHALPY(H2O,T=T_P) h_N2_P=ENTHALPY(N2,T=T_P) h_O2_P=ENTHALPY(O2,T=T_P)" "The mass flow rate of fuel is:" MM_fuel = 2*12 + 6*1+ 16 "[kg/kmol]" m_dot_fuel = N_dot_fuel*MM_fuel "[kg/s]" "The volume flow rate of fuel is:" rho_fuel = 790 "[kg/m^3]" V_dot_fuel = m_dot_fuel / rho_fuel "[m^3/s]" * convert(m^3/s,L/min) "[L/min]"

16-67 Propane gas is burned with 100% excess air. The combustion is incomplete. The balanced chemical reaction is to be written, and the dew-point temperature of the products and the heat transfer from the combustion chamber are to be determined. Assumptions 1. Combustion is incomplete. 2.

The combustion products contain CO2 , CO, H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 , N 2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , 28 kg / kmol , and

13-405

Analysis (a) The balanced reaction equation for stoichiometric air is

C3 H8 + ath [O2 + 3.76N 2 ] ¾ ¾® 3 CO2 + 4 H 2 O + ath ´ 3.76 N 2

The stoicihiometric coefficient ath is determined from an O2 balance:

ath = 3 + 2 = 5 Substituting,

C3 H8 + 5[O2 + 3.76N 2 ] ¾ ¾® 3 CO 2 + 4 H 2 O + 5´ 3.76 N 2 The reaction with 100% excess air and incomplete combustion can be written as

C3 H8 + 2´ 5[O 2 + 3.76N 2 ] ¾ ¾® 0.90´ 3 CO 2 + 0.10´ 3 CO + 4 H 2O + x O 2 + 2´ 5´ 3.76 N 2 The coefficient for O2 is determined from a mass balance,

O2 balance:

10 = 0.9´ 3 + 0.05´ 3 + 2 + x ¾ ¾ ® x = 5.15

Substituting,

C3 H8 + 10[O 2 + 3.76N 2 ] ¾ ¾® 2.7 CO 2 + 0.3 CO + 4 H 2 O + 5.15 O 2 + 37.6 N 2 (b) The partial pressure of water vapor is Pv =

N H2O 4 4 kmol Ptotal = (100 kPa) = (100 kPa) = 8.040 kPa N total 2.7 + 0.3 + 4 + 5.15 + 37.6 49.75 kmol

The dew point temperature of the product gases is the saturation temperature of water at this pressure:

Tdp = Tsat@ 8.04 kPa = 41.5°C (Table A-5) (c) The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Both the reactants and products are at 25°C. Assuming the air and the combustion products to be ideal gases, we have h = h(T). Also, since the temperature of products (25°C) is lower than the dew-po,nt temperature, some water vapor will condense. Noting from Table A-5 that Psat@ 25°C = 3.17 kPa , the molar amount of water that remain as vapor is determined from Pv =

N H2O,vapor N H2O Ptotal ¾ ¾ ® 3.17 kPa = (100 kPa) ¾ ¾ ® N H2O,vapor = 1.5 kmol N total N H2O,vapor + 2.7 + 0.3 + 5.15 + 37.6

Thus,

N H2O,liquid = 4 - 1.5 = 2.5 kmol Then, using the values given in the table,

- Qout = (2.7)(- 393,520) + (0.3)(- 110,530) + (1.5)(- 241,820) + (2.5)(- 285,830) - (1)(- 103,850) = - 2,069,120 kJ / kmol C3H8 or

Qout = 2,069,120 kJ / kmol C3 H8

Then the heat transfer for a 100 kmol fuel becomes

13-406

æm ö Qout = NQout = çç ÷ Q = (100 kmol fuel)(2,069,120 kJ/kmol fuel) = 2.069×108 kJ ÷ çè N ÷ ø out

EES (Engineering Equation Solver) SOLUTION "Given" P=100 [kPa] ex=1 "100 percent excess air" f_CO2=0.90 f_CO=0.10 T_reac=25 [C] T_prod=25 [C] N_fuel=100 [kmol] "Analysis" "Combustion equation is: C3H8 + 2a_th (O2+3.76N2) = 0.9x3 CO2 + 0.1x3 CO + 4 H2O + x O2 + 2a_thx3.76 N2" a_th=3+4/2 "O2 balance for stoichiometric equation" 2*a_th=0.9*3+(0.1/2)*3+2+x "O2 balance for actual equation" N_C3H8=1 [kmol] N_CO2=0.9*3 [kmol] N_CO=0.1*3 [kmol] N_H2O=4 [kmol] N_O2=x N_N2=2*a_th*3.76 N_total=N_CO2+N_CO+N_H2O+N_O2+N_N2 P_v=N_H2O/N_total*P T_dp=temperature(steam_iapws,P=P_v, x=1) P_g=pressure(steam_iapws,T=25, x=1) N_total_new=N_CO2+N_CO+N_H2O_v+N_O2+N_N2 P_g=N_H2O_v/(N_total_new)*P N_H2O_l=N_H2O-N_H2O_v h_CO2=enthalpy(CO2, T=T_prod) h_CO=enthalpy(CO, T=T_prod) h_H2O=enthalpy(H2O, T=T_prod) h_H2O_l=-285830 [kJ/kmol] h_O2=enthalpy(O2, T=T_prod) h_N2=enthalpy(N2, T=T_prod) h_C3H8=enthalpy(C3H8, T=T_reac) -Q_bar_out=N_CO2*h_CO2+N_CO*h_CO+N_H2O_l*h_H2O_l+N_H2O_v*h_H2O+N_O2*h_O2+N_N2*h_N2N_C3H8*h_C3H8 Q_out=N_fuel*Q_bar_out

16-68 A mixture of propane and methane is burned with theoretical air. The balanced chemical reaction is to be written, and the amount of water vapor condensed and the the required air flow rate for a given heat transfer rate are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 , N 2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , 28 kg / kmol , and

29 kg / kmol , respectively (Table A-1). Analysis (a) The balanced reaction equation for stoichiometric air is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-407

0.4 C3 H8 + 0.6 CH 4 + ath [O 2 + 3.76N 2 ]¾ ¾® 1.8 CO 2 + 2.8 H 2O + ath ´ 3.76 N 2 The stoicihiometric coefficient ath is determined from an O2 balance:

ath = 1.8 + 1.4 = 3.2 Substituting,

0.4 C3 H8 + 0.6 CH 4 + 3.2[O 2 + 3.76N 2 ]¾ ¾® 1.8 CO 2 + 2.8 H 2 O + 12.032 N 2 (b) The partial pressure of water vapor is Pv =

N H2O 2.8 2.8 kmol Ptotal = (100 kPa) = (100 kPa) = 16.84 kPa N total 1.8 + 2.8 + 12.032 16.632 kmol

The dew point temperature of the product gases is the saturation temperature of water at this pressure:

Tdp = Tsat@16.84 kPa = 56.2°C (Table A-5) Since the temperature of the product gases are at 398 K (125C), there will be no condensation of water vapor. (c) The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

The products are at 125°C, and the enthalpy of products can be expressed as

(h - h ) = c D T p

where D T = 125 - 25 = 100°C = 100 K . Then, using the values given in the table, - Qout = (1.8)(- 393,520 + 41.16´ 100) + (2.8)(- 241,820 + 34.28´ 100) + (12.032)(0 + 29.27´ 100) - (0.4)(- 103,850) - (0.6)(- 74,850)

= - 1, 246,800 kJ/kmol fuel

Qout = 1, 246,800 kJ/kmol fuel

For a heat transfer rate of 97,000 kJ / h , the molar flow rate of fuel is Nfuel =

Qout 97, 000 kJ/h = = 0.07780 kmol fuel/h Qout 1, 246,800 kJ/kmol fuel

The molar mass of the fuel mixture is

M fuel = 0.4´ 44 + 0.6´ 16 = 27.2 kg/kmol The mass flow rate of fuel is

mfuel = Nfuel M fuel = (0.07780 kmol/h)(27.2 kg/kmol) = 2.116 kg/h The air-fuel ratio is AF =

mair (3.2´ 4.76´ 29) kg = = 16.24 kg air/kg fuel mfuel (0.4´ 44 + 0.6´ 16) kg

The mass flow rate of air is then

mair = mfuel AF = (2.116 kg/h)(16.24) = 34.4 kg / h

13-408

EES (Engineering Equation Solver) SOLUTION "Given" P=100 [kPa] T_reac=(298-273) [C] T_prod=(398-273) [C] Q_dot_out=97000 [kJ/h] "Properties" MM_C3H8=44 [kg/kmol] MM_CH4=16 [kg/kmol] MM_air=29 [kg/kmol] "Analysis" "Combustion equation is: 0.4 C3H8 + 0.6 CH4 + a_th (O2+3.76N2) = 1.8 CO2 + 2.8 H2O + a_thx3.76 N2" a_th=1.8+2.8/2 "O2 balance" N_C3H8=0.4 [kmol] N_CH4=0.6 [kmol] N_CO2=1.8 [kmol] N_H2O=2.8 [kmol] N_N2=a_th*3.76 N_total=N_CO2+N_H2O+N_N2 P_v=N_H2O/N_total*P T_dp=temperature(steam_iapws,P=P_v, x=1) h_CO2=enthalpy(CO2, T=T_prod) h_H2O=enthalpy(H2O, T=T_prod) h_N2=enthalpy(N2, T=T_prod) h_C3H8=enthalpy(C3H8, T=T_reac) h_CH4=enthalpy(CH4, T=T_reac) -Q_bar_out=N_CO2*h_CO2+N_H2O*h_H2O+N_N2*h_N2-N_C3H8*h_C3H8-N_CH4*h_CH4 N_dot_fuel=Q_dot_out/Q_bar_out MM_fuel=0.4*MM_C3H8+0.6*MM_CH4 m_dot_fuel=N_dot_fuel*MM_fuel N_air=a_th*4.76 N_fuel=1 [kmol] AF=(N_air*MM_air)/(N_fuel*MM_fuel) m_dot_air=AF*m_dot_fuel

16-69 A mixture of methane and oxygen contained in a tank is burned at constant volume. The final pressure in the tank and the heat transfer during this process are to be determined. Assumptions 1. Air and combustion gases are ideal gases. 2. Combustion is complete. Properties The molar masses of CH4 and O2 are 16 kg / kmol and 32 kg / kmol , respectively (Table A-1). Analysis (a) The combustion is assumed to be complete, and thus all the carbon in the methane burns to CO2 and all of the hydrogen to H 2 O . The number of moles of CH4 and O2 in the tank are

13-409

N CH4 =

N O2 =

mCH4

M CH4

mO2 M O2

0.12 kg = 7.5´ 10- 3 kmol = 7.5 mol 16 kg/kmol

0.6 kg = 18.75´ 10- 3 kmol = 18.75 mol 32 kg/kmol

Then the combustion equation can be written as

7.5CH 4 + 18.75O2 ¾ ¾® 7.5CO2 + 15H 2 O + 3.75O2 At 1200 K, water exists in the gas phase. Assuming both the reactants and the products to be ideal gases, the final pressure in the tank is determined to be

æ öæ PRV = N R Ru TR ïü ÷ççTP ö÷ ÷ ïý P = P çç N P ÷ P Rç ÷ ÷ççTR ø÷ ÷ çè N R øè PPV = N P Ru TP ïïþ

Substituting,

æ26.25 mol öæ ÷çç1200 K ö÷ ÷ PP = (200 kPa )çç ÷ ÷ç 298 K ø÷= 805 kPa èç 26.25 mol øè which is relatively low. Therefore, the ideal gas assumption utilized earlier is appropriate. (b) The heat transfer for this constant volume combustion process is determined from the energy balance

Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to - Qout = å N P (h f + h - h - Pv ) - å N R (h f + h - h - Pv ) P

Since both the reactants and products are assumed to be ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT . It yields

- Qout = å N P (h f + h1200 K - h298 K - RuT ) - å N R (h f - RuT ) P

since the reactants are at the standard reference temperature of 25C. From the tables,

h 298 K

h1200 K

kJ / kmol

---

8682

38,447

H2 O ( g )

9904

44,380

CO2

9364

53,848

Substance

hf kJ / kmol

CH4 O2

Thus,

- Qout = (7.5)(- 393,520 + 53,848 - 9364 - 8.314´ 1200)+ (15)(- 241,820 + 44,380 - 9904 - 8.314´ 1200) + (3.75)(0 + 38, 447 - 8682 - 8.314´ 1200)- (7.5)(- 74,850 - 8.314´ 298)- (18.75)(- 8.314´ 298) = - 5,251,791 J = - 5252 kJ

Thus Qout = 5252 kJ of heat is transferred from the combustion chamber as 120 g of CH4 burned in this combustion chamber. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-410

EES (Engineering Equation Solver) SOLUTION "Input Data" T_reac = (25+273) [K] "reactant mixture temperature" P_reac = 200 [kPa] "reactant mixture pressure" T_prod = 1200 [K] "product mixture temperature" m_O2=0.600 [kg] "initial mass of O2" Mw_O2 = 32 [kg/kmol] m_CH4 = 0.120 [kg] "initial mass of CH4" Mw_CH4=(1*12+4*1) [kg/kmol] R_u = 8.314 [kJ/kmol-K] "universal gas constant" "For theoretical oxygen, the complete combustion equation is" "CH4 + A_th O2=1 CO2+2 H2O " 2*A_th=1*2+2*1"theoretical O balance" "now to find the actual moles of O2 supplied per mole of fuel" N_O2 = m_O2/Mw_O2/N_CH4 N_CH4= m_CH4/Mw_CH4 "The balanced complete combustion equation with Ex% excess O2 is" "CH4 + (1+EX) A_th O2=1 CO2+ 2 H2O + Ex( A_th) O2 " N_O2 = (1+Ex)*A_th "Apply First Law to the closed system combustion chamber and assume ideal gas behavior. (At 1200 K, water exists in the gas phase.)" E_in - E_out = DELTAE_sys E_in = 0 E_out = Q_out "kJ/kmol_CH4" "No work is done because volume is constant" DELTAE_sys = U_prod - U_reac "neglect KE and PE and note: U = H - PV = N(h - R_u T)" U_reac = 1*(enthalpy(CH4, T=T_reac) - R_u*T_reac) +(1+EX)*A_th*(enthalpy(O2,T=T_reac) - R_u*T_reac) U_prod = 1*(enthalpy(CO2, T=T_prod) - R_u*T_prod) +2*(enthalpy(H2O, T=T_prod) R_u*T_prod)+EX*A_th*(enthalpy(O2,T=T_prod) - R_u*T_prod) "The total heat transfer out, in kJ, is:" Q_out_tot=Q_out"kJ/kmol_CH4"/(Mw_CH4 "kg/kmol_CH4") *m_CH4"kg" "kJ" "The final pressure in the tank is the pressure of the product gases. Assuming ideal gas behavior for the gases in the constant volume tank, the ideal gas law gives:" P_reac*V =N_reac * R_u *T_reac P_prod*V = N_prod * R_u * T_prod N_reac = N_CH4*(1 + N_O2) N_prod = N_CH4*(1 + 2 + Ex*A_th)

16-70 Problem 16-69 is reconsidered. The effect of the final temperature on the final pressure and the heat transfer for the combustion process is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" T_reac = (25+273) [K] P_reac = 200 [kPa] T_prod = 1200 [K] m_O2=0.600 [kg] Mw_O2 = 32 [kg/kmol] m_CH4 = 0.120 [kg] Mw_CH4=(1*12+4*1) [kg/kmol] R_u = 8.314 [kJ/kmol-K] "For theoretical oxygen, the complete combustion equation is" "CH4 + A_th O2=1 CO2+2 H2O " 2*A_th=1*2+2*1 "theoretical O balance" "Now to find the actual moles of O2 supplied per mole of fuel" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-411

N_O2 = m_O2/Mw_O2/N_CH4 N_CH4= m_CH4/Mw_CH4 "The balanced complete combustion equation with Ex% excess O2 is" "CH4 + (1+EX) A_th O2=1 CO2+ 2 H2O + Ex( A_th) O2 " N_O2 = (1+Ex)*A_th "Apply First Law to the closed system combustion chamber and assume ideal gas behavior. (At 1200 K, water exists in the gas phase.)" E_in - E_out = DELTAE_sys E_in = 0 E_out = Q_out "No work is done because volume is constant" DELTAE_sys = U_prod - U_reac "neglect KE and PE and note: U = H - PV = N(h - R_u T)" U_reac = 1*(enthalpy(CH4, T=T_reac) - R_u*T_reac) +(1+EX)*A_th*(enthalpy(O2,T=T_reac) - R_u*T_reac) U_prod = 1*(enthalpy(CO2, T=T_prod) - R_u*T_prod) +2*(enthalpy(H2O, T=T_prod) R_u*T_prod)+EX*A_th*(enthalpy(O2,T=T_prod) - R_u*T_prod) Q_out_tot=Q_out/(Mw_CH4)*m_CH4 "The total heat transfer out" "The final pressure in the tank is the pressure of the product gases. Assuming ideal gas behavior for the gases in the constant volume tank, the ideal gas law gives:" P_reac*V =N_reac * R_u *T_reac P_prod*V = N_prod * R_u * T_prod N_reac = N_CH4*(1 + N_O2) N_prod = N_CH4*(1 + 2 + Ex*A_th)

Tprod

Qout,tot

Pprod

[K] 500 700 900 1100 1300 1500

[kJ] 5872 5712 5537 5349 5151 4943

[KPa] 335.6 469.8 604 738.3 872.5 1007

16-71 A stoichiometric mixture of octane gas and air contained in a closed combustion chamber is ignited. The heat transfer from the combustion chamber is to be determined. Assumptions 1. Both the reactants and products are ideal gases. 2. Combustion is complete. Analysis The theoretical combustion equation of C8 H18 with stoichiometric amount of air is

C8 H18 (g )+ ath (O2 + 3.76N 2 ) ¾ ¾® 8CO2 + 9H 2 O + 3.76ath N 2

13-412

where ath is the stoichiometric coefficient and is determined from the O2 balance,

ath = 8 + 4.5 = 12.5 Thus,

C8 H18 (g )+ 12.5(O 2 + 3.76N 2 ) ¾ ¾® 8CO2 + 9H 2 O + 47N 2

The heat transfer for this constant volume combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with Wother = 0 ,

- Qout = å N P (h f + h - h - Pv ) - å N R (h f + h - h - Pv ) P

For a constant pressure quasi-equilibrium process D U + Wb = D H . Then the first law relation in this case is

- Qout = å N P (h f + h1000 K - h298 K ) - å N R h f , R P

since the reactants are at the standard reference temperature of 25C. Since both the reactants and the products behave as ideal gases, we have h = h(T). From the tables, Substance

hf kJ / kmol

h 298 K

h1000 K

kJ / kmol

---

C8 H18 ( g ) O2

8682

31,389

8669

30,129

H2 O ( g )

9904

35,882

CO2

9364

42,769

Thus,

- Qout = (8)(- 393,520 + 42,769 - 9364)+ (9)(- 241,820 + 35,882 - 9904)+ (47 )(0 + 30,129 - 8669)- (1)(- 208, 450)- 0 - 0 = - 3,606, 428 kJ (per kmol of C8 H18 ) or

Qout = 3,606, 428 kJ (per kmol of C8 H18 ).

Total mole numbers initially present in the combustion chamber is determined from the ideal gas relation,

N1 =

(300 kPa )(0.5 m3 ) P1V1 = = 0.06054 kmol Ru T1 (8.314 kPa ×m3 /kmol ×K )(298 K )

Of these, 0.06054 / (1 + 12.5´ 4.76) = 1.001´ 10 kmol of them is C8 H18 . Thus the amount of heat transferred from the combustion chamber as 1.001´ 10- 3 kmol of C8 H18 is burned is -3

Qout = (1.001´ 10- 3 kmol C8 H18 )(3,606,428 kJ/kmol C8 H18 ) = 3610 kJ

13-413

EES (Engineering Equation Solver) SOLUTION "Given" P1=300 [kPa] V1=0.5 [m^3] T1=(25+273) [K] T2=1000 [K] T0=(25+273) [K] "Analysis" "The stoichiometric combustion equation is: C8H18 + a_th (O2+3.76N2) = 8 CO2 + 9 H2O + 3.76*a_th N2" a_th=8+4.5 "O2 balance" "Mol numbers of reactants and products in kmol" N_C8H18=1 N_O2=a_th N_N2=a_th*3.76 N_CO2=8 N_H2O=9 "Ideal gas enthalpies in kJ/kmol" h_f_C8H18=enthalpy(C8H18, T=T0) h_O2=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_N2_P=enthalpy(N2, T=T2) h_CO2=enthalpy(CO2, T=T2) h_H2O=enthalpy(H2O, T=T2) H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_N2*h_N2_P H_R=N_C8H18*h_f_C8H18+N_O2*h_O2+N_N2*h_N2_R -Q_out=H_P-H_R N1=(P1*V1)/(R_u*T1) R_u=8.314 [kPa-m^3/kmol-K] N_fuel=N1/(1+a_th*4.76) Q_out_kJ=N_fuel*Q_out

16-72 A high efficiency gas furnace burns gaseous propane C3 H8 with 140 percent theoretical air. The volume flow rate of water condensed from the product gases is to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The reaction equation for 40% excess air (140% theoretical air) is

C3 H8 + 1.4ath [O2 + 3.76N 2 ] ¾ ¾® B CO 2 + D H 2 O + E O 2 + F N 2 where ath is the stoichiometric coefficient for air. We have automatically accounted for the 40% excess air by using the factor 1.4ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance: Hydrogen balance:

B=3 2D = 8 ¾ ¾ ® D= 4

Oxygen balance:

2´ 1.4ath = 2B + D + 2E

0.4ath = E Nitrogen balance:

1.4ath ´ 3.76 = F

13-414

Solving the above equations, we find the coefficients (E = 2, F = 26.32, and ath = 5 ) and write the balanced reaction equation as

C3 H8 + 7 [O 2 + 3.76N 2 ] ¾ ¾® 3 CO 2 + 4 H 2 O + 2 O 2 + 26.32 N 2 The partial pressure of water in the saturated product mixture at the dew point is Pv ,prod = Psat@40°C = 7.3851 kPa

The vapor mole fraction is yv =

Pv ,prod Pprod

7.3851 kPa = 0.07385 100 kPa

The kmoles of water condensed is determined from yv =

N water N total,product

¾¾ ® 0.07385 =

4 - Nw ¾¾ ® N w = 1.503 kmol 3 + 4 - N w + 2 + 26.32

The steady-flow energy balance is expressed as

N fuel H R = Qfuel + N fuel H P where

Qfuel =

Qout 25, 000 kJ/h = = 26, 042 kJ/h hfurnace 0.96

H R = hfuel @ 25 ° C + 7hO2 @ 25°C + 26.32hN2 @ 25°C

= (- 103,850 kJ/kmol) + 7(0) + 26.32(0) = - 103,850 kJ/kmol H P = 3hCO2@25°C + 4hH2O@25°C + 2hO2@25°C + 26.32hN2@25°C + Nw (h fo H2O(liq) )

= 3(- 393,520 kJ/kmol) + 4( - 241,820 kJ/kmol) + 2(0) + 26.32(0) + 1.503(- 285,830 kJ/kmol) = - 2.577 ´ 106 kJ/kmol

Substituting into the energy balance equation,

N fuel H R = Qfuel + N fuel H P

Nfuel (- 103,850 kJ/kmol) = 26,042 kJ/h + Nfuel (- 2.577´ 106 kJ/kmol) ¾ ¾® Nfuel = 0.01053 kmol/h The molar and mass flow rates of the liquid water are

N w = N w Nfuel = (1.503 kmol/kmol fuel)(0.01053 kmol fuel/h) = 0.01582 kmol/h mw = N w M w = (0.01582 kmol/h)(18 kg/kmol) = 0.2850 kg/h The volume flow rate of liquid water is

Vw = (v f @25°C )mw = (0.001003 m3 /kg)(0.2850 kg/h) = 0.0002859 m3 /h = 6.86 L / day

EES (Engineering Equation Solver) SOLUTION "The reaction equation for 140% theroetical air is:" "C3H8+ 1.4*A_th (O2 +3.76N2)=B CO2 + D H2O + E O2 + F N2" "Carbon Balance:" 3=B "Hydrogen Balance:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-415

8=2*D "Oxygen Balance:" 1.4*A_th*2=B*2 + D + E*2 "Excess oxygen in the products is:" 0.4*A_th = E "Nitrogen Balance:" 1.4*A_th*3.76 = F "The product temperature is:" T_P = 25+273.15 "[K]" "The reactant temperature is:" T_R= 25+273.15 "[K]" "The combustion efficiency is:" eta_b = 0.96 Q_dot_fuel=Q_dot_out/eta_b "The required heat transfer rate is" Q_dot_out = 25000 "[kJ/h]" "The steady-flow energy balance is:" N_dot_fuel*H_R = Q_dot_fuel+N_dot_fuel*H_P "kmol water condensed:" "The product dew point temperature is:" T_DP = 40 +273.15"[K]" "The product pressure is:" P_prod = 100 "[kPa]" "The partial pressure of the water in the saturated product mixture at the dew point is:" P_v_prod = pressure(steam_iapws, T=T_DP, x=1) "[kPa]" P_v_prod = y_v * P_prod "The kmoles of water condensed are determined from the vapor mole fraction in the products:" y_v = (D-N_liq)/(B + D-N_liq+ E + F) h_bar_f_H2Oliq=-285830 H_R=1*h_C3H8_R+1.4*A_th*h_O2_R+1.4*A_th*3.76*h_N2_R H_P=B*h_CO2_P+D*h_H2O_P+N_liq*(h_bar_f_H2Oliq + h_steam_P-h_steam_0)+E*h_O2_P+F*h_N2_P h_C3H8_R=-103850 "ENTHALPY(C3H8,T=T_R)" h_O2_R=ENTHALPY(O2,T=T_R) h_N2_R=ENTHALPY(N2,T=T_R) h_CO2_P=ENTHALPY(CO2,T=T_P) h_H2O_P=ENTHALPY(H2O,T=T_P) h_steam_P=ENTHALPY(steam_iapws, T=T_P,x=0) h_steam_0=ENTHALPY(steam_iapws, T=298.15,x=0) h_O2_P=ENTHALPY(O2,T=T_P) h_N2_P=ENTHALPY(N2,T=T_P) N_dot_liq=N_liq*N_dot_fuel "The mass flow rate of fuel is:" MM_liq = MOLARMASS(H2O) m_dot_liq = N_dot_liq*MM_liq "The volume flow rate of liquid water is:" v_liq = VOLUME(Steam_iapws,T=T_P,x=0) V_dot_liq = N_dot_liq*v_liq"[m^3/h]" * convert(m^3/h,L/day) "[gal/day]"

16-73E A mixture of benzene gas and 60 percent excess air contained in a constant-volume tank is ignited. The heat transfer from the combustion chamber is to be determined. Assumptions 1. Both the reactants and products are ideal gases. 2. Combustion is complete.

13-416

Analysis The theoretical combustion equation of C6 H6 with stoichiometric amount of air is

C6 H 6 (g )+ ath (O2 + 3.76N 2 ) ¾ ¾® 6CO 2 + 3H 2 O + 3.76ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

ath = 6 + 1.5 = 7.5 Then the actual combustion equation with 60% excess air becomes

C6 H 6 (g )+ 12 (O 2 + 3.76N 2 ) ¾ ¾® 5.52CO 2 + 0.48CO + 3H 2O + 4.74O 2 + 45.12N 2 The heat transfer for this constant volume combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h - Pv ) - å N R (h f + h - h - Pv ) P

Since both the reactants and the products behave as ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT . It yields

- Qout = å N P (h f + h1800 R - h537 R - RuT ) - å N R (h f - RuT ) P

since the reactants are at the standard reference temperature of 77F. From the tables,

h 537

h 2100 R

Substance

hf Btu / Ibmol

Btu / Ibmol

C6 H6 ( g )

35,680

---

3725.1

16,011

3729.5

15,334

H2 O ( g )

−104,040

4258.0

18,467

−47,540

3725.1

15,463

CO2

−169,300

4027.5

22,353

Thus,

- Qout = (5.52)(- 169,300 + 22,353 - 4027.5 - 1.986 ´ 2100) + (0.48)(- 47,540 + 15,463 - 3725.1- 1.986´ 2100) + (3)(- 104,040 + 18,467 - 4258.0 - 1.986 ´ 2100) + (4.74)(0 + 16,011- 3725.1- 1.986´ 2100) + (45.12)(0 + 15,334 - 3729.5 - 1.986´ 2100) - (1)(35,680 - 1.986´ 537)- (12)(4.76)(- 1.986´ 537)

13-417

= - 757,400 Btu or

Qout = 757,400 Btu

EES (Engineering Equation Solver) SOLUTION "Given" T1=(77+460) [R] ex=1.6 "160% theoretical air" T2=2100 [R] T0=(77+460) [R] "Properties" R_u=1.986 [Btu/lbmol-R] "Analysis" "The combustion equation is: C6H6 + ex*a_th (O2+3.76N2) = 5.52 CO2 + 0.48 CO + 3 H2O + 2.49 O2 + ex*3.76*a_th N2" a_th=6+1.5 "O2 balance of stoichiometric equation" "Mol numbers of reactants and products in lbmol" N_C6H6=1 N_O2_R=ex*a_th N_O2_P=4.74 N_N2=ex*a_th*3.76 N_H2O=3 N_CO2=5.52 N_CO=0.48 "Enthalpy of formation data from Table A-26E in Btu/lbmol" h_f_C6H6=35680 "Ideal gas enthalpies in kJ/kmol" h_O2_R=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) h_CO=enthalpy(CO, T=T2) H_P=N_CO2*(h_CO2-R_u*T2)+N_CO*(h_CO-R_u*T2)+N_H2O*(h_H2O-R_u*T2)+N_O2_P*(h_O2_PR_u*T2)+N_N2*(h_N2_P-R_u*T2) H_R=N_C6H6*(h_f_C6H6-R_u*T1)+N_O2_R*(h_O2_R-R_u*T1)+N_N2*(h_N2_R-R_u*T1) -Q_out=H_P-H_R

Adiabatic Flame Temperature 16-74C For the case of stoichiometric amount of pure oxygen since we have the same amount of chemical energy released but a smaller amount of mass to absorb it.

16-75C Under the conditions of complete combustion with stoichiometric amount of air.

16-76E Hydrogen is burned with 20 percent excess air during a steady-flow combustion process. The exit temperature of product gases is to be determined. Assumptions

13-418

1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5 The combustion chamber is adiabatic. Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber under adiabatic conditions (Q = 0) with no work interactions (W = 0). Under steady-flow conditions the energy balance

Ein - Eout = D Esystem applied on the combustion chamber reduces to

å N (h + h - h ) = å N (h + h - h ) P

The combustion equation of H 2 with 20% excess air is

H 2 + 0.6 (O2 + 3.76N 2 ) ¾ ¾® H 2 O + 0.1O 2 + 2.256N 2 From the tables,

h 500

h 537

Substance

hf Btu / Ibmol

Btu / Ibmol

3386.1

3640.3

3466.2

3725.1

3472.2

3729.5

H2 O ( g )

−104,040

3962.0

4258.0

Thus,

(1)(- 104, 040 + hH O - 4258)+ (0.1)(0 + hO - 3725.1)+ (2.256)(0 + hN - 3729.5) 2

= (1)(0 + 3386.1- 3640.3)+ (0.6)(0 + 3466.2 - 3725.1)+ (2.256)(0 + 3472.2 - 3729.5) It yields

hH2 O + 0.1hO2 + 2.256hN2 = 116, 094 Btu The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 116,094 / (1 + 0.1 + 2.256) = 34,593 Btu / lbmol . This enthalpy value corresponds to about 4400 R for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 4400 R, but somewhat under it because of the higher specific heat of H2 O . At 4020 R:

hH2 O + 0.1hO2 + 2.256hN2 = (1)(40, 740)+ (0.1)(32,989)+ (2.256)(31,503)

= 115,110 Btu At 4100 R:

hH2 O + 0.1hO2 + 2.256hN2 = (1)(41, 745)+ (0.1)(33, 722)+ (2.256)(32,198)

= 117,756 Btu By interpolation,

(Lower than 116,094 Btu ) (Higher than 116,094 Btu )

TP = 4054 R

13-419

EES (Engineering Equation Solver) SOLUTION "Given" ex=1.2 "120% theoretical air" T1=40+460 "[R]" T0=77+460 "[R]" "Analysis" "The combustion equation is: H2 + ex*a_th (O2+3.76N2) = H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=0.5+(ex-1)*a_th "O2 balance" "Mol numbers of reactants and products in lbmol" N_H2=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_H2O=1 "Ideal gas enthalpies in Btu/lbmol" h_H2=enthalpy(H2, T=T1) h_O2_R=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) H_P=N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_H2*h_H2+N_O2_R*h_O2_R+N_N2*h_N2_R H_P=H_R

16-77 Methane is burned with 30 percent excess air. The adiabatic flame temperature is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with Q = W = 0 reduces to

å N (h + h - h ) = å N (h + h - h ) P

¾ ¾®

å N (h + h - h ) = å N h P

R f,R

since all the reactants are at the standard reference temperature of 25°C. Then,

CH 4 + 1.3´ ath (O2 + 3.76 N 2 ) ¾ ¾® CO2 + 2H 2O + 0.3O2 + 1.3´ ath ´ 3.76N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance, 1.3ath = 1 + 1 + 0.3ath

¾¾ ®

ath = 2

Thus, CH 4 + 2.6(O2 + 3.76 N 2 ) ¾ ¾ ® CO2 + 2H 2 O + 0.3O2 + 9.776N 2 From the tables,

13-420

Substance

hf kJ / kmol

kJ / kmol

H4 ( g )

−74,850

---

8682

8669

H2 O ( g )

−241,820

9904

CO2

−393,520

9364

h 298 K

Thus,

(1)(- 393,520 + hCO2 - 9364)+ (2)(- 241,820 + hH2O - 9904)+ (0.3)(0 + hO2 - 8682) + (9.776)(0 + hN2 - 8669) = (1)(- 74,850)+ 0 + 0

hCO2 + 2hH2O + 0.3hO2 + 9.776hN2 = 918,835 kJ

It yields

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 918,835 / (1 + 2 + 0.3 + 9.776) = 70,269 kJ / kmol . This enthalpy value corresponds to about 2150 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2150 K, but somewhat under it because of the higher specific heat of H 2 O . At 2000 K:

hCO2 + 2hH2O + 0.3hO2 + 9.776hN2 = 100,804 + 2´ 82,593 + 0.3´ 67,881 + 9.776´ 64,810 = 919,936 kJ (Higher than 918,835 kJ) At 1980 K:

hCO2 + 2hH2O + 0.3hO2 + 9.776hN2 = 99,606 + 2´ 81,573 + 0.3´ 67,127 + 9.776´ 64,090 = 909, 434 kJ (Lower than 918,835 kJ) By interpolation,

TP = 1998 K = 1725°C

16-78 Octane is burned with 40 percent excess air adiabatically during a steady-flow combustion process. The exit temperature of product gases is to be determined. Assumptions

13-421

1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis Under steady-flow conditions the energy balance E in  E out  E system applied on the combustion chamber with Q = W = 0 reduces to

 N h  h  h    N h  h  h   f



 f



since all the reactants are at the standard reference temperature of 25°C. Then,

C 8 H18 l   1.4a th O 2  3.76N 2    8CO 2  9H 2 O  0.4a th O 2  1.43.76 a th N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.4a th  8  4.5  0.4a th   a th  12.5 Thus, C 8 H18 g   17.5O 2  3.76N 2    8CO 2  9H 2 O  5O 2  65.8N 2 From the tables,

h 298 K

h 580K

Substance

hf kJ / kmol

kJ / kmol

C8 H18 ( )

−249,950

---

8682

17,290

8669

16,962

H2 O ( g )

−241,820

9904

---

CO2

−393,520

9364

---

Thus,

8  393,520  hCO2  9364    9   241,820  hH2O  9904   5  0  hO2  8682 





  65.8 0  hN2  8669  1 249,950   17.5  0  17, 290  8682   (65.8)(16,962  8669)

8hCO2  9hH2O  5hO2  65.8h N2  6,548,788 kJ

It yields

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 6,548,788 / (8 + 9 + 5 + 65.8) = 74,588 kJ / kmol . This enthalpy value corresponds to about 2250 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2250 K, but somewhat under it because of the higher specific heat of H 2 O . At 2100 K:

8hCO2  9hH2O  5hO2  65.8hN2  8106,864   9 87,735   5  71,668   65.8 68, 417 

 6,504,706 kJ

 Lower than 6,548,788 kJ 

13-422

At 2150 K:

8hCO2  9hH2O  5hO2  65.8hN2  8109,898  9 90,330   5 73,573  65.8 70, 226 

 6,680,890 kJ

 Higher than 6,548,788 kJ 

By interpolation,

TP = 2113 K = 1840°C EES (Engineering Equation Solver) SOLUTION "Given" ex=0.4 T_air=307 [C] P_air=600 [kPa] T_fuel=25 [C] "Analysis" "Combustion equation is: C8H18 (l) + (ex+1)a_th (O2+3.76N2) = 8 CO2 + 9 H2O + (ex)(a_th) O2 + (ex+1)a_thx3.76 N2" a_th=8+9/2 "O2 balance" N_C8H18=1 [kmol] N_CO2=8 [kmol] N_H2O=9 [kmol] N_O2=ex*a_th N_N2=(ex+1)*a_th*3.76 N_O2_R=(ex+1)*a_th h_CO2=enthalpy(CO2, T=T_af) h_H2O=enthalpy(H2O, T=T_af) h_O2=enthalpy(O2, T=T_af) h_N2=enthalpy(N2, T=T_af) h_O2_R=enthalpy(O2, T=T_air) h_N2_R=enthalpy(N2, T=T_air) h_C8H18=-249950 [kJ/kmol] H_P=H_R H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2*h_O2+N_N2*h_N2 H_R=N_C8H18*h_C8H18+N_O2_R*h_O2_R+N_N2*h_N2_R

16-79 Octane gas is burned with 30 percent excess air during a steady-flow combustion process. The exit temperature of product gases is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with Q = W = 0 reduces to

å N (h + h - h ) = å N (h + h - h ) ¾ ¾® å N (h + h - h ) = å N h P

R f,R

since all the reactants are at the standard reference temperature of 25°C. Then,

C8 H18 (g )+ 1.3ath (O 2 + 3.76N 2 ) ¾ ¾® 8CO 2 + 9H 2 O + 0.3ath O 2 + (1.3)(3.76)ath N 2

13-423

where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.3ath = 8 + 4.5 + 0.3ath ¾ ¾® ath = 12.5 ® 8CO 2 + 9H 2 O + 3.75O 2 + 61.1N 2 Thus, C8 H18 (g )+ 16.25 (O 2 + 3.76N 2 ) ¾ ¾

Therefore, 16.25  4.76 = 77.35 kmol of dry air will be used per kmol of the fuel. The partial pressure of the water vapor present in the incoming air is

Pv , in = f air Psat @ 25°C = (0.60)(3.1698 kPa ) = 1.902 kPa Assuming ideal gas behavior, the number of moles of the moisture that accompanies 77.35 kmol of incoming dry air is determined to be æPv , in ö æ 1.902 kPa ÷ ö ÷ ÷ N v , in = ççç ® N v , in = 1.48 kmol ÷ ÷N total = ççèç ÷(77.35 + N v , in ) ¾ ¾ 101.325 kPa ø èç Ptotal ÷ ø

The balanced combustion equation is obtained by adding 1.48 kmol of H 2 O to both sides of the equation,

C8 H18 (g )+ 16.25(O 2 + 3.76N 2 )+ 1.48H 2 O ¾ ¾® 8CO 2 + 10.48H 2 O + 3.75O 2 + 61.1N 2 From the tables,

hf kJ / kmol

Substance

C8 H18 ( g ) O2 N2 H2 O ( g ) CO2

h 298 K

−208,450

kJ / kmol ---

8682

8669

−241,820

9904

−393,520

9364

Thus,

(8)(- 393,520 + hCO - 9364)+ (10.48)(- 241,820 + hH O - 9904)+ (3.75)(0 + hO - 8682) 2

+ (61.1)(0 + hN2 - 8669) = (1)(- 208, 450)+ (1.48)(- 241,820)+ 0 + 0 It yields

8hCO2 + 10.48hH2 O + 3.75hO2 + 61.1hN2 = 5,857, 029 kJ

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the right-hand side of the equation by the total number of moles, which yields 5,857,029 / (8 + 10.48 + 3.75 + 61.1) = 70,287 kJ / kmol . This enthalpy value corresponds to about 2150 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2150 K, but somewhat under it because of the higher specific heat of H 2 O . At 2000 K: 8hCO2 + 10.48hH2 O + 3.75hO2 + 61.1hN2 = (8)(100,804)+ (10.48)(82,593)+ (3.75)(67, 881)+ (61.1)(64,810)

= 5,886, 451 kJ (Higher than 5,857,029 kJ ) At 1980 K: 8hCO2 + 10.48hH2O + 3.75hO2 + 61.1hN2 = (8)(99, 606)+ (10.48)(81,573)+ (3.75)(67,127)+ (61.1)(64, 090)

= 5,819,358 kJ (Lower than 5,857,029 kJ )

13-424

By interpolation,

TP = 1991 K

EES (Engineering Equation Solver) SOLUTION "Given" ex=1.3 "130% theoretical air" T1=25+273 "[K]" P1=101.325 "[kPa]" phi=0.60 T0=25+273 "[K]" "Analysis" "The combustion equation is: C8H18 + ex*a_th (O2+3.76N2) = 8 CO2 + 9 H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=8+4.5+(ex-1)*a_th "O2 balance" P_sat=pressure(steam_NBS, T=T1, x=1) P_v_in=phi*P_sat N_v_in=(P_v_in/P_total)*N_total "moles of H2O in air" P_total=P1 N_total=ex*a_th*4.76+N_v_in "Then the combustion equation becomes: C8H18 + ex*a_th (O2+3.76N2) + N_v_in H2O = 8 CO2 + (9+N_v_in) H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" "Mol numbers of reactants and products in kmol" N_C8H18=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_H2O_R=N_v_in N_H2O_P=9+N_v_in N_CO2=8 "Ideal gas enthalpies in kJ/kmol" h_f_C8H18=enthalpy(C8H18, T=T0) h_O2_R=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_H2O_R=enthalpy(H2O, T=T1) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O_P=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2+N_H2O_P*h_H2O_P+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C8H18*h_f_C8H18+N_O2_R*h_O2_R+N_N2*h_N2_R+N_H2O_R*h_H2O_R H_P=H_R

16-80 Problem 16-79 is reconsidered. The effect of the relative humidity on the exit temperature of the product gases is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input data" Fuel$ = 'Octane (C8H18)' PercentEX = 30 [%] RelHum=60 [%] T_fuel = (25+273) [K] Ex = PercentEX/100 P_air1 = 101.3 [kPa]

13-425

T_air1 = (25+273) [K] RH_1 = RelHum/100 M_air = 28.97 [kg/kmol] M_water = 18 [kg/kmol] M_C8H18=(8*12+18*1) [kg/kmol] "For theoretical dry air, the complete combustion equation is" "C8H18 + A_th(O2+3.76 N2)=8 CO2+9 H2O + A_th (3.76) N2 " A_th*2=8*2+9*1 "theoretical O balance" "Now to find the amount of water vapor associated with the dry air" w_1=HUMRAT(AirH2O,T=T_air1,P=P_air1,R=RH_1) N_w=w_1*((1+Ex)*A_th*4.76*M_air)/M_water "Moles of water in the atmoshperic air, kmol/kmol_fuel" "The balanced combustion equation with Ex% excess moist air is" "C8H18 + (1+EX)[A_th(O2+3.76 N2)+N_w H2O]=8 CO2+(9+N_w) H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " H_fuel = -208450 [kJ/kmol] "from Table A-26" HR=H_fuel+ (1+Ex)*A_th*enthalpy(O2,T=T_air1)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air1)+N_w*enthalpy(H2O,T=T_air1) HP=8*enthalpy(CO2,T=T_prod)+(9+N_w)*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) HP = HR "For Adiabatic Combustion" "This solution used the humidity ratio form psychrometric data to determine the moles of water vapor in atomspheric air. One should calculate the moles of water contained in the atmospheric air by the method shown in Chapter 14, which uses the relative humidity to find the partial pressure of the water vapor and, thus, the moles of water vapor. Explore what happens to the results as you vary the percent excess air, relative humidity, and product temperature. " RelHum [%] 0 10 20 30 40 50 60 70 80 90 100

TProd [kJ] 2024 2019 2014 2008 2003 1997 1992 1987 1981 1976 1971

13-426

16-81 Acetylene gas is burned with 30 percent excess air during a steady-flow combustion process. The exit temperature of product gases is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol of C2 H 2 , the combustion equation can be written as

C2 H 2 + 1.3ath (O2 + 3.76N 2 ) ¾ ¾® 2CO 2 + H 2 O + 0.3ath O 2 + (1.3)(3.76)ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.3ath = 2 + 0.5 + 0.3ath ¾ ¾® ath = 2.5 Thus,

C2 H 2 + 3.25(O2 + 3.76N 2 ) ¾ ¾® 2CO 2 + H 2 O + 0.75O 2 + 12.22N 2 Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

Substance

C2 H 2 O2 N2 H2 O ( g ) CO2

h 298 K

h 300 K

226,730

kJ / kmol ---

8682

8736

8669

8723

−241,820

9904

---

−393,520

9364

---

hf kJ / kmol

Thus,

- 75,000 = (2)(- 393,520 + hCO2 - 9364)+ (1)(- 241,820 + hH2O - 9904) + (0.75)(0 + hO2 - 8682)+ (12.22)(0 + hN2 - 8669)- (1)(226,730)

- (3.25)(0 + 8736 - 8682)+ (12.22)(0 + 8723 - 8669) It yields

2hCO2 + hH2 O + 0.75hO2 + 12.22hN2 = 1,321,184 kJ

13-427

The temperature of the product gases is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 1,321,184 / (2 + 1 + 0.75 + 12.22) = 82,729 kJ / kmol . This enthalpy value corresponds to about 2500 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2500 K, but somewhat under it because of the higher specific heats of CO2 and H2 O . At 2350 K: 2hCO2 + hH2 O + 0.75hO2 + 12.22hN2 = (2)(122, 091)+ (1)(100,846)+ (0.75)(81, 243)+ (12.22)(77, 496)

= 1,352,961 kJ (Higher than 1,321,184 kJ ) At 2300 K: 2hCO2 + hH2 O + 0.75hO2 + 12.22hN2 = (2)(119, 035)+ (1)(98,199)+ (0.75)(79,316)+ (12.22)(75, 676)

= 1,320,517 kJ (Lower than 1,321,184 kJ ) By interpolation,

TP = 2301 K

EES (Engineering Equation Solver) SOLUTION "Given" ex=1.3 "130% theoretical air" T1_fuel=25+273 "[K]" T1_air=27+273 "[K]" Q_out=75000 "[kJ] per kmol C2H2" T0=25+273 "[K]" "Analysis" "The combustion equation is: C2H2 + ex*a_th (O2+3.76N2) = 2 CO2 + H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=2+0.5+(ex-1)*a_th "O2 balance" "Mol numbers of reactants and products in kmol" N_C2H2=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_H2O=1 N_CO2=2 "Ideal gas enthalpies in kJ/kmol" h_f_C2H2=enthalpy(C2H2, T=T0) h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C2H2*h_f_C2H2+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R

16-82 Ethyl alcohol is burned with 200 percent excess air adiabatically in a constant volume container. The final pressure and temperature of product gases are to be determined. Assumptions 1. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible.

13-428

4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis The combustion equation is

C2 H5 OH + 3ath [O 2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3 H 2 O + 3ath ´ 3.76 N 2 + 2ath O 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

0.5 + 3ath = 2 + 1.5 + 2ath

¾ ¾®

ath = 3

Substituting,

C2 H5 OH + 9[O 2 + 3.76N 2 ] ¾ ¾® 2CO2 + 3 H 2 O + 33.84 N 2 + 6O 2 For this constant-volume process, the energy balance Ein - Eout = D Esystem applied on the combustion chamber with Q = W = 0 reduces to

å N (h + h - h - Pv ) = å N (h + h - h - Pv ) P

å N (h + h - h - R T ) = å N (h + h - h - R T ) P

From the tables, hf

h 298 K

Substance

kJ / kmol

C2 H5OH (g )

−235,310

---

8682

8669

H2 O ( g )

−241,820

9904

CO2

−393,520

9364

Thus,

(2)(- 393,520 + hCO2 - 9364 - 8.314´ Tp )+ (3)(- 241,820 + hH2O - 9904 - 8.314´ Tp ) + (6)(0 + hO2 - 8682 - 8.314´ Tp )

+ (33.84)(0 + hN2 - 8669 - 8.314´ Tp )

= (1)(- 235,310 - 8.314´ 298)+ (9)(0 - 8.314 ´ 298)+ (33.84)(- 8.314 ´ 298)

It yields 2hCO2 + 3hH2O + 6hO2 + 33.84hN2 - 372.8Tp = - 343,927 + 1,906,391 = 1,562, 464 kJ

The adiabatic flame temperature is obtained from a trial and error solution. A first guess may be obtained by assuming all the products are nitrogen and using nitrogen enthalpy in the above equation. That is, 44.84hN2 - 372.8Tp = 1,562, 464 kJ

An investigation of Table A-18 shows that this equation is satisfied at a temperature close to 1600 K but it will be somewhat under it because of the higher specific heat of H 2 O . © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-429

At 1500 K: (2)(71, 078) + (3)(57,999) + (6)(49, 292) + (33.84)(47, 073) - (372.8)(1500) = 1, 645, 655

(Higher than 1,562, 464 kJ) At 1400 K: (2)(65, 271) + (3)(53,351) + (6)(45, 648) + (33.84)(43, 605) - (372.8)(1400) = 1,518,156

(Lower than 1,562, 464 kJ) By interpolation,

TP = 1435 K The volume of reactants when 1 kmol of fuel is burned is

V = Vfuel + Vair = ( Nfuel + Nair )

RuT (8.314 kJ/kmol ×K)(298 K) = (1 + 42.84) kmol) = 1086 m3 P 100 kPa

The final pressure is then

P = N prod

RuT (8.314 kJ/kmol ×K)(1435 K) = (44.84 kmol) = 493 kPa V 1086 m3

16-83 Methane is burned with 300 percent excess air adiabatically in a constant volume container. The final pressure and temperature of product gases are to be determined. Assumptions 1. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis The combustion equation is CH 4 + 4ath [O 2 + 3.76N 2 ]¾ ¾ ® CO 2 + 2 H 2 O + 4ath ´ 3.76 N 2 + 3ath O 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

4ath = 1 + 1 + 3ath ¾ ¾ ® ath = 2 ® CO 2 + 2 H 2 O + 30.08 N 2 + 6 O 2 Substituting, CH 4 + 8[O 2 + 3.76N 2 ]¾ ¾

For this constant-volume process, the energy balance Ein - Eout = D Esystem applied on the combustion chamber with Q = W = 0 reduces to

å N (h + h - h - Pv ) = å N (h + h - h - Pv ) P

å N (h + h - h - R T ) = å N (h + h - h - R T ) P

From the tables,

13-430

h 298 K

Substance

hf kJ / kmol

kJ / kmol

CH4 ( g )

−74,850

---

8682

8669

H2 O ( g )

−241,820

9904

CO2

−393,520

9364

Thus,

(1)(- 393,520 + hCO2 - 9364 - 8.314´ Tp )+ (2)(- 241,820 + hH2O - 9904 - 8.314´ Tp ) + (6)(0 + hO2 - 8682 - 8.314´ Tp )+ (30.08)(0 + hN2 - 8669 - 8.314´ Tp ) = (1)(- 74,850 - 8.314´ 298)+ (8)(0 - 8.314´ 298)+ (30.08)(- 8.314 ´ 298)

It yields hCO2 + 2hH2O + 6hO2 + 30.08hN2 - 324.9Tp = - 171, 674 + 1, 219,188 = 1, 047,514 kJ

An investigation of Table A-18 shows that this equation is satisfied at a temperature close to 1200 K but it will be somewhat under it because of the higher specific heat of H 2 O . At 1140 K: (50, 484) + (2)(41, 780) + (6)(36,314) + (30.08)(34, 760) - (324.9)(1140) = 1, 027,123 (Lower than 1,047,514 kJ) At 1160 K: (51, 602) + (2)(42, 642) + (6)(37, 023) + (30.08)(35, 430) - (324.9)(1160) = 1, 047,874 (Higher than 1,047,514 kJ) By interpolation,

TP = 1159.7 @1160 K The volume of reactants when 1 kmol of fuel is burned is

V = Vfuel + Vair = ( Nfuel + Nair )

RuT (8.314 kJ/kmol ×K)(298 K) = (1 + 38.08) kmol) = 955.8 m3 P 101.3 kPa

The final pressure is then

P = N prod

RuT (8.314 kJ/kmol ×K)(1160 K) = (39.08 kmol) = 394 kPa V 955.8 m3

EES (Engineering Equation Solver) SOLUTION "Given" ex=3 T_reac=(25+273) [K] P_reac=101.325 [kPa] "Analysis" "Combustion equation is: CH4 + (ex+1)a_th (O2+3.76N2) = 1 CO2 + 2 H2O + (ex)(a_th) O2 + (ex+1)a_thx3.76 N2" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-431

a_th=1+2/2 "O2 balance" N_CH4=1 [kmol] N_CO2=1 [kmol] N_H2O=2 [kmol] N_O2=ex*a_th N_N2=(ex+1)*a_th*3.76 N_O2_R=(ex+1)*a_th N_air=(ex+1)*a_th*4.76 h_CO2=enthalpy(CO2, T=T_af) h_H2O=enthalpy(H2O, T=T_af) h_O2=enthalpy(O2, T=T_af) h_N2=enthalpy(N2, T=T_af) h_CH4=enthalpy(CH4, T=T_reac) h_O2_R=enthalpy(O2, T=T_reac) h_N2_R=enthalpy(N2, T=T_reac) R_u=8.314 [kJ/kmol-K] H_P=H_R H_P=N_CO2*(h_CO2-R_u*T_af)+N_H2O*(h_H2O-R_u*T_af)+N_O2*(h_O2-R_u*T_af)+N_N2*(h_N2-R_u*T_af) H_R=N_CH4*(h_CH4-R_u*T_reac)+N_O2_R*(h_O2_R-R_u*T_reac)+N_N2*(h_N2_R-R_u*T_reac) Vol=(N_CH4+N_air)*(R_u*T_reac)/P_reac P_prod=(N_CH4+N_air)*(R_u*T_af)/Vol

16-84 A certain coal is burned with 50 percent excess air adiabatically during a steady-flow combustion process. The temperature of product gases is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5 The combustion chamber is adiabatic. Properties The molar masses of C, H 2 , N 2 , O2 , S, and air are 12, 2, 28, 32, 32, and 29 kg / kmol , respectively (Table A1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are

mf C =

mC 79.61 kg 79.61 kg = = = 0.8712 mtotal (100-8.62) kg 91.38 kg

mf H2 =

mH2 4.66 kg = = 0.05100 mtotal 91.38 kg

mf O2 =

mO2 4.76 kg = = 0.05209 mtotal 91.38 kg

13-432

mf N2 =

mN2 1.83 kg = = 0.02003 mtotal 91.38 kg

mfS =

mS 0.52 kg = = 0.00569 mtotal 91.38 kg

We now consider 100 kg of this mixture. Then the mole numbers of each component are

NC =

mC 87.12 kg = = 7.26 kmol M C 12 kg/kmol

N H2 =

mH2 5.10 kg = = 2.55 kmol M H2 2 kg/kmol

N O2 =

mO2 5.209 kg = = 0.1628 kmol M O2 32 kg/kmol

N N2 =

mN2 2.003 kg = = 0.07154 kmol M N2 28 kg/kmol

NS =

mS 0.569 kg = = 0.01778 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are

Nm = 7.26 + 2.55 + 0.1628 + 0.07154 + 0.01778 = 10.062 kmol yC =

NC 7.26 kmol = = 0.7215 N m 10.06 kmol

yH2 =

N H2 2.55 kmol = = 0.2535 Nm 10.06 kmol

yO2 =

N O2 0.1628 kmol = = 0.01618 Nm 10.06 kmol

yN2 =

N N2 0.07154 kmol = = 0.00711 Nm 10.06 kmol

yS =

NS 0.01778 kmol = = 0.00177 Nm 10.06 kmol

Then, the combustion equation in this case may be written as

0.7215C + 0.2535H 2 + 0.01618O2 + 0.00711N 2 + 0.00177S + 1.5ath (O2 + 3.76N 2 ) ¾ ¾® xCO2 + yH 2 O + zSO2 + kN 2 + 0.5ath O2 According to the species balances,

C balance : x = 0.7215

H2 balance : y = 0.2535 S balance : z = 0.00177

13-433

0.01618 + 1.5ath = x + 0.5 y + z + 0.5ath 1.5ath - 0.5ath = 0.7215 + 0.5(0.2535) + 0.00177 - 0.01617 ¾ ¾ ® ath = 0.8339 0.00711 + 1.5´ 3.76ath = k ¾ ¾ ® k = 0.00711 + 1.5´ 3.76´ 0.8339 = 4.710

N 2 balance :

Substituting,

0.7215C + 0.2535H 2 + 0.01617O2 + 0.00711N 2 + 0.00177S + 1.2509(O2 + 3.76N 2 ) ¾ ¾® 0.7215CO2 + 0.2535H 2O + 0.0018SO2 + 4.71N2 + 0.4170O2 Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with Q = W = 0 reduces to

å N (h + h - h ) = å N (h + h - h ) P

¾¾ ®

å N (h + h - h ) = å N h P

f,R

From the tables, hf

h 298 K

h 400 K

Substance

kJ / kmol

8682

11,711

8669

11,640

H2 O ( g )

−241,820

9904

---

CO2

−393,520

9364

---

Thus,

(0.7215)(- 393,520 + hCO2 - 9364)+ (0.2535)(- 241,820 + hH2O - 9904)+ (0.4170)(0 + hO2 - 8682) + (4.71)(0 + hN2 - 8669) = (1.2509)(0 + 11, 711- 8682)+ (4.703)(11, 640 - 8669)

0.7215hCO2 + 0.2535hH2O + 0.4170hO2 + 4.71hN2 = 416,706 kJ

It yields

The product temperature is obtained from a trial and error solution. A first guess is obtained by dividing the right-hand side of the equation by the total number of moles, which yields 416,706 / (0.7215 + 0.2535 + 0.4170 + 4.71) = 68, 290 kJ / kmol This enthalpy value corresponds to about 2100 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2100 K, but somewhat under it because of the higher specific heat of H 2 O . At 2000 K: 0.7215hCO2 + 0.2535hH2O + 0.4170hO2 + 4.71hN2 = (0.7215)(100,804) + (0.2535)(82,593) + (0.4170)(67,881) + (4.71)(64,810)

= 427, 229 kJ (Higher than 416,706 kJ) At 1980 K: 0.7215hCO2 + 0.2535hH2O + 0.4170hO2 + 4.71hN2 = (0.7215)(99, 606) + (0.2535)(81,573) + (0.4170)(67,127) + (4.71)(64, 090)

= 422, 400 kJ (Higher than 416,706 kJ) By extrapolation,

TP = 1956 K = 1683°C

16-85 A certain coal is burned with 50 percent excess air adiabatically during a steady-flow combustion process. The combustion gases are expanded in an isentropic turbine. The work produced by this turbine is to be determined. Assumptions © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-434

1. 2. 3. 4.

Steady operating conditions exist. Combustion gases are ideal gases. Kinetic and potential energies are negligible. The effect of sulfur in the energy and entropy balances is negligible.

Properties The molar masses of C, O2 , H 2 , S, and air are 12, 32, 2, 32, and 29 kg / kmol , respectively (Table A-1). Analysis The balanced combustion equation from the previous problem is 0.7215C + 0.2535H 2 + 0.01617O2 + 0.00711N 2 + 0.00177S + 1.2509(O2 + 3.76N 2 ) ¾ ¾® 0.7215CO2 + 0.2535H 2O + 0.0018SO 2 + 4.71N2 + 0.4170O2

The change of the entropy of this mixture during the expansion is given by æ Pö ÷ D s = å y çççs2o - s1o - R ln 2 ÷ ÷ ÷ çè P1 ø

Since there is no entropy change during an isentropic process, this equation becomes P2

140 kPa

å y (s - s ) = R ln P å y = (8.314 kJ/kmol ×K) ln 1380 kPa (6.1038 kmol) = - 116.1 kJ/K o 2

o 1

Here, the total moles of the products are 6.1038 kmol. The exit temperature of the combustion gases may be determined by a trial-error method. The inlet temperature is 1956 K ( 1960 K) from the previous problem. Guessing the exit temperature to be 1000 K gives (Tables A-18 through A-23)

å y (s - s ) = (0.7215)(269.215 - 307.992) + (0.2535)(232.597 - 263.542) + (4.71)(228.057 - 251.242) + (0.417)(243.471- 267.891) o 2

o 1

= - 155.2 kJ/K For the exit temperature of 1200 K,

å y (s - s ) = (0.7215)(297.307 - 307.992) + (0.2535)(240.333 - 263.542) + (4.71)(234.115 - 251.242) + (0.417)(249.906 - 267.891) o 2

o 1

= - 101.8 kJ/K By interpolation, T2 = 1146 K The work produced during the isentropic expansion of combustion gases from 1956 K to 1146 K is (Tables A-18 through A-23)

wout = å y(h1 - h2 ) = (0.7215)(98,160 - 50,819) + (0.2535)(80,352 - 42,037) + (4.71)(63, 236 - 34,961) + (0.417)(66, 223 - 36,527) = 189, 428 kJ/kmol fuel

The molar mass of the product gases is Mm =

mm (0.7215´ 44 + 0.2535´ 18 + 0.0018´ 64 + 4.71´ 28 + 0.4170 ´ 32) kg = = 29.76 kg/kmol Nm (0.7215 + 0.2535 + 0.0018 + 4.71 + 0.417) kmol

The work produced per kg of the fuel is then wout =

wout 189,428 kJ/kmol fuel = = 6365 kJ / kg fuel Mm 29.76 kg/kmol

Entropy Change and Second Law Analysis of Reacting Systems 16-86C Assuming the system exchanges heat with the surroundings at T0 , the increase-in-entropy principle can be expressed as

13-435

Sgen = å N P sP - å N R sR +

Qout T0

16-87C By subtracting Rln( P / P0 ) from the tabulated value at 1 atm. Here P is the actual pressure of the substance and P0 is the atmospheric pressure.

16-88C It represents the reversible work associated with the formation of that compound.

16-89 Liquid octane is burned steadily with 50 percent excess air. The heat transfer rate from the combustion chamber, the entropy generation rate, and the reversible work and exergy destruction rate are to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis (a) The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol C8 H18 , the combustion equation can be written as C8 H18 ( )+ 1.5ath (O 2 + 3.76N 2 ) ¾ ¾ ® 8CO 2 + 9H 2 O + 0.5ath O 2 + (1.5)(3.76 )ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.5ath = 8 + 4.5 + 0.5ath ¾ ¾ ® ath = 12.5 Thus, C8 H18 ( )+ 18.75(O 2 + 3.76N 2 ) ¾ ¾ ® 8CO 2 + 9H 2 O + 6.25O 2 + 70.5N 2

Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) = å N P h f , P - å N R h f , R P

since all of the reactants are at 25°C. Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

13-436

Substance

kJ / kmol

C8 H18 ( )

−249,950

H2 O (l )

−285,830

CO2

−393,520

Substituting, - Qout = (8)(- 393,520)+ (9)(- 285,830)+ 0 + 0 - (1)(- 249,950)- 0 - 0 = - 5, 470, 680 kJ/kmol of C8 H18 or

Qout = 5, 470,680 kJ/kmol of C8 H18 The C8 H18 is burned at a rate of 0.25 kg / min or Thus,

m 0.25 kg/min = = 2.193´ 10- 3 kmol/min é(8)(12)+ (18)(1)ù kg/kmol M ë û

Qout = NQout = (2.193´ 10- 3 kmol/min )(5,470,680 kJ/kmol) = 11, 997 kJ / min The heat transfer for this process is also equivalent to the enthalpy of combustion of liquid C8 H18 , which could easily be de determined from Table A-27 to be hC = 5, 470, 740 kJ / kmolC8 H18 . (b) The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q ¾¾ ® Sgen = å N P sP - å N R sR + out Tsurr Tsurr

The C8 H18 is at 25°C and 1 atm, and thus its absolute entropy is sC8 H18 = 360.79 kJ / kmol.K (Table A-26). The entropy values listed in the ideal gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components which is equal to Pi = yi Ptotal , where yi is the mole fraction of component i. Also,

Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm )) The entropy calculations can be presented in tabular form as

sio (T, 1 atm)

R u ln (y i Pm )

Ni si

C8 H18

1.00

360.79

---

360.79

18.75

0.21

205.14

−12.98

4089.75

70.50

0.79

191.61

−1.96

13646.69

SR = 18,097.23 kJ / K

CO2

0.0944

213.80

−19.62

1867.3

H2 O ( )

---

69.92

---

629.3

6.25

0.0737

205.04

−21.68

1417.6

13-437

70.50

0.8319

191.61

−1.53

13,616.3

SP = 17,531 kJ / K Thus, and

Sgen = S P - S R +

Qsurr 5,470,523 kJ = 17,531- 18, 097 + = 17, 798 kJ/kmol ×K Tsurr 298 K

Sgen = NSgen = (2.193´ 10- 3 kmol/min )(17,798 kJ/kmol ×K ) = 39.03 kJ / min ×K (c) The exergy destruction rate associated with this process is determined from X destroyed = T0 Sgen = (298 K)(39.03 kJ/min ×K) = 11,632 kJ/min = 193.9 kW

EES (Engineering Equation Solver) SOLUTION "Given" ex=1.5 "150% theoretical air" T1=(25+273) [K] T2=(25+273) [K] P=101.325 [kPa] m_dot_fuel=0.25 [kg/min] T0=(25+273) [K] "Properties" MM_fuel=molarmass(C8H18) "Analysis" "The combustion equation is: C8H18 (l) + ex*a_th (O2+3.76N2) = 8 CO2 + 9 H2O (l) + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=8+4.5+(ex-1)*a_th "O2 balance" "(a)" "Mol numbers of reactants and products in kmol" N_C8H18=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_H2O=9 N_CO2=8 "Enthalpy of formation data from Table A-26 in kJ/kmol" h_f_C8H18=-249950 [kJ/kmol] h_f_O2=0 [kJ/kmol] h_f_N2=0 [kJ/kmol] h_f_H2O=-285830 [kJ/kmol] h_f_CO2=-393520 [kJ/kmol] H_P=N_CO2*h_f_CO2+N_H2O*h_f_H2O+N_O2_P*h_f_O2+N_N2*h_f_N2 H_R=N_C8H18*h_f_C8H18+N_O2_R*h_f_O2+N_N2*h_f_N2 -Q_out=H_P-H_R N_dot_fuel=m_dot_fuel/MM_fuel Q_dot_out=N_dot_fuel*Q_out "(b)" "Entropies of reactants" P_O2_R= 1/4.76*P "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T1, P=P_O2_R) P_N2_R= 3.76/4.76*P s_N2_R=entropy(N2, T=T1, P=P_N2_R) s_C8H18_R=entropy(C8H18, T=T1, P=P)-s_fg_fuel "Adjust the EES gaseous value by s_fg" s_fg_fuel = h_fg_fuel/T1 h_fg_fuel=41466 [kJ/kmol] "from Table A-27" S_R=N_C8H18*s_C8H18_R+N_O2_R*s_O2_R+N_N2*s_N2_R © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-438

"Entropies of products" N_prod =8+9+ (ex-1)*a_th+ex*a_th*3.76 "total kmol of products" P_O2_P=N_O2_P/N_prod*P "Partial pressure of O2 in products" s_O2_P=entropy(O2, T=T2, P=P_O2_P) P_N2_P=N_N2/N_prod*P s_N2_P=entropy(N2, T=T2, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T2, P=P_CO2_P) s_H2O_P=69.92 [kJ/kmol-K] "from Table A-26 for liquid H2O" S_P=N_CO2*s_CO2_P+N_H2O*s_H2O_P+N_N2*s_N2_P+N_O2_P*s_O2_P S_gen=S_P-S_R+Q_out/T0 S_dot_gen=N_dot_fuel*S_gen "(c)" X_dot_destroyed=T0*S_dot_gen*Convert(kJ/min, kW)

16-90 Liquid octane is burned steadily with 200 percent excess air in a automobile engine. The maximum amount of work that can be produced by this engine is to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol C8 H18 , the combustion equation can be written as

C8 H18 ( )+ 3ath (O2 + 3.76N 2 ) ¾ ¾® 8CO 2 + 9H 2 O + 2ath O 2 + (3)(3.76)ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

3ath = 8 + 4.5 + 2ath

¾ ¾®

ath = 12.5

Thus,

C8 H18 ( )+ 37.5(O 2 + 3.76N 2 ) ¾ ¾®

8CO 2 + 9H 2 O + 25O 2 + 141N 2

Under steady-flow conditions the energy balance

Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to - Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

13-439

h 298 K

h350 K

Substance

kJ / kmol

C8 H18 ( )

−249,950

---

8682

10,213

8669

10,180

H2 O ( g )

−241,820

9904

11,652

CO2

−393,520

9364

11,351

Thus,

- Qout = (8)(- 393,520 + 11,351- 9364)+ (9)(- 241,820 + 11,652 - 9904)+ (25)(0 + 10,213 - 8682)+ (141)(0 + 10,180 - 8669)- (1)(- 249,950)

= - 4,791,636 kJ/kmol of C8 H18 The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

The entropy values listed in the ideal gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components which is equal to

Pi = yi Ptotal , where yi is the mole fraction of component i. Then, Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln ( yi Pm ))

The entropy calculations can be presented in tabular form as

si (T, 1 atm)

R u ln (y i Pm )

Ni si

C8 H18

---

466.73

---

466.73

37.5

0.21

205.04

−12.98

8175.75

141

0.79

191.61

−1.960

27,293.37

SR = 35,936 kJ / K

CO2

0.0437

219.831

−26.03

1966.89

H2 O ( g )

0.0492

194.125

−25.04

1972.49

0.1366

209.765

−16.55

5657.88

141

0.7705

196.173

−2.17

27,966.36

SP = 37,565 kJ / K Thus, Sgen = S P - S R +

Qout 4, 791, 636 = 37,565 - 35,936 + = 17,708 kJ/K (per kmol C8 H18 ) Tsurr 298

The maximum work is equal to the exergy destruction

Wmax = X dest = T0 Sgen = (298)(17,708 kJ/K) = 5, 276,984 kJ/K (per kmol C8 H18 ) Per unit mass basis, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-440

Wmax =

5, 276,984 kJ/K ×kmol = 46, 290 kJ / kg fuel 114 kg/kmol

16-91 Methane is burned steadily with 200 percent excess air in a automobile engine. The maximum amount of work that can be produced by this engine is to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol CH4 , the combustion equation can be written as

CH 4 + 3ath (O2 + 3.76N 2 ) ¾ ¾® CO 2 + 2H 2 O + 2ath O 2 + (3)(3.76)ath N 2

where ath is the stoichiometric coefficient and is determined from the

O2 balance, 3ath = 1 + 1 + 2ath

¾ ¾®

ath = 2

Thus,

CH 4 + 6 (O2 + 3.76N 2 ) ¾ ¾® CO2 + 2H 2 O + 4O 2 + 22.56N 2 Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to - Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

h 298 K

h350 K

Substance

kJ / kmol

CH4

−74,850

---

8682

10,213

8669

10,180

H2 O ( g )

−241,820

9904

11,652

CO2

−393,520

9364

11,351

Thus,

13-441

- Qout = (1)(- 393,520 + 11,351- 9364)+ (2)(- 241,820 + 11,652 - 9904)+ (4)(0 + 10,213 - 8682)+ (22.56)(0 + 10,180 - 8669)- (1)(- 74,850)

= - 756,615 kJ/kmol of CH4 The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

Pi = yi Ptotal , where yi is the mole fraction of component i. Then, Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln ( yi Pm ))

The entropy calculations can be presented in tabular form as

si (T, 1 atm)

R u ln (y i Pm )

Ni si

CH4

---

186.16

---

186.16

0.21

205.04

−12.98

1308.12

22.56

0.79

191.61

−1.960

4366.94

SR = 5861.22 kJ / K CO2

0.0338

219.831

−28.16

247.99

H2 O ( g )

0.0677

194.125

−22.39

433.03

0.1353

209.765

−16.63

905.58

22.56

0.7632

196.173

−2.25

4476.42

SP = 6063.02 kJ / K Thus, Sgen = S P - S R +

Qout 756, 615 = 6063.02 - 5861.22 + = 2741 kJ/K (per kmol CH 4 ) Tsurr 298

The maximum work is equal to the exergy destruction

Wmax = X dest = T0 Sgen = (298)(2741 kJ/K) = 816,818 kJ/K (per kmol CH 4 ) Per unit mass basis, Wmax =

816,818 kJ/K ×kmol = 51, 050 kJ / kg fuel 16 kg/kmol

16-92 Liquid propane is burned steadily with 150 percent excess air. The mass flow rate of air, the heat transfer rate from the combustion chamber, and the rate of entropy generation are to be determined. Assumptions

13-442

1. 2. 3. 4.

Combustion is complete. Steady operating conditions exist. Air and the combustion gases are ideal gases. Changes in kinetic and potential energies are negligible.

Properties The molar masses of C3 H8 and air are 44 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis (a) The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol of C3 H8 , the combustion equation can be written as

C3H8   2.5ath O2  3.76N 2    3CO 2  4H 2O  1.5ath O2  2.53.76 ath N2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

2.5ath  3  2  1.5ath

 

ath  5

Substituting,

C3H8    12.5O2  3.76N 2    3CO 2  4H 2O  7.5O 2  47N 2 The air-fuel ratio for this combustion process is AF 

Thus,

m air 12.5  4.76 kmol29 kg/kmol   39.2 kg air/kg fuel m fuel 3 kmol12 kg/kmol  4 kmol2 kg/kmol

m air  AFm fuel   39.2 kg air/kg fuel0.4 kg fuel/min   15.7 kg air/min

(b) Under steady-flow conditions the energy balance E in  E out  E system applied on the combustion chamber with W = 0 reduces to  Qout 

 N h  h  h    N h  h  h  P

 f



 f



Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The h f of liquid propane is obtained by adding the h fg at 25°C to h f of gaseous propane). h 285 K

h 298 K

h1200 K

Substance

hf kJ / kmol

kJ / kmol

C3 H8 ( )

−118,910

---

8296.5

8682

38,447

8286.5

8669

36,777

H2 O ( g )

−241,820

---

9904

44,380

CO2

−393,520

---

9364

53,848

Thus, Qout   3 393,520  53,848  9364    4  241,820  44,380  9904    7.5  0  38, 447  8682    47  0  36,777  8669   1 118,910  h298  h298   12.5  0  8296.5  8682    47  0  8286.5  8669 

 190,464 kJ/kmol of C3H8

13-443

Thus 190,464 kJ of heat is transferred from the combustion chamber for each kmol (44 kg) of propane. This corresponds to 190, 464 / 44 = 4328.7 kJ of heat transfer per kg of propane. Then the rate of heat transfer for a mass flow rate of

0.4 kg / min for the propane becomes Q out  m q out  0.4 kg/min 4328.7 kJ/kg   1732 kJ/min

Qout  Tsurr

Qout

N s N s  T P P

R R

surr

The C3 H8 is at 25°C and 1 atm, and thus its absolute entropy for the gas phase is sC3 H8 = 269.91 kJ / kmol.K (Table A-26). Then the entropy of C3 H8 ( ) is obtained from s C3H8    s C3H8 g   s fg  s C3H8 g  

h fg T

 269 .91 

15,060  219 .4 kJ/kmol  K 298 .15





S i  N i s i T , Pi   N i s i T , P0   Ru ln  y i Pm 

The entropy calculations can be presented in tabular form as

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

C3 H8

---

219.40

---

219.40

12.5

0.21

203.70

−12.98

2708.50

0.79

190.18

−1.96

9030.58

SR = 11,958.48 kJ / K

CO2

0.0488

279.307

−25.112

913.26

H2 O ( g )

0.0650

240.333

−22.720

1052.21

7.5

0.1220

249.906

−17.494

2005.50

0.7642

234.115

−2.236

11108.50

SP = 15,079.47 kJ / K Thus, S gen  S P  S R 

Qout 190,464  15,079 .47  11,958 .48   3760 .1 kJ/K per kmol C 3 H 8  Tsurr 298

Then the rate of entropy generation becomes

 



 0.4  S gen  N S gen   kmol/min 3760.1 kJ/kmol  K   34.2 kJ/min  K  44 

EES (Engineering Equation Solver) SOLUTION Fuel$ = 'Propane (C3H8)_liq' T_fuel = (25 + 273.15) [K] P_fuel = 101.3 [kPa] m_dot_fuel = 0.4 [kg/min] Ex = 1.5 "Excess air" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-444

P_air = 101.3 [kPa] T_air = (12+273.15) [K] T_prod = 1200 [K] P_prod = 101.3 [kPa] MW_air = MolarMass(air) MW_C3H8=MolarMass(C3H8) T_surr = (25+273.15) [K] "For theoretical dry air, the complete combustion equation is" "C3H8 + A_th(O2+3.76 N2)=3 CO2+4 H2O + A_th (3.76) N2 " 2*A_th=3*2+4*1"theoretical O balance" "The balanced combustion equation with Ex%/100 excess moist air is" "C3H8 + (1+EX)A_th(O2+3.76 N2)=3 CO2+ 4 H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " "The air-fuel ratio on a mass basis is:" AF = (1+Ex)*A_th*4.76*Mw_air/(1*Mw_C3H8) "kg_air/kg_fuel" "The air mass flow rate is:" m_dot_air = m_dot_fuel * AF "Apply First Law SSSF to the combustion process per kilomole of fuel:" E_in - E_out = DELTAE_cv E_in =HR "Since EES gives the enthalpy of gasesous components, we adjust the EES calculated enthalpy to get the liquid enthalpy. Subtracting the enthalpy of vaporization from the gaseous enthalpy gives the enthalpy of the liquid fuel. h_fuel(liq) = h_fuel(gas) - h_fg_fuel" h_fg_fuel = 15060 "kJ/kmol from Table A-27" HR = 1*(enthalpy(C3H8, T=T_fuel) - h_fg_fuel)+ (1+Ex)*A_th*enthalpy(O2,T=T_air)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air) E_out = HP + Q_out HP=3*enthalpy(CO2,T=T_prod)+4*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) DELTAE_cv = 0 "Steady-flow requirement" "The heat transfer rate from the combustion chamber is:" Q_dot_out=Q_out"kJ/kmol_fuel"/(Mw_C3H8 "kg/kmol_fuel")*m_dot_fuel"kg/min" "[kJ/min]" "Entopy Generation due to the combustion process and heat rejection to the surroundings:" "Entopy of the reactants per kilomole of fuel:" P_O2_reac= 1/4.76*P_air "Dalton's law of partial pressures for O2 in air" s_O2_reac=entropy(O2,T=T_air,P=P_O2_reac) P_N2_reac= 3.76/4.76*P_air "Dalton's law of partial pressures for N2 in air" s_N2_reac=entropy(N2,T=T_air,P=P_N2_reac) s_C3H8_reac=entropy(C3H8, T=T_fuel,P=P_fuel) - s_fg_fuel "Adjust the EES gaseous value by s_fg" "For phase change, s_fg is given by:" s_fg_fuel = h_fg_fuel/T_fuel SR = 1*s_C3H8_reac + (1+Ex)*A_th*s_O2_reac + (1+Ex)*A_th*3.76*s_N2_reac "Entopy of the products per kmole of fuel:" "By Dalton's law the partial pressures of the product gases is the productof the mole fraction and P_prod" N_prod = 3 + 4 + (1+Ex)*A_th*3.76 + Ex*A_th "total kmol of products" P_O2_prod = Ex*A_th/N_prod*P_prod "Partial pressure O2 in products" s_O2_prod=entropy(O2,T=T_prod,P=P_O2_prod) P_N2_prod = (1+Ex)*A_th*3.76/N_prod*P_prod "Patrial pressure N2 in products" s_N2_prod=entropy(N2,T=T_prod,P=P_N2_prod) P_CO2_prod = 3/N_prod*P_prod "Partial pressure CO2 in products" s_CO2_prod=entropy(CO2, T=T_prod,P=P_CO2_prod) P_H2O_prod = 4/N_prod*P_prod "Patrial pressure H2O in products" s_H2O_prod=entropy(H2O, T=T_prod,P=P_H2O_prod) SP = 3*s_CO2_prod + 4*s_H2O_prod + (1+Ex)*A_th*3.76*s_N2_prod + Ex*A_th*s_O2_prod "Since Q_out is the heat rejected to the surroundings per kilomole fuel, the entropy of the surroundings is:" S_surr = Q_out/T_surr "Rate of entropy generation:" S_dot_gen = (SP - SR +S_surr)"kJ/kmol_fuel"/(Mw_C3H8 "kg/kmol_fuel")*m_dot_fuel"kg/min" "[kJ/min-K]"

13-445

16-93 Problem 16-92 is reconsidered. The effect of the surroundings temperature on the rate of exergy destruction is to be studied. Analysis The problem is solved using EES, and the solution is given below. Fuel$ = 'Propane (C3H8)_liq' T_fuel = (25 + 273.15) "[K]" P_fuel = 101.3 [kPa] m_dot_fuel = 0.4 [kg/min]*Convert(kg/min, kg/s) Ex = 1.5 "Excess air" P_air = 101.3 [kPa] T_air = (12+273.15) "[K]" T_prod = 1200 [K] P_prod = 101.3 [kPa] Mw_air = 28.97 "lbm/lbmol_air" Mw_C3H8=(3*12+8*1) "kg/kmol_C3H8" {TsurrC = 25 [C]} T_surr = TsurrC+273.15 "[K]" "For theoretical dry air, the complete combustion equation is" "C3H8 + A_th(O2+3.76 N2)=3 CO2+4 H2O + A_th (3.76) N2 " 2A_th=3*2+4*1"theoretical O balance" "The balanced combustion equation with Ex%/100 excess moist air is" "C3H8 + (1+EX)A_th(O2+3.76 N2)=3 CO2+ 4 H2O + (1+Ex) A_th (3.76) N2+ Ex( A_th) O2 " "The air-fuel ratio on a mass basis is:" AF = (1+Ex)*A_th*4.76*Mw_air/(1*Mw_C3H8) "kg_air/kg_fuel" "The air mass flow rate is:" m_dot_air = m_dot_fuel * AF "Apply First Law SSSF to the combustion process per kilomole of fuel:" E_in - E_out = DELTAE_cv E_in =HR "Since EES gives the enthalpy of gasesous components, we adjust the EES calculated enthalpy to get the liquid enthalpy. Subtracting the enthalpy of vaporization from the gaseous enthalpy gives the enthalpy of the liquid fuel. h_fuel(liq) = h_fuel(gas) - h_fg_fuel" h_fg_fuel = 15060 "kJ/kmol from Table A-27" HR = 1*(enthalpy(C3H8, T=T_fuel) - h_fg_fuel)+ (1+Ex)*A_th*enthalpy(O2,T=T_air)+(1+Ex)*A_th*3.76 *enthalpy(N2,T=T_air) E_out = HP + Q_out HP=3*enthalpy(CO2,T=T_prod)+4*enthalpy(H2O,T=T_prod)+(1+Ex)*A_th*3.76* enthalpy(N2,T=T_prod)+Ex*A_th*enthalpy(O2,T=T_prod) DELTAE_cv = 0 "Steady-flow requirement" "The heat transfer rate from the combustion chamber is:" Q_dot_out=Q_out"kJ/kmol_fuel"/(Mw_C3H8 "kg/kmol_fuel")*m_dot_fuel"kg/s" "kW" "Entopy Generation due to the combustion process and heat rejection to the surroundings:" "Entopy of the reactants per kilomole of fuel:" P_O2_reac= 1/4.76*P_air "Dalton's law of partial pressures for O2 in air" s_O2_reac=entropy(O2,T=T_air,P=P_O2_reac) P_N2_reac= 3.76/4.76*P_air "Dalton's law of partial pressures for N2 in air" s_N2_reac=entropy(N2,T=T_air,P=P_N2_reac) s_C3H8_reac=entropy(C3H8, T=T_fuel,P=P_fuel) - s_fg_fuel "Adjust the EES gaseous value by s_fg" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-446

"For phase change, s_fg is given by:" s_fg_fuel = h_fg_fuel/T_fuel SR = 1*s_C3H8_reac + (1+Ex)*A_th*s_O2_reac + (1+Ex)*A_th*3.76*s_N2_reac "Entopy of the products per kilomle of fuel:" "By Dalton's law the partial pressures of the product gases is the product of the mole fraction and P_prod" N_prod = 3 + 4 + (1+Ex)*A_th*3.76 + Ex*A_th "total kmol of products" P_O2_prod = Ex*A_th/N_prod*P_prod "Patrial pressure O2 in products" s_O2_prod=entropy(O2,T=T_prod,P=P_O2_prod) P_N2_prod = (1+Ex)*A_th*3.76/N_prod*P_prod "Patrial pressure N2 in products" s_N2_prod=entropy(N2,T=T_prod,P=P_N2_prod) P_CO2_prod = 3/N_prod*P_prod "Patrial pressure CO2 in products" s_CO2_prod=entropy(CO2, T=T_prod,P=P_CO2_prod) P_H2O_prod = 4/N_prod*P_prod "Patrial pressure H2O in products" s_H2O_prod=entropy(H2O, T=T_prod,P=P_H2O_prod) SP = 3*s_CO2_prod + 4*s_H2O_prod + (1+Ex)*A_th*3.76*s_N2_prod + Ex*A_th*s_O2_prod "Since Q_out is the heat rejected to the surroundings per kilomole fuel, the entropy of the surroundings is:" S_surr = Q_out/T_surr "Rate of entropy generation:" S_dot_gen = (SP - SR +S_surr)"kJ/kmol_fuel"/(Mw_C3H8 "kg/kmol_fuel")*m_dot_fuel"kg/s" "kW/K" X_dot_dest = T_surr*S_dot_gen"[kW]" TsurrC [C] 0 4 8 12 16 20 24 28 32 36 38

dest [kW] 157.8 159.7 161.6 163.5 165.4 167.3 169.2 171.1 173 174.9 175.8

13-447

177.5

Xdest [kW]

173.5

169.5

165.5

161.5

157.5 0

TsurrC [C] 16-94E Benzene gas is burned steadily with 90 percent theoretical air. The heat transfer rate from the combustion chamber and the exergy destruction are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and the combustion gases are ideal gases. 3. Changes in kinetic and potential energies are negligible. Analysis (a) The fuel is burned with insufficient amount of air, and thus the products will contain some CO as well as CO2 , H 2 O , and N 2 . The theoretical combustion equation of C6 H6 is

C6 H 6 + ath (O 2 + 3.76N 2 ) ¾ ¾ ® 6CO 2 + 3H 2 O + 3.76ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

ath = 6 + 1.5 = 7.5 Then the actual combustion equation can be written as C6 H 6 + (0.90)(7.5)(O 2 + 3.76N 2 ) ¾ ¾ ® xCO 2 + (6 - x)CO + 3H 2 O + 25.38N 2

The value of x is determined from an O2 balance,

(0.90)(7.5) = x + (6 - x)/2 + 1.5 ¾ ¾® x = 4.5 Thus, C6 H 6 + 6.75 (O 2 + 3.76N 2 ) ¾ ¾ ® 4.5CO 2 + 1.5CO + 3H 2O + 25.38N 2

13-448

Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to

- Qout = å N P (h fo + h - h o ) - å N R (h fo + h - h o ) = å N P (h fo + h - h o ) - å N R h fo, R P

since all of the reactants are at 77°F. Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, h537 R

h1900 R

Substance

hf Btu / Ibmol

Btu / Ibmol

C6 H6 ( g )

35,680

---

3725.1

14,322

3729.5

13,742

H2 O ( g )

−104,040

4258.0

16,428

−47,540

3725.1

13,850

CO2

−169,300

4027.5

19,698

Thus, - Qout = (4.5)(- 169,300 + 19, 698 - 4027.5)+ (1.5)(- 47,540 + 13,850 - 3725.1) + (3)(- 104, 040 + 16, 428 - 4258)+ (25.38)(0 + 13, 742 - 3729.5)- (1)(35, 680)- 0 - 0

= - 804,630 Btu / lbmol of C6H6 (b) The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

The C6 H6 is at 77°F and 1 atm, and thus its absolute entropy is sC6 H6 = 64.34 Btu / lbmol ×R (Table A-26E). The entropy values listed in the ideal gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components which is equal to Pi = yi Ptotal , where yi is the mole fraction of component i. Also,

Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm )) The entropy calculations can be presented in tabular form as

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

C6 H 6

1.00

64.34

---

64.34

6.75

0.21

49.00

−3.10

351.68

25.38

0.79

45.77

−0.47

1173.57

SR = 1589.59 Btu / R CO2

4.5

0.1309

64.999

−4.038

310.67

1.5

0.0436

56.509

−6.222

94.10

H2 O ( g )

0.0873

56.097

−4.843

182.82

25.38

0.7382

54.896

−0.603

1408.56

13-449

Thus, Sgen = S P - S R +

Qout 804, 630 = 1996.15 - 1589.59 + = 1904.9 Btu / R Tsurr 537

Then the exergy destroyed is determined from

X destroyed = T0 Sgen = (537 R )(1904.9 Btu/lbmol ×R ) = 1,022,950 Btu / R (per lbmol C6 H 6 )

EES (Engineering Equation Solver) SOLUTION "Given" ex=0.90 "90% theoretical air" T1=(77+460) [R] T2=1900 [R] T0=(77+460) [R] P=14.7 [psia] "Analysis" "The stoichiometric combustion equation is: C6H6 + a_th (O2+3.76N2) = 6 CO2 + 3 H2O + 3.76*a_th N2" a_th=6+1.5 "O2 balance" "The actual combustion equation is: C6H6 + ex*a_th (O2+3.76N2) = x CO2 + (6-x) CO + 3 H2O + ex*a_th*3.76 N2" ex*a_th=x+(6-x)/2+1.5 "O2 balance" "(a)" "Mol numbers of reactants and products in lbmol" N_C6H6=1 [lbmol] N_O2=ex*a_th N_N2=ex*a_th*3.76 N_CO2=x N_CO=6-x N_H2O=3 [lbmol] "Enthalpy of formation data from Table A-26E in Btu/lbmol" h_f_C6H6=35680 [Btu/lbmol] "Ideal gas enthalpies in Btu/lbmol" h_O2=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_N2_P=enthalpy(N2, T=T2) h_CO2=enthalpy(CO2, T=T2) h_CO=enthalpy(CO, T=T2) h_H2O=enthalpy(H2O, T=T2) H_P=N_CO2*h_CO2+N_CO*h_CO+N_H2O*h_H2O+N_N2*h_N2_P H_R=N_C6H6*h_f_C6H6+N_O2*h_O2+N_N2*h_N2_R -Q_out=H_P-H_R "(b)" "Entropies of reactants" P_O2_R= 1/4.76*P "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T1, P=P_O2_R) P_N2_R= 3.76/4.76*P s_N2_R=entropy(N2, T=T1, P=P_N2_R) s_C6H6_R=64.34 "from Table A-26E" S_R=N_C6H6*s_C6H6_R+N_O2*s_O2_R+N_N2*s_N2_R "Entropies of products" N_prod=x+6-x+3+ex*a_th*3.76 "total kmol of products" P_N2_P=N_N2/N_prod*P s_N2_P=entropy(N2, T=T2, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T2, P=P_CO2_P) P_CO_P=N_CO/N_prod*P © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-450

s_CO_P=entropy(CO, T=T2, P=P_CO_P) P_H2O_P=N_H2O/N_prod*P s_H2O_P=entropy(H2O, T=T2, P=P_H2O_P) S_P=N_CO2*s_CO2_P+N_CO*s_CO_P+N_H2O*s_H2O_P+N_N2*s_N2_P S_gen=S_P-S_R+Q_out/T0 X_destroyed=T0*S_gen

