SOLUTIONS MANUAL for Introduction to Chemical Engineering Thermodynamics 9th Edition By JSmith, Hend by ACADEMIAMILL

Solution 1.1 Problem Statement

Electric current is the fundamental SI electrical dimension, with the ampere (A) as its unit. Determine units for the following quantities as combinations of fundamental SI units. (a) Electric power (b) Electric charge (c) Electric potential difference (d) Electric resistance (e) Electric capacitance

Solution

(a) Power is power, whether it is electrical, mechanical, or otherwise. Thus, electric power has the usual units of power: power 

energy J N m kg m 2    time s s s3

(b) Electric current is by definition the time rate of transfer of electrical charge. Thus current 

charge time

charge  current*time  A s

(you probably recall that the Coulomb is the usual derived unit of charge, defined as 1 A s)

Solution continued on next page…

energy  current*electric potential time

electrical potential 

energy kg m2  current*time A s3

(you probably recall that this is defined as the volt)

(d) Because (by Ohm’s Law) current is electric potential divided by resistance, or

resistance 

electrical potential kg m2  2 3 current A s

(this is defined as the ohm)

(e) Because electric potential is electric charge divided by electric capacitance, electrical potential 

charge electrical capacitance

electrical capacitance 

charge As A 2 s4   electrical potential kg m 2 kg m2 3 As

Solution 1.2 Problem Statement Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the Antoine equation, which can be written in the form: log10 P sat / (torr )  a 

b t / C  c

Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be rewritten in the equivalent form: ln P sat / kPa  A 

B T / K C

Show how the parameters in the two equations are related.

Solution

We must convert both the units and the logarithm (between base 10 and natural logarithm). We know that t in degrees Celsius is equal to T in Kelvins minus 273.15. Also 1 kPa is equal to 7.50 torr (we might have to look up this conversion factor). So, we have a b   t /C c 

P sat / torr  10

  B  7.5 • P sat / kPa  7.5exp  A   T / K  C  

Next, we might recognize that 10 can be written as exp(ln(10)) or exp(2.303). That is how we convert from base 10 log to natural log in general. So,      b B exp2.303a    7.5exp A   T / K  273.15  c  T / K  C    

Here, I have also substituted T

273.15 for t. Taking the natural log of both sides gives

  b B 2.303a    ln7.5  A  T / K  273.15  c  T /K C 

For the two functions to be equal for all values of T, each part of the functions must be the same, so we must have A = 2.303a ln(7.5) or A = ln(10)*a

ln(7.5)

B = 2.303b or B = ln(10)*b C=c

273.15

Solution 1.3 Problem Statement

Table B.2 in Appendix B provides parameters for computing the vapor pressure of many substances by the Antoine equation (see Prob. 1.2). For one of these substances, prepare two plots of P

sat

versus T over the range of temperature

sat

for which the parameters are valid. One plot should present P on a linear scale and the other should present P

sat

a log scale.

Problem 1.2

Liquid/vapor saturation pressure Psat is often represented as a function of temperature by the Antoine equation, which can be written in the form: log10 P sat / (torr )  a 

b t / C  c

Here, parameters a, b, and c are substance-specific constants. Suppose this equation is to be rewritten in the equivalent form: ln P sat / kPa  A 

B T / K C

Show how the parameters in the two equations are related.

Solution

The point of this problem is just for you to practice evaluating and plotting a simple function. You will do many problems over the course of the semester (and many more over the course of your career) in which the results are best presented in graphical form. The only thing to be careful of in plotting the Antoine equation is to pay attention to the sat

units of T and P and to whether the constants are given for use with the base 10 logarithm or the natural logarithm.

Solution continued on next page…

sat

The parameters in Table B.2 are for use with T in °C, P in kPa, and the natural logarithm. Note that plotting P vs. T sat

means that we treat P as the dependent variable and plot it on the vertical axis and treat T as the independent variable and plot it on the horizontal axis. Note also that to put the vertical axis on a log scale, we do not actually compute the sat

logarithm of P , but change the scaling of the axis, historically by using log-scale graphing paper, but now by changing an option in the plotting software.

I used MS Excel to prepare the plots for diethyl ether. A portion of the spreadsheet (containing the plots) is shown below: 13.9891 B

223.24

Vapor pressure of Diethyl Ether

200 180

100

160

Vapor Pressure (kPa)

T (deg C) Psat (kPa) -30 3.450361508 -29 3.684455127 -28 3.931786307 -27 4.192943086 -26 4.468531566 -25 4.759176154 -24 5.065519787 -23 5.388224158 -22 5.727969934 -21 6.085456959 -20 6.461404464 -19 6.856551255 -18 7.271655907 -17 7.707496937 -16 8.16487298 -15 8.644602953 -14 9.147526209 -13 9.674502686 -12 10.22641305 -11 10.80415881 -10 11.40866246 -9 12.0408676 -8 12.70173902 -7 13.39226279 -6 14.11344639 -5 14.86631874 -4 15.65193033 -3 16.4713532 -2 17.3256811 -1 18.21602942 0 19.14353533 1 20.10935774

2463.93 C

Vapor Pressure (kPa)

140 120 100 80 60

40 20 0

1 -30

-20

-10

Temperature (°C)

-30

-10

Temperature (°C)

Solution 1.4 Problem Statement

At what absolute temperature do the Celsius and Fahrenheit temperature scales give the same numerical value? What is the value?

Solution

One way to solve this problem is to write both the Celsius and Fahrenheit temperatures in terms of the absolute temperature in Kelvin, then set these equal to each other and solve for the temperature in Kelvin. The Celsius temperature is T (°C) = T (K) – 273.15. The Fahrenheit temperature is T (°F) = 1.8*T (K) – 459.67. If we plot these two lines, they look like:

SVNAS, problem 1.4 300

T (deg C or deg F)

200 100

deg C

0 -100

deg F

-200 -300 -400 -500 0

100

150

200 T (K)

250

300

350

What we are asked to find is the point where these two lines cross. That is, where T (K) – 273.15 = 1.8*T (K) – 459.67 Solving this for T (K) gives T (K) = (459.67 – 273.15) / 0.8 = 233.15 K. At this temperature (in K) the Fahrenheit temperature is 1.8 * 233.15 – 459.67 = 40 °F. Likewise, the Celsius temperature is 233.15 – 273.15 = 40 °C. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 1.5 Problem Statement

The SI unit of luminous intensity is the candela (abbreviated cd), which is a primary unit. The derived SI unit of luminous flux is the lumen (abbreviated lm). These are based on the sensitivity of the human eye to light. Light sources are often evaluated based on their luminous efficacy, which is defined as the luminous flux divided by the 1

power consumed and is measured in lm⋅W . In a physical or online store, find manufacturer’s specifications for representative incandescent, halogen, high-temperature-discharge, LED, and fluorescent lamps of similar luminous flux and compare their luminous efficacy.

Solution

I found the following on www.bulbs.com in 2016. I chose all lamps that could be screwed into a regular medium-base socket Type

Watts Used

Lumens Produced

Luminous Efficacy

Incandescent

150 W

1735 lm

12 lm W

Halogen

120 W

1850 lm

15 lm W

High Pressure Sodium

35 W

2300 lm

66 lm W

Compact Fluorescent

23 W

1650 lm

72 lm W

LED

19 W

1680 lm

88 lm W

Solution continued on next page…

In this light-output range, the high pressure sodium lamp, compact fluorescent, and LED are comparable. All three high-efficiency options are about a factor of 4 better than incandescent or halogen bulbs. Eventually, LED’s have the potential to be much more efficient than fluorescents, but what is on the market now is not there yet. They are improving rapidly – 2016 was the first year in which the LED lamp was the most efficient when I did this exercise. Previously, the high-pressure sodium lamp was most efficient.

Solution 1.6 Problem Statement

Pressures up to 3000 bar are measured with a dead-weight gauge. The piston diameter is 4 mm. What is the approximate mass in kg of the weights required?

Solution

Pressure is force divided by area. So, the maximum force required (to be applied by the weights) is the maximum pressure to be measured times the area of the piston. The area of the piston is 2

A = d /4 = 4 mm = 12.57 mm = 1.257  10

The maximum pressure to be measured is 3000 bar = 3  10 Pa = 3  10 N m . Thus, the maximum force to be applied by the weights is F = 1.257  10

m * 3  10 N m

= 3770 N. If this force is to be applied by weights, and

the acceleration of gravity is 9.8 m s , then we have

F = mg = 9.8 m s

* m = 3770 N = 3770 kg m s

So, the mass of weights needed is approximately 3770/9.8 = 385 kg

Solution 1.7 Problem Statement

Pressures up to 3000(atm) are measured with a dead-weight gauge. The piston diameter is 0.17(in). What is the approximate mass in (lbm) of the weights required?

Solution

Pressure is force divided by area. So, the maximum force required (to be applied by the weights) is the maximum pressure to be measured times the area of the piston. The piston area is 2

A = d /4 = *0.17 /4 in = 0.0227 in

The maximum pressure to be applied is 3000 atm = 3000 * 14.7 psi/atm = 44090 psi = 44090 lbf in . For the standard acceleration of gravity (g numerically equal to gc) one lbm imparts one lbf to the piston. Thus, the mass of weights 2

required is 0.0227 in * 44090 lbf in

* (g/gc = 1 lbf/lbm)

= 1001 lbm.

Solution 1.8 Problem Statement

The reading on a mercury manometer at 25°C (open to the atmosphere at one end) is 56.38 cm. The local acceleration 2

of gravity is 9.832 m·s . Atmospheric pressure is 101.78 kPa. What is the absolute pressure in kPa being measured? 3

The density of mercury at 25°C is 13.534 g·cm .

Solution

The total pressure is equal to atmospheric pressure plus the weight per area (gh) of the mercury column. This is, in 3

SI units, p = 101780 Pa + 13534 kg/m * 9.832 m s

* 0.5638 m = 176800 Pa = 176.8 kPa.

Solution 1.9 Problem Statement

The reading on a mercury manometer at 70(°F) (open to the atmosphere at one end) is 25.62(in). The local 2

acceleration of gravity is 32.243(ft)·(s) . Atmospheric pressure is 29.86(in Hg). What is the absolute pressure in (psia) 3

being measured? The density of mercury at 70(°F) is 13.543 g·cm .

Solution

The total pressure is equal to atmospheric pressure plus the weight per area ((g/gc)h) of the mercury column. The only trick is to do this in the non-SI units given. Let us compute the pressure in psia (pounds-force per square inch, absolute). To do so, we should convert the density of mercury from g cm Thus,  = 13.543 g cm

to lbm in .

* (2.54 cm/in) / 453.59 g/lb = 0.4893 lbm in . We also need to convert atmospheric pressure

from in Hg to psia by multiplying by 0.4912 psi/(in Hg).

So, p = 29.86 in Hg * 0.4912 lbf in 27.2 lbf in

(in Hg)

+ 0.4893 lbm in

* 32.243 ft s

/ (32.1740 lbm ft lbf

s ) * 25.62 in =

= 27.2 psia.

Solution 1.10 Problem Statement

An absolute pressure gauge is submerged 50 m (1979 inches) below the surface of the ocean and reads P = 6.064 bar. This is P = 2434(inches of H2O), according to the unit conversions built into a particular calculator. Explain the apparent discrepancy between the pressure measurement and the actual depth of submersion.

Solution

The pressure unit “inches of water” is defined as the pressure exerted by one inch of water under standard 3

gravitational conditions (density of water = 1 g/cm , g = 9.8 m/s ). The pressure exerted on the gauge is given by P = Patm + Here P is the absolute pressure at the position of the gauge, is the density of the water, g is the acceleration of gravity 2

(~9.8 m/s ), and h is the depth of the water (50 m). We can take Patm = 1.013 bar = 1.013×10 Pa. The pressure difference caused by the mass of the column of water above the pressure gauge (excluding atmospheric pressure) is 5

then P – Patm = 6.064 – 1.013 = 5.051 bar = 5.051×10 Pa. Converting that value to inches of water using my calculator gives 2028 inches of water. This is much closer to the actual depth, but still about 2.5% higher. This remaining 2

discrepancy can be attributed to the density of seawater. If we take h = 50 m and g = 9.8 m/s , then we can use our 5

pressure measurement to estimate the density of the seawater: = (P – Patm)/(gh) = 5.051×10 N/m /(9.8 m/s *50 m) = 2

1030.8 kg*m/(m s )/(m /s ) = 1030.8 kg/m . This is a typical value for the density of seawater, which is higher than the density of fresh water because of its dissolved salt content.

Solution 1.11 Problem Statement

Liquids that boil at relatively low temperatures are often stored as liquids under their vapor pressures, which at ambient temperature can be quite large. Thus, n-butane stored as a liquid/vapor system is at a pressure of 2.581 bar 3

for a temperature of 300 K. Large-scale storage (>50 m ) of this kind is sometimes done in spherical tanks. Suggest two reasons why.

Solution

The advantages of spherical tanks stem from the fact that a spherical container has the minimum surface area for a given interior volume. Because of this: (a) A minimum quantity of metal is required for tank construction (for a given metal thickness). (b) The tensile stress within the tank wall is everywhere uniform, with no sites of stress concentration, and the stress is purely tensile (everywhere tangential to the tank surface). Moreover, the maximum stress within the tank wall is kept to a minimum. (c) The surface area for heat transfer is minimized, optimizing the efficiency of insulating the tank.

Solution 1.12 Problem Statement

The first accurate measurements of the properties of high-pressure gases were made by E. H. Amagat in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mineshaft and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height of manometer required.

Solution

We know that 1 bar is equal to 750 torr or 750 mm of Hg or 0.75 meters of mercury. So, to measure a pressure of 400 bar, he needed a column of mercury 400 bar * 0.75 mHg/bar = 300 m of mercury.

Solution 1.13 Problem Statement

An instrument to measure the acceleration of gravity on Mars is constructed of a spring from which is suspended a 2

mass of 0.40 kg. At a place on earth where the local acceleration of gravity is 9.81 m·s , the spring extends 1.08 cm. When the instrument package is landed on Mars, it radios the information that the spring is extended 0.40 cm. What is the Martian acceleration of gravity?

Solution

The spring constant can be determined from a force balance on the mass on earth. The gravitational force is F = mg = 0.40 kg * 9.81 m s

= 3.924 kg m s

= 3.924 N.

The spring force is F = kx, where k is the spring constant and x is the distance it is extended from its equilibrium length. These forces must be equal, so we have kx = 3.924 N = k * 1.08 cm = k * 0.0108 m. So,

k = 3.924/0.0108 = 363.3 N/m.

Solution continued on next page…

Now, we do the same force balance on Mars, with g (for Mars) as the unknown. F = kx = 363.3 N/m * 0.40 cm = 363.3 N/m * 0.004 m = 1.453 N. and

F = 1.453 N = mg = 0.40 kg * g

So,

g = 3.63 N/kg = 3.63 m s .

Solution 1.14 Problem Statement The variation of fluid pressure with height is described by the differential equation: dP   g dz

Here, is specific density and g is the local acceleration of gravity. For an ideal gas, =

P/RT, where

is molar

mass and R is the universal gas constant. Modeling the atmosphere as an isothermal column of ideal gas at 10°C, estimate the ambient pressure in Denver, where z = 1(mile) relative to sea level. For air, take

= 29 g·mol ; values

of R are given in App. A.

Solution

Substituting the ideal gas expression for density into the equation for variation of fluid pressure with height gives dP P  g dz RT

If T is constant (along with molar mass, , and g) then this is a separable equation that can be rearranged to give: dP g  dz P RT

Solution continued on next page…

This can be directly integrated to give P g ln     z  zo    P  RT

Or  g  P  Po exp z  z o    RT  2

The standard pressure at sea level is Po = 101325 Pa. If Denver is 1 mile = 1609 m above sea level, g = 9.8 m s , 2

 = 0.029 kg/mol, T = 10 °C = 283 K, and R = 8.314 kg m mol

K , then

 0.029 * 9.8  P  101325 exp  1609  83430 Pa  8.314 * 283 

The estimated pressure is 83430 Pa = 0.834 bar = 0.823 atm.

Solution 1.15 Problem Statement A group of engineers has landed on the moon, and they wish to determine the mass of some rocks. They have a spring 2

scale calibrated to read pounds mass at a location where the acceleration of gravity is 32.186(ft)(s) . One of the moon rocks gives a reading of 18.76 on this scale. What is its mass? What is its weight on the moon? 2

Take g(moon) = 5.32(ft)(s) .

Solution

The force on the scale, whether on earth or on the moon, can be written at F = mg/gc, and it is the force that is actually measured by the scale. The reading (displayed in lbm) is therefore proportional to the actual mass and proportional to the gravitational constant. It can be written as mread = A mtrue, where A, the calibration constant for 2

the scale, is numerically equal to g/gc. On earth, where g = 32.186 ft/s , A = 1. On the moon, where g = 5.32 ft/s , A needs to be corrected by the ratio of g on the moon to g on earth. A = 5.32/32.186 = 0.1653. So, on the moon, mread = 0.1653*mtrue. The true mass of the rock is mtrue = mread/0.1653 = 18.76/0.1653 = 113.5 lbm. Its weight on the moon is numerically equal to the reading. It is 18.76 lbf.

Solution 1.16 Problem Statement In medical contexts, blood pressure is often given simply as numbers without units. (a) In taking blood pressure, what physical quantity is actually being measured? (b) What are the units in which blood pressure is typically reported? (c) Is the reported blood pressure an absolute pressure or a gauge pressure? (d) Suppose an ambitious zookeeper measures the blood pressure of a standing adult male giraffe (18 feet tall) in its front leg, just above the hoof, and in its neck, just below the jaw. By about how much are the two readings expected to differ? (e) What happens to the blood pressure in a giraffe’s neck when it stoops to drink? (f) What adaptations do giraffes have that allow them to accommodate pressure differences related to their height?

Solution

(a) “Blood pressure” refers to the pressure exerted by the blood on the interior walls of the major arteries. The higher number (systolic pressure) represents the maximum pressure during the cardiac cycle, and the lower number (diastolic pressure) represents the minimum pressure during the cardiac cycle. In certain situations where patients require continuous, accurate, and rapid blood pressure monitoring, it is measured directly. This can be done by simply inserting a cannula (big needle) into a large artery and connecting it to a pressure transducer. This is much the same way that one measures pressure in a piece of chemical process equipment. Ordinarily, the blood pressure is measured indirectly using a sphygmomanometer. This is the familiar device with the inflatable cuff. What is measured is the pressure inside the cuff that is needed to stop the flow of blood through the artery it is squeezing (usually the brachial artery in the upper arm). The systolic pressure is the lowest pressure at which flow is completely stopped throughout the cardiac cycle, and the diastolic pressure is the pressure at which the flow is not stopped at any point in the cardiac cycle. Blood flow is detected by a health care practitioner listening for it using a stethoscope, or by a small transducer built into the inflatable cuff. (b) Although they are often omitted, the units for blood pressure are mm of mercury. In SI units, a blood pressure of 120 over 80 would be a systolic pressure of 16000 Pa over 10700 Pa. (c) It is a gauge pressure, measured and reported as the difference in pressure between the pressure in the blood vessels and the ambient pressure. For example, when a scuba diver descends 30 feet below the surface, her blood pressure remains the same, even though the ambient pressure has roughly doubled from its value at the surface. (d) To a first approximation, we expect the pressure difference to just be the hydrostatic pressure difference due the difference in elevation between the hoof and the head. Taking this to be 17 feet (5.2 m) and taking the density of 3

giraffe blood to be 1000 kg/m , the pressure difference should be roughly P=

= 1000 kg/m * 9.8 m/s * 5.2 m = 51,000 Pa = 0.5 bar. The pressure in the giraffe’s hooves is about half an

atmosphere higher than that in its head (when it is standing). Solution continued on next page…

(e) It will fairly quickly increase by half an atmosphere, at least relative to the pressure in the giraffe’s hooves. The actual increase is not quite so big, because the giraffe’s heart moves to a slightly lower level when it stoops, and there accompanying changes in the giraffe’s pulse rate, etc. that have some effect. (f) The biggest problem would occur when the giraffe raises its head. Without special adaptations, the rapid change in pressure would drain all of the blood out of its head and it would faint (compare to the head rush you may sometimes get when changing the elevation of your head by just 3 feet or so). However, giraffes have a collapsible jugular vein and unusually strong ability to constrict the veins in their neck. This allows them to dramatically increase the flow resistance in their veins as they raise their heads, which prevents all of the blood from draining out. It seems that through differential contraction of different arterial paths, they can divert blood flow around the brain, or force a larger fraction through the brain. Overall, they have a much more sophisticated arrangement for controlling cranial blood pressure than most other animals.

Solution 1.17 Problem Statement A 70 W outdoor security light burns, on average, 10 hours a day. A new bulb costs $5.00, and the lifetime is about 1000 hours. If electricity costs $0.10 per kW·h, what is the yearly price of “security,” per light?

Solution

70 Watts times 10 hours per day is 700 Watt-hours per day or 0.7 kW-hour/day. Multiplying this by 365 days per year gives 255.5 kW-hours/year of electricity use at $0.10 per kW-hour, for a total annual electricity cost of $25.55. If the light is on 3650 hours per year, one will use an average of 3.65 bulbs (each lasting 1000 hours, or 100 days at 10 hours per day) each year, at an average annual cost of $5*3.65 = $18.25/year. So the total annual cost to operate the light is about $18.25 + $25.55 = $44. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 1.18

Problem Statement A gas is confined in a 1.25(ft) diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight 2

together is 250(lbm). The local acceleration of gravity is 32.169(ft)(s) , and atmospheric pressure is 30.12 (in Hg). (a) What is the force in (lbf) exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in (psia)? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 1.7(ft), what is the work done by the gas in (ft)(lbf)? What is the change in potential energy of the piston and weight?

Solution

(a) The force exerted on the gas has two parts, the weight of the piston and the pressure of the atmosphere on the top of the piston. The force due to the weight of the piston is mg/gc where m is the piston mass and g is the acceleration of gravity. We have to use gc to make the units work out right. The force due to the pressure of the atmosphere is PA where P is the pressure and A is the cross-sectional area of the piston. The cross-sectional area 2

of the piston is A = d /4 = 0.7854*(1.25 ft) = 1.227 ft . So, the total force is:

F = 250 lbm * 32.169 ft s

/ (32.1740 lbm ft lbf

+ 30.12 (in Hg) * 0.49116 lbf in

/ (in Hg) * 1.227 ft * 144 in ft

F = 249.96 lbf + 2613.85 lbf = 2863.8 lbf Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(b) The pressure of the gas is the force divided by the area 2

P = F / A = 2863.8 lbf / (1.227 ft * 144 in ft ) = 16.21 lbf in

= 16.21 psia

(c) Because the weight of the piston, its cross-sectional area, and the pressure of the atmosphere on it all remain constant, the force on the gas and the gas pressure also remain constant. So, the work done by the gas can be computed as either F h or P V, using the pressure and force that we have already computed. Because we are given h = 1.7 ft, it is easiest to calculate it as

W = 2863.8 lbf * 1.7 ft = 4868 ft lbf Most of this work is done against the pressure of the atmosphere, so only a small part of it changes the potential energy of the piston. The potential energy of the piston changes by mg/gch = 249.96 lbf * 1.7 ft = 424.9 ft lbf

Solution 1.19 Problem Statement

A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and 2

weight together is 150 kg. The local acceleration of gravity is 9.813 m·s , and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa?

(c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?

Solution

(a) The force applied by the atmosphere (if the piston and weight were ‘massless’) would be the atmospheric 2

pressure (101.57 kPa) times the cross-sectional area of the piston (A = d /4 = (0.47) /4 m = 0.173 m . So, the 2

force due to the atmosphere is 0.173 m * 1.0157  10 N m piston and weight is F = mg = 150 kg * 9.813 m s

= 17600 N. The additional force applied by the

= 1472 N. Thus, the total force applied to the gas is about

19090 N. (b) The pressure of the gas is the total force found in (a) divided by the cross sectional area of the piston. 2

P = 19090 N / 0.173 m = 1.101  10 N m

= 1.101  10 Pa = 110.1 kPa.

(c) Because the mass of the piston and the pressure of the atmosphere don’t change in this process, the pressure of the gas inside the piston is constant as it expands, and the force applied by the piston and atmosphere to the gas is constant as the gas expands. We can compute the work either as the pressure times the change in volume or the force times the distance the piston moves. Because we’ve already computed the force in part (a), the simplest method is to multiply this force by the distance that the piston travels: W = Fl = 19090 N * 0.83 m = 15850 N m = 15850 J. The change in potential energy of the piston and weight is:

EP = mgz = 150 kg * 9.813 m s

* 0.83 m = 1222 kg m s

= 1222 J.

Thus, most of the work done by the gas goes toward ‘pushing back’ the atmosphere, not raising the piston and weight. This is a result of the fact that most of the pressure being applied to the gas is due to the atmosphere, and not to the weight and piston.

Solution 1.20 Problem Statement

Verify that the SI unit of kinetic and potential energy is the joule.

Solution

Kinetic energy is given by EK = ½ mv . Its fundamental units are therefore the fundamental unit of mass (kg) times those of velocity (length/time, m/s) squared: 2

EK [=] kg·(m/s) [=] kg·m /s [=] N·m [=] J Gravitational potential energy (EP = mgz) has units of mass·length·acceleration. Its fundamental units are therefore: 2

EP [=] kg·m·(m/s ) [=] kg·m /s [=] N·m [=] J

Solution 1.21 Problem Statement

An automobile having a mass of 1250 kg is traveling at 40 m·s . What is its kinetic energy in kJ? How much work must be done to bring it to a stop?

Solution

The kinetic energy is KE = ½ mv = ½ * 1250 kg * (40 m/s) = 1000000 kg m s

= 1000000 J. To bring this car to a

stop, a minimum of 1000000 J of work must be done on the car.

Solution 1.22 Problem Statement

The turbines in a hydroelectric plant are fed by water falling from a 50 m height. Assuming 91% efficiency for conversion of potential to electrical energy, and 8% loss of the resulting power in transmission, what is the mass flow rate of water required to power a 200 W light bulb?

Solution

We can either work forward, computing the energy output per mass of water, or work backward, starting from the energy requirement of the lightbulb. If we choose to work forward, then we compute that the potential energy change, per kilogram of water that falls is EP/m = gz = 9.8 m s

*50 m = 490 m s

= 490 J kg . We are told that

91% of this energy is converted to electrical energy, and 92% of that electrical energy is transmitted to the lightbulb (8% is lost). So, the energy transmitted to the lightbulb is 490 J/kg * 0.91 * 0.92 = 410 J/kg. The 200-Watt light bulb requires 200 J per second. Dividing this by the energy production per kilogram gives:

Required water flow rate = 200 J s

/ 410 J/kg = 0.488 kg s

= 29.3 kg per minute = 64.5 lbm per minute.

To get more familiar volumetric units for water, we can divide by the density of water (about 8.3 lbm per gallon) to get 7.7 gallons per minute.

Solution 1.23 Problem Statement

A wind turbine with a rotor diameter of 77 m produces 1.5 MW of electrical power at a wind speed of 12 m⋅s . What fraction of the kinetic energy of the air passing through the turbine is converted to electrical power? You may assume a density of 1.25 kg⋅m

for air at the operating conditions.

Solution

The kinetic energy per unit mass of the wind approaching the turbine is ½u = ½ (12) = 72 m s

= 72 J/kg. The

volumetric flow rate of air approaching the turbine is given by the velocity times the cross-sectional area: 2

q = uA = 12 m/s *  (77 m) /4 = 55880 m /s. If the density of the air is 1.25 kg m , then the mass flow rate is:

   q  1.25 kg m 3 * 55880 m 3 s 1 = 69850 kg s 1 m 1

If this has a kinetic energy of 72 J kg , then the total power available is 72*69850 = 5.03  10 J/s = 5.03 MW. The efficiency of the turbine (electrical energy output divided by kinetic energy of the approaching air) is 1.5/5.03 = 0.30.

Solution 1.24

Problem Statement

The annual average insolation (energy of sunlight per unit area) striking a fixed solar panel in Buffalo, New York, is 2

200 W⋅m , while in Phoenix, Arizona, it is 270 W⋅m . In each location, the solar panel converts 15% of the incident energy into electricity. Average annual electricity use in Buffalo is 6000 kW⋅h at an average cost of $0.15 kW⋅h, while in Phoenix it is 11,000 kW⋅h at a cost of $0.09 kW⋅h. (a) In each city, what area of solar panel is needed to meet the average electrical needs of a residence? (b) In each city, what is the current average annual cost of electricity? (c) If the solar panel has a lifetime of 20 years, what price per square meter of solar panel can be justified in each location? Assume that future increases in electricity prices offset the cost of borrowing funds for the initial purchase, so that you need not consider the time value of money in this analysis.

Solution

(a) One way to figure this out is to find the annual average usage by taking the total annual usage (in kW-h) and dividing by the number of hours in a year to get an average usage in kW. For Buffalo, we have (6000 kW2

h/yr)/(8766 h/yr) = 0.684 kW. In Pheonix, it is 11000/8766 = 1.255 kW. In Buffalo, we get 0.15*200 = 30 W m of 2

solar panel, while in Phoenix we get 0.15*270 = 40.5 W m . The total area needed in Buffalo is therefore 684/30 2

= 22.8 m , while the total area needed in Pheonix is 1255/40.5 = 31.0 m . (b) In Buffalo, it is 6000 kW h * 0.15 $/(kW h) = $900 In Pheonix, it is 11000 kW h * 0.09 $/(kW h) = $990

Solution continued on next page…

(c) Neglecting the time value of money, we need the solar panels in each case to cost less than 20 years worth of electricity. Thus, in Buffalo, we require the total cost to be less than 20*$900 = $18000 and in Phoenix we require it be less than 20*$990 = $19800. Dividing these by the total area needed in each location gives: 2

Buffalo: $18000/22.8 m = $790 m 2

Phoenix: $19800/31.0 m = $639 m

This must include the total installed cost in each case, not just the panels themselves. Support structures, power inverters, etc. will be required. Sizing the panels based on average insolation assumes that electricity can be stored and retrieved for indefinite periods of time at no additional cost. This might be the case if the local electricity supplier allows the homeowner to feed power into the grid and be fully credited against any power drawn from the grid at other times (net metering).

Solution 1.25 Problem Statement Following is a list of approximate conversion factors, useful for “back-of-the-envelope” estimates. None of them is exact, but most are accurate to within about ±10%. Use Table A.1 (App. A) to establish the exact conversions.

∙ 1(atm)  1 bar ∙ 1(Btu)  1 kJ ∙ 1(hp)  0.75 kW ∙ 1(inch)  2.5 cm ∙ 1(lbm)  0.5 kg ∙ 1(mile)  1.6 km ∙ 1(quart)  1 liter ∙ 1(yard)  1 m

Add your own items to the list. The idea is to keep the conversion factors simple and easy to remember.

Solution

The exact conversion factors are: 1 atm = 1.01325 bar 1 Btu = 1.055 kJ 1 hp = 0.7457 kW 1 inch = 2.54 cm 1 lbm = 0.45359 kg 1 mile = 1.6093 km Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

1 quart = 0.9464 liter 1 yard = 0.9144 m Some other useful ones to remember are: 1 ft = 30.48 cm  30 cm = 0.3 m, 1 bar = 14.50 psi  15 psi, and 1 mile/hour = 0.447 m/s  0.5 m/s.

Solution 1.26 Problem Statement Consider the following proposal for a decimal calendar. The fundamental unit is the decimal year (Yr), equal to the number of conventional (SI) seconds required for the earth to complete a circuit of the sun. Other units are defined in the following table. Develop, where possible, factors for converting decimal calendar units to conventional calendar units. Discuss pros and cons of the proposal.

Solution

There is a little ambiguity in the number of seconds per year, because of leap year, occasional addition of “leapseconds”, etc. Solution continued on next page…

A reasonable number is 365.25 days*24 hours/day*60 min/hour*60 s/min = 31557600 s. Using this value, the above decimal calendar units are 1 Sc = 31.6 s = 0.53 minutes 1 Mn = 316 s = 5.26 minutes 1 Hr = 3156 s = 52.6 minutes = 0.877 hour 1 Dy = 526 minutes = 8.77 hour = 0.365 days 1 Wk = 87.7 hour = 3.65 days 1 Mo = 36.5 days

An inherent problem with this, or any other, decimal calendar is that a calendar needs to accommodate multiple “natural” time scales – at least the period of rotation of the earth, so that day/night cycles are reflected in the time units and the period of the earth’s orbit around the sun, so that seasonal changes are reflected in the time units. Such “natural” scales that govern our everyday experience are less prevalent for other dimensions, such as mass, length, etc.

Solution 1.27 Problem Statement Energy costs vary greatly with energy source: coal @ $35.00/ton, gasoline @ a pump price of $2.75/gal, and electricity 1

@ $0.100/kW·h. Conventional practice is to put these on a common basis by expressing them in $⋅GJ . For this purpose, assume gross heating values of 29 MJ⋅kg

for coal and 37 GJ⋅m

for gasoline. 1

(a) Rank order the three energy sources with respect to energy cost in $⋅GJ . (b) Explain the large disparity in the numerical results of part (a). Discuss the advantages and disadvantages of the three energy sources.

Solution

(a) The cost of coal per GJ is $25/ton / (907.2 kg/ton * 0.029 GJ/kg ) = $0.95 /GJ. 3

The cost of gasoline is $2/gal * 264.2 gal/m / 37 GJ/m = $14.30 /GJ. The cost of electricity is $0.1 /(kW hr) / 0.0036 GJ/(kW hr) = $27.80 /GJ. Coal is least expensive, electricity is most expensive.

(b) One can explain this in a variety of ways, but the underlying factor is that energy that is more convenient and/or more efficiently usable costs more. Electricity can be converted to mechanical energy (work) with nearly 100% efficiency (at least >90% in a well-designed electric motor). In contrast, a gasoline-burning internal combustion engine might have a typical efficiency of 30 to 35%. Thus, with the costs in part (a), and all other things being equal, an electric car might be more cost effective to operate than a gasoline-powered one, because the higher efficiency of energy use would more than make up for the higher cost per unit of energy input. This is what makes electric or plugin hybrid vehicles (potentially) economical. On the other hand, coal is inexpensive because it is very inconvenient to use. It cannot readily be used on a small scale, and is mostly consumed in large-scale industrial electricity generation and heating processes. A coal-fired power plant can convert about 1/3 of the energy of the coal into electricity. If that were the only cost of generating electricity, then with the costs in part (a), generating electricity from coal would be a very profitable business. However, the consumer electricity cost of $0.10/kW*hr reflects (along with profits) the costs of building the coal-fired power plant (although most US plants are old, and fully amortized) the cost of crushing and cleaning up the coal for combustion, the cost of pollution mitigation, and the cost of transmitting the power from the power plant to the customer. In mid-2020, the actual average coal cost for power generation was about $2 /GJ, which leads to an electricity generation cost of something like $0.03 per kW-hr, without including transmission costs, costs of pollution/cleanup, or capital costs of building the plant. However, coal cost varies widely by location and quality, from below $1 /GJ for low-quality coal in a remote location, to above $2 /GJ for higher-quality coal in a more desirable location.

Solution 1.28 Problem Statement Chemical-plant equipment costs rarely vary in proportion to size. In the simplest case, cost C varies with size S according to the allometric equation C

The size exponent is typically between 0 and 1. For a wide variety of equipment types it is approximately 0.6. (a) For

, show that cost per unit size decreases with increasing size. (“Economy of scale.”) t

(b) Consider the case of a spherical storage tank. The size is commonly measured by internal volume Vi Show why one might expect that = 2/3. On what parameters or properties would you expect quantity

to depend?

Solution

(a) The cost per unit size is C/S =

/S =

. If is less than 1, then the exponent ( – 1) is negative, and the cost

per size decreases with increasing size (S). (b) To a good first approximation, the cost of the tank should be proportional to the amount (mass) of material required to construct the tank. If the wall thickness, t, is independent of the tank volume, and is small compared to the tank diameter, then the mass of material required to make the tank will be m =

, where d is the diameter of the spherical tank and is the

density of the material from which it is constructed. t

The tank volume is Vt =

/6. t

1/3

Or, writing the diameter in terms of the volume, d = (6 Vt ) . Solution continued on next page…

Substituting this into the expression for the mass of material gives m=

2/3

(6 Vt / )

1/3

(36/ )

t 2/3

This shows that the exponent should be approximately = 2/3. It also shows that the cost should depend on the thickness of the tank walls, which will, in turn, depend on the maximum pressure that the tank must be able to withstand. Other factors that will influence the quantity should include the material of construction and the fabrication techniques used. The material of construction will depend on the need for corrosion resistance and other such factors.

Solution 1.29 Problem Statement sat

A laboratory reports the following vapor-pressure (P ) data for a particular organic chemical: t∕°C

Psat∕kPa

18.5 9.5 0.2 11.8 23.1 32.7 44.4 52.1 63.3 75.5

3.18 5.48 9.45 16.9 28.2 41.9 66.6 89.5 129. 187.

Correlate the data by fitting them to the Antoine equation:

ln P sat / kPa  A 

B T / K C

That is, find numerical values of parameters A, B, and C by an appropriate regression procedure. Discuss the comparison of correlated with experimental values. What is the predicted normal boiling point (i.e. temperature at which the vapor pressure is 1(atm)) of this chemical?

Solution

This can be done using many different software packages. The Microsoft Excel spreadsheet shown below illustrates one way of doing such nonlinear regression: A 12.74

B 2017.87

C -80.87

T (°C) -18.5 -9.5 0.2 11.8 23.1 32.7 44.4 52.1 63.3 75.5

T(K) 254.65 263.65 273.35 284.95 296.25 305.85 317.55 325.25 336.45 348.65

Psat (kPa)

ln (Psat)

ln (Psat)fit

Error2

Psatfit (kPa)

3.18 5.48 9.45 16.9 28.2 41.9 66.6 89.5 129 187

1.157 1.701 2.246 2.827 3.339 3.735 4.199 4.494 4.860 5.231

1.125 1.697 2.253 2.849 3.368 3.767 4.211 4.479 4.841 5.201

1.02E-03 2.03E-05 4.77E-05 4.61E-04 7.96E-04 1.02E-03 1.43E-04 2.24E-04 3.50E-04 9.19E-04

3.08 5.46 9.52 17.27 29.01 43.26 67.40 88.17 126.61 181.42

Error Sum

5.01E-03

6.000

200 180

5.000

160

4.000

Pressure(kPa)

ln(Pressure(kPa))

Relative Error -3.2% -0.4% 0.7% 2.2% 2.9% 3.3% 1.2% -1.5% -1.9% -3.0%

3.000 2.000

140 120 100 80 60 40

1.000

20 0.000

0 200

250

300 Temperature (K)

350

400

-50

0 50 Temperature (°C)

100

Solution continued on next page…

The first and third columns have the data given in the problem statement. The second column has the temperature in th

K, and the 4 column has the natural logarithm of the vapor pressure data. The fifth column has the result of lnP sat / kPa  A 

B T /K C

With parameters A, B, and C as shown at the top of the spreadsheet. The sixth column has the square of the difference sat

between the actual and model values of ln(P ) (column 5 – column 4) . The cell labeled “Error Sum” has the sum of those differences. The Solver function in MS Excel was used to minimize that cell (the sum of the squares of the differences between the data and fit values) by simultaneously varying the values of A, B, and C.

The 6 column has the vapor pressure predicted by the fit, and the 7 column has the relative error in the vapor pressure at each point, which is (fit value – actual value)/actual value. The two plots show the data and fit in two formats, in each case with the data as discrete points and the fit as a smooth curve. The one on the left is what is actually used in the fitting. That is, the sum of the squares of the distances between the points and the curve was minimized to obtain the parameters. The one on the right shows the data and fit in the original format in which the data was given.

The boiling point is the temperature at which P

sat

= 101.325 kPa. If one already has the spreadsheet shown above,

then the simplest way to determine this is to vary one of the temperatures until the fitted pressure is equal to 101.325 kPa. This gives a value of 56.3 °C.

Solution 2.1 Problem Statement

A nonconducting container filled with 25 kg of water at 20°C is fitted with a stirrer, which is made to turn by gravity acting on a weight of mass 35 kg. The weight falls slowly through a distance of 5 m in driving the stirrer. Assuming 2

that all work done on the weight is transferred to the water and that the local acceleration of gravity is 9.8 m·s , determine: (a) The amount of work done on the water. (b) The internal energy change of the water. 1

Solution

(a) The work done on the water is equal to the work done by gravity on the weight, which is equal to force times distance. The force is mg = 35 kg·9.8 m s

= 343 N.

The distance is 5 m, so the work is 343 N·5 m = 1715 N m = 1715 J. (b) The internal energy change is equal to the work done, 1715 J. Solution continued on next page…

(U = mCpT) or 1715 J = 25 kg·4180 J kg

K ·T K, from which

T = 1715/(25·4180) = 0.016 K So, the final temperature of the water is 20.02°C (the initial temperature was 20°C). (d) To return the water to its original state, one must remove 1715 J as heat. (e) Zero, zero, zero! The total energy change of the universe is always zero (in the absence of nuclear reactions).

Solution 2.2 Problem Statement Rework Prob. 2.1 for an insulated container that changes in temperature along with the water and has a heat capacity equivalent to 5 kg of water. Work the problem with:

(a) The water and container as the system. (b) The water alone as the system. Problem 2.1 A nonconducting container filled with 25 kg of water at 20°C is fitted with a stirrer, which is made to turn by gravity acting on a weight of mass 35 kg. The weight falls slowly through a distance of 5 m in driving the stirrer. Assuming 2

(c) The final temperature of the water, for which CP = 4.18 kJ·kg ·°C . (d) The amount of heat that must be removed from the water to return it to its initial temperature. (e) The total energy change of the universe because of (1) the process of lowering the weight, (2) the process of cooling the water back to its initial temperature, and (3) both processes together. Solution (a) The work done on the water is equal to the work done by gravity on the weight, which is equal to force times distance. The force is mg = 35 kg·9.8 m s

= 343 N. The distance is 5 m, so the work is 343 N·5 m = 1715 N m =

1715 J. (b) The internal energy change of the system (water plus container) is equal to the work done, 1715 J. However, not all of this is change in the internal energy of the water. The heat capacity of the water is (25/30) = 5/6 of the total heat capacity. Thus, when the temperature of the water and container both increase, 5/6 of the energy goes into the water, and 1/6 goes into the container. So, the internal energy of the water increases by 5/6·1715 = 1429 J. The internal energy of the container changes by 286 J. (c) The internal energy change (1715 J) is equal to the total heat capacity times the temperature change t

(U = mCpT) or 1715 J = 30 kg·4180 J kg

·T K, from which

T = 1715/(30·4180) = 0.0137 K So, the final temperature of the water is 20.014°C (the initial temperature was 20 °C). (d) To return the water to its original state, one must remove 1715 J as heat (assuming the container as well as the water cools down). If only the water cools down, and not the container, then only 1429 J must be removed. (e) Zero, zero, zero! The total energy change of the universe is always zero (in the absence of nuclear reactions, of course).

Solution 2.3 Problem Statement

An egg, initially at rest, is dropped onto a concrete surface and breaks. With the egg treated as the system, (a) What is the sign of W? (b

EP ?

E K?

(e) What is the sign of Q? In modeling this process, assume the passage of sufficient time for the broken egg to return to its initial temperature. What is the origin of the heat transfer of part (e)?

Solution

(a) W is positive – work is done by gravity on the egg

(b) EP is negative – the potential energy of the egg decreases when it falls.

(c) EK is zero – the egg starts out with zero velocity (and therefore zero kinetic energy, relative to the earth) and ends with zero velocity. t

(d) If the egg returns to its original temperature, then U for the egg is zero. U is a state function, so if the egg starts and ends at the same T and P (and there are no chemical changes in the egg), then U doesn’t change. (e) Q is negative – the work done on the egg is converted to heat when the egg (irreversibly) breaks on the pavement. t

If the egg returns to its original temperature, this heat must all flow into the concrete, so that U = Q + W = 0. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 2.4 Problem Statement An electric motor under steady load draws 9.7 amperes at 110 volts, delivering 1.25(hp) of mechanical energy. What is the rate of heat transfer from the motor, in kW?

Solution At steady state, the electrical power input to the motor has to be balanced by the work and heat output of the motor. At steady-state, for a closed system, the internal energy is constant: dU t  Q  W  0 dt

The work is equal to the difference between the electrical work done on the system (positive) and the mechanical work done by the system (negative). Thinking way, way, back to first-year physics, we remember that electrical power (work per time) is the product of current and voltage (P = IV), and that the product of volts times amps is equal to watts. Thus, the energy input to the motor (electrical work) is P = 9.7 A·110 V = 1067 V·A = 1067 W. The work output is 1.25 hp, and from table A1 we see that 1 kW = 1000 W = 1.341 hp. So, the work output is 1.25 hp/ (1.341 kW/hp) = 0.932 kW = 932 W. The heat transfer rate (the energy removed from the motor as heat) is equal to the difference between the electrical energy input and the mechanical work output:

  (1067  932)  135 W Q  W

135 W is removed from the motor as heat.

Solution 2.5 Problem Statement An electric hand mixer draws 1.5 amperes at 110 volts. It is used to mix 1 kg of cookie dough for 5 minutes. After mixing, the temperature of the cookie dough is found to have increased by 5°C. If the heat capacity of the dough is 1

4.2 kJ⋅kg ⋅K , what fraction of the electrical energy used by the mixer is converted to internal energy of the dough? Discuss the fate of the remainder of the energy.

Solution The electrical power supplied to the mixer is 1.5 A·110 V = 165 V·A = 165 W. The total energy supplied over a period of 5 minutes is then 5·60·165 = 49500 J = 49.5 kJ. The increase in internal energy of the cookie dough is given by t

U = m Cv T = 1 kg·4.2 kJ·kg ·K ·5 °C = 21 kJ. Thus, the fraction of the electrical energy that went into heating the cookie dough is 21/49.5 = 0.42; 42% of the energy input has appeared as increased internal energy of the cookie dough. The rest of the energy input has been transferred as heat to the surroundings, or remains as internal energy of the electric mixer, which is probably also at a higher temperature at the end of the process than at the beginning. From the data given, we know the work input only if the system includes the mixer itself (because we know the current and voltage supplied to it, from which we compute electrical work). Taking the mixer AND the dough as the system, we know W = 49.5 kJ, but we do not separately know Q, the amount of heat transferred between the system U, the change in internal energy of the mixer and dough together. t

U = Q +W = 21 kJ, but we do not know Q and W individually. Here the work done by the mixer on the dough should be some amount smaller than the electrical work provided to the mixer, because the mixer must have some inherent irreversibilities that prevent it from fully converting the incoming electrical work into mechanical work done on the dough. From the given data, we cannot evaluate all terms in the energy balance for either the dough alone or the mixer and dough together as the system. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 2.6 Problem Statement One mole of gas in a closed system undergoes a four-step thermodynamic cycle. Use the data given in the following table to determine numerical values for the missing quantities indicated by question marks.

Step 12 23 34 41

? ?

12341

Ut/J

Q/J

W/J

200

6000 3800 800

4700

? 300 ?

1400

Solution t

For this problem, we must use two principles: U = Q + W , which gives us the 3 number in any row in which we t

already have 2 numbers, and U is a state function, so its value doesn’t change in a cyclic processl i.e., if we end up at t

the same conditions as where we started, U will return to its initial value.

Step

U (J)

200

Q (J)

W (J)

6000

3800

800

300

4700

12341

? 1400

So, we just go through and fill in the numbers by adding and subtracting things. Q for the first step must be Q = U

W = 5800 J. Likewise, U = Q + W = 500 for the third step. Also, U for the whole cycle is 0, as stated

above. Filling these in gives Step

U (J)

Q (J)

W (J)

200

5800

6000

500

4700

12341

3800

800

300 ? 1400

Now, we can fill in Q = 1400 J for the overall cycle (so Q + W = U ). Then we know all but one number in the t

columns for U and Q, and the sum of the 4 steps in each case has to add up to the value for the overall cycle, so we t

have U 23 = 0 + 200 + 500

4700 = 4000 J and Q41 = 1400 – 5800 + 3800 + 800 = 200. Filling these in gives: Step

U (J)

Q (J)

W (J)

200

5800

6000

4000

3800

500

800

300

4700

200

12341

1400

? 1400

Finally, we now know two of the three entries in the 2 t

case, we have W = U

and 4 rows, so we can fill in these last two numbers. In each

Q, which gives W = 200 J for step 23 and W = 4500 J for step 41. Filling these in completes

the table: Step

U (J)

Q (J)

W (J)

200

5800

6000

4000

3800

500

800

4700

200

12341

1400

200 300 4500 1400

As a check, we can add up the work values for the four steps and make sure the sum comes out to be

Solution 2.7 Problem Statement Comment on the feasibility of cooling your kitchen in the summer by opening the door to the electrically powered refrigerator.

Solution If the refrigerator is inside the kitchen, then the electrical energy entering the refrigerator (from outside the kitchen) must inevitably appear in the kitchen. The only mechanism is by heat transfer (from the condenser of the refrigerator, usually located behind the unit or in its walls). This raises, rather than lowers, the temperature of the kitchen. The only way to make the refrigerator double as an air conditioner is to place the condenser of the refrigerator outside the kitchen (outdoors). If, for example, one mounted the refrigerator (an older model with exposed coils on the back) in an exterior doorway, sealed the area around it, and left the refrigerator door open, one would have the (bulky, expensive, and relatively inefficient) equivalent of a window air conditioner.

Solution 2.8 Problem Statement A tank containing 20 kg of water at 20°C is fitted with a stirrer that delivers work to the water at the rate of 0.25 kW. How long does it take for the temperature of the water to rise to 30°C if no heat is lost from the water? For water, 1

CP = 4.18 kJ⋅kg ⋅°C .

Solution The total heat capacity of the water is 20 kg·4.18 kJ/(kg·C) = 83.6 kJ/C = 83600 J/C. So, to increase the temperature by 10C requires an energy input of 836 kJ. If we are doing work on the water at a rate of 0.25 kW = 0.25 kJ/s, then we will have to do so for 836/0.25 = 3344 s = 55.7 minutes = 0.929 hr.

Solution 2.9 Problem Statement Heat in the amount of 7.5 kJ is added to a closed system while its internal energy decreases by 12 kJ. How much energy is transferred as work? For a process causing the same change of state but for which the work is zero, how much heat is transferred?

Solution The first law of thermodynamics tells us that t

U = Q + W

So, if Q = 7.5 kJ and U = 12 kJ, we see that W = 19.5 kJ. That is, 19.5 kJ was transferred from the system to the surroundings as work. t

If the same change of state occurred (which tells us that U was the same), but no work was done (W = 0) we would have 12 kJ = Q + 0. That is, if no work was done, then 12 kJ would have to be removed from the system to cause the same change in the state of the system.

Solution 2.10 Problem Statement A steel casting weighing 2 kg has an initial temperature of 500°C; 40 kg of water initially at 25°C is contained in a perfectly insulated steel tank weighing 5 kg. The casting is immersed in the water and the system is allowed to come to equilibrium. What is its final temperature? Ignore the effects of expansion or contraction, and assume constant 1

specific heats of 4.18 kJ⋅kg ⋅K

for water and 0.50 kJ⋅kg ⋅K

for steel.

Solution An energy balance shows us that the internal energy decrease of the casting must equal the internal energy increase of the water and steel tank, because all of the heat that flows out of the casting flows into the water and tank (and none leaves the overall system, which is described as perfectly insulated). So, we have

mcasting C p, casting (Tcasting  Tfinal )  mwater C p, water (Tfinal  Twater )  mtank C p, tank (Tfinal  Ttank )

2  0.5  (500  Tfinal )  (40  4.18  5  0.5)(Tfinal  25) (1  2.5  167.2)Tfinal  500  4180  62.5 Tfinal  27.8C

Solution 2.11 Problem Statement An incompressible fluid ( = constant) is contained in an insulated cylinder fitted with a frictionless piston. Can energy as work be transferred to the fluid? What is the change in internal energy of the fluid when the pressure is increased from P1 to P2?

Solution If there is no change in volume of the fluid, no work can be done. Reversible work is dW = PdV. As long as V doesn't change, it doesn't matter what P does. If the work is zero and (because the piston is insulated) the heat transfer is also zero, the change in internal energy of the fluid is also zero. The internal energy of an incompressible fluid is independent of pressure. Note that if a mechanism for doing irreversible work were available (e.g., a stir-bar inside the fluid) then energy could be transferred to the fluid, but no such mechanism is available in the system as described.

Solution 2.12 Problem Statement 1

One kg of liquid water at 25°C, for which CP = 4.18 kJ·kg ·°C : t

U , in kJ?

U for part (a). What is

z, in meters? (c) Is accelerated from rest to final velocity u

U for part (a). What is

u, in m·s ? Compare and discuss the results of the three preceding parts.

Solution In the previous problems, the heat capacity of water was given as 4.18 kJ kg

K . For liquids, Cp and Cv are very

nearly the same, so it doesn’t matter whether this change occurs at constant P or at constant V. The change in internal energy is thus t

U = mCpT = 1 kg·4.18 kJ kg

K ·1 K = 4.18 kJ

(a) The change in gravitational potential energy is

EP = mgz Solution continued on next page…

Thus, to have the same change in potential energy as the change in internal energy in part (a), we need 2

4180 J = 1 kg·9.8 m s ·z

from which z = 427 m (or 1400 feet) 2

(b) This time, we need EK = ½mv = 4180 J, from which v = 8360 (m/s) or v = 91.4 m/s (or 205 miles per hour). What have we learned? We have discovered again that what corresponds, in our everyday experience, to a small amount of thermal energy corresponds to a relatively large amount of mechanical energy. The amount of energy required to heat a quart of water (1 kg of water is about 1 liter or 1 quart) from room temperature to its boiling point (increasing the temperature by about 75 K) would be enough to raise the water to a height of about 20 miles (in a frictionless world, or one in which we lifted the object very slowly so that the drag of the air on it was negligible).

Solution 2.13 Problem Statement An electric motor runs “hot” under load, owing to internal irreversibilities. It has been suggested that the associated energy loss be minimized by thermally insulating the motor casing. Comment critically on this suggestion.

Solution Electrical and mechanical irreversibilities cause an increase in the internal energy, and therefore the temperature, of the motor (they convert some of the input electrical energy to heat rather than mechanical work). The temperature of the motor rises until a steady state is established such that heat transfer from the motor to the surroundings is equal to the rate at which heat is produced within the motor by these irreversibilities. Insulating the motor does nothing to decrease the irreversibilities in the motor and merely causes the temperature of the motor to rise until steady-state heat-transfer is reestablished with the surroundings. Insulating the motor will just make it run hotter and probably damage it. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 2.14 Problem Statement A hydroturbine operates with a head of 50 m of water. Inlet and outlet conduits are 2 m in diameter. Estimate the 1

mechanical power developed by the turbine for an outlet velocity of 5 m⋅s .

Solution A general energy balance on the hydroturbine can be written as fs





d (mU ) cv   ( H  1 u 2  zg )m  Q  W 2 fs dt

The system can reasonably be assumed to operate at steady-state, so the first term is zero. The change in kinetic energy will also be negligible, because the inlet and outlet pipes have the same diameter, and therefore the water will have the same velocity at the inlet and outlet. A somewhat stronger assumption is that the change in enthalpy of the water is negligible. All real processes, including hydroturbines, have some degree of irreversibility. However, it turns out that a well-designed hydroturbine can be quite efficient, and can convert more than 90% of the potential energy of the water into mechanical energy (to be used to drive a generator, which will have its own inefficiencies, so that the final electrical energy output will be smaller). Thus, for the hydroturbine, it is reasonable to neglect the change in enthalpy of the water, as well as heat flow to or from the system. Doing so gives simply   mg  z W  mgz

This will be an upper limit for the mechanical power (rate of work) produced by the turbine. To compute it, we must know the mass flow rate of the water through the turbine (which is the same at the inlet and outlet, by a trivial mass 2

balance). The cross-

·(1 m) = 3.14 m . Multiplying this by the 1

water velocity of 5 m s gives a volumetric flow rate of 15.7 m /s. Multiplying this by the density of water (1000 kg/m ) gives the mass flow rate as 15,700 kg/s. Thus, the mechanical work is estimated as Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

W  15700 kg/s  9.8 m/s2  50 m  7.7  10 6 kg m 2 s3  7.7 MW

A crude estimate is that the average power consumption rate of a U.S. household is around 1 kW. Allowing for an overall efficiency (turbine plus generator) of 85%, this is still enough to power about 6500 households.

Solution 2.15 Problem Statement 1

A wind turbine with a rotor diameter of 40 m produces 90 kW of electrical power when the wind speed is 8 m⋅s . The 3

density of air impinging on the turbine is 1.2 kg⋅m . What fraction of the kinetic energy of the wind impinging on the turbine is converted to electrical energy?

Solution If we take a cylinder of 40 m diameter and wind speed of 8 m/s, then the volumetric flow rate of the air impinging on 2

the turbine is 8 m/s⋅(40 m) /4⋅ = 10050 m /s. The corresponding mass flow rate is 3

10050 m /s⋅1.2 kg/m = 12060 kg/s. The kinetic energy per unit mass is 8 /2 = 32 m /s = 32 J/kg. Thus, the total kinetic energy per time is 12060 kg/s⋅32 J/kg = 386000 J/s = 386 kW. The fraction of this converted to electrical power is 90/368 = 0.233. That is, 23.3% of the kinetic energy of the oncoming wind is captured.

Solution 2.16 Problem Statement The battery in a laptop computer supplies 11.1 V and has a capacity of 56 W⋅h. In ordinary use, it is discharged after 4 hours. What is the average current drawn by the laptop, and what is the average rate of heat dissipation from it? You may assume that the temperature of the computer remains constant.

Solution If the total capacity of the battery is 56 Watt-hours, and it is drained in 4 hours, then the average discharge rate is 56/4 = 14 W, or 14 J/s. Remembering that power is current times voltage, the average current drawn by the laptop is 14 W/11.1 V = 1.26 A. Taking the computer (not including the battery) as the system, and neglecting any change in internal energy of the computer, all of the electrical work supplied to the computer by the battery (at an average rate of 24.4 J/s) is ultimately converted to heat. That is, we can write the first law as dU t  Q  W  0 dt

If U remains constant, then Q  W  14 W . When the computer is at steady state (not changing temperature) 14 W must be dissipated into the surroundings as heat. If we instead considered a system that includes the battery as well as the rest of the computer, then the work would be zero (the battery is now part of the system, rather than the surroundings). We could write the electrical potential energy of the battery (when fully charged) as

EEP = 56 W-h Solution continued on next page…

Thus, the total energy of the system would include this as well as the internal energy of the computer and we could write the first law as t

U + EEP) = Q + W

Taking W

EEP = 56 W-h (when the battery is fully discharged, EEP = 0), we have 56 W-h = Q.

The total heat removal is 56 W-h over a period of 4 hours, so the average rate of heat flow is Q  56/4  14 W .

Solution 2.17 Problem Statement Suppose that the laptop of Prob. 2.16 is placed in an insulating briefcase with a fully charged battery, but it does not go into “sleep” mode, and the battery discharges as if the laptop were in use. If no heat leaves the briefcase, the heat capacity of the briefcase itself is negligible, and the laptop has a mass of 2.3 kg and an average specific heat of 1

0.8 kJ⋅kg ⋅°C , estimate the temperature of the laptop after the battery has fully discharged.

Problem 2.16 The battery in a laptop computer supplies 11.1 V and has a capacity of 56 W⋅h. In ordinary use, it is discharged after 4 hours. What is the average current drawn by the laptop, and what is the average rate of heat dissipation from it? You may assume that the temperature of the computer remains constant.

Solution If we take the laptop and bag as the system, then there is no exchange of heat or work between the system and t

U = 0. The internal energy does not change, but the chemical energy stored in the battery is released as sensible heat, which raises the temperature of the computer. The total amount of energy stored in the battery is 56 W·h = 56 J/s·h·3600 s/h = 201600 J = 201.6 kJ. The total heat capacity of the laptop is 2.3·0.8 = 1.84 kJ/K. Thus, energy release of 201.6 kJ would increase the temperature by 201.6 kJ/1.84 kJ/K = 110 K = 110°C (because this is a temperature difference, and not an absolute temperature, units of K and °C are the same). If the computer started off at a cool 20°C, the estimated final temperature is 130°C (about 270°F). At that point, the foam briefcase would probably start melting onto the surface of the laptop. Fortunately, in real life, this is unlikely to happen, and even if it did, the bag would not be perfectly insulating. Nonetheless, a typical laptop battery does store sufficient energy to cook the laptop in which it is installed if no heat could be removed.

Solution 2.18 Problem Statement In addition to heat and work flows, energy can be transferred as light, as in a photovoltaic device (solar cell). The energy content of light depends on both its wavelength (color) and its intensity. When sunlight impinges on a solar cell, some is reflected, some is absorbed and converted to electrical work, and some is absorbed and converted to heat. 2

Consider an array of solar cells with an area of 3 m . The power of sunlight impinging upon it is 1 kW⋅m . The array converts 17% of the incident power to electrical work, and it reflects 20% of the incident light. At steady state, what is the rate of heat removal from the solar cell array?

Solution At steady state, the net energy flow into the solar panel is zero. The rate of energy transfer to the panel as sunlight is 2

0.8⋅1 kW/m ⋅3 m = 2.4 kW (80% of the total, because 20% is reflected). The rate at which energy leaves as electrical 2

work is 0.17⋅1 kW/m ⋅3 m = 0.51 kW. At steady state, the remaining energy must leave as heat. Written as an energy balance, t

U = 0 = 2.4 kW

0.51 kW + Q = 0 from which Q = 1.89 kW. The rate of heat removal is 1.89 kW. Because solar

cells inherently convert a minority of the absorbed energy into electrical work, they inevitably require removal of substantial amounts of energy as heat. This can lead them to operate at high temperature, which may further reduce their efficiency.

Solution 2.19 Problem Statement

Liquid water at 180°C and 1002.7 kPa has an internal energy (on an arbitrary scale) of 762.0 kJ⋅kg 3

and a specific

volume of 1.128 cm ⋅g . (a) What is its enthalpy?

(b) The water is brought to the vapor state at 300°C and 1500 kPa, where its internal energy is 2784.4 kJ⋅kg 3

specific volume is 169.7 cm ⋅g

and its

H for the process.

Solution 3

(a) By definition, H = U + PV, so H = 762.0 kJ/kg + 1002.7 kPa⋅0.001128 m /kg = 763.1 kJ/kg.

(Note that we converted the specific volume to m per kg, and that Pa⋅m = N/m ⋅m = N⋅m = J, 3

so kPa⋅m = kJ).

(b) U = 2784.4

762 = 2022.4 kJ/kg. The enthalpy of the vapor is 3

H = 2784.4 kJ/kg + 1500 kPa⋅0.1697 m /kg = 3039.0 kJ/kg. So,

H = 3039.0 – 763.1 = 2275.9 kJ/kg. H is significantly larger than U because of the large increase in specific volume during the process.

Solution 2.20 Problem Statement

A solid body at initial temperature T0 is immersed in a bath of water at initial temperature Tw0. Heat is transferred from the solid to the water at a rate Q  K  (Tw – T ) , where K is a constant and Tw and T are instantaneous values of the temperatures of the water and solid. Develop an expression for T as a function of time . Check your result for the limiting cases, = 0 and = ∞. Ignore effects of expansion or contraction, and assume constant specific heats for both water and solid.

Solution

Let symbols without subscripts refer to the solid and symbols with subscript w refer to the water. Heat transfer from the solid to the water results in changes in internal energy of both. Because energy is conserved and there are no t

U = t

U w. If total heat capacity of the solid

is C (= mC) and total heat capacity of the water is C w (= mwCw), then: t

U = C (T

T0) = - U w = C w (Tw

Tw0)

Tw  Tw 0 

Ct (T  T0 ) Cwt

This equation relates instantaneous values of Tw and T. The heat-transfer rate is given as Q = K(Tw

T ). Thus, writing an energy balance (the first law) with the solid as the

system and W = 0, yields dU t  Q  K (Tw  T ) dt

This change in internal energy is related to the change in temperature by the heat capacity: dU t dT  Ct dt dt

Combining these equations, we have

dT  K (Tw  T ) dt

Solution continued on next page…

Substituting into this our expression for Tw in terms of T gives a single equation for T that is separable:

  dT Ct  K Tw 0  t (T  T0 )  T   dt Cw 

T 1 T dT 1     K  wt0  0t   t  t T   C dt Cw  C Cw  

dT  Kdt T 1    w 0 T0  1  t  t   t  t T  Cw  C Cw   C

Integrating from t = 0 (where T = T0) to t gives: T 1    w 0 T0  1    T   t C Cwt  C t C wt   1    Kt ln 1        1   Tw 0  T0   1  1 T   0  C t C t   C t Cwt  C t Cwt     w

Tw 0 T0  1 1      T Ct C wt  C t Cwt  Tw 0  T0 Ct

 1 1    expKt  t  t   Cw   C

 1  Tw 0 T0  1 Tw 0  T0 1     1      T  exp  Kt    C t C t  Ct Cwt  C t Cwt  Ct   w   1 1   T0C t  Cwt Tw 0  (C t  C wt )T  C wt (Tw 0  T0 ) exp Kt  t  t  Cw   C 

T

 1 1   T0C t  C wt Tw 0  Cwt (Tw 0  T0 ) exp Kt  t  t  C w   C  C t  Cwt

Solution 2.21 Problem Statement A list of common unit operations follows: (a) Single-pipe heat exchanger (b) Double-pipe heat exchanger (c) Pump (d) Gas compressor (e) Gas turbine (f) Throttle valve (g) Nozzle Develop a simplified form of the general steady-state energy balance appropriate for each operation. State carefully, and justify, any assumptions you make.

Solution

The general equation applicable here is Eq. (2.29): [( H  1 u2  gz ) m ] fs  Q  W s s 2

(a) We write this for the single stream flowing within the pipe, neglect potential and kinetic energy changes, and set the work term equal to zero. This yields:  )  m H  m ( H  (mH

out

 H )  Q in

In the second and third versions of the LHS, we have used the fact that there is a single inlet and outlet, and that mass conservation requires that the mass flow rate at the inlet and outlet are equal. Solution continued on next page…

(b) The equation is here written for the two streams (1 and 2) flowing in the two pipes, again neglecting any potential- and kinetic-energy changes. There is no work, and the heat transfer is internal to the system, between the two streams, making

= 0. Thus,

 )  m 1H1  m 2H 2  0 (mH

(c) For a pump operating on a single liquid stream, the assumption of negligible potential- and kinetic energy changes is reasonable, as is the assumption of negligible heat transfer to the surroundings. Whence,  )  m H  m ( H out  Hin )  W (mH

(d) For a properly designed gas compressor the result is the same as in Part (c).  )  m H  m ( H out  Hin )  W (mH

(e) For a properly designed turbine the result is the same as in Part (c).  )  m H  m ( H out  Hin )  W (mH

(f) The purpose of a throttle is to reduce the pressure on a flowing stream. One usually assumes adiabatic operation with negligible potential- and kinetic-energy changes. Because there is no work, the equation is:  )  m H  m ( H out  Hin )  0 (mH

(g) The whole purpose of a nozzle is to produce a stream of high velocity. The kinetic-energy change must therefore be taken into account. However, one usually assumes negligible potential-energy change. Then, for a single stream, adiabatic operation, and no work:   ( H  1 u2 )m   0 2  

Or ( H  1 u2 ) m  0 2 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The usual case is for a negligible inlet velocity. The equation then reduces to: H   1 u22 2

where u2 is the exit velocity.

Solution 2.22 Problem Statement The Reynolds number Re is a dimensionless group that characterizes the intensity of a flow. For large Re, a flow is turbulent; for small Re, it is laminar. For pipe flow, Re ≡

/ , where D is pipe diameter and

(a) If D and

 on Re? are fixed, what is the effect of increasing mass flow rate m

 and (b) If m

are fixed, what is the effect of increasing D on Re?

is dynamic viscosity.

Solution

(a) The mass flowrate, m  u

, is related to the density, diameter, and average velocity by

D2 4

Thus, uD 

4m D

And Re 

uD 4m    D

Thus, the Reynolds number is directly proportional to the mass flow rate. (b) The expression in part (a) shows that in this case the Reynolds number is inversely proportional to the diameter, D.

Solution 2.23 Problem Statement An incompressible ( = constant) liquid flows steadily through a conduit of circular cross-section and increasing 1

diameter. At location 1, the diameter is 2.5 cm and the velocity is 2 m⋅s ; at location 2, the diameter is 5 cm. (a) What is the velocity at location 2? 1

(b) What is the kinetic-energy change (J⋅kg ) of the fluid between locations 1 and 2?

Solution

(a) For an incompressible fluid flowing steadily, a mass balance reduces to flow in = flow out, or uinAin = uoutAout, or uinAin = uoutAout or uout = uin(Ain/Aout). In this case, the diameter increases by a factor of two, so the cross1

sectional area increases by a factor of 4, so Ain/Aout = 1/4, and uout = 2 m s /4 = 0.5 m s . 2

(b) The change in kinetic energy (per unit mass) is just the change in ½ u 2

(½ u ) = ½·(0.25 m s

4 m s ) = 1.875 m s

= 1.875 J kg .

Solution 2.24 Problem Statement

A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg·s with 0.8 kg·s

of cool water at 25°C 1

of hot water at 75°C. During mixing, heat is lost to the surroundings at the rate of 30 kJ·s . What is 1

the temperature of the warm water stream? Assume the specific heat of water is constant at 4.18 kJ·kg ·K .

Solution

Here it may be helpful to draw a very simple schematic:

1.8 kg/s at ?? °C

1 kg/s at 25°C 0.8 kg/s at 75°C 30 kJ/s

The above diagram assumes that you can do the mass balance ‘in your head’: 1 kg/s in plus 0.8 kg/s in = 1.8 kg/s out at steady state. We can write the enthalpy balance as H  Q  W

We are given that the heat removal rate is –30 kJ/s and that no work is done. So, we simply have H  30 kJ/s

Solution continued on next page…

The enthalpy change is the sum of the mass flow rates of the input streams each multiplied by their respective temperature change and by the heat capacity of water:

H = 1 kg/s · (Tout – 25) · 4.18 kJ/(kg K) + 0.8 kg/s · (Tout – 75) · 4.18 kJ/(kg K) = 30 kJ/s

H = 7.524 Tout – 355.3 = 30

from which Tout = 43.2 C Note that because only temperature differences were involved in the problem (and not absolute temperatures) using

C for the units of temperature worked fine. A temperature difference is the same in C as in K.

Solution 2.25 Problem Statement Gas is bled from a tank. Neglecting heat transfer between the gas and the tank, show that mass and energy balances produce the differential equation: dU dm  H  U m

Here, U and m refer to the gas remaining in the tank; H what conditions can one assume H

Solution

If we take the tank as our control volume, then the mass balance gives us dm  m  dt

where m is the mass of fluid still in the tank and ̇ ′ is the mass flow rate of the stream leaving the tank. Likewise, the energy balance gives d (mU )  m H  dt

where H' is the specific enthalpy of the fluid leaving the tank and U is the specific internal energy of the fluid still in the tank. In the energy balance we have neglected contributions of kinetic and gravitational potential energy. Q, the rate of heat transfer in the energy balance is zero because the tank is insulated. W is zero because no work is being done on or by the system other than the flow work that is already included in H'. We can expand the derivative in the energy balance to get

dU dm U  m H  dt dt

 ' from this equation to get We can use the mass balance to eliminate m m

dU dm dm U  H dt dt dt

Rearranging this gives

dU dm  (H  U ) dt dt

From which we can eliminate time as a variable and get the desired relationship dU dm   H U m

If conditions within the tank are uniform (no gradients in pressure, temperature, etc. between the main part of the tank and the tank exit) then H' = H. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 2.26 Problem Statement Water at 28°C flows in a straight horizontal pipe in which there is no exchange of either heat or work with the surroundings. Its velocity is 14 m⋅s

in a pipe with an internal diameter of 2.5 cm until it flows into a section where

the pipe diameter abruptly increases. What is the temperature change of the water if the downstream diameter is 3.8 cm? If it is 7.5 cm? What is the maximum temperature change for an enlargement in the pipe?

Solution Mass balance at steady state requires that the mass flow (and therefore volumetric flow for water with nearly constant density) be the same before and after the diameter change. The mass flowrate in the 2.5 cm diameter pipe is

m  H2o kg m3  14m s1  (0.025 m) 2 / 4 and the mass flow rate after the expansion is

m  H2o kg m3  out m s1  (dout m) 2 / 4 Setting these equal shows that the velocity in the outlet pipe is 2

vout = 14 m/s · (2.5 cm / dout cm)

So, if dout = 3.8 cm, then vout = 6.06 m/s The change in kinetic energy per unit mass is then 2

EK/m = ½ (6.06 – 14 ) = 79.6 m s

= 79.6 J kg

Solution continued on next page…

Because no heat is exchanged with the surroundings, there is no change in elevation (no change in gravitational potential energy), and no non-flow work is done on the surroundings, the energy balance is simply

 H +  EK = 0

From which H = 79.6 J kg

This enthalpy change corresponds to a temperature increase given by

H = CpT = 4180 J kg

K T =79.6 J kg

from which T = 0.019 K.

If the downstream diameter were 7.5 cm, then we would have vout = 1.556 m/s, and the change in kinetic energy per 1

unit mass would be 96.8 J kg . The corresponding temperature increase would be 96.8/4180 = 0.0232 K.

The maximum temperature increase would occur if the downstream velocity went to zero, so that all of the kinetic 2

energy of the flow were converted to internal energy. In this case, we would have EK/m = ½ (0 – 14 ) = 98 J kg , leading to a temperature increase of 98/4180 = 0.0234 K. For the downstream diameter of 7.5 cm, we had very nearly reached this limit.

Solution 2.27 Problem Statement Fifty (50) kmol per hour of air is compressed from P1 = 1.2 bar to P2 = 6.0 bar in a steady-flow compressor. Delivered mechanical power is 98.8 kW. Temperatures and velocities are:

T1 = 300 K

T2 = 520 K

u1 = 10 m·s 1

u2 = 3.5 m·s 1

7 Estimate the rate of heat transfer from the compressor. Assume for air that CP  R and that enthalpy is independent 2

of pressure.

Solution The energy balance, including kinetic energy contributions, for a steady flow system gives us  1 2 1  m  H out  Hin  uout  uin2   Q  W 2 2   1

The mass flow rate (in and out) is 50 kmol hr ·29 kg/kmol = 1450 kg/hr = 0.40278 kg/s

The enthalpy change is

H out  Hin  C p (Tout  Tin ) 

3.5  8.314 kJ kmol 1 K1 29 kg kmol1

(520K  300K)  220.75 kJ kg1

The kinetic energy change is 1 2 1 uout  uin2  0.5(12.25 m 2 s2  100 m 2 s2 )  43.9 m 2 s2  0.0439 kJ kg1 2 2

So, as is so often the case, the change in kinetic energy is negligible. The energy balance then becomes 0.40278 kg s

220.71 kJ kg

= 88.9 kJ s

= 88.9 kW =

+ 98.8 kW

from which

= –9.9 kW. That is, 9.9 kW of heat flows from the system (the air) to the surroundings, or 9.9 kW of the

input mechanical energy leaves the compressor as heat.

Solution 2.28 Problem Statement Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter of 1.5(in). A pressure drop results from flow through a partially opened valve. Just upstream from the valve the pressure is 100(psia), the 1

temperature is 120(°F), and the average velocity is 20(ft)·s . If the pressure just downstream from the valve is 20(psia), what is the temperature? Assume for air that PV/ T is constant, CV = (5/2)R, and CP = (7/2)R. (Values for R, the ideal gas constant, are given in App. A.)

Solution Because the pipe is insulated, we can assume Q is zero, and because the pipe and valve presumably have no moving parts, W is also zero. Furthermore, the pipe is horizontal, so there is no change in gravitational potential energy between the inlet and the outlet streams. Thus, we can write the energy balance for an open system with one inlet and one outlet like it is written in equation 2.31. Note that this equation, as written, is per unit mass.

H 

u 2  g z  Q  W 2

H 

u 2 0 2 gc

The difference in enthalpy between the inlet and outlet streams can be written in terms of the temperature change:

H = CpT. Substituting that in, we have

C p T 

u 2 0 2

The mass flow rates in and out must be equal (at steady state), min = mout, and the mass flow rate is the velocity times the cross-sectional area divided by the specific volume (volumetric flow rate divided by specific volume): m = uA/V. So, we have uin Ain uout Aout  Vin Vout

and

uout 

uin AinVout Aout Vin

Ain and Aout are the same, and because we have PV/T = a constant, we can write Vout Tout Pin  Vin Tin Pout

So, we can then write

uout 

uin PinTout PoutTin

Putting in the numbers, we have Pin/Pout = 5 and uin = 20 ft s

uout (m  s1 )  5 

= 6.096 m s , Tin = 120 F = 322.04 K, so

6.096(322.04  T )  30.48  0.09465T 322.04

Also, we have Cp = 3.5·R = 3.5·8.314 J/(mol K) = 29.10 J/(mol K). As given, this is the molar heat capacity. To get the specific heat, we divide by the molar mass of nitrogen (0.02801 kg/mol) to get Cp = 1039 J/(kg K). Putting this all together, we see that

1039  T 

(30.48  0.09465T )2  6.0962 0 2

where both terms have units of J/kg, and T has units of K or °C. Multiplying things out gives

0.004479  T 2  1041.8T  445.9  0 Applying the quadratic formula to this gives T = –0.428 K = –0.77F, so the downstream temperature is 119.23F.

Solution 2.29 Problem Statement Air flows at steady state through a horizontal, insulated pipe with inside diameter of 4 cm. A pressure drop results from flow through a partially opened valve. Just upstream from the valve, the pressure is 7 bar, the temperature is 1

45°C, and the average velocity is 20 m·s . If the pressure just downstream from the valve is 1.3 bar, what is the temperature? Assume for air that PV/T is constant, CV = (5/2)R, and CP = (7/2)R. (Values for R, the ideal gas constant, are given in App. A.)

H 

u 2  g z  Q  W 2

H 

u 2 0 2

m The difference in enthalpy between the inlet and outlet streams can be written in terms of the temperature change:

H = CpT. Substituting that in, we have

C p T 

u 2 0 2

The mass flow rates in and out must be equal (at steady state, with no chemical reactions),

, and the mass

out

flow rate is the velocity times the cross-sectional area divided by the specific volume (volumetric flow rate divided by specific volume):

= uA/V. So, we have

uin Ain uout Aout  Vin Vout

and

uout 

uin AinVout Aout Vin

Ain and Aout are the same, and because we have PV/T = a constant, we can write Vout Tout Pin   Vin Tin Pout 

So, we can then write

uout 

uin PinTout PoutTin

Solution continued on next page…

Putting in the numbers, we have Pin/Pout = 5.385 and uin = 20 m s , Tin = 45 C = 318.15 K, so

uout (m  s1 ) 

5.385  20(318.15  T ) 318.15

 107.69  0.3385T

T is in K or °C. Also, we have Cp = 3.5·R = 3.5·8.314 J/(mol K) = 29.10 J/(mol K). As given, this is the molar heat capacity. To get the specific heat, we divide by the molar mass of nitrogen (0.02897 kg/mol) to get Cp = 1004.5 J/(kg K). Putting this all together, we see that

1004.5  T 

(107.69  0.3385T )2  20 2 0 2

where both terms have units of J/kg, and T has units of K or °C. Multiplying things out gives

0.057291  T 2  1040.9T  5398.8  0 Applying the quadratic formula to this gives T = –5.19 K = –5.19C, so the downstream temperature is 39.8C.

Solution 2.30 Problem Statement Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil, the water changes state from liquid at 200 kPa and 80°C to vapor at 100 kPa and 125°C. Its entering velocity is 3 m⋅s

and its exit velocity is 200 m⋅s . Determine the heat transferred through the coil per unit mass of water.

Enthalpies of the inlet and outlet streams are: 1

Inlet: 334.9 kJ⋅kg ; Outlet: 2726.5 kJ⋅kg

Solution We can write the energy balance (per unit mass) as

H  EK  Q  W s 1

From the problem statement, we have H = 2726.5 – 334.9 = 2391.6 kJ kg , and no shaft work is done. Because there is a large change in velocity from inlet to outlet, we will take into account the change in kinetic energy, which is 2

EK/m = ½ (200 – 3 ) m s

= 19996 J kg

= 20.0 kJ kg . So, we have

Q  H  EK  2391.6  20.0  2411.6 kJ kg1 Even for this huge change in velocity, the change in kinetic energy is a very small part of the overall heat requirement.

Solution 2.31 Problem Statement Steam flows at steady state through a converging, insulated nozzle, 25 cm long and with an inlet diameter of 5 cm. At 1

the nozzle entrance (state 1), the temperature and pressure are 325°C and 700 kPa and the velocity is 30 m⋅s . At the nozzle exit (state 2), the steam temperature and pressure are 240°C and 350 kPa. Property values are: H1 = 3112.5 kJ⋅kg

H2 = 2945.7 kJ⋅kg

V1 = 388.61 cm3⋅g

V2 = 667.75 cm3⋅g

What is the velocity of the steam at the nozzle exit, and what is the exit diameter?

Solution Because the nozzle is insulated, Q is zero, and because it is a nozzle (with no moving parts) the nonflow work, Ws, is zero, so from the steady-state energy balance we have  1 2 1  m H out  Hin  uout  uin2   0  2 2 

or 1 2 H out  Hin  (uin2  uout ) 2

from which 1 2 H out  Hin  (uin2  uout ) 2 2 uout  2( Hin  Hout )  uin2  2(3112500 J kg1  2945700 J kg1 )  900 J kg 1

2 uout  334500 J kg1  334500 m 2 s2

uout  578 m s1

The mass balance simply tells us that the mass out equals the mass in, or

from which 667.75 cm3 g1  30 m s1 Aout Vout uin    0.0892 Ain Vin uout 388.61 cm3 g1  578 m s1

Therefore

dout  din

Aout  0.0892  0.299 Ain

so dout = 0.299·5 cm = 1.49 cm.

Solution 2.32 Problem Statement 1

In the following, take CV = 20.8 and CP = 29.1 J⋅mol ⋅°C

for nitrogen gas:

(a) Three moles of nitrogen at 30°C, contained in a rigid vessel, is heated to 250°C. How much heat is required if the 1

vessel has a negligible heat capacity? If the vessel weighs 100 kg and has a heat capacity of 0.5 kJ⋅kg ⋅°C , how much heat is required? (b) Four moles of nitrogen at 200°C is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 40°C if the heat capacity of the piston and cylinder is neglected?

Solution (a) At constant volume, the total heat input will be Q

U = n U = nCv T (for constant heat capacity – if the heat

capacity were not constant, we would have to do an integral). So, we have: Q = 3 moles⋅20.8 J/(mol K)⋅220 K = 13728 J, to heat only the gas. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

To heat the vessel, with a total heat capacity of 100 kg⋅500 J/(kg K) = 50000 J/K requires Q = 50000 J/K⋅220 K = 1.1  7

10 J. The energy required to heat up the vessel is roughly 1000 times larger than the energy required to heat only the gas. This is fairly typical for gases: the vessel that contains them has much greater mass, and therefore much greater total heat capacity, than the gas itself.

(b) At constant pressure, Q

H = n H = nCp T (again for constant heat capacity; if the heat capacity were not

constant, we would have to do an integral). So, we have Q = 4 moles⋅29.1 J/(mol K)⋅ –160 K = –18624 J. About 19 kJ has to be removed. Note that in both parts of this problem, we only needed to know the total heat capacity (number of moles and molar heat capacity) and did not need to know the actual pressure or volume.

Solution 2.33 Problem Statement 1

In the following take CV = 5 and CP = 7 (Btu)(lb mole) (°F)

for nitrogen gas:

(a) Three pound moles of nitrogen at 70(°F), contained in a rigid vessel, is heated to 350(°F). How much heat is required if the vessel has a negligible heat capacity? If it weighs 200(lbm) and has a heat capacity of 1

0.12(Btu)(lbm) (°F) , how much heat is required? (b) Four pound moles of nitrogen at 400(°F) is contained in a piston/cylinder arrangement. How much heat must be extracted from this system, which is kept at constant pressure, to cool it to 150(°F) if the heat capacity of the piston and cylinder is neglected?

Solution

(a) For this constant-volume process, we have (neglecting the heat capacity of the vessel) t

Q = U = nCvT = 3 (lb mole)⋅5(Btu)(lb mole) (F) ⋅280(F) = 4200 Btu If we add in the heat capacity of the vessel, then we have t

Q = U = ((nCv)gas + (mCv)vessel)T t

Q = U = (15 + 24) (Btu)(F) ⋅280(F) = 10920 Btu

(as is likely to be the case in real life, the heat capacity of the vessel is greater than that of the gas).

(b) For this constant-pressure process, we have (neglecting the heat capacity of the piston and cylinder) t

Q = H = nCpT = 4 (lb mole)⋅7(Btu)(lb mole) (F) ⋅–250(F) = –7000 Btu We must extract 7000 Btu from the system.

Solution 2.34 Problem Statement Find an equation for the work of reversible, isothermal compression of 1 mol of gas in a piston/cylinder assembly if the molar volume of the gas is given by V

RT b P

Solution

The work of a reversible expansion or compression is given by W   PdV

where the integral is over the path from the initial to final state. If the temperature is fixed and the pressure and volume are related by

V

RT b P

Then we can substitute P in terms of V or dV in terms of P into the integral (depending whether we want to specify the limits of the integral in terms of P or V). If we want substitute for dV in terms of P, we have to differentiate the relationship between V and P, to get

dV 

RT  dp P2

Substituting this into the integral for the work, we have P  RT dP  RT ln  2   P1  P P P2

W 1

Taking the alternative approach, we could substitute for P in terms of V, writing

P

RT V b

 V  b  RT  dV  RT ln  1 V b  V2  b  V V2

W  

Substitution for V1 andV2 in terms of P1 and P2 shows that this result is identical to the first one. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 2.35 Problem Statement 1

Steam at 200(psia) and 600(°F) [state 1] enters a turbine through a 3-inch-diameter pipe with a velocity of 10(ft)⋅s . The exhaust from the turbine is carried through a 10-inch-diameter pipe and is at 5(psia) and 200(°F) [state 2]. What is the power output of the turbine?

H1 = 1322.6(Btu)(lbm) H2 = 1148.6(Btu)(lbm)

V1 = 3.058(ft) (lbm)

V2 = 78.14(ft) (lbm)

Solution The volumetric flow rate into the pipe is the velocity (u = 10 ft s 2

= 3.048 m s ) times the cross-sectional area

(A = (0.25 ft) /4 = 0.04909 ft ), so the volumetric flowrate is 0.4909 ft s . Because the specific volume at these 3

conditions is 3.058 ft lbm , the mass flowrate in is 3

min = 0.4909 ft s

/ 3.058 ft lbm

= 0.1605 lbm s

= 0.07280 kg s .

At steady-state, the mass flow rate out must be the same as the mass flowrate in. So, if the specific volume at the 3

outlet conditions is 78.14 ft lbm , then the volumetric flowrate out is 78.14 ft lbm ·0.1605 lbm s 2

= 12.543 ft s .

The cross-sectional area of the 10-inch diameter exit pipe is (10/12 ft) /4 = 0.5454 ft , so the velocity is 3

uout = 12.543 ft s

/ 0.5454 ft = 23.00 ft s

= 7.0104 m s .

Now, we can use this in the energy balance for an open system as written in equation 2.30. 2   H  u  g z m  Q  W   2 

We will assume that the heat loss from the turbine is negligible (Q = 0) and that the change in elevation from inlet to outlet is negligible (z = 0), so the work output of the turbine is given by:  u2  W  H  m  2 

 (7.0104 m s1 )2  (3.048 m s1 )2   1 0.07280 kg s1 W  (1148.6  1322.6)Btu lbm    2  W  174.0 Btu lbm1  19.93 J kg1 0.07280 kg s1

W  (404724 J kg1  19.93 J kg1 )0.07280 kg s1  29462 J s1

W = –29.46 kJ s

= –29.46 kW = –27.92 Btu s . Our sign convention for W is that it is work done on the fluid, so the

work output from the system is 27.92 Btu s

= 39.5 hp.

Solution 2.36 Problem Statement Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 1

0.1 kg⋅s . The exhaust from the turbine is carried through a 25-cm-diameter pipe and is at 50 kPa and 100°C [state 2]. What is the power output of the turbine? H1 = 3150.7 kJ⋅kg H2 = 2682.6 kJ⋅kg

V1 = 0.20024 m3⋅kg V2 = 3.4181 m3⋅kg

Solution 1

The volumetric flow rate into the pipe is the mass flow rate times the specific volume (0.1 kg s · 0.20024 m kg 3

0.020024 m s ). The cross-sectional area of the inlet pipe is A = (0.08 m) /4 = 0.005027 m , so the average flow 1

velocity is 0.020024/0.005027 = 3.984 m s . At steady-state, the mass flow rate out must be the same as the mass 3

flowrate in. So, if the specific volume at the outlet conditions is 3.4181 m kg , then the volumetric flowrate out is 3

3.4181 m kg ·0.1 kg s

= 0.34181 m s . The cross-sectional area of the 25 cm diameter exit pipe is (0.25 m) /4 =

0.04909 m , so the velocity is uout = 0.34181 m s

/ 0.04909 m = 6.963 m s . Now, we can use this in the energy

balance for an open system as written in equation 2.30. 2   H  u  g z m  Q  W   2 

  (6.963 m s1 )2  (3.984 m s1 )2  W  (2682.6  3150.7 kJ kg1  0.1 kg s1  2   W  468.1 kJ kg1  16.31 J kg 1  0.1 kg s1  46.81 kJ s1

W = –46.81 kJ s

= –46.81 kW. Our sign convention for W is that it is work done on the fluid, so the work output

from the system is 46.8 kW.

Solution 2.37 Problem Statement

Carbon dioxide gas enters a water-cooled compressor at conditions P1 = 1 bar and T1 = 10°C, and is discharged at conditions P2 = 36 bar and T2 = 90°C. The entering CO2 flows through a 10-cm-diameter pipe with an average 1

velocity of 10 m⋅s , and is discharged through a 3-cm-diameter pipe. The power supplied to the compressor is 1

12.5 kJ·mol . What is the heat-transfer rate from the compressor?

H1 = 21.71 kJ⋅mol H2 = 23.78 kJ⋅mol

V1 = 23.40 L⋅mol

V2 = 0.7587 L⋅mol

Solution First, we need to determine the velocity of the discharge,

. To do this the molar flow rate must be determined by

using 

n  4

D12u1 cm3  mol  3354.70 V1 Ls

n  3.3547

mol . s

Using this, V u2  n  2 2  3.6026 m/s D2 4

Now that

has been determined, using Eqn. 2.31,

 u2  u 2  kJ kg m2 kJ 1  Q  H 2  H1  CO2  2  0.04402 87.02 2  12.5   Ws  2.07  2  mol mol mol s 2

and converting 87.02m /s by using 1 kJ/kg = 1000 m /s shows that the kinetic energy term is 0.0038 kJ/mol and Q = 10.43 kJ/mol. To obtain the heat transfer rate, Q  Q  n  35.0 kJ/s.

Solution 2.38 Problem Statement Carbon dioxide gas enters a water-cooled compressor at conditions P1 = 15(psia) and T1 = 50(°F), and is discharged at conditions P2 = 520(psia) and T2 = 200(°F). The entering CO2 flows through a 4-inch-diameter pipe with a velocity of 1

20(ft)⋅s , and is discharged through a 1-inch-diameter pipe. The shaft work supplied to the compressor is 1

5360(Btu)(lb mole) . What is the heat-transfer rate from the compressor in (Btu)·h ?

H1 = 307(Btu)(lbm) H2 = 330(Btu)(lbm)

V1 = 9.25(ft) (lbm) V2 = 0.28(ft) (lbm)

Solution

First, we need to determine the velocity of the discharge, u2. To do this, the mass flow rate must be determined by using 

m  4

Using this

D12u1 1 = 679.3 lbm hr . V1

V u2  m   2 2  9.686 ft/s. D2 4

Now that u2 has been determined, using Eqn. 2.31,

Q  H 2  H1 

u22  u12 W Btu ft 2 Btu  s  23  153.09 2  121.8 2 MCO2 lbm lbm s 2

and converting 153.09 ft /s by using 25036 lbm·ft /(s ·Btu), we obtain Q = –98.8 Btu/lbm. Thus, the heat transfer rate is

Solution 2.39 Problem Statement Show that W and Q for an arbitrary mechanically reversible nonflow process are given by: W   V dp 

 PV 

Q

H   V dp

Solution

Starting with W, we know that: d( PV )  VdP  PdV and that in Eqn. 1.3, dW  PdV . Substituting in for PdV we get: dW  VdP  d( PV ) and integrating we get W   VdP  ( PV ). From Eqn. 2.4, dQ  dU  dW and from Eqn. 2.10, U  H  PV . Taking the integral of U, we get dU  dH  PdV  VdP. and substituting this in for dU, we get

dQ  dH  PdV  VdP  dW . Knowing that dW  PdV from earlier and substituting we get dQ  dH  VdP. Lastly, by integrating this we obtain Q  H   VdP .

Solution 2.40 Problem Statement One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triples. Calculate W, Q 1

H for the process. Assume for air that PV / T = 83.14 bar⋅cm ⋅mol ⋅K

and

CP = 29 J⋅mol ⋅K .

Solution

If PV/T is constant then at constant pressure, the volume will triple when the temperature triples, so the final state will be 900 K and 1 bar. The enthalpy change for this constant volume process with constant heat capacity is given by

H  C p T  29 J mol1 K1  (900 K  300 K)  17400 J mol1  600 J g 1  600 kJ kg 1

where we've used the molar mass of air of about 29 g mol

to convert from molar enthalpy change to specific

enthalpy change. By the definition of U, we have

1

U  H  PV  17400 J mol  83.14 bar cm mol 3

1

Pa bar K (900 K  300 K) cm3 10 6 3 m 105

1

U  17400 J mol 1  4988 Pa m3 mol 1  26100 J mol1  4988 J mol1  12412 J mol1 U  12412 J mol 1  428 J g1  428 kJ kg1 The work is given by –PV = –PV at constant P, which we just computed to be

–PV = –4988 J mol

= –172 J g

= –172 kJ kg .

For a constant pressure process, Q is equal to the enthalpy change, so Q = 600 kJ/kg.

Solution 2.41 Problem Statement The conditions of a gas change in a steady-flow process from 20°C and 1000 kPa to 60°C and 100 kPa. Devise a U

H for

the process on the basis of 1 mol of gas. Assume for the gas that PV/T is constant, CV = (5/2)R, and CP = (7/2)R. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

This is a two-step process, going first from 20°C to 60°C, and then 1000 kPa to 100 kPa.

So for temperature: T1  293.16 K and T2  333.16 K, and using PV  RT and R  8.314

J to determine the mol K

intermediate temperature and the volume changes we have

T I  T1 

1 P2  293.16 K     29.316 K and  10  P1

T m3kPa 293.16 K m3 V1  R 1  8.314  103   2.437  103 P1 mol K 1000 k Pa mol T m3 V2  R 2  2.770  102 P2 mol

Using the intermediate temperature to determine the temperature change in both steps we have

Ta  T I  T1  263.844 K and Tb  T2  T I  303.844 K

Using R  8.314

J , and plugging the temperatures into Eqn. 2.16 and Eqn. 2.20, we have for Step A: mol K

H a  U a  V1 ( P2  P1 )  5483

U a  Cv Ta 

J m3 J  2.437  103 (100 kPa  1000 kPa)  7676 mol mol mol

5 1 J  8.314   263.844 K  5483 2 mol K mol

And for Step B: H b  C p Tb 

7 J J  8.314   303.844 K  8842 2 mol K mol

U b  H b  P2 (V2  V1 )  8842

 J m3 m 3  J   6315  100 k Pa 2.770  102  2.437  103  mol mol mol  mol

Now using these to determine the total

and

U  U a  U b  832

J mol

H  H a  H b  1165

J mol

Solution 2.42 Problem Statement

A flow calorimeter like that shown in Figure 2.6 is used with a flow rate of 20 g⋅min

of the fluid being tested and a

constant temperature of 0°C leaving the constant-temperature bath. The steady-state temperature at section two (T2) is measured as a function of the power supplied to the heater (P), to obtain the data shown in the table below. What is the average specific heat of the substance tested over the temperature range from 0°C to 10°C? What is the average specific heat from 90°C to 100°C? What is the average specific heat over the entire range tested? Describe how you would use this data to derive an expression for the specific heat as a function of temperature. T2 /°C

100

P/W

5.5

11.0

16.6

22.3

28.0

33.7

39.6

45.4

51.3

57.3

Solution

For this problem, we need to find the specific heat or the heat capacity over a range of temperatures. From the text:

Q  H  C p m T Given the power supplied at the Q values and the flow rate, solving for the heat capacity gives:

Cp 

Q m T

Given this the specific heat from 0°C to 10°C is

Cp 

5.5 J/s 0.333 g/s  10 K

 1.650

J gK

And going from 90°C to 100°C is

C p  1.715

J gK

The key here is that the initial temperature for each step is 0 C and that the and the new temperature. Over the whole range the T2

for each is the difference between 0°C

values are:

100

5.5

11.0

16.6

22.3

28.0

33.7

39.6

45.4

51.3

57.3

1.650

1.660

1.673

1.680

1.685

1.697

1.703

1.710

1.719

C p w

Cp J/(g K)

The average is

= 1.683 J/(g k).

Solution 2.43 Problem Statement Like the flow calorimeter of Figure 2.6, a particular single-cup coffee maker uses an electric heating element to heat a steady flow of water from 22°C to 88°C. It heats 8 fluid ounces of water (with a mass of 237 g) in 60 s. Estimate the power 1

requirement of the heater during this process. You may assume the specific heat of water is constant at 4.18 J⋅g ⋅°C .

Solution

The relevant steady-state energy balance is H=Q+W Whether we regard the energy input as heat or work depends on where we draw the system boundary. If we draw the system boundary to only contain the fluid, then W

H = Q, where Q is the flow of heat from the heating

element to the water. The total change in enthalpy of the 237 g of water is t

H = mCp T = 237 g⋅4.18 J g °C ⋅66 °C = 65384 J = 65.4 kJ Solution continued on next page…

This is also the total heat transferred to the water in a period of 60 s. The rate of heat transfer is therefore 65.4 kJ/60 s = 1.09 kJ/s = 1.09 kW. We can reasonably assume that the heating element fully converts electrical energy to heat, so at steady state, this would also be the power requirement for the heater. The actual power requirement might be higher, because of heat losses to the surroundings (not all of the heat generated in the heating element actually enters the water). If the coffee maker is plugged into a regular U.S. 110 V outlet, then it will draw at least 10 amps of current during the heating cycle. One wouldn’t want to run this along with much of anything else on a regular 15 A household circuit.

Solution 2.44 Problem Statement (a) An incompressible fluid ( = constant) flows through a pipe of constant cross-sectional area. If the flow is steady, show that velocity u and volumetric flow rate q are constant. (b) A chemically reactive gas stream flows steadily through a pipe of constant cross-sectional area. Temperature and

 , n , q, u? pressure vary with pipe length. Which of the following quantities are necessarily constant: m

Solution

(a) Based on Eqn. 2.23a (m = u A )

which means it

remains unchanged, then the velocity u is constant. Knowing that q = u A, that tells us that q is constant as well. (b) Due to the law of mass conservation,

must be constant. may change due to reaction in the stream. Due to the

temperature and pressure changes in the pipe, both q and u will also vary.

Solution 2.45 Problem Statement The mechanical-energy balance provides a basis for estimating pressure drop owing to friction in fluid flow. For steady flow of an incompressible fluid in a horizontal pipe of constant cross-sectional area, it may be written,

P 2  f u2  0 L D F where fF is the Fanning friction factor. Churchill gives the following expression for fF for turbulent flow:   

  

f F  0.3305 ln  0.27

Here, Re is the Reynolds number and

2

   7  ( ) 0.9   D Re  

/D is the dimensionless pipe roughness. For pipe flow, Re

/ , where D

is pipe diameter and is dynamic viscosity. The flow is turbulent for Re > 3000.

 (in kg⋅s ) and Consider the flow of liquid water at 25°C. For one of the sets of conditions given below, determine m P

L (in kPa·m ). Assume

= 9.0 × 10

/D = 0.0001. For liquid water at 25°C, = 996 kg·m , and

kg⋅m ⋅s . Verify that the flow is turbulent.

(a) D = 2 cm, u = 1 m·s (b) D = 5 cm, u = 1 m·s (c) D = 2 cm, u = 5 m·s (d) D = 5 cm, u = 5 m·s

Solution

(a) First, the Reynolds number must be determined: kg m Du 2.0 cm  996 m3  1.0 s Re    22133  9.0  104 mkgs

Next, using the equation for the fanning friction factor given, we get 2    0.9    7     fF  0.3305 ln  0.27  0.0001     0.00635  22133         

With the friction factor, the

and

m    u 

P / L 

can be determined:  kg m  kg  D2  996 3  1.0   0.004 m 2  0.313 4 s 4 s m

2 2 kg m2 kPa    fF  u2   996 3  0.00635  1.0 2  0.632 D 0.02 m m m s

(b) Re  55333, fF  0.00517, m  1.956

kg P kPa ,  0.206 s L m

kg P kPa ,  11.255 s L m

(d) Re  276667, fF  0.00390, m  9.778

kg P kPa ,  3.879 s L m

Solution 2.46 Problem Statement

Ethylene enters a turbine at 10 bar and 450 K, and exhausts at 1(atm) and 325 K. For

= 4.5 kg⋅s , determine the

cost C of the turbine. State any assumptions you make.

Data: H1 = 761.1

H2 = 536.9 kJ⋅ kg

C / $  (15, 200)( W / kW ) 0.573

Solution Making the assumption that the compressor is adiabatic, meaning that the Q  0 , then the

| W |  m  ( H2  H1 )  1.009 kW and

 |W | 0.573  Cost $  15200    $799,969  kW 

Solution 2.47 Problem Statement The heating of a home to increase its temperature must be modeled as an open system because expansion of the household air at constant pressure results in leakage of air to the outdoors. Assuming that the molar properties of air leaving the home are the same as those of the air in the home, show that energy and mole balances yield the following differential equation: dn dU Q  PV n dt dt

Here,

is the rate of heat transfer to the air in the home, and t is time. Quantities P, V, n, and U refer to the air in the

Solution

Performing an energy and material balance on a home, the only sources of energy loss are from the air, enthalpy, and internal energy and the only material loss is the air. This gives: d(nU ) dU dn Q  natr H   nalr H  n U dt dt dt



dn dt

By substituting, we obtain dn dU dn dn dU Q  H n U  ( H  U ) n dt dt dt dt dt

and we know that H = U + PV, or H – U = PV Substititing this, we obtain dn dU Q  PV n dt dt

Solution 2.48 Problem Statement

 in terms of line pressure P1, ambient (a) Water flows through the nozzle of a garden hose. Find an expression for m pressure P2, inside hose diameter D1, and nozzle outlet diameter D2. Assume steady flow, and isothermal, adiabatic operation. For liquid water modeled as an incompressible fluid, H2

H1 = (P2

P1)/ for constant temperature.

(b) In fact, the flow cannot be truly isothermal: we expect T2 > T1, owing to fluid friction. Hence, H2

H1 = C(T2

T1) + (P2

P1)/ , where C is the specific heat of water. Directionally, how would inclusion of the

 as found in Part (a)? temperature change affect the value of m

Solution

(a)

Starting with equation (2.31) we know that:

H

u2  g z  Q  WS 2

With the system being adiabatic, steady flow, assuming no height change, and that no work done by the shaft we get:

H 2  H1 

u22  u12 0 2

Determining the area for u gives:

u

4m  D2

Substituting this in for u1 and u2 gives: 2  4  m 2  1 1  u22  u12    2  4  4       D D1  2

We know that 1 H 2  H1  ( P2  P1 ) 

Solution continued on next page…

Substituting these both in for the second equation we obtain: 2 1 1  4  m 2  1 1  ( P2  P1 )    2  4  4   0  2      D2 D1 

And by solving for

we get : 1

2      D 4 D 4  2 m  2 ( P2  P1 )   4 1 2 4   4   D  D   1 2 

(b)

The only addition to the

equation from above would be a C(T2 – T1) term: 2      D 4 D 4  m  2[ ( P2  P1 )  C  2 (T2  T1 )]   4 1 2 4   4   D  D   1 2 

1/2

Solution 3.1 Problem Statement

How many phase rule variables must be specified to fix the thermodynamic state of each of the following systems?

(a) A sealed flask containing a liquid ethanol-water mixture in equilibrium with its vapor. (b) A sealed flask containing a liquid ethanol-water mixture in equilibrium with its vapor and nitrogen. (c) A sealed flask containing ethanol, toluene, and water as two liquid phases plus vapor.

Solution

(a) Using the phase rule F    N

 is the number of phases, N is the number of chemical species, and

F is the degrees of freedom or the number of variables that must be specified, we can determine that   2 for the liquid and vapor, and N = 2 for the ethanol and water, F     . 2 variables need to be specified. (b)   2 for liquid and vapor, and N = 3 for the ethanol, water, and nitrogen, leaving

F   

3 variables must be specified

F     . 2 variables must be specified

Solution 3.2 Problem Statement

A renowned laboratory reports quadruple-point coordinates of 10.2 Mbar and 24.1°C for four-phase equilibrium of allotropic solid forms of the exotic chemical -miasmone. Evaluate the claim.

Solution

Using the phase rule F     N

  for the phases and N = 1 for the one chemical,

gives F = –1, which is impossible, making their claim invalid.

Solution 3.3 Problem Statement

A closed, nonreactive system contains species 1 and 2 in vapor/liquid equilibrium. Species 2 is a very light gas, essentially insoluble in the liquid phase. The vapor phase contains both species 1 and 2. Some additional moles of species 2 are added to the system, which is then restored to its initial T and P. As a result of the process, does the total number of moles of liquid increase, decrease, or remain unchanged?

Solution

The total number of moles of liquid decreases. To see this, note that the phase rule allows only two intensive phase rule variables to be specified

F=2–+N=2–2+2=2

In both the initial state and the final state, T and P have been specified. Thus, the state of the system is fixed, and all intensive variables in each phase (molar volume, internal energy, and species mole fractions) are the same in the initial and final states. If we add more of species 2 to the vapor phase without adding more of species 1, then to keep the composition of the vapor the same, some of species 1 will have to evaporate from the liquid. As a result, less liquid will remain. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 3.4 Problem Statement

A system comprised of chloroform, 1,4-dioxane, and ethanol exists as a two-phase vapor/liquid system at 50°C and 55 kPa. After the addition of some pure ethanol, the system can be returned to two-phase equilibrium at the initial T and P. In what respect has the system changed, and in what respect has it not changed?

Solution

The phase rule tells us that the degrees of freedom of the system are F = 2 –  + N = 2 – 2 + 3 = 3. With only T and P fixed, one degree of freedom remains available. We can specify one more intensive variable, like the mole fraction of one of the components in one of the phases. Thus, changes in the phase compositions are possible for the given T and P. If ethanol is added in a quantity that allows T and P to be restored to their initial values while 2 phases remain present, the ethanol distributes itself between the phases so as to form new equilibrium phase compostions and different amounts of the vapor and liquid phases. Only T and P remain constant, whereas the phase compositions and the amount of each phase change.

Solution 3.5 Problem Statement

For the system described in Prob. 3.4: (a) How many phase-rule variables in addition to T and P must be chosen so as to fix the compositions of both phases? (b) If the temperature and pressure are to remain the same, can the overall composition of the system be changed (by adding or removing material) without affecting the compositions of the liquid and vapor phases?

Problem 3.4 A system comprised of chloroform, 1,4-dioxane, and ethanol exists as a two-phase vapor/liquid system at 50°C and 55 kPa. After the addition of some pure ethanol, the system can be returned to two-phase equilibrium at the initial T and P. In what respect has the system changed, and in what respect has it not changed?

Solution

(a) The phase rule tells us that the degrees of freedom of the system are

F=2–+N=2–2+3=3

Becausethere are 3 total degrees of freedom, one more phase-rule variable must be specified, in addition to T and P, to fix the values of all of the intensive variables in the system (including the compositions of both phases). Solution continued on next page…

(b) Adding or removing liquid with a composition equal to that of the liquid phase or adding or removing vapor with a composition equal to that of the vapor phase does not change the phase compositions, and does not alter the intensive state of the system.

Solution 3.6 Problem Statement

Express the volume expansivity and the isothermal compressibility as functions of density and its partial derivatives. For water at 50°C and 1 bar, = 44.18 × 10

bar

. To what pressure must water be compressed at 50°C to change

its density by 1%? Assume that is independent of P.

Solution

Starting with equations 3.3 and 3.4:



1  dV    V  dT P

1  dV       V  dP 

To put this in to terms of density, the amount of mass must be fixed in the system meaning

is constant. This allows

for the expression dV  Vd  and allows for the above equations to be changed to

1  d         dT P

1  d        dP T

Solution continued on next page…

◦

Now that these are in terms of density and assuming a constant temperature of 50 C, the equation reduces to

dP 

d 

And integrating yields   –1 P  ln 2  and value of k  44.18 * 106 bar   1 

From this the pressure that is need to change the density of water (1.00 g/mol) by 1% can be found by saying 2 

* 1

P 

P

ln 1.010





ln 1.010 44.18 * 10 6bar 1

= 225.22 bar. So the pressure would need to be 1.00 bar + 225.22 bar = 226.22 bar

to change the density by 1%.

Solution 3.7 Problem Statement Generally, volume expansivity and isothermal compressibility depend on T and P. Prove that:             P   T  T

Solution

      To prove that these two are equivalent,   must be integrated with respect to P and   must be integrated  T   P  P

with respect to T, as follows:       1  V  V  1  2V 1  V  V  1  2V    2     and     2      T   P   T  V PT V  P  T  V PT V  P  P

Substituting in equations 3.3 and 3.4, we obtain:

1  dV      V  dT 

1  dV       V  dP 

      1 2V 1 2V     and         P   T  V P T V P T T P

            P   T  T

Which proves the original statement.

Solution 3.8 Problem Statement The Tait equation for liquids is written for an isotherm as:  AP  V  V0 1    B  P 

where V is molar or specific volume, V0 is the hypothetical molar or specific volume at zero pressure, and A and B are positive constants. Find an expression for the isothermal compressibility consistent with this equation.

Solution

By definition, the isothermal compressibility is 1  V       V  P T

So, to find the isothermal compressibility, we just need to take the derivative of V with respect to P. For the Tait equation, we have  AP  V  V0 1   B  P      A B  P   AP  V0 AB V    V0  2 2  P   B  P    B  P  V0 AB 

1 V  V P

B  P 

 AP  V0 1    B  P 



AB  B  P  AP  B  P 

Solution 3.9 Problem Statement For liquid water the isothermal compressibility is given by:



c V ( P  b)

where c and b are functions of temperature only. If 1 kg of water is compressed isothermally and reversibly from 1 to 3

–1

500 bar at 60°C, how much work is required? At 60°C, b = 2700 bar and c = 0.125 cm ·g .

Solution

Using the definition of isothermal compressibility, we have (for an isothermal process) that dV  dP V

Using the given expression for the isothermal compressibility, this becomes dV c 0.125 cm3 g1  dP  dP  dP V V  P  b V P  2700 bar

or simply

dV 

0.125 cm3 g1 dP  P  2700 bar

The work done in the process is, as usual, given by dW = –PdV, or

dW  PdV  P

0.125 cm3 g1 dP  P  2700 bar

Integrating this from the initial pressure of 1 bar to the final pressure of 500 bar gives 500 bar

500 bar 0.125 cm 3 g1 3 1  P  2700 bar  dP  0.125 cm g  P  2700 bar ln P  2700 bar 1 bar 1 bar   3200 bar  W  0.125 cm3 g1 499 bar  2700 bar ln    5.1591 bar cm 3 g1   2701 bar 

W

W  0.51591 Pa m 3 g1  515.91 Pa m3 kg1  515.91 J kg 1

Solution 3.10 Problem Statement 3

Calculate the reversible work done in compressing 1(ft) of mercury at a constant temperature of 32(°F) from 1(atm) to 3000(atm). The isothermal compressibility of mercury at 32(°F) is: / ( atm )

= 3.9 × 10

0.1 × 10

P / (atm)

Solution 1  V  By definition,      , and we know that we can find the work done on the sample by integrating –PdV. V  P T

From our definition of  above, we see that at constant T, we have dV = –VdP. We can use this to first find V as a function of P, and then substitute that back in to do the integral of –PdV. To find the volume, we integrate the above relationship from the initial volume (Vo) at the initial pressure (Po) to an arbitrary pressure P, at which we have the volume V given by: 1 dV  dP  3.9 106  0.1109 P  dP V V  ln    3.9 106  Po  P   5 1011  P 2  Po2   Vo 





V  Vo exp 3.9 106  Po  P   5 1011  P 2  Po2 

In these expressions, the units of V are arbitrary, but the units of P must be atm, since that is the units in which the numerical value of  is given. If we only want to find the work required, we don’t actually have to solve for the final volume. Rather, we can simply substitute dV = –VdP into our expression for work, dW = –PdV = PVdP, which finally gives us





dW   PVdP  PVo exp 3.9  106 Po  P   5  1011  P 2  Po2  dP dW  3.9 10

6



 0.110 P  PVo exp 3.9  10 9

6

 Po  P   5 1011  P 2  Po2 dP

we can put into this Vo = 1 ft and Po = 1 atm, and then integrate it from P = 1 atm to P = 3000 atm, to get





dW  3.9 106  0.1 109 P  P exp 3.9 106 1  P   5 1011  P 2  1 dP dW  3.9 10 P  0.1 10 P exp 3.9 10 P  5 10 6

9

6

11

P dP 2

3000

W   3.9 106 P  0.1 109 P 2 exp 3.9 106 P  5 1011 P 2 dP 1

–6

–11

In the second expression above, we have factored out exp(1.9  10 ) = 0.9999961 and exp(5  10

) = 1.00000000005,

both of which we can take to be negligibly different from 1, and can therefore ignore. Doing the last integral with a 3

little help from Maple and/or an integral table gives us W = 16.53. We started out with P in atm and V in ft , so this 3

has units of W = 16.53 atm ft , which is not a particularly common unit for energy. We can find the conversion factor 3

–1

to get to more reasonable units by using the values of R (the ideal gas constant) as R = 0.7302 ft atm lbmol R = –1

–1

1.986 Btu lbmol R , so 0.7302 ft atm =1.986 Btu, or 1 ft atm = 2.720 Btu. So, we can multiply our result by 2.72 to get W = 45.0 Btu.

Solution 3. 11

Problem Statement Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar t

V , W, Q

U . The properties for

liquid carbon tetrachloride at 1 bar and 0°C may be assumed independent of temperature: = 1.2 × 10

= 0.84 kJ·kg

·K

, and = 1590 kg·m

Solution

For an isobaric process with a substance with constant coefficient of thermal expansion, the change in volume is given by

dV   dT  1.2 103 K1 dT V Integrating this over finite temperature gives V  ln  2    T2  T1   1.2 103 K 1 T2  T1   V1 

So, when the temperature increases by 20 degrees, the specific volume change is given by V  ln  2   1.2 103 K1 20 K   0.024  V1  –3

–4

–1

The initial specific volume is V1 = 1/ = 1/(1590 kg m ) = 6.29  10 m kg . Solution continued on next page…

The final specific volume is then V2  V1 exp 0.024   6.29  10  4 m 3 kg  1 * exp 0.024   6.44 10 4 m 3 kg 1 –5

–1

–5

So, the change in specific volume is 1.53  10 m kg , and the change in total volume is V = 7.65  10 m .

–5

–3

The work done is simply –PV = –10 Pa * 7.65  10 m = –7.65 Pa m = –7.65 J = –0.765 * 10 kJ (the carbon –5

tetrachloride does 7.65 J of work on its surroundings when it expands by 7.65  10 m ).

–1

The heat required is Q = mCpT = 5 kg * 0.84 kJ kg K * 20 K = 84 kJ = 84000 J (more than 1000 times greater than the work done).

The enthalpy change for the constant pressure process is equal to Q: H = 84 kJ

The internal energy change is U = H – PV = Q + W = 83.992 kJ

As is generally the case for liquids, which are nearly incompressible, the internal energy change and the enthalpy change are almost identical.

Solution 3.12

Problem Statement Various species of hagfish, or slime eels, live on the ocean floor, where they burrow inside other fish, eating them from the inside out and secreting copious amounts of slime. Their skins are widely used to make eelskin wallets and accessories. Suppose a hagfish is caught in a trap at a depth of 200 m below the ocean surface, where the water temperature is 10°C, then brought to the surface where the temperature is 15°C. If the isothermal compressibility and volume expansivity are assumed constant and equal to the values for water,

( = 10

and = 4.8 × 10

bar

)

what is the fractional change in the volume of the hagfish when it is brought to the surface? Table 3.2 provides the 25

specific volume, isothermal compressibility, and volume expansivity of several liquids at 20°C and 1 bar for use in Problems 3.13 to 3.15, where and may be assumed constant.

Solution

If the isothermal compressibility and volume expansivity are taken to be constant, then we can integrate

dV   dT  dP V From the initial state to the final state, to get V  ln    T   P  Vo 

Solution continued on next page…

The change in temperature is 5ºC = 5 K. The pressure changes with height in a column (or ocean) of fluid as P = 3

. So, if we take the density of water to be 1000 kg/m (actually, for sea water it is more like 1030), and g = 9.8 m/s , and h = –200 m (the fish ends up at the surface, where the pressure is lower), then 6

P = 1000*9.8*–200 = –2 × 10 Pa = –20 bar (about 20 atmopheres, or 300 psi). Thus, we have  V    (104 K 1 ) * (5 K)  (4.8 105 bar 1 ) * 20 bar  5 10 4  9.6 104  0.0015  V 

ln 

Thus, V/Vo = 1.0015. The hagfish expands by about 0.15%.

Solution 3.13

Problem Statement

For one of the substances in Table 3.2, compute the change in volume and work done when one kilogram of the substance is heated from 15°C to 25°C at a constant pressure of 1 bar.

Table 3.2 provides the specific volume, isothermal compressibility, and volume expansivity of several liquids at 20°C 25

and 1 bar , where and may be assumed constant.

Table 3.2: Volumetric Properties of Liquids at 20°C

Molecular

Chemical Name

Formula

Specific

Isothermal

Volume

Compressibility

Expansivity

–1

–5

V/L.kg

C2H4O2 C6H7N CS2 C6H5Cl C6H12 C4H10O C2H5OH C4H8O2 C8H10 CH3OH CCl4 C7H8 CHCl3

Acetic Acid Aniline Carbon Disulfide Chlorobenzene Cyclohexane Diethyl Ether Ethanol Ethyl Acetate m-Xylene Methanol Tetrachloromethane Toluene Trichloromethane

/10 bar

0.951 0.976 0.792 0.904 1.285 1.401 1.265 1.110 1.157 1.262 0.628 1.154 0.672

9.08 4.53 9.38 7.45 11. 18. 11. 11. 8.46 12. 10. 8.96 9.96

–1

/10–3·°C

–1

1.08 0.81 1.12 0.94 1.15 1.65 1.40 1.35 0.99 1.49 1.14 1.05 1.21

CRC Handbook of Chemistry and Physics, 90th Ed., pp. 6-140-6-141 and p. 15–25, CRC Press, Boca Raton, Florida,

2010.

Solution First ethanol will be chosen as the substance to be used. In the problem the pressure is held constant, then the volume expansively equation is used to determine the change in volume.



1  V    V  T P

Solving for dv and integrating gives: V  T   ln 2   V1 

Solving for V2 gives: V1e T   V2

Plugging in the specific volume and volume expansivity leads V2 = 1.283 L for 1 kilogram of ethanol. To determine the work we know that:

W   PV   bar * 

L

L = –0.018 bar L = –1.8 joules

Solution 3.14

Problem Statement

For one of the substances in Table 3.2, compute the change in volume and work done when one kilogram of the substance is compressed from 1 bar to 100 bar at a constant temperature of 20°C.

Table 3.2 provides the specific volume, isothermal compressibility, and volume expansivity of several liquids at 20°C 25

and 1 bar , where and may be assumed constant.

Table 3.2: Volumetric Properties of Liquids at 20°C

Molecular

Chemical Name

Formula

Specific

Isothermal

Volume

Compressibility

Expansivity

–1

V/L.kg

C2H4O2 C6H7N CS2 C6H5Cl C6H12 C4H10O C2H5OH C4H8O2 C8H10 CH3OH CCl4 C7H8 CHCl3

Acetic Acid Aniline Carbon Disulfide Chlorobenzene Cyclohexane Diethyl Ether Ethanol Ethyl Acetate m-Xylene Methanol Tetrachloromethane Toluene Trichloromethane

0.951 0.976 0.792 0.904 1.285 1.401 1.265 1.110 1.157 1.262 0.628 1.154 0.672

–5

–1

/10 bar

9.08 4.53 9.38 7.45 11. 18. 11. 11. 8.46 12. 10. 8.96 9.96

–3

/10 ·°C

–1

1.08 0.81 1.12 0.94 1.15 1.65 1.40 1.35 0.99 1.49 1.14 1.05 1.21

CRC Handbook of Chemistry and Physics, 90th Ed., pp. 6-140-6-141 and p. 15–25, CRC Press, Boca Raton, Florida,

2010.

Solution

First ethanol will be chosen as the substance to be used. In the problem the temperature is held constant, then the isothermal compressibility equation is used to determine the change in volume. 1  dV       V  dP 

Solution continued on next page…

Solving for dv and integrating gives:

V  – P  ln 2   V1 

Solving for V2 gives:

V1eP  V2

Plugging in the specific volume and isothermal compressibility factor leads V2 = 1.251 L for 1 kilogram of ethanol.

To determine the work we know that:

W   RTln

V2 V1



J * mol K

 1.251L   = 27.12 J/mol K * ln  L 

Solution 3.15

Problem Statement

For one of the substances in Table 3.2, compute the final pressure when the substance is heated from 15°C and 1 bar to 25°C at constant volume.

Table 3.2 provides the specific volume, isothermal compressibility, and volume expansivity of several liquids at 20°C 25

and 1 bar , where and may be assumed constant.

Table 3.2: Volumetric Properties of Liquids at 20°C

Molecular

Chemical Name

Formula

Specific

Isothermal

Volume

Compressibility

Expansivity

–1

V/L.kg

C2H4O2 C6H7N CS2 C6H5Cl C6H12 C4H10O C2H5OH C4H8O2 C8H10 CH3OH CCl4 C7H8 CHCl3

Acetic Acid Aniline Carbon Disulfide Chlorobenzene Cyclohexane Diethyl Ether Ethanol Ethyl Acetate m-Xylene Methanol Tetrachloromethane Toluene Trichloromethane

0.951 0.976 0.792 0.904 1.285 1.401 1.265 1.110 1.157 1.262 0.628 1.154 0.672

–5

–1

/10 bar

9.08 4.53 9.38 7.45 11. 18. 11. 11. 8.46 12. 10. 8.96 9.96

–3

/10 ·°C

–1

1.08 0.81 1.12 0.94 1.15 1.65 1.40 1.35 0.99 1.49 1.14 1.05 1.21

CRC Handbook of Chemistry and Physics, 90th Ed., pp. 6-140-6-141 and p. 15–25, CRC Press, Boca Raton, Florida,

2010.

Solution

First ethanol will be chosen as the substance to be used. In the problem the volume is held constant, both the temperature and pressure come into play here. We start with the equation dV   dT dP V

At constant volume, it becomes

  dT dP Integrating

  T2  T1   P2  P1  and solving for P2 gives  T2  T1  

 P1  P2

Plugging in the values gives 

3

C 1  5

C 

11.19 * 10 bar

1

C 

 bar  P2

Leading P2 = 126.11 bar

Solution 3.16

Problem Statement A substance for which is a constant undergoes an isothermal, mechanically reversible process from initial state (P1, V1) to final state (P2, V2), where V is molar volume. (a) Starting with the definition of , show that the path of the process is described by: V = A (T) exp (

(b) Determine an exact expression which gives the isothermal work done on 1 mol of this constant- substance. Solution

1  V  1 (a) If  is constant, then we can simply rearrange its definition,      to get dV  dP V  P  V T

V  (at constant T). This can then be integrated directly to get ln      P  P1  , or solving for V, we get  V1 

V  V1 exp  P  P1  . This is of the form V  A T exp  P  with A T   V1 exp  P1  which is a function only of T, since it is for a particular value of V and P.

(b) As in the previous problem, we have dW = –PdV = PVdP. Substituting into this the expression for V obtained above, we have dW  PVdP  PV1 exp  P  P1 dP   AT  P exp P dP P

W   AT   P exp P  dP  P1

AT  

 P  1exp P    P  1exp P  1

We could simplify this by recognizing that A(T)exp(–P1) = V1 and A(T)exp(–P) = V, so this is

W

V V 1  P1  1V1   P  1V   P1V1  PV  1   

for the work in going from (P1, V1) to (P, V).

Solution 3.17 Problem Statement

One mole of an ideal gas with CP = (7/2)R and CV = (5/2)R expands from P1 = 8 bar and T1 = 600 K to P2 = 1 bar by each of the following paths: (a) Constant volume; (b) Constant temperature; (c) Adiabatically. Assuming mechanical reversibility, calculate W, Q

H for each process. Sketch each path on a single PV

diagram.

Solution (a) Constant volume expansion from 8 bar and 600 K to 1 bar. Since PV/T is constant for an ideal gas, when P decreases by a factor of 8, T will also decrease by a factor of 8 to maintain constant volume, so in this case the final temperature is 600/8 = 75 K. As for any constant volume process, the work done is zero (W = 0). The heat transfer required is –1

–1

Q = Cv (T2 – T1) = 5/2 * R * –525 K = 2.5*8.3145 J mol K *(75 K–600 K) = –10912.78 J mol

–1

The change in internal energy is U = Q = –10910 J mol .

The change in enthalpy is –1

–1

H = Cp (T2 – T1) = 7/2 * R * –525 K = 3.5*8.3145 J mol K *(75 K–600 K) = –15277.89 J mol

–1

(b) For isothermal expansion, we know that H = U = 0. The heat flow required is –1

–1

Q = –RT ln(P2/P1) = –8.31451 J mol K * 600 K * ln(1/8) = 10373.71 J mol

–1

The work done is

W = –Q = RT ln(P2/P1) = –10373.71 J mol

–1

T2  P2  C p  1  7        0.5520 T1  P1   8 

So we have T2 = 0.5520*600 K = 331 K –1

–1

Then U = Cv (T2 – T1) = 5/2 * R * (331 K–600 K) = 2.5*8.3145 J mol K *–269 K = –5591.50 J mol –1

–1

And H = Cp (T2 – T1) = 7/2 * R * (331 K–600 K)= 3.5*8.3145 J mol K *–269 K = –7828.10 J mol

–1

By definition, Q = 0. –1

Finally, W = U = –5591.50 J mol

Solution continued on next page…

On a PV diagram, the three paths look like this:

8 bar

(a)

(b) (c) 1 bar

0.00624 m3 mol-1

0.0275 m 3 mol-1

0.0499 m 3 mol-1

Solution 3.18 Problem Statement

One mole of an ideal gas with CP = (5/2)R and CV = (3/2)R expands from P1 = 6 bar and T1 = 800 K to P2 = 1 bar by each of the following paths: (a) Constant volume (b) Constant temperature (c) Adiabatically Assuming mechanical reversibility, calculate W, Q

H for each process. Sketch each path on a single PV

diagram.

Solution

For each of these problems start with equations 3.13b, 3.14b, 3.16, and 3.17:

U   C v dT H   C p dT dQ  Cv dT  RT

dV dV and dW   RT V V

(a) at constant volume,

 U   C v dT  H   C p dT dQ  C v dT and dW 

So plugging the given values into each part gives:

3 5 3 U  RT H  RT Q  RT and dW  2 2 2

Now not knowing what T2 is, we must assume to stay at an ideal gas that since the pressure drops by a factor of 6, the temperature must drop by the same as well, leaving T2 = 800/6 = 133.33 K. Using this value gives the following values:

Q  U 

J  mol K

K

K  

J mol

J  mol K

K

K  

J mol

H 

dW 

(b) Using the same starting point as in part (a) and assuming constant temperature, reduces the equations to the following:

U  0, H  0, dQ  RT

dV dV , and dW   RT V V

Putting dQ and dW in terms of pressure (eqns 3.18 and 3.19) and integrating gives:

P  P  U  0, H  0, Q   RT * ln 2 , and W  RT * ln  2   P   P 

Putting in the known values and solving gives U  0

H  Q  

J * mol K

W  8.314

 bar   K * ln  bar 

J mol

 bar  J J     11917.35 * 800 K * ln  mol K  bar  mol

U  C v T H  C p T Cv dT  RT

dP  P

P   dQ and W  RT * ln  2   P 

However, we also have adiabatic expansion which changes T2. Taking Eqn 3.16 integrating and solving for T2 gives: 2

R /C p

P  T2  T1  2   P1 



 1bar 5 K *     6bar 

Plugging in given values and solving gives:

U  *

J  mol K

K

K 

J mol

J  mol K

K

K 

J mol

H 

Q 

W  U  

J mol

Solution 3.19 Problem Statement An ideal gas initially at 600 K and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 12, pressure decreases isothermally to 3 bar; in step 23, pressure decreases at constant volume to 2 bar; in step 34, volume decreases at constant pressure; and in step 41, the gas returns adiabatically to its initial state.

Take CP = (7/2)R and CV = (5/2)R. (a) Sketch the cycle on a PV diagram. (b) Determine (where unknown) both T and P for states 1, 2, 3, and 4. (c) Calculate Q, W

H for each step of the cycle.

Solution

(a) A rough sketch is given below. We could refine it after doing the calculations. 1

10 bar

2 3 bar 2 bar

4 0.00499 m3 mol-1

3 0.0166 m 3 mol-1

V 6

(b) For state 1, we are given that T1 = 600 K and P1 = 10 bar = 10 Pa. At these conditions, the molar volume is

V1 

RT1 P1



8.314 Pa m3mol1K1 * 600 K  0.00499 m3 /mol 10 6 Pa

Since step 12 is isothermal, we also have that T2 = 600 K, and it is specified that P2 is 3 bar = 0.3*10 Pa. At these conditions, the molar volume is Solution continued on next page…

V2 

RT2 P2



8.314 Pa m3mol 1K1 * 600 K  0.0166 m 3 /mol 0.3 * 10 6 Pa

In step 23, the pressure is reduced from 3 bar to 2 bar by cooling at constant volume. Decreasing the pressure by 2/3 6

requires decreasing the temperature by 2/3 at constant volume, so T3 = 400 K and P3 = 2 bar = 0.2*10 Pa, while 3

V3 = V2 = 0.0166 m /mol. Finally, we can find the conditions for state 4 from the fact that it can go back to state 1 via adiabatic compression. For (R/Cp)

an adiabatic process on an ideal gas, we have (T4/T1) = (P4/P1)

(2/7)

= (2 bar/10 bar)

= 0.6314.

So, T4 = 0.6314*600 K = 378.9 K. The molar volume is then

V4 

RT4 P4



8.314 Pa m3 mol 1K 1 * 400 K 0.2 * 10 6 Pa

V4 = 0.0157 m /mol.

0.00499m 3 /mol

) = 5995.89 J/mol.

For step 23, we can, as always, compute U and H from Solution continued on next page…

U = Cv T = 2.5 R*(400K–600 K) = 2.5*8.314 J mol1K1 *(–200 K) = –4157 J/mol, and H = Cp T = 3.5 R*(400K–600 K) = 3.5*8.314 J mol1K1 *(–200 K) = –5819.8 J/mol. At constant volume, W = 0, and Q = U = –4157 J/mol. For step 34, we again compute U and H from

U = Cv T = 2.5 R*(378.9 K – 400 K) = 2.5*8.314 J mol1K1 *(–21.1 K) = –438.56 J/mol, and

H = Cp T = 3.5 R*(378.9 K–400 K) = 3.5*8.314 J mol1K1 *(–21.1 K) = –613.98 J/mol. At constant pressure, Q = H = –613.98 J/mol.

Since U = Q + W (in general), we have W = U – Q = –438.56 J/mol – (–613.98 J/mol_= 175.42 J/mol.

For step 41, we again compute U and H from

U = Cv T = 2.5 R*(600 K–378.9 K) = 2.5*8.314 J mol1K1 *(221.1 K) = 4595.56 J/mol, and

H = Cp T = 3.5 R*(600 K–378.9 K) = 3.5*8.314 J mol1K1 *(221.1 K) = 6433.78 J/mol. This step is adiabatic, so Q = 0, and W = U = 4595.56 J/mol.

Solution 3.20 Problem Statement An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, pressure increases isothermally to 5 bar; in step 23, pressure increases at constant volume; and in step 31, the gas returns adiabatically to its initial state. Take CP = (7/2)R and CV = (5/2)R. (a) Sketch the cycle on a PV diagram. (b) Determine (where unknown) V, T, and P for states 1, 2, and 3. (c) Calculate Q, W

H for each step of the cycle.

Solution

(a)

For state 1, we are given that T1 = 300 K and P1 = 1 bar = 10 Pa. At these conditions, the molar volume is V1 

RT1 P1



8.314 Pa m 3mol 1K1 * 300 K 105 Pa

V1 = 0.02494 m /mol.

Solution continued on next page…

Since step 12 is isothermal, we also have that T2 = 300 K, and it is specified that P2 is 5 bar = 5*10 Pa. At these conditions, the molar volume is V1 

RT1 P1

1



1

5 * 10 Pa

0.00499 m /mol. In step 23, the pressure is increased from 5 bar to P3, which needs to be determined, at constant volume. Increasing the pressure by 5 in step 12 requires increasing the temperature by 5 at constant 3

volume, so T3 = 1500 K and P3 = 25 bar, while V3 = V2 = 0.00499 m /mol. Finally, the conditions for state 31 go back to state 1 via adiabatic expansion. No further calculations necessary. (b) Step 12 is isothermal so U = H = 0, and Q = –W = RT ln(V2/V1) = 3

–1

8.314 J mol1K1 *300 K *ln(0.00499 m mol /0.02494 m mol ) = –4013 J/mol. For step 23, we can, as always, compute U and H from U = Cv T = 2.5 R*(1500 K–300 K) = 2.5*8.314 J mol1K1 *(1200 K) = 24942 J/mol, and H = Cp T = 3.5 R*(1500 K–300 K) = 3.5*8.314 J mol1K1 *(1200 K) = 34918.8 J/mol. At constant volume, W = 0, and Q = U = 24942 J/mol. For step 31, we again compute U and H from U = Cv T = 2.5 R*(300 K–1500 K) = 2.5*8.314 J mol1K1 *(–1200 K) = –24942 J/mol, and H = Cp T = 3.5 R*(300 K–1500 K) = 3.5*8.314 J mol1K1 *(–1200 K) = –34918.8 J/mol. This step is adiabatic, so Q = 0, and W = U = –24942 J/mol.

Solution 3.21 Problem Statement 3

The state of an ideal gas with CP = (5/2)R is changed from P = 1 bar and V1t = 12 m to P2 = 12 bar and

V2t = 1 m

by the following mechanically reversible processes: (a) Isothermal compression. (b) Adiabatic compression followed by cooling at constant pressure. (c) Adiabatic compression followed by cooling at constant volume. (d) Heating at constant volume followed by cooling at constant pressure. (e) Cooling at constant pressure followed by heating at constant volume. t

Calculate Q, W, U , and H for each of these processes, and sketch the paths of all processes on a single PV diagram.

Solution

For all parts of the problem, the final temperature is the same as the initial temperature (we know this because the final PV is equal to the initial PV, and because one of the paths is specified to be isothermal). So, for all of the paths, H = U = 0. Furthermore, because for any path, U = Q + W, for each of the paths we have Q = –W. So, for each path, we just have to compute W (or Q) and we are done. (a)

For isothermal compression, t

W = P2V2 ln(P2/P1) = 12 bar *1 m * ln(12 bar/1 bar) = 29.8 bar m = 2.98  10 Pa m

W = 2980 kJ Q = –W = –2980 kJ Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(b)

For the adiabatic compression to the final pressure and some intermediate volume that we'll call Vi, we have Cv

Vi  P1  C p  1  5        0.225 V1  P2   12  3

so the intermediate volume is Vi = 0.225* 12 m = 2.70 m

The work for an adiabatic process on an ideal gas (with constant heat capacity) can be written as (see eq. 3.26)

W (adiabatic) 

P2Vi  P1V1  1



(12 bar * 2.70 m3 )  (1 bar *12 m 3 )  30.6 bar m3  30.60 *105 J  3060 kJ (5 / 3)  1 3

The work for the cooling at constant pressure to the final volume of 1 m is just

W (isobaric) = –PV = –12 bar * (1 m – 2.70 m ) = 20.4 bar m = 20.4 *10 J = 2040 kJ

The total work is therefore W = 3060 kJ + 2040 kJ = 5100 kJ, and Q = –5100 kJ

(c)

For adiabatic compression to the final volume and some intermediate pressure that we'll call Pi, we have Cp

5 Pi  V1  C v     12 3  62.90  P1  V2 

So, the intermediate pressure is 1 bar * 62.90 = 62.90 bar

Solution continued on next page…

The work for the adiabatic step is then

W (adiabatic) 

PV  P1V1 i 2  1



(62.9bar * 1 m3 )  (1 bar *12 m3 )  76.3 bar m3  76.30 *105 J  7630 kJ 5 / 3 1

No work is done in the constant volume step, so the total work is W = 7630 kJ, and Q = –7630 kJ

(d)

No work is done during the heating at constant volume to the final pressure of 12 bar. The work done during

the subsequent cooling at constant pressure is just 3

W (isobaric) = –PV = –12 bar * (1m – 12 m ) = 132 bar m = 132 *10 J = 13200 kJ

So, the total work is W = 13200 kJ, and Q = –13200 kJ

(e)

When the gas is cooled at constant pressure, the work done is 3

W (isobaric) = –PV = –1 bar * (1 m – 12 m ) = 11 bar m = 11 * 10 J = 1100 kJ

No work is done during the subsequent heating at constant volume, so the total work is W = 1100 kJ, and Q = –1100 kJ

Solution continued on next page…

The various paths are sketched on the following PV diagram:

(c)

12 bar

(b)

(d)

(a) (e)

1 bar 1 m3

12 m 3

Solution 3.22 Problem Statement The environmental lapse rate dT/dz characterizes the local variation of temperature with elevation in the earth’s atmosphere. Atmospheric pressure varies with elevation according to the hydrostatic formula, dP    g dz

where  is molar mass, is molar density, and g is the local acceleration of gravity. Assume that the atmosphere is an ideal gas, with T related to P by the polytropic formula: TP

)

= constant

Develop an expression for the environmental lapse rate in relation to , g, R, and .

Solution

First use the equation 3.23b to start: /

= constant

Differentiate this equation with respect to T: 1   1       1    1   1         dP      dT   1    P   dP    dT   * P  T   *   P      P  T           dz  dz   dz     P  dz 

 T  1    dP dT  1       P    dz  dz  P  0

T  1    dP dT   0  P    dz dz

Substituting the given equation for dP/dz and combining the terms reduces the equation to T  1    dT  * M  g     P    dz

We know that

/ = 1/ and substituting this allows the equation to be reduced to

Mg  1    dT   R    dz

Solution 3.23 Problem Statement An evacuated tank is filled with gas from a constant-pressure line. Develop an expression relating the temperature of the gas in the tank to the temperature T and ignore heat transfer between the gas and the tank.

Solution

The mass balance on the tank is

Inlet mass flow rate = m ' enthalpy = H’

Inside tank, m = 0 at t = 0

dm  m ' dt

and the energy balance on the tank is d mU  dt

 m ' H '

where m is the total mass in the tank and U is the specific internal energy of the gas in the tank. Q and W are both zero because the tank is both rigid (no non-flow work) and insulated (no heat flow through its boundaries). We can combine the two balance equations to get d mU  dt



dm H' dt

Integrating these, and using the fact that the enthalpy of the gas coming into the tank is constant, we get

mU t  mU t0  mH 't  mH 't0 Solution continued on next page…

where the subscript t denotes the value at some arbitrary time t and the subscript t = 0 denotes the value at t = 0. Since m is zero at t = 0, this is simply

mU t  mH 't or U = H' for any time. So, the specific internal energy of the gas in the tank is equal to the enthalpy of the gas entering the tank. The specific enthalpy of the gas in the tank is H = U + PV = U + RT = H' + RT where T is the temperature of the gas in the tank and H is the specific enthalpy of the gas in the tank. The enthalpy change when the gas goes from the temperature of the inlet line to the temperature of the tank is H – H' = Cp(T – T'). Substituting this into the above equation gives H – H' = Cp(T – T') = RT Or CpT' = (Cp – R)T = CvT Or T = CpT'/Cv = T'

The temperature of the gas inside the tank will be higher than that of the gas entering the tank.

Solution 3.24 Problem Statement 3

A tank of 0.1 m volume contains air at 25°C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at the constant conditions of 45°C and 1500 kPa. A valve in the line is cracked so that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly enough that the temperature in the tank remains at 25°C, how much heat is lost from the tank? Assume air to be an ideal gas for which CP = (7/2)R and CV = (5/2)R.

Solution

We can make a simple sketch of the situation as shown below.

Tsupply =318 K P supply =1500 kPa T= 298 K

Now, we need to write mass (or mole) and energy balances on the gas in the tank. We will use a control volume that includes the gas inside the tank and extends just upstream of the valve (so the pressure and temperature of the gas entering the system are 1500 kPa and 318 K, respectively). The mole balance on the gas in the tank is just dn  n dt

where n is the molar flow rate into through the valve (which need not be constant). The corresponding energy balance is: d nU  dt

 supply  Q  nH

where Q is the rate of heat addition to the system (it will be negative, since heat is removed from the system). The work term is zero, because no shaft work is done (there are no moving parts). We can expand the derivative on the left hand side using the product rule, and replace Hsupply by Usupply + PsupplyVsupply = Usupply + RTsupply. This gives

dU dn dn U  n U supply  RTsupply   Q  U supply  RTsupply   Q dt dt dt

Where in the last expression we have used the mole balance to substitute

dn for n . Now, since the temperature dt

inside the cylinder remains constant, U also remains constant (for an ideal gas, U is only a function of T). So, the first term on the LHS is zero. Solving for Q then leaves dn dn dn Q  U  U  RTsupply   U  U supply  RTsupply  dt dt supply dt

This says that the rate of heat addition is directly proportional to the rate at which gas flows into the cylinder. Note that everything inside the parenthesis on the RHS of the equation is constant. The total heat added is just the integral of this over the whole process.   U  U Q   Qdt  RTsupply  supply 

dn dt  U  U supply  RTsupply n final  ninitial  dt

We also recognize the U – Usupply = Cv(T – Tsupply) = 5/2 R*–20 K = (–50 K)*R, and RTsupply = (318 K)*R. So, the RT term is much bigger than the U–Usupply term. 3

–1

nfinal – ninitial = (V*(Pfinal – Pinitial) )/(RT) = (0.1 m *(1500000 Pa – 101330 Pa)) / (8.314 Pa*m mol K * 298 K) = 56.45 mol. –1

–1

So, finally, Q = –368 K * 8.314 J mol K * 56.45 mol = 172711.71 J = 172.71 kJ. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 3.25 Problem Statement Gas at constant T and P is contained in a supply line connected through a valve to a closed tank containing the same gas at a lower pressure. The valve is opened to allow flow of gas into the tank, and then is shut again. (a) Develop a general equation relating n1 and n2, the moles (or mass) of gas in the tank at the beginning and end of the process, to the properties U1 and U2, the internal energy of the gas in the tank at the beginning and end of the process, and H

Q, the heat transferred to the material in the tank

during the process. (b) Reduce the general equation to its simplest form for the special case of an ideal gas with constant heat capacities. (c) Further reduce the equation of (b) for the case of n1 = 0. (d) Further reduce the equation of (c) for the case in which, in addition, Q = 0. (e) Treating nitrogen as an ideal gas for which CP = (7/2)R, apply the appropriate equation to the case in which a 3

steady supply of nitrogen at 25°C and 3 bar flows into an evacuated tank of 4 m volume, and calculate the moles of nitrogen that flow into the tank to equalize the pressures for two cases: 1. Assume that no heat flows from the gas to the tank or through the tank walls. 2. Assume that the tank weighs 400 kg, is perfectly insulated, has an initial temperature of 25°C, has a 1

specific heat of 0.46 kJ·kg ·K , and is heated by the gas so as always to be at the temperature of the gas in the tank.

Solution

(a)

We can write a mole balance on the material in the tank as dn  n ' dt

where n ' is the molar flow rate of gas entering the tank. This is (for this non-reacting system) equivalent to the mass balance dm  m ' dt

and can be obtained from it by simply dividing by the molecular weight of the gas.

The corresponding energy balance is d nU  dt

 n ' H ' Q

where we have written this using the molar flow rate and the molar enthalpy and internal energy. This differs in form from problem 3.12 where we wrote it in terms of the mass flow rate and specific enthalpy and internal energy. Also, we have not assumed that the tank is adiabatic, so this equation has a heat flow term in it. However, we have assumed that the tank is rigid, so there is no non-flow work term. We can combine the energy and mole balances to get d nU  dt



dn H ' Q dt

Solution continued on next page…

and we can integrate this over an arbitrary time interval (from time t1 to time t2) to get n2U 2  n1U 1  n2  n1  H ' Q

where Q is the total amount of heat that flows into the tank during the time interval from t1 to t2, n1 and n2 are the number of moles of gas in the tank at times t1 and t2, and U1 and U2 are the molar internal energy of the gas in the tank at times t1 and t2. This can also be written as n2 U 2  H '  n1 U 1  H '  Q

(b) For an ideal gas with constant heat capacities, U2

H' can be written as

U 2  H '  U 2  U ' P ' V '  U 2  U ' RT '  C v T2  T '  RT '  C vT2  C pT '

and likewise for U1

H' , so

n2 CvT2  C pT '  n1 C vT1  C pT '  Q

n2 CvT2  C pT '  Q

(d) If Q is zero, then this is

n2 CvT2  C pT '  0 or CvT2  C pT '

Solution continued on next page…

or T2  T '

just as we found in problem 3.12.

(e) 1. This corresponds to the case in part (d) where the temperature in the tank is given by

T2   T ' with  = 1.4, so if the supply temperature is 298 K, the temperature of the gas in the tank will be 1.4*298 K = 417.2 K. When the pressure in the tank is equal to the supply pressure, gas will stop flowing. At 3 bar and 417.2 K the ideal gas law tells us that the number of moles in the tank will be

n

PV 300000 Pa  4 m3   346 mol RT 8.3145 Pa m3 mol1 K  417.2 K

(e) 2. The heat transferred to the tank is mtankCp,tank(T2 – 25 °C) or mtankCp,tank(T2 – 298 K). The heat transferred to the gas –1

–1

is the negative of this, so we have Q = –400 kg * 460 J kg K (T2 – 298 K), with T2 in K. Using this with the result from part (c) gives n2 C vT2  C pT '  mtank C p,tank T2  298 K n2 2.5T2  3.5  298 K  8.3145 J mol1 K  400 kg  460 J kg1 K 1 T2  298 K

We also know that

n2 

P2V2 300000 Pa  4 m 3  RT2 8.3145 Pa m3 mol1 K1 T2

Solution continued on next page…

Substituting this in the previous equation gives us an equation for T2 alone (without n2): 1200000 2.5T2  3.5 298 K  400  460 T2  298 K T2 3000000T2  1251600000  184000T22  54832000T2 T22  281.70T2  6802.2  0 T2  304.1 K or 44.7 K (of course only the first answer is physically meaningful)

So then if the final temperature is 304.1 K, the final number of moles is n2 = 1200000/(8.31451*304.1) = 475 mol

This is substantially more than in the previous part, because the temperature rise is small, because the heat capacity of the tank is very large compared to the heat capacity of the gas.

Solution 3.26 Problem Statement Develop equations that may be solved to give the final temperature of the gas remaining in a tank after the tank has been bled from an initial pressure P1 to a final pressure P2. Known quantities are initial temperature, tank volume, heat capacity of the gas, total heat capacity of the containing tank, P1, and P2. Assume the tank to be always at the temperature of the gas remaining in the tank and the tank to be perfectly insulated.

Solution

We can re-use the sketch from the last problem, with slight modifications:

T = f(P) ?

Our goal is to develop a relationship between the temperature in the tank and the pressure in the tank. As in the problem 3.14, we will write mole and energy balances on the tank. This time, however, the tank is insulated on the outside (so Q = 0) but the tank itself changes temperature with the gas. We will take as our control volume the tank and the gas inside the tank, up to a point just before the valve. Then the mole balance is dn  n dt

and the energy balance is dU t   nH dt t

in the energy balance, Q = W = 0. U is the total internal energy of the system, which includes both the tank and the t

gas inside. So, we can write U = Utank + nU, where Utank is the total internal energy of the tank and U is the molar internal energy of the gas. We can also write H = U + PV = U + RT. Subsituting these in gives Solution continued on next page…

dU d dU dn dn dn U tank  nU   tank  n  U  n U  RT   U  RT dt dt dt dt dt dt or dU tank dU dn n  RT dt dt dt

where I have gone ahead and expanded the derivative on the left hand side using the product rule, and have used the mole balance to substitute for n on the right-hand-side. Now, we can replace the internal energy terms with temperatures by noting that dUtank = Cv,tankdT and dU = CvdT, where Cv,tank is the total heat capacity of the tank, and Cv is the molar heat capacity of the gas. Canceling the term that appears on both sides of the equation above and then substituting these gives dT

Cv,tank  nCv  dt  RT dt

Rearranging this to put the T’s on one side and the n’s on the other, then integrating from the initial to final number of moles in the tank gives 1 dT R dn  T dt dt C  nC  v ,tank v  T  R  Cv ,tank  n2C v   ln  2   ln   T1  C v  Cv ,tank  n1Cv 

This gives us a relationship between the number of moles in the tank and the temperature, but we want a relationship between the pressure in the tank and the temperature. We can substitute for the number of moles in terms of the temperature and pressure using the ideal gas law: n = (PVtank)/(RT). This gives Solution continued on next page…

  PV  C  2 tank C v  v , tank  T  R   RT2  ln  2   ln      T1  C v  PV  Cv ,tank  1 tank Cv   RT1 

or R

 P2Vtank Cv  C  C  v , tank T2  RT2 v     P1Vtank  T1  C   Cv ,tank   RT1 v 

This could be solved numerically for T2 given all of the other information.

Solution 3.27

Problem Statement 3

A rigid, nonconducting tank with a volume of 4 m is divided into two unequal parts by a thin membrane. One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bar and 100°C, and the other side, representing 2/3 of the tank, is evacuated. The membrane ruptures and the gas fills the tank. (a) What is the final temperature of the gas? How much work is done? Is the process reversible? (b) Describe a reversible process by which the gas can be returned to its initial state. How much work is done?

Assume nitrogen is an ideal gas for which CP = (7/2)R and CV = (5/2)R.

Solution

(a) In the initial state, the gas has a volume of /3 m , at a pressure of 6 bar (600000 Pa) and a temperature of 100 ºC (373.15 K). Taking the system to be the entire contents of the tank, the total internal energy of the system is just the internal energy of the part that has the gas in it. The internal energy in the evacuated part is, by definition, zero. It is empty space. Since the tank is nonconducting (insulating) and rigid (constant volume) no heat or work flows enter or 3

leave the tank. Thus, the internal energy in the final state, when the gas has expanded to a volume of 4 m is the same as before the expansion. We can write the first law for the contents of the tank before and after the expansion as U = Q + W = 0. For an ideal gas, the internal energy is only a function of temperature. Therefore, if the internal energy does not change, the temperature of the gas does not change. The final temperature is 100ºC. No work is done. The process is irreversible. The gas expands across a finite pressure difference. Intuitively, this is obvious if one tries 1

to imagine the reverse process, in which the gas all spontaneously flows into /3 of the tank.

(b) One process to return the gas to its initial state would be reversible isothermal compression, from 4 m and a 4

pressure of 2 bar to /3 m and 6 bar. For isothermal compression, W = –Q = nRTln(V1/V2) = nRTln(3). Since nRT is 3

equal to PV and is constant, we have W = –Q = 4 m * 200000 Pa * ln(3) = 878900 Pa m = 878900 J.

Solution 3.28

Problem Statement

An ideal gas, initially at 30°C and 100 kPa, undergoes the following cyclic processes in a closed system: (a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30°C, and finally expanded isothermally to its original state. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(b) The cycle traverses exactly the same changes of state, but each step is irreversible with an efficiency of 80% compared with the corresponding mechanically reversible process. Note: The initial step can no longer be adiabatic.

Calculate Q, W

H for each step of the process and for the cycle. Take CP = (7/2)R and CV = (5/2)R.

Solution

State 1 is 303 K and 100 kPa, for which

V 

RT P



1

100 * 10 Pa

V = 0.0252 m /mol. For adiabatic compression from 100 kPa to 500 kPa, we have (R/Cp)

(T2/T1) = (P2/P1)

2/7

= (500 kPa/100 kPa)

= 1.584. So, the temperature after compression is

T2 = 1.584*303 K = 480 K. The final volume after compression will then be

V2 

RT2 P2



1

500 * 10 3 Pa

V2= 0.007981 m /mol.

Solution continued on next page…

As always for an ideal gas, U = Cv T = 2.5 R*(480 K – 303 K)) = 3679 J/mol, and H = Cp T = 3.5 R*(480 K – 303 K) = 5150 J/mol. For an adiabatic process, Q = 0, and W = U = 3679 J/mol.

If we cool the gas at constant pressure back to the original temperature, the enthalpy and internal energy go back to their original values, and we will have U = Cv T = 2.5 R*(303 K – 480 K) = –3679 J/mol, and H = Cp T = 3.5 R*(303 K – 480 K) = –5150 J/mol. During this process, the volume will decrease to 3

R*303 K / 500000 Pa = 0.005039 m /mol. The work done will be W = – PV = –500000 Pa * (0.005039 – 0.007981) m –1

–1

mol = 1471 Pa m mol = 1471 J/mol. The heat removed can be found from

Q = U W = –3679 J/mol

1471 J/mol = –5150 J/mol.

Finally, if we expand the gas back to its original state isothermally, we will have U = H = 0 (since they only depend on temperature, which doesn’t change). For the isothermal process, we will have Q = RT ln(P1/P2) = R*303 K * ln(5) = 4055 J/mol, and W = –Q = –4055 J/mol.

For the overall cycle, Q = 0 – 5150 + 4055 = –1095. W must then be 1095 for the cycle (so that the overall U is zero). To check, we can add up the three steps: W = 3679 + 1471 – 4055 = 1095. Of course U and H are zero for any cyclic process.

Solution continued on next page…

(b) In the first step, if we compress it to 500 kPa with a process that is 80% efficient, and still adiabatic, then we will put in W = 3679/0.8 = 4599 J of work to do the compression. If Q is still zero, then we will have U = 4599 J = Cv T, so T will be 221 K. In compressing the gas adiabiatically with an 80% efficient process, we will heat it to 524 K rather than 480 K. The enthalpy change will then be H = Cp T = 3.5 R*(221 K) = 6431 J/mol. 3

Because of the higher temperature, the molar volume will then be 0.008797 m /mol rather than 0.007981 m /mol that was found in (a).

If we cool the gas back to the original temperature, as in the reversible case, the enthalpy and internal energy will go back to their original values, and we will have U = Cv T = 2.5 R*(–221) = –4599 J/mol, and H = Cp T = 3.5 R*(–221) = –6431 J/mol. The volume after this step will be the same as in part (a): 3

R*303 K / 500000 Pa = 0.005039 m /mol. The reversible work required to cool the gas isobarically is then 3

–1

W = – PV = –500000 Pa * (0.005039 m /mol – 0.008797 m /mol) = 1879 Pa m mol = 1879 J/mol. So, if our process is 80% efficient, the total work required will be 1879/0.8 = 2349 J/mol. We still have U = –4599 J/mol = Q + W = Q + 2349, so Q = –6948 J/mol.

For the isothermal expansion, we will still have U = H = 0 (since they only depend on temperature, which doesn’t change). For a reversible isothermal process, we had Q = RT ln(P1/P2) = R*303 K * ln(5) = 4055 J/mol and W = Q = 4055 J/mol. If the process is only 80% efficient, then the work will be 0.8*(–4055) = –3244, and the heat requirement will be set by

Q = U W = 0 J/mol

(3244 J/mol) = 3244 J/mol.

For the overall cycle with 80% efficiency in each step, Q = 0 – 6948 +3244 = –3704 J/mol. W must then be 3704 for the cycle (so that the overall U is zero). To check, we can add up the three steps: W = 4599 + 2349 – 3244 = 3704 J/mol. Of course U and H are zero for any cyclic process, whether the steps are reversible or not. Note that the irreversibilities compound themselves so that the net work we put in here for the whole cycle is much more than 1.25 times (or 1/0.8) the net work that we put in for the reversible cycle in part (a).

Solution 3.29

Problem Statement

One cubic meter of an ideal gas at 600 K and 1000 kPa expands to five times its initial volume as follows: (a) By a mechanically reversible, isothermal process. (b) By a mechanically reversible, adiabatic process. (c) By an adiabatic, irreversible process in which expansion is against a restraining pressure of 100 kPa. 1

For each case calculate the final temperature, pressure, and the work done by the gas. Take CP = 21 J·mol ·K .

Solution

(a)

For isothermal expansion, the final temperature is, by definition, the initial temperature of 600 K. The final th

pressure is 1/5 of the initial pressure, or 200 kPa. The work done on the gas is W = –nRTln(V2/V1) = 6

–PVln(V2/V1) = –10 Pa*1m *ln(5) = –1.61  10 J = –1610 kJ. The work done by the gas is 1610 kJ, and the heat flow into the gas is 1610 kJ.

Solution continued on next page…

(b)



For the adiabatic process, the final pressure is given by P2/P1 = (V1/V2) . The heat capacity ratio is 1.66

 = 21/(21 – 8.3145) = 1.66, so P2 = P1 (1/5)

= 0.0696 P1 = 69.6 kPa = 69600 Pa R/Cv

The final temperature is given by T2 = T1 (V1/V2)

0.6554

where R/Cv = 8.3145/(21 – 8.3145) = 0.6554, so T2 = T1 (1/5)

= 0.348 T1 = 0.348*600 K = 208.9 K.

The work done on the gas can be computed as 3

(P2V2 – P1V1)/( – 1) = (5 m *69600 Pa – 10 Pa *1 m )/(1.66–1) = –987878 J = –987.87 kJ. The work done by the gas is equal to 987.87 kJ.

(c)

In this case, the work is done against a constant resisting pressure of 100 kPa, so the total work done on the t

gas is just –PV = –100000 Pa * (5 m – 1 m )= –400000 J = –400 kJ. This is a lot less than in the reversible t

case in part (b). In general, for a closed system, U = Q + W. In this case, the process is adiabatic, so Q = 0 t

and we have U = W = –400 kJ. We also know that for a closed system, U = nCvT, so we can get the final temperature from t

T = U /nCv We can get n from the initial pressure, volume, and temperature, to be n = PV/RT = (1000000*1)/(8.314*600) = 200.5 mol, so –1

–1

T = –400000 J/ (200.5 mol * (21 – 8.3145) J mol K ) = –157.3 K

Solution continued on next page…

T2 = 600 K – 157.3 K = 442.7 K The gas doesn’t cool off nearly as much as in (b) because it doesn’t do nearly as much work. The final pressure is P = nRT/V = 200.5*8.3145*442.7/5 = 147605.39 Pa = 147.61 kPa

Solution 3.30

Problem Statement

One mole of air, initially at 150°C and 8 bar, undergoes the following mechanically reversible changes. It expands isothermally to a pressure such that when it is cooled at constant volume to 50°C its final pressure is 3 bar. Assuming air is an ideal gas for which CP = (7/2)R and CV = (5/2)R, calculate W, Q

Solution

For the initial state of 150 °C and 8 bar, the molar volume is 3

–1

V = 8.314 Pa m mol K *423 K/800000 Pa = 0.004396 m mol .

If the final state is 50 °C and 3 bar, then the molar volume at the final state is 3

–1

V = 8.314 Pa m mol K *323 K/300000 Pa = 0.008951 m mol .

Solution continued on next page…

This is also the molar volume at the intermediate state, since the gas goes from the intermediate state to the final state at constant volume. The temperature at the intermediate state is 150 °C, since the gas goes from the initial state to the intermediate state isothermally. So, the pressure at the intermediate state is 3

–1

P = 8.314 Pa m mol K *423 K/0.008951 m mol = 392897 Pa = 3.929 bar.

For the isothermal expansion, U = H = 0 and Q = –RT ln(P2/P1) = –R*423 K *ln(3.929 bar/8 bar) = 2501 J/mol, and W = –Q = –2501 J/mol. For the isochoric cooling, U = Cv T = 2.5 R*(323 K–423 K) = –2079 J/mol, and H = Cp T = 3.5 R*(323 K – 423 K) = –2910 J/mol. At constant volume, W = 0, so Q = U = –2079 J/mol. For the overall 2-step process, we then have U = –2079 J/mol, H = –2910 J/mol, Q = 422 J/mol, and W = –2501 J/mol.

Solution 3.31

Problem Statement

An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The crosssectional area of the tube changes with length, and this causes the velocity to change. Derive an equation relating the 1

temperature to the velocity of the gas. If nitrogen at 150°C flows past one section of the tube at a velocity of 2.5 m·s , 1

what is its temperature at another section where its velocity is 50 m·s ?

Assume CP = (7/2)R.

Solution

So first let’s start with eqn 2.31 assuming no heat added and no work added:

H 

u22  u12 0 2

To get a relationship between temperature and velocity, we know that H  C p T

Substituting this in for H

T gives:

u2  u12 T   2 2C p

T2  T1  

T2  T1 

u22  u12 2C p

So now that the equation is derived, nitrogen is flowing at T1 = 150 C at u1 = 2.5 m/s to a point where the velocity reaches u2 = 50 m/s, we need to determine what T2 is. By also know that Cp = 7/2 R/MWN2

T2 

K

(50 m/s) 2  (2.5 m/s)2 * J 2 * 3.5 * 8.314 molK





Solution 3.32 Problem Statement One mole of an ideal gas, initially at 30°C and 1 bar, is changed to 130°C and 10 bar by three different mechanically reversible processes:  The gas is first heated at constant volume until its temperature is 130°C; then it is compressed isothermally until its pressure is 10 bar.  The gas is first heated at constant pressure until its temperature is 130°C; then it is compressed isothermally to 10 bar.  The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130°C.

Calculate Q, W

H in each case. Take CP = (7/2)R and CV = (5/2)R.

Alternatively, take CP = (5/2)R and CV = (3/2)R.

Solution

Of course, the overall U and H have to come out the same in every case, since the initial and final states are the same, but the total Q and W should, in general, be different for each case.

Solution continued on next page…

(a) heating at constant volume followed by compression at constant temperature: For the constant volume heating, W = 0, so Q = U = CvT = 5/2*R*(403 K – 303 K) = 2079 J/mol. As usual, H = CpT = 7/2*R*(403 K – 303 K) = 2910 J/mol. For the isothermal compression, U = H = 0, and Q = –W = –RT ln(Pfinal/Pinitial). The initial pressure for this second step is 1 bar * (403 K/303 K) = 1.330 bar. So, Q = –W = –R*303 K*ln(10/1.330) = –5082.438 J/mol. Adding the two steps gives, for the overall process, Q = 2079 J/mol – 5082.438 J/mol = –3003.438 J/mol, W = 5082.438 J/mol, U = 2079 J/mol, and H = 2910 J/mol. (b) Heating at constant pressure followed by compression at constant volume: For constant pressure heating, Q = H = CpT = 7/2*R*(403 K – 303 K) = 2910 J/mol. As usual, U = CvT = 5/2*R*(403 K – 303 K) = 2079 J/mol. Finally, W = U – Q = –RT = –831 J/mol. As in part (a), for the isothermal compression, we have U = H = 0, and Q = –W = –RT ln(Pfinal/Pinitial). So, Q = –W = –R*303 K*ln(10/1) = –5800.09 J/mol. Adding the two steps gives, for the overall process, Q = 2910 J/mol – 5800.09 J/mol = –2890.09 J/mol, W = 5800.09 J/mol – 831 J/mol = 4969.09 J/mol, U = 2079 J/mol, and H = 2910 J/mol.

(c) Isothermal compression followed by heating at constant pressure: For the isothermal compression, we have as in part (b) U = H = 0, and Q = –W = –RT /n(Pfinal/Pinitial). So, Q = –W = –R*303 K*ln(10/1) = –5800.09 J/mol. For the subsequent heating at constant pressure, Q = H = CpT = 7/2*R*(403 K – 303 K) = 2910 J/mol. As usual, U = CvT = 5/2*R*403 K – 303 K) = 2079 J/mol. Finally, W = U – Q = –RT = –831 J/mol. Adding the two steps gives, for the overall process, Q = 2910 J/mol– 5800.09 J/mol = –2890.09 J/mol, W = 5800.09 J/mol – 831 J/mol = 4969.09 J/mol, U = 2079 J/mol, and H = 2910 J/mol.

Solution 3.33 Problem Statement

One mole of an ideal gas, initially at 30°C and 1 bar, undergoes the following mechanically reversible changes. It is compressed isothermally to a point such that when it is heated at constant volume to 120°C its final pressure is 12 bar. Calculate Q, W

H for the process.

Solution

For the initial state of 30 °C and 1 bar, the molar volume is 3

–1

V = 8.314 Pa m mol K *303 K/100000 Pa = 0.02519 m mol .

If the final state is 120 °C and 12 bar, then the molar volume at the final state is 3

–1

V = 8.314 Pa m mol K *393 K/1200000 Pa = 0.002723 m mol .

This is also the molar volume at the intermediate state, since the gas goes from the intermediate state to the final state at constant volume. The temperature at the intermediate state is 30 °C, since the gas goes from the initial state to the intermediate state isothermally. So, the pressure at the intermediate state is 3

–1

P = 8.314 Pa m mol K *303 K/0.002723 m mol = 925134 Pa = 9.251 bar.

For the isothermal compression, U = H = 0 and Q = –RT ln(P2/P1) = –R*303 K*ln(9.251/1) = –5605 J/mol, and W = –Q = 5605 J/mol.

For the isochoric heating, U = Cv T = 2.5 R*(393 K – 303 K) = 1871 J/mol, and H = Cp T = 3.5 R*(393 K – 303 K) = 2619 J/mol. At constant volume, W = 0, so Q = U = 1871 J/mol. For the overall 2-step process, we then have U = 1871 J/mol, H = 2619 J/mol, Q = 1871 J/mol 5605 kJ/mol = –3734 J/mol, and W = 5605 J/mol.

Solution 3.34 Problem Statement One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochorically to reach a final state of 25°C and 1 bar. Assuming these processes are mechanically reversible, compute T and P after the adiabatic expansion, and compute Q, W

H for each step and for the overall process. Take

CP = (7/2)R and CV = (5/2)R.

Solution

This is a two step process. First state the given information,T1 = 298.15 K and P1 = 10 bar = 10 * 10 Pa. At these 3

conditions, the molar volume is V1 = 0.002479 m /mol (from V = RT/P). 5

If the final state is 25 °C = 298.15 K and 1 bar = 1 * 10 Pa, then the molar volume at the final state is 3

V3 = 0.02479 m /mol (from V = RT/P). Solution continued on next page…

This is also the molar volume at the intermediate state, since the gas goes from the intermediate state to the final state at constant volume. We know that Q = 0 due to it being adiabatic which leads to the relation R/Cv

T2 = T1 (V1/V2)

=118.7 K. Knowing T2, P2 = P1 (V1/V2)

So for the adiabatic expansion, Q

Cp /Cv

= 0.398 bar.

U = W = Cv T = 2.5 R*(118.7 K – 298.15 K) = –3729 J/mol, and

H =Cp T = 3.5 R*(118.7 K – 298.15 K) = –5221 J/mol.

For the isochoric heating, U = Cv T = 2.5 R*(298.15 K – 118.7 K)= 3729 J/mol, H =Cp T = 3.5 R*(298.15 K – 118.7 K)) = 5221 J/mol. At constant volume, W = 0, so Q = U = 3729 J/mol. For the overall 2-step process, we then have U = 3729 J/mol, H = 5221 J/mol, Q = 3729 J/mol, and

W = –3729 J/mol.

Solution 3.35 Problem Statement A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bar is cooled at constant volume to T = 350 K. (2) The air is then heated at constant pressure until its temperature reaches 800 K. If this two-step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to some final pressure P, what is the value of P that makes the work of the two processes the same? Assume mechanical reversibility and treat air as an ideal gas with CP = (7/2)R and CV = (5/2)R.

Solution

The initial volume of 1 mol of air at 800 K and 4 bar (400000 Pa) is V = RT/P = 8.3145 * 800 / 400000 = 0.01663 m . When it is cooled at constant volume to 350 K, the pressure drops to 4 bar * 350 K/800 K = 1.75 bar = 175000 Pa. No work is done during this constant volume cooling step. When it is heated back to 800 K at constant pressure, the 3

volume increases to is V = RT/P = 8.3145 * 800 / 175000 = 0.03801 m . The work done on the gas is 3

–PT = –175000 Pa * (0.03801 m – 0.01663 m ) = –3742 J. So, the work done by the gas is 3742 J.

For an isothermal expansion, the work will be W = RT ln(P2/P1), or P2 = P1 exp(W/RT), or –1

–1

P2 = 4 bar * exp(–3742 J mol /(8.3145 J mol K * 800 K)) = 2.27 bar. So, the isothermal expansion does the same amount of work with less expansion of the gas.

Solution 3.36 Problem Statement One cubic meter of argon is taken from 1 bar and 25°C to 10 bar and 300°C by each of the following two-step paths. For each path, compute Q, W

H for each step and for the overall process. Assume mechanical reversibility

and treat argon as an ideal gas with CP = (5/2)R and CV = (3/2)R. (a) Isothermal compression followed by isobaric heating. (b) Adiabatic compression followed by isobaric heating or cooling. (c) Adiabatic compression followed by isochoric heating or cooling. (d) Adiabatic compression followed by isothermal compression or expansion. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

(a) Argon goes from 1 bar and 25 C isothermally compressed to 10 bar, and isobaric heating to 300 C. First, the 3

volume at the initial state ( P 1= 1 bar and T1 = 25 C) is the V1 = 0.02479 m /mol (from V = RT/P). The volume at 3

the final state (P 2 = 10 bar and T 2 = 300 C)is the V2 = 0.004765 m /mol.

So for the isothermal compression, U = 0, Q = RT1 ln (P1/P2) = –5707 J/mol, W = –Q = –(–5707 J/mol) = 5707 J/mol, and H = 0.

For the isobaric heating, U = CvT = 1.5* R * (573.15 K – 298.15 K) = 3429 J/mol, H = Q = CpT = 2.5* R * (573.15 K – 298.15 K) = 5715 J/mol, W = –RT = R * (573.15 K – 298.15 K) = –2286 J/mol. Altogether, W =5707 J/mol – 2286 J/mol = 3421 J/mol, U =3429 J/mol, Q = –5707 J/mol + 5715 J/mol = 8 J/mol, H = 5715 J/mol. (b) Argon goes from 1 bar and 25 C adiabatically compressed to 10 bar, and isobaric heating to 300 C. Use the same volumes as in part (a).

So for the adiabatic compression, Q = 0,

 P  CP Tintermidiate  T1  2    P  1

U  W  CV T 

H

   

5    1



U W 

1



1





1



1

For the isobaric heating, U  CV T 



1







1

H  C P T 



1







1

H = Q = 4930.26 J/mol, 1

W  R T  

1





1

 

Altogether,

W = –1972.11 J/mol + 471.366 J/mol = –1500.74 J/mol, U = 471.366 J/mol + 2958.16 J/mol = 3429.52 J/mol, Q = 4930.26 J/mol, H = 4930.26 J/mol + 942.732 J/mol = 5873 J/mol.

(c) Argon goes from 1 bar and 25 C adiabatically compressed to 10 bar, and isochoric heating to 300 C. Use the same volumes as in part (a).

So for the adiabatic compression, Q = 0,

 P  CP Tintermidiate  T1  2    P  1

U  W  CV T 

H

   

5   

1



U W 

1



1





1



1

Solution continued on next page…

For the isochoric heating, Q 

U  CV T 

H  CP T 

1

 1



1



 

1

 1



W = 0 J/mol.

Altogether, W = 471.366 J/mol, U = 471.366 J/mol + 2958.16 J/mol = 3429.52 J/mol, H = 4930.26 J/mol + 942.732 J/mol = 5873 J/mol. Q = 2958.16 J/mol.

(d) Argon goes from 1 bar and 25 C adiabatically compressed to 300C, and isothermal compression to 10 bar. Use the same volumes as in part (a).

So for the adiabatic compression, Q = 0,

 P  CP Tintermidiate  T1  2    P  1

U  W  CV T 

H

U W 

   

5   

1



1







1



1

For the isothermal compression, H = U = 0, Q = RT1 ln (P1/P2) = –5707 J/mol, and W = –Q = 5707 J/mol

Altogether, W =

U = 471.366 J/mol, Q = –5707J/mol,

H = 942.732 J/mol.

Solution 3.37 Problem Statement t

A scheme for finding the internal volume VB of a gas cylinder consists of the following steps. The cylinder is filled with a gas to a low pressure P1, and connected through a small line and valve to an evacuated reference tank of t

known volume VA . The valve is opened, and gas flows through the line into the reference tank. After the system P in the t

cylinder. Determine the cylinder volume VB from the following data:  VAt

256 cm

 P/ P1

0.0639

Solution

Before the valve is opened, when all of the gas is in cylinder B, at pressure P1, we can write the number of moles of gas (assuming it to be ideal) as t

n = P1VB /(RT) After the gas is allowed to expand into the second tank and return to the initial temperature, we can write the total number of moles in terms of the final pressure (P1 n = (P1

P) and total volume (VA + VB ) as

P)*(VA + VB )/(RT)

Both the number of moles of gas and the temperature are the same before and after the expansion. So, we can write (P1

P)*(VA + VB ) = P1VB

Solving this for the unknown volume VB gives (P1

P) VA = – P VB

VB = (–(P1

P) VA = –( P1

P + 1) VA

Putting in the numbers, we have t

VB = –(1/–0.0639 + 1)*256 cm = 3750 cm

Solution 3.38

Problem Statement

A closed, nonconducting, horizontal cylinder is fitted with a nonconducting, frictionless, floating piston that divides the cylinder into Sections A and B. The two sections contain equal masses of air, initially at the same conditions, T1 = 300 K and P1 = 1(atm). An electrical heating element in Section A is activated, and the air temperatures slowly increase: TA in Section A because of heat transfer, and TB in Section B because of adiabatic compression by the slowly moving piston. Treat air as an ideal gas with CP =

7 R, and let nA be the number of moles of air in Section A. For the 2

process as described, evaluate one of the following sets of quantities: (a) TA, TB, and Q/nA, if P(final) = 1.25(atm) (b) TB, Q/nA, and P (final), if TA = 425 K (c) TA, Q/nA, and P (final), if TB = 325 K (d) TA, TB, and P (final), if Q/nA = 3 kJ·mol

Solution

We want to develop relationships between TA, TB, p, and Q/nA. Then, given any one of the quantities, we can find the other three. One relationship is that the total volume remains constant: VA + VB = VAo + VBo. We can rewrite this in terms of the number of moles in each chamber (nA), the pressure (which is the same in both chambers) and the temperature in each chamber: RTA RTB 2 RTo   p p po

Another relationship arises from the fact that the compression in chamber B is adiabatic, so R

TB  p  Cp  p  7       To  po   po 

The third relationship derives from the energy balances on the two closed systems. If we let WB be the work done by the gas in chamber A on the gas in chamber B, then nAUA = Q – WB nAUB = WB and

nA (UA + UB) = Q

Writing the internal energy changes in terms of the heat capacity and temperature changes: UA + UB = Cv (TA + TB) = Q/nA

/2R ( TA + TB – 2To) = Q/nA

Solution continued on next page…

Now, we can apply these three relationships to solve the problems posed in parts (a) through (d). You only had to do one part of your choice.

(a) If we know the final pressure, then we can first use: 2

TB  p  7    To  po 

from which TB = 300*1.25

2/7

= 319.75 K

Next, we can use TA TB 2 p   To To po

to get TA 2.5 319.75    1.4342 300 1 300

from which TA = 430.3 K

Finally, then, Q/nA = /2R ( TA + TB – 2To) = 3118 J mol

–1

Solution continued on next page…

 p 7 T T T 2p (b) This is a bit trickier. One way to approach it is to substitute B    into A  B  to get To  po  To To po 2

TA  p  7 2 p     To  po  po

We can then guess values of p/po until we find one that gives the specified value of TA/To = 425/300 = 1.417. If we want to iterate on the equation, we might write it as: 2   p 1  p  7 TA       po 2  po  To   

From part (a), we know that p/po will be close to 1.25. So, we might iterate on the above starting from a guess of p/po = 1.25. This quickly converges to p/po = 1.24, or p = 1.24 *1 atm = 1.24 atm.

Now, we can use this directly to get TB = 300*1.2400

2/7

= 319.0 K.

Finally, the two temperatures are used to get Q/nA = /2R ( TA + TB – 2To) = 2993 J mol

–1

(c)

 p 7 T This one is a bit more straightforward using our equations. First, we can use B    to get To  po  7/2

p = po(TB/To)

3.5

= 1 atm * (1.08333)

= 1.3233 atm.

Solution continued on next page…

Then, we can use TA TB 2 p   to get To To po TA 2 p TB 2 * 1.3233 325      1.563 To po To 1 300

TA = 1.563*T0 = 300*1.563 from which TA = 469 K. 5

Finally, the two temperatures are used to get Q/nA = /2R ( TA + TB – 2To) = 4032 J mol

(d)

–1

This time, we have to start with /2R ( TA + TB – 2To) = Q/nA = 3000 J mol , from which 5

TA + TB = 600 + 3000/( /2R) = 744.3 K, or TA = 744.3 – TB. Substituting this into

gives

–1

TA TB 2 p   To To po

744.3  TB T 2p  B  , or just 2p/po = 744.3/300 = 2.481, from which p = 1.2405 atm. TB happened to 300 300 po

conveniently cancel out. 7

 p 7 2 p T  2 Then we can use   B  to get TB  To    300 1.2405 7  319.05 K  po  po  To 

Finally, TA = 744.3 – TB = 425.2 K.

Solution 3.39

Problem Statement

One mole of an ideal gas with constant heat capacities undergoes an arbitrary mechanically reversible process. Show that:

U 

1  1

( PV )

Solution

For an ideal gas we know that: U  Cv T

PV  RT

 PV   RT

Substituting we obtain:

U 

Cv  PV  R

Knowing that R = Cp – Cv and substituting in we obtain

U 

Cv 1  PV   PV  C p  Cv  1

Solution 3.40

Problem Statement

Derive an equation for the work of mechanically reversible, isothermal compression of 1 mol of a gas from an initial pressure P1 to a final pressure P2 when the equation of state is the virial expansion [Eq. (3.33)] truncated to: Z=1+B P

How does the result compare with the corresponding equation for an ideal gas?

Eq. 3.33

+ C P

D P

...

Solution

We have, in general for a mechanically reversible process, dW = –PdV. If our equation of state is 2

V = RT/P (1 + B’P) = RT /P + RTB’, then dV = –RT/P dP. So, dW = –P (–RT/P ) dP = RT/P dP. Integrating this from P1 to P2 gives W = RT ln(P2/P1), which is identical to the result for an ideal gas. The second term in this two-term form of the virial expansion does not affect the work of an isothermal expansion, because when we write the volume nd

as an explicit function of pressure, the term containing the 2

virial coefficient is independent of pressure.

Solution 3.41

Problem Statement

A certain gas is described by the equation of state:

   PV  RT  b  P  RT 

Here, b is a constant and is a function of T only. For this gas, determine expressions for the isothermal compressibility and the thermal pressure coefficient (P /T ) V . These expressions should contain only T, P, , /dT, and constants.

Solution

For this problem, we simply need to use the definitions of the isothermal compressibility and the thermal pressure coefficient, and take the specified partial derivatives. If the equation of state is:

   PV  RT  b  P  RT  Then we can write V as a function of P and T as

V

RT     b   P  RT 

The definition of  is 1  V       V  P 

Solution continued on next page…

So, we just need to take the derivative of V with respect to P at constant T, which is  V  RT    2  P  P T

So,

RT 1 P2      RT    P b     b   P 1      P  RT  RT  RT   Similarly, to find the thermal pressure coefficient, we can solve the EOS for P in terms of T and V to get     P V  b    RT   RT  RT P  V b RT

So, taking the derivative with respect to T at constant V, we have       V  b    R  RT    1 d  V  b    R  d      RT 2 RT dT    P  RT RT dT T      2 2  T      V V  b    V  b    RT  RT   

If we want this only in terms of T, P,  and d/dT, we should then substitute into it

V b 

 RT  RT P

Solution continued on next page…

Note that we have to do this after taking the derivative, since we need to have V in the expression so that we can hold it constant while taking the derivative. Substituting this in gives RT d  2 R   P  P dT T   P    P     d        2  T V  T   RT  T dT   RT     P 

Solution 3.42 Problem Statement For methyl chloride at 100°C the second and third virial coefficients are: 3

B = 242.5 cm ·mol

C = 25,200 cm ·mol

Calculate the work of mechanically reversible, isothermal compression of 1 mol of methyl chloride from 1 bar to 55 bar at 100°C. Base calculations on the following forms of the virial equation:

(a) Z  1 

B C  V V2

(b) Z  1  B P  C  P 2

where B 

B C  B2 and C   RT ( RT )2

Why don’t both equations give exactly the same result?

Solution

In general, the work for the isothermal compression can be computed from dW = –PdV We will integrate this from the initial pressure and volume to the final pressure and volume. In (a) we will write P in terms of V, while in (b) we will write V in terms of P.

We will substitute P = RT (1/V + B/V + C/V ) into the expression for the work and integrate it from the initial volume to the final volume. This integral gives 2

W = –RT (ln(V2/V1) – B (1/V2 – 1/V1) – C/2 (1/V2 – 1/V1 ) )

To get the initial and final volumes, we have to solve 3

–1

1 bar = 83.145 cm bar mol K * 373 K *(1/V1 – 242.5/V1 + 25200/V1 ) 3

–1

where V is in cm mol .

We could write this as 2

V1 = 83.145*373*(1–242.5/V1 + 25200/V1 )/1 bar 3

–1

And iterate to get V1 = 30769 cm mol

Similarly, for V2, 3

–1

55 bar = 83.145 cm bar mol K * 373 K *(1/V2 – 242.5/V2 + 25200/V2 )

or 2

V2 = 83.145*373/55*(1–242.5/V2 + 25200/V2 )

From which 3

–1

V2 = 241.21 cm mol

Using these values, –1

W = –8.3145 J mol K *373 K * (ln(V2/V1) – 242.5 (1/V2 – 1/V1) – 25200/2 (1/V2 – 1/V1 ) )

W = 12615 J mol

–1

Now we will do the same thing, but substituting for dV instead of for P in the expression for the work. We have 2

V = RT (1/P + B’ + C’P), from which dV = RT (–1/P + C') dP 2

So dW = –PdV = RT (1/P – C’P)dP, from which W = RT (ln(P2/P1) – C'/2 (P2 – P1 ) ) or using the relationship between C’ and B and C 2

2 2

W = RT ( ln(P2/P1) – (C – B )* (P2 – P1 ) /(2R T ) ) –1

W = 8.3145 J mol K * 373 K * (ln(55) – (25200 – 242.5 )*(55 – 1 )/(2*83.145 *373 )) 3

–1

Note that the R in the denominator of the last term needs to be in bar cm mol K to make the units work out, –1

–1

while the R out front is most conveniently taken to be in J mol K so that the answer comes out in J.

–1

W = 12592 J mol K

The answers aren’t exactly the same in (a) and (b) because the two forms of the virial equation are only exactly the same if we keep all of the (infinite) number of terms.

Solution 3.43 Problem Statement

Any equation of state valid for gases in the zero-pressure limit implies a full set of virial coefficients. Show that the second and third virial coefficients implied by the generic cubic equation of state, Eq. (3.41), are:

B  b

a(T ) RT

C  b2 

(   )ba(T ) RT

Specialize the result for B to the Redlich/Kwong equation of state, express it in reduced form, and compare it numerically with the generalized correlation for B for simple fluids, Eq. (3.61). Discuss what you find. Eq. 3.41:

P

RT a (T )  V  b (V  b )(V   b )

Eq. 3.61:

B 0  0.083 

0.422 Tr1 .6

Solution

Writing the virial expansion in molar density (inverse molar volume) we have Z

PV B C D  1   2  3  ...  1  B   C  2  D 3  ... RT V V V

From this, we see that B

dZ d 0

C

1 d2 Z 2 d 2 0

The generic cubic equation of state P

a T  RT  V  b V  bV  b

can be written in terms of Z and as Z

aT V PV V   RT V  b RT V  bV   b 1

Z



1 b 



aT  1

 1   RT   b 1   b      aT 

1  1  b RT 1  b 1  b 

Now, we must take the derivative of this with respect to and evaluate it at = 0:   aT  1  b 1  b    b 1  b   b 1  b  dZ b     2 2  d 1  b 2 RT  1  b 1  b     aT   dZ b 1  b2  2     2 2 2 d 1  b  RT 1  b  1  b     aT  dZ  b d 0 RT

Solution continued on next page…

So, the second virial coefficient is B

aT  dZ  b d 0 RT

Similarly, we take a second derivative with respect to and evaluate the result at = 0 to get the third virial coefficient:





2 2 2 2 2 2 2    2b1  b1  b  2b1  b 1  b       2 3 4 4  d 1  b RT  1  b 1  b   2 2 2 2 2 a T   2b  1  b 1   b   1  b  2 b 1   b   2 b 1  b   dZ 2b      2 3 3 3  d 1  b RT  1  b 1  b   2  a T  a T     b  dZ 2  2b   2   b  2b 2  2 RT RT   d 2

a T   2b  1  b  1  b   1  b 

0

So, finally, we have

C

aT    b 1 d2 Z  b2  2 2 d 0 RT

Solution 3.44 Problem Statement Calculate Z and V for ethylene at 25°C and 12 bar by the following equations:

(a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients: B = 140 cm 3 ·mol

C = 7200 cm 6 ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)– (3.62)]

Eq. 3.36: Z

PV BP  1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

(a) the 3-term truncated virial equation is: PV B C  1  2 RT V V

Putting in the numbers from the problem statement gives

12 bar V 140 cm3 mol1 7200 cm6 mol2  1  3 1 1 V 83.145 bar cm mol K 298 K V2  140 cm3 mol1 7200 cm6 mol2   V  2064.8 cm3 mol1 1    V V2  3

Starting from the initial guess of V = P/RT = 2065 cm mol

1

and iterating gives V = 1918 cm mol .

(b) Looking in table B.1 on page 654, we see that for ethylene, Tc = 282.3 K, Pc = 50.40 bar, and  = 0.087. So, at 298 K, ethylene is above its critical temperature. We have Tr = 298/282.3 = 1.056 and Pr = 0.2381. Now, we find the second virial coefficient from the Pitzer correlations as 0.422  0.3040 Tr1.6 0.172 B 1  0.139  4.2  0.00218 Tr BPc  B 0   B1  0.3040  0.087  0.00218  0.3038 RTc  BP  P 0.2381 Z  1   c  r  1  0.3038  0.9315  1.056  RTc  Tr B 0  0.083 

Solution continued on next page…

1

So, V = ZP/RT = 0.9315*2064.8 = 1923.36 cm mol . This is just 0.3% greater than the value obtained from the virial equation with the experimental coefficients (which is roughly at the level of round-off error in the calculations, since we've only carried 4 digits through most of them). This is essentially exact agreement.

(c)

The Redlich/Kwong equation of state is

P

aT RT  V  b V V  b

with T 0.5 R2Tc2 1.05560.5  83.1452  282.32 a(T )  0.42748 r  0.42748  4548100 bar cm6 mol2 Pc 50.40 RT 83.145  282.3 b  0.08664 c  0.08664  40.35 cm 3 mol 1 Pc 50.40

To iteratively solve for this vapor-like root of the equation, we can rearrange it as suggested in SVNA aT  V  b RT b P P V V  b  4548100 V  40.35   cm3 mol1 V  2064.8  40.35   12V V  40.35   379008V  40.35   cm3 mol 1 V  2105.2   V V  40.35  V

Starting from the initial guess of V = P/RT = 2065 cm mol

1

and iterating gives V = 1915 cm mol . Again, the

agreement with the other methods is almost exact. The Redlich/Kwong equation of state does well for ethylene, in part, because ethylene had a very small acentric factor (which the Redlich/Kwong EOS doesn't use). Solution continued on next page…

(d)

The Soave/Redlich/Kwong EOS is the same as the Redlich/Kwong equation of state, except for the definition

of a(T), which is

1  0.480  1.574  0.176 1  T  R T a(T )  0.42748 2

0.5 r

2 c

  a(T )  0.42748

1  0.480  1.574 0.087  0.1760.087

a(T )  0.42748



1  1.0556

 83.145 282.3 2

50.40

0.9731 83.1452 282.32

50.40 a(T )  4547150 bar cm 6 mol 2

b  0.08664

RTc Pc

 0.08664

83.145 282.3  40.35 cm 3 mol1 50.40

So, we have aT  V  b RT b P P V V  b  4547150 V  40.35  3 1 V  2064.8  40.35   cm mol  12V V  40.35   378929V  40.35   cm3 mol 1 V  2105.2   V V  40.35  V

Starting from the initial guess of V = P/RT = 2065 cm mol

1

and iterating gives V = 1916 cm mol . Again, the

agreement with the other methods is almost exact.

(e)

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

Solution continued on next page…

with

1  0.37464  1.54226  0.2699 1  T  R T 2

a (T )  0.45724

0.5

1  0.37464  1.54226  0.087  0.26990.087 1  1.0556   83.145  282.3 a (T )  0.45724 2

50.40

a (T )  0.45724

0.9724  83.1452  282.32

50.40 6 2 a (T )  4860171 bar cm mol b  0.07779

RTc Pc

 0.07779

83.145  282.3 50.40

 36.23 cm mol 3

1

Rearranging to solve for V iteratively, this is a T  RT V b b P P V  0.4142b V  2.4142b  4860171V  36.23    cm 3 mol 1 V  2064.8  36.23  12V  15.005V  87.46    405014 V  36.23    cm 3 mol 1 V  2101.0   V  15.005V  87.46  V

Starting from the initial guess of V = P/RT = 2065 cm mol

1

and iterating gives V = 1899 cm mol . This is about

1% lower than the other values.

Solution 3.45 Problem Statement Calculate Z and V for ethane at 50°C and 15 bar by the following equations: (a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients: B

156.7 cm ·mol

9650 cm ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)– (3.62)] (c) The Redlich/Kwong equation (d) The Soave/Redlich/Kwong equation (e) The Peng/Robinson equation

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

(a) the 3-term truncated virial equation is: PV B C  1  2 RT V V

Putting in the numbers from the problem statement gives 15 bar  V 156.7 cm3 mol1 9650 cm 6 mol2 1  3 1 1 V 83.145 bar cm mol K  323.15 K V2 3  1 6  2  156.7 cm mol 9650 cm mol   V  1791.2 cm3 mol 1 1   V V2   3

Starting from the initial guess of V = RT/V = 1791 cm mol

1

and iterating gives V = 1625 cm mol . This

corresponds to a compressibility of Z = 1625/1791 = 0.9073.

(b) Looking in table B.1 on page 654, we see that for ethane, Tc = 305.3 K, Pc = 48.72 bar, and  = 0.100. So, at 323 K, ethane is above its critical temperature. We have Tr = 323.15/305.3 = 1.058 and Pr = 0.3079. Now, we find the second virial coefficient from the Pitzer correlations as 0.422  0.3026 Tr1.6 0.172 B 1  0.139  4.2  0.003266 Tr BPc  B 0   B 1  0.3026  0.100  0.003266  0.3023 RTc  BP  P 0.3079 Z  1   c  r  1  0.3023  0.9120  RTc  Tr 1.058 B 0  0.083 

Solution continued on next page…

1

So, V = ZRT /P= 0.9120*1791.8 = 1634 cm mol . This is just 0.6% greater than the value obtained from the virial equation with the experimental coefficients.

(c)

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 1.0580.5  83.1452  305.32 a(T )  0.42748 r  0.42748  5496600 bar cm 6 mol 2 Pc 48.72 RT 83.145  305.3 b  0.08664 c  0.08664  45.14 cm3 mol 1 Pc 48.72

To iteratively solve for this vapor-like root of the equation, we can rearrange it as suggested in SVNA aT  V  b RT b P P V V  b  5496600 V  45.14   cm3 mol1 V  1791.2  45.14   15V V  45.14   366440V  45.14    cm3 mol1 V  1836.3   V V  45.14  V

Starting from the initial guess of V = RT/P = 1791 cm mol

1

and iterating gives V = 1623 cm mol . Again, the

agreement with the other methods is almost exact. Solution continued on next page…

(d)

The Soave/Redlich/Kwong EOS is the same as the Redlich/Kwong equation of state, except for the definition

of a(T), which is

1  0.480  1.574  0.176 1  T  R T a(T )  0.42748 2

0.5 r

2 c

1  0.480  1.5740.100  0.1760.100 1  1.058  83.145 305.3 a(T )  0.42748 2

48.72

0.96398  83.1452 305.32 a(T )  0.42748 48.72 a(T )  5450090 bar cm 6 mol2

So, we have aT  V  b RT b P P V V  b  5450090 V  45.14   cm3 mol1 V  1791.2  45.14   15V V  45.14   363399V  45.14   cm3 mol1 V  1836.3   V V  45.14  V

Starting from the initial guess of V = RT/P = 1791 cm mol

1

and iterating gives V = 1625 cm mol . Again, the

agreement with the other methods is almost exact. (e)

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

Solution continued on next page…

with

1  0.37464  1.54226  0.2699 1  T  R T 2

a(T )  0.45724

0.5

1  0.37464  1.54226 0.100  0.26990.100 1  1.058  83.145 305.3 a(T )  0.45724 2

0.97014  83.1452  305.32 a(T )  0.45724 48.72 a(T )  5866378 bar cm6 mol 2 b  0.07779

48.72

RTc 83.145  305.3  0.07779  40.53 cm3 mol 1 Pc 48.72

Rearranging to solve for V iteratively, this is aT  RT V b b P P V  0.4142bV  2.4142b  5866378V  40.53    cm 3 mol1 V  1791.2  40.53   15V  16.78V  97.85  391091V  40.53    cm 3 mol1 V  1831.7   V  16.78V  97.85 V

Starting from the initial guess of V = RT/P = 1791 cm mol

1

and iterating gives V = 1606 cm mol . This is about

1% lower than the other values.

Solution 3.46

Problem Statement

Calculate Z and V for sulfur hexafluoride at 75°C and 15 bar by the following equations: (a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients:

194 cm ·mol

= 15,300 cm ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)] (c) The Redlich/Kwong equation (d) The Soave/Redlich/Kwong equation (e) The Peng/Robinson equation 3

For sulfur hexafluoride, Tc = 318.7 K, Pc = 37.6 bar, Vc = 198 cm ·mol , and

= 0.286.

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

(a) the 3-term truncated virial equation is: PV B C  1  2 RT V V

Putting in the numbers from the problem statement gives 15 bar  V 194 cm3 mol 1 15300cm 6 mol 2 1  3 1 1 V 83.145 bar cm mol K 348 K V2  194 cm 3 mol1 15300cm6 mol2   V  1929.0 cm3 mol1 1    V V2  3

Starting from the initial guess of V = RT/V = 1929 cm mol

1

and iterating gives V = 1722 cm mol .

(b) From the problem statement, we see that for SF6, Tc = 318.7 K, Pc = 37.6 bar, and  = 0.286. So, at 348 K, SF6 is above its critical temperature. We have Tr = 348/318.7 = 1.0919 and Pr = 0.3989. Now, we find the second virial coefficient from the Pitzer correlations as Solution continued on next page…

0.422  0.2836 Tr1.6 0.172 B 1  0.139  4.2  0.02011 Tr BPc  B 0   B1  0.2836  0.286  0.02011  0.2779 RTc  BP  P 0.3989 Z  1   c  r  1  0.2779  0.8985 1.0919  RTc  Tr B 0  0.083 

So, V = ZRT/V = 0.8985*1929.0 = 1733 cm mol

(c)

1

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 1.09190.5  83.1452  318.7 2 a(T )  0.42748 r  0.42748  7639640 bar cm6 mol 2 Pc 37.6 RT 83.145  318.7 b  0.08664 c  0.08664  61.06 cm3 mol1 Pc 37.6

To iteratively solve for this vapor-like root of the equation, we can rearrange it as suggested in SVNA aT  V  b RT b P P V V  b  7639640 V  61.06   cm 3 mol 1 V  1929.0  61.06   15V V  61.06   509309V  61.06   cm 3 mol 1 V  1990.1   V V  61.06  V

Solution continued on next page…

Starting from the initial guess of V = RT/V = 1929 cm mol

(d)

1

and iterating gives V = 1713 cm mol .

The Soave/Redlich/Kwong EOS is the same as the Redlich/Kwong equation of state, except for the definition

of a(T), which is

1  0.480  1.574  0.176 1  T  R T a(T )  0.42748 2

0.5 r

2 c

1  0.480  1.574 0.286  0.1760.286 1  1.0919  83.145 318.7 a(T )  0.42748 2

37.6

0.9193  83.145 2  318.7 2 a(T )  0.42748 36.7 a(T )  7519399 bar cm 6 mol2

So, we have aT  V  b RT b P P V V  b  7519399V  61.06   cm3 mol1 V  1929.0  61.06  15V V  61.06    501293V  61.06   cm 3 mol 1 V  1990.1   V V  61.06  V

Starting from the initial guess of VV = RT/V = 1929 cm mol

(e)

1

and iterating gives V = 1718 cm mol .

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

Solution continued on next page…

with

1  0.37464  1.54226  0.2699 1  T  R T a(T )  0.45724 2

0.5 r

2 c

  a(T )  0.45724

1  0.37464  1.54226  0.286  0.26990.286



1  1.0919

  83.145 318.7 2

37.6 0.92994  83.1452  318.72 a(T )  0.45724 37.6 a(T )  7940485 bar cm 6 mol 2 RT 83.145  318.7 b  0.07779 c  0.07779  54.82 cm3 mol 1 Pc 37.6

Rearranging to solve for V iteratively, this is a T  RT V b b P P V  0.4142b V  2.4142b  7940485V  54.82    cm 3 mol 1 V  1929.0  54.82   15 V  22.71V  132.35  529365 V  54.82    cm 3 mol 1 V  1983.8   V  22.71V  132.35  V

Starting from the initial guess of V = RT/V = 1929 cm mol

1

and iterating gives V = 1701 cm mol .

Solution 3.47

Problem Statement

Calculate Z and V for ammonia at 320 K and 15 bar by the following equations: (a) The truncated virial equation [Eq. (3.38)] with the following values of virial coefficients: 3

B = 208 cm ·mol

C = 4378 cm ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)] (c) The Redlich/Kwong equation (d) The Soave/Redlich/Kwong equation (e) The Peng/Robinson equation

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

From the back of the book: Ammonia – Tc = 405.7 K, Pc = 112.8 bar,

= 0.253

(a) the 3-term truncated virial equation is: PV B C 1  2 RT V V

Putting in the numbers from the problem statement gives

15bar * V bar cm3 * mol K

V

  K

cm 6 cm3 4378 mol  mol 2 V V2

208

3  cm6   208 cm  4378 2  cm   mol mol     2 mol  V V     3

Starting from the initial guess of V = RT/P = 1773 cm mol

1

and iterating gives V = 1568 cm mol . This

corresponds to a compressibility of Z = 1568/1773 =0.8841.

(b) From the back of the book, we see that ammonia Tc = 405.7 K, Pc = 112.8 bar, and

= 0.253. This shows that

ammonia is below the critical temperature. However, determining that its boiling point is approx. 40.5 C. We have Tr = 320/405.7 = 0.7888 and Pr = 0.1330. Now, we find the second virial coefficient from the Pitzer correlations as

B0 



0.422  Tr1.6

B1 



0.172  Tr4.2

BPc  B 0  B1   RTc  BP  P Z   c  r    RTc  Tr 3

So, V = ZRT/V = 0.8960*1773 = 1588.6 cm mol

(c)

0.1330  0.7888

1

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

With

aT  

Tr0.5 R2Tc2 Pc

b



RTc Pc



0.788760.5 * 83.1452 * 405.7 2  112.8

83.145 * 405.7  112.8

bar cm6mol 2

cm3mol 1

To iteratively solve for this vapor-like root of the equation, we can rearrange it as suggested in SVNA

V

aT  V  b RT b P P V V  b

Solution continued on next page…

V  1773  25.91 

4855305 V  25.91 15 V V  25.91

Starting from the initial guess of V = RT/V = 1773 cm mol

1

and iterating gives V = 1622.44 cm mol

1

and

Z = 1622.44/1773 = 0.9151.

(d)

The Soave/Redlich/Kwong EOS is the same as the Redlich/Kwong equation of state, except for the definition

of a(T), which is

 

aT  

 





 2   Tr0.5  R2Tc2 2

 Pc





 

 0.5



405.72

112.8

b

RTc Pc



V  1773  25.91 

83.145 * 405.7  112.8

bar cm6mol 2



cm3mol 1

5192738.352 V  25.91 15 V V  25.91

So, using the same equation as RK and starting from the initial guess of V = RT/V =1773 cm mol 3

1

and iterating

1

gives V = 1587.2 cm mol and Z = 1587.2/1773 = 0.8956.

Solution continued on next page…

(e)

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

 



aT  



bar cm 6mol 2



 

0.5

 83.145 405.7 2

112.8

and

b

RTc  Pc

cm3mol1

Rearranging to solve for V iteratively, this is

aT  V  b RT P V b P V  0.4142bV  2.4142b

Starting from the initial guess of V = RT/P= 1773 cm /mol

1

and iterating gives V = 1577.15 cm mol and Z = 1577.15/1773 = 0.8895

Due to this being below the critical temperature the values give less than accurate results and only apply to the gas phase.

Solution 3.48

Problem Statement

Calculate Z and V for boron trichloride at 300 K and 1.5 bar by the following equations: (a) The truncated virial equation [Eq. (3.38)] with the following values of virial coefficients: 3

B = 724 cm ·mol

C = 93,866 cm ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)] (c) The Redlich/Kwong equation (d) The Soave/Redlich/Kwong equation (e) The Peng/Robinson equation

For BCl3, Tc = 452 K, Pc = 38.7 bar, and

= 0.086.

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

For this problem follow the same steps and methods as in 3.46 and 3.47

(a) the 3-term truncated virial equation is: PV B C 1  2 RT V V

Putting in the numbers from the problem statement gives

1.5bar * V bar cm 3 * mol K

V

  K

cm6 cm3 93866 mol  mol 2 2 V V

724

3  cm6   724 cm  93866 cm  mol  mol 2     mol  V V2    3

Solution continued on next page…

Starting from the initial guess of V = RT/P = 16629 cm mol

1

and iterating gives V = 15899.36 cm mol . This

corresponds to a compressibility of Z = 15899.36/16629 = 0.9561.

(b) From the problem statment, we see that boron trichloride Tc = 452 K, Pc that boron trichloride is below the critical temperature.

We have Tr = 300/452 = 0.6637 and Pr = 1.5/38.7 = 0.03875. Now, we find the second virial coefficient from the Pitzer correlations as B0 



0.422  Tr1.6

B1 



0.172  Tr4.2

BPc  B 0  B1   RTc  BP  P Z   c  r    RTc  Tr

0.38750  0.6637 3

So, V = ZRT/V = 0.9532*16629 = 15850.76 cm mol

(c)

1

Using the Redlich/Kwong equation of state gives

P

aT  RT  V  b V V  b

Solution continued on next page…

With Tr0.5 R2Tc2

aT  

b  0.08664

RTc

 0.08664 *

0.66370.5 * 83.1452 * 4522  38.7



83.145 * 452  84.14 cm3 mol1 38.7

V

V  16629  84.14 

bar cm 6 mol 2

aT  V  b RT b P P V V  b

19149726 V  84.14 1.5 V V  84.14 

Starting from the initial guess of V = RT/V = 16629 cm mol

1

and iterating gives V = 15953.14 cm mol

1

and

Z = 15953.14/16629 = 0.9593. (d)

The Soave/Redlich/Kwong EOS is the same as the Redlich/Kwong equation of state, except for the definition

of a(T), which is

aT  

 





 2   Tr0.5  R 2Tc2 2

 Pc

1  0.48  1.574 * 0.086  0.176 * 0.0   a T  0.42748 2

 

0.5

 * 2

38.7

aT   19350625.74 bar cm 6 mol2

Solution continued on next page…

b  0.08664

RTc Pc

 0.08664 *

83.145 * 452  84.14 cm3 mol1 38.7

V  16629  84.14 

19350625.74 V  84.14  1.5 V V  84.14

Starting from the initial guess of V = RT/V =16629 cm mol

1

and iterating gives V = 15945.173 cm mol and

Z = 15945.173/16629 = 0.9588.

