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Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India)

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In This Chapter We Cover the Following Topics

Art. Content Page

5.1 Two Reversible Adiabatic Paths Cannot Intersect Each Other 3

5.2 Reversible Cycles and Clausius Theorem

Clausius Theorem

Clausius Inequality

4

4

5

5.3 Entropy 6

5.4 Principle of Entropy Increase 8

5.5 Temperature Entropy Diagram & 2nd Law of a Control Volume 11

5.6 Applications of Entropy Principle

Mixing of Two Fluids

Maximum Work Obtainable from Two Finite Bodies at Temperatures T1

and T2

Maximum Work Obtainable from a Finite Body and a TER

12

12

13

15

5.7 Entropy Generation in a Closed System 18

5.8 Entropy Generation in an Open System 20

5.9 First and Second Laws Combined 22

5.10 Entropy Change for Some Elementary Processes

Isothermal Process

Reversible Adiabatic Work in a Steady Flow System

Isentropic Process

For Ideal/Perfect Gases

Polytropic Process

23

23

24

25

26

27

5.11 Third Law of Thermodynamics 28

References:

1. M. J. Moran and H. N. Shapiro, Fundamentals of Engineering Thermodynamics, 6e,

John Wiley & Sons, Inc., New York, 2008.




2 Chapter 5: Entropy

2. G. J. Van Wylen, R. E. Sonntag, C. Borgnakke, Fundamentals of Thermodynamics,

John Wiley & Sons, Inc., New York, 1994.

3. J. P. Holman, Thermodynamics, 4e, McGraw-Hill, New York, 1988.

4. F. W. Sears, G. L. Salinger, Thermodynamics, Kinetic theory, and Statistical

Thermodynamics, 3e, Narosa Publishing House, New Delhi, 1998.

5. Y. A. Cengel and M. A. Boles, Thermodynamics: An Engineering Approach, 2e,

McGraw-Hill, New York, 1994.

6. E. Rathakrishnan, Fundamentals of Engineering Thermodynamics, 2e, PHI Learning

Private Limited, New Delhi, 2008.

7. P. K. Nag, Basic and Applied Thermodynamics, 1e, McGraw-Hill, New Delhi, 2010.

8. V Ganesan, Gas Turbine, 2e, Tata McGraw-Hill, New Delhi, 2003.

9. Y. V. C. Rao, An Introduction to Thermodynamics, 1e, New Age International (P)

Limited, Publishers, New Delhi, 1998.

10. Onkar Singh, Applied Thermodynamics, 2e, New Age International (P) Limited,

Publishers, New Delhi, 2006.

Please welcome for any correction or misprint in the entire manuscript and your

valuable suggestions kindly mail us [email protected].




3 Basic and Applied Thermodynamics By Brij Bhooshan

Till now the detailed explanation of Zeroth law, first law and second law of

thermodynamics have been made. Also we have seen that the first law of

thermodynamics defined a very useful property called internal energy. For overcoming

the limitations of first law, the second law of thermodynamics had been stated. Now we

need some mathematical parameter for being used as decision maker in respect of

feasibility of process, irreversibility, nature of process etc. The first law of

thermodynamics was stated in terms of cycles first and it was shown that the cyclic

integral of heat is equal to the cyclic integral of work. When the first law was applied for

thermodynamic processes, the existence of a property, the internal energy, was found.

Similarly, the second law was also first stated in terms of cycles executed by systems.

When applied to processes, the second law also leads to the definition of a new property,

known as entropy. ‘Entropy’ is the outcome of second law and is a thermodynamic

property. Entropy is defined in the form of calculus operation, hence no exact physical

description of it can be given. However, it has immense significance in thermodynamic

process analysis. In fact, thermodynamics is the study of three E's, namely, energy,

equilibrium and entropy.

5.1 TWO REVERSIBLE ADIABATIC PATHS CAN’T INTERSECT EACH OTHER

Let it be assumed that two reversible adiabatics AC and BC intersect each other at point

C (Diagram. 5.1). Let a reversible isotherm AB be drawn in such a way that it intersects

the reversible adiabatics at A and B. The three reversible processes AB, BC, and CA

together constitute a reversible cycle, and the area included represents the net work

output in a cycle. But such a cycle is impossible, since net work is being produced in a

cycle by a heat engine by exchanging heat with a single reservoir in the process AB,

which violates the Kelvin-Planck statement of the second law. Therefore, the

assumption of the intersection of the reversible adiabatics is wrong. Through one point,

there can pass only one reversible adiabatic.

