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Chapter 13

INTRODUCTION TO

THE SECOND LAW

There is good reason for wanting to violate the Second Law, for

if it could be done, all of man’s energy requirements could forever

be met without any depletion of resources or pollution of his sur-

roundings.

−H.C. Van Ness (Understanding Thermodynamics)

Learning thermodynamics even at an introductory level cannot be con-

sidered complete without being told what the second law is all about. This

chapter introduces the second law to a beginner in an unconventional way.

The objective of this chapter is to make the students know why it is impor-

tant to learn the second law, and what role it has in the thermodynamic

analyses of engineering systems.

306 Chapter 13

13.1 The Second Law

In this book, we adopt the most general approach to present the Second

Law of Thermodynamics, which is that of Clausius. He made the following

statement of the second law in 1865:

Total entropy change of any system and its sur-

roundings is positive for a real process, and it ap-

proaches zero when the real process approaches a

reversible process.

The mathematical equivalence of the above statement of the second

law is as follows:

∆Stotal

>0for a real (irreversible) process

=0for a reversible process

<0for an impossible process

(13.1)

The second law is a fundamental law, which means nobody has either

proved or disproved the second law yet. It contains in it information that

has been gathered over several hundred years of observations and experi-

mentations with what is possible and what is not possible in reality.

Student: Teacher, the ﬁrst law states energy is conserved. It is easy to under-

stand what the ﬁrst law is all about, and therefore to accept it as true.

But, the second law does not make any sense.

Teacher: Well, let me put it this way. The second law of thermodynamics

states that the total entropy of a system and its surroundings together

is conserved for a reversible process, and it increases for a real, I mean

physically realizable, process.

Student: Why should the total entropy of a system and its surroundings to-

gether increases for a real process? What is that physical entity that we

measure using this entropy?

Teacher: It is a bit diﬃcult to answer that question since it is impossible to

relate entropy to a single physical entity. Let me only tell you that entropy

is a property that provides a measure of the molecular disorder.

Introduction to the Second Law 307

Student: What is the connection between molecular disorder and the second

law?

Teacher: The Nature, it appears, prefers the direction of increasing molecular

disorder. Entropy could be used to measure the degree of molecular dis-

order. Thus, we could say that for any thermodynamic process to occur

in real life, the total entropy change of the system and its surroundings,

which is the measure of the total molecular disorder of the system and its

surroundings, must increase.

Student: Oh... I see.

13.2 Evaluation of Total Entropy Change

Since the value of ∆Scan be used to determine whether a process is

real, reversible or impossible, we need to learn about how to determine ∆S.

We shall do that in this section.

For a closed system,

∆Stotal =∆Ssys +∆Ssurr (13.2)

where the label ’sys’ stands for system and the label ’surr’ stands for

surroundings.

Since Sis a property, ∆Ssys can be evaluated using the following:

∆Ssys =msys ∆ssys =msys (sf−so)sys (13.3)

where the subscript fstands for the ﬁnal state and the subscript ostands

for the initial state.

If the surroundings has a ﬁnite mass then the entropy diﬀerence between

the ﬁnal and the initial states of this ﬁnite mass would give the entropy

change of the surroundings. We could therefore use

∆Ssurr =msurr ∆ssurr =msurr (sf−so)surr (13.4)

Expressions developed in Chapter 11 for ∆sof various substances could

be used to evaluate ∆ssys and ∆ssurr of (13.3) and (13.4), as will be

demonstrated in the worked examples of this chapter.

308 Chapter 13

The surroundings is, however, often taken as a thermal reservoir of

inﬁnite heat capacity. A thermal reservoir is assumed to remain at a con-

stant temperature regardless of the heat transferred to or from the reser-

voir. It is also assumed that no irreversibilities occur within the reservoir.

With these assumptions, the entropy change of the surroundings taken as

a thermal reservoir is written as

∆Ssurr =(Qin)surr

Tsurr

(13.5)

where (Qin)surr denotes the ﬁnite amount of heat transferred to the sur-

roundings from the system during the process, and Tsurr denotes the con-

stant temperature of the surroundings in the Kelvin scale.

If the system is thermally isolated from the surroundings during the

process, that is, in an adiabatic process, there is no heat exchange between

the system and the surroundings. Therefore,

∆Ssurr =0

which simpliﬁes (13.2) to the following:

∆Stotal =∆Ssys for an adiabatic process. (13.6)

Using (13.3), we could write (13.6) as

∆Stotal =msys (sf−so)sys for an adiabatic process. (13.7)

If the given adiabatic process takes place reversibly then, of course, we

know that the entropy of the system remains a constant (see Section 11.4).

Therefore, (13.7) reduces to

∆Stotal =0 for a reversible adiabatic process. (13.8)

To evaluate the entropy change for an open system, we need to incorpo-

rate the entropy of the matter that enters and leaves the system. Since we

will mostly be working with steady ﬂow processes, let us get the expression

for the steady ﬂow processes, which is the following:

d(∆Stotal)

dt =˙mese−˙misi+d(∆Ssurr)

dt (13.9)

where the subscript estands for the exit and the subscript istands for the

inlet.

Introduction to the Second Law 309

If the surroundings is taken as a thermal reservoir at a constant tem-

perature of Tsurr K, we get

d(∆Ssurr)

dt =(˙

Qin)surr

Tsurr

(13.10)

where (˙

Qin)surr denotes the rate at which heat is transferred from the

system to the surroundings during the process.

