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IMTS (ISO 9001-2008 Internationally Certified) THERMAL ENGINEERING

THERMAL ENGINEERING

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THERMAL ENGINEERING CONTENTS:UNIT –I

LOW TEMPERATURE PHYSICS

01-28

Joule – Thomson effect – porous plug experiment – Theory – Adiabatic demagnetization Liquefaction of air, nitrogen, Helium gases – Practical applications of low temperature – Refrigerators – Air conditioning machines – effect of chloro fluoro carbon on ozone layer.

UNIT II

THERMODYNAMICS

29-73

Zeroth, first, second and third laws of thermodynamics – Heat engines – Carnot, Otto and Diesel engines – Working and efficiency – entropy – Change in entropy in reversible and irreversible process – temperature – entropy diagram – Maxwell’s thermodynamic relationsapplications – Tds equations - Clausius - Clayperon latent heat equations.

UNIT III

CONDUCTION AND RADIATION

74-127

Thermal conducticity – Forbe’s method – Lee’s disc method – Black body radiation – Wien’s law – Rayleigh Jeans law – Planck’s law – Stefan’s

law

–

Determination

Stefan’s

constant

pyrometers

–

Pyrtheliometers – Solar constant – determination of temperature of sun.

UNIT IV

ELASTICITY

128-157

Bending of beams – expression for bending moment – depression of the loaded end of cantilever – uniform and nonuniform bending – theory and experiment – Koening’s method – theory and experiment – 1 form of grinders – Torsion – expression for couple per unit twist – Torsion pendulum theory and experiment – Static torsion method of determining rigidity modulus.

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UNIT V

VISCOSITY AND SURFACE TENSION

158-195

Coefficient of viscosity – Ostawals’s viscometer – Searle’s viscometer – Theory and experiment – Viscosity of gases – Meyer’s formula – Rankine’s method Surface tension – excess pressure inside a curved surface – surface tension ad interfacial surface tension – method of drops – Quincke’s method – surface tension and angle of contact of mercury – variation of surface tension with temperature.

UNIT QUESTIONS

196-204

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UNIT-1 LOW TEMPERATURE PHYSICS 1. Production of low temperature. The general principle involved in the production of low temperature is to devise means for removing from a body its heat content.

The following

methods have been employed, with what results we shall presently see. (i)

Freezing mixtures of salts in ice.

(ii)

Cooling due to evaporation of liquids under reduced pressure.

(iii)

Cooling due to Joule-Thomson effect.

(iv)

Cooking due to adiabatic expansion.

(v)

Cooling due to desorption.

(vi)

Adiabatic demagnetization, I.e., by demagnetizing certain crystals

adiabatically, magnetic energy is made to leave the system and thereby extreme cooling is produced under suitable conditions. Of these methods, the first two. viz., freezing mixtures and cooling by evaporation are of practical importance, as they are utilized in the modern refrigerating machines. The third and fourth methods have been used for the liquefaction of gases. These are very important, since these are intimately connected with the production of very low temperatures and are of great interest on account of the theoretical principles involved as well of the many practical applications both in science and industry. The method of cooling due to desorption has its use in aiding to a further cooling of an already very much cooled system. For example, o

charcoal absorbs helium at 5 atmospheres and 10 K and when it is allowed to escape at a lower pressure of 0.1 mm. cooling results due to the heat absorption by a process similar to o

the evaporation of the liquids and the temperature is lowered to 4 K which is sufficient to liquefy Helium. The last method of adiabatic demagnetization is of special interest to the physicist, as it has enabled him to reach almost the absolute zero, where matter, so to say, is born, and any information concerning the nature of substances in that region is bound to be of great value. It is to be noted that most of these methods are not exclusive of each other by complementary, in the sense that they can be used conductively reach the desired low temperature. 1.1 Joule-Thomson Effect If a gas initially at a constant high pressure is allowed to suffer throttle expansion through the porous plug of silk wool or cotton wool having a number of fine pores, to a region of constant

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lower pressure adiabatically, a change of temperature of the gas (either cooling or heating) is observed. This effect is called Joule-Thomson or Joule-Kelvin effect. Joule in collaboration with William Thomson [Lord Kelvin] devised a very sensitive technique known as porous plug experiment and performed number of experiments from 1852 to 1862 and established beyond doubt the existence of intermolecular attraction. In this effect, the total heat function H = U + PV remains constant. At ordinary temperatures, all gases except hydrogen show cooling effect on passing through the porous plug, but hydrogen shows a heating effect. At sufficiently low temperatures all gases show a cooling effect. 1.2 Joule-Thomson Porous Plug Experiment The experimental set up for the porous plug experiment to study the Joule-Thomson effect is shown in Fig.1.2. It consists of the following main parts: 1. The apparatus consists of a porous plug having two perforated brass discs D, D. 2. The space between D, D is packed with cotton wool or silk fibers. 3. The porous plug is fitted in a cylindrical box-wood W which is surrounded by a vessel containing cotton wool (Fig. 1.2). This is done to avoid loss or gain of heat from the surroundings. 4. T1 and T2 are two sensitive platinum resistance thermometers and they measure the temperatures of the incoming and outgoing gas. 5. The gas is compressed to a high pressure with the help of piston P and it is passed through a spiral tube immersed in water bath maintained at a constant temperature. If there is any heating of the gas due to compression, this heat is absorbed by the circulating water in the water bath.

Fig. 1.2.

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Experimental procedure By means of the compression pump P, the experimental gas is passed slowly and uniformly through the porous plug keeping the high pressure constant, read by pressure gauge. During its passage through the porous plug, the gas is throttled, i.e. the separation between the molecules increases. On passing through the porous plug, the volume of the gas increases against the atmospheric pressure. As there is no loss or gain of heat during the whole process, the expansion of the gas takes place adiabatically. The initial and final temperatures are noted by means of the platinum resistance thermometers T1 and T2. Experimental Results Inversion Temperature The behaviour of a large number of gases was studied at various inlet temperatures of the gas and the results obtained are as follows: (i)

At sufficiently low temperatures, all gases show a cooling effect.

(ii)

At ordinary temperatures, all gases except hydrogen and helium show cooling effect. Hydrogen and helium show heating instead of cooling at room temperature.

(iii)

The fall in temperature is directly proportional to the difference in pressure on the two sides of the porous plug.

(iv)

The fall in temperature for a given difference of pressure decreases with rise in the initial temperature of the gas. It was found that the cooling effect decreased with the increase of initial temperature and became zero at a certain temperature and at a temperature higher than this temperature, instead of cooling, heating was observed. This particular temperature at which the Joule-Thomson effect changes sign is called the temperature of inversion.

Definition The temperature at which Joule-Thomson effect is zero and changes sign is known as the temperature of inversion. It is denoted by Ti and at this temperature

2a ď&#x20AC;˝b RTi

Ti ď&#x20AC;˝

or when

2a Rb

T < Ti, cooling takes place, and T > Ti, heating takes place.

for

helium Ti = 35 K, and for hydrogen Ti = 193 K.

1.2.1 Regenerative Cooling

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In case of hydrogen and helium, heating was observed at room temperature because it was at a temperature far higher than its temperature of inversion. The temperature of inversion for hydrogen is –80°C and for helium is –258°C. If helium is passed through the porous plug at a temperature lower than –258°C, it will also show cooling effect. It means any gas below its temperature of inversion shows a cooling effect when it is passed through the porous plug or a throttle valve. This is called regenerative cooling or Joule-Kelvin cooling. This principle is used in the liquefaction of the so called permanent gases like nitrogen, oxygen, hydrogen and helium. 1.2.2 Estimates of J-T Cooling (Theory of Porous Plug Experiment) Theory The simple arrangement of the Joule-Thomson porous plug experiment is shown in Fig. 1.2.2. The gas is allowed to pass through the porous plug from the high pressure side to the low pressure side. Due to this large difference of pressure, the gas flows through the fine pores and becomes throttled or wire drawn, i.e. the molecules of the gas are further drawn apart from one another.

Fig. 1.2.2 Consider one gram molecule of a gas to the left and to right of the porous plug. Let P1, V1 and P2, V2 represent the pressure and volume on the two sides of the porous plug. When the piston A is moved through a certain distance dx, the piston B also moves through the same distance dx. The work done on the gas by the piston gas on the piston

B  P2 A2 dx  PV 2 2.

A  P1 A1dx  PV 1 1 . The work done by the

Thus, the net external work done by the gas is

PV 2 2  PV 1 1. If w is the work done by the gas in separating the molecules against their intermolecular attractions, the total amount of work done by the gas is

( PV 2 2  PV 1 1)  w No heat is gained or lost to the surroundings. There are three possible cases: (i) Below the Boyle temperature,

PV 1 1  PV 2 2

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and P2V2 – P1Vl is + ve, w must be either positive or zero. Thus a net + ve work is done by the gas and there must be cooling when the gas passes though the porous plug. (ii) At the Boyle temperature if P1 is not very high

PV 1 1  PV 2 2 PV 2 2  PV 1 1 =0

and

The total work done by the gas in this case is

Therefore cooling effect at this

temperature is only due to the work done by the gas in overcoming intermolecular attractions. (iii) Above the Boyle temperature

PV 1 1  PV 2 2 PV 2 2  PV 1 1 is – ve

Thus, the observed effect will depend upon whether ( PV 2 2  PV 1 1 ) is greater or less than w. If w > (PlVl - P2V2), cooling will be observed. If w < (PlVl - P2V2), heating will be observed. Thus, the cooling or heating of a gas due to free expansion through a porous plug from a high pressure to a low pressure side will depend on (i) the deviation from Boyle's law and (ii) work done in overcoming intermolecular attractions. 1.2.3 Joule-Kelvin Effect-Temperature of Inversion Assuming that the Van der Waals equation is obeyed, the attractive forces between the molecules are equivalent to an internal pressure

a . V2

V1 to V2 , the work done in overcoming intermolecular

When the gas expands from attractions

w   P.dV V1

a V2

But

P



w 

 If

a dV V2

a a  V2 V1

V1 and V2 represent the gram molecular volumes on the high and the low pressure

sides respectively, the external work done by the gas is

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Hence the total work done by the gas

W  ( PV 2 2  PV 1 1)  w  ( PV 2 2  PV 1 1) 

a a  V2 V1

Van der Waals equation of state for a gas is

a    P  2  (V  b)  RT V  

PV 

a ab  bP  2  RT V V PV  RT  bP 

a V

 ab  V 2 is negligible 

 a  a a a W   RT  bP2     RT  bP1     V2   V1  V2 V1  1 1   b  P2  P1   2a    V1 V2  RT P1

V1 



P   P W  b  P2  P1   2a  1  2   RT RT 

and

V2 

RT P2

But

2a  P1  P2  RT

W   b  P1  P2  

 2a  W   P1  P2   b  RT  Suppose the fall in temperature is

T

W  JH

 J [MCP T ] where M is the gram-molecular weight of the gas 

 2a  JMCP T   P1  P2   b  RT 

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…(i)

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 P1  P2   2a  b    JMCP   RT

T  

or (i) Since

T

…(ii)

P1  P2 is + ve

will be + ve if

 2a   b  is + ve   RT  2a  b or RT

i.e.,

T 

2a Rb

…(iii)

Therefore, cooling will take place if the temperature of the gas is less than (ii) For

T

2a . Rb

to be zero, from equation (ii),

2a b 0 RT T

2a Rb

This temperature is called the temperature of inversion and is represented by

Ti  (iii)

T

2a Rb

…(iv)

will be negative, if

 2a   b  is –ve   RT  i.e.,

b

2a RT

T

2a Rb

T  Ti Therefore, heating will take place if the temperature of the gas is more than the temperature of inversion. Results (i) If the gas is at the temperature of inversion, then no cooling or heating is observed when it is passed through the porous plug. (ii) If the gas is at a temperature lower than the temperature of inversion, cooling will take place when it is passed through the porous plug. This is called regenerative cooling or JouleKelvin cooling. This principle has been used in the liquefaction of the so-called permanent gases like nitrogen, oxygen, hydrogen and helium.

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(iii) If the gas is at a temperature higher than the temperature of inversion, instead of cooling, heating is observed when the gas s passed through the porous plug. Relation between Boyle Temperature, temperature of Inversion and Critical Temperature Temperature of inversion,

Ti 

2a Rb

…(1)

TB 

a Rb

…(2)

Tc 

8a 27 Rb

…(3)

Boyle temperature,

Critical temperature,

From (1) and (2), we have

Ti  2 TB From (1) and (3), we have

Ti 2a 27 Rb  . Tc Rb 8a  The experimental value of

27  6.75 4

…(4)

Ti for actual gases is just less than 6. Tc

It means that the temperature of inversion is very much higher than the critical temperature. For hydrogen

Ti 190 K and Tc  33 K. As Ti  Tc , the methods employing

regenerative cooling (Joule-Kelvin cooling) are preferred to those employing the initial cooling of the gas below the critical temperature. Since ratio

Ti is a number (i.e., 6.75) from equation (4), the ratio of temperature of Tc

inversion and critical temperature does not depend upon the nature of the gas. 1.3 Adiabatic Demagnetisation Debye, Giauque and Macdougall were able to produce temperatures below 1 K with the help of gadolinium sulphate. Haas and Kramers used the magnetic balance for a number of paramagnetic substances. They found that potassium and chromicum alum give a much lower temperature. Method. The apparatus used is shown in Fig. 1.3.. The paramagnetic specimen (salt) is suspended in a vessel a, which is surrounded by liquid helium. Liquid helium taken in Dewar

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flask D, is boiled under reduced pressure. It is surrounded by another Dewar flask D2 containing liquid hydrogen. The salt is in contact with the helium gas. A magnetic field of the order of 30,000 Gauss is applied. When the magnetic field is switched on the specimen (salt) is magnetized. The heat due to magnetization is removed by first introducing hydrogen gas into A and then pumping it off with high vacuum pump, so that the specimen is thermally isolated. In the mean time, the specimen (salt) picks up the temperature of the liquid helium i.e., the specimen and the liquid helium are at the same temperature. Now the magnetic field is switched off.

Fig. 1.3. Adiabatic demagnetization of the specimen takes place and its temperature falls. The temperature of the specimen is determined by fitting a co-axial solenoid coil round the tube A and measuring the self-inductance and hence susceptibility of the substance at the beginning and at the end of experiment. Then temperature T is calculated by Curie law ď Ł =

C . T

Haas, in 1944, was able to produce temperatures upto 0.002K using a double sulphate of potassium and aluminium.

Klerk, Stenland and Gorter used powdered mixed crystals of

chromium alum and aluminum alum and went down to a temperature of 0.0014K. Theory. When a paramagnetic material is placed in a magnetizing field H, its elementary magnetic dipoles get aligned parallel to the direction of the field. The magnetic moment per unit volume thus produced is called the intensity of magnetization (I). According to Curieâ&#x20AC;&#x2122;s law, this intensity of magnetization is directly proportional to the magnetizing field H and inversely proportional to temperature T of the paramagnetic substance. Thus

Iď Ą

H T

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H I C  T 

…(i)

where C is a constant, known as Curie constant. If V is the volume of 1 mole (i.e., molar volume), then intensity of magnetization of 1 mole of paramagnetic substance.

M  IV Substituting for I from eqn (i), we get

H M  CV   T 

…(ii)

In the experiment, let 1 mole of paramagnetic substance is place in magnetizing field H. Then its thermodynamic behaviour can be expressed in terms of thermodynamic variables P, V, T and S. In thermodynamic system, an increase in pressure P results in decrease in volume V, analogously in our case any increase in H results in an increase in M. Hence replacing P by – H and V by M in Maxwell’s third thermodynamic relation.

 T   V       P  S  S  P We have

 T   M        H S  S  H  M     T  H   S     T  H

Multiplying N and D by T, we get

 M  T   T  H  T       S   H  S T   T  H

 M  T    T  H   Q     T  H



T CH

  M     T  H

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T.S  Q 

…(iii)

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where

 Q    = CH, the specific heat of the substance at constant magnetic field H. The unit is  T  H

Joule/mole K. Since the process is carried out adiabatically (in which entropy S remains unchanged, i.e., constant), we may write.

T  M    dH CH  T  H

dT  

Therefore, when a field changes from H1 to H2, the change in temperature

T  

T CH



 M    .dH  T  H

…(iv)

Differentiating eqn. (ii) w.r. to T, we get

CVH  M     2 T  T  H

…(v)

Substituting in equation (iv), we have

T   =

T 

or,

T CH



CV CH .T



 CVH  2  T

  .dH 

H.dH

CV H 2 2  H12   2CH .T

…(vi)

If the magnetic field is reduced from H1 = H to H2 = 0, then the change in temperature will be

T 

CV H2 2CH .T

…(vii)

This is the desired relation. The following conclusions can be drawn form this equation (vii). Conclusions: (a)

The temperature of the paramagnetic substance decrease on decreasing the magnetizing field H (indicated by negative sign of ΔT), and

(b)

Greater is the initial field H and lower is the initial temperature T, greater is the temperature fall ΔT.

It should be noted that the quantity CV is the Curie constant per mole. If 1 gm of paramagnetic substance (salt) is taken, then CV would stand for Curie constant per gm. 1.3.1 T-S Diagram The cooling produced by adiabatic demagnetization is explained by using T-S i.e., temperature entropy diagram, as shown in Fig. 1.3.1.

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Fig. 1.3.1. In the diagram, the entropy of the substance is represented schematically as a function of temperature for two different magnetizing fields constants, i.e., H 2 = 0 and H1 = H. The arrow AB represents the change in entropy during isothermal magnetisaiton (at constant temperature). This arrow shows the decrease in entropy of the substance during magnetization (from H = 0 to H).

Now if the field is switched off, the substance demagnetizes adiabatically, so that the

temperature falls while during this process entropy remains constant, as represented by an arrow -4

VC. During this process, i.e., adiabatic demagnetization the temperature of the order of 10 can be reached. The main hurdle in this method is the measurement of such a low temperature. This problem is resolved by using the Curie law.

Assuming the validity of Curie law, and if the

susceptibility of the sample is measured, the Curie temperature

TC ď&#x20AC;˝

ď Ł

(also known as

magnetic temperature) can be calculated. Knowing the Curie temperature, the thermodynamical temperature T can be evaluated. The temperature measured by this method is of the order of 10 4

-6

K. We can still go down up to the order of 10 by making use of nuclear magnetism. But it

should be remembered that absolute zero can never be attained in actual practice.

LIQUEFACTION OF GASES 1.4 Introduction A gas goes into liquid and solid forms as the temperature is reduced. Thus the processes of liquefaction of gases and solidification of liquids are intimately involved in the production of low temperatures. Andrews experiment showed that if a gas is to be liquefied by merely applying pressure on it, it has to be cooled below its critical temperature. Critical temperatures of carbon dioxide, ammonia are sulphur dioxide are higher than room temperature. Hence these gases can be liquefied at room temperature without pre-cooling, simply by increasing the pressure. So, the simple process would be to cool the gas below its critical temperature by some coolant and then to liquefy it by applying a pressure on it. This method does not work always, in the case of some gases like oxygen, nitrogen, hydrogen and helium; no pre-cooling can bring the temperature below their critical temperatures. Thus, they cannot liquefied by this method.

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Jouleâ&#x20AC;&#x201C;Thomson expansion is a very important technique to liquefy gases. The cooling produced in J-T expansion of a gas depends on the difference of pressure on the two sides of the porous plug and the initial temperature. For most of the gases, the J-T cooling is very small. However, the cooling effect can be intensified by employing the process called regenerative cooling. 1.4.1 Regenerative Cooling This process consists in cooling the incoming gas by a portion of the gas which has already been cooled by J-T expansion. That is, the cooling effect is made cumulative. By maintaining a continuous cycle of such operations, the initial temperature can be rapidly and progressively lowered. This happens due to the fact that, lower the temperature the greater is the cooling in J â&#x20AC;&#x201C; T process.

Fig. 1.4.1 The schematic arrangement of regenerative principle is shown in Fig. 1.4.1. The gas is compressed to a high pressure by the compressor P. The heat of compression is removed by passing the gas in a water- cooled jacket (Heat exchanger No. 1). Then the gas undergoes J-T expansion at needle valve T so that it cools by a small amount. This cooled gas is allowed to go back to the pump P. while going up, it cools the incoming gas in heat exchanger No. 2. This gas gets further cooled after undergoing J-T expansion at T. This regenerative process continues till the gas gets liquefied. This process was first used by Linde to liquefy air.

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1.4.2 Liquefaction of air – Linde’s Process Linde liquefied air using Joule – Thomson effect and the principle of regenerative cooling. The apparatus used is shown in Fig. 1.4.2 Air is compressed to a pressure of about 25 atmospheres by the pump P1. It is then cooled by passing it thought the cold water bath. This compressed air is passed through KOH solution to remove CO 2 and water vapour. The air then enters the second pump P2. Here it is compressed to a pressure of about 200 atmospheres. The compressed air next passes through a cooling spiral immersed in a freezing mixture, where its o

temperature falls to about – 20 C. This cooled air at high pressure passes down through the tube 0

A and is allowed to come out of the nozzle N1. The temperature falls to about – 70 C. This cooled air flows upwards through the tube B, cools the incoming air through A and returns to P 2. The pump P2 again compresses the air to 200 atmospheres and puts it into circulation once more. It passes through the nozzle N1 and is further cooled.

Fig. 1.4.2 After the completion of a few cycles, the air is cooled to a sufficiently low temperature. At this stage, the second nozzle N2 is opened. The already much cooled air is allowed to expand to a pressure of one atmosphere when it is liquefied. The liquid air is collected in a Dewar flask D. the unliquefied air is led back through the tube C to the pump P1 and the process is repeated. The entire arrangement is packed in cotton wool to avoid any conduction or radiation. 1.4.3 Claude’s Process – Liquefaction of Air Fig. 1.4.3. shows the Claude’s liquefier with a two fold working. First, an adiabatic expansion is used to attain a temperature below the inversion temperature T i. Then J-T expansion is to complete the liquefaction. Pure, dry, air compressed by the pumps P1 and P2

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passes through heat exchanger E1 to cool itself. Then it is divided into two parts at A. 80% of the compressed air goes to the expansion cylinder and suffers adiabatic expansion and consequent cooling. The piston of the expansion cylinder is coupled to the compressors to drive them partially.

Fig. 1.4.3. Now the cooled gas flows upwards through the heat exchanger E 2. thereby it cools the rest 20% of the coming from A to E2 directly. This cooled 20% gas suffers a further cooling in the heat exchanger E3. Then it is subjected to J-T expansion at the throttle valve N. This gives a further cooling and a part of the air is liquefied the remaining cool air is made to pass through E 3, E2 and E1 to give extra cooling to the incoming air. Both the tubes inside the exchangers are in the form of spirals of coils for better cooling. The main problem in this process is that of solidification of lubricants at low temperature. o

1.5 Liquefaction of nitrogen. Liquid oxygen whose normal boiling point is -183 C, when o

allowed to boil under reduced pressure, a temperature of -218 C is reached.

The critical

temperature of nitrogen is -146 C. By adding a fourth unit containing liquid oxygen, nitrogen can be liquefied. Limitations. The limit of Pictet process is reached with nitrogen as there is no liquefied gas which on evaporation under reduced pressure will produce temperature low enough to bring

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16 o

about liquefaction of hydrogen, the critical temperature of which is -240 C. o

Neon critical

temperature -229 C and helium critical temperature -268 C also cannot be liquefied by this process. Apart from this limitation, the chief objection to Pictet process is that it is cumbersome and thus is not used nowadays in the commercial preparation of liquid air (oxygen). 1.6 Liquefaction of helium. Helium, the last remaining gas, could not be liquefied for a long time. The attempts made by Dewar and Olszewski by adiabatic expansion method proved unsuccessful. At last, in 1908, Kamerlingh Onnes, after a tenacious and systematic study of the problem for well over ten years, succeeded to liquefy helium by the Joule-Thomson process. He first cleared the ground by determining the various constants that were implicated in the experiment. Such were the critical o

temperature Tc = -268 C = 5 K, the critical pressure pc = 2.3 atmospheres, the normal boiling o

point = -268.8 C = 4.2 K, the inversion temperature = -238 C = 35 K and the Boyle temperature o

= - 256 C = 17 K. He then concluded that if helium was to be liquefied by the Joule-Thomson o

expansion it had to be pre-cooled to 17 K.

He succeeded in obtaining the temperature by

passing compressed helium through liquid hydrogen boiling at reduced pressure. The apparatus employed is very complicated due to the use of air and hydrogen liquefiers for pre-cooling processes. A simple form of the apparatus for final stages is as shown in Fig. 1.6. pre-cooled helium gas at a pressure of about 40 atmospheres enters

Fig. 1.6. the spiral tube at C and flows partly through the spiral tubes S1 and partly through S2. The spiral S1 is surrounded by hydrogen boiling under reduced pressure and portion passing through the

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spiral S2 is surrounded by cooled out going helium gas. Similar process takes place in the spirals S3 and S4 which are also surrounded by boiling liquid hydrogen and cooled helium respectively. The circulation process is repeated and when temperature of helium is sufficiently low it gets liquefied in passing through the nozzle N due to Joule-Thomson effect. The outgoing helium is compressed again by the compressor P and fed back to the spirals S1 and S2 and then S3 and S4. Liquid helium is collected in the Dewar flask through the tap T. To have perfect heat insulation the whole apparatus is surrounded by Dewar flasks. Kapteza has liquefied helium pre-cooled by passing through cooling liquid nitrogen. Simon using the fact that the thermal capacity of metal container is extremely small at temperatures near the absolute zero, has liquefied helium in small quantities. By adiabatically expanding helium (with subsequent cooling) contained in a small metal cylinder immersed in liquid hydrogen the gas of liquefied. Simon also making use of the fact that heat is evolved when a gas is adsorbed on charcoal and when it is desorbed the temperature of the gas falls, liquefied helium by using considerably large quantities of activated charcoal immersed in liquid hydrogen. To minimize heat exchanges with the surroundings the apparatus is enclosed in an evacuating vessel. On pumping off helium, the temperature considerably falls and thus it is liquefied. The storage of liquid hydrogen and liquid helium presents a more difficult problem than in the case of liquid air as these have a very low temperatures and low latent heats.

Liquid

hydrogen is usually stored in a double-Dewar vessel, the space between the evacuated containers is filled with liquid air. To store liquid helium a triple Dewar vessel is generally used, the liquid helium being surrounded by liquid hydrogen which in turn is surrounded by liquid air.

Liquid helium is a

colourless mobile liquid which boils at 4.2K. It has a density which is about one-eighth that of water and its very flat meniscus shows that it has a very small surface tension. 1.7 Practical application of low Temperatures Applications in Pure Science 1. Separation of the constituents of Air. A very important application of low temperatures is in the separation of the constitutes of air. The atmospheric air contains 21% of oxygen, 78% of nitrogen and 1% of the rare gases like helium, neon, krypton, etc. Liquid oxygen and liquid nitrogen are obtained by subjecting liquid air to fractional distillation. 0

Liquid nitrogen has a lower boiling point (-196 C) then liquid oxygen (-183 C). On evaporating liquid air, nitrogen first separates out and the residue gets richer in oxygen. Thus after subjecting liquid air to a few distillations, oxygen and nitrogen can be completely separated in the liquid form. By a similar process, the rare gases can also be separated. Liquid air is first separated into two parts, the more volatile parts containing hydrogen, helium and neon and the less volatile part

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containing oxygen, nitrogen, argon, krypton, xenon and carbon-dioxide. By cooling the more volatile portion with liquid hydrogen, all other components except hydrogen are removed. Hydrogen is subsequently removed by sparking with oxygen. Thus liquid helium is obtained. 2)

Production of high vacua. Both liquid air and oxygen are used in the production of high vacua. If a gas less volatile than air, like water vapour, is contained in a vessel enclosed in a bath of liquid air, the water vapour gets completely solidified producing a high degree of vacuum in the vessel. In the same way, a high vacuum is produced in a vessel by placing liquid air in the vessel and enclosing it in a bath of liquid hydrogen

Calorimetric work. Dewar used liquid hydrogen or liquid oxygen calorimeters to determine the specific heat capacities of substances at very low temperatures.

Study of the Properties of the Upper Atmosphere. The production of very low temperatures in the laboratory has given us a method of studying the properties of upper atmosphere. Conditions similar to those obtaining in the upper atmosphere can be produced in the laboratory and its effect on plant and animal life can be studied at leisure.

Study of properties of substances at low Temperatures. The effect of low temperature on properties of substances, like their thermal and electrical conductivities can be studied. The discovery and the study of phenomena like super-conductivity and super-fluidity have been possible due to production of very low temperatures.

Applications in Industry (1) Liquid oxygen stored up in cylinders are used in hospitals for artificial respiration. Liquid oxygen is used in the manufacture of explosives. Liquid air is kept in coal mines in order to supply large quantities of oxygen in the event of an explosion. Nitrogen is used for the production of synthetic ammonia and for the manufacture of fertilizers. Inert gases like argon, neon, krypton and xenon separated from liquid air, have good commercial value. Argon is used in â&#x20AC;&#x153;gas filledâ&#x20AC;? electric lamps. Neon is employed in discharge tubes required for illumination and advertising sign boards. Liquid air is of considerable importance in chemical industry, for separation and purification of gases. It is extensively used in high vacuum technique and also for calorimetric observations. (2) The most important application of liquefied gases is in the working of refrigerating and airconditioning machines.

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1.8 Refrigeration Definition Refrigeration is defined as the production of temperatures lower than those of the surroundings and maintaining the lower temperature within the boundary of a given space. A machine used for producing low temperature below surroundings and maintaining an enclosure at the temperature is called refrigerator. Actually, a refrigerator is a Carnot engine working in backward direction. In this case, certain amount of work is done on the engine which transfers heat from insulated chamber to the environment. Principle. The principle used in refrigeration is the production of low temperature by evaporation of a suitable liquid under reduced pressure.

Fig. 1.8. The liquid used for the purpose is called a refrigerant. Some of the common refrigerants in use are ammonia, sulphur dioxide, carbon dioxide, Freon (CCl 2, F2) and methyl chloride. In large refrigerating plants ammonia is used while in small plants used for domestic purposes sulphur dioxide and Freon are used. The principle of working of a refrigerator is shown in Fig. 1.8. The gas is compressed to a high pressure by the compressor P. It is cooled down by water circulating through a spiral pipe at C. The gas then gets liquefied. This high pressure liquid passes through a throttle valve T towards a low pressure side. It comes out as a mixture of liquid and vapour and its temperature gets lowered. At D this liquid â&#x20AC;&#x201C;vapour mixture takes heat from its surrounding material thus

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cooling this stuff. In this process, the remaining liquid also turns into vapour. This vapour is finally sucked by the pump P. 1.8.1 Carnot’s Cycle as Refrigerator When the carnot’s engine works as a refrigerator, it absorbs heat Q 2 from the sink at temperature T2. W amount of work is done on it by some external means and rejects heat Q 1 to the source at a temperature T1 (Fig. 1.8.1) Coefficient or performance

Q2 Q2 T2   W Q1  Q2 T1  T2

Fig. 1.8.1 The coefficient of performance (K) is defined as the ratio of the heat taken in from the cold body to the work needed to run the refrigerator. Refrigerant A refrigerant must fulfill the following requirements 1. The latent heat of vaporization of the refrigerant liquid must be large. When a small quantity of liquid evaporates, a larger amount of heat is removed. So the cooling will be more. 2. The refrigerant must be a vapour at normal temperature and pressure. 3. The refrigerant should have low boiling point and low freezing point. 4. The specific volume (volume occupied by I gm of refrigerant) should be small in order to reduce the size of compressors. 5. It should have high thermal conductivity.

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6. It should be non-flammable, non-explosive and should not have bad effect on the stored food material. 7. The pressure necessary to liquefy the vapour in condenser coils must not be very large. Otherwise, the apparatus will have to be very robust and heavy. Types of Refrigerating Machines Refrigerating machines are of two types: 1. Vapour compression machine, and 2. vapour absorption machine Here we shall discuss both the refrigerating machines. 1.8.2 Electrolux Refrigerator (Vapour Absorption Machine) Principle. The principle involved is the production of low temperature by making a liquid evaporate rapidly under reduced pressure. The liquid ammonia is made to evaporate under reduced pressure in an atmosphere of hydrogen. The Electrolux refrigerator is a modern form of vapour absorption machine. Fig. 1.8.2 shows the schematic diagram of an Electrolux refrigerator.

Fig. 1.8.2

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Construction: The whole unit consists of four essential parts. 1.

Generator. Here, ammonia gas is generated by heating the strong aqueous solution of ammonia in it by means of a heater H.

Condenser. It is cooled by water circulating round it. So the gaseous ammonia condenses to the liquid form.

Evaporator. The evaporator is inside the space desired to be cooled. Here, the liquid ammonia gets evaporated.

Absorber. The absorber is surrounded by a cold water jacket. Here, the gaseous ammonia from the evaporator gets absorbed by the weak aqueous solution from the generator.

Working the weak ammonia solution in the boiler is forced into the absorber. The strong ammonia solution goes into the boiler. The gaseous ammonia enters the condenser. It is cooled and is condensed. Liquid ammonia enter the evaporator and is mixed with hydrogen. Hydrogen reduces the partial pressure of ammonia below its saturation point and causes evaporation. This evaporator is surrounded by the compartment to be cooled. When ammonia evaporates under reduced pressure, the latent heat of vaporizations is taken from surrounding compartment. In this way, the compartment is cooled. Hydrogen and gaseous ammonia leave the evaporator and enter the absorber. Here they meet the weak ammonia solution. The ammonia gas is dissolved. Here hydrogen gas rises through the absorber and enters the evaporator. The Strong ammonia solution is forced up into the boiler again. The process continues and a sufficiently low temperature is produced in the compartment to be cooled. Advantage. No compressor is required and the circulation of the liquid and the gas is automatic. 1.8.3 Frigidaire (Vapour compression machine) Principle. A Frigidaire maintains an adiabatic enclosure at a low temperature. The principle involved is the production of low temperature by making a liquid evaporate rapidly under reduced pressure and circulating the evaporating liquid around the enclosure.