Diagram 5.1

Since two constant property lines can never intersect each other, it is inferred that a

reversible adiabatic path must represent some property, which is yet to be identified.

Diagram 5.2

P

Rev. isotherm

Rev. adiabatic

Rev. adiabatic

P

Rev. isotherm

Rev. adiabatics

C

B

A





5.2 REVERSIBLE CYCLES AND CLAUSIUS THEOREM

First of all we Substitution of a reversible process by the reversible isothermal and

reversible adiabatic processes. Let a system be taken from an equilibrium state i to

another equilibrium state f by following the reversible path i-f refer to Diagram 5.2. Let

a reversible adiabatic i-a and another reversible b-f be drawn in such a manner that a

and b can be joined by a reversible isotherm. Also, these reversible adiabatic and

reversible isothermal lines are such that the area under iabf is equal to the area under

i-f.

Applying the first law for i-f,

Applying the first law for iabf,

We know Wif = Wiabf (area under the curves are same).

From Eqn. (5.1) and (5.2)

Since Qia and Qbf are zero, then

Thus, any reversible path may be substituted by a reversible adiabatic, a reversible

isotherm and a reversible adiabatic between the same end states such that the heat

transferred during the isothermal process is the same as that transferred during the

original process.

Clausius Theorem

Let a smooth closed curve representing a reversible cycle (Diagram 5.3) be considered.

Let the closed cycle be divided into a large number of strips by means of reversible

adiabatics. Each strip may be closed at the top and bottom by reversible isotherms. The

original closed cycle is thus replaced by a zigzag closed path consisting of alternate

adiabatic and isothermal processes, such that the heat transferred during all the

isothermal processes is equal to the heat transferred in the original cycle. Thus the

original cycle is replaced by a large number of Carnot cycles. If the adiabatics are close

to one another and the number of Carnot cycles is large, the saw-toothed zigzag line will

coincide with the original cycle.

Diagram 5.3

P

Rev. isotherms

Rev. adiabatics





Consider a system undergoing a reversible cycle. The given cycle may be sub-divided by

drawing a family of reversible adiabatic lines. Every two adjacent adiabatic lines may be

joined by two reversible isotherms (refers to Diagram 5.3).

Now,

Also, a1-b1-c1-d1 is a Carnot cycle which receives heat dQ1 during the a1b1 process and

rejects heat dQ2 during the c1d1 process. Let the heat addition be at temperature T1 and

the heat rejection to be at temperature T2.

Then it is possible to write,

and

or,

Since dQ2 is negative, it reduces to

Similarly for the cycle e1, f1, g1, h1,

If similar equations are written for all the elementary cycles, then

or,

The cyclic integral of dQ/T for a reversible cycle is equal to zero. This is known as

Clausius' theorem. The letter R emphasizes the fact that the equation is valid only for a

reversible cycle.

Clausius Inequality

Let us go back to the cycle a1, b1, c1, d1, from Diagram 5.3,

Now, the efficiency of a general cycle will be equal to or less than the efficiency of a

reversible cycle. Now ηIR ≤ ηR where, ηIR = 1 ‒ dQ2/dQ1 and this is not equal to 1 ‒ T2/T1.

Or

Since





or

for any process a1, b1, c1, d1, reversible or irreversible.

For the reversible cycle,

Hence, for any process a1, b1, c1, d1,

Then for any cycle

Similarly, for the irreversible cycle e1, f1, g1, h1,

Summing up all elementary cycles

Since entropy is a property and the cyclic integral of any property is zero. The above two

conclusions about reversible and irreversible cycles can be generalized as

The equality holds good for a reversible cycle and the inequality holds good for an

irreversible cycles. The complete expression is known as Clausius Inequality.

If ∮dQ/T = 0, the cycle is reversible,

If ∮dQ/T < 0, the cycle is irreversible and possible

If ∮dQ/T > 0, the cycle is impossible, since it violates the second law.

5.3 ENTROPY

Clausius inequality can be used to analyze the cyclic process in a quantitative manner.

The second law became a law of wider applicability when Clausius introduced the

property called entropy. By evaluating the entropy change, one can explain as to why

spontaneous processes occur only in one direction.