(e)

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

 

aT   aT  





 

0.5

 * 3.145 * 452 2

38.7 6 2

bar cm6 mol 2

and

b  0.07779

RTc Pc

 75.54 cm3 mol 1

Rearranging to solve for V iteratively, this is

aT  V  b RT P V b P V  0.4142bV  2.4142b Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Starting from the initial guess of V = RT/P= 16629 cm /mol

1

and iterating gives V = 15878.5 cm mol and Z = 15878.5/16629 = 0.9548

Due to this being below the critical temperature the values give less than accurate results and only apply to the gas phase.

Solution 3.49 Problem Statement Calculate Z and V for nitrogen trifluoride at 300 K and 95 bar by the following equations: (a) The truncated virial equation [Eq. (3.38)] with the following values of virial coefficients: B

83.5 cm ·mol

5592 cm ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)] (c) The Redlich/Kwong equation (d) The Soave/Redlich/Kwong equation (e) The Peng/Robinson equation For NF3, Tc = 234 K, Pc = 44.6 bar, and

= 0.126.

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

For this problem follow the same steps and methods as in 3.46 and 3.47

(a) the 3-term truncated virial equation is: PV B C 1  2 RT V V

Putting in the numbers from the problem statement gives Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

bar * V bar cm3 * mol K

  K

cm6 cm3 5592 mol  mol 2 2 V V

83.5

3  cm 6   83.5 cm  5592 cm  mol  mol 2     mol  V V2     3

V

Using the truncated virial equation and starting from the initial guess of V = RT/P = 262.56 cm mol 3

1

and iterating

1

gives V = 193cm mol . This corresponds to a compressibility of Z = 193/262.56 = 0.7350.

(b) From the problem statement, we see that boron trichloride Tc = 234 K, Pc that nitrogen triflouride is above the critical temperature.

We have Tr = 300/234 = 1.282 and Pr = 95/44.6 = 2.13. Now, we find the second virial coefficient from the Pitzer correlations as

B0 



B1 



0.422  Tr1.6

0.172  Tr4.2

BPc  B 0  B1   RTc  BP  P Z   c  r    RTc  Tr

2.13  1.282

So, V = ZRT/V = 0.6833*262.56 = 179.4 cm mol

1

Solution continued on next page…

(c)

Using the Redlich/Kwong equation of state gives

P

aT  RT  V  b V V  b

With Tr0.5 R2Tc2

aT  



b  0.08664

RTc Pc

 0.08664 *

1.2820.5 * 83.145 2 * 234 2  44.6

83.145 * 234  37.80 cm3 mol1 44.6

V

V  262.56  37.80 

bar cm 6 mol 2

aT  V  b RT b P P V V  b

3204293 V  37.80 95 V V  37.80

Starting from the initial guess of V = RT/V = 262.56 cm mol

1

and iterating gives V = 176.784 cm mol

1

and

Z = 176.784/262.56 = 0.6733 (d)

The Soave/Redlich/Kwong EOS is the same as the Redlich/Kwong equation of state, except for the definition

of a(T),

aT  

 

aT   0.42748 *



   Tr



0.5

 R T 2

 





 

0.5



* 234

44.6

aT   3008454.99 bar cm 6 mol2

Solution continued on next page…

b  0.08664

V

RTc

 0.08664 *

83.145 * 234  37.80 cm3 mol1 44.6

a T  V  b RT b V P P V V  b





3008454.99 (V  37.80) 95 V (V  37.80)

Starting from the initial guess of V = RT/V =262.56 cm mol

(e)

1

and iterating gives V = 188.475 cm mol .

The Peng-Robinson EOS gives

P

aT  RT  V  b V  2.4142bV  0.4142b

with

aT  

  724

aT  

b

RTc  Pc





 

0.5

 * 2

.1452 * 234 2

44.6

2 bar cm6mol2

cm3mol 1

Rearranging to solve for V iteratively, this is

aT  V  b RT P V b P V  0.4142bV  2.4142b Solution continued on next page…

Starting from the initial guess of V = RT/P= 262.56 cm /mol

1

and iterating gives V = 178.388cm mol and Z = 178.388/262.56 = 0.6794

Due to this being below the critical temperature the values give less than accurate results and only apply to the gas phase.

Solution 3.50 Problem Statement Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the following experimental values of virial coefficients:

152.5 cm ·mol

C = 5800 cm ·mol

(b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.38: Z

PV B C 1  2 RT V V

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

(a) the 3-term truncated virial equation is: PV B C  1  2 RT V V

Putting in the numbers from the problem statement gives 18 bar  V 152.5 cm3 mol1 5800 cm 6 mol 2  1   V 83.145 bar cm 3 mol1 K 1  523 K V2 3 1 6 2   152.5 cm mol 5800 cm mol   V  2415.8 cm3 mol 1 1    V V2 

Solution continued on next page…

Starting from the initial guess of V = RT/V = 2416 cm mol

1

and iterating gives V = 2249 cm mol .

(b) From table B.1 in appendix B, we see that for water, Tc = 647.1 K, Pc = 220.55 bar, and  = 0.345. So, at 523 K, and 18 bar, have Tr = 0.8082 and Pr = 0.08161. Now, we find the second virial coefficient from the Pitzer correlations as

0.422  0.5103 Tr1.6 0.172 B1  0.139  4.2  0.2816 Tr BPc  B 0   B 1  0.5103  0.345 0.2816  0.6075 RTc  BP  P 0.08161 Z  1   c  r  1  0.6075  0.9387  RTc  Tr 0.8082 B 0  0.083 

1

So, V = ZRT/V = 0.9387*2415.8 = 2268 cm mol . This is within 1% of the value obtained in (a).

(c)

1

From the table for superheated steam (SI units) on p. 707, at 1800 kPa and 250 °C, V is 124.99 cm g 1

1

which

(multiplying by the molecular weight of water, 18.016 g mol ) is 2252 cm mol . This agrees almost exactly with the result from part (a).

Solution 3.51 Problem Statement With respect to the virial expansions, Eqs. (3.33) and (3.34), show that:  Z   Z  B    and B        P  T ,P 0 T ,0

where  1/V.

Eq. 3.33: 2

Z=1+B P+C P +D P +...

Eq. 3.34:

Z 1

B C D    V V2 V3

Solution

Starting with eqn 3.33 and differentiate Z   B P  C P 2  D P 3

Solution continued on next page…

 Z     B   C ' P  D' P 2  P  T

Making P = 0 leaves  Z     P 

 B

T ,P 0

Now, do the same for eqn 3.34

Z 1

B C D   V V2 V3

Substituted with V = Z  1  B  C  2  D 3

Differentiated  Z     B  C   D 2    T

And finally setting = 0  Z      

B

T ,0

Solution 3.52 Problem Statement

Equation (3.34) when truncated to four terms accurately represents the volumetric data for methane gas at 0°C with: 3

B = 53.4 cm ·mol

C = 2620 cm ·mol

D = 5000 cm ·mol

(a) Use these data to prepare a plot of Z vs. P for methane at 0°C from 0 to 200 bar. (b) To what pressures do Eqs. (3.36) and (3.37) provide good approximations? Eq. 3.34:

Z 1

B C D    V V2 V3

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.37:

Z

PV B 1 RT V

Solution

(a) Using the values given, determine Zi, Z1i , an Z2i using the following

PV B C D  1  2  3 RT V V V

i  to

Zi 

f Pi , Vi  Pi RT



Pi  

 Z 1i  

P (bar)

Z1i

Z2i

0.953

0.951

0.906

0.895

0.861

0.859

0.83

0.819

0.812

0.749

100

0.784

0.765

0.622

120

0.757

0.718

0.5 + 0.179i

140

0.74

0.671

0.5 + 0.281i

160

0.733

0.624

0.5 + 0.355i

180

0.735

0.577

0.5 + 0.416i

200

0.743

0.53

0.5 + 0.469i

10

BPi RT

* ibar Vi 





1 2

RT Pi

Z2i  

1 BPi  4 RT





Solution continued on next page…

Note that values of Z from Eq. (3.39) are not physically meaningful for pressures above 100 bar.

1 0.95 0.9

0.85 0.8

Z1i 0.75

0.7

Z1i

0.65

Z2i

0.6 0.55 0.5 0

100

150

200

P (bar)

(b) Eqs 3.36 and 3.37 apply to pressures below the critical pressure of the gas, so everything below 100 bar.

Solution 3.53 Problem Statement THIS SOLUTION IS MISSING EQUATIONS…………………………….. Calculate the molar volume of saturated liquid and the molar volume of saturated vapor by the Redlich/Kwong equation for one of the following and compare results with values found by suitable generalized correlations. (a) Propane at 40°C where P

(b) Propane at 50°C where P

sat

(d) Propane at 70°C where P

sat

= 13.71 bar

= 17.16 bar

= 21.22 bar = 25.94 bar

(e) n-Butane at 100°C where P

sat

(f ) n-Butane at 110°C where P

(g) n-Butane at 120°C where P

sat

(h) n-Butane at 130°C where P

(i) Isobutane at 90°C where P

sat

(j) Isobutane at 100°C where P

(m) Chlorine at 60°C where P

(n) Chlorine at 70°C where P

(o) Chlorine at 80°C where P

= 22.38 bar

= 26.59 bar

= 16.54 bar

sat

(p) Chlorine at 90°C where P

= 18.66 bar

sat

(k) Isobutane at 110°C where P (l) Isobutane at 120°C where P

= 15.41 bar

sat

= 20.03 bar

= 24.01 bar = 28.53 bar

= 18.21 bar

= 22.49 bar

= 27.43 bar

= 33.08 bar

(q) Sulfur dioxide at 80°C where P

(r) Sulfur dioxide at 90°C where P

sat

(s) Sulfur dioxide at 100°C where P

(t) Sulfur dioxide at 110°C where P

= 18.66 bar

= 23.31 bar

sat

= 28.74 bar

= 35.01 bar

(u) Boron trichloride at 400 K where P

sat

For BCl3, Tc = 452 K, Pc = 38.7 bar, and

= 17.19 bar = 0.086.

(v) Boron trichloride at 420 K where P

sat

(w) Boron trichloride at 440 K where P

(x) Trimethylgallium at 430 K where P

= 23.97 bar

sat

= 32.64 bar

= 13.09 bar

For Ga(CH3)3, Tc = 510 K, Pc = 40.4 bar, and (y) Trimethylgallium at 450 K where P

(z) Trimethylgallium at 470 K where P

sat

= 0.205.

= 18.27 bar

= 24.55 bar

Solution

This is not too bad, since we only have to find the vapor and liquid volumes at a given pressure (and don’t have to find the pressure itself – we’ll save that for later in the semester). First, we need to find the Redlich/Kwong parameters for propane from Appendix B. The critical temperature and pressure (all we need for the RK EOS) are Tc = 369.8 K and Pc = 42.48 bar.

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  369.82.5 1.8295  10 8 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol2 Pc Pc 42.48 T RTc 83.145  369.8 3 1 b  0.08664  0.08664  62.71 cm mol Pc 42.48 7

–2

(a) At 40°C = 313.15 K, we have a = 1.0338 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*313.15/13.71 = 1899.1 cm mol . To find the vapor phase volume, we iterate on aT  V  b RT b P P V V  b  10338000V  62.71    cm3 mol 1 V  1899.1  62.71   13.71  V V  62.71   754082V  62.71   cm3 mol 1 V  1961.8   V V  62.71  V

–1

Starting from V = 1961.8 cm mol and iterating gives V = 1499 cm mol for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  aT     83.14  313.15  62.17  13.71  13.71V  V  62.71  V V  62.71   10338000  1963.4  V  V  62.71  V V  62.71   754048  3

–1

Starting from V = 62.71 cm mol and iterating on this gives V = 108.2 cm /mol. The ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The Rackett equation is 0.2857

V sat  Vc Z c

1Tr 

–1

From appendix B, we have Vc = 200.0 cm mol , and Zc = 0.276. At 313.15 K, the reduced temperature is 0.8468, so we have 0.2857

V sat  200  0.276

10.8468

which gives V

sat

–1

= 94.2 cm mol . The R/K equation gave a value about 15% higher than this.

The Rackett equation is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8468, and the reduced pressure is Pr = 13.71/42.48 = 0.3227. The acentric factor for propane is 0.152. So, we have 0.422  0.4676 0.84681.6 0.172 B1  0.139   0.2068 0.8468 4.2 B 0  0.083 

Solution continued on next page…

and Z  1  0.4676  0.152  0.2068

0.3227  0.8098 0.8468 3

–1

From which V = ZRT/P = 0.8098*1899.1 = 1538 cm mol . 7

(b) Now, we do the same thing at 50°C = 323.15 K (hey, it’s PRACTICE). At 323.15 K, we have a = 1.0177 × 10 bar 6

–2

–1

cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*323.15/17.16 = 1565.8 cm mol . To find the vapor phase volume, we iterate on aT  V  b RT b P P V V  b  10177000 V  62.71    cm3 mol 1 V  1565.8  62.71  17.16  V V  62.71    593080V  62.71   cm3 mol 1 V  1628.5   V V  62.71  V

–1

Starting from V = 1566 cm mol

–1

and iterating gives V = 1175 cm mol for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  323.15  62.17  17.16  17.16V  V  62.71  V V  62.71   10177000   1627.8  V  V  62.71  V V  62.71   593065  3

–1

Starting from V = 62.71 cm mol and iterating on this gives V = 114.4 cm /mol. The ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The Rackett equation is 0.2857

V sat  Vc Z c

1Tr 

–1

From appendix B, we have Vc = 200.0 cm mol , and Zc = 0.276. At 323.15 K, the reduced temperature is 0.8739, so we have 0.2857

V sat  200  0.276

1 0.8739

which gives V

sat

–1

= 98.1 cm mol . The R/K equation gave a value about 17% higher than this.

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8739, and the reduced pressure is Pr = 17.16/42.48 = 0.4040. The acentric factor for propane is 0.152. So, we have 0.422  0.4406 0.87391.6 0.172 B1  0.139   0.1640 0.8739 4.2 B 0  0.083 

Solution continued on next page…

and Z  1  0.4406  0.152  0.1640 

0.4040  0.8014 0.8739 3

–1

From which V = ZRT/P = 0.8014*1565.8 = 1255 cm mol . 7

–2

–1

gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*333.15/21.22 = 1305.4 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT  b P P V V  b  10023300 V  62.71    cm3 mol 1 V  1305.4  62.71  21.22  V V  62.71   472351V  62.71   cm3 mol1 V  1368.1   V V  62.71  V

–1

Starting from V = 1305 cm mol and iterating gives V = 920 cm mol for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP   V  b  V V  b   aT    83.14  333.15  62.17  21.22  21.22V  V  62.71  V V  62.71   100233000   1367.5  V  V  62.71  V V  62.71   472351  3

Starting from V = 62.71 cm mol

–1

and iterating on this gives V = 122.5 cm /mol. The ‘appropriate generalized

correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. Solution continued on next page…

The Rackett equation is 0.2857

V sat  Vc Z c

1Tr 

–1

From appendix B, we have Vc = 200.0 cm mol , and Zc = 0.276. At 333.15 K, the reduced temperature is 0.9009, so we have 0.2857

V sat  200  0.276

10.9009

which gives V

sat

–1

= 103.3 cm mol . The R/K equation gave a value about 19% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. We also see that as the temperature goes up, the predictions from the R/K equation are getting slightly worse. The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.9009, and the reduced pressure is Pr = 21.22/42.48 = 0.4995. The acentric factor for propane is 0.152. So, we have 0.422  0.4157 0.90091.6 0.172 B1  0.139   0.1276 0.9009 4.2 B 0  0.083 

Solution continued on next page…

and Z  1  0.4157  0.152  0.1276

0.4995  0.7588 0.9009 3

–1

From which V = ZRT/P = 0.7588*1305.4 = 990 cm mol .

(d) This is not too bad, since we only have to find the vapor and liquid volumes at a given pressure (and don’t have to find the pressure itself – we’ll save that for later in the semester). First, we need to find the Redlich/Kwong parameters for propane from Appendix B. The critical temperature and pressure (all we need for the RK EOS) are Tc = 369.8 K and Pc = 42.48 bar.

The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  369.82.5 1.8295  10 8 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol2 Pc Pc 42.48 T RT 83.145  369.8 b  0.08664 c  0.08664  62.71 cm3 mol 1 Pc 42.48 6

–2

At 70°C = 343.15 K, we have a = 9.876 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*343.15/25.94 = 1099.9 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT  b P P V V  b  9876000 V  62.71   3 1 V  1099.9  62.71   cm mol 25.94  V V  62.71    380725V  62.71   cm3 mol1 V  1162.6   V V  62.71  V

–1

Starting from V = 1099.9 cm mol

–1

and iterating gives V = 717 cm mol for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  343.15  62.17  25.94  25.94V  V  62.71  V V  62.71   9876000   1162.0  V  V  62.71  V V  62.71   380725  3

Starting from V = 62.71 cm mol

–1

and iterating on this gives V = 133.3 cm /mol. The ‘appropriate generalized

correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. The Rackett equation is V sat  Vc Z c

0.2857

1Tr 

–1

From appendix B, we have Vc = 200.0 cm mol , and Zc = 0.276. At 343.15 K, the reduced temperature is 0.9279, so we have 0.2857

V sat  200  0.276

1 0.9279

which gives V

sat

–1

= 109.0 cm mol . The R/K equation gave a value about 22% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

In the problem assignment, you were told to use the Pitzer correlation for the vapor, which is:

Z  1  B0

Pr P   B1 r Tr Tr

Solution continued on next page…

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.9279, and the reduced pressure is Pr = 25.94/42.48 = 0.6106. The acentric factor for propane is 0.152. So, we have 0.422  0.3927 0.92791.6 0.172 B1  0.139   0.09652 0.92794.2 B 0  0.083 

and Z  1  0.3927  0.152  0.0965

0.6106  0.7319 0.9279 3

–1

From which V = ZRT/P = 0.7319*1099.9 = 805 cm mol . This is 12% higher than the value from the R/K equation of state. If you were paying attention, though, you may have noticed that the reduced pressure and temperature here are nd

outside the range where the Pitzer correlation for the 2

virial coefficient accurately represents the Lee/Kessler

correlation. Thus, if we want to know whether the R/K result is accurate, we should compare it to the full Lee/Kessler correlation. Unfortunately, we can’t readily do so using the tables in the text because the point of interest is on the saturation curve. In the L/K table on page 668 in appendix E, we would like to interpolate between Tr = 0.9 and Tr = 0.93 and between Pr = 0.6 and Pr = 0.8. Looking at the four entries that we would use in our interpolation, we see that 0

Tr = 0.93, Pr = 0.6 corresponds to a vapor with Z = 0.6635 and (from p. 669) Z = –0.1662, but that the other three points correspond to liquid propane. It doesn’t make any sense to interpolate between the liquid and vapor points to get the vapor volume. So, we can’t interpolate. At the point Tr = 0.93, Pr = 0.6, we have Z = 0.6635 – 0.152*0.1662 = 0.638. If we extrapolate the Lee/Kesler table entries using the values at (Tr = 0.95, Pr = 0.6), (Tr = 0.93, Pr = 0.4), and Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(Tr = 0.93, Pr = 0.6) to (Tr = 0.9279, Pr = 0.6106) we obtain Z = 0.6525 and Z = –0.1768, from which 3

Z = 0.6525 – 0.152*0.1768 = 0.6256. This compressibility gives V = 688 cm /mol. This value is 15% smaller than the one from the Pitzer correlation, and 4% smaller than the one from the R/K equation. Presumably, this means that the R/K equation of state did better than the Pitzer correlation, since the conditions were outside the range where the Pitzer correlation works well.

(e) Now, we switch from propane to n-butane, so we have to look up a new set of critical properties. For n-butane, they are Tc = 425.1 K, Pc = 37.96 bar, and  = 0.200. Otherwise, it is exactly like part (d) above. The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  425.12.5 2.9006  108 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol 2 Pc Pc 37.96 T RTc 83.145  425.1 3 1 b  0.08664  0.08664  80.67 cm mol Pc 37.96 7

–2

At 100°C = 373.15 K, we have a = 1.5016 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*373.15/15.41 = 2013.3 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  15016000 V  80.67    cm3 mol1 V  2013.3  80.67   15.41  V V  80.67   974430 V  80.67    cm3 mol1 V  2094.0   V V  80.67  V

–1

Starting from V = 2013.3 cm mol

–1

and iterating gives V = 1516 cm mol

for the vapor volume, corresponding to

Z = 0.7531. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  373.15  80.67  15.41  15.41V  V  80.67  V V  80.67   15016000   2094.1  V  V  80.67  V V  80.67   974430  3

Starting from V = 80.67 cm mol

–1

and iterating on this gives V = 148.9 cm /mol. The ‘appropriate generalized

correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. The Rackett equation is V sat  Vc Z c

0.2857

1Tr 

–1

From appendix B, we have Vc = 255 cm mol , and Zc = 0.274. At 373.15 K, the reduced temperature is 0.8778, so we have 0.2857

V sat  255  0.274

10.8778

which gives V

sat

–1

= 125.4 cm mol . The R/K equation gave a value about 19% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

Solution continued on next page…

In the problem assignment, you were told to use the Pitzer correlation for the vapor, which is:

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8778, and the reduced pressure is Pr = 15.41/37.96 = 0.4059. The acentric factor for n-butane is 0.200. So, we have 0.422  0.4369 0.87781.6 0.172 B1  0.139   0.1583 0.87784.2 B 0  0.083 

and Z  1  0.4369  0.200  0.1583

0.4059  0.7833 0.8778 3

–1

From which V = ZRT/P = 0.7833*2013.3 = 1577 cm mol . This is 4% higher than the value from the R/K equation of state. Again, if you were paying attention, you may have noticed that the reduced pressure and temperature here are nd

outside the range where the Pitzer correlation for the 2 virial coefficient accurately represents the Lee/Kessler correlation. Thus, if we want to know whether the R/K result is accurate, we should compare it to the full Lee/Kessler correlation. Unfortunately, we can’t readily do so using the tables in the text because the point of interest is on the saturation curve. In the L/K table on page 668 in appendix E, we would like to interpolate between Tr = 0.85 and Tr = 0.9 and between Pr = 0.4 and Pr = 0.6. Looking at the four entries that we would use in our interpolation, we see Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

that Tr = 0.9, Pr = 0.4 corresponds to a vapor with Z = 0.7800 and (from p. 669) Z = –0.1118, but that the other three points correspond to liquid rather than vapor. It doesn’t make any sense to interpolate between the liquid and vapor points to get the vapor volume. So, we can’t interpolate. At the point Tr = 0.9, Pr = 0.4, we have Z = 0.7800 – 0.200*0.1118 = 0.7576. If we extrapolate the Lee/Kesler table entries using the values at (Tr = 0.93, 0

Pr = 0.4), (Tr = 0.9, Pr = 0.2), and (Tr = 0.9, Pr = 0.4) to (Tr = 0.8778, Pr = 0.4059) we obtain Z = 0.7572 and 1

Z = –0.1361, from which Z = 0.7572 – 0.200*0.1361 = 0.7300. This compressibility gives V = 1470 cm /mol. This value is 7% smaller than the one from the Pitzer correlation, and 3% smaller than the one from the R/K equation. Presumably, this means that the R/K equation of state did better than the Pitzer correlation, since the conditions were outside the range where the Pitzer correlation works well. (f) The Redlich/Kwong equation of state is P

RT V b



aT  V V  b 

–2

At 110°C = 383.15 K, we have a = 1.4818 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*383.15/18.66 = 1707.2 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  14818000 V  80.67    cm3 mol 1 V  1707.2  80.67   18.66  V V  80.67 V

 794105V  80.67  3 1 V  1787.9   cm mol  V V  80.67 

–1

Starting from V = 1707.2 cm mol

–1

and iterating gives V = 1216 cm mol

for the vapor volume, corresponding to

Z = 0.7124.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  383.15  80.67  18.66  18.66V  V  80.67  V V  80.67   14818000   1787.9  V  V  80.67  V V  80.67   794105  3

Starting from V = 80.67 cm mol

–1

and iterating on this gives V = 158.3 cm /mol. The ‘appropriate generalized

correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. The Rackett equation is V sat  Vc Z c

0.2857

1Tr 

–1

From appendix B, we have Vc = 255 cm mol , and Zc = 0.274. At 383.15 K, the reduced temperature is 0.9013, so we have 0.2857

V sat  255  0.274

10.9013

which gives V

sat

–1

= 130.7 cm mol . The R/K equation gave a value about 21% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

In the problem assignment, you were told to use the Pitzer correlation for the vapor, which is:

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.9013, and the reduced pressure is Pr = 18.66/37.96 = 0.4916. The acentric factor for n-butane is 0.200. So, we have 0.422  0.4153 0.90131.6 0.172 B1  0.139   0.1271 0.90134.2 B 0  0.083 

and Z  1  0.4153  0.200  0.1271

0.4916  0.7596 0.9013 3

–1

From which V = ZRT/P = 0.7596*1707.2 = 1297 cm mol . This is about 7% higher than the value from the R/K equation of state. Again, if you were paying attention, you may have noticed that the reduced pressure and temperature here are outside the range where the Pitzer correlation for the 2

virial coefficient accurately represents

the Lee/Kessler correlation. Thus, if we want to know whether the R/K result is accurate, we should compare it to the full Lee/Kessler correlation. Unfortunately, we can’t readily do so using the tables in the text because the point of interest is on the saturation curve. In the L/K table on page 668 in appendix E, we would like to interpolate between Tr = 0.90 and Tr = 0.93 and between Pr = 0.4 and Pr = 0.6. Looking at the four entries that we would use in our interpolation, we see that Tr = 0.93, Pr = 0.4, Tr = 0.90, Pr = 0.4, and Tr = 0.93, Pr = 0.96 correspond to vapor, but that Tr = 0.90, Pr = 0.6 corresponds to a liquid. It doesn’t make any sense to interpolate between the liquid and vapor Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

points to get the vapor volume, but this time (unlike in parts (d) and (e)) we have 3 vapor points that we can use to interpolate. Doing this 3-point interpolation of the Lee/Kesler table entries using the values at (Tr = 0.93, Pr = 0.4), 0

(Tr = 0.93, Pr = 0.6), and (Tr = 0.9, Pr = 0.4) to (Tr = 0.9013, Pr = 0.4916) we obtain Z = 0.7159 and Z = –0.1514, from 3

which Z = 0.7159 – 0.200*0.1514 = 0.6856. This compressibility gives V = 1170 cm /mol. This value is about 10% smaller than the one from the Pitzer correlation, and 4% smaller than the one from the R/K equation. Presumably, this means that the R/K equation of state did better than the Pitzer correlation, since the conditions were outside the range where the Pitzer correlation works well.

(g)

The critical properties of n-butane are, from appendix B, Tc = 425.1 K, Pc = 37.96 bar, and  = 0.200. The

acentric factor isn’t used in the Redlich/Kwong equation of state, but we’ll use it in the next two problems.

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  425.12.5 2.9006  108 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol 2 Pc Pc 37.96 T RT 83.145  425.1 b  0.08664 c  0.08664  80.67 cm3 mol1 Pc 37.96 7

–2

At 120°C = 393.15 K, we have a = 1.4629 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*393.15/22.38 = 1460.6 cm mol .

Solution continued on next page…

To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  14628800 V  80.67     cm 3 mol 1 V  1460.6  80.67   22.38  V V  80.67  653655V  80.67   cm3 mol1 V  1541.3   V V  80.67  V

–1

Starting from V = 1460.6 cm mol and iterating gives V = 971.8 cm mol

–1

for the vapor volume, corresponding to

Z = 0.6653.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  aT     83.14  393.15  80.67  22.38  22.38V  V  80.67  V V  80.67   14628800   1541.3  V  V  80.67  V V  80.67   653655  3

Starting from V = 80.67 cm mol

–1

and iterating on this gives V = 170.4 cm /mol, corresponding to Z = 0.117. The

‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Lee/Kesler correlations for the vapor.

The Rackett equation is V sat  Vc Z c

0.2857

1Tr 

Solution continued on next page…

–1

From appendix B, we have Vc = 255 cm mol , and Zc = 0.274. At 393.15 K, the reduced temperature is 0.9248, so we have 0.2857

V sat  255  0.274

10.9248

which gives V

sat

–1

= 137.4 cm mol . The R/K equation gave a value about 24% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. Looking at the figure on p. 103 of the text, we see that we are outside the range of reduced temperature and pressure where the Pitzer correlation for the second virial coefficient will represent the Lee/Kesler correlation well. So, we should use the full version from the tables in the back of the book. The reduced temperature is 0.9248 and the reduced pressure is 22.38/37.96 = 0.5896. So, we would like to interpolate between reduced temperatures of 0.9 and 0.93, and between reduced pressures of 0.4 and 0.6. Since we are on the saturation curve, some of these points correspond to vapor and some to liquid. For the vapor volume, we only want to use the 3 that correspond to vapor (Tr = 0.9, Pr = 0.4; 0

Tr = 0.93, Pr = 0.4; and Tr = 0.93, Pr = 0.6). Doing so gives Z = 0.6635 – 0.0045 + 0.0074 = 0.6664. Likewise, 1

–1

Z = –0.1662 – 0.0062 + 0.0047 = –0.1677. So, Z = 0.6664 + 0.2*(–0.1677) = 0.6329 and V = 924.4 cm mol . The prediction from the Redlich/Kwong EOS is about 5% higher than this.

(h) Now, we’ll do this again for a different temperature, in exactly the same way. If, like me, you typed your homework solutions in MS Word, you could just cut and paste and change a few numbers, like this: The critical properties of n-butane are, from appendix B, Tc = 425.1 K, Pc = 37.96 bar, and  = 0.200. The acentric factor isn’t used in the Redlich/Kwong equation of state, but we’ll use it in the next two problems. The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

–2

At 130°C = 403.15 K, we have a = 1.4446 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*403.15/26.59 = 1260.6 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  14446000 V  80.67     cm 3 mol 1 V  1260.6  80.67   26.59  V V  80.67 V

 543287 V  80.67  3 1 V  1341.3   cm mol  V V  80.67  3

–1

Starting from V = 1260.6 cm mol and iterating gives V = 768.9 cm mol for the vapor volume, corresponding to Z = 0.6099. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  403.15  80.67  26.59  26.59V  V  80.67  V V  80.67   14446000   1341.3  V  V  80.67  V V  80.67   543287  3

–1

Starting from V = 80.67 cm mol and iterating on this gives V = 187.1 cm /mol, corresponding to Z = 0.1484. The ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Lee/Kesler correlations for the vapor. Solution continued on next page…

The Rackett equation is V sat  Vc Z c

0.2857

1Tr 

–1

From appendix B, we have Vc = 255 cm mol , and Zc = 0.274. At 403.15 K, the reduced temperature is 0.9484, so we have 0.2857

V sat  255  0.274

10.9484

which gives V

sat

–1

= 146.4 cm mol . The R/K equation gave a value about 28% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. We see that as we increase the temperature (approaching the critical temperature) the liquid and vapor compressibilities approach each other, and the R/K EOS overestimates the liquid compressibility. We might expect this, since we know that the R/K EOS predicts a critical compressibility of 1/3, which is ~22% higher than the actual critical compressibility of n-butane. Looking at the figure on p. 103 of the text, we see that we are outside the range of reduced temperature and pressure where the Pitzer correlation for the second virial coefficient will represent the Lee/Kesler correlation well. So, we should use the full version from the tables in the back of the book. The reduced temperature is 0.9484 and the reduced pressure is 26.59/37.96 = 0.7005. So, we would like to interpolate between reduced temperatures of 0.93 and 0.95, and between reduced pressures of 0.6 and 0.8. Since we are on the saturation curve, some of these points correspond to vapor and some to liquid. This time, both points at a reduced pressure of 0.6 correspond to vapor, but both points at a reduced pressure of 0.8 correspond to liquid. So, we will have to extrapolate the reduced pressure using the values at 0.4 and 0.6. The formula for doing so is exactly the same as for interpolating, the only difference is that the point of interest lies outside of the 4 points used in the extrapolation. If we are doing so repeatedly, we might want to type the interpolation formula into a spreadsheet. Solution continued on next page…

T (K) 403.15

P (bar) 26.59

Tc (K) 425.1

Pc (bar) 37.96



Tr 0.93 0.93 0.95 0.95

Pr 0.40 0.60 0.40 0.60

Z0 0.8059 0.6635 0.8206 0.6967

0.2

Tr 0.9484

Pr 0.7005

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

0.6310

Z1 -0.0763 -0.1662 -0.0589 -0.1110 -0.1432

0.6023 759.4

The R/K EOS predictions for compressibility and molar volume are only about 1% higher than these – nearly exact agreement. We see that the R/K EOS does much better for vapor-phase volume and compressibility than for the liquid phase. (i)

This is not too bad, since we only have to find the vapor and liquid volumes at a given pressure (and don’t

have to find the pressure itself – we’ll save that for later in the semester). First, we need to find the Redlich/Kwong parameters for isobutane from Appendix B. The critical temperature and pressure (all we need for the RK EOS) are Tc = 408.1 K and Pc = 36.48 bar. The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  408.12.5 2.7255  108 a(T )  0.42748 r  0.42748  0.42748  bar cm 6 mol 2 Pc Pc 36.48 T RTc 83.145  408.1 3 1 b  0.08664  0.08664  80.59 cm mol Pc 36.48

Solution continued on next page…

–2

At 90°C = 363.15 K, we have a = 1.4302 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*363.15/16.54 = 1825.5 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  14302000 V  80.59    cm3 mol 1 V  1825.5  80.59   16.54  V V  80.59 V

 864692V  80.59   cm3 mol1 V  1906.1  V V  80.59   3

–1

Starting from V = 1825.5 cm mol and iterating gives V = 1330 cm mol for the vapor volume. This corresponds to a compressibility of Z = 1330/1825.5 = 0.7286.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  363.15  80.59  16.54  16.54V  V  80.59  V V  80.59   14302000   1906.0  V  V  80.59  V V  80.59   864692  3

–1

Starting from V = 80.59 cm mol and iterating on this gives V = 153.2 cm /mol. The corresponding compressibility is Z = 153.2/1825.5 = 0.0839. The ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. Solution continued on next page…

The Rackett equation is 0.2857

V sat  Vc Z c

1Tr 

–1

From appendix B, we have Vc = 262.7 cm mol , and Zc = 0.282. At 363.15 K, the reduced temperature is 0.8899, so we have 0.2857

V sat  262.7  0.282

1 0.8899

which gives V

sat

–1

= 133.9 cm mol , corresponding to a compressibility of Z = 133.9/1825.5 = 0.0733. The R/K

equation gave a value about 14% higher than this. The Rackett equation is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8899, and the reduced pressure is Pr = 16.54/36.48 = 0.4534. The acentric factor for isobutane is 0.302. So, we have 0.422  0.4256 0.88991.6 0.172 B1  0.139   0.1417 0.88994.2 B 0  0.083 

and Z  1  0.4256  0.302  0.1417 

0.4534  0.7614 0.8899 3

–1

From which V = ZRT/P = 0.7614*1825.5 = 1390 cm mol . This is higher than the value from the R/K equation of state. However, this reduced pressure and temperature may be a bit outside the range where the Pitzer correlation using only the 2

virial coefficient accurately represents the full Lee/Kessler correlation. This is discussed in more

detail at the end of the solution to part (j), where the problem should be even more severe.

(j) Now, we do the same thing at 100°C = 373.15 K (hey, it’s PRACTICE). At 100°C = 373.15 K, we have 7

–2

a = 1.4109 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

–1

Vid = RT/P = 83.145*373.15/20.03 = 1549.0 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  14109000V  80.59    cm 3 mol 1 V  1549.0  80.59   20.03  V V  80.59 V

 704406V  80.59  3 1 V  1629.6   cm mol V V  80.59   3

–1

Starting from V = 1549 cm mol and iterating gives V = 1058 cm mol for the vapor volume. This corresponds to a compressibility of Z = 1058/1549 = 0.6830.

Solution continued on next page…

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  373.15  80.59  20.03  20.03V  V  80.59  V V  80.59   14109000   1629.6  V  V  80.59  V V  80.59   704406  3

–1

Starting from V = 80.59 cm mol and iterating on this gives V = 164.2 cm /mol, corresponding to a compressibility of Z = 164.2/1549 = 0.1060. The ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. The Rackett equation is V sat  Vc Z c

0.2857

1Tr 

–1

From appendix B, we have Vc = 262.7 cm mol , and Zc = 0.282. At 373.15 K, the reduced temperature is 0.9144, so we have 0.2857

V sat  262.7  0.282

1 0.9144

which gives V

sat

–1

= 140.3 cm mol , corresponding to Z = 0.0906. The R/K equation gave a value about 13% higher

than this. The Rackett equation is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. Solution continued on next page…

The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.9144, and the reduced pressure is Pr = 20.03/36.48 = 0.5491. The acentric factor for isobutane is 0.302. So, we have 0.422  0.4040 0.91441.6 0.172 B1  0.139   0.1115 0.9144 4.2 B 0  0.083 

and Z  1  0.4040  0.302  0.1115

0.5491  0.7372 0.9144 3

–1

From which V = ZRT/P = 0.7372*1549 = 1142 cm mol , almost 10% higher than the value from the R/K EOS.

Unfortunately, the reduced temperature and pressure for this part of the problem are probably a bit outside the range where the Pitzer correlation for the virial coefficients can accurately match the full Lee/Kesler correlation. It is also difficult to use the Lee/Kesler tables in the back of the book for the saturated vapor, because some of the data points from which one would like to do interpolation correspond to the liquid phase, and therefore cannot be used. Solution continued on next page…

Extrapolating to Tr = 0.9144 and Pr = 0.5491 using the four vapor phase points with Tr = 0.93 or 0.95, and Pr = 0.4 or 0.6 (interpolating in pressure, extrapolating in temperature) gives the following results:

T (K)

P (bar)

373.15

Tc (K)

20.03

Pc (bar)

408.1

36.48

0.302

0.9144

0.5491

Table Points

Tr (1)

Pr(1)

0.93

0.40

0.8059

–0.0763

Tr(1)

Pr(2)

0.93

0.60

0.6635

–0.1662

Tr(2)

Pr(1)

0.95

0.40

0.8206

–0.0589

Tr(2)

Pr(2)

0.95

0.60

0.6967

–0.1110

Interpolated Values

0.6775

–0.1789

Final Values Z

0.6234

V (cm /mol)

965.8

Solution continued on next page…

Extrapolating to Tr = 0.9144 and Pr = 0.5491 using the four vapor phase points with Tr = 0.90 or 0.93, and Pr = 0.2 or 0.4 (interpolating in temperature, extrapolating in pressure) gives the following results: T (K)

P (bar)

373.15

Tc (K)

20.03

Pc (bar)

408.1

36.48

0.302

0.9144

0.5491

Table Points

Tr (1)

Pr(1)

0.90

0.20

0.9015

–0.0442

Tr(1)

Pr(2)

0.90

0.40

0.7800

–0.1118

Tr(2)

Pr(1)

0.93

0.20

0.9115

–0.0326

Tr(2)

Pr(2)

0.93

0.40

0.8059

–0.0763

Interpolated Values

0.7075

–0.1367

Final Values Z

0.6662

V (cm /mol)

1032.1

Solution continued on next page…

We could also try using just the 3 closest vapor-phase points, at Tr = 0.9, Pr = 0.4; Tr = 0.93, Pr = 0.4; and Tr = 0.93, 0

Pr = 0.6. This would give Z = 0.8059 – 0.0134 – 0.1062 = 0.6857 and Z = –0.0763 – 0.0180 – 0.0670 = –0.1613, so Z = 0.6857 – 0.302*0.1613 = 0.6370 (in between the other two results). So, the Pitzer correlation predicted a volume and compressibility that were higher than the R/K prediction, but this extrapolation from the Lee/Kesler tables gives a value that is lower than the R/K prediction. The result based on the tables is probably the closest to the truth, though it also has significant uncertainty due to the difficulties associated with extrapolating to the vapor/liquid equilibrium line.

(k) This is not too bad, since we only have to find the vapor and liquid volumes at a given pressure (and don’t have to find the pressure itself – we’ll save that for later in the semester). First, we need to find the Redlich/Kwong parameters from Appendix B. The critical temperature and pressure (all we need for the RK EOS) are Tc = 408.1 K and Pc = 36.48 bar. The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

–2

At 110°C = 383.15 K, we have a = 1.392 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*383.15/24.01 = 1326.8 cm mol . To find the vapor phase volume, we iterate on aT  V  b RT b P P V V  b  13920000 V  80.59    cm3 mol 1 V  1326.8  80.59   24.01  V V  80.59  579922V  80.59   cm 3 mol1 V  1407.4   V V  80.59  V

–1

Starting from V = 1326.8 cm mol and iterating gives V = 835.3 cm mol for the vapor volume, corresponding to a compressibility of Z = 835.3/1326.8 = 0.6296. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  aT     83.14  383.15  80.59  24.01  24.01V  V  80.59  V V  80.59   13920000   1407.4  V  V  80.59  V V  80.59   579922  3

–1

Starting from V = 80.59 cm mol and iterating on this gives V = 179.1 cm /mol.

Some ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor.

Solution continued on next page…

The Rackett equation is 0.2857

V sat  Vc Z c

1Tr 

–1

From appendix B, we have Vc = 262.7 cm mol , and Zc = 0.282. At 383.15 K, the reduced temperature is 0.9389, so we have 0.2857

V sat  262.7  0.282

1 0.9389

which gives V

sat

–1

= 148.6 cm mol . The R/K equation gave a value about 20% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

The Pitzer correlation that we want to use is Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.9389, and the reduced pressure is Pr = 24.01/36.48 = 0.6582. The acentric factor is 0.181. So, we have 0.422  0.3838 0.93891.6 0.172 B1  0.139   0.0851 0.93894.2 B 0  0.083 

and Z  1  0.3838  0.181  0.0851

0.6582  0.7201 0.9389 3

–1

From which V = ZRT/P = 0.7201*1326.8 = 955.5 cm mol . Solution continued on next page…

In this case, we can go back and look at Figure 3.14 on p. 103, where we find that we are outside the range of reduced temperature and pressure where the Pitzer correlation is expected to work well. So, we cannot have much faith in either the R/K volume or the Pitzer correlation volume for the vapor. To obtain a better estimate of the vapor volume and compressibility, we could use the Lee/Kesler tables in appendix E. Because the point of interest is along the saturation curve, we must extrapolate to it using points from the ‘vapor’ side of the boundary between the vapor and liquid compressibilities.

One way to do the extrapolation is to set up a little spreadsheet like that shown below: T (K)

P (bar)

383.15

24.01

Tc (K)

Pc (bar)

408.1

36.48

0.181

0.9389

0.6582

Table Points

Tr (1)

Pr(1)

0.93

0.40

0.8059

–0.0763

Tr(1)

Pr(2)

0.93

0.60

0.6635

–0.1662

Tr(2)

Pr(1)

0.95

0.40

0.8206

–0.0589

Tr(2)

Pr(2)

0.95

0.60

0.6967

–0.1110

Interpolated Values

0.6392

–0.1630

Final Values Z

0.6097

V (cm /mol)

809.0

These results from the full Lee/Kesler correlation are close to those from the R/K equation of state. As expected, the Pitzer correlation does not work at these conditions. 7

–2

(l) Repeating part (k) at 393.15 K and 28.53 bar, we have a = 1.3746 × 10 bar cm mol . The ideal gas molar volume 3

–1

at this temperature and pressure is Vid = RT/P = 83.145*393.15/28.53 = 1145.8 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  13746000 V  80.59    cm3 mol1 V  1145.8  80.59   28.53  V V  80.59  V

 481800 V  80.59  3 1 V  1226.3   cm mol V V  80.59   3

–1

Starting from V = 1145.8 cm mol and iterating gives V = 645.8 cm mol for the vapor volume, corresponding to a compressibility of Z = 645.8/1145.8 = 0.5636.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  393.15  80.59  28.53  28.53V  V  80.59  V V  80.59   13746000   1226.3  V  V  80.59  V V  80.59   481800  3

–1

Starting from V = 80.59 cm mol and iterating on this gives V = 201.4 cm /mol (after very many iterations!).

Now, we again compare to generalized correlations. Solution continued on next page…

The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 262.7 cm mol , and Zc = 0.282. At 393.15 K, the reduced temperature is 0.9632, so we have V sat  262.7  0.282

0.2857

10.9632

which gives V

sat

–1

= 160.5 cm mol . The R/K equation gave a value about 25% higher than this. The Rackett equation

is usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. We also note that the agreement between the R/K equation and the Rackett equation is getting worse with increasing temperature. The Pitzer correlation that we want to use is Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.9632, and the reduced pressure is Pr = 28.53/36.48 = 0.7821. The acentric factor is 0.181. So, we have 0.422  0.3651 0.96321.6 0.172 B1  0.139   0.0623 0.9632 4.2 B 0  0.083 

and Z  1  0.3651  0.181  0.0623

0.7821  0.6944 0.9632 3

–1

From which V = ZRT/P = 0.6944*1145.8 = 795.6 cm mol .

Solution continued on next page…

Again, we are outside the range of reduced temperature and pressure where the Pitzer correlation is expected to work well. So, we cannot have much faith in either the R/K volume or the Pitzer correlation volume for the vapor. To obtain a better estimate of the vapor volume and compressibility, we will again use the Lee/Kesler tables in appendix E. T (K)

P (bar)

393.15

28.53

Tc (K)

Pc (bar)

408.1

36.48

0.181

0.9634

0.7821

Table Points

Tr (1)

Pr(1)

0.95

0.40

0.8206

–0.0589

Tr(1)

Pr(2)

0.95

0.60

0.6967

–0.1110

Tr(2)

Pr(1)

0.97

0.40

0.8338

–0.0450

Tr(2)

Pr(2)

0.97

0.60

0.7240

–0.0770

Interpolated Values

0.6107

–0.1235

Final Values Z

0.5884

V (cm /mol)

674.2

Solution continued on next page…

These results from the full Lee/Kesler correlation are close to those from the R/K equation of state. As expected, the Pitzer correlation does not work at these conditions. (m)

First, we need to find the Redlich/Kwong EOS parameters using the critical properties of chlorine from

Appendix B. The critical temperature and pressure of chlorine (all we need for the RK EOS) are Tc = 417.2 K and Pc = 77.10 bar.

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  417.22.5 1.362  10 8 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol 2 Pc Pc 77.10 T RT 83.145  417.2 b  0.08664 c  0.08664  38.98 cm 3 mol 1 Pc 77.10 6

–2

At 60°C = 333.15 K, we have a = 7.466 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*333.15/18.21 = 1521.1 cm mol . To find the vapor phase volume, we can iterate on aT  V  b RT b P P V V  b  7466000 V  38.98   3 1 V  1521.1  38.98   cm mol 18.21  V V  38.98    409980 V  38.98   cm 3 mol1 V  1560.1   V V  38.98  V

Solution continued on next page…

Starting from V = 1521 cm mol

–1

and iterating gives V = 1252.5 cm mol for the vapor volume, corresponding to a

compressibility of Z = 1252.5/1521.1 = 0.8234.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  333.15  38.98  18.21  18.21V  V  38.98  V V  38.98   7466000   1560.1  V  V  38.98  V V  38.98   409980  3

–1

Starting from V = 40 cm /mol cm mol and iterating on this gives V = 61.7 cm /mol.

Some ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor.

The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 124. cm mol , and Zc = 0.265. At 333.15 K, the reduced temperature is 0.7985, so we have V sat  124  0.265

0.2857

10.7985

Solution continued on next page…

which gives V

sat

–1

= 53.5 cm mol . The R/K equation gave a value about 15% higher than this. The Rackett equation is

usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.7985, and the reduced pressure is Pr = 18.21/77.10 = 0.2362. The acentric factor for Cl2 is 0.069. So, we have 0.422  0.5219 0.79851.6 0.172 B1  0.139   0.3036 0.7985 4.2 B 0  0.083 

and Z  1  0.5210  0.069  0.3036

0.2362  0.8397 0.7985 3

–1

From which V = ZRT/P = 0.8397*1521.1 = 1277 cm mol . This is within 2% of the value given by the R/K equation.

Solution continued on next page…

However, if we go back and look at Figure 3.14 on p. 103, we find that, because of the low reduced temperature, we are slilghtly outside the range of reduced temperature and pressure where the Pitzer correlation is expected to work well. So, we cannot have much faith in either the R/K volume or the Pitzer correlation volume for the vapor. To obtain a better estimate of the vapor volume and compressibility, we could use the Lee/Kesler tables in appendix E. Because the point of interest is along the saturation curve, we must extrapolate to it using points from the ‘vapor’ side of the boundary between the vapor and liquid compressibilities. One way to do the extrapolation is to set up a little spreadsheet like that shown below:

T (K) 333.15

P (bar) 18.21

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.7985

Pr 0.2362

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

Z0 0.9319 0.8539 0.9436 0.8810 0.8247

Z1 -0.0487 -0.1160 -0.0319 -0.0715 -0.1419

0.8149 1239.7

These results from the full Lee/Kesler correlation are close to those from the R/K equation of state and the Pitzer correlation. 6

–2

–1

(n) Repeating part (m) at 343.15 K and 22.49 bar, we have a = 7.353 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*343.15/22.49 = 1268.6 cm mol . To find the vapor phase volume, we iterate on aT  V  b RT b P P V V  b  7353000 V  38.98    cm3 mol 1 V  1268.6  38.98   22.49  V V  38.98  326900V  38.98   cm3 mol 1 V  1307.6   V V  38.98  V

Solution continued on next page…

–1

Starting from V = 1268 cm mol

–1

and iterating gives V = 1007.2 cm mol for the vapor volume, corresponding to a

compressibility of Z = 1007.6/1268.6 = 0.7940. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  343.15  38.98  22.49  22.49V  V  38.98  V V  38.98   7353000   1307.6  V  V  38.98  V V  38.98   326900  3

–1

Starting from V = 40 cm mol and iterating on this gives V = 64.1 cm /mol.

Now, we again compare to generalized correlations. The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 124. cm mol , and Zc = 0.265. At 343.15 K, the reduced temperature is 0.8225, so we have 0.2857

V sat  124  0.265

10.8225

which gives V

sat

–1

= 55.1 cm mol . Again, the R/K equation gave a value about 15% higher than this.

The Pitzer correlation is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8225, and the reduced pressure is Pr = 22.49/77.10 = 0.2917. The acentric factor is 0.069. So, we have 0.422  0.4939 0.82251.6 0.172 B1  0.139   0.2518 0.8225 4.2 B 0  0.083 

and Z  1  0.4939  0.069  0.2518

0.2917  0.8187 0.8225 3

–1

From which V = ZRT/P = 0.8187*1268.6 = 1038.6 cm mol , about 3% higher than the R/K result. Again, we are a little bit outside the region where the Pitzer correlation is a good approximation to the full Lee/Kesler correlation. Again, doing the extrapolation from entries for the vapor phase in the Lee/Kesler tables, we have:

T (K) 343.15

P (bar) 22.49

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.8225

Pr 0.2917

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

Z0 0.9319 0.8539 0.9436 0.8810 0.8009

Z1 -0.0487 -0.1160 -0.0319 -0.0715 -0.1462

0.7908 1003.4

This compressibility and volume are very close to the R/K values, and 3% smaller than those from the Pitzer correlation. (o)

First, we need to find the Redlich/Kwong EOS parameters using the critical properties of chlorine from

Appendix B. The critical temperature and pressure of chlorine (all we need for the RK EOS) are Tc = 417.2 K and Pc = 77.10 bar. The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  417.22.5 1.362  10 8 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol 2 Pc Pc 77.10 T RTc 83.145  417.2 3 1 b  0.08664  0.08664  38.98 cm mol Pc 77.10 6

–2

At 80°C = 353.15 K, we have a = 7.248 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*353.15/27.43 = 1070.5 cm mol . To find the vapor phase volume, we can iterate on aT  V  b RT b P P V V  b  7248000 V  38.98    cm3 mol 1 V  1070.5  38.98   27.43  V V  38.98  264224 V  38.98   cm3 mol 1 V  1109.4   V V  38.98  V

–1

Starting from V = 1070 cm mol and iterating gives V = 814.7 cm mol for the vapor volume, corresponding to a compressibility of Z = 814.7/1070.5 = 0.7611.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  353.15  38.98  27.43  27.43V  V  38.98  V V  38.98   7248000   1109.4  V  V  38.98  V V  38.98   264224  3

Starting from V = 50 cm /mol and iterating on this gives V = 67.0 cm /mol.

Some ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 124. cm mol , and Zc = 0.265. At 353.15 K, the reduced temperature is 0.8465, so we have 0.2857

V sat  124  0.265

10.8465

which gives V

sat

–1

= 57.0 cm mol . The R/K equation gave a value about 18% higher than this. The Rackett equation is

usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid. The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

Solution continued on next page…

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8465, and the reduced pressure is Pr = 27.43/77.10 = 0.3558. The acentric factor for Cl2 is 0.069. So, we have 0.422  0.4680 0.84651.6 0.172 B1  0.139   0.2073 0.8465 4.2 B 0  0.083 

and Z  1  0.4680  0.069  0.2073

0.3558  0.7973 0.8465 3

–1

From which V = ZRT/P = 0.7973*1070.5 = 853.5 cm mol . This is within 5% of the value given by the R/K equation.

Solution continued on next page…

saturation curve, we must extrapolate to it using points from the ‘vapor’ side of the boundary between the vapor and liquid compressibilities. One way to do the extrapolation is to set up a little spreadsheet like that shown below: T (K) 353.15

P (bar) 27.43

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.8465

Pr 0.3558

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z 3 V (cm /mol)

Z 0.9319 0.8539 0.9436 0.8810

Z -0.0487 -0.1160 -0.0319 -0.0715

0.7799

-0.1394

0.7703 824.6

These results from the full Lee/Kesler correlation fall between those from the R/K equation of state and the Pitzer correlation. 6

–2

–1

(p) Repeating part (o) at 363.15 K and 33.08 bar, we have a = 7.147×10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*363.15/33.08 = 912.8 cm mol . To find the vapor phase volume, we iterate on aT  V  b RT b P P V V  b  7147000 V  38.98    cm3 mol1 V  912.8  38.98   33.08  V V  38.98  216057 V  38.98    cm3 mol1 V  951.7   V V  38.98  V

Starting from V = 913 cm mol

–1

and iterating gives V = 661.4 cm mol for the vapor volume, corresponding to a

compressibility of Z = 661.4/912.8 = 0.7246.

Solution continued on next page…

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b   aT    83.14  363.15  38.98  33.08  33.08V  V  38.98  V V  38.98   7147000   951.7  V  V  38.98  V V  38.98   216057  3

–1

Starting from V = 50 cm mol and iterating on this gives V = 70.4 cm /mol.

Now, we again compare to generalized correlations.

The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 124. cm mol , and Zc = 0.265. At 363.15 K, the reduced temperature is 0.8704, so we have 0.2857

V sat  124  0.265

10.8704

which gives V

sat

–1

= 59.1 cm mol . The R/K equation gave a value about 19% higher than this.

The Pitzer correlation is

Z  1  B0

Pr P   B1 r Tr Tr

Solution continued on next page…

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8704, and the reduced pressure is Pr = 33.08/77.10 = 0.4291. The acentric factor is 0.069. So, we have 0.422  0.4439 0.87041.6 0.172 B1  0.139   0.1691 0.8704 4.2 B 0  0.083 

and Z  1  0.4439  0.069  0.1691

0.4291  0.7754 0.8704 3

–1

From which V = ZRT/P = 0.7754*912.8 = 707.8 cm mol , about 7% higher than the R/K result. Again, we are a little bit outside the region where the Pitzer correlation is a good approximation to the full Lee/Kesler correlation. Again, doing the extrapolation from entries for the vapor phase in the Lee/Kesler tables, we have:

T (K) 363.15

P (bar) 33.08

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.90 0.90 0.93 0.93

Pr 0.20 0.40 0.20 0.40

ω 0.069

Tr 0.8704

Pr 0.4291

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z 3 V (cm /mol)

Z 0.9015 0.7800 0.9115 0.8059 0.7346

Z -0.0442 -0.1118 -0.0326 -0.0763 -0.1600

0.7235 660.5

This compressibility and volume are very close to the R/K values, and smaller than those from the Pitzer correlation. We could also do the extrapolation from a different set of points (extrapolating further in Pr, but not in Tr).

T (K) 363.15

P (bar) 33.08

Tc (K) 417.2

Pc (bar) 77.1

Tr 0.85 0.85 0.90 0.90

Pr 0.10 0.20 0.10 0.20

ω 0.069

Tr 0.8704

Pr 0.4291

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z 3 V (cm /mol)

Z 0.9436 0.8810 0.9528 0.9015 0.7566

Z -0.0319 -0.0715 -0.0205 -0.0442 -0.1361

0.7472 682.1

This gives a slightly larger volume, but is still closer to the R/K values than the Pitzer correlation. (q)

First, we need to find the Redlich/Kwong EOS parameters using the critical properties of SO2 from Appendix

B. The critical temperature and pressure of SO2 (all we need for the RK EOS) are Tc = 430.8 K and Pc = 78.84 bar. The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  430.82.5 1.4439  10 8 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol 2 Pc Pc 78.84 T RTc 83.145  430.8 b  0.08664  0.08664  39.363 cm3 mol 1 Pc 78.84

–2

At 80°C = 353.15 K, we have a = 7.683 × 10 bar cm mol . The ideal gas molar volume at this temperature and 3

–1

pressure is Vid = RT/P = 83.145*353.15/18.66 = 1573.6 cm mol .

To find the vapor phase volume, we can iterate on aT  V  b RT  b P P V V  b  7683000 V  39.363    cm3 mol1 V  1573.6  39.363   18.66  V V  39.363  411736 V  39.363   cm3 mol 1 V  1613.0   V V  39.363  V

–1

Starting from V = 1570 cm mol and iterating gives V = 1318.9 cm mol for the vapor volume, corresponding to a compressibility of Z = 1318.9/1573.6 = 0.8382.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b    aT    83.14  353.15  39.363  18.66  18.66V  V  39.363  V V  39.363   7683000   1613.0  V  V  39.363  V V  39.363   411736  3

Starting from V = 50 cm /mol and iterating on this gives V = 64.7 cm /mol. This corresponds to Z = 64.7/1573.6 = 0.0411.

Solution continued on next page…

Some ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor.

The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 122. cm mol , and Zc = 0.269. At 353.15 K, the reduced temperature is 0.8198, so we have 0.2857

V sat  122  0.269

10.8198

which gives V

sat

–1

= 54.6 cm mol . The R/K equation gave a value about 19% higher than this. The Rackett equation is

usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

Solution continued on next page…

The reduced temperature is Tr = 0.8198, and the reduced pressure is Pr = 18.66/78.84 = 0.2367. The acentric factor for SO2 is 0.245. So, we have 0.422  0.4969 0.81981.6 0.172 B1  0.139   0.2572 0.81984.2 B 0  0.083 

and Z  1  0.4969  0.245  0.2572 

0.2367  0.8282 0.8198 3

–1

From which V = ZRT/P = 0.8282*1573.6 = 1303.2 cm mol . This is only about 1% different from the R/K result.

However, if we go back and look at Figure 3.14 on p. 103, we find that we are slightly outside the range of reduced temperature and pressure where the Pitzer correlation is expected to work well. So, we cannot have much faith in either the R/K volume or the Pitzer correlation volume for the vapor. To obtain a better estimate of the vapor volume and compressibility, we could use the Lee/Kesler tables in appendix E. Because the point of interest is along the saturation curve, we must extrapolate to it using points from the ‘vapor’ side of the boundary between the vapor and liquid compressibilities. One way to do the extrapolation is to set up a little spreadsheet like that shown below: T (K) 353.15

P (bar) 18.66

Tc (K) 430.8

Pc (bar) 78.84

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.245

Tr 0.8198

Pr 0.2367

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm 3/mol)

Z0 0.9319 0.8539 0.9436 0.8810 0.8382

Z1 -0.0319 -0.0715 -0.0205 -0.0442 -0.0729

0.8204 1291.0

These results from the full Lee/Kesler correlation are in good agreement with the other two predictions in this case. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

–2

–1

(r) Repeating part (q) at 363.15 K and 23.31 bar, we have a = 7.577 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*363.15/23.31 = 1295.3 cm mol . To find the vapor phase volume, we iterate on aT  V  b RT  b P P V V  b  7577000 V  39.363    cm3 mol1 V  1295.3  39.363   23.31  V V  39.363  325053V  39.363    cm3 mol1 V  1334.7   V V  39.363  V

–1

Starting from V = 1295 cm mol and iterating gives V = 1046.6 cm mol for the vapor volume, corresponding to a compressibility of Z = 1046.6/1295.3 = 0.8080.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b    aT    83.14  363.15  39.363  23.31  23.31V  V  39.363  V V  39.363   7577000   1334.7  V  V  39.363  V V  39.363   325053  3

Starting from V = 50 cm /mol and iterating on this gives V = 67.5 cm /mol. This corresponds to Z = 67.5/1295.3 = 0.0521.

Some ‘appropriate generalized correlations’ to which we can compare these are the Rackett equation for the liquid, and the Pitzer correlations for the vapor. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The Rackett equation is 0.2857

1Tr 

V sat  Vc Zc

–1

From appendix B, we have Vc = 122. cm mol , and Zc = 0.269. At 363.15 K, the reduced temperature is 0.8430, so we have 0.2857

V sat  122  0.269

10.8430

which gives V

sat

–1

= 56.3 cm mol . The R/K equation gave a value about 20% higher than this. The Rackett equation is

usually good to within a couple percent, so we expect that the R/K equation of state is overestimating the volume of the saturated liquid.

The Pitzer correlation that we want to use is

Z  1  B0

Pr P   B1 r Tr Tr

with 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

The reduced temperature is Tr = 0.8430, and the reduced pressure is Pr = 23.31/78.84 = 0.2957. The acentric factor for SO2 is 0.245. So, we have 0.422  0.4726 0.84301.6 0.172 B1  0.139   0.2134 0.8430 4.2 B 0  0.083 

Solution continued on next page…

and Z  1  0.4726  0.245  0.2134 

0.2957  0.8159 0.8430 3

–1

From which V = ZRT/P = 0.8159*1295.3 = 1056.8 cm mol . This is only about 1% different from the R/K result.

T (K) 363.15

P (bar) 23.11

Tc (K) 430.8

Pc (bar) 78.84

Tr 0.80 0.80 0.85 0.85

Pr 0.10 0.20 0.10 0.20

ω 0.245

Tr 0.8430

Pr 0.2931

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm 3/mol)

Z0 0.9319 0.8539 0.9436 0.8810 0.8169

Z1 -0.0319 -0.0715 -0.0205 -0.0442 -0.0722

0.7992 1044.3

These results from the full Lee/Kesler correlation are in good agreement with the other two predictions in this case.

Solution 3.54

Problem Statement

Use the Soave/Redlich/Kwong equation to calculate the molar volumes of saturated liquid and saturated vapor for the substance and conditions given by one of the parts of Prob. 3.53 and compare results with values found by suitable generalized correlations.

Problem 3.53 Calculate the molar volume of saturated liquid and the molar volume of saturated vapor by the Redlich/Kwong equation for one of the following and compare results with values found by suitable generalized correlations.

(a) Propane at 40°C where P (b) Propane at 50°C where P (c) Propane at 60°C where P

sat

= 13.71 bar

sat

= 17.16 bar

sat

(d) Propane at 70°C where P

= 21.22 bar

sat

(e) n-Butane at 100°C where P

= 25.94 bar

sat

(f ) n-Butane at 110°C where P (g) n-Butane at 120°C where P

sat

(h) n-Butane at 130°C where P (i) Isobutane at 90°C where P

sat

(j) Isobutane at 100°C where P

(m) Chlorine at 60°C where P (n) Chlorine at 70°C where P (o) Chlorine at 80°C where P

= 18.66 bar = 22.38 bar = 26.59 bar

= 16.54 bar

sat

(k) Isobutane at 110°C where P (l) Isobutane at 120°C where P

= 15.41 bar

sat

= 20.03 bar = 24.01 bar = 28.53 bar

= 18.21 bar

= 22.49 bar = 27.43 bar

(p) Chlorine at 90°C where P

sat

= 33.08 bar

(q) Sulfur dioxide at 80°C where P (r) Sulfur dioxide at 90°C where P

sat

(s) Sulfur dioxide at 100°C where P (t) Sulfur dioxide at 110°C where P

= 18.66 bar = 23.31 bar

sat

= 28.74 bar = 35.01 bar

(u) Boron trichloride at 400 K where P

sat

= 17.19 bar

For BCl3, Tc = 452 K, Pc = 38.7 bar, and (v) Boron trichloride at 420 K where P

sat

(w) Boron trichloride at 440 K where P (x) Trimethylgallium at 430 K where P

= 0.086. = 23.97 bar

sat

= 32.64 bar = 13.09 bar

For Ga(CH3)3, Tc = 510 K, Pc = 40.4 bar, and (y) Trimethylgallium at 450 K where P (z) Trimethylgallium at 470 K where P

sat

= 0.205.

= 18.27 bar = 24.55 bar

Solution

(a) – (c) not provided (d) For propane, the critical parameters are Tc = 369.8 K, Pc = 42.48 bar, and  = 0.152.

The Soave/Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with

a(T )  0.42748 b  0.08664

Tr  R2Tc2 Pc

 0.42748

 Tr 83.1452 369.82 42.48

 9.5134 106 Tr  bar cm6 mol2

RTc 83.145369.8  0.08664  62.71 cm3 mol1 Pc 42.48

Solution continued on next page…

At 70°C = 343.15 K, we have Tr = 343.15/369.8 = 0.9279. At this reduced temperature, we have

   T   1  0.480  1.574 * 0.152  0.176 * 0.152 1  0.9279   1.0532  Tr   1  0.480  1.574  0.176 2 1  Tr0.5 

0.5

–2

So, we have a = 1.0020 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*343.15/25.94 = 1099.9 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  10020000 V  62.71   cm 3 mol1 V  1099.9  62.71  25.94  V V  62.71   386259V  62.71   cm 3 mol1 V  1162.6   V V  62.71  V

1

Starting from V = 1099.9 cm mol

and iterating gives V = 703 cm mol

1

for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  a T     83.14  343.15  62.17  25.94  25.94V  V  62.71  V V  62.71   10020000  1162.0  V  V  62.71  V V  62.71   386259 

Starting from V = 62.71 cm mol

1

and iterating on this gives V = 128.3 cm /mol. Both the liquid and vapor volumes

are a little closer to the values from the generalized correlations than the R/K results. Solution continued on next page…

(e) Now, we switch from propane to n-butane, for which the critical properties are Tc = 425.1 K, Pc = 37.96 bar, and  = 0.200. Otherwise, it is exactly like part (d) above. The Soave/Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with

a(T )  0.42748 b  0.08664

Tr  R2Tc2 Pc

 0.42748

 Tr 83.1452  425.12 37.96

 1.4068107Tr  bar cm6 mol2

RTc 83.145 425.1  0.08664  80.67 cm3 mol1 Pc 37.96

At 100°C = 373.15 K, we have Tr = 373.15/425.1 = 0.8778. At this reduced temperature, we have

   T   1  0.480  1.574 * 0.200  0.176 * 0.200 1  0.8778   1.1019  Tr   1  0.480  1.574  0.176 2 1  Tr0.5 

0.5

2

So, we have a = 1.5502 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*373.15/15.41 = 2013.3 cm mol .

Solution continued on next page…

To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b   15502000 V  80.67    cm 3 mol 1 V  2013.3  80.67   15.41  V V  80.67   1005940 V  80.67    cm 3 mol 1 V  2094.0   V V  80.67   V

1

Starting from V = 2013.3 cm mol

and iterating gives V = 1487 cm mol

1

for the vapor volume, corresponding to Z

= 0.7387.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  a T     83.14  373.15  80.67  15.41  15.41V  V  80.67  V V  80.67   15502000   2094.1  V  V  80.67  V V  80.67   1005940  3

1

Starting from V = 80.67 cm mol

and iterating on this gives V = 142.1 cm /mol.

(f) The Soave/Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

with

a(T )  0.42748 b  0.08664

Tr  R2Tc2

RTc Pc

 0.42748

 0.08664

Tr   83.1452 425.12 37.96

 1.4068107Tr  bar cm6 mol2

83.145  425.1  80.67 cm3 mol1 37.96

At 110°C = 383.15 K, we have Tr = 383.15/425.1 = 0.9013.

At this reduced temperature, we have

   T   1  0.480  1.574 * 0.200  0.176 * 0.200 1  0.9013   1.0814  Tr   1  0.480  1.574   0.176 2 1  Tr0.5 

0.5

2

So, we have a = 1.5212 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*383.15/18.66 = 1707.2 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b   15212000 V  80.67    cm 3 mol1 V  1707.2  80.67  18.66  V V  80.67     815252 V  80.67    cm 3 mol 1 V  1787.9   V V  80.67   V

1

Starting from V = 1707.2 cm mol

and iterating gives V = 1190 cm mol

1

for the vapor volume, corresponding to

Z = 0.6968. Solution continued on next page…

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b    a T    83.14  383.15  80.67  18.66  18.66V  V  80.67  V V  80.67   15212000   1787.9  V  V  80.67  V V  80.67   815252  3

Starting from V = 80.67 cm mol

1

and iterating on this gives V = 150.7 cm /mol. Again, the answers move closer to

the best values from the generalized correlations when we go from the R/K to SRK equation of state. (g) The Soave/Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with a(T )  0.42748 b  0.08664

Tr  R2Tc2 Pc

 0.42748

 Tr 83.1452  425.12 37.96

 1.4068107Tr  bar cm6 mol2

RTc 83.145 425.1  0.08664  80.67 cm3 mol1 Pc 37.96

At 120°C = 393.15 K, we have Tr = 393.15/425.1 = 0.9248. At this reduced temperature, we have

   T   1  0.480  1.574 * 0.200  0.176 * 0.200 1  0.9248   1.0613  Tr   1  0.480  1.574  0.176 2 1  Tr0.5 

0.5

Solution continued on next page…

2

1

So, we have a = 1.493 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*393.15/22.38 = 1460.6 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b   14930000 V  80.67  3 1 V  1460.6  80.67   cm mol  22.38  V V  80.67   667113V  80.67   cm 3 mol 1 V  1541.3  V V  80.67   V

1

Starting from V = 1460.6 cm mol

and iterating gives V = 947.9 cm mol

1

for the vapor volume, corresponding to

Z = 0.6490. This is about 2.5% higher than the Lee/Kesler correlation. To find the liquid volume, we iterate on  RT  bP  VP   V  b  V V  b  a T     83.14  393.15  80.67  22.38  22.38V  V  80.67  V V  80.67   14930000   1541.3  V  V  80.67  V V  80.67   667113  3

1

Starting from V = 80.67 cm mol

and iterating on this gives V = 161.8 cm /mol. This is a little closer to the value from the

Rackett equation than what we got using the R/K equation of state, but still about 18% too high.

Solution continued on next page…

(h) At 130°C = 403.15 K, we have Tr = 403.15/425.1 = 0.9484. At this reduced temperature, we have

   T   1  0.480  1.574 * 0.200  0.176 * 0.200 1  0.9484   1.0416  Tr   1  0.480  1.574   0.176 2 1  Tr0.5 

0.5

1

The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*403.15/26.59 = 1260.6 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b   14654000 V  80.67   cm 3 mol 1 V  1260.6  80.67  26.59  V V  80.67    551101V  80.67    cm 3 mol1 V  1341.3   V V  80.67  V

2

So, we have a = 1.4654 × 10 bar cm mol . Starting from V = 1260.6 cm mol 1

mol

1

and iterating gives V = 748.0 cm

for the vapor volume, corresponding to Z = 0.5933. This is about 1.5% lower than the Lee/Kesler correlation (a

little better than the R/K EOS). To find the liquid volume, we iterate on  RT  bP  VP   V  b  V V  b  a T     83.14  403.15  80.67  26.59  26.59V  V  80.67  V V  80.67   14654000   1341.3  V  V  80.67  V V  80.67   551101 

Solution continued on next page…

Starting from V = 80.67 cm mol

1

and iterating on this gives V = 177.1 cm /mol. This is about 21% higher than the

value from the Rackett Equation. (s) The Soave/Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with

a(T )  0.42748 b  0.08664

Tr  R2Tc2

RTc Pc

Pc  0.08664

 0.42748

Tr   83.1452  430.82 78.84

 6.9565  106 Tr  bar cm6 mol2

83.145  430.8  39.363 cm3 mol1 78.84

At 100°C = 373.15 K, we have Tr = 373.15/430.8 = 0.8662. At this reduced temperature, we have

   T   1  0.480  1.574 * 0.245  0.176 * 0.245 1  0.8662   1.1220  Tr   1  0.480  1.574   0.176 2 1  Tr0.5 

0.5

2

1

So, we have a = 7.806 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*373.15/28.74 = 1079.5 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  7805500 V  39.36    cm 3 mol 1 V  1079.5  39.36   28.74  V V  39.36   271591V  39.36   cm 3 mol 1 V  1118.9  V V  39.36   V

Solution continued on next page…

1

Starting from V = 1079.5 cm mol

and iterating gives V = 817.1 cm mol

1

for the vapor volume, corresponding to Z

= 0.7569. This is about 2% lower than the R/K equation, and about 0.6% below the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  a T     83.14  373.15  39.36  28.74  28.74V  V  39.36  V V  39.36   7805500   1118.9  V  V  39.36  V V  39.36   271591  3

Starting from V = 50 cm /mol and iterating on this gives V = 66.9 cm /mol. This is a little closer to the value from the Rackett equation than what we got using the R/K equation of state, but still about 10% too high. (t) At 110°C = 383.15 K, we have Tr = 383.15/430.8 = 0.8894. At this reduced temperature, we have

   T   1  0.480  1.574 * 0.245  0.176 * 0.245 1  0.8894   1.0997  Tr   1  0.480  1.574   0.176 2 1  Tr0.5 

0.5

Using this, a = 7.650 × 10 bar cm mol

2

1

The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*383.15/35.01 = 909.9 cm mol . To find the vapor phase volume, we iterate on a T  V  b RT b P P V V  b  7650000 V  39.36    cm 3 mol 1 V  909.9  39.36  35.01  V V  39.36    218515 V  39.36   cm 3 mol 1 V  949.3   V V  39.36  V

Starting from V = 909 cm /mol and iterating gives V = 652.5 cm mol

1

for the vapor volume, corresponding to Z =

0.7171. This is about 2.8% lower than the Lee/Kesler correlation (a little worse than the R/K EOS).

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V V  b  a T     83.14  383.15  39.36  35.01  35.01V  V  39.36  V V  39.36   7650000   949.3  V  V  39.36  V V  39.36   218515  3

Starting from V = 50 cm /mol and iterating on this gives V = 70.5 cm /mol. This is about 16% higher than the value from the Rackett Equation.

Solution 3.55

Problem Statement

Use the Peng/Robinson equation to calculate the molar volumes of saturated liquid and saturated vapor for the substance and conditions given by one of the parts of Prob. 3.53 and compare results with values found by suitable generalized correlations. Problem 3.53 Calculate the molar volume of saturated liquid and the molar volume of saturated vapor by the Redlich/Kwong equation for one of the following and compare results with values found by suitable generalized correlations.

(a) Propane at 40°C where P (b) Propane at 50°C where P

sat

= 13.71 bar = 17.16 bar

sat

(d) Propane at 70°C where P

= 21.22 bar

sat

(e) n-Butane at 100°C where P

= 25.94 bar

sat

(f ) n-Butane at 110°C where P (g) n-Butane at 120°C where P

sat

(h) n-Butane at 130°C where P (i) Isobutane at 90°C where P

sat

(j) Isobutane at 100°C where P

(m) Chlorine at 60°C where P (n) Chlorine at 70°C where P (o) Chlorine at 80°C where P

= 22.38 bar = 26.59 bar

= 16.54 bar

sat

(p) Chlorine at 90°C where P

= 18.66 bar

sat

(k) Isobutane at 110°C where P (l) Isobutane at 120°C where P

= 15.41 bar

sat

= 20.03 bar = 24.01 bar = 28.53 bar

= 18.21 bar

= 22.49 bar = 27.43 bar = 33.08 bar

(q) Sulfur dioxide at 80°C where P (r) Sulfur dioxide at 90°C where P

sat

(s) Sulfur dioxide at 100°C where P (t) Sulfur dioxide at 110°C where P

= 18.66 bar = 23.31 bar

sat

= 28.74 bar = 35.01 bar

(u) Boron trichloride at 400 K where P

sat

= 17.19 bar

For BCl3, Tc = 452 K, Pc = 38.7 bar, and (v) Boron trichloride at 420 K where P

sat

(w) Boron trichloride at 440 K where P (x) Trimethylgallium at 430 K where P

= 0.086. = 23.97 bar

sat

= 32.64 bar = 13.09 bar

For Ga(CH3)3, Tc = 510 K, Pc = 40.4 bar, and (y) Trimethylgallium at 450 K where P (z) Trimethylgallium at 470 K where P

sat

= 0.205.

= 18.27 bar = 24.55 bar

Solution

(d) For propane, the critical parameters are Tc = 369.8 K, Pc = 42.48 bar, and  = 0.152.

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr  83.1452 369.82

42.48 a(T )   Tr 1.0176 10 7 bar cm6 mol 2 b  0.07779

RTc 83.145 369.8  0.07779  56.3 cm3 mol1 Pc 42.48

At 70°C = 343.15 K, we have Tr = 343.15/369.8 = 0.9279. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.152  0.2699 * 0.152 1  0.9279   1.0448  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.0632 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*343.15/25.94 = 1099.9 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   10632000 V  56.3   cm3 mol 1 V  1099.9  56.3   25.94 V  23.32V  135.92   409851V  56.3    cm 3 mol1 V  1156.2   V  23.32V  135.92 V

1

Starting from V = 1099.9 cm mol

and iterating gives V = 678 cm mol

1

for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b  aT     83.14  343.15  56.3  25.94  25.94V  V  56.3  V  23.32V  135.92   10632000   1156.2  V  V  56.3  V  23.32V  135.92   409851  3

Starting from V = 56.3 cm mol

1

and iterating on this gives V = 113.6 cm /mol.

(e) Now, we switch from propane to n-butane, for which the critical properties are Tc = 425.1 K, Pc = 37.96 bar, and  = 0.200. Otherwise, it is exactly like part (d) above. The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr  83.1452  425.12

37.96 a(T )   Tr 1.5048 107 bar cm 6 mol 2 b  0.07779

RTc 83.145  425.1  0.07779  72.43 cm3 mol1 Pc 37.96

At 100°C = 373.15 K, we have Tr = 373.15/425.1 = 0.8778. Solution continued on next page…

At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.8778   1.0866  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.6352 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*373.15/15.41 = 2013.3 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   16352000 V  72.43   cm3 mol1 V  2013.3  72.43   15.41V  30.00V  174.86  1061129V  72.43    cm 3 mol 1 V  2085.7   V  30.00V  174.86 V

1

Starting from V = 2013.3 cm mol

and iterating gives V = 1453 cm mol

1

for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b   aT    83.14  373.15  72.43 15.41  15.41V  V  72.43  V  30.00 V  174.96   16352000   2085.7  V  V  72.43  V  30.00 V  174.96   1061129  3

1

Starting from V = 72.43 cm mol

and iterating on this gives V = 125.3 cm /mol.

(f) The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

Solution continued on next page…

with  Tr  R Tc 2

a(T )  0.45724

 Tr  83.145  425.1 2

a(T )  0.45724

37.96 7 6 2 a(T )   Tr  1.5048  10 bar cm mol b  0.07779

RTc Pc

 0.07779

83.145  425.1 37.96

 72.43 cm mol 3

1

At 110°C = 383.15 K, we have Tr = 383.15/425.1 = 0.9013. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9013   1.0692  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.6090 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*383.15/18.66 = 1707.2 cm mol . To find the vapor phase volume, we iterate on a T  RT V b  b P P V  0.4142bV  2.4142b   16090000 V  72.43   cm3 mol 1 V  1707.2  72.43   18.66V  30.00V  174.86  862226 V  72.43    cm 3 mol1 V  1779.6   V  30.00 V  174.86    V

1

Starting from V = 1707.2 cm mol

and iterating gives V = 1156 cm mol

1

for the vapor volume.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b  aT     83.14  383.15  72.43 18.66  18.66V  V  72.43  V  30.00 V  174.96   16090000   1779.6  V  V  72.43  V  30.00 V  174.96   862226 

Starting from V = 72.43 cm mol

1

and iterating on this gives V = 133.0 cm /mol.

(g) The Peng-Robinson EOS is P

aT  RT  V  b V  2.4142bV  0.4142b

with a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr  83.1452  425.12

37.96 a(T )   Tr 1.5048 107 bar cm 6 mol 2 b  0.07779

RTc 83.145  425.1  0.07779  72.43 cm3 mol1 Pc 37.96

At 120°C = 393.15 K, we have Tr = 393.15/425.1 = 0.9248. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9248   1.0522  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.5833 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*393.15/22.38 = 1460.6 cm mol . To find the vapor phase volume, we iterate on a T  RT V b  b P P V  0.4142bV  2.4142b   15833000 V  72.43   cm 3 mol1 V  1460.6  72.43   22.38V  30.00V  174.86  707471V  72.43    cm3 mol1 V  1533.0   V  30.00V  174.86  V

1

Starting from V = 1460.6 cm mol

and iterating gives V = 915.0 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 0.6264. This is about 1% lower than the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP   V  b  V  0.4142b V  2.4142b  aT     83.14  393.15  72.43  22.38  22.38V  V  72.43  V  30.00V  174.96   15833000   1533.0  V  V  72.43  V  30.00V  174.96   707041  3

Starting from V = 72.43 cm mol

1

and iterating on this gives V = 143.3 cm /mol. This is only about 4% higher than

the Rackett equation (much better than the R/K or S/R/K EOS). (h)

At 130°C = 403.15 K, we have Tr = 403.15/425.1 = 0.9484.

At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9484   1.0355  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.5582 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*403.15/26.59 = 1260.6 cm mol . To find the vapor phase volume, we iterate on

a T  RT V b  b P P V  0.4142bV  2.4142b   15582000 V  72.43   cm 3 mol1 V  1260.6  72.43  26.59V  30.00V  174.86   586008V  72.43   3 1 V  1333.0   cm mol  V  30.00V  174.86  V

Solution continued on next page…

1

Starting from V = 1260.6 cm mol

and iterating gives V = 715.7 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 0.5677. This is almost 6% lower than the Lee/Kesler correlation. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142b V  2.4142b   aT    83.14  393.15  72.43  26.59  26.59V  V  72.43  V  30.00V  174.96   15582000   1333.0  V  V  72.43  V  30.00V  174.96   586008  3

Starting from V = 72.43 cm mol

1

and iterating on this gives V = 157.2 cm /mol. This is only about 7% higher than

the Rackett equation (much better than the R/K or S/R/K EOS).

(i) The critical properties for isobutane are Tc = 408.1 K, Pc = 36.48 bar, and

= 0.302.

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr  83.1452  408.12 36.48

a(T )  Tr 1.443110 7 bar cm 6 mol2 b  0.07779

RTc Pc

 0.07779

83.145  408.1  72.36 cm3 mol1 36.48

Solution continued on next page…

(i) At 90°C = 363.15 K, we have Tr = 363.15/408.1 = 0.8899. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.302  0.2699 * 0.302 1  0.8899   1.0946  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.5796 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 1

Vid = RT/P = 83.145*363.15/16.54 = 1825.5 cm mol . To find the vapor phase volume, we iterate on a T  RT V b  b P P V  0.4142bV  2.4142b   15796000 V  72.36   cm3 mol1 V  1825.5  72.36   16.54 V  29.97V  174.69   955018V  72.36    cm3 mol 1 V  1897.9   V  29.97 V  174.69    V

Starting from V = 1825.5 cm /mol and iterating gives V = 1252 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 1251.5/1825.5 = 0.6856. To find the liquid volume, we iterate on

 RT  bP  VP    aT  

V  b  V  0.4142b V  2.4142 b

 83.14  363.15  72.36  16.54  16.54V     15796000  1897.9  V  V  72.36  V  29.97 V  174.69  955018  V  72.36  V  29.97 V  174.69

Starting from V = 72.36 cm /mol and iterating on this gives V = 125.8 cm /mol, corresponding to Z = 125.8/1825.5 = 0.0689. The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the Redlich-Kwong EOS, and are closer to the generalized correlations. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

At 100ºC = 373.15 K and 20.03 bar, we have Tr = 0.9144 and:

   T   1  0.37464  1.54226 * 0.302  0.2699 * 0.302 1  0.9144   1.0727  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.5480 × 10 bar cm mol . b is the same as in the previous part, b = 72.36. The ideal gas molar 3

1

volume at this temperature and pressure is Vid = RT/P = 83.145*373.15/20.03 = 1549.0 cm mol . To find the vapor phase volume, we iterate on a T  RT V b  b P P V  0.4142bV  2.4142b   15480000 V  72.36   cm3 mol 1 V  1549.0  72.36   20.03V  29.97V  174.69  772841V  72.36    cm3 mol 1 V  1621.3   V  29.97V  174.69 V

Starting from V = 1549 cm /mol and iterating gives V = 984 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 983.9/1549.0 = 0.6352.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142b V  2.4142b  aT     83.14  373.15  72.36  20.03  20.03V  V  72.36  V  29.97V  174.69   15480000   1621.3  V  V  72.36  V  29.97V  174.69   772841 

Solution continued on next page…

Starting from V = 72.36 cm /mol and iterating on this gives V = 134.7 cm /mol, corresponding to Z = 134.7/1549.0 = 0.0869. The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the Redlich-Kwong EOS.

(k) The critical properties for isobutane are Tc = 408.1 K, Pc = 36.48 bar, and

= 0.181.

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr  83.145 2  408.12

36.48 a(T )   Tr  1.4431 10 7 bar cm 6 mol2 b  0.07779

RTc 83.145  408.1  0.07779  72.36 cm 3 mol1 Pc 36.48

At 110°C = 383.15 K, we have Tr = 383.15/408.1 = 0.9389. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.181  0.2699 * 0.181 1  0.9389   1.0405  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

Solution continued on next page…

2

So, we have a = 1.5015 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is 3

1

Vid = RT/P = 83.145*383.15/24.01 = 1326.8 cm mol . To find the vapor phase volume, we iterate on a T  RT V b  b P P V  0.4142bV  2.4142b   15015000 V  72.36   cm3 mol 1 V  1326.8  72.36   24.01V  29.97V  174.69  625364 V  72.36    cm 3 mol1 V  1399.2   V  29.97V  174.69 V

Starting from V = 1326.8 cm /mol and iterating gives V = 782.9 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 782.9/1326.8 = 0.5901. This is only about 3% below the prediction from the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142b V  2.4142b  aT     83.14  383.15  72.36  24.01  24.01V  V  72.36  V  29.97V  174.69   15015000   1399.2  V  V  72.36  V  29.97V  174.69   625364  3

Starting from V = 72.36 cm /mol and iterating on this gives V = 151.2 cm /mol, corresponding to Z = 151.2/1326.8 = 0.1140. This is about 2% higher than the volume predicted by the Rackett equation.

The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the RedlichKwong EOS, and are closer to the generalized correlations.

Solution continued on next page…

(l) At 120ºC = 393.15 K and 28.53 bar, we have Tr = 0.9634 and:

   T   1  0.37464  1.54226 * 0.181  0.2699 * 0.181 1  0.9634   1.0240  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

2

So, we have a = 1.4777 × 10 bar cm mol . b is the same as in the previous part, b = 72.36. The ideal gas molar 3

1

volume at this temperature and pressure is Vid = RT/P = 83.145*393.15/28.53 = 1145.8 cm mol . To find the vapor phase volume, we iterate on a T  RT V b  b P P V  0.4142bV  2.4142b   14777000 V  72.36   cm 3 mol1 V  1145.8  72.36   28.53V  29.97 V  174.69  517946V  72.36    cm 3 mol1 V  1218.1   V  29.97V  174.69 V

Starting from V = 1145.8 cm /mol and iterating gives V = 597.3 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 597.3/1145.8 = 0.5213.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142b V  2.4142b   aT    83.14  393.15  72.36  28.53  28.53V  V  72.36  V  29.97V  174.69   14777000   1218.1  V  V  72.36  V  29.97V  174.69   517946 

Solution continued on next page…

Starting from V = 72.36 cm /mol and iterating on this gives V = 170.3 cm /mol, corresponding to Z = 170.3/1145.8 = 0.1486. This is about 6% higher than the value from the Rackett equation.

Again, the predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the Redlich-Kwong EOS and closer to the predictions of the generalized correlation.

(m) The critical properties for chlorine are Tc = 417.2 K, Pc = 77.10 bar, and

= 0.069.

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr  83.145 2  417.2 2

77.10 a(T )   Tr  7.136 10 6 bar cm 6 mol 2 b  0.07779

RTc 83.145  417.2  0.07779  35.00 cm3 mol 1 Pc 77.10

At 60°C = 333.15 K, we have Tr = 333.15/417.2 = 0.7985.

At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.7985   1.1047  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

Solution continued on next page…

2

1

So, we have a = 7.883 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*333.15/18.21 = 1521.1 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   7883000 V  35.0   cm3 mol1 V  1521.1  35.00   18.21V  14.50V  84.49   432903V  35.00    cm3 mol 1 V  1556.1   V  14.50V  84.49 V

Starting from V = 1521.1 cm /mol and iterating gives V = 1233.1 cm mol

1

for the vapor volume, corresponding to a

compressibility of Z = 1233.1/1521.1 = 0.8106. This is within 1% of the prediction from the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b  a T     83.14  333.15  35.00  18.21  18.21V  V  35.00  V  14.50V  84.49   7883000   1556.1  V  V  35.00  V  14.50V  84.49   432903  3

Starting from V = 35 cm /mol and iterating on this gives V = 54.0 cm /mol, corresponding to Z = 54.0/1521.1 = 0.035. This is about 1% lower than the volume predicted by the Rackett equation.

The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the RedlichKwong EOS, and are closer to the generalized correlations.

Solution continued on next page…

(n) At 70ºC = 343.15 K and 22.49 bar, we have Tr = 0.8225 and:

   T   1  0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8225   1.091  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

–2

So, we have a = 7.788 × 10 bar cm mol . b is the same as in the previous part, b = 35.00. The ideal gas molar volume 3

–1

at this temperature and pressure is Vid = RT/P = 83.145*343.15/22.49 = 1268.6 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   7788000V  35.0   cm3 mol 1 V  1268.6  35.00   22.49V  14.50V  84.49   346265V  35.00    cm 3 mol 1 V  1303.6   V  14.50 V  84.49    V

–1

Starting from V = 1268.6 cm /mol and iterating gives V = 987.4 cm mol for the vapor volume, corresponding to a compressibility of Z = 1022.9/1268.6 = 0.8063.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b  aT     83.14  343.15  35.00  22.49  22.49V  V  35.00  V  14.50V  84.49   7788000   1303.6  V  V  35.00  V  14.50V  84.49   317293  3

Starting from V = 35 cm /mol and iterating on this gives V = 56.0 cm /mol, corresponding to Z = 56.0/1268.6 = 0.044. This is about 2% higher than the value from the Rackett equation. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(o) The critical properties for chlorine are Tc = 417.2 K, Pc = 77.10 bar, and

= 0.069.

The Peng-Robinson EOS is

P

a T  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R2Tc2 Pc  Tr   83.145 2  417.22

77.10 a(T )   Tr   7.136  10 6 bar cm 6 mol 2 RT 83.145  417.2 b  0.07779 c  0.07779  35.00 cm 3 mol 1 Pc 77.10

At 80°C = 353.15 K, we have Tr = 353.15/417.2 = 0.8465. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8465   1.078  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

–2

–1

So, we have a = 7.694 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*353.15/27.43 = 1070.5 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   7694000V  35.0  3 1 V  1070.5  35.00   cm mol  27.43V  14.50V  84.49   280490 V  35.00    cm 3 mol 1 V  1105.5   V  14.50V  84.49 V

–1

Starting from V = 1070 cm /mol and iterating gives V = 794.9 cm mol for the vapor volume, corresponding to a compressibility of Z = 794.9/1070.5 = 0.7426. This is almost 4% lower than the prediction from the Lee/Kesler correlation. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b  aT     83.14  353.15  35.00  27.43  27.43V  V  35.00  V  14.50V  84.49   7694000   1105.5  V  V  35.00  V  14.50V  84.49   280490  3

Starting from V = 50 cm /mol and iterating on this gives V = 58.5 cm /mol. This is about 2.5% higher than the volume predicted by the Rackett equation.

The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the RedlichKwong EOS, and are closer to the generalized correlations.

(p) At 80ºC = 363.15 K and 33.08 bar, we have Tr = 0.8705 and:

   T   1  0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8705   1.065  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

Solution continued on next page…

–2

So, we have a = 7.602 × 10 bar cm mol . b is the same as in the previous part, b = 35.00. The ideal gas molar volume 3

–1

at this temperature and pressure is Vid = RT/P = 83.145*363.15/33.08 = 912.8 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b    7602000 V  35.0   cm 3 mol 1 V  912.8  35.00   33.08V  14.50V  84.49  229807 V  35.00    cm3 mol1 V  947.8   V  14.50V  84.49 V

–1

Starting from V = 913 cm /mol and iterating gives V = 641.6 cm mol for the vapor volume, corresponding to a compressibility of Z = 641.6/912.8 = 0.7029. To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b   aT    83.14  363.15  35.00  33.08  33.08V  V  35.00  V  14.50V  84.49   7602000   047.8  V  V  35.00  V  14.50V  84.49   229807  3

Starting from V = 35 cm /mol and iterating on this gives V = 61.4 cm /mol. This is about 4% higher than the value from the Rackett equation.

(q) The critical properties for SO2 are Tc = 430.8 K, Pc = 78.84 bar, and

= 0.245.

The Peng-Robinson EOS is P

a T  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc  Tr   83.145 2  430.82

78.84 a(T )   Tr   7.441  10 6 bar cm 6 mol2 RT 83.145  430.8 b  0.07779 c  0.07779  35.34 cm 3 mol 1 Pc 78.84

At 80°C = 353.15 K, we have Tr = 353.15/430.8 = 0.8198.

At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8198   1.1441  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

–2

–1

So, we have a = 8.513 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*353.15/18.66 = 1573.6 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   8513000 V  35.34   cm3 mol 1 V  1573.6  35.34   18.66 V  14.64V  85.32   456217 V  35.34    cm 3 mol 1 V  1608.9   V  14.64V  85.32 V

–1

Starting from V = 1500 cm /mol and iterating gives V = 1280.3 cm mol for the vapor volume, corresponding to a compressibility of Z = 1280.3/1573.6 = 0.8136. This within 1% of the prediction from the Lee/Kesler correlation.

Solution continued on next page…

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b   aT    83.14  353.15  35.34  18.66  18.66V  V  35.34  V  14.64V  85.32   8513000   1608.9  V  V  35.34  V  14.64V  85.32   456217  3

Starting from V = 50 cm /mol and iterating on this gives V = 54.1 cm /mol. This within 1% of the volume predicted by the Rackett equation, which is much better than the Redlich/Kwong equation.

The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the RedlichKwong EOS, and are closer to the generalized correlations.

(p) At 80ºC = 363.15 K and 23.31 bar, we have Tr = 0.8430 and:

   T   1  0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8430   1.1242  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

–2

So, we have a = 8.365 × 10 bar cm mol . b is the same as in the previous part, b = 35.34. The ideal gas molar volume 3

–1

at this temperature and pressure is Vid = RT/P = 83.145*363.15/23.11 = 1306.5 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   8365000 V  35.34   cm3 mol 1 V  1306.5  35.34   23.11V  14.64V  85.32   361969V  35.34   3 1 V  1341.9   cm mol   V  14.64 V  85.32    V

–1

Starting from V = 1300 cm /mol and iterating gives V = 1021.6 cm mol for the vapor volume, corresponding to a compressibility of Z = 1021.6/1306.5 = 0.7820. This is about 2% below the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b   aT    83.14  363.15  35.34  23.11  23.11V  V  35.34  V  14.64V  85.32   8365000   1341.9  V  V  35.34  V  14.64V  85.32   361969  3

Starting from V = 50 cm /mol and iterating on this gives V = 56.3 cm /mol. This is identical to the value from the Rackett equation (to 3 digits).

(s) The critical properties for SO2 are Tc = 430.8 K, Pc = 78.84 bar, and

= 0.245.

The Peng-Robinson EOS is

P

aT  RT  V  b V  2.4142bV  0.4142b

with

a(T )  0.45724 a(T )  0.45724

 Tr  R 2Tc2 Pc  Tr   83.145 2  430.82

78.84 a(T )   Tr   7.441  10 6 bar cm 6 mol2 RT 83.145  430.8 b  0.07779 c  0.07779  35.34 cm 3 mol 1 Pc 78.84

At 100°C = 373.15 K, we have Tr = 373.15/430.8 = 0.8662. At this reduced temperature, we have

   T   1  0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8662   1.1047  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

–2

–1

So, we have a = 8.220 × 10 bar cm mol . The ideal gas molar volume at this temperature and pressure is Vid = RT/P = 83.145*373.15/28.74 = 1079.5 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b   8220000 V  35.34  3 1 V  1079.5  35.34   cm mol  28.74 V  14.64V  85.32   285997 V  35.34    cm 3 mol 1 V  1114.9   V  14.64V  85.32 V

–1

Starting from V = 1079 cm /mol and iterating gives V =800.6 cm mol for the vapor volume, corresponding to a compressibility of Z = 800.6/1079.5 = 0.7416. This is within 3% of the prediction from the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b   aT    83.14  373.15  35.34  28.74  28.74V  V  35.34  V  14.64V  85.32   8220000   1114.9  V  V  35.34  V  14.64V  85.32   285997  3

Starting from V = 50 cm /mol and iterating on this gives V = 58.9 cm /mol. This within 1% of the volume predicted by the Rackett equation, which is much better than the Redlich/Kwong equation. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The predicted molar volumes for both the liquid and vapor phases are smaller than those predicted by the RedlichKwong EOS.

(t) At 110ºC = 383.15 K and 35.01 bar, we have Tr = 0.8894 and:

   T   1  0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8894   1.0856  Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5 

0.5

–2

So, we have a = 8.078 × 10 bar cm mol . b is the same as in the previous part, b = 35.34 cm /mol. The ideal gas 3

–1

molar volume at this temperature and pressure is Vid = RT/P = 83.145*383.15/35.01 = 909.9 cm mol . To find the vapor phase volume, we iterate on aT  RT V b b P P V  0.4142bV  2.4142b    8078000 V  35.34   cm 3 mol 1 V  909.9  35.34   35.01V  14.64V  85.32  230772V  35.34    cm 3 mol1 V  945.2   V  14.64V  85.32 V

–1

Starting from V = 909 cm /mol and iterating gives V = 635.9 cm mol for the vapor volume, corresponding to a compressibility of Z = 635.9/909.9 = 0.6988. This is about 5% below the Lee/Kesler correlation.

To find the liquid volume, we iterate on  RT  bP  VP    V  b  V  0.4142bV  2.4142b  aT     83.14  383.15  35.34  35.01  35.01V  V  35.34  V  14.64V  85.32   8078000   945.2  V  V  35.34  V  14.64V  85.32   230772  3

Starting from V = 50 cm /mol and iterating on this gives V = 62.1 cm /mol. This is 2.5% higher than the value from the Rackett equation, but much closer than the R/K equation.

Solution 3.56

Problem Statement

Estimate the following: (a) The volume occupied by 18 kg of ethylene at 55°C and 35 bar. 3

(b) The mass of ethylene contained in a 0.25 m cylinder at 50°C and 115 bar.

Solution

You could use any of the equations of state or correlations that we've been using in the other problems to do this. For both parts (a) and (b) we want to find the specific volume given the temperature and pressure. For ethylene, Tc = 282.3 K, Pc = 50.40 bar, and  = 0.087 (from table B.1)

(a) At 55°C and 35 bar, we have Tr = 1.162 and Pr = 0.6944. Glancing at figure 3.14, which shows the Lee/Kesler 0

correlation for Z , we see that we are in the gas-like region of the supercritical fluid, and the compressibility should be in the ballpark of 0.8. This is at the edge of, or just outside, the region shown in figure 3.15 where the Pitzer virial coefficient correlation works well. So, we should either use a cubic equation of state or use the full Lee/Kesler correlation. Since we haven't tried the Lee/Kesler correlation yet, let's do that. From table E.1, on page 668, we see 0

that at Tr = 1.15 and Pr = 0.6, we have Z = 0.8576. At Tr = 1.15 and Pr = 0.8, we have Z = 0.8032. At Tr = 1.2 and 0

Pr = 0.6, we have Z = 0.8779. So, interpolating linearly between these to Tr = 1.162 and Pr = 0.6944 we get 0

Z = 0.8576 + 0.0944/0.2*(0.8032–0.8576) + 0.012/0.05*(0.8779–0.8576) = 0.8368

Solution continued on next page…

Similarly, from table E.2 on p. 669, we get 1

Z = 0.0237 + 0.0944/0.2*(0.0396 – 0.0237) + 0.012/0.05*(0.0326 – 0.0237) = 0.0333 0

So, Z = Z + Z = 0.8368 + 0.087*0.0333 = 0.8397 3

–1

So, V = 0.8397*RT/P = 0.8397*83.145*328/35 = 654 cm mol . –1

Returning to the question at hand, 18 kg of ethylene = 18000 g/(28.05 g mol ) = 642 mol of ethylene = 642 mol * 654 3

–1

cm mol = 419800 cm = 419.8 liters = 0.4198 m .

(b) At 50°C and 115 bar, we have Tr = 1.144 and Pr = 2.282. This is above the critical temperature, but looking at figure 3.14 on p. 101, we see that it is at conditions where the supercritical fluid is liquid-like (has a liquid-like compressibility near 0.4). We will again use the Lee/Kesler correlation. From table E.3, on page 670, we see that at 0

Tr = 1.15 and Pr = 2.0, we have Z = 0.4760. At Tr = 1.15 and Pr = 3.0, we have Z = 0.5042. At Tr = 1.1 and Pr = 2.0, 0

we have Z = 0.3953. So, interpolating linearly between these to Tr = 1.144 and Pr = 2.282 we get 0

Z = 0.4760 + 0.282/1.0*(0.5042 – 0.4760) + 0.006/0.05*(0.3953 – 0.4760) = 0.4743

Similarly, from table E.2 on p. 669, we get 1

Z = 0.1667 + 0.282/1.0*(0.0332 – 0.1667) + 0.006/0.05*(0.0698 – 0.1667) = 0.1174 0

So, Z = Z + Z = 0.4743 + 0.087*0.1174 = 0.4845 3

–1

So, V = 0.4845*RT/P = 0.4845*83.145*323/115 = 113.1 cm mol .

–1

–3

The density is then 28.05 g mol /113.1 cm mol = 0.2479 g cm = 247.9 kg m .

So, a 0.25 m cylinder contains 0.25*247.9 = 62.0 kg of ethylene.

Number of moles n =

P1Vtotal ZRT1

= 2171 mol

Mass = n * molwt = 2171mol *28.02 g/mol = 60831.42 g = 60.898 kg

Solution 3.57

Problem Statement 3

–1

The vapor-phase molar volume of a particular compound is reported as 23,000 cm ·mol at 300 K and 1 bar. No other data are available. Without assuming ideal-gas behavior, determine a reasonable estimate of the molar volume of the vapor at 300 K and 5 bar.

Solution Assume validity of Eq. 3.38 Z1 

B

RT1 P1

P1V1 RT1



 Z1  1  1.942 * 103

cm3 mol

Solution continued on next page…

With this recalculate at P2 = 5 bar Z2  

V2 

RT1 Z2 P2

BP2 RT1



cm3 mol

Solution 3.58

Problem Statement

To a good approximation, what is the molar volume of ethanol vapor at 480°C and 6000 kPa? How does this result compare with the ideal-gas value?

Solution

The critical properties of ethanol are Tc = 513.9 K, Pc = 61.48 bar, and  = 0.645. So, at 480°C = 753.15 K and 6000 kPa = 60 bar, we have Tr = 1.4656 and Pr = 0.9759. From figure 3.15 on p. 103 of SVNA, we see that these are nd

well within the range where the Pitzer correlation for the 2 P

virial coefficient performs well. So, we have

aT  RT  V  b V V  b

Solution continued on next page…

and Z  1  0.1459  0.645  0.1045

0.9759  0.9477 1.4656

So, the molar volume will be about 95% of the ideal gas volume. The ideal gas volume is id

–1

V = 83.145*753.15/60 = 1044 cm mol . A good estimate of the actual volume is then 3

–1

V = 0.9477*1044 = 989 cm mol .

Solution 3.59

Problem Statement 3

A 0.35 m vessel is used to store liquid propane at its vapor pressure. Safety considerations dictate that at a temperature of 320 K the liquid must occupy no more than 80% of the total volume of the vessel. For these conditions, determine the mass of vapor and the mass of liquid in the vessel. At 320 K the vapor pressure of propane is 16.0 bar.

Solution

If the total volume of the vessel is 0.35 m , and 80% of this is filled with liquid, then the total liquid volume is 3

0.28 m = 280000 cm , and the total vapor volume is 0.07 m = 70000 cm . Dividing the total volume of each phase by its molar volume would give the number of moles of each phase. Multiplying those results by the molecular weight of propane will give the mass of each phase. The quickest and most reliable estimates of these molar volumes will come Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

from suitable generalized correlations. The critical properties of propane are Tc = 369.8 K , Pc = 42.48 bar, Zc = 0.276, 3

Vc = 200.0 cm /mol, and 0.2857

1Tr 

V sat  Vc Zc

= 0.152. For the molar volume of the saturated liquid, we can use the Rackett equation:

 200.0 * 0.276(1320/369.8)

0.2857

 96.8 cm3 /mol

So, there are 280000 cm / 96.8 cm /mol = 2892.5619 mol of liquid, with a molar mass of 0.044097 kg/mol. So, the total mass of liquid is about 2893*0.044097 = 127.5 kg. For the vapor phase, we can use the Lee/Kesler correlations. The reduced temperature is 320/369.8 = 0.8653 and the reduced pressure is 16.0/42.48 = 0.3766. Interpolating using the vapor-phase data from Table E.1 at Tr = 0.85, 0

Pr = 0.2; Tr = 0.9, Pr = 0.2; and Tr = 0.9, Pr = 0.4, gives Z = 0.9015 – 0.0142 – 0.1073 = 0.7800 and 1

Z = –0.0442 – 0.0189 – 0.0597 = –0.1228, and Z = 0.7800 + 0.152* – 0.1228 = 0.7613. The ideal gas molar volume is id

–1

V = 83.14 bar cm mol K * 320 K/16.0 bar = 1662.8 cm /mol. So, the actual molar volume is about 3

V = 0.7613*1662.8 = 1266 cm /mol. Thus, the total number of moles of vapor is 70000/1266 = 55.3 mol. The total mass of vapor is 55.3*0.044097 = 2.341 kg. The vapor occupies 20% of the tank volume, but is less than 2% of the total mass of propane in the tank.

Solution 3.60

Problem Statement 3

A 30 m tank contains 14 m of liquid n-butane in equilibrium with its vapor at 25°C. Estimate the mass of n-butane vapor in the tank. The vapor pressure of n-butane at the given temperature is 2.43 bar.

Solution

First determine the reduced pressure and temperature from the given data:

Tr 

T1 Tc



Pr 1 

P1 Pc



Next determine the B values and the Volume:

V

B0 



0.422  Tr1.6

B1 



0.172  Tr4.2

T  RT     B 0   B1  R c   P  Pc 

cm3 mol

Lastly, determine the mass by Vvap V  MWnbutane

mvap 

Solution 3.61

Problem Statement 3

Estimate: (a) The mass of ethane contained in a 0.15 m vessel at 60°C and 14,000 kPa. (b) The temperature at which 3

40 kg of ethane stored in a 0.15 m vessel exerts a pres-sure 20,000 kPa.

Solution continued on next page…

Solution You could use any of the equations of state or correlations that we've been using in the other problems to do this. For both parts (a) and (b) we want to find the specific volume given the temperature and pressure. For ethane, Tc = 305.3 K, Pc = 48.72 bar, and  = 0.100 (from table B.1)

(a) At 60°C (333.15 K) and 14000 kPa (140 bar), we have Tr = 1.091 and Pr = 2.874. Glancing at figure 3.13, which 0

shows the Lee/Kesler correlation for Z , we see that these conditions correspond to a fairly dense supercritical fluid, and the compressibility should be in the ballpark of 0.4. This is far outside the region shown in figure 3.14 where the Pitzer virial coefficient correlation works well. So, we should either use a cubic equation of state or use the full Lee/Kesler correlation, and in general we expect the Lee/Kesler correlation to work better. In problem 3.38, we put together a little spreadsheet to do the interpolation, so let’s use that again. Putting in the numbers from tables E.3 and E.4 gives the following:

T (K) 333.15

P (bar) 140

Tc (K) 305.3

Pc (bar) 48.72



Tr 1.05 1.05 1.10 1.10

Pr 2.00 3.00 2.00 3.00

Z0 0.3452 0.4604 0.3953 0.4770

0.1

Tr 1.0912

Pr 2.8736

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

0.4630

Z1 -0.0432 -0.0838 0.0698 -0.0373 -0.0334

0.4597 91.0 3

Returning to the question at hand, a vessel of volume 0.15 m = 150000 cm will contain 150000/91.0 = 1649 mol of ethane at these conditions. Muliplying by the molecular weight of ethane (30.070 g/mol = 0.030070 kg/mol) gives 3

49.6 kg. The mass density is about 331 kg/m (about 1/3 the density of liquid water and almost 300 times greater than the density of air at standard conditions). Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(b) Neither our cubic equations of state nor our correlations can be explicitly solved for temperature. Thus, we will have to try different temperatures until we get one that gives us the desired molar volume (or compressibility) at the given pressure. 40 kg of ethane corresponds to 40/0.030070 = 1330 mol ethane. If this quantity is in a volume of 3

150000 cm , then the molar volume is 150000/1330 = 112.8 cm /mol. After a little trial and error using the Lee/Kesler interpolation spreadsheet, we come up with:

T (K) 388.8

P (bar) 200

Tc (K) 305.3

Pc (bar) 48.72



Tr 1.40 1.40 1.50 1.50

Pr 3.00 5.00 3.00 5.00

Z0 0.7202 0.7761 0.7887 0.8200

0.1

Tr 1.2735

Pr 4.1051

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

0.6816

Z1 0.2397 0.1737 0.2433 0.2309 0.1612

0.6978 112.8

This was a bit of a pain, because we have to have the appropriate points from tables E.3 and E.4, and we didn’t know in advance what the temperature would be.

Solution 3.62

Problem Statement

A size D compressed gas cylinder has an internal volume of 2.40 liters. Estimate the pressure in a size D cylinder if it contains 454 g of one of the following semiconductor process gases at 20°C:

(a) Phosphine, PH3, for which Tc = 324.8 K, Pc = 65.4 bar, and

= 0.045

(b) Boron trifluoride, BF3, for which Tc = 260.9 K, Pc = 49.9 bar, and

= 0.094

(d) Germane, GeH4, for which Tc = 312.2 K, Pc = 49.5 bar, and

(e) Arsine, AsH3, for which Tc = 373 K, Pc = 65.5 bar, and

= 0.434

= 0.151

= 0.011

(f) Nitrogen trifluoride, NF3, for which Tc = 234 K, Pc = 44.6 bar, and

= 0.120

Solution

The Redlich/Kwong equation will be used for each of these. (a) Calculate the moles of Phosphine using the formula, Moles of phosphine = mass/MW =

Molar volume =

454g = 13.353 mol 33.9975g / mol

total volume = 2.40 L /13.353mol = 0.1797 L /mol = 179.7 cm^3 / mol moles of gas

Also Tr = T/Tc = 0.9021 Calculate a(t) and b values for the RK equation. The Redlich / Kwong equation of state is

P

aT  RT  V  b V V  b

With 0.5

0.4278*Tr  *R 2 *Tc 2 a(t) = Pc 0.5

0.4278*0.9021 *83.1452 *324.82 a(t) =  65.4

b = 0.08664*

R*Tc  Pc

83.145*324.8  65.4

bar.cm^6. mol ^–2

cm 3 mol

substituting the value of a(t) and b in the RK equation.

P

aT  RT  V  b V V  b

83.145 * 293 – = 39.54 bar 179.7 * 35.776 179.7179.7  35.776

(b) Calculate the moles of BF3 using the formula,

Moles of BF3 = mass/MW =

Molar volume =

454g = 6.6941 mol 67.825g / mol

total volume = 2.40 L /6.6941 mol = 0.3585 L /mol = 358.5 cm^3 / mol moles of gas

Also Tr = T/Tc = 1.1230 Calculate a(t) and b values for the RK equation. The Redlich / Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

With 0.5

0.4278*Tr  *R 2 *Tc 2 a(t) = Pc 0.5

0.4278*1.1230 *83.1452 *260.92 a(t) =  3806699.54 bar.cm^6. mol ^–2 49.9

b = 0.08664*

R*Tc  Pc

83.145 *260.9  49.9

cm 3 mol

substituting the value of a(t) and b in the RK equation.

P

aT  RT  V  b V V  b

3806699.54 83.145 * 293 – = 49.128 bar 358.5  37.6641 358.5358.5  37.6641

Moles of SiH4 = mass/MW =

Molar volume =

454g

= 14.1358 mol

total volume = 2.40 L /14.1358 mol = 0.1698 L /mol = 169.78 cm^3 / mol moles of gas

Also Tr = T/Tc = 1.0864 Calculate a(t) and b values for the RK equation. The Redlich / Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

With 0.5

0.4278*Tr  *R 2 *Tc 2 a(t) = Pc 0.5

a(t) =

0.4278*1.0864

2 2 *83.145 *269.7  48.4

b = 0.08664*

R*Tc  Pc

bar.cm^6. mol ^–2

83.145 *269.7  48.4

cm 3 mol

substituting the value of a(t) and b in the RK equation.

P

aT  RT  V  b V V  b

83.145 * 293 – = 68.2717 bar 169.78  40.14 169.78169.78  40.14 

(d) Calculate the moles of GeH4 using the formula,

Moles of GeH4 = mass/MW =

Molar volume =

454g

= 5.925 mol

total volume = 2.40 L /5.925 mol = 0.4050 L /mol = 405.06 cm^3 / mol moles of gas

Also Tr = T/Tc = 0.9385 Calculate a(t) and b values for the RK equation. The Redlich / Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

With 0.5

0.4278*Tr  *R 2 *Tc 2 a(t) = Pc 0.5

a(t) =

0.4278*0.9385

2 2 *83.145 *312.2  49.5

b = 0.08664*

R*Tc  Pc

bar.cm^6. mol ^–2

83.145 *312.2  49.5

cm 3 mol

substituting the value of a(t) and b in the RK equation.

P

aT  RT  V  b V V  b

6011105.335 83.145 * 293 – = 34.7996 bar 405.06  45.434 405.06 405.06  45.434 

(e) Calculate the moles of AsH3 using the formula,

Moles of AsH3 = mass/MW =

Molar volume =

454g

= 5.8246 mol

total volume = 2.40 L /5.8246 mol = 0.4120 L /mol = 412.044 cm^3 / mol moles of gas

Also Tr = T/Tc = 0.7855 Calculate a(t) and b values for the RK equation. The Redlich / Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

With 0.5

0.4278*Tr  *R 2 *Tc 2 a(t) = Pc 0.5

0.4278*0.7855 *83.1452 *3732 a(t) =  65.5

b = 0.08664*

R*Tc  Pc

83.145 *373  65.5

bar.cm^6. mol ^–2

cm 3 mol

substituting the value of a(t) and b in the RK equation.

P

aT  RT  V  b V V  b

7087841.665 83.145 * 293 – = 27.693 bar 412.044  41.022 412.044 412.044  41.022

(f) Calculate the moles of NF3 using the formula,

Moles of NF3 = mass/MW =

Molar volume =

454g

= 6.394 mol

total volume = 2.40 L /6.394 mol = 0.3753 L /mol = 373.35 cm^3 / mol moles of gas

Also Tr = T/Tc = 1.2521 Calculate a(t) and b values for the RK equation. The Redlich / Kwong equation of state is

P

aT  RT  V  b V V  b

Solution continued on next page…

With 0.5

0.4278*Tr  *R 2 *Tc 2 a(t) = Pc 0.5

0.4278*1.2521 *83.1452 *234 2 a(t) =  44.6

b = 0.08664*

R*Tc  Pc

83.145 *234  44.6

bar.cm^6. mol ^–2

cm 3 mol

substituting the value of a(t) and b in the RK equation.

P

aT  RT  V  b V V  b

3244902.631 83.145 * 293 – = 51.2455 bar 375.35  37.795 375.35375.35  37.795

Solution 3.63

Problem Statement

For one of the substances in Prob. 3.62, estimate the mass of the substance contained in the size D cylinder at 20°C and 25 bar.

Problem 3.62

A size D compressed gas cylinder has an internal volume of 2.40 liters. Estimate the pressure in a size D cylinder if it contains 454 g of one of the following semiconductor process gases at 20°C:

(a) Phosphine, PH3, for which Tc = 324.8 K, Pc = 65.4 bar, and

= 0.045

(b) Boron trifluoride, BF3, for which Tc = 260.9 K, Pc = 49.9 bar, and

= 0.094

(d) Germane, GeH4, for which Tc = 312.2 K, Pc = 49.5 bar, and

(e) Arsine, AsH3, for which Tc = 373 K, Pc = 65.5 bar, and

= 0.434

= 0.151

= 0.011

(f) Nitrogen trifluoride, NF3, for which Tc = 234 K, Pc = 44.6 bar, and

= 0.120

Solution

(a) For Phosphine, first determine Tr = T/Tc = 0.903K Pr = P/Pc = 0.3822 Solution continued on next page…



0.422 Tr1.6



0.172 Tr4.2



BPc RTc

 B1



 BP  P  c  r  RT  T  c r









0.9021

1

MWphosphine

V = 2.4 L = 2400 cm

Using this,

Phosphine

PV ZRT

1 3

1

Solution continued on next page…

(b) For Boron, first determine Tr = T/Tc = 293/260.9 = 1.1230 Pr = P/Pc = 25/49.9 = 0.5010



0.422 Tr1.6



0.172 Tr4.2



BPc RTc

 B1



 BP  P  c  r  RT  T  c r









1.1230

1

MWBF3

V = 2.4 L = 2400 cm

Using this,

BF3

PV ZRT

1 3

1

Solution continued on next page…

c. For Silane, first determine Tr = T/Tc = 1.0864 Pr = P/Pc = 25/48.4 = 0.51652



0.422 Tr1.6



0.172 Tr4.2



BPc RTc

 B1



 BP  P  c  r   RTc  Tr









1.0864

1

MWSiH4

V = 2.4 L = 2400 cm

Using this,

SiH4

PV ZRT

1 3

1

Solution continued on next page…

(d) For germane first determine Tr = T/Tc = 0.9385



0.422 Tr1.6



0.172 Tr4.2



BPc RTc

 B1



 BP  P  c  r  RT  T  c r









0.9385 1

MWGeH 4

V = 2.4 L = 2400 cm

Using this,

GeH4

PV ZRT

1 3

1

Solution continued on next page…

(e) For Arsine, first determine Tr = T/Tc = 0.7855 Pr = P/Pc = 25/65.5 = 0.38168



0.422 Tr1.6



0.172 Tr4.2



BPc RTc

 B1



 BP  P  c  r  RT  T  c r









0.7855 1

MWAsH3

V = 2.4 L = 2400 cm

Using this,

AsH3

PV ZRT

1 3

1

Solution continued on next page…

f) For Nitrogen Triflouride, first determine Tr = T/Tc = 1.2521 Pr = P/Pc = 25/44.6 = 0.5605



0.422 Tr1.6



0.172 Tr4.2



BPc RTc

 B1



 BP  P  c  r  RT  T  c r



 2028

1.2521

1

MWNF3

V = 2.4 L = 2400 cm

Using this,

NF3

PV ZRT

1 3

1

Solution 3.64

Problem Statement

Recreational scuba diving using air is limited to depths of 40 m. Technical divers use different gas mixes at different depths, allowing them to go much deeper. Assuming a lung volume of 6 liters, estimate the mass of air in the lungs of:

(a) A person at atmospheric conditions.

(b) A recreational diver breathing air at a depth of 40 m below the ocean surface.

(c) A near-world-record technical diver at a depth of 300 m below the ocean surface, breathing 10 mol% oxygen, 20 mol% nitrogen, 70 mol% helium.

Solution

(a) Standard atmospheric conditions will be T = 298.16 K, P = 1.01325 bar 3

Vlung = 6 L = 6000 cm . Using PV = nRT, n = 0.2452 mol of air or 7.103 g of air (b) At 40 m in depth, temperature should stay relatively the same at T = 298.16 K. Pressure will be at 5 times what it was for P = 5.066 bar, this makes n = 1.226 mol of air or 35.52 g of air (c) At 300 m in depth, temperature should reduce to T = 280.16 K. Pressure will be

P  Patmosphere   gh  1.01325 bar  1027 kg / m3 * 9.8 m / s 2 * 300 m P = 31.16 bar.

Solution continued on next page…

The biggest change here is the gas composition, with reductions in the amount of oxygen and nitrogen and the addition of helium. A partial pressure must be used here which is Pp02 = 0.1*31.16 bar = 3.116 bar for O2, PpN2 = 0.2*31.16bar = 6.232 bar for N2, and PpHe = 0.7*31.16 bar = 21.812 bar for He this makes n = 0.8026 mol of O2 or 25.68 g of O2, n = 1.605 mol of N2 or 44.94g N2, and , n = 5.618 mol of He or 22.472 g of He

Solution 3.65

Problem Statement 3

To what pressure does one fill a 0.15 m vessel at 25°C in order to store 40 kg of ethylene in it?

Solution 40 kg of ethylene, with a molecular weight of 28.054 g/mol is 40*1000 g/28.054 g/mol = 1425.8 moles of 3

ethylene. If we want to put this into a 0.15 m vessel, then we need to have a molar volume or 5

1.5 × 10 cm /1425.8 moles = 105.2 cm /mol. This is a liquid-like molar volume. The critical temperature of ethylene is 282.3 K, so 25 °C = 298.15 K is above the critical point (Tr = 1.0561). Since the critical volume (from table B.1) is 3

131 cm /mol, and the needed volume is smaller, we must be well above the critical pressure of 50.4 bar. We can start by estimating the pressure using a cubic equation of state, and then refine the estimate using the Lee/Kesler correlation (in the tabular form in the back of the book). Since the Lee/Kesler correlation is in terms of reduced temperature and reduced pressure, we can’t use it directly (we don’t know the pressure). So, we can start with the pressure given by the Redlich/Kwong equation of state, and then iterate on pressure using the tables to get a better answer. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b 

with T 0.5 R2Tc2 1.05610.5  83.1452  282.32 a(T )  0.42748 r  0.42748  4547019 bar cm 6 mol 2 Pc 50.4 RTc 83.145  282.3 b  0.08664  0.08664  40.35 cm3 mol1 Pc 50.4

So, it predicts

P

83.145 * 298.15 4547019   85.3 bar 105.2  40.35 105.2105.2  40.35

If that is the pressure, then the reduced pressure would be 85.3/50.4 = 1.692, and the compressibility would be Z = PV/RT = 85.3*105.2/(83.145*298.15) = 0.3620. Now, we can see what compressibility is given by the Lee/Kesler correlation for Tr = 1.0561 and Pr = 1.692.

If we expect to have to interpolate in the Lee/Kesler tables (or steam tables or whatever) repeatedly, it is worthwhile to set up a small spreadsheet to do the interpolation.

Solution continued on next page…

T (K)

P (bar)

Tc (K)

Pc (bar)

298.15

85.3

282.3

50.4

0.0870

1.0561

1.6925

Table Points

Tr (1)

Pr(1)

1.05

1.50

0.3131

0.0451

Tr(1)

Pr(2)

1.05

2.00

0.3452

–0.0432

Tr(2)

Pr(1)

1.10

1.50

0.4580

0.1630

Tr(2)

Pr(2)

1.10

2.00

0.3953

0.0698

Interpolated Values 0.3388

0.0254

Final Value Z

0.3410

In this sheet, you enter T, P, and the critical properties, and the sheet then computes Tr and Pr. You then enter the table values of the four Tr, Pr pairs that surround the computed Tr and Pr that were computed (you read these from the table and then type them in). The spreadsheet then applies the 2-D interpolation formula to get the interpolated values and adds them up to get the final value of Z. So, we see that for P = 85.3 bar, we get Z = 0.3410, a little smaller than we needed. We can use this Z to compute a new P as P = ZRT/P = 0.3410*83.145*298.15/105.2 = 80.4 bar. Using Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

P = 80.4 bar in the above spreadsheet gives Z = 0.3385 (notice that we didn’t have to re-enter any of the table points, since the reduced pressure was still between 1.50 and 2.0, and the reduced temperature didn’t change. Using this value gives P = ZRT/P = 0.3385*83.145*298.15/105.2 = 79.8 bar. At 79.8 bar, we get Z = 0.3382, from which P = ZRT/P = 0.3382*83.145*298.15/105.2 = 79.7 bar. At 79.7 bar, Z is still 0.3382, so the process has converged, and our answer is P = 79.7 bar.

For a few well-studied substances like ethylene, accurate PVT data are available from the National Institute of Science and Technology (NIST) at http://webbook.nist.gov. After you enter the NIST WebBook, look for the “Thermophysical 3

Properties of Fluid Systems” link. The data available there gives P = 75.8 bar for T = 25 °C and V = 105.2 cm /mol. The Lee/Kesler correlation is off by 5%, and the R/K equation is off by almost 13%. In this case, the deviations from ideal gas behavior are quite large (a factor of 3) and thus accurate prediction using generalized correlations is relatively difficult.

Solution 3.66

Problem Statement 3

If 15 kg of H2O in a 0.4 m container is heated to 400°C, what pressure is developed?

Solution

If we have 15 kg in 0.4 m , then the specific volume is 0.4/15 = 0.02667 m kg

= 26.7 cm g . Looking in the steam 3

tables on pages 722 and 723, in the column corresponding to 400°C, we see that the specific volume is 27.056 cm g 3

at 9800 kPa and 26.408 cm g

at 10000 kPa. Interpolating, we get

P = 9800 + (26.7 – 27.056)/(26.408 – 27.056)*(10000 – 9800) = 9910 kPa = 99.1 bar

Solution 3.67

Problem Statement 3

A 0.35 m vessel holds ethane vapor at 25°C and 2200 kPa. If it is heated to 220°C, what pressure is developed?

Solution

This is much like the previous problem, except that we first have to find the molar volume at 25° C and 2200 kPa (298.15 K and 22.0 bar) and then find the pressure for which that molar volume is the same at 220 °C (493 K). For ethane, Tc = 305.3 K, Pc = 48.72 bar, and  = 0.100. So, at 298.15 K and 22.0 bar, we have Tr = 0.9766 and Pr = 0.4516. Interpolating in the Lee/Kesler tables for these values gives Z = 0.8056, from which

Solution continued on next page…

V = 929.7 cm mol , as shown in the table below:

T (K)

P (bar)

Tc (K)

Pc (bar)

298.15

305.3

48.72

0.1000

0.9766

0.4516

Table Points 0

Tr (1)

Pr(1)

0.97

0.40

0.8338

0.0450

Tr(1)

Pr(2)

0.97

0.60

0.7240

0.0770

Tr(2)

Pr(1)

0.98

0.40

0.8398

0.0390

Tr(2)

Pr(2)

0.98

0.60

0.7360

0.0641

Interpolated Values 0.8105

0.0481

Final Value Z

0.8056 3

V (cm /mol)

929.7

Now, we want to find the pressure that gives us the same molar volume at T = 220 °C (493 K). If we heated an ideal gas from 298 K to 493 K, the pressure would increase by a factor of 493/298 = 1.65. So, the pressure would increase to 1.65*22 bar = 36.4 bar. So, let’s use 36.4 bar as our initial guess, and iterate like we did in the last problem until we 3

arrive back at a molar volume of 929.7 cm mol . At 493.15 K and 36.4 bar, we have Solution continued on next page…

Tr = 1.6153 and Pr = 0.7471. Interpolating to these values in the Lee/Kesler tables gives Z = 0.9556 and 3

V = 1102.4 cm mol, as shown below:

T (K)

P (bar)

Tc (K)

Pc (bar)

493.15

36.4

305.3

48.72

0.1000

1.6153

0.7471

Table Points

Tr (1)

Pr(1)

1.60

0.60

0.9575

0.0501

Tr(1)

Pr(2)

1.60

0.80

0.9439

0.0677

Tr(2)

Pr(1)

1.70

0.60

0.9667

0.0497

Tr(2)

Pr(2)

1.70

0.80

0.9563

0.0667

Interpolated Values 0.9493

0.0629

Final Value Z

0.9556 3

V (cm /mol)

1102.4

Solution continued on next page…

For this compressibility, to get V = 929.7 cm mol , we would need P = ZRT/P = 0.9556*83.145*493.15/929.7 = 42.1 bar, at which Pr = 0.8641. This falls in a different pressure interval in the Lee/Kesler tables, so we have to put new table entries in the spreadsheet:

T (K)

P (bar)

Tc (K)

Pc (bar)

493.15

42.1

305.3

48.72

0.1000

1.6153

0.8641

Table Points 0

Tr (1)

Pr(1)

1.60

0.80

0.9439

0.0677

Tr(1)

Pr(2)

1.60

1.00

0.9308

0.0855

Tr(2)

Pr(1)

1.70

0.80

0.9563

0.0667

Tr(2)

Pr(2)

1.70

1.00

0.9463

0.0838

Interpolated Values 0.9417

0.0732

Final Value Z

0.9491 3

V (cm /mol)

929.1

Predicting P using this compressibility gives P = ZRT/P = 0.9491*83.145*493.15/929.7 = 41.9 bar, at which Pr = 0.8600. The interpolated Z at that pressure is 0.9493, from which P = ZRT/P = 0.9493*83.145*493.15/929.7 = 41.9 bar, and the process has converged to an answer of 41.9 bar (at which 3

the volume is 929.2 cm mol

– close enough).

Solution 3.68

Problem Statement 3

What is the pressure in a 0.5 m vessel when it is charged with 10 kg of carbon dioxide at 30°C?

Solution

In this problem, instead of knowing the temperature and pressure and computing the molar volume, we know the temperature and molar volume, and need to compute the pressure. To obtain the molar volume from the data given, 3

we simply divide the total volume (0.5 m = 500,000 cm ) by the number of moles of gas 3

(10,000 g / 44.01 g/mol = 227.2 mol). So, the molar volume is 500,000/227.2 = 2200 cm /mol.

We can get an initial estimate of the pressure using the ideal gas law: id

P = RT/V = 83.14*303.15/2200 = 11.45 bar

Solution continued on next page…

Using this pressure, we can estimate the reduced pressure. The critical properties of CO2 are Tc = 304.2 K, Pc = 73.83 bar, and

= 0.224. So, the reduced temperature is Tr = 0.9965 and the reduced pressure, if the gas were

ideal, would be Pr = 11.45/73.83 = 0.1551. Checking this reduced temperature and pressure against the Figure 3.14 on p. 103 of SVNA, we see that it is well within the range for which the Pitzer correlation performs well. So, we can use it for this problem. The reduced second virial coefficient is BP Bˆ  c  B 0   B1 RTc

With 0.422 Tr1.6 0.172 B1  0.139  4.2 Tr B 0  0.083 

At Tr = 0.9965, we get B = –0.3414 and B = –0.0356, from which B = –0.3414 – 0.224*.0356 = –0.3493. The dimensional virial coefficient is then 3

B = –0.3493RTc/Pc = –0.3493*83.14*304.2/73.83 = –119.7 cm /mol The 2-term virial equation can be written as

(PV)/(RT) = 1 + B/V, or

P = RT/V*(1 + B/V) We have already computed RT/V above, so we have P = 11.45*(1 – 119.7/2200) = 10.83 bar

Solution 3.69

Problem Statement

A rigid vessel, filled to one-half its volume with liquid nitrogen at its normal boiling point, is allowed to warm to 25°C. 3

–1

What pressure is developed? The molar volume of liquid nitrogen at its normal boiling point is 34.7 cm ·mol .

Solution

The normal boiling point of N2 (from appendix B) is 77.3 K. The critical properties are Tc = 126.2 K, 3

–1

Pc = 34.00 bar, and  = 0.038. The molar volume of the liquid at this temperature is given as 34.7 cm mol . At 1 bar, the molar volume of the vapor will be essentially the ideal gas value, since the reduced pressure will be about 0.03. 3

–1

The ideal gas molar volume at 77.3 K and 1 bar is 6427 cm mol (almost 200 times the liquid molar volume). So, if t

we have a volume V that is half full of liquid and half full of vapor, the total number of moles will be 0.5 V /34.7 + 0.5 t

V /6427 = 0.01449 V moles (where V is in cm ). When the N2 is allowed to warm to 25°C, it ends up as a single (supercritical) phase, since 25°C = 298.15 K is far above the critical temperature for N2. The molar volume of this t

–1

supercritical fluid is the total volume divided by the total number of moles, or V / (0.01449 V ) = 69.03 cm mol . So, 3

–1

we want to find the pressure for which the molar volume is 69.03 cm mol at 298.15 K. The easiest way to do this will be to use a cubic equation of state, since given temperature and molar volume, these give pressure explicitly. Since the acentric factor is very small, we expect the R/K EOS to perform well, and it is simpler than the SRK or P/R equations, so we might as well use it. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The Redlich/Kwong equation of state is

P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 T 0.5 R2Tc2.5 T 0.5  83.1452  126.22.5 1.5551  107 a(T )  0.42748 r  0.42748  0.42748  bar cm6 mol2 Pc Pc 34.00 T RT 83.145  126.2 b  0.08664 c  0.08664  26.74 cm3 mol1 Pc 34.00

–2

So, at 298.15 K, we have a = 900619 bar cm mol , and

P

RT 900619 83.145 * 298.15 900619     450.0 bar V  26.74 V V  26.74 69.03  26.74 69.0369.03  26.74

Nitrogen is another substance for which accurate PVT data are available from the National Institute of Science and Technology (NIST) at http://webbook.nist.gov. After you enter the NIST WebBook, look for the “Thermophysical Properties of Fluid Systems” link. The data available for nitrogen there gives P = 482.8 bar for T = 25 °C and 3

V = 69.03 cm /mol. This is 7% higher than the R/K prediction.

Solution 3.70

Problem Statement 3

–1

The specific volume of isobutane liquid at 300 K and 4 bar is 1.824 cm ·g . Estimate the specific volume at 415 K and 75 bar.

Solution

First find the reduced temperature and pressures for isobutane at both sets of conditions.

Tr 1 

Pr 1 

Tr 2 

Pr 2 



P1 Pc T2 Tc P2 Pc

 0.11



Using Fig. (3.14) determine the reduced density at each set of conditions: r 1 

The final T > Tc, and Fig. 3.16 probably should not be used. One can easily show that r 

PVc ZRT

Using Z and Eq (3.53) and the Tables D.3 and D.4, Z gives Z  Z0    Z1  0.322

r 2 

P2Vc ZRT2

 1.774

Lastly by using Eq. (3.71) Solve for the second volume: V V

r 1 r 2



Solution 3.71

Problem Statement –3

The density of liquid n-pentane is 0.630 g·cm at 18°C and 1 bar. Estimate its density at 140°C and 120 bar.

Solution

First find the reduced temperature and pressures for n-pentane at both sets of conditions.

Tr 1 

Pr 1 

Tr 2 

Pr 2 

T1 Tc P1 Pc T2 Tc P2 Pc

 0.62



Using Fig. (3.14) determine the reduced density at each set of conditions:

r 1 

and r 2 

Solution continued on next page…

Lastly by using Eq. (3.71) Solve for the second density:

 

r 2 r 1



Solution 3.72

Problem Statement

Estimate the density of liquid ethanol at 180°C and 200 bar.

Solution

The critical properties of ethanol are Tc = 513.9 K, Pc = 61.48 bar, and

= 0.645. The critical volume is

Vc = 167 cm /mol. Thus, at 180 ºC and 200 bar, the reduced temperature and pressure are Tr = 453.15/513.9 = 0.8818 and Pr = 200/61.48 = 3.25. The simplest estimate is obtained by reading the reduced density at these conditions directly from Figure 3.16 on p. 110. At Pr = 3.25 and Tr = 0.88, the reduced density is r = 2.2. The critical density is

3 c = 1/Vc = 0.00599 mol/cm . So, the estimated molar density is 2.2*0.00599 = 0.013

mol/cm . Multiplying by the molar mass of ethanol, 46.07 g/mol, gives = 46.07*0.013 =0.5989 g/cm .

Solution continued on next page…

An alternative approach would be to apply the Lee/Kesler correlation directly at these conditions. Interpolating in the Lee/Kesler tables gives: T (K)

P (bar)

Tc (K)

Pc (bar)

453.15

200

513.9

61.48

0.645

0.8818

3.2531

Table Points 0

Tr (1)

Pr(1)

0.85

3.00

0.4591

0.1602

Tr(1)

Pr(2)

0.85

5.00

0.7388

0.2439

Tr(2)

Pr(1)

0.90

3.00

0.4527

0.1463

Tr(2)

Pr(2)

0.90

5.00

0.7220

0.2195

Interpolated Values 0.4896

0.1611

Final Values Z

0.3857 3

V (cm /mol)

72.7

This molar volume corresponds to a molar density of = 1/72.7 = 0.0138 mol/cm , and multiplying by the molar 3

mass gives = 0.63 g/cm . The two approaches differ by about 4%, and thus we can be fairly confident in this estimate.

Solution 3.73

Problem Statement

Estimate the volume change of vaporization for ammonia at 20°C. At this temperature the vapor pressure of ammonia is 857 kPa.

Solution

For ammonia: Tc = 405.7 K, Pc = 112.8 bar, Vc = 72.5 cm /mol,

= 0.253, Zc = 0.242

Pr = 0.076, Tr = 0.723

Using Eq. (3.68) 0.2857   1Tr   

Vliquid  Vc Zc 

Vvapor 



B0 



0.422  Tr1.6

B1 



0.172  Tr4.2

cm3 mol

T  RT     B 0   B 1  R c   P  Pc 

V  Vvapor  Vliquid 

cm3 mol

Solution 3.74

Problem Statement PVT data may be taken by the following procedure: A mass m of a substance of molar mass

is introduced into a

thermostated vessel of known total volume V . The system is allowed to equilibrate, and the temperature T and pressure P are measured. (a) Approximately what percentage errors are allowable in the measured variables (m,

, V , T, and P) if the

maximum allowable error in the calculated compressibility factor Z is ±1%?

(b) Approximately what percentage errors are allowable in the measured variables if the maximum allowable error in calculated values of the second virial coefficient B is ±1%? Assume that Z

0.9 and that values of B are calculated by

Eq. (3.37).

Eq. 3.37:

Z

PV B 1 RT V

Solution

The compressibility factor is related to the measured quantities by:

Z

PV t MPV t  a nRT mRT

By Eq. (3.37) B   Z  1V 

 Z  1 MV t m

(b)

(a)By Eq.(a) dZ dM dP dV t dm dT    t   c  Z M P m T V

Thus Max  Z 

M 

 P  V t 

m  T

Assuming approximately equal error in the five variables, a ±1% maximum error in Z requires errors in the variables of <0.2%.

(b)By Eq.(b) dB Z dZ dM dV t dm    t  B Z Z M m V

By Eq(c) dB Z  B Z

 dP dT  2Z  1  dM dV t dm       t    P T  z   M m  V

Therefore Max  B 



2Z  1 z 1

Z Z

 P  T 

 V  m   M  t

Solution continued on next page…

For Z  0.9 and for approximately equal error in the five variables, a ±1% maximum error in B requires errors in the variables of less than about 0.02%. This is because the divisor Z 1  0.1. In the limit as Z 1, the error in B approaches infinity.

Solution 3.75

Problem Statement

For a gas described by the Redlich/Kwong equation and for a temperature greater than Tc, develop expressions for the two limiting slopes,  Z  limP 0    P  T

Note that in the limit as P

0, V

 Z  limP     P T

, and that in the limit as P

, V

Solution

The RK equation has two typical forms that are in terms of P and Z:

Z

V  V b

a 3 2

RT V  b 

P

RT  V b

a 1 2

VT V  b

Solution continued on next page…

Differentiating these with respect to V gives 3

aV  b  V 2 RT 2 V  b  Z     3  V T 2 2 RT 2 V  b V  b 2

aV  b  bRT 2 V  b  P     3  V T 2 2 RT 2 V  b V  b 2

In addition, we have the mathematical relation  Z     V 

 Z  T     P     P  T    V T

Combining these equations gives

 Z      P  T

For P 

aV 2 V  b  bV 2 RT 2 V  b 2

aRT 2V  bV  b  V 2 R 2T 2 V  b 2

V   the last equation becomes b  a / RT 3/2  Z  lim    P 0   P  RT T

And for P  

V  b the last equation becomes  Z  b lim     P  RT T

P 

Solution 3.76

Problem Statement 3

If 140(ft) of methane gas at 60(°F) and 1(atm) is equivalent to 1(gal) of gasoline as fuel for an automobile engine, what would be the volume of the tank required to hold methane at 3000(psia) and 60(°F) in an amount equivalent to 10(gal) of gasoline?

Solution

If 140 ft of methane at 60ºF and 1 atm is equivalent to 1 gallon of gasoline, then 1400 ft at those conditions will be 3

equivalent to 10 gallons of gasoline. So, the question is if we compress the methane from 1400 ft at 1 atm, to a pressure of 3000 psia (at constant T of 60ºF), what will the final volume be? At both pressures, we can write the volume as V = ZRT/P. In going from the initial to the final state, T stays the same, but Z and P change. So, 3

we can write V2 = V1*(Z2/Z1)*(P1/P2). In this expression, V1 is 1400 ft and P1/P2 = 14.696/3000 = 4.899 × 10 . 3

Multiplying these together, V2 = (Z2/Z1)*6.858 ft . Now, all that remains is to use an appropriate correlation to find the compressibility at the initial and final pressures. The critical properties of methane are Tc = 190.6 K, Pc = 45.99 bar, and

= 0.012. The given temperature of 60ºF corresponds to 288.7 K, or a reduced temperature of 1.515.

The reduced pressure initially is 1 atm/0.9869 (atm/bar)/45.99 bar = 0.0220. At this low reduced pressure and high reduced temperature, the compressibility should be very close to 1.0. Just for fun, we can estimate it using nd

the Pitzer correlation for the 2 4.2

0.139 – 0.172/Tr

1.6

virial coefficient. This gives B = 0.083 – 0.422/Tr

= 0.1342 and B =

= 0.1090.

So, Z = 1 – 0.1342*0.022/1.515 + 0.012*0.1090*0.022/1.515 = 0.9981. This is, in fact, very close to 1.0, and we could have assumed that the methane at 1 atm and 60ºF is an ideal gas with negligible error. At the final pressure of 300 psia, the reduced pressure is 3000 psia / (14.5038 psia/bar)/45.99 bar = 4.498. At this very high reduced pressure, we cannot reliably use the Pitzer correlation. We should interpolate in the full Lee/Kesler correlation tables. Doing so gives: T (K)

P (bar)

Tc (K)

Pc (bar)

288.7

206.8

190.6

45.99

0.012

1.5147

4.4966

Table Points 0

Tr (1)

Pr(1)

1.50

3.00

0.7887

0.2433

Tr(1)

Pr(2)

1.50

5.00

0.8200

0.2309

Tr(2)

Pr(1)

1.60

3.00

0.8410

0.2381

Tr(2)

Pr(2)

1.60

5.00

0.8617

0.2631

0.8186

0.2374

Interpolated Values Final Values Z

0.8215

Solution continued on next page…

So, Z2 = 0.8215, and the final volume is 6.858 ft * 0.8215/0.9981 = 5.64 ft

3

= 42.2 gallons. Even at this quite high

pressure, the methane still requires a much larger tank than gasoline. Perhaps more importantly, the tank has to be sturdy enough to handle the high pressure, which means it will be much, much heavier than an ordinary sheet-metal gas tank.

Solution 3.77

Problem Statement Determine a good estimate for the compressibility factor Z of saturated hydrogen vapor at 25 K and 3.213 bar. For comparison, an experimental value is Z = 0.7757.

Solution

First find the critical parameters for H2 by using the eqns 3.54 and 3.55

Tc 

43.6 21.8  T K

Tc 

1

43.6  21.8  K

Solution continued on next page…

Pc 

20.5 44.2 1 T K

Pc 

Assuming a simple fluid makes

20.5  10.92 bar 44.2  K

1

= 0, and gives

Tr 

T  Tc

Pr 

P  Pc

Now taking the Z form of the RK EoS, determining q and ,

1/2

 (Tr ) q Tr

3/2

 (Tr )  

1.5



0.42748  (0.821) 0.08664

 6.6326

P 0.294    r  0.08664   0.03102 Tr 0.821

Z  1    q 

Z  Z Z  

Z  1  0.03102  6.6326  0.03102 

Z  0.03102  0.7904 Z  Z  0.03102

This is reasonably close to Z = 0.7757, with a 2% difference from the experimental number.

Solution 3.78

Problem Statement The Boyle temperature is the temperature for which:  Z  lim    0 P 0   P T

(a) Show that the second virial coefficient B is zero at the Boyle temperature. (b) Use the generalized correlation for B, Eqs. (3.58)–(3.62), to estimate the reduced Boyle temperature for simple fluids.

Eq. 58: BP Bˆ  c RTc

Eq. 59: B̂  B0   B1

Eq. 60: Z 0  1  B0

Pr Tr

Eq. 61: B 0  0.083 

0.422 Tr1.6

Eq. 62: B1  0.139 

0.172 Tr4.2

Solution

(a) Differentiating equation 3.33 gives:  Z     B   C P  D P 2   P  T

Taking the limit as P goes to 0 yields  Z     B  P0   P  T

If the limiting value of the derivative is 0 then B 



(b) To estimate the Boyle temperature and knowing B = 0 from part (a), we can start from BP Bˆ   RT

Bˆ   B 0   B 1.  = 0 for simple fluids, which makes B 0  0.  0.422  0.422 . Setting this equal to 0 and solving for Tr gives: Tr    1.6  0.083  Tr

1/1.6

So we know from eqn 3.61 B 0  0.083 

= 2.763

Solution 3.79 Problem Statement Natural gas (assume pure methane) is delivered to a city via pipeline at a volumetric rate of 150 million standard cubic feet per day. Average delivery conditions are 50(°F) and 300(psia). Determine: (a) The volumetric delivery rate in actual cubic feet per day. (b) The molar delivery rate in kmol per hour. –1

The pipe is 24(in) schedule-40 steel with an inside diameter of 22.624(in). Standard conditions are 60(°F) and 1(atm). Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

(a) At standard conditions of 60 ºF (288.7 K) and 1 atm (1.01325 bar), the natural gas can be taken to be an ideal gas. The critical properties of methane are Tc = 190.6 K, Pc = 45.99 bar, and

= 0.012. Thus, standard conditions

correspond to a reduced temperature of Tr = 1.515 and Pr = 0.0220. A quick check of Figure 3.15 on p. 104 of SVNA and/or Figure 3.14 on p. 103 confirms that the compressibility will be within a fraction of a percent of 1.0. So, the 3

molar volume at standard conditions is V = RT/P = 23690 cm /mol (0.8366 ft /mol). At the pipeline conditions of

50 ºF (283.15 K) and 300 psia (20.684 bar), the reduced temperature and pressure are Tr = 1.4856 and Pr = 0.4497. Another check of Fig. 3.14 shows that the compressibility should be about 0.96, and that these conditions are well 0

within the range where the Pitzer correlation performs well. Applying it, we obtain B = 0.1410 , B = 0.1064, Z = 1 – (0.1410  0.0012*0.1064)*0.4497/1.4856 = 0.9573

So the molar volume at the delivery conditions is V = 0.9574*83.145*283.15/20.684 = 1089.7 cm /mol (0.0385 ft /mol). 8

Thus, the volumetric delivery rate in actual cubic feet per day will thus be 1.5 × 10 * 1089.7/23690 = 6.9 × 10 actual cubic feet per day. (b) The molar delivery rate is simply the volumetric delivery rate divided by the molar volume. For convenience, we can do this at standard conditions, where we have n  1.5 10 8 / 0.8366  1.79 10 8 mol/day  7470 kmol/hr

Solution continued on next page…

(c)

The velocity is simply the volumetric flow rate (at actual conditions) divided by the pipe cross-sectional area. 6

The volumetric flow rate (from part (a)) is 6.9 × 10 ft /day = 79.9 ft /s. The cross-sectional area of the pipe is 2

A = (22.624/2) *

= = 0.25935m = 2.79 ft . So, the average velocity is v = 79.9 ft /s / 2.79 ft = 28.62 ft/s

= 8.72 m/s.

Solution 3.80

Problem Statement

Some corresponding-states correlations use the critical compressibility factor Zc, rather than the acentric factor , as a third parameter. The two types of correlation (one based on Tc, Pc, and Zc, the other on Tc, Pc, and ) would be equivalent were there a one-to-one correspondence between Zc and . The data of App. B allow a test of this correspondence. Prepare a plot of Zc vs. (Zc = a +

to see how well Zc correlates with . Develop a linear correlation

) for nonpolar substances.

Solution

So first gather the data from appendix B, all the Zc and

values can be taken and directly used to develop a linear

equation: (The data below spans over several pages…)

Solution continued on next page…

Zc mass Methane

16.043

45.99

Ethane

30.07

48.72

Propane

44.097

42.48

n-Butane

58.123

37.96

n-Pentane

72.15

33.7

n-Hexane

86.177

30.25

n-Heptane

100.204

27.4

n-Octane

114.231

24.9

n-Nonane

128.258

22.9

n-Decane

142.285

21.1

Isobutane

58.123

36.48

Isooctane

114.231

25.68

Cyclopentane

70.134

45.02

Cyclohexane

84.161

40.73

Methylcyclopentane

84.161

37.85

Methylcyclohexane

98.188

34.71

Ethylene

28.054

50.4

Propylene

42.081

46.65

1-Butene

56.108

40.43

cis-2-Butene

56.108

42.43

trans-2-Butene

56.108

1-Hexene

84.161

31.4

Isobutylene

56.108

1,3-Butadiene

54.092

42.77

Cyclohexene

82.145

43.5

Acetylene

26.038

61.39

Benzene

78.114

48.98

Toluene

92.141

41.06

Ethylbenzene

106.167

36.06

Cumene

120.194

32.09

o-Xylene

106.167

37.34

m-Xylene

106.167

35.36

p-Xylene

106.167

35.11

Styrene

104.152

38.4

Naphthalene

128.174

40.51

Biphenyl

154.211

38.5

Formaldehyde

30.026

65.9

Acetaldehyde

44.053

55.5

Methyl acetate

74.079

47.5

Ethyl acetate

88.106

38.8

Acetone

58.08

47.01

Methyl ethyl ketone

72.107

41.5

Diethyl ether

74.123

36.4

Methyl t-butyl ether

88.15

34.3

Methanol

32.042

80.97

Ethanol

46.069

61.48

1-Propanol

60.096

51.75

1-Butanol

74.123

44.23

1-Hexanol

102.177

35.1

2-Propanol

60.096

47.62

Phenol

94.113

61.3

Ethylene glycol

62.068

Acetic acid

60.053

57.86

n-Butyric acid

88.106

40.64

Benzoic acid

122.123

44.7

Acetonitrile

41.053

48.3

Methylamine

31.057

74.6

Ethylamine

45.084

56.2

Nitromethane

61.04

63.1

Carbon tetrachloride

153.822

45.6

Chloroform

119.377

54.72

Dichloromethane

84.932

60.8

Methyl chloride

50.488

66.8

Ethyl chloride

64.514

52.7

Chlorobenzene

112.558

45.2

Tetrafluoroethane

102.03

40.6

Argon

39.948

48.98

Krypton

83.8

55.02

Xenon

131.3

58.4

Helium 4

4.003

2.28

Hydrogen

2.016

13.13

Oxygen

31.999

50.43

Nitrogen

28.014

Air†

28.851

37.45

Chlorine

70.905

77.1

Carbon monoxide

28.01

34.99

Carbon dioxide

44.01

73.83

Carbon disulfide

76.143

Hydrogen sulfide

34.082

89.63

Sulfur dioxide

64.065

78.84

Sulfur trioxide

80.064

82.1

Nitric oxide (NO)

30.006

64.8

Nitrous oxide (N2O)

44.013

72.45

Hydrogen chloride

36.461

83.1

Hydrogen cyanide

27.026

53.9

Water

18.015

220.55

Ammonia

17.031

112.8

Nitric acid

63.013

68.9

Solution continued on next page…

180 160 140 120

100 80

y = -0.4578x + 92.712

60 40 20 0 -20 0

100

150

200

250

This gives the line Zc = 

Solution 3.81 Problem Statement Figure 3.3 suggests that the isochores (paths of constant volume) are approximately straight lines on a P–T diagram. Show that the following models imply linear isochores. (a) Constant- , equation for liquids (b) Ideal-gas equation (c) Van der Waals equation

Solution

 P  An isochore requires that    constant because it is the intercept  T  V

 P   (a) Using equation 3.5 at constant volume     T V   P  R (b) For the ideal gas equation     T  V V  P  R (c) Differentiating Eqn 3.40 gives     T V V  b  P  With each part on the right being a constant, then the derivative of   is as well of each.  T  V

Solution 3.82

Problem Statement An ideal gas, initially at 25°C and 1 bar, undergoes the following cyclic processes in a closed system: (a) In mechanically reversible processes, it is first compressed adiabatically to 5 bar, then cooled at a constant pressure of 5 bar to 25°C, and finally expanded isothermally to its original pressure.

(b) The cycle is irreversible, and each step has an efficiency of 80% compared with the corresponding mechanically reversible process. The cycle still consists of an adiabatic compression step, an isobaric cooling step, and an isothermal expansion.

Calculate Q, W

H for each step of the process and for the cycle. Take CP = (7/2)R and CV = (5/2)R.

Solution

(a)

State 1 is 298 K and 100 kPa, for which V = 0.02479 m /mol. For adiabatic compression from 100 kPa to 500 (R/Cp)

kPa, we have (T2/T1) = (P2/P1)

2/7

= 1.584. So, the temperature after compression is 3

T2 = 1.584*298 = 472.2 K. The final volume after compression will then be 0.007853 m /mol. As always for an ideal gas, U = Cv T = 2.5 R*(472.2298) = 3618 J/mol, and H = Cp T = 3.5 R*(472.2298) = 5065 J/mol. For an adiabatic process, Q = 0, and W = U = 3618 J/mol.

Solution continued on next page…

If we cool the gas at constant pressure back to the original temperature, the enthalpy and internal energy go back to their original values, and we will have U = Cv T = 2.5 R*(174) = 3618 J/mol, and H = Cp T = 3.5 R*(174) = 3

5065 J/m. During this process, the volume will decrease to R*298 K / 500000 Pa = 0.004958 m /mol. The work done 3

will be W =  PV = 500000 Pa * (0.004958 – 0.007853) m mol

1

= 1447 Pa m mol

1

= 1447 J/mol. The heat

removed can be found from Q = U-W = 36181447 = 5065 J/mol

Finally, if we expand the gas back to its original state isothermally, we will have U = H = 0 (since they only depend on temperature, which doesn’t change). For the isothermal process, we will have Q = RT ln(P1/P2) = RT ln(0.2) = 4055 J/mol, and W = Q = 4055 J/mol.

For the overall cycle, Q = 0 – 5065  4055 = 9120 J/mol. W must then be 9120 for the cycle (so that the overall U is zero). To check, we can add up the three steps: W = 3618 + 1447 + 4055 = 9120. Of course U and H are zero for any cyclic process.

(b) In the first step, if we compress it to 500 kPa with a process that is 80% efficient, and still adiabatic, then we will put in W = 3618/0.8 = 4523 J of work to do the compression. If Q is still zero, then we will have U = 4523 J = Cv T, so T will be 217.6 K. In compressing the gas adiabiatically with an 80% efficient process, we will heat it to 515.7 K rather than 472.2 K. The enthalpy change will then be H = Cp T = 3.5 R*(217.6) = 6332 J/mol. Because of the 3

higher temperature, the molar volume will then be 0.008576 m /mol rather than 0.007853 m /mol that was found in (a). Solution continued on next page…

If we cool the gas back to the original temperature, as in the reversible case, the enthalpy and internal energy will go back to their original values, and we will have U = Cv T = 2.5 R*(217.6) = 4523 J/mol, and H = Cp T = 3.5 R*(217.6) = 6332 J/mol. The volume after this step will be the same as in part (a): 3

R*298 K / 500000 Pa = 0.004958 m /mol. The reversible work required to cool the gas isobarically is then 3

W =  PV = 500000 Pa * (0.004958 – 0.008576) m mol

1

= 1809 Pa m mol

1

= 1809 J/mol. So, if our process is

80% efficient, the total work required will be 1809/0.8 = 2261 J/mol. We still have U = 4523 J/mol = Q + W = Q + 2261, so Q = 6784 J/mol. For the isothermal expansion, we will still have U = H = 0 (since they only depend on temperature, which doesn’t change). For a reversible isothermal process, we had Q = RT ln(P1/P2) = RTln(0.2) = 4055 J/mol, and W = Q = 4055 J/mol. If the process is only 80% efficient, then the work will be 0.8*(4055) = 3244, and the heat requirement will be set by U = 0 = Q + W = Q – 3244, so Q = 3244 J/mol.

For the overall cycle with 80% efficiency in each step, Q = 0 – 6784  3244 = 10028 J/mol. W must then be 3540 for the cycle (so that the overall U is zero). To check, we can add up the three steps: W = 4523 + 2261 + 3244 = 10028 J/mol. Of course U and H are zero for any cyclic process, whether the steps are reversible or not. Note that the irreversibilities compound each other so that the net work we put in here for the whole cycle is much more than 1.25 times (or 1/0.8) the net work that we put in for the reversible cycle in part (a).

Solution 3.83

Problem Statement Show that the density-series second virial coefficients can be derived from isothermal volumetric data via the expression: B  lim ( Z  1 ) /   0

 molar density   1 / V

Solution

Using the equation 3.34, substitute V with

Z 

B C D   V V2 V3

Z   B  C  2  D 3

Subtract 1 from each side and divide by .

Z 1  B  C  D 2 

Lastly, take the limit as



lim B  C   D 2  B 0

Solution 3.84 Problem Statement Use the equation of the preceding problem and data from Table E.2 of App. E to obtain a value of B for water at one of the following temperatures: (a) 300°C (b) 350°C (c) 400°C

Solution (a) From table F.2 all of the values for specific gravity at the specific pressure for 300 C need to be tablulated. With this data, the density, Z values, and Z-

values can be determined.

P Kpa

V (cm^3/g)

Z-1/

125

2109.7

2.63188E-05

0.996686

125.9023437

150

1757

3.1602E-05

0.996072

124.2881683

175

1505.1

3.6891E-05

0.995477

122.6047371

200

1316.2

4.21856E-05

0.994901

120.8796292

225

1169.2

4.74895E-05

0.994258

120.9085917

250

1051.6

5.28002E-05

0.993616

120.9160244

275

955.45

5.81137E-05

0.993044

119.697015

300

875.29

6.34358E-05

0.992433

119.2916276

/cm^3)

With the values calculated, plot Z-

versus the density values and linearly fit the data. The intercept of the line is 3

Z-1/ρ

the B value. So for this data B = 129.29 cm /mol.

-118 -119 -120 -121 -122 -123 -124 -125 -126 -127

y = 166681x - 129.29

0.00002

0.00004

0.00006

0.00008

ρ (g/cm^3)

(a) Same as part (a), except use 350 C to tabulate the data. Fitting the data gives B = 106.77 cm /mol. P Kpa

V (cm^3/g)

125

2295.6

150

/cm^3)

Z-1/

2.41874E-05

0.997494

103.6016566

1912.2

2.90371E-05

0.997077

100.6644948

175

1638.3

3.38917E-05

0.996634

99.32289962

200

1432.8

3.87526E-05

0.996138

99.64691421

225

1273.1

4.36138E-05

0.995747

97.50696627

250

1145.2

4.84847E-05

0.995235

98.28634491

275

1040.7

5.33532E-05

0.994861

96.32176525

300

953.52

5.82313E-05

0.994386

96.4013883

Solution continued on next page…

-95 -96

y = 189330x - 106.77

-97

Z-1/ρ

-98 -99 -100 -101 -102 -103 -104 0

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

0.00007

ρ (g/cm^3)

(b) Same as part (a), except use 400 C to tabulate the data. Fitting the data gives B = 90.77 cm /mol.

P Kpa

V (cm^3/g)

125

2481.2

150

/cm^3)

Z-1/

2.23782E-05

0.998061

86.63762145

2066.9

2.68638E-05

0.997691

85.94703438

175

1771.1

3.13504E-05

0.997393

83.14166567

200

1549.2

3.58409E-05

0.997064

81.92792085

225

1376.6

4.03347E-05

0.996726

81.17728694

250

1238.5

4.48322E-05

0.996372

80.92926009

275

1125.5

4.93334E-05

0.99601

80.88365086

300

1031.4

5.38343E-05

0.995712

79.65046047

Z-1/ρ

-78 -79 -80 -81 -82 -83 -84 -85 -86 -87 -88

y = 216165x - 90.772

0.00001

0.00002

0.00003

0.00004

0.00005

0.00006

ρ (g/cm^3)

Solution 3.85 Problem Statement

Zc given in Table 3.1 for: (a) The Redlich/Kwong equation of state. (b) The Soave/Redlich/Kwong equation of state. (c) The Peng/Robinson equation of state.

Solution Three general equations for Zc in terms of

and

from the Vander Waals equation of state gives:

1  1       3 Zc  2       1    3Zc2  2   1    Zc3

(a,b)For Redlich/Kwong and Soave/Redlich/Kwong, = 0

= 1.

Substituting this in to the general equations and solving gives 1  (1  0  1)  3 Z c

Zc 

1 3

Also, 0  1  2  0  1 (  1)    3(0.333) 2 or   





and 0  1  2   1    (0.333)3   solving equations 1 and 2, we get   0.086640 

Solution continued on next page…

1  (1  (1  2)  (1  2) )  3Z c 1  (1  1  2  1  2))  3Z c   Z c  

Also,









(1  2) 1  2 2  1  2  1  2 (  1)    3 Zc2        

 Z

2 c

3  2    3 Zc  1   32  2    3   3   2      2   2

102  4  3  1



1  102  4 3

and





(1  2 ) 1  2 2   1    Zc3   3    

   2         3  3 

By solving equation (3), we get   0.33202

Solution continued on next page…

Also, from equation (1), we get 1  0.33202  3 Zc Z c  0.2227

and from equation (2), we get 2

   1.1435

 



Solution 3.86 Problem Statement

Suppose Z vs. Pr data are available at constant Tr . Show that the reduced density-series second virial coefficient can be derived from such data via the expression: Bˆ  lim ( Z  1 ) Z Tr / Pr Pr  0

Suggestion: Base the development on the full virial expansion in density, Eq. (3.34) Eq. 3.34:

Z  1

B C D  2  3  V V V

Solution

Starting with equation 3.34: Z   B  C  2

  P ZRT gives Z 1

BP CP 2  2 2 2 ZRT Z RT

From here substituting in

from eqns 3.58 and 3.64 and rearranging gives

 Z  1ZT

 Bˆ 

ˆ CP r ZTr

And taking the limit as Pr goes to 0, yields the original statement lim

Pr  0

 Z  1 ZTr Pr

 Bˆ

Solution 3.87 Problem Statement Use the result of the preceding problem and data from Table D.1 of App. D to obtain a value of B̂ for simple fluids at Tr = 1. Compare the result with the value implied by Eq. (3.61). Eq. 61: B 0  0.083 

0.422 Tr1.6

Solution First obtain the data from Table D.1 for simple fluids at Tr = 1: Pr

0.01

0.997

0.05

0.983

0.1

0.966

0.2

0.93

0.4

0.851

0.6

0.757

0.8

0.636

0.29

With this data, and using y =

 Z  1 ZTr



Pr ZTr

Bˆ Bˆ  

-0.2

(Z-1)*Z*Tr/ Pr

-0.22

y = 0.0364x - 0.333

-0.24 -0.26 -0.28 -0.3 -0.32 -0.34 0

0.5

1.5

2.5

3.5

Pr/Z*Tr

Using equations 3.61 and 3.62 to determine B̂ through a correlation gives

B0 



0.422  Tr1.6

B1 



0.172  Tr4.2

And assuming water for the acentric factor, for simple fluid omega = 0

Bˆ  

 *



This gives a difference of about 5% difference between the two calculations.

Solution 3.88 Problem Statement The following conversation was overheard in the corridors of a large engineering firm. New engineer: “Hi, boss. Why the big smile?” Old-timer: “I finally won a wager with Harry Carey, from Research. He bet me that I couldn’t come up with a quick but accurate estimate for the molar volume of argon at 30°C and 300 bar. Nothing to it; I used the ideal3

gas equation, and got about 83 cm ·mol . Harry shook his head, but paid up. What do you think about that?” New engineer (consulting his thermo text): “I think you must be living right.” Argon at the stated conditions is not an ideal gas. Demonstrate numerically why the old-timer won his wager. Solution

First obtain the necessary properties of pure argon from table B.1: Tc = 150.9 K, Pc =48.89 bar,

= 0.00.

Using this information, calculate the Z value from the RK method (any of the equations will work but it needs to be specified which one is used.) Use the values from table 3.1 and the equations 3.50 and 3.51. P   r  Tr

q

Z  1    q *

 Tr  Tr



Z 

Z   Z   

Plugging these in to equation 3.48 and a Z guess of 0.9 gives a Z value = 0.9866 which leads to the

V

ZRT 0.9866 * 83.145 * 303.15   P 300

cm3 mol

This about a 0.1% difference from the ideal gas value.

Solution 3.89

Problem Statement

Five mol of calcium carbide are combined with 10 mol of water in a closed, rigid, high-pressure vessel of 1800 cm

internal empty volume. Acetylene gas is produced by the reaction:

CaC2 ( s ) + 2 H2O ( l )

C2H2 ( g ) + Ca ( OH )2 ( s )

The vessel contains packing with a porosity of 40% to prevent explosive decomposition of the acetylene. Initial conditions are 25°C and 1 bar, and the reaction goes to completion. The reaction is exothermic, but owing to heat transfer, the final temperature is only 125°C. Determine the final pressure in the vessel. 3

Note: At 125°C, the molar volume of Ca(OH)2 is 33.0 cm ·mol . Ignore the effects of any gases (e.g., air) initially present in the vessel.

Solution

Since the reaction goes to completion, assume that there are 5 moles Ca(OH)2 and 5 moles C2H2. Now calculated the volume that will be available for the gas.

 V   

cm3    mol 

cm3  mol

cm3 mol

With this plug into RK equation (3.40) to find the pressure in the vessel.

Tr0.5 R2Tc2

aT  

b

P

RTc Pc

bar cm 6mol 2



cm3mol 1

aT  RT   V  b V V  B 

bar

Solution 3.90

Problem Statement Storage is required for 35,000 kg of propane, received as a gas at 10°C and 1(atm). Two proposals have been made: (a) Store it as a gas at 10°C and 1(atm). (b) Store it as a liquid in equilibrium with its vapor at 10°C and 6.294(atm). For this mode of storage, 90% of the tank volume is occupied by liquid. Compare the two proposals, discussing pros and cons of each. Be quantitative where possible.

Solution

So for both proposals let’s determine the amount of volume need to hold each of these first. (a) So for this problem estimate the volume using the RK equation: Tr0.5 R2Tc2

aT  

RTc

b

V

bar cm6mol2



cm3mol 1

aT  V  b RT b  P P V V  b

18084.02 m would require a propane cylinder with a 10 m height and a 46 m diameter, which is very large. This probably is not feasible. (b) Use the rackett equation 3.68 to determine the liquid volume (90%) and then the RK to determine the gas (only 10% is gas). This is at 6.249 atm or 6.377 bar and 283.16K. Tr0.5 R2Tc2

aT  

b

V

RTc Pc



aT  V  b RT b  P P V V  b

bar cm6mol2

cm3mol 1

cm3 of gas

Solution continued on next page…

For the liquid, 2/7

1Tr 

V sat  Vc Zc



cm3 * mol

10.7667



cm3 mol

To determine the total volume needed to house all of the propane:

nmol 

Vtank Vtank  Vliq Vvap

Solving for Vtank gives a required volume of 72.73 m . This would be cylinder with a 5 m height and 4.3 m diameter. This is more reasonable, however the tank would have to be able to hold the propane at 10 C and 6.294 atm.

Solution 3.91

Problem Statement The definition of compressibility factor Z, Eq. (3.32), may be written in the more intuitive form:

Z

V V (ideal gas )

where both volumes are at the same T and P. Recall that an ideal gas is a model substance comprising particles with no intermolecular forces. Use the intuitive definition of Z to argue that: (a) Intermolecular attractions promote values of Z < 1. (b) Intermolecular repulsions promote values of Z > 1.

(c) A balance of attractions and repulsions implies that Z = 1. (Note that an ideal gas is a special case for which there are no attractions or repulsions.)

Eq. 3.32:

Z

PV V  ig RT V

Solution

(a) For Intermolecular attractions, the distance between molecules would be smaller or having a smaller ig

volume, which would mean V < V . This would result in Z < 1. (b) For intermolecular repulsions, the distance between molecules would be larger or having a larger volume, ig

which would mean V > V . This would result in Z > 1. (c) A balance between attraction and repulsion would basically cancel out any distance between molecules or ig

any change in the volume, meaning V = V . This would result in Z = 1.

Solution 3.92

Problem Statement Write the general form of an equation of state as:

1 + Zrep ( )

Zattr ( T, )

where Zrep( ) represents contributions from repulsions, and Zattr(T, ) represents contributions from attractions. What are the repulsive and attractive contributions to the van der Waals equation of state?

Solution

From Van der Waals:

Z

V a  V  b VRT

Setting V = 1/ and substituting this in gives:

Z

1 a b a  1  1  b RT 1  b RT

This would leave

Zrep 

b a and Z atr  1  b RT

Solution 3.93

Problem Statement Given below are four proposed modifications of the van der Waals equation of state. Are any of these modifications reasonable? Explain carefully; statements such as, “It isn’t cubic in volume” do not qualify.

(a)

RT a  V b V

P

(b) P 

V  b



a V

RT a  2 V V  b V

(d) P 

RT a  2 V V

Solution

First, let’s put a and b equal to zero and check if it reduces to the ideal gas equation or not. If equation reduces to the ideal gas equation, then the modification is reasonable.

RT a  V  b V

when a = 0 and b = 0, then

RT V

The equation reduces to the ideal gas equation. Therefore, the modification is reasonable.

RT a  2 V V  b

Solution continued on next page…

when a = 0 and b = 0, then

RT V2

The equation does not reduce to the ideal gas equation. Therefore, the modification is not reasonable.

RT a  2 V V  b V

when a = 0 and b = 0, then

RT V2

The equation does not reduce to the ideal gas equation. Therefore, the modification is not reasonable.

RT a  2 V V

when a = 0 and b = 0, then

RT V

The equation reduces to the ideal gas equation. Therefore, the modification is reasonable.

Solution 3.94 Problem Statement With reference to Prob. 2.47, assume air to be an ideal gas, and develop an expression giving the household air temperature as a function of time.

Problem 2.47 The heating of a home to increase its temperature must be modeled as an open system because expansion of the household air at constant pressure results in leakage of air to the outdoors. Assuming that the molar properties of air leaving the home are the same as those of the air in the home, show that energy and mole balances yield the following differential equation:

dn dU Q  PV n dt dt Here, Q is the rate of heat transfer to the air in the home, and t is time. Quantities P, V, n, and U refer to the air in the home.

Solution

In Problem 2.47, the general equation was developed. dn dU Q   PV n dt dt

Solution continued on next page…

To get this as an expression for air temperature over time, first assume ideal gas and substitute in for dn/dt and dU/dt

n

PV dn PV dT dU dT and  and  Cv 2 RT dt dt dt RT dt PV dT PV dT Q  RT  Cv 2 RT dt RT dt

This reduces to PV dT Q  C p RT dt

Separating and integrating gives PV

 T 

 C ln    Qdt R  T  t2

Solution 3.95

Problem Statement A garden hose with the water valve shut and the nozzle closed sits in the sun, full of liquid water. Initially, the water is at 10°C and 6 bar. After some time the temperature of the water rises to 40°C. Owing to the increase in temperature and pressure and the elasticity of the hose, the internal diameter of the hose increases by 0.35%. Estimate the final pressure of the water in the hose.

Data: (ave) = 250 × 10

K ;

(ave) = 45 × 10

bar

Solution

For estimating PVT behavior of liquids, we will often assume constant coefficient of thermal expansion and isothermal compressibility. In that case, the differential relationship dV   dT  dP V

can be integrated to give the integral relationship: V  ln   T  P  Vo 

If the diameter of the hose increases by 0.35% (by a factor of 1.0035), then the cross-sectional area increases by a factor 2

of 1.0035 = 1.0070. Then ln(V/Vo) = ln(1.007) = 0.00699. Putting in the known values of , , and T, we have

0.00699 = 250 × 10

* 30 K – 45 × 10

bar

* P (bar)

from which P = 11.33 bar. The final pressure is 17.33 bar.

Solution 3.96 Problem Statement

Prepare a plot like that in example 3.13 comparing results from the Redlich/Kwong equation, Peng/Robinson equation, 2-term virial equation, 3-term virial equation, and NIST Webbook Thermophysical Properties of Fluids data for one of the following: (a) Water at 770 K and pressures up to 1000 bar. (b) Carbon monoxide at 160 K and pressures up to 200 bar. (c) Propylene at 460 K and pressures up to 250 bar. (d) Cyclohexane at 660 K and pressures up to 200 bar. (e) Argon at 200 K and pressures up to 300 bar. (f) Hydrogen sulfide at 450 K and pressures up to 450 bar. (g) Carbon dioxide at 370 K and pressures up to 400 bar. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

In each case, the solution follows exactly the method of example 3.13. For the Redlich/Kwong equation, the equation of state parameters for each case come out to be: 4

(a) (b) (c) (d) (e)

/P = 3.5531 × 10 , q = 3.8012 /P = 22.40 × 10 4, q = 3.7351 /P = 16.80 × 10 4, q = 3.4960 /P = 19.24 × 10 4, q = 3.7903 /P = 16.00 × 10 4, q = 3.2336

(f) (g)

/P = 8.743 × 10 4, q = 3.7257 /P = 10.61 × 10 4, q = 3.6782

For the Peng/Robinson equation, the parameters are: (a) /P = 2.965 × 10 4, q = 4.1855 (b) (c) (d) (e) (f) (g)

/P = 18.47 × 10 , q = 4.4656 /P = 13.25 × 10 4, q = 4.0293 /P = 16.02 × 10 4, q = 4.3273 4 /P = 11.98 × 10 , q = 3.9460 /P = 7.198 × 10 4, q = 4.3912 /P = 8.664 × 10 4, q = 4.1551

For the two-term virial equation with the generalized correlation for Bˆ , the parameters are: 0

B = 0.0561, B̂ =

(b) B =

B = 0.0601, B̂ =

B1 = 0.0735, B̂ =

(d) B0 =

B1 = 0.0568, B̂ =

(e) B0 =

B1 = 0.0863, B̂ =

(a) B =

1 1

(f) B =

B = 0.0607, B̂ =

(g) B0 =

B1 = 0.0634, B̂ =

Solution continued on next page…

For the three-term virial equation with the generalized correlations for B̂ and Cˆ , we also have values of: 0

(a) C = 0.0340, C = 0.0066, Ĉ = 0.03627 (b) C0 = 0.0338, C1 = 0.0055, Ĉ = 0.03409 (c) C0 = 0.0331, C1 = 0.0019, Ĉ = 0.03385 (d) C0 = 0.0340, C1 = 0.0064, Ĉ = 0.03532 (e) C0 = 0.0323, C1 = –0.0019, Ĉ = 0.03226 (f) C0 = 0.0338, C1 = 0.0054, Ĉ = 0.03431 (g) C0 = 0.0337, C1 = 0.047, Ĉ = 0.03471

The results from each of these, along with the compressibility computed from the data in the NIST Chemistry Webbook are presented in the plots below:

(a) 1.2

R/K P/R

2-term 3-term NIST

P/R

0.8

3-term R/K

0.6

0.4

NIST

2-term

0.2

200

400

600

800

1000

P /bar

(b) 1.2

R/K P/R

2-term 3-term NIST

0.8

3-term

R/K

0.6

NIST P/R 0.4

2-term

0.2

100

150

200

P /bar

1.2

0.8

3-term

0.6

R/K 0.4

P/R

NIST

R/K

2-term

P/R

0.2

2-term 3-term NIST

100

150

200

250

P /bar

(c)

(d) 1.2

R/K P/R

2-term 3-term NIST

0.8

P/R

3-term

0.6

R/K

NIST

0.4

2-term 0.2

100

150

P /bar

200

(e) 1.2

3-term R/K

NIST

0.8

P/R

0.6

0.4

R/K

2-term

P/R 2-term

0.2

3-term NIST

100

150

200

250

300

P /bar

1.2

R/K P/R

2-term 3-term NIST

0.8

3-term 0.6

NIST

R/K

P/R

0.4

2-term 0.2

100

200

P /bar

300

400

(g) 1.2

R/K P/R

2-term 3-term NIST

0.8

3-term 0.6

P/R

R/K

0.4

NIST

2-term 0.2

100

200

P /bar

300

400

Solution 4.1

Problem Statement

For steady flow in a heat exchanger at approximately atmospheric pressure, what is the heat transferred: (a) When 10 mol of SO2 is heated from 200 to 1100°C? (b) When 12 mol of propane is heated from 250 to 1200°C? (c) When 20 kg of methane is heated from 100 to 800°C? (d) When 10 mol of n-butane is heated from 150 to 1150°C? (e) When 1000 kg of air is heated from 25 to 1000°C? (f) When 20 mol of ammonia is heated from 100 to 800°C? (g) When 10 mol of water is heated from 150 to 300°C? (h) When 5 mol of chlorine is heated from 200 to 500°C? (i) When 10 kg of ethylbenzene is heated from 300 to 700°C?

Solution

 T  T   0   0

 R dT  AT  T   2 T  T   3 T  T   D TT 2

2 0

3 0

 R dT  ICPH T T A B C D 0

Q  H  n

ICPH

Solution continued on next page…

(a) If 10 mol of SO2 is heated from 200 to 1100°C, the heat requirement will be 10*R*ICPH, where ICPH is the heat capacity integral. Using our spreadsheet to evaluate ICPH, we get T1 (K)

T2 (K)

B (1/K)

473.15

1373.15

5.699

8.01E04 0

C (1/K )

D (K )

ICPH (K)

1.02E05

5.65E03

So, the heat requirement is 10 mol*8.314 J mol K*5650 K  469700 J  470 kJ

(b) Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives 2

C (1/K )

T1 (K)

T2 (K)

B (1/K)

D (K )

ICPH (K)

523.15

1473.15

1.213

2.88E02 8.82E06 0.00E00

1.95E04

So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(c) Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(d) Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(e) Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(f)

Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(g) Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(h) Similarly, if we heat 12 moles of propane from 250 to 1200 °C, the heat requirement is 12*R*ICPHPutting the heat capacity parameters for propane and these temperatures into the ICPH spreadsheet gives So, the total heat requirement is 12 mol*8.314 J mol K*19500 K  1.9510 J  1950 kJ 6

(i)

Solution 4.2 Problem Statement For steady flow through a heat exchanger at approximately atmospheric pressure, what is the final temperature, (a) When heat in the amount of 800 kJ is added to 10 mol of ethylene initially at 200°C? (b) When heat in the amount of 2500 kJ is added to 15 mol of 1-butene initially at 260°C? (c) When heat in the amount of 106(Btu) is added to 40(lb mol) of ethylene initially at 500(°F)? Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

If we have set up a spreadsheet to evaluate the heat capacity integral like the example spreadsheet from the lecture notes, we can use it to evaluate the heat capacity integral in each case, and simply vary the final temperature until we get the desired Q. (a) If 800 kJ is added to 10 mol of ethylene, then Q  H  800000 J/10 mol  80000 J mol So, H/R  80000 J mol /8.314 J mol K  9622 K, so the integral of Cp/R from 200°C (473 K) to the final temperature should be 9622 KUsing our spreadsheet to evaluate this, we get 2

C (1/K )

T1 (K)

T2 (K)

B (1/K)

D (K )

473

1374.4

1.424

1.44E02 4.39E06 0.00E00

ICPH (K) 9.622E03

So, the final temperature is 1374 K.

(b) Similarly, if we add 2500 kJ to 15 mol of 1-butene, we have H/R  2500000 J / (15 mol *8.314 J mol K)  20045 KPutting the heat capacity coefficients for 1-butene into the spreadsheet and trying final temperatures until we get this value, we get 2

C (1/K )

T1 (K)

T2 (K)

B (1/K)

D (K )

533

1414

1.967

3.16E02 9.87E06 0.00E00

ICPH (K) 2.005E04

So, the final temperature is 1414 K. (c) This is the same, but in English units, in which we can use R  1.986 Btu lbmol R, so H/R  10 Btu / (40 lbmol *1.986 Btu lbmol R)  12588 °R  6993 KThe initial temperature of 6

500 °F is 533 KSo, we get 2

C (1/K )

T1 (K)

T2 (K)

B (1/K)

D (K )

533

1202.7

1.424

1.44E02 4.39E06 0.00E00

ICPH (K) 6.994E03

And the final temperature is 1203 K  1705 °F.

Solution 4.3 Problem Statement For a steady-flow heat exchanger with a feed temperature of 100°C, compute the out-let stream temperature when heat in the amount of 12 kJ·mol is added to the following substances. 1

(a) methane, (b) ethane, (c) propane, (d) n-butane, (e) n-hexane, (f) n-octane, (g) propylene, (h) 1-pentene, (i) 1-heptene, (j) 1-octene, (k) acetylene, (l) benzene, (m) ethanol, (n) styrene, (o) formaldehyde, (p) ammonia, (q) carbon monoxide, (r) carbon dioxide, (s) sulfur dioxide, (t) water, (u) nitrogen, ( ) hydrogen cyanide

Solution

To compute the outlet temperature, use

Tout  H / C p  Tfeed

But first the heat capacity at 373.15 K must be determine for each compound using the Table C.1 and Eqn 4.5.

Cp R

 A  BT  CT 2  DT 2

This can easily be done in a spreadsheet. species

10 B

10 C

10 D 5

T out (K)

Methane

1.702

9.081

2.164

4.79

39.82

674.52

Ethane

1.131

19.225

5.561

7.53

62.61

564.82

Propane

1.213

28.785

8.824

10.73

89.17

507.72

n-Butane

1.935

36.915

11.402

14.12

117.41

475.35

n-Hexane

3.025

53.722

16.791

20.73

172.38

442.76

n-Octane

4.108

70.567

22.208

27.35

227.37

425.93

Propylene

1.637

22.706

6.915

9.15

76.05

530.95

1-Pentene

2.691

39.753

12.447

15.79

131.29

464.55

1-Heptene

3.768

56.588

17.847

22.40

186.22

437.59

1-Octene

4.324

64.96

20.521

25.71

213.72

429.30

Acetylene

6.132

1.952

1.299

6.86

57.04

583.54

Benzene

0.206

39.064

13.301

12.52

104.08

488.45

Ethanol

3.518

20.001

6.002

10.15

84.35

515.41

Styrene

2.05

50.192

16.662

18.46

153.47

451.34

Formaldehyde

2.264

7.022

1.877

4.62

38.43

685.37

Ammonia

3.578

3.02

0.186

4.70

39.12

679.92

Carbon monoxide

3.376

0.557

0.031

3.58

29.80

775.89

Carbon dioxide

5.457

1.045

1.157

5.85

48.61

620.01

Sulfur dioxide

5.699

0.801

1.015

6.00

49.87

613.79

Water

3.47

1.45

0.121

4.01

33.35

732.99

Nitrogen

3.28

0.593

0.04

3.50

29.11

785.38

Hydrogen cyanide

4.736

1.359

0.725

5.24

43.59

648.43

Solution 4.4 Problem Statement If 250(ft)3(s) of air at 122(°F) and approximately atmospheric pressure is preheated for a combustion process 1

to 932(°F), what rate of heat transfer is required?

Solution

Looking in table A.2, we are reminded that the gas constant is R  0.7302 ft atm (lb mol) R in the sort of 3

units used in this problem, so we can convert the volumetric flow rate to a molar flow rate using the ideal gas law (air at atmospheric conditions is very nearly an ideal gas) n  PV/RT  1 atm * 250 ft s / (10.7302 ft atm lbmol R * (122R  459.7R)  3

n  0.59 lbmol s To find the heat needed to heat the air at constant pressure from 122°F to 932 °F, we need to integrate the heat capacity over that temperature range 932  F

Q   nC p dT 122  F

The ideal gas heat capacity for air is given in table C.1 as C p  R  A  BT  DT 2 

T must be in Kelvin for use in this expression, so we convert 122°F to 323.15 K and 932 °F to 773.15 KWe then have 773 K

Q  nR  A  BT  DT 2 dT 323 K 773 K

 B D Q  nR  AT  T 2    2 T  323 K

Solution continued on next page…

Putting in n  0.59 lbmol s, R  1.986 Btu lbmol R, A  3.355, B  0.57510 K, D  1600 K gives 3

  1 1  Q  0.59 lbmol s1 *1.986 Btu lbmol1 R 1 3.355773  323  2.88 104 7732  3232   1600    K  1.8  R/K   773 323    Q  3478 Btu s1

Note that because the heat capacity integral comes out with units of K, while we are using R with units of °R, we have to multiply by 1.8 °R/K.

Solution 4.5 Problem Statement How much heat is required when 10,000 kg of CaCO3 is heated at atmospheric pressure from 50°C to 880°C?

Solution

How much heat is required when 10,000 kg of CaCO3 is heated at atmospheric pressure from 50°C to 880°C?

The number of moles of CaCO3 is 10 g/ 100.09 g mol  99910.08 mol CaCO3Evaluating the heat capacity 7

integral from 323 to 1153 K, we get H/R  11350 K T1 (K)

T2 (K)

323

1153

12.572

B (1/K)

C (1/K )

D (K )

ICPH (K)

2.64E03 0.00E00 3.12E05 1.182E04

So, 99910.08 mol * 8.314 J mol K * 11350 K  9.4310 J  9.4310 kJ. 1

Solution 4.6 Problem Statement If the heat capacity of a substance is correctly represented by an equation of the form, CP  A  BT  CT

show that the error resulting when ⟨CP⟩H is assumed equal to CP evaluated at the arithmetic mean of the initial and final temperatures is C(T2  T1) /12. 2

Solution

For consistency with the problem statement, we rewrite Eq. (4.8) as:

C   A  B2 T   1  C3 T     1 p

2 1

where  T2/T1. DefineCPam as the value of CP evaluated at the arithmetic mean temperature Tam.

Then: 2 C pm  A  BTam  CTam

Where

Tam 

T1   1 2

T2 2 and Tam  1  2     4

Whence, B

C   A  2 T   1  4 T   2  1 pm

2 1

Solution continued on next page…

Define  as the difference between the two heat capacities:

  2    1  2  2  1     CP  C pm  CT12    3 4 

This readily reduces to:  

CT12 2   1 12

Making the substitution  T2/T1 yields the required answer. CT12 2 2 C T 2 / T 1  1  T 2  T1 12 12



Solution 4.7 Problem Statement If the heat capacity of a substance is correctly represented by an equation of the form, CP  A  BT  D T

show that the error resulting when ⟨CP⟩H is assumed equal to CP evaluated at the arithmetic mean of the initial and final temperatures is:

D  T2  T1    T1T2  T2  T1 

Solution

For consistency with the problem statement, we rewrite Eq. (4.8) as

C   A  2 T   1  T p

where ≡ T2/ T1. De

2 1

CPam as the value of CP evaluated at the arithmetic mean temperature Tam.

C pm  A  BTam 

D 2 Tam

As in the preceding problem, Tam 

T1   1 2

T2 2 and Tam  1  2     4

Whence,

C   A  B2 T   1  pm

4D T   2  1 2 1

Define  as the difference between the two heat capacities:

  C P   C pm 

 D  1 4  2  2 T1     2  1 

This readily reduces to: 2 D   1  2  T1    1 

Making the substitution   T2/ T1 yields the required answer. 2 D T2  T1      T2 T1  T2  T1 

Solution 4.8 Problem Statement Calculate the heat capacity of a gas sample from the following information: The sample comes to equilibrium in a flask at 25°C and 121.3 kPa. A stopcock is opened briefly, allowing the pressure to drop to 101.3 kPa. With the stopcock closed, the flask warms, returning to 25°C, and the pressure is measured as 104.0 kPa. Determine CP in J·mol ·K assuming the gas to be ideal and the expansion of the gas remaining in the flask to be 1

reversible and adiabatic.

Solution

Let step 12 represent the initial reversible adiabatic expansion, and step 23, the final constant-volume heating.

T1 

T3 

K P1 

T2  T3 Cp 

J  mol K

P2 

kPa

P3 

kPa

P2 P3



T2 

Given

P  T2  T3  2   P3 

R /C p

Solve for Cp

C p  56.95

J mol K

Solution 4.9 Problem Statement A process stream is heated as a gas from 25°C to 250°C at constant P. A quick estimate of the energy requirement is obtained from Eq. (4.3), with CP taken as constant and equal to its value at 25°C. Is the estimate of Q likely to be low or high? Why?

Eq. 4.3

Q  H   C p dT T1

Solution

Except for the noble gases [Fig. (4.1)], Cp increases with increasing T. Therefore, the estimate is likely to be low.

Solution 4.10

Problem Statement

(a) For one of the compounds listed in Table B.2 of App. B, evaluate the latent heat of vaporization Hn by Eq. (4.13). How does this result compare with the value listed in Table B.2? (b) Handbook values for the latent heats of vaporization at 25°C of four compounds are given in the table. For one of these, calculate Hn using Eq. (4.14), and com-pare the result with the value given in Table B.2.

Eq. 4.13:

Hn RTn



1.092(lnPc  1.013) 0.930  Trn

Eq. 4.14:

0.38 H2  1  Tr2     H1  1  Tr1 

Solution

(a) In this part, we use equation 4.14 (Watson’s method) for computing the heat of vaporization at one reduced temperature at another reduced temperatureThe critical temperature for n-pentane (from appendix B) is 469.7 KSo, 25°C corresponds to a reduced temperature of Tr1  298.15/469.7  0.6348The normal boiling point for n-pentane (also from appendix B) is 309.2 K, which corresponds to a reduced temperature Tr2  309.2/469.7  0.6583So, using equation 4.14, we have 0.38

1  T  r2  H2  H1    1  Tr 1 

1  0.6583   366.3  1  0.6348 

0.38

 357.1 J g1

This is essentially exact agreement with the handbook value (it differs by 0.03%). Solution continued on next page…

(b) In this part, we estimate the heat of vaporization at the normal boiling point without using the information on the heat of vaporization at 25°C, using equation 4.13 (Reidel’s method)In this method, we also use the critical pressure, Pc  33.70 bar for n-pentaneThen, using equation 4.13:  1.092ln P  1.013  1.092ln33.7  1.013   c 1 H n  RTn    8.314 * 309.2    25876 J mol 0.930  Trn    0.930  0.6583 

Dividing by the molecular weight of n-pentane (72.15 g/mol) gives Hn  358.6 J gThis is just 0.4% higher than the handbook value, and it only required knowledge of the critical properties and the normal boiling point temperature.

Solution 4.11 Problem Statement Table 9.1 lists the thermodynamic properties of saturated liquid and vapor tetrafluoroethane. Making use of the vapor pressures as a function of temperature and of the saturated-liquid and saturated-vapor volumes, calculate the latent heat of vaporization by Eq. (4.12) at one of the following temperatures and compare the result with the latent heat of vaporization calculated from the enthalpy values given in the table.

(a) 16°C, (b) 0°C, (c) 12°C, (d) 26°C, (e) 40°C. Eq. 4.12:

H  TV

dP sat dT

Solution We want to evaluate the latent heat of vaporization using the Clapeyron equation:

H  TV

dP sat dT

(a) At 16°C, we see that the vapor volume is 0.12551 m kg and the liquid volume is 0.000743 m kg, 3

so V  0.124767 m kgNow, we can estimate the slope of the vapor pressure curve from the values 3

at 18, 16, and 14 °C, which are 1.446,1.573, and 1.708 bar, respectivelySo, we could estimate the slope as

sat



P T



1.708bar  1.446bar (273.15  (14)) K  (273.15  (18)) K

 00.0655bar.K 1  0.0655 * 10 2 kPa.K 1

So, we have

H  (273.15  ( 16))K * 0.124767m 3 .kg 1 * 0.0655 * 10 2 kPa.K 1  210.4kPa.m 3 .kg 1  210.4kJ.kg 1

(b) At 0°C, we see that the vapor volume is 0.069309 m kg and the liquid volume is 0.000722 m kg, so 3

V  0.068587 m kgNow, we can estimate the slope of the vapor pressure curve from the values at 3

2, 0, and 2 °C, which are 2.722,2.928, and 3.146 bar, respectivelySo, we could estimate the slope as

dP sat P 3.146bar  2.722bar    0.106bar.K 1  0.106 * 10 2 kPa.K 1 dT T (273.15  (2)) K  (273.15  (2)) K

So, we have H  (273.15  (0))K * 0.068587m 3 .kg 1 * 0.106 * 10 2 kPa.K 1  198.59kPa.m 3 .kg 1  198.59kJ.kg 1

Solution continued on next page…

V  0.045535 m kgNow, we can estimate the slope of the vapor pressure curve from the values at 3

10, 12, and 14 °C, which are 4.146,4.43, and 4.732 bar, respectivelySo, we could estimate the slope as dP

sat



P T



4.732bar  4.146bar (273.15  (14)) K  (273.15  (10)) K

 0.1465bar.K

1

 0.1465 * 10 2 kPa.K 1

So, we have H  (273.15  (12))K * 0.045535m 3 .kg 1 * 0.1465 * 10 2 kPa.K1  190.22kPa.m 3 .kg 1  190.22kJ.kg 1

(d) At 26°C, we see that the vapor volume is 0.029998 m kg and the liquid volume is 0.000831 m kg, so 3

V  0.029167 m kgNow, we can estimate the slope of the vapor pressure curve from the values at 3

24, 26, and 28 °C, which are 6.458,6.854, and 7.269 bar, respectivelySo, we could estimate the slope as dP sat P 7.269bar  6.458bar    0.20275bar.K 1  0.20275 * 10 2 kPa.K 1 dT T (273.15  (28)) K  (273.15  (24)) K

So, we have H  (273.15  (26)K * 0.029167m 3 .kg 1 * 0.20275 * 10 2 kPa.K 1  176.91kPa.m 3 .kg 1  176.91kJ.kg 1

(e) At 40°C, we see that the vapor volume is 0.019966 m kg and the liquid volume is 0.000872 m kg, so 3

V  0.019094 m kgNow, we can estimate the slope of the vapor pressure curve from the values at 3

35, 40, and 45 °C, which are 8.87, 10.166 and 11.599 bar, respectivelySo, we could estimate the slope as dP sat P 11.599bar  8.87bar    0.2729bar.K 1  0.2729 * 102 kPa.K 1 dT T (273.15  (45)) K  (273.15  (35)) K

So, we have H  (273.15  (40)K * 0.019094m 3 .kg 1 * 0.2729 * 10 2 kPa.K 1  163.17kPa.m 3 .kg 1  163.17kJ.kg 1

Solution 4.12 Problem Statement Handbook values for the latent heats of vaporization in J·g are given in the table for 1

three pure liquids at 0°C.

H at 0° C lv

Chloroform

270.9

Methanol

1189.5

Tetrachloromethane

217.8

For one of these substances, calculate: (a) The value of the latent heat at Tn by Eq. (4.14), given the value at 0°C. (b) The value of the latent heat at Tn by Eq. (4.13). By what percentages do these results differ from the value listed in Table B.2 of App. B?

Eq. 4.13:

Hn RTn



1.092(lnPc  1.013) 0.930  Trn

Eq. 4.14: 0.38 H2  1  Tr2     H1  1  Tr1 

By what percentages do these results differ from the value listed in Table B.2 of App. B?

Solution

(a) Equation 4.14 is used to estimate the latent heat at a temperature of interest from the (known) latent heat at some other temperature, and may be written as: 0.38

1  T  r2  H2  H1    1  Tr 1 

For chloroform, the critical temperature is 536.4 K and the normal boiling point is 334.3 KAt 273 K, Tr1  0.5089 and at 334.3 K, Tr2  0.6232So, 0.38

 1  0.6232   H 2  270.9 J g 1   1  0.5089 

 245.0 J g 1

This is 0.8% lower than the handbook value. Similarly for methanol, Tr1  0.5326 and Tr2  0.6592, so 0.38

 1  0.6592   H2  1189.5 J g   1  0.5326  1

 1055.0 J g1

This is 4.1% below the handbook value.

Similarly for tetrachloromethane (also known as carbon tetrachloride!), Tr1  0.4909 and Tr2  0.6287, so 0.38

 1  0.6287   H 2  217.8 J g   1  0.4909  1

 193.2 J g 1

This is 0.5% below the handbook value. Solution continued on next page…

(b) Equation 4.13 is used to estimate the heat of vaporization at the normal boiling point without knowing the heat of vaporization at any temperatureIt can be written as 1.092ln P  1.013  c  Hn  RTn    0.930  Tr  n

where Pc, the critical pressure, must be given in barFor chloroform, we get

Hn  8.314 J mol

1

 1.092ln54.72  1.013    29570 J mol1 334.3 K   0.930  0.6232  

dividing by the molecular weight of 119.4 g mol gives 247.7 J gThis is just 0.3% above the handbook value.

Likewise, for methanol:

 1.092ln 80.97  1.013    38302 J mol1 Hn  8.314 J mol1 K 1 * 337.9 K   0.930  0.6592   and dividing by the molecular weight of 32.042 g mol gives 1194.2 J gThis is 8.6% above the handbook value.

Finally, for carbon tetrachloride,

 1.092ln 45.6  1.013    29587 J mol1 Hn  8.314 J mol1 K1 349.8 K   0.930  0.6287   and dividing by the molecular weight of 153.822 g mol gives 192.3 J gThis is 1% below the handbook value.

Solution 4.13

Problem Statement

Table B.2 of App. B provides parameters for an equation that gives Psat as a function of T for a number of pure compounds. For one of them, determine the heat of vaporization at its normal boiling point by application of sat

Eq. (4.12), the Clapeyron equation. Evaluate dP /dT from the given vapor-pressure equation, and use generalized correlations from Chapter 3 to estimate V. Compare the computed value with the value of Hn listed in Table B.2. Note that normal boiling points are listed in the last column of Table B.2.

Eq. 4.12: H T V

dP sat dT

Solution

Again, we want to use the Clapeyron equation to find the latent heat of vaporization from the vapor pressure curve and molar volumes.

First, we should find the normal boiling point by setting P  101.3 kPa sat

Taking the derivative of the Antoine equation for the vapor pressure gives   dP sat d  B  B B  BP sat      exp A  exp  A  2  dT dT  T  C  T  C  T  C  T  C 2 

Solution continued on next page…

or, using the numbers for benzene dP sat 2772.78P sat  2 dT T  53.00

For T  353.2 K and P  101.325 kPa this gives sat

dP sat 2772.78101.325   3.118 kPa K1 2 dT 353.2  53.00

Now, we can use generalized correlations to find the liquid and vapor volumesThe critical properties of benzene are Tc  562.2 K, Pc  48.98 bar,   0.210, Zc  0.271, and Vc  259 cm mol. 3

For the volume of the saturated liquid, we can use the Rackett equation, which gives

sat

0.2857

1Tr 

 Vc Zc

0.2857 1353.2   562.2   

 96.8 cm3 mol1

 2590.271

Since the reduced pressure is only 1.013/48.98  0.0207, we expect that the vapor will not be far from ideal, even though the reduced temperature is only 353.2/562.2  0.6282We can get the compressibility from the Pitzer correlation as in the previous homework: 0.422  0.8049 0.62821.6 0.172 B 1  0.139   1.0730 0.6282 4.2 B 0  0.083 

and Z  1  0.8049  0.210 1.0730

0.02068  0.9661 0.6282

So, V  ZRT/P  0.9661*83.145*353.2/1  28370 cm mol. 3

So, V  28370 – 82.6  28288 cm molFinally, putting this all together, we have 3

H  353.2 K  28288 cm3 mol1 3.118 kPa K1  3.115 107 kPa cm3 mol1  3.11510 4 J/mol So, our final answer is H  31.15 kJ/mol. Solution continued on next page…

7th edition Prob. 4.12, choose ethylbenzene We want to use the Clapeyron equation to find the latent heat of vaporization from the vapor pressure curve and molar volumes.

First, we should find the normal boiling point by setting P  101.3 kPaThat gives sat

ln 101.325  13.9726  T

3259.93 T  212.3

3259.93  212.3  136.2 C 13.9726  ln 101.325

Reassuringly, this agrees with the value given in Table B.1 exactly (to the precision given).

Taking the derivative of the Antoine equation for the vapor pressure gives sat    dP sat d  B  B  A  B   BP    exp exp  A  dT dT  T  C  T  C 2 T  C  T  C 2  

or, using the numbers for ethylbenzene dP sat 3259.93P sat  2 dT T  212.3

For T  136.2 °C and P  101.325 kPa this gives sat

dP sat 3259.93101.325   2.720 kPa K1 2 dT 136.2  212.3

Now, we can use generalized correlations to find the liquid and vapor volumesThe critical properties of ethylbenzene are Tc  617.2 K, Pc  36.06 bar,   0.303, Zc  0.263, and Vc  374 cm mol. 3

For the volume of the saturated liquid, we can use the Rackett equation, which gives V

sat

0.2857

1Tr 

 Vc Z c

0.2857 1 409.3   617.2 

 374  0.263

 140.5 cm 3 mol1

Solution continued on next page…

Because the reduced pressure is only 1.013/36.06  0.0281, we expect that the vapor will not be far from ideal, even though the reduced temperature is only 353.2/562.2  0.6632We can get the compressibility from the Pitzer correlation: 0.422  0.7311 0.66321.6 0.172 B 1  0.139   0.8262 0.6632 4.2 B 0  0.083 

and Z  1  0.7311  0.303 0.8262

0.02809  0.9584 0.6632

So, V  ZRT/P  0.9584*83.145*409.3/1.013  32198 cm mol. 3

So, V  32198 – 140.5  32058 cm molFinally, putting this all together, we have 3

H  409.35 K 32058 cm3 mol1 2.720 kPa K1  3.569107 kPa cm3 mol1  3.56910 4 J/mol

So, our final answer is H  35.7 kJ/mol. This agrees with the value in the table to within a few tenths of a percent – essentially perfect agreement.

