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EGR 334 Thermodynamics Chapter 5: Sections 10-11 Lecture 22: Carnot Cycle Quiz Today?
EGR 334 ThermodynamicsChapter 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:
Homework Assignment: Read Chapter 6, Sections 1-5
Problems from Chap 5: 64, 79, 81,86
3
Energy Balance:
Recall from last time:
syssys sys
dEQ W
dt
sysdEdt
Q W
Entropy Balance:
sys genQST
Entropy Rate Balance:
sysgen
dS Qdt T
sys sys sysE Q W
Energy Rate Balance:
sysdSdt
QT
gen
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)
T
v
2 3
1 4
p
v
23
14
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 H
Cmax 1
TT
7Sec 5.10 : The Carnot Cycle
Gas only cycleThe Carnot cycle:
T
v
2 3
1 4QH
QC
Area = Work
QH
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.
8Sec 5.10 : The Carnot CycleAnalyzing the Carnot cycle:
Energy Balance: WUQ
VU c T VQ c T pdV
Process 2 -3 : Isothermal Expansion receiving QH.
First look at the two isothermal processes
23 VQ c T pdV 3 3 3
323 2 2 2
2
lnHH H
VRT dVQ pdV pdV dV RT RTV V V
Process 4 – 1 : Isothermal Compression, rejecting QC.
4
141 ln
VVRTQ C
and
41
23
41
23
41
23
lnln
lnln
VVTVVT
VVRTVVRT
C
H
C
H
C
H
W pdV
9Sec 5.10 : The Carnot CycleAnalyzing the Carnot cycle:
Energy Balance:
Then look at the two adiabatic processes (Q = 0)
12 0 VQ c dT pdV Process 1-2 : Adiabatic Compression.
Process 3 – 4 : Adiabatic Expansion.
VRTc dT pdV dVV
1
22
1ln
VV
VdV
TdT
RcH
C
T
TV
3
44
3lnVV
VdV
TdT
RcC
H
T
TV
The term
TdT
Rc
TdT
Rc C
H
H
C
T
TVT
TV
Thus,
3
4
2
1
1
2 lnlnlnVV
VV
VV
1 4 1 2
2 3 4 3
V V V VorV V V V
VQ c dT pdV
10Sec 5.10 : The Carnot Cycle
Analyzing the Carnot cycle:
With3
2
4
1
VV
VV
and 41
23
lnln
VVTVVT
C
H
C
H
Therefore,C
H
C
H
TT
We have now proven
max 1 1C C
H H
Q TQ T
11The Carnot Model of a Hurricane
Warm moist air
Warm air risesAdiabatic cooling
Cools & Expands as P
As T to dew point, vapor condenses, releasing hfg and warming air
Added heat causes further rising.
12The Carnot Model of a Hurricane
AB
CD
vcore>>v outer
Adiabatic
Adiabatic
Isothermal Expansion
Isothermal Compression
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.
Sec 5.10 : The Carnot Cycle
p
v
14
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.
Sec 5.10 : The Carnot Cycle
15
state 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.6988 0.6988x 0 1 0.643v (ft3/lb) 0.02363 0.2677 300.74u (Btu/lb) 609.9 1090.0 687.8
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.
Sec 5.10 : The Carnot Cycle
state 1 2 3 4T (°F/R) 600 600 90 90p (psi)x 0 1 0.643v (ft3/lb)u (Btu/lb)
Isothermal IsothermalAdiabatic Adiabatic
Using Table A-23
3 0.01610 0.643(467.7 0.01610) 300.74 / mv ft lb
3 58.07 0.643(1040.2 58.07) 687.8 / mu Btu lb
16
Example: (5.76) (b) Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
Isothermal IsothermalAdiabatic Adiabaticstate 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643v (ft3/lb) 0.0236
30.267
7300.7
4u (Btu/lb) 609.9 1090.