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Fig. 1.8.3 The refrigerant used is sulphur dioxide. It is now being increasingly replaced by Freon or F12. This refrigerator is generally used for domestic purposes. Fig. 1.8.3 shows the schematic diagram of a Frigidaire. Construction. C is an electric motor â&#x20AC;&#x201C; driven compressor. It is fitted with two spring â&#x20AC;&#x201C; loaded valves V1 and V2 in its bottom. During the down stroke of the piston, when the pressure on the gas or vapour in it increases, V1 is thrown open, V2 remaining closed. During the upstroke of the piston, when the pressure falls, V2 gets opened and V1 is closed. A steady stream of cold water is kept circulating round the condenser coils. The evaporating coils are surrounded by the space or the chamber to be cooled. The throttle valve has a small or orifice in it. The pressure on the liquid passing through it from the condenser coils on the high pressure side can be reduced to be equal to that prevailing in the evaporator coils on the low pressure side. Working. To start with, some vapour of the refrigerant is introduced into the compressor. The vapour is compressed by the compressor during its downward stroke. The valve V 1 opens while V2 closed. Now the vapour under high pressure enters into the condenser coils. The condenser coil is surrounded by an enclosure in which cold water is circulating. Here the heat of compression is removed by the cold water. So the vapour, under high pressure and low temperature, liquefies in the condenser coils. The liquid refrigerant passes through the regulator valve V into the evaporator coils. The regulator value reduces the pressure of liquid from high value (inside condenser) to low value (inside evaporator)

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Due to low pressure, the liquid evaporates in the evaporator coils. Here, when the liquid evaporates, it absorbs latent heat of vapourisation from surrounding cold storage space. The space or the chamber thus gets cooled. The low pressure vapour from the evaporator coils is then drawn into the compressor during the upstroke of its piston, through valve, V2. Now, valve V1 opens and V2 remains closed. The same cycle of operations is repeated again After some time, the temperature of the space around the evaporator coils gets cooled to the desired temperature. Uses for refrigeration are: 1.

Preservation of perishable products during storage or transportation;

The manufacture of ice and solid carbon dioxide (dry ice);

The control of air temperature and humidity in air-conditioning systems.

1.9 Air-Conditioning Definition: Air – conditioning may be defined as a method of controlling the weather conditions within the limited space of a room, or a hall, so that one may have the maximum comfort. The conditions required for a person to have maximum comfort are the following: 

The temperature of the air in the room should lie between 65 F and 70 F during o

summer and 65 F and 75 F during winter. 

Relative humidity should be maintained between 60-65%



Each individual in the room should at last have about 0.85m of air having a movement

of 15 metres per minute. At least 25% of the total air circulating must be fresh air. 

The air in the room should be free from dust, smoke, carbon dioxide, etc.

The essential requirements of an air-conditioning plant vary with the climate of the place. In cold countries, the chief requirement is the warming of the rooms. This is achieved by central heating using electrical power or gas heating. In hot countries, the requirement is to keep the rooms cool and fresh without the heat and dust. This is achieved by refrigeration.

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1.9.1 Air Conditioner Air conditioner cools the air, removers the dust and controls the humidity (Water vapour contents) of the atmosphere. Refrigeration is the process of removing heat from the space or materials. Air conditioners are of three types: window air conditioner, package air conditioner and centralized air conditioner. Window air conditioner is used for small rooms. Package air conditioner is used for comparatively larger rooms. Centralized air conditioners is used for theatres and lodges The important actions involved in the operation of an air conditioning system are: 1.

Temperature control

Humidity control

Air filtering, clearing and purification

Air movement and circulation A complete air conditioning system provides automatic control of these conditions for

both winter and summer Temperature control in winter means control of heating source and in summer means control of refrigerating system. Humidity control for winter requires addition of moisture (i.e use of humidifier) and in summer requires removal of moisture i.e. use of a dehumidifier. In general, air filtering is the same for both summer and winter air conditioning. 1.9.2

Window Air-Conditioner (Room Air-Conditioner) This is suitable for cooling rooms in summer. It consists of three basic parts; A hermatic compressor, a condenser and an evaporator. Construction. It consists of two units: (i) indoor unit and (ii) outdoor unit (fig. 1.9.2). The indoor unit consists of an evaporator, fan, air filter, control panel, trays, etc. The outdoor unit consists of a motor driven compressor, condenser, fan, trays, etc.

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Fig. 1.9.3 The condenser and evaporator are connected through a capillary tube. Working : The liquid refrigerant collects in the lower coils of the condenser and from there flows via capillary tube (refrigerant control) into the evaporator. Pressure inside the evaporator or cooling coil is very low. So the liquid refrigerant evaporates rapidly by picking up heat from evaporator surface which is consequently cooled. The motor- driven fan F2 draws air from inside the room through a suitable filter and forces it to flow over the evaporator. The air so cooled goes back to the room. An adjustable thermostat mounted on the control panel provides the necessary temperature control. The low-pressure refrigerant is sucked back into the compressor. From the compressor, the refrigerant is forced into the condenser to be cooled and condensed to a liquid. The cycle is repeated as described above Compressor and condenser are so mounted that fan F1 draws air in from outside, circulates it over the condenser where it is heated up (by the heat given up by the refrigerant during liquefaction) and then discharges it to outside. The moisture which collects on the evaporator is drained into a drip pan kept under the evaporator.

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1.9.3 Central heating system In cold countries, especially in winter, central hearting is employed to keep the rooms of a building warm by the principles of convection currents in air. Fig 1.9.3 illustrates the principle of central heating.

Fig 1.9.3 Water is heated in the boiler kept on the ground â&#x20AC;&#x201C;floor oh the building. The rooms are fitted with pipes having radiators fixed on the walls. Hot water from the boiler circulates at a controlled speed through the radiators of different rooms. These radiators absorb heat from the flowing hot water and radiate the heat to the air in the room. After losing heat to the radiators, the water becomes relatively cold. The cold water returns to the boiler to be heated up and used again. Convection currents are set up in the air in the rooms. Thus the rooms are kept continuously warm at the optimum steady desired temperature. 1.10 Effects CF2 Cl2 on Ozone layer In the atmosphere, about 50 km above the surface of the earth, the ozone molecules (O 3) from an umbrella. It prevents the penetration of harmful ultra violet radiation from the sun and thus protects the life on the earth. It is now feared that there is danger of holes appearing on the ozone umbrella. This is caused by the use of freons and other chlorine â&#x20AC;&#x201C; fluorine carbons as refrigerants in domestic refrigerators and other cold storage facilities. They destroy ozone

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molecules as a result of photochemical reactions (Fig.1.10) Fig 1.10 shows how breakdown products of CFCs attract atmospheric ozone molecules.

Fig.1.10 The compound CF2 Cl2, which is widely used as refrigerant, decomposes to form chlorine atoms when exposed to UV radiation at 200 nm.

CF 2 Cl2 ( g )  CF2 Cl ( g )  Cl ( g ) Chlorine atom strips an atom from the molecule of ozone turning it into ordinary oxygen.

Cl  O3 ClO  O2 The chlorine goes on to repeat the process, and in this way on CFC molecule can destroy thousands of molecules of ozone. Over the past 16 years, the density of the ozone layer has diminishing at an average rate of 3% The main cause of ozone depletion is the widespread use of chlorofluoro carbons (CFCs). About 30% of world CFC production is used in fridges, freezers and air conditions. The huge amount of these products being used and released to atmosphere are probably causing an irreparable damage to our environment.

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UNIT II THERMODYNAMICS 2. Thermodynamic System A system may be defined as a definite quantity of matter (solid, liquid or gases) bounded by some closed surface. The simplest example of a system is a gas contained in a cylinder with a movable piston. When a system is completely uniform throughout, such as gas, or mixture of gases, or a pure solid, or a liquid or a solution is called a homogeneous system. When a system consists of two or more phases, which are separated from one another by definite boundary surfaces, it is said to be a heterogeneous system. Examples of heterogeneous system are: a liquid and its vapour, two immiscible or partially miscible liquids. Anything outside this system which can exchange energy with it and has a direct bearing on the behaviour of the system is called as its surroundings. A system may be separated from its surroundings by a real or imaginary boundary through which heat or mechanical energy may pass. The existence of the boundary is, however, essential to visualize the system distinctly from rest of the universe. A thermodynamic system may contain no substance at all, but may consist of radiant energy or electric and magnetic field. The combination of a system and its surroundings is called an universe. Thermodynamic Variables and equation of state The thermodynamic state or macroscopic state of a system is determined by four observable properties. These properties are the composition, pressure, volume and temperature, and are called the variables of state. For a homogeneous system, consisting of a single substance, the composition is fixed. Hence the state of the system is determined completely by the three variables: pressure (P), volume (V) and temperature (T). For a system, in general, there is an equation of state f (P, V, T) = 0 Therefore, the pressure, volume and temperature are not independent variables. Consequently, the thermodynamic state of a homogeneous system is determined completely by knowing two out of three variables. For example: the equation of state for an ideal gas in a static condition is represented as PV = RT where R is universal gas constant. For a Van der Waals’ gas, the equation of state is

a    P  2  (V  b)  RT V   where a and b are Van der Waal’s constants.

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Limitations (i)

The equation of state is not applicable to systems which are not in thermodynamic equilibrium.

(ii)

Every thermodynamic system has its own equation of state independent of the others.

(iii)

An equation of state expresses the peculiar behaviour of one individual system which distinguishes it from the others.

(iv)

An equation of state is not a theoretical deduction from thermodynamics but is an experimental backing to it. The thermodynamic systems in engineering are gas, vapour, steam, mixture of gasoline

vapour and air, ammonia vapours and its liquid. In physics, thermodynamics includes besides the above, systems like stretched wires, thermocouples, magnetic materials, electrical condenser, electrical reversible cells, solid and surface films etc. There are three classes of system (i) Open system : A system which can exchange matter and energy with the surroundings is called an open system, e.g. Air compressor: Air at low pressure enters and air at high pressure leaves the system i.e. there is an exchange of matter and energy with the surroundings. (ii) Closed system : A system which can exchange only energy (and not matter) with the surroundings is called a closed system. e.g., Gas enclosed in a cylinder expands when heated and drives the piston outwards. The boundary of the system moves but the matter (here gas) in the system remains constant. (iii) Isolated system : A system which is thermally insulated and has no communication of heat or work with the surroundings is called isolated system.2.1.

2.1 Zeroth Law of thermodynamics The sense of touch is the simplest way to distinguish hot bodies from cold bodies. By touch we can arrange bodies in the order of their hotness. We speak of this as our temperature sense. This is a very subjective procedure for determining the temperature of a body and certainly not very useful for purposes of science. Further, the range of our temperature sense is limited. What we need is an objective, numerical, measure of temperature The first step toward attaining an objective measure of the temperature sense is to set up a criterion of equality of temperature. Let an object A which feels cold to the hand and an identical object B which feels hot be placed in contact with each other. After sufficient length of time, A and B give rise to the same temperature sensation. Then, A and B are said to be in thermal equilibrium with each other. â&#x20AC;&#x153;Two bodies are in thermal equilibriumâ&#x20AC;? means that the two bodies are in states such that, if the two were connected, the combined systems would be in thermal

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equilibrium. If two bodies are in thermal equilibrium when placed in contact, then by definition their temperatures are equal. Conversely, if the temperatures of two bodies are equal, they will be in thermal equilibrium when placed in contact. The logical and operational test for thermal equilibrium is to use a third or test body such as a thermometer. This is summarized in a postulate often called the zeroth law of thermodynamics. Statement. If A and B are in thermal equilibrium with a third body C (the thermometer), then A and B are in thermal equilibrium with each other. Explanation Thus if we want to know when two beakers of water are at the same temperature it is unnecessary to bring them into contact and see whether their properties change with time. We insert a thermometer (body C) in one beaker of water (body A) and wait until some property of the thermometer, such as the length of the mercury column in a glass capillary, becomes constant. Then by definition, the thermometer has the same temperature as the water in the beaker A. We next repeat the procedure with the other beaker of water (body B). If the lengths of the mercury columns are the same, the temperatures of A and B are equal. Experiment shows that if the two beakers are brought into contact, no changes in their properties take place. Significance. This law forms the basis of concept of temperature. All these three systems can be said to possess a property that ensures their being in thermal equilibrium with one another. This property is known as temperature. We may, therefore, define the temperature of a system as the property that determines whether or not the system is in thermal equilibrium, with the neighbouring systems. It is obvious that if two systems are not in thermal equilibrium, they will be at different temperatures. Speaking loosely, the essence of the zeroth law is : there exists a useful quantity called temperature.

2.2. First Law of thermodynamics Statement. The amount of heat supplied to a system is equal to the algebraic sum of the change in internal energy of the system and the amount of external work done by the system. Explanation. The differential form of the first law of thermodynamics is dQ

dU + dw.

amount of heat supplied to the system

increase in internal energy of the system

external work done by the system

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In all transformations, the energy due to heat units supplied must be balanced by the external work done plus the increase in internal energy. In the first law, a conversion between heat and mechanical energies (work) is set up and in the balancing process, internal energy U is defined. When a cold body is placed in contact with a hot body, first law says that the heat gained by the colder body is equal to heat lost by the hotter body when an equilibrium is attained. The first law does not specifically say that heat does not flow spontaneously from the colder to the hotter body. The fact that heat can only pass from a hot body to a colder body and not otherwise is not covered by first law. First law is, therefore, insufficient for the sake of completeness of thermodynamic studies. To cover the directional properties of heat, it is desired to define some law. This is the basis of the second law of thermodynamics. Significance of the first law. The first law of thermodynamics establishes the relation between heat and work. According to this law, heat can be produced only by the expenditure of energy in some form or the other. Hence it follows directly from this law that it is impossible to make a perpetual motion machine or to derive work without any expenditure of energy. Application of the first law. From the first law of thermodynamics, …(1)

dQ = dU + PdV

The thermal energy of a system U is a function of any two P, V and T. Choosing T and V.

U  f (T ,V ) We have,

…(2)

 U   U  dU    dT    dV  T v  V T

…(3)

Therefore, the first law becomes

 U    U  dQ    dT     P  dV  T v  V T 

…(4)

Dividing by dT, we get

dQ  U   U   dV       P dT  T v  V T  dT

…(5)

This equation is true for any process involving any temperature change dT and any volume change dV. (a) If V is constant, dV = 0 and

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 dQ   U       Cv  dT v  T v

…(6)

Here, Cv is the molar specific heat capacity at constant volume. (b) If P is constant, Eq. (5) becomes

  V   dQ   U   U          P    dT  P  T v  V T   T  P But, by definition,



 dQ     CP = Molar specific heat capacity at constant pressure.  dT  P

 U    V  CP  Cv     P    V T   T  P For a perfect gas,

…(7)

 U     0. Further  V T

PV  RT or P(V / T ) P  R

 (or)

Eq. (7) becomes,

CP  CV  R

…(8)

CP  CV  R

2.3 Second Law of Thermodynamics The first law of thermodynamics states the equivalence of heat and energy. It simply tells that whatever work is obtained, an equivalent amount of heat is used up, or vice versa. It does not say anything about the direction in which the change might occur or about the range or limit to which it can be possible. The first law shows that perpetual motion of the first kind is impossible i.e. energy can not be created out of nothing or production of energy without disappearance of an equivalent energy of another form is not possible. The first law has no answer why heat always flows from a body at high temperature to a body at lower temperature and does not flow in the reverse direction. It can not explain why vast amount of a variable heat cannot be converted into mechanical work. It we could control and make use of the limitless store of heat viz-solar energy, we could have an inexhaustible supply of useful energy. Thus, we could have a perpetual motion machine not forbidden by the first law. This is called perpetual motion of the second kind. In practice, however there is no engine which can convert the heat from the single source in to useful work without rejecting some heat to a heat sink at a lower temperature i.e. perpetual motion of the second kind i.e. production of useful energy from the internal energy of one body is impossible.

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It was the quest of several such questions which led to the formulation of second law of thermodynamics. The second law is a generalization of certain experiences and observations and is concerned with the direction in which energy transfer takes place. The law has been stated in a number of ways, which means the same thing. Lord Kelvin's Statement In a heat engine the working substance extracts heat from the source, converts a part of it into work and rejects the rest to a sink at a lower temperature. The temperature of the source must be higher than the surroundings and engine will not work when the temperatures of source and sink are the same. No engine has ever been constructed which converts all the heat absorbed from the source into work without rejecting a part of it to the cold body. As the engine absorbs more and more heat from the hot body, the latter suffers a continuous fall in temperature and if a continuous supply of work is desired, the hot body will in the long run become as cold as its surroundings. Then no heat flow will be possible, the engine will stop working and hence no mechanical work will be obtained. It means that we can not obtain a continuous supply of work from a single supply of it i.e. the presence of colder body is a must for the continuous conversion of heat into work. Such considerations led Lord Kelvin to state that “It is impossible to get a continuous supply of work from a body by cooling it to a temperature lower than that of its surroundings.” Planck's Statement “It is impossible to construct an engine which, working in a complete cycle, will produce no effect other than the raising of a weight and the cooling of a heat reservoir.” Thus, it is impossible to construct an engine which working in a complete cycle, will produce no effect other than the absorption of heat from a reservoir and its conversion into an equivalent amount of work i.e. perpetual motion of the second kind is impossible. The engine must reject a part of the heat absorbed to a sink at lower temperature. These two statements can be combined into one equivalent statement, known as the Kelvin-Planck's statement of the second law of thermodynamics. Kelvin-Planck Statement “It is impossible to construct an engine which, operating in a cycle, has the sole effect of extracting heat from a reservoir and performing an equivalent amount of work.” Clausius's Statement According to Clausius “It is impossible for a self-acting machine working in a cyclic process, unaided by external agency, to transfer heat from a body at a lower temperature to a body at a higher temperature.” In other words it may be stated as “Heat cannot flow of itself from a colder body to a hotter body.”

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This statement is based upon the performance of refrigerator - a heat engine working in the backward direction. This statement means that natural flow of heat is always from a hot body to a cold body. If heat is to be transferred from cold body to hot body, work will have to be done by external agency. A refrigerator is a device which transfers heat from a colder body to hotter body by doing external work on the working substance. The compression is brought about by an external agency i.e. ‘electricity’ by performing work on the working substance.

2.4 Third Law of Thermodynamics: Nernst's Heat Theorem In 1906, Nernst proposed a general principle supported by a series of experimental evidences on the problem of atomic heat at low temperature. Actually, it was proposed as 'the new heat theorem' but its importance is so great that today it is recognized as 'the third law of thermodynamics'. Statement: According to Nernst, "The heat capacities of all solids tend to zero as the absolute zero of temperature is approached and that the internal energies and entropies of all substances become equal there, approaching their common value asymptotically tending to zero". This ample statement of Nernst follows neither from the first law (law of conservation of energy) nor from the second law (law of transmutability of energy) and is thus of the nature of a new law, usually called as third law of thermodynamics. This theorem is useful in explaining the nature of bodies in the neighbourhood of absolute zero temperature. Its importance lies in the fact that it permits the calculations of absolute values of entropy, Helmholtz and Gibb's free energies, etc. In terms of entropy, the theorem may also be stated as: It states that “at absolute zero temperature, the entropy tends to zero and the molecules of a substance or a system are in perfect order (well arranged)”. In all heat engines, there is always loss of heat in the form of conduction, radiation and friction. Therefore, in actual heat engines



 Q1 Q2      T1 T2 

Q1 Q2 is not equal to . T1 T2

is not zero but it is a positive quantity. When cycle after cycle is

repeated, the entropy of the system increases and attains a maximum value. When the system has attained the maximum value, a stage of stagnancy is reached and no work can be done by the engine at this stage. In the universe; the entropy is increasing and ultimately the universe will also reach a maximum value of entropy when no work will be possible. With the increase in entropy, the disorder of the molecules of a substance increases. The entropy is also a measure of the disorder of the system. With decrease in entropy, the disorder deceases. At absolute zero

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temperature, the entropy tend to zero and the molecules of a substance or a system are in perfect order (well arranged). This is known as the third law of thermodynamics. Example: The molecules are more free to move in the gaseous state than in the liquid state. The entropy in the gaseous state is more than in the liquid state. The entropy in liquid state is more than in the solid state. Thus, when a substance is converted from solid to a liquid and then from the liquid to the gaseous state.

2.5 Heat Engines Any practical machine which converts heat into mechanical work is called a heat engine. Heat engines in their operation absorb heat at a higher temperature, converts part of it into mechanical work, and rejects the remaining heat at a low temperature. In this process, a working substance

used.

steam engines, the working substance is water vapour, and in all gas engines the working substance is a combustible mixture of gases.

Fig 2.5 In any heat engine, the working substance goes through certain changes of pressure, volume and temperature, and then returns to the initial state. The complete changes through which the working substance undergoes from its initial state and back to its starting state constitute one cycle of operation. Definition of Efficiency The efficiency,

 , of a heat engine is defined as the ratio of the mechanical work done by

the engine in one cycle to the heat absorbed from the high temperature source. Thus



Q1  Q2 Q1

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where Q1 is the heat absorbed from the source at-high temperature, Q2, is the heat rejected to a sink at low temperature and (Qt â&#x20AC;&#x201D; Q2) is the mechanical work done by the engine in one cycle. Since (Q1 - Q2) < Q1, the efficiency can never be 100%.

2.6 Carnotâ&#x20AC;&#x2122;s Ideal Heat Engine In 1824 the French engineer Sadi Carnot conceived a theoretical engine which is free from all practical imperfections. Such an engine can not be realised in practice. It has maximum efficiency and it is an ideal heat engine. Sadi Cornot's heat engine requires the following important parts: 1. A cylinder having perfectly non-conducting walls, a perfectly conducting base and is provided with a perfectly non-conducting piston which moves without friction in the cylinder. The cylinder contains one mole of perfect gas as the working substance. 2. Source. A reservoir maintained at a constant temperature Tx from which the engine can draw heat by perfect conduction. It has infinite thermal capacity and any amount of heat can be drawn from it at constant temperature T1. 3. Heat insulating stand. A perfectly non-conducting platform acts as a stand for adiabatic processes. 4. Sink. A reservoir maintained at a constant lower temperature T2 (T2 < T1) to which the heat engine can reject any amount of heat. The thermal capacity of sink is infinite so that its temperature remains constant at T2, no matter how much heat is given to it.

Fig. 2.6.

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2.6.1 Carnot's Cycle In order to obtain a continuous supply of work, the working substance is subjected to the following cycle of quasi-static operations known as Carnot's cycle. 1.

Isothermal expansion. The cylinder is first placed on the source, so that the gas acquires the temperature T1 of the source. It is then allowed to undergo quasi-static expansion. As the gas expands, its temperature tends to fall. Heat passes into the cylinder through the perfectly conducting base which is in contact with the source. The gas therefore, undergoes slow isothermal expansion at the constant temperature T1.

Fig 2.6.1 Let the working substance during isothermal expansion goes from its initial state

A ( P1 ,V1 , T1 ) to the state B ( P2 ,V2 , T1 ) at constant temperature T1 along AB. In this process, the substance absorbs heat V2

Q1 from the source at T1 and does work W1 given by

Q1  W1   P dV  RT1 log e V1

V2  area ABGEA V1

…(2)

2. Adiabatic expansion. The cylinder is now removed from the source and is placed on the insulating stand. The gas is allowed to undergo slow adiabatic expansion, performing external work at the expense of its internal energy, until its temperature falls to T2, the same as that of the sink.

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This operation is represented by the adiabatic BC, starting from the state to the state

B ( P2 ,V2 , T1 )

C ( P3 ,V3 , T2 ) . In this process, there is no transfer of heat, the temperature of the

substance falls to T2 and it does some external work W2 given by V3

W2   P . dV  K 

dV V

( Q During aciabatic process,

PV  = constant = K)

KV31  KV21  1  

PV 3 3  PV 2 2 1 



RT2  RT1 1 



R(T1  T2 )  Area BCHGB  1

  (Q PV 2 2  PV 3 3  K)

…(3)

3. Isothermal Compression. The cylinder is now removed from the insulating stand and is placed on the sink which is at a temperature T2. The piston is now very slowly moved inwards so that the work is done on the gas. The temperature tends to increase due to heat produced by compression since the conducting base of the cylinder is in contact with the sink, the heat developed passes to the sink and the temperature of the gas remains constant at T2. Thus the gas undergoes isothermal compression at a constant temperature T2 and gives up some heat to the sink. This operation is represented by the isothermal CD, starting from the state to the state

C ( P3 ,V3 , T2 )

D ( P4 ,V4 , T2 ) . In this process, the substance rejects heat Q2 to the sink at T2 and

work W3 is done on the substance given by V4

Q2  W3   PdV  RT2 log e V3

  RT2 log e

V4 V3

…(4)

V3  area CHFDC V4

(- ve sign indicates that work is done on the working substance) 4. Adiabatic Compression. The cylinder is now removed from sink and again placed on the insulating stand. The piston is slowly moved inwards so that the gas is adiabatically compressed and the temperature rises. The adiabatic compression is continued till the gas

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back

condition i.e. state

its

A ( P1 ,V1 , T1 ) , thus completing one full cycle.

This operation is represented by adiabatic DA, starting from state

original

D ( P4 ,V4 , T2 ) to the final

A ( P1 ,V1 , T1 ) . In this process, work W4 is done on the substance and is given by W4 





P.dV

R(T1  T2 )  Area DFEAD  1

…(5)

(- ve sign indicates that work is done on the working substance. Since W2 and W4 are equal and opposite, they cancel each other.)

Work done by the engine per cycle During the above cycle, the working substance absorbs an amount of heat Q1 from the source and rejects Q2 to the sink. Hence, the net amount of heat absorbed by the gas per cycle = Q1 – Q 2 The net work done by the engine per cycle = W1 + W2 + W3 + W4 ( Q W2 = –W4)

= W1 + W3 From the graph, the net work done per cycle

= area ABGEA + area BCHGB – area CHFDC – area DFEAD = area ABCDA Thus, the area enclosed by the Carnot’s cycle consisting of two isothermals and two adiabatics gives the net amount of work done per cycle. In the cyclic process, Net heat absorbed Q1 – Q2

= Net work done per cycle. = W1 + W3 =

RT1 log e

V V2  RT2 log e 3 V1 V4

…(6)

Since the points A and D lie on the same adiabatic DA  1 TV  T2V4 1 1 1

T2  V1    T1  V4 

 1

Similarly, points B and C lie on the same adiabatic BC

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 1 TV  T2V3 1 1 2

T2  V2    T1  V3 



 1

…(8)

From equations (7) and (8),  1

 V1     V4 

 1

V   2  V3 

V1 V2  V4 V3

V2 V3  V1 V4

Substituting in equation (6), we get Net work done

 Q1  Q2  RT1 log e

V2 V  RT2 log e 2 V1 V1

W  (Q1  Q2 )  R (T1  T2 ) log e

V2 V1

…(9)

Efficiency The efficiency of the heat engine is the rate of quantity of heat converted into work (Useful output) per cycle to the total amount of heat absorbed per cycle. Efficiency,

 

Useful Output W  Input Q1 (Q1  Q2 ) Q1 R (T1  T2 ) log e

 RT1 log e

 or

V2 V1

T1  T2 T1

  1

T2 T1

…(10)

…(11)

From equation (11), we conclude that the efficiency depends only upon the temperature of the source and sink and is always less than unity. The efficiency is independent of the nature

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of working substance. From equation,

  1

T2 , we get   1, if T2 = oK i.e. the temperature T1

of the sink is at absolute zero degrees. In practice, it is never possible to reach absolute zero and hence 100% conversion of heat energy into mechanical work is not possible. Again, the efficiency is minimum or zero when T1 = T2 i.e. the temperature of the source is equal to the temperature of sink, then

0

i.e. the engine does not work.

The Carnot's heat engine is perfectly reversible. It can be operated in the reverse direction also. Then it works as a refrigerator. The heat Q2 is taken from the sink and external work is done on the working substance and heat Q2 is rejected to the source at a higher temperature (principle of a refrigerator). Moreover in the Carnot's heat engine, the process of isothermal and adiabatic expansions and compressions are carried out very-very slowly i.e. quasi-static. This is an ideal case. Any practical engine can not satisfy these conditions. Therefore, all practical engines have an efficiency less than the Carnot's engine. 2.6.2 Effective Way to Increase Efficiency The expression for the efficiency of a Carnot's engine is

  1 The efficiency



T2 T1

can be increased by one of the following ways:

(i) Using a heat-source at constant temperature T1 and heat-sink at temperature as low as possible, or (ii) Using a heat-sink at constant temperature T2 and a heat-source at temperature as high as possible. In practice, it is not convenient to use heat sink at a temperature below that of the atmosphere. Therefore, the more effective way to increase the efficiency is to use a heat-source at temperature as high as possible. 2.6.3 Carnot's Engine and Refrigerator Carnot's cycle is perfectly reversible. It can work as a heat engine and also as a refrigerator. When it is operated as a heat engine, it absorbs heat Q1 from the source at a temperature T1, does an amount of work W and rejects heat Q2 to the sink at temperature T2, (T2< T1) as shown in [Fig 2.6.3.(a)].

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Fig 2.6.3 (a) Heat Engine

Fig. 2.6.3 (b) Refrigerator

When it is operated as a refrigerator, it absorbs heat Q2 from the sink at temperature T2. W amount of work is done on it by some external means and rejects heat Q1 to the source at temperature T1, (T1 > T2) as shown in [Fig 2.6.3 (b)]. In the second case, [Fig 2.6.3 (b)], heat flows from a body at a lower temperature T2 to a body at a higher temperature T1 with the help of external work done on the working substance. This action is that of a refrigerator. In every cycle, heat Q2 is extracted from the cold body. This will not be possible if the cycle is not completely reversible. Coefficient of Performance The amount of heat absorbed at the lower temperature is Q2. The amount of work done on the working substance by the external agency (input energy) = W and the amount of heat rejected = Q1. Here Q2 is the desired refrigerating effect in each cycle.



Coefficient of performance,

P

Q2 Q2  W Q1  Q2

…(1)

Suppose 200 joules of energy is absorbed at the lower temperature and 100 joules of work is done on it by external help. Then, 200 + 100 = 300 joules are rejected at the higher temperature. The coefficient of performance

P

Q2 W



Q2 Q1  Q2



200 2 300  200

Therefore, the coefficient of performance of a refrigerator is 200%.

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In the case of heat engine, the efficiency cannot be more than 100% but in case of a refrigerator, the coefficient of performance can be much higher than 100%. If the working substance is an ideal gas, then

Q1 Q2 Q  Q2   1 T1 T2 T1  T2 Q2 T2  Q1  Q2 T1  T2

Thus, equation can also be expressed as coefficient of performance,

P

T2 T1  T2

2.6.4 Carnot’s Theorem Statement From the second law of thermodynamics two important results are derived; these conclusions are taken together to constitute Carnot’s theorem which may be stated in the following forms. (a) ‘No engine can be more efficient than a perfectly reversible engine working between the same two temperatures.’ (b) ‘The efficiency of all reversible engines, working between the same two temperatures is the same, whatever the working substance.’ Proof: First Part: To prove the first part of the theorem, we consider two engines R and I working between the temperatures T1 and T2 where T1 > T2 [Fig. 2.6.4]. Of these two engines R is reversible and I is irreversible.

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Fig. 2.6.4 Suppose I is more efficient than R. Suppose in each cycle, R absorbs the quantity of heat Q1 from the source at T1 and rejects the quantity of heat Q2 to the sink at T2. Suppose in each cycle I absorbs the quantity of heat

Q1 from the source at T1 and gives up the quantity of heat

Q2 to the sink at T2. Let the two engines do the same amount of work W in each cycle. According to the assumption I is more efficient than R.



Q1  Q2 Q1  Q2  Q1 Q1

W W  Q1 Q1

Q1  Q1

and

Q1  Q2  Q1  Q2

Q2  Q2  Q1  Q1

Now because

Q1  Q1



Q2  Q2

Now suppose the two engines are coupled together so that I drives R backwards and suppose they use the same source and sink. The combination forms a self-acting machine in which I supplies external work W and R absorbs this amount of work in its reverse cycle. I in

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its cycle absorbs heat cycle, absorbs heat

Q1 from the source and gives up heat Q2 to the sink. R in its reverse

Q2 from the sink and gives up heat Q1 to the source.

The net result of the complete cycle of the coupled engines is given by Gain of heat by the source at

T1  Q1  Q1 T2  Q2  Q2

Loss of heat by the sink at External work done on the system = 0

Thus, the couple engines forming a self-acting machine unaided by any external agency transfer heat continuously from a body at low temperature to a body at a higher temperature. This conclusion is contrary to the second law of thermodynamics, according to which in a cyclic process heat cannot be transferred from one body to another at a higher temperature by a self-acting machine. Hence our assumption is incorrect and we conclude that no engine can be more efficient than a perfectly reversible engine working between the same temperatures. Second part: The second part of the theorem may be proved by the same arguments as before. For this purpose, we consider two reversible engines R1 and R2 and assume that R2 is more efficient than R1. Proceeding in the same way we can show that R2 cannot be more efficient than R1. Therefore, all reversible engines working between the same two temperatures have the same efficiency. Thus, the efficiency of a perfectly reversible engine depends only on the temperatures between which the engine work, and is independent of the nature of the working substance. 2.6.5 The internal combustion engine.

In this type of heat engine, the hot source is

produced by the explosion of a mixture of air and some combustible gas like petrol vapour, atmosphere forming the condense and air the working substance. It is called internal combustion engine, because in it heat is produced by the combustion of fuel inside the cylinder, whereas a steam engine, in which heat is produced outside the cylinder in a separate boiler, may be considered as external combustion engine. There is another essential difference between steam and internal combustion engines, viz., in the former heat is absorbed by the working substance at constant temperature, while in the latter at constant volume or constant pressure. The internal combustion engine is much more efficient than the steam engine, since the working substance can be heated to a much higher temperature.

There are two kinds of internal combustion

engines, known as the Otto engine and the Diesel engine; in the former petrol vapour is used as fuel, whereas in the latter a crude oil (Diesel oil) is used as the fuel. These two kinds are more scientifically distinguished by the different cycles employed, viz., the Otto cycle, in which the heat is absorbed by the working substance at constant volume and the Diesel cycle where the heat is absorbed at constant pressure. The majority of the internal combustion engines in use are of the

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Otto type. We shall now briefly study the two kinds of internal combustion engines and compare their action and relative merits.

2.7 The Otto engine. It consists of the usual cylinder and piston. The cylinder is provided with inlet valve I through which a mixture of petrol and air is sucked in and the exhaust valve E as shown in Fig. 2.7 (a). The operation of these valves are controlled by the motion of the piston.

