EGR 334 Thermodynamics
Chapter 5: Sections 10-11
Lecture 22:
Carnot Cycle
Quiz Today?
Today’s main concepts:
•
•
•
•
State what processes make up a Carnot Cycle.
Be able to calculate the efficiency of a Carnot Cycle
Be able to give the Classius Inequality
Be able to apply the Classisus Inequality to determine if a cycle is
reversible, irreversible, or impossible as predicted by the 2nd
Law.
Reading Assignment:
Read Chapter 6, Sections 1-5
Homework Assignment:
Problems from Chap 5: 64, 79, 81,86
3
Recall from last time:
Energy Balance:
 E sys  Q sys  W sys
dE sys
Energy Rate Balance:
d E sys
dt
W
Q
dt
 Q sys  W sys
Entropy Balance:
 S sys  
Q
T
Q
  g en
T
Entropy Rate Balance:
d S sys
dt
 
Q
T

d S sys
dt
gen
 g en
Carnot Cycle
►The Carnot cycle provides a specific example of a reversible cycle
that operates between two thermal reservoirs. Other examples
covered in Chapter 9 are the Ericsson and Stirling cycles.
►In a Carnot cycle, the system executing the cycle undergoes a series
of four internally reversible processes:
two adiabatic processes (Q = 0)
alternated with
two isothermal processes ( T = constant)
p
2
T
3
1
4
2
3
1
v
4
v
Carnot Power Cycles
The p-v diagram and schematic of a gas in a
piston-cylinder assembly executing a Carnot
cycle are shown below:
Carnot Power Cycles
The p-v diagram and schematic of water executing a Carnot cycle
through four interconnected components are shown below:
In each of these cases the
thermal efficiency is given by
 max  1 
TC
TH
Sec 5.10 : The Carnot Cycle
7
The Carnot cycle:
QH
Gas only cycle
QH
T
2
3
Area = Work
1
QC
4
v
QC
Process 1-2 : Adiabatic Compression.
Process 2 -3 : Isothermal Expansion
receiving QH.
Process 3 – 4 : Adiabatic Expansion.
Process 4 – 1 : Isothermal Compression,
rejecting QC.
Sec 5.10 : The Carnot Cycle
8
Analyzing the Carnot cycle:
Energy Balance: Q   U  W
 U  cV  T
W 
Q  cV  T 
 pdV

pdV
First look at the two isothermal processes
Process 2 -3 : Isothermal Expansion receiving QH.
Q 2 3  cV  T 
Q 23 
 pdV



pdV
3
pdV  
2
3
R TH
2
V
d V R TH

3
dV
2
V
 V3 
 R T H ln 

V
 2 
Process 4 – 1 : Isothermal Compression, rejecting QC.
Q 41  RT C
and
 V1
ln 
 V4
QH
QC





Q 23
Q 41

RT H ln V 3 V 2 
RT C ln V1 V 4 

T H ln V 3 V 2 
T C ln V1 V 4 
Sec 5.10 : The Carnot Cycle
9
Analyzing the Carnot cycle:
Energy Balance: Q  cV d T 


pdV
Then look at the two adiabatic processes (Q = 0)
Process 1-2 : Adiabatic Compression.
The term
Q 1 2  0   cV d T   p d V
TH c
TC c
dT
dT
V
V
TC R T   T H R T
RT
  cV d T   p d V  
dV
Thus,
V


TH
cV dT
R
TC
T


dV
2
V
1
 V2
 ln 
 V1
 V2
 ln 
 V1




Process 3 – 4 : Adiabatic Expansion.