16-95 Ethylene gas is burned steadily with 20 percent excess air. The temperature of products, the entropy generation, and the exergy destruction (or irreversibility) are to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis (a) The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol of C2 H 4 , the combustion equation can be written as C2 H 4 (g )+ 1.2ath (O 2 + 3.76N 2 ) ¾ ¾ ®

2CO 2 + 2H 2 O + 0.2ath O 2 + (1.2)(3.76)ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.2ath = 2 + 1 + 0.2ath ¾ ¾ ® ath = 3 Thus, C2 H 4 (g )+ 3.6 (O 2 + 3.76N 2 ) ¾ ¾ ®

2CO2 + 2H 2 O + 0.6O2 + 13.54N 2

Under steady-flow conditions, the exit temperature of the product gases can be determined from the steady-flow energy equation, which reduces to

å N (h + h - h ) = å N h P

f,R

= (Nh f )

C2 H 4

since all the reactants are at the standard reference state, and for O2 and N 2 . From the tables, hf

h 298

Substance

kJ / kmol

C2 H4 ( g )

52,280

---

8682

8669

H2 O ( g )

−241,820

9904

CO2

−393,520

9364

Substituting,

(2)(- 393,520 + hCO - 9364)+ (2)(- 241,820 + hH O - 9904)+ (0.6)(0 + hO - 8682)+ (13.54)(0 + hN - 8669) = (1)(52, 280) 2

or,

2hCO2 + 2hH2 O + 0.6hO2 + 13.54hN2 = 1, 484, 083 kJ

By trial and error,

13-451

(b) The entropy generation during this adiabatic process is determined from Sgen = S P - S R = å N P sP - å N R sR

The C2 H 4 is at 25C and 1 atm, and thus its absolute entropy is 219.83 kJ / kmol.K (Table A-26). The entropy values listed in the ideal gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components which is equal to Pi = yi Ptotal , where yi is the mole fraction of component i . Also,

Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm )) The entropy calculations can be presented in tabular form as

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

C2 H 4

1.00

219.83

---

219.83

3.6

0.21

205.14

−12.98

784.87

13.54

0.79

191.61

−1.96

2620.94

SR = 3625.64 kJ / K CO2

0.1103

316.881

−18.329

670.42

H2 O

0.1103

271.134

−18.329

578.93

0.6

0.0331

273.467

−28.336

181.08

13.54

0.7464

256.541

−2.432

3506.49

SP = 4936.92 kJ / K Thus,

Sgen = S P - S R = 4936.92 - 3625.64 = 1311.28 kJ / kmol ×K and (c) X destroyed = T0 Sgen = (298 K )(1311.28 kJ/kmol ×K C2 H 4 ) = 390, 800 kJ (per kmol C2 H 4 )

EES (Engineering Equation Solver) SOLUTION "Given" ex=1.2 "120% theoretical air" T1=(25+273) [K] P=101.325 [kPa] T0=(25+273) [K] "Analysis" "The combustion equation is: C2H4 + ex*a_th (O2+3.76N2) = 2 CO2 + 2 H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=2+1+(ex-1)*a_th "O2 balance" "(a)" "Mol numbers of reactants and products in kmol" N_C2H4=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_H2O=2 N_CO2=2 "Ideal gas enthalpies in kJ/kmol" h_f_C2H4=enthalpy(C2H4, T=T0) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-452

h_O2_R=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C2H4*h_f_C2H4+N_O2_R*h_O2_R+N_N2*h_N2_R H_P=H_R "(b)" "Entropies of reactants" P_O2_R= 1/4.76*P "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T1, P=P_O2_R) P_N2_R= 3.76/4.76*P s_N2_R=entropy(N2, T=T1, P=P_N2_R) s_C2H4_R=entropy(C2H4, T=T1, P=P) S_R=N_C2H4*s_C2H4_R+N_O2_R*s_O2_R+N_N2*s_N2_R "Entropies of products" N_prod =2+2+ (ex-1)*a_th+ex*a_th*3.76 "total kmol of products" P_O2_P=N_O2_P/N_prod*P "Partial pressure of O2 in products" s_O2_P=entropy(O2, T=T2, P=P_O2_P) P_N2_P=N_N2/N_prod*P s_N2_P=entropy(N2, T=T2, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T2, P=P_CO2_P) P_H2O_P=N_H2O/N_prod*P s_H2O_P=entropy(H2O, T=T2, P=P_H2O_P) S_P=N_CO2*s_CO2_P+N_H2O*s_H2O_P+N_N2*s_N2_P+N_O2_P*s_O2_P S_gen=S_P-S_R "(c)" X_destroyed=T0*S_gen

16-96 Liquid octane is burned steadily with 70 percent excess air. The entropy generation and exergy destruction per unit mass of the fuel are to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Properties The molar masses of C8 H18 and air are 144 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol C8 H18 , the combustion equation can be written as C8 H18 ( )+ 1.7ath (O 2 + 3.76N 2 ) ¾ ¾ ® 8CO 2 + 9H 2 O + 0.7 ath O 2 + (1.7 )(3.76)ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.7ath = 8 + 4.5 + 0.7ath ¾ ¾ ® ath = 12.5 Thus,

13-453

C8 H18 ( )+ 21.25(O 2 + 3.76N 2 ) ¾ ¾ ® 8CO 2 + 9H 2 O + 8.75O 2 + 79.9N 2

(b) Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, h 298 K

h 600 K

h1500 K

Substance

hf kJ / kmol

kJ / kmol

C8 H18 ( )

−249,950

---

8682

17,929

49,292

8669

17,563

47,073

H2 O ( g )

−241,820

9904

---

57,999

CO2

−393,520

9364

---

71,078

Thus, - Qout = (8)(- 393,520 + 71,078 - 9364)+ (9)(- 241,820 + 57,999 - 9904)+ (8.75)(0 + 49, 292 - 8682)+ (79.9)(0 + 47,073 - 8669)- (1)(- 249,950) - (21.25)(0 + 17,929 - 8682)- (79.9)(0 + 17,563 - 8669)

= - 1,631,335 kJ/kmol of C8 H18 The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

The entropy values listed in the ideal gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of Pm = 600 kPa(= 600 /101.325 = 5.92 atm) , but the entropies are to be calculated at the partial pressure of the components which is equal to Pi = yi Ptotal , where yi is the mole fraction of component i. Then,

Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm )) The entropy calculations can be presented in tabular form as

13-454

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

C8 H18

---

466.73

14.79

451.94

21.25

0.21

226.35

1.81

4771.48

79.9

0.79

212.07

12.83

15,919.28

SR = 21,142.70 kJ / K

CO2

0.0757

292.11

−6.673

2390.26

H2 O ( g )

0.0852

250.45

−5.690

2305.26

8.75

0.0828

257.97

−5.928

2309.11

79.9

0.7563

241.77

12.46

18,321.87

SP = 25,326.50 kJ / K Thus, Sgen = S P - S R +

Qout 1, 631,335 = 25,326.50 - 21,142.70 + = 9658.1 kJ/K (per kmol C8 H18 ) Tsurr 298

The exergy destruction is

X dest = T0 Sgen = (298)(9658.1 kJ/K) = 2,878,108 kJ/K (per kmol C8 H18 ) The entropy generation and exergy destruction per unit mass of the fuel are Sgen = X dest =

Sgen M fuel

9658.1 kJ/K ×kmol = 84.7 kJ / K ×kg C8H18 114 kg/kmol

X dest 2,878,108 kJ/K ×kmol = = 25,247 kJ / kg C8H18 M fuel 114 kg/kmol

EES (Engineering Equation Solver) SOLUTION "Given" ex=0.7 P_air=600 [kPa] T_air=327 [C] T_fuel=25 [C] P_prod=600 [kPa] T_prod=1227 [C] T0=(25+273) [K] P=P_air "Analysis" "Combustion equation is: C8H18 (l) + (ex+1)a_th (O2+3.76N2) = 8 CO2 + 9 H2O + (ex)(a_th) O2 + (ex+1)a_thx3.76 N2" a_th=8+9/2 "O2 balance" N_C8H18=1 [kmol] N_CO2=8 [kmol] N_H2O=9 [kmol] N_O2=ex*a_th N_N2=(ex+1)*a_th*3.76 N_O2_R=(ex+1)*a_th "Ideal gas enthalpies in kJ/kmol" h_f_C8H18_l=-249950 [kJ/kmol] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-455

h_O2_R=enthalpy(O2, T=T_air) h_N2_R=enthalpy(N2, T=T_air) h_O2_P=enthalpy(O2, T=T_prod) h_N2_P=enthalpy(N2, T=T_prod) h_H2O=enthalpy(H2O, T=T_prod) h_CO2=enthalpy(CO2, T=T_prod) -Q_out=H_P-H_R H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2*h_O2_P+N_N2*h_N2_P H_R=N_C8H18*h_f_C8H18_l+N_O2_R*h_O2_R+N_N2*h_N2_R "Entropies of reactants" P_O2_R= 1/4.76*P "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T_air, P=P_O2_R) P_N2_R= 3.76/4.76*P s_N2_R=entropy(N2, T=T_air, P=P_N2_R) s_C8H18_R=466.73-R_u*ln(P) R_u=8.314 [kJ/kmol-K] S_R=N_C8H18*s_C8H18_R+N_O2_R*s_O2_R+N_N2*s_N2_R "Entropies of products" N_prod =N_CO2+N_H2O+N_O2+N_N2 "total kmol of products" P_O2_P=N_O2/N_prod*P "Partial pressure of O2 in products" s_O2_P=entropy(O2, T=T_prod, P=P_O2_P) P_N2_P=N_N2/N_prod*P s_N2_P=entropy(N2, T=T_prod, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T_prod, P=P_CO2_P) P_H2O_P=N_H2O/N_prod*P s_H2O_P=entropy(H2O, T=T_prod, P=P_H2O_P) S_P=N_CO2*s_CO2_P+N_H2O*s_H2O_P+N_N2*s_N2_P+N_O2*s_O2_P S_bar_gen=S_P-S_R+Q_out/T0 X_bar_dest=T0*S_bar_gen MM_fuel=114 [kg/kmol] S_gen=S_bar_gen/MM_fuel X_dest=X_bar_dest/MM_fuel

16-97 CO gas is burned steadily with air. The heat transfer rate from the combustion chamber and the rate of exergy destruction are to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Properties The molar masses of CO and air are 28 kg / kmol and

29 kg / kmol , respectively (Table A-1). Analysis (a) We first need to calculate the amount of air used per kmol of CO before we can write the combustion equation,

13-456

v CO =

RT (0.2968 kPa ×m /kg ×K )(310 K ) = = 0.836 m3 /kg P 110 kPa

mCO =

VCO 0.4 m3 /min = = 0.478 kg/min v CO 0.836 m3 /kg

Then the molar air-fuel ratio becomes AF =

(1.5 kg/min )/ (29 kg/kmol) N air mair / M air = = = 3.03 kmol air/kmol fuel N fuel mfuel / M fuel (0.478 kg/min )/ (28 kg/kmol)

Thus the number of moles of O2 used per mole of CO is 3.03 / 4.76 = 0.637 . Then the combustion equation in this case can be written as CO + 0.637 (O 2 + 3.76N 2 ) ¾ ¾ ® CO 2 + 0.137O 2 + 2.40N 2

Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, h 298 K

h 310 K

h 900 K

Substance

hf kJ / kmol

kJ / kmol

−110,530

8669

9014

---

8682

---

27,928

8669

---

26,890

CO2

−393,520

9364

---

37,405

Substituting, - Qout = (1)(- 393,520 + 37, 405 - 9364)+ (0.137 )(0 + 27,928 - 8682)+ (2.4)(0 + 26,890 - 8669)- (1)(- 110,530 + 9014 - 8669)- 0 - 0 = - 208,929 kJ/kmol of CO

Thus 208,929 kJ of heat is transferred from the combustion chamber for each kmol (28 kg) of CO. This corresponds to 208,929 / 28 = 7462 kJ of heat transfer per kg of CO. Then the rate of heat transfer for a mass flow rate of 0.478 kg / min for CO becomes Qout = mqout = (0.478 kg/min )(7462 kJ/kg) = 3567 kJ / min

(b) This process involves heat transfer with a reservoir other than the surroundings. An exergy balance on the combustion chamber in this case reduces to the following relation for reversible work,

Wrev = å N R (h f + h - h - T0 s ) - å N P (h f + h - h - T0 s ) - Qout (1- T0 / TR ) R

The entropy values listed in the ideal gas tables are for 1 atm = 101.325 kPa pressure. The entropy of each reactant and the product is to be calculated at the partial pressure of the components which is equal to Pi = yi Ptotal , where yi is the mole fraction of component i, and Pm = 110 /101.325 = 1.0856 atm . Also,

13-457

Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm )) The entropy calculations can be presented in tabular form as

O2 N2

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

1 0.637

1.00 0.21

198.678 205.04

0.68 −12.29

198.00 138.44

2.400

0.79

191.61

−1.28

462.94

0.2827

263.559

−9.821

273.38

0.137

0.0387

239.823

−26.353

36.47

2.400

0.6785

224.467

−2.543

544.82

SR = 799.38 kJ / K

CO2 O2 N2

SP = 854.67 kJ / K The rate of exergy destruction can be determined from where

X destroyed = T0 Sgen = T0 m (Sgen / M ) Sgen = S P - S R +

Qout 208,929 kJ = 854.67 - 799.38 + = 316.5 kJ/kmol ×K Tres 800 K

Thus, X destroyed = (298 K )(0.478 kg/min )(316.5/28 kJ/kmol ×K ) = 1610 kJ / min

EES (Engineering Equation Solver) SOLUTION "Given" T1_fuel=(37+273) [K] P1_fuel=110 [kPa] V_dot_fuel=0.4 [m^3/min] T1_air=(25+273) [K] P1_air=110 [kPa] m_dot_air=1.5 [kg/min] T2=900 [K] T_reservoir=800 [K] T0=(25+273) [K] P=P1_fuel "Properties" R=R_u/MM_fuel R_u=8.314 [kJ/kmol-K] MM_air=molarmass(air) MM_fuel=molarmass(CO) "Analysis" "(a)" v_fuel=(R*T1_fuel)/P1_fuel m_dot_fuel=V_dot_fuel/v_fuel AF_bar=N_air/N_fuel N_air=m_dot_air/MM_air N_fuel=m_dot_fuel/MM_fuel a=AF_bar/4.76 "Then the combustion equation becomes: CO + a (O2+3.76N2) = CO2 + (a-0.5) O2 + 3.76*a N2" "Mol numbers of reactants and products in kmol" N_CO=1 N_O2_R=a N_O2_P=a-0.5 N_N2=a*3.76 N_CO2=1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-458

"Ideal gas enthalpies in kJ/kmol" h_CO_R=enthalpy(CO, T=T1_fuel) h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_CO2_P=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2_P+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_CO*h_CO_R+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R Q_dot_out=m_dot_fuel*(Q_out/MM_fuel) "(b)" "Entropies of reactants" P_O2_R= 1/4.76*P "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T1_air, P=P_O2_R) P_N2_R= 3.76/4.76*P s_N2_R=entropy(N2, T=T1_air, P=P_N2_R) s_CO_R=entropy(CO, T=T1_fuel, P=P) S_R=N_CO*s_CO_R+N_O2_R*s_O2_R+N_N2*s_N2_R "Entropies of products" N_prod =1+a-0.5+a*3.76 "total kmol of products" P_O2_P=N_O2_P/N_prod*P "Partial pressure of O2 in products" s_O2_P=entropy(O2, T=T2, P=P_O2_P) P_N2_P=N_N2/N_prod*P s_N2_P=entropy(N2, T=T2, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T2, P=P_CO2_P) S_P=N_CO2*s_CO2_P+N_N2*s_N2_P+N_O2_P*s_O2_P S_gen=S_P-S_R+Q_out/T_reservoir X_dot_destroyed=m_dot_fuel*T0*(S_gen/MM_fuel)

Review Problems 16-98 Propane is burned with stoichiometric amount of oxygen. The mass fractions of each of the products and the mass of water in the products per unit mass of fuel burned are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 and H 2 O .

Combustion gases are ideal gases.

Properties The molar masses of C, H 2 , O2 , and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The combustion equation in this case is

C3 H8 + 5O2 ¾ ¾ ® 3CO2 + 4H 2 O The mass of each product and the total mass are

mCO2 = NCO2 M CO2 = (3 kmol)(44 kg/kmol) = 132 kg

13-459

mH2O = NH2O M H2O = (4 kmol)(18 kg/kmol) = 72 kg mtotal = mCO2 + mH2O = 132 + 72 = 204 kg Then the mass fractions are mf CO2 =

mCO2 132 kg = = 0.6471 mtotal 204 kg

mf H2O =

mH2O 72 kg = = 0.3529 mtotal 204 kg

The mass of water in the products per unit mass of fuel burned is determined from mH2O 72 kg = = 1.64 kg H 2 O / kg C3 H 8 mC3H8 44 kg

16-99 Octane is burned steadily with 60 percent excess air. The dew-point temperature of the water vapor in the products is to be determined. Assumptions

1. 2. 3.

Combustion is complete. Steady operating conditions exist. Air and the combustion gases are ideal gases.

Properties The molar masses of C8 H18 and air are 144 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol C8 H18 , the combustion equation can be written as C8 H18 + 1.6ath (O 2 + 3.76N 2 ) ¾ ¾ ®

8CO 2 + 9H 2 O + 0.6ath O 2 + (1.6)(3.76 )ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

1.6ath = 8 + 4.5 + 0.6ath

¾¾ ®

ath = 12.5

Thus, C8 H18 + 20 (O 2 + 3.76N 2 ) ¾ ¾ ®

8CO 2 + 9H 2 O + 7.5O 2 + 75.2N 2

The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is,

æN ö æ ö 9 kmol ÷ ÷ ÷ Pv = ççç v ÷ (101.325 kPa) = 9.147 kPa ÷Pprod = ççç ÷ ÷ çè N prod ø ÷ è(8 + 9 + 7.5 + 75.2) kmol ÷ ø Thus,

Tdp = Tsat @ 9.147 kPa = 44.1°C

(from EES)

16-100E A coal from Utah is burned with 20% excess air. The mass of water in the products per unit mass of coal burned and the dew-point temperature of the water vapor in the products are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-460

Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , CO, H 2 O , SO2 , O2 , and N 2 .

Combustion gases are ideal gases.

Properties The molar masses of C, O2 , H 2 , S, and N 2 are 12, 32, 2, 32, and 28 kg / kmol , respectively (Table A-1E). Analysis The mass fractions of the constituent of the coal when the ash is substituted are mf C =

mC 61.40 lbm 61.40 lbm = = = 0.6463 mtotal (100 - 5.00) lbm 95.00 lbm

mf H2 =

mH2 5.79 lbm = = 0.06095 mtotal 95.00 lbm

mf O2 =

mO2 25.31 lbm = = 0.2664 mtotal 95.00 lbm

mf N2 =

mN2 1.09 lbm = = 0.01147 mtotal 95.00 lbm

mfS =

mS 1.41 lbm = = 0.01484 mtotal 95.00 lbm

We now consider 100 lbm of this mixture. Then the mole numbers of each component are

NC =

mC 64.63 lbm = = 5.386 lbmol M C 12 lbm/lbmol

N H2 =

mH2 6.095 lbm = = 3.048 lbmol M H2 2 lbm/lbmol

N O2 =

mO2 26.64 lbm = = 0.8325 lbmol M O2 32 lbm/lbmol

N N2 =

mN2 1.147 lbm = = 0.04096 lbmol M N2 28 lbm/lbmol

NS =

mS 1.484 lbm = = 0.04638 lbmol M S 32 lbm/lbmol

The mole number of the mixture and the mole fractions are

13-461

yC =

N C 5.386 lbmol = = 0.5758 N m 9.354 lbmol

yH2 =

N H2 3.048 lbmol = = 0.3258 Nm 9.354 lbmol

yO2 =

N O2 0.8325 lbmol = = 0.0890 Nm 9.354 lbmol

yN2 =

N N2 0.04096 lbmol = = 0.00438 Nm 9.354 lbmol

yS =

NS 0.04638 lbmol = = 0.00496 Nm 9.354 lbmol

Then, the combustion equation in this case may be written as

0.5758C + 0.3258H 2 + 0.0890O 2 + 0.00438N 2 + 0.00496S + 1.2ath (O2 + 3.76N 2 ) ¾¾ ® 0.5758CO2 + 0.3258H 2 O + 0.00496SO 2 + 0.2ath O 2 + 1.2ath ´ 3.76kN 2 According to the oxygen balance,

O2 balance : 0.0890 + 1.2ath = 0.5758 + 0.5´ 0.3258 + 0.00496 + 0.2ath ¾ ¾® ath = 0.6547 Substituting,

0.5758C + 0.3258H 2 + 0.0890O2 + 0.00438N 2 + 0.00496S + 0.7856(O2 + 3.76N 2 ) ¾¾ ® 0.5758CO2 + 0.3258H 2 O + 0.00496SO2 + 0.1309O 2 + 2.954N 2 The mass of water in the products per unit mass of fuel burned is determined from mH2O (0.3258´ 18) lbm = mcoal (0.5758´ 12 + 0.3258´ 2 + 0.0890´ 32 + 0.00438´ 28 + 0.00496´ 32) lbm

5.864 lbm = 0.5486 lbm H 2O / lbm coal 10.69 lbm The dew-point temperature of a gas-vapor mixture is the saturation temperature of the water vapor in the product gases corresponding to its partial pressure. That is, =

æN ÷ ö æ ö 0.3258 lbmol ÷ ÷ Pv = ççç v ÷ Pprod = çç (14.696 psia) = 1.20 psia ÷ ÷ ÷ ÷ ç ÷ è(0.5758 + 0.3258 + 0.00496 + 0.1309 + 2.954) lbmol ø èç N prod ø Thus,

Tdp = Tsat @ 1.20 psia = 108°F (from EES)

16-101 A coal from Colorado is burned with 10 percent excess air during a steady-flow combustion process. The fraction of the water in the combustion products that is liquid and the fraction that is vapor are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5 The combustion chamber is adiabatic. Properties The molar masses of C, H 2 , N 2 , O2 , S, and air are 12, 2, 28, 32, 32, and 29 kg / kmol , respectively (Table A-1). Analysis The mass fractions of the constituent of the coal when the ash is substituted are

13-462

mf C =

mC 79.61 kg 79.61 kg = = = 0.8712 mtotal (100-8.62) kg 91.38 kg

mf H2 =

mH2 4.66 kg = = 0.05100 mtotal 91.38 kg

mf O2 =

mO2 4.76 kg = = 0.05209 mtotal 91.38 kg

mf N2 =

mN2 1.83 kg = = 0.02003 mtotal 91.38 kg

mfS =

mS 0.52 kg = = 0.00569 mtotal 91.38 kg

We now consider 100 kg of this mixture. Then the mole numbers of each component are NC =

mC 87.12 kg = = 7.26 kmol M C 12 kg/kmol

N H2 =

mH2 5.10 kg = = 2.55 kmol M H2 2 kg/kmol

N O2 =

mO2 5.209 kg = = 0.1628 kmol M O2 32 kg/kmol

N N2 =

mN2 2.003 kg = = 0.07154 kmol M N2 28 kg/kmol

NS =

mS 0.569 kg = = 0.01778 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are

Nm = 7.26 + 2.55 + 0.1628 + 0.07154 + 0.01778 = 10.062 kmol yC =

NC 7.26 kmol = = 0.7215 N m 10.06 kmol

yH2 =

N H2 2.55 kmol = = 0.2535 Nm 10.06 kmol

yO2 =

N O2 0.1628 kmol = = 0.01618 Nm 10.06 kmol

13-463

yN2 =

N N2 0.07154 kmol = = 0.00711 Nm 10.06 kmol

yS =

NS 0.01778 kmol = = 0.00177 Nm 10.06 kmol

Then, the combustion equation in this case may be written as

0.7215C + 0.2535H 2 + 0.01618O2 + 0.00711N 2 + 0.00177S + 1.1ath (O2 + 3.76N 2 ) ¾¾ ® 0.7215CO2 + 0.2535H 2 O + 0.00177SO2 + 1.1ath ´ 3.76N 2 + 0.1ath O 2 According to the oxygen balance,

O2 balance : 0.01618 + 1.1ath = 0.7215 + (0.5)(0.2535) + (0.00177) + 0.1ath ¾ ¾ ® ath = 0.8339 Substituting,

0.7215C + 0.2535H 2 + 0.01617O 2 + 0.00711N 2 + 0.00177S + 0.9173(O2 + 3.76N 2 ) ¾ ¾® 0.7215CO2 + 0.2535H 2O + 0.0018SO2 + 3.449N 2 + 0.0834O 2 If the mixture leaves the boiler saturated with water, the partial pressure of water in the product mixture would be

Pv = Psat @ 50°C = 12.352 kPa and the water mole fraction would be

Pv 12.352 kPa = = 0.1219 P 101.325 kPa The actual mole fraction of water in the products is yv =

yv =

N H2O 0.2535 kmol = = 0.0562 N prod (0.7215 + 0.2535 + 0.0018 + 3.449 + 0.0834)kmol

Since the actual mole fraction of the water is less than that when the mixture is saturated, the mixture never becomes saturated and hence, no condensate is formed. All of the water in the mixture is then in the vapor form at this temperature.

16-102 A sample of a certain fluid is burned in a bomb calorimeter. The heating value of the fuel is to be determined. Properties The specific heat of water is 4.18 kJ / kg. C (Table A-3).

Analysis We take the water as the system, which is a closed system, for which the energy balance on the system Ein - Eout = D Esystem with W = 0 can be written as

Qin = D U or

Qin = mcD T

13-464

= (2 kg )(4.18 kJ/kg ×°C)(2.5°C) = 20.90 kJ (per gram of fuel)

Therefore, heat transfer per kg of the fuel would be 20, 900 kJ / kg fuel . Disregarding the slight energy stored in the gases of the combustion chamber, this value corresponds to the heating value of the fuel.

EES (Engineering Equation Solver) SOLUTION "Given" m_fuel=0.001 [kg] m_water=2 [kg] m_air=0.100 [kg] DELTAT=2.5 [C] "Properties" c_water=4.18 [kJ/kg-C] "at room temperature" "Analysis" Q=m_water*c_water*DELTAT Q_HV=Q/m_fuel

16-103E Hydrogen is burned with 100 percent excess air. The AF ratio and the volume flow rate of air are to be determined. Assumptions 1. Combustion is complete. 2. Air and the combustion gases are ideal gases. Properties The molar masses of H 2 and air are 2 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis (a) The combustion is complete, and thus products will contain only H 2 O , O2 and N 2 . The moisture in the air does not react with anything; it simply shows up as additional H 2 O in the products. Therefore, for simplicity, we will balance the combustion equation using dry air, and then add the moisture to both sides of the equation. The combustion equation in this case can be written as H 2 + 2ath (O 2 + 3.76N 2 ) ¾ ¾ ® H 2 O + ath O 2 + (2)(3.76)ath N 2

where ath is the stoichiometric coefficient for air. It is determined from

® ath = 0.5 O2 balance: 2ath = 0.5 + ath ¾ ¾ Substituting, H 2 + (O 2 + 3.76N 2 ) ¾ ¾ ® H 2 O + 0.5O 2 + 3.76N 2

Therefore, 4.76 lbmol of dry air will be used per kmol of the fuel. The partial pressure of the water vapor present in the incoming air is Pv , in = f air Psat @ 90° F = (0.60)(0.7063 psi) = 0.4238 psia

The number of moles of the moisture that accompanies 4.76 lbmol of incoming dry air ( N v, in ) is determined to be

13-465

æPv , in ÷ ö æ0.4238 psia ÷ ö ÷ ÷ N v , in = ççç N total = ççç 4.76 + N v , in ) ¾ ¾ ® N v , in = 0.1433 lbmol ÷ ÷ ÷( ÷ è 14.5 psia ø èç Ptotal ø

The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.142 lbmol of H 2 O to both sides of the equation, H 2 + (O2 + 3.76N 2 )+ 0.1433H 2 O ¾ ¾ ® 1.1433H 2 O + 0.5O 2 + 3.76N 2

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

(4.76 lbmol)(29 lbm/lbmol)+ (0.1433 lbmol)(18 lbm/lbmol) mair = = 70.31 lbm air / lbm fuel mfuel (1 lbmol)(2 lbm/lbmol)

(b) The mass flow rate of H 2 is given to be 10 lbm / h . Since we need 70.3 lbm air per lbm of H 2 , the required mass flow rate of air is mair = (AF)(mfuel ) = (70.31)(25 lbm/h ) = 1758 lbm/h

The mole fractions of water vapor and the dry air in the incoming air are yH2 O =

N H2 O N total

0.1433 = 0.02923 and ydry air = 1- 0.02923 = 0.9708 4.76 + 0.1433

Thus,

M = ( yM )H O + ( yM )dry air = (0.02923)(18)+ (0.9708)(29) = 28.68 lbm/lbmol 2

RT (10.73/28.68 psia ×ft /lbm ×R )(550 R ) = = 14.19 ft 3 /lbm P 14.5 psia 3

V = mv = (1758 lbm/h )(14.19 ft 3 /lbm) = 24, 946 ft 3 / h

EES (Engineering Equation Solver) SOLUTION "Given" ex=2 "200% theoretical air" T1=(90+460) [R] P1=14.5 [psia] phi=0.60 m_dot_fuel=25 [lbm/h] "Properties" R_u=10.73 [psia-ft^3/lbmol-R] MM_air=29 "molarmass(air)" MM_H2=2 "molarmass(H2)" MM_H2O=18 "molarmass(H2O)" "Analysis" "The combustion equation is: H2 + ex*a_th (O2+3.76N2) = H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=0.5+(ex-1)*a_th "O2 balance" "(a)" P_sat=pressure(steam_iapws, T=T1, x=1) P_v_in=phi*P_sat N_v_in=(P_v_in/P1)*N_total N_total=ex*a_th*4.76+N_v_in "The combustion equation becomes: H2 + ex*a_th (O2+3.76N2) + N_v_in H2O = (1+N_v_in) H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" AF=m_air/m_fuel m_air=ex*a_th*4.76*MM_air+N_v_in*MM_H2O m_fuel=N_fuel*MM_H2 N_fuel=1 [lbmol] "(b)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-466

m_dot_air=AF*m_dot_fuel y_H2O=N_v_in/N_total y_dryair=1-y_H2O MM=y_H2O*MM_H2O+y_dryair*MM_air v=(R_u/MM)*T1/P1 V_dot=m_dot_air*v

16-104 Butane is burned with stoichiometric amount of air. The fraction of the water in the products that is liquid is to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3 Air and the combustion gases are ideal gases. Analysis The fuel is burned completely with the air, and thus the products will contain only CO2 , H 2 O , and N 2 . Considering 1 kmol C4 H10 , the combustion equation can be written as C4 H10 + 6.5(O 2 + 3.76N 2 ) ¾ ¾ ® 4 CO 2 + 5H 2 O + 24.44N 2

The mole fraction of water in the products is

yv =

N H2O 5 kmol = = 0.1495 N prod (4 + 5 + 24.44)kmol

The saturation pressure for the water vapor is

Pv = Psat @ 40°C = 7.3851 kPa When the combustion gases are saturated, the mole fraction of the water vapor will be

Pv 7.3851 kPa = = 0.0729 P 101.325 kPa The mole fraction of the liquid water is then yg =

y f = yv - yg = 0.1495 - 0.0729 = 0.0766 Thus, the fraction of liquid water in the combustion products is f liquid =

yf yg

0.0766 = 0.512 0.1495

16-105 A gaseous fuel mixture of 60% propane, C3 H8 , and 40% butane, C4 H10 , on a volume basis is burned with an airfuel ratio of 19. The moles of nitrogen in the air supplied to the combustion process, the moles of water formed in the combustion process, and the moles of oxygen in the product gases are to be determined. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 only.

13-467

Properties The molar masses of C, H 2 , O2 and air are 12 kg / kmol , 2 kg / kmol , 32 kg / kmol , and 29 kg / kmol , respectively (Table A-1). Analysis The theoretical combustion equation in this case can be written as 0.6 C3 H8 + 0.4 C 4 H10 + ath [O 2 + 3.76N 2 ] ¾ ¾ ® B CO 2 + D H 2 O + F N 2

where ath is the stoichiometric coefficient for air. The coefficient ath and other coefficients are to be determined from the mass balances Carbon balance:

B = 3´ 0.6 + 4´ 0.4 = 3.4

Hydrogen balance:

8´ 0.6 + 10´ 0.4 = 2 D ¾ ¾ ® D = 4.4

Oxygen balance:

2ath = 2B + D ¾ ¾ ® 2ath = 2´ 3.4 + 4.4 ¾ ¾ ® ath = 5.6

Nitrogen balance:

3.76ath = F ¾ ¾ ® 3.76´ 5.6 = F ¾ ¾ ® F = 21.06

Then, we write the balanced theoretical reaction equation as 0.6 C3 H8 + 0.4 C4 H10 + 5.6 [O 2 + 3.76N 2 ] ¾ ¾ ® 3.4 CO 2 + 4.4 H 2O + 21.06 N 2

The air-fuel ratio for the theoretical reaction is determined from AFth =

mair (5.6´ 4.76 kmol)(29 kg/kmol) = = 15.59 kg air/kg fuel mfuel (0.6´ 44 + 0.4´ 58) kg

The percent theoretical air is PercentTH air =

AFactual 25 = ´ 100 = 160.4% AFth 15.59

The moles of nitrogen supplied is

PercentTHair 160.4 ´ ath ´ 3.76 = (5.6)(3.76) = 33.8 kmol per kmol fuel 100 100 The moles of water formed in the combustion process is N N2 =

NH2O = D = 4.40 kmol per kmol fuel The moles of oxygen in the product gases is æPercentTHair ö æ160.4 ÷ ö NO2 = çç - 1÷ ath = çç - 1÷ (5.6) = 3.38 kmol per kmol fuel ÷ ÷ ÷ ç çè è 100 ø ø 100

EES (Engineering Equation Solver) SOLUTION y_C3H8=0.6 y_C4H10=0.4 AF_act = 25 "[kg/kg_fuel]" "actual air-fuel ratio" MM_C3H8=44 [kg/kmol] MM_C4H10=58 [kg/kmol] MM_air=29 [kg/kmol] "The reaction equation for theroetical air is:" "0.3 C3H8 + 0.7 C4H10 + A_th (O2 +3.76N2)=B CO2 + D H2O + F N2" "Carbon Balance:" y_C3H8*3+y_C4H10*4=B "Hydrogen Balance:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-468

y_C3H8*8+y_C4H10*10=2*D "Oxygen Balance:" A_th*2=B*2 + D "Nitrogen Balance:" A_th*3.76 = F "The theoretical air-fuel ratio is:" MM_fuel=y_C3H8*MM_C3H8+y_C4H10*MM_C4H10 AF_th = A_th*4.76*MM_air/MM_fuel "Percent Theoretical air is:" PercentTH_air=AF_act/AF_th * 100 "[%]" "Mole of N2 supplied:" N_N2 = PercentTH_air/100*A_th*3.76 "[kmol_N2/kmol_fuel]" "Mole of H2O formed:" N_H2O = D "[kmol_H2O/kmol_fuel]" "Mole of O2 in products:" N_O2 = (PercentTH_air/100 - 1)*A_th "[kmol_O2/kmol_fuel]"

16-106 The higher and lower heating values of gaseous methane are to be determined and compared to the listed values. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 .

Combustion gases are ideal gases.

Properties The molar masses of C, O2 , H 2 , and air are 12, 32, 2, and 29 kg / kmol , respectively (Table A-1). Analysis The combustion reaction with stoichiometric air is CH 4 + 2 (O 2 + 3.76N 2 ) ¾ ¾ ®

CO 2 + 2H 2 O + 7.52N 2

Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values. The heat transfer for this process is equal to enthalpy of combustion. Note that N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then,

q = hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2O

- (Nh f )

CH4

For the HHV, the water in the products is taken to be liquid. Then, hC = (1 kmol)(- 393,520 kJ/kmol) + (2 kmol)(- 285,830 kJ/kmol) - (1 kmol)(- 74,850 kJ/kmol)

= - 890,330 kJ/kmol methane

The HHV of the methane is HHV =

- hC 890,330 kJ/kmol CH 4 = = 55, 500 kJ / kg CH 4 Mm 16.043 kg/kmol CH 4

The listed value from Table A-27 is 55, 530 kJ / kg . For the LHV, the water in the products is taken to be vapor. Then, hC = (1 kmol)(- 393,520 kJ/kmol) + (2 kmol)(- 241,820 kJ/kmol) - (1 kmol)(- 74,850 kJ/kmol)

= - 802,310 kJ/kmol methane

13-469

LHV =

- hC -802,310 kJ/kmol CH 4 = = 50, 010 kJ / kg CH 4 Mm 16.043 kg/kmol CH 4

The listed value from Table A-27 is 50, 050 kJ / kg . The calculated and listed values are practically identical.

16-107 CO gas is burned with air during a steady-flow combustion process. The rate of heat transfer from the combustion chamber is to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. Combustion is complete. Properties The molar masses of CO and air are 28 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis We first need to calculate the amount of air used per kmol of CO before we can write the combustion equation,

v CO =

RT (0.2968 kPa ×m /kg ×K )(310 K ) = = 0.836 m3 /kg P (110 kPa )

mCO =

VCO 0.4 m3 /min = = 0.478 kg/min v CO 0.836 m3 /kg

Then the molar air-fuel ratio becomes AF =

(1.5 kg/min )/ (29 kg/kmol) N air m / M air = air = = 3.03 kmol air/kmol fuel N fuel mfuel / M fuel (0.478 kg/min )/ (28 kg/kmol)

Thus the number of moles of O2 used per mole of CO is 3.03 / 4.76 = 0.637 . Then the combustion equation in this case can be written as

CO + 0.637 (O 2 + 3.76N 2 ) ¾ ¾® CO 2 + 0.137O 2 + 2.40N 2 Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to - Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

Substance

hf kJ / kmol

h 298 K

kJ / kmol

h 310 K kJ / kmol

h 900 K kJ / kmol

8669

9014

27,066

8682

---

27,928

8669

---

26,890

13-470

CO2

9364

---

37,405

Thus,

- Qout = (1)(- 393,520 + 37, 405 - 9364)+ (0.137)(0 + 27,928 - 8682) + (2.4)(0 + 26,890 - 8669)- (1)(- 110,530 + 9014 - 8669)- 0 - 0 = - 208,927 kJ/kmol of CO

Then the rate of heat transfer for a mass flow rate of 0.956 kg / min for CO becomes æ0.478 kg/min ö æm ö ÷ ç ÷ Qout = NQout = çç ÷ ÷ ÷(208,927 kJ/kmol) = 3567 kJ / min ÷Qout = ççè 28 kg/kmol ø çè N ø ÷

EES (Engineering Equation Solver) SOLUTION "Given" T1_fuel=(37+273) [K] P1_fuel=110 [kPa] V_dot_fuel=0.4 [m^3/min] T1_air=(25+273) [K] P1_air=110 [kPa] m_dot_air=1.5 [kg/min] T2=900 [K] T0=(25+273) [K] P=P1_fuel "Properties" R=R_u/MM_fuel R_u=8.314 [kJ/kmol-K] MM_air=molarmass(air) MM_fuel=molarmass(CO) "Analysis" v_fuel=(R*T1_fuel)/P1_fuel m_dot_fuel=V_dot_fuel/v_fuel AF_bar=N_air/N_fuel N_air=m_dot_air/MM_air N_fuel=m_dot_fuel/MM_fuel a=AF_bar/4.76 "Then the combustion equation becomes: CO + a (O2+3.76N2) = CO2 + (a-0.5) O2 + 3.76*a N2" "Mol numbers of reactants and products in kmol" N_CO=1 [kmol] N_O2_R=a N_O2_P=a-0.5 N_N2=a*3.76 N_CO2=1 [kmol] "Ideal gas enthalpies in kJ/kmol" h_CO_R=enthalpy(CO, T=T1_fuel) h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_CO2_P=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2_P+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_CO*h_CO_R+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R Q_dot_out=m_dot_fuel*(Q_out/MM_fuel)

13-471

16-108 Methane gas is burned steadily with dry air. The volumetric analysis of the products is given. The percentage of theoretical air used and the heat transfer from the combustion chamber are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. Analysis (a) Considering 100 kmol of dry products, the combustion equation can be written as xCH 4 + a [O2 + 3.76N 2 ]¾ ¾ ® 5.20CO2 + 0.33CO + 11.24O2 + 83.23N 2 + bH 2 O

The unknown coefficients x, a, and b are determined from mass balances, N2 :

3.76a = 83.23

¾¾ ®

a = 22.14

x = 5.20 + 0.33

¾¾ ®

x = 5.53

4 x = 2b

¾¾ ®

b = 11.06

(Check O2 :

a = 5.20 + 0.165 + 11.24 + b / 2

¾¾ ® 22.14 = 22.14)

Thus, 5.53CH 4 + 22.14 [O 2 + 3.76N 2 ]¾ ¾ ® 5.20CO 2 + 0.33CO + 11.24O 2 + 83.23N 2 + 11.06H 2 O

The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 5.53 CH 4 + 4[O 2 + 3.76N 2 ] ¾ ¾ ®

0.94CO 2 + 0.06CO + 2.03O 2 + 15.05N 2 + 2H 2 O

To find the percent theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of the fuel, CH 4 + ath [O2 + 3.76N 2 ]¾ ¾ ® CO 2 + 2H 2 O + 3.76ath N 2 O2 :

Then,

ath = 1 + 1 ¾ ¾ ® ath = 2.0

Percent theoretical air =

mair, act mair, th

N air, act N air, th

(4.0)(4.76) kmol = 200% (2.0)(4.76) kmol

(b) Under steady-flow conditions, energy balance applied on the combustion chamber reduces to - Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

13-472

Substance

hf kJ / kmol

h 290 kJ / kmol

h 298 kJ / kmol

h 700 kJ / kmol

CH4 ( g )

−74,850

---

8443

8682

21,184

8432

8669

20,604

H2 O ( g )

−241,820

---

9904

24,088

−110,530

---

8669

20,690

CO2

−393,520

---

9364

27,125

Thus, - Qout = (0.94)(- 393,520 + 27,125 - 9364)+ (0.06)(- 110,530 + 20, 690 - 8669) + (2)(- 241,820 + 24, 088 - 9904)+ (2.03)(0 + 21,184 - 8682 )+ (15.04 )(0 + 20, 604 - 8669) - (1)(- 74,850 + h298 - h298 )- (4)(0 + 8443 - 8682)- (15.04)(0 + 8432 - 8669)

= - 530,022 kJ / kmol CH4 or

Qout = 530,022 kJ / kmol CH4

EES (Engineering Equation Solver) SOLUTION "Given" T1_fuel=25+273 "[K]" T1_air=17+273 "[K]" T2=700 "[K]" T0=25+273 "[K]" "Analysis" "(a)" "Considering 100 kmol of dry products, the combustion equation is:" "x CH4 + a (O2+3.76N2) = 5.20 CO2 + 0.33 CO + 11.24 O2 + 83.23 N2 + b H2O" 3.76*a=83.23 "N2 balance" x=5.20+0.33 "C balance" 4*x=2*b "H balance" "Dividing the combustion equation by x, we obtain the combustion equation for 1 kmol of fuel" "(x/x) CH4 + (a/x) (O2+3.76N2) = (5.20/x) CO2 + (0.33/x) CO + (11.24/x) O2 + (83.23/x) N2 + (b/x) H2O" "Theoretical combustion equation is: CH4 + a_th (O2+3.76N2) = CO2 + 2 H2O + 3.76a_th N2" a_th=1+1 "O2 balance" PercentTheorAir=N_air_act/N_air_th*Convert(, %) N_air_act=a/x*4.76 N_air_th=a_th*4.76 "(b)" "Mol numbers of reactants and products in kmol" N_CH4=1 N_O2_R=a/x N_O2_P=11.24/x N_N2=a/x*3.76 N_CO2=5.20/x N_CO=0.33/x N_H2O=b/x "Ideal gas enthalpies in kJ/kmol" h_f_CH4=enthalpy(CH4, T=T0) h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-473

h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_CO2_P=enthalpy(CO2, T=T2) h_CO_P=enthalpy(CO, T=T2) h_H2O_P=enthalpy(H2O, T=T2) H_P=N_CO2*h_CO2_P+N_CO*h_CO_P+N_O2_P*h_O2_P+N_N2*h_N2_P+N_H2O*h_H2O_P H_R=N_CH4*h_f_CH4+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R

16-109 A mixture of hydrogen and the stoichiometric amount of air contained in a rigid tank is ignited. The fraction of

H2 O that condenses and the heat transfer from the combustion chamber are to be determined. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5 Combustion is complete. Analysis The theoretical combustion equation of H 2 with stoichiometric amount of air is

H 2 + ath (O2 + 3.76N 2 ) ¾ ¾® H 2 O + 3.76ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance, Thus,

ath = 0.5 H 2 + 0.5(O 2 + 3.76N 2 ) ¾ ¾ ® H 2 O + 1.88N 2

(a) At 25C part of the water (say, N w moles) will condense, and the number of moles of products that remains in the gas phase will be 2.88 - Nw . Neglecting the volume occupied by the liquid water and treating all the product gases as ideal gases, the final pressure in the tank can be expressed as Pf =

N f , gas RuT f V

(2.88 - N w kmol)(8.314 kPa ×m3 /kg ×K )(298 K ) 6 m3

= 412.9 (2.88 - N w ) kPa

Then, Nv P 1- N w 3.169 kPa = v ¾¾ ® = ¾¾ ® N w = 0.992 kmol Ngas Ptotal 2.88 - N w 412.9 (2.88 - N w )kPa

Thus 99.2% of the H 2 O will condense when the products are cooled to 25C. (b) The energy balance Ein - Eout = D Esystem applied for this constant volume combustion process with W = 0 reduces to

- Qout = å N P (h f + h - h - Pv ) - å N R (h f + h - h - Pv ) P

With the exception of liquid water for which the Pv term is negligible, both the reactants and the products are assumed to be ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT . It yields © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-474

- Qout = å N P (h f - RuT ) - å N R (h f - RuT ) P

= å N P h f , P - å N R h f , R - RuT (å N P , gas - å N R )

since the reactants are at the standard reference temperature of 25C. From the tables,

hf Substance

kJ / kmol

H2 O ( g ) H2 O ( ) Thus, - Qout = (0.008)(- 241,820)+ (0.992)(- 285,830)+ 0 - 0 - 0 - 0 - 8.314 ´ 298 (1.89 - 3.38) = - 281,786 kJ (per kmol H 2 )

Qout = 281,786 kJ (per kmol H2 )

EES (Engineering Equation Solver) SOLUTION "Given" V=6 [m^3] T=(25+273) [K] "Properties" R_u=8.314 [kPa-m^3/kmol-K] "Analysis" "Stoichiometric combustion equation is: H2 + a_th (O2+3.76N2) = H2O + 3.76a_th N2" a_th=0.5 "O2 balance" "(a)" P_f=(N_f_gas*R_u*T)/V N_v/N_f_gas=P_v/P_f N_f_gas=3.76*a_th+1-N_w N_v=1-N_w P_v=pressure(steam_iapws, T=T, x=1) "(b)" N_H2O_l=N_w N_H2O_g=1-N_w N_N2=3.76*a_th N_H2=1 N_O2=a_th "Enthalpy of formation data from Table A-26 in kJ/kmol" h_f_H2O_g=-241820 h_f_H2O_l=-285830 h_f_H2=0 h_f_O2=0 h_f_N2=0 -Q_out=H_P-H_R+R_u*T*(N_P_g-N_R) H_P=N_H2O_g*h_f_H2O_g+N_H2O_l*h_f_H2O_l+N_N2*h_f_N2 H_R=N_H2*h_f_H2+N_O2*h_f_O2+N_N2*h_f_N2 N_P_g=N_N2+N_H2O_g N_R=N_H2+N_O2+N_N2

13-475

16-110 Propane gas is burned with air during a steady-flow combustion process. The adiabatic flame temperature is to be determined for different cases. Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion chamber under adiabatic conditions (Q = 0) with no work interactions (W = 0). Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber reduces to

å N (h + h - h ) = å N (h + h - h ) ¾ ¾® å N (h + h - h ) = (Nh ) P

C3 H8

since all the reactants are at the standard reference temperature of 25C, and h f = 0 for O2 and N 2 . (a) The theoretical combustion equation of C3 H8 with stoichiometric amount of air is C3 H8 (g )+ 5(O 2 + 3.76N 2 ) ¾ ¾ ® 3CO 2 + 4H 2 O + 18.8 N 2

From the tables,

Substance

h 298 K

kJ / kmol

kJ / kmol ---

C3 H8 ( g ) O2

8682

8669

H2 O ( g )

9904

8669

CO2

9364

Thus,

(3)(- 393,520 + hCO - 9364)+ (4)(- 241,820 + hH O - 9904)+ (18.8)(0 + hN - 8669) = (1)(- 103,850) 2

It yields

3hCO2 + 4hH2 O + 18.8hN2 = 2, 274, 675 kJ The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 2, 274,675 / (3 + 4 + 18.8) = 88,165 kJ / kmol . This enthalpy value corresponds to about 2650 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2650 K, but somewhat under it because of the higher specific heats of CO2 and H 2 O . At 2400 K:

3hCO2 + 4hH2 O + 18.8hN2 = (3)(125,152)+ (4)(103,508)+ (18.8)(79,320)

= 2, 280, 704 kJ (Higher than 2,274,675 kJ )

13-476

At 2350 K:

3hCO2 + 4hH2 O + 18.8hN2 = (3)(122, 091)+ (4)(100,846)+ (18.8)(77, 496) = 2, 226,582 kJ (Lower than 2,274,675 kJ )

By interpolation,

TP = 2394 K (b) The balanced combustion equation for complete combustion with 200% theoretical air is C3 H8 (g )+ 10 (O 2 + 3.76N 2 ) ¾ ¾ ® 3CO 2 + 4H 2 O + 5O 2 + 37.6 N 2

Substituting known numerical values,

(3)(- 393,520 + hCO - 9364)+ (4)(- 241,820 + hH O - 9904)+ (5)(0 + hO - 8682)+ (37.6)(0 + hN - 8669)= (1)(- 103,850) 2

which yields

3hCO2 + 4hH2 O + 5hO2 + 37.6hN2 = 2, 481, 060 kJ The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 2, 481,060 / (3 + 4 + 5 + 37.6) = 50,021 kJ / kmol . This enthalpy value corresponds to about 1580 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 1580 K, but somewhat under it because of the higher specific heats of CO2 and H 2 O . At 1540 K:

3hCO2 + 4hH2O + 5hO2 + 37.6hN2 = (3)(73, 417)+ (4)(59,888)+ (5)(50,756)+ (37.6)(48, 470) = 2,536, 055 kJ (Higher than 2,481,060 kJ )

At 1500 K:

3hCO2 + 4hH2O + 5hO2 + 37.6hN2 = (3)(71,078)+ (4)(57,999)+ (5)(49, 292)+ (37.6)(47,073) = 2, 461, 630 kJ (Lower than 2,481,060 kJ )

By interpolation,

TP = 1510 K

(c) The balanced combustion equation for incomplete combustion with 90% theoretical air is C3 H8 (g )+ 4.5(O 2 + 3.76N 2 ) ¾ ¾ ® 2CO 2 + 1CO + 4H 2 O + 16.92 N 2

Substituting known numerical values,

(2)(- 393,520 + hCO - 9364)+ (1)(- 110,530 + hCO - 8669)+ 2

(4)(- 241,820 + hH O - 9904)+ (16.92)(0 + hN - 8669) = (1)(- 103,850) 2

which yields

2hCO2 + 1hCO + 4hH2 O + 16.92hN2 = 1,974, 692 kJ The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 1,974,692 / (2 + 1 + 4 + 16.92) = 82,554 kJ / kmol . This enthalpy value corresponds to about 2500 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2500 K, but somewhat under it because of the higher specific heats of CO2 and H 2 O . At 2250 K:

2hCO2 + 1hCO + 4hH2 O + 16.92hN2 = (2)(115,984)+ (1)(74,516)+ (4)(95,562)+ (16.92)(73,856) = 1,938,376 kJ (Higher than 1,974,692 kJ )

At 2300 K:

2hCO2 + 1hCO + 4hH2 O + 16.92hN2 = (2)(119,035)+ (1)(76,345)+ (4)(98,199)+ (16.92)(75,676) = 1,987, 649 kJ (Lower than 1,974,692 kJ )

13-477

TP = 2287 K

EES (Engineering Equation Solver) SOLUTION "Part a and b" "Given" ex=1 "Change this to zero to get the answer for the stoichiometric reaction" T_reac=25 [C] P_reac=101.325 [kPa] "Analysis" "Combustion equation is: C3H8 + (ex+1)a_th (O2+3.76N2) = 3 CO2 + 4 H2O + (ex)(a_th) O2 + (ex+1)a_thx3.76 N2" a_th=3+4/2 "O2 balance" N_C3H8=1 [kmol] N_CO2=3 [kmol] N_H2O=4 [kmol] N_O2=ex*a_th N_N2=(ex+1)*a_th*3.76 h_CO2=enthalpy(CO2, T=T_af) h_H2O=enthalpy(H2O, T=T_af) h_O2=enthalpy(O2, T=T_af) h_N2=enthalpy(N2, T=T_af) h_C3H8=enthalpy(C3H8, T=T_reac) H_P=H_R H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2*h_O2+N_N2*h_N2 H_R=N_C3H8*h_C3H8 "Part c" "Given" ex=0.90 T_reac=25 [C] P_reac=101.325 [kPa] "Analysis" "The stoichiometric combustion equation is: C3H8 + a_th (O2+3.76N2) = 3 CO2 + 4 H2O + 3.76*a_th N2" a_th=3+2 "O2 balance" "The actual combustion equation is: C3H8 + ex*a_th (O2+3.76N2) = x CO2 + (3-x) CO + 4 H2O + ex*a_th*3.76 N2" ex*a_th=x+(3-x)/2+2 "O2 balance" N_C3H8=1 N_N2=ex*a_th*3.76 N_CO2=x N_CO=3-x N_H2O=4 h_CO2=enthalpy(CO2, T=T_af) h_CO=enthalpy(CO, T=T_af) h_H2O=enthalpy(H2O, T=T_af) h_N2=enthalpy(N2, T=T_af) h_C3H8=enthalpy(C3H8, T=T_reac) H_P=H_R H_P=N_CO2*h_CO2+N_CO*h_CO+N_H2O*h_H2O+N_N2*h_N2 H_R=N_C3H8*h_C3H8

16-111 The highest possible temperatures that can be obtained when liquid gasoline is burned steadily with air and with pure oxygen are to be determined.

13-478

Assumptions 1. Steady operating conditions exist. 2. Air and combustion gases are ideal gases. 3. Kinetic and potential energies are negligible. 4. There are no work interactions. 5. The combustion chamber is adiabatic. Analysis The highest possible temperature that can be achieved during a combustion process is the temperature which occurs when a fuel is burned completely with stoichiometric amount of air in an adiabatic combustion chamber. It is determined from

å N (h + h - h ) = å N (h + h - h ) ¾ ¾® å N (h + h - h ) = (Nh ) P

C8 H18

since all the reactants are at the standard reference temperature of 25C, and for O2 and N 2 . The theoretical combustion equation of C8 H18 air is

C8 H18 + 12.5(O 2 + 3.76N 2 ) ¾ ¾® 8CO 2 + 9H 2 O + 47 N 2 From the tables,

h 298 K

Substance

hf kJ / kmol

kJ / kmol

C8 H18 ( )

−249,950

---

8682

8669

H2 O ( g )

−241,820

9904

CO2

−393,520

9364

Thus,

(8)(- 393,520 + hCO - 9364)+ (9)(- 241,820 + hH O - 9904)+ (47)(0 + hN - 8669) = (1)(- 249,950) 2

It yields

8hCO2 + 9hH2 O + 47hN2 = 5,646,081 kJ

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 5,646,081/ (8 + 9 + 47) = 88,220 kJ / kmol . This enthalpy value corresponds to about 2650 K for N 2 . Noting that the majority of the moles are N 2 , TP will be close to 2650 K, but somewhat under it because of the higher specific heat of H 2 O . At 2400 K:

8hCO2 + 9hH2 O + 47hN2 = (8)(125,152)+ (9)(103,508)+ (47)(79,320) = 5, 660,828 kJ (Higher than 5,646,081 kJ )

At 2350 K:

8hCO2 + 9hH2 O + 47hN2 = (8)(122, 091)+ (9)(100,846)+ (47)(77, 496)

= 5,526,654 kJ (Lower than 5,646,081 kJ ) By interpolation, TP = 2395 K

13-479

If the fuel is burned with stoichiometric amount of pure O2 , the combustion equation would be

C8 H18 + 12.5O2 ¾ ¾® 8CO2 + 9H 2 O Thus,

(8)(- 393,520 + hCO - 9364)+ (9)(- 241,820 + hH O - 9904) = (1)(- 249,950)

It yields

8hCO2 + 9hH2 O = 5, 238, 638 kJ

The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by dividing the righthand side of the equation by the total number of moles, which yields 5,238,638 / (8 + 9) = 308,155 kJ / kmol . This enthalpy value is higher than the highest enthalpy value listed for H 2 O and CO2 . Thus an estimate of the adiabatic flame temperature can be obtained by extrapolation. At 3200 K:

8hCO2 + 9hH2 O = (8)(174,695)+ (9)(147, 457) = 2,724,673 kJ

At 3250 K:

8hCO2 + 9hH2 O = (8)(177,822)+ (9)(150, 272) = 2,775,024 kJ

By extrapolation, we get TP = 3597 K . However, the solution of this problem using EES gives 5645 K. The large difference between these two values is due to extrapolation.

EES (Engineering Equation Solver) SOLUTION "Given" T1=(25+273) [K] P1=101.325 [kPa] T0=(25+273) [K] "Analysis" "The stoichiometric combustion equation is: C8H18 + a_th (O2+3.76N2) = 8 CO2 + 9 H2O + 3.76*a_th N2" a_th=8+4.5 "O2 balance" "Mol numbers of reactants and products in kmol" N_C8H18=1 N_O2=a_th N_N2=a_th*3.76 N_CO2=8 N_H2O=9 "Enthalpy of formation data from Table A-26 in kJ/kmol" h_f_C8H18=-249950 [kJ/kmol] "Ideal gas enthalpies in kJ/kmol" h_O2=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_N2_P=enthalpy(N2, T=T2_a) h_CO2_a=enthalpy(CO2, T=T2_a) h_H2O_a=enthalpy(H2O, T=T2_a) H_P_a=N_CO2*h_CO2_a+N_H2O*h_H2O_a+N_N2*h_N2_P H_R_a=N_C8H18*h_f_C8H18+N_O2*h_O2+N_N2*h_N2_R H_P_a=H_R_a "If pure oxygen is used instead of air, the stoichiometric combustion equation is: C8H18 + a_th O2 = 8 CO2 + 9 H2O" "Ideal gas enthalpies of products are, in kJ/kmol" h_CO2_b=enthalpy(CO2, T=T2_b) h_H2O_b=enthalpy(H2O, T=T2_b) H_P_b=N_CO2*h_CO2_b+N_H2O*h_H2O_b H_R_b=N_C8H18*h_f_C8H18+N_O2*h_O2 H_P_b=H_R_b

13-480

16-112 Liquid propane, C3 H8 (liq) is burned with 150 percent excess air. The balanced combustion equation is to be written and the mass flow rate of air, the average molar mass of the product gases, the average specific heat of the product gases at constant pressure are to be determined. Assumptions 1.

Combustion is complete.

The combustion products contain CO2 , H 2 O , O2 , and N 2 only.

Properties The molar masses of C, H 2 , O2 , N 2 , and air are 12, 2, 32, 28, and 29 kg / kmol , respectively (Table A-1). Analysis (a) The reaction equation for 150% excess air is C3 H8 (liq.) + 2.5ath [O 2 + 3.76N 2 ] ¾ ¾ ® B CO 2 + D H 2 O + E O 2 + F N 2

where ath is the stoichiometric coefficient for air. We have automatically accounted for the 150% excess air by using the factor 2.5ath instead of ath for air. The coefficient ath and other coefficients are to be determined from the mass balances

Carbon balance:

B=3

Hydrogen balance:

2D = 8 ¾ ¾ ® D= 4

Oxygen balance:

2´ 2.5ath = 2B + D + 2E

1.5ath = E Nitrogen balance:

2.5ath ´ 3.76 = F

Solving the above equations, we find the coefficients (E = 7.5, F = 47, and ath = 5 ) and write the balanced reaction equation as C3 H8 + 12.5 [O 2 + 3.76N 2 ] ¾ ¾ ® 3 CO 2 + 4 H 2 O + 7.5 O 2 + 47 N 2

The fuel flow rate is mfuel 0.4 kg/min = = 0.009071 kmol/min M fuel 44 kg/kmol

N fuel =

The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel, AF =

mair (12.5´ 4.76 kmol)(29 kg/kmol) = = 39.08 kg air/kg fuel mfuel (1 kmol)(44 kg/kmol)

Then, the mass flow rate of air becomes

mair = AFmfuel = (39.08)(0.4 kg/min) = 15.63 kg / min (b) The molar mass of the product gases is determined from M prod =

N CO2 M CO2 + N H2O M H2O + N O2 M O2 + N N2 M N2 N CO2 + N H2O + N O2 + N N2

3(44) + 4(18) + 7.5(32) + 47(28) 3 + 4 + 7.5 + 47

= 28.63 kg / kmol (c) The steady-flow energy balance is expressed as

13-481

N fuel H R = Qout + N fuel H P

where from Tables A-26 and A-27 hfuel = - 103,850 kJ/kmol

h fg = (335 kJ/kg)(44 kg/kmol)=14,740 kJ/kmol and

H R = h fuel@25°C - h fg + 12.5hO2@25°C + 47hN2@25°C

= (- 103,850 kJ/kmol - 14,740 kJ/kmol) + 12.5(0) + 47(0) = - 118,590 kJ/kmol fuel

H P = 3hCO2@ TP + 4hH2O@ TP + 7.5hO2@ TP + 47hN2@ TP Substituting into the energy balance equation,

N fuel H R = Qout + N fuel H P (0.009071 kmol/min)(- 118,590 kJ/kmol) = (53´ 60)kJ/min + (0.009071 kmol/min) H P

H P = - 469,158 kJ/kmol fuel Substituting this value into the H P relation above and by a trial-error approach with the values in ideal gas tables (Tables A-18 through A-26), we obtain the temperature of the products of combustion to be about

TP = 1140 K Checking this result with ideal gas tables (Tables A-18 through A-26):

H P = 3hCO2@ Tp + 4hH2O@ Tp + 7.5hO@ Tp + 47hN2@ Tp

= 3(- 393,520 + 50, 484 - 9364) + 4(- 241,820 + 41,780 - 9904) + 7.5(36,314 - 8682) + 47(34,760 - 8669) = - 463, 459 kJ / kmol fuel This is sufficiently close to the value calculated earlier ( - 469,158 kJ / kmol fuel ). Therefore, the result is confirmed.

EES (Engineering Equation Solver) SOLUTION "The reaction equation for 150% excess air is:" "C3H8(liq)+ 2.5*A_th (O2 +3.76N2)=B CO2 + D H2O + E O2 + F N2" "Carbon Balance:" 3=B "Hydrogen Balance:" 8=2*D "Oxygen Balance:" 2.5*A_th*2=B*2 + D + E*2 "Excess oxygen in the products is:" 1.5*A_th = E "Nitrogen Balance:" 2.5*A_th*3.76 = F "The fuel flow rate is:" m_dot_fuel = 0.4 "[kg/min]" N_dot_fuel = m_dot_fuel/molarmass(C3H8) "[kmol/min]" "The mass flow rate of air is:" AF = 2.5*A_th*4.76*molarmass(Air)/molarmass(C3H8) "[kg_air/kg_fuel]" m_dot_air=m_dot_fuel*AF"[kg/min]" "The molar mass of the product gases is:"

13-482

MM_prod = (B*molarmass(CO2) + D*molarmass(H2O) + E*molarmass(O2)+ F*molarmass(N2))/(B+D+E+F) "[kg/kmol]" "The average specific heat at constant pressure of the product gases is:" Cp_prod = (B*C_CO2 + D*C_H2O + E*C_O2+ F*C_N2)/(B+D+E+F) "[kJ/kmol]" C_CO2=specheat(CO2, T=T_P) C_H2O=specheat(H2O, T=T_P) C_O2=specheat(O2, T=T_P) C_N2=specheat(N2, T=T_P) "The product temperature is determined from the energy balance" "The reactant temperature is :" T_R= 25+273.15 "[K]" "The required heat transfer rate is:" Q_dot_out = 53*60 "[kJ/min]" "The steady-flow energy balance is:" N_dot_fuel*H_R = Q_dot_out+N_dot_fuel*H_P "From Table A-27 ,the enthalpy of vaporization of the fuel is:" MM_fuel =molarmass(C3H8)"[kg/kmol]" h_bar_fg=335 "[kJ/kg]"*MM_fuel "[kJ/kmol]" "Hint: when product temperature is unknown, a good guess value is 450 K and can be set under Options, Variable infromation." H_R=1*(h_fuel - h_bar_fg) +2.5*A_th*h_O2_R+2.5*A_th*3.76*h_N2_R "[kJ/kmol]" H_P=B*h_CO2_P+D*h_H2O_P +E*h_O2_P+F*h_N2_P "[kJ/kmol]" h_fuel=enthalpy(C3H8,T=T_R) h_O2_R=enthalpy(O2,T=T_R) h_N2_R=enthalpy(N2,T=T_R) h_CO2_P=enthalpy(CO2,T=T_P) h_H2O_P=enthalpy(H2O,T=T_P) h_O2_P=enthalpy(O2,T=T_P) h_N2_P=enthalpy(N2,T=T_P) "The dew point temperature of the products is (assume P_prod = 100 kPa):" P_prod = 100 "[kPa]" P_v = D/(B+D+E+F)*P_prod"[kPa]" T_dp_prod =temperature(Steam, P=P_v, x=1)"[K]" "Here T_P > T_dp_prod, and there is no condensation."

16-113 Octane is burned with stoichiometric amount of air. The maximum work that can be produced is to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis The combustion equation is

C8 H18 + 12.5(O 2 + 3.76N 2 ) ¾ ¾® 8CO2 + 9H 2 O + 47N 2

13-483

The reactants and products are at 25C and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products, Wrev = å N R g f , R - å N P g f , P = (1)(16,530)- (8)(- 394,360) - (9)(- 228,590)

= 5,228,720 kJ (per kmol of fuel) since the g f of stable elements at 25C and 1 atm is zero. Per unit mass basis, Wrev =

5,228,720 kJ/kmol = 45, 870 kJ / kg fuel 114 kg/kmol

16-114 Octane is burned with 100% excess air. The maximum work that can be produced is to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis The combustion equation is

C8 H18 + 25(O2 + 3.76N 2 ) ¾ ¾® 8CO2 + 9H 2 O + 12.5O 2 + 94N 2 The reactants and products are at 25C and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products,

Wrev = å N R g f , R - å N P g f , P = (1)(16,530)- (8)(- 394,360) - (9)(- 228,590)

= 5,228,720 kJ

(per kmol of fuel)

since the g f of stable elements at 25C and 1 atm is zero. Per unit mass basis, Wrev =

5,228,720 kJ/kmol = 45, 870 kJ / kg fuel 114 kg/kmol

16-115E Methane is burned with 100 percent excess air. The combustion is incomplete. The maximum work that can be produced is to be determined. Assumptions 1. Combustion is incomplete. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-484

2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Analysis The combustion equation is

CH 4 + 4 (O 2 + 3.76N 2 ) ¾ ¾ ® 0.90 CO 2 + 0.10CO + 2H 2 O + xO 2 + 15.04N 2

The coefficient for O2 is determined from its mass balance as

4 = 0.9 + (0.5)(0.10) + 1 + x ¾ ¾ ® x = 2.05 Substituting, CH 4 + 4 (O 2 + 3.76N 2 ) ¾ ¾ ® 0.90 CO 2 + 0.10CO + 2H 2 O + 2.05O 2 + 15.04N 2

The reactants and products are at 77F and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products, Wrev = å N R g f , R - å N P g f , P = (1)(- 21,860)- (0.9)(- 169, 680) - (0.10)(- 59, 010) - (2)(- 98,350)

= 333,453 Btu

(per lbmol of fuel)

since the g f of stable elements at 77F and 1 atm is zero. Per unit mass basis,

Wrev =

333,453 Btu/lbmol = 20, 840 Btu / lbm fuel 16 lbm/lbmol

16-116 Methane is burned steadily with 50 percent excess air in a steam boiler. The amount of steam generated per unit of fuel mass burned, the change in the exergy of the combustion streams, the change in the exergy of the steam stream, and the lost work potential are to be determined. Assumptions 1. 2. 3. 4.

Combustion is complete. Steady operating conditions exist. Air and the combustion gases are ideal gases. Changes in kinetic and potential energies are negligible.

Properties The molar masses of CH4 and air are 16 kg / kmol and 29 kg / kmol , respectively (Table A-1). Analysis (a) The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol CH4 , the combustion equation can be written as CH 4 + 3(O2 + 3.76N 2 ) ¾ ¾ ® CO2 + 2H 2 O + O 2 + 11.28N 2

13-485

Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to - Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, h 298 K

h500 K

Substance

hf kJ / kmol

kJ / kmol

CH4

−74,850

---

8682

14,770

8669

14,581

H2 O ( g )

−241,820

9904

16,828

CO2

−393,520

9364

17,678

Thus,

- Qout = (1)(- 393,520 + 17,678 - 9364)+ (2)(- 241,820 + 16,828 - 9904)+ (1)(0 + 14,770 - 8682)+ (11.28)(0 + 14,581- 8669)- (1)(- 74,850) = - 707, 373 kJ/kmol of fuel

The heat loss per unit mass of the fuel is Qout =

707,373 kJ/kmol of fuel = 44, 211 kJ/kg fuel 16 kg/kmol of fuel

The amount of steam generated per unit mass of fuel burned is determined from an energy balance to be (Enthalpies of steam are from tables A-4 and A-6) ms Q 44,211 kJ/kg fuel = out = = 18.72 kg steam / kg fuel mf D hs (3214.5 - 852.26) kJ/kg steam

(b) The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm )) The entropy calculations can be presented in tabular form as

Ni CH4 O2

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

---

0.21

186.16

205.04

−12.98

654.06

13-486

11.28

0.79

191.61

−1.960

2183.47

SR = 3023.69 kJ / K CO2 H2 O ( g ) O2 N2

0.0654

234.814

−22.67

257.48

0.1309

206.413

−16.91

446.65

0.0654

220.589

−22.67

243.26

11.28

0.7382

206.630

−2.524

2359.26

SP = 3306.65 kJ / K Thus, Sgen = S P - S R +

Qout 707,373 = 3306.65 - 3023.69 + = 2657 kJ/K (per kmol fuel) Tsurr 298

The exergy change of the combustion streams is equal to the exergy destruction since there is no actual work output. That is,

D X gases = - X dest = - T0 Sgen = - (298 K)(2657 kJ/K) = - 791,786 kJ/kmol fuel Per unit mass basis, D X gases =

- 791, 786 kJ/kmol fuel = - 49,490 kJ / kg fuel 16 kg/kmol

Note that the exergy change is negative since the exergy of combustion gases decreases. (c) The exergy change of the steam stream is

D X steam = D h - T0D s = (3214.5 - 852.26) - (298)(6.7714 - 2.3305) = 1039 kJ / kg steam (d) The lost work potential is the negative of the net exergy change of both streams:

æm ö ÷ X dest = - ççç s D X steam + D X gases ÷ ÷ ÷ ÷ èç m f ø = - [(18.72 kg steam/kg fuel)(1039 kJ/kg steam) + (- 49, 490 kJ/kg fuel) ]

= 30,040 kJ / kg fuel

EES (Engineering Equation Solver) SOLUTION "Given" T_w1=200 [C] x_w1=0 P_w2=4000 [kPa] T_w2=400 [C] ex=0.5 P=101.325 [kPa] T1=25 [C] T2=227 [C] T0=(25+273) [K] "Analysis" "(a)" "Combustion equation is: CH4 + (ex+1)a_th (O2+3.76N2) = 1 CO2 + 2 H2O + (ex)(a_th) O2 + (ex+1)a_thx3.76 N2" a_th=1+2/2 "O2 balance" N_CH4=1 [kmol] N_CO2=1 [kmol] N_H2O=2 [kmol] N_O2=ex*a_th N_O2_R=(ex+1)*a_th © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-487

N_N2=(ex+1)*a_th*3.76 "Enthalpies of reactants and products" h_CH4_R=enthalpy(CH4, T=T1) h_O2_R=enthalpy(O2, T=T1) h_N2_R=enthalpy(N2, T=T1) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) -Q_bar_out=H_P-H_R H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2*h_O2_P+N_N2*h_N2_P H_R=N_CH4*h_CH4_R+N_O2_R*h_O2_R+N_N2*h_N2_R MM_CH4=16 [kg/kmol] Q_out=Q_bar_out/MM_CH4 Fluid$='steam_iapws' h_w1=enthalpy(Fluid$, T=T_w1, x=x_w1) h_w2=enthalpy(Fluid$, T=T_w2, P=P_w2) MM_H2O=MolarMass(H2O) DELTAh_w=(h_w2-h_w1)/MM_H2O m_s\m_f=Q_out/DELTAh_w "(b)" "Entropies of reactants" P_O2_R= 1/4.76*P "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T1, P=P_O2_R) P_N2_R= 3.76/4.76*P s_N2_R=entropy(N2, T=T1, P=P_N2_R) s_CH4=entropy(CH4, T=T1, P=P) S_R=N_CH4*s_CH4+N_O2_R*s_O2_R+N_N2*s_N2_R "Entropies of products" N_prod =N_CO2+N_H2O+N_O2+N_N2 "total kmol of products" P_O2_P=N_O2/N_prod*P "Partial pressure of O2 in products" s_O2_P=entropy(O2, T=T2, P=P_O2_P) P_N2_P=N_N2/N_prod*P s_N2_P=entropy(N2, T=T2, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T2, P=P_CO2_P) P_H2O_P=N_H2O/N_prod*P s_H2O_P=entropy(H2O, T=T2, P=P_H2O_P) S_P=N_CO2*s_CO2_P+N_H2O*s_H2O_P+N_N2*s_N2_P+N_O2*s_O2_P S_bar_gen=S_P-S_R+Q_bar_out/T0 DELTAX_bar_gases=-T0*S_bar_gen DELTAX_gases=DELTAX_bar_gases/MM_CH4 "(c)" s_w1=entropy(Fluid$, T=T_w1, x=x_w1) s_w2=entropy(Fluid$, T=T_w2, P=P_w2) DELTAs_w=s_w2-s_w1 DELTAX_w=(h_w2-h_w1-T0*(s_w2-s_w1))/MM_H2O "(d)" X_dest=-(m_s\m_f*DELTAX_w+DELTAX_gases)

16-117 A coal from Utah is burned steadily with 50 percent excess air in a steam boiler. The amount of steam generated per unit of fuel mass burned, the change in the exergy of the combustion streams, the change in the exergy of the steam stream, and the lost work potential are to be determined. Assumptions 1. Combustion is complete. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-488

2. 3. 4. 5.

Steady operating conditions exist. Air and the combustion gases are ideal gases. Changes in kinetic and potential energies are negligible. The effect of sulfur on the energy and entropy balances is negligible.

Properties The molar masses of C, H 2 , N 2 , O2 , S, and air are 12, 2, 28, 32, 32, and 29 kg / kmol , respectively (Table A1). Analysis (a) We consider 100 kg of coal for simplicity. Noting that the mass percentages in this case correspond to the masses of the constituents, the mole numbers of the constituent of the coal are determined to be

NC =

mC 61.40 kg = = 5.117 kmol M C 12 kg/kmol

N H2 =

mH2 5.79 kg = = 2.895 kmol M H2 2 kg/kmol

N O2 =

mO2 25.31 kg = = 0.7909 kmol M O2 32 kg/kmol

N N2 =

mN2 1.09 kg = = 0.03893 kmol M N2 28 kg/kmol

NS =

mS 1.41 kg = = 0.04406 kmol M S 32 kg/kmol

The mole number of the mixture and the mole fractions are

Nm = 5.117 + 2.895 + 0.7909 + 0.03893 + 0.04406 = 8.886 kmol yC =

N C 5.117 kmol = = 0.5758 N m 8.886 kmol

yH2 =

N H2 2.895 kmol = = 0.3258 Nm 8.886 kmol

yO2 =

N O2 0.7909 kmol = = 0.0890 Nm 8.886 kmol

13-489

yN2 =

N N2 0.03893 kmol = = 0.00438 Nm 8.886 kmol

yS =

NS 0.04406 kmol = = 0.00496 Nm 8.886 kmol

0.5758C + 0.3258H 2 + 0.0890O 2 + 0.00438N 2 + 0.00496S + 1.5ath (O2 + 3.76N 2 ) ¾¾ ® 0.5758CO2 + 0.3258H 2 O + 0.00496SO2 + 0.5ath O2 + 1.5ath ´ 3.76N 2 According to the oxygen balance,

O2 balance: 0.0890 + 1.5ath = 0.5758 + 0.5´ 0.3258 + 0.00496 + 0.5ath ¾ ¾ ® ath = 0.6547 Substituting,

0.5758C + 0.3258H 2 + 0.0890O2 + 0.00438N 2 + 0.00496S + 0.9821(O2 + 3.76N2 ) ¾¾ ® 0.5758CO2 + 0.3258H2 O + 0.00496SO2 + 0.3274O2 + 3.693N 2 The apparent molecular weight of the coal is

Mm =

mm Nm

(0.5758´ 12 + 0.3258´ 2 + 0.0890´ 32 + 0.00438´ 28 + 0.00496´ 32) kg (0.5758 + 0.3258 + 0.0890 + 0.00438 + 0.00496) kmol

10.69 kg 1.0 kmol

= 10.69 kg/kmol coal Under steady-flow conditions the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0 reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

13-490

h 298 K

h500 K

Substance

hf kJ / kmol

kJ / kmol

8682

14,770

8669

14,581

H2 O ( g )

9904

16,828

CO2

9364

17,678

Thus,

- Qout = (0.5758)(- 393,520 + 17, 678 - 9364)+ (0.3258)(- 241,820 + 16,828 - 9904) + (0.3274)(0 + 14,770 - 8682)+ (3.693)(0 + 14,581 - 8669)- 0 = - 274,505 kJ/kmol of fuel

The heat loss per unit mass of the fuel is Qout =

274,505 kJ/kmol of fuel = 25, 679 kJ/kg fuel 10.69 kg/kmol of fuel

The amount of steam generated per unit mass of fuel burned is determined from an energy balance to be (Enthalpies of steam are from tables A-4 and A-6) ms Q 25,679 kJ/kg fuel = out = = 10.87 kg steam / kg fuel mf D hs (3214.5 - 852.26) kJ/kg steam

(b) The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

Pi = yi Ptotal , where yi is the mole fraction of component i. Then, Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln ( yi Pm ))

The entropy calculations can be presented in tabular form as

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

0.5758

5.74

−4.589

5.95

0.3258

130.68

−9.324

45.61

0.0890

205.04

−20.11

20.04

0.00438

191.61

−45.15

1.04

0.9821

0.21

205.04

−12.98

214.12

3.693

0.79

191.61

−1.960

714.85

SR = 1001.61 kJ / K CO2

0.5758

0.1170

234.814

−17.84

145.48

13-491

H2 O ( g )

0.3258

0.0662

206.413

−22.57

74.60

0.3274

0.0665

220.589

−22.54

79.60

3.693

0.7503

206.630

−2.388

771.90

SP = 1071.58 kJ / K Thus, Sgen = S P - S R +

Qout 274,505 = 1071.58 - 1001.61 + = 991.1 kJ/K (per kmol fuel ) Tsurr 298

The exergy change of the combustion streams is equal to the exergy destruction since there is no actual work output. That is,

D X gases = - X dest = - T0 Sgen = - (298 K)(991.1 kJ/K) = - 295,348 kJ/kmol fuel Per unit mass basis, D X gases =

- 295,348 kJ/K = - 27,630 kJ / kg fuel 10.69 kg/kmol

Note that the exergy change is negative since the exergy of combustion gases decreases. (c) The exergy change of the steam stream is

D X steam = D h - T0D s = (3214.5 - 852.26) - (298)(6.7714 - 2.3305) = 1039 kJ/ kg steam (d) The lost work potential is the negative of the net exergy change of both streams:

æm ö ÷ X dest = - ççç s D X steam + D X gases ÷ ÷ ÷ çè m f ÷ ø = - [(10.87 kg steam/kg fuel)(1039 kJ/kg steam) + (- 27, 630 kJ/kg fuel) ]

= 16,340 kJ / kg fuel

EES (Engineering Equation Solver) SOLUTION "Given" T_w1=200 [C] x_w1=0 P_w2=4000 [kPa] T_w2=400 [C] ex=0.5 P=101.325 [kPa] T1=25 [C] T2=227 [C] T0=(25+273) [K] m_C=61.40 [kg] m_H2=5.79 [kg] m_O2=25.31 [kg] m_N2=1.09 [kg] m_S=1.41 [kg] m_ash=5.00 [kg] "Properties" MM_C=12 [kg/kmol] MM_H2=2 [kg/kmol] MM_O2=32 [kg/kmol] MM_N2=28 [kg/kmol] MM_S=32 [kg/kmol] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-492

MM_CO2=44 [kg/kmol] MM_H2O=18 [kg/kmol] MM_SO2=64 [kg/kmol] MM_air=29 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" "(a)" N_C=m_C/MM_C N_H2=m_H2/MM_H2 N_O2=m_O2/MM_O2 N_N2=m_N2/MM_N2 N_S=m_S/MM_S N_m=N_C+N_H2+N_O2+N_N2+N_S y_C=N_C/N_m y_H2=N_H2/N_m y_O2=N_O2/N_m y_N2=N_N2/N_m y_S=N_S/N_m x=y_C "C balance" y=y_H2 "H2 balance" z=y_S "S balance" y_O2+a_th=x+0.5*y+z "O2 balance" k=y_N2+(1+ex)*3.76*a_th "N2 balance" m=ex*a_th N_C_R=y_C N_O2_a=(ex+1)*a_th N_N2_a=(ex+1)*a_th*3.76 N_H2_R=y_H2 N_S_R=y_S N_O2_R=y_O2 N_N2_R=y_N2 N_CO2=x N_H2O=y N_SO2=z N_O2_p=m N_N2_p=k MM_fuel=(N_C_R*MM_C+N_H2_R*MM_H2+N_S_R*MM_S+N_N2_R*MM_N2+N_O2_R*MM_O2)/N_tot N_tot=N_C_R+N_H2_R+N_S_R+N_O2_R+N_N2_R "Enthalpies of products" h_O2=enthalpy(O2, T=T2) h_N2=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2=enthalpy(CO2, T=T2) -Q_bar_out=H_P-H_R H_P=N_CO2*h_CO2+N_H2O*h_H2O+N_O2_p*h_O2+N_N2_p*h_N2 H_R=0 Q_out=Q_bar_out/MM_fuel Fluid$='steam_iapws' h_w1=enthalpy(Fluid$, T=T_w1, x=x_w1) h_w2=enthalpy(Fluid$, T=T_w2, P=P_w2) DELTAh_w=(h_w2-h_w1)/MM_H2O m_s\m_f=Q_out/DELTAh_w "(b)" "Entropies of reactants" s_C_R=N_C_R*(5.74-R_u*ln(y_C*1)) P_O2_a= 1/4.76*P s_O2_a=entropy(O2, T=T1, P=P_O2_a) P_N2_a= 3.76/4.76*P

13-493

s_N2_a=entropy(N2, T=T1, P=P_N2_a) P_N2_R=y_N2*P s_N2_R=entropy(N2, T=T1, P=P_N2_R) P_H2_R=y_H2*P s_H2_R=entropy(H2, T=T1, P=P_H2_R) P_O2_R=y_O2*P s_O2_R=entropy(O2, T=T1, P=P_O2_R) S_R=N_C_R*s_C_R+N_O2_a*s_O2_a+N_N2_a*s_N2_a+N_N2_R*s_N2_R+N_O2_R*s_O2_R+N_H2_R*s_H2 _R "Entropies of products" N_prod =N_CO2+N_H2O+N_O2_p+N_N2_p "total kmol of products" P_O2_P=N_O2_p/N_prod*P "Partial pressure of O2 in products" s_O2_P=entropy(O2, T=T2, P=P_O2_P) P_N2_P=N_N2_p/N_prod*P s_N2_P=entropy(N2, T=T2, P=P_N2_P) P_CO2_P=N_CO2/N_prod*P s_CO2_P=entropy(CO2, T=T2, P=P_CO2_P) P_H2O_P=N_H2O/N_prod*P s_H2O_P=entropy(H2O, T=T2, P=P_H2O_P) S_P=N_CO2*s_CO2_P+N_H2O*s_H2O_P+N_N2_p*s_N2_P+N_O2_p*s_O2_P S_bar_gen=S_P-S_R+Q_bar_out/T0 DELTAX_bar_gases=-T0*S_bar_gen DELTAX_gases=DELTAX_bar_gases/MM_fuel "(c)" s_w1=entropy(Fluid$, T=T_w1, x=x_w1) s_w2=entropy(Fluid$, T=T_w2, P=P_w2) DELTAs_w=s_w2-s_w1 DELTAX_w=(h_w2-h_w1-T0*(s_w2-s_w1))/MM_H2O "(d)" X_dest=-(m_s\m_f*DELTAX_w+DELTAX_gases)

16-118 Liquid octane is burned with 200 percent excess air during a steady-flow combustion process. The heat transfer rate from the combustion chamber, the power output of the turbine, and the reversible work and exergy destruction are to be determined. Assumptions 1. Combustion is complete. 2. Steady operating conditions exist. 3. Air and the combustion gases are ideal gases. 4. Changes in kinetic and potential energies are negligible. Properties The molar mass of C8 H18 is 114 kg / kmol (Table A-1).

13-494

Analysis (a) The fuel is burned completely with the excess air, and thus the products will contain only CO2 , H 2 O , N 2 , and some free O2 . Considering 1 kmol of C8 H18 , the combustion equation can be written as C8 H18 + 3ath (O 2 + 3.76N 2 ) ¾ ¾ ® 8CO 2 + 9H 2 O + 2ath O 2 + (3)(3.76ath )N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance,

3ath = 8 + 4.5 + 2ath ¾ ¾ ® ath = 12.5 Substituting, C8 H18 + 37.5(O 2 + 3.76N 2 ) ¾ ¾ ® 8CO 2 + 9H 2 O + 25O 2 + 141N 2

The heat transfer for this combustion process is determined from the energy balance

Ein - Eout = D Esystem applied on the combustion chamber with W = 0, - Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

hf kJ / kmol

C8 H18 ( )

h500 K

h 298 K

h1300 K

h 950 K

kJ / kmol

---

14,770

8682

42,033

26,652

14,581

8669

40,170

28,501

H2 O ( g )

---

9904

48,807

33,841

CO2

---

9364

59,552

40,070

Substituting, - Qout = (8)(- 393,520 + 59,522 - 9364)+ (9)(- 241,820 + 48,807 - 9904)+ (25)(0 + 42, 033 - 8682)+ (141)(0 + 40,170 - 8669) - (1)(- 249,950 + h298 - h298 )- (37.5)(0 + 14, 770 - 8682)- (141)(0 + 14,581- 8669)

= - 109,675 kJ/kmol C8 H18 The C8 H18 is burned at a rate of 0.8 kg / min or Thus,

m 0.8 kg/min = = 7.018´ 10- 3 kmol/min M ((8)(12)+ (18)(1)) kg/kmol

13-495

Qout = NQout = (7.018´ 10- 3 kmol/min )(109,675 kJ/kmol) = 770 kJ / min (b) Noting that no chemical reactions occur in the turbine, the turbine is adiabatic, and the product gases can be treated as ideal gases, the power output of the turbine can be determined from the steady-flow energy balance equation for nonreacting gas mixtures,

- Wout = å N P (he - hi ) ¾ ¾ ® Wout = å N P (h1300 K - h950 K ) Substituting, Wout = (8)(59,522 - 40, 070)+ (9)(48,807 - 33,841)+ (25)(42, 033 - 29, 652)+ (141)(40,170 - 28,501)

= 2,245,164 kJ/kmol C8 H18 Thus the power output of the turbine is

Wout = NWout = (7.018´ 10- 3 kmol/min )(2,245,164 kJ/kmol) = 15,756 kJ/min = 262.6 kW (c) The entropy generation during this process is determined from Sgen = S P - S R +

Qout Q = å N P sP - å N R sR + out Tsurr Tsurr

where the entropy of the products are to be evaluated at the turbine exit state. The C8 H18 is at 25C and 8 atm, and thus its absolute entropy is sC8 H18 = 360.79 kJ / kmol.K (Table A-26). The entropy values listed in the ideal gas tables are for 1 atm pressure. The entropies are to be calculated at the partial pressure of the components which is equal to Pi = yi Ptotal , where

yi is the mole fraction of component i. Also, Si = Ni si (T , Pi ) = Ni (si (T , P0 )- Ru ln (yi Pm ))

The entropy calculations can be presented in tabular form as

s i (T, 1 atm )

R u ln (y i Pm )

Ni si

C8 H18

1.00

360.79

17.288

343.50

37.5

0.21

220.589

4.313

8,110.34

141

0.79

206.630

15.329

26,973.44

SR = 35, 427.28 kJ / K CO2

0.0437

266.444

–20.260

2,293.63

H2 O

0.0490

230.499

–19.281

2,248.02

0.1366

241.689

–10.787

6,311.90

141

0.7705

226.389

3.595

31,413.93

SP = 42, 267.48 kJ / K Thus,

109,675 kJ = 7208.2 kJ/K (per kmol) 298 K Then the rate of entropy generation becomes Sgen = 42, 267.48 - 35, 427.28 +

Sgen = NSgen = (7.018´ 10- 3 kmol/min )(7208.2 kJ/kmol ×K ) = 50.59 kJ/min ×K and

X destruction = T0 Sgen = (298 K)(50.59 kJ/min ×K) = 15,076 kJ/min = 251.3 kW Wrev = W + X destruction = 262.6 + 251.3 = 513.9 kW

13-496

EES (Engineering Equation Solver) SOLUTION "Given" ex=3 "300% theoretical air" T1_fuel=(25+273) [K] P1_fuel=(8*101.325) [kPa] m_dot_fuel=0.8 [kg/min] T1_air=500 [K] P1_air=(8*101.325) [kPa] T2=1300 [K] P2=(8*101.325) [kPa] T3=950 [K] P3=(2*101.325) [kPa] T0=(25+273) [K] "Properties" MM_fuel=molarmass(C8H18) "Analysis" "(a)" "The combustion equation is: C8H18 (l) + ex*a_th (O2+3.76N2) = 8 CO2 + 9 H2O + (ex-1)*a_th O2 + ex*3.76*a_th N2" ex*a_th=8+4.5+(ex-1)*a_th "O2 balance" "Mol numbers of reactants and products in kmol" N_C8H18=1 N_O2_R=ex*a_th N_O2_P=(ex-1)*a_th N_N2=ex*a_th*3.76 N_CO2=8 N_H2O=9 "Enthalpy of formation data from Table A-26 in kJ/kmol" h_f_C8H18=-249950 "Enthalpies of reactants in kJ/kmol" h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) "Enthalpies of products in kJ/kmol" h_CO2_P=enthalpy(CO2, T=T2) h_H2O_P=enthalpy(H2O, T=T2) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) "The heat transfer rate" H_P=N_CO2*h_CO2_P+N_H2O*h_H2O_P+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C8H18*h_f_C8H18+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R N_dot_fuel=m_dot_fuel/MM_fuel Q_dot_out=N_dot_fuel*Q_out "(b)" "Enthalpies at the turbine exit in kJ/kmol" h_CO2_3=enthalpy(CO2, T=T3) h_H2O_3=enthalpy(H2O, T=T3) h_O2_3=enthalpy(O2, T=T3) h_N2_3=enthalpy(N2, T=T3) "Turbine power" W_out=N_CO2*(h_CO2_P-h_CO2_3)+N_H2O*(h_H2O_P-h_H2O_3)+N_O2_P*(h_O2_Ph_O2_3)+N_N2*(h_N2_P-h_N2_3) W_dot_out=N_dot_fuel*W_out*Convert(kJ/min, kW) "(c)" "Entropies of reactants" s_C8H18_R=360.79 "from Table A-26" P_O2_R= 1/4.76*P1_air "Partial pressure of O2 in reactants" s_O2_R=entropy(O2, T=T1_air, P=P_O2_R) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-497

P_N2_R= 3.76/4.76*P1_air s_N2_R=entropy(N2, T=T1_air, P=P_N2_R) S_R=N_C8H18*s_C8H18_R+N_O2_R*s_O2_R+N_N2*s_N2_R "Entropies of products at the turbine exit" N_prod =8+9+ (ex-1)*a_th+ex*a_th*3.76 "total kmol of products" P_CO2_P=N_CO2/N_prod*P3 "Partial pressure of CO2 in products" s_CO2_P=entropy(CO2, T=T3, P=P_CO2_P) P_H2O_P=N_H2O/N_prod*P3 s_H2O_P=entropy(H2O, T=T3, P=P_CO2_P) P_O2_P=N_O2_P/N_prod*P3 s_O2_P=entropy(O2, T=T3, P=P_O2_P) P_N2_P=N_N2/N_prod*P3 s_N2_P=entropy(N2, T=T3, P=P_N2_P) S_P=N_CO2*s_CO2_P+N_H2O*s_H2O_P+N_O2_P*s_O2_P+N_N2*s_N2_P "Entropy generation" S_gen=S_P-S_R+Q_out/T0 S_dot_gen=N_dot_fuel*S_gen "Exergy destruction" X_dot_destroyed=T0*S_dot_gen*Convert(kJ/min, kW) "Reversible work" W_dot_rev=W_dot_out+X_dot_destroyed

16-119 It is to be shown that the work output of the Carnot engine will be maximum when Tp =

æ ç shown that the maximum work output of the Carnot engine in this case becomes w = C Taf çç1çè

T0Taf . It is also to be 2

ö T0 ÷ ÷ . ÷ ÷ Taf ø÷

Analysis The combustion gases will leave the combustion chamber and enter the heat exchanger at the adiabatic flame temperature Taf since the chamber is adiabatic and the fuel is burned completely. The combustion gases experience no change in their chemical composition as they flow through the heat exchanger. Therefore, we can treat the combustion gases as a gas stream with a constant specific heat c p . Noting that the heat exchanger involves no work interactions, the energy balance equation for this single-stream steady-flow device can be written as Q = m (he - hi ) = mC (Tp - Taf )

where Q is the negative of the heat supplied to the heat engine. That is,

13-498

QH = - Q = mC (Taf - Tp ) Then the work output of the Carnot heat engine can be expressed as

æ T ö÷ æ T ö÷ W = QH ççç1- 0 ÷ = mC (Taf - Tp )ççç1- 0 ÷ ÷ ÷ ÷ ÷ çè Tp ø÷ èç Tp ø÷

(1)

Taking the partial derivative of W with respect to T p while holding Taf and

T0 constant gives æ T ö T ¶W ÷ = 0¾¾ ® - mC ççç1- 0 ÷ + mC (Tp - Taf ) 02 = 0 ÷ ÷ ç ÷ ¶ Tp Tp è Tp ø Solving for T p we obtain Tp =

T0Taf

which the temperature at which the work output of the Carnot engine will be a maximum. The maximum work output is determined by substituting the relation above into Eq. (1), æ T ö ÷ W = mC (Taf - Tp )ççç1- 0 ÷ = mC Taf ÷ ÷ çè Tp ø ÷

(

æ ç T0Taf çç1çè

)

ö ÷ ÷ ÷ ÷ ÷ T0Taf ø T0

It simplifies to

æ ç W = mC Taf çç1çè

ö T0 ÷ ÷ ÷ ÷ Taf ÷ ø

æ ç w = C Taf çç1çè

ö T0 ÷ ÷ ÷ ÷ Taf ÷ ø

which is the desired relation.

