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ME1201-ENGINEERING THERMODYNAMICS

S.K.AYYAPPAN, Lecturer, Department of mechanical engineering

UNIT II

SECOND LAW

Second law of thermodynamics – Kelvin’s and Clausius statements of

second law – Reversibility and irreversibility – Carnot theorem – Carnot

cycle – Reversed carnot cycle – efficiency – COP –Thermodynamic

temperature scale – Clausius inequality – concept of entropy – entropy of

ideal gas –principle of increase of entropy – availability.



Second Law

One form of energy can be converted into other form and any process is possible as long as it

does not create energy or destroy energy. Also, in a cyclic process, heat is converted in to

work and vice-versa.

But in actual practice, all forms of energy cannot be changed into work.

Let there be two thermodynamic system X and Y at temperature T1 and T2 respectively such

that T1›T2

T1›T2 T‹T1 T›T1

SECOND LAW OF THERMODYNAMICS:

i) Kelvin-planck statement and

ii) Clausius statement

Kelvin-Planck statement:

It is impossible to construct an operating device working on a cyclic process which

produces other effect than the extraction of energy as heat from a single thermal

reservoir and performs an equivalent amount of work.

Qs Qs

W=Qs- QR

W=Qs-QR

QR

Impossible heat engine by II law

Possible heat engine by II law

Otherwise, it is impossible to construct an engine working on a cyclic process which converts

all the heat energy supplied into equivalent amount of useful work.

Simply, we can say that all the heat energy given to an engine cannot be converted it into

useful work and some amount of heat energy will rejected to sink.

Clausius statement:

SYSTEM X

T1

SYSTEM Y

T2

SYSTEM

X

T

SYSTEM Y

T

RESERVOIR

HEAT

ENGINEE

Hot Reservoir

Cold reservoir

HEAT ENGINE



It is impossible to construct an operating device working on a cyclic process which produces

no other effect than the transfer of energy as heat from a low temperature body to a high

temperature body.

Otherwise, it may be stated as “heat cannot flow from cold reservoir to hot reservoir without

any external aid. But, heat can flow from hot reservoir to cold reservoir without any external

aid”

Q

Impossible system possible system

Clausious statement

Reversibility:

All the processes are divided into two types based on second law of thermodynamics.

1. Reversible process or ideal process

2. Irreversible process or actual process or natural process.

A

P B

V

A process is performed in such a way that it should trace the same path when the process is

reversed known as reversible process. It means it does not follow the different path without

producing any change in the universe when it is reversed.

Process taken from A to B. by a path A-B. Similarly, when the same process is reversed, it

attains the initial state by a path in the same manner as that of the process already carried out.

The work involved due to this process is known as reversible work or maximum work or

ideal work. A quasi-static process is also a reversible process.

Irreversibility:

Hot reservoir

Cold reservoir

Hot reservoir

Cold reservoir

Heat engine



All the processes are irreversible process. When any process is reversed, it definitely should

not the same path as the original process. At the same time, it produces a change in the

universe. So, the path traced by the reversal process is different. Therefore, the work involved

due to this process is called as irreversible work. Irreversibility is produced due to various

reasons.

Causes:

1. Lack of equilibrium

2. Heat transfer through a finite temperature difference

3. Lack of pressure equilibrium within the interior of the system.

4. Free expansion

5. Dissipative effects

Qualitative difference between heat and work:

The energy gained by one would be exactly equal to that lost by the other. It is the second law

of thermodynamics which provides the criterion as the probability of various processes.

When supplied to a system in the form of work, can be completely converted into heat (work

transfer internal energy increase heat transfer).

But the complete conversion of heat into work in a cycle is not possible. So heat and work are

not completely interchangeable forms of energy.

When work is converted into heat, we always have

W = Q

When heat is converted into work in a complete closed cycle process

Q ≥W

The arrow indicates the direction of energy transformation. A system is taken from state 1 to

state 2 by work transfer W1-2, and then by heat transfer Q2-1 the system is brought back

from state 2 to state1 to complete a cycle.

W1-2=Q2-1

Work is said to be high grade energy and heat low grade energy. The complete conversion of

low grade energy into high grade energy in a cycle is impossible.

Cyclic heat engine:

A heat engine cycle is a thermodynamic cycle in which there is a net heat transfer to the

system and a net work transfer from the system. The system which executes a heat engine

cycle is called a heat engine.

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

Qnet=Q1-Q2

And the net work transfer in a cycle

Wnet=WT-WP

(Or)

Wnet=WE-WC

By the first law of thermodynamics, we have

∑cycleQ=∑cycleW

Qnet=Wnet



Q1-Q2=WT-WP

η= net work output of the cycle/ total heat input to the cycle

=Wnet/Q1

η=Wnet/Q1=WT-Wp/Q1=Q1-Q2/Q1

η=1-Q2/Q1

This is also known as the thermal energy of heat engine cycle.

A heat engine is very often called upon to extract as much work (net) as possible from a

certain heat input, i.e., to maximize the cycle efficiency.

Heat engine:

Heat engine is a device which operates in a thermodynamics cycle and produces work by

supplying heat from hot reservoir.

Qs

W

η=W/Qs

W=Qs-QR

η=Qs-QR/QS=TH-TL/TH

=1-QR/QS=1-TL/TH

Energy reservoirs:

A thermal energy reservoir (TER) is defined as a large body of infinite heat capacity, which

is capable of absorbing or rejecting an unlimited quantity of heat without suffering

appreciable changes in its thermodynamic coordinates.

The thermal energy reservoir TERH from which heat Q1 is transferred to the system operating

in a heat engine cycle is called the source.