TC
TH
c V dT
R
T


4
3
dV
V
 V4
 ln 
 V3
V1




V2

V4
V3
 V4 

 V1 


  ln 


 V   ln  V 

 2 
 3 
or
V1
V4

V2
V3
Sec 5.10 : The Carnot Cycle
10
Analyzing the Carnot cycle:
With
V1

V4
Therefore,
V2
and
V3
QH
QC

QH

QC
T H ln V 3 V 2 
T C ln V1 V 4 
TH
TC
We have now proven
 m ax 
1
QC
QH

1
TC
TH
The Carnot Model of a Hurricane
11
Added heat causes
further rising.
As T to dew point, vapor
condenses, releasing hfg
and warming air
Cools & Expands as P
Adiabatic cooling
Warm air rises
Warm moist air
The Carnot Model of a Hurricane
12
Isothermal Compression
Adiabatic
vcore>>v outer
A
Adiabatic
Isothermal Expansion
B
D
C
Sec 5.10 : The Carnot Cycle
13
Example: (5.76) One-half pound of water executes a Carnot power
cycle. During the isothermal expansion, the water is heated at 600°F
from a saturated liquid to a saturated vapor. The vapor then expands
adiabatically to a temperature of 90°F and a quality of 64.3%
(a) Sketch the cycle on a P-v diagram.
(b) Evaluate the heat and work for each process in BTU
(c) Evaluate the thermal efficiency.
p
v
Sec 5.10 : The Carnot Cycle
Example: (5.76) One-half pound of water executes a Carnot power
cycle. During the isothermal expansion, the water is heated at 600°F
from a saturated liquid to a saturated vapor. The vapor then expands
adiabatically to a temperature of 90°F and a quality of 64.3%
(a) Sketch the cycle on a P-v diagram.
(b) Evaluate the heat and work for each process in BTU
(c) Evaluate the thermal efficiency.
14
Sec 5.10 : The Carnot Cycle
15
Example: (5.76) One-half pound of water executes a Carnot power
cycle. During the isothermal expansion, the water is heated at 600°F
from a saturated liquid to a saturated vapor. The vapor then expands
adiabatically to a temperature of 90°F and a quality of 64.3%
(a) Sketch the cycle on a P-v diagram.
(b) Evaluate the heat and work for each process in BTU
(c) Evaluate the thermal efficiency.
Isothermal
state
Adiabatic
Isothermal
Adiabatic
1
2
3
4
T (°F/R)
600
600
90
90
p (psi)
1541
1541
0.6988
0.6988
0
1
0.643
0.643
v (ft3/lb)
0.02363
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
x
Using Table A-2
v 3  0 .0 1 6 1 0  0 .6 4 3( 4 6 7 .7  0 .0 1 6 1 0 )  3 0 0 .7 4 ft / lb m
3
u 3  5 8 .0 7  0 .6 4 3(1 0 4 0 .2  5 8 .0 7 )  6 8 7 .8 B tu / lb m
Sec 5.10 : The Carnot Cycle
Example: (5.76)
16
(b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
T (°F/R)
600
600
90
90
1-2
240
p (psi)
1541
1541
0.6988
0.6988
2-3
0
1
0.643
3-4
0.02363
0.2677
300.74
4-1
609.9
1090.0
687.8
state
x
v (ft3/lb)
u (Btu/lb)
Process 1-2
Q
274.8 34.81
0
0
 U 12  m  u 2  u 1   0.5 lb m  1090  609.9  B tu / lb m  240 B tu
W12 
 pdV
 p  V 2  V1   p ( m v 2  m v1 )  m p ( v 2  v 1 )
 (1541lb f / in )  0.5 lb m   0.2677  0.02363  ft / lb m
2
3
144 in
1
ft
2
2
1
B tu
778 ft  lb f
 34.81 B tu
Q1 2   U 1 2  W 1 2  2 4 0  3 4 .8 1  2 7 4 .8 B tu
W
Sec 5.10 : The Carnot Cycle
Example: (5.76)
17
(b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
T (°F/R)
600
600
90
90
1-2
240
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
1
0.643
3-4
v (ft3/lb)
0.02363
0.2677
300.74
4-1
u (Btu/lb)
609.9
1090.0
687.8
state
x
Q
274.8 34.81
0
0
Process 2-3
Q 23  0
(adiabatic process)
 U 23  m  u 2  u 1   0.5 lb m  687.8  1090  B tu / lb m   201.1 B tu
W 2 3  Q 2 3   U 2 3  0  (  2 0 1 .1)  2 0 1 .1 B tu
W
201
Sec 5.10 : The Carnot Cycle
Example: (5.76)
18
(b) Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
3-4
-142.6
v (ft3/lb)
0.02363
0.2677
300.74
4-1
0
u (Btu/lb)
609.9
1090.0
687.8
state
x
For Process 3 – 4:
Q 34
for the Carnot cycle:
Q
where QH = Q12 = 274.8 Btu
QC
H

