Chatper 06 Second Law of Thermodynamics

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. . . .Engineering Thermodynamics132

6

6.1 Qualitative difference betweenheat and work

The first law of thermodynamics states that a certain energy balance will hold when asystem undergoes a change of state or a thermodynamic process. But it does not giveany information on whether that change of state or the process is at all feasible ornot. The first law cannot indicate whether a metallic bar of uniform temperature canspontaneously become warmer at one end and cooler at the other. All that the law canstate is that if this process did occur, the energy gained by one end would be exactlyequal to that lost by the other. I t is the second law of thermodynamics which providesthe cri teri on as to the probabil ity of vari ous processes.

Spontaneous processes in nature occur only in one direction. Heat always flowsfrom a body at a higher temperature to a body at a lower temperature, water alwaysflows downward, time always flows in the forward direction. The reverse of thesenever happens spontaneously. The spontaneity of the process is due to a finite driv-

ing potential, sometimes called the ‘force’ or the ‘cause’, and what happens is calledthe ‘flux’, the ‘current’ or the ‘effect’. The typical forces like temperature gradient,concentration gradient and electric potential gradient, have their respective conjugatefluxes of heat transfer, mass transfer, and flow of electric current. These transferprocesses can never spontaneously occur from a lower to a higher potential. Thisdirectional law puts a limitation on energy transformation other than that imposedby the first law.

Figure 6.1 demonstrates the various transport processes. The process 1-2 froma high to a low potential is spontaneous, whereas the process 2-1 requires energyaddition for that to happen.

The firm line indicates the spontaneous process and the dotted line indicates thatthe process can never happen spontaneously and it is made possible only by energyaddition.

Joule’s experiments (Section 4.1) amply demonstrate that energy, when

supplied to a system in the form of work, can be completely converted into heat(work transferÆ internal energy increaseÆ heat transfer). But the complete conver-sion of heat into work in a cycle is not possible.So heat and work are not completelyinterchangeable forms of energy.

C h a p t e r

SECOND LAW OF THERMODYNAMICS



Second Law of Thermodynamics. . . .133

When work is converted into heat, we always have

W = Q but when heat is converted into work in a complete closed cycle process

Q >

W

The arrow indicates the direction of energy transformation. This is illustrated inFig. 6.2. As shown in Fig. 6.2(a), a system is taken from state 1 to state 2 by worktransferW 1 – 2, and then by heat transferQ 2 – 1 the system is brought back from state 2to state 1 to complete a cycle. It is always found thatW 1 – 2 =Q 2 – 1. But if the systemis taken from state 1 to state 2 by heat transferQ 1 – 2, as shown in Fig. 6.2(b), then thesystem cannot be brought back from state 2 to state 1 by work transferW 2 – 1. Hence,heat cannot be converted completely and continuously into work in a cycle. Someheat has to be rejected. In Fig. 6.2(c),Q 1–2 is the heat supply to the system,W 2–3 isthe work done by the system,Q 3–4 is the heat rejected from the system andW 4–1 isthe work done upon the system to complete the cycle. The net work done in the cycle

is (W 2–3 – W 4–1) which is less thanQ 1–2 and some heatQ 3–4 must be rejected in thecycle. This underlies the work of Sadi Carnot, a French military engineer, who firststudied this aspect of energy transformation (1824). Work is said to be ahigh gradeenergy and heat a low grade energy. The complete conversion of low grade energyinto high grade energy in a cycle is impossible.

1 2 3 4 1Æ Æ Æ Æ

1 2 1 Æ

Æ 1 2 1

Æ Æ

Q1–2

W2–1

W Q1–2 2–1=

(a)

W Q2–1 1–2<

( – ) <W W Q2–3 4–1 1–2

W3–4 > 0

(b) (c)

Q2–1

W 2 – 1

Q1–2W

2 –3W4–1

Q3–4

Wnet < Q1–2

f. 6.2 Qualitative distinction between heat and work

6.2 cyclic heat engineFor engineering purposes, the second law is best expressed in terms of theconditions which govern the production of work by a thermodynamic systemoperating in a cycle.

(a) ( )b (c) (d) (e) (f)

1 1 1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2 2

Z1

Z2

p1

p2

T1

T2

C1

C2

E1

E2

W

f. 6.1 (a) Flow from a high to allow elevation (Z), (b) Fluid flow from a high to a

low pressure (p), (c) Heat transfer from a high to a low temperature (T),

(d) Mass transfer of a constituent in a mixture from a high to a low concentration

(C, gmol/m3 ), (e) Flow of electric current from a high to a low electric potential

(E), (f) Energy flux from work to heat transfer (W Æ Q)




A heat engine cycle is a thermodynamic cycle in which there is a net heat transferto the system and a net work transfer from the system. The system which executes aheat engine cycle is called aheat engine .

A heat engine may be in the form of mass of gas confined in a cylinder and pis-ton machine (Fig. 6.3a) or a mass of water moving in a steady flow through a steampower plant (Fig. 6.3b).

In the cyclic heat engine, as represented in Fig. 6.3(a), heatQ 1 is transferred to thesystem, workW E is done by the system, work W c is done upon the system, and thenheatQ 2 is rejected from the system. The system is brought back to the initial statethrough all these four successive processes which constitute a heat engine cycle. InFig. 6.3(b) heatQ 1 is transferred from the furnace to the water in the boiler to formsteam which then works on the turbine rotor to produce workW T , then the steam iscondensed to water in the condenser in which an amount of heatQ 2 is rejected fromthe system, and finally workW p is done on the system (water) to pump it to the boiler.

The system repeats the cycle. The net heat transfer in a cycle to either of the heat engines

Q net =Q 1 –Q 2 (6.1)

Q2

Q2

Wp

WT

Q1

H O2

H O2

Vapour

Boiler

Condenser

Sea, River or Atmosphere

Pump

Furnace

Turbine

(a)

(b)

WE

WC

System

Q1

f. 6.3 Cyclic heat engine

(a) Heat engine cycle performed by a closed system undergoing four

successive energy interactions with the surroundings, (b) Heat engine cycle performed by a steady flow system interacting with

the surroundings as shown




and the net work transfer in a cycle

W net =W T –W P (6.2)

(or W net =W E – W C )

By the first law of thermodynamics, we have

cycle

ÂQ =cycle

ÂW

\ Q net =W net

or Q 1 –Q 2 =W T –W P (6.3)

Figure 6.4 represents a cyclic heat engineintheformof a block diagram indicating thevarious energy interactions during a cycle.

Boiler (B ), turbine (T ), condenser (C ), andpump (P ), all four together constitute a heatengine. A heat engine is here a certain quan-tity of water undergoing the energy interac-tions, as shown, in cyclic operations to pro-duce net work from a certain heat input.

The function of a heat engine cycle is toproduce work continuously at the expenseof heat input to the system. So the net workW net and heat inputQ 1 referred to the cycleare of primary interest. The efficiency of a heat engine or a heat engine cycle isdefined as follows:

h =Net work output of thecycle

Total heat input to thecycle =

W

Q

net

1

(6.4)

From Eqs. (6.1), (6.2), (6.3) and (6.4),

h =W

Q

net

1

=W W

Q

T P -

1

=Q Q

Q

1 2

1

-

h =1 –Q

Q

2

1

(6.5)

This is also known as the thermal efficiency of a heat engine cycle. A heat engineis very often called upon to extract as much work (net) as possible from a certain heatinput, i.e., to maximise the cycle efficiency.

6.3 energy reservoirs

A thermal energy reservoir (TER) is defined as a large body of infinite heatcapacity, which is capable of absorbing or rejecting an unlimited quantity of heatwithout suffering appreciable changes in its thermodynamic coordinates. The chang-es that do take place in the large body as heat enters or leaves are so very slow and sovery minute that all processes within it are quasi-static.

WTWP

Q2

Q1H O( )2 l H O(g)2

P

C

B

T

H O( )2 l H O( )2 l

f. 6.4 Cyclic heat engine with energy

interactions represented in a

block diagram




The thermal energy reservoir TERH from which heat Q 1 is transferred to thesystem operating in a heat engine cycle iscalled the source . The thermal energy res-ervoir TERL to which heat Q 2 is rejectedfrom the system during a cycle is the sink .A typical source is a constant temperaturefurnace where fuel is continuously burnt,and a typical sink is a river or sea or theatmosphere itself.

A mechanical energy reservoir (MER)is a large body enclosed by an adiabaticimpermeable wall capable of storingwork as potential energy (such as a raisedweight or wound spring) or kinetic energy(such as a rotating flywheel). All processes of interest within an MER are essentiallyquasi-static. An MER receives and delivers mechanical energy quasi-statically.

Figure 6.5 shows a cyclic heat engine exchanging heat with a source and a sinkand deliveringW net in a cycle to an MER.

6.4 kelvin–Planck statement of second law

The efficiency of a heat engine is given by

h =W

Q

net

1

=1 –Q

Q

2

1

Experience shows that W net <Q 1, since heatQ 1 transferred to a system cannot be

completely converted to work in a cycle (Section 6.1). Therefore, h is less than unity.A heat engine can never be 100% efficient. Therefore,Q 2 >0, i.e., there has alwaysto be a heat rejection. To produce net work in a thermodynamic cycle, a heat enginehas thus to exchange heat with two reservoirs, the source and the sink.

TheKelvin–Planck statement of the second law states: It is impossible for a heat

engine to produce net work in a complete cycle if i t exchanges heat onl y with bodies

at a single fixed temperature.

If Q 2 =0 (i.e., W net =Q 1, or h =1.00), the heat engine will produce net workin a complete cycle by exchanging heat with only one reservoir, thus violating theKelvin-Planck statement (Fig. 6.6). Such a heat engineis called aperpetual motion

machine of the second kind , abbreviated to PMM2. A PMM2 is impossible.A heat engine has, therefore, to exchange heat with two thermal energy

reservoirs at two different temperatures to produce net work in a complete cycle(Fig. 6.7). So long as there is a difference in temperature, motive power (i.e., work)

can be produced. If the bodies with which the heat engine exchanges heat are of finiteheat capacities, work will be produced by the heat engine till the temperatures of thetwo bodies are equalised.

W P

W T

W net

MER

CHE

Q1

Q2

B

C

T P

TER

(Source)

H

TER

(Sink)

L

f. 6.5 Cyclic heat engine (CHE) with

source and sink




t 1

Q1

W net

= Q1

Q2

= O

H.E

Source at t 1

Sink at t 2

Q1

W net

Q2

H.E

f. 6.6 A PMM2 f. 6.7 Heat engine producing net work in a

cycle by exchanging heat at two differenttemperatures

6.5 clausius’ statement of the second law

Heat always flows from a body at a higher temperature to a body at a lower tempera-ture. The reverse process never occurs spontaneously.

