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# Thermodynamics for Beginners - Chapter 13 INTRODUCTION TO THE SECOND LAW

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Learning thermodynamics even at an introductory level cannot be considered 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 important to learn the second law, and what role it has in the thermodynamic analyses of engineering systems.
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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 (sfso)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 (sfso)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 (sfso)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.
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
Solution to Example 13.2
of an ideal gas through the nozzle, we get Te, the temperature at the nozzle
exit, as
Te=Tic2
ec2
i
2Cp
= 975 K9252
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(sesi)= ˙mCpln Te
TiRln Pe
Pi
mR 1.4
1.41ln 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
1.41ln 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(TiTe)=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
turbine at 60 bar and 400C 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 400C = 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.
d(∆Stotal )
dt m(sesi)= ˙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 30C is cooled until it freezes at 0C. Take
Cwater as 4.2 kJ/kg ·K, and the latent heat of freezing at 0Cas333
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 30Cto0
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 0C, during
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 0C 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 20C.
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 20C, 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 30Cto0
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 0Ctoiceat0
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 20Cin
working out the above example. What if I take the surroundings to be at
the room temperature, say, 30C. Will I still get Stotal >0?
Teacher: You can’t do that. How could the system, which is cooled from 30C
to 0C, 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 10C
?
Heat rejected to the
surroundings at 30C
work input
as electricity
superheated vapour
at 1.2 bar &20C
superheated vapour
at 10 bar &65oC
vapour-liquid
mixture at
1.2 bar &25.8C
subcooled
liquid at
10 bar &40C
Figure 13.1 Schematic of a typical refrigeration cycle.
Student: Teacher, the ﬁgure looks complicated except for the compressor,
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 10C. Heat must
be continuously removed from the freezer space to maintain such a low
temperature when the surrounding atmosphere is around 30C. Do you
agree with that?
Student: Yes, I do.
Teacher: Let us send a liquid, the refrigerant, at a temperature much lower
than 10C 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 10C
to the refrigerant at a temperature much lower than 10C.
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
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 =TLTH
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
THTL
THTLQin (13.20)
320 Chapter 13
For Stotal to be a positive quantity,
Win >THTL
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
THTL
(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
THTL
(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 25C. The atmospheric temperature is at 36C, 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 25C (which is the cooler reservoir) and rejecting
heat to the atmosphere at 36C (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
THTL
=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,andSsurr =(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
=THTL
THTLQin Wout
TL
(13.27)
For Stotal to be a positive quantity,
Wout <THTL
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
<1TL
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 =1TL
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 500C throughout
the operation. Take the cooling water removing heat from the condenser
as the cooler reservoir at a constant temperature of 27C. 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 =1TL
TH
=1273 + 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 500C
and 27C, 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, 23C, then
theworkoutputoftheheatenginecouldbeincreasedforthesame155
MJ/s of heat that the heat engine receives from the reservoir at 500C.
You therefore plan to use a cooler reservoir at 23C for the heat engine.
To maintain the temperature at 23C, you plan to use a heat pump that
operates between the 23C reservoir and the original cooler reservoir of
theheatengineofExample 13.12 at 27C. Assuming that all the heat
rejected by the heat engine to the reservoir at 23C 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 =1250
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 23C
reservoir at 27C
reservoir at 500C
Heat rejected by the heat engine to the reservoir at 23Cis
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
23C, 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
THTL
=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 =1500
1200 =58.3% and (Wout)gt =0.583 Q1
ηst =1300
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 =Q10.583 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=1300
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
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 =1300
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 =mssys =m(sfso)(13.3)
If the surroundings has a ﬁnite mass,
Ssurr =msurr ssurr =msurr (sfso)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.
Stotal =∆Ssys =m(sfso)(13.7)
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
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
THTL
(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
Carnot (13.29)
where
ηCarnot =1TL
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.