Solution 4.14 Problem Statement A method for determination of the second virial coefficient of a pure gas is based on the Clapeyron equation and measurements of the latent heat of vaporization H , the molar volume of saturated liquid V , and the lv

vapor pressure P . Determine B in cm ·mol for methyl ethyl ketone at 75°C from the following data at this sat

temperature: H

 31,600 J·mol

V  96.49 cm ·mol 

ln P / kPa  48.158  5623 / T  4.705 ln T sat

[T  K]

Solution

Using the equation above, solve for the pressure

P

5623  T



lnT 

kPa

Differentiate the above equation and determine dP/dT

 5623 4.705       (348.15)2 348.15 

 5623 4.705  dP  P   2    T dT T 

kPa  K

bar K

Using the Clapeyron equation, solve for V

V  Vliq 

H  dP T dT

cm 3 mol

cm  mol

 1



cm3 mol

Then using Eq. 3.37, Solve for B

 PV  B  V       RT 

       

kPa  kPa cm3 mol1 K 1 

cm3 mol

       K  

cm3 mol

Solution 4.15 Problem Statement

One hundred kmol per hour of subcooled liquid at 300 K and 3 bar is superheated to 500 K in a steady-flow heat exchanger. Estimate the exchanger duty (in kW) for one of the following: (a) Methanol, for which T  368.0 K at 3 bar. sat

(b) Benzene, for which T  392.3 K at 3 bar. sat

Solution To find the total heat exchanger duty, we must consider three process steps: (1) heating of the subcooled liquid from 300 K to its saturation temperature at 3 bar, (2) vaporization at the saturation temperature, and (3) heating of the vapor from the saturation temperature to 500 K. For the first step, we do a heat capacity integral from 300 K to 368.0 K, using the liquid phase heat capacity:

ICPH  T0 ,T;A,B,C,D  

R

T0(K) 300

T (K) 368.00

A 13.431

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K2) D (K2) B (1/K) ICPH (K) D Hig (J/mol) -5.13E-02 1.31E-04 0.00E+00 746.8 6209.3

Solution continued on next page…

For the second step, we need the heat of vaporization at 3 bar and 368.0 KIn table B.2, we find the heat of vaporization at the normal boiling point (337.9 K) is 35.21 kJ/molTo find the heat of vaporization at 368 K, we can apply the Watson equation (p. 134 of SVNA)This gives:  H2  RTn  

Pc   Tr 1

H 2 

1



          

 

     

1

H 2 

Finally, for the third step, we have: T2

ICPH  T0 ,T;A,B,C,D  

R

T0(K) 368

T (K) 500.00

A 2.211

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K2) D (K2) B (1/K) ICPH (K) D Hig (J/mol) 1.22E-02 -3.45E-06 0.00E+00 905.2 7526.6

Comparing these, we see that the vaporization step has the largest contribution to the overall heat requirementAdding them gives Q  H  62.64 kJ/mol/If the total methanol flow is 100 kmol/hr  27.78 mol/s, then the total heat exchanger duty is 62.64*27.78  1740.13 kW.

Solution 4.16 Problem Statement For each of the following substances, compute the final temperature when heat in the amount of 60 kJ·mol is 1

added to the subcooled liquid at 25°C at atmospheric pressure. (a) methanol (b) ethanol (c) benzene

(d) toluene (e) water

Solution

Need to gather several things for this problem, the boiling point, the latent heat of vaporization, and the heat capacities for each compound. Using the equation Htotal  C p, vap Tfinal  Tboiling  H latent  C p,liq Tboiling  Tinitial 

And solving it for Tfinal.

(a) Methanol – Tboiling  337.85 K, Cp,liq  81.46 J/mol K, Cp,vap  46.11 J/mol K, H latent  35210 J/mol So for the Liquid: H  C p,liq Tboiling  Tinitial  







J mol

Now plugging into the first equation and solving for Tfinal gives: Tfinal 

H

total

Hlatent Hliquid  C p vap

 Tboiling 

60000  35210  3233.96  46.11



(b) Ethanol – Tboiling  351.4 K, Cp,liq  111.77 J/mol K, Cp,vap  74.39 J/mol K, H latent  38560 J/mol So for the Liquid: H  C p,liq Tboiling  Tinitial  







J mol

Now plugging into the first equation and solving for Tfinal gives: Tfinal 

H

total

H latent H liquid  C p vap

 Tboiling 

60000  38560  5951.75  74.39



Solution continued on next page…

(c) Benzene – Tboiling  353.15 K, Cp,liq  134.33 J/mol K, Cp,vap  85.29 J/mol K, H latent  30720 J/mol So for the Liquid: H  C p,liq Tboiling  Tinitial  







J mol

Now plugging into the first equation and solving for Tfinal gives: Tfinal 

H

total

Hlatent H liquid  C p vap

 Tboiling 

60000  30720  7388.15  85.29



(d) Toluene – Tboiling  383.75 K, Cp,liq  154.73 J/mol K, Cp,vap  107.43 J/mol K, H latent  33180 J/mol So for the Liquid: H  C p,liq Tboiling  Tinitial  







J mol

Now plugging into the first equation and solving for Tfinal gives: Tfinal 

H

total

H latent H liquid  C p vap

 Tboiling 

60000  33180  13244.89  107.43



(e) Water – Tboiling  373.15 K, Cp,liq  75.40 J/mol K, Cp,vap  33.57 J/mol K, H latent  40660 J/mol So for the Liquid: H  C p ,liq Tboiling  Tinitial  







J mol

Now plugging into the first equation and solving for Tfinal gives: Tfinal 

H

total

H latent H liquid  C p vap

 Tboiling 

60000  40660  5655  33.57



Solution 4.17

Problem Statement

Saturated-liquid benzene at pressure P1  10 bar ( T1sat  451.7K) is throttled in a steady-flow process to a pressure P2  1.2 bar ( T2sat  358.7K), where it is a liquid/ vapor mixture. Estimate the molar fraction of the exit stream that is vapor. For liquid benzene, CP  162 J·mol ·K . Ignore the effect of pressure on the 1

enthalpy of liquid benzene.

Solution

Tc 

Pc 

bar Tn 

T1 sat 

J mol K

Cp 

Estimate Hv using Riedel equation (4.13) and Watson correction (4.14) Trn 

Tn Tc



Tr 2 sat 

T2 sat Tc



 P  1.092(ln  c   1.013)  bar  kJ H n  RTn  30.588 0.930  Trn mol

 1  T  r 2 sat  H v  H n    1  Trn 

0.38

 30.28

kJ mol

Assume the throttling process is adiabatic and isenthalpic. x

C p T1sat  T2 sat  H v

 0.498

Solution 4.18 Problem Statement

Estimate

H f298 for one of the following compounds as a liquid at 25°C.

(a) Acetylene, (b) 1,3-Butadiene, (c) Ethylbenzene, (d) n-Hexane, (e) Styrene.

Solution

In each case, we can look up the heat of formation of the compound as a gas in table C.4, where we find Hf,(acetylene(g))  227480 J mol, Hf(1,3-butadiene(g))  109240 J mol, Hf(ethylbenzene(g))  29920 J mol, Hf,(n-hexane(g))  166920 J mol, Hf(styrene(g))  147360 J molThe heat of formation of the liquid is equal to the heat of formation of the gas minus the heat of vaporization at 25°CIf we only know the critical properties and normal boiling point of each species, we could estimate this by first using equation 4.12 to estimate the heat of vaporization at the normal boiling point of each species, and then using equation 4.13 to estimate the heat of vaporization at 25°CWe could, if we like, combine the equations to get  1  T    r (25 C)  H25 C  Hn    1  Tr   

0.38

1.092ln P  1.013  1  Tr (25 C)   c  RTn     0.930  Tr  1  Trn    n

0.38

Solution continued on next page…

This is getting a bit cumbersome to type into the old calculator, so let's evaluate it in a spreadsheetDoing so gives:

Species

Hn

H25C

Hf (g)

Hf (l)

(J mol)

0.9671

16911

6638

227480

220842

0.6319

0.7012

22450

20740

109240

88500

409.4

0.6633

0.4831

35852

42196

29920

12276

30.25

341.9

0.6736

0.5874

29010

31712

166920

198632

38.4

418.3

0.6577

0.4688

36753

43434

147360

103926

(K)

(bar)

(K)

308.3

61.39

189.4

0.6143

1,3-butadiene 425.2

42.77

268.7

Ethylbenzene 617.2

36.06

n-hexane

507.6

Styrene

636

acetylene

Trn

Tr25C

In the final column above, we have subtracted the latent heat of vaporization at 25 °C from the heat of formation of each species in the gas phase to get the heat of formation of each species in the liquid phase.

Solution 4.19 Problem Statement Reversible compression of 1 mol of an ideal gas in a piston/cylinder device results in a pressure increase from 1 bar to P2 and a temperature increase from 400 K to 950 K. The path followed by the gas during compression is given by PV  const, and the molar heat capacity of the gas is given by: 1.55

CP /R

 3.85  0.57 × 10 T 3

[TK]

Determine the heat transferred during the process and the final pressure.

Solution

4.19 (7 edition Prob. 4.17)

1st law: dQ  dU  dW  CV·dT  P·dV

(A)

Ideal gas: P·V  R·T and P·dV  V·dP  R·dT

Whence V·dP  R·dT  P·dV

Since

P·V  const

then

(B)

P· V  ·dV   V ·dP 1

from which V·dP  P· ·d·V

Combines with (B) to yield:

PdV 

Rdt 1

Combines with (A) to give: dQ  C p dT  RdT 

RdT 1

Which reduces to: dQ  C p dT 

 RdT 1

Or Cp   dQ     RdT  R 1   

C 

Since CP is linear in T, the mean heat capacity is the value of CP at the arithmetic mean temperature. Thus

C p  3.85  0.57 * 103 * Tam

With T2 

T1 

K Tam 



, and integrating (C):

C p   dQ     R T1  T2    R 1   

J mol



P2  P3

T2 1 T1



bar

Solution 4.20 Problem Statement Hydrocarbon fuels can be produced from methanol by reactions such as the following, which yields 1-hexene: 6CH3OH(g)

→

C6 H12 (g)  6H2O(g)

Compare the standard heat of combustion at 25°C of 6 CH3OH(g) with the standard heat of combustion at 25°C of C6H12(g) for reaction products CO2(g) and H2O(g).

Solution The heat of combustion of 6 CH3OH(g) is the heat of reaction for 6 CH3OH(g)  9 O2(g)  6 CO2(g)  12 H2O(g) This is given by Hrxn  6*Hf(CO2(g))  12*Hf(H2O(g))  6*Hf(CH3OH(g))  6*(393.51)  12*(241.82) 6*(200.66)  4059 kJ mol. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The heat of combustion of C6H12(g) (assumed to be 1-hexene, though it doesn't really say and there are many other possible C6H12 isomers) is given by the heat of reaction for C6H12(g)  9 O2  6 CO2(g)  6 H2O(g) This is given by Hrxn  6*Hf(CO2(g)) 6*Hf(H2O(g))  Hf(C6H12(g))  6*(393.51)  6*(241.82)  (41.95)  3770 kJ mol.

The difference between these two heats of combustion is the heat of reaction for the condensation reaction given in the problem statementThat is, subtracting the second combustion reaction above from the first gives 6 CH3OH(g)  C6H12(g)  6 H2O(g) For which the heat of reaction is 4059  (3770)  289 kJ/mol.

The heat of combustion of ethylene at 25°C (with water vapor product) is the heat of reaction for C2H4(g) 3 O2(g)  2 CO2(g)  2 H2O(g) This is given by Hrxn  2*Hf(CO2(g))  2*Hf(H2O(g))  Hf(C2H4(g))  2*(393.51)  2*(241.82) (52.51)  1323 kJ mol  1.32310 J mol 6

The only difference in each case is how much excess O2 and N2 have to be heated up by the heat released by reaction. (a) The products (per mole of ethylene burned) will be 2 moles of CO2  2 moles of H2O  3*(79/21)  11.3 moles of N2 that was included with the 3 moles of O2So, we integrate the total heat capacity of this mixture from 298 K to some final temperature and vary the final temperature until we get H/R  1.32310 J mol/8.314 J mol K  159120 KDoing so, using our handy-dandy heat capacity 6

integrating spreadsheet gives 2533 K. (b) Similarly, here the products are 2 moles of CO2  2 moles of H2O  0.25*3  0.75 moles of O2  3*1.25*(79/21)  14.1 moles of N2, and integrating the heat capacity of this mixture to get get H/R  159120 K gives a final temperature of 2198 K. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(c) and (d) are the same as (a) and (b), but with different numbers that lead to final temperatures of 1951 and 1609 K, respectively (e) Here, we can add the heat required to heat the air from 298K to 773K to the heat of reactionWe can imagine a path in which we cool the air from 773K to 298K (removing heat H1  309424 J per mol C2H4 burned), carry out the reaction at 298K (releasing the standard heat of reaction, Hrxn  1.32310 ) and 6

then heat the products with both the heat removed to cool the air to 298K and the heat of reaction (Htot  1.66310 J per mol C2H4 burned), so the total Htot /R is 196330 KBoth the heat removed to cool the air 6

from 773 to 298 K and the heat required to heat the reaction products to the final temperature are again computed using the heat capacity integrating spreadsheetThis gives a final temperature of 2282 K.

NOTE: A problem with this homework problem is that several of the final temperatures are above 2000 K, whereas the heat capacity expressions used (from table C.1 in the book) are only valid up to 2000 KSo, the results for parts (a), (b), and (e) are probably all incorrect (they are the correct answers to the homework problem, but probably are not good values for the actual adiabaic flame temperature of ethylene).

Solution 4.21 Problem Statement Calculate the theoretical flame temperature when ethylene at 25°C is burned with:

(a) The theoretical amount of air at 25°C. (b) 25% excess air at 25°C. (c) 50% excess air at 25°C. (d) 100% excess air at 25°C. (e) 50% excess air preheated to 500°C. (f) The theoretical amount of pure oxygen.

Solution

C2H4  3O2  2CO2  2H2O(g)

H298 [2·(241818)  2·(393509)  52510]  1323164 J/mol

Parts (a) – (d) can be worked exactly as Example 4.7. However, with Mathcad capable of doing the iteration, it is simpler to proceed differently.

Solution continued on next page…

Index the product species with the numbers: 1  oxygen 2  carbon dioxide 3  water (g) 4  nitrogen (a) For the product species, no excess air:  0     2    n  2   11.286  

i

For the products

3.639   5.457    A  3.470   3.280   

A   i ni Ai 

H p  R 

 0.506   1.045  103   B  K 1.450    0.593  

B   i ni Bi 

T0 

1 K

0.227    1.157    D   0.121    0.040   

D   i ni Di  



The integral is given by Eq. (4.8). Moreover, by an energy balance,

H298  HP  0

 B D    1  Given H298  R  AT0   1  T02  2  1    2 T0    

Solving for  

T

Solution continued on next page…

Parts (b), (c), and (d) are worked the same way, the only change being in the numbers of moles of products. (b) nO2  0.75

nN2  14.107

T  2198.6·K Ans.

nN2  16.929

T  1950.9·K Ans.

(d) nO2  3.0

nN2  22.571

T  1609.2·K Ans.

(e) 50% excess air preheated to 500 degC. For this process,

Hair  H298  HP  0

Hair  MCPH·(298.15  773.15)

For one mole of air:

MCPH (773.15,298.15, 3.355, 0.575·10 , 0.0, 0.016·10 )  3.65606 3

For 4.5/0.21  21.429 moles of air: Hair  21.429·8.314·3.65606·(298.15  773.15)  309399 L/mol

The energy balance here gives: H298  Hair  HP  0

 1.5     2   n     2  16.929  

i

A   i ni Ai 

3.639   5.457   A    3.470  3.280   

B   i ni Bi 

 0.506   1.045  103  B    1.450  K  0.593  

1 K

0.227   1.157  D     0.121   0.040   

D   i ni Di  



Solution continued on next page…

 B D    1  Given H 298 H air  R  AT0     T02  2      2 T0    

Solving for  

T

(f) theoretical amount of pure oxygen,

H298  HP  0 1323164 J/mol  HP  0  3   n     2  

i

A   i ni Ai 

3.639    A    3.470   

 0.506   103  B    1.450  K  

1 K

B   i ni Bi 

H 298  R

Cp R

0.227    D     0.121   

D   i ni Di  

 10 5 K 2

* T  T0 

Starting with a guess and solving iteratively yields:

T



H 298 Cp



Solution 4.22 Problem Statement What is the standard heat of combustion of each of the following gases at 25°C if the combustion products are H2O(l) and CO2(g)? Compute both the molar and specific heat of combustion in each case. (a) methane, (b) ethane, (c) ethylene, (d) propane, (e) propylene, (f) n-butane, (g) 1-butene, (h) ethylene oxide, (i) acetaldehyde, (j) methanol, (k) ethanol.

Solution

(a) the heat of combustion of methane gas in the heat of reaction for CH4  2 O2 (g)  CO2 (g)  2 H2O (l) This is given by Hrxn  *Hf(CO2(g))  2*Hf(H2O(l))  Hf(CH4(g))  1*(393.51)  2*(285.83)(74.520)  890.65 kJ mol. (b) The heat of combustion of ethane gas is the heat of reaction for 2 C2H6  7 O2 (g) 4 CO2 (g)  6 H2O (l) This is given by Hrxn  4*Hf(CO2(g))  6*Hf(H2O(l)) – 2*Hf (C2H6 (g))  4*(393.51)  6*(285.83)2*(83.820)  3121.38 kJ mol. (c) The heat of combustion of ethylene gas is the heat of reaction for C2H4  3 O2 (g) 2 CO2 (g)  2 H2O (l) This is given by Hrxn  2*Hf(CO2(g))  2*Hf(H2O(l)) – *Hf (C2H6 (g))  2*(393.51)  2*(285.83)(52.510)  1306.17 kJ mol. (d) The heat of combustion of propane gas is the heat of reaction for C3H8  5 O2 (g) 3 CO2 (g)  4 H2O (l)

Solution continued on next page…

This is given by Hrxn  3*Hf(CO2(g))  4*Hf(H2O(l)) – *Hf (C3H8 (g))  3*(393.51)  4*(285.83) (104.680)  2219.47 kJ mol. (e) The heat of combustion of propylene, gas is the heat of reaction for 2 C3H6  9 O2 (g) 6 CO2 (g)  6 H2O (l) This is given by Hrxn  6*Hf(CO2(g))  6*Hf(H2O(l)) – 2*Hf (C3H6 (g))  6*(393.51)  6*(285.83)2*(19.710)  4036.62 kJ mol. (f) The heat of combustion of n-butane, gas is the heat of reaction for C4H10  13/2 O2 (g) 4 CO2 (g)  5 H2O (l) This is given by Hrxn  4*Hf(CO2(g))  5*Hf(H2O(l)) – *Hf (C4H10 (g))  4*(393.51)  5*(285.83)(125.790)  2877.4 kJ mol. (g) The heat of combustion of 1-butene, gas is the heat of reaction for C4H8  6 O2 (g) 4 CO2 (g)  4 H2O (l) This is given by Hrxn  4*Hf(CO2(g))  4*Hf(H2O(l)) – *Hf (C4H8 (g))  4*(393.51)  4*(285.83)(0.540)  2716.82 kJ mol. (h) The heat of combustion of ethylene oxide, gas is the heat of reaction for C4H8  6 O2 (g) 4 CO2 (g)  4 H2O (l) This is given by Hrxn  4*Hf(CO2(g))  4*Hf(H2O(l)) – *Hf (C4H8 (g))  4*(393.51)  4*(285.83) (0.540)  2716.82 kJ mol. Solution continued on next page…

(i) The heat of combustion of acetaldehyde gas is the heat of reaction for CH3CHO 2.5 O2 (g) 2 CO2 (g)  2 H2O (l) This is given by Hrxn  2*Hf(CO2(g))  2*Hf(H2O(l)) – *Hf (CH3CHO (g))  2*(393.51)  2*(285.83) (166.190)  1192.49 kJ mol. (j) The heat of combustion of methanol gas is the heat of reaction for 2 CH3OH  3 O2 (g) 2 CO2 (g)  4 H2O (l) This is given by Hrxn  2*Hf(CO2(g))  4*Hf(H2O(l)) – 2*Hf (CH3OH (g))  2*(393.51)  4*(285.83)2*(200.660)  1529.02 kJ mol. (k) The heat of combustion of ethanol gas is the heat of reaction for C2H6 O 6 O2 (g) 2 CO2 (g)  3 H2O (l) This is given by Hrxn  2*Hf(CO2(g))  3*Hf(H2O(l)) – *Hf (C2H6O (g))  2*(393.51)  4*(285.83)(235.100)  1695.24 kJ mol.

Solution 4.23 Problem Statement Determine the standard heat of each of the following reactions at 25°C:

Solution

(a) Hrxn  2*Hf(NH3(g))  Hf(N2(g)) – 3*Hf(H2(g))  2*(46.110) –00  92.22 kJ mol. (b) Hrxn  4*Hf(NO(g))  6*Hf(H2O(g)) – 4*Hf(NH3(g) – 5*Hf(O2(g)) Hrxn  4*(90.25)  6*(241.82) – 4*(46.110) – 5*(0) 905.5 kJ mol.

Solution continued on next page…

(c) Hrxn  2*Hf(HNO3(l))  Hf(NO(g)) – 3*Hf(NO2(g))  Hf(H2O(l)) Hrxn  2*(174.1)  (90.25) – 3*(33.18)  (285.830)  71.66 kJ mol. (d) Hrxn  Hf(CaO(s))  Hf(C2H2(g)) – Hf(CaC2(s))  Hf(H2O(l)) Hrxn  635.090  227.480  59.800  285.830  61.98 kJ mol. (e) Hrxn  2*Hf(NaOH(s))  Hf(H2(g))  2*Hf(Na(s)) – 2*Hf(H2O(g)) Hrxn  2*(425.609)  0 – 2*(0) – 2*(241.818) 367.6 kJ mol. (f) Hrxn  7*Hf(N2(g))  12*Hf(H2O(g)) – 6*Hf(NO2(g)) – 8*Hf(NH3(g)) Hrxn  7*(0)  12*(241.818) – 6*(33.18) – 8*(46.110)  2732 kJ mol. (g) Hrxn  Hf(‹(CH2)2›O(g))  Hf(C2H4(g)) – ½ Hf(O2(g)) Hrxn  52630 – 52510 – 0  105140 J mol (h) Hrxn  Hf(‹(CH2)2›O(g))  Hf(C2H2(g)) – Hf(H2O(g)) Hrxn  52630 – 227480 – 241818  38292 J mol (i) Hrxn  Hf(CO2(g))  4*Hf(H2(g))  Hf(CH4(g)) – 2*Hf(H2O(g)) Hrxn   4*(92307) 2*(241818)  1.144 * 10 J mol 5

(j) Hrxn  Hf(CH3OH(g))  Hf(H2O(g))  Hf(CO2(g)) – 3*Hf(H2(g)) Hrxn  200660  241818  393509 – 3*0  48969 J mol (k) Hrxn  Hf(HCHO(g))  Hf(H2O(g))  Hf(CH3OH(g)) – ½ *Hf(O2(g)) Hrxn  108570  241818  200660 – ½*0  149728 J mol (l) Hrxn  2*Hf(H2O(g))  2*Hf(SO2(g))  2*Hf(H2S(g)) – 3 *Hf(O2(g)) Hrxn  2*241818 2*296830 2*20630 – 3*0  1036036 J mol (m) Hrxn  3*Hf(H2(g))  Hf(SO2(g))  Hf(H2S(g)) – 2*Hf(H2O(g)) Hrxn  3*0  296830  20630 – 2*241818  207436 J mol (n) Hrxn  2*Hf(NO(g))  Hf(N2(g)) – Hf(O2(g)) Hrxn  2*90250 – 0 – 0  180500 J mol Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(o) Hrxn  Hf(CaO(s))  Hf(CO2(g)) – Hf(CaCO3(s)) Hrxn  635090 – 393509  1206920 178321 J mol (p) Hrxn Hf(H2SO4(l))  Hf(SO3(g)) – Hf(H2O(l)) Hrxn  813989  395720  285830  132439 J mol (q) Hrxn  Hf(C2H5OH(l))  Hf(C2H4(g)) – Hf(H2O(l)) Hrxn  277690 – 52510  285830  44370 J mol (r) Hrxn  Hf(C2H5OH(g))  Hf(CH3CHO(g)) – Hf(H2(g)) Hrxn  235100 166190 – 0  68910 J mol (s) Hrxn  Hf(CH3COOH(l))  Hf(H2O(l))  Hf(C2H5OH(l)) – Hf(O2(g)) Hrxn  484500 285830  277690 – 0  492640 J mol (t) Hrxn  Hf(CH2:CHCH:CH2(g))  Hf(H2(g))  Hf(C2H5CH:CH2(g)) Hrxn  109240  0  540  109780 J mol (u) Hrxn  Hf(CH2:CHCH:CH2(g))  2*Hf(H2(g))  Hf(C4H10(g)) Hrxn  109240  0  125790  235030 J mol (v) Hrxn  Hf(CH2:CHCH:CH2(g))  Hf(H2(g))  Hf(C2H5CH:CH2(g))  Hf(O2(g))  Hrxn  109240 (241818) (540)  132038 J mol (w) Hrxn  4*Hf(NH3(g)) 6* Hf(NO(g)) – 6*Hf(H2O(g))  5*Hf(N2(g)) Hrxn 4(46110) – 6(90250) – 6*(241818)  0  1807968 J mol (x) Hrxn  2*Hf(HCN(g))  Hf(C2H2(g))  2*(135.1)  (227.48)  42.72 kJ mol. (y) Hrxn  Hf(styrene)  Hf(ethylbenzene)  (147.36)  (29.92)  117.44 kJ mol. (z) Hrxn  Hf(CO(g))  Hf(H2O(l))  (110.525)  (285.830)  175.305 kJ mol.

Solution 4.24 Problem Statement Determine the standard heat for one of the reactions of Prob. 4.23: part (a) at 600°C, part (b) at 50°C, part (f) at 650°C, part (i) at 700°C, part (j) at 590(°F), part (l) at 770(°F), part (m) at 850 K, part (n) at 1300 K, part (o) at 800°C, part (r) at 450°C, part (t) at 860(°F), part (u) at 750 K, part (v) at 900 K, part (w) at 400°C, part (x) at 375°C, part (y) at 1490(°F).

Problem 4.23 Determine the standard heat of each of the following reactions at 25°C:

Solution

(a) A heat of reaction at a temperature other than standard temperature is given by the heat of reaction at standard temperature plus the integral of the difference in heat capacity between products and reactants: o o Hrxn,T  Hrxn ,298.15 K  

T 298.15 K

C

total p,products

 C ptotal ,reactants  dT

A handy spreadsheet for evaluating the integral was provided in the lecture notesPutting in the parameters for the reaction N2  3 H2  2 NH3 We have T

1 IDCPH  T0 ,T; A, B, C, D  R



C p dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









T0 (K) 298.15

T (K) 873.15

DA -5.871

D B (1/K) 4.18E-03

D C (1/K ) 0.00E+00

D D (K ) IDCPH (K) -6.61E+04 -2113.9

Species Name N2 H2 NH3

Stoichiometric coefficient -1 -3 2

A 3.28 3.249 3.578

B (1/K) 5.93E-04 4.22E-04 3.02E-03

C (1/K2) 0.00E+00 0.00E+00 0.00E+00

D (K2) n iAi 4.00E+03 -3.28E+00 8.30E+03 -9.75E+00 -1.86E+04 7.16E+00

Multiplying the result by R gives 17.58 kJ/mol for the integral of CpThe heat of reaction at 298.15 K is just twice the heat of formation of ammonia, because this is the formation reaction for ammonia, written with 2 NH3 as the productThus, H298  46110 J/mol *2  92.22 kJ/molAdding on the integral of Cp gives H600°C  109.80 kJ/mol for the reaction as written.

Solution continued on next page…

(b) Repeating this for the reaction 4 NH3  5 O2  4 NO  6 H2O gives T

1 IDCPH  T0 ,T; A, B, C, D  R

B

C

1

1 

 C dT  A T  T   2 T  T   3 T  T   D  T  T  p

2 0

3 0

T0 (K) 298.15

T (K) 773.15

DA 1.861

D B (1/K) -3.39E-03

D C (1/K ) 0.00E+00

2 D D (K ) IDCPH (K) 2.66E+05 568.8

Species Name NH3 O2 NO H2O

Stoichiometric coefficient -4 -5 4 6

A 3.578 3.639 3.387 3.47

B (1/K) 3.02E-03 5.06E-04 6.29E-04 1.45E-03

C (1/K2) 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) n iAi -1.86E+04 -1.43E+01 -2.27E+04 -1.82E+01 1.40E+03 1.35E+01 1.21E+04 2.08E+01

Multiplying the result by R gives 4.729 kJ/mol for the integral of CpThe heat of reaction at 298.15 K is given by: H298  6*H298(H2O)  4*H298(NO) – 4*H298(NH3) H298  6*(241.818)  4*90.250 – 4*(46.11)  905.47 kJ/mol. Adding on the heat capacity integral gives H500°C  900.74 kJ/mol

(f) A heat of reaction at a temperature other than standard temperature is given by the heat of reaction at standard temperature plus the integral of the difference in heat capacity between products and reactants: o o H rxn ,T  H rxn ,298.15 K  

298.15 K

C

total p ,products

 C ptotal ,reactants  dT

A handy spreadsheet for evaluating this was provided in the lecture notesPutting in the parameters for the reaction 6 NO2(g)  8 NH3(g)  7 N2(g)  12 H2O (g)

Solution continued on next page…

gives: Reference Temperature T0 (K) 298.15

Species Name

Temperature of Interest T (K) 923.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -2732.016 2.037 -2716.380 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T0 (kJ/mol)

1 RT

1

B

C

1

1 

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn,T0  R  A T  T0   2 3  T T0   





B (1/K)

C (1/K )

niCi

n iD i

NO 2

-6

33.18

4.982

1.20E-03

0.00E+00

-7.92E+04 -1.99E+02 -2.99E+01 -7.17E-03

0.00E+00

4.75E+05

NH3

-8

-46.11

3.578

3.02E-03

0.00E+00

-1.86E+04 3.69E+02 -2.86E+01 -2.42E-02

0.00E+00

1.49E+05

3.28

5.93E-04

0.00E+00

4.00E+03 0.00E+00

2.30E+01

4.15E-03

0.00E+00

2.80E+04

H2O

-241.818

3.47

1.45E-03

0.00E+00

1.21E+04 -2.90E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

4.16E+01 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.74E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.45E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DB (1/K) -9.78E-03

DC (1/K ) 0.00E+00

DA 6.084 Note: Light blue fields are inputs, pink fields are the final output.

D (K )



niDHf,i

niAi

niBi

DD (K ) 7.97E+05

The heat of reaction at 650°C is 2716 kJ/molThis is only slightly different from the heat of reaction at standard conditionsIt just happens that the average heat capacity of the reactants and products is about the same over this temperature range.

Solution 4.25

Problem Statement

Develop a general equation for the standard heat of reaction as a function of temperature for one of the reactions given in parts (a), (b), (e), (f), (g), (h), (j), (k), (l), (m), (n), (o), (r), (t), (u), (v), (w), (x), (y), and (z) of Prob. 4.23.

Problem 4.23 Determine the standard heat of each of the following reactions at 25°C:

Solution

This is a simple application of a combination of Eqs. (4.19) & (4.20) with evaluated parameters. In each case the value of H 298 is calculated in Pb. 4.23. The values of A, B, C and D are given for all cases except for 0

Solution continued on next page…

Parts (e), (g), (h), (k), and (z) in the preceding table. Those missing are as follows:

Part No.

A

103 B

106 C

105 D

7.425

20.778

3.737

3.629

8.816

4.904

0.114

9.987

20.061

9.296

1.178

1.704

3.997

1.573

0.234

3.858

1.042

0.18

0.919

Solution 4.26

Problem Statement

Compute the standard heat of reaction for each of the following reactions taking place at 298.15 K in dilute aqueous solution at zero ionic strength.

Solution

Use table C.5 to find the standard enthalpies. (a) Reactants  1262.2 kJ/mol  3627.9 kJ/mol  4890.1 kJ/mol Products  2274.6 kJ/mol  2638.5 kJ/mol  4913.1 kJ/mol Products – Reactants  4913.1 J/mol – (4881.4) J/mol  23 J/mol (b) Reactants  2274.6 kJ/mol Products  2265.9kJ/mol Products – Reactants  8.7 kJ/mol (c) Reactants  5893.8 kJ/mol Products  5956.8 kJ/mol Products – Reactants  63 kJ/mol (d) Reactants  9144.4 J/mol Products  9083.66J/mol Products – Reactants  60.74 kJ/mol (e) Reactants  9144.4 J/mol Products  9200.66 J/mol Products – Reactants  56.26 kJ/mol (f) Reactants  9144.4 kJ/mol Products  9231.66 kJ/mol Products – Reactants  87.26 J/mol (g) Reactants  75.5 kJ/mol Products  571.66 kJ/mol Products – Reactants  496.16 kJ/mol Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(h) Reactants  3941.1 kJ/mol Products  3913.73 kJ/mol Products – Reactants  27.37 kJ/mol (i) Reactants  7957 kJ/mol Products  8399.12 kJ/mol Products – Reactants  441.42 kJ/mol (j) Reactants  9141.6 kJ/mol Products  9231.66 kJ/mol Products – Reactants  90.06 kJ/mol (k) Reactants  9137.4 kJ/mol Products  9231.66 kJ/mol Products – Reactants  94.26 kJ/mol (l) Reactants  4703.8 kJ/mol Products  4707.2 kJ/mol Products – Reactants  3.4 kJ/mol

Solution 4.27 Problem Statement

The first step in the metabolism of ethanol is dehydrogenation by reaction with nicotinamide-adenine dinucleotide (NAD): C2H5 OH  NAD

→

C2H4O  NADH

What is the heat effect of this reaction upon metabolizing 10 g of ethanol from a typical cocktail? What is the total heat effect for complete metabolism of the 10 g of ethanol to CO2 and water? How, if at all, is the perception of warmth that accompanies ethanol consumption related to these heat effects? For computing heat effects, you may neglect the temperature, pH, and ionic strength dependence of the enthalpy of reaction (i.e. apply the enthalpies of formation from Table C.5 of App. C at physiological conditions).

Solution

First determine how many moles of ethanol we have, MWetoh  46.068 g/mol. Moles of EtOH  10 g/ 46.068 g/mol  0.217 mol.

Now, Balance the equation C2H5OH  NAD → C2H4O  NADH  H

H f 298    H 

1

 

H f 298 

1

 1



 

1





1



For the second part, write the equation C2H5OH  NAD → (2CO2  2H2O)  NADH  H

H f 298 

H  

 

1

H f 298 



1



 

1

 

1



1

  

1



1

 



This reaction gives off heat as it proceeds in the reaction.

J·mol

1

Solution 4.28 Problem Statement Natural gas (assume pure methane) is delivered to a city via pipeline at a volumetric rate of 150 million standard cubic feet per day. If the selling price of the gas is $5.00 per GJ of higher heating value, what is the expected revenue in dollars per day? Standard conditions are 60(°F) and 1(atm).

Solution q  150 * 10 6

ft 3 day

T  288.71 K

P 1 atm

The higher heating value is the negative of the heat of combustion with water as liquid product.

Calculate methane standard heat of combustion with water as liquid product:

CH4  2O2  CO2 2H2O

Standard Heats of Formation: H fCH 4  

J mol

H fO 2 

J mol

H fCO 2  

H fH 2Oliq  285830

J mol

H c H fCO   H fH Oliq H fCH   H fO 



J mol

Assuming methane is an ideal gas at standard conditions: nq*

n * Hc *

P mol  1.793 * 10 8 RT day

$5.00 dollar  7.985 * 105 GJ day

Solution 4.29 Problem Statement Natural gases are rarely pure methane; they usually also contain other light hydrocarbons and nitrogen. Determine an expression for the standard higher heat of combustion as a function of composition for a natural gas containing methane, ethane, propane, and nitrogen. Assume liquid water as a product of combustion. Which of the following natural gases has the highest heat of combustion?

(a) y CH 4



0.95, y C2H6



0.02, y C3H8 

0.02, y N2



0.01.

(b) y CH 4



0.90, y C2H6



0.05, y C3H8



0.03, y N2



0.02.



0.85, y C2H6



0.07, y C3H8



0.03, y N2



0.05.

Solution

Calculate methane standard heat of combustion with water as liquid product Standard Heats of Formation:CH4  2O2 > CO2 2H2O H fCH 4  

J mol

H fO 2 

J mol

H fCO 2  

H fH 2Oliq  285830

J mol

H cCH H fCO  H fH Oliq H fCH   H fO 

J mol

Calculate ethane standard heat of combustion with water as liquid product:

Solution continued on next page…

Standard Heats of Formation:C2H6  7/2O2 > 2CO2 3H2O H fC 2 H 6  

J mol

H fO2 

J mol

H fH 2Oliq  285830

H fCO 2  

J mol

7 H cC H  H fCO  H fH Oliq H fC H  H fO  2

J mol

Calculate propane standard heat of combustion with water as liquid product

Standard Heats of Formation:C3H8  5O2 > 3CO2 4H2O H fC 3 H 8  

J mol

H fO 2 

J mol

H fH 2Oliq  285830

H fCO2  

J mol

J mol kJ mol

H cC H   H fCO  H fH Oliq H fC H  H fO 

Calculate the standard heat of combustion for the mixtures (a) H cCH 

H cC H 

H cC H



kJ mol

 H cCH 

 H cC H 

 H cC H



kJ mol

H cCH 

 H cC H 

H cC H



kJ mol

(b)

(c)

Gas b) has the highest standard heat of combustion.

Solution 4.30 Problem Statement If the heat of combustion of urea, (NH2)2CO(s), at 25°C is 631,660 J·mol when the products are CO2(g), H2O(l), 1

and N2(g), what is

H f298 for urea?

Solution

The heat of combustion is defined as the heat of reaction for CO2(g)  2 H2O (l)  N2(g)  (NH2)2CO(s)  3/2 O2(g) Because the number given for the heat of combustion is positive, we know that it is for formation of urea plus oxygen from the combustion products (the reverse of the combustion reaction) So, the heat of combustion at 25°C is Hcomb(298)  ½*Hf(O2(g))  Hf((NH2)2CO(s))  Hf(CO2(g))  2*Hf(H2O(l))  Hf(N2(g)) We know all of the quantities in the above equation except the heat of formation of ureaSo, we can put the numbers in and solve for Hf((NH2)2CO(s))  Hf(CO2(g))  2*Hf(H2O(l))  Hf(N2(g)) –1.5*Hf(O2(g))  Hcomb(298) Hf((NH2)2CO(s))  393.509  2*(285.83)  0 – 1.5*0  631.66  333.5 kJ/mol.

Solution 4.31 Problem Statement The higher heating value (HHV) of a fuel is its standard heat of combustion at 25°C with liquid water as a product; the lower heating value (LHV) is for water vapor as product.

(a) Explain the origins of these terms. (b) Determine the HHV and the LHV for natural gas, modeled as pure methane. (c) Determine the HHV and the LHV for a home-heating oil, modeled as pure liquid n-decane. For n-decane as a liquid

H f298  249,700 J·mol 1.

Solution

(a) There is nothing profound hereIf we compute the heat of combustion with water as a liquid product, then the value we get is higher (larger in absolute value) than if we compute it with water as a vapor productThey differ by the value of the heat of vaporization of water at 25°C, which is 44012 J mol. (b) The heat of combustion for methane with liquid water as the product is the heat of reaction for CH4(g)  3 O2(g)  CO2(g)  2 H2O(l), which is Hrxn  Hf(CO2(g))  2*Hf(H2O(l))  Hf(CH4(g))  393.51  2*(285.83) (74.52)  890.7 kJ mol

Solution continued on next page…

With water vapor as the product, it is 802.6 kJ/molHeating values for gaseous fuels are more commonly expressed in Btu per standard cubic footIf we treat methane as an ideal gas and choose the commonly used 'standard conditions' for natural gas of 60°F and 1 atm, then at there will be n/V  P/RT  1 atm/(0.7302 SCF atm lbmol °R * 519.7 °R)  0.00264 lbmol /SCF 1.195 mol /SCFMultiplying this by the higher heating value in kJ/mol gives 1.195*890.7 1064 kJ/SCF  1009 Btu/SCF (Perry's handbook lists 1012 Btu ft , so we probably did it right)Multiplying it by the lower heating value gives 3

909 Btu/SCFSo, the HHV  1009 Btu/SCF and the LHV  909 Btu/SCF. (c) Similarly, the heat of combustion with liquid water product for n-decane is the heat of reaction for C10H22(l)  15.5 O2(g) 10 CO2(g)  11 H2O(l), which is Hrxn  10*Hf(CO2(g))  11*Hf(H2O(l))  Hf(C10H22(l))  10*(393.51)  11*(285.83) (249.7)  6829.5 kJ molSimilarly, the value with gas phase water product is 6345.4 kJ molFor liquid fuels, the heating value is usually expressed as Btu per gallon of the liquidThe density of n-decane (from Perry's handbook) is 0.73 g cm , and the molecular 3

weight is 142.3 g mol, so we get HHV  6829.5 kJ mol / 143 g mol * 0.73 g cm  34.86 kJ/cm * 3785 3

cm gal * 0.9478 Btu kJ  125000 Btu galLikewise, the lower heating value is 116000 Btu gal. 3

Solution 4.32 Problem Statement A light fuel oil with an average chemical composition of C10H18 is burned with oxygen in a bomb calorimeter. The heat evolved is measured as 43,960 J·g for the reaction at 25°C. Calculate the standard heat of 1

combustion of the fuel oil at 25°C with H2O(g) and CO2(g) as products. Note that the reaction in the bomb occurs at constant volume, produces liquid water as a product, and goes to completion.

Solution

On the basis of 1 mole of C10H18 (molar mass  162.27)

Q 



J 



This value is for the constant-volume reaction:

C10H18(l)  14.5O2(g)  10CO2(g)  9H2O(l)

Assuming ideal gases and with symbols representing total properties, Q U H  PV  H  RT ngas T

K ngas 

H  Q  RT ngas  

mol



This value is for the constant-V reaction, whereas the STANDARD reaction is at const. PHowever, for ideal gases H  f(T), and for liquids H is a very weak function of P. We therefore take the above value as the standard value, and for the specified reaction:

C10H18(l)  14.5O2(g)  10CO2(g)  9H2O(l)

9H2O(l)  9H2O(g)

H

396108

___________________________________________________ C10H18(l)  14.5O2(g)  10CO2(g)  9H2O(g)

H298 : H  Hvap  6748436 J

Solution 4.33 Problem Statement

Methane gas is burned completely with 30% excess air at approximately atmospheric pressure. Both the methane and the air enter the furnace at 30°C saturated with water vapor, and the flue gases leave the furnace at 1500°C. The flue gases then pass through a heat exchanger from which they emerge at 50°C. Per mole of methane, how much heat is lost from the furnace, and how much heat is transferred in the heat exchanger?

Solution

FURNACE: Basis is 1 mole of methane burned with 30% excess air. CH4  2O2  CO2  2H2O(g)

Entering: nme 

nO 2 

nN 2 

Total moles of dry gases entering nt  nme  nO 2  nN 2 

At 30 degC the vapor pressure of water is 4.241 kPa. Moles of water vapor entering:

nH 2 O 

Leaving:

CO2

1 mol

H2O

2.585 mol

2.6  2  0.6 mol

9.781 mol

4.241 * 13.381  0.585 101.325  4.241

By an energy balance on the furnace:

Q  H  H303  HP

From Example 4.7: H298  802625 J/mol

H303  H298  MCPH * R * (T – T0)  802279 J/mol

For evaluation of HP we number species as above.

 1     2 .585  n     0.6   9.781   

i

5.457    3.470   A     3.639  3.280   

1.045    1.450  103  B    0.506 K  0.593  

A   i ni Ai 

1.157     0.121   D    0.227   0.040   

B   i ni Bi 



3

1 D   i ni Di   K



The TOTAL value for MCPH of the product stream:

HP  R·MCPH(303.15K,1773.15K,A,B,C,D)·(1773.15

303.15)K

HP  731982 kJ/mol

Q HP  H303  70.30 kJ

Solution continued on next page…

HEAT EXCHANGER: Flue gases cool from 1500 C to 50 C. The partial pressure of the water in the flue gases leaving the furnace (in kPa) is

pp 

n2 n1  n2  n3  n4



kPa

The vapor pressure of water at 50 degC (exit of heat exchanger) is 12.34 kPa, and water must condense to lower its partial pressure to this value. Moles of dry flue gases:

n  n1  n3  n4 

Moles of water vapor leaving the heat exchanger:

n2 

12.34 * n  1.578 101.325  12.34

Moles of water condensing: n  2.585  1.578  1.007

Latent heat of water at 50 C in J/mol:

H50  2382.9 * 18.015  42927.94

J mol

Sensible heat of cooling the flue gases to 50 degC with all the water as vapor (we assumed condensation at 50 degC): Q  R  MCPH 

K

 K A B C D  



 K n H50 

Solution 4.34 Problem Statement Ammonia gas enters the reactor of a nitric acid plant mixed with 30% more dry air than is required for the complete conversion of the ammonia to nitric oxide and water vapor. If the gases enter the reactor at 75°C, if conversion is 80%, if no side reactions occur, and if the reactor operates adiabatically, what is the temperature of the gases leaving the reactor? Assume ideal gases. Solution

4NH3(g)  5O2(g)  4NO(g)  6H2O(g) BASIS: 4 moles ammonia entering reactor Moles O2 entering  (5)(1.3)  6.5 Moles N2 entering  (6.5)(79/21)  24.45 Moles NH3 reacting  moles NO formed  (4)(0.8)  3.2 Moles O2 reacting  (5)(0.8)  4.0 Moles water formed  (6)(0.8)  4.8

ENERGY BALANCE: H  HR  H298  HP  0 REACTANTS: 1NH3; 2O2; 3N2

 4     n    24.45  

i

3.578    A    3.280   

3.020    103  B     0.593 K  

A   i ni Ai 

B   i ni Bi 

0.186    D     0.040   

D   i ni Di  



TOTAL mean heat capacity of reactant stream: H R  R  MCPH 

K A B C D 

J mol

K 

K

The result of Pb. 4.23(b) is used to get H 298 



))  

m l

Products  0.8     2.5    n   3.2   4.8      24.45 i

3.578   3.639   A  3.387  3.470      3.280 

3.020    0.506   103 B   0.629  1.450  K      0.593

A   i ni Ai 

0.186    0.227    B   0.014  10 5 K 2  0.121      0.040   

B   i ni Bi 

D   i ni Di 

T0 



 B D    1   Given H 298 H R  R  AT0     T02  2     2 T0    



T

Solution 4.35 Problem Statement Ethylene gas and steam at 320°C and atmospheric pressure are fed to a reaction process as an equimolar mixture. The process produces ethanol by the reaction: C2H4(g)  H2O(g) → C2H5OH(l)

The liquid ethanol exits the process at 25°C. What is the heat transfer associated with this overall process per mole of ethanol produced?

Solution

The enthalpy balance for this steady-state flow process in which no non-flow work is done simply gives Q  H, where H is the enthalpy of the ethanol stream leaving the reactor minus the enthalpy of the ethylene and water entering the reactorTo compute the enthalpy change, we can use an imaginary 2-step path in which we first cool the ethane and water from 320°C to 25°C, and then carry out the reactionThe total enthalpy change will be the sum of the entropy change for the two steps:

H  

298.15 K

593.15 K

o C pfeed dT  Hrxn ,298.15 K

The choice of doing the heat capacity integral first (from 320 to 25°C) is particularly convenient because the final temperature happens to be the standard temperature for heats of formationIn general, if we have reactants and products at different temperatures, we would have to do two heat capacity integrals – one from the feed temperature to 298.15 K and one from 298.15 K back to the product temperature.

The heat of reaction is Hrxn  Hf(C2H5OH(l))  Hf(H2O(g)) – Hf(C2H4(g)) Hrxn  277690 – (241818) – (52510)  88382 J/mol

We can use the handy spreadsheet from the lecture notes to evaluate the heat capacity integral:

T1 (K)

T2 (K)

B (1/K)

C (1/K )

D (K )

ICPH (K)

593.15

298.15

4.894

1.58E02

4.39E06

1.21E04

3.28E03

Solution continued on next page…

ICPH  

dT  AT2  T1  

1

1

 2

T  T   3 T  T   D  T  T  2



Where for A, B, C, and D, we use the sum of the values for water and ethyleneMultiplying this by R gives



298.15 K 593.15 K

C pfeed dT  8.314 * (3820)  27270 J mol1

Thus, the total heat load of the process is Q  H  88380 – 27270  115560 J/mol  115.6 kJ/mol We must remove 115.6 kJ of heat from the reactor for each mole of ethanol produced.

Solution 4.36 Problem Statement A gas mixture of methane and steam at atmospheric pressure and 500°C is fed to a reactor, where the following reactions occur: CH4  H2O  CO  3H2 and CO  H2O  CO2  H2 The product stream leaves the reactor at 850°C. Its composition (mole fractions) is: y CO2



0.0275 yCO



0.1725 yH2O



0.1725 yH2 0.6275

Determine the quantity of heat added to the reactor per mole of product gas.

Solution

Let x1 be the extent of the first reaction and x2 be the extent of the second reaction, defined so that they are extensive extents of reaction with units of moles, so that the number of moles of each species coming out of the reactor are nCH4  nCH4,o  x1 nH2O  nH2O,o  x1  x2 nCO  x1  x2 nH2  3x1  x2 and

nCO2  x2

Given the reactor outlet composition in mole fractions, and doing the problem on a 'per mole of outlet gas' basis, the number of moles of each species coming out is just the given mole fractions, so we have nCH4  nCH4,o  x1  0 nH2O  nH2O,o  x1  x2  0.1725 nCO  x1  x2  0.1725 nH2  3x1  x2  0.6275 and

nCO2  x2  0.0275

Note that because the four given mole fractions sum to 1, the mole fraction of CH4 in the outlet (not explicitly given) is zeroAbove, we have 5 equations in 4 unknowns (x1, x2, nCH4,o, and nH2O,o), but fortunately they are consistent From them, we get x2  0.0275, x1  0.2000, nCH4,o  0.2000, and nH2O,o  0.4000That is, for each mole of the product stream, 0.4 moles of H2O and 0.2 moles of CH4 are fed to the reactor, and 0.2 moles of reaction 1 occur and 0.0275 moles of reaction 2 occur

Solution continued on next page…

The energy balance, for this steady flow process with no shaft work, is simply H  QThe heat input is equal to the enthalpy difference between the product stream and the reactant streamThis can be computed as the sum of three steps: (1) The change in enthalpy when 0.4 mol H2O and 0.2 mol CH4 are cooled from 500 ºC to 25 º C (2) 0.2 times the standard heat of reaction at 25ºC for CH4  H2O  CO  3 H2 plus 0.0275 times the standard heat of reaction for CO  H2O  CO2  H2 (3) The change in enthalpy when a mixture of 0.0275 mol CO2, 0.1725 mol CO, 0.1725 mol H2O, and 0.6275 mol H2 is heated from 25 ºC to 850 ºC.

For step (1), we can use one of our handy spreadsheetsA version adapted for gas mixtures is shown below: T

IDCPH  T0,T; A,B, C,D  

1 R

B

C

1

1 

 C dT  A T  T   2 T  T   3 T  T   D  T  T  p

2 0

3 0

T 0 (K) 773.15

T (K) 298.15

ΔA 1.7284

ΔB (1/K) 2.40E-03

ΔC (1/K2) -4.33E-07

ΔD (K2) IDCPH (K) 4.84E+03 -1377.787

ΔH (J) -11455

Species Name CH4 H2O CO H2 CO2

Amount (moles) 0.2 0.4 0 0 0

A 1.702 3.47 3.376 3.249 5.547

B (1/K) 9.08E-03 1.45E-03 5.57E-04 4.22E-04 1.05E-03

C (1/K2) -2.16E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) 0.00E+00 1.21E+04 -3.10E+03 8.30E+03 -1.16E+05

n iBi 1.82E-03 5.80E-04 0.00E+00 0.00E+00 0.00E+00

niAi 3.40E-01 1.39E+00 0.00E+00 0.00E+00 0.00E+00

niCi -4.33E-07 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niDi 0.00E+00 4.84E+03 0.00E+00 0.00E+00 0.00E+00

For step (3), we do the same thing, but with the composition of the product stream and the appropriate temperatures: T

IDCPH  T0,T; A,B, C,D  

1 R

B

C

1

1 

 C dT  A T  T   2 T  T   3 T  T   D  T  T  p

2 0

3 0

T 0 (K) 298.15

T (K) 1123.15

ΔA 3.372225

ΔB (1/K) 6.40E-04

ΔC (1/K2) 0.00E+00

ΔD (K2) IDCPH (K) 3.58E+03 3165.980

ΔH (J) 26322

Species Name CH4 H2O CO H2 CO2

Amount (moles) 0 0.1725 0.1725 0.6275 0.0275

A 1.702 3.47 3.376 3.249 5.547

B (1/K) 9.08E-03 1.45E-03 5.57E-04 4.22E-04 1.05E-03

C (1/K2) -2.16E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2) niAi 0.00E+00 0.00E+00 1.21E+04 5.99E-01 -3.10E+03 5.82E-01 8.30E+03 2.04E+00 -1.16E+05 1.53E-01

n iBi 0.00E+00 2.50E-04 9.61E-05 2.65E-04 2.87E-05

niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niDi 0.00E+00 2.09E+03 -5.35E+02 5.21E+03 -3.18E+03

Solution continued on next page…

For step (2), the relevant heats of formation are: Reaction 1: Species Name CH4 H2O CO H2

Reaction 2: Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -74.52 -1 -241.818 1 -110.53 3 0

Species Name CO H2O CO2 H2

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -110.53 -1 -241.818 1 -393.509 1 0

From which H 298(rxn 1)  205808 J/mol and H 298(rxn 2)  41161 J/mol o

Thus, for step (2), we have H  0.2*205808  0.0257*41161  40030 J.

Thus, the total for the three steps is 11455  40030  26322  54897 J.

Solution 4.37 Problem Statement A fuel consisting of 75 mol-% methane and 25 mol-% ethane enters a furnace with 80% excess air at 30°C. If 8 × 10 kJ·kmol fuel is transferred as heat to boiler tubes, at what temperature does the flue gas leave the 5

furnace? Assume complete combustion of the fuel.

Solution

Let’s take as a basis one mole of fuel (0.75 mol of methane and 0.25 mol of ethane)Because we use excess air, we can safely assume complete combustion of the fuel to give CO2 and waterSo, we have two combustion reactions: CH4  2 O2  CO2  2 H2O C2H6  7/2 O2  2CO2  3 H2O

Burning the 0.75 mol of CH4 requires 2*0.75  1.5 mol of O2, and burning the 0.25 mol of C2H6 requires 0.25*7/2  7/8  0.875 mol O2So, the total O2 required for stoichiometric combustion of the fuel is 2.375 mol O2 per mole of fuelIf we use 80% excess air, that means we have 1.8 times this much O2, and the O2 feed rate is 1.8*2.375  4.275 mol O2 per mole of fuelBecause we are using air, we get 79/21 moles of N2 for every mole of O2, and therefore the N2 feed rate is 79/21*4.275  16.08 mol N2 per mole of fuelSo, the feed to the furnace is (per mole of fuel) 0.75 mol CH4 0.25 mol C2H6 4.275 mol O2 16.08 mol N2

The flue gas has the same amount of N2 as the feed, but all of the CH4 and C2H6 (and some of the O2) have been converted to CO2 and H2OThe 0.75 mol of CH4 produces 0.75 mol of CO2 and 1.5 mol of H2OThe 0.25 mol C2H6 produces 0.5 mol CO2 and 0.75 mol H2OWe already figured out that the stoichiometric O2 (the amount used up in burning the fuel) is 2.375 mol O2, so we have 4.275 – 2.375  1.9 mol O2 remaining in the flue gas (this is the 80% excess O2)

So the flue gas leaving the furnace is (per mole of fuel) 1.9 mol O2 1.25 mol CO2 2.25 mol H2O 16.08 mol N2

The energy balance is simply Q  H, where H is the enthalpy change between the feed stream and the flue gas stream and Q is specified as 810 kJ per kgmol of fuel (or J per mol of fuel)We can compute the total 5

H from feed to flue gas using a hypothetical three-step process in which we cool the feed from 30°C to 25°C, Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

then react it at 25°C, then heat it to the flue gas outlet temperature (which we are supposed to find)So, we have

H  

298.15 K

303.15 K

o C pfeed dT Hrxn  ,298.15 K

Tflue 298.15 K

C pflue dT  Q

We know Q, and we can compute the first heat capacity integral and the standard enthalpy of reaction, then we can find the flue gas temperature (by trial and error) that satisfies the above energy balanceFor the integral of the feed heat capacity, we can use the same spreadsheet as always: T1 (K)

T2 (K)

B (1/K)

C (1/K )

D (K )

ICPH (K)

303.15

298.15

68.70413

0.023316

3E06

32722.5

3.75E02

ICPH   T1

Cp R

dT  AT2  T1  

1 B 2 C 1 T2  T12   T23  T13   D     T2 T1  2 3

CH4

1.702

9.08E03

2.16E06

C2H6

1.131

1.92E02

5.56E06

3.369

5.06E04

2.27E04

3.28

5.93E04

4.00E03

CO2

5.457

1.05E03

1.16E05

H2O

3.47

1.45E03

1.21E04

For convenience, I have entered the heat capacity coefficients of the individual species, which I then add up (multiplied by the number of moles of each species) to get the total heat capacity coefficients of the mixtureNote that I can type in the formula once (for A) and then fill right to get B, C, and DSo, this gives us



298.15 K 303.15 K

C pfeed dT  375R  3118 J mol1

The total heat of reaction is 0.75 times the enthalpy of reaction for CH4  2 O2  CO2  2 H2O, plus 0.25 times the enthalpy of reaction for CH4  2 O2  CO2  2 H2O, or Hrxn 0.75*(Hf(CO2) 2Hf(H2O) – Hf(CH4))  0.25*(2Hf(CO2) 3Hf(H2O) – Hf(C2H6)) Hrxn 1.25*Hf(CO2) 2.25Hf(H2O) –0.75Hf(CH4)) – 0.25Hf(C2H6)) Hrxn 1.25*(393509)2.25(241818) –0.75(74520) – 0.25(83820)  959132 J mol

So, our energy balance becomes

H 3118 J mol1  959132 J mol1   So, 

Tflue 298.15 K

C pflue dT 800000 J mol1

C pflue dT  162250 J molSo, we can put the flue gas composition into the heat capacity integral

and try different final temperatures until we get this value:

T1 (K)

T2 (K)

B (1/K)

C (1/K )

D (K )

ICPH (K)

ICPH*R (J/mol)

298.15

543.5519

73.77225

0.015066

96210

1.95E04

162250.00

ICPH   T1

Cp R

dT  AT2  T1  

1 B 2 C 1 T2  T12   T23  T13   D     T2 T1  2 3

CH4

1.702

9.08E03

2.16E06

C2H6

1.131

1.92E02

5.56E06

3.369

5.06E04

2.27E04

3.28

5.93E04

4.00E03

CO2

5.457

1.05E03

1.16E05

H2O

3.47

1.45E03

1.21E04

A flue gas temperature of 543.6 K  270°C satisfies the energy balance.

Solution 4.38 Problem Statement The gas stream from a sulfur burner consists of 15 mol-% SO2, 20 mol-% O2, and 65 mol-% N2. The gas stream at atmospheric pressure and 400°C enters a catalytic converter where 86% of the SO2 is further oxidized to SO3. On the basis of 1 mol of gas entering, how much heat must be added to or removed from the converter so that the product gases leave at 500°C?

Solution

On a basis of 1 mol of gas entering, we have 0.15 mol SO2, 0.2 mol O2, and 0.65 mol N2 in the inlet streamIf 86% of the SO2 is further oxidized to SO3, then 0.86*0.15  0.129 mol SO2 is oxidized, producing 0.129 mol SO3 by the reaction SO2  ½ O2  SO3This also consumes ½ * 0.129  0.0645 mol O2So, in the exit stream we have 0.021 mol SO2, 0.129 mol SO3, 0.1355 mol O2, and 0.65 mol N2Notice that for every 1 mol of gas entering, we only have a total of 0.9355 mol leaving.

Since the entrance and exit temperatures are different, to compute the net heat removal requirement we could either compute the heat required to heat the reactants from 400°C to 500°C and then compute the heat of reaction at 500°C, or we could compute the heat of reaction at 400 °C and then compute the heat required to heat the products from 400°C to 500°CThe net heat requirement (the sum of these 2 steps) will be the same either way. Of course, we could also cool the whole inlet stream to 298K, add the heat of reaction at that temperature, and then heat the whole product stream to 500°CThat would also give the same answer.

Solution continued on next page…

Using the handy spreadsheet from the notes, we can compute the heat of reaction at 400°C as follows: Reference Temperature T 0 (K) 298.15

Species Name SO2 O2 SO3

Temperature of Interest T (K) 673.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -98.890 0.053 -98.596 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T0 (kJ/mol) -1 -296.83 -0.5 0 1 -395.72

A 5.699 3.639 8.06

DA 0.5415 Note: Light blue fields are inputs, pink fields are the final output.

T   1 1 B C C dT   AT T0  T 2 T02   T 3  T03  D    T T0  RT T p T  2 3 o

   B 2 C o o Hrxn  Hrxn  R  AT  T0   T T02   3 T 3  T03  D T  T  ,T ,T0 2  0  

B (1/K) 8.01E-04 5.06E-04 1.06E-03

C (1/K2) 0.00E+00 0.00E+00 0.00E+00

D (K2) niDHf,i niAi -1.02E+05 2.97E+02 -5.70E+00 -2.27E+04 0.00E+00 -1.82E+00 -2.03E+05 -3.96E+02 8.06E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DB (1/K) 2.00E-06

DC (1/K2) 0.00E+00

DD (K2) -9.00E+04

niBi -8.01E-04 -2.53E-04 1.06E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niDi 1.02E+05 1.14E+04 -2.03E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

So, the heat of reaction at 400°C is 98.596 kJ per mol of SO2 reactedIf 0.129 mol reacts, then the heat released is 98.596*0.129  12.719 kJ.

Now, we want to find the heat input required to increase the temperature of the products from 400°C to 500°C We can make a modified heat capacity integral spreadsheet that adds up the total heat capacity of the mixture components to get the total heat capacity and then computes the integralThis is shown below.

T 1 (K) 673.15

T 2 (K) 773.15

Species SO2 O2 SO3 N2

Moles 0.021 0.1355 0.129 0.65

Total Values for Mixture T2   2 2 C (1/K ) D (K ) ICPH (K*mol) ICPH  C p dT  A T  T  B T 2  T 2  C T 3  T 3  D  1  1  A B (1/K)  2 1 2 1 2 1  T T  R 2 3  3.78E+00 6.07E-04 0.00E+00 -2.88E+04 416.822 2 1 T1





Species Values A 5.699 3.639 8.06 3.28

B (1/K) 0.000801 0.000506 0.001056 5.93E-04





Values * Moles 2

C (1/K ) D (K ) A mol B (mol/K) C (mol/K ) D (mol K ) 0.00E+00 -1.02E+05 1.20E-01 1.68E-05 0.00E+00 -2.13E+03 0.00E+00 -2.27E+04 4.93E-01 6.86E-05 0.00E+00 -3.08E+03 0.00E+00 -2.03E+05 1.04E+00 1.36E-04 0.00E+00 -2.62E+04 0.00E+00 4.00E+03 2.13E+00 3.85E-04 0.00E+00 2.60E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Solution continued on next page…



So, the heat capacity integral is 416.8 K molNotice that since we multiplied the polynomial coefficients by the number of moles of each species, the integral is now in terms of the total heat capacity for the number of moles specified and has units of K mol, rather than just KMultiplying by R  0.008314 kJ mol K gives the heat requirement as 3.465 kJSo, the reaction releases 12.719 kJ, and 3.465 kJ of this is required to heat the products to 500°CTherefore, 12.719  3.465  9.25 kJ has to be removed per mole of feed gas fed to the catalytic converter.

Solution 4.39 Problem Statement Hydrogen is produced by the reaction: CO(g)  H2O(g) →

CO2 (g)  H2(g).

The feed stream to the reactor is an equimolar mixture of carbon monoxide and steam, and it enters the reactor at 125°C and atmospheric pressure. If 60% of the H2O is converted to H2 and if the product stream leaves the reactor at 425°C, how much heat must be transferred to or from the reactor?

Solution

This problem is very much like the previous oneWe will work it on the basis of 1 mol of CO fed to the reactorThus, the feed is 1 mol CO and 1 mol H2OIf 65% of the H2O (and by stoichiometry therefore alse 65% of the CO) is converted, then the product is 0.35 mol CO, 0.35 mol H2O, 0.65 mol CO2, and 0.65 mol H2Just to be different than the previous problem, this time let’s first compute the heat required to heat the reactants from 125°C to the final temperature of 425°C (698.15 K), and then compute the heat of reaction at 425°CWe can use a spreadsheet just like the one used in the last problem: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

T 1 (K) 398.15

T 2 (K) 698.15

Species CO H2O

Moles

Total Values for Mixture 2 2 C (1/K ) D (K ) A B (1/K) 6.85E+00 2.01E-03 0.00E+00 9.00E+03

ICPH  

ICPH (K*mol) 2393.554

dT  A T2 T1  

  B 2 T  T12  C3 T23  T13  D T1  T1  2 2  2 1

Species Values Values * Moles 2 2 2 2 A B (1/K) C (1/K ) D (K ) A mol B (mol/K) C (mol/K ) D (mol K ) 1 3.376 5.57E-04 0.00E+00 -3.10E+03 3.38E+00 5.57E-04 0.00E+00 -3.10E+03 1 3.47 1.45E-03 0.00E+00 1.21E+04 3.47E+00 1.45E-03 0.00E+00 1.21E+04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Multiplying the heat capacity integral by R gives Q  2393.5*0.008314  19.90 kJ required to heat the feed from 398.15 K to 698.15 K

Now, we compute the heat of reaction at 698.15 K in the same way as in the lecture notes and the previous problem:

Reference Temperature

Temperature of Interest

T 0 (K) 298.15

T (K) 698.15

Species Name

Stoichiometric coefficient

CO H2O CO2 H2

-1 -1 1 1

Heat of Reaction at T 0 kJ/mol -41.166 Standard Heat of Formation at T 0 (kJ/mol) -110.525 -241.818 -393.509 0

Heat Capacity Integral

Heat of Reaction at T

(dimensionless) kJ/mol 0.591 -37.734 Heat Capacity Coefficients

T   1 1 B C C dT   A T  T0   T 2 T02   T 3  T03  D     T T0  RT T p T  2 3 o

   B 2 C o o Hrxn  Hrxn  R  AT T0   T T02  3 T 3  T03  D T  T  ,T ,T0 2  0   2

C (1/K )

A 3.376 3.47 5.457 3.249

B (1/K) 5.57E-04 1.45E-03 1.05E-03 4.22E-04

0.00E+00 0.00E+00 0.00E+00 0.00E+00

DA 1.86 Note: Light blue fields are inputs, pink fields are the final output.

DB (1/K) -5.40E-04

DC (1/K ) 0.00E+00

D (K ) n iDHf,i -3.10E+03 1.11E+02 1.21E+04 2.42E+02 -1.16E+05 -3.94E+02 8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iAi -3.38E+00 -3.47E+00 5.46E+00 3.25E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n i Bi -5.57E-04 -1.45E-03 1.05E-03 4.22E-04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n i Ci 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n i Di 3.10E+03 -1.21E+04 -1.16E+05 8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -1.16E+05

So, the heat of reaction is –37.734 kJ per mol of CO reactedIf 0.6 mol of CO reacts, then 37.734*0.6  22.64 kJ is released by the reactionSo, the net heat removal rate is 22.64 – 19.90  2.74 kJ per mol of CO fed to the reactor (1.37 kJ per mole of feed)The exothermicity of the reaction is greater than the heat needed for the temperature increase, so the net result is that heat has to be removed.

Solution 4.40 Problem Statement A direct-fired dryer burns a fuel oil with a lower heating value of 19,000(Btu) (lbm) . [Products of combustion 1

are CO2(g) and H2O(g).] The composition of the oil is 85% carbon, 12% hydrogen, 2% nitrogen, and 1% water by weight. The flue gases leave the dryer at 400(°F), and a partial analysis shows that they contain 3 mol-% CO2 and 11.8 mol-% CO on a dry basis. The fuel, air, and material being dried enter the dryer at 77(°F). If the entering air is saturated with water and if 30% of the net heating value of the oil is allowed for heat losses (including the sensible heat carried out with the dried product), how much water is evaporated in the dryer per (lbm) of oil burned?

Solution

BASIS: 100 lbmol DRY flue gases containing 3.00 lbmol CO2 and 11.80 lbmol CO x lbmol O2 and 100(14.8x) 85.2x lbmol N2. The oil therefore contains 14.80 lbmol carbon; a carbon balance gives the mass of oil burned:

12.011  0.85

lbm

209.133 * 0.01  18.015

lbmol



The oil also contains H2O:

Also H2O is formed by combustion of H2 in the oil in the amount:

209.133 * 0.12  2.016

lbmol

Solution continued on next page…

Find amount of air entering by N2 & O2 balances. N2 entering in oil:

209.133 * 0.02  28.013

lbmol

lbmol N2 entering in the air (85.2 x) 0.149 85.051 x

lbmol O2 in flue gas entering with dry air  3.00  11.8/2  x  12.448/2  15.124  x lbmol (CO2) (CO) (O2) (H2O from combustion)

Total dry air  N2 in air  O2 in air  85.051  x  15.124  x  100.175 lbmol Since air is 21 mol % O2,

x







lbmol

O2 in air  15.124  x  21.037 lbmol N2 in air  85.051  x  79.138 lbmol N2 in flue gas  79.138  0.149  79.287 lbmol [CHECK: Total dry flue gas  3.00  11.80  5.913  79.287  100.00 lbmol]

Humidity of entering air, sat. at 77 F in lbmol H2O/lbmol dry air, P(sat)0.4594(psia)

0.4594  0.03227 14.686  0.4594

lbmol H2O entering in air:

0.03227·100.175·lbmol  3.233 lbmol Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

If y  lbmol H2O evaporated in the drier, then lbmol H2O in flue gas: 0.116 12.448 3.233 y  15.797  y

Entering the process are oil, moist air, and the wet material to be dried, all at 77 F. The "products" at 400 F consist of: 3.00 lbmol CO2 11.80 lbmol CO 5.913 lbmol O2 79.287 lbmol N2 (15.797  y) lbmol H2O(g)

Energy balance: Q  H  H298  HP where Q  30% of net heating value of the oil:

Q 



BTU  lbm

lbm 



BTU



BTU

Reaction upon which net heating value is based:

OIL  (21.024)O2  (14.8)CO2  (12.448  0.116)H2O(g)  (0.149)N2

H298a  



 BTU  

To get the "reaction" in the drier, we add to this the following:

(11.8)CO2  (11.8)CO  (5.9)O2 Solution continued on next page…

H

 







 BTU

(y)H2O(l)  (y)H2O(g)

H

 y  44012  0.42993  y  BTU

[The factor 0.42993 converts from joules on the basis of moles to BTU on the basis of lbmol.]

Addition of these three reactions gives the "reaction" in the drier, except for some O2, N2, and H2O that pass through unchanged. Addition of the corresponding delta H values gives the standard heat of reaction at 298 K:

H

 y  H a  H b H c  y

For the product stream we need MCPH: T0 

  3    11.8    n   5.913   79.278      15.797  y 

i

5.457   3.376   A  3.639 3.280     3.470

A   i nyAi

 1.045     0.557    103 B   0.506   0.593 K     1.450 

T

1.157    0.031   D  0.227  10 5 K 2  0.040       0.121 

B   i nyi Bi D   i nyi Di



Cp 

   A  B  T     D  0 2  2 T0  

Given C p 



 BTU  Q H298

Solve for y  49.782 lbmol 49.782*18.015/209.13  4.288 lb H2O evap. per lb oil burned

Solution 4.41 Problem Statement An equimolar mixture of nitrogen and acetylene enters a steady-flow reactor at 25°C and atmospheric pressure. The only reaction occurring is: N2( g)  C2H2( g) → 2HCN( g). The product gases leave the reactor at 600°C and contain 24.2 mol-% HCN. How much heat is supplied to the reactor per mole of product gas?

Solution If we work from a basis of 1 mole of N2 and 1 mole of C2H2 entering the reactor, then we have 2 moles of gas entering the reactor, and (since there is no mole change in the reaction) 2 moles of gas leaving the reactorIf this case is 24.2 mol-% HCN, then the total number of moles of HCN is 2*0.242  0.484The number of moles of reaction that has occurred (as written) is 0.242 moles, and the number of moles of N2 and C2H2 remaining are (10.242)  0.758 moles eachThe total heat supplied to the reactor, on this basis, will therefore be 0.242 times the heat of reaction at 298 K (conveniently also the temperature at which the reactants enter the reactor) plus the heat required to heat 0.748 mol N2, 0.748 mol C2H2 and 0.484 mol HCN from 298 K to 873 KThe heat of reaction is Hrxn  2*Hf(HCN(g))  Hf(N2(g))  Hf(C2H2(g)) Hrxn  2*135100 – 0 – 227480  42720 J mol

Since 0.242 mol of reaction occurs, 0.242*42720  10338 J of heat input is required to carry out the reaction at 298.15 K

We can use a handy-dandy spreadsheet like the one provided in the notes to compute the heat capacity integral for the products: T2

ICPH  T0 ,T;A,B,C,D  

1

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  0

2 0

3 0

T0(K) 298.15

T (K) 873.15

A 9.42652

B (1/K) 0.002587

C (1/K2) 0

C2H2 N2 HCN

0.758 0.758 0.484

6.132 3.28 4.736

1.95E-03 5.93E-04 1.36E-03

0 0 0

ig D (K2) ICPH (K) D H (J/mol) -130522.2 6003.1 49912.6

-1.30E+05 4.00E+03 -7.25E+04

So, the total heat input is 10338  49913  60250 J per mole of N2 and C2H2 entering the reactorThis is 60250/2  30125 J per mole of total inlet or outlet stream, or 60250/0.484  124500 J per mole of HCN produced.

Solution 4.42 Problem Statement Chlorine is produced by the reaction: 4HCl( g)  O2 ( g) → 2 H2O( g)  2Cl2( g). The feed stream to the reactor consists of 60 mol-% HCl, 36 mol-% O2, and 4 mol-% N2, and it enters the reactor at 550°C. If the conversion of HCl is 75% and if the process is isothermal, how much heat must be transferred to or from the reactor per mole of the entering gas mixture?

Solution

Chlorine is produced by the reaction: 4 HC 1 ( g)  O 2 ( g) → 2 H 2 O ( g)  2 C 1 2 ( g). The feed stream to the reactor consists of 60 mol-% HCl, 36 mol-% O2, and 4 mol-% N2, and it enters the reactor at 550°C. If the conversion of HCl is 75% and if the process is isothermal, how much heat must be transferred to or from the reactor per mole of the entering gas mixture?

BASIS: 1 mole gas entering reactor, containing 0.6 mol HCl, 0.36 mol O2, and 0.04 mol N2.

HCl reacted  (0.6)(0.75)  0.45 mol

4HCl(g)  O2(g)  2H2O(g)  2Cl2(g) H298     

 

  



J mol

Evaluate H823 by Eq. (4.23) with T0 

2   2 n    4  1  

i

A   i ni Ai  

3.470     4.442  A    3.156  3.639   

T

 1.45     0.089 103  B     0.623 K  0.506   

B   i ni Bi 

 0.121    0.344  D     0.151  0.277  

D   i ni Di  

H 823 H298  MCPH  R  T  T0   

J mol

Solution continued on next page…

Heat transferred per mol of entering gas mixture:

Q

H823  4

mol 

Solution 4.43 Problem Statement A gas consisting only of CO and N2 is made by passing a mixture of flue gas and air through a bed of incandescent coke (assume pure carbon). The two reactions that occur both go to completion: CO2  C →

2CO and 2C  O2 → 2CO

The flue gas composition is 12.8 mol-% CO, 3.7 mol-% CO2, 5.4 mol-% O2, and 78.1 mol-% N2. The flue gas/air mixture is so proportioned that the heats of the two reactions cancel, and the temperature of the coke bed is therefore constant. If this temperature is 875°C, if the feed stream is preheated to 875°C, and if the process is adiabatic, what ratio of moles of flue gas to moles of air is required, and what is the composition of the gas produced?

Solution CO2  C  2CO H 298a  172459 



H 298a  

J mol J mol

Eq. (4.22) applies to each reaction: For (a) 2   n    1  

3.376     A    5.457   

0.557    103  B    1.045  K  

0.037     D    1.157   

i

A   i ni Ai  

B   i ni Bi  

H1148  H 298  MCPH * R * T  T0   1.696 * 105

4

D   i ni Di 

J mol

For (b) 2   n    2  

3.376    A    1.771  

 0.557    103  B     0.771 K  

i

A   i ni Ai  

0.031    D    0.867  

B   i ni Bi  

H1148  H 298  MCPH * R * T  T0   2.249 * 10

4

D   i ni Di 



J mol

The combined heats of reaction must be 0

nCO 2  H1148a  nO2  H1148b  0

r H1148b H1148a 

For 100 mol flue gas and x mol air, moles are:

Flue gas

Air

Feed mix

CO2

12.8

3.7

5.4

0.21x

5.4  0.21x

78.1

0.79x

78.1  0.79x

Solution continued on next page…

Whence in the feed mix: r 

12.8 5.4  0.21  x

x

mol

Flue gas to air ratio  100/20.218  4.946

Product composition

nCO : 3.7  2·(12.8  5.4  0.21·19.155)  48.145

nN2 78.1  0.79  19.155  93.232

mol % CO  34.07 % mol % N2  65.92 %

Solution 4.44 Problem Statement A fuel gas consisting of 94 mol-% methane and 6 mol-% nitrogen is burned with 35% excess air in a continuous water heater. Both fuel gas and air enter dry at 77(°F). Water is heated at a rate of 75(lbm)(s) from 77(°F) to 1

203(°F). The flue gases leave the heater at 410(°F). Of the entering methane, 70% burns to carbon dioxide and 30% burns to carbon monoxide. What volumetric flow rate of fuel gas is required if there are no heat losses to the surroundings?

Solution CH4  2O2  CO2  2H2O(g) H 298a 

J mol

CH4  (3/2)O2  CO  2H2O(g) H298 

J mol

BASIS: 1 mole of fuel gas consisting of 0.94 mol CH4 and 0.06 mol N2 Air entering contains

1.35·2·0.94  2.538 mol O2 2.538· 79/21 9.548 mol N2

Moles CO2 formed by reaction  0.94·0.7  0.658 Moles CO formed by reaction  0.94·0.3  0.282

H



 H

a

 H

b 



J mol

Moles H2O formed by reaction  0.94·2.0  1.88 Moles O2 consumed by reaction  2·0.658  (3/2)·0.282  1.739

Product gases contain the following numbers of moles: (1) CO2: 0.658 (2) CO: 0.282 (3) H2O: 1.880 (4) O2: 2.538  1.739  0.799 (5) N2: 9.548  0.060  9.608

Solution continued on next page…

 0.658    0.282   n  1.880   0.799     9.608  

5.457    3.376   A  3.470  3.639     3.280  

 1.045     0.557    103 B  1.450    0.506  K     0.593  

A   i ni Ai  

i

H P  R  MCPH 

1.157    0.031   D   0.121  10 5 K 2 0.227      0.04 0  

B   i ni Bi 

K A B C D  

3

D   i ni Di  

K

K



10 5 J/mol

KJ mol

Energy balance: H rx  H 298 H P  

H H 20  m H 2O Hrx  n fuel 

m H 2O 

From Table C.1: H H 20   n fuel 

H H 20 * m H 2O Hrx





Kg sec

kJ kg



Volumetric flow rate of fuel, assuming ideal gas: V

n fuel * R * 298.15 K Pa



m3 s

Solution 4.45

Problem Statement

A process for the production of 1,3-butadiene results from the catalytic dehydrogenation at atmospheric pressure of 1-butene according to the reaction: C4H8(g)

→

C4H6(g) 

H2(g)

To suppress side reactions, the 1-butene feed stream is diluted with steam in the ratio of 10 moles of steam per mole of 1-butene. The reaction is carried out isothermally at 525°C, and at this temperature 33% of the 1-butene is converted to 1,3-butadiene. How much heat is transferred to or from the reactor per mole of entering 1butene?

Solution

C4H8(g)  C4H6(g)  H2(g)

Using the basis of 1 mole C4H8 entering and only 33% reacts. The unreacted C4H8 and the diluent H2O pass through the reactor unchanged, and need not be included in the energy balance.

MCPH

tau

3.2638

4.016

0.004

0.000000991

8300

298.15

2.677008217

798.15

Then

H 798  H 298  MCPH  R  T  T0   Q

 H 798 



J mol

J J mol

Solution 4.46

Problem Statement

(a) An air-cooled condenser transfers heat at a rate of 12(Btu)·s to ambient air at 70(°F). If the air 1

temperature is raised 20(°F), what is the required volumetric flow rate of the air?

(b) Rework part (a) for a heat-transfer rate of 12 kJ·s , ambient air at 24°C, and a temperature rise of 13°C. 1

Solution

Assume Ideal Gas and P  1 atm

R

(a) T0 



3

BTU mol K

T

ICPH T



T n 

3



T



Q mol  38.899 R * ICPH s

n  R  T0 V   P (b) T0 



ft 3 sec

ICPH T T

 n 

3





Q mol  31.49 R  ICPH s

n  R  T0 V   P

m3 sec

Solution 4.47

Problem Statement

(a) An air-conditioning unit cools 50(ft) ·s of air at 94(°F) to 68(°F). What is the required heat-transfer rate in 3

(Btu)·s ? 1

(b) Rework part (a) for a flow rate of 1.5 m ·s , a temperature change from 35°C to 25°C, and units of kJ·s . 3

Solution

Assume Ideal Gas and P  1 atm

(a) T0  R

Rankine  3



atm ft 3 mol Rankine

ICPH T T



T

K V 

3

ft 3 sec

Rankine 

n 

P  V  R  T0

lbmol s



K 

m3 P  V n   sec R  T0

mol s





Q  R  ICPH  n  

(b) T0  R

K 

5

T atmm3 mol K

K V 

ICPH T T



3







Q  R  ICPH  n  

Solution 4.48

Problem Statement

A propane-fired water heater delivers 80% of the standard heat of combustion of the propane [at 25°C with CO2(g) and H2O(g) as products] to the water. If the price of propane is $2.20 per gallon as measured at 25°C, what is the heating cost in $ per million (Btu)? In $ per MJ?

Solution

First calculate the standard heat of combustion of propane

C3H8  5O2  3CO2(g)  4H2O (g)

H298 



J mol

Cost  2.20 Dollars/gal

Estimate the density of propane using the Rackett equation Tc  0.2857

1Tr 

Vsat  Vc Zc



cm3  mol

Zc 



Vc  0.2857

10.806

heating cost 



Vsat  cost  0.8  H 298

cm3 K T mol

K Tr 

cm3 mol

dollars  MJ

dollars BTU

Solution 4.49

Problem Statement

Determine the heat transfer (J·mol ) when one of the gases identified below is heated in a steady-flow process 1

from 25°C to 500°C at atmospheric pressure.

(a) Acetylene; (b) Ammonia; (c) n-Butane; (d) Carbon dioxide; (e) Carbon monoxide; (f) Ethane; (g) Hydrogen; (h) Hydrogen chloride; (i) Methane; (j) Nitric oxide; (k) Nitrogen; (l) Nitrogen dioxide; (m) Nitrous oxide; (n) Oxygen; (o) Propylene

Solution

The heat required to take an ideal gas from one temperature to another is given by T2

H   C p dT   R A  BT  CT 2  DT 2 dT   1 1  B C H  R  AT2  T1   T22  T12   T23  T13   D    T2 T1  2 3  

where the temperature dependence of the heat capacity has been expressed in the polynomial form used in SVNA. If we have set up a spreadsheet to evaluate the heat capacity integral like the example spreadsheet from the lecture notes, we can use it to evaluate the heat capacity integral in each case. (a) Using the coefficients for acetylene in the heat capacity polynomial, we have:

T 1 (K) 298.15

T 2 (K) 773.15 T2

ICPH 

A 6.132

C (1/K2) 0

B (1/K) 1.95E-03

D (K2) ICPH (K) DH (kJ/mol) -1.30E+05 3.14E+03 26.12

 1

1

 R dT  AT  T   2 T  T   3 T  T   D T  T  2

2 2

3 2

3 1

Solution continued on next page…

26.12 kJ/mol is required to heat acetylene from 25°C to 500 °C.

(b) Using the coefficients for ammonia in the heat capacity polynomial, we have: T 1 (K) 298.15

T 2 (K) 773.15 T2

ICPH 

A 3.578

C (1/K ) 0

B (1/K) 3.02E-03

D (K ) ICPH (K) DH (kJ/mol) -1.86E+04 2.43E+03 20.20

 1

1

 R dT  AT  T   2 T  T   3 T  T   D T  T  2

2 2

3 2

3 1

20.20 kJ/mol is required to heat ammonia from 25°C to 500 °C. (c) the heat required to take an ideal gas from one temperature to another is given by T2

H   C p dT   R A  BT  CT 2  DT 2 dT   1 1  B C H  R  AT2  T1   T22  T12   T23  T13   D    T2 T1  2 3  

Where the temperature dependence of the heat capacity has been expressed in the polynomial form. If we have set up a spreadsheet to evaluate the heat capacity integral, we can use it to evaluate the heat capacity integral in each cases. Using the coefficients for n-butane in the heat capacity polynomial, we have T0  298.15 K and T  773.15 K   R  ICPH T T



3



 0



6

J  mol

KJ mol

71.96 kJ/mol is required to heat n-butane from 25 C to 500 C.

(d) the heat required to take an ideal gas from one temperature to another is given by T2

Solution continued on next page…



3

 0

J  mol



6



KJ mol

71.96 kJ/mol is required to heat n-butane from 25 C to 500 C.

(e) Using the coefficients for carbon monoxide in the heat capacity polynomial, we have: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T 0(K) 298.15

T (K) 773.15

A 3.376

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 5.57E-04 0.00E+00 -3.10E+03 1738.9 14457.5

14.5 kJ/mol is required to heat carbon monoxide from 25°C to 500 °C. (f) Using the coefficients for ethane in the heat capacity polynomial, we have: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T 0(K) 298.15

T (K) 773.15

A 1.131

1 1  B 2 C T  T02  T 3  T03  D    2 3  T T0 









C (1/K2) D (K2) B (1/K) ICPH (K) DH (J/mol) 1.92E-02 -2.16E-06 0.00E+00 5114.5 42521.6

42.5 kJ/mol is required to heat ethane from 25°C to 500 °C.

Solution 4.50

Problem Statement

Determine the final temperature for one of the gases of the preceding problem if heat in the amount of 30,000 J·mol is transferred to the gas, initially at 25°C, in a steady-flow process at atmospheric pressure. 1

Solution

This is just like the previous problem, except that instead of being given the final temperature and computing Q, we are given Q and must compute the final temperatureTo do so, we can just try different values of the final temperature in the spreadsheet for evaluating the heat capacity integral until we get the desired QWe could use the ‘goal seek’ or ‘solver’ function in Excel to automate this trial and error. (a) For acetylene starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 835 K or 562°C: T 1 (K) 298.15

T 2 (K) 835.36 T2

ICPH 

R

A 6.132

dT  A T2  T1  

B (1/K) 1.95E-03

C (1/K ) 0

D (K ) ICPH (K) DH (kJ/mol) -1.30E+05 3.61E+03 30.00

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









(b) For ammonia starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 964 K or 691°C:

T 1 (K) 298.15

T 2 (K) 964.00 T2

ICPH 

R

A 3.578

dT  A T2  T1  

B (1/K) 3.02E-03

C (1/K2) 0

D (K2) ICPH (K) DH (kJ/mol) -1.86E+04 3.61E+03 30.00

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









T 1 (K) 298.15

T 2 (K) 534.4073 T2

ICPH 

R

A 1.935

dT  A  T2  T1  

C (1/K ) D (K ) B (1/K) ICPH (K) Q (J/mol) 3.69E-02 -1.14E-05 0.00E+00 3608 30000

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









Solution continued on next page…

(d) For CO2 starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 933 K or 660 °C:

T 1 (K) 298.15

T 2 (K) 932.9464 T2

ICPH 

R

C (1/K2) D (K2) B (1/K) ICPH (K) Q (J/mol) 1.05E-03 0.00E+00 -1.16E+05 3608 30000

A 5.457

dT  A  T2  T1  

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









(e) For carbon monoxide starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 1248 K or 975°C: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T 0(K) 298.15

T (K) 1248.143

A 3.376

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 5.57E-04 0.00E+00 -3.10E+03 3608.4 30000.0

(f) For ethane starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 664 K or 391°C: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T 0(K) 298.15

T (K) 664.2338

A 1.131

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 1.92E-02 -2.16E-06 0.00E+00 3608.4 30000.0

(g) For hydrogen starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 1298 K or 1025°C: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 298.15

T (K) 1298.38

A 3.249

B (1/K) 4.22E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 



C (1/K2) 0







ig D (K2) ICPH (K) D H (J/mol) 8.30E+03 3608.2 30000.0

Solution continued on next page…

(h) For hydrogen chloride starting at 25°C, the final temperature for a heat input of 30 kJ/mol is 1277 K or 1004°C: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 298.15

T (K) 1276.95

A 3.156

B (1/K) 6.23E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 



C (1/K2) 0







ig D (K2) ICPH (K) D H (J/mol) 1.51E+04 3608.2 30000.0

Solution 4.51 Problem Statement Quantitative thermal analysis has been suggested as a technique for monitoring the composition of a binary gas stream. To illustrate the principle, do one of the following problems.

(a) A methane/ethane gas mixture is heated from 25°C to 250°C at 1(atm) in a steady-flow process. If Q  11,500 J·mol , what is the composition of the mixture? 1

(b) A benzene/cyclohexane gas mixture is heated from 100°C to 400°C at 1(atm) in a steady-flow process. If Q  54,000 J·mol , what is the composition of the mixture? 1

(c) A toluene/ethylbenzene gas mixture is heated from 150°C to 250°C at 1(atm) in a steady-flow process. If Q  17,500 J·mol , what is the composition of the mixture? 1

Solution

Given T0 

T

J mol

Q

(a) Guess mole fraction of methane: y 0.5 Q  y  ICPH T T

ICPH T T

3

 3











6

 R

3



6

  R    y 



6

y  0.637 (b) T0 

T

J mol

Q

Guess mole fraction of benzene: y 0.5 Q  y  ICPH T T 

ICPH T T 

 3





6





 R  R    y 

y  0.245

T

J mol

Q

Guess mole fraction of benzene: y 0.5 Q  y  ICPH T T

ICPH T T

3

 

3





6

 6

 R

  R    y 

y  0.512

Solution 4.52 Problem Statement Saturated steam at 1(atm) and 100°C is continuously generated from liquid water at 1(atm) and 25°C by thermal contact with hot air in a counterflow heat exchanger. The air flows steadily at 1(atm). Determine values for m (steam)/ n (air) for two cases:

(a) Air enters the exchanger at 1000°C. (b) Air enters the exchanger at 500°C. For both cases, assume a minimum approach T for heat exchange of 10°C.

Solution

To solve the problem, apply an energy balance equation

TH 1

m c  H  n H  CP dT TH 2

m c  H  n H  R 

(a) TH 1 

air

K andTH 2 

Tin  25°C  298.15 K, Tboiling  100°C  373.15 K, and Tout  100°C  373.15 K

H  H

+ H

Solution continued on next page…

H liquid  R 



1



H vapor  R 



1



H boiling  H latent 

1



1

 1





1

B

C

1

H 

For air,

A

Assume as a basis

air

m c 

D 

 1 mol/s.



n H  R 

air

H

1



m c  n H

(b)

To solve the problem, apply an energy balance equation

TH 1

m c  H  n H  CP dT TH 2

m c  H  n H  R 

air

Solution continued on next page…

(a) TH 1 

K andTH 2 

Tin  25°C  298.15 K, Tboiling  100°C  373.15 K, and Tout  100°C  373.15 K

H  H

Hliquid  R 



1



H vapor  R 



1



H boiling  Hlatent 



1

 H

1

 1



 H

1



1

H 

For air,

A

Assume as a basis

air

m c 

B

C

D 

 1 mol/s.



n H  R  H

air



1

m c  n H

Solution 4.53 Problem Statement Saturated water vapor, i.e., steam, is commonly used as a heat source in heat-exchanger applications. Why saturated vapor? Why saturated water vapor? In a plant of any reasonable size, several varieties of saturated steam are commonly available; for example, saturated steam at 4.5, 9, 17, and 33 bar. But the higher the pressure the lower the useful energy content (why?), and the greater the unit cost. Why then is higher-pressure steam used?

Solution

Saturated because the large Hlv overwhelms the sensible heat associated with superheat. Water because it is cheap, available, non-toxic, and has a large Hlv. The lower energy content is a result of the decrease in Hlv with increasing T, and hence P. However, higher pressures allow higher temperature levels.

Solution 4.54 Problem Statement The oxidation of glucose provides the principal source of energy for animal cells. Assume the reactants are glucose [C6H12O6(s)] and oxygen [O2(g)]. The products are CO2(g) and H2O(l). (a) Write a balanced equation for glucose oxidation, and determine the standard heat of reaction at 298 K. (b) During a day an average person consumes about 150 kJ of energy per kg of body mass. Assuming glucose the sole source of energy, estimate the mass (grams) of glucose required daily to sustain a person of 57 kg.

(c) For a population of 275 million persons, what mass of CO2 (a greenhouse gas) is produced daily by mere respiration. Data: For glucose, H f 298 

1274.4 kJ· mol . Ignore the effect of temperature on the heat 1

of reaction.

Solution

(a) The reaction for oxidation of glucose to CO2 and water is the same as the combustion reaction: C6H12O6(s)  6 O2(g)  6 CO2(g)  6 H2O(l) The heat of reaction for this reaction is Hrxn  6*Hf(CO2(g))  6*Hf(H2O(l)) – Hf(C6H12O6(s)) – 6*Hf(O2(g)) Hrxn  6*(393.509)  6*(285.83) – (1274.4) – 6*(0)  2801.6 kJ/mol

(b) The total energy consumption for a 57 kg person (not a very big person) is 150 kJ/kg * 57 kg  8550 kJGiven that oxidation of glucose releases 2801.6 kJ/mol, this requires 8550/2802  3.05 moles of glucose per dayThe molecular weight of glucose is 180.16 g/mol, so this is 550 g of glucose

(c) Oxidation of 3.05 moles of glucose produces 18.3 moles of CO2, or 806 g of CO2Multiplying this by 275 million people gives 2.2210 g/day  2.2210 kg/day  222000 metric tons per day  81 million 11

metric tons per year.

Solution 4.55 Problem Statement A natural-gas fuel contains 85 mol-% methane, 10 mol-% ethane, and 5 mol-% nitrogen. (a) What is the standard heat of combustion (kJ·mol ) of the fuel at 25°C with H2O(g) as a product? 1

(b) The fuel is supplied to a furnace with 50% excess air, both entering at 25°C. The products leave at 600°C. If combustion is complete and if no side reactions occur, how much heat (kJ per mol of fuel) is transferred in the furnace? Solution Assume as a basis, 1 mole of fuel. 0.85 (CH4(g)  2 O2(g)  CO2(g)  2 H2O(g)) 0.10(C2H6 (g)  3.5 O2(g)  2 CO2(g)  3 H2O(g)) -----------------------------------------------------------------0.85 CH4(g)  0.10 C2H6(g)  2.05 O2(g)  1.05 CO2(g)  2 H2O(g) H

fCH 4

H

(a) H C 

 H

fCO 2

kJ H mol



fCO 2

  H



fN 2



fC 2 H 6

kJ H mol

 H

kJ H mol



fCH 4

fN 2





fO 2



kJ mol

 H fC 2 H 6 

 H

fO 2



kJ mol

(b) For complete combustion of 1 mole of fuel and 50% excess air, the exit gas will contain the following

numbers of moles:

nO 2 

mol nCO 2 

mol nH 2O 

mol

nN 2  11.618 mol

Solution continued on next page…

Air and fuel enter at 25 C and combustion products leave at 600 C.

T1 

A

B

C 2H 6

C2 H6



T1 





H 2O







mol



C

D



 



C2H 6

 





 H 2O  mol

  H 2O   mol

 





  H 2O 



 103

 106 



mol

Q H c  ICPH T T A B C D  R  

 10 5

kJ mol

Solution 5.1 Problem Statement Prove that it is impossible for two lines representing reversible, adiabatic processes on a PV diagram to intersect. (Hint: Assume that they do intersect, and complete the cycle with a line representing a reversible, isothermal process. Show that performance of this cycle violates the second law.)

Solution Shown at the bottom is a PV diagram with two adiabatic lines 1  2 and 2  3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3  1. An engine traversing this cycle would produce work. For the cycle U  0, and therefore by the first law, Q  W  0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is therefore false.

Solution 5.2 Problem Statement A Carnot engine receives 250 kJ·s of heat from a heat-source reservoir at 525°C and rejects heat to a heat-sink 1

reservoir at 50°C. What are the power developed and the heat rejected?

Solution

The work output of any heat engine can be given as the thermal efficiency times the heat input

W   QH For a Carnot engine, the thermal efficiency only depends on the high and low temperatures, and is given by   1

TC TH

For the problem at hand, QH is 250 kJ s, TC is 323 K, and TH is 798 K. So, the power developed is

 T   323 K  W  1  C  QH  1  250 kJ s1  148.8 kJ s1  798 K   TH  The heat rejected to the heat sink at 323 K is

QC  QH  W  250 kJ s1  148.8 kJ s1  101.2 kJ s1

Solution 5.3

Problem Statement

The following heat engines produce power of 95,000 kW. Determine in each case the rates at which heat is absorbed from the hot reservoir and discarded to the cold reservoir.

(a) A Carnot engine operates between heat reservoirs at 750 K and 300 K. (b) A practical engine operates between the same heat reservoirs but with a thermal efficiency   0.35.

Solution

(a) The thermal efficiency of the Carnot engine is   1

TC 300 K  1  0.6 TH 750 K

So, if 95,000 kW is produced, then 95,000/0.6  158333 kW of heat must be absorbed at 750 K, and 158333  95000  63333 kW must be rejected to the low temperature thermal reservoir.

(b) If the thermal efficiency of the real engine is 0.35, then the heat absorbed is 95000/0.35  271429 kW and the heat rejected is 271429  95000  176429 kW.

Solution 5.4

Problem Statement

A particular power plant operates with a heat-source reservoir at 350°C and a heat-sink reservoir at 30°C. It has a thermal efficiency equal to 55% of the Carnot-engine thermal efficiency for the same temperatures. (a) What is the thermal efficiency of the plant?

(b) To what temperature must the heat-source reservoir be raised to increase the thermal efficiency of the plant to 35%? Again  is 55% of the Carnot-engine value.

Solution

(a)

The thermal efficiency of a Carnot engine would be thermal  1 

QC QH

 1

TC TH

 1

303  0.514 623

If the thermal efficiency of the plant is 55% of this, it is thermal  0.55*0.514  0.283

(b)

As in part (a), the overall thermal efficiency is

 T   303    0.35 thermal  0.551  C   0.55 1   TH   TH  from which TH  833 K

Solution 5.5

Problem Statement

An egg, initially at rest, is dropped onto a concrete surface; it breaks. Prove that the process is irreversible. In modeling this process treat the egg as the system, and assume the passage of sufficient time for the egg to return to its initial temperature.

Solution

The energy balance for the over-all process is written: Q  U  EK  EP t

Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q. Thus heat is transferred to the surroundings. Solution continued on next page…

The total entropy change of the process is: Stotal  S  S surr t

Just as U for the egg is zero, so is S . Therefore, t

t  total   surr 

Qsurr T



Q T

Since Q is negative, Stotal is positive, and the process is irreversible.

Solution 5.6

Problem Statement

Which is the more effective way to increase the thermal efficiency of a Carnot engine: to increase TH with TC constant, or to decrease TC with TH constant? For a real engine, which would be the more practical way?

Solution

By Eq. (5.9) the thermal efficiency of a Carnot engine is:   

Differentiate:

   1    T   T  c TH H

Tc TH

   T 1   c and   TH  TH TH Tc

Since TC/TH is less unity, the efficiency changes more rapidly with TC than with TH. So in theory it is more effective to decrease TC. In practice, however, TC is fixed by the environment, and is not subject to control. The practical way to increase  is to increase TH. Of course, there are limits to this too.

Solution 5.7

Problem Statement

Large quantities of liquefied natural gas (LNG) are shipped by ocean tanker. At the unloading port, provision is made for vaporization of the LNG so that it may be delivered to pipelines as gas. The LNG arrives in the tanker at atmospheric pressure and 113.7 K, and represents a possible heat sink for use as the cold reservoir of a heat engine. For unloading of LNG as a vapor at the rate of 9000 m ·s , as measured at 25°C and 1.0133 bar, and assuming the 3

availability of an adequate heat source at 30°C, what is the maximum possible power obtainable and what is the rate of heat transfer from the heat source? Assume that LNG at 25°C and 1.0133 bar is an ideal gas with the molar mass of 17. Also assume that the LNG vaporizes only, absorbing only its latent heat of 512 kJ·kg at 113.7 K. 1

Solution

The maximum power would be obtained with a Carnot engine running between the heat source temperature (30 C) and the LNG temperature (113.7 K). The efficiency of a Carnot engine running between these two temperatures will be   1

TC 113.7 K  1  0.625 TH 303.15 K

So 62.5% of the heat absorbed will be converted to work, and the remaining 37.5% will be rejected into the LNG to vaporize it at 113.7 K. The heat absorbed will therefore be QH  QC/0.375 and the work will be W  0.625 QH  0.625/0.375 QC  1.667 QC. Solution continued on next page…

The rate at which heat can be rejected is equal to the flow rate of the natural gas multiplied by its heat of vaporization. Assuming that the 9000 m s of natural gas at 25 C and 1.0133 bar is an ideal gas with a molecular weight of 17, the 3

mass flow rate is m  17 g mol * 9000 m s * 101330 Pa / (8.3145 Pa m mol K * 298.15 K) 3

m  6254000 g s  6254 kg s Multiplying this by the heat of vaporization of 512 kJ kg gives QC  6254 kg s * 512 kJ kg  3.2 10 kJ s  3.2 10 kW 6

So, QH  3.2 10 kW/0.375  8.5 10 kW and W  1.667 * 3.2 10 kW  5.3 10 kW 6

Solution 5.8 Problem Statement With respect to 1 kg of liquid water: (a) Initially at 0°C, it is heated to 100°C by contact with a heat reservoir at 100°C. What is the entropy change of the water? Of the heat reservoir? What is Stotal? (b) Initially at 0°C, it is first heated to 50°C by contact with a heat reservoir at 50°C and then to 100°C by contact with a reservoir at 100°C. What is Stotal? (c) Explain how the water might be heated from 0°C to 100°C so that Stotal  0.

Solution

(a)

We can find the change in entropy and enthalpy of water from the steam tables. Because the properties of a

liquid are almost unaffected by pressure, we can use the values for saturated liquid water at each temperature. For water at 0C, S  0.0 kJ kg K and H  0.04 kJ kg (this is 0.01 degree below the reference state where all the properties are taken to be zero). At 100C, S  1.3069 kJ kg K and H  419.1 kJ kg. So, to heat the water from 0 to 100 C, the heat required is Q  H  419.1 kJ. The entropy change of the water is S  1.3069 kJ K. The entropy change of the reservoir is –Q/Tsurr  419.1/373.15 K  .123 kJ K. So, Stotal  1.307 – 1.123  0.184 kJ K.

(b)

The entropy change of the water is the same as in part (a), because the initial and final states are the same. If

Q1 is the heat required to heat the water to 50 C and Q2 is the heat required to heat the water the rest of the way to 100 C, then the entropy change of the reservoir is –Q1/323.15 – Q2/373.15. The enthalpy of water at 50C is 209.3 kJ kg, so Q1  209.3 kJ and Q2  419.1 – 209.3  209.8 kJ. So, the entropy change of the reservoir is 209.3/323.15  209.8/373.15  .210 kJ. So, Stotal  1.307  1.210  0.097 kJ K.

(c)

If we used an infinite number of heat reservoirs, starting infinitesimally above 0C and having each one

infinitesimally warmer than the last until the final one was infinitesimally above 100C, then the whole process would be reversible and Stotal would be zero.

Solution 5.9

Problem Statement A rigid vessel of 0.06 m volume contains an ideal gas, CV  (5/2)R, at 500 K and 1 bar. 3

(a) If heat in the amount of 15,000 J is transferred to the gas, determine its entropy change.

(b) If the vessel is fitted with a stirrer that is rotated by a shaft so that work in the amount of 15,000 J is done on the gas, what is the entropy change of the gas if the process is adiabatic? What is Stotal? What is the irreversible feature of the process?

Solution

(a)

The increase in entropy for an ideal gas with constant heat capacity is given by S  C p ln

P T  R ln    P0  T0

To apply this, we need to know the final temperature and pressure. When the gas is heated at constant volume, T2

Q  U t  n  Cv dT T1

where n is the number of moles of gas. For constant heat capacity, this is just

Q  U t  nCv T we can get n from the ideal gas law and the initial volume, temperature, and pressure as n  0.06 m *100000 Pa/(8.3145 Pa m mol K * 500 K)  1.443 mol 3

So, T 

Q nC v



15000 J 1.443 mol  2.5 8.31451J mol

1



 500 K

Solution continued on next page…

So, the final temperature is 500  500  1000 K. The final pressure is 2 bar (doubling the temperature at constant volume doubles the pressure). So, S  C p ln

P 7 T 5  R ln    R ln 2  R ln 2  R ln2  1.733R  14.4 J mol 1 K 1  P0  2 T0 2

Multiplying this by the number of moles gives S  1.443 mol *14.4 J mol K  20.8 J K. t

(b) Since the change in the state of the system is the same as in (a), the change in entropy is also the same, S  20.8 J K. The stirring process is irreversible - we put work into the stirring and it is converted irreversibly to t

heat.

Solution 5.10

Problem Statement An ideal gas, CP  (7/2)R, is heated in a steady-flow heat exchanger from 70°C to 190°C by another stream of the same ideal gas which enters at 320°C. The flow rates of the two streams are the same, and heat losses from the exchanger are negligible. (a) Calculate the molar entropy changes of the two gas streams for both parallel and countercurrent flow in the exchanger. (b) What is Stotal in each case? (c) Repeat parts (a) and (b) for countercurrent flow if the heating stream enters at 200°C.

Solution

Since the flow rates of the two streams are the same, and the heat capacities are the same (constant heat capacity) then the temperature change of the two streams is the same (in opposite directions). We can see this from the overall energy balance Hhot  Hcold  Q  0 Q is zero since no heat is lost to the surroundings. Each H can be written as CpT, so we have Hhot  Hcold  CpThot  CpTcold  0 from which Thot  Tcold  20 C. So, the outlet temperature of the hot stream is 320  120 C  200 C  473.15 K. The entropy change of the hot stream is Shot  Cp ln(Thot,out/Thot,in)  Cp ln(473.15/593.15)  3.5*R*0.2660  6.577 J mol K. The entropy change of the cold stream is Scold  Cp ln(Tcold,out/Tcold,in)  Cp ln(463.15/343.15)  3.5*R*0.2999  8.726 J mol K. This is independent of whether the flow is co-current or countercurrent.

(b) In either case, Stotal  8.726  6.577  2.15 J mol K.

(c) In this case, the outlet temperature of the hot stream is 200 C – 120 C  80 C  353.15 K. So, we have Shot  Cp ln(Thot,out/Thot,in)  Cp ln(353.15/473.15)  3.5*R*0.2925  8.512 J mol K. Scold is still the same, so Stotal  8.726 – 8.512  0.214 J mol K. This is much less, because the heat is always transferred across a small (10 C) temperature difference. This can only be the case for countercurrent flow, since with co-current flow and these inlet temperatures the two streams would come to the same temperature (135 C) and then their temperatures wouldn’t change any more.

Solution 5.11

Problem Statement

For an ideal gas with constant heat capacities, show that: (a) For a temperature change from T1 to T2, S of the gas is greater when the change occurs at constant pressure than when it occurs at constant volume. (b) For a pressure change from P1 to P2, the sign of S for an isothermal change is opposite that for a constant-volume change.

Solution

For an ideal gas with constant heat capacities, and for the changes T1  T2 and P1  P2, Eq. (5.14) can be rewritten as:

T  P  S  C p ln  2   Rln  2    T1   P1  T  (a) If P2  P1, SP  C p ln  2   T1 

and if V2  V1, then

T  2 P1 T1

Whence, SV  C p

 T     2   Rln T2   C  v    T1   T1 

 T   2     T1 

Since CP  CV, this demonstrates that SP  SV . P  (b) If T2  T1, ST  Rln  2   P1 

and if V2  V1, then

P2 P1

T  2 T1

Solution continued on next page…

Whence, SV  C p

 P     2   Rln P2   C  v  P     1  P1 

 P   2     P1 

This demonstrates that the signs for ST and SV are opposite.

Solution 5.12

Problem Statement

For an ideal gas prove that: T C ig dT S V  V  ln T R T V0 0 R

Solution

Start with the equation just preceding Eq. (5.10):

ig C ig dT dP dS CP dT   dlnP  P  R R T R T P

For an ideal gas PV  RT, and ln P  ln V  ln R  ln T. Therefore, dP dV dT   P V T

dP dT dV   P T V

Solution continued on next page…

Because

CPig C ig   V this reduces to R R

ig S CV dT  *  dlnV R R T

And integrating yields ig V  T C S dT V  *  ln   T0 R  V0  R T

********************** As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute dT/T  dP/P  dV/V:

ig S CP dP  R R P

Solution 5.13 Problem Statement A Carnot engine operates between two finite heat reservoirs of total heat capacity C Ht and CCt . (a) Develop an expression relating TC to TH at any time. (b) Determine an expression for the work obtained as a function of C Ht , CCt , TH and the initial temperatures TH0 and TC0 .

(c) What is the maximum work obtainable? This corresponds to infinite time, when the reservoirs attain the same temperature.

In approaching this problem, use the differential form of Carnot’s equation, dQH TH  dQC TC

and a differential energy balance for the engine, 

dQC  dQH  0

Here, QC and QH refer to the reservoirs. Solution As indicated in the problem statement the basic differential equations are: dW  dQH  dQC 

dQH dQC

T  H TC

(A) (B)

where QC and QH refer to the reservoirs.

(a) With dQH  CHt dTH and dQc  Cct dTc , Eq. (B) becomes:

C Ht dTH

T  H t TC Cc dTc

dTC Tc



C Ht dTH Cct dTC

Solution continued on next page…

Whence, d ln TC   d ln TH

where  

C Ht Cct

Integration from TH0 and TC0 to TH and TC yields:



TC  TH    TC 0  TH 0 

T  or TC  TC 0  H   TH 0 

(b) With dQH  CHt dTH and dQc  Cct dTc , Eq. (B) becomes dW  CHt dTH  Cct dTc

Integrating yields:

W  C Ht TH  TH 0  Cct Tc  TC 0  Eliminate TC by the boxed equation of Part (a) and rearrange slightly:

 T    T    W  C Ht TH 0  H    Cct TC 0  c     T    TH 0  C 0   (c) For infinite time, TH TC T, and the answer of Part (a) becomes: 

 T   T  TC 0  TH 0 



T   TC 0  H 0   T 

From which 

T   1  TC 0 TH 0 



T TC 0   TH 0  

and

1  1 T  TC 0   TH 0   TH 0

Because /(   1)  1  1/(   1), then: 1





  T   1  T    1  T  T  C0     C 0    and    TH 0  TH 0 TH 0   TH 0  

Solution continued on next page…

Because TH T, substitution of these quantities in the answer of Part (b) yields: 1       T    T        t  C 0  C 0     Cc TC 0   W  C TH 0      T  T     H0  H 0    t H

Solution 5.14 Problem Statement A Carnot engine operates between an infinite hot reservoir and a finite cold reservoir of total heat capacity C Ct . (a) Determine an expression for the work obtained as a function of C Ct , TH ( constant), TC, and the initial coldreservoir temperature TC0 . (b) What is the maximum work obtainable? This corresponds to infinite time, when TC becomes equal to TH.

The approach to this problem is the same as for Prob. 5.13.

Problem 5.13 A Carnot engine operates between two finite heat reservoirs of total heat capacity C Ht and CCt . (a) Develop an expression relating TC to TH at any time. (b) Determine an expression for the work obtained as a function of C Ht , CCt , TH and the initial temperatures TH0 and TC0 .

(c) What is the maximum work obtainable? This corresponds to infinite time, when the reservoirs attain the same temperature.

In approaching this problem, use the differential form of Carnot’s equation, dQH TH  dQC TC

and a differential energy balance for the engine, dW



dQC  dQH  0

Here, QC and QH refer to the reservoirs.

Solution

As indicated in the problem statement the basic differential equations are: dW  dQH  dQC 

dQH

T  H dQC TC

(A) (B)

where QC and QH refer to the reservoirs.

Solution continued on next page…

(a) With dQc  Cct dTc , Eq. (B) becomes: dQH Cct dTc

T  H TC

T or dQH  H Cct dTc TC

Substitute for dQH and dQC in Eq. (A):

dW  TH Cct

dTc TC

 Cct dTc

Integrate from TC0 to TC:

T  W  TH Cct ln c   Cct TC  TC 0   TC 0 

  T  or W  Cct TH ln c   TC  TC 0     TC 0  

(b) For infinite time, TC  TH, and the boxed equation above becomes:   T  W  Cct TH ln  c   TH  TC 0    TH  

Solution 5.15 Problem Statement A heat engine operating in outer space may be assumed equivalent to a Carnot engine operating between reservoirs at temperatures TH and TC . The only way heat can be discarded from the engine is by radiation, the rate of which is given (approximately) by: | Q C | kA TC4

where k is a constant and A is the area of the radiator. Prove that, for fixed power output W and for fixed temperature TH, the radiator area A is a minimum when the temperature ratio TC / TH is 0.75.

Solution

Write Eqs. (5.9) in rate form and combine to eliminate | Q H|:

W W  Q c

 

TC TH

  r or

W 1 r

 W  Q c

T where r  C TH

4 4 With | Q H| kATC   kArTH  this becomes:

 W   1   r  4 1   kArTH  or A   4  W     W   3   1  r 1  r   kTH  1  r r Differentiate, noting that the quantity in square brackets is constant:   W       dA  W   3 1     4r  31   4         dr  kTH   1  r r 4 1  r 2 r 3   kTH4   1  r 2 r 4         

Equating this equation to zero, leads immediately to: 4r  3 or r  0.75

Hence

TC TH

 0.75

Solution 5.16 Problem Statement Imagine that a stream of fluid in steady-state flow serves as a heat source for an infinite set of Carnot engines, each of which absorbs a differential amount of heat from the fluid, causing its temperature to decrease by a differential amount, and each of which rejects a differential amount of heat to a heat reservoir at temperature T . As a result of the operation of the Carnot engines, the temperature of the fluid decreases from T1 to T2. Equation (5.8) applies here in differential form, wherein  is defined as:   dW / dQ where Q is heat transfer with respect to the flowing fluid. Show that the total work of the Carnot engines is given by: W  Q  T S where S and Q both refer to the fluid. In a particular case, the fluid is an ideal gas, with CP  (7/2)R, and the operating temperatures are T1  600 K and T2  400 K. If T  300 K, what is the value of W in J·mol ? How much 1

heat is discarded to the heat reservoir at T ? What is the entropy change of the heat reservoir? What is Stotal? Eq. 5.8:



QC W

Solution

Starting with Eq. 5.8 T dW    dQ T

dW  dQ  T

dW yields

dQ T

Since dQ/T  dS, then

dW  dQ  T dS Integrating gives the required result.

T1 

T2 

Q  CP T2  T1  

W  Q  T S  

Sreservoir 



Q T



*

T  S  CP ln 2    T1 

Q  Q  W 

3539.7

T 

J mol

K 

K

 400     



*



J mol



J mol K

Stotal  S Sreservoir 

J mol K

 

J mol reservoir

j mol K

Solution 5.17 Problem Statement A Carnot engine operates between temperature levels of 600 K and 300 K. It drives a Carnot refrigerator, which provides cooling at 250 K and discards heat at 300 K. Determine a numerical value for the ratio of heat extracted by the refrigerator (“cooling load”) to the heat delivered to the engine (“heating load”).

Solution

The Carnot engine will have a thermal efficiency of thermal  1 

T 300  1 C  1  0.5 TH 600 QH

So, its work output will be W  0.5 QH.

For the Carnot refrigerator, we have W  QH QC  and QC  / QH   TC/TH, so we can write QH T W 300   1  H 1   1  0.2 TC 250 QC QC

So, the work requirement is W  0.2QC,refridgerator. If this is provided by the Carnot engine, we have W  0.5 QH,engine, so we have 0.5 QH,engine  0.2QC,refrigerator, or QH,engine  0.4QC,refrigerator. Because the engine operates over a bigger temperature difference than the refrigerator, we can ‘pump’ more heat with the refrigerator than we take into the engine.

Solution 5.18 Problem Statement An ideal gas with constant heat capacity undergoes a change of state from conditions T1, P1 to conditions T2, P2. Determine H (J·mol ) and S (J·mol ·K ) for one of the following cases. 1

(a) T1  300 K, P1  1.2 bar, T2  450 K, P2  6 bar, CP/R  7/2.

(b) T1  300 K, P1  1.2 bar, T2  500 K, P2  6 bar, CP/R  7/2. (c) T1  450 K, P1  10 bar, T2  300 K, P2  2 bar, CP/R  5/2. (d) T1  400 K, P1  6 bar, T2  300 K, P2  1.2 bar, CP/R  9/2. (e) T1  500 K, P1  6 bar, T2  300 K, P2  1.2 bar, CP/R  4.

Solution

 P  T (a) For an ideal gas with constant heat capacity, we have H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1 So, in this case, we have H  /2 R (T2 – T1)  3.5*8.314*(150 K)  4365 J/mol 7

and 7 T  P  S  R  ln 2  ln  2   8.314 3.5ln1.5  ln5  1.58 J mol 1 K 1  P1   2 T1

(b) In this case, we have H  /2 R (T2 – T1)  3.5*8.314*(200 K)  5820 J/mol 7

and 7 T  P  S  R  ln 2  ln  2   8.314 3.5ln1.667  ln5  1.48 J mol1 K 1  2 T1  P1 

 P  T (c) For an ideal gas with constant heat capacity, we have H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1 So, in this case, we have H  /2 R (T2 – T1)  3.5*8.314*(50 K)  3118 J/mol 5

Solution continued on next page…

and 5 T  P  S  R  ln 2  ln  2   8.314 2.5ln 0.667  ln 0.2  4.96 J mol 1 K 1  2 T1  P1 

 P  T (d) For an ideal gas with constant heat capacity, we have H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1 So, in this case, we have H  /2 R (T2 – T1)  4.5*8.314*(00 K)  3741 J/mol 9

and 9 T  P  S  R  ln 2  ln  2   8.314 4.5ln 0.75  ln 0.2  2.62 J mol1 K 1  P1   2 T1

 P  T (e) For an ideal gas with constant heat capacity, we have H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1 So, in this case, we have H  4R (T2 – T1)  4*8.314*(200 K)  6651 J/mol and   P  T S  R 3ln 2  ln  2   8.314 4 ln 0.6  ln 0.2  3.61 J mol 1 K 1   P1  T1

Solution 5.19

Problem Statement An ideal gas, CP  (7/2)R and CV  (5/2)R, undergoes a cycle consisting of the following mechanically reversible steps:  An adiabatic compression from P1, V1, T1 to P2, V2, T2.  An isobaric expansion from P2, V2, T2 to P3  P2, V3, T3.

 An adiabatic expansion from P3, V3, T3 to P4, V4, T4.  A constant-volume process from P4, V4, T4 to P1, V1  V4, T1. Sketch this cycle on a PV diagram and determine its thermal efficiency if T1  200°C, T2  1000°C, and T3  1700°C.

Solution

This cycle is shown below. Assuming constant specific heats, the efficiency is given by:



Wnet Qin



Qin  Qout Qin

1 T4  T1    1      T3  T2 

Temperature T4 is not given and must be calculated. The following equations are used to derive and expression for T4.

Solution continued on next page…

For adiabatic steps 1 to 2 and 3 to 4:

T1V1 1  T2 V2 1

and T3V3 1  T4 V4 1

For constant volume step 4 to 1:

V1 V4

For isobaric step 2 to 3: P2

P  3 T3 T3



T  Solving these 4 equations for T4 yields: T4 T1  2   T3 

7 CP  R 2

T1 

K 

T  T4 T1  2   T3 

5 Cv  R 2

T2 



CP Cv



K T3 

1.4



 1273.15   *   1973.15 



Plugging this into: 1 T4  T1  1 873.7855  473.15    1     1      T3  T2  1.4  1973.15  1273.15    0.591

Solution 5.20

Problem Statement

The infinite heat reservoir is an abstraction, often approximated in engineering applications by large bodies of air or water. Apply the closed-system form of the energy balance [Eq. (2.3)] to such a reservoir, treating it as a constantvolume system. How is it that heat transfer to or from the reservoir can be nonzero, yet the temperature of the reservoir remains constant?

Eq 2.3:

U t  Q  W

Solution

Because W 0, Eq. (2.3) here becomes:

Q  Ut  m CV T

A necessary condition for T to be zero when Q is non-zero is that m  . This is the reason that natural bodies (air and water) that serve as heat reservoirs must be massive (oceans) or continually renewed (rivers).

Solution 5.21

Problem Statement One mole of an ideal gas, CP  (7/2)R and CV  (5/2)R, is compressed adiabatically in a piston/cylinder device from 2 bar and 25°C to 7 bar. The process is irreversible and requires 35% more work than a reversible, adiabatic compression from the same initial state to the same final pressure. What is the entropy change of the gas?

Solution

A process that is both adiabatic and reversible is isentropic – that is it occurs at constant entropy. So, we can find the final temperature by setting S  0 in our equation for the entropy change of an ideal gas with constant heat capacity:

T  P  Sad,rev  C p ln rev   R ln 2   0  T1   P1  or T  P  C p ln  rev   R ln  2   T1   P1  R

Trev  P2  C p    T1  P1 

Of course, this is the equation that we obtained in chapter 3 when we considered mechanically reversible adiabatic processes (p. 75 in SVA). Thus, the final temperature (for reversible compression) is R

 P  Cp 7  7 Trev  T1  2   298.15   426.46 K  P1   2 

Solution continued on next page…

The work done can be found from the first law: U  Q  W, so W  U  CvT  5/2*R*(426.46 – 298.15)  2667 J/mol If the compression is irreversible, but still adiabatic, and requires 35% more work than this, then we have W  1.35*2667 J/mol  CvT  5/2*R*(T2 – 298.15), so the final temperature is T2  298.15  (1.35*2667)/(5/2*R)  471.4 K. Perhaps more intuitively, the temperature rise (471.4 – 298.15) is just 1.35 times the temperature rise in the adiabatic case (426.5 – 298.15).

Knowing the final temperature and pressure, we can now compute the change in entropy of the gas as

T  P   471.4  7 S  C p ln  2   R ln  2   7 / 2 * R ln   R ln    0.3506R   298.15   2   T1   P1  from which we obtain S  2.92 J/(mol K). If we subtracted from the above equation the equation for the reversible adiabatic compression:

T  P  Sad,rev  C p ln rev   R ln 2   0  T1   P1  then we would have a simpler equation for S:  T   P   T   P  S Sad ,rev  S  C p ln  2   R ln  2   C p ln  rev   R ln  2   T1   P1    T1   P1     T  T  T  S  C p ln  2   C p ln  rev   C p ln  2   T1   T1   Trev 

Solution 5.22

Problem Statement

A mass m of liquid water at temperature T1 is mixed adiabatically and isobarically with an equal mass of liquid water at temperature T2. Assuming constant CP, show

S t  Stotal  SG  2mCP ln

(T1  T2 ) / 2 (T1T2 )1/2

and prove that this is positive. What would be the result if the masses of the water were different, say, m1 and m2?

Solution

An appropriate energy balance here is: Q  Ht  0 Applied to the process described, with T as the final temperature, this becomes: m1CP T  T1   m2CP T  T2  

T

m1T1  m2T2 m1m2

The total entropy change as a result of temperature changes of the two masses of water: S t  m1CP ln

T T  m2CP ln T1 T2

Solution continued on next page…

These equations represent the general case. If m1  m2  m,

T2 T1T2

S t  mCP

or S t  mCP

S t  2mCP ln

T T1T2

T1 T2  T1T2

Since T T1  T2 

 T1T2

S is positive t

Solution 5.23

Problem Statement

Reversible adiabatic processes are isentropic. Are isentropic processes necessarily reversible and adiabatic? If so, explain why; if not, give an illustrative example.

Solution

Isentropic processes are not necessarily reversible and adiabatic. The term isentropic denotes a process for which the system does not change in entropy. There are two causes for entropy changes in a system: The process may be internally irreversible, causing the entropy to increase; heat may be transferred between system and surroundings, causing the entropy of the system to increase or decrease. Solution continued on next page…

For processes that are internally irreversible, it is possible for heat to be transferred out of the system in an amount such that the entropy changes from the two causes exactly compensate each other. One can imagine irreversible processes for which the state of the system is the same at the end as at the beginning of the process. The process is then necessarily isentropic, but neither reversible nor adiabatic.

More generally, the system conditions may change in such a way that entropy changes resulting from temperature and pressure changes compensate each other. Such a process is isentropic, but not necessarily reversible. Expansion of gas in a piston/cylinder arrangement is a case in point. It may be reversible and adiabatic, and hence isentropic. But the same change of state may be irreversible with heat transfer to the surroundings. The process is still isentropic, but neither reversible nor adiabatic. An isentropic process must be either reversible and adiabatic or irreversible and non-adiabatic.

Solution 5.24

Problem Statement

Prove that the mean heat capacitiesCPH and CP S are inherently positive, whether T  T0 or T  T0. Explain why they are well defined for T  T0.

Solution

By definition, C pH 

 C T0

dT P

T  T0

 C  T

dT P

T0 T

Solution continued on next page…

By inspection, one sees that for both T  T0 and T0  T the numerators and denominators of the above fractions have the same sign. Thus, for both cases CPH is positive. Similarly,

C pS 

 C T0

dT P

T  ln   T0 

 C  T

dT P

T  ln 0   T 

By inspection, one sees that for both T  T0 and T0  T the numerators and denominators of the above fractions have the same sign. Thus, for both cases CPS is positive. When T  T0, both the numerators and denominators of the above fractions become zero, and the fractions are indeterminate. Application of l’Hˆopital’s rule leads to the result: CPH  CPS  CP.

Solution 5.25 Problem Statement A reversible cycle executed by 1 mol of an ideal gas for which CP  (5/2)R and CV  (3/2)R consists of the following:  Starting at T1  700 K and P1  1.5 bar, the gas is cooled at constant pressure to T2  350 K.  From 350 K and 1.5 bar, the gas is compressed isothermally to pressure P2.  The gas returns to its initial state along a path for which PT  constant.

What is the thermal efficiency of the cycle?

Solution

Step 1-2: Volume decreases at constant P. Heat flows out of the system. Work is done on the system. W12    P V2  V1     R * T2  T1     

Step 2-3: Isothermal compression. Work is done on the system. Heat flows out of the system. W23  RT2

 P   3   RT 2  P   2

 P   3     P1 

Step 3-: Expansion process that produces work. Heat flows into the system. Since the PT product is constant,

PdT  TdP 

PV  RT

dP  dT P

 A

PdV  VdP  RdT

PdV  RdT  VdP  RdT  RT

dP P

Combine with (A) this becomes

PdV  RdT  RdT  2RdT

Moreover P3  P1

T1 T3

 P1

T1 T2

W31  PdV  RT1  T3    RT1  T2  V3

Q31  U 31  W31  CV T1  T3   2 RT1 T3  Solution continued on next page…

Q31  (CV  2R)T1  T3   (CP  R) T1  T2  

7 CP  R 2

T1 

W12    RT2  T1   

 P   3   P   1

Wnet Qin



T2 

W12  W23  W31 Q31

P1 

J * molK

W12  





W12  

J * molK

 3bar     1.5bar 

W31   RT1  T2    *

J  molK

K

K 

J mol



J mol

W23  RT2

W31  5.82 * 103

Q31  CP  R T1  T2    Q31 



T P3  P1 * 1  bar T2

bar

W12  W23  W31 Q31

* 

J  mol

* *

J mol

* R  R *

J mol



J mol

J  mol J 4 mol

J mol

 

Solution 5.26 Problem Statement One mole of an ideal gas is compressed isothermally but irreversibly at 130°C from 2.5 bar to 6.5 bar in a piston/cylinder device. The work required is 30% greater than the work of reversible, isothermal compression. The heat transferred from the gas during compression flows to a heat reservoir at 25°C. Calculate the entropy changes of the gas, the heat reservoir, and Stotal.

Solution If the gas was compressed reversibly, then the work required would be Wrev  nRTln(P2/P1)  1 mol *8.314 J/(mol K)*403 K * ln(6.5/2.5)  3203 J. If the actual work is 30% greater than this, it is W  1.3*3203  4163 J. Since the temperature of the gas does not change, its internal energy does not change either (for an ideal gas, U only depends on T and not on P or V). So, U  Q  W  0, and Q  W  4163 J. The entropy change of the gas is S  t

nRln(P2/P1)   mol * 8.314 J/(mol K) * ln(6.5/2.5)  7.94 J/K. The entropy change of the surroundings is Ssurr  Q/T  4163 J/298 K  13.96 J/K. So, the entropy change of the universe is Stotal  13.96 – 7.94  6.02 J/K.

Solution 5.27 Problem Statement For a steady-flow process at approximately atmospheric pressure, what is the entropy change of the gas: (a) When 10 mol of SO2 is heated from 200 to 1100°C? (b) When 12 mol of propane is heated from 250 to 1200°C?

(c) When 20 kg of methane is heated from 100 to 800°C? (d) When 10 mol of n-butane is heated from 150 to 1150°C? (e) When 1000 kg of air is heated from 25 to 1000°C? ( f ) When 20 mol of ammonia is heated from 100 to 800°C? (g) When 10 mol of water is heated from 150 to 300°C? (h) When 5 mol of chlorine is heated from 200 to 500°C? (i) When 10 kg of ethylbenzene is heated from 300 to 700°C?

Solution

(a) Here, we need to evaluate the entropy heat capacity integral: T

S   C p T0

P dT  R ln    P0  T

Which will give us the entropy change per mol. Multiplying that by the number of moles gives the total entropy change. For heat capacities expressed in the usual polynomial form: T P A S D     B  CT  3 dT  ln     R T T   P0  T 0

T  P S C D  A ln   B T  T0   T 2  T02   T 2  T02   ln   R 2 2  T0   P0 

and we have a spreadsheet from the lecture notes that is all set up to evaluate this. Putting the heat capacity coefficients for SO2 and the given temperatures into this expression gives: T1 (K) 473.15 T2

ICPS 

T2 (K) 1373.15

A 5.699

 T2 

C (1/K2) D (K2) B (1/K) 8.01E-04 0.00E+00 -1.02E+00

ICPS 6.793

S (J/(mol K)) 56.48

 RT dT  A ln  T   B T  T   2 T  T   2 T  T  T1

2 2

2 1

2 2

2 1

Solution continued on next page…

So, for 10 moles, the entropy change is 10*56.48  565 J/K.

(b) Putting the heat capacity parameters for propane and the given temperatures into the same expression as in part (a) gives T1 (K) 523.15 T2

ICPS 

T2 (K) 1473.15

A 1.213

 T2 

C (1/K ) D (K ) B (1/K) 2.88E-02 -8.82E-06 0.00E+00

ICPS 20.234

S (J/(mol K)) 168.23

 RT dT  A ln  T   B T  T   2 T  T   2 T  T  2

2 2

2 1

2 2

2 1

So, the total entropy change for 12 mol of gas is 12*168  2019 J/K.

(c) Following the same steps as in part (a) and covering to moles for methane the the total entropy change is 1250 mol of gas is 1250*59  73750 J/K.

(d) Following the same steps as in part (a) for n-butane the the total entropy change for 10 mol of gas is 10*248  2480 J/K.

(e) Following the same steps as in part (a) and covering to moles for air the the total entropy change is 34483 mol of gas is 34483*45  1.55 * 10 J/K. 6

(f) Following the same steps as in part (a) for ammonia the the total entropy change for 20 mol of gas is 20*49  980 J/K.

(g) Following the same steps as in part (a) for water the the total entropy change for 10 mol of gas is 10*11  110 J/K.

Solution continued on next page…

(h) Following the same steps as in part (a) for chlorine the the total entropy change for 5 mol of gas is 5*18  90J/K.

(i) Following the same steps as in part (a) and covering to moles for ethylbenzene the the total entropy change is 94 mol of gas is 94*142  13348 J/K.

Solution 5.28 Problem Statement What is the entropy change of the gas, heated in a steady-flow process at approximately atmospheric pressure, (a) When 800 kJ is added to 10 mol of ethylene initially at 200°C? (b) When 2500 kJ is added to 15 mol of 1-butene initially at 260°C? (c) When 106(Btu) is added to 40(lb mol) of ethylene initially at 500(°F)?

Solution

(a)

First, we have to find the final temperature, then we can find the entropy change from T

S   C p T0

dT T

(for an ideal gas at constant pressure).

For a steady-flow process, we have T2

Q  H  n  C p dT t

Solution continued on next page…

As in several of last week's homework problems, we want to do the heat capacity integral and try different values of the final temperature until we get the desired heat input. We can use our spreadsheet to find T2

Cp Q 800000 J   9622 K=  dT 1 1 nR 10 mol  8.314 J mol K R 473 K

So, we put the heat capacity parameters for ethylene into our spreadsheet and try different final temperatures until the above relationship is satisfied:

T1 (K) 473.15

T2 (K) 1374.5 T2

ICPH 

R

A 1.424

dT  A T2  T1  

C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPH (K) 9.622E+03

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









Doing so, we find a final temperature of 1374.5 K.

Now, we can evaluate 1374.5 K

C p dT S   R R T 473.15 K

using our similar entropy-integral-evaluating spreadsheet:

T1 (K) 473.15

T2 (K) 1374.5 T2

ICPS 

A 1.424

 T2 

C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPS 10.835

 RT dT  Aln  T   B T  T   2 T  T   2 T  T  2

2 2

2 1

2 2

2 1

so 1374.5 K

C p dT S    10.835 R R T 473.15 K

and S  10 mol * 8.31451 J mol K * 10.835  900.9 J K. t

Solution continued on next page…

(b) As in part (a), T2

Cp Q 2500000 J   20045 K=  dT 1 1 nR 15 mol  8.314 J mol K R 533.15 K

Putting the heat capacity parameters for 1-butene into the heat capacity integral and trying different final temperatures until this is satisfied gives: T1 (K) 533.15

T2 (K) 1413.7 T2

ICPH 

R

A 1.967

dT  A T2  T1  

C (1/K ) D (K ) B (1/K) 3.16E-02 -9.87E-06 0.00E+00

ICPH (K) 2.004E+04

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









So, the final temperature is 1413.7 K. Evaluating the entropy integral then gives T1 (K) 533.15

T2 (K) 1413.7 T2

ICPS 

 T1

A 1.967

C (1/K ) D (K ) B (1/K) 3.16E-02 -9.87E-06 0.00E+00

ICPS 21.307

T  C 2 D 2 dT  A ln  2   B T2  T1   T2  T12  T2  T12 RT T 2 2  1









so 1413.7 K

C p dT S    21.307 R R T 533.15 K

and S  15 mol * 8.31451 J mol K * 21.307  2657 J K. t

Cp Q 1000000 Btu   12588 R = 6993 K =  dT 1 1 nR 40 lbmol 1.986 Btu lbmol R R 533.15 K

Solution continued on next page…

Note that the initial temperature of 500 F has been converted to 533.15 K in the lower limit of the integral, and we have converted Q/(nR) to K as well, since our heat capacity parameters are in K. Putting the heat capacity parameters for ethylene into the heat capacity integral and trying different final temperatures until this is satisfied gives: T1 (K) 533.15

T2 (K) 1202.7 T2

ICPH 

R

A 1.424

dT  A T2  T1  

C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPH (K) 6.993E+03

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









So, the final temperature is 1202.7 K  1705.2 F. Evaluating the entropy integral then gives T1 (K) 533.15

T2 (K) 1202.7 T2

ICPS 

 T1

A 1.424

C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPS 8.244

T  C 2 D 2 dT  A ln  2   B T2  T1   T2  T12  T2  T12 RT T 2 2  1









so 1202.7 K

C p dT S    8.244 R R T 533.15 K

and S  40 lbmol * 1.986 Btu lbmol R * 8.244  654.9 Btu R. t

Solution 5.29 Problem Statement A device with no moving parts provides a steady stream of chilled air at 25°C and 1 bar. The feed to the device is compressed air at 25°C and 5 bar. In addition to the stream of chilled air, a second stream of warm air flows from the device at 75°C and 1 bar. Assuming adiabatic operation, what is the ratio of chilled air to warm air that the device produces? Assume that air is an ideal gas for which CP  (7/2)R.

Solution

A simple sketch of the system is as follows:

248.15 K 298.15 K

1 bar

5 bar

348.15 K 1 bar

A mole balance on the system gives nin  nhot  ncold An enthalpy balance (with Q  W  0, since the device has no moving parts and operates adiabatically) gives (H)fs  0, or ninHin  nhotHhot  ncoldHcold Substituting the mass balance into this, we have (nhot  ncold) Hin  nhotHhot  ncoldHcold nhot (Hhot – Hin)  ncold (Hin  Hcold) nhotCp(Thot – Tin)  ncold Cp(Tin  Tcold) 50nhot  50ncold so, finally, nhot  ncold  nin/2.ncold/nhot  0.94

Solution 5.30 Problem Statement An inventor has devised a complicated nonflow process in which 1 mol of air is the working fluid. The net effects of the process are claimed to be:  A change in state of the air from 250°C and 3 bar to 80°C and 1 bar.  The production of 1800 J of work.  The transfer of an undisclosed amount of heat to a heat reservoir at 30°C. Determine whether the claimed performance of the process is consistent with the second law. Assume that air is an ideal gas for which CP  (7/2)R.

Solution

First, we must use an energy balance (the first law) to find out how much heat is transferred out of the system. Then, nd

we need to write an entropy balance to find out whether the process is allowed by the 2 law. If the entropy nd

generation is positive, it is OK. If it is negative, then the process violates the 2 law and can’t be realized. The energy balance is just U  Q  W, where in this case the “delta” is after minus before. If it was out minus in for t

a continuous flow process at steady state, then we would use H instead. We can compute U from the change in t

temperature. Assuming constant heat capacity, it is just U  nCvT  1 mol * 5/2 R * (70 K)  3533 J. So, if t

W  800 J, then Q  U – W  3533  1800  733 J. The entropy balance can be written as t

Sgen  S – Q/T t

Solution continued on next page…

And the entropy change of the gas is S  n(Cpln(T2/T1) – Rln(P2/P1))  1 mol * R * (3.5*ln (353/523) – ln(1/3))  2.301 J/K t

So, we have Sgen  2.301 J/K  1733 J/ 303 K  3.42 J/K The entropy generation is positive. Hence, the claimed performance of the process is consistent with the second low.

Solution 5.31 Problem Statement Consider the heating of a house by a furnace, which serves as a heat-source reservoir at a high temperature TF. The house acts as a heat-sink reservoir at temperature T, and heat |Q| must be added to the house during a particular time interval to maintain this temperature. Heat |Q| can of course be transferred directly from the furnace to the house, as is the usual practice. However, a third heat reservoir is readily available, namely, the surroundings at temperature T , which can serve as another heat source, thus reducing the amount of heat required from the furnace. Given that TF  810 K, T  295 K, T  265 K, and |Q|  1000 kJ, determine the minimum amount of heat |QF| which must be extracted from the heat-source reservoir (furnace) at TF. No other sources of energy are available.

Solution The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the surroundings, a heat sink. Notation is as follows: |Q| Heat transfer to the house at temperature T

Solution continued on next page…

|QF| Heat transfer from the furnace at TF |Q | Heat transfer from the surroundings at T

The first and second laws provide the two equations: Q  QF  Q

and

Q T



Q T



QF TF



Combine these equations to eliminate |Q |, and solve for |QF|:  T  T  T   F QF  Q    TF  T  T

With T  295 K, TF  810 K, T  265 K, and |Q’|  1000 kJ

Q

Shown to the right is a scheme designed to accomplish this result. A Carnot heat engine operates with the furnace as heat source and the house as heat sink. The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat from the surroundings and discharges heat to the house. Thus the heat rejected by the Carnot engine (|Q1|) and by the Carnot refrigerator (|Q2|) together provide the heat |Q| for the house. The energy balances for the engine and refrigerator are:

Solution continued on next page…

|W|engine  |Q|  |Q 1| |W|refrig  |Q 2|  |Q’| Equation (5.7) may be applied to both the engine and the refrigerator:

Q 1

T   T Q

and

 T  T T Wengine  Q      Q T  T 

and

Q 2

T   T' Q

Combine these to give  T  T T Wrefrig  Q      Q  T'  T ' 

Since these are equivalent Q

T T T

Q

T  T T'

 T  T   T  Q  Q     T  T  T 

Solution 5.32 Problem Statement Consider the air conditioning of a house through use of solar energy. At a particular location, experiment has shown that solar radiation allows a large tank of pressurized water to be maintained at 175°C. During a particular time interval, heat in the amount of 1500 kJ must be extracted from the house to maintain its temperature at 24°C when the surroundings temperature is 33°C. Treating the tank of water, the house, and the surroundings as heat reservoirs, determine the minimum amount of heat that must be extracted from the tank of water by any device built to accomplish the required cooling of the house. No other sources of energy are available.

Solution

The process involves three heat reservoirs: the house, a heat source; the tank, a heat source; and the surroundings, a heat sink. Notation is as follows:

|Q| Heat transfer from the tank at temperature T |Q‘| Heat transfer from the house at T‘ |Q | Heat transfer to the surroundings at T

The first and second laws provide the two equations:

Q  Q   Q

and

Q T



Q T



Q T



Combine these equations to eliminate |Q |, and solve for |Q|:

 T  T   T  Q  Q     T  T  T 

With T  448.15 K , T’  297.15 K, T  306.15 K, and |Q’|  1500 kJ

Q

Shown below is a scheme designed to accomplish this result. A Carnot heat engine operates with the tank as heat source and the surroundings as heat sink.

The work produced by the engine drives a Carnot refrigerator (reverse Carnot engine) which extracts heat |Q_| from the house and discharges heat to the surroundings. The energy balances for the engine and refrigerator are: Solution continued on next page…

|W|engine  |Q|  |Q 1| |W|refrig  |Q 2|  |Q’| Equation (5.7) may be applied to both the engine and the refrigerator:

Q 1

T   T Q

and

 T  T T Wengine  Q      Q   T T

and

Q 2

T   T' Q

Combine these to give  T  T T Wrefrig  Q      Q   T '  T'

Since these are equivalent Q

T T T

Q

T  T T'

 T  T   T  Q  Q     T  T  T 

Solution continued on next page…

Solution 5.33 Problem Statement A refrigeration system cools a brine from 25°C to 15°C at a rate of 20 kg·s . Heat is discarded to the atmosphere at a 1

temperature of 30°C. What is the power requirement if the thermodynamic efficiency of the system is 0.27? The specific heat of the brine is 3.5 kJ·kg ·°C . 1

Solution

First, we should find the change in state of the brine – the change in enthalpy and entropy corresponding to the given change in temperature. Then, we can compute the ideal work, which is the minimum amount of work required to bring about the change of state by a reversible process. Finally, we divide the ideal work by the thermodynamic efficiency to get the actual work required. For the enthalpy change, we have H  CpT  3.5 kJ kg ºC * 40 ºC  40 kJ/kg For the entropy change, we have S  Cp ln(258.15/298.15)  0.5042 kJ kg K So, the ideal work is Wideal  H  T S  40 – 303.15*(0.5042)  12.85 kJ kg If the total flow rate is 20 kg/s, the the total ideal work rate is ideal

 20 kg/s* 12.85 kJ/kg  257 kW

The actual work is the ideal work divided by the thermodynamic efficiency: s

 257/0.27  952 kW.

Solution continued on next page…

Even with this modest efficiency, the amount of work required is still significantly smaller than the amount of heat removed from the brine (140 kJ/kg * 20 kg/s  2800 kW). The total heat removal rate from the system is the sum of these; 3752 kW is transferred to the surroundings at 30 ºC.

Solution 5.34 Problem Statement An electric motor under steady load draws 9.7 amperes at 110 volts; it delivers 1.25(hp) of mechanical energy. The temperature of the surroundings is 300 K. What is the total rate of entropy generation in W·K ? 1

Solution

The power input to the motor is P  IV  (9.7 A)*110 V  1067 Watts  1067 J s. Part of this is extracted as work, and the rest is converted to heat, which is dissipated to the surroundings. The work output is 1.25 hp  932 Watts. So, 1067 – 932  135 Watts (135 J s) is converted to heat. So, an entropy balance gives 

Q  S g T

All of the terms in the entropy balance are zero except for the one for heat flow to the surroundings. The electrical work in and mechanical work out don’t contribute to entropy directly. The system is at steady state, so the rate of change of the entropy of the system is zero, and there are no flow streams, so the flow stream terms are zero as well. So, the entropy generation is 1

135 J s S g   300 K

 0.45 J K1 s1

Solution continued on next page…

This entropy generation occurs when some of the electrical potential energy, which could be converted completely to work, is converted to heat by irreversibilities within the motor (friction, electrical resistance in the windings, etc.).

Solution 5.35 Problem Statement A 25-ohm resistor at steady state draws a current of 10 amperes. Its temperature is 310 K; the temperature of the surroundings is 300 K. What is the total rate of entropy generation SG ? What is its origin?

Solution

The power input to the resistor is P  I R  (10 A) *25-ohm  2500 Watts  2500 J s. This is completely converted 2

to heat, which is dissipated to the surroundings as heat. So, an entropy balance gives 

Q  S g T

All of the terms in the entropy balance are zero except for the one for heat flow to the surroundings. The system is at steady state, so the rate of change of the entropy of the system is zero, and there are no flow streams, so the flow stream terms are zero as well. So, the entropy generation is 2500 J s S g   300 K

1

 8.33 J K1 s1

This entropy generation occurs when the electrical potential energy, which could be converted completely to work, is converted to heat within the resistor.

Solution 5.36 Problem Statement Show how the general rate form of the entropy balance, Eq. (5.16), reduces to Eq. (5.2) for the case of a closed system.

Eq 5.2:  Stotal  0

Eq 5.16:

d (mS )CV dt

 ( Sm ) fs   j

Q j T , j

 SG  0

Solution

For a closed system the first term of Eq. (5.16) is zero, and it becomes:

d Sm cv dt

 j

Q j T , j

 SG  0

Solution continued on next page…

Where Q j is here redefined to refer to the system rather than to the surroundings. Nevertheless, the second t

term accounts for the entropy changes of the surroundings, and can be written simply as dS surr/dt:

d S cv dt



t d Ssurr 

 SG  0

Multiplication by dt and integration over finite time yields:

t t  cv  surr 

or  total 

Solution 5.37 Problem Statement A list of common unit operations follows: (a) Single-pipe heat exchanger; (b) Double-pipe heat exchanger; (c) Pump; (d) Gas compressor; (e) Gas turbine (expander); ( f ) Throttle valve; (g) Nozzle. Develop a simplified form of the general steady-state entropy balance appropriate to each operation. State carefully, and justify, any assumptions you make.

Solution

The general equation applicable here is Eq. (5.17):

Sm  fs  

Q j

j T , j

 SG  0

(a) Single-pipe heat exchanger : For a single stream flowing within the pipe and with a single heat source in the surroundings, this becomes:

S m 

Q j T , j

 S G  0

(b) Double-pipe heat exchanger: The equation is here written for two streams (1 and 2) flowing in two pipes. Heat transfer is internal, between the two streams, making Q  0. Thus,

S1 m 1 S2 m 2  SG  0 (c) Pump: For a pump operating on a single stream and with the assumption of negligible heat transfer to the surroundings (Q 0):

Sm  SG  0 (d) Gas compressor: For an adiabatic gas compressor the result is the same as for Part (c). (e) Gas turbine (expander): For an adiabatic turbine the result is the same as for Part (c). (f) Throttle valve: For an adiabatic throttle valve the result is the same as for Part (c). (g) Nozzle: For an adiabatic nozzle the result is the same as for Part (c).

Solution 5.38 Problem Statement Ten kmol per hour of air is throttled from upstream conditions of 25°C and 10 bar to a downstream pressure of 1.2 bar. Assume air to be an ideal gas with CP  (7/2)R.

(a) What is the downstream temperature? (b) What is the entropy change of the air in J·mol ·K ? 1

(d) If the surroundings are at 20°C, what is the lost work?

Solution

kmol T1  hr

m 

(a) Assuming an isenthalpic process: T1 T2 

(b)

T2 C 1 S P  dT  T1 R T R

P1 

P2 

 T   2   R  T1 

 P   2     P1 

bar

7 S  R 2

bar

S  17.628

J mol K

10 * 1000mol 17.628 mol K h

Wlost  T S

Wlost 

s h

W K

J mol

Solution 5.39 Problem Statement A steady-flow adiabatic turbine (expander) accepts gas at conditions T1, P1, and discharges at conditions T2, P2. Assuming ideal gases, determine (per mole of gas) W, Wideal, Wlost, and SG for one of the following cases. Take T  300 K.

(a) T1  500 K, P1  6 bar, T2  371 K, P2  1.2 bar, CP/R  7/2. (b) T1  450 K, P1  5 bar, T2  376 K, P2  2 bar, CP/R  4. (c) T1  525 K, P1  10 bar, T2  458 K, P2  3 bar, CP/R  11/2. (d ) T1  475 K, P1  7 bar, T2  372 K, P2  1.5 bar, CP/R  9/2. (e) T1  550 K, P1  4 bar, T2  403 K, P2  1.2 bar, CP/R  5/2.

Solution

(a)

First, we should find the enthalpy and entropy changes of the gas, from which we can compute the requested

quantities. For an ideal gas with constant heat capacity, we have P  T H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1

So, in this case, we have H  /2 R (T2 – T1)  3.5*8.314*(29 K)  3754 J/mol 7

Solution continued on next page…

and 7 T  P  S  R  ln 2  ln 2   8.314 3.5ln 0.742  ln 0.2  4.698 J mol 1 K 1  2 T1  P1 

The enthalpy balance for the expander is H  Q  W And, because the expander is adiabatic, Q  0.

So, the actual work is W  H  3754 J/mol Our sign convention is that W is negative when the system does work on the surroundings, so this is the right sign. The expander does 3754 J of work on the surroundings per mole of gas flowing through it.

The ideal work is Wideal  H  TS  3754 – 300*4.698  5163 J/mol.

Therefore, the lost work is Wlost  W – Wideal  1409 J/mol.

The entropy generation per unit mol can be found from the entropy balance: S   j

Qj Tj

 Sg

which for an adiabatic process is just S  Sg, because there are no flows of heat between the system and surroundings. Whatever increase in entropy the gas experiences is all due to entropy generation. Sg  4.698 J mol K. As a check, we know that the lost work should be given by Wlost  T S g , which for this case is Wlost  300 * 4.698  1409 J mol K (the same as our answer above). Solution continued on next page…

(b)

First, we should find the enthalpy and entropy changes of the gas, from which we can compute the requested

quantities. For an ideal gas with constant heat capacity, we have P  T H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1

So, in this case, we have H  4R (T2 – T1)  4*8.314*(74 K)  2461 J/mol and   P  T S  R 4ln 2  ln  2   8.314  4ln 0.8356  ln 0.4  1.643 J mol 1 K 1   P1  T1

The enthalpy balance for the expander is just H  Q  W and the expander is adiabatic, so Q  0.

Thus, the actual work is W  H  2461 J/mol Our sign convention is that W is negative when the system does work on the surroundings, so this is the right sign. The expander does 2461 J of work on the surroundings per mole of gas flowing through it.

The ideal work is Wideal  H  TS  2461  300*1.643  2954 J/mol.

Therefore, the lost work is Wlost  W  Wideal  493 J/mol.

The entropy generation per unit mol can be found from the entropy balance: S   j

Qj Tj

 Sg

Solution continued on next page…

which for an adiabatic process is just S  Sg, because there are no flows of heat between the system and surroundings. Whatever increase in entropy the gas experiences is all due to entropy generation. Sg  1.643 J mol K. As a check, we know that the lost work should be given by Wlost  T Sg , which for this case is Wlost  300 * 1.643  493 J mol K (the same as our answer above).

(c)

First, we should find the enthalpy and entropy changes of the gas, from which we can compute the requested

quantities. For an ideal gas with constant heat capacity, we have P  T H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1

So, in this case, we have H  /2 R (T2 – T1)  /2 *8.314*(67 K)  3064 J/mol 11

and  11 T  P  11  S  R  ln 2  ln  2   8.314  ln 0.8724  ln 0.3  3.767 J mol1 K 1  2 T1  P1   2 

The enthalpy balance for the expander is H  Q  W and the expander is adiabatic, so Q  0.

Thus, the actual work is W  H  3064 J/mol Our sign convention is that W is negative when the system does work on the surroundings, so this is the right sign. The expander does 3064 J of work on the surroundings per mole of gas flowing through it.

The ideal work is Wideal  H  TS  3064 – 300*3.767  4194 J/mol.

Solution continued on next page…

Therefore, the lost work is Wlost  W – Wideal  1130 J/mol.

The entropy generation per unit mol can be found from the entropy balance: S  

 Sg

which for an adiabatic process is just S  Sg, because there are no flows of heat between the system and surroundings. Whatever increase in entropy the gas experiences is all due to entropy generation. Sg  3.767 J mol K. As a check, we know that the lost work should be given by Wlost  T S g , which for this case is Wlost  300 * 3.767  1130 J mol K (the same as our answer above).

(d) First, we should find the enthalpy and entropy changes of the gas, from which we can compute the requested quantities. For an ideal gas with constant heat capacity, we have P  T H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1

So, in this case, we have H  /2 R (T2 – T1)  4.5*8.314*(03 K)  3854 J/mol 9

and 9 T  P  S  R  ln 2  ln  2   8.314 4.5ln 0.7832  ln 0.2143  3.663 J mol1 K1  2 T1  P1 

The enthalpy balance for the expander is just H  Q  W and since the expander is adiabatic, Q  0.

Solution continued on next page…

So, the actual work is W  H  3854 J/mol Our sign convention is that W is negative when the system does work on the surroundings, so this is the right sign. The expander does 3854 J of work on the surroundings per mole of gas flowing through it.

The ideal work is Wideal  H  TS  3854 – 300*3.663  4953 J/mol.

Therefore, the lost work is Wlost  W – Wideal  1099 J/mol.

The entropy generation per unit mol can be found from the entropy balance: S   j

Qj Tj

 Sg

which for an adiabatic process is just S  Sg, because there are no flows of heat between the system and surroundings. Whatever increase in entropy the gas experiences is all due to entropy generation. Sg  3.663 J mol K. As a check, we know that the lost work should be given by Wlost  T S g , which for this case is Wlost  300 * 3.663  1099 J mol K (the same as our answer above).

(e) First, we should find the enthalpy and entropy changes of the gas, from which we can compute the requested quantities. For an ideal gas with constant heat capacity, we have P  T H  Cp (T2 – T1), and S  C p ln 2  R ln  2   P1  T1

So, in this case, we have H  /2 R (T2 – T1)  2.5*8.314*(47 K)  3055 J/mol 5

Solution continued on next page…

and 5 T  P  S  R  ln 2  ln  2   8.314 2.5ln 0.7327  ln 0.3  3.546 J mol 1 K 1  2 T1  P1 

The enthalpy balance for the expander is H  Q  W and since the expander is adiabatic, Q  0.

So, the actual work is W  H  3055 J/mol Our sign convention is that W is negative when the system does work on the surroundings, so this is the right sign. The expander does 3055 J of work on the surroundings per mole of gas flowing through it.

The ideal work is Wideal  H  TS  3055 – 300*3.546  4119 J/mol.

Therefore, the lost work is Wlost  W – Wideal  1064 J/mol.

The entropy generation per unit mol can be found from the entropy balance: S   j

Qj Tj

 Sg

which for an adiabatic process is just S  Sg, because there are no flows of heat between the system and surroundings. Whatever increase in entropy the gas experiences is all due to entropy generation. Sg  3.546 J mol K. As a check, we know that the lost work should be given by Wlost  T S g , which for this case is Wlost  300 * 3.546  1064 J mol K (the same as our answer above).

Solution 5.40 Problem Statement Consider the direct heat transfer from a heat reservoir at T1 to another heat reservoir at temperature T2, where T1  T2  T . It is not obvious why the lost work of this process should depend on T , the temperature of the surroundings, because the surroundings are not involved in the actual heat-transfer process. Through appropriate use of the Carnot-engine formula, show for the transfer of an amount of heat equal to |Q| that Wlost  T Q

T1  T2  T SG T1T2

Solution

The figure below indicates the direct, irreversible transfer of heat |Q| from a reservoir at T1 to a reservoir at T2. The figure on the right depicts a completely reversible process to accomplish the same changes in the heat reservoirs at T1 and T2.

The entropy generation for the direct heat-transfer process is: 1 T  T2 1 SG  Q     Q 1   T1 T2  T1T2

Solution continued on next page…

For the completely reversible process the net work produced is Wideal: W1  Q

T1  T

W2  Q

and

Wideal  W1  W2  Q

T  T2 T2

T1  T2 T1T2

This is the work that is lost, Wlost, in the direct, irreversible transfer of heat |Q|. Therefore, Wlost  T Q

T1  T2 T1T2

 T SG

Note that a Carnot engine operating between T1 and T2 would not give the correct Wideal or Wlost, because the heat it transfers to the reservoir at T2 is not Q.

Solution 5.41 Problem Statement –1 An ideal gas at 2500 kPa is throttled adiabatically to 150 kPa at the rate of 20 mol·s . Determine SG and W lost

if T  300 K.

Solution T   S  R

P1 

 P   2     P   1

SG  m * S  W lost  T SG 

P2 

kPa m 

 150 KPa   * ln  kPa  

S 

kPa



mol sec J mol K

J sec K

J  W lost  sec K

Solution 5.42 Problem Statement An inventor claims to have devised a cyclic engine which exchanges heat with reservoirs at 25°C and 250°C, and which produces 0.45 kJ of work for each kJ of heat extracted from the hot reservoir. Is the claim believable?

Solution

One way to assess the claim would be to compare the claimed thermal efficiency (45%) to the efficiency of a Carnot engine operating between the same two temperatures. If the efficiency exceeds that of a fully reversible (Carnot) cycle, then it is not possible. An alternative way of assessing the claim is to write an entropy balance. If 0.45 kJ of work is extracted for each kJ of heat absorbed at 250ºC, then an enthalpy balance tells us that 0.55 kJ is transferred out of the system at 25 ºC. The only terms in the entropy balance are entropy generation and entropy changes of the surroundings due to heat transfer. There are no material flows in or out of the system, and no change in the entropy of the system (it is a cyclic process). So, we have Q Q 1 0.55 S g   H  C     6.7105 kJ/K TH TC 523.15 298.15

The amount of entropy generated per kJ of heat absorbed at 250 ºC is 6.7 × 10 kJ/K. The fact that this is negative 5

tells us that the process is not possible. It violates the second law of thermodynamics.

Solution 5.43 Problem Statement Heat in the amount of 150 kJ is transferred directly from a hot reservoir at TH  550 K to two cooler reservoirs at T1  350 K and T2  250 K. The surroundings temperature is T  300 K. If the heat transferred to the reservoir at T1 is half that transferred to the reservoir at T2, calculate: –1

(a) The entropy generation in kJ·K . (b) The lost work. How could the process be made reversible?

Solution

T 

(a) SG 

QH TH

T1 



Q1 T1

(b) Wlost T SG 



Q2 T2



T2 

K TH 

K QH  

150 50 100    SG  550 350 250



This could be reversible if SG 

Wlost 

Q1 

Q2 

kJ K

Wlost 

This could be done by the work lost

going back into the hot reservoir isentropically. If each reservoir were connected to two pistons and the work lost was going into the hot reservoir piston to compress and heat it up this could be achieved.

Solution 5.44 Problem Statement A nuclear power plant generates 750 MW; the reactor temperature is 315°C and a river with water temperature of 20°C is available. (a) What is the maximum possible thermal efficiency of the plant, and what is the minimum rate at which heat must be discarded to the river? (b) If the actual thermal efficiency of the plant is 60% of the maximum, at what rate must heat be discarded to the 3

–1

river, and what is the temperature rise of the river if it has a flow rate of 165 m ·s ?

Solution

(a) The maximum thermal efficiency is that of a Carnot cycle, which is 

T 293.15  1  C  1  0.5016 TH 588.15 QH

The amount of heat discarded to the river is QC  QH  W  750 MW*(1/0.5016  1)  745.3 MW. (b)

If the actual thermal efficiency is   0.6*0.5016  0.3009, then QC  750 MW*(1/0.3009  1)  1742 MW. If the volumetric flow rate of the river is 165 m /s, then the mass flow rate is 165 m /s * 1000 kg/m  165000 kg/s. The 3

energy input is 1.742×10 J/s, and therefore the energy input per kg of water is 1.742 × 10 / 1.65 × 10  10560 J/kg  10.56 kJ/kg. 9

The specific heat of water at 20 ºC is 4.18 kJ/(kg K), so the temperature rise is 10.56 kJ/kg/ 4.18 kJ/(kg K)  2.5 K. The water temperature should rise by about 2.5K, or 2.5 ºC, or 4.5 ºF. This is probably enough to have substantial ecological effects.

Solution 5.45 Problem Statement A single gas stream enters a process at conditions T1, P1, and leaves at pressure P2. The process is adiabatic. Prove that the outlet temperature T2 for the actual (irreversible) adiabatic process is greater than that for a reversible adiabatic process. Assume the gas is ideal with constant heat capacities.

Solution

Equation (5.14) can be written for both the reversible and irreversible processes:

Sirrev  

Tirrev T0

CPig *

Trev P P dT dT  ln  0  and Srev   CPig *  ln  0  T0 T  P  T  P 

The difference is that Srev  0, leaving Srev  

Trev T0

C Pig *

dT T

Since Sirrev must be greater than zero, Tirrev must be greater than Trev.

Solution 5.46 Problem Statement A Hilsch vortex tube operates with no moving mechanical parts and splits a gas stream into two streams: one warmer and the other cooler than the entering stream. One such tube is reported to operate with air entering at 5 bar and 20°C, and air streams leaving at 27°C and –22°C, both at 1(atm). The mass flow rate of warm air leaving is six times that of the cool air. Are these results possible? Assume air to be an ideal gas at the conditions given.

Solution

First, we need to do the energy balance to find out how much heat flows into or out of the vortex tube. Then, we can do the entropy balance and see whether the rate of entropy generation is positive or negative. We can use our usual heat capacity integrals to compute the enthalpy and entropy change of the gas streams. On a basis of one mole of gas th

entering the tube, we have 1/7 of a mole being cooled from 293 K to 251 K and we have 6/7 of a mole being heated from 293 K to 300 K. The enthalpy changes are: T 1 (K) 293.15

T 2 (K) 300.15 T2

ICPH 

R

A 3.355

dT  A  T2  T1  

T 1 (K) 293.15

T 2 (K) 251.15 T2

ICPH 

R

A 3.355

dT  A  T2  T1  

B (1/K) 5.75E-04

C (1/K ) 0

D (K ) ICPH (K) -1.60E+03 24.55

DH (J/mol) 204.12

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 



B (1/K) 5.75E-04







C (1/K ) 0

D (K ) ICPH (K) -1.60E+03 -146.57

DH (J/mol) -1218.58

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









So, the total enthalpy change is H  6/7*204.12  1/7*218.58  0.9 J, which is essentially zero compared to the change in enthalpy of the streams. Since the tube has no moving parts, it does no work on its surroundings. The energy balance is then just H  Q  W  Q 0 and the tube is adiabatic. Since there is no heat transfer to the surroundings, the only term left in the entropy balance is the S of the gas, and the entropy generated is just the entropy change of the gas.

Solution continued on next page…

For the two exiting gas streams, the entropy changes due to temperature changes are given by the heat capacity integrals:

T 1 (K) 293.15 T2

ICPS 

T 2 (K) 303.15

A 3.355

 T2 

C (1/K ) 0

D (K ) -1.60E+03

ICPS 0.118

 RT dT  A ln  T   B T  T   2 T  T   2 T  T 

T 1 (K) 293.15

T 2 (K) 251.15 T2



2 2

2 1

2 2

2 1

ICPS 

B (1/K) 5.75E-04

A 3.355

B (1/K) 5.75E-04

C (1/K ) 0

D (K ) -1.60E+03

ICPS -0.540

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT T 2 2  1









The total entropy change of the gas is S  R*(6/7*0.118  1/7*0.540 – ln(1.013/5))  13.5 J/(mol K)

This is positive, so the process is consistent with the 2 law.

Solution 5.47 Problem Statement (a) Air at 70(°F) and 1(atm) is cooled at the rate of 100,000(ft) (hr)1 to 20(°F) by refrigeration. For a surroundings 3

temperature of 70(°F), what is the minimum power requirement in (hp)? (b) Air at 25°C and 1(atm) is cooled at the rate of 3000 m ·hr1 to –8°C by refrigeration. For a surroundings 3

temperature of 25°C, what is the minimum power requirement in kW?

Solution

(a) First, we should probably convert the volumetric flow rate into a molar flow rate. We can use the ideal gas law with R  0.7302 ft atm lbmol R to obtain 3

n 

pV 1 * 100000   258.6 lbmol hr 1 RT 0.7302 * 529.67

For the heat capacity integrals, we need to use the temperatures in K, because the heat capacity coefficients in Appendix C are given for temperatures in K. So, the gas is being cooled from 294.3 K to 266.5 K. Using these temperatures with the spreadsheets for computing the entropy and enthalpy heat capacity integrals, we have: T2

R

ICPH  T0 ,T;A,B,C,D  

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









T0(K)

T (K)

B (1/K)

C (1/K )

D (K )

ICPH (K)

294.3

266.5

3.355

5.75E04

0.00E00

.60E03

97.2

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T0 (K)

T (K)

B (1/K)

C (1/K )

D (K )

ICPS

294.3

266.5

3.355

5.75E04

0.00E00

.60E03

0.347

Using R  1.986 Btu lbmol R, we have H  97.2 K * 1.8 R/K * 1.986 Btu lbmol R  347.4 Btu/lbmol H  0.347 * 1.986 Btu lbmol R  0.6889 Btu lbmol R Solution continued on next page…

Using these, with the surroundings temperature of 70 ºF  529.67 R, we have Wideal  347.5 – 529.67*(0.6889)  17.46 Btu/lbmol Multiplyijng by the total flow rate of 258.6 lbmol/hr gives ideal

 4514 Btu/hr  1.25 Btu/s  1.32 kW  0.986 hp

(b) First, we should probably convert the volumetric flow rate into a molar flow rate. We can use the ideal gas law with R  8.314 Pa m mol K to obtain 3

n 

pV 101325 * 3000   122629 mol hr 1  34.06 mol s1 RT 8.314 * 298.15

For the heat capacity integrals, we need to use the temperatures in K, because the heat capacity coefficients in Appendix C are given for temperatures in K. So, the gas is being cooled from 298.15 K to 265.15 K. Using these temperatures with the spreadsheets for computing the entropy and enthalpy heat capacity integrals, we have: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 298.15

T (K) 265.15

A 3.355 T

ICPS  T0 ,T;A,B,C,D  



T0 (K) 298.15

T (K) 265.15

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) ICPH (K)  H (J/mol) 0 -1.60E+03 -115.4 -959.4

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT 2 T T0   T0    

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS -0.410

S

(J/(mol K)) -3.4099

Using these, with the surroundings temperature of 25 ºC  298.15 K, we have Wideal  H  TS  959.4 – 298.15*3.4099  57.26 J/mol. Multiplying by the total flow rate of 34.06 mol/s gives ideal

 57.26 J/mol * 34.06 mol/s  1950 J/s  1.95 kW

Solution 5.48 Problem Statement A flue gas is cooled from 2000 to 300(°F), and the heat is used to generate saturated steam at 212(°F) in a boiler. The flue gas has a heat capacity given by: CP  3.83  0.000306T /  R R

Water enters the boiler at 212(°F) and is vaporized at this temperature; its latent heat of vaporization is –1

970.3(Btu)(lbm) .

–1

(a) With reference to a surroundings temperature of 70(°F), what is the lost work of this process in (Btu)(lb mole) of flue gas? –1

(b) With reference to a surroundings temperature of 70(°F), what is the maximum work, in (Btu)(lb mole) of flue gas, that can be accomplished by the saturated steam at 212(°F) if it condenses only, and does not subcool? (c) How does the answer to part (b) compare with the maximum work theoretically obtainable from the flue gas itself as it is cooled from 2000 to 300(°F)?

Solution

T1  H V 

Rankine T2  BTU lbm

MW 

g mol

 Rankine C p    T 

Rankine



0.000306T  R rankine 

Tsteam 

Rankine

Solution continued on next page…

(a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2

 C T  dT m

n gas

HV 

steam

m steam   T1C p T  dT  n gas H V

m steam  n gas

lb lbmol

Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas.

SG  SG steam  Sgas Calculate entropy generation per lbmol of gas: m SG  steam * Ssteam S gas n gas

Ssteam 

Sgas  

H V Tsteam

C p T 

 1.444

BTU lbm Rankine g BTU mol lbmol Rankine

dT 

m SG  steam * Ssteam Sgas  n gas

Wlost  SG * T

(b) H steam HV

Ssteam 

HV

BTU lbmol Rankine

BTU lbm Rankine

Ssteam 

Tsteam

BTU lbmol

Wlost 

Wideal H steam  T Ssteam 

BTU lbm

Calculate lbs of steam generated per lbmol of gas cooled. T2

m steam   T1C p T  dT  n gas H V

m steam  n gas

lb lbmol

Solution continued on next page…

Use ratio to calculate ideal work of steam per lbmol of gas

Wideal

m steam  n gas

BTU lbmol

Wideal H gas  T Sgas 

BTU lbmol

Solution 5.49 Problem Statement A flue gas is cooled from 1100 to 150°C, and the heat is used to generate saturated steam at 100°C in a boiler. The flue gas has a heat capacity given by: CP  3.83  0.000551T / K R –1

Water enters the boiler at 100°C and is vaporized at this temperature; its latent heat of vaporization is 2256.9 kJ·kg . –1

(a) With reference to a surroundings temperature of 25°C, what is the lost work of this process in kJ·mol of flue gas? –1

(b) With reference to a surroundings temperature of 25°C, what is the maximum work, in kJ·mol of flue gas, that can be accomplished by the saturated steam at 100°C if it condenses only, and does not subcool? (c) How does the answer to part (b) compare with the maximum work theoretically obtainable from the flue gas itself as it is cooled from 1100 to 150°C?

Solution

T1  H V 

K T2  kJ kg

MW 

 K C p   



T 

g mol

T  R 

Tsteam 

(a) First apply an energy balance on the boiler to get the ratio of steam flow rate to gas flow rate.: T2

n gas *  C p T  * dT  m steam * HV  T1

m steam   T1 C p T  * dT  n gas HV

m steam  n gas

g mol

Calculate the rate of entropy generation in the boiler. This is the sum of the entropy generation of the steam and the gas.

SG  SG steam  Sgas Calculate entropy generation per lbmol of gas: m SG  steam * Ssteam Sgas n gas

1000

H V Ssteam   Tsteam

* T2

C p T 

J kg K  K

* dT 

J mol K

m SG  steam * Ssteam Sgas  n gas

J mol K

Sgas  

Wlost  SG * T 

J * mol K

K  Wlost 

J kg K

kJ mol

Solution continued on next page…

(b) H steam HV

Ssteam 

HV

Ssteam 

Tsteam

J kg K

Wideal H steam  T Ssteam 

kJ kg

Calculate lbs of steam generated per lbmol of gas cooled. T2

m steam  T1 C p T  * dT  n gas HV

m steam  n gas

g mol

Use ratio to calculate ideal work of steam per lbmol of gas

Wideal *

m steam  n gas

kJ mol

Wideal H gas  T Sgas 

kJ mol

Solution 5.50 Problem Statement Ethylene vapor is cooled at atmospheric pressure from 830 to 35°C by direct heat transfer to the surroundings at a –1

temperature of 25°C. With respect to this surroundings temperature, what is the lost work of the process in kJ·mol ? Show that the same result is obtained as the work which can be derived from reversible heat engines operating with the ethylene vapor as heat source and the surroundings as sink. The heat capacity of ethylene is given in Table C.1 of App. C.

Solution

T1 

K T2 

(a) Sethylene  R ICPS T1 T2

3

K T 





6

Sethylene  0.0695 Qethylene  R ICPH T1 T2

3



6

kJ mol K



Qethylene  43.105

kJ mol

Wlost  T Sethylene  Qethylene 

kJ mol

Now place a heat engine between the ethylene and the surroundings. This would constitute a reversible process, therefore, the total entropy generated must be zero. calculate the heat released to the surroundings for Stotal  0. Sethylene 

Qc T



Solving for Qc Qc  T * Sethylene 

kJ mol

Now apply an energy balance around the heat engine to find the work produced. Note that the heat gained by the heat engine is the heat lost by the ethylene. QH  Qethylene

WHE  QH  QC

WHE 

kJ mol

The lost work is exactly equal to the work that could be produced by the heat engine.

Solution 6.1

Problem Statement

Starting with Eq. (6.9), show that isobars in the vapor region of a Mollier (HS) diagram must have positive slope and positive curvature.

Eq. 6.9:

dH = T dS + V dP

Solution

Starting from Eq 6.9 dH  TdS  VdP at constant pressure gives

 H    T  S  P

This shows that isobars have a positive slope. Differentiating this equation yields:

  2 H   T        S   S 2  P

Solution continued on next page…

Combine this with Eq 6.18 yields   2 H  T    S 2   C P

Giving isobars positive curvature.

Solution 6.2

Problem Statement

(a) Making use of the fact that Eq. (6.21) is an exact differential expression, show that: (CP / P )T =

T ( V / T )P

What is the result of application of this equation to an ideal gas?

(b) Heat capacities CV and CP are defined as temperature derivatives respectively of U and H. Because these properties are related, one expects the heat capacities also to be related. Show that the general expression connecting CP to CV is:  P   V  CP  CV  T      T   T  V

Show that Eq. (B) of Ex. 6.2 is another form of this expression.

Eq. 6.21:   V   dH  C P dT  V  T    dP     T   P 

Eq. B of Example 6.2:   CP  CV  TV     

Solution

(a) Application of Eq. (6.12) to Eq. (6.20) yields:

    V  T  V       T    C P    p     P      T T      

or  2V   V   C P   V             T  T   P   T 2   T  T

Finally this yields:  2V   CP       P   T  T 2  T P

Solution continued on next page…

For an ideal gas:

 V  R    P  T  P

 2V     T 2   P

(b) Equations (6.21) and (6.33) are both general expressions for dS, and for a given change of state both must give the same value of dS. They may therefore be equated to yield:

CP  CV 

 V  dT  P     dV    dP T  T V  T P

Restricting to constant P:  P   V  C P  CV  T      T V  T P

By Eqs. 3.3 and 6.34

 V  .    V  T P

 P       T V 

Combine this with the equation above yielding:   CP  CV  T V     

Solution 6.3 Problem Statement

If U is considered a function of T and P, the “natural” heat capacity is neither CV nor CP, but rather the derivative (U/T)P . Develop the following connections between (U/T )P , CP, and CV.:   P    V   U   V    C   T  P V    C P  P    C P   PV  CV  T    P   V      T  T   T P  T P     V P  

To what do these equations reduce for an ideal gas? For an incompressible liquid?

Solution

By the definition of H, U = H

PV. Differentiate with respect to T:

 U   H   V        P    T   T   T  P

 U   V     C P  P    T   T  P

Substituting for the final derivative with Eq. 3.3 yields

 U     C P  PV  T  P

Divide Eq 6.32 by dT and restrict to constant P yields

  P    V   U      Cv  T    P    T   T   T  P V P  

Solving for the two derivatives by substituting Eq 6.34 and 3.3 give:

 U      C v  T    P  V   T  P

Solution continued on next page…

These equations for an ideal gas, through some algebra where    U     Cp  R  T  P

1 1 and   reduce to T P

 U  and    Cv  T P

  and  

And these equations for an incompressible liquid,

reduce to

 U   U     C P and    CV  T   T  P P

Solution 6.4 Problem Statement The PVT behavior of a certain gas is described by the equation of state: P(V

b ) = RT

where b is a constant. If in addition CV is constant, show that:

(a) U is a function of T only. (b)

= const.

b) = const.

Solution

  P   (a) In general, dU  CV dT  T    P  dV     T V 

By the equation of state, P 

RT V b

(6.32)

 P     R  P  T  V b T V

(b) Substituting this derivative into Eq. (6.32) yields dU  CV dT

U  f T  only.

From the definition of H, dH  dU  d  PV  From the equation of state, d  PV   RdT  bdP Combining these two equations and the definition of part (a) gives: dH  CV dT  RdT  bdP  CV  R dT  bdP  H  Then,    CV  R  T P

By definition, this derivative is CP CP

C P  CV  R

  CP CV .

CV dT  PdV  

d V  b RT dV  RT V b V b

dT R   d lnV  b T CV

Solution continued on next page…

But from part (b),

C  CV R  P    1. Then CV CV d  1

From which: T V  b

T    d

V  b

T d

 1

V  b



 const.

Substitution for T by the equation of state gives  1

P V  bV  b  R



 const. or P V  b   const.

Solution 6.5 Problem Statement A pure fluid is described by the canonical equation of state: G

( T ) + RT ln P

T ) is a substance-

specific function of temperature. Determine for such a fluid expressions for V, S, H, U, CP, and CV. These results are consistent with those for an important model of gas-phase behavior. What is the model?

Solution

Knowing the Gibbs energy as a function of T and P, as given here, is sufficient information to determine all of the thermodynamic properties of a homogeneous fluid. Molar volume is related to Gibbs energy as    G  V      RT  P  RT T

So, if we have G/RT = (T)/RT + ln P, then taking the derivative with respect to P gives V/RT = 1/P, or V = RT/P.

Solution continued on next page…

Enthalpy is related to Gibbs energy as    G  H     2  T  RT  RT P

So, if we have G/RT = (T)/RT + lnP, then taking the derivative with respect to T gives T  H 1 dT    2 2 RT dT RT RT dT  H  T   T dT

Having H and G, we can get S from the definition of G as H

TS, or S = (G – H)/T :

 d T  1  S  T   T  T   RT ln P  T  dT  dT  S   R ln P dT

Likewise, we can get U from the definition of H as U + PV, or U = H

d T 

 RT   P    P  dT d T  U  T   T  RT dT U  T   T

Cp is, by definition, the derivative of H with respect to T at constant P, so dT   dT  d  T  dT  d    T   T   T  dT  dT  dT dT dT 2 2

Cp 

C p  T

d 2 T  dT 2

Solution continued on next page…

Likewise, Cv is the derivative of U with respect to T at constant V (since U is only a function of T in this model, it doesn’t matter whether we hold V constant, though).

Cv 

 d T  dT  d2 T  dT  d    T  T  RT   T  R     dT  dT dT dT dT 2 

Cv  T

d2  T  dT 2

 R  Cp  R

This model is that of an ideal gas.

Solution 6.6 Problem Statement A pure fluid, described by the canonical equation of state: G = F(T) + KP, where F(T) is a substance-specific function of temperature and K is a substance-specific constant. Determine for such a fluid expressions for V, S, H, U, CP, and CV. These results are consistent with those for an important model of liquid-phase behavior. What is the model?

Solution

It follows immediately from Eq.(6.11) that:  G  V     P T

 G  and S     T P

It follows immediately from Eq. (6.11) that:

V K

S 

dF T  dT

Solution continued on next page…

Once V and S (as well as G) are known, we can apply the equations: H  G T S

U  H  PV  H PK

These become: H  F T   K P T

dF T  dT

U  F T  T

dF T  dT

By Eqs. (2.16) and (2.20),  H  C P     T P

 U  and CV     T V

Because F is a function of temperature only, these become:

CP T

d2 F dT 2

and CV T

d2 F  CP dT 2

The equation for V shows it to be constant, independent of both T and P. This is the definition of an incompressible fluid. H is seen to be a function of both T and P, whereas U, S, CP, and CV are functions of T only. We also have the result that CP = CV. All of this is consistent with the model of an incompressible fluid, as discussed in Ex. 6.2.

Solution 6.7 Problem Statement Estimate the change in enthalpy and entropy when liquid ammonia at 270 K is compressed from its saturation l

pressure of 381 kPa to 1200 kPa. For saturated liquid ammonia at 270 K, V = 1.551 × 10 = 2.095 × 10

m ·kg , and

K .

Solution

For nearly incompressible liquids, we can assume the thermal expansivity (or coefficient of thermal expansion), , is constant. The dependence of entropy and enthalpy on pressure at constant temperature is given in a form appropriate for liquids in equations 6.25 and 6.26 on page 200 of SVNA:  dS     V  dP  T

 dH     1  T V  dP  T

If  is constant, then S  V P H  1  T V P

For  = 2.09510

K , V = 1.551  10

m kg , T = 270 K, and P = 819000 Pa, we get

S  2.095 103 K1  1.551103 m 3 kg1  819000 Pa =  2.66 Pa m 3 kg1 K1 =  2.66 J kg1 K1 H  1  2.095 103 K1  270 K  1.551  103 m 3 kg1  819000 Pa=552 J kg1

Solution 6.8 Problem Statement Liquid isobutane is throttled through a valve from an initial state of 360 K and 4000 kPa to a final pressure of 2000 kPa. Estimate the temperature change and the entropy change of the isobutane. The specific heat of 1

liquid isobutane at 360 K is 2.78 J·g ·°C . Estimates of V and may be found from Eq. (3.68).

Eq. 3.68:

Vsat = VcZc(1 Tr)

2/7

Solution

When a fluid is throttled through a valve, it doesn’t do any work on the surroundings and usually there is negligible heat flow, so the energy balance around the valve becomes: H = Q + W = 0 So, the enthalpy of the fluid leaving the valve is the same as the enthalpy of the fluid entering the valve. The dependence of enthalpy on pressure and temperature is given in a form appropriate for liquids in equation 6.27 on page 203 of SVNA: dH  C p dT  1  T VdP

Setting this equal to zero and rearranging, we have

dT  

V 1  T  Cp

Since the absolute changes in both temperature and molar volume are expected to be small (changing the pressure of a liquid doesn’t have a big effect) we can neglect the change in V, T, , and Cp and write

T  

V 1  T  Cp

P

Solution continued on next page…

We can estimate  and V from the Rackett equation if we neglect the effect of pressure on molar volume (since the Rackett equation only applies to the saturated vapor). The Rackett equation is: 2

sat

1Tr  7

 Vc Z c

1 T  7    Tc 

 Vc Z c

From which 2    1 V  ln V    T  7   ln Vc 1   ln Z c      Tc  V T T T     5

2ln Z c  T   1   7Tc  Tc 

For isobutene, we have Tc = 408.1 K, Vc = 262.7 cm mol , and Zc = 0.282. So, at 360 K 2

 360  7 1   408.1 

V  262.7 * 0.282

 132 cm 3 mol 1  0.000132 m 3 mol1

and 5

2ln0.282  360     1  7 * 408.1  408.1 

 0.00408 K 1

So, if the liquid isobutene is throttled from 4000 kPa to 2000 kPa, we have

T  

0.000132 m 3 mol1 1  0.00408 * 360 2.78 J g1 K1 * 58.123 g mol1

2000000 Pa  0.766 K 3

Note that we’ve carefully converted everything to SI units so that, if we recognize that 1 Pa m = 1 J, the units all work out.

Finding the entropy change is actually easier than the temperature change (maybe we should have done it first). The fundamental property relation for enthalpy is dH = TdS + VdP. So, if H is constant (dH = 0) we have dS = (V/T)dP. Again, neglecting changes in the absolute value of V and T, this can be integrated to get S = (V/T) P =

(0.000132 m mol )/(360 K)*( 2000000 Pa) = 0.733 J mol

K .

Solution 6.9 Problem Statement 3

One kilogram of water (V1 = 1003 cm ·kg ) in a piston/cylinder device at 25°C and 1 bar is compressed in a mechanically reversible, isothermal process to 1500 bar. Determine Q, W = 250 × 10

and = 45 × 10

bar

S given that

. A satisfactory assumption is that V is constant at its arithmetic

average value.

Solution

 dH     1  T V  dP  T

We find the volume as a function of pressure in the same way we did in homework 3. If  is constant, then we

1 1  V  can simply rearrange its definition,      to get dV  dP (at constant T). This can then be V  P  V T

V  integrated directly to get ln     P  P1  , or solving for V, V  V1 exp  P  P1  .  V1 

Solution continued on next page…

So, the final volume is V2  1003 cm3 kg1 exp45  106 bar1  1499 bar   937.6 cm3 kg1  9.376  104 m3 kg1 3

And the average volume is (1003 + 937.6)/2 = 970.3 cm kg . The reversible work is given by dW = PdV = PVdP. If we approximate V by the average of the initial and final volumes, then P2

W  Vave  PdP  P1

Vave 2

P  P  2 2

2 1

We can treat the expressions for dH and dS in the same way: S  Vave P H  1  T Vave P

For  = 250  10 K ,  = 45  10 bar , V = 1.003  10 m kg , T = 298.15 K, and P = 1499 bar 6

= 1.499  10 Pa, we get 8

V2  1003 cm3 kg1 exp45 106 bar1 1499 bar  937.5741 cm 3 kg1  9.375741  104 m 3 kg1 W  45 106 bar1 * 970.3 cm3 kg1 / 2 * 1500 2  12  bar 2  49120 bar cm3 kg1  4.91 kJ kg1 S  250  106 K1 * 970.3 cm3 kg1 * 1499 bar  363.6 bar cm3 kg1 K1  36.4 J kg1 K1 H  1  250  10

6

1

* 298.15 K * 970.3 cm kg 3

We can then find U from U = H

1

* 1499 bar  1.346 10 bar cm kg

1

 134.6 kJ kg

1

(PV) = 1.346  10 J kg – (1.5  10 Pa * 0.9376  10 m kg 5

 10 Pa * 1.003  10 m kg ) = 1.346  10 J kg 5

1.405  10 J kg = 5940 J kg = 5.94 kJ kg . 5

Finally, Q = U – W = 10800 J kg . 1

Solution 6.10 Problem Statement Liquid water at 25°C and 1 bar fills a rigid vessel. If heat is added to the water until its temperature reaches 50°C, what pressure is developed? The average value of between 25 and 50°C is 36.2 × 10 at 1 bar and 50°C is 4.42 × 10

K . The value of

bar , and may be assumed to be independent of P. The specific volume of 3

liquid water at 25°C is 1.0030 cm ·g .

Solution

The vessel is described as rigid, meaning that it is a constant and unchanging. This leaves eq. 3.6 as  T2  T1  

With T1 

T2 



  P2  P1  

5

K 1

5

K 1 * 



5

bar 1

P1  bar

Solving for P2 yields

P2 

 T2  T1 

 P1 

K 5

bar 1

K

 bar 

bar

Solution 6.11 Problem Statement R

Determine expressions for G , H , and S implied by the three-term virial equation in volume, Eq. (3.38).

Eq. 3.38:

Z

PV B C 1  2 RT V V

Solution

The three term virial equation is: Z  1  B  C 2

Plugging this into eqs. 6.58 , 6.59, and 6.60 yields The expression for the residual Gibbs energy in terms of molar density is P GR dp   Z    Z   0 RT p

Therefore, P GR dp    Bp  Cp2    Bp  Cp2   0 RT p

Or P GR    B  Cpdp   Bp  Cp2   0 RT

Or GR 3  B  C  2  RT 2

Solution continued on next page…

Or   3 G R  RT 2 Bp  Cp2  lnZ   2 

Also,  B dB   C 1 dC  2  HR  T           T dT  RT  T 2 dT     B dB   C 1 dC  2  SR  ln Z  T             R  T 2 dT    T dT 

Solution 6.12 Problem Statement R

Determine expressions for G , H , and S implied by the van der Waals equation of state, Eq. (3.39).

Eq. 3.39: P

RT a  V b V 2

Solution As we saw in class, to obtain the residual properties from a cubic equation of state, we must re-write the equation of state to give compressibility in terms of molar density. For the van der Waals equation of state, we have P

RT V b



a V



RT 1 



b

a 1 



 RT 1  b

 a 2

Solution continued on next page…

And

Z

P 1 a    RT 1  b RT

The expression for residual Gibbs energy in terms of molar density (derived in the notes and textbook) is G



   Z  1 0

d 

 Z  1  ln Z

Substituting the van der Waals equation of state into the integrand, we get G

RT G



 1  d a   1  Z  1  ln Z  1  b RT  

   0 



  



1    d  Z  1  ln Z RT   a

1  b   1  1  b a  G d  Z  1  ln Z      1  b  RT  RT 0   R  b G a       d  Z  1  ln Z  1  b RT  RT RT



  ln 1  b  

a RT

 Z  1  ln Z

It would probably be most convenient to put this all back in terms of Z by substituting



p ZRT

From which  GR bP  aP    ln1   Z  1  ln Z  RT ZRT  ZR2T 2

Finally, we could replace b and a by their dimensionless counterparts and q, via 

bP a aP , and q   RT bRT  R2T 2

Which leads to  GR   q  ln 1     Z  1  ln Z  RT Z  Z

Solution continued on next page…

This can be further simplified as follows:  Z     q GR   ln   Z  1  ln Z  Z  Z RT GR q  ln Z  ln  Z      Z  1  ln Z RT Z GR q  Z  1  ln  Z     RT Z

Similarly, for the residual enthalpy, we have   Z  d HR  Z  1  T     T   RT  0

The derivative that we need is  Z     a  T  RT 2 

So, 

HR a d a aP  Z 1T   Z 1  Z 1 2 RT RT RT  ZR 2T 0 HR q  Z 1 RT Z

Finally, SR H R GR   R RT RT R S q  q   Z  1    Z  1  ln  Z       R Z  Z  R S  ln  Z    R

Solution 6.13 Problem Statement R

Determine expressions for G , H , and S implied by the Dieterici equation:

P

 RT a   exp V b  VRT 

Here, parameters a and b are functions of composition only.

Solution

This equation does not fall within the compass of the generic cubic, Eq. (3.41); so we start anew. First, multiply the given equation of state by

V RT

 a  PV V   exp  VRT  RT V  b

Substitute:

Z

PV 1 V RT 

a q bRT

Then, Z

1 exp qb   b

Z

1 expq  1

With the definition,   b, this becomes:

Solution continued on next page…

Because



bP ZRT

Given T and P, these two equations may be solved iteratively for Z and . Because b is a constant, Eqs. (6.58) and (6.59) may be rewritten as:  GR d   Z    Z   0 RT 

  Z  d HR      Z 1 0 RT  T  

In these equations, Z is given by Eq. (A), from which is also obtained:

Z q  T  T 1   

ln Z ln    q

q 

The integrals in Eqs. (B) and (C) must be evaluated through the exponential integral, E(x), a special function whose values are tabulated in handbooks and are also found from such software packages as MAPLE® . The necessary equations, as found from MAPLE®, are:

d

 Z  1   expqE q1     E q  E q  lnq   

where is Euler’s constant, equal to 0.57721566. . . . and   Z  s T    q 0  T   

Once values for G R RT

H R RT





q E q      E q  

S R R come from Eq. (6.47). The difficulties of

integration here are one reason that cubic equations have found greater favor.

Solution 6.14 Problem Statement Estimate the entropy change of vaporization of benzene at 50°C. The vapor pressure of benzene is given by the equation:

ln P sat / kPa  13.8858 

2788.51 t / C  220.79

V .

(b) Use the Clausius/Clapeyron equation of Ex. 6.6.

Eq. 6.86:

dP sat H lv  dT TV lv Eq. 6.87 (Clausius/Clapeyron equation:

Z lv  Z v  Z l 

P sat V v P sat V l   1 0  1 RT RT

Solution







Solution continued on next page…

The pressure is the vapor pressure given by the Antoine equation:

  

P t  



2788.51   t  220.79 

kPa

dP kPa  1.382 dT K

(a) The entropy change of vaporization is equal to the latent heat divided by the temperature. For the Clapeyron equation, Eq. (6.85), we need the volume change of vaporization. For this we estimate the liquid volume by Eq. (3.59) and the vapor volume by the generalized virial correlation. For benzene:

 c



Pc 

Zc 



cm3 mol

 c



Pr 

Using 3.61, 3.62, 3.59, and 3.57 yields B0 

Vvap 

B1  

P RT    B0   * B1  r    P  Tr 

cm3 mol

Using Eq 3.68 yields

Vliq  Vc Z c

0.2857

1Tr 

 93.164

cm3 mol

Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change of vaporization: S 

dP J V  Vliq   100.94 mol K dT vap

Solution continued on next page…

(b) Here for the entropy change of vaporization: S 

RT dP  P dT

J mol K

Solution 6.15 Problem Statement

Let P1sat at and P2sat at be values of the saturation vapor pressure of a pure liquid at absolute temperatures T1 sat

and T2. Justify the following interpolation formula for estimation of the vapor pressure P at intermediate temperature T:

ln P sat  ln P1sat 

T2 T  T1 

P sat ln 2sat T T2  T1  P1

Solution

Assume the validity for purposes of interpolation of Eq. (6.89), and write it for T2, T, and T1: ln P2sat  A 

B T2

ln P sat  A 

B T

ln P1sat  A 

B T1

Solution continued on next page…

Subtract (C) from (A): 1 1  T  T  P sat 1 ln 2sat  B    B 2  T1 T2   T1T2  P1

Subtract (C) from (B):

 1 1  T  T  P sat 1  B    B  sat  P1  T1 T   T1T 

The ratio of these two equations, upon rearrangement, yields the required result.

Solution 6.16 Problem Statement Assuming the validity of Eq. (6.89), derive Edmister’s formula for estimation of the acentric factor: 3       log Pc  1 7  1   

where ≡ Tn/Tc , Tn is the normal boiling point, and Pc is in (atm).

Eq. 6.89:

ln P sat  A 

B T

Solution Write Eq. (6.89) in log10 form:

log P sat  A 

log PC  A 

Apply at the critical point:

By difference,

If P

sat

B T B TC

 1 1  T  1   log Prsat  B     B  r  TC T   T 

is in (atm), then application of (A) at the normal boiling point yields:

log1  A 

B or A  B / Tn Tn

With ≡ Tn/Tc, Eq. (B) can now be written:   T  T      n  Pc  B     B  c  B   TnTc   Tn  Tn Tc   T  Whence, B   n log Pc  1   

Equation (C) then becomes:  T  T  1     Tr  1   log Pc log Pc   log Prsat   n    r       T      Tr  3    Apply at Tr = 0.7: log( Prsat )T  0.7     logPc  1 r 7  1   

Solution 6.17 Problem Statement Very pure liquid water can be subcooled at atmospheric pressure to temperatures well below 0°C. Assume that 1 kg has been cooled as a liquid to subcooled liquid. If the subsequent change occurs adiabatically at atmospheric pressure, what fraction of the S total for the process, and what is its irreversible 1

feature? The latent heat of fusion of water at 0°C is 333.4 J  g , and the specific heat of subcooled liquid water is 4.226 J  g

 °C .

Solution

The process may be assumed to occur adiabatically and at constant pressure. It is therefore isenthalpic, and may for calculation purposes be considered to occur in two steps: (1) Heating of the water from 6 deg C to the final equilibrium temperature of 0 deg C. (2) Freezing of a fraction x of the water at the equilibrium T. Enthalpy changes for these two steps sum to zero:

C P  T  x H fusion  0

Cp 

J gK

H fusion 

J g

T  K

Solution continued on next page…

Solving for x yields x 

CP T  0.076 H fusion

The entropy change for both steps is:

 T  xH fusion  273  0.076 * 333.4 J S  CP ln  2    4.226  * ln  T2 gK    T1  

3

J gK

The freezing process itself is irreversible, because it does not occur at the equilibrium temperature of 0 deg C.

Solution 6.18 Problem Statement The state of 1(lbm) of steam is changed from saturated vapor at 20 (psia) to superheated vapor at 50 (psia) and 1000(°F). What are the enthalpy and entropy changes of the steam? What would the enthalpy and entropy changes be if steam were an ideal gas?

Solution

For any problem involving water, we probably want to use the steam tables. The needed data is on p. 762 of in appendix F.4. Solution continued on next page…

For saturated vapor at 20 psia, we find H = 1156.3 Btu/lbm and S = 1.7320 Btu/(lbm R) For superheated steam at 50 psia and 1000 ºF, H = 1533.4 Btu/lbm and S = 1.9977 Btu/(lbm R). H = 1533.4 –

S = 1.9977 –

1.7320 = 0.2657 Btu/(lbm R). To find the property changes for an ideal gas, one approach is to use the steam table entries for a pressure of 1 psia (low enough that the vapor will behave very nearly ideally). For the initial temperature of 227.96 ºF, we must interpolate between the entries for 200 ºF and 250 ºF. Doing so, we get H = 1150.2 + (27.96/50)*(1172.9 – 1150.2) = 1162.9 Btu/lbm S = 2.0509 + (27.96/50)*(2.0841 – 2.0509) = 2.0695 Btu/(lbm R) At 1 psia and 1000 ºF, we find H = 1534.9 Btu/lbm and S = 2.4296. The ideal gas enthalpy is independent of id

H for going from 20 psia and 227.96 ºF to 50 psia and 1000 ºF is simply id

H = 1534.9 – 1162.9 = 372.0 Btu/lbm. This is within 1.5% of the actual value. For the entropy change, we have to also include the ideal gas dependence of entropy on pressure, so we get S = 2.4296 – 2.0695 + R/MW*ln(20/50) S = 0.3601 + 1.986/18.02*ln(0.4) = 0.2591 Btu/(lbm R) Note that we use R = 1.986 Btu/(lbmol R) and we must divide by the molecular weight (18.02 lbm/lbmol) to get the entropy change per unit mass. This is within 2.5% of the actual value. Non-ideality is not terribly important here, because the reduced pressures are relatively low.

An alternative, and equally valid, means of computing the ideal gas changes would be to do the ideal gas heat capacity integrals for enthalpy and entropy. Using our usual spreadsheet, and doing the necessary unit conversions, we obtain: Solution continued on next page…

T 1 (K) 382.02

T 2 (K) 810.93 T2

ICPH 



A 3.47

C (1/K2) D (K2) B (1/K) ICPH (K) DH (J/mol) 1.45E-03 0.00E+00 1.21E+04 1876.0 15597.3

 1 B C 3 1 dT  A  T2  T1   T22  T12  T2  T13  D    R 2 3  T2 T1 









MW (g/mol) 18.02 DH (J/g) 865.6 DH (Btu/lbm) 372.1

Which agrees well with the above result. Likewise, for the change in entropy due to change in temperature, we have: T1 (K) 382.02

T 2 (K) 810.93 T2

ICPS 



A 3.47

C (1/K2) D (K2) B (1/K) 1.45E-03 0.00E+00 1.21E+04

ICPS 3.266

T  C D 2 dT  A ln  2   B  T2  T1   T22  T12  T2  T1 2 RT 2 2  T1 









DS (J/(mol K)) 27.15 MW (g/mol) 18.02 DS (J/(g K)) 1.507 DS (Btu/(lbm R)) 0.3599

We still have to add the effect of pressure changes to this, to get id

S = 0.3599 + 1.986/18.02*ln(0.4) = 0.2589 Btu/(lbm R) Again, this is in excellent agreement with the values we obtained using the steam table data at 1 psia.

Solution 6.19 Problem Statement A two-phase system of liquid water and water vapor in equilibrium at 8000 kPa consists of equal volumes of t

liquid and vapor. If the total volume V = 0.15 m , what is the total enthalpy H and what is the total entropy S ?

Solution Looking in the steam tables in the back of the book, we see that the vapor pressure is 7889.7 kPa at 567.15 K and 8118.9 at 569.15 K. Interpolating linearly between these, we get T = 567.15 + (569.15

567.15)*(8000

T = 567.15 + 0.4812*(569.15

7889.7)/(8118.9

7889.7)

567.15) = 568.11 K

Similarly, the liquid specific volume is l

V = 1.381 + 0.4812*(1.388

1.381) = 1.384 cm g

And the vapor specific volume is v

V = 23.90 + 0.4812*(23.13 3

23.90) = 23.53 cm g 3

So, if we have 0.075 m = 75000 cm each of liquid and vapor (so the total volume is 0.15 m ), then we have 75000/1.384 = 54190 g = 54.19 kg of liquid and 75000/23.53 = 3187 g = 3.187 kg of vapor.

Interpolating between the steam table entries, the specific enthalpy of the liquid is l

H = 1311.8 + 0.4812*(1322.8

1311.8) = 1317.1 kJ kg

2761.5) = 2759.9 kJ kg

And the specific enthalpy of the vapor is v

H = 2761.5 + 0.4812*(2758.2 So, the total enthalpy is t

H = 54.19*1317.1 + 3.187 * 2759.9 = 80169 kJ

Solution continued on next page…

Likewise, interpolating between the steam table entries, the specific entropy of the liquid is l

S = 3.1985 + 0.4812*(3.2173

3.1985) = 3.2075 kJ kg K

And the specific entropy of the vapor is v

S = 5.7545 + 0.4812*(5.7392

5.7545) = 5.7471 kJ kg K

So, the total entropy is t

S = 54.19*3.2075 + 3.187 * 5.7471 = 192.1 kJ K

Solution 6.20 Problem Statement

A vessel contains 1 kg of H2O as liquid and vapor in equilibrium at 1000 kPa. If the vapor occupies 70% of the volume of the vessel, determine H and S for the 1 kg of H2O.

Solution

We know that the overall molar volume, in terms of the fractions of the water in the liquid and vapor phases, is V = xlVl + xvVv. The fraction of the total volume that is liquid is xlVl/V, and the fraction of the volume that is vapor is xvVv/V. So, if the vapor occupies 70% of the total volume, then xvVv/V = 0.7, and xlVl/V = 0.3. Dividing one of these by the other gives (xvVv)/(xlVl)=7/3. Finally, using the fact that xv = 1 – xl, we have ((1 – xl)Vv)/(xlVl)=7/3, or Solution continued on next page…

xl*(7/3Vl + Vv) = Vv, or xl = Vv/(7/3Vl + Vv) 3

For saturated steam at 1000 kPa (180 C), we find in the steam tables that Vl = 1.128 cm /g, and Vl = 193.8 3

cm /g. Using these values in the above expression gives xl = 0.987. Even though the vapor occupies 70% of the volume, only 1.3% of the mass is actually vapor. The fraction of the mass that is vapor (the quality of the steam) is 0.013. The enthalpy of the liquid is 763.1 kJ/kg and the enthalpy of the vapor is 2776.3 kJ/kg, so the enthalpy of the mixture is H = 0.987*763.1 + 0.013*2776.3 = 789.3 kJ/kg. The entropy of the liquid is 2.1393 kJ/(kg K), and the entropy of the vapor is 6.5819 kJ/(kg K). Thus, the entropy of the mixture is S = 0.987*2.1393 + 0.013*6.5819 = 2.197 kJ/(kg K).

Solution 6.21 Problem Statement A pressure vessel contains liquid water and water vapor in equilibrium at 350(°F). The total mass of liquid and vapor is 3(lbm). If the volume of vapor is 50 times the volume of liquid, what is the total enthalpy of the contents of the vessel?

Solution

The problem involves pure water, so we should get properties from the steam tables. For saturated liquid at 3

350 ºF, we have V = 0.01799 ft /lbm and H = 321.76 Btu/lbm. For the saturated vapor, V = 3.342 ft /lbm and H = 1192.3 Btu/lbm. The total mass is equal to the mass ov vapor plus the mass of liquid, and is 3 lbm. In terms v

of the volume of vapor and volume of liquid, this is V /3.342 ft /lbm + V /0.01799 ft /lbm = 3 lbm. That is, the mass of vapor is the volume of vapor divided by its specific mass, and the mass of liquid is the volume of liquid divided by its specific mass, and these add up to the total mass. If the vapor volume is 50 times l

the liquid volume, then this is 50V /3.342 ft /lbm + V /0.01799 ft /lbm = 70.547 V = 3 lbm, from which l

V = 0.04253 ft and V = 2.126 ft . The corresponding masses are m = 0.04253/0.01799 = 2.364 lbm and v

m = 2.126/3.342 = 0.636 lbm. So, the total enthalpy of the liquid and vapor together is H = 2.364*321.76 + 0.636*1192.3 = 1519 Btu. Solution 6.22 Problem Statement 3

Wet steam at 230°C has a density of 0.025 g·cm . Determine x, H, and S.