13-499

16-120 The combustion of a mixture of an alcohol fuel ( Cn Hm Ox ) and a hydrocarbon fuel ( Cw Hz ) with excess air and incomplete combustion is considered. The coefficients of the reactants and products are to be written in terms of other parameters. Analysis The balanced reaction equation for stoichiometric air is y1 Cn H m O x + y2 Cw H z + Ath [O 2 + 3.76N 2 ]¾ ¾ ® ( y1n + y2 w) CO 2 + 0.5( y1m + y2 z ) H 2 O + 3.76 Ath N 2

The stoichiometric coefficient Ath is determined from an O2 balance:

y1 x / 2 + Ath = ( y1n + y2 w) + ( y1m + y2 z ) / 4 ¾ ¾ ® Ath = ( y1n + y2 w) + ( y1m + y2 z) / 4 - y1 x / 2 The reaction with excess air and incomplete combustion is

y1 Cn H m O x + y2 Cw H z + (1 + B) Ath [O 2 + 3.76N 2 ] ¾¾ ® a( y1n + y2 w) CO 2 + b( y1n + y2 w) CO + 0.5( y1m + y2 z ) H 2 O + G O 2 + 3.76(1 + B) Ath N 2 The given reaction is y1 Cn H m O x + y2 Cw H z + (1 + B) Ath [O 2 + 3.76N 2 ]¾ ¾ ® D CO 2 + E CO + F H 2 O + G O 2 + J N 2

Thus,

D = a( y1n + y2 w) E = b( y1n + y2 w) F = 0.5( y1m + y2 z ) J = 3.76(1+ B) Ath The coefficient G for O2 is determined from a mass balance,

O2 balance:

0.5 y1 x + (1+ B) Ath = a( y1n + y2 w) + 0.5b( y1n + y2 w) + 0.25( y1m + y2 z ) + G 0.5 y1 x + ( y1n + y2 w) + 0.25( y1m + y2 z ) - 0.5 y1 x + BAth = a( y1n + y2 w) + 0.5b( y1n + y2 w) + 0.25( y1m + y2 z ) + G G = ( y1n + y2 w) - a( y1n + y2 w) - 0.5b( y1n + y2 w) + BAth = ( y1n + y2 w)(1- a) - 0.5b( y1n + y2 w) + BAth = b( y1n + y2 w) - 0.5b( y1n + y2 w) + BAth

= 0.5b( y1n + y2 w) + BAth

16-121 An expression for the HHV of a gaseous alkane Cn H 2n + 2 in terms of n is to be developed. Assumptions 1. Combustion is complete. 2.

The combustion products contain CO2 , H 2 O , and N 2 .

3. Combustion gases are ideal gases. Analysis The complete reaction balance for 1 kmol of fuel is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-500

3n + 1 3n + 1 (3.76)N 2 (O2 + 3.76N 2 ) ¾ ¾® nCO 2 + ( n + 1)H 2 O + 2 2 Both the reactants and the products are taken to be at the standard reference state of 25C and 1 atm for the calculation of heating values. The heat transfer for this process is equal to enthalpy of combustion. Note that N 2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then, Cn H 2 n+ 2 +

q = hC = H P - H R = å N P h f , P - å N R h f , R = (Nh f )

CO2

+ (Nh f )

H2O

- (Nh f )

fuel

For the HHV, the water in the products is taken to be liquid. Then,

hC = n(- 393,520) + ( n + 1)(- 285,830) - (h f° )

fuel

The HHV of the fuel is

HHV =

n(- 393,520) + (n + 1)(- 285,830) - (h f ) - hC fuel = M fuel M fuel

For the LHV, the water in the products is taken to be vapor. Then,

LHV =

n(- 393,520) + (n + 1)(- 241,820) - (h f )

fuel

M fuel

16-122 A general program is to be written to determine the adiabatic flame temperature during the complete combustion of a hydrocarbon fuel Cn Hm at 25°C in a steady-flow combustion chamber when the percent of excess air and its temperature are specified. Analysis The problem is solved using EES, and the solution is given below. "Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) <--> xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) <--> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + x-z/2) (Theo_air/100 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K ." Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-501

h_fuel = 226730 "Table A.26" else If fuel$='C3H8(l)' then x=3; y=8; z=0 Name$='propane(liq)' h_fuel = -103850-15060 "Tables A.26 and A.27" else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane(liq)' h_fuel = -249950 "Table A.26" else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 "Table A.26" endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END "Input data" T_air = 298 [K] Theo_air = 120 [%] Fuel$='CH4(g)' T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-502

HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 SOLUTION for the sample calculation A_th=2 Fuel$='CH4(g)' HP=-74651 [kJ/kg] HR=-74651 [kJ/kg] h_fuel=-74601 Moles_CO=0.000 Moles_CO2=1.000 Moles_H2O=2 Moles_N2=9.024 Moles_O2=0.400 MolO2=0.4 Name$='methane' SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=120 [%] Th_air=1.200 T_air=298 [K] T_fuel=298 [K] T_prod=2069 [K] w=0 x=1 y=4 z=0

16-123 The effect of the amount of air on the adiabatic flame temperature of liquid octane ( C8 H18 ) is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Adiabatic Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) <--> xCO2 + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + (y/4 + x-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Adiabatic, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (y/4 + x-z/2) (Theo_air/100) (O2 + 3.76 N2) <--> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (y/4 + x-z/2) (Theo_air/100) N2 + ((y/4 + x-z/2) (Theo_air/100 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450 K"

13-503

Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='Acetylene' h_fuel = 226730 "Table A.26" else If fuel$='C3H8(l)' then x=3; y=8; z=0 Name$='Propane(liq)' h_fuel = -103850-15060 "Tables A.26 and A.27" else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='Octane(liq)' h_fuel = -249950 "Table A.26" else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='Methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='Methyl alcohol' h_fuel = -200670 "Table A.26" endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' {MolCO = 0 MolCO2 = x} w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END "Input data" T_air = 298 [K] Theo_air = 200 Fuel$='C8H18(l)' T_fuel = 298 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-504

Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=HR "Adiabatic" HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2 Theo air

TProd

[%] 75 90 100 120 150 200 300 500 800

[k] 2077 2287 2396 2122 1827 1506 1153 840.1 648.4

16-124 The fuel among CH4 ( g ) , C2 H2 ( g ) , C2 H6 ( g ) , C3 H8 ( g ) , and C8 H18 (1) that gives the highest temperature when burned completely in an adiabatic constant-volume chamber with the theoretical amount of air is to be determined. Analysis The problem is solved using EES, and the solution is given below. "Adiabatic Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-505

"Adiabatic, Incomplete Combustion of fuel CnHm with Stoichiometric Air at T_fuel =T_air=T_reac in a constant volume, closed system: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the adiabatic combustion temperature, assuming no dissociation. Theo_air is the % theoretical air. " "The initial guess value of T_prod = 450K" Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 "Table A.26" else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 "Table A.26" else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 "Table A.26" endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: END © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-506

"Input data" Theo_air = 100 [%] Fuel$='CH4(g)' T_reac = 298 [K] T_air = T_reac T_fuel = T_reac R_u = 8.314 [kJ/kmol-K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) UR=(h_fuel-R_u*T_fuel)+ (x+y/4-z/2) *(Theo_air/100) *(enthalpy(O2,T=T_air)-R_u*T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *(enthalpy(N2,T=T_air)-R_u*T_air) UP=(x-w)*(enthalpy(CO2,T=T_prod)-R_u*T_prod)+w*(enthalpy(CO,T=T_prod)R_u*T_prod)+(y/2)*(enthalpy(H2O,T=T_prod)-R_u*T_prod)+3.76*(x+y/4-z/2)* (Theo_air/100)*(enthalpy(N2,T=T_prod)-R_u*T_prod)+MolO2*(enthalpy(O2,T=T_prod)-R_u*T_prod) UR =UP "Adiabatic, constant volume conservation of energy" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

SOLUTION for CH4 A_th=2 Moles_CO=0.000 Moles_N2=7.520 Name$='methane' SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] T_fuel=298 [K] UP=–100981 x=1 SOLUTION for C2H2 A_th=2.5 Moles_CO=0.000 Moles_N2=9.400 Name$=' acetylene ' SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] T_fuel=298 [K] UP=194717 x=2

fuel$='CH4(g)' Moles_CO2=1.000 Moles_O2=0.000 R_u=8.314 [kJ/kmol–K]

h_fuel=–74875 Moles_H2O=2 MolO2=0

Th_air=1.000 T_prod=2824 [K] UR=–100981 y=4

T_air=298 [K] T_reac=298 [K] w=0

fuel$=' C2H2(g)' Moles_CO2=2.000 Moles_O2=0.000 R_u=8.314 [kJ/kmol–K]

h_fuel=226730 Moles_H2O=1 MolO2=0

Th_air=1.000 T_prod=3535 [K] UR=194717 y=2

T_air=298 [K] T_reac=298 [K] w=0 z=0

13-507

SOLUTION for CH3OH A_th=1.5 Moles_CO=0.000 Moles_N2=5.640 Name$='methyl alcohol' SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] T_fuel=298 [K] UP=−220869 x=1 SOLUTION for C3H8 A_th=5 Moles_CO=0.000 Moles_N2=18.800 Name$='propane' SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] T_fuel=298 [K] UP=−165406 x=3 SOLUTION for C8H18 A_th=12.5 Moles_CO=0.000 Moles_N2=47.000 Name$=' octane ' SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] T_fuel=298 [K] UP=−400104 x=8

16-125

fuel$=' CH3OH(g)' Moles_CO2=1.000 Moles_O2=0.000 R_u=8.314 [kJ/kmol–K]

h_fuel=226730 Moles_H2O=1 MolO2=0

Th_air=1.000 T_prod=2817 [K] UR=−220869 y=4

T_air=298 [K] T_reac=298 [K] w=0 z=1

fuel$=' C3H8(g)' Moles_CO2=3.000 Moles_O2=0.000 R_u=8.314 [kJ/kmol–K]

h_fuel=226730 Moles_H2O=1 MolO2=0

Th_air=1.000 T_prod=2909 [K] UR=−165406 y=8

T_air=298 [K] T_reac=298 [K] w=0 z=0

fuel$='C8H18(g)' Moles_CO2=8.000 Moles_O2=0.000 R_u=8.314 [kJ/kmol–K]

h_fuel=−249950 Moles_H2O=9 MolO2=0

Th_air=1.000 T_prod=2911 [K] UR=−400104 y=18

T_air=298 [K] T_reac=298 [K] w=0 z=0

The rate of heat transfer is to be determined for the fuels CH4 ( g ) , C2 H2 ( g ) , CH3OH ( g ) , C3 H8 ( g ) , and

C8 H18 (1) when they are burned completely in a steady-flow combustion chamber with the theoretical amount of air. Analysis The problem is solved using EES, and the solution is given below. "Steady-floe combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Steady-flow, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-508

--> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the product gas temperature, assuming no dissociation. Theo_air is the % theoretical air. " Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) "This procedure takes the fuel name and returns the moles of C ,H and O and molar mass" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 "Table A.26" MM=2*12+2*1 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) MM=molarmass(C3H8) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 "Table A.26" MM=8*12+18*1 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) MM=molarmass(CH4) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670"Table A.26" MM=1*12+4*1+1*16 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif 10: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-509

END "Input data" m_dot_fuel = 0.1 [kg/s] T_air = 298 [K] Theo_air = 200 [%] Fuel$='CH4(g)' T_fuel = 298 [K] T_prod = 1200 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) HR =Q_out+HP Q_dot_out=Q_out/MM*m_dot_fuel "The heat transfer rate" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

SOLUTION for a sample calculation A_th=1.5 HR=–200701 [kJ/kg] Moles_CO=0.000 Moles_N2=5.640 m_dot_fuel=1 [kg/s] Q_out=404241.1 [kJ/kmol_fuel] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=100 [%] T_fuel=298 [K] x=1 16-126

fuel$='CH3OH(g)' h_fuel=–200670 Moles_CO2=1.000 Moles_O2=0.000 Name$='methyl alcohol'

HP=–604942 [kJ/kg] MM=32 Moles_H2O=2 MolO2=0 Q_dot_out=12633 [kW]

Th_air=1.000 T_prod=1200 [K] y=4

T_air=298 [K] w=0 z=1

The rates of heat transfer are to be determined for the fuels CH4 ( g ) , C2 H2 ( g ) , CH3OH ( g ) , C3 H8 ( g ) , and

C8 H18 (1) when they are burned in a steady-flow combustion chamber with for 50, 100, and 200 percent excess air. Analysis The problem is solved using EES, and the solution is given below. "Steady-flow combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Steady-flow, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2)

13-510

--> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the product gas temperature, assuming no dissociation. Theo_air is the % theoretical air. " Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='acetylene' h_fuel = 226730 "Table A.26" MM=2*12+2*1 else If fuel$='C3H8(g)' then x=3; y=8; z=0 Name$='propane' h_fuel = enthalpy(C3H8,T=T_fuel) MM=molarmass(C3H8) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='octane' h_fuel = -249950 "Table A.26" MM=8*12+18*1 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='methane' h_fuel = enthalpy(CH4,T=T_fuel) MM=molarmass(CH4) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='methyl alcohol' h_fuel = -200670 "Table A.26" MM=1*12+4*1+1*16 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 endif; endif © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-511

10: END "Input data" T_air = 298 [K] m_dot_fuel=1 [kg/s] Theo_air = 200 [%] Fuel$='CH4(g)' T_fuel = 298 [K] T_prod = 1200 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) HR =Q_out+HP Q_dot_out=Q_out/MM*m_dot_fuel "The heat transfer rate" Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

SOLUTION for a sample calculation A_th=12.5 HR=–250472 [kJ/kg] Moles_CO=0.000 Moles_N2=94.000 m_dot_fuel=1 [kg/s] Q_out=1390433.6 [kJ/kmol_fuel] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=200 [%] T_fuel=298 [K] x=8

fuel$='C8H18(l)' fuel$='C8H18(l)' Moles_CO2=8.000 Moles_O2=12.500 Name$='octane'

HP=–1.641E+06 [kJ/kg] MM=114 [kg/kmol] Moles_H2O=9 MolO2=12.5 Q_dot_out=12197 [kW]

Th_air=2.000 T_prod=1200 [K] y=18

T_air=298 [K] w=0 z=0

16-127 A general program is to be written to determine the heat transfer during the complete combustion of a hydrocarbon fuel Cn Hm at 25°C in a steady-flow combustion chamber when the percent of excess air and the temperatures of air and the products are specified. Analysis The problem is solved using EES, and the solution is given below. "Steady-flow combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> xCO2 + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + (x+y/4-z/2) (Theo_air/100 - 1) O2" "For theoretical oxygen, the complete combustion equation for CH3OH is" "CH3OH + A_th O2=1 CO2+2 H2O " "1+ 2*A_th=1*2+2*1""theoretical O balance" "Steady-flow, Incomplete Combustion of fuel CnHm entering at T_fuel with Stoichiometric Air at T_air: © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-512

Reaction: CxHyOz + (x+y/4-z/2) (Theo_air/100) (O2 + 3.76 N2) --> (x-w)CO2 +wCO + (y/2) H2O + 3.76 (x+y/4-z/2) (Theo_air/100) N2 + ((x+y/4-z/2) (Theo_air/100 - 1) +w/2)O2" "T_prod is the product gas temperature, assuming no dissociation. Theo_air is the % theoretical air. " Procedure Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) "This procedure takes the fuel name and returns the moles of C and moles of H" If fuel$='C2H2(g)' then x=2;y=2; z=0 Name$='Acetylene' h_fuel = 226730 "Table A.26" MM=2*12+2*1 else If fuel$='C3H8(l)' then x=3; y=8; z=0 Name$='Propane(liq)' h_fuel = -103850-15060 "Tables A.26 and A.27" MM=molarmass(C3H8) else If fuel$='C8H18(l)' then x=8; y=18; z=0 Name$='Octane(liq)' h_fuel = -249950 "Table A.26" MM=8*12+18*1 else if fuel$='CH4(g)' then x=1; y=4; z=0 Name$='Methane' h_fuel = enthalpy(CH4,T=T_fuel) MM=molarmass(CH4) else if fuel$='CH3OH(g)' then x=1; y=4; z=1 Name$='Methyl alcohol' h_fuel = -200670 "Table A.26" MM=1*12+4*1+1*16 endif; endif; endif; endif; endif end Procedure Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) ErrTh =(2*x + y/2 - z - x)/(2*A_th)*100 IF Th_air >= 1 then SolMeth$ = '>= 100%, the solution assumes complete combustion.' w=0 MolO2 = A_th*(Th_air - 1) GOTO 10 ELSE w = 2*x + y/2 - z - 2*A_th*Th_air IF w > x then Call ERROR('The moles of CO2 are negative, the percent theoretical air must be >= xxxF3 %',ErrTh) Else SolMeth$ = '< 100%, the solution assumes incomplete combustion with no O_2 in products.' MolO2 = 0 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-513

endif; endif 10: END "Input data" T_air = 298 [K] Theo_air = 150 [%] Fuel$='C3H8(l)' T_fuel = 298 [K] T_prod = 1800 [K] Call Fuel(Fuel$,T_fuel:x,y,z,h_fuel,Name$,MM) A_th =x + y/4 - z/2 Th_air = Theo_air/100 Call Moles(x,y,z,Th_air,A_th:w,MolO2,SolMeth$) HR=h_fuel+ (x+y/4-z/2) *(Theo_air/100) *enthalpy(O2,T=T_air)+3.76*(x+y/4-z/2) *(Theo_air/100) *enthalpy(N2,T=T_air) HP=(x-w)*enthalpy(CO2,T=T_prod)+w*enthalpy(CO,T=T_prod)+(y/2)*enthalpy(H2O,T=T_prod)+3.76*(x+y/4z/2)* (Theo_air/100)*enthalpy(N2,T=T_prod)+MolO2*enthalpy(O2,T=T_prod) Q_out=(HR-HP)/MM Moles_O2=MolO2 Moles_N2=3.76*(x+y/4-z/2)* (Theo_air/100) Moles_CO2=x-w Moles_CO=w Moles_H2O=y/2

SOLUTION for a sample calculation A_th=5 fuel$='C3H8(l)' HP=–149174 [kJ/kg] HR=–119067 [kJ/kg] h_fuel=–118910 MM=44.1 [kg/kmol] Moles_CO=0.000 Moles_CO2=3.000 Moles_H2O=4 Moles_N2=28.200 Moles_O2=2.500 MolO2=2.5 Name$='Propane(liq)' Q_out=682.8 [kJ/kg_fuel] SolMeth$='>= 100%, the solution assumes complete combustion.' Theo_air=150 [%] Th_air=1.500 T_air=298 [K] T_fuel=298 [K] T_prod=1800 [K] w=0 x=3 y=8

z=0

Fundamentals of Engineering (FE) Exam Problems 16-128 A fuel is burned steadily in a combustion chamber. The combustion temperature will be the highest except when (a) the fuel is preheated. (b) the fuel is burned with a deficiency of air. (c) the air is dry. (d) the combustion chamber is well insulated. (e) the combustion is complete. Answer (b) the fuel is burned with a deficiency of air.

13-514

16-129 A fuel is burned with 70 percent theoretical air. This is equivalent to (a) 30% excess air (b) 70% excess air (c) 30% deficiency of air (d) 70% deficiency of air (e) stoichiometric amount of air Answer (c) 30% deficiency of air Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). air_th=0.7 air_th=1–air_deficiency

16-130 Propane ( C3 H8 ) is burned with 125 percent theoretical air. The air-fuel mass ratio for this combustion process is (a) 12.3 (b) 15.7 (c) 19.5 (d) 22.1 (e) 23.4 Answer (c) 19.5 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). coeff=1.25 "coeff=1 for theoretical combustion, 1.25 for 25% excess air" n_C=3 n_H=8 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel

16-131 Benzene gas ( C6 H6 ) is burned with 90 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products is (a) 1.7% (b) 2.3% (c) 3.6% (d) 4.4% (e) 14.3% Answer (d) 4.4% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-515

coeff=0.90 "coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_C=6 n_H=6 a_th=n_C+n_H/4 "Assuming all the H burns to H2O, the combustion equation is C6H6+coeff*a_th(O2+3.76N2)----- (n_CO2) CO2+(n_CO)CO+(n_H2O) H2O+(n_N2) N2" n_O2=coeff*a_th n_N2=3.76*n_O2 n_H2O=n_H/2 n_CO2+n_CO=n_C 2*n_CO2+n_CO+n_H2O=2*n_O2 "Oxygen balance" n_prod=n_CO2+n_CO+n_H2O+n_N2 "Total mole numbers of product gases" y_CO=n_CO/n_prod "mole fraction of CO in product gases" "Some Wrong Solutions with Common Mistakes:" W1_yCO=n_CO/n1_prod; n1_prod=n_CO2+n_CO+n_H2O "Not including N2 in n_prod" W2_yCO=(n_CO2+n_CO)/n_prod "Using both CO and CO2 in calculations"

16-132 One kmol of methane ( CH4 ) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are 1 kmol of free O2 in the products, the air-fuel mass ratio is (a) 34.6 (b) 25.7 (c) 17.2 (d) 14.3 (e) 11.9 Answer (b) 25.7 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). (coeff-1)*a_th=1 "O2 balance: coeff=1 for theoretical combustion, 1.5 for 50% excess air" n_C=1 n_H=4 m_fuel=n_H*1+n_C*12 a_th=n_C+n_H/4 n_O2=coeff*a_th n_N2=3.76*n_O2 m_air=n_O2*32+n_N2*28 AF=m_air/m_fuel "Some Wrong Solutions with Common Mistakes:" W1_AF=1/AF "Taking the inverse of AF" W2_AF=n_O2+n_N2 "Finding air-fuel mole ratio" W3_AF=AF/coeff "Ignoring excess air"

16-133 The higher heating value of a hydrocarbon fuel Cn Hm with m = 8 is given to be 1560 MJ / kmol of fuel. Then its lower heating value is (a) 1384 MJ / kmol (b) 1208 MJ / kmol

13-516

(c) 1402 MJ / kmol (d) 1540 MJ / kmol (e) 1550 MJ / kmol Answer (a) 1384 MJ / kmol Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). n_H=8 HHV=1560 [MJ/kmol] h_fg=2.4423 [MJ/kg] "Enthalpy of vaporization of water at 25C" n_water=n_H/2 m_water=n_water*18 LHV=HHV-h_fg*m_water "Some Wrong Solutions with Common Mistakes:" W1_LHV=HHV - h_fg*n_water "Using mole numbers instead of mass" W2_LHV= HHV - h_fg*m_water*2 "Taking mole numbers of H2O to be m instead of m/2" W3_LHV= HHV - h_fg*n_water*2 "Taking mole numbers of H2O to be m instead of m/2, and using mole numbers"

16-134 Methane ( CH4 ) is burned completely with 80% excess air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is (a) 890 MJ / kg (b) 802 MJ / kg (c) 75 MJ / kg (d) 56 MJ / kg (e) 50 MJ / kg Answer (d) 56 MJ / kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). EXCESS=0.8 T= 25 [C] P=1 [atm] "Heat transfer in this case is the HHV at room temperature," HHV_CH4 =55.53 [MJ/kg] LHV_CH4 =50.05 [MJ/kg] "Some Wrong Solutions with Common Mistakes:" W1_Q=LHV_CH4 "Assuming lower heating value" W2_Q=EXCESS*hHV_CH4 "Assuming Q to be proportional to excess air"

13-517

16-135 Acetylene gas ( C2 H 2 ) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at 25°C, and the products leave at 1500 K. If the enthalpy of the products relative to the standard reference state is - 404 MJ / kmol of fuel, the heat transfer from the combustion chamber is (a) 177 MJ / kmol (b) 227 MJ / kmol (c) 404 MJ / kmol (d) 631 MJ / kmol (e) 751 MJ / kmol Answer (d) 631 MJ / kmol Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_reac=25 [C] T_prod=1500 [K] H_prod=-404 [MJ/kmol] hf_fuel=226730/1000 [MJ/kmol] H_react=hf_fuel Q_out=H_react-H_prod "Some Wrong Solutions with Common Mistakes:" W1_Qout= -H_prod "Taking Qout to be H_prod" W2_Qout= H_react+H_prod "Adding enthalpies instead of subtracting them"

16-136 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and 60C. The entropy change of carbon dioxide in the dehumidifying section is (a) - 2.8 kJ / kg ×K (b) - 0.13 kJ / kg ×K (c) 0 (d) 0.13 kJ / kg ×K (e) 2.8 kJ / kg ×K Answer (b) - 0.13 kJ / kg ×K Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=(60+273) [K] P1= 1 [atm] T2=T1 P2=P1 cp_CO2=0.846 [kJ/kg-K] R_CO2=0.1889 [kJ/kg-K] y1_CO2=0.5 P1_CO2=y1_CO2*P1 y2_CO2=1 P2_CO2=y2_CO2*P2 Ds_CO2=cp_CO2*ln(T2/T1)-R_CO2*ln(P2_CO2/P1_CO2) "Some Wrong Solutions with Common Mistakes:" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-518

W1_Ds=0 "Assuming no entropy change" W2_Ds=cp_CO2*ln(T2/T1)-R_CO2*ln(P1_CO2/P2_CO2) "Using pressure fractions backwards"

16-137 A fuel is burned during a steady-flow combustion process. Heat is lost to the surroundings at 300 K at a rate of 1120 kW. The entropy of the reactants entering per unit time is 17 kW / K and that of the products is 15 kW / K . The total rate of exergy destruction during this combustion process is (a) 520 kW (b) 600 kW (c) 1120 kW (d) 340 kW (e) 739 kW Answer (a) 520 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). To=300 [K] Q_out=1120 [kW] S_react=17 [kW/K] S_prod= 15 [kW/K] S_react-S_prod-Q_out/To+S_gen=0 "Entropy balance for steady state operation, Sin-Sout+Sgen=0" X_dest=To*S_gen "Some Wrong Solutions with Common Mistakes:" W1_Xdest=S_gen "Taking Sgen as exergy destruction" W2_Xdest=To*S_gen1; S_react-S_prod-S_gen1=0 "Ignoring Q_out/To"

16-138 … 16-143 Design and Essay Problems 16-139 A certain industrial process generates a liquid solution of ethanol and water as the waste product. The solution is to be burned using methane. A combustion process is to be developed to accomplish this incineration process with minimum amount of methane. Analysis The mass flow rate of the liquid ethanol-water solution is given to be 10 kg / s . Considering that the mass fraction of ethanol in the solution is 0.2, methanol = (0.2)(10 kg/s) = 2 kg/s mwater = (0.8)(10 kg/s) = 8 kg/s

Noting that the molar masses M ethanol = 46 and M water = 18 kg / kmol and that mole numbers N = m / M , the mole flow rates become N ethanol =

methanol 2 kg/s = = 0.04348 kmol/s M ethanol 46 kg/kmol

N water =

mwater 8 kg/s = = 0.44444 kmol/s M water 18 kg/kmol

Note that N water 0.44444 = = 10.222 kmol H 2 O/kmol C2 H5 OH Nethanol 0.04348

13-519

That is, 10.222 moles of liquid water is present in the solution for each mole of ethanol. Assuming complete combustion, the combustion equation of C2 H5OH ( ) with stoichiometric amount of air is C2 H5 OH ( )+ ath (O 2 + 3.76N 2 ) ¾ ¾ ® 2CO 2 + 3H 2 O + 3.76ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance, Thus,

1 + 2ath = 4 + 3 ¾ ¾ ® ath = 3

C2 H5 OH ( )+ 3(O 2 + 3.76N 2 ) ¾ ¾ ® 2CO 2 + 3H 2 O + 11.28N 2

Noting that 10.222 kmol of liquid water accompanies each kmol of ethanol, the actual combustion equation can be written as C2 H5 OH ( )+ 3(O 2 + 3.76N 2 )+ 10.222H 2 O ( ) ¾ ¾ ® 2CO 2 + 3H 2 O (g )+ 11.28N 2 + 10.222H 2O (l )

The heat transfer for this combustion process is determined from the steady-flow energy balance equation with W = 0,

Q = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). We assume all the reactants to enter the combustion chamber at the standard reference temperature of 25C. Furthermore, we assume the products to leave the combustion chamber at 1400 K which is a little over the required temperature of 1100C. From the tables, h 298 K

h1400 K

Substance

hf kJ / kmol

kJ / kmol

C2 H5OH ( )

−277,690

---

CH4

−74,850

---

8682

45,648

8669

43,605

H2 O ( g )

−241,820

9904

53,351

H2 O ( )

−285,830

---

CO2

−393,520

9364

65,271

Thus, Q = (2)(- 393,520 + 65, 271- 9364)+ (3)(- 241,820 + 53,351- 9904)+ (11.28)(0 + 43, 605 - 8669)- (1)(- 277, 690)- 0 - 0 + (10.222)(- 241,820 + 53,351- 9904)- (10.222)(- 285,830)

= 295,409 kJ/kmol of C2 H5OH The positive sign indicates that 295,409 kJ of heat must be supplied to the combustion chamber from another source (such as burning methane) to ensure that the combustion products will leave at the desired temperature of 1400 K. Then the rate of heat transfer required for a mole flow rate of 0.04348 kmol C2 H5OH/s CO becomes Q = NQ = (0.04348 kmol/s)(295,409 kJ/kmol) = 12,844 kJ/s

Assuming complete combustion, the combustion equation of CH4 ( g ) with stoichiometric amount of air is CH 4 + ath (O 2 + 3.76N 2 ) ¾ ¾ ® CO2 + 2H 2 O + 3.76ath N 2

where ath is the stoichiometric coefficient and is determined from the O2 balance, Thus, ath = 1 + 1 ¾ ¾ ® ath = 2

CH 4 + 2 (O 2 + 3.76N 2 ) ¾ ¾ ® CO 2 + 2H 2 O + 7.52N 2

13-520

The heat transfer for this combustion process is determined from the steady-flow energy balance Ein - Eout = D Esystem equation as shown above under the same assumptions and using the same mini table: Q = (1)(- 393,520 + 65, 271- 9364)+ (2)(- 241,820 + 53,351 - 9904)+ (7.52)(0 + 43, 605 - 8669)- (1)(- 74,850)- 0 - 0

= - 396,790 kJ/kmol of CH4 That is, 396,790 kJ of heat is supplied to the combustion chamber for each kmol of methane burned. To supply heat at the required rate of 12,844 kJ / s , we must burn methane at a rate of or,

NCH4 =

Q 12,844 kJ/s = = 0.03237 kmol CH 4 /s Q 396,790 kJ/kmol

mCH4 = M CH4 NCH4 = (16 kg/kmol)(0.03237 kmol CH4 /s) = 0.5179 kg / s Therefore, we must supply methane to the combustion chamber at a minimum rate 0,5179 kg / s in order to maintain the temperature of the combustion chamber above 1400 K.

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

Chapter 17 CHEMICAL AND PHASE EQUILIBRIUM

13-521

to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw Hill LLC.

13-522

CYU - Check Your Understanding CYU 17-1 ■ Check Your Understanding CYU 17-1.1 Select the correct statement(s) regarding chemical equilibrium. I A chemical reaction at a specified temperature and pressure proceeds in the direction of a decreasing Gibbs function. II A chemical reaction in an adiabatic chamber stops when the entropy reaches a minimum. III A chemical reaction in an adiabatic chamber proceeds in the direction of increasing entropy. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III CYU 17-1.2 Select the incorrect statement below regarding chemical equilibrium. (a) The changes in the number of moles of the components during a chemical reaction are proportional to the stoichiometric coefficients. (b) A chemical reaction at a specified temperature and pressure stops when the Gibbs function attains a minimum value. (c) Statement (b) above is valid for chemical reactions in both adiabatic and non-adiabatic chambers. (d) None of the above Answer: (d) None of the above.

CYU 17-2 ■ Check Your Understanding CYU 17-2.1 Select the correct statement(s) below regarding the equilibrium constant. (a) The equilibrium constant K P of an ideal-gas mixture at a specified temperature can be determined from a knowledge of the standard-state Gibbs function change at the same temperature. (b) Once the equilibrium constant is available, it can be used to determine the equilibrium composition of reacting idealgas mixtures. (c) The Gibbs function of an ideal gas depends on both temperature and pressure. (d) All of the above Answer: (d) All of the above CYU 17-2.2 Select the correct equilibrium constant relation(s) for reacting ideal-gas mixtures. Reactants: A and B, Products: C and D. Dn

P nC P n D æ P ö÷ ÷ K p = Cn Dn ççç ÷ PA A PB B çè N total÷ ø

K p = e- D G *(T )/RuT

III

Kp =

ö N CnC N Dn D æ çç P ÷ ÷ ÷ nA nB ç N A N B çè N total÷ ø

13-523

(a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (d) II and III

CYU 17-3 ■ Check Your Understanding CYU 17-3.1 Select the incorrect statement regarding the equilibrium constant of ideal-gas mixtures. (a) The larger the K P , the more complete the reaction. (b) The K P of the reverse reaction is 1 / K P . (c) The K P of a reaction depends on temperature only. (d) The mixture pressure affects the equilibrium composition. (e) The mixture pressure affects the equilibrium constant K P . Answer: (e) The mixture pressure affects the equilibrium constant K P . CYU 17-3.2 Select the incorrect statement regarding the equilibrium constant of ideal-gas mixtures. (a) The presence of inert gases does not affect the equilibrium composition. (b) The presence of inert gases does not affect the equilibrium constant K P . (c) When the stoichiometric coefficients are doubled, the value of K P is squared. (d) Free electrons in the equilibrium composition can be treated as an ideal gas. (e) Equilibrium calculations do not provide information on the reaction rate. Answer: (a) The presence of inert gases does not affect the equilibrium composition. CYU 17-3.3 If the equilibrium constant for the reaction CO + 12 O 2 Û CO 2 at 2000 K and 200 kPa is K P , what is the 1

equilibrium constant of the reaction CO 2 Û CO + 2 O 2 at 2000 K and 200 kPa? (a) 0.5K P (b) K P (c) 2K P (d) ( K P )2 (e) 1/ K P Answer: (e) 1/ K P CYU 17-3.4 If the equilibrium constant for the reaction CO + 12 O 2 Û CO 2 at 1000 K is K P , what is the equilibrium constant of the reaction 2CO + O 2 Û 2CO 2 at 1000 K? (a) 0.5K P (b) K P (c) 2K P © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-524

(d) ( K P )2 (e) 1/ K P Answer: (d) ( K P )2

CYU 17-4 ■ Check Your Understanding CYU 17-4.1 How many K P relations are needed to determine the equilibrium composition of the mixture which consists of

H 2 O , OH, O2 , and H 2 ? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer: (b) 2 The number of chemical species = 4 The number of elements = 2 Number of KP relations = 4 −2 = 2

CYU 17-5 ■ Check Your Understanding CYU 17-5.1 Select the correct statement(s) below regarding the effect of temperature on the equilibrium composition. I Exothermic reactions are less complete at higher temperatures. II The equilibrium constant increases with temperature for combustion reactions. III van't Hoff equation is an expression of the variation of K P with temperature in terms of hR (T ). (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III

CYU 17-6 ■ Check Your Understanding CYU 17-6.1 Select the correct statement(s) regarding phase equilibrium. (a) The Gibbs function difference is the driving force for phase change. (b) The two phases of a pure substance are in equilibrium when each phase has the same value of specific Gibbs function. (c) At the triple point, the specific Gibbs functions of all three phases are equal to each other. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-525

(d) All of the above Answer: (d) All of the above CYU 17-6.2 Select the correct statement(s) regarding the phase rule? I A single-component, three-phase system has no independent variable. II A single-component, one-phase system has one independent variable. III A single-component, two-phase system has two independent variables. (a) Only I (b) Only II (c) I and II (d) I and III (e) II and III Answer: (a) Only I IV = the number of independent variables C = the number of components PH = the number of phases present in equilibrium IV = C − PH + 2 I. IV = 1 − 3 + 2 = 0 II. IV = 1 − 1 + 2 = 2 III. IV = 1 − 2 + 2 = 1 CYU 17-6.3 Select the incorrect statement regarding Henry's law. (a) The amount of gas dissolved in a liquid can be increased by increasing the pressure of the gas. (b) The larger the Henry’s constant, the larger the concentration of dissolved gases in the liquid. (c) The fraction of a dissolved gas in the liquid decreases with increasing temperature. (d) None of the above. Answer: (b) The larger the Henry’s constant, the larger the concentration of dissolved gases in the liquid. CYU 17-6.4 How many independent variables does a saturated liquid-vapor mixture of water have? (a) 0 (b) 1 (c) 2 (d) 3 (e) 4 Answer: (b) 1 IV = the number of independent variables C = the number of components PH = the number of phases present in equilibrium IV = C − PH + 2 IV = 1 − 2 + 2 = 1 CYU 17-6.5 The solubility of calcium bicarbonate in water at 360 K in 100 kg of water is 18.1 kg. What is the mass fraction calcium bicarbonate in the saturated brine? (a) 0 (b) 1 (c) 0.181 (d) 18.1/ 118.1 (e) 18.1 Answer: (d) 18.1 / 118.1

13-526

mfcb  mcb / mm  (18.1 kg) / (100  18.1 kg)  18.1/118.1

13-527

PROBLEMS K p and Equilibrium Composition of Ideal Gases 17-1C Because when a reacting system involves heat transfer, the increase-in-entropy principle relation requires a knowledge of heat transfer between the system and its surroundings, which is impractical. The equilibrium criteria can be expressed in terms of the properties alone when the Gibbs function is used.

17-2C They are

P vC P n D K p = Cn A DvB , K p = e- D G*(T )/ RuT PA PB

ö N nC N n D æ P ÷ ÷ and K p = Cn A Dn B ççç ÷ N A N B çè N total ÷ ø

where D n = nC + n D - n A - n B . The first relation is useful in partial pressure calculations, the second in determining the

K P from gibbs functions, and the last one in equilibrium composition calculations.

17-3C No, the wooden table is NOT in chemical equilibrium with the air. With proper catalyst, it will reach with the oxygen in the air and burn.

17-4C (a) The equilibrium constant for the reaction CO + 12 O 2 Û CO 2 can be expressed as æ P çç n O2 ç n CO N N çè N n

Kp =

N COCO22

ö CO2 ÷ ÷ ÷ ÷ total ø

- n CO - n O2 )

Judging from the values in Table A-28, the K P value for this reaction decreases as temperature increases. That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of moles of CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases. (b) The value of the exponent in this case is 1 – 1 – 0.5 = – 0.5, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of K p depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (CO2 ) must increase, and the number of moles of the reactants (CO, O2 ) must decrease.

17-5C (a) The equilibrium constant for the reaction N2 Û 2N can be expressed as ( n - n N2 )

ö N N nN æ P ÷ ÷ K p = nNN ççç ÷ ø N N 2 çè N total ÷ 2

Judging from the values in Table A-28, the K p value for this reaction increases as the temperature increases. That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number of moles of N will increase and the number moles of N 2 will decrease as the temperature increases. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-528

(b) The value of the exponent in this case is 2 – 1 = 1, which is positive. Thus as the pressure increases, the term in the brackets also increases. The value of K p depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (N) must decrease, and the number of moles of the reactants (N2 ) must increase.

17-6C The equilibrium constant for the reaction CO + 12 O 2 Û CO 2 can be expressed as æ P çç Kp = n n O2 ç CO N N çè N n

N COCO22

ö CO2 ÷ ÷ ÷ ÷ total ø

- n CO - n O2 )

Adding more N 2 (an inert gas) at constant temperature and pressure will increase N total but will have no direct effect on other terms. Then to keep the equation balanced, the number of moles of the products (CO2 ) must increase, and the number of moles of the reactants (CO, O2 ) must decrease.

17-7C The values of the equilibrium constants for each dissociation reaction at 3000 K are, from Table A-28,

N 2 Û 2N Û ln K p = - 22.359 H 2 Û 2H Û ln K p = - 3.685

(greater than - 22.359)

Thus H 2 is more likely to dissociate than N 2 .

17-8C (a) No, because K p depends on temperature only. (b) In general, the total mixture pressure affects the mixture composition. The equilibrium constant for the reaction N2 + O2 Û 2NO can be expressed as Kp =

æ P çç n N2 n O2 ç N N çè N n NO N NO

NO ö ÷ ÷ ÷ ÷ total ø

- n N 2 - n O2 )

The value of the exponent in this case is 2 – 1 – 1 = 0. Therefore, changing the total mixture pressure will have no effect on the number of moles of N 2 , O2 and NO.

17-9C (a) K p , (b) 1/ K p , (c) K p2 , (d) K p , (e) 1 / K 3p .

13-529

(b) 1/ K p , (c) K p , (d) K p , (e) K p2 .

17-11C (a) This reaction is the reverse of the known CO reaction. The equilibrium constant is then

1/ K P (b) This reaction is the reverse of the known CO reaction at a different pressure. Since pressure has no effect on the equilibrium constant,

1/ K P (c) This reaction is the same as the known CO reaction multiplied by 2. The quilibirium constant is then

K P2 (d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant,

K P2

17-12 The temperature at which 10 percent of H 2 dissociates into 2H is to be determined. Assumptions 1

The equilibrium composition consists of H 2 and H only.

The constituents of the mixture are ideal gases.

Analysis This is a dissociation process that is significant at very high temperatures only. For simplicity we consider 1 kmol of H 2 , as shown in Fig. 16–8. The stoichiometric and actual reactions in this case are as follows: Stoichiometric:

H2 Û 2H (thus nH2 = 1 and nH = 2) H 2 Û 0.9 H 2 + 0.2 H

Actual:

reactants (leftover)

products

A double-headed arrow is used for the stoichiometric reaction to differentiate it from the actual reaction. This reaction involves one reactant (H 2 ) and one product (H). The equilibrium composition consists of 0.9 kmol of H 2 (the leftover reactant) and 0.2 kmol of H (the newly formed product). Therefore, N H 2  0.9 and NH  0.2 and the equilibrium constant

K P is determined from Eq. 16–15 to be n - n H2

Kp =

öH N HnH æ çç P ÷ ÷ ÷ n H2 ç N H2 èç N total ø÷

2- 1

ö 0.22 æ çç 10 ÷ ÷ ÷ 0.9 èç0.9 + 0.2 ø

= 0.404

From Table A–28, the temperature corresponding to this K P value is T = 3535 K

13-530

Discussion We conclude that 10 percent of H 2 dissociates into H when the temperature is raised to 3535 K. If the temperature is increased further, the percentage of H 2 that dissociates into H will also increase.

17-13E The temperature at which 15 percent of diatomic oxygen dissociates into monatomic oxygen at two pressures is to be determined. Assumptions 1

The equilibrium composition consists of O2 and O.

The constituents of the mixture are ideal gases.

Analysis (a) The stoichiometric and actual reactions can be written as Stoichiometric:

O2 Û 2O (thus nO2 = 1 and nO = 2)

O2 Û 0.85O2 + 0.3O

Actual:

prod.

react.

The equilibrium constant K p can be determined from n - nO2

Kp =

öO N OnO æ çç P ÷ ÷ nO2 ç ÷ çè N total ÷ N O2 ø

2- 1

ö 0.32 æ çç 3 /14.696 ÷ ÷ 0.85 èç0.85 + 0.3 ø÷

= 0.01880

and

ln K p = - 3.974 From Table A-28, the temperature corresponding to this ln K P value is T = 3060 K = 5508 R (b) At 100 psia, n - nO2

öO N OnO æ P ÷ ç ÷ K p = nO2 çç ÷ N O2 çè N total ÷ ø

2- 1

ö 0.32 æ çç100 /14.696 ÷ = ÷ ÷ ç è ø 0.85 0.85 + 0.3

= 0.6265

ln K p = - 0.4676 From Table A-28, the temperature corresponding to this ln K P value is T = 3701 K = 6662 R

17-14 The mole fractions of the constituents of an ideal gas mixture is given. The Gibbs function of the CO in this mixture at the given mixture pressure and temperature is to be determined.

13-531

Analysis From Tables A-21 and A-26, at 1 atm pressure,

g *(800 K, 1 atm) = g of + D éêëh (T ) - Ts o (T )ùúû = - 137,150 + (23,844 - 800´ 227.162) - (8669 - 298´ 197.543) = - 244,837 kJ/kmol The partial pressure of CO is

PCO = yCO P = (0.30)(10 atm) = 3 atm The Gibbs function of CO at 800 K and 3 atm is

g (800 K, 3 atm) = g *(800 K, 1 atm) + RuT ln PCO = - 244,837 kJ/kmol + (8.314 kJ/kmol)(800 K)ln(3 atm) = - 237, 530 kJ / kmol

17-15E The equilibrium constant of the reaction H 2 O Û H 2 + 12 O 2 is to be determined using Gibbs function.

Analysis (a) The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * D G *(T ) = n H2 g H2 (T ) + nO2 gO2 (T ) - n H2O g H2O (T )

At 1440 R, * * * D G *(T ) = n H2 g H2 (T ) + nO2 gO2 (T ) - n H2O g H2O (T )

= n H2 (h - Ts ) H2 + n O2 (h - Ts )O2 - n H2O (h - Ts )H2O

= nH2 [(h f + h1440 - h537 ) - Ts ]H2 + nO2 [(h f + h1440 - h537 ) - Ts ]O2 - nH2O [(h f + h1440 - h537 ) - Ts ]H2O = 1´ (0 + 9956.9 - 3640.3 - 1440´ 38.079) + 0.5´ (0 + 10,532.0 - 3725.1- 1440´ 56.326) - 1´ (- 104,040 + 11,933.4 - 4258 - 1440´ 53.428) = 87, 632 Btu/lbmol Substituting, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-532

ln K p = - (87,632 Btu/lbmol)/[(1.986 Btu/lbmol.R)(1440 R)]= - 30.64 or

K p = 4.93×10- 14 (Table A-28: ln K p = - 34.97 by interpolation) At 3960 R, * * * D G *(T ) = n H2 g H2 (T ) + nO2 gO2 (T ) - n H2O g H2O (T )

= n H2 (h - Ts )H2 + nO2 (h - Ts )O2 - n H2O (h - Ts )H2O = nH2 [(h f + h3960 - h537 ) - Ts ]H2 + nO2 [(h f + h3960 - h537 ) - Ts ]O2 - n H2O [(h f + h3960 - h537 ) - Ts ]H2O = 1´ (0 + 29,370.5 - 3640.3 - 3960´ 45.765) + 0.5´ (0 + 32, 441- 3725.1- 3960´ 65.032) - 1´ (- 104,040 + 39,989 - 4258 - 3960´ 64.402) = 53, 436 Btu/lbmol Substituting,

ln K p = - (53, 436 Btu/lbmol)/[(1.986 Btu/lbmol.R)(3960 R)]= - 6.79 or

K p = 1.125×10- 3 (Table A-28: ln K p = - 6.768)

17-16 The reaction 2H2 O Û 2H2 + O2 is considered. The mole fractions of the hydrogen and oxygen produced when this reaction occurs at 4000 K and 10 kPa are to be determined. Assumptions 1

The equilibrium composition consists of H 2 O , H 2 , and O2 .

2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

2H2 O Û 2H2 + O2 (thus nH2O = 2, nH2 = 2, and nO2 = 1)

2H 2 O ¾ ¾ ® xH 2 O + yH 2 +zO2

Actual:

react.

products

H balance:

4 = 2x + 2 y

O balance:

2 = x + 2z ¾ ¾ ® z = 1- 0.5 x

Total number of moles:

¾¾ ®

y = 2- x

Ntotal = x + y + z = 3 - 0.5x

The equilibrium constant relation can be expressed as (n

H2 N nH2 N nO2 æ P ö÷ ÷ K p = H2 nH2OO2 ççç ÷ N H2O èç N total ø÷

+ n O2 - n H2O )

From Table A-28, ln K p = - 0.542 at 4000 K . Since the stoichiometric reaction being considered is double this reaction,

K p = exp(- 2´ 0.542) = 0.3382

13-533

Substituting, 2+ 1- 2

0.3382 =

ö (2 - x) 2 (1- 0.5 x) æ çç10 /101.325 ÷ ÷ 2 ÷ çè 3 - 0.5 x ø x

Solving for x, x = 0.4446 Then, y = 2  x = 1.555 z = 1  0.5x = 0.7777 Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is

0.4446 H2 O+1.555 H2 + 0.7777 O2 The mole fractions of hydrogen and oxygen produced are yH2 =

N H2 1.555 1.555 = = = 0.560 N total 3 - 0.5´ 0.4446 2.778

yO2 =

N O2 0.7777 = = 0.280 N total 2.778

17-17 Hydrogen is burned with 150 percent theoretical air. The temperature at which 98 percent of H 2 will burn to H 2 O is to be determined. Assumptions 1

The equilibrium composition consists of H 2 O , H 2 , O2 , and N 2 .

The constituents of the mixture are ideal gases.

Analysis Assuming N 2 to remain as an inert gas, the stoichiometric and actual reactions can be written as Stoichiometric:

H 2 + 21 O2  H 2 O (thus  H 2O  1, H 2  1, and  O2  21 )

H 2 + 0.75(O 2  3.76 N 2 )

Actual:

 

0.98 H 2 O  0.02 H 2 + 0.26O 2  2.82 N 2          product

reactants

inert

The equilibrium constant K p can be determined from 

Kp 

 P    H2  O2  N H2 N O2  N total  N HH22OO

( H 2O  H 2  O2 )

11.5

0.98 1     0.5  0.02  0.26  0.98  0.02  0.26  2.82   194.11

From Table A-28, the temperature corresponding to this K p value is T = 2472 K.

13-534

EES (Engineering Equation Solver) SOLUTION "Given" P=1 [atm] ex=0.5 f_H2O=0.98 "Analysis" "Stoichiometric reaction: H2 + 1/2 O2 = H2O" nu_H2=1 nu_O2=1/2 nu_H2O=1 "Actual reaction: H2 + (ex+1)a_th (O2 + 3.76 N2) = f_H2O H2O + f_H2 H2 + x O2 + (ex+1)a_thx3.76 N2" a_th=0.5 f_H2=1-f_H2O 2*(ex+1)*a_th=f_H2O+2*x "O balance" N_H2O=f_H2O N_H2=f_H2 N_O2=x N_N2=(ex+1)*a_th*3.76 N_total=N_H2O+N_H2+N_O2+N_N2 K_p=N_H2O^nu_H2O/(N_H2^nu_H2*N_O2^nu_O2)*(P/N_total)^(nu_H2O-nu_H2-nu_O2) lnK_p=ln(K_p) "Find the temperature crresponding to this K_p value from Table A-28"

17-18 The reaction 2H2 O Û 2H2 + O2 is considered. The mole fractions of hydrogen gas produced is to be determined at 100 kPa and compared to that at 10 kPa. Assumptions 1

The equilibrium composition consists of H 2 O , H 2 , and O2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

2H2 O Û 2H2 + O2 (thus nH2O = 2, nH2 = 2, and nO2 = 1)

2H 2 O ¾ ¾ ® xH 2 O + yH 2 +zO2

Actual:

react.

products

H balance:

4 = 2x + 2 y

O balance:

2 = x + 2z ¾ ¾ ® z = 1- 0.5 x

Total number of moles:

¾¾ ®

y = 2- x

Ntotal = x + y + z = 3 - 0.5x

The equilibrium constant relation can be expressed as (n

H2 N nH2 N nO2 æ P ö÷ ÷ K p = H2 nH2OO2 ççç ÷ N H2O èç N total ø÷

+ n O2 - n H2O )

13-535

From Table A-28, ln K p = - 0.542 at 4000 K . Since the stoichiometric reaction being considered is double this reaction,

K p = exp(- 2´ 0.542) = 0.3382 Substituting, 2+ 1- 2

0.3382 =

ö (2 - x)2 (1- 0.5 x) æ çç100 /101.325 ÷ ÷ çè 3 - 0.5 x ÷ ø x2

Solving for x, x = 0.8870 Then, y = 2  x = 1.113 z = 1  0.5x = 0.5565 Therefore, the equilibrium composition of the mixture at 4000 K and 100 kPa is

0.8870 H2 O+1.113 H2 + 0.5565 O2 That is, there are 1.113 kmol of hydrogen gas. The mole number of hydrogen at 10 kPa reaction pressure was obtained in the previous problem to be 1.555 kmol. Therefore, the amount of hydrogen gas produced has decreased.

17-19 The reaction 2H2 O Û 2H2 + O2 is considered. The mole number of hydrogen gas produced is to be determined if inert nitrogen is mixed with water vapor is to be determined and compared to the case with no inert nitrogen. Assumptions 1

The equilibrium composition consists of H 2 O , H 2 , O2 , and N 2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

2H2 O Û 2H2 + O2 (thus nH2O = 2, nH2 = 2, and nO2 = 1)

2H 2 O + 0.5N 2 ¾ ¾ ® xH 2 O + yH 2 + zO2 + 0.5N 2

Actual:

react.

H balance:

4 = 2x + 2 y

O balance:

2 = x + 2z ¾ ¾ ® z = 1- 0.5 x

Total number of moles:

¾¾ ®

products

inert

y = 2- x

Ntotal = x + y + z + 0.5 = 3.5 - 0.5x

The equilibrium constant relation can be expressed as (n

H2 N nH2 N nO2 æ P ö ÷ ÷ K p = H2 nH2OO2 ççç ÷ ÷ N H2O çè N total ø

+ n O2 - n H2O )

From Table A-28, ln K p = - 0.542 at 4000 K . Since the stoichiometric reaction being considered is double this reaction,

K p = exp(- 2´ 0.542) = 0.3382

13-536

Substituting, 2+ 1- 2

0.3382 =

ö (2 - x) 2 (1- 0.5 x) æ çç10 /101.325 ÷ ÷ 2 ÷ çè 3.5 - 0.5 x ø x

Solving for x, x = 0.4187 Then, y = 2  x = 1.581 z = 1  0.5x = 0.7907 Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is

0.4187 H2 O+1.581 H2 + 0.7907 O2 That is, there are 1.581 kmol of hydrogen gas. The mole number of hydrogen without inert nitrogen case was obtained in an earlier problem to be 1.555 kmol. Therefore, the amount of hydrogen gas produced has increased.

17-20 The equilibrium constant of the reaction H2  1/ 2O2  H2 O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data.

Analysis (a) The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where

D G *(T ) = nH2O gH* 2O (T ) - nH2 gH* 2 (T ) - nO2 gO* 2 (T ) At 25C,

D G *(T ) = 1(- 228,590) - 1(0) - 0.5(0) = - 228,590 kJ/kmol Substituting,

ln K p = - (- 228,590 kJ/kmol)/[(8.314 kJ/kmol ×K)(298 K)]=92.26 or

K p = 1.12×1040 (Table A-28: ln K p = 92.21) (b) At 2000 K,

D G *(T ) = nH2O gH* 2O (T ) - nH2 gH* 2 (T ) - nO2 gO* 2 (T )

= n H2 O (h - Ts )H2 O - n H2 (h - Ts )H2 - nO2 (h - Ts )O2 = n H2O [(h f + h2000 - h298 ) - Ts ]H2 O - n H2 [(h f + h2000 - h298 ) - Ts ]H2 - nO2 [(h f + h2000 - h298 ) - Ts ]O2 = 1´ (- 241,820 + 82,593 - 9904 - 2000´ 264.571) - 1´ (0 + 61, 400 - 8468 - 2000´ 188.297) - 0.5´ (0 + 67,881- 8682) - 2000´ 268.655)

13-537

= - 135,556 kJ/kmol Substituting,

ln K p = - (- 135,556 kJ/kmol)/[(8.314 kJ/kmol ×K)(2000 K)]=8.152 or

K p = 3471 (Table A-28: ln K p = 8.145)

EES (Engineering Equation Solver) SOLUTION "Given" T=298 "[K]" "or 2000 [K]" P=101.325 "[kPa], assumed" "Reaction: H2+1/2 O2 = H2O" "Properties" R_u=8.314 "[kJ/kmol-K]" "Analysis" DELTAG=nu_H2O*g_H2O-nu_H2*g_H2-nu_O2*g_O2 K_p=exp(-DELTAG/(R_u*T)) "Stoichiometric coefficients" nu_H2=1 nu_O2=1/2 nu_H2O=1 "Gibbs functions" g_H2=h_H2-T*s_H2 g_O2=h_O2-T*s_O2 g_H2O=h_H2O-T*s_H2O "Enthalpies" h_H2=enthalpy(H2, T=T) h_O2=enthalpy(O2, T=T) h_H2O=enthalpy(H2O, T=T) "Entropies" s_H2=entropy(H2, T=T, P=P) s_O2=entropy(O2, T=T, P=P) s_H2O=entropy(H2O, T=T, P=P)

17-21 The reaction C  O2  CO2 is considered. The mole fraction of the carbon dioxide produced when this reaction occurs at a1 atm and 3800 K are to be determined. Assumptions 1

The equilibrium composition consists of CO2 , C and O2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are

13-538

C + O2 Û CO2 (thus nC = 1, nO2 = 1, and nCO2 = 1)

Stoichiometric:

C + O2 ¾ ¾ ® xC + yO 2 + zCO 2

Actual:

products

react.

C balance:

1= x + z ¾ ¾ ® z = 1- x

O balance:

2 = 2 y + 2z ¾ ¾ ® y = 1- z = 1- (1- x) = x

Total number of moles:

N total = x + y + z = 1+ x

The equilibrium constant relation can be expressed as (n

ö CO2 N nCO2 æ P ÷ ÷ K p = nCCO2nO2 ççç ÷ N C N O2 çè N total ÷ ø

- nC - nO2 )

From the problem statement at 3800 K, ln K p = - 0.461 . Then,

K p = exp(- 0.461) = 0.6307 Substituting, 1- 1- 1

0.6307 =

ö (1- x) æ çç 1 ÷ ÷ ÷ ç ( x)( x) è1 + x ø

Solving for x, x = 0.7831 Then, y = x = 0.7831 z = 1  x = 0.2169 Therefore, the equilibrium composition of the mixture at 3800 K and 1 atm is

0.7831 C + 0.7831 O2 + 0.2169 CO2 The mole fraction of carbon dioxide is yCO2 =

N CO2 0.2169 = = 0.1216 N total 1 + 0.7831

EES (Engineering Equation Solver) SOLUTION "Given" P=1 [atm] T=3800 [K] lnK_p=-0.461 "Analysis" "Stoichiometric reaction: C + O2 = CO2" nu_C=1 nu_O2=1 nu_CO2=1 "Actual reaction: C + O2 = x C + y O2 + z CO2" N_C=x N_O2=y N_CO2=z N_total=N_C+N_O2+N_CO2 1=x+z "C balance" 2=2*y+2*z "O balance" K_p=N_CO2^nu_CO2/(N_C^nu_C*N_O2^nu_O2)*(P/N_total)^(nu_CO2-nu_C-nu_O2) lnK_p=ln(K_p) y_CO2=N_CO2/N_total © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-539

17-22 The reaction N2  O2  2NO is considered. The equilibrium mole fraction of NO 1600 K and 1 atm is to be determined. Assumptions 1

The equilibrium composition consists of N 2 , O2 , and NO.

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

N2 + O2 Û 2NO (thus n N2 = 1, nO2 = 1, and n NO = 2)

N 2 + O2 ¾ ¾ ® xN 2 + yO2 + zNO

Actual:

products

react.

N balance:

2 = 2x + z ¾ ¾ ® z = 2 - 2x

O balance:

2 = 2y + z ¾ ¾ ® y= x

Total number of moles:

N total = x + y + z = 2

The equilibrium constant relation can be expressed as (n

ö NO N n NO æ P ÷ ÷ K p = n N2NO nO2 ççç ÷ N N2 N O2 çè N total ÷ ø

- n N2 - nO2 )

From Table A-28, at 1600 K, ln K p = - 5.294 . Since the stoichiometric reaction being considered is double this reaction,

K p = exp(- 2´ 5.294) = 2.522´ 10- 5 Substituting, 2- 1- 1

2.522´ 10- 5 =

ö (2 - 2 x) 2 æ çç1 ÷ ÷ 2 ÷ çè 2 ø x

Solving for x, x = 0.9975 Then, y = x = 0.9975 z = 2  2x = 0.005009 Therefore, the equilibrium composition of the mixture at 1600 K and 1 atm is

0.9975 N2 + 0.9975 O2 + 0.005009 NO The mole fraction of NO is then yNO =

N NO 0.005009 = = 0.002505 N total 2

13-540

"Analysis" "Stoichiometric reaction: N2+O2 = 2NO" "Stoichiometric coefficients" nu_N2=1 nu_O2=1 nu_NO=2 "Actual reaction: N2+O2 = xN2+yO2+zNO" 2=2*x+z "N balance" 2=2*y+z "O balance" N_total=x+y+z N_N2=x N_O2=y N_NO=z "From Table A-28, lnK_p = -5.294 at 1600 K. Thus" K_p=exp((-2)*5.294) K_p=N_NO^nu_NO/(N_N2^nu_N2*N_O2^nu_O2)*(P/N_total)^(nu_NO-nu_N2-nu_O2) "The mole fraction of NO is" y_NO=N_NO/N_total

17-23 The dissociation reaction CO2  CO + O is considered. The composition of the products at given pressure and temperature is to be determined when nitrogen is added to carbon dioxide. Assumptions 1

The equilibrium composition consists of CO2 , CO, O, and N 2 .

The constituents of the mixture are ideal gases.

Analysis For the stoichiometric reaction CO 2 Û CO + 12 O 2 , from Table A-28, at 2500 K

ln K p = - 3.331 For the oxygen dissociation reaction 0.5O2 Û O , from Table A-28, at 2500 K,

ln K p = - 8.509 / 2 = - 4.255 For the desired stoichiometric reaction

CO2 Û CO + O (thus nCO2 = 1, nCO = 1 and nO = 1) ,

ln K p = - 3.331- 4.255 = - 7.586 and

K p = exp(- 7.586) = 0.0005075 Actual: C balance:

CO 2 + 3N 2 ¾ ¾ ® xCO 2 + yCO+zO + 3 N 2

1= x + y

react.

products

¾ ¾®

y = 1- x

inert

13-541

O balance:

2 = 2x + y + z ¾ ¾ ® z = 1- x

Total number of moles:

Ntotal = x + y + z + 3 = 5 - x

The equilibrium constant relation can be expressed as n

Kp =

n CO ö CO N CO N OnO æ çç P ÷ ÷ nCO2 çèç N ÷ ÷ N CO2 total ø

+ nO - nCO2

Substituting, 1+ 1- 1

0.0005075 =

ö (1- x)(1- x) æ çç 1 ÷ ÷ ÷ ç è5 - x ø x

Solving for x, x = 0.9557 Then, y = 1  x = 0.0443 z = 1  x = 0.0443 Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is

0.9557 CO2 + 0.0443 CO + 0.0443 O + 3N 2

EES (Engineering Equation Solver) SOLUTION "Given" P=1 [atm] T=2500 [K] "Analysis" "Stoichiometric reaction: CO2 = CO + 1/2 O2" lnK_p1=-3.331 "from Table A-28" "Stoichiometric reaction: 1/2 O2 = O" lnK_p2=-8.509/2 "from Table A-28" "Desired stoichiometric reaction: CO2 = CO + O" nu_CO2=1 nu_CO=1 nu_O=1 lnK_p=lnK_p1+lnK_p2 K_p=exp(lnK_p) "Actual reaction: CO2 + 3 N2 = x CO2 + y CO + z O + 3 N2" N_CO2=x N_CO=y N_O=z N_N2=3 N_total=N_CO2+N_CO+N_O+N_N2 1=x+y "C balance" 2=2*x+y+z "O balance" K_p=(N_CO^nu_CO*N_O^nu_O)/N_CO2^nu_CO2*(P/N_total)^(nu_CO+nu_O-nu_CO2)

17-24 The equilibrium constant of the reaction CO  1/ 2O2  CO2 at 298 K and 2000 K are to be determined, and compared with the values listed in Table A-28.

13-542

Analysis (a) The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * D G *(T ) = nCO2 gCO2 (T ) - nCO g CO (T ) - n O2 g O2 (T )

At 298 K,

D G *(T ) = 1(- 394,360) - 1(- 137,150) - 0.5(0) = - 257, 210 kJ/kmol where the Gibbs functions are obtained from Table A-26. Substituting, ln K p = -

(- 257, 210 kJ/kmol) = 103.81 (8.314 kJ/kmol ×K)(298 K)

From Table A-28:

ln K p = 103.76 (b) At 2000 K, * * * D G *(T ) = nCO2 gCO2 (T ) - n CO g CO (T ) - n O2 g O2 (T )

= n CO2 (h - T s )CO2 - n CO (h - T s )CO - n O2 (h - T s )O2

= 1[(- 302,128) - (2000)(309.00) ]- 1[(- 53,826) - (2000)(258.48) ]- 0.5[(59,193) - (2000)(268.53) ] = - 110, 409 kJ/kmol The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa (1 atm) are obtained from EES. Substituting, ln K p = -

(- 110, 409 kJ/kmol) = 6.64 (8.314 kJ/kmol ×K)(2000 K)

From Table A-28:

ln K p = 6.635

EES (Engineering Equation Solver) SOLUTION Tprod_1 = 298 [K] Tprod_2 = 2000 [K] R_u=8.314 [kJ/kmol-K] "The following equations provide the molar specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, Tprod, at 1 atm pressure, 101.3 kPa" "For Tprod_1:" g_CO=h_CO_1-Tprod_1 *s_CO_1 h_CO_1=Enthalpy(CO,T=Tprod_1 ) s_CO_1=Entropy(CO,T=Tprod_1 ,P=101.3) g_O2=h_O2_1-Tprod_1*s_O2_1 h_O2_1=Enthalpy(O2,T=Tprod_1 ) s_O2_1=Entropy(O2,T=Tprod_1 ,P=101.3) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-543

g_CO2=h_CO2_1-Tprod_1*s_CO2_1 h_CO2_1=Enthalpy(CO2,T=Tprod_1 ) s_CO2_1=Entropy(CO2,T=Tprod_1 ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO2-1*g_CO-0.5*g_O2 "The equilibrium constant is given by:" K_p_1 = exp(-DELTAG /(R_u*Tprod_1 )) lnK_p_1=ln(k_p_1) "For Tprod_2:" g_CO_2=h_CO_2-Tprod_2 *s_CO_2 h_CO_2=Enthalpy(CO,T=Tprod_2) s_CO_2=Entropy(CO,T=Tprod_2 ,P=101.3) g_O2_2=h_O2_2-Tprod_2 *s_O2_2 h_O2_2=Enthalpy(O2,T=Tprod_2 ) s_O2_2=Entropy(O2,T=Tprod_2 ,P=101.3) g_CO2_2=h_CO2_2-Tprod_2 *s_CO2_2 h_CO2_2=Enthalpy(CO2,T=Tprod_2 ) s_CO2_2=Entropy(CO2,T=Tprod_2 ,P=101.3) "The standard-state Gibbs function is" DELTAG_2 =1*g_CO2_2-1*g_CO_2-0.5*g_O2_2 "The equilibrium constant is given by:" K_p_2 = exp(-DELTAG_2 /(R_u*Tprod_2 )) lnK_p_2= ln(k_P_2) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text." "Note: The equilibrium equation is the reverse of that in Table A-28. Therefore, K_p is the inverse of the value obtained from Table A-28 and the ln(K_p) is the negative of the value in Table A-28" lnK_p_1_TableA28 = 103.762 lnK_p_2_TableA28 = 6.635

17-25 The effect of varying the percent excess air during the steady-flow combustion of hydrogen is to be studied. Analysis The combustion equation of hydrogen with stoichiometric amount of air is

H 2 + 0.5[O 2 + 3.76N 2 ]¾ ¾ ® H 2 O + 0.5(3.76) N 2 For the incomplete combustion with 100% excess air, the combustion equation is

H 2 + (1 + Ex)(0.5) [O 2 + 3.76N 2 ]¾ ¾ ® 0.97 H 2 O + a H 2 + b O 2 + c N 2 The coefficients are to be determined from the mass balances Hydrogen balance:

2 = 0.97 ´ 2 + a ´ 2 ¾ ¾ ® a = 0.03

Oxygen balance:

(1+ Ex)´ 0.5´ 2 = 0.97 + b´ 2

Nitrogen balance: (1 + Ex)´ 0.5´ 3.76´ 2 = c´ 2 Solving the above equations, we find the coefficients (Ex = 1, a = 0.03 b = 0.515, c = 3.76) and write the balanced reaction equation as

H 2 + [O 2 + 3.76N 2 ]¾ ¾ ® 0.97 H 2 O + 0.03 H 2 + 0.515 O 2 + 3.76 N 2 Total moles of products at equilibrium are

Ntot = 0.97 + 0.03 + 0.515 + 3.76 = 5.275 The assumed equilibrium reaction is

H 2 O ¬ ¾® H 2 + 0.5O2 The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

13-544

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * D G *(T ) = n H2 g H2 (Tprod ) + n O2 g O2 (Tprod ) - n H2O g H2O (Tprod )

and the Gibbs functions are defined as * gH2 (Tprod ) = (h - Tprod s )H2 * gO2 (Tprod ) = (h - Tprod s )O2 * gH2O (Tprod ) = (h - Tprod s )H2O

The equilibrium constant is also given by 1+ 0.5- 1

æP ö ÷ ÷ K p = ççç ÷ èç Ntot ÷ ø

0.5

ö (0.03)(0.515)0.5 ab0.5 æ çç 1 ÷ = = 0.009664 ÷ ø 0.97 0.971 èç5.275 ÷

and ln K p = ln(0.009664) = - 4.647 The corresponding temperature is obtained solving the above equations using EES to be

Tprod = 2600 K This is the temperature at which 97 percent of H 2 will burn into H 2 O . The copy of EES solution is given next.

EES (Engineering Equation Solver) SOLUTION "To solve this problem, we need to give EES a guess value for T_prop other than the default value of 1. Set the guess value of T_prod to 1000 K by selecting Variable Infromation in the Options menu. Be sure to select the molar basis for properties." "Input data" PercentEx = 10 Ex = PercentEx/100 P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The combustion equation of H2 with stoichiometric amount of air is H2 + 0.5(O2 + 3.76N2)=H2O +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is H2 + (1+EX)(0.5)(O2 + 3.76N2)=0.97 H2O +aH2 + bO2+cN2" "Specie balance equations give the values of a, b, and c." 2 = 0.97*2 + a*2 "H, hydrogen" (1+Ex)*0.5*2=0.97 + b*2 "O, oxygen" (1+Ex)*0.5*3.76 *2 = c*2 "N, nitrogen" N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is H2O=H2+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each H2mponent in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) DELTAG =1*g_H2+0.5*g_O2-1*g_H2O "The standard-state Gibbs function" K_P = exp(-DELTAG /(R_u*T_prod )) "The equilibrium constant" P=P_prod /101.3 "The equilibrium constant is also given by" "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text."

13-545

ln K p

PercentEx [%]

Tprod

10 20 30 40 50 60 70 80 90 100

2440 2490 2520 2542 2557 2570 2580 2589 2596 2602

[K]

17-26 The equilibrium constant of the reaction CH4  2O2  CO2  2H2 O at 25C is to be determined.

Analysis The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * D G *(T ) = nCO2 gCO (T ) + nH2O gH* 2O (T ) - nCH4 gCH (T ) - nO2 gO* 2 (T ) 2 4

At 25C,

D G *(T ) = 1(- 394,360) + 2(- 228,590) - 1(- 50,790) - 2(0) = - 800, 750 kJ/kmol Substituting, or

K p = 1.96´ 10140

EES (Engineering Equation Solver) SOLUTION "Given" T=298 [K] P=101.325 [kPa] "assumed" "Reaction: CH4 + 2O2 = CO2 + 2H2O" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" DELTAG=nu_CO2*g_CO2+nu_H2O*g_H2O-nu_CH4*g_CH4-nu_O2*g_O2 K_p=exp(-DELTAG/(R_u*T)) "Stoichiometric coefficients" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-546

nu_CH4=1 nu_O2=2 nu_CO2=1 nu_H2O=2 "Gibbs functions" g_CH4=h_CH4-T*s_CH4 g_CO2=h_CO2-T*s_CO2 g_H2O=h_H2O-T*s_H2O g_O2=h_O2-T*s_O2 "Enthalpies" h_CH4=enthalpy(CH4, T=T) h_CO2=enthalpy(CO2, T=T) h_H2O=enthalpy(H2O, T=T) h_O2=enthalpy(O2, T=T) "Entropies" s_CH4=entropy(CH4, T=T, P=P) s_CO2=entropy(CO2, T=T, P=P) s_H2O=entropy(H2O, T=T, P=P) s_O2=entropy(O2, T=T, P=P)

17-27 Carbon dioxide is heated to a high temperature at a constant pressure. The percentage of CO2 that will dissociate into CO and O2 is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, and O2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

CO2 Û CO + 12 O2 (thus nCO2 = 1, nCO = 1, and nO2 = 12 )

Actual:

CO2 ¾ ¾ ® x CO2 + y CO + z O2 react.

products

C balance:

1= x + y ¾ ¾ ® y = 1- x

O balance:

2 = 2x + y + 2z ¾ ¾ ® z = 0.5 - 0.5x

Total number of moles:

Ntotal = x + y + z = 1.5 - 0.5x

The equilibrium constant relation can be expressed as Kp =

n n CO N CO N OO2 2 æ çç P n CO2 ççè N N CO 2

CO ö ÷ ÷ ÷ ÷ total ø

+ n O2 - n CO2 )

From Table A-28, ln K p = - 3.860 at 2400 K. Thus K p = 0.02107 Substituting,

13-547

1.5- 1

ö (1- x)(0.5 - 0.5 x)1/ 2 æ 3 ÷ çç 0.02107 = ÷ ÷ ç è1.5 - 0.5 x ø x Solving for x, x = 0.936

Thus the percentage of CO2 which dissociates into CO and O2 is 1 0.936 = 0.064 or 6.4%

EES (Engineering Equation Solver) SOLUTION "Given" T=2400 [K] P=3 [atm] P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "Stoichiometric reaction: CO2 = CO + 1/2 O2" "Stoichiometric coefficients" nu_CO2=1 nu_CO=1 nu_O2=1/2 "Actual reaction: CO2 = N_CO2 CO2 + N_CO CO + N_O2 O2" 1=N_CO2+N_CO "C balance" 2=2*N_CO2+N_CO+2*N_O2 "O balance" N_total=N_CO2+N_CO+N_O2 K_p=(N_CO^nu_CO*N_O2^nu_O2)/N_CO2^nu_CO2*(P/N_total)^(nu_CO+nu_O2-nu_CO2) "The equilibrium constant can also be expressed as" K_p=exp(-DELTAG/(R_u*T)) DELTAG=nu_CO*g_CO+nu_O2*g_O2-nu_CO2*g_CO2 "Gibbs functions" g_CO2=h_CO2-T*s_CO2 g_CO=h_CO-T*s_CO g_O2=h_O2-T*s_O2 "Enthalpies" h_CO2=enthalpy(CO2, T=T) h_CO=enthalpy(CO, T=T) h_O2=enthalpy(O2, T=T) "Entropies" s_CO2=entropy(CO2, T=T, P=P0) s_CO=entropy(CO, T=T, P=P0) s_O2=entropy(O2, T=T, P=P0) "The percentage of CO2 that will dissociate into CO and O2 is" Percent_dissociate=(1-N_CO2)*Convert(, %)

17-28 The equilibrium constant of the reaction CO2  CO  1/ 2O2 is listed in Table A-28 at different temperatures. It is to be verified using Gibbs function data.

13-548

Analysis (a) The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where

* * D G *(T ) = nCO gCO (T ) + nO2 gO* 2 (T ) - nCO2 gCO (T ) 2

At 298 K,

D G *(T ) = 1(- 137,150) + 0.5(0) - 1(- 394,360) = 257, 210 kJ/kmol Substituting,

ln K p = - (257, 210 kJ/kmol)/[(8.314 kJ/kmol ×K)(298 K)] = - 103.81 or

K p = 8.24´ 10- 46 (Table A-28: ln K p = - 103.76)

(b) At 1800 K, * * D G *(T ) = nCO gCO (T ) + nO2 gO* 2 (T ) - nCO2 gCO (T ) 2

= nCO (h - T s )CO + nO2 (h - T s )O2 - nCO2 (h - T s )CO2 = nCO [(h f + h1800 - h298 ) - T s ]CO + nO2 [(h f + h1800 - h298 ) - T s ]O2 - nCO2 [(h f + h1800 - h298 ) - T s ]CO2 = 1´ (- 110,530 + 58,191- 8669 - 1800´ 254.797) + 0.5´ (0 + 60,371- 8682 - 1800´ 264.701) - 1´ (- 393,520 + 88,806 - 9364 - 1800´ 302.884) = 127, 240.2 kJ/kmol Substituting, ln K p = - (127, 240.2 kJ/kmol)/[(8.314 kJ/kmol ×K)(1800 K)] = - 8.502 or

K p = 2.03´ 10- 4 (Table A-28: ln K p = - 8.497)

EES (Engineering Equation Solver) SOLUTION "Given" T=298 [K] "or 1800 [K]" P=101.325 [kPa] "assumed" "Reaction: CO2 = CO + 1/2O2" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" DELTAG=nu_CO*g_CO+nu_O2*g_O2-nu_CO2*g_CO2 K_p=exp(-DELTAG/(R_u*T)) "Stoichiometric coefficients" nu_CO2=1 nu_CO=1 nu_O2=1/2 "Gibbs functions" g_CO2=h_CO2-T*s_CO2 g_CO=h_CO-T*s_CO g_O2=h_O2-T*s_O2 "Enthalpies" h_CO2=enthalpy(CO2, T=T) h_CO=enthalpy(CO, T=T) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-549

h_O2=enthalpy(O2, T=T) "Entropies" s_CO2=entropy(CO2, T=T, P=P) s_CO=entropy(CO, T=T, P=P) s_O2=entropy(O2, T=T, P=P)

17-29 Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, O2 , and N 2 .

The constituents of the mixture are ideal gases.

Analysis Assuming N 2 to remain as an inert gas, the stoichiometric and actual reactions can be written as Stoichiometric:

CO + 12 O2 Û CO2 (thus nCO2 = 1, nCO = 1, and nO2 = 12 )

CO + 1(O2 + 3.76 N 2 ) ¾ ¾ ® 0.97 CO2 + 0.03 CO + 0.515 O2 + 3.76 N 2

Actual:

product

reactants

inert

The equilibrium constant K p can be determined from æ P çç n O2 ç n CO N N çè N n

Kp =

N COCO22

ö CO2 ÷ ÷ ÷ ÷ total ø

- n CO - n O2 )

1- 1.5

æ ö 0.97 1 ÷ ç = ÷ 0.5 ç ÷ ç è 0.03´ 0.515 0.97 + 0.03 + 0.515 + 3.76 ø = 103.5 and

ln K p = 4.639 From Table A-28, the temperature corresponding to this K p value is T = 2276 K

EES (Engineering Equation Solver) SOLUTION "Given" P=1 [atm] ex=1 f_CO2=0.97 "Analysis" "Stoichiometric reaction: CO + 1/2 O2 = CO2" nu_CO=1 nu_O2=1/2 nu_CO2=1 "Actual reaction: CO + (ex+1)a_th (O2 + 3.76 N2) = f_CO2 CO2 + f_CO CO + x O2 + (ex+1)a_thx3.76 N2" a_th=0.5 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-550

f_CO=1-f_CO2 1+2*(ex+1)*a_th=2*f_CO2+f_CO+2*x "O balance" N_CO2=f_CO2 N_CO=f_CO N_O2=x N_N2=(ex+1)*a_th*3.76 N_total=N_CO2+N_CO+N_O2+N_N2 K_p=N_CO2^nu_CO2/(N_CO^nu_CO*N_O2^nu_O2)*(P/N_total)^(nu_CO2-nu_CO-nu_O2) lnK_p=ln(K_p) "Find the temperature crresponding to this K_p value from Table A-28"

17-30 Problem 17-29 is reconsidered. The effect of varying the percent excess air during the steady-flow process from 0 to 200 percent on the temperature at which 97 percent of CO burn into CO2 is to be studied. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, we need to give EES a guess value for T_prop other than the default value of 1. Set the guess value of T_prod to 1000 K by selecting Variable Information in the Options menu." "Input Data" PercentEx = 100 Ex = PercentEx/100 P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] f=0.97 "The combustion equation of CO with stoichiometric amount of air is CO + 0.5(O2 + 3.76N2)=CO2 +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is CO + (1+EX)(0.5)(O2 + 3.76N2)=0.97 CO2 +aCO + bO2+cN2" "Specie balance equations give the values of a, b, and c." 1 = f + a "C, Carbon" 1 +(1+Ex)*0.5*2=f*2 + a *1 + b*2 "O, oxygen" (1+Ex)*0.5*3.76 *2 = c*2 "N, nitrogen" N_tot =f+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The standard-state Gibbs function" K_P = exp(-DELTAG /(R_u*T_prod )) "The equilibrium constant" P=P_prod /101.3 "The equilibrium constant is also given by K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *f lnK_p = ln(k_P) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text."

13-551

Tprod

PercentEx [%]

[K]

0 20 40 60 80 100 120 140 160 180 200

2065 2193 2230 2250 2263 2272 2279 2285 2289 2293 2296

17-31E Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, O2 , and N 2 .

The constituents of the mixture are ideal gases.

Analysis Assuming N 2 to remain as an inert gas, the stoichiometric and actual reactions can be written as Stoichiometric: Actual:

CO + 12 O2 Û CO2 (thus nCO2 = 1, nCO = 1, and nO2 = 12 )

CO + 1(O2 + 3.76 N 2 ) ¾ ¾ ® 0.97 CO2 + 0.03CO + 0.515O2 + 3.76 N 2 product

reactants

inert

The equilibrium constant K p can be determined from æ P çç Kp = n n N CO N O2 ççè N n

N COCO22

ö CO2 ÷ ÷ ÷ ÷ total ø

- n CO - n O2 )

1- 1.5

æ ö 0.97 1 ÷ ç ÷ 0.5 ç ÷ 0.03´ 0.515 çè0.97 + 0.03 + 0.515 + 3.76 ø = 103.5

and

ln K p = 4.639 From Table A-28, the temperature corresponding to this K p value is T = 2276 K = 4097 R

13-552

EES (Engineering Equation Solver) SOLUTION "Given" P=1 [atm] ex=1 f_CO2=0.97 "Analysis" "Stoichiometric reaction: CO + 1/2 O2 = CO2" nu_CO=1 nu_O2=1/2 nu_CO2=1 "Actual reaction: CO + (ex+1)a_th (O2 + 3.76 N2) = f_CO2 CO2 + f_CO CO + x O2 + 3.76 N2" a_th=0.5 f_CO=1-f_CO2 1+2*(ex+1)*a_th=2*f_CO2+f_CO+2*x "O balance" N_CO2=f_CO2 N_CO=f_CO N_O2=x N_N2=3.76 N_total=N_CO2+N_CO+N_O2+N_N2 K_p=N_CO2^nu_CO2/(N_CO^nu_CO*N_O2^nu_O2)*(P/N_total)^(nu_CO2-nu_CO-nu_O2) lnK_p=ln(K_p) "Find the temperature crresponding to this K_p value from Table A-28"

17-32 Air is heated to a high temperature. The equilibrium composition at that temperature is to be determined. Assumptions 1

The equilibrium composition consists of N 2 , O2 , and NO.

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

1 2

N2 + 12 O2 Û NO (thus n NO = 1, n N2 = 12 , and nO2 = 12 )

3.76 N 2 + O 2 ¾ ¾ ® x NO + y N 2 + z O 2

Actual:

prod.

N balance: O balance:

7.52 = x + 2y or y 2 = x + 2z or z

Total number of moles:

reactants

x x

Ntotal  x  y  z  x  4.76 x  4.76

The equilibrium constant relation can be expressed as Kp =

æ P çç n N2 n O2 ç N N çè N n NO N NO

NO ö ÷ ÷ ÷ ÷ total ø

- n N 2 - n O2 )

From Table A-28, ln K p  3.931 at 2000 K. Thus K p  0.01962 . Substituting,

13-553

1- 1

æ 2 ÷ ö x ç 0.01962 = ÷ 0.5 0.5 ç ÷ ç è ø 4.76 (3.76 - 0.5 x) (1- 0.5 x)

Solving for x, x = 0.0376 Then, y = 3.76−0.5x = 3.7412 z = 1−0.5x = 0.9812 Therefore, the equilibrium composition of the mixture at 2000 K and 2 atm is

0.0376NO + 3.7412N2 + 0.9812O2 The equilibrium constant for the reactions O2 Û 2O (ln K p  14.622 ) and N2 Û 2N (ln K p  41.645 ) are much smaller than that of the specified reaction (ln K p  3.931 ). Therefore, it is realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture. Also the equilibrium composition is in this case is independent of pressure since D n = 1- 0.5 - 0.5 = 0.

EES (Engineering Equation Solver) SOLUTION "Given" T=2000 [K] P=2 [atm] P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "Stoichiometric reaction: 1/2 N2 + 1/2 O2 = NO" "Stoichiometric coefficients" nu_N2=1/2 nu_O2=1/2 nu_NO=1 "Actual reaction: 3.76 N2 + O2 = N_NO NO + N_N2 N2 + N_O2 O2" 3.76*2=N_NO+2*N_N2 "N balance" 2=N_NO+2*N_O2 "O balance" N_total=N_NO+N_N2+N_O2 K_p=N_NO^nu_NO/(N_N2^nu_N2*N_O2^nu_O2)*(P/N_total)^(nu_NO-nu_N2-nu_O2) lnK_p=-3.931 "from Table A-28 at 2000 K" lnK_p=ln(K_p) "Compare this lnK_p value to those for the reactions O2=2O and N2=2N from Table A-28 to see whether any monatomic oxygen or nitrogen will likely be present in the equilibrium mixture. Also, double the pressure and see whether the results will change"

17-33 The equilibrium constant, K p is to be estimated at 2500 K for the reaction CO  H2 O  CO2  H2 . Analysis (a) The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * * D G *(T ) = nCO2 gCO2 (T ) + n H2 g H2 (T ) - nCO gCO (T ) - n H2O g H2O (T )

13-554

* * * * D G *(T ) = nCO2 gCO2 (T ) + n H2 g H2 (T ) - nCO g CO (T ) - n H2O g H2O (T )

= n CO2 (h - T s )CO2 + n H2 (h - T s ) H2 - n CO (h - T s )CO - n H2O (h - T s ) H2O = 1[(- 393,520 + 131, 290 - 9364) - (2500)(322.808) ] = 1[(- 393,520 + 131, 290 - 9364) - (2500)(322.808) ]+ 1[(0 + 78,960 - 8468) - (2500)(196.125) ]

- 1[(- 110,530 + 83, 692 - 8669) - (2500)(266.755 ]- 1[(- 241, 820 + 108,868 - 9904) - (2500)(276.286 ] = 37,531 kJ/kmol Substituting, ln K p = -

37,531 kJ/kmol = - 1.8057 ¾ ¾ ® K p = 0.1644 (8.314 kJ/kmol ×K)(2500 K)

The equilibrium constant may be estimated using the integrated van't Hoff equation:

æK p , est ÷ ö h æ1 1 ö ÷ ÷ ln ççç = R ççç - ÷ ÷ ÷ ÷ ÷ èç K p1 ÷ ø Ru èçTR T ø æ K p , est ö - 26,176 kJ/kmol æ 1 ö ÷ ÷ ç 1 ÷ ln ççç = ® K p , est = 0.1612 ÷¾ ¾ ÷ ÷ ÷ 8.314 kJ/kmol.K çèç 2000 K 2500 K ø çè0.2209 ø

EES (Engineering Equation Solver) SOLUTION T_HR=2000 [K] "Temperature for enthalpy of reaction" h_bar_R=-26176 [kJ/kmol] "Enthalpy of reaction at 2000 K" K_P_1=0.2209 "Equilibrium constant at 2000 K" "The reaction temperature for Kp calculation is:" T_reac= 2500 [K] R_u=8.314 [kJ/kmol-K] "The assumed equilibrium reaction is CO + H2O =CO2 + H2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_reac, at 1 atm pressure, 101.3 kPa" "Estimate the equilibrium constant K_P_est by using the integrated van't Hoff equation." ln(K_P_est/K_P_1)=h_bar_R/R_u*(1/(T_HR)-1/(T_reac)) "Second find K_P_calculated at T_reac" g_CO2_2=h_CO2-(T_reac)*s_CO2 h_CO2=Enthalpy(CO2,T=T_reac ) s_CO2=Entropy(CO2,T=T_reac ,P=101.3) g_CO_2=h_CO-(T_reac)*s_CO h_CO=Enthalpy(CO,T=T_reac ) s_CO=Entropy(CO,T=T_reac ,P=101.3) g_H2_2=h_H2-(T_reac)*s_H2 h_H2=Enthalpy(H2,T=T_reac ) s_H2=Entropy(H2,T=T_reac ,P=101.3) g_H2O_2=h_H2O-(T_reac)*s_H2O h_H2O=Enthalpy(H2O,T=T_reac ) s_H2O=Entropy(H2O,T=T_reac ,P=101.3) "The standard-state Gibbs function is" DELTAG_2 =1*g_CO2_2+1*g_H2_2-1*g_CO_2-1*g_H2O_2 "The equilibrium constant K_P is given by Eq. 15-14." K_P_calculated = exp(-DELTAG_2 /(R_u*T_reac ))

13-555

17-34 A gaseous mixture consisting of methane and nitrogen is heated. The equilibrium composition (by mole fraction) of the resulting mixture is to be determined. Assumptions 1

The equilibrium composition consists of CH4 , C, H 2 , and N 2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric: C + 2H2 Û CH4 (thus nC = 1, nH2 = 2, and nCH4 = 1)

CH 4 + N 2 ¾ ¾ ® xCH 4 + yC+ zH 2 + N 2

Actual:

react.

