Physics Program, Lecture Notes

1. HEAT ENGINES AND REFRIGERATORS
THERMODYNAMICS
Department of Physics, Faculty of Science
Jazan University, KSA
Part-6
222 PHYS

2. Heat engine: Is a device that transforms heat partly
into work (mechanical energy) by a working
substance undergoing a cyclic process.
Several hundreds atmospheres and
water vaporizes at about 500 ªC
Schematic drawing of a steam engine.

3. Various statements of the second law
So far we have presented three statements of the second law of
thermodynamics.
1. The entropy of the system never decreases,
2. You can not change heat energy into work with
100% efficiency. That is, there are no perfect engines.
3. You can not transfer heat energy from a low- temperature reservoir to a
higher temperature reservoir without doing work. That is, there are no
perfect refrigerators.
0S
---Kelvin version
---Clausius version

4. We assume that all thermodynamic processes are reversible.
Carnot engine
CARNOT CYCLE
- Heat engine with the maximum possible efficiency
consistent with 2nd law.
- All thermal processes in the cycle must be reversible
- All heat transfer must occur (isothermally) because
conversion of work into heat is irreversible.
- When the temperature of the working substances
changes, there must be zero heat exchange
(adiabatic process).
- Carnot cycle – consists of two reversible isotherms
and two reversible adiabats
 Qh = heat taken from heat
source
 Qc = heat removed to heat sink
 W = Work done
 Th = temperature of heat source
 TC = temperature of heat sink
Sadi Carnot (1824) Frenchman

5. Tc
Th
V
P Isothermal Expansion
Adiabatic Expansion
Isothermal
Compression
Adiabatic Compression
The Carnot Cycle

6. The purpose of an engine is to transform as much of the extracted heat
into work as possible.
We define “thermal efficiency as
h cW Q Q 
HQ
. .
. . . h
Wenergy you get
energy you pay for Q
  
1 1c c
h h h
Q TW
Q Q T
     
Efficiency of a Carnot engine
In practice, the Carnot cycle cannot be used for a heat engine because the
slopes of the adiabatic and isothermal lines are very similar and the net work
output (area enclosed by PV diagram) is too small to overcome friction & other
losses in a real engine.
Efficiency of Carnot engine - max possible efficiency
for a heat engine operating between Tc and Th

7. The Carnot Cycle: Derivation of Maximum efficiency for a heat engine.
Carnot cycle is a reversible cycle between two heat reservoirs
1
2
ln
2
1 V
V
TRndVPWQ h
V
V
gasbyh  
4
3
ln
4
3 V
V
TRndVPWQ c
V
V
gasonc  
Isothermal processes
Adiabatic processes
1
4
1
1
1
3
1
2






VTVT
VTVT
ch
ch
4
3
1
2
1
4
1
3
1
1
1
2
V
V
V
V
V
V
V
V
 







2
1
2
1
ln
ln
c
c c
h h
h
V
T
Q TV
VQ TT
V
  1 1h c c c
h h h h
Q Q Q TW
Q Q Q T


     
V
p
2
1
3
4
1 and 2: “cold”
3 and 4: “hot”

8. h
h
h
Q
S
T
 
C
C
C
T
Q
S 
hQ
CQ
W
entropy
heat
work
heat
hot reservoir, Th
cold reservoir, TC
Perfect Engines (no extra S generated)
The condition of continuous operation:
c ch
h c c h
h c h
Q TQ
S S Q Q
T T T
      
The work generated during one cycle of a
reversible process:
h c
h c h
h
T T
W Q Q Q
T

  
Carnot efficiency
the highest possible value of the energy conversion
efficiency
max 1 1c
h h
TW
Q T
    
This is the absolute max efficiency of a heat engine

9. Carnot Engine
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 200 400 600 800 1000 1200 1400 1600 1800 2000
Temperature TH (o
C)
efficiency

10. Perpetualmotionmachine

11. Real Engines
Real heat engines have lower efficiencies because the processes within the
devices are not perfectly reversible
The entropy will be generated by irreversible processes:
max1 c
h h
TW
Q T
    
η = ηmax only in the limit of reversible operation.