Solution l

Looking in the steam tables at 230 C, we see that the saturated liquid properties are V = 1.209 cm g , l

H = 990.3 kJ kg , and S = 2.6102 kJ kg

K . The vapor properties are V = 71.45 cm g ,

Solution continued on next page…

H = 2802.0 kJ kg , and S = 6.2107 kJ kg 3

40 cm g

= V + x (V

K . The specific volume of the wet steam is V = 1/(0.025 g cm ) =

V ) = 1.209 + x (71.45

1.209), from which x = 0.5523 (55.2 % of the water is l

saturated vapor, and 44.8% of it is saturated liquid). Then H = H + 0.5523*(H 990.3) = 1991 kJ kg

and S = S + 0.5523*(S

S ) = 2.6102 + 0.5523*(6.2107

H ) = 990.3 + 0.5523*(2802.0 2.6102) = 4.5988 kJ kg

Solution 6.23 Problem Statement 3

A vessel of 0.15 m volume containing saturated-vapor steam at 150°C is cooled to 30°C. Determine the final volume and mass of liquid water in the vessel.

Solution

We know that: Vtotal  mtotal Vliq  mvap Vlv

Vtotal 

Solution continued on next page…

And from Table E.1 at 150 C and 30 C

Vvap 

cm 3 g

Vliq 

cm 3 g

Vlv 

First figure out the total mass and the mass of vapor: mtotal 

mvap 

Vtotal  Vvap

Vtotal mtotal Vliq Vliq



3

Now determine the mass and volume of the liquid:

mliq  mtotal  mvap 

Vliq total  mliq Vliq 

cm3

Solution 6.24 Problem Statement Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. What is the quality of the steam in its initial state?

Solution

Table F.2, 1100 kPa:

liq





J gm



vap

Interpolate @101.325 kPa & 105 degC:

Const.-H throttling:



liq

 

vap

liq



J gm



liq

 vap

liq



J gm

 J  gm J   gm



 2



J gm  J  gm





Solution 6.25 Problem Statement Steam at 2100 kPa and 260°C expands at constant enthalpy (as in a throttling process) to 125 kPa. What is the temperature of the steam in its final state, and what is its entropy change? What would be the final temperature and entropy change for an ideal gas?

Solution Using the steam tables, at 2100 kPa and 260 °C (interpolating between 250 °C and 275 °C), the properties of steam are H = 2897.9 + 10/25*(2961.9 – 2897.9) = 2923.5 kJ/kg S = 6.5162 + 10/25*(6.6356 – 6.5162) = 6.5640 kJ/(kg K) Solution continued on next page…

At a pressure of 125 kPa, this enthalpy almost exactly matches the value at 225 °C. We could just take the final temperature to be 225 °C, or we can interpolate as follows, using values at 200 °C and 225 °C: H = 2923.5 = 2874.2 + (T – 200)/25*(2923.9 – 2874.2), from which T = 200 + 25*(2923.5 – 2874.2)/(2923.9 – 2874.2) = 224.8 °C At this temperature, we have S = 7.7300 + 24.8/25*(7.8324 – 7.7300) = 7.8316 kJ/(kg K) The final temperature is 224.8 °C, and the entropy change is S = 7.8316 – 6.5640 = 1.2676 kJ/(kg K) For an ideal gas, enthalpy is only a function of temperature, so expanding the gas at constant enthalpy would give no temperature change. The final temperature for an ideal gas would be 260 °C. The entropy change would be given by S = R/MW * ln(P1/P2) = (23.46 J/mol) / (0.01802 kg/mol) = 1301.8 J/kg = 1.3018 kJ/kg Even though the outlet temperature is much different for an ideal gas than the actual steam, the entropy change only differs by about 3%.

Solution 6.26 Problem Statement Steam at 300 (psia) and 500(°F) expands at constant enthalpy (as in a throttling process) to 20 (psia). What is the temperature of the steam in its final state, and what is its entropy change? What would be the final temperature and entropy change for an ideal gas?

Solution

Data, Table F.4 at 300(psia) and 500 degF: 





BTU lb m



BTU  lbm 1



BTU lbm  rankine

Final state is at this enthalpy and a pressure of

20(psia). By interpolation at these conditions, the final temperature is 438.87 degF and





BTU  lbm  rankine





BTU lbm  rankine



For steam as an ideal gas, there would be no temperature change and the entropy change would be given by:



P  R  ln  2   P1  molwt

 



1



1

 1

  







  

  

lb lbmol

BTU lbm  rankine

Solution 6.27 Problem Statement Superheated steam at 500 kPa and 300°C expands isentropically to 50 kPa. What is its final enthalpy?

Solution

Collect the data from Table E.2 at 500 kPa and 300 degC

S1  7.4614

J gK

The final stat at this entropy and pressure of 50 kPa is a wet steam which leads Sliq 

J gK

H liq 

J g

Svap 

J gK

And the enthalpies are H vap 

J g

Then solve for the final enthalpy

S2  S2  Sliq  x Svap  Sliq 

Solve for the mole fraction

x

S2  Sliq Svap  Sliq

 0.98

H2  Hliq  x  H vap  Hliq  

J g

Solution 6.28 Problem Statement What is the mole fraction of water vapor in air that is saturated with water at 25°C and 101.33 kPa? At 50°C and 101.33 kPa?

Solution

Vapor pressures of water from Table E.1: At 25 C:

Psat 

x water  At 50 C:

kPa

Psat P



Psat 

x water 



kPa

3.166  101.33 kPa

Psat

P

kPa

12.34 KPa  0.122 101.33KPa

Solution 6.29 Problem Statement 3

A rigid vessel contains 0.014 m of saturated-vapor steam in equilibrium with 0.021 m3 of saturated-liquid water at 100°C. Heat is transferred to the vessel until one phase just disappears, and a single phase remains. Which phase (liquid or vapor) remains, and what are its temperature and pressure? How much heat is transferred in the process?

Solution

Consider at constant total volume

Vtotal 





Data from table E.1 at 100 C: J g

U liq 

mliq 

J Vliq  g

U vap 



mvap 

cm 3 g

Vvap 



cm 3 g

mtotal 

Mole fraction of vapor and volume at the phase change is

x

m vap mtotal



4

V2 

Vtotal mtotal

cm 3 g



This V2 is first reached as saturated liquid at 349.83 C For this state, P = 16,500.1 kPa, and U2 

J U 1  U liq  x U vap  U liq   g

J  g

Q U 2  U 1 

J g

4

  

J  g

J   g 

J g

The amount of heat transferred is then

Solution 6.30 Problem Statement 3

A vessel of 0.25 m capacity is filled with saturated steam at 1500 kPa. If the vessel is cooled until 25% of the steam has condensed, how much heat is transferred, and what is the final pressure?

Solution

For saturated steam at 1500 kPa, the steam tables tell us that T = 471.44 K, V = 131.66 cm /g, U = 2592.4 3

kJ/kg, and H = 2789.9 kJ/kg. So, if the vessel has a volume of 0.25 m = 250000 cm , then the mass of steam in the vessel is m = 250000/131.66 = 1898.8 g = 1.899 kg. After cooling, the same volume is occupied by ml = 0.25*1.899 = 0.4747 kg of liquid and mv = 1.424 kg of vapor. The total volume remains the same, and is equal to t

V = 474.7 Vl + 1424.1 Vv = 250000 cm

with the liquid and vapor specific volumes in cm /g. Using the liquid specific volume from the initial conditions will be a very good approximation because (a) liquid specific volume is relatively insensitive to 3

pressure, and (b) the liquid will occupy a tiny fraction of the total volume. So, using Vl = 1.154 cm /g, we can solve the above equation to get Vv = 175.2. This is almost exactly 4/3 of the initial specific volume, because ¾ of the substance has expanded to fill essentially all of the container (only 0.2% is occupied by the liquid). Finding this specific volume in the steam tables (and interpolating between the entries for 184 ºC and 186 ºC) we find that the final temperature is T = 458.5 K.

U = Q, because no work is done. So, to find the heat removed, we need the change in internal energy. The internal energy of the vapor and liquid at this temperature are Uv = 2780.7 kJ/kg and Ul = 785.5 kJ/kg. So, the total internal energy is t

U = 0.4747 Ul + 1.4241 Uv = 0.4747*785.5 + 1.4241*2585.3 = 4046.6 kJ t

In the original state, the total internal energy was U = 1.899*2592.4 = 4923.0 kJ. Thus, the heat added is Q

U = 4046.6 – 4923.0 = 876.5 kJ. A total of 876.5 kJ must be removed to condense 25% of the steam.

Solution 6.31 Problem Statement 3

A vessel of 2 m capacity contains 0.02 m of liquid water and 1.98 m of water vapor at 101.33 kPa. How much heat must be added to the contents of the vessel so that the liquid water is just evaporated?

Solution

Table F.2,101.325 kPa:

liq



liq





cm 3 gm



vap

J gm



vap





cm 3 gm



J gm

Solution continued on next page…

mtotal 

x

m vap m total



liq





 liq

vap

liq



vap



0.02  m3 Vliq



1.98  m3  Vvap

106 cm3  cm3 g 1

10 6 cm 3

1



= 19157.08  1183.5 = 20340.58 g

20340.58 g = 0.0581



 vap

liq



cm3 gm



liq

 

vap



liq





J gm

Since the total volume and the total mass do not change during the process, the initial and final specific volumes are the same. The final state is therefore the state for which the specific volume of saturated vapor is 98.326 cu cm/gm. By interpolation in Table F.1, we find t = 213.0 degC and



total











J gm



Solution 6.32 Problem Statement 3

A rigid vessel of 0.4 m volume is filled with steam at 800 kPa and 350°C. How much heat must be transferred from the steam to bring its temperature to 200°C?

Solution Data, Table F.2 at 800 kPa and 350 degC:





cm3 gm





J gm

total





The final state at 200 degC has the same specific volume as the initial state, and this occurs for superheated steam at a pressure between 575 and 600 kPa. By interpolation, we find P = 596.4 kPa and U2 

Q 

Vtotal  U 2  U 1 Q  V1

0.4 * 10 6 * cm 3 gm 1



J gm 1 

J gm J gm1   

Solution 6.33 Problem Statement One kilogram of steam is contained in a piston/cylinder device at 800 kPa and 200°C. (a) If it undergoes a mechanically reversible, isothermal expansion to 150 kPa, how much heat does it absorb? (b) If it undergoes a reversible, adiabatic expansion to 150 kPa, what is its final temperature and how much work is done? Solution First obtain the data from table E.2 for steam at 800 kPa and 200 C

U1 

J g

S1 

J gK

Solution continued on next page…

(a) Collect the data for isothermal expansion for steam at 150 kPa and 200 C J g

U2 

Q  kg * T * S2  S1  

S2 

J gK

 K *  

J  gK

T

J   g K 

kJ heat absorbed

(b) Constant-entropy expansion to 150 kPa. The final state is wet steam: U liq 

J g

Sliq 

J gK

U vap 

J g

Svap 

J gK

Solve for the mole fraction:

x

S2  Sliq Svap  Sliq

 0.929

Solve for the internal energy U 2  U liq  x U vap  U liq   2.367 * 10 3

J g

W  kg * U 2 U 1   

Now solve for the work

Since wet steam at 150 kPa, the temperature is 111.4 C.

Solution 6.34 Problem Statement Steam at 2000 kPa containing 6% moisture is heated at constant pressure to 575°C. How much heat is required per kilogram?

Solution x= 0.94

J g

H vap 

Hliq 

J g

Determine the enthalpy at the first state J  g

H1  Hliq  x  H vap  Hliq  

  

J   g 

J  g

J g

For superheated vapor at 2000 kPa and 575 C, by interpolation: H 2  3634.65

J g

Now determine the heat that is required for 1 kg

 Q  kg *  H2  H1   *  

J  g

J   g 

Solution 6.35 Problem Statement Steam at 2700 kPa and with a quality of 0.90 undergoes a reversible, adiabatic expansion in a nonflow process to 400 kPa. It is then heated at constant volume until it is saturated vapor. Determine Q and W for the process.

Solution

First step:



Second step:



For process:



Table F.2,

liq



2700 kPa:

liq



liq

J gm





liq











liq





liq



liq



vap

liq

J gm





vap







J gm  K



J  gm

  



J  gm



J   gm 

J gm

  vap 

liq



J J J  0.9(6.2244   2.5924  )  S1 gm  K gm  K gm  K 3

m2 s 2K

J gm  K





vap

J gm  K

 2.5924 

Table F.2, 400 kPa:











J gm

cm3 gm

vap





J gm  K

vap





J gm





cm 3 gm

Solution continued on next page…

Since step 1 is isentropic,



liq



  vap 



liq







 2 

 liq

vap

liq



 vap

S1  Sliq





liq



 2

liq

S vap  Sliq







J gm

cm 3 gm

and the final state is sat. vapor with this specific volume.

Interpolate to find that this V occurs at T = 509.23 degC and.



J gm





J gm









J gm



Solution 6.36 Problem Statement Four kilograms of steam in a piston/cylinder device at 400 kPa and 175°C undergoes a mechanically reversible, isothermal compression to a final pressure such that the steam is just saturated. Determine Q and W for the process.

Solution

Table F.2, 400 kPa &





J gm





J gm  K

Solution continued on next page…

175 degC:





J gm





J gm  K

   

  2 







 = 4000 g * 448.15 K* (









J  gm  K





J  gm  K

= 4000g * (2578.8 J/g – 2605.8 J/g ) ( 775.66KJ) = 667.66 kJ

Solution 6.37 Problem Statement Steam undergoes a change from an initial state of 450°C and 3000 kPa to a final state of 140°C and 235 kPa. H

(a) From steam-table data. (b) By equations for an ideal gas. (c) By appropriate generalized correlations.

Solution (a) From the steam tables, steam at 450C and 3000 kPa has H = 3344.6 kJ/kg and S = 7.0854 kJ/(kg K). At 140 C and 235 kPa, interpolation from the three nearest points in the superheated steam tables gives. H = 2766.5 + (10/25)*(2764.5 2766.5) + (10/25)*(2713.8 2766.5) = 2744.6 kJ/kg S = 7.2213 + (10/25)*(7.1689 7.2213) + (10/25)*(7.0928 7.2213) = 7.1489 kJ/(kg K) H = 600.0 kJ/kg =

S = 0.06354 kJ/(kg K) = 63.5 J/(kg K) = 1.14 J/(mol K)

Solution continued on next page…

(b) For an ideal gas, we would use the ideal gas heat capacity integrals, with heat capacity polynomial coefficients for water from Table C.1. For enthalpy, we have

ICPH  T0 ,T;A,B,C,D  

1

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  2

2 0

3 0

T0(K) 723.15

T (K) 413.15

A 3.47

B (1/K) 1.45E-03

C (1/K2) 0

ig D (K2) ICPH (K) D H (J/mol) 1.21E+04 -1343.6 -11171.7 Enthalpy of

an ideal gas is independent of pressure, so we have ig

H = ( 11172 J/mol) / (18.02 g/mol) = 620.0 J/g = 620 kJ/kg. This is about 3% larger than the steamtable-based value. For entropy, we have the temperature dependence as:

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T0 (K) 723.15

T (K) 413.15

A 3.47

B (1/K) 1.45E-03

C (1/K2) 0

D (K2) 1.21E+04

ICPS -2.416

(J/(mol K)) -20.0858

Adding the pressure dependence gives ig

S = ( 20.086 – 8.314*ln(235/3000)) = 1.089 J/(mol K)/(0.01802 kg/mol) = 60.5 J/kg. This is about 5% smaller than the steam-table-based value. (c) To use a generalized correlation, we will compute the residual properties in the initial and final state, and then add those to the ideal gas values. We will thus have R

H = H (450 ºC,3000 kPa) + H + H (140 ºC,235 kPa)

S = S (450 ºC,3000 kPa) + S + S (140 ºC,235 kPa)

Solution continued on next page…

The critical properties of steam are Tc = 647.1 K, Pc = 220.55 bar = 22055 kPa,  = 0.345. So, at the initial conditions, the reduced T and P are Tr = 1.118 and Pr = 0.1360. This is well within the range where we can use the Pitzer correlation for the 2

virial coefficient. At the outlet conditions, Tr = 0.6385, Pr = 0.0107.

At the inlet conditions, we have:

T (K) 723.15

Tr 1.1175

P (bar) 30

Pr 0.1360

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) 647.1

B0 -0.2703

Pc (bar) 220.55

B1 0.0311

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.345

dB0 dTr

dB1 dTr

0.5056

0.4051

HR /RTc -0.1334

HR (J mol-1) -717.78

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 723.15

Tr 1.1175

P (bar) 30

Pr 0.1360

Tc (K) 647.1

Pc (bar) 220.55

dB0 dTr

dB1 dTr

0.5056

0.4051

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.345

S R/R -0.0878

SR (J mol-1 K-1) -0.73

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output.

Solution continued on next page…

and at the outlet conditions:

T (K) 413.15

Tr 0.6385

P (bar) 2.35

Pr 0.0107

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) 647.1

B0 -0.7822

Pc (bar) 220.55

B1 -0.9933

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.345

dB0 dTr

dB1 dTr

2.1675

7.4444

HR /RTc -0.0442

HR (J mol-1) -237.82

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output.

T (K)

P (bar)

Tc (K)

Pc (bar)

413.15

2.35

647.1

220.55

0.345

dB0 dTr

0.6385

0.0107

dB1 dTr R

2.1675

7.4444

-1

S /R

S (J mol K )

0.0505

0.42

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output. Solution continued on next page…

So, finally: R

H = –H (450 ºC,3000 kPa) + H + H (140 ºC,235 kPa) = 771 – 11172 – 238 = –10639 J/mol = –590 kJ/kg.

S = –S (450 ºC,3000 kPa) + S + S (140 ºC,235 kPa) = 0.73 + 1.09 – 0.42 = 1.40 J/(mol K) = 77.7 kJ/(kg K).

Solution 6.38 Problem Statement A piston/cylinder device operating in a cycle with steam as the working fluid executes the following steps: • Steam at 550 kPa and 200°C is heated at constant volume to a pressure of 800 kPa. • It then expands, reversibly and adiabatically, to the initial temperature of 200°C. • Finally, the steam is compressed in a mechanically reversible, isothermal process to the initial pressure of 550 kPa. What is the thermal efficiency of the cycle?

Solution

Data, Table F.2 superheated steam at 550 kPa and 200 degC:





cm3 gm





J  gm 1



J gm  K

Step 1--2: Const.-V heating to 800 kPa. At the initial specific volume and this P, interpolation gives t = 401.74 degC, and





J gm





J gm  K

Solution continued on next page…





J  gm



J Step 2--3: Isentropic expansion to initial T. gm



 1

 

cycle









J gm

J gm  K

Step 3--1: Constant-T compression to initial P. 



  31  

J gm

For the cycle, the internal energy change = 0. cycle



 

Q 31 Q12



Wcycle Q12



Solution 6.39 Problem Statement A piston/cylinder device operating in a cycle with steam as the working fluid executes the following steps: • Saturated-vapor steam at 300 (psia) is heated at constant pressure to 900(°F). • It then expands, reversibly and adiabatically, to the initial temperature of 417.35(°F). • Finally, the steam is compressed in a mechanically reversible, isothermal process to the initial state. What is the thermal efficiency of the cycle?

Solution

Table F.4, sat.vapor, 300(psi): 









BTU lb m

BTU lb m  rankine

Superheated steam at 300(psi) & 900 degF







BTU lb m





 1 

BTU  lbm  rankine 3

 31  

BTU lbm

For the cycle, the internal energy change = 0.

cycle

 

 

cycle

Q 31 Q12

 





Wcycle Q12

Whence



Solution 6.40 Problem Statement Steam entering a turbine at 4000 kPa and 400°C expands reversibly and adiabatically. (a) For what discharge pressure is the exit stream a saturated vapor? (b) For what discharge pressure is the exit stream a wet vapor with quality of 0.95?

Solution If the steam expands reversibly and adiabatically, that means it expands isentropically. For this problem, I used the properties from the NIST Webbook, to avoid manual interpolation. For the inlet conditions of 400 °C and 4000 kPa, the properties are H = 3214.5 kJ/kg and S = 6.7714 kJ/(kg K). These agree with the values in the steam tables in the back of the book to within a fraction of a percent. (a) If the discharge stream is a saturated vapor, then we simply want to look at properties of saturated vapor and find the temperature and pressure for which S = 6.7714 kJ/(kg K). Trying different pressure ranges, we find that for P = 579 kPa, S = 6.7713 kJ/(kg K), T = 157.4 °C, and H = 2754.6 kJ/kg. Although the question doesn’t explicitly ask, another item of interest is the work output, which is W = H = 459.9 kJ/kg. (b) If the exit stream is a wet vapor with a quality of 0.95, then we have 0.05 Sl + 0.95 Sv = 6.7714 kJ/(kg K) We can choose a range of pressures from which to get the data from the NIST webbook and then see where this is satisfied. To a very rough approximation, we could neglect the entropy of the liquid, so that we would have Sv = 6.7714/0.95 = 7.13 kJ/(kg K). Trying some different pressures, we find that for P = 200 kPa, Sv = 7.1269 kJ/(kg K). So, the pressure should be a bit above 200 kPa. At 200 kPa, Sl = 1.5302 kJ/(kg K). Using that value in our equation above would give Sv = (6.7714 – 0.05*1.5302)/0.95 = 7.047 kJ/(kg K). At 260 kPa, Sv = 7.0394, so the pressure should be close to that value. Now, the simplest thing to do is probably to download properties for a range of pressures near 260 kPa and see which best satisfies the equation. Going from 250 kPa to 270 kPa in 1 kPa increments gives: Solution continued on next page…

T ( °C)

P (MPa) H l (kJ/kg) S l (kJ/(kg K)) H v (kJ/kg) S v (kJ/(kg K)) H (kJ/kg) S (kJ/(kg K))

127.41 127.54 127.67 127.80 127.94 128.07 128.19 128.32 128.45 128.58 128.71 128.84 128.96 129.09 129.22 129.34 129.47 129.59 129.72 129.84 129.97

0.250 0.251 0.252 0.253 0.254 0.255 0.256 0.257 0.258 0.259 0.260 0.261 0.262 0.263 0.264 0.265 0.266 0.267 0.268 0.269 0.270

535.34 535.91 536.46 537.02 537.58 538.13 538.68 539.23 539.78 540.33 540.87 541.42 541.96 542.50 543.04 543.57 544.11 544.64 545.18 545.71 546.24

1.6072 1.6086 1.6100 1.6114 1.6128 1.6142 1.6155 1.6169 1.6183 1.6196 1.6210 1.6223 1.6237 1.6250 1.6264 1.6277 1.6290 1.6303 1.6317 1.6330 1.6343

2716.5 2716.7 2716.9 2717.0 2717.2 2717.4 2717.6 2717.8 2717.9 2718.1 2718.3 2718.5 2718.6 2718.8 2719.0 2719.2 2719.3 2719.5 2719.7 2719.9 2720.0

7.0524 7.0511 7.0498 7.0485 7.0472 7.0458 7.0445 7.0432 7.0419 7.0407 7.0394 7.0381 7.0368 7.0355 7.0343 7.0330 7.0318 7.0305 7.0293 7.0280 7.0268

2607.4 2607.7 2607.9 2608.0 2608.2 2608.4 2608.7 2608.9 2609.0 2609.2 2609.4 2609.6 2609.8 2610.0 2610.2 2610.4 2610.5 2610.8 2611.0 2611.2 2611.3

6.7801 6.7790 6.7778 6.7766 6.7755 6.7742 6.7731 6.7719 6.7707 6.7696 6.7685 6.7673 6.7661 6.7650 6.7639 6.7627 6.7617 6.7605 6.7594 6.7583 6.7572

The last column gives the total entropy (0.05 times the liquid entropy plus 0.95 times the vapor entropy). The equation is most nearly satisfied at a discharge pressure of 257 kPa, corresponding to T = 128.3 °C and H = 2608.9 kJ/(kg K). In this case, the work output would be W

H = 605.6 kJ/kg. By condensing just 5% of the steam, we get 32% more work

out of the turbine.

Solution 6.41 Problem Statement A steam turbine, operating reversibly and adiabatically, takes in superheated steam at 2000 kPa and discharges at 50 kPa. (a) What is the minimum superheat required so that the exhaust contains no moisture? Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(b) What is the power output of the turbine if it operates under these conditions and the steam rate is 5 kg⋅s ?

Solution

(a) Table F.2 at the final conditions of saturated vapor at 50 kPa:





kJ kg  K





kJ  kg 1

Find the temperature of superheated vapor at 2000 kPa with this entropy. It occurs between 550 and 600 degC. By interpolation



Superheat:

(b)

 











 1 



kJ kg



kg sec









Solution 6.42 Problem Statement An operating test of a steam turbine produces the following results. With steam supplied to the turbine at 1350 kPa and 375°C, the exhaust from the turbine at 10 kPa is saturated vapor. Assuming adiabatic operation and negligible changes in kinetic and potential energies, determine the turbine efficiency, i.e., the ratio of actual work of the turbine to the work of a turbine operating isentropically from the same initial conditions to the same exhaust pressure.

Solution

If the turbine operates adiabatically, and changes in kinetic and potential energy are negligible, then the H = Ws. According to the steam tables, at 1350 K and 375 °C, H = 3205.4 kJ/kg and S = 7.2410 kJ/(kg K). For saturated vapor at 10 kPa, H = 2584.8 kJ/kg and S = 8.1511 kJ/(kg K). Thus, for the actual process, Ws

H = 2584.8 3205.4 = 620.6 kJ/kg.

If the turbine operated isentropically, then it would discharge a mixture of saturated liquid and saturated vapor with the sam entropy as the steam entering the turbine. For saturated liquid at 10 kPa, H = 191.832 kJ/kg and l

S = 0.6493 kJ/(kg K). The entropy of the two-phase stream at 10 kPa is x *0.6493 + (1 x )*8.1511 = 7.2410 (for l

isentropic operation). Solving this gives x = (8.1511 7.2410)/(8.1511 0.6493) = 0.1213. The fraction vapor is v

therefore x = 0.8787. The enthalpy of this two-phase stream is H = 0.1213*191.832 + 0.8787*2584.8 = 2294.5 kJ/kg. Thus, if the turbine operated isentropically, the work would be Ws

H = 2294.5 3205.4 = 910.9 kJ/kg.

The isentropic efficiency is = 620.6/910.9 = 0.681.

Solution 6.43 Problem Statement 1

A steam turbine operates adiabatically with a steam rate of 25 kg·s . The steam is supplied at 1300 kPa and 400°C and discharges at 40 kPa and 100°C. Determine the power output of the turbine and the efficiency of its operation in comparison with a turbine that operates reversibly and adiabatically from the same initial conditions to the same final pressure.

Solution H = W. From the steam tables, we find that at 1300 kPa and 400 ºC, H = 3259.7 kJ/kg, and S = 7.3404 kJ/(kg K). At 40 kPa and 100ºC, H = 2683.8 kPa and S = 7.8009 kJ/(kg K). So, W

H = (2683.8 – 3259.7) = –575.9 kJ/kg. So, for a steam rate of 25 kg/s, the

turbine produces 25*575.9 = 14398 kJ/s = 14398 kW = 14.4 MW. A turbine that operates reversibly and adiabatically would produce no entropy change in the gas (adiabatic + reversible = isentropic). So, we want to find the temperature at which the entropy of steam is 7.3404 kJ/(kg K) at a pressure of 40 kPa. A check of the steam tables shows that at 40 kPa, the entropy of saturated steam is 7.6709 kJ/(kg K), which is higher than the entropy of the steam entering the turbine. So, a reversible turbine would produce wet steam (a mixture of saturated liquid and vapor). The entropy of saturated liquid at 40 kPa is 1.0261 kJ/kg. So, we have S = 7.3404 = 1.0261 xl + 7.6709 xv = 1.0261 xl + 7.6709 (1 – xl) From which xl = (7.6709 – 7.3404)/(7.6709 – 1.0261) = 0.0497. A little less than 5% of the steam condenses. So, the enthalpy of the wet steam leaving a reversible turbine would be H = xlHl + xvHv = 0.0497*317.65 + 0.9503*2636.9 = 2521.5 kJ/(kg K) Thus, the work done by a reversible turbine with the same discharge pressure would be W

H = (2521.5 – 3259.7) = –738.2 kJ/kg, which for a steam rate of 25 kg/s provides an output of 18.46 MW.

The efficiency of the turbine, relative to a reversible adiabatic turbine with the same discharge pressure, is = 575.9/738.2 = 0.780, or 78%. While the actual turbine discharges superheated steam at 100 ºC, a reversible one would discharge a saturated liquid/vapor mixture, with a quality of 0.9503 at a temperature of 75.9 ºC.

Solution 6.44 Problem Statement R

From steam-table data, estimate values for the residual properties V , H , and S for steam at 225°C and 1600 kPa, and compare with values found by a suitable generalized correlation.

Solution

P 

Table F.2 at 1600 kPa and 225 degC:

V 



 kPa

cm3 H  gm



J S  gm



J gm  K

Table F.2 (ideal-gas values, 1 kPa and 225 degC) H ig 



T 

J S  gm ig



 KT 



J P   kPa gm  K 0

KVR  V 

R T  molwt P

The enthalpy of an ideal gas is independent of pressure, but the entropy

DOES depend on P: H R  H  H ig

Sig  

cm3 H  gm R

VR  

Reduced conditions:

P R  ln   SR  S Sig  Sig   P0  molwt J S  gm R



Tc 

 KPc 

Tr 

T T Tc r

Pr 

J gm  K

 bar

P P Pc r

The generalized virial-coefficient correlation is suitable here B0 



0.422 B0   Tr1.6

B1 



0.172 B1   Tr4.2

Solution continued on next page…

By Eqs. (3.61) + (3.62) & (3.63) along with Eq. (6.40) Z    B0    B1  

HR : 

R  TC molwt

VR  

Pr Z Tr

VR 

 HRB Tr , Pr ,  SR : 

cm3 H  gm R

R T P  molwt

 Z  

R  SRB Tr , Pr ,  molwt

J S  gm R

J gm  K

Solution 6.45 Problem Statement From data in the steam tables: l

(a) Determine values for G and G for saturated liquid and vapor at 1000 kPa. Should these be the same? lv

H /T R

S at 1000 kPa. Should these be the same? R

(d) Estimate a value for dP /dT

S at 1000 kPa.

Does this result agree with the steam-table value? R

Apply appropriate generalized correlations for evaluation of V , H , and S for saturated vapor at 1000 kPa. Do these results agree with the values found in (c)?

Solution First collect all the data from table E.2 for saturated vapor and liquid at 1000 kPa and 453.03 K V1 

cm 3 g

Vv 

cm 3 g

H1 

J g

Hv 

J g

S1 

J gK

Sv 

J gK

The difference between liquid and vaper for each is Vlv 

cm 3 g

J g

Gv  H v  TSv 

(a) G1  H1  TS1 

J g

Hlv 

J g

Slv 

J gK

yes these should be the same when they

are at equlibrium Hlv  T

(b) Slv 

J gK

r

R T * MW P

VR  

J gK

yes they are the same value

cm 3 g

For enthalpy and entropy, assume that steam at 179.88 degC and 1 kPa is an ideal gas. By interpolation in Table E.2 at 1 kPa: J g

Hig 

J gK

Sig 

The enthalpy of an ideal gas is independent of pressure; the entropy DOES depend on P: H R  H v  H ig  

J g

Sig 

SR  Sv  Sig Sig  

R * Mw

P  P0

J gK

Solution continued on next page…

(d) Assume ln P vs. 1/T linear and fit three data pts @ 975, 1000, & 1050 kPa.  975     kPa Pi     1050   xi 

178.79   C t    182.02   

1 ti  273.15

yi 

slope  

dP P  2 * slope * K  dt T

kPa K

Slv Slv *

Reduced conditions:  

i

Tc 

K

Tr 

dP  dt

j gK

Pc 

 bar

Pr 

The generalized virial-coefficient correlation is suitable here B0 

B1 

By Eqs. (3.58) + (3.59) & (3.60) along with Eq. (6.40) P Z    B0   B1  r  Tr HR 

RTc HRB Tr Pr     MW

VR 

J g

RT * Z     P MW

SR 

R SRB Tr Pr     MW

cm 3 g J gK

These are different from those found in part (c)

Solution 6.46 Problem Statement From data in the steam tables: l

(a) Determine values for G and G for saturated liquid and vapor at 150(psia). Should these be the same? Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

H /T R

S at 150(psia). Should these be the same? R

(d) Estimate a value for dP /dT

S at 150(psia).

Does this result agree with the steam-table value? R

Apply appropriate generalized correlations for evaluation of V , H , and S for saturated vapor at 150(psia). Do these results agree with the values found in (c)?

Solution By definition, G = H – TS, and we can look up H, T, and S for saturated liquid and water vapor at 150 psia in the steam tables. The tables for superheated steam also give the values for saturated vapor and liquid, and are conveniently arranged by pressure so we can use the 150 psia entries and we do not have to interpolate. From l

page 739 of SVNA, we find that T = 358.43 F = 818.1 R, H = 330.65 Btu lbm , H = 1194.1 Btu lbm l

S = 0.5141 Btu lbm

R , and S = 1.5695 Btu lbm

 1

R . So, G = 330.65 – 818.1*0.5141 = 89.9 Btu lbm ,

and G = 1194.1 – 818.1*1.5695 = 89.9 Btu lbm . These are equal (to within the accuracy of the table entries) as they must be since the vapor and liquid in question are in equilibrium with each other.

(a) H /T = (1194.1 – 330.65)/818.1 = 1.055 Btu lbm lbm

R . S = 1.5695

0.5141 = 1.0554 Btu

R . These should be equal, since the phase transition takes place at constant temperature, and lv

Q = H , so S =Q/T = H /T .

Solution continued on next page…

(b) The ideal gas specific volume is V = RT/MP = 10.73*818.1/(18.02*150) = 3.248 ft lbm . The volume 3

given ion the steam tables is 3.014 ft lbm . So, the residual, volume is V = 3.014 – 3.248 = 0.234 ft 1

lbm . To find the ideal gas enthalpy and entropy, we can use values from the steam tables at low pressure. We can assume that steam at 358.43 F and 1 psia is an ideal gas. Interpolating to this ig

temperature using the entries for superheated steam at 1 psia gives H = 1222.6 Btu lbm . The enthalpy of an ideal gas is independent of pressure, so this is also the value at 150 psia. The entropy of an ideal gas depends on pressure as S = R/M ln(P2/P1), so at 150 psia, the ideal gas entropy will be 1.986/18.02*ln(150) = 0.552 Btu lbm

R

different than at 1 psia. Using this with the

interpolated value from the table gives S = 2.149

0.552 = 1.597 Btu lbm

R . Subtracting these

ideal gas values from the actual values given in the steam tables gives H = 1194.1 Btu lbm

and S = 1.5695 – 1.597 = 0.028 Btu lbm

sat

1222.6 = 28.5

R .

sat

= 143.57 psia at 355 F and 153.01 psia at 360 F. So, we can estimate dP /dT as 1

(153.01 – 143.57)/5 = 1.888 psia R . From the steam tables, we have V = 3.014 – 0.0181 = 2.996 ft 1

lbm . So, the Clayperon equation gives us

H  TV

dP sat  818.1 * 2.996 * 1.888  4628 ft 3 psia lbm1  856.5 Btu lbm1 dT lv

So, we estimate that S =856.5 / 818.1 = 1.047 Btu lbm

R , which agrees with the answer from

sat

part (b) to within the accuracy of the approximation to dP /dT Solution continued on next page…

If we assume the propane is an ideal gas in the initial state, then we only have to worry about residual properties in the final state. The critical properties of propane are Tc = 369.8 K, Pc = 42.48 bar, and  = 0.152. So, in the final state, Tr = 1.2660 and Pr = 3.178. For these conditions, we should use the full Lee/Kesler tables.

Doing the needed interpolations gives: T (K) 468.15

P (bar) 135

Tc (K) 369.8

Pc (bar) 42.48

Tr 1.20 1.20 1.30 1.30

Pr 3.00 5.00 3.00 5.00

w 0.152

Tr 1.2660

Pr 3.1780

Z1 0.1095 -0.0141 0.2079 0.0875

(HR )0/(RTc) -2.801 -3.166 -2.274 -2.825

0.1636

-2.4968

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Z0 0.5425 0.7069 0.6344 0.7358

Interpolated Values Final Values Z

HR /(RTc)

0.6389

V (cm /mol)

0.6140

184.2

H (J/mol)

(HR )1/(RTc) -0.934 -1.84 -0.3 -1.066

(SR )0/R -1.727 -1.827 -1.299 -1.554

(S R )1/R -0.991 -1.767 -0.29 -0.73

-1.4627

-0.5780

-0.5883

-2.58623

S R /R

-1.55056

-7951.4

S R (J/(mol K))

-12.8914

The estimated final volume is 184.2 cm /mol. To compute the enthalpy and entropy changes for the process, we need to add the residual properties shown above to the heat capacity integrals for the ideal gas enthalpy and entropy changes:

ICPH  T0 ,T;A,B,C,D  

R

T0(K) 308.15

T (K) 468.15

A 1.213

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) B (1/K) ICPH (K) D H (J/mol) 2.88E-02 -8.82E-06 0.00E+00 1766.0 14683.6

Solution continued on next page…

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T0 (K)

T (K)

B (1/K)

308.15

468.15

1.213

2.88E-02

C (1/K )

D (K )

ICPS

-8.82E-06 0.00E+00 4.565

S (J/(mol K)) 37.9524

For the entropy, there is also the pressure-dependent ideal gas term –R ln (135)= –40.785 J/(mol K). Putting this all together, we have H = 14684 –

S = 37.95 – 40.785 – 12.891 = –15.73 J/(mol K).

Solution 6.47 Problem Statement Propane gas at 1 bar and 35°C is compressed to a final state of 135 bar and 195°C. Estimate the molar volume of the propane in the final state and the enthalpy and entropy changes for the process. In its initial state, propane may be assumed an ideal gas. Solution The critical values for propane are as follows: Tc 

Pc 

bar



Tr 

T

K P

bar P0  bar

Pr 

Use the Lee/Kessler correlation; by interpolation, to determine the volume of the propane in its final state: Z0 

Z1 

Z  Z0   Z1 

V

ZRT cm3  184.2 P mol

Now for the enthalpy and entropy changes, estimate the values for each at the initial state and final state using tables D.5 – D.12 H R0  

* R * Tc  

J H R1   mol

SR 0  

* R

J S  mol K R 1

J mol K

* R 

J S  S R 0   SR 1   mol R

H R  H R 0   H R1  

J mol

* R * Tc  

J mol K

The change in enthalpy is H  R * ICPH 

3



6

 H 

J mol

And the change in entropy is  S  R ICPS  

3



6



 ln PP   S   0



J mol K

Solution 6.48 Problem Statement Propane at 70°C and

S for the process

kPa P0 

kPa

by suitable generalized correlations.

Solution The critical values for propane are as follows: Tc 

Pc 

bar

 Tr 

T

K P

Pr 

Assume propane an ideal gas at the initial conditions. Use generalized virial correlation at final conditions. H  RTc HRB Tr Pr    

J mol

 P S  R SRB Tr Pr    ln     P0 

J mol K

Solution 6.49 Problem Statement A stream of propane gas is partially liquefied by throttling from 200 bar and 370 K to 1 bar. What fraction of the gas is liquefied in this process? The vapor pressure of propane is given by Eq. (6.91) with parameters: A = –6.72219, B = 1.33236, C = –2.13868, D = –1.38551.

Eq. 6.91:

ln Prsat 

A  B 1.5  C 3  D 6 1 

Solution For propane: Tc 

Pc 

bar

Zc 



Vc 

cm3 mol

If the final state is a two-phase mixture, it must exist at its saturation temperature at 1 bar. This temperature is found from the vapor pressure equation:

P  bar A 

B

C    1

D 

T Tc

P  exp

A *   B *  1.5  C *  3  D *  6 1

Solving for T gives

T

The latent heat of vaporization at the final conditions will be needed for an energy balance. It is found by the Clapeyron equation. We proceed exactly as in Pb. 6.14. P T   Pc * exp

A *   B *  1.5  C *  3  D *  6 1 

dP T  dT

Vvap 

 4.428

kPa K

Pr 

Tr 

B0  

B1  

Pr  RT   B   B    P  Tr  Vliq  Vc Zc(1Tr )

0.2857

Hlv  T * Vvap  Vliq  *



dP T  dT

cm3 mol

 1.879 * 10 4

J mol

ENERGY BALANCE: For the throttling process there is no enthalpy change. The calculation path from the initial state to the final is made up of the following steps: (1) Transform the initial gas into an ideal gas at the initial T & P. (2) Carry out the temperature and pressure changes to the final T & P in the ideal-gas state. (3) Transform the ideal gas into a real gas at the final T & P. (4) Partially condense the gas at the final T & P. Solution continued on next page…

The sum of the enthalpy changes for these steps is set equal to zero, and the resulting equation is solved for the fraction of the stream that is liquid.

For Step (1), use the generalized correlation of Tables D.7 & D.8, and let 0

 H R   r0   RTc  T1 

 H R   r1   RTc 

P1 

bar Tr 

r0 

r1 

Pr 

By interpolation, find

For Step (2) the enthalpy change is given by H2  R * ICPH T1 T

3



Pr 

1bar  Pc

6

 

J mol

For Step (3) the enthalpy change is given by Tr 

H3  R * Tc * HRB Tr Pr   

For step (4):

230.703  Tc

232.291 J/mol

 H 4   x * H lv

For the process: H1  H2  H3  xH lv  0 Solving for x x

H1  H2  H3 H lv

x

Solution 6.50 Problem Statement Estimate the molar volume, enthalpy, and entropy for 1,3-butadiene as a saturated vapor and as a saturated liquid at 380 K. The enthalpy and entropy are set equal to zero for the ideal-gas state at 101.33 kPa and 0°C. The vapor pressure of 1,3-butadiene at 380 K is 1919.4 kPa. Solution For 1,3-butadiene: Tc 

Pc 

P0 

bar

kPa

Zc 



T0 

Tr 

T

K P

Pr 

Tn 

kPa K

Use Lee/Kesler correlation. HOWEVER, the values for a saturated vapor lie on the very edge of the vapor region, and some adjacent numbers are for the liquid phase. These must NOT be used for interpolation. Rather, EXTRAPOLATIONS must be made from the vapor side. There may be some choice in how this is done, but the following values are as good as any:

Z0 

V

Z1 

Z  Z0  Z1 

5 3 1 1 ZRT 0.7079 * 83.14 * 10 Pa. cm . mol . K * 380 K cm3   1164.7 P mol 1919.4 * 10 3 Pa

HR0  

* R * Tc  

J H R1   mol

SR 0  

*R

J S  mol K R 1

H R  H R 0   H R1  

* R * Tc  

J mol

*R

J mol K

J S  S R 0   SR 1   mol R

J mol K

Solution continued on next page…

The change in enthalpy is H vap  R * ICPH 

3



 H 

6

J mol

And the change in entropy is  Svap  R  ICPS  

3



6

P

  ln P   S   0



J mol K

By Eqs. (3.68), (4.13) & (4.14) Vliq  Vc Zc(1Tr )

0.2857

cm3 mol



     1.092ln Pc   1.013   Hn  R * Tn   Tn   0.930    T c        1  T  r    H  H n   T   1  n   Tc 

J mol

0.38



H liq  H vap  H  

Sliq  Svap 

H  T

J mol

J mol J mol K

Solution 6.51 Problem Statement Estimate the molar volume, enthalpy, and entropy for n-butane as a saturated vapor and as a saturated liquid at 370 K. The enthalpy and entropy are set equal to zero for the ideal-gas state at 101.33 kPa and 273.15 K. The vapor pressure of n-butane at 370 K is 1435 kPa.

Solution The critical values for n-butane are as follows: Tc 

Pc 

P0 

Zc 

bar

kPa



T0 

T

Tr 

Pr 

K P

kPa

Tn 

Z1   V

H R0  

* R * Tc  

SR 0  

* R 

Z  Z0   Z1 

ZRT cm3  1579.5 P mol J H R1   mol J S  mol K R 1

H R  H R 0   H R1  

J mol

* R * Tc  

* R 

J mol K

J S  SR 0   SR 1   mol R

J mol K

The change in enthalpy is H vap  R * ICPH 

3



6

 H 

J mol

P

J mol K

And the change in entropy is  Svap  R  ICPS  

3



6

  ln P   S  0



Solution continued on next page…

By Eqs. (3.68), (4.13) & (4.14) Vliq  Vc Z c(1Tr )

0.2857

cm3 mol



     1.092ln Pc   1.013   Hn  R * Tn   Tn   0.930    Tc        1  T   r   H  H n    Tn   1    Tc 

J mol

0.38



J mol

Hliq  H vap H  

J mol

H  T

J mol K

Sliq  Svap 

Solution 6.52 Problem Statement The total steam demand of a plant over the period of an hour is 6000 kg, but instantaneous demand fluctuates 1

from 4000 to 10,000 kg⋅h . Steady boiler operation at 6000 kg⋅h is accommodated by inclusion of an accumulator, essentially a tank containing mostly saturated liquid water that “floats on the line” between the boiler and the plant. The boiler produces saturated steam at 1000 kPa, and the plant operates with steam at 700 kPa. A control valve regulates the steam pressure upstream from the accumulator and a second control valve regulates the pressure downstream from the accumulator. When steam demand is less than boiler output, steam flows into and is largely condensed by liquid residing in the accumulator, in the process increasing the pressure to values greater than 700 kPa. When steam demand is greater than boiler output, water in the

accumulator vaporizes and steam flows out, thus reducing the pressure to values less than 1000 kPa. What accumulator volume is required for this service if no more that 95% of its volume should be occupied by liquid?

Solution Under the stated conditions the worst possible cycling of demand can be represented as follows:

10,000 kg/hr 2/3 hr

1/3 hr

Demand 1 hr

(kg/hr)

time

6,000 4,000 kg/hr

Net storage

Net depletion

of steam

This situation is also represented by the equation: 4000 

Solution gives

 





2 3

The steam stored during this leg is:

prime

   

kg  hr

prime

kg   hr 



Solution continued on next page…

We consider this storage leg, and for this process of steam addition to a tank the equation developed in Problem 6-74 is applicable:  Hfg2    m1  Hprime  H1   Vtank  P2  P1  Vfg2   m2  Hfg2 H prime  Hf 2  Vf 2  Vfg2

We can replace Vtank by m2V2, and rearrange to get  Hfg2 Hfg2  m2    H prime  Hf 2  Vf 2   V2  P2  P1    Hprime  H1 m1  Vfg2 Vfg2    

However

 1





m 2 V1  m1 V2

tank

Eq. (A)

Making this substitution and rearranging we get Hprime  Hf 2  Vf 2 

H fg2 Vfg2

 P2  P1 

H fg2 Vfg2



Hprime  H1 V1

In this equation we can determine from the given information everything except Hprime and Vprime. These quantities are expressed by

H1  H f 1  x1  H fg 1

V1 V f 1  x1  V fg 1

Therefore our equation becomes (with Hprime = Hg2) H



H  H   V   V  g2



fg2

fg 2



 P2  P1 

Hfg2 fg2



Hg 2  Hf 1  x 1  Hfg1 f1





Eq. (B)

fg1

Solution continued on next page…

In this equation only x1 is unknown and we can solve for it as follows. First

we need V2: From the given information we can write:

 

Therefore

19 



 f 2

x 2  Vg2

V2 

Vg2

    

 f 2



Vf 2  g2

Then



 20    19 1 f2   Vf 2 Vg2

Vf 2  g2

Eq. (c)

Now we need property values: 

Initial state in accumulator is wet steam at 700 kPa. We find from the steam tables



kJ kg



cm3 gm



kJ kg



cm 3 gm

Final state in accumulator is wet steam at 1000 kPa.

fg1



fg1



 g1

 f 1 fg1



fg1



kJ kg

cm 3 gm

Solution continued on next page…

From the steam tables



kJ kg



cm3 gm



kJ kg



cm3 gm



kJ kg

fg2



cm3 gm

3



fg2





fg2







Solve Eq. (C) for V2

 2

    

Vg2

Vf 2  g2

    2 f2 

Guess: x1 :  0.1

Next solve Eq. (B) for x1 Given







 H   fg 2      Vfg2 

Thus



Eq. (A) gives





fg1



m2 V1  m1 V2

 P2  P1 

 1 1 



fg2

Vfg 2









fg1

Vf 1  x 1  Vfg1 4

cm3 gm 



Solve for m1 and m2 using a Mathcad Solve Block: Guess: m1 : 

m prime 2

m2 :  m1

Solution continued on next page…

m2 V1  m1 V2

Given



 m   1    m 2 







Finally, find the tank volume

tank







4 2

tank





Note that just to store 1333.3 kg of saturated vapor at 1000 kPa would require a volume of: 



One can work this problem very simply and almost correctly by ignoring the vapor present. By first equation of problem 3-15

 



prime



 prime





 prime

prime



prime





kJ kg

Given m2 1



 m 2  Vf 2 0.95



prime



prime



f1 2 f2



 m   1      2

 1





Solution 6.53 Problem Statement Propylene gas at 127°C and 38 bar is throttled in a steady-state flow process to 1 bar, where it may be assumed to be an ideal gas. Estimate the final temperature of the propylene and its entropy change.

Solution When the gas is throttled through a valve, there is no work extracted, and negligible heat flow to or from the gas. Thus, the energy balance (around the valve) for the

H = Q + W = 0. The enthalpy of the

gas in the final state is equal to its enthalpy in the initial state. As usual, we can write the overall entropy change in three parts: H  H R 400 K,38 bar   

Tf 400 K

C igp dT  H R Tf ,1 bar 

Where Tf is the final temperature. If we assume that the gas behaves as an ideal gas at the final pressure and temperature, then the last term in the above equation is zero, and we simply have:

H  H R  400 K,38 bar  



Tf 400 K

C igp dT  0

C igp dT  H R 400 K,38 bar

So, we should first compute the residual enthalpy at the initial temperature and pressure, and then do the heat capacity integral, trying different final temperatures until the above equation is satisfied. For propylene, the critical properties are Tc = 365.6 K, Pc = 46.65, and

= 0.140. So, 400 K and 38 bar correspond to Tr = 1.094 and Pr = 0.815.

Checking figure 3.14, we see that the Pitzer correlation is not expected to be accurate at this reduced temperature and pressure, and we should interpolate in the Lee/Kesler tables in the back of the book. Using a spreadsheet to do so yields the following results: T (K) 400.15

P (bar) 38

Tc (K) 365.6

Pc (bar) 46.65

Tr 1.0945

Pr 0.8146

Tr 1.05 1.05 1.10 1.10

Pr 0.80 1.00 0.80 1.00

Z0 0.7130 0.6026 0.7649 0.6880

Z1 -0.0032 0.0220 0.0236 0.0476

(HR)0/(RTc ) -0.995 -1.359 -0.827 -1.120

0.7533

0.0224

-0.8674

HR/(RTc) HR (J/mol)

-0.94242 -2864.57

0.14

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Interpolated Values Final Values Z V (cm3/mol)

0.7565 662.4

(SR)0/R -0.656 -0.965 -0.537 -0.742

(SR)1/R -0.642 -0.820 -0.470 -0.577

-0.5359

-0.5659

-0.4973

SR (J/(mol K))

-0.63548 -5.28337

(HR)1/(RTc) -0.691 -0.877 -0.507 -0.617

SR/R

Solution continued on next page…

So, the residual enthalpy is 2864.6 J/mol. Now, we can use our old friend the heat capacity integral spreadsheet to find a final temperature that satisfies the above equation (a final temperature that corresponds to removing 2864.6 J/mol). The result is as follows: T 1 (K) 400

T 2 (K) 362.9176 T2

ICPH 

A 1.637

C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 2.27E-02 -6.92E-06 0.00E+00 -344.5 -2864.6

 1

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T  2

2 2

2 1

3 2

3 1

So, the final temperature is 363 K. Like the enthalpy change, the entropy change is the sum of three parts: S  S R  400 K,38 bar   

C igp

400 K

dT  R ln

1  S R 363 K,1 bar  38

Once again, the last term can be neglected. The first term was already found by interpolating in the Lee/Kesler tables, R

and is S = 5.28 J/(mol K). To compute the second term, we do an entropy heat capacity integral (cutting and pasting the heat capacity coefficients from our enthalpy spreadsheet). The result is:

T 1 (K) 400 T2

ICPS 

T 2 (K) 362.9176

A 1.637

 T2 

C (1/K ) D (K ) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

ICPS -0.903

DS (J/(mol K)) -7.51

 RT dT  A ln  T   B T  T   2 T  T   2 T  T 

2 2

2 1

2 2

2 1

S = 5.28 – 7.51 + 30.24 = 28.0 J/(mol K).

Solution 6.54 Problem Statement Propane gas at 22 bar and 423 K is throttled in a steady-state flow process to 1 bar. Estimate the entropy change of the propane caused by this process. In its final state, propane may be assumed to be an ideal gas.

Solution When the gas is throttled through a valve, there is no work extracted, and negligible heat flow to or from the gas. H = Q + W = 0. The enthalpy of the gas in the final state is equal to its enthalpy in the initial state. As usual, we can write the overall entropy change in three parts: H  H R 423 K,22 bar   

Tf 423 K

C igp dT  H R Tf ,1 bar 

Where Tf is the final temperature. If we assume that the gas behaves as an ideal gas at the final pressure and temperature, then the last term in the above equation is zero, and we simply have:

H  H R  423 K,22 bar  



Tf 423 K

C igp dT  0

C igp dT  H R 423 K,22 bar

For propane, Tc = 369.8, Pc = 42.48, and

= 0.152. Thus, at 423 K and 22 bar, we have Tr = 1.1439,

Pr = 0.5179. Checking Fig. 3.14 on p. 103 of SVNA, we see that this is within the range where the Pitzer correlation for the 2

virial coefficient can be expected to work well. So, we can use our handy spreadsheets for calculating

residual enthalpy and entropy from the 2

virial coefficient:

Solution continued on next page…

T (K) 423

Tr 1.1439

P (bar) 22

Pr 0.5179

B0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) Pc (bar) 369.8 42.48

B -0.2573

w 0.152

dB0 dTr

dB1 dTr

0.4759

0.3589

B 0.0412

dB0 0.675  2.6 dTr Tr

H /RTc -0.4443

-1

(J mol ) -1366.04

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

So, the residual enthalpy at the initial state is H (423K, 22 bar) = 1366 J/mol. Now, we want to find the final temperature for which the heat capacity integral will balance this. Using our handy heat capacity integral spreadsheet, we have: T 1 (K) 423

T 2 (K) 408.9078 T2

ICPH 

A 1.213

C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 2.88E-02 -8.82E-06 0.00E+00 -164.3 -1366.0

 1

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T  2

2 2

2 1

3 2

3 1

In this spreadsheet, the final temperature (T2) was varied until

H was achieved. The resulting final

temperature was 408.9 K. Now we can use that temperature to find the entropy change (which is what we’re supposed to be finding). Like the enthalpy change, the entropy change is the sum of three parts: S  S 423 K,22 bar    R

408.9

C igp

423 K

dT  R ln

1  S R 408.9 K,1 bar  22

Solution continued on next page…

Once again, the last term can be neglected. The first term can be found using the handy residusl entropy spreadsheet, just like the enthalpy:

T (K) 423

P (bar) 22

Tr 1.1439

Pr 0.5179

Tc (K) Pc (bar) 369.8 42.48

dB0 dTr

dB1 dTr

0.4759

0.3589

dB0 0.675  2.6 dTr Tr

w 0.152

SR/R -0.2747

SR (J mol-1 K-1) -2.28

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. The heat capacity integral is calculated using its handy spreadsheet as well: T 1 (K) 423

T 2 (K) 408.9078 T2

ICPS 



A 1.213

C (1/K ) D (K ) B (1/K) 2.88E-02 -8.82E-06 0.00E+00

ICPS -0.395

T  C 2 D 2 dT  A ln  2   B T2  T1   T2  T12  T2  T12 RT T 2 2  1







DS (J/(mol K)) -3.28



S = 2.28 – 3.28 +8.314*ln(22) = 24.70 J/(mol K). The entropy increase due to the decrease in pressure is much larger than either the change due to cooling or the residual entropy in the initial state.

Solution 6.55 Problem Statement Propane gas at 100°C is compressed isothermally from an initial pressure of 1 bar to a final pressure of 10 bar. H

Solution For each property, we can write the overall change as H= H

H + H (10 bar, 373 K)

S= S

S + S (10 bar, 373 K)

The ideal gas enthalpy change is zero for an isothermal process, while the ideal gas entropy change is ig

S = Rln(P2/P1) = Rln(10) = 19.145 J/(mol K). We can use the Pitzer correlation for the residual properties (conditions are such that it is applicable):

T (K) 373.15

Tr 1.0091

P (bar) 1

Pr 0.0235

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) 369.8

B0 -0.3330

Pc (bar) 42.48

B1 -0.0266

dB0 0.675  2.6 dTr Tr

w 0.152

dB0 dTr

dB1 dTr

0.6594

0.6889

HR /RTc -0.0261

HR (J mol-1) -80.20

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 373.15

Tr 1.0091

P (bar) 10

Pr 0.2354

B  0.083  B1  0.139 

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 369.8

B -0.3330

Pc (bar) 42.48

B -0.0266

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.152

dB0 dTr

dB1 dTr

0.6594

0.6889

HR /RTc -0.2608

HR (J mol-1) -801.97

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 373.15

Tr 1.0091

P (bar) 1

Pr 0.0235

Tc (K) 369.8

Pc (bar) 42.48

dB0 dTr

dB1 dTr

0.6594

0.6889

dB0 0.675  2.6 dTr Tr

w 0.152

SR /R -0.0180

SR (J mol-1 K-1) -0.1496

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. T (K) 373.15

Tr 1.0091

P (bar) 10

Pr 0.2354

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

Tc (K) 369.8

Pc (bar) 42.48

dB0 dTr

dB1 dTr

0.6594

0.6889

w 0.152

SR /R -0.1799

SR (J mol-1 K-1) -1.4955

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output.

Using these results: H = 80.20 + 0 – 801.97 = 721.8 J/mol S = 0.1496 – 19.145 – 1.4955 = 20.491 J/(mol K) According to the NIST webbook, at 100 °C and 1 bar, H = 33.927 kJ/mol and S = 143.83 J/(mol K) and at 100 °C and 10 bar, H = 33.178 kJ/mol and S = 123.28 J/(mol K). Thus, the actual enthalpy and entropy changes are: HNIST = 33.178 – 33.927 = 0.749 kJ/mol = 749 J/mol SNIST = 123.28 – 143.83 = 20.55 J/(mol K) These are in good agreement with the values we got using the Pitzer correlation.

Solution 6.56 Problem Statement Hydrogen sulfide gas is compressed from an initial state of 400 K and 5 bar to a final state of 600 K and 25 bar. H

Solution The critical values for H2S are as follows:

Tc 

P2 

bar

T2 

Pc 

bar

Tr 1 



T1 

Pr 1 

K P1  bar

Tr 2 

Pr 2 

Use generalized virial-coefficient correlation for both sets of conditions. Eqs. (6.72) & (6.73) are written H  R * ICPH T1 T2  7821.41

3



  RT  HRBT P c

r 2,





  HRB Tr 1, Pr 1  

J mol

 S  R ICPS T1 T2  J  1.551 mol K

3



P2 

 ln P   R SRBT P  SRB T P  1



r 2, r 2

r 1, r 1

Solution 6.57 Problem Statement Carbon dioxide expands at constant enthalpy (as in a throttling process) from 1600 kPa and 45°C to 101.33 kPa. S for the process.

Solution H = 0 (the throttling process is adiabatic and does not produce or consume work). As usual, we can break the change up into 3 parts: R

H = H (45 ºC,1600 kPa) + H + H (T2, 101.33 kPa) = 0

Carbon dioxide’s critical properties are Tc = 304.2K, Pc = 73.83 bar, and  = 0.224. So, at the inlet conditions, Tr = 1.046, Pc = 0.2167. The Pitzer correlation for the 2

T (K) 318.15

Tr 1.0459

P (bar) 16

Pr 0.2167

B 0  0.083 

0.422

B1  0.139 

0.172

Tc (K) 304.2

Pc (bar) 73.83

B0 -0.3098

B1 -0.0035

dB0 0.675  2.6 dTr Tr

Tr1.6

dB1 0.722  5.2 dTr Tr

Tr4.2

virial coefficient is applicable, and we have:

w 0.224

dB0 dTr

dB1 dTr

0.6007

0.5718

HR /RTc -0.2325

HR (J mol-1) -588.04

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output.

At the outlet conditions (1.01 bar) the gas will be nearly ideal and the residual properties will be near zero. So, to a R

good approximation, we have H (45 ºC,

H = 588 J/mol.

So, we do the heat capacity integral and vary the final temperature until the enthalpy change is 588 J/mol:

ICPH  T0 ,T;A,B,C,D  

R

T0(K) 318.15

T (K) 302.71

A 5.457

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) B (1/K) ICPH (K) D H (J/mol) 1.05E-03 0.00E+00 -1.16E+05 -70.7 -588.0

Solution continued on next page…

So, the estimated outlet temperature is 302.7 K = 29.6 ºC. At these outlet conditions, the residual enthalpy is: T (K)

P (bar)

Tc (K)

Pc (bar)

302.7

1.0133

304.2

73.83

0.224

dB0 dTr 0

0.9951

0.0137

B 0  0.083  1

B  0.139 

dB1 dTr

0.3424

H /RTc

0.0366

0.6837

0.7408

H (J mol )

0.0164

41.52

0.422 Tr1.6

dB0 0.675  2.6 dTr Tr

0.172 Tr4.2

dB1 0.722  5.2 dTr Tr

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output. Adding this back to the overall equation gives: R

H (45 ºC,1600 kPa)

H (29.6 ºC,101.33 kPa) = H = 546 J/mol. Re-doing the heat capacity integral:

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 318.15

T (K) 303.83

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) B (1/K) ICPH (K) D H (J/mol) 1.05E-03 0.00E+00 -1.16E+05 -65.7 -546.0

A 5.457

This gives an improved estimate of the outlet temperature to be 303.8 K. Using that value, we compute the entropy change as R

S = S (45 ºC,1600 kPa) + S + S (20.7ºC, 101.33 kPa)

The inlet residual entropy is: T (K) 318.15

Tr 1.0459

P (bar) 16

Pr 0.2167

Tc (K) 304.2

Pc (bar) 73.83

dB0 dTr

dB1 dTr

0.6007

0.5718

dB0 0.675  2.6 dTr Tr

w 0.224

S R/R -0.1579

SR (J mol-1 K-1) -1.31

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr The outlet residual entropy is:

T (K) 303.8

Tr 0.9987

P (bar) 1.0133

Pr 0.0137

Tc (K) 304.2

Pc (bar) 73.83

dB0 dTr

dB1 dTr

0.6773

0.7270

dB0 0.675  2.6 dTr Tr

w 0.224

S R/R -0.0115

SR (J mol-1 K-1) -0.10

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr The heat capacity integral is:

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T0 (K) 318.15

T (K) 303.83

A 5.457

C (1/K2) D (K2) B (1/K) 1.05E-03 0.00E+00 -1.16E+05

ICPS -0.211

(J/(mol K)) -1.7557

Finally, the ideal gas pressure dependence is –R ln (101.33/1600) = 22.94 J/(mol K) The total entropy change is: S = 1.31 – 1.76 + 22.94 – 0.10 = 22.39 J/(mol K)

Solution 6.58 Problem Statement A stream of ethylene gas at 250°C and 3800 kPa expands isentropically in a turbine to 120 kPa. Determine the temperature of the expanded gas and the work produced if the properties of ethylene are calculated by: (a) Equations for an ideal gas; (b) Appropriate generalized correlations.

Solution (a) For an ideal gas, we have the entropy change, set equal to zero, as S  



C igp

523 K

C igp

523 K

dT  R ln

120 0 3800

dT  28.73 J/(mol K)

Evaluating the heat capacity integral using our handy spreadsheet, and varying the final temperature to achieve the desired entropy change, we get: T 1 (K) 523.15 T2

ICPS 

T 2 (K) 308.167

A 1.424

 T2 

C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

ICPS -3.456

DS (J/(mol K)) -28.73

 RT dT  A ln  T   B T  T   2 T  T   2 T  T 

2 2

2 1

2 2

2 1

H = Q + W, shows that if the expander is adiabatic (which it must be to be isentropic) then the work done on the gas is equal to its enthalpy change. Still treating it as an ideal gas, the enthalpy change is given by the usual heat capacity integral (enthalpy is independent of pressure for an ideal gas): Solution continued on next page…

T 1 (K) 523.15

T 2 (K) 308.167 T2

ICPH 

R

C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 1.44E-02 -4.39E-06 0.00E+00 -1425.6 -11852.5

A 1.424

dT  A  T2  T1  

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









The work extracted from the gas is 11.85 kJ/mol. (b) If we include the residual enthalpy and entropy of the ethane, then we will have Tf

C igp

523 K

S  S R 523 K,38 bar   

 

C igp

T C igp

523 K

dT  R ln

dT  S R 523 K,38 bar   R ln

1.2  S R Tf ,1.2 bar   0 38

1.2  S R Tf ,1.2 bar  38

dT  S R 523 K,38 bar   28.73 J/(mol K)  S R Tf ,1.2 bar 

For ethylene, the critical properties are Tc = 282.3 K, Pc = 50.40 K, and 1.8532, Pr = 0.7540. For these conditions, the Pitzer correlation for the 2

= 0.087. So, in the initial state, we have Tr = nd

virial coefficient is valid, and again we can

use our handy spreadsheets. For the residual entropy, we have:

T (K) 523.15

Tr 1.8532

P (bar) 38

Pr 0.7540

Tc (K) Pc (bar) 282.3 50.4

dB0 dTr

dB1 dTr

0.1357

0.0292

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.87

SR/R -0.1215

SR (J mol-1 K-1) -1.01

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output. If we neglect the residual entropy in the final (low pressure) state, then we have



C igp

523 K

dT  1.01  28.73  29.74 J/(mol K)

Solution continued on next page…

Modifying the heat capacity integral spreadsheet to achieve this entropy change gives: T 1 (K) 523.15

T 2 (K) 301.3147 T2

ICPS 



C (1/K ) D (K ) B (1/K) 1.44E-02 -4.39E-06 0.00E+00

A 1.424

ICPS -3.577

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T12 RT 2 2  T1 







DS (J/(mol K)) -29.74



Now, the predicted outlet temperature is 301.3 K, about 7 K lower than that predicted using the ideal gas properties. Using this final temperature in the heat capacity integral for enthalpy gives: T 1 (K) 523.15

T 2 (K) 301.3147 T2

ICPH 

R

dT  A  T2  T1  

C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 1.44E-02 -4.39E-06 0.00E+00 -1462.6 -12160.3

A 1.424

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









However, we also need to include the residual enthalpy in the initial state, which is:

T (K) 523.15

Tr 1.8532

P (bar) 38

Pr 0.7540

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) Pc (bar) 282.3 50.4

B -0.0743

B 0.1261

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.87

dB0 dTr 0.1357

dB1 dTr

0.0292

HR/RT c -0.1984

HR (J mol-1) -465.77

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output.

Still neglecting the residual enthalpy in the final state, this gives us the total enthalpy change as: H  H R 523 K,38 bar   

301 K

523 K

C igp dT  465.8  12160.3  11695 J/mol

Including the residual properties in the initial state, the work extracted from the gas is 11.70 kJ/mol. This is about 1.3% lower than what we predicted using the ideal gas properties. Finally, we should check that the residual R

properties in the final state really are negligible. At 1.2 bar and 301 K, S = 0.20 J/(mol K) (about 20% of the value at the initial state). Adjusting the entropy by this amount would change the final temperature to 302.7 K. Using this final temperature and adding in the residual enthalpy at this final temperature (

H to 11709 J/mol.

The estimated work extracted from the gas is 11.71 kJ/mol. Including residual properties in the final state changed the computed work by less than 0.1%. Neglecting them was a good approximation.

Solution 6.59 Problem Statement A stream of ethane gas at 220°C and 30 bar expands isentropically in a turbine to 2.6 bar. Determine the temperature of the expanded gas and the work produced if the properties of ethane are calculated by: (a) Equations for an ideal gas; (b) Appropriate generalized correlations.

Solution In this case, we want to calculate the final temperature from the specification that the process is isentropic. The overall entropy change is: S  S R 493 K,30 bar   

C pig

493 K

dT  R ln

2.6  S R Tf K,2.6 bar   0 30

(a) If we assume that the ethane is an ideal gas, in both the initial and final state, then the first and last terms are zero. The third term evaluates to 8.314 ln(11.54) = 20.33 J/(mol K). So, if ethane were an ideal gas, we would have



C igp

493 K

dT  20.33 J/(mol K)

Solution continued on next page…

Using our handy-dandy entropy integral spreadsheet, we have T 1 (K) 493

T 2 (K) 367.472 T2

ICPS 



C (1/K ) D (K ) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

A 1.131

ICPS -2.445

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T12 RT T 2 2  1







DS (J/(mol K)) -20.33



So, using the ideal gas equations, the final temperature is 367.5 K. Since the process is adiabatic (Q = 0), the work done on the system (negative, since the system does work on the surroundings) is just equal to the enthalpy change of H = W. So, to compute the work done, we need to do an enthalpy heat capacity integral: T 1 (K) 493

T 2 (K) 367.472 T2

ICPH 

R

C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 1.92E-02 -5.56E-06 0.00E+00 -1050.1 -8730.7

A 1.131

dT  A  T2  T1  

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









The work produced is about 8.7 kJ per mole of gas expanded through the turbine. (b) In this case, we should take into account the residual entropy in the initial and final state. For ethane, the critical properties are Tc = 305.3 K, Pc = 48.72 bar, and

= 0.100. Thus, in the initial state, the reduced temperature and

pressure are Tr = 1.600 and Pr = 0.616. A quick check of figure 3.14 shows that this is well within the range where the Pitzer correlation for the 2

virial coefficient performs well. So, we can use the simple spreadsheets provided in the

lecture notes for computing the residual entropy and enthalpy from the Pitzer correlation. In the final state, the reduced pressure is 0.0534, and the reduced temperature is not known exactly, since we don’t yet know the final temperature. For the final temperature from the ideal gas calculations, the reduced temperature would be 1.2. For such a low reduced pressure, non-idealities should be small (hopefully negligible) so at least initially we can neglect the residual entropy at the final conditions. Then we have S  S R 493 K,30 bar   

C igp

493 K

dT  R ln

2.6 0 30

Solution continued on next page…



C igp

493 K

dT  S R 493 K,30 bar   20.33 J/(mol K)

Using the handy-dandy Pitzer correlation spreadsheet, the residual entropy at 493 K and 30 bar is:

T (K) 493.15

P (bar) 30

Tr 1.6153

Pr 0.6158

Tc (K) Pc (bar) 305.3 48.72

dB0 dTr

dB1 dTr

0.1940

0.0597

dB0 0.675  2.6 dTr Tr

w 0.1

SR/R -0.1231

SR (J mol-1 K-1) -1.02

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

The residual entropy is just 1.02 J/(mol K). So, we have



C igp

493 K

dT  21.35 J/(mol K)

Just as in part (a), we can evaluate the entropy heat capacity integral and try different final temperatures until we S. T 1 (K) 493

T 2 (K) 361.4254 T2

ICPS 



A 1.131

C (1/K ) D (K ) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

ICPS -2.568

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T12 RT T 2 2  1







DS (J/(mol K)) -21.35



Solution continued on next page…

So, we predict the final temperature to be 361.4 K, neglecting the residual entropy in the final state. The residual entropy at 361.4 K and 2.6 bar is 0.21 J/(mol K) as shown below:

T (K) 361

P (bar) 2.6

Tr 1.1824

Pr 0.0534

Tc (K) Pc (bar) 305.3 48.72

dB0 dTr

dB1 dTr

0.4366

0.3021

dB0 0.675  2.6 dTr Tr

w 0.1

SR/R -0.0249

SR (J mol-1 K-1) -0.21

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Including this (previously neglected) contribution, we have

 

C pig

T C pig

493 K

dT  S R  493 K,30 bar  20.33 J/(mol K)  S R Tf ,2.6 bar  dT  1.02  20.33  0.21  21.14 J/(mol K)

Using this value to compute the final temperature gives: T 1 (K) 493

T 2 (K) 362.6735 T2

ICPS 



A 1.131

C (1/K ) D (K ) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

ICPS -2.543

T  C 2 D 2 dT  A ln  2   B T2  T1   T2  T12  T2  T12 RT T 2 2  1







DS (J/(mol K)) -21.14



And the predicted final temperature is 363 K. The residual entropy at 363 K is negligibly different from that at 361 K, and we do not need to iterate further. Now, the enthalpy change is H  H R 493 K,30 bar   

363 K 493 K

C igp dT  H R 363 K,2.6 bar 

Solution continued on next page…

We use the usual heat capacity integral for the second term, and the Pitzer correlation for the first and third terms. This gives:

T (K) 493

P (bar) 30

Tr 1.6148

Pr 0.6158

B  0.083 

Tc (K) Pc (bar) 305.3 48.72

B -0.1130

Tr1.6

B 0.1160

B1  0.139 

0.172 Tr4.2

dB1 0.722  5.2 dTr Tr

T 1 (K) 493

T 2 (K) 363

A 1.131

ICPH 

0.1

dB0 dTr

dB0 0.675  2.6 dTr Tr

0.422

0.1942

dB1 dTr

0.0597

Tr 1.1890

-1

(J mol ) -663.72

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 1.92E-02 -5.56E-06 0.00E+00 -1083.3 -9006.2

 1

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T  2

2 2

2 1

3 2

3 1

T (K) 363

H /RT c -0.2615

P (bar) 2.6

Pr 0.0534

B  0.083  B1  0.139 

Tc (K) Pc (bar) 305.3 48.72

B0 -0.2369

0.422 Tr1.6 0.172 Tr4.2

So, the final result is that W

B1 0.0559

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.1

dB0 dTr 0.4304

dB1 dTr

0.2935

HR/RT c -0.0415

HR (J mol-1) -105.38

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

H = 663.7 – 9006.2 – 105.4 = 8448 J/mol. The predicted work extracted is about

8.4 kJ per mole of gas expanded through the turbine, which is about 3% lower than we predicted using the ideal gas equations.

Solution 6.60 Problem Statement Estimate the final temperature and the work required when 1 mol of n-butane is compressed isentropically in a steady-flow process from 1 bar and 50°C to 7.8 bar.

Solution

If the compression is isentropic, then we want to use the entropy balance to find the outlet conditions. We can write it as: R

S= S

S + S (7.8 bar, T2) = 0

The ideal gas part of the entropy change is T2

2 Cp  7.8  dT  R ln     dT  17.08 J mol1 K1   T 1 T   323.15 323.15

S ig  

We can use the Pitzer correlation to obtain the residual entropy at the inlet conditions (which we expect will be small): T (K) 323.15

Tr 0.7602

P (bar) 1

Pr 0.0263

Tc (K) 425.1

Pc (bar) 37.96

dB0 dTr

dB1 dTr

1.3770

3.0046

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.2

SR /R -0.0521

SR (J mol-1 K-1) -0.4332

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output.

Solution continued on next page…

The entropy balance then gives us: T2

 T dT  17.08  0.43  S T ,7.8 bar  16.65  S T ,7.8 bar J mol R

1

K 1

323.15

If the n-butane were an ideal gas, then we would simply have had T2

 T dT  17.08 J mol

1

K 1

323.15

We can use that equation to get a good initial estimate of the actual outlet temperature. Evaluating the entropy heat capacity integral and varying the outlet temperature to satisfy this equation gives:

ICPS  T0 ,T;A,B,C,D  



T0 (K) 323.15

T (K) 376.56

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT 2 T T0   T0    

A 1.935

C (1/K2) D (K2) B (1/K) 3.69E-02 -1.14E-05 0.00E+00

ICPS 2.054

(J/(mol K)) 17.0803

Evaluating the residual entropy at 376.56 K and 7.8 bar gives: T (K) 376.56

Tr 0.8858

P (bar) 7.8

Pr 0.2055

Tc (K) 425.1

Pc (bar) 37.96

dB0 dTr

dB1 dTr

0.9251

1.3563

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.2

SR /R -0.2458

SR (J mol-1 K-1) -2.0440

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output.

Using that value in the entropy balance gives: T2

 T dT  17.08  0.43  S T ,7.8 bar  16.65  2.044  18.69 J mol R

1

K 1

323.15

Varying the final temperature in the heat capacity integral to satisfy the entropy balance then gives:

ICPS  T0 ,T;A,B,C,D  



T0 (K) 323.15

T (K) 381.69

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT 2 T T0   T0    

C (1/K2) D (K2) B (1/K) 3.69E-02 -1.14E-05 0.00E+00

A 1.935

ICPS 2.248

(J/(mol K)) 18.6893

Re-evaluating the residual entropy at 381.69 K gives: T (K) 381.69

Tr 0.8979

P (bar) 7.8

Pr 0.2055

Tc (K) 425.1

Pc (bar) 37.96

dB0 dTr

dB1 dTr

0.8932

1.2641

dB0 0.675  2.6 dTr Tr

w 0.2

SR /R -0.2355

SR (J mol-1 K-1) -1.9579

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output.

Using that value in the entropy balance gives: T2

 T dT  17.08  0.43  S T ,7.8 bar  16.65  1.958  18.604 J mol R

1

K1

323.15

Varying the final temperature in the heat capacity integral to satisfy the entropy balance then gives: T

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T0 (K) 323.15

T (K) 381.42

C (1/K2) D (K2) B (1/K) 3.69E-02 -1.14E-05 0.00E+00

A 1.935

ICPS 2.238

(J/(mol K)) 18.6041

Further iteration would have no significant effect, so we can say that the outlet temperature is 381.4 K = 108.3 °C. We obtain the work requirement from the enthalpy balance for this adiabatic process: Ws = H = H

H + H (7.8 bar, 381.4 K)

The ideal gas enthalpy change is given by:

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 323.15

T (K) 381.42

A 1.935

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) B (1/K) ICPH (K) D H (J/mol) 3.69E-02 -1.14E-05 0.00E+00 787.9 6550.6

The residual enthalpies are

T (K) 323.15

Tr 0.7602

P (bar) 1

Pr 0.0263

B  0.083  B1  0.139 

0.422 Tr1.6 0.172 Tr4.2

Tc (K) 425.1

B -0.5714

Pc (bar) 37.96

B -0.4051

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.2

dB0 dTr

dB1 dTr

1.3770

3.0046

HR /RTc -0.0568

HR (J mol-1) -200.75

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

and

T (K) 381.42

Tr 0.8972

P (bar) 7.8

Pr 0.2055

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) 425.1

B0 -0.4189

Pc (bar) 37.96

B1 -0.1322

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.2

dB0 dTr

dB1 dTr

0.8948

1.2688

HR /RTc -0.3033

HR (J mol-1) -1071.92

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

Adding these up gives: Ws = H = 200.75 + 6550.6

1071.92 = 5679 J/mol

We could also treat this problem using data from the NIST WebBook, because n-butane is one of the substances for which properties are tabulated. First, we would look up the properties at 50 °C and 1 bar, where we find they are H = 39.064 kJ/mol, S = 157.3 J/(mol K). Next, we look for the temperature where S = 157.3 J/(mol K) at P = 7.8 bar. At 109.0 °C, the WebBook gives an entropy value of 157.32 J/(mol K). Using the Pitzer correlation and heat capacity integral, we underpredicted the temperature by about 0.7 °C, which is pretty negligible in most contexts. The enthalpy at 7.8 bar and 109 °C is 44.745 kJ/mol. Thus, the work using the NIST data would be Ws,NIST = 44745 – 39064 = 5681 J/mol, which is in essentially exact agreement with the result obtained using the Pitzer correlation and heat capacity integral.

Solution 6.61 Problem Statement Determine the maximum amount of work obtainable in a flow process from 1 kg of steam at 3000 kPa and 450°C for surrounding conditions of 300 K and 101.33 kPa.

Solution The maximum work results when the 1 kg of steam is reduced in a completely reversible process to the conditions of the surroundings, where it is liquid at 300 K (26.85 degC). This is the ideal work. From Table E.2 for the initial state of superheated steam: H1 

kJ S  kg 1

kJ kg K

Solution continued on next page…

From Table E.1, the state of sat. liquid at 300 K is essentially correct: H2 

kJ S  kg 2

kJ T  kg K

By Eq. (5.22), kJ kg

Wideal   H2  H1   T S2  S1  

Solution 6.62 Problem Statement

Liquid water at 325 K and 8000 kPa flows into a boiler at a rate of 10 kg·s

and is vaporized, producing saturated

vapor at 8000 kPa. What is the maximum fraction of the heat added to the water in the boiler that can be converted into work in a process whose product is water at the initial conditions, if T = 300 K? What happens to the rest of the heat? What is the rate of entropy change in the surroundings as a result of the work-producing process? In the system? Total?

Solution Sat. liquid at 325 K (51.85 C), Table E.1: H liq 

kJ kg

kJ kg K

Sliq 

P1 

kPa

cm3 g

Vliq 

T

Psat 

kPa

For the compressed liquid at 325 K and 8000 kPa, apply Eqs. (6.27) and (6.28) with   460 * 106

1 K kj kg

H1  Hliq  Vliq   T  *  P1  Psat   S1  Sliq  Vliq  *  P1  Psat  

kj kg K

For sat. vapor at 8000 kPa, from Table E.2: kJ kg

H2 

Heat added in boiler:

S2 

kJ kg K

T 

kJ kg

Q  H2  H1 

Maximum work from steam, by Eq. (5.22): Wideal  H1  H 2  T S1  S2  

kJ kg

Work as a fraction of heat added:

Frac 

Wideal Q

 0.4058

The heat not converted to work ends up in the surroundings. Q  Wideal kg kW SG, surr  * 10  50.234 T s K SG , system  S1  S2  *

kg  s

kW K

Obviously the TOTAL rate of entropy generation is zero. This is because the ideal work is for a completely reversible process.

Solution 6.63 Problem Statement Suppose the heat added to the water in the boiler in the preceding problem comes from a furnace at a temperature of 600°C. What is the total rate of entropy generation as a result of the heating process? What is W lost ?

Solution Treat the furnace as a heat reservoir, for which Q 

kJ s

T

The rate of entropy lost is Q kW kW SG   50.234  21.19 T K K And by Eq 5.29

T 

W lost T SG 

Solution 6.64 Problem Statement 1

An ice plant produces 0.5 kg·s of flake ice at 0°C from water at 20°C (T ) in a continuous process. If the latent heat 1

of fusion of water is 333.4 kJ·kg and if the thermodynamic efficiency of the process is 32%, what is the power requirement of the plant?

Solution For sat. liquid water at 20 C, Table E.1: H1 

kJ S  kg 1

kJ kg K

H0  

kJ S  kg 0

kJ kg K

For sat. liquid water at 0 C, Table E.1:

Solution continued on next page…

For ice at 0 C: H2  

kJ kg

kJ kg K

S2  

T 

kg s

m 

t 

By Eqs. (5.21) and (5.23): W ideal  m  H1  H2   T  S1  S2   W W  ideal  t

Solution 6.65 Problem Statement An inventor has developed a complicated process for making heat continuously available at an elevated temperature. Saturated steam at 100°C is the only source of energy. Assuming that there is plenty of cooling water available at 0°C, what is the maximum temperature level at which heat in the amount of 2000 kJ can be made available for each kilogram of steam flowing through the process?

Solution This is a variation on Example 5.6., pp. 175-177, where all property values are given. We approach it here from the point of view that if the process is completely reversible then the ideal work is zero. We use the notation of Example 5.6: 





kJ  kg 1

kJ kg  K

 



kJ kg  K 

kJ kg





kJ kg



Solution continued on next page…

The system consists of two parts: the apparatus and the heat reservoir at elevated temperature, and in the equation for ideal work, terms must be included for each part.

 

ideal



apparatus.reservoir

 





 apparatus.reservoir





  apparatus.reservoir 



   

Q    T  



Q T

ideal





kJ kg

(Guess)

Given



 

   

kJ  kg







(136.64 degC)

Solution 6.66 Problem Statement Two boilers, both operating at 200(psia), discharge equal amounts of steam into the same steam main. Steam from the first boiler is superheated at 420(°F) and steam from the second is wet with a quality of 96%. Assuming adiabatic mixing and negligible changes in potential and kinetic energies, what is the equilibrium condition after mixing and what is SG for each (lbm) of discharge steam?

Solution From Table F.4 at 200(psi):





BTU lb m

liq





BTU lbm



vap





BTU lb m  rankine



BTU lbm

(at 420 degF)

(Sat. liq. And vapor)

liq







BTU lb m  rankine

liq







vap





vap

 BTU lb m

liq



BTU lbm  rankine





liq



  vap 



liq



BTU lbm  ranknine



Neglecting kinetic- and potential-energy changes, on the basis of 1 pound mass of steam after mixing, Eq. (2.30) yields for the exit stream: 







BTU lbm



(wet stream)



liq



H  Hliq H vap  H liq

  vap 

liq



BTU lbm  rankine

 

By Eq. (5.17) on the basis of 1 pound mass of exit steam,





 1

 2 G



4

BTU lbm  rankine

Solution 6.67 Problem Statement 3

A rigid tank of 80(ft) capacity contains 4180(lbm) of saturated liquid water at 430(°F). This amount of liquid almost completely fills the tank, the small remaining volume being occupied by saturated-vapor steam. Because a bit more vapor space in the tank is wanted, a valve at the top of the tank is opened, and saturated-vapor steam is vented to the atmosphere until the temperature in the tank falls to 420(°F). Assuming no heat transfer to the contents of the tank, determine the mass of steam vented.

Solution From Table F.3 at 430 degF (sat. liq. and vapor): 

liq





BTU lb m



vap

liq





liq





liq



tank





vap

liq



 vap



liq

ft 3 lb m



vap

Vvap 

vap



tank

BTU lbm

liq



liq

VOL vap

ft 3 lbm

vap









vap

m liq  m vap



BTU lb m



By Eq. (2.28) multiplied through by dt, we can write,

 t  t 





(Subscript t denotes the contents of the tank. H and m refer to the exit stream.)

Integration gives:











From Table F.3 we see that the enthalpy of saturated vapor changes from 1203.9 to 1203.1(Btu/lb) as the temperature drops from 430 to 420 degF. This change is so small that use of an average value for H of 1203.5(Btu/lb) is fully justified. Then 2









liq



ave

vap







ave







BTU lbm



Solution continued on next page…

Property values below are for sat. liq. and vap. at 420 degF



 



liq





liq





V2 mass : 

Vtank

  vap 



liq



ft3 lb m

BTU lbm

m 2 mass

vap





ft 3 lbm

vap





BTU lbm

x mass : 

V2 mass  Vliq vap



liq



(Guess)

mass 

Given

liq



ave



2 2





 







Solution 6.68 Problem Statement 3

A tank of 50 m capacity contains steam at 4500 kPa and 400°C. Steam is vented from the tank through a relief valve to the atmosphere until the pressure in the tank falls to 3500 kPa. If the venting process is adiabatic, estimate the final temperature of the steam in the tank and the mass of steam vented.

Solution The steam remaining in the tank is assumed to have expanded isentropically. Data from Table E.2 at 4500 kPa and 400 degC: S1 

J gK

V1 

cm3 g

Vtank 

Solution continued on next page…

Interpolate using this entropy and 3500 kPa to find V2 S2  S1 

m1 

Vtank V1



J gK m2 

V2 

Vtank V2



cm3 T2  g kg

m  m2  m1 

Solution 6.69 Problem Statement 3

A tank of 4 m capacity contains 1500 kg of liquid water at 250°C in equilibrium with its vapor, which fills the rest of the tank. A quantity of 1000 kg of water at 50°C is pumped into the tank. How much heat must be added during this process if the temperature in the tank is not to change?

Solution This problem is similar to Example 6.8, where it is shown that  







Here, the symbols with subscript t refer to the contents of the tank, whereas H refers to the entering stream. We illustrate here development of a simple expression for the first term on the right. The1500 kg of liquid initially in the tank is unchanged during the process. Similarly, the vapor initially in the tank that does NOT condense is unchanged. The only two enthalpy changes within the tank result from: 1.

Addition of 1000 kg of liquid water. This contributes an enthalpy change of

Hliq mt Solution continued on next page…

Condensation of y kg of sat. vapor to sat. liq. This contributes an enthalpy change of 

liq



vap



Thus





liq





liq







Similarly, Whence

   



liq













 lv  t

Required data from Table F.1 are: 

At 50 degC:

At 250 degC:





kJ kg



liq









kJ kg





Vliq  m t  Vlv liq



liq





kJ  lv  kg

cm3 gm 

cm3 gm



 



Solution 6.70 Problem Statement 3

Liquid nitrogen is stored in 0.5 m metal tanks that are thoroughly insulated. Consider the process of filling an evacuated tank, initially at 295 K. It is attached to a line containing liquid nitrogen at its normal boiling point of 1

77.3 K and at a pressure of several bars. At this condition, its enthalpy is –120.8 kJ·kg . When a valve in the line is opened, the nitrogen flowing into the tank at first evaporates in the process of cooling the tank. If the tank has a mass 1

of 30 kg and the metal has a specific heat capacity of 0.43 kJ·kg ·K , what mass of nitrogen must flow into the tank just to cool it to a temperature such that liquid nitrogen begins to accumulate in the tank? Assume that the nitrogen

and the tank are always at the same temperature. The properties of saturated nitrogen vapor at several temperatures are given as follows:

T /K P /bar

80 85 90 95 100 105 110

0.1640 0.1017 0.06628 0.04487 0.03126 0.02223 0.01598

78.9 82.3 85.0 86.8 87.7 87.4 85.6

1.396 2.287 3.600 5.398 7.775 10.83 14.67

Solution At the point when liquid nitrogen starts to accumulate in the tank, it is filled with saturated vapor nitrogen at the final temperature and having properties mvap ,Tvap ,Vvap ,Hvap ,and Uvap By Eq. (2.28) multiplied through by dt d nt  U t   H  dm  dQ Subscript t denotes the contents of the tank; H and m refer to the inlet stream. Since the tank is initially evacuated, integration gives

mvap  U vap  H in  mvap  Q  mtank  C  Tvap  T1 

mvap 

Vtank Vvap

(a)

(b)

Solution continued on next page…

Calculate internal-energy values for saturated vapor nitrogen at the given values of T: v

T /K

P /bar

V /m ⋅kg

1.396

0.164

78.9

56.0056

2.287

0.1017

82.3

59.04121

3.6

0.06628

61.1392

5.398

0.04487

86.8

62.579174

100

7.775

0.03126

87.7

63.39535

105

10.83

0.02223

87.4

63.32491

110

14.67

0.01598

85.6

62.15734

H /kJ⋅kg

U kJ/kg

Fit tabulated data with cubic spline: U s  lspline T U  U vap t   inter p U s T U t 

Vs  lspline T U  Vvap t   inter p Vs T V t 

Combining Eqs. (A) & (B) gives: U vap Tvap   H in 

mtank C * T1  Tvap  * Vvap Tvap  Vtank

Solving for Tvap and mvap Tvap 

mvap  Vtank Vvap Tvap 

Solution 6.71 Problem Statement 3

A well-insulated tank of 50 m volume initially contains 16,000 kg of water distributed between liquid and vapor phases at 25°C. Saturated steam at 1500 kPa is admitted to the tank until the pressure reaches 800 kPa. What mass of steam is added?

Solution







Whence Also







  1 



liq.2





liq.2







lv.2

 lv .2



Vtank m2

Eliminating x2 from these equations gives      2    

   2    lv .2   liq.2  Vlv .2   Vtank

 



which is later solved for m2

tank





3 1







Vtank m1

 1

Data from Table F.1



3

m3 kg

@ 25 degC:  liq.1



cm3  lv .1  gm



cm 3 gm

liq.1

 



kJ  kg



liq.1

Vlv .1



lv .1

liq.1

5









cm3 gm





cm3  gm

kJ kg







lv .1



kJ kg





Data from Table F.2 @ 800 kPa:

liq.2

 lv.2  

m3  kg

 lv .2 



Data from Table F.2 @ 1500 kPa:

1 2

 



liq.2

lv.2



kJ kg





lv .2

 

kJ kg

 U

 

liq.2

   Vlv.2   U  lv.2     liq.2   V 

 tank  

lv.2





lv.2

steam





steam





Solution 6.72 Problem Statement 3

An insulated evacuated tank of 1.75 m volume is attached to a line containing steam at 400 kPa and 240°C. Steam flows into the tank until the pressure in the tank reaches 400 kPa. Assuming no heat flow from the steam to the tank, prepare graphs showing the mass of steam in the tank and its temperature as a function of pressure in the tank.

Solution Assume no heat flows from the gas to the tank or through the tank walls, this leaves n1  Q 

U2  H

From Table F.2 at 400 kPa and 240 C H  2943.9

kJ kg

Interpolation in Table F.2 will produce values of T and V for a given P where U = 2943.9 kJ/kg. Vtank 

mass 

Vtank V2

Determine the mass for each P and T

P2 (kPa)

T2 (K)

V2 (cm^3/g)

mass (kg)

384.09

303316

0.00576956

100

384.82

3032.17

0.577144421

200

385.57

1515.61

1.154650603

300

386.31

1010.08

1.732536037

400

387.08

757.34

2.310719096

Solution continued on next page…

Finally, plot the mass and T versus P:

2.5

387.5 387 386.5 386

1.5

385.5 1

T2 (K)

Mass (kg)

385 384.5

0.5

384 0

383.5 0

100

150

200

250

300

350

400

P2 (kPa)

T rises very slowly as P increases

Solution 6.73 Problem Statement 3

A 2 m tank initially contains a mixture of saturated-vapor steam and saturated-liquid water at 3000 kPa. Of the total mass, 10% is vapor. Saturated-liquid water is bled from the tank through a valve until the total mass in the tank is 40% of the initial total mass. If during the process the temperature of the contents of the tank is kept constant, how much heat is transferred?

Solution Data from Table F.2 @ 3000 kPa: cm3 g

Vliq 

H liq 

Vvap 

kJ kg

H vap 

cm3 g

Vtank  m3

kJ kg

x1 

V1  Vliq  x1 Vvap  Vliq   7.757 * 103 m1 

Vtank V1



m3 kg

The process is the same as that of Example 6.8, except that the stream flows out rather than in. The energy balance is the same, except for a sign: Q mt * H t   H * mtank

where subscript t denotes conditions in the tank, and H is the enthalpy of the stream flowing out of the tank. The only changes affecting the enthalpy of the contents of the tank are:

1. Evaporation of y kg of sat. liq.: 2. Exit of 0.6 m1 kg of liquid from the tank: Leading to the equation mt * H t   y  H vap  Hliq   0.6m1Vliq

Similarly, since the volume of the tank is constant 0  y Vvap  Vliq   0.6m1Vliq

Solution continued on next page…

Solving for y yields:

y

0.6m1Vliq

V  V  vap

Q

0.6m1Vliq

V  V  vap

We know that Hliq  H

mtank 

liq

*  Hvap  Hliq  

m1Vliq  H * mtank

liq

m1 , this leaves those two terms in the last equation to cancel out,

yielding

Q

0.6m1Vliq

V  V  vap

*  H vap  Hliq  

liq

Solution 6.74 Problem Statement

A stream of water at 85°C, flowing at the rate of 5 kg⋅s

is formed by mixing water at 24°C with saturated steam at

400 kPa. Assuming adiabatic operation, at what rates are the steam and water fed to the mixer?

Solution Obtain the data from table E.1 for saturate liquid at 24 C and 85 C: H1 

kJ kg

H3 

kJ kg

Data from Table E.2 for sat. vapor at 400 kPa: H 2  2737.6

kJ kg

By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3m 3  H1m 1  H2m 2  0

Also m 1  m 3  m 2

m 3 

kg s

Whence

Solving for m 2

m 2  m 3 H3  H1

H1  H 2

kg s

m 1 

m 1 yields m 2 

kg s

Solution 6.75 Problem Statement In a desuperheater, liquid water at 3100 kPa and 50°C is sprayed into a stream of superheated steam at 3000 kPa and 375°C in an amount such that a single stream of saturated-vapor steam at 2900 kPa flows from the desuperheater at 1

the rate of 15 kg·s . Assuming adiabatic operation, what is the mass flow rate of the water? What is SG for the process? What is the irreversible feature of the process?

Solution Obtain the data from Table E.2 for sat. vapor at 2900 kPa (3),3000 kPa at 375 C (2), and from table E.1 saturated liquid at 50 C (liq): kJ kg

H3  H2 

Vliq 

kJ kg

cm3 H liq  g

Psat 

kJ kg K

S3  S2 

m 3  kJ kg K

kJ S  kg liq

kPa T 

kg s

kJ kg K

Find changes in H and S caused by pressure increase from 12.34 to 3100 kPa. First estimate the volume expansivity from sat. liq, data at 45 and 55 C: 3

V  *



cm3 g

T 

1 V  * Vliq T

P

4

kPa

1 K

Apply Eqs. (6.27) & (6.28) at constant T: kJ kg

H1  Hliq  Vliq   T  P  Psat   kJ kg K

S1  Sliq  Vliq   P  Psat  

By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms equal to zero: H3m 3  H1m 1  H 2m 2 

Also

m 2  m 3  m 1

Whence m 1 

m 3  H3  H 2  H1  H 2



kg s m 2  m 3  m 1  13.11

kg s

For adiabatic conditions, Eq. (5.17) becomes SG  S3m 3  S1m 1  S2m 2 

kJ sK

The mixing of the two streams at different temperatures is irreversible.

Solution 6.76 Problem Statement 1

Superheated steam at 700 kPa and 280°C flowing at the rate of 50 kg·s

is mixed with liquid water at 40°C to produce

steam at 700 kPa and 200°C. Assuming adiabatic operation, at what rate is water supplied to the mixer? What is SG for the process? What is the irreversible feature of the process?

Solution Obtain the data from Table E.2 for sat. vapor at 700 kPa at 200 C (3),700 kPa at 280 C (2), and from table E.1 saturated liquid at 40 C (liq): kJ kg

H3  kJ kg

H2 

H liq 

kJ kg K

S3 

S2 

kJ kg K

kJ S  kg liq

m 1 

kg s

kJ kg K

By Eq. (2.29), neglecting kinetic and potential energies and setting the heat and work terms equal to zero:

H2  Hliq Also

H3m 3  H1m 1  H2m 2 

m 3  m 2  m 1

Whence m 2 

m 1  H1  H 3  H3  H2



kg s m 2  m 3  m 1  13.11

kg s

For adiabatic conditions, Eq. (5.17) becomes

S2  Sliq SG  S3m 3  S1m 1  S2m 2 

kJ sK

The mixing of the two streams at different temperatures is irreversible. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 6.77 Problem Statement A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2.5 times the mass flow rate. If this process were accomplished reversibly and adiabatically, what would be the temperature and pressure of the resulting air stream? Assume air to be an ideal gas for which CP = (7/2)R.

Solution Basis: 1 mol air at 12 bar and 900 K (1) + 2.5 mol air at 2 bar and 400 K (2) = 3.5 mol air at T and P. T1 

T2 

K P1 

bar P2  bar n1  mol

7 CP  R 

n2 

mol

J mol K

1 law: given n1C p T  T1   n2C p T  T2   J Solving for T gives: 2

T

law: Given   T   P  T   P  J n1 C p ln   R ln   n2 C p ln   R ln    0       K  T1   P1   T2   P2   

Solving for P gives:

P

bar

Solution 6.78 Problem Statement 1

Hot nitrogen gas at 750(°F) and atmospheric pressure flows into a waste-heat boiler at the rate of 40(lbm)s , and transfers heat to water boiling at 1(atm). The water feed to the boiler is saturated liquid at 1(atm), and it leaves the boiler as superheated steam at 1(atm) and 300( F). If the nitrogen is cooled to 325( F) and if heat is lost to the

surroundings at a rate of 60(Btu) for each (lbm) of steam generated, what is the steam-generation rate? If the surroundings are at 70(°F), what is SG for the process? Assume nitrogen to be an ideal gas for which CP = (7/2)R.

Solution MW  28.104

M n  40

lb mol

Cp 

7 R BTU  0.248 2 MW lbm rankine

lbm T3  1209.67 Rankine T4  784.67 Rankine s

Ms = steam rate in lbm/sec Mn = nitrogen rate in lbm/sec

(1) = sat. liq. water @ 212 degF entering (2) = exit steam at 1 atm and 300 degF (3) = nitrogen in at 750 degF (4) = nitrogen out at 325 degF BTU lbm

H 2  1192.6

BTU lbm rankine

S2  1.8158

H1  180.17

S1  0.3121

BTU lbm

BTU lbm rankine

Eq. (2.30) applies with negligible kinetic and potential energies and with the work term equal to zero and with the heat transfer rate given by Given: M s * ( H 2  H1 )  M n CP (T4  T3 )  60

BTU s

Solution continued on next page…

Solving for Ms M s  3.931

lbm s

Eq. (5.17) here becomes Q SG  M s ( S2  S1 )  M n ( S4  S3 )  T

T  S4  S3  CP ln 4   T3 

Q  60 * M s  235.967

T  529.67 rankine Substituting in for ( S4  S3 ) yields   T  Q SG  M s S2  S1   M n * CP ln  4     T3  T

Solving for SG SG  2.064

BTU sec Rankine

Solution 6.79

Problem Statement 1

Hot nitrogen gas at 400°C and atmospheric pressure flows into a waste-heat boiler at the rate of 20 kg·s , and transfers heat to water boiling at 101.33 kPa. The water feed to the boiler is saturated liquid at 101.33 kPa, and it leaves the boiler as superheated steam at 101.33 kPa and 150°C. If the nitrogen is cooled to 170°C and if heat is lost to the surroundings at a rate of 80 kJ for each kilogram of steam generated, what is the steam-generation rate? If the surroundings are at 25°C, what is SG for the process? Assume nitrogen to be an ideal gas for which CP = (7/2)R.

Solution MW  28.104

M n  20

lb mol

CP 

7 R J  1.039 2 MW gK

kg T3  673.15 K s

T4  443.15 K

Ms = steam rate in kg/sec Mn = nitrogen rate in kg/sec

(1) = sat. liq. water @ 101.33 kPa entering (2) = exit steam at 101.33 kPa and 150 C (3) = nitrogen in at 400 C (4) = nitrogen out at 170 C

H1  419.064 S1  1.3069

kJ kg

H2  2776.2

kJ kg

kJ kJ S2  7.6075 kg K kg K

Eq. (2.30) applies with negligible kinetic and potential energies and with the work term equal to zero and with the heat transfer rate given by Given: M s * ( H2  H1 )  M nC P (T4  T3 )  80

kJ s

Solving for Ms M s  1.961

lbm s

Solution continued on next page…

Eq. (5.17) here becomes Q SG  M s ( S2  S1 )  M n ( S4  S3 )  T

T  S4  S3  CP ln 4   T3 

Q  60 * M s

T  298.15 K Substituting in for ( S4  S3 ) yields   T  Q SG  M s S2  S1   M n * CP ln  4    T3  T 

Solving for SG kJ SG  4.194 sec K

Solution 6.80

Problem Statement Show that isobars and isochores have positive slopes in the single-phase regions of a TS diagram. Suppose that CP = a + bT, where a and b are positive constants. Show that the curvature of an isobar is also positive. For specified T and S, which is steeper: an isobar or an isochore? Why? Note that CP > CV.

Solution The slopes of isobars and isochores on a T S diagram are given by Eqs. (6.18) and (6.30):  T   T  T T    and     S P C P   S V CV

Both slopes are necessarily positive. With CP > CV , isochores are steeper. Solution continued on next page…

An expression for the curvature of isobars results from differentiation of the first equation above:    2T  1  T  T  C  T T  C P   T  T  CP   2         P       T / C 1     P  C  T     S 2 P CP   S P CP2   S P CP2 CP2  T P   S P  P P 

 C  With CP  a  bT  P   b  T P



T  CP  bT a      CP  T P a  bT a  bT

Because this quantity is positive, so then is the curvature of an isobar.

Solution 6.81

Problem Statement Starting with Eq. (6.9), show that isotherms in the vapor region of a Mollier (HS) diagram have slopes and curvatures given by:

 2 H   H  1 1       T  1 2    3    S    S   V  P  T

Here, is volume expansivity. If the vapor is described by the two-term virial equation in P, Eq. (3.36), what can be said about the signs of these derivatives? Assume that, for normal temperatures, B is negative and dB/dT is positive.

Eq. 6.9: dH = T dS + V dP

Eq. 3.36: Z

PV BP 1 RT RT

Solution Division of Eq. (6.9) by dS and restriction to constant T yields:  H   P     T  V    S   S  T



 P    V  S 

 

Solution continued on next page…

 H  1 1    T   T  1  S   

Therefore,

Also,

 2 H          P                S 2    2   S    2   P    S    2   P   V  T

Whence,

 2 H  1           S 2   3V   P 

By Eqs. (3.3) and (3.37):  

Whence,

1  V  RT  andV  B  V  T P P

 V  R dB 1  R dB     and      V  P dT   T P P dT

Differentiation of the second preceding equation yields:     R dB  1  V  R R 1  V         2      V  2   2 2      P T   P dT  V  P T VP VP V   P T  V  RT    2   P  P

From the equation of state,

Whence,

   R  RT R       T   2 2 V   P  VP P V P2 T

Clearly, the signs of quantity ( relation of and V to B and dB/dT: RT dB dB T T B T  R dB  dT   dT T        P RT RT V  P dT  B B P P

In this equation dB/dT is positive and B is negative. Because RT/P is greater than |B|, the quantity This makes the derivative in the first boxed equation positive, and the second derivative in the second boxed equation negative.

Solution 6.82

Problem Statement

The temperature dependence of the second virial coefficient B is shown for nitrogen in Fig. 3.8. Qualitatively, the shape of B(T ) is the same for all gases; quantitatively, the temperature for which B = 0 corresponds to a reduced temperature of about Tr = 2.7 for many gases. Use these observations to show by Eqs. (6.54) through (6.56) that the R

residual properties G , H , and S are negative for most gases at modest pressures and normal temperatures. What can R

you say about the signs of V and CPR ?

Eq. 6.54: GR BP  RT RT

Eq. 6.55: H R P  B dB      RT R  T dT 

Eq. 6.56: SR P dB  R R dT

Solution Since a reduced temperature of Tr = 2.7 is well above “normal” temperatures for most gases, we expect on the basis of 2

Fig. 3.10 that B is (

B/dT is (

By Eqs. (6.54) and (6.56), G = BP and S = R

Whence, both G and S are (

, H = G +T S , and H is (

By Eqs. (3.37) and (6.40), V = B, and V is ( R

Combine the equations above for G , S , and H :  H R   dB d2 B dB  d2 B    P   T   PT   T  dT 2 dT  dT 2  dT

 dB  H R  P  B  T   dT 

Therefore,

 H R   CPR    T 

is(+).

(See Fig. 6.5.)

Solution 6.83

Problem Statement An equimolar mixture of methane and propane is discharged from a compressor at 5500 kPa and 90°C at the rate of 1

1.4 kg·s . If the velocity in the discharge line is not to exceed 30 m·s , what is the minimum diameter of the discharge line?

Solution Methane = 1; propane = 2



















8  bar

The elevated pressure here requires use of either an equation of state or the Lee/Kesler correlation with pseudocritical parameters. We choose the latter.



 c1 



 c2 pc 



T Tpc



 c1 

 c2



P Ppc



By interpolation in Tables E.3 and E.4: 























For the molar mass of the mixture, we have:   1



ZRT P  molwt





cm3 gm



 4A

gm mol



cm 3 sec

kg sec



Vdot u







gm mol





m sec 2



Solution 6.84 Problem Statement R

Estimate V , H , and S for one of the following by appropriate generalized correlations: (a) 1,3-Butadiene at 500 K and 20 bar. (b) Carbon dioxide at 400 K and 200 bar. (c) Carbon disulfide at 450 K and 60 bar. (d) n-Decane at 600 K and 20 bar. (e) Ethylbenzene at 620 K and 20 bar. (f) Methane at 250 K and 90 bar. (g) Oxygen at 150 K and 20 bar. (h) n-Pentane at 500 K and 10 bar. (i) Sulfur dioxide at 450 K and 35 bar. (j) Tetrafluoroethane at 400 K and 15 bar.

Solution

For these, we should first try the Pitzer correlation, and then if the reduced temperature and pressure are outside the range of its validity, we should use the Lee/Kesler tables. (a) For 1,3-butadiene at 500 K and 20 bar, the reduced temperature and pressure are Tr = 1.176 and Pr = 0.468. These conditions are well within the range of validity of the Pitzer correlation. Adding a cell for computing Z to the residual enthalpy spreadsheet we have been using, we get: Solution continued on next page…

T (K) 500

Tr 1.1759

P (bar) 20

Tc (K) 425.2

B0 -0.2426

Pr 0.4676

B 0  0.083 

0.422

B1  0.139 

0.172

Pc (bar) 42.77

B1 0.0519

dB0 0.675  2.6 dTr Tr

Tr1.6

dB1 0.722  5.2 dTr Tr

Tr4.2

w 0.19

0.9074

dB0 dTr

dB dTr

0.4429

0.3109



Z  1  B 0   B1

HR /RTc -0.3849

HR (J mol-1) -1360.65

  HR dB 0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

and for the entropy, we get T (K) 500

Tr 1.1759

P (bar) 20

Pr 0.4676

Tc (K) 425.2

Pc (bar) 42.77

dB0 dTr

dB1 dTr

0.4429

0.3109

dB0 0.675  2.6 dTr Tr

w 0.19

SR /R -0.2347

SR (J mol-1 K-1) -1.9517

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

The residual volume is V = (Z – 1) RT/P = 192.5 cm /mol. Summarizing: R

V =

192.5 cm /mol, H = 1360.7 J/mol, and S = 1.952 J/(mol K)

Solution continued on next page…

 TP

(b) For CO2 at 400 K and 200 bar, we have a reduced temperature and reduced pressure of Tr = 1.3149 and Pr = 2.7089. These conditions are well outside the range where the Pitzer correlation is applicable. Thus, we should interpolate in the full Lee/Kesler tables. Using the entries for Pr = 2 and 3 and Tr = 1.3 and 1.4, we get the following: T (K) 400

P (bar) 200

Tc (K) 304.2

Pc (bar) 73.83

Tr 1.30 1.30 1.40 1.40

Pr 2.00 3.00 2.00 3.00

w 0.224

Tr 1.3149

Pr 2.7089

Z1 0.1991 0.2079 0.1894 0.2397

(HR )0/(RTc) -1.56 -2.274 -1.253 -1.857

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Z0 0.6908 0.6344 0.7753 0.7202

Interpolated Values Final Values Z 3

V (cm /mol)

0.6636 HR /(RTc)

0.7102

118.1

H (J/mol) R

0.2083

(HR )1/(RTc) -0.178 -0.3 -0.07 -0.044

-2.0087

(SR )0/R -0.891 -1.299 -0.663 -0.99

(S R )1/R -0.3 -0.481 -0.22 -0.29

-1.1376

-0.4046

-0.2327

-2.06084

S R /R

-1.22828

-5212.12

-10.2119

S (J/(mol K)) 3

The full Lee/Kesler correlation gives V = (Z – 1) RT/P = 48.2 cm /mol, H = 5212 J/mol, and R

S = 10.21 J/(mol K).

Solution 6.85 Problem Statement R

Estimate Z, H , and S for one of the following equimolar mixtures by the Lee/Kesler correlations: (a) Benzene/cyclohexane at 650 K and 60 bar. (b) Carbon dioxide/carbon monoxide at 300 K and 100 bar. (c) Carbon dioxide/n-octane at 600 K and 100 bar. (d) Ethane/ethylene at 350 K and 75 bar. (e) Hydrogen sulfide/methane at 400 K and 150 bar. (f) Methane/nitrogen at 200 K and 75 bar.

(g) Methane/n-pentane at 450 K and 80 bar. (h) Nitrogen/oxygen at 250 K and 100 bar.

Solution Vectors containing T, P, Tc1, Tc2, Pc1, Pc2, 1, and 2 for Parts (a) through (h) T

Tc1

Tc2

Pc1

Pc2

650

562.2

553.6

48.98

40.73

0.21

300

100

304.2

132.9

73.83

34.99

0.224

0.048

600

100

304.2

568.7

73.83

24.9

0.224

0.4

350

305.3

282.3

48.72

50.4

0.1

0.087

400

150

373.5

190.6

89.63

45.99

0.094

0.012

200

190.6

126.2

45.99

0.012

0.038

450

190.6

469.7

45.99

33.7

0.012

0.252

250

100

126.2

154.6

50.43

0.038

0.022

Tpc

Ppc

Tpr

Ppr

557.9

44.855

0.21

1.165083

1.337644

218.55

54.41

0.136

1.372684

1.837897

436.45

49.365

0.312

1.374728

2.025727

293.8

49.56

0.0935

1.191287

1.513317

282.05

67.81

0.053

1.418188

2.212063

158.4

39.995

0.025

1.262626

1.875234

330.15

39.845

0.132

1.363017

2.00778

140.4

42.215

0.03

1.780627

2.368826

Lee/Kesler Correlation --- By linear interpolation in Tables D.1--D.12: Z0

0.6543

0.1219

1.395

0.461

0.89

0.466

0.7706

0.1749

1.214

0.116

0.658

0.235

0.7527

0.1929

1.346

0.097

0.729

0.242

0.6434

0.1501

1.51

0.4

0.944

0.43

0.7744

0.199

1.34

0.049

0.704

0.224

0.6631

0.1853

1.623

0.254

0.965

0.348

0.7436

0.1933

1.372

0.11

0.75

0.25

0.9168

0.1839

0.82

0.172

0.361

0.095

 HR

H0 

RTpc

 HR

H1 

RTpc

HR H RTpc

SR

S0 

SR

S1 

S

SR R

Z   Z   Z  H   H   H  S  S   S 

HR  H * Tpc * R

SR  S * R

The Z, HR, SR Z

HR (J/mol)

SR (J/mol K)

0.680

6919.58

8.21307

0.794

2234.53

5.73633

0.813

4993.97

6.68865

0.657

3779.76

8.18268

0.785

3148.34

5.95176

0.668

2145.75

8.09534

0.769

3805.81

6.50986

0.922

951.151

3.02505

Solution 6.86 Problem Statement For the reversible isothermal compression of a liquid for which and may be assumed independent of pressure, show that:

W  P1 V1  P2 V2 

S 

H 

V2  V1 

 V  V1   2 1  T V2  V1  

Do not assume that V is constant at an average value, but use Eq. (3.6) for its P dependence (with V2 replaced by V). Apply these equations to the conditions stated in Prob. 6.9. What do the results suggest with respect to use of an average value for V?

Eq. 3.6:

V2   T2  T1     P2  P1  V1

Problem 6.9 3

One kilogram of water (V1 = 1003 cm ·kg ) in a piston/cylinder device at 25°C and 1 bar is compressed in a mechanically reversible, isothermal process to 1500 bar. Determine Q, W = 250 × 10

and = 45 × 10

bar

S given that

. A satisfactory assumption is that V is constant at its arithmetic average

value.

Solution 1 V P  ln  P1  V1

By Eq. (3.6) at constant T :

(a) Work

1 V    1 1 dW P dV   ln  P1 dV  ln V dV   P1  ln V1 dV        V1  W

  1 V2 1 ln V dV   P1  ln V1 V2  V1     V1   

1 W  V2 

V2  V2  V1

1 V1 V1   P1 V2 V1  V2  

V1 V1

V1 

 V 1  V2 ln 2  V1  V2   P1 V2 V1   V1  V2

(b) Entropy

By Eq. (6.28),

By Eq. (A), P 

dS V dP

lnV lnV1 1   P1and  dP  d ln V    dS 

V  V1 W  P1V1  P2V2  2 

   P2  P1 

By Eq. (3.6),

V   d ln V  and S  V2  V1    

dH 

By Eq. (6.27), dH   

Substitute for dP:

T V

H 



T V dP

V 

T dV 

T V1  V2  

These equations are so simple that little is gained through use of an average V. For the conditions given in Pb. 6.9, calculations give: W

kJ kg 1S  

kJ kg 1 K 1H 

kJ kg 1

Solution 6.87 Problem Statement In general for an arbitrary thermodynamic property of a pure substance, M = M(T, P); whence  M   M  dM    dT    dP  T   P  P

For what two distinct conditions is the following equation true? T2  M   dT M    T1   T  P

Solution  M  The given equation will be true if and only if   dP  0  P T  M  The two circumstances for which this condition holds are when    0 or when dP = 0.  P  T

The former is a property feature and the latter is a process feature.

Solution 6.88 Problem Statement The enthalpy of a pure ideal gas depends on temperature only. Hence, Hig is often said to be “independent of ig

pressure,” and one writes (H /P ) T = 0. Determine expressions for (H /P)V and (H /P)S . Why are these quantities not zero?

Solution  H ig   H ig   H ig   T    ig  T            P     P    T    P   CP   P  V

Neither CP nor (T/ P)V is in general zero for an ideal gas.  H ig   H ig   H ig   T    ig  T            P     P    T    P   CP   P  S

 T   T    S ig  T   S ig          ig      P    S    P  C ig   P  S

 H ig    S ig     T     P    P  S

Neither T nor ( S / P)T is in general zero for an ideal gas. The difficulty here is that the expression independent of pressure is imprecise.

Solution 6.89 Problem Statement Prove that

dS 

CV  T  CP  T   dV   dP   T  P V T  V P

For an ideal gas with constant heat capacities, use this result to derive Eq. (3.23c). Eq. 3.23c: ig

P (V )

const

Solution  S   S  For S  S  P, V  : dS    dP    dV  P V  V P

By the chain rule for partial derivatives,  S   T   S   T  dS      dP      dV  T   P   T   V  V

With Eqs. (6.30) and (6.18), this becomes: dS 

C v  T  C P  T   d   dP   T  P  T  V  V

Solution 6.90 Problem Statement The derivative (U / V )T is sometimes called the internal pressure and the product T (P / T )V the thermal pressure. Find equations for their evaluation for:

(a) An ideal gas; (b) A van der Waals fluid; (c) A Redlich/Kwong fluid.

Solution By Eq. (6.31),

 P   U  P  T       T   V T V

Solution continued on next page…

(a) For an ideal gas, P 

Therefore

RT V

 U  RT RT  U     and  0   V V  V T  V T

(b) For a van der Waals gas, P 

Therefore

 P  R     T V V

 P     R  T V V  b

RT a  V b V 2

 U   U  RT a RT a  2    and   V b V V  b  V r  V r V 2

 U      V  T

 32  A 1

T 2 V V  b

Where A  aTc   Tc1/2

Solution 6.91 Problem Statement

(a) A pure substance is described by an expression for G(T, P). Show how to determine Z, U, and CV, in relation to G, T, and P and/or derivatives of G with respect to T and P.

(b) A pure substance is described by an expression for A(T, V). Show how to determine Z, H, and CP, in relation to A, T, and V and/or derivatives of A with respect to T and V.

Solution  G   G  S    andV     T P   P T

Solution continued on next page…

Combining the definition of Z with the second of these gives:

Z

PV P  G     RT RT   P T

U=G+TS

PV.

Replacing S and V by their derivatives gives:

 G   G  U  G  T    P    T P   P T

Developing an equation for CV is much less direct. First differentiate the above equation for U with respect to T and then with respect to P: The two resulting equations are:  2G    2G   U       T  2   P   T   T   T  P  P

  2G    2G   U     P      T     P   T  P    P 2  T

From the definition of CV and an equation relating partial derivatives:  U   U   U    P  CV             T   T    P   T   V P T V

Combining the three equations yields:  2G    2G     2G    2G     P     T    P    CV   T  2   P   2    T P  T  P    T  P   P T   T V

Evaluate ( P/T )V through use of the chain rule:  V     T    P    P   V  P           V   T V  V T  T P     P  T

Solution continued on next page…

The two derivatives of the final term come from differentiation of V = (G/ P)T :   2G   2G   V   V  and       2    T    P    PT    P  P

  2G     T    P  P    2   T   V   G   P 2  T

Then

And

  2G      2G    2G     2G    2G    PT        T    P  2   CV  T  2  P   T   T  P    T  P   P T    2G  P   2   P T

Some algebra transforms this equation into a more compact form:   2G     T P 

  2G  CV   T  2   T   2G   T P    P 2 

(b) The solution here is analogous to that of part (a), but starting with the derivatives inherent in Eq. (6.9).

Solution 6.92 Problem Statement Use steam tables to estimate a value of the acentric factor

for water. Compare the result with the value given in

Table B.1.

Solution Tc := 647.1K

Pc := 220.55bar

At Tr = 0.7:

T := 0.7·Tc

T = 452.97K

Solution continued on next page…

Find Psat in the Saturated Steam Tables at T = 452.97 K

T1 := 451.15K

P1 := 957.36kPa

T2 := 453.15K

P sat 

P 2  P1  T  T  P Psat = 998.619 kPa T 2  T1

Psatr:=

Psat Psatr := 0.045 Pc

log(Psatr)

P2 := 1002.7kPa

Psat = 9.986 bar

= 0.344 Ans.

This is very close to the value reported in Table B.1 ( = 0.345).

Solution 6.93 Problem Statement The critical coordinates for tetrafluoroethane (refrigerant HFC-134a) are given in Table B.1, and Table 9.1 shows saturation properties for the same refrigerant. From these data determine the acentric factor

for HFC-134a, and

compare it with the value given in Table B.1.

Solution Tc := 374.2K

Pc := 40.60bar

At Tr = 0.7:

T := 0.7·Tc

T := T

459.67rankine

T = 471.492 rankine T = 11.822 degF

Find Psat in Table 9.1 at T = 11.822 F

T1 := 10degF

Psat:=

P1 := 26.617psi

P 2  P1  (T  T 1)  P1 T 2  T1

Psatr :=

Psat

Psatr = 0.047

T2 := 15degF

P2 := 29.726psi

Psat = 27.75 psi

:= 1

Psat = 1.913 bar

log(Psatr)

= 0.327 Ans.

This is exactly the same as the value reported in Table B.1.

Solution 6.94 Problem Statement lv

H is not independent of T; in fact, it becomes zero at the critical point. Nor may saturated vapors in general be considered ideal gases. Why is it then that Eq. (6.89) provides a reasonable approximation to vapor-pressure behavior over the entire liquid range?

Eq. 6.89: ln P sat  A 

B T

Solution

Equation (6.88) is exact:

d ln P sat H lv  1 RZ lv d   T 

The right side is approximately constant owing to the qualitatively similar behavior of H lv

Z lv . Both decrease

monotonically as T increases, becoming zero at the critical point.

Solution 6.95 Problem Statement Rationalize the following approximate expressions for solid/liquid saturation pressures: sat

(a) Psl = A + BT ; sat

(b) Psl

= A + BlnT

Solution

By the Clapeyron equation:

If the ratio S sl

dP sat S sl H sl   dT V sl T V sl

S sl is assumed approximately constant, then

P sat  A  BT

If the ratio H sl

V sl is assumed approximately constant, then

P sat  A  BlnT

Solution 6.96 Problem Statement As suggested by Fig. 3.1, the slope of the sublimation curve at the triple point is generally greater than that of the vaporization curve at the same state. Rationalize this observation. Note that triple-point pressures are usually low; hence assume for this exercise that Z SV  Z lv  1.

Solution

By Eq, (6.87) and its analog for sv equilibrium: sv sv  dP sat   sv   Pt Ht  Pt Ht  dT  2 sv 2 RTt Zt RTt  t

lv lv  dP sat   lv   Pt H t  Pt Ht   dT  RTt2 Ztlv RTt2  t

 dP sat   sat   sv    dPlv   Pt H sv H lv  t   t   dT   RTt2  t  dT t

Because Htsv H tlv   Htsl is positive, then so is the left side of the preceding equation.

Solution 6.97 Problem Statement Show that the Clapeyron equation for liquid/vapor equilibrium may be written in the reduced form:

d ln Prsat dTr

lv  lv  H H lv   2 where  H  RTc Tr Z lv

Solution

By Eq. (6.87):

dP sat H lv  dT T V lv

But

V lv 

RT dlnP sat H lv lv  Z whence  dT P sat RT 2 Z lv dlnPrsat dTr



Tc H lv RT 2Z

 lv

lv  H lv 1  H  2  RTc Tt Z lv Tr2 Z lv

Solution 6.98 Problem Statement Use the result of the preceding problem to estimate the heat of vaporization at the normal boiling point for one of the sat

substances listed below. Compare the result with the value given in Table B.2 of App. B. Ground rules: Represent Pr with Eqs. (6.92), (6.93), and (6.94), with

given by Eq. (6.95). Use Eqs. (3.57), (3.58), (3.59), (3.61), and (3.62) for Z ,

and Eq. (3.69) for Z . Critical properties and normal boiling points are given in Table B.1. (a) Benzene; (b) iso-Butane;

Solution

(a)for benzene 



6.09648  Tr

ln Tr 

Tr6

15.6875  Tr

ln Tr 

Tr6

Trn 

Psat 

Using eqs. 6.89–6.95 to determine the acentric factor ln Pr 0 



ln Pr 1 





ln Psat rn   ln Pr  ln Pr 1 



ln  Psatr   ln Pr 0   ln Pr 1

Z sat liq 

Psat rn Trn

(1 (1Tr n )0.2857

 0.00334

Solution continued on next page…

Determine the B values from Eqs 3.61, 3.62, 3.60 B0 

B1  

Z0 

Z1 

Determine the Zsat for the vapor

Zsat vap  Z   Z  Determine the change in Z from liquid to vapor

Zlv  Zsat vap  Zsat liq   Hˆ lv 

dPsat r dTrn

* Trn * Z lv  6.59 2

Lastly determine the heat of vaporization H lv  RTc Hˆ lv

H lv 

kJ mol

This compares well with the value in Table B.2 of 30.19 kJ/mol The results for the other species are given in the table below. Estimated Value

Table B.2

(kJ/mol)

Benzene

30.8

30.72

iso-butane

21.39

21.3

tetrachloride

29.81

29.82

cyclohexane

30.03

29.97

n-Decane

39.97

38.75

n-Hexane

29.27

28.85

n-Ocatane

34.7

34.41

Toluene

33.72

33.18

o-Xylene

37.23

36.24

Carbon

Solution 6.99 Problem Statement Riedel proposed a third corresponding-states parameter

c, related to the vapor-pressure curve by:

 d ln P sat   c    d lnT    T Tc

For simple fluids, experiment shows that

5.8; for non-simple fluids,

c increases with increasing molecular

complexity. How well does the Lee/Kesler correlation for Prsat accommodate these observations?

Solution

Convert

to reduced conditions:  d ln P sat   d ln P sat   d ln P sat   d ln P sat  r r r        c     T        r  d ln T d ln T dT dT   T T    T 1   T 1 r  r r Tr 1 c r r

From the Lee/Kesler equation, find that  d ln P sat  r    5.8239  4.8300    dT r   T 1 r

Thus,

(L/K) = 5.82 for

= 0, and increases with increasing molecular complexity as quantified by .

Solution 6.100 Problem Statement

Triple-point coordinates for carbon dioxide are Tt = 216.55 K and Pt = 5.170 bar. Hence, CO2 has no normal boiling point. (Why?) Nevertheless, one can define a hypothetical normal boiling point by extrapolation of the vapor-pressure curve.

(a) Use the Lee/Kesler correlation for Prsat in conjunction with the triple-point coordinates to estimate

for CO2.

Compare it with the value in Table B.1.

(b) Use the Lee/Kesler correlation to estimate the hypothetical normal boiling point for CO2. Comment on the likely reasonableness of this result.

Solution

For CO2:

:= 0.224 Tc := 304.2K Pc := 73.83bar

At the triple point:

a) At Tr = 0.7

Ttr :=

Tt Tc

lnPr Tr  

Tt := 216.55K

T := 0.7Tc

Ttr = 0.712



Pt := 5.170bar

T = 212.94K

Ptr :=

6.09648  Tr

Pt Pc

Ptr = 0.07

 lnTr  

 Tr 6 Eqn. (6.93)

Solution continued on next page…

lnPr Tr  





15.6875  Tr

ln Ptr   lnPr0 Ttr 



lnPr1Ttr 



 

 

 Tr 6

Eqn. (6.94)

Ans.

This is exactly the same value as given in Table B.1

Psatr 

Given

1atm Pc



ln(Psatr) = lnPr0(Trn) + ·lnPr1(Trn)

Trn = 0.609 Tn := Trn·Tc

Trn := Find(Trn)

Tn = 185.3K Ans.

This seems reasonable; a Trn of about 0.6 is common for triatomic species.

Solution 7.1 Problem Statement 1

Air expands adiabatically through a nozzle from a negligible initial velocity to a final velocity of 325 m · s . What is the temperature drop of the air, if air is assumed to be an ideal gas for which CP = (7/2) R?

Solution

u2 

m s

J mol K

R

MW 

g mol

Cp 

7 R MW

Knowing that the heat, work, potential energy, and initial velocity are equal to 0, Eq 2.31 reduces to

H 

u22  2

H  C pT

Substitution yields

T  

u22 2CP

T  

Solution 7.2 Problem Statement

In Ex. 7.5 an expression is found for the Joule/Thomson coefficient, = (T/P )H, that relates it to a heat capacity and equation-of-state information. Develop similar expressions for the derivatives: (a) (T/P)S

(b) (T/V)U

Solution Apply the mathematical identity in section 7.1 to the particular derivative of interest here:  T   T   S          P   S   P  S

The two partials on the right are found from Eqs. 6.17, and 6.18, with substitution yielding:  T  T  V        P  C  T  S

For gases this derivative is positive. It applies to reversible adiabatic expansions and compressions in turbines and compressors.

Apply the same mathematical identity in section 7.1 to the second derivative:  T   T   U          V   U V  V T U

Solution continued on next page…

The right sides is found from Eqs, 2.15 and 6.30 yielding:  T   P   1   P  T       V U CV   T V 

For gases, this may be positive or negative, depending on conditions. Note that it is zero for an ideal gas. It applies directly to the Joule expansion, an adiabatic expansion of gas confined in a portion of a container to fill the entire container.

Solution 7.3 Problem Statement The thermodynamic sound speed c is defined in Sec. 7.1. Prove that:

c

where V is molar volume and

VCP

 CV 

is molar mass. To what does this general result reduce for: (a) an ideal gas?

(b) an incompressible liquid? What do these results suggest qualitatively about the speed of sound in liquids relative to gases?

Solution The equation giving the thermodynamic sound speed appears in section 7.1. As written, it implicitly requires that V represent specific volume. This is easily confirmed by a dimensional analysis. If V is to be molar volume, then the right side must be divided by molar mass:

c2  

V 2  P    M  V S

Applying the mathematical identity from Pb. 7.3 to the derivative yields:

 P   P   S          V   S   V  S

Separating further, this can be written as:

 P   T    S   T       P       T   S    P   T                       S   T    T   V    V      S V P V P  T V  S V   T P  V P  

Division of Eq. (6.16) by Eq. (6.29) shows that the first product in square brackets on the far right is the ratio CP/CV . Reference again to the mathematical identity from Pb. 7.3 shows that the second product in square brackets on the far right is ( P / V )T , which is given by Eq. (3.4).

Therefore:

 P  C  P  C  1     P    P   C  V  C  V   V  S

c2 

V 2 CP M CV 

Substitute into our first equation gives:

So for an ideal gas, V = RT/P and = 1/P. Therefore

c ig 

RT CP M CV

For an incompressible liquid. V is constant, and = 0. This leads to the result that c = . This leads to the conclusion that the speed of sound is much greater in liquids that gases.

Solution 7.4 Problem Statement Steam enters a nozzle at 800 kPa and 280°C at negligible velocity and discharges at a pressure of 525 kPa. Assuming isentropic expansion of the steam in the nozzle, what is the exit velocity and what is the cross1

sectional area at the nozzle exit for a flow rate of 0.75 kg·s ?

Solution

First obtain values from table e.2 for steam at 800 kPa and 280 C.

H1 

kJ kg

S1 

kJ kg K

Interpolation in Table E.2 at 525 kPa and S = 7.1595 kJ/kg K yields:

H2 

kJ kg

cm3 g

V2 

m 

kg s

Knowing that the heat, work, potential energy, and initial velocity are equal to 0, Eq 2.31 reduces to

H 

u22  2

Solving for u2 yields u2    H2  H1 

u2  565.2

m s

Solving for the area using eq 2.26 yields:

A2 

 2 mV u2

A2 

cm2

Solution 7.5 Problem Statement Steam enters a converging nozzle at 800 kPa and 280°C with negligible velocity. If expansion is isentropic, what is the minimum pressure that can be reached in such a nozzle, and what is the cross-sectional area at the 1

nozzle throat at this pressure for a flow rate of 0.75 kg·s ? Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

The calculations of the preceding problem may be carried out for a series of exit pressures until a minimum cross-sectional area is found. The corresponding pressure is the minimum obtainable in the converging nozzle. Initial property values are as in the preceding problem.

H1 

kJ kg

S1 

kJ kg K

S2  S1

Interpolations in Table E.2 at several pressures and at the given entropy, as well as solving for the velocity and area yield the following: u2    H2  H1 

A2 

P (kPa)

(kJ/kg)

(cm3/g)

(m/sec)

(cm2)

400

2855.2

531.21

565.15

7.05

425

2868.2

507.12

541.66

7.02

450

2880.7

485.45

518.07

7.03

475

2892.5

465.69

494.77

7.06

500

2903.9

447.72

471.17

7.13

 2 mV u2

Solution continued on next page…

Fit the P vs. A2 data with cubic spline and find the minimum P at the point where the first derivative of the spline is zero.

7.14 7.12

Area (cm^2)

7.10 7.08 7.06 7.04 7.02 7.00 400

420

440

460

480

500

Pressure (kPa)

The minimum here is

P

kPa

A

cm2

Solution 7.6 Problem Statement

A gas enters a converging nozzle at pressure P1 with negligible velocity, expands isentropically in the nozzle, and discharges into a chamber at pressure P2. Sketch graphs showing the velocity at the throat and the mass flow rate as functions of the pressure ratio P2/P1.

Solution

As P2 decreases from an initial value of P2 = P1, both u2

 steadily increase until the critical pressure m

ratio is reached. At this value of P2, u2 equals the speed of sound in the gas, and further reduction in P2 does not affect u2

 . m

Solution 7.7 Problem Statement

For isentropic expansion in a converging/diverging nozzle with negligible entrance velocity, sketch graphs of mass flow rate m , velocity u, and area ratio A/A1 versus the pressure ratio P/P1. Here, A is the cross-sectional area of the nozzle at the point in the nozzle where the pressure is P, and subscript 1 denotes the nozzle entrance.

Solution

The mass-flow rate m is of course constant throughout the nozzle from entrance to exit. The velocity u rises monotonically from nozzle entrance (P/P1 = 1) to nozzle exit as P and P/P1 decrease. The area ratio decreases from A/A1 = 1 at the nozzle entrance to a minimum value at the throat and thereafter increases to the nozzle exit.

Solution 7.8 Problem Statement An ideal gas with constant heat capacities enters a converging/diverging nozzle with negligible velocity. If it expands isentropically within the nozzle, show that the throat velocity is given by:

2 uthroat 

 RT 1  2        1 

where T1 is the temperature of the gas entering the nozzle,

is the molar mass, and R is the molar gas

constant.

Solution

Substitution of Eq. (7.12) into (7.11), with u1  0 gives:

2 uthroat 

 2   P1 V1  2     P1V1              

where V1 is specific volume in m • kg

2 and P1 is in Pa. The units of uthroat

are then:

Pa  m 3  kg 1  

N  m 3  kg 1  N  m  kg 1  kg  m  s2  m  kg 1  m 2  s 2 m2

With respect to the final term in the preceding equation, note that P1V1 has the units of energy per unit mass. 1

Because 1 N · m = 1 J, equivalent units are J·kg . Moreover, P1V1 = RT1/M; whence

2 uthroat 

 RT1  2    m    

With R in units of J·(kg mol) ·K , RT1/M has units of J·kg

or m ·s .

Solution 7.9 Problem Statement Steam expands isentropically in a converging/diverging nozzle from inlet conditions of 1400 kPa, 325°C, and 2

negligible velocity to a discharge pressure of 140 kPa. At the throat, the cross-sectional area is 6 cm . Determine the mass flow rate of the steam and the state of the steam at the exit of the nozzle.

Solution

From Table E.2 at 1400 kPa and 325 degC:

H1 

kJ kg

kJ kg K

S1 

S2  S1

Interpolate in Table E.2 at a series of downstream pressures and at S = 7.0499 kJ/(kg*K) to find the minimum cross-sectional area.

u2    H2  H1 

V  A2   2 m  u2 

Since m is constant, the quotient V2/u2 is a measure of the area. Its minimum value occurs very close to the value at vector index i = 3. See the table below for the A2 values P (kPa)

H2 (kJ/kg)

V2 (cm3/g)

u2 (m/sec)

A2 (cm2 s/kg)

800

2956

294.81

530.09

5.5615

775

2948.5

302.12

544.06

5.5531

750

2940.8

309.82

558.03

5.5520

725

2932.8

317.97

572.19

5.5571

700

2924.9

326.69

585.83

5.5765

Solution continued on next page…

Using the value of A2  6 cm 2

m 

A2 u2



m 

kg s

At the nozzle exit, P = 140 kPa and S = S1, the initial value. From Table E.2 we see that steam at these conditions is wet. By interpolation,

Sliq 

x

kJ kg K S1  Sliq Svap  Sliq

Svap 

kJ kg K

x

State is saturated liquid-vapor mixture

Solution 7.10 Problem Statement Steam expands adiabatically in a nozzle from inlet conditions of 130(psia), 420(°F), and a velocity of 230(ft)(s)

to a discharge pressure of 35(psia) where its velocity is 2000(ft)(s) . What is the state of the steam

at the nozzle exit, and what is SG for the process?

Solution ft s

u1 

ft s

u2 

From Table E.4 at 130 psia and 420 degF:

H1 

BTU lbm

BTU lbm Rankine

S1 

H u2  u22 H  1 2

H  

BTU lbm

H 2  H 1  H

H2 

BTU lbm

H determine what H2 is:

From Table E.4 at 35(psi), we see that the final state is wet steam:

H liq 

Sliq 

BTU lbm

H vap 

BTU lbm Rankine

x

H2  H liq H vap  H liq

Svap 

BTU lbm

BTU lbm Rankine

x

Solution continued on next page…

State is saturated liquid-vapor mixture With this x value determine S2 S2  Sliq  x Svap  Sliq 

S2 

BTU lbm Rankine

Now solve for SG

SG  S2  S1

SG 

BTU lbm Rankine

Solution 7.11 Problem Statement 1

Air discharges from an adiabatic nozzle at 15°C with a velocity of 580 m · s . What is the temperature at the entrance of the nozzle if the entrance velocity is negligible? Assume air to be an ideal gas for which CP = (7/2)R.

Solution

u2 

m T2  s

MW 

g 7 R CP  mol 2 MW

Solution continued on next page…

Knowing that work and potential energy are zero. Using Eq 2.31 set u1 = 0: u2  u22 u22 H  1  2 2

Now substitute H  CP T , this yields

T 

u22 2C p

T  

Now determine the initial temperature:

T1 T  T2 

Solution 7.12 Problem Statement Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What is the lost work per kilogram of water for this everyday household happening? At 15°C and 1(atm), the volume expansivity for liquid water is about 1.5 × 10

K . The surroundings temperature T is

20°C. State carefully any assumptions you make. The steam tables are a source of data.

Solution

Values from the steam tables for saturated-liquid water:

At 15 C:

V

cm3 g

T

Enthalpy difference for saturated liquid for a temperature change from 14 to 15 C:

H  



J g



T  K



CP 

H T

1.5 * 104 K

J gK

CP 

P  atm

Apply Eq. (7.25) to the constant-enthalpy throttling process. Assumes very small temperature change and property values independent of P.

T 

V 1  T P  1  J   3  C  9.86923 cm atm 

T 

The entropy change for this process is given by Eq. (7.26):

S  CP

 T  T      V P  T 

S 

3

J gK

Apply Eq. (5.31) with Q = 0:

T 

Wlost  T S

Wlost 

kJ kg

Solution 7.13 Problem Statement For a pressure-explicit equation of state, prove that the Joule/Thomson inversion curve is the locus of states for which:

 Z   Z  T        T    T  Apply this equation to (a) the van der Waals equation; (b) the Redlich/Kwong equation. Discuss the results. Solution

It is shown at the end of Ex. 7.5 that the Joule/Thomson inversion curve is the locus of states for which

(Z /T ) p  0. We apply the following general equation of differential calculus:

 x   x           x   w      y   y   w   y  z

 Z            Z    Z       T      T   T  

 Z   Z   Z                T   T      T 

Whence,



Because P  ZRT  

P ZRT

            P  1  Z  T  Z    T P R  ZT 2   T P     

Setting (Z /T ) p  0 in each of the two preceding equations reduces them to:

     Z        Z     and      P          T  T  T   T  ZRT 2 

Combining these two equations yields:

 Z   Z  T           T  

(a) Equation (3.42) with van der Waals parameters becomes:

P

RT a  2 V b V

Multiply through by V/RT, substitute Z = PV/RT, V = 1/ , and rearrange:

Z

1 a   b RT

In accord with Eq. (3.51), define q  a/bRT. In addition, define 

Z





Tr) = 1 for the van der Waals equation, q   / Tr . Whence,

dq  dT

Then,

1  q 1 

 Z   Z         dq  T   T  dT

Differentiate:

By Eq. (3.54) with

. Then,

  dT  q   r      2  T 2  dT  Tr Tc  TTr T  r 

 Z         q   q   T   T  T 

Solution continued on next page…

In addition,

 Z       b Z         T





 qb

Substitute for the two partial derivatives in the boxed equation:

q b    qb or q   q 2 2 T     

By Eq. (3.46), Pc  RTc b

 1

Whence,

P  Z RT

Pr  ZbT/ Tc Whence ZTr

Pr 



These equations allow construction of a Tr vs. Pr inversion curve as in Fig. 7.2. For a given value of Tr, calculate q. Equation (B) then gives , Eq. (A) gives Z, and Eq. (C) gives Pr.

(b) Proceed exactly as in Part (a), with exactly the same definitions. This leads to a new Eq. (A): 1 q   

Z

By Eq. (3.54) with  Tr   Tr0.5

q

Tr1.5 . This leads to:

 Z  dq 1.5q 1.5q  and    dT T  T  T    

Moreover,

 Z        T







 

Solution continued on next page…

Substitution of the two derivatives into the boxed equation leads to a new Eq. (B): 2

     q       



   

As in Part (a), for a given Tr, calculate q, and solve Eq. (B) for , by trial or by a computer routine. As before, Eq. (A) then gives Z, and Eq. (C) of Part (a) gives Pr.

Solution 7.14 Problem Statement Two nonconducting tanks of negligible heat capacity and of equal volume initially contain equal quantities of the same ideal gas at the same T and P. Tank A discharges to the atmosphere through a small turbine in which the gas expands isentropically; tank B discharges to the atmosphere through a porous plug. Both devices operate until discharge ceases. (a) When discharge ceases, is the temperature in tank A less than, equal to, or greater than the temperature in tank B? (b) When the pressures in both tanks have fallen to half the initial pressure, is the temperature of the gas discharging from the turbine less than, equal to, or greater than the temperature of the gas discharging from the porous plug?

(c) During the discharge process, is the temperature of the gas leaving the turbine less than, equal to, or greater than the temperature of the gas leaving tank A at the same instant?

(d) During the discharge process, is the temperature of the gas leaving the porous plug less than, equal to, or greater than the temperature of the gas leaving tank B at the same instant? Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(e) When discharge ceases, is the mass of gas remaining in tank A less than, equal to, or greater than the mass of gas remaining in tank B?

Solution

(a) Equal to. (b) Less than. (c) Less than. (d) Equal to. (e) Equal to.

Solution 7.15 Problem Statement A steam turbine operates adiabatically at a power level of 3500 kW. Steam enters the turbine at 2400 kPa and 500°C and exhausts from the turbine as saturated vapor at 20 kPa. What is the steam rate through the turbine, and what is the turbine efficiency?

Solution

W  

kJ kg

H2 

kJ kg

S1 

m 

W H 2  H1

m 

kg s

Data from Table E.2:

H1 

kJ kg K

Using Eq 7.13

For isentropic expansion, exhaust is wet steam: kJ kg K

Sliq 

x

Hliq 

kJ kg K

Svap 

S2  Sliq Svap  Sliq kJ kg

S2  S1

x

kJ kg

H vap 

H 2  Hliq  x  H vap  Hliq 

H2

kJ kg

The efficiency becomes



H2  H1 H '2  H 1



Solution 7.16 Problem Statement A turbine operates adiabatically with superheated steam entering at T1 and P1 with a mass flow rate m . The exhaust pressure is P2 and the turbine efficiency is

For one of the following sets of operating conditions,

determine the power output of the turbine and the enthalpy and entropy of the exhaust steam. 1

(a) T1 = 450°C, P1 = 8000 kPa, m = 80 kg · s , P2 = 30 kPa, = 0.80. 1

(b) T1 = 550°C, P1 = 9000 kPa, m = 90 kg · s , P2 = 20 kPa, = 0.77. 1

(d) T1 = 400°C, P1 = 7000 kPa, m = 65 kg · s , P2 = 50 kPa, = 0.75. 1

(e) T1 = 200°C, P1 = 1400 kPa, m = 50 kg · s , P2 = 200 kPa, = 0.75. 1

(f) T1 = 900(°F), P1 = 1100 (psia), m = 150(lbm)(s) , P2 = 2(psia), = 0.80. 1

(g) T1 = 800(°F), P1 = 1000 (psia), m = 100(lbm)(s) , P2 = 4(psia), = 0.75.

Solution

(a) At the inlet conditions, from the steam tables, we have H = 3274.3 kJ/kg and S = 6.5597 kJ/(kg K). First we find the outlet conditions and work produced if this were expanded to the outlet pressure isentropically. At the l

outlet pressure of 30 kPa, saturated liquid has S = 0.9441 kJ/(kg K) and saturated vapor has v

S = 7.7695 kJ/(kg K). Thus, steam at 30 kPa with the inlet entropy of 6.5597 kJ/(kg K) is a mixture of vapor v

and liquid. The vapor phase fraction (or quality) can be found as x = (6.5597 – 0.9441)/(7.7695 – 0.9441) = l

0.8228. The enthalpy of saturated liquid at 30 kPa is H = 289.302 kJ/kg and for the vapor it is Solution continued on next page…

H = 2625.4 kJ/kg. Thus, the enthalpy of the mixture leaving an isentropic expander would be H = 289.302 + 0.8228*(2625.4 – 289.302) = 2211.3 kJ/kg. The energy balance across the isentropic expander would then give Wisentropic

H = 2211.3

3274.3 = 1063.0 kJ/kg

If the turbine efficiency is 0.80, then the actual work is W = 0.8* 1063.0 = 850.4 kJ/kg The actual enthalpy of the stream leaving the turbine is H = 3274.3 – 850.4 = 2423.9 kJ/kg. This is less than the enthalpy of saturated vapor at 30 kPa, so the real turbine also discharges a vapor/liquid mixture. The vapor v

fraction is x = (2423.9 – 289.302)/(2625.4 – 289.302) = 0.9138. The corresponding entropy of this stream is S = 0.9441 + 0.9138*(7.7695 – 0.9441) = 7.1808 kJ/(kg K). For a steam rate of 80 kg/s, the total power output will be P = 80 * 850.4 = 68032 kJ/s = 68.0 MW.

(b) At the inlet conditions (550 °C, 9000 kPa), from the steam tables, we have H = 3509.8 kJ/kg and S = 6.8143 kJ/(kg K). First we find the outlet conditions and work produced if this were expanded to the outlet l

pressure isentropically. At the outlet pressure of 20 kPa, saturated liquid has S = 0.8321 kJ/(kg K) and v

saturated vapor has S = 7.9094 kJ/(kg K). Thus, steam at 20 kPa with the inlet entropy of 6.8143 kJ/(kg K) is a v

mixture of vapor and liquid. The vapor phase fraction (or quality) can be found as x = (6.8143 – 0.8321)/ l

(7.9094 – 0.8321) = 0.8453. The enthalpy of saturated liquid at 20 kPa is H = 251.453 kJ/kg and for the v

vapor it is H = 2609.9 kJ/kg. Thus, the enthalpy of the mixture leaving an isentropic expander would be H = 251.453 + 0.8453*(2609.9 – 251.453) = 2245.0 kJ/kg. The energy balance across the isentropic expander would then give Wisentropic

H = 2245.0 – 3509.8 = –1264.8 kJ/kg

Solution continued on next page…

If the turbine efficiency is 0.77, then the actual work is W = 0.77*–1264.8 = –973.9 kJ/kg The actual enthalpy of the stream leaving the turbine is H = 3509.8 – 973.9 = 2535.9 kJ/kg. This is less than the enthalpy of saturated vapor at 20 kPa, so the real turbine also discharges a vapor/liquid mixture. The vapor v

fraction is x = (2535.9 – 251.453)/(2609.9 – 251.453) = 0.9686. The corresponding entropy of this stream is S = 0.8321 + 0.9686*(7.9094 – 0.8321) = 7.687 kJ/(kg K). For a steam rate of 90 kg/s, the total power output will be P = 90 * 973.9 = 87650 kJ/s = 87.7 MW.

(c) At the inlet conditions (600 °C, 8600 kPa), from the steam tables, we have H = 3634.5 kJ/kg and S = 6.9813 kJ/(kg K). First we find the outlet conditions and work produced if this were expanded to the outlet l

pressure isentropically. At the outlet pressure of 10 kPa, saturated liquid has S = 0.6493 kJ/(kg K) and v

saturated vapor has S = 8.1511 kJ/(kg K). Thus, steam at 10 kPa with the inlet entropy of 6.9813 kJ/(kg K) is a v

mixture of vapor and liquid. The vapor phase fraction (or quality) can be found as x = (6.9813 – l

0.6493)/(8.1511 – 0.6493) = 0.8441. The enthalpy of saturated liquid at 10 kPa is H = 191.832 kJ/kg and for the v

vapor it is H = 2584.8 kJ/kg. Thus, the enthalpy of the mixture leaving an isentropic expander would be H = 191.832 + 0.8441*(2584.8 – 191.832) = 2211.7 kJ/kg. The energy balance across the isentropic expander would then give Wisentropic

H = 2211.7 – 3634.5 = –1422.8 kJ/kg

If the turbine efficiency is 0.82, then the actual work is W = 0.82*–1422.8 = –1166.7 kJ/kg The actual enthalpy of the stream leaving the turbine is H = 3634.5 – 1176.7 = 2467.8 kJ/kg. This is less than the enthalpy of saturated vapor at 10 kPa, so the real turbine also discharges a vapor/liquid mixture.

Solution continued on next page…

The vapor fraction is x = (2467.8 – 191.832)/(2584.8 – 191.832) = 0.9511. The corresponding entropy of this stream is S = 0.6493 + 0.9511*(8.1511 – 0.6493) = 7.784 kJ/(kg K). For a steam rate of 70 kg/s, the total power output will be P = 70 * 1166.7 = –81669 kJ/s = 81.7 MW.

(d) At the inlet conditions (400 °C, 7000 kPa), from the steam tables, we have H = 3161.2 kJ/kg and S = 6.4536 kJ/(kg K). First we find the outlet conditions and work produced if this were expanded to the outlet l

pressure isentropically. At the outlet pressure of 50 kPa, saturated liquid has S = 1.0912 kJ/(kg K) and v

saturated vapor has S = 7.5947 kJ/(kg K). Thus, steam at 50 kPa with the inlet entropy of 6.4536 kJ/(kg K) is a mixture of vapor and liquid. The vapor phase fraction (or quality) can be found as v

x = (6.4536 – 1.0912)/(7.5947 – 1.0912) = 0.8245. The enthalpy of saturated liquid at 50 kPa is l

H = 340.564 kJ/kg and for the vapor it is H = 2646.0 kJ/kg. Thus, the enthalpy of the mixture leaving an isentropic expander would be H = 340.564 + 0.8245*(2646.0 – 340.564) = 2241.4 kJ/kg. The energy balance across the isentropic expander would then give Wisentropic

H = 2241.4 – 3161.2 = –919.8 kJ/kg

If the turbine efficiency is 0.75, then the actual work is W = 0.75*–919.8 = –689.9 kJ/kg The actual enthalpy of the stream leaving the turbine is H = 3161.2 – 689.9 = 2471.3 kJ/kg. This is less than the enthalpy of saturated vapor at 50 kPa, so the real turbine also discharges a vapor/liquid mixture. The vapor v

fraction is x = (2471.3 – 340.564)/(2646.0 – 340.564) = 0.9242. The corresponding entropy of this stream is S = 1.0912 + 0.9242*(7.5947 – 1.0912) = 7.102 kJ/(kg K). For a steam rate of 65 kg/s, the total power output will be P = 65 * 689.9 = 44840 kJ/s = 44.8 MW.

Solution continued on next page…

(e) At the inlet conditions (200 °C, 1400 kPa), from the steam tables, we have H = 2801.4 kJ/kg and S = 6.4941 kJ/(kg K). First we find the outlet conditions and work produced if this were expanded to the outlet l

pressure isentropically. At the outlet pressure of 200 kPa, saturated liquid has S = 1.5301 kJ/(kg K) and v

saturated vapor has S = 7.1268 kJ/(kg K). Thus, steam at 200 kPa with the inlet entropy of 6.4941 kJ/(kg K) is a mixture of vapor and liquid. The vapor phase fraction (or quality) can be found as v

x = (6.4941 – 1.5301)/(7.1268 – 1.5301) = 0.8870. The enthalpy of saturated liquid at 200 kPa is l

H = 504.701 kJ/kg and for the vapor it is H = 2706.3 kJ/kg. Thus, the enthalpy of the mixture leaving an isentropic turbine would be H = 504.701 + 0.8870*(2706.3 – 504.701) = 2457.4 kJ/kg. The energy balance across the isentropic expander would then give Wisentropic

H = 2457.4 – 2801.4 = –344 kJ/kg

If the turbine efficiency is 0.75, then the actual work is W = 0.75*–344 = –258 kJ/kg The actual enthalpy of the stream leaving the turbine is H = 2801.4 – 258 = 2543.4 kJ/kg. This is less than the enthalpy of saturated vapor at 200 kPa, so the real turbine also discharges a vapor/liquid mixture. The vapor v

fraction is x = (2543.5 – 504.701)/(2706.3 – 504.701) = 0.926. The corresponding entropy of this stream is S = 1.5301 + 0.926*(7.1268 – 1.5301) = 6.717 kJ/(kg K). For a steam rate of 50 kg/s, the total power output will be P = 50 * 258 = 12900 kJ/s = 12.9 MW.

(f) At the inlet conditions, from the steam tables, we have H = 1444.7 Btu/lbm and S = 1.6000 Btu/(lbm R). First we find the outlet conditions and work produced if this were expanded to the outlet pressure l

isentropically. At the outlet pressure of 2 psia, saturated liquid has S = 0.1749 Btu/(lbm R)and saturated vapor Solution continued on next page…

has S = 1.9201 Btu/(lbm R). Thus, steam at 2 psia with the inlet entropy of 1.6000 Btu/(lbm R) is a mixture of v

vapor and liquid. The vapor phase fraction (or quality) can be found as x = (1.6000 – 0.1749)/(1.9201 – 0.1749) = l

0.8166. The enthalpy of saturated liquid at 2 psia is H = 93.96 Btu/lbm and for the vapor it is H = 1116.1 Btu/lbm. Thus, the enthalpy of the mixture leaving an isentropic turbine would be H = 93.96 + 0.8166*(1116.1 – 93.96) = 928.6 Btu/lbm. The energy balance across the isentropic expander would then give

Wisentropic

H = 928.6 – 1444.7 = –516.1 Btu/lbm

If the turbine efficiency is 0.80, then the actual work is W = 0.8*–516.1 = –412.8 Btu/lbm The actual enthalpy of the stream leaving the turbine is H = 1444.7 – 412.8 = 1031.9 Btu/lbm. This is less than the enthalpy of saturated vapor at 2 psia, so the real turbine also discharges a vapor/liquid mixture. The vapor v

fraction is x = (1031.9 – 93.96)/(1116.1 – 93.96) = 0.9176. The corresponding entropy of this stream is S = 0.1749 + 0.9176*(1.9201 – 0.1749) = 1.7763 Btu/(lbm R). For a steam rate of 100 lbm/s, the total power output will be P = 100 * 412.8 = 41280 Btu/s = 58405 hp = 43553 kW.

(g) At the inlet conditions, from the steam tables, we have H = 1289.6 Btu/lbm and S = 1.5677 Btu/(lbm R). First we find the outlet conditions and work produced if this were expanded to the outlet pressure l

isentropically. At the outlet pressure of 4 psia, saturated liquid has S = 0.2200 Btu/(lbm R) and saturated vapor v

has S = 1.8625 Btu/(lbm R). Thus, steam at 4 psia with the inlet entropy of 1.5677 Btu/(lbm R) is a mixture v

of vapor and liquid. The vapor phase fraction (or quality) can be found as x = (1.5677 – 0.2200)/(1.8625 – l

0.2200) = 0.8205. The enthalpy of saturated liquid at 4 psia is H = 120.99 Btu/lbm and for the vapor it is Solution continued on next page…

H = 1127.3 Btu/lbm. Thus, the enthalpy of the mixture leaving an isentropic turbine would be H = 120.99 + 0.8205*(1127.3 – 120.99) = 946.7 Btu/lbm. The energy balance across the isentropic expander would then give

Wisentropic

H = 946.7 – 1389.6 = –442.9 Btu/lbm

If the turbine efficiency is 0.75, then the actual work is W = 0.75*–442.9 = –332.2 Btu/lbm The actual enthalpy of the stream leaving the turbine is H = 1389.6 – 332.2 = 1057.4 Btu/lbm. This is less than the enthalpy of saturated vapor at 4 psia, so the real turbine also discharges a vapor/liquid mixture. The vapor v

fraction is x = (1057.4 – 120.99)/(1127.3 – 120.99) = 0.9305. The corresponding entropy of this stream is S = 0.2200 + 0.9305*(1.8625 – 0.2200) = 1.7483 Btu/(lbm R). For a steam rate of 100 lbm/s, the total power output will be P = 100 * 332.2 = 33220 Btu/s = 47001 hp = 35049 kW.

Solution 7.17 Problem Statement Nitrogen gas initially at 8.5 bar expands isentropically to 1 bar and 150°C. Assuming nitrogen to be an ideal gas, calculate the initial temperature and the work produced per mole of nitrogen.

Solution

T

K P0 

bar P  bar

Isentropic expansion means that  S  0 Solution continued on next page…

So to determine the S, first we need the values for the heat capacity of nitrogen:

A

B

0.593 * 103 K

D

Then for the entropy change of an ideal gas, combine Eqs. (5.10) & (5.11) with C = 0. Substitute:    T D    1     1  ln P  S  R  A * ln    B *  2  *      T  2  P0    Now solving for  S = 0 gives the initial temperature as

T0 

T 

T0 

This leaves the initial temperature at 489.27 C. Now that the initial temperature is calculated, let’s determine the work done.

First determine the mean heat capacity





CPig  R * MCPH 762.42, 423.15, 3.280,  0.593 * 103 , 0,  0.040 * 105 

CPig  30.85

J mol K

From here we know that: Wisentropic  CPig * T  T0 

Wisentropic  10466.48

kJ mol

Solution 7.18 Problem Statement Combustion products from a burner enter a gas turbine at 10 bar and 950°C and discharge at 1.5 bar. The turbine operates adiabatically with an efficiency of 77%. Assuming the combustion products to be an ideal-gas 1

mixture with a heat capacity of 32 J · mol · K , what is the work output of the turbine per mole of gas, and what is the temperature of the gases discharging from the turbine?

Solution

T1 

P1 

bar

P2 

bar C p 

J mol K



Eqs. (7.18) and (7.19) derived for isentropic compression apply equally well for isentropic expansion. They combine to give:

R    P2 CP   WS  C PT1       P1    

WS  W S

J mol

WS  

WS  

J mol

Knowing that H WS , eq. (7.21) can be used to determine the temperature of the discharge:

T2  T1 

H CP

T2 

Solution 7.19 Problem Statement 1

Isobutane expands adiabatically in a turbine from 5000 kPa and 250°C to 500 kPa at a rate of 0.7 kg mol · s . If the turbine efficiency is 0.80, what is the power output of the turbine and what is the temperature of the isobutane leaving the turbine?

Solution

The critical values for isobutene are as follows:

Tc 

Pc 

bar

T0 

P

kPa



S  0

kJ kg K

37.853 * 103 K

C

Pr 0 

Pr 

P0 

kPa

For the heat capacity of isobutene A

B

Tr 0 

11.945 * 106 K2

The entropy change is given by Eq. (6.73) combined with Eq. (5.11) with D = 0:  S  R  A * 

 

   1   *      2 



T   * T0

    B * T0  CT0 2 

T

  * T  P 0  SRB  Pr    SRBTr 0 Pr 0    TC P0  

K Tr 

Solution continued on next page…

The enthalpy change for the final T is given by Eq. (6.72):





Hig  R * ICPH T0 , T , 1.677, 37.853 * 103 ,  11.945 * 106 , 0

Hig  

J mol

H   Hig  RTc HRB Tr , Pr    RTc HRB Tr 0 Pr 0  

H   

J mol

The actual enthalpy change from Eq. (7.17):



n 

mol s

H  H   

W  n H  

J mol

The actual final temperature is now found from Eq. (6.72) combined with Eq (4.8), written:     * T  B D    1  0   Tc  HRB  H  R  AT0     * T02  2    *  Pr    HRB Tr 0 Pr 0      TC 2 T0      

Using this to solve for yields



T   * T0 

Solution 7.20 Problem Statement The steam rate to a turbine for variable output is controlled by a throttle valve in the inlet line. Steam is supplied to the throttle valve at 1700 kPa and 225°C. During a test run, the pressure at the turbine inlet is 1

1000 kPa, the exhaust steam at 10 kPa has a quality of 0.95, the steam flow rate is 0.5 kg · s , and the power output of the turbine is 180 kW. (a) What are the heat losses from the turbine? (b) What would be the power output if the steam supplied to the throttle valve were expanded isentropically to the final pressure?

Solution

From Table E.2 @ 1700 kPa & 225 degC:

H1 

kJ kg

kJ kg K

S1 

The values of the exhaust steam at 10 kPa are:

x2 

Sliq 

Hliq 

kJ kg

kJ kg K

H vap 

kJ kg K

Svap 

kJ kg

m 

kg s

Solution continued on next page…

Solving for H2 H 2  Hliq  x 2  H vap  H liq 

H  H 2  H 1

Part(a): the heat losses for the turbine with W 

H2 

H  

kJ kg

kW are:

Q  m * H  W

Q  

kJ s

Part(b): The power output with the isentropic expansion would be:

x2 

S1  Sliq Svap  Sliq

x 2 

H 2  Hliq  x 2  H vap  H liq 

H2

W  m  H2  H1 

W  

kJ kg

Solution 7.21 Problem Statement Carbon dioxide gas enters an adiabatic expander at 8 bar and 400°C and discharges at 1 bar. If the turbine efficiency is 0.75, what is the discharge temperature and what is the work output per mole of CO2? Assume CO2 to be an ideal gas at these conditions.

Solution

First, we compute the outlet conditions if the expansion were isentropic. Treating CO2 as an ideal gas, the total entropy change is given by:

S  

1 dT  R ln    0 673 K T  8  Tf

C igp

The entropy increase due to the pressure decrease is –Rln(1/8) = 17.29 J/(mol K). Thus,



C igp

673 K

dT  17.29 J/(mol K)

Evaluating the entropy integral and varying the final temperature to get this value gives: T

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2     RT T0  2 T T  0    T



T0 (K)

673.15

T (K)

466.44

5.457

B (1/K)

1.05E–03

C (1/K )

D (K )

ICPS

S (J/(mol K))

0.00E+00

1.16E+05

–2.080

–17.29

Thus, the predicted final temperature is 466.3 K, for isentropic expansion. The enthalpy change for going from 673 K and 8 bar to 466.4 K and 1 bar is (still treating CO2 as an ideal gas):

H  

465.9 K

673.15 K

C igp dT

Solution continued on next page…

Evaluating this gives: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 673.15

T (K) 466.40

A 5.457

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) 1.05E-03 0.00E+00 -1.16E+05 -1175.1

If the expander has an isentropic efficiency of 75%, then the Ws

DHig (J/mol) -9770.7

H) is:

Hactual = 0.75* 9770.7 = 7328.1 J/mol H gives T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 673.15

T (K) 520.00

A 5.457

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) 1.05E-03 0.00E+00 -1.16E+05 -880.6

DHig (J/mol) -7321.8

So, the predicted outlet temperature is 520 K.

Solution 7.22 Problem Statement

Tests on an adiabatic gas turbine (expander) yield values for inlet conditions (T1, P1) and outlet conditions (T2, P2). Assuming ideal gases with constant heat capacities, determine the turbine efficiency for one of the following:

(a) T1 = 500 K, P1 = 6 bar, T2 = 371 K, P2 = 1.2 bar, CP /R = 7/2.

(b) T1 = 450 K, P1 = 5 bar, T2 = 376 K, P2 = 2 bar, CP /R = 4.

(c) T1 = 525 K, P1 = 10 bar, T2 = 458 K, P2 = 3 bar, CP /R = 11/2. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(d) T1 = 475 K, P1 = 7 bar, T2 = 372 K, P2 = 1.5 bar, CP /R = 9/2.

(e) T1 = 550 K, P1 = 4 bar, T2 = 403 K, P2 = 1.2 bar, CP /R = 5/2.

Solution

(a) For the inlet and outlet temperatures given, and an adiabatic expander, assuming an ideal gas with constant heat capacity, the energy balance gives: W

H = Cp T = /2R*(371 – 500) = 3754 J/mol

If the turbine was isentropic, then we could get the outlet temperature from the entropy balance: S = 0 = Cpln(Tout/Tin) – Rln(Pout/Pin) R/Cp

From which Tout = Tin*(Pout/Pin)

2/7

= 500 K * (1.2/6)

= 315.69 K. If that were the outlet temperature, we

would have 7

Wisentropic = /2R*(315.69 – 500) = 5363 J/mol Thus, the turbine efficiency is = (3754/5363) = 0.700

(b) For the inlet and outlet temperatures given, and an adiabatic expander, assuming an ideal gas with constant heat capacity, the energy balance gives: W

H = Cp T = 4R*(376 – 450) = 2461 J/mol

If the turbine was isentropic, then we could get the outlet temperature from the entropy balance: S = 0 = Cpln(Tout/Tin) – Rln(Pout/Pin) R/Cp

From which Tout = Tin*(Pout/Pin)

0.25

= 450 K * (2/5)

= 357.9 K. If that were the outlet temperature, we

would have Wisentropic = 4R*(357.9 – 450) = 3064 J/mol Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Thus, the turbine efficiency is = (2461/3064) = 0.803

(c) For the inlet and outlet temperatures given, and an adiabatic expander, assuming an ideal gas with constant heat capacity, the energy balance gives: W

H = Cp T = 5.5*R*(458 – 525) = –3064 J/mol

If the turbine was isentropic, then we could get the outlet temperature from the entropy balance: S = 0 = Cpln(Tout/Tin) – Rln(Pout/Pin) R/Cp

From which Tout = Tin*(Pout/Pin)

(2/11)

= 525 K * (3/10)

= 421.8 K. If that were the outlet temperature, we

would have Wisentropic = 5.5*R*(421.8 – 525) = –4719 J/mol Thus, the turbine efficiency is = (3064/4719) = 0.649

(d) For the inlet and outlet temperatures given, and an adiabatic expander, assuming an ideal gas with constant heat capacity, the energy balance gives: W

H = Cp T = 4.5*R*(372 – 475) = –3854 J/mol

If the turbine was isentropic, then we could get the outlet temperature from the entropy balance: S = 0 = Cpln(Tout/Tin) – Rln(Pout/Pin) R/Cp

From which Tout = Tin*(Pout/Pin)

(2/9)

= 475 K * (1.5/7)

= 337.3 K. If that were the outlet temperature, we

would have Wisentropic = 4.5*R*(337.3 – 475) = –5152 J/mol Thus, the turbine efficiency is = (3854/5152) = 0.748

Solution continued on next page…

(e) For the inlet and outlet temperatures given, and an adiabatic expander, assuming an ideal gas with constant heat capacity, the energy balance gives: W

H = Cp T = 2.5*R*(403 – 550) = –3056 J/mol

If the turbine was isentropic, then we could get the outlet temperature from the entropy balance: S = 0 = Cpln(Tout/Tin) – Rln(Pout/Pin) R/Cp

From which Tout = Tin*(Pout/Pin)

(2/5)

= 550 K * (1.2/4)

= 339.8 K. If that were the outlet temperature, we

would have Wisentropic = 2.5*R*(339.8 – 550) = –4369 J/mol Thus, the turbine efficiency is = (3056/4369) = 0.699

Solution 7.23 Problem Statement The efficiency of a particular series of adiabatic gas turbines (expanders) correlates with power output according to the empirical expression: = 0.065 + 0.080 ln W . Here, W is the absolute value of the actual power output in kW. Nitrogen gas is to be expanded from inlet conditions of 550 K and 6 bar to an outlet 1

pressure of 1.2 bar. For a molar flow rate of 175 mol·s , what is the delivered power in kW? What is the efficiency of the turbine? What is the rate of entropy generation SG? Assume nitrogen to be an ideal gas with CP = (7/2)R.

Solution

7 Cp  R 2

mol s

n 

T1 

K P1  bar

P2 

bar

Determine the power delivered:

R    P2 CP     pT1    1 W 0.065  0.08 * ln W nC  P    1   

Guessing for W

W





Then solving to the efficiency

W



Now to solve for SG. For an expander operating with an ideal gas with constant Cp, one can show that:

T2  T

R    P2 CP            P1    

T2 

Solution continued on next page…

Then using eq 5.10

C T P S  R *  P * ln 2  ln 2   R T1 P1 

J mol K

S 

Lastly using eq 5.32 to solve for SG for adiabatic operation yields:

SG  n S

SG 

J Ks

Solution 7.24 Problem Statement A turbine operates adiabatically with superheated steam entering at 45 bar and 400°C. If the exhaust steam must be “dry,” what is the minimum allowable exhaust pressure for a turbine efficiency, = 0.75? Suppose the efficiency were 0.80. Would the minimum exhaust pressure be lower or higher? Why?

Solution

Properties of superheated steam at 4500 kPa and 400 C from Table E.2:

H1 

kJ kg

S1 

kJ kg K

Solution continued on next page…

If the exhaust steam (Point 2, Fig. 7.4) is "dry," i.e., saturated vapor, then Isentropic expansion to the same pressure (Point 2', Fig. 7.4) must produce "wet" steam, with entropy:

S2  S1  637093  x * Svap  1  x  * Sliq

A second relation follows from Eq. (7.16), written:

H  H vap  3207.1  H s  0.75[ x * H vap  1  x  * Hliq  3207.1

Each of these equations may be solved for x. Given a final temperature and the corresponding vapor pressure, values for Svap, Sliq, Hvap, and Hliq are found from the table for saturated steam, and substitution into the equations for x produces two values. The required pressure is the one for which the two values of x agree. This is clearly a trial process. For a final trial temperature of 120 C, the following values of H and S for saturated liquid and saturated vapor are found in the steam table:

HL 

kJ kg

HV 

kJ kg

SL 

kJ kg K

SV  .1293

kJ kg K

Solving for x in both equations yields:

xH 

Solving for both x’s gives: x H 

H V  801.7  0.75 H L 0.75 *  HV  H L 

xS 

6.7093  SL SV  SL

xS 

These are sufficiently close, and we conclude that:

T

P

kPa

same work and H.

Solution 7.25 Problem Statement Turbines can be used to recover energy from high-pressure liquid streams. However, they are not used when the high-pressure stream is a saturated liquid. Why? Illustrate by determining the downstream state for isentropic expansion of saturated liquid water at 5 bar to a final pressure of 1 bar.

Solution

When a saturated liquid is expanded in a turbine some of the liquid vaporizes. A turbine properly designed for expansion of liquids cannot handle the much larger volumes resulting from the formation of vapor. For example, if saturated liquid at 5 bar expands isentropically to 1 bar, the fraction of the original liquid that vaporizes is found as follows:

S2  S2l  x 2v S2V  S2l   S1

Solution continued on next page…

And solving for x2v

x2v 

S1  S2l

S2V  S2l 



1.8604  1.3027  0.0921 7.3598  1.3027

Were the expansion irreversible, the fraction of liquid vaporized would be even greater.

Solution 7.26 Problem Statement Liquid water enters an adiabatic hydroturbine at 5(atm) and 15°C, and exhausts at 1(atm). Estimate the power output of the turbine in J · kg

of water if its efficiency is = 0.55. What is the outlet temperature of the water?

Assume water to be an incompressible liquid.

Solution

P1  atm

P2  atm

T1 



Data in Table E.1 for saturated liquid water at 15 C give:

V

cm3 kg

CP 

kJ kg C

Solution continued on next page…

Eqs. (7.16) and (7.24) combine to give:

H  V  P2  P1 

WS  H

kJ kg

WS  

Eq. (7.25) with = 0 is solved for T:

T 

H  V  P2  P1  Cp

T 

So final temperature is: 14.956 C

Solution 7.27 Problem Statement An expander operates adiabatically with nitrogen entering at T1 and P1 with a molar flow rate n . The exhaust pressure is P2, and the expander efficiency is

Estimate the power output of the expander and the

temperature of the exhaust stream for one of the following sets of operating conditions. 1

(a) T1 = 480°C, P1 = 6 bar, n = 200 mol · s , P2 = 1 bar, = 0.80. 1

(b) T1 = 400°C, P1 = 5 bar, n = 150 mol · s , P2 = 1 bar, = 0.75. 1

(d) T1 = 450°C, P1 = 8 bar, n = 100 mol · s , P2 = 2 bar, = 0.85. 1

(e) T1 = 900(°F), P1 = 95(psia), n = 0.5(lb mol)(s) , P2 = 15(psia), = 0.80.

Solution

Assume nitrogen an ideal gas. First find the temperature after isentropic expansion from a combination of Eqs. (5.14) & (5.15) with C = 0. Then find the work (enthalpy change) of isentropic expansion by a combination of Eqs. (4.2) and (4.8) with C = 0. The actual work (enthalpy change) is found from Eq. (7.20). From this value, the actual temperature is found by a second application of the preceding equation, this time solving it for the temperature. The following vectors contain values for Parts (a) through (e):

All of this can be done iteratively in excel, the equations needed are as follows:    T D    1     1  ln P  S  R  A * ln    B *  2  *    T  2  P0   





H '  R * ICPH T0 , T , 1.677, 37.853 * 103 ,  11.945 * 106 , 0

 B D    1  H  R  AT0   1  * T02  2  1  *   2 T0     W  n * H

Solution continued on next page…

The solutions for this are as follows:

T P0

ndot T (K)

final

Wdot

(K)

(kW)

Part

T0 (K)

(bar)

P (bar)

(mol/sec)

753.15

200

0.8

459.72

9042.15

7233.72

515.85

1446.74

673.15

150

0.75

430.41

7394.54

5545.91

489.74

831.886

773.15

175

0.78

452.45

9911.01

7730.59

520.02

1352.85

723.15

100

0.85

493.82

7056.97

5998.42

526.34

599.842

755.37

95 psi

15 psi

226.795

0.8

454.13

9285.79

7428.63

511.64

1684.78

Solution 7.28 Problem Statement What is the ideal-work rate for the expansion process of Ex. 7.6? What is the thermodynamic efficiency of the process? What is the rate of entropy generation SG ? What is W lost ? Take T = 300 K.

Solution

Property values and data from Example 7.6: kJ S  kg 1

H1 

m 

kJ H2  kg K

kJ kg

kg s

W  

kJ kg K

S2 

Using Eq. 5.21 to calculate the ideal work

W ideal  m *  H 2  H1  T S2  S1   

W ideal  

This leaves the efficiency as



W W



ideal

Since the process is adiabatic the SG and W lost are

SG  m S2  S1 

SG 

kW K

W lost  T SG  17685 kW

Solution 7.29 Problem Statement

Exhaust gas at 400°C and 1 bar from internal-combustion engines flows at the rate of 125 mol · s

into a waste-

heat boiler where saturated steam is generated at a pressure of 1200 kPa. Water enters the boiler at 20°C (T ), and the exhaust gases are cooled to within 10°C of the steam temperature. The heat capacity of the exhaust gases is CP /R = 3.34 + 1.12 × 10

T/K. The steam flows into an adiabatic turbine and exhausts at a pressure of 25 kPa.

If the turbine efficiency is 72%, (a) What is W S the power output of the turbine? (b) What is the thermodynamic efficiency of the boiler/turbine combination? (c) Determine SG for the boiler and for the turbine. (d) Express W lost (boiler) and W lost (turbine) as fractions of W ideal , the ideal work of the process.

Solution

For sat. vapor steam at 1200 kPa, Table E.2:

H2 

kJ kg

S2 

kJ kg K

The saturation temperature is 187.96 C. The exit temperature of the exhaust gas is therefore 197.96 C, and the temperature change of the exhaust gas is 202.04 K. Solution continued on next page…

For the water at 20 C from Table E.1, kJ kg

H1 

kJ kg K

S1 

The turbine exhaust will be wet vapor steam. For sat. liquid and sat. vapor at the turbine exhaust pressure of 25 kPa, the best property values are found from Table E.1 by interpolation between 64 and 65 degC: kJ kg

Hliq 

kJ kg

Hlv 

kJ kg K

Sliq 

kJ kg K

Slv 



For isentropic expansion of steam in the turbine:

S3  S2

S3 

x 3 

kJ kg K

S3  Sliq

H3  H liq  x 3 H lv

Slv

x 3 

H3 

H23    H3  H 2 

x3 

H3  H liq Hlv

x3 

kJ kg

H3  H 2  H 23

kJ kg

H 23  

H3 

kJ kg

S3  Sliq  x 3 Slv

S3 

kJ kg K

For the exhaust gases: mol s

n 

T1 

T2 



MW 

g mol



H gas  R * MCPH T1 , T2 , 3.34, 1.12 * 103 , 0, 0 * T2  T1  T S gas  R * MCPS T1 , T2 , 3.34, 1.12 * 103 , 0, 0 * ln 2 T1



H gas  



J mol

S gas  

kJ mol K

The energy balance on the boiler is:

m 

n H gas H 2  H1

kg s

m 

T 

Now that all the setup has been done

Part(a): Solve for W  m *  H3  H 2 

W  

Part(b): Solve for the ideal work using Eq 5.25 W ideal  n H gas  m ( H3  H1 )  T  n Sgas  m ( S3  S1   W ideal  

t 

W W



ideal

Part(c): For both the boiler and the turbine, Eq. (5.28) applies with Q = 0. For the boiler: SG  n * Sgas  m S2  S1

For the boiler SG 

kW K

For the turbine SG  m S2  S1

SG 

Part(d): Now for the work lost

W lost boiler 

kW *T K

W lost boiler 

W lost turbine 

kW *T K

W lost boiler 

Fractionboiler 

W lost boiler  W ideal

Fractionturbine 

W lost turbine  W ideal

Solution 7.30 Problem Statement 3

A small adiabatic air compressor is used to pump air into a 20-m insulated tank. The tank initially contains air at 25°C and 101.33 kPa, exactly the conditions at which air enters the compressor. The pumping process continues until the pressure in the tank reaches 1000 kPa. If the process is adiabatic and if compression is isentropic, what is the shaft work of the compressor? Assume air to be an ideal gas for which CP = (7/2)R and CV = (5/2)R.

Solution

Apply Eq. (2.29) to this non-steady-state process, with n replacing m, with the tank as control volume, and with a single inlet stream. Since the process is adiabatic and the only work is shaft work, this equation may be multiplied by dt to give: d nU tank  Hdn  dWs

Because the inlet stream has constant properties, integration from beginning to end of the process yields: Ws  n2U 2  n1U 1  nH

where the subscripted quantities refer to the contents of the tank and n and H refer to the inlet stream. Solution continued on next page…

Substitute n = n2

n1 and H = U + PV = U + RT:

Ws  n2U 2  n1U 1  n2  n1  

  n2  U 2  U  RT   n1 U1  U  RT 

With U CvT for an ideal gas with constant heat capacities, this becomes:

Ws  n2 Cv T2  T   RT   n1 Cv T1  T   RT 

However, T = T1, and therefore: Ws  n2 C v T2  T1   RT1   n1 RT1

By Eq. 3.23b 1/

P  T2  T1  2   P1 

Moreover

n1 

With  

T2 

P1Vtank RT1

n2 

P2 Vtank RT2

m3 kPa kmol K

R

n1 

and

n2 

Substitution of numerical values into the boxed equation, with R = 8.314 kJ kmol 1 K 1, gives:

Ws  15633 kJ

Solution 7.31

Problem Statement Saturated steam at 125 kPa is compressed adiabatically in a centrifugal compressor to 700 kPa at the rate of 1

2.5 kg · s . The compressor efficiency is 78%. What is the power requirement of the compressor and what are the enthalpy and entropy of the steam in its final state?

Solution

From Table E.2 for sat. vap. at 125 kPa: kJ kg

H1 

For isentropic expansion, S '2  S1  7.2847

S1 

kJ kg K

Interpolation in Table E.2 at 700 kPa for the enthalpy of steam with this entropy gives

H 2 

kJ kg



H2  H1 

H 

H2 

kJ kg

H 

H 2  H1  H

kj kg

Solution continued on next page…

Interpolation in Table E.2 at 700 kPa for the entropy of steam with this enthalpy gives

S2  7.4586

With m 

kJ kg K

kg the power requirement is s

W  m H

W 

Solution 7.32

Problem Statement A compressor operates adiabatically with air entering at T1 and P1 with a molar flow rate n . The discharge pressure is P2 and the compressor efficiency is

Estimate the power requirement of the compressor and the

temperature of the discharge stream for one of the following sets of operating conditions. 1

(a) T1 = 25°C, P1 = 101.33 kPa, n = 100 mol · s , P2 = 375 kPa, = 0.75. 1

(b) T1 = 80°C, P1 = 375 kPa, n = 100 mol · s , P2 = 1000 kPa, = 0.70. 1

(d) T1 = 100°C, P1 = 500 kPa, n = 50 mol · s , P2 = 1300 kPa, = 0.75. 1

(e) T1 = 80(°F), P1 = 14.7 (psia), n = 0.5 (lb mol)(s) , P2 = 55 (psia), = 0.75. 1

(f) T1 = 150(°F), P1 = 55 (psia), n = 0.5 (lb mol)(s) , P2 = 135 (psia), = 0.70. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

(a) We can start by computing the work required for isentropic compression. Then, we divide that work by the isentropic efficiency to get the actual work, which is also the actual enthalpy change of the gas. Finally, we compute the outlet temperature that gives that enthalpy change. So, we start from:

S  S R 298.15 K,1.0133 bar   



Tf 298.15 K

 3.75    S R Tf ,3.75 bar   0 dT  R ln  1.0133  298.15 K T Tf

C pig

 3.75  dT  R ln   S R 298.15 K,1.0133 bar   S R Tf ,3.75 bar   1.0133  T

C igp

We can get our first estimate for Tf by neglecting the residual entropies, and solving



 3.75  dT  R ln    10.880 J/(mol K) 1.0133  T

C igp

Tf 298.15 K

Using the handy entropy integral spreadsheet, we get

ICPS  T0,T;A,B,C,D  



T0 (K) 298.15

T (K) 431.1

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT 2 T T0   T0    

A 3.355

C (1/K ) D (K ) B (1/K) 5.75E-04 0.00E+00 -1.60E+03

ICPS 1.309

DS (J/(mol K)) 10.880

So, our first estimate for Tf is 431.1 K. For air, with pseudo-critical properties of Tc = 132.2 K and Pc = 37.45 bar, both the inlet and outlet conditions correspond to relatively low reduced pressure and high reduced temperature, so we expect the residual entropies to be small. We can compute them using the Pitzer correlation, with the spreadsheet shown here for the outlet temperature:

Solution continued on next page…

T (K) 431.1

Tr 3.2610

P (bar) 3.75

Tc (K) Pc (bar) 132.2 37.45

Pr 0.1001

dB0 dTr

dB1 dTr

0.0312

0.0015

dB0 0.675  2.6 dTr Tr

w 0.035

SR/R -0.0031

SR (J mol-1 K-1) -0.026

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

At the inlet conditions, S = 0.018 J/(mol K). So, if we added in these corrections, we would have:



C igp

298.15 K

dT  10.880  0.026  0.018  10.872 J/(mol K)

Which changes the computed temperature to 431.0 K – a negligible difference. Now for the work of isentropic compression, we have:

Ws (isentropic )  H S  H R 298.15 K,1.013 bar   

431.0 K

298.15 K

C pig dT  H R 431.0 K,3.75 bar 

Evaluating the three terms in this equation gives:

ICPH  T0,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 298.15

T (K) 431.0

A 3.355

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 5.75E-04 0.00E+00 -1.60E+03 471.7 3922.1

Solution continued on next page…

T (K) 431.1

Tr 3.2610

P (bar) 3.75

Pr 0.1001

B0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) Pc (bar) 132.2 37.45

B 0.0193

dB0 dTr

B 0.1378

dB0 0.675  2.6 dTr Tr

w 0.035

0.0312

dB1 dTr

0.0015

H /RTc -0.0078

-1

(J mol ) -8.57

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output.

The residual enthalpy at the feed conditions (using the same spreadsheet) is 6.30 J/(mol K). Thus, we have Ws (isentropic )  H S  6.30  3922.1  8.57  3919.8 J/mol

Again, the residual properties make a negligible difference.

If the isentropic efficiency is 0.75, then the actual work is Ws = 3918.8/0.75 = 5266.4 J/mol. So, for the actual process, we have

Ws  H  H R 298.15 K,1.013 bar  

298.15 K

C igp dT  H R Tf ,3.75 bar   5266.4 J/mol

Neglecting the residual enthalpies gives



298.15 K

C igp dT  5266.4 J/mol

Solution continued on next page…

ICPH  T0,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 298.15

T (K) 475.8

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









C (1/K ) D (K ) B (1/K) ICPH (K) DH (J/mol) 5.75E-04 0.00E+00 -1.60E+03 633.4 5266.4

A 3.355

The residual enthalpy at this temperature is 5.9 J/mol, which almost perfectly cancels the residual enthalpy at the feed conditions ( 6.3 J/mol) so there is certainly no need to include the difference of those terms. Thus, the predicted discharge temperature is 476 K. The power requirement for the compressor, for a flow rate of 100 mol/s is

 S  100 mol/s  5266 J/mol = 526.6 kW W s  nW

(b) We can start by computing the work required for isentropic compression. Then, we divide that work by the isentropic efficiency to get the actual work, which is also the actual enthalpy change of the gas. Finally, we compute the outlet temperature that gives that enthalpy change. So, we start from:

S  S R 353.15 K,3.75 bar   



 10  dT  R ln    S R Tf ,10 bar   0 353.15 K T  3.75  Tf

C igp

 10  dT  R ln    S R 353.15 K,3.75 bar   S R Tf ,10 bar   3.75  353.15 K T Tf

C igp

We can get our first estimate for Tf by neglecting the residual entropies, and solving



 10  dT  R ln   8.155 J/(mol K)  3.75  353.15 K T Tf

C igp

Solution continued on next page…

Using the handy entropy integral spreadsheet, we get

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T0 (K) 353.15

T (K) 464.51

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 0.981

(J/(mol K)) 8.1551

So, our first estimate for Tf is 464.51 K. For air, with pseudo-critical properties of Tc = 132.2 K and Pc = 37.45 bar, both the inlet and outlet conditions correspond to relatively low reduced pressure and high reduced temperature, so we expect the residual entropies to be small. We can compute them using the Pitzer correlation, with the spreadsheet shown here for the outlet conditions:

T (K) 464.51

Tr 3.5137

P (bar) 10

Pr 0.2670

dB0 0.675  2.6 dTr Tr

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0257

0.0010

w 0.035

S R/R -0.0069

SR (J mol-1 K-1) -0.0572

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

At the inlet conditions, S = 0.0438 J/(mol K). So, if we added in these corrections, we would have:



C igp

353.15 K

dT  8.1551  0.0572  0.0438  8.1685 J/(mol K)

Solution continued on next page…

Which changes the computed temperature to 464.7 K – a negligible difference. Now for the work of isentropic compression, we have:

Ws (isentropic )  H S  H R 353.15 K,3.75 bar  

464.7 K

353.15 K

C pig dT  H R 464.7 K,10 bar

Evaluating the three terms in this equation gives:

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 353.15

T (K) 464.72

A 3.355

B (1/K) 5.75E-04

T (K) 464.72

P (bar) 10

Tc (K) 132.2

Pc (bar) 37.45

Tr 3.5153

Pr 0.2670

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

B0 0.0265

B1 0.1381

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 399.5 3321.2

w 0.035

dB0 dTr

dB1 dTr

0.0257

0.0010

HR /RTc -0.0158

HR (J mol-1) -17.34

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output.

The residual enthalpy at the feed conditions (using the same spreadsheet) is 15.45 J/(mol K). Thus, we have Ws (isentropic )  H S  15.45  3321.2  17.34  3349.3 J/mol

Solution continued on next page…

Again, the residual properties make a negligible difference.

If the isentropic efficiency is 0.70, then the actual work is Ws = 3349.3/0.7 = 4784.7 J/mol. So, for the actual process, we have

Ws  H  H R 353.15 K,3.75 bar   

353.15 K

C pig dT  H R Tf ,10 bar  4784.7 J/mol

Neglecting the residual enthalpies gives



353.15 K

C igp dT  4784.7 J/mol

Using the ICPH spreadsheet, we find the final

ICPH  T0 ,T;A,B,C,D  

2 0

T (K) 513.21

A 3.355

3 0

1

1  0

T0(K) 353.15

 R dT  AT  T   2 T  T   3 T  T   D  T  T  B (1/K) 5.75E-04

ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 575.5 4784.7

The residual enthalpy at this temperature is 11.0 J/mol, which slightly smaller than the residual enthalpy at the feed conditions ( 15.45 J/mol). Including these would give



353.15 K

C igp dT  4789.2 J/mol

Which leads to an outlet temperature of 513.4 K. If the total flow rate is 100 mol/s, then the total power requirement for the compressor is

 S  100 mol/s  4784 J/mol = 478.4 kW W s  nW

Solution continued on next page…

(c) We can start by computing the work required for isentropic compression. Then, we divide that work by the isentropic efficiency to get the actual work, which is also the actual enthalpy change of the gas. Finally, we compute the outlet temperature that gives that enthalpy change. So, we start from:

S  S R 303.15 K,1.00 bar   



C igp

303.15 K

Tf 303.15 K

 500  dT  R ln    S R Tf ,5 bar   0  100  T

C igp

dT  R ln 5  S R 303.15 K,1.00 bar   S R Tf ,5 bar 

We can get our first estimate for Tf by neglecting the residual entropies, and solving



C igp

303.15 K

dT  R ln 5  13.382 J/(mol K)

Using the handy entropy integral spreadsheet, we get

ICPS  T0 ,T;A,B,C,D  



T0 (K) 303.15

T (K) 476.21

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT 2 T T0   T0    

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 1.610

(J/(mol K)) 13.3820

So, our first estimate for Tf is 476.2 K. For air, with pseudo-critical properties of Tc = 132.2 K and Pc = 37.45 bar, both the inlet and outlet conditions correspond to relatively low reduced pressure and high reduced temperature, so we expect the residual entropies to be small. We can compute them using the Pitzer correlation, with the spreadsheet shown here for the outlet temperature: Solution continued on next page…

T (K) 476.21

P (bar) 5

Tr 3.6022

Pr 0.1335

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0241

0.0009

dB0 0.675  2.6 dTr Tr

w 0.035

SR /R -0.0032

SR (J mol-1 K-1) -0.0268

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

At the inlet conditions, S = 0.0174 J/(mol K). So, if we added in these corrections, we would have:



C pig

298.15 K

dT  13.382  0.0268  0.0174  13.372 J/(mol K)

Which changes the computed temperature to 476.1K – a negligible difference. Now for the work of isentropic compression, we have:

Ws (isentropic )  H S  H R 303.15 K,1.00 bar  

476.1 K

303.15 K

C pig dT  H R  476.1 K,5.00 bar

Evaluating the three terms in this equation gives:

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 303.15

T (K) 476.06

A 3.355

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 616.9 5129.3

Solution continued on next page…

T (K) 303.15

P (bar) 1

Tr 2.2931

Pr 0.0267

0.422

B  0.083  B1  0.139 

Tr1.6 0.172 Tr4.2

Tc (K) 132.2

B -0.0288

Pc (bar) 37.45

B 0.1337

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.035

dB0 dTr

dB1 dTr

0.0780

0.0096

HR /RTc -0.0054

HR (J mol-1) -5.98

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

The residual enthalpy at the outlet conditions (using the same spreadsheet) is 7.85 J/(mol K). Thus, we have Ws (isentropic )  H S  7.85  5129.3  5.98  5131 J/mol

Again, the residual properties make a negligible difference.

If the isentropic efficiency is 0.80, then the actual work is Ws = 5131/0.80 = 6414 J/mol. So, for the actual process, we have

Ws  H  H R 303.15 K,1.00 bar   

303.15 K

C igp dT  H R Tf ,5.00 bar   6413.96 J/mol

Neglecting the residual enthalpies gives



303.15 K

C igp dT  6413.96 J/mol

Solution continued on next page…

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 303.15

T (K) 518.57

A 3.355

B (1/K) 5.75E-04

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 771.4 6414.0

The residual enthalpy at this temperature is 5.2 J/mol, which almost perfectly cancels the residual enthalpy at the feed conditions ( 5.98 J/mol) so there is certainly no need to include the difference of those terms. Thus, the predicted discharge temperature is 519 K. The power requirement for the compressor, for a flow rate of 150 mol/s is

 S  150 mol/s  6414 J/mol = 962.1 kW W s  nW

(d) We can start by computing the work required for isentropic compression. Then, we divide that work by the isentropic efficiency to get the actual work, which is also the actual enthalpy change of the gas. Finally, we compute the outlet temperature that gives that enthalpy change. So, we start from:

S  S R 373.15 K,5.00 bar   



C igp

373.15 K

13  dT  R ln    S R Tf ,13 bar   0 373.15 K T  5  Tf

C pig

dT  R ln 13 / 5  S R 373.15 K,5.00 bar   S R Tf ,13 bar 

We can get our first estimate for Tf by neglecting the residual entropies, and solving



C igp

373.15 K

dT  R ln 13 / 5  7.945 J/(mol K)

Solution continued on next page…

Using the handy entropy integral spreadsheet, we get

ICPS  T0,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T



T0 (K) 373.15

T (K) 486.89

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 0.956

D Sig (J/(mol K)) 7.9453

So, our first estimate for Tf is 486.9 K. For air, with pseudo-critical properties of Tc = 132.2 K and Pc = 37.45 bar, both the inlet and outlet conditions correspond to relatively low reduced pressure and high reduced temperature, so we expect the residual entropies to be small. We can compute them using the Pitzer correlation, with the spreadsheet shown here for the outlet temperature:

T (K) 486.89

P (bar) 13

Tr 3.6830

Pr 0.3471

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0228

0.0008

dB0 0.675  2.6 dTr Tr

w 0.035

SR /R -0.0079

SR (J mol-1 K-1) -0.0658

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

At the inlet conditions, S = 0.0506 J/(mol K). So, if we added in these corrections, we would have:



C igp

298.15 K

dT  7.945  0.0506  0.0658  7.960 J/(mol K)

Solution continued on next page…

Which changes the computed temperature to 487.1K – a negligible difference. Now for the work of isentropic compression, we have:

Ws (isentropic )  H S  H R 373.15 K,6.00 bar  

487.1 K

373.15 K

C pig dT  H R 487.1 K,13.00 bar 

Evaluating the three terms in this equation gives:

ICPH T0,T;A,B,C,D  

1

2 0

3 0

T0(K) 373.15

T (K) 487.13

A 3.355

B (1/K) 5.75E-04

T (K) 487.1

P (bar) 13

Tc (K) 132.2

Pc (bar) 37.45

Tr 3.6846

Pr 0.3471

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T 

B0 0.0306

B1 0.1383

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 409.6 3405.5

w 0.035

dB0 dTr

dB1 dTr

0.0227

0.0008

HR /RTc -0.0168

HR (J mol-1) -18.47

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output.

The residual enthalpy at the inlet conditions (using the same spreadsheet) is 17.77 J/(mol K). Thus, we have Ws (isentropic )  H S  18.47  3405.5  17.77  3406.2 J/mol

Solution continued on next page…

Again, the residual properties make a negligible difference.

If the isentropic efficiency is 0.75, then the actual work is Ws = 3406.2/0.75 = 4542 J/mol. So, for the actual process, we have

Ws  H  H R 373.15 K,5.00 bar  

373.15 K

C igp dT  H R Tf ,13.00 bar   4541.6 J/mol

Neglecting the residual enthalpies gives



373.15 K

C igp dT  4541.6 J/mol H:

ICPH T0,T;A,B,C,D  

2 0

T (K) 524.67

A 3.355

3 0

1

1  0

T0(K) 373.15

 R dT  AT  T   2 T  T   3 T  T   D  T  T  B (1/K) 5.75E-04

ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 546.2 4541.6

The residual enthalpy at this temperature is 12.63 J/mol. Including this and the residual enthalpy at the inlet gives.



373.15 K

C igp dT  4541.6  17.77  12.63 = 4536.5 J/mol

Using this value gives an outlet temperature of 524.5 K. The power requirement for the compressor, for a flow rate of 50 mol/s is

 S  50 mol/s  4541.6 J/mol = 227.1 kW W s  nW

Solution continued on next page…

(e) We can start by computing the work required for isentropic compression. Then, we divide that work by the isentropic efficiency to get the actual work, which is also the actual enthalpy change of the gas. Finally, we compute the outlet temperature that gives that enthalpy change. In SI units, the initial conditions are 299.82 K and 1.013 bar, and the outlet pressure is 3.791 bar. So, we start from:

S  S R 299.82 K,1.013 bar   



C igp

299.82 K

 3.791  dT  R ln    S R Tf ,3.791 bar   0 299.82 K T  1.013  Tf

C igp

dT  R ln 3.791 / 1.013  S R 298.82 K,1.013 bar   S R Tf ,3.791 bar 

We can get our first estimate for Tf by neglecting the residual entropies, and solving



C igp

299.82 K

dT  R ln3.791 / 1.013  10.973 J/(mol K)

Using the handy entropy integral spreadsheet, we get

ICPS  T0,T;A,B,C,D  



T0 (K) 299.82

T (K) 434.78

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT 2  T T0   T0   

A 3.355

B (1/K) 5.75E-04

C (1/K2) D (K2) 0 -1.60E+03

ICPS 1.320

(J/(mol K)) 10.9730

So, our first estimate for Tf is 434.8 K. For air, with pseudo-critical properties of Tc = 132.2 K and Pc = 37.45 bar, both the inlet and outlet conditions correspond to relatively low reduced pressure and high reduced temperature, so we expect the residual entropies to be small. We can compute them using the Pitzer correlation, with the spreadsheet shown here for the outlet temperature: Solution continued on next page…

T (K) 434.78

P (bar) 3.791

Tr 3.2888

Pr 0.1012

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0305

0.0015

dB0 0.675  2.6 dTr Tr

w 0.035

SR /R -0.0031

SR (J mol-1 K-1) -0.0258

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output.

At the inlet conditions, S = 0.0181 J/(mol K). So, if we added in these corrections, we would have:



C igp

299.82 K

dT  10.973  0.0181  0.0258  10.980 J/(mol K)

Using this as the target entropy change in the entropy integral spreadsheet changes the computed temperature to 434.88 K – a negligible difference. Now for the work of isentropic compression, we have:

Ws (isentropic )  H S H R 299.82 K, 1.013 bar  

434.88 K 299.82 K

C pig dT  H R  434.88 K,3.791 bar

Solution continued on next page…

Evaluating the three terms in this equation gives:

ICPH T0,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 299.82

T (K) 434.88

A 3.355

B (1/K) 5.75E-04

T (K) 299.82

P (bar) 1.013

Tc (K) 132.2

Pc (bar) 37.45

Tr 2.2679

Pr 0.0270

B  0.083  B1  0.139 

0.422 Tr1.6 0.172 Tr4.2

B -0.0308

B 0.1335

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 









ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 480.0 3990.9

w 0.035

dB0 dTr

dB1 dTr

0.0803

0.0102

HR /RTc -0.0057

HR (J mol-1) -6.22

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 434.88

Tr 3.2896

P (bar) 3.791

Pr 0.1012

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) 132.2

B0 0.0202

Pc (bar) 37.45

B1 0.1378

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.035

dB0 dTr

dB1 dTr

0.0305

0.0015

HR /RTc -0.0076

HR (J mol-1) -8.41

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

Thus, we have Ws (isentropic )  H S  6.22  3990.9  8.41  3988.7 J/mol

Again, the residual properties make a negligible difference.

If the isentropic efficiency is 0.75, then the actual work is Ws = 3988.7/0.75 = 5318 J/mol. So, for the actual process, we have

Ws  H  H R 299.82 K,1.013 bar  

299.82 K

C pig dT  H R Tf ,3.791 bar   5318.3 J/mol

Neglecting the residual enthalpies gives



299.82 K

C igp dT  5318.3 J/mol

Using the ICPH

ICPH T0,T;A,B,C,D  

2 0

3 0

T (K) 479.10

1  0

T0(K) 299.82

1

 R dT  AT  T   2 T  T   3 T  T   D  T  T 

A 3.355

B (1/K) 5.75E-04

ig C (1/K2) D (K2) ICPH (K) D H (J/mol) 0 -1.60E+03 639.6 5318.3

The residual enthalpy at this temperature is 5.79 J/mol. Including this and the residual enthalpy at the inlet gives.



299.82 K

C igp dT  5318.3  6.22  5.79 = 5317.9 J/mol

Using this value does not change the predicted outlet temperature. The power requirement for the compressor, for a flow rate of 0.5 lbmol/s = 226.8 mol/s is

 S  226.8 mol/s 5318.3 J/mol = 1.206 106 W  1.21 MW W s  nW Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(f) We can start by computing the work required for isentropic compression. Then, we divide that work by the isentropic efficiency to get the actual work, which is also the actual enthalpy change of the gas. Finally, we compute the outlet temperature that gives that enthalpy change. In SI units, the initial conditions are 338.71 K and 3.792 bar, and the outlet pressure is 9.308 bar. So, we start from:

S  S R 338.71 K,3.792 bar   



C pig

338.71 K

 9.308  dT  R ln    S R Tf ,9.308 bar   0 338.71 K T  3.792  Tf

C igp

dT  R ln 9.308 / 3.792  S R 338.71 K,3.792 bar   S R Tf ,9.308 bar 

We can get our first estimate for Tf by neglecting the residual entropies, and solving



C igp

338.71 K

dT  R ln9.308 / 3.792  7.466 J/(mol K)

Using the handy entropy integral spreadsheet, we get

ICPS  T0,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2 T T0      T



T 0 (K) 338.71

T (K) 435.72

A 3.355

B (1/K) 5.75E-04

C (1/K2) 0

D (K2) -1.60E+03

ICPS 0.898

ig DS (J/(mol K)) 7.4660

So, our first estimate for Tf is 435.72 K. For air, with pseudo-critical properties of Tc = 132.2 K and Pc = 37.45 bar, both the inlet and outlet conditions correspond to relatively low reduced pressure and high reduced temperature, so we expect the residual entropies to be small. We can compute them using the Pitzer correlation, with the spreadsheet shown here for the outlet temperature:

Solution continued on next page…

T (K) 338.71

P (bar) 3.792

Tr 2.5621

Pr 0.1013

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0585

0.0054

dB0 0.675  2.6 dTr Tr

w 0.032

S /R -0.0059

-1

(J mol K ) -0.0494

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. T (K) 435.72

P (bar) 9.308

Tr 3.2959

Pr 0.2485

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0304

0.0015

dB0 0.675  2.6 dTr Tr

w 0.032

S /R -0.0076

(J mol K ) -0.0629

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output. R

At the inlet conditions, S = 0.0494 J/(mol K) and at the outlet conditions, it is 0.0629. So, if we added in these corrections, we would have:



C igp

338.71 K

dT  7.466  0.0494  0.0629  7.480 J/(mol K)

Using this as the target entropy change in the entropy integral spreadsheet changes the computed temperature to 435.93 K – a negligible difference. Now for the work of isentropic compression, we have:

Ws (isentropic )  H S  H R 338.71 K,3.792 bar  

435.93 K

338.71 K

C pig dT  H R 435.93 K,9.308 bar

Solution continued on next page…

Evaluating the terms in his equation gives:

ICPH  T0,T;A,B,C,D  

R

dT  A T  T0  

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 



T0(K) 338.71

T (K) 435.93

A 3.355

B (1/K) 5.75E-04

C (1/K2) 0

T (K) 338.71

P (bar) 3.792

Tc (K) 132.2

Pc (bar) 37.45

w 0.032

Tr 2.5621

Pr 0.1013

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

dB0 dTr

-0.0107

0.1357

0.0585

dB0 0.675  2.6 dTr Tr







D (K2) ICPH (K) -1.60E+03 346.8

dB1 dTr 0.0054

HR/RTc -0.0159

DH (J/mol) 2883.1

HR (J mol-1) -17.43

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output.