The energy reservoir TERL to which heat Q2 is rejected from the system during a cycle is the

sink.

A mechanical energy reservoir (MER) is a large body enclosed by an adiabatic impermeable

wall capable of storing work as potential energy or kinetic energy.

Q1

Wp WT

Wnet

CHE

Q2

TERH (source)

B

P T C

TERL (sink)

MER

Hot temperature reservoir,TH

Low temperature reservoir, TL

Heat engine

QR



Cyclic heat engine (CHE) with source and sink.

Refrigerator:

Refrigerator is a device which is used to remove heat from a cold system. In other words, it is

used to maintain the temperature of the body lower than that of surroundings.

Qs

Refrigerator

W

Performance of a refrigerator or a heat pump is measured with a term called coefficient of

performance. It is the ratio of heat extracted to the work input.

COP of a refrigerator is given by

(C.O.P) ref= QR/W=QR/QS-QR=TL/TH-TL

Where, TH and TL are higher lower temperature respectively.

Heat pump:

Heat pump is a device which is used to supply heat to a hotter system. In other words, it is

used to maintain the temperature of the body higher than the surroundings.

Qs

W

(C.O.P)HP=QS/W=QS/QS-QL=TH/TH-TL

Hot temperature reservoir,TH (atm)

Low temperature reservoir, TL

Ref

QR

Hot temperature reservoir,TH

Low temperature reservoir, TL(Atm)

HP

QR



Where, TH and TL are higher lower temperature respectively.

Corollaries of the second law of thermodynamics:

The statements which can be proved with the help of reversible cycles are called as

corollaries of the second law of thermodynamics.

Corollary-1

It is impossible to construct a system which will operate in a cycle and transfer heat from a

cooler to a hotter body without supplying work on the system by surroundings.

Proof:

Suppose the above statement is not true, the system could be represented by a heat pump for

which W=0.

If it takes Qs units of heat from the cold reservoir, it must deliver QS=QR units to the hot

reservoir to satisfy the first law.

A heat engine could also be operated between two reservoirs. It delivers QR units of heat to

the cold reservoir while performing W units of work. Then the first law states that the engine

must be supplied with (W+QR) or (W+Qs) units of heat from the hot reservoir.

QS1=QR1 QS2+W

W=0

W

QS1 QR2

If we combine both the plants then the cold reservoir becomes unnecessary and the heat

rejected from the heat engine is directed to heat pump as heat intake. The combined plant

represents a heat engine extracting (W+QR)-QR=W units of heat from a reservoir, and

delivering an equivalent amount of work. This is impossible according to second law. Thus,

corollary-1 must true.

Perpetual motion machine of second kind (PMM-II)

PMM-II is the machine which receives heat energy from hot reservoir and converts it into

equivalent amount of work. PMM-II gives 100% efficiency. Therefore, it is impossible to

construct. This violates the second law of thermodynamics.

Carnot cycle:

HOT RESERVOIR

COLD RESERVOIR

HP HE



It is also called as constant temperature cycle. It consists of four processes such as two

isentropic or reversible adiabatic and two isothermal process. The p-v and T-s diagram for

cycle as follows.

2 2 3

3 T2=T3

P W Q

1 4 T1=T4

1 4

S1=S2

S3=S4

V

p-V diagram T-s diagram

Process 1-2

Air is compressed isentropically from stage 1 to 2. During this process, both the pressure and

temperature increase from p1 to p2 and T1 to T2 resp. volume decreases from V1 to V2.

There is no heat added or rejected during this process.

Process 2-3:

During this process, heat is supplied to the fluid at constant temperature. Therefore, there is

no change in temperature T3=T2 but volume and entropy increase from V2 to V3 and S2 to

S3 resp. pressure decreases from p2 to p3.

Heat supplied Qs2-3=T2ds=T3ds ds=dQ/T&T2=T3

dQ=T.ds

Process 3-4:

Air is expanded isentropic ally from 3-4. During this process, both pressure and temperature

decrease from p3 to p4 and T3 to T4 resp. but, entropy remains constant (S3=S4).

Process 4-1;

During this process, heat is rejected isothermally from the fluid and attains its initial position.

Thus, the fluid completes one full cycle.

Heat rejected QR4-1=T1.dS=T4.dS [T1=T4]

Work done during the cycle W=heat supplied-heat rejected

W=T2.dS-T1.dS

W= (T2-T1) dS

Efficiency η=W/Qs= (T2-T1) ds/T2Ds

η Carnot=T2-T1/T2=TH-TL/TH

Where T1&T2 are minimum temperature resp.

Therefore T2=TH and T1=TL

Four reversible process, it is a reversible cycle.

A Carnot gas cycle operating in a given temperature range is shown in the T-s diagram

In Fig. One way to carry out the processes of this cycle is through the use of

Steady-state, steady-flow devices as shown in Fig. The isentropic expansion



process 2-3 and the isentropic compression process 4-1 can be simulated quite well by

A well-designed turbine and compressor respectively, but the isothermal expansion

process 1-2 and the isothermal compression process 3-4 are most difficult to achieve.

Because of these difficulties, a steady-flow Carnot gas cycle is not practical.

The Carnot gas cycle could also be achieved in a cylinder-piston apparatus (a

reciprocating engine) as shown in Fig. The Carnot cycle on the p-v diagram is as

shown in Fig, in which processes 1-2 and 3-4 are isothermal while processes 2-3

and 4-1 are isentropic. We know that the Carnot cycle efficiency is given by the

expression.