For an adiabatic steady ﬂow process, the system does not thermally

interact with the surroundings. Therefore,

d(∆Stotal)

dt =˙mese−˙misi(13.11)

13.3 Worked Examples

Example 13.1

A thermally insulated rigid box of negligible

heat capacity is divided into equal halves by a partition of negligible mass.

Initially, one compartment contains air at 2 bar and 400 K, and the other is

evacuated. When the dividing partition is raptured, air will rush to ﬁll the

entire box. Show that this process is a irreversible process. It is a common

assumption that air at low or moderate pressures behaves like an ideal gas.

Solution to Example 13.1

Let us evaluate the total entropy change of the system and its surroundings

to determine if the given process is reversible or not. The system is isolated from

the surroundings, and therefore

∆Stotal =∆Ssys =mair ∆sair

Since air is assumed to behave as an ideal gas and since air volume has doubled

during the process, we could use (11.23) or (11.25) to determine ∆sair as follows:

∆sair =Cvln (Tf/To)+Rln (2) = Cvln (Pf/Po)+Cpln (2)

310 Chapter 13

Let us now determine the temperature or pressure ratio between the ﬁnal

and the initial equilibrium states of the process. Since the entire system is

isolated from its surroundings, the internal energy of the system remains constant

throughout the process. Since air is assumed to behave as an ideal gas, and

since the internal energy of an ideal gas is a function of temperature alone, the

temperature at the ﬁnal equilibrium state is also 400 K. Thus, ln (Tf/To)=0.

We therefore get

∆Stotal =mair ∆sair =mair Rln (2) >0

Thus, according to the second law given by (13.1), the given process is a

irreversible process. We, of course, know that the expansion of air into vacuum

is unrestrained, and therefore the expansionprocessisfarfromreversible. The

second law has just proven it.

Example 13.2

Hot gases leaving the turbine of a turbojet

engine is reported to enter the nozzle of the turbojet engine at 6 bar and

975 K and exit the nozzle at 0.9 bar and 925 m/s. Determine if such a ﬂow

is possible in reality. Ignore the speed of the gases at the nozzle entrance.

Assume that the gases behave as ideal gas and that Cp= 1.005 kJ/kg ·K

and γ= 1.4. Also, assume steady adiabatic ﬂow through the nozzle.

Solution to Example 13.2

Using the steady ﬂow energy equation applied to the steady adiabatic ﬂow

of an ideal gas through the nozzle, we get Te, the temperature at the nozzle

exit, as

Te=Ti−c2

e−c2

i

2Cp

= 975 K−9252

2×1005 K= 549.3K

Since the ﬂow through the adiabatic nozzle exchanges no heat with its sur-

roundings, using (13.11) and (11.24), we get

d(∆Stotal )

dt =˙m(se−si)= ˙mCpln Te

Ti−Rln Pe

Pi

=˙mR 1.4

1.4−1ln 549.3

975 −ln 0.9

6

=−0.11 ˙mR< 0

Introduction to the Second Law 311

Thus, according to the second law given by (13.1), the given process is an

impossible process.

Example 13.3

Determine the maximum possible speed re-

alizable at the nozzle exit of Example 13.2 for the given inlet condition

and exit pressure.

Solution to Example 13.3

At the reversible limit, d(∆Stotal)/dt)= 0 for the ﬂow through the adiabatic

nozzle. Therefore, we get the temperature at the nozzle exit, Te,as

˙mR 1.4

1.4−1ln Te

975 −ln 0.9

6=0

which gives Te= 567 K.

Using this exit temperature in the steady ﬂow energy equation applicable for

the ﬂow through adiabatic nozzle, we get the speed at the nozzle exit, ce,as

ce=2×Cp(Ti−Te)=2×1005 ×(975 −567) = 905.5m/s

which is the maximum possible speed realizable at the nozzle exit for the given

conditions.

Example 13.4

A steady ﬂow of steam enters an adiabatic

turbine at 60 bar and 400◦C with negligible velocity. It is reported to leave

the nozzle at 3 bar as saturated steam. Is that possible?

Solution to Example 13.4

The speciﬁc entropy of steam at the entrance of the turbine can be found

from a Steam Table as se=sat 60 bar and 400◦C = 6.541 kJ/kg ·K. The

312 Chapter 13

speciﬁc entropy of steam at the exit of the turbine is found from a Steam Table

as si=sof saturated steam at 3 bar = 6.993 kJ/kg ·K.

Therefore, for steady ﬂow through an adiabatic turbine,

d(∆Stotal )

dt =˙m(se−si)= ˙m(6.541 −6.993) <0

Thus, according to the second law given by (13.1), the given process is not

possible in reality.

Example 13.5

Calculate the entropy change for the process

where 3 kg of liquid water at 30◦C is cooled until it freezes at 0◦C. Take

Cwater as 4.2 kJ/kg ·K, and the latent heat of freezing at 0◦Cas−333

kJ/kg.

