Transcript 열역학 제5장 강의노트

Lecture Notes on Thermodynamics 2008
Chapter 7 Entropy
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Prof. Man Y. Kim, Autumn 2008, ⓒ[email protected], Aerospace Engineering, Chonbuk National University, Korea
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Entropy (1/3)
• The Inequality of Clausius

Q
T
0
 The inequality of Clausius is a corollary or a consequence of the 2nd law of
thermodynamics.
 It is valid for all possible cycles, including both reversible and irreversible ones
 The entropy is defined from this formulation, i.e.,
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 Q 
dS  

 T rev
and
2
S2  S1 

2
1
 Q 


 T rev
Entropy (2/3)
• Proof of the Inequality of Clausius

Consider first a reversible (Carnot) heat engine cycle :  Q  QH  QL  0
From the definition of absolute temperature scale ( QH QL  TH TL )
 QH TH
 Q QH QL
Q
Q 


0 

 H  L
T
TH TL
TH TL 
 QL TL
 Q QH QL


0
If TH  TL,
 Q  QH  QL and
0
T
TH TL
Finally, we conclude that for all reversible heat engines,
Q
0
 Q  0 an
T
d
Now consider an irreversible cycle heat engine :





Wirr  Wrev  QH  QL ,irr  QH  QL ,rev QL ,irr  QL ,rev
Consequently, for the irreversible cycle engine,
 Q QH QL ,irr
 Q  QH  QL ,irr  0 and


0
T
TH
TL


If we make the engine become more and more irreversible, but keepQH T, H , andTL fixed,
Q
 Q  0 and
0
T
Finally, we conclude that for all irreversible heat engine cycles,
Q
0
 Q  0 an
T
d be applied for both reversible and irreversible refrigeration cycles.
Similarly, the same procedure can
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



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Entropy (3/3)
• Entropy – A Property of a System
Reversible process along path A-B

Q
T
0

2
1
 Q 

 
 T A
 Q 


2  T B

1
Reversible process along path C-B

Q
T
0

2
1
 Q 

 
 T C
 Q 


2  T B

1
Subtracting the second equation from the first, we have

2
1
 Q 

 
 T A

2
1
 Q 


 T C
 
Q
T
is independent of the path → point function → property
This property is called entropy
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 Q 
dS  

 T rev
ds 
and
S2  S1 

1
2
1  Q 


m  T rev
 Q 


 T rev
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Principle of the Increase of Entropy
•(1/2)
Increase of Entropy Principle
From the Clausius Inequality

Q
T
0
or

2
1
Here, you can find that
T

 Q 
0


2  T  int rev

1
Q
 Q 
 

0

 S1  S2  0  S1  S2 

1 T
2  T  int rev
1 T
Entropy generation
2
Q
S2  S1 
 Sgen
1 T

2
Q
Q

1
• Entropy Generation


 dS 
2
Q
T

2
1
Q
T
 dS 
Q
T
  S gen
where,  Sgen  0 :entropy generation due to irreversibility occurring inside the system ( because of
friction, unrestricted expansion, internal energy transfer over a finite temp. difference, etc.)
Reversible process : Q  TdS and  W  PdV
Irreversible process : Qirr  T  dS   Sgen   TdS  T Sgen
1st law :  Qirr  dU   Wirr
Thermodynamic property relation : TdS  dU  PdV
 S2  S1 

2
dS 

2
  Wirr  PdV  T Sgen
Q
T
 1 S2 gen
Lost Work → Exergy (Chapter 8)
1
1
Thus we have an expression for the change of entropy for an irreversible process as an equality,
whereas in the last slide we had an inequality.
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Principle of the Increase of Entropy
(2/2)
• Discussions on Entropy Generation
dS 
Q
T
  Sgen
 Sgen  0
 Wirr  PdV  T Sgen
Discussion 1 : There are 2 ways in which the entropy of a system can be increased by
(1) transferring heat to the system
(2) having an irreversible process
Note : There is only one way in which entropy can be decreased by transferring heat from the
system
Discussion 2 : For an adiabatic system, the increase of entropy is always associated with the
irreversibility
Discussion 3 : The presence of irreversibility will cause the work to be smaller than the reversible work
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Entropy Change of a Pure Substance
• see Examples 7–3 (p.326) and 7–4 (p.327)
• Isentropic Process
s  0
or
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s 2  s1
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Isentropic Relations
Consider the case of an ideal gas undergoing an isentropic process,
s2  s1  C p 0 ln
T  R
P 
T2
P
T P 
 R ln 2  0  ln  2  
ln  2   2   2 
T1
P1
T1  P1 
 T1  C p 0  P1 
However,
R
Cp 0
C p0
R C p0  Cv0 k  1
, where k 
: specific heat ratio


C p0
C p0
k
Cv0
Finally we can obtain
T2  P2 
 
T1  P1 
k1
k
,
T2  v1 
 
T1  v2 
k1
and
P2  v1 
 
P1  v2 
k
: Isentropic Relation
Note : Pv  constant is a special case of a polytropic process in which the polytropic
exponent n is equal to the specific heat ratio k
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k
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T–s Diagram of the Carnot Cycle
Consider the Carnot cycle, i.e.,
① → ② : reversible isothermal heat addition process
1 2
Q
 Q 