Reversed Carnot Cycle:

Reversed Carnot cycle is shown in Fig.. It consists of the following processes.

Process a-b: Absorption of heat by the working fluid from refrigerator at constant low

temperature T2 during isothermal expansion.

Process b-c: Isentropic compression of the working fluid with the aid of external work.

The temperature of the fluid rises from T2 to T1.



Process c-d: Isothermal compression of the working fluid during which heat is rejected

at constant high temperature T1.

Process d-a: Isentropic expansion of the working fluid. The temperature of the working

fluid falls from T1 to T2.

Practically, the reversed Carnot cycle cannot be used for refrigeration purpose as the

isentropic process requires very high speed operation, whereas the isothermal process

requires very low speed operation.

Thermodynamic Temperature Scale

To define a temperature scale that does not depend on the thermometric property of a

substance, Carnot principle can be used since the Carnot engine efficiency does not depend

on the working fluid. It depends on the temperatures of the reservoirs between which it

operates.



Consider the operation of three reversible engines 1, 2 and 3. The engine 1 absorbs energy Q1

as heat from the reservoir at T1, does work W1 and rejects energy Q2 as heat to the reservoir

at T2.

Let the engine 2 absorb energy Q2 as heat from the reservoir at T2 and does work W2 and

rejects energy Q3 as heat to the reservoir at T3.

The third reversible engine 3, absorbs energy Q1as heat from the reservoir at T1, does work

W3 and rejects energy Q3 as heat to the reservoir at T3.

1 = W1 / Q1 = 1- Q2/Q1 = f(T1,T2)

or, Q1/Q2 = F(T1,T2)

2 = 1- Q3/Q2 = f(T2,T3)

or, T2/T3 = F(T2,T3)

3 = 1- Q3/Q1 = f(T1,T3)

T1/T3 = F(T1,T3)

Then , Q1/Q2 = (Q1/Q3)/(Q2/Q3)

Or, F(T1,T2) = F(T1,T3) /F(T2,T3)

Since T3 does not appear on the left side, on the RHS also T3 should cancel out. This is

possible if the function F can be written as

F(T1, T2) = (T1) (T2)

(T1) (T2) = { (T1) (T3)} / { (T2) (T3)}

= (T1) (T2)



Therefore, (T2) = 1 / (T2)

Hence, Q1 / Q2 = F(T1,T2) = (T1)/ (T2)

Now, there are several functional relations that will satisfy this equation. For the

thermodynamic scale of temperature, Kelvin selected the relation

Q1/Q2 = T1/T2

That is, the ratio of energy absorbed to the energy rejected as heat by a reversible engine is

equal to the ratio of the temperatures of the source and the sink.

The equation can be used to determine the temperature of any reservoir by operating a

reversible engine between that reservoir and another easily reproducible reservoir and by

measuring efficiency (heat interactions). The temperature of easily reproducible thermal

reservoir can be arbitrarily assigned a numerical value (the reproducible reservoir can be at

triple point of water and the temperature value assigned 273.16 K).

The efficiency of a Carnot engine operating between two thermal reservoirs the temperatures

of which are measured on the thermodynamic temperature scale, is given by

1 = 1- Q2/Q1 = 1 – T2/T1

The efficiency of a Carnot engine, using an ideal gas as the working medium and the

temperature measured on the ideal gas temperature scale is also given by a similar

expression.

(COP)R = QL /(QH – QL) = TL / (TH – TL)

(COP)HP= QH /(QH – QL) = TH / (TH – TL)

The Carnot cycle uses only two thermal reservoirs – one at high temperature T1 and the other

at two temperature T2.



If the process undergone by the working fluid during the cycle is to be reversible, the heat

transfer must take place with no temperature difference, i.e. it should be isothermal.

The Carnot cycle consists of a reversible isothermal expansion from state 1 to 2, reversible

adiabatic expansion from state 2 to 3, a reversible isothermal compression from state 3 to 4

followed by a reversible adiabatic compression to state 1.

The thermal efficiency, is given by

= Net work done / Energy absorbed as heat

During processes 2-3 and 4-1, there is no heat interaction as they are adiabatic.

)/ln( 121

2

1

1

2

1

21 vvRTv

dvRTPdvQ

Similarly for the process 3-4,

)/ln( 342

4

3

2

4

3

43 vvRTv

dvRTPdvQ

Net heat interaction = Net work done

= RT1ln(v2/v1) + RT2ln(v4/v3)

= RT1ln(v2/v1) - RT2ln(v3/v4)

The processes 2-3 and 4-1 are reversible, adiabatic and hence

T1v2-1

= T2v3-1

Or, v2/v3 = (T2/T1)1/( -1)

And T2v4-1

= T1v1-1

Or, v1/v4 = (T2/T1)1/( -1)

v2/v3 = v1/v4 or v2/v1 = v3/v4

= {RT1ln(v2/v1) - RT2ln(v3/v4)} / RT1ln(v2/v1)

= (T1 – T2)/T1

= 1- T2/T1

The Carnot Principles

1. The efficiency of an irreversible heat engine is always less than the efficiency of a

reversible one operating between same two thermal reservoirs.



2. The efficiencies of all reversible heat engines operating between the same two

thermal reservoirs are the same.

Lets us assume it is possible for an engine I to have an efficiency greater than the efficiency

of a reversible heat engine R.

I > R

Let both the engines absorb same quantity of energy Q1. Let Q and Q2 represent the energy

rejected as heat by the engines R, and I respectively.