Solution to Example 13.5

Considering water as an incompressible substance, we could use (11.22) to

determine the entropy change for the process, in which the temperature of water

is reduced from 30◦Cto0

◦C, as follows:

∆s=(4.2kJ/kg ·K)×ln (273/303) = −0.44 kJ/kg ·K

Conversion of water to ice occurs at a constant temperature of 0◦C, during

which vchanges but Premains constant. Thus, let us start with Tds=dh −

vdP, which reduces to Tds=dh for a constant-pressure process such as the

one considered here. Therefore, we get

∆s=dh

T=∆h

T

for the given constant-pressure and constant-temperature phase change process.

Since ∆hfor the process of water turning into ice at 0◦C is the latent heat

of freezing, we get

∆s=−333 kJ/kg

273 K=−1.22 kJ/kg ·K

The total change in the entropy for the given process is therefore

∆Ssys =3kg ×(−0.44 −1.22) kJ/kg ·K=−4.98 kJ/K (13.12)

Introduction to the Second Law 313

Student: Teacher, the total entropy change of the system is negative in the

above example. According the second law, it is then an impossible process.

How could that be? We know that water can be frozen to get ice. What

is wrong?

Teacher: Nothing is wrong, dear Student. Observe that it is the total en-

tropy change of the system that takes a negative value, not the total

entropy change of the system and the surroundings. The given system

has interacted with its surroundings during cooling, by losing heat to the

surroundings. Therefore, we need to evaluate the entropy change of the

surroundings, and add that value to the entropy change of the system to

get the total entropy change of the system and its surroundings. It is this

total that should take a positive value for the process to be a real process.

Example 13.6

Calculate the total entropy change of the sys-

tem and its surroundings for the process given in Example 13.5.Take

the surroundings to be a thermal reservoir at −20◦C.

Solution to Example 13.6

The entropy change of the system is calculated in the Solution to Exam-

ple 13.5, and is given by (13.12). Now, let us evaluate the entropy change of

the surroundings. Since the surroundings is assumed to be a thermal reservoir

at −20◦C, we could use (13.5) to get ∆Ssurr as follows:

∆Ssurr =(Qin)surr

253 K(13.13)

where (Qin)surr is the heat lost by the system to the surroundings.

The heat lost by water when it is cooled from 30◦Cto0

◦C is evaluated using

(3 kg) ×(4.2 kJ/kg ·K) ×(303 −273) K, which becomes 378 kJ. Heat lost

during the conversion of water at 0◦Ctoiceat0

◦C is evaluated using (3 kg) ×

333 kJ/kg, which becomes 999 kJ.

314 Chapter 13

The total heat lost to the surroundings is therefore 1377 kJ. Substituting it

in (13.13), we get

∆Ssurr =1377 kJ

253 K=5.44 kJ/K (13.14)

Combining (13.12) and (13.14), we get

∆Stotal =0.46 kJ/K >0

Therefore, according to the second law, the process considered is a real

(irreversible) process.

Student: Teacher, you have assumed that the surroundings was at −20◦Cin

working out the above example. What if I take the surroundings to be at

the room temperature, say, 30◦C. Will I still get ∆Stotal >0?

Teacher: You can’t do that. How could the system, which is cooled from 30◦C

to 0◦C, lose heat to a surroundings that is at a higher temperature than

the system?

Student: Yes, that is true.... But, Teacher, it happens with the refrigerator.

Doesn’t it?

Teacher: Yes, it does. Heat is transferred from the inside of the refrigerator,

which is at a much lower temperature than the atmospheric temperature,

to the surroundings, which is at atmospheric temperature. That is correct.

Student: How does that happen, Teacher?

Teacher: A refrigerator cannot lose heat to its surroundings unless we provide

work in the form of electricity to operate its compressor. Do you agree

with that?

Student: Yes, I do. I know about the compressor in the refrigerator. If the

compressor fails then refrigerator does not function. Well, how does a

refrigerator work?

Teacher: Each refrigerator has an engine, which consists of a ﬂuid known as

refrigerant. Are you familiar with that?

Student: I know about the refrigerant used in a refrigerator. I know that

CFCs, the chemicals that make a hole in the ozone layer, are used as

refrigerants. But, nowadays, it is being replaced by other chemicals that

does not damage the ozone layer.

Introduction to the Second Law 315

Teacher: It’s very good that you know so much about refrigerants. Now, let

me tell you about the cyclic process executed by the refrigerant in the

refrigerator. The schematic diagram of a typical refrigeration cycle is

given in Figure 13.1.

evaporator

condenser

reciprocating

compressor

throttling

valve

?

heat removed from the

refrigerated space at −10◦C

?

Heat rejected to the

surroundings at 30◦C

work input

as electricity

superheated vapour

at 1.2 bar &−20◦C

superheated vapour

at 10 bar &65oC

vapour-liquid

mixture at

1.2 bar &−25.8◦C

subcooled

liquid at

10 bar &40◦C

Figure 13.1 Schematic of a typical refrigeration cycle.

Student: Teacher, the ﬁgure looks complicated except for the compressor,

throttling valve and the condenser. We have already learned about them

in Chapter 10. Haven’t we?

Teacher: Yes, you have. Let me explain the workings of the refrigeration

cycle shown in the ﬁgure. First of all, let’s take the freezer compartment

of a refrigerator, which must be maintained, say, at −10◦C. Heat must

be continuously removed from the freezer space to maintain such a low

temperature when the surrounding atmosphere is around 30◦C. Do you

agree with that?

Student: Yes, I do.