C balance:

1= x + y

H balance:

4 = 4x + 2z ¾ ¾ ® z = 2 - 2x

Total number of moles:

¾ ¾®

products

inert

y = 1- x

Ntotal = x + y + z + 1 = 4 - 2 x

The equilibrium constant relation can be expressed as n

CH4 N nCH4 æ P ö ÷ ÷ K p = nCCH4nH2 ççç ÷ ÷ N C N H2 çè N total ø

- n C - n H2

From the problem statement at 1000 K, ln K p = 2.328 . Then,

K p = exp(2.328) = 10.257 Substituting, 1- 1- 2

10.257 =

æ 1 ÷ ö x çç ÷ ÷ (1- x)(2 - 2 x) 2 çè 4 - 2 x ø

Solving for x, x = 0.5194 Then, y = 1  x = 0.4806 z = 2  2x = 0.9612 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is

0.5194 CH4 +0.4806 C + 0.9612 H2 + 1 N2 The mole fractions are N 0.5194 0.5194 yCH4 = CH4 = = = 0.1754 N total 4 - 2´ 0.5194 2.961 yC =

NC 0.4806 = = 0.1623 N total 2.961

yH2 =

N H2 0.9612 = = 0.3246 N total 2.961

13-556

yN2 =

N N2 1 = = 0.3377 N total 2.961

17-35 A mixture of N 2 , O2 , and Ar is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1

The equilibrium composition consists of N 2 , O2 , Ar, and NO.

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

1 2

N2 + 12 O2 Û NO (thus n NO = 1, n N2 = 12 , and nO2 = 12 )

3 N 2 + O2 + 0.1Ar ¾ ¾ ®

Actual:

x NO + y N 2 + z O 2 + 0.1Ar prod.

N balance:

6 = x + 2y ¾ ¾ ®

y = 3 - 0.5 x

O balance:

2 = x + 2z ¾ ¾ ®

z = 1- 0.5 x

Total number of moles:

reactants

inert

Ntotal = x + y + z + 0.1 = 4.1

The equilibrium constant relation becomes, æ P K p = νN νO ççç 2 2 çN N N è νNO N NO

(ν

NO ö ÷ ÷ ÷ ÷ total ø

- νN 2 - νO2 )

1- 0.5- 0.5

x æ P ö ÷ ÷ = 0.5 0.5 ççç ÷ ÷ y z èç N total ø

From Table A-28, ln K p = - 3.019 at 2400 K. Thus K p = 0.04885. Substituting, 0.04885 =

x ´1 (3 - 0.5 x) (1- 0.5 x) 0.5 0.5

Solving for x, x = 0.0823 Then, y x = 2.9589 z x = 0.9589 Therefore, the equilibrium composition of the mixture at 2400 K and 10 atm is

0.0823NO + 2.9589N 2 + 0.9589O2 + 0.1Ar

13-557

P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "Stoichiometric reaction: 1/2 N2 + 1/2 O2 = NO" "Stoichiometric coefficients" nu_N2=1/2 nu_O2=1/2 nu_NO=1 "Actual reaction: 3 N2 + O2 + 0.1 Ar = N_NO NO + N_N2 N2 + N_O2 O2 + N_Ar AR" 6=N_NO+2*N_N2 "N balance" 2=N_NO+2*N_O2 "O balance" N_Ar=0.1 "N2 balance" N_total=N_NO+N_N2+N_O2+N_Ar K_p=N_NO^nu_NO/(N_N2^nu_N2*N_O2^nu_O2)*(P/N_total)^(nu_NO-nu_N2-nu_O2) "The equilibrium constant can also be expressed as" K_p=exp(-DELTAG/(R_u*T)) DELTAG=nu_NO*g_NO-nu_N2*g_N2-nu_O2*g_O2 "Gibbs functions" g_N2=h_N2-T*s_N2 g_O2=h_O2-T*s_O2 g_NO=h_NO-T*s_NO "Enthalpies" h_N2=enthalpy(N2, T=T) h_O2=enthalpy(O2, T=T) h_NO=enthalpy(NO, T=T) "Entropies" s_N2=entropy(N2, T=T, P=P0) s_O2=entropy(O2, T=T, P=P0) s_NO=entropy(NO, T=T, P=P0)

17-36 The mole fraction of sodium that ionizes according to the reaction Na  Na   e at 2000 K and 1.5 atm is to be determined. Assumptions All components behave as ideal gases.

Analysis The stoichiometric and actual reactions can be written as Stoichiometric: Actual:

Na Û Na + + e-

(thus n Na = 1, n Na+ = 1 and ne- = 1)

Na ¾ ¾ ® x Na + y Na + + y ereact.

products

Na balance: 1 = x + y or y = 1- x Total number of moles:

Ntotal = x + 2 y = 2 - x

The equilibrium constant relation becomes,

13-558

n Na ö Na+ N Na N e-e- æ çç P ÷ ÷ Kp = ÷ n Na çèç N ø ÷ N Na total

+ n - - n Na )

1+ 1- 1

ö y2 æ çç P ÷ ÷ ÷ ç ÷ x èç N total ø

Substituting, 0.668 =

ö (1- x) 2 æ çç 1.5 ÷ ÷ ç è ø x 2- x÷

Solving for x, x = 0.4449 Thus the fraction of Na which dissociates into Na  and e  is 1 − 0.4449 = 0.5551 or 55.5%

EES (Engineering Equation Solver) SOLUTION "Given" T=2000 [K] P=1.5 [atm] K_p=0.668 "Analysis" "Stoichiometric reaction: Na = Na(+) + e(-)" "Stoichiometric coefficients" nu_Na=1 nu_e=1 "Actual reaction: Na = x Na + y Na(+) + y e(-)" 1=x+y "Na balance" N_total=x+2*y K_p=(y^nu_Na*y^nu_e)/x^nu_Na*(P/N_total)^(nu_Na+nu_e-nu_Na) "The percentage of Na which dissociates into Na(+) and e(-) is" Percent_dissociate=(1-x)*Convert(, %)

17-37 The partial pressures of the constituents of an ideal gas mixture is given. The Gibbs function of the nitrogen in this mixture at the given mixture pressure and temperature is to be determined.

Analysis The partial pressure of nitrogen is

PN2 = 110 kPa = (110 /101.325) = 1.086 atm The Gibbs function of nitrogen at 293 K and 1.086 atm is

g (293 K, 1.086 atm) = g *(293 K, 1 atm) + RuT ln PN2 = 0 + (8.314 kJ/kmol.K)(293 K)ln(1.086 atm) = 200 kJ / kmol

13-559

EES (Engineering Equation Solver) SOLUTION "Given" T=(20+273) [K] P_N2=110 [kPa] P_CO2=230 [kPa]-P_N2 P_NO=(350-230) [kPa] "Analysis" P_N2_atm=P_N2*Convert(kPa, atm) R_u=8.314 [kJ/kmol-K] g=R_u*T*ln(P_N2_atm)

17-38E A steady-flow combustion chamber is supplied with CO and O2 . The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, and O2 .

The constituents of the mixture are ideal gases.

Analysis (a) We first need to calculate the amount of oxygen used per lbmol of CO before we can write the combustion equation,

v CO =

RT (0.3831 psia ×ft 3 /lbm ×R)(560 R) = = 13.41 ft 3 /lbm P 16 psia

mCO =

VCO 12.5 ft 3 /min = = 0.932 lbm/min v CO 13.41 ft 3 /lbm

Then the molar air-fuel ratio becomes (it is actually O2 -fuel ratio)

AF =

N O2 Nfuel

mO2 /M O2 mfuel /M fuel

(0.7 lbm/min)/(32 lbm/lbmol) = 0.657 lbmol O2 /lbmol fuel (0.932 lbm/min)/(28 lbm/lbmol)

Then the combustion equation can be written as

CO + 0.657O2 ¾ ¾ ®

xCO2 + (1- x)CO + (0.657 - 0.5x)O2

The equilibrium equation among CO2 , CO, and O2 can be expressed as

CO2 Û CO + 12 O2 (thus nCO2 = 1, nCO = 1, and nO2 = 12 ) and Kp =

n n CO N CO N OO2 2 æ çç P n CO2 ççè N N CO 2

ö÷ CO ÷ ÷ ÷ total ø

+ n O2 - n CO2 )

where N total = x + (1- x) + (0.657 - 0.5x) = 1.657 - 0.5 x

13-560

P = 16 /14.7 = 1.088 atm

From Table A-28, ln K p = - 6.635 at T = 3600 R = 2000 K. Thus K p = 1.314´ 10- 3. Substituting, 1.5- 1

ö (1- x)(0.657 - 0.5 x)0.5 æ çç 1.088 ÷ 1.314´ 10 = ÷ ç è1.657 - 0.5 x ø÷ x - 3

Solving for x, x = 0.9966 Then the combustion equation and the equilibrium composition can be expressed as

CO + 0.657O2 ¾ ¾ ® 0.9966CO2 + 0.0034CO + 0.1587O 2 and

0.9966CO2 + 0.0034CO + 0.1587O2 (b) The heat transfer for this combustion process is determined from the steady-flow energy balance Ein - Eout = D Esystem on the combustion chamber with W = 0,

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

h537 R

h560 R

h3600 R

Btu / lbmol

CO O2

−47,540 0

3725.1 3725.1

3889.5 ---

28,127 29,174

CO2

−169,300

4027.5

---

43,411

Substituting, - Qout = 0.9966(- 169,300 + 43, 411- 4027.5) + 0.0034(- 47,540 + 28,127 - 3725.1)

+ 0.1587(0 + 29,174 - 3725.1) - 1(- 47,540 + 3889.5 - 3725.1) - 0 = - 78,139 Btu/lbmol of CO or

Qout = 78,139 Btu/lbm of CO

The mass flow rate of CO can be expressed in terms of the mole numbers as

m 0.932 lbm/min = = 0.0333 lbmol/min M 28 lbm/lbmol

Thus the rate of heat transfer is

Qout = N ´ Qout = (0.0333 lbmol/min)(78,139 Btu/lbmol) = 2602 Btu / min

EES (Engineering Equation Solver) SOLUTION "Given" T1_CO=560 [R] P1_CO=16 [psia] V_dot_CO=12.5 [ft^3/min] T1_O2=537 [R] P1_O2=16 [psia] m_dot_O2=0.7 [lbm/min] T2=3600 [R] P2=16 [psia] T0=(77+460) [R]

13-561

P0=14.7 [psia] "Properties" R=R_u/MM_CO*Convert(Btu, psia-ft^3) R_u=1.986 [Btu/lbm-R] MM_O2=molarmass(O2) MM_CO=molarmass(CO) "Analysis" "(a)" v_CO=(R*T1_CO)/P1_CO m_dot_CO=V_dot_CO/v_CO AF_bar=N_dot_O2/N_dot_CO N_dot_O2=m_dot_O2/MM_O2 N_dot_CO=m_dot_CO/MM_CO a=AF_bar "Then the combustion equation becomes: CO + a O2 = x CO2 + (1-x) CO + (a-0.5*x) O2" "The equilibrium equation is: CO2 = CO + 1/2 O2" "Stoichiometric coefficients" nu_CO2=1 nu_CO=1 nu_O2=1/2 N_total=x+1-x+a-0.5*x P2_atm=P2*Convert(psia, atm) K_p=((1-x)^nu_CO*(a-0.5*x)^nu_O2)/x^nu_CO2*(P2_atm/N_total)^(nu_CO+nu_O2-nu_CO2) "The equilibrium constant can also be expressed as" K_p=exp(-DELTAG/(R_u*T2)) DELTAG=nu_CO*g_CO+nu_O2*g_O2-nu_CO2*g_CO2 "Gibbs functions" g_CO2=h_CO2-T2*s_CO2 g_CO=h_CO-T2*s_CO g_O2=h_O2-T2*s_O2 "Enthalpies" h_CO2=enthalpy(CO2, T=T2) h_CO=enthalpy(CO, T=T2) h_O2=enthalpy(O2, T=T2) "Entropies" s_CO2=entropy(CO2, T=T2, P=P0) s_CO=entropy(CO, T=T2, P=P0) s_O2=entropy(O2, T=T2, P=P0) "The equilibrium composition of the products" N_CO2_P=x N_CO_P=1-x N_O2_P=a-0.5*x "(b)" "Moles of reactants" N_CO_R=1 N_O2_R=a "Ideal gas enthalpies" h_CO_R=enthalpy(CO, T=T1_CO) h_O2_R=enthalpy(O2, T=T1_O2) h_CO_P=enthalpy(CO, T=T2) h_O2_P=enthalpy(O2, T=T2) h_CO2_P=enthalpy(CO2, T=T2) H_P=N_CO2_P*h_CO2_P+N_CO_P*h_CO_P+N_O2_P*h_O2_P H_R=N_CO_R*h_CO_R+N_O2_R*h_O2_R -Q_out=H_P-H_R Q_dot_out=m_dot_CO/MM_CO*Q_out

13-562

17-39 Liquid propane enters a combustion chamber. The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined. Assumptions 1

The equilibrium composition consists of CO2 , H 2 O , CO, N 2 , and O2 .

The constituents of the mixture are ideal gases.

Analysis (a) Considering 1 kmol of C3 H8 , the stoichiometric combustion equation can be written as

C3 H8 ( ) + ath (O2 + 3.76 N 2 ) ¾ ¾ ® 3CO2 + 4 H 2O+3.76ath N 2 where ath is the stoichiometric coefficient and is determined from the O2 balance,

2.5ath = 3 + 2 + 1.5ath ¾ ¾ ® ath = 5 Then the actual combustion equation with 150% excess air and some CO in the products can be written as

C3 H8 ( ) + 12.5(O2 + 3.76N 2 ) ¾ ¾ ® xCO2 + (3 - x)CO + (9 - 0.5x)O2 + 4H 2 O + 47N 2 After combustion, there will be no C3 H8 present in the combustion chamber, and H 2 O will act like an inert gas. The equilibrium equation among CO2 , CO, and O2 can be expressed as

CO2 Û CO + 12 O2 (thus nCO2 = 1, nCO = 1, and nO2 = 12 ) and n n CO N CO N OO2 2 æ çç P Kp = n CO2 ççè N N CO 2

CO ö ÷ ÷ ÷ ÷ total ø

+ n O2 - n CO2 )

where

Ntotal = x + (3 - x) + (9 - 0.5x) + 4 + 47 = 63 - 0.5 x From Table A-28, ln K p = - 17.871 at 1200 K. Thus K p = 1.73´ 10- 8. Substituting, 1.5- 1

1.73´ 10- 8 =

ö÷ (3 - x)(9 - 0.5 x)0.5 æ 2 çç ÷ ç è x 63 - 0.5 x ø÷

Solving for x,

x = 2.9999999 @3.0 Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss in accuracy. Then the combustion equation and the equilibrium composition can be expressed as

C3 H8 ( ) + 12.5(O2 + 3.76N 2 ) ¾ ¾ ® 3CO2 + 7.5O2 + 4H 2O + 47N 2 and

3CO2 + 7.5O2 + 4H 2O + 47N 2 (b) The heat transfer for this combustion process is determined from the steady-flow energy balance Ein - Eout = D Esystem on the combustion chamber with W = 0,

- Qout = å N P (h fo + h - h o ) - å N R (h fo + h - h o ) P

13-563

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The h f of liquid propane is obtained by adding the h fg at 25°C to h f of gaseous propane).

13-564

Substance

C3 H8 ( ) O2 N2 H2O ( g ) CO2

h285 K

h298 K

h1200 K

kJ / kmol

−118,910

---

8696.5

8682

38,447

8286.5

8669

36,777

−241,820

---

9904

44,380

−393,520

---

9364

53,848

Substituting,

- Qout = 3(- 393,520 + 53,848 - 9364) + 4(- 241,820 + 44,380 - 9904) + 7.5(0 + 38, 447 - 8682) + 47(0 + 36,777 - 8669) - 1(- 118,910 + h298 - h298 ) - 12.5(0 + 8296.5 - 8682) - 47(0 + 8186.5 - 8669) = - 185,764 kJ/kmol of C3 H8 or

Qout = 185,764 kJ/kmol of C3 H8 The mass flow rate of C3 H8 can be expressed in terms of the mole numbers as N=

m 1.2 kg/min = = 0.02727 kmol/min M 44 kg/kmol

Thus the rate of heat transfer is

Qout = N ´ Qout = (0.02727 kmol/min)(185,746 kJ/kmol) = 5066 kJ / min The equilibrium constant for the reaction 12 N 2 + 12 O 2 Û NO is ln K p  7.569 , which is very small. This indicates that the amount of NO formed during this process will be very small, and can be disregarded.

EES (Engineering Equation Solver) SOLUTION "Given" T1_fuel=(25+273) [K] m_dot_fuel=1.2 [kg/min] T1_air=(12+273) [K] ex=2.5 "250% theoretical air" T2=1200 [K] P2=2 [atm] T0=(25+273) [K] "standard temperature" P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] MM_air=molarmass(air) MM_C3H8=molarmass(C3H8) "Analysis" "(a)" "The stoichiometric combustion equation is: C3H8 + a_th (O2+3.76N2) = 3 CO2 + 4 H2O + 3.76*a_th N2" a_th=3+2 "O2 balance" "The actual combustion equation is: C3H8 + ex*a_th (O2+3.76N2) = x CO2 + (3-x) CO + 4 H2O + (9-0.5*x) O2 + ex*3.76*a_th N2" "The equilibrium equation is: CO2 = CO + 1/2 O2" "Stoichiometric coefficients" nu_CO2=1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-565

nu_CO=1 nu_O2=1/2 N_total=x+3-x+9-0.5*x+4+3.76*ex*a_th K_p=((3-x)^nu_CO*(9-0.5*x)^nu_O2)/x^nu_CO2*(P2/N_total)^(nu_CO+nu_O2-nu_CO2) "The equilibrium constant can also be expressed as" K_p=exp(-DELTAG/(R_u*T2)) DELTAG=nu_CO*g_CO+nu_O2*g_O2-nu_CO2*g_CO2 "Gibbs functions" g_CO2=h_CO2-T2*s_CO2 g_CO=h_CO-T2*s_CO g_O2=h_O2-T2*s_O2 "Enthalpies" h_CO2=enthalpy(CO2, T=T2) h_CO=enthalpy(CO, T=T2) h_O2=enthalpy(O2, T=T2) "Entropies" s_CO2=entropy(CO2, T=T2, P=P0) s_CO=entropy(CO, T=T2, P=P0) s_O2=entropy(O2, T=T2, P=P0) "The equilibrium composition is" N_CO2=x N_CO=3-x N_O2=9-0.5*x N_H2O=4 N_N2=3.76*ex*a_th "(b)" "Remaining mol numbers of reactants and products" N_C3H8=1 N_O2_R=ex*a_th N_O2_P=N_O2 "Enthalpy of formation data from Table A-26" h_f_C3H8_gas=enthalpy(C3H8, T=T0) h_f_C3H8=h_f_C3H8_gas-h_fg h_fg=h_g-h_f h_f=enthalpy(propane, T=T0, x=0) h_g=enthalpy(propane, T=T0, x=1) "Ideal gas enthalpies" h_O2_R=enthalpy(O2, T=T1_air) h_N2_R=enthalpy(N2, T=T1_air) h_O2_P=enthalpy(O2, T=T2) h_N2_P=enthalpy(N2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_CO2_P=enthalpy(CO2, T=T2) H_P=N_CO2*h_CO2_P+N_H2O*h_H2O+N_O2_P*h_O2_P+N_N2*h_N2_P H_R=N_C3H8*h_f_C3H8+N_O2_R*h_O2_R+N_N2*h_N2_R -Q_out=H_P-H_R Q_dot_out=m_dot_fuel/MM_C3H8*Q_out

17-40 Problem 17-39 is reconsidered. It is to be investigated if it is realistic to disregard the presence of NO in the product gases. Analysis The problem is solved using EES, and the solution is given below. "To solve this problem, the Gibbs function of the product gases is minimized. Click on the Min/Max icon." © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-566

"The adiabatic flame temperature for the complete combustion of C3H8(liq) for 250% theoretical air (150% excess) is 1286 K. For this problem at 1200 K the moles of CO are 0.000 and moles of NO are 0.000, thus we can disregard both the CO and NO. However, try some product temperatures above 1286 K and observe the sign change on the Q_out and the amout of CO and NO present as the product temperature increases." "The reaction of C3H8(liq) with excess air can be written: C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO The coefficients A_th and EX are the theoretical oxygen and the percent excess air on a decimal basis. Coefficients a, b, c, d, e, and f are found by minimiming the Gibbs Free Energy at a total pressure of the product gases P_Prod and the product temperature T_Prod. The equilibrium solution can be found by applying the Law of Mass Action or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are six compounds present in the products subject to four specie balances, so there are two degrees of freedom. Minimize the Gibbs function of the product gases with respect to two molar quantities such as coefficients b and f. The equilibrium mole numbers a, b, c, d, e, and f will be determined and displayed in the Solution window." PercentEx = 150 [%] Ex = PercentEx/100 P_prod =2*P_atm T_Prod=1200 [K] m_dot_fuel = 0.5 [kg/s] Fuel$='C3H8' T_air = (12+273) [K] T_fuel = (25+273) [K] P_atm = 101.325 [kPa] R_u=8.314 [kJ/kmol-K] "Theoretical combustion of C3H8 with oxygen: C3H8 + A_th O2 = 3 C02 + 4 H2O " 2*A_th = 3*2 + 4*1 "Balance the reaction for 1 kmol of C3H8" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" b_max = 3 f_max = (1+Ex)*A_th*3.76*2 e_guess=Ex*A_th 1*3 = a*1+b*1 "Carbon balance" 1*8=c*2 "Hydrogen balance" (1+Ex)*A_th*2=a*2+b*1+c*1+e*2+f*1 "Oxygen balance" (1+Ex)*A_th*3.76*2=d*2+f*1 "Nitrogen balance" "Total moles and mole fractions" N_Total=a+b+c+d+e+f y_CO2=a/N_Total; y_CO=b/N_Total; y_H2O=c/N_Total; y_N2=d/N_Total; y_O2=e/N_Total; y_NO=f/N_Total "The following equations provide the specific Gibbs function for each component as a function of its molar amount" g_CO2=Enthalpy(CO2,T=T_Prod)-T_Prod*Entropy(CO2,T=T_Prod,P=P_Prod*y_CO2) g_CO=Enthalpy(CO,T=T_Prod)-T_Prod*Entropy(CO,T=T_Prod,P=P_Prod*y_CO) g_H2O=Enthalpy(H2O,T=T_Prod)-T_Prod*Entropy(H2O,T=T_Prod,P=P_Prod*y_H2O) g_N2=Enthalpy(N2,T=T_Prod)-T_Prod*Entropy(N2,T=T_Prod,P=P_Prod*y_N2) g_O2=Enthalpy(O2,T=T_Prod)-T_Prod*Entropy(O2,T=T_Prod,P=P_Prod*y_O2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-567

g_NO=Enthalpy(NO,T=T_Prod)-T_Prod*Entropy(NO,T=T_Prod,P=P_Prod*y_NO) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance" Gibbs=a*g_CO2+b*g_CO+c*g_H2O+d*g_N2+e*g_O2+f*g_NO "For the energy balance, we adjust the value of the enthalpy of gaseous propane given by EES:" h_fg_fuel = 15060 [kJ/kmol] "Table A.27" h_fuel = enthalpy(Fuel$,T=T_fuel)-h_fg_fuel "Energy balance for the combustion process:" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" HR =Q_out+HP HR=h_fuel+ (1+Ex)*A_th*(enthalpy(O2,T=T_air)+3.76*enthalpy(N2,T=T_air)) HP=a*enthalpy(CO2,T=T_prod)+b*enthalpy(CO,T=T_prod)+c*enthalpy(H2O,T=T_prod)+d*enthalpy(N2,T=T_pro d)+e*enthalpy(O2,T=T_prod)+f*enthalpy(NO,T=T_prod) Q_dot_out=Q_out/molarmass(Fuel$)*m_dot_fuel "heat transfer rate" SOLUTION a=3.000 [kmol] A_th=5 b=0.000 [kmol] b_max=3 c=4.000 [kmol] d=47.000 [kmol] e=7.500 [kmol] Ex=1.5 e_guess=7.5 f=0.000 [kmol] Fuel$='C3H8' f_max=94 Gibbs=−17994897 [kJ] g_CO=−703496 [kJ/kmol]

g_CO2=−707231 [kJ/kmol] g_H2O=−515974 [kJ/kmol] g_N2=−248486 [kJ/kmol] g_NO=−342270 [kJ/kmol] g_O2=−284065 [kJ/kmol] HP=−330516.747 [kJ/kmol] HR=−141784.529 [kJ/kmol] h_fg_fuel=15060 [kJ/kmol] h_fuel=−118918 [kJ/kmol] m_dot_fuel=0.5 [kg/s] N_Total=61.5 [kmol/kmol_fuel] PercentEx=150 [%] P_atm=101.3 [kPa] P_prod=202.7 [kPa]

Q_dot_out=2140 [kW] Q_out=188732 [kJ/kmol_fuel] R_u=8.314 [kJ/kmol-K] T_air=285 [K] T_fuel=298 [K] T_Prod=1200.00 [K] y_CO=1.626E-15 y_CO2=0.04878 y_H2O=0.06504 y_N2=0.7642 y_NO=7.857E-08 y_O2=0.122

17-41 Oxygen is heated during a steady-flow process. The rate of heat supply needed during this process is to be determined for two cases. Assumptions 1

The equilibrium composition consists of O2 and O.

All components behave as ideal gases.

Analysis (a) Assuming some O2 dissociates into O, the dissociation equation can be written as

O2 ¾ ¾ ® x O2 + 2(1- x)O The equilibrium equation among O2 and O can be expressed as

13-568

O 2 Û 2O (thus n O2 = 1 and n O = 2) Assuming ideal gas behavior for all components, the equilibrium constant relation can be expressed as n - n O2

Kp =

öO N OnO æ çç P ÷ ÷ ÷ n ÷ N OO2 ççè N total ø 2

where

Ntotal = x + 2(1- x) = 2 - x

From Table A-28, ln K p = - 4.357 at 3000 K. Thus K p = 0.01282. Substituting, 2- 1

0.01282 =

ö (2 - 2 x)2 æ çç 1 ÷ ÷ ÷ ç è x 2- xø

Solving for x gives x = 0.943 Then the dissociation equation becomes

O2 ¾ ¾ ® 0.943O2 + 0.114O The heat transfer for this combustion process is determined from the steady-flow energy balance Ein - Eout = D Esystem on the combustion chamber with W = 0,

Qin = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables, Substance O O2

h298 K

h3000 K

kJ / kmol

249,190 0

6852 8682

63,425 106,780

Substituting,

Qin = 0.943(0 + 106,780 - 8682) + 0.114(249,190 + 63, 425 - 6852) - 0 = 127,363 kJ/kmol O2 The mass flow rate of O2 can be expressed in terms of the mole numbers as N=

m 0.5 kg/min = = 0.01563 kmol/min M 32 kg/kmol

Thus the rate of heat transfer is

Qin = N ´ Qin = (0.01563 kmol/min)(127,363 kJ/kmol) = 1990 kJ / min (b) If no O2 dissociates into O, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be

Qin = m(h2 - h1 ) = N (h2 - h1 ) = (0.01563 kmol/min)(106,780-8682) kJ/kmol = 1533 kJ / min

EES (Engineering Equation Solver) SOLUTION "Given" T1=298 [K] T2=3000 [K] P=1 [atm] m_dot=0.5 [kg/min] T0=(25+273) [K] "standard temperature" P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-569

MM_O2=molarmass(O2) "Analysis" "(a)" "The dissociation equation is: O2 = x O2 + 2(1-x) O" "The equilibrium equation is: O2 = 2O" "Stoichiometric coefficients" nu_O2=1 nu_O=2 N_total=x+2*(1-x) "From Table A-28, ln K_p = -4.357 at 3000 K. Thus" K_p=exp(-4.357) K_p=(2-2*x)^nu_O/x^nu_O2*(P/N_total)^(nu_O-nu_O2) "The equilibrium composition is" N_O2_P=x N_O_P=2-2*x N_O2_R=1 "Enthalpy of formation data from Table A-26" h_f_O=249190 h_f_O2=0 "Ideal gas enthalpies from the text tables" h_O2_R=8682 "at T1" h_O2_P=106780 "at T2" h_O_P=63425 "at T2" h_o_O2=8682 "at T0" h_o_O=6852 "at T0" H_P=N_O2_P*(h_f_O2+h_O2_P-h_o_O2)+N_O_P*(h_f_O+h_O_P-h_o_O) H_R=N_O2_R*(h_f_O2+h_O2_R-h_o_O2) Q_a=H_P-H_R Q_dot_a=m_dot/MM_O2*Q_a "(b)" Q_b=N_O2_R*(h_O2_P-h_O2_R) Q_dot_b=m_dot/MM_O2*Q_b

17-42 A constant volume tank contains a mixture of H 2 and O2 . The contents are ignited. The final temperature and pressure in the tank are to be determined. Analysis The reaction equation with products in equilibrium is

H 2 + O2 ¾ ¾ ® a H 2 + b H 2 O + c O2 The coefficients are determined from the mass balances Hydrogen balance:

2 = 2a + 2b

Oxygen balance:

2 = b + 2c

The assumed equilibrium reaction is

H 2 O ¬ ¾® H 2 + 0.5O2 The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * D G *(T ) = n H2 g H2 (Tprod ) + n O2 g O2 (Tprod ) - n H2O g H2O (Tprod )

and the Gibbs functions are given by * gH2 (Tprod ) = (h - Tprod s )H2

13-570

* gO2 (Tprod ) = (h - Tprod s )O2 * gH2O (Tprod ) = (h - Tprod s )H2O

The equilibrium constant is also given by 1+ 0.5- 1

Kp =

ö a1c 0.5 æ çç P ÷ ÷ ÷ 1 ç b çè N tot ÷ ø

0.5

ö ac 0.5 æ çç P2 /101.3÷ ÷ ÷ b èç a + b + c ø

An energy balance on the tank under adiabatic conditions gives

UR = UP where U R = 1(hH2@25°C - RuTreac ) + 1(hO2@25°C - RuTreac )

= 0 - (8.314 kJ/kmol.K)(298.15 K) + 0 - (8.314 kJ/kmol.K)(298.15 K) = - 4958 kJ/kmol U P = a(hH2@ Tprod - RuTprod ) + b(hH2O@ Tprod - RuTprod ) + c(hO2@ Tprod - RuTprod )

The relation for the final pressure is P2 =

ö æa + b + c ÷ öæ N tot Tprod çç Tprod ÷ ÷ P1 = çç (101.3 kPa) ÷ ÷ ÷ç ç è øèç 298.15 K ÷ N1 Treac 2 ø

Solving all the equations simultaneously using EES, we obtain the final temperature and pressure in the tank to be

Tprod = 3857 K P2 = 1043 kPa

EES (Engineering Equation Solver) SOLUTION "The product temperature isT_prod and pressure is P_2" "The reactant temperature is:" T_reac= (25+273) [K] "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 [kJ] P_1=101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The reaction equation with products in equilibrium:" "H2+ O2 =a H2 + b H2O + c O2 " "Hydrogen Balance:" 2=2*a + 2*b "Oxygen Balance:" 2=b + c*2 N_tot =a +b + c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is H2O=H2+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_H2+0.5*g_O2-1*g_H2O "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_2/101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*c^0.5)/(b^1)" sqrt(P/N_tot )*a*sqrt(c )=K_P *b © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-571

"Closed system, constant volume, adiabatic energy balance is:" U_R = Q_out+U_P U_R=1*(ENTHALPY(H2,T=T_reac)-R_u*T_reac) +1*(ENTHALPY(O2,T=T_reac)-R_u*T_reac) U_P=a*(ENTHALPY(H2,T=T_prod)-R_u*T_prod)+b*(ENTHALPY(H2O,T=T_prod)R_u*T_prod)+c*(ENTHALPY(O2,T=T_prod)-R_u*T_prod) P_2=N_tot/N_1*T_prod/T_reac*P_1 N_1=1+1

Simultaneous Reactions 17-43C It can be expresses as “ (dG )T , P  0 for each reaction.” Or as “the K p relation for each reaction must be satisfied.”

17-44C The number of K p relations needed to determine the equilibrium composition of a reacting mixture is equal to the difference between the number of species present in the equilibrium mixture and the number of elements.

17-45 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined.

Assumptions 1

The equilibrium composition consists of H 2 O , OH, O2 , and H 2 .

2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as

H2O ¾ ¾ ® x H 2 O + y H 2 + z O2 +w OH Mass balances for hydrogen and oxygen yield H balance: 2 = 2 x + 2 y + w (1) O balance:

1 = x + 2z + w

(2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the K p relations) to determine the equilibrium composition of the mixture. They are

H 2 O Û H 2 + 12 O 2

(reaction 1)

H 2 O Û 12 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be

ln K P1 = - 1.891 ¾ ¾ ®

K P1 = 0.15092

ln K P 2 = - 1.576 ¾ ¾ ®

K P 2 = 0.20680

13-572

The K p relations for these two simultaneous reactions are (n

K P1 =

n n ö H2 N HH2 2 N OO2 2 æ çç P ÷ ÷ ÷ n ÷ N HHO2O çèç N total ø

+ n O2 - n H 2O )

and

KP2 =

where

n n OH H2 æ P ö N HH2 2 N OH ÷ çç ÷ ÷ n ÷ N HHO2O èçç N total ø

+ n OH - n H 2O )

N total = N H2 O + N H2 + N O2 + N OH = x + y + z + w

Substituting, 1/ 2

0.15092 =

ö ( y )( z )1/ 2 æ 1 ÷ çç ÷ ÷ ç x è x + y + z + w÷ ø

0.20680 =

ö (w)(y )1/ 2 æ 1 ÷ çç ÷ ÷ ç x è x + y + z + w÷ ø

(3)

1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.574 y = 0.308 z = 0.095 w = 0.236 Therefore, the equilibrium composition becomes

0.574H2O + 0.308H2 + 0.095O2 + 0.236OH

EES (Engineering Equation Solver) SOLUTION "Given" T=3400 [K] P=1 [atm] "Analysis" "Actual reaction: H2O = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_OH OH" 2=2*N_H2O+2*N_H2+N_OH "H balance" 1=N_H2O+2*N_O2+N_OH "O balance" N_total=N_H2O+N_H2+N_O2+N_OH "Stoichiometric reaction 1: H2O = H2 + 1/2 O2" "Stoichiometric coefficients for reaction 1" nu_H2O_1=1 nu_H2_1=1 nu_O2_1=1/2 "Stoichiometric reaction 2: H2O = 1/2 H2 + OH" "Stoichiometric coefficients for reaction 2" nu_H2O_2=1 nu_H2_2=1/2 nu_OH_2=1 "The equilibrium constant for these two reactions at 3400 K are determined from Table A-28" K_p1=exp(-1.891) K_p2=exp(-1.576) "K_p relations are" K_p1=(N_H2^nu_H2_1*N_O2^nu_O2_1)/N_H2O^nu_H2O_1*(P/N_total)^(nu_H2_1+nu_O2_1-nu_H2O_1) K_p2=(N_H2^nu_H2_2*N_OH^nu_OH_2)/N_H2O^nu_H2O_2*(P/N_total)^(nu_H2_2+nu_OH_2-nu_H2O_2)

17-46E Two chemical reactions are occurring in air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1

The equilibrium composition consists of O2 , N 2 , O, and NO.

The constituents of the mixture are ideal gases.

13-573

Analysis The reaction equation during this process can be expressed as

¾¾ ®

O2 +3.76 N 2

x N 2 + y NO + z O2 + w O

Mass balances for nitrogen and oxygen yield N balance:

7.52 = 2x + y

(1)

O balance:

2 = y + 2z + w

(2)

N 2 + 12 O 2 Û NO

(reaction 1)

O2 Û 2O

(reaction 2)

1 2

The equilibrium constant for these two reactions at T = 5400 R = 3000 K are determined from Table A-28 to be

ln K P1 = - 2.114 ¾ ¾ ®

K P1 = 0.12075

ln K P 2 = - 4.357 ¾ ¾ ®

K P 2 = 0.01282

The K P relations for these two simultaneous reactions are K P1 =

æ P çç n N2 n O2 ç N N çè N n NO N NO

ö NO ÷ ÷ ÷ ÷ total ø

- n N 2 - n O2 )

n O - n O2

KP2 =

ö N OnO æ çç P ÷ ÷ ÷ n O2 ç ÷ N O èç N total ø 2

where

N total = N N2 + N NO + N O2 + N O = x + y + z + w

Substituting, 1- 0.5- 0.5

0.12075 =

ö y æ 1 ÷ ç ÷ ÷ 0.5 0.5 ç ÷ ç x z è x + y + z + wø

(3)

2- 1

ö w2 æ 1 ÷ çç ÷ 0.01282 = ÷ ÷ ç z è x + y + z + wø

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.658 y = 0.2048 z = 0.7868 Thus the equilibrium composition is

w = 0.2216

3.658N2 + 0.2048NO + 0.7868O2 + 0.2216O The equilibrium constant of the reaction N2 Û 2N at 5400 R is ln K P  22.359 , which is much smaller than the K P values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

13-574

EES (Engineering Equation Solver) SOLUTION T=5400 [R] P=1 [atm] "To solve without the diagram window, press F4 and minimize the variable Gibbs with respect to b and d." AO2=0.21; BN2=0.79 "Composition of air" AO2*2=a*2+b+d "Oxygen balance" BN2*2=c*2+d "Nitrogen balance" "The total moles at equilibrium are" N_tot=a+b+c+d y_O2=a/N_tot; y_O=b/N_tot; y_N2=c/N_tot; y_NO=d/N_tot "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T)-T*Entropy(O2,T=T,P=P*y_O2) g_N2=Enthalpy(N2,T=T)-T*Entropy(N2,T=T,P=P*y_N2) g_NO=Enthalpy(NO,T=T)-T*Entropy(NO,T=T,P=P*y_NO) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ so we must convert h and s to English units." T_K=T*Convert(R,K) "Convert R to K" Call JANAF('O',T_K:Cp`,h`,S`) "Units from JANAF are SI" S_O=S`*Convert(kJ/kgmole-K, Btu/lbmole-R) h_O=h`*Convert(kJ/kgmole, Btu/lbmole) "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_O=h_O-T*(S_O-R_u*ln(Y_O)) R_u=1.9858 [Btu/lbmol-R] "The universal gas constant" "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=a*g_O2+b*g_O+c*g_N2+d*g_NO

17-47E Problem 17-46E is reconsidered. Using appropriate software, the equilibrium solution is to be obtained by minimizing the Gibbs function by using the optimization capabilities built into the softare. This solution technique is to be compared with that used in the previous problem. Analysis The problem is solved using EES, and the solution is given below. "This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. 0.21 O2+0.79 N2 = a O2+b O + c N2 + d NO Two of the four coefficients, a, b, c, and d, are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 5400 R. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, select MinMax from the Calculate menu. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficients b and d. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses."

13-575

"Data input" T=3000 [R] P=1 [atm] AO2=0.21; BN2=0.79 "Composition of air" AO2*2=a*2+b+d "Oxygen balance" BN2*2=c*2+d "Nitrogen balance" "The total moles at equilibrium are" N_tot=a+b+c+d y_O2=a/N_tot; y_O=b/N_tot; y_N2=c/N_tot; y_NO=d/N_tot "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T)-T*Entropy(O2,T=T,P=P*y_O2) g_N2=Enthalpy(N2,T=T)-T*Entropy(N2,T=T,P=P*y_N2) g_NO=Enthalpy(NO,T=T)-T*Entropy(NO,T=T,P=P*y_NO) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ so we must convert h and s to English units." T_K=T*Convert(R,K) Call JANAF('O',T_K:Cp`,h`,S`) "Units from JANAF are SI" S_O=S`*Convert(kJ/kgmole-K, Btu/lbmole-R) h_O=h`*Convert(kJ/kgmole, Btu/lbmole) "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_O=h_O-T*(S_O-R_u*ln(Y_O)) R_u=1.9858 [Btu/lbmole-R] "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=a*g_O2+b*g_O+c*g_N2+d*g_NO d [lbmol]

b [lbmol]

0.002698 0.004616 0.007239 0.01063 0.01481 0.01972 0.02527 0.03132 0.03751 0.04361

0.00001424 0.00006354 0.0002268 0.000677 0.001748 0.004009 0.008321 0.01596 0.02807 0.04641

Gibbs

yO 2

y NO

yN 2

[Btu / lbmol] −162121 −178354 −194782 −211395 −228188 −245157 −262306 −279641 −297179 −314941

T [R]

0.2086 0.2077 0.2062 0.2043 0.2015 0.1977 0.1924 0.1849 0.1748 0.1613

0.0000 0.0001 0.0002 0.0007 0.0017 0.0040 0.0083 0.0158 0.0277 0.0454

0.0027 0.0046 0.0072 0.0106 0.0148 0.0197 0.0252 0.0311 0.0370 0.0426

0.7886 0.7877 0.7863 0.7844 0.7819 0.7786 0.7741 0.7682 0.7606 0.7508

3000 3267 3533 3800 4067 4333 4600 4867 5133 5400

13-576

Discussion The equilibrium composition in the above table are based on the reaction in which the reactants are 0.21 kmol O2 and 0.79 kmol N 2 . If you multiply the equilibrium composition mole numbers above with 4.76, you will obtain equilibrium composition for the reaction in which the reactants are 1 kmol O2 and 3.76 kmol N 2 .

17-48 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, O2 , and O.

The constituents of the mixture are ideal gases.

Analysis The reaction equation during this process can be expressed as

2CO2 + O2 ¾ ¾ ® x CO2 + y CO + z O2 + w O Mass balances for carbon and oxygen yield C balance: 2 = x + y (1) O balance:

6 = 2 x + y + 2 z + w (2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the K P relations) to determine the equilibrium composition of the mixture. They are

CO 2 Û CO + 12 O 2

(reaction 1)

O2 Û 2O

(reaction 2)

The equilibrium constant for these two reactions at 2800 K are determined from Table A-28 to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-577

ln K P1 = - 1.894 ¾ ¾ ® K P1 = 0.1505

ln K P 2 = - 5.826 ¾ ¾ ® K P 2 = 0.002950 The K P relations for these two simultaneous reactions are (n

K P1 =

n n CO ö CO N CO N OO2 2 æ çç P ÷ ÷ ÷ n ÷ N COCO2 çèç N total ø

+ n O2 - n CO2 )

n O - n O2

KP2 =

ö N OnO æ çç P ÷ ÷ ÷ n O2 ç ÷ N O èç N total ø 2

where

N total = N CO2 + N O2 + N CO + N O = x + y + z + w Substituting, 1/ 2

0.1505 =

ö ( y )( z )1/ 2 æ 4 ÷ çç ÷ ÷ ç x è x + y + z + w÷ ø

(3)

2- 1

ö w2 æ 4 ÷ çç ÷ 0.002950 = ÷ ÷ ç z è x + y + z + wø

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously using an equation solver such as EES for the four unknowns x, y, z, and w yields x = 1.773 y = 0.2266 z = 1.088 w = 0.05019 Thus the equilibrium composition is

1.773CO2 + 0.2226CO + 1.088O2 + 0.05019O

EES (Engineering Equation Solver) SOLUTION "Given" T=2800 [K] P=4 [atm] "The equilibrium constant for these two reactions at 2800 K are determined from Table A-28" K_p1=exp(-1.894) K_p2=exp(-5.826) "Analysis" "Actual reaction: 2 CO2 + O2 = N_CO2 CO2 + N_CO CO + N_O2 O2 + N_O O" 2=N_CO2+N_CO "C balance" 6=2*N_CO2+N_CO+2*N_O2+N_O "O balance" N_total=N_CO2+N_CO+N_O2+N_O "Stoichiometric reaction 1: CO2 = CO + 1/2 O2" "Stoichiometric coefficients for reaction 1" nu_CO2_1=1 nu_CO_1=1 nu_O2_1=1/2 "Stoichiometric reaction 2: O2 = 2 O" "Stoichiometric coefficients for reaction 2" nu_O2_2=1 nu_O_2=2 "K_p relations are" K_p1=(N_CO^nu_CO_1*N_O2^nu_O2_1)/N_CO2^nu_CO2_1*(P/N_total)^(nu_CO_1+nu_O2_1-nu_CO2_1) K_p2=N_O^nu_O_2/N_O2^nu_O2_2*(P/N_total)^(nu_O_2-nu_O2_2)

13-578

17-49 Water vapor is heated during a steady-flow process. The rate of heat supply for a specified exit temperature is to be determined for two cases. Assumptions 1

The equilibrium composition consists of H 2 O , OH, O2 , and H 2 .

The constituents of the mixture are ideal gases.

Analysis (a) Assuming some H 2 O dissociates into H 2 , O2 , and O, the dissociation equation can be written as

H2 O ¾ ¾®

x H2 O + y H 2 + z O2 + w OH

Mass balances for hydrogen and oxygen yield H balance: 2 = 2 x + 2 y + w (1) O balance:

1 = x + 2z + w

(2)

H 2 O Û H 2 + 12 O 2

(reaction 1)

H 2 O Û 12 H 2 + OH

(reaction 2)

The equilibrium constant for these two reactions at 3000 K are determined from Table A-28 to be

ln K P1 = - 3.086 ¾ ¾ ® K P1 = 0.04568 ln K P 2 = - 2.937 ¾ ¾ ® K P 2 = 0.05302 The K P relations for these three simultaneous reactions are K P1 =

n n N HH2 2 N OO2 2 æ çç P n H 2O ççè N N

ö H2 ÷ ÷ ÷ ÷ total ø

+ n O 2 - n H 2O )

n OH æ P N OH çç n H 2O ççè N N

+ n OH - n H 2O )

H2 O

KP2 =

n H2 H2

H2 O

ö H2 ÷ ÷ ÷ ÷ total ø

where

N total = N H2 O + N H2 + N O2 + N OH = x + y + z + w Substituting, 1/ 2

0.04568 =

ö ( y )( z )1/ 2 æ 1 ÷ çç ÷ ÷ x èç x + y + z + w ÷ ø

0.05302 =

ö ( w)( y )1/ 2 æ 1 ÷ çç ÷ ÷ ç x è x + y + z + w÷ ø

(3)

1/ 2

(4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 0.7835 y = 0.1622 z = 0.05396 w = 0.1086 Thus the balanced equation for the dissociation reaction is

13-579

The heat transfer for this dissociation process is determined from the steady-flow energy balance Ein - Eout = D Esystem with W = 0,

Qin = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables, Substance

H2 O H2 O2 OH

h298 K

h3000 K

kJ / kmol

−241,820

9904

136,264

8468

97,211

8682

106,780

39,460

9188

98,763

Substituting,

Qin = 0.7835(- 241,820 + 136, 264 - 9904) + 0.1622(0 + 97, 211- 8468) + 0.05396(0 + 106,780 - 8682) + 0.1086(39, 460 + 98,763 - 9188) - (- 241,820) = 185,058 kJ/kmol H2 O The mass flow rate of H 2 O can be expressed in terms of the mole numbers as N=

m 0.2 kg/min = = 0.01111 kmol/min M 18 kg/kmol

Thus,

Qin = N ´ Qin = (0.01111 kmol/min)(185,058 kJ/kmol) = 2056 kJ / min (b) If no dissociates takes place, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be

Qin = m(h2 - h1 ) = N (h2 - h1 ) = (0.01111 kmol/min)(136,264 - 9904) kJ/kmol = 1404 kJ / min

EES (Engineering Equation Solver) SOLUTION "Given" T1=298 [K] T2=3000 [K] P=1 [atm] m_dot=0.2 [kg/min] T0=298 [K] "The equilibrium constant for these two reactions at 2400 K are determined from Table A-28" K_p1=exp(-3.086) K_p2=exp(-2.937) "Properties" MM_H2O=molarmass(H2O) "Analysis" "(a)" "Actual reaction: H2O = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_OH OH" 2=2*N_H2O+2*N_H2+N_OH "H balance" 1=N_H2O+2*N_O2+N_OH "O balance"

13-580

N_total=N_H2O+N_H2+N_O2+N_OH "Stoichiometric reaction 1: H2O = H2 + 1/2 O2" "Stoichiometric coefficients for reaction 1" nu_H2O_1=1 nu_H2_1=1 nu_O2_1=1/2 "Stoichiometric reaction 2: H2O = 1/2 H2 + OH" "Stoichiometric coefficients for reaction 2" nu_H2O_2=1 nu_H2_2=1/2 nu_OH_2=1 "K_p relations are" K_p1=(N_H2^nu_H2_1*N_O2^nu_O2_1)/N_H2O^nu_H2O_1*(P/N_total)^(nu_H2_1+nu_O2_1-nu_H2O_1) K_p2=(N_H2^nu_H2_2*N_OH^nu_OH_2)/N_H2O^nu_H2O_2*(P/N_total)^(nu_H2_2+nu_OH_2-nu_H2O_2) "Enthalpy of formation data from Table A-26" h_f_OH=39460 "Enthalpies of products" h_H2O_R=enthalpy(H2O, T=T1) h_H2O_P=enthalpy(H2O, T=T2) h_H2=enthalpy(H2, T=T2) h_O2=enthalpy(O2, T=T2) h_OH=98763 "at T2 from the ideal gas tables in the text" "Standard state enthalpies" h_o_OH=9188 "at T0 from the ideal gas tables in the text" "Heat transfer" H_P=N_H2O*h_H2O_P+N_H2*h_H2+N_O2*h_O2+N_OH*(h_f_OH+h_OH-h_o_OH) H_R=N_H2O_R*h_H2O_R N_H2O_R=1 Q_in_a=H_P-H_R Q_dot_in_a=(m_dot/MM_H2O)*Q_in_a "(b)" Q_in_b=N_H2O_R*(h_H2O_P-h_H2O_R) Q_dot_in_b=(m_dot/MM_H2O)*Q_in_b

17-50 studied.

Problem 17-49 is reconsidered. The effect of the pressure on the rate of heat supplied for the two cases is to be

Analysis The problem is solved using EES, and the solution is given below. "Given" T1=298 [K] T2=3000 [K] P=1 [atm] m_dot=0.2 [kg/min] T0=298 [K] MM_H2O=molarmass(H2O) "The equilibrium constant for these two reactions at 3000 K are determined from Table A-28" K_p1=exp(-3.086) K_p2=exp(-2.937) "Actual reaction: H2O = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_OH OH" 2=2*N_H2O+2*N_H2+N_OH "H balance" 1=N_H2O+2*N_O2+N_OH "O balance" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-581

N_total=N_H2O+N_H2+N_O2+N_OH "Stoichiometric reaction 1: H2O = H2 + 1/2 O2" "Stoichiometric coefficients for reaction 1" nu_H2O_1=1 nu_H2_1=1 nu_O2_1=1/2 "Stoichiometric reaction 2: H2O = 1/2 H2 + OH" "Stoichiometric coefficients for reaction 2" nu_H2O_2=1 nu_H2_2=1/2 nu_OH_2=1 "K_p relations are" K_p1=(N_H2^nu_H2_1*N_O2^nu_O2_1)/N_H2O^nu_H2O_1*(P/N_total)^(nu_H2_1+nu_O2_1-nu_H2O_1) K_p2=(N_H2^nu_H2_2*N_OH^nu_OH_2)/N_H2O^nu_H2O_2*(P/N_total)^(nu_H2_2+nu_OH_2-nu_H2O_2) "Enthalpy of formation data from Table A-26" h_f_OH=39460 "Enthalpies of products" h_H2O_R=enthalpy(H2O, T=T1) h_H2O_P=enthalpy(H2O, T=T2) h_H2=enthalpy(H2, T=T2) h_O2=enthalpy(O2, T=T2) h_OH=98763 "at T2 from the ideal gas tables in the text" h_o_OH=9188 "at T0 from the ideal gas tables in the text" "Heat transfer" H_P=N_H2O*h_H2O_P+N_H2*h_H2+N_O2*h_O2+N_OH*(h_f_OH+h_OH-h_o_OH) H_R=N_H2O_R*h_H2O_R N_H2O_R=1 Q_in_a=H_P-H_R Q_dot_in_a=(m_dot/MM_H2O)*Q_in_a Q_in_b=N_H2O_R*(h_H2O_P-h_H2O_R) Q_dot_in_b=(m_dot/MM_H2O)*Q_in_b

P [atm] 1 2 3 4 5 6 7 8 9 10

Qin,Dissoc

Qin,NoDissoc

[kJ/min]

2066 1946 1885 1846 1818 1796 1778 1763 1751 1740

1417 1417 1417 1417 1417 1417 1417 1417 1417 1417

13-582

17-51 Ethyl alcohol C2 H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted. Analysis The complete combustion reaction in this case can be written as

C2 H5 OH (gas) + (1 + Ex)ath [O 2 + 3.76N 2 ]¾ ¾ ® 2 CO 2 + 3 H 2 O + ( Ex)( ath ) O 2 + f N 2 where ath is the stoichiometric coefficient for air. The oxygen balance gives

1 + (1 + Ex)ath ´ 2 = 2´ 2 + 3´ 1 + ( Ex)(ath )´ 2 The reaction equation with products in equilibrium is

C2 H5 OH (gas) + (1 + Ex)ath [O 2 + 3.76N 2 ]¾ ¾ ® a CO 2 + b CO + d H 2 O + e O 2 + f N 2 + g NO The coefficients are determined from the mass balances Carbon balance:

2= a+ b

Hydrogen balance:

6 = 2d ¾ ¾ ® d= 3

Oxygen balance:

1+ (1 + Ex)ath ´ 2 = a´ 2 + b + d + e´ 2 + g

Nitrogen balance:

(1+ Ex)ath ´ 3.76´ 2 = f ´ 2 + g

Solving the above equations, we find the coefficients to be Ex = 0.4, ath  3 , a = 1.995, b = 0.004712, d = 3, e = 1.17, f = 15.76, g = 0.06428 Then, we write the balanced reaction equation as C2 H 5 OH (gas) + 4.2 [O 2 + 3.76N 2 ] ¾¾ ® 1.995 CO 2 + 0.004712 CO + 3 H 2 O + 1.17 O 2 + 15.76 N 2 + 0.06428 NO

Total moles of products at equilibrium are

Ntot = 1.995 + 0.004712 + 3 + 1.17 + 15.76 = 21.99 The first assumed equilibrium reaction is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-583

CO2 ¬ ¾® CO + 0.5O2 The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

æ- D G1 *(Tprod ) ÷ ö ÷ K p1 = exp ççç ÷ ÷ ÷ RuTprod èç ø Where

* * * D G1 *(Tprod ) = n CO g CO (Tprod ) + n O2 g O2 (Tprod ) - n CO2 g CO2 (Tprod )

and the Gibbs functions are defined as * gCO (Tprod ) = (h - Tprod s )CO * gO2 (Tprod ) = (h - Tprod s )O2 * gCO2 (Tprod ) = (h - Tprod s )CO2

The equilibrium constant is also given by 1+ 0.5- 1

K p1 =

ö be0.5 æ çç P ÷ ÷ ÷ ç ÷ a èç N tot ø

0.5

ö (0.004712)(1.17)0.5 æ çç 1 ÷ = 0.0005447 ÷ ÷ ç è 21.99 ø 1.995

The second assumed equilibrium reaction is

0.5N 2 + 0.5O2 ¬ ¾® NO Also, for this reaction, we have * g NO (Tprod ) = (h - Tprod s ) NO * g N2 (Tprod ) = (h - Tprod s ) N2 * gO2 (Tprod ) = (h - Tprod s )O2

* * * D G2 *(Tprod ) = n NO g NO (Tprod ) - n N2 g N2 (Tprod ) - n O2 g O2 (Tprod )

æ- D G2 *(Tprod ) ÷ ö ÷ K p 2 = exp ççç ÷ ÷ Ru Tprod èç ø÷ 1- 0.5- 0.5

æP ÷ ö ÷ K p 2 = ççç ÷ çè N tot ÷ ø

æ 1 ÷ ö g 0.06428 = çç = 0.01497 ÷ ø (1.17)0.5 (15.76)0.5 eo.5 f 0.5 èç 21.99 ÷

A steady flow energy balance gives

HR = HP where H R = hfuel@25°C + 4.2hO2@25°C + 15.79hN2@25°C

= (- 235,310 kJ/kmol) + 4.2(0) + 15.79(0) = - 235,310 kJ/kmol H P = 1.995hCO2@Tprod + 0.004712hCO@Tprod + 3hH2O@Tprod + 1.17 hO2@Tprod + 15.76hN2@Tprod + 0.06428hNO@Tprod

Solving the energy balance equation using EES, we obtain the adiabatic flame temperature

Tprod = 1901 K The copy of entire EES solution including parametric studies is given next:

EES (Engineering Equation Solver) SOLUTION T_reac= (25+273) [K] "The reactant temperature" Q_out = 0 [kJ] "For adiabatic combustion of 1 kmol of fuel" PercentEx = 40 "Percent excess air" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-584

Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O + Ex*A_th O2 + f N2 " 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "Oxygen Balance for complete combustion" "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2 + g NO" 2=a + b "Carbon Balance" 6=2*d "Hydrogen Balance" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 +g "Oxygen Balance" (1+Ex)*A_th*3.76 *2= f*2 + g "Nitrogen Balance" N_tot =a +b + d + e + f +g "Total kilomoles of products at equilibrium" "The first assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 "The standard-state Gibbs function" K_P_1 = exp(-DELTAG_1 /(R_u*T_prod )) "The equilibrium constant" P=P_prod /101.3 "The equilibrium constant is also given by K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot) *b *sqrt(e) =K_P_1*a "The second assumed equilibrium reaction is 0.5N2+0.5O2=NO" g_NO=Enthalpy(NO,T=T_prod )-T_prod *Entropy(NO,T=T_prod ,P=101.3) g_N2=Enthalpy(N2,T=T_prod )-T_prod *Entropy(N2,T=T_prod ,P=101.3) DELTAG_2 =1*g_NO-0.5*g_O2-0.5*g_N2 "The standard-state Gibbs function" K_P_2 = exp(-DELTAG_2 /(R_u*T_prod )) "The equilibrium constant" "The equilibrium constant is also given by K_ P_2 = (P/N_tot)^(1-0.5-0.5)*(g^1)/(e^0.5*f^0.5)" g=K_P_2 *sqrt(e*f) H_R = Q_out+H_P "The steady-flow energy balance" h_bar_f_C2H5OHgas=-235310 [kJ/kmol] H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) +g*ENTHALPY(NO,T=T_prod)

a th

PercentEx [%]

3 3 3 3 3 3 3 3

1.922 1.971 1.988 1.995 1.998 1.999 2 2

0.07779 0.0293 0.01151 0.004708 0.001993 0.0008688 0.0003884 0.0001774

3 3 3 3 3 3 3 3

0.3081 0.5798 0.8713 1.17 1.472 1.775 2.078 2.381

12.38 13.5 14.63 15.76 16.89 18.02 19.15 20.28

0.0616 0.06965 0.06899 0.06426 0.05791 0.05118 0.04467 0.03867

10 20 30 40 50 60 70 80

Tprod [K] 2184 2085 1989 1901 1820 1747 1682 1621

13-585

3 3

2 2

Variations of

0.00008262 0.00003914

3 3

2.683 2.986

21.42 22.55

0.0333 0.02856

90 100

1566 1516

with Temperature

17-52C It enables us to determine the enthalpy of reaction hR from a knowledge of equilibrium constant K P .

17-53C At 2000 K since combustion processes are exothermic, and exothermic reactions are more complete at lower temperatures.

17-54 The hR value for the dissociation process O2  2O at a specified temperature is to be determined using enthalpy and K P data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The dissociation equation of O2 can be expressed as

O2 Û 2O The hR of the dissociation process of O2 at 3100 K is the amount of energy absorbed or released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 3100 K, and can be determined from

hR = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables,

13-586

Substance

O O2

h298 K

h2900 K

kJ / kmol

249,190 0

6852 8682

65,520 110,784

Substituting,

hR = 2(249,190 + 65,520 - 6852) - 1(0 + 110, 784 - 8682) = 513, 614 kJ / kmol

(b) The hR value at 3100 K can be estimated by using K P values at 3000 K and 3200 K (the closest two temperatures to 3100 K for which K P data are available) from Table A-28, ln

KP2 h æ1 1 ö h æ1 1 ö ÷ ÷ ÷ ÷ @ R ççç or ln K P 2 - ln K P1 @ R ççç ÷ ÷ ÷ ÷ K P1 Ru çèT1 T2 ø Ru èçT1 T2 ø

- 3.072 - (- 4.357) @

æ 1 ö hR 1 ÷ çç ÷ ÷ 8.314 kJ/kmol ×K çè3000 K 3200 K ø

hR @512, 808 kJ / kmol

EES (Engineering Equation Solver) SOLUTION "Given" T=3100 [K] T0=298 [K] "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "(a)" "The combustion equation: O2 = 2 O" "Mol numbers" N_O2=1 N_O=2 "Enthalpy of formation data from Table A-26" h_f_O2=0 [kJ/kmol] h_f_O=249190 [kJ/kmol] "Enthalpies at T" h_O2=enthalpy(O2, T=T) h_O=65520 "at T from ideal gas tables in the text" "Standard state enthalpies" h_o_O2=enthalpy(O2, T=T0) h_o_O=6852 [kJ/kmol] "at T0 from ideal gas tables in the text" "Enthalpy of reaction" h_bar_R_a=N_O*(h_f_O+h_O-h_o_O)-N_O2*(h_f_O2+h_O2-h_o_O2) "(b)" "The equilibrium constants at T1 = 2800 K and T2 = 3000 K, from Table A-28" T1=3000 [K] T2=3200 [K] K_p1=exp(-4.357) K_p2=exp(-3.072) ln(K_p2/K_p1)=h_bar_R_b/R_u*(1/T1-1/T2) 17-55E The hR at a specified temperature is to be determined using the enthalpy and K P data. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-587

Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of CO can be expressed as

CO + 12 O 2 Û CO 2 The hR of the combustion process of CO at 3960 R is the amount of energy released as one kmol of H 2 is burned in a steady-flow combustion chamber at a temperature of 3960 R, and can be determined from

hR = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables,

Substance

CO2 CO O2

h537 R

h3960 R

Btu / lbmol

−169,300

4027.5

48,647

−47,540 0

3725.1 3725.1

31,256.5 32,440.5

Substituting,

hR = 1(- 169,300 + 48, 647 - 4027.5) - 1(- 47,540 + 31, 256.5 - 3725.1)

- 0.5(0 + 32, 440.5 - 3725.1) = - 119, 030 Btu / lbmol

(b) The hR value at 3960 R can be estimated by using K P values at 3600 R and 4320 R (the closest two temperatures to 3960 R for which K P data are available) from Table A-28, ln

KP2 h æ1 1 ö h æ1 1 ö ÷ ÷ ÷ ÷ @ R ççç or ln K P 2 - ln K P1 @ R ççç ÷ ÷ ÷ ÷ K P1 Ru çèT1 T2 ø Ru èçT1 T2 ø

3.860 - 6.635 @

æ 1 ö hR 1 ÷ çç ÷ 1.986 Btu/lbmol ×R çè3600 R 4320 R ø÷

hR @- 119, 041 Btu / lbmol

EES (Engineering Equation Solver) SOLUTION "Given" T=3960 [R] T0=537 [R] "Properties" R_u=1.986 [Btu/lbmol-R] "Analysis" "(a)" "The combustion equation: CO + 1/2 O2 = CO2" "Mol numbers" N_CO=1 N_O2=1/2 N_CO2=1 "Enthalpies at T" h_CO=enthalpy(CO, T=T) h_O2=enthalpy(O2, T=T) h_CO2=enthalpy(CO2, T=T) "Enthalpy of reaction" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-588

h_bar_R_a=N_CO2*h_CO2-N_CO*h_CO-N_O2*h_O2 "(b)" "The equilibrium constants at T1 = 3600 R and T2 = 4320 R, from Table A-28" T1=3600 [R] T2=4320 [R] K_p1=exp(6.635) K_p2=exp(3.860) ln(K_p2/K_p1)=h_bar_R_b/R_u*(1/T1-1/T2)

17-56 The K P value of the combustion process H2  1/ 2O2  H2 O is to be determined at a specified temperature using

hR data and K P value. Assumptions Both the reactants and products are ideal gases. Analysis The hR and K P data are related to each other by ln

ö KP2 h æ1 1 ö hR æ ÷ ç1 1 ÷ ÷ ÷ @ R ççç ÷ or ln K P 2 - ln K P1 @ ççç ÷ ÷ ÷ ç K P1 Ru èT1 T2 ø Ru èT1 T2 ø

The hR of the specified reaction at 3000 K is the amount of energy released as one kmol of H 2 is burned in a steady-flow combustion chamber at a temperature of 3000 K, and can be determined from

hR = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the H 2 O , H 2 and O2 to be ideal gases, we have h = h (T). From the tables, Substance

H2 O H2 O2

h298 K

kJ / kmol

−241,820

9904

136,264

8468

97,211

8682

106,780

h3000 K

Substituting,

hR = 1(- 241,820 + 136, 264 - 9904) - 1(0 + 97, 211- 8468)

- 0.5(0 + 106,780 - 8682)

= - 253, 252 kJ/kmol The K P value at 3200 K can be estimated from the equation above by using this hR value and the K P value at 2800 K which is ln K P1  3.812 or K P1  45.24 , ln( K P 2 / 45.24) @

ö - 253, 252 kJ/kmol æ 1 ÷ çç 1 ÷ ÷ ç 8.314 kJ/kmol ×K è 2800 K 3200 K ø

ln K P 2 = 2.452

(Table A-28: ln K P 2 = 2.451)

K P 2 = 11.61

13-589

T=3000 [K] T2=3200 [K] T0=298 [K] "The equilibrium constant at T1 = 2800 K, from Table A-28" T1=2800 [K] K_p1=exp(3.812) "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "The combustion equation: H2 + 1/2 O2 = H2O" "Mol numbers" N_H2=1 N_O2=1/2 N_H2O=1 "Enthalpies at T" h_H2O=enthalpy(H2O, T=T) h_H2=enthalpy(H2, T=T) h_O2=enthalpy(O2, T=T) "Enthalpy of reaction" h_bar_R=-253252 "N_H2O*h_H2O-N_H2*h_H2-N_O2*h_O2" ln(K_p2/K_p1)=h_bar_R/R_u*(1/T1-1/T2)

17-57 The hR value for the dissociation process CO2  CO  1/ 2O2 at a specified temperature is to be determined using enthalpy and K p data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The dissociation equation of CO2 can be expressed as

CO 2 Û CO+ 12 O 2 The hR of the dissociation process of CO2 at 2200 K is the amount of energy absorbed or released as one kmol of CO2 dissociates in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from

hR = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables, Substance

CO2 CO O2

h298 K

h2200 K

kJ / kmol −393,520

kJ / kmol 9364

kJ / kmol 112,939

−110,530 0

8669 8682

72,688 75,484

Substituting,

hR = 1(- 110,530 + 72, 688 - 8669) + 0.5(0 + 75, 484 - 8682) - 1(- 393,520 + 112,939 - 9364) = 276, 835 kJ / kmol

(b) The hR value at 2200 K can be estimated by using K P values at 2000 K and 2400 K (the closest two temperatures to 2200 K for which K P data are available) from Table A-28, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-590

KP2 h æ1 1 ö h æ1 1 ö ÷ ÷ ÷ ÷ @ R ççç or ln K P 2 - ln K P1 @ R ççç ÷ ÷ ÷ ÷ K P1 Ru çèT1 T2 ø Ru èçT1 T2 ø

- 3.860 - (- 6.635) @

æ 1 ö hR 1 ÷ çç ÷ ÷ 8.314 kJ/kmol ×K çè 2000 K 2400 K ø

hR @276, 856 kJ / kmol

EES (Engineering Equation Solver) SOLUTION "Given" T=2200 [K] T0=298 [K] "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "(a)" "The combustion equation: CO2 = CO + 1/2 O2" "Mol numbers" N_CO2=1 N_CO=1 N_O2=1/2 "Enthalpies at T" h_CO2=enthalpy(CO2, T=T) h_CO=enthalpy(CO, T=T) h_O2=enthalpy(O2, T=T) "Enthalpy of reaction" h_bar_R_a=N_CO*h_CO+N_O2*h_O2-N_CO2*h_CO2 "(b)" "The equilibrium constants at T1 = 2000 K and T2 = 2400 K, from Table A-28" T1=2000 [K] T2=2400 [K] K_p1=exp(-6.635) K_p2=exp(-3.860) ln(K_p2/K_p1)=h_bar_R_b/R_u*(1/T1-1/T2)

17-58

The enthalpy of reaction for the equilibrium reaction CH4  2O2  CO2  2H2 O at 2500 K is to be estimated

using enthalpy data and equilibrium constant, K p data. Analysis For this problem we use enthalpy and Gibbs function properties from EES. The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * * D G *(T ) = nCO2 gCO2 (T ) + n H2O g H2O (T ) - nCH4 gCH4 (T ) - n O2 g O2 (T )

At T1  2500  10  2490 K : * * * * D G1 *(T ) = nCO2 gCO2 (T1 ) + n H2O g H2O (T1 ) - nCH4 gCH4 (T1 ) - nO2 gO2 (T1 )

= 1(- 1,075, 278) + 2(- 831,031) - 1(- 722,303) - 2(- 611,914)

13-591

At T2  2500  10  2510 K : * * * * D G2 *(T ) = nCO2 gCO2 (T2 ) + n H2O g H2O (T2 ) - nCH4 gCH4 (T2 ) - n O2 g O2 (T2 )

= 1(- 1,081,734) + 2(- 836,564) - 1(- 728,962) - 2(- 617, 459) = - 790,981 kJ/kmol The Gibbs functions are obtained from enthalpy and entropy properties using EES. Substituting,

æ ö - 791, 208 kJ/kmol ÷ ÷ K p1 = exp çç= 3.966´ 1016 ÷ çè (8.314 kJ/kmol ×K)(2490 K) ø÷

æ ö - 790,981 kJ/kmol ÷ ÷ K p 2 = exp çç= 2.893´ 1016 ÷ èç (8.314 kJ/kmol ×K)(2510 K) ÷ ø The enthalpy of reaction is determined by using the integrated van't Hoff equation:

æK p 2 ö÷ h æ1 1 ö ÷ ÷= R çç ÷ ln ççç ÷ ÷ çè K p1 ø÷ ÷ Ru çèçT1 T2 ÷ ø æ2.893´ 1016 ö æ 1 ö hR 1 ÷ ÷ çç ÷= ln ççç ¾¾ ® hR = - 819,472 kJ / kmol ÷ 16 ÷ ÷ è3.966´ 10 ÷ ø 8.314 kJ/kmol.K èç 2490 K 2510 K ø The enthalpy of reaction can also be determined from an energy balance to be

hR = H P - H R where

H R = 1hCH4 @ 2500 K + 2hO2 @ 2500 K = 106,833 + 2(78, 437) = 263,708 kJ/kmol H P = 1hCO2 @ 2500 K + 2hH2O @ 2500 K = (- 271,531) + 2(- 142,117) = - 555,765 kJ/kmol The enthalpies are obtained from EES. Substituting,

hR = H P - H R = (- 555, 765) - (263, 708) = - 819,472 kJ / kmol which is identical to the value obtained using K p data.

EES (Engineering Equation Solver) SOLUTION T_reac= 2500 [K] DELTAT=10 [K] R_u=8.314 [kJ/kmol-K] "The assumed equilibrium reaction is CH4 + 2 O2 = CO2 + 2 H2O" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_reac, at 1 atm pressure, 101.3 kPa" "First find K_P_1 at TmDT" TmDT=T_reac-DELTAT g_CO2_1=Enthalpy(CO2,T=TmDT )-(TmDT) *Entropy(CO2,T=TmDT ,P=101.3) g_O2_1=Enthalpy(O2,T=TmDT )-(TmDT) *Entropy(O2,T=TmDT ,P=101.3) g_CH4_1=Enthalpy(CH4,T=TmDT )-(TmDT) *Entropy(CH4,T=TmDT ,P=101.3) g_H2O_1=Enthalpy(H2O,T=TmDT )-(TmDT) *Entropy(H2O,T=TmDT ,P=101.3) DELTAG_1 =1*g_CO2_1+2*g_H2O_1-1*g_CH4_1-2*g_O2_1 "The standard-state Gibbs function" K_P_1 = exp(-DELTAG_1 /(R_u*TmDT )) "The equilibrium constant" "Second find K_P_2 at TpDT" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-592

TpDT=T_reac+DELTAT g_CO2_2=Enthalpy(CO2,T=TpDT )-(TpDT) *Entropy(CO2,T=TpDT ,P=101.3) g_O2_2=Enthalpy(O2,T=TpDT )-(TpDT) *Entropy(O2,T=TpDT ,P=101.3) g_CH4_2=Enthalpy(CH4,T=TpDT )-(TpDT) *Entropy(CH4,T=TpDT ,P=101.3) g_H2O_2=Enthalpy(H2O,T=TpDT )-(TpDT) *Entropy(H2O,T=TpDT ,P=101.3) DELTAG_2 =1*g_CO2_2+2*g_H2O_2-1*g_CH4_2-2*g_O2_2 K_P_2 = exp(-DELTAG_2 /(R_u*TpDT )) "Estimate the enthalpy of reaction by using the integrated van't Hoff equation." ln(K_P_2/K_P_1)=h_bar_Rkp/R_u*(1/(TmDT)-1/(TpDT)) "The enthalpy of reaction is obtained from the energy balance" h_bar_Reng = H_P-H_R H_R= 1*h_CH4+2*h_O2 H_P=1*h_CO2+2*h_H2O h_CH4=ENTHALPY(CH4,T=T_reac) h_O2=ENTHALPY(O2,T=T_reac) h_CO2=ENTHALPY(CO2,T=T_reac) h_H2O=ENTHALPY(H2O,T=T_reac) SOLUTION DELTAG_1=−791208 [kJ/kmol] DELTAT=10 [K] g_CH4_2=−728962 [kJ/kmol] g_CO2_2=−1081734 [kJ/kmol] g_H2O_2=−836564 [kJ/kmol] g_O2_2=−617459 [kJ/kmol] h_bar_Rkp=−819472 [kJ/kmol] h_CO2=−271531 [kJ/kmol] h_O2=78437 [kJ/kmol] H_R=263708 [kJ/kmol] K_P_2=2.893E+16 TmDT=2490 [K] T_reac=2500 [K]

DELTAG_2=−790981 [kJ/kmol] g_CH4_1=−722303 [kJ/kmol] g_CO2_1=−1075278 [kJ/kmol] g_H2O_1=−831031 [kJ/kmol] g_O2_1=−611914 [kJ/kmol] h_bar_Reng=−819472 [kJ/kmol] h_CH4=106833 [kJ/kmol] h_H2O=−142117 [kJ/kmol] H_P=−555765 [kJ/kmol] K_P_1=3.966E+16 R_u=8.314 [kJ/kmol-K] TpDT=2510 [K]

Phase Equilibrium 17-59C Yes. Because the number of independent variables for a two-phase (PH = 2), two-component (C = 2) mixture is, from the phase rule,

IV = C - PH + 2 = 2 - 2 + 2 = 2 Therefore, two properties can be changed independently for this mixture. In other words, we can hold the temperature constant and vary the pressure and still be in the two-phase region. Notice that if we had a single component (C = 1) two phase system, we would have IV=1, which means that fixing one independent property automatically fixes all the other properties.

17-60C No. Because the specific gibbs function of each phase will not be affected by this process; i.e., we will still have g f  g g .

17-61C Using solubility data of a solid in a specified liquid, the mass fraction w of the solid A in the liquid at the interface at a specified temperature can be determined from © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-593

mf A =

msolid msolid + mliquid

where msolid is the maximum amount of solid dissolved in the liquid of mass mliquid at the specified temperature.

17-62C The molar concentration Ci of the gas species i in the solid at the interface Ci, solid side (0) is proportional to the partial pressure of the species i in the gas Pi, gas side (0) on the gas side of the interface, and is determined from

Ci, solid side (0)  S  Pi, gas side (0) (kmol / m3 ) where S is the solubility of the gas in that solid at the specified temperature.

17-63C Using Henry’s constant data for a gas dissolved in a liquid, the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from Henry’s law expressed as

yi, liquid side (0) =

Pi, gas side (0) H

where H is Henry’s constant and Pi, gas side (0) is the partial pressure of the gas i at the gas side of the interface. This relation is applicable for dilute solutions (gases that are weakly soluble in liquids).

17-64E Water is sprayed into air, and the falling water droplets are collected in a container. The mass and mole fractions of air dissolved in the water are to be determined. Assumptions 1 2 3

Both the air and water vapor are ideal gases. Air is saturated since water is constantly sprayed into it. Air is weakly soluble in water and thus Henry’s law is applicable.

Properties The saturation pressure of water at 80F is 0.5075 psia (Table A-4E). Henry’s constant for air dissolved in water at 80F (300 K) is given in Table 17-2 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 lbm/ lbmol , respectively (Table A-1). Analysis Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 80F,

Pvapor = Psat@80°F = 0.5075 psia Then the partial pressure of dry air becomes

Pdry air = P - Pvapor = 14.3 - 0.5075 = 13.793 psia From Henry’s law, the mole fraction of air in the water is determined to be

ydry air, liquid side =

Pdry air, gas side H

13.793 psia(1 atm /14.696 psia) = 1.29´ 10- 5 74,000 bar (1 atm/1.01325 bar)

which is very small, as expected. The mass and mole fractions of a mixture are related to each other by mf i =

mi Ni M i M = = yi i mm Nm M m Mm

where the apparent molar mass of the liquid water - air mixture is

13-594

@1´ 18.0 + 0´ 29.0 @18.0 kg/kmol Then the mass fraction of dissolved air in liquid water becomes

mf dry air, liquid side = ydry air, liquid side (0)

M dry air Mm

= (1.29´ 10- 5 )

29 = 2.08´ 10- 5 18

EES (Engineering Equation Solver) SOLUTION "Given" T=80 [F] P=14.3 [psia] "Properties" P_sat=pressure(steam_iapws, T=T, x=1) H=74000 "[bar], from Table 15-2 in the text" MM_air=molarmass(air) MM_water=molarmass(H2O) "Analysis" P_vapor=P_sat P_dryair=P-P_vapor y_dryair_liquid=(P_dryair*Convert(psia, atm))/(H*Convert(bar, atm)) y_water_liquid=1-y_dryair_liquid MM_m=y_dryair_liquid*MM_air+y_water_liquid*MM_water "molar mass of mixture" w_dryair_liquid=y_dryair_liquid*MM_air/MM_m

17-65 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 300 kPa satisfies the criterion for phase equilibrium. Analysis The saturation temperature at 300 kPa is 406.7 K. Using the definition of Gibbs function and enthalpy and entropy data from Table A-5,

g f = h f - Ts f = (561.43 kJ/kg) - (406.7 K)(1.6717 kJ/kg ×K) = - 118.5 kJ/kg g g = hg - Tsg = (2724.9 kJ/kg) - (406.7 K)(6.9917 kJ/kg ×K) = - 118.6 kJ/kg which are practically same. Therefore, the criterion for phase equilibrium is satisfied.

EES (Engineering Equation Solver) SOLUTION "Given" P=300 [kPa] "Analysis" Fluid$='steam_iapws' g_f=h_f-T*s_f g_g=h_g-T*s_g T=temperature(Fluid$, P=P, x=1) h_f=enthalpy(Fluid$, P=P, x=0) s_f=entropy(Fluid$, P=P, x=0) h_g=enthalpy(Fluid$, P=P, x=1) s_g=entropy(Fluid$, P=P, x=1)

17-66 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 100C satisfies the criterion for phase equilibrium. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-595

Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-4,

g f = h f - Ts f = (419.17 kJ/kg) - (373.15 K)(1.3072 kJ/kg ×K) = - 68.61 kJ/kg g g = hg - Tsg = (2675.6 kJ/kg) - (373.15 K)(7.3542 kJ/kg ×K) = - 68.62 kJ/kg which are practically same. Therefore, the criterion for phase equilibrium is satisfied.

EES (Engineering Equation Solver) SOLUTION "Given" T=100+273.15 "[K]" "Analysis" Fluid$='steam_iapws' g_f=h_f-T*s_f g_g=h_g-T*s_g h_f=enthalpy(Fluid$, T=T, x=0) s_f=entropy(Fluid$, T=T, x=0) h_g=enthalpy(Fluid$, T=T, x=1) s_g=entropy(Fluid$, T=T, x=1)

17-67 The values of the Gibbs function for saturated refrigerant-134a at 0C as a saturated liquid, saturated vapor, and a mixture of liquid and vapor are to be calculated. Analysis Obtaining properties from Table A-11, the Gibbs function for the liquid phase is,

g f = h f - Ts f = 51.83 kJ/kg - (273.15 K)(0.20432 kJ/kg ×K) = - 3.98 kJ / kg

For the vapor phase,

g g = hg - Tsg = 250.50 kJ/kg - (273.15 K)(0.93158 kJ/kg ×K) = - 3.96 kJ / kg For the saturated mixture with a quality of 30%,

h = h f + xh fg = 51.83 kJ/kg + (0.30)(198.67 kJ/kg) = 111.43 kJ/kg s = s f + xs fg = 0.20432 kJ/kg ×K + (0.30)(0.72726 kJ/kg ×K) = 0.42250 kJ/kg ×K g = h - Ts = 111.43 kJ/kg - (273.15 kJ/kg)(0.42250 kJ/kg ×K) = - 3.98 kJ / kg The results agree and demonstrate that phase equilibrium exists.

17-68 The values of the Gibbs function for saturated refrigerant-134a at −10C are to be calculated.

13-596

Analysis Obtaining properties from Table A-11, the Gibbs function for the liquid phase is,

g f = h f - Ts f = 38.53 kJ/kg - (263.15 K)(0.15496 kJ/kg ×K) = - 2.25 kJ / kg For the vapor phase,

g g = hg - Tsg = 244.55 kJ/kg - (263.15 K)(0.93782 kJ/kg ×K) = - 2.24 kJ / kg The results agree and demonstrate that phase equilibrium exists.

17-69 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the nitrogen and the mass fraction of the oxygen at this temperature is to be determined. Analysis From the equilibrium diagram (Fig. 17-21) we read T = 88 K. For the liquid phase, from the same figure,

y f ,O2 = 0.90

and y f ,N2 = 0.10

Then,

mf f ,O2 =

m f ,O2 m f ,total

y f ,O2 M O2 y f ,O2 M O2 + y f ,N2 M N2

(0.90)(32) = 0.911 (0.90)(32) + (0.10)(28)

17-70 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the liquid phase. Analysis From the equilibrium diagram (Fig. 17-21) we read T = 84 K.

17-71 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the mass of the oxygen in the liquid and gaseous phases is to be determined for a specified composition of the mixture. Properties The molar masses of O2 is 32 kg/ kmol and that of N 2 is 28 kg/ kmol (Table A-1). Analysis From the equilibrium diagram (Fig. 17-21) at T = 84 K, the oxygen mole fraction in the vapor phase is 34% and that in the liquid phase is 70%. That is,

y f ,O2 = 0.70

and yg ,O2 = 0.34

The mole numbers are N O2 =

mO2 30 kg = = 0.9375 kmol M O2 32 kg/kmol

N N2 =

mN2 40 kg = = 1.429 kmol M N2 28 kg/kmol

Ntotal = 0.9375 + 1.429 = 2.366 kmol The total number of moles in this system is

N f + N g = 2.366

(1)

The total number of moles of oxygen in this system is

0.7 N f + 0.34 N g = 0.9375

(2)

Solving equations (1) and (2) simultneously, we obtain

N f = 0.3696 N g = 1.996 Then, the mass of oxygen in the liquid and vapor phases is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-597

m f ,O2 = y f ,O2 N f M O2 = (0.7)(0.3696 kmol)(32 kg/kmol) = 8.28 kg

mg ,O2 = yg ,O2 N g M O2 = (0.34)(1.996 kmol)(32 kg/kmol) = 21.72 kg

EES (Engineering Equation Solver) SOLUTION "Given" m_O2=30 [kg] m_N2=40 [kg] T=84 [K] P=100 [kPa] "Properties" MM_O2=32 [kg/kmol] MM_N2=28 [kg/kmol] "Analysis" "From Fig. 17-21 at 84 K" y_f_O2=0.70 y_g_O2=0.34 N_O2=m_O2/MM_O2 N_N2=m_N2/MM_N2 N_total=N_O2+N_N2 N_f+N_g=N_total y_f_O2*N_f+y_g_O2*N_g=N_O2 m_f_O2=y_f_O2*N_f*MM_O2 m_g_O2=y_g_O2*N_g*MM_O2

17-72 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the total mass of the liquid phase is to be determined. Properties The molar masses of O2 is 32 kg/ kmol and that of N 2 is 28 kg/ kmol (Table A-1). Analysis From the equilibrium diagram (Fig. 17-21) at T = 84 K, the oxygen mole fraction in the vapor phase is 34% and that in the liquid phase is 70%. That is,

y f ,O2 = 0.70

and yg ,O2 = 0.34

y f ,N2 = 0.30

and yg ,N2 = 0.66

Also,

The mole numbers are N O2 =

mO2 30 kg = = 0.9375 kmol M O2 32 kg/kmol

N N2 =

mN2 40 kg = = 1.429 kmol M N2 28 kg/kmol

Ntotal = 0.9375 + 1.429 = 2.366 kmol The total number of moles in this system is

N f + N g = 2.366

(1)

The total number of moles of oxygen in this system is

0.7 N f + 0.34 N g = 0.9375

(2)

Solving equations (1) and (2) simultneously, we obtain

13-598

N g = 1.996 The total mass of liquid in the mixture is then

m f ,total = m f ,O2 + m f ,O2 = y f ,O2 N f M O2 + y f ,N2 N f M N2 = (0.7)(0.3696 kmol)(32 kg/kmol) + (0.3)(0.3696 kmol)(28 kg/kmol) = 11.38 kg

17-73 The number of independent properties needed to fix the state of a mixture of oxygen and nitrogen in the gas phase is to be determined. Analysis In this case the number of components is C = 2 and the number of phases is PH = 1. Then the number of independent variables is determined from the phase rule to be IV = C − PH + 2 = 2 − 1 + 2 = 3 Therefore, three independent properties need to be specified to fix the state. They can be temperature, the pressure, and the mole fraction of one of the gases.

EES (Engineering Equation Solver) SOLUTION "Given" C=2 "number of components" PH=1 "number of phases" "Analysis" IV=C-PH+2

13-599

17-74 A rubber wall separates O2 and N 2 gases. The molar concentrations of O2 and N 2 in the wall are to be determined. Assumptions The O2 and N 2 gases are in phase equilibrium with the rubber wall. Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 kg/ kmol , respectively (Table A-1). The solubility of 3

oxygen and nitrogen in rubber at 298 K are 0.00312 and 0.00156 kmol/ m .bar , respectively (Table 17-3).

Analysis Noting that 300 kPa = 3 bar, the molar densities of oxygen and nitrogen in the rubber wall are determined to be

CO2 , solid side (0) = S´ PO2 , gas side

= (0.00312 kmol/m3 ×bar)(3 bar) = 0.00936 kmol / m 3 CN2 , solid side (0) = S´ PN2 , gas side

= (0.00156 kmol/m3 ×bar)(3 bar) = 0.00468 kmol / m 3 That is, there will be 0.00936 kmol of O2 and 0.00468 kmol of N 2 gas in each m 3 volume of the rubber wall.

EES (Engineering Equation Solver) SOLUTION "Given" T=(25+273) [K] P=300 [kPa]*Convert(kPa, bar) "Properties" S_O2=0.00312 [kmol/m^3-bar] S_N2=0.00156 [kmol/m^3-bar] "from Table 17-3 in the text" "Analysis" C_O2_solidside=S_O2*P C_N2_solidside=S_N2*P

17-75 A rubber plate is exposed to nitrogen. The molar and mass density of nitrogen in the iron at the interface is to be determined. Assumptions Rubber and nitrogen are in thermodynamic equilibrium at the interface. Properties The molar mass of nitrogen is M  28.0 kg/ kmol (Table A-1). The solubility of nitrogen in rubber at 298 K is

0.00156 kmol/ m3 .bar (Table 17-3). Analysis Noting that 250 kPa = 2.5 bar, the molar density of nitrogen in the rubber at the interface is determined to be

13-600

= (0.00156 kmol/m3 ×bar)(2.5 bar) = 0.0039 kmol / m 3 It corresponds to a mass density of r N2 , solid side (0) = CN2 , solid side (0) M N2

=(0.0039 kmol/m3 )(28 kg/kmol) = 0.1092 kg / m 3 That is, there will be 0.0039 kmol (or 0.1092 kg) of N 2 gas in each m 3 volume of iron adjacent to the interface.

EES (Engineering Equation Solver) SOLUTION "Given" T=298 [K] P_N2_gasside=250[kPa]*Convert(kPa, bar) "Properties" MM_N2=molarmass(N2) S=0.00156 [kmol/m^3-bar] "from Table 15-3 in the text" "Analysis" C_N2_solidside=S*P_N2_gasside rho_N2_solidside=C_N2_solidside*MM_N2

17-76 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The pressure of ammonia is to be determined for two compositions of the liquid phase. Assumptions The mixture is ideal and thus Raoult’s law is applicable.

Analysis According to Raoults’s law, when the mole fraction of the ammonia liquid is 20%,

PNH3 = y f ,NH3 Psat,NH3 (T ) = 0.20(615.3 kPa) = 123.1kPa When the mole fraction of the ammonia liquid is 80%,

PNH3 = y f ,NH3 Psat,NH3 (T ) = 0.80(615.3 kPa) = 492.2 kPa

EES (Engineering Equation Solver) SOLUTION "Given" T=10 [C] y_f_NH3=0.20 "or 0.80" P_sat=615.3 [kPa] "Analysis" P_NH3=y_f_NH3*P_sat

13-601

17-77 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the liquid phase is given. The composition of the vapor phase is to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable.

Properties At 25°C, Psat,H2 O = 3.170 kPa and

Psat,NH3 = 1003.5 kPa .