Diagram 5.4

Let a system be taken from an initial equilibrium state i to a final equilibrium state f by

following the reversible path R1, (Diagram 5.4). The system is brought back from f to i

by following another reversible path R2. Then the two paths R1 and R2 together

constitute a reversible cycle. From Clausius' theorem

P





The above integral may be replaced as the sum of two integrals, one for path R1 and the

other for path R2,

or

Since R2 is a reversible path

Since R1 and R2 represent any two reversible

Is independent of the reversible path connecting i and f. Therefore, there exists a

property of a system whose value at the final state f minus its value at the initial state i

is equal to

This property is called entropy, and is denoted by S. lf Si is the entropy at the initial

state i, and Sf is the entropy at the final state f then

When the two equilibrium states are infinitesimally near

where dS is an exact differential because S is a point function and a property. The

subscript R in dQ indicates that heat dQ is transferred reversibly.

The word 'entropy' was first used by Clausius, taken from the Greek word 'tropee'

meaning 'transformation'. It is an extensive property, and has the unit J/K. The specific

entropy

If the system is taken from an initial equilibrium state i to a final equilibrium state f by

an irreversible path, since entropy is a point or state function, and the entropy change is

independent of the path followed, the non-reversible path is to be replaced by a

reversible path to integrate for the evaluation of entropy change in the irreversible

process (Diagram 5.5).





Integration can be performed only on a reversible path.

Diagram 5.5

Calculation of Entropy Change

The following facts should be kept in mind while calculating the change in entropy for a

process

1. ds = (dQ/T)R for a reversible process

2. Entropy is a state function. The entropy change of a system is determined by its

initial and final states only, irrespective of how the system has changed its state.

3. In analyzing irreversible processes, it is not necessary to make a direct analysis

of the actual process. One can substitute the actual process by a reversible

process connecting the final state to the initial state, and the entropy change for

the imaginary reversible process can be evaluated.

Diagram 5.6

Mathematical formulation for entropy (dQrev = T·dS) can be used for getting property

diagrams between “temperature and entropy” (T – S), “enthalpy and entropy” (h – S).

Area under process curve on T–S diagram (Diagram 5.6) gives heat transferred, for

internally reversible process

5.4 PRINCIPLE OF ENTROPY INCREASE

Refer to Diagram 5.7. Let a system change from state 1 to state 2 by a reversible process

A and return to state 1 by another reversible process B. Then 1A2B1 is a reversible

cycle.

Diagram 5.7

C

1

P 2

B

A

2

1

Rev. path which replaces

the irrev. path Actual irrev. path





Therefore, the Clausius inequality gives

If the system is restored to the initial state 1 from state 2 by an irreversible process C,

then 1A2C1 is an irreversible cycle. Then the Clausius inequality gives

Subtracting (5.15) from (5.16)

Since the process 2B1 is reversible

or,

Thus, for any irreversible process,

whereas for a reversible process

Therefore, for the general case, we can write

Where the equality sign holds good for a reversible process and the inequality sign holds

good for an irreversible process.

Diagram 5.8

Now let us apply the above results to evaluate the entropy of the universe when a

system interacts with the surroundings. Refer to Diagram 5.8.

Let Temperature of the surroundings = Tsur

Temperature of the system = Tsys and,

If the energy exchange takes place, dQ will be the energy transfer from the

surroundings to the system.

Surrounding





Since Tsur > Tsys, then

For any infinitesimal process undergone by a system, we have from Eq. (5.17) for the

total mass

If the system is isolated, that is, when there is no energy interaction between the system

and the surroundings, dQ = 0, then

For a reversible process,

For a irreversible process,

Through generalization we can write for an isolated system

Examine the statement critically. For a reversible process it is equal to zero. For an

irreversible process it is greater than zero.

It is thus proved that the entropy of an isolated system can never decrease. It always

increases and remains constant only when the process is reversible. This is known as

the principle of increase of entropy, or simply the entropy principle. It is the quantitative

general statement of second law from the macroscopic viewpoint.

An isolated system can always be formed by including any system and its surroundings

within a single boundary (Diagram 5.9). Sometimes the original system which is then

only a part of the isolated system is called a 'subsystem'.

Diagram 5.9 Isolated system

The system and the surroundings together (the universe or the isolated system) include

everything which is affected by the process. For all possible processes that a system in

the given surroundings can undergo

The equality sign holds good if the process undergone is reversible; the inequality sign

holds good if the process undergone is irreversible.

