1. The document discusses the Carnot cycle, which includes the Carnot engine and Carnot refrigerator. It uses the principles of the first and second laws of thermodynamics. 2. A Carnot engine operates between a high temperature TH and low temperature TL. Heat is added at TH and rejected at TL to produce net work. The efficiency depends only on TH and TL. 3. A Carnot refrigerator operates in reverse, absorbing heat at TL and rejecting it at TH, requiring net work. Its coefficient of performance depends only on TH and TL.

1. CARNOT CYCLE
By: Engr. Yuri G. Melliza
• Second Law of Thermodynamics
• Kelvin – Planck Statement
• Carnot engine
• Carnot Refrigerator
• Sample Problems

2. Second Law of Thermodynamics:
Whenever energy is transferred, the level of
energy cannot be conserved and some
energy must be permanently reduced to a
lower level.
When this is combined with the first law of
thermodynamics, the law of energy conservation, the
statement becomes:
Whenever energy is transferred, energy
must be conserved, but the level of energy
cannot be conserved and some energy must
be permanently reduced to a lower level.

3. Kelvin-Planck statement of the Second Law:
No cyclic process is possible whose sole result
is the flow of heat from a single heat reservoir
and the performance of an equivalent amount
of work.
For a system undergoing a cycle:
The net heat is equal to the net work.
Q
W 

 
 dW
dQ Where:
W - net work
Q - net heat

4. CARNOT CYCLE
Nicolas Leonard Sadi Carnot 1796-1832
1.Carnot Engine
Processes:
1 to 2 - Heat Addition (T = C)
2 to 3 - Expansion (S = C)
3 to 4 - Heat Rejection (T = C)
4 to 1 - Compression (S = C)
PV Diagram
TS Diagram

5. P
V
2
1
3
4
T = C
S = C
S = C
T = C

6. T
S
2
1
3
4
T H
T L
Q A
Q R
TH = T1 = T2
TL = T3 = T4

7. Heat Added (T = C)
QA = TH(S)  1
Heat Rejected (T = C)
QR = TL(S) 2
S = S2 - S1 = S3 – S4  3
Net Work
W = Q = QA - QR  4
W = (TH - TL)(S)  5
PV Diagram
TS Diagram

8. %
x
Q
Q
e
%
x
Q
Q
Q
e
%
x
Q
W
e
A
R
A
R
A
A
100
1
100
100




  6
 7
 8
PV Diagram
TS Diagram

9. Substituting eq.1 and eq. 5 to eq 6
%
x
T
T
e
%
x
T
T
T
e
H
L
H
L
H
100
1
100



  9
 10
PV Diagram
TS Diagram

10. TH
TL
W
QA
QR
E
Carnot Engine

11. 2. Carnot Refrigerator: Reversed
Carnot Cycle
Processes:
1 to 2 - Compression (S =C)
2 to 3 - Heat Rejection (T = C)
3 to 4 - Expansion (S = C)
4 to 1 - Heat Addition (T = C)
TS Diagram

12. QR
QA
1
2
4
3
S
T
TH
TL
TH = T2 = T3
TL = T1 = T2

13. Heat Added (T = C)
QA = TL(S)  1
Heat Rejected (T = C)
QR = TH(S) 2
S = S1 - S4 = S2 - S3  3
Net Work
W = Q 4
W = QR - QA  5
W = (TH - TL)(S)  6
TS Diagram

14. Coefficient of Performance
(COP)
L
H
L
A
T
T
T
COP
W
Q
COP


  7
 8
TS Diagram

15. Tons of Refrigeration
211 KJ/min = 1 TR
3. Carnot Heat Pump:A heat pump
uses the same components as the
refrigerator but its purpose is
to reject heat at high energy
level.

16. Performance Factor
(PF)
A
R
R
R
Q
Q
Q
PF
W
Q
PF


  10
 11

17. 13
1
12






COP
PF
L
T
H
T
H
T
PF

18. TH
TL
W
QA
QR
R
Carnot Refrigerator

19. A Carnot engine operating between 775 K and
305K produces 54 KJ of work. Determine the
change of entropy during heat addition.
TH = 775 K ; TL = 305 K
W = 54 KJ
TS Diagram

20. TH
TL
W
QA
QR
E

21. K
KJ
0.015
775
89.04
T
Q
S
-
S
)
S
-
(S
T
Q
KJ
89.04
0.606
54
e
W
Q
Q
W
e
0.606
775
305
775
T
T
T
e
H
A
1
2
1
2
H
A
A
A
H
L
H















22. A Carnot heat engine rejects 230 KJ of
heat at 25C. The net cycle work is 375 KJ.
Determine the cycle thermal efficiency and
the cycle high temperature .
Given:
QR = 230 KJ
TL = 25 + 273 = 298K
W = 375 KJ

23. TL = 298K
TH
W
E
QR = 230 KJ
QA
K
87
.
783
772
.
0
605
)
S
-
(S
Q
T
K
KJ/
-0.772
)
S
-
(S
K
KJ/
772
.
0
)
S
S
(
)
S
S
(
298
230
)
S
S
(
S
S
)
S
S
(
T
Q
)
S
S
(
T
Q
62
.
0
605
375
QA
W
e
KJ
605
QA
)
230
375
(
Q
W
Q
Q
Q
W
1
2
A
H
1
2
3
4
3
4
1
2
3
4
3
4
L
R
1
2
H
A
R
A
R
A




