Fig. 2.7 (a) Principle of working. The Otto Cycle first-employed successfully by Otto, a German engineer. in 1876, consists of four operations known as strokes. (i)

The charging stroke. The inlet valves are open and suitable mixture of air and petrol vapour is sucked into the cylinder C by the forward motion of the piston P.

(ii)

The compression stroke. All the valves are closed and the mixture is compressed adiabatically to about 1/5 of its original volume by the backward motion of piston. The o

temperature of the mixture is thereby raised to about 600 C. The process is called pre-heating. At the end of the compression stroke, the mixture is fired by a series of sparks from a sparking plug. The combustion produces a large amount of heat which o

raises the temperature of the gas to about 2000 C and a corresponding high pressure of about 15 atmospheres is produced instantaneously.

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48 Fig. 2.7.(b)

(iii)

The working stroke. The piston is now driven forward with great force owing to the high pressure of the gas, and the gas expands adiabatically to the original volume with a corresponding drop of temperature.

(iv)

The exhaust stroke. At the end of the third stroke, the cylinder is filled with a mixture of gases which is useless for further work. The exhaust valves are then opened and the spent up mixture is ejected out of the cylinder, the piston moving backward. After this scavenging of useless gases is completed, a fresh charge of air and petrol vapour is sucked in and a new cycle starts.

The four operations described above are represented on the indicator diagram in Fig. 2.7.(b) EA represents the charging stroke in which the mixture is sucked in at atmospheric pressure. The mixture at A is at a temperature T 1K, not mush higher than that of the surrounding. Let the specific volume of the working substance (which is air, not the petrol vapour, whose function is to merely heat the air by its combustion) at A be v2 say. AB represents the compression stroke. At o

B (pressure about 5 atmospheres, temperature T 2 about 600 C) the mixture is fired by a spark; the specific volume of the working substance remaining practically constant, say at v1, there is an o

instantaneous large increase of pressure and temperature to T 3 (15 atmospheres and 2000 C). This change is represented by BC. CD represents the working stroke during which the gas expands adiabatically to the original volume v2 with a consequent drop of temperature to T 1. At D the exhaust value is opened and the pressure falls to the atmospheric pressure as shown by DA. AE represents the exhaust stroke. The efficiency of this engine can be obtained as follows:Consider 1 gram molecule of the working substance. The heat Q 1 supplied to it by combustion is given by Q1 = Cv(T3-T2) Where Cv is the specific heat at constant volume. Q2 = Cv(T4-T1) Hence the efficiency

 =1-

Q2 T T  1 4 1 Q1 T3  T2

…(i)

since AB is an adiabatic,

T1v2 1  T2v1 1 where  is the ration of the two specific heats of the working substance. 

T1  v1    T2  v2 

 1

Similarly, since CD is an adiabatic

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49 T3 v1 1  T4 v2 1 T4  v1    T3  v2 

 1

…(iii)  1



T4 T1 T4  T1  v1      T3 T2 T3  T2  v2 

Using this result in (i), we get

v    1  1   v2  Since

 1

v2 is the adiabatic expansion ration , v1  1

1   1    where

…(iv)

1 is called the adiabatic compression ratio.  Thus we obtain for the efficiency of the Otto cycle an expression similar to that obtained

in the Carnot cycle. Then why not use the Carnot cycle, since it has maximum efficiency ? But in designing an engine, other practical considerations must be taken into account, besides theoretical efficiency. In the first place, it is desirable to have a machine, which is fast-working, i.e., in which the cycle of operations is completed as quickly as possible. This rules out the Carnot engine since it absorbs the required heat by a comparatively slow isothermal process, while the Otto engine acquires the heat by a much quicker adiabatic process. Secondly, the maximum pressure developed in the Carnot engine also forbids its use in practice. For instance, o

between the same two temperatures 2040 K and 340 K, the Carnot engine develops a maximum pressure of 1178 atmospheres, while the Otto engine develops only 27 atmospheres, although the efficiency of the former will be 83%, while that of the latte only 45%. This reduction in efficiency in the Otto engine is more than compensated by its design being much more compact, less heavy and stout than the Carnot engine, which has necessarily to be very bulky and heavy to withstand the very high pressure developed, so that its power output will be extremely small compared to bulk.

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2.8 The Diesel engine. Diesel, another German engineer, not satisfied with low efficiency of the Otto engine and investigating the possibility of arriving at the efficiency of the Carnot cycle by some contrivance or other, produced another engine, which was more efficient than the Otto type, though is could not reach that of the Carnot type. The efficiency depends on the expansion ratio  given by the relation.  1

1   1   

If  is increase also will increase. But in the Otto engine  cannot be increased beyond a certain value (about 5), otherwise the mixture will be spontaneously ignited during compression before the spark passes. Hence Diesel designed an engine in which no fuel is introduced until the compression is completed.

(a)

(b)

(c)

(d)

Fig. 2.8(i) The Diesel engine consists of a cylinder and piston, the cylinder being much more robust than in the Otto engine, since it has to withstand a higher maximum pressure (about 35 atmospheres). The cylinder is provided with inlet valve I for air, the working substance, another inlet N through which a heavy crude oil acting as fuel can be sprayed into the cylinder at the right moment and exhaust valves. The operations of these inlets and outlets are controlled by the motion of the piston. Principle of working. The working of the diesel engine is shown in Fig. 2.8.(i) The diesel cycles consists of five operations.

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Fig. 2.8(ii) (i)

The suction stroke. The air-valve I is opened and pure air is sucked into the

cylinder at constant pressure (atmospheric). This is represented by p 1A on the indicator diagram (Fig. 2.8(ii)). (ii)

The compression stroke. All the valve are closed and air is compressed to th

about 1/17 of its original volume. The temperature rises to about 1000 C and the pressure to about 35 atmospheres. AB on the indicator diagram represents this operation. (iii)

The injection of oil. The supply valve is opened and oil under pressure is

sprayed into the cylinder. The oil burns spontaneously as the temperature in the cylinder is much above its ignition point. The pressure during the process is maintained constant by regulating the oil supply. This represented by BC. Due to combustion the temperature rises o

to about 2000 C. At this stage the oil supply is cut off. (iv)

The working stroke. All the valves are closed. The gas mixture is allowed

to expand adiabatically, the piston moving forward, until the curve CD is described. Strictly speaking the mixture must be allowed to expand freely until the pressure drops to the original value represented by F; but as this would involve a large volume for the cylinder, in actual practice, the exhaust valve is opened at a stage represented by D and the pressure drops to A. (v)

The exhaust stroke. The useless spent up gas mixture is forced out. This is

represented by Ap1. The machine then become ready for the next cycle. The efficiency of this engine may be calculated as follows. We shall first consider the ideal cycle CFAB. Let the specific volumes and pressures of the working substance at C and F be v1 and v2 and p2 and p1 respectively. Let T1, T2, T3 and T4 be the temperatures at A, B, C and F respectively. Considering 1 gm. Molecule of the working substance the heat supplied by combustion of the fuel along BC is given by

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52 Q1 = Cv(T3-T2)

Where Cv is the specific heat at constant pressure. The heat rejected outside along FA is given by Q2 = Cv(T4-T1)

 =1-

The efficiency

Q2 T T  1 4 1 Q1 T3  T2

…(i)

Since A and B lie on the same adiabatic, 

p11 T3  p21 T2

T1  p2    T2  p1 

1



…(ii)

Similarly, since C and F lie on the same adiabatic 

p21 T3  p11 T4

T4  p2    T3  p1 

1



…(iii)

Comparing (ii) and (iii), we have  1

T1 T4 T4  T1  p2       T2 T3 T3  T2  p1  Again since C and F lie on the same adiabatic,

p2 v1  p1 v2 

p2  v2    p1  v1  

T4  T1  p2    T3  T2  p1 

1



1

v   2   v1 

Substituting this in (i), we get  1

  1  ( )

1

1  1   

…(iv)

Thus we have for the efficiency the same formula as for the Carnot or the Otto cycle. Relation (iv) gives the efficiency for the ideal cycle, where that could be rejected at constant pressure along FA, Actually, how-even, the exhaust valve is opened at D and heat is rejected at constant volume along DA. Hence Q2 = Cv(T4-T1),

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Taking temperatures at D and F to be practically the same. For the actual Diesel cycle,  1

C (T  T ) 1 1    1 v 4 1  1   Cv (T3  T2 )  



…(v)

The efficiency for the ideal cycle can be calculated and is found to be about 63%. The efficiency for the actual cycle is, therefore, somewhat lower, about 55%. Even with this lower value, the Diesel engine is more efficient than the Otto engine; also it consumes less and cheaper fuel. But it must be more robustly built to withstand higher pressure and is beset with much greater mechanical difficulties, which have been, however successfully overcome. All things considered, it must be said that the Diesel engine is a decided improvement on the Otto engine. During the past fifty years the internal combustion engines have been greatly developed and have replaced the steam engines, which consume much more fuel, in several cases. Some of the modern means of transport such as the automobile and aeroplane would not have been possible, but for these internal combustion engines. The Otto engine is usually employed for motor cars and aeroplanes while the Diesel engine for driving ships. Now-a-days, motor trucks and buses as well as locomotives also employ Diesel engines. The following table shows the differences between the Otto and Diesel engines. The Otto engine 1.

The diesel engine

The fuel vapours are introduced into the

cylinder, along with air

The fuel vapours are introduced into the cylinder after the air has first been sufficiently compressed.

Heart is taken in by the air at constant

Heat is taken in by the air at constant pressure.

volume 3. The adiabatic compression ratio is low

The adiabatic compression ratio is high

4. Its efficiency is not high.

Its efficiency is comparatively higher.

2.9 Entropy Definition: The entropy of a substance is that physical quantity which remains constant when the substance undergoes a reversible adiabatic process.

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Explanation: Consider two adiabatics AF and BE (Fig. 2.9) crossed by a number of isothermals at temperatures T1, T2, T3. Consider the Carnot cycle ABCD. Let Q1 be the heat absorbed from A to B at temperature T1, Let Q2 be the heat rejected from C to D at temperature T2.

Fig. 2.9 Then, from the theory of Carnot engine,

Q1 Q2 .  T1 T2 Similarly, consider the Carnot cycle DCEF. Q2 is the heat absorbed at temperature T 2 and Q3 heat rejected at temperature T3. Then,

Q2 Q3  T2 T3 

Q1 Q2 Q3    constant. T1 T2 T3 Thus, if Q is the amount of heat absorbed or rejected in going from one adiabatic to

another along any isothermal at temperature T, then

Q = constant T This constant ratio is called the change in entropy in going from the adiabatic AF to the adiabatic BE. If a system absorbs a quantity of heat dQ at constant temperature T during a reversible process, the entropy increases by

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dQ . T

Similarly, if a substance gives out a quantity of heat dQ at termperature T, during a reversible change, then its entropy decreases by

ds  Unit of entropy is JK

dQ T

-1

For an adiabatic change, we have dQ=0.

 ds = 0

Thus, there is no change of entropy during a reversible adiabatic process.

2.10 Change of Entropy in a Reversible Process (Carnot’s Cycle) Consider a reversible Carnot cycle ABCD (Fig. 2. 10). (i)

In the isothermal expansion from A to B, the working substance absorbs an amount of heat Q1 at a constant temperature T1.

increase in entropy of working  Q1  . substance from A to B  T1 (ii)

During the adiabatic expansion from B to C, there is no change in entropy.

Fig. 2.10 (iii)

During the isothermal compression from C to D, the working substance gives out a quantity of heat Q2 at a constant temperature T2.

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Decrease in entropy of working substance from C to D =

(iv)

Q2 . T2

During the adiabatic compression from D to A, there is no change in entropy.

The net change in entropy of the working  Q1 Q2   . substance during the cycle ABCD  T1 T2



For a reversible cycle,

Q1 Q2  T1 T2 Q1 Q2 =0  T1 T2



Thus, the total change of entropy is zero during a Carnot’s cycle. Entropy change in a reversible cycle is zero. Change in Entropy in an Irreversible Process. Consider an irreversible process like conduction or radiation of heat. Suppose a body at a higher temperature T1 conducts away a small quantity of heat dQ to another body at a lower temperature T2. Then, decrease in entropy of the hot body = dQ/T 1 Increase in entropy of the cold body = dQ/T 2 

the net increase in the entropy of the system

 dS 

dQ dQ  T2 T1

dS is always positive since T 1 > T2. Hence there is an increase of entropy. Similarly, there is an increase of entropy during the loss of heat by radiation. Therefore, generally, the entropy of a system increases in all irreversible processes. This is called the law of increase of entropy.

2.11. Temperature – Entropy diagram. The state of a substance may be represented by points plotted with temperature as ordinates and entropies as abscissae. This is the TS diagram. Here the isothermals are horizontal straight lines (parallel to S – axis). The adiabatics are vertical straight lines (parallel to T – axis)

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Consider the Carnot cycle ABCD (Fig. 2.11a)

Fig. 2.11 (i)

From A to B, heat energy Q1 is absorbed at temperature T1. The increase in entropy S1 takes place from A to B (Fig 2.11 (b))

S1 

Q1 T1

…(1)

(ii)

From B to C there is no change in entropy. Fall of temperature if T 1-T2.

(iii)

From C to D, there is decrease in entropy (S2) at constant temperature T2.

S2  (iv)

Q2 T2

…(2)

From D to A, there is no change in entropy but the temperature increases from T 2 to T1. The area ABCD in the T-S diagram

S1  S2  

 S1 (T1  T2 )  S2 (T1  T2 )

Q1 Q2 Q1  Q2   T1 T2 T1  T2

Area of figure ABCD in T-S diagram



Q1  Q2 (T1  T2 )  Q1  Q2 T1  T2

The area ABCD represents the energy converted to work. The efficiency of the engine,

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…(3)

…(4)

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 

Heat energy converted int o work Total heat absorbed

S1 (T1  T2 ) T1  T2  T1S1 T1

2.12 Thermodynamic variables The thermodynamic state of a substance is specified by some of its properties like pressure, volume, temperature, internal energy and entropy. These properties undergo a change when the system passes from one state to another.

These variables are known as

thermodynamic variables or co-ordinates. These are called macroscopic co-ordinates. They require the specification of few measurable properties of the system and do not require the knowledge of microscopic structure of matter composing the system. 2.12.1 Extensive and intensive variables An extensive variable of a system is a macroscopic parameter which describes a system in equilibrium and which has a value equal to the sum of its values in each part of the system. The extensive variable depends upon the mass or the size of the substance present in the system. Examples Mass, volume, internal energy, entropy, length, area, heat-capacity, electric charge, magnetization etc. An intensive variable of a substance is a macroscopic parameter which describes the system in equilibrium and has the same value in any part of the system. It is independent of mass or size of ht system. It is a characteristic of the substance present tin the system. Examples Pressure, temperature, viscosity, refractive index, density, specific volume, magnetic induction, surface tension, electromotive force etc. Distinction between Extensive and Intensive variables Consider a homogenous system in equilibrium. Suppose the system is divided into many parts. If a macroscopic variable x of the system has the values x 1, x2, x3,……. in each of these parts of the system respectively, then (i)

x is said to be extensive variable, if x = x1 + x2 + x3 + …….., and

(ii)

x is said to be intensive variable, if x = x1 = x2 = x3 = …………..

An extensive variable may become an intense variable by specifying unit amount of substance.

Thus, mass and volume are extensive variables by density

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M    and V 

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V    are intensive variables. Similarly heat capacity is extensive M 

specific volume

variable but specific heat is an intensive variable. The extensive and intensive variables of some systems are given in Table 1. Table 1 System

Extensive Variable

Intensive Variable

Chemical system

Volume

Pressure

Stretched wire

Length

Tension

Surface film

Area

Surface tension

Electric cell

Charge

E.M.F.

Paramagnetic solid

Magnetisation

Magnetic intensity H

2.12.2 Maxwell’s Thermodynamical Relation (General Relationship) From the first and second law of thermodynamics, Maxwell was able to derive six fundamental thermodynamical relations. The state of a system can be specified by any pair of quantities, viz.

pressure (P), volume (V), temperature (T), and entropy (S).

In solving any

thermodynamical problem, the most suitable pair is chosen and the quantities constituting the pair are taken as independent variables. From the first law of thermodynamics.

 Q  dU   W  Q  dU  PdV dU   Q  PdV For the second law of thermodynamics,

dS 

Q

T  Q  TdS or Substituting this value of Q, we get U = TdS – PdV.

(1)

Considering U, S and V to be function of two independent variables x and y [ here, in general, x and y can be any two variables out of P.V.T. and S].

 U   U  dU    dy  dx    x  y  y  x  S   S  dS    dx    dy  x  y  y  x

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and

 V   V  dV    dy  dx    x  y  y  x

Substituting these values in equation (1), we get

 S    V   U   S   V    U   dy  T   dx    dy   P   dy    dx    dx    y  x  y  x  y  x  y       x   y     y y     S    S   V    V    T    P    dy   dy  T    P   y  y  x  x    x  y    x  x  Comparing the coefficients of dx and dy, we get

 U   S   V     T    P   x  y  x  y  x  y

…(2)

 U   S   V     T    P   y  x  y  x  y  x

…(3)

Differentiating equation (2) with respect to y and equation (3) with respect to ġ,

 2U  T   S   2 S  P   V   2V   T   P        y.x  y  x  x  y y.x  y  x  x  y V .x and

 2U  T   S   2 S  P   V   2V      P    T x.y  x  y  y  x x.y  x  y  y  x x.y

The change in internal energy brought about by changing V and T, whether V is changed by dV first and dT later or vice versa is the same. It means dU is a perfect differential. 

 2U  2U and  y.x x.y  T   S   2 x  P   V   2V  T   P         y.x  y  x  x  y y.x  y  x  x  y  2 S  P   V   2V  T   S    T   P        x.y  x  y  y  x y.x  x  y  y  x

Since dS and dV are also perfect differentials, we have

 2S  2S V V  and  x.y y.x x.y y.x Equation (4), therefore reduces to

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…(4)

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 T   S   P   V   T   S   P   V                   y  x  x  y  y  x  x  y  x  y  y  x  x  y  y  x

…(5)

This is the general expression for Maxwell’s thermodynamical relations. In place of the independent variables ġ and y, any two of the four variables S, T, P and V can be substituted so that there may be one mechanical variable (P or V) an one thermal variable (S or T). Thus, there may be four sets of possible substitutions (S, V), (T, V), (S, P) and (T, P), providing the four Maxwell’s thermodynamical relations. First Relation: Put x = S and y = V in equation (5), so that

S V  1, 1 x y S V  0, 0 y x

and

Substituting in equations (5), we get

 T   P        x  y  y  x But

y  V (y = V) and s  S (as x = S). Hence

 T   P        V  S  S V

…(6)

This is Maxwell’s first thermodynamical relation. Second Relation: Put x = T and y = V in equation (5), then

T V  1, 1 x y

and

T V  0, 0 y x

Substituting in equation (5), we get

 S   P       V T  T V

…(7)

This is Maxwell’s second thermodynamical relation Third Relation: Put x = S and y = P, in equation (5) then

S P S P  1,  1,  0, 0 x y y x

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Substituting these in equation (5), we get

 T   V       P  S  S  P

…(8)

This is Maxwell’s third thermodynamical relation Fourth Relation: Put x = T and y = P, in equation (5) gives

T P T P  1,  1,  0and 0 x y y x Substituting these values in equation (5), we get

 S   V        P T  T  P This is Maxwell’s fourth thermodynamical relation.

…(9) These are the four Maxwell’s

fundamental thermodynamic relations. Further there are two more relations within the mechanical variables (P,V) and thermal variables (T,S). Fifth Relation: Put x = P and y = V

P V P V  1,  1,  0, and 0 x y y x Substituting these values in equation (5), we get

 T   S   T   S          1  P V  V  P  V  P  P V

…(10)

Sixth Relation: Put x = T and y = S

T S T S  1,  1,  0and 0 x y y x Substituting in equation (5), we get

 P   V   P   V          1  T  S  S T  S T  T  S

…(11)

Out of these six thermodynamic relations, the one suited for a particular problem is used and the problem is solved. Let us see, some of the important applications of these Maxwell’s thermodynamic relations.

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2.12.3 Applications of Maxwell’s Thermodynamic Relations 2.12.3.1 Specific Heat Equation The specific heat at constant pressure is given as

 Q   S  CP    T    T  P  T  P

(

Q=T.S)

and the specific heat at constant volume is

 Q   S  CV    T    T V  T V Now, if the entropy S is regarded as a function of T and V and since dS is a perfect differential,

 S   S  dS    dT+   dV  T V  V T 

 S   S   T   S   V             T  P  T V  T  P  V T  T  P

 S   S   S   V  T  T   T      T  P  T V  V T  T  P

But

 S   P      , from Maxwell’s second relation (Eqn. 7)  V T  T V



 S   S   P   V  T  T   T      T  P  T V  T V  T  P  P   V  CP  CV  T      T V  T  P

…(12)

(a) For a perfect gas: The equation of state is PV = RT 

R R  P   V     and     T V V  T  P P

Hence, equation (12) becomes 2 2  R  R  R T R T CP  CV  T      V  P  PV RT



(

PV =RT)

CP  CV  R (b) Van der Waals: The equation of state is

a    P  2  (V  b)  RT V   where a and b are constants.

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…(13)

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a  RT  P 2   V  (V  b)  Differentiating with respect to T at constant volume, we get

R  P      T V (V  b) and differentiating with respect to T at constant pressure we get

0

2a  V  RT  V  R     3  2  V  T  P (V  b)  T  P (V  b)

2a  R  V   RT  3    2  T  P  (V  b) V  (V  b)

 R   (V  b)   V       2a   T  P  RT  (V  b)2  V 3   

Substituting these values of

 P   V    and   in equation 12), we get  T V  T  P

 R  R  T (V  b)  (V  b)  R   CP  CV    2a (V  b) 2  RT 2a  1 3 .  (V  b)2  V 3  V RT   Neglecting b in comparison to V,

R R 2a   CP  CV    R 1   2 2a 2a V  VRT  1 1 3 . VRT V RT

1

Expanding binomially and neglecting a in higher powers, as a is also small as compared to V, we have

2a   CP  CV  R 1    VRT 

…(14)

[Here R is in heat units]

2.13 The T. dS Equations 1. First T.dS Equation. The entropy S of a pure substance can be taken as a function of temperature and volume.

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

 S   S  dS    dT    dV  T V  V T

Multiplying both sides by T

 S   S  T .dS  T   dV  T   dV  T V  V T  But

 S  CV  T    T V

and from Maxwell’s relations

 S   P       V T  T V Substituting these values in equation (ii)

 P  TdS  CV dT  T   dV  T V

…(1)

Equation (2) is called the first T,dS equation. 2. Second T.dS Equation. The entropy S of a pure substance can also be regarded as a function of temperature and pressure.



 S   S  dS    dT    dP  T  P  P T

Multiplying both sides by T

 S   S  T .dS  T   dT  T   dP  T  P  P T But

 S  CP  T    T  P

and from Maxwell’s relations

 S   V        P T  T  P Substituting these values in equation (v)

 V  T .dS  CP dT  T   dP  T  P

…(2)

Equation (2) is called the second T dS equation. 2.13.1 Clapeyron’s Latent Heat Equation using Maxwell’s Thermodynamical Relations

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From Maxwell’s second thermodynamical relation

 S   P       V T  T V Multiplying by T, we get

 S   P  T  T    V T  T V  Q   P    T    V T  T V Here

( T S  Q)

 Q    represents the quantity of heat absorbed per unit increase in volume at constant  V T

temperature. This quantity of heat absorbed at constant temperature is the latent heat (L). Thus,

Q = L and V  V2  V1 , for unit mass of a substance.

Substituting,

 L   P    T    T V  V2  V1 T

L dP T V2  V1 dT

dP L  dT T V2  V 1

…(3)

This is Clapeyron’s latent heat equation. 2.13.2 Clapeyron Latent Heat Equation using Carnot’s Cycle Consider the isothermals FBAE at temperature T+dT and GCDH at temperature T. Here EA and HD show the liquid state of the substance. At A and D, the substance is purely in the liquid state (Fig. 2.13.2) From A to B or D to C, the substance is in transition from the liquid to the gaseous state and vice versa. At B and C, the substance is purely I the gaseous state. From B to F or C to G, the substance is in the gaseous state. Join A to D and B to C by dotted lines.

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Fig. 2.13.2 The cycle ABCD represents a complete cycle and Carnot’s theorem can be applied. Suppose the volume at the point A is V1 and temperature is T + dT. The pressure is just below its saturation pressure and the liquid begins to evaporate and at the point B the volume is V 2. The substance is in the vapour state. Suppose the mass of the liquid at B is one gram the amount of heat absorbed is Q1, here Q1 = L + dL, where L + dL is the latent heat of the liquid at temperature (T + dT). At the point B, the pressure is decreased by dP.

The vapour will expand and its

temperature falls. The temperature at C is T. At this pressure and temperature T, the gas beings to condense and is converted into the liquid state. At the point D, the substance is in the liquid state. From C to D, the amount of heat rejected (given out) is Q2. Here Q2 = L where L is latent heat at temperature T. By increasing the pressure a little, the original point A is restored. The cycle ABCDA is completely reversible. Applying the principle of the Carnot’s reversible cycle.

Q1 Q2  T1 T2 or

Q1 T1  Q2 T2 Q1  Q2 T1  T2  Q2 T1

Here,

Q1  L  dL, Q2  L, T1  T  dT , T2  T Q1  Q2  L  dL  L  dL T1  T2  T  dT  T  dT

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dL dT  L T The area of fig.

ABCD  Q1  Q2  dL dP(V2  V1 )

dP(V2  V1 ) dT  L T dP L  T T (V2  V1 )

…(4)

This is called the Clapeyron’s latent heat equation. Applications 1. Effect of change of pressure on the melting point. When a solid is converted into a liquid, there is change in volume. (i)

If V2 is greater than V1.

(ii)

dP is a possible quantity. It means that the rate of change of pressure with respect T to temperature is positive. In such cases, the melting point of the substance will increase with increase in pressure and vice versa.

(iii)

If V2 is less than V1.

dP is a negative quantity. It means that the rate of change of pressure with respect to T temperature is negative. In such cases, the melting point of the substance will decrease with increase in pressure and vice versa. In the case of melting ice, the volume of water formed is less than the volume of ice taken, Hence V2 < V1. Therefore, the melting point of ice decreases with increase in pressure. Hence ice will melt at a temperature lower than zero degree centigrade at a pressure higher than the normal o

pressure. Ice melts at 0 C only at a pressure of 76 cm of Hg. 2. Effect of change of pressure on the boiling point. When the liquid is converted in to a gaseous state, the volume V 2 of the gas is always greater than the corresponding volume V1 of the liquid i.e., V2>V1. Therefore,

dP is a +ve quantity. T

With increase in pressure, the boiling points of a substance increases and vice versa. The liquid will boil at a temperature under reduced pressure. In the case of water, the boiling o

points increases with increase in pressure and vice versa. Water boils at 100 C only at 76 cm of the Hg pressure. In the laboratories, while preparing steam, the boiling point is less

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than 100 C because the atmospheric pressure is less than 76 cm of Hg.

In pressure

cookers, the liquid boils at a higher temperature because the pressure inside is more than the atmospheric pressure. 2.14 Clausius – Clapeyron’s Equation (First Latent Heat Equation) The equation

dP L was first by Clapeyron using Carnot’s reversible cycle.  dT T (V2  V1 )

Therefore, it is some times called Clapeyron’s equation. Whenever there is a change of state, either from a solid state to liquid state or from liquid state to vapour state, the temperature remains constant; as far as the change takes place. This temperature (either melting point or boiling point) depends upon the pressure and is a characteristic of each substance. The above relation shows how melting or boiling points vary with pressure and was derived by Clapeyron by applying the second law of thermodynamics. Later Clausius obtained the same relation by using the Maxwell’s thermodynamic relations, as under: Derivation: From second thermodynamical relation

 S   P       V T  T V Multiplying both sides by T, we have

 S   P  T  T    V T  T V But 

T S  Q

 Q   P    T    V T  T V The quantity

…(1)

 Q    represents the quantity of heat absorbed or liberated per unit  V T

change in volume at constant temperature. As there is a change in volume due to the heat absorbed at constant temperature, the heat represents the latent heat used when a substance changes from solid to liquid (melting) or liquid to vapour (boiling) state when the temperature remains constant, during the change of state. If L is the quantity of heat required to change the state of a unit mass of the substance, V2 and V1 the corresponding specie volumes (Volume per unit mass) then

Q  Land V  V2  V1.

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L  Q      V T V2  V1

Hence

Therefore,

L  P  T   V2  V1  T V

dP L  dT T (V2  V1 )

…(2)

Where dT is the change in melting point or boiling due to a change in pressure dP. This is Clausius – Clapeyron latent heat equation.

Solved Problems: Example 1: Find the efficiency of the Carnot’s engine working between the steam point and the ice point. o

Sol. T1 = 100 C = 373K, T2 = 0 C = 273K, η = ?



T1  T2 373  273   0.2681  26.81% T1 373

Example 2: A Carnot’s engine is operated between two reservoirs at temperatures of 450 K and 350 K. If the engine receives 4200 joules of heat from the source in each cycle, calculate the amount of heat rejected to the sink in each cycle. Calculate the efficiency of the engine and the work done by the engine in each cycle. Sol: T1 = 450K;

T2 = 350K;

Q1= 4200J;

Q2 = ?

Q1 Q2  T1 T2 

Q  4200 Q2   1  T2  X 350  3267 J . 450  T1 

  1

T2  350   1    0.2222  22.22% T1  450 

Work done by the engine in each cycle

 Q1  Q2  4200  3267  933J Example 3: A Carnot’s engine whose temperature of the source is 400 K takes 840 J of heat at this temperature and rejects 630 J of heat to the sink. What is the temperature of the sink? Also calculate the efficiency of the engine. Sol: Q1= 840 J,

Q2 = 630 J,

T1= 400 K,

Q1 Q2  T1 T2

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T2 =?

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T   400  T2   1  Q2    630  300 K .  840   Q1 

  1

T2  300   1    0.25 T1  400 

%efficiency  25% o

Example 4: A Carnot’s whose low temperature reservoir is at 7 C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased? o

Sol: I Case, η = 50% = 0.5; Now,

  1

or 0.5 = 1 -

or 0.7  1 

T1 = ?

T2 T1 280 or T1 =560K T1

II Case, η = 70% = 0.7;

  1 

T2 = 7 C = 273 + 7 = 280K;

T2 = 280K;

T1 = ?

T2 T1 

280 orT1  933.3K T1

Increase in temperature

 T1  T1  933.3  560  373.3K th

Example 5: A Carnot engine converts 1/6 of heat input into work. When the temperature of the o

sink is reduced by 62 C, the efficiency is doubled. Find the temperature of the source and the sink. Sol. I Case. 



II Case   

T1  T2 1  orT1  1.2T2 T1 6 T1  (T2  62) 1  2X T1 6

T1  T2  62 1  T1 3

2T1  3T2  186

2 X 1.2T2  3T2  186



T2  310 K  37o C T1  1.2T2  1.2 X 310  372 K  99o C. IMTSINSTITUTE.COM

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Example 6: A Carnot’s engine working as a refrigerator between 260 K and 300 K receives 2100 J of heat from the reservoir at a lower temperature. Calculate the amount of heat rejected to the reservoir at a higher temperature. Calculate also the amount of work done in each cycle to operate the refrigerator. Sol:

T1 = 300K,



Q1 Q2  T1 T2



Q1 

T2 = 260 K,

Q2 = 2100J;

Q1 = ?

T1 300 XQ2  X 2100  2423J T2 260

Work done = W =Q1 – Q2 = 2423 – 2100 = 323 J. -2 o Example 7: Calculate the change in entropy when 10 kg of ice at 0 C is converted into water at 5

-1

the same temperature. Given that the specific latent heat of fusion of ice is 3.36 X 10 Jkg . -2

Sol. Quantity of heat given to 10 kg of ice to convert it into water without any change of temperature is -2

dQ = mL = 10 X (3.36 X 10 ) = 3.36 X10 J. This process takes place at a constant temperature T = 273 K.

dQ 3.36 X 103 Increase in entropy = dS =   12.31JK 1 T 273 o

Example 8: Calculate the change in entropy when 5 kg of water at 100 C is converted into steam at the same temperature. o

Sol: Heat absorbed by 5 kg of water at 100 C when it is converted into steam at 100 C = 4

dQ = 5 X (226 X 10 ) J=1.13 X 10 J.

dQ 1.13 X 107 Increase in entropy = dS =   30295 JK 1. T 373 Example 9: Calculate the change in temperature when helium gas suffers Joule-Thomson o

expansion at -173 C, the pressure difference between the two sides of the plug being 20 atmospheres. Does the gas show a heating effect or a cooling effect in this expansion at – o

-1

173 C? Given : R = 8.3 JK mol and for helium Van der Waals constants are a = 0.0341 atm 2

lit /mole and b = 0.02357 litre/mole. 5

Sol: Here, P1 – P2 = 20 atmospheres = 20 X 1.01325 X 10 Pa; o

T = -173 C = 100K; 2

-6

a = 0.00341 atm lit /mole = 0.0341 X [(1.01325 X 10 )X10 ] -4

-2

Nm mole = 0.003455 Nm mole ; 3

-1

b = 0.02357 litre/mole = 0.00002357 m mol .

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5 R  20.75 Jk 1mole 1 ,  T  ? 2  P  P   2a  T   1 2    b  C   RT  p   (20X1.01325X105 )  2 X 0.003455  =  0.00002357   20.75  8.3 X 100  = -1.489K. Cp 

Since T is negative, there is a rise in temperature equal to 1.489 K. Now,

Ti 

2a 2 X 0.003455   35.43K  237.6o C Rb 8.3 X (0.00002357) o

Since the initial temperature is -173 C which is higher than Ti, a heating effect will take place. -2

Example 10. The Van der Waals constants for hydrogen are a = 2.45 X 10 Nm /mole and b = -5

2.67 X 10 m /mole.

Calculate temperature of inversion, critical temperature and Boyle

temperature. Sol:

Temperature  2a 2 X (2.45 X 102 )  T    220 K .  i of inversion  Rb 8.314 X (2.67 X 105 )  8a 8 X (2.45 X 102 )   32.7 K   TC  temperature  27 Rb 27 X (2.67 X 105 )8.314 Boyle  a 2.45 X 102  T    110.4 K .  B temperature  Rb 8.314 X (2.67 X 105 )

Critical

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UNIT III CONDUCTION AND RADIATION 3 Introduction Conduction of heat is defined as the flow of heat through an unequally heated body from places of higher temperature to places of lower temperature without the actual motion of the particles. Conduction of heat takes place in solids, liquids and gases. When one end of a metal rod is heated, the other end soon becomes war because heat is conducted through the rod. When one end of a metal rod is heated, the molecules at the heated end vibrate with higher amplitude and hence with greater kinetic energy. They transmit the heat energy from one particle to the next and so on. However, the particles remain in their mean equilibrium position. 3.1 Coefficient of Thermal Conductivity Consider a slab of material of length x and area of cross section A (fig 3.1). The opposite faces of the slab are maintained at temperatures 1 and 2 where 1>2. Assume that no heat is lost from the sides of the slab.