Entropy may decrease locally at some region within the isolated system, but it must be

compensated by a greater increase of entropy somewhere within the system so that the

net effect of an irreversible process is an entropy increase of the whole system. The

entropy increase of an isolated system is a measure of the extent of irreversibility of the

process undergone by the system.

Surrounding isolated

(composite) system System





Clausius summarized these as:

Die Energie der Welt ist konstant.

Die Entropie der Welt strebt einem Maximum zu.

This means:

The energy of the universe (world) in constant (first law).

The entropy of the universe (world) tends towards a maximum (second law).

The entropy of an isolated system always increases and becomes a maximum at the

state of equilibrium. If the entropy of an isolated system varies with some parameter x,

then there is a certain value of xe which maximizes the entropy (when dS/dx = 0) and

represents the equilibrium state (Diagram 5.10). The system is then said to exist at the

peak of the entropy hill, and dS = 0. When the system is at equilibrium, any conceivable

change in entropy would be zero.

Diagram 5.10

5.5 TEMPERATURE ENTROPY DIAGRAM AND 2ND LAW OF A CONTROL

VOLUME

Entropy change of a system is given by dS = (dQ/T)R. During the reversible process, the

energy transfer as heat to the system from the surroundings is given by

Refer to Diagram 5.11. Here T and S are chosen as independent variables. The ∮TdS is

the area under the curve.

Diagram 5.11

The first law of thermodynamics gives dU = dQ ‒ dW. Also for a reversible process, we

can write,

Therefore,

For a cyclic process, the above equation reduces to

For a cyclic process ∮T.dS represents the net heat interaction which is equal to the net

work done by the system. Hence the area enclosed by a cycle on a T−S diagram

1

2

3 4

Equilibrium state

Possible

Impossible





represents the net work done by a system. For a reversible adiabatic process, we know

that

or,

or,

Hence a reversible adiabatic process is also called an isentropic process. On a T−S

diagram, the Carnot cycle can be represented as shown in Diagram 5.11. The area under

the curve 1-2 represents the energy absorbed as heat Q1 by the system during the

isothermal process. The area under the curve 3-4 is the energy rejected as heat Q2 by the

system. The shaded area represents the net work done by the system.

We have already seen that the efficiency of a Carnot cycle operating between two

thermal reservoirs at temperatures T1 and T2 (T2 < T1) is given by

This was derived assuming the working fluid to be an ideal gas. The advantage of T−S

diagram can be realized by a presentation of the Carnot cycle on the T−S diagram. Let

the system change its entropy from SA to SB during the isothermal expansion process 1-

2. Then,

and,

and,

or,

This demonstrates the utility of T-S diagram.

5.6 APPLICATIONS OF ENTROPY PRINCIPLE

The principle of increase of entropy is one of the most important laws of physical

science. It is the quantitative statement of the second law of thermodynamics. Every

irreversible process is accompanied by entropy increase of the universe, and this entropy

increase quantifies the extent of irreversibility of the process. The higher the entropy

increase of the universe, the higher will be the irreversibility of the process. A few

applications of the entropy principle are illustrated in the following.

Application 5.1: (Mixing of Two Fluids): A mass m of water at T1 is isobarically and

adiabatically mixed with an equal mass of water T2.

Show that and prove that this is non-negative.





Diagram 5.12 Mixing of two fluids

Solution: Consider a general case, Subsystem 1 having a fluid of mass m1, specific heat

c1, and temperature T1, and subsystem 2 consisting of a fluid of mass m2, specific heat c2,

and temperature T2, comprise a composite system in an adiabatic enclosure (Diagram

5.12). When the partition is removed, the two fluids mix together, and at equilibrium, let

Tf be the final temperature, and T2 < Tf < T1. Since energy interaction is exclusively

confined to the two fluids, the system being isolated

Entropy change for the fluid in subsystem 1

This will be negative, since Tf < T1.

Entropy change for the fluid in subsystem 2

This will be positive, since T2 < Tf.

ΔSuni will be positive definite, and the mixing process is irreversible.

Although the mixing process is irreversible, to evaluate the entropy change for the

subsystems, the irreversible path was replaced by a reversible path on which the

integration was performed.

If m1 = m2 = m and c1 = c2 = c.

This is always positive, since the arithmetic mean of any two numbers is always greater

than their geometric mean.