12. Refrigerators, Air conditioners and Heat Pumps
Refrigerators or air conditioners
Coefficient of
Performance
W
Q
COP c


13. Second Law of Thermodynamics. Refrigerators. Heat Pumps
Schematic representation of a
refrigerator.

14. Entropy and the performance of refrigerators
The Carnot refrigerator
Adjacent fig., shows the basic elements of a
refrigerator. We call it a Carnot refrigerator. W
hT
cT
cQ
hQ
A
B
C
D
0 AV BV CVDV
AP
BP
CP
DP
hT
cT
cQ
hQ
ch
h c
QQ
T T
 are satisfied.
h cW Q Q 


15. Let’s define the “Coefficient of Performance”
cQwhat you want
COP
what you pay for W
 
 
  
h cW Q Q Q
c
h c
Q
COP
Q Q


ch
h c
QQ
T T
Q
c
h c
T
COP
T T


for Carnot refrigerator
HW

16. Example-1
A “perfect” heat engine with η = 0.4 is used as a refrigerator
(the heat reservoirs remain the same). How much heat QC can
be transferred in one cycle from the cold reservoir to the hot
one if the supplied work in one cycle is W =10 kJ?
H C
H H
Q QW
Q Q


 
H
C
H
C
CH
cc
Q
Q
Q
Q
QQ
Q
W
Q
COP




1
1C
H
Q
Q
  1 1 0.6
1.5
1 1 0.4
COP
 
 
 
    
 
kJ15cycle)one(in  COPWQC

17. Example-2 A household refrigerator, COP =4.5, extracts heat from the
food chamber at the rate of .
(a) how much work per cycle is required to operate the refrigerator?
(b) how much heat per cycle is discharged to the room?
cycleJ /250
Solution:
(a)
(b)
4.5 cQ
COP
W
  cycleJ
cycleJ
W /53
5.4
)/250(

303 /h cQ Q W J cycle  

18. ch QQW 
Work done by engine each cycle
h
W
Q
 
J470J890W  J420
The efficiency of the engine
J908
J420
 %2.47472.0 
Th= 550 K
Tc
Engine
Qh
Qc
W
= 470 J
= 890 J
For the engine shown in the figureExample-3a:

19. Example-3b: The engine ejects heat at 10 oC . Find the temperature of the hot
reservoir?
h
c
h
c
T
T
Q
Q

c
h
ch
Q
Q
TT 







J550
J900
K283
h
T K463 C190 o

Tc
Engine
Qh
Qc
W
= 550J
= 900 J
For the engine shown in the figure

20. Example-4: A Carnot engine operates
between a hot reservoir at 650 K and a cold
reservoir at 300 K. If it absorbs 400 J of heat
at the hot reservoir, how much work does it
deliver?
Th= 650 K
Tc= 300 K
Engine
Qh
Qc
W = ?
= 400 J
h
W
Q
 
h
ch
T
TT 






 

h
ch
h
T
TT
QW





 

K650
K300K650
J400W J215

21. Example-5: If the theoretical efficiency of a Carnot engine is to be 100%, the heat
sink must be
(A) at absolute zero.
(B) at 0°C.
(C) at 100°C.
(D) infinitely hot.
Example-6: A Carnot cycle consists of :
(A) two adiabats and two isobars.
(B) two isobars and two isotherms.
(C) two isotherms and two isomets.
(D) two adiabats and two isotherms.

22. Example-7: The temperature inside the engine of a helicopter is 20000C, the
temperature of the exhaust gases is 9000C. The mass of the helicopter is M =
2×103 kg, the heat of combustion of gasoline is qcomb = 47×103 kJ/kg, and the
density of gasoline is  = 0.8 g/cm3. What is the maximum height that the
helicopter can reach by burning V = 1 liter of gasoline?
The work done on lifting the helicopter: MgHW 
1 C
H H
TW
Q T
   For the ideal Carnot cycle (the maximum efficiency):
VqQ combH  The heat released in the gasoline combustion:
m
K
K
m/skg
literkg/liter0.8kJ/kg
2
928
2273
1173
1
8.92000
11047
1
1
3

























H
Ccomb
comb
H
C
T
T
Mg
Vq
H
Vq
T
T
MgH


Thus, H
H
C
Q
T
T
W 





 1