Clausius’ statement of the second law gives: I t is impossible to construct a devicewhich, operating in a cycle, wil l produce no effect other than the transfer of heat from

a cooler to a hotter body.

Heat cannot flow of itself from a body at a lower temperature to a body at a highertemperature. Some work must be expended to achieve this.

6.6 refrigerator and heat PumPA refrigerator is a device which, operating in a cycle, maintains a body at atemperature lower than the temperature of the surroundings. Let the bodyA (Fig. 6.8)be maintained at t 2, which is lower than the ambient temperaturet 1. Even thoughA isinsulated, there will always be heat leakageQ 2 into the body from the surroundingsby virtue of the temperature difference. In order to maintain bodyA at the constanttemperature t 2, heat has to be removed from the body at the same rate at which heatis leaking into the body. This heat (Q 2) is absorbed by a working fluid, called therefrigerant , which evaporates in the evaporator E 1 at a temperature lower than t 2 absorbing the latent heat of vaporisation from the body A which is cooled orrefrigerated (Process 4–1). The vapour is first compressed in the compressor C 1,driven by a motor, which absorbs workW C (Process 1–2), and is then condensed inthe condenserC 2 rejecting the latent heat of condensationQ 1 at a temperature higherthan that of the atmosphere (at t 1) for heat transfer to take place (Process 2–3). The

condensate then expands adiabatically through an expander (an engine or turbine)producing workW E , when the temperature drops to a value lower than t 2 such thatheatQ 2 flows from the bodyA to make the refrigerant evaporate (Process 3–4). Sucha cyclic device of flow throughE 1–C 1–C 2–E 2 is called a refrigerator . In a refrigerator




cycle, attention is concentrated on the bodyA.Q 2 andW are of primary interest. Justlike efficiency in a heat engine cycle, there is a performance parameter in a refrigeratorcycle, called the coefficient of performance , abbreviated to COP, which is defined as

COP =Desiredeffect

Work input

=Q

W

2

\ [COP]ref =Q

Q Q

2

1 2- (6.6)

A heat pump is a device which, operat-ing in a cycle, maintains a body, sayB

(Fig. 6.9), at a temperature higherthan the temperature of the surround-ings. By virtue of the temperature dif-ference, there will be heat leakageQ 1 from the body to the surroundings. Thebody will be maintained at the constanttemperature t 1, if heat is dischargedinto the body at the same rate at which

heat leaks out of the body. The heatis extracted from the low temperaturereservoir, which is nothing but the at-mosphere, and discharged into the hightemperature body B , with the expen-

Bodyat

B

t1

Refrigerant

Atmosphereat t2

3

4

2

1

WE

WC

Wnet = W WC E

–

= Q Q1 2 –

Q1

Q1

Q2

E2

C2

E1

C1

t2 < t1

f. 6.9 A cyclic heat pump

Atmospheret1

Condenser

Evaporator

Body A

t2

(a) (b)

Bodyat

A

t2

Expander

Compressor Refrigerant

Atmosphereat t1

3

2

4

1

3

4

2

1

C1

C2

E2

E1

WE

WC

WE WC

Q2Q2

Q2

Q1

Q1

Q2

E2

C2

E1

C1

t1 > t2

f. 6.8 A cyclic refrigeration plant




diture of workW in a cyclic device called a heat pump . The working fluid operatesin a cycle flowing through the evaporator E 1, compressorC 1, condenserC 2 and ex-panderE 2, similar to a refrigerator, but the attention is here focussed on the high tem-perature bodyB . HereQ 1 andW are of primary interest, and the COP is defined as

COP =Q

W

1

\ [COP]H.P . =Q

Q Q

1

1 2- (6.7)

From Eqs. (6.6) and (6.7), it is found that

[COP]H.P . =[COP]ref +1 (6.8)

The COP of a heat pump is greater than the COP of a refrigerator by unity. Equation(6.8) expresses a very interesting feature of a heat pump.Since

Q 1 =[COP]H.P . W

=[COPref +1]W (6.9)

Q 1 is always greater thanW .For an electrical resistance heater, ifW is the electrical energy consumption, then

the heat transferred to the space at steady state isW only, i.e.,Q 1 =W .A 1 kW electric heater can give 1 kW of heat at steady state and nothing more. In

other words, 1 kW of work (high grade energy) dissipates to give 1 kW of heat (lowgrade energy), which is thermodynamically inefficient.

However, if this electrical energy W is used to drive the compressor of a heatpump, the heat suppliedQ 1 will always be more than W , or Q 1 >W . Thus, a heatpump provides a thermodynamic advantage over direct heating.

For heat to flow from a cooler to a hotter body, W cannot be zero, and hence,the COP (both for refrigerator and heat pump) cannot be infinity. Therefore,

W >0, and COP <•.It may be noted that the energy conversion efficiency from energy input (work) to

desired energy output (heat) taking place in a refrigerator or heat pump and expressedby Eqs. (6.6) and (6.7), the denominator being less than the numerator, exceeds100%, in which no law is violated but results from the way be which the efficiencyis defined. To eliminate this confusion, the expression ‘coefficient of performance’(CUP) is used instead of ‘efficiency’ in such situations.

6.7 eQuivalence of kelvin–Planck

and clausius statements

At first sight, Kelvin–Planck’s and Clausius’ statements may appear to be uncon-nected, but it can easily be shown that they are virtually two parallel statements ofthe second law and are equivalent in all respects.

The equivalence of the two statements will be proved if it can be shown that theviolation of one statement implies the violation of the second, and vice versa. 1. Let us first consider a cyclic heat pump P which transfers heat from a low

temperature reservoir (t 2) to a high temperature reservoir (t 1) with no othereffect, i.e., with no expenditure of work, violating Clausius statement(Fig. 6.10).




Hot Reservoir at t 1

Cold Reservoir at t 2

W = 0 W net

= – Q Q1 2

Q1

H.P.

P E

Q1

Q1

Q2

H.E.

f. 6.10 Violation of the Clausius statement

Let us assume a cyclic heat engine E operating between the same thermalenergy reservoirs, producingW net in one cycle. The rate of working of the heatengine is such that it draws an amount of heatQ 1 from the hot reservoir equalto that discharged by the heat pump. Then the hot reservoir may be eliminatedand the heatQ 1 discharged by the heat pump is fed to the heat engine. So wesee that the heat pumpP and the heat engineE acting together constitute aheat engine operating in cycles and producing net work while exchangingheat only with one body at a single fixed temperature(t 2). This violates theKelvin–Planck statement.

2. Let us now consider a perpetual motion machine of the second kind (E ) whichproduces net work in a cycle by exchanging heat with only one thermal energyreservoir (at t 1) and thus violates the Kelvin–Planck statement (Fig. 6.11).

t 1

t 2

W = Q1

Q1

H.E.P (E)

PMM2 H.P.

Q1 + Q2

Q2 = 0 Q2

f. 6.11 Violation of the Kelvin–Planck statement




Let us assume a cyclic heat pump (P ) extracting heatQ 2 from a low temperaturereservoir at t 2 and discharging heat to the high temperature reservoir at t 1 with theexpenditure of work W equal to what the PMM2 delivers in a complete cycle. SoE andP together constitute a heat pump working in cycles and producing the soleeffect of transferring heat from a lower to a higher temperature body, thus violatingthe Clausius statement.

6.8 reversibility and irreversibility

The second law of thermodynamics enables us to divide all processes into two class-es: 1. Reversible or ideal process 2. Irreversible or natural process

A reversible process is one which is performed insuch a way that at the conclusion of the process, boththe system and the surroundings may be restored totheir initial states, without producing any changesin the rest of the universe. Let the state of a systembe represented by A (Fig. 6.12), and let the systembe taken to state B quasistatically by following thepathA–B . If the system and also the surroundings arerestored to their initial states on reverse, and nochange in the universe is produced, then the processA–B will be a reversible process. In the reverse process, the system has to be takenfrom stateB toA by following the same path B –A. A reversible process should notleave any trace or relic behind to show that the process had ever occurred.

A reversible process is carr ied out i nfinitely slowly with an infinitesimal gradient,

so that every state passed through by the system is an equil ibr ium state . So a revers-ible process coincides with a quasi-static process.

Any natural process carried out with a finite gradient is an irreversible process.A reversible process, which consists of a succession of equil ibr ium states , is an ide-alised hypothetical process, approached only as a limit. It is said to be anasymptoteto reali ty . All spontaneous processes are irreversible.

Time has an important effect on reversibility. If the time allowed for a process tooccur is infinitely large, even though the gradient is finite, the process becomes re-versible. However, if this time is squeezed to a finite value, the finite gradient makesthe process irreversible.

6.9 causes of irreversibility

The irreversibility of a process may be due to either one or both of the following:

1. Lack of equilibrium during the process 2. Involvement of dissipative effects

B

A

y

x

f. 6.12 A reversible process




6.9.1 i l eq

The lack of equilibrium (mechanical, thermal or chemical) between the system andits surroundings, or between two systems, or two parts of the same system, causesa spontaneous change which is irreversible. The following are specific examples inthis regard:

1. Heat transfer through a finite temperature difference A heat transfer

process approaches reversibility as the temperature difference between two bodies

approaches zero. We define a reversible heat transfer process as one in which heat

is transferred through an infinitesimal temperature difference. So to transfer a finite

amount of heat through an infinitesimal temperature difference would require an

infinite amount of time, or infinite area. All actual heat transfer processes are through

a finite temperature difference and are, therefore, irreversible, and greater the tem-perature difference, the greater is the irreversibility.

We can demonstrate by the second law that heat transfer through a finite tem-

perature difference is irreversible. Let us assume that a source at t A and a sink at

t B (t A >t B ) are available, and letQ A – B be the amount of heat flowing from A to B

(Fig. 6.13). Let us assume an engine operating betweenA andB , taking heatQ 1 from

A and discharging heatQ 2 toB . Let the heat transfer process be reversed, andQ B – A

be the heat flowing fromB toA, and let the rate of working of the engine be such that

Q 2 =Q B –A

Q A B – Q

2

Q1

W netE

Source A, t A

Sink B, t B

QB A – Q

2

Q1

W net

Source A, t A

Sink B, t B

E

f. 6.13 Heat transfer f. 6.14 Heat transfer through

through a finite a finite temperature

temperature difference difference is irreversible

(Fig. 6.14). Then the sink B may be eliminated. The net result is thatE produces net-workW in a cycle by exchanging heat only withA, thus violating the Kelvin–Planckstatement. So the heat transfer processQ A – B is irreversible, andQ B – A is not possible.




2. Lack of pressure equilibrium within the interior of the system or

between the system and the surroundings When there exists a difference in

pressure between the system and the surroundings, or within the system itself, then

both the system and its surroundings or the system alone, will undergo a change of

state which will cease only when mechanical equilibrium is established. The reverse

of this process is not possible spontaneously without producing any other effect. That

the reverse process will violate the second law becomes obvious from the following

illustration.