Application 5.2: (Maximum Work Obtainable from Two Finite Bodies at Temperatures

T1 and T2): A heat engine is working between the starting temperature limits of T1 and

T2 of two bodies. Working fluid flows at rate ‘m’ kg/s and has specific heat at constant

pressure as Cp. Determine the maximum obtainable work from engine.

Solution: Let us consider two identical finite bodies of constant heat capacity at

temperatures T1 and T2 respectively, T1, being higher than T2. If the two bodies are

Adiabatic enclosure

Partition

Subsystem 1

Subsystem 2





merely brought together into thermal contact, delivering no work, the final temperature

Tf reached would be the maximum

Diagram 5.13

If a heat engine is operated between the two bodies acting as thermal energy reservoirs

(Diagram 5.13), part of the heat withdrawn from body 1 is converted to work W by the

heat engine, and the remainder is rejected to body 2. The lowest attainable final

temperature Tf corresponds to the delivery of the largest possible amount of work, and is

associated with a reversible process.

As work is delivered by the heat engine, the temperature of body 1 will be decreasing

and that of body 2 will be increasing. When both the bodies attain the final temperature

Tf, the heat engine will stop operating. Let the bodies remain at constant pressure and

undergo no change of phase.

Total heat withdrawn from body 1

where Cp is the heat capacity of the two bodies at constant pressure.

Total heat rejected to body 2

Amount of total work delivered by the heat engine

For given values of Cp, T1, and T2, the magnitude of work W depends on Tf. Work

obtainable will be maximum when Tf is minimum.

Now, for body 1, entropy change ΔS1 is given by

For body 2, entropy change ΔS2 would be

Since the working fluid operating in the heat engine cycle does not undergo any entropy

change, ΔS of the working fluid in heat engine = ∮dS = 0.

Applying the entropy principle

HE

Body 1

Body 2





From equation (5.25), for Tf to be a minimum

Since Cp ≠ 0,

For W to be a maximum, Tf will be . From equation (5.24)

The final temperatures of the two bodies, initially at T1, and T2, can range from (T1 +

T2)/2 with no delivery of work to with maximum delivery of work.

Application 5.3: (Maximum Work Obtainable from a Finite Body and a TER)

Suppose that one of the bodies considered in the previous application be a thermal

energy reservoir. The finite body has a thermal capacity Cp and is at temperature T and

the TER is at temperature T0, such that T > T0. Let a heat engine operate between the

two (Diagram 5.14).

Diagram 5.14

As heat is withdrawn from the body, its temperature decreases. The temperature of the

TER would, however, remain unchanged at T0. The engine would stop working, when

the temperature of the body reaches T0. During that period, the amount of work

delivered is W, and the heat rejected to the TER is (Q ‒ W). Then

Applying the entropy principle

or

TER T0

HE

Body T





or

Application 5.4: Two identical bodies of constant heat capacity are at the same initial

temperature T1. A refrigerator operates between these two bodies until one body is

cooled to the temperature T2. If the bodies remain at constant pressure and undergo no

change of phase, obtain an expression for the minimum amount of work required to

achieve this.

[Engg. Services 1987]

Diagram 5.15

Solution: Suppose Cp be Specific heat of the body at constant pressure.

For minimum work, the refrigerator should operate on a reversible cycle. If it is so, then

entropy change for the total cycle is equal to zero.

If Tf is the final temperature of the high level reservoir, then the above equation can be

written as

Cyclic work done = Cyclic heat added

work done = heat rejected at the higher temperature − heat picked up at the lower

temperature.

HP

Body B

Body A





Utilizing Eqn. (i) and (ii), then we have

Application 5.5: An ideal gas is heated from temperature T1 to T2 by keeping its

volume constant. The gas is expanded back to it's initial temperature according to the

law pvn = constant. If the entropy changes in the two processes are equal, find the value

of ‘n’ in terms of adiabatic index γ.

[Engg. Services-1997]

Diagram 5.16

Solution: During constant volume process change in entropy

Change in entropy during polytropic process,

Since the entropy change is same, so

Application 5.6: A cool body at temperature T1 is brought in contact with high

temperature reservoir at temperature T2. Body comes in equilibrium with reservoir at

constant pressure. Considering heat capacity of body as C, show that entropy change of

universe can be given as;

Solution: Since body is brought in contact with reservoir at temperature T2, the body

shall come in equilibrium when it attains temperature equal to that of reservoir, but

there shall be no change in temperature of the reservoir.