Analysis The vapor pressures are

PH2 O = y f ,H2 O Psat,H2 O (T ) = 0.50(3.170 kPa) = 1.585 kPa PNH3 = y f ,NH3 Psat,NH3 (T ) = 0.50(1003.5 kPa) = 501.74 kPa Thus the total pressure of the mixture is

Ptotal = PH2 O + PNH3 = (1.585 + 501.74) kPa = 503.33 kPa Then the mole fractions in the vapor phase become

yg ,H2 O =

yg ,NH3 =

PH2 O Ptotal

PNH3 Ptotal

1.585 kPa = 0.0031 or 0.31% 503.33 kPa

501.74 kPa = 0.9969 or 99.69% 503.33 kPa

EES (Engineering Equation Solver) SOLUTION "Given" T=25 "[C]" y_f_NH3=0.50 y_f_H2O=0.50 "Properties" P_sat_H2O=pressure(steam_iapws, T=T, x=1) P_sat_NH3=pressure(ammonia, T=T, x=1) "Analysis" P_H2O=y_f_H2O*P_sat_H2O P_NH3=y_f_NH3*P_sat_NH3 P_total=P_H2O+P_NH3 y_g_H2O=P_H2O/P_total y_g_NH3=P_NH3/P_total

17-78 A mixture of water and ammonia is considered. The mole fractions of the ammonia in the liquid and vapor phases are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 25C, Psat, H2O  3.1698 kPa and Psat, NH3  1003 .5 kPa (Tables A-4).

13-602

Analysis According to Raoults’s law, the partial pressures of ammonia and water in the vapor phase are given by

Pg,NH3  yf ,NH3 Psat,NH3  Pg,H2O  yf ,H2O Psat,H2O 

N f ,NH3 Nf ,NH3  Nf ,H2O N f ,H2O N f ,H2O  N f ,H2O

(1003.5 kPa)

(3.1698 kPa)

The sum of these two partial pressures must equal the total pressure of the vapor mixture. In terms of x 

N f ,H2O N f , NH3

, this sum

is 1003 .5 3.1698 x   100 x 1 x 1 Solving this expression for x gives

x  9.331 kmol H2O / kmol NH3 In the vapor phase, the partial pressure of the ammonia vapor is 1003 .5 1003 .5   97.13 kPa x 1 9.331  1 The mole fraction of ammonia in the vapor phase is then Pg , NH3 

y g ,NH3 

Pg ,NH3 P



97.13 kPa  0.9713 100 kPa

According to Raoult’s law,

y f , NH3 

Pg , NH3 Psat, NH3



97.13 kPa  0.0968 1003 .5 kPa

EES (Engineering Equation Solver) SOLUTION "Given" N_NH3=1 [kmol] N_H2O=2 [kmol] P=100 [kPa] T=25 [C] P_sat_NH3=1003.5 [kPa] "Analysis" P_sat_H2O=pressure(steam_iapws, T=T, x=0) P_sat_NH3/(x+1)+(P_sat_H2O*x)/(x+1)=P "x=N_f_H2O/N_f_NH3" P_g_NH3=P_sat_NH3/(x+1) y_g_NH3=P_g_NH3/P y_f_NH3=P_g_NH3/P_sat_NH3

17-79 An ammonia-water absorption refrigeration unit is considered. The operating pressures in the generator and absorber, and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-603

Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 0°C, Psat, H2O = 0.6112 kPa and at 46°C, Psat, H2O = 10.10 kPa (Table A-4). The saturation pressures of ammonia at the same temperatures are given to be 430.6 kPa and 1830.2 kPa, respectively. Analysis According to Raoults’s law, the partial pressures of ammonia and water are given by

Pg, NH3 = y f , NH3 Psat, NH3 Pg, H2O = y f , H2O Psat, H2O = (1- y f , NH3 ) Psat, H2O Using Dalton’s partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is

yg , NH3 = 0.96 =

y f , NH3 Psat, NH3 y f , NH3 Psat, NH3 + (1- y f , NH3 Psat, H2O ) 430.6 y f , NH3 430.6 y f , NH3 + 0.6112(1- y f , NH3 )

¾¾ ® y f , NH3 = 0.03294

Then,

P = y f , NH3 Psat, NH3 + (1- y f , NH3 ) Psat, H2O = (0.03294)(430.6) + (1- 0.03294)(0.6112) = 14.78 kPa Performing the similar calculations for the regenerator,

0.96 =

1830.2 y f , NH3 1830.2 y f , NH3 + 10.10(1- y f , NH3 )

¾¾ ® y f , NH3 = 0.1170

P = (0.1170)(1830.2) + (1- 0.1170)(10.10) = 223.1kPa

EES (Engineering Equation Solver) SOLUTION "Given" T_abs=0 [C] T_gen=46 [C] y_g_NH3=0.96 P_sat_abs_NH3=430.6 [kPa] P_sat_gen_NH3=1830.2 [kPa] "Analysis" P_sat_abs_H2O=pressure(steam_iapws, T=T_abs, x=1) P_sat_gen_H2O=pressure(steam_iapws, T=T_gen, x=1) y_g_NH3=(y_f_NH3_abs*P_sat_abs_NH3)/((y_f_NH3_abs*P_sat_abs_NH3)+(1y_f_NH3_abs)*P_sat_abs_H2O) P_abs=(y_f_NH3_abs*P_sat_abs_NH3)+(1-y_f_NH3_abs)*P_sat_abs_H2O y_g_NH3=(y_f_NH3_gen*P_sat_gen_NH3)/((y_f_NH3_gen*P_sat_gen_NH3)+(1y_f_NH3_gen)*P_sat_gen_H2O) P_gen=(y_f_NH3_gen*P_sat_gen_NH3)+(1-y_f_NH3_gen)*P_sat_gen_H2O

17-80 An ammonia-water absorption refrigeration unit is considered. The operating pressures in the generator and absorber, and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 6°C, Psat, H2O = 0.9353 kPa and at 40°C, Psat, H2O = 7.3851 kPa (Table A-4 or EES). The saturation pressures of ammonia at the same temperatures are given to be 534.8 kPa and 1556.7 kPa, respectively. Analysis According to Raoults’s law, the partial pressures of ammonia and water are given by

13-604

Pg, NH3 = y f , NH3 Psat, NH3

Pg, H2O = y f , H2O Psat, H2O = (1- y f , NH3 ) Psat, H2O Using Dalton’s partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is

yg ,NH3 = 0.96 =

y f , NH3 Psat, NH3 y f , NH3 Psat, NH3 + (1- y f , NH3 Psat, H2O ) 534.8 y f ,NH3 534.8 y f , NH3 + 0.9353(1- y f , NH3 )

¾¾ ® y f , NH3 = 0.04028

Then,

P = y f , NH3 Psat, NH3 + (1- y f , NH3 ) Psat, H2O = (0.04028)(534.8) + (1- 0.04028)(0.9353) = 22.44 kPa Performing the similar calculations for the regenerator,

0.96 =

1556.7 y f , NH3 1556.7 y f , NH3 + 7.3851(1- y f , NH3 )

¾¾ ® y f , NH3 = 0.1022

P = (0.1022)(1556.7) + (1- 0.1022)(7.3851) = 165.7 kPa

EES (Engineering Equation Solver) SOLUTION "Given" T_abs=6 [C] T_gen=40 [C] y_g_NH3=0.96 P_sat_abs_NH3=534.8 [kPa] P_sat_gen_NH3=1556.7 [kPa] "Analysis" P_sat_abs_H2O=pressure(steam_iapws, T=T_abs, x=1) P_sat_gen_H2O=pressure(steam_iapws, T=T_gen, x=1) y_g_NH3=(y_f_NH3_abs*P_sat_abs_NH3)/((y_f_NH3_abs*P_sat_abs_NH3)+(1y_f_NH3_abs)*P_sat_abs_H2O) P_abs=(y_f_NH3_abs*P_sat_abs_NH3)+(1-y_f_NH3_abs)*P_sat_abs_H2O y_g_NH3=(y_f_NH3_gen*P_sat_gen_NH3)/((y_f_NH3_gen*P_sat_gen_NH3)+(1y_f_NH3_gen)*P_sat_gen_H2O) P_gen=(y_f_NH3_gen*P_sat_gen_NH3)+(1-y_f_NH3_gen)*P_sat_gen_H2O

17-81 A glass of water is left in a room. The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100 percent. 3 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 27C is 3.568 kPa (Table A-4). Henry’s constant for air dissolved in water at 27C (300 K) is given in Table 17-2 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 kg/ kmol , respectively (Table A-1).

13-605

Analysis (a) Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27C,

Pvapor = Psat@27°C = 3.568 kPa (Table A-4 or EES) Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air is determined to be yvapor =

Pvapor P

3.568 kPa = 0.0388 92 kPa

(b) Noting that the total pressure is 92 kPa, the partial pressure of dry air is

Pdry air = P - Pvapor = 92 - 3.568 = 93.43 kPa = 0.8843 bar From Henry’s law, the mole fraction of air in the water is determined to be ydry air, liquid side =

Pdry air, gas side H

0.8843 bar = 1.20´ 10- 5 74,000 bar

Discussion The amount of air dissolved in water is very small, as expected.

EES (Engineering Equation Solver) SOLUTION "Given" T=27 [C] P=92[kPa]*Convert(kPa, bar) phi=1 "Properties" P_sat=pressure(steam_iapws, T=T, x=1) H=74000[bar] "from Table 17-2 in the text" "Analysis" "(a)" P_vapor=phi*P_sat y_vapor=P_vapor/P "(b)" P_dryair=P-P_vapor y_dryair_liquid=P_dryair/H

17-82 A carbonated drink in a bottle is considered. Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as a water, determine the mole fraction of the water vapor in the CO2 gas

13-606

and the mass of dissolved CO2 in a 300 ml drink are to be determined when the water and the CO2 gas are in thermal and phase equilibrium. Assumptions 1 The liquid drink can be treated as water. 2

Both the CO2 and the water vapor are ideal gases.

The CO2 gas and water vapor in the bottle from a saturated mixture.

The CO2 is weakly soluble in water and thus Henry’s law is applicable.

Properties The saturation pressure of water at 27C is 3.568 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 27C (300 K) is given in Table 17-2 to be H = 1710 bar. Molar masses of CO2 and water are 44 and 18 kg/ kmol , respectively (Table A-1). Analysis (a) Noting that the CO2 gas in the bottle is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27C,

Pvapor = Psat@27°C = 3.568 kPa

(more accurate EES value compared to interpolation value from Table A-4)

Assuming both CO2 and vapor to be ideal gases, the mole fraction of water vapor in the CO2 gas becomes yvapor =

Pvapor P

3.568 kPa = 0.0310 115 kPa

(b) Noting that the total pressure is 115 kPa, the partial pressure of CO2 is

PCO2 gas = P - Pvapor = 115 - 3.568 = 111.4 kPa = 1.114 bar From Henry’s law, the mole fraction of CO2 in the drink is determined to be yCO2, liquid side =

PCO2, gas side H

1.114 bar = 6.52´ 10- 4 1710 bar

Then the mole fraction of water in the drink becomes

ywater, liquid side = 1- yCO2, liquid side = 1- 6.52´ 10- 4 = 0.9993 The mass and mole fractions of a mixture are related to each other by mf i =

mi Ni M i M = = yi i mm Nm M m Mm

where the apparent molar mass of the drink (liquid water - CO2 mixture) is

M m = å yi M i = yliquid water M water + yCO2 M CO2 = 0.9993´ 18.0 + (6.52´ 10- 4 ) ´ 44 = 18.03 kg/kmol Then the mass fraction of dissolved CO2 gas in liquid water becomes mf CO2, liquid side = yCO2, liquid side (0)

M CO2 44 = 6.52´ 10- 4 = 0.00159 Mm 18.03

Therefore, the mass of dissolved CO2 in a 300 ml  300 g drink is

mCO2 = mfCO2 mm = (0.00159)(300 g) = 0.477 g

EES (Engineering Equation Solver) SOLUTION "Given" T=27 [C] P=115[kPa]*Convert(kPa, bar) V_m=300 [mL] "Properties" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-607

P_sat=pressure(steam_iapws, T=T, x=1) H=1710 [bar] "from Table 17-2 in the text" MM_CO2=molarmass(CO2) MM_water=molarmass(H2O) rho_m=1 [g/mL] "Analysis" "(a)" P_vapor=P_sat y_vapor=P_vapor/P "(b)" P_CO2_gas=P-P_vapor y_CO2_liquid=P_CO2_gas/H y_water_liquid=1-y_CO2_liquid MM_m=y_CO2_liquid*MM_CO2+y_water_liquid*MM_water "molar mass of mixture" w_CO2_liquid=y_CO2_liquid*MM_CO2/MM_m m_m=rho_m*V_m m_CO2=w_CO2_liquid*m_m

17-83E A mixture of water and R-134a is considered. The mole fractions of the R-134a in the liquid and vapor phases are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 77°F, Psat,H2O = 0.4597 psia and Psat,R = 96.56 psia (Tables A-4E, A-11E or EES).

Analysis According to Raoults’s law, the partial pressures of R-134a and water in the vapor phase are given by

Pg ,R = y f ,R Psat,R =

N f ,R N f ,R + N f ,H2O

Pg ,H2O = y f ,H2O Psat,H2O =

(96.56 psia)

N f ,H2O N f ,H2O + N f ,H2O

(0.4597 psia)

The sum of these two partial pressures must equal the total pressure of the vapor mixture. In terms of x =

N f ,H2O N f ,R

, this sum

96.56 0.4597 x + = 14.7 x+ 1 x+ 1 Solving this expression for x gives

x  5.748 kmol H2O / kmol R  134a In the vapor phase, the partial pressure of the R-134a vapor is

Pg ,R =

96.56 96.56 = = 14.31 psia x + 1 5.748 + 1

The mole fraction of R-134a in the vapor phase is then

13-608

yg ,R =

Pg ,R

14.31 psia = 0.9735 14.7 psia

According to Raoult’s law,

y f ,R =

Pg ,R Psat,R

14.31 psia = 0.1482 96.56 psia

EES (Engineering Equation Solver) SOLUTION "p=pressure(r134a, T=77, x=1) 0.96=1556.7*y/(1556.7*y+(1-y)*7.3851)" y_fR*94.48=N_fr/(N_fr+N_fw) y_fw*0.460=N_fr/(N_fr+N_fw) (1-y_fR)*0.4597 P_gR+P_gw=14.7

17-84 A liquid mixture of water and R-134a is considered. The mole fraction of the water and R-134a vapor are to be determined. Assumptions The mixture is ideal and thus Raoult’s law is applicable. Properties At 20C, Psat, H2O  2.3392 kPa and Psat, R  572 .07 kPa (Tables A-4, A-11). The molar masses of water and R134a are 18.015 and 102.03 kg/ kmol , respectively (Table A-1).

Analysis The mole fraction of the water in the liquid mixture is

yf ,H2O 

N f ,H2O N total 



mf f ,H2O / M H2O (mf f ,H2O / M H2O )  (mf f ,R / M R )

0.9 / 18.015  0.9808 (0.9 / 18.015)  (0.1/ 102.03)

According to Raoults’s law, the partial pressures of R-134a and water in the vapor mixture are

Pg ,R  y f ,R Psat, R  (1  0.9808 )(572 .07 kPa)  10.98 kPa Pg ,H2O  y f ,H2O Psat, H2O  (0.9808 )(2.3392 kPa)  2.294 kPa The total pressure of the vapor mixture is then

Ptotal  Pg ,R  Pg ,H2O  10.98  2.294  13.274 kPa Based on Dalton’s partial pressure model for ideal gases, the mole fractions in the vapor phase are

y g ,H2O 

y g ,R 

Pg ,H2O Ptotal

Pg ,R Ptotal



2.294 kPa  0.1728 13.274 kPa

10.98 kPa  0.8272 13.274 kPa

13-609

EES (Engineering Equation Solver) SOLUTION "Given" mf_w=0.90 T=20 [C] "Properties" MM_w=18.015 [kg/kmol] MM_R=102.03 [kg/kmol] "Analysis" P_sat_w=pressure(steam_iapws, T=T, x=0) P_sat_R=pressure(R134a, T=T, x=0) mf_R=1-mf_w N_f_w=mf_w/MM_w N_f_R=mf_R/MM_R N_total=N_f_w+N_f_R y_f_w=N_f_w/N_total y_f_R=1-y_f_w P_g_R=y_f_R*P_sat_R P_g_w=y_f_w*P_sat_w P_total=P_g_R+P_g_w y_g_w=P_g_w/P_total y_g_R=P_g_R/P_total

Review Problems 17-85 The mole fraction of argon that ionizes at a specified temperature and pressure is to be determined. Assumptions All components behave as ideal gases.

Analysis The stoichiometric and actual reactions can be written as

Ar Û Ar + +e-

Stoichiometric:

(thus nAr = 1, nAr+ = 1 and ne- = 1)

Ar ¾ ¾ ® x Ar + y Ar + +y e-

Actual:

react.

products

Ar balance: 1 = x + y or y = 1- x Total number of moles:

Ntotal = x + 2 y = 2 - x

The equilibrium constant relation becomes n

Kp =

n Ar N Ar N e-e- æ çç P N nAr ççè N Ar

ö÷ Ar+ ÷ ÷ ÷ total ø

+ n - - n Ar )

1+ 1- 1

ö y2 æ çç P ÷ ÷ ÷ x èçç N total ø÷

Substituting, 0.00042 =

ö (1- x) 2 æ çç 0.35 ÷ ÷ ÷ ç x è2 - x ø

13-610

Thus the fraction of Ar which dissociates into Ar  and e  is 1  0.965 = 0.035 or 3.5%

EES (Engineering Equation Solver) SOLUTION "Given" T=10000 [K] P=0.35 [atm] K_p=0.00042 "Analysis" "Stoichiometric reaction: Ar = Ar(+) + e(-)" "Stoichiometric coefficients" nu_Ar=1 nu_e=1 "Actual reaction: Ar = x Ar + y Ar(+) + y e(-)" 1=x+y "Ar balance" N_total=x+2*y K_p=(y^nu_Ar*y^nu_e)/x^nu_Ar*(P/N_total)^(nu_Ar+nu_e-nu_Ar) "The percentage of Ar which sissociates into Ar(+) and e(-) is" Percent_dissociate=(1-x)*Convert(, %)

17-86 The equilibrium constant of the dissociation process O2  2O is given in Table A-28 at different temperatures. The value at a given temperature is to be verified using Gibbs function data.

Analysis The K P value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where

D G *(T ) = nO gO* (T ) - nO2 gO* 2 (T )

= n O ( h - T s ) O - n O2 ( h - T s ) O2 = nO [(h f + h 2000 - h 298 ) - T s ]O - nO2 [(h f + h 2000 - h 298 ) - T s ]O2 = 2´ (249,190 + 42,564 - 6852 - 2000´ 201.135) - 1´ (0 + 67,881- 8682 - 2000´ 268.655) = 243,375 kJ/kmol Substituting,

ln K p = - (243,375 kJ/kmol)/[(8.314 kJ/kmol ×K)(2000 K)] = - 14.636 or

K p = 4.4´ 10- 7

(Table A-28: ln K P  14.622 )

13-611

EES (Engineering Equation Solver) SOLUTION "Given" T=2000 [K] T0=298 [K] "standard temperature" P0=101.325 [kPa] "standard pressure" "Reaction: O2 = 2 O" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" DELTAG=nu_O*g_O-nu_O2*g_O2 K_p=exp(-DELTAG/(R_u*T)) "Stoichiometric coefficients" nu_O2=1 nu_O=2 "Gibbs functions" g_O2=h_f_O2+h_O2-h_o_O2-T*s_O2 g_O=h_f_O+h_O-h_o_O-T*s_O "Enthalpy of formation data from Table A-26" h_f_O2=0 [kJ/kmol] h_f_O=249190 [kJ/kmol] "Enthalpies at T" h_O2=enthalpy(O2, T=T) h_O=42564 [kJ/kmol] "from ideal gas tables in the text" "Enthalpies at T0" h_o_O2=enthalpy(O2, T=T0) h_o_O=6852 [kJ/kmol] "from ideal gas tables in the text" "Entropies at T" s_O2=entropy(O2, T=T, P=P0) s_O=201.135 [kJ/kmol-K] "from ideal gas tables in the text"

17-87 A mixture of H 2 O , O2 , and N 2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined. Assumptions 1

The equilibrium composition consists of H 2 O , O2 , N 2 and H 2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric: Actual:

H2 O Û H2 + 12 O2 (thus nH2O = 1, nH2 = 1, and nO2 = 12 )

H 2 O+2O2 + 5 N 2

¾¾ ®

x H 2 O + y H 2 +z O2 + 5 N 2 react.

H balance:

2 = 2x + 2 y ¾ ¾ ®

products

inert

y = 1- x

13-612

O balance:

5 = x + 2z

Total number of moles:

¾ ¾®

z = 2.5 - 0.5x

Ntotal = x + y + z + 5 = 8.5 - 0.5x

The equilibrium constant relation can be expressed as (n

Kp =

n n ö H2 N HH2 2 N OO2 2 æ çç P ÷ ÷ ÷ n ÷ N HHO2O çèç N total ø 2

- n O 2 - n H 2O )

1+ 0.5- 1

ö y z 0.5 æ çç P ÷ ÷ ÷ ç ÷ x èç N total ø

From Table A-28, ln K P   6.768 at 2200 K. Thus K P  0.00115 . Substituting, 0.5

0.00115 =

ö÷ (1- x)(1.5 - 0.5 x)0.5 æ 5 çç ÷ çè8.5 - 0.5 x ø÷ x

Solving for x, x = 0.9981 Then, y = 1 − x = 0.0019 z = 2.5 −0.5x = 2.00095 Therefore, the equilibrium composition of the mixture at 2200 K and 5 atm is

0.9981H2O + 0.0019H2 + 2.00095O2 + 5N2 The equilibrium constant for the reaction H 2 O Û OH+ 12 H 2 is ln K P   7.418 , which is very close to the K P value of the reaction considered. Therefore, it is not realistic to assume that no OH will be present in equilibrium mixture.

EES (Engineering Equation Solver) SOLUTION "Given" T=2200 [K] P=5 [atm] P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "Stoichiometric reaction: H2O = H2 + 1/2 O2" "Stoichiometric coefficients" nu_H2O=1 nu_H2=1 nu_O2=1/2 "Actual reaction: H2O + 2 O2 + 5 N2 = N_H2O H2O + N_H2 H2 + N_O2 O2 + N_N2 N2" 2=2*N_H2O+2*N_H2 "H balance" 5=N_H2O+2*N_O2 "O balance" N_N2=5 "N2 balance" N_total=N_H2O+N_H2+N_O2+N_N2 K_p=(N_H2^nu_H2*N_O2^nu_O2)/N_H2O^nu_H2O*(P/N_total)^(nu_H2-nu_O2-nu_H2O) "The equilibrium constant can also be expressed as" K_p=exp(-DELTAG/(R_u*T)) DELTAG=nu_H2*g_H2+nu_O2*g_O2-nu_H2O*g_H2O "Gibbs functions" g_H2O=h_H2O-T*s_H2O g_H2=h_H2-T*s_H2 g_O2=h_O2-T*s_O2 "Enthalpies" h_H2O=enthalpy(H2O, T=T) h_H2=enthalpy(H2, T=T) h_O2=enthalpy(O2, T=T) "Entropies" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-613

s_H2O=entropy(H2O, T=T, P=P0) s_H2=entropy(H2, T=T, P=P0) s_O2=entropy(O2, T=T, P=P0)

17-88 Methane gas is burned with stoichiometric amount of air during a combustion process. The equilibrium composition and the exit temperature are to be determined. Assumptions 1

The product gases consist of CO2 , H 2 O , CO, N 2 , and O2 .

2 3

The constituents of the mixture are ideal gases. This is an adiabatic and steady-flow combustion process.

Analysis (a) The combustion equation of CH4 with stoichiometric amount of O2 can be written as

CH 4 + 2(O2 + 3.76N 2 ) ¾ ¾ ® xCO2 + (1- x)CO+(0.5 - 0.5x)O2 + 2H 2 O + 7.52N 2 After combustion, there will be no CH4 present in the combustion chamber, and H 2 O will act like an inert gas. The equilibrium equation among CO2 , CO, and O2 can be expressed as

CO2 Û CO + 12 O2 (thus nCO2 = 1, nCO = 1, and nO2 = 12 ) and K p =

n n CO N CO N OO2 2 æ çç P n CO2 ççè N N CO2

ö CO ÷ ÷ ÷ ÷ total ø

+ n O2 - n CO2 )

where

Ntotal = x + (1- x) + (1.5 - 0.5x) + 2 + 7.52 = 12.02 - 0.5x Substituting, 1.5- 1

Kp =

ö÷ (1- x)(0.5 - 0.5 x)0.5 æ 1 çç ÷ çè12.02 - 0.5 x ø÷ x

The value of K P depends on temperature of the products, which is yet to be determined. A second relation to determine K P and x is obtained from the steady-flow energy balance expressed as

0 = å N P (h fo + h - h o ) - å N R (h fo + h - h o ) ¾ ¾ ® 0 = å N P (h fo + h - h o ) - å N R h Ro P

since the combustion is adiabatic and the reactants enter the combustion chamber at 25C. Assuming the air and the combustion products to be ideal gases, we have h = h (T). From the tables, Substance

hf , kJ / kmol

h 298 K , kJ / kmol

CH4 ( g ) N2 O2 H2 O (g) CO

−74,850

8669

8682

−241,820

9904

−110,530

8669

13-614

CO2

−393,520

9364

Substituting,

0 = x(- 393,520 + hCO2 - 9364) + (1- x)(- 110,530 + hCO - 8669)

+ 2(- 241,820 + hH2O - 9904) + (0.5 - 0.5 x)(0 + hO2 - 8682) + 7.52(0 + hN2 - 8669) - 1(- 74,850 + h 298 - h 298 ) - 0 - 0 which yields xhCO2 + (1- x)hCO + 2hH2 O + (0.5 - 0.5 x)hO2 + 7.52hN2 - 279,344 x = 617,329

Now we have two equations with two unknowns, TP and x. The solution is obtained by trial and error by assuming a temperature TP , calculating the equilibrium composition from the first equation, and then checking to see if the second equation is satisfied. A first guess is obtained by assuming there is no CO in the products, i.e., x = 1. It yields TP  2328 K . The adiabatic combustion temperature with incomplete combustion will be less.

Take Tp = 2300 K ¾ ¾ ®

ln K p = - 4.49 ¾ ¾ ®

x = 0.870 ¾ ¾ ®

RHS = 641, 093

Take Tp = 2250 K ¾ ¾ ®

ln K p = - 4.805 ¾ ¾ ® x = 0.893 ¾ ¾ ®

RHS = 612, 755

By interpolation, Tp = 2258 K and x = 0.889 Thus the composition of the equilibrium mixture is

0.889CO2 + 0.111CO + 0.0555O2 + 2H2 O + 7.52N2

EES (Engineering Equation Solver) SOLUTION "Input Data" P_prod =101.3 [kPa] R_u=8.134 [kJ/kmol-K] T_fuel=298 [K] T_air=298 [K] "The combustion equation of CH4 with stoichiometric amount of air is CH4 + 2(O2 + 3.76N2)=CO2 +2H2O+2(3.76)N2" "For the incomplete combustion process in this problem, the combustion equation is CH4 + 2(O2 + 3.76N2)=aCO2 +bCO + cO2+2H2O+2(3.76)N2" "Specie balance equations" "O" 4=a *2+b +c *2+2 "C" 1=a +b N_tot =a +b +c +2+2*3.76 "Total kilomoles of products at equilibrium" "We assume the equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by:" K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod*Convert(kPa, Atm) "The equilibrium constant is also given by:

13-615

K_ P = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1) which can be written as:" sqrt(P/N_tot )*b *sqrt(c )=K_P *a "Conservation of energy for the reaction, assuming SSSF, neglecting work , ke, and pe:" E_in - E_out = DELTAE_cv E_in = Q_net + HR DELTAE_cv = 0 "Steady-flow requirement" "The enthalpy of the reactant gases is" HR=enthalpy(CH4,T=T_fuel)+ 2 *enthalpy(O2,T=T_air)+2*3.76 *enthalpy(N2,T=T_air) E_out = HP "The enthalpy of the product gases is" HP=a *enthalpy(CO2,T=T_prod )+b *enthalpy(CO,T=T_prod ) +2*enthalpy(H2O,T=T_prod)+2*3.76*enthalpy(N2,T=T_prod ) + c *enthalpy(O2,T=T_prod) {Q_net=0} "For an adiabatic reaction the net heat added is zero."

17-89 Problem 17-88 is reconsidered. The effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 200 percent is to be studied. Analysis The problem is solved using EES, and the solution is given below. "Often, for nonlinear problems such as this one, good guesses are required to start the solution. First, run the program with zero percent excess air to determine the net heat transfer as a function of T_prod. Just press F3 or click on the Solve Table icon. From Plot Window 1, where Q_net is plotted vs T_prod, determnine the value of T_prod for Q_net=0 by holding down the Shift key and move the cross hairs by moving the mouse. Q_net is approximately zero at T_prod = 2269 K. From Plot Window at T_prod = 2269 K, a, b, and c are approximately 0.89, 0.10, and 0.056, respectively." "For EES to calculate a, b, c, and T_prod directly for the adiabatic case, remove the '{ }' in the last line of this window to set Q_net = 0.0. Then from the Options menu select Variable Info and set the Guess Values of a, b, c, and T_prod to the guess values selected from the Plot Windows. Then press F2 or click on the Calculator icon." "Input Data" PercentEx = 0 Ex = PercentEX/100 P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] T_fuel=298 [K] T_air=298 [K] "The combustion equation of CH4 with stoichiometric amount of air is CH4 + (1+Ex)(2)(O2 + 3.76N2)=CO2 +2H2O+(1+Ex)(2)(3.76)N2" "For the incomplete combustion process in this problem, the combustion equation is CH4 + (1+Ex)(2)(O2 + 3.76N2)=aCO2 +bCO + cO2+2H2O+(1+Ex)(2)(3.76)N2" "Specie balance equations" 4=a *2+b +c *2+2 "O" 1=a +b "C" N_tot =a +b +c +2+(1+Ex)*(2)*3.76 "Total kilomoles of products at equilibrium" "We assume the equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-616

g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3 "The equilibrium constant is also given by K_ P = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )=K_P *a "Conservation of energy for the reaction, assuming SSSF, neglecting work, ke, and pe:" E_in - E_out = DELTAE_cv E_in = Q_net + HR HR=enthalpy(CH4,T=T_fuel)+ (1+Ex)*(2) *enthalpy(O2,T=T_air)+(1+Ex)*(2)*3.76 *enthalpy(N2,T=T_air) E_out = HP HP=a *enthalpy(CO2,T=T_prod )+b *enthalpy(CO,T=T_prod ) +2*enthalpy(H2O,T=T_prod )+(1+Ex)*(2)*3.76*enthalpy(N2,T=T_prod ) + c *enthalpy(O2,T=T_prod ) DELTAE_cv = 0 Q_net=0 PercentEx

Tprod

0 20 40 60 80 100 120 140 160 180 200

[K] 2260 2091 1940 1809 1695 1597 1511 1437 1370 1312 1259

13-617

17-90 Solid carbon is burned with a stoichiometric amount of air. The number of moles of CO2 formed per mole of carbon is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, O2 , and N 2 .

The constituents of the mixture are ideal gases.

Analysis Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2 , O2 , and N 2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by

C + (O2 + 3.76N 2 ) ¾ ¾ ® CO2 + 3.76N 2 The number of moles of CO2 in the products is then

N CO2 =1 NC

EES (Engineering Equation Solver) SOLUTION "Given" P=1[atm] T_air=298 [K] T_prod=(967+273) [K] "Analysis" "Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2, O2, and N2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by C + (O2+ 3.76 N2) = CO2 + 3.76 N2" N_CO2\N_C=1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-618

17-91 Solid carbon is burned with a stoichiometric amount of air. The amount of heat released per kilogram of carbon is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , CO, O2 , and N 2 .

The constituents of the mixture are ideal gases.

C + (O2 + 3.76N 2 ) ¾ ¾ ® CO2 + 3.76N 2 The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

hf kJ/kmol

kJ/kmol

N2 CO2

8669

38,129

9364

56,108

h298 K

h1240 K

Substituting,

- Qout = (1) (- 393,520 + 56,108 - 9364)+ (3.76) (0 + 38,129 - 8669) = - 236,006 kJ/kmol C or

Qout =

236,006 kJ/kmol = 19,670 kJ / kg carbon 12 kg/kmol

EES (Engineering Equation Solver) SOLUTION "Given" P=1 [atm] T_air=298 [K] T_prod=(967+273) [K] "Analysis" "Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2, O2, and N2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by C + (O2+ 3.76 N2) = CO2 + 3.76 N2" N_CO2=1 N_N2=3.76 h_N2=enthalpy(N2, T=T_prod) h_CO2=enthalpy(CO2, T=T_prod) -Q_bar_out=N_CO2*h_CO2+N_N2*h_N2 MM_C=12 [kg/kmol]

13-619

Q_out=Q_bar_out/MM_C

17-92 Propane gas is burned with 30% excess air. The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined. Assumptions 1

The equilibrium composition consists of CO2 , H 2 O , O2 , NO, and N 2 .

2 The constituents of the mixture are ideal gases. Analysis (a) The stoichiometric and actual reactions in this case are Stoichiometric:

N2 + O2 Û 2NO (thus n N2 = 1, nO2 = 1, and n NO = 2)

C3 H8 + 1.3´ 5(O2 + 3.76N 2 ) ¾ ¾ ® 3CO2 + 4H 2 O + xNO + yO2 + zN 2

Actual: N balance:

48.88 = x + 2 z ¾ ¾ ® z = 24.44 - 0.5 x

O balance:

13 = 6 + 4 + x + 2 y ¾ ¾ ® y = 1.5 - 0.5 x

Total number of moles:

Ntotal = 3 + 4 + x + y + z = 32.94

The equilibrium constant relation can be expressed as (n

ö NO N n NO æ P ÷ ÷ K p = n N2NO nO2 ççç ÷ N N2 N O2 çè N total ÷ ø

- n N2 - nO2 )

From Table A-28, at 1600 K, ln K p = - 5.294 . Since the stoichiometric reaction being considered is double this reaction,

K p = exp(- 2´ 5.294) = 2.522´ 10- 5 Substituting, 2- 1- 1

2.522´ 10

- 5

æ 1 ÷ ö x2 çç = ÷ ø (1.5 - 0.5 x)(24.44 - 0.5 x) çè32.94 ÷

Solving for x, x = 0.03024 Then, y = 1.5  0.5x = 1.485 z = 24.44  0.5x = 24.19 Therefore, the equilibrium composition of the products mixture at 1600 K and 1 atm is

C3 H8 + 6.5(O2 + 3.76N 2 ) ¾ ¾ ® 3CO 2 + 4H 2O + 0.03024NO +1.485O 2 + 24.19N 2

13-620

(b) The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

C3 H8 O2 N2 H2 O CO2

h298 K

h1600 K

kJ/kmol

−103,850

---

8682

52,961

8669

50,571

−241,820

9904

62,748

−393,520

9364

76,944

Neglecting the effect of NO in the energy balance and substituting,

- Qout = (3) (- 393,520 + 76,944 - 9364)+ (4)(- 241,820 + 62, 748 - 9904) + 1.485(52,961- 8682)

+ (24.19)(50,571- 8669) - (- 103,850) = - 654,360 kJ/kmol C3 H8 or Qout =

654,360 kJ/kmol = 14, 870 kJ / kg C3 H 8 44 kg/kmol

17-93 Methane gas is burned with 30 percent excess air. The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined. Assumptions 1

The equilibrium composition consists of CO2 , H 2 O , O2 , NO, and N 2 .

2 The constituents of the mixture are ideal gases. Analysis Inspection of the equilibrium constants of the possible reactions indicate that only the formation of NO need to be considered in addition to other complete combustion products. Then, the stoichiometric and actual reactions in this case are Stoichiometric: N2 + O2 Û 2NO (thus n N2 = 1, nO2 = 1, and n NO = 2) Actual:

CH 4 + 2.6(O2 + 3.76N 2 ) ¾ ¾ ® CO2 + 2H 2 O + xNO + yO2 + zN 2

N balance:

2´ 9.776 = x + 2 z ¾ ¾ ® z = 9.776 - 0.5 x

O balance:

5.2 = 2 + 2 + x + 2 y ¾ ¾® y = 0.6 - 0.5 x

Total number of moles:

Ntotal = 1+ 2 + x + y + z = 13.38

The equilibrium constant relation can be expressed as

13-621

Kp =

n NO æ P ÷ ö NO N NO çç ÷ n N2 nO2 ç ÷ çè N total ÷ N N2 N O2 ø

- n N2 - nO2 )

From Table A-28, at 1600 K, ln K p = - 5.294. Since the stoichiometric reaction being considered is double this reaction,

K p = exp(- 2´ 5.294) = 2.522´ 10- 5 Substituting, 2- 1- 1

2.522´ 10

- 5

æ 1 ÷ ö x2 çç = ÷ ÷ ç è ø (0.6 - 0.5 x)(9.766 - 0.5 x) 13.38

Solving for x, x = 0.0121 Then, y = 0.6  0.5x = 0.594 z = 9.776  0.5x = 9.77 Therefore, the equilibrium composition of the products mixture at 1600 K and 1 atm is

CH4 + 2.6(O2 + 3.76N2 ) ¾ ¾ ® CO2 + 2H2O + 0.0121NO + 0.594O2 + 9.77N2 The heat transfer for this combustion process is determined from the energy balance Ein - Eout = D Esystem applied on the combustion chamber with W = 0. It reduces to

- Qout = å N P (h fo + h - h o ) - å N R (h fo + h - h o ) P

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

CH4 O2 N2 H2 O CO2

h298 K

h1600 K

kJ/kmol

−74,850

---

8682

52,961

8669

50,571

−241,820

9904

62,748

−393,520

9364

76,944

Neglecting the effect of NO in the energy balance and substituting,

- Qout = (1) (- 393,520 + 76,944 - 9364)+ (2)(- 241,820 + 62, 748 - 9904) + 0.594(52,961- 8682)

+ (9.77)(50,571- 8669) - (- 74,850) = - 193,500 kJ/kmol CH4 or

Qout = 193,500 kJ / kmol CH4

EES (Engineering Equation Solver) SOLUTION "Given" ex=0.3 P=1 [atm] T1=298 [K] T2=1600 [K] "Analysis" "Stoichiomettric reaction N2 + O2 = 2 NO" nu_N2=1 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-622

nu_O2=1 nu_NO=2 "Actual reaction: CH4 = (ex+1)(a_th) (O2 + 3.76 N2) = CO2 + 2 H2O + x NO + y O2 + z N2" a_th=1+2/2 2*(ex+1)*a_th*3.76=x+2*z "N balance" 2*(ex+1)*a_th=2+2+x+2*y "O balance" N_total=1+2+x+y+z N_CO2=1 N_H2O=2 N_NO=x N_O2=y N_N2=z N_CH4=1 lnK_p=-5.294 "Table A--28 at 1600 K" K_p=exp(-2*5.294) K_p=N_NO^nu_NO/(N_N2^nu_N2*N_O2^nu_O2)*(P/N_total)^(nu_NO-nu_N2-nu_O2) h_CH4=enthalpy(CH4, T=T1) h_CO2=enthalpy(CO2, T=T2) h_H2O=enthalpy(H2O, T=T2) h_O2=enthalpy(O2, T=T2) h_N2=enthalpy(N2, T=T2) -Q_out=N_CO2*h_CO2+N_H2O*h_H2O+N_O2*h_O2+N_N2*h_N2-N_CH4*h_CH4

17-94E Gaseous octane gas is burned with 40% excess air. The equilibrium composition of the products of combustion is to be determined. Assumptions 1

The equilibrium composition consists of CO2 , H 2 O , O2 , NO, and N 2 .

The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are Stoichiometric:

N2 + O2 Û 2NO (thus n N2 = 1, nO2 = 1, and n NO = 2)

Actual:

C8 H18 + 1.4´ 12.5(O2 + 3.76N 2 ) ¾ ¾ ® 8CO2 + 9H 2 O + xNO + yO2 + zN 2

N balance:

131.6 = x + 2 z ¾ ¾ ® z = 65.8 - 0.5 x

O balance:

35 = 16 + 9 + x + 2 y ¾ ¾ ® y = 5 - 0.5x

Total number of moles:

Ntotal = 8 + 9 + x + y + z = 87.8

The equilibrium constant relation can be expressed as (n

ö NO N n NO æ P ÷ ÷ K p = n N2NO nO2 ççç ÷ N N2 N O2 çè N total ÷ ø

- n N2 - nO2 )

From Table A-28, at 2000 K (3600 R), ln K p = - 3.931. Since the stoichiometric reaction being considered is double this reaction,

13-623

Substituting, 2- 1- 1

3.851´ 10- 4 =

æ600 /14.7 ÷ ö x2 çç ÷ ÷ ç (5 - 0.5 x)(65.8 - 0.5 x) è 87.8 ø

Solving for x, x = 0.3492 Then, y = 5  0.5x = 4.825 z = 65.8  0.5x = 65.63 Therefore, the equilibrium composition of the products mixture at 2000 K and 4 MPa is

C8H18 + 17.5(O2 + 3.76N2 ) ¾ ¾ ® 8CO2 + 9H2O + 0.3492NO + 4.825O2 + 65.63N2

EES (Engineering Equation Solver) SOLUTION "Given" ex=0.4 P=600 [psia]*Convert(psia,atm) T=3600 [R] "Analysis" "Stoichiomettric reaction N2 + O2 = 2 NO" nu_N2=1 nu_O2=1 nu_NO=2 "Actual reaction: C8H18 = (ex+1)(a_th) (O2 + 3.76 N2) = 8 CO2 + 9 H2O + x NO + y O2 + z N2" a_th=8+9/2 2*(ex+1)*a_th*3.76=x+2*z "N balance" 2*(ex+1)*a_th=16+9+x+2*y "O balance" N_total=8+9+x+y+z N_CO2=8 N_H2O=9 N_NO=x N_O2=y N_N2=z N_C8H18=1 lnK_p=-3.931 "Table A--28 at 2000 K = 3600 R" K_p=exp(-2*3.931) K_p=N_NO^nu_NO/(N_N2^nu_N2*N_O2^nu_O2)*(P/N_total)^(nu_NO-nu_N2-nu_O2)

17-95 A mixture of H 2 and O2 in a tank is ignited. The equilibrium composition of the product gases and the amount of heat transfer from the combustion chamber are to be determined. Assumptions 1

The equilibrium composition consists of H 2 O , H 2 , and O2 .

The constituents of the mixture are ideal gases.

13-624

Analysis (a) The combustion equation can be written as

H 2 + 0.5O2 ¾ ¾ ®

x H 2 O + (1- x)H 2 + (0.5 - 0.5x)O2

The equilibrium equation among H 2 O , H 2 , and O2 can be expressed as

H2 O Û H2 + 12 O2 (thus nH2O = 1, nH2 = 1, and nO2 = 12 ) Total number of moles:

Ntotal = x + (1- x) + (0.5 - 0.5x) = 1.5 - 0.5x

The equilibrium constant relation can be expressed as (n

n n ö H2 N HH2 2 N OO2 2 æ çç P ÷ ÷ ÷ n ø N HHO2O ççè N total ÷

Kp =

+ n O 2 - n H 2O )

From Table A-28, ln K P  3.812 at 2800 K. Thus K P  0.02210 . Substituting, 1+ 0.5- 1

ö÷ (1- x)(0.5 - 0.5 x)0.5 æ 5 çç 0.0221 = ÷ ç è x 1.5 - 0.5 x ø÷

Solving for x, x = 0.944 Then the combustion equation and the equilibrium composition can be expressed as

H 2 + 0.5O2 ¾ ¾ ® 0.944H 2 O + 0.056H 2 + 0.028O2 and 0.944H2O + 0.056H2 + 0.028O2 (b) The heat transfer can be determined from

- Qout = å N P (h fo + h - h o - Pv ) - å N R (h fo + h - h o - Pv ) P

Since W = 0 and both the reactants and the products are assumed to be ideal gases, all the internal energy and enthalpies depend on temperature only, and the Pv terms in this equation can be replaced by RuT . It yields

- Qout = å N P (h fo + h2800 K - h298 K - RuT ) - å N R (h fo - RuT ) P

since reactants are at the standard reference temperature of 25C. From the tables, Substance hf h h 298 K

H2 O2 H2 O

2800 K

kJ / kmol

8468

89,838

8682

98,826

9904

125,198

Substituting,

- Qout = 0.944(- 241,820 + 125,198 - 9904 - 8.314´ 2800) + 0.056(0 + 89,838 - 8468 - 8.314´ 2800) + 0.028(0 + 98,826 - 8682 - 8.314´ 2800) - 1(0 - 8.314´ 298) - 0.5(0 - 8.314´ 298) = - 132,574 kJ/kmol H2 or

Qout = 132,600 kJ / mol H2

13-625

The equilibrium constant for the reaction H 2 O Û OH+ 12 H 2 is ln K P   3.763 , which is very close to the K P value of the reaction considered. Therefore, it is not realistic to assume that no OH will be present in equilibrium mixture.

EES (Engineering Equation Solver) SOLUTION "Given" T1=(25+273) [K] P1=1 [atm] T2=2800 [K] P2=5 [atm] T0=(25+273) [K] "standard temperature" P0=101.325 [kPa] "standard pressure" "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "(a)" "The combustion equation is: H2 + 0.5 O2 = x H2O + (1-x) H2 + (0.5-0.5*x) O2" N_H2O=x N_H2=1-x N_O2=0.5-0.5*x "The equilibrium equation is: H2O = H2 + 1/2 O2" "Stoichiometric coefficients" nu_H2O=1 nu_H2=1 nu_O2=1/2 N_total=N_H2O+N_H2+N_O2 K_p=(N_H2^nu_H2*N_O2^nu_O2)/N_H2O^nu_H2O*(P2/N_total)^(nu_H2+nu_O2-nu_H2O) "The equilibrium constant can also be expressed as" K_p=exp(-DELTAG/(R_u*T2)) DELTAG=nu_H2*g_H2+nu_O2*g_O2-nu_H2O*g_H2O "Gibbs functions" g_H2O=h_H2O-T2*s_H2O g_H2=h_H2-T2*s_H2 g_O2=h_O2-T2*s_O2 "Enthalpies" h_H2O=enthalpy(H2O, T=T2) h_H2=enthalpy(H2, T=T2) h_O2=enthalpy(O2, T=T2) "Entropies" s_H2O=entropy(H2O, T=T2, P=P0) s_H2=entropy(H2, T=T2, P=P0) s_O2=entropy(O2, T=T2, P=P0) "(b)" "Enthalpies of reactants" h_H2_R=enthalpy(H2, T=T1) h_O2_R=enthalpy(O2, T=T1) "Enthalpies of products" h_H2O_P=enthalpy(H2O, T=T2) h_H2_P=enthalpy(H2, T=T2) h_O2_P=enthalpy(O2, T=T2) H_P=N_H2O*(h_H2O_P-R_u*T2)+N_H2*(h_H2_P-R_u*T2)+N_O2*(h_O2_P-R_u*T2) H_R=N_H2_R*(h_H2_R-R_u*T1)+N_O2_R*(h_O2_R-R_u*T1) N_H2_R=1 N_O2_R=0.5 -Q_out=H_P-H_R © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-626

17-96 Methane is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Properties The molar mass and gas constant of methane are 16.043 kg/ kmol and 0.5182 kJ/ kg.K (Table A-1).

Assumptions 1

The equilibrium composition consists of O2 and O.

2 The constituents of the mixture are ideal gases. Analysis (a) An energy balance for the process gives

Ein - Eout Net energy transfer by heat, work, and mass

D Esystem Change in internal, kinetic, potential, etc. energies

Qin = N (u2 - u1 )

= N éëêh2 - h1 - Ru (T2 - T1 )ùûú Using the empirical coefficients of Table A-2c, 2

h2 - h1 = ò c p dT = a (T2 - T1 ) + 1

= 19.89(1000 - 298) + +

b 2 c d (T2 - T12 ) + (T23 - T13 ) + (T24 - T14 ) 2 3 4

0.05024 1.269´ 10- 5 (10002 - 2982 ) + (10003 - 2983 ) 2 3

- 11.01´ 10- 9 (10004 - 2984 ) 4

= 38, 239 kJ/kmol Substituting,

Qin = (10 kmol) [38,239 kJ/kmol - (8.314 kJ/kmol ×K)(1000 - 298)K ]= 324,000 kJ (b) The stoichiometric and actual reactions in this case are Stoichiometric: CH4 Û C + 2H2 (thus nCH4 = 1, nC = 1 and nH2 = 2)

CH 4 ¾ ¾ ® xCH 4 + yC + zH 2

Actual:

react.

products

C balance:

1= x + y ¾ ¾ ® y = 1- x

H balance:

4 = 4x + 2z ¾ ¾ ® z = 2 - 2x

Total number of moles:

Ntotal = x + y + z = 3 - 2 x

The equilibrium constant relation can be expressed as n + n H2 - n CH4

Kp =

n H2 æ öC N CnC N H2 çç P ÷ ÷ ÷ n CH4 ÷ ççè N total ø N CH4

From the problem statement, at 1000 K, ln K p = - 2.328. Then,

13-627

Substituting, 1+ 2- 1

0.09749 =

ö (1- x)(2 - 2 x)2 æ çç 1 ÷ ÷ ÷ çè3 - 2 x ø x

Solving for x, x = 0.6414 Then, y = 1  x = 0.3586 z = 2  2x = 0.7172 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is

0.6414 CH4 + 0.3586 C + 0.7172 H2 The mole fractions are yCH4 =

N CH4 0.6414 0.6414 = = = 0.3735 N total 0.6414 + 0.3586 + 0.7172 1.7172

yC =

NC 0.3586 = = 0.2088 N total 1.7172

yH2 =

N H2 0.7172 = = 0.4177 N total 1.7172

The heat transfer can be determined from

Qin = N ( yCH4 cv , CH4T2 + yH2 cv , H2T2 + yC cv ,CT2 ) - Ncv , CH4T1 = (10) [(0.3735)(63.3)(1000) + (0.4177)(21.7)(1000) + (0.2088)(0.711)(1000) ]- (10)(27.8)(298) = 245,700 kJ

EES (Engineering Equation Solver) SOLUTION "Given" N=10 [kmol] P=1[atm] T1=298 [K] T2=1000 [K] lnK_p=2.328 c_v_CH4=63.3 [kJ/kmol-K] c_v_H2=21.7 [kJ/kmol-K] c_v_C=0.711 [kJ/kmol-K] c_v_CH4_room=27.8 [kJ/kmol-K] "Analysis" "(a)" DELTAh=a*(T2-T1)+b/2*(T2^2-T1^2)+c/3*(T2^3-T1^3)+d/4*(T2^4-T1^4) a=19.89 "Coefficients are from Table A-2c" b=0.05024 c=1.269E-5 d=-11.01E-9 R_u=8.314 [kJ/kmol-K] Q_in_a=N*(DELTAh-R_u*(T2-T1)) "(b)" "Stoichiomettric reaction CH4 = C + 2 H2" nu_CH4=1 nu_C=1 nu_H2=2 "Actual reaction: CH4 = x CH4 + y C + z H2" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-628

1=x+y "C balance" 4=4*x+2*z "H balance" N_total=x+y+z N_CH4=x N_C=y N_H2=z K_p=(N_C^nu_C*N_H2^nu_H2)/N_CH4^nu_CH4*(P/N_total)^(nu_C+nu_H2-nu_CH4) K_p=exp(-lnK_p) y_CH4=N_CH4/N_total y_C=N_C/N_total y_H2=N_H2/N_total Q_in_b=N*(y_CH4*c_v_CH4*T2+y_H2*c_v_H2*T2+y_C*c_v_C*T2)-N*c_v_CH4_room*T1

17-97 The equilibrium constant for the reaction CH4  2O2  CO2  2H2 O at 100 kPa and 2000 K is to be determined. Assumptions 1 The constituents of the mixture are ideal gases.

Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide,

CH4 Û C + 2H2

K P = e- 7.847

C + O2 Û CO2

K P = e23.839

When these two reactions are summed and the common carbon term cancelled, the result is

CH4 + O2 Û CO2 + 2H2 K P = e

(23.839- 7.847)

= e15.992

Next, we include the water dissociation reaction (Table A-28), 2(8.145) = e16.29 2H2 + O2 Û 2H2 O K P = e

which when summed with the previous reaction and the common hydrogen term is cancelled yields 15.992 + 16.29 = e32.282 CH4 + 2O2 Û CO2 + 2H2 O K P = e

Then,

ln K P = 32.282

17-98 The equilibrium mole fraction of the water vapor for the reaction CH4  2O2  CO2  2H2 O at 100 kPa and 2000 K is to be determined. Assumptions 1

The equilibrium composition consists of CH4 , O2 , CO2 , and H 2 O .

The constituents of the mixture are ideal gases.

13-629

Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide,

CH4 Û C + 2H2

K P = e- 7.847

C + O2 Û CO2

K P = e23.839

When these two reactions are summed and the common carbon term cancelled, the result is

CH4 + O2 Û CO2 + 2H2 K P = e

(23.839- 7.847)

= e15.992

Next, we include the water dissociation reaction, 2(8.145) = e16.29 2H2 + O2 Û 2H2 O K P = e

which when summed with the previous reaction and the common hydrogen term is cancelled yields 15.992 + 16.29 = e32.282 CH4 + 2O2 Û CO2 + 2H2 O K P = e

Then,

ln K P = 32.282 Actual reeaction:

CH 4 + 2O2 ¾ ¾ ® xCH 4 + yO2 + zCO2 +mH 2 O react.

C balance:

1= x + z

H balance:

4 = 4 x + 2m ¾ ¾ ® m = 2 - 2x

O balance:

4 = 2 y + 2z + m ¾ ¾ ® y = 2x

Total number of moles:

¾¾ ®

products

z = 1- x

Ntotal = x + y + z + m = 3

The equilibrium constant relation can be expressed as nCO2 + n H2O - nCH4 - nO2

Kp =

nCO2 n H2O æ ö N CO2 N H2O çç P ÷ ÷ ÷ nCH4 nO2 ç ÷ N CH4 N O2 çè N total ø

Substituting, 1+ 2- 1- 2

e32.282 =

ö (1- x)(2 - 2 x) 2 æ çç100 /101.325 ÷ ÷ 2 ÷ ç è ø 3 x(2 x)

Solving for x, x = 0.00002122 Then, y = 2x = 0.00004244 z = 1  x = 0.99997878 m = 2  2x = 1.99995756 Therefore, the equilibrium composition of the mixture at 2000 K and 100 kPa is

0.00002122 CH4 +0.00004244 O2 + 0.99997878 CO2 + 1.99995756 H2 O The mole fraction of water vapor is then

yH2O =

1.99995756 = 0.6667 3

13-630

17-99 The hR at a specified temperature is to be determined using enthalpy and K p data. Assumptions Both the reactants and products are ideal gases. Analysis (a) The complete combustion equation of H 2 can be expressed as

H 2 + 12 O 2 Û H 2 O The hR of the combustion process of H2 at 2400 K is the amount of energy released as one kmol of H 2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from

hR = å N P (h fo + h - h o ) - å N R (h fo + h - h o ) P

Assuming the H 2 O , H 2 , and O2 to be ideal gases, we have h = h (T). From the tables, Substance

H2 O H2 O2

h298 K

h2400 K

kJ / kmol

−241,820

9904

103,508

8468

75,383

8682

83,174

Substituting,

hR = 1(- 241,820 + 103,508 - 9904) - 1(0 + 75,383 - 8468) - 0.5(0 + 83,174 - 8682) = - 252, 377 kJ / kmol

(b) The hR value at 2400 K can be estimated by using K P values at 2200 K and 2600 K (the closest two temperatures to 2400 K for which K P data are available) from Table A-28, ln

ö KP2 h æ1 1 ö ÷ or ln K - ln K @ hR æ ÷ çç 1 - 1 ÷ ÷ @ R ççç P2 P1 ÷ ÷ ÷ ÷ K P1 Ru çèT1 T2 ø Ru èççT1 T2 ø

4.648 - 6.768 @

æ 1 hR 1 ö÷ çç ÷ ç è 8.314 kJ/kmol ×K 2200 K 2600 K ø÷

hR @- 252, 047 kJ / kmol

EES (Engineering Equation Solver) SOLUTION "Given" T=2400 [K] T0=298 [K] "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "(a)" "The combustion equation: H2 + 1/2 O2 = H2O" "Mol numbers" N_H2=1 N_O2=1/2 N_H2O=1 "Enthalpies at T" h_H2=enthalpy(H2, T=T) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-631

h_O2=enthalpy(O2, T=T) h_H2O=enthalpy(H2O, T=T) "Enthalpy of reaction" h_bar_R_a=N_H2O*h_H2O-N_H2*h_H2-N_O2*h_O2 "(b)" "The equilibrium constants at T1 = 2200 K and T2 = 2600 K, from Table A-28" T1=2200 T2=2600 K_p1=exp(6.768) K_p2=exp(4.648) ln(K_p2/K_p1)=h_bar_R_b/R_u*(1/T1-1/T2)

17-100 Problem 17-99 is reconsidered. The effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 2000 to 3000 K is to be investigated. Analysis The problem is solved using EES, and the solution is given below. T_prod=2400 [K] DELTAT_prod =25 [K] R_u=8.314 [kJ/kmol-K] T_prod_1 = T_prod - DELTAT_prod T_prod_2 = T_prod + DELTAT_prod "The combustion equation is 1 H2 + 0.5 O2 =>1 H2O" "The enthalpy of reaction H_bar_R using enthalpy data is:" h_bar_R_Enthalpy = HP - HR HP = 1*Enthalpy(H2O,T=T_prod ) HR = 1*Enthalpy(H2,T=T_prod ) + 0.5*Enthalpy(O2,T=T_prod ) "The enthalpy of reaction H_bar_R using enthalpy data is found using the following equilibruim data:" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_H2O_1=Enthalpy(H2O,T=T_prod_1 )-T_prod_1 *Entropy(H2O,T=T_prod_1 ,P=101.3) g_H2_1=Enthalpy(H2,T=T_prod_1 )-T_prod_1 *Entropy(H2,T=T_prod_1 ,P=101.3) g_O2_1=Enthalpy(O2,T=T_prod_1 )-T_prod_1 *Entropy(O2,T=T_prod_1 ,P=101.3) g_H2O_2=Enthalpy(H2O,T=T_prod_2 )-T_prod_2 *Entropy(H2O,T=T_prod_2 ,P=101.3) g_H2_2=Enthalpy(H2,T=T_prod_2 )-T_prod_2 *Entropy(H2,T=T_prod_2 ,P=101.3) g_O2_2=Enthalpy(O2,T=T_prod_2 )-T_prod_2 *Entropy(O2,T=T_prod_2 ,P=101.3) DELTAG_1 =1*g_H2O_1-0.5*g_O2_1-1*g_H2_1 DELTAG_2 =1*g_H2O_2-0.5*g_O2_2-1*g_H2_2 K_p_1 = exp(-DELTAG_1/(R_u*T_prod_1)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod_2)) "From EES data" "The entahlpy of reaction is estimated from the equilibrium constant K_p" ln(K_P_2/K_P_1)=h_bar_R_Kp/R_u*(1/T_prod_1 - 1/T_prod_2) PercentError = ABS((h_bar_R_enthalpy - h_bar_R_Kp)/h_bar_R_enthalpy)*Convert(, %)

Percent Error [%]

Tprod

h REnthalpy

h RKp

[K]

[kJ/ kmol]

0.0002739

2000

−251723

−251722

13-632

0.0002333 0.000198 0.0001673 0.0001405 0.0001173 0.00009706 0.00007957 0.00006448 0.00005154 0.0000405

2100 2200 2300 2400 2500 2600 2700 2800 2900 3000

−251920 −252096 −252254 −252398 −252532 −252657 −252778 −252897 −253017 −253142

−251919 −252095 −252254 −252398 −252531 −252657 −252777 −252896 −253017 −253142

17-101 The K P value of the dissociation process O2  2O at a specified temperature is to be determined using the hR data and K P value at a specified temperature. Assumptions Both the reactants and products are ideal gases. Analysis The hR and K P data are related to each other by ln

KP2 h æ1 1 ö h æ1 1 ö ÷ ÷ ÷ ÷ @ R ççç or ln K P 2 - ln K P1 @ R ççç ÷ ÷ ÷ ÷ K P1 Ru èçT1 T2 ø Ru èçT1 T2 ø

The hR of the specified reaction at 2800 K is the amount of energy released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 2800 K, and can be determined from

hR = å N P (h f + h - h ) - å N R (h f + h - h ) P

Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables,

Substance O O2

h298 K

h2800 K

kJ / kmol

249,190 0

6852 8682

59,241 98,826

Substituting,

hR = 2(249,190 + 59, 241- 6852) - 1(0 + 98,826 - 8682) = 513, 014 kJ/kmol © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-633

The K P value at 3000 K can be estimated from the equation above by using this hR value and the K P value at 2600 K which is ln K P1  7.521 , ln K P 2 - (- 7.521) =

513, 014 kJ/kmol æ 1 ö ÷ çç 1 ÷ ÷ ç 8.314 kJ/kmol ×K è 2600 K 3000 K ø

ln K P 2 = - 4.357

(Table A-28: ln K P 2 = - 4.357)

K P 2 = 0.01282

EES (Engineering Equation Solver) SOLUTION "Given" T=2800 [K] T2=3000 [K] T0=298 [K] "Properties" R_u=8.314 [kJ/kmol-K] "Analysis" "The combustion equation: O2 = 2 O" "Mol numbers" N_O2=1 N_O=2 "Enthalpy of formation data from Table A-26" h_f_O2=0 h_f_O=249190 "Enthalpies at T" h_O2=98826 "enthalpy(O2, T=T)" h_O=59241 "from ideal gas tables in the text" "Standard state enthalpies" h_o_O2=8682 "enthalpy(O2, T=T0)" h_o_O=6852 "from ideal gas tables in the text" "Enthalpy of reaction" h_bar_R=N_O*(h_f_O+h_O-h_o_O)-N_O2*(h_f_O2+h_O2-h_o_O2) "The equilibrium constant at T1 = 2000 K, from Table A-28" T1=2600 K_p1=exp(-7.521) ln(K_p2/K_p1)=h_bar_R/R_u*(1/T1-1/T2)

17-102 A glass of water is left in a room. The mole fraction of the water vapor in the air at the water surface and far from the surface as well as the mole fraction of air in the water near the surface are to be determined when the water and the air are at the same temperature. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 25C is 3.170 kPa (Table A-4). Henry’s constant for air dissolved in water at 25C (298 K) is given in Table 17-2 to be H = 71,600 bar. Molar masses of dry air and water are 29 and 18 kg/ kmol , respectively (Table A-1).

13-634

Analysis (a) Noting that the relative humidity of air is 70%, the partial pressure of water vapor in the air far from the water surface will be

Pv , room air = f Psat @ 25°C = (0.7)(3.170 kPa) = 2.219 kPa Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the room air is yvapor =

Pvapor P

2.219 kPa = 0.0222 100 kPa

(or 2.22%)

(b) Noting that air at the water surface is saturated, the partial pressure of water vapor in the air near the surface will simply be the saturation pressure of water at 25C, Pv , interface = Psat@25°C = 3.170 kPa. Then the mole fraction of water vapor in the air at the interface becomes

yv , surface =

Pv , surface P

3.170 kPa = 0.0317 100 kPa

(or 3.17%)

Pair, surface = P - Pv, surface = 100 - 3.170 = 96.83 kPa From Henry’s law, the mole fraction of air in the water is determined to be

ydry air, liquid side =

Pdry air, gas side H

(96.83 /100) bar = 1.35´ 10- 5 71,600 bar

Discussion The water cannot remain at the room temperature when the air is not saturated. Therefore, some water will evaporate and the water temperature will drop until a balance is reached between the rate of heat transfer to the water and the rate of evaporation.

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] P=(100/100) [bar] phi=0.70 "Properties" P_sat=pressure(steam_iapws, T=T, x=1) H=71600 [bar] "from Table 15-2 in the text" "Analysis" "(a)" P_v=phi*P_sat y_v=P_v/P "(b)" P_v_surface=P_sat y_v_surface=P_v_surface/P "(c)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-635

P_air_surface=P-P_v_surface y_dryair_liquid=P_air_surface/H

17-103 A glass of water is left in a room. The mole fraction of the water vapor in the air at the water surface and far from the surface as well as the mole fraction of air in the water near the surface are to be determined when the water and the air are at the same temperature. Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is weakly soluble in water and thus Henry’s law is applicable. Properties The saturation pressure of water at 25C is 3.170 kPa (Table A-4). Henry’s constant for air dissolved in water at 25C (298 K) is given in Table 17-2 to be H = 71,600 bar. Molar masses of dry air and water are 29 and 18 kg/ kmol , respectively (Table A-1). Analysis (a) Noting that the relative humidity of air is 25%, the partial pressure of water vapor in the air far from the water surface will be

Pv , room air = f Psat @ 25°C = (0.25)(3.170 kPa) = 0.7925 kPa Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the room air is yvapor =

Pvapor P

0.7925 kPa = 0.0079 100 kPa

(or 0.79%)

Pv , surface P

3.170 kPa = 0.0317 100 kPa

(or 3.17%)

Pair, surface = P - Pv, surface = 100 - 3.170 = 96.83 kPa From Henry’s law, the mole fraction of air in the water is determined to be ydry air, liquid side =

Pdry air, gas side H

(96.83 / 100) bar = 1.35´ 10- 5 71,600 bar

EES (Engineering Equation Solver) SOLUTION "Given" T=25 [C] P=100[kPa]*Convert(kPa, bar) phi=0.25 "Properties" P_sat=pressure(steam_iapws, T=T, x=1) H=71600 [bar] "from Table 15-2 in the text" "Analysis" "(a)" P_v=phi*P_sat y_v=P_v/P "(b)" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-636

P_v_surface=P_sat y_v_surface=P_v_surface/P "(c)" P_air_surface=P-P_v_surface y_dryair_liquid=P_air_surface/H

17-104 A 2-L bottle is filled with carbonated drink that is fully charged (saturated) with CO2 gas. The volume that the

CO2 gas would occupy if it is released and stored in a container at room conditions is to be determined. Assumptions 1

The liquid drink can be treated as water.

Both the CO2 gas and the water vapor are ideal gases.

The CO2 gas is weakly soluble in water and thus Henry’s law is applicable.

Properties The saturation pressure of water at 17C is 1.938 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 17C (290 K) is H = 1280 bar (Table 17-2). Molar masses of CO2 and water are 44.01 and 18.015 kg/ kmol , 3

respectively (Table A-1). The gas constant of CO2 is 0.1889 kPa.m /kg.K . Also, 1 bar = 100 kPa. Analysis In the charging station, the CO2 gas and water vapor mixture above the liquid will form a saturated mixture. Noting that the saturation pressure of water at 17C is 1.938 kPa, the partial pressure of the CO2 gas is

PCO2 , gas side = P - Pvapor = P - Psat@17°C = 600 - 1.938 = 598.06 kPa = 5.9806 bar From Henry’s law, the mole fraction of CO2 in the liquid drink is determined to be yCO2 , liquid side =

PCO2 , gas side H

5.9806 bar = 0.00467 1280 bar

Then the mole fraction of water in the drink becomes

ywater, liquid side = 1- yCO2 , liquid side = 1- 0.00467 = 0.99533 The mass and mole fractions of a mixture are related to each other by wi =

mi Ni M i M = = yi i mm Nm M m Mm

where the apparent molar mass of the drink (liquid water - CO2 mixture) is M m = å yi M i = yliquid water M water + yCO2 M CO2 = 0.99533´ 18.015 + 0.00467 ´ 44.01 = 18.14 kg/kmol

Then the mass fraction of dissolved CO2 in liquid drink becomes

wCO2 , liquid side = yCO2 , liquid side (0)

M CO2 Mm

= 0.00467

44.01 = 0.0113 18.14

Therefore, the mass of dissolved CO2 in a 2 L  2 kg drink is

mCO2 = wCO2 mm = 0.0113(2 kg) = 0.0226 kg Then the volume occupied by this CO2 at the room conditions of 20C and 100 kPa becomes V=

mRT (0.0226 kg)(0.1889 kPa ×m3 / kg ×K)(293 K) = = 0.0125 m 3 = 12.5 L P 100 kPa

13-637

Discussion Note that the amount of dissolved CO2 in a 2-L pressurized drink is large enough to fill 6 such bottles at room temperature and pressure. Also, we could simplify the calculations by assuming the molar mass of carbonated drink to be the same as that of water, and take it to be 18 kg/ kmol because of the very low mole fraction of CO2 in the drink.

EES (Engineering Equation Solver) SOLUTION "Given" T1=17 [C] P1=600[kPa]*Convert(kPa, bar) V_m=2 [L] T2=20 [C] P2=100[kPa]*Convert(kPa, bar) "Properties" P_sat=pressure(steam_iapws, T=T1, x=1) H=1280 [bar] "from Table 15-2 in the text at T1" MM_CO2=molarmass(CO2) MM_water=molarmass(H2O) R_CO2=R_u/MM_CO2 R_u=8.314 [kPa-m^3/kmol-K] rho_m=1 [kg/L] "Analysis" P_vapor=P_sat P_CO2_gasside=P1-P_vapor y_CO2_liquidside=P_CO2_gasside/H y_water_liquidside=1-y_CO2_liquidside MM_m=y_CO2_liquidside*MM_CO2+y_water_liquidside*MM_water "molar mass of mixture" w_CO2_liquidside=y_CO2_liquidside*MM_CO2/MM_m m_m=rho_m*V_m m_CO2=w_CO2_liquidside*m_m V=(m_CO2*R_CO2*(T2+273))/(P2*Convert(bar, kPa))

17-105 The natural log of the equilibrium constant as a function of temperature between 298 to 3000 K for the equilibrium reaction CO  H2 O  CO2  H2 is to be tabulated and compared to those given in Table A-28 Analysis The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * * D G *(T ) = nCO2 gCO2 (T ) + n H2 g H2 (T ) - nCO gCO (T ) - n H2O g H2O (T )

and the Gibbs functions are defined as * gCO (Tprod ) = (h - Tprod s )CO * gH2O (Tprod ) = (h - Tprod s )H2O * gCO2 (Tprod ) = (h - Tprod s )CO2 * gH2 (Tprod ) = (h - Tprod s )H2

The copy of entire EES solution with resulting parametric table is given next: T_prod = 298 [K] R_u=8.314 [kJ/kmol-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-638

g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) DELTAG =1*g_CO2+1*g_H2-1*g_CO-1*g_H2O K_p = exp(-DELTAG /(R_u*T_prod )) lnK_p=ln(k_p)

Tprod [K]

ln K p

298 500 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000

11,58 4,939 0,3725

17-106 Ethyl alcohol C2 H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 90 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted. Analysis The complete combustion reaction in this case can be written as

C2 H5 OH (gas) + (1 + Ex)ath [O 2 + 3.76N 2 ]¾ ¾ ® 2 CO 2 + 3 H 2 O + ( Ex)( ath ) O 2 + f N 2 where ath is the stoichiometric coefficient for air. The oxygen balance gives

1 + (1 + Ex)ath ´ 2 = 2´ 2 + 3´ 1 + ( Ex)(ath )´ 2 The reaction equation with products in equilibrium is

C2 H5 OH (gas) + (1 + Ex) ath [O 2 + 3.76N 2 ]¾ ¾ ® a CO 2 + b CO + d H 2 O + e O 2 + f N 2 The coefficients are determined from the mass balances Carbon balance: 2= a+ b Hydrogen balance: 6 = 2d ¾ ¾ ® d= 3 Oxygen balance:

1+ (1 + Ex)ath ´ 2 = a´ 2 + b + d + e´ 2

Nitrogen balance: (1 + Ex)ath ´ 3.76 = f Solving the above equations, we find the coefficients to be Ex = 0.9, ath  3 , a = 2, b = 0.00008644, d = 3, e = 2.7, f = 21.43 Then, we write the balanced reaction equation as

C2 H5 OH (gas) + 5.7 [O 2 + 3.76N 2 ]¾ ¾ ® 2 CO 2 + 0.00008644 CO + 3 H 2 O + 2.7 O2 + 21.43 N 2 Total moles of products at equilibrium are

13-639

The assumed equilibrium reaction is

CO2 ¬ ¾® CO + 0.5O2 The K p value of a reaction at a specified temperature can be determined from the Gibbs function data using

K p = e- D G*(T )/ RuT or ln K p = - D G *(T ) / RuT where * * * D G *(T ) = n CO g CO (Tprod ) + n O2 g O2 (Tprod ) - n CO2 g CO2 (Tprod )

and the Gibbs functions are defined as * gCO (Tprod ) = (h - Tprod s )CO * gO2 (Tprod ) = (h - Tprod s )O2 * gCO2 (Tprod ) = (h - Tprod s )CO2

The equilibrium constant is also given by 1+ 0.5- 1

ö be0.5 æ çç P ÷ ÷ Kp = ÷ a ççè N tot ÷ ø

0.5

ö (0.00008644)(2.7)0.5 æ çç 1 ÷ ÷ = 0.00001316 ç è 29.13ø÷ 2

A steady flow energy balance gives

HR = HP where o H R = hfuel@25 ° C + 5.7hO2@25° C + 21.43hN2@25° C

= (- 235,310 kJ/kmol) + 5.7(0) + 21.43(0) = - 235,310 kJ/kmol H P = 2hCO2@Tprod + 0.00008644hCO@Tprod + 3hH2O@Tprod + 2.7hO2@Tprod + 21.43hN2@Tprod

Solving the energy balance equation using EES, we obtain the adiabatic flame temperature to be

Tprod = 1569 K The copy of entire EES solution including parametric studies is given next: T_reac= (25+273.15) [K] "The reactant temperature is" Q_out = 0 [kJ] "For adiabatic combustion of 1 kmol of fuel" PercentEx = 90 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The complete combustion reaction equation for excess air is: C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O +Ex*A_th O2 + f N2" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "Oxygen Balance for complete combustion" "The reaction equation for excess air and products in equilibrium is: C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" 2=a + b "Carbon Balance" 6=2*d "Hydrogen Balance" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 "Oxygen Balance" (1+Ex)*A_th*3.76 = f "Nitrogen Balance" N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-640

each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3 "atm" "The equilibrium constant is also given by K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 [kJ/kmol] H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) PercentEx [%]

a th

Tprod

10 20 30 40 50 60 70 80 90 100

1.921 1.97 1.988 1.995 1.998 1.999 2 2 2 2

3 3 3 3 3 3 3 3 3 3

0.07868 0.03043 0.01212 0.004983 0.002111 0.0009184 0.0004093 0.0001863 0.00008644 0.00004081

3 3 3 3 3 3 3 3 3 3

0.3393 0.6152 0.9061 1.202 1.501 1.8 2.1 2.4 2.7 3

12.41 13.54 14.66 15.79 16.92 18.05 19.18 20.3 21.43 22.56

2191 2093 1996 1907 1826 1752 1685 1625 1569 1518

[K]

17-107 The percent theoretical air required for the combustion of octane such that the volume fraction of CO in the products is less than 0.1% and the heat transfer are to be determined. Also, the percent theoretical air required for 0.1% CO in the products as a function of product pressure is to be plotted. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-641

Analysis The complete combustion reaction equation for excess air is

C8 H18  Pth a th O 2  3.76N 2    8 CO 2  9 H 2 O  ( Pth  1)a th O 2  f N 2 The oxygen balance is

Pth a th  2  8  2  9  1  ( Pth  1)a th  2 The reaction equation for excess air and products in equilibrium is

C8 H18  Pth a th O 2  3.76N 2    a CO 2  b CO  d H 2 O  e O 2  f N 2 The coefficients are to be determined from the mass balances Carbon balance:

8ab

Hydrogen balance:

18  2d   d  9

Oxygen balance:

Pth a th  2  a  2  b  d  e  2

Nitrogen balance:

Pth a th  3.76  f

Volume fraction of CO must be less than 0.1%. That is, y CO 

b b   0.001 N tot a  b  d  e  f

The assumed equilibrium reaction is

CO 2  CO  0.5O 2 The K p value of a reaction at a specified temperature can be determined from the Gibbs function data:  gCO (Tprod )  (h  Tprod s)CO  (53,826)  (2000)(258.48)  570,781 kJ/kmol  gO2 (Tprod )  (h  Tprod s)O2  (59,193)  (2000)(268.53)  477,876 kJ/kmol  gCO2 (Tprod )  (h  Tprod s)CO2  (302,128)  (2000)(309.00)  920,121 kJ/kmol

The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa are obtained from EES. Substituting,    G *(Tprod )   CO gCO (Tprod )   O2 gO2 (Tprod )  CO2 gCO2 (Tprod )

 1(570,781)  0.5(477,876)  (920,121)  110, 402 kJ/kmol

  G * (Tprod )    exp  110,402   0.001308 K p  exp  (8.314 )(2000 )   Ru Tprod      The equilibrium constant is also given by be 0.5  P    Kp  a  N tot 

1 0.5 1

be 0.5  Pprod / 101 .3   a  a  b  d  e  f 

1 0.5 1

The steady flow energy balance gives

H R  Qout  H P where

HR  1hC8H18 @ 298 K  Pth a th h O2 @ 298 K  (Pth a th  3.76)h N2 @ 298 K  (208, 459)  Pth a th (0)  (Pth a th  3.76)(0)  208, 459 kJ/kmol

HP  ah CO2 @ 2000 K  bh CO @ 2000 K  dh H2O @ 2000 K  eh O2 @ 2000 K  f h N2 @ 2000 K  a(302,128)  b(53,826)  d(169,171)  e(59,193)  f (56,115) The enthalpies are obtained from EES. Solving all the equations simultaneously using EES, we obtain

Pth  1.024, ath  12.5, a  7.935, b  0.06544, d  9, e  0.3289, f  48.11 PercentTh  Pth 100  1.024 100  102.4% © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-642

Qout  995,500 kJ / kmol C8H18 The copy of entire EES solution including parametric studies is given next: "The product temperature is:" T_prod = 2000 "[K]" "The reactant temperature is:" T_reac= 25+273 "[K]" "PercentTH is Percent theoretical air" Pth= PercentTh/100 "Pth = % theoretical air/100" P_prod = 5 "[atm]" *convert(atm,kPa)"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C8H18+ Pth*A_th (O2 +3.76N2)=8 CO2 + 9 H2O +(Pth-1)*A_th O2 + f N2" "Oxygen Balance for complete combustion:" Pth*A_th*2=8*2+9*1 + (Pth-1)*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C8H18+ Pth*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 8=a + b "Hydrogen Balance:" 18=2*d "Oxygen Balance:" Pth*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" Pth*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The volume faction of CO in the products is to be less than 0.1%. For ideal gas mixtures volume fractions equal mole fractions." "The mole fraction of CO in the product gases is:" y_CO = 0.001 y_CO = b/N_tot "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P H_R=1*ENTHALPY(C8H18,T=T_reac)+Pth*A_th*ENTHALPY(O2,T=T_reac)+Pth*A_th*3.76*ENTHALPY(N2,T= T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]"

Pprod [kPa]

PercentTh [%]

13-643

100 300 500 700 900 1100 1300 1500 1700 1900 2100 2300

112 104.1 102.4 101.7 101.2 101 100.8 100.6 100.5 100.5 100.4 100.3

112

PercentTh %

110 108 106 104 102 100 0

500

1000

1500

2000

2500

Pprod [kPa]

17-108 It is to be shown that as long as the extent of the reaction, , for the disassociation reaction X2  2X is smaller KP 4 + KP

than one,  is given by a =

Assumptions The reaction occurs at the reference temperature. Analysis The stoichiometric and actual reactions can be written as Stoichiometric:

X2 Û 2X (thus nX2 = 1 and nX = 2)

Actual: X 2 Û (1- a ) X 2 + 2a X prod.

react.

The equilibrium constant K p is given by n - n X2

X N XnX æ P ö ÷ ç ÷ K p = nX2 çç ÷ ÷ N X2 çè N total ø

2- 1

ö (2a ) 2 æ çç 1 ÷ = ÷ ÷ ç è (1- a ) a + 1ø

4a 2 (1- a )(1 + a )

Solving this expression for  gives

13-644

KP 4 + KP

17-109 It is to be shown that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same.

Analysis The total Gibbs function of the three phase mixture of a pure substance can be expressed as

G = ms g s + m g + mg g g where the subscripts s, , and g indicate solid, liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions, g) constant yields

dG = g s dms + g dm + g g dmg From conservation of mass,

dms + dm + dmg = 0 ¾ ¾ ® dms = - dm - dmg Substituting,

dG = - g s (dm + dmg ) + g dm + g g dmg Rearranging,

dG = ( g - g s ) dm + ( g g - g s ) dmg For equilibrium, dG = 0. Also dm and dmg can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. It yields

g = g s and g g = g s Combining these two conditions gives the desired result,

g = gs = gs

17-110 It is to be shown that when the two phases of a two-component system are in equilibrium, the specific Gibbs function of each phase of each component is the same.

13-645

Analysis The total Gibbs function of the two phase mixture can be expressed as

G = (m 1 g 1 + mg1 g g1 ) + (m 2 g 2 + mg 2 g g 2 ) where the subscripts  and g indicate liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions) constant yields

dG = g 1dm 1 + g g1dmg1 + g 2 dm 2 + g g 2 dmg 2 From conservation of mass,

dmg1 = - dm 1 and dmg 2 = - dm 2 Substituting,

dG = ( g 1 - g g1 )dm 1 + ( g 2 - g g 2 )dm 2 For equilibrium, dG = 0. Also dm 1 and dm 2 can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. Then we have

g 1 = g g1 and g 2 = g g 2 which is the desired result.

17-111 Using Henry’s law, it is to be shown that the dissolved gases in a liquid can be driven off by heating the liquid. Analysis Henry’s law is expressed as

yi,liquid side (0) =

Pi, gas side (0) H

Henry’s constant H increases with temperature, and thus the fraction of gas i in the liquid yi,liquid side decreases. Therefore, heating a liquid will drive off the dissolved gases in a liquid.

Fundamentals of Engineering (FE) Exam Problems

17-112 Of the reactions given below, the reaction whose equilibrium composition at a specified temperature is not affected by pressure is (a) H2  1/ 2O2  H2 O (b) CO  1/ 2O2  CO2 (c) N2  O2  2NO (d) N2  2N (e) all of the above. Answer (c) N2  O2  2NO

13-646

17-113 Of the reactions given below, the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and temperature is (a) H2  1/ 2O2  H2 O (b) CO  1/ 2O2  CO2 (c) N2  O2  2NO (d) N2  2N (e) none of the above. Answer (d) N2  2N

17-114 If the equilibrium constant for the reaction H2  1/ 2O2  H2 O is K, the equilibrium constant for the reaction 2H2 O  2H2  O2 at the same temperature is (a) 1/ K (b) 1/ (2 K ) (c) 2K (d) K 2 (e) 1 / K 2 Answer (e) 1 / K 2

17-115 If the equilibrium constant for the reaction CO  1/ 2O2  CO2 is K, the equilibrium constant for the reaction CO2  3N2  CO  1/ 2O2  3N2 at the same temperature is (a) 1/ K (b) 1/ ( K + 3) (c) 4K (d) K (e) 1 / K 2 Answer (a) 1/ K

17-116 The equilibrium constant for the reaction H2  1/ 2O2  H2 O at 1 atm and 1500C is given to be K. Of the reactions given below, all at 1500C, the reaction that has a different equilibrium constant is (a) H2  1/ 2O2  H2 O at 5 atm , (b) 2H2  O2  2H2 O at 1 atm , (c) H2  O2  H2 O  1/ 2O2 at 2atm , (d) H2  1/ 2O2  3N2  H2 O  3N2 at 5 atm , (e) H2  1/ 2O2  3N2  H2 O  3N2 at 1 atm , Answer (b) 2H2  O2  2H2 O at 1 atm ,

13-647

17-117 Moist air is heated to a very high temperature. If the equilibrium composition consists of H 2 O , O2 , N 2 , OH,

H 2 , and NO, the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (c) 3

17-118 Propane C3 H8 is burned with air, and the combustion products consist of CO2 , CO, H 2 O , O2 , N 2 , OH, H 2 , and NO. The number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (d) 4

17-119 Consider a gas mixture that consists of three components. The number of independent variables that need to be specified to fix the state of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (d) 4

17-120 The value of Henry’s constant for CO2 gas dissolved in water at 290 K is 12.8 MPa. Consider water exposed to air at 100 kPa that contains 3 percent CO2 by volume. Under phase equilibrium conditions, the mole fraction of

CO2 gas dissolved in water at 290 K is (a) (b) (c) (d) (e)

2.3  104 3.0  104

0.80  104 2.2  104 5.6  104

Answer (a) 2.3  104 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). H=12.8 [MPa]

13-648

P=0.1 [MPa] y_CO2_air=0.03 P_CO2_air=y_CO2_air*P y_CO2_liquid=P_CO2_air/H "Some Wrong Solutions with Common Mistakes:" W1_yCO2=P_CO2_air*H "Multiplying by H instead of dividing by it" W2_yCO2=P_CO2_air "Taking partial pressure in air"

17-121 The solubility of nitrogen gas in rubber at 25C is 0.00156 kmol/ m3 .bar . When phase equilibrium is established, the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at 300 kPa is (a) 0.005 kg/ m3 (b) 0.018 kg/ m3 (c) 0.047 kg/ m3 (d) 0.13 kg/ m3 (e) 0.28 kg/ m3 Answer (d) 0.13 kg/ m3 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=25 [C] S=0.00156 [kmol/bar-m^3] P_N2=3 [bar] MM_N2=28 [kg/kmol] S_mass=S*MM_N2 rho_solid=S_mass*P_N2 "Some Wrong Solutions with Common Mistakes:" W1_density=S*P_N2 "Using solubility per kmol"

17-122 … 17-127 Design and Essay Problems

SOLUTIONS MANUAL Thermodynamics: An Engineering Approach 10th Edition Yunus A. Çengel, Michael A. Boles, Mehmet Kanoğlu McGraw-Hill, 2023

13-649

Chapter 18 COMPRESSIBLE FLOW PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGraw Hill LLC and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw Hill LLC: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of McGraw Hill LLC.

13-650

CYU - Check Your Understanding CYU 18-1 ■ Check Your Understanding CYU 18-1.1 What is the correct expression for the stagnation enthalpy of a fluid? (a) h + V 2 / 2 (b) h + V 2 / 2 + gz (c) u + P / r (d) u + P / r + gz Answer: (a) h + V 2 / 2 CYU 18-1.2 What is the correct expression for the stagnation temperature of an ideal gas? (a) T + V 2 / 2 (b) T + V 2 / (2c p ) (c) T + V 2 / (2c p )+ gz / c p (d) T / c p + V 2 / (2c p ) Answer: (b) T + V 2 / (2c p )

CYU 18-1.3 What is the dynamic temperature of air flowing at 200 m / s ? Take the specific heat of air to be c p = 1kJ / kg ×K . (a) 5 K (b) 10 K (c) 15 K (d) 20 K (e) 25 K Answer: (d) 20 K Tdynamic = V 2 / (2cp ) = (200 m / s) 2 / [(2)(1 kJ / kgK)](1/1000) = 20 K

since 1kJ / kg = 1000 m2 / s2

CYU 18-2 ■ Check Your Understanding CYU 18-2.1 Select the incorrect statement regarding the speed of sound and the Mach number. (a) The Mach number of an aircraft cruising at constant velocity in still air may be different at different locations. (b) The Mach number is the ratio of the actual speed of the fluid to the speed of sound in the same fluid at the same state. (c) The sonic speed is defined as the speed at which an infinitesimally small pressure wave travels through a medium. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-651

(d) The flow of a fluid is called supersonic when Mach number is greater than 1. (e) The speed of sound in a specified ideal gas depends on temperature and pressure. Answer: (e) The speed of sound in a specified ideal gas depends on temperature and pressure. CYU 18-2.2 For an ideal gas, the speed of sound is not a function of (a) Gas constant (b) Specific heat ratio (c) Pressure (d) Temperature Answer: (c) Pressure CYU 18-2.3 The speed of a fluid is equal to speed of sound when (a) Ma < 1 (b) Ma = 1 (c) Ma > 1 (d) Ma  1 (e) Ma >> 1 Answer: (b) Ma = 1

CYU 18-3 ■ Check Your Understanding CYU 18-3.1 Consider the supersonic flow of a gas through a converging-diverging nozzle. The velocity of the gas increases after passing the throat although the flow area increases. This is due to the (a) rapid increase in the fluid density (b) rapid decrease in the fluid density (c) rapid decrease in the fluid temperature (d) rapid increase in the fluid temperature (e) rapid increase in the fluid pressure Answer: (b) rapid decrease in the fluid density CYU 18-3.2 During subsonic flow of a gas through a duct, the pressure decreases in _______ ducts. (a) converging (b) converging-diverging (c) diverging (d) converging (e) diverging Answer: (d) converging

13-652

CYU 18-3.3 During subsonic flow of a gas through a duct, the pressure increases in _______ ducts. (a) converging (b) diverging (c) converging-diverging (d) diverging (e) converging Answer: (d) diverging CYU 18-3.4 During supersonic flow of a gas through a duct, the pressure decreases in _______ ducts. (a) converging (b) converging-diverging (c) diverging (d) converging (e) diverging Answer: (e) diverging CYU 18-3.5 During supersonic flow of a gas through a duct, the pressure increases in ________ ducts. (a) converging (b) diverging (c) converging-diverging (d) diverging (e) converging Answer: (e) converging CYU 18-3.6 Select the incorrect statement for a converging subsonic nozzle. (a) Velocity increases (b) Pressure decreases (c) Density increases (d) Temperature decreases (e) Mach number increases Answer: (c) Density increases CYU 18-3.7 Select the incorrect statement for a converging supersonic diffuser. (a) Velocity decreases (b) Pressure increases (c) Density increases (d) Temperature increases (e) Mach number increases Answer: (e) Mach number increases

13-653

CYU 18-4 ■ Check Your Understanding CYU 18-4.1 The maximum mass flow rate of a particular ideal gas through a nozzle cannot be controlled by changing the (a) Back pressure (b) Stagnation temperature at the inlet (c) Stagnation pressure at the inlet (d) Throat area Answer: (a) Back pressure CYU 18-4.2 If the back pressure Pb during flow of a gas in a converging nozzle is lower than the critical pressure P* , select the incorrect statement below. (a) The Mach number at the exit plane is unity. (b) The pressure at the exit plane Pe is less than the critical pressure P* . (c) The mass flow rate is the maximum flow rate. (d) None of these Answer: (b) The pressure at the exit plane Pe is less than the critical pressure P* . CYU 18-4.3 Select the correct statement(s) below regarding flow through nozzles. (a) The highest velocity to which a fluid can be accelerated in a converging nozzle is limited to the sonic velocity. (b) Accelerating a fluid to supersonic velocities can be accomplished only by a converging-diverging nozzle. (c) A converging-diverging nozzle is standard equipment in supersonic aircraft and rocket propulsion. (d) All of the above Answer: (d) All of the above

CYU 18-5 ■ Check Your Understanding CYU 18-5.1 Select the incorrect statement below regarding shock waves during flow in a duct. (a) The stagnation temperature of an ideal gas remains constant across the shock. (b) The entrance and exit flow areas for the control volume of a shock wave are approximately equal. (c) Gas flow through the shock wave can be approximated as being isentropic. (d) None of the above. Answer: (c) Gas flow through the shock wave can be approximated as being isentropic. CYU 18-5.2 For gas flow through an adiabatic duck, combining the conservation of mass and conservation of energy equations into a single equation and plotting it on the h-s diagram yield a curve called (a) Fanno line (b) Oblique shock line (c) Velocity inversion line (d) Rayleigh line

13-654

Answer: (a) Fanno line CYU 18-5.3 Select the incorrect statement below for the variation of flow properties across a normal shock during gas flow in a duct. (a) Velocity decreases (b) Pressure increases (c) Density increases (d) Temperature increases (e) Mach number increases Answer: (e) Mach number increases CYU 18-5.4 Select the correct statement(s) regarding oblique shocks. I. The Mach number increases across an oblique shock. II. Oblique shocks are possible only if the upstream flow is supersonic. III. Downstream of an oblique shock can be subsonic, sonic, or supersonic. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (d) II and III

CYU 18-6 ■ Check Your Understanding CYU 18-6.1 Select the correct statement(s) regarding Rayleigh flow. I. The stagnation enthalpy and stagnation temperature increase during Rayleigh flow when heat is transferred to the fluid. II. The entropy of a fluid always increase or remain constant during Rayleigh flow. III. The mass balance through a constant-area duct during Rayleigh flow can be expressed as r 1V1 = r 2V2 . (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (c) I and III CYU 18-6.2 Which statement(s) below are correct regarding Rayleigh flow? (a) During heating, fluid temperature always increases if the Rayleigh flow is supersonic, but the temperature may actually drop if the flow is subsonic. (b) Heating increases the flow velocity in subsonic flow but decreases it in supersonic flow. (c) Density decreases with heat transfer to the fluid in subsonic flow but increases with heat gain in supersonic flow. (d) All of the above

13-655

Answer: (d) All of the above

CYU 18-7 ■ Check Your Understanding CYU 18-7.1 During flow through a nozzle, steam which propagates in the wet region without any condensation taking place is called (a) Superheated steam (b) Supersaturated steam (c) Saturated steam (d) Subcooled steam (e) Compressed steam Answer: (b) Supersaturated steam CYU 18-7.2 Select the correct statement(s) below regarding steam nozzles. I. Steam flowing through a nozzle at a high speed is assumed to begin condensation when the 4 percent moisture line is crossed. II. The critical-pressure ratio P* / P0 for steam depends on the nozzle inlet state as well as on whether the steam is superheated or saturated at the nozzle inlet. III. During the expansion of steam in a nozzle, steam may temporarily exist in the wet region without containing any liquid. (a) Only I (b) I and II (c) I and III (d) II and III (e) I, II, and III Answer: (e) I, II, and III

13-656

PROBLEMS Stagnation Properties 18-1C We are to discuss the temperature change from an airplane’s nose to far away from the aircraft. Analysis The temperature of the air rises as it approaches the nose because of the stagnation process. Discussion In the frame of reference moving with the aircraft, the air decelerates from high speed to zero at the nose (stagnation point), and this causes the air temperature to rise.