TH
TC
 TC 
 550 
  QC   Q H 
   274.85 B tu 
   142.6 B tu
 1060 
 TH 
 U 34  m ( u 4  u 3 )
W 34 
 pdV
 p 3 (V 4  V 3 )  p 3 m ( v 4  v 3 )
Sec 5.10 : The Carnot Cycle
Example: (5.76)
19
(b) Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic Isothermal Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
3-4
-142.6
v (ft3/lb)
0.02363
0.2677
300.74
4-1
0
u (Btu/lb)
609.9
1090.0
687.8
state
x
continuing for Process 3 – 4:
 U 34  Q 34  W 34
m ( u 4  u 3 )  Q 34  m ( p 4 v 4  p 3 v 3 )
recalling h = u + pv
Q 34  m ( u 4  p 4 v 4 )  m ( u 3  p 3 v 3 )
h4  ( u 4  p 4 v 4 ) 
h4 
 142.6 B tu
0.5 lb m
Q 34
m
 (u 3  p 3v3 )
 (687.8 B tu / lb m  (0.6988 lb f / in )(300.74 ft / lb m ))
2
3
144 in
1 ft
2
2
1B tu
778 lb f  ft
 441.5 B tu / lb m
Sec 5.10 : The Carnot Cycle
Example: (5.76)
20
(b) Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic Isothermal Adiabatic
11
22
33
44
Process
U
Q
W
TT (°F/R)
(°F/R)
600
600
600
600
90
90
90
90
1-2
240
274.85
34.81
pp (psi)
(psi)
1541
1541
1541
1541
0.6988
0.6988 0.6988
0.6988
2-3
-201
0
201
00
11
0.643
0.643
0.368
3-4
-142.6
0.02363
0.02363 0.2677
0.2677 300.74
300.74
172.1
4-1
0
state
state
xx
vv (ft
(ft33/lb)
/lb)
uu (Btu/lb)
(Btu/lb)
609.9
609.9
1090.0
1090.0
687.8
687.8
419.5
continuing for Process 3 – 4:
Then using Table A2 at h4 = 441.5 Btu/lbm and T4 = 90 deg.
x4 
h4  h4 f
h 4 fg

4 4 1 .5  5 8 .0 7
 0 .3 6 8
1 0 4 2 .7
which let the state 4 intensive properties be found:
v 4  0 .0 1 6 1 0  0 .3 6 8( 4 6 7 .7  0 .0 1 6 1 0 )  1 7 2 .1 ft / lb m
3
u 4  5 8 .0 7  0 .3 6 8(1 0 4 0 .2  5 8 .0 7 )  4 1 9 .5 B tu / lb m
Sec 5.10 : The Carnot Cycle
21
Example: (5.76) (b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal
Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
0.368
3-4
-134.2
-142.6
-8.32
v (ft3/lb)
0.02363
0.2677
300.74
172.1
4-1
u (Btu/lb)
609.9
1090.0
687.8
419.5
state
x
0
and for Process 4-1:
 U 34  m  u 4  u 3   0.5 lb m (419.5  687.8) B tu / lb m   134.2 B tu
W 34 

p d V  p 3 V 4  V3   p 3 m  v 4  v3 
W 3 4  (0 .6 9 8 8 lb f / in )(0 .5 lb m ) 1 7 2 .1  3 0 0 .7 4  ft / lb m
2
3
1 4 4 in
1
ft
2
2
1
B tu
7 7 8 ft  lb f
  8.32 B tu
Sec 5.10 : The Carnot Cycle
22
Example: (5.76) (b) Evaluate the heat and work for each process in BTU
Isothermal Adiabatic Isothermal
state
Adiabatic
1
2
3
4
T (°F/R)
600
600
90
90
p (psi)
1541
1541
0.6988
0.6988
0
1
0.643
0.368
v (ft3/lb)
0.02363
0.2677
300.74
172.1
u (Btu/lb)
609.9
1090.0
687.8
419.5
x
Process
1-2
2-3
3-4
4-1
U
240
Q
274.85
W
34.81
-201
-134.2
0
-142.6
201
-8.32
95.2
0
-95.2
Finally, process 4 – 1:
 U 41  m  u 1  u 4   0.5 lb m (609.9  419.5) B tu / lb m  95.2 B tu
W 4 1  Q 4 1   U 4 1  0  9 5 .2   9 5 .2 B tu
Sec 5.10 : The Carnot Cycle
23
Example: (5.76) (c) Evaluate the thermal efficiency.
Isothermal Adiabatic Isothermal
state
Adiabatic
1
2
3
4
Process
U
Q
W
T (°F/R)
600
600
90
90
1-2
240
274.85
34.81
p (psi)
1541
1541
0.6988
0.6988
2-3
-201
0
201
0
1
0.643
0.368
3-4
-134.2
-142.6
-8.32
v (ft3/lb)
0.02363
0.2677
300.74
172.1
4-1
95.2
0
-95.2
u (Btu/lb)
609.9
1090.0
687.8
419.5
x
Thermal efficiency of the cycle:
 m ax  1 
 