Teacher: Let us send a liquid, the refrigerant, at a temperature much lower

than −10◦C through the evaporator coil that is attached to the inner walls

of the freezer compartment. Tell me what happens to the refrigerant then.

316 Chapter 13

Student: Of course, heat will be transferred from the freezer space at −10◦C

to the refrigerant at a temperature much lower than −10◦C.

Teacher: That’s correct. The refrigerant entering the evaporator is usually

maintained as a mixture of saturated liquid and vapour at a temperature

at least 10 degrees below the freezer space temperature, and at a pressure

slightly above the atmospheric pressure. The heat transferred from the

freezer space to the refrigerant heats the refrigerant to a slightly super-

heated vapour state. This vapour is compressed to a pressure about 8 to

10 times the atmospheric pressure in a reciprocating type of compressor

powered by electricity.

Student: Why are we compressing the refrigerant, Teacher?

Teacher: When the refrigerant vapour is compressed, its temperature would

increase. We need to increase the temperature of the refrigerant well

above the temperature of the surroundings, so that the refrigerant could

lose all the heat that it has gained in the evaporator to the surroundings,

as it passes through the condenser coil that is exposed to the atmosphere.

Student: I see, that is why the air around the refrigerator is warm. It explains

why my cat loves to lie down on the refrigerator during the cold days.

Teacher: Yes, that is right.

Student: Okay, Teacher, I understand how the refrigeration cycle works. But,

tell me why there is a throttling valve in the cycle.

Teacher: The temperature of the refrigerant leaving the condenser could not

be reduced below the atmospheric temperature. Do you agree with that?

Student: Yes, I do. We could not lower the temperature of the refrigerant

below the temperature of the atmosphere to which it is losing heat.

Teacher: That’s correct. The refrigerant leaving the condenser at a temper-

ature above the atmospheric temperature is throttled using a throttling

valve to reduce the temperature of the refrigerant passing through it, to

a value that is much below the atmospheric temperature.

Student: Well, I see why a throttling valve is needed.

Teacher: The refrigeration cycle is so designed that the temperature and the

pressure reductions across the throttling valve, which is simply a capillary

tube, will bring the refrigerant to the temperature and the pressure at

which it enters the evaporator. In this way, the refrigeration cycle is

completed and the same refrigerant is used over and over again.

Introduction to the Second Law 317

Student: Oh, I see. That’s neat. But... I have a question. It appears that the

refrigerant is used and reused. Then, how it enters the atmosphere and

destroys the ozone layer.

Teacher: Oh, well, the discarded refrigerators are the main culprits who let

the refrigerant into the atmosphere. Now, I want you to take note of the

fact that heat cannot be transferred from a lower temperature body to a

higher temperature body unless work is provided to the engine, as with

the refrigerators.

Student: Yes, I have noted that.

Teacher: In the following examples, we will prove that fact using the second

law given by (13.1).

Example 13.7

Is it possible for an engine, whose working

ﬂuid operating in a cyclic process as in the refrigeration cycle discussed

above, to transfer heat from a cooler reservoir to a hotter reservoir without

producing any other eﬀects on the surroundings?

Solution to Example 13.7

Let us consider the schematic of an engine shown in Figure 13.2, whose

working ﬂuid describe is said to describe a cyclic process. It receives Qin amount

of heat from the cold reservoir at TLK and rejects Qout amount of heat to the

hot reservoir at THK, where TH>T

L.

Since the system describes a cyclic process, the internal energy of the system

remains a constant. Therefore, according to the ﬁrst law, all heat removed from

the cold reservoir could be transferred to the hot reservoir. That is,

Qin =Qout (13.15)

Let us now calculate the total entropy change of the engine of Figure 13.2.

The system undergoes a cyclic process and entropy is a property, and therefore

the entropy of the system would not change. That is,

∆Ssys =0 (13.16)

318 Chapter 13

hot reservoir at THK

cold reservoir at TLK

Qout

Qin

Figure 13.2 Heat transfer from a cooler reservoir to a hotter reservoir.

The surroundings consists of the hot reservoir at THK and the cold reservoir

at TLK. The entropy change of the surroundings can therefore be evaluated

using (13.5) as follows:

∆Ssurr =Qout

TH

−Qin

TL

(13.17)

where the hot reservoir gains Qout amount of heat from the system and the cold

reservoir loses Qin amount of heat to the system.

Combining (13.16) and (13.17), we get the total entropy change of the

engine of Figure 13.2 as

∆Stotal =Qout

TH

−Qin

TL

(13.18)

Using (13.15), the above could be rewritten as

∆Stotal =TL−TH

THTLQin <0

since TL, the temperature of the cold reservoir, is less than TH, the temperature

of the hot reservoir.

According to the second law, given by (13.1), a process for which ∆Stotal <

0 is an impossible process. Thus, no engine, whose working ﬂuid operating in a

cyclic process, could transfer heat from a cooler reservoir to a hotter reservoir,

without producing any other eﬀects.

Introduction to the Second Law 319

Example 13.8

Prove using the second law that it is possible

for an engine, whose working ﬂuid operating in a cyclic process, to remove

heat from a cooler reservoir and rejects heat to a hotter reservoir, if work

is supplied to the engine?

Solution to Example 13.8

The engine given in this example is what is known as the refrigerator or,

as the heat pump in general, and the schematic of which is shown in Figure

13.3.

hot reservoir at THK

cold reservoir at TLK

Qout

Qin

Figure 13.3 Schematic of a heat pump.