3. Free expansion Let us consider an in-

sulated container (Fig. 6.15) which is divided

into two compartments A andB by a thin dia-

phragm. Compartment A contains a mass ofgas, while compartmentB is completely evacu-

ated. If the diaphragm is punctured, the gas inA

will expand into B until the pressures in A and

B become equal. This is known asfree or unre-

strained expansion . We can demonstrate by the

second law, that the process of free expansion

is irreversible.

To prove this, let us assume that free expansion is reversible, and that the gas inB returns into A with an increase in pressure, and B becomes evacuated as before(Fig. 6.16). There is no other effect. Let us install an engine (a machine, not a cyclic

heat engine) between A and B , and permit thegas to expand through the engine fromA toB . The engine develops a work outputW at the expense of the internal energyof the gas. The internal energy of the gas (system) inB can be restored to its ini-tial value by heat transfer Q (=W ) froma source. Now, by the use of the reversedfree expansion, the system can be restoredto the initial state of high pressure inA andvacuum in B . The net result is a cycle, inwhich we observe that net work outputW is accomplished by exchanging heat with asingle reservoir. This violates the Kelvin–Planck statement. Hence, free expansion isirreversible.

The same argument will hold if thecompartmentB is not in vacuum but at apressure lower than that in compartmentA (case b).

6.9.2 i dp e

The irreversibility of a process may be due to thedissipati ve effects in which workis done without producing an equivalent increase in the kinetic or potential energyof any system. The transformation of work into molecular internal energy either of

Engine

Heatsource, t

A

W

B

Q = W

f. 6.16 Second law demonstrates

that free expansion is

irreversible

Gas

Diaphragm Insulation

Vacuum

A B

f. 6.15 Free expansion




the system or of the reservoir takes place through the agency of such phenomena asfriction, viscosity, inelasticity, electrical resistance, and magnetic hysteresis. Theseeffects are known as dissipative effects, and work is said to be dissipated.

1. Friction Friction is always present in moving devices. Friction may be reducedby suitable lubrication, but it can never be completely eliminated. If this were pos-sible, a movable device could be kept in continual motion without violating either ofthe two laws of thermodynamics. The continual motion of a movable device in thecomplete absence of friction is known as perpetual motion of the third kind.

That friction makes a process irrevers-ible can be demonstrated by the second law.Let us consider a system consisting of aflywheel and a brake block (Fig. 6.17). The

flywheel was rotating with a certain rpm,and it was brought to rest by applying thefriction brake. The distance moved by thebrake block is very small, so work transferis very nearly equal to zero. If the brakingprocess occurs very rapidly, there is littleheat transfer. Using suffix 2 after brakingand suffix 1 before braking, and applyingthe first law, we have

Q 1 – 2 =E 2 –E 1 +W 1 – 2

0 =E 2 –E 1 +0

\ E 2 =E 1 (6.10)

The energy of the system (isolated) remains constant. Since the energy may exist inthe forms of kinetic, potential, and molecular internal energy, we have

U 2 +m V2

2

2 +m Z2g =U 1 +

m V12

2 +m Z1g

Since the wheel is brought to rest, V 2 =0, and there is no change in P.E.

U 2 =U 1 +m V1

2

2 (6.11)

Therefore, the molecular internal energy of the system (i.e., of the brake and thewheel) increases by the absorption of the K.E. of the wheel. The reverse process,i.e., the conversion of this increase in molecular internal energy into K.E. within thesystem to cause the wheel to rotate is not possible. To prove it by the second law, letus assume that it is possible, and imagine the following cycle with three processes:

Process A Initially, the wheel and the brake are at high temperature as a result ofthe absorption of the K.E. of the wheel, and the flywheel is at rest. Let the flywheelnow start rotating at a particular rpm at the expense of the internal energy of thewheel and brake, the temperature of which will then decrease.

System boundary

Flywheel Brake block

F

f. 6.17 Irreversibility due to

dissipative effect like friction




Process B Let the flywheel be brought to rest by using its K.E. in raising weights,with no change in temperature.

Process C Now let heat be supplied from a source to the flywheel and the brake, torestore the system to its initial state.

Therefore, the processesA, B , andC together constitute a cycle producing workby exchanging heat with a single reservoir. This violates the Kelvin–Planck state-ment, and it will become a PMM2. So the braking process, i.e., the transformation ofK.E. into molecular internal energy, is irreversible.

2. Paddle-Wheel work transfer Work may be transferred into a system in aninsulated container by means of a paddle wheel (Fig. 6.18) which is also known asstirring work. Here work transferred is dissipated adiabatically into an increase in

the molecular internal energy of the system. To prove the irreversibility of the pro-cess, let us assume that the same amount of workis delivered by the system at the expense of itsmolecular internal energy, and the temperatureof the system goes down (Fig. 6.19). The systemis brought back to its initial state by heat trans-fer from a source. These two processes togetherconstitute a cycle in which there is work outputand the system exchanges heat with a single res-ervoir. It becomes a PMM2, and hence the dis-sipation of stirring work to internal energy is ir-reversible.

Adiabatic

Diathermic

Heat

source

W Q W =

f. 6.19 Irreversibility due to dissipation of stirring work into internal energy

3. Transfer of electricity through a resistor The flow of electric currentthrough a wire represents work transfer, because the current can drive a motor whichcan raise a weight. Taking the wire or theresistor as the system (Fig. 6.20) and writingthe first law

Q 1 – 2 =U 2 –U 1 +W 1 – 2

Here bothW 1 – 2 andQ 1 – 2 are negative.

W 1 – 2 =U 2 –U 1 +Q 1 – 2 (6.12)

Insulation

System

W

f. 6.18 Adiabatic work transfer

W

Q

I I

Resistor (system)

f. 6.20 Irreversibility due to

dissipation of electrical work into internal energy




A part of the work transfer is stored as an increase in the internal energy of the wire

(to give an increase in its temperature), and the remainder leaves the system as heat.

At steady state, the internal energy and hence the temperature of the resistor become

constant with respect to time and

W 1 – 2 =Q 1 – 2 (6.13)

The reverse process, i.e., the conversion of heatQ 1 – 2 into electrical workW 1 – 2

of the same magnitude is not possible. Let us assume that this is possible. Then heat

Q 1 – 2 will be absorbed and equal workW 1 – 2 will be delivered. But this will become

a PMM2. So the dissipation of electrical work into internal energy or heat is irrevers-ible.

6.10 conditions for reversibilityA natural process is irreversible because the conditions for mechanical, thermal andchemical equilibrium are not satisfied, and the dissipative effects, in which workis transformed into an increase in internal energy, are present. For a process to bereversible, it must not possess these features. If a process is performed quasi-stati-cally, the system passes through states of thermodynamic equilibrium, which may betraversed as well in one direction as in the opposite direction. I f there are no dissipative effects, all the work done by the system duri ng the performance of a process in

one direction can be returned to the system duri ng the reverse process.

A process will be reversible when it is performed in such a way that the system

is at al l times infinitesimall y near a state of thermodynamic equil ibr ium and i n the

absence of di ssipat ive effect of any form . Reversible processes are, therefore, purelyideal, limiting cases of actual processes.

6.11 carnot cycle

A reversible cycle is an ideal hypothetical cycle in which all the processes consti-tuting the cycle are reversible. Carnot cycle is a reversible cycle. For a stationarysystem, as in a piston and cylinder machine, the cycle consists of the following foursuccessive processes (Fig. 6.21):

1. A reversible isothermal process in which heat Q 1 enters the system at t 1 reversibly from a constant temperature source at t 1 when the cylinder coveris in contact with the diathermic cover A. The internal energy of the systemincreases.

From first law, Q 1 =U 2 –U 1 +W 1–2 (6.14)

(for an ideal gas only,U 1 =U 2)




Sink, t2

Source, t1

System Adiabatic

WP

WE

Adiabatic cover ( )B

Diathermic cover ( ) A

Q2

Q1

f. 6.21 Carnot heat engine—Stationary system

2. A reversible adiabati c process in which the diathermic coverA is replaced bythe adiabatic cover B , and workW E is done by the system adiabatically andreversibly at the expense of its internal energy, and the temperature of thesystem decreases from t 1 to t 2.

Using the first law,

0 =U 3 –U 2 +W 2 – 3 (6.15)

3. A reversible isothermal process in whichB is replaced byA and heatQ 2 leavesthe system at t 2 to a constant temperature sink at t 2 reversibly, and the internalenergy of the system further decreases.

From the first law,

– Q 2 =U 4 –U 3 –W 3 – 4 (6.16)

only for an ideal gas, U 3 =U 4

4. A reversibl e adiabatic process in which B again replacesA, and workW p isdone upon the system reversibly and adiabatically, and the internal energy ofthe system increases and the temperature rises from t 2 to t 1.

Applying the first law,

0 =U 1 –U 4 –W 4 – 1 (6.17)

Two reversible isotherms and two reversible adiabatics constitute a Carnotcycle, which is represented inp -v coordinates in Fig. 6.22.




W P Q1

Rev. adiabatics

2

3

4

v

Rev. isotherm ( )t 2

Rev. isotherm ( )t 1

1

p

Q2

W E

f. 6.22 Carnot cycle

Summing up Eqs. (6.14) to (6.17), Q 1 –Q 2 =(W 1 – 2 +W 2 – 3) – (W 3 – 4 +W 4 – 1)

or

cycle

ÂQnet =cycle

ÂW net

A cyclic heat engineoperating on theCarnot cycleis called aCarnot heat engine.For a steady flow system, the Carnot cycle is represented as shown in Fig. 6.23.

Here heatQ 1 is transferred to the system reversibly and isothermally at t 1 in the heatexchangerA, workW T is done by the system reversibly and adiabatically in the tur-bine (B ), then heatQ 2 is transferred from the system reversibly and isothermally at t 2 in the heat exchanger (C ), and then workW p is done upon the system reversibly andadiabatically by the pump (D ). To satisfy the conditions for the Carnot cycle, theremust not be any friction or heat transfer in the pipelines through which the workingfluid flows.

Flow

Flow

Sink, t

2

Heat exchanger ( ) A

Heat exchanger ( )C

Turbine ( )BPump ( )D

t 2

W P W T

Source, t 1

Systemboundary

t 1 t 1

t 1

t 2

t 2

Q2

Q1

W W W P net = -t

= Q Q1 2-

4

f. 6.23 Carnot heat engine—Steady flow system




6.12 reversed heat engine

Since all the processes of the Carnot cycle are reversible, it is possible to imaginethat the processes are individually reversed and carried out in reverse order. When a

reversible process is reversed, all the energy transfers associated with the process arereversed in direction, but remain the same in magnitude. The reversed Carnot cyclefor a steady flow system is shown in Fig. 6.24. The reversible heat engine and the

reversed Carnot heat engine are represented in block diagrams in Fig. 6.25. If E is areversible heat engine (Fig. 6.25a), and if it is reversed (Fig. 6.25b), the quantitiesQ 1,Q 2 andW remain the same in magnitude, and only their directions are reversed. The

reversed heat engine$ takes heat from a low temperature body, dis charges heat to ahigh temperature body, and receives an inward flow of network.