Fig. 3.1 The quantity of heat Q conducted from one face to the other is (i)

Directly proportional to the area of cross- section A.

(ii)

Directly proportional to the difference of temperature between the ends (1-2),

(iii)

Directly proportional to the time of conduction t, and

(iv)

Inversely proportional to the length x.

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(or)

Q

A(1  2 )t     or Q  KA  1 2  t x  x 

Here K is a constant called the coefficient of thermal conductivity of the material of the slab. The quantity

(1  2 ) / x represents the rate of fall of temperature with respect to

distance. It is called the temperature gradient. Consider a slab of infinitesimal thickness dx, across which there is a temperature difference

d

Then the temperature gradient is

d . dx

d is negative, since it represents rate of fall of temperature with distance. dx Q   KA(d / dx)t

 If

 1m2 ,

d  1and t  1 second, then Q = K. dx

Definition. The coefficient of thermal conductivity of a mater is defined as numerically equal to the quantity of heat conducted per second normally across unit area of cross- section of the material per unit temperature gradient. The unit of K is W

m-1

-1

K .

Thermal Diffusivity. Let C be the specific heat capacity of the material and p its density. Then the ratio K/p C is called the thermal diffusivity. When one end of a metal bar is heated, diffusivity determines the rate at which the temperature of any part of the material changes before a steady state is reached. Thermal conductivity of good conductors.

The coefficient of thermal conductivity can be

determined by the direct application of the relation.

Q

KA(1   2 )t r

3.2 Forbes’ method. This is one of the earliest methods used for determining the absolute conductivity of metals devised by Forbes. Principle. A long bar of the metal of uniform area of cross-section under test is heated at one end till a steady state is reached. The amount of heat passing per second across the crosssection at any point B at a distance x from the hot end is equal to the heat lost by radiation by the bar beyond the section B and is given by

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Fig. 3.2(a)

 d   KA    dx  where K is the conductivity, A the cross-sectional area and

 d    , the temperature gradient at  dx  B

B. The amount of heat lost in radiation by the bar beyond the section B 

end

 B

 and s are the density and specific heat of the material of the bar and of temperature.



 d   KA     dx  B

end

d

  As dt dx, B

d dx dt B  d     dx  B end

K

s 

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 As

d dx, where dt

d is the rate of charge dt

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 d  Hence, in order to determine K,   , and  dx  B

end

 B

d dx, are to be found, since  and s dt

are known. The experiment, therefore, consists of two parts, viz., static to determine

and dynamic to determine

 d    and then  dt 

end

 B

 d    ,  dx  B

d dx. dt

Static experiment. The apparatus consists of a long metallic bar of uniform section, the length being such that when one end of it is heated steadily, the other end will be practically at the room temperature. Forbes used a bar of wrought iron 8 ft. long 1

1 inch square section. 4

One end of the bar is heated by immersing it in a high constant temperature bath of molten lead or silver. In the laboratory experiment a steam chamber is used to heat the end. The bar is provided with a series of holes at regular intervals into which thermometers are inserted with a little mercury in the holes for good contact, in order to record the temperature at different points along the bar when the steady state is reached.

A non-conducting screen S protects the

thermometers from the direct heat of the bath. After a long time of heating the steady state is reached when the thermometers indicate constant readings. Forbes took nearly six hours to obtain the steady state. The temperatures indicated by the thermometers and their distances from the hot end are recorded.

The

temperature distribution is indicated by the dotted curve in Fig. 3.2 (a) and obeys the relation  = 0e . A graph is plotted between the excess of temperature  above that of the surrounding and -x

the distance x from the hot end. A curve of the form shown in Fig. 3.2 (b). similar to the dotted curve mentioned above is obtained. This is known as the static experiment, since it deals with the steady state of heat flow.

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78 Fig. 3.2 (b)

Drawing a tangent to the curve at a point corresponding to B near the hot end, the value of

 d    is obtained, since  dx  B AB  d   tan    =  dx  B BC (i)

Dynamic experiment. A short piece of the original bar, on of a similar one with the same cross-section, is heated to a temperature attained by the hot end

Fig. 3.2(i)

Fig. 3.2(ii)

in the static experiment and then exposed to air to cool in exactly the same surroundings as the static bar. When the temperature of the bar is a little higher than that of B chosen in the static part, a thermometer T is inserted in a hole in the bar with a little mercury. The readings of the temperature at regular intervals are noted and a cooling curve is plotted as shown in Fig. 3.2(ii).

This is known as the dynamic experiment as it deals with the variation of

temperature with time. By drawing tangents to this curve corresponding to the temperatures indicated by the various thermometers in different holes along the bar in the static experiment, the values of

d d at different distance x along the bar are obtained. A graph is plotted between and x dt dt as shown in Fig. 3.2 (iii). The curve is prolonged to meet and corresponding to the point B a point is located on the curve. The area of the shaded portion, gives the value of

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79 end

 B

d .dx dt

The area is measured either with the help of the planimeter or from the graph directly. Hence the value of K is obtained from the equation.

Fig. 3.2 (iii)

 s X area of shaded portion tan 

It may be noted that different points can be taken as B and an average value of K can be found. The diffusivity K/s is also determined in this experiment for which it is not necessary to assume the values of  and s. This method is not only tedious on account of the long hours required for the static part and of the data to be gathered for drawing the three graphs, but also fails to given accurate results, due to several sources of error, such as: (i) the specific heat does not remain constant at different temperatures as assumed in the method, (ii) the distribution of temperature in the bar in the static and dynamic parts are different. The one great merit of the method is that it seeks to determine the conductivity absolutely, basing itself on the fundamental definition of the phenomenon, instead of using any theoretical equations founded on the hypothesis of constant conductivity and Newton’s law of cooling. 3.3 Lee’s and Charltons method for bad conductors. The apparatus designed is used in the laboratory experiment and consists of a thick metallic disc A of copper or brass nickel-polished which is suspended by means of three strings from a heavy retort-stand so that its top face is horizontal. On this is placed the disc D of the material having the same diameter as that of A. Above this rests a hollow cylindrical steam chamber C with a heavy base so that the disc D is well pressed between A and C. A thermometer T 1 is placed near the bottom of the cylindrical steam vessel C and a thermometer T 2 is placed in a hole bored in the disc A. To ensure food contact with the thermometers a little of mercury is placed in the hole.

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Fig. 3.3. Steam is passed through the cylindrical vessel for a pretty long time so that a steady state is reached, i.e., the thermometers T 1 and T2 show constant temperatures. In this condition the rate at which heat is conducted across the specimen disc is equal to the rate at which the heat is radiated through the exposed surface of the lower disc. Let 1 and 2 be the temperatures registered by the thermometer T 1 and T2 respectively in the steady state. If Q is the heat conducted per second across the disc of the material, then

Q

KA(1   2 ) cals/sec. d

…(i)

If m is the mass of the disc A, s the specific heat of the material, then rate of radiation

 d    from the disc A at temperature 2 is given by  dt  Q  m.s.

d  r  2l    dt  2r  2l 

…(ii)

where r is the radius of the slab and l its thickness. The quantity within the bracket gives the fraction of the total radiation which refers to the side and top surface of the slab. 

m.s.

K . A(1   2 ) d  r  2l    = Q d dt  2r  2l 

K . r 2 (1   2 )  d

To measure

…(iii)

d the experimental disc is removed and the steam chamber C is directly dt

placed over the disc A. Steam is passed till the temperature of the disc is about 10 above 2 . o

The chamber C is then removed and the disc A is allowed to cool and the temperature is noted at

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regular intervals of time say 30 seconds, till the temperature falls to about 10 below 2 . A graph o

is then plotted between temperature and time. A tangent is drawn at the point P corresponding to 2 . The slope of the curves gives the value of

d corresponding to the temperature 2 . dt

Thus K can be determined from the relation (iii). 3.4 Thermal Radiation Introduction If a hot body is placed in vacuum, it cannot lose heat by conduction or convection. It is found that this body still loses heat by radiating energy from it. These radiations are called thermal radiations. The average wavelength of thermal radiation is greater than that of visible light. Radiation is the process in which heat is transferred from one place to another directly without the aid of intervening medium. Heat from the sun reaches earth due to radiation without affecting the intervening medium. Total emissive power. This is the radiant energy of all wavelengths emitted per unit time -2

-1

from unit surface area of a radiating body. It is denoted by E. Unit : Jm s . Spectral emissive power. The spectral emissive power of a body at a particular wavelength is the radiant energy emitted per unit time per unit surface area of the body within a unit wavelength range. It is denoted by E. 

E   E d  0

Black body. A perfectly block body is one which absorbs all the heat radiations, of whatever wavelength, incident on it. When such a body is heated it emits radiations of all possible wavelengths. It is found that inside a black body, the nature of radiation becomes independent of the shape, size and material of the body and depends only upon the temperature of the body. This radiation is called black body radiation. Also the black body completely absorbs heat radiations of all wavelengths. So a perfectly black body is a goods absorber as well as a good radiator. Characteristics of a black body radiation A perfectly black body absorbs radiations of all wavelengths which fall on it. According to kirchhoff’s law, such a body will emit all wavelengths when heated to a suitable high temperature. The radiation emitted by a black body is known as black radiation of full radiation or total radiation.

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The wavelengths of the radiation emitted depends only on the temperature and is independent of the material of the body. Absorptive power. It is the ratio of the amount of heat absorbed in a given time by time by the surface to the amount of heat incident on the surface in the same time. Emissive power. It is the ratio of the amount of heat radiations emitted by unit area of a surface in one second to the amount of heat radiated by a perfectly black body per unit area in one second under identical conditions. Kirchhoff’s law. The ratio of emissive power and the absorptive power of all bodies is the same and is equal to the emissive power of a perfectly black body. 3.4.1 Fery’s black body: Fery’s black body is shown in Fig.3.4.1. it is in the form of double walled conducting sphere. The inter space between the walls is evacuated to prevent loss of heat by conduction and convection. It has a small hole O and a conical projection P just opposite the hole O. It is lampblacked inside and nickel polished outside.

Fig.3.4.1 When the radiation is incident on the hole, it passes inside the enclosure. The radiation suffers multiple reflections inside the enclosure and is completely absorbed. There is hardly any possibility of the radiation to get out except by direct reflection from the surface opposite to hole. This is eliminated by the projection P. In this way, any radiation that enters the body does not escape and is absorbed. Therefore, the body appears as perfectly black. Here the hole behaves very closely like a black body. If the enclosure is heated to a definite temperature, it is filled with black radiation having all possible wavelengths. 3.4.2 Wien’s black body: Wien’s black body is shown in Fig 3.4.2. It consists of a long metallic true C blackened inside and surrounded by concentric porcelain tubes P, P. The tube is heated by electric current

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passing in the coil (Heating coil) wound around it. The temperature of the central part of the tube is measured by thermocouple T. Heat radiations emerge out of the hole O. The radiation form the inner chamber can be limited with the help of diaphragms shown by dotted lines. The hold acts as a black body radiator.

Fig 3.4.2 3.4.3 Energy Distribution in Black Body Radiation: If the radiation emitted by a black body at a fixed temperature is analysed by means of a suitable spectroscopic arrangement, it is found to spread up into a continuous spectrum. The total energy is not distributed uniformly over the entire range of the spectrum. The distribution of energy in black body radiation for different wavelengths and at various temperatures was determined experimentally by Lummer and Pringsheim. The Experimental arrangement is shown in Fig. 3.4.3 (a). The radiation from the black body is rendered into a parallel beam by the concave mirror M1. It is then allowed to fall on a fluorspar prism to resolve it into a spectrum. The spectrum is brought to focus by another concave mirror M2 on to a linear bolometer. The bolometer is connected to a galvanometer. The deflections in the galvanometer corresponding to different  are noted by rotating the prism table. Then curves are plotted for E  versus . The experiment is done with the black body at different temperatures.

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Fig. 3.4.3 (a).

Fig. 3.4.3 (b).

The curves obtained are shown in Fig. 3.4.3 (b). Results (i) At any given temperature E first increases with , reaching a maximum value corresponding to a particular wavelength m and then decreases for longer wavelengths. (ii)

The value of E for any  increases as temperature increases.

(iii)

The wavelength corresponding to the maximum energy shifts to shorter wavelength side as the temperature increases. This confirms Wien’s displacement law, m T= constant.

(iv)

Total energy emitted per unit area of the source per second at a given temperature is 

 E

d  . It is represented by the total area between the curve for that temperature

and the

 axis.

This area is found to be proportional to the fourth power of the

absolute temperature. This verifies Stefan’s law. 3.4.4 Planck’s hypothesis: According to the classical theory of radiation, energy changes of radiators taken place continuously. The classical theory failed to explain the experimentally observed distribution of energy in the spectrum of a black body Planck succeeded in deriving a formula which agrees extremely well with experimental results. He discarded both the idea of radiation being a continuous stream as we as the law of equipartition of energy. He suggested the quantum theory of radiation. His assumptions are:

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A black-body radiation chamber is filled up not only with radiation, but also with simple harmonic oscillators or resonators of molecular dimensions. They can vibrate with all possible frequencies.

(2)

The oscillators or resonators cannot radiate or absorb energy continuously. But an oscillator of frequency υ can only radiate or absorb energy in units or quanta of magnitude h υ. H is universal constant called Planck’s constant. The emission of radiation corresponds to a decrease and absorption to an increase in the energy and amplitude of an oscillator,

3.4.5 Planck’s law of radiation Statement : The energy density of radiation in an enclosure at temperature T having wavelengths in the range

E d   Here,

 to 

+ d  is

8 hc 5 d (ehc / kT  1)

h = Planck’s constant; C= speed of light; K= Boltzmann’s constant; T= temperature of the enclosure.

Explanation. Planck proposed the quantum theory of radiation. Planck’s law is derived on the basis of quantum theory. According to quantum theory, energy is emitted in the from of quanta called photons. The energy of a single photon of frequency υ is E=hυ. Planck’s law curve agrees well with the experiment in the entire range of wavelengths. Planck’s formula reduces to Wien’s formula for small wavelengths. Planck’s formula reduces to Rayleigh Jean’s formula for longer wavelengths. So, Rayleigh Jean’s law and Wien’s law are special cases of Planck’s law. Hence Planck’s formula is the correct formula for thermal radiation. Fig. 3.4.5. shows energy vs.

 curves according to different laws of black body radiation.

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Fig. 3.4.5. 3.5 Wien’s Law (i) Wien’s Displacement Law Statement. The wavelength

(m ) of the strongly emitted radiation in the continuous

spectrum of a black body is inversely proportional to the absolute temperature (T) of that body.

m T  b -3

Here, b is Wien’s constant = 2.898 x 10 m K Explanation.

m is the wavelength at which the emissive power E is a maximum at the

absolute temperature T. This relation implies an inverse variation of

m 

m

with T, i.e.,

1 T

This relation shows that as the temperature increases, the maximum intensity of radiation emitted shifts (or gets displaced) towards shorter wavelengths. Hence it is called wien’s displacement law. This law is confirmed by Lummer- Pringsheim experiments. This law is used in measurement of high temperature. The temperature of the sun can be calculated using wien’s displacement law. (ii)

Wien’s Fifth Power Law : Wien also showed that if Em is the maximum value of

E corresponding to m at temperature T, we have

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Em T 5 (or)

Em = Constant x T



Em  Constant. T5

This is called the fifth – power law. (iii)

Wien’s Law of Energy Distribution Statement. The energy density of radiation in an enclosure at temperature T having wavelengths in the range

 to   d  is E d   8 hc  5 e hckT d 

Here, h = Planck’s constant; C= speed of light ; K = Boltzmann’s constant; T = temperature of the enclosure. Explanation. Wien’s law may be written in the form

E  C1 5 ec 2/ T . Here,

E is the energy emitted by a black- body at a given wavelength  and at a

temperature T. C1 and C2 are called the first and second radiation constants. Wien’s formula agrees with experimental results in the region of shorter wavelengths. It fails in the longer wavelength region. Derivation of wien’s Law from Plank’s Law Planck’s radiation Law is

E d   When

 is small ehc / kT Hence

8 hc 5 d (ehc / kT  1)

…(1)

is large when compared to 1.

ehc / kT  1  ehc / kT

So Eq. (1) reduce to

E d   8 hc 5 e hc / kT d  This is when’s law

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Thus Planck;s law reduces to Wien’s law for shorter wavelengths.

3.6. Rayleigh-Jean’s Law Statement. The energy density of radiation in an enclosure at temperature T having wavelengths in the range

 to   d  is

E d   8 kT  4d  Explanation. Rayleigh- Jean’s law agrees with the experimental results in the longer wavelength region. If fails in the shorter wavelength region. Derivation of Rayleigh-Jean’s Law from Planck’s Law Planck’s radiation law is

E d  

When

8 hc 5 d (ehc / kT  1)

…(1)

hc   .  is large, ehc / kT  1    kT 

Hence Planck’s law reduces to

8 hc 5 E d  d   8 hc   1   1   kT  

hc 5 d  hc      kT 

E d   8 kT  4d 

…(2)

This is Rayleigh-Jean’s formula This Planck’s formula reduces to Rayleigh Jean’s formula for longer wavelengths. 3.6.1 Derivation of Wien’s Law Consider a spherical enclosure B with perfectly reflecting walls and capable of expansion radially outwards (Fig.3.6.1 (a))

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Fig. 3.6.1 (a) Let it be filled with black body radiation of energy density u at a temperature T. Let V be the volume of the enclosure. Total internal energy U of radiation is …(1)

U=uV

Let us imagine that the walls of enclosure move outward such that radiation inside it expands adiabatically. Let dV be the change in volume and p, the pressure of radiation on the walls of enclosure. Work done by the pressure of radiation on the walls of enclosure is dW=pdV. From first law of thermodynamics, we have dQ = dU + dW (or)

d(uV) + p dV = 0 ( dQ  0 for adiabatic change)

1 d (uV )  udV  0 3 (or)

1 udV  Vdu  udV  0 3

(or)

4 udV  V du  0 3

(or)

4 dV du  0 3 V u

1 ( p  u) 3

Integrating the above expression, we have

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4 log e V  log e u  0 3 4

V 3 u  constant

(or)

u  T 4 .

Form Stefan’s law,



V 3 ( T 4 )  constant 1

(or)

V 3T  constant

…(2)

Let us calculate the change in wavelength when a radiation wave is reflected by slowly moving reflecting wall of the enclosure, due to Doppler’s effect. Let OA be a ray of wavelength  incident at an angle  on the wall in position S1 (Fig. 3.6.1 (b)). A particular wave crest strikes the wall at A and is reflected along AC. Let AC =. As the reflected wave crest reaches C, the next crest will reach A in time T. Here, T is the period of wave motion. When the next crest reaches A, the wall has moved a distance AM, equal to vT. Here, v is the velocity of the expansion of the wall.

Fig. 3.6.1 (b) Now the crest is reflected from the point B of the new position S 2 of the wall. The crest if reflected in the direction BD. Let AB + BD = 1.

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The change in wavelength is

d   1    ( AB  BD)  AC  ( AB  BN  ND )  AC  AB  BN (

ND  AC )

 A' B  BN (

AB  A' B )

 A ' N  AA' cos   2 AM cos   2vT cos  2

v  cos   c 

T 

 c 

Here, c is velocity of radiation Every ray inside that spherical enclosure undergoes repeated reflections. The path of a single ray is shown in Fig. 3.6.1 (c). Between two successive reflections, say at A and B, the wave travels a distance 2r cos .



number of reflections per sec. =

Number of reflection in time

c 2r cos 

dt 

c dt 2r cos 

The change in wavelength in time dt is d = change in wavelength in one reflection X number of reflections in time dt



(or)

d 

d





2v cos c dt v X  . dt c 2r cos r

v dt dr    r r 

v

dr   dt 

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Fig. 3.6.1 (c). Integrating this expression,

loge   loge r  loge C Here, C is a constant

  rC

(or)

( / r )  constant From Eq. (2) V

1/3

T= constant

1/ 3

(or)

4 3  3  r 

(or)

…(4)

T = constant

4 3   V  3  r 

r T = constant

…(5)

From Eqs. (4) and (5), we get

 r (or)

XrT = constant

T  constant

 'T '

…(6)

This is called Wien’s displacement law. This is the first of the two socalled wien’s displacement laws. Let us now suppose that the waves of wavelengths lying between  and +d in the spherical chamber are isolated and are allowed to undergo adiabatic expansion. Its energy will be

U  d  . Now  as well as d will change. Form the first law of thermodynamics,

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d (U  d  )  p dV  0 (or)

1 d (U V d  )  u d V  0 3

(or)

1 Vd (u d  )  u d dV  u d  dV  0 3

(or)

1 Vd (u d  )  u d  0 3

(or)

d (u d  ) 4 dV dr   4 u d  3 V r Integrating this expression, we get

loge (u d  )  4loge r  loge C (constant) (or)

u d  r 4 =constant

(or)

u dr r 4 =constant

(  r or d dr )

Integrating over r, we get

u r 5 = constant uT 5 = constant

(or)

(see Eq.(5))

We know that spectral radiancy

E is proportional to energy density U 

Therefore

ET 5 = constant = E ' (T )5

…(7)

E T 5  E ' T '5



This is the second of Wien’s displacement laws. Combining the two relations viz.,

T  a constant and E (T )5  a constant, we get E  5  a

constant. Wien supposed that

E  5 should be some function of (T ) , i.e.,

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94 E  5  Af (T )

Here, A is an absolute constant.

E  A 5 f (T )



…(8)

Eq. (8) is wien’s general law for distribution of energy in a black body spectrum. With certain assumptions, Wien was able to deduce the relation

f (T )  Ce



T

Here, C is a constant.



E  C1 5eC2 / T

Here, C1 and C2 are constants Eq. (9) represents the wien’s law for distribution of energy in the spectrum of black body radiation. 3.6.2 Derivation of Rayleigh-Jean’s Law Consider a hollow cubic enclosure of side l with perfectly reflecting walls. Let us place a black particle inside the enclosure. The radiations emitted by the particle will be reflected by the walls. According to electromagnetic theory, the radiation is supposed to consists of number of waves. The waves in the enclosure travel in all possible directions. They undergo multiple reflections from the various walls of the enclosure. In course of time, the enclosure will be filled with the stationary waves of all wavelengths. The reason is that reflected wave interferes with the corresponding incident wave to form stationary wave.

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Fig. 3.6.2 (a) When a wire of length l is fixed at both ends and plucked at its mid point, the stationary waves are formed with nodes at the fixed points. If the wire vibrates in n loops due to formation of stationary waves, then

 2

Here, n = 1,2,3,…..

 l or   (2l / n)

…(1)

.

Similar is the case with an enclosure filled with radiation. Let / be the distance between the walls. The corresponding allowed frequencies (overtones) are

v





cn 2l

…(2)

Here, c is velocity of light. The number of overtones may be referred to as the several degrees of freedom with which the system is capable of vibrating. Moreover, every allowed frequency is called a mode of vibration. Let the three edges of the cube from the three axes in the space. The number of loops, in each direction nx, ny and nz are given by

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

from Eq. (1)

 2   n  2   n  2 

l  nx 

…(3)

The waves can be inclined at any angle to the rectangular axes.

 ,  and 

For waves making an angle

with the three axes (fig. 3.6.2(a)), we have

 2 l cos   n    2 l cos   n    2 l cos   nx 

….(4)

Here the direction cosines obey the relation

cos2   cos2   cos2   1

…(5)

From Eqs. (3) and (4), we can write

 2l  nx  ny  nz     2

(or)

….(6)

 2l  nx  ny  nz  r , then r     2

r



Eq. (6) represents an ellipsoid with nx,ny and nz directions as coordinate axes. Each set of values of nx, ny and nz satisfying the above equation corresponds to one mode of vibration. The total number of modes of vibration are the total number of possible set of (nx,ny,nz). The number of modes of vibration in wavelength interval

 and   d 

can be found using the above

equation. To make the picture clear, let us consider the number of modes of vibration in two dimensional plane. For two dimensional plane, Eq. (6) has the form

 2l  nx  ny     2

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Fig. 3.6.2(b). This equation represents a circle when we plot n x on X axis and ny on Y-axiz as shown in Fig. 3.6.2(b). Every point of intersection gives a mode of vibration. Since positive values of n x and ny are allowed, we have to consider intersections in positive quadrant only. The area of each square is unity. Hence the number of squares in the positive quadrant is

1  2l   l 2    2 4   2

So the number of modes of vibration with wavelength lying between

and   d  can



be obtained by differentiating the above equation. If we now extend the above idea to three dimensional space, the circle will be sphere and each unit square will be a unit cube. Now the total number of modes of vibrations f upto wavelength

 will be those in the octant of the spherical shell of radii (2l /  ) . So,

3 1 4  1 4  2l  4 l f    r 3   . .   8 3 3 3  8 2  3

The number of modes of vibration i.e. degree of freedom between

…(7)

 and   d  is

obtained by differentiating the above expression.

df 

4 l 3 d  1  4 l 3  3  .  3   4 d  3 d    3   

…(8)

Neglecting negative sign as d  is positive, we have

df 

4V

4

d

…(9)

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Here, V is the volume of enclosure. The above number should be multiplied by 2 as transverse electromagnetic waves have two polarisations for each mode. Thus the total number of modes of vibration is given by

df 

8V

4

d

Number of modes of vibration per unit volume



8 d 

…(11)

4

Rayleigh and jeans assumed that the law of equipartition of

energy holds good in case of

radiation also. According to this law, the average energy per mode of vibration is kT. Energy density within wavelength

 and   d 

is given by

E d  = Total number of modes of vibration x average energy per mode 

8 d 

4

X kT

E d   8 kT  4d  This is Rayleigh – Jean’s formula for energy distribution. Example -1. The cavity of a black body radiator is in the shape of a cube. Find the number of modes of vibration per unit volume in the wavelength range between 499.5 nm and 500.5 nm. Sol. Number of modes of vibration per unit volume in the wavelength range

  d 

Here,



8

4

d

499.5  500.5 nm  500nm  5 X 107 m 2

d   (500.5  499.5)nm  1nm  109 m 

Number of modes of vibration =



8X 3.14 X 109  4.022 X 1017 7 4 (5 X 10 )

3.7 Derivation of planck’s law

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 and

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Planck’s Hypothesis. Planck introduced the quantum theory of radiation. His assumptions are : 1.

A black – body radiation chamber is filled up not only with radiation but also with simple harmonic oscillators or resonators or molecular dimensions. They can vibrate with all possible frequencies.

The oscillators or resonators cannot radiate or absorb energy continuously. But an oscillator of frequency v can only radiate or absorb energy in units or quanta of magnitude hυ. The emission of radiation corresponds to a decrease and absorption to an increase in the energy and amplitude of an oscillator.

In general for an oscillator of frequency v, the possible values of the energy are given by.

 nh Here, n is any positive integer and h is planck’s constant. Let N be the total number of Planck’s resonators and E their total energy. Then, average energy per oscillator

  E

. Now,

N  N0  N0e  Here

N 0e



 N0e 2

 ...  N0e r

 ..

N0 = number of resonators having 0 energy. kT

= number of resonators having energy

,

N0e2 / kT = number of resonators having energy 2

N0e r / kT = number of resonators having energy r and so on. Putting

 / kT  x , N  N0  N0e x  N0e2 x  .........N0e rx

N

N0 1  e x

The total energy of placnk’s resonators is

E  0 X N0   X N0e x  2 X N0e2 x  ....  r X N0e rx  ... Ee x   N0e2 x  2 N0e3 x  ...  r N0e( r 1) x

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…(1)

THERMAL ENGINEERING Subtracting,

100

E (1  e x )   N0e x   N0e2 x   N0e3 x  ...



 N 0e  x 1  e x

E

 N 0e  x (1  e  x ) 2 ...(2)

Average energy of a resonator

 

According to Planck’s hypothesis,





and x 

e

E  e x    x x N 1 e e 1

  hv. Further v  c 



hc kT  kT



 kT

…(3)

 1

Number of oscillators per unit volume in the wavelength range  d -4

 and   d  = …(4)

Fig. 3.7 Hence, energy density of radiation between wavelengths



and

  d

(average energy of a Planck’s oscillator) X (number of oscillators per unit volume).

E d  



ehc   1

X 8 4 d 

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E d  

101

8 hc  5 d ehc kT  1

…(5)

Eq. (5) represents Planck’s radiation law. Planck’s formula agrees well with the experimental curves throughout the whole range of wavelengths (Fig. 3.7). Planck’s formula reduces to Wien’s formula for small wavelengths. When

 kT

 is

small,

is large when compared to 1. Hence Eq. (5) reduces to

E d   8 hc  5e

 hc

 kT

d

…(7)

This is Wien’s law Planck’s formula reduces to Rayleigh Jean’s formula for longer wavelengths. When

 kT



 1  hc

 kT



is large,



Hence Planck’s law reduces to

E d   8



hc 5 d   8 kT  4 d  hc  kT



…(8)

This is Rayleigh-Jeans formula Example 1. Using Planck’s radiation law, prove that the total energy density

Etot  aT 4 , where a  8 5k 4 /(15h3 c3 )

Etot   E d   8 hc   5  e 

Sol.



Put

 kT

 1 d  1

x



 x (e 3

 kT

 KT  Etot  8 hc    hc  Now,

 1)dx 

4 

 x (e 3

 1)1 d 

4 15

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Etotal is given by

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

Etot  aT 4 ,

a

102

8 5 k 4 15 h3 c3

This is Stefan’s law. Example 2. Deduce Wien’s displacements law from planck’s radiation formula Sol. Planck’s radiation formula is

The value of

E  8 hc( 5 )  e

 kT

 1 . 1

max is obtained by solving the equation d ( E ) d   0

1 2 hc hc hc   hc   8 hc  5 6  e kT  1   5 (1)  e kT  1 e kT  2    0   kT   

hc e kT 5  kT (ehc / kT  1)

(or)

Putting

 kT  x,

we get

 x5  e

x

1

This equation has a single root given by x=4.965. that is.

maxT 

hc b 4.965 k

This is Wien’s displacement law. Example .3 calculate the average energy of an oscillator of frequency 5.6x 10 T=330 K, treating it as (i) classical oscillator, (ii) planck;s oscillator. Sol. The average energy of a classical oscillator is

E  kT  (1.380 X 1023 ) X 330 4.554 X 1021 J The average energy of Planck’s oscillator is

E



e

hv hv

 1

(6.626 X 1034 ) X (5.6 X 10 12)   6.626 X 1034 X 5.6 X 1012   exp.    1 23  1.380 X 10 X 300   

= 2.9450 X 10

-21

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103

3.8 Stefan’s Law Statement. The total amount of heat radiated by a perfectly black body per second per unit area is directly proportional to the fourth power of its absolute temperature, i.e.,

E  T 4 or E   T 4 Here,  is a constant called stefan’s constant -8

-2

-1

-4

Its value is experimentally found to be 5.67 X 10 Jm S K . It the body is not perfectly black and its relative emittance is e, then

E  e T 4 E varies between 0 and 1, depending on the nature of the surface. For a perfectly black body e=1 Boltzmann gave a theoretical proof of Stafan’s law on the basis of therimodynamics. Therefore, this law is also called Stefan- Boltzmann law. Explanation. This law implies that the emissive power of a black surface depends on temperature alone. If the temperature of a black body is doubled, total radiation emitted by the 4

surface becomes 2 = 16 times. The above equation refers to emission only and not to the net loss of heat by the body after exchange with the surroundings. But the law can be extended to represent the net loss of heat by a body. It may be stated as follows: If a black-body at absolute temperature T is surrounded by another black body at absolute temperature T0, the amount of heat lose by the former per second per unit area is given by

E   (T 4  T 4o ) This form of the law is a law of cooling: If the body has an emissivity, (i.e., relative emittance) e, then

E   (T 4  T 4o ) 3.8.1 Derivation of Newton’s Law of cooling from stefan’s Law Let T and T0 be the absolute temperatures of the hot body and the surroundings. According to Stefan’s law, rate of loss of heat from the hot body is given by

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104

R   (T 4  T 4o )   (T  To )(T 3  T 2 T0  TT02  T03 )

T  T0

When

R   (T  To )4T03 R  (T  To )

(or)

(since

4 T03 = K = a constant)

Thus the rate of cooling of the body is directly proportional to the excess of temperature of the body over that of the surroundings, if this excess temperature is small. The is Newton’s law of cooling. 5

Example 1. An iron furnace radiates 6.405 X 10 joules per hour through an opening of cross-4

section 10 m . If the relative emittance of the furnace is 0.8, calculate the temperature of the furnace. Given

  5.669X 108 Wm2 K 4

Sol. Energy radiated by the furnace in one hour is Q

e T 4 x surface area x time

e T 4 At 5

Here, energy radiated per hour = Q = 6.405 X 10 J

A  104 m2 ; e  0.8, t  3600s;

  5.669 X 108 Wm2 K 4 ;T  ?  Q  T   e At 

  6.405 x105   8 4  0.8 X (5.669X 10 )X 10 X 3600 

 2504 K

Example 2. Calculate the radiant emittance of a black body at a temperate of (i) 400 K (ii) 4000K.

[  5.672 X 108 J m2 s 1 K 4 Sol.

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105

E  T 4 8

Sol. E= (i ) E  (5.672 X 10 )(400)  1452 watts / m 4

(ii) E  (5.672 X 108 )(4000) 4  1.452 x107 Wm2 Example 3. Two large closely spaced concentric spheres (both are black body radiators) are maintained at temperatures of 200 K and 300 K respectively. The space in between the two spheres is evacuated. Calculate the net rate of energy transfer between the two spheres.