Entropy change of universe ΔSuni = ΔSbody + ΔSreservoir

1 2 3

T

3

2

1





as, heat gained by body = Heat lost by reservoir

Thus, ΔSuni is

or, rearranging the terms,

5.7 ENTROPY GENERATION IN A CLOSED SYSTEM

The entropy of any closed system can increase in two ways:

(a) by heat interaction in which there is entropy transfer

(b) internal irreversibilities or dissipative effects in which work (or K.E.) is

dissipated into internal energy increase.

If dQ is the infinitesimal amount of heat transferred to the system through its boundary

at temperature T, the same as that of the surroundings, the entropy increase dS of the

system can be expressed as

where deS is the entropy increase due to external heat interaction and diS is the entropy

increase due to internal irreversibility. From Eq. (5.28),

The entropy increase due to internal irreversibility is also called entropy production or

entropy generation, Sgen.

In other words, the entropy change of a system during a process is greater than the

entropy transfer (dQ/T) by an amount equal to the entropy generated during the

process within the system (diS), so that the entropy balance gives:

Entropy change = Entropy transfer + Entropy generation

which is a verbal statement of Eq. (5.28).

It may so happen that in a process (e.g., the expansion of a hot fluid in a turbine) the

entropy decrease of the system due to heat loss to the surroundings (‒∫dQ/T) is equal to

the entropy increase of the system due to internal irreversibilities such as friction, etc.

(‒∫diS), in which case the entropy of the system before and after the process will remain

the same (∫dS = 0). Therefore, an isentropic process need not be adiabatic or reversible.

But if the isentropic process is reversible, it must be adiabatic. Also, if the isentropic

process is adiabatic. it cannot but be reversible. An adiabatic process need not be





isentropic, since entropy can also increase due to friction etc. But if the process is

adiabatic and reversible, it must be isentropic.

For an infinitesimal reversible process by a closed system,

If the process is irreversible,

Since U is a property,

The difference (pdV ‒ dW) indicates the work that is lost due to irreversibility, and is

called the lost work d(LW), which approaches zero as the process approaches

reversibility as a limit. Equation (5.30) can be expressed in the form

Thus the entropy of a closed system increases due to heat addition (deS) and internal

dissipation (diS ).

Diagram 5.17

In any process executed by a system, energy is always conserved, but entropy is

produced internally. For any process between equilibrium states 1 and 2 (Diagram 5.17),

the first law can be written as

Energy transfer Energy change

By second law,

It is only the transfer of energy as heat which is accompanied by entropy transfer, both

of which occur at the boundary where the temperature is T. Work interaction is not

accompanied by any entropy transfer. The entropy change of the system (S2 ‒ S1)

exceeds the entropy transfer

The difference is produced internally due to irreveersibility. The amount of entropy

generation Sgen is given by

Surrounding

(Work Transfer)

(Entropy Transfer)

(Heat Transfer)

T

(Boundary

temperature)

Boundary System 1→2





The second law states that, in general, any thermodynamic process is accompanied by

entropy generation.

Diagram 5.18

Process 1-2, which does not generate any entropy (Sgen = 0), is a reversible process

(Diagram 5.18). Paths for which Sgen > 0 are considered irreversible. Like heat transfer

and work transfer during the process 1-2, the entropy generation also depends on the

path the system follows. Sgen is, therefore, not a thermodynamic property and dSgen is an

inexact differential, although (S2 ‒ S1) depends only on the end states. In the differential

form, Eq. (5.31) can be written as

The amount of entropy generation quantifies the intrinsic irreversibility of the process. If

the path A causes more entropy generation than path B (Diagram 5.18), i.e.

the path A is more irreversible than path B and involves more 'lost work'.

If heat transfer occurs at several locations on the boundary of a system, the entropy

transfer term can be expressed as a sum, so Eq. (5.31) takes the form

where Qj/Tj is the amount of entropy transferred through the portion of the boundary at

temperature Tj.

On a time rate basis, the entropy balance can be written as

where dS/dτ is the rate of change of entropy of the system, is the rate of entropy

transfer through the portion of the boundary whose instantaneous temperature is Tj,

and is the rate of entropy generation due to irreversibilities within the system.

5.8 ENTROPY GENERATION IN AN OPEN SYSTEM

In an open system, there is transfer of three quantities: mass, energy and entropy. The

control surface can have one or more openings for mass transfer (Diagram 5.19). It is

rigid, and there is shaft work transfer across it.

