lecture 17 heat engines and refrigerators

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Lecture 17 heat engines and refrigerators

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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

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