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1
Entropy:
A measure of Disorder
7CHAPTER
7.1. Entropy and the
Clausius Inequality
2
3
Introduction
The second law often leads to
expressions that involve inequalities.
,
,
th rev
th
th rev
η
η
η
<

=
,th th revη η≤
4
The Inequality of Clausius
 The inequality of Clausius is a consequence of
the second law of thermodynamics.
 Q is the heat transfer to or from the system.
 T is the absolute temperature at the boundary.
 The symbol is the cyclic integral
0≤∫ T
Qδ
∫
5
The cyclic integral
 The cyclic integral
indicates that the integral
should be performed
over the entire cycle and
over all parts of the
boundary.
Q
T
δ
∫
2 3 4 1
1 2 3 4
Q Q Q Q
T T T T
δ δ δ δ
= + + +∫ ∫ ∫ ∫
6
The cyclic integral
0 0H L
H L
Q Q
T T
= + − +
Q
T
δ
∫
2 3 4 1
1 2 3 4
Q Q Q Q
T T T T
δ δ δ δ
= + + +∫ ∫ ∫ ∫
H L
H L
Q Q
T T
= −
7
The cyclic integral of
Reversible Heat Engine
0 0H L
H L
Q Q
T T
= + − +
Q
T
δ
∫
2 3 4 1
1 2 3 4
Q Q Q Q
T T T T
δ δ δ δ
= + + +∫ ∫ ∫ ∫
H L
H L
Q Q
T T
= −
H H
L L
Q T
Q T
=
0=
Since
8
The cyclic integral of
Irreversible Heat Engine
irr revW W<
H L
H L
Q Q
T T
=
Q
T
δ
∫
H L
H L
Q Q
T T
= − We cannot use this
( ) ( )H L H Lirr rev
Q Q Q Q− < −
It is Irreversible
H L irr H L revQ Q Q Q− < −
L irr L revQ Q>
H Lirr
H L
Q Q
T T
− 0<
9
The cyclic integral of
Reversible Refrigeration
0 0L H
L H
Q Q
T T
= + + −
Q
T
δ
∫
2 3 4 1
1 2 3 4
Q Q Q Q
T T T T
δ δ δ δ
= + + +∫ ∫ ∫ ∫
L H
L H
Q Q
T T
= −
H H
L L
Q T
Q T
=
0=
Since
10
The cyclic integral of
Irreversible Refrigeration
irr revW W>
H L
H L
Q Q
T T
=
Q
T
δ
∫
H L
H L
Q Q
T T
=− +
We cannot use this
( ) ( )H L H Lirr rev
Q Q Q Q− > −
It is Irreversible
H irr L H rev LQ Q Q Q− > −
H irr H revQ Q>
H irr L
H L
Q Q
T T
− + 0<
11
IrreversibleReversible
< 00=Heat Engine
< 00=Refrigeration
0≤∫ T
Qδ
Q
T
δ
∫
Derivation of Clausius Inequality
12
All paths are arbitrary
0
Q
T
δ
=∫
Derivation of Entropy (Reversible Process)
2 2
1 1A C
Q Q
T T
δ δ   
=   
   
∫ ∫Subtracting gives
2 1
1 2
0
C B
Q Q
T T
δ δ   
= + =   
   
∫ ∫
For reversible cycle A-B
2 1
1 2
0
A B
Q Q
T T
δ δ   
= + =   
   
∫ ∫
For reversible cycle C-B
0
Q
T
δ
=∫
Q
the quantity is independent of the path and dependent on the end states only
T
δ
∴ ∫
Since paths A and C are arbitrary, it follows that the integral of δQ/T
has the same value for ANY reversible process between the two sates.
13
work & heat are dependent on path Path functions
Recall are independent of path
properties Point functions
and depend on state only





( )
is a thermodynamic property
we call it entropy S
δQ
T
⇒ ∫
Entropy (the unit)
S = entropy (kJ/K); s = specific entropy (kJ/kg K)
∫ 