W C ycle
Q in

TC
 1
TH
550
1060
34.81  201  8.32  95.2
274.85
 0 .4 8 1  4 8 %

131.99
274.85
 0.480  48%
Clausius Inequality
►The Clausius inequality is developed from the KelvinPlanck statement of the second law and can be expressed
as:

Q 

    cycle
 T b
The nature of the cycle executed is indicated by
the value of cycle:
cycle = 0 no irreversibilities present within the system
cycle > 0 irreversibilities present within the system
cycle < 0 impossible
Sec 5.11 : The Clausius Inequality
We have shown that: 
QH
25

QC
Therefore: Q H
TH

QC
TH
TC
and thus
QH
 
TH
QC
TC
For an ideal/reversible process
0
TC
Now consider a general process
Each part of the cycle is divided into
an infinitesimally small process
dQ
p
H

dQ C
0
TH
TC
Then sum (integrate) the entire process
v
dQ = heat transfer at
 Q 
boundary
  T   0
b
T = absolute T at that
part of the cycle.
Sec 5.11 : The Clausius Inequality
26
For a real process, Qreal > Qreversible
Therefore:
 Q 
  T   0
b
 Q 
  T     cycle
b
We can then define σ, where
and σcycle = 0  reversible process
σcycle > 0  irreversible process
σcycle < 0  impossible process
P
We now also have the mathematical
definition of enthalpy.
dQ
 dS  dQ  TdS
T
v
More on this in Chapter 6
Sec 5.11 : The Clausius Inequality
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 500 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
TH= 500 K
W=?
Q = 50 kJ
TH= 300 K
27
Sec 5.11 : The Clausius Inequality
28
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 1000 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
QC
  1
 Q C  1   Q H
TH= 500 K
QH
 Q 
W=?
and  cycle    

 T b
Q = 50 kJ
TH= 300 K
 cycle
Therefore:
 cycle
QH
QC 
 

  QH
TC 
 TH
 1
1    



T
T
C
 H

1  0 . 6  
kJ
 1
  1000 kJ 



0
.
667
300 K 
K
 500 K
Since σcycle is negative, the cycle is impossible.
Sec 5.11 : The Clausius Inequality
29
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 1000 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
TH= 500 K
W=?
Q = 50 kJ
 cycle
TH= 300 K
 cycle
QH
QC 
 

  QH
TC 
 TH
 1
1    



T
T
C
 H

1  0 . 4  
kJ
 1
  1000 kJ 

0

300 K 
K
 500 K
Since σcycle is zero, the cycle is internally reversible.
Sec 5.11 : The Clausius Inequality
30
Example: (5.81) A system executes a power cycle while receiving heat
transfer at a temperature of 1000 K and discharging 1000 kJ by heat
transfer at a temperature of 300 K. There are no other heat transfers.
Appling the Clausius Inequality, determine σcycle if the thermal
efficiency is (a) 60% (b) 40% (c) 20 %. In each case is the cycle
reversible, or impossible?
TH= 500 K
W=?
Q = 50 kJ
TH= 300 K
 cycle
 cycle
QH
QC 
 

  QH
TC 
 TH
 1
1    



T
T
C
 H

1  0 . 2  
kJ
 1
  1000 kJ 

  0 . 667

300 K 
K
 500 K
Since σcycle is positive, the cycle is irreversible.
31
End of Lecture 21 Slides