18-2C We are to define dynamic temperature. Analysis Dynamic temperature is the temperature rise of a fluid during a stagnation process. Discussion When a gas decelerates from high speed to zero speed at a stagnation point, the temperature of the gas rises.

18-3C We are to discuss the measurement of flowing air temperature with a probe – is there significant error? Analysis No, there is not significant error, because the velocities encountered in air-conditioning applications are very low, and thus the static and the stagnation temperatures are practically identical. Discussion If the air stream were supersonic, however, the error would indeed be significant.

18-4C Solution We are to define and discuss stagnation enthalpy. Analysis Stagnation enthalpy combines the ordinary enthalpy and the kinetic energy of a fluid, and offers convenience when analyzing high-speed flows. It differs from the ordinary enthalpy by the kinetic energy term. Discussion Most of the time, we mean specific enthalpy, i.e., enthalpy per unit mass, when we use the term enthalpy.

18-5 Air at 320 K is flowing in a duct. The temperature that a stationary probe inserted into the duct will read is to be determined for different air velocities. Assumptions The stagnation process is isentropic. Properties

The specific heat of air at room temperature is

Analysis The air which strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature, T0 . It is determined from T0  T  (a) T0  320 K+

V2 . The results for each case are calculated below: 2c p

 1 kJ/kg  (1 m/s)2    320.0 K 2  1.005 kJ/kg  K  1000 m2 / s2 

(b) T0  320 K +

(10 m/s) 2  1 kJ/kg     320.1 K 2  1.005 kJ/kg  K  1000 m 2 / s 2 

(100 m/s) 2  1 kJ/kg     325.0 K 2  1.005 kJ/kg  K  1000 m 2 / s 2 

13-657

(d) T0  320 K +

(1000 m/s) 2  1 kJ/kg     817.5 K 2  1.005 kJ/kg  K  1000 m 2 / s 2 

Discussion Note that the stagnation temperature is nearly identical to the thermodynamic temperature at low velocities, but the difference between the two is significant at high velocities.

EES (Engineering Equation Solver) SOLUTION "Given" T=320 [K] Vel_a=1 [m/s] Vel_b=10 [m/s] Vel_c=100 [m/s] Vel_d=1000 [m/s] "Properties" c_p=1.005 [kJ/kg-K] "Analysis" T_0_a=T+Vel_a^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) T_0_b=T+Vel_b^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) T_0_c=T+Vel_c^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) T_0_d=T+Vel_d^2/(2*c_p)*Convert(m^2/s^2, kJ/kg)

18-6 Air flows through a device. The stagnation temperature and pressure of air and its velocity are specified. The static pressure and temperature of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas. Properties The properties of air at an anticipated average temperature of 600 K are c p = 1.051kJ / kg ×K and k = 1.376. Analysis The static temperature and pressure of air are determined from T = T0 -

æ 1 kJ/kg ÷ ö V2 (520 m/s) 2 çç = 673.2 ÷= 544.6 K @ 545 K ø 2c p 2´ 1.051 kJ/kg ×K çè1000 m 2 / s 2 ÷

and k / ( k - 1)

æT ö ÷ P = P0 ççç ÷ ÷ çèT0 ÷ ø

1.376/ (1.376- 1)

æ544.6 K ö ÷ = (0.4 MPa) çç ÷ ÷ èç673.2 K ø

= 0.184 MPa

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

EES (Engineering Equation Solver) SOLUTION "Given" P_0=600 [kPa] T_0=(400+273) [K] Vel=570 [m/s] "Properties" c_p=CP(air, T=600) "at the anticipated average temperature of 600 K" c_v=CV(air, T=600) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-658

k=c_p/c_v "Analysis" T=T_0-Vel^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P=P_0*(T/T_0)^(k/(k-1))

18-7E Steam flows through a device. The stagnation temperature and pressure of steam and its velocity are specified. The static pressure and temperature of the steam are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Steam is an ideal gas. Properties Steam can be treated as an ideal gas with c p = 0.4455 Btu / lbm ×R and k = 1.329. Analysis T = T0 -

The static temperature and pressure of steam are determined from æ 1 Btu/lbm ÷ ö V2 (900 ft/s)2 çç = 700°F ÷= 663.7°F 2 2÷ ç 2c p 2´ 0.4455 Btu/lbm ×°F è 25,037 ft / s ø k /( k - 1)

æT ö ÷ P = P0 ççç ÷ ÷ çèT ø÷ 0

1.329/(1.329- 1)

æ1123.7 R ÷ ö = (120 psia) çç ÷ èç 1160 R ÷ ø

= 105.5 psia

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

EES (Engineering Equation Solver) SOLUTION "Given" P_0=120 [psia] T_0=(700+460) [R] Vel=900 [ft/s] "Properties" c_p=0.445 [Btu/lbm-R] k=1.329 "Analysis" T=T_0-Vel^2/(2*c_p)*Convert(ft^2/s^2, Btu/lbm) P=P_0*(T/T_0)^(k/(k-1))

18-8 The states of different substances and their velocities are specified. The stagnation temperature and stagnation pressures are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Helium and nitrogen are ideal gases. Analysis (a) Helium can be treated as an ideal gas with c p = 5.1926 kJ / kg ×K and k = 1.667. Then the stagnation temperature and pressure of helium are determined from T0 = T +

æ 1 kJ/kg ÷ ö V2 (240 m/s) 2 çç = 50°C + = 55.5°C 2 2÷ ÷ ç è ø 2c p 2´ 5.1926 kJ/kg ×°C 1000 m / s k / (k - 1)

æT ö P0 = P çç 0 ÷ ÷ çè T ø÷

1.667/ (1.667- 1)

æ328.7 K ö÷ = (0.25 MPa) çç ÷ çè323.2 K ÷ ø

= 0.261 MPa

(b) Nitrogen can be treated as an ideal gas with c p = 1.039 kJ / kg ×K and k =1.400. Then the stagnation temperature and pressure of nitrogen are determined from T0 = T +

æ 1 kJ/kg ÷ ö V2 (300 m/s)2 çç = 50°C + = 93.3°C 2 2÷ ÷ ç è 2c p 2´ 1.039 kJ/kg ×°C 1000 m / s ø

13-659

k / ( k - 1)

æT ö P0 = P çç 0 ÷ ÷ çè T ø÷

1.4/ (1.4- 1)

æ366.5 K ö÷ = (0.15 MPa) çç ÷ çè323.2 K ø÷

= 0.233 MPa

(c) Steam can be treated as an ideal gas with c p = 1.865 kJ / kg ×K and k =1.329. Then the stagnation temperature and pressure of steam are determined from T0 = T +

æ 1 kJ/kg ö V2 (480 m/s) 2 ÷ çç = 350°C + ÷ ÷= 411.8°C = 685 K 2c p 2´ 1.865 kJ/kg ×°C çè1000 m 2 / s 2 ø k / ( k - 1)

æT ö P0 = P çç 0 ÷ ÷ çè T ø÷

1.329/ (1.329- 1)

æ 685 K ö÷ = (0.1 MPa) çç ÷ çè623.2 K ÷ ø

= 0.147 MPa

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

EES (Engineering Equation Solver) SOLUTION "Given" P_a=250 [kPa] T_a=(50+273) [K] Vel_a=240 [m/s] P_b=150 [kPa] T_b=(50+273) [K] Vel_b=300 [m/s] P_c=100 [kPa] T_c=(350+273) [K] Vel_c=480 [m/s] "Properties" c_p_He=5.1926 [kJ/kg-K] k_He=1.667 c_p_N2=1.039 [kJ/kg-K] k_N2=1.4 c_p_s=1.865 [kJ/kg-K] k_s=1.329 "Analysis" "(a)" T_0_a=T_a+Vel_a^2/(2*c_p_He)*Convert(m^2/s^2, kJ/kg) P_0_a=P_a*(T_0_a/T_a)^(k_He/(k_He-1)) "(b)" T_0_b=T_b+Vel_b^2/(2*c_p_N2)*Convert(m^2/s^2, kJ/kg) P_0_b=P_b*(T_0_b/T_b)^(k_N2/(k_N2-1)) "(c)" T_0_c=T_c+Vel_c^2/(2*c_p_s)*Convert(m^2/s^2, kJ/kg) P_0_c=P_c*(T_0_c/T_c)^(k_s/(k_s-1))

18-9 The state of air and its velocity are specified. The stagnation temperature and stagnation pressure of air are to be determined. Assumptions 1 The stagnation process is isentropic. 2 Air is an ideal gas. Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4. Analysis The stagnation temperature of air is determined from T0 = T +

æ 1 kJ/kg ÷ ö V2 (325 m/s) 2 çç = 238 K + = 290.5 @ 291 K 2 2÷ ÷ ç è 2c p 2´ 1.005 kJ/kg ×K 1000 m /s ø

Other stagnation properties at the specified state are determined by considering an isentropic process between the specified state and the stagnation state, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-660

k /( k - 1)

æT ö P0 = P çç 0 ÷ ÷ çè T ÷ ø

1.4/(1.4- 1)

æ290.5 K ö÷ = (36 kPa) çç ÷ çè 238 K ø÷

= 72.37 kPa @ 72.4 kPa

Discussion Note that the stagnation properties can be significantly different than thermodynamic properties.

EES (Engineering Equation Solver) SOLUTION "Given" P=36 [kPa] T=238 [K] Vel=470 [m/s] "Properties" c_p=CP(air, T=T) c_v=CV(air, T=T) k=c_p/c_v "Analysis" T_0=T+Vel^2/(2*c_p)*Convert(m^2/^s^2, kJ/kg) P_0=P*(T_0/T)^(k/(k-1))

18-10 The inlet stagnation temperature and pressure and the exit stagnation pressure of air flowing through a compressor are specified. The power input to the compressor is to be determined. Assumptions 1 The compressor is isentropic. 2 Air is an ideal gas.

Properties The properties of air at room temperature are c p = 1.005 kJ / kg ×K and k = 1.4. Analysis The exit stagnation temperature of air T02 is determined from ( k - 1)/ k

æP ÷ ö T02 = T01 ççç 02 ÷ ÷ ÷ çè P ø 01

(1.4- 1)/1.4

æ900 ö ÷ = (308.2 K) çç ÷ ÷ èç100 ø

= 577.4 K

From the energy balance on the compressor, Win = m(h20 - h01 )

or,

Win = mc p (T02 - T01 ) = (0.04 kg/s)(1.005 kJ/kg ×K)(577.4 - 308.2)K = 10.8 kW Discussion Note that the stagnation properties can be used conveniently in the energy equation.

13-661

EES (Engineering Equation Solver) SOLUTION "Given" P_01=100 [kPa] T_01=(35+273) [K] P_02=900 [kPa] m_dot=0.04 [kg/s] "Properties" c_p=CP(air, T=T_01) c_v=CV(air, T=T_01) k=c_p/c_v "Analysis" T_02=T_01*(P_02/P_01)^((k-1)/k) W_dot_in=m_dot*C_p*(T_02−T_01)

18-11 The inlet stagnation temperature and pressure and the exit stagnation pressure of products of combustion flowing through a gas turbine are specified. The power output of the turbine is to be determined. Assumptions 1 The expansion process is isentropic. 2 Products of combustion are ideal gases. Properties The properties of products of combustion are c p = 1.157 kJ / kg ×K , R = 0.287 kJ / kg ×K , and k = 1.33. Analysis The exit stagnation temperature T02 is determined to be ( k - 1)/ k

æP ö ÷ T02 = T01 ççç 02 ÷ ÷ çè P ø÷ 01

(1.33- 1)/1.33

æ 0.1 ÷ ö = (1113.2 K) çç ÷ çè0.90 ÷ ø

= 645.4 K

Also,

c p = kcv = k (c p - R) ¾ ¾ ® cp = =

kR k- 1

1.33(0.287 kJ/kg ×K ) 1.33 - 1

= 1.157 kJ/kg ×K

From the energy balance on the turbine,

- wout = (h20 - h01 ) or, wout = c p (T01 - T02 ) = (1.157 kJ/kg ×K)(1113.2 - 645.4)K = 541.2 kJ/kg @541kJ / kg Discussion Note that the stagnation properties can be used conveniently in the energy equation. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-662

EES (Engineering Equation Solver) SOLUTION "Given" P_01=750 [kPa] T_01=(690+273) [K] P_02=100 [kPa] k=1.33 R=0.287 [kJ/kg-K] "Analysis" T_02=T_01*(P_02/P_01)^((k-1)/k) c_p=(k*R)/(k-1) w_out=c_p*(T_01-T_02)

Speed of Sound and Mach Number 18-12C We are to define and discuss sound and how it is generated and how it travels. Analysis Sound is an infinitesimally small pressure wave. It is generated by a small disturbance in a medium. It travels by wave propagation. Sound waves cannot travel in a vacuum. Discussion Electromagnetic waves, like light and radio waves, can travel in a vacuum, but sound cannot.

18-13C We are to discuss whether sound travels faster in warm or cool air. Analysis Sound travels faster in warm (higher temperature) air since c =

kRT .

Discussion On the microscopic scale, we can imagine the air molecules moving around at higher speed in warmer air, leading to higher propagation of disturbances.

18-14C We are to compare the speed of sound in air, helium, and argon. Analysis Sound travels fastest in helium, since c = kRT and helium has the highest kR value. It is about 0.40 for air, 0.35 for argon, and 3.46 for helium. Discussion We are assuming, of course, that these gases behave as ideal gases – a good approximation at room temperature.

18-15C We are to compare the speed of sound in air at two different pressures, but the same temperature. Analysis Air at specified conditions will behave like an ideal gas, and the speed of sound in an ideal gas depends on temperature only. Therefore, the speed of sound is the same in both mediums. Discussion If the temperature were different, however, the speed of sound would be different.

18-16C We are to discuss sonic velocity – specifically, whether it is constant or it changes. Analysis The sonic speed in a medium depends on the properties of the medium, and it changes as the properties of the medium change. Discussion The most common example is the change in speed of sound due to temperature change.

18-17C We are to examine whether the Mach number remains constant in constant-velocity flow. Analysis In general, no, because the Mach number also depends on the speed of sound in gas, which depends on the temperature of the gas. The Mach number remains constant only if the temperature and the velocity are constant. Discussion It turns out that the speed of sound is not a strong function of pressure. In fact, it is not a function of pressure at all for an ideal gas. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-663

18-18C We are to state whether the propagation of sound waves is an isentropic process. Analysis Yes, the propagation of sound waves is nearly isentropic. Because the amplitude of an ordinary sound wave is very small, and it does not cause any significant change in temperature and pressure. Discussion No process is truly isentropic, but the increase of entropy due to sound propagation is negligibly small.

18-19 Carbon dioxide flows through a nozzle. The inlet temperature and velocity and the exit temperature of CO2 are specified. The Mach number is to be determined at the inlet and exit of the nozzle. Assumptions 1 CO2 is an ideal gas with constant specific heats at room temperature. 2 This is a steady-flow process. Properties The gas constant of carbon dioxide is R = 0.1889 kJ / kg ×K . Its constant pressure specific heat and specific heat ratio at room temperature are c p = 0.8436 kJ / kg ×K and k = 1.288. Analysis (a) At the inlet

c1 =

k1 RT1 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.288)(0.1889 kJ/kg ×K)(800 K) ççç = 441.2 m/s ÷ è 1 kJ/kg ÷ ø

Thus, Ma1 =

V1 50 m/s = = 0.113 c1 441.2 m/s

(b) At the exit,

c2 =

k2 RT2 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.288)(0.1889 kJ/kg ×K)(400 K) çç = 312.0 m/s ÷ çè 1 kJ/kg ø÷

The nozzle exit velocity is determined from the steady-flow energy balance relation,

0 = h2 - h1 +

V2 2 - V12 V 2 - V12  0 = c p (T2 - T1 ) + 2 2 2

0 = (0.8439 kJ/kg ×K)(400 - 800 K) +

ö V2 2 - (50 m/s) 2 æ çç 1 kJ/kg ÷ ¾¾ ® V2 = 823.2 m/s 2 2÷ ÷ ç è ø 2 1000 m / s

Thus, Ma 2 =

V2 823.2 m/s = = 2.64 c2 312.0 m/s

Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using Table A-2:

13-664

At 800 K : c p = 1.169 kJ / kg ×K, k = 1.193 ®

c1 = 424.6 m / s, V1 = 50 m / s, Ma1 = 0.118

At 400 K : c p = 0.939 kJ / kg ×K, k = 1.252 ®

c1 = 307.6 m / s, V2 = 868 m / s, Ma 2 = 2.82

Therefore, the constant specific heat assumption results in an error of 4.4% at the inlet and 6.8% at the exit in the Mach number, which are significant.

EES (Engineering Equation Solver) SOLUTION "Given" T_1=800 [K] Vel_1=50 [m/s] T_2=400 [K] "Properties" R=0.1889 [kJ/kg-K] c_p=0.8439 [kJ/kg-K] k=1.288 "Analysis" "(a)" C_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) Ma_1=Vel_1/C_1 "(b)" C_2=sqrt(k*R*T_2*Convert(kJ/kg, m^2/s^2)) DELTAh=c_p*(T_2−T_1) 0=DELTAh+(Vel_2^2−Vel_1^2)/2*Convert(m^2/s^2, kJ/kg) "energy balance" Ma_2=Vel_2/C_2 "Discussion" "(a)" k_a=1.193 C_1_a=sqrt(k_a*R*T_1*Convert(kJ/kg, m^2/s^2)) Ma_1_a=Vel_1/C_1_a "(b)" c_p_b=0.939 k_b=1.252 C_2_b=sqrt(k_b*R*T_2*Convert(kJ/kg, m^2/s^2)) DELTAh_b=c_p_b*(T_2−T_1) 0=DELTAh_b+(Vel_2_b^2-Vel_1^2)/2*Convert(m^2/s^2, kJ/kg) "energy balance" Ma_2_b=Vel_2_b/C_2_b

18-20 Nitrogen flows through a heat exchanger. The inlet temperature, pressure, and velocity and the exit pressure and velocity are specified. The Mach number is to be determined at the inlet and exit of the heat exchanger. Assumptions 1 N 2 is an ideal gas. 2 This is a steady-flow process. 3 The potential energy change is negligible. Properties The gas constant of N 2 is R = 0.2968 kJ / kg ×K . Its constant pressure specific heat and specific heat ratio at room temperature are c p = 1.040 kJ / kg ×K and k = 1.4. Analysis For the inlet,

c1 =

k1 RT1 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.400)(0.2968 kJ/kg ×K)(283 K) ççç = 342.9 m/s ÷ è 1 kJ/kg ÷ ø

13-665

Thus, Ma1 =

V1 100 m/s = = 0.292 c1 342.9 m/s

From the energy balance on the heat exchanger,

qin = c p (T2 - T1 ) +

V2 2 - V12 2

120 kJ/kg = (1.040 kJ/kg.°C)(T2 - 10°C) +

ö (200 m/s) 2 - (100 m/s) 2 æ çç 1 kJ/kg ÷ 2 2÷ ÷ ç è1000 m / s ø 2

It yields

T2 = 111°C = 384K c2 =

k2 RT2 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.4)(0.2968 kJ/kg ×K)(384 K) ççç = 399 m/s ÷ ÷ è 1 kJ/kg ø

Thus, Ma 2 =

V2 200 m/s = = 0.501 c2 399 m/s

Discussion The specific heats and their ratio k change with temperature, and the accuracy of the results can be improved by accounting for this variation. Using EES (or another property database): At 10°C : c p = 1.038 kJ / kg ×K, k = 1.400 ®

c1 = 343 m / s, V1 = 100 m / s, Ma1 = 0.292

At 111°C : c p = 1.041 kJ / kg ×K, k = 1.399 ®

c2 = 399 m / s, V2 = 200 m / s, Ma 2 = 0.501

Therefore, the constant specific heat assumption results in an error of 4.5% at the inlet and 15.5% at the exit in the Mach number, which are almost identical to the values obtained assuming constant specific heats.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=150 [kPa] T_1=(10+273) [K] Vel_1=100 [m/s] P_2=100 [kPa] Vel_2=200 [m/s] q_in=120 [kJ/kg] "Properties" R=0.2968 [kJ/kg-K] c_p=1.040 [kJ/kg-K] k=1.4 "Analysis" C_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) Ma_1=Vel_1/C_1

13-666

q_in=c_p*(T_2−T_1)+(Vel_2^2-Vel_1^2)/2*Convert(m^2/s^2, kJ/kg) "energy balance" C_2=sqrt(k*R*T_2*Convert(kJ/kg, m^2/s^2)) Ma_2=Vel_2/C_2

18-21 The speed of sound in refrigerant-134a at a specified state is to be determined. Assumptions R-134a is an ideal gas with constant specific heats at room temperature. Properties The gas constant of R-134a is R = 0.08149 kJ / kg ×K . Its specific heat ratio at room temperature is k = 1.108. Analysis From the ideal-gas speed of sound relation,

c= Discussion

kRT =

æ1000 m 2 / s 2 ÷ ö ÷ (1.108)(0.08149 kJ/kg ×K)(60 + 273 K) çç = 173 m / s ÷ çè 1 kJ/kg ÷ ø

Note that the speed of sound is independent of pressure for ideal gases.

EES (Engineering Equation Solver) SOLUTION "Given" P=900 [kPa] T=(60+273) [K] "Properties" R=0.08149 [kJ/kg-K] k=1.108 "Analysis" c=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2))

18-22 The Mach number of an aircraft and the speed of sound in air are to be determined at two specified temperatures. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ / kg ×K . Its specific heat ratio at room temperature is k = 1.4. Analysis From the definitions of the speed of sound and the Mach number, (a) At 300 K,

c= and Ma =

kRT =

æ1000 m 2 / s 2 ö÷ ÷ = 347 m / s (1.4)(0.287 kJ/kg ×K)(300 K) çç ÷ çè 1 kJ/kg ø÷

V 330 m/s = = 0.951 c 347 m/s

(b) At 800 K,

kRT =

æ1000 m 2 / s 2 ö÷ ÷ = 567 m / s (1.4)(0.287 kJ/kg ×K)(800 K) çç ÷ çè 1 kJ/kg ø÷

V 330 m/s = = 0.582 c 567 m/s Discussion Note that a constant Mach number does not necessarily indicate constant speed. The Mach number of a rocket, for example, will be increasing even when it ascends at constant speed. Also, the specific heat ratio k changes with temperature. and Ma =

13-667

EES (Engineering Equation Solver) SOLUTION "Given" T_a=300 [K] T_b=800 [K] Vel=330 [m/s] "Properties" R=0.287 [kJ/kg-K] k=1.4 "Analysis" "(a)" C_a=sqrt(k*R*T_a*Convert(kJ/kg, m^2/s^2)) Ma_a=Vel/C_a "(b)" C_b=sqrt(k*R*T_b*Convert(kJ/kg, m^2/s^2)) Ma_b=Vel/C_b

18-23E Steam flows through a device at a specified state and velocity. The Mach number of steam is to be determined assuming ideal gas behavior. Assumptions Steam is an ideal gas with constant specific heats. Properties The gas constant of steam is R = 0.1102 Btu / lbm ×R . Its specific heat ratio is given to be k = 1.3. Analysis From the ideal-gas speed of sound relation,

kRT =

æ25,037 ft 2 / s 2 ÷ ö ÷ (1.3)(0.1102 Btu/lbm ×R)(1160 R) çç ÷ = 2040 ft/s çè 1 Btu/lbm ÷ ø

Thus,

V 900 ft/s = = 0.441 c 2040 ft/s Discussion Using property data from steam tables and not assuming ideal gas behavior, it can be shown that the Mach number in steam at the specified state is 0.446, which is sufficiently close to the ideal-gas value of 0.441. Therefore, the ideal gas approximation is a reasonable one in this case. Ma =

EES (Engineering Equation Solver) SOLUTION "Given" P=120 [psia] T=(700+460) [R] Vel=900 [ft/s] k=1.3 "Properties" R=0.1102 [Btu/lbm-R] "Analysis" c=sqrt(k*R*T*Convert(Btu/lbm, ft^2/^s^2)) Ma_max=Vel/c

18-24E Problem 2-23E is reconsidered. The variation of Mach number with temperature as the temperature changes between 350 and 700F is to be investigated, and the results are to be plotted. Analysis The EES Equations window is printed below, along with the tabulated and plotted results.

13-668

P=120 [psia] T=700 [F] Vel=900 [ft/s] k=1.3 R=0.1102 [Btu/lbm-R] c=sqrt(k*R*(T+460)*Convert(Btu/lbm, ft^2/^s^2)) Ma=Vel/c Temperature, T, F 350 375 400 425 450 475 500 525 550 575 600 625 650 675 700

Mach number Ma 0.528 0.520 0.512 0.505 0.498 0.491 0.485 0.479 0.473 0.467 0.462 0.456 0.451 0.446 0.441

Discussion Note that for a specified flow speed, the Mach number decreases with increasing temperature, as expected.

18-25 The inlet state and the exit pressure of air are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.287 kJ / kg ×K and k = 1.4. The specific heat ratio k varies with temperature, but in our case this change is very small and can be disregarded. Analysis The final temperature of air is determined from the isentropic relation of ideal gases, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ çè P1 ÷ ø

(1.4- 1)/1.4

æ0.4 MPa ÷ ö = (350.2 K) çç ÷ èç 2.2 MPa ÷ ø

= 215.2 K

Treating k as a constant, the ratio of the initial to the final speed of sound can be expressed as

Ratio =

c2 = c1

k1 RT1 k2 RT2

T1 T2

350.2 215.2

= 1.28

Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-669

EES (Engineering Equation Solver) SOLUTION "Given" P_1=2200 [kPa] T_1=(77+273) [K] P_2=400 [kPa] "Properties" c_p=CP(air, T=T_1) c_v=CV(air, T=T_1) k=c_p/c_v "Analysis" T_2=T_1*(P_2/P_1)^((k-1)/k) Ratio=sqrt(T_1)/sqrt(T_2)

18-26 The inlet state and the exit pressure of helium are given for an isentropic expansion process. The ratio of the initial to the final speed of sound is to be determined. Assumptions Helium is an ideal gas with constant specific heats at room temperature. Properties The properties of helium are R = 2.0769 kJ / kg ×K and k = 1.667. Analysis The final temperature of helium is determined from the isentropic relation of ideal gases, ( k - 1)/ k

æP ö ÷ T2 = T1 ççç 2 ÷ ÷ èç P ÷ ø 1

(1.667- 1)/1.667

æ0.4 ö = (350.2 K) çç ÷ ÷ ÷ çè 2.2 ø

= 177.0 K

The ratio of the initial to the final speed of sound can be expressed as

Ratio =

c2 = c1

k1 RT1 k2 RT2

T1 T2

350.2 177.0

= 1.41

Discussion Note that the speed of sound is proportional to the square root of thermodynamic temperature.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=2200 [kPa] T_1=(77+273) [K] P_2=400 [kPa] "Properties" k=1.667 "Analysis" T_2=T_1*(P_2/P_1)^((k-1)/k) Ratio=sqrt(T_1)/sqrt(T_2)

18-27 The Mach number of a passenger plane for specified limiting operating conditions is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ / kg ×K . Its specific heat ratio at room temperature is k = 1.4. Analysis From the speed of sound relation

kRT =

æ1000 m 2 / s 2 ö÷ ÷ = 293 m/s (1.4)(0.287 kJ/kg ×K)(-60 + 273 K) çç ÷ çè 1 kJ/kg ø÷

Thus, the Mach number corresponding to the maximum cruising speed of the plane is © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-670

Vmax (945 / 3.6) m/s = = 0.897 c 293 m/s

Ma =

Discussion Note that this is a subsonic flight since Ma < 1. Also, using a k value at -60C would give practically the same result.

EES (Engineering Equation Solver) SOLUTION "Given" Mass_plane=260000 [kg] L=64 [m] Wingspan=60 [m] Vel_max=945 [km/h] Passenger=271 Altitude=14000 [m] Range=12000 [km] T=(-60+273) [K] "Properties" c_p=CP(air, T=T) c_v=CV(air, T=T) k=c_p/c_v R=c_p-c_v "Analysis" C=sqrt(k*R*T*Convert(kJ/kg, m^2/^s^2)) Ma_max=(Vel_max*Convert(km/h, m/s))/C

18-28 The expression for the speed of sound for an ideal gas is to be obtained using the isentropic process equation and the definition of the speed of sound. Analysis The isentropic relation Pv k = A where A is a constant can also be expressed as k

æ1 ÷ ö P = Açç ÷ = Ar k ÷ çèv ø Substituting it into the relation for the speed of sound,

ö æ¶ P ö æ¶ ( Ar ) k ÷ ÷ ÷ c 2 = çç ÷ = çç = kAr k - 1 = k ( Ar k ) / r = k ( P / r ) = kRT ÷ ÷ èç ¶r ø÷s çè ¶r ø÷s since for an ideal gas P = r RT or RT = P / r . Therefore, c =

kRT , , which is the desired relation.

Discussion Notice that pressure has dropped out; the speed of sound in an ideal gas is not a function of pressure.

One Dimensional Isentropic Flow 18-29C We are to determine if it is possible to accelerate a gas to supersonic velocity in a converging nozzle. Analysis No, it is not possible. Discussion The only way to do it is to have first a converging nozzle, and then a diverging nozzle.

18-30C We are to discuss what happens to several variables when a subsonic gas enters a diverging duct. Analysis (a) The velocity decreases. (b), (c), (d) The temperature, pressure, and density of the fluid increase. Discussion The velocity decrease is opposite to what happens in supersonic flow.

13-671

18-31C We are to discuss the pressure at the throats of two different converging-diverging nozzles. Analysis The pressures at the two throats are identical. Discussion Since the gas has the same stagnation conditions, it also has the same sonic conditions at the throat.

18-32C We are to discuss what happens to several variables when a supersonic gas enters a converging duct. Analysis (a) The velocity decreases. (b), (c), (d) The temperature, pressure, and density of the fluid increase. Discussion The velocity decrease is opposite to what happens in subsonic flow.

18-33C We are to discuss what happens to several variables when a supersonic gas enters a diverging duct. Analysis (a) The velocity increases. (b), (c), (d) The temperature, pressure, and density of the fluid decrease. Discussion The velocity increase is opposite to what happens in subsonic flow.

18-34C We are to discuss what happens to the exit velocity and mass flow rate through a converging nozzle at sonic exit conditions when the nozzle exit area is reduced. Analysis (a) The exit velocity remains constant at sonic speed, (b) the mass flow rate through the nozzle decreases because of the reduced flow area. Discussion Without a diverging portion of the nozzle, a converging nozzle is limited to sonic velocity at the exit.

18-35C We are to discuss what happens to several variables when a subsonic gas enters a converging duct. Analysis (a) The velocity increases. (b), (c), (d) The temperature, pressure, and density of the fluid decrease. Discussion The velocity increase is opposite to what happens in supersonic flow.

18-36 Helium enters a converging-diverging nozzle at specified conditions. The lowest temperature and pressure that can be obtained at the throat of the nozzle are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of helium are k = 1.667 and c p = 5.1926 kJ / kg ×K . Analysis The lowest temperature and pressure that can be obtained at the throat are the critical temperature T * and critical pressure P*. First we determine the stagnation temperature T0 and stagnation pressure P0 , T0 = T +

æ 1 kJ/kg ö V2 (100 m/s)2 ÷ çç = 800 K + ÷ ÷= 801 K 2c p 2´ 5.1926 kJ/kg ×°C çè1000 m 2 / s 2 ø k / ( k - 1)

æT ö P0 = P çç 0 ÷ ÷ èç T ø÷

1.667/ (1.667- 1)

æ801 K ö÷ = (0.7 MPa) çç ÷ èç800 K ø÷

= 0.702 MPa

13-672

Thus, æ 2 ö æ 2 ö ÷ ÷ ç T * = T0 çç ÷ ÷ ÷= (801 K) èçç1.667 + 1ø ÷= 601 K èç k + 1ø

and k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ çè k + 1ø

1.667/(1.667- 1)

æ 2 ö ÷ = (0.702 MPa) çç ÷ ÷ çè1.667 + 1ø

= 0.342 MPa

Discussion These are the temperature and pressure that will occur at the throat when the flow past the throat is supersonic.

EES (Engineering Equation Solver) SOLUTION "Given" P=700 [kPa] T=800 [K] Vel=100 [m/s] "Properties" c_p=5.1926 [kJ/kg-K] k=1.667 "Analysis" T0=T+Vel^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P0=P*(T0/T)^(k/(k-1)) T_cr=T0*2/(k+1) P_cr=P0*(2/(k+1))^(k/(k-1))

18-37 The speed of an airplane and the air temperature are give. It is to be determined if the speed of this airplane is subsonic or supersonic. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ / kg ×K . Its specific heat ratio at room temperature is k = 1.4. Analysis The temperature is – 50 + 273.15 = 223.15 K. The speed of sound is

æ1000 m 2 / s 2 ÷ öæ3.6 km/h ö ÷ ç ÷ (1.4)(0.287 kJ/kg ×K)(223.15 K) çç ÷= 1077.97 km/h ÷ççè 1 m/s ø÷ çè 1 kJ/kg ÷ ø

kRT =

Ma =

V 1050 km/h = = 0.9741 km/h @0.974 c 1077.97 km/h

and

The speed of the airplane is subsonic since the Mach number is less than 1. Discussion Subsonic airplanes stay sufficiently far from the Mach number of 1 to avoid the instabilities associated with transonic flights.

13-673

EES (Engineering Equation Solver) SOLUTION "Given" V=1050 [km/h] T=(-50+273.15) [K] "Properties" R=0.287 [kJ/kg-K] k=1.4 "Analysis" c=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) Ma=V/(c*Convert(m/s, km/h))

18-38 The critical temperature, pressure, and density of air and helium are to be determined at specified conditions. Assumptions Air and Helium are ideal gases with constant specific heats at room temperature. Properties The properties of air at room temperature are R = 0.287 kJ / kg ×K , k = 1.4, and c p = 1.005 kJ / kg ×K . The properties of helium at room temperature are R = 2.0769 kJ / kg ×K , k = 1.667, and c p = 5.1926 kJ / kg ×K . Analysis (a) Before we calculate the critical temperature T*, pressure P*, and density *, we need to determine the stagnation temperature T0 , pressure P0 , and density r 0 . T0 = 100°C +

æ 1 kJ/kg ö V2 (325 m/s)2 ÷ çç = 100 + ÷ ÷= 152.5°C=425.7 K 2c p 2´ 1.005 kJ/kg ×°C çè1000 m 2 / s 2 ø

k /( k - 1)

æT ö P0 = P çç 0 ÷ ÷ çè T ÷ ø r0 =

1.4/(1.4- 1)

æ425.7 K ö÷ = (200 kPa) çç ÷ çè373.2 K ø÷

= 317.0 kPa

P0 317.0 kPa = = 2.595 kg/m3 RT0 (0.287 kPa ×m3 /kg ×K)(425.7 K)

Thus, æ 2 ö æ 2 ö ÷ ÷ ç T * = T0 çç ÷ ÷ ÷= (425.7 K) èçç1.4 + 1ø ÷= 355 K çè k + 1ø k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ èç k + 1ø

1/( k - 1)

æ 2 ö ÷ r * = r 0 çç ÷ ÷ çè k + 1ø

1.4/(1.4- 1)

æ 2 ö ÷ = (317.0 kPa) çç ÷ ÷ èç1.4 + 1ø

= 167 kPa 1/(1.4- 1)

æ 2 ö ÷ = (2.595 kg/m3 ) çç ÷ ÷ èç1.4 + 1ø

= 1.65 kg / m3

(b) For helium, T0 = T +

æ 1 kJ/kg ÷ ö V2 (300 m/s)2 çç = 60°C + ÷= 68.7°C=341.9 K ø 2c p 2´ 5.1926 kJ/kg ×°C çè1000 m 2 / s 2 ÷ k / ( k - 1)

æT ö P0 = P çç 0 ÷ ÷ èç T ø÷ r0 =

1.667/ (1.667- 1)

æ341.9 K ö÷ = (200 kPa) çç ÷ èç333.2 K ø÷

= 213.3 kPa

P0 213.3 kPa = = 0.3004 kg/m3 RT0 (2.0769 kPa ×m3 /kg ×K)(341.9 K)

Thus, æ 2 ö æ 2 ö ÷ ÷ ç T * = T0 çç ÷ ÷ ÷= (341.9 K) ççè1.667 + 1ø ÷= 256 K èç k + 1ø

13-674

k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ èç k + 1ø

1/( k - 1)

æ 2 ÷ ö r * = r 0 çç ÷ çè k + 1÷ ø

1.667/(1.667- 1)

æ 2 ö ÷ = (213.3 kPa) çç ÷ ÷ èç1.667 + 1ø

= 104 kPa 1/(1.667- 1)

æ 2 ö ÷ = (0.3004 kg/m3 ) çç ÷ çè1.667 + 1÷ ø

= 0.195 kg / m3

Discussion These are the temperature, pressure, and density values that will occur at the throat when the flow past the throat is supersonic.

EES (Engineering Equation Solver) SOLUTION "Given" P_air=200 [kPa] T_air=(100+273.15) [K] Vel_air=250 [m/s] P_He=200 [kPa] T_He=(40+273.15) [K] Vel_He=300 [m/s] "Properties" c_p_air=1.005 [kJ/kg-K] k_air=1.4 R_air=0.287 [kJ/kg-K] c_p_He=5.1926 [kJ/kg-K] k_He=1.667 R_He=2.0769 [kJ/kg-K] "Analysis" "(a)" T0_air=T_air+Vel_air^2/(2*c_p_air)*Convert(m^2/s^2, kJ/kg) P0_air=P_air*(T0_air/T_air)^(k_air/(k_air-1)) rho_0_air=P0_air/(R_air*T0_air) T_cr_air=T0_air*2/(k_air+1) P_cr_air=P0_air*(2/(k_air+1))^(k_air/(k_air-1)) rho_cr_air=rho_0_air*(2/(k_air+1))^(1/(k_air-1)) "(b)" T0_He=T_He+Vel_He^2/(2*c_p_He)*Convert(m^2/s^2, kJ/kg) P0_He=P_He*(T0_He/T_He)^(k_He/(k_He-1)) rho_0_He=P0_He/(R_He*T0_He) T_cr_He=T0_He*2/(k_He+1) P_cr_He=P0_He*(2/(k_He+1))^(k_He/(k_He-1)) rho_cr_He=rho_0_He*(2/(k_He+1))^(1/(k_He-1))

18-39E Air flows through a duct at a specified state and Mach number. The velocity and the stagnation pressure, temperature, and density of the air are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.06855 ×Btu / lbm ×R = 0.3704 psia ft 3 / lbm ×R and k = 1.4. Analysis First, T = 320 + 459.67 = 779.67 K. The speed of sound in air at the specified conditions is

kRT =

æ25,037 ft 2 / s 2 ÷ ö ÷ (1.4)(0.06855 Btu/1bm ×R)(779.67 R) ççç = 1368.72 ft/s ÷ è 1 Btu/1bm ÷ ø

Thus,

V = Ma ´ c = (0.7)(1368.72 ft/s) = 958.10 @958 ft / s Also,

13-675

P 25 psia = = 0.086568 1bm/ft 3 RT (0.3704 psia ×ft 3 /lbm ×R)(779.67 R)

Then the stagnation properties are determined from

æ (k - 1)Ma 2 ö÷ æ (1.4 - 1)(0.7) 2 ö÷ ÷ ÷ T0 = T çç1 + = (779.67 R) çç1 + ÷ ÷= 856.08 R @ 856 R çè 2 2 ø÷ èç ø÷ k / ( k - 1)

æT ö P0 = P çç 0 ÷ ÷ çè T ÷ ø

1/( k - 1)

æT ö r 0 = r çç 0 ÷ ÷ çè T ÷ ø

1.4/ (1.4- 1)

æ856.08 R ÷ ö = (25 psia) çç ÷ çè779.67 R ÷ ø

= 34.678 psia @ 34.7 psia 1/(1.4- 1)

æ856.08 R ö÷ = (0.08656 1bm/ft 3 ) çç ÷ çè779.67 R ÷ ø

= 0.10936 lbm/ft 3 @ 0.109 lbm / ft 3

Discussion Note that the temperature, pressure, and density of a gas increases during a stagnation process.

EES (Engineering Equation Solver) SOLUTION "Given" P=25 [psia] T=(320+459.67) [R] Ma=0.7 "Properties" c_p=CP(air, T=T) c_v=CV(air, T=T) k=c_p/c_v R=c_p-c_v "Analysis" c=sqrt(k*R*T*Convert(Btu/lbm, ft^2/s^2)) Vel=Ma*c rho=P/(R*Convert(Btu, psia-ft^3)*T) T0=T*(1+((k-1)*Ma^2)/2) P0=P*(T0/T)^(k/(k-1)) rho_0=rho*(T0/T)^(1/(k-1))

18-40 Air enters a converging-diverging nozzle at specified conditions. The lowest pressure that can be obtained at the throat of the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air at room temperature is k = 1.4. Analysis The lowest pressure that can be obtained at the throat is the critical pressure P*, which is determined from k / ( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ èç k + 1ø

1.4/ (1.4- 1)

æ 2 ö ÷ = (1200 kPa) çç ÷ ÷ èç1.4 + 1ø

= 634 kPa

Discussion This is the pressure that occurs at the throat when the flow past the throat is supersonic.

EES (Engineering Equation Solver) SOLUTION "Given" P0=1200 [kPa] "Properties" k=1.4 "Analysis" P_cr=P0*(2/(k+1))^(k/(k-1)) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-676

18-41 An aircraft is designed to cruise at a given Mach number, elevation, and the atmospheric temperature. The stagnation temperature on the leading edge of the wing is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The properties of air are R = 0.287 kPa ×m3 / kg.K , c p = 1.005 kJ / kg ×K , and k = 1.4. Analysis The speed of sound in air at the specified conditions is

kRT =

æ1000 m 2 / s 2 ÷ ö ÷ = 308.0 m/s (1.4)(0.287 kJ/kg ×K)(236.15 K) çç ÷ çè 1 kJ/kg ø÷

Thus,

V = Ma ´ c = (1.1)(308.0 m/s) = 338.8 m/s Then, T0 = T +

ö V2 (338.8 m/s) 2 æ çç 1 kJ/kg ÷ = 236.15 + = 293 K 2 2÷ ÷ ç 2c p 2´ 1.005 kJ/kg ×K è1000 m / s ø

Discussion Note that the temperature of a gas increases during a stagnation process as the kinetic energy is converted to enthalpy.

EES (Engineering Equation Solver) SOLUTION "Given" Ma=1.1 H=12000 [m] T=236.15 [K] "Properties" c_p=CP(air, T=T) c_v=CV(air, T=T) k=c_p/c_v R=c_p-c_v "Analysis" C=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) Vel=Ma*C T0=T+Vel^2/(2*c_p)*Convert(m^2/s^2, kJ/kg)

18-42 Quiescent carbon dioxide at a given state is accelerated isentropically to a specified Mach number. The temperature and pressure of the carbon dioxide after acceleration are to be determined. Assumptions Carbon dioxide is an ideal gas with constant specific heats at room temperature. Properties The specific heat ratio of the carbon dioxide at room temperature is k = 1.288. Analysis The inlet temperature and pressure in this case is equivalent to the stagnation temperature and pressure since the inlet velocity of the carbon dioxide is said to be negligible. That is, T0 = Ti = 500 K and P0 = Pi = 900 kPa . Then,

æ ö æ ö 2 2 ÷ ÷ ÷= (500 K) çç ÷ T = T0 çç = 475.4 K @ 475 K ÷ 2÷ 2 ÷ ÷ èç 2 + (k - 1)Ma ø èç 2 + (1.288 - 1)(0.6) ø and k / ( k - 1)

æT ö ÷ P = P0 ççç ÷ ÷ çèT0 ÷ ø

1.288/ (1.288- 1)

æ475.4 K ÷ ö = (900 kPa) çç ÷ ÷ èç 500 K ø

= 717.9 K @ 718 kPa

Discussion Note that both the pressure and temperature drop as the gas is accelerated as part of the internal energy of the gas is converted to kinetic energy.

13-677

EES (Engineering Equation Solver) SOLUTION "Given" P0=900 [kPa] T0=500 [K] M=0.6 "Properties" k=1.288 "Analysis" T=T0*2/(2+(k-1)*M^2) P=P0*(T/T0)^(k/(k-1))

18-43 The Mach number of scramjet and the air temperature are given. The speed of the engine is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.287 kJ / kg ×K . Its specific heat ratio at room temperature is k = 1.4. Analysis The temperature is – 20 + 273.15 = 253.15 K. The speed of sound is

kRT =

æ1000 m2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(253.15 K) ççç = 318.93 m/s ÷ è 1 kJ/kg ÷ ø

and æ3.6 km/h ÷ ö V = cMa = (318.93 m/s)(7) çç = 8037 km/h @ 8040 km / h ÷ çè 1 m/s ÷ ø

Discussion Note that extremely high speed can be achieved with scramjet engines. We cannot justify more than three significant digits in a problem like this.

EES (Engineering Equation Solver) SOLUTION "Given" Ma=7 T=(-20+273.15) [K] "Properties" R=0.287 [kJ/kg-K] k=1.4 "Analysis" c=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) V=c*Ma*Convert(m/s, km/h)

18-44E The Mach number of scramjet and the air temperature are given. The speed of the engine is to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The gas constant of air is R = 0.06855 Btu / lbm ×R . Its specific heat ratio at room temperature is k = 1.4 . Analysis The temperature is 0 + 459.67 = 459.67 R. The speed of sound is

kRT =

æ25,037 ft 2 / s 2 ÷ ö ÷ (1.4)(0.06855 Btu/lbm ×R)(459.67 R) çç ÷ = 1050.95 ft/s çè 1 Btu/lbm ø÷

and

13-678

æ 1 mi/h ö ÷= 5015.9 mi/h @ 5020 mi / h V = cMa = (1050.95 ft/s)(7) çç ÷ çè1.46667 ft/s ÷ ø

Discussion Note that extremely high speed can be achieved with scramjet engines. We cannot justify more than three significant digits in a problem like this.

EES (Engineering Equation Solver) SOLUTION "Given" Ma=7 T=(0+459.67) [R] "Properties" R=0.06855 [Btu/lbm-R] k=1.4 "Analysis" c=sqrt(k*R*T*Convert(Btu/lbm, ft^2/s^2)) V=c*Ma*Convert(ft/s, mph)

Isentropic Flow Through Nozzles 18-45C We are to analyze if it is possible to accelerate a fluid to supersonic speeds with a velocity that is not sonic at the throat. Analysis No, if the flow in the throat is subsonic. If the velocity at the throat is subsonic, the diverging section would act like a diffuser and decelerate the flow. Yes, if the flow in the throat is already supersonic, the diverging section would accelerate the flow to even higher Mach number. Discussion In duct flow, the latter situation is not possible unless a second converging-diverging portion of the duct is located upstream, and there is sufficient pressure difference to choke the flow in the upstream throat.

18-46C We are to discuss what would happen if we add a diverging section to supersonic flow in a duct. Analysis The fluid would accelerate even further, as desired. Discussion This is the opposite of what would happen in subsonic flow.

18-47C We are to discuss the difference between Ma* and Ma. Analysis Ma* is the local velocity non-dimensionalized with respect to the sonic speed at the throat, whereas Ma is the local velocity non-dimensionalized with respect to the local sonic speed. Discussion The two are identical at the throat when the flow is choked.

18-48C We are to discuss what would happen if we add a diverging section to supersonic flow in a duct. Analysis The fluid would accelerate even further instead of decelerating. Discussion This is the opposite of what would happen in subsonic flow.

18-49C We are to consider subsonic flow through a converging nozzle with critical pressure at the exit, and analyze the effect of lowering back pressure below the critical pressure. Analysis (a) No effect on velocity. (b) No effect on pressure. (c) No effect on mass flow rate. Discussion In this situation, the flow is already choked initially, so further lowering of the back pressure does not change anything upstream of the nozzle exit plane.

13-679

18-50C We are to compare the mass flow rates through two identical converging nozzles, but with one having a diverging section. Analysis If the back pressure is low enough so that sonic conditions exist at the throats, the mass flow rates in the two nozzles would be identical. However, if the flow is not sonic at the throat, the mass flow rate through the nozzle with the diverging section would be greater, because it acts like a subsonic diffuser. Discussion Once the flow is choked at the throat, whatever happens downstream is irrelevant to the flow upstream of the throat.

18-51C We are to discuss the hypothetical situation of hypersonic flow at the outlet of a converging nozzle. Analysis Maximum flow rate through a converging nozzle is achieved when Ma = 1 at the exit of a nozzle. For all other Ma values the mass flow rate decreases. Therefore, the mass flow rate would decrease if hypersonic velocities were achieved at the throat of a converging nozzle. Discussion Note that this is not possible unless the flow upstream of the converging nozzle is already hypersonic.

18-52C We are to consider subsonic flow through a converging nozzle, and analyze the effect of setting back pressure to critical pressure for a converging nozzle. Analysis (a) The exit velocity reaches the sonic speed, (b) the exit pressure equals the critical pressure, and (c) the mass flow rate reaches the maximum value. Discussion In such a case, we say that the flow is choked.

18-53C We are to discuss what happens to several variables in the diverging section of a subsonic converging-diverging nozzle. Analysis (a) The velocity decreases, (b) the pressure increases, and (c) the mass flow rate remains the same. Discussion Qualitatively, this is the same as what we are used to (in previous chapters) for incompressible flow.

18-54 Nitrogen enters a converging-diverging nozzle at a given pressure. The critical velocity, pressure, temperature, and density in the nozzle are to be determined. Assumptions 1 Nitrogen is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of nitrogen are k = 1.4 and R = 0.2968 kJ / kg ×K . Analysis The stagnation pressure in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle,

P0 = Pi = 700 kPa T0 = Ti = 400 K r0 =

P0 700 kPa = = 5.896 kg/m3 RT0 (0.2968 kPa ×m3 /kg ×K)(400 K)

13-680

Critical properties are those at a location where the Mach number is Ma = 1. From Table A-32 at Ma = 1, we read T / T0 = 0.8333 , P / P0 = 0.5283 , and r / r 0 = 0.6339 . Then the critical properties become T * = 0.8333T0 = 0.8333(400 K) = 333 K P* = 0.5283P0 = 0.5283(700 kPa) = 370MPa r * = 0.6339r 0 = 0.6339 (5.896 kg / m3 ) = 3.74kg / m 3

Also,

V * = c* =

kRT * =

æ1000 m 2 /s 2 ö÷ ÷ = 372 m / s (1.4)(0.2968 kJ/kg ×K)(333 K) çç ÷ çè 1 kJ/kg ø÷

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=700 [kPa] T_i=400 [K] Vel_i=0 "negigible inlet velocity" "Properties" R=0.2968 [kJ/kg-K] k=1.4 "Analysis" P0=P_i T0=T_i rho_0=P0/(R*T0) Ma=1 T_star=T0/(1+(k-1)/2*Ma^2) P_star=P0*(1+(k-1)/2*Ma^2)^(-k/(k-1)) rho_star=rho_0*(1+(k-1)/2*Ma^2)^(-1/(k-1)) c_star=sqrt(k*R*T_star*Convert(kJ/kg, m^2/s^2)) Vel_star=c_star

18-55 Air enters a converging-diverging nozzle at a specified pressure. The back pressure that will result in a specified exit Mach number is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is k = 1.4.

Analysis The stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. It remains constant throughout the nozzle since the flow is isentropic, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-681

P0 = Pi = 1.2 MPa From Table A-32 at Ma e = 1.8 , we read Pe / P0 = 0.1740 . Thus,

Pe = 0.1740P0 = 0.1740(1.2 MPa) = 0.209 MPa = 209 kPa Discussion If we solve this problem using the relations for compressible isentropic flow, the results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=1200 [kPa] Vel_i=0 "negligible inlet velocity" Ma_e=1.8 "Properties" k=1.4 "Analysis" P0=P_i "since inlet velocity is negligible" P_e=P0*(1+(k-1)/2*Ma_e^2)^(-k/(k-1))

18-56 For subsonic flow at the inlet, the variation of pressure, velocity, and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions. Assumptions 1 The gas is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The flow is choked at the throat. Analysis Using EES and CO2 as the gas, we calculate and plot flow area A, velocity V, and Mach number Ma as the pressure drops from a stagnation value of 1400 kPa to 200 kPa. Note that the curve for A is related to the shape of the nozzle, with horizontal axis serving as the centerline. The EES equation window and the plot are shown below.

P=600 [kPa] k=1.289 cp=0.846 [kJ/kg-K] R=0.1889 [kJ/kg-K] P0=1400 [kPa] T0=473 [K] m=3 [kg/s] rho_0=P0/(R*T0) rho=P/(R*T) rho_norm=rho/rho_0 "Normalized density" T=T0*(P/P0)^((k-1)/k) Tnorm=T/T0 "Normalized temperature" V=SQRT(2*cp*(T0-T)*1000) V_norm=V/500 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-682

A=m/(rho*V)*500 c=SQRT(k*R*T*1000) Ma=V/c

Discussion We are assuming that the back pressure is sufficiently low that the flow is choked at the throat, and the flow downstream of the throat is supersonic without any shock waves. Mach number and velocity continue to rise right through the throat into the diverging portion of the nozzle, since the flow becomes supersonic.

18-57 We repeat Prob. 18-56, but for supersonic flow at the inlet. The variation of pressure, velocity, and Mach number along the length of the nozzle are to be sketched for an ideal gas under specified conditions. Analysis Using EES and CO2 as the gas, we calculate and plot flow area A, velocity V, and Mach number Ma as the pressure rises from 200 kPa at a very high velocity to the stagnation value of 1400 kPa. Note that the curve for A is related to the shape of the nozzle, with horizontal axis serving as the centerline.

13-683

13-684

Discussion Note that this problem is identical to the proceeding one, except the flow direction is reversed. In fact, when plotted like this, the plots are identical.

18-58 For an ideal gas, an expression is to be obtained for the ratio of the speed of sound where Ma = 1 to the speed of sound based on the stagnation temperature, c* / c0 . Analysis For an ideal gas the speed of sound is expressed as c =

c* = c0

kRT . Thus,

1/ 2

1/2 æT *ö æ 2 ÷ ö ç ÷ = ççç ÷ = ÷ çç ÷ èk +1÷ ø èç T0 ø÷ kRT0

kRT *

Discussion Note that a speed of sound changes the flow as the temperature changes.

18-59 It is to be explained why the maximum flow rate per unit area for a given ideal gas depends only on P0 / T0 . Also

(

)

for an ideal gas, a relation is to be obtained for the constant a in mmax /A* = a P0 / T0 . Properties The properties of the ideal gas considered are R = 0.287 kPa ×m3 / kg.K and k = 1.4. Analysis The maximum flow rate is given by ( k + 1)/ 2( k - 1)

æ 2 ö ÷ mmax = A*P0 k / RT0 çç ÷ ÷ çè k + 1ø

( k + 1)/ 2( k - 1)

(

mmax / A* = P0 / T0

)

æ 2 ö ÷ k / R çç ÷ ÷ çè k + 1ø

For a given gas, k and R are fixed, and thus the mass flow rate depends on the parameter P0 / T0 . Thus, mmax /A* can be

(

)

expressed as mmax /A* = a P0 / T0 where

13-685

( k + 1)/ 2( k - 1)

æ 2 ö ÷ k / R çç ÷ ÷ èç k + 1ø

2.4/ 0.8

æ 2 ö 1.4 ÷ çç ÷ ÷ æ1000 m 2 / s 2 ö èç1.4 + 1ø ÷ ç ÷ (0.287 kJ/kg.K) çç ÷ ÷ è 1 kJ/kg ø

= 0.0404 (m / s) K

Discussion Note that when sonic conditions exist at a throat of known cross-sectional area, the mass flow rate is fixed by the stagnation conditions.

EES (Engineering Equation Solver) SOLUTION "Given" R=0.287 [kJ/kg-K] k=1.4 "Analysis" a=sqrt(k/R*Convert(m^2/s^2, kJ/kg))*(2/(k+1))^((k+1)/(2*(k-1)))

18-60 An ideal gas is flowing through a nozzle. The flow area at a location where Ma = 1.6 is specified. The flow area where Ma = 0.8 is to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio is given to be k = 1.4. Analysis The flow is assumed to be isentropic, and thus the stagnation and critical properties remain constant throughout the nozzle. The flow area at a location where Ma 2 = 0.8 is determined using A / A* data from Table A-32 to be Ma1 = 1.6 :

A1 A1 45 cm 2 = 1.2502 ¾ ¾ ® A* = = = 35.99 cm 2 A* 1.2502 1.2502

Ma 2 = 0.8 :

A2 = 1.0382 ¾ ¾ ® A2 = (1.0382) A* = (1.0382)(35.99 cm 2 ) = 37.4 cm2 A*

Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" M1=2.4 A1=36 [cm^2] M2=1.2 k=1.4 "Analysis" Ratio_A1=1/M1*((2/(k+1))*(1+(k-1)/2*M1^2))^((k+1)/(2*(k-1))) "Ratio_A1=A1/A_star" A_star=A1/Ratio_A1 Ratio_A2=1/M2*((2/(k+1))*(1+(k-1)/2*M2^2))^((k+1)/(2*(k-1))) "Ratio_A2=A2/A_star" A2=Ratio_A2*A_star

18-61 An ideal gas is flowing through a nozzle. The flow area at a location where Ma = 1.6 is specified. The flow area where Ma = 0.8 is to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Analysis The flow is assumed to be isentropic, and thus the stagnation and critical properties remain constant throughout the nozzle. The flow area at a location where Ma 2 = 0.9 is determined using the A / A* relation,

13-686

( k + 1)/ 2( k - 1)

öïüï A 1 ïíï æ 2 öæ ÷ çç1 + k - 1 Ma 2 ÷ = ÷ ÷ ì ççç ý ÷ ÷ ç ï øïïþ A* Ma ïî è k + 1øè 2 For k = 1.33 and Ma1 = 1.8 :

2.33/ 2´ 0.33

öïüï A1 1 ïíï æ 2 öæ ÷ çç1 + 1.33 - 11.62 ÷ = ÷ ÷ý ì ççç ÷ ç ø÷ïþï A* 1.6 ïîï è1.33 + 1øè 2 and, A* =

= 1.2646

A1 45 cm 2 = = 35.58 cm 2 1.2646 1.2646

For k = 1.33 and Ma 2 = 0.8 : 2.33/ 2´ 0.33

öüïï A2 1 íïï æ 2 öæ ÷ çç1 + 1.33 - 1 0.82 ÷ = ÷ ÷ý ì ççç ÷ç ø÷ïþï A* 0.8 ïîï è1.33 + 1øè 2

= 1.0391

and A2 = (1.0391) A* = (1.0391)(35.58 cm2 ) = 37.0 cm2 Discussion Note that the compressible flow functions in Table A-32 are prepared for k = 1.4, and thus they cannot be used to solve this problem.

EES (Engineering Equation Solver) SOLUTION "Given" M1=2.4 A1=36 [cm^2] M2=1.2 k=1.33 "Analysis" Ratio_A1=1/M1*((2/(k+1))*(1+(k-1)/2*M1^2))^((k+1)/(2*(k-1))) "Ratio_A1=A1/A_star" A_star=A1/Ratio_A1 Ratio_A2=1/M2*((2/(k+1))*(1+(k-1)/2*M2^2))^((k+1)/(2*(k-1))) "Ratio_A2=A2/A_star" A2=Ratio_A2*A_star

18-62E Air enters a converging-diverging nozzle at a specified temperature and pressure with low velocity. The pressure, temperature, velocity, and mass flow rate are to be calculated in the specified test section. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and R = 0.06855 Btu / lbm ×R = 0.3704 psia ×ft 3 / lbm ×R . Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic.

P0 = Pi = 150 psia

and T0 = Ti = 100°F » 560 R

13-687

æ ö æ ö 2 2 ÷ ÷ ÷= (560 R) çç ÷ Te = T0 ççç = 311R ÷ 2÷ 2 ç ÷ è 2 + (k - 1)Ma ÷ ø è 2 + (1.4 - 1)2 ø k / ( k - 1)

æT ö ÷ Pe = P0 ççç ÷ ÷ èçT ÷ ø 0

re =

1.4/ 0.4

æ311÷ ö = (150 psia) çç ÷ çè560 ÷ ø

= 19.1 psia

Pe 19.1 psia = = 0.166 1bm/ft 3 RTe (0.3704 psia.ft 3 /1bm ×R)(311 R)

The nozzle exit velocity can be determined from Ve = Ma e ce , where ce is the speed of sound at the exit conditions,

æ25,037 ft 2 / s 2 ÷ ö ÷ = 1729 ft/s @1730 ft / s Ve = Ma e ce = Ma e kRTe = (2) (1.4)(0.06855 Btu/1bm ×R )(311 R )çç ÷ çè 1 Btu/1bm ÷ ø Finally, m = r e AeVe = (0.166 1bm / ft 3 )(5 ft 2 )(1729 ft / s) = 1435 lbm / s @1440 lbm / s

Discussion Air must be very dry in this application because the exit temperature of air is extremely low, and any moisture in the air will turn to ice particles.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=150 [psia] T_i=(100+460) [R] Vel_i=0 [ft/s] "low velocity" A_e=5 [ft^2] M_e=2 "Properties" C_p=CP(air, T=T_i) C_v=CV(air, T=T_i) k=C_p/C_v R=C_p-C_v "Analysis" P0=P_i T0=T_i T_e=T0/(1+(k-1)/2*M_e^2) P_e=P0/(T0/T_e)^(k/(k-1)) rho_e=P_e/(R*Convert(Btu, psia-ft^3)*T_e) C_e=sqrt(k*R*T_e*Convert(Btu/lbm, ft^2/s^2)) Vel_e=M_e*C_e m_dot=rho_e*A_e*Vel_e

18-63 Air enters a nozzle at specified temperature, pressure, and velocity. The exit pressure, exit temperature, and exit-toinlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4 and c p = 1.005 kJ / kg ×K . Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic.

13-688

T0 = Ti +

æ 1 kJ/kg ÷ ö Vi 2 (110 m/s) 2 çç = 420 K + = 426.02 2 2÷ ÷ ç 2c p 2´ 1.005 kJ/kg ×K è1000 m /s ø

and k / ( k - 1)

æT ö P0 = P çç 0 ÷ ÷ èç T ø÷

1.4/ (1.4- 1)

æ426.02 K ö÷ = (0.5 MPa) çç ÷ èç 420 K ÷ ø

= 0.52554 MPa

From Table A-32 (or from Eqs. 12-18 and 12-19) at Ma = 1, we read T / T0 = 0.8333 , P / P0 = 0.5283 . Thus,

T = 0.8333T0 = 0.8333(426.02 K) = 355.00 K » 355 K and P = 0.5283P0 = 0.5283(0.52554 MPa) = 0.27764 MPa » 0.278 MPa = 278 kPa Also,

ci =

kRTi =

æ1000 m2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(420 K) çç = 410.799 m/s ÷ çè 1 kJ/kg ÷ ø

and Ma i =

Vi 110 m/s = = 0.2678 ci 410.799 m/s

Ma i =

Vi 150 m/s = = 0.3651 ci 410.799 m/s

From Table A-32 at this Mach number we read Ai / A* = 2.3343 . Thus the ratio of the throat area to the nozzle inlet area is

A* 1 = = 0.42839 @0.428 A 2.3343 Discussion We can also solve this problem using the relations for compressible isentropic flow. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=500 [kPa] T_i=420 [K] Vel_i=110 [m/s] "Properties" R=0.287 [kJ/kg-K] c_p=1.005 [kJ/kg-K] k=1.4 "Analysis" T0=T_i+Vel_i^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P0=P_i*(T0/T_i)^(k/(k-1)) Ma=1 T=T0/(1+(k-1)/2*Ma^2) P=P0/(T0/T)^(k/(k-1)) C_i=sqrt(k*R*T_i*Convert(kJ/kg, m^2/s^2)) Ma_i=Vel_i/C_i "From Table A-15 of the text at this Ma_i = 0.4, we read A_i/A_cr = 1.7452. Therefore, A_cr/A_i is" Ratio=1/1.7452 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-689

18-64 Air enters a nozzle at specified temperature and pressure with low velocity. The exit pressure, exit temperature, and exit-to-inlet area ratio are to be determined for a Mach number of Ma = 1 at the exit. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The specific heat ratio of air is k = 1.4. Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic.

T0 = Ti = 420 K

and

P0 = Pi = 0.5 MPa

From Table A-32 (or from Eqs. 12-18 and 12-19) at Ma = 1, we read T / T0 = 0.8333 , P / P0 = 0.5283 . Thus,

T = 0.8333T0 = 0.8333(420 K) = 350 K

and

P = 0.5283P0 = 0.5283(0.5 MPa) = 0.264 MPa

The Mach number at the nozzle inlet is Ma = 0 since Vi @ 0 . From Table A-32 at this Mach number we read Ai / A* = ¥ . Thus the ratio of the throat area to the nozzle inlet area is

A* 1 = = 0. Ai ¥

Discussion If we solve this problem using the relations for compressible isentropic flow, the results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=500 [kPa] T_i=420 [K] Vel_i=0 "negligible inlet velocity" "Properties" k=1.4 "Analysis" T0=T_i P0=P_i Ma=1 T=T0/(1+(k-1)/2*Ma^2) P=P0/(T0/T)^(k/(k-1)) Ma_i=0.00001 "since Vel_i=0" Ratio_A=1/Ma_i*((2/(k+1))*(1+(k-1)/2*Ma_i^2))^((k+1)/(2*(k-1))) "Ratio_A=A/A_star" Ratio=1/Ratio_A "For Ma_i=0, Ratio_A is infinity, and Ratio is zero. For EES to be able do calculations, we used a very small Ma_i value instead of zero, and obtained the same result."

18-65 Air enters a converging nozzle at a specified temperature and pressure with low velocity. The exit pressure, the exit velocity, and the mass flow rate versus the back pressure are to be calculated and plotted. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-690

Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of air are k = 1.4, R = 0.287 kJ / kg ×K , and c p = 1.005 kJ / kg ×K . Analysis The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic,

P0 = Pi = 900 kPa

T0 = Ti = 400 K

The critical pressure is determined to be k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ èç k + 1ø

1.4/0.4

æ 2 ö ÷ = (900 kPa) çç ÷ ÷ èç1.4 + 1ø

= 475.5 kPa

Then the pressure at the exit plane (throat) will be

Pe = Pb

for Pb ³ 475.5 kPa

Pe = P* = 475.5 kPa for Pb < 475.5 kPa (choked flow)

13-691

Thus the back pressure will not affect the flow when 100 < Pb < 475.5 kPa . For a specified exit pressure Pe , the temperature, the velocity and the mass flow rate can be determined from ( k - 1)/ k

æP ö ÷ Temperature Te = T0 ççç e ÷ ÷ çè P0 ÷ ø

0.4/1.4

æP ö = (400 K) çç e ÷ ÷ ÷ èç900 ø

æ1000 m 2 /s 2 ÷ ö ÷ 2(1.005 kJ/kg ×K)(400 - Te ) çç ÷ çè 1 kJ/kg ÷ ø

Velocity

2c p (T0 - Te ) =

Density

re =

Pe Pe = RTe (0.287 kPa ×m 3 / kg ×K )Te

Mass flow rate

m = r eVe Ae = r eVe (0.001 m2 )

EES SOLUTION Procedure ExitPress(P_back,P_crit : P_exit, Condition$) If (P_back>=P_crit) then P_exit:=P_back "Unchoked Flow Condition" Condition$:='unchoked' else P_exit:=P_crit "Choked Flow Condition" Condition$:='choked' Endif End

13-692

"Data input" Gas$='Air' A_cm2=10 [cm^2] "Throat area" P_inlet = 900 [kPa] T_inlet= 400 [K] P_back =475.5 [kPa] A_exit = A_cm2*Convert(cm^2,m^2) c_p=specheat(Gas$,T=T_inlet) c_p-c_v=R k=c_p/c_v M=MOLARMASS(Gas$) "Molar mass of Gas$" R=R_u/M "Gas constant for Gas$" R_u= 8.314 [kJ/kmol-K] "Since the inlet velocity is negligible, the stagnation temperature = T_inlet; and, since the nozzle is isentropic, the stagnation pressure = P_inlet." P_o=P_inlet "Stagnation pressure" T_o=T_inlet "Stagnation temperature" P_crit /P_o=(2/(k+1))^(k/(k-1)) "Critical pressure " Call ExitPress(P_back,P_crit : P_exit, Condition$) T_exit /T_o=(P_exit/P_o)^((k-1)/k) "Exit temperature for isentopic flow, K" V_exit ^2/2=c_p*(T_o-T_exit)*Convert(kJ, J) "Exit velocity, m/s" Rho_exit=P_exit/(R*T_exit) "Exit density, kg/m3" m_dot=Rho_exit*V_exit*A_exit "Nozzle mass flow rate, kg/s"

Pb , kPa

Pe , kPa

Te , k

Ve , m / s

r e , kg / m3

m, kg/s

100 200 300 400 475.5 600 700 800 900

475.5 475.5 475.5 475.5 475.5 600 700 800 900

333.3 333.3 333.3 333.3 333.3 356.2 372.3 386.8 400

366.1 366.1 366.1 366.1 366 296.6 236 163.1 0

4.97 4.97 4.97 4.97 4.97 5.868 6.551 7.207 7.839

1.819 1.819 1.819 1.819 1.819 1.74 1.546 1.176 0

900 850 800

Pexit, kPa

750 700 650 600 550 500 450 100

200

300

400

500

600

700

800

900

Pback, kPa

13-693

400 350

Vexit m/s

300 250 200 150 100 50 0 100

200

300

400

500

600

700

800

900

600

700

800

900

600

700

800

900

Pback, kPa 2.0

m, kg/s

1.6

1.2

0.8

0.4

0.0 100

200

300

400

500

Pback, kPa 8.0 7.5

r exit kg/m3

7.0 6.5 6.0 5.5 5.0 4.5 100

200

300

400

500

Pback, kPa

13-694

400 390

Texit, K

380 370 360 350 340 330 100

200

300

400

500

600

700

800

900

Pback, kPa

Discussion We see from the plots that once the flow is choked at a back pressure of 475.5 kPa, the mass flow rate remains constant regardless of how low the back pressure gets.

18-66 We are to reconsider Prob. 18-65. Using appropriate software, we are to solve the problem for the inlet conditions of 0.8 MPa and 1200 K. Analysis We use EES for this problem. Air at 800 kPa, 1200 K enters a converging nozzle with a negligible velocity. The throat area of the nozzle is 10 cm2. Assuming isentropic flow, calculate and plot the exit pressure, the exit velocity, and the mass flow rate versus the back pressure Pb for 0.8 > = Pb > = 0.1 MPa . Procedure ExitPress(P_back,P_crit : P_exit, Condition$) If (P_back>=P_crit) then P_exit:=P_back "Unchoked Flow Condition" Condition$:='unchoked' else P_exit:=P_crit "Choked Flow Condition" Condition$:='choked' Endif End "Input data" Gas$='Air' A_cm2=10 [cm^2] P_inlet = 800 [kPa] T_inlet= 1200 [K] P_back =422.7 [kPa] A_exit = A_cm2*Convert(cm^2,m^2) c_p=specheat(Gas$,T=T_inlet) c_p-c_v=R k=c_p/c_v M=MOLARMASS(Gas$) R=R_u/M R_u=8.314 [kJ/kmol-K] "Since the inlet velocity is negligible, the stagnation temperature = T_inlet; and, since the nozzle is isentropic, the stagnation pressure = P_inlet." P_o=P_inlet "Stagnation pressure" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-695

T_o=T_inlet "Stagnation temperature" P_crit /P_o=(2/(k+1))^(k/(k-1)) "Critical pressure" Call ExitPress(P_back,P_crit : P_exit, Condition$) T_exit /T_o=(P_exit/P_o)^((k-1)/k) "Exit temperature for isentopic flow, K" V_exit ^2/2=c_p*(T_o-T_exit)*1000 "Exit velocity, m/s" Rho_exit=P_exit/(R*T_exit) "Exit density, kg/m3" m_dot=Rho_exit*V_exit*A_exit "Nozzle mass flow rate, kg/s" The table of results and the corresponding plot are provided below. EES SOLUTION A_cm2=10 A_exit=0.001 Condition$='choked' c_p=1.208 c_v=0.9211 Gas$='Air' k=1.312 M=28.97 m_dot=0.9124 P_back=422.7 P_crit=434.9 P_exit=434.9 P_inlet=800 P_o=800 R=0.287 Rho_exit=1.459 T_exit=1038 T_inlet=1200 T_o=1200 V_exit=625.2

Pback [kPa] 100 200 300 400 422.7 500 600 700 800

Pexit [kPa] 434.9 434.9 434.9 434.9 434.9 500 600 700 800

Vexit [m / s] 625.2 625.2 625.2 625.2 625.2 553.5 437.7 300.9 0.001523

m [kg / s]

0.9124 0.9124 0.9124 0.9124 0.9124 0.8984 0.8164 0.6313 0.000003538

Texit

ρexit

[K]

ékg / m 3 ù ëê ûú 1.459 1.459 1.459 1.459 1.459 1.623 1.865 2.098 2.323

1038 1038 1038 1038 1038 1073 1121 1163 1200

13-696

13-697

Discussion We see from the plot that once the flow is choked at a back pressure of 422.7 kPa, the mass flow rate remains constant regardless of how low the back pressure gets.

Shock Waves and Expansion Waves 18-67C We are to discuss the applicability of the isentropic flow relations across shocks and expansion waves. Analysis The isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves. Discussion Flow across any kind of shock wave involves irreversible losses – hence, it cannot be isentropic.

18-68C We are to discuss the states on the Fanno and Rayleigh lines. Analysis The Fanno line represents the states that satisfy the conservation of mass and energy equations. The Rayleigh line represents the states that satisfy the conservation of mass and momentum equations. The intersections points of these lines represent the states that satisfy the conservation of mass, energy, and momentum equations. Discussion T-s diagrams are quite helpful in understanding these kinds of flows.

18-69C We are to analyze a claim about oblique shock analysis. Analysis Yes, the claim is correct. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is b = p / 2 , or 90°. Discussion The component of flow in the direction normal to the oblique shock acts exactly like a normal shock. We can think of the flow parallel to the oblique shock as “going along for the ride” – it does not affect anything.

18-70C We are to discuss the effect of a normal shock wave on several properties. Analysis (a) velocity decreases, (b) static temperature increases, (c) stagnation temperature remains the same, (d) static pressure increases, and (e) stagnation pressure decreases. Discussion In addition, the Mach number goes from supersonic (Ma > 1) to subsonic (Ma < 1).

18-71C We are to discuss the formation of oblique shocks and how they differ from normal shocks. Analysis Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. Discussion In addition, while a normal shock must go from supersonic (Ma > 1) to subsonic (Ma < 1), the Mach number downstream of an oblique shock can be either supersonic or subsonic.

18-72C We are to discuss whether the flow upstream and downstream of an oblique shock needs to be supersonic. Analysis Yes, the upstream flow has to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. Discussion The latter is not true for normal shocks. For a normal shock, the flow must always go from supersonic (Ma > 1) to subsonic (Ma < 1).

13-698

18-73C We are to determine if Ma downstream of a normal shock can be supersonic. Analysis No, the second law of thermodynamics requires the flow after the shock to be subsonic. Discussion A normal shock wave always goes from supersonic to subsonic in the flow direction.

18-74C We are to discuss if a shock wave can develop in the converging section of a converging-diverging nozzle. Analysis No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. Discussion A normal shock (if it is to occur) would occur in the supersonic (diverging) section of the nozzle.

18-75C We are to discuss shock detachment at the nose of a 2-D wedge-shaped body. Analysis When the wedge half-angle  is greater than the maximum deflection angle θmax , the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is  = 90°. Discussion When  is less than qmax , the oblique shock is attached to the nose.

18-76C We are to discuss the shock at the nose of a rounded body in supersonic flow. Analysis When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle  at the nose is 90°, and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether two-dimensional, axisymmetric, or fully three-dimensional. Discussion Since  = 90° at the nose,  is always greater than qmax , regardless of Ma or the shape of the rest of the body.

18-77 Air flowing through a nozzle experiences a normal shock. Various properties are to be calculated before and after the shock. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air at room temperature are k = 1.4, R = 0.287 kJ / kg ×K , and c p = 1.005 kJ / kg ×K . Analysis The stagnation temperature and pressure before the shock are T01 = T1 +

æ 1 kJ/kg ÷ ö V12 (815 m/s) 2 çç = 230 + ÷= 560.5 K ø 2c p 2(1.005 kJ/kg ×K) çè1000 m 2 / s 2 ÷ k / ( k - 1)

æT ö ÷ P01 = P1 ççç 01 ÷ ÷ èç T1 ÷ ø

1.4/ (1.4- 1)

æ560.5 K ÷ ö = (26 kPa) çç ÷ çè 230 K ÷ ø

= 587.3 kPa

13-699

The velocity and the Mach number before the shock are determined from

æ1000 m 2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(230 K) ççç = 304.0 m / s ÷ è 1 kJ/kg ÷ ø

c1 =

kRT1 =

Ma1 =

V1 815 m/s = = 2.681 c1 304.0 m/s

and

The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-33. For Ma1 = 2.681 we read Ma 2 = 0.4972,

P02 = 9.7330, P1

P2 = 8.2208, P1

and

T2 = 2.3230 T1

Then the stagnation pressure P02 , static pressure P2 , and static temperature T2 , are determined to be

P02 = 9.7330P1 = (9.7330)(26 kPa) = 253.1 kPa

P2 = 8.2208P1 = (8.2208)(26 kPa) = 213.7 kPa T2 = 2.3230T1 = (2.3230)(230 K) = 534.3 K The air velocity after the shock can be determined from V2 = Ma 2 c2 , where c 2 is the speed of sound at the exit conditions after the shock,

æ1000 m 2 / s 2 ÷ ö ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.4972) (1.4)(0.287 kJ/kg.K)(534.3 K) ççç = 230.4 m / s ÷ è 1 kJ/kg ø÷ Discussion This problem could also be solved using the relations for compressible flow and normal shock functions. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" Px=26 [kPa] Tx=230 [K] Velx=815 [m/s] "Properties" c_p=CP(air, T=300) c_v=CV(air, T=300) k=c_p/c_v R=c_p-c_v "Analysis" T0x=Tx+Velx^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P0x=Px*(T0x/Tx)^(k/(k-1)) Cx=sqrt(k*R*Tx*Convert(kJ/kg, m^2/s^2)) Mx=Velx/Cx © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-700

My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) P0y=Px*(1+k*Mx^2)/(1+k*My^2)*(1+My^2*(k-1)/2)^(k/(k-1)) Py=Px*(1+k*Mx^2)/(1+k*My^2) Ty=Tx*(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) Cy=sqrt(k*R*Ty*Convert(kJ/kg, m^2/s^2)) Vely=My*Cy

18-78 Air flowing through a nozzle experiences a normal shock. The entropy change of air across the normal shock wave is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air at room temperature are R = 0.287 kJ / kg ×K and c p = 1.005 kJ / kg ×K . Analysis The entropy change across the shock is determined to be T P s2 - s1 = c p ln 2 - R ln 2 T1 P1

= (1.005 kJ/kg ×K)ln(2.3230) - (0.287 kJ/kg ×K)ln(8.2208) = 0.242 kJ / kg×K Discussion A shock wave is a highly dissipative process, and the entropy generation is large during shock waves.

EES (Engineering Equation Solver) SOLUTION "Given" Px=26 [kPa] Tx=230 [K] Velx=815 [m/s] "Properties" c_p=CP(air, T=300) c_v=CV(air, T=300) k=c_p/c_v R=c_p-c_v "Analysis" T0x=Tx+Velx^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P0x=Px*(T0x/Tx)^(k/(k-1)) Cx=sqrt(k*R*Tx*Convert(kJ/kg, m^2/s^2)) Mx=Velx/Cx My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) P0y=Px*(1+k*Mx^2)/(1+k*My^2)*(1+My^2*(k-1)/2)^(k/(k-1)) Py=Px*(1+k*Mx^2)/(1+k*My^2) Ty=Tx*(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) Cy=sqrt(k*R*Ty*Convert(kJ/kg, m^2/s^2)) Vely=My*Cy DELTAs=c_p*ln(Ty/Tx)-R*ln(Py/Px)

18-79 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-701

2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic,

P01 = Pi = 2.4 MPa

It is specified that A / A* = 3.5 . From Table A-32, Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 = 2.80 and P1 / P01 = 0.0368 . The pressure ratio across the shock for this Ma1 value is, from Table A-33,

P2 / P1 = 8.98 . Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be P2 = 8.98P1 = 8.98´ 0.0368P01 = 8.98´ 0.0368´ (2.4 MPa) = 0.793 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=2000 [kPa] T_i=(100+273) [K] Ratio_A=3.5 "Ratio_A=A/A_star" "Properties" c_p=CP(air, T=T_i) c_v=CV(air, T=T_i) k=c_p/c_v R=c_p-c_v "Analysis" P0x=P_i Ratio_A=1/Mx*(2/(k+1)*(1+(k-1)/2*Mx^2))^((k+1)/(2*(k-1))) Px=P0x*(1+(k-1)/2*Mx^2)^(-k/(k-1)) My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) Py=Px*(1+k*Mx^2)/(1+k*My^2)

18-80 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic,

P0x = Pi = 2.4 MPa

13-702

It is specified that A / A* = 2 . From Table A-32, the Mach number and the pressure ratio which corresponds to this area ratio are the Ma1 = 2.20 and P1 / P01 = 0.0935 . The pressure ratio across the shock for this M1 value is, from Table A-33,

P2 / P1 = 5.48 . Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be

P2 = 5.48P1 = 5.48´ 0.0935P01 = 5.48´ 0.0935´ (2.4 MPa) = 1.23 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=2000 [kPa] T_i=(100+273) [K] Ratio_A=2 "Ratio_A=A/A_star" "Properties" c_p=CP(air, T=T_i) c_v=CV(air, T=T_i) k=c_p/c_v R=c_p-c_v "Analysis" P0x=P_i Ratio_A=1/Mx*(2/(k+1)*(1+(k-1)/2*Mx^2))^((k+1)/(2*(k-1))) Px=P0x*(1+(k-1)/2*Mx^2)^(-k/(k-1)) My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) Py=Px*(1+k*Mx^2)/(1+k*My^2)

18-81E Air flowing through a nozzle experiences a normal shock. Effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.06855 Btu / lbm ×R , and the properties of helium are k = 1.667 and

R = 0.4961 Btu / lbm ×R . Analysis The air properties upstream the shock are Ma1 = 2.5 , P1 = 10 psia , and T1 = 440.5 R

13-703

Fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-33. For Ma1 = 2.5 , Ma 2 = 0.513,

P02 P T = 8.5262, 2 = 7.125, and 2 = 2.1375 P1 P1 T1

Then the stagnation pressure P02 , static pressure P2 , and static temperature T2 , are determined to be

P02 = 8.5262P1 = (8.5262)(10 psia ) = 85.3 psia P2 = 7.125P1 = (7.125)(10 psia ) = 71.3 psia T2 = 2.1375T1 = (2.1375)(440.5 R) = 942 R The air velocity after the shock can be determined from V2 = Ma 2 c2 , where c 2 is the speed of sound at the exit conditions after the shock,

æ25,037 ft 2 / s 2 ÷ ö ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.513) (1.4)(0.06855 Btu/1bm ×R)(941.6 R) ççç = 772 ft / s ÷ è 1 Btu/1bm ÷ ø We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-33 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, 1/ 2

1/ 2 æ Ma12 + 2 / (k - 1) ÷ ö æ ö 2.52 + 2 / (1.667 - 1) ÷ ç ç ÷ ÷ Ma 2 = çç ÷ = çç ÷ = 0.553 2 2 è 2´ 2.5 ´ 1.667 / (1.667 - 1) - 1÷ ø èç 2Ma1 k / (k - 1) - 1÷ ø

P2 1 + kMa12 1 + 1.667´ 2.52 = = = 7.5632 2 P1 1 + kMa 2 1 + 1.667´ 0.5532 T2 1 + Ma12 (k - 1) / 2 1 + 2.52 (1.667 - 1) / 2 = = = 2.7989 T1 1 + Ma 22 (k - 1) / 2 1 + 0.5532 (1.667 - 1) / 2 k /( k - 1) P02 æ 1 + kMa12 ö÷ ÷ 1 + (k - 1)Ma 22 / 2) = ççç 2 ÷( ÷ ç P1 è1 + kMa 2 ø

æ 1 + 1.667 ´ 2.52 ÷ ö 1.667/ 0.667 ÷(1 + (1.667 - 1)´ 0.5532 / 2) = çç = 9.641 2÷ çè1 + 1.667´ 0.553 ø÷ Thus,

P02 = 11.546P1 = (11.546)(10 psia ) = 115 psia

P2 = 7.5632P1 = (7.5632)(10 psia ) = 75.6 psia T2 = 2.7989T1 = (2.7989)(440.5 R) = 1233 R æ25,037 ft 2 / s 2 ÷ ö ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.553) (1.667)(0.4961 Btu/1bm.R)(1232.9 R) ççç = 2794 ft / s ÷ ÷ è 1 Btu/1bm ø Discussion This problem could also be solved using the relations for compressible flow and normal shock functions. The results would be identical.

13-704

EES (Engineering Equation Solver) SOLUTION Procedure NormalShock(M_x, k : M_y, PyOPx, TyOTx, RhoyORhox, PoyOPox, PoyOPx) If M_x < 1 Then M_y = -1000;PyOPx=-1000;TyOTx=-1000;RhoyORhox=-1000 PoyOPox=-1000;PoyOPx=-1000 else M_y=sqrt( (M_x^2+2/(k-1)) / (2*M_x^2*k/(k-1)-1) ) PyOPx=(1+k*M_x^2)/(1+k*M_y^2) TyOTx=( 1+M_x^2*(k-1)/2 )/(1+M_y^2*(k-1)/2 ) RhoyORhox=PyOPx/TyOTx PoyOPox=M_x/M_y*( (1+M_y^2*(k-1)/2)/ (1+M_x^2*(k-1)/2) )^((k+1)/(2*(k-1))) PoyOPx=((1+k*M_x^2)/(1+k*M_y^2))*(1+M_y^2*(k-1)/2)^(k/(k-1)) Endif End Function ExitPress(P_back,P_crit) If P_back>=P_crit then ExitPress:=P_back "Unchoked Flow Condition" If P_back<P_crit then ExitPress:=P_crit "Choked Flow Condition" End Procedure GetProp(Gas$ : Cp, k, R) "Cp and k data are from Text Table A.2E. A more general GetProp function could be written but care must be excercised since the fluid Gas$ can be an ideal gas (e.g., air, C02) or a real fluid (e.g., helium) which differ in the number of parameters required for property evaluations. The function IfIdealGas can be used to test if Gas$ is an ideal gas." M=MOLARMASS(Gas$) "Molar mass of Gas$" R_u=1545 "[ft-lbf/lbmol-R]" R= 1545[ft-lbf/lbmol-R]/M "[ft-lbf/lbm-R] Particular gas constant for Gas$" if Gas$='Air' then Cp=0.24"Btu/lbm-R"; k=1.4 endif if Gas$='CO2' then Cp=0.203"Btu/lbm_R"; k=1.289 endif if Gas$='Helium' then Cp=1.25"Btu/lbm-R"; k=1.667 endif End "Variable Definitions:" "M = flow Mach Number" "P_ratio = P/P_o for compressible, isentropic flow" "T_ratio = T/T_o for compressible, isentropic flow" "Rho_ratio= Rho/Rho_o for compressible, isentropic flow" "A_ratio=A/A* for compressible, isentropic flow" "Fluid properties before the shock are denoted with a subscript x" "Fluid properties after the shock are denoted with a subscript y" "M_y = Mach Number down stream of normal shock" "PyOverPx= P_y/P_x Pressue ratio across normal shock" "TyOverTx =T_y/T_x Temperature ratio across normal shock" "RhoyOverRhox=Rho_y/Rho_x Density ratio across normal shock" "PoyOverPox = P_oy/P_ox Stagantion pressure ratio across normal shock" "PoyOverPx = P_oy/P_x Stagnation pressure after normal shock ratioed to pressure before shock" "Input Data"

13-705

{P_x = 10 "psia" "Vaules of P_x, T_x, and M_x are set in the Parametric Table" T_x = 440.5 "R" M_x = 2.5} Gas$='Air' "This program has been written for the gases Air, CO2, and Helium" Call GetProp(Gas$:Cp,k,R) Call NormalShock(M_x,k:M_y,PyOverPx, TyOverTx,RhoyOverRhox, PoyOverPox, PoyOverPx) P_oy_air=P_x*PoyOverPx "Stagnation pressure after the shock" P_y_air=P_x*PyOverPx "Pressure after the shock" T_y_air=T_x*TyOverTx "Temperature after the shock" M_y_air=M_y "Mach number after the shock" "The velocity after the shock can be found from the product of the Mach number and speed of sound after the shock." C_y_air = sqrt(k*R*T_y_air*Convert(ft-lbf/lbm, ft^2/s^2) ) V_y_air=M_y_air*C_y_air DELTAs_air=entropy(air,T=T_y_air, P=P_y_air) -entropy(air,T=T_x,P=P_x) Gas2$='Helium' "Gas2$ can be either Helium or CO2" Call GetProp(Gas2$:Cp_2,k_2,R_2) Call NormalShock(M_x,k_2:M_y2,PyOverPx2, TyOverTx2,RhoyOverRhox2, PoyOverPox2, PoyOverPx2) P_oy_he=P_x*PoyOverPx2 "Stagnation pressure after the shock" P_y_he=P_x*PyOverPx2 "Pressure after the shock" T_y_he=T_x*TyOverTx2 "Temperature after the shock" M_y_he=M_y2 "Mach number after the shock" "The velocity after the shock can be found from the product of the Mach number and speed of sound after the shock." C_y_he = sqrt(k_2*R_2*T_y_he*Convert(ft-lbf/lbm, ft^2/s^2)) V_y_he=M_y_he*C_y_he DELTAs_he=entropy(helium,T=T_y_he, P=P_y_he) -entropy(helium,T=T_x,P=P_x)

18-82E We are to reconsider Prob. 12-81E. Using appropriate software, we are to study the effects of both air and helium flowing steadily in a nozzle when there is a normal shock at a Mach number in the range 2 < Mx < 3.5. In addition to the required information, we are to calculate the entropy change of the air and helium across the normal shock, and tabulate the results in a parametric table. Analysis We use EES to calculate the entropy change of the air and helium across the normal shock. The results are given in the Parametric Table for 2 < M_x < 3.5. Procedure NormalShock(M_x,k:M_y,PyOPx, TyOTx,RhoyORhox, PoyOPox, PoyOPx) If M_x < 1 Then M_y = -1000;PyOPx=-1000;TyOTx=-1000;RhoyORhox=-1000 PoyOPox=-1000;PoyOPx=-1000 else M_y=sqrt( (M_x^2+2/(k-1)) / (2*M_x^2*k/(k-1)-1) ) PyOPx=(1+k*M_x^2)/(1+k*M_y^2) TyOTx=( 1+M_x^2*(k-1)/2 )/(1+M_y^2*(k-1)/2 ) RhoyORhox=PyOPx/TyOTx PoyOPox=M_x/M_y*( (1+M_y^2*(k-1)/2)/ (1+M_x^2*(k-1)/2) )^((k+1)/(2*(k-1))) PoyOPx=(1+k*M_x^2)*(1+M_y^2*(k-1)/2)^(k/(k-1))/(1+k*M_y^2) Endif End Function ExitPress(P_back,P_crit) If P_back>=P_crit then ExitPress:=P_back "Unchoked Flow Condition" If P_back<P_crit then ExitPress:=P_crit "Choked Flow Condition" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-706

End Procedure GetProp(Gas$:cp,k,R) "cp and k data are from Text Table A.2E" M=MOLARMASS(Gas$) R= 1545/M if Gas$='Air' then cp=0.24 [Btu/lbm-R] k=1.4 endif if Gas$='CO2' then cp=0.203 [Btu/lbm-R] k=1.289 endif if Gas$='Helium' then cp=1.25 [Btu/lbm-R] k=1.667 endif End "Variable Definitions:" "M = flow Mach Number" "P_ratio = P/P_o for compressible, isentropic flow" "T_ratio = T/T_o for compressible, isentropic flow" "Rho_ratio= Rho/Rho_o for compressible, isentropic flow" "A_ratio=A/A* for compressible, isentropic flow" "Fluid properties before the shock are denoted with a subscript x" "Fluid properties after the shock are denoted with a subscript y" "M_y = Mach Number down stream of normal shock" "PyOverPx= P_y/P_x Pressue ratio across normal shock" "TyOverTx =T_y/T_x Temperature ratio across normal shock" "RhoyOverRhox=Rho_y/Rho_x Density ratio across normal shock" "PoyOverPox = P_oy/P_ox Stagantion pressure ratio across normal shock" "PoyOverPx = P_oy/P_x Stagnation pressure after normal shock ratioed to pressure before shock" "Input Data" {P_x = 10 [psia]} "Vaules of P_x, T_x, and M_x are set in the Parametric Table" {T_x = 440.5 [R]} {M_x = 2.5} Gas$='Air' "This program has been written for the gases Air, CO2, and Helium" Call GetProp(Gas$:cp,k,R) Call NormalShock(M_x,k:M_y,PyOverPx, TyOverTx,RhoyOverRhox, PoyOverPox, PoyOverPx) P_oy_air=P_x*PoyOverPx "Stagnation pressure after the shock" P_y_air=P_x*PyOverPx "Pressure after the shock" T_y_air=T_x*TyOverTx "Temperature after the shock" M_y_air=M_y "Mach number after the shock" "The velocity after the shock can be found from the product of the Mach number and speed of sound after the shock." c_y_air = sqrt(k*R*T_y_air*32.2) V_y_air=M_y_air*c_y_air DELTAs_air=entropy(air,T=T_y_air, P=P_y_air) -entropy(air,T=T_x,P=P_x) Gas2$='Helium' "Gas2$ can be either Helium or CO2" Call GetProp(Gas2$:cp_2,k_2,R_2) Call NormalShock(M_x,k_2:M_y2,PyOverPx2, TyOverTx2,RhoyOverRhox2, PoyOverPox2, PoyOverPx2) P_oy_he=P_x*PoyOverPx2 "Stagnation pressure after the shock" P_y_he=P_x*PyOverPx2 "Pressure after the shock" T_y_he=T_x*TyOverTx2 "Temperature after the shock" M_y_he=M_y2 "Mach number after the shock" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-707

"The velocity after the shock can be found from the product of the Mach number and speed of sound after the shock." c_y_he = sqrt(k_2*R_2*T_y_he*32.2) V_y_he=M_y_he*c_y_he DELTAs_he=entropy(helium,T=T_y_he, P=P_y_he) -entropy(helium,T=T_x,P=P_x) The parametric table and the corresponding plots are shown below.