WI = Q1 - Q

WR= Q1 – Q2

I = WI / Q1 = (Q1 - Q)/Q1 = 1-Q/Q1

R = WR/Q1 = (Q1 - Q2)/Q1 = 1-Q2/Q1

Since I > R,

1-Q/Q1 > 1-Q2/Q1

or, Q < Q2



Therefore, WI (= Q1-Q) > WR (=Q1 – Q2)

Since the engine R is reversible, it can be made to execute in the reverse order. Then, it will

absorb energy Q2 from the reservoir at T2 and reject energy Q1 to the reservoir at T1 when

work WR is done on it.

If now engines I and R are combined, the net work delivered by the combined device is given

by

WI – WR = Q1 – Q – (Q1 – Q2) = Q2 – Q

The combined device absorbs energy (Q2 – Q) as heat from a single thermal reservoir and

delivers an equivalent amount of work, which violates the second law of thermodynamics.

Hence, R I

Carnot principle 2

Consider two reversible heat engines R1 and R2 , operating between the two given thermal

reservoirs at temperatures T1 and T2.

Let R1 > R2

Q1= energy absorbed as heat from the reservoir at T1 by the engines R1 and R2, separately.

Q = energy rejected by reversible engine R1 to the reservoir at T2

Q2 = energy rejected by reversible engine R2 to the reservoir at T2.

WR1 = Q1 - Q = work done by a reversible engine R1 .

WR2 = Q1 –Q2 = work done by a reversible engine R2

According to assumption, R1 > R2 Or, 1 – Q/Q1 > 1- Q2/Q1

Q1 –Q >Q1-Q2 or WR1 >WR2

WR1 – WR2 = (Q1 –Q) – (Q1- Q2) = Q2 – Q

Since the engine R2 is reversible, it can be made to execute the cycle in the reverse by

supplying WR2.



Since WR1 > WR2 the reversible engine R2 can be run as a heat pump by utilizing part of the

work delivered by R1. For the combined device,

WR1 – WR2 = Q2 – Q, by absorbing energy Q2 – Q from a single thermal reservoir which

violates the second law of thermodynamics.

Hence R1 > R2 is incorrect.

By similar arguments, if we assume that R2 > R1 then,

R1 R2

Therefore, based on these two equations,

R1 = R2

The efficiency of a reversible heat engine is also independent of the working fluid and

depends only on the temperatures of the reservoirs between which it operates.

Carnot theorem:

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

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

(or)

“no heat engine operating in a cycle process between two fixed temperatures can be more

efficient than a reversible engine while operating between the same temperatures limit”Let

two heat engines EA and EB operate between the given source at temperature t1 and the

given sink at temperature t2 as shown in fig.

Source t1

Q1A Q1B

WA WB

Let EA be any heat engine and EB be any reversible heat engine. We have to prove that the

efficiency of EB is more than that of EA. Let us assume that this is not true and ηA>ηB

Q1A=Q1B=Q1

ηA>ηB

WA/Q1A>WB/Q1B

WA>WB

Let EB be reversed. Since EB is a reversible heat engine, the magnitude of heat and work

transfer quantities will remain the same, but their directions will be reversed. Since WA>WB

Some part of WA (Equal to WB ) may be fed to drive the reversed heat engine B. Since

Q1A=Q1B=Q1, the heat discharged by B may be supplied to A. the source may, therefore,

EA EB

Q2B Q2A

Sink t2

Two cyclic heat engines EA and EB operating between the same source and sink, of which EB is reversible.



be eliminated. The net result is that A and B together constitute a heat engine which,

operating in a cycle, produces a net work WA-WB, while exchanging heat with a single

reservoir at t2. This violates the Kelvin-planck statement of the second law. Hence the

assumption that ηA>ηB is wrong.

Therefore ηB≥ηA.

Corollaries of Carnot theorem:

Corollary I

All reversible engines operating between the two given temperature limits have the efficiency

Corollary II

Efficiency of all the reversible engine depends only on the temperature limit of the reservoirs

and is independent of the nature of working fluid.

Clausius Inequality

For a Carnot cycle

Q1/Q2=T1/T2

Or Q1/T1-Q2/T2=0 for a reversible engine.

With the usual sign convention, that is, heat flow into a system taken as positive and heat

outflow of the system taken as negative

Q1/T1+Q2/T2=0 or Qi/Ti=0

For an irreversible engine absorbing Q1 amount of heat from a reservoir at T1 and rejecting

Q21 to a reservoir at T2, then

1-Q21/Q1 1-Q2/Q1

or 1-Q21/Q1 1-T2/T1

or Q21/Q1 T2/T1

or Q21/T2 Q1/T1

making use of the sign convention, we get

Q21/T2+Q1/T1 0

Or Q/T 0 for an irreversible engine



Replacement of a Reversible process by an equivalent process

Let us consider cyclic changes in a system other than heat engines. If the cycle can be split up

into a large number of heat engine cycles then the above observation can be made use of in

relating the heat interactions with the absolute temperatures.

Any reversible process can be approximated by a series of reversible, isothermal and

reversible, adiabatic processes.

Consider a reversible process 1-2. The same change of a state can be achieved by process 1-a

(reversible adiabatic process), isothermal process a-b-c and a reversible adiabatic process c-2.