T (K) 435.93

Tr 3.2975

P (bar) 9.308

Pr 0.2485

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6 Tr4.2

Tc (K) 132.2

Pc (bar) 37.45

dB0 dTr

dB1 dTr

0.0205

0.1379

0.0303

0.0015

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.032

HR/RTc -0.0187

HR (J mol-1) -20.58

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output.

Thus, we have Ws (isentropic )  H S  17.43  2883.1  20.58  2880.0 J/mol

Again, the residual properties make a negligible difference.

If the isentropic efficiency is 0.70, then the actual work is Ws = 2880.0/0.70 = 4114.2 J/mol. So, for the actual process, we have

Ws  H H R 338.71 K, 3.791 bar  

338.71 K

C pig dT  H R Tf , 9.308 bar  4114.2 J/mol

Neglecting the residual enthalpies gives



338.71 K

C igp dT  4114.2 J/mol H:

ICPH  T0,T;A,B,C,D  

1

2 0

3 0

T0(K) 338.71

T (K) 476.94

1 

 R dT  A T  T   2 T  T   3 T  T   D  T  T 

A 3.355

B (1/K) 5.75E-04

C (1/K2) 0

D (K2) ICPH (K) -1.60E+03 494.8

DH (J/mol) 4114.2

The residual enthalpy at this temperature is 14.61 J/mol. Including this and the residual enthalpy at the inlet gives.



338.71 K

C igp dT  17.43  4114.2  14.61 = 4117.0 J/mol

Using this value increases the outlet temperature to 476.85 K (a negligible difference). The power requirement for the compressor, for a flow rate of 0.5 lbmol/s = 226.8 mol/s is

 S  226.8 mol/s  4114.2 J/mol = 9.33110 5 W  0.933 MW W s  nW

Solution 7.33 Problem Statement Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the final temperature, the work required, and the entropy change of the ammonia.

Solution Ammonia gas critical values are:

Pc 

P0 

bar

T0 

kPa

Tc 

S  0

P

kPa

K 

J mol k

For the heat capacity of ammonia

A

B

Tr 0 

3.020 * 103 K

D

Pr 0 

Pr 

Use generalized second-virial correlation: The entropy change is given by Eq. (6.73) combined with Eq. (5.11) with D = 0:

     T D    1     1  ln P  SRB  * T0 , P ,   SRB T , P ,   S  R  A * ln     B *  2    *  r r 0 r 0         T  2  P0  TC    Solution continued on next page…

Solving for 



T   * T0

T

K Tr 

The enthalpy change for the final T is given by Eq. (6.72), with HRB for this T:





Hig  R * ICPH T0 , T , 1.702, 9.081 * 103 ,  2.164 * 106 , 0

H ig 

J mol

H   H ig  RTc HRB Tr , Pr    RTc HRB Tr 0 Pr 0  

H  

J mol

The actual enthalpy change from Eq. (7.17): 

H 

H   

J mol

The actual final temperature is now found from Eq. (6.72) combined with Eq (4.8), written:

    T  B D    1    Tc  HRB  * 0 , Pr ,    HRB Tr 0 , Pr 0 ,   H  R  AT0   1  * T02  2  1  *  2 T0     TC    

Using this to solve for  yields



T   * T0 

Tr 

The entropy change is:

       1    D P   S  R  Aln    B * T0      ln  SRB T P   SRB T P   r r   r 0 r 0   *  2 *   2 P   0 T0     

S  2.449

J mol K

Lastly the work required is

W  n H

n 

mol sec

W 

Solution 7.34 Problem Statement 1

Propylene is compressed adiabatically from 11.5 bar and 30°C to 18 bar at a rate of 1 kg mol · s . If the compressor efficiency is 0.8, what is the power requirement of the compressor, and what is the discharge temperature of the propylene?

Solution

As usual, we first consider the isentropic compression, find the work requirement, multiply that by the efficiency, and then use the energy balance to find the actual outlet conditions.

S  S R 303.15 K, 11.5 bar  

Tf, ientropic

303.15 K

 18  dT  Rln   S R Tf , isentropi, 18 bar  0  11.5  T

C igp





The third term is Rln(18/11.5) = 3.725 J/(mol K). If we treated the gas as ideal, then we would have simply:



Tf, ientropic, ig 303.15 K

 18  dT  R ln    3.725 J/(mol K)  11.5  T

C igp

Doing the heat capacity integral and varying the final temperature to achieve this value gives:

T0 (K) 303.15

T (K) 320.52

A 1.637

C (1/K2) D (K2) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

ICPS 0.448

DSig (J/(mol K)) 3.7250

The predicted outlet temperature for isentropic compression of an ideal gas is 320.5 K. Now, we can compute the residual entropy at the inlet conditions and at this estimated outlet condition: T (K) 303.15

Tr 0.8501

P (bar) 11.5

Pr 0.2465

Tc (K) 356.6

Pc (bar) 46.65

dB0 dTr

dB1 dTr

1.0296

1.6798

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.14

S /R -0.3118

-1

(J mol K ) -2.59

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output.

Solution continued on next page…

T (K) P (bar) 320.51 18

Tr 0.8988

Tc (K) 356.6

Pc (bar) 46.65

dB0 dTr

dB1 dTr

w 0.14

Pr 0.3859

0.8908

dB0 0.675  2.6 dTr Tr

1.2575

S /R -0.4117

-1

(J mol K ) -3.42

 dB 0 SR dB1    Pr     dT R dTr   r

dB1 0.722  5.2 dTr Tr

Pale blue boxes are input fields, pink boxes are the final output.

Using these in the entropy balance gives: 2.59  



Tf ,ientropic

C igp

303.15 K ig Tf ,ientropic p

303.15 K

dT  3.725  3.42  0

dT  4.555

Varying the outlet temperature to satisfy this gives: T0 (K) 303.15

T (K) 324.42

A 1.637

C (1/K ) D (K ) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

ICPS 0.548

DS (J/(mol K)) 4.5547

Re-evaluating the residual entropy at this temperature gives: T (K) 324.41

Tr 0.9097

P (bar) 18

Pr 0.3859

Tc (K) 356.6

Pc (bar) 46.65

dB0 dTr

dB1 dTr

0.8632

1.1808

dB0 0.675  2.6 dTr Tr dB1 0.722  5.2 dTr Tr

w 0.14

S /R -0.3969

-1

(J mol K ) -3.30

 dB 0 SR dB1    Pr     dT R dTr   r

Pale blue boxes are input fields, pink boxes are the final output.

Solution continued on next page…

Substituting this back into the entropy balance gives:



Tf ,ientropic

C igp

303.15 K

dT  4.435 J/(mol K)

Varying the outlet temperature to satisfy this gives: T

ICPS  T0 ,T;A,B,C,D  

 T  D  T 2  T02  dT  A ln    B T  T0    C  2 2    RT T0  2  T T0     T



T0 (K) 303.15

T (K) 323.86

A 1.637

C (1/K ) D (K ) B (1/K) 2.27E-02 -6.92E-06 0.00E+00

ICPS 0.533

 S (J/(mol K)) 4.4351

Further iteration does not give a significant change in the outlet temperature, so we can take the isentropic outlet temperature to be 323.9 K. Next, we apply the energy balance, assuming that the process is adiabatic and results in negligible change in kinetic or potential energy: Ws,isentropic  H isentropic  H R 303.15 K, 11.5 bar   

323.9 K 303.15 K

C pig dT  H R 323.9 K, 18 bar 

The three pieces of this evaluate to:

T (K) 303.15

Tr 0.8501

P (bar) 11.5

Pr 0.2465

B 0  0.083 

0.422

B1  0.139 

0.172

Tc (K) 356.6

B -0.4642

Pc (bar) 46.65

B -0.2012

dB0 0.675  2.6 dTr Tr

Tr1.6

dB1 0.722  5.2 dTr Tr

Tr4.2

w 0.14

dB0 dTr

dB1 dTr

1.0296

1.6798

H /RTc -0.3864

-1

(J mol ) -1145.75

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr      RTc dT dT r r   

Pale blue boxes are input fields, pink boxes are the final output. T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 303.15

T (K) 323.95

A 1.637

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 







C (1/K ) D (K ) B (1/K) 2.27E-02 -6.92E-06 0.00E+00



ICPH (K) 168.0

DHig (J/mol) 1396.8

Solution continued on next page…

T (K) 323.9

Tr 0.9083

P (bar) 18

Pr 0.3859

B 0  0.083 

0.422

B1  0.139 

0.172

Tc (K) 356.6

B -0.4092

Pc (bar) 46.65

B -0.1186

dB0 0.675  2.6 dTr Tr

Tr1.6

dB1 0.722  5.2 dTr Tr

Tr4.2

w 0.14

dB0 dTr

dB1 dTr

0.8668

1.1905

H /RTc -0.5265

-1

(J mol ) -1561.03

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

Adding these gives Ws,isentropic  Hisentropic  1145.8  1396.8  1561.0  981.5 J/mol Dividing this by the compressor efficiency using it in the energy balance gives Ws,actual  H actual  H R 303.15 K, 11.5 bar   

Tf, actual

303.15 K

C pig dT  H R Tf , actual , 18 bar   1226.9 J/mol

Now, we proceed by using the residual enthalpy at the initial state (which we’ve already calculated) approximating the residual enthalpy at the final state by the value we had for the isentropic final state, and finding the final temperature that satisfies the energy balance:

Ws, actual  Hactual  1145.8  



Tf, actual 303.15 K

Tf , actual 303.15 K

C igp dT 1561.0  1226.9 J/mol

C igp dT  1642.1 J/mol

Varying the final temperature in the heat capacity integral to satisfy this gives: T2

ICPH  T0 ,T;A,B,C,D  

R

dT  A T  T0  

T0(K) 303.15

T (K) 327.51

A 1.637

1 1  B 2 C 3 T  T02  T  T03  D    2 3  T T0 







C (1/K ) D (K ) B (1/K) 2.27E-02 -6.92E-06 0.00E+00



ICPH (K) 197.5

DHig (J/mol) 1642.1

Solution continued on next page…

Thus, our new best guess for the outlet temperature is 327.5 K. Evaluating the residual enthalpy at that temperature gives:

T (K) 327.51

Tr 0.9184

P (bar) 18

B -0.4006

Pr 0.3859

B 0  0.083 

0.422

B1  0.139 

0.172

Tc (K) 356.6

Pc (bar) 46.65

B -0.1069

dB1 0.722  5.2 dTr Tr

Tr4.2

dB0 dTr

dB1 dTr

0.8422

1.1239

dB0 0.675  2.6 dTr Tr

Tr1.6

w 0.14

H /RTc -0.5145

-1

(J mol ) -1525.54

  HR dB0 dB1    Pr  B 0  Tr    B1  Tr    RTc dTr dTr    

Pale blue boxes are input fields, pink boxes are the final output.

Substituting this back into the energy balance gives:

Ws ,actual  Hactual  1145.8  

Tf ,actual

303.15 K



Tf ,actual 303.15 K

C igp dT  1525.5  1226.9 J/mol

C igp dT  1606.6 J/mol

Varying the final temperature in the heat capacity integral to satisfy this gives: T2

ICPH  T0 ,T;A,B,C,D  

1

2 0

3 0

T0(K) 303.15

T (K) 326.99

1 

 R dT  AT  T   2 T  T   3 T  T   D  T  T  A 1.637

B (1/K) 2.27E-02

C (1/K2) -6.92E-06

D (K2) 0.00E+00

ICPH (K) 193.2

DHig (J/mol) 1606.6

So, our best estimate for the discharge temperature is 327 K = 53.9 °C. The work required is 1226.9 J/mol. If 1

the flow rate is 1 mol s , then the power requirement for the compressor is 1226.9 kJ/s = 1227 kW. At the outlet conditions, we were actually a bit outside the range where the Pitzer correlation works well. A slightly more reliable approach would have been to use the Lee/Kesler tables for the residual properties, though it would have been a little more tedious. Another, more practical, option for a well-studied substance like propylene is to use tabulated data like that in the NIST WebBook: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

At the initial state (303.15 K, 11.5 bar) the NIST fluid properties database gives S = 100.00 J/(mol K) and H = 25.606 kJ/mol. To analyze the isentropic compression, we find the temperature at which the entropy is 100.00 J/mol at a pressure of 18 bar. This turns out to be 326.16 K, where the enthalpy is 26.565 kJ/mol. Thus, the enthalpy change for isentropic compression would be 26.565 – 25.606 = 0.959 kJ/mol. This is about 2% lower than the value we obtained using the Pitzer correlation for the residual properties with the heat capacity integral. Dividing this by the given efficiency of 0.8 gives Ws,actual

Hactual = 1.199 kJ/mol,

and the corresponding work requirement for the compressor is 1.199 kJ/mol * 1000 mol/s = 1199 kW. The enthalpy of the gas leaving the compressor is 25.606 + 1.199 = 26.805 kJ/mol. At 18 bar, the enthalpy reaches this value at 329.1 K. The outlet temperature predicted using the NIST data is 2 K higher than that predicted using the Pitzer correlation and heat capacity integral.

Solution 7.35 Problem Statement Methane is compressed adiabatically in a pipeline pumping station from 3500 kPa and 35°C to 5500 kPa at a 1

rate of 1.5 kg mol · s . If the compressor efficiency is 0.78, what is the power requirement of the compressor and what is the discharge temperature of the methane?

Solution

Methane:

Pc 

P0 

bar

T0 

kPa

Tc 

S  0

P

kPa

K  J mol k

For the heat capacity of methane:

A

Tr 0 

T0  Tc

B

9.081 * 103 2.164 * 106 C K K2

Pr 0 

P0  Pc

Pr 

P  Pc

Use generalized second-virial correlation: The entropy change is given by Eq. (6.73) combined with Eq. (5.11) with D = 0:

   T      1  P 0 S  R  A  ln    B  T0  CT02        ln  SRB  Pr    SRB Tr 0 Pr 0    2   TC P0     



T   * T0

T

K Tr 

Solution continued on next page…

The enthalpy change for the final T is given by Eq. (6.72), with HRB for this T:





Hig  R * ICPH T0 , T , 1.702, 9.081 * 103 ,  2.164 * 106 , 0

H ig 

J mol

H   Hig  RTc HRB Tr , Pr    RTc HRB Tr 0 Pr 0  

H  

J mol

H   

J mol

W  n H 

The actual enthalpy change from Eq. (7.17):



H 

n 

mol s

The actual final temperature is now found from Eq. (6.72) combined with Eq (4.8), written:

    T  B C 0 H  R  AT0      T02  2    T03  3    Tc  HRB  Pr    HRB Tr 0 Pr 0   2 3  TC    

Using this to solve for  yields



T    T0 

Solution 7.36

Problem Statement What is the ideal work for the compression process of Ex. 7.9? What is the thermodynamic efficiency of the process? What are SG and W lost? Take T = 293.15 K.

Solution

From the data and results of Example 7.9,

P1 

kPa

P2 

kPa

J mol

W

Determine

T1 

T2 

T 

and

H  R  ICPH T1 T2 H  5288.196  S  R   ICPS T1 T2 

3







6



J mol

3





6

P2 

 ln P  1

S  3.366



J mol K

Solution continued on next page…

Since the process is adiabatic SG  S

SG  3.366 Wideal  H  T S

J mol K

Wlost  T S

Wlost 

Wideal W

t 

J mol

Wideal  J mol

Solution 7.37

Problem Statement A fan is (in effect) a gas compressor which moves large volumes of air at low pressure across small (1 to 15 kPa) pressure differences. The usual design equation is: R T1 W  n P  P1

where subscript 1 denotes inlet conditions and is the efficiency with respect to isentropic operation. Develop this equation. Show also how it follows from the usual equation for compression of an ideal gas with constant heat capacities.

Solution

Combine Eqs. (7.13) and (7.17):

W s  n H 

n H s 

H s   V dP  V P

By Eq. (6.8),

Assume now that P is small enough that V , an average value, can be approximated by V1 = RT1/P1. Then

H s  n

RT1 P P1

RT W s  n 1 P  P1

Equation (7.22) is the usual equation for isentropic compression of an ideal gas with constant heat capacities. For irreversible compression it can be rewritten:

W s 

R    pT1  P CP  nC  2    1      P1    

For P sufficiently small, the quantity in square brackets becomes:

   P CP  C  2   1  1  P  P  1  1  R P   1      P   P1  C p P1   1 The boxed equation is immediately recovered from this result.

Solution 7.38 Problem Statement For an adiabatic gas compressor, the efficiency with respect to isentropic operation is a measure of internal irreversibilities; so is the dimensionless rate of entropy generation

SG R



SG .  nR

Assuming that the gas is ideal with constant heat capacities, show that and SG /R are related through the expression: SG CP   1  ln( ) R R 

where   ( P2 /P1 )

R /CP

Solution

The equation immediately preceding Eq. (7.22) gives T '2  T1. With this substitution, Eq. (7.23) becomes:

T2  T1 

T1   T1 

   1    T1     

The entropy generation SG is simply S for the compression process, for which Eq. (5.14) may be rewritten: R

T P C T C P CP S C P   ln 2  ln 2  P  ln 2  P  ln 2 R R T1 P1 R T1 R P1

Combine these two equations:

   1    1       C C S  1      ln   P  ln  P ln1    R   R R           Which reduces to: SG R



     1    ln     R

Solution 7.39 Problem Statement Air at 1(atm) and 35°C is compressed in a staged reciprocating compressor (with intercooling) to a final pressure of 50(atm). For each stage, the inlet gas temperature is 35°C and the maximum allowable outlet temperature is 200°C. Mechanical power is the same for all stages, and isentropic efficiency is 65% for each 3

stage. The volumetric flow rate of air is 0.5 m ·s

at the inlet to the first stage.

(a) How many stages are required? (b) What is the mechanical-power requirement per stage? (c) What is the heat duty for each intercooler? (d) Water is the coolant for the intercoolers. It enters at 25°C and leaves at 45°C. What is the cooling-water rate per intercooler?

Assume air is an ideal gas with CP = (7/2)R.

Solution

P1  atm

P2 

atm

T1 

CP 

T2 

V

RT1 P1

n 

V V



V 

m3 s

mol s

n 

With compression from the same initial conditions (P1,T1) to the same final conditions (P2,T2) in each stage, the same efficiency in each stage, and the same power delivered to each stage, the applicable equations are:

 P N r   2  (where r is the pressure ratio in each stage and N is r the number of stages.)  P1 

Eq. (7.23) may be solved for T’2:

T 2  T2  T1   T1   

T2

Eq. (7.18) written for a single stage is:

 P  N CP T '2  T1  2   P  1

Solve this for N

P  ln 2   P1  R N CP T '2  ln    T1 

N

Part(a): Although any number of stages greater than this would serve, design for 4 stages. Solution continued on next page…

 P N Part(b): Calculate r for 4 stages: N = 4 r   2   P1 

r

Power requirement per stage follows from Eq. (7.22). In kW/stage:

 R    P T1 r CP  1 nC      Wr  

W r 

Part(c): Because the gas (ideal) leaving the intercooler and the gas entering the compressor are at the same temperature (308.15 K), there is no enthalpy change for the compressor/interchanger system, and the first law yields:

Q r  W r

Q r  

Heat duty = 87.94 kW/interchanger

Part(d): Energy balance on each interchanger (subscript w denotes water):

With data for saturated liquid water from the steam tables:

H W  





kJ kg

HW 

kJ kg

Solve for the cooler water rate: m W 

Q r H W

m W 

kg s

Solution 7.40

Problem Statement Demonstrate that the power requirement for compressing a gas is smaller the more complex the gas. Assume fixed values of n , , T1, P1, and P2, and that the gas is ideal with constant heat capacities.

Solution

The relevant fact here is that CP increases with increasing molecular complexity. Isentropic compression work on a mole basis is given by Eq. (7.22), which can be written: R

Ws  CPT1   

 P CP    2   P1 

This equation is a proper basis, because compressor efficiency and flowrate

are fixed. With all other

variables constant, differentiation yields:  dWs d   T1     CP dCP dCP  

From the definition of ,

ln  

Then,

P R ln 2 CP P1

P d ln  d R    2 ln 2 dCP  dC P P1 CP

d R P   2 ln 2 dCP CP P1

Solution continued on next page…

And

 dWs R P2   T1    ln   T1    ln    dCP C P P1 

When  = 1, the derivative is zero; for  > 1, the derivative is negative (try some values). Thus, the work of compression decreases as Cp increases and as the molecular complexity of the gas increases.

Solution 7.41

Problem Statement

Tests on an adiabatic gas compressor yield values for inlet conditions (T1, P1) and outlet conditions (T2, P2). Assuming ideal gases with constant heat capacities, determine the compressor efficiency for one of the following:

(a) T1 = 300 K, P1 = 2 bar, T2 = 464 K, P2 = 6 bar, CP /R = 7/2.

(b) T1 = 290 K, P1 = 1.5 bar, T2 = 547 K, P2 = 5 bar, CP /R = 5/2.

(d) T1 = 300 K, P1 = 1.1 bar, T2 = 505 K, P2 = 8 bar, CP /R = 11/2.

(e) T1 = 305 K, P1 = 1.5 bar, T2 = 496 K, P2 = 7 bar, CP /R = 4.

Solution

This is one can easily be done in excel. Use the given values to determine

H  CP T2  T1 

R      P CP   2   H S  CP T1       P     1     

s, and



H (J/mol)

H S H

Part

T1 (K)

P1 (bar)

T2 (K)

P2 (bar)

Hs (J/mol)

300

464

29.099

4772.236

3218.973

0.675

290

1.5

547

20.785

5341.745

3728.973

0.698

295

1.2

455

37.413

5986.08

4745.499

0.793

300

1.1

505

45.727

9374.035

5959.231

0.636

305

1.5

496

33.256

6351.896

4764.994

0.750

Solution 7.42

Problem Statement Air is compressed in a steady-flow compressor, entering at 1.2 bar and 300 K and leaving at 5 bar and 500 K. Operation is nonadiabatic, with heat transfer to the surroundings at 295 K. For the same change in state of the air, is the mechanical-power requirement per mole of air greater or less for nonadiabatic than for adiabatic operation? Why?

Solution

The appropriate energy balance can be written: W  H  Q Since Q is negative (heat transfer is out of the system), the work of non-adiabatic compression is greater than for adiabatic compression, where Q = 0. Note that in order to have the same change in state of the air, i.e., the same H, the irreversibilities of operation would have to be quite different for the two cases.

Solution 7.43

Problem Statement A boiler house produces a large excess of low-pressure [50(psig), 5(°F)-superheat] steam. An upgrade is proposed that would first run the low-pressure steam through an adiabatic steady-flow compressor, producing medium-pressure [150(psig)] steam. A young engineer expresses concern that compression could result in the formation of liquid water, damaging the compressor. Is there cause for concern? Suggestion: Refer to the Mollier diagram of Fig. F.3 of App. F.

Solution

There is in fact no cause for concern, as adiabatic compression sends the steam further into the superheat region. As long as it stays in the superheated region it will remain there. Once it reaches a point where it is saturated then problems can arise.

Solution 7.44

Problem Statement A pump operates adiabatically with liquid water entering at T1 and P1 with a mass flow rate m . The discharge pressure is P2, and the pump efficiency is

For one of the following sets of operating conditions, determine

the power requirement of the pump and the temperature of the water discharged from the pump. 1

(a) T1 = 25°C, P1 = 100 kPa, m = 20 kg · s , P2 = 2000 kPa, = 0.75, = 257.2 × 10

(b) T1 = 90°C, P1 = 200 kPa, m = 30 kg · s , P2 = 5000 kPa, = 0.70, = 696.2 × 10 1

K .

(d) T1 = 70(°F), P1 = 1(atm), m = 50(lbm)(s) , P2 = 20(atm), = 0.70, = 217.3 × 10 1

K .

(e) T1 = 200(°F), P1 = 15(psia), m = 80(lbm)(s) , P2 = 1500(psia), = 0.75, = 714.3 × 10

K .

Solution

(a) If the pump operated reversibly and therefore the process was isentropic (since it is also adiabatic) then we could write

T  S  C p ln  2   V P  0  T1  Solution continued on next page…

Taking , Cp, and V to be approximately constant, we can get the isentropic outlet temperature as  V P    T2  T1 exp  C p  3

For water at 25 ºC, we have Cp = 4184 J/(kg K) and V = 0.001003 m /kg. With the value of  given in the problem statement, we get

 257.2 106 K 1  0.001003 m3 kg1  1900 kPa    298.185 T2  298.15exp   4.184 kJ kg1 K1  The temperature rise of 0.035 K is entirely negligible.

The enthalpy change for isentropic compression is then H  C p T  V 1  T P

H  4.184 * 0.035  0.001003 * 1  257.2 106 * 298.15 * 1900  1.906 kJ kg 1

So, the work for isentropic pumping would be W

H = 1.906 kJ kg . If, instead of including  in the

calculation, we had just assumed an incompressible fluid, we would have had W = V P = 0.001003*1900 = 1.906 kJ kg

(agreeing with the more rigorous result to 4 digits, and thus

showing that compressibility is completely negligible here). If the pump efficiency is = 0.75, then the actual work is W = 1.906/0.75 = 2.541 kJ/kg. For a total flow of 20 kg/s, the power requirement is 20*2.541 = 50.82 kJ/s = 50.8 kW. The temperature rise can then be found from the energy balance:

H  C pT  V 1  T P  2.541 T 

2.541  0.001003 * 1  257.2106 * 298.15 * 1900 4.184

 0.187 K

The expected actual outlet temperature is therefore T2 = 298.34 K.

Solution continued on next page…

(b) If the pump operated reversibly and therefore the process was isentropic (since it is also adiabatic) then, taking , Cp, and V to be approximately constant, we could write

T  S  C p ln  2   V P  0  T1  Thus, we can get the isentropic outlet temperature as  V P    T2  T1 exp  C p  For water at 90 ºC and 200 kPa (from the NIST WebBook), we have Cp = 4205 J/(kg K) and 3

V = 0.001036 m /kg. Using these with the value of  given in the problem statement, we get

 696.2106 K1 * 0.001036 m 3 kg1 * 4800 kPa    363.45 T2  363.15 exp  4.205 kJ kg1 K 1  The temperature rise of 0.30 K is probably negligible. The enthalpy change for isentropic compression is then H  C p T  V 1  T P

H  4.205 * 0.30  0.001036 * 1  696.2 106 * 363.15 * 4800  4.976 kJ kg 1

So, the work for isentropic pumping would be W

H = 4.976 kJ kg . If, instead of including  in the

calculation, we had just assumed an incompressible fluid, we would have had W = V P = 0.001036*4800 = 4.973 kJ kg

(agreeing with the more rigorous result to within 0.1%, and thus

showing that compressibility is completely negligible here). If the pump efficiency is = 0.70, then the actual work is W = 4.976/0.70 = 7.108 kJ/kg. For a total flow of 30 kg/s, the power requirement is 30*7.108 = 213.2 kJ/s = 213.2 kW. The temperature rise can then be found from the energy balance: H  C p T  V 1  T P  7.108 T 

7.108  0.001036 * 1  696.2 106 * 363.15 * 4800 4.205

 0.807 K

The expected actual outlet temperature is therefore T2 = 363.96 K. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(c) If the pump operated reversibly and therefore the process was isentropic (since it is also adiabatic) then, taking , Cp, and V to be approximately constant, we could write

T  S  C p ln  2   V P  0  T1  Thus, we can get the isentropic outlet temperature as  V P    T2  T1 exp  C p  For water at 60 ºC, the heat capacity polynomial with coefficients from table C.3 gives Cp = 75.73 J/(mol K) = 4204 J/(kg K). We can take the volume to be that of the saturated liquid at 60 °C, 3

where the vapor pressure is 19.92 kPa. For those conditions V = 0.001017 m /kg. Using these with the value of  given in the problem statement, we get  523.110 6 K1 * 0.001017 m3 kg1 * 4980 kPa    333.36 1 1  4.204 kJ kg K 

T2  333.15 exp

The temperature rise of 0.21 K is probably negligible. The enthalpy change for isentropic compression is then H  C p T  V 1  T P

H  4.204 * 0.21  0.001017 * 1  523.1106 * 333.15  * 4980  5.065 kJ kg 1

So, the work for isentropic pumping would be W

H = 5.065 kJ kg . If, instead of including  in the

calculation, we had just assumed an incompressible fluid, we would have had W = V P = 0.001017*4980 = 5.065 kJ kg

(agreeing with the more rigorous result to four digits, and thus

showing that compressibility is completely negligible here, at least for calculating work).

Solution continued on next page…

If the pump efficiency is = 0.75, then the actual work is W = 5.065/0.75 = 6.753 kJ/kg. For a total flow of 15 kg/s, the power requirement is 15*6.753 = 101.3 kJ/s = 101.3 kW. The temperature rise can then be found from the energy balance:

H  C pT  V 1  T P  6.753 T 

6.753  0.001017 * 1  523.1106 * 333.15 * 4980 4.204

 0.612 K

The expected actual outlet temperature is therefore T2 = 333.76 K.

(d) If the pump operated reversibly and therefore the process was isentropic (since it is also adiabatic) then, taking , Cp, and V to be approximately constant, we could write

T  S  C p ln  2   V P  0  T1  Thus, we can get the isentropic outlet temperature as  V P    T2  T1 exp  C p  For water at 70 ºF = 294.26 K = 21.11 °C, the heat capacity polynomial with coefficients from table C.3 gives Cp = 75.36 J/(mol K) = 4182 J/(kg K). We can take the volume to be that of the saturated liquid at 70 ºF, 3

where the vapor pressure is 0.3629 psia = 2.501 kPa. For those conditions V = 0.010605 ft /lbm = 0.001002 3

m /kg. Using these with the value of  given in the problem statement, we get

 217.3106 K1 * 0.001002 m 3 kg1 * 1925 kPa    294.29 K T2  294.26 exp  4.182 kJ kg1 K1 

The temperature rise of 0.029 K is negligible. The enthalpy change for isentropic compression is then H  C p T  V 1  T P

H  4.182 * 0.029  0.001002 * 1  217.3106 * 294.26 * 1925  1.927 kJ kg 1

So, the work for isentropic pumping would be W

H = 1.931 kJ kg . If, instead of including  in the

calculation, we had just assumed an incompressible fluid, we would have had 1

W = V P = 0.001002*1925 = 1.929 kJ kg (showing that compressibility is completely negligible here, at least for calculating work).

If the pump efficiency is = 0.70, then the actual work is W = 1.927/0.70 = 2.753 kJ/kg. For a total flow of 50 lbm/s = 22.68 kg/s, the power requirement is 22.68*2.753 = 62.4 kJ/s = 62.4 kW. The temperature rise can then be found from the energy balance:

H  C pT  V 1  T P  2.753 T 

2.753  0.001002 * 1  217.3106 * 294.26 * 1925 4.182

 0.226 K

The expected actual outlet temperature is therefore T2 = 294.49 K = 70.4 °F.

(e) If the pump operated reversibly and therefore the process was isentropic (since it is also adiabatic) then, taking , Cp, and V to be approximately constant, we could write

T  S  C p ln  2   V P  0  T1  Thus, we can get the isentropic outlet temperature as  V P    T2  T1 exp  C p  For water at 200 ºF = 366.5 K the heat capacity polynomial with coefficients from table C.3 gives Cp = 76.04 J/(mol K) = 4221 J/(kg K). We can take the volume to be that of the saturated liquid at 200 ºF, where the vapor pressure is 11.526 psia = 79.47 kPa. For those conditions 3

V = 0.01664 ft /lbm = 0.001039 m /kg. Using these with the value of  given in the problem statement, we get Solution continued on next page…

 714.3106 K 1 * 0.001039 m3 kg1 * 10239 kPa    367.16 K T2  366.5 exp  4.221 kJ kg1 K1  The temperature rise of 0.66 K is pretty minor. The enthalpy change for isentropic compression is then H  C p T  V 1  T P

H  4.221 * 0.66  0.001039 * 1  714.3106 * 366.5 * 10239  10.64 kJ kg 1

So, the work for isentropic pumping would be W

H = 10.64 kJ kg . If, instead of including  in the

calculation, we had just assumed an incompressible fluid, we would have had 1

W = V P = 0.001039*10239 = 10.64 kJ kg (showing that compressibility is completely negligible here, at least for calculating work).

If the pump efficiency is = 0.75, then the actual work is W = 10.64/0.75 = 14.19 kJ/kg. For a total flow of 80 lbm/s = 36.29 kg/s, the power requirement is 36.29*14.19 = 515.0 kJ/s = 515 kW. The temperature rise can then be found from the energy balance:

H  C pT  V 1  T P  14.19 T 

14.19  0.001039 * 1  714.3106 * 366.5 * 10239 4.221

 1.50 K

The expected actual outlet temperature is therefore T2 = 368.0 K = 202.7 °F.

Solution 7.45

Problem Statement What is the ideal work for the pumping process of Ex. 7.10? What is the thermody-namic efficiency of the process? What is SG? What is Wlost? Take T = 300 K.

Solution

Results from Example 7.10: H 

kJ kg

W

Wideal  H  T S Wideal 

kJ kg K

S 

t 

kJ kg K

T 

Wideal W

t 

Since the process is adiabatic, SG is just the change in entropy SG  



3

kJ kg K

And the work lost is Wlost  T S  2.7

kJ kg

Solution 7.46 Problem Statement Show that the points on the Joule/Thomson inversion curve [for which = (T/P)H = 0] are also characterized by each of the following:  Z  (a)    0;  T  P

 H  (b)    0;  P  T

 V  V (c)  P  ;  T  T  Z  (d)    0;  V  P

 P   P  (e) V    T    0  V   T  T

Solution

Part(a):  

RT 2  Z    CP P  T P

 Z      T 



 H  RT 2  Z  Part(b):        P  P  T P T

 H      P 

 H   V  Part(c):    V  T    P   T 

 V  V    T  T 

  

RT 2  Z    P  T P

 H  V     V  T     P   T 

 Z   Z   T  Part(d):        but by part (a) this is 0  V   T   V  P P P

Part(e): rearranging the given equation yields:  P     T 

 V   P   V  V V                T  P T  T V  T P  P   V  T

This leaves the solution exactly the same as part (c).

Solution 7.47

Problem Statement According to Prob. 7.3, the thermodynamic sound speed c depends on the PVT equation of state. Show how isothermal sound-speed measurements can be used to estimate the second virial coefficient B of a gas. Assume that Eq. (3.36) applies, and that the ratio CP /CV is given by its ideal-gas value. Eq. 3.36: Z

PV BP 1 RT RT

Problem 7.3 The thermodynamic sound speed c is defined in Sec. 7.1. Prove that: c

V CP

 CV 

where V is molar volume and is molar mass. To what does this general result reduce for: (a) an ideal gas? (b) an incompressible liquid? What do these results suggest qualitatively about the speed of sound in liquids relative to gases? Solution From the result of Pb. 7.3:

c

1  V       V  P T

V CP 1  * M CV 

With V

RT B P

 V  RT    2  P  P



CP CV

This leads to c  PV

   BP   RT   RT  BP       MRT MRT RT  M 

c

 RT B  M RT

 RT *P M

A value for B at temperature T may be extracted from a linear fit of c vs. P.

Solution 7.48 Problem Statement Real-gas behavior for turbomachinery is sometimes empirically accommodated through the expression W  Z W ig, where W ig is the ideal-gas mechanical power and Z is some suitably defined average value of the compressibility factor. (a) Rationalize this expression. (b) Devise a turbine example incorporating real-gas behavior via residual properties, and determine a numerical value of Z for the example.

Solution Part(a): On the basis of Eq. (6.8), write: RT dP P ZRT H S   V dP   dP P H Sig   V ig dP  

ZRT dP P   Z RT H Sig  P dP



H S

By extension, and with equal turbine efficiencies, H S H

ig S



W  Z W ig

Part(b): For this problem, you can use the lee/keseler correlation to determine a Z for a real gas. Let’s say we use the following values for water that is being isothermally expanded at T

P2 

Pr 2 

P1 

bar

Tr 

bar

Pr 1 

Then our Z values for this are

Z0 

Z1 

With   0.345, Z then is

Z  Z0  Z1  Use this Z for the equation in part A to calculate Z P  ZRT * ln  2   P1  Z  P  RT * ln 2   P  1

Z  0.535

Solution 7.49

Problem Statement Operating data are taken on an air turbine. For a particular run, P1 = 8 bar, T1 = 600 K, and P2 = 1.2 bar. However, the recorded outlet temperature is only partially legible; it could be T2 = 318, 348, or 398 K. Which must it be? For the given conditions, assume air to be an ideal gas with constant CP = (7/2)R.

Solution By Eq. (7.16), H   H S

For CP = constant, T2  T1   T2 S  T1   

For an ideal gas with constant CP, T2 S is related to T1 by: R

 P CP T2 S  T1  2   P1 

Combine the last two equations, and solve for T2: R     P CP    T2  T1 1    2   1  P     1  

From which



T2 1 T1 R

 P CP  2   1  P  1

Results: For T2 = 318 K, = 1.123 For T2 = 348 K, = 1.004 For T2 = 398 K, = 0.805 Only T2 = 398 K is possible since efficiency must be less than 100%

Solution 7.50

Problem Statement Liquid benzene at 25°C and 1.2 bar is converted to vapor at 200°C and 5 bar in a two-step steady-flow process: compression by a pump to 5 bar, followed by vaporization in a counterflow heat exchanger. Determine the 1

power requirement of the pump and the duty of the exchanger in kJ·mol . Assume a pump efficiency of 70%, 1

and treat benzene vapor as as an ideal gas with constant CP = 105 J·mol ·K . Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

T1 

P1 

bar P2  bar T3 

kJ mol



H lv 

P3  bar

C pv 

J mol K

First estimate the specific volume of liquid benzene using the rackett equation: Tc 

Zc 

cm 3 mol

Vc 

Tn 

Trn 

Tn Tc



Assume that Vliq  Vsat : 0.2857

V  VC ZC

1Trn 

cm3 mol



Calculate the pump power Ws 

V  P2  P1  

 0.053

kJ mol

Assume that no temperature change occurs during the liquid compression. This means T2  T1 Estimate the saturation temperature at P = 5 bar using the Antoine Equation and values from Table B.2 For benzene from Table B.2: A 

B

Tsat 

B C A  lnP2

C Tsat 

C

Estimate the heat of vaporization using Watson’s method, obtain the heat of vaporization starting at the boiling point. H lv  30.72

Tr 1 

353.15 K  Tc

kJ mol Tr 2 

1  T  r2  Hlv 2  H lv    1  Tr 1 

Tsat Tc



0.38

 26.810

kJ mol

Solution continued on next page…

Calculate the heat exchanger duty: Q  R * ICPH T2 , Tsat , 0.747, 67.96 * 103 , 37.78 * 106   H lv 2  C p T3  Tsat  Q  51.1

kJ mol

Solution 7.51

Problem Statement Liquid benzene at 25°C and 1.2 bar is converted to vapor at 200°C and 5 bar in a two-step steady-flow process: vaporization in a counterflow heat exchanger at 1.2 bar, followed by compression as a gas to 5 bar. Determine 1

the duty of the exchanger and the power requirement of the compressor in kJ · mol . Assume a compressor 1

efficiency of 75%, and treat benzene vapor as as an ideal gas with constant CP = 105 J · mol · K .

Solution

T1 

P1 

bar P2 

bar T3 

P3  bar

C pv 

J mol K

  0.75

Calculate the compressor inlet temperature. T2 

R    1  P3 C PV 1     1       P2    

T2 

K or

Calculate the compressor power: Ws  CP T3  T2   6.834

kJ mol

Calculate the heat exchanger duty. Note that the exchanger outlet temperature, T2, is equal to the compressor inlet temperature. The benzene enters the exchanger as a subcooled liquid. In the exchanger the liquid is first heated to the saturation temperature at P1, vaporized and finally the vapor is superheated to temperature T2. Estimate the saturation temperature at P = 1.2 bar using the Antoine Equation and values from Table B.2 Solution continued on next page…

For benzene from Table B.2:

A

B

Tsat 

B C A  lnP2

C

Tsat 

C

Estimate the heat of vaporization using Watson’s method, obtain the heat of vaporization starting at the boiling point. H lv  30.72

TC 

Tr 1 

kJ mol

353.15 K  Tc

 1  T  r2  Hlv 2 Hlv   1  Tr 1 

Tr 2 

0.38



Tsat Tc



kJ mol

Calculate the heat exchanger duty: Q  R * ICPH T2 , Tsat,  0.747, 67.96 * 103,  37.78 * 106   H lv 2  C p T3  Tsat Q  44.393

kJ mol

Solution 7.52

Problem Statement Of the processes proposed in Probs. 7.50 and 7.51, which would you recommend? Why?

Problem 7.50 Liquid benzene at 25°C and 1.2 bar is converted to vapor at 200°C and 5 bar in a two-step steady-flow process: compression by a pump to 5 bar, followed by vaporization in a counterflow heat exchanger. Determine the 1

power requirement of the pump and the duty of the exchanger in kJ · mol . Assume a pump efficiency of 70%, 1

and treat benzene vapor as as an ideal gas with constant CP = 105 J · mol · K .

Problem 7.51 Liquid benzene at 25°C and 1.2 bar is converted to vapor at 200°C and 5 bar in a two-step steady-flow process: vaporization in a counterflow heat exchanger at 1.2 bar, followed by compression as a gas to 5 bar. Determine 1

the duty of the exchanger and the power requirement of the compressor in kJ · mol . Assume a compressor 1

efficiency of 75%, and treat benzene vapor as as an ideal gas with constant CP = 105 J · mol · K .

Solution

The proposal of Pb. 7.50, i.e., pumping of liquid followed by vaporization. The reason is that pumping a liquid is much less expensive than vapor compression.

Solution 7.53

Problem Statement Liquids (identified below) at 25°C are completely vaporized at 1(atm) in a countercurrent heat exchanger. Saturated steam is the heating medium, available at four pressures: 4.5, 9, 17, and 33 bar. Which variety of steam is most appropriate for each case? Assume a minimum approach T of 10°C for heat exchange. (a) Benzene; (b) n-Decane; (c) Ethylene glycol; (d) o-Xylene

Solution

What is required here is the lowest saturated steam temperature that satisfies the T constraint. Data from Tables E.2 and B.2 lead to the following: part (a): Benzene/4.5 bar part (b): n-Decane/17 bar part (c): Ethylene glycol/33 bar part (d): o-Xylene/9 bar Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 7.54

Problem Statement One hundred (100) kmol · h

of ethylene is compressed from 1.2 bar and 300 K to 6 bar by an electric-motor-

driven compressor. Determine the capital cost C of the unit. Treat ethylene as an ideal gas with constant 1

CP = 50.6 J · mol · K .

(compressor) = 0.70 C(compressor)/$ = 3040 (W S /kW ) 0.952 where W S  isentropic power requirement for the compressor. 0.855

C(motor)/$ = 380( W e /kW)

where W e  delivered shaft power of motor.

Solution

T1 

P1 

kmol hr

bar P2  bar n 



Cp 

J mol K

Assuming the compressor is adiabatic, solve for T2 R

 P CP T2  T1  2   P1 

T2 

 P T2  T1  W s  nC

W s 

W W e  s  

K kW kW

With the work calculated determine the capital cost for each unit  W  Ccompressor  $3040  s   kW 

0.952

 W  Cmotor  $380  e   kW 

 $307,451.24

0.855

 $32,572.26

Solution 7.55

Problem Statement Four different types of drivers for gas compressors are: electric motors, gas expanders, steam turbines, and internal-combustion engines. Suggest when each might be appropriate. How would you estimate operating costs for each of these drivers? Ignore such add-ons as maintenance, operating labor, and overhead.

Solution

Electric motors can be used as a driver for a gas compressor when electricity is available or in an environment that may be more hazardous for an internal combustion engine. Operating cost for this can be estimated by determine the amount of energy that it takes to compress the gas and the efficiency of the motor.

Steam turbines are used usually to expand steam or gas to provide energy or work to move something. An example of this is using a turbine to drive conveyors to move product down a line. The operating cost for this can be estimated by knowing the pressure of the steam coming into it and the pressure coming out and how efficient the turbine is.

Gas expander is similar to the steam turbine but can be used to drive something with a high pressure gas instead of steam. Typically natural gas or another fuel can be used here. The operating cost would depend on the pressure of the gas coming in, what pressure it is expanded to, and the efficiency of the expander.

Internal combustion engine uses a combustion of a fuel to compress gas, combust it, and repeat. A car is a good example of this. It is typically used to provide work in the form of movement, such as provide power to rotate the wheels of a car. The operating costs would depend on how much the fuel costs, how much it needs to be compressed, and the amount of work needed to drive the process, as well as the efficiency of the engine. On a side note, typically combustion engines are the least efficient when it comes to the rest of these types of drivers.

Solution 7.56

Problem Statement Two schemes are proposed for the reduction in pressure of ethylene gas at 375 K and 18 bar to 1.2 bar in a steady-flow process: (a) Pass it through a throttle valve. (b) Send it through an adiabatic expander of 70% efficiency. For each proposal, determine the downstream temperature, and the rate of entropy generation in 1

J · mol · K . What is the power output for proposal (b) in kJ · mol ? Discuss the pros and cons of the two proposals. Do not assume ideal gases.

Solution

T1 

P1 

bar P2 

bar

The characteristic properties for ethylene are

Tr 1 

A



TC 

T1  TC

Pr 1 

B

K PC 

P1  PC

3

bar Pr 2 

C 

P2  PC

6

D0

Part (a): For throttling process, assume the process is adiabatic. Find T2 such that H = 0. H  CP , mig T2  T1   HR2  HR1 Use the MCPH function to calculate the mean heat capacity and the HRB function for the residual enthalpy.

T  J  MCPH T1 , T2 , A, B, C , D * R T3  Tsat   RTc HRB  2 , Pr 2 ,    RTc HRB Tr 1 , Pr 1 ,   TC mol 

T2 

Calculate change in entropy using Eq. (6-94) along with MCPS function for the mean heat capacity and SRB function for the residual entropy.  T   P  S   R * MCPS T1 , T2 , A, B, C , D * ln  2   R * ln  2   R * SRB Tr 2 , Pr 2 ,   R * SRB Tr 1 , Pr 1 ,   T1   P1   Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

J mol K

S 

Part(b);

= 0.7

First find T2 for isentropic expansion. Solve Eq. (6-94) with S = 0.

 T   P  J  R * MCPS T1 , T2 , A, B, C, D * ln 2   R * ln  2   R * SRB Tr 2 , Pr 2 ,    R * SRB Tr 1 , Pr 1 ,  mol K   T1   P1 

T2 

Tr 2 

Now calculate the isentropic enthalpy change, HS. H S  R * MCPH T1 , T2 , A, B, C, D * R T2  T1   RTc HRB Tr 2 , Pr 2 ,    RTc HRB Tr 1 , Pr 1 ,   J mol Calculate actual enthalpy change using the expander efficiency. H S  

J mol

H   H S 

Find T2 such that H matches the value above. 4709.88

J  MCPH T1 , T2 , A, B, C , D * R T2  T1   RTc HRB Tr 2 , Pr 2 ,    RTc HRB Tr 1 , Pr 1 ,   mol

T2 

Now recalculate S at calculated T2.  T   P  S   R * MCPS T1 , T2 , A, B, C , D * ln  2   R * ln  2   R * SRB Tr 2 , Pr 2 ,   R * SRB Tr 1 , Pr 1 ,   T1   P1  

S 

J mol K

Calculate power produced by expander kJ mol The advantage of the expander is that power can be produced in the expander which can be used in the plant. The disadvantages are the extra capital and operating cost of the expander and the low temperature of the gas leaving the expander compared to the gas leaving the throttle valve. P   * H

P 

Solution 7.57 Problem Statement A stream of hydrocarbon gas at 500°C is cooled by continuously combining it with a stream of light oil in an adiabatic tower. The light oil enters as a liquid at 25°C; the combined stream leaves as a gas at 200°C. (a) Draw a carefully labeled flow diagram for the process. (b) Let F and D denote respectively the molar flow rates of hot hydrocarbon gas and light oil. Use data given below to determine a numerical value for the oil-to-gas ratio D/F. Explain your analysis. (c) What is the advantage to quenching the hydrocarbon gas with a liquid rather than with another (cooler) gas? Explain. 1

Data: CPv (ave) = 150 J · mol · K

CPv (ave) = 200 J · mol · K

H lv (oil) = 35,000 J · mol

for the hydrocarbon gas.

for the oil vapor. at 25°C.

Solution

Part(a):

Hydrocarbon gas stream: T 

Light oil stream: C p oil 

J mol K

C CP gas  H lv 

J T  mol K

J mol

Solution continued on next page…

Combined Stream: T3  200 C

Part(b): Assume that the oil vaporizes at 25 C. For an adiabatic column, the overall energy balance is as follows.

FC P gas T  T   D Hlv  CP oil T  T   0   Solving for the D/F ratio yields:

 CP gas T  T  D    H  C T  T  F lv P oil  

D  F

Part (c): Using liquid oil to quench the gas stream requires a smaller oil flow rate. This is because a significant portion of the energy lost by the gas is used to vaporize the oil.

Solution 8.1 Problem Statement The basic cycle for a steam power plant is shown in Fig. 8.1. The turbine operates adiabatically with inlet steam at 6800 kPa and 550°C, and the exhaust steam enters the condenser at 50°C with a quality of 0.96. Saturated liquid water leaves the condenser and is pumped to the boiler. Neglecting pump work and kinetic- and potential-energy changes, determine the thermal efficiency of the cycle and the turbine efficiency.

Solution

With reference to Fig. 8.1, SI units, Point 2: enthalpy and entropy are H2 

kJ S  kg 2

kJ kg K

Point 4: enthalpy is H 4  209.3

kJ kg

Point 1: H1  H 4

Solution continued on next page…

Point 3: going into the condenser kJ x  kg 3

Hliq  H 4 Hlv 

H3  H liq  x3H lv H3 

kJ kg

kJ Slv  kg K

kJ kg K

Sliq 

For isentropic expansion:

S 3  S2 x 3 

S 3  Sliq Slv

x 3  kJ kg

H 3  Hliq  x 3H lv H 3 

So the turbine efficiency is turbine 

H3  H 2   H 3  H2 turbine

Now for the efficiency on the cycle Ws  H3  H 2 Ws  

kJ kg

QH  H2  H1 QH  3.322 * 103 kJ/kg cycle 

Ws QH

cycle 

Solution 8.2 Problem Statement A Carnot engine with H2O as the working fluid operates on the cycle shown in Fig. 8.2. The H2O circulation 1

rate is 1 kgs . For TH = 475 K and TC = 300 K, determine: (a) The pressures at states 1, 2, 3, and 4. (b) The quality xv at states 3 and 4.

(c) The rate of heat addition. (d) The rate of heat rejection. (e) The mechanical power for each of the four steps. ( f ) The thermal efficiency of the cycle.

Solution First find the property values for the first 4 states by interpolation from table E.1: State 1: Sat liquid at TH: H1 

kJ S  kg 1

kJ P  kg K 1

bar

State 2: Sat liquid at TH: H 2 

kJ S  kg 2

kJ P  kg K 2

bar

State 3: Sat liquid at TC: H liq 

kJ H  kg vap

kJ P  kg K 1

bar

State 4: Sat liquid at TC: Sliq 

kJ S  kg vap

kJ P  kg K 1

bar

Part(a): The pressures are in bar above. Part(b): Steps 2-3 and 4-1 (Fig. 8.2) are isentropic, for which S3 = S2 and S1 = S4. Thus by Eq. 6.96a):

x3 

S2  Sliq Svap  Sliq

x3 

x4 

S1  Sliq

x4 

Svap  Sliq

Part(c): The rate of heat addition, Step 1-2: Q 12  m  H 2  H1 Q 12 

kJ s

Part(d): The rate of heat rejection, Step 3-4:

H3  Hliq  x3  H vap  Hliq  H 4  Hliq  x 4  Hvap  Hliq  H3 

kJ H  kg 4

Q 34  m  H 4  H3 Q 34  

kJ kg

kJ s

Part(e): The mechanical power for each step: W 23  m  H 3  H2 W 23   W 41  m  H1  H 4 W 41 

kJ s kJ s

Part(f): The thermal efficiency of the cycle 

W 23  W 41  Q 12

Note this matches Carnot efficiency:

   TC TH   

Solution 8.3 Problem Statement A steam power plant operates on the cycle of Fig. 8.4. For one of the following sets of operating conditions, determine the steam rate, the heat-transfer rates in the boiler and condenser, and the thermal efficiency of the plant.

(a) P1 = P2 = 10,000 kPa; T2 = 600°C; P3 = P4 = 10 kPa; (turbine) = 0.80; (pump) = 0.75; power rating = 80,000 kW. (b) P1 = P2 = 7000 kPa; T2 = 550°C; P3 = P4 = 20 kPa; (turbine) = 0.75; (pump) = 0.75; power rating = 100,000 kW. (c) P1 = P2 = 8500 kPa; T2 = 600°C; P3 = P4 = 10 kPa; (turbine) = 0.80; (pump) = 0.80; power rating = 70,000 kW. (d) P1 = P2 = 6500 kPa; T2 = 525°C; P3 = P4 = 101.33 kPa; (turbine) = 0.78; (pump) = 0.75; power rating = 50,000 kW. (e) P1 = P2 = 950(psia); T2 = 1000(°F); P3 = P4 = 14.7(psia); (turbine) = 0.78; (pump) = 0.75; power rating = 50,000 kW. ( f ) P1 = P2 = 1125(psia); T2 = 1100(°F); P3 = P4 = 1(psia); (turbine) = 0.80; (pump) = 0.75; power rating = 80,000 kW.

Solution (a) For reference, here is Figure 8.4, which sketches out the cycle:

Essentially, what we want to know is the heat or work requirement for each step in the cycle. So, we can go around the cycle, on a basis of 1 kg of water. We can start at point 4, because its state is fully known (saturated liquid at 10 kPa). From the steam tables, we find that at 10 kPa, the saturation tmperature is 45.83 ºC = 318.98 K. The enthalpy and entropy of the saturated liquid are H4 = 191.832 kJ/kg and 3

S4 = 0.6493 kJ/(kg K). The specific volume is V4 = 1.010 cm /g = 1.01 × 10

m /kg. If the pump compressed

this liquid isentropically to the boiler pressure of 10000 kPa, and the volume remained approximately constant, then we could estimate the work required as:

WS isentropic   H S 

10000 kPa



VdP  V4 9990 kPa

10 kPa

Solution continued on next page…

Thus Ws(isentropic) = 1.01 × 10

m /kg*9990 kPa = 10.09 kJ/kg. The actual work input is obtained by dividing

this by the pump efficiency of 0.75. Thus, Ws H41 = 10.09/0.75 = 13.45 kJ/kg. Adding this to the enthalpy at point 4, we obtain the enthalpy at point 1 as H1 = 191.832 + 13.453 = 205.285 kJ/kg. We could also compute the temperature and entropy at point 1 from equations in chapter 7, but do not need to do so for this problem. The properties at point 2, for which the temperature and pressure are specified as 600 ºC and 10000 kPa, respectively, can be obtained directly from the steam tables. They are H2 = 3622.7 kJ/kg and S2 = 6.9013 kJ/kg. The enthalpy change from point 1 to point 2 gives the heat input in the boiler QH H12 = 3622.7 – 205.3 = 3417.4 kJ/kg To find the conditions at point 3, we first consider isentropic expansion through the turbine, with wet steam l

leaving it at 10 kPa and 45.83 ºC. At this temperature, the saturated liquid has H = 191.832 kJ/kg and l

S = 0.6493 kJ/(kg K), while the saturated vapor has H = 2584.8 kJ/kg and S = 8.1511 kJ/(kg K). If the turbine is isentropic, then the entropy of the wet steam leaving it is the same as that of the superheated steam entering it: l

S3’ = x *0.6493 + (1 – x )*8.1511 = S2 = 6.9013 kJ/kg/K, from which l

x = (8.1511 – 6.9013)/(8.1511 – 0.6493) = 0.1666 v

and therefore x = 0.8334. The enthalpy of this wet steam is then: H3’ = 0.1666*191.832 + 0.8334*2584.8 = 2186.1 kJ/kg Thus, for isentropic expansion, the work would be Ws(isentropic

H)S = 2186.1 – 3622.7 = –1436.6 kJ/kg

If the turbine efficiency is 0.80, then Ws

H will be 80% of this value

H23 = 0.8*–1436.6 = –1149.3 kJ/kg

Thus, the enthalpy at point 3 is H3 = 3622.7 – 1149.3 = 2473.4 kJ/kg The heat removed in the condenser is equal to the enthalpy difference between point 4 and point 3: QC

H34 = 191.832 – 2473.4 = –2281.6 kJ/kg

The net work produced by the cycle is the sum of the compressor work and the pump work: Wnet = 13.45 – 1149.3 = –1135.9 kJ/kg Thus, the thermal efficiency of the cycle is

thermal

= |Wnet|/QH = 1135.9/3417.4 = 0.332

The specified power rating of 80000 kW will require a steam rate of 80000/1135.9 = 70.4 kg/s of steam. Thus, the total heat transfer rate in the boiler will be 3417.4 kJ/kg * 70.4 kg/s = 240690 kW = 240.7 MW. The heat transfer rate in the condenser will be 2281.6 kJ/kg*70.4 kg/s = –160690 kW = –160.7 MW. The total heat and work input for the cycle is zero, as it must be. 240.7 MW enters the cycle in the boiler, 80 MW leaves as work from the turbine, and the remaining 160.7 MW leaves as heat in the condenser.

Solution continued on next page…

(b) For reference, here is Figure 8.4, which sketches out the cycle:

Essentially, what we want to know is the heat or work requirement for each step in the cycle. So, we can go around the cycle, on a basis of 1 kg of water. We can start at point 4, because its state is fully known (saturated liquid at 20 kPa). From the steam tables, we find that at 20 kPa, the saturation tmperature is 60.09 ºC = 333.24 K. The enthalpy and entropy of the saturated liquid are H4 = 251.453 kJ/kg and S4 = 0.8321 kJ/(kg K). The specific 3

–3

volume is V4 = 1.017 cm /g = 1.017 × 10 m /kg. If the pump compressed this liquid isentropically to the boiler pressure of 7000 kPa, and the volume remained approximately constant, then we could estimate the work required as: WS isentropic   H S 

7000 kPa

 VdP  V 6980 kPa  4

10 kPa

–3

Thus Ws(isentropic) = 1.017 × 10 m /kg*6980 kPa = 7.10 kJ/kg. The actual work input is obtained by dividing this by the pump efficiency of 0.75. Thus, Ws H41 = 7.10/0.75 = 9.46 kJ/kg. Adding this to the enthalpy at point 4, we obtain the enthalpy at point 1 as H1 = 260.918 kJ/kg. We could also compute the temperature and entropy at point 1 from equations in chapter 7, but do not need to do so for this problem. The properties at point 2, for which the temperature and pressure are specified as 550 ºC and 7000 kPa, respectively, can be obtained directly from the steam tables. They are H2 = 3529.6 kJ/kg and S2 = 6.9485 kJ/kg. The enthalpy change from point 1 to point 2 gives the heat input in the boiler QH

H12 = 3529.6 – 260.9 = 3268.7 kJ/kg

Solution continued on next page…

To find the conditions at point 3, we first consider isentropic expansion through the turbine, with wet steam l

leaving it at 20 kPa and 60.09 ºC. At this temperature, the saturated liquid has H = 251.453 kJ/kg and l

S = 0.8321 kJ/(kg K), while the saturated vapor has H = 2609.9 kJ/kg and S = 7.9094 kJ/(kg K). If the turbine is isentropic, then the entropy of the wet steam leaving it is the same as that of the superheated steam entering it: l

S3’ = x *0.8321 + (1 – x )*7.9094 = S2 = 6.9485 kJ/kg/K, from which l

x = (7.9094 – 6.9485)/(7.9094 – 0.8321) = 0.1358 v

and therefore x = 0.8642. The enthalpy of this wet steam is then: H3’ = 0.1358*251.453 + 0.8642*2609.9 = 2289.6 kJ/kg Thus, for isentropic expansion, the work would be Ws(isentropic H)S = 2289.6 – 3529.6 = –1240.0 kJ/kg H will be 75% of this value Ws H23 = 0.75*–1240.0 = –930.0 kJ/kg Thus, the enthalpy at point 3 is H3 = 3529.6 – 930.0 = 2599.6 kJ/kg The heat removed in the condenser is equal to the enthalpy difference between point 4 and point 3: QC H34 = 251.453 – 2599.6 = –2348.2 kJ/kg The net work produced by the cycle is the sum of the turbine work and the pump work: Wnet = 9.5 – 930.0 = –920.5 kJ/kg Thus, the thermal efficiency of the cycle is thermal = |Wnet|/QH = 920.5/3268.7 = 0.282 The specified power rating of 100,000 kW will require a steam rate of 100,000/920.5 = 108.6 kg/s of steam. Thus, the total heat transfer rate in the boiler will be 3268.7 kJ/kg * 108.6 kg/s = 355000 kW = 355 MW. The heat transfer rate in the condenser will be –2348.2 kJ/kg*108.6 kg/s = –255000 kW = –255 MW. The total heat and work input for the cycle is zero, as it must be. 355 MW enters the cycle in the boiler, 100 MW leaves as work from the turbine, and the remaining 255 MW leaves as heat in the condenser. (c) For reference, here is Figure 8.4, which sketches out the cycle:

Essentially, what we want to know is the heat or work requirement for each step in the cycle. So, we can go around the cycle, on a basis of 1 kg of water. We can start at point 4, because its state is fully known (saturated liquid at 10 kPa). From the steam tables, we find that at 10 kPa, the saturation temperature is 45.83 ºC = 318.98 K. The enthalpy and entropy of the saturated liquid are H4 = 191.832 kJ/kg and 3

–3

S4 = 0.6493 kJ/(kg K). The specific volume is V4 = 1.010 cm /g = 1.01 × 10 m /kg. If the pump compressed this liquid isentropically to the boiler pressure of 8500 kPa, and the volume remained approximately constant, then we could estimate the work required as:

WS isentropic   H S 

10000 kPa



VdP  V4 8490 kPa

10 kPa

–3

Thus Ws(isentropic) = 1.01 × 10 m /kg*8490 kPa = 8.575 kJ/kg. The actual work input is obtained by dividing this by the pump efficiency of 0.80. Thus, Ws H41 = 8.575/0.80 = 10.72 kJ/kg. Adding this to the enthalpy at point 4, we obtain the enthalpy at point 1 as H1 = 191.832 + 10.72 = 202.55 kJ/kg. We could also compute the temperature and entropy at point 1 from equations in chapter 7, but do not need to do so for this problem. The properties at point 2, for which the temperature and pressure are specified as 600 ºC and 8500 kPa, respectively, can be obtained directly from the steam tables (interpolating between values at 8400 kPa and 8600 kPa). They are H2 = 3635.4 kJ/kg and S2 = 6.9875 kJ/kg. The enthalpy change from point 1 to point 2 gives the heat input in the boiler QH H12 = 3635.2 – 202.6 = 3432.8 kJ/kg To find the conditions at point 3, we first consider isentropic expansion through the turbine, with wet steam l

leaving it at 10 kPa and 45.83 ºC. At this temperature, the saturated liquid has H = 191.832 kJ/kg and l

S3’ = x *0.6493 + (1 – x )*8.1511 = S2 = 6.9875 kJ/kg/K, from which l

x = (8.1511 – 6.9875)/(8.1511 – 0.6493) = 0.1551 v

and therefore x = 0.8449. The enthalpy of this wet steam is then: H3’ = 0.1551*191.832 + 0.8449*2584.8 = 2213.7 kJ/kg Thus, for isentropic expansion, the work would be Ws(isentropic H)S = 2213.7 – 3635.4 = –1421.7 kJ/kg If the turbine efficiency is 0.80, then the H will be 80% of this value Ws H23 = 0.8*–1421.7 = –1137.4 kJ/kg Thus, the enthalpy at point 3 is H3 = 3635.4 – 1137.4 = 2498.0 kJ/kg The heat removed in the condenser is equal to the enthalpy difference between point 4 and point 3: QC H34 = 191.832 – 2498.0 = –2306.2 kJ/kg The net work produced by the cycle is the sum of the compressor work and the pump work: Wnet = 10.7 – 1137.4 = –1126.6 kJ/kg Solution continued on next page…

Thus, the thermal efficiency of the cycle is thermal = |Wnet|/QH = 1126.6/3432.8 = 0.328 The specified power rating of 70000 kW will require a steam rate of 70000/1126.6 = 62.13 kg/s of steam. Thus, the total heat transfer rate in the boiler will be 3432.8 kJ/kg * 62.13 kg/s = 213285 kW = 213.3 MW. The heat transfer rate in the condenser will be –2306.2 kJ/kg*62.13 kg/s = –143285 kW = 143.3 MW. The total heat and work input for the cycle is zero, as it must be. 213.3 MW enters the cycle in the boiler, 70 MW leaves as work from the turbine, and the remaining 143.3 MW leaves as heat in the condenser. (d) For reference, here is Figure 8.4, which sketches out the cycle:

Essentially, what we want to know is the heat or work requirement for each step in the cycle. So, we can go around the cycle, on a basis of 1 kg of water. We can start at point 4, because its state is fully known (saturated liquid at 101.33 kPa). From the steam tables (or in this case from our general knowledge), we find that at 101.33 kPa, the saturation tmperature is 100.00 ºC = 373.15 K. The enthalpy and entropy of the saturated liquid are H4 = 419.064 kJ/kg and S4 = 1.3069 kJ/(kg K). The specific 3

–3

volume is V4 = 1.044 cm /g = 1.044 × 10 m /kg. If the pump compressed this liquid isentropically to the boiler pressure of 6500 kPa, and the volume remained approximately constant, then we could estimate the work required as:

WS isentropic   H S 

6500 kPa



VdP  V4 6398.67 kPa

101.33 kPa

–3

Thus Ws(isentropic) = 1.044 × 10 m /kg*6398.67 kPa = 6.680 kJ/kg. The actual work input is obtained by dividing this by the pump efficiency of 0.75. Thus, Ws H41 = 6.680/0.75 = 8.907 kJ/kg. Adding this to the enthalpy at point 4, we obtain the enthalpy at point 1 as H1 = 419.064 + 8.907 = 427.97 kJ/kg. We could also compute the temperature and entropy at point 1 from equations in chapter 7, but do not need to do so for this problem. The properties at point 2, for which the temperature and pressure are specified as 525 ºC and 6500 kPa, respectively, can be obtained directly from the steam tables. They are H2 = 3475.6 kJ/kg and S2 = 6.9145 kJ/kg.

Solution continued on next page…

The enthalpy change from point 1 to point 2 gives the heat input in the boiler QH

H12 = 3475.6 – 427.97 = 3047.6 kJ/kg

To find the conditions at point 3, we first consider isentropic expansion through the turbine, with wet steam l

leaving it at 101.33 kPa and 100 ºC. At this temperature, the saturated liquid has H = 419.064 kJ/kg and S = v

1.3069 kJ/(kg K), while the saturated vapor has H = 2676.0 kJ/kg and S = 7.3554 kJ/(kg K). If the turbine is isentropic, then the entropy of the wet steam leaving it is the same as that of the superheated steam entering it: v

S3, = x *7.3554 + (1 – x )*1.3069 = S2 = 6.9145 KJ/kg/K, from which l

x = (6.9145 – 1.3069)/(7.3554 – 1.3069) = 0.9271 v

and therefore x = 0.0729. The enthalpy of this wet steam is then: H3, = 0.0729*419.064 + 0.9271*2676.0 = 2511.5 kJ/kg Thus, for isentropic expansion, the work would be Ws(isentropic

H)S = 2511.5 – 3475.6 = –964.1 kJ/kg H will be 78% of this value

H23 = 0.78*–964.1 = –752.0 kJ/kg

Thus, the enthalpy at point 3 is H3 = 3475.6 – 752.0 = 2723.6 kJ/kg The heat removed in the condenser is equal to the enthalpy difference between point 4 and point 3: QC

H34 = 419.064 – 2723.6 = –2304.5 kJ/kg

The net work produced by the cycle is the sum of the turbine work and the pump work: Wnet = 8.91 – 752.0 = –743.1 kJ/kg Thus, the thermal efficiency of the cycle is

thermal

= |Wnet|/QH = 743.1/3047.6 = 0.244

The specified power rating of 50000 kW will require a steam rate of 50000/743.1 = 67.29 kg/s of steam. Thus, the total heat transfer rate in the boiler will be 3047.6 kJ/kg * 67.29 kg/s = 205059 kW = 205.1 MW. The heat transfer rate in the condenser will be –2304.5 kJ/kg*67.29 kg/s = –155059 kW = 155.1 MW. The total heat and work input for the cycle is zero, as it must be. 205.1 MW enters the cycle in the boiler, 50 MW leaves as work from the turbine, and the remaining 155.1 MW leaves as heat in the condenser.

Solution 8.4 Problem Statement Steam enters the turbine of a power plant operating on the Rankine cycle (Fig. 8.3) at 3300 kPa and exhausts at 50 kPa. To show the effect of superheating on the performance of the cycle, calculate the thermal efficiency of the cycle and the quality of the exhaust steam from the turbine for turbine-inlet steam temperatures of 450, 550, and 650°C. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution Subscripts refer to Fig. 8.3. Saturated liquid at 50 kPa (point 4) kJ H4  V  cm3 P1  kPa and P4  kPa kg 4 Saturated liquid and vapor at 50 kPa: H liq  H 4 , H vap 

kJ S  kg vap

kJ and Sliq  kg K

Wpump  V4  P4  P1 

kJ kg

H1  H 4  Wpump 

kJ kg

kJ kg K

By Eq. 7.24, the pump work is

The following vectors give values for temperatures of 450, 550, and 650 C: H2 (kJ/kg) 3340.6 3565.3 3792.9 Determine the turbine work and the heat added by

S2 (kJ/kg K) 7.0373 7.3282 7.5891

S 3  S2 x 3 

S 3  Sliq Svap  Sliq

H 3  Hliq  x 3  Hvap  Hliq  Solution continued on next page…

Wturbine  H 3  H2 QH   H2  H1  

Wturbine  Wpump QH

The values for the quality and thermal efficiency are: x'3 0.914 0.959

0.297 0.314

0.999

0.332

Solution 8.5 Problem Statement Steam enters the turbine of a power plant operating on the Rankine cycle (Fig. 8.3) at 600°C and exhausts at 30 kPa. To show the effect of boiler pressure on the performance of the cycle, calculate the thermal efficiency of the cycle and the quality of the exhaust steam from the turbine for boiler pressures of 5000, 7500, and 10,000 kPa.

Solution

Subscripts refer to Fig. 8.3. Saturated liquid at 30 kPa (point 4) H4 

kJ V  kg 4

cm3 and P4  g

kPa

Solution continued on next page…

Saturated liquid and vapor at 30 kPa: H liq  H 4 H vap 

kJ S  kg vap

kJ and Sliq  kg K

kJ kg K

P1 (kPa) 5000 7500 10000 By Eq. 7.24, the pump work is

Wpump  V4  P4  P1  Wpump (kJ/kg) 5.079 7.634 10.189

H1  H 4  Wpump

H1 (kJ/kg) 294.381 296.936 299.491 The following vectors give values for temperatures of 5000, 7500, and 10000 kPa at 600 C: H2 (kJ/kg) 3664.5

S2 (kJ/kg K) 7.2578

3643.7 3622.7

7.0526 6.9013

Determine the turbine work and the heat added by

S 3  S2 x 3 

S 3  Sliq Svap  Sliq

Solution continued on next page…

H 3  Hliq  x 3  Hvap  Hliq  Wturbine  H 3  H2 QH   H2  H1   

Wturbine  Wpump QH

The values for the quality and thermal efficiency are: x'3 0.925 0.895

0.359 0.375

0.873

0.386

Solution 8.6 Problem Statement A steam power plant employs two adiabatic turbines in series. Steam enters the first turbine at 650°C and 7000 kPa and discharges from the second turbine at 20 kPa. The system is designed for equal power outputs from the two turbines, based on a turbine efficiency of 78% for each turbine. Determine the temperature and pressure of the steam in its intermediate state between the two turbines. What is the overall efficiency of the two turbines together with respect to isentropic expansion of the steam from the initial to the final state? Solution

From Table E.2 at 7000 kPa and 650 C: H1 

kJ S  kg 1

kJ S 2  S1 kg K

For sat. liq. and sat. vap. at 20 kPa: H liq 

kJ H vap  kg

kJ S  kg vap

kJ S  kg K liq

kJ kg K

The following enthalpies are interpolated in Table E.2 at four values for intermediate pressure P2: P2 (kPa) 725

H'2 (kJ/kg) 3023.9

750 775

3032.5 3040.9

800

3049

Solution continued on next page…

Solve the work from step 1 to 2, and for H2:



W12   H 2  H1



H2  H1  W12 W12 (kJ/kg) –579.15

H2 (kJ/kg) 3187.25

S2 (kJ/kg K) 7.4939

–572.442 –565.89

3193.958 3200.51

7.4898 7.4851

–559.572

3206.828

7.4797

The entropy values are by interpolation in E.2. Now solving for the Work between steps 2 and 3 and the work difference:

x 3 

S2  Sliq Svap  Sliq

H 3  Hliq  x 3  H vap  Hliq 



W23   H 3  H2



W  W12  W23 X'3

H'3 (kJ/kg)

W23 (kJ/kg)

W (kJ/kg)

0.941 0.941

2471.438 2470.072

–558.333 –564.631

–20.817 –7.811

0.940 0.939

2468.506 2466.706

–570.963 –577.295

5.073 17.723

The work difference is essentially linear in P2, and we interpolate linearly to find the value of P2 for which the work difference is zero:

P2 

Also needed are values of H2 and S2 at this pressure. Again we do linear interpolations: H2 

kJ S  kg 2

kJ kg K

We can now find the temperature at this state by interpolation in Table E.2. This gives an intermediate steam temperature T2 of 366.6 C.

Solution continued on next page…

The work calculations must be repeated for THIS case:

W12  H2  H1 x 3 

S2  Sliq Svap  Sliq kJ kg

W12  

x 3  0.94 H 3  H liq  x 3  H vap  Hliq   2.469 * 10 3





W23   H 3  H2  

kJ kg

Work  W12  W23 

kJ kg

For a single isentropic expansion from the initial pressure to the final pressure, which yields a wet exhaust:

x 3 

S1  Sliq Svap  Sliq

H 3  Hliq  x 3  H vap  Hliq  x 3 

H 3 

W   H 3  H1  

kJ kg

This leaves the overall efficiency as: overall 

Work  W

Solution 8.7 Problem Statement A steam power plant operating on a regenerative cycle, as illustrated in Fig. 8.5, includes just one feedwater heater. Steam enters the turbine at 4500 kPa and 500°C and exhausts at 20 kPa. Steam for the feedwater heater is extracted from the turbine at 350 kPa, and in condensing raises the temperature of the feedwater to within 6°C of its condensation temperature at 350 kPa. If the turbine and pump efficiencies are both 0.78, what is the thermal efficiency of the cycle, and what fraction of the steam entering the turbine is extracted for the feedwater heater?

Solution

Solution continued on next page…

From Table E.2 for steam at 4500 kPa and 500 C: kJ S  kg 2

H2 

kJ S 3  S2 kg K

By interpolation at 350 kPa and this entropy kJ  kg

H 3 



WI   * H 3  H 2



H3  H2  WI H3 

kJ kg kJ kg

WI 

Isentropic expansion to 20 kPa:

S 4  S2 Exhaust is wet: for sat. liq. & vap.: H liq 

kJ S  kg liq

kJ H vap  kg K

x 4 

S 4  Sliq Svap  Sliq

kJ S  kg vap



H 4  H liq  x  4  H vap  H liq  





H 4  H 2   H 4  H2  H5  Hliq V5 

cm3 P5  g

Wpump 

kJ kg K

* *

kPa P6 

kJ kg

kJ kg kPa

V5  P6  P5  

H 6  H5  Wpump Wpump 

kJ kg

H 6  257.294

kJ kg

Solution continued on next page…

For sat. liq. at 350 kPa (Table E.2): kJ T  kg 7

H7 

C

We need the enthalpy of compressed liquid at point 1, where the pressure is 4500 kPa and the temperature is:

T1 

 

At this temperature, 132.87 degC or 406.2 K, interpolation in Table E.1 gives: kJ P  kg sat

H sat liq 

cm3 g

kPa Vsat liq 

Also by approximation, the definition of the volume expansivity yields:



1.083  1.063  cm3 1    9.32 * 104 *   Vsat liq  K gK 1

P1  P6 By Eq. (7.25), kJ kg

H1  H sat .liq  Vsat .liq   T1  *  P1  Psat  

By an energy balance on the feedwater heater:

mass 

H1  H6 * kg  H3  H7

i.e. 13% of the steam is extracted from the high-pressure turbine Work in 2nd section of turbine:

WII   kg  mass  H 4  H3   

Wnet  WI  Wpump  * kg  WII 

QH   H2  H1  * kg QH 

And lastly the thermal efficiency is



Wnet QH



Solution 8.8 Problem Statement A steam power plant operating on a regenerative cycle, as illustrated in Fig. 8.5, includes just one feedwater heater. Steam enters the turbine at 650(psia) and 900(°F) and exhausts at 1(psia). Steam for the feedwater heater is extracted from the turbine at 50(psia) and in condensing raises the temperature of the feedwater to within 11(°F) of its condensation temperature at 50(psia). If the turbine and pump efficiencies are both 0.78, what is the thermal efficiency of the cycle, and what fraction of the steam entering the turbine is extracted for the feedwater heater?

Solution Refer to figure in preceding problem. From Table E.4 for steam at 650 psia & 900 F: BTU lbm BTU S2  1.6671 lbm Rankine S 3  S2 H2 

Solution continued on next page…

By interpolation at 50 psia and this entropy BTU lbm

H 3  1180.4 



WI   * H 3  H 2



H3  H2  WI H3 

WI 

BTU lbm BTU lbm

Isentropic expansion to 1 psia:

S 4  S2 Exhaust is wet: for sat. liq. & vap.: BTU lbm BTU lbm Rankine BTU lbm BTU lbm Rankine

H liq  69.73 Sliq  H vap  Svap 

x 4 

S 4  Sliq Svap  Sliq



H 4  H liq  x  4  H vap  H liq  





H 4  H 2   H 4  H 2 

BTU lbm BTU lbm

H5  Hliq

ft 3 lbm P5  psia

V5 

P6  Wpump 

V5  P6  P5  

psia H 6  H5  Wpump

Solution continued on next page…

BTU lbm BTU H 6  72.209 lbm

Wpump 

For sat. liq. at 50 psia (Table E.4): BTU lbm BTU lbm Rankine F Rankine

H 7  250.21 S7 

T7 

We need the enthalpy of compressed liquid at point 1, where the pressure is 650 psia and the temperature is:

T1 





F

Rankine

At this temperature, 270.01 F, interpolation in Table E.3 gives: BTU lbm BTU Ssat .liq  0.3960 lbm Rankine Psat  psia H sat .liq 

Vsat liq 

ft 3 lbm

The definition of the volume expansivity yields:



 0.01726  0.01709  ft 3 1    4.95 * 105 * Vsat liq  Rankine  lbm Rankine 1

P1  P6 By Eq. 7.25 and 7.26 BTU lbm BTU lbm Rankine

H1  H sat .liq  Vsat .liq   T1  * P1  Psat  S1  Ssat .liq  Ssat .liq *  *  P1  Psat 

By an energy balance on the feedwater heater: mass 

H1  H6 * lbm  H3  H7

lbm

i.e. 30.83% of the steam is extracted from the high-pressure turbine

Solution continued on next page…

Work in 2nd section of turbine:

WII   lbm  mass H4  H3 WII  

BTU

Wnet  WI  Wpump  * lbm  WII  

BTU

QH   H2  H1  * 1 lbm  1.102 * 103 BTU And lastly the thermal efficiency is 

Wnet QH



Solution 8.9 Problem Statement A steam power plant operating on a regenerative cycle, as illustrated in Fig. 8.5, includes two feedwater heaters. Steam enters the turbine at 6500 kPa and 600°C and exhausts at 20 kPa. Steam for the feedwater heaters is extracted from the turbine at pressures such that the feedwater is heated to 190°C in two equal increments of temperature rise, with 5°C approaches to the steam-condensation temperature in each feedwater heater. If the turbine and pump efficiencies are each 0.80, what is the thermal efficiency of the cycle, and what fraction of the steam entering the turbine is extracted for each feedwater heater?

Solution

Steam at 6500 kPa & 600 degC (point 2) Table E.2: kJ S  kg 2

H2 

kJ P  kg K 2

kPa

At point 3 the pressure must be such that the steam has a condensation temperature in feedwater heater I of 195 C, 5 deg higher than the temperature of the feed water to the boiler at point 1. Its saturation pressure, corresponding to 195 C, from Table E.1, is 1399.0 kPa. The steam at point 3 is superheated vapor at this pressure, and if expansion from P2 to P3 is isentropic,

S 3  S2 By interpolation at 1399 kPa and this entropy H 3  3142.6

kJ kg



WI   *  H 3  H2  H3  H2  WI 

WI  

kJ kg

H10  829.9

kJ kg

From table E.1:

Solution continued on next page…

Similar calculations are required for feedwater heater II. At the exhaust conditions of 20 kPa, the properties of sat. liq. and sat.vap. are: H liq  251.453

kJ kg

Sliq 

kJ kg K

Vliq 

cm3 g

H vap 

kJ kg

Svap 

kJ kg K

If we find t7, then t8 is the mid-temperature between t7 and t1(190 C), and that fixes the pressure of stream 4 so that its saturation temperature is 5 C higher. At point 6, we have saturated liquid at 20 kPa, and its properties from Table E.2 are:

t sat 

Tsat  t sat 



H 6  Hliq V6  Vliq

P6  Wpump 

kPa V6  P2  P6  

Wpump  8.238

kJ kg

H 67  Wpump

We apply Eq. (7.25) for the calculation of the temperature change from point 6 to point 7. For this we need values of the heat capacity and volume expansivity of water at about 60 C. They can be estimated from data in Table E.1:  1.083  1.063  cm3 1 4 1     *  * V K   g K sat liq

CP 

272.0  230.2 kJ  kg K

kJ kg K

Solving Eq. (7.25) for T gives:

T67 

H67  Vliq 1  Tsat P2  P6  CP



t7  t sat  t9 

190  t7 2

T67 K

 60.768 C

 t7 

C

t8  t9  

From table E.1 kJ kg

H8 

kJ kg

H 7  H liq  H 67 

At points 9 and 1, the streams are compressed liquid (P = 6500 kPa), and we find the effect of pressure on the liquid by Eq. (7.25). Values by interpolation in Table E.1 at saturation temperatures t9 and t1: H sat 

kJ kg

H sat  807.5

kJ kg

Vsat 

cm3 g

Vsat 

cm3 g

Psat  234.9 kPa P6 

kPa

T1 

V  0.019

cm3 g

V1 

cm3 g

T 

K   1  V   9  * Vsat  T  1 

1 Vsat

 V1    *   T 

1 K 1 1  1.226 * 103 K 9 

4

H  H sat  Vsat    T  *  P  Psat  

kJ kg

H  H sat  Vsat    T  *  P  Psat  

kJ kg

Now we can make an energy balance on feedwater heater I to find the mass of steam condensed: mass 

H1  H9 * kg  H3  H10

i.e. 11.6% of the steam is extracted The temperature at point 8, t8 = 130.38 (see above) is the saturation temperature in feedwater heater II. The saturation pressure by interpolation in Table E.1 is 273.28 kPa. Isentropic expansion of steam from the initial conditions to this pressure results in a slightly superheated vapor, for which by double interpolation in Table E.2: kJ kg

H 4 

H 4  H2    H 4  H2  

kJ kg

We can now make an energy balance on feedwater heater II to find the mass of steam condensed:

mII 

 H9  H7 1 kg  mI *  H10  H8  H4  H8



i.e. ~10% of the remaining steam is extracted The final stage of expansion in the turbine is to 20 kPa, where the exhaust is wet. For isentropic expansion,

x 5 

S2  Sliq Svap  Sliq

 0.889

H 5  Hliq  x 5  H vap  H liq   2.349 * 103





H5  H2   H 5  H 2  2609.4

kJ kg

The work of the turbine

Wturbine  WI * kg   kg  mI  *  H 4  H3    kg  mI  mII  *  H5  H4  Wturbine  

QH   H2  H1  * kg QH 

And lastly the thermal efficiency is 

Wturbine  Wpump * 1 kg QH



and the total heat extracted is m = 0.11563 + 0.09971 = 0.21534 kg

Solution 8.10 Problem Statement A power plant operating on heat recovered from the exhaust gases of internal combustion engines uses isobutane as the working medium in a modified Rankine cycle in which the upper pressure level is above the critical pressure of isobutane. Thus the isobutane does not undergo a change of phase as it absorbs heat prior to its entry into the turbine. Isobutane vapor is heated at 4800 kPa to 260°C and enters the turbine as a supercritical fluid at these conditions. Isentropic expansion in the turbine produces a superheated vapor at 450 kPa, which is cooled and condensed at constant pressure. The resulting saturated liquid enters the pump for return to the heater. If the power output of the modified Rankine cycle is 1000 kW, what are the isobutane flow rate, the heat-transfer rates in the heater and condenser, and the thermal efficiency of the cycle? The vapor pressure of isobutane can be computed from data given in Table B.2 of App. B.

Solution continued on next page…

Solution The critical values for isobutane are as follows:

Tc 

K Pc 

bar  

For isentropic expansion in the turbine, let the initial state be represented by symbols with subscript zero and the final state by symbols with no subscript. Then

T0 

K P0  S  0

kPa P 

kPa

kJ kg K

For the heat capacity of isobutene A

B

37.853 * 103 11.945 * 106 C K K2

Tr 0 

Pr 0 

Pr 

Using generalized second-virial correlation, the entropy change is given by Eq. (6.73) combined with Eq. (5.11) with D = 0, solving for  yields:    *T      1  P 0 S  R  A * ln    B * T0  CT02   *     ln  SRB  Pr    SRB Tr 0 Pr 0     2  P0  TC     



T   * T0 T 

K Tr 

The enthalpy change for the final T is given by Eq. (6.72), with HRB at the above T:



H ig  R * ICPH T0 T

Hig  1.151 * 10 4

3



6



J mol

Hturbine  Hig  RTc HRBTr , Pr    RTc HRB Tr 0 Pr 0   Wturbine  H turbine  8946.0

J mol

Solution continued on next page…

The work of the pump is given by Eq. (7.24), and for this we need an estimate of the molar volume of isobutane as a saturated liquid at 450 kPa. This is given by Eq. (3.68). The saturation temperature at 450 kPa is given by the Antoine equation solved for t degC: VP 

kPa Avp 

t sat 

Bvp 

Cvp 

Bvp  C vp   VP  Avp  ln   kPa 

Tsat  t sat 



cm3 mol

VC 

ZC  0.282 Trsat 

Tsat TC



cm3 mol

1Trsat 7

Vliq  VC ZC



J mol

Wpump  Vliq  P0  P 

The flow rate of isobutane can now be found: 1000 kW

m 

Wturbine  Wpump

mol s



The enthalpy change of the isobutane in the cooler/condenser is calculated in two steps: a. Cooling of the vapor from 454.49 to 307.15 K b. Condensation of the vapor at 307.15 K Enthalpy change of cooling: HRB at the initial state has already been calculated. For saturated vapor at 307.15 K:



H ig  R * ICPH T Tsat

3

Hig  1.774 * 10 4



6



J mol

Ha  Hig  RTc HRB Trsat , Pr    RTc HRBTr Pr  H a  18106.8

J mol

For the condensation process, we estimate the latent heat by Eqs. (4.13) and (4.14):

Tn  Trn 

H n 

Tn Tc

K 

  P   R * Tn * 1.092ln c   1.013   bar   0.930  Trn

 1  T  r sat  H b  Hrn    1  Trn 

 2.118 * 10 4

0.38

J mol



Q out  m * Ha  Hb 

Q in  Wturbine  Wpump m  Q out Q out  

kW Q in 

kW  Q in



Solution 8.11 Problem Statement A power plant operating on heat from a geothermal source uses isobutane as the working medium in a Rankine cycle (Fig. 8.3). Isobutane is heated at 3400 kPa (a pressure just below its critical pressure) to a temperature of 140°C, at which conditions it enters the turbine. Isentropic expansion in the turbine produces superheated vapor at 450 kPa, which is cooled and condensed to saturated liquid and pumped to the 1

heater/boiler. If the flow rate of isobutane is 75 kgs , what is the power output of the Rankine cycle, and what are the heat-transfer rates in the heater/ boiler and cooler/condenser? What is the thermal efficiency of the cycle? The vapor pressure of isobutane is given in Table B.2 of App. B.

Repeat these calculations for a cycle in which the turbine and pump each have an efficiency of 80%.

Solution The critical values for isobutane are as follows:

Tc 

K Pc 

bar  

For isentropic expansion in the turbine, let the initial (inlet) state be represented by symbols with subscript zero and the final (exit) state by symbols with no subscript. Then T0 

K P0 

kPa P 

S  0

kPa MW 

g mol

kJ kg K

For the heat capacity of isobutene A

B

37.853 * 103 11.945 * 106 C K K2

Tr 0  1.0124, Pr 0  0.932, Pr  0.123 Use Lee/Kesler correlation for turbine-inlet state, designating values by HRLK and SRLK:

HRLK 0  

SRLK 0 

The entropy change is given by Eq. (6.73) combined with Eq. (5.11) with D = 0, solving for  yields:    * T      1  P 0 S  R  A * ln    B * T0  CT02       ln  SRB  Pr    SRLK 0    2  *  P0  TC     



T   * T0 

K Tr 

The enthalpy change for the final T is given by Eq. (6.72), with HRB at the above T:





H ig  R * ICPH T0 , T , 1.677, 37.853 * 103 ,  11.945 * 106 , 0 Hig  9.231 * 10 3

J mol

Hturbine  Hig  RTc HRB Tr , Pr ,    RTc HRLK 0 Wturbine  Hturbine  4863

J mol

The work of the pump is given by Eq. (7.24), and for this we need an estimate of the molar volume of isobutane as a saturated liquid at 450 kPa. This is given by Eq. (3.68). The saturation temperature at 450 kPa is the value calculated in Problem 8.10: VP AVP  13.8254, BVP  2181.79, CVP  248.870

T sat

     BVP    CVP    VP       AVP      

      

2181.79    

  



      

Solution continued on next page…

1

Vc Zc Trsat

Vliq

T sat Tc

Vc Z c

 2     

 Trsat  7 



 2    7

l J mol

Wpump  Vliq  P0  P 

For the cycle, the net power output is: m 

kg  s

mol  W  m Wturbine  W pump     s

The enthalpy change of the isobutane in the cooler/condenser is calculated in two steps: a. Cooling of the vapor from 359.53 to 307.15 K b. Condensation of the vapor at 307.15 K Enthalpy change of cooling: HRB at the initial state has already been calculated. For saturated vapor at 307.15 K was found in Problem 8.10 as:



K Hig  R * ICPH T Tsat

Tsat 

Hig  2.897

3



6



kJ mol

Ha  Hig  RTc HRBTrsat , Pr   RTc HRB Tr Pr  H a  253.91

J mol

Solution continued on next page…

For the condensation process, the enthalpy change was found in Problem 8.10:

Tn  261.4 K

Trn

Tn Tc RTn

H n H n

 Pc      0.930  Trn 4

1

  

0.930  0.641

·mol1

 1  T sat  r  Hn   1  Trn 

0.38

H b

  

1

1  0.753     1  0.641 

0.38



1



Q out  m * Ha  Hb  



1

For the heater/boiler:

Q in  W  Q out  

W  Q in

We now recalculate results for a cycle for which the turbine and pump each have an efficiency of 0.8. The work of the turbine is 80% of the value calculated above, i.e., W turbine 

J mol

* Wturbine  

The work of the pump is: W pump 



Wpump 0.8

 414.3



W  m W turbine  W  pump W 

J mol

Solution continued on next page…

The decrease in the work output of the turbine shows up as an increase in the heat transferred out of the cooler condenser. Thus





Q out  Q out  m Wturbine  W turbine  

The increase in pump work shows up as a decrease in the heat added in the heater/boiler. Thus





Q in  Q in  m W pump  Wpump  29740 kW

Lastly the thermal efficiency is: 

W  Q in

Solution 8.12 Problem Statement For comparison of Diesel and Otto-engine cycles: (a) Show that the thermal efficiency of the air-standard Diesel cycle can be expressed as

 1

 1   1     r 

rc  1  (rc  1)

where r is the compression ratio and rc is the cutoff ratio, defined as rc = VA/VD. (See Fig. 8.10.) (b) Show that for the same compression ratio the thermal efficiency of the air-standard Otto engine is greater than the thermal efficiency of the air-standard Diesel cycle. Hint: Show that the fraction which multiplies (1/r)

in the preceding equation for is greater than unity by expanding rc in a Taylor series with the

remainder taken to the first derivative. (c) If = 1.4, how does the thermal efficiency of an air-standard Otto cycle with a compression ratio of 8 compare with the thermal efficiency of an air-standard Diesel cycle with the same compression ratio and a cutoff ratio of 2? How is the comparison changed if the cutoff ratio is 3?

Solution a) Because Eq. (8.7) for the efficiency Diesel

re  VB VA , we relate

this quantity to the compression ratio, r  VC VD

rc  VA VD . Since

VC  VB re  VC VA . Whence, VC V V r 1 r  D  A  rc or  c re VC VD re r VA

Equation (8.7) can therefore be written:    r      1    c    1             r   1   r  1  r   rc  1   Diesel  1     1    rc  1  rc  1  1     r  r r     1



1       r 

rc  1

 rc  1

b) We wish to show that: rc  1

 rc  

xa a x







Taylor’s theorem with remainder, taken to the 1st derivative, is written:

g  g 1  g  1 ·  x  1  R Where, Then,

g  1    x  1   x 2   2! x a   a x   12 a a

R



     x



a 2

x



Note that the final term is R

x a   a x

a

x

R

Therefore:

 xa   a rc  1

And

 rc  1

c) If  

r

x  



then by Eq. (8.6) 1      8 

0.4

Otto 

 Desiel 

      8 

Desiel 

1.4         8  1.4 3  1

0.4



rc 

Otto 

   

1.4

0.4



rc 

Solution 8.13 Problem Statement

An air-standard Diesel cycle absorbs 1500 Jmol

of heat (step DA of Fig. 8.10, which simulates combustion).

The pressure and temperature at the beginning of the compression step are 1 bar and 20°C, and the pressure at the end of the compression step is 4 bar. Assuming air to be an ideal gas for which CP = (7/2)R and CV = (5/2)R, what are the compression ratio and the expansion ratio of the cycle?

Solution

7 CP  R Pc  bar Tc  2

K PD  bar  

Starting from Eq 3.23c, the compression ratio is: PC VC  PDVD 1

 P   P    D  or r   D    PC  VD  PC  VC

Starting from Eq. 3.23b, the expansion ratio is:  1

P   TD  TC  D    PC 

J mol

QDA 

QDA  C P TA  TD  TA 

QDA  TD  CP

RTC V V P re  B  C  C V A V A RTA PA

PA  PD re 

TC PA TA PC

re 

Solution 8.14 Problem Statement Calculate the efficiency for an air-standard gas-turbine cycle (the Brayton cycle) operating with a pressure ratio of 3. Repeat for pressure ratios of 5, 7, and 9. Take = 1.35.

Solution The efficiency of an air-standard gas turbine cycle (Brayton cycle) is  1

P    1   A   PB 



If = 1.35, this becomes

 P 0.25926   1   A   PB  For PA/PB = 1/3, this gives   1  0.333330.25926  0.248

Similarly, compression rations of 5, 7, and 9 give efficiencies of 0.341, 0.396, and 0.434, respectively.

Solution 8.15 Problem Statement An air-standard gas-turbine cycle is modified by the installation of a regenerative heat exchanger to transfer energy from the air leaving the turbine to the air leaving the compressor. In an optimum countercurrent exchanger, the temperature of the air leaving the compressor is raised to that of point D in Fig. 8.12, and the temperature of the gas leaving the turbine is cooled to that of point B in Fig. 8.12. Show that the thermal efficiency of this cycle is given by (1)/ 

T P    1  A  B  TC  PA 

Solution See the figure below. In the regenerative heat exchanger, the air temperature is raised in step B 

D

D . Heat addition (replacing combustion) is in step B*  C . *

WAB WCD QB*C

By definition,



Where,

WAB   H B  H A   CP TB  TA  WCD   H D

HC   CP TD

TC 

QB* c  CP TC TB*   CP TC TD  

Whence,

TA  TB  TC  TD TC  TD

1

TB  TA TC  TD

By Eq. (3.23b),  1

P   TB  TA  B   PA 

Then,

 1

 P   P   TD  TC  D   TC  A   PB   PC 

 1    P     B   TA    1  P    A        1    P      A  TC 1       PB      

 1

P   Multiplication of numerator and denominator by  B  gives:  PA   1



T P    A  B  TC  PA 

Solution 8.16 Problem Statement Consider an air-standard cycle for the turbojet power plant shown in Fig. 8.13. The temperature and pressure of the air entering the compressor are 1 bar and 30°C. The pressure ratio in the compressor is 6.5, and the temperature at the turbine inlet is 1100°C. If expansion in the nozzle is isentropic and if the nozzle exhausts at 1 bar, what is the pressure at the nozzle inlet (turbine exhaust), and what is the velocity of the air leaving the nozzle?

Solution

Figure shows the air-standard turbojet power plant on a PV diagram.

TA 

K TC 

7 K CP  R 2

By Eq. 7.22: R    2   P CP   B WAB  CP TA    1  CP TA cr 7  1     PA     R    2   P CP   D  WCD  CP TC    1  CP TC er 7  1     PC     where cr is the compression ratio and er is the expansion ratio. Since the two work terms are equal but of opposite signs,

WAB  WCD

 2   2  TC er 7  1   TA cr 7  1    

Solving for er yields er  0.552 R CP

2  P  By Eq. 7.18: TD  TC  D   TC er 7   PC 

By Eq. 7.11:  1        2  P V P D D   E    A uE2  u2D      1   PD    

Solution continued on next page…

Knowing the relation: er 

P PD P P cr  B  C cr * er  D PC PA PE PE

Make the following substitutions into (A) uD 

P  1 R 2 1   PDVD  RTD E   CP 7 PD cr * er

Then (A) becomes uE 

2     1 7  R    * * T   MW D   cr * er    

With MW 

g mol

uE  331.8

m s

Finally with the nozzle exhaust at 1 bar the nozzle inlet is PD  cr * er * PE 

bar

Solution 8.17 Problem Statement Air enters a gas-turbine engine (see Fig. 8.11) at 305 K and 1.05 bar and is compressed to 7.5 bar. The fuel is methane at 300 K and 7.5 bar; compressor and turbine efficiencies are each 80%. For one of the turbine inlet temperatures TC given below, determine: the molar fuel-to-air ratio, the net mechanical power delivered per mole of fuel, and the turbine exhaust temperature TD. Assume complete combustion of the methane and expansion in the turbine to 1(atm).

(a) TC = 1000 K (b) TC = 1250 K (c) TC = 1500 K

Solution

TA 

K PA 

bar PB 

bar  

Assume air to be an ideal gas with mean heat capacity (final temperature by iteration):

Cpmair  R * MCPH 

Cpmair  29.231

3



* 10  5

J mol K

Compressor: R    Cpmair TA  PB Cpmair    Ws air      P    A     Ws air TB  TA  T  Cpmair B

J mol K

Combustion: CH4 + 2O2 = CO2 + 2H2O Basis: Complete combustion of 1 mol CH4. Reactants are N mol of air and 1 mol CH4. Because the combustion is adiabatic, the basic equation is:

H R H298 H P  Solution continued on next page…

For H_R, the mean heat capacities for air and methane are required. The value for air is given above. For methane the temperature change is very small; use the value given in Table C.1 for 298 K: 4.217*R. The solution process requires iteration for N. Assume a value for N until the above energy balance is satisfied. Part(a): TC 

K N HR  Cpmair * N * 





H R  8723

J mol

R





The product stream contains: 1 mol CO2, 2 mol H2O, 0.79N mol N2, and (0.21N-2) mol O2 n A B D 1 2

5.457 3.47

0.001045 0.00145

–115700 12100

45.53402 10.10398

3.28 3.639

0.000593 0.000506

4000 –22700

Sum

58.638

198.52

0.04

–138724.27

Cpmp  R * MCPH 

H P  Cpm p TC 

From Ex 4.7: H298  802625





J mol

K 

J mol H R H 298 H P  

Solving for this iteratively, it is found that N moles of air per mole of methane or

J mol

moles of methane per moles of air

Assume expansion of the combustion products in the turbine is to 1(atm), i.e., to 1.0133 bar:

PD 

bar PC 

bar

The pertinent equations are analogous to those for the compressor. The mean heat capacity is that of the combustion gases, and depends on the temperature of the exhaust gases from the turbine, which must therefore be found by iteration. For an initial calculation use the mean heat capacity already determined. This calculation yields an exhaust temperature of about 390 K. Thus iteration starts with this value. Parameters A, B, and D have the final values determined above.

Cpm  R * MCPH 1000 K , 343.12 K , 198.517, 0.036, 0.0,  1.3872 * 10 5  Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Cpm 

J mol K

R    Cpm * 57.795 * TC  PD Cpm  Ws             PC     Ws TD  TC   K Cpm

Ws net  Ws  Ws air * N  

J mol

J  J per mole of methane mol

Parts (b) and (c) are solved in exactly the same way, with the following results: Part(b): TC 

K N

Ws net  

Part(c): TC 

K N

Ws net  

J T  mol D

Solution 8.18 Problem Statement

Most electrical energy in the United States is generated in large-scale power cycles through conversion of thermal energy to mechanical energy, which is then converted to electrical energy. Assume a thermal efficiency of 0.35 for conversion of thermal to mechanical energy, and an efficiency of 0.95 for conversion of mechanical to electrical energy. Line losses in the distribution system amount to 20%. If the cost of fuel for the 1

power cycle is $4.00 GJ , estimate the cost of electricity delivered to the customer in $ per kWh. Ignore operating costs, profits, and taxes. Compare this number with that found on a typical electric bill. Solution The key question is how many kW-h of delivered electrical energy can be provided per GJ of primary energy. To answer this, we simply multiply all of the efficiencies: 1 GJ primary energy *0.35*0.95*0.8 = 0.266 GJ energy reaching a household Where 0.35 is the fraction of primary energy converted to mechanical energy 0.95 is the fraction of mechanical energy converted to electrical energy 0.8 is the fraction of electrical energy that reaches the consumer. Thus, the cost of electrical energy reaching –1

the consumer would be $4/0.266 = $15.04 (GJ energy reaching a household) Solution continued on next page…

Now, we need to convert this to a price per kwh. One kwh is equal to 3600 kJ, or 3.6 MJ, or 0.0036 GJ. Thus, the cost per kwh is –1

$15.04(GJ energy reaching a household) *0.0036 GJ/kwh = $0.054 kwh

–1

My most recent bill from National Grid shows “electricity supply” at a rate of $0.046 kwh , plus an optional –1

$0.016 kwh for using a “renewable energy” supplier. However, after adding delivery fees taxes, and a wide –1

variety of other surcharges, the effective rate (total bill divided by kwh used) comes out to be $0.1706 kwh .

Solution 8.19 Problem Statement

Liquefied natural gas (LNG) is transported in very large tankers, stored as liquid in equilibrium with its vapor at approximately atmospheric pressure. If LNG is essentially pure methane, the storage temperature then is about 111.4 K, the normal boiling point of methane. The enormous amount of cold liquid can in principle serve as a heat sink for an onboard heat engine. Energy discarded to the LNG serves for its vaporization. If the heat source is ambient air at 300 K, and if the efficiency of a heat engine is 60% of its Carnot value, estimate the 1

vaporization rate in moles vaporized per kJ of power output. For methane, Hnlv = 8.206 kJmol . Solution TC 

K TH 

K Hnlv 

kJ mol

First determine the efficiency of the heat engine Carnot  

TC  TH

 HE 

Carnot 

Assume as a basis of W = 1 kJ and determine the vaporization rate QH 

W   HE

kJ QC  QH    HE  

The vaporization rate is

QC  1 kJ * Hnlv

kJ mol kJ

Solution 8.20 Problem Statement The oceans in the tropics have substantial surface-to-deep-water temperature gradients. Depending on location, relatively constant temperature differences of 15 to 25°C are observed for depths of 500 to 1000 m. This provides the opportunity for using cold (deep) water as a heat sink and warm (surface) water as a heat source for a power cycle. The technology is known as OTEC (Ocean Thermal Energy Conversion). (a) Consider a location where the surface temperature is 27°C and the temperature at a depth of 750 m is 6°C. What is the efficiency of a Carnot engine operating between these temperature levels? (b) Part of the output of a power cycle must be used to pump the cold water to the surface, where the cycle hardware resides. If the inherent efficiency of a real cycle is 0.6 of the Carnot value, and if 1/3 of the generated power is used for moving cold water to the surface, what is the actual efficiency of the cycle? (c) The choice of working fluid for the cycle is critical. Suggest some possibilities. Here you may want to consult a handbook, such as Perry’s Chemical Engineers’ Handbook. Solution

TH 

K TC 

Part(a): the efficiency for a carnot engine would be carnot  

TC  TH

Part(b): The actual efficiency would be actual  carnot *

2 *  3

Part(c): The thermal efficiency is low and high fluid rates are required to generate reasonable power. This argues for working fluids that are relatively inexpensive. Candidates that provide reasonable pressures at the required temperature levels include ammonia, n-butane, and propane.

Solution 8.21 Problem Statement Air-standard power cycles are conventionally displayed on PV diagrams. An alternative is the PT diagram. Sketch air-standard cycles on PT diagrams for the following: (a) Carnot cycle (b) Otto cycle (c) Diesel cycle Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(d) Brayton cycle Why would a PT diagram not be helpful for depicting power cycles involving liquid/ vapor phase changes? Solution We give first a general treatment of paths on a PT diagram for an ideal gas with constant heat capacities undergoing reversible polytropic processes. Equation (3.35c), p. 78, may be rewritten as 

P  KT 1 ln P  ln K 

 ln T  1

dP  dT  P  1 T dP  P   A dT   1 T Constant P

dP dT





   dP dT  

By Eq. (A),

 

 dP dT 

Constant T

d2 P   1 dP P   1   P P    2       2   1  T dT T    1 T    1 T T  dT

d2 P  P   B dT 2   2 T 2

d 2 P dT 2

 , i.e., +

For a constant-V process, P varies with T in accord with the ideal-gas law: P = RT/V or P = KT 

With respect to the initial equation, P  KT 1

   Moreover, dP/dT = K and d2 P dT 2 

Thus a constant-V process is represented on a PT diagram as part of a straight line passing through the origin. The slope K is determined by the initial PT coordinates. For a reversible adiabatic process (an isentropic process),   In this case Eqs. (A) and (B) become: dP  P d2 P  P   2 2 2 dT   1 T dT   1 T

We note here that

   



  

are both > 1. Thus in relation to a constant-V process the isentropic

process is represented by a line of greater slope and greater curvature for the same T and P.

Solution continued on next page…

Lines characteristic of the various processes are shown on the following diagram.

The required sketches appear on the following page. (Courtesy of Prof. Mark T. Swihart, State University of New York at Buffalo.)

Figure 1: The Carnot cycle

Figure 2: The Otto cycle

Figure 3: The Diesel cycle

Figure 4: The Brayton cycle

Solution 8.22 Problem Statement A steam plant operates on the cycle of Fig. 8.4. The pressure levels are 10 kPa and 6000 kPa, and steam leaves the turbine as saturated vapor. The pump efficiency is 0.70, and the turbine efficiency is 0.75. Determine the thermal efficiency of the plant.

Solution P2  P1 

kPa P4  P3 

kPa  pump 

turbine 

First let’s determine state 4 T 

kJ S liq  kg

K H liq 

kJ v  kg K

cm3 g

The isentropic work is:

Wisentropic  V4 





kJ kg

This then means that the actual work is: Ws 

Wisentropic  pump

Ws 

kJ kg

Adding this to the enthalpy at state 4 gives the enthalpy at state 1: H1  200.47

kJ kg

First let’s determine state 3, due to it being a saturated vapor: T3 

K H3 

kJ S  kg 3

kJ kg K

We know that the entropy at S2  S3 ’, so guess an x =0.2 

S3  x l *

xl *

 S2 

kJ kg K



kJ kg

Following the same guess for H3 ’ is 

H3  x l *

xl *

Now to figure out H2 . Knowing the entropy at state 2, determine the temperature that matches that entropy. Then use that temperature and pressure to determine the enthalpy through linear interpolation kJ T2  K H2  kg So to determine if our guess is correct, we use the turbine efficiency to iterate and get the correct values turbine 

H 2  H3 ,  0.75 H 2  H3

So through iteration the values for everything are as follows xl 

H2 

kJ T  kg 2

C S2 

kJ H3  kg K

kJ kg

So now that we have all the enthalpies, the thermal efficiency is: thermal 

 H 2  H3    H1  H 4  H 2  H1

thermal  0.2941

Solution 8.23 Problem Statement Devise a general scheme for analyzing four-step air-standard power cycles. Model each step of the cycle as a polytropic process described by P V = constant which implies that TP

)

= constant

with a specified value of . Decide which states to fix, partially or completely, by values of T and/or P. Analysis here means determination of T and P for initial and final states of each step, Q and W for each step, and the thermal efficiency of the cycle. The analysis should also include a sketch of the cycle on a PT diagram. Solution This is a challenging and open-ended problem for which we offer no solution. Problem 8.21 may offer some insight. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 9.1

Problem Statement

An easy way to rationalize definitions of cycle performance is to think of them as:

Measure of performance 

What you get What you pay for

Thus, for an engine, thermal efficiency is = |W| / |Q H|; for a refrigerator, the coefficient of performance is = |QC| / |W|. Define a coefficient of performance

for a heat pump. What is

for a Carnot heat pump?

Solution

Since the object of doing work |W| on a heat pump is to transfer heat |QH| to a heat sink, then:

What you get  QH What you pay for  W This leaves the coefficient of performance for a heat pump as:



QH W

For a Carnot heat pump:



QH QH  QC



TH TH  TC

Solution 9.2

Problem Statement

The contents of the freezer in a home refrigerator are maintained at leaks amount to 125,000 kJ per day, and if electricity costs $0.08/kWh, estimate the yearly cost of running the refrigerator. Assume a coefficient of performance equal to 60% of the Carnot value.

Solution

The hot and cold temperatures are:

TH 

TC 

The heat transferred is kJ Q H  125000 day

First find the coefficient of performance for the Carnot refrigerator carnot 

TH  6.329 TH  TC

The actual carnot coefficient is 60% of this value which is 

* carnot 

125000

kJ day

kW * 



The total work it takes to run the refrigerator is: Q W  C  



So the cost of running it is:

Cost 

0.08 0.075 * W  * kWhr kWhr

 hr year 1 

dollars year

Solution 9.3

Problem Statement

Consider the startup of a refrigerator. Initially, the contents are at the same temperature as the surroundings:

TC0  TH , where TH is the (constant) surroundings temperature. With the passage of time, owing to work input, the contents’ temperature is reduced from TC0 to its design value TC. Modeling the process as a Carnot refrigerator operating between an infinite hot reservoir and a finite cold reservoir of total heat capacity Ct, determine an expression for the minimum work required to decrease the contents temperature from TC0 to TC.

Solution

Because the temperature of the finite cold reservoir (contents of the refrigerator) is a variable, use differential forms of Carnot’s equations, Eqs. (5.4) and (5.5), with some rearrangement:

dQH T  H dQC TC

 T  dW    C dQH  TH 

In these equations QC and QH refer to the reservoirs. With dQC  C t dTC , the first of Carnot’s equations becomes: dQH  TH C t

dTC TC

Combine this with the other equation: dW  C t dTC  TH C t

dTC TC

Integration from TH toTC yields:

T  W  C t TH ln C   C t TC  TH   TH  Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Or simplified:

T  W  C t TH ln C   C t TC  TH   TH 

 T  T  W  C t TH * ln C   C  1   TH  TH 

Solution 9.4

Problem Statement

A Carnot refrigerator has tetrafluoroethane as the working fluid. The cycle is the same as that shown by Fig. 8.2, except the directions are reversed. For TC =

TH = 40°C, determine:

(a) The pressures at states 1, 2, 3, and 4. (b) The quality xv at states 3 and 4. (c) The heat addition per kg of fluid. (d) The heat rejection per kg of fluid. (e) The mechanical power per kg of fluid for each of the four steps. (f) The coefficient of performance

for the cycle.

Solution

Basis: 1 lbm of tetrafluoroethane Part(a): The following property values are found from Table 9.1:

State 1, Sat. Liquid at TH: H1 

kJ kg

S1 

kJ kgK

P1 

bar

State 2, Sat. Vapor at TH: H 2 

kJ kg

S2 

kJ kgK

P2 

bar

State 3, Sat. Vapor at TC: H liq 

kJ kg

H vap 

kJ kg

State 4, Sat. Vapor at TC: Sliq 

Svap 

kJ kgK kJ kgK

P3  P4 

bar

Part (b): Steps 3-2 and 1-4 (Fig. 8.2) are isentropic, for which S3 = S2 and S1 = S4. Thus by Eq. 6.82): x3 

S2  Sliq

x3 

Svap  Sliq

S1  Sliq

x4 

Svap  Sliq

x4 

Part (c): The heat addition is between steps 4-3 H3  Hliq  x3   H vap  H liq 

H3 

kJ kg

H 4  Hliq  x 4   H vap  H liq 

H4 

kJ kg

Q43  H3  H 4

Q43 

kJ kg

Q21  

kJ kg

Part (d): The heat rejection is between steps 2-1 Q21  H1  H2

Solution continued on next page…

Part (e): the mechanical power for each step is as follows:

W21 

W43 

W32  H 2  H3

W32 

kJ kg

W14  H 4  H1

W14  

kJ kg

Part (f): the coefficient of performance is: 

Q43 W14  W32



Solution 9.5

Problem Statement

Which is the more effective way to increase the coefficient of performance of a Carnot refrigerator: to increase TC with TH constant, or to decrease TH with TC constant? For a real refrigerator, does either of these strategies make sense?

Solution

Differentiation of Eq. (9.3) for constant TH yields:    TC TH 1    T   T  T  (T  T )2  (T  T )2  C TH H C H C H C

for constant TC yields:    TC   2  T     H TC TH  TC 

Because TH  TC

Solution continued on next page…

However, for a real refrigeration system increasing TC TH TH is usually the ambient temperature and usually not subject to control. In colder climates though this could be

done by placing the refrigerator outside.

Solution 9.6

Problem Statement

In comparing the performance of a real cycle with that of a Carnot cycle, one has in principle a choice of temperatures to use for the Carnot calculation. Consider a vapor-compression refrigeration cycle in which the average fluid temperatures in the condenser and evaporator are TH and TC, respectively. Corresponding to TH and TC, the heat transfer occurs with respect to surroundings at temperature T H and T C. Which provides the more conservative estimate of

Carnot: a calculation based on TH and TC, or one based on T H and T C?

Solution

For a Carnot refrigerator,

is given by Eq. (9.3). Write this equation for the two cases: 

TH TH  TC

 

T H T H  TC

Because the directions of heat transfer require that TH > T H and TC < T C, a comparison shows that and therefore that

is the more conservative value.

Solution 9.7

Problem Statement

A Carnot engine is coupled to a Carnot refrigerator so that all of the work produced by the engine is used by the 1

refrigerator in extraction of heat from a heat reservoir at 0°C at the rate of 35 kJ s . The source of energy for the Carnot engine is a heat reservoir at 250°C. If both devices discard heat to the surroundings at 25°C, how much heat does the engine absorb from its heat-source reservoir?

If the actual coefficient of performance of the refrigerator is

= 0.6

Carnot and if the thermal efficiency of the engine

is = 0.6 Carnot, how much heat does the engine absorb from its heat-source reservoir?

Solution

The hot temperature and cold temperature for the engine is:

TH 

TC 

The hot temperature and cold temperature for the refrigerator is:

TH 

TC 

carnot  

TC TH

 carnot 

TC TH  TC

 carnot 

The carnot efficiency is found by eq 5.7:

Using eq 9.3, the coefficient of performance is:  carnot 

So we also know that 

Wengine QH



QC Wrefrig

And we know that QC 

Wengine  Wrefrig

kJ s

Then the heat absorbed is: kJ 35 QC s QH    carnot * carnot 0.43 * 10.926

Now given that = 0.6

Carnot

and

Q QH  C  *

= 0.6

Carnot

QC **



kJ s

QH 

then the actual heat absorbed is: 35



kJ s *



QH 

kJ kg

Solution 9.8

Problem Statement

A refrigeration system requires 1.5 kW of power for a refrigeration rate of 4 kJ s . (a) What is the coefficient of performance? (b) How much heat is rejected in the condenser? (c) If heat rejection is at 40°C, what is the lowest temperature the system can possibly maintain?

Solution (a): Using Eq. 9.2, the coefficient of performance is: 4 kJ QC s    2.667 W 1.5 kW

(b): The heat rejected can be found by rearranging Eq. 9.1:

QH  QC  W

QH 

(c): To find the lowest temperature the system can maintain, start with Eq. 9.3 

TH TH  TC

Rearrange to solve for TC    TC  TH      1 

With TH 



TC is TC 

 2.667     2.667  1 

K or 

Solution 9.9

Problem Statement

A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is tetrafluoroethane (Table 9.1, Fig. F.2). For one of the following sets of operating conditions, determine the circulation rate of the refrigerant, the heat-transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle, and the coefficient of performance of a Carnot refrigeration cycle operating between the same temperature levels. 1

(a) Evaporation T = 0°C; condensation T = 26°C; (compressor) = 0.79; refrigeration rate = 600 kJ·s . 1

(b) Evaporation T = 6°C; condensation T = 26°C; (compressor) = 0.78; refrigeration rate = 500 kJ·s .

T = 26°C; (compressor) = 0.77; refrigeration rate = 400 kJ·s .

(d) Evaporation T =

T = 26°C; (compressor) = 0.76; refrigeration rate = 300 kJ·s .

(e) Evaporation T =

T = 26°C; (compressor) = 0.75; refrigeration rate = 200 kJ·s .

See Table 9.1 and Fig. F.2 on next pages…

Solution at end of Tables…

Solution

The data below is the from the table 9.1 and is for parts (a) through (e). The H2 and S2 values are for saturated vapor. The H4 value is for saturated liquid.

T4 (K)

299.15

H4 (kJ/kg)

235.97

Part

T2 (K)

QdotC (kJ/s)

H2 (kJ/kg)

S2 (kJ/kg K)

273.15

0.79

600

398.6

1.727

267.15

0.78

500

395.06

1.731

261.15

0.77

400

391.46

1.735

255.15

0.76

300

387.79

1.74

248.15

0.75

200

383.45

1.746

The saturation pressure at Point 4 from Table 9.1 is 6.854 bar. For isentropic compression, from Point 2 to Point 3', we must read values for the enthalpy at Point 3' from Fig. F.2 at this pressure and at the entropy values S2. This cannot be done with much accuracy. The most satisfactory procedure is probably to read an enthalpy at S = 1.70 (H = 405) and at S = 1.80 (H = 440) and interpolate linearly for intermediate values of H. This leads to the following values (rounded to 1 decimal): H3' (kJ/kg)

H3 (kJ/kg)

414.5

20.12658

418.7266

415.9

26.71795

421.7779

417.3

33.55844

425.0184

419

41.06579

428.8558

421.1

50.2

433.65

Solution continued on next page…

The values for H23 and H3 can be found through the equations below. This values have been calculated above. H 23 

H 3  H2 

H3  H 2  H 23

kJ kg

H1  H 4 

 Q H At this point the values for m

W can be found by:

m 

Q C

Q H  m *  H 4  H3 

H 2  H1

W  m * H 23

mdot (kg/s)

QdotH (kJ/s)

Wdot (kJ/s)

3.69

–674.25

74.25

3.14

–583.97

83.97

2.57

–486.33

86.33

1.98

–381.15

81.15

1.36

–268.08

68.08

Lastly the coefficient of performance of the cycle and the coefficient of performance for the carnot cycle is:



QC W

TC  T2

TH  T4

8.080358

10.50577

5.954424

8.348438

4.633409

6.872368

3.696995

5.798864

2.937849

4.865686

carnot 

TC TH  TC

Solution 9.10

Problem Statement

A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is water. Given that the 1

evaporation T = 4°C, the condensation T = 34°C, (compressor) = 0.76, and the refrigeration rate = 1200 kJ s , determine the circulation rate of the refrigerant, the heat-transfer rate in the condenser, the power requirement, the coefficient of performance of the cycle, and the coefficient of performance of a Carnot refrigeration cycle operating between the same temperature levels.

Solution Subscripts in the following refer to Fig. 9.1. All property values come from Tables E.1 and E.2.

T2 

T4 

Q C 

kJ s

H4 

kJ kg

H2  S2  S2 

K kJ kg

 S2 

kJ kg K on

Solution continued on next page…

The saturation pressure at Point 4 from Table E.1 is 5.318 kPa. We must find in Table E.2 the enthalpy (Point 3') at this pressure and at the entropy S2. This requires double interpolation. The pressure lies between entries for pressures of 1 and 10 kPa, and linear interpolation with P is unsatisfactory. Steam is here very nearly an ideal gas, for which the entropy is linear in the logarithm of P, and interpolation must be in accord with this relation. The enthalpy, on the other hand, changes very little with P and can be interpolated linearly. Linear interpolation with temperature is satisfactory in either case. The result of interpolation is kJ kg

H3 

H23 

H3  H 2

H23  402.368 kJ  kg

H3  H 2  H23 

H1  H 4

 kJ kg

kJ  kg

H3 

 m 

kg s

kJ kg

So the circulation rate of the refrigerant is:

m 

Q C H 2  H1



1200

kJ s

  2508.9 kJ  142.4 kJ  kg kg  

The heat transferred into the condenser is: kg  * s 

Q H  m *  H 4  H 3  

kJ  kg

kJ    Q H   kg 

kJ kg

The power requirement for this is;

W  m * H23

W 

The coefficient of performance for the cycle is:



Q C W



Lastly the coefficient of performance of a carnot cycle is: Carnot 

T2 T4  T2

Carnot 

Solution 9.11

Problem Statement

A refrigerator with tetrafluoroethane (Table 9.1, Fig. F.2) as refrigerant operates with an evaporation temperature of rated liquid refrigerant from the condenser flows through an expansion valve into the evaporator, from which it emerges as saturated vapor. 1

(a) For a cooling rate of 5 kJ·s , what is the circulation rate of the refrigerant? (b) By how much would the circulation rate be reduced if the throttle valve were replaced by a turbine in which the refrigerant expands isentropically? (c) Suppose the cycle of (a) is modified by the inclusion of a countercurrent heat exchanger between the condenser and the throttle valve in which heat is transferred to vapor returning from the evaporator. If liquid from the condenser enters the exchanger at 26°C and if vapor from the evaporator enters the exchanger at 20°C, what is the circulation rate of the refrigerant? (d) For each of (a), (b), and (c), determine the coefficient of performance for isentropic compression of the vapor.

See Table 9.1 and Fig. F.2 on next pages…

Solution after Tables…

See Fig. F.2 on next page…

Solution

Parts (a) & (b): subscripts refer to Fig. 9.1

At the conditions of Point 2 [t = –25 C and P = 1.064 bar] for sat. liquid and sat. vapor from Table 9.1: kJ kg

Hliq 

kJ kg K

Sliq 

kJ kg

H vap 

Svap 

H 2  H vap kJ kg K

For sat. liquid at Point 4 (26 C): kJ kg

H4 

S4 

kJ kg K

(a): Isenthalpic expansion: H1  H4

kJ Q C  s

m 

Q C H 2  H1

 383.45

kJ s

kJ kJ  235.97 kg kg

 m 

kg s

Solution continued on next page…

(b): Isentropic expansion: x1 

S1  S4 S1  Sliq Svap  Sliq

H1  Hliq  x1  H vap  H liq 

x1 

Q C H 2  H1

m 

H1 

kJ kg

kg s

(c): The sat. vapor from the evaporator is superheated in the heat exchanger to 20 C at a pressure of 5.717 bar. Property values for this state are read (with considerable uncertainty) from Fig. F.2: H2 A 

kJ kg

S2 A 

kJ kg K

m 

Q C H2 A  H 4

m 

kg s

(d): For isentropic compression of the sat. vapor at Point 2

S3  Svap and from Fig. F.2 at this entropy and P = 6.854 bar H3 

kJ kg

In the first case H1 has the value of H4 using Eq 9.4: a 

H2  H 4 H3  H 2



383.45  235.97  490  383.45

a 

In the second case H1 has its last calculated value [Part (b)]: b 

H 2  H1 H3  H 2



383.45  229.29  490  383.45

a 

In Part (c), compression is at constant entropy of 1.95 kJ/kg*K to the final pressure. Again from Fig. F.2: H3 

kJ kg

W   H3  H 2 A m

c 

Q C W

W  

kJ s

c 

Solution 9.12

Problem Statement

A vapor-compression refrigeration system is conventional except that a countercurrent heat exchanger is installed to subcool the liquid from the condenser by heat exchange with the vapor stream from the evaporator. The minimum temperature difference for heat transfer is 5°C. Tetrafluoroethane is the refrigerant (Table 9.1, Fig. F.2), evaporating at

0 kJ·s . If the compressor efficiency is 75%,

what is the power requirement? How does this result compare with the power required by the compressor if the system operates without the heat exchanger? How do the refrigerant circulation rates compare for the two cases?

See Table 9.1 and Fig. F.2 on next pages…

Solution after Tables…

Solution

Subscripts: see figure of the preceding problem.

At the conditions of Point 2 [sat. vapor, t = –6 C and P = 2.343 bar] from Table 9.1: kJ kg

H2 

kJ kg K

S2 

At Point 2A we have a superheated vapor at the same pressure and at 21 C. From Fig. F.2: kJ kg

H2 A 

S2 A 

kJ kg K

S4 

kJ kg K

For sat. liquid at Point 4 (26 C): kJ kg

H4 

Energy balance, heat exchanger: kJ  kg

H1  H 4  H 2 A  H 2 

Q C 

kJ s

kJ  kg

kJ  kg kJ s

Q C m   H2  H1

kJ  s

kJ kg

H1 

m 

kJ s

kg s

For compression at constant entropy of 1.92 kJ/kg*K to the final pressure of 6.854 bar, by Fig. F.2: H3 

kJ kg



H comp 

W  m H comp

H comp 

W 

H 3  H2 A 

kJ kg

If the heat exchanger is omitted, then H1 = H4. Points 2A & 2 coincide, and compression is at a constant entropy of 1.731 to P = 6.854 bar m 

Q C H2  H 4

m 

kg H 3  s

W  m Hcomp

kJ kg

H comp 

W 

H3  H 2 

H comp 

kJ kg

Solution 9.13

Problem Statement

Consider the vapor-compression refrigeration cycle of Fig. 9.1 with tetrafluoroethane as refrigerant (Table 9.1, Fig. F.2). If the evaporation temperature is of performance by making calculations for condensation temperatures of 16, 28, and 40°C. (a) Assume isentropic compression of the vapor.

(b) Assume a compressor efficiency of 75%.

See Table 9.1 and Fig. F.2 on next pages…

Solution after Tables…

Solution

Subscripts refer to Fig. 9.1.

At Point 2 [sat. vapor @ –12 C] from Table 9.1: kJ kg

H2 

kJ kg K

S2 

H values for sat. liquid at Point 4 come from Table 9.1 and H values for Point 3` come from Fig. F.2. The vectors following give values for condensation temperatures of 16, 28, 40 C at pressures of 5.043, 7.269, & 10.166 bar respectively. 221.87    kJ  H 4    kg   256.41

404    kJ  H3    kg K   420

H1  H 4

(a): H  H1  2 H3  H2

(b): H 

H3  H 2 0.75

13.52          4.73 

H  H 3  H2 , then Eq 9.4 now becomes



H2  H1 H

10.14          3.55 

Solution 9.14

Problem Statement

A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a low-temperature heat source; during the summer, it acts as a high-temperature heat sink. The heattransfer rate through the walls and roof of the house is 0.75 kJ s

for each °C of temperature difference between the

inside and outside of the house, summer and winter. The heat-pump motor is rated at 1.5 kW. Determine the minimum outside temperature for which the house can be maintained at 20°C during the winter and the maximum outside temperature for which the house can be maintained at 25°C during the summer.

Solution

TH 

W 

We know that the heat lost is:

Q H  

TH  TC 

And the work coefficient of performance: T  TC W  H  T Q H

Combining these yields:

T  TC W  H TH 0.75TH  TC  Giving a guess for TC and iterating gives

TC 

K or 

TC 

W 

We know that the heat lost is:

Q C 

TH  TC 

And the work coefficient of performance: T  TC W  H  T Q C

Combining these yields:

T  TC W  H TC 0.75TH  TC  Giving a guess for TC and iterating gives

TH 

K or

Solution 9.15

Problem Statement Dry methane is supplied by a compressor and precooling system to the cooler of a Linde liquid-methane system (Fig. 9.6) at 180 bar and 300 K. The low-pressure methane leaves the cooler at a temperature 6°C lower than the Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

temperature of the incoming high-pressure methane. The separator operates at 1 bar, and the product is saturated liquid at this pressure. What is the maximum fraction of the methane entering the cooler that can be liquefied? The NIST Chemistry WebBook (http://webbook.nist. gov/chemistry/fluid/) is a source of data for methane.

Solution

Data comes from Perry's Handbook, 7th ed. H4 

z

kJ kg

H 4  H15  H9  H15

H9 

kJ kg

H15 

kJ kJ  1186.7 kg kg kJ kJ  kg kg

1033.5

z

Solution 9.16

Problem Statement

Rework Problem 9.15 for methane entering at 200 bar and precooled to 240 K by external refrigeration.

Problem 9.15

Dry methane is supplied by a compressor and precooling system to the cooler of a Linde liquid-methane system (Fig. 9.6) at 180 bar and 300 K. The low-pressure methane leaves the cooler at a temperature 6°C lower than the temperature of the incoming high-pressure methane. The separator operates at 1 bar, and the product is saturated liquid at this pressure. What is the maximum fraction of the methane entering the cooler that can be liquefied? The NIST Chemistry WebBook (http://webbook.nist. gov/chemistry/fluid/) is a source of data for methane.

Solution

Data comes from Perry’s Handbook, 7th ed. H4 

kJ kg

H9 

H  H15 z 4  H9  H15

kJ kg

H15 

kJ kJ  1056.4 kg kg kJ kJ 284.7  1056.4 kg kg 785.3

z

Solution 9.17

Problem Statement

An advertisement is noted in a rural newspaper for a dairy-barn unit that combines a milk cooler with a water heater. Milk must, of course, be refrigerated, and hot water is required for washing purposes. The usual barn is equipped with a conventional air-cooled electric refrigerator and an electric-resistance water heater. The new unit is said to provide both the necessary refrigeration and the required hot water at a cost for electricity about the same as the cost of running just the refrigerator in the usual installation. To assess this claim, compare two refrigeration units: The advertised unit takes 15 kJ·s

from a milk cooler at

to raise the

temperature of water from 13 to 63°C. The conventional unit takes the same amount of heat from the same milk cooler at

-cooled condenser at 50°C; in addition, the same amount of water is

heated electrically from 13 to 63°C. Estimate the total electric power requirements for the two cases, assuming that the actual work in both is 50% greater than required by Carnot refrigerators operating between the given temperatures.

Solution

The advertised combination unit says: TH 

TC 

QC 

kJ s

So first find the Carnot work: Wcarnot  QC *

TH  TC TC



kJ 338.15  271.15 * s 271.15

Wcarnot 

kJ s

We know that the actual work is 1.5 times greater than the carnot refrigerator, so the actual work is WI  1.5 * Wcarnot  1.5 * 3.706  5.559

kJ s

This is the TOTAL power requirement for the advertized combination unit. The amount of heat rejected at the higher temperature of 65 C is: QH  WI  QC

kJ S

QH 

For the conventional water heater, this amount of energy must be supplied by resistance heating, which requires power in this amount. For the conventional cooling unit,

TH  Wcarnot  QC *

TC 

TH  TC

K kJ s

Wcarnot 

W  1.5 * Wcarnot  4.3155

kJ s

The amount of heat rejected at the higher temperature of 50 C is QH  WI  QC

QH 

kJ S

So the total power required by the conventional unit is WII  QH  W 

kJ  S

kJ  WII  s

kJ S

Solution 9.18

Problem Statement

A two-stage cascade refrigeration system (see Fig. 9.3) operates between TC = 210 K and TH = 305 K. Intermediate temperatures are TC = 255 K and TH = 260 K. Coefficients of performance corresponding values for a Carnot refrigerator. Determine

of each stage are 65% of the

for the real cascade, and compare it with that for a

Carnot refrigerator operating between TC and TH.

Solution

TC 

TC 

TH 

TH 

Starting from Eq. 9.3, determine each stage’s coefficient of performance: 

TC TH  TC

I 



TC *  TH  TC

I 

 II 

TC TH  TC

 II 

The work for each stage is Wcarnot 

QC 



WI 

QC I



WII 

QC  II



QC  WI  II



QC 

QC 3.315



Even though we don’t know QC , we can still define the ratio of the actual work to the carnot work as: 1 1    r      I  II  

QC QC 

QC 

r

Solution 9.19

Problem Statement

The condenser of a home refrigerator is commonly underneath the appliance; thus, the condensing refrigerant exchanges heat with household air, which has an average temperature of about 21°C. It is proposed to reconfigure a refrigerator so that the condenser is outside the home, where the average yearly temperature is about 10°C. Discuss the pros and cons of this proposal. Assume a freezer temperature of 60% that of a Carnot refrigerator.

Solution

This problem is just a reworking of Example 9.3 with different values of x. It could be useful as a group project.

Solution 9.20

Problem Statement

Solution

On average, the coefficient of performance will increase, thus providing savings on electric costs. On the other hand, installation costs would be higher. The proposed arrangement would result in cooling of the kitchen, as the refrigerator would act as an air conditioner. This would be detrimental in the winter, but beneficial in the summer, at least in temperate climates.

Solution 9.21

Problem Statement

A common misconception is that the coefficient of performance of a refrigerator must be less than unity. In fact, this is rarely the case. To see why, consider a real refrigerator for which in order for

= 0.6

Carnot. What condition must be satisfied

< 1? Assume that TH is fixed.

Solution

First the relation for the coefficient of performance is

 T  H   0.6 * carnot  0.6 *    TH  TC  Now for  

TC 

TH 1.6

TH 

K

C

TC 

K 

C  This is not likely when most

refrigerators run at their coldest –20 C.

Solution 9.22

Problem Statement

A furnace fails in a home in the winter. Mercifully, the electric power remains on. The resident engineer tells her spouse not to worry; they’ll move into the kitchen, where the heat discarded from the refrigerator may provide for a temporarily comfortable living space. However (the engineer is reminded), the kitchen loses heat to the outdoors. Use the following data to determine the allowable rate of heat loss (kW) from the kitchen for the engineer’s proposal to make sense.

Data: Desired kitchen temperature = 290 K. Refrigerator freezer temperature = 250 K. Average mechanical power input to refrigerator = 0.40 kW. Performance: Actual

= 65% of Carnot

Solution

TH 

TC 

Ws 

To get the actual coefficient of performance use the carnot coefficient of performance to determine this: carnot 

TC TH  TC

carnot 



* carnot



Now use this to determine QC

QC   * Ws 

kW 

QH 

Now for the heat loss

QH  Ws  QC 

kW 

Solution 9.23

Problem Statement Fifty (50) kmol·h

of liquid toluene at 1.2 bar is cooled from 100 to 20°C. A vapor-compression refrigeration cycle is

used for the purpose. Ammonia is the working fluid. Condensation in the cycle is effected by an air-cooled fin/fan heat exchanger for which the air temperature may be assumed essentially constant at 20°C. Determine: (a) The low and high pressure levels (bar) in the refrigeration cycle. 1

(b) The circulation rate of ammonia (mol·s ).

Assume 10°C minimum approach temperature differences for heat exchange. Data for ammonia: H nlv  23.34 kJ mol 1 ln P sat  45.327 

where P

sat

4104.67 P sat  5.146 ln T  615.0 2 T T

is in bars and T is in kelvins.

Solution

Follow the notation from Fig. 9.1. With air at 20 C and the specification of a minimum approach T = 10 C:

T1 

T4 

T2  T1

Calculate the high and low operating pressures using the given vapor pressure equation PL  PL   T1  4104.67   45.327  ln   5.146 * ln    615.0 * bar2 T1  bar   K   T1    K  K 

PL 

bar

PH  PH   T4  4104.67   45.327  ln   5.146 * ln    615.0 * bar2 T4  bar   K   T4    K  K 

PH 

bar

Calculate the heat load kmol hr

n toluene 

T1 

T2 

Using values from Table C.3 Q C  n toluene * R * ICPH T1 T2

Q C 

Hliq4  Hliq1  x1Hlv 1

And:

Hliq 4  Hliq1  Vliq P4  P1   

3

6



H4  H1

Since the throttling process is adiabatic:

But:

Hliq 4  Hliq1  x1Hlv 1 T4 T1

C pliq T dT

Solution continued on next page…

Estimate Vliq using the Rackett Eqn.

 ZC 

VC 

TC 

PC 

bar

cm3 mol

Tn 

H lvn  3.34

Tr 

Vliq  Vc Z c

293.15 K TC

kJ mol

Tr 

0.2857

1Tr 

cm3 mol

Vliq 

Estimate Hlv at 10C using Watson correlation

Trn 

Tr 1 

 1  T  r1  Hlv  H lvn    1  Trn 

0.38

Hliq 41  Vliq  PH  PL   R * ICPH T1 T2

kJ mol

Hliq 41 

kJ mol

Hlv 



x1 

H liq 41 Hlv

3

6



x1 

H12  H 2  H1  H1 vap   H1liq  x1Hlv   1  x1 Hlv

H12  n 

Q C H12

kJ n 

mol s

Solution 9.24 Problem Statement

Cascade refrigeration is often applied in freeze dryers, where water is sublimated from a frozen sample under vacuum and collects on the evaporator coil of the low-temperature stage of a two-stage refrigeration cascade. A particular freeze dryer is specified as being able to collect (freeze from the low-pressure gas phase) 4 kg of water per hour while maintaining a coil temperature of –50 °C. The condenser of the hightemperature stage is designed to operate at 40 °C. A 10 °C temperature difference is required for heat transfer from the condenser of the low-temperature stage to the evaporator of the high-temperature stage. Propane (R-290) is used as refrigerant in the high-temperature stage, while ethane (R-170) is used in the Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

low-temperature stage. Properties data for these are available in the NIST WebBook. The heat of –1 sublimation of ice at –50 °C is 2838 kJ · kg . For both stages, the isentropic efficiency of the compressor is 75%. Compute the work requirement and refrigeration recirculation rate for each stage, if the hightemperature evaporator and low-temperature condenser operate at: (a) –15 °C and –5 °C (b) –10 °C and 0 °C (c) –5 °C and 5 °C Solution

(a) With the interchanger temperature specified, we simply analyze each vapor-compression cycle separately. Using the notation of Fig. 9.1, we obtain the properties of the saturated vapor leaving the evaporator of the high-temperature side (point 2) and of the saturated liquid leaving the condenser of the high-temperature cycle (point 4) directly from the NIST WebBook: –1

–1

H2,h = 558.08 kJ·kg , S2,h = 2.3924 kJ·kg ·K , H4,h = H1,h = 307.50 kJ·kg , S4,h = 1.3606 kJ·kg ·K

The evaporator pressure is P2,h = 2.9164 bar, and the condenser pressure is P4,h = 13.697 bar. Next, we want to find the work for isentropic compression of the vapor from the evaporator to the condenser pressure. At 13.697 bar, we find that the entropy is 2.3924 kJ·kg–1·K–1 for a temperature of 47.25 °C, where the enthalpy is –1

–1

H3’,h = 630.80 kJ·kg . Thus, the isentropic work is (Ws)S = 630.80 – 558.08 = 72.72 kJ·kg . The actual work is then Ws,h = 72.72/0.75 = 96.96 kJ·kg–1, and the enthalpy at point 3, leaving the compressor, is H3,h = 558.08 + 96.96 = 655.04 kJ·kg–1. The heat effect in the condenser is QH,h = 307.50 – 655.04 = –347.54 kJ·kg–1, and that –1

in the evaporator is QC,h = 558.08 – 307.50 = 250.58 kJ·kg . Following the same process for the low-temperature cycle with properties of ethane: H2,l = 527.07 kJ·kg–1, S2,l = 2.4000 kJ·kg–1·K–1, H4,l = H1,l = 230.58 kJ·kg–1, S4,l = 1.0028 kJ·kg–1·K–1 The evaporator pressure is P2,l = 5.5216 bar, and the condenser pressure is P4,l = 21.115 bar. At 21.115 bar, we find that the entropy is 2.4000 kJ·kg–1·K–1 for a temperature of 18.95 °C, where the enthalpy is H3’,l = 607.62 kJ·kg–1. Thus, the isentropic work is (Ws)S = 607.62 – 527.07 = 80.55 kJ·kg–1. The actual work is then Ws,l = 80.55/0.75 = 107.40 kJ·kg–1, and the enthalpy at point 3, leaving the compressor, is H3,l = 558.08 + 96.96 = 634.47 kJ·kg–1. The heat effect in the condenser is QH,h = 230.58 – 634.47 = –403.89 kJ·kg–1, and that in the evaporator is QC,h = 527.07 – 230.58 = 296.49 kJ·kg–1. Solution continued on next page…

Now, we want to relate these values per kg of refrigerant in each stage to the actual cooling requirement of –1 –1 –1 –1 Q C ,l = 2838 kJ·kg · 4 kg·h = 11352 kJ·h = 3.1533 kJ·s = 3.1533 kW to freeze 4 kg per hour of saturated

water vapor at –50 °C. The actual duty would be a bit higher due to heat transfer to the condenser coils from their surroundings, but we neglect that here. The refrigerant circulation rate in the low-temperature cycle is then m l = 3.1533 kJ·s–1/ 296.49 kJ·kg–1 = 0.01064 kg·s–1 = 10.6 g·s–1. The rate of work input is then –1 –1 –1 W s ,l  0.01064 kg·s * 107.40 kJ·kg = 1.142 kW. The interchanger duty is Q H ,l = 0.01064 kg·s * –403.89

kJ·kg–1 = –4.296 kW. This is equal in magnitude to the heat absorption rate of the high-temperature cycle, Q C ,h = 4.296 kW, from which the refrigerant circulation rate in the high-temperature cycle is m h = 4.2956

kJ·s–1/250.58 kJ·kg–1 = 0.01714 kg·s–1 = 17.1 g·s–1. The rate of work input is then W s,h  0.01714 kg·s–1 * 96.96 kJ·kg = 1.662 kW. The total work input is W s ,t = 1.142 + 1.662 = 2.804 kW. –1

(b) Doing the same thing for interchanger temperatures of –10 °C and 0 °C, we have: –1

–1

H2,h = 563.84 kJ·kg , S2,h = 2.3853 kJ·kg ·K , H4,h = H1,h = 307.50 kJ·kg , S4,h = 1.3606 kJ·kg ·K P2,h = 3.4530 bar, P4,h = 13.697 bar, T3’,h = 46.22 °C, H3’,h = 628.53 kJ·kg , (Ws)S = 64.69 kJ·kg –1

–1

Ws,h = 86.25 kJ·kg , H3,h = 650.09 kJ·kg , QH,h = –342.59 kJ·kg , QC,h = 256.34 kJ·kg –1

–1

H2,l = 527.07 kJ·kg , S2,l = 2.4000 kJ·kg ·K , H4,l = H1,l = 247.57 kJ·kg , S4,l = 1.0630 kJ·kg ·K P2,l = 5.5216 bar, P4,l = 23.875 bar, T3’,l = 26.10 °C, H3’,l = 615.63 kJ·kg , (Ws)S = 88.56 kJ·kg –1

–1

Ws,l = 118.08 kJ·kg , H3,l = 650.09 kJ·kg , QH,l = –397.58 kJ·kg , QC,l = 279.50 kJ·kg

–1 –1 m l = 0.01128 kg·s , W s ,l  1.3322 kW, Q H ,l = –4.486 kW, Q C ,h = 4.486 kW, m h = 0.01750 kg·s

W s ,h  1.503 kW, W s ,t = 2.841 kW.

(c) Doing the same thing for interchanger temperatures of –5 °C and 5 °C, we have: H2,h = 569.52 kJ·kg–1, S2,h = 2.3789 kJ·kg–1·K–1, H4,h = H1,h = 307.50 kJ·kg–1, S4,h = 1.3606 kJ·kg–1·K–1 P2,h = 4.0606 bar, P4,h = 13.697 bar, T3’,h = 45.30 °C, H3’,h = 626.50 kJ·kg–1, (Ws)S = 56.98 kJ·kg–1 Ws,h = 75.97 kJ·kg–1, H3,h = 645.49 kJ·kg–1, QH,h = –337.99 kJ·kg–1, QC,h = 262.02 kJ·kg–1 Solution continued on next page…

H2,l = 527.07 kJ·kg–1, S2,l = 2.4000 kJ·kg–1·K–1, H4,l = H1,l = 265.35 kJ·kg–1, S4,l = 1.1247 kJ·kg–1·K–1 P2,l = 5.5216 bar, P4,l = 26.891 bar, T3’,l = 33.18 °C, H3’,l = 23.48 kJ·kg–1, (Ws)S = 96.41 kJ·kg–1 Ws,l = 128.55 kJ·kg–1, H3,l = 655.62 kJ·kg–1, QH,l = –390.27 kJ·kg–1, QC,l = 261.72 kJ·kg–1 –1 –1 m l = 0.01205 kg·s , W s ,l  1.5488 kW, Q H ,l = –4.702 kW, Q C ,h = 4.702 kW, m h = 0.01795 kg·s

W s ,h 

1.363 kW, W s ,t = 2.912 kW.

Solution 10.1 Problem Statement 3

What is the change in entropy when 0.7 m of CO2 and 0.3 m of N2, each at 1 bar and 25°C, blend to form a gas mixture at the same conditions? Assume ideal gases. Solution For ideal gases, the partial molar entropy is Siig T , P   Siig T , pi   Siig T , P   R ln yi

So the entropy change on mixing is N

i 1

ig Smix  S ig   yi Siig  R  yi ln yi  R  yi ln

1 yi

In this case, that is mix

S

–1

= 8.314*(0.3 ln (1/0.3) + 0.7 ln (1/0.7)) = 5.08 J mol K

The volume change of mixing for ideal gases is zero, so both before and after mixing, we have 1.0 m of gas. At 1 bar and 25 degrees C, this is n = PV/RT = 100000*1/(8.3145*298.15) = 40.33 moles –1

So, the total entropy change on mixing is 5.08 * 40.33 = 204.9 J K .

Solution 10.2 Problem Statement A vessel, divided into two parts by a partition, contains 4 mol of nitrogen gas at 75°C and 30 bar on one side and 2.5 mol of argon gas at 130°C and 20 bar on the other. If the partition is removed and the gases mix adiabatically and completely, what is the change in entropy? Assume nitrogen to be an ideal gas with CV = (5/2)R and argon to be an ideal gas with CV = (3/2)R. Solution One way to approach this problem is to first find the final temperature and pressure of the system, then compute the entropy change of the pure components in going from their respective initial states to the final temperature and pressure, and then finally compute the entropy of mixing at the final temperature. There will be some entropy increase due to the mixing of different gases, and some additional entropy increase due to mixing of fluids at different temperature (both processes are irreversible). Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The overall process is adiabatic and occurs at constant (total) volume (so no work or heat, including flow work, is exchanged with the surrounding). So, the internal energy change for the whole process is zero, and the internal energy change for mixing of ideal gases is zero. So, the internal energy increase of the nitrogen (which is initially at lower temperature) must be equal to the internal energy decrease of the argon (which is initially –1

–1

at higher temperature). The total heat capacity of the nitrogen is 4 mol * 2.5 * R J mol K . The total heat –1

–1

capacity of the argon is 2.5 * mol * 1.5 * R J mol K . So, we have (4*2.5*R)*(T–75) = (2.5*1.5*R)*(130–T) 5.5*T = 1.5*130 + 4*75 T = 90 C Now, we want to find the final pressure. The total volume is equal to the total initial volume, which is –1

–1

V = 4 mol * R J mol K * 348.15 K / 3000000 Pa + 2.5 mol * R J mol K * 403.15 K / 2000000 Pa = 3

0.00805 m . So, at the final temperature, the pressure will be –1

–1

P = 6.5 mol * R J mol K * 363.15 K / 0.00805 m = 2438000 Pa = 24.38 bar The change in entropy for the nitrogen going from 75C and 30 bar to 90C and 24.38 bar is S = n(Cp ln (T/To) – R ln (P/Po)) = 4*(3.5 R ln (363.15/348.15) – R ln(24.38/30)) –1

S = 11.809 J K

The change in entropy for the argon going from 130C and 20 bar to 90C and 24.38 bar is S = n(Cp ln (T/To) – R ln (P/Po)) = 2.5*(2.5 R ln (363.15/403.15) – R ln(24.38/20)) –1

S = –9.546 J K

So, the net change in entropy for taking the pure components from their initial temperatures and pressures to –1

the final temperature and pressure is 11.809 – 9.546 = 2.26 J K . The entropy change of mixing is then computed in the same way as in problem 1. The mole fraction nitrogen is yN2 = 4/6.5 = 0.615 and yAr = 0.385. The entropy change of mixing is then mix

S

–1

= 6.5*8.314*(0.385 ln (1/0.385) + 0.615 ln (1/0.615)) = 36.02 J K . –1

Thus, finally, the total change in entropy is 36.02 + 2.26 = 38.28 J K .

Solution 10.3 Problem Statement

A stream of nitrogen flowing at a rate of 2 kg·s

and a stream of hydrogen flowing at a rate of 0.5 kg·s

mix

adiabatically in a steady-flow process. If the gases are assumed ideal, what is the rate of entropy increase as a result of the process? Solution For ideal gases, the enthalpy of mixing is zero, so if the gases come in at the same temperature, they will also leave at that temperature (there will be no heat effects). For ideal gases, the partial molar entropy is Siig T , P   Siig T , pi   Siig T , P   R ln yi

So the entropy change of mixing is N

i 1

ig Smix  S ig   yi Siig  R  yi ln yi  R  yi ln

1 yi

We need to evaluate the molar flow rates to get the mole fractions. 2 kg/s N2 with a molecular weight of 0.028 kg/mol corresponds to a molar flow rate of 71.4 moles/s. 0.5 kg/s of H2 with a molecular weight of 0.002 kg/mol corresponds to a molar flow rate of 250 moles/s. So, the total molar flow rate is 321.4 moles per second and the mole fractions are xH2 = 0.778, xN2 = 0.222. So, the entropy of mixing per mole is mix

S

–1

= 8.314*(0.222 ln (1/0.222) + 0.778 ln (1/0.778)) = 4.40 J mol K –1 –1

So, the total rate of entropy change is 4.40 * 321.4 = 1415 J K s .

Solution 10.4 Problem Statement

What is the ideal work for the separation of an equimolar mixture of methane and ethane at 175°C and 3 bar in a steady-flow process into product streams of the pure gases at 35°C and 1 bar if T = 300 K?

Solution We learned in chapter 5 that the ideal work (minimum work input required or maximum work that can be extracted) to achieve a change of state of a fluid is Wideal   H  T S . So, what we need to do is compute H and S for going from a mixture of 50% methane/50% ethane at 175C and 3 bar to the separate components at 35C and 1 bar. We can use a 3-step hypothetical process to achieve this change, in which we (1) unmix the components to get the pure species, then (2) expand them isothermally from 3 bar to 1 bar, then (3) cool them at constant pressure from 175C to 35C. Let’s do this all on a basis of 1 mole of the mixture (0.5 mol ethane, 0.5 mol methane). Of course, we will assume that methane and ethane are ideal gases at these relatively low pressures (reduced pressures around 0.1) and high temperatures (reduced temperatures above 1.0). So, for these 3 steps, we have (1) Unmixing. The enthalpy of mixing for ideal gases is zero, so H1 = 0. The entropy of mixing is N

ig Smix  R  yi ln i 1

1 . Since we are unmixing the gases, we want the negative of the entropy of mixing, which yi

for this equimolar binary mixture is just S1 = –R*(0.5*ln(2) + 0.5*ln(2)) = –R ln(2) = –5.763 J mol K –1

–1

(2) Isothermal expansion. The enthalpy of ideal gases is independent of pressure, so H2 = 0. The entropy change is S2 = R ln(P1/P2) = R ln(3) = 9.134 J mol K . The ½ mole of ethane and ½ mole of methane each –1

–1

have the same entropy change in this step, so the total entropy change is that for 1 mole of gas. (3) Isobaric cooling. To compute the changes in enthalpy and entropy when the two pure substances are cooled from 175C to 35C, we need to do the usual heat capacity integrals. Putting the appropriate temperatures and ideal gas heat capacity parameters into the spreadsheets we have already set up for evaluating the integrals, we get: For methane: T1 (K) 448.15

T2 (K) 308.15 T2

ICPH 

R

A 1.702

dT  A T2  T1  

C (1/K ) D (K ) B (1/K) 9.08E-03 -2.16E-06 0.00E+00

ICPH (K) -6.75E+02

H (J/mol) -5614

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









Solution continued on next page…

T1 (K) 448.15

T2 (K) 308.15 T2

ICPS 

 T1

A 1.702

C (1/K2) D (K2) B (1/K) 9.08E-03 -2.16E-06 0.00E+00

ICPS -1.794

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT 2 2  T1 







S (J/(mol K)) -14.92



And for ethane: T1 (K) 448.15

T2 (K) 308.15 T2

ICPH 

R

A 1.131

dT  A T2  T1  

T1 (K) 448.15 T2

ICPS 

T2 (K) 308.15

A 1.131

 T2 

C (1/K ) D (K ) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

ICPH (K) -1.06E+03

H (J/mol) -8843

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









C (1/K2) D (K2) B (1/K) 1.92E-02 -5.56E-06 0.00E+00

ICPS -2.821

S (J/(mol K)) -23.45

 RT dT  A ln  T   B T  T   2 T  T   2 T  T  2

2 2

2 1

2 2

2 1

So, for ½ mole of methane and ½ mole of ethane (to give the result per mole of mixture entering the process) we have H3 = ½ * –5614 + ½ * –8843 = –7229 J/mol, and S3 = ½ * –14.92 + ½ * –23.45 = –19.19 J mol K . –1

–1

Adding the three steps together, we have for the overall process: H = –7229 J/mol and S = –5.76 + 9.13 – –1

–1

19.19 = –15.82 J mol K . So, finally, the ideal work is Wideal = H – TS = –7229 – 300*(–15.82) = –2485 J/mol. In fact, the enthalpy decrease more than makes up for the entropy decrease, so the ideal work is negative. We can actually take work out of the overall process (in an ideal world with no entropy generation). The maximum work that could be extracted from the combined separation, cooling, and expansion is 2485 J per mol of mixed gases entering the process.

Solution 10.5 Problem Statement

What is the work required for the separation of air (21-mol-% oxygen and 79-mol-% nitrogen) at 25°C and 1 bar in a steady-flow process into product streams of pure oxygen and nitrogen, also at 25°C and 1 bar, if the thermodynamic efficiency of the process is 5% and if T = 300 K? Solution We learned in chapter 5 that the ideal work, the minimum work required for a steady-flow process that requires work input to achieve a change in state of the process stream is given by Wideal   H  T S

If the process has a thermodynamic efficiency of 5%, then the actual work can be found from the definition of thermodynamic efficiency: t  work required 

W ideal  0.05 W s

From which the actual work is Ws  20Wideal  20 H  T S

Air at 1 bar can be treated as an ideal gas, and the pressure and temperature of the outlet streams are the same as that of the inlet stream. Therefore, the enthalpy and entropy changes are only those due to the “unmixing” of the nitrogen and oxygen. H = 0. The entropy of mixing for ideal gases is N

i 1

ig Smix  S ig   yi Siig  R  yi ln yi  R  yi ln

1 yi

In this case, it is mix

S

–1

= 8.314*(0.21 ln (1/0.21) + 0.79 ln (1/0.79)) = 4.27 J mol K . –1

The process we are looking at is unmixing –

–1

S = –4.27 J mol K . Using this in the

expression for the work required, we obtain –1

–1

Ws = 20*(0 – 300 K * (–4.27 J mol K ) = 25640 J/mol = 25.6 kJ/mol About 25.6 kJ is required to separate one mole of air into nitrogen and oxygen streams with a process that has a thermodynamic efficiency of 5%.

Solution 10.6 Problem Statement What is the partial molar temperature? What is the partial molar pressure? Express results in relation to the T and P of the mixture. Solution Using Eq 10.7, the partial molar temperature is:   nT    n    Ti    T  T   ni   ni  P T n P, T , n j j And the partial molar pressure using Eq. 10.7 is:   nP    n    Pi    P  P   ni   ni  P T n P, T , n j j

Solution 10.7 Problem Statement

Show that: (a) The “partial molar mass” of a species in solution is equal to its molar mass. (b) A partial specific property of a species in solution is obtained by division of the partial molar property by the molar mass of the species. Solution Part(a): Let m be the mass of the solution, and define the partial molar mass by:   nT    mi     ni  P , T , n j

Let Mk be the molar mass of species k. This leads

m  mk M k  ni M i  n j M j k

 j  i

Solution continued on next page…

And   n M    m  i i       Mi   n   n  i P T n j   P T n i j

Which yields

mi  Mi Part(b): Define a partial specific property as:  M t   M i    mi 

P, T , mj

If Mi is the molar mass of species i then

ni 

 M t      ni 

 n   i   mi   P, T, mj P, T , mj  n   i   mi 

mi Mi

P, T, mj



1 Mi

Because constant mj implies constant nj , the initial equation may be written Mi 

Mi Mi

Solution 10.8 Problem Statement If the molar density of a binary mixture is given by the empirical expression: = a0 + a1 x1 + a2 x 12 find the corresponding expressions for V1 and V2 . Solution For any partial molar property of a binary mixture, we derived in class (and it is derived in the text) that M1  M  x 2

dM dM  M  1  x1  dx1 dx1

M 2  M  x1

dM dx1

and

Solution continued on next page…

So, if we write the molar volume as a function only of x1, then we can apply these equations to get the partial molar volume of each species. The molar volume is just the inverse of the molar density. So, in this case, we have 1 1 V   a0  a1 x 1  a2 x 12 Taking the derivative of this with respect to x1 gives a1  2a2 x1 dV  2 dx1 a  a x  a x2



1 1



Substituting this into the relationships for the partial molar volumes gives V1  V  1  x1  V1 

1  x1 a1  2a2 x1  dV 1   2 2 dx1 a0  a1 x1  a2 x1 a  a x  a x2



1 1

a0  a1 x1  a x  a1  2a2 x1  a1 x 1  2a2 x



2 1

a  a x  a x  0

V1 

2 2 1

1 1

2 1

a0  a1  2a1 x1  2a2 x 1  3a2 x12

a  a x  a x  0

1 1

2 2 1

and V2  V  x1 V2 

x1 a1  2a2 x1  dV 1   2 2 dx1 a0  a1 x1  a2 x1 a  a x  a x2



a0  2a1 x2  3a2 x



2 1 2

a  a x  a x  0

1 1

2 1

Solution 10.9 Problem Statement For a ternary solution at constant T and P, the composition dependence of molar property M is given by: M = x1 M1 + x2 M2 + x3 M3 + x1x2x3 C where M1, M2, and M3 are the values of M for pure species 1, 2, and 3, and C is a parameter independent of composition. Determine expressions for M 1 , M 2 , and M 3 by application of Eq. (10.7). As a partial check on your results, verify that they satisfy the summability relation, Eq. (10.11). For this correlating equation, what are the Mi at infinite dilution?

Eq. 10.7:   nM    Mi     ni  P , T , n j

Eq. 10.11:

M   xi M i i

Solution The definition of a partial molar property of species i in a mixture is   nM    M i    ni  T , P ,n

ji

Multiplying the given composition dependence by the total number of moles n gives (writing the mole fractions in terms of the numbers of moles (x1 = n1/n = n1/(n1 + n2 + n3), etc.) 2

nM = n1M1 + n2M2 + n3M3 + (n1n2n3/n )C 2

nM = n1M1 + n2M2 + n3M3 + n1n2n3C/(n1 + n2 + n3)

Taking the partial derivative of this with respect to n1 (holding n2 and n3 constant) gives   nM   2 3   M1   = M1 + n2n3C/(n1 + n2 + n3) – 2n1n2n3C/(n1 + n2 + n3)  n1  T, P,n , n 2

Similarly, taking the partial derivatives with respect to n2 and n3 gives 2

M2 = M2 + n1n3C/(n1 + n2 + n3) – 2n1n2n3C/(n1 + n2 + n3)

M3 = M3 + n1n2C/(n1 + n2 + n3) – 2n1n2n3C/(n1 + n2 + n3)

Or, writing these in terms of the mole fractions, M1 = M1 + x2x3C– 2x1x2x3C = M1 + (1– 2x1) x2x3C M2 = M2 + x1x3C – 2x1x2x3C = M2 + (1– 2x2) x1x3C M3 = M3 + x1x2C – 2x1x2x3C = M3 + (1– 2x3) x1x2C

Summing up the three mole fractions times the partial molar properties gives: M = x1 M 1  x 2 M 2  x 3 M 3 M = x1 (M1 + x2x3C – 2x1x2x3C) + x2 (M2 + x1x3C – 2x1x2x3C) + x3 (M3 + x1x2C – 2x1x2x3C) M = x1 M1 + x2M2 + x3M3 + 3x1x2x3C – 2(x1 + x2 + x3) x1x2x3C M = x1 M1 + x2M2 + x3M3 + x1x2x3C Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The infinite dilution values are the values of the partial molar properties for the component as the mole fraction of that component goes to zero: lim x1  0 M1 = M1 + x2x3C lim x2  0 M 2 = M2 + x1x3C lim x3  0 M 3 = M3 + x1x2C

Solution 10.10 Problem Statement A pure-component pressure pi for species i in a gas mixture may be defined as the pressure that species i would exert if it alone occupied the mixture volume. Thus, pi 

yi Zi RT V

where yi is the mole fraction of species i in the gas mixture, Zi is evaluated at pi and T, and V is the molar volume of the gas mixture. Note that pi as defined here is not a partial pressure yiP, except for an ideal gas. Dalton’s “law” of additive pressures states that the total pressure exerted by a gas mixture is equal to the sum of the pure-component pressures of its constituent species: P   pi . Show that Dalton’s “law” implies that i

Z   yi Z i , where Zi is the compressibility factor of pure species i evaluated at the mixture temperature but at i

its pure-component pressure. Solution With the given equation and the Dalton’s-law requirement that P   pi . Then: i

P

RT y Z V i i i

Knowing that P = ZRT/V. these two equations combine to yield Z   yi Zi i

Solution 10.11 Problem Statement

If for a binary solution one starts with an expression for M (or M or M ) as a function of x1 and applies Eqs. (10.15) and (10.16) to find M1 and M2 (or M1R and M2R or M1E and M2E ) and then combines these expressions by Eq. (10.11), the initial expression for M is regenerated. On the other hand, if one starts with expressions for M 1 and M 2 , combines them in accord with Eq. (10.11), and then applies Eqs. (10.15) and (10.16), the initial

expressions for M1 and M 2 are regenerated if and only if the initial expressions for these quantities meet a specific condition. What is the condition? Eq. 10.11: M   xi M i i

Eq. 10.15: M1  M  x 2

dM dx1

Eq. 10.16: M 2  M  x1

dM dx1

Solution

The general principle is simple enough: Given equations that represent partial properties M i , M iR , M iE as functions of composition, one may combine them by the summability relation to yield a mixture property. Application of the defining (or equivalent) equations for partial properties then regenerates the given equations if and only if the given equations obey the Gibbs/Duhen equation.

Solution 10.12 Problem Statement With reference to Ex. 10.4, (a) Apply Eq. (10.7) to Eq. (A) to verify Eqs. (B) and (C). (b) Show that Eqs. (B) and (C) combine in accord with Eq. (10.11) to regenerate Eq. (A). (c) Show that Eqs. (B) and (C) satisfy Eq. (10.14), the Gibbs/Duhem equation. (d) Show that at constant T and P,

 dH   1   dx  1

xi 1

 dH    2   dx1 

 0

x1  0

(e) Plot values of H, H1 , and H2, calculated by Eqs. (A), (B), and (C), vs. x1. Label points

H1, H2 , H1 , and H2, and show their values. Example 10.4 Eq. A:

H  600  180 x1  20 x13 Example 10.4 Eq. B:

H1  420  60 x12  40 x13 Example 10.4 Eq. C:

H2  600  40 x13 Eq. 10.7:   nM    Mi     ni  P ,T ,n j

Eq. 10.11: M   xi M i i

Eq. 10.14:

 x dM  0 i

Solution

(a): Multiply Eq. (A) of Ex. 10.4 by n  n1  n2  and eliminate Multiply Eq. (A) of Ex. 10.4 by n  n1  n2  and eliminate x1 by x 1 

nH 



n1 n1  n2

  2 2n13   3n1     n  n 2 n  n 3  2 1 2  1 



n12

 2

n1  n2 

n13

n1  n2 

Further simplification yields:

Hi  420  60 x12  40 x13 Form the partial derivative of nH with respect to n2 at constant n1: Hi 



n13

n1  n2 

Hi 



x i3

(b): In accord with eq. 10.11: H  x1 420  60 x12  40 x13   1  x2 600  40 x13 

Simplifying

H  600  180 x1  20 x13  dH   dH  (c): Write Eq. 10.14 for a binary system and divide by dx1: x1  1   x2  2   0  dx1   dx1 

Differentiate the boxed equations of part (a):  dH   1     dx   1

x1 

x12 x2 

 dH   2    dx   1

x12

Multiply each derivative by the appropriate mole fraction and add:

120 x12 x 2  120 x12 x 2  0 (d): Substitute x1 

x 2  in the first derivative expression of part (c) and substitute x1  0

in the second derivative expression of part (c). The results are:  dH   dH   1   2   0    dx    1  x1   dx1 x1 

Solution continued on next page…

(e):

Solution 10.13 Problem Statement 3

The molar volume (cm ·mol ) of a binary liquid mixture at T and P is given by: V = 120 x1 + 70 x2 + (15 x1 + 8 x2 ) x1 x2 (a) Find expressions for the partial molar volumes of species 1 and 2 at T and P. (b) Show that when these expressions are combined in accord with Eq. (10.11) the given equation for V is recovered. (c) Show that these expressions satisfy Eq. (10.14), the Gibbs/Duhem equation.

 dV   dV    2   0. (d) Show that  1   dx1   dx1 x1  0 x1  1 (e) Plot values of V , V1 , and V2 calculated by the given equation for V and by the equations developed in (a) vs. x1. Label points V1 , V2 , V1 , and V2 and show their values.

Eq. 10.11: M   xi M i i

Eq. 10.14:

 x dM  0 i

Solution

(a) The definition of a partial molar volume of species i in a mixture is   nV   Vi     ni  T , P ,n

ji

Multiplying the given composition dependence of the total volume by the total number of moles n gives (writing the mole fractions in terms of the numbers of moles (x1 = n1/n = n1/(n1 + n2), etc.) 2

nV = 120n1 + 70n2 + (15n1 + 8n2) n1n2/(n1 + n2) (cm ) Taking the partial derivative of this with respect to n1 (holding n2 constant) gives 3

V1  120 – 2*(15n1 + 8n2) n1n2/(n1 + n2) + (15n1 + 8n2) n2/(n1 + n2) + 15n1n2/(n1 + n2) 2

V1  120 – 30x1 x2 – 16x1x2 + 30x1x2 + 8x2

or, in terms of x1 alone: 2

V1  120 – 30x1 (1–x1) – 16x1(1–2x1 + x1 ) + 30x1(1–x1) + 8(1–2x1 + x1 ) 3

V1  120 + 30x1 – 30x1 – 16x1 + 32x1 – 16x1 + 30x1 – 30x1 + 8 – 16x1 + 8x1 2

V1  128 – 2x1 – 20 x1 + 14x1

Similarly, taking the partial derivative with respect to n2 gives 3

V2  70 – 2*(15n1 + 8n2) n1n2/(n1 + n2) + (15n1 + 8n2) n1/(n1 + n2) + 8n1n2/(n1 + n2) 2

V2  70 – 30x1 x2 – 16x1x2 + 16x1x2 + 15x1

or, in terms of x1 alone: 2

V2  70 – 30x1 (1–x1) – 16x1(1–2x1 + x1 ) + 16x1(1–x1) + 15x1 3

V2  70 + 30x1 – 30x1 – 16x1 + 32x1 – 16x1 + 16x1 – 16x1 + 15x1 2

V2  70 + x1 + 14x1

Solution continued on next page…

(b) Adding the two mole fractions times the partial molar volumes gives: V = x1V1  x2V2 2

V = x1 (120 – 30x1 x2 – 16x1x2 + 30x1x2 + 8x2 ) + x2 (70 – 30x1 x2 – 16x1x2 + 16x1x2 + 15x1 ) 2

V = 120x1 + 70x2 – 30 x1 x2 (x1 + x2) – 16x1x2 (x1 + x2) + 30x1 x2 + 8x1x2 + 16x1x2 + 15x1 x2 2

V = 120x1 + 70x2 – 30 x1 x2 – 16x1x2 + 30x1 x2 + 8x1x2 + 16x1x2 + 15x1 x2 2

V = 120x1 + 70x2 + 8x1x2 + 15x1 x2 V = 120x1 + 70x2 + (15x1 + 8x2) x1 x2 (c) The Gibbs-Duhem equation says that at constant T and P the partial molar volumes have to satisfy N

 x dV  0 i1

For a binary mixture, this is just x1 dV1  x 2 dV2  0

we can re-write the expressions for the partial molar volumes to eliminate x2 using x2 = 1 – x1. This will make it easier to cancel things out so that we end up with zero, as we expect. Eliminating x2, we have 2

V1  120 – 30x1 (1 – x1) – 16x1(1 – x1) + 30x1(1 – x1) + 8(1 – x1) 2

V1  120 – 30x1 + 30x1 – 16 x1 + 32 x1 – 16x1 + 30x1 – 30x1 + 8 – 16x1 + 8x1 2

V1  128 – 2x1 – 20 x1 + 14x1 2

V2  70 – 30x1 (1 – x1) – 16x1(1 – x1) + 16x1(1 – x1) + 15x1 2

V2  70 – 30x1 + 30x1 – 16 x1 + 32 x1 – 16x1 + 16x1 – 16x1 + 15x1 2

V2  70 + x1 + 14x1

and

dV1  (–2 – 40 x1 + 42x1 ) dx1 2

dV2  (2x1 + 42x1 ) dx1

Substituting these into x1 dV1  x 2 dV2  0 2

x1dV1  x2 dV2 = x1(–2 – 40 x1 + 42x1 ) dx1 + (1 – x1) (2x1 + 42x1 ) dx1 2

x1dV1  x 2 dV2 = (–2x1 – 40 x1 + 42x1 + 2x1 + 42x1 – 2 x1 – 42x1 ) dx1 = 0

Everything cancels out, and the Gibbs-Duhem relation is satisfied, as it must be.

Solution continued on next page…

(d) We’ve basically taken the necessary derivatives in the previous part, where we had 2

dV1  (–2 – 40 x1 + 42x1 ) dx1 2

dV2  (2x1 + 42x1 ) dx1

so dV1  2  40 1  421  0 dx1 x  1 1

dV2  20  420  0 dx1 x  0 1

(e) A plot of all this looks like:

Total or Partial Molar Volume (cm 3/mol)

SVA problem 11.13(e) 140

V1,  128 130

V1  120

120 110 100

V2,  85

90 80

V2  70 V2

70 60 0

0.2

0.4

0.6

0.8

Solution 10.14 Problem Statement For a particular binary liquid solution at constant T and P, the molar enthalpies of mixtures are represented by the equation H = x1 ( a1 + b1 x1) + x2 ( a2 + b2x2 ) where the ai and bi are constants. Because the equation has the form of Eq. (10.11), it might be that Hi  ai  bi x i . Show whether this is true.

Eq. 10.11: M   xi M i i

Solution

Once again, we want to apply the definition of a partial molar property and then take the required derivatives. The partial molar enthalpy is, by definition:   nH   H i     ni  T , P ,n

ji

Multiplying the given expression for H by n and writing it in terms of the mole numbers instead of mole fractions, we have   n1  n2    n2 a2  b2  nH  n1 a1  b1   n1  n2  n1  n2   Taking the derivative of this with respect to n1 gives

 nH     n1n2 n22  a  b n1   b  H1     b  1 2 2  1  n1  n1  n2  1 n  n 2  n1  n2  1 2 T , P ,n 2

Or, rewriting it in terms of mole fractions again H 1  a1  b1 x 1   b1 x 1 x 2  b2 x 22

Since the original expression is symmetric with respect to x1 and x2, the result for component 2 will be the same, but with the indices switched: H 2  a2  b2 x 2   b2 x 1 x 2  b1 x12

These are not of the form proposed in the problem, but they do (of course) satisfy the summability relationship. If we combine them, we have x1 H1  x 2 H2  x1 a1  b1 x 1   b1 x 12 x 2  b2 x 1 x 22  x 2 a2  b2 x 2   b2 x 1 x 22  b1 x 12 x 2 x1 H1  x 2 H2  x1 a1  b1 x 1   x 2 a2  b2 x 2 

This last expression is the expression for H from which we started. A quicker way to do this one would be to directly use the expressions specific to binary systems dH dx1 dH H 2  H  x1 dx 1 H1  H  x 2

But in that case, one must be sure to write the original expression for H in terms of x1 alone before taking the derivative.

Solution 10.15 Problem Statement  : Analogous to the conventional partial property Mi , one can define a constant-T, V partial property M i

      (nM )  M i  n  i   T ,V ,n j  and M are related by the equation: Show that M i i

    M  (V  V ) M  M i i i   V 

T, x

 satisfy a summability relation, M   x M  . Demonstrate that the M i i i i

Solution

Apply the following general equation of differential calculus:  dx            dx    dx   dw   dy   dy   dw  y  dy  z

  nM     nM     nM    V             n     n   n  V   P T n   V T n   T n  i P , T , n j i i j j

Substitution yields:  V     ni  T, n 

    n M  Mi  M i  V 

P , T , nj

 V     ni  T, n 

    M  n M  M i i  V 

P , T , nj

By definition   nV    V   Vi    n  V    ni P , T , n j  ni  P T n j

 V  n    ni 

 Vi  V

P, T , n j

This leads us to     M  V  V  M  M i i i  V T , n

Solution 10.16 Problem Statement From the following compressibility-factor data for CO2 at

P/bar

150°C, prepare plots of the fugacity and fugacity coefficient of CO2 vs. P for pressures up to 500 bar. Compare results with those found from the generalized correlation represented by Eq. (10.68). Eq. 10.68: P    exp  r  B 0   B1   Tr 

10 20 40 60 80 100 200 300 400 500

0.985 0.970 0.942 0.913 0.885 0.869 0.765 0.762 0.824 0.910

Solution

The fugacity coefficient is related to compressibility by P

Zi  1 dP P 0

ln i  

So, to get the fugacity coefficient, we can simply compute (Z – 1)/P from the numbers in the table and integrate this vs. P using the trapezoid rule or other simple rule. We basically just add up the values of (Z – 1)/P*P. This is shown in the table below, along with a plot of (Z – 1)/P vs. P. The area under this curve gives ln.

0.000000 -0.000200

100

200

300

400

500

600

-0.000400 (Z - 1)/P (1/bar)

P (Z-1)/P f (bar) Z (1/bar) ln(Phi) Phi (bar) 10 0.9850 -0.001500 -0.0150 0.9851 9.85 20 0.9700 -0.001500 -0.0300 0.9704 19.41 40 0.9420 -0.001450 -0.0595 0.9422 37.69 60 0.9130 -0.001450 -0.0885 0.9153 54.92 80 0.8850 -0.001438 -0.1174 0.8893 71.14 100 0.8690 -0.001310 -0.1449 0.8652 86.52 200 0.7650 -0.001175 -0.2691 0.7641 152.81 300 0.7620 -0.000793 -0.3675 0.6925 207.74 400 0.8240 -0.000440 -0.4292 0.6510 260.42 500 0.9100 -0.000180 -0.4602 0.6312 315.58

-0.000600 -0.000800 -0.001000 -0.001200 -0.001400 -0.001600 Pressure (bar)

Now, we wish to compare this to the predictions from generalized correlations. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

For CO2, the critical properties are Tc = 304.2 K, Pc = 73.83 bar, and  = 0.224. So, at 150C, Tr = (423.15/304.2) = 1.3910. By checking figure 3.15 on p. 103, we see that for a reduced temperature of about 1.4, we can use the second virial coefficient version of the Lee/Kesler correlation up to a reduced pressure of about 1.5. So, we can safely use the second virial coefficient version for this problem for the pressures up to 100 bar. To get the values at 200, 300, 400, and 500 bar, we will have to use the tables. So, for pressures up to 100 bar, we can use the spreadsheet from the lecture notes. This is shown below for a pressure of 10 bar. The fugacity coefficients for 20, 40, 60, 80, and 100 bar are obtained in the same way. T (K)

P (bar)

Tc (K)

Pc (bar)

304.2

73.83

423.15

 0.224



1.3910

0.1354

-0.1659

0.0960

0.9860

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 P    exp  B 0   B 1 r  T r  





Tr4.2

Pale blue boxes are input fields, pink box is the final output.

For the last four points in the table, the reduced pressure is outside the range where the 2

virial coefficient

representation of the generalized fugacity correlations is valid. So, we will use the Lee/Kesler correlation as presented in the tables in the back of the book. This is shown belopw for the point at 200 bar and the values at 300, 400, and 500 bar are obtained in the same way. T (K)

P (bar)

423.15

200

Tc (K) Pc (bar)



304.2

73.83

0.224 1.391026 2.708926

0



2.00 3.00 2.00 3.00

0.7345 0.6383 0.7925 0.7145

1.1776 1.2853 1.1858 1.2942

Interpolated for Pr=2 Interpolated for Pr=3 Interpolated to Tr, Pr

0.7873 0.7077 0.7308 0.7699

1.1851 1.2934 1.2619

Table Points Tr (1) Tr (1) Tr (2) Tr (2)

Pr(1) Pr(2) Pr(1) Pr(2)

1.30 1.30 1.40 1.40

Final Value of 

Solution continued on next page…

The table below summarizes the results from the generalized correlations. Two cases are shown: one where the values are obtained as described above, and a second case where the 2

virial coefficient method is used

for the points at 200 – 500 bar as well as for the lower pressures. The columns labeled ‘percent difference’ refer to the percent difference of the value of f shown here and that computed from the compressibility data given in the problem statement. If the full form of the generalized correlation is used at the higher pressures, agreement is good for all pressures. If we use equation 11.65 as suggested by SVNA in the problem statement (the 2

virial coefficient formulation) then the errors are substantial for the highest pressures.

Using appropriate Using 2nd virial method coefficient for all P f Percent f Percent (bar) Phi (bar) difference Phi (bar) difference 10 0.9860 9.86 0.09% 0.9860 9.86 0.09% 20 0.9723 19.45 0.19% 0.9723 19.45 0.19% 40 0.9453 37.81 0.33% 0.9453 37.81 0.33% 60 0.9191 55.15 0.41% 0.9191 55.15 0.41% 80 0.8936 71.49 0.49% 0.8936 71.49 0.49% 100 0.8689 86.89 0.43% 0.8689 86.89 0.43% 200 0.7699 153.98 0.76% 0.7549 150.98 -1.20% 300 0.7074 212.22 2.16% 0.6559 196.77 -5.28% 400 0.6626 265.04 1.78% 0.5699 227.96 -12.46% 500 0.6422 321.10 1.75% 0.4952 247.60 -21.54%

Solution 10.17 Problem Statement R

For SO2 at 600 K and 300 bar, determine good estimates of the fugacity and of G /RT.

Solution

Using the Lee/Kesler correlation, we can adapt the spreadsheet we had set up for interpolating in the Lee/Kesler tables for the chapter 6 homework.

Solution continued on next page…

T (K) 600

P (bar) 300

Tc (K) 430.8

Pc (bar) 78.84

Tr 1.30 1.30 1.40 1.40

Pr 3.00 5.00 3.00 5.00

 0.245

Tr 1.3928

Z 0.2079 0.0875 0.2397 0.1737

Pr 3.8052

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

Z 0.6344 0.7358 0.7202 0.7761

Interpolated Values Final Values Z 3 V (cm /mol)

0.7378 HR/(RTc) R H (J/mol)

0.7891 131.2

0.2092 -2.20103 -7883.38

R 0

(H ) /(RT c) -2.274 -2.825 -1.857 -2.486

R 1

R 0

R 1

(S ) /R -1.299 -1.554 -0.99 -1.303

(S ) /R -0.481 -1.147 -0.29 -0.73

-0.2567

-1.1367

-0.4876

S (J/(mol K))

-1.25615 -10.4436

(H ) /(RT c) -0.3 -1.066 -0.044 -0.504

-2.1382 R

S /R R

0 0.6383 0.5383 0.7145 0.6237

1 1.2853 1.3868 1.2942 1.4488

0.6722

1.3542

 f (bar)

0.723998 217.1995

-0.32419 0.723111 216.9334

G /(RT)  f (bar)

Again, we can compute  in two different ways. We can interpolate directly in the tables for , or we can use the values of residual enthalpy and entropy and take GiR H R  TS R  RT RT In the above spreadsheet, both ways are applied, and they agree through 3 digits. The small disagreement is ln i 

due to the difference in interpolation. Since  is not a linear function of G , linear interpolation in H and S R

which gives a linear interpolation of G is not equivalent to linear interpolation in . If we had done linear interpolation on ln() the answers would have come out exactly the same.

Solution 10.18 Problem Statement Estimate the fugacity of isobutylene as a gas: (a) At 280°C and 20 bar; (b) At 280°C and 100 bar. Solution Estimate the fugacity of isobutylene as a gas: (a) At 280°C and 20 bar; (b) At 280°C and 100 bar. For isobutylene, the critical properties are Tc = 417.9 K, Pc = 40.00 bar, and  = 0.194. So, at 280C, Tr = (553.15/417.9) = 1.32. Solution continued on next page…

(a) At 20 bar, the reduced pressure is 0.500, and looking at the chart on page 115, we see that conditions are well within the range for which the 2

virial coefficient formulation of the generalized correlations is

accurate. We can use the handy spreadsheet that was included in the lecture notes to evaluate the fugacity coefficient:

T (K)

P (bar)

Tc (K)

Pc (bar)

417.9

553.15

 0.194



1.3236

0.5000

-0.1865

0.0860

0.9379

0.422

B0  0.083 

Tr1.6

 P    exp  B 0   B1 r  Tr  



0.172

B  0.139 



Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

The fugacity is f = P = 18.76 bar (b) At 100 bar, the reduced pressure is Pr = 2.5, and we are outside the range where the 2

virial

coefficient representation of the generalized fugacity correlations is valid. So, we will use the Lee/Kesler correlation as presented in the tables in the back of the book.

T (K)

P (bar)

553.15

Tc (K) Pc (bar)

100

417.9



0.194 1.323642

Pr 2.5

Table Points

0



2.00 3.00 2.00 3.00

0.7345 0.6383 0.7925 0.7145

1.1776 1.2853 1.1858 1.2942

Interpolated for Pr=2 Interpolated for Pr=3 Interpolated to Tr, Pr

0.7482 0.6563 0.7023 0.7314

1.1795 1.2874 1.2335

Tr Tr (1) Tr (1) Tr (2) Tr (2)

Pr(1) Pr(2) Pr(1) Pr(2)

1.30 1.30 1.40 1.40

Final Value of 

Interpolating to Tr = 1.32 and Pr = 2.5 in the table in the back of SVNA, we get the above results for  and  , 0

1 

which we combine to get  = ( )( ) = 0.7314. So, f = 0.7314*100 bar = 73.14 bar. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 10.19 Problem Statement Estimate the fugacity of one of the following: (a) Cyclopentane at 110°C and 275 bar. At 110°C the vapor pressure of cyclopentane is 5.267 bar. (b) 1-Butene at 120°C and 34 bar. At 120°C the vapor pressure of 1-butene is 25.83 bar. Solution (a) Here we are asked to find the fugacity of a compressed (subcooled) liquid. One way to find this is to directly use the Lee/Kesler tables, which include values for subcooled liquids: T (K)

P (bar)

Tc (K)

Pc (bar)

383.15

275

511.8

45.02

0.196

0.7486

6.1084

Table Points 0

Tr(1)

Pr(1)

0.70

5.00

0.0406

0.0851

Tr(1)

Pr(2)

0.70

7.00

0.0402

0.0752

Tr(2)

Pr(1)

0.75

5.00

0.0625

0.1552

Tr(2)

Pr(2)

0.75

7.00

0.061

0.1387

Interpolated Values 0.0611

0.1442 0.041795

f (bar)

11.5

Another approach is to find the fugacity of the saturated liquid (equal to that of the saturated vapor) and then apply the Poynting factor to get to the specified pressure:

fi   P

sat sat i i

 V sat  P  P sat   i i  exp  RT  

Solution continued on next page…

sat

Remember that the volume in this equation is the volume of the liquid. We could get Vi

from the Rackett

equation. We have Tr = 0.7486, and for cyclopentane, 3

–1

Vc = 258 cm mol and Zc = 0.273, so V 

sat

using the Pitzer correlation for the 2

T (K)

0.2857

= 258 * 0.273^(1 – 0.7486)

virial coefficient:

P (bar)

Tc (K)

Pc (bar)

383.15

5.267

511.8

45.02



0.7486

0.1170

-0.5876

-0.4412

0.9000

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

= 107.5 cm /mol. We can estimate

 0.196

 P    exp  B0   B1 r  T r  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

Thus, we finally get:  107.5 cm 3 mol 1 275  5.267 bar    fi  0.9000 * 5.267 bar* exp 3 1 1  83.145 cm bar mol K * 383.15 K  fi  11.78 bar

This agrees quite well with the result we obtained directly from the Lee/Kesler tables. We could also have done the same thing getting the fugacity and volume of the saturated liquid from a cubic equation of state.

Solution 10.20 Problem Statement Justify the following equations:  lnˆi  VR    i  P  RT T ,x

 lnˆ  H iR  i  T    RT 2  P , x

GR =  xi ln ˆi RT i

 x d ln ˆ i

(const T , P )

Solution Equation (11.59) demonstrates that ln ˆ i

G R RT

ˆi  Gi RT The

partial-property analogs of Eqs. (13.8) and (13.9) are:   ln ˆ  Vi R  i      P   T , x RT

  ln ˆ  i d    T 



P,x

HiR RT 2

The summability and Gibbs/Duhem equations take on the following forms: GR  x i ln ˆi RT i

x d i

ˆi 

 ns T P

Solution 10.21 Problem Statement From data in the steam tables, determine a good estimate for f/f f

sat

for liquid water at 150°C and 150 bar, where

is the fugacity of saturated liquid at 150°C.

Solution The ratio f/f fi fi

 sat

sat

is given by the Poynting factor:

 V sat  P  P sat   i i   exp   sat sat RT i Pi   fi

From the steam tables, we see that for saturated steam at 150C, the pressure is 476 kPa = 4.76 bar, and the 3

specific volume of the liquid is 1.091 cm /g. Multiplying the specific volume by the molecular weight of water sat

gives the molar volume as V

= 1.091 cm /g * 18.015 g/mol = 19.65 cm /mol. So, we have

 19.65 cm3 mol1 150  4.76 bar    exp   1.085  3 1 1 fi sat  83.145 cm bar mol K * 423.15 K  fi

The fugacity of the liquid is, as expected, not very sensitive to pressure.

Solution 10.22 Problem Statement For one of the following, determine the ratio of the fugacity in the final state to that in the initial state for steam undergoing the isothermal change of state: (a) From 9000 kPa and 400°C to 300 kPa. (b) From 1000(psia) and 800(°F) to 50(psia). Solution For one of the following, determine the ratio of the fugacity in the final state to that in the initial state for steam undergoing the isothermal change of state: (a) From 9,000 kPa and 400°C to 300 kPa. (b) From 1,000(psia) and 800(°F) to 50(psia). (a) The fugacity, for both the initial and final state, is defined by Gi  i T   RT ln fi

So, if we take the difference in Gibbs energy between the final and intial state, we get G final  Ginitial  i T   RT ln f final  i T   RT ln finitial  f final   G final  Ginitial  RT ln   finitial 

Or, solving for the ratio of the fugacity in the final state to that in the initial state,  G final  Ginitial        exp G   exp H  S   exp      finitial  RT  RT   RT R   f final

–1

From the steam tables, at 400C and 300 kPa, H = 3275.2 kJ kg and S = 8.0338 kJ kg K . At 400 C –1

–1

and 9000 kPa, H = 3121.2 kJ kg and S = 6.2915 kJ kg K . So, H = 154 kJ kg = 154 J g and –1

–1

S = 1.7423 kJ kg K = 1.7423 J g K . Multiplying these by the molecular weight of water (18.02 g/mol) gives –1

–1

H = 2775.08 J mol and S = 31.396 J mol K . So,  H S   2775.08 31.396   exp    exp      finitial R   RT  8.3145 * 673.15 8.3145  f final  exp0.4958  3.7761  0.0376 finitial f final

(b) The fugacity, for both the initial and final state, is defined by Gi  i T   RT ln fi

Or, solving for the ratio of the fugacity in the final state to that in the initial state,  G final  Ginitial        exp G   exp H  S   exp      finitial RT  RT   RT R   f final

–1

From the steam tables, at 800F and 1000 psia, H = 1389.6 Btu lbm and S = 1.5677 Btu lbm R . At 800 F –1

–1

and 50 psia, H = 1431.7 Btu lbm and S = 1.9227 Btu lbm R . So, H = 42.1 Btu lbm and S = 0.355 –1

–1

Btu lbm R . Multiplying these by the molecular weight of water (18.02 lbm/lbmol) gives H = 758.6 Btu –1

–1

lbmol and S = 6.397 Btu lbmol R . So,  H S   758.6 6.397   exp    exp       finitial R   RT  1.986 * 1259.7 1.986  f final  exp0.30323  3.2210  0.0541 finitial f final

Solution 10.23 Problem Statement Estimate the fugacity of one of the following liquids at its normal-boiling-point temperature and 200 bar: (a) n-Pentane; (b) Isobutylene; (c) 1-Butene. Solution (a) In this case, the vapor pressure is low, and it makes sense to use the Pitzer correlation to get the fugacity coefficient for the saturated vapor (at 1 atm and the normal boiling point) and then apply the Poynting correction:  V sat  P  P sat   i i  sat sat fi  i Pi exp  RT   Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

First, we can estimate the fugacity of the saturated vapor using the Pitzer correlation. The normal boiling point of n-pentane is 36C = 309.2 K. The critical properties are 3

 = 0.252, Tc = 469.7 K, Pc = 33.7 bar, Zc = 0.270, and Vc = 313 cm /mol. T (K)

P (bar)

Tc (K)

Pc (bar)

309.2

1.013

469.7

33.7



0.6583

0.0301

-0.7408

-0.8568

0.9573

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 0.252

 P    exp  B0   B1 r  Tr  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

So, we estimate the fugacity coefficient of the saturated liquid as 0.957. We can find the volume of the saturated liquid from the Rackett equation: V

sat

0.2857

= Vc * Zc^(1 – Tr)

0.2857

= 313*0.270^(1–0.6583)

= 119.4 cm /mol.

Thus, we finally get:  119.4 cm 3 mol1 200  1.01 bar    fi  0.9573 * 1.013 bar* exp  83.145 cm3 bar mol 1 K1 * 309.2 K  fi  2.4 bar

(b) In this case, the vapor pressure is low, and it makes sense to use the Pitzer correlation to get the fugacity coefficient for the saturated vapor (at 1 atm and the normal boiling point) and then apply the Poynting correction:  V sat  P  P sat   i i  fi  isat Pi sat exp  RT   First, we can estimate the fugacity of the saturated vapor using the Pitzer correlation. The normal boiling point of isobutylene is 266.3 K. The critical properties are  = 0.194, Tc = 417.9 K, Pc = 40.0 bar, Zc = 0.267, 3

and Vc = 220.4 cm /mol. Using these values in the Pitzer correlation gives: T (K)

P (bar)

266.3

Tc (K) Pc (bar)



417.9

0.194



0.6372

0.0250

-0.7848

-1.0025

0.9623

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 P    exp  B0   B1 r  T  r 





Tr4.2

So, we estimate the fugacity coefficient of the saturated liquid as 0.9623. We can find the volume of the saturated liquid from the Rackett equation: V

sat

0.2857

= Vc * Zc^(1 – Tr)

2/7

= 220.4*0.267^(1–0.6372)

= 82.0 cm /mol.

Thus, we finally get:  V sat  P  P sat   i i  fi   P exp   RT    82.0 cm3 mol1 200  1.01 bar    fi  0.9623 * 1.013 bar * exp  83.145 cm3 bar mol1 K1 * 266.3 K  fi  2.04 bar sat sat i i

(c) In this case, the vapor pressure is low, and it makes sense to use the Pitzer correlation to get the fugacity coefficient for the saturated vapor (at 1 atm and the normal boiling point) and then apply the Poynting correction: fi   P

sat sat i i

 V sat  P  P sat   i i  exp  RT  

First, we can estimate the fugacity of the saturated vapor using the Pitzer correlation. The normal boiling point of 1-butene is 266.9 K. The critical properties are  = 0.191, Tc = 420.0 K, Pc = 40.43 bar, Zc = 0.277, and 3

Vc = 239.3 cm /mol. Using these values in the Pitzer correlation gives:

T (K)

P (bar) Tc (K) Pc (bar)



266.9

1.01325

420

0.191

0.6355

0.0251

-0.7887

-1.0158

B 0  0.083  1

B  0.139 

40.43

 0.9620

0.422 Tr1.6 0.172

 P    exp  B 0   B1 r  Tr  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

So, we estimate the fugacity coefficient of the saturated liquid as 0.9620. We can find the volume of the saturated liquid from the Rackett equation: V

sat

0.2857

= Vc * Zc^(1 – Tr)

2/7

= 239.3*0.277^(1–0.6355)

= 91.43 cm /mol.

Thus, we finally get:  V sat  P  P sat   i i  fi  isat Pi sat exp  RT    91.43 cm3 mol1 200  1.01 bar    fi  0.9620 * 1.013 bar * exp  83.145 cm 3 bar mol1 K1 * 266.9 K  fi  2.21 bar

Solution 10.24 Problem Statement Assuming that Eq. (10.68) is valid for the vapor phase and that the molar volume of saturated liquid is given by Eq. (3.68), prepare plots of f vs. P and of vs. P for one of the following: (a) Chloroform at 200°C for the pressure range from 0 to 40 bar. At 200°C the vapor pressure of chloroform is 22.27 bar. (b) Isobutane at 40°C for the pressure range from 0 to 10 bar. At 40°C the vapor pressure of isobutane is 5.28 bar. Eq. 3.68: 2/7

V sat  Vc Zc (1Tr )

Eq. 10.68:

P    exp  r  B 0   B1   Tr 

Solution

(a) In this case, we can adapt the spreadsheet given in the notes to evaluate  at many pressures, but a single temperature. Because the virial coefficient only depends on temperature, we only have to compute it once. We can then apply eq. 11.68 (the Pitzer correlation) for pressures up to the vapor pressure. Above the vapor pressure, we can assume that the volume of the liquid is equal to the volume of the saturated liquid, as estimated from the Rackett equation. The modified spreadsheet looks like this:

T (K)

sat

(bar)

Tc (K)

22.27

536.4

473.15

0.8821

-0.4328

-0.1523

Pc (bar) 54.72

sat

(cm /mol) 122.7

Below P sat

B 0  0.083 

0.422 Tr1.6

 P    exp  B0   B1 r  Tr  0.172  B1  0.139  4.2 Tr





Above P sat



Vi f i  isat Pi sat exp   

sat

 P  P   sat

 

 0.222

Vc (cm /mol)

239.0

P (bar)

0.293

f (bar)

f 0.0 0.5 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

1.000 0.995 0.990 0.986 0.981 0.976 0.971 0.967 0.962 0.957 0.953 0.948 0.944 0.939 0.935 0.930 0.926 0.921 0.917 0.912

0.00 0.50 0.99 1.48 1.96 2.44 2.91 3.38 3.85 4.31 4.76 5.22 5.66 6.10 6.54 6.98 7.40 7.83 8.25 8.67

Solution continued on next page…

The resulting plot looks like this:

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

Fugacity Coefficient

Fugacity (bar)

Fugacity and Fugacity Coefficient of Chloroform at 200 °C

0.0 0

Pressure (bar)

Part (b) is solved by using the same method used in part (a).

Solution 10.25 Problem Statement For the system ethylene(1)/propylene(2) as a gas, estimate fˆ1 , fˆ2 , ˆ1 , and ˆ2 at t = 150°C, P = 30 bar, and y1 = 0.35: (a) Through application of Eqs. (10.63). (b) Assuming that the mixture is an ideal solution. Eq. 10.63a: P lnˆ1  ( B  y22 12 ) RT 11 Eq. 10.63b: P lnˆ2  ( B  y1212 ) RT 22

Solution (a) This is exactly like example 11.9, and we can do it using the same spreadsheet (putting in the critical parameters for this problem): Spreadsheet for computing fugacity coefficients in a binary mixture (Applied to Problem 11.25) T (K) P (bar) 423.15 30 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc Species 1 Ethylene 0.35 282.3 50.4 131 0.281 Species 2 Propylene 0.65 365.6 46.65 188.4 0.289

Cross-Parameters

kij

Vc Tc (K) Pc (bar) (cm3/mol) 0 321.26139 48.19 158.0

i   j 2 Z cij RTcij Pcij  Vcij

2nd Virial Coefficients (cm /mol) B11 -60 B22 -159 B12 -99 3

12 (cm /mol)

ln ˆ1 ln ˆ 2

ˆ1 ˆ2

ij 

20.83

Pr 0.5952 0.6431

B0 -0.1378 -0.2510

B1 0.1076 0.0459

0.1135

Tr 1.3172

Pr 0.6225

B0 -0.1886

B1 0.0849



Tcij  TciTcj 1  kij 

B0  0.083 

0.422

Z ci  Z cj

B1  0.139 

0.172

Zcij 

Tr1.6 Tr4.2

 Vci1/ 3  Vcj1/ 3  Vcij     2  

-0.04351 -0.13371 0.957 0.875

Zc 0.2850

0.087 0.14

Tr 1.4989 1.1574



Bij 

RTcij Pcij

B  B  0

P P  B11  y22 12   RT  B11  y 22  2 B12  B11  B22  RT P P ln ˆ2   B22  y1212   RT  B22  y12  2 B12  B11  B22   RT ln ˆ1 

These calculations confirm that we are at low enough reduced pressure and high enough reduced temperature that the 2-term virial equation expressions used here are applicable. The fugacities are just the fugacity coefficients times the pressure: fˆethylene  ˆethylene yethylene P  0.957 * 0.35 * 30 bar  10.05 bar fˆpropylene  ˆ propylene y propylene P  0.875 * 0.65 * 30 bar  17.06 bar

(b) For an ideal solution, the fugacity of a component in the mixture is its pure component fugacity times its partial pressure:

fˆiid  yi fi  yii P

Solution continued on next page…

So, for this part, we want to find the pure species fugacity coefficients. We can do that using the other spreadsheet from the lecture notes. For ethylene: T (K)

P (bar)

Tc (K)

Pc (bar)

282.3

50.4

423.15

1.4989

0.5952

-0.1378

0.1076

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 0.087

 0.9503

 P    exp  B0   B1 r  Tr  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

and for propylene: T (K)

P (bar)

Tc (K)

Pc (bar)

365.6

46.65

423.15

1.1574

0.6431

-0.2510

0.0459

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 0.14

 0.8729

 P    exp  B0   B1 r  Tr  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

So, putting in this fugacity coefficient, id fˆethylene  yethylene ethylene P  0.35 * 0.9503 * 30 bar  9.98 bar ˆf id  y propylene  propylene P  0.65 * 0.8729 * 30 bar  17.02 bar propylene

These aren’t much different from what we got with the full treatment in part (a).

Solution 10.26 Problem Statement Rationalize the following expression, valid at sufficiently low pressures, for estimating the fugacity coefficient: ln

Z

Solution For a pressure low enough that Z and ln

are given approximately by Eqs. (3.36) and (10.36):

Z 

BP RT

and

ln  

BP RT

The ln   Z  1

Solution 10.27 Problem Statement For the system methane(1)/ethane(2)/propane(3) as a gas, estimate fˆ1 , fˆ2 , fˆ3 , ˆ1 , ˆ2 , and ˆ3 at t = 100°C, P = 35 bar, y1 = 0.21, and y2 = 0.43: (a) Through application of Eq. (10.64). (b) Assuming that the mixture is an ideal solution. Eq. 10.64: ln ˆk 

 P  1 Bkk   yi y j (2ik  i j )  RT  2 i j 

Solution (a) As we discussed in recitation, for this problem we need to extend the spreadsheet for binary systems from the lecture notes to apply it to a ternary system. The general equation is

ln ˆk 

 P  1 N N  Bkk   yi y j 2ik  ij  RT  2 i1 j 1 

where ik  ki  2 Bik  Bii  Bkk ij   ji  2 Bij  Bii  B jj ii   jj  kk  0

As we saw in recitation, if we carefully write out the terms in the double sum and simplify the result, we end up with P  B11  y2212  y32 13  y2 y3 12  13  23  RT P ln ˆ2  B  y3223  y12 12  y3 y1 23  12  13  RT 22 P ln ˆ3  B  y12 13  y2223  y1 y2 13  23  12  RT 33 ln ˆ1 

Putting these equations into the spreadsheet gives the results shown below. Spreadsheet for computing fugacity coefficients in a binary mixture (problem 11.27 in SVA) T (K) P (bar) 373.15 35

Species 1 Species 2 Species 3

Name Methane Ethane Propane

Mole Fraction 0.21 0.43 0.36

Cross-Parameters 1,2 1,3 2,3

kij

Tc (K) Pc (bar) 0 241.22641 47.01 0 265.488 43.26 0 336.00586 45.26

2nd Virial Coefficients (cm /mol) B11 -21 B22 -113 B33 -244 B12 -52 B13 -78 B23 -167 3

12 (cm /mol) 3 13 (cm /mol) 3 23 (cm /mol)

ln ˆ1 ln ˆ2

30.33 107.52 23.10

ln ˆ3

0.01895 -0.12320 -0.25478

ˆ1 ˆ2 ˆ

1.019 0.884 0.775

Tc (K) Pc (bar) 190.6 45.99 305.3 48.72 369.8 42.48

Vc 3 (cm /mol) 98.6 145.5 200 Vc (cm3/mol) 120.5 143.4 171.3

i   j 2 Z cij RTcij Pcij  Vcij

Tr 1.9578 1.2222 1.0091

Pr 0.7610 0.7184 0.8239

0.012 0.1 0.152

B -0.0610 -0.2231 -0.3330

B 0.1288 0.0650 -0.0266

Tr 1.5469 1.4055 1.1105

Pr 0.7446 0.8090 0.7734

0.0560 0.0820 0.1260

B -0.1270 -0.1618 -0.2738

B 0.1115 0.0978 0.0283

 0.286 0.279 0.276

Zc 0.2825 0.2810 0.2775



Tcij  TciTcj 1  kij 

ij 

Z cij 

 V 1/ 3  Vcj1/ 3  Vcij   ci   2  

Z ci  Z cj

B 0  0.083  1

B  0.139 

0.422 Tr1.6 0.172 Tr4.2

Bij 

RTcij Pcij

B   B  0

P  B11  y2212  y3213  y2 y3 12  13   23   RT P ln ˆ2   B22  y32 23  y1212  y3 y1  23  12  13   RT P ln ˆ3   B33  y1213  y22 23  y1 y2 13   23  12   RT ln ˆ1 

f1 (bar) f2 (bar) f3 (bar)

7.49 13.31 9.77

(b) For an ideal solution, the fugacity of a component in the mixture is its pure component fugacity times its partial pressure:

fˆiid  yi fi  yii P

Solution continued on next page…

So, for this part, we want to find the pure species fugacity coefficients. We can do that using the other spreadsheet from the lecture notes. For methane: T (K)

P (bar)

373.15

Tc (K) Pc (bar) 190.6

45.99

 0.012



1.9578

0.7610

-0.0610

0.1288

0.9771

B 0  0.083 

0.422 Tr1.6

B1  0.139 

0.172

 P    exp  B 0   B1 r  T  r 





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

For ethane: T (K)

P (bar)

373.15

Tc (K) Pc (bar) 305.3

48.72

 0.1



1.2222

0.7184

-0.2231

0.0650

0.8805

B0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 P    exp  B0   B1 r  Tr  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

and for propane: T (K)

P (bar)

Tc (K)

Pc (bar)

369.8

42.48

373.15

 0.152



1.0091

0.8239

-0.3330

-0.0266

0.7594

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 P    exp  B0   B1 r  Tr  





Tr4.2

Pale blue boxes are input fields, pink boxes are the final output.