Win

This example is similar to Example 13.7,exceptforthefactthatthe

system is provided with work. Therefore, according to the ﬁrst law,

Qin +Win =Qout (13.19)

The total entropy change of the heat pump of Figure 13.3 would be the

same as that is given by (13.18). Eliminating Qout from (13.18) using (13.19),

we get

∆Stotal =Qin +Win

TH

−Qin

TL

=Win

TH

−TH−TL

THTLQin (13.20)

320 Chapter 13

For ∆Stotal to be a positive quantity,

Win >TH−TL

TLQin (13.21)

Thus, it is possible to construct an engine, whose working ﬂuid operating

in a cyclic process, capable of removing heat from a cooler reservoir and rejects

heat to a hotter reservoir, if the work done on the working ﬂuid of the engine

satisﬁes (13.21).

Comment:FromExample 13.7 and Example 13.8, we shall conclude

that it is impossible to construct an engine, whose working ﬂuid operating in a

cyclic process, capable of transferring heat from a cooler body to a hotter body,

without producing no other eﬀect. This is the famous Clausius Statement

of the Second Law.

The ratio of heat removed from the cooler reservoir by the heat pump

to the work supplied to the heat pump is known as the coeﬃcient of

performance, and is denoted by COP. From (13.21), we can determine

the upper limit of the COP as follows:

COP =Qin

Win

<TL

TH−TL

(13.22)

Any heat pump that works at the upper limit of the COP will have

∆Stotal = 0, which means that such a heat pump operates as a reversible

heat pump. A reversible heat pump is known as the Carnot heat pump,

and its COP is known at the Carnot COP,givenby

COPCarnot =TL

TH−TL

(13.23)

Example 13.9

Determine the minimum amount of work re-

quired to operate an air-conditioner which will maintain an indoor tem-

perature of 25◦C. The atmospheric temperature is at 36◦C, and the heat

Introduction to the Second Law 321

generated indoor from the people and other heat generating devices are esti-

mated to be 26 MJ/h. Determine also the heat rejected to the atmosphere

by the air-conditioner.

Solution to Example 13.9

The air-conditioner functions as a heat pump removing 26 MJ/h of heat from

the indoor space maintained at 25◦C (which is the cooler reservoir) and rejecting

heat to the atmosphere at 36◦C (which is the hotter reservoir). Combining

(13.22) and (13.23), we can write the coeﬃcient of performance of an air-

conditioner, as

COP =Qin

Win

≤COPCarnot

which gives

Win =Qin

COP ≥Qin

COPCarnot

The minimum work required by the air-conditioner is therefore given by

(Win)min =Qin

COPCarnot

We can calculate the COP using (13.23) as

COPCarnot =TL

TH−TL

=273 + 25

(273 + 36) −(273 + 25) =27.1

It is given that Qin = 26 MJ/h, and therefore

(Win)min =26 MJ/h

27.1=0.27 kW

The minimum amount of heat rejected to the atmosphere by the air-conditioner

is calculated as follows:

(Qout)min =Qin +(Win)min

=26MJ/h +26

27.1MJ/h ≈27 MJ/h

Comment: Note that making the living space cool and comfortable by air-

conditioning results in additional amount of heat being rejected into the envi-

ronment. At least 1 MJ/h of waste heat is added in the case considered in this

example.

322 Chapter 13

Example 13.10

Is it possible to convert all heat provided to

an engine, whose working ﬂuid operating in a cycling process, into useful

work?

Solution to Example 13.10

Let us consider the engine shown in Figure 13.4. The working ﬂuid of the

engine, taken as the system, is said to operate a cyclic process. It receives Qin

amount of heat from the surroundings and does Wout amount of work on the

surroundings.

hot reservoir at THK

Qin

Figure 13.4 Schematic of an engine converting all heat into work.

Wout

Since the system describes a cyclic process, the internal energy of the system

remains a constant. Therefore, accordingtotheﬁrstlaw,allheatprovidedto

the system is converted into work. That is, Wout =Qin.

Well, what does the second law says about such a 100% conversion of heat

into work? To answer this question, let us calculate the total entropy change of

the engine of Figure 13.4. Following the methods used in the previous examples,

we could write that ∆Ssys =0,and∆Ssurr =−(Qin/TH).

The total entropy change of the engine of Figure 13.4 is therefore

∆Stotal =−Qin

TH

<0(13.24)

since Qin and THare positive quantities.

According to the second law, given by (13.1), a process for which ∆Stotal

<0 is an impossible process. Therefore, it is not possible to convert all heat

provided to an engine, whose working ﬂuid operating in a cycling process, into

work.

Introduction to the Second Law 323

Example 13.11

Prove using the second law that it is pos-

sible for an engine, whose working ﬂuid operating in a cyclic process, to

convert part of the heat that it receives from a hotter reservoir into work

done on the surroundings, provided the remaining heat is rejected to a

cooler reservoir.

Solution to Example 13.11

The engine given in this example is what is known as the heat engine,and

the schematic of which is shown in Figure 13.5.

hot reservoir at THK

cold reservoir at TLK

Qin

Qout

Figure 13.5 Schematic of a heat engine.

