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Lec 17: Carnot principles, entropy

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• For next time:– Read: § 6-9 to 6-14 and 7-1– HW 9 due October 29, 2003

• Outline:– Carnot’s corollaries– Kelvin temperature scale– Clausius inequality and definition of entropy

• Important points:– Do not forget the first law of thermodynamics and

the conservation of mass – we still need these to solve problems

– Kelvin temperature scale helps us find maximum efficiencies for power cycles

– Understand how entropy is defined as a system property

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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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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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Carnot’s second corollary

• And

• so

1R,thH

1,R

H

2,R2R,th Q

W

Q

W

1R,th2R,th

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Refrigerators and heat pumps

• We can show in a manner parallel to that for the Carnot corollaries:

1. The COP of an actual, or irreversible, refrigeration cycle is always less than the COP for a reversible cycle operating between the same two reservoirs.

2. The COP’s of two reversible refrigerators or heat pumps operating between the same two reservoirs are the same.

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Kelvin Temperature Scale

• The thermal efficiency of all reversible power cycles operating between the same two thermal energy reservoirs are the same.

• It does not depend on the cycle or the mechanism.

• What can depend upon?

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Kelvin Temperature Scale

= (L, H)

• where the ’s are temperatures.

• It is also true that =1 -(QL/QH)rev, so it must be true that

),( HL

REVH

L

Q

Q

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Kelvin Temperature Scale

• The previous equation provides the basis for a thermodynamic temperature scale--that is, one independent of the working fluid’s properties, of the cycle type, or any machine.

• We are free to pick the function any way we wish. We will go with the following simple choice:

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Kelvin Temperature Scale

H

LHL T

T),(

•So, H

L

REVH

L

T

T

Q

Q

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• This only assigns the T ratio. We proceed by assigning Ttp 273.16 K to the triple point of water. Then, if that is one reservoir,

Kelvin Temperature Scale

REVtpQ

QK16.273T

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(Reversible) Power Cycles’ Efficiency

• With Carnot’s first corollary, the maximum efficiency one can expect from a power-producing cycle is that of a reversible cycle.

• Also, recall that for any cycle (reversible or irreversible)

H

L

in

outth Q

Q

Q

Q 11

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(Reversible) Power Cycles’ Efficiency

revH

L

revin

outrevth Q

Q

Q

Q

11maxmaximum,

•But H

L

revH

L

T

T

Q

Q

•SoH

L

revH

Lrevmax T

T1

Q

Q1

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TEAMPLAYTEAMPLAY

• Many power cycles for electricity supply operate between a steam supply reservoir of about 1,000 °F and a heat rejection reservoir of about 70 °F.

• What is the maximum thermal efficiency you can expect from such a system?

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Carnot Efficiency

• This maximum thermal efficiency for a power cycle is called the “Carnot Efficiency.”

H

L

revH

Lrevmax T

T1

Q

Q1

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Efficiencies

• So, if an efficiency is obtained that is too large, it may be an impossible situation:

engineheatimpossible

engineheatreversible

engineheat(real)le,irreversib

,

,

,

revthth

revthth

revthth

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TEAMPLAYTEAMPLAY

Problem 6-90E

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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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Coefficient of Performance

Refrigerators/Air conditioners

InputWork

EffectCoolingCOPac

input

L

input

Lac W

Q

W

QCOP

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For a Carnot refrigerator or air conditioner

LH

L

LH

Lac Q-Q

Q

Q-Q

QCOP

LH

L

T-T

TacCOP

Substituting for temperatures

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Coefficient of Performance for Heat Pumps

InputWork

EffectHeatingCOPhp

input

H

input

Hhp W

Q

W

QCOP

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For a Carnot heat pump

LH

H

LH

Hhp Q-Q

Q

Q-Q

QCOP

LH

H

T-T

ThpCOP

Substituting for temperatures

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TEAMPLAYTEAMPLAY

Problem 6-104

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This is going to seem pretty abstract..so hang on for the ride!

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Clausius Inequality

Another corollary of the 2nd Law.

Now we will deal with increments of heat and work, Q and W, rather than Q and W.

We will employ the symbol , which means to integrate over all the parts of the cycle.

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Clausius Inequality

The cyclic integral of Q/T for a closed system is always equal to or less than zero.

0T cycle

Q

.integralcyclicasignifiestermthe Remember,

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Look at a reversible power cycle

Hot reservoir

Cold reservoir

System

HQ

LQ

outRW

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Look at a reversible cycle:

LHcycleQQQ

We know:

And:

0WoutRcycleQ

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For the reversible cycle

cycleT

Q

L

L

H

H

T

Q

T

Q Look at Q/T:

Since the heat transfer occurs at constant temperature, we can pull T out of integrals:

cycleT

Q

L

L

H

H

T

Q

T

Q

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For the reversible cycle

H

L

revH

L

T

T

Q

Q

or

L

L

H

H

T

Q

T

Q

This allows us to write:

cycleT

Q

0T

Q

T

Q

H

H

H

H

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What were signs of irreversibilities?

• Friction

• unrestrained expansion

• mixing

• heat transfer across a temperature difference

• inelastic deformation

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For an irreversible cycle

Hot reservoir

Cold reservoir

System

HQ

LIQ

outIW

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For an irreversible cycle

outRoutI WW For the same heat input:

For both cycles we can write:

LHoutR QQW LIHoutI QQW and

Apply inequality:

LHLIH QQQQ or LLI QQ LLI QQ or

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Apply cyclic integral

cycleT

Q

L

LI

H

H

T

Q

T

Q 0

T

Q

T

Q

L

L

H

H

For the irreversible cycle:

0T

Q

cycle

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Clausius Inequality

0T

Q

cycle

And so, we have

Where the goes with an irreversible cycle and the = goes with a reversible one.

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Clausius Inequality

Why do we go through the “proof”?…The inequality will lead to a new property.

0T

Q

cycle

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Let’s look at a simple reversible cycle on a p-v digram with two

processes

P

1

2

.

.A

B

Let A and B both be reversible

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Find the cyclic integral:

2

1 B

2

1 A

1

2 B

2

1 Acycle T

Q

T

Q

T

Q

T

Q0

T

Q

2

1 B

2

1 A T

Q

T

Q

We can then write:

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What can we conclude?

T

QThe integral is the same for all reversible

Paths between points (states) 1 and 2. This integral is only a function of the end states and is therefore a property of the system.

RevInt T

Q=dS

We’ll define a new property, entropy as:

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Entropy

2

1 revintT

q

Rlb

Btuor

Kkg

kJ

m

s2 - s1 =

Units are