 Q  1 2  0  1 Q2  0 


TH 1
TH
1  T rev
Area 1-2-b-a-1 : heat transferred to the working fluid during the process
② → ③ : reversible adiabatic process
 Q 
dS  
  0 → isentropic process
 T rev
③ → ④ : reversible isothermal heat rejection process
4
Q
1 4
 Q 
S4  S3  
 Q  3 4  0  3 Q4  0 
 
TL 3
TL
3  T rev
Area 3-4-a-b-3 : heat transferred from the working fluid to the low-temperature reservoir.
④ → ① : reversible adiabatic process
 Q 
dS  
  0 → isentropic process
 T rev
Area 1-2-3-4-1 : net work of the cycle
Efficiency
 W  W
area 1  2  3  4  1
th   out   net 
area 1  2  b  a  1
 Qin  QH
S2  S1 


2


Comments on efficiency : TH  th , TL  th ,
• see Example 7–6
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TL  0  th  100%
What is Entropy ?
Figure 7–23
Figure 7–22
Figure 7–20
Figure 7–24
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Figure 7–21
Figure 7–27
Figure 7–25
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Figure 7–26
Thermodynamic Property Relations
• Gibbs Equations (T–ds Relations)
TdS  dU  PdV
and
TdS  dH  VdP
For the simple compressible substance with no motion or gravitational effects, the 1st law becomes
 Q  dU   W
For a reversible process of a simple compressible substance,
 Q  TdS
and
 W  PdV
TdS  dU  PdV
Since enthalpy is defined asH
 U  PV
dH  dU  PdV  VdP  dU  TdS  dU  VdP
TdS  dH  VdP
For a unit mass,
Tds  du  Pdv


du Pdv
 ds  T  T
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and
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Tds  dh  vdP


dh vdP
 ds 


T
T

Entropy Change during Irreversible
Process
Reversible cycle : reversible process along path A-B

Q
T


2
1
 Q 

 
 T A
 Q 

 0
2  T B

1
Irreversible cycle : irreversible path C and reversible path B
2
1
Q
 Q 
 Q 
 



 0
T
1  T C
2  T B



Subtracting the second equation from the first, we have
2
2
2
2
2
 Q 
 Q 
 Q 
 Q 


dS








 
C
1  T A
1  T C
1
1  T C
1  T A






2
1
dSA 
2
 dS
1
C
As path C was arbitrary, the general result is (both reversible and irreversible cases)
dS 
Q
T
and
S2  S1 

2
1
Q
T
This is one of the most important equations of thermodynamics !
Q
dS  

 T rev
and
Q
dS  

 T irr
  dS 
Q
T
Therefore, we can find that the entropy change for an irreversible process is larger than the change in a
reversible process for the same Q and T.
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Entropy Change for a Solid(Liquid) and Ideal
Gas
• For a Solid or Liquid
specific volume is very small,
 ds
• For an Ideal Gas
Tds  du  Pdv duand
du
T
We know that Tds  du  Pdv,
C
dT
T
Similarly, Tds  dh  vdP
,
 s2  s1
and
R
s2  s1 
v

T P
dh  C p 0 dT
and
dT
dP
 ds  C p 0
R
T
P
and
C ln
P R

T v
and
du  C v 0 dT
dT Rdv
 ds  C v 0

T
v
dh du CdT
s2  s1 
2
1
Cv0

2
1
T2
T1
dT
v
 R ln 2
T
v1
C p0
dT
P
 R ln 2
T
P1
If we assume that the specific heat is constant,
s2  s1  C v 0 ln
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T2
v
 R ln 2
T1
v1
and
s2  s1  C p 0 ln
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T2
P
 R ln 2
T1
P1
Reversible Polytropic Process for an Ideal
• Polytropic Process
Gas
d ln P
d ln V
 n  d ln P  nd ln V  0
If n is a constant, PV
P V 
 2  1 
P1  V2 
n
n
n
 constant  PV
1 1  P2V2
n
T2  P2 
 
T1  P1 
and
n 1
n
V 
 1 
 V2 
n 1
: Polytropic Relation
• Work done during a reversible polytropic process
1W2 
 PdV
1W2 

2
1

2
1
and PV n  constant

PdV  constant
2
1
P2V2  P2V2 mR T2  T1 

1n
1n
dV
Vn
 Isobaric process (P=constant) : n=0
 Isothermal process (T=constant) : n=1
 Isentropic process (s=constant) : n=k
 Isochoric process (v=constant) : n=∞
for any value of n except n=1
•The reversible isothermal process
dV
V2
P1
 PV
 PV
1 1 ln
1 1 ln
V1
P2
1
1 V
V2
P
 mRT ln 1
1W2  mRT ln
V1
P2
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PV  constant  PV
1 1  P2V2  1W2 
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or

2
PdV  constant

2
Heat Transfer and Entropy Generation
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Examples (1/3)
• Turbine : Example 7–14
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Examples (2/3)
• Compressor : Example 7–14
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Examples (3/3)
• Nozzle : Example 7–16
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Saemangum @
Jellabukdo
Homework #7
Solve the Examples 7–1 ~ 7–23
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