Vy,he

Vy,air Ty,he Ty,air

[ft / s] [ft / s] [R]

[R]

Py,he

Py,air

Poy,he

Poy,air M y,he M y,air Mx

[R] [psia] [psia] [psia] [psia] [psia]

2644 771.9 915. 743. 440. 6 3 5 2707 767.1 1066 837. 440. 6 5 2795 771.9 1233 941. 440. 6 5 3022 800.4 1616 1180 440. 5 3292 845.4 2066 1460 440. 5

47.5

60.79 57.4

75.63 71.25

110

103.3

150.6 141.3

ΔShe

ΔSair

[Btu / lbm – R] [Btu / lbm – R]

63.46 56.4 0.607 0.577 2 4 79.01 70.02 0.575 0.540 2.25 9 6 96.41 85.26 0.553 0.513 2.5

0.1345

0.0228

0.2011

0.0351

0.2728

0.04899

136.7 120.6 0.522 0.475 3 3 2 184.5 162.4 0.503 0.451 3.5 2 2

0.4223

0.08

0.5711

0.1136

13-708

Discussion In all cases, regardless of the fluid or the Mach number, entropy increases across a shock wave. This is because a shock wave involves irreversibilities.

13-709

18-83 Air flowing through a converging-diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The properties of air are k = 1.4 and R = 0.287 kJ / kg ×K . Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Then, P01 = Pi = 1 MPa

T01 = Ti = 300 K

Then,

æ ö÷ æ ö÷ 2 2 ÷= (300 K) çç ÷= 139.4 K T1 = T01 ççç 2÷ 2÷ çè 2 + (1.4 - 1)2.4 ÷ çè 2 + (k - 1)Ma1 ÷ ø ø and k / ( k - 1)

æT ö ÷ P1 = P01 ççç 1 ÷ ÷ èçT0 ÷ ø

1.4/ 0.4

æ139.4 ÷ ö = (1 MPa) çç ÷ ÷ èç 300 ø

= 0.06840 MPa

The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-33. For Ma1 = 2.4 we read Ma 2 = 0.5231 @ 0.523,

P02 P T = 0.5401, 2 = 6.5533, and 2 = 2.0403 P01 P1 T1

Then the stagnation pressure P02 , static pressure P2 , and static temperature T2 , are determined to be

P02 = 0.5401P01 = (0.5401)(1.0 MPa) = 0.540 MPa = 540 kPa P2 = 6.5533P1 = (6.5533)(0.06840 MPa) = 0.448 MPa = 448 kPa T2 = 2.0403T1 = (2.0403)(139.4 K) = 284 K The air velocity after the shock can be determined from V2 = Ma 2 c2 , where c 2 is the speed of sound at the exit conditions after the shock,

æ1000 m 2 / s 2 ÷ ö ÷ V2 = Ma2 c2 = Ma 2 kRT2 = (0.5231) (1.4)(0.287 kJ/kg ×K)(284 K) ççç = 177 m / s ÷ è 1 kJ/kg ÷ ø Discussion We can also solve this problem using the relations for normal shock functions. The results would be identical.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=1000 [kPa] T_i=300 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-710

Vel_i=0 "negligible" Mx=2.4 "Properties" k=1.4 R=0.287 [kJ/kg-K] "Analysis" P0x=P_i T0x=T_i Tx=T0x*2/(2+(k-1)*Mx^2) Px=P0x*(Tx/T0x)^(k/(k-1)) My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) P0y=P0x*Mx/My*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^((k+1)/(2*(k-1))) Py=Px*(1+k*Mx^2)/(1+k*My^2) Ty=Tx*(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) Cy=sqrt(k*R*Ty*Convert(kJ/kg, m^2/s^2)) Vely=My*Cy

18-84 The entropy change of air across the shock for upstream Mach numbers between 0.5 and 1.5 is to be determined and plotted. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4, R = 0.287 kJ / kg ×K , and c p = 1.005 kJ / kg ×K . Analysis The entropy change across the shock is determined to be T P s2 - s1 = c p ln 2 - R ln 2 T1 P1

1/ 2

where and

æ Ma12 + 2 / (k - 1) ÷ ö P2 1 + kMa12 ÷ = Ma 2 = ççç , , ÷ 2 P1 1 + kMa 22 èç 2Ma1 k / (k - 1) - 1÷ ø

T2 1 + Ma12 (k - 1) / 2 = T1 1 + Ma 22 (k - 1) / 2

13-711

EES (Engineering Equation Solver) SOLUTION k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) RatioT=(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) "RatioT=Ty/Tx" RatioP=(1+k*Mx^2)/(1+k*My^2) "RatioP=Py/Tx" DELTAs=c_p*ln(RatioT)-R*ln(RatioP) Mx

RatioT

RatioP

D s, kg / kg ×K

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5

2.6458 1.8778 1.5031 1.2731 1.1154 1.0000 0.9118 0.8422 0.7860 0.7397 0.7011

0.4375 0.6286 0.7563 0.8519 0.9305 1.0000 1.0649 1.1280 1.1909 1.2547 1.3202

0.1250 0.2533 0.4050 0.5800 0.7783 1.0000 1.2450 1.5133 1.8050 2.1200 2.4583

-0.2340 -0.0724 -0.0213 -0.0048 -0.0005 -0.0000 0.0003 0.0021 0.0061 0.0124 0.0210

0.05 0

sy - sx

-0.05 -0.1 -0.15 -0.2 -0.25 0.4

0.6

0.8

1.2

1.4

1.6

Mx Discussion The total entropy change is negative for upstream Mach numbers Ma1 less than unity. Therefore, normal shocks cannot occur when Ma1 < 1 .

18-85 Supersonic airflow approaches the nose of a two-dimensional wedge and undergoes a straight oblique shock. For a specified Mach number, the minimum shock angle and the maximum deflection angle are to be determined. Assumptions Air is an ideal gas with a constant specific heat ratio of k = 1.4 (so that Fig. 18-43 is applicable).

13-712

Analysis For Ma = 3, we read from Fig. 12-41 Minimum shock (or wave) angle:

b min = 19°

Maximum deflection (or turning) angle:

qmax = 34°

Discussion Note that the minimum shock angle decreases and the maximum deflection angle increases with increasing Mach number Ma1 .

18-86 Air flowing at a specified supersonic Mach number undergoes an expansion turn. The Mach number, pressure, and temperature downstream of the sudden expansion along a wall are to be determined. Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin. 3 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k = 1.4. Analysis On the basis of Assumption #2, we take the deflection angle as equal to the wedge half-angle, i.e.,   = 15°. Then the upstream and downstream Prandtl-Meyer functions are determined to be ö k + 1 - 1æ k- 1 ÷ - 1 n (Ma) = tan ççç (Ma 2 - 1) ÷ Ma 2 - 1 ÷- tan ÷ çè k + 1 k- 1 ø

(

)

Upstream:

n (Ma1 ) =

ö 1.4 + 1 - 1 æ 1.4 - 1 ÷ tan ççç (3.62 - 1) ÷ - tan - 1 ÷ ÷ çè 1.4 + 1 1.4 - 1 ø

( 3.6 - 1)= 60.09° 2

13-713

Then the downstream Prandtl-Meyer function becomes

n (Ma 2 ) = q + n (Ma1 ) = 15°+ 60.09° = 75.09°

Ma 2 is found from the Prandtl-Meyer relation, which is now implicit: Downstream: n (Ma 2 ) =

ö÷ 1.4 + 1 - 1 æ 1.4 - 1 tan ççç Ma 22 - 1) ÷ - tan - 1 ÷ ÷ ç 1.4 - 1 è 1.4 + 1 ø

( Ma - 1)= 75.09° 2 2

Solution of this implicit equation gives Ma 2 = 4.81 . Then the downstream pressure and temperature are determined from the isentropic flow relations:

P2 =

P2 / P0 [1 + Ma 22 (k - 1) / 2]- k /( k - 1) [1 + 4.812 (1.4 - 1) / 2]- 1.4/0.4 P1 = P = (32 kPa) = 6.65 kPa 1 P1 / P0 [1 + Ma12 (k - 1) / 2]- k /( k - 1) [1 + 3.62 (1.4 - 1) / 2]- 1.4/0.4

T2 =

T2 / T0 [1 + Ma 22 (k - 1) / 2]- 1 [1 + 4.812 (1.4 - 1) / 2]- 1 T1 = T = (240 K) = 153 K 1 T1 / T0 [1 + Ma12 (k - 1) / 2]- 1 [1 + 3.62 (1.4 - 1) / 2]- 1

Note that this is an expansion, and Mach number increases while pressure and temperature decrease, as expected. Discussion There are compressible flow calculators on the Internet that solve these implicit equations that arise in the analysis of compressible flow, along with both normal and oblique shock equations.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=32 [kPa] T_1=240 [K] Ma_1=3.6 theta=15 [degrees] "Properties" k=1.4 "Analysis" nu_Ma1=sqrt((k+1)/(k-1))*Arctan(sqrt((k-1)/(k+1)*(Ma_1^2-1)))-Arctan(sqrt(Ma_1^2-1)) nu_Ma2=theta+nu_Ma1 nu_Ma2=sqrt((k+1)/(k-1))*Arctan(sqrt((k-1)/(k+1)*(Ma_2^2-1)))-Arctan(sqrt(Ma_2^2-1)) P_2=P_1*P2\P0/P1\P0 P2\P0=(1+Ma_2^2*(k-1)/2)^((-k)/(k-1)) P1\P0=(1+Ma_1^2*(k-1)/2)^((-k)/(k-1)) T_2=T_1*T2\T0/T1\T0 T2\T0=(1+Ma_2^2*(k-1)/2)^(-1) T1\T0=(1+Ma_1^2*(k-1)/2)^(-1)

13-714

18-87E Air flowing at a specified supersonic Mach number is forced to undergo a compression turn (an oblique shock)., The Mach number, pressure, and temperature downstream of the oblique shock are to be determined. Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin. 3 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k = 1.4. Analysis On the basis of Assumption #2, we take the deflection angle as equal to the wedge half-angle, i.e.,    = 15°. Then the two values of oblique shock angle  are determined from

tan  

2(Ma 12 sin 2   1) / tan  Ma 12 (k  cos 2 )  2

tan 15 



2(2 2 sin 2   1) / tan  2 2 (1.4  cos 2 )  2

which is implicit in . Therefore, we solve it by an iterative approach or with an equation solver such as EES. It gives βweak = 45.34° and βstorng = 79.83° . Then the upstream “normal” Mach number Ma1,n becomes Weak shock:

Ma 1,n  Ma 1 sin   2 sin 45.34  1.423

Strong shock:

Ma 1,n  Ma 1 sin   2 sin 79.83  1.969

Also, the downstream normal Mach numbers Ma 2,n become Weak shock:

Strong shock:

Ma 2, n 

(k  1)Ma 1,2 n  2 2kMa 1,2 n  k  1 (k  1)Ma 1,2 n  2 2kMa 1,2 n  k  1



(1.4  1)(1.423) 2  2 2(1.4)(1.423) 2  1.4  1 (1.4  1)(1.969 ) 2  2 2(1.4)(1.969 ) 2  1.4  1

 0.7304

 0.5828

The downstream pressure and temperature for each case are determined to be Weak shock:

P2  P1

2kMa 1,2 n  k  1 k 1

 (6 psia )

2(1.4)(1.423) 2  1.4  1  17.57  17.6 psia 1.4  1

13-715

P2 1 P 2  (k  1)Ma 1,n 17.57 psia 2  (1.4  1)(1.423) 2  T1 2  ( 480 R )  609 .5 R  610 R P1  2 P1 (k  1)Ma 1,2 n 8 psia (1.4  1)(1.423) 2 2

T2  T1

Strong shock:

P2  P1

2kMa 1,2 n  k  1 k 1

 (8 psia )

2(1.4)(1.969 ) 2  1.4  1  34.85  34.9 psia 1.4  1

P2 1 P 2  (k  1)Ma 1,n 34.85 psia 2  (1.4  1)(1.969 ) 2  T1 2  ( 480 R )  797 .9 R  798 R P1  2 P1 (k  1)Ma 1,2 n 8 psia (1.4  1)(1.969 ) 2 2

T2  T1

The downstream Mach number is determined to be Ma 2, n 0.7304 Weak shock: Ma 2    1.45 sin(    ) sin( 45.34  15) Strong shock:

Ma 2 

Ma 2, n sin(    )



0.5828  0.644 sin(79.83  15)

Discussion Note that the change in Mach number, pressure, temperature across the strong shock are much greater than the changes across the weak shock, as expected. For both the weak and strong oblique shock cases, Ma1,n is supersonic and Ma 2,n is subsonic. However, Ma 2 is supersonic across the weak oblique shock, but subsonic across the strong oblique

shock.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=8 [psia] T_1=480 [R] Ma_1=2.0 theta=15 [degrees] "Properties" k=1.4 "Analysis" tan(theta)=(2*(Ma_1^2*sin(beta_weak)^2-1)/tan(beta_weak))/(Ma_1^2*(k+cos(2*beta_weak))+2) tan(theta)=(2*(Ma_1^2*sin(beta_strong)^2-1)/tan(beta_strong))/(Ma_1^2*(k+cos(2*beta_strong))+2) Ma_1_n_weak=Ma_1*sin(beta_weak) Ma_1_n_strong=Ma_1*sin(beta_strong) Ma_2_n_weak=sqrt(((k-1)*Ma_1_n_weak^2+2)/(2*k*Ma_1_n_weak^2-k+1)) Ma_2_n_strong=sqrt(((k-1)*Ma_1_n_strong^2+2)/(2*k*Ma_1_n_strong^2-k+1)) P_2_weak=P_1*(2*k*Ma_1_n_weak^2-k+1)/(k+1) P_2_strong=P_1*(2*k*Ma_1_n_strong^2-k+1)/(k+1) T_2_weak=T_1*P_2_weak/P_1*(2+(k-1)*Ma_1_n_weak^2)/((k+1)*Ma_1_n_weak^2) T_2_strong=T_1*P_2_strong/P_1*(2+(k-1)*Ma_1_n_strong^2)/((k+1)*Ma_1_n_strong^2) Ma_2_weak=Ma_2_n_weak/sin(beta_weak-theta) Ma_2_strong=Ma_2_n_strong/sin(beta_strong-theta)

18-88 Air flowing at a specified supersonic Mach number undergoes an expansion turn over a tilted wedge. The Mach number, pressure, and temperature downstream of the sudden expansion above the wedge are to be determined. Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin. 3 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k = 1.4.

13-716

Analysis On the basis of Assumption #2, the deflection angle is determined to be    = 25 – 10 = 15°. Then the upstream and downstream Prandtl-Meyer functions are determined to be

n (Ma) =

ö k + 1 - 1æ k- 1 ÷ tan ççç (Ma 2 - 1) ÷ - tan - 1 ÷ ÷ çè k + 1 k- 1 ø

( Ma - 1) 2

Upstream:

n (Ma1 ) =

ö÷ 1.4 + 1 - 1 æ 1.4 - 1 - 1 tan ççç (2.42 - 1) ÷ ÷- tan ÷ 1.4 - 1 èç 1.4 + 1 ø

( 2.4 - 1)= 36.75° 2

Then the downstream Prandtl-Meyer function becomes

n (Ma 2 ) = q + n (Ma1 ) = 15°+ 36.75° = 51.75° Now Ma 2 is found from the Prandtl-Meyer relation, which is now implicit: Downstream: n (Ma 2 ) =

ö 1.4 + 1 - 1 æ 1.4 - 1 ÷ tan ççç (Ma 22 - 1) ÷ - tan - 1 ÷ ÷ çè 1.4 + 1 1.4 - 1 ø

( Ma - 1)= 51.75° 2 2

It gives Ma 2 = 3.105 . Then the downstream pressure and temperature are determined from the isentropic flow relations

P2 = T2 =

P2 / P0 [1 + Ma 22 (k - 1) / 2]- k /( k - 1) [1 + 3.1052 (1.4 - 1) / 2]- 1.4/0.4 P1 = P = (70 kPa) = 23.8 kPa 1 P1 / P0 [1 + Ma12 (k - 1) / 2]- k /( k - 1) [1 + 2.42 (1.4 - 1) / 2]- 1.4/0.4

T2 / T0 [1 + Ma 22 (k - 1) / 2]- 1 [1 + 3.1052 (1.4 - 1) / 2]- 1 T1 = T = (260 K) = 191 K 2 - 1 1 T1 / T0 [1 + Ma1 (k - 1) / 2] [1 + 2.42 (1.4 - 1) / 2]- 1

EES (Engineering Equation Solver) SOLUTION "Given" P_1=70 [kPa] T_1=260 [K] Ma_1=2.4 theta=(25-10) [degrees] "Properties" k=1.4 "Analysis" nu_Ma1=sqrt((k+1)/(k-1))*Arctan(sqrt((k-1)/(k+1)*(Ma_1^2-1)))-Arctan(sqrt(Ma_1^2-1)) nu_Ma2=theta+nu_Ma1 nu_Ma2=sqrt((k+1)/(k-1))*Arctan(sqrt((k-1)/(k+1)*(Ma_2^2-1)))-Arctan(sqrt(Ma_2^2-1)) P_2=P_1*P2\P0/P1\P0 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-717

P2\P0=(1+Ma_2^2*(k-1)/2)^((-k)/(k-1)) P1\P0=(1+Ma_1^2*(k-1)/2)^((-k)/(k-1)) T_2=T_1*T2\T0/T1\T0 T2\T0=(1+Ma_2^2*(k-1)/2)^(-1) T1\T0=(1+Ma_1^2*(k-1)/2)^(-1)

18-89 Air flowing at a specified supersonic Mach number undergoes a compression turn (an oblique shock) over a tilted wedge. The Mach number, pressure, and temperature downstream of the shock below the wedge are to be determined. Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin. 3 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k = 1.4. Analysis On the basis of Assumption #2, the deflection angle is determined to be    = 25 + 10 = 35°. Then the two values of oblique shock angle  are determined from

tan q =

2(Ma12 sin 2 b - 1) / tan b Ma12 (k + cos 2b ) + 2

tan12° =

2(3.42 sin 2 b - 1) / tan b 3.42 (1.4 + cos 2b ) + 2

which is implicit in . Therefore, we solve it by an iterative approach or with an equation solver such as EES. It gives βweak = 49.86° and βstorng = 77.66° . Then for the case of strong oblique shock, the upstream “normal” Mach number Ma1,n becomes Ma1,n = Ma1 sin b = 5sin 77.66° = 4.884

Also, the downstream normal Mach numbers Ma 2,n become

Ma 2,n =

2 (k - 1)Ma1,n +2 2 2kMa1,n - k+1

(1.4 - 1)(4.884) 2 + 2 = 0.4169 2(1.4)(4.884) 2 - 1.4 + 1

The downstream pressure and temperature are determined to be P2 = P1

2 2kMa1,n - k+1

k+1

= (70 kPa)

2(1.4)(4.884) 2 - 1.4 + 1 = 1940 kPa 1.4 + 1 2

T2 = T1

P2 r 1 P 2 + (k - 1)Ma1,n 1940 kPia 2 + (1.4 - 1)(4.884) 2 = T1 2 = (260 K) = 1450 K 2 P1 r 2 P1 (k + 1)Ma1,n 70 kPa (1.4 + 1)(4.884) 2

The downstream Mach number is determined to be

Ma 2 =

Ma 2,n sin(b - q)

0.4169 = 0.615 sin(77.66°- 35°)

13-718

Discussion Note that Ma1,n is supersonic and Ma 2,n and Ma 2 are subsonic. Also note the huge rise in temperature and pressure across the strong oblique shock, and the challenges they present for spacecraft during reentering the earth’s atmosphere.

EES (Engineering Equation Solver) SOLUTION "Given" P_1=70 [kPa] T_1=260 [K] Ma_1=5 theta=(25+10) [degrees] "Properties" k=1.4 "Analysis" tan(theta)=(2*(Ma_1^2*sin(beta_strong)^2-1)/tan(beta_strong))/(Ma_1^2*(k+cos(2*beta_strong))+2) Ma_1_n_strong=Ma_1*sin(beta_strong) Ma_2_n_strong=sqrt(((k-1)*Ma_1_n_strong^2+2)/(2*k*Ma_1_n_strong^2-k+1)) P_2_strong=P_1*(2*k*Ma_1_n_strong^2-k+1)/(k+1) T_2_strong=T_1*P_2_strong/P_1*(2+(k-1)*Ma_1_n_strong^2)/((k+1)*Ma_1_n_strong^2) Ma_2_strong=Ma_2_n_strong/sin(beta_strong-theta)

18-90E Air flowing at a specified supersonic Mach number is forced to turn upward by a ramp, and weak oblique shock forms. The wave angle, Mach number, pressure, and temperature after the shock are to be determined. Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin. 3 Air is an ideal gas with constant specific heats.

Properties The specific heat ratio of air is k = 1.4. Analysis On the basis of Assumption #2, we take the deflection angle as equal to the ramp, i.e.,    = 8°. Then the two values of oblique shock angle  are determined from

tan q =

2(Ma12 sin 2 b - 1) / tan b 2(22 sin 2 b - 1) / tan b tan 8 ° =  Ma12 (k + cos 2b ) + 2 22 (1.4 + cos 2b ) + 2

which is implicit in . Therefore, we solve it by an iterative approach or with an equation solver such as EES. It gives βweak = 37.21° and βstrong = 85.05° . Then for the case of weak oblique shock, the upstream “normal” Mach number Ma1,n becomes Ma1, n = Ma1 sin b = 2sin 37.21° = 1.209

Also, the downstream normal Mach numbers Ma 2,n become

13-719

Ma 2, n =

(k - 1)Ma1,2 n + 2 2kMa1,2 n - k + 1

(1.4 - 1)(1.209) 2 + 2 = 0.8363 2(1.4)(1.209) 2 - 1.4 + 1

The downstream pressure and temperature are determined to be P2 = P1

2kMa1,2 n - k + 1 k+1

= (12 psia)

2(1.4)(1.209) 2 - 1.4 + 1 = 18.5 psia 1.4 + 1 2

T2 = T1

P2 r 1 P 2 + (k - 1)Ma1, n 18.5 psia 2 + (1.4 - 1)(1.209)2 = T1 2 = (490 R) = 556 R P1 r 2 P1 (k + 1)Ma1,2 n 12 psia (1.4 + 1)(1.209)2

The downstream Mach number is determined to be

Ma 2 =

Ma 2,n sin(b - q)

0.8363 = 1.71 sin(37.21°- 8°)

Discussion Note that Ma1,n is supersonic and Ma 2,n is subsonic. However, Ma 2 is supersonic across the weak oblique shock (it is subsonic across the strong oblique shock).

EES (Engineering Equation Solver) SOLUTION "Given" P_1=12 [psia] T_1=(30+460) [R] Ma_1=2 theta=8 [degrees] "Properties" k=1.4 "Analysis" tan(theta)=(2*(Ma_1^2*sin(beta_weak)^2-1)/tan(beta_weak))/(Ma_1^2*(k+cos(2*beta_weak))+2) Ma_1_n_weak=Ma_1*sin(beta_weak) Ma_2_n_weak=sqrt(((k-1)*Ma_1_n_weak^2+2)/(2*k*Ma_1_n_weak^2-k+1)) P_2_weak=P_1*(2*k*Ma_1_n_weak^2-k+1)/(k+1) T_2_weak=T_1*P_2_weak/P_1*(2+(k-1)*Ma_1_n_weak^2)/((k+1)*Ma_1_n_weak^2) Ma_2_weak=Ma_2_n_weak/sin(beta_weak-theta)

18-91 Air flowing at a specified supersonic Mach number impinges on a two-dimensional wedge, The shock angle, Mach number, and pressure downstream of the weak and strong oblique shock formed by a wedge are to be determined.

13-720

Assumptions 1 The flow is steady. 2 The boundary layer on the wedge is very thin. 3 Air is an ideal gas with constant specific heats. Properties The specific heat ratio of air is k = 1.4. Analysis On the basis of Assumption #2, we take the deflection angle as equal to the wedge half-angle, i.e.,    = 8°. Then the two values of oblique shock angle  are determined from

tan q =

2(Ma12 sin 2 b - 1) / tan b Ma12 (k + cos 2b ) + 2



tan 8° =

2(3.42 sin 2 b - 1) / tan b 3.42 (1.4 + cos 2b ) + 2

which is implicit in . Therefore, we solve it by an iterative approach or with an equation solver such as EES. It gives weak  23.15 and strong  87.45 . Then the upstream “normal” Mach number Ma1,n becomes Weak shock:

Ma1, n = Ma1 sin b = 3.4sin 23.15° = 1.336

Strong shock:

Ma1, n = Ma1 sin b = 3.4sin 87.45° = 3.397

Also, the downstream normal Mach numbers Ma 2,n become Weak shock: Ma 2, n =

Strong shock:

(k - 1)Ma1,2 n + 2 2kMa1,2 n - k + 1

(k - 1)Ma1,2 n + 2

Ma 2, n =

2kMa

2 1, n

- k+1

(1.4 - 1)(1.336) 2 + 2 = 0.7681 2(1.4)(1.336) 2 - 1.4 + 1 (1.4 - 1)(3.397) 2 + 2 = 0.4553 2(1.4)(3.397) 2 - 1.4 + 1

The downstream pressure for each case is determined to be Weak shock: P2 = P1 Strong shock:

2kMa1,2 n - k + 1 k+1

P2 = P1

= (40 kPa)

2kMa1,2 n - k + 1 k+1

2(1.4)(1.336) 2 - 1.4 + 1 = 76.6 kPa 1.4 + 1

= (40 kPa)

2(1.4)(3.397) 2 - 1.4 + 1 = 532 kPa 1.4 + 1

The downstream Mach number is determined to be Weak shock: Ma 2 =

Strong shock:

Ma 2, n sin(b - q)

Ma 2 =

0.7681 = 2.94 sin(23.15°- 8°)

Ma 2, n sin(b - q)

0.4553 = 0.463 sin(87.45°- 8°)

Discussion Note that the change in Mach number and pressure across the strong shock are much greater than the changes across the weak shock, as expected. For both the weak and strong oblique shock cases, Ma1,n is supersonic and Ma 2,n is subsonic. However, Ma 2 is supersonic across the weak oblique shock, but subsonic across the strong oblique shock.

13-721

EES (Engineering Equation Solver) SOLUTION "Given" P_1=60 [kPa] T_1=240 [K] Ma_1=3.4 theta=8 [degrees] "Properties" k=1.4 "Analysis" tan(theta)=(2*(Ma_1^2*sin(beta_weak)^2-1)/tan(beta_weak))/(Ma_1^2*(k+cos(2*beta_weak))+2) tan(theta)=(2*(Ma_1^2*sin(beta_strong)^2-1)/tan(beta_strong))/(Ma_1^2*(k+cos(2*beta_strong))+2) Ma_1_n_weak=Ma_1*sin(beta_weak) Ma_1_n_strong=Ma_1*sin(beta_strong) Ma_2_n_weak=sqrt(((k-1)*Ma_1_n_weak^2+2)/(2*k*Ma_1_n_weak^2-k+1)) Ma_2_n_strong=sqrt(((k-1)*Ma_1_n_strong^2+2)/(2*k*Ma_1_n_strong^2-k+1)) P_2_weak=P_1*(2*k*Ma_1_n_weak^2-k+1)/(k+1) P_2_strong=P_1*(2*k*Ma_1_n_strong^2-k+1)/(k+1) Ma_2_weak=Ma_2_n_weak/sin(beta_weak-theta) Ma_2_strong=Ma_2_n_strong/sin(beta_strong-theta)

18-92 Air flowing through a nozzle experiences a normal shock. The effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium under the same conditions. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, onedimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.287 kJ / kg ×K , and the properties of helium are k = 1.667 and

R = 2.0769 kJ / kg ×K .

Analysis The air properties upstream the shock are

Ma1  2.6 , P1  58kPa , and T1  270 K Fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions in Table A-33. For Ma1  2.6 , Ma 2 = 0.5039,

P02 = 9.1813, P1

P2 = 7.7200, P1

and

T2 = 2.2383 T1

We obtained these values using analytical relations in Table 14. Then the stagnation pressure P02 , static pressure P2 , and static temperature T2 , are determined to be

P02  9.1813P1  (9.1813)(58kPa)  532.5kPa P2  7.7200P1  (7.7200)(58kPa)  447.8kPa

13-722

The air velocity after the shock can be determined from V2  Ma 2 c2 , where c 2 is the speed of sound at the exit conditions after the shock, æ1000 m 2 / s 2 ö ÷ ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.5039) (1.4)(0.287 kJ/kg ×K)(604.3 K) çç = 248.3 m / s ÷ ÷ çè 1 kJ/kg ø

We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-33 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, 1/ 2

1/ 2 æ Ma12 + 2 / (k - 1) ö æ ö 2.62 + 2 / (1.667 - 1) ÷ ÷ çç ÷ ÷ Ma 2 = ççç = = 0.5455 ÷ 2 çè 2´ 2.62 ´ 1.667 / (1.667 - 1) - 1÷ ÷ ÷ ø èç 2Ma1 k / (k - 1) - 1ø

P2 1 + kMa12 1 + 1.667´ 2.62 = = = 8.2009 P1 1 + kMa 22 1 + 1.667´ 0.54552 T2 1 + Ma12 (k - 1) / 2 1 + 2.62 (1.667 - 1) / 2 = = = 2.9606 T1 1 + Ma 22 (k - 1) / 2 1 + 0.54552 (1.667 - 1) / 2 k /( k - 1) P02 æ 1 + kMa12 ö÷ ÷ 1 + (k - 1)Ma 22 / 2) = ççç 2 ÷( ÷ ç P1 è1 + kMa 2 ø

æ 1 + 1.667 ´ 2.62 ÷ ö 1.667/0.667 2 ÷ = çç = 10.389 ÷(1 + (1.667 - 1) ´ 0.5455 / 2) çè1 + 1.667 ´ 0.54552 ø÷ Thus,

P02  10.389P1  (10.389)(58kPa)  602.5kPa

P2  8.2009P1  (8.2009)(58kPa)  475.7kPa T2  2.9606T1  (2.9606)(270 K)  799.4K æ1000 m 2 / s 2 ö ÷ ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.5455) (1.667)(2.0769 kJ/kg ×K)(799.4 K) ççç = 907.5 m / s ÷ ÷ è 1 kJ/kg ø

Discussion The velocity and Mach number are higher for helium than for air due to the different values of k and R.

EES (Engineering Equation Solver) SOLUTION "Given" Mx=2.6 Px=58 [kPa] Tx=270 [K] "Properties" c_p_a=CP(air, T=300) c_v_a=CV(air, T=300) k_a=c_p_a/c_v_a R_a=c_p_a-c_v_a c_p_b=CP(Helium, T=300, P=100) c_v_b=CV(Helium, T=300, P=100) k_b=c_p_b/c_v_b R_b=c_p_b-c_v_b "Analysis" "for air" My_a=sqrt((Mx^2+2/(k_a-1))/(2*Mx^2*k_a/(k_a-1)-1)) P0y_a=Px*(1+k_a*Mx^2)/(1+k_a*My_a^2)*(1+My_a^2*(k_a-1)/2)^(k_a/(k_a-1)) Py_a=Px*(1+k_a*Mx^2)/(1+k_a*My_a^2) Ty_a=Tx*(1+Mx^2*(k_a-1)/2)/(1+My_a^2*(k_a-1)/2) Cy_a=sqrt(k_a*R_a*Ty_a*Convert(kJ/kg, m^2/s^2)) Vely_a=My_a*Cy_a "for helium" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-723

My_b=sqrt((Mx^2+2/(k_b-1))/(2*Mx^2*k_b/(k_b-1)-1)) P0y_b=Px*(1+k_b*Mx^2)/(1+k_b*My_b^2)*(1+My_b^2*(k_b-1)/2)^(k_b/(k_b-1)) Py_b=Px*(1+k_b*Mx^2)/(1+k_b*My_b^2) Ty_b=Tx*(1+Mx^2*(k_b-1)/2)/(1+My_b^2*(k_b-1)/2) Cy_b=sqrt(k_b*R_b*Ty_b*Convert(kJ/kg, m^2/s^2)) Vely_b=My_b*Cy_b

18-93 Air flowing through a nozzle experiences a normal shock. The entropy change of air across the normal shock wave is to be determined. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, onedimensional, and isentropic before the shock occurs. Properties The properties of air are R = 0.287 kJ / kg ×K and c p = 1.005 kJ / kg ×K , and the properties of helium are

R = 2.0769 kJ / kg ×K and c p = 5.1926 kJ / kg ×K . Analysis For air, the entropy change across the shock is determined to be T P s2 - s1 = c p ln 2 - R ln 2 = (1.005 kJ/kg ×K)ln(2.2383) - (0.287 kJ/kg ×K)ln(7.7200) = 0.223 kJ / kg ×K T1 P1

For helium, the entropy change across the shock is determined to be T P s2 - s1 = c p ln 2 - R ln 2 = (5.1926 kJ/kg ×K)ln(2.9606) - (2.0769 kJ/kg ×K)ln(8.2009) = 1.27 kJ / kg ×K T1 P1

Discussion Note that shock wave is a highly dissipative process, and the entropy generation is large during shock waves.

EES (Engineering Equation Solver) SOLUTION "Given" Mx=2.6 Px=58 [kPa] Tx=270 [K] "Properties" c_p_a=CP(air, T=300) c_v_a=CV(air, T=300) k_a=c_p_a/c_v_a R_a=c_p_a-c_v_a c_p_b=CP(Helium, T=300, P=100) c_v_b=CV(Helium, T=300, P=100) k_b=c_p_b/c_v_b R_b=c_p_b-c_v_b "Analysis " "for air" My_a=sqrt((Mx^2+2/(k_a-1))/(2*Mx^2*k_a/(k_a-1)-1)) P0y_a=Px*(1+k_a*Mx^2)/(1+k_a*My_a^2)*(1+My_a^2*(k_a-1)/2)^(k_a/(k_a-1)) Py_a=Px*(1+k_a*Mx^2)/(1+k_a*My_a^2) Ty_a=Tx*(1+Mx^2*(k_a-1)/2)/(1+My_a^2*(k_a-1)/2) Cy_a=sqrt(k_a*R_a*Ty_a*Convert(kJ/kg, m^2/s^2)) Vely_a=My_a*Cy_a "for helium" My_b=sqrt((Mx^2+2/(k_b-1))/(2*Mx^2*k_b/(k_b-1)-1)) P0y_b=Px*(1+k_b*Mx^2)/(1+k_b*My_b^2)*(1+My_b^2*(k_b-1)/2)^(k_b/(k_b-1)) Py_b=Px*(1+k_b*Mx^2)/(1+k_b*My_b^2) Ty_b=Tx*(1+Mx^2*(k_b-1)/2)/(1+My_b^2*(k_b-1)/2) Cy_b=sqrt(k_b*R_b*Ty_b*Convert(kJ/kg, m^2/s^2)) Vely_b=My_b*Cy_b

13-724

DELTAs_a=c_p_a*ln(Ty_a/Tx)-R_a*ln(Py_a/Px) DELTAs_b=c_p_b*ln(Ty_b/Tx)-R_b*ln(Py_b/Px)

18-94 For an ideal gas flowing through a normal shock, a relation for V2 / V1 in terms of k, Ma1 , and Ma 2 is to be developed. Analysis The conservation of mass relation across the shock is r 1V1 = r 2V2 and it can be expressed as

ö V2 r 1 P / RT1 æ P öæ ççT2 ÷ ÷ ÷ = = 1 = ççç 1 ÷ ÷ ÷çç T1 ø÷ ÷ V1 r 2 P2 / RT2 çè P2 øè From Eqs. 12-35 and 12-38,

V2 æ 1 + kMa 22 öæ 1 + Ma12 (k - 1) / 2 ö÷ ÷ ÷çç ÷ = ççç 2 ÷ç ÷ç1 + Ma 22 ( k - 1) / 2 ø÷ ÷ V1 çè1 + kMa1 øè Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known.

Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 18-95C We are to discuss the characteristic aspect of Rayleigh flow, and its main assumptions. Analysis The characteristic aspect of Rayleigh flow is its involvement of heat transfer. The main assumptions associated with Rayleigh flow are: the flow is steady, one-dimensional, and frictionless through a constant-area duct, and the fluid is an ideal gas with constant specific heats. Discussion Of course, there is no such thing as frictionless flow. It is better to say that frictional effects are negligible compared to the heating effects.

18-96C We are to discuss the effect of heating on the flow velocity in subsonic Rayleigh flow. Analysis Heating the fluid increases the flow velocity in subsonic Rayleigh flow, but decreases the flow velocity in supersonic Rayleigh flow. Discussion These results are not necessarily intuitive, but must be true in order to satisfy the conservation laws.

18-97C We are to discuss what the points on a T-s diagram of Rayleigh flow represent. Analysis The points on the Rayleigh line represent the states that satisfy the conservation of mass, momentum, and energy equations as well as the property relations for a given state. Therefore, for a given inlet state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. Discussion The T-s diagram is quite useful, since any downstream state must lie on the Rayleigh line.

18-98C We are to discuss the effect of heat gain and heat loss on entropy during Rayleigh flow. Analysis In Rayleigh flow, the effect of heat gain is to increase the entropy of the fluid, and the effect of heat loss is to decrease the entropy. Discussion You should recall from thermodynamics that the entropy of a system can be lowered by removing heat.

18-99C We are to discuss how temperature and stagnation temperature change in subsonic Rayleigh flow.

13-725

Analysis In Rayleigh flow, the stagnation temperature T0 always increases with heat transfer to the fluid, but the temperature T decreases with heat transfer in the Mach number range of 0.845 < Ma < 1 for air. Therefore, the temperature in this case will decrease. Discussion This at first seems counterintuitive, but if heat were not added, the temperature would drop even more if the air were accelerated isentropically from Ma = 0.92 to 0.95.

18-100C We are to examine the Mach number at the end of a choked duct in Rayleigh flow when more heat is added. Analysis The flow is choked, and thus the flow at the duct exit remains sonic. Discussion There is no mechanism for the flow to become supersonic in this case.

18-101 Argon flowing at subsonic velocity in a constant-diameter duct is accelerated by heating. The highest rate of heat transfer without reducing the mass flow rate is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Mass flow rate remains constant. Properties We take the properties of argon to be k = 1.667, c p = 0.5203 kJ / kg ×K , and R = 0.2081 kJ / kg ×K .

Analysis Heat transfer stops when the flow is choked, and thus Ma 2 = V2 / c2 = 1 . The inlet stagnation temperature is

æ k- 1 ö æ 1.667-1 2 ö÷ T01 = T1 çç1 + Ma12 ÷ = (400 K) çç1 + 0.2 ÷ = 405.3 K ÷ ÷ çè ø èç ø÷ 2 2 The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are

T02 / T0* = 1 (since Ma 2  1 )

T01 (k + 1)Ma12 [2 + (k - 1)Ma12 ] (1.667 + 1)0.22 [2 + (1.667 - 1)0.22 ] = = = 0.1900 T0* (1 + kMa12 ) 2 (1 + 1.667 ´ 0.22 ) 2 Therefore,

T0 2 T02 / T0* 1 = = ® T02 = T01 / 0.1900 = (405.3 K) / 0.1900 = 2133 K * T0 1 T01 / T0 0.1900 Then the rate of heat transfer becomes

Q = mair c p (T02 - T01 ) = (0.85 kg/s)(0.5203 kJ/kg ×K)(2133 - 400) K = 766 kW Discussion It can also be shown that T2  1600 K , which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. Also, in the solution of this problem, we cannot use the values of Table A-34 since they are based on k = 1.4.

13-726

EES (Engineering Equation Solver) SOLUTION "Given" Ma_1=0.2 T_1=400 [K] P_1=320 [kPa] m_dot=1.2 [kg/s] Ma_2=1 "Properties" k=1.667 c_p=0.5203 [kJ/kg-K] R=0.2081 [kJ/kg-K] "Analysis" T_01=T_1*(1+(k-1)/2*Ma_1^2) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01)

18-102 Air is heated in a duct during subsonic flow until it is choked. For specified pressure and velocity at the exit, the temperature, pressure, and velocity at the inlet are to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K . Analysis Noting that sonic conditions exist at the exit, the exit temperature is

c2 = V2 /Ma 2 = (680 m/s)/1 = 680 m/s c2 =

kRT2 ®

æ1000 m2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)T2 çç = 680 m/s ÷ çè 1 kJ/kg ÷ ø

It gives

T2  1151K . Then the exit stagnation temperature becomes T02 = T2 +

æ 1 kJ/kg ÷ ö V22 (680 m/s) 2 çç = 1151 K + = 1381 K ÷ 2 2 ÷ 2c p 2´ 1.005 kJ/kg ×K çè1000 m /s ø

The inlet stagnation temperature is, from the energy equation q = c p (T02 - T01 ),

T01 = T02 -

q 67 kJ/kg = 1381 K = 1314 K cp 1.005 kJ/kg ×K

13-727

The maximum value of stagnation temperature T0* occurs at Ma = 1, and its value in this case is T02 since the flow is choked. Therefore, T0* = T02 = 1381 K . Then the stagnation temperature ratio at the inlet, and the Mach number corresponding to it are, from Table A-34,

T01 1314 K = = 0.9516 T0* 1381 K

Ma1  0.7779  0.778



The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1  0.7779 : Ma 2  1 :

T1 / T * = 1.018 ,

P2 / P* = 1.301 , V1 / V * = 0.7852

T2 / T * = 1 , P2 / P* = 1 , V2 / V * = 1

Then the inlet temperature, pressure, and velocity are determined to be

T2 T2 / T * 1 = = * T1 T1 / T 1.018

 T1 = 1.018T2 = 1.018(1151 K) = 1172 K

P2 P2 / P* 1 = = P1 P1 / P* 1.301



V2 V2 / V * 1 = = * V1 V1 / V 0.7852

 V1 = 0.7852V2 = 0.7852(680 m/s) = 533.9 m / s

P1 = 1.301P2 = 1.301(270 kPa) = 351.3 kPa

Discussion Note that the temperature and pressure decreases with heating during this subsonic Rayleigh flow while velocity increases. This problem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.

EES (Engineering Equation Solver) SOLUTION "Given" q=67 [kJ/kg] V_2=680 [m/s] P_2=270 [kPa] Ma_2=1 "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" c_2=V_2/Ma_2 c_2=sqrt(k*R*T_2*Convert(kJ/kg, m^2/s^2)) T_02=T_2+V_2^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) q=c_p*(T_02-T_01) T_01/T_02=T01\T0star T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 V1\Vstar=((1+k)*Ma_1^2)/(1+k*Ma_1^2) P1\Pstar=(1+k)/(1+k*Ma_1^2) T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 V2\Vstar=((1+k)*Ma_2^2)/(1+k*Ma_2^2) P2\Pstar=(1+k)/(1+k*Ma_2^2) T_2/T_1=T2\Tstar/T1\Tstar V_2/V_1=V2\Vstar/V1\Vstar P_2/P_1=P2\Pstar/P1\Pstar

13-728

18-103 Air enters the combustion chamber of a gas turbine at a subsonic velocity. For a specified rate of heat transfer, the Mach number at the exit and the loss in stagnation pressure to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 The crosssectional area of the combustion chamber is constant. 3 The increase in mass flow rate due to fuel injection is disregarded. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K . Analysis The inlet stagnation temperature and pressure are

æ k- 1 ö æ 1.4 - 1 2 ö÷ T01 = T1 çç1 + Ma12 ÷ = (700 K) çç1 + 0.2 ÷ = 705.6 K ÷ ÷ çè ø èç ø÷ 2 2 k /( k - 1)

æ k- 1 ö P01 = P1 çç1 + Ma12 ÷ ÷ èç ø÷ 2

1.4/0.4

æ 1.4 - 1 2 ö÷ = (600 kPa )çç1 + 0.2 ÷ ÷ èç ø 2 = 617.0 kPa

The exit stagnation temperature is determined from

Q = mair c p (T02 - T01 )  150 kJ/s = (0.3 kg/s)(1.005 kJ/kg ×K)(T02 - 705.6) K It gives

T02  1203 K At Ma1  0.2 we read from T02 /T0* = 0.1736 (Table A-34). Therefore, T0* =

T01 705.6 K = = 4064.5 K 0.1736 0.1736

Then the stagnation temperature ratio at the exit and the Mach number corresponding to it are (Table A-34)

T02 1203 K = = 0.2960 * 4064.5 K T0



Ma 2  0.2706  0.271

Also,

Ma1 = 0.2 ®

P01 / P0* = 1.2346

Ma 2 = 0.2706 ®

P02 / P0* = 1.2091

Then the stagnation pressure at the exit and the pressure drop become

P0 2 P02 / P0* 1.2091 = = = 0.9794  P02 = 0.9794P01 = 0.9794(617 kPa) = 604.3 kPa * P0 1 P01 / P0 1.2346 and

D P0 = P01 - P02 = 617.0 - 604.3 = 12.7 kPa Discussion This problem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.

13-729

EES (Engineering Equation Solver) SOLUTION "Given" T_1=700 [K] P_1=600 [kPa] Ma_1=0.2 m_dot=0.3 [kg/s] Q_dot=150 [kJ/s] "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" T_01=T_1*(1+(k-1)/2*Ma_1^2) P_01=P_1*(1+(k-1)/2*Ma_1^2)^(k/(k-1)) Q_dot=m_dot*c_p*(T_02-T_01) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T_01/T_0_star=T01\T0star T_02/T_0_star=T02\T0star T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 P01\Pstar=(k+1)/(1+k*Ma_1^2)*((2+(k-1)*Ma_1^2)/(k+1))^(k/(k-1)) P02\Pstar=(k+1)/(1+k*Ma_2^2)*((2+(k-1)*Ma_2^2)/(k+1))^(k/(k-1)) P_02/P_01=P02\Pstar/P01\Pstar DELTAP_0=P_01-P_02

18-104 Air enters the combustion chamber of a gas turbine at a subsonic velocity. For a specified rate of heat transfer, the Mach number at the exit and the loss in stagnation pressure to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 The crosssectional area of the combustion chamber is constant. 3 The increase in mass flow rate due to fuel injection is disregarded. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K . Analysis The inlet stagnation temperature and pressure are

æ k- 1 ö æ 1.4 - 1 2 ö÷ T01 = T1 çç1 + Ma12 ÷ = (700 K) çç1 + 0.2 ÷ = 705.6 K ÷ ÷ çè ø èç ø÷ 2 2 k /( k - 1)

æ k- 1 ö P01 = P1 çç1 + Ma12 ÷ ÷ çè ø÷ 2

1.4/0.4

æ 1.4 - 1 2 ö÷ = (600 kPa )çç1 + 0.2 ÷ èç ø÷ 2 = 617.0 kPa

The exit stagnation temperature is determined from

13-730

Q = mair c p (T02 - T01 )  300 kJ/s = (0.3 kg/s)(1.005 kJ/kg ×K)(T02 - 705.6) K It gives

T02  1701K At Ma1  0.2 we read from T01 / T0* = 0.1736 (Table A-34). Therefore, T0* =

T01 705.6 K = = 4064.5 K 0.1736 0.1736

Then the stagnation temperature ratio at the exit and the Mach number corresponding to it are (Table A-34)

T02 1701 K = = 0.4185 T0* 4064.5 K

Ma 2  0.3393  0.339



Also,

Ma1 = 0.2 ®

P01 / P0* = 1.2346

Ma 2 = 0.2706 ®

P02 / P0* = 1.2091

Then the stagnation pressure at the exit and the pressure drop become

P0 2 P02 / P0* 1.1820 = = = 0.9574 P0 1 P01 / P0* 1.2346



P02 = 0.9574P01 = 0.9574(617 kPa) = 590.7 kPa

and

D P0 = P01 - P02 = 617.0 - 590.7 = 26.3 kPa Discussion This problem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.

EES (Engineering Equation Solver) SOLUTION "Given" T_1=700 [K] P_1=600 [kPa] Ma_1=0.2 m_dot=0.3 [kg/s] Q_dot=300 [kJ/s] "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" T_01=T_1*(1+(k-1)/2*Ma_1^2) P_01=P_1*(1+(k-1)/2*Ma_1^2)^(k/(k-1)) Q_dot=m_dot*c_p*(T_02-T_01) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T_01/T_0_star=T01\T0star T_02/T_0_star=T02\T0star T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 P01\Pstar=(k+1)/(1+k*Ma_1^2)*((2+(k-1)*Ma_1^2)/(k+1))^(k/(k-1)) P02\Pstar=(k+1)/(1+k*Ma_2^2)*((2+(k-1)*Ma_2^2)/(k+1))^(k/(k-1)) P_02/P_01=P02\Pstar/P01\Pstar DELTAP_0=P_01-P_02

13-731

18-105E Air flowing with a subsonic velocity in a round duct is accelerated by heating until the flow is choked at the exit. The rate of heat transfer and the pressure drop are to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 The flow is choked at the duct exit. 3 Mass flow rate remains constant. Properties We take the properties of air to be k = 1.4, c p = 0.2400 Btu / lbm ×R , and

R = 0.06855 Btu / lbm ×R = 0.3704 psia ×ft 3 / lbm ×R .

Analysis The inlet density and velocity of air are r1 =

P1 30 psia = = 0.1012 lbm/ft 3 RT1 (0.3704 psia ×ft 3 /lbm ×R)(800 R)

V1 =

mair 9 lbm/s = = 452.9 ft/s r 1 Ac1 (0.1012 lbm/ft 3 )[p (6/12 ft) 2 / 4]

The stagnation temperature and Mach number at the inlet are T01 = T1 +

æ 1 Btu/lbm ÷ ö V12 (452.9 ft/s)2 çç = 800 R + ÷= 817.1 R 2 2÷ ç 2c p 2´ 0.2400 Btu/lbm ×R è 25,037 ft /s ø æ25,037 ft 2 / s 2 ö ÷ ÷ (1.4)(0.06855 Btu/lbm ×R)(800 R) ççç = 1386 ft/s ÷ ÷ è 1 Btu/lbm ø

c1 =

kRT1 =

Ma1 =

V1 452.9 ft/s = = 0.3268 c1 1386 ft/s

The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1  0.3268 :

T1 / T * = 0.4642 ,

P1 / P* = 2.0857 ,

T01 / T0* = 0.3957

Ma 2  1 :

T2 / T * = 1 ,

P2 / P* = 1 ,

T02 / T0* = 1

Then the exit temperature, pressure, and stagnation temperature are determined to be

T2 T2 / T * 1 = = * T1 T1 / T 0.4642



T2 = T1 / 0.4642 = (800 R) / 0.4642 = 1723 R

P2 P2 / P* 1 = = * P1 P1 / P 2.0857



P2 = P1 / 2.0857 = (30 psia) / 2.0857 = 14.4 psia

T0 2 T02 / T * 1 = = T0 1 T01 / T * 0.3957

 T02 = T01 / 0.3957 = (817.1 R) / 0.3957 = 2065 R

Then the rate of heat transfer and the pressure drop become

Q = mair c p (T02 - T01 ) = (9 lbm/s)(0.2400 Btu/lbm ×R)(2065 - 817.1) R = 2695 Btu / s

D P = P1 - P2 = 30 - 14.4 = 15.6 psia Discussion Note that the entropy of air increases during this heating process, as expected.

13-732

EES (Engineering Equation Solver) SOLUTION "Given" D=(4/12) [ft] m_dot=5 [lbm/s] T_1=800 [R] P_1=30 [psia] Ma_2=1 "Properties" k=1.4 c_p=0.240 [Btu/lbm-R] R=0.06855 [Btu/lbm-R] "Analysis" rho_1=P_1/(R*Convert(Btu/lbm-R, psia-ft^3/lbm-R)*T_1) A_c=pi*D^2/4 V_1=m_dot/(rho_1*A_c) T_01=T_1+V_1^2/(2*c_p)*Convert(ft^2/s^2, Btu/lbm) c_1=sqrt(k*R*T_1*Convert(Btu/lbm, ft^2/s^2)) Ma_1=V_1/c_1 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 P1\Pstar=(1+k)/(1+k*Ma_1^2) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 P2\Pstar=(1+k)/(1+k*Ma_2^2) T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_2/T_1=T2\Tstar/T1\Tstar P_2/P_1=P2\Pstar/P1\Pstar T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01) DELTAP=P_1-P_2

18-106 Air flowing with a subsonic velocity in a duct. The variation of entropy with temperature is to be investigated as the exit temperature varies from 600 K to 5000 K in increments of 200 K. The results are to be tabulated and plotted. Analysis We solve the probem using EES. P1=350 [kPa] T1=600 [K] V1=70 [m/s] k=1.4 cp=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] C1=sqrt(k*R*T1*1000) Ma1=V1/C1 T01=T1*(1+0.5*(k-1)*Ma1^2) P01=P1*(1+0.5*(k-1)*Ma1^2)^(k/(k-1)) F1=1+0.5*(k-1)*Ma1^2 T01Ts=2*(k+1)*Ma1^2*F1/(1+k*Ma1^2)^2 P01Ps=((1+k)/(1+k*Ma1^2))*(2*F1/(k+1))^(k/(k-1)) T1Ts=(Ma1*((1+k)/(1+k*Ma1^2)))^2 P1Ps=(1+k)/(1+k*Ma1^2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-733

V1Vs=Ma1^2*(1+k)/(1+k*Ma1^2) F2=1+0.5*(k-1)*Ma2^2 T02Ts=2*(k+1)*Ma2^2*F2/(1+k*Ma2^2)^2 P02Ps=((1+k)/(1+k*Ma2^2))*(2*F2/(k+1))^(k/(k-1)) T2Ts=(Ma2*((1+k)/(1+k*Ma2^2)))^2 P2Ps=(1+k)/(1+k*Ma2^2) V2Vs=Ma2^2*(1+k)/(1+k*Ma2^2) T02=T02Ts/T01Ts*T01 P02=P02Ps/P01Ps*P01 T2=T2Ts/T1Ts*T1 P2=P2Ps/P1Ps*P1 V2=V2Vs/V1Vs*V1 Delta_s=cp*ln(T2/T1)-R*ln(P2/P1) Exit temperature T2 , K 600 800 1000 1200 1400 1600 1800 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800 4000 4200 4400 4600 4800 5000

Exit Mach number, Ma 2 0.143 0.166 0.188 0.208 0.227 0.245 0.263 0.281 0.299 0.316 0.333 0.351 0.369 0.387 0.406 0.426 0.446 0.467 0.490 0.515 0.541 0.571 0.606

Exit entropy relative to inlet, s2 , kJ / kg ×K 0.000 0.292 0.519 0.705 0.863 1.001 1.123 1.232 1.331 1.423 1.507 1.586 1.660 1.729 1.795 1.858 1.918 1.975 2.031 2.085 2.138 2.190 2.242

13-734

Discussion Note that the entropy of air increases during this heating process, as expected.

18-107 Fuel is burned in a rectangular duct with compressed air. For specified heat transfer, the exit temperature and Mach number are to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid.

Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K . Analysis The stagnation temperature and Mach number at the inlet are c1 =

kRT1 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(300 K) ççç = 347.2 m/s ÷ ÷ è 1 kJ/kg ø

V1 = Ma1c1 = 2(347.2 m/s) = 694.4 m/s T01 = T1 +

ö V12 (694.4 m/s) 2 æ çç 1 kJ/kg ÷ = 300 K + = 539.9 K 2 2÷ ÷ ç 2c p 2´ 1.005 kJ/kg ×K è1000 m /s ø

The exit stagnation temperature is, from the energy equation q = c p (T02 - T01 ),

T02 = T01 +

q 55 kJ/kg = 539.9 K + = 594.6 K cp 1.005 kJ/kg ×K

The maximum value of stagnation temperature T0* occurs at Ma = 1, and its value can be determined from Table A-34 or from the appropriate relation. At Ma1  2 we read T01 / T0* = 0.7934 . Therefore,

13-735

T0* =

T01 539.9 K = = 680.5 K 0.7934 0.7934

The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-34,

T02 594.6 K = = 0.8738  Ma 2  1.642  1.64 T0* 680.5 K Also,

Ma1 = 2 ® T1 / T * = 0.5289 Ma 2 = 1.642 ® T2 / T * = 0.6812 Then the exit temperature becomes

T2 T2 /T * 0.6812 = = = 1.288 * T1 T1 /T 0.5289



T2 = 1.288T1 = 1.288(300 K) = 386 K

Discussion Note that the temperature increases during this supersonic Rayleigh flow with heating. This problem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.

EES (Engineering Equation Solver) SOLUTION "Given" T_1=300 [K] P_1=420 [kPa] Ma_1=2 q=55 [kJ/kg] "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) V_1=Ma_1*c_1 T_01=T_1+V_1^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) q=c_p*(T_02-T_01) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T_01/T_0_star=T01\T0star T_02/T_0_star=T02\T0star T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 T_2/T_1=T2\Tstar/T1\Tstar

18-108 Compressed air is cooled as it flows in a rectangular duct. For specified heat rejection, the exit temperature and Mach number are to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K .

13-736

Analysis The stagnation temperature and Mach number at the inlet are c1 =

kRT1 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(300 K) ççç = 347.2 m/s ÷ ÷ è 1 kJ/kg ø

V1 = Ma1c1 = 2(347.2 m/s) = 694.4 m/s T01 = T1 +

ö V12 (694.4 m/s) 2 æ çç 1 kJ/kg ÷ = 300 K + = 539.9 K 2 2÷ ÷ ç 2c p 2´ 1.005 kJ/kg ×K è1000 m /s ø

The exit stagnation temperature is, from the energy equation q = c p (T02 - T01 ),

T02 = T01 +

q - 55 kJ/kg = 539.9 K + = 485.2 K cp 1.005 kJ/kg ×K

T01 539.9 K = = 680.5 K 0.7934 0.7934

The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-34,

T02 485.2 K = = 0.7130  Ma 2  2.479  2.48 T0* 680.5 K Also,

Ma1 = 2 ® T1 / T * = 0.5289

Ma 2 = 1.642 ® T2 / T * = 0.6812 Then the exit temperature becomes

T2 T2 /T * 0.3838 = = = 0.7257  T2 = 0.7257T1 = 0.7257(300 K) = 218 K T1 T1 /T * 0.5289 Discussion Note that the temperature decreases and Mach number increases during this supersonic Rayleigh flow with cooling. This problem can also be solved using appropriate relations instead of tabulated values, which can likewise be coded for convenient computer solutions.

EES (Engineering Equation Solver) SOLUTION "Given" T_1=300 [K] P_1=420 [kPa] Ma_1=2 q=-55 [kJ/kg] "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-737

"Analysis" c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) V_1=Ma_1*c_1 T_01=T_1+V_1^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) q=c_p*(T_02-T_01) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T_01/T_0_star=T01\T0star T_02/T_0_star=T02\T0star T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 T_2/T_1=T2\Tstar/T1\Tstar

18-109 Air flowing at a supersonic velocity in a duct is decelerated by heating. The highest temperature air can be heated by heat addition and the rate of heat transfer are to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Mass flow rate remains constant. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K . Analysis Heat transfer will stop when the flow is choked, and thus Ma 2 = V2 / c2 = 1 . Knowing stagnation properties, the static properties are determined to be - 1

- 1

æ k- 1 ö÷ æ 1.4 - 1 2 ö÷ T1 = T01 çç1 + Ma12 ÷ = (600 K) çç1 + 1.8 ÷ = 364.1 K ÷ èç ø èç ø÷ 2 2 - k /( k - 1)

æ k- 1 ö P1 = P01 çç1 + Ma12 ÷ ÷ èç ø÷ 2

- 1.4/0.4

æ 1.4 - 1 2 ö÷ = (140 K) çç1 + 1.8 ÷ èç ø÷ 2

= 24.37 kPa

r1 =

P1 24.37 kPa = = 0.2332 kg/m 3 RT1 (0.287 kJ/kgK)(364.1 K)

Then the inlet velocity and the mass flow rate become c1 =

kRT1 =

æ1000 m 2 / s 2 ö ÷ ÷ (1.4)(0.287 kJ/kg ×K)(364.1 K) çç = 382.5 m/s ÷ ÷ çè 1 kJ/kg ø

V1 = Ma1c1 = 1.8(382.5 m/s) = 688.5 m/s mair = r 1 Ac1V1 = (0.2332 kg/m3 )[p (0.07 m) 2 / 4](688.5 m/s) = 0.6179 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34):

13-738

Ma1  1.8 : T1 / T * = 0.6089 , T01 / T0* = 0.8363

Ma 2  1 :

T2 / T * = 1 , T02 / T0* = 1

Then the exit temperature and stagnation temperature are determined to be

T2 T2 /T * 1 = = T1 T1 /T * 0.6089



T2 = T1 / 0.6089 = (364.1 K) / 0.6089 = 598 K

T0 2 T02 /T0* 1 = = * T0 1 T01 /T0 0.8363



T02 = T01 / 0.8363 = (600 K ) / 0.8363 = 717.4 K @717 K

Finally, the rate of heat transfer is

Q = mair c p (T02 - T01 ) = (0.6179 kg/s)(1.005 kJ/kg ×K)(717.4 - 600) K = 72.9 kW Discussion Note that this is the highest temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature will cause the mass flow rate to decrease. Also, once the sonic conditions are reached, the thermodynamic temperature can be increased further by cooling the fluid and reducing the velocity (see the T-s diagram for Rayleigh flow).

EES (Engineering Equation Solver) SOLUTION "Given" D=0.07 [m] Ma_1=1.8 P_01=140 [kPa] T_01=600 [K] Ma_2=1 "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" T_1=T_01*(1+(k-1)/2*Ma_1^2)^(-1) P_1=P_01*(1+(k-1)/2*Ma_1^2)^(-k/(k-1)) rho_1=P_1/(R*T_1) c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) V_1=Ma_1*c_1 A_c=pi*D^2/4 m_dot=rho_1*A_c*V_1 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_2/T_1=T2\Tstar/T1\Tstar T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01)

18-110E Air flowing with a subsonic velocity in a square duct is accelerated by heating until the flow is choked at the exit. The rate of heat transfer and the entropy change are to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 The flow is choked at the duct exit. 3 Mass flow rate remains constant. Properties We take the properties of air to be k = 1.4, c p = 0.2400 Btu / lbm ×R , and

13-739

Analysis The inlet density and mass flow rate of air are r1 =

P1 80 psia = = 0.3085 lbm/ft 3 RT1 (0.3704 psia ×ft 3 /lbm ×R)(700 R)

mair = r 1 Ac1V1 = (0.3085 lbm/ft 3 )(6´ 6/144 ft 2 )(260 ft/s) = 20.06 lbm/s The stagnation temperature and Mach number at the inlet are T01 = T1 +

æ 1 Btu/lbm ÷ ö V12 (260 ft/s) 2 çç = 700 R + ÷= 705.6 R 2 2÷ ç 2c p 2´ 0.2400 Btu/lbm ×R è 25,037 ft /s ø æ25,037 ft 2 / s 2 ö ÷ ÷ (1.4)(0.06855 Btu/lbm ×R)(700 R) ççç = 1297 ft/s ÷ ÷ è 1 Btu/lbm ø

c1 =

kRT1 =

Ma1 =

V1 260 ft/s = = 0.2005 c1 1297 ft/s

The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1  0.2005 :

T1 / T * = 0.2075 , P1 / P* = 2.272 , T01 / T0* = 0.1743

Ma 2  1 :

T2 / T * = 1 ,

P2 / P* = 1 ,

T02 / T0* = 1

Then the exit temperature, pressure, and stagnation temperature are determined to be

T2 T2 /T * 1 = = T1 T1 /T * 0.2075

 T2 = T1 / 0.2075 = (700 R) / 0.2075 = 3374 R

P2 P2 /P* 1 = = * P1 P1 /P 2.272



P2 = P1 / 2.272 = (80 psia) / 2.272 = 35.2 psia

T0 2 T02 /T * 1 = =  T02 = T01 / 0.1743 = (705.6 R) / 0.1743 = 4048 R * T0 1 T01 /T 0.1743 Then the rate of heat transfer and entropy change become

Q = mair c p (T02 - T01 ) = (20.06 lbm/s)(0.2400 Btu/lbm ×R)(4048 - 705.6) R = 16,090 Btu / s T P 3374 R 35.2 psia D s = c p ln 2 - R ln 2 = (0.2400 Btu/lbm ×R) ln - (0.06855 Btu/lbm ×R) ln = 0.434 Btu / lbm ×R T1 P1 700 R 80 psia

Discussion Note that the entropy of air increases during this heating process, as expected.

EES (Engineering Equation Solver) SOLUTION "Given" L=(6/12) [ft] V_1=260 [ft/s] T_1=700 [R] P_1=80 [psia]

13-740

Ma_2=1 "Properties" k=1.4 c_p=0.240 [Btu/lbm-R] R=0.06855 [Btu/lbm-R] "Analysis" rho_1=P_1/(R*Convert(Btu/lbm-R, psia-ft^3/lbm-R)*T_1) A_c=L^2 m_dot=rho_1*A_c*V_1 T_01=T_1+V_1^2/(2*c_p)*Convert(ft^2/s^2, Btu/lbm) c_1=sqrt(k*R*T_1*Convert(Btu/lbm, ft^2/s^2)) Ma_1=V_1/c_1 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 P1\Pstar=(1+k)/(1+k*Ma_1^2) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 P2\Pstar=(1+k)/(1+k*Ma_2^2) T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_2/T_1=T2\Tstar/T1\Tstar P_2/P_1=P2\Pstar/P1\Pstar T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01) DELTAs=c_p*ln(T_2/T_1)-R*ln(P_2/P_1)

Steam Nozzles 18-111C The delay in the condensation of the steam is called supersaturation. It occurs in high-speed flows where there isn’t sufficient time for the necessary heat transfer and the formation of liquid droplets.

18-112 Steam enters a converging nozzle with a low velocity. The exit velocity, mass flow rate, and exit Mach number are to be determined for isentropic and 94 percent efficient nozzle cases. Assumptions 1 Flow through the nozzle is steady and one-dimensional. 2 The nozzle is adiabatic. Analysis (a) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h01  h1 . At the inlet,

P1 = P01 = 5 MPa üïï h1 = h01 = 3196.7 kJ/kg ý T1 = T01 = 400°C ïïþ s1 = s2 s = 6.6483 kJ/kg ×K At the exit,

üïï h2 = 3057.0 kJ/kg ý s2 = 6.6483 kJ/kg ×K ïïþ v 2 = 0.08601 m3 /kg

P2 = 3 MPa

13-741

Then the exit velocity is determined from the steady-flow energy balance Ein = Eout with q = w = 0, h1 + V12 /2 = h2 + V22 /2

¾¾ ®

0 = h2 - h1 +

V22 - V12Z 0 2

Solving for V2 , V2 =

2(h1 - h2 ) =

æ1000 m 2 /s 2 ÷ ö ÷ 2(3196.7 - 3057.0) kJ/kg ççç = 528.5 m / s ÷ è 1 kJ/kg ÷ ø

The mass flow rate is determined from m=

1 1 A2V2 = (75´ 10- 4 m 2 )(528.5 m/s) = 46.08 kg / s v2 0.08601 m3 /kg

The velocity of sound at the exit of the nozzle is determined from 1/ 2

1/ 2

æ¶ P ö æ D P ö÷ ç ÷ c = çç ÷ ÷ @ççD (1/ v ) ÷ ÷ çè ¶r ø÷ è ø÷s s

The specific volume of steam at s2 = 6.6483 kJ / kg ×K and at pressures just below and just above the specified pressure (2.5 and 3.5 MPa) are determined to be 0.09906 and 0.07632 m3 / kg . Substituting,

æ1000 m 2 / s 2 ö (3500 - 2500) kPa ÷ çç ÷ = 576.6 m/s ÷ çè 1 kPa.m3 ø ÷ æ 1 ö 1 3 ÷ çç ÷ ÷ kg/m çè0.07632 0.09906 ø

c2 =

Then the exit Mach number becomes Ma 2 =

V2 528.5 m/s = = 0.917 c2 576.6 m/s

(b) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h01  h1 . At the inlet,

P1 = P01 = 5 MPa ïüï h1 = h01 = 3196.7 kJ/kg ý T1 = T01 = 400°C ïïþ s1 = s2 s = 6.6483 kJ/kg ×K At state 2s,

ïüï ý h2 s = 3057.0 kJ/kg s2 s = 6.6483 kJ/kg ×K ïïþ

P2 s = 3 MPa

The enthalpy of steam at the actual exit state is determined from hN =

h01 - h2 3196.7 - h2 ¾¾ ® 0.94 = ¾¾ ® h2 = 3065.4 kJ/kg h01 - h2 s 3196.7 - 3057.0

13-742

Therefore,

üïï v 2 = 0.08666 m3 /kg ý h2 = 3065.4 kJ/kg ïïþ s2 = 6.6622 kJ/kg ×K

P2 = 3 MPa

Then the exit velocity is determined from the steady-flow energy balance Ein = Eout with q = w = 0, h1 + V12 /2 = h2 + V22 /2 ¾ ¾ ® 0 = h2 - h1 +

V22 - V12Z 0 2

Solving for V2 , V2 =

2(h1 - h2 ) =

æ1000 m 2 /s 2 ÷ ö ÷ 2(3196.7 - 3065.4) kJ/kg çç ÷ = 512.4 m / s çè 1 kJ/kg ÷ ø

The mass flow rate is determined from m=

1 1 A2V2 = (75´ 10- 4 m 2 )(512.4 m/s) = 44.34 kg / s v2 0.08666 m3 /kg

The velocity of sound at the exit of the nozzle is determined from 1/ 2

1/ 2

æ¶ P ö æ D P ö÷ ÷ ÷ c = çç ÷ @çç ÷ ÷ çè ¶r ø÷ èçD (1 / v ) ø÷s s

The specific volume of steam at s2 = 6.6622 kJ / kg ×K and at pressures just below and just above the specified pressure (2.5 and 3.5 MPa) are determined to be 0.09981 and 0.07689 m3 / kg . Substituting,

c2 =

æ1000 m2 /s 2 ö (3500 - 2500) kPa ÷ çç ÷ ÷ = 578.7 m/s 3 ÷ æ 1 ö ç 1 1 kPa.m è ø 3 ÷ kg/m çç ÷ ÷ çè0.07689 0.09981ø

Then the exit Mach number becomes Ma 2 =

V2 512.4 m/s = = 0.885 c2 578.7 m/s

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=5000 [kPa] T[1]=400 [C] Vel1=0 "negligible inlet velocity" P[2]=3000 [kPa] A2=75E-4 [m^2] Eta_N=0.94 "Analysis" Fluid$='steam_iapws' "(a)" h[1]=enthalpy(Fluid$, T=T[1], P=P[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) v[2]=volume(Fluid$, P=P[2], s=s[2]) Vel2a=sqrt(2*(h[1]-h[2])*Convert(kJ/kg, m^2/s^2)) "energy balance" m_dot_a=1/v[2]*A2*Vel2a P2a=P[2]-500 "selected" P2b=P[2]+500 "selected" v1=volume(Fluid$, P=P2a, s=s[2]) v2=volume(Fluid$, P=P2b, s=s[2]) C2a=sqrt((P2b-P2a)/(1/v2-1/v1)*Convert(kJ/kg, m^2/s^2)) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-743

M2a=Vel2a/C2a "(b)" Eta_N=(h[1]-h_b[2])/(h[1]-h[2]) v_b[2]=volume(Fluid$, P=P[2], h=h_b[2]) s_b[2]=entropy(Fluid$, P=P[2], h=h_b[2]) Vel2b=sqrt(2*(h[1]-h_b[2])*Convert(kJ/kg, m^2/s^2)) m_dot_b=1/v_b[2]*A2*Vel2b v1b=volume(Fluid$, P=P2a, s=s_b[2]) v2b=volume(Fluid$, P=P2b, s=s_b[2]) C2b=sqrt((P2b-P2a)/(1/v2b-1/v1b)*Convert(kJ/kg, m^2/s^2)) M2b=Vel2b/C2b

18-113E Steam enters a converging nozzle with a low velocity. The exit velocity, mass flow rate, and exit Mach number are to be determined for isentropic and 90 percent efficient nozzle cases. Assumptions 1 Flow through the nozzle is steady and one-dimensional. 2 The nozzle is adiabatic. Analysis (a) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h01  h1 .

At the inlet,

P1  P01  450 psia  h1  h01  1468 .6 Btu/1bm   s1  s 2s  1.7117 Btu/1bm  R

T1  T01  900 F At the exit,

P2  275 psia

 h2  1400 .5 Btu/1bm  s2 s  1.7117 Btu/1bm  R  v 2  2.5732 ft 3/1bm Then the exit velocity is determined from the steady-flow energy balance Ein  Eout with q = w = 0, 0

h1  V12 / 2  h2  V22 / 2   0  h2  h1 

V22  V12 2

Solving for V2 ,  25,037 ft 2 /s 2    1847 ft/s V2  2(h1  h2 )  2(1468 .6  1400 .5) Btu/1bm   1 Btu/1bm   

Then, m 

A2V2 

1 2.5732 ft 3 / 1bm

(3.75 / 144 ft 2 )(1847 ft/s)  18.7 1bm/s

13-744

1/ 2

 P  c       s

1/ 2

 P      (1 / v )  s

The specific volume of steam at s2 = 1.7117 Btu / lbm ×R and at pressures just below and just above the specified pressure (250 and 300 psia) are determined to be 2.7709 and 2.4048 ft 3 / lbm . Substituting,  25,037 ft 2 / s 2   (300  250 ) psia 1 Btu     2053 ft/s 3     1 1   3  1 Btu/1bm  5.4039 ft  psia     1bm/ft  2.4048 2.7709 

c2 

Then the exit Mach number becomes Ma 2 

V2 1847 ft/s   0.900 c2 2053 ft/s

(b) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h01  h1 . At the inlet,

P1  P01  450 psia  h1  h01  1468 .6 Btu/1bm  T1  T01  900 F  s1  s 2s  1.7117 Btu/1bm  R At state 2s,

P2 s  275 psia

  h2 s  1400 .5 Btu/lbm s 2 s  1.7117 Btu/lbm  R 

The enthalpy of steam at the actual exit state is determined from

N 

h01  h2 1468 .6  h2   0.90    h2  1407 .3 Btu/1bm h01  h2 s 1468 .6  1400 .5

Therefore,

P2  275 psia

 v 2  2.6034 ft 3/1bm  h2  1407 .3 Btu/lbm  s 2  1.7173 Btu/1bm  R

Then the exit velocity is determined from the steady-flow energy balance E in  E out with q = w = 0, 0

h1  V12 / 2  h2  V22 / 2   0  h2  h1 

V22  V12 2

Solving for V2 ,  25,037 ft 2 /s 2    1752 ft/s V2  2(h1  h2 )  2(1468 .6  1407 .3) Btu/1bm   1 Btu/1bm   

Then, m 

A2V2 

1 2.6034 ft 3 / 1bm

(3.75 / 144 ft 2 )(1752 ft/s)  17.53 1bm/s

The velocity of sound at the exit of the nozzle is determined from 1/ 2

 P  c       s

1/ 2

 P      (1 / v )  s

The specific volume of steam at s2 = 1.7173 Btu / lbm ×R and at pressures just below and just above the specified pressure (250 and 300 psia) are determined to be 2.8036 and 2.4329 ft 3 / lbm . Substituting,

13-745

c2 

 25,037 ft 2 / s 2   (300  250 ) psia 1 Btu     2065 ft/s 3    1   1 5.4039 ft  psia  3  1 Btu/lbm      lbm/ft  2.4329 2.8036 

Then the exit Mach number becomes Ma 2 

V2 1752 ft/s   0.849 c2 2065 ft/s

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=450 [psia] T[1]=900 [F] Vel1=0 "negligible inlet velocity" P[2]=275 [psia] A2=3.75[in^2]*Convert(in^2, ft^2) Eta_N=0.90 "Analysis" Fluid$='steam_iapws' "(a)" h[1]=enthalpy(Fluid$, T=T[1], P=P[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) v[2]=volume(Fluid$, P=P[2], s=s[2]) Vel2a=sqrt(2*(h[1]-h[2])*Convert(Btu/lbm, ft^2/s^2)) "energy balance" m_dot_a=1/v[2]*A2*Vel2a P1=250 P2=300 v1=volume(Fluid$, P=P1, s=s[2]) v2=volume(Fluid$, P=P2, s=s[2]) C2a=sqrt((P2-P1)/(1/v2-1/v1)*Convert(psia-ft^3/lbm, ft^2/s^2)) M2a=Vel2a/C2a "(b)" Eta_N=(h[1]-h_b[2])/(h[1]-h[2]) v_b[2]=volume(Fluid$, P=P[2], h=h_b[2]) s_b[2]=entropy(Fluid$, P=P[2], h=h_b[2]) Vel2b=sqrt(2*(h[1]-h_b[2])*Convert(Btu/lbm, ft^2/s^2)) "energy balance" m_dot_b=1/v_b[2]*A2*Vel2b v1b=volume(Fluid$, P=P1, s=s_b[2]) v2b=volume(Fluid$, P=P2, s=s_b[2]) C2b=sqrt((P2-P1)/(1/v2b-1/v1b)*Convert(psia-ft^3/lbm, ft^2/s^2)) M2b=Vel2b/C2b

18-114 Steam enters a converging-diverging nozzle with a low velocity. The exit area and the exit Mach number are to be determined. Assumptions Flow through the nozzle is steady, one-dimensional, and isentropic. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h01  h1 .

13-746

At the inlet,

P1 = P01 = 1 MPa üïï h1 = h01 = 3479.1 kJ/kg ý T1 = T01 = 500°C ïïþ s1 = s2s = 7.7642 kJ/kg ×K At the exit,

ïüï h2 = 3000.0 kJ/kg ý s2 = 7.7642 kJ/kg ×K ïïþ v 2 = 1.2325 m3 /kg

P2 = 0.2 MPa

Then the exit velocity is determined from the steady-flow energy balance Ein = Eout with q = w = 0, h1 + V12 /2 = h2 + V22 /2 ¾ ¾ ® 0 = h2 - h1 +

V22 - V12Z 0 2

Solving for V2 , V2 =

2(h1 - h2 ) =

æ1000 m 2 /s 2 ö ÷ ÷ 2(3479.1- 3000.0) kJ/kg çç = 978.9 m/s ÷ ÷ çè 1 kJ/kg ø

The exit area is determined from A2 =

mv 2 (2.5 kg/s)(1.2325 m3 /kg) = = 31.5´ 10- 4 m 2 = 31.5 cm 2 V2 (978.9 m/s)

The velocity of sound at the exit of the nozzle is determined from 1/ 2

1/ 2

æ¶ P ö æ D P ö÷ ÷ @çç ÷ c = çç ÷ ÷ ÷ çè ¶r ø÷ èçD (1 / v ) ø÷s s

The specific volume of steam at s2 = 7.7642 kJ / kg ×K and at pressures just below and just above the specified pressure (0.1 and 0.3 MPa) are determined to be 2.0935 and 0.9024 m3 / kg . Substituting,

c2 =

æ1000 m 2 /s 2 ö÷ (300 - 100) kPa çç ÷ = 563.2 m/s 3 ÷ ç ÷ æ 1 ö÷ 1 3 è 1 kPa ×m ø çç ÷ kg/m çè0.9024 2.0935 ø÷

Then the exit Mach number becomes Ma 2 =

V2 978.9 m/s = = 1.738 c2 563.2 m/s

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=1000 [kPa] T[1]=500 [C] Vel1=0 "negligible inlet velocity" m_dot=2.5 [kg/s] P[2]=200 [kPa] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-747

Fluid$='steam_iapws' h[1]=enthalpy(Fluid$, T=T[1], P=P[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) v[2]=volume(Fluid$, P=P[2], s=s[2]) Vel2=sqrt(2*(h[1]-h[2])*Convert(kJ/kg, m^2/s^2)) "energy balance" A2=(m_dot*v[2])/Vel2 P1=100 P2=300 v1=volume(Fluid$, P=P1, s=s[2]) v2=volume(Fluid$, P=P2, s=s[2]) C2=sqrt((P2-P1)/(1/v2-1/v1)*Convert(kJ/kg, m^2/s^2)) M2=Vel2/C2

18-115 Steam enters a converging-diverging nozzle with a low velocity. The exit area and the exit Mach number are to be determined. Assumptions Flow through the nozzle is steady and one-dimensional. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h01  h1 .

At the inlet,

P1 = P01 = 1 MPa ïüï h1 = h01 = 3479.1 kJ/kg ý T1 = T01 = 500°C ïïþ s1 = s2s = 7.7642 kJ/kg ×K

üïï ý h2 s = 3000.0 kJ/kg s2 = 7.7642 kJ/kg ×K ïïþ The enthalpy of steam at the actual exit state is determined from At state 2s,

hN =

P2 = 0.2 MPa

h01 - h2 3479.1- h2 ¾¾ ® 0.90 = ¾¾ ® h2 = 3047.9 kJ/kg h01 - h2 s 3479.1- 3000.0

Therefore,

üïï v 2 = 1.2882 m3 /kg ý h2 = 3047.9 kJ/kg ïïþ s2 = 7.7642 kJ/kg ×K

P2 = 0.2 MPa

Then the exit velocity is determined from the steady-flow energy balance Ein = Eout with q = w = 0, h1 + V12 /2 = h2 + V22 /2 ¾ ¾ ® 0 = h2 - h1 +

V22 - V12Z 0 2

Solving for V2 , V2 =

2(h1 - h2 ) =

æ1000 m 2 /s 2 ö ÷ ÷ 2(3479.1- 3047.9) kJ/kg ççç = 928.7 m/s ÷ ÷ è 1 kJ/kg ø

13-748

A2 =

mv 2 (2.5 kg/s)(1.2882 m3 /kg) = = 34.7´ 10- 4 m2 = 34.7 cm 2 V2 928.7 m/s

The velocity of sound at the exit of the nozzle is determined from 1/ 2

1/ 2

æ¶ P ö÷ æ D P ö÷ ÷ c = çç ÷ @ççç ÷ ÷ çè ¶r ø÷ èD (1 / v ) ø÷s s

The specific volume of steam at s2 = 7.7642 kJ / kg ×K and at pressures just below and just above the specified pressure (0.1 and 0.3 MPa) are determined to be 2.1903 and 0.9425 m3 / kg . Substituting,

c2 =

æ1000 m 2 /s 2 ÷ ö (300 - 100) kPa çç ÷ = 575.2 m/s 3 ÷ ç æ 1 ö 1 ÷ è 1 kPa ×m ÷ ø çç kg/m3 ÷ çè0.9425 2.1903 ÷ ø

Then the exit Mach number becomes Ma 2 =

V2 928.7 m/s = = 1.61 c2 575.2 m/s

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=1000 [kPa] T[1]=500 [C] Vel1=0 "negligible inlet velocity" m_dot=2.5 [kg/s] P[2]=200 [kPa] eta_N=0.85 "Analysis" Fluid$='steam_iapws' h[1]=enthalpy(Fluid$, T=T[1], P=P[1]) s[1]=entropy(Fluid$, T=T[1], P=P[1]) s_s[2]=s[1] h_s[2]=enthalpy(Fluid$, P=P[2], s=s_s[2]) Eta_N=(h[1]-h[2])/(h[1]-h_s[2]) v[2]=volume(Fluid$, P=P[2], h=h[2]) s[2]=entropy(Fluid$, P=P[2], h=h[2]) Vel2=sqrt(2*(h[1]-h[2])*Convert(kJ/kg, m^2/s^2)) "energy balance" A2=(m_dot*v[2])/Vel2 P1=100 [kPa] P2=300 [kPa] v1=volume(Fluid$, P=P1, s=s[2]) v2=volume(Fluid$, P=P2, s=s[2]) C2=sqrt((P2-P1)/(1/v2-1/v1)*Convert(kJ/kg, m^2/s^2)) M2=Vel2/C2

Review Problems 18-116 A Pitot-static probe measures the difference between the static and stagnation pressures for a subsonic airplane. The speed of the airplane and the flight Mach number are to be determined. Assumptions 1 Air is an ideal gas with a constant specific heat ratio. 2 The stagnation process is isentropic. Properties The properties of air are R = 0.287 kJ / kg ×K and k = 1.4. Analysis The stagnation pressure of air at the specified conditions is

P0 = P + D P = 54 + 16 = 70 kPa

13-749

Then, k / k- 1

P0 æ (k - 1)Ma 2 ö ÷ ÷ = çç1 + ÷ ÷ P çè 2 ø

1.4/ 0.4

70 æ (1.4 - 1)Ma 2 ö ÷ ÷ ¾¾ ® = çç1 + ÷ ÷ 54 èç 2 ø

It yields Ma = 0.620 The speed of sound in air at the specified conditions is c=

kRT =

æ1000 m 2 / s 2 ö ÷ ÷ (1.4)(0.287 kJ/kg ×K)(256 K) çç ÷ = 320.7 m/s ÷ çè 1 kJ/kg ø

Thus,

V = Ma ´ c = (0.620)(320.7 m/s) = 199 m / s Discussion Note that the flow velocity can be measured in a simple and accurate way by simply measuring pressure.

EES (Engineering Equation Solver) SOLUTION "Given" H=5000 [m] P=54 [kPa] T=256 [K] DELTAP=16 [kPa] "Properties" k=1.4 R=0.287 [kJ/kg-K] "Analysis" P0=P+DELTAP P0/P=(1+((k-1)*M^2)/2)^(k/(k-1)) C=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) Vel=M*C

18-117 The thrust developed by the engine of a Boeing 777 is about 380 kN. The mass flow rate of gases through the nozzle is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of combustion gases through the nozzle is isentropic. 3 Choked flow conditions exist at the nozzle exit. 4 The velocity of gases at the nozzle inlet is negligible. Properties The gas constant of air is R = 0.287 kPa ×m3 / kg ×K , and it can also be used for combustion gases. The specific heat ratio of combustion gases is k = 1.33. Analysis The velocity at the nozzle exit is the sonic speed, which is determined to be V = c=

kRT =

æ1000 m 2 / s 2 ÷ ö ÷ (1.33)(0.287 kJ/kg ×K) çç (215 K) = 286.5 m/s ÷ ÷ çè 1 kJ/kg ø

Noting that thrust F is related to velocity by F = mV , the mass flow rate of combustion gases is determined to be

2ö F 380, 000 N æ çç1 kg ×m / s ÷ ÷ = ÷= 1326 kg/s @1330 kg / s V 286.5 m/s çè 1 N ø÷

Discussion The combustion gases are mostly nitrogen (due to the 78% of N 2 in air), and thus they can be treated as air with a good degree of approximation.

13-750

F=380000 [N] T=220 [K] P=40 [kPa] "Properties" R=0.287 [kJ/kg-K] "taken same as that of air" k=1.33 "from Table 18-2 of the text" "Analysis" C=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2)) Vel=C "Choked flow is assumed" m_dot=F/Vel

18-118 A stationary temperature probe is inserted into an air duct reads 85C. The actual temperature of air is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The stagnation process is isentropic. Properties The specific heat of air at room temperature is c p = 1.005 kJ / kg ×K . Analysis The air that strikes the probe will be brought to a complete stop, and thus it will undergo a stagnation process. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature. The actual air temperature is determined from T = T0 -

æ 1 kJ/kg ÷ ö V2 (190 m/s) 2 çç = 85°C = 67.0°C 2 2÷ ÷ ç 2c p 2´ 1.005 kJ/kg ×K è1000 m / s ø

Discussion Temperature rise due to stagnation is very significant in high-speed flows, and should always be considered when compressibility effects are not negligible.

EES (Engineering Equation Solver) SOLUTION "Given" Vel=190 [m/s] T0=85 [C] "Properties" c_p=1.005 [kJ/kg-K] "Analysis" T=T0-Vel^2/(2*c_p)*Convert(m^2/s^2, kJ/kg)

18-119 Nitrogen flows through a heat exchanger. The stagnation pressure and temperature of the nitrogen at the inlet and the exit states are to be determined. Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 Flow of nitrogen through the heat exchanger is isentropic.