[  5.672 X 108 J m2 s 1 K 4 ] Sol.

E   (T 4  T04 )  (5.673 X 108 ) (300) 4  (200) 4   368.68Wm2

Example 4. Establish the value of Stefan’s constant if the temperature of the filament of a 40 watt o

-5

tungsten lamp is 2170 C and the effective surface area of the filament is 6.6 X 10 m . You are to assume that the energy radiated is 0.31 of that from a black body in similar conditions and that any effect due to radiation from the glass envelope is negligible. Sol.

Q  eA T 4 40  0.31 X (6.6 X 105 ) X  X (2443) 4



40 0.31 X (6.6 X 105 )(2443) 4

 5.489 X 108 J m2 S 1 K 4 . -3

Example 5. The relative emittance of tungsten is 0.35. A tungsten sphere of surface area 10 m

is suspended inside a large evacuated enclosure whose walls are at 300 K. What power input is required to maintain the sphere at a temperate of 3000 K? Sol. The metal sphere will radiate out energy to the enclosure and its temperature will therefore fall. Hence if its temperature is to be maintained constant at 3000 K, it must obviously be supplied with as must energy per second as it radiates out to the enclosure. Energy radiated by the sphere per second is

Q  eA (T 4  T04 )  0.35 X (5.67 X 108 ) X 103 [(3000) 4  (300) 4 ] 1607W  The power input = 1607 W

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106

Example 6. An aluminium foil of relative emittance 0.1 is placed in between two concentric spheres at temperatures 300 K and 200 K respectively. Calculate the temperature of the foil after the steady state is reached. Assume that the spheres are perfect black body radiators. Also calculate the rate of energy transfer between one of the spheres and the foil. Sol. Here

e  0.1, T1  300 K , T2  200 K

  5.67 X 108 J m2 s 1 K 4 (i)

Let x be the temperature of the foil in the steady state.

e (T14  x 4 )  e ( x4  T24 )

[(300)4  x4 ] [ x4  (200)4 ] 

x  258.2K

(ii) Rate of energy transfer

e (T14  x 4 )

 0.1 X (5.67 X 108 )[(300)4  (258.2) 4 ]  20.98Wm2

Example 7. At what temperature a black body will radiate thermal energy at the rate of 1 watt per square cm?

E  1W / cm2  1 X 104 Wm 2 Sol. Here,

  5.67 X 108 J m2 s 1 K 4 E  T 4



E T    

 1 x104    648 K 8  5.67 X 10  

Example 8. Calculate the energy radiated per minute from the filament of an incandescent lamp at 2000 K if the surface area is 5 x 10

-5

and its relative emittance is 0.85. Given

  5.7 X 108 W m2 K 4 Sol. Total energy radiated by the filament per minute

Q  eA T 4 x surface area x time = e T 4 At

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107

t  60s;T  2000 K ; A  5 X 105 m2 e  0.85;  5.7 X 108 Wm2 K 4 Q  e  T 4 At  (0.85) X (5.7 X 108 )(2000) 4 X (5 X 105 )(60) = 2325 J.

3.8.2 Derivation of Stefan Boltzamann law or radiation Radiation exerts a pressure on the surface on which it is incident. For normal incidence on a surface, the pressure of radiation is equal to the energy density. If the radiation is diffuse, the pressure of radiation will be one-third of the energy density. Pressure of diffuse radiation



P

(or)

1 total energy density 3

Black – body radiation behaves like a gas, as it exerts pressure and possesses energy. Hence thermodynamic laws can be applied to black body radiation. Let

 denote

the energy density or radiation inside a uniform temperature enclosure at

absolute temperature T. Let V be the volume and P the pressure of radiation. Both

 and P are

simple functions of T. Energy of the radiation

 U  V and P  

From first law of thermodynamics,

dQ  dU  P dV From Maxwell’s thermodynamical relations,

 S   P       V T  T V (or)

 dQ   dP    T   dV T  dT V



 dU  P dV   dP    T  dV  T  dT V

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…(1)

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108

(or)

 dU   dP     P T   dV T  dT V



 dU   dP    T   P  dV T  dT V Now

U  V or

( dU

dV )T

…(2)



1  d   dP  P   or     3  dT T 3  dT 

T d  2 dT 3





(or)

4 T d  3 3 dT d





Integrating,

dT T

loge   4loge T  constant

  T 4

(or) Here,

4

…(3)

…(4)

 is Stefan’s constant.

3.8.3 Experimental verification of Stefan’s law Lummer are Pringsheim experimentally verified Stefan’s law over a wide range of o

temperature (100 C to 1,300 C). Apparatus. The apparatus is shown in Fig. 3.8.3

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109 Fig. 3.8.3. o

The black body for the range 200 to 600 C consists of a hollow copper sphere C coated with platinum black. It is placed in a bath containing a mixture of fused sodium and potassium o

nitrates. The bath can be maintained at any temperature between 200 C to 600 C. The temperature can be measured by a thermocouple T. For higher temperatures, the copper sphere is replaced by an iron cylinder enclosed in a double-walled gas furnace. The radiation coming out of the operating O is measured by the bolometer B designed by Lummer and Kurlbaum. The radiation could be stopped when required by a water-cooled shutter S. o

Experiment. The bolometer is calibrated by means of a Fery’s black body kept at 100 C. The bolometer is kept at different distances from this black body and the deflections are observed. The deflections are found to be inversely proportional to the square of the square of the distances. From this it is concluded that the deflection of the galvanometer in the bolometer circuit is directly proportional to the quantity of radiation received by it. The black body C is next heated to any desired temperature and the water-cooled shutter is removed. The radiation is allowed to fall on the receiving face of the bolometer. The steady deflection of the galvanometer is noted. Keeping the bolometer at a particular distance, the deflection  in the galvanometer is found for various steady temperatures of the black body. Calculation Let T be the temperature of ht back body and To, the temperature of the shutter. Then

  a(T 4  T04 ) The coefficient a is found to be constant. This verifies the law. Note: The constant a is not Stefan’s constant . a is some other constant which depends on the calibration of the bolometer.

3.9 Determination of Stefan’s constant Apparatus. The apparatus is shown in Fig. 3.9. B is a metal hemisphere with its inner surface blackened. The hemisphere is enclosed in a steam chamber (W). The entire arrangement rests on a wooden platform P with a hole at the centre of the hemisphere. B is heated to a uniform temperature measured by the two thermometers TT. The blackened surface of B with a small hole at its centre thus constitutes the black body radiator. A small silver disc (D), blackened on the top surface is fixed at the centre of the hemisphere. The ebonite covering E is used to shield

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110

the disc D from or expose the disc to the black body radiation from the enclosure. This can be done by turning the handle H form outside.

Fig. 3.9. One junction (J1) of a thermocouple is fixed to the bottom face of the disc. The other junction (J2) is placed in a constant temperature bath C. A galvanometer G is included in the thermocouple circuit. The wires leading to G are packed in cotton wool. Theory. When steam is passed through the steam chamber, let the steady temperature of the hemispherical black-body be T1 K. If now, the silver disc is introduced in the central hole, the disc absorbs the radiation emitted by the black body. The disc also emits radiation. But its rate of emission per unit area is less than its rate of absorption. As a result, there is a net gain of heat energy by the disc. Let R1 be the rate of absorption of heat energy per unit area by the disc.

R1   T14 At any instant let T2 be the temperature of the disc. Let R2 the rate of emission of heat energy per unit area by the disc.

R2   T24



Net rate of gain of energy by the disc per unit area

 R1  R2   (T14  T24 )

Let A be the surface area of the disc.

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Heat energy gained by the disc per second

  (T14  T24 ) A

This absorption of heat results in a rise in the temperature of the disc. Let

m = mass of the disc, C = specific heat capacity of its material and

dT = rate of change of temperature of the disc. dt mC (dT dt )   A(T14  T24 )



mC (dT

) dt A(T14  T24 )

…(1)

Here, mC and A for the disc are known. T1, the temperature of the steam chamber is also known. Hence, if T2 and

(dT

) are determined, we can calculate  . The actual experiment

consists of two parts:Experiment (i) Calibration of the thermocouple: Before passing steam in W, the temperature of the black body B is the same as the room temperature. The silver disc is arranged at the central hole. It is allowed to attain this temperature. The water bath C is heated. Now, the disc (J1) is the cold junction and J2 is the hot junction. The deflection in the galvanometer () is noted at regular intervals, corresponding to the difference of temperature (T) between J 1 and J2. A graph is the plotted between  and T (Fig. 4.17). The graph is a straight line passing through the origin. The slope of the line gives

dT BC  tan   d AB

(ii) Determination of

d

and T2:

The silver disc is removed from the hole. Steam is passed in W for a long time until B has attained a steady temperature as indicated by the two thermometers. Now, the silver disc is again introduced in the central hole. The temperature of the silver disc increases. The deflections of the galvanometer are noted are regular intervals.

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A graph is plotted between deflection and time (Fig. 4.18). The slope of the curve at a convenient point P is obtained. Rate of change of deflection in the galvanometer at point P



d EF .  dt DE

dT dT d BC EF  X  X dt d dt AB DE



T2 for the point P can be found from the first (T,) graph. Calculation of Stefan’s constant   is calculated using Eq. (1)

 dT  mC    dt   A(T14  T24 ) Hence  is calculated Example 1 A body of mass 10

-2

kg is kept in an enclosure at temperature 27 C. If the

-1

temperature of the body is 127 C, its specific heat capacity is 418 J kg K and area of emitting surface

body

-3

find

out

the

rate

cooling

the

body.

(  5.72 X 108 Jm2 s 1 K 4 ) Sol. Energy radiated from the whole body in one second is

Q   (T 4  T04 ) A Let m and C be the mass and specific heat capacity of the body. If the temperature of the body falls by an amount dT in one second,

Q  mC dT

mC dT   (T 4  T04 ) A (or)

dT 



 (T 4  T04 ) A mC

(5.72 X 108 ) (4004  3004 )10 3 102 X 418

 0.239o C Per second.

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3.10 Pyrometry o

The physics of measurement of very high temperatures above 1600 C is called pyrometry. Pyrometers are instruments used for measurement of high temperatures. The fact that the radiation emitted by a black body depends solely upon its temperature provides radiation methods for measuring high temperatures. Thee pyrometers based on the principles of radiation are termed as radiation pyrometers. Both stefan’s law and plank’s law are used for the determination of temperature. There are two types of radiation pyrometers: (i)

Total Radiation Pyrometers. In these types of instruments, the total energy of the radiation emitted by the body under test is measured. The temperature is determined by applying Stefan’s law ( ( E

(ii)

  T 4 ) assuming that the body is black.

Optical or spectral Pyrometers. These are based on the Wien’s distribution law. The intensity of radiation of a certain wavelength emitted by the body is compared with that of the radiation of same wavelength emitted by a standard body at a known temperature. The temperature of the body is deduced form Wien’s distribution law, or more accurately from planck’s law.

Advantages. (i)

They can be used to measure any high temperatures even when the hot body is inaccessible.

(ii)

They need not be put in contact with the hot body nor raised to the temperature of the body.

(iii)

There is no difficulty in extrapolation because radiation laws are valid at all temperatures.

Disadvantages. (i)

Their range roughly begins at 600 C. These pyrometers cannot be used to measure o

temperatures below about 600 C. Below this limit, measurement will be inaccurate due to the low emission of radiation from the bodies.

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The temperature obtained by these pyrometers is always less than the actual temperature of the source because they use radiation laws which are true only for black bodies. So the measurements are accurate for black bodies only.

3.10.1 Fery’s Total Radiation Pyrometer Principle. The total energy of the radiation emitted by the body under test is measured. The temperature is determined by applying Stefan’s law

( E   T 4 ) assuming that the body is black.

Construction. The total radiation pyrometer is shown in Fig. 3.10.1.

It consists of a large

concave mirror C of nickel-plated copper. The mirror has a small hole at the centre. An eyepiece (E) is fitted into the hole. The mirror can be moved by a rack and pinion arrangement. Two plane o

semi-circular mirrors M, M inclined to each other at about 5 with an opening of about 1.5 mm at the centre are placed in front of the hole in the mirror C. The opening serves as a limiting diaphragm. The opening is backed by a blackened strip S. This strip is connected to one junction of the thermocouple.

The leads of thermocouple are connected to a

millivoltmeter. The

millivoltmeter measures the e.m.f developed. The cold junction is shielded from direct radiation. The strip S is properly protected from external radiation.

Fig. 3.10.1. Working. The radiations from the hot body for which the temperature is to be determined are allowed to fall on the concave mirror C. By moving the concave mirror C suitably, the radiations are focused on the blackened strip S. The focusing can be observed through the eyepiece E. the focusing is perfect (in focus) when the images in the two mirrors overlap exactly to form a complete circle. When the two halves appear to be displaced (out of focus), the focusing is not perfect. By moving C, perfect focusing is obtained. Now strip S is heated and produces a thermo e.m.f the thermo emf is read by the milli-voltmeter.

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The distance of the pyrometer form the hot body should be such that the image formed by concave mirror C is larger then the opening of two mirrors M, M. the can be done by the geometry of the apparatus. By doing so, the readings of the milli-voltmeter are independent of the distance of the hot body from the pyrometer. This is very essential for the accurate measurement of temperatures. Calculation of Temperature. According to Stefan–Boltzmann law, the milli-voltmeter deflection  will be proportional to Here,

(T 4  T04 ).

T = absolute temperature of the source, T0 = absolute temperature of the strip

As T0 is very small compared to T, we have

 T 4 In practice, the power of T varies 3.8 to 4.2 due to stray reflections from the walls etc.



 T k Taking logarithm, log  = K log T.

Using bodies whose temperatures are accurately known and measuring the deflections, a graph drawn between log  and log T. The graph will be a straight line. This is known as calibration graph. Any unknown temperature can be obtained with the help of this graph, provided the corresponding deflection is known. The milli-voltmeter may also directly be calibrated in degrees. Modification for Measurement of Very high Temperatures. o

To measure temperature above 1400 C, a rapidly rotating sector (Fig. 3.10.2) is interposed the hot body and the pyrometer.

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116 Fig. 3.10.2.

Let  be the angle of the sector. The rotating sector allows only a fraction (/360) of the incident radiation to enter the pyrometer. Now the pyrometer indicates a temperature which is less than the true temperature of the body. Let T be the true temperature of the source and T 1 the temperature measured with the instrument. Then

T 4 360  T14 

T4 

T14 (360)



1/ 4

 360  T  T1     



Advantage. It can be easily adapted for a continuous recording of temperature e.g., the roof of an open-hearth. Drawbacks. (i)

It is not suitable for the measurement of temperatures of smaller bodies, whose heat image is likely to be too small to cover the whole of the disc.

(ii)

Calibrated in the manner as described above, its utility is confined to the limited o

range, 800 C to 1600 C. (iii)

The readings of the apparatus, while theoretically quite independent of the distance from the hot body, are not really so in actual practice.

3.10.3 Disappearing Filament Optical Pyrometer Principle. It is based on the Wien’s distribution law. The intensity of radiation of a certain wavelength emitted by the body is compared with that of the radiation of same wavelength emitted by a standard body at a known temperature. The temperature of the body is deduced from wien’s distribution law, or more accurately from Planck’s law. Construction. It consists of a telescope fitted with an objective O at one end and an eyepiece E at the other end (Fig. 3.10.3). The distance between objective and eyepiece can be adjusted by a rack and pinion arrangement. The cross wires of the telescope are replaced by an electric lamp S (serving as standard body) D1 and D2 are two diaphragms which limit the cone of radiation entering the telescope. A red glass filter G is placed before eye piece E. The filament F of the lamp is connected to a battery B, rheostat Rh and ammeter A in series. The filament is heated by

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the electric current flowing through it and its temperature can be changed by adjusting the current in the circuit. The current is recorded by ammeter.

Fig. 3.10.3. Working. The telescope is directed towards the hot body whose temperature is to be measured. By adjusting the position of the objective of telescope L, an image of black body is focused on the filament F. The temperature of the filament is adjusted by means of the rheostat Rh, until the filament F disappears against the image of the black body. When the brightness of filament F and that of the black body have been exactly matched, the reading of the ammeter A is noted. Since both filament and image are equally bright, they must be emitting equal amounts of energy per unit area per second. Hence they must be at the same temperature. Initially the instrument is calibrated by taking sources at known temperature and the corresponding currents passing through the filament are noted. A graph is plotted between the current and the temperature. From the graph, for a particular value of current passing through the filament (for disappearing position) the corresponding unknown temperature can be determined. The ammeter can also be calibrated to measure the temperatures directly. o

The instrument is suitable for measuring temperatures form 600 C to 1500 C. The range o

can be extended to 2700 C by using a rotating sector. Advantages (i)

It is portable

(ii)

It can be used to measure the temperature of even small bodies, since the tip to the filament can be matched with any part of the image seen through the telescope.

(iii)

It can be easily constructed and this makes it highly precise and accurate.

(iv)

The theory underlying it is relatively less affected by a departure of the body from a perfectly black body. So the corrections involved are comparatively much smaller.

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Drawback. It requires some wavelengths in the visible part of the spectrum for its working. It cannot, therefore, be used for measuring temperatures of bodies which are not actually glowing. 3.10.4 Polarising Optical Pyrometer In this optical pyrometer, a ray of a particular colour from the hot body is compared with a ray of the same colour from a standard electric lamp, by means of a polarizing device. Its essential parts are shown in Fig. 3.10.4 Construction. The radiation beam from the hot body whose temperature is to be measured after passing through a diaphragm D1 falls on a calcite prism P. Calcite is a doubly refracting crystal. It breaks the incident beam into two plane polarized components in planes perpendicular to each other. The two components are shown by vertical and horizontal arrows in the figure. The two components now pass through a biprism B. The biprism bends the components is such a way that the horizontal component passes through diaphragm D 2 while the vertical component (↕) is cut off. Similarly, a radiation beam from the standard body at a steady temperature is passed through the system.

In this case, the vertical component is passed

through diaphragm D2 while the horizontal component  is cut off. These two mutually perpendicular polarized components pass through Nicol prism N and then through a red filter F. On looking through an eyepiece E, two semi-circular patches of light are observed side by side.

Fig. 3.10.4. Working. To measure the temperature of hot body, the Nicol N is rotated in its own plane. It is observed that in a particular position of N, the light from the given hot body is transmitted fully while the light from the standard body is completely cut off. On the other hand, if o

the Nicol N is rotated through 90 from this position, the light from the standard body is transmitted while light from hot body is cut off. In between these two positions, there is a position when light from both the bodies is equally bright. The angle  through which the Nicol N is rotated from the first position to equally bright position is measured.

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Calculation of temperature. The unknown temperature T can be calculated by the following formula:

b log tan   a  . T Here, a and b, are constants. They are obtained by calibrating the pyrometer against known temperatures. So measuring  with the help of pyrometer, the unknown temperature T is calculated. Derivation of the formula. Let

E be the intensity of bright red colour emitted from the

hot body under test at unknown temperature T K and

S be the intensity of radiation of the same

colour emitted from standard body. Then

E  tan 2  S

…(1)

Now keeping the standard body same, the experiment is repeated with another body at a known absolute temperature T. Let

 be the angle of rotation of Nicol in this case. Then

E '  tan 2  ' S Here

…(2)

E ' = intensity of radiation emitted from the second body.

Form Eqs. (1) and (2),

E tan 2   E ' tan 2  '

…(3)

Form Wien’s distribution law,

E  C1 5 eC2 / T and

…(4)

E '  C1 5 eC2 / T

…(5)

Here C1 and C2 are constants. From Eqs. (4) and (5),

E  eC2 /  [1/ T '1/ T ] ' E Comparing Eqs. (3) and (6), we get

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…(6)

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tan 2   eC2 /  [1/ T '1/ T ] 2 ' tan  Taking log of both sides, we get

 tan 2   C2  1 1  log   2     tan  '    T ' T  2[log tan   log tan  '] 

(or)

C2  1 1     T ' T 

…(7)

This equation is of the from

b log tan   a  . T Here a and b are constant. We can obtain them by calibrating the pyrometer against known temperatures. Knowing a, b and , we can easily evaluate the temperature of the given hot body. o

The normal range of this pyrometer is from 700 C to 1400 C. The range can be raised to o

4000 C by putting absorption glasses in the path of the incident radiation. 3.11 Determination of Solar Constant- Water flow Pyrherliometer. Apparatus: The water flow pyrheliometer consists of a double walled cylindrical vessel V (Fig. 3.11). The lower side is conical in shape and is coated with platinum black on its inner side. Water circulates in spiral channels through the space between the walls. T 1 and T2 are the platinum resistance thermometers to record the temperatures of incoming and out going water. The vessel is kept inside a Dewar flask to avoid any heat loss due to conduction and radiation. D is a diaphragm which allows the heat radiations from the sun through a known area of cross-section. Solar radiation is absorbed by the blackened inner walls and the heat is transferred to the circulating water.

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Fig. 3.11. Working: The rate of flow of water is adjusted so that the thermometers show constant temperatures. After some time, when the steady state is reached, the constant temperatures, 1and 2 of the inflowing and out flowing water are noted. The solar radiation is now screened off. C is a manganin coil wound over the conical bottom. Now a suitable current is passed through the manganin coil C, so that the thermometers T1 and T2 show the same constant temperatures 1and 2. The rate of flow of water is left undisturbed. Let V be the potential difference applied between the ends of the manganin coil and I the constant current passed through it. Calculation Let A be the area of cross-section of the diaphragm S is the solar constant. Heat energy received per second from the sun = SA. Heat energy produced per second by passing electric current = VI The heat energy received from the sun is exactly equal to the heat energy generated electrically. Hence SA = VI



S

VI A

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Thus, the solar constant can be calculated. 3.11.1 Angstrom’s Pyrheliometer Construction. Angstrom’s pyrheliometer is shown in Fig. 3.11.1. It consists of two identical strips of blackened platinum foil S1 and S2. The two strips are arranged such that one is open to receive radiation from sun normally while the other is protected by a double walled shield H. S 1 and S2 are connected to the junctions of a thermocouple circuit containing a sensitive galvanometer G. The shielded strip S2 is heated by an electric current.

Fig. 3.11.1 Working. When both the foils S1 and S2 are shielded from radiation, the galvanometer G shows no deflection. When S1 is exposed to radiation from the sun, its temperature rises and galvanometer shows deflection. Now current is passed through strips S 2 and its strength is so adjusted that the galvanometer shows no deflection. Under this condition, the temperatures of A and B are the same i.e., the rate at which heat is supplied to both is same. The heat energy supplied to strip S2 can be calculated from the known values of the current and potential difference in the electric circuit. Calculation. Let A be the area of cross-section of the strip and a its absorption coefficient. Let S be the solar constant. …(1)

Energy absorbed by the strip S1 per second = SAa. Let I be the current flowing through S2 and E the potential difference across it. Heat produced in one second in the strip S2 = EI Equating Eqs. (1) and (2),

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…(2)

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SAa  EI 

S

…(3)

EI Aa

Here, V is voltmeter reading in volts and I is ammeter reading in amperes. Hence S can be calculated To calculate the True Value of Solar Constant S0 We know that the atmosphere absorbs a part of solar radiation. The experiment is performed several times on the same day under constant sky conditions for different elevations of the sun.

Fig(i) The observed value of solar constant S and the true of solar constant S0 are connected by the relation

S  S0t s ec z Here, t is the transmission coefficient of the atmosphere and Z is the zenith distance of the sun (or angular altitude) Taking logarithms, we have

log S  log S0  sec z.log t From the experimental data, a graph is drawn between log S (on Y-axis) and sec Z (on Xaxis). We get a straight line graph (Fig.(i)). The intercept of the straight line on Y-axis gives the value of log S0. From this the value of S0 can be calculated.

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3.11.2 Water Stir Pyrheliometer Principle. In this apparatus, solar radiation is absorbed by a vessel and is transferred to water. The water is continuously stirred so that the temperature throughout its volume is the same. The rise in temperature depends on the energy absorbed. The incident radiation is estimated in terms o electrical energy necessary to produce the same effect as due to the absorbed solar radiation. From this, the solar constant is calculated. Construction. The apparatus is shown in fig.3.11.2 Construction. A cylindrical vessel C with a conical bottom is blackened inside. It is kept in a calorimeter containing water. The heat radiations are received from the sun by the vessel C. the water is kept vigorously stirred by a stirrer driven by an electric motor. A coil of manganin wire W is wound around C. it is used as a platinum resistance thermometer to measure the temperature of water.

Fig. 3.11.2. Working. Heat radiations from the sun are allowed to enter the vessel C. the rise in temperature of water in one second is calculated. The heat radiations are cut off. A suitable current is passed through the manganin coil W so that the rise in temperature of water in one second is there same. Let E volts be the potential difference and I amperes be the current passing through the manganin coil W.

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Electrical energy spent per second in W=EI Therefore, the amount of heat radiations received from the sun in one second by the exposed surface of the vessel C = EI From this value, the solar constant is calculated.

3.12 Solar Constant Definition. Solar constant is defined as the amount of solar energy received per second by unit area of a perfectly black surface, held perpendicular to the sun’s rays and placed at a mean distance of the earth from the sun, in the absence of the atmosphere. Explanation. The sun is a source of thermal energy. It emits radiant energy continuously in all directions, but only a part of the energy thus radiated is received by the earth. From the amount of energy received by the earth, the temperature of the sun can be estimated. For this purpose, a constant called solar constant has to be defined. The value of the solar constant S is -2 -1

1400 J m s . The instruments used to measure the solar constant are called pyrheliometers.

3.13 Temperature of the sun The surface temperature of the sun may be estimated by assuming that the sun radiates energy like a black body. The temperature so estimated is known as the black body temperature of the sun. Let R be the radius of the sun (Fig. 3.13). Then its surface area is 4 R 2 . Let T be the absolute temperature of the sun. According to Stefan’s law, total energy emitted by the sun per second Here,



is stefan’s constant.

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 4 R2 X  T 4 .

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Fig (3.13) This energy is spread in all directions. Let us consider a sphere of radius r concentric with the sun. r is the mean distance of the earth from the sun. Then this radiated energy will be spread over the surface 4 r . 2

Let S be value of the solar constant. Total energy received by the surface



4 r 2 per second  4 r 2 S

4 R2 X  T 4  4 r 2 X S

Sr 2 R 2



T4 

(or)

 r  2 1 T    X SX    R 

1/ 4

Thus temperature of sun can be calculated be determining values of S, r, R and Example. 1. Calculate the temperature of the sun from the following date: Solar constant = 1400 Wm

-2

Radius of the sun = 6.96 X 10 m 11

Distance of the sun from earth = 1.496 X 10 m 8

-2

-4

Stefan’s constant = 5.6697 X 10- Wm K . 11

Sol. Here, r = 1.496 X 10 m, R= 6.96 X 10 m, S = 1400 Wm

-2

 = 5.667 x 10-8 Wm-2 K-4 The black body temperature of the sun is given by

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

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127 1/ 4

 r 2 S  T      R   

1/ 4

 (1.496 X 1011 )2  1400 T  X 8 2 8  (5.6697 X10 )   (6.96 X 10 ) = 5812 K. Temperature of the sun using Wien’s Displacement Law

The temperature of the sun can also be calculated using Wien’s displacement law. According to Wien’s displacement law, the product of wavelength corresponding to maximum energy

m and the absolute temperature T is constant i.e.,

m T  constant = b. -3

Here, b is Wien’s constant = 2.898 x 10 mK Abbot’s measurements show that

m

surface temperature of the sun is

T

m



-9

for the solar radiation is 475.3 x 10 m. Hence the

2.898 X 103  6097 K . 475.3 X 109

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4. Introduction A body can be deformed (i.e., changed in shape or size) by the suitable application of external forces on it. A body is said to be perfectly elastic, if it regains its original shape or size, when the applied forces are removed. This property of a body to regain its original state or condition on removal of the applied forces is called elasticity. A body which does not tend to regain its original shape or size, even when the applied forces are removed, is called a perfectly plastic body. No body, in nature, is either perfectly elastic or perfectly plastic. Quartz fibre is the nearest approach to a perfectly elastic body. When an external force is applied on body, there will be relative displacement of the particles and due to the property of elasticity, the particles tend to regain their original positions. Stress is defined as the restoring force per unit area. If a force F is applied normally to the area -1

-2

of cross-section A of a wire, then stress = F/A. Its dimensions are ML T . 4.1 Definitions Beam: A beam is defined as a rod or bar of uniform cross-section (circular or rectangular) whose length is very much greater than its thickness. Bending Couple: If a beam is fixed at one end and loaded at the other end, it bends. The load acting vertically downwards at its free end and the reaction at the support acting vertically upwards, constitute the bending couple.

Fig. 4.1 This couple tends to bend the beam clockwise. Since there is no rotation of the beam, the external bending couple must be balanced by another equal and opposite couple which comes into play inside the body due to elastic nature of the body. The moment of this elastic couple is called the internal bending moment. When the beam is in equilibrium, the external banding moment = the internal bending moment. Plane of Bending : The plane of bending is the plane is which the bending takes place and the bending couple acts in this plane. In Fig. 4.1, the plane of paper is the plane of bending.

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Neutral Axis: When a beam is bent as in Fig. 4.1, filaments like ab in the upper part of the beam are elongated and filament like cd in the lower part are compressed. Therefore, there must be filament like ef in between, which is neither elongated nor compressed. Such a filament is known as the neutral filament and the axis of the beam lying on the neutral filament is the neutral axis. The change in length of any filament is proportional to the distance of the filament from the neutral axis.

4.2 Expression for the bending moment Consider a portion of the beam to be bent into a circular arc, as shown in Fig. 4.2. ef is the neutral axis. Let R be the radius of curvature of the neutral axis and  the angle subtended by it at its centre of curvature C. Filaments above ef are elongated while filaments below ef are compressed. The filament ef remains unchanged in length. Let ab be a filament at a distance z from the neutral axis. The length of this filament ab before bending is equal to that of the corresponding filament on the neutral axis ab. We have, original length = ab = R

Fig 4.2. Its extended length = ab = (R+z)  Increase in its length = ab -ab = (R+z)  - R = z. .  Linear strain =

increase in length z. z   original length R. R

If E is the Young’s modulus of the material, E = Stress / Linear strain ie.,

Stress = E X Linear strain = E(z/R)

If A is the area of cross-section of the filament,

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the tensile force on the area A = stress X area =

E.z  A. R

Moment of this force about the neutral axis ef =

E. z E  A.z   A.z 2 . R R

 E 2     A.z . forces acting on all the filaments  R The sum of the moments of



 A.z

E  A.z 2  R

is called the geometrical moment of inertia of the cross-section of the beam about an 2

axis through its centre perpendicular to the plane of bending. It is written as equal to Ak . i.e.,

 A.z

= Ak . (A = Area of cross-section and k = radius of gyration).

But the sum of moments of forces acting on all the filaments is the internal bending moment which comes into play due to elasticity. 2

Thus, bending moment of a beam = E Ak / R

. 2

Notes : (i) For a rectangular beam of breadth b, and depth (thickness) d, A = bd and k = 2

d / 12. 

Ak = bd / 12. (ii)

For a beam of circular cross-section of radius r, A = r and k = r / 4. 2



Ak = r / 4. 2

(iii)

E Ak is called the flexural rigidity of the beam.

4.3 Depression of the loaded end of a cantilever Cantilever: A cantilever is a beam fixed horizontally at one end and loaded at the other end.

Fig. 4.3 Let OA be a cantilever of length l fixed at O and loaded with a weight W at the other end. OA is the unstrained position of the beam. Let the depression AA of the free end be y (Fig. 4.3).

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Let us consider an element PQ of the beam of length dx at a distance (QA = x) from the loaded end. C is the centre of curvature of the element PQ and R its radius of curvature. The load W at A and the force of reaction W at Q constitute the external couple, so that, the external bending moment = W.x. The internal bending moment =

E. Ak 2 R

E. Ak 2 E. Ak 2 For equilibrium, Wx = or R = R Wx

…(1)

Draw tangent at P and Q meeting the vertical line at T and S respectively. Let TS = dy and d = Angle between the tangents. Then, PCQ also = d. Now, PQ = dx = R d or d =

dx Wx (From Eq. 1)  dx. R EAk 2

Wx 2 dx Wxdx We have, dy = x d = x . = EAk 2 EAk 2 

…(2)

1 the total depression of  Wx 2 Wl 3  y  dx   0 EAk 2 the end of the cantilever  3EAk 2

Angle between the tangents at the ends of a cantilever: Since the beam is fixed horizontally at O, the tangent at O is horizontal. If a tangent is drawn at A (the free end of the bent bar), it makes an angle  with the horizontal.

Angle between the  Wx dx.   d  tangents at P and Q  EAk 2 1 Angle between the  Wx dx.     tangents at O and A  EAk 2 0



=

Wl 2 . 2 EAk 2

Work done in uniform bending. Consider a beam bent uniformly by an external couple. Let A be the area of cross-section of the beam. Consider a filament of area of cross-section A at a distance z from the neutral axis (Fig. 1.13). Then, the tensile force on the area A =

Ez  A. . R

The linear strain of this filament = z / R. If l is the length of the filament, then, the extension of the filament = zl / R.

 1   forceXextension bending the filament  2

The work done in

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1 Ez zl 1 El 2  AX  Xz . A. 2 R R 2 R2

For uniform bending R is constant. Hence, the work done in bending the whole beam is

W Here,

1 El 1 El 1 EAk 2 1 2 2 z  A  XAk  X  2 R2 2 R2 2 R R

EAk 2 = the bending moment and l / R = the angle subtended by the bent beam at its R

centre of curvature. 

The work done in uniform bending =

1 (bending moment) X (Angle subtended by the 2

bent beam at its centre of curvature).