0687.8
Process
U Q W
1 - 2 240 274.8
34.81
2 - 3 0 3 - 44 - 1 0
12 2 1 0.5 1090 609.9 / 240m mU m u u lb Btu lb Btu
12 2 1 2 1 2 1( ) ( )W pdV p V V p mv mv mp v v
22 3
2
144 1(1541 / ) 0.5 0.2677 0.02363 /1 778f m m
f
in Btulb in lb ft lbft ft lb
Process 1-2
34.81 Btu
Process
U Q W
1 - 22 - 3 03 - 44 - 1 0
12 12 12 240 34.81 274.8Q U W Btu
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Example: (5.76) (b) Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
Isothermal IsothermalAdiabatic Adiabaticstate 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643v (ft3/lb) 0.0236
30.267
7300.7
4u (Btu/lb) 609.9 1090.
0687.8
Process
U Q W
1 - 2 240 274.8
34.81
2 - 3 -201
0 201
3 - 44 - 1 0Process 2-3
23 2 1 0.5 687.8 1090 / 201.1m mU m u u lb Btu lb Btu
23 23 23 0 ( 201.1) 201.1W Q U Btu
23 0Q (adiabatic process)
Process
U Q W
1 - 2 240 274.8
34.81
2 - 3 03 - 44 - 1 0
18
Example: (5.76) (b) Evaluate the heat and work for each process in Btu
Sec 5.10 : The Carnot Cycle
Isothermal IsothermalAdiabatic Adiabaticstate 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643v (ft3/lb) 0.0236
30.267
7300.7
4u (Btu/lb) 609.9 1090.
0687.8
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201
0 201
3 - 4 -142.64 - 1 0
H H
C C
Q TQ T
For Process 3 – 4: for the Carnot
cycle:
34550274.85 142.6
1060C
C HH
TQ Q Q Btu Btu
T
where QH = Q12 = 274.8 Btu
34 3 4 3 3 4 3( ) ( )W pdV p V V p m v v 34 4 3( )U m u u
19
Example: (5.76) (b) Evaluate the heat and work for each process in Btu
Sec 5.10 : The Carnot Cycle
Isothermal IsothermalAdiabatic Adiabaticstate 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643v (ft3/lb) 0.0236
30.267
7300.7
4u (Btu/lb) 609.9 1090.
0687.8
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201
0 201
3 - 4 -142.64 - 1 0continuing for Process 3 – 4:
recalling h = u + pv
4 3 34 4 4 3 3( ) ( )m u u Q m p v p v 34 34 34U Q W
34 4 4 4 3 3 3( ) ( )Q m u p v m u p v 34
4 4 4 4 3 3 3( ) ( )Q
h u p v u p vm
2
2 34 2
142.6 144 1(687.8 / (0.6988 / )(300.74 / ))0.5 1 778m f m
m f
Btu in Btuh Btu lb lb in ft lblb ft lb ft
441.5 / mBtu lb
20
Example: (5.76) (b) Evaluate the heat and work for each process in Btu
Sec 5.10 : The Carnot Cycle
Isothermal IsothermalAdiabatic Adiabaticstate 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643 0.368v (ft3/lb) 0.0236
30.267
7300.7
4172.1
u (Btu/lb) 609.9 1090.0
687.8 419.5
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201
0 201
3 - 4 -142.64 - 1 0continuing for Process 3 – 4:
Then using Table A2 at h4 = 441.5 Btu/lbm and T4 = 90 deg.
4 44
4
441.5 58.07 0.3681042.7
f
fg
h hx
h
which let the state 4 intensive properties be found:3
4 0.01610 0.368(467.7 0.01610) 172.1 / mv ft lb
4 58.07 0.368(1040.2 58.07) 419.5 / mu Btu lb
state 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643v (ft3/lb) 0.0236
30.267
7300.7
4u (Btu/lb) 609.9 1090.