The names heat pump and refrigerator are applied to the reversed heat engine,

which have already been discussed in Section 6.6, where the working fluid flowsthrough the compressor (B ), condenser (A), expander (D ), and evaporator (C ) to

complete the cycle.

Flow

Flow Systemboundry

Heat exchanger ( ) A

Heat exchanger ( )C

Turbine( )B

Pump( )D

WP WT /WC

t1

t1

t1

t2

t2

Q2

Q1

t2t2

t1

Expander ( )WE

Compressor ( )WC

(a)

v

p

3

2

4

Q1

t1

Q2 1 t2

WT

WF

WC

(b)

f. 6.24 Reversed Carnot heat engine—Steady flow process




W P W P W T

W T D DE E

C C

A A

B B

t 1 t 1

t 2 t 2

(a) (b)

Q1 Q1

Q2 Q2

W W W = -net T P

W W W = -net T P

f. 6.25 Carnot heat engine and reversed Carnot heat engine shown in block diagrams

6.13 carnot’s theorem

It states that of all heat engines operating between a given constant temperature

source and a given constant temperature sink, none has a higher efficiency than a

reversibl e engine.

Let two heat enginesE A andE B operate between the given source at temperaturet 1 and the given sink at temperature t 2 as shown in Fig. 6.26.

Q1 A

Q2 A

Sink t 2

Source, t 1

Q2B

E A W A W BE B

Q1B

f. 6.26 Two cyclic heat engines E A and E B operating between the same source and sink,

of which E B is reversible

LetE A beany heat engine andE B beany reversible heat engine. We have to provethat the efficiency of

E B is more than that of

E A. Let us assume that this is not true

andhA >hB . Let the rates of working of the engines be such that

Q 1A =Q 1B =Q 1

Since hA >hB




W

Q

A

A1

>W

Q

B

B 1

\ W A >W B

Now, let E B be reversed. SinceE B is a reversible heat engine, the magnitudes ofheat and work transfer quantities will remain the same, but their directions will bereversed, as shown in Fig. 6.27. SinceW A >W B , some part ofW A (equal toW B ) maybe fed to drive the reversed heat engine$ B .

Since Q 1A =Q 1B =Q 1, the heat discharged by $B may be supplied toE A. Thesource may, therefore, be eliminated (Fig. 6.28). The net result is that E A and $ B together constitute a heat engine which, operating in a cycle, produces net workW A – W B , while exchanging heat with a single reservoir at t 2. This violates the Kel-

vin–Planck statement of the second law. Hence the assumption thathA >hB is wrong. Therefore, hB ≥ hA

Q1 A

Q2 A

Sink, t 2

Source, t 1

Q2B

E A B EW A

W B

Q1B

Q2 A

Q1 = A Q1 =BQ1 Q1

Sink, t 2

E A EB

W A

W A

W B

W B

Q2B

–

f. 6.27 E B is reversed f. 6.28 E A and $ B together violate the

K-P statement

6.14 corollary of carnot’s theorem

The efficiency of all reversible heat engines operati ng between the same temperature

levels is the same.

Let both the heat enginesE A and E B (Fig. 6.26) be reversible. Let us assumehA >hB . Similar to the procedure outlined in the preceding article, if E B is reversedto run, say, as a heat pump using some part of the work output (W A) of engineE A, weseethat thecombined systemof heat pump E B and engineE A, becomes a PMM2. So

hA cannot be greater thanhB . Similarly, if we assumehB >hA and reverse the engineE A, we observe thathB cannot be greater thanhA.

Therefore, hA =hB




Since the efficiencies of all reversible heat engines operating between the sameheat reservoirs are the same, the efficiency of a reversible engine i s independent of thenature or amount of the working substance undergoing the cycle.

6.15 absolute thermodynamic temPerature scale

The efficiency of any heat engine cycle receiving heat Q 1 and rejecting heatQ 2 isgiven by

h =W

Q

net

1

=Q Q

Q

1 2

1

- =1 –

Q

Q

2

1

(6.18)

By the second law, it is necessary to have a temperature difference (t 1 – t 2) to ob-tain work for any cycle. We know that the efficiency of all heat engines operatingbetween the same temperature levels is the same, and it is independent of the work-

ing substance. Therefore, for a reversible cycle (Carnot cycle), the efficiency willdepend solely upon the temperaturest 1 and t 2, at which heat is transferred, or

hrev = f (t 1, t 2) (6.19)where f signifies some function of the temperatures. From Eqs. (6.18) and (6.19)

1 –Q

Q

2

1

=f (t 1, t 2)

In terms of a new function F

Q

Q

1

2

=F (t 1, t 2) (6.20)

If some functional relationship is assigned between t 1, t 2 andQ 1/Q 2, the equationbecomes the definition of a temperature scale.

Let us consider two reversible heat engines, E 1 receiving heat from the sourceat t 1, and rejecting heat at t 2 to E 2 which, in turn, rejects heat to the sink at t 3(Fig. 6.29a).

Q2

t 2

Q1

Heat reservoir, t 1

Heat reservoir, t 3

Q1

Q2

Q3 Q3

E 1

E 3

E 2

W 1

W 3

W 2

Q Q1 2-

Q Q1 3-

Q Q2 3-

=

=

=

f. 6.29 () Three Carnot engines




NowQ

Q

1

2

=F (t 1, t 2);Q

Q

2

3

=F (t 2, t 3)

E 1and E 2together constituteanother heat engineE 3 operating between t 1 and t 3.

\ Q

Q

1

3

=F (t 1, t 3)

NowQ

Q

1

2

=Q Q

Q Q

1 3

2 3

/

/

orQ

Q

1

2

=F (t 1, t 2) =F t t

F t t

( , )

( , )

1 3

2 3

(6.21)

The temperatures t 1, t 2 and t 3 are arbitrarily chosen. The ratio Q 1/Q 2 dependsonly on t 1 and t 2, and is independent of t 3. So t 3 will drop out from the ratio on

the right in Eq. (6.21). After it has been cancelled, the numerator can be written asf (t 1), and the denominator as f (t 2), wheref is another unknown function. Thus,

Q

Q

1

2

=F (t 1, t 2) =f

f

( )

( )

t

t

1

2

Heat transfer (Kelvin)(J)

Q1 600 K

500 K

300 KQ2

Q1

Q2

600 K

E Wnet

300 K

f. 6.29 ()

Since f (t ) is an arbitrary function, the simplest possible way to define the abso-

lute thermodynamic temperature T is to let f (t ) =T , as proposed by Kelvin. Then, bydefinition

Q

Q

1

2

=T

T

1

2

(6.22)

The absolute thermodynamic temperature scale is also known as theKelvin scale . Two temperatures on the Kelvin scale bear the same relationship to each other as dothe heats absorbed and rejected respectively by a Carnot engine operating betweentwo reservoirs at these temperatures Fig. 6.29 (b). The Kelvin temperature scale is,therefore, independent of the peculiar characteristics of any particular substance.

The heat absorbedQ 1 and the heat rejectedQ 2 during the two reversible isother-

mal processes bounded by two reversible adiabatics in a Carnot engine can be mea-sured. In defining the Kelvin temperature scale also, the triple point of water is takenas the standard reference point. For a Carnot engine operating between reservoirs attemperatures T and T t , T t being the triple point of water (Fig. 6.30), arbitrarily as-signed the value 273.16 K,




Q

Q t

=T

T t

\ T =273.16Q

Q t

(6.23)

Q

Q t

T t

T

E W net

273.16K=

f. 6.30 Carnot heat engine with sink

at triple point of water

Q1

Q2

Q2

Q3

Q3

Q4

Q4

T 1

T 2

T 3

T 4

E 1

E 2

E 3

W 1

W2

W 3

Q Q1 2-

Q Q2 3-

Q Q3 4-

=

=

=

f. 6.31 Heat engines operating in series

If this equation is compared with the equations given in Section 2.3, it is seenthat in the Kelvin scale , Q plays the role of thermometr ic property . The amount ofheat supply Q changes with change in temperature, just like the thermal emf in athermocouple.

That the absolute thermodynamic temperature scale has a definite zero point can

be shown by imagining a series of reversible engines, extending from a source at T 1 to lower temperatures (Fig. 6.31).Since

T

T

1

2

=Q

Q

1

2

\ T T

T

1 2

2

- =

Q Q

Q

1 2

2

-

or T 1 – T 2 = (Q1 – Q2)T

Q

2

2Similarly,

T 2 – T 3 =(Q 2 –Q 3)T

Q

3

3

=(Q 2 –Q 3)T

Q

2

2

T 3 – T 4 =(Q 3 –Q 4)T

Q

2

2and so on.

If T 1 – T 2 =T 2 – T 3 =T 3 – T 4 =..., assuming equal temperature intervals

Q 1 –Q 2 =Q 2 –Q 3 =Q 3 –Q 4 =...




or W 1 =W 2 =W 3 =...

Conversely, by making the work quantities performed by the engines in series equal(W 1 =W 2 =W 3 =...), we will get

T 1 – T 2 =T 2 – T 3 =T 3 – T 4 =...

at equal temperature intervals. A scale having one hundred equal intervals betweenthe steam point and the ice point could be realised by a series of one hundred Carnotengines operating as in Fig. 6.31. Such a scale would be independent of the workingsubstance.

If enough engines are placed in series to make the total work output equal toQ 1,then by the first law the heat rejected from the last engine will be zero. By the secondlaw, however, the operation of a cyclic heat engine with zero heat rejection cannotbe achieved, although it may be approached as a limit. When the heat rejected ap-

proaches zero, the temperature of heat rejection also approaches zero as a limit.Thusit appears that a definite zero point exists on the absolute temperature scale but thi s

point cannot be reached without a viol ation of the second law.

It follows from Eq. (6.23),

T =273.16Q

Q t that the heat transferred isothermally between two given adiabatics decreases as thetemperature decreases. Conversely, the smaller the value ofQ , the lower the tempera-tureT . The smallest possible value ofQ is zero, and the correspondingT is zero.Thusif a system undergoes a reversible isothermal process without transfer of heat, the

temperature at which this process takes place is call ed absolute zero . In other words,at absolute zero, an isotherm and an adiabatic are identical.

Also, since Q 2 =0, the isothermal process atT 2 =0 would also be adiabatic ren-dering Carnot cycle ambiguous.

Thus any attainable value of absolute temperature is always greater than zero. Thisis also known as theThird law of thermodynamics which may be stated as follows: I tis impossibl e by any procedure, no matter how ideal ized, to reduce any system to the

absolute zero of temperature in a finite number of operati ons.

