lecture 17 heat engines and refrigerators
DESCRIPTION
Lecture 17 heat engines and refrigeratorsTRANSCRIPT
Lecture 17Heat engines and
refrigerators.
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
Hot and cold reservoirs
Stages of the cycle–Absorb heat from hot reservoir (QH)
–Perform mechanical work (W )–Dump excess heat into cold reservoir (QC < 0)
Reservoir = large body whose temperature does not change when it absorbs or releases heat.
Energy flow
Working substance in engine completes a cycle, so ΔU = 0:
H C 0Q Q W
H C H CW Q Q Q Q
This relation follows naturally from the diagram (QH “splits”). Draw it every time!
Energy flow diagrams
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 QH = W, or QC = 0)
Efficiency
what you useEffi ciency
what you pay f or
For a heat engine: H
We
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
eQ
C C
30 J
70 J 70 J
W
Q Q
H C 100 JQ W Q
0 1e
ACT: Two engines
Two engines 1 and 2 with efficiencies e1 and e2 work in series as shown. Let e be the efficiency of the combination. Which of the following is true?
A. e > e1 + e2
B. e = e1 + e2
C. e < e1 + e2
11
1
22
2
We
Q
We
Q
1 2
1
W We
Q
1 2 1 2Q 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
TC
TH
Q3
Q1
Q2
W2
W1e1
e2
The Stirling engine
d
a
b
c
DEMO: Stirling engine
1:
isoch
ori
c
1
hot water 100°C
Gas warms up
2: isotherm
2
hot water 100°C
ΔV
Hot gas
3: iso
choric
3
Room temperature 20°C
Gas cools down
4: isotherm
ΔV
Room temperature 20°C
4
Cold gas
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
Otto cycle
Idealization of the four-stroke gasoline engine
StartQC
Intake: • mix of air and fuel enter• at patm
• n increase QC
Compression:
Adiabatic compression:• temperature increase• no heat exchange• work done on the gas (small because of small pressure)
QC
Combustion: Heating at constant volume• No work
QC
Power stroke:
Adiabatic expansion•Temperature decrease• No heat exchange •Work done by the gas (large because of large pressure)
QC
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 patm
QC
Exhaust:
n decreaseQC
Compression ratio
QC
max
min
compression ratioV
rV
1 12 2 1 1
Compression:
T V TV
1
1 2T rV
1
2 1T T r
1 13 3 4 4
Expansion:
T V T V
1
4 3T rV
13 4T T r
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 QWe
Q Q
QC
3 2 1 4
3 2
T T T T
T T
12 1
13 4
T T r
T T r
1 14 1 1 4
1 14 1
T r T r T T
T r T r
14 1
14 1
1T T r
T T r
1
11e
r
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
11
10 101.27
15 15
e r
e r
Diatomic gas:
772 1.4
5 52
P
V
RC
CR
Why not simply use a higher compression ratio?
• V2 big huge, heavy engine
• V1 small temp. gets too high premature ignition need to use octane in gas to raise combustion temperature
00.10.20.30.40.50.60.70.80.91
1 5 9 13 17
Conmpression Ratio (V2/V1)
Eff
icie
nc
y o
f O
tto
Cy
cle
Monatomic
Diatomic
Nonlinear'
Compression
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 = 1.4 (air), e = 0.56
Realistic cycle of 4-stroke engine
e ~ 0.3
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
Refrigerators
• Absorb heat from cold reservoir (QC > 0) • Work done on engine (W < 0)
• Dump heat into hot reservoir (QH > 0)
(We want as much QC while paying for the smallest possible W .)
Energy balance:
H C
H C
W Q Q
W Q Q
CQK
W
Coefficient of performance (refrigerator)
0 K
Heat pumps
A very efficient way to warm a house: bring heat from the colder outside.
Same energy diagram as refrigerator
Outside of house TC
Inside of house TH
Heat pump Coefficient of performance (heat pump)
HQK
W
This time we are interested in QH :
1 K