The continuity equation gives

Rev

1

2

B A





Diagram 5.19

Net mass transfer rate = rate of mass accumulation in the CV

The energy equation gives

Net rate of energy transfer = rate of energy accumulation in the CV

The second law inequality or the entropy principle gives

Net rate of entropy transfer = rate of increase of entropy of the CV

Here represents the rate of heat transfer at the location of the boundary where the

instantaneous temperature is T. The ratio /T accounts for the entropy transfer along

with heat. The terms ṁisi and ṁese account, respectively, for rates of entropy transfer

into and out of the CV accompanying mass flow. The rate of entropy increase of the

control volume exceeds, or is equal to, the net rate of entropy transfer into it. The

difference is the entropy generated within the control volume due to irreversibility.

Hence, the rate of entropy generation is given by

By the second law,

If the process is reversible, = 0. For an irreversible process, > 0.

The magnitude of quantifies the irreversibility of the process. If systems A and B

operate so that

it can be said that the system A operates more irreversibly than system B.

At steady slate, the continuity equation gives

the energy equation becomes

CS

(Entropy Transfer Rate)

Surrounding

Shaft Work Transfer Rate

(Heat Transfer Rate)

Surface temperature, T

S

M E

Control volume





and the entropy equation reduces to

These equations often must be solved simultaneously, together with appropriate

property relations.

Mass and energy are conserved quantities, but entropy is not generally conserved. The

rate at which entropy is transferred out must exceed the rate at which entropy enters

the CV, the difference being the rate of entropy generated within the CV owing to

irreversibilities.

For one-inlet and one-exit control volumes, the entropy equation becomes

5.9 FIRST AND SECOND LAWS COMBINED

By the second law

dQ = T.dS

and by the first law, for a closed non-flow system,

dQ = dU + p.dV

T.dS = dU + p.dV [5.43]

Again, the enthalpy

H = U + pV

dH = dU + p.dV + V.dp = T.dS + V.dp

T.dS = dH – V.dp [5.44]

Equations (5.43) and (5.44) are the thermodynamic equations relating the properties of

the system.

Let us now examine the following equations as obtained from the first and second laws:

(a) dQ = dE + dW→ This equation holds good for any process, reversible or

irreversible, and for any system.

(b) dQ = dU + pdW→ This equation holds good for any process undergone by a closed

stationary system.

(c) dQ = dU + pdV→ This equation holds good for a closed system when only pdV-

work is present. This is true only for a reversible (quasi-static) process.

(d) dQ = TdS→ This equation is true only for a reversible process.

(e) TdS = dU + pdV→ This equation holds good for any process reversible or

irreversible, undergone by a closed system, since it is a relation among properties

which are independent of the path,

(f) TdS = dH ‒ Vdp→ This equation also relates only the properties of a system.

There is no path function term in the equation. Hence the equation holds good for

any process.





The use of the term 'irreversible process' is doubtful, since no irreversible path or

process can be plotted on thermodynamic coordinates. It is more logical to state that 'the

change of state is irreversible, rather than say 'it is an irreversible process'. A natural

process which is inherently irreversible is indicated by a dotted line connecting the

initial and final states, both of which are in equilibrium. The dotted line has no other

meaning, since it can be drawn in any way. To determine the entropy change for a real

process, a known reversible path is made to connect the two end states, and integration

is performed on this path using either equation (e) or equation (f'), as given above.

Therefore, the entropy change of a system between two identifiable equilibrium states is

the same whether the intervening process is reversible or the change of state is

irreversible.

For constant pressure process, dh = T·ds

or

which means slope of constant pressure line on enthalpy – entropy diagram (h – s) is

given by temperature.

Also from above two relations

For a constant pressure process above yields

It gives the slope of constant pressure line on T – s diagram.

Similarly, for a constant volume process,

It gives the slope of constant volume line on T – s diagram.

It can be concluded from the above mathematical explanations for slope that slope of

constant volume line is more than the slope of constant pressure line as cp > cv.

Diagram 5.20

5.10 ENTROPY CHANGE FOR SOME ELEMENTARY PROCESSES

Isothermal Process

Let us find out entropy change for isothermal heat addition process. As isothermal

process can be considered internally reversible, therefore entropy change shall be;

Constant pressure line

Constant volume line





where Qa‒b is total heat interaction during state change a – b at temperature T.