Wout

This example is similar to Example 13.10,exceptforthefactthatthe

system rejects heat into a cooler reservoir. Therefore, according to the ﬁrst law,

Qin =Wout +Qout (13.25)

Let us calculate the total entropy change of the engine of Figure 13.5 fol-

lowing the methods used in the previous examples as follows: ∆Ssys =0,and

∆Ssurr =−(Qin/TH)+(Qout/TL).

The total entropy change of the engine of Figure 13.5 is therefore

∆Stotal =−Qin

TH

+Qout

TL

(13.26)

324 Chapter 13

Eliminating Qout from (13.26) using (13.25), we get

∆Stotal =−Qin

TH

+Qin −Wout

TL

=TH−TL

THTLQin −Wout

TL

(13.27)

For ∆Stotal to be a positive quantity,

Wout <TH−TL

THQin (13.28)

Thus, it is possible to construct an engine, whose working ﬂuid operating in

a cyclic process, to convert part of the heat it receives from a hotter reservoir

into work done on the surroundings, and to reject the remaining heat to a cooler

reservoir, provided (13.28) is satisﬁed.

Comment:FromExample 13.10 and Example 13.11, we shall conclude

that it is impossible to construct an engine, whose working ﬂuid operating in a

cyclic process, capable of converting all heat it receives into useful work, without

producing no other eﬀect in its surroundings. This is the famous Kelvin-Plank

Statement of the Second Law.

The ratio of work obtained from the heat engine to the heat provided to

theheatengineisknownasthethermal eﬃciency, and is denoted by η.

From (13.28), we can determine the upper limit of the thermal eﬃciency

as follows:

η=Wout

Qin

<1−TL

TH

(13.29)

Any heat engine that works at the upper limit of the thermal eﬃciency

will have ∆Stotal = 0, which means that such a heat engine operates as a

reversible heat engine. A reversible heat engine is known as the Carnot

heat engine, and its thermal eﬃciency is known at the Carnot eﬃ-

ciency, denoted by

ηCarnot =1−TL

TH

(13.30)

No heat engine can have a thermal eﬃciency higher than the Carnot

eﬃciency, which is a function of the maximum and minimum temperatures

across which the heat engine operates.

Introduction to the Second Law 325

Increasing the temperature of the hotter reservoir and/or decreasing the

temperature of the cooler reservoir are the only means by which the Carnot

eﬃciency of a heat engine could be increased.

Example 13.12

Consider the steam turbine of Example

12.4, whose working ﬂuid (water/steam) operates in a cyclic process. Take

the hot gases providing heat to the steam generator as the hotter reservoir,

and assume that it remains at a constant temperature of 500◦C throughout

the operation. Take the cooling water removing heat from the condenser

as the cooler reservoir at a constant temperature of 27◦C. If the heat input

to the steam engine from the hotter reservoir is 155 MJ/s, determine the

maximum work output possible from the heat engine, and the amount of

heat rejected to the cooler reservoir.

Solution to Example 13.12

Combining (13.29) and (13.30), we can write that the thermal eﬃciency of

aheatengineas

η=Wout

Qin

≤ηCarnot

which gives

Wout =η×Qin ≤ηCarnot ×Qin

The maximum work obtainable is therefore given by

(Wout)max =ηCarnot ×Qin

We can calculate ηCarnot using (13.30) as

ηCarnot =1−TL

TH

=1−273 + 27

273 + 500 =61.2%

It is given that Qin = 155 MJ/s, and therefore

(Wout)max =0.612 ×155 MJ/s =95MW

326 Chapter 13

The minimum amount of heat rejected by the heat engine can be calculated

using the ﬁrst law as follows:

(Qout)min =Qin −(Wout )max = 155 MJ/s −95 MW =60MJ/s

Comment: For any real heat engine working between the reservoirs at 500◦C

and 27◦C, the thermal eﬃciency would be less than 61.2%, which is the Carnot

eﬃciency, the work output would be less than 95 MW, and the heat rejected by

the engine would be more than 60 MJ/s.

Example 13.13

If the temperature of the cooler reservoir

of the heat engine of Example 13.12 is reduced to, say, −23◦C, then

theworkoutputoftheheatenginecouldbeincreasedforthesame155

MJ/s of heat that the heat engine receives from the reservoir at 500◦C.

You therefore plan to use a cooler reservoir at −23◦C for the heat engine.

To maintain the temperature at −23◦C, you plan to use a heat pump that

operates between the −23◦C reservoir and the original cooler reservoir of

theheatengineofExample 13.12 at 27◦C. Assuming that all the heat

rejected by the heat engine to the reservoir at −23◦C is removed by the heat

pump, determine the maximum net work output and the overall thermal

eﬃciency of the combined system.

Solution to Example 13.13

Figure 13.6 shows the combined system. The heat pump could be operated

only if we provide work to it, which is denoted as Win is the ﬁgure. The net

work output is therefore Wout −Win ,whereWout is the work output of the heat

engine.

For the heat engine, following the procedure adopted in the Solution to

Example 13.12,weget

ηCarnot =1−250

773 =0.677

and therefore the maximum work obtainable from the heat engine is given by

(Wout)max =ηCarnot ×Q1=0.677 ×155 MJ/s = 105 MW

Introduction to the Second Law 327

Q1= 155 MJ/s

Q2

Figure 13.6 Schematic for Example 13.13.