So, putting in these fugacity coefficients, f1 (bar) 7.18 f2 (bar) 13.25 f3 (bar) 9.57 These aren’t much different from what we got with the full treatment in part (a). Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 10.28 Problem Statement E

Given below are values of G /J·mol , H /J·mol , and CPE /J ·mol1 · K 1 for some equimolar binary liquid 1

mixtures at 298.15 K. Estimate values of G , H , and S at 328.15 K for one of the equimolar mixtures by two procedures: (I) Use all the data; (II) Assume CPE = 0. Compare and discuss your results for the two procedures. E

(a) Acetone/chloroform: G = 622, H = 1920, CPE = 4.2. E

(b) Acetone/n-hexane: G = 1095, H = 1595, CPE = 3.3. (c) Benzene/isooctane: G = 407, H = 984, CPE = 2.7. E

(d) Chloroform/ethanol: G = 632, H = 208, CPE = 23.0. E

(e) Ethanol/n-heptane: G = 1445, H = 605, CP = 11.0. E

(f ) Ethanol/water: G = 734, H = 416, C CPE = 11.0. E

(g) Ethyl acetate/n-heptane: G = 759, H = 1465, CPE = 8.0. Solution For CPE  const . , the following equations are readily developed from those given in the last column of Table 10.1:

H E  CPE T

 G E  T   CPE S E    T  T P x

Working equations are then:

S1E 

H1E  G1E T1

H2E  H1E  CPE T

S2E  S1E  CPE

T T

G2E  H2E  T2 S2E

Solution continued on next page…

For T1 

K T2 

K T 

T 

results for all parts of the problem are

given in the following table: I

Part

G1E

H1E

S1E

CPE

S2E

H2E

G2E

S2E

H2E

G2E

a b

–622.0 1095.0

–1920.0 1595.0

–4.354 1.677

4.2 3.3

–3.951 1.993

–1794 1694

–497.4 1039.9

–4.354 1.677

–1920.0 1595.0

–491.4 1044.7

c d

407.0 632.0

984.0 –208.0

1.935 –2.817

–2.7 23.0

1.677 –0.614

903 482

352.8 683.5

1.935 –2.817

984.0 –208.0

348.9 716.5

e f

1445.0 734.0

605.0 –416.0

–2.817 –3.857

11.0 11.0

–1.764 –2.803

935 –86

1513.7 833.9

–2.817 –3.857

605.0 –416.0

1529.5 849.7

759.0

1465.0

2.368

–8.0

1.602

1225

699.5

2.368

1465.0

688.0

Solution 10.29 Problem Statement

The data in Table 10.2 are experimental values of VE for binary liquid mixtures of 1,3-dioxolane(1) and isooctane(2) at 298.15 K and 1(atm). (a) Determine from the data numerical values of parameters a, b, and c in the correlating equation:

V E  x1 x 2 (a  bx1  cx12 ) E

(b) Determine from the results of part (a) the maximum value of V . At what value of x1 does this occur? (c) Determine from the results of part (a) expressions for V1E and V2E. Prepare a plot of these quantities vs. x1, and discuss its features.

Table 10.2: Excess Volumes for 1,3-Dioxolane(1)/Isooctane(2) at 298.15 K VE/10 3 cm3·mol 1

x1 0.02715 0.09329 0.17490 0.32760 0.40244 0.56689 0.63128 0.66233 0.69984 0.72792 0.77514 0.79243 0.82954 0.86835 0.93287 0.98233

87.5 265.6 417.4 534.5 531.7 421.1 347.1 321.7 276.4 252.9 190.7 178.1 138.4 98.4 37.6 10.0

R. Francesconi et al., Int. DATA Ser., Ser. A, Vol. 25, No. 3, p. 229, 1997.

Solution (a) To correlate the data, a difference analysis between the given and equation values can be found and minimized. This will allow adjustments to be made to the equation in part(a) to obtain a good fit for the data. x1 0.02715

VE 87.5

0.09329 0.1749

265.6 417.4

0.3276 0.40244

534.5 531.7

0.56689 0.63128

421.1 347.1

0.66233 0.69984

321.7 276.4

0.72792 0.77514

252.9 190.7

0.79243 0.82954

178.1 138.4

0.86835 0.93287

98.4 37.6

0.98233

600 500

400

VE actual

300 200 100 0 0

0.2

0.4

0.6

0.8

This yields the values for a, b, and c as: a

(b): To find the maximum, set

b 

c

dV E  dx1

3 2 dV E  4c  x1   3c  b *  x1   2b  a x1  a dx1

Setting

dV E  dx1

x1 

Plugging this back into the equation in part(a) yields: E Vmax  x1   x1  * a  bx1  cx12 

E Vmax 

(c): 2 V1E  1  x1   a  2bx1  3cx12    2 E  V2   x 1   a  b  2b  c  x1  3cx 12   

Plotting these two equations gives 3500 3000 2500

VE1 VE2

2000 1500 1000 500 0 -500

0.2

0.4

0.6

0.8

Discussion: Part (a): Partial property for species i goes to zero WITH ZERO SLOPE as xi > 1. Part (b): Interior extrema come in pairs: VEbar min for species 1 occurs at the same x1 as VEbar max for species 2, and both occur at an inflection point on the VE vs. x1 plot. Part (c): At the point where the VEbar lines cross, the VE plot shows a maximum.

Solution 10.30 Problem Statement R

For an equimolar vapor mixture of propane(1) and n-pentane(2) at 75°C and 2 bar, estimate Z, H , and S . 3 1 Second virial coefficients, in cm ·mol are:

t/°C

B11

50 75 100

331 276 235

B22

B12 980 809 684

558 466 399

Equations (3.36), (6.55), (6.56), and (10.62) are pertinent. Eq. 3.36: Z

PV BP 1 RT RT

Eq. 6.55: H R P  B dB      RT R  T dT 

Eq. 6.56: SR P dB  R R dT

Eq. 10.62:

B  y12 B11  2 y1 y2 B12  y22 B22

Solution

To compute Z, we just need the 2 virial coefficient of the mixture (B), but to compute H and S we also need the derivative of B with respect to temperature. So, let’s compute B for the mixture at each of the three B T to approximate the derivative. The rule for combining the B’s to get the overall B of the mixture is:

B  y12 B11  2 y1 y2 B12  y22 B22 For an equimolar mixture (y1 = y2 = 0.5), this is just B = 0.25B11 + 0.5B12 + 0.25B22. Applying this at the 3 3 3 3 temperatures gives B (50 ºC) = –606.75 cm /mol, B (75 ºC) = –504.25 cm /mol, and B (100 ºC) = –429.25 cm /mol. 3 –1 –1 We can approximate dB/dT at 75 ºC as (–429.25 + 606.75)/50 = 3.55 cm mol K . Using these values, we have Z 1

BP 504.25 * 2  1  0.965 RT 83.14 * 348.15

H R BP P dB 504.25 * 2 2     * 3.55  0.1202 RT RT R dT 83.14 * 348.15 83.14 SR P dB 2   * 3.55  0.08540 R R dT 83.14 R

Or, finally, putting H and S in dimensional form, we have Z = 0.965, H = –348 J/mol, and S = –0.710 J/(mol K).

Solution 10.31 Problem Statement Use the data of Prob. 10.30 to determine ˆ1 and ˆ2 as functions of composition for binary vapor mixtures of propane(1) and n-pentane(2) at 75°C and 2 bar. Plot the results on a single graph. Discuss the features of this plot. Problem 10.30 R

For an equimolar vapor mixture of propane(1) and n-pentane(2) at 75°C and 2 bar, estimate Z, H , and S . 3

Second virial coefficients, in cm ·mol

are:

t/°C

B11

50 75 100

331 276 235

B22

B12 980 809 684

558 466 399

Equations (3.36), (6.55), (6.56), and (10.62) are pertinent. Eq. 3.36: Z

PV BP 1 RT RT

Eq. 6.55: H R P  B dB      RT R  T dT 

Eq. 6.55: SR P dB  R R dT

Eq. 10.62:

B  y12 B11  2 y1 y2 B12  y22 B22 Solution T

P  bar

y1 

276 466 cm3  B   466 809 mol

i , j  Bi, j  Bi , i  B j , j

n

i

j

Using equations 10.63a and 10.63b  P  2 ˆ1  exp  B  1  y1  1, 2   RT 1, 1   P  ˆ2  exp  B  y121,2   RT 2, 2 









Set y1  0,0.1, . .10, and plot the data 1 0.95

ɸhat1 ɸhat2

0.9 0.85 0.8 0.75

ɸhat1

0.7

ɸhat2

0.65 0.6 0.55 0.5 0

0.2

0.4

0.6

0.8

ˆ 1

ˆ 2 has a positive trend from a plateau.

Solution 10.32 Problem Statement For a binary gas mixture described by Eqs. (3.36) and (10.62), prove that: d12 Py1 y2 dT

G E  12 P y1 y2

SE  

 d  H E  12  T 12  Py1 y2  dT 

CPE   T

See also Eq. (10.87), and note that

d 2 12 dT 2

Py1 y2

12 = 2 B12

B11

B22.

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 10.62:

B  y12 B11  2 y1 y2 B12  y22 B22 Eq. 10.87: M E  M R   x i M iR i

Solution By Eq. (10.87), written with M  G and with x replaced by y: G E  G R   i yiGiR Equations (10.33) and (10.36) together give GiR  Bii P. Then for a binary mixture: G E  BP  y1 B11 P  y2 B22 P

G E  P  B  y1 B11  y2 B22 

Combine this equation with eq 10.53: G E  12 Py1 y2  dG E   From the last column of Table 10.1: S E    dT  P, x

Because 12

SE  

d12 Py y dT 1 2

By the definition of G E H E  G E  TS E

 d  H E  12  T 12  Py1 y2  dT 

 dH E   Again from the last column of Table 10.1: CPE    dT  P, x

With this and the last equation, it directly leads to: CPE  T

d 2 12 dT 2

Py1 y2

Solution 10.33 Problem Statement The data in Table 10.3 are experimental values of HE for binary liquid mixtures of 1,2-dichloroethane(1) and dimethyl carbonate(2) at 313.15 K and 1(atm). (a) Determine from the data numerical values of parameters a, b, and c in the correlating equation: E

H = x1 x2 (a + b x1 + cx1 ) E

(b) Determine from the results of part (a) the minimum value of H . At what value of x1 does this occur? (c) Determine from the results of part (a) expressions for H1E and H2E. Prepare a plot of these quantities vs. x1, and discuss its features.

Table 10.3 : HE Values for 1,2-Dichloroethane(1)/Dimethylcarbonate(2) at 313.15 K x1

HE /J·mol 1

0.0426

23.3

0.0817

45.7

0.1177

66.5

0.1510

86.6

0.2107

118.2

0.2624

144.6

0.3472

176.6

0.4158

195.7

0.5163

204.2

0.6156

191.7

0.6810

174.1

0.7621

141.0

0.8181

116.8

0.8650

85.6

0.9276

43.5

0.9624

22.6 R. Francesconi et al., Int. Data Ser., Ser. A, 5, No. 3, p. 225, 1997.

Solution

(a) To correlate the data, a difference analysis between the given and equation values can be found and minimized. This will allow adjustments to be made to the equation in part(a) to obtain a good fit for the data. x1 0.0426

HE –23.3

0.0817 0.1177

–45.7 –66.5

0.151 0.2107

–86.6 –118.2

0.2624 0.3472

–144.6 –176.6

0.4158 0.5163

–195.7 –204.2

0.6156 0.681

–191.7 –174.1

0.7621 0.8181

–141 –116.8

0.865 0.9276

–85.6 –43.5

0.9624

–22.6

0 0

0.2

0.4

0.6

0.8

-50

-100 -150 HE actual

-200

equation -250

This yields the values for a, b, and c as: a 

(b): To find the maximum, set

b 

c

dH E  dx1

3 2 dH E  4 c  x1   3c  b *  x1  2 b  a x1  a dx1

Setting

dV E  dx1

x1 

Plugging this back into the equation in part(a) yields: E Hmin  x1  x1  * a  bx1  cx12 

E Hmin 

(c): H  H  x1    x1 E 1

H 2E  H E  x1  x1

dH E x1 dx1

dH  x1 E

dx1

Plotting these two equations gives:

100 0 0

0.2

0.4

0.6

0.8

-100 HEbar1

HEbar

-200

HEbar2 -300 -400 -500 -600 -700

Discussion: Part (a) Partial property for species i goes to zero WITH ZERO SLOPE as xi > 1. Part (b) Interior extrema come in pairs: HEbar min for species 1 occurs at the same x1 as HEbar max for species 2, and both occur at an inflection point on the HE vs. x1 plot. Part (c) At the point where the HEbar lines cross, the HE plot shows a minimum. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 10.34 Problem Statement Make use of Eqs. (3.36), (3.61), (3.62), (6.54), (6.55), (6.56), (6.70), (6.71), (10.62), and (10.69)–(10.74), to estimate V, R

H , S , and G for one of the following binary vapor mixtures: (a) Acetone(1)/1,3-butadiene(2) with mole fractions y1 = 0.28 and y2 = 0.72 at t = 60°C and P = 170 kPa. (b) Acetonitrile(1)/diethyl ether(2) with mole fractions y1 = 0.37 and y2 = 0.63 at t = 50°C and P = 120 kPa. (c) Methyl chloride(1)/ethyl chloride(2) with mole fractions y1 = 0.45 and y2 = 0.55 at t = 25°C and P = 100 kPa. (d) Nitrogen(1)/ammonia(2) with mole fractions y1 = 0.83 and y2= 0.17 at t = 20°C and P = 300 kPa. (e) Sulfur dioxide(1)/ethylene(2) with mole fractions y1 = 0.32 and y2 = 0.68 at t = 25°C and P = 420 kPa. Note: Set kij = 0 in Eq. (10.71). Eq. 3.36: Z

PV BP 1 RT RT

Eq. 3.61: B 0  0.083 

0.422 Tr1.6

Eq. 3.62: B1  0.139 

0.172 Tr4.2

Eq. 6.54: GR BP  RT RT

Eq. 6.55: H R P  B dB      RT R  T dT 

Eq. 6.56: SR P dB  R R dT

Eq. 6.70: dB 0 0.675  2.6 dTr Tr

Eq. 6.71: dB 1 0.172  4.2 dTr Tr

Eq. 10.62:

B  y12 B11  2 y1 y2 B12  y22 B22 Eq. 10.69 a and b: Bˆ ij  B 0  ij B1 Bi j 

Bi j Pcij RTcij

Eq. 10.70–74:

(10.70) : i j  (10.72) : Pcij 

i   j 2 Z cij RTcij Vcij

(10.71): Tcij = TciTcj 

1/2

(10.73): Z cij =

Zci  Z cj 2

1  k  ij

 V 1/3 _ V 1/3   ci cj  (10.74): Vcij =    2 

Solution

(a) To find the volume and residual properties, we need to find the second virial coefficient of the mixture, N

which is B   yi y j Bij in general, or B  y12 B11  2 y1 y2 B12  y22 B22 for a binary mixture. B11 and B22 are i 1 j 1

the pure component virial coefficients, which we find from the Pitzer correlation, as we have done many times in the past: B

RTc

0.422

 B   B , with B  0.083  T 0

1.6 r

and B1  0.139 

0.172 Tr4.2

For the cross-coefficient, we use the same equations, but with the critical temperature and pressure based on the mixing rules given in lecture 17: ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj

2 3  V 1/3  V 1/3   ci cj  Vcij     2 

The above equations give us everything we need to find B for the mixture. However, to find the residual enthalpy and entropy, we also need the derivative of B with respect to temperature. Taking the derivative of the expression above for B, we have dB dB dB dB  y12 11  2 y1 y2 12  y22 22 dT dT dT dT

and the derivatives of the individual virial coefficients are dBij dT



0 dBij1  R  dBij0 dBij1  dBij0 0.675 dBij1 0.722 RTc  dBij    , with      and  5.2  Pc  dT dT  Pc  dTr dTr  dT dT Tr2.6 Tr

Solution continued on next page…

Finally, once we have evaluated these, we can get the volume, residual Gibbs energy, residual entropy, and residual enthalpy from V

RT  B, P

G R  BP,

H R  BP  TP

dB , dT

and S R  P

dB dT

The mixing rules and equations for the components of B are already included in the spreadsheet from the lecture notes for computing the fugacity coefficients in a binary mixture. So, we can just modify that spreadsheet to compute the above properties instead of the fugacity coefficients. The results of doing so are shown below: Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(a) T (K) P (bar) 333.15 1.7 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 Acetone 0.28 508.2 47.01 209 0.233 0.307 Species 2 1,3-butadiene 0.72 425.2 42.77 220.4 0.267 0.190

Cross-Parameters

kij 0

Vc Tc (K) Pc (bar) (cm3/mol) 464.8512 45.02 214.6

2nd Virial Coefficients (cm /mol) B11 -912 -500 B22 B12 -665

Zc 0.2500

i   j 2 Z cij RTcij Pcij  Vcij

Derivatives of 2nd Virial Coefficients (cm /((mol K)) 7.10 dB11/dT 3.42 dB22/dT dB12/dT 4.84

1/3 ci

V Vcij    

Zcij  1/ 3 cj

V 2

Pr 0.0362 0.0397

B -0.7464 -0.5405

B -0.8744 -0.3402

Tr 0.7167

Pr 0.0378

B -0.6361

B -0.5579

Tcij  TciTcj 1  kij 

ij 

 0.2485

Tr 0.6555 0.7835

  

Z ci  Z cj 2

B 0  0.083  1

B  0.139 

dB dT 6.4892 2.5674

dB dT 4.0817

dB dT 2.0236 1.2729 dB dT 1.6049

0.422 Tr1.6 0.172 Tr4.2

Bij 

RTcij Pcij

B   B  0

dBij RTc  dBij0 dBij1  dBij0 0.675 dBij1 0.722   , with    2.6 and  5.2 3 Mixture 2nd Virial Coefficient (cm /mol)   dT Pc dT dT dT dT Tr Tr   -599 B 3 2 2 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) B  y1 B11  2y1 y2 B12  y2 B22 4.28 dB/dT

Mixture Properties: V (cm3/mol) Z R G (J/mol) R H (J/mol) R S (J/(mol K))

dB 2 dB11 dB dB  y1  2 y1 y2 12  y22 22 dT dT dT dT

15695 0.9632 -101.8 -344.3 -0.728

V

RT B, P

G R  BP ,

H R  BP  TP

dB , dT

and S R  P

dB dT

(b) To find the volume and residual properties, we need to find the second virial coefficient of the mixture, N

which is B   yi y j Bij in general, or B  y12 B11  2 y1 y2 B12  y22 B22 for a binary mixture. B11 and B22 are the i 1 j 1

pure component virial coefficients, which we find from the Pitzer correlation, as we have done many times in the past: B

RTc Pc

0.422

 B   B , with B  0.083  T 0

1.6 r

and B1  0.139 

0.172 Tr4.2

Solution continued on next page…

For the cross-coefficient, we use the same equations, but with the critical temperature and pressure based on the mixing rules given in lecture 17: ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj

2 3  V 1/3  V 1/3   ci cj  Vcij     2 

and the derivatives of the individual virial coefficients are dBij dT



0 dBij1  R  dBij0 dBij1  dBij0 0.675 dBij1 0.722 RTc  dBij     2.6 and  5.2    , with Pc  dT dT  Pc  dTr dTr  dT dT Tr Tr

Finally, once we have evaluated these, we can get the volume, residual Gibbs energy, residual entropy, and residual enthalpy from V

RT  B, P

G R  BP,

H R  BP  TP

dB , dT

and S R  P

dB dT

Solution continued on next page…

which is B   yi y j Bij in general, or B  y12 B11  2 y1 y2 B12  y22 B22 for a binary mixture. B11 and B22 are the i 1 j 1

pure component virial coefficients, which we find from the Pitzer correlation, as we have done many times in the past: RTc

0.422

 B   B , with B  0.083  T P

B

1.6 r

and B1  0.139 

0.172 Tr4.2

For the cross-coefficient, we use the same equations, but with the critical temperature and pressure based on the mixing rules given in lecture 17: ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj

2 3  V 1/3  V 1/3   ci cj  Vcij     2 

and the derivatives of the individual virial coefficients are dBij dT



0 dBij1  dBij0 0.675 dBij1 0.722 RTc  dBij , with    and  5.2 Pc  dT dT  dT dT Tr2.6 Tr

Finally, once we have evaluated these, we can get the volume, residual Gibbs energy, residual entropy, and residual enthalpy from V

RT  B, P

G R  BP,

H R  BP  TP

dB , dT

and S R  P

dB dT

Solution continued on next page…

The mixing rules and equations for the components of B are already included in the spreadsheet from the lecture notes for computing the fugacity coefficients in a binary mixture. So, we can just modify that spreadsheet to compute the above properties instead of the fugacity coefficients. The results of doing so are shown below: Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(c) T (K) P (bar) 298.15 1 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Methyl chloride 0.45 416.3 66.8 143 0.276 0.153 Species 1 Species 2 Ethyl chloride 0.55 460.4 52.7 200 0.275 0.190

Cross-Parameters kij

Vc Pc (bar) (cm3/mol) 59.02 169.9

Tc (K) 437.7951

0 3

2nd Virial Coefficients (cm /mol) B11 -374 B22 -682 -507 B12

i  j

ij  Pcij  3

Pr 0.0150 0.0190

B -0.6369 -0.7627

B -0.5599 -0.9278

Tr 0.6810

Pr 0.0169

B -0.6973

B -0.7245

Tcij  TciTcj 1  kij 

2 Z cij RTcij

Zcij 

Vcij

 V 1/ 3  Vcj1/3  Vcij   ci   2  

Derivatives of 2nd Virial Coefficients (cm /((mol K)) dB11/dT 2.78 dB22/dT 5.37 dB12/dT 3.87

 0.1715

Zc 0.2755

Tr 0.7162 0.6476

Z ci  Z cj 2

0.422

B1  0.139 

0.172

dB dT 4.0963 6.9148

dB dT 5.3221

dB dT 1.6078 2.0889 dB dT 1.8326

Tr1.6 Tr4.2

Bij 

dBij RTc  dBij0 dBij1  dBij0 0.675   , with    2.6 and 3 Mixture 2nd Virial Coefficient (cm /mol) dT Pc  dT dT  dT Tr   B -533 3 2 2 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) B  y1 B11  2 y1 y2 B12  y2 B22 dB/dT 4.10

Mixture Properties: V (cm3/mol) Z R G (J/mol) R H (J/mol) R S (J/(mol K))

B 0  0.083 

dBij1 dT



RTcij Pcij

B  B  0

0.722 Tr5.2

dB dB dB dB  y12 11  2y1 y2 12  y22 22 dT dT dT dT

24257 0.9785 -53.3 -175.6 -0.410

V

RT B, P

G R  BP ,

H R  BP  TP

dB , dT

and S R  P

dB dT

(d) To find the volume and residual properties, we need to find the second virial coefficient of the mixture, N

which is B   yi y j Bij in general, or B  y12 B11  2 y1 y2 B12  y22 B22 for a binary mixture. B11 and B22 are the i 1 j 1

pure component virial coefficients, which we find from the Pitzer correlation, as we have done many times in the past: B

RTc Pc

0.422

 B   B , with B  0.083  T 0

1.6 r

and B1  0.139 

0.172 Tr4.2

Solution continued on next page…

For the cross-coefficient, we use the same equations, but with the critical temperature and pressure based on the following mixing rules: ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj

2 3  V 1/3  V 1/3   ci cj  Vcij     2 

and the derivatives of the individual virial coefficients are dBij dT



0 dBij1  dBij0 0.675 dBij1 0.722 RTc  dBij , with    and  5.2 Pc  dT dT  dT dT Tr2.6 Tr

Finally, once we have evaluated these, we can get the volume, residual Gibbs energy, residual entropy, and residual enthalpy from V

RT  B, P

G R  BP,

H R  BP  TP

dB , dT

and S R  P

dB dT

Solution continued on next page…

Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(d) T (K) P (bar) 293.15 3 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 Nitrogen 0.83 126.2 34 89.2 0.289 0.038 Species 2 Ammonia 0.17 405.7 112.8 72.5 0.242 0.253

Cross-Parameters kij 0

Vc Tc (K) Pc (bar) (cm3/mol) 226.2727 62.00 80.6

2nd Virial Coefficients (cm 3/mol) B11 -7 B22 -228 B12 -56

i   j 2 Z cij RTcij Pcij  Vcij 1/ 3 ci

Derivatives of 2nd Virial Coefficients (cm /((mol K)) dB11/dT 0.19 dB22/dT 1.89 dB12/dT

V Vcij    

Z cij  1/ 3 cj

V 2

 0.1455

Pr 0.0882 0.0266

B0 -0.0266 -0.6267

B1 0.1340 -0.5343

dB0 dT 0.0754 1.5711

dB1 dT 0.0090 3.9114

Tr 1.2956

Pr 0.0484

B0 -0.1959

B1 0.0810

dB0 dT 0.3443

dB1 dT 0.1878

Tcij  TciTcj 1  kij 

 ij 

Zc 0.2655

Tr 2.3229 0.7226

Zci  Zcj 2

B0  0.083  1

B  0.139 

0.422 Tr1.6 0.172 Tr4.2

  

Bij 

RTcij Pcij

B   B  0

0.50

dBij RTc  dBij0 dBij1  dBij0 0.675 dBij1 0.722   , with    2.6 and  5.2 Mixture 2nd Virial Coefficient (cm 3/mol) dT Pc  dT dT  dT dT T Tr r   B -27 3 Derivative of Mixture 2nd Virial Coefficient (cm /((mol K)) B  y12 B11  2y1 y2 B12  y22 B22 dB/dT 0.32 Mixture Properties: V (cm3/mol) Z GR (J/mol) HR (J/mol) S R (J/(mol K))

dB dB dB 2 dB11  y1  2y1 y2 12  y22 22 dT dT dT dT

8098 0.9967 -8.1 -36.5 -0.097

V

RT B, P

GR  BP ,

H R  BP  TP

dB , dT

and S R  P

dB dT

(e) To find the volume and residual properties, we need to find the second virial coefficient of the mixture, N

which is B   yi y j Bij in general, or B  y12 B11  2 y1 y2 B12  y22 B22 for a binary mixture. B11 and B22 are i 1 j 1

the pure component virial coefficients, which we find from the Pitzer correlation, as we have done many times in the past: B

RTc

0.422

 B   B , with B  0.083  T P 0

1.6 r

and B1  0.139 

0.172 Tr4.2

Solution continued on next page…

For the cross-coefficient, we use the same equations, but with the critical temperature and pressure based on the following mixing rules: ij 

i   j

2 Tcij  TciTcj 1  kij  Pcij  Z cij 

Z cij RTcij Vcij Z ci  Z cj

2 3  V 1/3  V 1/3   ci cj  Vcij     2 

and the derivatives of the individual virial coefficients are dBij dT



0 dBij1  dBij0 0.675 dBij1 0.722 RTc  dBij , with    and  5.2 Pc  dT dT  dT dT Tr2.6 Tr

Finally, once we have evaluated these, we can get the volume, residual Gibbs energy, residual entropy, and residual enthalpy from V

RT  B, P

G R  BP,

H R  BP  TP

dB , dT

and S R  P

dB dT

Solution continued on next page…

Spreadsheet for computing residual properties of a binary mixture, for SVA problem 11.37(e) T (K) P (bar) 293.15 4.2 Vc Mole 3 Fraction Name Tc (K) Pc (bar) (cm /mol) Zc  Species 1 sulfur dioxide 0.32 430.8 78.84 122 0.269 0.245 Species 2 ethylene 0.68 282.3 50.4 131 0.281 0.087

Cross-Parameters kij 0

Vc Tc (K) Pc (bar) (cm3/mol) 348.7332 63.06 126.4

2nd Virial Coefficients (cm /mol) B11 -398 B22 -147 B12 -235

i   j 2 Z cij RTcij Pcij  Vcij

Derivatives of 2nd Virial Coefficients (cm /((mol K)) dB11/dT 3.32 dB22/dT 1.09 dB12/dT

 0.1660

Pr 0.0533 0.0833

Tr 0.8406

Pr 0.0666

Tcij  TciTcj 1  kij 

 ij 

Zc 0.2750

Tr 0.6805 1.0384

Z cij 

Zci  Zcj 2

-0.6983 -0.3143

-0.7274 -0.0078

-0.4741

-0.2176

B0  0.083 

0.422

B1  0.139 

0.172

dB0 dTr 1.8365 0.6120

dB1 dTr 5.3445 0.5934

dB0 dTr 1.0601

dB1 dTr 1.7809

Tr1.6 Tr4.2

 Vci1/ 3  Vcj1/ 3  Vcij     2  

Bij 

RTcij Pcij

B   B  0

1.79

dBij R  dBij0 dBij1  dB 0 0.675  , with ij  2.6 and    Mixture 2nd Virial Coefficient (cm 3/mol)   dT Pc dTr dTr dTr Tr   B -211 2 2 Derivative of Mixture 2nd Virial Coefficient (cm 3/((mol K)) B  y1 B11  2y1 y2 B12  y2 B22 dB/dT 1.62 Mixture Properties: V (cm3/mol) Z GR (J/mol) HR (J/mol) S R (J/(mol K))

dBij1 dTr



0.722 Tr5.2

dB dB dB 2 dB11  y1  2y1 y2 12  y22 22 dT dT dT dT

5593 0.9637 -88.5 -288.4 -0.682

V

RT B, P

GR  BP ,

H R  BP  TP

dB , dT

and S R  P

dB dT

Solution 10.35

Problem Statement E

Laboratory A reports the following results for equimolar values of G for liquid mixtures of benzene(1) with 1-hexanol(2): E

G = 805 J·mol

at T = 298 K G = 785 J·mol

at T = 323 K E

Laboratory B reports the following result for the equimolar value of H for the same system: E

H = 1060 J·mol

at T = 313 K

Are the results from the two laboratories thermodynamically consistent with one another? Explain.

Solution

 (G E /RT )  E     H From Eq. 13.6 :    T RT 2 

 (G E/ RT )  E     H   T T2  

 (G E/RT )  (G E/ RT ) HE    To an excellent approximation, write:      T T T 2mean P

From the given data: (G E/ RT ) T

785 805  0.271  323 298    0.01084 323  298 25

And 

HE 1060   0.0182 2 T mean 3132

The data is thermodynamically consistent.

Solution 10.36 Problem Statement The following expressions have been proposed for the partial molar properties of a particular binary mixture: M1  M1  A x 2

M 2  M 2  A x1

Here, parameter A is a constant. Can these expressions possibly be correct? Explain.

Solution

By Eq. (10.14), the Gibbs/Duhem equation, x1 Given that M1  M1  Ax2 Then x1

dM1 ds1

 x2

dM2 ds1

dM1 ds1

M 2  M 2  Ax2

 x2

dM 2 ds1

dM1 ds1

0

 A

dM2 ds1

A

  Ax1  Ax2  0

Solution 10.37 Problem Statement 1

Two (2) kmol·hr

of liquid n-octane (species 1) are continuously mixed with 4 kmol·hr

of liquid iso-octane

(species 2). The mixing process occurs at constant T and P; mechanical power requirements are negligible. (a) Use an energy balance to determine the rate of heat transfer. 1

(b) Use an entropy balance to determine the rate of entropy generation (W·K ). State and justify all assumptions.

Solution n 1 

x1 

kmol hr

n 1 n 3

kmol hr

n 2 

x1 

n 3  n 1  n 2

x 2   x2

x2 

(a): Assume an ideal solution since n-octane and iso-octane are non-polar and very similar in chemical structure. For an ideal solution, there is no heat of mixing therefore the heat transfer rate is zero. (b): St  R x1 * ln  x1   x 2ln  x2  * n 3

St 

W K

Solution 10.38 Problem Statement Fifty (50) mol·s

of enriched air (50 mol-% N2, 50 mol-% O2) are produced by continuously combining air

(79 mol-% N2, 21 mol-% O2) with a stream of pure oxygen. All streams are at the constant conditions T = 25°C and P = 1.2(atm). There are no moving parts. 1

(a) Determine the rates of air and oxygen (mol·s ). (b) What is the rate of heat transfer for the process? (c) What is the rate of entropy generation S ∙ G (W·K

State all assumptions. Suggestion: Treat the overall process as a combination of demixing and mixing steps.

Solution For air entering the process: xO21 

x N 21 

For the enhanced air leaving the process:

xO22 

Total moles leaving:

n 2  50

x N 21 

mol s

(a): Apply mole balances to find rate of air and O2 fed to process xO21 * n air  n O 2  xO22 * n 2 x N 21 * n air  x N 22 * n 2

Solving for n air

n O 2

mol mol n O 2  s s (b): Assume ideal gas behavior. For an ideal gas there is no heat of mixing, therefore, the heat transfer rate is zero. n air 

(c): To calculate the entropy change, treat the process in two steps: 1. Demix the air to O2 and N2 2. Mix the N2 and combined O2 to produce the enhanced air Entropy change of demixing: S12  R  x O 21 * ln  x O 21   x N 21 * ln x N 21 

S12  

J mol K

Entropy change of mixing: S23  R  x O 22 * ln x O 22   x N 22 * ln x N 22 

S23 

J mol K

Total rate of entropy generation: SG  n air * S12  n 2 * S23

SG 

W K

Solution 10.39 Problem Statement E

A simple expression for M of a symmetrical binary system is M = Ax1x2. However, countless other empirical expressions can be proposed which exhibit symmetry. How suitable would the two following expressions be for general application? (a) M E  A x12 x22 E

(b) M = A sin(

Suggestion: Look at the implied partial properties M1E and M2E.

Solution M1E  Ax 1 x 22   x 1 

(a): For M E  Ax12 x 22

 x2  

Note that both x1  

M 2E  Ax 2 x 12   x 2

x1   x 2   M1E  M 2E 



In particular,  M1E    M 2E   0 Although M E

M1E

M 2E change sign, which is

unusual behavior. Find also that dM1E  Ax2   x1 x 2  dx1

dM2E  Ax1   x 1 x 2  dx 1

The two slopes are thus of opposite sign, as required; they also change sign, which is unusual.

(b): For M E  A

x1 

dM1E  A dx1

x1 

dM1E  dx1

dM 2E  dx1 dM 2E  A dx1

 x1  find M1E  A

 x1  A x2 cos  x1 dM1E   A 2 x2 dx1

 x1

M2E  A

 x1  A x1 cos  x1

dM 2E   A 2 x 1 dx 1

 x1

The two slopes are thus of opposite sign, as required. But note the following, which is unusual: x1 

and x1 

dM1E  dx1

dM2E  dx1

and

Part(a) graph 2

ME MEbar 1

ME values

1.5 1 0.5 0 -0.5 0

0.2

0.4

0.6

0.8

Part(b) graph 35 30

ME values

25 ME 20

MEbar1

MEbar2

10 5 0 0

0.2

0.4

0.6

0.8

Solution 10.40 Problem Statement For a multicomponent mixture containing any number of species, prove that  M   M     xk   M i  M   x i   x k  k T ,P T,P where the summation is over all species. Show that for a binary mixture this result reduces to Eqs. (10.15) and (10.16). Eq. 10.15: M1  M  x 2

dM dx1

Eq. 10.16: M 2  M  x1

dM dx1

Solution

By Eq. 10.7

 nM    M    M i    M  n  ni   ni T , P , n j  T , P , n j

At constant T and P,  M   M   dx k   k  x k T , P , x j

Divide by dni with restriction to constant n j  j  i :  M   M   x    k       n   n     x  i T , P , n j    k k T,P, x i n j j

With xk 

nk n

 nk    x  2   k    n  1 ni  ni n j   2  n n

and

 k  i k  i

 M   M   M  1 1       x k   1  xi   n   xi  n k i  xk T , P , x n  i T , P , n j  T , P , nj j  M  1  M  1       k x k   xk  n  x i T , P , n n  T , P, x j j  M   M i  M   x i 

T , P, n j

 M     k xk   x k 

T, P, x j

For species 1 of a binary mixture (all derivatives at constant T and P):   M   M   M   M   M    x 1    x 2    M  x 2       M i  M    x1   x1   x2   x1   x2        x2 x2 x1 x2 x1  

Because x1  x 2 

the partial derivatives in this equation are physically unrealistic; however, they

do have mathematical significance. Because M  M x1 x2 ), we can quite properly write:  M   M   dx1    dx 2 dM   x 1   x 2   x2 x1

Division by dx1 yields:  M  dx  M   M  dM  M  2              dx 1  x 1 x  x 2 x dx 1  x1 x  x 2 x 2

wherein the physical constraint on the mole fractions is recognized. Therefore M i  M  x2

The expression for

M  x1

is found the same way.

Solution 10.41 Problem Statement The following empirical two-parameter expression has been proposed for correlation of excess properties of symmetrical liquid mixtures:

  1 1  M E  A x1 x2   x 2  Bx1   x1  Bx2 Here, quantities A and B are parameters that depend at most on T. (a) Determine from the given equation the implied expressions for M1E and M 2E . (b) Show that the results of part (a) satisfy all necessary constraints for partial excess properties. 



a) Apply Eq. (10.7) to species 1:

  nM E     M1E     n   1   n2

Multiply the given equation by n and eliminate the mole fractions in favor of mole numbers:   1 1  nM E  an1n2    n1  Bn2 n2  Bn1        1 1 1 B     n1  M  An2     2 2    n1  Bn2 n2  Bn1  n  Bn  n  Bn     1 2 2 1   E 1

Conversion back to mole fractions yields:      1 1 1 B    x1  M1E  Ax2     2 2   x1  Bx 2 x2  Bx1   x1  Bx 2   x2  Bx1   

Solution continued on next page…

The first term in the first parentheses is combined with the first term in the second parentheses and the second terms are similarly combined:      1 1 1  x 1   1  Bx1  M1E  Ax 2     x 1  Bx 2  x 1  Bx 2  x 2  Bx 1  x 2  Bx1  

Reduction yields:    B 1  M  Ax      x  Bx 2  x  Bx 2  2 2 1  1  E 1

2 2

Similarly,    1 B  M  Ax      x  Bx 2  x  Bx 2  2 2 1  1  E 2

2 1

b) The excess partial properties should obey the Gibbs/Duhem equation, Eq. (10.14)when written for excess properties in a binary system at constant T and P: x1

dM1E dM 2E  x2 0 dx1 dx1

If the answers to part (a) are mathematically correct, this is inevitable, because they were derived from a proper expression for M E

M iE

xi 









 M   A B1  1 M   A1  B1  E 1



E 2



Solution 10.42 Problem Statement E

Commonly, if M for a binary system has a single sign, then the partial properties M1E and M2E have the same E

sign as M over the entire composition range. There are occasions, however, where the M1E may change sign E

even though M has a single sign. In fact, it is the shape of the M vs. x1 curve that determines whether the M1E E

change sign. Show that a sufficient condition for M1E and M2E to have single signs is that the curvature of M vs. x1 have a single sign over the entire composition range.

Solution

By Eqs. (10.15) and (10.16), written for excess properties, find: dM1E d2 M E  x2 dx1 dx12

At x1 

dM 2E d2 M E  x1 dx1 dx12

dM1E  and by continuity can only increase or decrease for dx1

x1 

dM1E dx1

d2 M E dx12 dM2E dx1

x1 

dM 2E  dx1

d2 M E dx12

Solution 10.43 Problem Statement

An engineer claims that the volume expansity of an ideal solution is given by  id   x i i i

Is this claim valid? If so, show why. If not, find a correct expression for

id.

Solution

The claim is not in general valid. 

 id 

1  V    V  T P

V id  x i Vi i

 V  1 1 x i  i    x iVi i    T  xiVi i  P  xiVi i i

The claim is valid only if all the Vi are equal.

Solution 10.44 Problem Statement E

Following are data for G and H (both in J·mol ) for equimolar mixtures of the same organic liquids. Use all E

of the data to estimate values of G , H , and TS for the equimolar mixture at 25°C. E



At T = 10°C: G = 544.0, H = 932.1



At T = 30°C: G = 513.2, H = 893.4



At T = 50°C: G = 494.2, H = 845.9

Suggestion: Assume CPE is constant and use material developed in Example 10.10.

Solution

T (K)

GE (J/mol)

HE (J/mol)

283.15 303.15

544 513

932.1 893.4

323.15

494.2

845.9

Assuming Cp is constant, H can be in the form of a line: H E  c  a * T Find the constants a and c, which are the slope and intercept of a line, respectively. a

J mol K

c

J mol

 T   E G takes the form: G E  a * Tln    T   bT  c     K  

Rearrange this to solve for b:

b

 T   G E  aTln    T   c   K  

T Find values for each b, and then average them together to obtain one b value b  13.549

J mol K

Solution continued on next page…

Using the constants found above re calculate H , G , and T*S at 25 C: H E  1544

J J J  2.155 * 298.15 K  901.487 mol mol K mol

G E  2.155 * 298.15 * ln298.15  298.15  13.549 * 298.15  1544  522.394 S E  H E  G E  378.848

J mol

Solution 11.1

Problem Statement

At 25°C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation V = x1 x2 (45 x1 + 25 x2) 3

V is in cm ·mol . At these conditions, V1 = 110 and V2 = 90 cm ·mol . Determine the partial molar volumes

V1 and V2 in a mixture containing 40-mol-% of species 1 at the given conditions. Solution One way to do this is to go back to the equations from chapter 11 (p. 376 of SVNA) that relate the partial molar properties of species in a binary mixture to the total property:

dM dx1 dM M 2  M  x1 dx1 M1  M  x 2

These are for any property M and corresponding partial molar property V, so in this case, they are

dV dx1 dV V2  V  x1 dx1 V1  V  x 2

Now, we need to write V in terms of x1 (eliminating x2): V  x 1V1  x 2V2  V E  x1V1  x 2V2  V V  110 x 1  90 x 2  x 1 x 2 45 x 1  25 x 2 

V  110 x 1  90 1  x1   x 1 1  x1  45 x 1  251  x 1  V  20 x1  90   x1  x 12 20 x 1  25 V  90  45 x1  5 x12  20 x13

Solution continued on next page…

So, dV  45  10 x1  60 x12 dx1 3

–1

And finally, at x1 = 0.4, V = 105.92 cm mol and

3 –1 dV = 31.4 cm mol , so dx1

dV  105.92  0.6 * 31.4  124.76 cm3 mol1 dx1 dV V2  V  x1  105.92  0.4 * 31.4  93.36 cm3 mol1 dx1 V1  V  x 2

Notice that both of the partial molar volumes are greater than the corresponding pure species volumes. This is to be expected, since the excess volume and volume change of mixing are positive. One could also do this problem by writing the total volume in terms of the number of moles of each species, and using the definition of the partial molar properties,   nV    Vi    ni  T , P ,n

ji

In that more general approach, we have V  x1V1  x2V2  V E  x1V1  x 2V2  V

V  110 x1  90 x 2  x 1 x 2 45 x1  25 x 2  n1n2 nV  110n1  90n2  45n1  25n2  2 n1  n2 

Taking the derivative of this with respect to n1 gives     nV   n1n2 n2 2n1n2     V1    110  45  45 n  25 n     1 2  2  n  n 2 n  n 3   n1  n  n    1  1 2 2 1 2 T , P ,n 2

V  110  45 x1 x2   45 x1  25 x 2  x2  2 x1 x2  V  110  90 x1 x2 1  x1   25 x 22  50 x1 x 22 V  110  40 x1 x 22  25 x 22

Solution continued on next page…

–1

Evaluating this for x1 = 0.4, x2 = 0.6 gives V1 = 124.76 cm mol (reassuringly identical to the answer we got before). We can get the partial molar volume of component 2 in the same way, or we can use the summability relation: V  x 1V1  x 2V2 V2 

V  x1V1 105.92  0.4 * 124.76   93.36 cm3 mol1 x2 0.6

The solutions manual has a problem for this one. I think that in MathCad, when the derivative of V w.r.t. x1 was taken, V had already been defined as a number, so the derivative was zero.

Solution 11.2

Problem Statement

The volume change of mixing (cm ·mol ) for the system ethanol(1)/methyl butyl ether(2) at 25°C is given by the equation

V = x1 x2 [ 1.026 + 0.0220 (x1 3

x2)]

Given that V1 = 58.63 and V2 = 118.46 cm ·mol , what volume of mixture is formed when 750 cm of pure species 1 3

is mixed with 1500 cm of species 2 at 25°C? What would be the volume if an ideal solution were formed?

Solution

The molar volume of the mixture is V = x1V1 + x2V2 + V = x1V1 + x2V2 + x1x2(–1.026 + 0.220(x1 – x2))

Solution continued on next page…

–1

So, first we need to find the mole fractions. We have 750 cm /58.63 cm mol = 12.79 mol of species 1 and 3

1500 cm /118.46 cm /mol = 12.66 mol of species 2. So, the mole fractions are x1 = 12.79/(12.79 + 12.66) = 0.503 and x2 = 0.497. So, 3

V = x1x2(–1.026 + 0.220(x1 – x2)) = –0.256 cm /mol and 3

V = x1V1 + x2V2 + V = 88.11 cm /mol id

The molar volume for an ideal solution would be V = x1V1 + x2V2 = 88.37 cm /mol. 3

The total volume is then 88.11*(12.66 + 12.79) = 2242.4 cm . For an ideal solution, the total volume would be the sum 3

of the initial volumes, 750 + 1500 = 2250 cm . This is consistent with the molar volume of the ideal solution 3

computed above, since 88.37*(12.66 + 12.79) = 2249 cm , which is within round-off error of 2250. The V of mixing for this system is quite small.

Solution 11.3

Problem Statement

If LiCl⋅2H2O(s) and H2O(l) are mixed isothermally at 25°C to form a solution containing 10 mol of water for each mole of LiCl, what is the heat effect per mole of solution?

Solution

If we write this mixing process as a chemical reaction, then we have: LiCl2H2O(s) + 8 H2O(l)  LiCl in 10 mol H2O Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

We have the heats of formation for LiCl•2H2O(s) and LiCl in 10 mol H2O on page 457 of the book. The thing that is a little bit tricky is that the heat of formation for LiCl•2H2O(s) includes the heat of formation of water in it. That is, it is the heat of reaction for Li(s) + ½ Cl2(g) + 2H2(g) + O2(g)

LiCl•2H2O(s), H = –1012650 J/mol

While the heat of formation for LiCl in 10 mol H2O does not include the heat of formation of the water. That is, it is the heat of reaction for Li(s) + ½ Cl2(g) + 10H2O

LiCl in 10 mol H2O, H = –441579 J/mol

So, to get the heat effect for LiCl•2H2O(s) + 8H2O(l)

LiCl in 10 mol H2O

we also have to add in the heat of formation for 2 mols of H2O. From table C.4 on p. 687, we see that the heat of formation of liquid water is –285830 J/mol. So, we want to add together the 3 reactions, in the appropriate directions: Li(s) + ½ Cl2(g) + 10H2O

LiCl in 10 mol H2O, H = –441579 J/mol

Li(s) + ½ Cl2(g) + 2H2(g) + O2(g), H = 1012650 J/mol

LiCl•2H2O(s)

2*( H2(g) + ½ O2(g)

H2O(l)), H = 2*(–285830) J/mol

LiCl•2H2O(s) + 8 H2O(l)

LiCl in 10 mol H2O, H = –589 J/mol

The net result is that there is almost no heat effect. 589 joules is released per mole of LiCl•2H2O(s) dissolved. The heat effect per mole of solution is –589 J/(mol solute) / 11 mols of solution per mol solute = –53.5 J/mol solution. One +

might also reasonably count it as 12 moles of solution, since the LiCl is presumably present as 1 mole of Li ions and –

1 mole of Cl ions.

Solution 11.4

Problem Statement

If a liquid solution of HCl in water, containing 1 mol of HCl and 4.5 mol of H2O, absorbs an additional 1 mol of HCl(g) at a constant temperature of 25°C, what is the heat effect?

Solution

The heat effect of making a mixture of 2 mol HCl in 4.5 mol water from a mixture of 1 mol HCl in 4.5 mol water plus 1 mol of HCl gas is equal to the heat effect of mixing 2 mol HCl with 4.5 mol water, minus the heat effect of mixing 1 mol of HCl with 4.5 mol water. So, it is twice the heat of solution of HCl at a concentration of 2.25 moles of water per mol HCl minus the heat of solution of HCl at a concentration of 4.5 moles of water per mol HCl. From figure 12.14 on p. 445, we see that the heat of solution of HCl at 2.25 moles of water per mol HCl is about –50.5 kJ/mol HCl, and the heat of solution at 4.5 moles of water per mol HCl is about –62 kJ/mol HCl. Thus, the heat effect of adding 1 mol of HCl(g) to a solution of 1 mol HCl in 4.5 mol water is H = 2*(–50.5) + 62 = –39 kJ.

Solution 11.5

Problem Statement What is the heat effect when 20 kg of LiCl(s) is added to 125 kg of an aqueous solution containing 10-wt-% LiCl in an isothermal process at 25°C?

Solution

In this problem, we mix a solution containing 12.5 kg of LiCl and 112.5 kg H2O with 20 kg of LiCl(s). The result will be a solution containing 32.5 kg of LiCl and 112.5 kg H2O. We want to find the heat effect (heat released or absorbed) when this process is carried out isothermally. This heat effect is given by the heat of mixing of the solution containing 12.5 kg of LiCl and 112.5 kg H2O with 20 kg of LiCl(s). We can obtain that heat of mixing as the difference between the heat of mixing for 32.5 kg LiCl(s) + 112.5 kg H2O, and the heat of mixing for 12.5 kg LiCl(s) + 112.5 kg H2O. We could write this as the sum of two steps:

12.5 kg LiCl in 112.5 kg H2O

32.5 kg LiCl in 112.5 kg H2O

with H1

12.5 kg LiCl(s) + 112.5 kg H2O(l)

with H2

32.5 kg LiCl(s) + 112.5 kg H2O(l)

12.5 kg LiCl in 112.5 kg H2O + 20 kg LiCl(s) t

32.5 kg LiCl in 112.5 kg H2O

with H = H1 + H2 t

where H1 is the heat of solution for mixing 32.5 kg LiCl(s) with 112.5 kg H2O(l) and –H2 is the heat of solution for mixing 12.5 kg LiCl(s) with 112.5 kg H2O(l). Heats of solution are given per mole of solute, so we need to convert the masses given to moles. The molecular weights are 42.39 g/mole for LiCl and 18.015 g/mole for water, so we have 32.5 kg LiCl(s) = 767 moles LiCl(s) 12.5 kg LiCl(s) = 295 moles LiCl(s) 112.5 kg H2O(l) = 6245 moles H2O(l) Solution continued on next page…

So, we have

295 mol LiCl in 6245 mol H2O

767 mol LiCl in 6245 mol H2O

with H1

295 mol LiCl(s) + 6245 mol H2O(l)

with H2

767 mol LiCl(s) + 6245 mol H2O(l)

295 mol LiCl in 6245 mol H2O + 472 mol LiCl(s)

767 mol LiCl in 6245 mol H2O

with H = H1 + H2

The molar ratio of water to LiCl in the concentrated solution is 6245/767 = 8.15. The molar ratio of water to LiCl in the dilute solution is 6245/295 = 21.2. From figure 12.14, the heat of solution for 1 mole of LiCl in 8.15 moles water is about –32 kJ/mole LiCl(s). The heat of solution for 1 mole of LiCl in 21.2 moles of water is about –35 kJ/mole LiCl(s). t

So, H1 = 767 moles LiCl(s) * –32 kJ/mole LiCl(s) = –24540 kJ and –H2 = 295 moles LiCl(s) * –35 kJ/mole LiCl(s) = –10320 kJ. The net heat of mixing is then H = –25540 + 10320 = –14220 kJ. If the mixing process is isothermal, then 14200 kJ of heat must be removed from the system.

Solution 11.6

Problem Statement

An LiCl/H2O solution at 25°C is made by adiabatically mixing cool water at 10°C with a 20-mol-% LiCl/H2O solution at 25°C. What is the composition of the solution formed?

Solution Basis: 1 mole of 20% LiCl solution entering the process. Assume 3 steps in the process: 1. Heat M1 moles of water from 10 C to 25 C 2. Unmix 1 mole (0.8 moles water + 0.2 moles LiCl) of 20 % LiCl solution 3. Mix (M1 + 0.8) moles of water and 0.2 moles of LiCl Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Step 1: From Steam Tables  kJ kJ  kg H1  104.8  41.99  * 18.015   kg kg  kmol H 1  1.132

kJ mol

Step 2: From Fig. 11.7 with n = 4 moles H2O/mole solute: H 2  25.5

kJ mol

Step 3: Guess M1 and find H3

H

for process. Continue to guess M1 until H  M1 

mol

for adiabatic process. mol  M1

n3 

mol

H  M1  H1 

H 3  

 mol  H2 

H   x



0.2 mol M1  1 mol

kJ mol

 mol  H3

kJ x

Solution 11.7

Problem Statement

A 20-mol-% LiCl/H2O solution at 25°C is made by mixing a 25-mol-% LiCl/H2O solution at 25°C with chilled water at 5°C. What is the heat effect in joules per mole of final solution?

Solution

The mixing occurs in three steps:

C  H2O

H2O



LiCl  H2O      LiCl  H2O   LiCl  H2O       LiCl  H2O  

H2O

C  LiCl  H 2       LiCl  H2O

 H1   

kj  kg

H 2 

kJ mol

kJ  * kg 

H3  

g mol

H1 

kJ mol

H  

kJ mol

H  H1  H2  H3  *

mol

Solution 11.8

Problem Statement

A 20-mol-% LiCl/H2O solution is made by six different mixing processes:

(a) Mix LiCl(s) with H2O(l). (b) Mix H2O(l) with a 25-mol-% LiCl/H2O solution. (c) Mix LiCl ·H2O(s) with H2O(l). (d) Mix LiCl(s) with a 10-mol-% LiCl/H2O solution.

(e) Mix a 25-mol-% LiCl/H2O solution with a 10-mol-% LiCl/H2O solution. (f) Mix LiCl·H2O(s) with a 10-mol-% LiCl/H2O solution.

Mixing in all cases is isothermal, at 25°C. For each part determine the heat effect in J·mol

of final solution.

Solution

(a) This is a straightforward computation of the heat of mixing. The heats of mixing are tabulated per mole of solute (LiCl). So, the heat effect per mole of solute when we mix 1 mol of LiCl(s) with 4 moles of H2O to get 5 moles, total, of 20% LiCl in H2O will be given by the heat of solution of LiCl(s) in 4 moles H2O at 25 °C. We can read  = –25.5 kJ per mole LiCl. This gives us this directly from figure 12.14 on p. 445 of SVNA to be H

5 moles of solution total, so the heat effect per mole of solution is –25.5/5 = –5.1 kJ per mole of solution.

(b) Again, we will do this on the basis of 1 mole of LiCl. If we want to end up with 4 moles of water mixed with 1 mole of LiCl, we need 4 moles of the 25% LiCl solution (which will contain 1 mole of LiCl) plus 1 mole of water. The overall heat release in going from 1 mole of LiCl(s) plus 4 moles of water will be –25.5 kJ/mol, as computed in part (a). The total change in enthalpy is independent of the path by which the process occurs. So, we can compute the change in enthalpy to make the 4 moles of 25% LiCl solution, and then the change in enthalpy when we mix the 4 moles of 25% LiCl solution with 1 mole of water to get 5 moles of 20% LiCl solution will be equal to the change in enthalpy for making 5 moles of 20% LiCl solution from LiCl plus water, directly, minus the change in enthalpy for making the 4 moles of 25% LiCl solution. We can read the heat of solution for 1 mole of LiCl in 3 moles of water from figure 12.14, or we can compute it as the difference between the heat of formation of LiCl in 3 mol H2O and LiCl(s) –1

–1

using the numbers given on page 444 of SVNA: –429366 J mol – –408610 J mol = –20800 J mol = –20.8 kJ.

The heat effect when the 4 moles of 25% LiCl solution is mixed with 1 mole of water is then given by –25.5 kJ – –20.8 kJ = –4.7 kJ. This is illustrated schematically below. This was computed for 5 moles (total) of –1

solution, so per mole of solution this is –4.7/5 = –0.94 kJ (mol solution) .

4 moles water

Enthalpy Change = -25.5 kJ

1 mol LiCl Enthalpy Change = -20.8 kJ

3 moles water

5 moles 20% LiCl in water

1 mol LiCl

5 moles 20% LiCl in water

Enthalpy Change = -25.5 kJ +20.8 kJ = -4.7 kJ 5 moles 20% LiCl in water

1 mole water

(c) In this case, we are dealing with the hydrated salt, LiCl•H2O(s), rather than the pure salt or solutions of it. This case is most easily treated using the heats of formation of the hydrated salt and the final solution. We need to mix 1 mole of LiCl•H2O(s) with 3 moles of water to get our 5 moles of 20% LiCl solution that we have been using as our basis. This overall process can be written as the sum of the following formation reactions: (1) LiCl•H2O(s)

Li + ½ Cl2(g) + H2(g) + ½ O2(g)

(2) Li + ½ Cl2(g)

LiCl(4 H2O)

(3) H2(g) + ½ O2(g)

H2O(l)

Net Reaction: LiCl•H2O(s)

LiCl(4 H2O)

Reaction (1) is the reverse of the formation reaction of the hydrated salt. By definition, the heat of reaction for this reaction includes the heat of formation of the water. Reaction 2 is the formation reaction for LiCl in 4 moles of H2O. By definition, the heat of reaction for this reaction does not include the heat of formation of water. Finally, reaction (3) is the formation reaction for water. We only need to include this for the 1 mole of water for which the heat of formation was included in reaction (1). The heat of reaction for (1), from p. 444 is –712.58 kJ/mol. For reaction (2), the heat of reaction isn’t given on p. 444. However, we can compute it using our result from part (a) to be –408.61 – 25.5 = –434.11 kJ/mol (the heat of formation of LiCl(s) plus the heat of solution for dissolving it in 4 moles H2O. Finally, the heat of reaction for (3) is the heat of formation of (liquid) water from table C.4 on p. 660, which is –285.83 kJ/mol. So, the heat effect is 712.58 + –434.11 + –285.83 = –7.4 kJ/mol. Dividing this by 5 total moles of solution gives –1.47 kJ/mole of solution.

(d) For each mole of LiCl in the 10% LiCl/H2O solution, we will get 9 moles of water. So, to get 4 moles of water, we 4

will need 4 /9 moles (total) of the 10% LiCl/H2O solution, which will contain 4 moles of water plus /9 of a mole of 5

LiCl. We will have to mix it with /9 of a mole of LiCl(s) to get 5 moles (total) of 20% LiCl/H2O.

The heat effect of mixing /9 of a mole of LiCl(s) with 4 moles of H2O to get the 4 /9 moles (total) of the 10% LiCl/H2O 4

solution will be /9 times the heat of solution of LiCl(s) in 9 moles of H2O. From Figure 12.14, this heat of solution is 4

about –32.4 kJ/mole solute. Multiplying this by /9 gives –14.4 kJ for the change in enthalpy when /9 of a mole of LiCl(s) is mixed with 4 moles of H2O at constant temperature of 25°C. From part (a), the overall change in enthalpy 4

when 1 mole of LiCl(s) is mixed with 4 moles of H2O is –25.5 kJ. The enthalpy change when the 4 /9 moles of the 5

10% LiCl/H2O solution is mixed with /9 of a mole of LiCl(s) is then the difference between these two –25.5 – (–14.4) = –11.1 kJ. Dividing this by 5 moles of solution total gives a heat effect of –2.2 kJ per mole of solution. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(e) First, we need to find out how much of each solution we need. If n1 is the number of moles of 25% LiCl solution needed and n2 is the number of moles of 10% LiCl solution needed, then we have 0.25 n1 + 0.10 n2 = 1 mole LiCl 0.75 n1 + 0.90 n2 = 4 moles H2O and

n1 + n2 = 5 moles total

Of course, only two of these mole balances are independent. Solving the last one for n2 and substituting it into the first one gives 0.25 n1 + 0.10 (5 – n1) = 1 mole LiCl 0.15 n1 = 0.5 10

from which n1 = /3 moles and n2 = /3 moles.

The /3 moles of 25% LiCl solution contain /6 mole LiCl and 2.5 moles water. The /3 moles of 10% LiCl solution 1

contain /6 mole LiCl and 1.5 moles water. The change in enthalpy when /6 mole LiCl is mixed with 2.5 moles water 5

is /6 times the heat of solution of LiCl(s) in 3 moles water, which is –20.8 kJ/(mol solute). Multiplying by /6 gives 1

–17.3 kJ. The enthalpy change when the /6 mole LiCl is mixed with 1.5 moles water is /6 of the heat of solution of 1

LiCl(s) in 9 moles water, which is –32.4 kJ/(mol solute), as in part (d). Multiplying by /6 gives –5.4 kJ. The total enthalpy change for mixing 1 mole LiCl(s) with 4 moles H2O is –25.5 kJ. The enthalpy change for mixing the 10

/3 moles of 25% LiCl solution with the /3 moles of 10% LiCl solution is then the difference between these values:

–25.5 – (–17.3 + –5.4) = –2.8 kJ for 5 total moles of solution. Dividing by 5 gives –0.56 kJ per mole of solution. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(f) Since this one involves the solid hydrate of the salt, it is again most easily treated using formation reactions. First, however, we need to see how much of the LiCl•H2O(s) and how much 10% LiCl in H2O we need to mix to get our final mixture of 1 mole LiCl in 4 moles H2O. If n1 is the number of moles of LiCl•H2O(s) and n2 is the number of moles of 10% LiCl solution, then we have n1 + 0.1 n2 = 1 n1 + 0.9 n2 = 4 and

2n1 + n2 = 5

Substituting n2 = 5 – 2n1 into the first equation gives n1 + 0.5 – 0.2 n1 = 1 0.8 n1 = 0.5 5

n1 = /8 and n2 = 3.75 = /8 3

The 3.75 moles of 10% LiCl in H2O contains /8 mole of LiCl. So, we can write the overall process of mixing /8 mole of LiCl•H2O(s) with 3.75 moles of 10% LiCl in H2O as the sum of the following formation reactions: (1)

(2)

(3)

(4)

/8(LiCl•H2O(s)

Li(s) + ½ Cl2(g) + H2(g) + ½ O2(g))

/8(LiCl/(9 H2O)

Li(s) + ½ Cl2(g))

Li(s) + ½ Cl2(g)

LiCl/(4 H2O)

/8(H2(g) + ½ O2(g) 5

H2O (l))

Overall reaction: /8 LiCl•H2O(s) + /8 LiCl/(9 H2O)

LiCl/(4 H2O)

The heat of reaction for (1) is the negative of the heat of formation of LiCl•H2O(s), which is 712.58 kJ/mol. The heat of reaction for reaction (2) is the negative of the heat of formation of LiCl/(9 H2O), which is –(–408.6 + –32.4) = 441.0 kJ/mol. The heat of reaction for (3) is the heat of formation of LiCl/(4 H2O), which is –434.1 kJ/mol. Finally, reaction (4) is the formation reaction for water, for which the heat of reaction is –285.8 kJ/mol. Adding these up gives: 5

/8(712.6) + /8(441.0) + –434.1 + /8(–285.8) = –2.0 kJ

Dividing this by 5 total moles of solution gives a heat release of 0.4 kJ per mole of solution.

Solution 11.9

Problem Statement

A stream of 12 kg·s

of Cu(NO3)2⋅6H2O and a stream of 15 kg·s

of water, both at 25°C, are fed to a tank where

mixing takes place. The resulting solution passes through a heat exchanger that adjusts its temperature to 25°C. What is the rate of heat transfer in the exchanger? • For Cu (NO3)2, H f 298 = • For Cu (NO3)2 ⋅ 6H2O, H f 298 = 2110.8 kJ.

• The heat of solution of 1 mol of Cu(NO3)2 in water at 25°C is

pendent of ñ for the concentrations of

interest here.

Solution

BASIS: 1 second, during which the following are mixed: (1) 12 kg hydrated (6 H2O) copper nitrate (2) 15 kg H2O k mol s

n1 

Mole ratio of the final solution:

6n1  n2 n1

k mol s

n2 



The mixing occurs in four steps:

O    H O l 

H 



Cu  N  O     Cu NO 



Cu  NO3 

H2O    Cu  N 2  O2  H2

Cu NO3  

H2O    Cu NO3  

Cu  NO3 

H1  * 

H2O 

H 2 O

H2Ol  Cu  NO3  

kJ   H3 



kJ kJ

H 2 O

H2  

H 4  

H  H1  H2  H3  H 4  



This value is for 1 mol of the hydrated copper nitrate. On the basis of 1 second, Q  n1

H mol

Q

kJ mol

Solution 11.10

Problem Statement

A liquid solution of LiCl in water at 25°C contains 1 mol of LiCl and 7 mol of water. If 1 mol of LiCl⋅3H2O(s) is dissolved isothermally in this solution, what is the heat effect?

Solution

The mixing occurs in four steps:

LiCl H2O    Li 

Cl2  H2 



H 

O    H O l 



Li 

Cl2  H2O    LiCl  H2O



LiCl  H2O     Li 



Cl2  H2O

LiCl  H2O  LiCl H2O    LiCl  H2O Values were pulled from the heats of formation for LiCl in Ch 11 and from table C.4

H1 

H2  

kJ H3  

H  H1  H2  H3  H 4  

Q  H

Q

H4 

Solution 11.11

Problem Statement

You need to produce an aqueous LiCl solution by mixing LiCl⋅2H2O(s) with water. The mixing occurs both adiabatically and without change in temperature at 25°C. Determine the mole fraction of LiCl in the final solution.

Solution

The mixing occurs in three steps:

Li 

Cl  n  H2 

H O    LiCl n  H O

O2    H2O



Cl2  H2  O2  

LiCl H2O    Li 

LiCl H2O  n H2O    LiCl n  H2O

H2  

H2 

H  H1  H2  H3 

Since it is also adiabatic,  H  0

This leads to H1  H2  H3

H1  

Interpolation in the table for the heats of formation for LiCl in Ch 11 shows that the LiCl is dissolved in 8.878 mol H2O.

x LiCl 

1 9.878

x LiCl 

Solution 11.12

Problem Statement

Data from the Bureau of Standards (J. Phys. Chem. Ref. Data, vol. 11, suppl. 2, 1982) include the following heats of formation for 1 mol of CaCl2 in water at 25°C: CaCl2 in 10 mol H2O

862.74 kJ

CaCl2 in 15 mol H2O

867.85 kJ

CaCl2 in 20 mol H2O CaCl2 in 25 mol H2O CaCl2 in 50 mol H2O CaCl2 in 100 mol H2O CaCl2 in 300 mol H2O CaCl2 in 500 mol H2O CaCl2 in 1000 mol H2O

 From these data prepare a plot of  H, the heat of solution at 25°C of CaCl2 in water, vs. n, the mole ratio of

water to CaCl2.

Solution

Hf (kJ)

– HfCaCl2

–862.74

–66.94

–867.85

–72.05

–870.06

–74.26

–871.07

–75.27

–872.91

–77.11

100

–873.82

–78.02

300

–874.79

–78.99

500

–875.13

–79.33

1000

–875.54

–79.74

To obtain the graph, first determine the heats of formation for CaCl2 + H2O

Ca  Cl2  n H2O    CaCl n H 2O CaCl2 s     Ca  Cl2

Hf  HfCaCl2

CaCl2 s   n H2O     CaCl2 n H2O  Obtain the heat of formation for CaCl2 from table C.4

H fCaCl2  

 H

Solution continued on next page…

Plotted on a logarithmic scale:

-66 -68

Hf -HfCaCl2

-70 -72 -74 -76 -78 -80 10

100

1000

Solution 11.13

Problem Statement

A liquid solution contains 1 mol of CaCl2 and 25 mol of water. Using data from Prob. 11.12, determine the heat effect when an additional 1 mol of CaCl2 is dissolved isothermally in this solution.

Problem 11.12 Data from the Bureau of Standards (J. Phys. Chem. Ref. Data, vol. 11, suppl. 2, 1982) include the following heats of formation for 1 mol of CaCl2 in water at 25°C: CaCl2 in 10 mol H2O

862.74 kJ

CaCl2 in 15 mol H2O CaCl2 in 20 mol H2O CaCl2 in 25 mol H2O

871.07 kJ

CaCl2 in 50 mol H2O CaCl2 in 100 mol H2O CaCl2 in 300 mol H2O CaCl2 in 500 mol H2O CaCl2 in 1000 mol H2O

 From these data prepare a plot of  H , the heat of solution at 25°C of CaCl2 in water, vs. n, the mole ratio of

water to CaCl2.

Solution The mixing occurs in three steps:



CaCl2    Ca  Cl2 Ca  Cl2  CaCl2 

H2O    CaCl2 

H2O     Ca  Cl2 

CaCl2 

H 2 O H 2O

H2O  CaCl2    CaCl2 

  H2O

Values were pulled from the heats of formation for CaCl2 from table C.4

H1 

H2  

kJ H3 

H  H1  H2  H3  H 4   

Q  H

Q

Solution 11.14

Problem Statement

Solid CaCl2⋅6H2O and liquid water at 25°C are mixed adiabatically in a continuous process to form a brine of 15-wt-% CaCl2. Using data from Prob. 11.12, determine the temperature of the brine solution formed. The specific 1

heat of a 15-wt-% aqueous CaCl2 solution at 25°C is 3.28 kJ·kg ·°C .

Problem 11.12

Data from the Bureau of Standards (J. Phys. Chem. Ref. Data, vol. 11, suppl. 2, 1982) include the following heats of formation for 1 mol of CaCl2 in water at 25°C: CaCl2 in 10 mol H2O

862.74 kJ

CaCl2 in 15 mol H2O CaCl2 in 20 mol H2O CaCl2 in 25 mol H2O CaCl2 in 50 mol H2O Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

CaCl2 in 100 mol H2O CaCl2 in 300 mol H2O CaCl2 in 500 mol H2O CaCl2 in 1000 mol H2O

 From these data prepare a plot of  H , the heat of solution at 25°C of CaCl2 in water, vs. n, the mole ratio of

water to CaCl2.

Solution

The process may be considered in two steps: Mix at 25 degC, then heat/cool solution to the final temperature. The two steps together are adiabatic and the overall enthalpy change is 0.

Calculate moles H2O needed to form solution: 85 18.015 n 15 110.986

n

Moles of water added per mole of CaCl2.6H2O:

n 

mol

Basis: 1 mol of CaCl2.6H2O dissolved

CaCl

H O s     Ca  Cl  H  O

H2O   CaCl2 

Ca  Cl2  H 

O    H2O

CaCl2 H2O 

H2O    CaCl2 

 H2O

  H2 O 

Values were pulled from the heats of formation for CaCl2 from table C.4

H1 

H2  

kJ H3  

H298  H1  H2  H3  H 4   

H298 

Determine the mass of the solution:

msoln  



g

msoln 

kJ kg C

H 298  msoln C p T 

T 

H298 msoln C p

T  

T

C

T

Cp 

Now determine the change in temperature

The initial temperature of the brine is: C

Solution 11.15

Problem Statement  Consider a plot of  H , the heat of solution based on 1 mol of solute (species 1), vs. n, the moles of solvent per mole

of solute, at constant T and P. Figure 11.4 is an example of such a plot, except that the plot considered here has a  linear rather than logarithmic scale along the abscissa. Let a tangent drawn to the  H vs. n curve intercept the ordinate at point I. (a) Prove that the slope of the tangent at a particular point is equal to the partial excess enthalpy of the solvent in a solution with the composition represented by n ; i.e., prove that:  d H  H2E dn

(b) Prove that the intercept I equals the partial excess enthalpy of the solute in the same solution; i.e., prove that:

I  H1E

Solution x1 

(a) As shown on page 458,

1  and  H  H (1  n)  n

Eliminating  n

H   H x1

Differentiation yields:

 d H 1 dH H dx1  1 dH H  dx1   2   2  dn x 1 dn x 1 dn  x 1 dx 1 x 1  dn

Where

dx1 1   x12 2  dn   n

Whence,

 d H dH dH E  H  x 1  H E  x1 dn dx1 dx1

Comparison with Eq. (11.16) written with M  H E H2E  H E  x1

dH E dx1

 d H  H 2E dn

Shows that

  d H  HI   dn n

(b) By geometry, with reference to the following figure,

Combining this with the result of Part (a) gives:

H2E 

From which,

  2E I  H  nH

Substitute:

x H HE   H  and n  2 x1 x1 x1

Whence,

I

  HI n

x H E  x 2 H2E HE  2 H 2E  x1 x1 x1

However, by the summability equation,

H E  x 2 H2E  x1 H1E I  H1E

Then,

Solution 11.16

Problem Statement

H for a particular solute(1)/solvent(2) system is represented by the equation:

H = x1 x2 (A21 x1 + A12 x2)  Relate the behavior of a plot of  H vs. n to the features of this equation. Specifically, rewrite Eq. (A) in the form   H ( n ) , and then show that:  (a) lim  H =0 n 0

 (b) lim  H = A12 n

 (c) lim d  H /d n = A21 n 0

Solution Combine the given equation with Eq. (A) of the preceding problem:   H  x 2  A21 x 1  A12 x 2 

With x 2   x 1

x1 

  n 

n

 x  1  n 2

The preceding equations combine to give:

  H

A n  n  A21  12   1  n  1  n 1  n 

(a) It follows immediately from the preceding equation that:

(b) Because n   n 

 lim  H0 n 0

 lim  H  A12

n  

n 

(c) Analogous to Eq. (12.10b), page 438, we write: H2E  x12  A21  2 A12  A21  x2    Eliminate the mole fractions in favor of n :  1   n     A  2 A12  A21  H 2E   1  n   21 1  n   2

In the limit as n 

A21. From the result of Part (a) of the preceding problem, it follows that  d H  A21 n  dn

lim

Solution 11.17

Problem Statement

If the heat of mixing at temperature t0

H0 and if the heat of mixing of the same solution at temperature t

show that the two heats of mixing are related by: t

H  H 0   CP dt t0

CP is the heat-capacity change of mixing, defined by Eq. (11.1).

Eq. 11.1:

M  M   x i M i i

Solution

By Eq. (11.1) with M  H H  H xi Hi . Differentiate: i

 H     t 

P,x

 H  With    t 

 H     t 

 CP P,x

Therefore,

P,x



H  H0

 H   xi  i   t  i P, x

 H      t 

P,x

 C P   x i CPi  C P i

d H    CP dt H  H 0   C P dt

Solution 11.18

Problem Statement Use Figure 11.4 to obtain heat of solution data for this problem.

What is the heat effect when 75 kg of H2SO4 is mixed with 175 kg of an aqueous solution containing 25-wt-% H2SO4 in an isothermal process at 300 K?

Solution

Enthalpy values pulled from Figure 11.4 m1 

m2 

H1 

kJ kg

H2  

kJ kg

The new weight percent for the mixture is: 100% * m1  25% * m2 m1  m2 m3  m2  m1

m3 

 H3  

kJ kg

The heat effect is then:

Q  m3 H3  m1 H1  m2 H 2 

Q

Solution 11.19

Problem Statement Use Figure 11.4 to obtain heat of solution data for this problem.

For a 50-wt-% aqueous solution of H2SO4 at 350 K, what is the excess enthalpy H in kJ·kg ?

Solution

Enthalpy values pulled from Figure 11.4 x1  H 

x2   x1 kJ kg

H SO

H 

kJ kg

H 

kJ kg

The excess enthalpy: H E  H   x 1 H1  x 2 H2 

HE  

kJ kg

Solution 11.20

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

A single-effect evaporator concentrates a 20-wt-% aqueous solution of H2SO4 to 70-wt-%. The feed rate is 15 kg·s , and the feed temperature is 300 K. The evaporator is maintained at an absolute pressure of 10 kPa, at which pressure the boiling point of 70-wt-% H2SO4 is 102°C. What is the heat-transfer rate in the evaporator?

Solution

Enthalpy values pulled from Figure 11.4 x1 

m1  x2 

kg s

H1  

H3 

kJ kg

H 2 SO4

H2  

m3  m1  m2

m3 

kJ kg

H 2O

First determine the mass flows: m2 

x1m1

m2 

kg s

Then determine the heat transfer rate: Q  m2 H 2  m3 H 3  m1 H 1

Q

kJ s

Solution 11.21

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

What is the heat effect when sufficient SO3(l) at 25°C is reacted with H2O at 25°C to give a 50-wt-% H2SO4 solution at 60°C?

Solution

First react 1 mol SO3(l) with 1 mol H2O(l) to form 1 mol H2SO4(l): SO3(l) + H2O(l) ---> H2SO4(l) With data from Table C.4:

H298   

 

 J



H298  

Mix 1 mol or 98.08 g H2SO4(l) with mg H2O to form a 50% solution.

mH2 SO4 

msoln 

mH2 SO4

mH2O  msoln  mH2 SO4

msoln 

0.5

mH2O 

Enthalpy values pulled from Figure 11.4 H H2 SO4 

kJ kg

pure H2 SO4 H soln  

H H 2O 

C kJ kg

kJ kg

pure H2O

Hmix  

mixture

Determine the Change in enthalpy of mixing

Hmix  msoln H soln  mH2 SO4 H H2SO4  mH2O H H2O And the heat transfer rate is: Q

H298  Hmix msoln

Q 

kJ kg

Solution 11.22

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

A mass of 70 kg of 15-wt-% solution of H2SO4 in water at 70°C is mixed at atmospheric pressure with 110 kg of 80-wt-% H2SO4 at 38°C. During the process heat in the amount of 20,000 kJ is transferred from the system. Determine the temperature of the product solution.

Solution

x1 

m1 

x2 

m2 

Enthalpy values pulled from Figure 11.4 with interpolation: H1 

kJ kg

H2  

kJ kg

m3  m1  m2

m3 

H 2 SO4

Solution continued on next page…

x3 

Q

m1 x1  m2 x2

H3 

x3 

Q  m1 H1  m2 H2  m3

H3  

kJ kg

Solution 11.23

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

An insulated tank, open to the atmosphere, contains 750 kg of 40-wt-% sulfuric acid at 290 K. It is heated to 350 K by injection of saturated steam at 1 bar, which fully condenses in the process. How much steam is required, and what is the final concentration of H2SO4 in the tank?

Solution

Enthalpy values pulled from Figure 11.4 with interpolation: x1 

m1 

H1  

kJ kg

H 2  2675.4

kJ kg

H 2 SO4

Saturated steam at 1 bar; Table F.2:

By knowing that H3 is equal to: H3 

m1 H1  m2 H 2 m3

And that: x3 

x1m1 m1  m2

and

m3  m1  m2

Determine what m2, x3, and H3 by guessing an m2 H3  

kJ kg

x3 

m2 

The question now is whether this result is in agreement with the value read from Fig. 11.4 at 37.0% and 350 K. It is close and about as close as you can get.

Solution 11.24

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

Saturated steam at 3 bar is throttled to 1 bar and mixed adiabatically with (and condensed by) 45-wt-% sulfuric acid at 300 K in a flow process that raises the temperature of the acid to 350 K. How much steam is required for each pound mass of entering acid, and what is the concentration of the hot acid?

Solution

Enthalpy values pulled from Figure 11.4: x1 

kg  lbm

m1 

H1  

kJ kg

H2 SO4

Saturated steam at 1 bar; Table F.2: H 2  2724.7

kJ kg

By knowing that H3 is equal to: H3 

m1 H1  m2 H 2 m3

And that: x3 

x1m1 m1  m2

and

m3  m1  m2

Determine what m2, x3, and H3 by guessing an m2 H3  

kJ kg

x3 

m2 

The question now is whether this result is in agreement with the value read from Fig. 11.4 at 37.5% and 350 K. It is close and about as close as you can get.

Solution 11.25

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

For a 35-wt-% aqueous solution of H2SO4

H in kJ·kg ?

Solution

Enthalpy values pulled from Figure 11.4: H

kJ kg

pure water

H1 

x2   x1

x2 

kJ kg

pure acid

H2  

kJ kg

x 1  .35

H  H  x1 H1  x2 H2

The enthalpy of mixing is: H  162.5

kJ kg

Solution 11.26

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

If pure liquid H2SO4 at 300 K is added adiabatically to pure liquid water at 300 K to form a 40-wt-% solution, what is the final temperature of the solution?

Solution

Enthalpy values pulled from Figure 11.4: H1 

x2   x1

kJ kg

x2 

kJ kg

H2  

x1 

Q  H  H  x1 H1  x2 H2 

The enthalpy of mixing is: H  110

kJ kg

From Fig. 11.4, for a 40% soln. to have this enthalpy the temperature is well above 350 K, probably about 475 K.

Solution 11.27

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

A liquid solution containing 1 kg mol H2SO4 and 7 kg mol H2O at 300 K absorbs 0.5 kg mol of SO3(g), also at 300 K, forming a more concentrated sulfuric acid solution. If the process occurs isothermally, determine the heat transferred.

Solution x1 

Final solution:

x2 

98.08  98.08  7 * 18.015

1.5 * 98.08  0.5568 1.5 * 98.08  6.5 * 18.015

For these values the enthalpies @ 300 K are pulled from figure 11.4: H H2 O 

kJ kg

H H2 SO4 

kJ kg

H1  

kJ kg

H2  

Unmix the initial solution:

Hunmix   x1 HH2 SO4  1  x1  H H2O   H1   Hunmix  290

kJ kg

React 1 mol SO3(g) with 1 mol H2O(l) to form 1 mol H2SO4(l). We neglect the effect of Ton the heat of reaction, taking the value at 300 K equal to the value at 25 C.

H f SO3  

J mol

H f H2O  

J mol

Hrx  H f H2 SO4  H f H2O  H f SO3

H f H2 SO4  

Hrx  

J mol

Finally, mix the constituents to form the final solution (assume 1kg basis):

Hmix  H2   x2 H H2 SO4    x2  * H H2O   

Hmix  

kJ kg

Q  Hunmix 98.08  7 * 18.015  1 kg * Hrx  Hmix 1.5 * 98.08  6.5 * 18.015

Q 

Solution 11.28

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

H of sulfuric acid in water and the partial specific enthalpies of H2SO4 and H2O for a solution containing 65-wt-% H2SO4 at 300 K.

Solution

For these values the enthalpies @ 300 K are pulled from figure 11.4: H 

kJ kg

H1 

x2   x1

kJ pure acid kg

H2 

x2 

H  H  x1 H1  x2 H2 H  340

kJ pure water kg

x1 

kJ kg

From the intercepts of a tangent line drawn to the 300 K curve of Fig. 11.4 at 65%, find the approximate values: H1  

kJ kg

H2  

kJ kg

Solution 11.29

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

It is proposed to cool a stream of 75-wt-% sulfuric acid solution at 330 K by diluting it with chilled water at 280 K. Determine the amount of water that must be added to 1 kg of 75-wt-% acid before cooling below 330 K actually occurs.

Solution

Graphical solution: If the mixing is adiabatic and water is added to bring the temperature to 330 K, then the point on the H-x diagram of Fig. 11.4 representing the final solution is the intersection of the 330 K isotherm with a straight line between points representing the 75 wt % Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

solution at 330 K and pure water at 280 K. This intersection gives x3, the wt % of the final solution at 330 K:

x3 

m1  kg

By a mass balance: x3 

0.75 * m1 m1  m2

m2 

0.75 m1 x3

 m1

m2 

Solution 11.30

Problem Statement

Use Figure 11.4 to obtain heat of solution data for this problem.

The following liquids, all at atmospheric pressure and 300 K, are mixed: 25 kg of pure water, 40 kg of pure sulfuric acid, and 75 kg of 25-wt-% sulfuric acid. (a) How much heat is liberated if mixing is isothermal at 300 K? (b) The mixing process is carried out in two steps: First, the pure sulfuric acid and the 25-wt-% solution are mixed, and the total heat of part (a) is extracted; second, the pure water is added adiabatically. What is the temperature of the intermediate solution formed in the first step?

Solution

(a): m1 

kg m2 

kg m3 

x1 

x2 

kJ kg

H2 

x3 

Enthalpy values pulled from Figure 11.4: H1 

kJ kg

H3  

m4  m1  m2  m3 x4 

x1m1  x 2m2  x3m3 m4 H 4  285

x4 

kJ kg

Q  m4 H4  m1 H1  m2 H2  m3 H3 

Q

(b) Determine the enthalpy of the intermediate to determine the temperature: H1 

kJ kg

H2 

m3  m1  m2

kJ m  kg 1 x3 

x3 

m2 

x1 m1  x 2 m2 m3

H3  

x1 

H3 

x2 

Q  x1m1  x 2 m2 m3

kJ kg

Refering to Figure 11.4 at that concentration and enthalpy puts the temperature approximately at 350 K.

Solution 11.31

Problem Statement

A large quantity of very dilute aqueous NaOH solution is neutralized by addition of the stoichiometric amount of a 10-mol-% aqueous HCl solution. Estimate the heat effect per mole of NaOH neutralized if the tank is maintained at 25°C and 1(atm) and the neutralization reaction goes to completion. Data:

• For NaCl,

n 

• For NaOH,

  H

n 

  H 

kJ mol1 kJ mol 1

Solution

BASIS: 1 mol NaOH neutralized. For following reaction; data from Table C.4:

NaOH  s  HCl  g     NaCl  s   H2O l

H298   



 

H298  

 J

 5

NaOH  s  HCl  g     NaCl  s  H2O l 



NaOH inf H2O    NaOH  s   inf H2O



HCl  H2O    HCl  g  



H2Ol 

NaCl s   inf H2O    NaCl inf H2O

 

NaOH inf H2O  HCl  H2O    NaCl inf H2O

H1  H298 

H2 

H3 

H4 

H  H1  H2  H3  H4

Q  H

Q

Solution 11.32

Problem Statement

A large quantity of very dilute aqueous HCl solution is neutralized by the addition of the stoichiometric amount of a 10-mol-% aqueous NaOH solution. Estimate the heat effect per mole of HCl neutralized if the tank is maintained at 25°C and 1(atm) and the neutralization reaction goes to completion.  • For NaOH(9H2O),  H  45.26 kJ ·mol1

• For NaCl,

n 

  H

kJ mol1

Solution

With the heat of solution given, this problem is simple H soln 

kJ mol

Now, on the BASIS of 1 mol of HCl neutralized: NaCl   inf H 2O     NaCl inf H 2O 



NaOH  H2O     NaOH  s  

H 2O



HCl inf H 2O    HCl  g   inf H 2O



NaOH  s  HCl  g     NaCl  s  H2O l   

HCl inf H 2O  NaOH  H 2O     NaCl inf H 2O 

Solution continued on next page…

H 1  

H 2 

H 3 

H 4 

H  H1  H2  H 3  H 4

ange

Q H

Q

Solution 11.33

Problem Statement

(a) Making use of Eqs. (10.15) and (10.16), written for excess properties, show for a binary system that:

  dX  dX   and M2E  x12  X  x2  M1E  x22  X  x1    dx1  dx1 

where X

ME x1 x2

Eq. 10.15: M 2E  M E  x1

dM E dx1

Eq. 10.16: H  H   x i H i i

(b) Plot on a single graph the values of H /(x1x2), H1E , and H2E determined from the following heat-of-mixing data for the H2SO4(1)/H2O(2) system at 25°C:

x1 = mass fraction H2SO4

H/kJ·kg

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.85 0.90 0.95

73.27 144.21 208.64 262.83 302,84 323.31 320.98 279.58 237.25 178.87 100.71

Solution

(a) From the definition of M: Differentiate:

 A

M E  x1 x2 M

dM E dM  M  x 2  x1   x 1 x 2 dx1 dx1

 B

Substitution of Eqs. (A) & (B) into Eqs. (10.15) & (10.16), written for excess properties, yields the required result.

In order to take the necessary derivatives of H, we will fit the data to a third order polynomial of the form H

HE  a  bx1  cx12  dx13 x1 x2 rd

Taking the data, plugging this into an excel spreadsheet, and plotting a 3 order trendline yields the following values for a, b, c, and d

a

b 

c

d 

A new H(x1) can be calculated by:

H  x1   a  bx1  cx12  dx13 Solution continued on next page…

Using the equations given in the problem statement and taking the derivatives of the polynomial analytically: 2 HEbar    x1  *  H  x1   x1  b  cx1  dx12      2 HEbar 2   x1  *  H  x1   1  x1   b  2cx1  3dx12     

(b) 0 0

0.2

0.4

0.6

0.8

H (kJ/kg)

-500

-1000 Hebar1 Hebar2 -1500

H(x1)

-2000

-2500

By adding acid to water, a steady decrease in the enthalpy of mixing is seen, and this allows for easy measuring and can less error when trying to attain the correct pH. Whereas, adding water to acid there is a sharp increase up to 100 acid, this makes it difficult to get the correct pH.

Solution 11.34

Problem Statement

A 90-wt-% aqueous H2SO4 solution at 25°C is added over a period of 6 hours to a tank containing 4000 kg of pure water also at 25°C. The final concentration of acid in the tank is 50-wt-%. The contents of the tank are cooled continuously to maintain a constant temperature of 25°C. Because the cooling system is designed for a constant rate Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

of heat transfer, this requires the addition of acid at a variable rate. Determine the instantaneous 90-%-acid rate as a 1

function of time, and plot this rate (kg·s ) vs. time. The data of the preceding problem may be fit to a cubic equation E

expressing H /(x1x2) as a function of x1, and the equations of the preceding problem then provide expressions for

H1E and H2E .

Solution

In order to take the necessary derivatives of H, we will fit the data to a third order polynomial of the form H

HE  a  bx1  cx12  dx13 x1 x2 rd

Taking the data, plugging this into an excel spreadsheet, and plotting a 3 order trendline yields the following values for a, b, c, and d

a

b 

c

d 

A new H(x1) can be calculated by:

H  x1   a  bx1  cx12  dx13 Using the equations given in the problem statement and taking the derivatives of the polynomial analytically: 2 HEbar    x1  *  H  x1   x1  b  cx1  dx12      2 HEbar 2   x1  *  H  x1   1  x1   b  2cx1  3dx12     

At time , let:

x1 = mass fraftion of H2SO4 in tank m = total mass of 90% H2SO4 added up to time H = enthalpy of H2SO4 solution in tank at 25 C

H2 = enthalpy of pure H2O at 25 C H1 = enthalpy of pure H2SO4 at 25 C H3 = enthalpy of 90% H2SO4 at 25 C Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Material and energy balances are then written as: x1 * 

 M 

 x1  

4000 * x1

 A

0.9  x1

Q  Ht  4000  m * H  4000 * H 2  mH3 Since H  H  x1 H1  x2 H2

H1  H 2

H H The energy

balance then becomes:

Q 

 mH  m * H3

 B

Applying these equations to the overall process, for which: 

H3  H 

x1 

H  H 



H3  

H  

kJ kg

Define quantities as a function of x1 Q  x1   

m x 1   Qt  x 1   

 m  x 1  * H  x 1   m  x 1  * H 3

4000 * x1 0.9  x1

m

 m  x1  * H  m  x 1  H3

 Qt 



Since the heat transfer rate q is constant: q

Qt  x1 

  x1  



Qt  x1  q

C 

The following is probably the most elegant solution to this problem, and it leads to the direct calculation of the required rates, r

dm d

Solution continued on next page…

When 90% acid is added to the tank it undergoes an enthalpy change equal to: Hbar 

Hbar  H

Hbar and

Hbar

x1 existing in the tank at the

instant of addition. This enthalpy change equals the heat required per kg of 90% acid to keep the temperature at 25 C. Thus, r  x1  

q 0.9 Hbar1  0.1Hbar 2  H3

Lastly, plot the rate as a function of time. 3600 3100

r(x1) (kg/h)

2600 2100 1600 1100 600 0

Θ (x1)

Solution 11.35

Problem Statement id

S by appropriate application of Eqs. (5.36) and (5.37) to a mixing process. Eq. 11.12: S id  R  x i ln x i i

Eq. 5.36: 

N! ( N 1 !)( N 2 !)( N 3 !)

Eq. 5.37: S=k

Solution

In this application the microscopic “state” of a particle is its species identity, i.e., 1, 2, 3, . . . . By assumption, this label is the only thing distinguishing one particle from another. For mixing, t t t S t  Smixed  Sunmixed  Smixed   Sit i

where the total emtropies are given by Eq. (5.42). Thus, for an unmixed species i, and for the mixed system of particles,

Sit  k ln i  k ln

Combining the last three equations gives:

From which:

Ni ! Ni

S t  k ln

S S t S t 1 N!    ln  N  R kN N N1 N 2 N 3  R    N A 

t  0 Smixed  k ln

N! N1 N 2 N 3



1 (ln N !   ln N i !) i N

 1  S 1   N ln N  N  N i ln N i   N i    N ln N   x i N ln x i N  R N   N   i i i 

 1   N ln N   xi N ln xi   x i N ln N    x i ln x1 N   i i i

Solution 11.36

Problem Statement

Ten thousand (10,000) kg·h

of an 80-wt-% H2SO4 solution in water at 300 K is continuously diluted with chilled

water at 280 K to yield a stream containing 50-wt-% H2SO4 at 330 K. 1

(a) What is the mass flow rate of chilled water in kg·h ? (b) What is the rate of heat transfer in kJ·h

for the mixing process? Is heat added or removed?

(c) If the mixing occurred adiabatically, what would be the temperature of the product stream? Assume here the same inlet conditions and the same product composition as for part (b).

Heat of solution data is available in Fig. 11.4.

Solution kg h

m 1 

x1 

T1 

H1  

x2 

T2 

H2  

kJ kg

x3 

T3 

H3  

kJ kg

(a): Use mass balances to find feed rate of cold water and product rate. Given :

 1  m 2  m 3 m m 1 x1  m 2 x2  m 3 x3

Assume:

 1  m 2 m

2 Solve for m

H 2 SO4 balance

 3  * m 1 m

3 m m 2 

kg h

kg hr

m 3 

(b): Apply an energy balance on the mixer Q  m 3 H3  m 1 H1  m 2 H2 

kJ h

Q  

Q is negative, heat is removed from the mixer.

(c): For an adiabatic process, Q is zero. Solve the energy balance to find H3 H3 

m 1 H1  m 2 H 2 m 3

H3  

kJ kg

From figure 11.4 this corresponds to a temperature of approximately 375 K

Solution 12.1 Problem Statement Consider a closed vessel of fixed volume containing equal masses of water, ethanol, and toluene at 70°C. Three phases (two liquid and one vapor) are present. (a) How many variables, in addition to the mass of each component and the temperature, must be specified to fully determine the intensive state of the system? (b) How many variables, in addition to the mass of each component and the temperature, must be specified to fully determine the extensive state of the system? (c) The temperature of the system is increased to 72°C. What, if any, intensive or extensive coordinates of the system remain unchanged?

Solution (a) Using phase rule equation 3.1, F  2    N , F = 2 Since temperature is specified, the number of variables to fully determine the intensive state of the system is: 2–1=1 (b) Duhem’s theorem states, for any closed system formed from known amounts of prescribed chemical species, the equilibrium state is completely determined when any two independent variables are fixed. The number of independent variables is given by phase rule. For this case F = 2, therefore the number of variables to fully determine the extensive state of the system is zero.

(c) As the system is closed, extensive coordinates remain unchanged. The intensive variables will change value as temperature changes.

Solution 12.2 Problem Statement Consider a binary (two-species) system in vapor/liquid equilibrium. Enumerate all of the combinations of intensive variables that could be fixed to fully specify the intensive state of the system.

Solution Using phase rule equation 3.1, F  2    N , F = 2 We can specify the following combinations of intensive variables: Temperature and pressure Temperature and mole fraction Pressure and mole fraction Problems 12.3 through 12.8 refer to the Pxy diagram for Ethanol(1)/Ethyl acetate(2) at 70°C shown in Fig. 12.19.

FIGURE 12.19 Pxy diagram for vapor-liquid equilibrium of Ethanol(1)/Ethyl Acetate(2) at 70°C.

Solution 12.3 Problem Statement For this problem, refer to the Pxy diagram for ethanol(1)/ethyl acetate(2) at 70°C shown in Fig. 12.19.

The pressure above a mixture of ethanol and ethyl acetate at 70°C is measured to be 86 kPa. What are the possible compositions of the liquid and vapor phases? Solution The high pressure corresponds to liquid phase and low pressure corresponds to vapor phase. Draw a horizontal line at 86 kPa. From the point of intersection of this horizontal line and the upper curve, draw a vertical line, thus the mole fraction of liquid ethanol, x1 = 0.12 and mole fraction of liquid ethyl acetate, x2 = 1–x1 = 0.88 Similarly, the composition of vapor phases can be determined from the intersection of the horizontal line and the lower curve. Thus the mole fraction of vapor ethanol, y1 = 0.18 and mole fraction of vapor ethyl acetate, y2 = 1–y1 = 0.82 As you see from figure 12.19, the horizontal line intersects the upper and lower curves twice, so the remaining compositions are the following: x1 = 0.79, x2 = 0.21 and y1 = 0.69, y2 = 0.31

Solution 12.4 Problem Statement For this problem, refer to the Pxy diagram for ethanol(1)/ethyl acetate(2) at 70°C shown in Fig. 12.19.

The pressure above a mixture of ethanol and ethyl acetate at 70°C is measured to be 78 kPa. What are the possible compositions of the liquid and vapor phases? Solution Draw a horizontal line at 78 kPa. From the point of intersection of this horizontal line and the upper curve, draw a vertical line, thus the mole fraction of liquid ethanol, x1 = 0.94 and mole fraction of liquid ethyl acetate, x2 = 1–x1 = 0.06 Similarly, the composition of vapor phases can be determined from the intersection of the horizontal line and the lower curve. Thus the mole fraction of vapor ethanol, y1 = 0.875 and mole fraction of vapor ethyl acetate, y2 = 1–y1 = 0.125

Solution 12.5 Problem Statement For this problem, refer to the Pxy diagram for ethanol(1)/ethyl acetate(2) at 70°C shown in Fig. 12.19.

Consider an ethanol(1)/ethyl acetate(2) mixture with x1 = 0.70, initially at 70°C and 100 kPa. Describe the evolution of phases and phase compositions as the pressure is gradually reduced to 70 kPa. Solution As we can see from the figure, for x1 = 0.70 at pressures > 89 kPa, the entire mixture is liquid. So for 89 kPa < P < 100 kPa, y1 = 0, y2 = 0, where P is pressure Now, for pressures 85.8 kPa < P < 89 kPa we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given pressure and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.3, 12.4. As we continue to lower pressure from 85.8 kPa, the mixture exists only in vapor phase, so for 70 kPa < P < 85.8 kPa x1 = 0, x2 = 0

Solution 12.6 Problem Statement For this problem, refer to the Pxy diagram for ethanol(1)/ethyl acetate(2) at 70°C shown in Fig. 12.19.

Consider an ethanol(1)/ethyl acetate(2) mixture with x1 = 0.80, initially at 70°C and 80 kPa. Describe the evolution of phases and phase compositions as the pressure is gradually increased to 100 kPa. Solution As we can see from the figure, for x1 = 0.80 at pressures < 81.5 kPa, the entire mixture is vapor. So for 80 kPa < P < 81,5 kPa, x1 = 0, x2 = 0, where P is pressure Now, for pressures 81.5 kPa < P < 85.9 kPa we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given pressure and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.3, 12.4. As we continue to gradually increase pressure from 85.9 kPa to 100 kPa, the mixture condenses and turns to liquid, so for 85.9 kPa < P < 100 kPa y1 = 0, y2 = 0

Solution 12.7 Problem Statement For this problem, refer to the Pxy diagram for ethanol(1)/ethyl acetate(2) at 70°C shown in Fig. 12.19.

What is the composition of the azeotrope for the ethanol(1)/ethyl acetate(2) system? Would this be called a high-boiling or low-boiling azeotrope? Solution For binary mixture, azeotrope exists when the components have same composition in liquid and vapor phases, so x1 = y1, x2 = y2 In this case, the composition of the azeotrope: x1 = y1= 0.45, x2 = y2 = 0.55 The boiling point of ethanol is 173.1°F (78.37°C) and the boiling point of ethyl acetate is 170.8°F (77.1°C). Since the azeotrope occurs at 70°C, this is a positive deviation or minimum boiling azeotrope or low-boiling azeotrope.

Solution 12.8 Problem Statement For this problem, refer to the Pxy diagram for ethanol(1)/ethyl acetate(2) at 70°C shown in Fig. 12.19.

Consider a closed vessel initially containing 1 mol of pure ethyl acetate at 70°C and 86 kPa. Imagine that pure ethanol is slowly added at constant temperature and pressure until the vessel contains 1 mol ethyl acetate and 9 mol ethanol. Describe the evolution of phases and phase compositions during this process. Comment on the practical feasibility of carrying out such a process. What sort of device would be required? How would the total system volume change during this process? At what composition would the system volume reach its maximum value? Solution Initially, there is no ethanol in the vessel, and figure 12.19, shows that the mixture at 70°C and 86 kPa is liquid, thus the mole fraction of liquid ethanol, x1,initial = 0 and mole fraction of liquid ethyl acetate, x2,initial = 1–x1,initial = 1 The final mole fraction of ethanol = 9/(9 + 1) = 0.9. Again as shown in figure 12.19, at 70°C and mole fraction 0.9, the mixture is liquid, so x1,final = 0.9 and x2,final = 0.1

Solution continued on next page…

During the evolution of this process, the mixture has 2 phases at the following compositions: x1 = 0.11, x2 = 1–x1 = 0.88; y1 = 0.18, y2 = 1–y1 = 0.82 x1 = 0.79, x2 = 0.21 and y1 = 0.69, y2 = 0.31 Practical feasibility: It might be difficult to do such a process as we are keeping temperature and pressure constant. Device: Volume Change: The volume increases with increasing number of moles and with increase in vapor phase. The system reaches its maximum volume when the mixture has the maximum number of moles, this is the final stage when the total number of moles = 10 Problems 12.9 through 12.14 refer to the Txy diagram for Ethanol(1)/Ethyl acetate(2) shown in Fig. 12.20.

Solution 12.9 Problem Statement For this problem, refer to the Txy diagram for ethanol(1)/ethyl acetate(2) shown in Fig. 12.20.

A mixture of ethanol and ethyl acetate is heated in a closed system at 100 kPa to a temperature of 74°C, and two phases are observed to be present. What are the possible compositions of the liquid and vapor phases? Solution The high temperature corresponds to vapor phase and low temperature corresponds to liquid phase. Draw a horizontal line at 74°C. From the point of intersection of this horizontal line and the lower curve (bubble point curve), draw a vertical line, thus the mole fraction of liquid ethanol, x1 = 0.14 and mole fraction of liquid ethyl acetate, x2 = 1–x1 = 0.86 Similarly, the composition of vapor phases can be determined from the intersection of the horizontal line and the upper curve (dew point curve). Thus the mole fraction of vapor ethanol, y1 = 0.2 and mole fraction of vapor ethyl acetate, y2 = 1–y1 = 0.8 As you see from figure 12.20, the horizontal line intersects the upper and lower curves twice, so the remaining compositions are the following: x1 = 0.79, x2 = 0.21 and y1 = 0.70, y2 = 0.30

Solution 12.10 Problem Statement For this problem, refer to the Txy diagram for ethanol(1)/ethyl acetate(2) shown in Fig. 12.20.

A mixture of ethanol and ethyl acetate is heated in a closed system at 100 kPa to a temperature of 77°C, and two phases are observed to be present. What are the possible compositions of the liquid and vapor phases? Solution Draw a horizontal line at 77°C. From the point of intersection of this horizontal line and the upper curve (dew point curve), draw a vertical line, thus the mole fraction of vapor ethanol, y1 = 0.92 and mole fraction of vapor ethyl acetate, y2 = 1–y1 = 0.08 Similarly, the composition of liquid phases can be determined from the intersection of the horizontal line and the lower curve (bubble point curve). Thus the mole fraction of liquid ethanol, x1 = 0.96 and mole fraction of vapor ethyl acetate, x2 = 1–x1 = 0.04

Solution 12.11 Problem Statement For this problem, refer to the Txy diagram for ethanol(1)/ethyl acetate(2) shown in Fig. 12.20.

Consider an ethanol(1)/ethyl acetate(2) mixture with x1 = 0.70, initially at 70°C and 100 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually increased to 80°C. Solution As we can see from the figure 12.20, for x1 = 0.70 at temperatures < 73.2°C, the entire mixture is liquid. So for 70°C < T < 73.2°C, y1 = 0, y2 = 0, where T is temperature. Now, for temperatures 73.2°C < T < 74°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.9, 12.10. As we continue to increase temperature to 80°C, the mixture exists only in vapor phase, so for 74°C < T < 80°C x1 = 0, x2 = 0

Solution 12.12 Problem Statement For this problem, refer to the Txy diagram for ethanol(1)/ethyl acetate(2) shown in Fig. 12.20.

Consider an ethanol(1)/ethyl acetate(2) mixture with x1 = 0.20, initially at 70°C and 100 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually increased to 80°C. Solution As we can see from the figure 12.20, for x1 = 0.20 at temperatures < 73.5°C, the entire mixture is liquid. So for 70°C < T < 73.5°C, y1 = 0, y2 = 0, where T is temperature. Now, for temperatures 73.5°C < T < 74°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.9, 12.10. As we continue to increase temperature to 80°C, the mixture exists only in vapor phase, so for 74°C < T < 80°C x1 = 0, x2 = 0

Solution 12.13 Problem Statement For this problem, refer to the Txy diagram for ethanol(1)/ethyl acetate(2) shown in Fig. 12.20.

Consider an ethanol(1)/ethyl acetate(2) mixture with x1 = 0.20, initially at 80°C and 100 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually reduced to 70°C. Solution As we can see from the figure 12.20, for x1 = 0.20 at temperatures > 74°C, the entire mixture is vapor. So for 74°C < T < 80°C, x1 = 0, x2 = 0, where T is temperature Now, for temperatures 73.5°C < T < 74°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.9, 12.10. As we continue to gradually decrease temperature from 73.5°C to 70°C, the mixture condenses and turns to liquid, so for 70°C < T < 73.5°C, y1 = 0, y2 = 0, where T is temperature.

Solution 12.14 Problem Statement For this problem, refer to the Txy diagram for ethanol(1)/ethyl acetate(2) shown in Fig. 12.20.

Consider an ethanol(1)/ethyl acetate(2) mixture with x1 = 0.80, initially at 80°C and 100 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually reduced to 70°C. Solution As we can see from the figure 12.20, for x1 = 0.80 at temperatures > 75.4°C, the entire mixture is vapor. So for 75.4°C < T < 80°C, x1 = 0, x2 = 0, where T is temperature Now, for temperatures 74.1°C < T < 75.4°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.9, 12.10. As we continue to gradually decrease temperature to 70°C, the mixture condenses and turns to liquid, so for 70°C < T < 74.1°C, y1 = 0, y2 = 0, where T is temperature.

Solution 12.15 Problem Statement Consider a closed vessel initially containing 1 mol of pure ethyl acetate at 74°C and 100 kPa. Imagine that pure ethanol is slowly added at constant temperature and pressure until the vessel contains 1 mol ethyl acetate and 9 mol ethanol. Describe the evolution of phases and phase compositions during this process. Comment on the practical feasibility of carrying out such a process. What sort of device would be required? How would the total system volume change during this process? At what composition would the system volume reach its maximum value? Solution Initially, there is no ethanol in the vessel, and figure 12.20, shows that the mixture at 74°C and 100 kPa is liquid, thus the mole fraction of liquid ethanol, x1,initial = 0 and mole fraction of liquid ethyl acetate, x2,initial = 1–x1,initial = 1 The final mole fraction of ethanol = 9/(9 + 1) = 0.9. Again as shown in figure 12.20, at 74°C and mole fraction 0.9, the mixture is liquid, so x1,final = 0.9 and x2,final = 0.1 During the evolution of this process, the mixture has 2 phases at the following compositions: x1 = 0.14, x2 = 1–x1 = 0.86; y1 = 0.2, y2 = 1–y1 = 0.8 x1 = 0.79, x2 = 0.21 and y1 = 0.70, y2 = 0.30 The system reaches its maximum volume when the mixture has the maximum number of moles. This is the final stage when the total number of moles is 10. Fig. 12.21.

FIGURE 12.21 Pxy diagram for vapor-liquid equilibrium of Chloroform(1)/Tetrahydrofuran(2) at 50°C. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 12.16 Problem Statement For this problem, refer to the Pxy diagram for chloroform(1)/tetrahydrofuran(2) at 50°C shown in Fig. 12.21.

The pressure above a mixture of chloroform and tetrahydrofuran at 50°C is measured to be 62 kPa. What are the possible compositions of the liquid and vapor phases? Solution Draw a horizontal line at 62 kPa. From the point of intersection of this horizontal line and the upper curve (bubble point curve), draw a vertical line, thus the mole fraction of liquid chloroform, x1 = 0.88 and mole fraction of liquid tetrahydrofuran, x2 = 1–x1 = 0.12 Similarly, the composition of vapor phases can be determined from the intersection of the horizontal line and the lower curve (dew point curve). Thus the mole fraction of vapor ethanol, y1 = 0.96 and mole fraction of vapor ethyl acetate, y2 = 1–y1 = 0.04

Solution 12.17 Problem Statement For this problem, refer to the Pxy diagram for chloroform(1)/tetrahydrofuran(2) at 50°C shown in Fig. 12.21.

The pressure above a mixture of chloroform and tetrahydrofuran at 50°C is measured to be 52 kPa. What are the possible compositions of the liquid and vapor phases? Solution Draw a horizontal line at 52kPa. From the point of intersection of this horizontal line and the upper curve (bubble point curve), draw a vertical line, thus the mole fraction of liquid chloroform, x1 = 0.20 and mole fraction of liquid tetrahydrofuran, x2 = 1–x1 = 0.80 Similarly, the composition of vapor phases can be determined from the intersection of the horizontal line and the lower curve (dew point curve). Thus the mole fraction of vapor chloroform, y1 = 0.14 and mole fraction of vapor tetrahydrofuran, y2 = 1–y1 = 0.86 As you see from figure 12.21, the horizontal line intersects the upper and lower curves twice, so the remaining compositions are the following: x1 = 0.68, x2 = 0.32 and y1 = 0.79, y2 = 0.21

Solution 12.18 Problem Statement For this problem, refer to the Pxy diagram for chloroform(1)/tetrahydrofuran(2) at 50°C shown in Fig. 12.21.

Consider a chloroform (1)/tetrahydrofuran(2) mixture with x1 = 0.80, initially at 50°C and 70 kPa. Describe the evolution of phases and phase compositions as the pressure is gradually reduced to 50 kPa. Solution As we can see from the figure 12.21, for x1 = 0.80 at pressures > 58 kPa, the entire mixture is liquid. So for 58 kPa < P < 70 kPa, y1 = 0, y2 = 0, where P is pressure Now, for pressures 52.1 kPa < P < 58 kPa we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given pressure and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.16, 12.17. As we continue to lower pressure to 50kPa, the mixture exists only in vapor phase, so for 50 kPa < P < 52.1 kPa, x1 = 0, x2 = 0

Solution 12.19 Problem Statement For this problem, refer to the Pxy diagram for chloroform(1)/tetrahydrofuran(2) at 50°C shown in Fig. 12.21.

Consider an chloroform(1)/tetrahydrofuran(2) mixture with x1 = 0.90, initially at 50°C and 50 kPa. Describe the evolution of phases and phase compositions as the pressure is gradually increased to 70 kPa. Solution As we can see from the figure 12.21, for x1 = 0.90 at pressures < 57 kPa, the entire mixture is vapor. So for 50 kPa < P < 57 kPa, x1 = 0, x2 = 0, where P is pressure Now, for pressures 57 kPa < P < 63.6 kPa we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given pressure and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.16, 12.17. As we continue to gradually increase pressure from 63.6 kPa to 70 kPa, the mixture condenses and turns to liquid, so for 63.6 kPa < P < 70 kPa y1 = 0, y2 = 0

Solution 12.20 Problem Statement For this problem, refer to the Pxy diagram for chloroform(1)/tetrahydrofuran(2) at 50°C shown in Fig. 12.21.

What is the composition of the azeotrope for the chloroform(1)/tetrahydrofuran (2) system? Would this be called a high-boiling or low-boiling azeotrope? Solution For binary mixture, azeotrope exists when the components have same composition in liquid and vapor phases, so x1 = y1, x2 = y2 In this case, the composition of the azeotrope: x1 = y1 = 0.46, x2 = y2 = 0.54 The boiling point of chloroform is 61.2°C and the boiling point of tetrahydrofuran is 66°C. Since the azeotrope occurs at 50°C, this is a positive deviation or minimum boiling azeotrope or low-boiling azeotrope.

Solution 12.21 Problem Statement For this problem, refer to the Pxy diagram for chloroform(1)/tetrahydrofuran(2) at 50°C shown in Fig. 12.21.

Consider a closed vessel initially containing 1 mol of tetrahydrofuran at 50°C and 52 kPa. Imagine that pure chloroform is slowly added at constant temperature and pressure until the vessel contains 1 mol tetrahydrofuran and 9 mol chloroform. Describe the evolution of phases and phase compositions during this process. Comment on the practical feasibility of carrying out such a process. What sort of device would be required? How would the total system volume change during this process? At what composition would the system volume reach its maximum value? Solution Initially, there is no chloroform in the vessel, and figure 12.21, shows that the mixture at 50°C and 52 kPa is vapor, thus the mole fraction of vapor chloroform, y1,initial = 0 and mole fraction of liquid tetrahydrofuran, y2,initial = 1–y1,initial = 1 The final mole fraction of chloroform = 9/(9 + 1) = 0.9. Again as shown in figure 12.21, at 50°C and mole fraction 0.9, the mixture is vapor, so y1,final = 0.9 and y2,final = 0.1 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

During the evolution of this process, the mixture has 2 phases at the following compositions: x1 = 0.20, x2 = 1–x1 = 0.80; y1 = 0.14, y2 = 1–y1 = 0.86 x1 = 0.68, x2 = 0.32: y1 = 0.79, y2 = 0.21 Practical feasibility: It might be difficult to do such a process as we are keeping temperature and pressure constant. Device: Volume Change: The volume increases with increasing number of moles and with increase in vapor phase. The system reaches its maximum volume when the mixture is vapor and has the maximum number of moles, this is the final stage when the total number of moles = 10 Problems 12.22 through 12.28 refer to the Txy diagram for Chloroform(1)/Tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

FIGURE 12.22 Txy diagram for vapor-liquid equilibrium of Chloroform(1)/Tetrahydrofuran (2) at 120 kPa.

Solution 12.22 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

A mixture of chloroform and tetrahydrofuran is heated in a closed system at 120 kPa to a temperature of 75°C, and two phases are observed to be present. What are the possible compositions of the liquid and vapor phases? Solution The high temperature corresponds to vapor phase and low temperature corresponds to liquid phase. Draw a horizontal line at 75°C. From the point of intersection of this horizontal line and the lower curve (bubble point curve), draw a vertical line, thus the mole fraction of liquid chloroform, x1 = 0.23 and mole fraction of liquid tetrahydrofuran, x2 = 1–x1 = 0.77 Similarly, the composition of vapor phases can be determined from the intersection of the horizontal line and the upper curve (dew point curve). Thus the mole fraction of vapor chloroform, y1 = 0.16 and mole fraction of vapor tetrahydrofuran, y2 = 1–y1 = 0.84 As you see from figure 12.22, the horizontal line intersects the upper and lower curves twice, so the remaining compositions are the following: x1 = 0.68, x2 = 0.32 and y1 = 0.76, y2 = 0.24 Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 12.23 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

A chloroform and tetrahydrofuran mixture is heated in a closed system at 120 kPa to a temperature of 70°C, and two phases are observed to be present. What are the possible compositions of the liquid and vapor phases? Solution Draw a horizontal line at 70°C. From the point of intersection of this horizontal line and the upper curve (dew point curve), draw a vertical line, thus the mole fraction of vapor chloroform, y1 = 0.95 and mole fraction of vapor tetrahydrofuran, y2 = 1–y1 = 0.05 Similarly, the composition of liquid phases can be determined from the intersection of the horizontal line and the lower curve (bubble point curve). Thus the mole fraction of liquid chloroform, x1 = 0.86 and mole fraction of vapor tetrahydrofuran, x2 = 1–x1 = 0.14

Solution 12.24 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

Consider a chloroform(1)/tetrahydrofuran(2) mixture with x1 = 0.80, initially at 70°C and 120 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually increased to 80°C. Solution As we can see from the figure 12.22, for x1 = 0.80 at temperatures < 72°C, the entire mixture is liquid. So for 70°C < T < 72°C, y1 = 0, y2 = 0, where T is temperature. Now, for temperatures 72°C < T < 74.4°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.22, 12.23. As we continue to increase temperature to 80°C, the mixture exists only in vapor phase, so for 74.4°C < T < 80°C x1 = 0, x2 = 0

Solution 12.25 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

Consider a chloroform(1)/tetrahydrofuran(2) mixture with x1 = 0.20, initially at 70°C and 120 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually increased to 80°C. Solution As we can see from the figure 12.22, for x1 = 0.20 at temperatures < 74.6°C, the entire mixture is liquid. So for 70°C < T < 74.6°C, y1 = 0, y2 = 0, where T is temperature. Now, for temperatures 74.6°C < T < 75.5°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.22, 12.23. As we continue to increase temperature to 80°C, the mixture exists only in vapor phase, so for 75.5°C < T < 80°C x1 = 0, x2 = 0

Solution 12.26 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

Consider a chloroform(1)/tetrahydrofuran(2) mixture with x1 = 0.10, initially at 80°C and 120 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually reduced to 70°C. Solution As we can see from the figure 12.22, for x1 = 0.10 at temperatures > 74°C, the entire mixture is vapor. So for 74°C < T < 80°C, x1 = 0, x2 = 0, where T is temperature Now, for temperatures 73°C < T < 74°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.22, 12.23. As we continue to gradually decrease temperature from 74°C to 70°C, the mixture condenses and turns to liquid, so for 70°C < T < 74°C, y1 = 0, y2 = 0, where T is temperature.

Solution 12.27 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

Consider a chloroform(1)/tetrahydrofuran(2) mixture with x1 = 0.90, initially at 76°C and 120 kPa. Describe the evolution of phases and phase compositions as the temperature is gradually reduced to 66°C. Solution As we can see from the figure 12.22, for x1 = 0.90 at temperatures > 72°C, the entire mixture is vapor. So for 72°C < T < 80°C, x1 = 0, x2 = 0, where T is temperature Now, for temperatures 69.1°C < T < 72°C we have two phase region, the compositions of different phases could be determined by drawing a horizontal line at the given temperature and finding the points of intersection between this line and the bubble point and dew point curves as done in problems 12.22, 12.23.

Solution 12.28 Problem Statement For this problem, refer to the Txy diagram for chloroform(1)/tetrahydrofuran(2) at 120 kPa shown in Fig. 12.22.

Consider a closed vessel initially containing 1 mol of pure tetrahydrofuran at 74°C and 120 kPa. Imagine that pure chloroform is slowly added at constant temperature and pressure until the vessel contains 1 mol tetrahydrofuran and 9 mol chloroform. Describe the evolution of phases and phase compositions during this process. Comment on the practical feasibility of carrying out such a process. What sort of device would be required? How would the total system volume change during this process? At what composition would the system volume reach its maximum value? Solution Initially, there is no chloroform in the vessel, and figure 12.22, shows that the mixture at 74°C is vapor, thus the mole fraction of vapor chloroform, y1,initial = 0 and mole fraction of liquid tetrahydrofuran, y2,initial = 1–y1,initial = 1 The final mole fraction of chloroform = 9/(9 + 1) = 0.9. Again as shown in figure 12.22, at 74°C and mole fraction 0.9, the mixture is vapor, so y1,final = 0.9 and y2,final = 0.1 During the evolution of this process, the mixture has 2 phases at the following compositions: y1 = 0.1, y2 = 1–y1 = 0.9; x1 = 0.16, x2 = 1–x1 = 0.84 y1 = 0.82, y2 = 1–y1 = 0.18; x1 = 0.72, x2 = 1–x1 = 0.28 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Practical feasibility: It might be difficult to do such a process as we are keeping temperature and pressure constant. Device: Volume Change: The volume increases with increasing number of moles and with increase in vapor phase. The system reaches its maximum volume when the mixture is vapor and has the maximum number of moles, this is the final stage when the total number of moles = 10 Problem 12.29 through 12.33 refer to the xy diagram provided in Figure 12.23. This diagram shows xy curves for both Ethanol(1)/Ethyl acetate(2) and for Chloroform(1)/Tetrahydrofuran(2), both at a constant pressure of 1 bar. The curves are intentionally unlabeled. Readers should refer to Figures 12.19 through 12.22 to deduce which curve is for which pair of substances.

FIGURE 12.23 xy diagram for Ethanol(1)/Ethyl acetate(2) and for Chloroform(1)/ Tetrahydrofuran(2), both at a constant pressure of 1 bar. Note that the curves are intentionally unlabeled, but can be identified based on information presented in Fig. 12.19 through 12.22.

Solution 12.29 Problem Statement For this problem, refer to the xy diagram provided in Fig. 12.23. This diagram shows xy curves both for ethanol(1)/ethyl acetate(2) and for chloroform(1)/tetrahydrofuran(2), both at a constant pressure of 1 bar. The curves are intentionally unlabeled. Readers should refer to Figs. 12.19 through 12.22 (see next page) to deduce which curve is for which pair of substances.

What is the composition of the vapor phase in equilibrium with a liquid-phase ethanol(1)/ethyl acetate(2) mixture of the following compositions at P = 1 bar? (a) x1 = 0.1 (b) x1 = 0.2 (c) x1 = 0.3 (d) x1 = 0.45 (e) x1 = 0.6 (f) x1 = 0.8 (g) x1 = 0.9 See other figures(12.19 thru 12.22) on next page…

See Solution after other figures…

Solution (a) By using figure 12.20, we can quickly identify the upper curve before azeotrope (x1 = 0.45, y1 = 0.45) corresponds to ethanol/ethyl acetate and the lower curve after azeotrope corresponds to ethanol/ethyl acetate. Now using figure 12.23, y1 = 0.17, y2 = 0.83 (b) y1 = 0.26, y2 = 0.74 (c) y1 = 0.34, y2 = 0.66 (d) y1 = 0.45, y2 = 0.55 (e) y1 = 0.55, y2 = 0.45 (f) y1 = 0.70, y2 = 0.30 (g) y1 = 0.83, y2 = 0.17

Solution 12.30 Problem Statement For this problem, refer to the xy diagram provided in Fig. 12.23. This diagram shows xy curves both for ethanol(1)/ethyl acetate(2) and for chloroform(1)/tetrahydrofuran(2), both at a constant pressure of 1 bar. The curves are intentionally unlabeled. Readers should refer to Figs. 12.19 through 12.22 (see next page) to deduce which curve is for which pair of substances.

What is the composition of the liquid phase in equilibrium with a vapor-phase ethanol(1)/ethyl acetate(2) mixture of the following compositions at P = 1 bar? (a) y1 = 0.1 (b) y1 = 0.2 (c) y1 = 0.3 (d) y1 = 0.45 (e) y1 = 0.6 (f) y1 = 0.8 (g) y1 = 0.9

See other figures (12.19 thru 12.22) on next page… See Solution after other figures…

Solution (a) x1 = 0.05, x2 = 0.95 (b) x1 = 0.14, x2 = 0.86 (c) x1 = 0.24, x2 = 0.76 (d) x1 = 0.45, x2 = 0.55 (e) x1 = 0.68, x2 = 0.32 (f) x1 = 0.88, x2 = 0.12 (g) x1 = 0.95, x2 = 0.05

Solution 12.31 Problem Statement For this problem, refer to the xy diagram provided in Fig. 12.23. This diagram shows xy curves both for ethanol(1)/ethyl acetate(2) and for chloroform(1)/tetrahydrofuran(2), both at a constant pressure of 1 bar. The curves are intentionally unlabeled. Readers should refer to Figs. 12.19 through 12.22 (see next page) to deduce which curve is for which pair of substances.

What is the composition of the vapor phase in equilibrium with a liquid-phase chloroform(1)/tetrahydrofuran(2) mixture of the following compositions at P = 1 bar? (a) x1 = 0.1 (b) x1 = 0.2 (c) x1 = 0.3 (d) x1 = 0.45 (e) x1 = 0.6 (f) x1 = 0.8 (g) x1 = 0.9 See other figures (12.19 thru 12.22) on next page…

See Solution after other figures…

Solution (a) y1 = 0.05, y2 = 0.95 (b) y1 = 0.14, y2 = 0.86 (c) y1 = 0.24, y2 = 0.76 (d) y1 = 0.45, y2 = 0.55 (e) y1 = 0.67, y2 = 0.33 (f) y1 = 0.90, y2 = 0.10 (g) y1 = 0.97, y2 = 0.03

Solution 12.32 Problem Statement For this problem, refer to the xy diagram provided in Fig. 12.23. This diagram shows xy curves both for ethanol(1)/ethyl acetate(2) and for chloroform(1)/tetrahydrofuran(2), both at a constant pressure of 1 bar. The curves are intentionally unlabeled. Readers should refer to Figs. 12.19 through 12.22 (see next page) to deduce which curve is for which pair of substances.

What is the composition of the liquid phase in equilibrium with a vapor-phase chloroform(1)/tetrahydrofuran(2) mixture of the following compositions at P = 1 bar? (a) y1 = 0.1 (b) y1 = 0.2 (c) y1 = 0.3 (d) y1 = 0.45 (e) y1 = 0.6 (f) y1 = 0.8 (g) y1 = 0.9 See other figures (12.19 thru 12.22) on next page… See Solution after other figures…

Solution (a) x1 = 0.16, x2 = 0.84 (b) x1 = 0.25, x2 = 0.75 (c) x1 = 0.34, x2 = 0.66 (d) x1 = 0.45, x2 = 0.55 (e) x1 = 0.55, x2 = 0.45 (f) x1 = 0.70, x2 = 0.30 (g) x1 = 0.80, x2 = 0.20

Solution 12.33 Problem Statement For this problem, refer to the xy diagram provided in Fig. 12.23. This diagram shows xy curves both for ethanol(1)/ethyl acetate(2) and for chloroform(1)/tetrahydrofuran(2), both at a constant pressure of 1 bar. The curves are intentionally unlabeled. Readers should refer to Figs. 12.19 through 12.22 (see next page) to deduce which curve is for which pair of substances.

Consider a binary liquid mixture for which the excess Gibbs energy is given by G /RT = Ax1x2. What is the minimum value of A for which liquid/liquid equilibrium is possible?

See other figures (12.19 thru 12.22) on next page… See Solution after other figures…

Solution

A single phase binary mixture is stable when

 2 G E RT   2 x1

0

GE  Ax1 x 2 RT GE  Ax1 1  x1  RT

G E RT  x 1

 A  2 Ax1

 2 G E RT  2 x1

 2 A

Using equation 12.5,  2 G E RT   x1 2



1 x1 x 2

Therefore, 2A 

1 x1 x 2

When x1  x2  0.5, the right hand side of this inequality has the minimum value of 4. So the minimum value of A for which liquid/liquid equilibrium is possible is 2.

Solution 12.34 Problem Statement E

Consider a binary liquid mixture for which the excess Gibbs energy is given by G /RT = Ax1x2 (x1 + 2x2). What is the minimum value of A for which liquid/liquid equilibrium is possible? Solution  2 G E RT   2 x1



GE  Ax1 x2  x1  2 x 2  RT

GE  Ax1   x1  x1    x1   A  x12  x13  x1  RT

d G E RT  dx1

 A * 6 x1  3x12  2

d 2 G E /RT  dx12

 A * 6  6 x1 

Using equation 12.5,  2 G E RT   x1 2



1 x1 x 2

Therefore, – 6A

1 x1 x 22

Or A

1 6 x1 x 22

To find the minimum value of A, take the derivative of A with respect to x1 and set the equal to 0. Based on this, the minimum will occur at x1 = 1/3 and x2 = 2/3. The minimum value of A for which liquid/liquid equilibrium is possible is 1.125.

Solution 12.35 Problem Statement E

Consider a binary mixture for which the excess Gibbs energy is given by G /RT = 2.6x1x2. For each of the following overall compositions, determine whether one or two liquid phases will be present. If two liquid phases will be present, find their compositions and the amount of each phase present (phase fractions). (a) z1 = 0.2 (b) z1 = 0.3 (c) z1 = 0.5 (d) z1 = 0.7 (e) z1 = 0.8

Solution In general the Gibbs energy change of mixing equation is written as G GE  x1 ln x1  x2 ln x 2  RT RT

To determine if the mixture is of one or two phases at a fixed temperature and pressure the following must be true:  G  d 2    RT  dx12

0

Now this tells us if each phase is stable or not. To determine if an overall composition at which a single phase is not stable, the compositions of the two phases in equilibrium can be found by:  G  d 2    RT  dx12

0

This now can allow us to determine if the overall compositions will split into one or two phases. So taking the derivative twice of the first equation yields: G  x1 ln x1  x2 ln x2  2.6 x1 x2  x1 ln x1  1  x1  ln1  x1   2.6 x1 1  x1  RT  G  d    RT   1  ln x1  1  ln 1  x1   2.6  5.2 x1 dx1

 G  d 2    RT  1 1    5.2 2 x dx1 1  x1  1

Setting this equal to 0, and simplification, yields 5.2 x12  5.2 x1  1  0

With this equation apply the quadratic formula giving the roots of x1  0.2598 and x1  0.7402. Thus Part (a): the solution is stable, single phase Parts (b), (c), (d): The solution is not stable, and the phase the compositions for phase 

x1 



x1 

Part(e): the solution is stable, single phase

Solution 12.36 Problem Statement

Consider a binary mixture for which the excess Gibbs energy is given by G /RT = 2.1x1x2 (x1 + 2x2). For each of the following overall compositions, determine whether one or two liquid phases will be present. If two liquid phases will be present, find their compositions and the amount of each phase present (phase fractions). (a) z1 = 0.2 (b) z1 = 0.3 (c) z1 = 0.5 (d) z1 = 0.7 (e) z1 = 0.8

Solution

The solution is the same as 12.35, save for the following: G  x1 ln x1  x 2 ln x 2  2.1 x1 x2  x1  2 x2   x1 ln x1  1  x1 ln1  x1   2.1x1 1  x1  x1  21  x1  RT  G  d    RT   1  ln x1  1  ln1  x1   2.1 x12  6 x1  2 dx1

 G  d 2    RT  1 1    12.6 x1  12.6 x1 1  x1  dx12

Setting this equal to 0, and simplifying, yields: 12.6 x13  20.8 x12  8.2 x1  1  0

This cubic equation has three roots, at x1 = 0.0974, x1 = 0.6508, and x1 = 1.2518. The physically meaningful ones are x1 = 0.0974 and x1 = 0.6508. From here we can determine that for: Parts (a), (b), (c): The solution splits into

x1 = 0.6508.

Solution 13.1

Problem Statement

Assuming the validity of Raoult’s law, do the following calculations for the benzene(1)/toluene(2) system: (a) Given x1 = 0.33 and T = 100°C, find y1 and P. (b) Given y1 = 0.33 and T = 100°C, find x1 and P. (c) Given x1 = 0.33 and P = 120 kPa, find y1 and T. (d) Given y1 = 0.33 and P = 120 kPa, find x1 and T. (e) Given T = 105°C and P = 120 kPa, find x1 and y1. (f) For part (e), if the overall mole fraction of benzene is z1 = 0.33, what molar fraction of the two-phase system is vapor? (g) Why is Raoult’s law likely to be an excellent VLE model for this system at the stated (or computed) conditions? Note: Solutions to some of the problems of this chapter require vapor pressures as a function of temperature. Table B.2, Appendix B, lists parameter values for the Antoine equation, from which these can be computed.

Solution (a) This is a bubble point pressure calculation – the easiest of the types of VLE calculations we have just learned how sat

to do. At 100°C, P1

= exp(13.7819 – 2726.81/(100 + 217.572)) = 180.45 kPa (using the Antoine parameters for

benzene from table B.2 on p. 682). Similarly, using the parameters for toluene, sat

= exp(13.9320 – 3056.96/(100 + 217.625)) = 74.26 kPa. So, the total pressure is

Solution continued on next page…

P   xi Pi sat  x1 P1sat  x2 P2sat i 1

P  0.33  180.45  0.67  74.26  109.3 kPa The mole fractions are then obtained from yi 

xi Pi sat

P x1 P1sat 0.33  180.45 y1    0.545 P 109.3 y2  1  y1  0.455

(b) This is a dew point pressure calculation at the same temperature. The saturation pressures are the same as in (a). The total pressure can be found from P

1 N

P i 1

P

yi sat



1 y1 sat 1



y2 P2sat

1  92.16 kPa 0.33 0.67  180.45 74.26

Then the liquid phase mole fractions are found from 0.33  92.16  0.169 180.45 P x 2  1  x1  0.831 x1 

y1 P sat 1



(c) This is a bubble point temperature calculation. One way to do it is to do a series of bubble point pressure calculations until we find the one that gives us the specified pressure. We can write the total pressure (as in (a)) as N

P   xi Pi sat  x1 P1sat  x 2 P2sat i 1

  B1  B2    x2 exp A2   P  x1 exp A1    T  C1  T  C2     2726.81  3056.96  120  0.33exp13.7819    0.67 exp13.9320      T  217.572  T  217.625  

Solution continued on next page…

We can’t solve this explicitly for T. From the result of part (a), we know that T should be a little bit above 100 °C. By trial and error, we can find that a value of T = 103.31 °C, P = 120.00 kPa. At this temperature, sat

sat

= 197.15 kPa and P2

= 82.00 kPa. So, the vapor phase mole fractions are

x1 P1sat 0.33  197.15   0.542 P 120 y2  1  y1  0.458 y1 

(d) This is a dew point temperature calculation. One way to do it is to do a series of dew point pressure calculations until we find the one that gives us the specified pressure. We can write the total pressure (as in (b)) as P

1 N

P i 1

sat



y1 P1sat



y2 P2sat 1

120 

0.33  2726.81  exp 13.7819   T  217.572  



0.67  3056.96  exp13.9320   T  217.625  

Once again, we aren’t able to solve explicitly for T. We know that T should be a bit higher than in part (d). By trial sat

and error, we can find that at T = 109.13 °C, P = 120.00 kPa. At this temperature, P1 sat

= 229.40 kPa and

= 97.17 kPa. So, the liquid phase mole fractions are 0.33  120.00  0.173 229.40 P x 2  1  x 1  0.827 x1 

y1 P sat 1



(e) This isn’t one of our four standard calculations, but for a binary system it is pretty straightforward. At 105 °C, sat

sat

= 206.14 kPa and P2

= 86.20 kPa. So, we can write the total pressure as

P   xi Pi sat  x1 P1sat  x2 P2sat i 1

P  x1  206.14  1  x1   86.20  120 kPa 120  86.20  0.282 206.14  86.20 x2  1  x1  0.718 x1 

P1sat 206.14  0.282  0.484 P 120 y2  1  y1  0.516 y1  x1

Solution continued on next page…

(f) The overall mole fraction of species 1 is given by z1 = x1L + y1V where L is the fraction of the system that is liquid and V is the fraction of the system that is vapor. Also, L + V = 1, so we can substitute L = 1 – V to get z1 = x1(1 – V) + y1V = x1 + V(y1 – x1) and finally,

z1  x 1 0.33  0.282   0.238 y1  x1 0.484  0.282 L  1  V  0.762

V

(e) Benzene and toluene are very chemically similar (they differ only by 1 methyl group), so interactions between a benzene molecule and a toluene molecule should be very similar to interactions between two benzene molecules or between two toluene molecules. This means that the solution will be nearly an ideal solution. The pressure is only a few percent of the critical pressures of the species, while the temperature is more than half the critical temperatures of the components, so the vapor will not be far from being an ideal gas.

Solution 13.2

Problem Statement Assuming Raoult’s law to be valid, prepare a Pxy diagram for a temperature of 90°C and a txy diagram for a pressure of 90 kPa for one of the following systems: (a) Benzene(1)/ethylbenzene(2). (b) 1-Chlorobutane(1)/chlorobenzene(2). Note: Solutions to some of the problems of this chapter require vapor pressures as a function of temperature. Table B.2, Appendix B, lists parameter values for the Antoine equation, from which these can be computed.

Solution (a) While we could do the individual data points with a hand calculator, to construct a whole P-x-y or T-x-y diagram, it makes sense to set up a spreadsheet or a computer program. Here is what the spreadsheet for generating the P-x-y diagram looks like:

Spreadsheet to compute a P-x-y diagram for SVA problem 10.2(a) Species 1 is benzene, species 2 is ethylbenzene Antoine Coefficients (for T in deg C and P in kPa) T (deg C) A1 B1 C1 A2 B2 C2 90 13.8594 2773.78 220.07 14.0045 3279.47 P (kPa)

x2 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1.00 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

24.25 29.84 35.44 41.03 46.63 52.22 57.82 63.41 69.01 74.60 80.20 85.79 91.39 96.98 102.58 108.17 113.77 119.36 124.96 130.55 136.15

sat

(kPa) P2 (kPa) 136.148 24.24732

P (kPa)

y2 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

1.00 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

24.25 25.29 26.42 27.66 29.02 30.52 32.18 34.04 36.12 38.48 41.16 44.25 47.84 52.06 57.10 63.21 70.80 80.45 93.16 110.62 136.15

P-x-y Diagram for Benzene(1)/Ethylbenzene(2) at 90 deg C 140.00 120.00 100.00 P (kPa)

P1 213.2

80.00 60.00 40.00 20.00 0.00

0.20

0.40 0.60 x1 or y1

0.80

1.00

For the T-x-y diagram, we have to iteratively solve for T for each point. In the spreadsheet shown below, this was 2

done using Solver in Microsoft Excel. For each cell in the (P-Pspec) columns, the corresponding number in the T 2

(deg C) was varied until the (P-Pspec) cell was zero. This was done by varying all of the temperatures until the sum 2

of the whole (P-Pspec) column was minimized (to zero).

Solution continued on next page…

Spreadsheet to compute a T-x-y diagram for SVNA problem 10.2(a) Species 1 is 1-benzene, species 2 is ethylbenzene Antoine Coefficients (for T in deg C and P in kPa) P (kPa) A1 B1 C1 A2 B2 C2 90 13.7819 2726.81 217.572 13.9726 3259.93 x2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29

P (kPa) (P-Pspec) y 1 T (deg C) P 1 sat P 2 sat 1.00 131.84 394.58 90.00 90.00 0.00 0.99 130.62 383.98 87.03 90.00 0.00 0.98 129.44 373.84 84.21 90.00 0.00 0.97 128.28 364.14 81.52 90.00 0.00 0.96 127.15 354.85 78.96 90.00 0.00 diagram for 0.95 126.04 345.95 T-x-y 76.53 90.00 0.00 Benzene1)/Ethylbenzene(2) kPa 0.94 124.97 337.41 74.21 90.00 at 90 0.00 0.93 123.91 329.23 71.99 90.00 0.00 0.92140.00 122.88 321.37 69.88 90.00 0.00 0.91 121.88 313.83 67.86 90.00 0.00 0.90 120.89 306.58 65.94 90.00 0.00 130.00 0.89 119.93 299.62 64.09 90.00 0.00 0.88 118.99 292.92 62.33 90.00 0.00 120.00 0.87 118.07 286.48 60.64 90.00 0.00 0.86 117.16 280.27 59.03 90.00 0.00 110.00 0.85 116.28 274.30 57.48 90.00 0.00 0.84 115.42 268.54 55.99 90.00 0.00 100.00 0.83 114.57 262.99 54.57 90.00 0.00 0.82 113.74 257.64 53.20 90.00 0.00 90.00 0.81 112.93 252.48 51.89 90.00 0.00 0.80 112.13 247.50 50.63 90.00 0.00 80.00 0.79 111.35 242.68 49.41 90.00 0.00 0.78 110.58 238.03 48.25 90.00 0.00 0.77 70.00 109.83 233.54 47.12 90.00 0.00 0.00 229.190.20 46.04 0.4090.00 0.60 0.76 109.10 0.00 0.75 108.37 224.99 45.00 90.00 x 1 or y1 0.00 0.74 107.66 220.92 44.00 90.00 0.00 0.73 106.97 216.98 43.03 90.00 0.00 0.72 106.28 213.17 42.10 90.00 0.00 0.71 105.61 209.47 41.20 90.00 0.00

Temperature (deg C)

212.3

SUM((P-Pspec)2) 0.00 y2 0.00 0.04 0.08 0.12 0.16 0.19 0.22 0.26 0.29 0.31 0.34 0.37 0.39 0.41 0.44 0.46 0.48 0.50 0.52 0.53 0.55 0.57 0.58 0.60 0.80 0.61 0.62 0.64 0.65 0.66 0.67

1.00 0.96 0.92 0.88 0.84 0.81 0.78 0.74 0.71 0.69 0.66 0.63 0.61 0.59 0.56 0.54 0.52 0.50 0.48 0.47 0.45 0.43 0.42 0.40 1.00 0.39 0.38 0.36 0.35 0.34 0.33

(b) While we could do the individual calculations in problem 1 with a hand calculator, to construct a whole P-x-y or T-x-y diagram, it makes sense to set up a spreadsheet or a computer program. Here is what the spreadsheet for generating the P-x-y diagram looks like (only showing the top portion, and not the entire columns of numbers):

Solution continued on next page…

Spreadsheet to compute a P-x-y diagram for SVA problem 10.2(b) Species 1 is 1-chlorobutane, species 2 is chlorobenzene Antoine Coefficients (for T in deg C and P in kPa) T (deg C) A1 B1 C1 A2 B2 C2 P1sat (kPa) P2sat (kPa) 90 13.96 2826.26 224.1 13.9926 3295.12 217.55 142.8846 26.53606 P (kPa)

x2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22

1.00 0.99 0.98 0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.90 0.89 0.88 0.87 0.86 0.85 0.84 0.83 0.82 0.81 0.80 0.79 0.78

26.54 27.70 28.86 30.03 31.19 32.35 33.52 34.68 35.84 37.01 38.17 39.33 40.50 41.66 42.82 43.99 45.15 46.32 47.48 48.64 49.81 50.97 52.13

y2 0.00 0.05 0.10 0.14 0.18 0.22 0.26 0.29 0.32 0.35 0.37 0.40 0.42 0.45 0.47 0.49 0.51 0.52 0.54 0.56 0.57 0.59 0.60

1.00 0.95 0.90 0.86 0.82 0.78 0.74 0.71 0.68 0.65 0.63 0.60 0.58 0.55 0.53 0.51 0.49 0.48 0.46 0.44 0.43 0.41 0.40

P-x-y Diagram for 1-Chlorobutane(1)/Chlorobenzene(2) at 90 deg C 140.00 120.00 100.00 P (kPa)

80.00 60.00 40.00 20.00 0.00

0.20

0.40 0.60 x1 or y1

0.80

1.00

For the T-x-y diagram, we have to iteratively solve for T for each point. In the spreadsheet shown below, this was 2

done using the “Solver” function in Microsoft Excel. A single cell, labeled SUM((P-Pspec) ) was set up as the sum of 2

all the numbers in the column labeled (P-Pspec) . Then the “solver” function was used to minimize that sum by changing the whole column of T values. The sum goes to zero only when each of the P-Pspec values goes to zero. That is, it goes to zero only when each temperature has been varied to give the desired pressure. Note that the “goal seek” function will only change one cell at a time, but the “solver” function will change many cells (multivariable optimization). This is a particularly easy case of multivariable optimization, because each of the temperatures acts independently of the others (each temperature only affects one term in the sum that is being minimized).

Solution continued on next page…

Spreadsheet to compute a T-x-y diagram for SVNA problem 10.2(b) Species 1 is 1-chlorobutane, species 2 is chlorobenzene Antoine Coefficients (for T in deg C and P in kPa) P (kPa) A1 B1 C1 A2 B2 C2 90 13.96 2826.26 224.1 13.9926 3295.12 217.55 x2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.32 0.33

sat

P (kPa) (P-Pspec) y 1 P2 y2 T (deg C) P 1 1.00 129.57 391.01 90.00 90.00 0.00 0.00 0.99 128.36 380.45 87.07 90.00 0.00 0.04 0.98 127.18 370.35 84.28 90.00 for 0.00 0.08 T-x-y diagram 0.97 126.03 360.71 81.63 90.00 0.00 1-Chlorobutane(1)/Chlorobenzene(2) at 90 0.12 kPa 0.96 124.91 351.48 79.11 90.00 0.00 0.16 0.95 123.82 342.64 76.70 90.00 0.00 0.19 140.00 0.94 122.75 334.17 74.41 90.00 0.00 0.22 0.93 121.71 326.06 72.23 90.00 0.00 0.25 130.00 0.92 120.69 318.28 70.15 90.00 0.00 0.28 0.91 119.69 310.82 68.16 90.00 0.00 0.31 120.00 0.90 118.72 303.65 66.26 90.00 0.00 0.34 0.89 117.77 296.76 64.44 90.00 0.00 0.36 110.00 0.88 116.84 290.15 62.71 90.00 0.00 0.39 0.87 115.93 283.78 61.04 90.00 0.00 0.41 100.00 0.86 115.04 277.66 59.45 90.00 0.00 0.43 0.85 114.17 271.76 57.92 90.00 0.00 0.45 0.84 90.00 113.32 266.09 56.46 90.00 0.00 0.47 0.83 112.48 260.61 55.06 90.00 0.00 0.49 0.82 80.00 111.66 255.34 53.71 90.00 0.00 0.51 0.81 110.86 250.25 52.41 90.00 0.00 0.53 0.80 70.00 110.08 245.34 51.17 90.00 0.00 0.55 0.00 0.20 49.97 0.4090.00 0.600.00 0.80 0.79 109.31 240.59 0.56 0.78 108.55 236.01 48.82 90.00 0.00 0.58 x1 or y1 0.77 107.81 231.59 47.71 90.00 0.00 0.59 0.76 107.09 227.31 46.64 90.00 0.00 0.61 0.75 106.38 223.17 45.61 90.00 0.00 0.62 0.74 105.68 219.16 44.62 90.00 0.00 0.63 0.73 104.99 215.28 43.66 90.00 0.00 0.65 0.72 104.32 211.52 42.74 90.00 0.00 0.66 0.71 103.66 207.89 41.85 90.00 0.00 0.67 0.70 103.01 204.36 40.99 90.00 0.00 0.68 0.69 102.37 200.94 40.16 90.00 0.00 0.69 0.68 101.74 197.62 39.35 90.00 0.00 0.70 0.67 101.13 194.40 38.58 90.00 0.00 0.71

Temperature (deg C)

SUM((P-Pspec) ) 0.00

1.00 0.96 0.92 0.88 0.84 0.81 0.78 0.75 0.72 0.69 0.66 0.64 0.61 0.59 0.57 0.55 0.53 0.51 0.49 0.47 0.45 1.00 0.44 0.42 0.41 0.39 0.38 0.37 0.35 0.34 0.33 0.32 0.31 0.30 0.29

Solution 13.3

Problem Statement

Assuming Raoult’s law to apply to the system n-pentane(1)/n-heptane(2), (a) What are the values of x1 and y1 at t = 55°C and P  system that is vapor

1 sat (P1  P2sat )? For these conditions plot the fraction of 2

vs. overall composition z1.

(b) For t = 55°C and z1 = 0.5, plot P, x1, and y1 vs. .

Note: Solutions to some of the problems of this chapter require vapor pressures as a function of temperature. Table B.2, Appendix B, lists parameter values for the Antoine equation, from which these can be computed.

Solution

(a) First, we need to evaluate the vapor pressures for species 1 (n-pentane) and species 2 (n-heptane) using the sat

Antoine equation and the coefficients given in table 10.2. For n-pentane, we have P1 = exp(13.8183 – 2477.07/(55 + sat

233.21)) = 185.6 kPa. For n-heptane, we have P2

= exp(13.8587 – 2911.32/(55 + 216.64) = 23.13 kPa. Using these, sat

we can compute the specified pressure as P = ½ (P1

sat

+ P2 ) = 104.4 kPa. Specifying the temperature (T = 55 ºC)

and pressure (P = 104.4 kPa) is sufficient to specify the state of the 2-phase system, since the phase rule tells us that F=2

+N=2 sat

x1P1

2 + 2 = 2. We can write Raoult’s law for each species as sat

= y1P and x2P2

= y2P

Solution continued on next page…

Substituting x2 = 1 – x1 and y2 = 1 – y1, and putting in the numbers for the vapor pressures and total pressure, we have 185.6 x1 = 104.4 y1 and 23.13 (1 – x1) = 104.4 (1 – y1) Adding these equations gives (185.6 – 23.13) x1 + 23.13 = 104.4, or x1 = (104.4 – 23.13)/(185.6 – 23.13) = 0.5

More generally (without substituting in the numbers), we would have sat

sat

x1 = (P – P2 ) / (P1

sat

– P2 )

Thinking about this a little more generally, we realize that for a binary mixture that obeys Raoult’s law, the total pressure is a linear function of the liquid phase mole fractions. In fact, it is a straight line connecting the two pure component vapor pressures: sat

P = x1P1

sat

+ (1 – x1)P2

sat

= P2

sat

+ x1 (P1

sat

– P2 )

Thus, the total pressure will be halfway between the two vapor pressures for x1 = 0.5. sat

The corresponding vapor phase mole fraction is given by y1 = x1P1 /P = 0.5*185.6/104.4 = 0.889.

The fraction of the system that is vapor is related to the overall mole fraction of species 1 by the equation x1L + y1V = z1 For the specified T and P, x1 and y1 remain constant. So, we can substitute L = 1 – V and then solve this for V (L and V are the fractions of the total number of moles that are liquid and vapor, respectively, so they have to add up to one). x1 (1 – V) + y1V = z1 V = (z1 – x1)/(y1 – x1) Solution continued on next page…

This is only physically meaningful for V ranging from 0 to 1 (since we can’t have a fraction vapor that is negative or greater than one). V is zero when z1 = x1 = 0.5 and V is one when z1 = y1 = 0.889. So, the range of z1 for which we can have two phases is 0.5 < z1 < 0.889. Plotting V vs. z1 for this range gives:

Problem 10.3(a), V vs. z1

Vapor phase fraction, V

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.5

0.6

0.7

0.8

0.9

Overall mole fraction z1

(b) While we are asked to plot P, x1, and y1 vs. V, as a practical matter it will be simpler to specify a range of values of P and compute x1, y1, and V for each value of P, in basically the same way that we did for part (a). The range of pressures for which there can be two phases is delimited by the dew point pressure and bubble point pressure for the given overall composition. For a mixture with z1 = 0.5, the bubble point pressure is given by sat

Pbub = x1P1

sat

+ x2P2 = 0.5 P1 + 0.5 P2

= 0.5*185.61 + 0.5*23.13 = 104.37 kPa

This is the same as for part (a), where we found that the corresponding y1 value was y1 = 0.889. Solution continued on next page…

sat

The dew point is given by Pdew = 1/(y1/ P1

sat

+ y2/P2 ) = 1/(0.5/185.61 + 0.5 / 23.13) = 41.13 kPa sat

The corresponding liquid phase mole fraction is x1 = y1P/P1

= 0.5*185.61/41.13 = 0.111

At the bubble point, V = 0, and at the dew point V = 1. To generate a plot, we can select several values of P ranging from Pdew to Pbub, and for each pressure compute the liquid phase mole fraction (as we did in part (a)) as sat

sat

x1 = (P – P2 ) / (P1

sat

– P2 )

The corresponding y1 is then sat

y 1 = x1 P 1

Finally, the corresponding vapor phase fraction is (again using the equation developed in part (a) V = (z1 – x1)/(y1 – x1) This is implemented in the spreadsheet shown below.

Solution continued on next page…

A1 13.8183

B1 2477.07

C1 233.21

P1sat (kPa)P2sat (kPa) 185.61 23.13

A2 13.8587

B2 2911.32

C2 T (deg C) 216.64 55

Pbub (kPa) Pdew (kPa) 104.37 41.13

z1 0.5 P vs. V for Problem 10.3(b)

110.0000

x1 0.1108 0.1303 0.1497 0.1692 0.1886 0.2081 0.2276 0.2470 0.2665 0.2859 0.3054 0.3249 0.3443 0.3638 0.3832 0.4027 0.4222 0.4416 0.4611 0.4805 0.5000

y1 0.5000 0.5458 0.5856 0.6204 0.6511 0.6783 0.7027 0.7247 0.7446 0.7627 0.7792 0.7943 0.8082 0.8211 0.8330 0.8440 0.8543 0.8639 0.8729 0.8813 0.8892

V 1.0000 0.8897 0.8036 0.7332 0.6733 0.6207 0.5733 0.5296 0.4884 0.4490 0.4107 0.3731 0.3356 0.2979 0.2596 0.2205 0.1801 0.1383 0.0945 0.0486 0.0000

100.0000

Pressure (kPa)

P (kPa) 41.1340 44.2957 47.4575 50.6192 53.7809 56.9426 60.1043 63.2660 66.4278 69.5895 72.7512 75.9129 79.0746 82.2363 85.3980 88.5598 91.7215 94.8832 98.0449 101.2066 104.3683

90.0000 80.0000 70.0000 60.0000 50.0000 40.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Vapor phase fraction (V)

x1 and y1 vs. V for Problem 10.3(b) 1.0000 0.9000 0.8000

x1 or y1

0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Vapor phase fraction (V)

The solution given here differs slightly from that in the online solutions manual, because the solutions manual uses 2991.32 for the Antoine coefficient B, instead of 2911.32, which is the value in table 10.2.

Solution 13.4

Problem Statement

Rework Prob. 13.3 for one of the following: (a) t = 65°C; (b) t = 75°C; (c) t = 85°C; (d) t = 95°C.

Problem 13.3 Assuming Raoult’s law to apply to the system n-pentane(1)/n-heptane(2), (a) What are the values of x1 and y1 at t = 55°C and P 

1 sat (P  P2sat )? For these conditions plot the fraction of 2 1

system that is vapor vs. overall composition z1.

(b) For t = 55°C and z1 = 0.5, plot P, x1, and y1 vs. .

Solution (a) First, we need to evaluate the vapor pressures for species 1 (n-pentane) and species 2 (n-heptane) using the sat

Antoine equation. For n-pentane, we have P1 = exp(13.7667 – 2451.88/(65 + 232.014)) = 247.55 kPa. For Solution continued on next page…

sat

n-heptane, we have P2

= exp(13.8622 – 2910.26/(65 + 216.432) = 33.83 kPa. Using these, we can compute the sat

specified pressure as P = ½ (P1

sat

+ P2 ) = 140.69 kPa. Specifying the temperature (T = 65 ºC) and pressure (P =

140.69 kPa) is sufficient to specify the state of the 2-phase system, since the phase rule tells us that F = 2

+N=

2 – 2 + 2 = 2. We can write Raoult’s law for each species as sat

x1P1

sat

= y1P and x2P2

= y2P

Substituting x2 = 1 – x1 and y2 = 1 – y1, and putting in the numbers for the vapor pressures and total pressure, we have 247.55 x1 = 140.69 y1 and 33.83 (1 – x1) = 140.69 (1 – y1) Adding these equations gives (247.55 – 33.83) x1 + 33.83 = 140.69, or x1 = (140.69 – 33.83)/(247.55 – 33.83) = 0.5

More generally (without substituting in the numbers), we would have sat

sat

x1 = (P – P2 ) / (P1

sat

– P2 )

P = x1P1

sat

+ (1 – x1)P2

sat

= P2

sat

+ x1 (P1

sat

– P2 )

Thus, the total pressure will be halfway between the two vapor pressures for x1 = 0.5. sat

The corresponding vapor phase mole fraction is given by y1 = x1P1 /P = 0.5*247.55/140.69 = 0.880.

The fraction of the system that is vapor is related to the overall mole fraction of species 1 by the equation x1L + y1V = z1 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

For the specified T and P, x1 and y1 remain constant. So, we can substitute L = 1 – V and then solve this for V (L and V are the fractions of the total number of moles that are liquid and vapor, respectively, so they have to add up to one). x1 (1 – V) + y1V = z1 V = (z1 – x1)/(y1 – x1) This is only physically meaningful for V ranging from 0 to 1 (since we can’t have a fraction vapor that is negative or greater than one). V is zero when z1 = x1 = 0.5 and V is one when z1 = y1 = 0.880. So, the range of z1 for which we can have two phases is 0.5 < z1 < 0.880. Plotting V vs. z1 for this range gives:

Problem 10.4(a), V vs. z 1 1

Vapor phase fraction, V

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.5000

0.6000

0.7000

0.8000

0.9000

Overall mole fraction z 1

Solution continued on next page…

Pbub = x1P1

sat

+ x2P2 = 0.5 P1 + 0.5 P2

= 0.5*247.55 + 0.5*33.83 = 140.69 kPa

This is the same as for part (a), where we found that the corresponding y1 value was y1 = 0.880. sat

The dew point is given by Pdew = 1/(y1/ P1

sat

+ y2/P2 ) = 1/(0.5/247.55 + 0.5 / 33.83) = 59.52 kPa sat

The corresponding liquid phase mole fraction is x1 = y1P/P1

= 0.5*247.55/59.52 = 0.120

sat

x1 = (P – P2 ) / (P1

sat

– P2 )

The corresponding y1 is then sat

y 1 = x1 P 1

Finally, the corresponding vapor phase fraction is (again using the equation developed in part (a) V = (z1 – x1)/(y1 – x1)

Solution continued on next page…

This is implemented in the spreadsheet shown below: A1 13.7667

B1 2451.88

C1 232.014

P1sat (kPa)P2sat (kPa) 247.55 33.83

A2 13.8622

B2 2910.26

C2 T (deg C) 216.43 65

Pbub (kPa) Pdew (kPa) 140.69 59.52

z1 0.5 P vs. V for Problem 10.4

160.0000

x1 0.1202 0.1392 0.1582 0.1772 0.1962 0.2152 0.2342 0.2531 0.2721 0.2911 0.3101 0.3291 0.3481 0.3671 0.3861 0.4051 0.4240 0.4430 0.4620 0.4810 0.5000

y1 0.5000 0.5420 0.5790 0.6118 0.6411 0.6674 0.6911 0.7127 0.7323 0.7503 0.7669 0.7821 0.7962 0.8093 0.8215 0.8328 0.8434 0.8534 0.8627 0.8715 0.8798

V 1.0000 0.8957 0.8123 0.7428 0.6829 0.6299 0.5818 0.5372 0.4951 0.4549 0.4157 0.3772 0.3390 0.3006 0.2617 0.2219 0.1811 0.1388 0.0948 0.0486 0.0000

140.0000

Pressure (kPa)

P (kPa) 59.5227 63.5810 67.6393 71.6975 75.7558 79.8140 83.8723 87.9305 91.9888 96.0470 100.1053 104.1635 108.2218 112.2800 116.3383 120.3965 124.4548 128.5130 132.5713 136.6295 140.6878

120.0000 100.0000 80.0000 60.0000 40.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Vapor phase fraction (V)

x1 and y1 vs. V for Problem 10.4 1.0000 0.9000 0.8000

x1 or y1

0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

Vapor phase fraction (V)

Parts (c) and (d) were solved using the same method.

Solution 13.5

Problem Statement

Prove: An equilibrium liquid/vapor system described by Raoult’s law cannot exhibit an azeotrope.

1.0000

Solution

For a binary system, the next equation following Eq. (13.17) sat

P = P2

sat

+ x1 (P1

sat

– P2 )

shows that P is linear in x1. Thus no maximum or minimum can exist in this relation. Since such an extremum is required for the existence of an azeotrope, no azeotrope is possible.

Solution 13.6

Problem Statement

Of the following binary liquid/vapor systems, which can be approximately modeled by Raoult’s law? For those that cannot, why not? Table B.1 (App. B) may be useful.

(a) Benzene/toluene at 1(atm). (b) n-Hexane/n-heptane at 25 bar. (c) Hydrogen/propane at 200 K. (d) Iso-octane/n-octane at 100°C. (e) Water/n-decane at 1 bar.

Solution

(a): Because benzene and toluene are chemically similar and the pressure is only 1(atm), this system can be modeled by Raoult’s law to a good approximation.

(b): Although n-hexane and n-heptane are chemically similar, a pressure of 25 bar is too high for modeling this system by Raoult’s law.

(c): At 200 K, hydrogen is supercritical, and modeling the hydrogen/propane system at this temperature by Raoult’s law is out of the question, because no value of P sat for hydrogen is known.

(d): Because isooctane and n-octane are chemically similar and at a temperature (373.15 K) close to their normal boiling points, this system can be modeled by Raoult’s law to a good approximation

(e): Water and n-decane are much too dissimilar to be modeled by Raoult’s law, and are in fact only slightly soluble in one another at 300 K.

Solution 13.7

Problem Statement

A single-stage liquid/vapor separation for the benzene(1)/ethylbenzene(2) system must produce phases of the following equilibrium compositions. For one of these sets, determine T and P in the separator. What additional information is needed to compute the relative amounts of liquid and vapor leaving the separator? Assume that Raoult’s law applies. (a) x1 = 0.35, y1 = 0.70. (b) x1 = 0.35, y1 = 0.725. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(c) x1 = 0.35, y1 = 0.75. (d) x1 = 0.35, y1 = 0.775. Note: Solutions to some of the problems of this chapter require vapor pressures as a function of temperature. Table B.2, Appendix B, lists parameter values for the Antoine equation, from which these can be computed.

Solution

sat

(a) Applying Raoult’s law to both species, we have x1P1

sat

= y1P and x2P2

= y2P. Given the compositions

(x1, y1, x2, and y2) we want to find the pressure and temperature. The vapor pressures are known functions of temperature, which we can write out explicitly as sat

= exp(Ai – Bi/(T + Ci))

Substituting this into Raoult’s law, we have x1 exp(A1 – B1/(T + C1)) = y1P x2 exp(A2 – B2/(T + C2)) = y2P We now have two equations in two unknowns. Multiplying the second equation by y1/y2 and subtracting it from the first gives us a single equation for T x1 exp(A1 – B1/(T + C1)) – y1x2/y2 exp(A2 – B2/(T + C2)) = 0 Taking the logarithm of this equation gives ln x1 + A1 – B1/(T + C1) = ln(y1x2/y2) + A2 – B2/(T + C2) (ln((y1x2)/(y2x1)) + A2 – A1)(T + C1)(T + C2) – B2 (T + C1) + B1 (T + C2) = 0 2

T + (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) T + C1C2 + (B1C2 – B2C1)/(ln((y1x2)/(y2x1)) + A2 – A1) = 0

Solution continued on next page…

Although this is a little ugly looking, it is just a quadratic equation. If we plug in the numbers, it will look more manageable. We have x1 = 0.35, x2 = 0.65, y1 = 0.7, y2 = 0.3, so ln((y1x2)/(y2x1) = 1.466. The Antoine coefficients are (in consistent units) A1 = 13.8594, B1 = 2773.78, C1 = 220.07, A2 = 14.0045, B2 = 3279.47, and C2 = 213.20. So, A2 – A1 = 0.1451, and (ln((y1x2)/(y2x1)) + A2 – A1) = 1.6114. B1 – B2 = –505.69, and C1 + C2 = 433.27. Thus, (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) = 433.27 – 505.69/1.6114 = 119.45. Also, C1C2 = 46918 and (B1C2 – B2C1) = –130343. So, the final quadratic equation becomes 2

T + 119.45 T – 33969 = 0 Applying the quadratic formula to this gives T = 134.0 °C or T = –253.5 °C. Although the negative temperature is above absolute zero and therefore a realizable temperature, our experience with benzene and ethylbenzene tells us that they would freeze far above that temperature at any pressure, so we have T = 134.0 °C. At this temperature, the sat

sat

vapor pressures are P1 = 413.8 kPa and P2 sat

P = x1P1

sat

+ x2P2

= 95.5 kPa. Thus, the total pressure is

= 0.35*413.8 + 0.65*95.5 = 206.9 kPa.

We can double-check these results by computing the vapor phase mole fraction sat

y1 = x1P1 /P = 0.35 * 413.8 / 206.9 = 0.700. In order to calculate the relative amounts of liquid and vapor leaving the separator, we would have to know the composition of the feed. In order to have two phases leaving the separator, it would have to have a composition between the specified liquid and vapor phase compositions.

sat

(b) Applying Raoult’s law to both species, we have x1P1

sat

= y1P and x2P2

= y2P. Given the compositions

(x1, y1, x2, and y2) we want to find the pressure and temperature. The vapor pressures are known functions of temperature, which we can write out explicitly as sat

= exp(Ai – Bi/(T + Ci))

Solution continued on next page…

T + (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) T + C1C2 + (B1C2 – B2C1)/(ln((y1x2)/(y2x1)) + A2 – A1) = 0 Although this is a little ugly looking, it is just a quadratic equation. If we plug in the numbers, it will look more manageable. We have x1 = 0.35, x2 = 0.65, y1 = 0.725, y2 = 0.275, so ln((y1x2)/(y2x1) = 1.5884. The Antoine coefficients are (in consistent units) A1 = 13.8594, B1 = 2773.78, C1 = 220.07, A2 = 14.0045, B2 = 3279.47, and C2 = 213.20. So, A2 – A1 = 0.1451, and (ln((y1x2)/(y2x1)) + A2 – A1) = 1.7335. B1 – B2 = –505.69, and C1 + C2 = 433.27. Thus, (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) = 433.27 – 505.69/1.7335 = 141.56. Also, C1C2 = 46918 and (B1C2 – B2C1) = –130343. So, the final quadratic equation becomes 2

T + 141.56 T – 28270 = 0 Applying the quadratic formula to this gives T = 111.6 °C or T = –253.2 °C. Although the negative temperature is above absolute zero and therefore a realizable temperature, our experience with benzene and ethylbenzene tells us that they would freeze far above that temperature at any pressure, so we have

sat

T = 111.6 °C. At this temperature, the vapor pressures are P1 = 244.1 kPa and P2

= 49.9 kPa. Thus, the total

pressure is sat

P = x1P1

sat

+ x2P2

= 0.35*244.1 + 0.65*49.9 = 117.8 kPa.

We can double-check these results by computing the vapor phase mole fraction sat

y1 = x1P1 /P = 0.35 * 244.1 / 117.8 = 0.725. In order to calculate the relative amounts of liquid and vapor leaving the separator, we would have to know the composition of the feed. In order to have two phases leaving the separator, it would have to have a composition between the specified liquid and vapor phase compositions.

sat

= y1P and x2P2

= y2P. Given the compositions

(x1, y1, x2, and y2) we want to find the pressure and temperature. The vapor pressures are known functions of temperature, which we can write out explicitly as sat

= exp(Ai – Bi/(T + Ci))

T + (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) T + C1C2 + (B1C2 – B2C1)/(ln((y1x2)/(y2x1)) + A2 – A1) = 0 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Although this is a little ugly looking, it is just a quadratic equation. If we plug in the numbers, it will look more manageable. We have x1 = 0.35, x2 = 0.65, y1 = 0.75, y2 = 0.25, so ln((y1x2)/(y2x1) = 1.7177. The Antoine coefficients are (in consistent units) A1 = 13.8594, B1 = 2773.78, C1 = 220.07, A2 = 14.0045, B2 = 3279.47, and C2 = 213.20. So, A2 – A1 = 0.1451, and (ln((y1x2)/(y2x1)) + A2 – A1) = 1.8628. B1 – B2 = –505.69, and C1 + C2 = 433.27. Thus, (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) = 433.27 – 505.69/1.7335 = 161.80. Also, C1C2 = 46918 and (B1C2 – B2C1) = –130343. So, the final quadratic equation becomes 2

T + 161.8 T – 28270 = 0 Applying the quadratic formula to this gives T = 87.4 °C or T = –249.2 °C. Although the negative temperature is above absolute zero and therefore a realizable temperature, our experience with benzene and ethylbenzene tells us that they would freeze far above that temperature at any pressure, so we have T = 87.4 °C. At this temperature, the sat

sat

vapor pressures are P1 = 126.1 kPa and P2 sat

P = x1P1

sat

+ x2P2

= 22.0 kPa. Thus, the total pressure is

= 0.35*126.1 + 0.65*22.0 = 58.5 kPa.

We can double-check these results by computing the vapor phase mole fraction sat

y1 = x1P1 /P = 0.35 * 126.1 / 58.5 = 0.75. In order to calculate the relative amounts of liquid and vapor leaving the separator, we would have to know the composition of the feed. In order to have two phases leaving the separator, it would have to have a composition between the specified liquid and vapor phase compositions.

sat

(d) Applying Raoult’s law to both species, we have x1P1

sat

= y1P and x2P2

= y2P. Given the compositions

(x1, y1, x2, and y2) we want to find the pressure and temperature. The vapor pressures are known functions of temperature, which we can write out explicitly as sat

= exp(Ai – Bi/(T + Ci))

Solution continued on next page…

Substituting this into Raoult’s law, we have x1 exp(A1 – B1/(T + C1)) = y1P x2 exp(A2 – B2/(T + C2)) = y2P We now have two equations in two unknowns. Multiplying the second equation by y1/y2 and subtracting it from the first gives us a single equation for T (eliminating P, the other unknown). x1 exp(A1 – B1/(T + C1)) – y1x2/y2 exp(A2 – B2/(T + C2)) = 0 Taking the logarithm of this equation gives ln x1 + A1 – B1/(T + C1) = ln(y1x2/y2) + A2 – B2/(T + C2) (ln((y1x2)/(y2x1)) + A2 – A1)(T + C1)(T + C2) – B2 (T + C1) + B1 (T + C2) = 0 2

T + (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) T + C1C2 + (B1C2 – B2C1)/(ln((y1x2)/(y2x1)) + A2 – A1) = 0 Although this is a little ugly looking, it is just a quadratic equation. If we plug in the numbers, it will look more manageable. We have x1 = 0.35, x2 = 0.65, y1 = 0.775, y2 = 0.225, so ln((y1x2)/(y2x1) = 1.8558. The Antoine coefficients are (in consistent units) A1 = 13.8594, B1 = 2773.78, C1 = 220.07, A2 = 14.0045, B2 = 3279.47, and C2 = 213.20. So, A2 – A1 = 0.1451, and (ln((y1x2)/(y2x1)) + A2 – A1) = 2.0009. B1 – B2 = –505.69, and C1 + C2 = 433.27. Thus, (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) = 433.27 – 505.69/2.0009 = 180.54. Also, C1C2 = 46918 and (B1C2 – B2C1) = –130343. So, the final quadratic equation becomes 2

T + 180.54 T – 18224 = 0 Applying the quadratic formula to this gives T = 72.1 °C or T = –252.7 °C. Although the negative temperature is above absolute zero and therefore a realizable temperature, our experience with benzene and ethylbenzene tells us that they would freeze far above that temperature at any pressure, so we have T = 72.1 °C. At this temperature, the sat

sat

vapor pressures are P1 = 78.8 kPa and P2 sat

P = x1P1

sat

+ x2P2

= 12.31 kPa. Thus, the total pressure is

= 0.35*78.8 + 0.65*12.3 = 35.6 kPa.

We can double-check these results by computing the vapor phase mole fraction sat

y1 = x1P1 /P = 0.35 * 78.8 / 35.6 = 0.775. In order to calculate the relative amounts of liquid and vapor leaving the separator, we would have to know the composition of the feed. In order to have two phases leaving the separator, it would have to have a composition between the specified liquid and vapor phase compositions.

Solution 13.8

Problem Statement Do all four parts of Prob. 13.7, and compare the results. The required temperatures and pressures vary significantly. Discuss possible processing implications of the various temperature and pressure levels.

Problem Solution To increase the relative amount of benzene in the vapor phase, the temperature and pressure of the process must be lowered. For parts (c) and (d), the process must be operated under vacuum conditions. The temperatures are well within the bounds of typical steam and cooling water temperatures.

Problem 13.7 A single-stage liquid/vapor separation for the benzene(1)/ethylbenzene(2) system must produce phases of the following equilibrium compositions. For one of these sets, determine T and P in the separator. What additional information is needed to compute the relative amounts of liquid and vapor leaving the separator? Assume that Raoult’s law applies. (a) x1 = 0.35, y1 = 0.70. (b) x1 = 0.35, y1 = 0.725. (c) x1 = 0.35, y1 = 0.75. (d) x1 = 0.35, y1 = 0.775.

Solution

sat

Applying Raoult’s law to both species, we have x1P1

sat

= y1P and x2P2

= y2P. Given the compositions

(x1, y1, x2, and y2) we want to find the pressure and temperature. The vapor pressures are known functions of temperature, which we can write out explicitly as sat

= exp(Ai – Bi/(T + Ci))

C2 = 212.320. So, A2 – A1 = 0.1907, and (ln((y1x2)/(y2x1)) + A2 – A1) =1.6570. B1 – B2 = –533.12, and C1 + C2 = 429.872. Thus, (C1 + C2 + (B1 – B2)/(ln((y1x2)/(y2x1)) + A2 – A1)) = 433.27 – 533.12/1.65704 = 180.1411. Also, C1C2 = 46190.54 and (B1C2 – B2C1) = –130367.73. So, the final quadratic equation becomes 2

T + 180.54 T – 32484.67 = 0 Applying the quadratic formula to this gives T = 134.10 °C or T = –242.7 °C. Although the negative temperature is above absolute zero and therefore a realizable temperature, our experience with benzene and ethylbenzene tells us that they would freeze far above that temperature at any pressure, so we have T = 134.10 °C. At this temperature, the sat

sat

vapor pressures are P1 = 414.91 kPa and P2 sat

P = x1P1

sat

+ x2P2

= 95.75 kPa. Thus, the total pressure is

= 0.35*414.91 + 0.65*95.75 = 207.45 kPa.

We can double-check these results by computing the vapor phase mole fraction sat

y1 = x1P1 /P = 0.35 * 414.91 / 207.45 = 0.700. In order to calculate the relative amounts of liquid and vapor leaving the separator, we would have to know the composition of the feed. In order to have two phases leaving the separator, it would have to have a composition between the specified liquid and vapor phase compositions.

Solution 13.9

Problem Statement

A mixture containing equimolar amounts of benzene(1), toluene(2), and ethylbenzene(3) is flashed to conditions T and P. For one of the following conditions, determine the equilibrium mole fractions {xi} and {yi} of the liquid and vapor phases formed and the molar fraction of the vapor formed. Assume that Raoult’s law applies. (a) T = 110°C, P = 90 kPa. (b) T = 110°C, P = 100 kPa. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

(a) First, we need to compute the saturation pressures at T = 110 °C. For benzene, this is sat

sat

= exp(13.8594 – 2773.78/(110 + 220.07)) = 237.89 kPa. For toluene, this is = exp(14.0098 – 3103.01/(110 + 219.79)) = 99.56 kPa. For ethylbenzene, this is = exp(14.0045 – 3279.47/(110 + 213.20)) = 47.35 kPa. Because we also know the pressure in this problem, we sat

can compute the K-values for each of the species from Raoult’s law as Ki = yi/xi = P1 /P. So, K1 = 237.89/90 = 2.642, K2 = 1.106, K3 = 0.5261. Know we can combine this information with the known overall mole fractions z1 = z2 = z3 = 0.333 and use equation

z1 K1

1  V  K1  1



z2 K 2

1  V  K 2  1



z3 K 3

1  V  K 3  1

1

1  V  K 1  11  V  K 2  11  V  K 3  1  z1 K1 1  V  K 2  11  V  K 3  1  1  V  K1  1 z2 K 2 1  V  K 3  1  1  V  K1  11  V  K 2  1 z3 K 3  0 or, substituting in the K’s and z’s

V 3  4.288587V 2  10.43753V  5.148752  0 Solving for V gives V = –6.129, V = 0.8400, or V = 1.0001 and of course the only one that is physically meaningful is the one that is between zero and one. So the fraction of the system that is vapor is V = 0.8400 and the fraction that is liquid is L = 0.1600.

Solution continued on next page…

Now, we can find the vapor phase mole fractions from zi K i L  VK i 0.3333  2.642 y1   0.3701 0.1600  0.8400  2.642 0.3333  1.106 y2   0.3385 0.1600  0.8400  1.106 0.3333  0.5261 y3   0.2913 0.1600  0.8400  0.5261 yi 

Then, the liquid phase mole fractions are found from xi = yi/Ki, and we get x1 = 0.1401, x2 = 0.3061, x3 = 0.5537.

(b) First, we need to compute the saturation pressures at T = 110 °C. For benzene, this is sat

sat

= exp(13.8594 – 2773.78/(110 + 220.07)) = 237.89 kPa. For toluene, this is = exp(14.0098 – 3103.01/(110 + 219.79)) = 99.56 kPa. For ethylbenzene, this is = exp(14.0045 – 3279.47/(110 + 213.20)) = 47.35 kPa. Since we also know the pressure in this problem, we can sat

compute the K-values for each of the species from Raoult’s law as Ki = yi/xi = P1 /P. So, K1 = 237.89/100 = 2.3789, K2 = 0.9956, K3 = 0.4735. Now we can combine this information with the known overall mole fractions z1 = z2 = z3 = 0.333 and use equation z1 K 1

1  V  K1  1



z2 K 2

1  V  K 2  1



z3 K 3

1  V  K 3  1

1

or, substituting in the K’s and z’s

0.3333 * 2.3789 0.3333 * 0.9956 0.3333 * 0.4735   1 1  1.3789V 1  0.0044V 1  0.5265V Solving for V by trial and error gives V = 0.5833 So the fraction of the system that is vapor is V = 0.5833 and the fraction that is liquid is L = 0.4167.

Solution continued on next page…

Now, we can find the vapor phase mole fractions from yi 

zi K i

1  V  K i  1

y1 

0.3333 * 2.3789  0.4395 1  1.3789 * 0.5833

y2 

0.3333 * 0.9956  0.3327 1  0.0044 * 0.5833

y3 

0.3333 * 0.4735  0.2278 1  0.5265 * 0.5833

Then, the liquid phase mole fractions are found from xi = yi/Ki, and we get x1 = 0.1847, x2 = 0.3342, x3 = 0.4811.

sat

we can compute the K-values for each of the species from Raoult’s law as sat

Ki = yi/xi = P1 /P. So, K1 = 237.89/110 = 2.1626, K2 = 0.9051, K3 = 0.4305. Now we can combine this information with the known overall mole fractions z1 = z2 = z3 = 0.333 and use equation

z1 K1

1  V  K1  1



z2 K 2

1  V  K 2  1



z3 K 3

1  V  K 3  1

1

or, substituting in the K’s and z’s

0.3333 * 2.1626 0.3333 * 0.9051 0.3333 * 0.4305   1 1  1.1626V 1  0.0949V 1  0.5695V Solving for V by trial and error gives V = 0.3641

Solution continued on next page…

So the fraction of the system that is vapor is V = 0.3641 and the fraction that is liquid is L = 0.6359. Now, we can find the vapor phase mole fractions from yi 

zi K i

1  V  K i  1

y1 

0.3333 * 2.1626  0.5064 1  1.1626 * 0.3641

y2 

0.3333 * 0.9051  0.3125 1  0.0949 * 0.3641

y3 

0.3333 * 0.4305  0.1810 1  0.5695 * 0.3641

Then, the liquid phase mole fractions are found from xi = yi/Ki, and we get x1 = 0.2342, x2 = 0.3453, x3 = 0.4204.

(d) First, we need to compute the saturation pressures at T = 110 °C. For benzene, this is sat

sat

= exp(13.7819 – 2726.81/(110 + 217.572)) = 234.5 kPa. For toluene, this is = exp(13.9320 – 3056.96/(110 + 217.650)) = 99.7 kPa. For ethylbenzene, this is = exp(13.9726 – 3259.93/(110 + 212.30)) = 47.4 kPa. Since we also know the pressure in this problem, we can sat

compute the K-values for each of the species from Raoult’s law as Ki = yi/xi = P1 /P.

So, K1 = 234.5/120 = 1.9544, K2 = 0.8307, K3 = 0.3948. Now we can combine this information with the known overall mole fractions z1 = z2 = z3 = 0.333 and use equation

z1 K1

1  V  K1  1



z2 K 2

1  V  K 2  1



z3 K 3

1  V  K 3  1

1

or, substituting in the K’s and z’s

0.3333 * 1.9544 0.3333 * 0.8307 0.3333 * 0.3948   1 1  0.9544V 1  0.1693V 1  0.6052V Solving for V by trial and error gives V = 0.145 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

So the fraction of the system that is vapor is V = 0.145 and the fraction that is liquid is L = 0.855. Now, we can find the vapor phase mole fractions from yi 

zi K i

1  V  K i  1 0.3333 * 1.9544 y1   0.572 1  0.9544 * 0.145 0.3333 * 0.8307 y2   0.284 1  0.1693 * 0.145 0.3333 * 0.3948 y3   0.144 1  0.6052 * 0.145

Then, the liquid phase mole fractions are found from xi = yi/Ki, and we get x1 = 0.293, x2 = 0.342, x3 = 0.365.

Solution 13.10

Problem Statement

Do all four parts of Prob. 13.9, and compare the results. Discuss any trends that appear.

Problem 13.10

As the pressure increases, the fraction of vapor phase formed (V) decreases, the mole fraction of benzene in both phases increases and the the mole fraction of ethylbenzene in both phases decreases.

Problem 13.9

(a) T = 110°C, P = 90 kPa. (b) T = 110°C, P = 100 kPa. (c) T = 110°C, P = 110 kPa. (d) T = 110°C, P = 120 kPa.

Solution

First, we need to compute the vapor pressures at T = 110 °C. For benzene, this is sat

sat

Since we also know the pressure in this problem, we can compute the K-values for each of the species from Raoult’s sat

law as Ki = yi/xi = P1 /P. So, K1 = 234.5/90= 2.6432, K2 =1.1062, K3 = 0.5261. Now we can combine this information with the known overall mole fractions z1 = z2 = z3 = 0.333 and use equation

z1 K 1

1  V  K1  1



z2 K 2

1  V  K 2  1



z3 K 3

1  V  K 3  1

1

or, substituting in the K’s and z’s. Solving for V by trial and error gives V = 0.8400 So the fraction of the system that is vapor is V = 0.8400 and the fraction that is liquid is L = 0.1600. Now, we can find the vapor phase mole fractions from Eq 13.105

yi 

zi Ki

1  V  K i  1

Y1 = 0.3702 Y2 = 0.3385 Y3 = 0.2913 Then, the liquid phase mole fractions are found from xi = yi/Ki, and we get x1 = 0.1400, x2 = 0.3060 x3 = 0.5538.

Solution 13.11

Problem Statement

A binary mixture of mole fraction z1 is flashed to conditions T and P. For one of the following, determine the equilibrium mole fractions x1 and y1 of the liquid and vapor phases formed, the molar fraction of the vapor formed, and the fractional recovery

of species 1 in the vapor phase (defined as the ratio for species 1 of moles in the vapor

to moles in the feed). Assume that Raoult’s law applies. (a) Acetone(1)/acetonitrile(2), z1 = 0.75, T = 340 K, P = 115 kPa. (b) Benzene(1)/ethylbenzene(2), z1 = 0.50, T = 100°C, P = 0.75(atm). (c) Ethanol(1)/1-propanol(2), z1 = 0.25, T = 360 K, P = 0.80(atm). (d) 1-Chlorobutane(1)/chlorobenzene(2), z1 = 0.50, T = 125°C, P = 1.75 bar.

Solution

(a) At the specified temperature of 340 K = 66.85 ºC, the vapor pressures of acetone (component 1) and acetonitrile (component 2) are:  2756.22  P1sat (kPa)  exp14.3145    143.8 kPa  66.85  228.06    3413.10 P2sat (kPa)  exp14.8950    62.9 kPa 66.85  250.523  

Solution continued on next page…

The bubble point pressure at this temperature and overall composition (z1 = 0.75) is Pbub = 0.75*143.8 + 0.25*62.9 = 123.6 kPa. The corresponding dew point pressure is Pdew = 1/(0.75/143.8 + 0.25/62.9) = 108.8 kPa. The specified pressure is between the bubble point and dew point pressures, so 2 phases will be present and we can do a flash calculation. In general, for a flash calculation, we can write equation 13.106 N

zi K i

i 1

 1  V  K  1  1 For a 2-component system, this is

z1 K 1

1  V  K 1  1



z2 K2

1  V  K 2  1

1

Which can be rearranged to give z1 K 1 1  V  K 2  1  z2 K 2 1  V  K 1  1  1  V  K 2  11  V  K 1  1 z1 K 1  z2 K 2   z1 K 1 K 2  z2 K 2 K 1  z1 K 1  z2 K 2 V  1   K 1  K 2  2 V   K 2  1 K 1  1V

 K 2  1 K 1  1V   K 2 K 1  z1 K 1  z2 K 2  K 1  K 2  2  z1 K 1  z 2 K 2  1  0 2

If the system obeys Raoult’s law, then the K-values can be computed without knowing the phase compositions, and therefore this is a single equation for a single unknown, V, the vapor phase fraction. In the present case, z1 = 0.75, sat

z2 = 0.25, K1 = P1 /P = 143.8/115 = 1.250, and K2 = 62.9/115 = 0.547. So, using these numbers, we have

0.11325V 2  0.1875V  0.07425  0 Solving this quadratic equation gives V = 0.656 (as well as the trivial solution, V = 1). So, L = 1 – 0.656 = 0.344. The vapor phase mole fractions are given by 13.105: yi 

zi K i

1  V  K i  1

So, y1 = 0.75*1.25/(1 + 0.656*0.25) = 0.805, and y2 = 0.195. x1 = y1/K1 = 0.644 and x2 = 0.356.

Solution continued on next page…

The recovery of component 1 is the total moles of component 1 in the vapor phase per mole of component 1 in the feed: R = 0.805*0.656/0.75 = 0.704. That is, 70.4% of the acetone in the feed ends up in the vapor stream. A little more than 50% of the acetonitrile also ends up in the vapor stream.

(b) At the specified temperature of 100 ºC, the vapor pressures of benzene (component 1) and ethylbenzene (component 2) are:  

P1 (kPa)  exp 13.7819  sat

 

P2 (kPa)  exp 13.9726  sat

   180.45 kPa 100  217.572  2726.81

3259.93 

  34.27 kPa

100  212.3 

The bubble point pressure at this temperature and overall composition (z1 = 0.50) is Pbub = 0.50*180.45 + 0.50*34.27 = 107.36 kPa. The corresponding dew point pressure is Pdew = 1/(0.50/180.45 + 0.50/34.27) = 57.60 kPa. The specified pressure (0.75 atm = 75.99 kPa) is between the bubble point and dew point pressures, so 2 phases will be present and we can do a flash calculation. In general, for a flash calculation, we can write equation 13.106 : N

zi K i

 1  V  K  1  1 i 1

For a 2-component system, this is z1 K 1

1  V  K 1  1



z2 K 2

1  V  K 2  1

1 sat

In the present case, z1 = 0.50, z2 = 0.50, K1 = P1 /P = 180.45/75.99 = 2.3747, and K2 = 34.27/75.99 = 0.4510. So, using these numbers, we have

1.1874 0.2255  1 1  1.3747V 1  0.5491V 1.1874  0.6519V  0.2255  0.3100V  1  0.8256V  0.7873V 2 0.7548V 2  1.1675V  0.4129  0 Solving this quadratic equation gives V = 0.5474 (as well as the trivial solution, V = 1). Solution continued on next page…

So, L = 1 – 0.5474 = 0.4526. The vapor phase mole fractions are given by:

yi 

So, y1 

zi Ki

1  V  K i  1

1.1874  0.6775 1  1.3747V

y2 

0.2255  0.3224 1  0.5491V

The recovery of component 1 is the total moles of component 1 in the vapor phase per mole of component 1 in the feed: R = 0.6775*0.5474/0.50 = 0.742. That is, 74.2% of the benzene in the feed ends up in the vapor stream. The liquid phase mole fractions are x1 = 0.6775/2.3747 = 0.2853 and x2 = 0.3224/0.4510 = 0.7149. These add up to 1.0002, which is close enough to 1.0 to reassure us that we have done things correctly.

(c) At the specified temperature of 360 K = 86.85 K, the vapor pressures of ethanol (component 1) and 1-propanol (component 2) are: 16.1154, 3483.67, 205.807   3795.17 P1sat (kPa)  exp16.8958    141.54 kPa 86.85  230.918     3483.67 P2sat (kPa)  exp16.1154    67.48 kPa  86.85  205.807 

The bubble point pressure at this temperature and overall composition (z1 = 0.25) is Pbub = 0.25*141.54 + 0.75*67.48 = 85.99 kPa. The corresponding dew point pressure is Pdew = 1/(0.25/141.54 + 0.75/67.48) = 77.63 kPa. The specified pressure (0.80 atm = 81.06 kPa) is between the bubble point and dew point pressures, so 2 phases will be present and we can do a flash calculation. In general, for a flash calculation, we can write equation 13.106 N

zi K i

i 1

 1  V  K  1  1 For a 2-component system, this is

z1 K 1

1  V  K 1  1



z2 K2

1  V  K 2  1

1

sat

In the present case, z1 = 0.25, z2 = 0.75, K1 = P1 /P = 141.54/81.06 = 1.746, and K2 = 67.48/81.06 = 0.8324. So, using these numbers, we have

0.4365 0.6243  1 1  0.7461V 1  0.1676V 0.4365  0.0731V  0.6243  0.4658V  1  0.5786V  0.1250V 2 0.1250V 2  0.1859V  0.061  0 Solving this quadratic equation gives V = 0.4869 (as well as the trivial solution, V = 1). So, L = 1 – 0.4869 = 0.5131. The vapor phase mole fractions are given by:

yi 

So, y1 

zi Ki

1  V  K i  1

0.4365  0.3202 1  0.7461V

y2 

0.6243  0.6798 1  0.1676V

The recovery of component 1 is the total moles of component 1 in the vapor phase per mole of component 1 in the feed: R = 0.3202*0.4869/0.25 = 0.624. That is, 62.4% of the ethanol in the feed ends up in the vapor stream. The liquid phase mole fractions are x1 = 0.3202/1.7641 = 0.1834 and x2 = 0.6798/0.8324 = 0.8166. These add up to 1.0000, which provides some reassurance that we have done things correctly. Part (d) is solved using the same method as parts (a)–(c).

Solution 13.12

Problem Statement

Humidity, relating to the quantity of moisture in atmospheric air, is accurately given by equations derived from the ideal-gas law and Raoult’s law for H2O.

(a) The absolute humidity h is defined as the mass of water vapor in a unit mass of dry air. Show that it is given by: h

where

H2 O

pH2O

air P  pH2O

represents a molar mass and pH2O is the partial pressure of the water vapor, i.e., pH2O = yH2O P . sat

(b) The saturation humidity h

is defined as the value of h when air is in equilibrium with a large body of pure

water. Show that it is given by: h sat 

H2 O

pHsat2 O

air P  pHsatO 2

where pHsat2 O is the vapor pressure of water at the ambient temperature. (c) The percentage humidity is defined as the ratio of h to its saturation value, expressed as a percentage. On the other hand, the relative humidity is defined as the ratio of the partial pressure of water vapor in air to its vapor pressure, expressed as a percentage. What is the relation between these two quantities?

Solution

PV t  ni RT Multiply both sides by yi , the mole fraction of

For a total volume V t species i in the mixture: yi PV t  ni RT

pi V t 

mi RT Mi

Solution continued on next page…

where mi is the mass of species i, Mi is its molar mass, and pi is its partial pressure, defined as pi  yi P

mi :

mi 

M i pi V t RT

Applied to moist air, considered a binary mixture of air and water vapor, this gives:

mH 2O 

M H 2 O pH 2 O V t RT

mair 

M air pair V t RT

(a): By definition,

h

m H 2O mair

h

M H 2 O pH 2 O M air pair

Since the partial pressures must sum to the total pressure, pair  P  pH 2O : Whence, h

M H 2 O pH 2 O M air P  pH 2 O

(b): If air is in equilibrium with liquid water, then the partial pressure of water vapor in the air equals the vapor pressure of the water, and the preceding equation becomes:

h sat 

M H 2O pH 2O sat M air P  pH 2O sat

(c): Percentage humidity and relative humidity are defined as follows:

hpc 

pH 2O P  pH 2O sat h   h sat pH 2O sat P  pH 2O



and hrel 

pH 2 O pH 2O sat





Combining these two definitions to eliminate pH2O gives: hpc  hrel

P  pH 2O sat h  P  pH 2O sat  rel   100 

Solution 13.13

Problem Statement

A concentrated binary solution containing mostly species 2 (but x2  1) is in equilibrium with a vapor phase containing both species 1 and 2. The pressure of this two-phase system is 1 bar; the temperature is 25°C. At this temperature,

1 = 200 bar and

p2sat = 0.10 bar. Determine good estimates of x1 and y1. State and justify all

assumptions.

Solution

H1 

bar

Psat 2 

bar P  bar

Assume at 1 bar that the vapor is an ideal gas. The vapor-phase fugacities are then equal to the partial presures. Assume the Lewis/Randall rule applies to concentrated species 2 and that Henry’s law applies to dilute species 1. Then:

y1 P  H1 x1

y2 P  x 2 Psat 2

P  H1 x1    x1  Psat 2

x1  x 2  Solve for x 1

P  y1 P  y2 P

y1 x1 

P  Psat 2 H1  Psat 2

y1 

H1 x 1 P

y1 

x1 

3

Solution 13.14

Problem Statement

Air, even more than carbon dioxide, is inexpensive and nontoxic. Why is it not the gas of choice for making soda water and (cheap) champagne effervescent? Table 13.2 may provide useful data. Table 13.2: Henry’s Constants for Gases Dissolved in Water at 25°C Gas

/bar

Acetylene Air Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Hydrogen sulfide Methane Nitrogen Oxygen

1,350 72,950 1,670 54,600 30,600 11,550 126,600 71,600 550 41,850 87,650 44,380

Solution

Because the vapor space above the liquid phase is nearly pure gas, Eq. P = xiHi . For the same mole fraction of gas dissolved in the liquid phase, P is then proportional to Hi . Values given in Table 10.1 indicate that were air used rather than CO2, P would be about 44 times greater, much too high a pressure to be practical.

Solution 13.15

Problem Statement

Helium-laced gases are used as breathing media for deep-sea divers. Why? Table 13.2 may provide useful data.

Table 13.2: Henry’s Constants for Gases Dissolved in Water at 25°C Gas

/bar

Acetylene Air Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Hydrogen sulfide Methane Nitrogen Oxygen

1,350 72,950 1,670 54,600 30,600 11,550 126,600 71,600 550 41,850 87,650 44,380

Solution

Because Henry’s constant for helium is very high, very little of this gas dissolves in the blood streams of divers at approximately atmospheric pressure

Solution 13.16

Problem Statement

A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium at temperature T. The 2

overall mole fraction of species 1 in the system is z1 = 0.65. At temperature T, ln 1 = 0.67 x 22 ; ln 2 = 0.67 x1 ;

P1sat = 32.27 kPa ; and P2sat = 73.14 kPa. Assuming the validity of Eq. (13.19), (a) Over what range of pressures can this system exist as two phases at the given T and z1? (b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar fraction of the system is vapor? (c) Show whether or not the system exhibits an azeotrope.

Eq. 13.19:

yi P  x i i Pi sat

i  1, 2, . . . , N 

Solution

Psat 1 

kPa

Psat 2 

1  x1 x2   exp A * x22 

kPa

A

z1  0.65

2  x1 x2  

A* x  2 1

P  x1 x2   x1 1  x1 x2  Psat1  x2 2  x1 x2  Psat 2

(a): BuBL P calculation:

x1  z1

x 2   x1

Pbubl  P  x1 x2 

Pbubl 

kPa

Solution continued on next page…

DEW P calculation:

y1  z1

y2   y1

P 

Psat 1  Psat 2 2

y1 P   x1 1  x1 , 1  x1  Psat1 P   x1 1  x1 , 1  x1  Psat 1  1  x1  * 2  x1 , 1  x1  Psat 2 You will have to solve iteratively for x1

Pdew

Pdew  43.864 kPa

The pressure range for two phases is from the dewpoint to the bubblepoint: From 43.864 to 56.745 kPa

(b): BUBL Calculation at x1  0.75

y1  x1  

x1 1  x1 , 1  x1  Psat1 P  x 1 , 1  x1 

The fraction vapor, by material balance is:

V

z1  x1

y1  x1   x1

V

P  x1

 x1  

kPa

(c): Refer to Ex 13.1

12.0 

1 0, 1 Psat 1 Psat 2 12.0 

12.1 

Psat1

2 1, 0 Psat 2

12.1 

Since alpha does not pass through 1.0 for 0<x1<1, there is no azeotrope.

Solution 13.17

Problem Statement

For the system ethyl ethanoate(1)/n-heptane(2) at 343.15 K, ln 1 = 0.95 x 22 ; ln 2 = 0.95 x12 ; P1sat = 79.80 kPa; and P1sat = 40.50 kPa. Assuming the validity of Eq. (13.19), (a) Make a BUBL P calculation for T = 343.15 K, x1 = 0.05. (b) Make a DEW P calculation for T = 343.15 K, y1 = 0.05. (c) What are the azeotrope composition and pressure at T = 343.15 K?

Eq. 13.19:

yi P  xi  i Pi sat

i  1, 2, . . . , N 

Solution

The bubble point pressure calculation (and the dew point pressure calculation) are basically the same as in earlier problems using Raoult’s law, but with the addition of the activity coefficients.

Solution continued on next page…

(a) For the bubble point calculation, the total pressure is given by P  x 1 1 P1sat  x 2  2 P2sat

P  x 1 exp0.95 x 22  P1sat  x 2 exp0.95 x 12  P2sat









P  0.05exp 0.950.95 79.80  0.95exp 0.95 0.05 40.50 2

P  47.97 kPa

and then we have 2 sat x1 1 P1sat x1 exp0.95 x 2  P1 y1   P P 2 0.05exp 0.950.95 79.80 y1   0.196 47.97 y2  1  0.196  0.804





(b) For the dew point calculation, the total pressure is given by P

P

1 y1

 1 P1sat

y  2

 2 P2sat 1

exp0.95 x 22  P1sat



exp 0.95 x12  P2sat

We can’t immediately compute the pressure, since we can’t compute the activity coefficients without knowing the liquid phase mole fractions. So, let’s write the whole thing in terms of x1, and then solve iteratively for it: x1 

y1 P  1 P1sat y1

x1 

exp0.95 x22  P1sat

exp0.95 x 22  P1sat y1

x1 



exp 0.95 x 12  P2sat





exp 0.951  x1  P1sat





exp 0.95 1  x1  P 2

sat 1



exp0.95 x12  P2sat

To get an initial guess for x1, we can set the activity coefficients to 1, to get y1

x1, guess 

P1sat y2 sat 

0.05 

0.05

sat 2

79.80

79.80  0.95

 0.026 40.50

Starting from this value, we can iterate on: 0.05 x1 

0.05





exp 0.951  x 1  79.80





exp 0.951  x1  79.80 2

 0.95

exp0.95 x 12  40.50

until we converge to x1 = 0.0104. So, x2 = 0.9896, and putting this into the expression for the total pressure, P

1 0.05

exp0.95  0.98962 79.80

 0.95

exp0.95  0.0104 40.50

P  42.19 kPa

x1 P  x1 1 P1sat x2 P  x2  2 P2sat which is the modified Raoult’s law, with y1 and y2 replaced by x1 and x2, respectively. So, at the azeotrope,

P  1 P1sat P  2 P2sat so 1 P1sat   2 P2sat

exp0.95 x22 79.80  exp0.95 x12  40.50





exp 0.95 x22  x12   40.50 / 79.80

 



exp 0.95 1  x1   x12  0.50752 2





0.95 1  2 x1  x12   x12  ln0.50752 1  2 x1  ln 0.50752 / 0.95 x1 

1  ln0.50752 / 0.95 2

 0.8570

So, at the azeotrope, x1 = y1 = 0.8570, and x2 = y2 = 0.1430. The total pressure is

P  exp0.950.14302 79.80  81.37 kPa Note that the pressure of the azeotropic mixture is higher than the saturation pressure of either component. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 13.18

Problem Statement

A liquid mixture of cyclohexanone(1)/phenol(2) for which x1 = 0.6 is in equilibrium with its vapor at 144°C. Determine the equilibrium pressure P and vapor composition y1 from the following information:  ln 1 = A x22

= A x12

 At 144°C, P1sat = 75.20 kPa and P2sat = 31.66 kPa  The system forms an azeotrope at 144°C for which x1az  y1az  0.294.

We have an azeotrope when x1 = y1 and x2 = y2. We can therefore write x1 P  x1  1 P1sat x2 P  x2  2 P2sat

which is the modified Raoult’s law, with y1 and y2 replaced by x1 and x2, respectively. So, at the azeotrope, P  1 P1sat P   2 P2sat

Solution continued on next page…

so 1 P1sat   2 P2sat exp Ax22  P1sat  exp Ax12  P2sat





exp A x 22  x12   P2sat / P1sat





exp A 1  x1   x12  P2sat / P1sat 2





A 1  2 x1  x12   x12  ln P2sat / P1sat  A1  2 x1   ln  P2sat / P1sat  A

ln  P2sat / P1sat 

1  2 x1 



ln31.66 / 75.20

1  20.294

 2.0998

Now that we have the value of A, we just need to do a bubble point pressure calculation to find the pressure and vapor phase mole fractions. This is basically the same as in problem 10.1, but with the addition of the activity coefficients.

For the bubble point pressure calculation, the total pressure is given by P  x1 1 P1sat  x2  2 P2sat









P  0.6 exp 2.09980.4 75.20  0.4 exp 2.09980.6 31.66 2

P  38.18 kPa

and then we have y1 

y1 

x1  1 P1sat P





0.6 exp 2.09980.4 75.20 2

38.18

 0.845

y2  1  0.845  0.155

Solution 13.19

Problem Statement

A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium at temperature T, for which ln  1  1.8 x 22 , ln  2  1.8 x12 , P1sat  1.24 bar, and P2sat = 40.50 kPa. Assuming the validity of Eq. (13.19),

(a) For what range of values of the overall mole fraction z1 can this two-phase system exist with a liquid mole fraction x1 = 0.65? (b) What are the pressure P and vapor mole fraction y1 within this range? (c) What are the pressure and composition of the azeotrope at temperature T?

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Solution

(a) The phase rule tells us that F = 2 for two components with two phases, so specifying the liquid phase mole fractions and the temperature fully specifies the state of the system (provided two phases remain present). So, as the overall composition changes, the liquid and vapor phase compositions and the total pressure must remain constant. The vapor fraction will change with z1 such that these can remain constant and still satisfy the overall mass balance. The vapor fraction and the overall composition are related by x1L + y1V = z1 or

x1 (1 – V) + y1V = z1

V = (z1 – x1)/(y1 – x1)

At fixed compositions, the fraction vapor is a linear function of z1. V ranges from 0 to 1 (its allowable range) as z1 ranges from x1 (for which V = 0) to y1 (for which V = 1). We specify that x1 = 0.65. We will find y1 in part (b), and it will turn out to be y1 = 0.601. Surprisingly, even though species 1 has a higher pure component vapor pressure, its concentration in the vapor is lower than its concentration in the liquid. Therefore, we can have two phases present with x1 = 0.65 for overall compositions in the range 0.601 < z1 < 0.65.

