Entropy

1. Entropy
Dr. Rohit Singh Lather

2. Introduction
• It has already been proved that for a reversible cycle working between source and sink
temperatures, the heat supplied and rejected to the engine is related with source and sink
temperature as
𝑄1
𝑇1
=
𝑄2
𝑇2
'(
)(
+
'+
)+
= 0
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.

3. Replacement of a Reversible process by an equivalent process
Pressure
Volume
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.
The same change of a state can be achieved by process
Reversible Process 1-2
1-a
Reversible adiabatic
process
a-b-c
Isothermal process
c-2
Reversible adiabatic
process
The areas 1-a-b and b-c-2 are equal
U2-U1=Q1-a-b-c-2-W1-a-b-c-2

4. • 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
• 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
∮ 𝑑𝑊 = W1-a-b-c-2 + W2-b-1 = 0
W1-a-b-c-2 = - W2-b-1 = W1-b-2

5. .Hence,
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.
heat transferred in the process 1 – 2 is equal to heat
transferred in the isothermal process a – b
• So any reversible path can be replaced by a zig-zag path between the same end states
• The zig-zag path consists of a reversible adiabatic followed by a reversible isotherm and then by
a reversible adiabatic in such a way that will satisfy above equation i.e., the heat transferred
during the isothermal process is the same as that transferred during the original process
U2 - U1 = Qa-b-c - W1-b-2

6. Introduction to Clausius Theorem and Clausius Inequality
• When a reversible engine uses more than two reservoirs, the third or higher numbered reservoirs
will not be equal in temperature to the original two
• Consideration of expression for efficiency of the engine indicates that for maximum efficiency,
all the heat transfer should take place at maximum or minimum reservoir temperatures
- Any intermediate reservoir used will, therefore, lower the efficiency of the heat engine
• Practical engine cycles often involve continuous changes of temperature during heat transfer
• A relationship among processes in which these sort of changes occur is necessary
• The ideal approach to a cycle in which temperature continually changes is to consider the system
to be in communication with a large number of reservoirs in procession
• Each reservoir is considered to have a temperature differing by a small amount from the
previous one

7. • 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
Reversible Cycle
Adiabatic Lines
Pressure
Volume
Isothermal LinesThe original reversible cycle thus is a split into a family of
Carnot cycles
a1-b1-d1-c1
is a Carnot cycle
Qa-b = Qa1-b1 and
Qc-d = Qc1-d1
Clausius Theorem

8. For a1-b1-d1-c1 Carnot cycle
- aQ1 is added at contact temperature T1 and aQ2 is rejected at temperature T2
So we can write
𝑄1
𝑇1
=
𝑄2
𝑇2
'(
)(
+
'+
)+
= 0
'(
)(
+
'+
)+
+ ….......= 0
'/
)/
+
'0
)0
= 0
∑
'
)2 = 0
∮
3'
)
= 0 Clausius Theorem
• The work interaction along the reversible path is equal to the work interaction along the
reversible adiabatic and the reversible isothermal path

9. Temperature Volume
Let us Consider cycle A-B-C-D
- AB is a general process, either reversible or irreversible
- Let the cycle be divided into number of elementary cycles as shown
𝜂 = 1 -
𝜹𝑸 𝟐
𝜹𝑸
𝜹𝑸 𝒊𝒔 𝒉𝒆𝒂𝒕 𝒔𝒖𝒑𝒑𝒍𝒊𝒆𝒅 𝒂𝒕 𝑻
𝜹𝑸2 𝒊𝒔 𝒉𝒆𝒂𝒕 𝒓𝒆𝒋𝒆𝒄𝒕𝒆𝒅 𝒂𝒕 𝑻 𝟐
𝜂 ≤ 𝜂rev
The efficiency of a general cycle is always less
than the efficiency of a reversible cycle
Source: P K Nag, “Engineering Thermodynamics", Mc Graw Hill Eductaion,5th Edition, 2016
1 -
𝜹𝑸 𝟐
𝜹𝑸
≤ (1 -
𝜹𝑸 𝟐
𝜹𝑸
)rev 𝜹𝑸2
𝜹𝑸
Clausius Inequality