Example 1: Obtain an expression for the depression at the free end of a heavy beam clamped horizontally at end and loaded at the other end. Consider an element PQ of the beam of length dx at a distance x from the fixed end O (Fig. 1.14). Now, in addition to the load W acting at A, a weight equal to that of the portion (l – x) of the beam also acts at is mid – point. Let W 1 be the weight of the beam. Then, the weight per unit length of the beam = W 1/l. Now, we have an additional weight W 1(l-x)/l acting at a distance (l – x)/2 from Q. Therefore,

 W1 (l  x)   W (l  x)  (l  x) external couple applied  l 2 total moment of the

 W (l  x) 

W1 (l  x) 2 2l

The beam being in equilibrium, this must be balanced by the bending moment Therefore,

W (l  x) 

 d2y  W1 2 EAk 2 (l  2l.x  x 2 )   EAk 2  2  2l R  dx 

Integrating,

 x2  W  x3  dy W  lx    1  l 2 .x  l.x 2    EAk 2 C 2  2l  3 dx  where C is constant of integration. Since at x = 0, dy/dx =0, we have C = 0. Integrating once again,

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EAk

133

2  dy  W  (lx  x / 2)dx 

 l3  W  l4  EAk 2 y  W    1    3  2l  4 

Wl 3 W1l 3 EAk y   . 3 8

3 l3 y  (W  W1 ) 8 3EAk 2

W1 (l 2 x  lx 2  x3 / 3)dx 2l 0

4.3.1 Measurement of E (1) Cantilever depression : The given beam is clamped rigidly at on end (Fig. 4.3.1(a)). A weight – hanger (H) is suspended at the free end of the beam. A pin (P) is fixed vertically by some wax at the free end of the beam. A traveling microscope (M) is focused on the pin. The microscope is adjusted so that the horizontal cross-wire coincides with the tip of the pin and the reading on the vertical scale is noted. Then weights m, 2m, 3 m, 4 m, etc., are added to the weight-hanger. The microscope is adjusted each time to make the horizontal cross-wire coincide with the tip of the pin and the reading on the vertical scale of the microscope is noted in each case. Observations are made for decreasing loads also. The results are tabulated as follows:

Fig. 4.3.1(a) Load in kg

Microscope Reading Load Increasing

Load Decreasing

Mean

Depression For M(= 4 m)

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Mean depression for a load of M kg

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W W+m W + 2m W + 3m

x x1 x2 x3

W + 4m W + 5m W + 6m W + 7m

x4 x5 x6 x7

x4 - x x5 - x1 x6 - x2 x7 - x3

The mean depression(y) for a load M kg is found. The length of the beam (l0ween the clamped end and the loaded end is measured. The mean breadth (b) of the beam and its mean thickness (d) are determined. If y is the depression produced for a load of Mg, then,

y= 

Now,

Mgl3 Mgl3 orE  . 3EAk 2 3Ak 2 y 2

Ak = bd / 12 for a rectangular beam.

Mgl3 4Mgl 3 Hence, E =  3(bd3 /12)y bd 3 y. The young’s modulus of the material of the beam is calculated using this relation. (2) E - by measuring the tilt in a loaded cantilever. The given rectangular beam is rigidly clamped at one end and a small plane mirror M is fixed at the free end [Fig. 4.3.1(b)]. A weight hanger (H) is attached at the free end of the beam. A vertical scale (S) and telescope (T) are arranged in front of the mirror. The telescope is focused so that the image of the vertical scale due to reflection in the mirror is obtained in the telescope. The reading on the scale which coincides with the horizontal cross-wire is noted. Then weights m, 2m, 4m etc., are added to the weight-hanger and the readings of the scale as observed in the telescope, are noted in each case.

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Fig 4.3.2(b) Observations are made for decreasing loads also.

The results are tabulated as

follows: Load in kg

Readings on the scale Load increasing Load decreasing

Mean

Change in scale reading for M kg

The average of the readings in the last column gives the mean change in scale reading s for a load of M kg. The distance D between the mirror and the scale is found. The breadth (b) and the thickness (d) of the beam are accurately measured.

 s    of the cantilever for a load of M kg  2D The angle between the two ends



But,

Mg.l2 Mg.l2  2EAk 2 2E.bd3 /12

…(1)

 bd 3  2 SinceAk    12  

6Mgl 2  . bd 3 E From (1) and (2),

…(2)

s 6Mg.l 2 12Mg.l 2 D  orE  . 2 D bd 3 E bd 3 s

4.3.2 Oscillations of a cantilever Let OA be a cantilever of length l, of negligible mass fixed at O. Let a mass M be attached at the other end A (Fig.4.3.2). If the mass is slightly depressed and then released, the cantilever will execute simple harmonic motion about its original depressed position.

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Fig.4.3.2 The depression of the loaded end of the cantilever is

y

Wl 3 3EAk 2

W

3EAk 2 y. l3

This must be equal to the elastic reaction of the cantilever balancing it and hence directed oppositely to it. 2

If M is the mass of the weight W and d y/dt , the acceleration (upwards), we have, elastic reaction =

d2y dt 2

d 2 y 3EAk 2  y dt 2 l3



M

d 2 y 3EAk 2  y. dt 2 Ml 3

But,

3EAk 2 = A constant Ml 3 The acceleration of mass M or the free end of the cantilever is thus proportional to its

displacement and is directed opposite to it. It, therefore, executes a S.H.M. of time period T, given by

Displacement y Ml 3 T  2  2  2 Acceleration 3EAk 2  3EAk 2 y    3  M .l  If the mass of the cantilever is not negligible, it can be shown that,

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137

1 ( M  m)l 3 3 where m = mass of the cantilever. T  2 3EAk 2 The mass of the cantilever can be eliminated by finding the periods T 1 and T2 for two different masses M1 and M2 attached to the cantilever at the same length. Then,

1 1 ( M1  m)l 3 ( M 2  m)l 3 2 2 3 3 and T2  4  T12  4 2 2 3EAk 3EAk 2

T2 2  T12 

4 2 ( M 2  M1 )l 3 . 3EAk 2

4 2 ( M 2  M1 )l 3 E . 3 Ak 2 (T2 2  T12 )



Experiment: The given beam is rigidly clamped at O. A certain load of M1 kg is suspended from the other end A. The beam is set in transverse oscillations and the time for 25 oscillations is found. From this the period of oscillation T 1 is calculated. Similarly, the period T 2 with at load M2 is found.

E

We have, For a rectangular bar,

E

Hence,

4 2 ( M 2  M1 )l 3 . 3 Ak 2 (T2 2  T12 ) 2

Ak = bd / 12.

16 2l 3 ( M 2  M1 ) bd 3 (T2 2  T12 )

The length of the cantilever l, the breadth b and depth d are measured. E is calculated using the above formula. -2

-3

Example 1: A steel bar, 0.3 m long, 2 X 10 m broad and 2 X 10 m thick is clamped at one end and loaded at the other with a mass of 0.01 kg. Calculate the period of vibration of the bar, neglecting the effect of weight of the bar. (E for steel = 20 X 10

For a cantilever of length l, loaded  Ml 3   T  2 with a mass M, period of vibraiton  3EAk 2 10

Here, l = 0.3m, M = 0.01 kg; E = 20 X 10 Nm 2

-2

-3 3

-2

Ak = bd / 12 = (2 X 10 )(2 X 10 ) /12 =



4 X 1011 3

(0.01) X (0.3)3 T  2  0.0365sec onds. 4 10 11  3 X (20 X 10 ) X  X 10  3 

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Nm ).

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138

4.3.3 Depression at the mid-point of a beam loaded at the middle Let AB represent a beam of length l, supported on two knife-edges at A and B and loaded with a weight W at the centre C.

The reaction at each knife-edge is W/2 acting vertically

upwards. The beam bends as shown in Fig. 4.3.3, the depression being maximum at the centre. The bending is non-uniform. Let CD = y. The portion DA of the beam may be considered as a cantilever of length l/2, fixed at D and bending upwards under a load W/2. Hence the elevation of A above D or, the depression of D below A = y =

(W / 2)(l / 2)3 Wl 3  3EAk 2 48EAk 2

Note: The inclination of the tangent at the points A and B is given by

Fig. 4.3.3

tan  

dy Wl 2  dx 16 EAK 2

Since  is small, tan    . 

Wl 2  16 EAK 2

4.3.4 Uniform bending of a beam Consider a beam of negligible mass supported symmetrically on two knife-edges A and B in a horizontal level (Fig. 4.3.4 (a)). Let AB = 1. Let equal weights W, W be added to the beam at its ends C and D. Let AC = BD = a. Then the beam is bent into an arc of a circle. The reactions on the knife-edges will then be W and W, acting vertically upwards. Consider the cross-section of the beam at any point P. The only forces acting on the part PC of the beam are the forces W at C and the reaction W at A. The external bending moment with respect to P

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139

 W .Cp  W .AP  W (CP  AP)  W .AC  Wa. 2

This must be balanced by the internal bending moment EAk /R. 2

Hence,

…(1)

Wa = EAk /R

Fig. 4.3.4 (a) 2

Since for a given load W, E, a and Ak are constant, R is a constant. The bending is then said to be uniform. If y is the elevation of the mid-point of AB above its normal position (Fig. 4.3.4 (b)),

Fig. 4.3.4 (b) EF(2R – EF) = AF

y(2R – y) = (l / 2)

y.2R = l / 4

(

y is negligible) 2

y = l / 8R From (1),

1 Wa  R EAk 2

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y



140

Wal 2 8EAk 2

4.4 Measurement of Young’s modulus – By bending of a beam (1) Non-uniform Bending: The given beam is symmetrically supported on two knife-edges (Fig. 4.4 (1)). A weighthanger is suspended by means of a loop of thread from the point C exactly midway between the knife-edges. A pin is fixed vertically at C by some wax. A traveling microscope is focused on the tip of the pin such that the horizontal cross-wire coincides with the tip of the pin. The reading in the vertical traverse scale of microscope is noted. Weights are added in equal steps of m kg and the corresponding readings are noted. Similarly, readings are noted while unloading. The results are tabulated as follows: -

Fig. 4.4 (1) Load in kg

Readings on the microscope Load increasing Load decreasing

Mean

y for M kg

The mean depression y is found for a load of M kg. The length of the beam (l) between the knife-edges is measured. The breadth b and the thickness d of the beam are measured with a vernier calipers and screw gauge respectively. Then,

y

Wl 3 Wl 3 orE  48EAk 2 48 Ak 2 y

E

Mgl 3 48 X (bd 3 /12) Xy

( W  Mg and A k 2  bd 3 /12)

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E



141

Mgl 3 4bd 3 y

Example 1: In an experiment a rod of diameter 0.0126 m was supported on two knife-edges, placed 0.7 metre apart. On applying a load of 0.9 kg exactly midway between the knife-edges, the depression on the middle point was observed to be 0.00025 m.

Calculate the Young’s

modulus of the substance.

E

Mgl 3 (0.9)(9.8)(0.7)3  12 y r 4 12(0.00025) (0.0063) 4

E  2.039 X 1011 Nm2



(2) Uniform bending: The given beam is supported symmetrically on two knife-edges A and B (Fig. 4.4 (2)). Two equal weight-hangers are suspended, so that their distances from the knife-edges are equal. The elevations of the centre of the beam may be measured accurately by using a single optic level (L). The front leg of the single optic lever rests on the centre of the loaded beam and the hind legs are supported on a separate stand. A vertical scale (S) and telescope (T) are arranged in front of the mirror. The telescope is focused on the mirror and adjusted so that the reflected image of the scale in the mirror is seen through the telescope. The load on each hanger is increased in equal steps of m kg and the corresponding readings on the scale are noted. Similarly, readings are noted while unloading. The results are tabulated as follows:

Fig. 4.4 (2) Load in kg

Readings of the scale as seen in the telescope Load increasing Load decreasing Mean

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Shift in reading for M kg

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The shift in scale reading for M kg is found from the table. Let it be S. If D = The distance between the scale and the mirror, x = the distance between the front leg and the plane containing the two hind legs of the optic lever, then

y = Sx / 2D.

The length of the beam l between the knife-edges, and a, the distance between the point of suspension of the load and the nearer knife-edge (AC = BD = a) are measured. The breadth b and the thickness d of the beam are also measured.

y

Then, 2

Wal 2 Sx Mgal 2 or  8EAk 2 2 D 8E (bd 3 /12)

[Since W = Mg and Ak = bd / 12]

3Mgal 2 D E Sxbd 3



Pin and Microscope Method: The given beam is supported symmetrically on two knife-edges A and B. Two equal weight-hangers are suspended so that their distances from the knife-edges are equal. A pin is placed vertically at the centre of the beam. The tip of the pin is viewed by a microscope. The load on each hanger is increased in equal steps of m kg and the corresponding microscope readings are noted. Similarly, readings are noted while unloading. The results are tabulated as follows: Load in kg

Readings of the microscope Load increasing Load decreasing

Mean

y for M kg

The mean elevation (y) of the centre for M kg is found. The length of the beam l between the knife-edges and a, the distance between the point of suspension of the load and the nearer knife-edge (AC = BD = a) are measured. The breadth b and the thickness d of the beam are also measured.

y



E

Wal 2 Mgal 2  8EAk 2 8E (bd 3 /12)

 bd3  2 W  Mg and Ak =   12  

Mgal 2 2bd 3 y

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Using the above formula we can calculate the Young’s modulus of the material of the beam. Example 1: Distinguish between uniform and non-uniform bending. In uniform bending every element of the beam is bent with the same radius of curvature (R). In non-uniform bending, R is not the same for all the elements in the beam. 4.5 Koenig’s Method The beam is supported on two knife-edges K1 and K2 separated by a distance l. Two plane mirrors m1 and m2 are fixed near the two ends of the beam at equal distances beyond the knife-edges. {Fig. 4.5 (a)}. The two plane mirrors face each other and they are inclined slightly outwards from the vertical.

Fig .4.5 (a) An illuminated translucent scale and a telescope (T) are arranged as shown.

The

reading of a point C on the scale as reflected first by m 2 and then by m1 is viewed in the telescope. Let the load suspended at the mid-point of the beam be M. The beam is then bent and the bending is non-uniform. The mirrors at the ends are turned towards each other [Fig. 4.5. (b)]. Let the shift in the scale reading be s. The Young’s modulus of the material of the beam is then calculated from the relation

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144

Fig. 4.5 (b)

E where

Mgl 2 (2 D  L) 2bd 3 s

l = Distance between the knife-edges D = Distance between the scale and the remote mirror, m 2 L = Distance between the two mirrors. s = Shift in scale reading for a load of M kg b = Breadth of the beam d = Thickness of the beam



Wl 2 16 EAK 2

The mirrors m1 and m2 also turn through the same angle  due to loading. In fig. 4.5 (b), m1 and m2 represent the initial and m 1 and m2 the displaced positions of the mirrors. Originally, the image of the scale division at C coincides with the cross-wire and finally when the load is applied, H is seen to be in coincidence with the cross-wire. For convenience in evaluating , consider the rays of light to be reversed in their path. TQEC will be the original path. When m 1 is turned through an angle  to the position m 1, QE is turned through 2 and strikes m2 at G. Then EG = L2. The ray GH is turned through an angle 4, since, in addition to QE having moved through 2, m2 itself has turned through . Draw GK parallel to EC. Then, KGH = 4 and CK = EG. KH = D4 

The total shift in scale reading = s = CK + KH = EG +KH (

CK = EG)

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145 = L2 + D4 = (L+2D)2



But

Wl 2 16 EAK 2

Hence,

Wl 2 16 EAK 2 Wl 2 ( L  2 D) E 8 AK 2 s

s  ( L  2 D)2

Now Ak = bd /12 for a beam of rectangular cross-section and W = Mg. 

E

Mgl 2 ( L  2 D) 3Mgl 2 (2D  L)  8(bd 3 /12) s 2bd 3 s

4.6 I Section of girders: A girder supported at its two ends as on the opposite walls of a room, bends under its own weight and, or, under the load place above it. The middle portion gets 3

depressed. The depression (y) at the mid-point of a rectangular beam is proportional to Wl /Ebd . For the depression (y) to be small for a given load (W), the length of the girder (l) should be small and its breadth (b), depth (d) and Young’s modulus for its material (E) must be large. Due to depression, the upper parts of the beam above the neutral surface contract, while those below the neutral surface extend. Hence the stresses have a maximum value at the top and bottom and progressively decrease to zero as we approach the neutral surface from either face.

Therefore, the upper and the lower surfaces of the beam must be stronger than the

intervening part. That is why the two surfaces of a girder or iron rails (for railway tracks etc.) are made much broader than the rest of it, thus giving its cross-section the shape of the letter I. In this manner, material will be saved, without appreciably impairing its strength. Torsion 4.7 Torsion of a body When a body fluid at one end and twisted about its axis by means of a torque at the other end, the body is said to be under torsion. Torsion involves shearing strain and so the modulus involved is the rigidity modulus. 4.8 Torsion of a cylinder-Expression for torque per unit twist Consider a cylindrical wire of length L and radius a fixed at its upper end and twisted through an angle  by applying a torque at the lower end. Consider the cylinder to consist of an infinite number of hollow co-axial cylinders. Consider one such cylinder of radius x and thickness dx. [Fig. 4.8(1)].

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Fig. 4.8. (1)

Fig. 4.8. (2)

A line such as AB initially parallel to the axis OO' of the cylinder is displaced to the position AB′ through an angle  due to the twisting torque [Fig. 4.8(2)]. The result of twisting the cylinder is a shear strain. The angle of shear = BAB' = . Now

BB′ = x. = L  or  = x./L

We have, rigidity modulus = G = 

Shearing stress = G. = Gx/L But,



Shearing stress angle of shear (  )

Shearing stress =

Shearing force Area on which the force acts

Shearing force = Shearing stress X Area on which the force acts. The area over which the shearing force acts = 2x dx Hence, the shearing force = F =

Gx X 2 xdx L

The moment of this  2 G 3  Gx force about the axis   2 xdx.x  x dx L L OO' of the cylinder   

a Twisting torque on the  2 G 3  C  x dx   whole cylinder L  0

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C

147

 Ga 4 2L

The torque per unit twist (i.e.,   Ga 4  c   the torque when  = 1 radian)  2L Note 1: When an external torque is applied on the cylinder to twist it, at once an internal torque, due to elastic forces comes into play. In the equilibrium position, these two torques will be equal and opposite. Note 2: If the material is in the form of a hollow cylinder of internal radius a and external b, then, b The torque acting  2 G 3  G 4 4  C  x dx  (b  a )   on the cylinder  L 2L a

Torque per unit twist = c = G(b -a )/(2L) 4

Example 1: What torque must be applied to a wire one metre long, 10

-3

metre in diameter in

order to twist one end of it through 90 , the other end remaining fixed? The rigidity of the material of the wire is 2.8 X 10

-2

Nm .

Here, L = 1m; G = 2.8 X 10

-2

Nm ; a =

103 m  0.5 X 103 m; 2

 = 90 = /2 radian; o



C

 Ga 4 2L



 (2.8 X 1010 )(0.5 X 103 )4 2 X1

 2

-3

= 4.318 X 10 Nm. -3

Example 2: A circular bar one metre long and 8 X 10 metre diameter is rigidly clamped at one end in a vertical position. A torque of magnitude 2.5 Nm is applied at the other end. As a result, a mirror fixed at this end deflects a spot of light by 0.15 metres on a scale one metre away. Calculate the modulus of rigidity of the bar. For a twist , the mirror turns through , and the reflected beam through 2. If the deflection is d on a scale D away, 2D = d or

=

d 0.15   0.075radians; 2D 2 X 1

Here, C= 2.5 Nm; a = 4 X 10 m;  = 0.075 radians; L = 1m; G = ? -3

Hence,

i.e.,

G

C

 Ga 4 2L

G

C.2 L  a 4

2.5 X 2 X 1  8.290 X 1010 Nm2 3 4  (4 X 10 ) (0.075) -4

-3

Example 3: A steel wire of diameter 3.6 X 10 m and length 4 m extends by 1.8 X 10 m under -5

a load of 1 kg and twists by 1.2 radians when subjected to a total torsional torque of 4 X 10 Nm at one end. Find the values of E, G and v for steel.

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We have,

E

148

F / A FL  l/L A.l

Here, F = mg = 1 X 9.8 = 9.8N: L = 4m; A = a = (1.8 X 10 ) m and l = 1.8 X 10 m. 2

-3

9.8 X 4  2.139 X 1011 Nm2 4 2 3  (1.8 X 10 ) X 1.8 X 10

E



-4 2

Torque which must be applied   Ga 4  to twist one end of the wire   C  2L  through an angle  radians  G

C.2 L  a 4 -5

-4

Here, C = 4 X 10 Nm; L = 4m; a = 1.8 X 10 m;  = 1.2 radians 

G

v

(4 X 105 ) X 2 X 4 C.2 L =  0.8083 X 1011 Nm2 4 4 4  a   (1.8 X 10 ) X 1.2

E 2.139 X 1011 1   1  1.323  1  0.323. 2G 2 X 0.8083 X 1011

Example 4: Explain why a hollow rod is a better shaft than a solid one of the same mass, length and material. Consider a solid cylinder of length L, radius r and shear modulus G.

The torque required to twist the   Gr 4  C   1 solid cylinder through an angle   2L

…(1)

Let r1 and r2 be the inner and outer radii of the hollow cylinder of the same length, mass and material. Then the torque required to twist it through the same angle  is

C2  Hence,

 G(r2 4  r14 ) 2L

…(2)

C2 r2 4  r14 (r2 2  r12 )(r2 2  r12 )   C1 r4 r4

Since the two cylinders have the same mass,

 (r22  r12 )l    r 2l. (where  is the density of the material of the cylinders). or

r2 2  r12  r 2

…(3)

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Adding 2r1 to both sides, Hence,

or 

149

r22  r12  r 2  2r12

…(4)

C2 (r 2  2r12 )r 2 r 2  2r12   C1 r4 r2

C2 2r 2  1  12 C1 r C2>C1 i.e., the twisting torque for a hollow cylinder is greater than that for a solid cylinder of the same mass, length and material. Hence a hollow cylinder is stronger and a better shaft than a solid one of the same mass, length and material.

Torsional oscillations of a body Suppose a wire is clamped vertically at one end and the other end carries a body (i.e., a disc, bar or a cylinder) of moment of inertia I about the wire as the axis. Let the length, radius and rigidity modulus of the wire be respectively l, a and G. When the body is given a slight rotation by applying a torque, say by the hand, the wire is twisted. If the body is released, the body oscillates in the horizontal plane about the wire as axis. These oscillations are called Torsional oscillations and the arrangement is known as a Torsion pendulum. Let us consider the energy of the vibrating system when the angle of twist is . Let  be the angular velocity of the body. The potential energy of the wire due to the twist =

1 2 c. . 2

The kinetic energy of the  1 2 1  d 2   I  I   body due to its rotation  2 2  dt  The total energy  1  d 2 1 2  c  cons tan t  I of the system  2  dt  2 Differentiating this with respect to t,

1 d d 2 1 d I .2 . 2  c2  0. 2 dt dt 2 dt or

d 2 d 2 c I 2  c or 2    0 dt dt I

The body has simple harmonic motion and its period is given by

T  2

I c

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4.9 Rigidity modulus by Torsion pendulum (Dynamic torsion method): The torsion pendulum consists of a wire with one end fixed in a split chuck and the other end to the centre of a circular disc as in Fig. 4.9. Two equal symmetrical masses (each equal to m) are placed along a diameter of the disc at equal distances d1 on either side of the centre of the disc. The disc is rotated through an angle and is then released. The system executes torsional oscillations about the axis of the wire. The period of oscillations T1 is determined.

Fig. 4.9. Then

T1  2

T12 

I1 c

4 2 I1 . c

Here, I1 = Moment of inertia of the whole system about the axis of the wire and c= torque per unit twist. Let I0 = M.I. of the disc alone about the axis of the wire. i = M.I. of each mass about a parallel axis passing through its centre of gravity. Then by the parallel axes theorem,

I1  I 0  2i  2md12 

4 2  I 0  2i  2m.d12  . T  c 2 1

…(1)

The two masses are now kept at equal distances d2 from the centre of the disc and the corresponding period T2 is determined. Then,

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T2 2  

151

4 2  I 0  2i  2m.d 2 2  . c

…(2)

4 2 .2m.  d 2 2  d12  c

…(3)

T2 2  T12 

 Ga 4

But

Hence

T2  T 

G

2L 2 1

4 2 .2m.  d 2 2  d12  2 L

 Ga 4

16 Lm(d 2 2  d12 ) a 4 (T2 2  T12 )

Using this relation, G is determined. M.I. of the disc by torsional oscillations. The two equal masses are removed and the period T0 is found when the disc alone is vibrating. Then,

cT0 2 4 2 T0  I 0orI 0  c 4 2 2

…(4)

4 2 .2m.  d 2 2  d12 

From (3),

c

Hence

4 2 .2m.  d 2 2  d12  T0 2 2m.  d 2 2  d12  T0 2 I0  . 2  T2 2  T12 4 T2 2  T12

T2 2  T12

From this relation, the moment of inertia of the disc about the axis of the wire is calculated. Maxwell’s needle: Maxwell’s needle consists of a hollow metal tube suspended from a torsion head T by a wire whose rigidity modulus is required. Two solid and two hollow cylinders of equal lengths exactly fit into the tube. The length of each cylinder is equal to one quarter of the length of the needle. The solid cylinders (S and S) are first placed in the inner position and the hollow cylinders (H and H) at the end as shown in Fig. (i). The needle is then rotated through a small angle about the wire as axis and let go. The system oscillates torsionally about the wire as axis and the time period T1 is found. The positions of the solid and hollow cylinders are then interchanged [Fig. (ii)] and the period of torsional oscillations T 2 is found. The length of the wire (L) and the radius of the wire (a) are also measured.

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Let I0 = M.I. of the hollow tube wire as axis, i1 = M.I. of the solid cylinder about an axis through its C.G. parallel to the axis of the wire. i2 = M.I. of the hollow cylinder about an axis through its C.G. parallel to the axis of the wire, m1 = mass of the solid cylinder, m2 = mass of the hollow cylinder. d = the length of each cylinder. Then, the M.I. I1 of the system about the axis of the wire in the first case is given by, 2

I1 =I0 + 2i1 +2i2 +2m1 (d/2) + 2m2 (3d/2) Further,

T1  2 I1 / c or

…(1)

T12  4 2 I1 / c

…(2)

Similarly, the M.I. I2 of the system about the axis of the wire in the second case is given by, 2

I2 =I0 + 2i1 +2i2 +2m2 (d/2) + 2m1 (3d/2) and

T2 2  4 2 I 2 / c

…(4)

4 2 Subtracting (2) from (4), T2  T   I 2  I1  c 2

…(3)

2 1

Now, subtracting (1) from (3),

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I 2  I1  4d 2  m1  m2  and c = Hence,

T2 2  T12  G

 Ga 4 2L

4 2 X 2 L 2 4d  m1  m2   Ga 4 32 Ld 2 X  m1  m2 

T

…(5)

 T12  a 4

Example 1: A metal disc of 0.1 m radius and mass 1 kg is suspended in a horizontal plane by a -3

vertical wire attached to its centre. If the diameter of the wire is 10 m, its length 1 m, and the period of torsional vibrations is 4 seconds, find the rigidity modulus of the wire. The metal disc executes torsional oscillations about the axis of the wire. Hence

T  2 I / c or T 2  4 2 I / c , where I = moment of inertia of the disc about the axis of the wire and c = torque per unit twist =

 Ga 4 2L

T 2  4 2



8 IL 1 or G  2 4 2 T a ( Ga / 2 L)

Here, I = MR /2 [where M = mass of the disc = 1 kg and R = radius of the disc = 0.1m]. 

I = 1(0.1) /2 = 0.005 kg m . L = Length of the wire = 1m; a = 0.5  10 m; T = 4s. -3



G

8 IL 8 (0.005) 1 = 2  1.256 1011 Nm2 2 4 3 4 T a 4 (0.5 10 )

Example 2: A steel bar is suspended in a horizontal position by a vertical wire attached to its o

centre. A horizontal torque of moment 5 Nm twists the bar horizontally through an angle of 12 . When the bar is released, it oscillates as a torsion pendulum with a period of ½ s. Determine the moment of inertia. A torque of moment 5 Nm produces a twist of 12 or 12  /180 rad = 0.2090 rad. o

Torque per unit twist = c = 5/0.2090 = 23.88 Nm.

T  2 I / c or T 2  4 2 I / c

Period of oscillation = 

Here, 

Moment of inertia = I = cT /4 . c = 23.88 Nm, T = 0.5 s 2

I = (23.88)(0.5) /4 = 0.1513 Kg m .

Example 3: A body, suspended symmetrically from the lower end of a wire, 1 m long, 1.22  10

-3

m is diameter, oscillates about the wire as axis with a period of 1.25 seconds. If the modulus of

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rigidity of the material of the wire is 8  10

-2

Nm , Calculate the moment of inertia of the body

about the axis of rotation. The body executes torsional oscillations about the axis of the wire with a time-period

T  2 I / c Where I = M.I. of the body about the about the axis of the wire or the axis of rotation and c = torque per unit twist =

T 2  4 2 I / c .

Hence or

 Ga 4

I T2

c 4 2



I

T2

 Ga 4 2 L4 2

T 2Ga 4 8 L

Here, T = 1.25s; G = 8  10 Nm ; a = 0.61  10 m; L = 1 m. 10

Hence

I

-2

-3

(1.25)2 (8X1010 )(0.61X 103 )4 8 X 1 -4

= 6.885 X 10 Kgm . 4.10 Determination of rigidity modulus – Static torsion method Searle’s apparatus: The experimental rod is rigidly fixed at one end A and fitted into the axle of a wheel W at the other end B (Fig. 4.10.). The wheel is provided with a grooved edge over which passes a tape. The tape carries a weight hanger at its free end. The rod can be twisted by adding weights to the hanger. The angle of twist can be measured by means of two pointers fixed at Q and R which move over circular scales S1 and S2. The scales are marked in degrees with centre zero.

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155 Fig. 4.10.

With no weights on the hanger, the initial readings of the pointers on the scales are adjusted to be zero. Loads are added in steps of m kg (conveniently 0.2 kg). The readings on the two scales are noted for every load, both while loading and unloading. The experiment is repeated after reversing the twisting torque by winding the tape over the wheel in the opposite way. The observations are tabulated. The readings in the last column give the twist for a load of M kg for the length QR (=L) of the rod. Reading on S1 Torque Clockwise

Torque Anticlockwise

Torque Clockwise

Load Load increasin g

Load decreasi ng

Load increasin g

= 2 1

Reading on S2

Mean 1

Load decreasi ng

Load increa sing

Load decr easin g

Torque Anticlockwise Load increasin g

Loa d dec rea sin g

M ea n 2

The radius a of the rod and the radius R of the wheel are measured. If a load of M kg is suspended from the free end of the tape, the twisting torque = MgR. The angle of twist =  degrees = ./180 radians. 

The restoring torque =

For equilibrium,

 Ga 4 

MgR 

180

 Ga 4  . 2 L 180

. orG 

360MgRL  2 a 4

Since a occurs in the fourth power in the relation used, it should be measured very accurately. Notes: (1) We eliminate the error due to the eccentricity of the wheel by applying the torque in both clockwise and anticlockwise directions. (3) We eliminate errors due to any slipping at the clamped end by observing readings at two points on the rod.

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4.10.1 Determination of rigidity modulus – Static torsion method. (Searle’s apparatus – Scale and Telescope) A plane mirror strip is fixed to the rod at a distance L from the fixed end of the rod [Fig. 4.10.1]. A vertical scale (S) and telescope (T) are arranged in front of the mirror. The telescope is focused on the mirror and adjusted so that the reflected image of the scale in the mirror is seen through the telescope. With some dead load W on the weight-hanger, the reading of the scale division coinciding with the horizontal cross-wire is taken. Weights are added in steps of m kg and the corresponding scale readings are taken. Weights are then decreased continuously in steps of m kg and the readings taken again. The torque is reversed now, by passing the tape anticlockwise on the wheel. The readings are taken as before. From these readings, the shift in scale reading (s) for a load m kg is found.

Fig 4.10.1 The length L of the rod from the fixed end to the mirror is measured. The mean radius a of the rod is accurately measured with a screw gauge. The radius (R) of the wheel is found by measuring its circumference with a thread. The distance (D) between the scale and the mirror is measured with a metre scale. G is calculated using the formula G =

4mgRLD  a4s

Telescope Reading Load in kg

Torque clockwise Loading

Unloadin g

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Mean (Y)

Mean (X)

Unloadin g

X Y 2

Torque anticlockwise

Shift in scale reading for 4m kg

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W W+m W+2m W+3m W+4m W+5m W+6m W+7m Work done in twisting a wire Consider a cylindrical wire of length L and radius a fixed at its upper end and twisted through an angle  by applying a torque at the lower end. If c is the torque per unit angular twist of the wire, then the torque required to produce a twist  in the wire is C = c. The work done in twisting the wire through a small angle d is Cd = c d.  The total work done in twisting   W   0 c. d the wire through an angle  



1 W  c. 2 2

The work done in twisting the wire is stored up in the wire as potential energy.

-3

Example 1: Find the amount of work done in twisting a steel wire of radius 10 m and of length o

0.25 m through an angle of 45 . Given G for steel = 8 X 10 We have, Here,

1 1  Ga 4 2   Ga 4  W  c. 2    c  2 2 2L 2L  

G = 8 X 10

Nm ; a = 10 m;  = 45 = /4 rad; -2

-3

L = 0.25 m 

10 -3 4 1  (8X10 )(10 )  /4  W 2 2 X 0.25

-2

Nm .

 0.1550 J .

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158 UNIT V

VISCOSITY AND SURFACE TENSION 5. Introduction When two parallel layers of a liquid are moving with different velocities, they experience tangential forces which tend to retard the faster layer and accelerate the slower layer. These forces are called forces of viscosity. Consider two layers of liquid separated by a distance dz (Fig. 2.1). Let  and +d be the velocities of two layers. So the velocity gradient is d/dz. Let A be the surface area of the layer. The viscous force is directly proportional to the surface area A and velocity gradient d/dz i.e.,

F A

d d orF   A dz dz

...(1)

Where η is a constant for the liquid and called coefficient of viscosity. If A = 1. and d/dz = 1, we have F= η. The coefficient of viscosity is defined as the tangential force per unit area required to maintain a unit velocity gradient. -2

Unit of η is Nsm , It is called the pascal second. Dimensions of

[ ] 

[F ] MLT 2  2  ML1T 1 1 [ A][(dv / dz )] L ( LT / L)

Streamline Flow and Turbulent Flow Consider a liquid flowing in a pipe. Let the velocity of flow be 1 at A, 2 at B and 3 at C (Fig. 5). If as time passes, the velocities at A, B, and C are constant in magnitude and direction, then the flow is said to be steady. In a steady flow, each particle follows exactly the same path and has exactly the same velocity as its predecessor. In such a case, the liquid is said to have an orderly or streamline flow.