The areas 1-a-b and b-c-2 are equal. From the first law

U2-U1=Q1-a-b-c-2-W1-a-b-c-2

Consider the cycle 1-a-b-c-2-b-1. The net work of the cycle is zero. Then

01221 bcba WWdW

or

211221 bbcba WWW

The heat interaction along the path 1-a-b-c-2 is

Q1-a-b-c-2=Q1-a+Qa-b-c+Qc-2=Qa-b-c



Since 1-a and c-2 are reversible adiabatic paths. Hence

U2-U1=Qa-b-c-W1-b-2

Application of the first law of the thermodynamics to the process 1-b-2 gives

U2-U1=Q1-b-2-W1-b-2

Comparing the two equations

Qa-b-c=Q1-b-2

The heat interaction along the reversible path 1-b-2 is equal to that along the isothermal path

a-b-c. Therefore a reversible process can be replaced by a series of reversible adiabatic and

reversible isothermal processes.

Clausius Inequality

A given cycle may be subdivided by drawing a family of reversible, adiabatic lines. Every

two adjacent adiabatic lines may be joined by two reversible isotherms.

The heat interaction along the reversible path is equal to the heat interaction along the

reversible isothermal path.

The work interaction along the reversible path is equal to the work interaction along the

reversible adiabatic and the reversible isothermal path.

That is,



Qa-b=Qa1-b1 and Qc-d=Qc1-d1

a1-b1-d1-c1 is a Carnot cycle.

The original reversible cycle thus is a split into a family of Carnot cycles. For every Carnot

cycle

0/TdQ . Therefore for the given reversible cycle,

0/TdQ

If the original cycle is irreversible

0/TdQ

so the generalized observation is

0/TdQ

Whenever a system undergoes a cyclic change, however complex the cycle may be( as long

as it involves heat and work interactions), the algebraic sum of all the heat interactions

divided by the absolute temperature at which heat interactions are taking place considered

over the entire cycle is less than or equal to zero (for a reversible cycle).

Clausius inequality:

Clausius inequality states that “when a system undergoes a cyclic process, the summation of

dQ/T around a closed cycle is less than or equal to zero

Consider an engine operating between two fixed temperature reservoirs TH and TL. Let dQs,

units of heat be supplied at temperature TH and dQR units of heat be rejected at temperature

TL during a cycle.

Thermal efficiency,η=dQs-dQR/dQs

Thermal efficiency of any reversible engine working on the same temperature limit is given

by

Thermal efficiency for reversible engine=TH-TL/TH

The efficiency of an actual engine cycle must be less than that of a reversible cycle.

Since no engine can be more efficient than that of a reversible engine

dQs-dQR/dQs≤TH-TL/TH

dQR/dQs≤ TL/TH

dQR/TL≤ dQs/TH

dQR/TL- dQs/TH ≤0

entire cycle ∫ dQ/T≤0



this equation is known as clausius inequality. It provides the criterion of the reversibility of a

cycle.

If ∫ dQ/T=0, The cycle is revesible.

If ∫ dQ/T<0, the cycle is irrerversible and possible.

If ∫ dQ/T>0, the cycle is impossible.

Since the cyclic integral ∫ dQ/T<0, the cycle violates the second law of thermodynamics. So,

it is impossible

∫ dQ/T=0,

Concept of entropy:

Entropy is an index of unavailability or degradation of energy. Heat always flow from hot

body to cold body and thus become lesser value available. This unavailability of energy is

measured by entropy. It is an important thermodynamics property of the working substance.

It increases with the addition of heat and decreases with its removal. It is the function of

quantity of heat with respect to temperature.

We are usually interested in change in entropy. The change in entropy for reversible process

is mathematically given by

Change in entropy, dS= change of heat transfer/absolute temperature=( dQ/T)

Unit of entropy is kj/K or j/K

For reversible adiabatic process change in entropy is zero.

Entropy

1. T

dQ has the same value irrespective of path as long as path is reversible

2. RT

dQ is an exact differential of some function which is identical as entropy

3. 2

1

2

1

12RT

dQdSSSS

4. RT

dQdS for reversible process only

Calculation of Entropy change

1. Entropy is a state function. The entropy change is determined by its initial and final

states only

2. In analyzing irreversible process, it is not necessary to make a direct analysis of

actual reversible process.



Substitute actual process by an imaginary reversible process. The entropy change for

imaginary reversible process is same as that of an irreversible process between given

final and initial states.

(a) Absorption of energy by a constant temperature reservoir

Energy can be added reversibly or irreversibly as heat or by performing work.

RT

dQS

Example:-

The contents of a large constant-temperature reservoir maintained at 500 K are continuously

stirred by a paddle wheel driven by an electric motor. Estimate the entropy change of the

reservoir if the paddle wheel is operated for two hours by a 250W motor.

Paddle wheel work converted into internal energy- an irreversible process. Imagine a

reversible process with identical energy addition

kJT

Q

T

dQS

R

6.0500

)3600(225.0

(b) Heating or cooling of matter

UQ for constant volume heating

HQ for constant pressure heating

2

11

2ln

T

T

ppT

TmC

T

dTCm

T

dQS

, for constant pressure

2

11

2ln

T

T

vvT

TmC

T

dTCm

T

dQS

, for constant volume process

Example: -

Calculate entropy change if 1kg of water at 300 C is heated to 80

0C at 1 bar pressure. The

specific heat of water is 4.2kJ/kg-K

Kkg

kJ

T

TCS p

.6415.0

30273

80273ln102.4ln 3

1

2

(c) Phase change at constant temperature and pressure

sf

sf

sfT

h

T

dQS



T

h

T

dQS

fg

fg

Example:-

Ice melts at 00C with latent heat of fusion= 339.92 kJ/kg. Water boils at atmospheric

pressure at 1000C with hfg= 2257 kJ/kg.