This is what is called the Fowler–Guggenheim statement of the third law. The thirdlaw itself is an independent law of nature, and not an extension of the second law. Theconcept of heat engine is not necessary to prove the non-attainability of absolute zeroof temperature by any system in a finite number of operations.

6.16 efficiency of the reversible heat engine

The efficiency of a reversible heat engine in which heat is re ceived solely at T 1 isfound to be

hrev

=hmax

=1 –Q

Q

2

1

Ê

Ë Á

ˆ

¯ ˜ rev =1 –

T

T

2

1

or hrev =T T

T

1 2

1

-




It is observed here that asT 2 decreases, andT 1 increases, the efficiency of the revers-ible cycle increases.

Sinceh is always less than unity, T 2 is always greater than zero and positive. The COP of a refrigerator is given by

(COP)refr =Q

Q Q

2

1 2- =

1

11

2

Q

Q -

For a reversible refrigerator, using

Q

Q

1

2

=T

T

1

2

[COPrefr]rev =T

T T

2

1 2- (6.24)

Similarly, for a reversible heat pump

[COPH.P .]rev =T

T T

1

1 2- (6.25)

6.17 eQuality of ideal gas temPerature and kelvin

temPerature

Let us consider a Carnot cycle executed by an ideal gas, as shown in Fig. 6.32. The two isothermal processesa –b andc –d atq 1andq 2are represented by equilat-

eral hyperbolas whose equations are respectively

pV =nR q 1and pV =nR q 2For any infinitesimal reversible process of an ideal gas, the first law may be written as

d -Q = C v d q + pdV = pdV

Applying this equation to the isothermal processa –b , the heat absorbed is found tobe

Q1 = pdV V

V

a

b

Ú =nR

V V

V

a

b q 1Ú dV = nRq 1 lnV

V

b

a

Reversible

Reversible

Adiabatics

Isotherms

a

q 2

q 1

p

d

v

Q2

Q1

W C

W E

c

b

f. 6.32 Carnot cycle of an ideal gas




Similarly, for the isothermal process c –d , the heat rejected is

Q 2 =nR q 2 lnV

V

c

d

\ Q

Q

1

2

=

q

q

1

2

ln

ln

V

V

V

V

b

a

c

d

(6.26)

Since the processb –c is adiabatic, the first law gives

–C v d q =pdV =

nR

V

q dV

1

2

1

nR q

q

Ú C v

d q

q =lnV

V

c

b

Similarly, for the adiabatic process d –a

1

2

1

nR q

q

Ú C v d q

q =lnV

V

d

a

\ lnV

V

c

b

=lnV

V

d

a

orV

V

c

b

=V

V

d

a

orV

V

b

a

=V

V

c

d

(6.27)

Equation (6.26) thus reduces to

Q

Q

1

2

=q

q 1

2

(6.28)

Kelvin temperature was defined by Eq. (6.22)

Q

Q

1

2

=T

T

1

2

If q andT refer to any temperature, andq t andT t refer to the triple point of water,

q

q t =

T

T t

Sinceq T =T t =273.16 K, it follows that

q =T (6.29)

The Kelvin temperature is, therefore, numerically equal to the ideal gas temperatureandmay be measured by means of a gas thermometer




6.18 tyPes of irreversibility

It has been discussed in Section 6.9 that a process becomes irreversible if it occurs

due to a finite potential gradient like the gradient in temperature or pressure, or if

there is dissipative effect like friction, in which work is transformed into internal

energy increase of the system. Two types of irreversibility can be distinguished:

1. Internal irreversibility

2. External irreversibility

The internal ir reversibili ty is caused by the internal dissipative effects like friction,

turbulence, electrical resistance, magnetic hysteresis, etc., within the system. Theexternal ir reversibi li ty refers to the irreversibility occurring at the system boundarylike heat interaction with the surroundings due to a finite temperature gradient.

Sometimes, it is useful to make other distinctions. I f the irreversibility of a process

is due to the dissipation of work into the increase in internal energy of a system, ordue to a finite pressure gradient, it is calledmechanical irreversibi li ty . If the processoccurs on account of a finite temperature gradient, it is thermal ir reversibili ty , and ifit is due to a finite concentration gradient or a chemical reaction, it is called chemical

irreversibility .A heat engine cycle in which there is a temperature difference (i ) between the

source and the working fluid during heat supply, and (ii ) between the working fluidand the sink during heat rejection, exhibits external thermal irreversibility. If the realsource and sink are not considered and hypothetical reversible processes for heatsupply and heat rejection are assumed, the cycle can be reversible. With the inclusion

of the actual source and sink, however, the cycle becomes externally irreversible.

solved examPles

ep 6.1 A cycl ic heat engine operates between a source temperatur e of

800°C and a sink temperature of 30°C. What is the least rate of heat rejecti on per kW

net output of the engine?

s For a reversible engine, the rate of heat rejection will be minimum(Fig. 6.33).

Q1

Q2

W = Q Q =1 2

– 1 kW

T 1

= 1073 K

Source

Sink

= 303 KT 2

HE

f. 6.33




hmax =hrev =1 –T

T

2

1

=1 –30 273

800 273

+

+

=1 – 0.282 =0.718

NowW

Q

net

1

=h max =0.718

\ Q 1 =1

0718. =1.392 kW

Now Q 2 =Q 1 –W net =1.392 – 1 = 0.392 kW

This is the least rate of heat rejection.

ep 6.2 A domestic food freezer maintains

a temperature of – 15°C. The ambient air tempera-

ture is 30°C. I f heat l eaks into the freezer at the

continuous rate of 1.75 kJ/s what is the least power

necessary to pump this heat out cont inuously?

s Freezer temperature,T 2 =– 15 +273 =258 K

Ambient air temperature,T 1 =30 +273 =303 K

The refrigerator cycle removes heat from thefreezer at the same rate at which heat leaks into it

(Fig. 6.34).For minimum power requirement

Q

T

2

2

=Q

T

1

1

\ Q 1 =175

28

.

. ¥ 303 =2.06 kJ/s

\ W =Q 1 –Q 2

=2.06 – 1.75 =0.31 kJ/s= 0.31 kW

ep 6.3 An ideal gas cycle is represented by a rectangle on a p-V diagram. If

p 1 and p 2 are the lower and higher pressures; and V 1 and V 2 , the smaller and larger

volumes, respectively, then (a) cal culate the work done per cycle, (b) indicate which

par ts of the cycle invol ve heat flow into the gas. (c) Show that

h =g

g

-

- -

1

p

p p+

V

V V

2

2 1

1

2 1

if heat capaci ties are constant.

Q2

Q1

Ambient air T 1

= 303 K

Freezer T 2

= 258 K

Q2

= 1.75 kJ/s

R W

f. 6.34




s

p2

p1

b

a

c

d

p

V 1 V 2V

f. 6.35

(a) W =area of the cycle (Fig. 6.35)

=(p 2 – p 1) (V 2 –V 1) Ans.

(b) Processesab andbc

Heat absorbed by 1 mole of gas in one cycle,

Q =Q ab +Q bc =C v (T b – T a ) +C p (T c – T b )

Now, T a =T b p

p

1

2

andp 2 V 1 =R T b

T c =T b V

V

2

1

, T b = p V

R

2 1

Q =C v T b 1 1

2

-Ê Ë Á ˆ ¯ ˜ p p

+C p T b V V 2

11-Ê Ë Á ˆ ¯ ˜

= p V

RC

p p

pC

V V

V p p

2 1 2 1

2

2 1

1

-+

-Ê Ë Á

ˆ ¯ ˜ Ans.

(c) Q =W

Q =

( ) ( ) p p V V

p V

RC

p p

pC

V V

V p v

2 1 2 1

2 1 2 1

2

2 1

1

- -

-+

-È

ÎÍ

˘

˚˙

= R p p V V

C V p p C p V V v p

( )

( ) ( )

2 1 2 1

1 2 1 2 2 1

- -( )- + -

=C C

C V

V V C

p

p p

p v

v p

-

-

+

-

1

2 1

2

2 1

=g

g

-

-

+

-

1

1

2 1

2

2 1

V

V V

p

p p

Ans.




ep 6.4 A Carnot engine absorbs 200 J of heat from a reservoir at the

temperature of the normal boil ing point of water and rejects heat to a reservoir at the

temperature of the tri ple point of water. Find the heat rejected, the work done by the

engine and the thermal efficiency.

sQ 1 =200 J atT 1 =373.15 K

T 2 =273.16 K

Q 2 =Q 1 T

T

2

1

=200¥ 273 16

373 15

.

. =146.4 J

W =Q 1 –Q 2 =53.6 J

h =

W

Q1 =

53 6

200

.

=0.268

ep 6.5 A reversibl e heat engine operates between two reservoi rs at

temperatures of 600°C and 40°C. The engi ne dr ives a reversible refr igerator which

operates between reservoi rs at temperatures of 40°C and – 20°C. The heat t ransfer to

the heat engine is 2000 kJ and the net work output of the combined engine refr igerator

plant is 360 kJ.

(a) Evaluate the heat transfer to the refri gerant and the net heat transfer to the

reservoi r at 40°C.

(b) Reconsider (a) given that the efficiency of the heat engine and the COP of the

refr igerator are each 40% of their maximum possible values.

s (a) Maximum efficiency of the heat engine cycle (Fig. 6.36) is given by

hmax =1 –T

T

2

1

=1 –313

873

=1 – 0.358 =0.642

AgainW

Q

1

1

=0.642

\ W 1 =0.642¥ 2000 =1284 kJ

Q1

= 2000 kJ Q4

Q2

W1

W2

Q Q W3 4 2= +

T1= 873 K T

3= 253 K

W = 360 kJ

T2

= 313 K

HE R

f. 6.36




Maximum COP of the refrigerator cycle

(COP)max =T

T T

3

2 3- =

253

313 253- =4.22

Also COP =Q

W

4

2

=4.22

Since W 1 –W 2 =W =360 kJ

\ W 2 =W 1 –W =1284 – 360 =924 kJ

\ Q 4 =4.22¥ 924 =3899 kJ

\ Q 3 =Q 4 +W 2 =924 +3899 =4823 kJ

Q 2 =Q 1 –W 1 =2000 – 1284 =716 kJ

Heat rejection to the 40°C reservoir =Q 2 +Q 3 =716 +4823= 5539 kJ

(b) Efficiency of the actual heat engine cycle

h =0.4 h max =0.4¥ 0.642

\ W 1 =0.4¥ 0.642¥ 2000

=513.6 kJ

\ W 2 =513.6 – 360 =153.6 kJ

COP of the actual refrigerator cycle

COP =Q

W

4

2

=0.4¥ 4.22 =1.69

Therefore, Q 4 =153.6¥ 1.69 =259.6 kJ

Q 3 =259.6 +153.6 =413.2 kJ

Q 2 =Q 1 –W 1 =2000 – 513.6 =1486.4 kJ

Heat rejected to the 40°C reservoir

=Q 2 +Q 3 =413.2 +1486.4 =1899.6 kJ

ep 6.6 Which i s the more effective way to increase the efficiency of a Carnot

engine: to increase T 1 , keeping T 2 constant; or to decrease T 2 , keeping T 1 constant?

s The efficiency of a Carnot engine is given by

h =1 –T

T

2

1

If T 2 is constant

∂∂

Ê Ë Á

ˆ ¯ ˜

h

T T 12

=T

T

2

12

AsT 1 increases, h increases, and the slope∂∂

Ê Ë Á

ˆ ¯ ˜

h

T T 12

decreases (Fig. 6.37). If T 1 isconstant,




∂∂

Ê Ë Á

ˆ ¯ ˜

h

T T 21

=-1

1T

AsT 2 decreases, h increases, but the slope∂∂

Ê Ë Á

ˆ ¯ ˜

h

T T 21

remains constant (Fig. 6.38).