10. (
𝜹𝑸
𝜹𝑸 𝟐
)rev =
𝑻
𝑻 𝟐
𝜹𝑸
𝜹𝑸 𝟐
≤
𝑻
𝑻 𝟐
𝜹𝑸
𝑻
≤
𝜹𝑸 𝟐
𝑻 𝟐
𝒅𝑺 =
𝜹𝑸
𝑻
=
𝜹𝑸 𝟐
𝑻 𝟐
For any process A-B
For reversible process A-B
𝜹𝑸
𝑻
≤ 𝑑𝑆 For any process AB
I
𝛿𝑄
𝑇
≤ I 𝑑𝑆
I
𝜹𝑸
𝑻
≤ 𝟎
dS = 0 for a cyclic process
Clausius Inequality
For cyclic process
We know that
Replacing for
reversible
-
𝜹𝑸 𝟐
𝜹𝑸
≤ - (
𝜹𝑸 𝟐
𝜹𝑸
)rev
𝜹𝑸 𝟐
𝜹𝑸
≥ (
𝜹𝑸 𝟐
𝜹𝑸
)rev
𝜹𝑸
𝜹𝑸 𝟐
≤ (
𝜹𝑸
𝜹𝑸 𝟐
)rev

11. 𝜕Q
Reservoir at Variable Temperature T
Reversible
Heat
Engine
Reservoir at Temperature Tres
Heat rejected
𝜕Q’
𝜕W’
𝜕W
engine delivers a
small amount of work
Thermal reservoir at Tres which delivers a small amount
of heat δQ′ to a reversible cyclic engine
reservoir itself delivers a different
small amount of work δW to the
surroundings
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA

12. The first law in differential form for the combined system is
δQ is internal and so does not cross the boundary of the combined system and is not present in our first law formulation
Rearrange
Replace δQ’
Let this configuration undergo a thermodynamic cycle
Apply the Kelvin-Planck form of the second law of thermodynamicsW + 𝑊O ≤ 0
We cannot convert all the heat to work, but we can convert all the work to heat
Q'O
Q'
=
)RST
)
Q'O
)RST
=
Q'
)
dE = (𝛿𝑄′) − (𝛿𝑊 + 𝛿𝑊′)
(𝛿𝑊 + 𝛿𝑊′) = (𝛿𝑄′) − dE
(𝛿𝑊 + 𝛿𝑊′) = (Tres
Q'
)
) − dE
(∮ 𝛿𝑊 + ∮ 𝛿𝑊′= ∮Tres
Q'
)
− ∮dE
W + 𝑊′= Tres ∮
Q'
)
Tres ∮
Q'
)
≤ 0
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA

13. This is known as Clausius Inequality
And since Tres > 0, we can divide above equation by it, without changing the sense of the inequality
to get a mathematical representation of the second law of thermodynamics
This is known as Clausius Inequality for a reversible process
This is known as Clausius Inequality for a irreversible process
∮
Q'
)
≤ 0
I
𝛿𝑄
𝑇
≤ 0
I
𝛿𝑄
𝑇
= 0
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition
For every Carnot cycleI
𝛿𝑄
𝑇
= 0

14. 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)
Source: P K Nag, “Engineering Thermodynamics", Mc Graw Hill Eductaion,5th Edition, 2016
• The integral symbol with a circle in the middle is used to indicate that the integration is to be
performed over the entire cycle
• Any heat transfer to or from a system can be considered to consist of differential amounts of
heat transfer
• Then the cyclic integral of dQ/T can be viewed as the sum of all these differential amounts of
heat transfer divided by the temperature at the boundary

15. Entropy Path-Independent Thermodynamic Property
Sketch of P − V diagram for various
combinations of processes forming cyclic
integrals
• Consider starting from 1, proceeding on path A to 2,
and returning to 1 via path B.
• The cyclic integral δQ/T =0 decomposes to
• Perform the same exercise going from 1 to 2 on path
A and returning on path C, yielding
Volume
Pressure
1
2
C
B
A
A, B, & C are arbitrary processes
between state 1 and state 2
(∫
Q'
)
+
(
)A + (∫
Q'
)
)
(
+ B = 0
(∫
Q'
)
+
(
)A + (∫
Q'
)
)
(
+ C = 0
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA

16. We can reverse direction and recover the same result
• Since paths B and C are different and arbitrary, but ∫ δQ/T is the same on either path, the
integral must be path-independent.
• It therefore defines a thermodynamic property of the system.
• We define that property as entropy, S, an extensive thermodynamic property
(∫
Q'
)
)
(
+ B - (∫
Q'
)
)
(
+ C = 0
(∫
Q'
)
)
(
+ B = (∫
Q'
)
)
(
+ C
(∫
Q'
)
)
+
( B = (∫
Q'
)
)
+
( C
S2 – S1 = ∫
Q'
)
+
(
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA

17. scale by the constant mass m to get the
corresponding intensive variable s = S/m
Integrating
Differential Form
This is the heat transfer equivalent to
s2 – s1= ∫
Q'
)
+
(
𝛿𝑠 =
𝛿𝑞
𝑇
𝛿𝑞 = 𝑇. 𝛿𝑠
] 𝛿𝑞
+
(
= ] 𝑇. 𝛿𝑠
+
(
1q2 = ∫ 𝑇. 𝛿𝑠
+
(
1w2= ∫ 𝑇. 𝛿𝑠
+
(
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA

18. The entropy change between two specific
states is the same whether the process is
reversible or irreversible
Isentropic = Adiabatic + Reversible
The heat transfer for a process from 1 to 2 is given
by the area under the curve in the T − s plane
• If a process lies on a so-called Isentrope: a line on
which entropy s is constant, then, 1q2 = 0; thus, the
process is adiabatic
• Above equation only applies for a reversible process
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

19. Example : For a particular power plant, the heat added and rejected both occur at constant
temperature and no other processes experience any heat transfer. The heat is added in the amount
of 3150 kJ at 440oC and is rejected in the amount of 1950 kJ at 20oC. Is the Clausius inequality
satisfied and is the cycle reversible or irreversible?
• The Clausius inequality is satisfied
• Since the inequality is less than zero, the cycle has at least one irreversible process and the cycle
is irreversible
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

20. Example: For a particular power plant, the heat added and rejected both occur at constant
temperature; no other processes experience any heat transfer. The heat is added in the amount of
3150 kJ at 440oC and is rejected in the amount of 1294.46 kJ at 20oC. Is the Clausius inequality
satisfied and is the cycle reversible or irreversible?
• The Clausius inequality is satisfied
• Since the cyclic integral is equal to zero, the cycle is made of reversible processes
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

21. Second Law in terms of Entropy – Entropy Change in an Irreversible Process
Consider the cycle in the T − S diagram as shown
- We start at 1, and proceed to 2 along path I, which
represents an irreversible process
- We return from 2 to 1 along path R, which represents a
reversible process
Entropy
Temperature
1
2
I
R
I
𝛿𝑄
𝑇
≤ 0
S2 – S1 = ∫
Q'
)
+
(
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
Statement of second law valid for reversible and irreversible heat transfer
Definition of Entropy provided the heat transfers are reversible
S2 – S1 = ∫
Q'
)
+
(
For a reversible process

22. S1 – S2 = ∫
Q'
)
(
+
Since the process is reversible, we reverse and get
Now, apply the second law
0 ≥ (∫
Q'
)
+
(
)I + (∫
Q'
)
)
(
+ R
Now, substitute, to eliminate the integral along R to get 0 ≥ (∫
Q'
)
+
(
)I + S1 –S2
S2 – S1 ≥ (∫
Q'
)
+
(
)I
More generally, we can write the second law of thermodynamics as, S2 – S1 ≥ ∫
Q'
)
+
(
If 2 → 1 is reversible, the equality holds; if 1 → 2 is irreversible, the inequality holds

23. • For an isolated system, there can be no heat transfer interactions and 𝜕𝑄 = 0,
So S2 − S1 ≥ 0,
- This implies 2 occurs later in time than 1
- Thus, for isolated systems, the entropy increases as time moves forward
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
Isolated system
Example: Two large thermal reservoirs, one at TA and the other at TB, exchange a finite amount of
heat Q with no accompanying work exchange. The reservoirs are otherwise isolated and thus form
their own universe, when considered as a combined system. Consider the implications for entropy and
the second law.
A
TA
Heat transfer from A to B
B
TB
Q