(b) Because we are specifying T and x1, we need to do a bubble point pressure calculation to find P and y1. For the bubble point pressure calculation, the total pressure is given by

P  x1 1 P1sat  x2 2 P2sat

Solution continued on next page…

The activity coefficients are given by 1 = exp (1.8*0.35 ) = 1.2467 and 2 = exp (1.8*0.65 ) = 2.1393. So, P = 0.65*1.2467*1.24 + 0.35*2.1393*0.89 = 1.67 bar and then we have

y1 

x1 1 P1sat

P 0.65 * 1.2467 * 1.24 y1   0.601 1.67

Surprisingly, even though species 1 has a higher pure component vapor pressure, its concentration in the vapor is lower than its concentration in the liquid. (c) We have an azeotrope when x1 = y1 and x2 = y2. We can therefore write

x1 P  x1 1 P1sat x2 P  x2  2 P2sat which is the modified Raoult’s law, with y1 and y2 replaced by x1 and x2, respectively. So, at the azeotrope,

P  1 P1sat P  2 P2sat so 1 P1sat   2 P2sat

exp1.8 x22  P1sat  exp1.8 x12  P2sat





exp 1.8 x22  x12   P2sat / P1sat

 



exp 1.8 1  x1   x12  P2sat / P1sat



1.8 1  2 x1  x   x 2 1

2 1

  ln P / P  sat 2

sat 1

1.81  2 x1   ln P2sat / P1sat  sat sat   1  ln  P2 / P1  1  ln 0.89 / 1.24    1  x1  1    0.592   2  1.8 1.8   2  2

The azeotrope occurs at x1 = 0.592. At this composition, the activity coefficients are 1 = exp(1.8*0.408 ) = 2

1.3491 and 2 = exp (1.8*0.592 ) = 1.8797. So, P = 0.592*1.3491*1.24 + 0.408*1.8797*0.89 = 1.673 bar

Solution 13.20

Problem Statement

For the acetone(1)/methanol(2) system, a vapor mixture for which z1 = 0.25 and z2 = 0.75 is cooled to temperature T in the two-phase region and flows into a separation chamber at a pressure of 1 bar. If the composition of the liquid product is to be x1 = 0.175, what is the required value of T, and what is the value of y1? For liquid mixtures of this system, to a good approximation, ln

1 = 0.64

x22 and ln 2 = 0.64 x12.

Solution

Since the liquid phase composition (x1 = 0.175) and pressure (P = 1 bar) for the separator are specified, we want to do a bubble point temperature calculation. As in the earlier problems, the easiest way to do this may be to set up a bubble point pressure calculation and then try different temperatures until we get the desired pressure. In this case, the equilibrium is described by the modified Raoult’s law: for all species i  1, 2, ..., N  . So, for a bubble point calculation, we have:

yi P  x i  i Pi sat N

i 1

 yi P   xi  i Pi sat N

P   xi i Pi sat i 1

Solution continued on next page…

sat

For a 2-component system, this is just P = x11P1

sat

+ x22P2

. So, using the known liquid-phase mole fractions we

compute the activity coefficients from 1 = exp(0.64 x2 ) and 2 = exp(0.64 x1 ), and compute the vapor pressures from Raoult’s law (at a guessed temperature). Then, we put it all together to calculate P, and try different temperatures until we get P = 1 bar = 100 kPa. The result of doing this is shown in the spreadsheet below, where I have tried different temperatures until I found that 72.9 °C gives a total pressure of 100 kPa. The vapor phase mole fractions are y1 = 0.47, y2 = 0.53. Acetone (1) A B 14.3145

C 2756.22

Methanol(2) A B C 228.06 16.5785 3638.27

T (deg C) P1sat (kPa) P2sat (kPa) 59.5 113.4 82.4

x1 0.175

x2 0.825

239.5 g1 1.55

P (kPa) g2 1.02 100

y1 0.307

y2 0.693

Oddly, the problem statement doesn’t ask about the fraction of the stream leaving the separator as a liquid and as a vapor. We can find the vapor fraction by requiring that the overall mole fraction of acetone be 0.25: x1V + y1(1-V) = z1, or 0.175V + 0.307 – 0.307 V = 0.25, or V = (0.307 – 0.25)/(0.307 – 0.175) = 0.431. 43% of the stream leaves as a vapor, and the other 57% leaves as a liquid.

Solution 13.21

Problem Statement

The following is a rule of thumb: For a binary system in VLE at low pressure, the equilibrium vapor-phase mole fraction y1 corresponding to an equimolar liquid mixture is approximately

y1 

P1sat P1sat  P2sat

where Pi sat is a pure-species vapor pressure. Clearly, this equation is valid if Raoult’s law applies. Prove that it is also valid for VLE described by Eq. (13.19), with ln 1 = A x 22 and ln

2 = A x1 .

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Solution

By Eq. (13.19) and the given equations for ln 1

 Ax  P

y1 P  x1

2 2

sat 1

2

y2 P  x2

 Ax  P

 x2

 Ax  P

2 1

sat 2

These equations sum to give: P  x1

 Ax P 2 2

sat 1

2 1

sat 2

Dividing the equation for y1 P by the preceding equation yields:

y1 

x1 exp Ax 22  P1sat x1 exp Ax 22  P1sat  x 2 exp  Ax 12  P2sat

For x1  x2 this equation obviously reduces to:

P

P1sat P1sat  P2sat

Solution 13.22

Problem Statement

A process stream contains light species 1 and heavy species 2. A relatively pure liquid stream containing mostly 2 is desired, obtained by a single-stage liquid/vapor separation. Specifications of the equilibrium composition are: x1 = 0.002 and y1 = 0.950. Use data given below to determine T (K) and P (bar) for the separator. Assume that Eq. (13.19) applies; the calculated P should be used to validate this assumption. Data: For the liquid phase, ln

1 = 0.93

x22

2 = 0.93

x12

ln Pi sat B  Ai  i bar T /K

A1 = 10.08, B1 = 2572.0, A2 = 11.63, B2 = 6254.0 Eq. 13.19:

i  1, 2, . . . , N 

yi P  x i  i Pi sat

Solution

x1 

y1 

T

Psat 1 T   e

x2   x1

A1  A1

B1 T /K

y2   y1

B1 

A2 

Psat 2 T   e 2

1  e 0.93 x2

B2 

B2 T/K

2  e 0.93 x1

Given that Psat 1 T  Psat 2 T 



x 2 y1 1 x 1 y2  2

Determine what T is: T  376.453 K

Now determine what P is

P

x1 1 Psat 1 T  y1

P

bar

Solution 13.23

Problem Statement

If a system exhibits VLE, at least one of the K-values must be greater than 1.0 and at least one must be less than 1.0. Offer a proof of this observation.

Solution

A little reflection should convince anyone that there is no other way that BOTH the liquid-phase and vapor-phase mole fractions can sum to unity.

Solution 13.24

Problem Statement

Flash calculations are simpler for binary systems than for the general multicomponent case because the equilibrium compositions for a binary are independent of the overall composition. Show that, for a binary system in VLE,

x1  

1 K2 K1  K2

y1 =

K1 1  K 2  K1  K 2

z1 ( K 1  K 2 )  (1  K 2 ) ( K1  1)(1  K 2 )

Solution

By the definition of a K-value, y1  K 1 x 1

Substitute for x1

y1  y2 

These equations combine to yield:

K1 x  K 2   x 1  

K 1 x1  K 2 x 2  Solve for x1 x 1 

y2  K 2 x 2

1  K2 K1  K 2

y1  K1 x1:

y1  Note that when two phases exist both x1

K1   K 2  K1  K 2 z 1.

Solution continued on next page…

By a material balance on the basis of 1 mole of feed,

x1   y1  z1 or x1 1–    y1   z1 Substitute for both x 1

y1 by the equations derived above:

K 1– K 2  1 – K2   z1 1     1 K1 – K 2 K1 – K 2 Solve this equation for  :  

z1  K 1  K 2   K 2 

 K1    K 2 

Note that the relative amounts of liquid and vapor phases do depend on z1.

Solution 13.25

Problem Statement

The NIST Chemistry WebBook reports critically evaluated Henry’s constants for selected chemicals in water at 25°C. Henry’s constants from this source, denoted here by kHi, appear in the VLE equation written for the solute in the form: mi  kHi yi P

where mi is the liquid-phase molality of solute species i, expressed as mol i / kg solvent. (a) Determine an algebraic relation connecting kHi to

i, Henry’s constant in Eq. (13.26). Assume that xi is “small.” 1

(b) The NIST Chemistry WebBook provides a value of 0.034 mol·kg ·bar implied value of

for kHi of CO2 in H2O at 25°C. What is the

i in bar? Compare this with the value given in Table 13.2, which came from a different source.

Eq. 13.26:

fˆil  x i

Table 13.2: Henry’s Constants for Gases Dissolved in Water at 25°C

Gas

/bar

Acetylene Air Carbon dioxide Carbon monoxide Ethane Ethylene Helium Hydrogen Hydrogen sulfide Methane Nitrogen Oxygen

1,350 72,950 1,670 54,600 30,600 11,550 126,600 71,600 550 41,850 87,650 44,380

Solution

(a)

For the formulation of Henry’s law used in the NIST WebBook, kiyiP = Mi, we can write yiP = Mi/ki.

For the formulation of Henry’s law used in SVNA, we have yiP = xiMi. Comparing these, we see that xiMi = Mi/ki, or Mi = Mi/(kixi). If ni is the number of moles of solute, and n is the total number of moles, then xi = ni/n, and Mi = ni/((n – ni)Ms), where Ms is the molar mass of the solvent and the number of moles of solvent is equal to the total number of moles minus the number of moles of solute. Substituting these into the above relationship for Mi gives

 i  i  ki x i

M s n  ni  ki ni



n 1  ki M s n  ni  ki M s 1  x i 

So, if xi is “small” (much less than 1), we have

i 

1 ki M s

(b) Applying the above formula for CO2 in H2O, with ki = 0.034 mol kg

i 

0.034 mol kg

1

bar

and Ms = 0.01802 kg mol , we have

1  1632 bar bar 1 0.01802 kg mol1

This is about 2% lower than the value in table 10.1, which I would consider to be very good agreement.

Solution 13.26

Problem Statement

(a) A feed containing equimolar amounts of acetone(1) and acetonitrile(2) is throttled to pressure P and temperature T. For what pressure range (atm) will two phases (liquid and vapor) be formed for T = 50°C? Assume that Raoult’s law applies. (b) A feed containing equimolar amounts of acetone(1) and acetonitrile(2) is throttled to pressure P and temperature T. For what temperature range (°C) will two phases (liquid and vapor) be formed for P = 0.5(atm)? Assume that Raoult’s law applies.

Solution 

Acetone: Psat 1 T   e

2756.22 T 228.06 deC



Acetonitrile: Psat 2 T   e

3413.1 T 250.523 deC

(a): Find BUBL P and DEW P values T

x1 

y1 

BUBL P  x1 Psat 1 T     x1  Psat 2 T  DEW P 

BUBL P 

DEW P 

1  y1   Psat 1 T  Psat 2 T  y1

atm atm

At T = 50 C two phases will form between P = 0.478 atm and 0.573 atm

(b): Find BUBL T and DEW T values

P

atm

x1 

y1 

Iteratively solved for the temperature until the BUBL P = 0.5

x1 Psat 1 T   1  x1  Psat 2 T   P BUBL T  46.316 C

Iteratively solve for the temperature to where DEW P = 0.5

x1 Psat 1 T     x1  Psat 2 T   P

  x1  Psat 2 T     y1 P

DEW T  51.238 C

At P = 0.5 atm, two phases will form between T = 46.3 C and 51.2 C

Solution 13.27

Problem Statement A binary mixture of benzene(1) and toluene(2) is flashed to 75 kPa and 90°C. Analysis of the effluent liquid and vapor streams from the separator yields: x1 = 0.1604 and y1 = 0.2919. An operator remarks that the product streams are “off-spec,” and you are asked to diagnose the problem. (a) Verify that the exiting streams are not in binary equilibrium. (b) Verify that an air leak into the separator could be the cause.

Solution

Calculate x and y at T 

13.7819

Benzene: Psat 1 T   e

13.932 

Toluene: Psat 2 T   e

P

kPa

2726.81 T  217.572 deC

3056.96 T  217.625 deC

(a): Calculate the equilibrium composition of the liquid and vapor at the flash T and P T

P

kPa

Using the equations

x1 Psat 1 T   y1 P

  x1  Psat 2 T     y1  P

Iteratively find x1 and y1 x1 

y1 

The equilibrium compositions do not agree with the measured values.

(b): Assume that the measured values are correct. Since air will not dissolve in the liquid to any significant extent, the mole fractions of toluene in the liquid can be calculated. x1 

y1 

x 2   x1

x 2  8396

Now calculate the composition of the vapor. y3 represents the mole fraction of air in the vapor.

  x1  Psat 2 T     y1  y3  P

y1  y2  y2 

Solve this iteratively for y2 and y3 y2 

y3 

Conclusion: An air leak is consistent with the measured compositions.

Solution 13.28

Problem Statement

Ten (10) kmol·hr

of hydrogen sulfide gas is burned with the stoichiometric amount of pure oxygen in a special unit.

Reactants enter as gases at 25°C and 1(atm). Products leave as two streams in equilibrium at 70°C and 1(atm): a phase of pure liquid water, and a saturated vapor stream containing H2O and SO2. (a) What is the composition (mole fractions) of the product vapor stream? 1

(b) What are the rates (kmol·hr ) of the two product streams?

Solution

H2S + 3/2 O2 –> H2O + SO2 By a stoichiometric balance, calculate the following total molar flow rates Feed: n H S  2

kmol h

Products: n SO2  n H2 S

3 n O  n H S  2 2 2

kmol h

n H2 O  n H2 S

Exit conditions: 16.3872

P  atm

T

Psat H2O  e

3885.7 T 230.17 deC

Solution continued on next page…

(a): Calculate the mole fraction of H2O and SO2 in the exiting vapor stream assuming vapor is saturated with H2O

yH2O vap 

Psat H2O T2 

y H2O vap 

ySO2   yH2O vap

ySO2 

(b): Calculate the vapor stream molar flow rate using balance on SO2 n vap 

n SO2

kmol h

n vap 

ySO2

Calculate the liquid H2O flow rate using balance on H2O n H O vap  n vap * y H O vap 2

n H O vap 

n H O liq  n H O  n H O vap 2

n H O liq  2

kmol h kmol h

Solution 13.29 Problem Statement

Physiological studies show the neutral comfort level (NCL) of moist air corresponds to an absolute humidity of about 0.01 kg H2O per kg of dry air.

(a) What is the vapor-phase mole fraction of H2O at the NCL? (b) What is the partial pressure of H2O at the NCL? Here, and in part (c), take P = 1.01325 bar. (c) What is the dewpoint temperature (°F) at the NCL?

Solution

(a) 0.01 kg of H2O / 0.01802 kg/mol = 0.5549 mol H2O and 1 kg dry air / 0.02885 kg/mol = 34.662 mol dry air. So, the mole fraction H2O is xH O = 0.5549/(34.662 + 0.5549) = 0.0158. 2

(b) The partial pressure, is just the mole fraction times the total pressure: PH O = 0.0158*1.01325 = 0.0160 bar = 2

16 kPa = 0.232 psia. (c) The dew point temperature is just the temperature where the vapor pressure of water is 0.232 psia. Checking the steam tables on p. 754, we see that this is between 56 and 58 ºF. Interpolating between the two corresponding vapor pressures of 0.2218 psia at 56 ºF and 0.2384 at 58 ºF, we find Tdew = 56 + 2*(0.232 – 0.2218)/(0.2384 – 0.2218) = 57.2 ºF

Solution 13.30

Problem Statement

An industrial dehumidifier accepts 50 kmol·hr

of moist air with a dewpoint of 20°C. Conditioned air leaving the 1

dehumidifier has a dewpoint temperature of 10°C. At what rate (kg·hr ) is liquid water removed in this steady-flow process? Assume P is constant at 1(atm).

Solution

kmol h

n 1 

Tdp1  

PsatH O T   e

3885.7 T  230.17 deC

y1 

PsatH2 O Tdp1  P

Tdp 2 

P  atm

g mol

MWH O  2

PsatH2O Tdp 2 

y1 

y2 

n 2liq  n1

n 2 vap  n1

y2 

By a mole balance on the process

n1 y1  n 2 vap y2  n 2liq

H2O balance

n1  n 2 vap  n 2liq

Overall balance n 2 vap 

kmol h

m 2liq  n 2liq * MWH O 2

n 2liq 

kmol h

m 2liq 

kg h

Solution 13.31

Problem Statement

Vapor/liquid-equilibrium azeotropy is impossible for binary systems rigorously described by Raoult’s law. For real systems (those with i

1), azeotropy is inevitable at temperatures where the P is at are equal. Such a temperature is

called a Bancroft point. Not all binary systems exhibit such a point. With Table B.2 of App. B as a resource, identify three binary systems with Bancroft points, and determine the T and P coordinates. Ground rule: A Bancroft point must lie in the temperature ranges of validity of the Antoine equations.

Solution

Start with Benzene and Cyclohexane:

Benzene: A1 

B1 

Cyclohexane: A2 

C1  B1 

C1 

A1 

Psat 1 T   e

B1 T C1 deC

A2 

Psat 2 T   e

kPa

T

B2 T C2 deC

kPa

Psat 1 T   Psat 2 T  Solution continued on next page…

The Bancroft point for this system is:

Psat 1 T  

kPa

Component 1

Component 2

T (deg C)

P (kPA)

Benzene

Cyclohexane

52.3

39.6

2-butanol

water

87.7

64.2

Acentonitrile

Ethanol

65.8

60.6

T

Solution 13.32

Problem Statement

The following is a set of VLE data for the system methanol(1)/water(2) at 333.15 K:

P/kPa x1 19.953 39.223 42.984 48.852 52.784 56.652

0.0000 0.1686 0.2167 0.3039 0.3681 0.4461

y1 0.0000 0.5714 0.6268 0.6943 0.7345 0.7742

P/kPa x1

60.614 63.998 67.924 70.229 72.832 84.562

0.8085 0.8383 0.8733 0.8922 0.9141 1.0000

0.5282 0.6044 0.6804 0.7255 0.7776 1.0000

Extracted from K. Kurihara et al., J. Chem. Eng. Data, vol. 40, pp. 679–684, 1995.

(a) Basing calculations on Eq. (13.24), find parameter values for the Margules equation that provide the best fit of E

(d) Using Barker’s method, find parameter values for the Margules equation that provide the best fit of the P–x1 data. Prepare a diagram showing the residuals

and

1 plotted vs. x1.

(e) Repeat (d) for the van Laar equation. (f) Repeat (d) for the Wilson equation.

Eq. 13.24: i 

yi P x i fi



yi P x i Pi sat

Solution

In parts (a), (b), and (c), we find the best fit to the excess Gibbs energy using each of three models, while in parts (d), (e), and (f) we find the best fit to the pressure vs. x1 curve. Links will be provided here to directly download spreadsheets for each of the problem parts.

(a) We compute the activity coefficients from the pressure and composition, then compute the excess Gibbs energy E

Solution continued on next page…

Problem 12.1(a) Data Fit 0.7000

y = -0.2075x + 0.6828 R² = 0.828

0.6500

GE/(x1x2RT)

0.6000

0.5500

0.5000

0.4500

0.4000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

Using the parameters determined from the above plot, we can go back and look at the activity coefficients and the Gibbs energy, which are given by ln 1  x 22  A12  2 A21  A12  x1  ln  2  x 12  A21  2 A12  A21  x 2 

and

GE  x 1 x 2  A21 x1  A12 x 2  RT

Problem 12.1(a), Fit Results 0.7000

ln(g 1),ln(g 2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

Finally, we can construct a P-x-y diagram from the Margules equation using P  x1 1 P1sat  x 2  2 P2sat y1 

x 1 1 P1sat x 1 1 P1sat  x 2  2 P2sat

The diagram constructed in this way, and compared with the original data, is shown below:

P-x-y diagram with Margules Fit for Problem 12.1a 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1 E

(b) Here, we will do the same thing with the Van Laar model, for which we fit a straight line through (RTx1x2)/G . The plot for doing the fitting is shown here:

Problem 12.1(b) Data Fit 2.2000

y = 0.6414x + 1.4185 R² = 0.8148 2.0000

(x1x2RT)/GE

1.8000 1.6000 1.4000 1.2000 1.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

Methanol Mole Fraction

In this case, the Van Laar plot doesn’t look much better or worse than the Margules plot. From this fit, we get the Van Laar parameters as A 21= 0.4855 and A 12 = 0.7050. The activity coefficients are then given by 2

 A x  ln 1  A21 1  12 1   A21 x2  2  A21 x 2     ln  2  A12 1    A12 x1  E

Computing and plotting these, as well as G /(RT) gives

Problem 12.1(b), Fit Results 0.8000

ln(g 1),ln(g2), and GE/(RT)

0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

Finally, using these activity coefficients and the same equations as in part (a) to construct the P-x-y diagram gives the following.

Solution continued on next page…

P-x-y diagram with Van Laar Fit for Problem 12.1(b) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

This isn’t easily converted to a form for which we can fit a straight line or other simple function (like a polynomial) using linear regression. That is, we can’t readily write it in a form that is linear in the coefficients that we want to determine (21 and 12). So, instead, we need to do nonlinear regression to the original functional form. For a given E

set of parameter values, we compute G /(RT) at each experimental value of x1 using the above equation.

Solution continued on next page…

We then minimize the sum of the squares of the differences between the values computed and the experimental E

values of G /(RT). The resulting fit of G /(RT) is shown here:

Data Fit for Problem 12.1(c) 0.1600 0.1400 0.1200

GE/(RT)

0.1000 0.0800 0.0600 0.0400 0.0200 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

The best fit values for the Wilson parameters obtained from this fit are 12 = 0.4763 and 21= 1.0262.

The activity coefficients for the Wilson equation are given by   12 21  ln 1   ln  x1  x2 12   x 2    x1  x2 12 x2  x1 21    12 21  ln  2   ln  x2  x1  21   x1    x1  x2 12 x2  x121 

Solution continued on next page…

Computing these and comparing them to the data gives

Problem 12.1(c), Fit Results 0.7000

ln(g1),ln(g2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Methanol Mole Fraction

Finally, the fit to the P-x-y diagram is shown here:

P-x-y diagram with Wilson Fit for Problem 12.1(c) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Solution continued on next page…

(d) Here, we repeat the fit to the Margules equation, but instead of plotting G /(x1x2RT) and fitting a straight line through it, we directly fit the P vs. x1 data. We guess values of A21 and A12, then use them to compute the activity coefficients and the pressure for each experimental value of x1. Then we take the difference between the experimental and computed pressure at each value of x1 and vary the model parameters to minimize the sum of the squares of the differences. Doing so using the spreadsheet gives A21 = 0.4346 and A12 = 0.7577 (in part (a), we got A21 = 0.4573 and A12 = 0.6828).

Solution continued on next page…

The new values of the parameters give the following fits to the activity coefficients and the P-x-y diagram:

Problem 12.1(d), Fit Results 0.7000

ln(g 1),ln(g 2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

P-x-y diagram with Margules Fit for Problem 12.1d 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Solution continued on next page…

For this part, we were asked to plot the residuals of P and y1 as well. Those plots are shown below:

Problem 12.1 (d) - Pressure Residuals 0.400

0.300

d (P) kPa

0.200

0.100

0.000

-0.100

-0.200

-0.300

-0.400 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Problem 12.1 (d) - Vapor Composition Residuals 0.0080

0.0060

d (y1)

0.0040

0.0020

0.0000

-0.0020

-0.0040

-0.0060 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Solution continued on next page…

(e) As in part (d), in this part we minimize the sum of the squares of the difference between the computed and measured pressure, this time using the van Laar equation for the activity coefficients. When this is done using the spreadsheet the parameter values that give the best fit are A 21 = 0.4677 and A 12 = 0.8301. As was the case for the Margules equation, these differ a bit from the values we obtained in part (b) when fitting the excess Gibbs energy rather than the pressure. The resulting fits are shown here:

Problem 12.1(e), Fit Results 0.9000

ln(g1),ln(g2), and GE/(RT)

0.8000 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol Mole Fraction

P-x-y diagram with Van Laar Fit for Problem 12.1(e) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

The residuals of pressure and vapor composition are shown below. These are a little higher than for the Margules equation. So, while all of the models provide a good fit, this (along with the higher sum of squares of the errors) shows that we cannot fit the data quite as well with the Van Laar equation as with the Margules equations.

Problem 12.1 (e) - Pressure Residuals 0.400

0.300

d (P) kPa

0.200

0.100

0.000

-0.100

-0.200

-0.300

-0.400 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Problem 12.1 (e) - Vapor Composition Residuals 0.0100

0.0080

0.0060

d (y1)

0.0040

0.0020

0.0000

-0.0020

-0.0040

-0.0060 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Solution continued on next page…

(f) Finally, here we fit the Wilson equation parameters to minimize the sum of the squares of the differences between computed and experimental pressures rather than excess Gibbs energies. Using the spreadsheet, the resulting Wilson parameters are 12 = 0.3483 and 21= 1.1982. Again these differ a bit from the values obtained by fitting the excess Gibbs energy. The resulting fits to the activity coefficients and P-x-y diagram are shown below.

Problem 12.1(f), Fit Results 0.7000

ln(g 1),ln(g 2), and GE/(RT)

0.6000 0.5000 0.4000 0.3000 0.2000 0.1000 0.0000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

Methanol Mole Fraction

P-x-y diagram with Wilson Fit for Problem 12.1(f) 90.000 80.000 70.000

P(kPa)

60.000 50.000 40.000 30.000 20.000 10.000 0.000 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

x1 or y1

The plots of pressure and vapor mole fraction residuals are shown below. The fit is about the same as the Van Laar equation, and not as good as the Margules equation.

Problem 12.1 (f) - Pressure Residuals 0.600

0.400

d (P) kPa

0.200

0.000

-0.200

-0.400 0.0000

0.2000

0.4000

0.6000

0.8000

1.0000

1.2000

-0.600

Problem 12.1 (f) - Vapor Composition Residuals 0.0120 0.0100 0.0080

d (y1)

0.0060 0.0040 0.0020 0.0000 -0.0020 -0.0040 -0.0060 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Note: Spreadsheets were also used in solving this problem.

Solution 13.33

Problem Statement

If Eq. (13.24) is valid for isothermal VLE in a binary system, show that:

 dP   dP      P2sat   P1sat  dx   1  x1  0  dx 1 x1 1

Eq. 13.24: i 

yi P x i fi



yi P x i Pi sat

Solution

Equation (13.19)) may be written: yi P  xi i Pi sat. Summing for i 

P  x1 1 P1sat  x2 2 P2sat.

Differentiate at constant T :

 d   d  dP  P1sat  x1 1   P2sat  x2 2  2   dx1  dx1  dx1 

Apply this equation to the limiting conditions:

For

x1 

x2   1  1  2 

For

x1 

x2  1   2   2

d 2 dx1 d1 dx1



Solution continued on next page…

Then,  dP   dP      P1sat 1  P2sat or   P2sat  P1sat 1  dx    dx  1 x1 0  1 x1 0  dP   dP      P1sat  P2sat  2 or   P1sat  P2sat  2  dx    dx  1 x1 1  1 x1 1

Since both Pi sat

i are always positive definite, it follows that:  dP     P2sat  dx   1 x1 0

 dP     P1sat   dx1 x1 1

Solution 13.34

Problem Statement

The following is a set of VLE data for the system acetone(1)/methanol(2) at 55°C:

P/kPa x1 68.728 72.278 75.279 77.524 78.951 82.528 86.762 90.088 93.206 95.017 96.365

0.0000 0.0287 0.0570 0.0858 0.1046 0.1452 0.2173 0.2787 0.3579 0.4050 0.4480

P/kPa

0.0000 0.0647 0.1295 0.1848 0.2190 0.2694 0.3633 0.4184 0.4779 0.5135 0.5512

97.646 98.462 99.811 99.950 100.278 100.467 100.999 101.059 99.877 99.799 96.885

x1 0.5052 0.5432 0.6332 0.6605 0.6945 0.7327 0.7752 0.7922 0.9080 0.9448 1.0000

y1 0.5844 0.6174 0.6772 0.6926 0.7124 0.7383 0.7729 0.7876 0.8959 0.9336 1.0000

D. C. Freshwater and K. A. Pike, J. Chem. Eng. Data, vol. 12, 79–183, 1967.

(a) Basing calculations on Eq. (13.24), find parameter values for the Margules equation that provide the best fit of E

G RT to the data, and prepare a Pxy diagram that compares the experimental points with curves determined from the correlation. (b) Repeat (a) for the van Laar equation. (c) Repeat (a) for the Wilson equation. (d) Using Barker’s method, find parameter values for the Margules equation that provide the best fit of the P–x1 data. Prepare a diagram showing the residuals

and

1 plotted vs. x1.

(e) Repeat (d) for the van Laar equation. (f) Repeat (d) for the Wilson equation.

Eq. 13.24: i 

yi P x i fi



yi P x i Pi sat

Solution

In parts (a), (b), and (c), we find the best fit to the excess Gibbs energy using each of three models, while in parts (d), (e), and (f) we find the best fit to the pressure vs. x1 curve.

(a) We compute the activity coefficients from the pressure and composition, then compute the excess Gibbs energy E

from the activity coefficients, then plot G /(x1x2RT) and fit a line through it. The plot of G /(x1x2RT) vs. x1 with the line fit through it is shown below. The Margules coefficients determined from this are A21 = 0.7077 – 0.0182 = 0.6895 and A12 = 0.7077. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Problem 12.3(a) Data Fit 1.2000

y = -0.0182x + 0.7077 2 R = 0.0034 1.0000

GE/(x1x2RT)

0.8000

0.6000

0.4000

0.2000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

and

GE  x 1 x 2  A21 x1  A12 x 2  RT

Solution continued on next page…

Problem 12.3(a), Fit Results 0.8000

ln(g1),ln(g2), and G /(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

Finally, we can construct a P-x-y diagram from the Margules equation using P  x1 1 P1sat  x 2  2 P2sat y1 

x 1 1 P1sat x 1 1 P1sat  x 2  2 P2sat

The diagram constructed in this way, and compared with the original data, is shown below:

P-x-y diagram with Margules Fit for Problem 12.3a 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

As you can see, the fit is reasonable, but not great. We might improve it by either fitting a different model to the excess Gibbs energy (in parts (b) and (c)) or doing the fit directly to the P vs. x1 data (in part (d)) or both (in parts (e) and (f)).

(b) Here, we will do the same thing with the Van Laar model, for which we fit a straight line through (RTx1x2)/G . The plot for doing the fitting is shown here:

Problem 12.3(b) Data Fit 1.8000 1.6000 1.4000

(x1x2RT)/GE

1.2000

y = 0.015x + 1.4424 2 R = 0.0008

1.0000 0.8000 0.6000 0.4000 0.2000 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

In this case, the Van Laar plot doesn’t look much better or worse than the Margules plot. From this fit, we get the Van Laar parameters as A 21= 0.6862 and A 12 = 0.6933. The activity coefficients are then given by 2

 A x  ln 1  A21 1  12 1  A21 x 2   2  A x  ln  2  A12 1  21 2  A12 x1  

Solution continued on next page…

Computing and plotting these, as well as G /(RT) gives

Problem 12.3(b), Fit Results 0.8000

ln(g1),ln(g2), and G /(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

Finally, using these activity coefficients and the same equations as in part (a) to construct the P-x-y diagram gives the following.

P-x-y diagram with Van Laar Fit for Problem 12.3(b) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Solution continued on next page…

The fit isn’t much different than what we obtained with the Margules equations (neither much better nor much worse). (c) For the Wilson model, we have GE  x1 ln  x1  x 2 12   x2 ln  x 2  x121  RT

set of parameter values, we compute G /(RT) at each experimental value of x1 using the above equation. We then minimize the sum of the squares of the differences between the values computed and the experimental values of E

G /(RT). The resulting fit of G /(RT) is shown here:

Data Fit for Problem 12.3(c) 0.1800 0.1600 0.1400

GE/(RT)

0.1200 0.1000 0.0800 0.0600 0.0400 0.0200 0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

The best fit values for the Wilson parameters obtained from this fit are 12 = 0.7100 and 21 = 0.6806. The activity coefficients for the Wilson equation are given by   12 21  ln 1   ln  x1  x2 12   x 2    x1  x2 12 x2  x1 21    12 21  ln  2   ln  x2  x1  21   x1    x1  x2 12 x2  x121 

Computing these and comparing them to the data gives

Problem 12.3(c), Fit Results 0.7000

0.5000

ln(g1),ln(g2), and G /(RT)

0.6000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

Finally, the fit to the P-x-y diagram is shown here:

P-x-y diagram with Wilson Fit for Problem 12.3(c) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Solution continued on next page…

This looks slightly, but not dramatically, better than the fits obtained with the Margules and van Laar equations. E

(d) Here, we repeat the fit to the Margules equation, but instead of plotting G /(x1x2RT) and fitting a straight line through it, we directly fit the P vs. x1 data. We guess values of A21 and A12, then use them to compute the activity coefficients and the pressure for each experimental value of x1. Then we take the difference between the experimental and computed pressure at each value of x1 and vary the model parameters to minimize the sum of the squares of the differences. Doing so using the spreadsheet gives A21 = 0.6443 and A12 = 0.6716. These are lower than the values obtained in part (a) by about 5% and 7%, respectively. They give the following fits to the activity coefficients and the P-x-y diagram: Problem 12.3(d), Fit Results 0.8000

ln(g1),ln(g2), and G /(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

P-x-y diagram with Margules Fit for Problem 12.3d 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

These fits look a little better than the ones obtained in part (a). (e) As in part (d), in this part we minimize the sum of the squares of the difference between the computed and measured pressure, this time using the van Laar equation for the activity coefficients. When this is done, using the spreadsheet, the parameter values that give the best fit are A 21= 0.6725 and A 12 = 0.6438. As was the case for the Margules equation, this are similar to, but slightly lower than, the values we obtained in part (b) when fitting the excess Gibbs energy rather than the pressure. The resulting fits are shown here:

Problem 12.3(e), Fit Results 0.8000

ln(g1),ln(g2), and GE/(RT)

0.7000

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

P-x-y diagram with Van Laar Fit for Problem 12.3(e) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

These look slightly better than the fits obtained in part (b).

(f) Finally, here we fit the Wilson equation parameters to minimize the sum of the squares of the differences between computed and experimental pressures rather than excess Gibbs energies. Using the spreadsheet, the resulting Wilson parameters are 12 = 0.7316 and 21=0.6632. Again these differ by a few percent from the values obtained by fitting the excess Gibbs energy. The resulting fits to the activity coefficients and P-x-y diagram are shown below.

Problem 12.3(f), Fit Results 0.7000

ln(g1 ),ln(g2), and GE/(RT)

0.6000

0.5000

0.4000

0.3000

0.2000

0.1000

0.0000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone Mole Fraction

P-x-y diagram with Wilson Fit for Problem 12.3(f) 105.000

100.000

95.000

P(kPa)

90.000

85.000

80.000

75.000

70.000

65.000

60.000 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

x1 or y1

Solution 13.35

Problem Statement The excess Gibbs energy for binary systems consisting of liquids not too dissimilar in chemical nature is represented to a reasonable approximation by the equation: E

G /RT = A x1 x2 where A is a function of temperature only. For such systems, it is often observed that the ratio of the vapor pressures of the pure species is nearly constant over a considerable temperature range. Let this ratio be r, and determine the range of values of A, expressed as a function of r, for which no azeotrope can exist. Assume the vapor phase to be an ideal gas.

Solution

By Eqs. (13.37 and 13.38),

Therefore, ln

By Eq 13.24

1 2

ln 1  Ax22

ln 2  Ax12

 A x22  x12   A x2  x1   A1  2 x1 

y1 / x1 P2sat 1 y x P sat  1 2 2sat   12 r 2 y2 / x2 P1sat y2 x1 P1

Solution continued on next page…

x 1 ln r  A  x 1az 

If an azeotrope exists, 12  at  x 1az  The quantity A  x1 

x1 A  ln r

A  ln r

Solution 13.36 Problem Statement For the ethanol(1)/chloroform(2) system at 50°C, the activity coefficients show interior extrema with respect to composition [see Fig. 13.4(e)]. (a) Prove that the van Laar equation cannot represent such behavior. (b) The two-parameter Margules equation can represent this behavior, but only for particular ranges of the ratio A21A12. What are they?

Solution

(a) Perhaps the easiest way to proceed here is to note that an extremum in ln 1 is accompanied by the opposite extremum in ln 2. Thus the difference ln 1

ln 2 is also an extremum, and Eq. (13.56)

becomes useful: d G /RT   ln 1  ln  2  ln 1  2 dx1 E

Thus, given an expression for

GE  g  x1  , we locate an extremum through: RT     G E   dln 1  d 2   RT    2   0 2 dx1 dx 1

For the van Laar equation, write Eq. (13.42), omitting the primes ( ‘): x x GE  A12 A21 1 2 RT A

A  A12 x 1  A21 x 2

Moreover, dA  A12  A21 dx 1

d2 A  dx 12

Then

dG E /RT  dx1

x  x x x dA  1   A12 A21  2  1 22  A A dx1 

Solution continued on next page…

 G E   d 2   RT  dx12

 2 x  x dA x 1 x 2 d 2 A dA  2 x1 x 2 dA x  x   A12 A21   2 2 1  2   3  2 2 1  2 dx1 dx 1  A A dx1 A dx1 A   A 2   2 2 x 2  x 1  dA 2 x 1 x 2  dA    A12 A21      dx1 A2 A3  dx1    A   2   dA   2 A12 A21  dA 2    x 1 x 2   A   x 2  x1  A   dx1 A3   dx1    



2A12 A21  dA   x dA  A  A  x2  3  1 dx1 dx1  A  

This equation has a zero value if either A12 or A21 is zero. However, this makes GE/RT everywhere zero, and no extremum is possible. If either quantity in parentheses is zero, substitution for A and dA dx1

A12 

A21 

G E RT everywhere zero. We conclude that no values

   of the parameters exist that provide for an extremum in ln 1   2 

(b) The Margules equation is given by: GE   A12 x 1  A21 x1  x1 x 2 RT GE  Ax 1 x 2 RT

A  A12 x1  A21 x 2

dA  A12  A21 dx 1

d2 A  dx 12

Then,  G E   d    RT  dA    A x 2  x1   x1 x 2  dx1 dx1   G E   d 2   RT  dA dA d2 A   2 A  x  x  x  x  x x     2 1 2 1 1 2 dx1 dx1 dx12 dx12

 2 A  2 x2  x1 

  dA dA  2  x 2  x1   A dx1 dx1  

This equation has a zero value when the quantity in square brackets is zero. Then:

 x2  x1 

dA  A   x2  x1  A12  A21   A21 x 1  A12 x 2  A21 x 2  A12 x1   A21 x1  A12 x 2   dx1

Substituting x2  1  x1 and solving for x1 yields:

x1 

A21  2 A12

3 A21  A12 

x1 

r 2 3r  1

r

A21 A12

When r = 2, x1 = 0, and the extrema in ln 1 and ln 2 occur at the left edge of a diagram such as those of Fig. 12.9. For values of r > 2, the extrema shift to the right, reaching a limiting value for r =  at x1 = 1/3. For positive values of the parameters, in all of these cases A21 > A12, and the intercepts of the ln 2 curves at x1 = 1 are larger than the intercepts of the ln 1 curves at x1 = 0. When r = 1/2, x1 = 1, and the extrema in ln 1 and ln 2 occur at the right edge of a diagram such as those of Fig. 12.9. For values of r < 1/2, the extrema shift to the left, reaching a limiting value for r = 0 at x1 = 2/3. For positive values of the parameters, in all of these cases A21 < A12, and the intercepts of the ln 1 curves at x1 = 0 are larger than the intercepts of the ln 2 curves at x1 = 1.

No extrema exist for values of r between 1/2 and 2.

Solution 13.37

Problem Statement

VLE data for methyl tert-butyl ether(1)/dichloromethane(2) at 308.15 K are as follows: P/kPa x1 85.265 83.402 82.202 80.481 76.719 72.422 68.005 65.096

0.0000 0.0330 0.0579 0.0924 0.1665 0.2482 0.3322 0.3880

y1 0.0000 0.0141 0.0253 0.0416 0.0804 0.1314 0.1975 0.2457

P/kPa x1

59.651 56.833 53.689 51.620 50.455 49.926 49.720 49.624

0.3686 0.4564 0.5882 0.7176 0.8238 0.9002 0.9502 1.0000

0.5036 0.5749 0.6736 0.7676 0.8476 0.9093 0.9529 1.0000

Mato, C. Berro, and A. Péneloux, J. Chem. Eng. Data, vol 36, pp. 259–262, 1991. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The data are well correlated by the three-parameter Margules equation [an extension of Eq. (13.39)]: GE  ( A21 x 1  A12 x 2  Cx1 x 2 ) x1 x 2 RT

Implied by this equation are the expressions:

ln1  x 22 [ A12  2( A21  A12  C ) x1  3Cx12 ] ln2  x12 [ A21  2( A12  A21  C ) x2  3Cx 22 ]

(a) Basing calculations on Eq. (13.24), find the values of parameters A12, A21, and C that provide the best fit of G /RT to the data. (b) Prepare a plot of ln 1, ln 2, and GE/(x1x2RT ) vs. x1 showing both the correlation and experimental values. (c) Prepare a Pxy diagram [see Fig. 13.8(a)] that compares the experimental data with the correlation determined in (a). (d) Prepare a consistency-test diagram like Fig. 13.9. (e) Using Barker’s method, find the values of parameters A12, A21, and C that provide the best fit of the P–x1 data. Prepare a diagram showing the residuals

and

1 plotted vs. x1.

Eq. 13.24: i 

yi P x i fi



yi P x i Pi sat

See Figure 13.8(a) and Figure 13.9 on next page… Solution after Figures…

Solution

(a) This is very similar to problems 12.1(a) and 12.3(a), except that this time we will use a slightly extended version of the Margules equations for which GE  x 1 x 2  A21 x 1  A12 x 2  Cx 1 x 2  RT

And ln 1  x22  A12  2 A21  A12  C  x1  3Cx12  ln  2  x12  A21  2 A12  A21  C  x2  3Cx22 

As in the earlier problems, we compute the activity coefficients from the pressure and composition, then compute the E

excess Gibbs energy from the activity coefficients, then plot G /(x1x2RT) and fit a curve through it. The difference is that this time instead of fitting a straight line through the data (as for the Margules equations) we will fit a quadratic equation through the data. To see how the coefficients in the above equation are related to the coefficients obtained E

by fitting a quadratic polynomial through G /(x1x2RT) we can re-write this as a function of only x1: GE  A21 x1  A12 x 2  Cx1 x2 x1 x2 RT GE  A21 x1  A12 1  x1   Cx1 1  x1  x1 x2 RT GE  Cx12   A21  A12  C  x1  A12 x1 x2 RT E

So, if we fit G /(x1x2RT) vs. x1 to a function of the form y = ax + bx + c, then we will have a = C, b = A21 – A12 – C, and c = A12, or A21 = a + b + c, A12 = c, and C = a.

Solution continued on next page…

Doing this fitting yields the following plot and result:

-0.35

-0.4

GE/(x1x2RT)

y = 0.0760x - 0.2393x - 0.3771 R2 = 0.9216 -0.45

-0.5

-0.55 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

x1 Using the above relationship between the coefficients in the fit above and the parameters in the excess Gibbs energy model gives: A21 = A12 = C = 0.076

Solution continued on next page…

(b) For this part, we simply evaluate the model and plot it along with the experimentally derived values to see how well it fits. The result is shown below: 0

GE/RT or ln(g1) or ln(g2)

-0.1

-0.2

-0.3

-0.4

-0.5

-0.6 0

0.2

0.4

0.6

0.8

Solution continued on next page…

(c) Finally, we use the activity coefficients shown above to compute the P-x-y diagram for comparison to the original data, as shown below: 90 85 80

P (kPa)

75 70 65 60 55 50 45 40 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

x1 or y1

(d) Consistency Test:





GERTi :  GeRT x1i x2i  GERTi    i 

       x 1i x 2 i      ln  1i   2 x x    2i  1i 2i  





Calculate mean absolute deviation of residuals

   GERT



4

   | ln  1  2|

(e) Barker’s Method by non-linear least squares: Margules Equation   A  2   A  A  C   x1  2 12 21 12  1  x1, x 2, A12 , A21, C  :  exp  x 2   2   3  C  x 1       A  2   A  A  C   x 2  2 21 12 21   2  x1, x 2, A12, A21, C exp:   x1   2    C  x    

Minimize sum of the squared errors using the Mathcad Minimize function.

A12 

Guesses:

C :  0.2

A21 

  x    x x , A , A , C   Psat    1i 1 1i 2i 12 21  1  SSE  A12 A21 C     Pi      x 2i   2  x1i , x2i , A12, A21, C   Psat 2  i     A   12   A  :  Minimize SSE , A , A , C   12 21  21   C   

 A  0.364  12     A   0.521   21    C   0.23     

Note: Spreadsheets were also used in solving this problem.

Solution 13.38

Problem Statement

Equations analogous to Eqs. (10.15) and (10.16) apply for excess properties. Because ln i is a partial property with E

respect to G RT, these analogous equations can be written for ln

1 and ln 2 in a binary system.

(a) Write these equations, and apply them to Eq. (13.42) to show that Eqs. (13.43) and (13.44) are indeed obtained. (b) The alternative procedure is to apply Eq. (13.7). Show that by doing so Eqs. (13.43) and (13.44) are again reproduced.

dM dx1

Eq. 10.16: M 2  M  x1

dM dx1

Eq. 13.7:   (nG E / RT )   ln  i    ni   P ,T ,n j

Eq. 13.42: A12 A21 GE  x1 x 2 RT A12 x 1  A21 x 2

Eq. 13.43: 2

 A x  ln  1  A12 1  12 1   A21 x2 

Eq. 13.44:

 A x  ln 2  A21 1  21 2   A12 x1 

Solution

Equations 13.03) and (11.16) here become: d G / RT  d G / RT  GE GE  x2 and ln  2   x1 RT dx1 RT dx1 E

ln 1 

(a) For simplicity of notation, omit the primes that appear on the parameters in Eqs. (12.16) and (12.17), and write Eq. (12.16) as: x x GE  A12 A21 1 2 where D  A12 x1  A21 x2 RT D

d G E / RT 

Then,

And

dx1

 x  x1 x1 x2   A12 A21  2  2  A12  A21   D  D  

x x  x  x1 x1 x2  ln 1  A12 A21  1 2  x2  2  2  A12  A21   D D   D 



 A12 A21  x1 x 22 2  x x  x  x x  A  A   1 2 2 1 2 12 21  D  D 

A12 A21 x22 D2 2 2 A12 A21 x2

 D  A12 x1  A21 x1  

A12 A21 x22 D2 2

2  D   A21 x2      A12   A12    D   A x  21

 A21 x2  A21 x1  2

 A x  A x  21 2   A12  12 1  A21 x2  

2

 A x  ln 1  A12 1  12 1  A21 x2  

The equation for ln 2 is derived in analogous fashion. Solution continued on next page…

(b) With the understanding that T

  (nG E / RT   1    n1   n 2

and Eq. (12.16) may be written: A A nn nG E  12 21 1 2 RT nD

n D  A12 n1  A21n2

Differentiation in accord with the first equation gives:    1 n  nD  ln 1  A12 A21n2   1 2     nD nD  n1   n2  

ln 1 

 A A x  A12 A21n2  n A x  1  1 A12   12 21 2 1  12 1  nD  nD D D   



A12 A21 x 2 D2

 D  A12 x1  

A12 A21 x2 D2

A21 x2 

A12 A212 x22 D2

The remainder of the derivation is the same as in Part (a).

Solution 13.39

Problem Statement The following is a set of activity-coefficient data for a binary liquid system as determined from VLE data:

0.0523 0.1299 0.2233 0.2764 0.3482 0.4187 0.5001

1.202 1.307 1.295 1.228 1.234 1.180 1.129

1.002 1.004 1.006 1.024 1.022 1.049 1.092

0.5637 0.6469 0.7832 0.8576 0.9388 0.9813

1 1.120 1.076 1.032 1.016 1.001 1.003

2 1.102 1.170 1.298 1.393 1.600 1.404

Inspection of these experimental values suggests that they are noisy, but the question is whether they are consistent, and therefore possibly on average correct. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(a) Find experimental values for GE∕RT and plot them along with the experimental values of ln

1 and ln 2 on a single

graph. E

(b) Develop a valid correlation for the composition dependence of G RT and show lines on the graph of part (a) that represent this correlation for all three of the quantities plotted there. (c) Apply the consistency test described in Ex. 13.4 to these data, and draw a conclusion with respect to this test. Note: Solutions to some of the problems of this chapter require vapor pressures as a function of temperature. Table B.2, Appendix B, lists parameter values for the Antoine equation, from which these can be computed.

Solution (a): Experimental values for

GE by the eqn: RT GE  x1 ln1   x 2 ln  2  RT

GE/RT

0.0523

1.202

1.002

0.0115

0.1299

1.307

1.004

0.0383

0.2233

1.295

1.006

0.0624

0.2764

1.228

1.024

0.0739

0.3482

1.234

1.022

0.0874

0.4187

1.18

1.049

0.0971

0.5001

1.129

1.092

0.1047

0.5637

1.12

1.102

0.1063

0.6469

1.076

1.17

0.1028

0.7832

1.032

1.298

0.0812

0.8576

1.016

1.393

0.0608

0.9388

1.001

1.6

0.0297

0.9813

1.003

1.404

0.0093

(b): Fit GE/RT data to Margules eqn. by linear least-squares procedure: GE Yi  RT x1 x 2

X i  x1

Determine the slope and intercept from this:

Slope = 0.247

intercept = 0.286

This corresponds the to A12 and A21 values by

A12  intercept 

A21  slope  A12 

1  x1 , x2   exp  x22  A12  2 A21  A12  x1      2  x1 , x2   exp  x12  A21  2 A12  A21  x2     









GE/RT  x1 , x 2   x1 ln 1  x1 , x2   x 2 ln 2  x1 , x2 

1.8000 1.6000

ϒ's and GE/RT

1.4000 1.2000

GE/RT

1.0000

GE/RT(x1,x2)

0.8000

ϒ1(x1,x2)

0.6000

ϒ2(x1,x2)

0.4000

ϒ1 ϒ2

0.2000 0.0000 0

0.2

0.4

0.6

0.8

Solution continued on next page…

(c): Calculate and plot the residuals for consistency test

GERT  GeRT  x1 , x2   GERT    x , x       1 1 2   1  ln 1  2  ln    ln    2    2  x1 , x 2  

del GE/RT

del ln

0.0033

0.098001

0.0023

0.00017

0.0032

0.0206

0.0030

0.026484

0.0029

0.01917

0.0022

0.006071

0.0022

0.028597

0.0015

0.00945

0.0008

0.00924

0.0003

0.00057

0.0001

0.01079

0.0001

0.027934

0.0004

0.16868

Solution continued on next page…

0.1

del lnϒ1/ϒ2

0.05 0 -0.05 -0.1 -0.15 -0.2 0

0.2

0.4

0.6

0.8

The average del GE/RT = 0.0011 and average del ln

0.0025

Based on the graph and mean absolute deviations, the data show a high degree of consistency

Solution 13.40

Problem Statement

Following are VLE data for the system acetonitrile(1)/benzene(2) at 45°C:

P/kPa

29.819 31.957 33.553 35.285 36.457 36.996 37.068

0.0000 0.0455 0.0940 0.1829 0.2909 0.3980 0.5069

0.0000 0.1056 0.1818 0.2783 0.3607 0.4274 0.4885

36.978 36.778 35.792 34.372 32.331 30.038 27.778

0.5458 0.5946 0.7206 0.8145 0.8972 0.9573 1.0000

0.5098 0.5375 0.6157 0.6913 0.7869 0.8916 1.0000

Extracted from I. Brown and F. Smith, Austral. J. Chem., vol. 8, p. 62, 1955.

The data are well correlated by the three-parameter Margules equation (see below from Prob. 13.37). E

(a) Basing calculations on Eq. (13.24), find the values of parameters A12, A21, and C that provide the best fit of G RT to the data. E

(b) Prepare a plot of ln 1, ln 2, and G x1x2 RT vs. x1 showing both the correlation and experimental values. (c) Prepare a Pxy diagram [see Fig. 13.8(a)] that compares the experimental data with the correlation determined in (a). (d) Prepare a consistency-test diagram like Fig. 13.9. (e) Using Barker’s method, find the values of parameters A12, A21, and C that provide the best fit of the P–x1 data. Prepare a diagram showing the residuals

and

1 plotted vs. x1.

Eq. 13.24: i 

yi P x i fi



yi P x i Pi sat

Prob 13.37: the three-parameter Margules equation [an extension of Eq. (13.39)]: GE  ( A21 x1  A12 x 2  Cx1 x 2 ) x1 x2 RT

Solution

Using the data above, determine experimental values of activity coefficients and excess Gibbs energy of the mixture of acetonitrile and benzene.

1 

y1 P x 1 Psat 1

2 

y2 P x 2 Psat 2

GE /RTx1 x 2 

GE  x1 RT

1   x 2  2 

GE /RT x1 x 2

(a) Fit GE/RT data to Margules eqn. by nonlinear least squares. Minimize sum of the squared errors using excel solver.

 GE  SSE  A12 , A21 , C       A21 x1  A12 x2  Cx1 x2  x1 x2    i  RT

Minimizing this by solving for A12 A21 and C yields

A12 

A21 

C

(b): Plot and fit the data

GE /RTx1 x 2  x1 , x 2   A21 x 1  A12 x 2  Cx 1 x 2 GE RT  x1 x 2   GE RTx 1 x 2  x1 x 2   x 1 x 2

ln 1  x1 , x2   x22  A12  2  A21  A12  C  x1  3Cx12    ln  2  x1 , x 2   x12  A21  2 A12  A21  C  x 2  3Cx 22   

Solution continued on next page…

(c): Plot Pxy diagram with fit and data









1  x1 , x2   exp ln 1  x1 , x2 

2  x1 , x2   exp ln 2  x1 , x 2 

Pcalc  X 1  1  X 1 , X 2   Psat1  X 2   2  X 1 , X 2   Psat 2 y1calc 

X 1  1  X 1 , X 2   Psat 1 Pcalc

(d): Consistency test: GERT  GeRT  x1 , x2   GERT    x , x      1 1 2    ln 1   ln 1 2  ln   2    2  x 1 , x 2 

The average del GE/RT = –0.0002 and average del ln

–0.0091

Solution continued on next page…

(e): Barker’s Method by non-linear least squares: Margules Equation

1  x1 , x 2   exp  x 22  A12  2 A21  A12  C  x1  3Cx12      2  x1 x 2  

 x 2  A   A  A  C  x  Cx 2   12 21 2 2    1  21

Minimize sum of the squared errors using excel solver.





SSE  A12 , A21 , C     Pi  x1 1  x1 , x2 , A12 , A21 , C   Psat 1  x2  2  x1 , x2 , A12 , A21 , C   Psat 2  .   i 2

Solve for A12 , A21 , C

A12 

A21 

C

Plot P-x,y diagram for Margules Equation with parameters from Barker’s Method.

Pcalc  X 1  1  X 1 , X 2 , A12 , A21 , C   Psat1  X 2  2  X 1 , X 2 , A12 , A21 , C   Psat 2 y1calc 

X 1  1  X 1 , X 2 , A12 , A21 , C   Psat 1 Pcalc

Solution continued on next page…

38 37 36

P and Pcalc

35 34

P/kPa

Pcalc

Y1calc

30 29 28 0

0.2

0.4

0.6

0.8

x1,y1,X1,Y1calc

0.06

P and Y1 Residuals

0.04 0.02 0 -0.02

Y1 residual

-0.04

P residual

-0.06 -0.08 -0.1 0

0.2

0.4

0.6

0.8

The RMS deviations is P is:

 P  Pcalc 

RMS 

 i



kPa

Solution 13.41

Problem Statement

An unusual type of low-pressure VLE behavior is that of double azeotropy, in which the dew and bubble curves are S-shaped, thus yielding at different compositions both a minimum-pressure and a maximum-pressure azeotrope. Assuming that Eq. (13.57) applies, determine under what circumstances double azeotropy is likely to occur.

Eq. 13.57:

P  x1 1 P1sat  x2 2 P2sat

Solution

This behavior requires positive deviations from Raoult’s law over part of the composition range and negative deviations over the remainder. Thus a plot of GE vs. x1 starts and ends with GE  0 at x 

x 

and shows positive values over part of the composition range and negative values

over the remainder, with an intermediate crossing of the x1 axis. Because these deviations are usually quite small, the vapor pressures Psat 1

Psat 2 must not be too different, otherwise the dewpoint and

bubblepoint curves cannot exhibit extrema.

Solution 13.42 Problem Statement Rationalize the following rule of thumb, appropriate for an equimolar binary liquid mixture: GE 1 (equimolar )  ln ( 1  2 ) RT 8

Solution

Assume the Margules equation, eq. 13.39, applies:

GE   A12 x1  A21 x1  x1 x2 RT

GE 1 equimolar   A12  A21  RT 8

But

1  A12

2  A21

This yields

GE 1 equimolar   1  RT 8

2 

GE 1 equimolar  ln1  2  RT 8

Problems 13.43 through 13.54 require parameter values for the Wilson or NRTL equation for liquid-phase activity coef pressure are given in Table B.2, Appendix B.

Solution continued on next page…

Table 13.10 Parameter Values for the Wilson and NRTL Equations 3

Parameters a12, a21, b12, and b21 have units of cal mol , and V1 and V2 have units of cm mol . Values are those recommended by Gmehling et al., Vapor-Liquid Equilibrium Data Collection, Chemistry Data Series, vol. I, parts 1a, 1b, 2c and 2e, DECHEMA, Frankfurt/Main, 1981–1988. V1

Wilson Equation

NRTL Equation

System

a12

a21

b12

b21

Acetone(1)

74.05

291.27

1,448.01

631.05

1.197.41

0.5343

Water(2)

18.07

Methanol(1)

40.73

107.38

469.55

845.21

0.2994

Water(2)

18.07

1-Propanol(1)

75.14

775.48

1,351.90

500.40

1,636.57

0.5081

Water(2)

18.07

Water(1)

18.07

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1)

40.73

196.75

343.70

314.59

0.2981

Acetonitrile(2)

66.30

Acetone(1)

74.05

583.11

184.70

222.64

0.3084

Methanol(2)

40.73

Methyl acetate(1)

79.84

813.18

381.46

346.54

0.2965

Methanol(2)

40.73

Methanol(1)

40.73

183.04

730.09

1,175.41

0.4743

1,696.98

504.31

1,734.42

Solution continued on next page

Benzene(2)

89.41

Ethanol(1)

58.68

Toluene(2)

106.85

1,556.45

210.52

713.57

1,147.86

0.5292

Values are those recommended by Gmehling et al., Vapor-Liquid Equilibrium Data Collection, Chemistry Data Series, vol. I, parts 1a, 1b, 2c and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution 13.43

Problem Statement

For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the Wilson equation, prepare a Pxy diagram for t = 60°C.

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

Acetone(1)/Water(2) System: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table 10.2 on p. 362):

  B1  2795.82    exp14.3916  P1sat  exp  A1    115.68 kPa    T  C1  60  230.00     B2  3799.89    exp16.2620  P2sat  exp  A2    19.924 kPa  T  C  60  226.35   2

Then, we want to use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C:

12 

 291.27   a  18.07 V2    0.1572 exp 12   exp  RT  74.05 1.987333.15  V1

21 

 1448.01   a  74.05 V1    0.4598 exp 21   exp V2  RT  18.07 1.987333.15 

Now that we’ve got the parameters, we can compute the activity coefficients, pressure, and vapor mole fraction at specified liquid mole fraction using the Wilson equation exactly as in problem 12.3, part (c). Doing so gives the P-x-y plot shown below.

P-x-y Plot for Acetone (1) / Water (2) 120.00

Total Pressure (kPa)

100.00

80.00

60.00

40.00

20.00

0.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone mole fraction (x1 or y1)

Methanol(1)/Water(2) system: First, we need to get the vapor pressures at 60 C from the Antoine equation (using the values for the coefficients from table 10.2 on p. 362):

  B1  3644.3    exp16.5938  P1sat  exp  A1    121.07 kPa  T  C1  60  239.76     B2  3799.89    exp16.2620  P2sat  exp  A2    19.924 kPa  T  C2  60  226.35    Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Then, we want to use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C: 12 

 107.38   a  18.07 V2    0.3772 exp 12   exp V1  RT  40.73 1.987 333.15 

21 

 469.55   a  40.73 V1    1.1089 exp 21   exp  RT  18.07 1.987333.15  V2

Now that we’ve got the parameters, we can compute the activity coefficients, pressure, and vapor mole fraction at specified liquid mole fraction using the Wilson equation exactly as in problem 12.1 parts (c) and (f). Doing so gives the P-x-y plot shown below.

P-x-y Plot for Methanol (1) / Water (2) 140.00

Total Pressure (kPa)

120.00

100.00

80.00

60.00

40.00

20.00

0.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol mole fraction (x1 or y1)

Propanol(1)/Water(2) system: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  3483.67    exp16.1154  P1sat  exp A1    20.275 kPa    T  C1  60  205.807     B2  3885.70    exp16.3872  P2sat  exp A2    20.007 kPa  T  C  60  230.17   2

Then, we want to use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C: 12 

 775.48  a  18.07 V2    0.07453 exp  12   exp  V1  RT  75.14 1.987333.15 

21 

 1351.90   a  75.14 V1    0.5395 exp 21   exp  RT  18.07 1.987333.15  V2

P-x-y Plot for 1-Propanol (1) / Water (2) 32.00

Total Pressure (kPa)

30.00

28.00

26.00

24.00

22.00

20.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

1-propanol mole fraction (x1 or y1)

Water(1)/Dioxane(2) System: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  3885.70    exp16.3872  P1sat  exp A1    20.007 kPa    T  C1  60  230.17     B2  3579.78    exp 15.0967  P2sat  exp A2    23.986 kPa  T  C  60  240.337   2

Then, we want to use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C:

12 

 1696.98   a  85.71 V2    0.3654 exp 12   exp  RT  18.07  1.987333.15  V1

21 

   a  18.07 V1 219.39    0.2937 exp 21   exp  V2  RT  85.71 1.987333.15 

P-x-y Plot for Water (1) / 1,4-Dioxane (2) 34.00

32.00

Total Pressure (kPa)

30.00

28.00

26.00

24.00

22.00

20.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Water mole fraction (x1 or y1)

Solution continued on next page…

Ethanol(1)/Toluene(2) System: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  3795.17    exp16.8598  P1sat  exp A1    45.345 kPa   T  C1  60  230.918      B2  3056.96    exp13.932  P2sat  exp A2    18.558 kPa    T  C2  60  217.625   Then, we want to use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C:

12 

 1556.45  a  106.85 V2    0.1734 exp  12   exp  V1  RT  58.68 1.987333.15 

21 

 210.52   a  58.68 V1    0.3996 exp 21   exp  RT  106.85 1.987 333.15  V2

Now that we’ve got the parameters, we can compute the activity coefficients, pressure, and vapor mole fraction at specified liquid mole fraction using the Wilson equation exactly as in problem 12.1 from last week. Doing so gives the P-x-y plot shown below.

P-x-y Plot for Ethanol (1) / Toluene (2) 55.00

50.00

Total Pressure (kPa)

45.00

40.00

35.00

30.00

25.00

20.00

15.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Ethanol mole fraction (x1 or y1)

Methanol(1)/Benzene(2) System: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  3638.27    exp16.5785  P1sat  exp A1    88.989 kPa   T  C1  60  239.50      B2  2726.81    exp13.7819  P2sat  exp A2    52.377 kPa    T  C2  60  215.582   Then, we use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C:

12 

 1734.42   a  89.41 V2    0.1598 exp 12   exp V1  RT  40.73 1.987 333.15 

21 

 183.04   a  40.73 V1    0.3455 exp 21   exp V2  RT  89.41 1.987333.15 

Now that we’ve got the parameters, we can compute the activity coefficients, pressure, and vapor mole fraction at specified liquid mole fraction using the Wilson equation. Doing so gives the P-x-y plot shown below.

P-x-y Plot for methanol(1)/benzene2) 110.00

Total Pressure (kPa)

100.00

90.00

80.00

70.00

60.00

50.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone mole fraction (x1 or y1)

Methanol(1)/Acetonitrile(2) System: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  3638.27    exp16.5785  P1sat  exp  A1    83.989 kPa  T  C1  60  239.50     B2  3413.10    exp14.895  P2sat  exp  A2    49.578 kPa  T  C  60  250.523    2

Then, we use the values of V1, V2, a12, and a21 with equation 13.53 to get the Wilson parameters at 60 °C:

12 

 504.31   a  66.30 V2    0.7599 exp 12   exp  RT  40.73 1.987 333.15  V1

21 

 196.75   a  40.73 V1    0.4564 exp 21   exp   1.987333.15  V2  RT  66.30

Total Pressure (kPa)

80.00 75.00 70.00 65.00 60.00 55.00 50.00 45.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol mole fraction (x1 or y1)

Note: Spreadsheets were also used to solve this problem. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 13.44

Problem Statement

For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the Wilson equation, prepare a txy diagram for P = 101.33 kPa.

Eq. 13.19: yi P  xi  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73

Methanol(1) Benzene(2)

40.73

Ethanol(1) Toluene(2)

58.68

1734.42

183.04

730.09

1175.41

0.4743

1556.45

210.52

713.57

1147.86

0.5292

89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

Water(1)/Dioxane(2) System: First we look up and enter the Antoine equation parameters and the Wilson equation parameters. For each composition (x1 value) for which we want to plot a point, we guess a temperature, compute the vapor pressures from the Antoine equation, compute the activity coefficient parameters and activity coefficients from the Wilson equation, then use those activity coefficients to compute a pressure and vapor-phase mole fraction. A bit of the spreadsheet used to do this, and the resulting T-x-y diagram are shown below: Antoine Parameters A1 B1 16.3872

3885.7

Wilson Parameters V1

(cm 3/mol) 18.07

(cm3/mol) 85.71

230.17

A2 B2 C2 15.0967 3579.78 240.337

a 12 a 21 (cal/mol) (cal/mol) 1696.98 -219.39

T (°C) P1sat (kPa) P 2sat (kPa) x1 x2 101.3 106.1 101.3 0.0000 1.0000 99.3 98.8 95.2 0.0200 0.9800 97.6 93.1 90.5 0.0400 0.9600 96.3 88.7 86.7 0.0600 0.9400 95.2 85.2 83.7 0.0800 0.9200 94.3 82.3 81.3 0.1000 0.9000 93.5 80.0 79.3 0.1200 0.8800 92.8 78.1 77.7 0.1400 0.8600 92.3 76.5 76.3 0.1600 0.8400

ln(g 1) ln(g 2) P/kPa y1 L 12 L 21 Wilson Wilson Wilson Wilson 0.4848 0.2831 1.4410 0.0000 101.31 0.0000 0.4788 0.2836 1.3920 0.0006 101.33 0.0784 0.4740 0.2840 1.3417 0.0025 101.33 0.1406 0.4700 0.2843 1.2907 0.0055 101.33 0.1909 0.4668 0.2845 1.2398 0.0098 101.32 0.2323 0.4641 0.2847 1.1893 0.0152 101.33 0.2668 0.4618 0.2849 1.1395 0.0218 101.33 0.2960 0.4599 0.2851 1.0907 0.0296 101.33 0.3211 0.4583 0.2852 1.0429 0.0384 101.34 0.3427

Error2 2.66E-04 3.65E-06 2.17E-06 1.74E-05 4.45E-05 2.10E-05 3.86E-08 1.34E-05 2.89E-05

Solution continued on next page…

T-x-y Plot for water(1)/1,4-dioxane(2) 103.0 101.0

Temperature (deg. C)

99.0 97.0 95.0 93.0 91.0 89.0 87.0 85.0 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Methanol mole fraction (x1 or y1)

Solution 13.45

Problem Statement

For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the NRTL equation, prepare a Pxy diagram for t = 60°C.

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1.

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

methanol(1)/benzene(2) system: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  3638.27    exp16.5785  P1sat  exp A1    88.989 kPa    T  C1  60  239.50      B2  2726.81    exp13.7819  P2sat  exp A2    52.377 kPa  T  C  60  215.582   2

Then, we use the values of , b12, and b21 with the appropriate equations from p. 448-449 of SVNA, to get the NRTL equation parameters at 60 °C: 12  21 

b12 RT b21 RT



730.09  1.1029 1.987333.15



1175.41  1.7756 1.987333.15

G12  exp12   exp0.4743 * 1.1029  0.5927

G21  exp21   exp 0.4743 * 1.7756  0.4308 Now that we’ve got the parameters, we can compute the activity coefficients, pressure, and vapor mole fraction at specified liquid mole fraction using the NRTL equation. The activity coefficients are computed from Eq. 13.49 and 13.50 of SVNA:

  Ln 1 = x22   

2     G        x 1  x 2G21  x  x G

   2  

2       G21 G12    ln 1  x  21     x1  x2G21   x  x G 2  2 1 12 2      G12 G21    ln  2  x12 12  2   x  x G   2 1 12   x1  x 2G21   2 2

Solution continued on next page…

Doing all this gives the P-x-y plot shown below.

P-x-y Plot for methanol(1)/benzene2) 110.00

Total Pressure (kPa)

100.00

90.00

80.00

70.00

60.00

50.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

Acetone mole fraction (x1 or y1)

methyl acetate(1)/methanol(2) system: First, we need to get the vapor pressures at 60 °C from the Antoine equation (using the values for the coefficients from table B.2 on p. 682):

  B1  2662.78    exp14.2456  P1sat  exp A1    112.745 kPa   T  C1  60  219.69    B2  3638.27    exp16.5785  P2sat  exp A2    84.748 kPa  T  C  60  239.50   2

Solution continued on next page…

Then, we use the values of , b12, and b21 with the appropriate equations from p. 448-449 of SVNA, to get the NRTL equation parameters at 60 °C: 12   21 

b12 RT b21

 

381.46

1.987333.15 346.54

 0.5763  0.5235

1.987333.15 G  exp   exp0.2965 * 0.5763  0.8429 G  exp   exp0.2965 * 0.5235  0.8562 RT

Now that we’ve got the parameters, we can compute the activity coefficients, pressure, and vapor mole fraction at specified liquid mole fraction using the NRTL equation. The activity coefficients are computed from Eq. 13.49 and 13.50 of SVNA: Ln 1 = 2       G21 G12    ln 1  x  21     x1  x2G21   x  x G 2  2 1 12 2       G12 G21  2  ln  2  x1 12    2     x 2  x1G12   x1  x 2G21   2 2

Doing all this gives the P-x-y plot shown below. P-x-y Plot for methanol(1)/benzene2) 130.00 125.00

Total Pressure (kPa)

120.00 115.00 110.00 105.00 100.00 95.00 90.00 85.00 80.00 0.0000

0.1000

0.2000

0.3000

0.4000

0.5000

0.6000

0.7000

0.8000

0.9000

1.0000

methyl acetate mole fraction (x1 or y1)

Note: Spreadsheets were also used in solving this problem. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 13.46

Problem Statement

This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B. For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the NRTL equation, prepare a txy diagram for P = 101.33 kPa. Eq. 13.19: yi P  xi  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73

Methanol(1) Benzene(2)

40.73

Ethanol(1) Toluene(2)

58.68

1734.42

183.04

730.09

1175.41

0.4743

1556.45

210.52

713.57

1147.86

0.5292

89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

We present the solution for the system 1-propanol(1)/water(2). Antoine Coefficients: 1-Propanol: Water:

A1 

B1 

A2 

B1 

C1 

K K

C1 

 Psat T   exp  A  T  

 Psat T   exp  A  T  

  kPa K   C1     kPa K   C2  

Parameters for the NRTL equation:



T  

cal mol



T  

b12 RT

   T 

cal mol



T  

 G

 b21 RT

T  

   T 

Solution continued on next page…

2      2 G12T   G21T  21T         1  x 1 , x 2 , T   exp x 2  12T    2     x 1  x 2 G12T  x  x G 21 T    1 2      2      2 G12 T   G 21T  21T         2  x1 , x 2 , T   exp x 1 12T    2     x 2  x1G12T  x  x G 21 T    1 2     

Make a T-x, y diagram at P = 101.33 kPa Given the equations P  x11  x1 , x 2 , T   Psat 1 T   1  x1  2  x1 , x2 , T  Psat 2 T 

Solution continued on next page…

T(eq) can be solved iteratively y eq 

y(eq)

T(eq)

373.149

0.05

0.32

363.606

0.1

0.377

361.745

0.15

0.394

361.253

0.2

0.402

361.066

0.25

0.408

360.946

0.3

0.415

360.843

0.35

0.424

360.757

0.4

0.434

360.697

0.45

0.447

360.676

0.5

0.462

360.709

0.55

0.48

360.807

0.6

0.5

360.985

0.65

0.524

361.262

0.7

0.552

361.66

0.75

0.586

362.215

0.8

0.629

362.974

0.85

0.682

364.012

0.9

0.754

365.442

0.95

0.853

367.449

370.349

x11  x1 , x2 , T   Psat 1 T  / P

Solution continued on next page…

T(eq)

374 372

370

y(eq)

368 366 364 362 360 0

0.2

0.4

0.6

0.8

x1, y(eq)

Solution 13.47

Problem Statement

This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B. For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL P : t = 60° C, x1 = 0.3. (b) DEW P : t = 60° C, y1 = 0.3. (c) P, T

flash: t = 60° C, P =

1 (P + Pdew), z 1 = 0.3. 2 bubble az

(d) If an azeotrope exists at t = 60° C, find P and x1

 y1az

Eq. 13.19: yi P  x i i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

Acetone(1)/Water(2) system: If we’ve already succeeded in doing problem 12.12, this one is pretty trivial. (a)

We already have the calculated P and y at 60°C and x1 = 0.3 in our solution to problem 12.12: P = 92.88 kPa and y1 = 0.8117.

Solution continued on next page…

(b) Rather than do a conventional dew point calculation, we can use our spreadsheet from problem 12.12 and vary x1 at 60°C until we get y1 = 0.3. Doing so gives x1 = 0.00730 and P = 28.28 kPa. (c)

The bubble point for z1 = 0.3 is our answer to part (a), and the dew point for z1 is our answer to part (b), so 0.5*(Pbubble + Pdew) = 0.5*(92.88 + 28.28) = 60.58 kPa. We can find the vapor and liquid composition at P = 60.58 kPa by varying x1 at 60°C until we get this pressure. We see from the plot we made in problem 12.12 that x1 should be about 0.06 to get a pressure near 60 kPa. Varying x1 to match the desired P gives x1 = 0.0551 and y1 = 0.6849. (so x2 = 0.9449 and y2 = 0.3151). We can then get the fraction of the system in the vapor phase from x1L + y1V = x1( 1 – V ) + y1V = z1, or z1 – x1 = (y1 – x1)V, or V = (z1 – x1 )/(y1 – x1) = (0.3 – 0.0551)/(0.6849 – 0.0551) = 0.389. So, L = 0.611.

(d) From the plot we made for problem 12.12, it looks like there might be an azeotrope very near x1 = 1.0. However, if we plot that region more carefully and look closely, we see that there is no azeotrope. Here is the region from x1 = 0.99 to x1 = 1.0:

Methanol(1)/Water(2) system: If we’ve already succeeded in doing problem 12.12, this one is pretty trivial. (a)

We already have the calculated P and y at 60°C and x1 = 0.3 in our solution to problem 12.12: P = 63.75 kPa and y1 = 0.7608.

(b)

Rather than do a conventional dew point calculation, we can use our spreadsheet from problem 12.12 and vary x1 at 60°C until we get y1 = 0.3. Doing so gives x1 = 0.0315 and P = 27.60 kPa.

Solution continued on next page…

(c)

The bubble point for z1 = 0.3 is our answer to part (a), and the dew point for z1 is our answer to part (b), so 0.5*(Pbubble + Pdew) = 0.5*(63.75 + 27.60) = 45.68 kPa. We can find vapor and liquid composition at P = 45.68 kPa by varying x1 at this 60°C until we get P = 45.68. We see from the plot we made in problem 12.12 that x1 should be about 0.13 to get a pressure near 45 kPa. Varying x1 to match the desired P gives x1 = 0.1359 and y1 = 0.6144. (so x2 = 0.8641 and y2 = 0.3856). We can then get the fraction of the system in the vapor phase from x1L + y1V = x1( 1 – V ) + y1V = z1, or z1 – x1 = (y1 – x1)V, or V = (z1 – x1 )/(y1 – x1) = (0.3 – 0.1359)/(0.6144 – 0.1359) = 0.343. So, L = 0.657.

(d)

We see from the plot we made for problem 12.12 that there is no azeotrope for this system.

(e) Propanol(1)/Water(2) system: If we’ve already succeeded in doing problem 12.12, this one is pretty trivial. (a) We already have the calculated P and y at 60°C and x1 = 0.3 in our solution to problem 12.12: P = 31.32 kPa and y1 = 0.4126. (b) Rather than do a conventional dew point calculation, we can use our spreadsheet from problem 12.12 and vary x1 at 60°C until we get y1 = 0.3. Doing so gives x1 = 0.0422 and P = 27.79 kPa. (c) The bubble point for z1 = 0.3 is our answer to part (a), and the dew point for z1 is our answer to part (b), so 0.5*(Pbubble + Pdew) = 0.5*(31.32 + 27.79) = 29.56 kPa. We can find the vapor and liquid composition at P = 29.56 kPa by varying x1 at 60°C until we get this pressure. We see from the plot we made in problem 12.12 that x1 should be about 0.1 to get a pressure near 29.6 kPa. Varying x1 to match the desired P gives x1 = 0.0805 and y1 = 0.3506. (so x2 = 0.9195 and y2 = 0.6494). We can then get the fraction of the system in the vapor phase from x1L + y1V = x1( 1 – V ) + y1V = z1, or z1 – x1 = (y1 – x1)V, or V = (z1 – x1 )/(y1 – x1) = (0.3 – 0.0805)/(0.3506 – 0.0805) = 0.813. So, L = 0.187. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(d) From the plot we prepared for problem 12.12, we see that the azeotrope is near x1 = 0.44, where the pressure is about P = 31.5 kPa. We can try different values of x1 until the computed y1 is equal to x1. That gives x1 = y1 = 0.4385 and P = 31.50 kPa.

Methyl acetate(1)/Methanol(2) system: If we’ve already succeeded in doing problem 12.12, this one is pretty trivial. (a) We already have the calculated P and y at 60°C and x1 = 0.3 in our solution to problem 12.12: P = 120.45 kPa and y1 = 0.4577. (b) Rather than do a conventional dew point calculation, we can use our spreadsheet from problem 12.12 and vary x1 at 60°C until we get y1 = 0.3. Doing so gives x1 = 0.1524 and P = 107.15 kPa. (c) The bubble point for z1 = 0.3 is our answer to part (a), and the dew point for z1 is our answer to part (b), so 0.5*(Pbubble + Pdew) = 0.5*(120.45 + 107.15) = 113.80 kPa. We can find the vapor and liquid composition at P = 113.80 kPa by varying x1 at 60°C until we get this pressure. We see from the plot we made in problem 12.12 that x1 should be about 0.20. Varying x1 to match the desired P gives x1 = 0.1995 and y1 = 0.3771. (so x2 = 0.8005 and y2 = 0.6229). We can then get the fraction of the system in the vapor phase from x1L + y1V = x1( 1 – V ) + y1V = z1, or z1 – x1 = (y1 – x1)V, or V = (z1 – x1 )/(y1 – x1) = (0.3 – 0.1995)/(0.3771 – 0.1995) = 0.566. So, L = 0.434. (d) From the plot we made for problem 12.12, an azeotrope exists near x1 = y1 = 0.64. Varying the liquid phase mole fraction until the vapor-phase mole fraction computed from it is exactly the same (to 4 digits) as the liquid-phase mole fraction gives the precise location of the azeotrope at x1 = y1 = 0.6418.

Solution 13.48

Problem Statement

Work Prob. 13.47 for the NRTL equation.

Problem 13.47 This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B. For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL P : t = 60° C, x1 = 0.3. (b) DEW P : t = 60° C, y1 = 0.3. (c) P, T

flash: t = 60° C, P =

1 ( P bubble + P dew ), z 1 = 0.3. 2

(d) If an azeotrope exists at t = 60° C, find P

and x1az  y1az

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

NRTL Equation

a21

b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

We present the solution for the system 1-propanol(1)/water(2). Antoine Coefficients: 1-Propanol: A1  Water:

A2 

B1  B1 

K K

C1  C1 

K K

  B1   Psat T   exp  A   kPa  T  273.15 K   C1    

Solution continued on next page…

 Psat T   exp  A  T  

  kPa  K   C2  

Parameters for the NRTL equation:



cal mol



T  

cal mol



b12 RT

T  



T   exp *  T 



b21 RT

T   exp *  T 

2      2 G12 T   G21T   21T       1  x1 , x2 , T   exp  x 2   12T       x1  x 2G12 T    x  x G21T 2     1 2    2     G12T    2 G21T  21T          2  x1 , x 2 , T   exp x1  12T    2     x 2  x1G12T     x1  x 2G21T     

(a): Determine the BUBL Pressure:

T

x1 

x2   x1

y1 

y2   y1

P

kPa

P

kPa

Find Pbubl, y1, and y2 y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Subject to y1  y2  1

Pbubl 

kPa

y1 

y2 

(b): Find the DEW P

T

x1 

x 2   x1

y1 

y2   y1

Find Pdew, x1, and x2 y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Pdew 

x1 

kPa

x2 

(c): P, T flash calculation P

Pdew  Pbubl

T

z1 

y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T  x1 1  V   y1 V  z1

Determine x1, x2, y1, y2, and V

x1 

x2 

y1 

y2 

V

(d): Azeotrope calculation: test for the azeotrope at T = 333.15 K 1  



T  

2 

1 0, 1, T  Psat 1 T  Psat 2 T 



Psat 1 T 

 2 0, 1, T  Psat 2 T 

T 







Since one of these values is >1 and the other is <1, an azeotrope exists. Let’s determine where that point is.

x1  y1 for anazeotrope

x1  x2 

y1  y2 

y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Solve for P, x1, and y1

Pazeo 

kPa

x1 

y1 

Solution 13.49

Problem Statement

This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B. For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL T : P = 101.33 kPa, x1 = 0.3. (b) DEW T : P = 101.33 kPa, y1 = 0.3. (c) P, T

flash : P = 101.33 kPa, T = 1/2 ( T bubble + T dew ), z 1 = 0.3. az

(d) If an azeotrope exists at P = 101.33 kPa, find T and x1az  y1az .

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1.

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

NRTL Equation

a21

b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

We present the solution for the system 1-propanol(1)/water(2). Antoine Coefficients: 1-Propanol: Water:

A1 

A2 

B1  B1 

K K

C1 

K K

  B1   Psat T   exp  A   kPa  T  273.15 K   C1      Psat T   exp  A  T  

  kPa  K   C2  

Parameters for the Wilson equation:

V1 

cm 3 mol

V2 

 12T  

cm 3 mol

a12 

 a  V2 exp 12   RT  V1

cal mol

 21T  

cal mol

a21 

 a  V1 exp 21   RT  V2

   12T   21T     exp  x 2      x1  x 2  12T  x 2  x1  21T  1  x 1 , x 2 , T   x1  x 2  12T     12T   21T    exp x 1     x1  x 2  12T  x 2  x1  21T      2  x1 , x 2 , T   x 2  x1  21T 

(a): Determine the BUBL Temperature:

T

x1 

x 2   x1

y1 

y2   y1

P

kPa

Find Tbubl, y1, and y2 y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Subject to y1  y2  1

Tbubl 

y1 

y2 

Solution continued on next page…

(b): Find the DEW T

T

x1 

x 2   x1

y1 

y2   y1

P

kPa

Find Tdew, x1, and x2 y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Pdew 

P

kPa

x1 

x2 

(c): P, T flash calculation T

Tdew  Tbubl 2

z1 

V

x1 

y1 

x   x , x , T  * Psat T 

y1 

y2 

x   x , x , T  * Psat T  P

x1 1  V   y1 V  z1

Determine x1, x2, y1, y2, and V

x1 

x2 

y1 

y2 

V

Part(d): Azeotrope calculation: test for the azeotrope at P = 101.33 kPa 1 

Tb  

2 

Tb  

 B   1 Tb1    C1    A1  ln P 

 K  

Tb1 

 B   2 Tb2    C2    A2  ln P 

 K  

Tb2 









1 0, 1, T  Psat 1 Tb2 P P

 2 0, 1, T  Psat 2 Tb1









Solution continued on next page…

Since one of these values is >1 and the other is <1, an azeotrope exists. Let’s determine where that point is.

x1  y1 for anazeotrope

x1  x2 

y1  y2 

y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Solve for T, x1, and y1

Tazeo 

x1 

y1 

Solution 13.50

Problem Statement

Work Prob. 13.49 for the NRTL equation.

Problem 13.49

This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B. For one of the binary systems listed in Table 13.10, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL T : P = 101.33 kPa, x1 = 0.3. (b) DEW T : P = 101.33 kPa, y1 = 0.3. (c) P, T

flash : P = 101.33 kPa, T = 1/2 ( T bubble + T dew ), z 1 = 0.3. az

(d) If an azeotrope exists at P = 101.33 kPa, find T and x1az  y1az . Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Eq. 13.19:

i  1, 2, . . . , N 

yi P  x i i Pi sat

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

NRTL Equation

a21

b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

We present the solution for the system 1-propanol(1)/water(2). Antoine Coefficients: 1-Propanol:

A1 

B1 

C1 

Water:

A2 

B1 

C1 

 Psat T   exp  A  T  

  kPa  K   C1     kPa  K   C2  

Parameters for the NRTL equation:



cal mol



T  

cal mol



b12 RT

T   exp *  T 

T  

 G



b21 RT

T   exp *  T 

2     G12T    2 G21T  21T        1  x1 , x 2 , T   exp  x 2  12T  2     x1  x2G12 T  x  x G 21 T     1 2      2     G12T    2 G21T  21T        2  x1 , x2 , T   exp  x 1  12T       x2  x1G12T   x  x G21T 2     1 2   

(a): Determine the BUBL Temperature:

T

x1 

x2   x1

y1 

y2   y1

P

kPa

Find Tbubl, y1, and y2 y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Solution continued on next page…

Subject to y1  y2  1

Tbubl 

y1 

y2 

(b): Find the DEW T

T

x1 

x2   x1

y1 

y2   y1

P

kPa

Find Tdew, x1, and x2 y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Tdew 

x1 

x2 

(c): P, T flash calculation

T

Tdew  Tbubl

P

2 y1 

y2 

z1 

kPa

x   x , x , T  * Psat T  P x   x , x , T  * Psat T  P

x1 1  V   y1V  z1

Determine x1, x2, y1, y2, and V

x1 

x2 

y1 

y2 

V

(d): Azeotrope calculation: test for the azeotrope at P = 101.33 kPa 1 

Tb  

 B   1 Tb1    C1    A1  ln P 

2   K  

Tb   Tb1 

Solution continued on next page…

 B   2 Tb2    C2    A2  ln P   



 K  

1 0, 1, T  Psat 1 Tb2



P P

 2 0, 1, T  Psat 2 Tb1

Tb2 









Since one of these values is >1 and the other is <1, an azeotrope exists. Let’s determine where that point is.

x1  y1 for anazeotrope

x1  x2 

y1  y2 

y P  x   x , x , T  * Psat T  y P  x   x , x , T  * Psat T 

Solve for T, x1, and y1

Tazeo 

x1 

y1 

Solution 13.51

Problem Statement

For the acetone(1)/methanol(2)/water(3) system, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL P : t = 65° C, x1 = 0.3, x2 = 0.4.

(b) DEW P : t = 65° C, y1 = 0.3, y2 = 0.4. (c) P, T

flash : t = 65° C, P = 1/2 ( P bubble + P dew ), z1 = 0.3, z2 = 0.4.

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

First, gather all necessary data: Molar Volumes and Antoine Coefficients V

74.05

14.3145

2756.22

228.06

40.73

16.5785

3638.27

239.5

18.07

16.3872

3885.7

230.17 T  338.15 K

 Psat i T   exp  Ai  T  

  kPa  K   Ci  

Psati 135.575914 102.5303094 25.10135396

The Wilson parameters are: Wilson Parameters (cal/mol) 1 2 3 0 583.11

–161.88 0

291.27 107.38

1448.01

469.55

0 i , j T  

 ai , j   exp Vi, j  RT  Vi, j

i

j

p

(a): Determine the BUBL Pressure:

x1 

x2 

x3   x1  x 2

       x p p , i T       i, x , T   exp 1  ln   x j i , j T  ]      p x  T      j     j p, j j    









Find Pbubl, y1, y2, and y3 Pbubl   xi  i x T  Psat i T 

yi 

Pbubl 

kPa

y2 

x2 

x3   x1  x2

x i  i, x , T  Psat i T 

y1 

Pbubl

y3 

(b): Find the DEW P

x1 

y1 

y2 

y3   y1  y2

P  117.1 kPa Find Pdew, x1, x2, and x3 y P  x  1, x , T  * Psat T  y P  x  2, x , T  * Psat T  y P  x  3, x , T  * Psat T 

x1  x2  x3  1

Pdew 

kPa

x1 

x2 

x3 

(c): P, T flash calculation

P

Pdew  Pbubl 2

T

V y1 

y2 

z1 

x1 

z2 

z3   z1  z 2

y1 

x  1, x , T  * Psat T  P x  2, x , T  * Psat T  P x  3, x , T  * Psat T  P

Solution continued on next page…

x1 1  V   y1 V  z1 x 2 1  V   y2V  z2 x 3 1  V   y3V  z3

Determine x1, x2, x3, y1, y2, y3 and V

x1 

x2 

x3 

y1 

y2 

y3 

V

Solution 13.52

Problem Statement Work Prob. 13.51 for the NRTL equation. Problem 13.51

This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B. For the acetone(1)/methanol(2)/water(3) system, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL P : T = 65° C, x1 = 0.3, x2 = 0.4. (b) DEW P : T = 65° C, y1 = 0.3, y2 = 0.4. (c) P, T

flash : T = 65° C, P = 1/2 ( P bubble + P dew ), z1 = 0.3, z2 = 0.4.

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

First, gather all necessary data: Molar Volumes and Antoine Coefficients V

74.05

14.3145

2756.22

228.06

40.73

16.5785

3638.27

239.5

18.07

16.3872

3885.7

230.17 T  338.15 K

 Psat i T   exp  Ai  T  

  kPa  K   Ci  

Psati 135.575914 102.5303094 25.10135396

The NRTL parameters are: NRTL Parametes a 1

0.3084

0.5343

0.3084

0.2994

0.5343

0.2994

NRTL Parametes b (cal/mol) 0

184.7

222.64

1197.41

845.21

 i, j 

bi, j RT

i

631.05 0

j

l

k

Gi, j 

   i, j

i, j

(a): Determine the BUBL Pressure:

x1 

x2 





x3   x1  x2





    k , j * Gk , j * x k   Gi , j * x j    j i , j * Gi , j * x j  k  i , j    i, x , T   exp      j G x G x   l Gl , j * xl  l l, j * l   l l, j * l   













Find Pbubl, y1, y2, and y3 Pbubl   xi  i x T  Psat i T 

yi 

Pbubl 

y2 

x i  i, x , T  Psat i T 

kPa

y1 

Pbubl

y3 

(b): Find the DEW P

x1 

x2 

x3   x1  x2

y1 

y2 

y3   y1  y2

P  117.1 kPa Find Pdew, x1, x2, and x3 y P  x  1, x , T  * Psat T  y P  x  2, x , T  * Psat T  y P  x  3, x , T  * Psat T 

x1  x2  x3  1

Pdew 

kPa

x1 

x2 

x3 

(c): P, T flash calculation P

Pdew  Pbubl 2

T

V y1 

z1 

x1 

z2 

z3   z1  z2

y1 

x  1, x , T  * Psat T  P

Solution continued on next page…

y2 

x  2, x , T  * Psat T  P x  3, x , T  * Psat T  P

x1 1  V   y1V  z1 x 2 1  V   y2V  z 2 x 3 1  V   y3V  z3

Determine x1, x2, x3, y1, y2, y3 and V

x1 

x2 

x3 

y1 

y2 

y3 

V

Solution 13.53

Problem Statement

For the acetone(1)/methanol(2)/water(3) system, based on Eq. (13.19) and the Wilson equation, make the following calculations: (a) BUBL T : P = 101.33 kPa, x1 = 0.3, x2 = 0.4. (b) DEW T : P = 101.33 kPa, y1 = 0.3, y2 = 0.4. (c) P, T

flash : P = 101.33 kPa, T = 1/2 ( T bubble + T dew ), z 1 = 0.3, z2 = 0.2.

Eq. 13.19: yi P  x i  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

First, gather all necessary data: Molar Volumes and Antoine Coefficients V

74.05

14.3145

2756.22

228.06

40.73

16.5785

3638.27

239.5

18.07

16.3872

3885.7

230.17

P  101.33 kPa   Bi   Psat i T   exp  Ai   kPa  T  273.15 K   Ci    

Psati 135.575914 102.5303094 25.10135396

The Wilson parameters are: Wilson Parameters (cal/mol) 1

3 161.88

291.27

583.11

107.38

1448.01

469.55

0 i, j T  

 ai, j    exp  RT  Vi, j   Vi, j

i

j

p

(a): Determine the BUBL temperature:

x1 

x2 

x3   x1  x 2

       x p p, i T        i, x , T   exp 1  ln   x j i, j T  ]      p    j  j x j p, j T       









Find Tbubl, y1, y2, and y3 P  x i  i x T  Psat i T 

yi P  x i  i x T  Psat i T 

Tbubl 

y2 

y1 

y3 

(b): Find the DEW T

x1 

x2 

x3   x1  x2

y1 

y2 

y3   y1  y2

P  334.08 K

Find Tdew, x1, x2, and x3 y P  x  1, x , T  * Psat T  y P  x  2, x , T  * Psat T  y P  x  3, x , T  * Psat T 

x1  x2  x3  1

Tdew 

x1 

x2 

x3 

(c): P, T flash calculation

T

Tdew  Tbubl 2

T

V

x1 

y1 

y2 

z1 

z2 

z3   z1  z2

y1 

x  1, x , T  * Psat T  P x  2, x , T  * Psat T  P x  3, x , T  * Psat T  P

Solution continued on next page…

x1 1  V   y1V  z1 x 2 1  V   y2V  z2 x 3 1  V   y3V  z3

Determine x1, x2, x3, y1, y2, y3 and V

x1 

x2 

x3 

y1 

y2 

y3 

V

Solution 13.54

Problem Statement

Work Prob. 13.53 for the NRTL equation.

Problem 13.53 This problem requires parameter values for the Wilson or NRTL equation for liquid-phase activity coefficients. Table 13.10 gives parameter values for both equations. Antoine equations for vapor pressure are given in Table B.2, Appendix B.

flash : P = 101.33 kPa, T = 1/2 ( T bubble + T dew ), z 1 = 0.3, z2 = 0.2.

Eq. 13.19: yi P  xi  i Pi sat

i  1, 2, . . . , N 

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1. V1

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07 18.07 18.07

66.30 40.73 40.73 89.41 106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution

First, gather all necessary data: Molar Volumes and Antoine Coefficients V

74.05

14.3145

2756.22

228.06

40.73

16.5785

3638.27

239.5

18.07

16.3872

3885.7

230.17 T  338.15 K

  Bi   Psat i T   exp  Ai   kPa  T  273.15 K   Ci    

Psati 135.575914 102.5303094 25.10135396 The NRTL parameters are: NRTL Parametes a 1

0.3084

0.5343

0.3084

0.2994

0.5343

0.2994

NRTL Parametes b (cal/mol) 0

184.7

222.64

1197.41

845.21  i, j 

631.05

0 bi, j RT

i

j

l

k

Gi, j 

   i, j

i, j

(a): Determine the BUBL temperature:

x1 

x2 



x3   x1  x2







    k, j, T * Gk, j, T * xk   Gi, j, T * x j    j  i, j, t * Gi, j, T * x j  k  i, j, T    i, x , T   exp     j Gl, j, T * xl  Gl, j, T * xl   l Gl, j, T * xl     l l   













Find Tbubl, y1, y2, and y3 P  x i  i x T  Psat i T 

yi P  x i  i x T  Psat i T 

Tbubl 

y2 

y1 

y3 

(b): Find the DEW T

x1 

x2 

x3   x1  x 2

y1 

y2 

y3   y1  y2

T  334.08 K

Find Tdew, x1, x2, and x3 y P  x  1, x , T  * Psat T  y P  x  2, x , T  * Psat T  y P  x  3, x , T  * Psat T 

x1  x2  x3  1

Tdew 

x1 

x2 

x3 

(c): P,T flash calculation T

Tdew  Tbubl 2

T

V y1  y2  y2 

z1 

x1 

z2 

z3   z1  z2

y1 

x  1, x , T  * Psat T  P x  2, x , T  * Psat T  P x  3, x , T  * Psat T  P

x1 1  V   y1V  z1 x2 1  V   y2V  z2 x3 1  V   y3V  z3

Determine x1, x2, x3, y1, y2, y3 and V

x1 

x2 

x3 

y1 

y2 

y3 

V

Solution 13.55

Problem Statement

The following expressions have been reported for the activity coefficients of species 1 and 2 in a binary liquid mixture at given T and P:

ln 1  x 22 (0.273  0.096 x1 )

ln  2  x 12 (0.273  0.096 x 2 )

(a) Determine the implied expression for G RT. (b) Generate expressions for ln 1 and ln

2 from the result of (a).

(c) Compare the results of (b) with the reported expressions for ln 1 and ln 2. Discuss any discrepancy. Can the reported expressions possibly be correct?

Solution

(a) For this part, we need to apply the general relationship between G and the activity coefficients: GE  x1 ln 1  x 2 ln  2 RT

Substituting in the given expressions for ln 1 and ln 2 gives

GE  x1 x22 a  bx1   x2 x12 a  bx 2  RT GE  x1 x2 ax2  bx1 x 2  ax1  bx1 x2  RT GE  x1 x2 a x1  x 2   ax1 x2  0.273 x1 x 2 RT





where I used a and b rather than the numbers during the derivation. In this case, b has cancelled out E

of the expression for G /RT. It seems unlikely that it would re-appear when we derive the activity coefficient expressions from this in part (b), but let’s see if it does. E

(b) The activity coefficients are obtained from G through the relationship  d  nG E   d  nG E       ln 1   and ln     2  dn  RT   dn1  RT   T , P, n  2 T , P, n 2

Using the result from part (a),

an1n2 nG E  nax1 x 2  RT n1  n2 Taking the derivative of this with respect to n1 at constant n2 gives  d  an n   1 2  ln 1      dn1  n1  n2  ln 1 

an22

n1  n2 

   n2 n1  n2   n1n2    a 2   n1  n2    T , P, n2

 ax22  0.273 x22

Similarly,    d  an n   n1 n1  n2   n1n2    1 2      ln  2     a   2   dn2  n1  n2   n  n      T , P, n 1 2 2

ln  2 

an12

n1  n2 

 ax12  0.273 x12

These are not the same as the expressions from which we derived G /RT in part (a). That is because the expressions given in the problem statement violate the Gibbs-Duhem equation, as we will show in part (c).

(c) The reported expressions cannot possibly be correct. They violate the Gibbs-Duhem relationship. The Gibbs-Duhem equation requires that, at constant T and P, the activity coefficients be related by x1 d ln 1   x 2 d ln  2   0

Substituting the given expressions into this, we get









x1 d x22 a  bx1   x 2 d x12 a  bx 2   0 Substituting x2 = 1 – x1 everywhere in this expression gives





 



x1 d 1  x1  a  bx1   1  x1  d x12 a  b1  x1   0 2

Now, the expression is a function of a single variable, x1. We can write the differentials as derivatives with respect to x1 as x1









2 d d x 2 a  b 1  x1   0 1  x1  a  bx1   1  x1  dx1 dx1 1

Taking the derivatives gives









x1 21  x1 a  bx1   b 1  x1   1  x1  2 x1 a  b 1  x1   x12 b  0 2

Now that we are done taking the derivatives, we can put x2 back in for (1 – x1) x1 2 x2 a  bx1   bx22   x 2 2 x1 a  bx 2   x12 b  0 2 x1 x2 a  2 x12 x2 b  x1 x22 b  2 x 1 x 2 a  2 x 1 x 22 b  x12 x 2 b  0 x12 x2 b  x1 x22 b  0

x 1 x 2  x 1  x 2  b  0 x 1 x 2 b  0

so, we see that this can only satisfy the Gibbs-Duhem equation if b = 0. The given expressions violate the Gibbs-Duhem equation, while the expressions derived in part (b) do not.

Solution 13.56

Problem Statement

Possible correlating equations for ln

1 in a binary liquid system are given here. For one of these cases,

determine by integration of the Gibbs/Duhem equation [Eq. (13.11)] the corresponding equation for ln 2. E

What is the corresponding equation for G RT? Note that by its definition,

i = 1 for xi = 1.

(a) ln

1 = A x2 ;

(b) ln

1 = x2

( A + Bx2 );

( A + B x2 + C x22 );

Eq. 13.11:

 x d ln   0 i

Solution

Write Eq. (13.10) for a binary system, and divide through by dx1:

dln 1  dx1

 x2

dln 2  dx2



dln 1 x dln 1 x1 dln 1  1  dx1 x2 dx1 x2 dx2

whence

Integrate, recalling that ln 2 = 1 for x1 = 0:

ln  2  ln  

x1 dln1

x 2 dx 2

dx1  

x1 dln1

x 2 dx 2

dx1

(a): For ln1  Ax22

Whence:

ln  2  A  x1dx1

ln 2  Ax12

By eq. 13.10 GE  Ax1 x 2 RT

(b): For ln 1   A  Bx 2  x 22

dln 1  dx2

 2 x 2  A  Bx2   Bx22  2 Ax 2  3Bx22  2 Ax2  3Bx 2 1  x1 

This leads to: x1

ln  2  A  x1 dx1  B  x1dx1  B  x12 dx1 Integration yields: 3 ln  2  Ax 12  Bx12  Bx 13 2

    3B B ln 2  x12  A   Bx1   x12  A    x 1      2 2  

Solution continued on next page…

Apply Eq. 13.10:   GE 3B  x1 x 22  A  Bx 2   x 2 x 12  A   Bx1   RT 2  

Reducing this can lead to multiple forms:   GE B  x1 x 2  A  1  x 2    RT 2 

(c): For ln 1   A  Bx2  Cx 22  x22 d ln 1  dx 2

 2 x 2  A  Bx 2  Cx 22   x 22  B  2Cx 2   2 Ax 2  3 Bx 22  4Cx 23

 2 Ax 2  3 Bx 2 1  x1   4Cx 2 1  x1 

This leads to: x1

ln  2  A x1dx1  B  x1   x1  dx1  C  x1   x1  dx1 2

Simplifying yields: x1

ln  2   A  B  C   x1 dx1   B  C   x12 dx1  C  x13 dx1 Integrating: ln  2 

2 A  3B  4C  2 3B  8C  3 2

x1 

x1  Cx14

Simplifying further yields:

   3B 8C  ln 2  x12  A   C   B   x1  Cx12      2 3     Or   B C ln  2  x12  A    x1     x 2  x 22    2 3

Applying eq 13.10 yeilds   GE B C  x 1 x 2  A  1  x 2  1  x 2  x 22    RT 2 3

Solution 13.57

Problem Statement

A storage tank contains a heavy organic liquid. Chemical analysis shows the liquid to contain 600 ppm (molar basis) of water. It is proposed to reduce the water concentration to 50 ppm by boiling the contents of the tank at constant atmospheric pressure. Because the water is lighter than the organic, the vapor will be rich in water; continuous removal of the vapor serves to reduce the water content of the system. Estimate the percentage loss of organic (molar basis) in the boil-off process. Comment on the reasonableness of the proposal. Suggestion: Designate the system water(1)/organic(2) and do unsteady-state molar balances for water and for water + organic. State all assumptions. Data: Tn2 = normal boiling point of organic = 130°C.

1 = 5.8 for water in the liquid phase at 130°C.

Solution

Let L = total moles of liquid at any point in time and Vdot = rate at which liquid boils and leaves the system as vapor. An unsteady state mole balance yields:

dL  V dt

An unsteady state species balance on water yields:

Expanding the derivative gives: L

d  Lx1  dt

  y1V

dx1 dL  x1   y1V dt dt

Substituting -Vdot for dL/dt: L

dx1  x 1V   y1V dt

Rearranging this equation gives: L

Substituting -dL/dt for Vdot: : L

Eliminating dt and rearranging:

dx 1   x 1  y1 V dt

dx 1 dL   y1  x 1  dt dt

dx1

 y1  x1 



dL L

At low concentrations y1 and x1 can be related by:

  P y1   in f 1 sat1 x1   K 1 x1  P 

where K1  in f 1

Psat 1 P

Substitution yields:

dx1

 K1  1 x1



dL L

Integrating yields: ln

Lf L0



x1 f 1 ln  K1  1 x10

where L0 and x10 are the initial conditions of the system

For this problem the following applies:

L0  mol

x10 

 Psat  exp  

600 10 6

x1 f 



50 10 6

T

3885.7  kPa T   230.17 

P  atm

Psat 

 in f 1 

kPa

Solution continued on next page…

K 1   in f 1

Psat 1 P

 15.459

 1 x1 f  L f  L0 * exp  ln    K 1  1 x10 

Lf 

mol

norg 0  L0   x 10 

norg 0 

mol

norgf  Lf   x1 f 

norgf 

mol

The % loss is lossorg 

norg 0  norgf norg 0

lossorg 

The water can be removed but almost 16% of the organic liquid will be removed with the water.

Solution 13.58

Problem Statement

Binary VLE data are commonly measured at constant T or at constant P. Isothermal data are much preferred E

for determination of a correlation for G for the liquid phase. Why?

Solution

Isobaric data reduction is complicated by the fact that both composition and temperature vary from point to point, whereas for isothermal data composition is the only significant variable. (The effect of pressure on liquid-phase properties is assumed negligible.) Because the activity coefficients are strong functions of both liquid composition and T , which are correlated, it is quite impossible without additional information to separate the effect of composition from that of T . Moreover, the Pi

sat

values

depend strongly on T , and one must have accurate vapor-pressure data over a temperature range.

Solution 13.59

Problem Statement

Consider the following model for G RT of a binary mixture:

GE k  ( x1 A21  x 2 A12k )1 k x1 x 2 RT E

This equation in fact represents a family of two-parameter expressions for G RT; specification of k leaves A12 and A21 as the free parameters.

(a) Find general expressions for ln

1 and ln 2, for any k.

(b) Show that ln 1 = A12 and ln2  A21, for any k.

∞ . Two of the cases should generate

familiar results. What are they?

Solution

(a): Written for G E, Eqs.10.15 and 10.16, become: G1E  G E  x 2

Divide through by RT, and note that

dG E dx 1

and

G2E  G E  x 1

dG E dx 1

GiE  ln i , this yields RT

GE G ln 1   x 2 RT RT dx1 E

GE G ln 2   x1 RT RT dx1 E

and

Given that 1

GE  Ak x1 x2 RT

k A  x 1 A21  x 2 A12k

with

Whence E

G  x1 x2 A RT

1 k

dA 1  A dx1 k

1 1 k

GE 1 1 k RT  x x dA  A k  x  x  1 2 2 1 dx1 dx1

1 k

and

GE 1 1 k RT  x x dA Ak  A k  A k  x  x   1 2 21 12  2 1 dx1 dx1

1 k

dA 1 A   Ak  A12k  dx1 k A 21

d and

Finally simplifying yields:

  A k  Ak  x  12 1  21  ln 1  x A    kA     2 1

1 k

and

  Ak  Ak  x  21 12 2  ln  2  x A    kA     1 k

2 1

(b): Appropriate substitution in the preceding equations of x1 = 1 and x1 = 0 yields:

ln 1  A k   A12k k  A12

and

ln 2  A k   A21k k  A21

(c): Let 1

g

If k = 1 then

If k = –1 then

g  x1 A21  x2 A12

GE k  A k   x 1 A21  x2 A12k  x1 x2 RT

which is the Margules equation

1 1 g  x1 A21  x 2 A12 

A21 A12 x1 A21  x 2 A12

which is the van Laar equation

If k  0,  , or  , then indeterminate forms appear , most easily resolved by working with the logarithm: 1

1 ln g  ln x1 A21k  x 2 A12k k  ln x 1 A21k  x 2 A12k  k

Apply L’Hopital’s rule to the final term

d ln x1 A21k  x2 A12k  dk



x1 A21k ln  A21   x2 A12k ln  A12  x1 A21k  x2 A12k

Consider the limits of the quantity on the right as k approaches several limiting values:

For k 

ln g  x 1 ln A21  x 2 ln A12  ln A21x1  ln A12x2

A12

For k   

A21

and g  A21x1 A12x2



A  x1 ln A21   x 2  12  ln  A12   A21  k

A  x1  x 2  12   A 

A  limk    12    A21  k

For k  

This yields g  A21

and

x1 

limk   ln g  ln A21

 A12

 A   12     A  k     21  k

For k    This yields

A12 A21



g  A12

x1 

lim k   ln g  ln A12

and

 A21

A21 A12

Solution 13.60

Problem Statement

A breathalyzer measures volume-% ethanol in gases exhaled from the lungs. Calibration relates it to volume-% ethanol in the bloodstream. Use VLE concepts to develop an approximate relation between the two quantities. Numerous assumptions are required; state and justify them where possible.

Solution Assume that Eq. (12.1) is the appropriate equilibrium relation, written as x e  e Pesat  x e  e Pesat  ye Pe 

Because P is low, we have assumed ideal gases, and for small xe let e  e . For volume fraction e in the vapor, the ideal

gas assumption provides ev  ye , and for the liquid phase, with xe small

el 

Then

xe Vel xe Vel  x bVb

Vb l  sat   P  ev P Ve e e e



x e Vel x Vl  e e b xb Vb Vb



Ve P Vb  e Pesat

Solution 13.61

Problem Statement Table 13.10 gives values of parameters for the Wilson equation for the acetone(1)/ methanol(2) system. Estimate values of ln 1 and ln2 at 50°C. Compare with the values suggested by Fig. 13.4(b). Repeat the exercise with the NRTL equation. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Table 13.10: Parameters a12, a21, b12, and b21 have units of cal·mol 1, and V1 and V2 have units of cm3·mol 1.

Wilson Equation

System

a12

Acetone(1) Water(2)

74.05

291.27

1448.01

631.05

1197.41

0.5343

Methanol(1) Water(2)

40.73

107.38

469.55

253.88

845.21

0.2994

1-Propanol(1) Water(2)

75.14

775.48

1351.90

500.40

1636.57

0.5081

Water(1)

18.07

1696.98

219.39

715.96

548.90

0.2920

1,4-Dioxane(2)

85.71

Methanol(1) Acetonitrile(2)

40.73

504.31

196.75

343.70

314.59

0.2981

Acetone(1) Methanol(2)

74.05

161.88

583.11

184.70

222.64

0.3084

Methyl acetate(1) Methanol(2)

79.84

31.19

813.18

381.46

346.54

0.2965

Methanol(1) Benzene(2)

40.73

1734.42

183.04

730.09

1175.41

0.4743

Ethanol(1) Toluene(2)

58.68

1556.45

210.52

713.57

1147.86

0.5292

a21

NRTL Equation b12

b21

18.07

66.30

40.73

89.41

106.85

Vapor-Liquid Equilibrium Data Collection, Chemistry parts 1a, 1b, 2c, and 2e, DECHEMA, Frankfurt/Main, 1981–1988.

Solution Acetone – 1 Methanol – 2

T = 323.15 K

Starting with the Wilson equation:

a12  

cal mol

12 

 a  V2 exp 12  V1  RT 

With these values the ln  1

From Figure 13.4(b)

cal mol

a21 

12 

cm 3 mol

V1 

21 

 a  V1 exp 21  V2  RT 

V2 

cm 3 mol

21 

ln  2 are

ln 1  ln 12   21 

ln 1 

ln  2  ln 21   12 

ln  2 

ln 1 

ln  2 

For the NRTL equation: b12  12 

b12 RT

G12  exp 12 

cal mol  21 

cal mol

b21  b21 RT

12 

G21  exp 21 



21  G12 

ln 1   21  12 exp 12 

ln 1 

ln  2  12   21 exp  21 

ln  2 

G21 

Both estimates are in close agreement with the values from Figure 13.4(b)

Solution 13.62

Problem Statement

For a binary system derive the expression for H implied by the Wilson equation for G RT. Show that the implied excess heat capacity CPE is necessarily positive. Recall that the Wilson parameters depend on T, in accord with Eq. (13.53).

Eq. 13.53:

i j 

Vj Vi

exp

aij RT

(i  j)

Solution

By Eq. 13.6   G E       RT  E H    T   RT  T     P , x

And knowing GE   x1 ln( x1  x2 12 )  x2 ln( x2  x1 21 ) RT

Solution continued on next page…

Then   G E      d d x 1 x 2 12 x 2 x1 21   RT     dT dT    T    x  x   x  x  1 2 12 2 1 21     x

and

 d12 d21     HE  dT dT  x 1 x 2T    RT x 2  x1 21   x1  x 2 12  

Knowing that ij  dij dT



 aij   exp  Vi  RT 

i  j

 aij  aij aij  exp  ij 2   Vi RT 2  RT  RT

Substitution yields

  a 21a21  12 12  H E  x1 x 2   x2  x121   x1  x2 12 Because CPE  dH E /dT differentiate the preceding expression and reduce to get:

2 2   a12   a21         x  x     2 21   1 12  RT  CPE  RT     x1 x 2   2  x  x  2 R  x2  x121   2 12   1   

Because 12

21

CPE must always be positive

Solution 13.63

Problem Statement

A single P-x1-y1 data point is available for a binary system at 25°C. Estimate from the data: (a) The total pressure and vapor-phase composition at 25°C for an equimolar liquid mixture. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

(b) Whether azeotropy is likely at 25°C. sat

Data: At 25°C, P1

sat

at = 183.4 and P2

at = 96.7 kPa

For x1 = 0.253, y1 = 0.456 and P = 139.1 kPa

Solution

Psat 1 

Psat 2 

kPa

x1 

y1 

P

kPa

Check whether or not the system is ideal using Raoult’s Law (RL) PRL  x 1 Psat 1    x 1  Psat 2

Since PRL < P, 



PRL 

kPa

are not equal to 1. Therefore, we need a model for GE/RT. A two parameter model will

work. Assume a model of the form from the Margules equation:

GE  x1 x 2  A21 x1  A12 x2  RT 1  exp( x22  A21  2 A21  A12  x1  2  exp( x12  A12  2 A12  A21  x2 

Find 



x1 

from the given data 1 

Use the values of 



y1 P x 1 Psat 1

1 

2 

y2 P x 2 Psat 2

x1 

1 

A12

A21

ln 1  ( x22  A21  2 A21  A12  x1  ln  2  ( x12  A12  2  A12  A21  x 2 

Determining A12

A21 yields A12 

A21 

1  exp( x22  A21  2 A21  A12  x1  2  exp( x12  A12  2 A12  A21  x2 

Part(a): x1 

y1 

x1 1  x1  Psat 1 P

y1 

P  x 1 1  x 1  Psat1  x 2  2  x1  Psat 2

P

kPa

Part(b): ln 1  exp A12 





1 * Psat1 Psat 2

ln 1 



ln  2  exp  A21 







* Psat1 2 Psat 2

ln  2 





Since 

Solution 13.64

Problem Statement

A single P–x1 data point is available for a binary system at 35°C. Estimate from the data:

(a) The corresponding value of y1. (b) The total pressure at 35°C for an equimolar liquid mixture. (c) Whether azeotropy is likely at 35°C.

sat

Data: At 25°C, P1

sat

at = 120.2 and P2

at = 73.9 kPa

For x1 = 0.389, P = 108.6 kPa

Solution

P

kPa

x 

T

Psat 

kPa

PRL 

kPa

Psat 

kPa

Check whether or not the system is ideal using Raoult’s Law (RL)

PRL  x1 Psat 1    x1  Psat 2

Since PRL < P, 1 and 2 are not equal to 1. Therefore, we need a model for GE/RT. A one parameter model will work.

Assume a model of the form:

GE  Ax1 x 2 RT 1  exp  Ax22  2  exp Ax12 

Since we have no y1 value, we must use the following equation to find A:

P  x1 1 Psat1  x2 2 Psat 2

Solution continued on next page…

Use the data to find the value of A





P  x1 exp A  x1  Psat 1   x1  exp  Ax12  Psat 2 2

A  0.677



1  x1   exp A  x1 



 2  x1  

 Ax  2 1

Part(a): The corresponding value of y1 is P y1  x1 * 1  x1  sat 1 P

y1 

Part(b): The total pressure for an equimolar liquid mixture is P  x 1 1  x 1  Psat1   x1   2  x1  Psat 2

P

kPa

Part(c): If there is an azeotrope for the binary system at 35 C  in f  exp A

 in f  



 in f Psat Psat

 in f  exp A

 in f 





Psat 1  in f Psat





Since 12 ranges from less than 1 to greater than 1, an azeotrope is likely based on the assumption that our model is reliable.

Solution 13.65

Problem Statement

The excess Gibbs energy for the system chloroform(1)/ethanol(2) at 55°C is well represented by the Margules E

equation, G RT = (1.42 x1 + 0.59 x2)x1x2. The vapor pressures of chloroform and ethanol at 55°C are P1sat at = 82.37 and P2sat at = 37.31 kPa. (a) Assuming the validity of Eq. (13.19), make BUBL P calculations at 55°C for liquid-phase mole fractions of 0.25, 0.50, and 0.75. (b) For comparison, repeat the calculations using Eqs. (13.13) and (13.14) with virial coefficients: B11 = B22 =

B12 = 52 cm ·mol .

Eq. 13.19:

i  1, 2, . . . , N 

yi P  x i i Pi sat

Solution

(a) This is the usual modified Raoult’s law (with ideal vapor phase) calculation like we did many times in chapter 12. sat

We have P = x11P1

sat

+ x22P2

sat

and then y1 = x11P1 /P. For the Margules equation, we have









1  exp x 22  A12  2 A21  A12  x1 

and

2  exp x12  A21  2 A12  A21  x2 

Solution continued on next page…

Applying these equations for the three specified values of x1 gives: SVNA Problem 14.1, part (a) P1

sat

(kPa)

sat

(kPa)

82.37

A21

37.31

A12

1.42

0.59

 1  exp  x22  A12  2  A21  A12  x1   x1

x2 0.25 0.5 0.75

 2  exp  x12  A21  2  A12  A21  x2  

g1 g2 P (kPa) y1 y2 0.75 1.759998 1.010998 64.53299 0.561616 0.438384 0.5 1.426181 1.158933 80.35715 0.730952 0.269048 0.25 1.121523 1.759998 85.70126 0.808446 0.191554

(b) Now, we want to extend what we did in part (a) to include nonideality in the vapor phase. So, we have

yii P  i xi Pi sat , with N N    B  P  P sat   P y j yk 2 ji   jk   i  ii  2 j1 k 1  i  sat  exp   RT i     

ˆi

For a binary system, this simplifies to  B  P  P sat   Py 2    11 1 2 12   1  exp   RT    B  P  P sat   Py 2    22 2 1 12   2  exp  RT  

So, we can start by calculating

1 and

2 using the P, y1, and y2 calculated in part (a), then compute a new value

of P from P

1 x1 P1sat  x P sat  2 2 2 1 2

Solution continued on next page…

and a new value of y1 from y1 

1 x1 P1sat 1 P

Then, we compute new values of

1 and

2 using the new P, y1, and y2 and repeat until the values stop changing

significantly. The first time we compute

2, P, y1, and y2 using the values of the virial coefficients given in the problem

statement and the P, y1, and y2 calculated in part (a), we get

P (kPa)

1.0182

1.0041

63.77

0.5582

0.4418

1.0063

1.0169

79.63

0.7330

0.2670

1.0018

1.0265

85.15

0.8122

0.1878

After repeating this a few times, we converge to 1

P (kPa)

1.0186

1.0041

63.76

0.5581

0.4419

1.0064

1.0171

79.62

0.7330

0.2670

1.0019

1.0270

85.14

0.8123

0.1877

The pressure is relatively low (< 1 atmosphere) and therefore the effects of vapor-phase non-ideality are small. Note: A spreadsheet was also used in solving this problem.

Solution 13.66

Problem Statement

Find expressions for ˆ1 and ˆ2 for a binary gas mixture described by Eq. (3.38). The mixing rule for B is given by Eq. (10.62). The mixing rule for C is given by the general equation:

C   yi y j yk Cijk i

where Cs with the same subscripts, regardless of order, are equal. For a binary mixture, this becomes:

C  y13C111  3 y12 y2C112  3 y1 y22C122  y23C222

Eq. 3.38:

ln 2  A0 x12

Eq. 10.62:

B  y12 B11  2 y1 y2 B12  y22 B22

Solution

Start with the equation immediately following Eq. (14.49), which can be modified slightly to read:

ln ˆi 

 nG R     RT  ni



 nZ  ni

n

 ln Z 1 ni

where the partial derivatives written here and in the following development without subscripts are understood to be at constant T n  

 n





n can be written:

 3  nG R  2nnB   n 2 nC    n ln Z   RT n 2  n  2

Differentiate:  nG R /RT  ni

By definition,

 nG R /RT  ni

 ln Z  3      nB  nBi     n 2C  n Ci   n  ln Z 2  n  n  n   ln Z 3    B  Bi    2  C  Ci   n  ln Z 2 ni

 nB     andC    nC  Bi   i     ni  T ,n  ni  T ,n j j

The equation of state, Eq. (3.40), can be written:   nZ  n  nnB   n 2 nC    n   n 

Z   B  C

Differentiate:

nZ  ni  nZ  ni

      nB  n Bi    n2C  n2 Ci   n   n  2

    B  Bi    2  C  Ci 

When combined with the two underlined equations, the initial equation reduces to: 1 ˆi     B  Bi    2  C  Ci  2

The two mixing rules are:

B  y12 B11  2 y1 y2 B12  y22 B22 C  y13C111  3 y12 y2C112  3 y1 y22C122  y23C222 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Application of the definitions of Bi

Ci to these mixing rules yields:

B1  y1 2  y1  B11  2 y22 B12  y22 B22 C1  y12 3  2 y1 C111  6 y1 y22C112  3 y22 1  2 y1 C122  2 y23C222 B2   y12 B11  2 y12 B12  y2 2  y2  B22 C2  2 y13C111  3 y12 1  2 y2 C112  6 y1 y22C122  2 y22 3  2 y2 C222 In combination with the mixing rules, these give:

B  B1  2 y1 B11  y2 B12  2C  C1  3 y12C111  2 y1 y2C112  y22C122 

B  B2  2 y2 B22  y1 B12  2C  C2  3 y22C222  2 y1 y2C122  y12C112 

In combination with the boxed equation these expressions along with Eq. (3.40) allow calculation of

ˆ1

ln ˆ2 .

Solution 13.67

Problem Statement A system formed of methane(1) and a light oil(2) at 200 K and 30 bar consists of a vapor phase containing 95 mol-% methane and a liquid phase containing oil and dissolved methane. The fugacity of the methane is given by Henry’s law, and at the temperature of interest Henry’s constant is

1 = 200 bar. Stating any assumptions, estimate

the equilibrium mole fraction of methane in the liquid phase. The second virial coefficient of pure methane at 200 K is

·mol .

Solution continued on next page…

Solution

T

P

y1 

bar

H1 

bar

B 150

cm3 mol

Assume Henry’s law applies to methane(1) in the liquid phase, and that the Lewis/Randall rule applies to the methane in the vapor:

fˆ1l  H1 x1

fˆ1v  y11 P

Determine 1 by:  BP     RT 

1 

Set the liquid- and vapor-phase fugacities equal to each other and solve for x1:

x1 

y11 P H1

x1 

Solution 13.68

Problem Statement

Use Eq. (13.13) to reduce one of the following isothermal data sets, and compare the result with that obtained by E

application of Eq. (13.19). Recall that reduction means developing a numerical expression for G RT as a function of composition. (a) Methylethylketone(1)/toluene(2) at 50°C: Table 13.1. (b) Acetone(1)/methanol(2) at 55°C: Prob. 13.34. (c) Methyl tert-butyl ether(1)/dichloromethane(2) at 35°C: Prob. 13.37. (d) Acetonitrile(1)/benzene(2) at 45°C: Prob. 13.40. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Second-virial-coefficient data are as follows:

B11/cm3·mol 1 B12/cm3·mol 1 B22/cm3·mol 1

Part (a)

Part (b)

Part (c)

Part (d)

1840

1440

2060

4500

1800

1150

860

1300

1150

1040

790

1000

Eq. 13.13:

yii P  x i i Pi sat

Eq. 13.19:

i  1, 2, . . . , N 

yi P  x i i Pi sat Problem 13.34

The following is a set of VLE data for the system acetone(1)/methanol(2) at 55°C:

P/kPa x1 68.728 72.278 75.279 77.524 78.951 82.528 86.762 90.088 93.206 95.017 96.365

0.0000 0.0287 0.0570 0.0858 0.1046 0.1452 0.2173 0.2787 0.3579 0.4050 0.4480

P/kPa

0.0000 0.0647 0.1295 0.1848 0.2190 0.2694 0.3633 0.4184 0.4779 0.5135 0.5512

97.646 98.462 99.811 99.950 100.278 100.467 100.999 101.059 99.877 99.799 96.885

x1 0.5052 0.5432 0.6332 0.6605 0.6945 0.7327 0.7752 0.7922 0.9080 0.9448 1.0000

y1 0.5844 0.6174 0.6772 0.6926 0.7124 0.7383 0.7729 0.7876 0.8959 0.9336 1.0000

D. C. Freshwater and K. A. Pike, J. Chem. Eng. Data, vol. 12, 79–183, 1967.

Table 13.1: VLE Data for Methyl Ethyl Ketone(l)/Toluene(2) at 50°C P/kPa

x1 sat

12.30 (P ) 15.51 18.61 21.63 24.01 25.92 27.96 30.12 31.75 34.15 36.09 (P sat)

fˆl = y 1P

y1 0.0000 0.0895 0.1981 0.3193 0.4232 0.5119 0.6096 0.7135 0.7934 0.9102 1.0000

0.0000 0.2716 0.4565 0.5934 0.6815 0.7440 0.8050 0.8639 0.9048 0.9590 1.0000

0.000 4.212 8.496 12.835 16.363 19.284 22.508 26.021 28.727 32.750 36.090 (P sat )

fˆl = y 2P

sat

12.300 (P ) 11.298 10.114 8.795 7.697 6.636 5.542 4.099 3.023 1.400 0.000

1.000 1.009 1.026 1.050 1.078 1.105 1.135 1.163 1.189 1.268

1.304 1.188 1.114 1.071 1.044 1.023 1.010 1.003 0.997 1.000

Problem 13.37 VLE data for methyl tert-butyl ether(1)/dichloromethane(2) at 308.15 K are as follows:

P/kPa x1 85.265 83.402 82.202 80.481 76.719 72.422 68.005 65.096

0.0000 0.0330 0.0579 0.0924 0.1665 0.2482 0.3322 0.3880

0.0000 0.0141 0.0253 0.0416 0.0804 0.1314 0.1975 0.2457

P/kPa x1

59.651 56.833 53.689 51.620 50.455 49.926 49.720 49.624

0.3686 0.4564 0.5882 0.7176 0.8238 0.9002 0.9502 1.0000

0.5036 0.5749 0.6736 0.7676 0.8476 0.9093 0.9529 1.0000

Mato, C. Berro, and A. Péneloux, J. Chem. Eng. Data, vol 36, pp. 259–262, 1991. Problem 13.40 Following are VLE data for the system acetonitrile(1)/benzene(2) at 45°C:

P/kPa

29.819 31.957 33.553 35.285 36.457 36.996 37.068

0.0000 0.0455 0.0940 0.1829 0.2909 0.3980 0.5069

0.0000 0.1056 0.1818 0.2783 0.3607 0.4274 0.4885

36.978 36.778 35.792 34.372 32.331 30.038 27.778

0.5458 0.5946 0.7206 0.8145 0.8972 0.9573 1.0000

0.5098 0.5375 0.6157 0.6913 0.7869 0.8916 1.0000

Extracted from I. Brown and F. Smith, Austral. J. Chem., vol. 8, p. 62, 1955.

Solution

(a): P(kPA)

15.51

0.0895

0.9105

0.2716

0.7284

1.304

1.009

18.61

0.1981

0.8019

0.4565

0.5435

1.188

1.026

21.63

0.3193

0.6807

0.5934

0.4066

1.114

1.05

24.01

0.4232

0.5768

0.6815

0.3185

1.071

1.078

25.92

0.5119

0.4881

0.744

0.256

1.044

1.105

27.96

0.6096

0.3904

0.805

0.195

1.023

1.135

30.12

0.7135

0.2865

0.8639

0.1361

1.01

1.163

31.75

0.7934

0.2066

0.9048

0.0952

1.003

1.189

34.15

0.9102

0.0898

0.959

0.041

0.997

1.268

Data reduction with the Margules equation and 13.19

GERT  x1 ln 1   x 2 ln 2  1 

y1 P

2 

x 1 Psat 1

y2 P x 2 Psat 2

Determine A12 and A21 values by minimizing F  A12 A21   GERT   A21 x1  A12 x 2  x 1 x 2   

This yields values of A12 

 RMS Error 

and n i 1

A21 

GERT   A x  A x  x x  i 21 1i 12 2 i 1i 2 i   n

RMS Error 

3

Solution continued on next page…

0.080

GERT and GERTcalc

0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 0

0.2

0.4

0.6

0.8

Data reduction with the Margules equation and Eq 13.13 B11  

cm3 mol

B22  

cm3 mol

B12  

12  2B12  B11  B22

  B  P  P   Py 2    sat 1 2 * 12     11 * 1  exp    RT    

1 

  B P  P   Py 2    sat 2 1 * 12     22 * 2  exp    RT    

2 

y11 P x1 Psat 1

y22 P x2 Psat 2

Redetermine the values for A12 and A21 using the preceding equations F  A12 A21   GERT   A21 x 1  A12 x 2  x1 x 2    A12 

A21 

Solution continued on next page…

0.080

GERT and GERTcalc

0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 0

0.2

0.4

0.6

0.8

The RMS error with Eqn. (13.13) is about 11% lower than the RMS error with Eqn. (13.19).

Parts (b)-(d) are solved using the same method as part (a).

Solution 13.69

Problem Statement

For one of the following substances, determine Psat∕bar from the Redlich/Kwong equation at two temperatures: T = Tn (the normal boiling point), and T = 0.85Tc. For the second temperature, compare your result with a value from the literature (e.g., Perry’s Chemical Engineers’ Handbook). Discuss your results. (a) Acetylene; (b) Argon; (c) Benzene; (d) n-Butane; (e) Carbon monoxide; (f) n-Decane; (g) Ethylene; (h) n-Heptane; (i) Methane; (j) Nitrogen

Solution

(a) The critical properties for acetylene are Tc = 308.3 K and Pc = 61.39 bar. The normal boiling point is Tn = 189.4 K. So, the reduced temperature at the normal boiling point is Tr = 0.6143. The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 0.61430.5  83.1452  308.32 a(T )  0.42748 r  0.42748  5837599 bar cm 6 mol2 Pc 61.39 RT 83.145  308.3 b  0.08664 c  0.08664  36.18 cm3 mol1 Pc 61.39

The dimensionless equation of state parameters are



aT  bP 36.18 * P (bar) 5837599   0.002297 P and q    10.247 RT 83.145 * 189.4 bRT 36.18 * 83.145 * 189.4

To solve iteratively for the compressibility at the ‘vapor-like’ root of this equation, we write it as Zv  1   

q  Z v    Zv Zv   

and to solve iteratively for the compressibility at the ‘liquid-like’ root of this equation, we write it as Zl    Zl  Zl   

1    Zl q

Then, for either the liquid or the vapor, we can compute the fugacity coefficient as  Z    ln   Z  1  ln  Z     q ln    Z 

Solution continued on next page…

The criterion for phase equilibrium is that the chemical potential be the same for both the liquid and the vapor, which is satisfied if ln  is the same for both the liquid and the vapor. So, what we must do is try different pressures, computing a new value of  for each pressure, then solve for the compressibility of each phase, and then use the compressibility to get the fugacity coefficient in each phase. We should keep trying different pressures until the fugacity coefficient is the same for both the liquid and the vapor. Since we have to do the same calculations over and over again, let’s use a spreadsheet, as shown below. The successive rows in the columns containing the compressibilities and the fugacity coefficients correspond to successive iterations. Each compressibility calculation uses the compressibility from the row above it. To find the vapor pressure, P was varied until the difference between the liquid and vapor fugacity coefficients (in column E) was near zero (in the last row, where the compressibilities are well-converged). T

P (bar) 189.4 1.603982

a(T)

Tc (K)

Pc (K)

308.3 beta

61.39

Pr 0.6143

0.0261

q T 0.5 R 2 Tc2 a(T )  0.42748 r Pc

5837599 36.17683 0.003685 10.24679 Zv

Zl Phi(v) 1 0.003685

Phi(l)

q

-0.04614 -0.0041

0.964852 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485 0.96485

-0.00045 -5.3E-05 -6.3E-06 -7.7E-07 -9.6E-08 -1.6E-08 -5.9E-09 -4.7E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09 -4.5E-09

0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992

0.966445 0.966045 0.965999 0.965993 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992 0.965992

RTc Pc

Phi(v)-Phi(l)

0.966205 0.004404 0.965993 1.012135 0.964904 0.004628 0.965992 0.970095 0.004703 0.004728 0.004737 0.00474 0.004741 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742 0.004742

b  0.08664

a T  bRT

Zv  1   



q  Zv    Zv  Z v   

bP RT

Zl    Z l  Zl   

1    Zl q

Z  ln   Z  1  ln  Z     q ln    Z 

The computed vapor pressure at the normal boiling point (where it should be 1.013 bar) is 1.603 bar, which is pretty far off. Changing the temperature to find the temperature where the vapor pressure is 1.013 (the predicted normal boiling point) gives T = 181.3 K – again, pretty far off. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

At T = 0.85Tc = 262.06 K gives P = 22.89 bar. Table 2-8 of Perry’s handbook gives vapor pressure curves in the form C5

P (kPa) = exp (C1 + C2/T + C3 ln(T) + C4 T ) with T in K. –16

For acetylene, the coefficients in this are given as C1 = 39.63, C2 = –2552.20, C3 = –2.78, C4 = 2.3930  10

, and

C5 = 6. Evaluating this at T = 262.1 K gives P = 1907000 Pa = 19.07 bar. Our value from the R/K equation of state is 20% higher than this. In general, we cannot expect the Redlich/Kwong equation to do a great job of predicting vapor pressures. It is parameterized to fit the critical point, but does not have any other information about the vapor pressure. In contrast, the Soave-Redlich-Kwong and Peng-Robinson equations use the acentric factor, which is based on a vapor pressure measurement at a reduced temperature of 0.7. Including this third parameter allows them to do much better at predicting vapor pressure curves than the Redlich-Kwong equation.

(c) First, we’ll do this one by hand. The critical properties for benzene are Tc = 562.2 K and Pc = 48.98 bar. The normal boiling point is Tn = 353.2 K. So, the reduced temperature at the normal boiling point is Tr = 0.6282. The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 0.62820.5  83.1452  562.22 a(T )  0.42748 r  0.42748  24060400 bar cm 6 mol2 Pc 48.98 RT 83.145  562.2 b  0.08664 c  0.08664  82.68 cm3 mol1 Pc 48.98

The dimensionless equation of state parameters are



aT  bP 82.68 * P (bar) 24060400   0.002815P and q    9.909 RT 83.145 * 353.2 bRT 82.68 * 83.145 * 353.2

Solution continued on next page…

To solve iteratively for the compressibility at the ‘vapor-like’ root of this equation, we write it as Zv  1   

q  Z v    Zv Zv   

and to solve iteratively for the compressibility at the ‘liquid-like’ root of this equation, we write it as Zl    Zl  Zl   

1    Zl q

Then, for either the liquid or the vapor, we can compute the fugacity coefficient as  Z    ln   Z  1  ln  Z     q ln    Z 

P (bar) Tc (K) Pc (K) Tr Pr 353.2 1.600138 562.2 48.98 0.6282 0.0327

a(T)

beta

24059470 82.68498 0.004505 9.908378 Zv 1 0.960265 0.958452 0.958365 0.958361 0.958361 0.958361 0.958361 0.958361 0.958361 0.958361 0.958361

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.004505 0.005415 0.959958 1.009294 -0.049336 0.005708 0.959957 0.964648 -0.004691 0.00581 0.959957 0.960514 -0.000558 0.005846 0.959957 0.960027 -7.04E-05 0.005859 0.959957 0.959966 -8.79E-06 0.005864 0.959957 0.959958 -8.29E-07 0.005866 0.959957 0.959957 2.09E-07 0.005866 0.959957 0.959956 3.45E-07 0.005866 0.959957 0.959956 3.63E-07 0.005866 0.959957 0.959956 3.65E-07 0.005866 0.959957 0.959956 3.65E-07

T 0.5 R 2Tc2 a (T )  0.42748 r Pc

q

a T  bRT

Zv  1   



q   Zv    Zv  Z v   

b  0.08664

RTc Pc

bP RT

Zl    Zl  Zl   

1    Zl q

Z ln   Z  1  ln  Z     q ln    Z 

The computed vapor pressure at the normal boiling point (where it should be 1.013 bar) is 1.600 bar, which is pretty far off. Changing the temperature to find the temperature where the vapor pressure is 1.013 (the predicted normal boiling point) gives T = 337.6 K – again, pretty far off. At T = 0.85Tc = 477.9 K gives P = 18.26 bar. Table 2-6 of Perry’s handbook gives vapor pressure curves in the form C5

P (kPa) = exp (C1 + C2/T + C3 ln(T) + C4 T ) with T in K. –6

For benzene, the coefficients in this are given as C1 = 83.918, C2 = –6517.7, C3 = –9.3453, C4 = 7.12  10 , and C5 = 2. Evaluating this at T = 477.9 K gives P = 1545000 Pa = 15.45 bar. Our value from the R/K equation of state is 18% higher than this.

(d) First, we’ll do this one by hand. The critical properties for n-butane are Tc = 425.1 K and Pc = 37.96 bar. The normal boiling point is Tn = 272.7 K. So, the reduced temperature at the normal boiling point is Tr = 0.6415. The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 0.64150.5  83.1452  425.12 a(T )  0.42748 r  0.42748  17565000 bar cm 6 mol2 Pc 37.96 RT 83.145  425.1 b  0.08664 c  0.08664  80.67 cm3 mol1 Pc 37.96

The dimensionless equation of state parameters are



aT  bP 80.67 * P (bar) 17565000   0.003558P and q    9.603 RT 83.145 * 272.7 bRT 80.67 * 83.145 * 272.7

To solve iteratively for the compressibility at the ‘vapor-like’ root of this equation, we write it as Zv  1   

q  Z v    Zv Zv   

and to solve iteratively for the compressibility at the ‘liquid-like’ root of this equation, we write it as Zl    Zl  Zl   

1    Zl q

Then, for either the liquid or the vapor, we can compute the fugacity coefficient as  Z    ln   Z  1  ln  Z     q ln    Z 

P (bar) Tc (K) Pc (K) Tr Pr 272.7 1.517526 425.1 37.96 0.6415 0.0400

a(T)

beta

17564956 80.67138 0.005399 9.602986 Zv 1 0.954107 0.951668 0.951532 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524 0.951524

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.005399 0.006524 0.953679 1.006166 -0.05249 0.006898 0.953676 0.958988 -0.00531 0.007033 0.953676 0.954353 -0.00068 0.007083 0.953676 0.953768 -9.2E-05 0.007101 0.953676 0.953689 -1.3E-05 0.007109 0.953676 0.953677 -1.8E-06 0.007111 0.953676 0.953676 -2.8E-07 0.007112 0.953676 0.953676 -5.7E-08 0.007113 0.953676 0.953676 -2.6E-08 0.007113 0.953676 0.953676 -2.1E-08 0.007113 0.953676 0.953676 -2.1E-08 0.007113 0.953676 0.953676 -2E-08 0.007113 0.953676 0.953676 -2E-08 0.007113 0.953676 0.953676 -2E-08 0.007113 0.953676 0.953676 -2E-08

T 0.5 R 2Tc2 a(T )  0.42748 r Pc

q

a T  bRT

Zv  1   



q  Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT

Zl    Zl  Z l   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

Solution continued on next page…

The computed vapor pressure at the normal boiling point (where it should be 1.013 bar) is 1.52 bar, which is pretty far off. Changing the temperature to find the temperature where the vapor pressure is 1.013 (the predicted normal boiling point) gives T = 261.7 K – low by 11 K. At T = 0.85Tc = 361.3 K gives P = 14.15 bar. Table 2-6 of Perry’s handbook gives vapor pressure curves in the form C5

P (kPa) = exp (C1 + C2/T + C3 ln(T) + C4 T ) with T in K. –6

For n-butane, the coefficients in this are given as C1 = 66.343, C2 = –4363.2, C3 = –7.046, C4 = 9.4509  10 , and C5 = 2. Evaluating this at T = 361.3 K gives P = 12.04 bar. Our value from the R/K equation of state is 18% higher than this.

(e) The critical properties for CO are Tc = 132.9 K and Pc = 34.99 bar. The normal boiling point is Tn = 81.7 K. So, the reduced temperature at the normal boiling point is Tr = 0.6147. The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 0.61470.5  83.145 2  132.92 a(T )  0.42748 r  0.42748  1902592 bar cm 6 mol 2 Pc 34.99 RT 83.145  132.9 b  0.08664 c  0.08664  27.36 cm 3 mol 1 Pc 34.99

The dimensionless equation of state parameters are



aT  bP 27.36 * P (bar) 1902592   0.004028P and q    10.237 RT 83.145 * 81.7 bRT 27.36 * 83.145 * 81.7

To solve iteratively for the compressibility at the ‘vapor-like’ root of this equation, we write it as Zv  1   

q  Z v    Zv Zv   

and to solve iteratively for the compressibility at the ‘liquid-like’ root of this equation, we write it as Zl    Zl  Zl   

1    Zl q

Then, for either the liquid or the vapor, we can compute the fugacity coefficient as  Z    ln   Z  1  ln  Z     q ln    Z 

The criterion for phase equilibrium is that the chemical potential be the same for both the liquid and the vapor, which is satisfied if ln  is the same for both the liquid and the vapor. So, what we must do is try different pressures, computing a new value of  for each pressure, then solve for the compressibility of each phase, and then use the compressibility to get the fugacity coefficient in each phase. We should keep trying different pressures until the fugacity coefficient is the same for both the liquid and the vapor. Because we have to do the same calculations over and over again, we should use a spreadsheet, as shown below, or a computer program. The successive rows in the columns containing the compressibilities and the fugacity coefficients correspond to successive iterations. Each compressibility calculation uses the compressibility from the row above it. To find the vapor pressure, P was varied until the difference between the liquid and vapor fugacity coefficients (in column E) was near zero (in the last row, where the compressibilities are well-converged). T

P (bar) Tc (K) Pc (K) Tr Pr 81.7 0.920433 132.9 34.99 0.6147 0.0263

a(T)

beta

1902592 27.36123 0.003707 10.23651 0.004028 Zv 1 0.966037 0.964723 0.96467 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668 0.964668

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.003707 0.004432 0.965823 1.012058 -0.04623 0.004657 0.965822 0.969942 -0.00412 0.004733 0.965822 0.966278 -0.00046 0.004759 0.965822 0.965876 -5.4E-05 0.004768 0.965822 0.965829 -6.5E-06 0.004771 0.965822 0.965823 -8.8E-07 0.004772 0.965822 0.965822 -2E-07 0.004773 0.965822 0.965822 -1.2E-07 0.004773 0.965822 0.965822 -1.1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07 0.004773 0.965822 0.965822 -1E-07

T 0.5 R 2 Tc2 a(T )  0.42748 r Pc q

a T  bRT

Zv  1   



q   Zv    Zv  Zv   

b  0.08664

RTc Pc

bP RT

Zl    Zl  Zl   

1    Zl q

Z  ln   Z  1  ln  Z     q ln    Z 

The computed vapor pressure at the normal boiling point (where it should be 1.013 bar) is 0.920 bar, which is significantly different. Changing the temperature to find the temperature where the vapor pressure is 1.013 (the predicted normal boiling point) gives T = 82.48 K, which is high by less than 1 K.

At T = 0.85Tc = 112.97 K, the R/K equation gives P = 13.04 bar. Fluid property data for CO is available from the NIST Webbook. It gives P

sat

= 1.020 bar at 81.7 K and P

sat

= 12.74 bar

at 112.97 K. The R/K prediction at 0.85Tc is reasonable. In general, the R/K equation will work best closer to the critical point (because it is fit to match exactly at that point) and for substances like CO that have a relatively small acentric factor.

(h) The critical properties for n-heptane are Tc = 540.2 K and Pc = 27.40 bar. The normal boiling point is Tn = 371.6 K. So, the reduced temperature at the normal boiling point is Tr = 0.6879. The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 0.68790.5  83.1452  540.22 a(T )  0.42748 r  0.42748  37947574 bar cm 6 mol2 Pc 27.40 RT 83.145  540.2 b  0.08664 c  0.08664  142.02 cm3 mol 1 Pc 27.40

The dimensionless equation of state parameters are



aT  bP 142.02 * P (bar) 37947600   0.004597 P and q    8.648 RT 83.145 * 371.6 bRT 142.02 * 83.145 * 371.6

To solve iteratively for the compressibility at the ‘vapor-like’ root of this equation, we write it as Zv  1   

q  Z v    Zv Zv   

Solution continued on next page…

and to solve iteratively for the compressibility at the ‘liquid-like’ root of this equation, we write it as Zl    Zl  Zl   

1    Zl q

Then, for either the liquid or the vapor, we can compute the fugacity coefficient as  Z    ln   Z  1  ln  Z     q ln    Z 

P (bar) Tc (K) Pc (K) Tr Pr 371.6 2.064241 540.2 27.4 0.6879 0.0753

a(T) b beta q 37947757 142.023 0.009489 Zv 1 0.928973 0.922943 0.922389 0.922338 0.922333 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332 0.922332

8.648

Zl Phi(v) Phi(l) Phi(v)-Phi(l) 0.009489 0.011683 0.927772 0.99209 -0.06432 0.012496 0.927752 0.935764 -0.00801 0.012827 0.927752 0.92904 -0.00129 0.012965 0.927752 0.927977 -0.00022 0.013024 0.927752 0.927792 -4E-05 0.013049 0.927752 0.927759 -7.4E-06 0.01306 0.927752 0.927753 -1.4E-06 0.013065 0.927752 0.927752 -2.6E-07 0.013067 0.927752 0.927752 -5.5E-08 0.013067 0.927752 0.927752 -1.8E-08 0.013068 0.927752 0.927752 -1.1E-08 0.013068 0.927752 0.927752 -9.5E-09 0.013068 0.927752 0.927752 -9.3E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09 0.013068 0.927752 0.927752 -9.2E-09

T 0.5 R 2Tc2 a (T )  0.42748 r Pc

q

a T  bRT

Zv  1   



q   Zv    Zv Zv   

b  0.08664

RTc Pc

bP RT

Z l    Zl  Z l   

1    Zl q

Z   ln   Z  1  ln  Z     q ln    Z 

The computed vapor pressure at the normal boiling point (where it should be 1.013 bar) is 2.06 bar, which is pretty far off. Changing the temperature to find the temperature where the vapor pressure is 1.013 (the predicted normal boiling point) gives T = 343.7 K, which is low by 28 K.

At T = 0.85Tc = 459.2 K, the R/K equation gives P = 10.2 bar.

sat

Using the Antoine equation coefficients from Table B.2 of SVNA, we get P = 1.015 bar at 371.6 K and P

sat

= 7.59 bar

at 459.2 K. However, we notice that the second temperature (459.2 K = 186 ºC) is outside the stated range of validity for the Antoine coefficients given in the table (4 to 123 ºC for n-heptane)

(j) The critical properties for nitrogen are Tc = 126.2 K and Pc = 34.00 bar. The normal boiling point is Tn = 77.3 K. So, the reduced temperature at the normal boiling point is Tr = 0.6125. The Redlich/Kwong equation of state is P

aT  RT  V  b V V  b

with T 0.5 R2Tc2 0.61250.5  83.1452  126.22 a(T )  0.42748 r  0.42748  1768755 bar cm 6 mol 2 Pc 34.00 RT 83.145  126.2 b  0.08664 c  0.08664  26.74 cm3 mol1 Pc 34.00

The dimensionless equation of state parameters are



aT  bP 26.74 * P (bar) 1768755   0.00416P and q    10.292 RT 83.145 * 77.3 bRT 26.74 * 83.145 * 77.3

To solve iteratively for the compressibility at the ‘vapor-like’ root of this equation, we write it as Zv  1   

q  Z v    Zv Zv   

Solution continued on next page…

and to solve iteratively for the compressibility at the ‘liquid-like’ root of this equation, we write it as Zl    Zl  Zl   

1    Zl q

Then, for either the liquid or the vapor, we can compute the fugacity coefficient as  Z    ln   Z  1  ln  Z     q ln    Z 

P (bar) 77.3 0.861996

a(T)

Tc (K)

Pc (K)

126.2 beta

Tr 34

Pr 0.6125

q T 0.5 R 2Tc2 a (T )  0.42748 r Pc

1768755 26.73838 0.003586 10.29241 Zv

Zl Phi(v) 1 0.003586

Phi(l)

q

-0.04573 -0.00403

0.965648 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646 0.965646

-0.00044 -5E-05 -5.1E-06 1.84E-07 8.17E-07 8.93E-07 9.02E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07 9.03E-07

0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738 0.966738

0.967178 0.966788 0.966743 0.966738 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737 0.966737

b  0.08664

RTc Pc

Phi(v)-Phi(l)

0.96694 0.004283 0.966739 1.012471 0.965696 0.004499 0.966738 0.970768 0.004571 0.004595 0.004604 0.004607 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608 0.004608

0.0254

a T  bRT

Zv  1   



q   Zv    Zv  Zv   

bP RT

Z l    Zl  Z l   

1    Zl q

Z  ln   Z  1  ln  Z     q ln    Z 

The computed vapor pressure at the normal boiling point (where it should be 1.013 bar) is 0.862 bar, which is significantly different. Changing the temperature to find the temperature where the vapor pressure is 1.013 (the predicted normal boiling point) gives T = 79.23 K, which is high by about 2 K.

At T = 0.85Tc = 107.27 K, the R/K equation gives P = 12.68 bar.

Antoine equation parameters for nitrogen are available from the NIST Webbook. Using them, in the appropriate sat

form of the Antoine equation, gives P

= 0.961 bar at 77.3 K and P

sat

= 12.66 bar at 107.27 K. The R/K prediction at

0.85Tc is quite good. In general, the R/K equation will work best closer to the critical point (because it is fit to match exactly at that point) and for substances like N2 that have a very small acentric factor.

Solution 13.70

Problem Statement

Work Prob. 13.69 for one of the following: (a) The Soave/Redlich/Kwong equation; (b) the Peng/Robinson equation.

Problem 13.69 For one of the following substances, determine Psat/bar from the Redlich/Kwong equation at two temperatures: T = Tn (the normal boiling point), and T = 0.85Tc. For the second temperature, compare your result with a value from the literature (e.g., Perry’s Chemical Engineers’ Handbook). Discuss your results.

(a) Acetylene; (b) Argon; (c) Benzene; (d) n-Butane; (e) Carbon monoxide; (f) n-Decane; (g) Ethylene; (h) n-Heptane; (i) Methane; (j) Nitrogen

Solution

(a) For Soave/Redlich/Kwong: Acetylene

Tc 



Pc 

T  Tn 





  Tr       



bar Tn 

Tr  

First layout the equations needed:

 Tr   

 Tr ,   R 2T 2 Pc



q Tr  

1     2   Tr2   

 Tr ,    Tr

 Tr , Pr  

 Pr Tr

Define z for the vapor zv    Tr , Pr   q Tr  *  Tr , Pr  *

zv   Tr , Pr 

zv   T P  * zv   T P  r, r

r, r

Define z for the liquid zl   Tr , Pr    zl   Tr , Pr  *  zl   Tr , Pr 

  Tr , Pr   zl qTr   Tr , Pr 

Solution continued on next page…

To find liquid root, restrict search for zl to values less than 0.2, zl 

Iv Tr , Pr  

 zv T P    T P   1  r, r r, r  ln       zv Tr , Pr    Tr , Pr  

Il Tr , Pr  

 zl T P    T P   1  r, r r, r  ln        zl Tr , Pr    Tr , Pr  

ln v Tr , Pr   zv Tr , Pr    ln  zv Tr , Pr    Tr , Pr   q Tr  * Iv Tr , Pr 

ln l Tr , Pr   zl Tr , Pr    ln zl Tr , Pr    Tr , Pr   q Tr  * Il Tr , Pr  Psatr 

Given

l Tr , Psatr   Determine zv Tr , Pr  zl Tr , Pr  Psatr

Psatr 

l Tr , Psatr 

v Tr , Psatr 

v Tr , Psatr  and  Tr , Pr 

zl Tr , Pr  

3

zv Tr , Pr  

Psat  1.073 bar Now solve for 

    log  Psatr 

 

Solution continued on next page…

The following table lists answers for all parts. Literature values are interpolated from tables in Perry’s Chemical Engineers’ Handbook, 6th ed. The last column shows the percent difference between calculated and literature values at 0.85Tc. These range from less than 0.1 to 2.5%. For the normal boiling point (Tn), Psat should be 1.013 bar. Tabulated results for Psat agree well with this value. Differences range from near 0 to 6%.

Psat(bar) @

0.85Tc

Psat (bar) @0.85

Psat (bar) Lit

Tn(K)

(K)

value

Difference

Acetylene

1.894

1.073

262.1

20.016

19.78

1.2

Argon

87.3

0.976

128.3

18.79

18.7

0.5

Benzene

353.2

1.007

477.9

15.658

15.52

0.9

n-Butane

272.7

1.008

361.3

12.239

12.07

1.4

monoxide

81.7

1.019

113

12.871

12.91

–0.3

n-Decane

447.3

1.014

525

5.324

5.21

2.1

Ethylene

169.4

1.004

240

17.918

17.69

1.3

n-Heptane

371.6

1.011

459.2

7.779

7.59

2.5

Methane

111.4

0.959

162

17.46

17.33

0.8

Nitrogen

77.3

0.992

107.3

12.617

12.57

0.3

Carbon

Solution 13.71

Problem Statement

Departures from Raoult’s law are primarily from liquid-phase nonidealities ( i

1). But vapor-phase nonidealities

( ˆi  1 ) also contribute. Consider the special case where the liquid phase is an ideal solution, and the vapor phase a nonideal gas mixture described by Eq. (3.36). Show that departures from Raoult’s law at constant temperature are likely to be negative. State clearly any assumptions and approximations.

Eq. 3.36: Z

PV BP 1 RT RT

Solution

For the case described, Eqs. (13.13) and (13.14) combine to give:

isat ˆ

yi P  x i Pi sat

If the vapor phase is assumed an ideal solution, ˆi  i

yi P  xi Pi sat

isat i

When Eq. (3.38) is valid, the fugacity coefficient of pure species i is given by Eq. (10.36): i 

Bii P RT

and

isat 

Bii Pi sat RT

Therefore, Bii  Pi  P   sat B P sat B P ln i  ln isat  ln i  ii i  ii  i RT RT RT sat

For small values of the final term, this becomes approximately: Bii  Pi sat  P  isat 1 i RT

Whence, yi P  xi Pi

sat

 Bii Pi sat  P      RT    

yi P  xi Pi

sat



xi Pi sat Bii  Pi sat  P  RT

Write this equation for species 1 and 2 of a binary mixture, and sum. This yields on the left the difference between the actual pressure and the pressure given by Raoult’s law: P  P  RL 

x1 P1sat B11  P1sat  P   x 2 P2sat B22  P2sat  P  RT

Because deviations from Raoult’s law are presumably small, P on the right side may be replaced by its Raoult’s-law value. For the two terms, P1sat  P  P1sat  x1 P1sat  x 2 P2sat  P1sat  1  x2  P1sat  x2 P2sat  x2  P1sat  P2sat  P2sat  P  P2sat  x1 P1sat  x2 P2sat  P2sat  x1 P1sat  1  x1  P2sat  x1 P2sat  P1sat 

Combine the three preceding equations: P  P  RL  x 2 x P B11  P sat 1 1

sat 1

P

sat 2

  x x P B P 2

sat 1 2

sat 2

P

sat 1



x 2 x1  P2sat  P1sat  RT

P B  P B  sat 1

sat 2

Rearrangement yields the following:

x 2 x1  P1sat  P2sat   P sat B  P sat B  11 2 22   1 sat P  P  RL   sat RT  P1  P2  2

sat  x 2 x1  P1sat  P2sat    B   B11  B22  P2    11  RT P1sat  P2sat    2

x 2 x1  P1sat  P2sat 











 B  P sat  B22  P1  P2sat 

 B11  1  1  11  sat 2

Solution continued on next page…

Clearly, when B22 = B11, the term in square brackets equals 1, and the pressure deviation from the Raoult’s-law value has the sign of B11; this is normally negative. When the virial coefficients are not equal, a reasonable assumption is that species 2, taken here as the “heavier” species (the one with the smaller vapor pressure) has the more negative second virial coefficient. This has the effect of making the quantity in parentheses negative and the quantity in square brackets < 1. However, if this latter quantity remains positive (the most likely case), the sign of B11 still determines the sign of the deviations.

Parts (b), (f), (g), and (i) are solved using the same method.

Solution 13.72

Problem Statement

Determine a numerical value for the acentric factor

implied by:

(a) The van der Waals equation; (b) The Redlich/Kwong equation.

Solution

(a) For van der Waals Eqn Tr  0.7









 Tr  

qTr  

 Tr   Tr

 Tr , Pr  

 Pr Tr

Solution continued on next page…

First layout the equations needed:

zv    Tr , Pr   q Tr  *  Tr , Pr  *

zl   Tr , Pr    zl  * 2

Iv Tr , Pr  

 Tr , Pr 

zv   Tr , Pr  2

  Tr , Pr   zl qTr   Tr , Pr 

Il Tr , Pr  

zv Tr , Pr 

 zv

 Tr , Pr  zl Tr , Pr 

ln v Tr , Pr   zv Tr , Pr    ln  zv Tr , Pr    Tr , Pr   q Tr  * Iv Tr , Pr 

ln l Tr , Pr   zl Tr , Pr    ln zl Tr , Pr    Tr , Pr   q Tr  * Il Tr , Pr  Psatr 

Given

l Tr , Psatr   Determine zv Tr , Pr  zl Tr , Pr  Psatr

l Tr , Psatr 

zv Tr , Pr  

v Tr , Psatr  

v Tr , Psatr  and  Tr , Pr 

zl Tr , Pr  

l Tr , Psatr   

Psatr 

v Tr , Psatr   

 Tr , Pr  

    log  Psatr 

 

Now solve for 

(b): For Redlich/Kwong 





 Tr   Tr5



 Tr , Pr  

q Tr  

 Tr   Tr

 Pr Tr

Solution continued on next page…

First layout the equations needed: zv    Tr , Pr   qTr  *  Tr , Pr  *

zl   Tr , Pr   zl  zl   Tr , Pr  *

Iv Tr , Pr  

 zv T P    T P   r, r r, r    zv Tr , Pr   

zv   Tr , Pr  zv  zv   Tr , Pr 

  Tr , Pr   zl q Tr   Tr , Pr 

Il Tr , Pr  

 zl T P    T P   r, r r, r    zl Tr , Pr   

ln v Tr , Pr   zv Tr , Pr    ln  zv Tr , Pr    Tr , Pr   q Tr  * Iv Tr , Pr 

ln l Tr , Pr   zl Tr , Pr    ln zl Tr , Pr    Tr , Pr   q Tr  * Il Tr , Pr  Psatr 

Given

l Tr , Psatr   Determine zv Tr , Pr  zl Tr , Pr  Psatr

zv Tr , Pr  

l Tr , Psatr 

v Tr , Psatr 

v Tr , Psatr  and  Tr , Pr 

zl Tr , Pr  

Psatr 

v Tr , Psatr   

 Tr , Pr  

    log  Psatr 

 

l Tr , Psatr   

Now solve for 

Solution 13.73

Problem Statement

The relative volatility

12 is commonly used in applications involving binary VLE. In particular (see Ex. 13.1), it serves

as a basis for assessing the possibility of binary azeotropy. (a) Develop an expression for

12 based on Eqs. (13.13) and

(13.14). (b) Specialize the expression to the composition limits x1 = y1 = 0 and x1 = y1 = 1. Compare with the result obtained from modified Raoult’s law, Eq. (13.19). The difference between the results reflects the effects of vapor-phase nonidealities. (c) Further specialize the results of part (b) to the case where the vapor phase is an ideal solution of real gases.

Eq. 13.13:

yii P  x i i Pi sat

Eq. 13.14: i 

ˆiv isat

Eq. 13.19:

yi P  x i i Pi sat

i  1, 2, . . . , N 

Solution

(a): In accord with Eqs. (13.13) and (13.14),

ˆi P  x i  i Pi sat sat ˆ 



Ki 

12 

K1 K2



yi xi



 i Pi sat ˆi P ˆ sat i

1 P1sat 1sat ˆi *  P sat ˆ ˆ sat 2 2

(b):

12  x1  0 

12  x1  1 

1 P1sat P2sat

1  P1sat 

2 P2sat 

1 P1sat

 * ˆ  sat * P2sat i  P2  2 P2sat 

1  P1sat 

* ˆ  sat i  P2 

1  P1sat  ˆ2  P1sat    P sat ˆ2  P1sat   1 sat1 *  sat * sat * sat 2 P2 P2 1  P1  2 P2  2  P2sat  P1sat

The final fractions represent corrections to modified Raoult’s law for vapor nonidealities

Part(c): If the vapor phase is an ideal solution of gases, then ˆi  i for all compositions

Solution 13.74

Problem Statement

Although isothermal VLE data are preferred for extraction of activity coefficients, a large body of good isobaric data exists in the literature. For a binary isobaric T-x1-y1 data set, one can extract point values of i via Eq. (13.13): yi i (Tk , P , y ) P

i ( x ,Tk ) 

x i Pi sat (Tk )

Here, the variable list for i recognizes a primary dependence on x and T; pressure dependence is normally negligible. The notation Tk emphasizes that temperature varies with data point across the composition range, and the calculated activity coefficients are at different temperatures. However, the usual goal of VLE data reduction and correlation is to E

develop an appropriate expression for G RT at a single temperature T. A procedure is needed to correct each activity E

coefficient to such a T chosen near the average for the data set. If a correlation for H (x) is available at or near this T, show that the values of i corrected to T can be estimated by the expression:  H E  T  i   i ( x ,T )   i ( x ,Tk )exp    1   RT  Tk

Solution

Equation (1113.9) applies:  ln i     T 

P,x



HiE RT 2

Assume that H E

HiE are functions of composition only. Then integration from Tk to T gives:

   x , T     i H E T dT H iE  1 HE  T 1     i   1   i  ln     R Tk T 2 R  T Tk  RT  Tk    i  x , Tk 

HE  T   i  x , T    i  x , Tk  * exp  i   1    RT  Tk

Solution 13.75

Problem Statement

What are the relative contributions of the various terms in the gamma/phi expression for VLE? One way to address the question is through calculation of the activity coefficients for a single binary VLE data point via Eq. (13.19):

i 

yi P ˆi fi sat  sat  fi xi Pi sat  i 

 ( A)

( B)

Term (A) is the value that would follow from modified Raoult’s law; term (B) accounts for vapor-phase nonidealities; term (C) is the Poynting factor [see Eq. (10.44)]. Use the following single-point data for the butanenitrile(1)/benzene(2) system at 318.15 K to evaluate all terms for i = 1 and i = 2. Discuss the results. VLE data: P = 0.20941 bar, x1 = 0.4819, y1 = 0.1813.

Ancillary data:

P1sat = 0 . 07287 and P2sat = 0.29871 bar B 11 =

B22 =

B12 =

V1l = 90, V2l = 92 cm ·mol

V1 

cm3 mol

·mol

Eq. 13.19:

i  1, 2, . . . , N 

yi P  x i i Pi sat

Eq. 10.44

fi  isat Pi sat exp

Vi l  P  Pi sat  RT

Solution

1-butanenitrile

2-benzene

Psat 1  Psat 2 

bar

V2 

cm3 mol

B11  

cm3 mol

T

K P

i

j

B22  

cm3 mol

B11  

bar x1  k

x 2   x1

cm3 mol

y1  y2   y1

Solution continued on next page…

The equations for each term are as follows: Term A is calculated using the given data. Term Ai 

yi P x i Psati

Term B is calculated from Eq 13.60 and 13.61

i, j  2 B ji  B jj  Bii      P  1      ˆi  exp  B  y y 2    j , k       ii 2      j k  i , j RT        j  k

isat 

Bii Pi sat RT

TermBi 

ˆi isat

Lastly, Term C is calculated from Eq 10.44

fi sat  isat Pi sat

  V P  P sat      i *  i fi  isat Pi sat * exp     RT      Term A1 

Term A2 

Term B1 

Term B2 

Term C1 

Term Ci 

fi sat fi

Term C2 

Solution 13.76

Problem Statement

Generate P-x1-y1 diagrams at 100°C for one of the systems identified below. Base activity coefficients on the Wilson equation, Eqs. (13.45) to (13.47). Use two procedures: (i) modified Raoult’s law, Eq. (13.19), and (ii) the gamma/phi i given by Eq. (13.14). Plot the results for both procedures on the same graph. Compare

and discuss them. Data sources: For Pi sat at use Table B.2. For vapor-phase nonidealities, use material from Chap. 3; assume that the vapor phase is an (approximately) ideal solution. Estimated parameters for the Wilson equation are given for each system. (a

21 = 0.8637

21 = 0.7100

21 = 0.5011

(d) Benzene(1)/n-

21 = 0.4530

(e) Carbon tetrachloride(

21 = 0.7757

(f) Carbon tetrachloride(1)/n-

21 = 0.5197

(g) Carbon tetrachloride(1)/n-

21 = 0.6011

(h) Cyclohexane(1)/n(i) Cyclohexane(1)/n-

12 12

21 = 0.7046 21 = 0.5901

Eq. 13.13:

yii P  x i i Pi sat Eq. 13.14: ˆiv i  sat i Eq. 13.19: yi P  x i i Pi sat

i  1, 2, . . . , N 

Eq. 13.45: GE   x1 ln( x1  x2 12 )  x2 ln( x2  x121 ) RT

Eq. 13.46:

  12 21  ln 1   ln( x1  x2 12 )  x2    x1  x2 12 x 2  x121  Eq. 13.47:

  12 21 ln 2  ln( x2  x1 21 )  x1     x1  x2 12 x2  x1 21 

Solution

The solution is presented for one of the systems given. The solutions for the other systems follow in the same manner.

A1 

f) 1-Carbon Tetrachloride:

2-n-heptane:

A2 

B1 

C1 

1 

Tc1 

B1 

2 

Tc 2 

K C1  K

  Psat T   exp  A   T  

Using Wilson’s equation:

K 12 

Tr 1 

Tr 2 

bar

  Psat T   exp  A   T  

T

Pc1 

Pc1  B1

bar

   kPa K   C1      kPa K   C2  

Psat 1r 

Psat 2r  .039

21 

    12 21   1  x1   exp ln  x1  x2 12   x2      x1  x 2 12 x2  x121       12 21    2  x1   exp ln  x 2  x1 21   x1      x  x  x  x    2 12 2 1 21    1 

For part i, use the modified Raoult’s Law. Define the pressure and vapor mole fraction y1 as functions of the liquid mole fraction, x1.

Pi  x1   x1 1  x1  Psat 1 T   x 2  2  x 1  Psat 2 T 

yi  x 1  

x1 1  x 1  Psat 1 T  Pi  x 1 

For part ii, assume the vapor phase is an ideal solution. Use Eqn. (10.68) and the PHIB function to calculate ˆ and sat

sat 1  PHIB Tr 1 Psat 1r 1 

sat 1 

ˆ1  P   PHIB Tr 1 P Pc1 1 

1  P 

sat 2  PHIBTr 2 Psat 2r 2 

sat 2 

ˆ2  P   PHIB Tr 2 P Pc 2 2 

2  P  

ˆ1  P  sat 1

ˆ2  P  sat 2

Solve Eqn. (14.1) for y1 and P given x1 Given y1 1  P  P  x 1 1  x 1  Psat1 T  y2 2 P  P  x 2  2  x1  Psat 2 T 

fii is a vector containing the values of P and y1. Extract the pressure, P and vapor mole fraction, y1 as functions of the liquid mole fraction. Pii  x1   fii  x1 

yii  x1   fii  x 1 

Solution 13.77 Problem Statement

Construct plots like those in Fig. (13.8) using activity coefficients predicted by UNIFAC and compare the result to the data of Table 13.5. You may use the pure component vapor pressures from Table 13.5, but should not use any other data.

Solution

Using the UNIFAC method to predict activity coefficients starts by identifying the groups present in each molecule, as delineated in Table G.1 of Appendix G. In this case, species 1, diethyl ketone (or 3-pentanone) comprises 2 CH3, 1 CH2CO, and 1 CH2 groups, while species 2, n-hexane, comprises 4 CH2 and 2 CH3 groups. The corresponding values of Rk and Qk and the values of ri and qi for the species are then looked up and computed as shown in the table below. These also allow the calculation of the eki: Group

ν k (1)

ν k (2)

ek1

e k2

CH3 CH2 CH2CO

1 2 20

0.9011 0.6744 1.4457

0.848 0.540 1.180

2 1 1

2 4 0

3.9223 q1 3.416

4.4998 q2 3.856

0.4965 0.1581 0.3454

0.4398 0.5602 0.0000

CH3 and CH2 both fall in main group 1, while CH2CO falls in main group 9. Thus, the interaction parameters amk, which we find in table G.2 of appendix G. There, we find (using k subgroup values as the indices) a1,1 = a1,2 = a2,1 = a2,2 = a20,20 = 0 K, and a1,20 = a2,20 = 476.4 K and a20,1 = a20,2 = 26.76 K. For the given temperature of 65 °C = 338.15 K, the corresponding values of mk are 1,1 = 1,2 = 2,1 = 2,2 = 20,20 = 1, and 1,20 = 2,20 = 0.2444, and 20,1 = 20,2 = 0.9239. Using these in Eq. G.18 gives the following values of ik: β ik 1 2

1 0.9737 1.0000

2 0.9737 1.0000

20 0.5054 0.2444

and sk for each of the three groups, at each of the compositions Table 13.5 (with which we will compare our final answers). We compute Li and Ji for each of the two species at each composition from Eqs. G.10 and G.11. Finally, we combine these according to Eq. G.24 and G.14 to get the results shown in the table below for the combinatorial and residual components of k

the activity coefficients: x1 0.000 0.063 0.248 0.372 0.443 0.508 0.561 0.640 0.702 0.763 0.834 0.874 1.000

x2 1.000 0.937 0.752 0.628 0.557 0.492 0.439 0.360 0.298 0.237 0.166 0.126 0.000

θ1 0.4398 0.4430 0.4526 0.4593 0.4633 0.4669 0.4699 0.4745 0.4781 0.4818 0.4861 0.4886 0.4965

θ2 θ 20 0.5602 0.5376 0.4693 0.4218 0.3940 0.3681 0.3467 0.3142 0.2883 0.2625 0.2318 0.2144 0.1581

0.0000 0.0194 0.0781 0.1189 0.1428 0.1650 0.1834 0.2113 0.2335 0.2558 0.2821 0.2971 0.3454

s1 1.0000 0.9985 0.9941 0.9910 0.9891 0.9874 0.9860 0.9839 0.9822 0.9805 0.9785 0.9774 0.9737

s2 1.0000 0.9985 0.9941 0.9910 0.9891 0.9874 0.9860 0.9839 0.9822 0.9805 0.9785 0.9774 0.9737

s 20 0.2444 0.2591 0.3034 0.3343 0.3523 0.3691 0.3830 0.4041 0.4209 0.4377 0.4575 0.4689 0.5054

J1 0.8717 0.8788 0.9003 0.9154 0.9242 0.9325 0.9393 0.9497 0.9580 0.9663 0.9761 0.9818 1.0000

J2 1.0000 1.0082 1.0329 1.0501 1.0603 1.0697 1.0776 1.0895 1.0990 1.1086 1.1199 1.1263 1.1472

L1 0.8859 0.8923 0.9117 0.9252 0.9331 0.9404 0.9465 0.9557 0.9630 0.9704 0.9791 0.9840 1.0000

L2 1.0000 1.0072 1.0291 1.0443 1.0532 1.0615 1.0684 1.0788 1.0871 1.0954 1.1052 1.1108 1.1288

ln γ 1C -0.0060 -0.0040 -0.0028 -0.0023 -0.0018 -0.0014 -0.0010 -0.0007 -0.0004 -0.0002 -0.0001

ln γ 2C 0.0000 -0.0004 -0.0009 -0.0013 -0.0018 -0.0022 -0.0028 -0.0034 -0.0041 -0.0050 -0.0055

ln γ 1R

ln γ 2R

0.7524 0.4427 0.2931 0.2244 0.1710 0.1337 0.0877 0.0590 0.0367 0.0177 0.0101

0.0043 0.0603 0.1272 0.1744 0.2227 0.2655 0.3347 0.3932 0.4543 0.5295 0.5739

Solution continued on next page…

We then add the components to get the activity coefficients for each species at each composition and compute a P-x-y diagram in the usual way. Results are shown in tabular and graphical form below: UNIFAC predictions x1 0.000 0.063 0.248 0.372 0.443 0.508 0.561 0.640 0.702 0.763 0.834 0.874 1.000

x2 1.000 0.937 0.752 0.628 0.557 0.492 0.439 0.360 0.298 0.237 0.166 0.126 0.000

ln γ 1 0.746 0.439 0.290 0.222 0.169 0.132 0.087 0.058 0.036 0.017 0.010

ln γ 2 0.004 0.060 0.126 0.173 0.221 0.263 0.332 0.390 0.450 0.525 0.568

G E /(x 1x 2RT ) 0.865 0.825 0.802 0.790 0.779 0.771 0.759 0.751 0.743 0.734 0.729

γ1

γ2

2.1093 1.5507 1.3368 1.2487 1.1844 1.1415 1.0906 1.0600 1.0369 1.0176 1.0100

1.0043 1.0617 1.1346 1.1890 1.2473 1.3013 1.3935 1.4766 1.5685 1.6897 1.7654

P (kPa) 90.15 88.69 83.13 78.66 75.75 72.77 70.07 65.47 61.25 56.46 49.90 45.65 29.00

y1 0.000 0.043 0.134 0.183 0.212 0.240 0.265 0.309 0.352 0.406 0.493 0.561 1.000

Experimental results from table 13.5 P (kPa) y1 ln γ 1 ln γ 2 G E /(x 1x 2RT ) 90.15 0.000 0.000 91.78 0.049 0.901 0.033 1.481 88.01 0.131 0.472 0.121 1.114 81.67 0.182 0.321 0.166 0.955 78.89 0.215 0.278 0.210 0.972 76.82 0.248 0.257 0.264 1.043 73.39 0.268 0.190 0.306 0.977 66.45 0.316 0.123 0.337 0.869 62.95 0.368 0.129 0.393 0.993 57.70 0.412 0.072 0.462 0.909 50.16 0.490 0.016 0.536 0.740 45.70 0.570 0.027 0.548 0.844 29.00 1.000 0.000

In each case, the solid lines show the UNIFAC predictions, while the filled circles are the results from Table 13.5. The UNIFAC method does quite a good job of predicting the activity coefficients without any input from the data (purely from the structure of the molecules). The Pxy diagram is also reproduced well with the UNIFAC-predicted activity coefficients. The agreement with the data is comparable to what is obtained from fitting the data to a simple activity coefficient model.

Solution 14.1

Problem Statement

Develop expressions for the mole fractions of reacting species as functions of the reaction coordinate for: (a) A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4 NH3(g)

+ 5O2(g)



4NO(g)

+ 6

(g)

(b) A system initially containing 3 mol H2S and 5 mol O2 and undergoing the reaction: 2H2S(g)

3O2(g)



2H2O(g)

+ 2SO2(g)

8NH3(g)



7N2(g)

12H2O(g)

Solution

5 + 4 + 6 = 1. The total number of moles and number of moles of each species are n 7 nNH  2  4 3

nO  5  5 2

nNO  4  n H O  6 2

The mole fractions are then 2  4 7 5  5 yO  2 7 4 y NO  7 6 yH O  2 7 y NH  3

Solution continued on next page…

(b) The overall stoichometric number (change in number of moles)

3+2+2=

The total number of moles and number of moles of each species are n  8 nH S  3  2 2

nO  5  3 2

n H O  2 2

nSO  2 2

The mole fractions are then

3  2 8 5  3 yO  2 8 2 yH O  2 8  2 ySO  2 8  yH S  2

The solutions manual has 8 + in the denominator, rather than 8

The total number of moles and number of moles of each species are n  8  5 nNO  3  6 2

nNH  4  8 3

nN  1  7 2

nH O  12 2

The mole fractions are then 3  6 8  5 4  8 y NH  3 8  5 1  7 yN  2 8  5 12 yH O  2 8  5 y NO  2

Solution 14.2

Problem Statement

A system initially containing 2 mol C2H4 and 3 mol O2 undergoes the reactions: C2H4(g) C2H4(g)

1/2O2(g)

3O2(g)



(CH2)2O(g)

2CO2(g)

2H2O(g)

Develop expressions for the mole fractions of the reacting species as functions of the reaction coordinates for the two reactions.

Solution

The two reactions occurring are: (1)

C2H4(g) + ½ O2(g)  (CH2)2O, with reaction coordinate 1, and

(2)

C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O (g), with reaction coordinate 2

Initially, there are 2 moles of C2H4 (nC2H4,o = 2 mol) and 3 moles of O2 (nO2,o = 3 mol). There is none of the other species present initially (n(CH2)2O,o = nCO ,o = nH O,o = 0). We write the number of moles of each species 2

as a function of the extents of reaction by simply multiplying each extent of reaction by the stoichiometric coefficient of each species. nC2H4 = 2 nO = 3 2

½ 1

n(CH )2O = 1 2

nCO = 2 2 2

Solution continued on next page…

nH2O = 2 2 The total number of moles can be obtained by adding all of these, or by taking the initial total number of moles plus the sum of the stoichiometric coefficients for each reaction times each reaction coordinate n=5

½ 1

Each mole fraction is the number of moles of that species divided by the total number of moles: yC2H = (2

yO = (3

½ 1

y(CH )2O = 1/(5 2

2)/(5

½ 1)

yH O = 2 2/(5

½ 1)

yCO = 2 2/(5 2

½ 1)

Solution 14.3

Problem Statement

A system formed initially of 2 mol CO2, 5 mol H2, and 1 mol CO undergoes the reactions: CO2(g)

CO2(g)

3H2(g)



CH3OH(g)

+ H2O(g)



CO(g)

H2O(g)

H2(g)

Develop expressions for the mole fractions of the reacting species as functions of the reaction coordinates for the two reactions.

Solution

The reactions to be considered are: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

CO2 + 2H2  CH3OH + H2O And

CO2 + H2  CO + H2O 2 is the extent of reaction 2, and we initially have 2 moles of CO2, 5 moles of

H2, and 1 mole of CO then the number of moles of each species is nCO = 2 2

nH = 5

nCH3OH nH O 2

nCO

The total number of moles is ntotal = 8 yCO = (2

yH = (5 –

1 – 2)/(8 –

yCH3OH yH O 2

yCO

2)/(8

1/(8 –

2)/(8 –

1. So, the mole fractions are

Solution 14.4

Problem Statement

Consider the water-gas-shift reaction: H2(g)

CO2(g)



H2 (g)

CO(g)

At high temperatures and low to moderate pressures the reacting species form an ideal-gas mixture. By Eq. (10.27):

G   yiGi  RT  yi ln yi i

When the Gibbs energies of the elements in their standard states are set equal to zero, Gi  G °fi for each species, and then:

G   yi G°fi  RT  yi ln yi i

(A)

At the beginning of Sec. 14.2 we noted that Eq. (12.3) is a criterion of equilibrium. Applied to the water-gasshift reaction with the understanding that T and P are constant, this equation becomes:

dGt  dnG  ndG  Gdn  0 n Here, however,

dG dn G 0 d d

= 0. The equilibrium criterion therefore becomes:

dG 0 d

(B)

Once the yi are eliminated in favor of , Eq. (A) relates G to . Data for G °fi for the compounds of interest are given with Ex. 14.13. For a temperature of 1000 K (the reaction is unaffected by P) and for a feed of 1 mol H2 and 1 mol CO2: (a) Determine the equilibrium value of by application of Eq. (B). (b) Plot G vs. , indicating the location of the equilibrium value of determined in (a).

Solution

H2(g) + CO2(g) = H2O(g) + CO(g)

v  vi 1  1  1  1  0

n0  1  1  2

By EQ 14.5:

yH 2  yCO  2

1  2

y H O  yCO  2

 2

T  1000 K

By the equation below and with data from Example 14.13 at 1000 K:    1     395790 J     192420  200240  J  RT 2  1   ln  1     2  ln   G             2   mol  2 mol 2 2 2   2  

dG e  de

0

J mol

(a): Determine the equilibrium point for the reaction. This can easily be seen if we graph this, but it can also be solved iteratively.

e  0.45308 (b): -208200 -208250 -208300 -208350

Equlibrium Point at 0.45308

-208400 -208450 -208500 -208550 -208600 -208650 -208700 0.3

0.35

0.4

0.45

0.5

0.55

0.6

εe

Solution 14.5

Problem Statement

Rework Prob. 14.4 for a temperature of:

(a) 1100 K; (b) 1200 K; (c) 1300 K.

Problem 14.4 Consider the water-gas-shift reaction: H2(g)

CO2(g)



H2 (g)

CO(g)

At high temperatures and low to moderate pressures the reacting species form an ideal-gas mixture. By Eq. (10.27):

G   yiGi  RT  yi ln yi i

When the Gibbs energies of the elements in their standard states are set equal to zero, Gi  G °fi for each species, and then:

G   yi G°fi  RT  yi ln yi i

(A)

At the beginning of Sec. 14.2 we noted that Eq. (12.3) is a criterion of equilibrium. Applied to the water-gasshift reaction with the understanding that T and P are constant, this equation becomes:

dGt  dnG  ndG  Gdn  0 n Here, however,

dG dn G 0 d d

= 0. The equilibrium criterion therefore becomes:

dG 0 d

(B)

Solution

This problem has the same setup as the 14.4. Just rework the problem for the different Temperatures. This will also change the numbers that were taken from Ex 14.13 Part(a) T  1100 K H2(g) + CO2(g) = H2O(g) + CO(g)

v  vi 1  1  1  1  0

n0  1  1  2

By EQ 14.5:

yH  yCO  2

1  2

yH O  yCO  2

 2

By the equation below and with data from Example 14.13:  1   1     1    J   J    395960    187000  209110    2 ln  G      RT 2  ln    2   2   mol  2 mol 2 2 2 

dG e  de



J mol

Determine the equilibrium point for the reaction. This can easily be seen if we graph this, but it can also be solved iteratively.

Solution continued on next page…

e  0502

-209800 -209900 -210000 -210100

-210200 -210300 -210400 -210500 -210600 -210700 -210800 0

0.2

0.4

0.6

0.8

εe

Part(b): T 

H2(g) + CO2(g) = H2O(g) + CO(g)

v  vi 1  1  1  1  0

n0  1  1  2

By EQ 14.5:

yH  yCO  2

1  2

yH O  yCO  2

 2

By the equation below and with data from Example 14.13 :

   1     396020 J     181380  217830  J  RT 2  1   ln  1     2  ln   G               mol  2 mol 2 2 2   2   2  

dG e  de



J mol

Determine the equilibrium point for the reaction. This can easily be seen if we graph this, but it can also be solved iteratively.

e  053988

-212100 -212200

-212300 -212400 -212500 -212600 -212700 0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

εe

Part(c): T 

H2(g) + CO2(g) = H2O(g) + CO(g)

v  vi 1  1  1  1  0

n0  1  1  2

By EQ 14.5:

yH  yCO  2

1  2

yH O  yCO  2

 2

By the equation below and with data from Example 14.13 :

   1     36080 J     175720  226530  J  RT 2  1   ln  1     2  ln   G           2   2   mol  2 mol 2 2 2 

dG e  de

0

J mol

Determine the equilibrium point for the reaction. This can easily be seen if we graph this, but it can also be solved iteratively.

e  057088

-214250 -214300 -214350

-214400 -214450 -214500 -214550 -214600 -214650 -214700 0.4

0.45

0.5

0.55

0.6

0.65

0.7

0.75

εe

Solution 14.6

Problem Statement

Use the method of equilibrium constants to verify the value of found as an answer in one of the following: (a) Prob. 14.4; (b) Prob. 14.5(a); (c) Prob. 14.5(b); (d) Prob. 14.5(c).

Problem 14.5 Rework Prob. 14.4 for a temperature of: (a) 1100 K; (b) 1200 K; (c) 1300 K.

Problem 14.4

Consider the water-gas-shift reaction: H2(g)

CO2(g)



H2 (g)

CO(g)

At high temperatures and low to moderate pressures the reacting species form an ideal-gas mixture. By Eq. (10.27):

G   yiGi  RT  yi ln yi i

When the Gibbs energies of the elements in their standard states are set equal to zero, Gi  G °fi for each species, and then:

G   yi G°fi  RT  yi ln yi i

(A)

At the beginning of Sec. 14.2 we noted that Eq. (12.3) is a criterion of equilibrium. Applied to the water-gasshift reaction with the understanding that T and P are constant, this equation becomes:

dGt  dnG  ndG  Gdn  0 Here, however,

dG dn G 0 d d

= 0. The equilibrium criterion therefore becomes:

dG 0 d

(B)

Solution

H2(g) + CO2(g) = H2O(g) + CO(g)

v  vi 1  1  1  1  0

n0  1  1  2

By EQ 14.5:

yH  yCO  2

1  2

yH O  yCO  2

 2

With data from Example 14.13, the following table represent values for Parts (a) through (d):

T (K) 1000

3130

1100 1200 1300

Combining Eqs. (14.5), (14.11a), and (14.28) gives          2   2   1     1      2   2 



 G    K  exp  RT  (1  ) 2

 G     exp   RT 



 1

0.4531 0.5021 0.5399

Solution 14.7

Problem Statement

Develop a general equation for the standard Gibbs-

G° as a function of temperature

for one of the reactions given in parts (a), ( f ), (i), (n), (r), (t), (u), (x), and ( y ) of Prob. 4.23. Problem 4.23 Determine the standard heat of each of the following reactions at 25°C:

Solution

The equation for G ° , appearing just above Eq. (14.16) is:

G°  H °0 

C °P C °P dT T H °0 G°0   R  dT  RT   T0 R R T T T



To calculate values of G







H 0 H 298

A B C G 0 G 298

D

J mol1 are:

(a)

Solution 14.8

Problem Statement

For ideal gases, exact mathematical expressions can be developed for the effect of T and P on e. For 

conciseness, let  ( yi ) i  K y . i

Then:

          e    K y  de and  e   K y  de     T     T  dK y  P T  P T dK y P P Use Eqs. (14.28) and (14.14), to show that:    K de (a)  e   H ° 2 dK  T  RT y P    K y de (b)  e   ( )  P dK y  P T

(c)

e /dKy is always positive. (Note: It is equally valid and perhaps easier to show that the reciprocal is

positive.) Eq. 14.14:

d ln K H o  dT RT 2 Eq. 14.28:  P   K  P o 

 ( yi ) i   i

Solution

The relation of K y

K is given by Eq. (14.28), which may be concisely written:

 P  v K y    K  P ° 

(a) Differentiate this equation with respect to T and combine with Eq. (14.14):

 K   P v dK K y dK d ln K K y H °  y       Ky        K dT dT  P °  dT  T  RT 2 P    Substitute into the given equation for  e  :  T  P

    e   K y de H °  T  RT 2 dK y P (b) The derivative of K y

P is:

 K  vK y  P v 1 1  P v  P 1 1  y      v K   vK               P° P  P°   P°   P°  P °  P T    Substitute into the given equation for  e  :  P  T     e   K y de v   P  P dK y T 

Because yi  ni n

dyi de



1 dni ni dn 1  dn dn   2   i  yi  n de n de n  de de 

But

ni  ni  vi e and n  n0  ve 0

Solution continued on next page…

Whence,

dni de

 vi

and

dn v de dyi

Therefore,

de



vi  yi v n0  e

Substitution into Eq. (A) gives

 v2  v    yi   1 dK y 1  i      i  i  v v  i    n    y   K y de y n     i 0 e 0 e i  i  i 

m  2 m  1  vi   v v  i k  n0  e  y  i1  i k 1 

In this equation, both K y and n0  e ( n) are positive. It remains to show that the summation term

is positive. If m  2, this term becomes

v12 y1

 1 1  2  

v22 y2

 y21  y12 

 2 1  2  

y1 y2

where the expression on the right is obtained by straight-forward algebraic manipulation. One can proceed by induction to find the general result, which is

m  2  vi

 m m  y   y  2 k i i k  y  vi vk    y y i k i1  i k 1  i k m

i  k 

All quantities in the sum are of course positive.

Solution 14.9

Problem Statement

For the ammonia synthesis reaction written: 1 3 N  g   H2  g   NH3  g  2 2 2

with 0.5 mol N2 and 1.5 mol H2 as the initial amounts of reactants and with the assumption that the 1/2  P   equilibrium mixture is an ideal gas, show that: e  1  1  1299 K   P ° 

Solution

1 3 N 2  g   H2  g   NH3  g  2 2

For the given reaction,  

By Eq. (14.5),

By Eq. (14.28),

n0 

1 3 1  e  1  e   2 yN  yH  2 yNH  e 2 2 3 2  e 2  e 2  e y NH

1 3 y N2 y H2 2 2

1 2 3

3 2

1  1  e   1  e   2   2 

e   e 

K

P P°

 1 2  3 2 P P      K  1299 K 2  2   2  P° P°

Whence,

  e 

r e 2  re  r   

This may be written:

Where,

e   e 



r  1  1299 K

P P°

e  1 

The roots of the quadratic are:



1r 2 1

Because e  1, e  1  r



1  1  P  2 2 ,   1  1  1299 K   e 



P° 

Solution 14.10

Problem Statement

Peter, Paul, and Mary, members of a thermodynamics class, are asked to find the equilibrium composition at a particular T and P and for given initial amounts of reactants for the following gas-phase reaction: 5 2 NH3  3 NO  3 H2O  N2 2

(A)

Each solves the problem correctly in a different way. Mary bases her solution on reaction (A) as written. Paul, who prefers whole numbers, multiplies reaction (A) by 2:

4 NH3

+ 6NO



6 H2O

+ 5 N2

(B)

Peter, who usually does things backward, deals with the reaction:

3 H2O

5 N 2 2



2 NH3

+ 3NO

(C)

Write the chemical-equilibrium equations for the three reactions, indicate how the equilibrium constants are related, and show why Peter, Paul, and Mary all obtain the same result.

Solution

If we write the reaction as (A)

2NH3 + 3NO  3H2O + /2N2

Then the change in standard state Gibbs energy will be 5 o o GAo  3GHo O  GNo  2GNH  3GNO 2 3 2 2

The equilibrium constant will then be

 o  5 o o   K A  exp GAo  exp3GH  GNo  2GNH  3GNO  2O 2 3 2 





The equilibrium constant is related to the fugacities by 5

KA 

 fˆ 3  fˆ  2  H2O   N2   o   o   f H O   f N  2 2 3  fˆ 2  NH3   fˆNO      o   o   fNH   fNO   3 

If we write the reaction as (B)

4NH3 + 6NO  6H2O + 5N2

Then the change in standard state Gibbs energy will be o o  GBo  6GHo O  5GNo  4GNH  6GNO 2

The equilibrium constant will then be







o o K B  exp GBo  exp 6GHo O  5GNo  4GNH  6GNO 2



The equilibrium constant is related to the fugacities by Solution continued on next page…

KB 

 fˆ 6  fˆ 5  H2O   N2   o   o   f H O   f N  2 2 6  fˆ 4  NH3   fˆNO      o   o   f NH   f NO   3 

If we write the reaction as (C)

3H2O + /2N2  2NH3 + 3NO

Then the change in standard state Gibbs energy will be 5 o o GCo  2GNH  3GNO  3GHo O  GNo 3 2 2 2

The equilibrium constant will then be

 o  5 o KC  exp GCo  exp2GNH  3GNO  3GHo O  GNo   3 2 2 2 





The equilibrium constant is related to the fugacities by

KC 

3  fˆ 2  NH3   fˆNO   o   o    fNH3   fNO  5

 fˆ 3  fˆ  2  H2O   N2   o   o     f H2O   f N2 

Comparing these, we see that



o o K B  exp 6GHo O  5GNo  4GNH  6GNO 2



   5 o o   K B  exp23GHo O  GNo  2GNH  3GNO 2 3 2 2    2   o  2 5 o o o    K B  exp3GH  G  2 G  3 G   K A  NH3 NO  2O   2 N2 

and

 o  5 o K C  exp 2GNH  3GNO  3GHo O  GNo   3 2 2 2    o  5 o o   K C  exp 3GH O  GNo  2GNH  3GNO 2 3   2 2  1 1 KC    o  KA 5 o o   exp3GH O  GNo  2GNH  3GNO 2 3  2 2 

and we also see that the changes in the equilibrium constants come with corresponding changes in the relationship between the fugacities and the equilibrium constant. 2

So, if we substitute KB = (KA) into the relationship between KB and the fugacities, we get

K B  K A   2

 fˆ 6  fˆ 5  H2O   N2   o   o     f H2O   f N2  6  fˆ 4  NH3   fˆNO    o   o   f NH   fNO   3 

Taking the square root of both sides gives 5

KA 

 fˆ 3  fˆ  2  H2O   N2   o   o   f H O   f N  2 2 3  fˆ 2  NH3   fˆNO      o   o    fNH3   fNO 

The same relationship we got when we wrote version (A) of the reaction. Likewise, if we substitute KC = 1/KA into the relationship between KC and the fugacities, we get

KC 

 fˆ 2  ˆ 3  NH3   f NO   o   o    f NH3   f NO 

1  K A  ˆ 3  ˆ  5 2  fH2O   fN2   o   o     fH O   fN  2

Inverting both sides gives Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

KA 

 fˆ 3  fˆ  2  H2O   N2   o   o     f H2O   f N2  3  fˆ 2  NH3   fˆNO   o   o   fNH   fNO   3 

The same relationship we got when we wrote version (A) of the reaction.

Solution 14.11

Problem Statement

The following reaction reaches equilibrium at 500°C and 2 bar:

4HCl (g)

O2 (g)



2H2O (g)

+ 2Cl2 (g)

If the system initially contains 5 mol HCl for each mole of oxygen, what is the composition of the system at equilibrium? Assume ideal gases.

Solution

The reaction to be considered is 4 HCl(g) + O2(g)  2 H2O (g) + 2 Cl2(g) at 500 °C and 2 bar. First, we will calculate the equilibrium constant at that temperature. We can evaluate the equilibrium constant in the format

K  Ko K1 K2 Solution continued on next page…

where

 G o T   o   K o  exp RTo   is the value of the equilibrium constant at the reference temperature

  T Cpo  1 T  o K2  exp  C dT  dT p  RT   RT  To To   is the smaller temperature dependence due to the fact that the heat capacity of reactants and products is (in general) different, and therefore the enthalpy of reaction changes with temperature. To do that, we’ll first look up the enthalpy and Gibbs energy of formation of the species. From table C.4, we find the following:

Compound

Hf(298 K) J/mol

HCl(g) O2(g)

92307

95299

0 241818

H2O(g) Cl2(g)

Gf(298 K) J/mol

228572

So, for the reaction, we have H (298 K) = 2* 1

and G (298 K) =

So, we have

  75948   2.02051013 Ko  exp  8.3145  298.15  Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

and    114408 1  773.15   4.8527 1013 K 1  exp  298.15   8.3145  773.15 

Finally, we need to evaluate the heat capacity integrals from 298.15 to 773.15 K to evaluate K2. We can do this using our handy spreadsheets and the heat capacity polynomials given in table C.1. Doing so, we have for the first integral:

IDCPH  T0 ,T; A, B, C, D  

1 R

B

C

1

1 

 C dT  A T  T   2 T  T   3 T  T   D  T  T  p

2 0

3 0

T0 (K) 298.15

T (K) 773.15

DA -0.439

DB (1/K) 8.00E-05

DC (1/K2 ) 0.00E+00

DD (K2) IDCPH (K) -8.23E+04 -357.8

Species Name HCl O2 H2O Cl2

Stoichiometric coefficient -4 -1 2 2

A 3.156 3.639 3.47 4.442

B (1/K) 6.23E-04 5.06E-04 1.45E-03 8.90E-05

C (1/K2) 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2 ) niAi n iBi 1.51E+04 -1.26E+01 -2.49E-03 -2.27E+04 -3.64E+00 -5.06E-04 1.21E+04 6.94E+00 2.90E-03 -3.44E+04 8.88E+00 1.78E-04

IDCPH/T -0.463

niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi -6.04E+04 2.27E+04 2.42E+04 -6.88E+04

niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi -6.04E+04 2.27E+04 2.42E+04 -6.88E+04

The dimensionless heat capacity integral is T

IDCPS  T0 ,T; A, B, C, D  



 T  D  T 2  T02  dT  A ln    B  T  T0    C  2 2     RT 2 T T0   T0    

C p

T0 (K) 298.15

T (K) 773.15

DA -0.439

DB (1/K) 8.00E-05

DC (1/K2 ) 0.00E+00

DD (K2) -8.23E+04

Species Name HCl O2 H2O Cl2

Stoichiometric coefficient -4 -1 2 2

A 3.156 3.639 3.47 4.442

B (1/K) 6.23E-04 5.06E-04 1.45E-03 8.90E-05

C (1/K2) 0.00E+00 0.00E+00 0.00E+00 0.00E+00

D (K2 ) niAi n iBi 1.51E+04 -1.26E+01 -2.49E-03 -2.27E+04 -3.64E+00 -5.06E-04 1.21E+04 6.94E+00 2.90E-03 -3.44E+04 8.88E+00 1.78E-04

IDCPS -0.774

and this integral is

K 2  exp0.463  0.774   0.7327 13

Thus, at 500 °C, we have K = 2.0205×10 * 4.8527×10

* 0.7327 = 7.18.

Now, we need to relate this to the composition at equilibrium. For an ideal gas mixture, our general relationship: Solution continued on next page…

 N  fˆ  i  i  K     fo i1   i 



reduces to simply:  P   P  i     y  K  K  o   i   1 bar   P  N

i1

and for this problem, at a pressure of 2 bar, this is 2 2 yH O yCl 2

4 yHCl yO

 2 bar 1   14.37  7.184  1 bar 

We can write the number of moles of species and the total number of moles in terms of a single reaction coordinate (using a basis of 5 moles HCl and 1 mole O2) as n  6 n HCl  5  4  nO  1   2

n H O  2 2

nCl  2  2

and

5  4 6 1 yO  2 6 2 yH O  2 6 2 yCl  2 6 yHCl 

Substituting these into the equilibrium relationship, we have  2 4    6     5  4  4  1         6     6   

 14.368

 2 4  1       14.368  0  5  4    6   

I found it rather difficult to find a form of this that converged when I iterated on it. We see from the mole fractions that, to have all positive mole fractions, we must have between zero and 1. Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

 2 4  1      14.368  for between zero and 1 gives: Plotting   5  4    6    Problem 13.11

zero at equilibrium

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

-5

extent of reaction

From which the solution is near = 0.8. Trying values near 0.8 shows that the solution is = 0.793. Using this, we obtain the mole fractions at equilibrium as:

5  4  0.35 6 1  yO   0.040 2 6 2 yH O   0.305 2 6 2 yCl   0.305 2 6 yHCl 

Solution 14.12

Problem Statement

The following reaction reaches equilibrium at 650°C and atmospheric pressure: N2(g)

C2H2(g)



2HCN(g)

If the system initially is an equimolar mixture of nitrogen and acetylene, what is the composition of the system at equilibrium? What would be the effect of doubling the pressure? Assume ideal gases.

Solution

We have the reaction N2(g) + C2H2(g)  2HCN(g) reaching equilibrium at 650 °C and 1.013 bar (1 atm.), starting from equal amounts of N2 and C2H2. To find the composition, we need to compute the equilibrium constant at 650 °C, write the species mole fractions in terms of a single reaction coordinate, and then solve 2

(yHCN) /(yN2 yC2H2) = K for the reaction coordinate. To find K, we first look up the enthalpy and Gibbs energy of the reactants and products, from which we can compute the equilibrium constant at 298.15 K (K0), and the main part of the temperature dependence of the equilibrium constant (K1). We can also look up the heat capacity coefficients, which we will use to find the small temperature dependence that arises from changes in the enthalpy of reaction with temperature (K2) From appendix C in the back of the book, we find:

G°f(298)

H°f(298)

(J/mol)

3.280

0.593 × 10

…

C2H2

209,970

227,480

6.132

1.952 × 10

…

1.299 × 10

HCN

124,700

135,100

4.736

1.359 × 10

…

0.725 × 10

Species

B K

C 1

D 2

0.040 × 10

Using these, we can compute the change in each of these quantities upon reaction as twice the value for HCN minus the value for N2 minus the value for C2H2. So, for the reaction we have: G°rxn(298)

H°rxn(298)

(J/mol)

39,430

42,720

A

B K

0.06

C

0.173 × 10

K 3

…

D K

0.191 × 10

The equilibrium constant at 298 K is then K0 = exp(

*298.15)) = 1.2365 × 10 . The main

part of change in the equilibrium constant from 298 K to 650°C (923 K) is:

 H o  To   T   42720  923   5 K1  exp   1     exp    1     1.1668  10   RT  To    8.314 * 923  298    Finally, for the minor part of the temperature dependence, we have

 T Cpo   1 T Cpo K2  exp  dT   dT    T R RT To To   Evaluating the two heat capacity integrals using our spreadsheets ICPH.xls and ICPS.xls and the delta Cp coefficients above gives:

T 1 (K) 298.15

T 2 (K) 923.15 T2

ICPH 

R

A 0.06

dT  A T2  T1  

C (1/K ) D (K ) B (1/K) 1.73E-04 0.00E+00 1.91E-06

ICPH (K) 1.04E+02

 1 B 2 C 3 1 T2  T12  T2  T13  D    2 3  T2 T1 









and

T 1 (K) 298.15

T 2 (K) 923.15 T2

ICPS 

 T1

A 0.06

C (1/K2) D (K2) B (1/K) 1.73E-04 0.00E+00 1.91E-06

ICPS 0.176

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT T 2 2  1









 104  K2  exp  0.176  1.066  923 

Putting this all together, K = 1.2365 × 10

* 1.1668×10 * 1.066 = 0.01538.

Now, we need to write the mole fraction of each species in terms of a single reaction coordinate. The reaction was N2(g) + C2H2(g)  2HCN(g), starting from equal amounts of N2 and C2H2. So, we can use as a basis initial number of moles 0.5 mol N2 and 0.5 mol C2H2. Then the total number of moles will be 1 initially, and will remain 1 because there is no change in mole number for this reaction. Thus, the number of moles of each species and its mole fraction are the same, and we have yN2 = 0.5

, yC2H2 = 0.5

, and

yHCN = 2 . So, our final equation for is 2

(yHCN) /(yN2 yC2H2) = 4 /(0.5

) = K = 0.01538.

Taking the square root of both sides, we have 2 /(0.5

) = 0.124

and 2.124 = 0.062 and = 0.0292 The resulting mole fractions are yN2 = yC2H2 = 0.471 and yHCN = 0.058. Doubling the pressure would not change the equilibrium composition, because there is no change in total number of moles for this reaction and therefore the equilibrium is independent of pressure.

Solution 14.13

Problem Statement

The following reaction reaches equilibrium at 350°C and 3 bar: CH3CHO(g)

+ H2(g)



C2H5OH(g)

If the system initially contains 1.5 mol H2 for each mole of acetaldehyde, what is the composition of the system at equilibrium? What would be the effect of reducing the pressure to 1 bar? Assume ideal gases.