Wout

Q4

Q3

Win heat

engine

heat

pump

reservoir at −23◦C

reservoir at 27◦C

reservoir at 500◦C

Heat rejected by the heat engine to the reservoir at −23◦Cis

Q2=Q1−(Wout)max = 155 MJ/s −105 MW =50MJ/s

This amount of heat is removed by the heat pump from the reservoir at

−23◦C, and therefore

Q3=Q2=50MJ/s

Combining (13.22) and (13.23), we can write the coeﬃcient of performance

of a heat pump as

COP =Q3

Win

≤COPCarnot

which gives

Win =Q3

COP ≥Q3

COPCarnot

The minimum work required by the heat pump is therefore given by

(Win)min =Q3

COPCarnot

We can calculate the COP using (13.23) as

COPCarnot =TL

TH−TL

=250

300 −250 =5

We know that Q3= 50 MJ/s, and therefore

(Win)min =50 MJ/s

5=10MW

328 Chapter 13

The net work output of the combined system is calculated as follows:

(Wout)net =(Wout )max −(Win )min = 105 MW −10 MW =95MW

which is the maximum net work output obtainable from the combined system.

The overall thermal eﬃciency of the combined system is determined as fol-

lows:

ηoverall =(Wout )net

Q1

=95 MW

155 MJ/s =61.2

Comment: Observe that the maximum net work output and the overall thermal

eﬃciency of the combined system are the same as the maximum work output and

the thermal eﬃciency of the heat engine alone in Example 13.12. Therefore,

it is of no advantage to use the combined system proposed in this example to

generate the work required.

Example 13.14

A metal block A of 70 kg is at 800 K and

a metal block B of 200 kg is at 300 K. A heat engine, the working ﬂuid of

which operating in a cyclic process, is to be operated using the two given

metal blocks as the heat source and heat sink, respectively. It is reported

that during a trial run, the temperature of the metal block A is reduced to

470 K and that of the metal block B is increased to 370 K. The speciﬁc

heat of the metal is given as 0.45 kJ/kg·K. Verify the report by carrying

out a second law analysis.

If the system satisﬁes the second law, determine the work output and

the thermal eﬃciency of the heat engine.

Solution to Example 13.14

Consider the heat engine as the system and the metal blocks A and B as

the surroundings. Since the heat engine describes a cyclic process, ∆Ssys =0.

Therefore,

∆Stotal =∆Ssurr =∆SA+∆SB

Taking the metal blocks to be incompressible substances, we could use

Introduction to the Second Law 329

(11.15) to calculate the entropy changes of blocks A and B as follows:

∆SA=(70kg)×(0.45 kJ/kg ·K)×ln 470

800=−16.75 kJ/K

∆SB= (200 kg)×(0.45 kJ/kg ·K)×ln 370

300=18.88 kJ/K

Therefore,

∆Stotal =−16.75 kJ/K +18.88 kJ/K =2.12 kJ/K >0

Since ∆Stotal >0, according to the second law, the given system is physically

realizable.

The work output of the engine, according to the ﬁrst law, is given by the

diﬀerence between the heat received by the engine from block A and the heat

rejected by the engine to block B. Heat received by the engine from block A is

given by 70 ×0.45 ×(800 - 470) kJ = 10,395 kJ. Heat rejected by the engine

to block B is given by 200 ×0.45 ×(370 - 300) kJ = 6,300 kJ. The work output

of the engine is therefore 4,095 kJ. And, the thermal eﬃciency of the engine is

39.4%.

Example 13.15

What should be the ﬁnal temperature of

block B of Example 13.14 for the work output of the heat engine to

reach its maximum? Assume all other data of Example 13.14 remains

unchanged. Determine also the value of the maximum work output and

the corresponding thermal eﬃciency.

Solution to Example 13.15

The maximum work output could be obtained, in theory, when the system of

Example 13.14 reaches its reversible limit. That is, when ∆Stotal =∆SA+

∆SB=0,whichgives

70 ×0.45 ×ln 470

800+ 200 ×0.45 ×ln TBf

300 =0

where TBf is the ﬁnal temperature of block B.

330 Chapter 13

Solving the above, we get TBf = 361.4 K. Therefore, heat rejected by the

engine to block B will become 200 ×0.45 ×(361.4 - 300) kJ = 5,526 kJ. Heat

received by the engine from block A remains the same as in the Solution to

Example 13.14, which is 10,395 kJ. The maximum work output will therefore

be 4,869 kJ, and the corresponding thermal eﬃciency will be 46.8%.

Example 13.16

A reversible gas turbine, whose working

ﬂuid is considered to operate in a cyclic process, works between two thermal

reservoirs, say A and B. The reservoir A is at 1200 K and the reservoir B

is at 500 K. A reversible steam turbine is operated between the reservoir B

and the atmosphere at 300 K. Determine the overall thermal eﬃciency of

this idealized combined gas turbine - steam turbine plant.

Compare this overall thermal eﬃciency to the thermal eﬃciency of a

reversible heat engine that would operate between the reservoirs at 1200 K

and 300 K.