24. • Assume for now that positive Q leaves A and enters B
• Both A and B are so massive that the respective loss and gain of thermal energy does not alter
their respective temperatures. Consider the entropy changes for each system:
Now, our universe is the combination of A and B, so the entropy change of the universe is found by
adding the entropy changes of the components of the universe
SA2 – SA1 = ∫
Q'
)
+
(
=
(
)^
∮ 𝛿𝑄 = −
'
)^
+
(
SB2 – SB1 = ∫
Q'
)
+
(
=
(
)_
∮ 𝛿𝑄 =
'
)_
+
(
(SA2 + SB2) – (SA1 + SB1) =
'
)_
+
'
)^
= SU2 = SU1
Source: Joseph M. Powers, “Lecture notes on thermodynamics", University of Notre Dame, Notre Dame, Indiana, USA
With the universe entropy SU as SU = SA + SB , we get SU2 – SU1 = Q (
(
)_
-
(
)^
)

25. The universe is isolated, so the second law holds that SU2 − SU1 ≥ 0; thus,
Now, we have assumed Q > 0; therefore, we can divide by Q without changing the sense of the
inequality:
• We have thus confirmed that our mathematical formulation of the second law in terms of entropy
yields a result consistent with the Clausius statement of the second law.
• We must have TA ≥ TB in order to transfer positive heat Q from A to B
Q (
(
)_
-
(
)^
) ≥ 0
(
)_
-
(
)^
≥ 0
)^` )_
)^)_
≥ 0
𝑇 𝐴 − 𝑇 𝐵 ≥ 0
𝑇 𝐴 ≥ 𝑇 𝐵
Since TA > 0 and TB > 0, we can multiply both sides by TATB without changing the sense of the
inequality to get

26. 6-3
The entropy change of an isolated system is the sum of the entropy changes of its components, and
is never less than zero
Entropy Change of an Isolated System
“The entropy of an isolated system either increases or, in the limit, remains constant.”
The principle of entropy increase: The entropy of an isolated system can never decrease. It always
increases with every irreversible process and remains constant, only when the process is reversible.
dSsys =
`Q'
)TcT
dSsur =
Q'
)TdR
dSiso = dSsys + dSsur
= 𝛿𝑄 [
(
)TdR
-
(
)TcT
] > 0
The entropy of an isolated system either increases if it undergoes an irreversible process
System
Tsys
Surroundings
Tsur
SQ
If, Tsys = Tsur, the process would occur reversibly and entropy will remain constant
Source: D. S. Kumar, “Engineering Thermodynamics", S.K. Kataria & Sons, 2nd Edition

27. • Second law of thermodynamics while applied to a process give rise to a property, called entropy
• Absolute entropy (S) or entropy is a measure of energy dispersion in a system
• Following are basic features of entropy:
- Entropy transfer is associated with heat transfer
- Entropy is proportional to mass flow in an open system
- Entropy does not transfer with work
Entropy
Heat to Work
Work to Work
Work to Heat
Entropy Transfer
No Entropy Transfer
No Entropy Transfer; Entropy Generates

28. • If no irreversibilities occur within the system as well as the reversible cyclic device,
- Then the cycle undergone by the combined system will be internally reversible
- As such, it can be reversed
• In the reversed cycle case, all the quantities will have the same magnitude but the opposite
sign
- Therefore, the work WC, which could not be a positive quantity in the regular case, cannot be a
negative quantity in the reversed case
- Then it follows that WC,int rev = 0 since it cannot be a positive or negative quantity, and therefore
for internally reversible cycles
• Thus we conclude that the equality in the Clausius inequality holds for totally or just internally reversible
cycles and the inequality for the irreversible ones.
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

29. • Entropy change of the system will have the same sign as the heat transfer in a reversible
process
• Processes can occur in a certain direction only, not in just any direction, such that Sgen ≥ 0
• Entropy is a non-conserved property, and there is no such thing as the conservation of entropy
principle
• The entropy of the universe is continuously increasing, as all the processes happening in the
universe are irreversible
• The performance of engineering systems is degraded by the presence of irreversibilities
• Entropy generation is a measure of the magnitudes of the irreversibilities present during that
process
• Entropy change is caused by heat transfer and irreversibilities
Some Remarks about Entropy
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

