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Ch19 Heat Engines and Refrigerators

講者： 許永昌 老師

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ContentsTurning Heat into WorkEnergy transfer diagrams

Heat engine: Thermal EfficiencyRefrigerators:

Coefficient of performance.

Practical applications:Ideal-gas Heat EnginesIdeal-gas Refrigerators

The Limits of EfficiencyThe Carnot Cycle.

Entropy

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Turning Heat into Work ( 請預讀 P567)

Example:

The gas do positive work on the piston.W<0 (work done by the environment on the

system)Ws=-W (work done by the system)W+Q=DEth.

Q=Ws+DEth.

Based on the 2nd LawTflame > Tf > Ti.

SystemEmech+Eth

Q

W

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Energy Transfer diagram ( 請預讀 P568)

Three diagrams:Phase diagrams: tell us the phase of a matterPV diagram for ideal gas:

tell us the change of state of the described system If n is provided, we can get P, V, T and W.

If CV is provided, we can get DEth and Q.

Energy Transfer diagram for all substance: Emphasize the ideas about the transfer and

transformation of energy for a CYCLIC PROCESS. Cyclic process: A system will return to its initial state at

the end of process.

P

V

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Energy Transfer diagram (continue)ENERGY RESERVOIR:

An object or a part of the environment so large that its temperature does not change when heat is transferred between the system and the reservoir.

Hot reservoir: TH > Ts.Cold reservoir: TC < Ts.

Where Ts is the temperature of the system and it is changeable during the process.

QH & QC:Because we already use the arrow to

describe the direction of energy transfers, therefore, QH & QC >0 in this diagram .

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Energy Transfer Diagram (continue)Properties:

The process described in this diagram must be a cyclic process for REUSE ( 不然就不是引擎了 ).

Work can be transformed into heat with 100% efficiency.

There are no perfect engines that turn heat into work with 100% efficiency.Th

Tc

Th

Tc

STOP TO THINK: In Ch17, we get Q=Wout of an isothermal process, i.e. the heat is 100% transformed into work. Is the 3rd property wrong?

No perfect engine

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Heat engine: Thermal Efficiency ( 請預讀 P570~P572)

Heat engine:It must be a closed-cycle device

for reuse. DEth=0.

Because of the 1st Law of thermodynamics (energy conservation), we get QH=Wout+QC.

Thermal efficiency:

h~0.1~0.5 for real engine.

What you get1 .

What you had to payout C

H H

W Q

Q Q

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A Heat-Engine ExampleDescription:

PV diagram:

Energy Transfer diagram:

1 2 23 3112

STOP TO THINK: How to calculate its h?TH & TC?

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RefrigeratorsSimilar to the heat engine.It pumps energy from the cold

reservoir to the hot one.The coefficient of performance (COP):

STOP TO THINK:Since the 2nd Law says that heat is not

spontaneously transferred from a colder to a hotter object, how can we transfer energy from the cold reservoir to the hot reservoir?

What you get.

What you had to payC

in

QK

W

ThTc

Adiabaticprocess

Tc’

Th’

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No perfect Heat EngineProof:

Suppose we had a perfect engineWe could use its output to provide the work input to

the refrigerator.The combined system becomes a perfect refrigerator .

It conflicts with the 2nd Law of thermodynamics.

In fact, we can get the upper limit of the efficiency.

, 1.K

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HomeworkStudent Workbook

19.219.319.519.6

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Ideal-Gas Heat Engines( 請預讀P575~P579)Strategy:

Model: Identify each process in the cycle.

Visualize: Draw the PV diagram of the cycle.

Solve: Use the ideal-gas law to complete your knowledge of n, P, V, and T at one

point in the cycle. Calculate Q, Ws, and DEth for each process by W+Q=DEth. Add just the positive values of Q to find QH (What you pay) Wout=SiWi. Calculate the quantities you need to complete the solution.

Assess: Is (DEth)net=0? Do all the signs of Ws and Q make sense? Does h have a reasonable value?

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Ideal-gas summaryTable 19.1

Table 19.2

Process Gas Law Work W Heat Q DEth

Isochoric DV=0 0 Q=nCVDT DEth=Q

Isobaric DP=0 -PDV Q=nCpDT DEth=Q+W

Isothermal DT=0 nRTln(Vf/Vi) Q=-W DEth=0

Adiabatic nCVDT Q=0 DEth=W

ANY PV=nRT Area under curve

DEth=Q+W

i i f fPV P V

,P

V

C

C

Monatomic Diatomic

Eth 3/2*nRT 5/2*nRT

CV 3/2*R 5/2*R

CP 5/2*R 7/2*R

g 5/3~1.67 7/5=1.4

這些表上除了 PV=nRT外，都是各位可以推得的，請務必會推導。這些乃是 Ch 16~Ch 18的重點。

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Work done during one full cycleFor a heat engine, the gas must go around

the PV trajectory in a clockwise direction for Wout to be positive.Besides, a refrigerator uses a

counterclockwise(ccw) cycle.