1.0

0

h

T 1

T 2

1.0

0

Slope = – 1/T

h

T 1

T 2

f. 6.37 f. 6.38

Also∂∂

Ê Ë Á

ˆ ¯ ˜

h

T T 11

=T

T

2

12 and

∂∂

Ê Ë Á

ˆ ¯ ˜

h

T T 21

=-T

T

1

12

Since T 1 >T 2,∂∂

Ê Ë Á

ˆ ¯ ˜ >

∂∂

Ê Ë Á

ˆ ¯ ˜

h h

T T T T 2 1

1 2

So, the more effective way to increase the efficiency is to decreaseT 2. Alterna-

tively, letT 2 be decreased byDT withT 1 remaining the same

h1 =1 –T T

T

2

1

- D

If T 1 is increased by the sameDT , T 2 remaining the same

h2 =1 –T

T T

2

1+ D Then

h1 – h2 =T

T T

T T

T

2

1

2

1+-

-

D

D

=( ) ( )

( )

T T T T

T T T

1 22

1 1

- +

+

D D

D

Since T 1 >T 2, (h1 – h2) >0 The more effective way to increase the cycle efficiency is to decreaseT 2.




ep 6.7 Kelvi n was the first to poi nt out the

thermodynamic wastefulness of burning fuel for the

direct heating of a house. It is much more economi-

cal to use the high temperature heat produced by

combustion in a heat engine and then to use the work

so developed to pump heat from outdoors up to the

temperature desired in the house. In Fig. 6.39 a boil -

er fur nishes heat Q 1 at the high temperature T 1 . This

heat is absorbed by a heat engine, which extracts

work W and rejects the waste heat Q 2 into the house

at T 2 . Work W is in tur n used to operate a mechanical

refr igerator or heat pump, which extracts Q 3 from

outdoors at temperature T 3 and reject Q ¢

2 (where Q ¢

2

= Q 3 + W) into the house. As a result of thi s cycle of

operations, a total quantity of heat equal to Q 2 + Q ¢ 2 is li berated in the house, against Q 1 which would be

provided directly by the ordinary combustion of the

fuel. Thus the ratio (Q 2 + Q ¢ 2 )/Q 1 represents the heatmultipl icati on factor of thi s method. Determine this

multi plication factor if T 1 = 473 K, T 2 = 293 K, and

T 3 = 273 K.

s For the reversible heat engine (Fig. 6.39)

Q

Q

2

1

=T

T

2

1

\ Q 2 =Q

1

T

T

2

1

Ê

Ë Á

ˆ

¯ ˜

Also h =W

Q 1

=T T

T

1 2

1

-

or W =T T

T

1 2

1

- ◊ Q 1

For the reversible heat pump

COP =¢Q

W

2 =T

T T

2

2 3-

\ Q ¢2 =T

T T

T T

T

2

2 3

1 2

1-

◊

-

◊ Q 1

\ multiplication factor (M.F.)

=Q Q

Q

2 2

1

+ ¢ =

Q T

T Q

T

T T

T T

T

Q

1

2

1

1

2

2 3

1 2

1

1

+ ◊

-

◊-

W

W

Boiler

Outdoors

House

Q1

T 3

Q1

Q2

Q3

Q Q W ¢ 2 3

= +

T 1

Q3

T 2

CHE

H.P.

f. 6.39




or M.F. =T T T T T T

T T T

22

2 3 2 1 22

1 2 3

- + -

-( )

or M.F. =T T T

T T T

2 1 3

1 2 3

( )

( )

-

-

Here T 1 =473 K, T 2 =293 K andT 3 =273 K

\ M.F. =293 473 273

473 293 273

( )

( )

-

- =

2930

473 =6.3

which means that every kg of coal burned would deliver the heat equivalent to over6 kg. Of course, in an actual case, the efficiencies would be less than Carnotefficiencies, but even with a reduction of 50%, the possible savings would be quite

significant.

ep 6.8 It is proposed that solar energy be used to warm a large coll ec-

tor plate. This energy woul d, in tur n, be transferred as heat to a fluid within a heat

engine, and the engine would reject energy as heat to the atmosphere. Experiments

indicate that about 1880 kJ/m 2 h of energy can be coll ected when the plate is operat-

ing at 90°C. Estimate the minimum collector area that would be required for a pl ant

producing 1 kW of useful shaft power. The atmospheri c temperature may be assumed

to be 20°C.

s The maximum efficiency for the heat engine operating between the col-lector plate temperature and the atmospheric temperature (Fig. 6.40) as follows:

hmax =1 2

1

- T

T =1

293

363- =0.192

The efficiency of any actual heat engine operating between these temperatures wouldbe less than this efficiency.

\ Qmin =W

hmax

=1 kJ /s

0 192. = 5.21 kJ/s

= 18,800 kJ/h

\ minimum area required for the collector plate

=18 800

1880

, = 10 m

2

ep 6.9 A reversible heat engine in a satell it e operates between a hot

reservoir at T 1 and a radiati ng panel at T 2 . Radiati on from the panel i s proporti onal

to i ts area and to T 4 2 . For a gi ven work output and value of T 1 show that the area of

the panel wil l be minimum when T

T

2

1

= 0.75.

Determine the minimum area of the panel for an output of 1 kW if the constant ofproportionality is 5.67¥ 10–8 W/m2 K 4 andT 1 is 1000 K.




s For the heat engine (Fig. 6.40), the heat rejectedQ 2 to the panel (atT 2) is equal to the energy emitted fromthe panel to the surroundings by radiation. If A is the area ofthe panel,Q 2 µ AT 42, orQ 2 =KAT 42, whereK is a constant.

Now, h =W

Q1

=T T

T

1 2

1

-

orW

T T 1 2- =

Q

T

Q

T

1

1

2

2

= = KAT

T

2

4

2

=KAT 32

\ A =W

KT T T 23 1 2( )-

=W

K T T T ( )1 23 24-For a givenW andT 1, A will be minimum when

dA

dT 2

= -W

K (3T 1T

22 – 4T 32)◊ (T 1T

32 – T

42)

–2 =0

Since (T 1T 32 – T

42)

–2 π 0, 3T 1T 22 =4T 32

\ T

T

2

1

=0.75 Proved.

∵ Amin =W

K T T T ( . ) ( . )0 75 0 753

13

1 1-

=W

K T

27

256 1

4

=256

27 1

4

W

KT

Here W =1 kW,K =5.67¥ 10–8W/m2 K 4, andT 1 =1000 K

\ Amin =256 1

27 5 67 10 1000

2 4

8 4 4

¥ ¥

¥ ¥ ¥-

kW m K

W K . ( )

=25 6 10

27 5 67 10 10

3

8 12

¥

¥ ¥ ¥-

. m2 = 0.1672 m2

summary

A heat engine cycle is a thermodynamic cycle in which there is a net flow of heat to the system

and a net flow of work from the system. The system which executes a heat engine cycle is a

heat engine. The efficiency of a heat engine, or of its cycle is

h =W

Q

net

1

whereQ 1 is the heat transferred to the system in a cycle, andW net is the net work of the cycle.

This efficiency is called thermal efficiency.

Panel

HE

T 2

Q KAT =2

42

Q1

Q2

W

T 1

f. 6.40




A heat pump or a refrigerator is a system to which there is a net flow of work and from

which there is a net flow of heat in a cycle.

The second law is stated as follows (after Kelvin and Planck): It is impossible for a heat

engine to produce net work in a complete cycle if it exchanges heat only with bodies at a single

fixed temperature.

An equivalent statement (after Clausius) follows: It is impossible for a system working in

a complete cycle to accomplish as its sole effect the transfer of heat from a body at a given

temperature to a body at a higher temperature.

A process is reversible if, after the process has been carried out, it is possible by any means

whatsoever to restore both the system and the entire surroundings to exactly the same states

that were in before the process. A process that is not reversible is irreversible. A reversible

process is a process that can be undone in such a way that no trace remains anywhere of the

fact that the process had occurred. The causes of irreversibility of a process are:

1. Finite potential gradient causing the process like temperature, pressure, concentration,etc.

2. Presence of dissipative effects like friction in which macroscopic work dissipates into

an increase of internal energy and then heat.

A reversible process would require infinite time.

A reversible cycle is a cycle composed entirely of reversible processes. The classical

example is the Carnot cycle which consists of two reversible isothermal processes and two

reversible adiabatic processes.

No heat engine operating between fixed temperature levels can be more efficient than a

reversible engine operating between the same temperatures. The efficiency of all reversible

heat engines operating between the same temperature levels is the same. The efficiency of

a reversible heat engine is independent of the nature or amount of the working substance

undergoing the cycle. The absolute thermodynamic scale is defined by the relation,

T

T

Q

Q

1

2

1

2

=

whereQ 1 is the heat received from a source atT 1 andQ 2 is the heat rejected to a sink atT 2 by

a reversible heat engine. The efficiency of a reversible heat engine receiving heat solely atT 1

and rejecting heat solely atT 2 is given by

hmax =hrev =T T

T

1 2

1

- =1 –

T

T

2

1

The COPs of a Carnot refrigerator and a Carnot heat pump are

(COPmax)Ref =T

T T

2

1 2-

, (COPmax)HP=T

T T

1

1 2-

The absolute thermodynamic temperature scale or the Kelvin scale is equivalent to the ideal

gas temperature scale, and the Kelvin temperature can be measured by a gas thermometer.

review Questions

6.1 What is the qualitative difference between heat and work? Why are heat and work not

completely interchangeable forms of energy?