Kkg

kJSsf

.2261.1

15.273

92.334

Kkg

kJS fg

.0485.6

15.373

2257

(d) Adiabatic mixing

Example:-

A lump of steel of mass 30kg at 4270 C is dropped in 100kg oil at 27

0C.The specific heats

of steel and oil is 0.5kJ/kg-K and 3.0 kJ/kg-K respectively. Calculate entropy change of

steel, oil and universe.

T= final equilibrium temperature.

oilpsteelp TmCTmC

)300(3100)700(5.0300 TT

or T=319K

KkJS universe /6343.64226.187883.11)(

Tds relations

From the definition of entropy,

dQ = Tds

From the first law of thermodynamics,

dW = PdV

Therefore,

TdS = dU + PdV

Or, Tds = du + Pdv

This is known as the first Tds or, Gibbs equation.

KkJ

T

TmCS

oil

poil

/4226.18300

319ln3100

1

2ln)(

KkJ

T

TmC

T

dTmC

T

dQS

steel

p

p

steel

/7883.11700

319ln5.030

1

2ln)(

2

1

2

1



The second Tds equation is obtained by eliminating du from the above equation using

the definition of enthalpy.

h = u + Pv dh = du + vdP

Therefore, Tds = dh – vdP

The two equations can be rearranged as

ds = (du/T) + (Pdv/T)

ds = (dh/T) – (vdP/T)

Change of state for an ideal gas

If an ideal gas undergoes a change from P1, v1, T1 to P2, v2, T2 the change in entropy can be

calculated by devising a reversible path connecting the two given states.

Let us consider two paths by which a gas can be taken from the initial state, 1 to the final

state, 2.

The gas in state 1 is heated at constant pressure till the temperature T2 is attained and then it

is brought reversibly and isothermally to the final pressure P2.

Path 1-a: reversible, constant-pressure process.

Path a-2: reversible, isothermal path

s1-a = dq/T = Cp dT/T = Cp ln(T2/T1)

sa-2 = dq/T = (du+Pdv)/T = (Pdv)/T = Rln(v2/va)

(Since du = 0 for an isothermal process)

Since P2v2 = Pava = P1va

Or, v2/va = P1/P2

Or, sa-2 = -Rln(P2/P1)

Therefore, s = s1-a + sa-2

= Cp ln(T2/T1) – Rln(P2/P1)

Path 1-b-2: The gas initially in state 1 is heated at constant volume to the final temperature T2

and then it is reversibly and isothermally changed to the final pressure P2.

1-b: reversible, constant volume process

b-2: reversible, isothermal process

s1-b = Cv ln(T2/T1)

sb-2 =Rln(v2/v1)

or, s = Cv ln(T2/T1)+ Rln(v2/v1)

The above equation for s can also be deduced in the following manner:

ds = (dq/T)R = (du + Pdv)/T = (dh – vdP)/T

or,

Entropy- a property of a system

Let a thermodynamic system undergoes a change of state from 1 to 2 by reversible process 1-

A-2 and returns to its original state 1 by another reversible process 2-B-1 and completing a

cycle 1-2-1.



For this cyclic reversible process, the entropy equation is

∫rev dQ/T=0=2A∫1A dQ/T+1A∫2B

dQ/T

Not let us consider the cycle 1-2-1 is completed by another reversible process 2-C-1

1

A

B

P C

2

V

Then, ∫ dQ/T=0=2A∫1A dQ/T+

1c∫2c dQ/T

Subtracting equation

1B∫2B dQ/T=1C∫2C

dQ/T

We conclude that

the ∫ dQ/T is the

same for all

reversible

paths between

states 1 and 2. It is independent of the path and a function of end states only. Hence, entropy

is a property of a system.

T-s diagram:

We can represent a state of system by selecting properties i.e. temperature (T) and entropy (s)

as co-ordinates.

1

T1 1 (T1, S1)

2

T

S S1 ds

2 2

1 1

2 2

1 1

2

2 2

1 11

( )

ln ln

Similarly,

( )ln ln

v

v

p

C dTdu pdv Rdvs

T T v

T vC R

T v

T Pdh vdps C R

T T P



dQ=Tds

when a system undergoes a change of states from state 1 to state 2, then the area under the

process 1-2 represents heat transfer from state 1 to state 2

(Q1-2) rev=∫Tds

The quantity of heat transfer (Q1-2) rev represented by the area under the T-s diagram ( area 1-

2-s2-s1)

Calculations

Let us invoke the I law for a process namely dq=dw+du

Substitute for dq=Tds and dw = p dv Tds = pdv +du

For a constant volume process we have Tds = du… (1)

We have by definition

h = u+ pv

Differentiating

dh=du+pdv+vdp

dh= Tds +vdp For a constant pressure process Tds = dh…. (2)

For a perfect gas

perfect gas du=cvdT and dh=cpdT

Substitute for du in (1) and dh in (2)

for v=const Tds = cvdT or dT/ds v=const= T / cv

for p=const Tds = cpdT or dT/ds p=const= T / cp

1. Since cp > cv a constant pressure line on T-s plane will be

flatter than a constant volume line.

2. The both (isobars and isochores) will have +ve slopes and

curve upwards because the slope will be larger as the

temperature increases

Const V line, Const. P line

1-2 Isothermal expansion

1-3 Isothermal compression

1.4 Isentropic compression



1-5 Isentropic expansion

1-6 Isochoric heating

1-7 Isochoric cooling

1-8 Isobaric heating/expansion

1-9 Isobaric cooling/compression

Comparison Between

P-v and T-s Planes

A similar comparison can be made for processes going in

the other direction as well.

Note that n refers to general index in pvn=const.