=−





≡
2
1
12gintegratin
revrev T
Q
SS
T
Q
dS
δδ S2 – S1 depends on the end
states only and not on the path,
∴ it is same for any path
reversible or irreversible
Derivation of Entropy (Reversible Process)
14
Consider 2 cycles AB is reversible and CB is irreversible
2 1
1 2
for cycle A-B (reversible)
0
A B
Q Q Q
T T T
δ δ δ   
= + =   
   
∫ ∫ ∫
2 1
1 2
for path C-B (irreversible)
0
C B
Q Q Q
T T T
δ δ δ   
= + <   
   
∫ ∫ ∫
2 2
1 1
comparing gives
A C
Q Q
T T
δ δ



    
>   
   
∫ ∫
 
2 2 2
1 1 1
reversible it is a
property
but A C
A
δQ
dS dS
T
 
=== === 
 
∫ ∫ ∫
in general
δQ
dS
T
⇒ ≥
2 2
1 1C
C
δQ
dS
T
 
∴ >  
 
∫ ∫
2
2 1 1
or
δQ
S S
T
− ≥ ∫
equality for reversible
inequality for irreversible
Derivation of Entropy (Irreversible Process)
15
Example (7-1) Entropy change during
isothermal process.
Solution:
 This is simple problem.
 No irreversibilities occur within the system
boundaries during the heat transfer process.
A friction-less piston-cylinder device contains a liquid-
vapor mixture of water at 300 K. During a constant
pressure process, 750 kJ of heat is transferred to the
water. As a result, part of the liquid in the cylinder
vaporizes. Determine the entropy change of the water
during this process.
16
kkJ
k
kJ
T
Q
S
sys
/5.2
300
750
===∆
( )
2 2
1 1
1
rev
rev
Q Q
S Q
T T T
δ
δ
 
∆= = = 
 
∫ ∫
Q
S
T
∆ =
 We computed the entropy change for a
system using the RHS of the equation.
 But we can not get easy form each time.
 So, we need to know how to evaluate the
LHS which is path independent.
7.3. Heat Transfer as the
Area Under the T-s
Curve
17
18
Entropy change for different
substances (∆S = S2-S1)
 We need to find how to compute the
left hand side of the entropy balance
for the following substances:
1. Pure substance like water, R-134,
Ammonia etc..
2. Solids and liquids
3. Ideal gas
19
1- ∆S for Pure Substances
The entropy of a pure substance is determined from the
tables, just as for any other property
These values were tabulated after
conducting a tedious integration.
These values are given relative to
an arbitrary reference state.
)( 12 ssmS −=∆
Entropy change for a closed system
with mass m is given as expected:
For water its assigned zero at 0.01 C.
For R-134 it is assigned a zero at -40 C.
20
Property diagrams involving
entropy
 Recall the definition of entropy
 Property diagrams serves as great
visual aids in the thermodynamic
analysis of process.
 We have used P-v and T-v diagrams
extensively in conjunction with the
first law of thermodynamics.
T
Q
dS revδ
= TdSQrev =δ
In the second-law analysis, it is very helpful to plot the processes on T-s
and h-s diagrams for which one of the coordinates is entropy.
W PdVδ =
21
∫=
2
1
TdSQrev
This area has no meaning
for irreversible processes!
Thus
It can be done only for a reversible
process for which you know the
relationship between T and s during a
process. Let us see some of them.
22
Isothermal reversible process.
Entropy
Isothermal Process
Q
1 2
0 0T S T m s= ∆ = ∆
23
Adiabatic reversible process
 In this process Q =0, and
therefore the area under
the process path must be
zero.
 This process on a T-s
diagram is easily
recognized as a vertical-
line.
Entropy
Isentropic Process
1
2
Q=0
24
T-s Diagram for the Carnot
Cycle
Temperature
Entropy
1
Isentropic
compression
Qin
W=Qin-Qout
Qout
2
Isothermal
expansion
3
Isentropic
expansion
Isothermal
compression
4
25
Another important diagram is
the h-s Diagram
 This diagram is important in the
analysis of steady flow devices
such as turbines.
 In analyzing the steady flow of
steam through an adiabatic
turbine, for example,
 The vertical distance between
the inlet and the exit states (∆h)
is a measure of the work output
of the turbine,
 The horizontal distance (∆s) is a
measure of the irreversibilities
associated with the process.

