Lecture 17 heat engines and refrigerators

Lecture 17 heat engines and refrigerators

1. Lecture 17
Heat engines and refrigerators.

2. Heat engine
= device with a working substance (eg. gas) that operates in
a thermodynamic cycle. In each cycle, the net result is that
the system absorbs heat (Q > 0) and does work (W > 0).
Examples:
- Car engine: burns fuel, heats air inside piston. Piston expands,
does mechanical work to move car
- Animal: burns “food” to be able to move

3. Hot and cold reservoirs
Stages of the cycle
–Absorb heat from hot
reservoir (Q
H
)
–Perform mechanical work (W )
–Dump excess heat into cold
reservoir (Q
C
< 0)
Reservoir = large body whose temperature does not
change when it absorbs or releases heat.

4. Energy flow
Working substance in engine completes
a cycle, so ΔU = 0:
( )
H C
0Q Q W+ - =
H C H C
W Q Q Q Q= + = -
This relation follows naturally from the diagram
(Q
H
“splits”). Draw it every time!

5. Energy flow diagrams

6. Limitations
We are not saying that you can absorb 10 J of heat from a hot
source (a burning fuel) and produce 10 J of mechanical work...
You can absorb 10 J of heat from a hot source (a burning fuel) and
produce 7 J of mechanical work and release 3 J into a cold source
(cooling system).
… so at the end you absorbed 10 J but used (= converted to
work) only 7 J.
(We’ll see later that it is impossible to make Q
H
= W, or Q
C
= 0)

7. Efficiency
what you use
Efficiency
what you pay for
=
For a heat engine:
H
W
e
Q
=
Example: A heat engine does 30 J of work and exhausts 70 J by heat
transfer. What is the efficiency of the engine?
H
0.3 (or 30%)
W
e
Q
= =
C C
30 J
70 J 70 J
W
Q Q
=
= Þ = -
H C
100 JQ W Q= - =
0 1e< <

8. ACT: Two engines
Two engines 1 and 2 with efficiencies e
1
and e
2
work in series as
shown. Let e be the efficiency of the combination. Which of the
following is true?
A. e > e
1
+ e
2

B. e = e
1
+ e
2

C. e < e
1
+ e
2

1
1
1
2
2
2
W
e
Q
W
e
Q
=
=
1 2
1
W W
e
Q
+
=
1 2 1 2
Q Q W Q= + >
1 2
1 1
Q Q
<
1 2
1 1
W W
Q Q
= +
1 2
1 2
W W
Q Q
< +
T
C
T
H
Q
3
Q
1
Q
2
W
2
W
1e
1
e
2

9. The Stirling engine
d
a
b
c
DEMO:
Stirling
engine
1
:

i
s
o
c
h
o
r
i
c
1
hot water 100°C
Gas warms
up
2: isotherm
2
hot water 100°C
ΔV
Hot gas
3
:

i
s
o
c
h
o
r
i
c
3
Room temperature 20°C
Gas cools
down
4: isotherm
ΔV
Room temperature 20°C
4
Cold gas

10. Internal combustion engines
The heat source (fuel combustion) is inside the engine and
mixed with the working substance (air)
Note: No real cold and hot reservoirs.
- Otto (4 stroke gasoline)
- Diesel

11. Otto cycle
Idealization of the four-stroke gasoline engine
Start
Q
C

12. Intake:
• mix of air
and fuel enter
• at p
atm
• n increase Q
C
Compression:
Adiabatic compression:
• temperature increase
• no heat exchange
• work done on the gas
(small because of small
pressure)
Q
C

13. Combustion:
Heating at constant
volume
• No work
Q
C
Power stroke:
Adiabatic expansion
•Temperature decrease
• No heat exchange
•Work done by the gas
(large because of large
pressure)
Q
C

14. Heat reject:
When piston at the bottom,
very fast cooling, i.e. at
constant volume
• Excess heat absorbed by
water jacket
• Valve opens ® Pressure drops
to p
atm
Q
C
Exhaust:
n decrease
Q
C

15. Compression ratio
Q
C
max
min
compression ratio
V
r
V
=
1 1
2 2 1 1
Compression:
TV TV
g g- -
=
( )
1
1 2
T rV
g-
=
1
2 1
T Tr
g-
=
1 1
3 3 4 4
Expansion:
TV TV
g g- -
=
( )
1
4 3
T rV
g-
=
1
3 4
T T r
g-
=

16. Efficiency of the Otto cycle
( )
( )
H 3 2
C 1 4
V
V
Q nC T T
Q nC T T
= -
= -
0
0
>
<
H C
H H
Q QW
e
Q Q
+
= =
Q
C
3 2 1 4
3 2
T T T T
T T
- + -
=
-
1
2 1
1
3 4
T Tr
T T r
g
g
-
-
=
= 1 1
4 1 1 4
1 1
4 1
T r Tr T T
T r Tr
g g
g g
- -
- -
- + -
=
-
( )( )
( )
1
4 1
1
4 1
1T T r
T T r
g
g
-
-
- -
=
-
1
1
1e
r
g-
= -

17. In-class example: Otto’s engine efficiency
Two idealized Otto cycles have a compression ratio of 5 and 10,
respectively. What is the ratio of their efficiencies? Take the
gas mixture to be a diatomic gas.
A.1.27
B.1.33
C.1.50
D.1.67
E.2.00
( )
( )
10
?
5
e r
e r
=
=
=
( )
( )
1.4 1
1.4 1
1
1
10 10
1.27
15
1
5
e r
e r
-
-
-
=
= =
=
-
Diatomic gas:
7
72
1.4
5 5
2
P
V
R
C
C
R
g= = = =

18. Why not simply use a higher compression ratio?
• V
2
big  huge, heavy engine
• V
1
small  temp. gets too high  premature ignition  need
to use octane in gas to raise combustion temperature
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 5 9 13 17
Conmpression Ratio (V2/V1)
E
f
f
i
c
i
e
n
c
y

o
f

O
t
t
o

C
y
c
l
e
Monatomic
Diatomic
Nonlinear'
Compression

19. Real four-stroke engine
The Otto cycle is an idealization:
• assumes ideal gas
• neglects friction, turbulence, loss of
heat to walls
For r = 8 and g = 1.4 (air), e = 0.56
Realistic cycle of 4-stroke engine
e ~ 0.3

20. Diesel engine
• Intake and compression happen without fuel.
• Fuel is injected after compression, and keeps pressure constant.
• Compression rate r is 15-20
® Larger temperatures
® Fuel ignites spontaneously
• Ideal efficiency e ~ 0.65-0.70

21. Refrigerators
• Absorb heat from cold reservoir (Q
C
> 0)
• Work done on engine (W < 0)
• Dump heat into hot reservoir (Q
H
> 0)
(We want as much Q
C

while paying for the
smallest possible W .)
Energy balance:
H C
H C
W Q Q
W Q Q
= +
= -
C
Q
K
W
=
Coefficient of
performance
(refrigerator)
0K< < ¥

22. Heat pumps
A very efficient way to warm a house: bring heat from the colder outside.
Same energy diagram as refrigerator
Outside of house T
C
Inside of house T
H
Heat pump
Coefficient of
performance
(heat pump)
H
Q
K
W
=
This time we are interested in Q
H
:
1K< < ¥