Reversible Adiabatic Work in a Steady Flow System

In the differential form, the steady flow energy equation per unit mass is given by,

For a reversible process, dQ = Tds

Using the property relation, Eq. (5.45), per unit mass,

in Eq. (5.45), we have

On integration

If the changes in KE and P.E. are neglected, Eq. (5.47) reduces to

If dQ = 0, implying ds = 0, the property relation gives

or

From Eqs (5.48) and (5.49),

The integral

represents an area on the p-v plane (Diagram 5.21a). To make the

integration, one must have a relation between p and v such as pvn = constant

Diagram 5.21 Reversible work interaction

(a) Steady flow

2

1

(b) A closed system

2

1





Equation (5.50) holds good for a steady flow work-producing machine like an engine or

turbine as well as for a work-absorbing machine like a pump or a compressor, when the

fluid undergoes reversible adiabatic expansion or compression.

For closed system (Diagram 5.21b), the reversible work done would be

Isentropic Process

It is the process during which change in entropy is zero and entropy remains constant

during process.

It indicates that when ΔSa‒b = 0, then Qa‒b = 0

which means there is no heat interaction during such process and this is adiabatic

process.

Hence, it can be said that "a reversible isentropic process shall be adiabatic, where as if

isentropic process is adiabatic then it may or may not be reversible’’.

Thus, adiabatic process may or may not be reversible. It means in reversible adiabatic

process all states shall be in equilibrium and no dissipative effects are present along

with no heat interaction whereas in adiabatic process there is no heat interaction but

process may be irreversible.

Finally, it can be concluded that an adiabatic process may or may not be isentropic

whereas a reversible adiabatic process is always isentropic.

An adiabatic process of non isentropic type is shown in Diagram 5.22 where

irreversibility prevails, say due to internal friction.

Diagram 5.22

Here a – b is reversible adiabatic expansion of isentropic type.

Non-isentropic or adiabatic expansion is shown by a – b'.

Isentropic expansion efficiency may be defined as ratio of actual work to ideal work

available during expansion.

Similarly, isentropic and non-isentropic compression process are shown as c–d and c–d'

respectively.

Isentropic compression efficiency can be defined on same lines as,

Absorption of energy by a constant temperature reservoir: A certain amount of heat is

added to a constant temperature reservoir. The actual process can be replaced by a





reversible path in which an equivalent amount of energy is added to the reservoir. Then,

the entropy change of the reservoir is given by

Heating or Cooling of matter: The heating can be carried out either at constant pressure

or at constant volume. From the first law of thermodynamics

Q = ΔU for constant volume heating/cooling process

Q = ΔH for constant pressure heating/cooling process

or,

Similarly,

or,

Phase change at constant temperature and pressure

Melting:

Evaporation:

For Ideal/Perfect Gases

Suppose a certain quantity of a perfect gas being heated by any thermodynamics

process. Suppose

m = mass of gas,

p1 = initial pressure of a gas,

v1 = initial volume of a gas,

T1 = initial temperature of a gas,

p2, v2, T2 are corresponding final values of the gas,

dT = small change in temperature, and

dv = small change in volume.

Now from first law of thermodynamics

Dividing by T

On integration





Now from general equation of gas

Using equation (5.51a), then we have

Now we know that Cp ‒ Cv = R, then

Again from general equation of gas

Using equation (5.51a), then we have

Polytropic Process

Entropy change in a polytropic process having governing equation as pvn = constant, can

be obtained as below,

For polytropic process between 1 and 2,

Also, from gas laws,

Above two pressure ratios give,

Now using equation (5.61a), then we have

For perfect gas R = cp – cv, then R = cv (γ ‒1)

Substituting R in entropy change





5.11 THIRD LAW OF THERMODYNAMICS

‘Third law of thermodynamics’, an independent principle uncovered by ‘Nernst’ and

formulated by ‘Planck’, states that the “Entropy of a pure substance approaches zero at

absolute zero temperature.” This fact can also be corroborated by the definition of

entropy which says it is a measure of molecular disorderness. At absolute zero

temperature substance molecules get frozen and do not have any activity, therefore it

may be assigned zero entropy value at crystalline state. Although the attainment of

absolute zero temperature is impossible practically, however theoretically it can be used

for defining absolute entropy value with respect to zero entropy at absolute zero

temperature. Second law of thermodynamics also shows that absolute zero temperature

can’t be achieved. Third law of thermodynamics is of high theoretical significance for the

sake of absolute property definitions and has found great utility in thermodynamics.

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