Solution

The reaction to be considered is CH3CHO(g) + H2(g)  C2H5OH(g) at 300 °C and 3 bar. First, we will calculate the equilibrium constant at that temperature. We can evaluate the equilibrium constant in the format

K  Ko K1 K2 where

 G o T   o   K o  exp  RT  o  is the value of the equilibrium constant at the reference temperature

 H o T     T  o  1   K 1  exp  RT  To    is the main part of the temperature dependence, and is the change in the equilibrium constant that we would have if the enthalpy change of reaction were constant, and Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

  T Cpo  1 T  o K2  exp Cp dT   dT     RT RT To To   is the smaller temperature dependence due to the fact that the heat capacity of reactants and products is (in general) different, and therefore the enthalpy of reaction changes with temperature. To do that, we’ll first look up the enthalpy and Gibbs energy of formation of the species. From table C.4, we o

find that for acetaldehyde (CH3CHO ), Hf (298 K) =

H2(g) is an element in its standard state, so by definition, it has Hf (298 K) = 0 J mol 1

mol . Finally, for ethanol (as a gas) we find Hf (298 K) = 1

G (298 K) =

and Gf (298 K) = 0 J

and Gf (298 K) = 1

mol . So, for the reaction, we have H (298 K) =

and Gf (298 K) =

and

So, we have

  39630   8.7669106 Ko  exp  8.3145  298.15  and    68910 1  623.15   5.0546 107 K 1  exp   8.3145  623.15  298.15 

Finally, we need to evaluate the heat capacity integrals from 298.15 to 623.15 K to evaluate K2. We can do this using our handy spreadsheets and the heat capacity polynomials given in table C.1. Doing so, we have for the first integral:

Solution continued on next page…

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 623.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -0.363 Heat Capacity Coefficients

Standard Heat Species Stoichiometric of Formation Name coefficient at T 0 (kJ/mol) acetaldehyde -1 H2 -1 ethanol 1

A 1.693 3.249 3.518

DA -1.424 Note: Light blue fields are inputs, pink fields are the final output.

1 RT

1

B

C

1 

1

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

  1 1  B 2 C 3 o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R  A T  T0   2 3  T T0  



D (K ) n iDHf,i n iAi 0.00E+00 0.00E+00 -1.69E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 0.00E+00 3.52E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -8.30E+03

B (1/K) 1.80E-02 4.22E-04 2.00E-02

C (1/K ) -6.16E-06 0.00E+00 -6.00E-06

DB (1/K) 1.60E-03

DC (1/K ) 1.56E-07





n iBi -1.80E-02 -4.22E-04 2.00E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



niCi 6.16E-06 0.00E+00 -6.00E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 1073.15

Species Stoichiometric Name coefficient acetaldehyde -1 H2 -1 ethanol 1

Entropy Change of Reaction at To J/(mol K)

Standard Entropy at T0 (J/(mol K))

Entropy Heat Capacity Change of Integral Reaction at T (dimensionless) J/(mol K) -0.543 Heat Capacity Coefficients

1 R



T 

C

1 

D  1

 T dT   Aln  T   B T  T   2 T  T   2  T  T  2

2 o

 T  C 2 D  1 1  o o S rxn A ln    B T  To   T  To2   2  2   ,T   S rxn ,T0  R    T 2 2 T T o   o   



D (K ) niDHf,i niAi 0.00E+00 0.00E+00 -1.69E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 0.00E+00 3.52E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -8.30E+03

A 1.693 3.249 3.518

B (1/K) 1.80E-02 4.22E-04 2.00E-02

C (1/K ) -6.16E-06 0.00E+00 -6.00E-06

DA -1.424 Note: Light blue fields are inputs, pink fields are the final output.

DB (1/K) 1.60E-03

DC (1/K ) 1.56E-07

n iBi -1.80E-02 -4.22E-04 2.00E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



niCi 6.16E-06 0.00E+00 -6.00E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

and this integral is

K 2  exp0.363  0.543  0.8353 6

So, at 350 °C, we have K = 8.7669 × 10 * 5.0546 × 10

* 0.8353 = 3.701.

Now, we need to relate this to the composition at equilibrium. For an ideal gas mixture, our general relationship: Solution continued on next page…

 N  fˆ  i  i  K     fo i1   i 



reduces to simply: N

i

 yi  i1

 P   P      K  o   K   P   1 bar 

and for this problem, at a pressure of 3 bar, this is yC H OH 2

y H yCH CHO 2

 3 bar    11.104  3.701  1 bar 

We can write the number of moles of species and the total number of moles in terms of a single reaction coordinate as n  no   nCH CHO  nCH CHO ,o   3

nH  nH ,o   2

nC H OH  nC H OH ,o   2

and yCH CHO  3

yH 

nCH CHO,o   3

no   nH ,o   2

no   nC H OH ,o   yC H OH  2 5 2 5 no   2

If we initially have H2 and acetaldehyde in a 1.5 to 1 ratio, and we do the calculation on the basis of 1 mole of acetaldehyde, we have 1  2.5   1.5   yH  2 2.5    yC H OH  2 5 2.5   yCH CHO  3

Substituting these into the equilibrium relationship, we have

 2.5   

1.5  1  

 11.104



2.5  2  11.104 1.5  2.5   2



12.104   30.26  16.656  0

Applying the quadratic formula to this last expression gives = 0.8182 or = 1.682 Looking back at the relationships between the mole fractions and the reaction coordinate, we see that a reaction coordinate greater than 1 would lead to a negative mole fraction for acetaldehyde, which is not allowed. So, the physically meaningful solution is = 0.8182, which give equilibrium mole fractions of 1  0.8182  0.1081 2.5  0.8182 1.5  0.8182 yH   0.4054 2 2.5  0.8182 0.8182 yC H OH   0.4865 2 5 2.5  0.8182 yCH CHO  3

If we reduced the pressure to 1 bar, then we would shift the equilibrium toward the reactants, because the change in mole number on reaction is negative. More quantitatively, we would have  1 bar    3.701  3.701  1 bar  y H yCH CHO yC H OH 2

2.5  

1.5  1  

 3.701



2.5  2  3.701 1.5  2.5  2



4.701  11.7525  5.5515  0

Applying the quadratic formula to this last expression gives = 0.6323 or = 1.868 so at 1 bar, we would have 1  0.6323  0.1969 2.5  0.6323 1.5  0.6323 yH   0.4646 2 2.5  0.6323 0.6323 yC H OH   0.3385 2 5 2.5  0.6323 yCH CHO  3

Solution 14.14

Problem Statement

The following reaction, hydrogenation of styrene to ethylbenzene, reaches equilibrium at 650°C and atmospheric pressure:

C6H5CH:CH2(g)

+ H2(g)



C6H5.C2H5(g)

If the system initially contains 1.5 mol H2 for each mole of styrene (C6H5CH:CH2), what is the composition of the system at equilibrium? Assume ideal gases.

Solution

We want to find the equilibrium composition at 650 ºC and 1 atm (1.013 bar) resulting from the gas phase reaction Styrene + H2  Ethylbenzene with 50% excess hydrogen in the feed (1.5 mol H2 per mol styrene). If our basis is 1 mol of styrene, then the mole numbers are related to the extent of reaction by nC8H8 = 1 nH2 = 1.5 nC8H10 = n = 2.5 The mole fractions are Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

yC8H8 = (1

)/(2.5

)

yH2 = (1.5

)/(2.5

)

yC8H10 = /(2.5

)

So, assuming an ideal gas mixture, the extent of reaction or reaction coordinate is related to the equilibrium constant by K

yC H 8



2.5   

1  1.5   2.5     K 1   1.5    0 ( K  1)2  2.5 K  1  1.5 K  0 yC H y H 8

2.5 K  1  6.25 K  1  6 K  K  1 2

 

2 K  1 2.5 K  1  0.25 K 2  6.5 K  6.25

  1.25 

2  K  1 0.25 K 2  6.5 K  6.25 2 K  1

In the above solution, we chose the negative sign in the quadratic formula, because it is clear that the positive sign would give a value of greater than 1.25. Since the number of moles of styrene is 1

, we require that

1 so that the styrene mole fraction remains positive. Now, the second part of solving the problem is to compute the value of K under the specified conditions. We can write it as

K  Ko K1 K2 where

 G o T   o   K o  exp RTo   is the value of the equilibrium constant at the reference temperature (To

 H o T     T  o  1   K 1  exp  RT  To  

is the main part of the temperature dependence, and is the change in the equilibrium constant that we would have if the enthalpy change of reaction were constant, and

For this reaction, we have º

Gf 298 (ethylbenzene)

Gf 298 (styrene)

Gf 298 (H2)

G (298 K) = 130890 H

Hf 298 (ethylbenzene)

Hf 298 (styrene)

Hf 298 (H2)

H (298 K) = 29920 So, we have

 83010    3.4961014 Ko  exp  8.314  298  And

 117440   1  923   1.174 1014 K 1  exp   8.314  923  298 

The heat capacity integrals for K2 can be evaluated using the handy-dandy spreadsheets, which end up looking like this:

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 923.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -117.440 -1.375 -127.990 Heat Capacity Coefficients

Standard Heat Species Stoichiometric of Formation Name coefficient at T 0 (kJ/mol) styrene -1 147.36 H2 -1 0 ethylbenzene 1 29.92

A 2.05 3.249 1.124

DA -4.175 Note: Light blue fields are inputs, pink fields are the final output.

1 RT

1

B

C

1 

1

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R  A T  T0   2 3  T T0  





D (K ) n iDHf,i n iAi 0.00E+00 -1.47E+02 -2.05E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 2.99E+01 1.12E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -8.30E+03

B (1/K) 5.02E-02 4.22E-04 5.54E-02

C (1/K ) -1.67E-05 0.00E+00 -1.85E-05

DB (1/K) 4.77E-03

DC (1/K ) -1.81E-06





n iBi -5.02E-02 -4.22E-04 5.54E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niCi 1.67E-05 0.00E+00 -1.85E-05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

And Temperature of Interest T (K) 923.15

Entropy Change of Reaction at To J/(mol K) -115.48

Species Stoichiometric Name coefficient styrene -1 H2 -1 ethylbenzene 1

Standard Entropy at T0 (J/(mol K)) -223.2 0.0 -338.7

Reference Temperature T0 (K) 298.15

Entropy Heat Capacity Change of Integral Reaction at T (dimensionless) J/(mol K) -2.474 -136.048 Heat Capacity Coefficients

1 R



T 

C

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   2

2 o

 T  C 2 D  1 1  o o  S rxn T  To2     B  T  To   ,T  S rxn ,T0  R  A ln   T 2 T 2    T 2 2  o  o   



D (K ) niDHf,i n iAi 0.00E+00 2.23E+02 -2.05E+00 8.30E+03 0.00E+00 -3.25E+00 0.00E+00 -3.39E+02 1.12E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -8.30E+03

A 2.05 3.249 1.124

B (1/K) 5.02E-02 4.22E-04 5.54E-02

C (1/K ) -1.67E-05 0.00E+00 -1.85E-05

DA -4.175

DB (1/K) 4.77E-03

DC (1/K ) -1.81E-06

n iBi -5.02E-02 -4.22E-04 5.54E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



n iCi 1.67E-05 0.00E+00 -1.85E-05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 0.00E+00 -8.30E+03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

So, the third piece of K is   T Cpo  1 T  o K 2  exp C p dT   dT     RT RT To To   K 2  exp1.375  2.474   0.3332 14

Finally, K = 3.496 × 10 *1.174 × 10

*0.3332 = 1.37. Substituting this into the expression for gives

= 0.416, from which the mole numbers are nC8H8 = 1

= 0.584

nH2 = 1.5

= 1.084

n = 2.5

= 2.084

and the mole fractions are yC8H8 = (1

)/(2.5

) = 0.280

yH2 = (1.5

)/(2.5

) = 0.520

yC8H10 = /(2.5

) = 0.200

Solution 14.15

Problem Statement

The gas stream from a sulfur burner is composed of 15-mol-% SO2, 20-mol-% O2, and 65-mol-% N2. This gas stream at 1 bar and 480°C enters a catalytic converter, where the SO2 is further oxidized to SO3. Assuming that the reaction reaches equilibrium, how much heat must be removed from the converter to maintain isothermal conditions? Base your answer on 1 mol of entering gas.

Solution

The amount of heat that must be removed is just the (negative of the) enthalpy of reaction, times the number of moles of reaction that occur. So, we need to (1) find the equilibrium constant at the specified reaction temperature, (2) use it to find the amount of reaction that occurs, and (3) multiply that amount of reaction (per mole of feed) by the heat of reaction (at the reaction temperature) to find the heat removal per mole of feed. If we start from 1 mol of gas of the specified composition, and the reaction that occurs is written as SO2 + ½ O2  SO3 Solution continued on next page…

Then nSO2 = 0.15 nO2 = 0.20

nSO3 = nN2 = 0.65 and

n=1

The mole fractions are then ySO2 = (0.15

)/(1

yO2 = (0.20

½ )/(1

ySO3 = /(1

½ )

yN2 = 0.65/(1

½ ) ½ )

½ )

The equilibrium relationship (assuming ideal gases) is then  P 0.5 K  o   K   P 

ySO

 

ySO yO 2

0.5



 1  0.5

0.15   0.20  0.5

where we have noted that the pressure is the same as the reference pressure (P/P = 1). Now, the second part of solving the problem is to compute the value of K under the specified conditions. We can write it as

K  Ko K1 K2 where

 G o T   o   K o  exp  RT  o  is the value of the equilibrium constant at the reference temperature (To

 H o T     T  o  1   K 1  exp  RT  To   is the main part of the temperature dependence, and is the change in the equilibrium constant that we would have if the enthalpy change of reaction were constant, and

  T Cpo  1 T  o  K2  exp Cp dT   dT    RT  RT To To  is the smaller temperature dependence due to the fact that the heat capacity of reactants and products is (in general) different, and therefore the enthalpy of reaction changes with temperature. For this reaction, we have º

Gf 298 (SO3)

Gf 298 (SO2)

Gf 298 (O2)

G (298 K) = H

Hf 298 (SO3)

Hf 298 (SO2)

Hf 298 (O2)

H (298 K) = So, we have

  70866   2.6051012 K o  exp  8.314  298.15  And    98890 1  753.15   3.4121011 K 1  exp   298.15   8.314  753.15 

The heat capacity integrals for K2 can be evaluated using the handy-dandy spreadsheets, which end up looking like this:

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 753.15

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -98.890 0.086 -98.353 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -296.83 -0.5 0 1 -395.72

Species Name SO2 O2 SO3

A 5.699 3.639 8.06

DA 0.5415 Note: Light blue fields are inputs, pink fields are the final output.

Reference Temperature T0 (K) 298.15

Species Name SO2 O2 SO3

Temperature of Interest T (K) 753.15

Entropy Change of Reaction at To J/(mol K) -93.99

Stoichiometric coefficient -1 -0.5 1

Standard Entropy at T0 (J/(mol K)) 11.28 0 -82.71

1 RT

1

C

2 0

1 

1

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R  A T  T0   2 3  T T0  



D (K ) n iDHf,i -1.02E+05 2.97E+02 -2.27E+04 0.00E+00 -2.03E+05 -3.96E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -9.00E+04

B (1/K) 8.01E-04 5.06E-04 1.06E-03

C (1/K ) 0.00E+00 0.00E+00 0.00E+00

DB (1/K) 2.00E-06

DC (1/K ) 0.00E+00

Entropy Heat Capacity Change of Integral Reaction at T (dimensionless) J/(mol K) 0.076 -93.361 Heat Capacity Coefficients

B

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T  

1 R

n iAi -5.70E+00 -1.82E+00 8.06E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00





n iBi -8.01E-04 -2.53E-04 1.06E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 1.02E+05 1.14E+04 -2.03E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00





T 

C

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   2

2 o

 T  C 2 D  1 1  o o S rxn T  To2       B  T  To   ,T  S rxn ,T0  R  A ln   2 2  T 2 To2    To  



D (K ) niDHf,i -1.02E+05 -1.13E+01 -2.27E+04 0.00E+00 -2.03E+05 -8.27E+01 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -9.00E+04

A 5.699 3.639 8.06

B (1/K) 8.01E-04 5.06E-04 1.06E-03

C (1/K ) 0.00E+00 0.00E+00 0.00E+00

DA 0.5415 Note: Light blue fields are inputs, pink fields are the final output.

DB (1/K) 2.00E-06

DC (1/K ) 0.00E+00

n iAi -5.70E+00 -1.82E+00 8.06E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iBi -8.01E-04 -2.53E-04 1.06E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



n iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 1.02E+05 1.14E+04 -2.03E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

So, we have K2 = exp( 0.086 + 0.076) = 0.990, and K = 2.605 × 10 *3.412 × 10

*.990 = 88.0.

Note that we could also get K directly using the enthalpy and entropy change of reaction at the actual reaction temperature (from the above spreadsheets): K = exp(

G (753 K)/(RT)) = exp((

H (753 K) + T S (753 K))/(RT)

K = exp ((98353 753.15*93.361)/(8.314*753.15)) = 88.0

These two approaches to finding the equilibrium constant are, of course, equivalent, and give the same answer to within the round-off error. Now, we can return to finding the equilibrium extent of reaction. We had Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

 1  0.5

0.15   0.20  0.5

K

My first thought was to rearrange this as 

K 0.15   0.20  0.5 1  0.5



88.0 0.15    0.20  0.5 1  0.5

And iterate to find a solution. However, that didn’t converge. Another reasonable rearrangement is  1  0.5

 0.15    K 0.20  0.5  1  0.5   0.15  K 0.20  0.5

Since K is significantly greater than one, we expect the majority of the limiting reactant to be consumed. So, we might start with a guess of = 0.1 (two-thirds of the SO2 is converted). Iterating on this version of the equation gives rapid convergence to = 0.1455. The final answer to the question is that the heat released by reaction is Q=

H (753 K) = 0.1455* 98353 =

oved per mole of gas

fed to the catalytic converter.

Solution 14.16

Problem Statement

For the cracking reaction,

C3H8(g)



C2H4(g)

CH4(g)

the equilibrium conversion is negligible at 300 K, but it becomes appreciable at temperatures above 500 K. For a pressure of 1 bar, determine:

(a) The fractional conversion of propane at 625 K. (b) The temperature at which the fractional conversion is 85%.

Solution

(a) The reaction to be considered is C3H8(g)  C2H4(g) + CH4 First at 625 K and 1 bar, and then as a function of temperature. First, we will calculate the equilibrium constant at that temperature. We can evaluate the equilibrium constant in the format

K  Ko K1 K2 where

 G o T   o   K o  exp RTo   is the value of the equilibrium constant at the reference temperature

Compound

Hf(298 K) J/mol

Gf(298 K) J/mol

C3H8(g)

104680

24290

C2H4(g) CH4(g)

52510

68460

74520

50460

H (298 K) = 52510 o

G (298 K) = 68460

and

So, we have

 42290    3.901108 Ko  exp  8.3145  298.15  and  82670      1  T   exp 82670 1  625   3.750 10 7 K 1  exp     8.3145  T   8.3145  625  298.15  298.15 

Finally, we need to evaluate the heat capacity integrals from 298.15 to 625 K to evaluate K2. We can do this using our handy spreadsheets and the heat capacity polynomials given in table C.1. Doing so, we have for the first integral: Reference Temperature T 0 (K) 298.15

Species Name Propane Ethylene Methane

Temperature of Interest T (K) 625

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T 0 kJ/mol (dimensionless) kJ/mol 0.000 -0.018 -0.094 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 1 1

A 1.213 1.424 1.702

DA 1.913 Note: Light blue fields are inputs, pink fields are the final output.

1 RT

1

 C dT  T  AT  T  p

T  T0 

T  T0  D

 o o H rxn ,T   H rxn ,T0  R  A T  T0  

B (1/K) 2.88E-02 1.44E-02 9.08E-03

C (1/K ) -8.82E-06 -4.39E-06 -2.16E-06

D (K ) 0.00E+00 0.00E+00 0.00E+00

DB (1/K) -5.31E-03

DC (1/K ) 2.27E-06

DD (K ) 0.00E+00

T  T0 

1 1    T T 

T  T0  D



Notice that I’ve just left the heat of reaction and heat of formation entries blank, since we don’t need them for this problem. The heat capacity integral is Likewise for the second integral, we have Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) 0.00 0.023 0.187 Heat Capacity Coefficients Standard Stoichiometric Entropy at T 0 coefficient (J/(mol K)) A B (1/K) -1 1.213 2.88E-02 1 1.424 1.44E-02 1 1.702 9.08E-03

Reference Temperature T 0 (K) 298.15

Temperature of Interest T (K) 625

Species Name Propane Ethylene Methane

DA 1.913 Note: Light blue fields are inputs, pink fields are the final output.

1 R



T 

 T dT   A ln  T   B T  T 

 o o S rxn ,T  S rxn ,T0  R  A  

DB (1/K) -5.31E-03

T T

C (1/K ) -8.82E-06 -4.39E-06 -2.16E-06

D (K ) 0.00E+00 0.00E+00 0.00E+00

DD (K ) 0.00E+00

DC (1/K ) 2.27E-06

 B T  To 

and this integral is 0.023. So, we have

K 2  exp0.018  0.023  1.042 This contribution to the temperature dependence is pretty negligible. Putting it all together, we have, at 625 K, K = 3.901 × 10

* 3.750 × 10 * 1.042 = 1.524.

Now, we need to relate this to the composition at equilibrium. For an ideal gas mixture, our general relationship:  N  fˆ  i  i  K     fo i1   i 



reduces to simply:  P   P  i     y  K  K  o   i   1 bar   P  N

i1

and for this problem, at a pressure of 1 bar, this is

yC H yCH 2

yC H 3



 1 bar 1   1.524  K   1 bar 

T  To 

   



We can write the number of moles of species and the total number of moles in terms of a single reaction coordinate (using a basis of 1 mole of propane) as n  1  nC H  1   3

nC H   nCH   4

and 1  1  yC H  2 4 1  yCH  4 1 yC H  3

Substituting these into the equilibrium relationship, we have   2    1     1       1   



2 2  K 1  1   1  2





2 1  K   K K 1.524   0.777 1 K 2.524



Putting this into the equations for the mole fractions gives 1  0.736  0.152 1  0.736 0.736 yC H   0.424 2 4 1  0.736 0.736 yCH   0.424 4 1  0.736 yC H  3

Since the fractional conversion is defined as the fraction of the propane that has reacted, and we start with 1 mole of propane, the fractional conversion is the same as the extent of reaction in this case (fractional conversion = 77.7%). (b) A fractional conversion of 85% means that 85% of the propane that was initially present has reacted. Since we started with 1 mole of propane, 85% has reacted when = 0.85. We saw in part (a) that

Solution continued on next page…

  2    1     1       1   



2 2  K 1  1   1  2





So, to get = 0.85, we need K = 0.85 / (1

) = 2.604. At 625 K we had K = 1.182, so the temperature

needs to be just a little higher. From part (a), we have

 42920    3.901108 Ko  exp  8.3145  298.15  and   82670   1  T  K 1  exp   298.15   8.3145  T 

and K2 was pretty negligible (perhaps 1.04 or so). So, we can write:  82670   1  T   1.04  2.604 K  3.901108 exp    8.3145  T  298.15   82670  1    1   6.41810 7 exp   8.3145  T 298.15  82670  1 1    17.977   8.3145  T 298.15  1 1   0.001808  0.001546 T 298.15

From which T = 646.8 K. Re-evaluating K2 at this temperature gives

K 2  exp0.036  0.002  1.035 which is within half a percent of what we used based on the result at 625 K, so we do not need to change it and re-calculate T.

Solution 14.17

Problem Statement

Ethylene is produced by the dehydrogenation of ethane. If the feed includes 0.5 mol of steam (an inert diluent) per mole of ethane and if the reaction reaches equilibrium at 1100 K and 1 bar, what is the composition of the product gas on a water-free basis?

Solution

C2H6(g) = H2(g) + C2H4(g)

On a water free basis.

n0  By Eq. 14.5:

yC 2 H 6 

1  1 

yH 

 1

yC 2 H 4 

 1

From data in Table C.4

H298  136330

J J G298  100315 mol mol

The following table represents the species of the reaction in the order in which they appear:

1.131

0.019225

3.249

0.000422

1.424

0.014394

D 0

8300 0

Solution continued on next page…

A   vi Ai B   ( vi Bi ) C   viCi  D   vi Di  i

A  3542 B  4409  103 C  1169  106 D  83  103 T  1100 K T  29815 K

G  H298 

T H298 G298   R  IDCPH T0 ,T ,A,B,C,D RTIDCPS T0 ,T ,A,B,C,D T0 G  5429  103

 G  J  K  181048 K  exp  RT  mol

By Eq 14.28, determine  2  K   08026 1  1  

By Eq. 14.4, determine that concentrations of the gases yC 2 H 6  0109 y H  04452

yC 2 H 4  04452

Solution 14.18

Problem Statement

The production of 1,3-butadiene can be carried out by the dehydrogenation of 1-butene:

C2H5CH:CH2(g)



H2C:CHHC:CH2(g)

+ H2(g)

Side reactions are suppressed by the introduction of steam. If equilibrium is attained at 950 K and 1 bar and if the reactor product contains 10-mol-% 1,3-butadiene, find: (a) The mole fractions of the other species in the product gas. (b) The mole fraction of steam required in the feed.

Solution

C2H5CH:CH2(g) = CH2:CHCH:CH2(g) + H2(g) Number the species as shown. Basis is 1 mol species 1 + x mol steam. n0 = 1 = x y1 

By Eq. (14.5),

1   , y2  y3  = 0.10 1 x  1 x  

From data in Table C.4

H 298  109780

J J G298  79455 mol mol

The following vectors represent the species of the reaction in the order in which they appear:

1    V =  1     1 

1967   A  2734   3249

A  

B   

31630   B   26786 103    0422 



C  

9873   C  8882106    00 



G  H

H  G

T T

 RTIDCPS T

G  4896 x10

J mol

By Eq. (13.28),

X :=

 1   010

K K  010

T  A B C D

K = 0.53802

01011  x   1 

  R.IDCPH T T  A B C D

K

= 0.843

x = 6.5894

1  YH 2O  1– 02 – y1 1 x 

Y 1  00186 (b)

 G   K  exp  R.T 

   x      

(a) Y1 

D   x

T 0  298.15kelvin

T = 950

Since

 00    D   00 105   0083

Ysteam 

YH2O  07814

65894 Ysteam  08682 75894

Solution 14.19

Problem Statement

The production of 1,3-butadiene can be carried out by the dehydrogenation of n-butane:

C4H10(g)



H2C:CHHC:CH2(g)

2H2(g)

Side reactions are suppressed by the introduction of steam. If equilibrium is attained at 925 K and 1 bar and if the reactor product contains 12-mol-% 1,3-butadiene, find: (a) The mole fractions of the other species in the product gas. (b) The mole fraction of steam required in the feed.

Solution

C4H10(g) = CH2:CHCH:CH2(g) + 2H2(g) (1)

(2)

(3)

v=2

Number the species as shown. Basis 1 mol species 1 + x mol steam entering. By Eq. (13.5),

Y1 

N0 = 1 + x

  , Y2  = 0.12 ,  x    x  

Y3 = 2

2 = 0.24

From data in Table C.4

H





J G mol





J mol

The following vectors represent the species of the reaction in the order in which they appear:

1 1.935 36.915 11.402  0.0                 3   6 v :  1  A : 2.734B : 26.78610 C :  8.882 10 D :  0.0 105            2  3.249  0.422   0.0  0.083

A   vi  Ai 

B   vi  Bi 

C   vi Ci

D   vi  Di

A 

B 

T = 925

3

C 

6

D 

T0= 298.15

G  H298 

T H298 G298   R IDCPH T0 T A B C D  RT  T0

IDCPS T0 ,T , A, B, C , D

G 

J mol



 G     RT 

K = 0.30066

0.12(0.24) 1  x  2

By Esq. (14.28),



  x    

Because

x 

    0.12

(a):

Y1 



x = 4.3151

 YH O   1  x  2 

Ysteam 

4.3151 5.3151

K K  (0.24)2

  0.839

 Y1

Y1 

(b):

K



Ysteam 

Solution 14.20

Problem Statement

For the ammonia synthesis reaction,

1/2N2(g)+3/2H2



NH3(g)

the equilibrium conversion to ammonia is large at 300 K, but it decreases rapidly with increasing T. However, reaction rates become appreciable only at higher temperatures. For a feed mixture of hydrogen and nitrogen in the stoichiometric proportions, (a) What is the equilibrium mole fraction of ammonia at 1 bar and 300 K? (b) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure of 1 bar? (c) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal gas? (d) At what temperature does the equilibrium mole fraction of ammonia equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal solution of gases?

Solution

For this one, let’s work out a general relationship between the mole fractions and the equilibrium constant in advance, then answer the questions in parts (a) through (d). If we have a mixture of ideal gases with, initially 1 mole of N2 and 3 moles of H2 (so I can use whole numbers) then n  4 nN  1  0.5 2

nH  3  1.5 2

nNH   3

Solution continued on next page…

and

1  0.5 4 3  1.5 yH  2 4  yNH  3 4 yN  2

and the mole fractions and extent of reaction are related to the equilibrium constant by  P   P  i     y  K  K  o  i   P   1 bar  N

i1

y NH

y  y  0.5

1.5

 P    K   1 bar 

 4  

1  0.5 3  1.5 0.5

1.5

 KP

For these particular initial concentrations, we can simplify this by recognizing that (3

) = 3*(1

), so

we get  4   

1  0.5 3 2

1.5

 KP

For given values of the equilibrium constant and pressure, we could solve this iteratively for . Or, if we want to find the equilibrium constant or pressure required to get a certain value of , we could do that directly. So, on to the questions: (a) To find the equilibrium composition at 300 K, we need to compute the standard enthalpy and Gibbs energy changes for reaction. Conveniently, this reaction is the formation reaction for ammonia, so the enthalpy and Gibbs energy changes are just the enthalpy of formation and Gibbs energy of formation of ammonia. At 298.15 K, these are o

H (298 K) =

and G (298 K) =

So, at 298.15 K, the equilibrium constant is

  16450   761.9 K  Ko  exp  8.3145  298.15  The change in K in going from 298.15 to 300 K is

 46110   1  300   0.8833 K 1  exp   298.15   8.3145  300 

The temperature dependence is strong – the 0.67% increase in temperature leads to a 12% decrease in the rate constant! So, at 300 K, K = 0.8833*761.9 = 673. (we can neglect the secondary temperature dependence for this small change in temperature). So, at 300 K and 1 bar, we have  4  

1  0.5 3 2

1.5

 673

This is rather tricky to solve. The allowable range of the reaction coordinate is from zero to 2 (when = 2, the reactants are used up). Since the equilibrium constant is much greater than 1, equilibrium at these conditions lies toward the product, and the reaction coordinate will be near its maximum value (near 2). When = 2, the denominator goes zero. To avoid having a zero in the denominator, let’s rearrange it as  4    6733

1.5

 1  0.5

      4     4        21    21  1.5     3497   673 3      

Starting with an initial guess of 2 and iterating on this gives = 1.932. So, the mole fractions at equilibrium at 300 K are yN  2

yH 

1  0.51.932 4  1.932 3  1.51.932

 0.016

 0.049 4  1.932 1.932 yNH   0.934 3 4  1.932 2

(b) To have y NH  3

 4    4   0.5 we require = /3 = 1.333. Substituting this into  KP gives 2 1.5 4 1  0.5  3  

1.3334  1.333

1  0.51.333 3 2

1.5

 6.158  KP. So, we are looking for the temperature where K = 6.158 (since P = 1 bar).

Neglecting, for the moment, the secondary dependence of K on temperature, we would have Solution continued on next page…

 46110   1  T  K  K 0 K1  761.9exp  8.3145  T  298.15     46110  46110  exp  K  761.9exp  8.3145  298.15   8.3145  T   46110   K  6.365 106 exp  8.3145  T 

So, to find the T where K = 6.158, we take

 6.158  46110  ln  6.365106  8.3145  T T

46110  402 K  6.158    8.3145  ln   6.365106 

1  0.51.333 3 2

1.5

 6.158  KP  100 K .

So, now we are looking for the temperature where K = 0.06158.

Neglecting, for the moment, the secondary dependence of K on temperature, we again have

 46110   K  6.365106 exp  8.3145  T  So, the temperature is given by

 0.06158  46110  ln  6.365106  8.3145  T 46110 T  604 K  0.06158   8.3145  ln   6.365106 

To include the secondary temperature dependence, we have to do the heat capacity integrals. For the first one, we get

Solution continued on next page…

Reference Temperature T0 (K) 298.15

Temperature of Interest T (K) 604

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -0.997 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T0 (kJ/mol) -0.5 -1.5 1

Species Name N2 H2 NH3

1 RT

1

B

C

1

1 

 C dT  T  AT  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R A  T  T 0   2 3  T T0  





D (K ) n iDHf,i n iAi 4.00E+03 0.00E+00 -1.64E+00 8.30E+03 0.00E+00 -4.87E+00 1.86E+04 0.00E+00 3.58E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) 4.15E+03

A 3.28 3.249 3.578

B (1/K) 5.93E-04 4.22E-04 3.02E-03

C (1/K ) 0.00E+00 0.00E+00 0.00E+00

DA -2.9355

DB (1/K) 2.09E-03

DC (1/K ) 0.00E+00





n iBi -2.97E-04 -6.33E-04 3.02E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi -2.00E+03 -1.25E+04 1.86E+04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

For the second one, we get Reference Temperature T0 (K) 298.15

Species Name N2 H2 NH3

Temperature of Interest T (K) 604

Stoichiometric coefficient -0.5 -1.5 1

Entropy Change of Reaction at To J/(mol K)

Entropy Heat Capacity Change of Integral Reaction at T (dimensionless) J/(mol K) -1.415 Heat Capacity Coefficients

Standard Entropy at T0 (J/(mol K))

1 R



T 

C

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   2

2 o

 T  C 2 D  1 1  o o S rxn T  To2      B  T  To   ,T  S rxn ,T0  R  A ln   T 2 T 2    T 2 2  o   o  



D (K ) niDHf,i n iAi 4.00E+03 0.00E+00 -1.64E+00 8.30E+03 0.00E+00 -4.87E+00 1.86E+04 0.00E+00 3.58E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) 4.15E+03

A 3.28 3.249 3.578

B (1/K) 5.93E-04 4.22E-04 3.02E-03

C (1/K ) 0.00E+00 0.00E+00 0.00E+00

DA -2.9355

DB (1/K) 2.09E-03

DC (1/K ) 0.00E+00

n iBi -2.97E-04 -6.33E-04 3.02E-03 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



n iCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi -2.00E+03 -1.25E+04 1.86E+04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

So, we have K2 = exp(0.997 1.415) = 0.6584 So, we then have

 46110   46110    0.6584  4.1905106 exp  K  K0 K1 K2  6.365106 exp  8.3145  T   8.3145  T  T

46110  578.0 K  0.06158  8.3145  ln    4.1905 106 

Re-doing the heat capacity integrals with this final temperature gives K2 = exp(0.966 1.341) = 0.6873, and

 46110   46110    0.6873  4.3747106 exp K  K0 K1 K2  6.365106 exp   8.3145  T   8.3145  T 

T

46110  580.6 K  0.06158  8.3145  ln    4.3747 106 

Re-calculating again wouldn’t lead to any significant change in T.

(d) If we (more nearly correctly) treat the gas mixture as an ideal mixture of non-ideal gases, then instead of N

i

 yi  i1

 P   P     K    K   P o   1 bar   P     K 1.5  1 bar 

y NH

y  y  0.5

 4  

1  0.5 3  1.5 0.5

1.5

 KP

We have N

 P 

i

i yi   K  o  P  NH y NH

i1

 y   y  0.5

 P    K   1 bar 

1.5

 P    K   1 bar 

     NH  3   KP 0.5 1.5  0.5 1.5   1  0.5 3  1.5  N2 H2   4   

   

So now we need to evaluate the pure species fugacity coefficients at the (unknown) final temperature and 4

(known) pressure. For a NH3 mole fraction of 0.5, we still need = /3, so now we have         NH NH  3 3   6.158   KP    2 0.5 1.5  0.5 1.5  1.5      H 1  0.51.333 3  N2 H2   N2  2 1.3334  1.333

   

and we need to find the temperature where     NH 6.158  3   K 0.5 1.5   P  H  N2  2

   

A reasonable first guess for the temperature is the answer we found in part (c), where we assumed that the fugacity coefficients were 1. So, we will (1) evaluate the fugacity coefficients at 580.6 K, then (2) use them to find the K that we need, then (3) find the T that gives us the needed K, just like in parts (b) and (c), and (4) repeat this, evaluating the fugacity coefficients at the newly found temperature, until the temperature converges. We can use the Lee/Kesler correlation to find the fugacity coefficients. The critical properties of the species are: NH3:

Tc = 405.7 K, Pc = 112.8 bar,

= 0.253

N2:

Tc = 126.2 K, Pc = 34.0 bar,

= 0.038

H2:

Tc = 33.19 K, Pc = 13.13 bar,

So, at 580.6 K and 100 bar, the reduced temperature and pressure for each species are: NH3:

Tr = 1.431, Pr = 0.8865

N2:

Tr = 4.601, Pr = 2.941

H2:

Tr = 17.49, Pr = 7.616

For these values, we can use the 2-term virial equation version of the Lee/Kesler correlation, because for all three species these reduced conditions fall within the range where it fits the full (tabular) correlations well. Note that the 2-term virial equation form always works if the reduced temperature is above about 3. Luckily, NH3 has a very high critical pressure, so the reduced pressure is low enough that it also works for NH3. For each species, we want to evaluate  P  i  exp  Bi0   Bi1 r   Tr 





and conveniently, we have a handy-dandy spreadsheet for doing this, from the Lecture 17 notes. For NH3, we get

T (K)

P (bar)

580.6

100

Tc (K) Pc (bar) 405.7

112.8

 0.253



1.4311

0.8865

-0.1548

0.1008

0.9230

0.422

B 0  0.083 

Tr1.6 0.172

B  0.139 

 P    exp  B0   B1 r  Tr  





Tr4.2

For N2 we get

T (K)

P (bar)

Tc (K)

Pc (bar)

100

126.6

580.6

 0.038



4.5861

2.9412

0.0461

0.1387

1.0335

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 P    exp  B 0   B1 r  Tr  





Tr4.2

and for H2, we get

T (K)

P (bar)

580.6

100

Tc (K) Pc (bar) 33.19

13.13

 -0.216



17.4932

7.6161

0.0787

0.1390

1.0214

B 0  0.083 

0.422

B1  0.139 

0.172

Tr1.6

 P    exp  B0   B1 r  Tr  





Tr4.2

Using these values in our equilibrium expression, we have     NH 6.158  3   K 0.5 1.5   P  H  N 2  2   6.158  0.9230   K   0.06158 0.8795  0.05416 0.5 1.5 100  1.0335 1.0214    

   

Solution continued on next page…

Putting this into the equation for T that we had in part (d), which already had in it the value of K2 at 580.6 K, we get T

46110  588.5 K  0.05416   8.3145  ln    4.3747 106 

Re-evaluating the fugacity coefficients and K2 at this temperature causes negligible changes (less than a degree), so we can consider this our final answer. In the solutions manual, it is assumed that H2 is ideal, but here we compute a fugacity coefficient of 1.021. This enters the calculation to the 3/2 power, and leads to a difference of a few K in the final answer.

Solution 14.21

Problem Statement

For the methanol synthesis reaction,

CO(g)

2H2(g)



CH3OH(g)

the equilibrium conversion to methanol is large at 300 K, but it decreases rapidly with increasing T. However, reaction rates become appreciable only at higher temperatures. For a feed mixture of carbon monoxide and hydrogen in the stoichiometric proportions,

(a) What is the equilibrium mole fraction of methanol at 1 bar and 300 K? (b) At what temperature does the equilibrium mole fraction of methanol equal 0.50 for a pressure of 1 bar? (c) At what temperature does the equilibrium mole fraction of methanol equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal gas? (d) At what temperature does the equilibrium mole fraction of methanol equal 0.50 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal solution of gases?

Solution

(a) For the reaction CO(g) + 2H2(g)  CH3OH(g) The enthalpies, entropies, and Gibbs energies of formation of the reactants and products are o

Gf (298 K)

Sf (298 K)

J/mol

J/(mol K)

CO (g) H2 (g)

Hf (298 K)

110,525 0

CH3OH (g)

137,169 0

200,660

89.36 0

161,960

129.80

So, for the reaction, we have º

H (298 K) = º

G (298 K) = º

S (298 K) = At 298.15 K, the equilibrium constant is K = exp(24791/(8.314*298.15)) = 22050 º

G (300 K) =

*300 =

K = exp(24387/(8.314*300)) = 17632. The

equilibrium constant decreased by 20% due to just a 1.85 K change in temperature. This reminds us of how strongly equilibrium constants can depend on T. For a stoichiometric feed, the mole numbers and mole fractions are nCO = 1 nH2 = 2 nCH3OH = Solution continued on next page…

n=3 yCO = (1

)/(3

yH2 = 2(1

)/(3

yCH3OH = /(3

) ) )

So, the equilibrium relationship is yCH OH 3

2 yCO y H

 3  2



 P 2    K 3  P o  4 1   

At 1 bar, P = P , and at 300 K, we found that K = 17632. So, we have  3  2

4 1   

 17632

Since K is much greater than 1, we expect the equilibrium to lie very close to the products, so (1

) is nearly

zero. Given that expectation, a good way to rearrange the above equation for iterative solution is 1

2 3   3  2     1    4  17632   

Starting from a guess of = 1 and iterating, this converges rapidly to = 0.9752 So, at 300 K and 1 bar, the equilibrium mole fraction of methanol is yCH3OH = /(3

) = 0.929

Like the problem statement says – the equilibrium conversion is high at 300 K. (b) To have a methanol mole fraction of 0.5, we require yCH3OH = /(3

) = 0.500

from which = 0.75. Substituting this into the equilibrium relationship, we have

yCH OH 3

2 yCO y H

3  2

0.753  1.5

41  

4 1  0.75 



 3

 P 2  K  o   P 

 P 2 K  o   27  P  Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

So, at P = 1 bar, we need to find the temperature where K = 27. The value of K is related to the enthalpy, entropy, and Gibbs energies of reaction as R ln K =

G (T)/T =

H (T)/T + S (T)

Solving this for T gives º

H (T

S (T) H

R ln K) º

S at 298 K in this expression gives 219.16 8.314ln(27)) = 366 K H

S at 366 K, then use them to re-calculate T, and

then repeat if necessary. Using our handy heat capacity integral spreadsheets, we get: Reference Temperature T 0 (K) 298.15

Species Name CO H2 CH3OH

Temperature of Interest T (K) 366

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -90.135 -0.849 -92.717 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -110.53 -2 0.00 1

-200.66

1 RT

1

B

C

1 

1

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R  A T  T0   2 3  T T0  









A 3.376 3.249

B (1/K) 5.57E-04 4.22E-04

C (1/K ) 0.00E+00 0.00E+00

D (K ) n iDHf,i n iAi n iBi -3.10E+03 1.11E+02 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

niCi 0.00E+00 0.00E+00

n iDi 3.10E+03 -1.66E+04

2.211

1.22E-02

-3.45E-06

0.00E+00 -2.01E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DB (1/K) 1.08E-02

DC (1/K ) -3.45E-06

DA -7.663 Note: Light blue fields are inputs, pink fields are the final output.

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -1.35E+04

Temperature of Interest T (K) 366

Entropy Change of Reaction at To J/(mol K) -219.16

Species Name CO H2

Stoichiometric coefficient -1 -2

Standard Entropy at T0 (J/(mol K)) 89.36 0.00

A 3.376 3.249

B (1/K) 5.57E-04 4.22E-04

C (1/K ) 0.00E+00 0.00E+00

D (K ) niDHf,i n iAi n iBi -3.10E+03 -8.94E+01 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

n iCi 0.00E+00 0.00E+00

n iDi 3.10E+03 -1.66E+04

CH3OH

-129.80

2.211

1.22E-02

-3.45E-06

0.00E+00 -1.30E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DA -7.663 Note: Light blue fields are inputs, pink fields are the final output.

DB (1/K) 1.08E-02

DC (1/K ) -3.45E-06

Reference Temperature T0 (K) 298.15

Entropy Heat Capacity Change of Integral Reaction at T (dimensionless) J/(mol K) -0.941 -226.986 Heat Capacity Coefficients

2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1 R



T 

C

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   2

2 o



2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



DD (K ) -1.35E+04

Solution continued on next page…

From which T=

226.99

Re-

8.314 ln(27)) = 364.5 K

S at 364.5 K results in negligible change, so this is our answer.

3  2

0.753  1.5

41  

4 1  0.75 

yCH OH



2 yCO y H

 3

 P 2  K  o   P 

 P 2 K  o   27  P  o

But with (P/P ) = 100, this gives K = 0.0027. With that, we proceed just as in part (b): º

R ln K)

8.314ln(0.0027)) = 530 K

Re-

Reference Temperature T 0 (K) 298.15

Species Name CO H2 CH3OH

T = H (T) / ( S (T) 219.16

Temperature of Interest T (K) 530

S at 530 K gives

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol -90.135 -1.696 -97.609 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -110.53 -2 0.00 1

-200.66

1 RT

1

B

C

1

1 

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R  A T  T0   2 3  T T0  









A 3.376 3.249

B (1/K) 5.57E-04 4.22E-04

C (1/K ) 0.00E+00 0.00E+00

D (K ) n iDHf,i n iAi n iBi -3.10E+03 1.11E+02 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

niCi 0.00E+00 0.00E+00

n iDi 3.10E+03 -1.66E+04

2.211

1.22E-02

-3.45E-06

0.00E+00 -2.01E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DB (1/K) 1.08E-02

DC (1/K ) -3.45E-06

DA -7.663 Note: Light blue fields are inputs, pink fields are the final output.

2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) -1.35E+04

Temperature of Interest T (K) 530

Entropy Change of Reaction at To J/(mol K) -219.16

Species Name CO H2

Stoichiometric coefficient -1 -2

Standard Entropy at T0 (J/(mol K)) 89.36 0.00

A 3.376 3.249

B (1/K) 5.57E-04 4.22E-04

C (1/K ) 0.00E+00 0.00E+00

D (K ) niDHf,i n iAi n iBi -3.10E+03 -8.94E+01 -3.38E+00 -5.57E-04 8.30E+03 0.00E+00 -6.50E+00 -8.44E-04

n iCi 0.00E+00 0.00E+00

n iDi 3.10E+03 -1.66E+04

CH3OH

-129.80

2.211

1.22E-02

-3.45E-06

0.00E+00 -1.30E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-3.45E-06 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DA -7.663 Note: Light blue fields are inputs, pink fields are the final output.

DB (1/K) 1.08E-02

DC (1/K ) -3.45E-06

Reference Temperature T0 (K) 298.15

H T=

Entropy Heat Capacity Change of Integral Reaction at T (dimensionless) J/(mol K) -2.284 -238.155 Heat Capacity Coefficients

1 R



T 

C

1 

D  1

 T dT   A ln  T   B T  T   2 T  T   2  T  T   2

2 o

 T  C 2 D  1 1  o o S rxn T  To2   2  2     B  T  To   ,T  S rxn ,T0  R  A ln    T 2 2 T T  o  o  



2.21E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.22E-02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



DD (K ) -1.35E+04

S gives

238.155

8.314ln(0.0027)) = 516.5 K H

S at 516.5 K, then using the results to compute a new value of T

gives T = 516.5 K again, so the process has converged and this is our final answer.

(d) If we consider the gases to be an ideal mixture of non-ideal gases, then we have

CH OH yCH OH 3



CO yCO H y H 2

 2



CH OH 3

 3  2

 P 2    K 2 3  P o  4 1  

 

CO H

The fugacity coefficients in this equation are the pure component fugacity coefficients, which are independent of the mixture composition. Thus, for a given temperature and pressure, we can compute them without knowing the composition. So, we can start by computing them at the temperature found in part (c), then iterating once again. To obtain the fugacity coefficients from generalized correlations, we need the critical properties of each reactant and product. For CO, we can use either the Pitzer correlation or interpolate in the Lee/Kesler tables. Using the Pitzer correlation we get:

T (K)

P (bar)

516.5

100

Tc (K) Pc (bar)



132.9

34.99

0.048



3.8864

2.8580

0.0349

0.1384

1.0310

B 0  0.083  1

B  0.139 

0.422 Tr1.6 0.172

 P    exp  B0   B1 r  Tr  





Tr4.2

Interpolating directly in the Lee/Kesler tables gave

CO = 1.0402 (less than 1% different). For hydrogen, the

reduced temperature is so high that it falls outside the range of the Lee/Kesler tables, as given in the book, so we will use the Pitzer correlation:

T (K)

P (bar)

516.5

100

Tc (K) Pc (bar)



33.19

-0.216

13.13



15.5619

7.6161

0.0778

0.1390

1.0236

B 0  0.083  1

B  0.139 

0.422 Tr1.6 0.172

 P    exp  B0   B1 r  Tr  





Tr4.2

At such high reduced temperature, the gas is nearly ideal, even at very high reduced pressure. For methanol, we must interpolate in the Lee/Kesler tables, since the reduced temperature and pressure are outside the range of validity of the Pitzer correlation:

T (K) 516.5

P (bar) 100

Tc (K) 512.6

Pc (bar) 80.97

Tr 1.00 1.00 1.01 1.01

Pr 1.20 1.50 1.20 1.50

 0.564

Tr 1.0076

Pr 1.2350

Table Points Tr (1) Tr(1) Tr(2) Tr(2)

Pr(1) Pr(2) Pr(1) Pr(2)

0.5781 0.4875 0.5970 0.5047

0.9204 0.9078 0.9462 0.9333

Interpolated Values

0.5818

0.9385

Final Value

0.5613

So, finally, we have: CH OH yCH OH 3



CO yCO H y H 2

 2



3  2

27  0.5613

  41  

1.04021.0236

CH OH 3

CO H

 3

 P 2    10 4 K  K 2  P o 

From which K = 0.001391. Using the enthalpy and entropy of reaction at 516.5 K that were found in part (c) with this value of the equilibrium constant gives: T=

238.155

8.314 ln(0.001391)) = 532.0 K º

Re-evaluating the enthalpy and entropy of

H =

J/(mol K). Re-evaluating the fugacity coefficients at 532.0 K gives

CH3OH = 0.6268,

S = CO = 1.0417, and

H2 =

1.0231. Using these values, we have CH OH yCH OH 3



CO yCO H y H 2

 2



3  2

CH OH 3

 3

  41  

CO H

 P 2    10 4 K  K 2  P o  1.0417 1.0231 27  0.6268

From which K = 0.001641 and then T=

238.248 8.314*ln(0.001641)) = 528.1 K.

An additional iteration might still change the result slightly, but would probably not be worth the effort required.

Solution 14.22

Problem Statement

Limestone (CaCO3) decomposes upon heating to yield quicklime (CaO) and carbon dioxide. At what temperature is the decomposition pressure of limestone 1(atm)?

Solution

For the reaction CaCO3 (s)  CaO (s) + CO2 (g), the CaCO3 and CaO can be considered to exist as pure solid phases, while the CO2 exists only as a pure ideal gas. Thus, when we write the equilibrium relationship, we get  fˆ  ˆ    CO2  fCaO   yCO P  2 1  o  o   i  fCO  fCaO   P o  N  fˆ   P  2 K   io     o    ˆ  1  fi   P i1   fCaCO3    o  fCaCO   3 

The vapor phase mole fraction of CO2 is 1, and the fugacities of the solids can just be taken to be equal to their pure component fugacities, since they exist as two separate solid phases, each of which is a pure component. This is usually, but not always, the case for solid phase reactions like this. From the equation above, we see that the equilibrium pressure exerted by the CO2 from the decomposing is just proportional to the equilibrium constant for the reaction. The decomposition pressure will be one atmosphere when K = 1 atm/1 bar = 1.013. So, we have K = exp( H

G /(RT)) = 1.013. Taking the natural

S , we have

Solution continued on next page…

H /T

S = R ln(1.013) = 0.107 o

Initially taking H and S to be constant and solving for T gives T o

S ). For the reactant and

S for the reaction at 298.15 K. From these, we can get an initial answer. If we want to refine that answer, H

S for the reaction at the temperature

found using the values from 298K. If necessary, we could iterate again to further refine the answer. We have: o

Gf (298 K)

Sf (298 K)

J/mol

J/(mol K)

CaCO3 (s)

1,206,920

1,128,790

CO2 (g)

393,509

394,359

CaO (s)

635,090

604,030

Hf (298 K)

262.05 2.85 104.18

S (298) = 160.7 J/(mol K). So, our first estimate for T is T =

178321/160.8 = 1109 K. This is pretty far from the reference temperature, so we had better do the heat capacity integrals to get an improved estimate. Putting the relevant information into the handy spreadsheets, we have Reference Temperature T 0 (K) 298.15

Species Name CaCO3 CO2 CaO

Temperature of Interest T (K) 1109

Heat Capacity Heat of Heat of Integral Reaction at T Reaction at T0 kJ/mol (dimensionless) kJ/mol 178.321 -1.128 167.923 Heat Capacity Coefficients

Standard Heat Stoichiometric of Formation coefficient at T 0 (kJ/mol) -1 -1206.92 1 -393.509 1 -635.09

A 12.572 5.457 6.104

DA -1.011 Note: Light blue fields are inputs, pink fields are the final output.

1 RT

1

B

C

1

1 

 C dT  T  A T  T   2 T  T   3 T  T   D  T  T   p

2 0

3 0

 1 B 2 C 3 1  o o H rxn T  T02  T  T03  D    ,T   H rxn ,T0  R  A T  T0   2 3  T T0  



C (1/K )

B (1/K) 2.64E-03 1.05E-03 4.43E-04

0.00E+00 0.00E+00 0.00E+00

DB (1/K) -1.15E-03

DC (1/K ) 0.00E+00

D (K ) n iDHf,i -3.12E+05 1.21E+03 -1.16E+05 -3.94E+02 -1.05E+05 -6.35E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iAi -1.26E+01 5.46E+00 6.10E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00





n iBi -2.64E-03 1.05E-03 4.43E-04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00



niCi 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

n iDi 3.12E+05 -1.16E+05 -1.05E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

DD (K ) 9.16E+04

Solution continued on next page…

Reference Temperature T0 (K) 298.15

Species Name CaCO3

Entropy Entropy Change of Heat Capacity Change of Reaction at To Integral Reaction at T J/(mol K) (dimensionless) J/(mol K) 160.72 -1.782 145.910 Heat Capacity Coefficients Standard Stoichiometric Entropy at T0 coefficient (J/(mol K)) A B (1/K) -1 -262.05 12.572 2.64E-03 Temperature of Interest T (K) 1109

CO2 CaO

1 1

2.85 -104.18

1 R



1 

D  1

2 o

 T  C 2 D  1 1  o o Srxn T  To2   2  2     B  T  To   ,T  Srxn,T0  R  A ln    T 2 2 T T  o o   



C (1/K ) 0.00E+00

1.05E-03 4.43E-04

0.00E+00 0.00E+00

A -1.011 Note: Light blue fields are inputs, pink fields are the final output.

B (1/K) -1.15E-03

C (1/K ) 0.00E+00

C

5.457 6.104

T 

 T dT    A ln  T    B T  T   2 T  T   2  T  T  



D (K ) iHf,i iAi iBi -3.12E+05 2.62E+02 -1.26E+01 -2.64E-03

iCi 0.00E+00

iDi 3.12E+05

-1.16E+05 2.85E+00 -1.05E+05 -1.04E+02 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

-1.16E+05 -1.05E+05 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

5.46E+00 6.10E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

1.05E-03 4.43E-04 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00 0.00E+00

 D (K ) 9.16E+04

S (1109 K) = 145.9 J/(mol K) gives T = o

167923/(146.0) = 1150 K. Re-

H (1150 K) =

S (1109 K) = 145.235 J/(mol K), which give back the same temperature of 1150 K. So, the result is consistent, and this is our final answer. To have a decomposition pressure of 1 atm, we have to heat limestone to about 1150 K.

Solution 14.23

Problem Statement

Ammonium chloride [NH4Cl(s)] decomposes upon heating to yield a gas mixture of ammonia and hydrochloric acid. At what temperature does ammonium chloride exert a decomposition pressure of 1.5 bar? For NH4Cl(s),

H °f 298 =

and G °f 298 =

Solution

NH4Cl(s) = NH3(g) + HCl(g)

The NH4Cl exists PURE as a solid phase, for which the activity is f/f0. Since f and f0 are for practical purposes the same, the activity is unity. If the equimolar mixture of NH3 and HCl is assumed an ideal gas mixture at 1.5 bar, then with f0 = 1 bar the activity of each gas species is its partial pressure, (0.5)(1.5) = 0.75. As a result, Eq. (13.10) becomes K = (0.75)(0.75) = 0.5625, and we must find the T for which K has this value. From the given data and the data of Table C.4,

H298  176013

J J , G298  91121 mol mol

The following vectors represents the species of the reaction in the order in which they appear:

1    v   1     1 

5939   A  3578   3156

i  1..3

A   i ( vi  Ai )

A  

B   

16105   B   3020 103    0623 

 00    D  0186105    0151 

B   i ( vi  Bi )

C 

D   i ( vi  Di )

D   x

T  62397 K G  H 298 

T0  29815 K

T H298 G298   R IDCPH T0 , T , A, B, C, D  RT T0  IDCPS T0 T A B C D 

G  

Thus

J mol

 G   K  K  exp   RT 

T  62397 K

Although a number of trials were required to reach this result, only the final trial is shown.

Solution 14.24

Problem Statement

A chemically reactive system contains the following species in the gas phase: NH3, NO, NO2, O2, and H2O. Determine a complete set of independent reactions for this system. How many degrees of freedom does the system have?

Solution

Formation reactions:

1 3 N2  H2  NH3 2 2

(1)

1 1 N  O  NO 2 2 2 2

(2)

1 N  O2  NO2 2 2

(3)

1 H2  O2  H2 O 2

(4)

Combine Eq. (3) with Eq. (1) and with Eq. (2) to eliminate N2 : 3 NO2  H2  NH3  O2 2

(5)

1 NO2  O2  NO 2

(6)

The set now comprises Eqs. (4), (5), and (6); combine Eq. (4) with Eq. (5) to eliminate H2 : 3 3 NO2  H2O  NH3  1 O2 2 4

(7)

Equations (6) and (7) represent a set of independent reactions for which r  2. Other equivalent sets of two reactions may be obtained by different combination procedures. By the phase rule, F  2 –   N –r – s  2 –15 – 2 – 0  4

Solution 14.25

Problem Statement

The relative compositions of the pollutants NO and NO2 in air are governed by the reaction, NO

1/2O2



NO2

For air containing 21-mol-% O2 at 25°C and 1.0133 bar, what is the concentration of NO in parts per million if the total concentration of the two nitrogen oxides is 5 ppm?

Solution

NO(g) + (1/2)O2(g) = NO2(g) yNO 2 y NO  yO 2 

0.5

G298  35240

From Table C.4



y NO 2 y NO 0.21

0.5

 K T  29815 K

J mol  G  6 298  K  exp  K  1493  10  RT 

yNO2  

05

 K  yNO and yNO2  yNO  

05

K

Determine yNO

yNO  7307  1012 This about 7*10

ppm, which is negligible,

Solution 14.26

Problem Statement

Consider the gas-phase oxidation of ethylene to ethylene oxide at a pressure of 1 bar with 25% excess air. If the reactants enter the process at 25°C, if the reaction proceeds adiabatically to equilibrium, and if there are no side reactions, determine the composition and temperature of the product stream from the reactor.

Solution

C2H4(g) + (1/2) O2(g) = <(CH2)2>O(g)

H298  105140

J J G298  81470 mol mol

Basis: 1 mol C2H4 entering reactor. Moles O2 Entering:

nO2  125  05

Moles N2 entering:

nN 2  nO 2

79 21

n0 : 1  nO2  nN 2n0  3976 Index the product species with the numbers: 1 = ethylene 2 = oxygen 3 = ethylene oxide 4 = nitrogen The numbers of moles in the product stream are given by Eq. (14.5). For the product stream, data from Table C.1: Guess:

  08

Solution continued on next page…

 1      n  05  O2  n           nN 2   1424  14394      3639   0506  3   10  B   A       0  385 23  463   K         3280   0593  4392    00  6  10 C    2  9  296  K     00 

i := 1 .. 4

 00   1      0227 05   5 2 10  K v   D      0  0 1          0040   0 

A   (n)i  Ai  B   (n)i  Bi  i

C    (n)i Ci  D  (n)i  Di  i

y   

n  n0  05

K   ( y i ) vi K   15947 i

The energy balance for the adiabatic reactor is:

H298 Hp  0

For the second term, we combine Eqs. (4.3) & (4.8).

The three equations together provide the energy balance. A  3629, B  8816

103 106 , C  4904 kelvin kelvin 2

D  0114105  K 2 ,T0  29815 K  B 2 2 C 3 3 D    1   IDCPH  AT0   1  T0    1  T0    1   2 3 T0    









   D     1    IDCPS  A In   BT0  C T0        2   2   (   T )   0     IDCPH  



KIDCPS   

Given

Solution continued on next page…

 B   2 2 C   3 3 D      1   H298  R  AT0     T0     T0      2 3 T0     









 H G  H298  298 298   IDCPS  1  IDCPH  K   exp    RT0 RT0   T0    Determine  and    088244

  318374

00333    0052    y(0.88244) =   02496    06651

Knowing  , the temperature is

T :  T0

T  94923K

Solution 14.27

Problem Statement

Carbon black is produced by the decomposition of methane: CH4(g)



C(s)

+ 2H2(g)

For equilibrium at 650°C and 1 bar, (a) What is the gas-phase composition if pure methane enters the reactor, and what fraction of the methane decomposes? (b) Repeat part (a) if the feed is an equimolar mixture of methane and nitrogen.

Solution

CH4(g) = C(s) + 2H2(g) V = 1 (gases only)

The carbon exists PURE as an individual phase, for which the activity is unity. Thus we leave it out of consideration. From the data of Table C.4,

H298 : 74520 

J J G298 : 50460  mol mol

The following vectors represent the species of the reaction in the order in which they appear: 1 1702 9081         v   1  A  1771 B  0771        2  3249 0422

3

2164   C   00     00 

6

 00    D  0867    0083 

A   vi  Ai B   vi  Bi C   vi  Ci D   vi  Di  i

A  

B   

T

G H298 

G 

By Eq. (14.5),





3

C  

 kelvin T0 





6

D  



 kelvin

T H298 G298   R  IDCPH T0 T A B C DR  T T0  IDCPS T0 T A B C D

J mol

 G   = 4.2392  exp  RT 

n0 = 1

YCH4 =

1  2  ,YH2  1 1

(a) By Eq. (14.28),

 :

K   07173 4K

YCH 

(2)2 4  2  K 1  1   1  2

(fraction decomposed)

1  2  YH  YCH   1 1

Yh  

(b) For a feed of 1 mol CH4 and 1 mol N2, n0 = 2 By Eq. (14.28), Given

(guess)

(2)2  K   Find   2  1  

  07893

YCH4 =



1  2

YH2 = 0.5659

(fraction decomposed)

YH2 =

2  2

YN2 = 1-YCH4-YH2

YCH4 = 0.0756 YN2 = 0.3585

Solution 14.28

Problem Statement

Consider the reactions 1/2N2(g) + 1/2O2(g)  NO(g) 1/2N2(g) + O2(g)  NO2(g)

If these reactions come to equilibrium after combustion in an internal-combustion engine at 2000 K and 200 bar, estimate the mole fractions of NO and NO2 present for mole fractions of nitrogen and oxygen in the combustion products of 0.70 and 0.05.

Solution

In problems like this, I like to use integer stoichiometric coefficients. The final answer will not depend on how we write the reactions. So, let’s write them as (1)

N2 + O2  2 NO

(2)

N2 + 2 O2  2 NO2

It may be slightly tricky to write the number of moles of each species, because the mole fractions of the two reactants (N2 and O2) do not sum to 1.0. Of course, in the engine, there are also CO, CO2, H2O, and perhaps small amounts of other species as well. However, if we only consider the equilibrium of these two reactions, then the identity of the other species doesn’t matter. The fact that they are present does matter, since it affects how the total number of moles changes. We can write:

nN2 = 0.70

nO2 = 0.05

nNO = 2 1 nNO2 = 2 2 n=1

and the mole fractions are yN2 = (0.70

yO2 = (0.05

2)/( 1

2) /( 1

yNO = 2 1/( 1

yNO2 = 2 2/( 1

The equilibrium constants for these reactions are given in Figure 13.2, but we can also compute them fairly easily from the usual approach. Since both reactions are formation reactions, the enthalpy, entropy, and Gibbs energies of reaction are just the corresponding enthalpies, entropies, and Gibbs energies of formation of the products. That is º

H 1 = 2 H f,,NO = 2 (90250) = 180500 J/mol H 2 = 2 H f,,NO2 = 2 (33180) = 66360 J/mol G 1 = 2 G f,,NO = 2 (86550) = 173100 J/mol G 2 = 2 G f,,NO2 = 2 (51310) = 102620 J/mol So, we have K0,1 = exp( 173100/8.314/298.15) = 4.724 × 10 And K0,2 = exp( 102620/8.314/298.15) = 1.049 × 10 K1,1 = exp( 180500/8.314*(1/2000

× 10

Solution continued on next page…

K2,1 = exp( 66360/8.314*(1/2000

× 10

Evaluating the heat capacity integrals in the usual way, we get K2,1 = exp( 0.102 + 0.264) = 1.175 K2,2 = exp( 0.095 So, K1 = 4.724 × 10

0.438) = 0.587 31

And K2 = 1.049 × 10

* 8.122 × 10 * 1.175 = 0.000451

* 7.820 × 10 * 0.587 = 4.813 × 10

Although the pressure is very high, the temperature is also very high, and therefore we can treat the gases as an ideal gas mixture. So, the equilibrium relationships are

 yNO 

21   K yO y N 070  1  2 005  1  22  1 2

y   P 2  1     K    200 K  P  070     0.05    2  y y  2

NO2

We could rearrange these for iterative solution as 1  K1 070  1  2 005  1  22  / 4 200 K 2 070  1  2 005  1  22 

2 

41  2 

Since both K1 and K2 are much less than one, both extents of reaction will be small, and we can initially guess 0.5 = 1 = 2 = 0, then iterate on the above two equations. The first iteration will just give 1 = (0.035K1/4) 0.5

0.00199 and 2 = (0.00175K2/4)

= 2.05 × 10 . Putting these values back into the above equations and 5

repeating quickly gives convergence to 1 = 0.00194 and 2 = 1.97 × 10 . The corresponding mole fractions of 5

NO and NO2 are yNO = 0.0039 and yNO2 = 3.9 × 10 . These might also be expressed as 3900 ppm of NO and 39 ppm of NO2. These may seem like small mole fractions, but they are far above the legal limits for NOx emissions – which is part of the reason that our cars have catalytic converters that help to re-establish the equilibrium among these species at lower temperatures where the NOx concentrations are acceptable. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 14.29

Problem Statement

Oil refineries often have both H2S and SO2 to dispose of. The following reaction suggests a means of getting rid of both at once:

2H2S(g)

SO2(g)



3S(s)

2H2O(g)

For reactants in the stoichiometric proportion, estimate the percent conversion of each reactant if the reaction comes to equilibrium at 450°C and 8 bar.

Solution

2H2S(g) + SO2(g) = 3S(s) + 2H2O(g)

The sulfur exists PURE as a solid phase, for which the activity is f/f0. Since f and f0 are for practical purposes the same, the activity is unity, and it is omitted from the equilibrium equation. Thus for the gases only, v= From the given data and the data of Table C.4,

H298 



J G298  mol



J mol

The following vectors represent the species of the reaction in the order in which they appear:

2   1  v =    3     2 

3.931   5.699    A=   4.114   3.470 

i= 1 .. 4 A   vi  Ai

 1.490     0.801    B =   1.728    1.450 

B   vi  Bi

D    vi  Di 

A  T = 723.15K,

B 



3

C  D 



T0 = 298.15 K G H298 

G   

T H298 G298   R  IDCPH T0 T A B C D  R  T T0  IDCPS T0 T A B C D

 G  J  = 12.9169 K = exp   RT  mol

By Eq. (14.5), gases only: n0 = 3

YH2S =

0.232   1.015  5  D =  10 0.783    0.121 

(basis)

2  2  1  2 , YSO2  ,YH 2O  3  3  3

By Eq. (14.28)

(2)2 3  

Given

(2  2)2 1   

 8 K   

Percent conversion of reactants = PC PC =

ni0  ni ni0

100 

vi 3 ni 0

100

[By Eq. (14.4)]

Since the reactants are present in the stoichiometric proportions, for each reactant,

ni0  vi

Whence PC :=   100

PC = 76.667

Solution 14.30

Problem Statement

Species N2O4 and NO2 as gases come to equilibrium by the reaction: N2O4  2NO2. (a) For T = 350 K and P = 5 bar, calculate the mole fractions of these species in the equilibrium mixture. Assume ideal gases.

(b) If an equilibrium mixture of N2O4 and NO2 at conditions of part (a) flows through a throttle valve to a pressure of 1 bar and through a heat exchanger that restores its initial temperature, how much heat is exchanged, assuming chemical equilibrium is again attained in the final state? Base the answer on an amount of mixture equivalent to 1 mol of N2O4, i.e., as though the NO2 were present as N2O4.

Solution

N2O4(g) = 2NO2(g) (a)

(b)

Data from Tables C.4 and C.1 provide the following values:

H298  T0 = 298.15·K



J G298  mol



J mol

T = 350·K

A   G H298 

B  



3

C 

D  



T  H298 G298   R  IDCPH T0 T A B C D  R  T T0  IDCPS T0 T A B C D

G 



J mol

 G   = 3.911 K = exp   RT 

Basis: 1 mol species (a) initially. Then

Ya  Yb 

1  1 2 1

 P 1    K 1    1    P 0  (2  )2

(a) P =5



P0 =1

K   04044 4P  K

Ya 

1  1

Ya  0.4241

(b) P =1



P0 =1

K  4 P  K

By Eq. (4.19), at 350 K:

H  H298  R IDCPH T0 T A B C DH 

J mol

This is Q per mol of reaction, which is

  07031  04044  0299 Whence Q  H  

J mol

Solution 14.31

Problem Statement

The following isomerization reaction occurs in the liquid phase: A  B, where A and B are miscible liquids for E

° which: G RT = 0.1xAxB. If G298 =

much error is introduced if one assumes that A and B form an ideal solution?

Solution

As we have seen in previous homeworks, the activity coefficients are obtained from G through the relationship  d  nG E     ln 1     dn  RT   1 

and T ,P ,n2

 d  nG E     ln  2     dn  RT   2 

T , P ,n1

Using the given equation for G (letting a represent the number 0.1) an1n2 nG E  nax1 x 2  RT n1  n2

and taking the derivative of this with respect to n1 at constant n2 gives  d  an n    1 2  ln 1     dn1  n1  n2  ln 1 

an22

n1  n2 

   n2 n1  n2   n1n2    a 2   n  n  1 2   T , P ,n2

 ax22  0.1 x22

Similarly, Solution continued on next page…

 d  an n    1 2  ln 2     dn2  n1  n2  ln 2 

an12

n1  n2 

   n1 n1  n2   n1n2     a 2   n  n  1 2   T , P ,n2

 ax12  0.1 x12

Our general equilibrium relationship is  N  fˆ  i  i  K     fo i1   i 



and for liquids at low to moderate pressures with fugacities represented by activity coefficients, this becomes N

i

i x i   K i 1

For the simple isomerization reaction, this is B xB AxA

K

The equilibrium constant at 298.15 K is K = exp( G/(RT)) = exp( 1000/(8.314*298.15)) = 1.497. If we assumed an ideal solution, then the activity coefficients would both be 1.0, and we would have xB

 1.497 1  xB 1.497 xB   0.600 2.497 xA



So, if we assume an ideal solution, we get xB = 0.600 and xA = 0.400. If we use the activity coefficients derived above, we have

B x B AxA



   exp 0.1 x  x x  exp 0.1 x  x x  x x  K   x    x   exp0.1x  x exp 0.1x 2A x B

Substituting xA =1 B xB AxA

2 B

2 A

2 B

xB gives

 exp 0.11  2 x B 

xB 1 xB

 1.497

Solving this by trial and error gives xB = 0.6045, so xA = 0.3955. In this case, the answer assuming an ideal mixture is negligibly different from the answer using the activity coefficients.

Solution 14.32

Problem Statement

Hydrogen gas can be produced by the reaction of steam with “water gas,” an equimolar mixture of H2 and CO obtained by the reaction of steam with coal. A stream of “water gas” mixed with steam is passed over a catalyst to convert CO to CO2 by the reaction:

H2O(g)

CO(g) 

H2(g)

CO2(g)

Subsequently, unreacted water is condensed and carbon dioxide is absorbed, leaving a product that is mostly hydrogen. The equilibrium conditions are 1 bar and 800 K. (a) Is any advantage gained by carrying out the reaction at pressures above 1 bar? (b) Would increasing the equilibrium temperature increase the conversion of CO? (c) For the given equilibrium conditions, determine the molar ratio of steam to “water gas” (H2 + CO) required to produce a product gas containing only 2-mol-% CO after cooling to 20°C, where the unreacted H2O has been virtually all condensed. (d ) Is there any danger that solid carbon will form at the equilibrium conditions by the reaction

2CO(g)



CO2(g)

C(s)

Solution

H2O(g) + CO(g) = H2(g) + CO2(g)

v0

From the data of Table C.4, Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

H298 : 41166

T0 = 298.15·kelvin

J J , G298  28618 mol mol

T = 800·kelvin

A  1860  B   0540 103

C   D  1164 105

G  H298 

G   



T H298 G298   R IDCPH T0 T A B C DRT T0  IDCPS T0 T A B C D

 G   =4.27837 K = exp   RT 

J mol

(a) No. Since v  0 , at low pressures P has no effect (b) No. K decreases with increasing T. (The standard heat of reaction is negative.). (c) Basis: 1 mol CO, 1 mol H2, w mol H2O feed. From the problem statement, nCO nCO  nH 2  nCO 2

By Eq. (14.4),

= 0.02

nCO  1   nH 2  1   nCO   2

1  1  096   002     0941 1  1     2   102 Let z = w/2 = moles H2O/mole “Water gas”. By Eq. (14.5),

yH 2 O 

w   2z    2  w 2  2z

yCO 

1 2  2z

yH 2 

1 2  2z

yCO2 

 2  2z

Solution continued on next page…

Given  1  

2 z   1   

K

Determine z z  41 (d)

2CO(g) = CO2(g) + C(s)

V  1

From the data of Table C.4 and C.1,

H298 : 172459 G298  120021

J mol

A  0476  B  0702103

C   D  1962105

G  H 298 

G   

T H 298  G298   R IDCPH T0 ,T , A, B, C , D  RT T0  IDCPS T0 T A B C D  4



J mol

 G   K  1017 K = exp   RT 

By Eq. (14.28), gases only, with P = P0 = 1 bar yCO2

 yCO 

 K  1017 for the reaction at equilibrium

If the ACTUAL value of this ratio is GREATER than this value, the reaction tries to shift left to reduce the ratio. But if no carbon is present, no reaction is possible, and certainly no carbon is formed. The actual value of the ratio in the equilibrium mixture of Part (c) is Solution continued on next page…

yCO  2

 2  2z

1  2  2z

yCO  

yCO

02424

yCO 

yCO  

The ratio of these two is

Ratio 

yCO



001522

 104916

No carbon can be deposited from the equilibrium mixture.

Solution 14.33

Problem Statement The feed gas to a methanol synthesis reactor is composed of 75-mol-% H2, 15-mol-% CO, 5-mol-% CO2, and 5-mol-% N2. The system comes to equilibrium at 550 K and 100 bar with respect to the reactions: 2H2(g)

CO(g) 

CH3OH(g)H2(g)

CO2(g)



CO(g)

+ H2O(g)

Assuming ideal gases, determine the composition of the equilibrium mixture.

Solution

The two reactions we are considering are (1)

2H2(g) + CO(g)  CH3OH(g)

(2)

H2(g) + CO2(g)  CO(g) + H2O(g)

We can choose as a basis whatever we like, so let’s choose a basis of 100 moles of feed gas. That way the mole percents given simply become our initial mole numbers. So, we have nH2,o = 75 moles, nCO,o = 15 moles, nCO ,o 2

= 5 moles, and nN2,o = 5 moles for the initial mole numbers. If 1 is the reaction coordinate for reaction 1 and 2 is the reaction coordinate for reaction 2, then the numbers of moles of each species, as a function of the reaction coordinates, are nH = 75

nCO = 15

1+ 2

nCH3OH = 1 nCO = 5 2

nH O = 2 2

nN = 5 2

and the total number of moles is n = 100

yH = (75

2)/( 100

yCO = (15

1 + 2) /( 100

yCH3OH = 1/( 100 yCO = (5 2

2) /( 100

yH O = 2/( 100

yN = 5/( 100

1. So, the corresponding mole fractions are

Our equilibrium relationships for the two reactions (assuming ideal gases) are 2

o 2

yCH3OH/((yH2) *yCO) = Kr1(P/P ) and

(yCO*yH O)/(yH *yCO ) = Kr2 2

Where Kr1 and Kr2 are the equilibrium constants for reaction 1 and 2, respectively, at 550 K. The thermodynamic properties needed for computing the equilibrium constants (enthalpy and Gibbs energy of formation and coefficients for heat capacity polynomials) are as follows: Species

G°f(298)

H°f(298)

(J/mol)

B K

C 1

D 2

3.249

0.422 × 10

…

137169

110525

3.376

0.557 × 10

CH3OH

161960

200660

2.211

12.216 × 10

CO2

394359

393509

5.547

1.045 × 10

…

H2O

228572

241818

3.470

1.450 × 10

…

0.083 × 10

0.031 × 10

3.450 × 10

… 1.157 × 10 0.121 × 10

The changes in each of these quantities for each of the reactions are then: Reaction

(1) (2)

G°rxn(298)

H°rxn(298)

(J/mol)

24791 28618

90135 41166

A

B K

C

7.663

10.815 × 10

1.950

0.540 × 10

3.45 × 10 …

D K 6

1.350 × 10 0.116 × 10

So, for reaction 1, the equilibrium constant at 298 K is K0,1 = exp(24791/(8.314*298.15)) = 22038. The main part of change in the equilibrium constant from 298 K to 550 K is: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

 H o T     90135    T  o  1  550   5878108 1    exp K1,1  exp    83145  550  29815  RT  To   Finally, for the minor part of the temperature dependence, we have

  T Cpo  1 T C po  K 2  exp  dT   dT   T R RT To To   Evaluating the two heat capacity integrals using our spreadsheets ICPH.xls and ICPS.xls and the delta Cp coefficients above gives:

T 1 (K) 298.15

T 2 (K) 550 T2

ICPH 

R

A -7.663

dT  A T2  T1  

C (1/K2) D (K2) B (1/K) 1.08E-02 -3.45E-06 -1.35E+04

ICPH (K) -9.56E+02

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









and

T 1 (K) 298.15

T 2 (K) 550 T2

ICPS 

 T1

A -7.663

C (1/K2) D (K2) B (1/K) 1.08E-02 -3.45E-06 -1.35E+04

ICPS -2.391

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT T 2 2  1









so  956  K 21  exp  2391  0521  550 

Putting this all together, Kr1 = 22038 * 5.878 × 10

* 0.521 = 6.74 × 10

Likewise, for the second reaction, the equilibrium constant at 298 K is 6

K0,2 = exp( 28618/(8.314*298.15)) = 9.691 × 10 . The main part of change in the equilibrium constant from 298 K to 550 K is: Solution continued on next page…

 H o T     41166    T  o  1  550   2006 1    exp  K12  exp     83145  550  29815  RT  To   Finally, for the minor part of the temperature dependence, we have

T 1 (K) 298.15

T 2 (K) 550 T2

ICPH 

R

C (1/K2) D (K2) B (1/K) 5.40E-04 0.00E+00 1.16E+05

A -1.95

dT  A T2  T1  

ICPH (K) -2.55E+02

 1 1 B 2 C 3 T2  T12  T2  T13  D    2 3  T2 T1 









and

T 1 (K) 298.15

T 2 (K) 550 T2

ICPS 

 T1

C (1/K2) D (K2) B (1/K) 5.40E-04 0.00E+00 1.16E+05

A -1.95

ICPS -0.596

T  C D 2 dT  A ln  2   B T2  T1   T22  T12  T2  T1 2 RT T 2 2  1









so  255  K 22  exp  0596  0876  550 

Putting this all together, Kr2 = 9.691 × 10

* 2006 * 0.876 = 0.01703.

So, at 100 bar and 550 K, the equilibrium relationships become: 2

o 2

yCH3OH/((yH ) *yCO) = Kr1(P/P ) = 6.74 × 10 2

and

* 10 = 6.74

(yCO*yH O)/(yH *yCO ) = Kr2 = 0.01703 2

Substituting in the expressions for the mole fractions in terms of the two reaction coordinates gives: 1 100  21 

75  21  2  15  1  2  2

and

 674

15  1  2 2  001703 75  21  2 5  2 

So, now we have two equations in two unknowns, and it is simply a matter of solving them. From the expressions for the number of moles of each species, we see that 2 must be between 0 and 5 (if it was less than zero, the water concentration would be negative, and if it was greater than 5, the CO2 concentration would be negative). Likewise, 1 has to be between 0 and 20 (so that the methanol and CO concentrations are positive). So, we have a relatively small range of values over which we need to look for a solution. We can use Maple to solve these, as follows: > > eq1:=e1*(100 eq2:=(15-e1+e2)*e2/(75

1-e2)^2/(15 -e2)/(5-e2)=0.01703;

solve({eq1,eq2},{e1,e2});

eq1 :=

e1 ( 100 2 e1 ) 2 6.74 ( 75 2 e1e2 ) 2 ( 15e1 e2 )

eq2 :=

( 15e1e2 ) e2 .01703 ( 752 e1e2 ) ( 5e2 )

{ e2-50.32722105 , e1-31.79049524 }, { e139.96789395 4.768453918 I, e2.02238269229 .02815501014 I }, { e139.96789395 4.768453918 I, e2.02238269229 .02815501014 I }, { e2.8787346421 , e111.85184823 }, { e143.71142266, e229.30072336 }, { e1109.7205911 , e298.65062054 }

So, Maple found 6 solutions, but only 1 of them ( 1 = 11.85, 2 = 0.879) has values of the reaction coordinates that lead to positive values for all of the species concentrations. Though we won’t prove it, it can be proven that the equilibrium composition is always unique. Substituting these values back into the expressions for the mole fractions gives: > e1:= 11.85184823; e2:= .8787346421; yH2 := (75 yCO := (15 yCH3OH := e1/( 100 yCO2 := (5 yH2O := e2/( 100 yN2 := 5/( 100

2*e1);

e1 := 11.85184823 e2 := .8787346421 yH2 := .6608127335 yCO := .05277957418 yCH3OH := .1553397436 yCO2 := .05401657966 yH2O := .01151739470 yN2 := .06553397435 So, the equilibrium composition is 66.1 mol % H2, 5.3 mol % CO, 15.5 mol % methanol, 5.4 mol % CO2, 1.2 mol % H2O, and 6.6 mol % N2. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 14.34

Problem Statement

“Synthesis gas” can be produced by the catalytic re-forming of methane with steam. The reactions are: CH4(g) + H2O(g)  CO(g) + 3H2(g)CO(g) + H2O(g)  CO2(g) + H2(g) Assume equilibrium is attained for both reactions at 1 bar and 1300 K. (a) Would it be better to carry out the reaction at pressures above 1 bar? (b) Would it be better to carry out the reaction at temperatures below 1300 K? (c) Estimate the molar ratio of hydrogen to carbon monoxide in the synthesis gas if the feed consists of an equimolar mixture of steam and methane. (d) Repeat part (c) for a steam-to-methane mole ratio in the feed of 2. (e) How could the feed composition be altered to yield a lower ratio of hydrogen to carbon monoxide in the synthesis gas than is obtained in part (c)? ( f ) Is there any danger that carbon will deposit by the reaction 2CO  C + CO2 under conditions of part (c)? Part (d)? If so, how could the feed be altered to prevent carbon deposition?

Solution

CH4(g) + H2O(g) = CO(g) + 3H2(g) v=2 From the given data and the data of Table C.4,

H298 



J G298  mol



J mol

Solution continued on next page…

The following vectors represent the species of the reaction in the order in which they appear: 1   1  v =   1      3 

1702  9.081     3470 1450     B  A=     3  376 0  557         3249 0422



2.164     0   6 C =  10 0      0 

 0     0.121   5 D =  10  0.031     0.083 

i= 1 .. 4 A   i ( vi  Ai )

B   i ( vi  Bi )

A   T = 1300K



3

C  

D   i ( vi  Di ) 

6

D 



T0 = 298.15 K G  H298 

G   

B  

C   i ( vi Ci )



J  K mol

T H298 G298   R IDCPH T0 T A B C DRT T0  IDCPS T0 T A B C D

 G   K  13845 = exp   RT 

H2O + CO = H2 + CO2

H298 : 41166

J mol

G298 : 28618

J mol

A    B   0540  103

C    D   1164  105

Solution continued on next page…

G  H 298 

G  



J and K mol

T H298 G298   R IDCPH T0 T A B C D RT T0  IDCPS T0 T A B C D

 G   = 0.5798 = exp   RT 

Part(a): No. Primary reaction (1) shifts left with increasing P. Part(b): The higher temperature would favor the reaction because the reaction is endothermic. Therefore, it would not be better to carry out the reaction at temperatures below 1300 K. Par(c): The value of K1 is so large compared with the value of K2 that for all practical purposes reaction (1) may be considered to go to completion. With a feed equimolar in CH4 and H2O, no H2O then remains for reaction (2). In this event the ratio, moles H2/moles CO is very nearly equal to 3.0. Part(d): With H2O present in an amount greater than the stoichiometric ratio, reaction (2) becomes important. However, reaction (1) for all practical purposes still goes to completion, and may be considered to provide the feed for reaction (2). On the basis of 1 mol CH4 and 2 mol H2O initially, what is left as feed for reaction (2) is: 1 mol H2O, 1 mol CO, and 3 mol H2; n0 = 5. Thus, for reaction (2) at equilibrium by Eq. (14.5):

yCO  y H O  2

1 5

yCO  2

 5

yH  2

3 5

Determine   3   

1  

Ratio 

yH 2 yCO

 K 2  05798

  01375

3 1 

Ratio  

Ratio 

Part(e): One practical way is to add CO2 to the feed. Some H2 then reacts with the CO2 by reaction (2) to form additional CO and to lower the H2/CO ratio.

Part(f): 2CO(g) = CO2(g) + C(s)

V= 1 (gases)

This reaction is considered in the preceding problem, Part (d), from which we get the necessary parameter values:

H298 : 172459

J mol

G298 : 120021

J mol

For T = 1300 K T0  29815 K

A    B   0702  103

C : 0  D   1962  105

G  H 298 

T H298 G298   R IDCPH T0 T A B C D RT T0  IDCPS T0 T A B C D 

G  5673  10 4 

J mol

 G   K  52557  103 K = exp   RT  Ratio 

yCO

 yCO 

When for ACTUAL compositions the value of this ratio is greater than the equilibrium value as given by K, there can be no carbon deposition. Thus in Part (c), where the CO2 mole fraction approaches zero, there is danger of carbon deposition. However, in Part (d) there can be no carbon deposition, because Ratio > K: Ratio 

5

1  

 0924

Solution 14.35

Problem Statement

Consider the gas-phase isomerization reaction: A B. (a) Assuming ideal gases, develop from Eq. (14.28) the chemical-reaction equilibrium equation for the system. (b) The result of part (a) should suggest that there is one degree of freedom for the equilibrium state. Upon verifying that the phase rule indicates two degrees of freedom, explain the discrepancy. Eq. 14.28:

i

 P  ( yi )   o  K   i

Solution

(a) Equation (14.28) here becomes:

Whence,

0 y B  P     K  K y A  P ° 

yB  K T  1  yB

(b) The preceding equation indicates that the equilibrium composition depends on temperature only. However, application of the phase rule, Eq. (14.36), yields: F  22 –1–1 2

This result means in general for single-reaction equilibrium between two species A and B that two degrees of freedom exist, and that pressure as well as temperature must be specified to fix the equilibrium state of the system. However, here, the specification that the gases are ideal removes the pressure dependence, which in the general case appears through the ˆi s.

Solution 14.36

Problem Statement

A low-pressure, gas-phase isomerization reaction, A B, occurs at conditions such that vapor and liquid phases are present. (a) Prove that the equilibrium state is univariant. (b) Suppose T is specified. Show how to calculate xA, yA, and P. State carefully, and justify, any assumptions.

Solution

For the isomerization reaction in the gas phase at low pressure, assume ideal gases. Equation (13.28) then becomes: 0 y B  P     K  K y A  P  

whence

1 yA yA

 K T 

Assume that vapor/liquid phase equilibrium can be represented by Raoult’s law, because of the low pressure and the similarity of the species:

x A PAsat T   y A P (a) Application of Eq. (13.36) yields:

  x A PBsat T     y A P F

N r





(b) Given T, the reaction-equilibriuum equation allows solution for yA. The two phase–equilibrium equations can then be solved for xA and P. The equilibrium state therefore depends solely on T.

Solution 14.37

Problem Statement

Set up the equations required for solution of Ex. 14.14 by the method of equilibrium constants. Verify that your equations yield the same equilibrium compositions as given in the example.

Solution

Formation reactions: C + 2H2 = CH4 H2 + (1/2)O2 = H2O C + (1/2)O2 = CO C + O2 = CO2 Elimination first of C and then of O2 leads to a pair of reactions: CH4 + H2O = CO + 3H2 (1) CO + H2O = CO2 + H2 (2) There are alternative equivalent pairs, but for these: The stoichiometric numbers are: i

CH4

H2O

CO2

j 1 2

Solution continued on next page…

For initial amounts: 2 mol CH4 and 3 mol H2O, n0  5, and by Eq. (14.7): yCH 4 

2  1

y H 2O 

5  21 yCO 2 

3  1  2

yCO 

5  21

2

yH 2 

5  21

1  2 5  21

31  2 5  21

By Eq. (14.40), with P = P0 = 1 bar

  k

yCO  yH 2

yCH yH O 4

yCO  yH 2

yCH yH O 4

 k2

From the data given in Example 14.14,

G1  27540

J J G2  3130 T  1000 K mol mol

 G1   G2   K 2  exp  K 1  27453 K 2  1457 K 1  exp   RT   RT 

Given

1  2 31  2  2 31  2   K K 1 2 1  2 3  1  2  2 2  1 3  1  2 5  21  3

Determine 1 and 2 iteratively

1  18304

2  03211

Plug this back into the y’s

yCH  00196 yH O  0098 yCO  01743 4

yCO  00371 yH  06711 2

These results are in agreement with those of Example 14.14.

Solution 14.38

Problem Statement

Reaction-equilibrium calculations may be useful for estimation of the compositions of hydrocarbon feedstocks. A particular feedstock, available as a low-pressure gas at 500 K, is identified as “aromatic C8.” It could in principle contain the C8H10 isomers: o-xylene (OX), m-xylene (MX), p-xylene (PX), and ethylbenzene (EB). Estimate how much of each species is present, assuming the gas mixture has come to equilibrium at 500 K and low pressure. The following is a set of independent reactions (why?): OX  MX

(I)

OX  PX

(II)

OX  EB

(III)

(a) Write reaction-equilibrium equations for each equation of the set. State clearly any assumptions. (b) Solve the set of equations to obtain algebraic expressions for the equilibrium vapor-phase mole fractions of the four species in relation to the equilibrium constants, KI, KII, KIII. (c) Use the following data to determine numerical values for the equilibrium constants at 500 K. State clearly any assumptions. (d) Determine numerical values for the mole fractions of the four species. Species

H of 298 / J mol1

G of 298 / J mol1

OX(g) MX(g) PX(g) EB(g)

19,000 17,250 17,960 29,920

122,200 118,900 121,200 130,890

Solution

(a): For low pressure and a temperature of 500 K, the system is assumed to be a mixture of ideal gases, for which Eq. (14.28) is appropriate. Therefore,  P 0 y MX  K I    K I  P0  yOX

 P 0 yPX  K II    K II yOX  P0 

 P 0 yEB  K III    K III yOX  P0 

(b): These equations lead to the following set:

yMX  K I yOX

yPX  K II yOX

yEB  K III yOX

The mole fractions must sum to unity, and therefore:

yOX  K I yOX  K II yOX  K III yOX  yOX 1  K I  K II  K III   1 yOX 

(c): With the assumption that CP° 

1 1  K I  K II  K III 

K 2  Eq 14.20, 14.21, and 14.22 combine to give

 G°   H °  T   298  298  K  K 0 K 1  exp 1  0  exp   RT0  T   RT0   Whence  H °   29815  °   298   RT 1  500   G298    0  K  exp   8314  29815        

The data provided lead to the following property changes of reaction and equilibrium constants at 500 K:

Reaction

2.847

1.2637

III

10920

8690

0.1778

(d): Substitution of numerical values into Eqs. (1), (2), (3), and (4) yields the following values for the mole fractions:

yOX  01891 yMX  05383 yPX  02390 yEB  00336

Solution 14.39

Problem Statement

Ethylene oxide as a vapor and water as liquid, both at 25°C and 101.33 kPa, react to form a liquid solution containing ethylene glycol (1,2-ethanediol) at the same conditions: (CH2)2O

H2O



CH2OH.CH2OH

If the initial molar ratio of ethylene oxide to water is 3.0, estimate the equilibrium conversion of ethylene oxide to ethylene glycol. At equilibrium the system consists of liquid and vapor in equilibrium, and the intensive state of the system is fixed by the specification of T and P. Therefore, one must first determine the phase compositions, independent of the ratio of reactants. These results may then be applied in the material-balance equations to find the equilibrium conversion. Choose as standard states for water and ethylene glycol the pure liquids at 1 bar and for ethylene oxide the pure ideal gas at 1 bar. Assume any water present in the liquid phase has an activity coefficient of unity and that the vapor phase is an ideal gas. The partial pressure of ethylene oxide over the liquid phase is given by: p i /kPa

The vapor pressure of ethylene glycol at 25°C is so low that its concentration in the vapor phase is negligible.

Solution

Phase-equilibrium equations:

Solution continued on next page…

Ethylene oxide(1): p1  y1 P 

Water(2): x2 Psat 2  y2 P Psat 2  

P



kPa x2 

kPa x 1 

y1 P 415kPa

y2 P Psat 2

Ethylene glycol(3): Psat 3  0 y3  0 Therefore y2  1  y1 and x3  1  x2  x1 For the specified standard states: (CH2)2O(g) + H2O(l) = CH2OH.CH2OH(l) By Eq. (14.40) and the stated assumptions, k

Data from Table C.4:

G298  72941

 3 x3 x  3  P  y1 x 2  y   1 P  2 x 2   0

T  29815 K

J mol

 G  298  k  exp k  6018  1012  RT 

So large a value of k requires either y1 or x2 to approach zero. If y1 approaches zero, y2 approaches unity, and the phase-equilibrium expression for water(2) makes x2 = 32, which is impossible. Thus x2 must approach zero, and the phase-equilibrium equation requires y2 also to approach zero. This means that for all practical purposes the reaction goes to completion. For initial amounts of 3 moles of ethylene oxide and 1 mole of water, the water present is entirely reacted along with 1 mole of the ethylene oxide. Conversion of the oxide is therefore 33.3 %.

Solution 14.40

Problem Statement

In chemical-reaction engineering, special measures of product distribution are sometimes used when multiple 12

reactions occur. Two of these are yield Yj and selectivity Sj/k. We adopt the following definitions :

Yj 

moles formed of desired product j moles of j that would be formed with no side reactions and with complete consumption of the limiting reactant species

S j /k 

moles formed of desired product j moles formed of undesired product k

For any particular application, yield and selectivity can be related to component rates and reaction coordinates. For two-reaction schemes the two reaction coordinates can be found from Yj and Sj/k, allowing the usual material-balance equations to be written. Consider the gas-phase reactions: A



C (I)

A+C



D (II)

Here, C is the desired product, and D is the undesired by-product. If the feed to a steady-flow reactor contains 10 kmol·h

of A and 15 kmol·h

of B, and if YC = 0.40 and SC/D = 2.0, determine complete product rates and

the product composition (mole fractions), using reaction coordinates. 12

R. M. Felder, R. W. Rousseau, and L. G. Bullard, Elementary Principles of Chemical Processes, 4th ed., Sec.

4.6d, Wiley, New York, 2015.

Solution

For the given flowrates, nA0 

nB0 

nA0 the limiting reactant without (II) n A  nA0  I  II nB  nB0  I nC  I  II nD  II n  n0  I  II

Use given values of YC and SC / D

I and II YC 

Solve for I

I  II n A0

SC / D 

I  II II

II S    1  n Y   2  1   10  04  6 I   C / D A 0 C   SC / D   2 

II 

nA 

n A0YC SC / D

  



10  040 2 2

yA 

2   17

nB  15  6  9

yB 

9  05295 17

nC  6  2  4

yC 

4  02353 17

nD  2  2

yD 

n  17

2  01176 17 y 1

Solution 14.41

Problem Statement

The following problems involving chemical-reaction stoichiometry are to be solved through the use of reaction coordinates. (a) Feed to a gas-phase reactor comprises 50 kmol·h 1 of species A, and 50 kmol·h 1 of species B. Two independent reactions occur: A



(I)

A+C

 D

(II)

Analysis of the gaseous effluent shows mole fractions yA = 0.05 and yB = 0.10. 1

(i) What is the reactor effluent rate in kmol·h ?

(ii) What are the mole fractions yC and yD in the effluent? (b) Feed to a gas-phase reactor comprises 40 kmol·h

of species A, and 40 kmol·h

of species B. Two

independent reactions occur: A



(I)

A + 2B



(II) 1

Analysis of the gaseous effluent shows mole fractions: yC = 0.52 and yD = 0.04. Determine the rates (kmol·h ) of all species in the effluent stream. (c) Feed to a gas-phase reactor is 100 kmol·h A



B+C

(I)

of pure species A. Two independent reactions occur: A+B

 D

(II)

Reaction (I) produces valuable species C and coproduct B. The side reaction (II) produces by-product D. 1

Analysis of the gaseous effluent shows mole fractions yC = 0.30 and yD = 0.10. Determine the rates (kmol·h ) of all species in the effluent stream. 1

(d) The feed to a gas-phase reactor is 100 kmol·h , containing 40 mol-% species A and 60 mol-% species B. Two independent reactions occur: A+B



(I)

A+B



D+E

(II)

Analysis of the gaseous effluent shows mole fractions yC = 0.25 and yD = 0.20. Determine: (i) Rates in kmol·h

of all species in the effluent stream.

(ii) Mole fractions of all species in the effluent stream.

Solution

    1 1    (a): Stoichiometric Coefficients: v      1  1    1   0

Number of components: I =1..4

Number of reactions: j = 1..2

v j   v i, j i

Given values: y A  005

    50   n0       0     0 

1 v    1

n0  n0 i i

n0  100

kmol h

yB  010

Solution continued on next page…

Given yA 

n01  1  2

yB 

n0  1  2

n0 2  1 n0  1  2

Make guesses for yC, yD, 1 and 2

yC 

n03  1  2

yD 

n0  1  2

n0 4  2 n0  1  2

1 and 2 1  44737

kmol h

2  2632

kmol hr

Now determine the n values Part(i)

n A  n 01  1  2

n A  2632

kmol h

nB  n02  1

nB  5263

kmol h

nC  n 03  1  2

nC  42105

nD  n0 4  2

n A  2632

n  n A  nB  nC  nD

n  52632

kmol h

kmol h kmol h

Part(ii): yC  08 yD  005     1 1     (b) Stoichiometric Coefficients: v       0   1   1   0

Number of components: I =1..4

    40   n0       0     0 

Number of reactions: j = 1..2 v j  vi, j i

1 v    n0  n0 i 1 i

n0  80

kmol h

Solution continued on next page…

Given values: yC  005

yD  004

Given yA 

n01  1  2

yB 

n0  1  22

n02  1  22 n0  1  22

Make guesses for yA, yB, 1 and 2

yC 

n03  1

yD 

n0  1  22

n0 4  2 n0  1  22

1 and 2 1  26

kmol h

2  2

kmol hr

y A  024

yB  02

n A  n 01  1  2

n A  12

nB  n02  1  22

nB  10

Now determine the n values

nC  26

kmol h

nD  n04  2

nA  2

kmol h     100   n0       0     0 

Number of reactions: j = 1..2 v j  vi, j i

Given values: yC  003

kmol h

nC  n03  1

    1 1       (c): Stoichiometric Coefficients: v     0   1   1   0

Number of components: I =1..4

kmol h

 1  v    n0  n0 i 1 i

n0  100

kmol h

yD  01

Solution continued on next page…

Given yA 

n01  1  2

yB 

n0  1  22

n0 2  1  22

yC 

n0  1  22

Make guesses for yA, yB, 1 and 2

n03  1

yD 

n0  1  22

n0 4  2 n0  1  22

1 and 2 kmol h

2  125

y A  0

yB  02

1  375

kmol hr

Now determine the n values

n A  50

nB  n02  1  22

nB  25

nC  375

kmol h

nD  n04  2

n A  125

kmol h          40   n0       0     0     0 

Number of reactions: j = 1..2

v j   v i, j i

Given values: yC  025

kmol h

nC  n03  1

        1 1   (d): Stoichiometric Coefficients: v        1 0     0 1    1   0

Number of components: I =1..5

kmol h

n A  n01  1  2

1 v    0

n0  n0 i i

n0  100

kmol h

yD  02

Solution continued on next page…

Given yA 

n01  1  2 n0  1

yB 

n0 2  1  22

yC 

n0  1 yE 

Make guesses for yA, yB, yE, 1 and 2

n03  1 n0  1

yD 

n0 4  2 n0  1

n05  2 n0  1

1 and 2 1  20

kmol h

yA  

2  16

kmol hr

yB  

yE  

Now determine the n values

kmol h

n A  n01  1  2

nA  4

nB  n02  1  22

nB  24

nC  n03  1

nC  20

nD  n0 4  2

n A  16

kmol h

nE  n05  2

n A  16

kmol h

Solution 14.42

Problem Statement

The following is an industrial-safety rule of thumb: compounds with large positive G °f must be handled and stored carefully. Explain.

Solution

A compound with large positive G °f has a disposition to decompose into its constituent elements. Moreover, large positive G °f often implies large positive H °f . Thus, if any decomposition product is a gas, high pressures can be generated in a closed system owing to temperature increases resulting from exothermic decomposition.

Solution 14.43

Problem Statement

Two important classes of reactions are oxidation reactions and cracking reactions. One class is invariably endothermic; the other, exothermic. Which is which? For which class of reactions (oxidation or cracking) does equilibrium conversion increase with increasing T?

Solution

Oxidation is typically the loss of electrons from a chemical substance, being replaced by oxygen. This loss of energy would mean that the process is exothermic, as that energy is release typically as heat. This would leave cracking as endothermic. To prove this, cracking is the breaking of long chain hydrocarbons, but to break these chains a lot of energy is needed before the bond will break. This addition of energy means it is an endothermic reaction

Looking at the gibbs free energy equation at equilibrium we have

0  H  TS For cracking, entropy dominates this equation which would mean if temperature increased then the conversion would increase. For oxidation, enthalpy dominates this equation which would mean if temperature increased, and enthalpy is dependent on temperature, the conversion would also increase.

Solution 14.44

Problem Statement

H° for gas-phase reactions is independent of the choice of standard-state pressure P

G° for such reactions does depend on P°. Two choices

of P° are conventional: 1 bar (the basis adopted in this text), and 1.01325 b

G° for gas-

phase reactions from values based on P° = 1 bar to those based on P° = 1.01325 bar.

Solution

By Eq. (14.12), G °   i viGi°



G P  V ° i

° i

 G °   G°     v  i   v V °   i  P   i i °   P T  T i i

Solution continued on next page…

For the ideal gas state, Vi°  RT /P ° . Therefore  G °     v RT  vRT   i P° P° °   P T i

 

and G ° P2° G ° P1°  vRT

P2° P1°

Solution 14.45

Problem Statement

Ethanol is produced from ethylene via the gas-phase reaction C2H4(g)

H2O(g)



C2H5OH(g)

Reaction conditions are 400 K and 2 bar. (a) Determine a numerical value for the equilibrium constant K for this reaction at 298.15 K. (b) Determine a numerical value for K for this reaction at 400 K. (c) Determine the composition of the equilibrium gas mixture for an equimolar feed containing only ethylene and H2O. State all assumptions. (d) For the same feed as in part (c), but for P = 1 bar, would the equilibrium mole fraction of ethanol be higher or lower? Explain.

Solution

C2H4(g) + H2O(g) -> C2H5OH(g)

T0 



K P0  bar T 

K P  bar

Solution continued on next page…

J mol

G 0 f 1  68460

J mol

G0 f 1  228572

H 0 f 1  52500

1= C2H4(g)

2= H2O(g) H 0 f 2  241818

3= C2H5OH(g)

H 0 f 3  235100

J mol

J mol J mol

G0 f 3  168490

J mol

H 0  H 0 f 1 H 0 f 2  H 0 f 3

H 0  45782

G 0  G0 f 1  G 0 f 2  G 0 f 3

G 0  8378

A  1424  3470  3518

A  1376

 B  14394   1450  20001  103    C   4392  0  6002  106    D  0  0121  0  105  

kJ mol

B  4157  10 4  C  161  106

 D  121  10 4

(a): The equilibrium constant at 298 Kelvin is:  G 0   K 298  K 0 K 0  exp  RT 

K 298  29366

(b): the equilibrium constant at 400K is  H 0  T 0   K  907  103 1  K 1  exp  1  RT  T 

K2 

 1   IDCPH T T A B C D   IDCPS T T A B C D  K    2  T 

Solution continued on next page…

K 400  K0 K1 K2

K e  0263

(c): Assume as a basis there is initially 1 mol of C2H4 and 1 mol of H2O y1 

1  e 2  e

y2 

1  e 2  e

y3 

e 2  e

Assuming Ideal gas behavior  P   K    P0  y1 y2 y3

Substitution results in the following expression: e  P  2  e  K 400    1    1  e   P0   e     2  e   2  e 

Solve for e iteratively with either mathcad or excel

e  0191 Plug this back into the relations for y1-y3

y1  0447 y2  0447 y3  0105 (d): Since v = 1< 0, a decrease in the pressure will cause a shift on the reaction to the left and the mole fraction of ethanol will decrease.

Solution 14.46

Problem Statement

A good source for formation data for compounds is the NIST Chemistry WebBook site. Values of H °f , but not of G °f , are reported. Instead, values of absolute standard entropies S° are listed for compounds and elements. To illustrate the use of NIST data, let H2O2 be the compound of interest. Values provided by the Chemistry WebBook are as follows:  H °f [H2O2(g)] = 136.1064 J·mol 1

 S°[H2O2(g)] = 232.95 J·mol ·K 1

 S°[H2(g)] = 130.680 J·mol ·K  S°[O2(g)]= 205.152 J·mol ·K

All data are for the ideal-gas state at 298.15 K and 1 bar. Determine a value of G °f 298 for H2O2(g). Solution

Using the givens in the problem statement, we can determine that: H2(g) + O2(g) –> H2O2(g)

S fH O  S0 H  S0O  S0 H 0 2 2

2 2

S fH O  102882 2 2

J mol K

And

G0 f  H fH O  T S fH O 2 2

2 2

G0 f 



kJ mol

Solution 14.47

Problem Statement

Reagent-grade, liquid-phase chemicals often contain as impurities isomers of the nominal compound, with a consequent effect on the vapor pressure. This can be quantified by phase-equilibrium/reaction-equilibrium analysis. Consider a system containing isomers A and B in vapor/liquid equilibrium, and also in equilibrium with respect to the reaction A  B at relatively low pressure. (a) For the reaction in the liquid phase, determine an expression for P (the “mixture vapor pressure”) in terms l

of PAsat , PBsat , and K l , the reaction equilibrium constant. Check the result for the limits K = 0 and K = . v

(b) For the reaction in the vapor phase, repeat part (a). Here, the relevant reaction equilibrium constant is K . (c) If equilibrium prevails, then whether the reaction is assumed to occur in one phase or the other makes no difference. Thus the results for parts (a) and (b) must be equivalent. Use this idea to show the connection l

between K and K through the pure-species vapor pressures. (d ) Why is the assumption of ideal gases and ideal solutions both reasonable and prudent? (e) Results for parts (a) and (b) should suggest that P depends on T only. Show that this is in accord with the phase rule.

Solution

(a) For isomers at low pressure Raoult’s law should apply:



P  x A PAsat  x B PBsat  PBsat  x A PAsat  PBsat



For the given reaction with an ideal solution in the liquid phase, the equation becomes: Kl 

xB xA



1 xA

xA 

1 l

K 

The preceding equation now becomes,  1  sat  1  sat P  1  l  PB   l  PA   K  1  K  1   K l      P sat  1  P sat P   l B  K l  1  A  K  1 

(A)

For K l  0 P  PAsat For K l   P  PBsat

(b) Given Raoult’s law:

 x A  xB  y A

P

1 yA

PBsat

 sat

 yB sat



 y yB   A  P     P sat P sat  PBsat A B P

PAsat PBsat

 sat

y A PBsat  yB PA

PAsat PBsat



PAsat  y A PBsat  PAsat



For the given reaction with ideal gases in the vapor phase, Eq. (14.28) becomes: yB yA

 K v whence y A 

1 v

K 1

Elimination of y A from the preceding equation and reduction gives:

 K  1 P P P v

sat sat A B

K v PAsat  PBsat

(B)

For K v  0 P  PAsat ForK v   P  PBsat (c) Equations (A) and (B) must yield the same P. Therefore Solution continued on next page…





K v  1 PAsat PBsat  K l      P sat   1  P sat   l  B  K l  1  A K v PAsat PBsat  K  1 

Some algebra reduces this to:

PBsat

PAsat

 l

(d) As mentioned already, the species (isomers) are chemically similar, and the low pressure favors ideal-gas behavior. (e) F  N     r      

T should suffice.

Solution 14.48

Problem Statement

Cracking propane is a route to light olefin production. Suppose that two cracking reactions occur in a steadyflow reactor: C3H8(g)  C3H6(g) + H2(g)

(I)

C3H8(g)  C2H4(g) + CH4(g) (II) Calculate the product composition if both reactions go to equilibrium at 1.2 bar and (a) 750 K; (b) 1000 K; (c) 1250 K

Solution

C3H8(g) –> C3H6(g) + H2(g) (I) C3H8( (g) –> C2H4(g) + CH4(g) (II)

T0  1 = C3H8 (g)

H 0 f 1  104680

2 = C3H6 (g)

H 0 f 2  19710

3 = H2 (g)

H 0 f 1  0

4 = C2H4 (g)

H 0 f 2  52510

5 = CH4 (g)

H 0 f 1  74520



K P0  bar T 

K P   bar

J J G 0 f 1  24290 mol mol

J J G 0 f 1  62205 mol mol

J J G 0 f 1  0 mol mol J J G 0 f 1  68460 mol mol J J G 0 f 1  50460 mol mol

Calculate equilibrium constant for reaction I:

H 0 I  H 0 f 1  H 0 f 2  H 0 f 3 H 0 I  12439

G 0 I  G 0 f 1  G 0 f 2  G 0 f 3 G 0 I  86495

kJ mol

AI  1213  1637  3249 AI  3673  BI  28785  22706  0422   103    CI  8824    6915  0   10 6    DI  0   0  0083  105  

BI   5657  10 3  CI  1909  10 6

DI  83  10 3

 G 0 I   KI 0  0 KI 0  exp  RT 

 H 0 I  T 0   KI  1348  1013 1  KI1  exp  1 T   RT 

KI 2 

 1   IDCPH T T AI BI CI DI   IDCPS T T AI BI CI DI    T 

KI2  1714 KI  KI 0 KI1 KI2

KI  0016

Calculate equilibrium constant for reaction II:

H 0 II  H 0 f 1  H 0 f 4  H 0 f 5 H 0 II  8267

G0 II  G0 f 1  G 0 f 4  G 0 f 5 G0 II  4229

kJ mol

AII  1213  1424  1702 AII  1913  BII  28785  14394  9081  103  

BII  531  10 3

 CII  8824    4392   2164   10 6  CII  2268  106    DII  0  0  0  105  

 DII  0

 G 0 II   KII 0  3897  108 KII 0  exp  RT   H 0 II  T 0   KII  5322  1013 1  KII1  exp  1  T   RT

KII2 

 1   IDCPH T T AI BI CI DI   IDCPS T T AI BI CI DI    T 

KII2  1028 KII  KII 0 KII1 KII2

KII  21328

Part(a) Assume an ideal gas and 1 mol of C3H8 initially.

y1 

1  I  II

y2 

1  I  II y4 

I

y3 

1  I  II

II 1  I  II

y3 

I 1  I  II

II 1  I  II

The equilibrium relationships are: y2 y3 y1

P   KI  0   P 

y4 y5 y1

P   KII  0   P 

Substitution yields the following equations:   I I     1      1      P   I II  I II   KI  0  1  I  II  P  1  I  II

and

 II II     1      1      P   I II  I II   KII  0  1  I  II  P  1  I  II

Solve iteratively for I II , this can be done in matlab, Mathcad, or excel.

I  0026 II  0948 Plug this into the gas compositions to determine y1-y5

y1  001298

y2  00132

y3  00132

y4  04803

y3  04803

A summary of the values for the other temperatures is given in the table below. Part

750

1000

1250

0.013

0.00047

0.00006

0.0132

0.034

0.0593

0.0132

0.034

0.0593

0.4803

0.4658

0.4407

0.4803

0.4658

0.4407

T= (K) y1 y2 y3 y4 y5

Solution 14.49

Problem Statement

Equilibrium at 425 K and 15 bar is established for the gas-phase isomerization reaction 

n-C4H10(g)

iso-C4

10(g)

Estimate the composition of the equilibrium mixture by two procedures: (a) Assume an ideal-gas mixture. (b) Assume an ideal solution with the equation of state given by Eq. (3.36). Compare and discuss the results. 1

Data: For iso-butane, H °f 298 =

; G °f 298 =

Eq. 3.36: Z

PV BP 1 RT RT

Solution

n-C4H10(g) –> iso-C4H10(g)

T0  1= n-C4H10(g)

H 0 f 1  125790

2= iso-C4H10(g) H 0 f 2  134180

J mol



K P0  bar T  G0 f 1  16570

G0 f 1  20760

K P

bar

J mol

Solution continued on next page…

H  H

f 1  H f 2

kJ mol

H   

G0  G 0 f 1  G0 f 2 G0  419

kJ mol

A  1935  1677 A  0258  B  36915  37853   10 3  B  938  104    C  11402   11945   106    D        

C   543  10 7

D 

 G 0   K 0  5421 K 0  exp  RT   H 0  T 0  1   K  0364 K 1  exp   1  RT  T 

K2 

 1   IDCPH T T A B C D  IDCPS T T A B C D   T 

K2 

Ke  K0 K1 K2 K e  1974 Assume as a basis there is initially 1 mol of n-C4H10(g)

y1  1  e

(a): Assuming Ideal gas behavior

y2 y1

y2  e

 Ke

Substitution results in the following expression: K e 

e 1  e

e

Solving for Ke e 

y1  1  e

1 1  Ke

e  0336

y1  0664 y2  e

y2  0336

Solution continued on next page…

(b): Assume the gas is an ideal solution. In this case Eqn. (14.27) applies.

  P v    ( yii   P  K  0 i   

Substituting for yi yields:

Solve for e

e 

1  e 2 e 1

K

2 2  K e 1

Calculate i for each pure component using the PHIB function For n-C4H10

1  0200 Tc1  4251K Pc1  3796 bar Tr 1  1 Pr 1  0395 1  PHIBTr1 Pr1 1  1  

For iso-C4H10:

2  0200 Tc 2  4251 K Pc 2  3796 bar Tr 2  1041 Pr 2  0411 2  PHIB Tr 2 , Pr 2 , 2  2  0884

Solving for e yields e 

y1  1  e

2 2  K e 1

e  0339

y1  0661 y2  e

y2  0339

The values of y1 and y2 calculated in parts a) and b) differ by less than 1%. Therefore, the effects of vapor-phase nonidealities is here minimal.

Solution 14.50

Problem Statement

Compute the emf and work output per mole of fuel for a reversible fuel cell that uses methanol as fuel and air as the source of oxygen, operating at 50°C.

Solution

The reaction in this case is CH3OH + /2 O2  CO2 + 2 H2O. The work output is simply equal to the Gibbs 3

energy change of reaction (at 50°C and accounting for the use of air rather than pure O2 as the oxygen source). We will follow the approach illustrated in the text and not account for the dilution of the products in nitrogen, which is slightly inconsistent with accounting for the dilution of the oxidant, but has a small effect in any case. We assume that the methanol is delivered as a vapor and the water is produced as a vapor. At 298.15 K, we have G298

Gf,298(CO2

G298 = H298

Gf,298(H2O)

Gf,298(CH3OH)

228572 Hf,298(CO2

H298 =

Hf,298(H2O)

Hf,298(CH3OH)

241818

G at 50°C, we also integrate the heat capacity difference from 298.15 to 323.15 K, for both enthalpy and entropy: 323.15 Cpo  323.15   323.15  323.15 o o    H 298.151     Cp dT  T  G 323.15  G 298.15 dT  298.15   298.15  T o

298.15

For enthalpy, we use the ICPDH spreadsheet, and for entropy the ICPDS spreadsheet: T

IDCPH  T0 ,T; A, B, C, D  

1 R

B

C

1

1 

 C dT  A T  T   2 T  T   3 T  T   D  T  T  p

2 0

3 0

T0 (K) 298.15

T (K) 323.15

DB (1/K)

DC (1/K )

DD (K )

IDCPH (K)

4.7275

-9.03E-03

3.45E-06

-5.75E+04

41.5

Species Name

Stoichiometric coefficient

B (1/K)

C (1/K )

CH3OH

-1

2.211

1.22E-02

-3.45E-06

0.00E+00 -2.21E+00

D (K )

n iAi

-1.5

3.639

5.06E-04

0.00E+00

-2.27E+04 -5.46E+00

CO2

5.457

1.05E-03

0.00E+00

-1.16E+05 5.46E+00

H2O

3.47

1.45E-03

0.00E+00

1.21E+04

C p

T 

IDCPS  T0 ,T; A, B, C, D  

6.94E+00

D   T 2  T02   2 2  2 0 



 RT dT  A ln  T   B T  T    C  T T   0

T0 (K) 298.15

T (K) 323.15

DB (1/K)

DC (1/K )

DD (K )

4.7275

-9.03E-03

3.45E-06

-5.75E+04

Species Name

Stoichiometric coefficient

B (1/K)

C (1/K )

CH3OH

-1

2.211

1.22E-02

-3.45E-06

0.00E+00 -2.21E+00

D (K )

IDCPS 0.134

n iAi

-1.5

3.639

5.06E-04

0.00E+00

-2.27E+04 -5.46E+00

CO2

5.457

1.05E-03

0.00E+00

-1.16E+05 5.46E+00

H2O

3.47

1.45E-03

0.00E+00

1.21E+04

6.94E+00

Thus, IDCPH = 35.2 K and IDCPS = 0.113. Using these, we get  323.15   323.15    6764851   G o 323.15  689543  298.15   8.3145  41.5  8.3145  323.15  0.134  298.15 

From which G323 = Finally, if we account for the dilution of oxygen in air, noting that we use 1.5 mol O2 per mol methanol or mol reaction, we have G323,air = Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

That is the electrical work available from a reversible methanol fuel cell operating at 50°C. To find the voltage, we note that 6 electrons are produced per mole of methanol oxidized. Note that each oxygen atom coming from O2 takes two electrons to form CO2 or H2O, but the oxygen atom in methanol is already reduced and does not take any electrons from the fuel. The emf of the fuel cell is then E=

G/6F = 696223/(6*96485) = 1.203 V

Where F is Faraday’s constant (96485 C/mol) and one J/C equals 1 V.

Solution 14.51

Problem Statement

Compute the emf and work output per mole of Zn for a zinc-air battery operating at 30°C. Treat it as a steadystate fuel cell using Zn as the fuel and find the needed thermodynamic data in the NIST Chemistry WebBook.

Solution

Here, the reaction is simply Zn(s) + ½O2(g)  ZnO(s), which is the formation reaction for zinc oxide. Thus, the enthalpy and Gibbs energy of reaction are just the enthalpy and Gibbs energy of formation of ZnO. We get the enthalpy of formation from the NIST

Hf =

of formation are not tabulated there, so we have to use the entropy values provided there to get an entropy of formation and use that to get the Gibbs energy of formation. We have Sf = 43.65 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

The large negative entropy of reaction is associated with a reduction in gas-phase species (gaseous O2 ends up in solid ZnO). Gf = To get the value at 30°C, we can reasonably neglect the effect of changes in enthalpy and entropy of reaction over a 5 degree interval (skip the heat capacity integrals) and use the values at 298.15 K. Doing so: Gf(30°C) = Finally, if we account for the dilution of oxygen in air, noting that we use 0.5 mol O2 per mol methanol or mol reaction, we have G323,air = Thus, the work output per mole of Zn oxidized is oxidized, the emf of the fuel cell is then E=

G/2F = 318010/(2*96485) = 1.65 V

Where F is Faraday’s constant (96485 C/mol) and one J/C equals 1 V.

Solution 14.52

Problem Statement Solid oxide fuel cells use a ceramic membrane to separate the anode and cathode and operate at elevated temperature. Compute the emf and work output per mole of H2 for a reversible solid-oxide fuel cell operating with air as the oxygen source at a temperature of 800°C.

Solution

As shown in the text, for a hydrogen fuel cell, the Gibbs energy and enthalpy changes of reaction are just the Gibbs energy and enthalpy of formation of gaseous water: G298 =

H298 =

We want to use these, along with the enthalpy and entropy heat capacity integrals, to compute the Gibbs energy of hydrogen oxidation at 800°C, according to: 1073.15 Cpo  1073.15   1073.15  1073.15  H o 298.151     Cpo dT  T  G o 1073.15  G o 298.15 dT  298.15  298.15  298.15 T  298.15

We use the IDCPH and IDCPS spreadsheets to evaluate the heat capacity integrals as shown below:

IDCPH  T0 ,T; A, B, C, D  

1 1  1 B 2 C 3 C p dT  A  T  T0   T  T02  T  T03  D    R 2 3  T T0 











T0 (K) 298.15

T (K) 1073.15

DB (1/K)

DC (1/K )

DD (K )

IDCPH (K)

-1.5985

7.75E-04

0.00E+00

1.52E+04

-790.3

Species Name

Stoichiometric coefficient

B (1/K)

C (1/K )

-1

3.249

4.22E-04

0.00E+00

8.30E+03 -3.25E+00

-0.5

3.639

5.06E-04

0.00E+00

-2.27E+04 -1.82E+00

H2O

3.47

1.45E-03

0.00E+00

1.21E+04

C p

T 

IDCPS  T0 ,T; A, B, C, D  

D (K )



n iAi

3.47E+00

2 2 D   T  T0   2 2  2 0 

 RT dT  A ln  T   B T  T    C  T T   0

T0 (K) 298.15

T (K) 1073.15

DB (1/K)

DC (1/K )

DD (K )

-1.5985

7.75E-04

0.00E+00

1.52E+04

Species Name

Stoichiometric coefficient

B (1/K)

C (1/K )

D (K )

-1

3.249

4.22E-04

0.00E+00

8.30E+03 -3.25E+00

-0.5

3.639

5.06E-04

0.00E+00

-2.27E+04 -1.82E+00

H2O

3.47

1.45E-03

0.00E+00

1.21E+04

IDCPS -1.368

n iAi

3.47E+00

Solution continued on next page…

Using these, we get  1073.15   1073.15    2418181    8.3145 790.3  8.3145  1073.15  1.368 G o 1073.15  228572  298.15   298.15  

From which G1073 = Finally, if we account for the dilution of oxygen in air, noting that we use 0.5 mol O2 per mol methanol or mol reaction, we have G1073,air = That is the electrical work available from a reversible hydrogen fuel cell operating at 800°C. To find the voltage, we note that 2 electrons are produced per mole of reaction. The emf of the fuel cell is then E=

G/6F = 186409/(2*96485) = 0.966 V

Where F is Faraday’s constant (96485 C/mol) and one J/C equals 1 V.

Solution 15.1

Problem Statement

An absolute upper bound on G for stability of an equimolar binary mixture is G = RT ln 2. Develop this result. What is the corresponding bound for an equimolar mixture containing N species?

Solution

Stability requires that G  





G 

in which event Eq. (12.30)

becomes: G E  RT xi lnxi i

For an equimolar solution x i 

N is the number of species. Therefore,

GE 

  N i

1 1  RT  N N i

N  RT

For the special case of a binary solution, N  2,andG E max   RT ln 2

Solution 15.2

Problem Statement

A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for parameters A12 and A21 in the Margules equation at 25°C: (a) x1  0.10,

x1  0.90; (b) x1  0.20,

x1  0.90; (c) x1  0.10,

x1  0.80.

Solution

(a) The criterion for liquid-liquid equilibrium (as for any phase equilibrium) is that the fugacity of each component is the same in both phases. For liquids, we can write the fugacities using an activity coefficient model as: x i  i fi  x i  i fi

Or, because the pure component fugacity is the same for both phases (it is not a function of composition, but just a property of the component at given T and P), simply x i  i  x i i

For a binary mixture, using the Margules equations for the activity coefficients, this becomes: 2 2     x1 exp x2 A12  2  A21  A12  x1   x1 exp x 2 A12  2 A21  A12  x1       2 2     x2 exp x1 A21  2  A12  A21  x 2   x 2 exp x1 A21  2 A12  A21  x 2      

 



 



 



 



Putting in numerical values for the mole fractions in each phase gives us two equations for two unknowns (the Margules parameters): 2 2     0.1exp0.9  A12  2  A21  A12 0.1  0.9exp0.1  A12  2 A21  A12 0.9     2 2     0.9exp0.1  A21  2  A12  A21 0.9  0.1exp0.9  A21  2 A12  A21 0.1    

Rearranging these and taking the natural logarithm gives 0.656 A12 + 0.144 A21 = ln(9) = 2.1972 0.144 A12 + 0.656 A21 = ln(9) = 2.1972 Eliminating A21 gives 0.6244 A12 = 1.71491, or A12 = 2.746 Substituting this into either of the equations allow one to solve for A21 = 2.746. As we might expect from the symmetry of the mole fractions and of the phases, the two parameters are the same. (b) The criterion for liquid-liquid equilibrium (as for any phase equilibrium) is that the fugacity of each component is the same in both phases. For liquids, we can write the fugacities using an activity coefficient model as: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

x i  i fi  x i  i fi

For a binary mixture, using the Margules equations for the activity coefficients, this becomes: 2 2     x1 exp x2 A12  2  A21  A12  x1   x1 exp x 2 A12  2 A21  A12  x1      2 2     x2 exp x1 A21  2  A12  A21  x 2   x 2 exp x1 A21  2 A12  A21  x 2     

 



 



 



 



Putting in numerical values for the mole fractions in each phase gives us two equations for two unknowns (the Margules parameters): 2 2     0.2exp0.8  A12  2  A21  A12 0.2  0.9exp0.1  A12  2 A21  A12 0.9     2 2     0.8exp0.2  A21  2  A12  A21 0.8  0.1exp0.9  A21  2  A12  A21 0.1    

Rearranging these and taking the natural logarithm gives 0.392 A12 + 0.238 A21 = ln(4.5) = 1.50408 0.098 A12 + 0.672 A21 = ln(8) = 2.148 Eliminating A21 gives 0.3573 A12 = 0.76761, or A12 = 2.746 Substituting this into either of the equations allow one to solve for A21 = 2.781. (c) The criterion for liquid-liquid equilibrium (as for any phase equilibrium) is that the fugacity of each component is the same in both phases. For liquids, we can write the fugacities using an activity coefficient model as: x i  i fi  x i  i fi

For a binary mixture, using the Margules equations for the activity coefficients, this becomes: 2 2     x1 exp x2 A12  2  A21  A12  x1   x1 exp x 2 A12  2 A21  A12  x1      2 2     x2 exp x1 A21  2  A12  A21  x 2   x 2 exp x1 A21  2 A12  A21  x 2     

 



 



 



 



Putting in numerical values for the mole fractions in each phase gives us two equations for two unknowns (the Margules parameters): 2 2     0.1exp0.9  A12  2 A21  A12 0.1  0.8exp 0.2  A12  2 A21  A12 0.8     2 2     0.9exp0.1  A21  2 A12  A21 0.9  0.2exp0.8  A21  2  A12  A21 0.2    

Rearranging these and taking the natural logarithm gives 0.672 A12 + 0.098 A21 = ln(8) = 2.0794 0.238 A12 + 0.392 A21 = ln(4.5) = 1.504 Eliminating A21 gives -2.45 A12 = -6.814, or A12 = 2.78 Substituting this into either of the equations allow one to solve for A21 = 2.15.

Solution 15.3

Problem Statement

Work Prob. 15.2 for the van Laar equation.

Problem 15.2

A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for parameters A12 and A21 in the Margules equation at 25°C: (a ) x1  010, x1  0 90; (b ) x1  020, x1  0 90; (c ) x1  010, x1  0 80 . Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution

x1  01 x 2  1  x1 x 2  09 x 2  1  x1

Guess

a12  2a21  2

Given 2  2          a12 x1    a12 x1    exp  a12 1     x1  exp a12 1     x1     a21 x2   a21 x2           2  2          a21 x2    a21 x2    exp  a21 1     x2  exp a21 1     x2     a12 x2   a12 x1          

a   12   Find a , a a  2747 a  2747  12 21  12 21 a   21 

x1  02 x 2  1  x1 x 2  09 x 2  1  x1

Guess

a12  2a21  2

Given 2  2          a12 x1    a12 x1         exp a12 1    x1  exp a12 1     x1         a x a x     21 2 21 2         2  2          a21 x2    a21 x2         exp a21 1    x2  exp a21 1     x2            a x a x     12 2 12 1       

a   12   Find a , a a  2199a  281  12 21  12 21 a   21 

x1  01 x 2  1  x1 x 2  08 x 2  1  x1

Guess

a12  2a21  2

Given

2  2          a12 x1    a12 x1         exp a12 1    x1  exp a12 1     x1           a x a x 21 2   21 2           2  2          a21 x2    a21 x2         exp a21 1    x2  exp a21 1     x2           a x a x 12 2   12 1          

a   12   Find a , a  12 21  a12  281a21  2199 a   21 

Solution 15.4

Problem Statement

Consider a binary vapor-phase mixture described by Eqs. (3.36) and (10.62). Under what (highly unlikely) conditions would one expect the mixture to split into two immiscible vapor phases?

Eq. 3.36: Z

PV BP 1 RT RT

Eq. 10.62: B  y12 B11  2 y1 y2 B12  y22 B22

Solution

We know that,

G E  12 Py1 y2 or

This equation has the form:

G E 12 P  y y RT RT 1 2 GE  Ax1 x 2 RT

Solution continued on next page…

for which phase-splitting occurs for A

12 P RT 

Suppose T = 300 K and P = 5 bar. The preceding condition then requires : 12 

cm 3

1

For vapor – phase

immiscibility. Such large postive values for 12 are unknown for real mixtures. (Examples of gas/gas equilibria are known, but at conditions outside the range of applicability of the two-term virial EOS.)

Solution 15.5

Problem Statement

Figures 15.1, 15.2, and 15.3 are based on Eqs. (A) and (F) of Ex. 15.3 with CPE assumed to be positive and given by CPE / R = 3x 1x 2. Graph the corresponding figures for the following cases, in which CPE is assumed to be negative:

(a) A 

975  184  3 lnT T

(b) A 

540  171  3 lnT T

1500  199  3 lnT T

Eq. A of Ex. 15.3: GE  Ax1 x 2 RT

Eq. F of Ex. 15.3: A

a  b  c ln T T

Solution

a  975b 184c 3 T  250450 AT  

a  b  c  lnT  T

Parameter A = 2 at two temperatures. The lower one is an UCST, because A decreases to 2 as T increases. The higher one is a LCST, because A decreases to 2 as T decreases. Guess: x = 0.25

Given

 1  x   AT   1  2 x   ln   x  x

x 

Eq. E, Ex. 15.3 x1 T   Find  x  x 2 T    x1 T 

UCST = 300 (guess) Given

A (UCST) =2

UCST = Find (UCST)

UCST= 272.93

LCST = Find (LCST)

LCST= 391.21

LCST = 400 (guess) Given

A (LCST) =2

Plot phase diagram as a function of T T1= 225, 225.1 ..UCST

T2= LCST..450

(b) a 

b    c  T  250450 AT  

a  b  c  lnT  T

Parameter A = 2 at a single temperature. It is an LCST, because A decreases to 2 as T decreases. Guess: x = 0.25

Given

 1  x   AT   1  2 x   ln   x 

Eq. E, Ex. 15.3

x  0 x  05 x1 T   Find x 

LCST = 350 (guess) Given

A (UCST) =2

LCST = Find (LCST)

Plot phase diagram as a function of T

LCST= 346

T= LCST..450

b

 c 

T  250450 AT  

a  b  c  lnT  T

Parameter A = 2 at a single temperature. It is a UCST, because A decreases to 2 as T increases. Guess: x = 0.25

Given

 1  x   AT   1  2 x   ln   x 

Eq. E, Ex. 15.3

x  0 x  05 x1 T   Find x 

UCST = 350 (guess) Given

A (UCST) =2

UCST = Find (UCST)

UCST= 339.66

Plot phase diagram as a function of T

T= UCST..250

Solution 15.6

Problem Statement

It has been suggested that a value for G of at least 0.5 RT is required for liquid/liquid phase splitting in a binary system. Offer some justification for this statement.

Solution

Consider a quadratic mixture, described by:

GE  Ax1 x 2 RT

It is shown in Example 15.4 that phase splitting occurs for such a mixture if A  A

x1  x 2  5. Thus, for a quadratic mixture, phase-splitting obtains if:

1 1 G E  2    RT  05 RT 2 2

This is a model-dependent result. Many liquid mixtures are known which are stable as single phases, even though G E  05 RT for equimolar composition.

Solution 15.7

Problem Statement

Pure liquid species 2 and 3 are for practical purposes immiscible in one another. Liquid species 1 is soluble in both liquid 2 and liquid 3. One mole each of liquids 1, 2, and 3 are shaken together to form an equilibrium mixture of two liquid phases: an -phase containing species 1 and 2, and a -phase containing species 1 and 3. What are the mole fractions of species 1 in the

and phases, if at the temperature of the experiment, the excess Gibbs energies of the

phases are given by: (G E )  0.4 x1 x 2 RT

and

(G E )  0.8 x1 x3 RT

Solution

The criterion for species 1 to be in equilibrium between the two phases is the its fugacity is the same in both phases, which implies that x1 1  x1 1 . Since we specify that the mole fraction of species 2 in phase and of species 3 in phase

are zero, we do not have any similar equilibrium relationship for those two species. We have seen in chapter

12 and the previous homework the activity coefficients corresponding to the excess Gibbs energy model GE  ax1 x 2 RT

are ln 1  ax 22 and ln 2  ax12

So, in phase , we have 2 2   1  exp0.4 x2   exp0.4 1  x1      

 





and in phase we have 2 2   1  exp0.8 x3   exp0.8 1  x1     

 





Substituting these into the equilibrium relationship for species 1, we have Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

2 2   x1 exp 0.4 1  x1   x1 exp0.8 1  x1      









This gives us 1 equation and 2 unknowns. The additional equation that we need comes from a balance on the total amount of species 1 present. We know that phase

contains 1 mole of species 2, and phase contains one mole of

species 3, and that a total of 1 mole of species 1 is split between the two phases. So, we want to write an equation that relates x1 and x1 based on the fact that the total number of moles of species 1 is 1. These mole fractions, in terms of the number of moles of species 1 in each phase, are x1 

n1 n1  n2



n1 n1  1

and

x1 

n1 n1  n3



n1 n1  1

Solving these for n1 and n1 gives n1 

x1

n1 

and

1  x1

x1 1  x1

So, we can write the requirement that the total number of moles of species 1 is 1 as n1  n1 

x1 1  x1



x1 1  x1

1

Solving this for x1 in terms of x1 gives x1 

2 x1  1

1  x1 

and

3 x1  2

x1  1 3 x1  2

Now, we can substitute this into the equilibrium relationship for species 1 to get a single equation for x1 : 



x1 exp0.4 1  x1  

2      x  1   1     exp0.8   3 x   2    3 x1  2   1  

2



2 x1  1

This, of course, must finally be solved numerically. One can do so using Maple as follows: > eq:=xa*exp(0.4*(1-xa)^2)=(2*xa-1)/(3*xa-2)*exp(0.8*((xa-1)/(3*xa-2))^2);

eq := xa e

( .4 ( 1 xa )2 )



2  ( xa  1 )   .8   2   ( 3 xa  2 ) 

( 2 xa1 ) e 3 xa2

> fsolve(eq,xa=0..1);

.3711378130 So, we have x1 = 0.371, x1 = 0.291, n1 = 0.590, n1 = 0.410. As we would expect, species 1 partitions preferentially into phase

where the excess Gibbs energy due to its presence is lower.

Solution 15.8

Problem Statement

It is demonstrated in Ex. 15.5 that the Wilson equation for G is incapable of representing LLE. Show that the simple modification of Wilson’s equation given by: E

G RT = C[x1ln(x1 + x2

12) + x2 ln(x2 + x1

21)]

can represent LLE. Here, C is a constant.

Solution

Comparison of the Wilson equation, with the modified Wilson equation shows that

 G E   G E       RT   C  RT   m  



g

GE ; then RT

gm  Cgngm  Cng

Addition and subtraction of

 ngm  n1

C

 ng  n1

 1 m  C C

1

x 1 on the right side yields:

ln  x1  1   ln x1  C ln  x1  1   C ln x1 m

ln  x1 1   C ln  x1  1   C  1ln x 1 m

d ln x1 1 

Differentiate:

dx1

C

d ln  x1 1  dx1



C 1 x1

The derivative on the right side of this equation is always positive. However, for C sufficiently greater than unity, the contribution of the second term on the right can make d ln  x 1 1 

dx1

0

over part of the composition range, thus violating the stability condition and implying the formation of two liquid phases.

Solution 15.9

Problem Statement

Vapor sulfur hexafluoride SF6 at pressures of about 1600 kPa is used as a dielectric in large primary circuit breakers for electric transmission systems. As liquids, SF6 and H2O are essentially immiscible, and one must therefore specify a low enough moisture content in the vapor SF6 so that if condensation occurs in cold weather, a liquid water phase will not form first in the system. For a preliminary determination, assume the vapor phase can be treated as an ideal gas and prepare the phase diagram [like Fig. 12.18(a)] for H2O(1)/SF6(2) at 1600 kPa in the composition range up to 1000 parts per million of water (mole basis). The following approximate equations for vapor pressure are adequate: ln P1sat  kPa  191478 

536370 TK

ln P2sat  kPa  146511 

204897 TK

Solution

 53637   P1sat T   exp191478   T 

P

Water

kPa

 20897   P2sat T   exp146511   T 

SF6

Find 3-phase equilibrium temperature and vapor-phase composition: Guess: T= 300 K Given

P  P1sat T   P2sat T  T   Find T T * 

y1 



  y  10  695

P1sat T  P

 1

Find saturation temperatures of pure species 2: Guess: T = 300 K Solution continued on next page…

P2sat T   P

Given II

T2  FindT T2 

T = T , T +0.0001 ..T2 I

T = T , T +0.01 ..T +6

y1II T   1  y1I T  



P2sat T  P

P1sat T  P

Because of the very large difference in scales appropriate to regions I and II, the txy diagram is presented on the following page in two parts, showing regions I and II separately.

Solution 15.10

Problem Statement

In Ex. 15.2 a plausibility argument was developed from the LLE equilibrium equations to demonstrate that positive deviations from ideal-solution behavior are conducive to liquid/liquid phase splitting. (a) Use one of the binary stability criteria to reach this same conclusion. (b) Is it possible in principle for a system exhibiting negative deviations from ideality to form two liquid phases?

Solution

(a) Refer to the stability requirement of Eq. (12.5). For instability, i.e., for the formation of two liquid phases,

 G E   d2   RT  dx12



1 x1 x2

over part of the composition range. The second derivative of G E must be sufficiently negative so as to satisfy this E

condition for some range of x1. Negative curvature is the norm for mixtures for which G is positive; see,e.g., E

the sketches of G vs. x1 for systems (a),(b),(d),(e),and (f) in Fig.11.4. Such systems are candidates for liquid/ E

liquid phase splitting,although it does not in fact occur for the cases shown.Rather large values of G are usually required. (b) Nothing in principle precludes phase-splitting in mixtures for which G E  0; one merely requires that the curvature be sufficiently negative over part of the composition range. However, positive curvature is the norm for such mixtures. We know of no examples of liquid/liquid phase splitting in systems exhibiting negative deviations from ideal-solution behavior.

Solution 15.11

Problem Statement

Toluene(l) and water(2) are essentially immiscible as liquids. Determine the dew-point temperatures and the compositions of the first drops of liquid formed when vapor mixtures of these species with mole fractions z 1 = 0.2 and z 1 = 0.7 are cooled at a constant pressure of 101.33 kPa. What is the bubblepoint temperature and the composition of the last drop of vapor in each case? See Table B.2 for vapor-pressure equations.

Solution

Temperatures in °C; pressures in kPa  305696   P1sat T   exp 139320  T  217625  

Toluene

P  10133  388570   P2sat T   exp 163872   T  230170 

Water

Find the 3-phase equilibrium T and y: Guess: T = 25 Given

P  P1sat T   P2sat T  T   Find T T * 

y1 



  y  0444

P1sat T *

* 1

For z1< y1 , first liquid is pure species 2: y1  02 Guess : Tdew  T 

Given

y1  

P2sat Tdew  P

Tdew  Find Tdew  Tdew 



For z1> y1 , first liquid is pure species 1: Solution continued on next page…

y1  07 Guess : Tdew  T 

Given

y1 

P1sat Tdew  P

Tdew  Find Tdew  Tdew 



In both cases the bubble point temperature is T , and the mole fraction of the last vapor is y1 .

Solution 15.12

Problem Statement

n-Heptane(l) and water(2) are essentially immiscible as liquids. A vapor mixture containing 65-mol-% water at 100°C and 101.33 kPa is cooled slowly at constant pressure until condensation is complete. Construct a plot for the process showing temperature vs. the equilibrium mole fraction of heptane in the residual vapor. See Table B.2 for vaporpressure equations.

Solution

Temperatures in °C; pressures in KPa  291026   P1sat T   exp 138622  T  216432  

n-heptane

P  10133  388570   P2sat T   exp 163872   T  230170 

Water

Find the 3-phase equilibrium T and y: Guess: T = 50 Given

P  P1sat T   P2sat T  T   FindT T  

y1* 



  y  0548

P1sat T  P

 1

For 0.35 < y1 , first liquid is pure species 2:

y1  1 

P2sat T  P

Find temperature of initial condensation at y1= 0.35:

y1  035 Guess : Tdew  T  0

Given

y1  Tdew   y1

Tdew  Find Tdew  Tdew 



Define the path of vapor mole fraction above and below the dew point.



y1 path  T   if T  Tdew y1 y1  T  0



T= 100, 99.9…T

Path of mole fraction heptane in residual vapor as temperature is decreased. No vapor exists below T

Solution 15.13

Problem Statement

Consider a binary system of species 1 and 2 in which the liquid phase exhibits partial miscibility. In the regions of miscibility, the excess Gibbs energy at a particular temperature is expressed by the equation: GE/RT = 2.25 x 1x2

In addition, the vapor pressures of the pure species are:

P1sat = 75 kPa and P2sat = 110 kPa Making the usual assumptions for low-pressure VLE, prepare a Pxy diagram for this system at the given temperature.

Solution

Pressures in KPa P1sat T   2   A   x1    2  x1    

1  x1  

P2sat T  

P1sat T  

A x  2 1

Find the solubility limits: Guess:

x1  0.1

Given

 x1 

 1  x    1      x1 

x1  0224 x 1  1  x1

 

x1  Find x1

x1  0 776

Find the conditions for VLLE: Guess:

P   P1sat

y1  05

Given

P   x1 1 x1  P1sat  1  x1  2 x1  P2sat

 



  

 

y1 P  x1 1 x1  P1sat  P           Find P y1  y1 



 P 



y1  

Calculate VLE in two-phase region. Modified Raoult’s law; vapor in an ideal gas. Guess:

x1  01 P  50

Given P  x1 1  x1   P1sat  1  x 1    2  x1   P2sat

Solution continued on next page…

P  x1   Find P

y1  x1  

x1 1  x1   P1sat P

Plot the phase diagram. Define liquid equilibrium line: PL(x1):= if (P(x1) < P*, P(x1), P*) Define vapor equilibrium line: PV(x1):= if (P(x1) < P*, P(x1), P*) Define pressures for liquid phases above P*: Pliq := P*..P*+10 X1:= 0, 0.01 ..1

x1:= 0, 0.05 ..0.2 x1= 0 0.05 0.1 0.15 0.2 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

x1:= 1, 0.95 ..0.2 PL(x1) = 110 133.66 147.658 155.523 159.598 y1 (x1) = 0 0.214 0.314 0.368 0.397 x1= 1 0.95 0.9 0.85 0.8 PL(x1)= 75 113.556 137.096 150.907 158.506 Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

y1 (x1) = 1 0.631 0.504 0.444 0.414 x1  0 224

x1  0 776

y1  0 405

Solution 15.14

Problem Statement

The system water(l)/n-pentane(2)/n-heptane(3) exists as a vapor at 101.33 kPa and 100°C with mole fractions z 1 = 0.45, z 2 = 0.30, z 3 = 0.25. The system is slowly cooled at constant pressure until it is completely condensed into a water phase and a hydrocarbon phase. Assuming that the two liquid phases are immiscible, that the vapor phase is an ideal gas, and that the hydrocarbons obey Raoult’s law, determine: (a) The dewpoint temperature of the mixture and composition of the first condensate. (b) The temperature at which the second liquid phase appears and its initial composition. (c) The bubblepoint temperature and the composition of the last bubble of vapor. See Table B.2 for vapor-pressure equations.

Solution

Temperatures in °C; pressures in kPa  388570   P1sat T   exp 163872  T  230170  

Water

 245188   P2sat T   exp 137667  T  232014  

n-Pentane

 291026   P3sat T   exp 138622  T  216432  

n-Heptane

P  10133 z1  045 z2  030 z3  1  z1  z2 Guess: T1dew  100 x 2  z2 Given



x3  1  x 2





P  x2 P2sat T1dew  x3 P3sat T1dew



 

1 z3 P  x3 P3sat Tdew

x 2  x 3  1

 x    2       1   x3   Find x2 , x3 , Tdew T 1   dew 



 1 Tdew  66 602 z1  0 45 x3  0 706 x 2  0 294

Calculate dew point temperature assuming the water layer forms first: 2 x1  0 706 Guess : Tdew  100

Given

  T

2 z1 P  x1 P1sat Tdew

  T

2 2 dew  Find Tdew

2 dew 



2 1 Since Tdew  Tdew the water layer forms first. 3 Tdew 

(b) Guess:

x 2  z2

x 3   x 2

y1  z1









y2  z2







 y z

Given P  P1sat T3dew  x2 P2sat T3dew  x3 P3sat T3dew y1 P  P1sat T3dew



y2 P  x2 P2sat T3dew

y3  z3

y1  y2  y3  1

 x  x 1  2

 3

 y   1   y   2   y   3  dew   T dew   Find y1 , y2 , y3 ,T3 , x2 , x2  3   x    2      x3 





y1  0288 y2  0388 y3  0324 T3dew  68 437 3 (c) Tbubble  Tdew x2 

Given



z2 z2  z3



 0545 x 3 



z3 z2  z3



x 2  0 1446 x 3  0 8554

 0455



P  P1sat T bubble  x2 P2sat T bubble  x3 P3sat T bubble



Tbubble  Find Tbubble  Tbubble 

y1 

y2 

y3 



P1sat T bubble P

  0111



x 2 P2sat T bubble P



x 3 P3sat T bubble P



  081

  0078

Solution 15.15

Problem Statement

Work the Problem 15.14 for mole fractions z1 = 0.32, z2 = 0.45, z3 = 0.23.

Problem 15.14

water phase and a hydrocarbon phase. Assuming that the two liquid phases are immiscible, that the vapor phase is an ideal gas, and that the hydrocarbons obey Raoult’s law, determine: (a) The dewpoint temperature of the mixture and composition of the first condensate. (b) The temperature at which the second liquid phase appears and its initial composition. (c) The bubblepoint temperature and the composition of the last bubble of vapor. See Table B.2 for vapor-pressure equations.

Solution

Temperatures in °C; pressures in kPa  388570   P1sat T   exp 163872  T  230170  

Water

 245188   P2sat T   exp 137667   T  232014 

n-Pentane

 291026   P3sat T   exp 138622   T  216432 

n-Heptane

P  10133 z1  032 z2  045 z3  1  z1  z2 a)

Guess: T1dew  70 x 2  z2

Given





x3  1  x 2



P  x2 P2sat T1dew  x3 P3sat T1dew



 

1 z3 P  x3 P3sat Tdew

x 2  x 3  1

 x    2       1  x3   Find x2 , x3 , Tdew  1  Tdew 





1 Tdew  65 122 x3  0686 x 2  0 314

Calculate dew point temperature assuming the water layer forms first: 2 x1  1 Guess : Tdew  170

Given

  T

2 z1 P  x1 P1sat Tdew

  T

2 2 dew  Find Tdew

2 dew  70854

1 2 Since Tdew  Tdew a hydrocarbon layer forms first.

x 2  z2

3 Tdew  100

(b) Guess:

x 3  1  x 2

y1  z1











Given P  P1sat T3dew  x2 P2sat T3dew  x3 P3sat T3dew



y1 P  P1sat T3dew

y2  z2

y3  z3



y2 y3





y2 P  x2 P2sat T3dew

y1  y2  y3  1

 x  x 1  2

 3

 y   1   y   2   y   3  dew   T dew   Find y1 , y2 , y3 , T3 , x2 , x2  3   x    2    x3 





y1  024 y2  0503 y3  0257 T3dew  64 298 x 2  0 2099 x 3  0 7901 3 (c) Tbubble  Tdew

Given

x2 



z2 z2  z3

 0662 x3 







z3 z2  z3

 0338



P  P1sat T bubble  x2 P2sat T bubble  x3 P3sat T bubble



Tbubble  Find Tbubble  Tbubble  49939

y1 

y2 

y3 



P1sat T bubble P

  009



x 2 P2sat T bubble P



x 3 P3sat T bubble P

  0861   0049

Solution 15.16

Problem Statement

The Case I behavior for SLE (Sec. 15.4) has an analog for VLE. Develop the analogy.

Solution

The analogy is Raoult’s law, Eq. (13.16), applied at constant P: yi P  x i Pi sat If the vapor phase in VLE is ideal and the liquid molar volumes are negligible (assumptions inherent in

Raoult’s law), then the Clausius/Clapeyron equation applies: d lnPisat dT



Hilv RT 2

Integration from the boiling temperature



Pi sat  P



Pisat  Pisat ) gives:

lv T H P sat i ln i   Tb RT 2 P i

Solution continued on next page…

Combination with Eq. (10.1) yields:

yi  xi exp 

T H

lv i dT 2

which is an analog of the Case I SLE equations.

Solution 15.17

Problem Statement

An assertion with respect to Case II behavior for SLE (Sec. 15.4) was that the condition zi  si  1 corresponds to complete immiscibility for all species in the solid state. Prove this.

Solution

Consider binary (two-species) equilibrium between two phases of the same kind. Equation (15.16 a) applies: x i  i  x i  i i  1, 2 

If phase 



Hence,

x1  1  x 1  1  1 and x 2  2  x 2  2  1

The reasoning applies generally to (degenerate) N 

x 1  1 

x 2   2 

N mutually immiscible species.

Whence the cited result for solids.

Solution 15.18

Problem Statement

Use results of Sec. 15.4 to develop the following (approximate) rules of thumb: (a) The solubility of a solid in a liquid solvent increases with increasing T. (b) The solubility of a solid in a liquid solvent is independent of the identity of the solvent species. (c) Of two solids with roughly the same heat of fusion, that solid with the lower melting point is the more soluble in a given liquid solvent at a given T. (d) Of two solids with similar melting points, that solid with the smaller heat of fusion is the more soluble in a given liquid solvent at a given T.

Solution

The rules of thumb are based on Case II binary SLE behavior. For concreteness, let the solid be pure species 1 and the solvent be liquid species 2. Then Eqs. (15.18 )) and (15.19.a)) apply:

x1  1  exp

H1sl  T  Tm1    RTm  T  1

(a) Differentiate:

Thus x1 dT

dx1 dT

 1 

H1sl RT 2

(b) Equation (15.19a) contains no information about species 2. Thus, to the extent that Eqs. (15.18) and (15.19.a) are valid, the solid solubility x 1 is independent of the identity of species 2. Solution continued on next page…













are related by:





  sl  H TmB  TmA  xA   exp    xB RTm Tm   A B  

where by assumption, Accordingly, x A x B 

TA  TB , thus validating the rule of thumb.

(d) Identify the solid species as in Part (c). Then x A

x B are related by:





  sl sl  H B H A Tm  T   xA   exp    xB RTmT    

where by assumption, Notice that Tm  T

Tm A  Tm  Tm B

x A xB 

H Asl  H Bsl , in accord with the rule of thumb.

Solution 15.19

Problem Statement

Estimate the solubility of naphthalene(l) in carbon dioxide(2) at a temperature of 80°C at pressures up to 300 bar. Use the procedure described in Sec. 15.4, with l12 = 0.088. Compare the results with those shown in Fig. 15.6. Discuss any differences. P1sat = 0.0102 bar at 80°C.

Solution 0302 7484  TC       3042 K 0224  

P

bar

4051 bar PC   7383

bar

T  35315 K Tr 

bar T TC

Use SRK EOS Solution continued on next page…

From Table 3.1,   1   0   008664





 



 2   1  0480  1574    0176   2  1  Tr05    a

6842 kg m 5  a   0325 s 2 mol 2

    R 2  Tc2

b

Ω  R  TC

1331104  3   m b   2968105  mol  

2  P  

b2  P R T

q2 

a2 b2  R  T

z2  1 (guess)

z2  1  2 P  q2  2 P  

z2  2  P 

z2    2 P   z2    2 P 

Eq 3.48

z2  P   Find z 2

 z  P     P   2 2  I2 P   ln  z2  P    For simplicity, let

1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution.

I12  0088

     a 05 b   b   1  P   exp   1 *  Z2  P   1  ln z2  P   2  P  q2 2  1  I12    1   1   I2  P    b2      a b  2 2       P1sat  

Eqs. 15.28, 15.29 with

bar V1 



cm3 mol

sat = 1 and P  P  P1sat, combine to give: 1

Solution continued on next page…

y1  P  

 P  V  1 exp   R  T P  1 P    P1sat

Solution 15.20

Problem Statement

Estimate the solubility of naphthalene(l) in nitrogen(2) at a temperature of 35°C at pressures up to 300 bar. Use the procedure described in Sec. 15.4, with l12 = 0. Compare the results with those shown in Fig. 15.6 for the naphthalene/CO2 system at 35°C with l12 = 0. Discuss any differences.

Solution 0302 7484  TC    K      0038 1262 

4051 bar PC   3400

P

bar

T



K Tr 

T TC

Use SRK EOS From Table 3.1,   1   0   008664   0 42748







 2   1  0480  1574    0176   2  1  Tr05    a

7298 kg m 5  a   0067 s 2mol2

    R 2  Tc2

b

Ω  R  TC

1331104  3   m b   2674 105  mol  

2  P  

b2  P R T

q2 

b2  R  T

z2  1 (guess)

z2  1  2 P  q2  2 P  

z2  2  P 

z2    2 P   z2    2 P 

Eq 3.48

z2  P   Find z 2

 z  P     P   2 2  I2 P   ln    z P     2 For simplicity, let

Eq 13.72

1 represent the infinite-dilution value of the fugacity coefficient of species 1 in solution.

I12  00

     a 05 b    b1    1 1 1  P   exp    Z2  P  1  ln  z2  P  2  P  q2 2  1  I12        I2  P  Eq 13.99   b2    a2  b2        P1sat   

Eqs. 15.28, 15.29 with

y1  P  

4

bar

V1 

cm3 mol

sat = 1 and P  P  P1sat, combine to give: 1

 P  V  1 exp   R  T  P  1  P  P1sat

Note: y axis is log scale

Solution 15.21

Problem Statement

The qualitative features of SVE at high pressures shown in Fig. 15.6 are determined by the equation of state for the gas. To what extent can these features be represented by the two-term virial equation in pressure, Eq. (3.36)? Eq. 3.36: Z

PV BP 1 RT RT

Solution

The shape of the solubility curve is characterized in part by the behavior of the derivative

dyi  dP

T  A general

expression is found from the equation (15.28), y1  P1sat P / F1 , where the enhancement factor F1 depends (at constant T) on P and y1. Thus,   F  dy  dy1 P1sat P1sat  F1   1 1   F1       dP P  P  y  y1 P dP  P2   1    ln F  dy  y   ln F1  1 1      1  y1    P   P  y  y1 P dP    1

Solution continued on next page…

  1   ln F1     y1  P   P  y dy1  1      ln F  dP 1 1  y1    y1 

Whence,

This is a general result. An expression for F1 is given by Eq. (15.29):



V1s P  P1sat 1sat F1  exp RT ˆ1



From this, after some reduction:   ln ˆ    ln ˆ    ln F    ln F  V1s   1 1 1 1     and         P     P RT  y  y    y   y  1 P 1  P 1

    ln ˆ1  V1s 1    y1        P  RT P   y1  dy1    ln ˆ  dP  1 1  y1    y1  P

Whence,

This too is a general result. If the two-term virial equation in pressure applies, then ln ̂1 is given by Eq. (10.63a), from which:   ln ˆ    ln ˆ  y2 12 P 1   2 1 1  B  y  and .      11 2 12  RT RT   P   y1 



Whence,



 V s  B  y2 1   11 2 12 y1  1   RT P   dy1  2y y  P dP 1  1 2 12 RT

The denominator of this equation is positive at any pressure. Hence, the sign of dy1/dP is determined by the sign of the group in parentheses.For very low pressures the 1/P term dominates and dy1/dP is negative.For very high pressures,1/P is small,and dy1/dP can be positive.If this is the case,then dy1/dP is zero for some intermediate pressure,and the solubility y1 exhibits a minimum with respect to pressure. However, the two-term virial equation is only valid for low to moderate pressures, and is unable to mimic the change in curvature and “flattening” of the y1

P curve observed for high pressures for the naphthalene/CO2 system.

Solution 15.22

Problem Statement

The UNILAN equation for pure-species adsorption is: n

m c  P es ln( ) 2s c  P e s 1

z  1  bn

where m, s, and c are positive empirical constants. (a) Show that the UNILAN equation reduces to the Langmuir isotherm for s = 0. (Hint: Apply l’Hôpital’s rule.) (b) Show that Henry’s constant k for the UNILAN equation is: k (UNILAN) 

m sinh s cs

Solution

(a) Rewrite the UNILAN equation: n









m s s  ln c  Pe  ln c  Pe    2s

As s  0, this expression becomes indeterminate. Application of l’Hôpital’s rule gives: m  Pe s Pes     s s 0 2  c  Pes   c  Pe

lim n  lim s 0



m  P P      2  c  P c  P 

lim n  s 0

mP cP

which is the Langmuir isotherm. (b) Henry’s constant, by definition:

dn P 0 dP

k  lim

dn m  e s es      dP 2 s  c  Pe s c  Pes 

Differentiate Eq. (A):

Whence,

k

m  e s es  m  e s  es  m      or k  sinh s 2 s  c c  cs  2  cs

P are well-behaved in the zero-pressure limit: dn m  sinh s P 0 dP cs lim

d2 n

m   2 sinh2s P 0 dP c s lim

lim

d3 n

P 0 dP



2m c3 s

sinh3s

Etc. Numerical studies show that the UNILAN equation, although providing excellent overall correlation of adsorption data at low-to-moderate surface coverage, tends to underestimate Henry’s constant.

Solution 15.23

Problem Statement

In Ex. 15.8, Henry’s constant for adsorption k, identified as the intercept on a plot of n/P vs. n, was found from a polynomial curve-fit of n/P vs. n. An alternative procedure is based on a plot of ln(P/n) vs. n. Suppose that the 2

adsorbate equation of state is a power series in n: z = 1 + Bn + C n + . . . . Show how from a plot (or a polynomial curve-fit) of ln(P/n) vs. n one can extract values of k and B. [Hint: Start with Eq. (15.39).] Eq. 15.39: n  dn   n  kP exp   (1  z )  1 z n 0   

Solution

Start with Eq. (15.39), written as: P n dn ln    ln k    z  1  z  1  n  0 n

With z  1  Bn  Cn 2  , this becomes: P  3 ln     ln k  2 Bn  Cn2   2 n 

Thus a plot of ln(P/n) vs.n produces -ln k as the intercept and 2B as the limiting slope (for n  0). Alternatively, a polynomial curve fit of ln(P/n) in n yields -ln k and 2B as the first two coefficients.

Solution 15.24

Problem Statement

It was assumed in the development of Eq. (15.39) that the gas phase is ideal, with Z = 1. Suppose for a real gas phase that Z = Z ( T, P ). Determine the analogous expression to Eq. (15.39) appropriate for a real (nonideal) gas phase. [Hint: Start with Eq. (15.35).] Eq. 15.35

a d   xi di  0 i

Eq. 15.39: n  dn   n  kP exp   (1  z )  1 z n 0   

Solution

For species i in a real-gas mixture, Eqs. (10.46) and (10.52) give: ig 

T   RT

y i ˆi P

At constant temperature, dig  RTd ln yiˆi P With i  dig, Eq. (14.105) then becomes: 

a d  dlnP  xid lnyiˆi  RT i



s T

For pure-gas adsorption, this simplifies to: a d  d lnP  dln RT



T

On the left side of Eq. (A), introduce the adsorbate compressibility factor z

  z

a A  : RT nRT

a dn d  dz  z RT n

 

where n is moles adsorbed. On the right side of Eq. (A), make the substitution: d

  Z  

dP P

C 

which follows from Eq. (10.35). Combination of Eqs. (A), (B), and (C) gives on rearrangement d ln

n dn dP  1  z   dz   Z  1 P n P

which yields on integration and rearrangement: P

n  kP·exp   Z  1 0

n  dP dn   ·exp   1  z   1  z  P n 0   

Solution 15.25

Problem Statement

n and z vs. n for ethylene adsorbed on a carbon molecular sieve. Discuss the plots.

Solution

400 300 200 100 0 0

n 2.5

2 1.5

1 0

Both charts show a similar trend of reaching an asymptote at about 4. The first chart is on a much larger scale than the second.

Solution 15.26

Problem Statement

Suppose that the adsorbate equation of state is given by z = (1

bn )

, where b is a constant. Find the implied

adsorption isotherm, and show under what conditions it reduces to the Langmuir isotherm.

Solution

1

With z  1  bm , one obtains the isotherm:  bn   n  kP 1  bnexp  1  bn 

For bn

   bn    bn  1  bn  1  bn

Whence, by Eq. (A),

n  kP   bn n 

kP 1  2bkP

which is the Langmuir isotherm. With z    n

n  kP

 n

from which, for n sufficiently small, the Langmuir isotherm is again recovered.

Solution 15.27

Problem Statement

Suppose that the adsorbate equation of state is given by z = 1 +

, where is a function of T only. Find the implied

adsorption isotherm, and show under what conditions it reduces to the Langmuir isotherm.

Solution

1

With z  1  bm , one obtains the isotherm:  bn   n  kP 1  bnexp  1  bn 

For bn

   bn    bn  1  bn  1  bn

Whence, by Eq. (A),

n  kP   bn  n 

kP 1  2bkP

which is the Langmuir isotherm. With z    n

n  kP

 n 

from which, for n sufficiently small, the Langmuir isotherm is again recovered.

Solution 15.28

Problem Statement

Derive the result given in the third step of the procedure for predicting adsorption equilibria by ideal-adsorbedsolution theory at the end of Sec. 15.5.

Solution

a

A Ad dP , n n RT P

The definition of  and its derivative are: 

A Ad and d  RT RT d  n

Whence,

dP P

 

By Eq. (13.16), the Raoult’s law analogy, x i  yi P Pi 

P yields:

 

xi  P  P i i

By general differentiation, d x i  P d  i

yi Pi 

 i

yi Pi 

C 

The equation,  i x i  1 , is an approximation that becomes increasingly accurate as the solution procedure converges. Thus, by rearrangement of Eq. (B),

 P i 

 ixi  1

With P fixed, Eq. (C) can now be written in the simple but approximate form: d xi  i

dP P

Equation (A) then becomes: d  n d x i

    n   x i   i 

where we have replaced differentials by deviations. The deviation in  i x i is known, since the true value must be unity. Therefore,  x i  P  i

yi Pi 

1

and 1

n

 i( n i ) i

Combine the three preceding equations:

 

P

yi 1 i P i  x 

 i n i  i

When x i  yi P Pi  , the Raoult’s law analogy, is substituted the required equation is reproduced:

 

P

yi 1 iP i

P

yi i P n  i i

Solution 15.29

Problem Statement

Consider a ternary system comprising solute species 1 and a mixed solvent (species 2 and 3). Assume that: GE _  A12 x 1 x 2  A13 x1 x3  A23 x 2 x 3 RT

Show that Henry’s constant 1 for species 1 in the mixed solvent is related to Henry’s constants 1, 2 and 1, 3 for species 1 in the pure solvents by:

ln 1  x2 ln1,2  x3 ln1,3  A23 x2 x3 Here x 2 and x 3 are solute-free mole fractions: x2 

x2 x 2  x3

x3 

x3 x 2  x3

Solution

Multiply the given equation for G E RT

n and convert all mole fractions to mole numbers:

nn nn nn nG E  A12 1 2  A13 1 3  A23 2 3 RT n n n

for i :

1 n  1 n  nn ln 1  A12 n2   12   A13n3   12   A23 2 2 3  n n   n n  n  A12 x 2 1  x 1   A13 x 3 1  x1   A23 x 2 x 3

Introduce solute-free mole fractions: x2 

Whence,

x2 x2  x3



x2 1  x1

and x3 

x3 1  x1

ln 1  A12 x2 1  x1   A13 x3 1  x1   A23 x2 x3 1  x1  2

For x  ln1  A12 x 2  A13 x3  A23 x 2 x3 Apply this equation to the special case of species 1 infinitely dilute in pure solvent 2. In this case, x 2 

x3 

   also ln1,3  ln1,2  A12  A13

Whence,

  ln 1  x 2 ln 1,2  x 3ln 1,3  A23 x 2 x3

In logarithmic form the equation may be applied to the several infinite-dilution cases:   lnH 1  lnf1  ln 1 lnH 1,2  lnf1  ln 1,2 lnH 1,3  lnf1  ln 1,3

Whence, Or

lnH1  lnf1  x2 (lnH1,2  lnf1 )  x2 (lnH1,3  lnf1 )  A23 x2 x3 ln H1  x2 ln H1,2  x3 ln H1,3  A23 x 2 x3

and

Solution 15.30

Problem Statement

It is possible in principle for a binary liquid system to show more than one region of LLE for a particular temperature. For example, the solubility diagram might have two side-by-side “islands” of partial miscibility separated by a G vs. x1 diagram at constant T look like for this case? Suggestion: See Fig. 12.13 for a mixture showing normal LLE behavior.

Solution

For the situation described, the two regions like the one shown from 

 probably one on either side of the

minimum in curve II.

Solution 15.31

Problem Statement

With V2  V2 , Eq. (15.66) for the osmotic pressure may be represented as a power series in x1: V2 x1 RT

 1  B x1  C x12  . . .

Reminiscent of Eqs. (3.33) and (3.34), this series is called an osmotic virial expansion. Show that the second osmotic virial coefficient B is:   2  1   d ln2   B  1    2     2   d x1 x1 0  E

What is B for an ideal solution? What is B if G = A x1x2? Eq. 3.33: Z

+ BP

+ C P

+ D P

...

Eq. 3.34: Z 1

B C D    2 V V V3

Eq. 15.66: 

RT ln( x2 2 ) V2

Solution

By Eq. (15.65) with V2  V2 : Represent

V2 RT

 ln x2  2 

 2 by a Taylor series: ln  ln

x1 



dln 2 dx1

x1 

x 

1 d ln2 x 2  2 dx12 x1 

But at x1 

 x2  

2 and its first derivative are zero. Therefore,

 2  1  d ln 2   ln2   x12  2  dx12  x 0 1

x12

x13

x14

Also,

ln x2  ln 1  x1    x1 

Therefore,

   2   2 1  1  d ln  2   x   ln x 2  2   ln x 2  ln  2  x1  1    1 2  2 2  dx1   x1 0  

And

  2  1  1  d ln 2   x   1  1   2   1  x1 RT 2 2  dx1  x1  0   





V2

Comparison with the given equation shows that:

  2  1  1  d ln  2   B  1     2  2  dx12  x1  0   

Solution 15.32

Problem Statement

A liquid-process feed stream F contains 99 mol-% of species 1 and 1 mol-% of impurity, species 2. The impurity level is to be reduced to 0.1 mol-% by contacting the feed stream with a stream S of pure liquid solvent, species 3, in a mixer/settler. Species 1 and 3 are essentially immiscible. Owing to “good chemistry,” it is expected that species 2 will selectively concentrate in the solvent phase. (a) With the equations given below, determine the required solvent-to-feed ratio nS/nF. (b) What is mole fraction x2 of impurity in the solvent phase leaving the mixer/settler? (c) What is “good” about the chemistry here? With respect to liquid-phase nonidealities, what would be “bad” chemistry for the proposed operation? dE Given: G12  RT  1 5 x1 x 2

E G23  RT   08 x 2 x 3

Solution

Define the values given in the problem statement. Assume as a basis a feed rate, nF= 1mol/s nF  1

mol s

x F 1  099 x F 2  001 x S 3  1

x2  0001 x1  1  x2 Apply mole balances around the process as well as an equilibrium relationship

A12  15 A23 08 2 2   2  x2   exp  A12  1  x2   2  x2   exp  A23  1  x2      

Material Balances nS  nF  nE  nR

(Total)

nS  x 3  nE

(Species 3)

x F 1  nF  x 1  nR

(Species 1)

Substituting the species balances into total balance yields nS  nF 

xF 1 nE  1 nF x 3 x1

Solving for the ratio of solvent to feed (nS /nF ) gives: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

 x  xF   x   3 1    1    nF  1  x 3   x 1  nS

We need 3 . Assume exiting streams are at equilibrium. Here, the only distributing species is 2. Then x 2  2  x 2  2

Substituting for 2

2 2 2   x2  exp  A12  1  x2    x 2  exp  A23  1  x2      

Solve for using Mathcad Solve Block Guess: x2  05 Given 2 2   x2  exp  A12  1  x2    x 2  exp  A23  1  x2      

x 2  Find  x 2  x 2  

From above, the equation for the ration

nS nF

x 3   x 2

x 3  

is:

 x  xF   x   3 1    1     09112 nF  1  x 3   x 1  nS

(b) x2  000979

E G23





E G12

Solution 15.33

Problem Statement

At 25°C the solubility of n-hexane in water is 2 ppm (molar basis), and the solubility of water in n-hexane is 520 ppm. Estimate the activity coefficients for the two species in the two phases.

Solution

N-hexane

2- Water Since this is a dilute system in both phases, Eqns (C) and (D) from Example 15.2 can be used to find 1 x 1 

520 10

, x 2  1  x1

1  2 

x 1 x1

x 2 

2 106

2

x 1  1  x 2

 1923103

1  x1 1  x 1

 4997 105

Solution 15.34

Problem Statement

A binary liquid mixture is only partially miscible at 298 K. If the mixture is to be made homogeneous by increasing E

the temperature, what must be the sign of H ?

Solution

Equation (10.58) applies:

  E     G          RT      T     



HE RT 2

P, x

For the partially miscible system GE/RT is necessarily “large,” and if it is to decrease with increasing T, the derivative must be negative. This requires that HE be positive.

Solution 15.35

Problem Statement

The spinodal curve for a binary liquid system is the locus of states for which

d2 G  RT  d x12

 0 const T , P 

Thus it separates regions of stability from instability with respect to liquid/liquid phase splitting. For a given T, there are normally two spinodal compositions (if any). They are the same at a consolute temperature. On curve II of Fig. 12.13 they are a pair of compositions between x1 and x1 , corresponding to zero curvature. Suppose a liquid mixture is described by the symmetrical equation GE  AT  x1 x 2 RT

(a) Find an expression for the spinodal compositions as a function of A(T). (b) Assume that A(T) is the expression used to generate Fig. 15.2. Plot on a single graph the solubility curve and the spinodal curve. Discuss.

Solution

   d   RT  2  G 

(a) Equivalent to

For

dx12



GE  Ax1 x 2  A x1  x 12 RT



 G E   d2   RT  dx12



1 x1 x2



 G E   d   RT   A1  2 x1  dx1

 G E   d2   RT  dx12 Thus,

 2 A

 A 

1 x1 x 2

Ax1 x2  .

Substituting for x2 x1  x12 

1 2A

x12  x1 

1  2A 1  1

The solution to this equation yields two roots:

x1 

2 A

2 1  1 x1 

And

The two roots are symmetrical around x1 

2 A

1 2

Note that for:

A  2 : No real roots A

x1 

1 (consolute point) 3

A

x1 

x1 

(b) Plot the spinodal curve along with the solubility curve

From Fig. 15.2(a):

AT   

T  540 K  211  3ln   K  T

Both curves are symmetrical around x1 

1 Create functions to represent the left and right halves of the curves. 2

Solution continued on next page…

From above, the equations for the spinodal curves are: 1 1 AT  1 1 AT  xspr1 T  :    xspl1 T  :   2 2 2 2 AT  AT 

xr :  07 xl :  03 From Eq. (E) in Example 15.3, the solubility curves are solved using a Solve Block:

Given

 1  xr   xr  0.5 xr1 T  : Find  xr  AT  1  2 xr   ln   xr 

Given

 1  xl   xl  05 xl1 T  : Find xl  AT  1  2 xl   ln   xl 

Find the temperature of the upper consolute point. T :  300 KGivenAT   2Tu : FindT Tu  345998 K

T :  250 K ..346K

360 340 320 300 280 260 240

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

xl1 xr1 xspl1

Solution 15.36

Problem Statement

Two special models of liquid-solution behavior are the regular solution, for which S = 0 everywhere, and the E

athermal solution, for which H = 0 everywhere.

(a)

Ignoring the P-dependence of G , show that for a regular solution,

FR  x  GE  RT RT

(b)

Ignoring the P-dependence of G , show that for an athermal solution,

GE  FA  x  RT

(c)

Suppose that G /RT is described by the symmetrical equation

GE  AT  x1 x 2 RT

From parts (a) and (b), we conclude that GE   x x RT RT 1 2 GE   x1 x 2 RT

where

(regular)

(A)

(athermal)

(B)

and are constants. What are the implications of Eqs. (A) and (B) with respect to the shapes of predicted

solubility diagrams for LLE? Find from Eq. (A) an expression for the consolute temperature, and show that it must be an upper consolute temperature. Suggestion: See Ex. 15.3 for numerical guidance.

Solution

 G E    (a) find:   T 

 S E 

P,x

FR  x  GE  RT rt

Therefore

  E     G         RT   HE     2 RT  0  T      

(b) By Eq. (10.58),



GE  FA  x  RT





A A With respect to Eq. (A), increasing T

A

A

LLE

G E RT smaller. thus, the consolute point is an upper consolute point. Its

value follows from:    2  TU  RTU 2R

The shape of the solubility curve is as shown on Fig. 15.2(a).

Solution 15.37

Problem Statement

Many fluids could be used as solvent species for supercritical separation processes (Sec. 15.4). But the two most popular choices seem to be carbon dioxide and water. Why? Discuss the pros and cons of using CO2 vs. H2O as a supercritical solvent.

Solution

Why? Because they are both nontoxic, relatively inexpensive, and readily available. For CO2, its Tc is near room temperature,making it a suitable solvent for temperature-sensitive materials.It is considerably more expensive than water,which is probably the cheapest possible solvent. Tc

Pc for water are high, which increases heating and pumping costs.

Solution 16.1

Problem Statement

A plant takes in water at 21°C, cools it to 0°C, and freezes it at this temperature, producing 0.5 kg·s

of ice. Heat

rejection is at 21°C. The heat of fusion of water is 333.5 kJ·kg . (a) What is W ideal for the process? (b) What is the power requirement of a single Carnot heat pump operating between 0 and 21°C? What is the thermodynamic efficiency of this process? What is its irreversible feature? (c) What is the power requirement if an ideal tetrafluoroethane vapor-compression refrigeration cycle is used? Ideal here implies isentropic compression, infinite cooling-water rate in the condenser, and minimum heat-transfer driving forces in evaporator and condenser of 0°C. What is the thermodynamic efficiency of this process? What are its irreversible features? (d ) What is the power requirement of a tetrafluoroethane vapor-compression cycle for which the compressor efficiency is 75%, the minimum temperature differences in evaporator and condenser are 5°C, and the temperature rise of the cooling water in the condenser is 10°C? Make a thermodynamic analysis of this process.

Solution

(a) Initial state: Liquid water at 21°C H1 

kJ kg 1 S1 

kJ kg . K

(Table F.3)

Final state: Ice at 0°C H2  

 S2   

T  



kJ kg 1  

333.5  kJ   273.15  kg. K

kJ kg 1

kJ kg.K

 K = 294.15 K



Point A: saturated vapor at 0°C Point B: Superheated vapor at 5.92 bar and the entropy of Point A. Point C: saturated liquid at 21°C, P = 5.92 bar Point D: Mix of saturated liquid and saturated vapor at 21°C with the enthalpy of Point C. Data for Points A, C, & D from Table 9.1. Data for Point B from Fig. F.2 Wideal  H 2  H1  T  S2  S2   

kJ kg

1



W ideal  m  Wideal 

kJ kg

kg s1 

1



K 



kJ kg 1 K 1 

m 

kg s  1

kJ kg 1 



kJ kg 1 K 1 

kJ kg 1

Solution continued on next page…

(b) For the Carnot heat pump, heat equal to the enthalpy change of the water is extracted from a cold reservoir at 0°C, with heat rejection to the surroundings at 21°C. TC 

TH  T

kJ kg1 

QC  H2  H1  

kJ kg 1  

 T  T  C Work  QC   H   TC  kg s1

m 

t 

W

ideal

W

W  m  Work 





kJ kg 1

t 

The only irreversibility is the transfer of heat from the water as it cools from 21°C to 0°C to the cold reservoir of the Carnot heat pump at 21°C. (c) Conventional refrigeration cycle under ideal conditions of operation: Isentropic compression, infinite flow rate of cooling water, and minimum temperature difference for heat transfer = 0 For saturated liquid and vapor 0°C, by interpolation in the table: kJ kg 1 S A 

HA 

kJ kg . K

For saturated liquid at 21°C: HC 

kJ kg 1 H D  HC

For superheated vapor at 5.92 bar and S = 0.9307:

HB 

kJ kg 1

Refrigerant circulation rate:

m 

 H2  H1  

kg s1

H A  HD

W  m   H B  H A  

t 





kJ kg

kg s1 



kJ kg 1 





1

kJ kg1 

kJ kg 1  

kJ kg

kg s1

1



kJ kg 1 

kg s1



Wideal  0.780 W

The irreversibilities are in the throttling process and in heat transfer in both the condenser and evaporator, where there are finite temperature differences. (d) Practical cycle

= 0.75

Point A: saturated vapor at 5°C Point B: Superheated vapor at 9.29 bar. Point C: saturated liquid at 36°C. Point D: Mix of saturated liquid and saturated vapor at 5°C with the enthalpy of Point C. For saturated liquid and vapor at 5°C: Hliq 

Sliq 

kJ kg 1 H vap  kJ S  kg K vap

kJ kg kJ kg K

H A  H vap

S A  Svap

For saturated liquid at 36°C, P = 9.29 bar: HC 

kJ kg 1 SC 

kJ kg . K

For isentropic compression, the entropy of point B is 0.93 at P = 9.29 bar. From Fig.F.2, H B 

kJ kg 1 H B  H A 

H B  H A 



kJ kg 1

Solution continued on next page…

SB 

kJ kg 1

H D  HC

xD 



H D  Hliq H vap  Hliq



kJ kg . K



kg s1

SD  Sliq  x D  Svap  Sliq  0.39



Refrigerant circulation rate: m 

W  m   H B  H A  

 H2  H1  

kg s1

H A  HD

k t 

W ideal W

t 

T 

Thermodynamic Analysis

W lost .comp  T  m  SB  S A  Q condenser  m   HC  H B  W lost.condenser  T  m  SC  SB   Q condenser W lost.throttle  T  m   SD  SC    W lost .evap  T  m 1  S A  SD    W ideal 

 kg  S2  S1  s 

27.82%

W lost.comp 

kW 13.87%

W lost .cond 

kW 36.09%

W lost . throttle 

kW 12.14%

W lost .evap 

10.09%

The percent values above express each quantity as a percentage of the actual work, to which the terms sum. Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

Solution 16.2

Problem Statement

Consider a steady-flow process in which the following gas-phase reaction takes place: CO + 1/2O2  CO2. The surroundings are at 300 K. (a) What is Wideal when the reactants enter the process as pure carbon monoxide and as air containing the stoichiometric amount of oxygen, both at 25°C and 1 bar, and the products of complete combustion leave the process at the same conditions? (b) The overall process is exactly the same as in (a), but the CO is here burned in an adiabatic reactor at 1 bar. What is Wideal for the process of cooling the flue gases to 25°C? What is the irreversible feature of the overall process? What is its thermodynamic efficiency? What has increased in entropy, and by how much?

Solution

Assume ideal gases. Data from Table C.4 H298  S298 

H 298 G298 K

BASIS: 1 mol CO and 0.5 mol O2 entering with accompanying N 2  nCO  mol

nair 

mol

G298 

S298 

J K

 79       21 

mol

nN 2 

mol

nCO 2  mol

Solution continued on next page…

Isothermal process at 298.15 K:

Since the enthalpy change of mixing for ideal gases is zero, the overall enthalpy of change for the process is H  H 298 For unmixing the air, define

y1 

nair

y1 

y2   y1

By Eq. (11.4) (original 12.35) with no minus sign: Sunmixing  nair  R   y1ln  y1   y2 ln  y2  

J mol1 K 1  

mol 

ln



ln

J K

  

For mixing the products of reaction, define y1 



nCO 2

y1 

nN 2  nCO 2

y2   y1



Smixing  nN  nCO  R   y1ln y1   y2 ln  y2  

mol  mol  

S  Sunmixing  S298  Smixing  



ln





J K 1 

J   K T 

ln 



 

J K

J K 1  

Solution continued on next page…

J K

Wideal  H  T S Wideal  

(b) Adiabatic combustion:

Heat-capacity data for the product gases from Table C.1:

A

B

nCO  5.457  nN  3.280 2

mol nCO  1.045  nN  0.593 2

 11.627

3

mol





D

3

nCO 1.157  nN  0.040 2

mol





For the products, T

H P  R  T0

CP R

T0 

Integrating the above equation and by an energy balance, H298  H P  0

Guess

2

A

B



3

K 1

D



Solution continued on next page…

Given  2 B D    1   H298  R  A  T0       T0    2     2 T0    





T  T0   



For the cooling process from this temperature to the final temperature of 298.15 K, the entropy change is calculated by

ICPS





ICPS  

3





 05 29.701 J K

S  R  ICPS  

 H   H 298 Wideal.cooling   H  T   S

H  

t 

Wideal.cooling Wideal



Wideal.cooling  

t 

The surroundings increase in entropy in the amount:



Q   H298  Wideal.cooling

 S  T 

S 

J K

The irreversibility is in the combustion reaction.

Solution 16.3

Problem Statement

A plant has saturated steam available at 2700 kPa, but there is little use for this steam. Rather, steam at 1000 kPa is required. Also available is saturated steam at 275 kPa. A suggestion is that the 275 kPa steam be compressed to 1000 kPa using the work of expanding the 2700 kPa steam to 1000 kPa. The two streams at 1000 kPa would then be mixed. Determine the rates at which steam at each initial pressure must be supplied to provide enough steam at 1000 kPa so that upon condensation to saturated liquid, heat in the amount of 300 kJ·s

is released

(a) if the process is carried out in a completely reversible manner. (b) if the higher-pressure steam expands in a turbine of 78% efficiency and the lower-pressure steam is compressed in a machine of 75% efficiency. Make a thermodynamic analysis of this process.

Solution

For the saturated steam at 2700 kPa, Table F.2: Solution continued on next page…

H1 

kJ kg

S1 

kJ kg K

H2 

kJ kg

S2 

kJ kg K

kJ kg

Sliquid 

For the saturated steam at 275 kPa, Table F.2:

For saturated liquid and vapor at 1000 kPa, Table F.2: Hliquid 

H vapor 

kJ kg

Svapor 

kJ kg K

kJ kg K Tsat 

(a) Assume no heat losses, no shaft work and negligible changes in kinetic and potential energy. Then for a completely reversible process:

 fs  H  m  

 fs S  m  

We can also write a material balance, a quantity requirement, and relation between H3 and S3 which assumes wet steam at point 3. The five equations and five unknowns are as follows: m 3  m 1  m 2 (A)

H3m 3  H1m 1  H2m 2  kJ s1 (B) S3m 3  S1m 1  S2m 2  kJ  s1  K 1 (C)

 H3  Hliq m 3  S3  Sliq 

H3  Hliq Tsat

kJ  s1 (D)

E

Defining x1  m 1 / m 3, equations B and C can be re-written:

H3  H2  x1  H1  H2  S3  S2  x1 S1  S2 

F 

G 

Rearranging equation E yields: H3  Hliq  Tsat S3  Sliq 

H 

Substituting Equation G into H and equation to F yields: H3  H 2  x 1  H1  H2   Hliq  Tsat  S2  x1 S1  S2   Sliq   

x1 

I 

 Hliq  H2  Tsat Sliq  S2   0.574  H1  H2   Tsat S1  S2 

H 3  H 2  x 1  H1  H 2  

kJ  kg 1

S3  S2  x1 S1  S2  

kJ kg 1  K 1

m 3 



kJ s1 H3  Hliq

m 1  x1m 3 





kJ s1

2767.2  762.6kJ kg 1



kg  s1

kg s1 and m 2  m 3  m 1 

kg  s1

The steam at Point 3 is wet. (b) Turbine: Constant-S expansion of steam from Point 1 to 1000 kPa results in wet steam of quality kJ kJ  2.1382 kg K kg K   xturb   kJ kJ Svap  Sliquid 6.5828  2.1382 kg K kg K S1  Sliquid





  Hliquid  xturb   H vap  Hliq  Hturb

6.2244

kJ kg

turb  0.78   H1   Hturb  H1  turb   Hturb

kJ kg

Solution continued on next page…

xturb 

Hturb  Hliquid H vap  Hliquid







Sturb  Sliquid  xturb  Svap  Sliq 

kJ kg K

Compressor: Constant-S compression of steam from Point 2 to 10000 kPa results in superheated steam. Interpolation in Table F.2 yields kJ kg

 H comp   H   H   comp 2 H comp  H2      comp 

By interpolation:

comp 

kJ kg

Scomp  7.1083

kJ kg K

The energy balance, mass balance, and quantity requirement equations of Part (a) are still valid. In addition, the work output of the turbine equals the work input of the compressor. Thus we have 4 equations (4 unknowns): Guesses: m 1  H3 

kg s1

m 2 

kg s1

m 3 

kg s 1

kJ kg

Given

 H comp  H2   m 2   Hturb  H1   m 1  H3  m 3    H1  m 1    H2  m 2   kJ s1 m 3  m 1  m 2

H3  Hliquid   m 3 

kJ s 1

 1 m 2 m 3 H3 Find m m 1 

kg s1 m 2 

kg s1

m 3 

kg s1

H3  2.77844 103

kJ kg

Steam at Point 3 is slightly superheated By interpolation, S3  6.5876

Thermodynamic Analysis

T 

kJ kg K

(assumed)

By Eq. (5.20), with the enthalpy term equal to zero:

W ideal  T  m 3  S3  m 1  S1  m 2  S2  

W lost.turb  T  m 1 Sturb  S1 



W lost .comp  T  m 2 Scomp  S2



W lost .mixing  T  m 3  S3  m 1  Sturb   m 2  Scomp    W lost .turb 

48.2815 %

W lost.comp 

34.1565 %

W lost .mixing 

kW 17.5620%

The percent values above express each quantity as a percentage of the absolute value of the ideal work, to which the quantities sum.

Solution 16.4

Problem Statement

Make a thermodynamic analysis of the refrigeration cycle of Ex. 9.1(b).

Solution

Some property values with reference to Fig.9.1 are given in Ex.9.1. Others come from Table 9.1 or Fig.F.2. For saturated liquid and vapor at the evaporator temperature of 25°C: Hliquid 

kJ kg

H vapor 

kJ kg

Sliquid 

kJ kg K

Svapor 

kJ kg K

For saturated liquid at the condenser outlet temperature of 26°C: kJ kg

H4 

kJ kg K

S4 

H 2  H vap S2  Svap H 1  H 4

x1 

H1  Hliq H vap  H liq







S1  Sliquid  x1  Svap  Sliq 

kJ kg K

From Ex.9.1(b) for the compression step: kJ kg

H 

H3  H2  H 

kJ kg

From Fig.F.2 at H3 and P = 6.854 bar: S3 

kJ kg K

W  m  H 

m 

kg hr kJ hr

The purpose of the condenser is to transfer heat to the surroundings. Thus the heat transferred in the condenser is Q in the sense of Chapter 16; i.e., it is heat transfer to the surroundings, taken here to be at a temperature of 26°C.

Internal heat transfer (within the system) is not Q. The heat transferred in the evaporator comes from a space maintained at 21°C, which is part of the system, and is treated as an internal heat reservoir. The ideal work of the process is that of a Carnot engine operating between the temperature of the refrigerated space and the temperature of the surroundings. T  TC  

 



T  TH kJ hr

Q  

T  TC W ideal  Q C  H  TC

kJ hr

W lost .comp  T  m S3  S2  Q  m   H 4  H3     

kJ hr

W lost .cond  T m S4  S3   Q    W lost .throttle  T  m  S1  S4   

  H  H2    m  W lost .evap  T  m   S2  S1   T  1    TC   

The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir). W ideal 

kJ hr

22.82%

W lost .comp 

kJ hr

26.99%

W lost .cond 

kJ hr

39.38%

W lost .throttle 

kJ hr

7.60%

Solution continued on next page…

W lost .evap 

kJ hr

3.21%

The percent values above express each quantity as a percentage of the actual work, to which they sum: W 

kJ hr

Solution 16.5

Problem Statement

Make a thermodynamic analysis of the refrigeration cycle described in one of the parts of Prob. 9.9. Assume that the refrigeration effect maintains a heat reservoir at a temperature 5°C above the evaporation temperature and that T is 5°C below the condensation temperature.

Problem 9.9

(a) Evaporation T = 0°C; condensation T = 26°C; (compressor) = 0.79; refrigeration rate = 600 kJ·s . 1

(b) Evaporation T = 6°C; condensation T = 26°C; (compressor) = 0.78; refrigeration rate = 500 kJ·s .

(d) Evaporation T = 18°C; condensation T = 26°C; (compressor) = 0.76; refrigeration rate = 300 kJ·s . 1

(e) Evaporation T = 25°C; condensation T = 26°C; (compressor) = 0.75; refrigeration rate = 200 kJ·s .

Solution

Here

T 



T  TH

The following vectors refer to part a to e:  5  600       1  500      kJ tC   7  Q C  400     s 13 300      20 200  TC  tC 



T  TC W ideal  Q C  H TC

For saturated liquid and vapor at the evaporator temperature, Table 9.1:  200    198.66    kJ Hliq  184.07   kg 176.23   167.19

 398.6    398.015    kJ H vap   391.46    kg  387.79      383.45 

H2  H vap

 1    0.995    kJ Sliq   0.941   kg K  0.91     0.875

 1.727    1.7275    kJ Svap   1.735    kg K  1.74      1.746 

S2  Svap

For saturated liquid at the condenser temperature, Table 9.1: H4 

x1 

kJ kg

H1  H liq H vap  Hliq

S4 

kJ kg

H1  H 4





S1   Sliq  x1  Svap  Sliq   

From the results of Pb.9.9:  418.7266    421.7779    kJ H3  425.0184   kg  428.8558    433.65 

From the above H3 values, we must find corresponding entropies from Fig. F.2. They are read at the vapor pressure for 21°C. The flow rates come from Problem 9.9: 1.741   1.752    kJ S3   1.77    kg K 1.785    1.80 

3.69   3.14     kg m  2.57   s 1.98    1.36 

Solution continued on next page…

W lost .comp  T  m S3  S2  Q  m   H 4  H3    W lost .cond  T  m  S4  S3   Q    W lost .throttle  T  m  S1  S4   

  H  H2    m  W lost .evap  T  m   S2  S1   T  1    TC   

The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir) 58.176    40.419    kJ W ideal  40.082   s 39.208     32.392 

15.196   22.629    kJ W lost .comp  26.459   s 26.209    21.602

5.757    4.320    kJ W lost .cond  1.746    s 2.481   1.185 

 7.243     6.549     kJ W lost .throttle  11.113   s 11.284   10.807

12.12    0.014     kJ W lost .evap  8.376    s 7.012    4.562

74.267   74.616    kJ W  86.245   s 81.310    68.272

In each case the ideal work and the lost work terms sum to give the actual work, and each term may be expressed as a percentage of the actual work.

Solution 16.6

Problem Statement

Make a thermodynamic analysis of the refrigeration cycle described in the first paragraph of Prob. 9.12. Assume that the refrigeration effect maintains a heat reservoir at a temperature 5°C above the evaporation temperature and that T is 5°C below the condensation temperature.

Problem 9.12

at 6°C and condensing at 26°C. The heat load on the evaporator is 2000 kJ·s . If the compressor efficiency is 75%, what is the power requirement? How does this result compare with the power required by the compressor if the system operates without the heat exchanger? How do the refrigerant circulation rates compare for the two cases?

Solution

Here

T 



T  TH TC  

Q C  



T  TC W ideal  Q C  H  TC

kJ s kJ s

For saturated liquid and vapor at the evaporator temperature, Table 9.1: Hliq 

kJ kg

H vap 

kJ kg

H2  H vap

Sliq 

kJ kg K

Svap 

kJ kg K

S2  Svap

For saturated liquid at the condenser temperature, Table 9.1: Solution continued on next page… Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education.

kJ kg

H4 

kJ kg

S4 

H1  H 4

From the results of Pb.9.12: kJ kg

H2 A 

kJ kg K

S2 A 

H3  H2 A  Hcomp 





S3  1.84

kJ kg K

kJ kg

From Fig. F.2 at this enthalpy and 2.343 bar:

Energy balance on heat exchanger: kJ kg

H1  H 4  H2 A  H2 

x1 

H1  Hliq H vap  Hliq







S1   Sliq  x1  Svap  Sliq    

kJ kg K

Upstream from the throttle (Point 4A) the state is subcooled liquid with the enthalpy: H 4 A  H1

The entropy at this point is essentially that of saturated liquid with this enthalpy; by interpolation in Table 9.1: S4 A  0.82876

kJ kg K

From Prob. 9.12: m  12.57

kg s

W lost.comp  T  m S3  S2 A  Q  m   H 4  H3    W lost .cond  T  m  S4  S3   Q    W lost .throttle  T  m  S1  S4 A   

  H  H2    m  W lost .evap  T  m   S2  S1   T  1    TC   

Solution continued on next page…

The final term accounts for the entropy change of the refrigerated space (an internal heat reservoir). W lost .exchanger  T  m  S2 A  S2  S4 A  S4   

W  m   H3  H 2 A  kJ s

W ideal 

11.65%

W lost .comp 

kJ s

21.32%

W lost .cond 

kJ s

32.06%

W lost .throttle 

kJ s

2.04%

kJ s

W lost .evap  W lost .exchanger  W 

4.34% kJ s

28.58%

kJ s

The figures on the right are percentages of the actual work, to which terms sum.

Solution 16.7

Problem Statement

A colloidal solution enters a single-effect evaporator at 100°C. Water is vaporized from the solution, producing a more concentrated solution and 0.5 kg·s

of steam at 100°C. This steam is compressed and sent to the heating coils of the

evaporator to supply the heat required for its operation. For a minimum heat-transfer driving force across the evaporator coils of 10°C, for a compressor efficiency of 75%, and for adiabatic operation, what is the state of the steam leaving the heating coils of the evaporator? For a surroundings temperature of 300 K, make a thermodynamic analysis of the process.

Solution

Compression to a pressure at which condensation in coils occurs at 110°C. Table F.1 gives this saturated pressure as 143.27 kPa.  comp  0.75

H1 

kJ kg

S1 

kJ kg K

(Saturated liquid)

H2 

kJ kg

S2 

kJ kg K

(Saturated vapor)

For isentropic compression to 143.27 kPa, we find by double interpolation Table F.2:

H3 

 H   H  kJ  2 H3  H2   3   comp  kg

kJ kg

By more double interpolation in Table F.2 at 143.27 kPa: S3 

kJ kg K

By an energy balance, assuming the slurry passes through unchanged, H 4  H1  H3  H 2 H 4 

kJ kg

This enthalpy is a bit larger than that of saturated liquid at 110°C; find quality and then the entropy: Hliq 

Slv 

kJ kg

Hlv 

kJ kg K

x4 

kJ kg

H 4  Hliq H lv kJ kg K

S4  Sliq  x 4  Slv S4 

T 

kJ kg K

Sliq 

x4 

m 

kg s

W ideal  m 2   H 4  H1  T  S4  S1   

W lost.evap T  m 1  S4  S3  S2  S1  W lost.comp  T  m 2  S3  S2  W lost  m  H3  H2  W ideal 

W lost .evap 

W lost .comp  W 

21.16 %

kW kW

60.62 % 18.22%

The figures on the right are percentages of the actual work, to which the terms sum.

Solution 16.8

Problem Statement

Make a thermodynamic analysis of the process described in Prob. 8.9. T = 27°C.

Problem 8.9 A steam power plant operating on a regenerative cycle, as illustrated in Fig. 8.5, includes two feedwater heaters. Steam enters the turbine at 6500 kPa and 600°C and exhausts at 20 kPa. Steam for the feedwater heaters is extracted from the turbine at pressures such that the feedwater is heated to 190°C in two equal increments of temperature rise, with 5°C approaches to the steam-condensation temperature in each feedwater heater. If the turbine and pump efficiencies are each 0.80, what is the thermal efficiency of the cycle, and what fraction of the steam entering the turbine is extracted for each feedwater heater?

Solution

A thermodynamic analysis requires an exact definition of the overall process considered, and in this case we must therefore specify the source of the heat transferred to the boiler. Since steam leaves the boiler at 900°F, the heat source may be considered a heat reservoir at some higher temperature. We assume this temperature as 950°F. The assumption of a different temperature would provide a variation in the solution. The ideal work of the process in this case is given by a Carnot engine operating between this temperature and that of the surroundings, here specified to be 80°F. Assume a basis of 1 lbm of water passing through the boiler. Required property values come from Pb. 8.8

TH  



rankine TC  



rankine

T  TC

Subscripts below correspond to points on figure of Pb.8.7. H1 

BTU lbm

S1 

BTU lbmrankine H2 

BTU lbm

S2 

BTU lbm rankine

H3 

BTU lbm

S3 

BTU lbmrankine

H4 

BTU lbm

S4 

BTU lbmrankine

H5 

BTU lbm

S5 

BTU lbmrankine

H7 

BTU lbm

S7 

BTU lbmrankine

 T  QH  H2  H1 lbm Sideal  QH    C   TH 

For purposes of thermodynamic analysis, we consider the following 4 parts of the process: Solution continued on next page…

The boiler/heat reservoir combination The turbine The condenser and throttle valve The pump and feed water heater  Q  Wlost .boiler .reservoir  T  S2  S1   lbm  H  TH   m

(From Pb.8.8)

lbm

Wlost .turbine T  m   S3  S2    lbm  m   S4  S2   

The purpose of the condenser is to transfer heat to the surroundings. The amount of heat is Q  lbm  H5    lbm  m   H 4   m  H7  

BTU

Wlost .cond .valve  T   lbm  S5    lbm  m   S4  m  S7   Q  

Wlost. pump.heater T   lbm   S1  S5   m  S7  S3    The absolute value of the actual work comes from Pb.8.8: Wabs. value 

50.43%

BTU

Wlost .boiler .reservoir  Wlost .turbine 

BTU BTU

30.24%

13.30%

Wlost .cond. valve 

BTU

4.9%

Wlost . pump.heater 

BTU

1.13%

Wideal 

BTU

The numbers on the right are percentages of the absolute value of the ideal work, to which they sum.

Solution 16.9

Problem Statement

Make a thermodynamic analysis of the process described in Ex. 9.3. T = 295 K.

Example 9.3 Natural gas, assumed here to be pure methane, is liquefied in a Claude process. Compression is to 60 bar and precooling is to 300 K. The expander and throttle exhaust to a pressure of 1 bar. Recycle methane at this pressure leaves the exchanger system (point 15, Fig. 9.7) at 295 K. Assume no heat leaks into the system from the surroundings, an expander efficiency of 75%, and an expander exhaust of saturated vapor. For a draw-off to the expander of 25% of the methane entering the exchanger system (x = 0.25), what fraction z of the methane is liquefied, and what is the temperature of the high-pressure stream entering the throttle valve?

Solution Refer to Fig.9.7, the analysis presented here is for the liquefaction section to the right of the dashed line. Enthalpy and entropy values are those given in example 9.3 plus additional values from the reference cited at conditions given in example 9.3. Property values: H4 

kJ kg

S4 

kJ kg K

H5 

kJ kg

S5 

kJ kg K

H7 

kJ kg

S7 

kJ kg K

H9 

kJ kg

S9 

kJ kg K

H10 

kJ kg

S10 

kJ kg K

H14 

kJ kg

S14 

kJ kg K

H15 

kJ kg

S15 

kJ kg K

H6  H5 S6  S5 H11  H5 H12  H10 S12  S10 H13  H10 T 

S11  S5 S13  S10

The basis for all calculations is 1 kg of methane entering at Point 4. All work quantities are in kJ. Results given in Ex. 9.3 on this basis are: Fraction of entering methane that is liquefied: Solution continued on next page…

Fraction of entering methane passing through the expander: On this basis also Eq. 5.20 for ideal work, Eq.5.17 for entropy generation, and Eq.5.25 for lost work can be written:

z

x Wideal   H  m  T  S  m fs

SG  S  m  fs

Q T

Wlost  T  SG

Wideal   H15  1  z    H 9  z   H 4   T   S15  1  z   S9  z  S4     

Wideal  

kJ kg

Wout   H12  H11  * x a)

Heat Exchanger I: SG .a   S5  S4    S15  S14   1  z    SG .a 

kJ Wlost .a  T  SG.a  kg K

kJ kg

b) Heat Exchanger II: SG .b   S7  S6   1  x   S14  S13   1  z    SG .b 

kJ Wlost .b  T  SG .b  kg K

kJ kg

c) Expander: SG .c  S12  S11   x SG .c 

kJ Wlost .c  T  SG.c  kg K

kJ kg

d) Throttle: SG .d   S9  z  S10  1  z  x  S7  1  x    SG .d 

kJ Wlost .d  T  SG .d  kg K

kJ kg

Entropy-generation analysis:

kJ kg

Percent of 

SG.a

0.044

2.98%

SG.b

0.313

21.18%

SG.c

0.157

10.62%

SG.d

0.964

65.22%



1.478

100.00%

Work analysis, Eq.(15.3):

kJ kg

Percent of 

|Wout|

53.20

10.88%

Wlost.a

13.02

2.66%

Wlost.b

92.24

18.86%

Wlost.c

46.24

9.46%

Wlost.d

284.30

58.14%



489

100.00%

Note that:  = |Wideal|