6.2 What is a cyclic heat engine?

6.3 Explain a heat engine cycle performed by a closed system.




6.4 Explain a heat engine cycle performed by a steady flow system.

6.5 Define the thermal efficiency of a heat engine cycle. Can this be 100%?

6.6 Draw a block diagram showing the four energy interactions of a cyclic heat engine.

6.7 What is a thermal energy reservoir? Explain the terms ‘source’ and ‘sink’.

6.8 What is a mechanical energy reservoir?

6.9 Why can all processes in a TER or an MER be assumed to be quasi-static?

6.10 Give the Kelvin–Planck statement of the second law.

6.11 To produce net work in a thermodynamic cycle, a heat engine has to exchange heat with

two thermal reservoirs. Explain.

6.12 What is a PMM2? Why is it impossible?

6.13 Give the Clausius’ statement of the second law.

6.14 Explain the operation of a cyclic refrigerator plant with a block diagram.

6.15 Define the COP of a refrigerator.

6.16 What is a heat pump? How does it differ from a refrigerator?

6.17 Can you use the same plant as a heat pump in winter and as a refrigerator in summer?Explain.

6.18 Show that the COP of a heat pump is greater than the COP of a refrigerator by unity.

6.19 Why is direct heating thermodynamically wasteful?

6.20 How can a heat pump upgrade low grade waste heat?

6.21 Establish the equivalence of Kelvin–Planck and Clausius statements.

6.22 What is a reversible process? A reversible process should not leave any evidence to

show that the process had ever occurred. Explain.

6.23 How is a reversible process only a limiting process, never to be attained in practice?

6.24 All spontaneous processes are irreversible. Explain.

6.25 What are the causes of irreversibility of a process?

6.26 Show that heat transfer through a finite temperature difference is irreversible.

6.27 Demonstrate, using the second law, that free expansion is irreversible.

6.28 What do you understand by dissipative effects? When is work said to be dissipated?

6.29 Explain perpetual motion of the third kind.

6.30 Demonstrate using the second law how friction makes a process irreversible. 6.31 When a rotating wheel is brought to rest by applying a brake, show that the molecular

internal energy of the system (of the brake and the wheel) increases.

6.32 Show that the dissipation of stirring work to internal energy is irreversible.

6.33 Show by second law that the dissipation of electrical work into internal energy or heat

is irreversible.

6.34 What is a Carnot cycle? What are the four processes which constitute the cycle?

6.35 Explain the Carnot heat engine cycle executed by (a) a stationary system, and (b) a

steady flow system.

6.36 What is a reversed heat engine?

6.37 Show that the efficiency of a reversible engine operating between two given constant

temperatures is the maximum.

6.38 Show that the efficiency of all reversible heat engines operating between the same tem-

perature levels is the same.

6.39 Show that the efficiency of a reversible engine is independent of the nature or amount

of the working substance going through the cycle. 6.40 How does the efficiency of a reversible cycle depend only on the two temperatures at

which heat is transferred?

6.41 What is the absolute thermodynamic temperature scale? Why is it called absolute?

6.42 How is the absolute scale independent of the working substance?




6.43 How doesQ play the role of thermometric property in the Kelvin scale?

6.44 Show that a definite zero point exists on the absolute temperature scale but that this

point cannot be reached without a violation of the second law.

6.45 Give the Fowler–Guggenheim statement of the third law.

6.46 Is the third law an extension of the second law? Is it an independent law of nature?

Explain.

6.47 How does the efficiency of a reversible engine vary as the source and sink temperatures

are varied? When does the efficiency become 100%?

6.48 For a givenT 2, show that the COP of a refrigerator increases asT 1 decreases.

6.49 Explain how the Kelvin temperature can be measured with a gas thermometer.

6.50 Establish the equality of ideal gas temperature and Kelvin temperature.

6.51 What do you understand by internal irreversibility and external irreversibility?

6.52 Explain mechanical, thermal and chemical irreversibilities.

6.53 A Carnot engine with a fuel burning device as source and a heat sink cannot be treated

as a reversible plant. Explain.

Problems with hints

6.1 A reversible power cycle operates between a reservoir at temperatureT and a lower

temperature reservoir at 200 K. At steady state, the cycle develops 40 kW of powerwhile rejecting 1000 kJ /min of energy by heat transfer to the cold reservoir.

Determine T .

H int: W cycle = Q Q

1 2- where W =40 kW andQ 2 =

1000

60 kW

Find Q1 . For a reversible cycle,

Q

Q

1

2

=T

T

1

2

where

T 2 =200 K, getT 1 =T =951.9 K Ans.

6.2 At steady state, a refrigerator whose COP is 2.5 removes energy by heat transfer from afreezer cabinet at 0°C at the rate of 8000 kJ/h and discharges energy by heat transfer to

the surroundings, which are at 20°C. Determine the power input to the refrigerator andcompare with the power required by a reversible refrigerator cycle operating between

reservoirs at these two reservoirs.

H int: Find W =Q2

COP which is 3200 kJ/hAns.

COPrev =T

T T

2

1 2-

would give 13.65.

Find W min =586.1 kJ/h, which is 5 ½ times less.Ans.

6.3 When the outside temperature is –10°C, a residential heat pump must provide 3.5 ¥ 106 kJ per day to a dwelling to maintain its temperature at 20°C. (Fig. 6.41). If electric-

ity cost Rs. 5.0 per kWh, determine the minimum theoretical operating cost for each

day of operation.

H int:

Surroundingsat –10°C

Dwellingat 20°C

1Q2

.Q1

.

Wcyel

f. 6.41




COPmax =T

T T

1

1 2-

=293

293 263. =9.77

\ W min

=Q1

COP =

3 5 10

9 77

6.

.

¥ kJ

h

\ Operating cost per day = W min ¥ 24 h¥

5

3600 rupees.Ans.

6.4 Air of mass 0.5 kg as an ideal gas executes a Carnot cycle having a thermal efficiency of

50%. The heat transfer to the air during the isothermal expansion is 40 kJ. At the begin-

ning of isothermal expansion the pressure is 7 bar and the volume is 0.12 m3. Determine

(a) the maximum and minimum temperatures for the cycle, (b) the volume at the end of

isothermal expansion, (c) the work and heat transfer for each of the four processes.

H int:

p

1

4

v

2

Q 4 – 1

= 0

Q1

Q2

T1

T2

Q 2 – 3

= 0

f. 6.42 Carnot cycle

T 1 = PV

RT

1 1

1

=585.4 K. Ans.

T T

2

1

=1 – h =0.5

\ T 2 =292.7 K. Ans.

Process 1–2

W 1–2 =mRT 1 lnV

V

2

1

\ lnV

V

2

1

=0.4761

FindV 2 =0.193 m3 Ans.

Process 2–3

Q 2–3 =0,W 23 =U 2 –U 3 =mc u (T 2 – T 3)

=107.5 kJ Ans.

Process 3–4

Q

Q

2

1

=T

T

2

1

Q 2 =1

2 Q 1 =20 kJ

=W 3–4 Ans.




Process 4–1

Q 41 =0,W 41 =U 4 –U 1 =107.5 kJ Ans.

hcycle =40 107 5 20 107 5

40

+ - -. . =0.5 (Proved).

6.4 Before introducing the temperature scale, now known as the Kelvin scale, Kelvin

suggested a logarithmic scale in which the functionQ 1/Q 2 takes the form

Q

Q

1

2

=exp

exp

q

q 1

2

where q 1 and q 2 denote the temperatures of the source and sink respectively on this

scale.

(a) Show that the relation between the Kelvin temperatureT and the temperatureq on the

logarithmic scale is

q =ln T +C . whereC is a constant.

(b) On the Kelvin scale, temperatures vary fromO to +µ. Determine the range of tempera-

tures on the logarithmic scale.

(c) Obtain an expression for the thermal efficiency of any system undergoing a reversible

power cycle operating between reservoirs at temperaturesq 1 andq 2 on the logarithmic

scale.

H int: The two scales are

Q

Q

2

1

=T

T

2

1

Kelvin scale

Q

Q

2

1

=exp

exp

q

q 2

1

logarithmic scale.

By comparison,

T

T 2

1

=exp

exp

q

q 2

1 =exp (q 2 – q 1)

lnT

T

2

1

=q 2 – q 1

or, ln T 2 – lnT 1 =q 2 – q 1

\ q =lnT +C Proved.Ans.

(b) Temperature on the logarithmic scale varies from –• to +•.

(c)Q

Q

2

1

=exp

exp

q

q 2

1

\ hcycle =1 –exp

exp

q

q 2

1

Ans.

6.5 A heat engine (Fig. 6.43). Operating between two reservoirs at temperatures 600°C and

40°C drives a refrigerator operating between reservoirs at temperatures of 40°C and

–15°C. The heat transfer to the heat engine is 2500 kJ and the net work output of the

combined engine and refrigerator plant is 400 kJ. The efficiency of the heat engine and

the COP of the refrigerator are each 40% of the maximum possible values. Estimate the

heat transfer to the refrigerant and the net heat transfer to the reservoir at 40°C.




T1

= 600 + 273

= 273 K

T1

= –15+ 273

= 258 K

T1

= 40 + 273

= 313 K

Q1

= 2500kJ

Q Q W3 4 2

= +

HE R

Q A

W1

W2

Q2

400 kJ

f. 6.43

H int:

For the heat engine,hmax =1 –T

T

2

1

=0.642

h =0.4hmax,W 1 =hQ 1 =642 kJ

W 2 =642 – 400 =202 kJ

COP =0.4¥ 258

3 10 2 58- =1.376 =

Q

W

4

2

\ Q 4 =333 kJ

\ Q 3 =Q 4 +W 2, Q 2 =Q 1 –W 1

\ Q 2 +Q 3 =2433 kJ Ans.

Problems

6.1 An inventor claims to have developed an engine that takes in 105 MJ at a temperature

of 400 K, rejects 42 MJ at a temperature of 200 K, and delivers 15 kWh of mechanical

work. Would you advise investing money to put this engine in the market? (Ans. No)

6.2 If a refrigerator is used for heating purposes in winter so that the atmosphere be-

comes the cold body and the room to be heated becomes the hot body, how

much heat would be available for heating for each kW input to the driving

motor? The COP of the refrigerator is 5, and the electromechanical efficiency of the

motor is 90%. How does this compare with resistance heating? (Ans. 5.4 kW)

6.3 Using an engine of 30% thermal efficiency to drive a refrigerator having a COP of 5,

what is the heat input into the engine for each MJ removed from the cold body by the

refrigerator? (Ans. 666.67 kJ)

If this system is used as a heat pump, how many MJ of heat would be available for heat-

ing for each MJ of heat input to the engine? (Ans. 1.8 MJ ) 6.4 An electric storage battery which can exchange heat only with a constant tem-

perature atmosphere goes through a complete cycle of two processes. In pro-

cess 1–2, 2.8 kWh of electrical work flow into the battery while 732 kJ of heat

flow out to the atmosphere. During process 2–1, 2.4 kWh of work flow out




of the battery. (a) Find the heat transfer in process 2–1. (b) If the process 1–2

has occurred as above, does the first law or the second law limit the maximum

possible work of process 2–1? What is the maximum possible work? (c) If the maxi-

mum possible work were obtained in process 2–1, what will be the heat transfer in the

process? (Ans. (a) – 708 kJ (b) Second law,W 2 – 1 =9348 kJ (c)Q 2 – 1 =0)

6.5 A household refrigerator is maintained at a temperature of 2°C. Every time

the door is opened, warm material is placed inside, introducing an average of

420 kJ, but making only a small change in the temperature of the refrigerator. The door

is opened 20 times a day, and the refrigerator operates at 15% of the ideal COP. The

cost of work is Rs. 2.50 per kWh. What is the monthly bill for this refrigerator? The

atmosphere is at 30°C. (Ans. Rs. 118.80)

6.6 A heat pump working on the Carnot cycle takes in heat from a reservoir at 5°C and

delivers heat to a reservoir at 60°C. The heat pump is driven by a reversible heat engine

which takes in heat from a reservoir at 840°C and rejects heat to a reservoir at 60°C.