Solution to Example 13.16

The schematic of the combined power plant is shown in Figure 13.7. The

overall thermal eﬃciency of the combined power plant would be

ηoverall =(Wout )gt +(Wout)st

Q1

From the data given for the reversible gas and steam turbines, we get

ηgt =1−500

1200 =58.3% and (Wout)gt =0.583 Q1

ηst =1−300

500 =40.0% and (Wout)st =0.4Q3

Assuming that all the heat rejected by the gas turbine to the reservoir at

500 K is taken by the steam turbine, we get

Q3=Q2=Q1−(Wout)gt =Q1−0.583 Q1=0.417 Q1

which gives

(Wout)st =0.4×0.417 Q1=0.167 Q1

Introduction to the Second Law 331

The overall thermal eﬃciency of the combined power plant therefore becomes

ηoverall =0.583 Q1+0.167 Q1

Q1

= 75%

Q1

Q2

Figure 13.7 Schematic for Example 13.16.

(Wout)gt

B: reservoir at 500 K

C: reservoir at 300 K

(Wout)st

Q3

Q4

gas turbine

steam turbine

A: reservoir at 1200 K

For a reversible heat engine operating between the reservoirs at 1200 K and

300 K, the thermal eﬃciency is

η=1−300

1200 = 75%

The results show that regardless of whether we operate a single heat engine

or a combined power plant, the thermal eﬃciency remains unchanged as far

as the temperatures of the hotter reservoir and the cooler reservoir remain the

same.

In reality, however, there are certain constraints in operating a gas turbine or

a steam turbine alone between the two given temperature extremes to reach the

kind of eﬃciencies that could be achieved by the combined power plant. The

details of which is beyond the scope of this text book, and therefore will not be

discussed here.

332 Chapter 13

Example 13.17

The combined power plants are known for their

improved thermal eﬃciency. The thermal eﬃciency of a newly installed com-

bined power plant is about 48%. The plant operates between the maximum

temperature of 1000 K and the atmospheric temperature of 300 K. A company

that comes with foreign aid claims that they will be able to increase the thermal

eﬃciency of the power plant to 70% by installing energy saving devices at a cost.

You have been asked to advice the Minister of Energy on that. What will be

your advice?

Solution to Example 13.17

A single (or even a combined) reversible heat engine operating between 1000

K and 300 K will have the following Carnot eﬃciency:

ηCarnot =1−300

1000 = 70%

No engine, however cleverly built, could have an eﬃciency that is higher

than the Carnot eﬃciency, which is 70% in this case, as long as it is operated

between the temperature extremes of 1000 K and 300 K.

Reaching the 70% thermal eﬃciency means that the combined power plant

must be operated under reversible conditions. Even though, in theory, it is

possible to improve the thermal eﬃciency to achieve the Carnot eﬃciency, it

would be impossible to reach such eﬃciency in real life situations.

My advice to the Minister of Energy would be that the company’s claim is

not physically realizable.

13.4 Summary

•The mathematical equivalence of the second law is as follows:

∆Stotal

>0for a real (irreversible) process

=0for a reversible process

<0for an impossible process

(13.1)

Introduction to the Second Law 333

•For a closed system,

∆Stotal =∆Ssys +∆Ssurr (13.2)

where

∆Ssys =m∆ssys =m(sf−so)(13.3)

•If the surroundings has a ﬁnite mass,

∆Ssurr =msurr ∆ssurr =msurr (sf−so)surr (13.4)

•Athermal reservoir is assumed to remain at a constant temperature

regardless of the heat transferred to or from the reservoir. It is also

assumed that no irreversibilities occur within the thermal reservoir.

•If the surroundings is taken as a thermal reservoir at a temperature of

Tsurr Kthen

∆Ssurr =(Qin)surr

Tsurr

(13.5)

where (Qin)surr denotes the ﬁnite amount of heat transferred to the sur-

roundings from the system during the process.

•For an adiabatic process,

∆Stotal =∆Ssys =m(sf−so)(13.7)

•For a reversible adiabatic process,

∆Stotal =0 (13.8)

•For an op en system,

d(∆Stotal )

dt =˙mese−˙misi+d(∆Ssurr)

dt (13.9)

•If the surroundings of the open system is taken as a thermal reservoir at

a constant temperature of Tsurr K, then

d(∆Ssurr )

dt =(˙

Qin)surr

Tsurr

(13.10)

where (˙

Qin)surr denotes the rate at which heat is transferred from the

system to the surroundings during the process.

334 Chapter 13

•For an adiabatic steady ﬂow process,

d(∆Stotal )

dt =˙mese−˙misi(13.11)

•A heat pump is a device, whose working ﬂuid operating in a cyclic process,

receives heat from a cooler reservoir and work from the surroundings, and

rejects all that it received, as heat to a hotter reservoir.

•The coeﬃcient of performance of a heat pump is given by

COP =Qin

Win

<COP

Carnot (13.22)

where

COPCarnot =TL

TH−TL

(13.23)

•No heat pump can have a coeﬃcient of performance higher than the

Carnot coeﬃcient of performance, which is a function of the maximum

and minimum temperatures across which the heat pump operates.

•Heat engine is a device, whose working ﬂuid operating in a cyclic process,

receives heat from a hotter reservoir, does work on the surroundings, and

rejects the remaining heat to a cooler reservoir.

•The thermal eﬃciency of a heat engine is given by

η=Wout

Qin

<η

Carnot (13.29)

where

ηCarnot =1−TL

TH

(13.30)

•No heat engine can have a thermal eﬃciency higher than the Carnot

eﬃciency, which is a function of the maximum and minimum temperatures

across which the heat engine operates.