Fig. 5. The line ABC is called a stream-line, which is the path followed by an orderly procession of particles. The tangent to the streamline at any point gives the velocity of the liquid at that point. The flow is steady or streamlined only as long as the velocity of the liquid does not exceed a limiting value, called the critical velocity. When the external pressure causing the flow of the liquid is excessive, the motion of the liquid takes place with a velocity greater than the

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critical velocity and the motion becomes unsteady or turbulent.

This caused eddies and

whirlpools in the motion of the liquid. This turbulent motion is also known as vortex motion. The distinction between stream-line flow and turbulent flow can be demonstrated by injecting a jet of ink axially in a wider tube in which water is made to flow axially. When the velocity of the liquid is small, the ink will move in a straight line. As the speed of flow is increased beyond the critical velocity, the link will spread out, the showing that the motion has become turbulent. Definition of critical velocity. Critical velocity of a liquid is the velocity below which the motion of the liquid is orderly and above which the motion of the liquid becomes turbulent. Expression for the critical velocity. The critical velocity of a liquid may depend upon (i) the coefficient of viscosity of the liquid (η) (ii) the density of the (ρ) and (iii) the radius r of the tube through which the liquid is flowing. We may write c = K η ρ r a

b c

Where k is constant called Reynolds’ number. Writing the dimensions of these quantities,

[ LT 1 ]  [ML1T 1 ]a [ML3 ]b [ L]c [ LT 1 ]  [M ab L a3b  cT  a ] 

a+b = 0; -a-3b+c = 1 and –a = -1 From these equations we have, a = 1, b = -1 and c = -1.

vc 



k . r

Significance of Reynold’s number :

k

vc  r



The significance of the Reynold’s number k is that its value determines the nature of flow of flow of a liquid through a tube. In the case of apparatus, geometrically similar, whatever their actual dimensions, turbulence in at the same constant value of Reynold’s number in all cases of liquid flow. The flow will be steady and stream-line in each individual case, until the number is not exceeded. After exceeding this number, the flow becomes turbulent. Eventhough the values of r, ρ and η may all vary from each other, but so long as k remains the same, the liquid-flow will be similar in all the cases. Poiseuille’s formula for the flow of a liquid through a capillary tube

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Fig. 5(a)

Fig. 5 (b)

Suppose a constant pressure difference p is maintained between the two ends of the capillary tube of length l and radius a, as shown in Fig. 5. (a). Consider the steady flow of a liquid of coefficient of viscosity η through the tube. The velocity of the liquid is a maximum along the axis and is zero at the walls of the tube. Assume that there is no radial flow. Consider a cylindrical shell of the liquid co-axial with the tube of inner radius r and outer radius r+dr [Fig. 5 (b)]. Let the velocity of the liquid on the inner surface of the shell be  and that on the outer surface be  -d. (d/dr) is the velocity gradient. The surface area of the shell = A = 2πrl. According to Newton’s law of viscous flow, the backward dragging tangential force exerted by the outer layer on the inner layer, opposite to the direction of motion.

F1   A

d dv   2 rl dr dr

The driving force on the liquid shell, accelerating it forward

F2  p r 2 2

where p= pressure difference across the two ends of the tube and πr = Area of cross-section of the inner cylinder. When the motion is steady, Backward dragging force (F1) = The driving force (F2)

 2 rl

Integrating,

dv p   r 2ordv  rdr. dr 2l

v

 p r2  C. 2l 2

Where C is a constant of integration. When r = a,  = 0, Hence, 0 =



v

 p a2 pa 2  C or C  2 l 2 4 l

p 2 2 (a  r ). 4l

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This gives us the average velocity of the liquid flowing through the cylindrical shell. Hence the volume of the liquid that flows out per second through this shell.

dv =

 Area of cross-section of the shell of    X Velocity of flow  radius r and thickness dr  =

2 rdr

p 2 2 p 2 (a  r )  (a r  r 3 ) dr 4l 2l

The volume of the liquid that flows out per second is obtained by integrating the expression for dV between the limits r = 0 to r = a. a

p 2  p  2 r2 r4  V  (a r  r 3 )dr  a  2l 2l  2 4  0 0 a

 V

 p a4 2 l 4

 pa 4 8 l

Corrections to Poiseuille’s formula Two important corrections are to be applied in the Poiseuille’s equation. (ii)

Correction for pressure head: The outgoing liquid acquires K.E. due to its velocity after passing through the tube. Hence the pressure-head maintained is utilized not only for overcoming viscous resistance but also in imparting considerable K.E. to emergent liquid. So the effective pressure is less and is given by

p1  p 

V 2

 2 4

This can be deduced as follows: The K.E. given to the liquid of density ρ per second a

E' =

1 2 3 0 2 (2 r dr  ) =  0 r dr

p 2 (a  r 2 ) 4 l

But

v



 p  2 2 3 E '    r   (a  r ) dr 4 l  0  a

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 p  a8   pa 4        2 4   4 l  8  8 l   a



V 3  2a4

The work done in overcoming viscosity is p1V whereas total work done per unit volume is pV. Here p1 is the effective pressure. 

V 3 pV  p1V  2 4  a

p1  p 



 V2  p1  g   h  2 4   a g  2

V 2  2a4

2 4

Thus [V /(π a g)] is the correction factor to the pressure head for gain of kinetic energy by the emergent liquid. (iii)

Correction for length of tube: At the inlet end of the tube, the flow of the liquid is not stream-line for some distance.

Consequently the liquid is accelerated.

The

effective length of the tube is thus increased from l to l + 1.64 a. Thus, the corrected relation for η becomes



 a4

 V2  h    g 8V (l  1.64a)   2a4 g 

Poiseuille’s method for determining coefficient of viscosity of a liquid. The liquid is taken in the constant tank upto a height h (Fig (i)). A capillary tube AB is fixed to the bottom of the tank. A weighted beaker is placed below the free end B of the capillary tube. The mass m of the liquid collected in it in time t is found out.

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Fig (i). Volume of liquid flowing per second = V =m/(ρ.t) where ρ is the density of the liquid. The length l of the capillary tube is measured by a meter rod. The radius a of the capillary tube is determined very accurately, using the traveling microscope. Then from the relation



 pa 4

(where p = hg),

8Vl

The value of η for the liquid can be easily calculated. 5.1 Comparison of viscosities: The liquid whose viscosity is η1 is first used in the constant level tank and the volume of liquid flowing per second =V 1 = m1/ ρ1.t is determined with a capillary tube. The tube is then taken out and cleaned well. The experiment is repeated for the other liquid whose viscosity is η2 and the volume of liquied flowing per second = =V2 = m2/ ρ1.t is determined for the same pressure head and with the same capillary tube. If l is the length of the tube,  its radius and ρ1 and ρ2 the densities of the two liquids,

1 

 h1 ga 4 8V1l

and

2 

 h2 ga 4 8V2l



1 1V2  2  2V1

ρ1/ ρ2 can be determined with a Hare’s apparatus. Thus the viscosities of two liquids can be compared. 5.2 Ostwald’s Viscometer This instrument is used to compare the viscosities of two liquids. Its is also used to study the variation of viscosity of a liquid with temperature.

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Fig. 5.2 The apparatus consists of two glass bulbs A and B joined by a capillary tube DE bent into a U-form (Fig. 5.2).

The bulb A is connected to a funnel F. The bulb B is connected to an

exhaust pump through a stop-cocks S. K, L, and M are fixed marks, as shown in figure. The whole apparatus is placed inside a constant temperature bath. The liquid is then introduced into the apparatus through the funnel and its volume its adjusted, so that the liquid occupies the portion between the marks K and M, When the stop-cock is closed. The stop-cock is now opened and with the help of the exhaust pump the liquid is sucked up above the mark K. The stop-cock is closed and the exhaust pump is removed. The stop-cock is again opened. The liquid is allowed to flow through the capillary tube. The time (t1) taken the liquid to fall from the mark K to the mark L is noted.

The

experiment is then repeated with the second liquid and the time (t 2) taken by it to fall from K to L is noted. Theory: Let η1 and η2 be the coefficients of viscosity and ρ2 the densities of the two liquids respectively. Let the volume of liquid between K and L be V. Then, the rate of flow of the first liquid = V1 = V/t1 and the rate of flow of the second liquid = V2 = V/t2 Now,

1 

 .1 .a 4 8V1 .l

and

2 

…(1) …(2)

 .2 .a 4 8V2 .l

1 V2 p1  X 2 V1 p2

…(3)

But the pressure P is proportional to the density of the liquid used (P = h ρ g) Hence,

Also, dividing (2) by (1),

p1 1  p2  2

…(4)

V2 t1  V1 t2

…(5)

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1 t1.1  2 t2 .2

Hence,

…(6)

From equation (6), η1/ η2 can be calculated. 5.2.1 Poiseuille’s method for determining coefficient of viscosity of a liquid. [Variable pressure head] The given liquid is poured into a graduated burette. The capillary tube is fixed as shown in figure (Fig. 5.2.1). The clip is opened fully. The liquid is allowed to flow slowly through the capillary tube. When the liquid-level in the burette crosses the zero marking, a stop clock is started. The readings of the stop clock are noted when the liquid-level crosses the 10 cc, 20cc, 30cc. etc., markings. The vertical height h between the capillary tube and midpoints of the range 0-10cc, 10-20cc, 20-30cc etc., are measured.

Fig. 5.2.1 The length of the capillary tube (l) is measured. The radius a of the capillary tube is measured thread or microscope.

The density of the liquid ρ is determined using Hare’s

apparatus. The readings are tabulated as follows: Burette

Stop

Burette

Volume

Mean

Time of

Reading

Clock

Reading

of liquid

Pressure

Flow

Reading

range

flowing

Head

seconds

m 0 10

0 to 10cc

-6

10 x 10

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h.t V

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166

10 to 20cc

“

20 to 30cc

“

Mean h.t/V = The coefficient of viscosity is calculated using the formula,



 ga 4  h.t  8l

  V 

Example 1: Water flow through a horizontal tube of length 0.2 metres and internal radius 8.1 x -4

-4

10 metre under a constant head of the liquid 0.2 metres high. In 12 minutes 8.64 x 10 m of liquid issues from the tube. Calculate the coefficient of viscosity of water. (The density of water = -3

-2

1000 kg m and g= 9.81 ms ). From Poiseuille’s formula,

V

 p.a 4 8 .l

-2

-4

Here, p = h.ρ.g = 0.2 x 1000 x 9.8 Nm ; a = 8.1 x 10 m

8.64 X 104 -6 3 l = 0.2m, V = = 1.2 X 10 m ; = ? 12 X 60



Now,

 p.a 4 8V .l

(3.14)(0.2 X 1000 X 9.8)(8.1X 104 ) 4 8(1.2 X 106 )0.2 -3

= 1.38 X10 Nsm

-2 2

Example 2: A vessel of cross-section 0.002m has at the bottom a horizontal capillary tube of length 0.1 m and internal radius 0.0005m. It is initially filled with water upto a height of 0.2m above the capillary tube. Find the time taken by the vessel to empty one-half of its contents, -2

given the viscosity of water = 10-3 Nsm . According to Poiseuille’s formula,

Volume of liquid flowing   pa 4  Out of a tube per second  = V  8 l  

…(1)

Let A be the area of cross-section of the vessel and h the height of water above the capillary tube. Then, ρ = hρg. Suppose in a small time flowing out in time dt = A dh. 

Rate of flow of water = -A dh/dt

From (1) and (2),

dt =



dh  (h g )a 4 A  dt 8l

8Al dh  (h g ).a 4 X  ga 4 h 8 .l IMTSINSTITUTE.COM

…(2)

THERMAL ENGINEERING

Integrating, t





167

8 Al h log e hh2 4  1  ga

h  8 Al log e  1  4  ga  h2 

Here, A = 0.002 m ; = 10 Nsm , l = 01 m;  = 1000 Kg 2

-3

-2

-3

-2

g = 9.8ms , a = 0.0005m; h1 = 0.2m and h2 = 0.1m.

t



8 X 0.002 X 103 X 0.1 log e 2  576.2s. 4  X 1000 X 9.8(0.0005)

Example 3: Calculate the mass of water flowing in 10 minutes through a tube 0.001 m diameter and 0.4 m long if there is a constant pressure head of water. The coefficient of viscosity of water -2

is 0.00082 Nsm .

  pa 4t  M  VXtX     8l 

Mass of water flowing in 10 minutes Here,

-2

p = hg = 0.2 X 1000 X 9.8 = 1960 Nm ; a = 0.0005 m. t = 10X60 = 600s; = 0.00082 Nsm and l = 0.4 m. -2

M



 X 1960 X (0.0005) 4 X 600 8 X 0.00082 X 0.4

X 1000  0.08795kg

Example 4: Calculate the maximum velocity with which water of coefficient of viscosity 0.001 -2

-4

Nsm flows through a tube of radius 6 x 10 m without turbulence being produced. Reynolds number is 1000. Find the rate of flow of water through the tube at this velocity and the pressure head required to maintain it if the length of the tube is 0.2 m. Critical velocity =

c 

k 1000 X 0.001   1.667ms 1 4  r 1000 X 6 X 10

Rate of flow of water through the tube V = Area X Velocity = .r X c = (6 X 10 ) 1.667 2

-4 2

-6

= 1.884 X 10 m /sec.

V

 p.a 4 8 .l

h

8 X 0.001X 0.2 X (1.884 X 10 6 )  0.7559m.  X 1000 X 9.8(6 X 104 )4

 (h g ).a 4 8l.V or h   g.a 4 8 .l

5.3 Searle’s Viscometer: Rotating cylinder method of finding .

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This method is suitable for finding the coefficient of viscosity of a very viscous liquid like castor oil, glycerin etc. Theory: Consider two coaxial cylinders, one suspended inside the other. The space between the two cylinders is filled with the liquid whose coefficient of viscosity () is to be determined. The outer cylinder B (of radius a) is fixed and the inner cylinder A (of radius b) is rotated with a constant angular velocity 1 [Fig. 5.3.]. The layer of liquid in contact with the inner cylinder will have an angular velocity 1 and the layer of liquid in contact with the outer cylinder will be at rest. The velocity of the liquid increases from zero at the outer cylinder to a maximum at the inner cylinder and thus a velocity gradient is set up. Let us consider a co-axial cylindrical layer of liquid of radius r, thickness dr and length l. If  be its angular velocity, the linear velocity, of the layer  = r.

Fig. 5.3 Now, velocity gradient



dv d d  (r )    r . dr dr dr

The angular velocity  of the layer is necessary to prevent slipping of the layer and is not responsible for viscous drag. Thus the effective velocity gradient is r(d/dr); According to Newton’s formula,

The viscous force acting on the  dv   F  A Cylindrical layer of radius r dr 

  2 rlr

d dr

…(1)

The moment of this force (or torque)  d Xr   C   2 rlr about the axis of rotation dr 

 2 l r 3

d dr

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169 dr  2 l d. r3

Integrating,

C



1 dr  2  l  0 d r3

 1  1 1 C   2   2 l1 i.e., C  2  2   4 l1  2r  b b a  C=

41a 2b 2 l (a 2  b 2 )

…(2)

Here we have omitted the torque due to viscosity between the bottoms of the two cylinders. Let it be c'. This effect can be eliminated by performing the experiment with two different heights of the liquid l1 and l2 without changing the distance between the bottoms of the cylinders. Then, the total torque when the height of the liquid is l1.

41 a 2 b 2 C1= l1  c' (a 2  b 2 )

The total torque when the  41a 2 b2  C  l2  c '  2 height of the liquid l2 (a 2  b 2 )  

C1-C2 =

41a 2b2 (l1  l2 ) (a 2  b 2 )

…(3)

Note: This expression is valid for gases also. Only change in such a case is that we have to use separate inner cylinders of lengths l1 - l2. Experiment: The apparatus consists of an inner cylinder I fixed to a spindle AB (Fig. 5.3(i)). The inner cylinder I is surrounded by the liquid contained in the fixed outer cylinder R. The spindle is pivoted at its ends A and B so that it can be free to rotate about its axis AB. The disc F and the drum D are also attached to the same spindle. The disc F, in combination with the index E, measures the period rotation of the cylinder. A string is wound over the drum D. The ends of the string are passed over pulleys P1 and P2. Two weights mg, mg are suspended from the ends of the string. As the weights fall, a steady condition will be reached when the couple due to the weights will be balanced by the couple due to the viscous forces and the inner cylinder rotates with uniform angular velocity. If the period of rotation of the inner cylinder is T, then

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Fig. 5.3 (i) 1 =

2 T

C(a 2 -b 2 ) 1 = 4 a 2b 2l

From (2),

…(4)

Here C = mgd, where d is the diameter of the drum. Substituting for 1and C in Eq. (4), we get,

gd (a 2  b2 ) mT 2 mgd (a 2  b 2 ) or   X  8 2 a 2b2 l T 4 a 2b 2l The experiment is repeated for different heights of the liquid and correspondingly adjusting m to get the same period of rotation T. A graph is plotted between mT and l. The graph be a straight line.

The intercept on the negative side of the length axis gives the bottom

correction k. Thus the bottom correction can be accounted for by taking l as l + k.



Hence

gd (a 2  b2 ) mT X 2 2 2 (l  k ) 8 a b

5.4 VISCOSITY OF GASES Modifications of Poiseuille’s formula for Gases - Meyer’s Formula Poseuille’s formula does not apply to gases since gases are; unlike liquids, highly compressible.

The density of a gas varies directly with pressure (  P).

Under steady

conditions, there us no accumulation of gas anywhere and so the mass (and not the volume) of the gas crossing any section of the tube per second must be constant. i.e., PV = constant or V constant (

  P).

Consider the flow of a gas through a capillary tube AB of length l and radius a [Fig. 5.4].

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Fig. 5.4 Consider an intermediate element CD of the tube of length dx. Let dp be the difference of pressure between C and D. Let p be the average pressure of the gas in CD.

The volume of gas flowing   a 4 dp  V   through CD per second 8 dx 

…(1)

The –ve sign indicates that the pressure decreases as x increases. Let p1 and p2 be the pressures at the inlet and outlet of the tube. Let V 1 and V2 be the volumes of the gas entering and leaving the tube respectively. Now 

p1V1 = pV =

…(2)

p1V1 = p2V2 = pV

  a 4 dp  p   8 dx  p1V1dx =

 a 4 pdp 8

Integrating over the whole length of the tube,

 a4 p1 V1  dx   8 0 1



pdp

 a4  p2  p1V1 ( x)     8  2  p 1 0

p1V1l 

Hence,

 a 4 [ p2 2  p12 ] 16

p1V1 

 a4 2 [ p1  p2 2 ] 16l

p1V1 

 a4 ( p12  p2 2 )  p2V2 16l

[From Eq. (2)].

This equation is called Meyer’s Formula for the rate of flow of a gas through a capillary tube.

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5.5 Rankine’s method for determination of of a gas His apparatus consists of a closed vessel ABCD [Fig. 5.5]. Between A and B there is a capillary tube of length l and radius a. In the opposite branch, there is a mercury pellet of mass m. There are two fixed marks C and D such that the volume  of the upper portion BC is equal to the volume of the lower portion AD. Let the volume of the gas in the whole tube ABCD be V.T1 and T2 are taps for filling the vessel with the experimental gas. Let α be the area of cross-section of the tube CD.

Fig. 5.5 Then the pressure difference between the ends of the capillary tube is caused by the mercury pellet and the excess of pressure =



. When the apparatus is held vertical, the mercury pellet

descends down the tube, forcing a certain amount of gas through the capillary tube AB. A stop clock is started just when the upper end of the pellet crosses the mark C and stopped just when the lower end of the pellet crosses the mark D. The time interval t is noted. Then is calculated using the formula,



 a 4 mgt 8 l (V  2 )

Theory of the Experiment: When the apparatus is placed horizontally, the pressure p is the same throughout and the density of the gas is p, where  is the density per unit pressure. The mass of the gas in the whole tube =pV and remains constant. At the beginning of the Experiment: Pressure at the outlet B = Pressure above mercury pellet at C = p1 (say) Pressure at the inlet A = Pressure below the mercury pellet at C

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THERMAL ENGINEERING  p1 



173

 P1

At the end of the experiment: Pressure at the outlet B = Pressure above D = p2 (say). Pressure at the inlet A = Pressure below mercury pellet at D.

 p1 



 P2 .

Since the total mass of the gas remains constant, we can write two equations representing the beginning and the end of the experiment.

mg   p V  1   p1   (V   )   

…(1) at the beginning.

mg   p V  p2  (V   )   p2     

…(2) at the end

From (1),

pV  p1  p1V  p1  or

p1V  pV 

p1  p 



p1  p1 



V





.

mg   V





V

 p

mg   V

From (2),

pV  p2V  p2  p2  or

p2V  pV 

p2  p 



.

mg  .  V

p2  p2 

 p



 p

mg   1   V

mg  mg   V 

 . 

At the beginning of the experiment, the mass of the gas enclosed between the first position of the mercury pellet and the inlet of the capillary = P1(V-).

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At the end of the experiment, the mass of the gas enclosed between the inlet and the second position of the mercury pellet = P1. Hence the mass of the gas that goes through the capillary tube in time t = P 1(V-)- P2.

mg    p  V  =

mg    1   (V   )  [ p    V 

 ] 

p (V  2 ).

The average rate of flow  p  (V  2 ) .  in kilogram per second  t Let us find the average rate of flow from Meyer’s formula. At the beginning of the experiment, pressure at the inlet A = P1. Let V1 be the volume of the gas flowing through the capillary tube per second. From Meyer’s formula,

P1 V1 

 a4 2 [ P1  p12 ] 16.l

Similarly, at the end of the experiment, let V2 be the volume of the gas flowing through the capillary tube per second. Then,

P2 V2 

 a4 [ P2 2  p2 2 ] 16.l

  PV 1 1   PV 2 2  flow in kilogram per second  2 The average rate of



 a 4 2  a4 P1  P12    P2 2  P2 2    16l 16l 2

 a 4 [ P12  P12    P2 2  P2 2 ] 32l

Now,

mg 2mg   mg  P12  p12  ( P1  p1 )( P1  p1 )   2 p     V   

mg 2mg   mg  P2 2  p2 2  ( P2  p2 )( P2  p2 )   2 p     V    

Average rate of flow =

 a 4 mg 4p 32l 

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p

 a 4 mg 8 l 

Equating the two expressions for the rate of flow, we have

p  (V  2 )  a 4 mg  p t 8l 





 a 4 mgt 8 l (V  2 )

Notes: (1) However pure the mercury of the pellet may be, some of it does stick to the wall of the tube and the effect of sticking is to reduce the effective weight of the pellet. In the actual experiment a correction for the sticking of the pellet is necessary. (2)

Rankine’s method is used to find the variation of  with pressure and temperature. Rankine determined the coefficient of viscosity of gases under various pressures and found that coefficient of viscosity is quite independent of pressure.

It was

found that the coefficient of viscosity of a gas increases with temperature. (3)

According to the kinetic theory of gases, the coefficient of viscosity of a gas is given by

1 3

  m.nc where m is the mass,

c the mean speed and  the mean free path of the molecules of the gas. n

is the number of molecules per unit volume. Determination of enables us to estimate the mean free path. Further,



1 2 2 n

where σ is the molecular diameter. By measuring , the molecular diameter σ can be estimated.

Surface Tension 5.6 Introduction Any liquid in small quantity, so that gravity influence is negligibly small, will always assume the form of a spherical drop – e.g., rain drops, small quantities of mercury placed on a clean glass plate etc.

So a liquid must experience some kind of force, so as to occupy a

minimum surface area. This contracting tendency of a liquid surface is known as surface tension of liquid. This is a fundamental property of every liquid.

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Fig. 5.6(a)

Fig. 5.6(b)

The following experiment illustrates the tendency of a liquid to decrease its surface area. When a camel hair brush is dipped into water, the bristles spread out [Fig. 5.6 (a)]. When the brush is taken out, the bristles cling together on account of the films of water between them contracting [Fig. 5.6 (b)]. This experiment clearly shows that the surface of a liquid behaves like an elastic membrane under tension with a tendency to contract. This tension or pull in the surface of a liquid is called its surface tension. Definition: It may be defined as the force per unit length of a line drawn in the liquid surface, acting perpendicular to it at every point and tending to pull the surface apart along the line. Unit of surface tension.

Surface tension being force per unit length, its SI unit is

-1

Newton per meter (Nm ). Dimensions of surface Tension: Since it is the ratio of a force to a length, its -2

-2

dimensions are MLT /L = MT . Molecular forces: There are two kinds of molecular forces: (i)

adhesive forces and (ii) cohesive forces.

(ii)

Forces of attraction between molecules of different substances are known as adhesive forces.

For example, the force of attraction between the glass

molecules of a beaker and molecules of water contained in it is an adhesive force. Adhesive force is different for different pairs of substances. (iii)

Force of attraction between molecules of the same substance is called cohesive force. This force varies inversely probably as the eighth power of the distance between two molecules.

Hence, it is very appreciable when the distance

between two molecules is small. It is the greatest in solids, less in liquids and the least in gases. Therefore, a solid has a definite shape, a liquid has a definite free surface and a gas has neither. The maximum distance upto which a molecule exerts a force of attraction on another is -9

called the range of molecular attraction and is generally of the order of 10 m. A sphere with the molecule as centre and the range of molecular attraction as radius is called the

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sphere of influence of the molecule. The molecule attracts and is, in turn, attracted by the molecules present inside this sphere. 5.6.1 Explanation of surface tension on kinetic theory Consider three molecules A, B and C of a liquid [Fig. 5.6.1]. The circles around them indicate their respective spheres of influence.

Fig. 5.6.1 (i)

The molecule A is well within the liquid and I is attracted equally in all directions by the other molecules lying within its sphere of influence. Therefore, it does not experience any resultant force in any direction. This happens only as long as the sphere of influence is well within the liquid.

(ii)

The sphere of influence of molecule B lies partly outside the liquid. The upper half of the sphere contains fewer molecules attracting the molecule B upwards, than the lower half attracting it downwards. Hence, there is a resultant downward force acting on B.

(iii)

The molecule C lies on the surface of the liquid. Half of its sphere of influence lies above the surface of the liquid and contains only a few vapour molecules. Whereas there are many liquid molecules in its entire lower half. Thus the resultant downward force in this case is the maximum. If a plane RS is drawn parallel to the free surface PQ of the liquid at a distance equal to the molecular range, then the layer of the liquid between the planes PQ and RS is called the surface film. Hence all the molecules in the surface film are pulled downward due to the cohesive force between molecules.

If a molecule is to be brought from the interior of the liquid to the surface of the liquid, work has to be done against the downward cohesive force acting upon it. Hence molecules in the surface film have greater potential energy than the molecules inside the liquid. Since the potential energy of a system tends towards a minimum, the surface film tends to contract, so as to contain minimum number of molecules in it. Thus the surface of the liquid is under tension and behaves like a stretched elastic membrane. Surface energy: The potential energy per unit area of the surface film is called its surface energy. 5.6.2 Forms of liquid drops When a quantity of liquid rests upon a horizontal solid plats, which it does not wet, the shape of the drop us determined by surface tension and gravity. For extremely small drops, the S.T. effects are great and the gravitational effects small. So S.T. determines the shape of the

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drop. It is therefore spherical. Rain drops, a small quantity of mercury on a glass plate, water drops on leaves, all assume spherical shapes on account of this. On increasing the size of the drop, the effect of gravitation becomes greater and that of S.T. less. Now, the effect of gravitation alone would be to make the drop spread out, so that its centre of gravity may be the lowest. Hence, a large drop of a heavy liquid spreads out when placed on a glass plate. Therefore, a large drop of mercury is always flat. 5.6.3 Angle of Contact When a glass plate is dipped in water, the water molecules cling to the surface of glass and the water molecules rise along the plate. The shape of water is as shown in Fig. 5.6.3 (i). When the glass plate is dipped in mercury molecules cling to the surface and the liquid is depressed along the plate as shown in Fig. 5.6.3 (ii)

Fig. 5.6.3 (i)

Fig. 5.6.3 (ii)

The angle of contact  is defined as the angle made by the tangent at the point of contact of the liquid surface with the glass surface inside the liquid. o

This angle may have any value between 0 and 180 . For most liquids and glass, it is o

less than 90 ; for mercury and glass, it is about 140 . It really depends upon the nature of the liquid and the solid. It is quite independent of the angle of inclination of the solid to the liquid surface. 5.6.4 Spreading of one liquid over another Let a liquid (liquid A) be in contact with another liquid (liquid B) as shown in Fig. 5.6.4. The free surfaces of both liquids are in contact with air. Three forces are acting at the point of contact P of the three substances. (1) S.T. σ1 between liquid A and air, (2) S.T. σ2 between liquid B and air and (3) S.T. σ3 between liquids A and B. The directions of these tensions are along the tangents at the lines of common contact.

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Fig. 5.6.4 If equilibrium is possible, we should be able to represent the three forces acting at the point P by the three sides of a triangle, taken in order. Such a triangle is known as the Neumann’s triangle. The sum of any two sides of a triangle is always greater than the third side. Therefore, the sum of any two of the three surface tensions σ1, σ2 and σ3 should be greater than the third. No two pure liquids satisfy this condition. Thus it is not possible to have a drop of one liquid remaining in equilibrium over another liquid. The liquid B, if it is lighter than the liquid A, will spread over the surface of A. Quincke showed that pure water spreads over pure mercury. But, if the mercury surface is contaminated with grease, the water will form a drop on the mercury surface. For contaminated surfaces of liquids, the construction of Neumann’s triangle can be possible. 5.6.5 Pressure difference across a liquid surface (a) If the free surface of the liquid is plane, as in Fig. 5.6.5 (a), the resultant force due to S.T. on a molecule on the surface is zero. (b) IF the free surface of the liquid is concave, as if Fig. 5.6.5 (b), the resultant force due to S.T. on a molecule on the surface acts vertically upwards. (c) If the free surface of the liquid is convex, as in Fig. 5.6.5 (c), the resultant force due to S.T. on a molecule on the surface acts vertically downwards (into the liquid).

Fig. 5.6.5 Excess pressure inside a liquid drop: A spherical liquid drop has a convex surface, as in Fig. (i).

The molecules near the surface of the drop experience a resultant force, acting

inwards due to surface tension. Therefore, the pressure inside the drop must be greater than the pressure outside it. Let this excess pressure inside the liquid drop over the pressure outside it be p.

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Imagine the drop to be divided into two exactly equal halves. Consider the equilibrium of the upper-half (or the upper hemisphere) of the drop as shown in Fig. (ii). IF r is the radius of the drop, and σ its S.T.,

The upward force on the plane face  2   p r ABCD due to the excess pressure p  The downward force due to surface tension acting     2 r along the circumference of the circle ABCD  Since the hemisphere is in equilibrium, the two forces are equal.

p r 2   2 r or p  2 / r



Excess pressure inside a soap bubble: A soap bubble has two liquid surfaces in contact with air, one inside the bubble and the other outside the bubble. The force due to S.T. in this case =

2 X  2 r  4 r .

Therefore, for equilibrium of the hemisphere,

p r 2  4 r or p  4 / r . Thus the excess pressure inside a drop or a bubble is inversely proportional to its radius (i.e.,

p1/ r ). Since p1/ r , the pressure needed to form a very small bubble is high. This

explains why one needs to blow hard to start a balloon growing. Once the balloon has grown, less air pressure is needed to make it expand more. 5.7 Excess pressure inside a curved liquid surface When the pressure on both sides of a liquid surface is same, then the surface is flat, without any curvature. But when it is curved convex upwards, then the pressure inside must be greater than the pressure outside and the excess pressure inside is balanced by the force of S.T. To find the excess pressure, consider a small curvilinear rectangular element A1B1C1D1 of a liquid surface [Fig. 5.7]. A1B1 has a radius of curvature R1 with centre at O1. B1C1 has a radius of curvature R2 with centre at O2. Let p be the excess of pressure inside the surface over that outside. Then the outward thrust on the surface A1B1C1D1 = p X Area of the element A1B1C1D1 = p X A1B1 X B1C1.

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Fig. 5.7. Now, let the surface be moves outward through a very small distance x. Let the new position of the surface be A2B2C2D2. Work done in the displacement = p.A1B1.B1C1. x

Now, increase   = Area of A2B2C2D2 – Area of A1B1C1D1 in surface area  = A2B2.B2C2 - A1B1.B1C1 From similar triangles, A1B1O1 and A2B2O1,

A1 B1 A1O1 AB R1 or 1 1   A2 B2 A2O1 A2 B2 ( R1   x)

  x A1B1 ( R1   x)  A1B1 1   R1  R1 



A2 B2 

Similarly,

B2C2 = B1C1

Hence

 x  x  A 2 B2 .B2 C2 =A1B1 1+  B1C1 1+   R1   R2 



 x  1   R2  

Increase in   x x     A1B1.B1C1 1+   A1B1.B1C1 surface area   R1 R 2   1 1     R1 R 2 

= A1B1.B1C1 δx 

Work done in increasing  Surface tension X increase =   the area of a surface  in surface area 

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182

 1 1     R1 R 2 

= σ. A1B1.B1C1.x 

…(2)

Equating (1) and (2) we have,

 1 1  p.A1B1.B1C1. x= . A1B1.B1C1. x     R1 R 2   1 1     R1 R 2 

p = σ

0or

The excess of pressure inside any particular surface can be deduced from the above expression.

(1)

Spherical liquid drop (an air bubble in a liquid).

It has only one surface and the radius of curvature is the same everywhere, i.e., R 1 = R2 = r. p = 2σ/r. (2)

Spherical soap bubble. Here there are two surfaces having the same radius of curvature p = 4σ/r. R1 = r = Radius of the cylinder and R2 = , p = σ/r

(3)

Cylindrical drop.

(4)

Cylindrical bubble. p = 2σ/r since it has two surfaces.

1 1    . The expression holds good for surfaces such as  R1 R2 

Note: We have, p = σ 

Fig. 5.7(a) & (b) spherical or ellipsoidal, for which the principal radii of curvature are on the same side. Such surfaces are called synclastic surfaces [Fig. 5.7 (a)]. But in cases where the two radii are in opposite directions, the surface is called anticlastic [Fig. 5.7 (b)]. surfaces is

 1 1   .  R1 R2 

p = σ

Combining the two cases, the general relation is

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183

 1 1   .  R1 R2 

p = σ

-3

Example 1: Calculate the work done in spraying a spherical drop of water of 10 m radius into -3

-1

million droplets, all of the same size, the surface tension of water being 72 X10 Nm . Breaking the liquid drop means an increase in surface area. Therefore work has to be done for this purpose. Work done = Surface tension X Increase in surface area. Let us calculate the increase in surface area. Let R be the radius of the larger drop and r the radius of the smaller droplets. 6

Volume of the original drop = Volume of 10 droplets.