13-751

Properties The properties of nitrogen are c p = 1.039 kJ / kg ×K and k = 1.4. Analysis The stagnation temperature and pressure of nitrogen at the inlet and the exit states are determined from T01 = T1 +

æ 1 kJ/kg ÷ ö V12 (100 m/s) 2 çç = 10°C + = 14.8°C 2 2÷ ÷ ç è ø 2c p 2´ 1.039 kJ/kg ×°C 1000 m /s k / ( k - 1)

æT ö ÷ P01 = P1 ççç 01 ÷ ÷ èç T ÷ ø 1

1.4/ (1.4- 1)

æ288.0 K ÷ ö = (150 kPa) çç ÷ çè 283.2 K ÷ ø

= 159 kPa

From the energy balance relation Ein - Eout = D Esystem with w = 0 qin = c p (T2 - T1 ) +

V22 - V12 + D peZ 0 2

150 kJ/kg = (1.039 kJ/kg ×°C)(T2 - 10°C) +

ö (200 m/s) 2 - (100 m/s) 2 æ çç 1 kJ/kg ÷ ÷ ÷ çè1000 m 2 /s 2 ø 2

T2 = 139.9°C and T02 = T2 +

æ 1 kJ/kg ÷ ö V2 2 (200 m/s) 2 çç = 139.9°C + ÷= 159°C ø 2c p 2´ 1.039 kJ/kg ×°C çè1000 m 2 / s 2 ÷ k / ( k - 1)

æT ö ÷ P02 = P2 ççç 02 ÷ ÷ çè T ÷ ø 2

1.4/ (1.4- 1)

æ432.3 K ÷ ö = (100 kPa) çç ÷ çè 413.1 K ÷ ø

= 117 kPa

Discussion Note that the stagnation temperature and pressure can be very different than their thermodynamic counterparts when dealing with compressible flow.

EES (Engineering Equation Solver) SOLUTION "Given" P1=150 [kPa] T1=(10+273.15) [K] Vel1=100 [m/s] P2=100 [kPa] Vel2=180 [m/s] q_in=125 [kJ/kg] "Properties" c_p=1.039 [kJ/kg-K] k=1.4 "Analysis" T01=T1+Vel1^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) T01_C=T01-273.15 P01=P1*(T01/T1)^(k/(k-1)) q_in=c_p*(T2-T1)+(Vel2^2-Vel1^2)/2*Convert(m^2/s^2, kJ/kg) "energy balance" T02=T2+Vel2^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-752

T02_C=T02-273.15 P02=P2*(T02/T2)^(k/(k-1))

18-120 The equivalent relation for the speed of sound is to be verified using thermodynamic relations. æ ö çè ¶r ÷ ø

Analysis The two relations are c 2 = çç¶ P ÷ ÷ ÷

and

From

æ¶ P ö ÷ c 2 = k çç ÷ ÷ ÷ çè ¶r ø T

æ¶ P ö ö ö ö æ¶ T ö 2æ 2æ 2æ ç¶ P ÷ ç¶ P ¶ T ÷ ç¶ P ÷ ç ÷ r = 1/ v ¾ ¾ ® dr = - d v /v 2 . Thus, c 2 = çç ÷ ÷ ÷ ÷ ÷ ÷ ÷ = - v èçç ¶ v ø ÷ = - v èçç ¶ T ¶ v ø ÷ = - v èçç ¶ T ø ÷ èçç ¶ v ø ÷ çè ¶ r ø s s s s s

From the cyclic rule, æ¶ P ö æ¶ T ÷ ö æ¶ s ö æ¶ P ö æ¶ s ö æ¶ P ö÷ ç çç ÷ ç ( P, T , s ) : çç ÷ = - 1¾¾ ® çç ÷ = - çç ÷ ÷ ÷ ÷ ÷ ÷ èçç ¶ s ÷ ÷ ÷ ÷ ÷ èçç ¶ s ø÷ ÷ çè ¶ T ø ç ç øP è¶ P øT è ¶ T øs èç¶ T ø s P T

æ¶ T ö æ¶ v ö æ¶ s ö æ¶ T ö æ¶ s ö÷ æ¶ T ö÷ ç ÷ ç ÷ ç ç (T , v , s) : çç ÷ ® çç ÷ ÷ ÷ ÷ ÷ èçç ¶ s ø ÷ èçç¶ T ø ÷ = - 1¾¾ ÷ = - èçç¶ v ø÷ ÷ èçç ¶ s ø÷ ÷ çè ¶ v ø çè ¶ v ø÷ s T v s T v

Substituting, æ¶ s ö æ¶ P ö æ¶ s ö æ¶ T ö ö æ¶ T ö æ¶ P ö 2æ ç ÷ ç ÷ ç ÷ ç¶s ÷ ç ÷ ç ÷ c 2 = - v 2 çç ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ ÷ èçç ¶ s ø ÷ èçç¶ v ø ÷ èçç ¶ s ø ÷ = - v èçç¶T ø ÷ èçç ¶ s ø ÷ èçç ¶ v ø ÷ çè¶ T ø P T T v P v T

Recall that æ¶ s ö c æ¶ s ö and v = çç ÷ = çç ÷ ÷ ÷ ÷ ÷ ç T çè¶ T ø T è¶T øP v

Substituting,

æc p öæT ö÷æ¶ P ö æ¶ P ö çç ÷ç ÷ ÷ c 2 = - v 2 ççç ÷ = - v 2 k çç ÷ ÷ ÷ ç ÷ ÷ ÷ ç èç ¶ v ø÷T øèç ¶ v øT èç T ø÷èçcv ÷ Replacing - d v / v 2 by d, we get æ¶ P ö ÷ c 2 = k ççç ÷ ÷ ÷ è ¶r ø T

which is the desired expression Discussion Note that the differential thermodynamic property relations are very useful in the derivation of other property relations in differential form.

18-121 For ideal gases undergoing isentropic flows, expressions for P / P* , T / T * , and r / r * as functions of k and Ma are to be obtained. T* 2 T0 2 + (k - 1)Ma 2 = Analysis Equations 12-18 and 12-21 are given to be and = T0 k+1 T 2

æT T *÷ ö æ2 + (k - 1)Ma 2 ÷ öæ 2 ö ÷ çç ÷ ÷ = ççç Multiplying the two, ççç 0 ÷ ÷ ÷ ø 2 ø÷çè k + 1÷ èç T T0 ø÷ è Simplifying and inverting, k / ( k - 1)

From

æT ö P = çç ÷ ÷ ø P * çèT *÷

T k+1 = T * 2 + (k - 1)Ma 2

(1) k / ( k - 1)

ö÷ P æ k+1 ÷ = çç ¾¾ ® 2÷ ç P * è 2 + (k - 1)Ma ÷ ø

(2)

13-753

k / ( k - 1)

From

ær ö r ÷ = ççç ÷ ÷ r * è r *÷ ø

k / ( k - 1)

æ ö÷ r k+1 ÷ = ççç ¾¾ ® 2÷ r * è 2 + (k - 1)Ma ø÷

(3)

Discussion Note that some very useful relations can be obtained by very simple manipulations.

18-122 It is to be verified that for the steady flow of ideal gases dT0 / T = dA / A + (1- Ma 2 )dV / V . The effect of heating and area changes on the velocity of an ideal gas in steady flow for subsonic flow and supersonic flow are to be explained. V2 = c p (T0 - T ) 2

Analysis We start with the relation Differentiating, We also have

V dV = c p (dT0 - dT )

(2)

d r dA dV + + = 0 r A V

(3)

dP + V dV = 0 r

and

(4) P = RT,

Differentiating the ideal gas relation

dP d r dT = + = 0 P r T

From the speed of sound relation, c 2 = kRT = (k - 1)c pT = kP / r Combining Eqs. (3) and (5),

dP dT dA dV + + = 0 P T A V

Combining Eqs. (4) and (6),

dP dP = = - VdV r kP / c 2

dP k V 2 dV dV = - 2 V dV = - k 2 = - kMa 2 P V C C V

or,

Combining Eqs. (2) and (6), or, dT = T

dT = dT0 - V

(5) (6) (7)

(8)

dV cp

dT0 dT0 dT0 V 2 dV dT V2 dV dV = = - 2 = - (k - 1)Ma 2 T C pT V T T T V C /(k - 1) V

Combining Eqs. (7), (8), and (9), or,

(1)

- (k - 1)Ma 2

(9)

dV dT0 dV dA dV + (k - 1)Ma 2 + + = 0 V T V A V

dT0 dA é dV = + êë- kMa 2 + (k - 1)Ma 2 + 1ù ú û T A V

Thus,

dT0 dA dV = + (1- Ma 2 ) T A V

(10)

Differentiating the steady-flow energy equation q = h02 - h01 = c p (T02 - T01 )

dq = c p dT0

(11)

Eq. (11) relates the stagnation temperature change dT0 to the net heat transferred to the fluid. Eq. (10) relates the velocity changes to area changes dA, and the stagnation temperature change dT0 or the heat transferred. (a) When Ma < 1 (subsonic flow), the fluid accelerates if the duct converges (dA < 0) or the fluid is heated ( dT0  0 or q > 0). The fluid decelerates if the duct diverges (dA > 0) or the fluid is cooled ( dT0  0 or q < 0).

13-754

(b) When Ma > 1 (supersonic flow), the fluid accelerates if the duct diverges (dA > 0) or the fluid is cooled ( dT0  0 or q < 0). The fluid decelerates if the duct converges (dA < 0) or the fluid is heated ( dT0  0 or q > 0). Discussion Some of these results are not intuitively obvious, but come about by satisfying the conservation equations.

18-123 An expression for the ratio of the stagnation pressure after a shock wave to the static pressure before the shock wave as a function of k and the Mach number upstream of the shock wave is to be found. Analysis The relation between P1 and P2 is

æ1 + kMa12 ö÷ P2 1 + kMa 22 ÷ = ¾¾ ® P2 = P1 ççç 2 ÷ çè1 + kMa 22 ÷ P1 1 + kMa1 ø We substitute this into the isentropic relation k / (k - 1) P02 = (1 + (k - 1)Ma 22 / 2) P2

which yields k /( k - 1) P02 æ 1 + kMa12 ö÷ ÷ 1 + (k - 1)Ma 22 / 2) = ççç 2 ÷( P1 çè1 + kMa 2 ø÷

where Ma 22 =

Ma12 + 2 / (k - 1) 2kMa 22 / (k - 1) - 1

Substituting, k / ( k - 1)

2 ö P02 æ (1 + kMa12 )(2kMa12 - k + 1) öæ ÷ çç1 + (k - 1)Ma1 / 2 + 1 ÷ ÷ ÷ = ççç ÷ ÷ 2 2 ÷çç 2kMa1 / (k - 1) - 1ø ÷ P1 çè kMa1 (k + 1) - k + 3 øè

Discussion Similar manipulations of the equations can be performed to get the ratio of other parameters across a shock.

18-124 An expression for the speed of sound based on van der Waals equation of state is to be derived. Using this relation, the speed of sound in carbon dioxide is to be determined and compared to that obtained by ideal gas behavior. Properties The properties of CO2 are R = 0.1889 kJ / kg ×K and k = 1.279 at T = 50C = 323.2 K. Analysis Van der Waals equation of state can be expressed as P = Differentiating,

RT a - 2. v- b v

æ¶ P ö RT 2a çç ÷ = + 3 2 ÷ çè ¶ v ÷ øT (v - b) v

® d r = - dv /v 2 , the speed of sound relation becomes Noting that r = 1/v ¾ ¾ æ¶ P ö ö 2 æ ç¶ P ÷ Substituting, c 2 = k çç ÷ ÷ ÷ ÷ = v k èçç ¶ v ø ÷ èç ¶ r ø T T

c2 =

v 2 kRT 2ak v (v - b) 2

Using the molar mass of CO2 ( M = 44 kg / kmol ), the constant a and b can be expressed per unit mass as

a = 0.1882 kPa ×m6 /kg 2 and b = 9.70´ 10- 4 m3 / kg The specific volume of CO2 is determined to be © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-755

(0.1889 kPa ×m3 / kg ×K)(323.2 K) 2´ 0.1882 kPa ×m 6 / kg 2 v - 0.000970 m3 /kg v2

200 kPa=

v = 0.300 m3 / kg

Substituting, 1/ 2

ææ(0.300 m3 /kg) 2 (1.279)(0.1889 kJ/kg ×K)(323.2 K) 2(0.1882 kPa ×m 6 /kg 3 )(1.279) öæ1000 m 2 /s 2 öö ÷ ÷çç ÷ ÷ ÷÷ c = ççççç ÷ ÷ ÷ ÷ç1 kPa ×m3 /kg ø ÷ø ÷ (0.300 - 0.000970 m3 /kg) 2 (0.300 m3 /kg) 2 øè èççè = 271 m / s

If we treat CO2 as an ideal gas, the speed of sound becomes c=

kRT =

æ1000 m 2 /s 2 ÷ ö ÷ (1.279)(0.1889 kJ/kg ×K)(323.2 K) çç = 279 m / s ÷ ÷ çè 1 kJ/kg ø

Discussion Note that the ideal gas relation is the simplest equation of state, and it is very accurate for most gases encountered in practice. At high pressures and/or low temperatures, however, the gases deviate from ideal gas behavior, and it becomes necessary to use more complicated equations of state.

EES (Engineering Equation Solver) SOLUTION "Given" T=(80+273.15) [K] P=320 [kPa] a_m=364.3 [kPa-m^6/kmol^2] b_m=0.0427 [m^3/kmol] "Properties" R=0.1889 [kJ/kg-K] k=1.279 "at 50 C" M=44 [kg/kmol] "Analysis" a=a_m/M^2 b=b_m/M "a and b are expressed per unit mass basis" P=(R*T)/(v-b)-a/v^2 C=sqrt((v^2*k*R*T)/(v-b)^2*Convert(kJ/kg, m^2/s^2)-(2*a*k)/v*Convert(kJ/kg, m^2/s^2)) "derived by hand" C_ideal=sqrt(k*R*T*Convert(kJ/kg, m^2/s^2))

18-125 Helium gas is accelerated in a nozzle. The pressure and temperature of helium at the location where Ma = 1 and the ratio of the flow area at this location to the inlet flow area are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Properties The properties of helium are R = 2.0769 kJ / kg ×K , c p = 5.1926 kJ / kg ×K , and k = 1.667. Analysis The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. We first determine the stagnation temperature and pressure, which remain constant throughout the nozzle since the flow is isentropic. T0 = Ti +

æ 1 kJ/kg ÷ ö Vi 2 (120 m/s) 2 çç = 600 K + = 601.4 K 2 2÷ ÷ ç 2c p 2´ 5.1926 kJ/kg ×K è1000 m / s ø

and k / ( k - 1)

æT ö ÷ P0 = Pi ççç 0 ÷ ÷ çè T ÷ ø i

1.667/ (1.667- 1)

æ601.4 K ÷ ö = (0.5 MPa) çç ÷ ÷ çè 600 K ø

= 0.5029 MPa

13-756

The Mach number at the nozzle exit is given to be Ma = 1. Therefore, the properties at the nozzle exit are the critical properties determined from

æ 2 ö÷ æ 2 ö÷ T * = T0 çç = (601.4 K) çç ÷ ÷= 451.0 K = 451K ÷ èç k + 1ø èç1.667 + 1ø÷ k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ èç k + 1ø

1.667/(1.667- 1)

æ 2 ö ÷ = (0.5027 MPa) çç ÷ ÷ èç1.667 + 1ø

= 0.2450 MPa=0.245 MPa

The speed of sound and the Mach number at the nozzle inlet are æ1000 m2 /s 2 ö ÷ ÷ (1.667)(2.0769 kJ/kg ×K)(600 K) ççç = 1441 m/s ÷ ÷ è 1 kJ/kg ø

ci =

kRT i =

Ma i =

Vi 120 m/s = = 0.08328 ci 1441 m/s

The ratio of the entrance-to-throat area is ( k + 1)/[2( k - 1)]

öù Ai 1 éæ 2 öæ k- 1 ÷ êçç ú = Ma i2 ÷ ÷ ÷ ççç1 + * ÷ ÷ú ç ê ø Ma i ëè k + 1øè 2 A û

2.667/(2´ 0.667)

éæ 2 öæ öù 1 ÷ çç1 + 1.667 - 1 (0.08328) 2 ÷ êçç ÷ ÷ú ÷ ø÷úû 0.08328 êëèç1.667 + 1øèç 2 = 6.785 Then the ratio of the throat area to the entrance area becomes A* 1 = = 0.1474 @0.147 Ai 6.785 =

Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=600 [kPa] T_i=560 [K] Vel_i=120 [m/s] "Properties" R=2.0769 [kJ/kg-K] c_p=5.1926 [kJ/kg-K] k=1.667 "Analysis" T0=T_i+Vel_i^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P0=P_i*(T0/T_i)^(k/(k-1)) T_star=T0*(2/(k+1)) P_star=P0*(2/(k+1))^(k/(k-1)) C_i=sqrt(k*R*T_i*Convert(kJ/kg, m^2/s^2)) M_i=Vel_i/C_i Ratio_A=1/M_i*((2/(k+1))*(1+(k-1)/2*M_i^2))^((k+1)/(2*(k-1))) "Ratio_A=A_i/A_star" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-757

Ratio=1/Ratio_A

18-126 Helium gas enters a nozzle with negligible velocity, and is accelerated in a nozzle. The pressure and temperature of helium at the location where Ma = 1 and the ratio of the flow area at this location to the inlet flow area are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The entrance velocity is negligible. Properties The properties of helium are R = 2.0769 kJ / kg ×K , c p = 5.1926 kJ / kg ×K , and k = 1.667. Analysis We treat helium as an ideal gas with k = 1.667. The properties of the fluid at the location where Ma = 1 are the critical properties, denoted by superscript *. The stagnation temperature and pressure in this case are identical to the inlet temperature and pressure since the inlet velocity is negligible. They remain constant throughout the nozzle since the flow is isentropic.

T0  Ti  600 K P0  Pi  0.5MPa

The Mach number at the nozzle exit is given to be Ma = 1. Therefore, the properties at the nozzle exit are the critical properties determined from

æ 2 ö÷ æ 2 ö ÷ ç T * = T0 çç = 450 K ÷= (600 K) èçç1.667 + 1÷ ÷ çè k + 1ø÷ ø k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ èç k + 1ø

1.667/(1.667- 1)

æ 2 ö ÷ = (0.5 MPa) çç ÷ ÷ èç1.667 + 1ø

= 0.244 MPa

The ratio of the nozzle inlet area to the throat area is determined from ( k + 1)/[2( k - 1)]

öù Ai 1 éæ 2 öæ ÷ çç1 + k - 1 Ma i2 ÷ êçç = ÷ ÷ú * ÷ ç ç ø÷úû Ma i êëè k + 1øè 2 A

But the Mach number at the nozzle inlet is Ma = 0 since Vi  0 . Thus the ratio of the throat area to the nozzle inlet area is A* 1 = = 0 Ai ¥

Discussion The compressible flow functions are essential tools when determining the proper shape of the compressible flow duct.

EES (Engineering Equation Solver) SOLUTION "Given" P_i=600 [kPa] T_i=560 [K] Vel_i=0 [m/s] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-758

"Properties" R=2.0769 [kJ/kg-K] c_p=5.1926 [kJ/kg-K] k=1.667 "Analysis" T0=T_i P0=P_i M=1 T_star=T0*(2/(k+1)) P_star=P0*(2/(k+1))^(k/(k-1)) M_i=0.00001 "since Vel_i=0" Ratio_A=1/M_i*((2/(k+1))*(1+(k-1)/2*M_i^2))^((k+1)/(2*(k-1))) "Ratio_A=A_i/A_star" Ratio=1/Ratio_A "For M_i=0, Ratio_A is infinity, and Ratio is zero. For EES to be able do calculations, we used a very small M_i value instead of zero, and obtained the same result."

18-127 Nitrogen gas enters a converging nozzle. The properties at the nozzle exit are to be determined. Assumptions 1 Nitrogen is an ideal gas with k = 1.4. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. Analysis The schematic of the duct is shown in Fig. 12–25. For isentropic flow through a duct, the area ratio A / A* (the flow area over the area of the throat where Ma = 1) is also listed in Table A –32. At the initial Mach number of Ma = 0.3, we read A1 = 2.0351, A*

T1 = 0.9823, T0

and

P1 = 0.9395 P0

With a 20 percent reduction in flow area, A2  0.8 A1 , and A2 A A = 2 1* = (0.8)(2.0351) = 1.6281 A1 A A*

For this value of A2 / A* from Table A–32, we read T2 = 0.9791, T0

P2 = 0.8993, and Ma 2 = 0.391 P0

Here we chose the subsonic Mach number for the calculated A2 / A* instead of the supersonic one because the duct is converging in the flow direction and the initial flow is subsonic. Since the stagnation properties are constant for isentropic flow, we can write

æT /T ö æ0.9791ö÷ T2 T2 /T0 ÷ = ® T2 = T1 ççç 2 0 ÷ = (400 K) çç ÷ ÷= 399 K çè0.9823 ø÷ T1 T1 /T0 èç T1 /T0 ÷ ø æP /P ö æ0.8993 ö÷ P2 P2 /P0 ÷ = ® P2 = P1 ççç 2 0 ÷ = (100 kPa) çç ÷= 95.7 K ÷ çè P1 /P0 ø÷ èç0.9395 ÷ ø P1 P1 /P0 which are the temperature and pressure at the desired location. Discussion Note that the temperature and pressure drop as the fluid accelerates in a converging nozzle.

EES (Engineering Equation Solver) SOLUTION "Given" T1=400 [K] P1=100 [kPa] M1=0.3 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-759

A2\A1=0.8 "Analysis" k=1.4 RatioA1=1/M1*((2/(k+1))*(1+(k-1)/2*M1^2))^((k+1)/(2*(k-1))) "Ratio_A=A/A_star" RatioT1=(1+(k-1)/2*M1^2)^(-1) RatioP1=(1+(k-1)/2*M1^2)^(-k/(k-1)) RatioA2=A2\A1*RatioA1 RatioA2=1/M2*((2/(k+1))*(1+(k-1)/2*M2^2))^((k+1)/(2*(k-1))) RatioT2=(1+(k-1)/2*M2^2)^(-1) RatioP2=(1+(k-1)/2*M2^2)^(-k/(k-1)) T2=T1*RatioT2/RatioT1 P2=P1*RatioP2/RatioP1

18-128 Nitrogen gas enters a converging nozzle. The properties at the nozzle exit are to be determined. Assumptions 1 2

Nitrogen is an ideal gas with k = 1.4. Flow through the nozzle is steady, one-dimensional, and isentropic.

Analysis The schematic of the duct is shown in Fig. 12–25. For isentropic flow through a duct, the area ratio A / A* (the flow area over the area of the throat where Ma = 1) is also listed in Table A –32. At the initial Mach number of Ma = 0.5, we read A1 T P = 1.3398, 1 = 0.9524, and 1 = 0.8430 * T0 P0 A

With a 20 percent reduction in flow area, A2  0.8 A1 , and A2 A A = 2 1* = (0.8)(1.3398) = 1.0718 A1 A A*

For this value of A2 / A* from Table A–32, we read T2 P = 0.9010, 2 = 0.6948, and Ma 2 = 0.740 T0 P0

T2 T2 /T0 = T1 T1 /T0

æT /T ö æ0.9010 ö÷ ÷ ® T2 = T1 ççç 2 0 ÷ = (400 K) çç ÷ ÷= 378 K çè0.9524 ø÷ èç T1 /T0 ÷ ø

P2 P2 /P0 = P1 P1 /P0

æP /P ö æ0.6948 ö÷ ÷= (100 kPa) çç ® P2 = P1 ççç 2 0 ÷ ÷= 82.4 K ÷ çè P1 /P0 ÷ èç0.8430 ÷ ø ø

which are the temperature and pressure at the desired location. Discussion Note that the temperature and pressure drop as the fluid accelerates in a converging nozzle.

EES (Engineering Equation Solver) SOLUTION "Given" T1=400 [K] P1=100 [kPa] M1=0.5 A2\A1=0.8

13-760

"Analysis" k=1.4 RatioA1=1/M1*((2/(k+1))*(1+(k-1)/2*M1^2))^((k+1)/(2*(k-1))) "Ratio_A=A/A_star" RatioT1=(1+(k-1)/2*M1^2)^(-1) RatioP1=(1+(k-1)/2*M1^2)^(-k/(k-1)) RatioA2=A2\A1*RatioA1 RatioA2=1/M2*((2/(k+1))*(1+(k-1)/2*M2^2))^((k+1)/(2*(k-1))) RatioT2=(1+(k-1)/2*M2^2)^(-1) RatioP2=(1+(k-1)/2*M2^2)^(-k/(k-1)) T2=T1*RatioT2/RatioT1 P2=P1*RatioP2/RatioP1

18-129 Nitrogen entering a converging-diverging nozzle experiences a normal shock. The pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock are to be determined. The results are to be compared to those of air under the same conditions. Assumptions 1 2 3

Nitrogen is an ideal gas with constant specific heats. Flow through the nozzle is steady, one-dimensional, and isentropic. The nozzle is adiabatic.

Properties The properties of nitrogen are R = 0.297 kJ / kg ×K and k = 1.4. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Assuming the flow before the shock to be isentropic,

P01 = Pi = 620 kPa

T01 = Ti = 310 K

Then,

æ ö÷ æ ö÷ 2 2 ÷= (310 K) çç ÷= 110.7 K T1 = T01 ççç 2÷ 2÷ çè 2+(1.4 - 1)3 ø÷ çè 2 + (k - 1)Ma1 ø÷ and k / ( k - 1)

æT ö ÷ P1 = P01 ççç 1 ÷ ÷ ÷ çèT ø 01

1.4/ 0.4

æ110.7 ÷ ö = (620 kPa) çç ÷ çè 310 ÷ ø

= 16.88 kPa

The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-33. For Ma1  3.0 we read Ma 2 = 0.4752 @ 0.475,

P02 = 0.32834, P01

P2 = 10.333, P1

and

T2 = 2.679 T1

13-761

Then the stagnation pressure P02 , static pressure P2 , and static temperature T2 , are determined to be

P02 = 0.32834P01 = (0.32834)(620 kPa) = 203.6 kPa @204 kPa P2 = 10.333P1 = (10.333)(16.88 kPa) = 174.4 kPa @174 kPa T2 = 2.679T1 = (2.679)(110.7 K) = 296.6 K @297 K The velocity after the shock can be determined from V2  Ma 2 c2 , where c 2 is the speed of sound at the exit conditions after the shock, æ1000 m 2 /s 2 ÷ ö ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.4752) (1.4)(0.297 kJ/kg ×K)(296.6 K) ççç = 166.9 m/s @167 m / s ÷ ÷ è 1 kJ/kg ø

Discussion For air at specified conditions k = 1.4 (same as nitrogen) and R = 0.297 kJ / kg ×K . Thus the only quantity which will be different in the case of air is the velocity after the normal shock, which happens to be 164.0 m / s .

EES (Engineering Equation Solver) SOLUTION "Given" P_i=620 [kPa] T_i=310 [K] Vel_i=0 "negligible" Mx=3 "Properties" R=0.2968 [kJ/kg-K] k=1.4 "Analysis" P0x=P_i T0x=T_i Tx=T0x*2/(2+(k-1)*Mx^2) Px=P0x*(Tx/T0x)^(k/(k-1)) My=sqrt((Mx^2+2/(k-1))/(2*Mx^2*k/(k-1)-1)) P0y=P0x*Mx/My*((1+My^2*(k-1)/2)/(1+Mx^2*(k-1)/2))^((k+1)/(2*(k-1))) Py=Px*(1+k*Mx^2)/(1+k*My^2) Ty=Tx*(1+Mx^2*(k-1)/2)/(1+My^2*(k-1)/2) Cy=sqrt(k*R*Ty*Convert(kJ/kg, m^2/s^2)) Vely=My*Cy

18-130 The diffuser of an aircraft is considered. The static pressure rise across the diffuser and the exit area are to be determined. Assumptions 1 2 3

Air is an ideal gas with constant specific heats at room temperature. Flow through the diffuser is steady, one-dimensional, and isentropic. The diffuser is adiabatic.

Properties Air properties at room temperature are R = 0.297 kJ / kg ×K , c p = 1.005 kJ / kg ×K , and k = 1.4. Analysis The inlet velocity is æ1000 m 2 / s 2 ö ÷ = 281.0 m/s ÷ V1 = Ma1c1 = M1 kRT1 = (0.9) (1.4)(0.287 kJ/kg ×K)(242.7 K) çç ÷ ÷ çè 1 kJ/kg ø

Then the stagnation temperature and pressure at the diffuser inlet become

13-762

T01 = T1 +

ö V12 (281.0 m/s) 2 æ çç 1 kJ/kg ÷ = 242.7 + = 282.0 K 2 2÷ ÷ ç 2c p 2(1.005 kJ/kg ×K) è1000 m / s ø k / ( k - 1)

æT ö ÷ P01 = P1 ççç 01 ÷ ÷ èç T ÷ ø 1

1.4/ (1.4- 1)

æ282.0 K ö ÷ = (41.1 kPa) çç ÷ ÷ çè 242.7 K ø

= 69.50 kPa

For an adiabatic diffuser, the energy equation reduces to h01  h02 . Noting that h  c pT and the specific heats are assumed to be constant, we have T01 = T02 = T0 = 282.0 K The isentropic relation between states 1 and 02 gives k / ( k - 1)

æT ö ÷ P02 = P02 = P1 ççç 02 ÷ ÷ èç T1 ÷ ø

1.4/ (1.4- 1)

æ282.0 K ö ÷ = (41.1 kPa) çç ÷ ÷ èç 242.7 K ø

= 69.50 kPa

The exit velocity can be expressed as æ1000 m 2 /s 2 ÷ ö ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (0.3) (1.4)(0.287 kJ/kg ×K) T2 çç ÷ = 6.01 T2 ÷ çè 1 kJ/kg ø

Thus T2 = T02 -

ö V2 2 6.012 T2 m 2 /s 2 æ çç 1 kJ/kg ÷ = (282.0) = 277.0 K 2 2÷ ÷ ç 2c p 2(1.005 kJ/kg ×K) è1000 m / s ø

Then the static exit pressure becomes k / ( k - 1)

æT ö ÷ P2 = P02 ççç 2 ÷ ÷ ÷ èçT02 ø

1.4/ (1.4- 1)

æ277.0 K ö ÷ = (69.50 kPa) çç ÷ ÷ èç 282.0 K ø

= 65.28 kPa

Thus the static pressure rise across the diffuser is D P = P2 - P1 = 65.28 - 41.1 = 24.2 kPa Also, r 2 =

P2 65.28 kPa = = 0.8211 kg/m3 RT2 (0.287 kPa ×m 3 /kg ×K)(277.0 K)

V2 = 6.01 T2 = 6.01 277.0 = 100.0 m/s Thus A2 =

m 38 kg/s = = 0.463 m2 r 2V2 (0.8211 kg/m3 )(100.0 m/s)

Discussion The pressure rise in actual diffusers will be lower because of the irreversibilities. However, flow through well-designed diffusers is very nearly isentropic.

EES (Engineering Equation Solver) SOLUTION "Given" M1=0.9 H=7000 [m] P1=41.1 [kPa] T1=242.7 [K] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-763

M2=0.3 m_dot=38 [kg/s] "Properties" c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] k=1.4 "Analysis" C1=sqrt(k*R*T1*Convert(kJ/kg, m^2/s^2)) Vel1=M1*C1 T01=T1+Vel1^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P01=P1*(T01/T1)^(k/(k-1)) P02=P1*(T02/T1)^(k/(k-1)) C2=sqrt(k*R*T2*Convert(kJ/kg, m^2/s^2)) Vel2=M2*C2 T02=T01 T2=T02-Vel2^2/(2*c_p)*Convert(m^2/s^2, kJ/kg) P2=P02*(T2/T02)^(k/(k-1)) DELTAP=P2-P1 rho_2=P2/(R*T2) A2=m_dot/(rho_2*Vel2)

18-131 The critical temperature, pressure, and density of an equimolar mixture of oxygen and nitrogen for specified stagnation properties are to be determined. Assumptions Both oxygen and nitrogen are ideal gases with constant specific heats at room temperature. Properties The specific heat ratio and molar mass are k = 1.395 and M = 32 kg / kmol for oxygen, and k = 1.4 and

M = 28 kg / kmol for nitrogen. Analysis The gas constant of the mixture is M m = yO M O + yN M N = 0.5´ 32 + 0.5´ 28 = 30 kg/kmol 2

Rm =

Ru 8.314 kJ/kmol ×K = = 0.2771 kJ/kg ×K Mm 30 kg/kmol

The specific heat ratio is 1.4 for nitrogen, and nearly 1.4 for oxygen. Therefore, the specific heat ratio of the mixture is also 1.4. Then the critical temperature, pressure, and density of the mixture become æ 2 ö÷ æ 2 ö÷ ç T * = T0 çç ÷= (550 K) èçç1.4 + 1ø÷ ÷= 458.3 K @ 458 K çè k + 1ø÷ k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ çè k + 1ø

r*=

1.4/(1.4- 1)

æ 2 ö ÷ = (350 kPa) çç ÷ ÷ èç1.4+1ø

= 184.9 kPa @185 kPa

P* 184.9 kPa = = 1.46 kg / m3 RT * (0.2771 kPa ×m 3 /kg ×K)(458.3 K)

Discussion If the specific heat ratios k of the two gases were different, then we would need to determine the k of the mixture from k = c p , m / cv , m where the specific heats of the mixture are determined from c p , m = mfO2 c p , O2 + mf N2 c p , N2 = ( yO2 M O2 /M m )c p , O2 + ( yN2 M N2 /M m )c p , N2 cv , m = mfO2 cv , O2 + mf N2 cv , N2 = ( yO2 M O2 /M m )cv , O2 + ( y N2 M N2 /M m )cv , N2

where mf is the mass fraction and y is the mole fraction. In this case it would give c p , m = (0.5´ 32 / 30)´ 0.918 + (0.5´ 28 / 30) ´ 1.039 = 0.974 kJ/kg.K

c p , m = (0.5´ 32 / 30)´ 0.658 + (0.5´ 28 / 30) ´ 0.743 = 0.698 kJ/kg.K and k = 0.974 / 0.698 = 1.40

13-764

EES (Engineering Equation Solver) SOLUTION "Given" y_O2=0.5 y_N2=0.5 T0=550 [K] P0=350 [kPa] "Properties" Fluid1$='O2' Fluid2$='N2' C_p_O2=0.918 [kJ/kg-K] C_v_O2=0.658 [kJ/kg-K] k_O2=1.395 M_O2=32 [kg/kmol] C_p_N2=1.039 [kJ/kg-K] C_v_N2=0.743 [kJ/kg-K] k_N2=1.4 M_N2=28 [kg/kmol] R_u=8.314 [kJ/kmol-K] "Analysis" M_m=y_O2*M_O2+y_N2*M_N2 R_m=R_u/M_m C_p_m=(y_O2*M_O2/M_m)*C_p_O2+(y_N2*M_N2/M_m)*C_p_N2 C_v_m=(y_O2*M_O2/M_m)*C_v_O2+(y_N2*M_N2/M_m)*C_v_N2 k_m=C_p_m/C_v_m T_star=T0*(2/(k_m+1)) P_star=P0*(2/(k_m+1))^(k_m/(k_m-1)) rho_star=P_star/(R_m*T_star)

18-132E Helium gas is accelerated in a nozzle. For a specified mass flow rate, the throat and exit areas of the nozzle are to be determined for the case of isentropic nozzle. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of helium are R = 0.4961 Btu / lbm ×R = 2.6809 psia ×ft 3 / lbm ×R , c p = 1.25 Btu / lbm× R, and k = 1.667. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible,

T01 = T1 = 740 R P01 = P1 = 220 psia

The flow is assumed to be isentropic, thus the stagnation temperature and pressure remain constant throughout the nozzle, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-765

T02 = T01 = 740 R

P02 = P01 = 220 psia The critical pressure and temperature are determined from

æ 2 ö÷ æ 2 ö÷ T * = T0 çç ÷= (740 R) ççç ÷= 554.9 R èç k + 1ø÷ è1.667 + 1ø÷ k /( k - 1)

æ 2 ö ÷ P* = P0 çç ÷ ÷ çè k + 1ø

r*=

= 107.2 psia

P* 107.2 psia = = 0.07203 1bm/ft 3 RT * (2.6809 psia ×ft 3 /lbm ×R)(554.9 R)

V * = c* =

and A* =

1.667/(1.667- 1)

æ 2 ö ÷ = (220 psia) çç ÷ ÷ çè1.667 + 1ø

æ25,037 ft 2 / s 2 ÷ ö ÷ (1.667)(0.4961 Btu/lbm ×R)(554.9 R) ççç = 3390 ft/s ÷ ÷ è 1 Btu/1bm ø

kRT * =

m 0.2 1bm/s = = 8.19´ 10- 4 ft 2 r *V * (0.07203 1bm/ft 3 )(3390 ft/s)

At the nozzle exit the pressure is P2  15psia . Then the other properties at the nozzle exit are determined to be k / ( k - 1)

ö p0 æ k- 1 = çç1 + Ma 22 ÷ ÷ ÷ ç ø p2 è 2

1.667/ 0.667

¾¾ ®

ö 220 psia æ 1.667 - 1 = çç1 + Ma 22 ÷ ÷ ÷ ç è ø 15 psia 2

It yields Ma 2  2.405 , which is greater than 1. Therefore, the nozzle must be converging-diverging.

æ ö æ ö÷ 2 2 ÷ ÷= (740 R) çç ÷= 252.6 R T2 = T0 ççç 2÷ 2÷ ç è 2 + (1.667 - 1) ´ 2.405 ø÷ èç 2 + (k - 1)Ma 2 ø÷ r2 =

P2 15 psia = = 0.02215 1bm/ft 3 RT2 (2.6809 psia ×ft 3 /lbm ×R)(252.6 R)

æ25,037 ft 2 / s 2 ÷ ö ÷ V2 = Ma 2 c2 = Ma 2 kRT2 = (2.405) (1.667)(0.4961 Btu/lbm ×R)(252.6 R) ççç = 5500 ft/s ÷ ÷ è 1 Btu/1bm ø

Thus the exit area is m 0.2 lbm/s A2 = = = 0.00164 ft 2 r 2V2 (0.02215 lbm/ft 3 )(5500 ft/s) Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities.

EES (Engineering Equation Solver) SOLUTION "Given" P1=220 [psia] T1=740 [R] Vel1=0 "negligible" P2=15 [psia] m_dot=0.2 [lbm/s] "Properties" R=0.4961 [Btu/lbm-R] c_p=1.25 [Btu/lbm-R] k=1.667 "Analysis" T01=T1 P01=P1 "since inlet velocity is negligible" T02=T01 P02=P01 "since the flow is isentropic" T_star=T01*2/(k+1) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-766

P_star=P01*(2/(k+1))^(k/(k-1)) rho_star=P_star/(R*Convert(Btu, psia-ft^3)*T_star) C_star=sqrt(k*R*T_star*Convert(Btu/lbm, ft^2/s^2)) Vel_star=C_star A_star=m_dot/(rho_star*Vel_star) P02/P2=(1+(k-1)/2*M2^2)^(k/(k-1)) "If M2 is greater than 1, the nozzle is converging-diverging" T2=T02*2/(2+(k-1)*M2^2) rho_2=P2/(R*Convert(Btu, psia-ft^3)*T2) C2=sqrt(k*R*T2*Convert(Btu/lbm, ft^2/s^2)) Vel2=M2*C2 A2=m_dot/(rho_2*Vel2)

18-133 Helium gas is accelerated in a nozzle isentropically. For a specified mass flow rate, the throat and exit areas of the nozzle are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The properties of helium are R = 2.0769 kJ / kg ×K , c p = 5.1926 kJ / kg ×K , and k = 1.667. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible,

T01 = T1 = 500 K

P01 = P1 = 0.8 MPa

The flow is assumed to be isentropic, thus the stagnation temperature and pressure remain constant throughout the nozzle,

T02 = T01 = 500 K P02 = P01 = 0.8 MPa The critical pressure and temperature are determined from

æ 2 ö÷ æ 2 ö÷ T * = T0 çç ÷= (500 K) ççç ÷= 375.0 K èç k + 1ø÷ è1.667+1ø÷ k / ( k - 1)

æ 2 ÷ ö P* = P0 çç ÷ çè k + 1÷ ø

r*=

1.667/ (1.667- 1)

æ 2 ÷ ö = (0.8 MPa) çç ÷ çè1.667+1÷ ø

= 0.3897 MPa

P* 389.7 kPa = = 0.5003 kg/m 3 RT * (2.0769 kPa ×m 3 /kg ×K)(375 K)

V * = c* =

kRT * =

æ1000 m 2 /s 2 ÷ ö ÷ (1.667)(2.0769 kJ/kg ×K)(375 K) ççç = 1139.4 m/s ÷ ÷ è 1 kJ/kg ø

13-767

Thus the throat area is A* =

m 0.34 kg/s = = 5.964´ 10- 4 m 2 = 5.96 cm2 3 r *V * (0.5003 kg/m )(1139.4 m/s)

At the nozzle exit the pressure is P2  0.1MPa . Then the other properties at the nozzle exit are determined to be k / ( k - 1)

ö P0 æ k- 1 = çç1 + Ma 22 ÷ ÷ ÷ ø P2 èç 2

1.667/ 0.667

¾¾ ®

ö 0.8 MPa æ 1.667 - 1 = çç1 + Ma 22 ÷ ÷ ÷ ø 0.1 MPa èç 2 å

EES code for this equation: 0.8 / 0.1 = (1 + (1.667 - 1) / 2 Ma 2) (1.667 / 0.667) It yields Ma 2  1.973 , which is greater than 1. Therefore, the nozzle must be converging-diverging.

æ ö÷ æ ö÷ 2 2 ÷= (500 K) çç ÷ T2 = T0 ççç ÷= 217.6 K 2÷ 2÷ ç è 2 + (1.667 - 1) ´ 1.973 ø èç 2 + (k - 1)Ma 2 ÷ ø r2 =

P2 100 kPa = = 0.2213 kg/m3 RT2 (2.0769 kPa ×m3 /kg ×K)(217.6 K)

æ1000 m 2 / s 2 ö ÷ ÷ = 1712.5 m/s V2 = Ma 2 c2 = Ma 2 kRT2 = (1.973) (1.667)(2.0769 kJ/kg ×K)(217.6 K) çç ÷ ÷ çè 1 kJ/kg ø

Thus the exit area is A2 =

m 0.34 kg/s = = 8.972´ 10- 4 m 2 @ 8.97 cm2 r 2V2 (0.2213 kg/m 3 )(1712.5 m/s)

Discussion Flow areas in actual nozzles would be somewhat larger to accommodate the irreversibilities.

EES (Engineering Equation Solver) SOLUTION "Given" P1=1000 [kPa] T1=500 [K] Vel1=0 "negligible" P2=100 [kPa] m_dot=0.46 [kg/s] "Properties" c_p=5.1926 [kJ/kg-K] R=2.0769 [kJ/kg-K] k=1.667 "Analysis" T01=T1 P01=P1 "since inlet velocity is negligible" T02=T01 P02=P01 "since the flow is isentropic" T_star=T01*2/(k+1) P_star=P01*(2/(k+1))^(k/(k-1)) rho_star=P_star/(R*T_star) C_star=sqrt(k*R*T_star*Convert(kJ/kg, m^2/s^2)) Vel_star=C_star A_star=m_dot/(rho_star*Vel_star)*Convert(m^2, cm^2) P02/P2=(1+(k-1)/2*M2^2)^(k/(k-1)) "If M2 is greater than 1, the nozzle is converging-diverging" T2=T02*2/(2+(k-1)*M2^2) rho_2=P2/(R*T2) C2=sqrt(k*R*T2*Convert(kJ/kg, m^2/s^2)) Vel2=M2*C2 A2=m_dot/(rho_2*Vel2)*Convert(m^2, cm^2) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-768

18-134 Air flowing at a subsonic velocity in a duct is accelerated by heating. The highest rate of heat transfer without affecting the inlet conditions is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant.

Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K . Analysis Heat transfer will stop when the flow is choked, and thus Ma 2 = V2 / c2 = 1 . The inlet density and stagnation temperature are r1 =

P1 350 kPa = = 2.904 kg/m 3 RT1 (0.287 kJ/kgK)(420 K)

æ k- 1 ö æ 1.4 - 1 2 ö÷ T01 = T1 çç1 + Ma12 ÷ = (420 K) çç1 + 0.6 ÷ = 450.2 K ÷ ÷ èç ø èç ø÷ 2 2 Then the inlet velocity and the mass flow rate become c1 =

kRT1 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(420 K) çç ÷ = 410.8 m/s çè 1 kJ/kg ÷ ø

V1 = Ma1c1 = 0.6(410.8 m/s) = 246.5 m/s mair = r 1 Ac1V1 = (2.904 kg/m3 )(0.1´ 0.1 m 2 )(246.5 m/s) = 7.157 kg/s The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02 / T0* = 1 (since Ma 2  1 ) . T01 (k + 1)Ma12 [2 + (k - 1)Ma12 ] (1.4 + 1)0.62 [2 + (1.4 - 1)0.62 ] = = = 0.8189 T0* (1 + kMa12 )2 (1 + 1.4´ 0.62 ) 2

Therefore, T0 2 T02 / T0* 1 = = T0 1 T01 / T0* 0.8189

 T02 = T01 / 0.8189 = (450.2 K) / 0.8189 = 549.8 K

Then the rate of heat transfer becomes

Q = mair c p (T02 - T01 ) = (7.157 kg/s)(1.005 kJ/kg ×K)(549.8 - 450.2) K = 716 kW Discussion It can also be shown that T2  458K , which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. We can also solve this problem using the Rayleigh function values listed in Table A-34.

13-769

P_1=350 [kPa] T_1=420 [K] "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" Ma_2=1 rho_1=P_1/(R*T_1) T_01=T_1*(1+(k-1)/2*Ma_1^2) c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) V_1=Ma_1*c_1 A_c=L^2 m_dot=rho_1*A_c*V_1 T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01) T2/T_02=(1+(k-1)/2*Ma_2^2)^(-1)

18-135 Helium flowing at a subsonic velocity in a duct is accelerated by heating. The highest rate of heat transfer without affecting the inlet conditions is to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of helium to be k = 1.667, c p = 5.193 kJ / kg ×K , and R = 2.077 kJ / kg ×K . Analysis Heat transfer will stop when the flow is choked, and thus Ma 2 = V2 / c2 = 1 . The inlet density and stagnation temperature are r1 =

P1 350 kPa = = 0.4012 kg/m3 RT1 (2.077 kJ/kgK)(420 K)

æ k- 1 ö æ 1.667 - 1 2 ö÷ T01 = T1 çç1 + Ma12 ÷ 0.6 ÷ = 470.4 K ÷= (420 K) ççç1 + èç ø÷ è ø÷ 2 2

Then the inlet velocity and the mass flow rate become c1 =

kRT1 =

æ1000 m 2 / s 2 ÷ ö ÷ (1.667)(2.077 kJ/kg ×K)(420 K) ççç = 1206 m/s ÷ ÷ è 1 kJ/kg ø

V1 = Ma1c1 = 0.6(1206 m/s) = 723.5 m/s mair = r 1 Ac1V1 = (0.4012 kg/m3 )(0.1´ 0.1 m2 )(723.5 m/s) = 2.903 kg/s © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-770

The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are T02 / T0* = 1 (since Ma 2  1 ) T01 (k + 1)Ma12 [2 + (k - 1)Ma12 ] (1.667 + 1)0.62 [2 + (1.667 - 1)0.62 ] = = = 0.8400 T0* (1 + kMa12 ) 2 (1 + 1.667 ´ 0.62 ) 2

Therefore, T0 2 T02 / T0* 1 = = * T0 1 T01 / T0 0.8400

 T02 = T01 / 0.8400 = (470.4 K) / 0.8400 = 560.0 K

Then the rate of heat transfer becomes

Q = mair c p (T02 - T01 ) = (2.903 kg/s)(5.193 kJ/kg ×K)(560.0 - 470.4) K = 1350 kW Discussion It can also be shown that T2  420 K , which is the highest thermodynamic temperature that can be attained under stated conditions. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease. Also, in the solution of this problem, we cannot use the values of Table A-34 since they are based on k = 1.4.

EES (Engineering Equation Solver) SOLUTION "Given" l=0.10 [m] Ma_1=0.6 P_1=350 [kPa] T_1=420 [K] "Properties" k=1.667 c_p=5.193 [kJ/kg-K] R=2.077 [kJ/kg-K] "Analysis" Ma_2=1 rho_1=P_1/(R*T_1) T_01=T_1*(1+(k-1)/2*Ma_1^2) c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) V_1=Ma_1*c_1 A_c=L^2 m_dot=rho_1*A_c*V_1 T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01) T2/T_02=(1+(k-1)/2*Ma_2^2)^(-1)

18-136 Air flowing at a subsonic velocity in a duct is accelerated by heating. For a specified exit Mach number, the heat transfer for a specified exit Mach number as well as the maximum heat transfer are to be determined. Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. 2 Inlet conditions (and thus the mass flow rate) remain constant. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K .

13-771

Analysis The inlet Mach number and stagnation temperature are æ1000 m 2 / s 2 ÷ ö ÷ (1.4)(0.287 kJ/kg ×K)(400 K) ççç = 400.9 m/s ÷ è 1 kJ/kg ÷ ø

c1 =

kRT1 =

Ma1 =

V1 100 m/s = = 0.2494 c1 400.9 m/s

æ k- 1 ö æ 1.4 - 1 ö T01 = T1 çç1 + Ma12 ÷ = (400 K) çç1 + 0.24942 ÷ ÷ ÷= 405.0 K ÷ çè ç ø è ø÷ 2 2 The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1  0.2494 :

T01 / T * = 0.2559

Ma 2  0.8 :

T02 / T * = 0.9639

Then the exit stagnation temperature and the heat transfer are determined to be T0 2 T02 / T * 0.9639 = = = 3.7667 * T0 1 T01 / T 0.2559



T02 = 3.7667T01 = 3.7667(405.0 K) = 1526 K

q = c p (T02 - T01 ) = (1.005 kJ/kg ×K )(1526 - 405) K = 1126 kJ/kg @1130 kJ / kg Maximum heat transfer will occur when the flow is choked, and thus Ma 2  1 and thus T02 / T * = 1 . Then, T0 2 T02 / T * 1 = = T0 1 T01 / T * 0.2559



T02 = T01 / 0.2559 = (405 K) / 0.2559 = 1583 K

qmax = c p (T02 - T01 ) = (1.005 kJ/kg ×K)(1583 - 405) K = 1184 kJ/kg @1180 kJ / kg

Discussion This is the maximum heat that can be transferred to the gas without affecting the mass flow rate. If more heat is transferred, the additional temperature rise will cause the mass flow rate to decrease.

EES (Engineering Equation Solver) SOLUTION "Given" V_1=100 [m/s] T_1=400 [K] P_1=35 [kPa] Ma_2=0.8 "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) Ma_1=V_1/c_1 T_01=T_1*(1+(k-1)/2*Ma_1^2) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-772

T_02/T_01=T02\T0star/T01\T0star q=c_p*(T_02-T_01) "For maximum heat transfer, the flow is choked: Ma_2 = 1" T02\T0star_max=1 T_02_max/T_01=T02\T0star_max/T01\T0star q_max=c_p*(T_02_max-T_01)

18-137 Air flowing at sonic conditions in a duct is accelerated by cooling. For a specified exit Mach number, the amount of heat transfer per unit mass is to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K .

Analysis Noting that Ma1  1 , the inlet stagnation temperature is

æ k- 1 ö æ 1.4 - 1 2 ö÷ T01 = T1 çç1 + Ma12 ÷ = (340 K) çç1 + 1÷ = 408 K ÷ ÷ çè ø èç ø÷ 2 2 The Rayleigh flow functions T0 / T0* corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1  1 :

T01 / T0* = 1

Ma 2  1.6 :

T02 / T0* = 0.8842

Then the exit stagnation temperature and heat transfer are determined to be T0 2 T02 / T0* 0.8842 = = = 0.8842 * T0 1 T01 / T0 1

 T02 = 0.8842T01 = 0.8842(408 K) = 360.75 K

q = c p (T02 - T01 ) = (1.005 kJ/kg ×K)(360.75 - 408) K = - 47.49 kJ/kg @- 47.5 kJ / kg

Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated.

EES (Engineering Equation Solver) SOLUTION "Given" Ma_1=1 T_1=340 [K] P_1=250 [kPa] Ma_2=1.6 "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-773

T_01=T_1*(1+(k-1)/2*Ma_1^2) T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_02/T_01=T02\T0star/T01\T0star q=c_p*(T_02-T_01)

18-138 Air flowing at a supersonic velocity in a duct is accelerated by cooling. For a specified exit Mach number, the rate of heat transfer is to be determined. Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) are valid. Properties We take the properties of air to be k = 1.4, c p = 1.005 kJ / kg ×K , and R = 0.287 kJ / kg ×K .

Analysis Knowing stagnation properties, the static properties are determined to be - 1

- 1

æ k- 1 ö æ 1.4 - 1 2 ö T1 = T01 çç1 + Ma12 ÷ = (350 K) çç1 + 1.2 ÷ ÷ ÷ ÷ ÷ = 271.7 K èç ø èç ø 2 2 - k /( k - 1)

æ k- 1 ö P1 = P01 çç1 + Ma12 ÷ ÷ ÷ çè ø 2

r1 =

- 1.4/0.4

æ 1.4 - 1 2 ö = (240 kPa) çç1 + 1.2 ÷ ÷ ÷ èç ø 2

= 98.97 kPa

P1 98.97 kPa = = 1.269 kg/m3 RT1 (0.287 kJ/kgK)(271.7 K)

Then the inlet velocity and the mass flow rate become c1 =

kRT1 =

æ1000 m 2 / s 2 ö ÷ ÷ (1.4)(0.287 kJ/kg ×K)(271.7 K) çç = 330.4 m/s ÷ ÷ çè 1 kJ/kg ø

V1 = Ma1c1 = 1.2(330.4 m/s) = 396.5 m/s mair = r 1 Ac1V1 = (1.269 kg/m3 )[p (0.30 m) 2 / 4](396.5 m/s) = 35.57 kg/s The Rayleigh flow functions T0 / T0* corresponding to the inlet and exit Mach numbers are (Table A-34):

Ma1  1.8 : T01 / T0* = 0.9787 Ma 2  2 :

T02 / T0* = 0.7934

Then the exit stagnation temperature is determined to be T0 2 T02 / T0* 0.7934 = = = 0.8107 T0 1 T01 / T0* 0.9787

 T02 = 0.8107T01 = 0.8107(350 K) = 283.7 K

Finally, the rate of heat transfer is

Q = mair c p (T02 - T01 ) = (35.57 kg/s)(1.005 kJ/kg ×K )(283.7 - 350) K= - 2370 kW

13-774

Discussion The negative sign confirms that the gas needs to be cooled in order to be accelerated. Also, it can be shown that the thermodynamic temperature drops to 158 K at the exit, which is extremely low. Therefore, the duct may need to be heavily insulated to maintain indicated flow conditions.

EES (Engineering Equation Solver) SOLUTION "Given" D=0.20 [m] Ma_1=1.2 T_01=350 [K] P_01=240 [kPa] Ma_2=2 "Properties" k=1.4 c_p=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Analysis" T_1=T_01*(1+(k-1)/2*Ma_1^2)^(-1) P_1=P_01*(1+(k-1)/2*Ma_1^2)^(-k/(k-1)) rho_1=P_1/(R*T_1) c_1=sqrt(k*R*T_1*Convert(kJ/kg, m^2/s^2)) V_1=Ma_1*c_1 A_c=pi*D^2/4 m_dot=rho_1*A_c*V_1 T1\Tstar=((Ma_1*(1+k))/(1+k*Ma_1^2))^2 T01\T0star=((k+1)*Ma_1^2*(2+(k-1)*Ma_1^2))/(1+k*Ma_1^2)^2 T2\Tstar=((Ma_2*(1+k))/(1+k*Ma_2^2))^2 T02\T0star=((k+1)*Ma_2^2*(2+(k-1)*Ma_2^2))/(1+k*Ma_2^2)^2 T_2/T_1=T2\Tstar/T1\Tstar T_02/T_01=T02\T0star/T01\T0star Q_dot=m_dot*c_p*(T_02-T_01)

18-139 Saturated steam enters a converging-diverging nozzle with a low velocity. The throat area, exit velocity, mass flow rate, and exit Mach number are to be determined for isentropic and 90 percent efficient nozzle cases. Assumptions 1 Flow through the nozzle is steady and one-dimensional. 2 The nozzle is adiabatic. Analysis (a) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h10  h1 . At the inlet,

h1 = (h f + x1h fg )@ 1.75 MPa = 878.16 + 0.90´ 1917.1 = 2603.5 kJ/kg s1 = ( s f + x1s fg )@ 1.75 MPa = 2.3844 + 0.90´ 4.0033 = 5.9874 kJ/kg ×K

13-775

At the exit, P2  1.2 MPa and s2 = s2s = s1 = 5.9874 kJ / kg ×K . Thus, s2 = s f + x2 s fg ® 5.9874 = 2.2159 + x2 (4.3058) ® x2 = 0.8759

h2 = h f + x2 h fg = 798.33 + 0.8759´ 1985.4 = 2537.4 kJ/kg v 2 = v f + x2v fg = 0.001138 + 0.8759´ (0.16326 - 0.001138) = 0.14314 m 3 / kg Then the exit velocity is determined from the steady-flow energy balance to be V12 V2 = h2 + 2 2 2 Solving for V2 , h1 +

V2 =

0 = h2 - h1 +

V22 - V12 2

æ1000 m 2 / s 2 ÷ ö ÷ 2(2603.5 - 2537.4)kJ/kg ççç = 363.7 m / s ÷ è 1 kJ/kg ÷ ø

2(h1 - h2 ) =

The mass flow rate is determined from m=

1 1 A2V2 = (25´ 10- 4 m 2 )(363.7 m/s) = 6.35 kg / s v2 0.14314 m3 / kg

The velocity of sound at the exit of the nozzle is determined from 1/ 2

1/ 2 æ DP ÷ ö æ¶ P ö ÷ c = çç ÷ @ççç ÷ ÷ çè ¶ r ÷ ÷ øs èD (1 / v ) øs

The specific volume of steam at s2 = 5.9874 kJ / kg ×K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1547 and 0.1333 m3 / kg . Substituting,

c2 =

(1300 - 1100) kPa æ 1 1 ö÷ çç kg/m ÷ çè0.1333 0.1547 ÷ ø

æ1000 m 2 / s 2 ö÷ çç ÷ = 438.9 m/s 3 ÷ è 1 kPa ×m ÷ ø 3ç

Then the exit Mach number becomes V 363.7 m/s Ma 2 = 2 = = 0.829 c2 438.9 m/s The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be Pt = P* = 0.576´ P01 = 0.576´ 1.75 = 1.008 MPa Then at the throat, Pt = 1.008 MPa

and

st = s1 = 5.9874 kJ/kg ×K

Thus, ht = 2507.7 kJ/kg

v t = 0.1672 m3 / kg Then the throat velocity is determined from the steady-flow energy balance,

13-776

Z 0

V2 V2 h1 + 1 = ht + t 2 2 Solving for Vt , Vt =

2(h1 - ht ) =

0 = ht - h1 +

Vt 2 2

æ1000 m 2 / s 2 ÷ ö ÷ 2(2603.5 - 2507.7)kJ/kg ççç = 437.7 m/s ÷ è 1 kJ/kg ÷ ø

Thus the throat area is

mv t (6.35 kg/s)(0.1672 m3 /kg) = = 24.26´ 10- 4 m2 = 24.26 cm 2 Vt 437.7 m/s

At =

(b) The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Thus h10  h1 . At the inlet,

h1 = (h f + x1h fg )@1.75 MPa = 878.16 + 0.90´ 1917.1 = 2603.5 kJ/kg s1 = ( s f + x1s fg )@1.75 MPa = 2.3844 + 0.90´ 4.0033 = 5.9874 kJ/kg ×K

At state 2s, P2  1.2 MPa and s2 = s2s = s1 = 5.9874 kJ / kg ×K . Thus,

s2 s = s f + x2 s fg ® 5.9874 = 2.2159 + x2 (4.3058) ® x2 s = 0.8759 h2 s = h f + x2 h fg = 798.33 + 0.8759´ 1985.4 = 2537.4 kJ/kg The enthalpy of steam at the actual exit state is determined from hN =

h01 - h2 2603.5 - h2 ¾¾ ® 0.92 = ¾¾ ® h2 = 2542.7 kJ/kg h01 - h2 s 2603.4 - 2537.4

Therefore at the exit, P2  1.2 MPa and h2 = 2542.7 kJ / kg ×K . Thus,

h2 = h f + x2 h fg ¾ ¾ ® 2542.7 = 798.33 + x2 (1985.4) ¾ ¾ ® x2 = 0.8786

s2 = s f + x2 s fg = 2.2159 + 0.8786´ 4.3058 = 5.9989 v 2 = v f + x2v fg = 0.001138 + 0.8786´ (0.16326 - 0.001138) = 0.1436 kJ / kg Then the exit velocity is determined from the steady-flow energy balance to be h1 +

V12 V2 = h2 + 2 2 2

0 = h2 - h1 +

V22 - V12 2

Solving for V2 , V2 =

2(h1 - h2 ) =

æ1000 m2 / s 2 ÷ ö ÷ 2(2603.5 - 2542.7)kJ/kg ççç = 348.9 m / s ÷ è 1 kJ/kg ÷ ø

The mass flow rate is determined from m=

1 1 A2V2 = (25´ 10- 4 m 2 )(348.9 m/s) = 6.07 kg / s v2 0.1436 m3 /kg

13-777

The velocity of sound at the exit of the nozzle is determined from 1/ 2

1/ 2

æ¶ P ö æ D P ö÷ ÷ @çç ÷ c = çç ÷ ÷ ÷ çè ¶r ø÷ èçD (1 / v ) ø÷s s

The specific volume of steam at s2 = 5.9989 kJ / kg ×K and at pressures just below and just above the specified pressure (1.1 and 1.3 MPa) are determined to be 0.1551 and 0.1337 m3 / kg . Substituting,

(1300 - 1100) kPa

c2 =

æ 1 1 ö ÷ kg/m çç ÷ çè0.1337 0.1551÷ ø

æ1000 m 2 / s 2 ö ÷ çç ÷ = 439.7 m/s 3 ÷ ç ÷ ø 3 è 1 kPa ×m

Then the exit Mach number becomes Ma 2 =

V2 348.9 m/s = = 0.793 c2 439.7 m/s

The steam is saturated, and thus the critical pressure which occurs at the throat is taken to be

Pt = P* = 0.576´ P01 = 0.576´ 1.75 = 1.008 MPa At state 2ts, Pts  1.008 MPa and sts = s1 = 5.9874 kJ / kg ×K . Thus, hts = 2507.7 kJ / kg . The actual enthalpy of steam at the throat is hN =

h01 - ht 2603.5 - ht ¾¾ ® 0.92 = ¾¾ ® ht = 2515.4 kJ/kg h01 - hts 2603.5 - 2507.7

Therefore at the throat, P2 = 1.008 MPa and ht = 2515.4 kJ/kg. Thus, vt = 0.1679 m3 / kg . Then the throat velocity is determined from the steady-flow energy balance, Z 0

V2 V2 h1 + 1 = ht + t 2 2

0 = ht - h1 +

Vt 2 2

Solving for Vt , Vt =

2(h1 - ht ) =

æ1000 m2 / s 2 ÷ ö ÷ 2(2603.5 - 2515.4)kJ/kg ççç = 419.9 m/s ÷ è 1 kJ/kg ÷ ø

Thus the throat area is

At =

mv t (6.07 kg/s)(0.1679 m3 / kg) = = 24.30´ 10- 4 m2 = 24.30 cm2 Vt 419.9 m/s

EES (Engineering Equation Solver) SOLUTION "Given" P[1]=1750 [kPa] x[1]=0.90 Vel1=0 "negligible inlet velocity" P[2]=1200 [kPa] A2=25E-4 [m^2] Eta_N=0.92 P1=1100 P2=1300 "Analysis" Fluid$='steam_iapws' "In the equations below, a stands for part (a), b stands for part (b) and t stands for throat" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-778

"(a)" h[1]=enthalpy(Fluid$, P=P[1], x=x[1]) s[1]=entropy(Fluid$, P=P[1], x=x[1]) s[2]=s[1] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) v[2]=volume(Fluid$, P=P[2], s=s[2]) x[2]=quality(Fluid$, P=P[2], s=s[2]) Vel2a=sqrt(2*(h[1]-h[2])*Convert(kJ/kg, m^2/s^2)) m_dot_a=1/v[2]*A2*Vel2a v1=volume(Fluid$, P=P1, s=s[2]) v2=volume(Fluid$, P=P2, s=s[2]) C2a=sqrt((P2-P1)/(1/v2-1/v1)*Convert(kJ/kg, m^2/s^2)) M2a=Vel2a/C2a P_ta=0.576*P[1] "0.576 from Sec. 18-7 of the text" s_ta=s[1] h_ta=enthalpy(Fluid$, P=P_ta, s=s_ta) v_ta=volume(Fluid$, P=P_ta, s=s_ta) Vel_ta=sqrt(2*(h[1]-h_ta)*Convert(kJ/kg, m^2/s^2)) "energy balance" A_ta=(m_dot_a*v_ta)/Vel_ta*Convert(m^2, cm^2) "(b)" Eta_N=(h[1]-h_b[2])/(h[1]-h[2]) v_b[2]=volume(Fluid$, P=P[2], h=h_b[2]) s_b[2]=entropy(Fluid$, P=P[2], h=h_b[2]) x_b[2]=quality(Fluid$, P=P[2], h=h_b[2]) Vel2b=sqrt(2*(h[1]-h_b[2])*Convert(kJ/kg, m^2/s^2)) "energy balance" m_dot_b=1/v_b[2]*A2*Vel2b v1b=volume(Fluid$, P=P1, s=s_b[2]) v2b=volume(Fluid$, P=P2, s=s_b[2]) C2b=sqrt((P2-P1)/(1/v2b-1/v1b)*Convert(kJ/kg, m^2/s^2)) M2b=Vel2b/C2b Eta_N=(h[1]-h_tb)/(h[1]-h_ta) v_tb=volume(Fluid$, P=P_ta, h=h_tb) Vel_tb=sqrt(2*(h[1]-h_tb)*Convert(kJ/kg, m^2/s^2)) "energy balance" A_tb=(m_dot_b*v_tb)/Vel_tb*Convert(m^2, cm^2)

18-140 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane.

Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-779

P01  Pi  2.0 MPa It is specified that A / A* = 2.896 . From Table A-13, the Mach number and the pressure ratio which corresponds to this area ratio are Ma1  2.60 and P1 / P01 = 0.0501 . The pressure ratio across the shock for this Ma1 value is, from Table A-14,

P2 / P1 = 7.72 . Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be P2  7.72P1  7.72  0.0501P01  7.72  0.0501 (2.0MPa)  0.774MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results should be the same.

18-141 Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for an ideal gas with k = 1.667. Properties The specific heat ratio of the ideal gas is given to be k = 1.667. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. 0.5( k + 1)/( k - 1)

Ma * = Ma

öù A 1 éæ 2 öæ ÷ çç1 + k - 1 Ma 2 ÷ êçç ú = ÷ ÷ * ÷ ÷ ç ç øúû Ma êëè k + 1øè 2 A

k+1 2 + (k - 1)Ma 2 - k / ( k - 1)

- 1/ ( k - 1)

ö P æ k- 1 = çç1 + Ma 2 ÷ ÷ ÷ ç ø P0 è 2

æ k- 1 ö r = çç1 + Ma 2 ÷ ÷ ÷ ç ø r0 è 2

- 1

ö T æ k- 1 = çç1 + Ma 2 ÷ ÷ ÷ ç ø T0 è 2

k=1.667 PP0=(1+(k–1)*M^2/2)^(–k/(k–1)) TT0=1/(1+(k–1)*M^2/2) DD0=(1+(k–1)*M^2/2)^(–1/(k–1)) Mcr=M*SQRT((k+1)/(2+(k–1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k–1)*M^2))^(0.5*(k+1)/(k–1))/M Ma 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 5.0 

Ma * 0 0.1153 0.2294 0.3413 0.4501 0.5547 0.6547 0.7494 0.8386 0.9222 1.0000 1.1390 1.2572 1.3570 1.4411 1.5117 1.5713 1.6216 1.6643 1.7007 1.7318 1.8895 1.9996

A / A*  5.6624 2.8879 1.9891 1.5602 1.3203 1.1760 1.0875 1.0351 1.0081 1.0000 1.0267 1.0983 1.2075 1.3519 1.5311 1.7459 1.9980 2.2893 2.6222 2.9990 9.7920 

P / P0 1.0000 0.9917 0.9674 0.9288 0.8782 0.8186 0.7532 0.6850 0.6166 0.5501 0.4871 0.3752 0.2845 0.2138 0.1603 0.1202 0.0906 0.0686 0.0524 0.0403 0.0313 0.0038 0

r / r0 1.0000 0.9950 0.9803 0.9566 0.9250 0.8869 0.8437 0.7970 0.7482 0.6987 0.6495 0.5554 0.4704 0.3964 0.3334 0.2806 0.2368 0.2005 0.1705 0.1457 0.1251 0.0351 0

T / T0 1.0000 0.9967 0.9868 0.9709 0.9493 0.9230 0.8928 0.8595 0.8241 0.7873 0.7499 0.6756 0.6047 0.5394 0.4806 0.4284 0.3825 0.3424 0.3073 0.2767 0.2499 0.1071 0

13-780

Discussion The tabulated values are useful for quick calculations, but be careful – they apply only to one specific value of k, in this case k = 1.667.

18-142 Using the normal shock relations, the normal shock functions are to be evaluated and tabulated as in Table A33 for an ideal gas with k = 1.667. Properties The specific heat ratio of the ideal gas is given to be k = 1.667. Analysis The normal shock relations listed below are expressed in EES and the results are tabulated.

Ma 2 =

(k - 1)Ma12 + 2 2kMa12 - k + 1

T2 2 + Ma12 (k - 1) = T1 2 + Ma 22 (k - 1)

P2 1 + kMa12 2kMa12 - k + 1 = = P1 1 + kMa 22 k+1

r 2 P2 / P1 (k + 1)Ma12 V = = = 1, r 1 T2 / T1 2 + (k - 1)Ma12 V2 k+1

2( k - 1) P02 Ma1 éê1 + Ma 22 (k - 1) / 2 ù ú = P01 Ma 2 êë1 + Ma12 (k - 1) / 2 ú û

P02 (1 + kMa12 )[1 + Ma 22 (k - 1) / 2]k /( k - 1) = P1 1 + kMa 22

k=1.667 My=SQRT((Mx^2+2/(k–1))/(2*Mx^2*k/(k–1) –1)) PyPx=(1+k*Mx^2)/(1+k*My^2) TyTx=(1+Mx^2*(k–1)/2)/(1+My^2*(k–1)/2) RyRx=PyPx/TyTx P0yP0x=(Mx/My)*((1+My^2*(k–1)/2)/(1+Mx^2*(k–1)/2))^(0.5*(k+1)/(k–1)) P0yPx=(1+k*Mx^2)*(1+My^2*(k–1)/2)^(k/(k–1))/(1+k*My^2)

Ma1

Ma 2

P2 / P1

r 2 / r1

T2 / T1

P02 / P01

P02 / P1

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 4.0 5.0 

1.0000 0.9131 0.8462 0.7934 0.7508 0.7157 0.6864 0.6618 0.6407 0.6227 0.6070 0.5933 0.5814 0.5708 0.5614 0.5530 0.5455 0.5388 0.5327 0.5273 0.5223 0.4905 0.4753 0.4473

1.0000 1.2625 1.5500 1.8626 2.2001 2.5626 2.9501 3.3627 3.8002 4.2627 4.7503 5.2628 5.8004 6.3629 6.9504 7.5630 8.2005 8.8631 9.5506 10.2632 11.0007 19.7514 31.0022 

1.0000 1.1496 1.2972 1.4413 1.5805 1.7141 1.8415 1.9624 2.0766 2.1842 2.2853 2.3802 2.4689 2.5520 2.6296 2.7021 2.7699 2.8332 2.8923 2.9476 2.9993 3.3674 3.5703 3.9985

1.0000 1.0982 1.1949 1.2923 1.3920 1.4950 1.6020 1.7135 1.8300 1.9516 2.0786 2.2111 2.3493 2.4933 2.6432 2.7989 2.9606 3.1283 3.3021 3.4819 3.6678 5.8654 8.6834 

1 0.999 0.9933 0.9813 0.9626 0.938 0.9085 0.8752 0.8392 0.8016 0.763 0.7243 0.6861 0.6486 0.6124 0.5775 0.5442 0.5125 0.4824 0.4541 0.4274 0.2374 0.1398 0

2.0530 2.3308 2.6473 2.9990 3.3838 3.8007 4.2488 4.7278 5.2371 5.7767 6.3462 6.9457 7.5749 8.2339 8.9225 9.6407 10.3885 11.1659 11.9728 12.8091 13.6750 23.9530 37.1723 

Discussion The tabulated values are useful for quick calculations, but be careful – they apply only to one specific value of k, in this case k = 1.667.

13-781

18-143 Using appropriate software, the shape of a converging-diverging nozzle is to be determined for specified flow rate and stagnation conditions. The nozzle and the Mach number are to be plotted. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The specific heat ratio of air at room temperature is 1.4. Analysis The problem is solved using EES, and the results are tabulated and plotted below. k=1.4 cp=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] P0=1400 [kPa] T0=(200+273) [K] m=3 [kg/s] rho_0=P0/(R*T0) rho=P/(R*T) T=T0*(P/P0)^((k-1)/k) V=SQRT(2*cp*(T0-T)*1000) A=m/(rho*V)*10000 c=SQRT(k*R*T*1000) Ma=V/c Pressure P, kPa

Flow area

A,cm2

Mach number Ma

1400 1350 1300 1250 1200 1150 1100 1050 1000 950 900 850 800 750 700 650 600 550 500 450 400 350 300 250 200 150 100

 30.1 21.7 18.1 16.0 14.7 13.7 13.0 12.5 12.2 11.9 11.7 11.6 11.5 11.5 11.6 11.8 12.0 12.3 12.8 13.3 14.0 15.0 16.4 18.3 21.4 27.0

0 0.229 0.327 0.406 0.475 0.538 0.597 0.655 0.710 0.766 0.820 0.876 0.931 0.988 1.047 1.107 1.171 1.237 1.308 1.384 1.467 1.559 1.663 1.784 1.929 2.114 2.373

13-782

Discussion The shape is not actually to scale since the horizontal axis is pressure rather than distance. If the pressure decreases linearly with distance, then the shape would be to scale.

18-144 Steam enters a converging nozzle. The exit pressure, the exit velocity, and the mass flow rate versus the back pressure for a specified back pressure range are to be plotted. Assumptions 1 Steam is to be treated as an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic. 3 The nozzle is adiabatic. Properties The ideal gas properties of steam are R = 0.462 kJ / kg ×K , c p = 1.872 kJ / kg ×K , and k = 1.3.

13-783

Analysis We use EES to solve the problem. The stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Since the flow is isentropic, they remain constant throughout the nozzle,

P0  Pi  6MPa

and

T0  Ti  700 K

The critical pressure is determined to be k /( k - 1)

æ 2 ö÷ P* = P0 çç ÷ çè k + 1ø÷

1.3/0.3

æ 2 ö÷ = (6 MPa) çç ÷ èç1.3 + 1ø÷

= 3.274 MPa

Then the pressure at the exit plane (throat) is

Pe  Pb

for

Pb  3.274MPa

Pe  P  3.274MPa

for

Pb  3.274MPa (choked flow)

Thus the back pressure does not affect the flow when 3  Pb  3.274 MPa . For a specified exit pressure Pe , the temperature, velocity, and mass flow rate are © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-784

( k - 1)/ k

æP ö ÷ Temperature Te = T0 ççç e ÷ ÷ èç P ÷ ø

æP ö = (700 K) çç e ÷ ÷ ÷ çè 6 ø

Velocity V =

2c p (T0 - Te ) =

æ1000 m 2 /s 2 ÷ ö ÷ 2(1.872 kJ/kg ×K)(700 - Te ) ççç ÷ ÷ è 1 kJ/kg ø

Density

re =

Mass flow rate

0.3/1.3

Pe Pe = RTe (0.462 kPa ×m3 /kg ×K)Te

m = r eVe Ae = r eVe (0.0008 m2 )

EES Solution P_i=6000 [kPa] T_i=700 [K] Vel_i=0 [m/s] "negligible" A_e=8E-4 [m^2] R=0.462 [kJ/kg-K] cp=1.872 [kJ/kg-K] k=1.3 T0=T_i P0=P_i P_star=P0*(2/(k+1))^(k/(k-1)) T_e=T0*(P_e/P0)^((k-1)/k) Vel_e=sqrt(2*cp*(T0-T_e)*Convert(kJ/kg, m^2/s^2)) rho_e=P_e/(R*T_e) m_dot=rho_e*Vel_e*A_e "P_e=P_b for P_b >= P_star and P_e=P_star for P_b < P_star

Pb , MPa

Pe , MPa

Te , K

Ve , m / s

ρe, kg / m 3

m, kg / s

6000 5500 5000 4500 4000 3500 3274 3000

6000 5500 5000 4500 4000 3500 3274 3274

700.0 686.1 671.2 655.0 637.5 618.1 608.7 608.7

0 228.3 328.6 410.3 483.8 553.6 584.7 584.7

18.55 17.35 16.13 14.87 13.58 12.26 11.64 11.64

0 3.168 4.239 4.881 5.257 5.428 5.446 5.446

6000 5500

5000 4500 4000 3500 3000 3000

3500

4000

4500

5000

5500

6000

13-785

600 500

Vele

400 300 200 100 0 3000

3500

4000

4500

5000

5500

6000

Pb 6

m [kg/s]

5 4 3 2 1 0 3000

3500

4000

4500

5000

5500

6000

Pb Discussion Once the back pressure drops below 3.274 MPa, the flow is choked, and m remains constant from then on.

18-145 Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for air. Properties The specific heat ratio is given to be k = 1.4 for air. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. 0.5( k + 1)/( k - 1)

k+1 Ma = Ma 2 + (k - 1)Ma 2 *

- k / ( k - 1)

ö P æ k- 1 = çç1 + Ma 2 ÷ ÷ ÷ ç ø P0 è 2

öù A 1 éæ 2 öæ ÷ çç1 + k - 1 Ma 2 ÷ êçç ú = ÷ ÷ * ÷ ÷ ç ç ê øúû Ma ëè k + 1øè 2 A - 1/ ( k - 1)

æ k- 1 ö r = çç1 + Ma 2 ÷ ÷ ÷ ç ø r0 è 2

- 1

ö T æ k- 1 = çç1 + Ma 2 ÷ ÷ ÷ ø T0 çè 2

13-786

TT0=1/(1+(k–1)*M^2/2) DD0=(1+(k–1)*M^2/2)^(–1/(k–1)) Mcr=M*SQRT((k+1)/(2+(k–1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k–1)*M^2))^(0.5*(k+1)/(k–1))/M Ma

Ma * 1.0000 1.3646 1.6330 1.8257 1.9640 2.0642 2.1381 2.1936 2.2361 2.2691 2.2953 2.3163 2.3333 2.3474 2.3591 2.3689 2.3772 2.3843 2.3905

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

A / A* 1.0000 1.1762 1.6875 2.6367 4.2346 6.7896 10.7188 16.5622 25.0000 36.8690 53.1798 75.1343 104.1429 141.8415 190.1094 251.0862 327.1893 421.1314 535.9375

P / P0 0.5283 0.2724 0.1278 0.0585 0.0272 0.0131 0.0066 0.0035 0.0019 0.0011 0.0006 0.0004 0.0002 0.0002 0.0001 0.0001 0.0000 0.0000 0.0000

r / r0 0.6339 0.3950 0.2300 0.1317 0.0762 0.0452 0.0277 0.0174 0.0113 0.0076 0.0052 0.0036 0.0026 0.0019 0.0014 0.0011 0.0008 0.0006 0.0005

T / T0 0.8333 0.6897 0.5556 0.4444 0.3571 0.2899 0.2381 0.1980 0.1667 0.1418 0.1220 0.1058 0.0926 0.0816 0.0725 0.0647 0.0581 0.0525 0.0476

Discussion The tabulated values are useful for quick calculations, but be careful – they apply only to one specific value of k, in this case k = 1.4.

18-146 Using the compressible flow relations, the one-dimensional compressible flow functions are to be evaluated and tabulated as in Table A-32 for methane. Properties The specific heat ratio is given to be k = 1.3 for methane. Analysis The compressible flow functions listed below are expressed in EES and the results are tabulated. 0.5( k + 1)/( k - 1)

öù A 1 éæ 2 öæ ÷ çç1 + k - 1 Ma 2 ÷ êçç ú = ÷ ÷ * ÷ç ÷ øúû Ma êëèç k + 1øè 2 A

k+1 Ma = Ma 2 + (k - 1)Ma 2 *

- k /( k - 1)

- 1/( k - 1)

ö P æ k- 1 = çç1 + Ma 2 ÷ ÷ ÷ ø P0 çè 2

æ k- 1 ö r = çç1 + Ma 2 ÷ ÷ ÷ ø r 0 çè 2

- 1

ö T æ k- 1 = çç1 + Ma 2 ÷ ÷ ÷ ç ø T0 è 2

Methane: k=1.3 PP0=(1+(k–1)*M^2/2)^(–k/(k–1)) TT0=1/(1+(k–1)*M^2/2) DD0=(1+(k–1)*M^2/2)^(–1/(k–1)) Mcr=M*SQRT((k+1)/(2+(k–1)*M^2)) AAcr=((2/(k+1))*(1+0.5*(k–1)*M^2))^(0.5*(k+1)/(k–1))/M Ma

Ma *

A / A*

P / P0

r / r0

T / T0

13-787

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0

1.0000 1.3909 1.6956 1.9261 2.0986 2.2282 2.3263 2.4016 2.4602 2.5064 2.5434 2.5733 2.5978 2.6181 2.6350 2.6493 2.6615 2.6719 2.6810

1.0000 1.1895 1.7732 2.9545 5.1598 9.1098 15.9441 27.3870 45.9565 75.2197 120.0965 187.2173 285.3372 425.8095 623.1235 895.5077 1265.6040 1761.2133 2416.1184

0.5457 0.2836 0.1305 0.0569 0.0247 0.0109 0.0050 0.0024 0.0012 0.0006 0.0003 0.0002 0.0001 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000

0.6276 0.3793 0.2087 0.1103 0.0580 0.0309 0.0169 0.0095 0.0056 0.0033 0.0021 0.0013 0.0008 0.0006 0.0004 0.0003 0.0002 0.0001 0.0001

0.8696 0.7477 0.6250 0.5161 0.4255 0.3524 0.2941 0.2477 0.2105 0.1806 0.1563 0.1363 0.1198 0.1060 0.0943 0.0845 0.0760 0.0688 0.0625

Discussion The tabulated values are useful for quick calculations, but be careful – they apply only to one specific value of k, in this case k = 1.3.

Fundamentals of Engineering (FE) Exam Problems 18-147 Consider a converging nozzle with a low velocity at the inlet and sonic velocity at the exit plane. Now the nozzle exit diameter is reduced by half while the nozzle inlet temperature and pressure are maintained the same. The nozzle exit velocity will (a) remain the same (b) double (c) quadruple (d) go down by half (e) go down to one-fourth Answer (a) remain the same 18-148 An aircraft is cruising in still air at 5C at a velocity of 400 m / s . The air temperature at the nose of the aircraft where stagnation occurs is (a) 5C (b) 25C (c) (c) 55 (d) (d) 80C (e) (e) 85C Answer (e) 85C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=5 [C] Vel1= 400 [m/s] k=1.4 cp=1.005 [kJ/kg-K] T1_stag=T1+Vel1^2/(2*cp*1000) © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-788

"Some Wrong Solutions with Common Mistakes:" W1_Tstag=T1 "Assuming temperature rise" W2_Tstag=Vel1^2/(2*cp*1000) "Using just the dynamic temperature" W3_Tstag=T1+Vel1^2/(cp*1000) "Not using the factor 2"

18-149 Air is flowing in a wind tunnel at 25C, 95 kPa, and 250 m / s . The stagnation pressure at a probe inserted into the flow stream is (a) 184 kPa (b) 98 kPa (c) 161 kPa (d) 122 kPa (e) 135 kPa Answer (e) 135 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=25 [C] P1=95 [kPa] Vel1= 250 [m/s] k=1.4 cp=1.005 [kJ/kg-K] T1_stag=(T1+273)+Vel1^2/(2*cp*1000) T1_stag/(T1+273)=(P1_stag/P1)^((k-1)/k) "Some Wrong Solutions with Common Mistakes:" T11_stag/T1=(W1_P1stag/P1)^((k-1)/k); T11_stag=T1+Vel1^2/(2*cp*1000) "Using deg. C for temperatures" T12_stag/(T1+273)=(W2_P1stag/P1)^((k-1)/k); T12_stag=(T1+273)+Vel1^2/(cp*1000) "Not using the factor 2" T13_stag/(T1+273)=(W3_P1stag/P1)^(k-1); T13_stag=(T1+273)+Vel1^2/(2*cp*1000) "Using wrong isentropic relation"

18-150 Air is flowing in a wind tunnel at 12C and 66 kPa at a velocity of 190 m / s . The Mach number of the flow is (a) 0.56 (b) 0.65 (c) 0.73 (d) 0.87 (e) 1.7 Answer (a) 0.56 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=(12+273) [K] P1=66 [kPa] Vel1=190 [m/s] k=1.4 Cp=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] VS1=SQRT(k*R*T1*1000)

13-789

Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_Mach=Vel1/VS2; VS2=SQRT(k*R*(T1-273)*1000) "Using C for temperature" W2_Mach=VS1/Vel1 "Using Mach number relation backwards" W3_Mach=Vel1/VS3; VS3=k*R*T1 "Using wrong relation"

18-151 An aircraft is reported to be cruising in still air at -20C and 40 kPa at a Mach number of 0.86. The velocity of the aircraft is (a) 91 m / s (b) 220 m / s (c) 186 m / s (d) 274 m / s (e) 378 m / s Answer (d) 274 m / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=(-20+273) [K] P1=40 [kPa] Mach=0.86 k=1.4 cp=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] VS1=SQRT(k*R*T1*1000) Mach=Vel1/VS1 "Some Wrong Solutions with Common Mistakes:" W1_vel=Mach*VS2; VS2=SQRT(k*R*T1) "Not using the factor 1000" W2_vel=VS1/Mach "Using Mach number relation backwards" W3_vel=Mach*VS3; VS3=k*R*T1 "Using wrong relation"

18-152 Air is approaching a converging-diverging nozzle with a low velocity at 12C and 200 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of air at the throat of the nozzle is (a) 338 m / s (b) 309 m / s (c) 280 m / s (d) 256 m / s (e) 95 m / s Answer (b) 309 m / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=(12+273) [K] P1=200 [kPa] © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-790

Vel1=0 [m/s] k=1.4 cp=1.005 [kJ/kg-K] R=0.287 [kJ/kg-K] "Properties at the inlet" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(To-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation"

18-153 Argon gas is approaching a converging-diverging nozzle with a low velocity at 20C and 150 kPa, and it leaves the nozzle at a supersonic velocity. If the cross-sectional area of the throat is 0.015 m2 , the mass flow rate of argon through the nozzle is (a) 0.47 kg / s (b) 1.7 kg / s (c) 2.6 kg / s (d) 6.6 kg / s (e) 10.2 kg / s Answer (d) 6.6 kg / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=(20+273) [K] P1=150 [kPa] Vel1=0 [m/s] A=0.015 [m^2] k=1.667 cp=0.5203 [kJ/kg-K] R=0.2081 [kJ/kg-K] "Properties at the inlet" To=T1 "since velocity is zero" Po=P1 "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) rho_throat=P_throat/(R*T_throat) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) m=rho_throat*A*V_throat "Some Wrong Solutions with Common Mistakes:" W1_mass=rho_throat*A*V1_throat; V1_throat=SQRT(k*R*T1_throat*1000); T1_throat=2*(To-273)/(k+1) "Using C for temp" W2_mass=rho2_throat*A*V_throat; rho2_throat=P1/(R*T1) "Using density at inlet" © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-791

18-154 Carbon dioxide enters a converging-diverging nozzle at 60 m / s , 310C, and 300 kPa, and it leaves the nozzle at a supersonic velocity. The velocity of carbon dioxide at the throat of the nozzle is (a) 125 m / s (b) 225 m / s (c) 312 m / s (d) 353 m / s (e) 377 m / s Answer (d) 353 m / s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T1=(310+273) [K] P1=300 [kPa] Vel1=60 [m/s] k=1.289 cp=0.846 [kJ/kg-K] R=0.1889 [kJ/kg-K] "Properties at the inlet" To=T1+Vel1^2/(2*cp*1000) To/T1=(Po/P1)^((k-1)/k) "Throat properties" T_throat=2*To/(k+1) P_throat=Po*(2/(k+1))^(k/(k-1)) "The velocity at the throat is the velocity of sound," V_throat=SQRT(k*R*T_throat*1000) "Some Wrong Solutions with Common Mistakes:" W1_Vthroat=SQRT(k*R*T1*1000) "Using T1 for temperature" W2_Vthroat=SQRT(k*R*T2_throat*1000); T2_throat=2*(T_throat-273)/(k+1) "Using C for temperature" W3_Vthroat=k*R*T_throat "Using wrong relation"

18-155 Consider gas flow through a converging-diverging nozzle. Of the five statements below, select the one that is incorrect: (a) The fluid velocity at the throat can never exceed the speed of sound. (b) If the fluid velocity at the throat is below the speed of sound, the diversion section will act like a diffuser. (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic. (d) There will be no flow through the nozzle if the back pressure equals the stagnation pressure. (e) The fluid velocity decreases, the entropy increases, and stagnation enthalpy remains constant during flow through a normal shock. Answer (c) If the fluid enters the diverging section with a Mach number greater than one, the flow at the nozzle exit will be supersonic.

18-156 Combustion gases with k = 1.33 enter a converging nozzle at stagnation temperature and pressure of 350C and 400 kPa, and are discharged into the atmospheric air at 20C and 100 kPa. The lowest pressure that will occur within the nozzle is (a) 13 kPa © MCGRAW HILL LLC. ALL RIGHTS RESERVED. NO REPRODUCTION OR D ISTRIBUTION WITHOUT THE PRIOR WRITTEN CONSENT OF M CGRAW HILL LLC.

13-792

(b) 100 kPa (c) 216 kPa (d) 290 kPa (e) 315 kPa Answer (c) 216 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.33 Po=400 [kPa] P_throat=Po*(2/(k+1))^(k/(k-1)) "The critical pressure" "The lowest pressure that will occur in the nozzle is the higher of the critical or atmospheric pressure." "Some Wrong Solutions with Common Mistakes:" W2_Pthroat=Po*(1/(k+1))^(k/(k-1)) "Using wrong relation" W3_Pthroat=100 "Assuming atmospheric pressure"

18-157 … 18-160 Design and Essay Problems 18-160 A summary of the differences between incompressible flow, subsonic flow, and supersonic flow is to be written. Analysis Students discussions will vary greatly, but we would expect them to mention things like: 

Mach number being < about 0.3 for “incompressible flow”, between about 0.3 and 1 for compressible but subsonic flow, and > 1 for supersonic flow.  How density is nearly constant in incompressible and low Mach number flows, but highly variable in compressible flows.  How disturbances in incompressible flow are felt everywhere else instantaneously (the speed of sound is “infinite”), but in subsonic flow disturbances travel at the speed of sound, and for supersonic flow disturbances like shock waves can travel faster than the speed of sound.  How shock waves are possible only in supersonic flow.  Etc. Discussion There are many more differences than those listed here.