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Ch07a entropy

  • 1. 1 Entropy: A measure of Disorder 7CHAPTER
  • 2. 7.1. Entropy and the Clausius Inequality 2
  • 3. 3 Introduction The second law often leads to expressions that involve inequalities. , , th rev th th rev η η η <  = ,th th revη η≤
  • 4. 4 The Inequality of Clausius  The inequality of Clausius is a consequence of the second law of thermodynamics.  Q is the heat transfer to or from the system.  T is the absolute temperature at the boundary.  The symbol is the cyclic integral 0≤∫ T Qδ ∫
  • 5. 5 The cyclic integral  The cyclic integral indicates that the integral should be performed over the entire cycle and over all parts of the boundary. Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫
  • 6. 6 The cyclic integral 0 0H L H L Q Q T T = + − + Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫ H L H L Q Q T T = −
  • 7. 7 The cyclic integral of Reversible Heat Engine 0 0H L H L Q Q T T = + − + Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫ H L H L Q Q T T = − H H L L Q T Q T = 0= Since
  • 8. 8 The cyclic integral of Irreversible Heat Engine irr revW W< H L H L Q Q T T = Q T δ ∫ H L H L Q Q T T = − We cannot use this ( ) ( )H L H Lirr rev Q Q Q Q− < − It is Irreversible H L irr H L revQ Q Q Q− < − L irr L revQ Q> H Lirr H L Q Q T T − 0<
  • 9. 9 The cyclic integral of Reversible Refrigeration 0 0L H L H Q Q T T = + + − Q T δ ∫ 2 3 4 1 1 2 3 4 Q Q Q Q T T T T δ δ δ δ = + + +∫ ∫ ∫ ∫ L H L H Q Q T T = − H H L L Q T Q T = 0= Since
  • 10. 10 The cyclic integral of Irreversible Refrigeration irr revW W> H L H L Q Q T T = Q T δ ∫ H L H L Q Q T T =− + We cannot use this ( ) ( )H L H Lirr rev Q Q Q Q− > − It is Irreversible H irr L H rev LQ Q Q Q− > − H irr H revQ Q> H irr L H L Q Q T T − + 0<
  • 11. 11 IrreversibleReversible < 00=Heat Engine < 00=Refrigeration 0≤∫ T Qδ Q T δ ∫ Derivation of Clausius Inequality
  • 12. 12 All paths are arbitrary 0 Q T δ =∫ Derivation of Entropy (Reversible Process) 2 2 1 1A C Q Q T T δ δ    =        ∫ ∫Subtracting gives 2 1 1 2 0 C B Q Q T T δ δ    = + =        ∫ ∫ For reversible cycle A-B 2 1 1 2 0 A B Q Q T T δ δ    = + =        ∫ ∫ For reversible cycle C-B 0 Q T δ =∫ Q the quantity is independent of the path and dependent on the end states only T δ ∴ ∫ Since paths A and C are arbitrary, it follows that the integral of δQ/T has the same value for ANY reversible process between the two sates.
  • 13. 13 work & heat are dependent on path Path functions Recall are independent of path properties Point functions and depend on state only      ( ) is a thermodynamic property we call it entropy S δQ T ⇒ ∫ Entropy (the unit) S = entropy (kJ/K); s = specific entropy (kJ/kg K) ∫       =−      ≡ 2 1 12gintegratin revrev T Q SS T Q dS δδ S2 – S1 depends on the end states only and not on the path, ∴ it is same for any path reversible or irreversible Derivation of Entropy (Reversible Process)
  • 14. 14 Consider 2 cycles AB is reversible and CB is irreversible 2 1 1 2 for cycle A-B (reversible) 0 A B Q Q Q T T T δ δ δ    = + =        ∫ ∫ ∫ 2 1 1 2 for path C-B (irreversible) 0 C B Q Q Q T T T δ δ δ    = + <        ∫ ∫ ∫ 2 2 1 1 comparing gives A C Q Q T T δ δ         >        ∫ ∫   2 2 2 1 1 1 reversible it is a property but A C A δQ dS dS T   === ===    ∫ ∫ ∫ in general δQ dS T ⇒ ≥ 2 2 1 1C C δQ dS T   ∴ >     ∫ ∫ 2 2 1 1 or δQ S S T − ≥ ∫ equality for reversible inequality for irreversible Derivation of Entropy (Irreversible Process)