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Stop to Think ( 請預讀 P577)

Stop to think 19.3What is the thermal efficiency of this heat

engine?a. 0.10b. 0.50c. 0.25d. 4e. Can’t tell without knowing QC. V(m3)

P(Pa)

QH=4000J40,000

20,000

0.1 0.2

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Some Examples of Heat Engines

Otto Cycle:Diesel Cycle:Brayton Cycle:

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The Brayton Cycle ( 請預讀 P578~P579)

12 & 34: adiabaticPVg=constant.

23 & 41: isobaric

Thermal Efficiency:

max 1 4

min 2 3P

P V Vr

P V V

.PP

CQ nC T P V

R

min 4 1 1/

1 /max 3 2

1 11 1 1 * 1 .out C

pH H P p

P V VW Qr

Q Q P V V r r

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The conditions of TH & TC ( 請預讀 P580)

Key concept: The heat is always transferred from a hotter object to a colder object.Heat Engine:

THMax(T of Q>0 processes). TCMin(T of Q<0 processes).

Refrigerator: THMin(T of Q<0 processes). TCMax(T of Q>0 processes).

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HomeworkStudent Workbook

19.819.1119.1219.14

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Ideal-Gas Refrigerators ( 請預讀P579~P581)

To make a Brayton refrigerator you must both reverse the cycle and change the hot and cold reservoirs.Example 19.3 is a very nice example.

TH T2=108oC TCT4=-23oC

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HomeworkStudent Workbook

19.1519.16

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THE LIMITS OF EFFICIENCY ( 請預讀P582~P588)

很多課本都是由此推導出 entropy 的概念，可見得有多重要。

What is a perfectly reversible engine?It has the maximum efficiency.

Q:Compare to what?A perfectly reversible engine must use only

two types of processes.Q: How about the other processes?

The Carnot Cycle and its efficiency.Entropy.

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What is a perfectly reversible engine?A perfectly reversible engine:

A device that can be operated as either a heat engine or a refrigerator between the same two energy reservoirs and with the same energy transfers, with only their direction changed.

A perfectly reversible engine has maximum efficiency.

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A perfectly reversible engine must use only two types of processes ( 請預讀 P584)

A perfectly reversible engine must use only two types of processes:Adiabatic Process:

Frictionless mechanical interactions with no heat transfer (Q=0).

Isothermal Process: During this process, the heat should be transferred

infinitely slowly. reversible.

Any engine that uses only these two types of processes is called a Carnot engine.

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The Carnot Cycle ( 請預讀 P585~P588)

Any Carnot engine operating between TH and TC must have exactly the same efficiency as any other Carnot engine operating between the same two energy reservoirs.

The Carnot Cycle is an ideal-gas cycle.Find the thermal efficiency Wout/QH:

Known: TH and TC. Ideal gas: PV=nRT. Adiabatic Process:

TVg-1=C=constant. Q=0. W=DE-Q=nCVDT.

Isothermal Process: W=-PdV=-nRTD(lnV) Q=DE-W=-W.

------------------------------ We get 1 C

H

T

T 5 10 15 20

10

20

30

40

50

60

70

QH

QC 1

2

3

4

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The maximum efficiencyThe thermal efficiency:

The coefficient of performance:

dQrevTdS.

S is entropy and it is a state variable. (Why?)

Carnot 1 1C

H

T

T

CarnotC

H C

TK K

T T

carnot 0dQ

T

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EntropyIn microscopic scale, entropy S a macrostate is

related to the number of microstates W.S kBln W .

dQrev=TdS is a statistic result.It is related to the Lagrange multiplier.dQ is the real energy transfer by heat and dQdQrev

usually.For a reversible process, dQ=TdS which is a statistic

result.For an isolated system, DS 0.

2nd Law of Thermodynamics. e.g. free-expansion of ideal gas.

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For an isolated process, DS 0Macroscopic:

h hcarnot.

Microscopic:The system tends to have the maximum

number of microstates.

0.C C H C

H H H C

Q T Q Q

Q T T T

0.dQ

T

a

b

0a

rev

bdashed

dQ dQ

T T

0 if 0 for this irreversible process.b

rev

a

dQS dQ

T

用虛線是因為不可逆過程在過程中，一般我們無法決定它所有的巨觀量如 P&V 。

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HomeworkStudent Workbook:

19.1719.18

Student Textbook:19.4319.54

請自行製作 terms and notation 的卡片。