30. • Heat transfer to a system increases the entropy; heat transfer from a system decreases it
• The effect of irreversibilities is always to increase the entropy
- In fact, a process in which the heat transfer is out of the system may be so irreversible that
the actual entropy change is positive (Friction is one source of irreversibilities in a system)
• The entropy change during a process is obtained by integrating the dS equation over the process
• The inequality is to remind us that the entropy change of a system during an irreversible process
is always greater than ∮ 𝛿𝑄/𝑇
+
(
, called the entropy transfer
- That is, some entropy is generated or created during an irreversible process
- Generation is due entirely to the presence of irreversibilities.
• The entropy generated during a process is called entropy generation and is denoted as Sgen
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

31. • We can remove the inequality by noting the following
- Sgen is always a positive quantity or zero
- Its value depends upon the process and thus it is not a property
- Sgen is zero for an internally reversible process
• The integral ∮ 𝛿𝑄/𝑇
+
(
is performed by applying the first law to the process to obtain the heat
transfer as a function of the temperature. The integration is not easy to perform, in general
∆𝑆 𝑠𝑦𝑠 = 𝑆2 − 𝑆1 = I
𝛿𝑄 𝑛𝑒𝑡
𝑇
+
+
(
𝑆 𝑔𝑒𝑛
Source: Yunus A. Cengel and Michael A. Boles Thermodynamics: An Engineering Approach, McGraw Hill, 8th Edition

32. 6-9
Entropy
Temperature
1
2
dA = TdS = 𝜕Q
Internally Reversible Process
Heat Transfer as the Area under a T-S Curve
For the reversible process, the equation for dS implies that or the incremental heat transfer in a process is the
product of the temperature and the differential of the entropy, the differential area under the process curve
plotted on the T-S diagram

33. 6-9
Heat Transfer for a Internally Reversible Process
On a T-S diagram, the area under the process curve represents the heat transfer for internally reversible
processes
Entropy
Temperature
1
2
dA = TdS = 𝜕Q
Internally Reversible Process
Adiabatic Process
Temperature
Entropy
Isothermal Process
Temperature
Entropy
𝛿𝑄𝑟𝑒𝑣 = T∫ 𝑑𝑆 = 𝑇
+
(
(S2 − S1)

34. Volume
Pressure
Reversible Adiabatic Processes
Heat exchange with a single
reservoir in the process AB
Violates Kelvin Plank Statement
• For a reversible adiabatic process, heat interactions are zero so
change in entropy is zero
• Two adiabatic reversible paths never intersect each other
• Entropy is a property and a point function
• Through one point, there can only pass one reversible
adiabatic line
• If, two adiabatic reversible paths intersect each other, then
entropy will have two values at that point, this violates Kalvin
Planks statement (engine which exchanges heat with a single
reservoir)

35. 6-11
For adiabatic steady-flow devices, the vertical distance is ∆h on an h-s diagram, and the horizontal
distance is ∆s
h-s Diagram for Adiabatic Steady Flow Devices
Entropy
Enthalpy(h)
1
2
∆h
∆s
measure of work
measure of irreversibilities

36. • Temperature-entropy and enthalpy-entropy diagrams for water
• The h-s diagram, called the Mollier diagram, is a useful aid in solving steam power plant problems.
Constant Temperature
Lines
Steam Quality
Lines
Super-heated Vapor Steam Region
Mixed Region
of vapor and liquid
• The Mollier diagram shows only two
regions, the mixed region of vapor and
liquid and the super-heated vapor steam
region.
• The two regions are separated by the
downward sloping saturation line, where
steam quality is equal to 1

37. Constant Volume and Constant Pressure Lines on T-S Diagram
3)
3o
=
)
pq
3)
3o
=
)
pr
ds= Cv
3)
)
ds= Cp
3)
)
As Cp > Cv, Constant volume lines on T-s diagram is steeper than constant pressure line, working
between the same temperature limits
3)
3o
𝛼 𝑇 For both Constant volume and constant pressure
Constant volume
Constant pressure
Slope of constant volume line on T-S diagram
Slope of constant pressure line on T-S diagram

38. The level of molecular disorder (entropy) of a substance increases as it melts and evaporates
Level of Molecular Disorder (Entropy)
During a heat transfer process, the net disorder (entropy) increases
Net disorder decreases during heat transfer
the increase in the disorder of the cold body more than offsets the decrease in the disorder in the hot body

39. • The third law of thermodynamics states that “The entropy of a pure crystalline substance at
absolute zero temperature is zero.”
• This law provides an absolute reference point for the determination of entropy
• The entropy determined relative to this point is called absolute entropy
Third Law of Thermodynamics