  • 15. 15 Example (7-1) Entropy change during isothermal process. Solution:  This is simple problem.  No irreversibilities occur within the system boundaries during the heat transfer process. A friction-less piston-cylinder device contains a liquid- vapor mixture of water at 300 K. During a constant pressure process, 750 kJ of heat is transferred to the water. As a result, part of the liquid in the cylinder vaporizes. Determine the entropy change of the water during this process.
  • 16. 16 kkJ k kJ T Q S sys /5.2 300 750 ===∆ ( ) 2 2 1 1 1 rev rev Q Q S Q T T T δ δ   ∆= = =    ∫ ∫ Q S T ∆ =  We computed the entropy change for a system using the RHS of the equation.  But we can not get easy form each time.  So, we need to know how to evaluate the LHS which is path independent.
  • 17. 7.3. Heat Transfer as the Area Under the T-s Curve 17
  • 18. 18 Entropy change for different substances (∆S = S2-S1)  We need to find how to compute the left hand side of the entropy balance for the following substances: 1. Pure substance like water, R-134, Ammonia etc.. 2. Solids and liquids 3. Ideal gas
  • 19. 19 1- ∆S for Pure Substances The entropy of a pure substance is determined from the tables, just as for any other property These values were tabulated after conducting a tedious integration. These values are given relative to an arbitrary reference state. )( 12 ssmS −=∆ Entropy change for a closed system with mass m is given as expected: For water its assigned zero at 0.01 C. For R-134 it is assigned a zero at -40 C.
  • 20. 20 Property diagrams involving entropy  Recall the definition of entropy  Property diagrams serves as great visual aids in the thermodynamic analysis of process.  We have used P-v and T-v diagrams extensively in conjunction with the first law of thermodynamics. T Q dS revδ = TdSQrev =δ In the second-law analysis, it is very helpful to plot the processes on T-s and h-s diagrams for which one of the coordinates is entropy. W PdVδ =
  • 21. 21 ∫= 2 1 TdSQrev This area has no meaning for irreversible processes! Thus It can be done only for a reversible process for which you know the relationship between T and s during a process. Let us see some of them.
  • 22. 22 Isothermal reversible process. Entropy Isothermal Process Q 1 2 0 0T S T m s= ∆ = ∆
  • 23. 23 Adiabatic reversible process  In this process Q =0, and therefore the area under the process path must be zero.  This process on a T-s diagram is easily recognized as a vertical- line. Entropy Isentropic Process 1 2 Q=0
  • 24. 24 T-s Diagram for the Carnot Cycle Temperature Entropy 1 Isentropic compression Qin W=Qin-Qout Qout 2 Isothermal expansion 3 Isentropic expansion Isothermal compression 4
  • 25. 25 Another important diagram is the h-s Diagram  This diagram is important in the analysis of steady flow devices such as turbines.  In analyzing the steady flow of steam through an adiabatic turbine, for example,  The vertical distance between the inlet and the exit states (∆h) is a measure of the work output of the turbine,  The horizontal distance (∆s) is a measure of the irreversibilities associated with the process.