The reversible heat engine also drives a machine that absorbs 30 kW. If the heat pumpextracts 17 kJ/s from the 5°C reservoir, determine (a) the rate of heat supply from the

840°C source, and (b) the rate of heat rejection to the 60°C sink.

(Ans. (a) 47.61 kW, (b) 34.61 kW)

6.7 A refrigeration plant for a food store operates as a reversed Carnot heat engine cycle.

The store is to be maintained at a temperature of – 5°C and the heat transfer from the

store to the cycle is at the rate of 5 kW. If heat is transferred from the cycle to the atmo-

sphere at a temperature of 25°C, calculate the power required to drive the plant.

(Ans. 0.56 kW)

6.8 A heat engine is used to drive a heat pump. The heat transfers from the heat engine and

from the heat pump are used to heat the water circulating through the radiators of a

building. The efficiency of the heat engine is 27% and the COP of the heat pump is 4.

Evaluate the ratio of the heat transfer to the circulating water to the heat transfer to the

heat engine. (Ans. 1.81)

6.9 If 20 kJ are added to a Carnot cycle at a temperature of 100°C and 14.6 kJ are rejected

at 0°C, determine the location of absolute zero on the Celsius scale. (Ans. – 270.37°C)

6.10 Two reversible heat enginesA andB are arranged in series, A rejecting heat directly to

B . EngineA receives 200 kJ at a temperature of 421°C from a hot source, while engine

B is in communication with a cold sink at a temperature of 4.4°C. If the work output

of A is twice that of B , find (a) the intermediate temperature betweenA andB , (b) the

efficiency of each engine, and (c) the heat rejected to the cold sink.

(Ans. 143.4°C, 40% and 33.5%, 80 kJ)

6.11 A heat engine operates between the maximum and minimum temperatures of 671°C

and 60°C respectively, with an efficiency of 50% of the appropriate Carnot efficiency.

It drives a heat pump which uses river water at 4.4°C to heat a block of flats in which

the temperature is to be maintained at 21.1°C. Assuming that a temperature difference

of 11.1°C exists between the working fluid and the river water, on the one hand, and

the required room temperature on the other, and assuming the heat pump to operate

on the reversed Carnot cycle, but with a COP of 50% of the ideal COP, find the heat

input to the engine per unit heat output from the heat pump. Why is direct heatingthermodynamically more wasteful? (Ans. 0.79 kJ/kJ heat input)

6.12 An ice-making plant produces ice at atmospheric pressure and at 0°C from

water. The mean temperature of the cooling water circulating through the condenser




of the refrigerating machine is 18°C. Evaluate the minimum electrical work in kWh

required to produce 1 tonne of ice. (The enthalpy of fusion of ice at at mospheric pres-

sure is 333.5 kJ/kg). (Ans. 6.11 kWh)

6.13 A reversible engine works between three thermal reservoirs, A, B and C . The

engine absorbs an equal amount of heat from the thermal reservoirs A and B

kept at temperatures T A and T B respectively, and rejects heat to the thermal

reservoir C kept at temperature T C . The efficiency of the engine is a times the

efficiency of the reversible engine, which works between the two reservoirsA andC .

Prove that

T

T

A

B

= (2a – 1) + 2 (1 – a )T

T

A

C

6.14 A reversible engine operates between temperatures T 1 and T (T 1 > T ). The

energy rejected from this engine is received by a second reversible engine

at the same temperature T . The second engine rejects energy at temperatureT 2 (T 2 <T 1). Show that (a) temperature T is the arithmetic mean of temperatures T 1

and T 2 if the engines produce the same amount of work output, and (b) temperature

T is the geometric mean of temperaturesT 1 andT 2 if the engines have the same cycle

efficiencies.

6.15 Two Carnot enginesA andB are connected in series between two thermal reservoirs

maintained at 1000 K and 100 K respectively. EngineA receives 1680 kJ of heat from

the high-temperature reservoir and rejects heat to the Carnot engineB . EngineB takes

in heat rejected by engineA and rejects heat to the low-temperature reservoir. If engines

A andB have equal thermal efficiencies, determine (a) the heat rejected by engineB , (b)

the temperature at which heat is rejected by engine, A, and (c) the work done during the

process by engines,A andB respectively. If enginesA andB deliver equal work, deter-

mine (d) the amount of heat taken in by engineB , and (e) the efficiencies of enginesA

andB .

(Ans. (a) 168 kJ, b) 316.2 K, (c) 1148.7, 363.3 kJ , (d) 924 kJ, (e) 45%, 81.8%)

6.16 A heat pump is to be used to heat a house in winter and then reversed to cool the housein summer. The interior temperature is to be maintained at 20°C. Heat transfer through

the walls and roof is estimated to be 0.525 kJ/s per degree temperature difference be-

tween the inside and outside. (a) If the outside temperature in winter is 5°C, what is the

minimum power required to drive the heat pump? (b) If the power output is the same as

in part (a), what is the maximum outer temperature for which the inside can be main-

tained at 20°C? (Ans. (a) 403 W, (b) 35.4°C)

6.17 Consider an engine in outer space which operates on the Carnot cycle. The

only way in which heat can be transferred from the engine is by radiation. The

rate at which heat is radiated is proportional to the fourth power of the absolute

temperature and to the area of the radiating surface. Show that for a given power output

and a givenT 1, the area of the radiator will be a minimum when

T

T

2

1

=3

4

6.18 It takes 10 kW to keep the interior of a certain house at 20°C when the outside tempera-ture is 0°C. This heat flow is usually obtained directly by burning gas or oil. Calculate

the power required if the 10 kW heat flow were supplied by operating a reversible




engine with the house as the upper reservoir and the outside surroundings as the lower

reservoir, so that the power were used only to perform work needed to operate the

engine. (Ans. 0.683 kW)

6.19 Prove that the COP of a reversible refrigerator operating between two given tempera-

tures is the maximum.

6.20 A house is to be maintained at a temperature of 20°C by means of a heat pump pump-

ing heat from the atmosphere. Heat losses through the walls of the house are estimated

at 0.65 kW per unit of temperature difference between the inside of the house and the

atmosphere. (a) If the atmospheric temperature is – 10°C, what is the minimum power

required to drive the pump? (b) It is proposed to use the same heat pump to cool the

house in summer. For the same room temperature, the same heat loss rate, and the same

power input to the pump, what is the maximum permissible atmospheric temperature?

(Ans. 2 kW, 50°C)

6.21 A solar-powered heat pump receives heat from a solar collector atT h , rejects heat to the

atmosphere atT a , and pumps heat from a cold space atT c . The three heat transfer ratesareQ h ,Q a , andQ c respectively. Derive an expression for the minimum ratioQ h /Q c , in

terms of the three temperatures.

If T h =400 K, T a =300 K, T c =200 K,Q c =12 kW, what is the minimumQ h ? If the

collector captures 0.2 kW/m 2, what is the minimum collector area required?

(Ans. 26.25 kW, 131.25 m2)

6.22 A heat engine operating between two reservoirs at 1000 K and 300 K is used to drive a

heat pump which extracts heat from the reservoir at 300 K at a rate twice that at which

the engine rejects heat to it. If the efficiency of the engine is 40% of the maximum

possible and the COP of the heat pump is 50% of the maximum possible, what is the

temperature of the reservoir to which the heat pump rejects heat? What is the rate of

heat rejection from the heat pump if the rate of heat supply to the engine is 50 kW?

(Ans. 326.5 K, 86 kW)

6.23 A reversible power cycle is used to drive a reversible heat pump cycle. The power cycle

takes inQ 1 heat units atT 1 and rejectsQ 2 at T 2. The heat pump abstractsQ 4 from the

sink at T 4 and dischargesQ 3 at T 3. Develop an expression for the ratioQ 4/Q 1 in termsof the four temperatures.

(Ans.Q

Q

4

1

=T T T

T T T

4 1 2

1 3 4

( )

( )

-

- ) 6.24 Prove that the following propositions are logically equivalent: (a) A PMM2 is impos-

sible, (b) A weight sliding at constant velocity down a frictional inclined plane executes

an irreversible process.

6.25 A heat engine receives half of its heat supply at 1000 K and half at 500 K while rejecting

heat to a sink at 300 K. What is the maximum thermal efficiency of the heat engine?

(Ans. 55%)

6.26 A heat pump provides 3¥ 104 kJ /h to maintain a dwelling at 23°C on a day when the

outside temperature is 0°C. The power input to the heat pump is 4 kW. Determine the

COP of the heat pump and compare it with the COP of a reversible heat pump operating

between the reservoirs at the same two temperatures.

(Ans. 2.08, 12.87)

6.27 When the outside temperature is – 10°C, a residential heat pump must provide 3.5¥

106 kJ per day to a dwelling to maintain its temperature at 20°C. If the electricity costs

Rs 2.10 per kWh, find the minimum operating cost for each day of operation.

(Ans. Rs 208.83)




6.28 A reversible power cycle receives energy Q H from a reservoir at temperature T H and

rejectsQ C to a reservoir at temperatureT C . The work developed by the power cycle is

used to drive a reversible heat pump that removesQ ¢ C from a reservoir atT ¢C and rejects

energy Q ¢H to a reservoir at temperature T ¢H . (a) Develop an expression for the ratio

Q ¢H /Q H in terms of the temperatures of the four reservoirs. (b) What must be the

relationship of the temperaturesT H , T C , T ¢C andT ¢H for¢Q

Q

H

H

to exceed a value of unity?

(Ans. (a)¢

= ¢ -

¢ - ¢

Q

Q

T T T

T T T

H

H

H H C

H H C

( )

( ), (b)

T

T

T

T

C

C

H

H ¢<

¢ )

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Chatper 06 Second Law of Thermodynamics