4 4  R3  106 X  r 3 3 3 4 4  (103 )3  106 X  r 3 3 3 3 -15 or r = 10 m or r=10-5m Area of original large drop

R=10 -3m]

 4 R2  4 (103 )2  4 X 106 m2

Area of 10 droplets  10 6

[

X 4 r 2  106 X 4 (105 )2 -4

= 4X10 m

Hence increase in surface area is

 4 X 104  4 X 106  1.244 X 103 m2 

Work done = Surface tension X Increase in surface area.

 (72 X 103 ) X (1.244 X 103 )  8.956 X 105 joules. Example 2: Calculate the amount of work done if a soap bubble is slowly enlarged from a radius of 0.1 m to a radius of 0.2m.  = 30 X10 Nm . -3

-1

Increase in surface area = 2[4(r2 -r1 )] 2

= 8[(0.2) -(0.1) ] = 0.7536 m 

Work done = Increase in surface area X  -3

= 0.7536 X (30 X 10 ) = 0.02261J. Example 3: What is the work done in blowing a soap bubble of radius 0.1 m? What additional work will be performed in further blowing it, so that its radius becomes 0.15 m?  = 30 X 10 Nm -3

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Work done in blowing a soap    surface tension X increase in surface area bubble of radius 0.1 m  W = X8r1 = (30 X 10 ) 8 (0.1) = 7.536 X 10 J. 2

-3

Work done in increasing the radius of  2 2    X 8 (r2  r1 ) the soap bubble from 0.1m to 0.15m  i.e.,

W = (30X10-3 ) 8 [(0.15) 2 -(0.1) 2 ]=9.42X10-3J.

Example 4: There is a minute circular hole at the bottom of a small hollow vessel. The vessel has to be immersed in water to a depth of 0.4m, before any water penetrates inside. Find the -3

-1

radius of the hole, if the surface tension and density of water be 73 X 10 Nm and 1000 kg m

-3

respectively. Water cannot penetrate till the hydrostatic pressure is greater that the excess of pressure that comes into play due to surface tension. At the limiting condition, when water just penetrates, the two pressures should be equal; i.e., hg = 2/r where r is the radius of the hole. 

r = 2. /( hg)

Here,  = 73 X 10 Nm ; -3

-1

r

=1000kgm ; -3

h = 0.4m;

r =?

2 X (73 X 103 )  3.724 X 105 m 0.4 X 1000 X 9.8 -4

Example 5: What would be the pressure inside a small air bubble of 10 m radius, situated just below the surface of water?

S.T. of water may be taken to be 70 X 10 5

-3

-1

and the

-2

atmospheric pressure to be 1.012 X 10 Nm .

Excess of pressure inside the spherical  2  p air bubble over that of the atmosphere  r Here,  = 70 X 10 Nm ; -3

-1

-4

r = 10 m.

Excess  2 X (70 X 10 3 ) 2 =  1400 Nm2  =p= 4 pressure  10 r 

Total pressure inside the  Atmospheric pressure   =  air bubble  +Excess pressure  5

-2

= 1.012 X 10 +1400 = 1.026 X 10 Nm . -3

-3

Example 6: The pressure of air in a soap bubble of 7 X 10 m diameter is 8 X 10 m of water above the atmospheric pressure. Calculate the S.T. of the soap solution.

Excess of pressure inside a soap  4 .  p bubble over that outside it r  Here,

-3

-2

p = 8 X 10 m of water = (8 X 10 ) X 1000 X 9.81 Nm .

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185 -2

-3

r = (7 X 10 )/2 = 3.5 X 10 m. 

=

pr 78.48 X (3.5 X 103 )   68.67 X 103 Nm1 4 4

Example 7: A spherical bubble of radius 0.001m is blown in an atmosphere whose pressure is 5

-2

-1

10 Nm . If the S.T. of the liquid comprising the film is 0.05 Nm , to what pressure must the surrounding atmosphere be brought in order that the radius of the bubble may be doubled? The pressure inside the bubble initially is

pi 

4 4 X 0.05 P  105  1.002 X 105 Nm2 r 0.001

The volume of  4 3 4 9 3  =Vi =  r =  X 10 m the bubble 3 3  In the second case, the radius of the bubble is 0.002m. Let the required pressure p f. Then the total pressure inside the bubble,

Pf 

4 4 X 0.05  Pf   Pf  (100  Pf ) Nm2 0.002 0.002

In the second case, volume of the bubble =

4 V f   (0.002)3 3

Applying Boyle’s law, piVi = pfVf,

4 4  (1.002 X 105 )   X 109   (100  Pf )  (0.002)3 3 3  4 2 Simplifying , Pf  1.252 X 10 Nm Example 8: Two soap bubbles of radii r1 and r2 coalesce to form a single bubble of radius r. If the external pressure is P, prove that the S.T. of the solution from which the soap bubble is formed is given by

1 4

  P(r 3  r13  r23 ) /(r12  r2 2  r 2 ).

Pressure inside the first bubble = p1 = (4/r1)+P Pressure inside the second bubble = p2 = (4/r2)+P Let the bubbles coalesce into a larger bubble of radius r. Then, pressure inside this larger bubble = p3 = (4/r)+P. Let V1, V2 be the volumes of the two bubbles before they coalesce and V3 the volume of the large bubble formed. Then by Boyle’s law, p1V1+ p2V2 = p3V3

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 4 4  4 4  4 4  P   r13    P   r23    P   r 3.   r1 3  r2 3  r 3 or 4 ( r12  r2 2  r 2 )  P(r 3  r13  r23 )

1 4

  P(r 3  r13  r23 ) /(r12  r2 2  r 2 ).

5.8 Drop-weight method of determining the surface tension of a liquid Experiment:

A short glass tube is connected to the lower end of a burette (or funnel) clamped

vertically, by means of a rubber tube [Fig. 5.8.]. The funnel is filled with the liquid whose S.T. is to be determined. A beaker is arranged under the glass tube to collect the liquid dropping from the funnel. The stopcock is adjusted so that the liquid drops are formed slowly. In a previously weighed beaker a known number of drops, (say 50) are collected.

Fig. 5.8 The beaker is again weighed. The difference between this weight and the weight of the empty beaker gives the weight of 50 drops of the liquid. From this the mass m of each drop is calculated. The inner radius r of the tube is determined using a vernier calipers. The S.T. of the liquid at the room temperature is calculated using the formula,



m.g 3.8r

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Here, we consider the vertical forces that keep a small drop of liquid in

equilibrium, just before it gets detached from the end of a vertical glass tube of circular aperture. At the instant the drop gets detached, it assumes a cylindrical shape at the orifice of the tube (Fig. 5.8). Let σ = S.T. of the liquid and r = radius of the orifice.

Excess pressure (p) inside the drop over    the outside atmospheric pressure  r The area of the section is r . Therefore, 2

Downward forces on the drop  2  r due to this excess of pressure  r The weight mg of the drop also acts vertically downwards. 

Total force on the drop = (r σ/r)+mg

This downward force is balanced by the upward pull due to surface tension 2rσ acting along a circle of radius r. Therefore 2 rσ =

 r 2 r





 mg or 2 rσ = rσ+mg m.g  .r

But the equilibrium of the drop at the instant of its detachment is dynamic and not static. Lord Rayleigh, taking dynamical aspect into account, showed that



m.g 3.8r

Example 1: In a drop weight method for the determination of S.T. between water and air, a glass tube of external diameter 2mm is used, and 100 drops of water are collected. The mass of these drops is 2.8 gms. Find the S.T. of water in air. Here, 

-3

-5

r = 10 m.m = 2.8 X 10 /100 = 2.8 X 10 kg



m.g (2.8 X 10-5 ) X 9.8 =  0.07221Nm1 3.8 X 103 3.8r

Interfacial Tension: At the surface of separation between two immiscible liquids there is a tension similar to surface tension. It is called the interfacial tension. Definition: When one liquid rests on another without mixing with it, the interface between the two liquids possesses energy just like the surface of a liquid.

The

interfacial tension is the value of the force acting per metre normal to a line drawn on the interface. 5.8.1 Experiment to determine the interfacial tension between water and kerosene

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Sufficient amount of the lighter liquid (kerosene) is taken in a beaker. The weight w 1 of the beaker with kerosene is determined. The heavier liquid (water) is taken in the burette [Fig 5.8.1]. The glass tube is fixed vertically with its end under the surface of kerosene. The flow of water is regulated so that drops of water detach themselves into kerosene one by one. After collecting 50 drops, the beaker is again weighed. Let this weight be w2. Then w2-w1 gives the mass of 50 drops.

Fig 5.8.1 From this the average mass m of each drop is calculated. The interfacial tension σ between water and kerosene is calculated using the formula



 2  1    1 

Mass of beaker +

Mass of 50 drops

lighter liquid

=(w2-w1) kg

No Trial I

Theory:

m.g 3.8r

Trial II

Mean

+50 drops

w1 kg

Trial I

Mean Trial II

Let 1 and 2 be the densities of water and kerosene respectively. Let m be the

mass of water drop in air. 

w2 kg

Volume of water drop = m/1

Volume of kerosene displaced  m  by the water drop  1

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mass of kerosene displaced  m2  by the water drop  1 

Apparent weight of the  m 2 g   mg  water drop in kerosene  1

Let σ be the S.T. at the interface between the two liquids. Then,

2 r 





 r 2 r

 m.g 

m  2 .g

1

m.g   2  1    r  1 

Again the more accurate equation will be



m.g 3.8r

 2  1    1 

5.9 Quincke’s method The shape of a drop depends on the combined action of S.T. and gravity. When a large drop of mercury is placed on a clean horizontal glass plate, the drop flattens out until its top becomes perfectly horizontal as shown in Fig. 5.9 (a).

Fig. 5.9 (a). Imagine that the drop is cut into two halves by a vertical plane ABCD [Fig. 5.9 (b)]. Let the drop the further cut by two vertical planes BCG and ADH at a distance l from each other and perpendicular to ABCD. GH is the most protruding portion of the drop. Let h1 be the height of the flat top AB above the horizontal plane EFGH and h2 the total height BC of the drop.

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Fig. 5.9 (b). The portion of the drop lyin0g above the horizontal plane EFGH is in equilibrium under the action of the following forces: (1) Force due to S.T. acting at right angles to AB from left to right horizontally = σ.l. (2) Hydrostatic thrust acting horizontally from right to left on the plane ABEF of the drop due to part of the liquid on the right. The hydrostatic pressure is zero at AB and increases to h1g at EF.

1 h1g. 2



Average pressure = (0 + h1g)/2 =



Total hydrostatic thrust on the area ABEF =

1 h1g X h1l 2 =

1 2 h1 gl 2

(3) The S.T. at G, which acts vertically upwards has to component along the horizontal. Since the drop is in equilibrium, these horizontal forces must balance. Hence σ.l =

1 2 1 2 h1 gl or h1 g. 2 2

Determination of the angle of contact : Consider the equilibrium of the whole drop. The forces acting on the face ABCD are: Pull due to S.T. acting horizontally from left to right and perpendicular to AB = σ.l

Hydrostatic thrust acting horizontally  1 2  1 from right to left on the face ABCD  = h2g X h2l= h2 gl 2 2

The S.T. pull due to mercury on glass at K acting   = σ.l tangentially to the slice in the direction KS 

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The glass plate exerts an equal and opposite reaction along KQ.

The horizontal component of this   reactional force due to glass along    .lcos   KC, acting from left to right  where α = (180 - ) In equilibrium, σ.l + σ.l cos α =

1 2 h2  .g.l 2 1 2

 (1+cos )= h2 2 .g

1 2 1 h1 .g (1+cos )= h2 2 .g 2 2

(1+cos )=

cos   (h22  h12 ) / h12 .

1 2     2 h1  .g 

h2 2 h2 2 or cos  = 1 h12 h12

The value of α can be determined form this relation. The angle of contact  180 – α. Experiment:

Place a clean glass plate on a small table provided with leveling screws.

Arrange it horizontally with the help of a spirit level. Form a large drop of mercury on the glass plate so that its upper surface is perfectly plane. Sprinkle a fine layer of lycopodium powder at the top of the drop. Adjust the microscope so that the horizontal cross-wire is in line with the lycopodium powder. Lower the microscope and focus it on the line of separation of mercury and glass. The difference between the two readings gives h2.

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192 Fig. 5.9.(i)

Move the microscope to one edge of the drop.

Focus the microscope so that the

horizontal cross-wire is in line with the most protruding part of the drop. To find this position very accurately, light from an incandescent lamp S is focused by a lens L and the plane glass plate P, acting as a mirror, on to the edge of the drop (Fig. 5.9.(i)). A bright, then, horizontal line is seen at G, where the drop protrudes out most. The horizontal cross-wire placed coinciding with this bright horizontal line and the reading is noted. The difference between the top surface reading and this reading gives h1. Using the relations given above, S.T. of mercury and angle of contact are determined. 5.9.1 Vapour pressure over flat and curved surfaces. When a liquid is contained in a closed vessel, the space above the liquid is filled with vapour and a stage is reached when evaporation and condensation go on at equal rates. At this stage, the vapour is in dynamic equilibrium with the liquid and the pressure exerted by it is called the saturation vapour pressure of the liquid at that temperature. Let a capillary tube of radius r be dipped vertically in a liquid (which wets the tube) of S.T. σ and density  (fig. 5.9.1). Let the whole arrangement be enclosed in an exhausted bell jar, so that the effect of atmosphere may be neglected. The liquid evaporates till a steady state is reached and the space within the bell-jar is saturated with the vapour. The liquid rises to a height h given by σ = rhg/2

…(i)

If p1 is the vapour pressure at the horizontal surface A, p2 is the vapour pressure above the concave surface B (Fig. 5.9.1) of the liquid in the tube and ' is the density of the vapour supposed to be uniform between A and B of height h, then …(ii)

p1 - p2 = h'g Dividing (ii) by (i)

p1  p2





2  2   orp1  p2  X . r r 

Therefore, the saturation vapour pressure over the concave surface is less than that over a flat surface by

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193

Fig. 5.9.1 In the case of liquid like mercury, there is a capillary depression and the meniscus is convex. Then the saturation vapour pressure over the convex surface is more than that over the flat surface by,

2   X . t 

Formation of drops. over a flat surface

The saturated vapour pressure over a convex surface is greater than that

2   X . Let us place a drop of water in a space in which the vapour t 

pressure is at the saturation value for a plane surface. The maximum vapour pressure over a convex surface must be more than that on a plane surface. Hence the drop evaporates in order to increase the vapour pressure to its own saturation value. This will result in a further decrease in the radius of the drop and the value of saturation vapour pressure over it must rise further. Therefore, the drop evaporates more and more rapidly to increase the vapour pressure. The space then contains more vapour than that required for Saturation at that

temperature, without condensation taking place and the vapour becomes

super-saturated. The dust particles or charged ions are present in the atmosphere, vapour condenses on them and the size of the drop formed, even in the beginning is sufficiently large. The surface of a large drop is almost flat and it has no tendency to evaporate. As its size grows, its tendency to evaporate becomes smaller still. The dust particles and smoke particles play an important role in the condensation of water vapour and formation of clouds. That is why the weather looks foggy in large industrial towns, where smoke and dust particles provide nuclei for condensation to set in. Even in a super-saturated cloud, condensation will not take place if dust particles (which act as nuclei for the vapour to condense on) are not present in the atmosphere.

Artificial rain is sometimes

brought about by injecting solid carbondioxide into a saturated cloud to act as nuclei for the water vapour to condense on.

5.10 Variation of Surface Tension with Temperature Liquids are of two types, v.z., (i) unassociated liquid and (ii) associated liquid.

unassociated liquid contains the individual molecules of that liquid. Example: Benzene and carbon tetrachloride. An associated liquid contains groups of molecules of quite another type. These groups, however, tend to break up into single molecules with a rise in temperature. At the ordinary temperature, water is known to consist of groups, consisting of two H 2O molecules, in addition to ordinary H2O molecules. Thus water is an associated liquid at these temperatures.

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The S.T. of an unassociated liquid is found to decrease with rise of temperature, o

according to the simple formula σt = σo(1-αt) where σt is the S.T. at t C, σo at 0 C and α, the temperature coefficient of S.T. for the liquid. Van der Waals and Ferguson suggested other relations from which it could be easily deduced that the S.T. is zero at the critical temperature. The best relation connecting S.T. and temperature, for both associated and unassociated liquids, is due to Eotvos. This formula was later modified by Ramsay and shields. This is represented by σ(Mx)

3/2

= k(c -  - d) where σ = Surface tension at  K, c =

Critical temperature, d = a constant, varying from 6 to 8 for most of the liquids, k = another constant having the value 2.12 for associated liquids and 2.22 for unassociated liquids. x = Coefficient of association =

effective molecular weight of associate d liquid mol wt of the unassociated liquid with the same molecules

M = molecular weight of the unassociated liquid and  its specific volume. This shows that the S.T. is zero, when  = (c – d) i.e., at a temperature a little below the critical temperature. Note 1:

At higher temperature, the molecules themselves possess greater average

kinetic energy and are moving about more rapidly.

This reduces the effect of molecular

attractions. So less work is needed to bring a molecule from the interior of the liquid to the surface. Hence the S.T. of all liquids decreases as the temperature rises. Note 2: Impurities, contaminations and dissolves substances all lower the S.T. of a liquid. 5.10.1 Experimental study of the variation of surface tension with temperature: A capillary glass tube AB is connected to another glass tube CD, of a comparatively wider bore, so as to form a complete U tube (Fig.5.10.1). The experimental liquid is poured into o

this U tube. If the angle of contact () for the liquid and glass is less than 90 , the liquid stands much higher in the narrower limb AB of the tube than in its wider limb CD. If >90 (for example, o

in the case of mercury and glass), the liquid column in AB will be depressed into the tube to a position h below the liquid column in CD. Let σ be the S.T. of the liquid and , its angle of contact with glass. Let r 1 and r2 be the radii of the bores of the two limbs respectively. Let p be the atmospheric pressure.

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195

Fig. 5.10.1 Then,

Pressure just below the  2 cos   P liquid meniscus in AB  r1 Pressure just below the  2 cos   P liquid meniscus in CD  r2 Difference of pressure   = 2 σ cos  in the two limbs 



1 1   .  r1 r2 

This is balanced by the hydrostatic pressure hg due to the liquid column in the U-tube. Here, h is the difference of liquid levels in the two limbs.  is the density of the liquid. 

2 σ cos 

σ=

1 1    r1 r2

  . =hg 

h g 1 1 2    cos   r1 r2 

The experiment is repeated with the liquid maintained at different temperatures by means of suitable temperature baths.

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UNIT QUESTIONS

UNIT -I QUESTIONS Part A 1. Explain Joule-Thomson effect. 2. Explain the principle of regenerative cooling. How is the principle applied in the liquefaction of gases? 3. Explain the principle involved in the liquefaction of liquid helium. 4. Explain the principle of cooling by adiabatic demagnetisation 5. Explain the principle of refrigeration. 6. Name two refrigerants normally used. 7. Explain briefly the principle and working of a refrigerator. 8. Describe briefly the principles and practice of air conditioning. 9. What are the effects of CF2 Cl2 on ozone layer? Part B 10. Describe the porous plug experiment. Sketch a liquid air plant and explain its action. 11. Describe Linde’s method of liquefying air. 12. Describe Claude’s process for the liquefaction of air. 13. Describe the methods for the liquefaction of hydrogen and helium using Joule – Thomson effect. 14. Describe, with necessary theory, the method of producing very low temperatures by adiabatic demagnetisation. 15. Write an essay on the industrial and scientific applications of “Low Temperatures”. 16. What is refrigeration? Explain the principle of working of a vapour compression machine. 17. Describe, in detail, the principle and working of the Electrolux refrigerator. 18. Describe, in detail, the principle and working of the Frigidaire. 19. Explain, with the help of a neat sketch, the working of a window air conditioner. 20. With a help of a neat sketch, explain the working of a room air-conditioner. Part-C 21. Describe Joule-Kelvin effect and give its theory.

How is it utilized in the liquefaction of

gases?

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22. What is Joule-Thomson effect?

Obtain an expression for the cooling produced in this

process in case of van der Waal’s gas. Why do hydrogen and helium show a heating effect at ordinary temperature? 23. Describe Joule-Thomson effect and give its theory.

How has it been utilised in the

liquefaction of gases? 24. Describe Joule-Thomson porous plug experiment. What are the important inferences from this experiment? 25. Give an account of the methods of liquefying gases and discuss the principles on which they depend. 26. Define Joule-Thomson effect for liquefaction of gases. Prove that any gas below its inversion temperature will cool on suffering Joule-Thomson expansion. What is regenerative cooling. 27. Explain the phenomenon of adiabatic demagnetisation.

How will you employ this

phenomenon to produce and measure very low temperatures?

UNIT-II QUESTIONS PART-A

28. State and explain Zeroth law of thermodynamics. 29. State and explain first law of thermodynamics. 30. Write down the efficiency of an Otto cycle and Diesel cycle. Of these two which is preferred and why? 31. Explain the reversible process with an example? 32. Explain the change of entropy in a reversible process. 33. Distinguish between reversible and irreversible processes, giving examples. 34. Explain reversible and irreversible process. 35. State and explain II law of thermodynamics. 36. State different statements of second law of thermodynamics. 37. Define and explain ‘Entropy’. 38. Define “Entropy”. What is its physical significance? What happens to entropy in the case of reversible and irreversible processes? 39. What is the change of entropy in a Carnot’s cycle? 40. Obtain an expression for change in entropy in an irreversible process. 41. Explain what is T-S diagram. What information does it provide? 42. Derive the expression for the efficiency of a Carnot’s engine directly from T-S diagram. 43. State the third law of thermodynamics.

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Part B 44. State the second law of thermodynamics. Explain a Carnot cycle and prove Carnot theorem. 45. Describe with diagram the working of an Otto engine and deduce expression for its efficiency. 46. Describe with a neat diagram the construction and working an internal combustion petrol engine. 47. Which of the two cycles Otto or Diesel would you prefer in the design of internal combustion engine and why? 48. Bring out the differences between an Otto cycle and Diesel cycle. 49. Show that the change of entropy in a reversible cycle is zero. 50. State and explain the Third law of Thermodynamics.

Part C 51. Derive an expression for the efficiency of a Carnot’s engine in terms of the temperatures of the source and the sink. Show how an absolute scale of temperature can be defined with the help of the ideal Carnot’s engine. 52. Explain the concepts of reversible and irreversible processes. Obtain an expression for the efficiency of a reversible Carnot engine with a perfect gas as the working substance. Show that for the same pair of working temperatures, the efficiency of all reversible engines is the same and is the maximum value attainable. 53. Explain the working of Otto and Diesel engines with relevant diagrams.

Work out their

efficiencies. 54. Enunciate the second law of thermodynamics. Define a temperature scale without making use of the peculiarities of any selected thermometric substance. Show that Kelvin’s work scale is such a scale and that the ratio of the two temperatures as measured on the Kelvin scale is identical with the ratio of the same two temperatures on a perfect gas scale. 55. Define Entropy. What is its physical significance? Show that the entropy of a perfect gas remains constant in a reversible process but increases in an irreversible process.

Problems o

56. Find the efficiency of the Carnot’s engine working between 150 C and 50 C. o

57. Find the efficiency of a Carnot’s engine working between 227 C and 27 C. 58. A refrigerator has to transfer an average energy of 200 joules of heat per second from a o

temperature – 10 C to 25 C. Calculate the average power consumed assuming an ideal reversible engine.

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Sol:

Q1  Q2

199 T1 298  200 X  226.6 J / sec. T2 263

 Average power consumed, W = Q1 – Q2. = 226.6 – 200 = 26.6 J/sec =

26.6 H .P.  0.0357 H .P. 746

UNIT-III QUESTIONS

Part A 1. Define coefficient of thermal conductivity and give its units. 2. Distinguish between thermal conductivity and thermal diffusity of a substance. 3. Explain clearly the terms: temperature gradient, steady state and thermal conductivity. 4. What is the principle of operation of Lee’s disc method? Ans: In the steady state, the rate at which heat is conducted across the cardboard is

equal

to the rate at which heat is radiated from the surface of the slab.

K r 2

(1   2 ) d  r  2l   mc   d dt  2r  2l 

5. Explain what is meant by black body. 6. What is a black body? What are the salient features of black body radiation? 7. What is a black body? Describe Fery’s black body. 8. What so you understand by a black body? How can it be realized in practice? Explain with the help of a diagram. 9. Explain the distribution of energy in a black body spectrum. 10. State the three formulae of radiation, viz., Rayleigh- Jeans, Wien’s and Planck’s. 11. State Wien’s displacement law and Rayleigh Jean’s law of radiation. 12. State and explain Wien’s displacement law of radiation. What is its use in Astrophysics? 13. State and explain Planck’s law of radiation. 14. State and explain Rayleigh Jean’s law of radiation. 15. Deduce Rayleigh – Jeans law from Planck’s law of radiation. 16. Deduce Wien’s law of radiation from Planck’s radiation formula. 17. State and explain Stefan’s law of radiation. 18. Define Stefan’s constant. Give one practical application. Application.: If Stefan’s constant () is known, the rate of emission of heat energy by a hot body can be found by using Stefan’s law,

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200

E = T . 4

19. What is Pyrometer. 20. Define solar constant. State its value. Obtain an expression for the surface temperature of the sun in terms of the solar constant. 21. What are Pyrotheliometers? State the different types of Pyrotheliometers. Ans: The instruments used for the determination of Solar constant are called pyrotheliometers Different Types of pyrotheliometers (i)

The Water-flow Pyrotheliometers

(ii)

The Water-stir Pyrotheliometers

(iii)

Angstrom’s Compensation Pyrotheliometers

22. Describe how the surface temperature of the sun can be estimated. 23. Explain the origin of solar energy. Part B 24. Explain the method of Lee’s disc to determine the coefficient of thermal conductivity of a bad conductor. 25. State Wien’s displacement law. Prove the law thermodynamically. 26. Derive Wien’s distribution law thermodynamically and discuss its importance. 27. Derive Rayleigh – Jean’s law of radiation. 28. Derive Planck’s formula for the distribution of energy in black body radiation. 29. State Planck’s hypothesis and deduce Planck’s law. Show that Rayleigh-Jeans law and Wien’s law are special cases of Planck’s law. 30. Derive Planck’s formula for the spectral distribution of energy in the black body radiation. 31. Discuss the distribution of energy in black body radiation. limitations of the classical theory in explaining it.

Discuss the success and

Derive Planck’s radiation formula,

explaining how the shortcomings of classical theory were overcome. 32. State and explain Stefan’s law. Describe an experiment to verify the law. 33. State Stefan’s law of radiation and deduce Newton’s law of cooling from it. Describe and experiment to determine Stefan’s constant. 34. State Stefan’s law and describe a method of determining Stefan’s constant. 35. Deduce Stefan’s law of radiation from the principles of thermodynamics. 36. What is Stefan’s law of radiation?

Give a laboratory method for determining Stefan’s

constant. 37. Describe the construction and working of a total radiation Pyrometer. 38. Describe the construction and working of a “Disappearing filament optical pyrometer”. 39. State the principle underlying the measurement of very high temperature by means of optical pyrometer. Describe fully and clearly a practical form of apparatus of this type.

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40. Define solar constant. Describe the construction of water flow pyrotheliometer. Explain how it is used to determine solar constant. 41. Define solar constant. Explain how the solar constant is determined experimentally. 42. How can the temperature of the sun be calculated from the solar constant? Comment on the source of energy in the sun. Part C 43. Describe with necessary theory, how the thermal conductivity of a bad conductor, such as ebonite, may be measured. 44. Deduce Wien’s wavelength – temperature, displacement law form thermodynamical considerations. 45. Obtain the Wien’s displacement laws. 46. Find the number of modes of vibration in unit volume of an enclosure. Deduce Rayleigh – Jeans formula. 47. Describe the circumstances which led Planck to propose the quantum theory of radiation. What are the assumptions he made? Derive Planck’s formula and show that Wien’s and Rayleigh-Jeans formula are special cases of the same. 48. Describe the energy distribution in the black body spectrum. How is it experimentally verified and theoretically explained? 49. Explain what you understand by a black body. State Stefan’s law of radiation and prove it from thermodynamical considerations. Indicate how it can be verified. How is this law distinguished from Newton’s law of cooling? 50. What is meant by a black body? How has it been experimentally realized? Describe how the Stefan’s law has been verified with it. 51. Deduce Stefan’s law of radiation from the principles of thermodynamics.

Describe an

experiment to determine Stefan’s constant. 52. Discuss fully the methods that have been used to measure high temperatures in the o

neighbourhood of 2000 C. 53. What are radiation pyrometers? Describe the construction and working of the total radiation pyrometer. 54. What is the principle of optical pyrometer? With the help of a diagram, explain the construction and working of an optical pyrometer. 55. Define constant and explain how it can be determined experimentally.

How do you

determine the temperature of the sun using Stefan’s law? Problems o

56. What at 100 C is boiling in an iron saucepan whose bottom is 150 sq. cm and thickness is o

0.3 cm. This is placed on a hot plate at 150 C. How much water is evaporated per minute? The loss of heat from the sides and the effect due to surface film are to be neglected.

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202 -1

-1

Thermal conductivity of iron = 67.2 Wm K , specific latent heat of steam = 226 X 10 Jkg . (Ans: 0.444 kg) 2

57. One face of a metal plate 1m in area and 0.02 m thick is in contact with melting ice while the o

other face is in contact with boiling water at 100 C. If thermal conductivity of the metal be 56 -1

-1

Wm k , find the mass of ice melted in one hour. (Ans: 3000 kg) 7

58. In an atomic explosion, the maximum temperature reached was of the order of 10 K. Calculate the wavelength of maximum energy.

-10

(Ans: 2.93 X 10

UNIT-IV QUESTIONS Part-A 59. Calculate the work done in twisting a wire. 60. Derive an expression for the period of oscillation of a torsion pendulum. 61. What is meant by a beam? Explain the terms neutral surface, neutral axis, plane of bending, and bending moment of a beam. 62. Explain why steel girders and rails are made in the form of I – section Part-B 63. Derive an expression for the moment of the couple required to twist one end of a cylinder when the other is fixed. 64. Describe, with necessary theory, how the rigidity modulus of the material of a rod is determined by the static torsion method. 65. Describe, with necessary theory, how you would determine the rigidity modulus of a wire experimentally by using the torsion pendulum. 66. What is a cantilever? Obtain an expression for the depression at the free end of a thin light beam clamped horizontally at one end and loaded at the other. 67. Derive an expression for the bending of a bar supported at the two ends and loaded in the middle. Describe an experiment to determine E by bending. 68. Describe, with necessary theory, the oscillation method to determine E for the material of a cantilever. 69. Describe Konig’s method for the determination of Young’s modulus of a beam. Part-C 70. A wire of length lm and diameter 10

-3

m is fixed at one end and twisted at the other end

through an angle of 70 by applying a couple of value 0.01 Nm. Evaluate the rigidity modulus of the wire.

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203 UNIT-V QUESTIONS

71. Derive Poiseuille’s formula for the rate of flow of a liquid through a capillary tube. Describe a laboratory method for determining the coefficient of viscosity of a liquid at room temperature. 72. Describe Rankine’s experiment for the measurement of the viscosity of a gas. 73. Water is conveyed through a horizontal tube 0.08 m in diameter and 4000m in length at the rate of 20 litres per second. Assuming only viscous resistance. Calculate the pressure required to maintain the flow. η = 0.001 SI units. Hint. p =

8Vl 8 X 0.02 X 0.001X 4000   7.962 X 104 Nm2 4 4 a  (0.04)

74. Define S.T. and angle of contact. 75. What are the units and dimensions of surface tension? 76. Explain S.T. from the point of view of surface tension? 77. What is surface energy? How is it related to S.T.? 78. It is easier to spray water when soap is added to it that when it is pure. Why? [Hint. When soap is added to water, the surface tension decreases. Therefore less work is required to spray water.] 79. Obtain the relation between the vapour pressure over a curved surface and that over a flat surface. 80. Obtain an expression for the excess of pressure inside (i) a spherical soap bubble and (ii) a spherical liquid drop. 81. Prove that the excess or pressure on one side of a soap film of surface tension  over that on the other side is given by

1 1  p  2    .  R1 R2 

82. Describe Quincke’s method of finding S.T. Derive the formula employed. -4

83. A drop of water of radius 10 m is split into 1000 equal tiny droplets. Find the mechanical work done. Calculate the excess of pressure inside each droplet, if the S.T. of water is 75 X -3

-1

10 Nm . -8

-2

[Ans. W = 8.482 X 10 J; P = 1.5 X 10 Nm ]. -3

84. Calculate the excess pressure inside a soap bubble of radius 3 X 10 m. Surface tension of -3

soap solution is 20 X 10 N/m. Calculate also the surface energy of the soap bubble. 2

-6

[Ans. 26.66 Nm . 4.525 X 10 J]. 85. A liquid drop of radius R breaks into 64 tiny drops. Find the resulting change in energy. Surface tension of the liquid is  . [Ans.2.12 R ]. 2

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204

86. If a number of little droplets of water, all of the same radius r coalesce to form a single drop of radius R, show that the rise in temperature of water will be given by

1 r

  3  

1  where R

 is the S.T. of water. 87. Calculate the amount of energy evolved when 1000 droplets of water of radius 0.0002 m -3

-1

-5

combine to form a single drop. S.T. of water = 72 X 10 Nm . [Ans. 3.257 X 10 J]. 88. Two spherical soap bubbles of diameter 0.1 m and 0.06 m respectively are formed at the ends of a narrow horizontal tube. What is the difference of pressure between the two ends of -2

-1

the tube? (S.T. of soap solution = 3 X 10 Nm ). 89. A spherical soap bubble of radius 0.01 m is formed inside another soap bubble of radius 0.02 m. Calculate the radius of a single soap bubble which will have an excess of pressure equal to the difference in pressure between the inside of the inner bubble and the outside of the larger bubble.

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