1 lec 16: refrigerators, heat pumps, and the carnot cycle

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Lec 16: Refrigerators, heat pumps, and the Carnot cycle

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• For next time:– Read: § 6-5 to 6-6 and 6-8 to 6-9

• Outline:– Refrigerators– Heat pumps– Carnot cycle

• Important points:– Understand the different performance

measures for cyclic devices.– Realize that COPHP and COPR are different– Start learning to recognize systems that violate

the 2nd Law of Thermodynamics

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Second Law of Thermodynamics

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Review--Heat Engine or Cycle Efficiency

InputEnergyHeat

OutputWorkNet

H

L

H

LH

H

output

H

output

Q

Q1

Q

QQ

Q

W

Q

W

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Refrigerators, air conditioners and heat pumps

Hot reservoir at TH

Cold reservoir at TL

System

HQ

LQ

inputW

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Refrigerators/‘air conditioners’

Remember: the purpose of a refrigerator or ‘air conditioner’ is to remove heat QL from a cold region at TL.

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Refrigerator

Basic components and typical operating conditions

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Heat Pump

Remember: the purpose of a heat pump is to add heat QH to a warm region at TH.

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Coefficient of Performance

Refrigerators/Air conditioners

InputWork

EffectCoolingCOP AC/R

input

L

input

LAC/R W

Q

W

QCOP

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Coefficient of Performance

Refrigerators/Air conditioners

LHinput QQW

LH

LAC/R QQ

QCOP

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Coefficient of Performance for Heat Pumps

InputWork

EffectHeatingCOPhp

input

H

input

Hhp W

Q

W

QCOP

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Coefficient of Performance for Heat Pumps

LH

Hhp QQ

QCOP

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TEAMPLAYTEAMPLAY

Problem 6-52

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Perpetual Motion Machines (PMM)

• PMM1--A perpetual motion machine of the first kind violates the first law or the law of conservation of energy. An example would be an adiabatic system that supplies work with no change in internal energy, kinetic energy or potential energy.

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Perpetual Motion Machines (PMM)--Teamplay

• PMM2--A perpetual motion machine of the second kind violates the second law of thermodynamics.

• Your book has Figure 6-34 and goes into a correct explanation of why it violates the first law.

• The contraption in Figure 6-34 also violates the second law, as does the machine in Figure 6-35. Why?

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Carnot Cycle

• Composed of four internally reversible processes.– Two isothermal processes– Two adiabatic processes

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Carnot cycle for a gas.

TL=const

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The Carnot cycle for a gas might occur as visualized below.

TL

QL

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•This is a Carnot cycle involving two phases--it is still two adiabatic processes and two isothermal processes.

•It is always reversible--a Carnot cycle is reversible by definition.

TL

TL

TL

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Carnot refrigeration cycle for a gas.

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Analytical form of KP Statement:

• Conservation of Energy for a cycle says

• E = 0 = Qcycle - Wcycle, or

Qcycle = Wcycle

• We have not limited the number of heat reservoirs (or work interactions, for that matter). Qcycle could be QH - QC, for example.

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Analytical form of KP statement.

• Let us limit ourselves to the special case of one TER (thermal energy reservoir):

TER

HE W

Q

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TEAMPLAYTEAMPLAY

• Can the system on the previous slide do work while operating in a cycle? If not, what does it violate?

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Analytical form of the KP statement.

• However, it would not violate the KP statement if work were done on the system during the cycle, or if work were zero.

)reservoirgle(sin0Wcycle

)reservoirgle(sin0Q,Also cycle

These are analytical forms of the KP statement.

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Analytical forms of the KP statement.

• Both the equations may be regarded as analytical forms of the KP statement.

• It can be shown that the equality applies to reversible processes and that the inequality applies to irreversible processes.

• Consider a cycle for which the equality applies, that is Qcycle = Wcycle.

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Carnot’s first corollary

• The thermal efficiency of an irreversible power cycle is always less than the thermal efficiency of a reversible power cycle when each operates between the same two reservoirs.

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R I

QHQH

WI

WR

QC=QH-WR IHC WQQ

Hot reservoir

Cold reservoir

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Carnot’s first corollary

• Each engine receives identical amounts of heat QH and produces WR or WI.

• Each discharges an amount of heat Q to the cold reservoir equal to the difference between the heat it receives and the work it produces.

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R I

QH

QH

QH

WI

WR

WR

QC

QC=QH-WR IHC WQQ

Hot reservoir

Cold reservoir

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Carnot’s first corollary.

• Taken together,

• Now reverse the reversible engine.CHCHIR QQQQWW

CHCHIR QQQQWW

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R I

QHQH

QH

WI

WR

QC

QC=QH-WR IHC WQQ

Cold reservoir

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Carnot’s first corollary

•If WI WR, the system puts out net work and exchanges heat with one reservoir. This violates KP. So, WI cannot be WR.

CCIR QQWW

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Carnot’s first corollary

•If WI = WR, QC = Q′C

And the irreversible engine is identical to the reversible engine, i.e., it is just the reversible engine.

CCIR QQWW

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Carnot’s first corollary

• So, WI WR, and

H

II,th Q

W

H

RR,th Q

W

•So th,I th,R

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Carnot’s second corollary

• All reversible power cycles operating between the same two thermal reservoirs have the same thermal efficiencies.

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R1 R2

QH

QH

QH

WR,2

WR,1

WR,1

QC

QC

Hot reservoir

Cold reservoir

QC

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Carnot’s second corollary

• Both engines receive QH, and Qcycle = 0 and Wcycle= 0 for both engines with one reversed because they are both reversible.

• Now, with engine 1 reversed. Wcycle = 0 = WR,1-WR,2

and WR,1 = WR,2

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Carnot’s second corollary

• And

• so

1R,thH

1,R

H

2,R2R,th Q

W

Q

W

1R,th2R,th