1. EGR 334 Thermodynamics Chapter 5: Sections 10-11
Lecture 22:
Carnot Cycle
Quiz Today?

2. 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:
Energy Rate Balance:
Entropy Balance:
Entropy Rate Balance:

4. 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 v 2 3 1 4 T v 2 3 1 4

5. Carnot Power Cycles
The p-v diagram and schematic of a gas in a piston-cylinder assembly executing a Carnot cycle are shown below:

6. In each of these cases the thermal efficiency is given by
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

7. The Carnot cycle: QH Gas only cycle Area = Work QH QC QC
Sec 5.10 : The Carnot Cycle T v 2 3 1 4
The Carnot cycle: QH
Gas only cycle
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. QC

8. Analyzing the Carnot cycle:
Sec 5.10 : The Carnot Cycle
Analyzing the Carnot cycle:
Energy Balance:
First look at the two isothermal processes
Process 2 -3 : Isothermal Expansion receiving QH.
Process 4 – 1 : Isothermal Compression, rejecting QC.
and

9. Analyzing the Carnot cycle:
Sec 5.10 : The Carnot Cycle
Analyzing the Carnot cycle:
Energy Balance:
Then look at the two adiabatic processes (Q = 0)
Process 1-2 : Adiabatic Compression.
The term
Thus,
Process 3 – 4 : Adiabatic Expansion.

10. Analyzing the Carnot cycle:
Sec 5.10 : The Carnot Cycle
Analyzing the Carnot cycle:
With
and
Therefore,
We have now proven

11. Added heat causes further rising.
The Carnot Model of a Hurricane
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

12. The Carnot Model of a Hurricane
Isothermal Compression
Adiabatic
vcore>>v outer A
Adiabatic B D
Isothermal Expansion C

13. Sketch the cycle on a P-v diagram.
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%
Sketch the cycle on a P-v diagram.
Evaluate the heat and work for each process in BTU
Evaluate the thermal efficiency. p v

14. Sketch the cycle on a P-v diagram.
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%
Sketch the cycle on a P-v diagram.
Evaluate the heat and work for each process in BTU
Evaluate the thermal efficiency.

15. Sketch the cycle on a P-v diagram.
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%
Sketch the cycle on a P-v diagram.
Evaluate the heat and work for each process in BTU
Evaluate the thermal efficiency.
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi) x
0.643
v (ft3/lb)
u (Btu/lb)
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
v (ft3/lb)
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
Using Table A-2

16. Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in BTU
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
v (ft3/lb)
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
Process U Q W
1 - 2
240
274.8
34.81
2 - 3
3 - 4
4 - 1
Process U Q W
1 - 2
2 - 3
3 - 4
4 - 1
Process 1-2

17. Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in BTU
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
v (ft3/lb)
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
Process U Q W
1 - 2
240
274.8
34.81
2 - 3
3 - 4
4 - 1
Process U Q W
1 - 2
240
274.8
34.81
2 - 3
-201
201
3 - 4
4 - 1
Process 2-3
(adiabatic process)

18. Evaluate the heat and work for each process in Btu
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
v (ft3/lb)
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
Process U Q W
1 - 2
240
274.85
34.81
2 - 3
-201
201
3 - 4
-142.6
4 - 1
For Process 3 – 4:
for the Carnot cycle:
where QH = Q12 = Btu

19. Evaluate the heat and work for each process in Btu
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
v (ft3/lb)
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
Process U Q W
1 - 2
240
274.85
34.81
2 - 3
-201
201
3 - 4
-142.6
4 - 1
continuing for Process 3 – 4:
recalling h = u + pv

20. Evaluate the heat and work for each process in Btu
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in Btu
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
0.368
v (ft3/lb)
0.2677
300.74
172.1
u (Btu/lb)
609.9
1090.0
687.8
419.5
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
v (ft3/lb)
0.2677
300.74
u (Btu/lb)
609.9
1090.0
687.8
Process U Q W
1 - 2
240
274.85
34.81
2 - 3
-201
201
3 - 4
-142.6
4 - 1
continuing for Process 3 – 4:
Then using Table A2 at h4 = Btu/lbm and T4 = 90 deg.
which let the state 4 intensive properties be found:

21. Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in BTU
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
0.368
v (ft3/lb)
0.2677
300.74
172.1
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
201
3 - 4
-142.6
4 - 1
Process U Q W
1 - 2
240
274.85
34.81
2 - 3
-201
201
3 - 4
-134.2
-142.6
-8.32
4 - 1
and for Process 4-1:

22. Evaluate the heat and work for each process in BTU
Sec 5.10 : The Carnot Cycle
Example: (5.76)
Evaluate the heat and work for each process in BTU
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
0.368
v (ft3/lb)
0.2677
300.74
172.1
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
201
3 - 4
-134.2
-142.6
-8.32
4 - 1
Process U Q W
1 - 2
240
274.85
34.81
2 - 3
-201
201
3 - 4
-134.2
-142.6
-8.32
4 - 1
95.2
-95.2
Finally, process 4 – 1:

23. (c) Evaluate the thermal efficiency.
Sec 5.10 : The Carnot Cycle
Example: (5.76)
(c) Evaluate the thermal efficiency.
Isothermal
Adiabatic
Isothermal
Adiabatic
state 1 2 3 4
T (°F/R)
600 90
p (psi)
1541
0.6988 x
0.643
0.368
v (ft3/lb)
0.2677
300.74
172.1
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
201
3 - 4
-134.2
-142.6
-8.32
4 - 1
95.2
-95.2
Thermal efficiency of the cycle:

24. scycle = 0 no irreversibilities present within the system
Clausius Inequality
The Clausius inequality is developed from the Kelvin- Planck statement of the second law and can be expressed as:
The nature of the cycle executed is indicated by
the value of scycle:
scycle = 0 no irreversibilities present within the system
scycle > 0 irreversibilities present within the system
scycle < 0 impossible

25. For an ideal/reversible process
Sec 5.11 : The Clausius Inequality
We have shown that:
and thus
Therefore:
For an ideal/reversible process
Now consider a general process
Each part of the cycle is divided into an infinitesimally small process p v
Then sum (integrate) the entire process
dQ = heat transfer at boundary
T = absolute T at that part of the cycle.

26. For a real process, Qreal > Qreversible
Sec 5.11 : The Clausius Inequality
For a real process, Qreal > Qreversible
Therefore:
We can then define σ, where
and σcycle = 0  reversible process
σcycle > 0  irreversible process
σcycle < 0  impossible process P v
We now also have the mathematical definition of enthalpy.
More on this in Chapter 6

27. 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?
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ

28. Since σcycle is negative, the cycle is impossible.
Sec 5.11 : The Clausius Inequality
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?
W=?
TH= 500 K
TH= 300 K
and
Therefore:
Q = 50 kJ
Since σcycle is negative, the cycle is impossible.

29. Since σcycle is zero, the cycle is internally reversible.
Sec 5.11 : The Clausius Inequality
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?
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ
Since σcycle is zero, the cycle is internally reversible.

30. Since σcycle is positive, the cycle is irreversible.
Sec 5.11 : The Clausius Inequality
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?
W=?
TH= 500 K
TH= 300 K
Q = 50 kJ
Since σcycle is positive, the cycle is irreversible.

31. End of Lecture 21 Slides