For 1 < n < g the end point will lie between 2 and 5

For n > g the end point will lie between 5 and 7

Note: All work producing cycles will have a clockwise direction even on

the T-s plane

Consider the Clausius inequality

dQ /T= 0



In the cycle shown let A be a reversible process (R) and B an

Irreversible one (ir), such that 1A2B1 is an irreversible cycle.

Entropy of ideal gas:

Consider an ideal gas heated from state 1 to state 2 and thus temperature is increased from T1

to T2. During the heating process, there should be some change in entropy on the gas. If we

consider a change in heat transfer dQ to the gas at an absolute temperature T, then change in

entropy is given by ds= dQ/T

Expression for change in entropy of ideal gas:

i) In terms of temperature and volume.

ii) In terms of pressure and temperature.

iii) In terms of pressure and volume by gas equation.

Application of change in entropy for different processes

a) Constant volume process

b) Constant pressure process

c) Constant temperature process

d) Reversible adiabatic or isentropic process

e) Polytropic process

W = cdv/ vn

w = (P1v1- P2v2)/ (n-1)

du = dq – dw

u2 – u1 = q - (P1v1- P2v2)/(n-1)

u2 – u1 = Cv (T2 – T1) = q – w

q = R (T2 – T1)/ ( -1) + (P1v1- P2v2)/ (n-1)

= R (T1 – T2) {1/ (n-1) – 1/ ( -1)}

= (P1v1- P2v2)/ (n-1) {( -n)/ ( -1)}

=w. {( -n)/ ( -1)}

Problem: Air (ideal gas with = 1.4) at 1 bar and 300K is compressed till the final volume is

one-sixteenth of the original volume, following a polytropic process Pv1.25

= const. Calculate

(a) the final pressure and temperature of the air, (b) the work done and (c) the energy

transferred as heat per mole of the air.

Solution: (a) P1v11.25

= P2v21.25

P2 = P1(v1/v2)1.25

= 1(16)1.25

= 32 bar

T2 = (T1P2v2)/(P1v1) = (300 x 32 x 1)/(1x16)

= 600K

(b) w = (P1v1- P2v2)/(n-1)

= Ru (T1 – T2)/(n-1)

= 8.314 (300 – 600)/ (1.25-1) = -9.977 kJ/mol



(c) q = w.{ ( -n)/( -1)}

= -9.977 (1.4 – 1.25)/(1.4-1)

= -3.742 kJ/mol

Unresisted or Free expansion

In an irreversible process, w Pdv

Vessel A: Filled with fluid at pressure

Vessel B: Evacuated/low pressure fluid

Valve is opened: Fluid in A expands and fills

both vessels A and B. This is known as unresisted expansion or free expansion.

No work is done on or by the fluid.

No heat flows (Joule‟s experiment) from the

boundaries as they are insulated.

U2 = U1 (U = UA + UB)

Problem: A rigid and insulated container of 2m3 capacity is divided into two equal

compartments by a membrane. One compartment contains helium at 200kPa and 127oC while

the second compartment contains nitrogen at 400kPa and 227oC. The membrane is punctured

and the gases are allowed to mix. Determine the temperature and pressure after equilibrium

has been established. Consider helium and nitrogen as perfect gases with their Cv as 3R/2 and

5R/2 respectively.

Solution: Considering the gases contained in both the compartments as the system, W= 0

and Q = 0. Therefore, U = 0 (U2 = U1)

Amount of helium = NHe = PAVA/RuTA

= 200 x 103 x 1/(8.314 x400)

= 60.14 mol.

Amount of nitrogen = NN2 = PBVB/RuTB

= 400 x 103 x 1/(8.314x500)

= 96.22 mol.



Let Tf be the final temperature after equilibrium has been established. Then,

[NCv(Tf-400)]He + [NCv(Tf-500)]N2 = 0

Ru[60.14(Tf-400)3 + 96.22(Tf-500)5 ] /2 = 0

Or, Tf = 472.73 K

The final pressure of the mixture can be obtained by applying the equation of state:

PfVf = (NHe + NN2)Ru Tf

2Pf = (60.14 + 96.22) 8.314 (472.73)

or, Pf = 307.27 kPa

Principle of increase of entropy:

Application of entropy principle:

Transfer of heat through a finite temperature difference

Mixing of two fluids

Maximum work obtainable from two finite bodies at temperature TH and TL

Absolute entropy

The entropy of a pure crystalline substance at absolute zero temperature is zero. This is the

statement of third law of thermodynamics. This law provides an absolute reference point is

called as absolute entropy.

It is very much useful in thermodynamic analysis of chemical reactions.

Consider a system being at state 1

The absolute entropy, ∆S=S1-So

S1- Entropy of system at state 1

So- entropy of system at standard state denoted „o‟

Solved problems

Availability

a) The first law of thermodynamics does note deal the same grade of energy and

its feasibility but it says that all the energy absorbed is converted into work

output without losses.

Η1=work output/heat supplied=W/Q

[Work is high grade energy and heat is a low

η1=100% where Q=W]

b) The second law of thermodynamics says that all the energy absorbed as heat

by an engine cannot be converted into work. Some part of the energy must be

rejected to sink.

Q1=W+Q2 [W- useful work, Q2-losses]

Losses occur due to friction called irreversibility.

ηII=100%

Available energy and unavailable energy

The second law of thermodynamics tells us that it is not possible to convert all the heat

absorbed by a system into work.

Suppose a certain quantity of energy Q as heat can be received from a body at temperature T.



The maximum work can be obtained by operating a Carnot engine (reversible engine) using

the body at T as the source and the ambient atmosphere at T0 as the sink.

Where s is the entropy of the body supplying the energy as heat.