0687.8
21
Example: (5.76)(b) Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
state 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643 0.368v (ft3/lb) 0.0236
30.267
7300.7
4172.1
u (Btu/lb) 609.9 1090.0
687.8 419.5
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201 0 2013 - 4 -
134.2-142.6 -8.32
4 - 1 0
Isothermal IsothermalAdiabatic Adiabatic
34 4 3 0.5 (419.5 687.8) / 134.2m mU m u u lb Btu lb Btu and for Process 4-1:
34 3 4 3 3 4 3W pdV p V V p m v v
22 3
34 2
144 1(0.6988 / )(0.5 ) 172.1 300.74 /1 778f m m
f
in BtuW lb in lb ft lbft ft lb
8.32 Btu
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201 0 2013 - 4 -142.64 - 1 0
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Example: (5.76)(b) Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
state 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643 0.368v (ft3/lb) 0.0236
30.267
7300.7
4172.1
u (Btu/lb) 609.9 1090.0
687.8 419.5
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201 0 2013 - 4 -
134.2-142.6 -8.32
4 - 1 95.2 0 -95.2
Isothermal IsothermalAdiabatic Adiabatic
41 1 4 0.5 (609.9 419.5) / 95.2m mU m u u lb Btu lb Btu
41 41 41 0 95.2 95.2W Q U Btu
Finally, process 4 – 1:
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201 0 2013 - 4 -
134.2-142.6 -8.32
4 - 1 0
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Example: (5.76)
Sec 5.10 : The Carnot Cycle
state 1 2 3 4T (°F/R) 600 600 90 90p (psi) 1541 1541 0.698
80.698
8x 0 1 0.643 0.368v (ft3/lb) 0.0236
30.267
7300.7
4172.1
u (Btu/lb) 609.9 1090.0
687.8 419.5
Process
U Q W
1 - 2 240 274.85
34.81
2 - 3 -201 0 2013 - 4 -
134.2-142.6 -8.32
4 - 1 95.2 0 -95.2
Isothermal IsothermalAdiabatic Adiabatic
Thermal efficiency of the cycle:
(c) Evaluate the thermal efficiency.
max5501 1 0.481 48%
1060C
H
TT
34.81 201 8.32 95.2 131.99 0.480 48%274.85 274.85
Cycle
in
WQ
Clausius Inequality►The Clausius inequality is developed from the
Kelvin-Planck statement of the second law and can be expressed as:
cycleb
QT
The nature of the cycle executed is indicated by the value of cycle:
cycle = 0 no irreversibilities present within the systemcycle > 0 irreversibilities present within the systemcycle < 0 impossible
25Sec 5.11 : The Clausius Inequality
We have shown that: and thusC
H
C
H
TT
For an ideal/reversible process
C
C
H
H
TQ
TQ
Therefore:0
C
C
H
H
TQ
TQ
p
v
Now consider a general processEach part of the cycle is divided into an infinitesimally small process
0C
C
H
H
TdQ
TdQ
Then sum (integrate) the entire process
0
bTQ dQ = heat transfer
at boundaryT = absolute T at that
part of the cycle.
26Sec 5.11 : The Clausius Inequality
For a real process, Qreal > Qreversible
Therefore:
P
v
We can then define σ, where
0
bTQ
cyclebT
Q
and σcycle = 0 reversible processσcycle > 0 irreversible processσcycle < 0 impossible process
We now also have the mathematical definition of enthalpy.
TdSdQdSTdQ
More on this in Chapter 6
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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?
Sec 5.11 : The Clausius Inequality
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ
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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?
Sec 5.11 : The Clausius Inequality
bcycle T
Q
HCH
C QQQQ 11
and
Therefore:
CHH
C
C
H
Hcycle TT
QTQ
TQ 11
KkJ
KKkJcycle 667.0
3006.01
50011000
Since σcycle is negative, the cycle is impossible.
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ
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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?
Sec 5.11 : The Clausius Inequality
CHH
C
C
H
Hcycle TT
QTQ
TQ 11
KkJ
KKkJcycle 0
3004.01
50011000
Since σcycle is zero, the cycle is internally reversible.
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ
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?
Sec 5.11 : The Clausius Inequality
CHH
C
C
H
Hcycle TT
QTQ
TQ 11
KkJ
KKkJcycle 667.0
3002.01
50011000
Since σcycle is positive, the cycle is irreversible.
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ
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End of Lecture 21 Slides