The Carnot cycle and the available energy is shown in figure.The area 1-2-3-4 represents the

available energy.

The shaded area 4-3-B-A represents the energy, which is discarded to the ambient

atmosphere, and this quantity of energy cannot be converted into work and is called

Unavailable energy.

Suppose a finite body is used as a source. Let a large number of differential Carnot engines

be used with the given body as the source.

If the initial and final temperatures of the

source are T1 and T2 respectively, the total work done or the available energy is given by

||

1

0

00

2

1

2

1

sTQorW

T

dQTQ

T

TdQdQW

T

T

T

T

Loss in Available Energy

Suppose a certain quantity of energy Q is transferred from a body at constant temperature T1

to another body at constant temperature T2 (T2<T1).

Initial available energy, with the body at T1,

T

TQ 01

Final available energy, with the body at T2,

2

01T

TQ

Loss in available energy

||1 00 sTQ

T

TQQW

T

TdQdQdW 01



unisTT

Q

T

QT

T

TQ

T

TQ 0

12

0

2

0

1

0 11

Where suni is the change in the entropy of the universe.

Availability Function

The availability of a given system is defined as the maximum useful work that can be

obtained in a process in which the system comes to equilibrium with the surroundings or

attains the dead state.

(a) Availability Function for Non-Flow process:-

Let P0 be the ambient pressure, V1 and V0 be the initial and final volumes of the system

respectively.

If in a process, the system comes into equilibrium with the surroundings, the work done in

pushing back the ambient atmosphere is P0(V0-V1).

Availability= Wuseful=Wmax-P0(V0-V1)

Consider a system which interacts with the ambient at T0. Then,

Wmax=(U1-U0)-T0(S1-S0)

Availability= Wuseful=Wmax-P0(V0-V1)

= ( U1-T0 S1) - ( U0-T0 S0)- P0(V0-V1)

= (U1+ P0V1-T0 S1)- ( U0+P0V0-T0 S0)

= 1- 0

Where =U+P0V-T0S is called the availability function for the non-flow process. Thus, the

availability: 1- 0

If a system undergoes a change of state from the initial state 1 (where the availability is ( 1-

0) to the final state 2 (where the availability is ( 2- 0), the change in the availability or the

change in maximum useful work associated with the process, is 1- 2.

(b) Availability Function for Flow process:-

The maximum power that can be obtained in a steady flow process while the control volume

exchanges energy as heat with the ambient at T0, is given by:

)()(

)()(

001101(max)

01001(max)

STHSTHW

SSTHHW

sh

sh

Sometimes the availability for a flow process is written as:

STHBwhere

BBWuseful

0

01

,

Which is called the Darrieus Function.

Second Law Efficiency

The second law efficiency ( 2) of a process,



2= Change in the available energy of the system

------------------------------------------------------

Change in the available energy of the source

(a) Compressors and Pumps:-

Change in the availability of the system is given by:

)()( 1201212 ssThhBBWrev

Where T0 is the ambient temperature

The second law efficiency of a compressor or pump is given by,

W

ssThh

W

WrevPC

)()( 12012/2

(b) Turbines and Expanders:-

The change in the available energy of the system=W

The change in the available energy of the source= Wrev=B1-B2

The second law efficiency of the turbine 2T/E is given by,

Work Potential Associated with Internal Energy

The total useful work delivered as the system undergoes a reversible process from the given

state to the dead state (that is when a system is in thermodynamic equilibrium with the

environment), which is Work potential by definition.

Work Potential = Wuseful= Wmax- P0(V0-V1)

= (U1-T0 S1)- ( U0-T0 S0)- P0(V0-V1)

21

/2BB

W

W

W

rev

ET



= (U1+ P0V1-T0 S1)- ( U0+P0V0-T0 S0)

= 1- 0

The work potential of internal energy (or a closed system) is either positive or zero. It is never

negative.

Work Potential Associated with Enthalpy, h

The work potential associated with enthalpy is simply the sum of the energies of its

components.

)()(

)()(

001101(max)

01001(max)

STHSTHW

SSTHHW

sh

sh

The useful work potential of Enthalpy can be expressed on a unit mass basis as:

)()( 01001 ssThhwsh

Here h0 and s0 are the enthalpy and entropy of the fluid at the dead state. The work potential

of enthalpy can be negative at sub atmospheric pressures.

Availability

A thermodynamic system undergoes a charge of state through a reversible process until it

comes to equilibrium with the atmosphere. Then, the work done by the system on the

atmosphere is a maximum. The maximum work obtained is called as the availability of

system.

The system transfers heat to the atm..at pressure p0 and temperature T0. here, the equilibrium

will exist between the system and atm (p0=1.01325 bar, T0=25°C)

It is denoted by Φ for closed system and χ (or) B for open system.

Availability, Φ=Q-To∆S for any type of closed system

Availability, χ (or) B=(h1-h2)-To(S1-S2) for any type of open system

[Wmax= (or) Φ (or) χ]

Irreversibility (I)

It is defined as the difference between maximum works to the actual work obtained in a

process.

I=Wmax-Wact

=To∆S (energy loss)

Second law efficiency:

The ratio between the change in the available energy the system and the change in the

available energy of the source.

Otherwise. It may also be defined as ratio between the availability of output to the availability

of input.

ηII=Aout/Ain

Availability analysis to closed system:

Case: A constant volume process

Case: B constant pressure process

Case: C constant temperature or isothermal process



Case: D polytropic process

Availability analysis to open system:

Case: A turbine

Case: B compressor/pump

Case: C heat exchanger

Case: D throttling process

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