Ch20 Young Freedman2

8/12/2019 Ch20 Young Freedman2

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The Carnot Cycle (The Most EfficientHeat Engine)

A reversible cycle described by Sadi Carnot in1824.

The Carnot Theorem gives the theoretical limit

to the thermal efficiency of any heat engine. The Carnot cycle consists of:

Operates between two temperatures: 2 reversible isothermal processes in which

2 reversible adiabatic processes in which

An Ideal Gas as its working substance

andH C

T T

H CT T

Q


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Steps of the Carnot Cycle animation


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Details on the Carnot Cycle

The isothermal expansion (ab) and compression (cd):

0isothermalU (Tis constant and U(T) is a functionof Tonly for an Ideal Gas.)

ln bH ab H

a

VQ W nRT V

(ab : isothermal expansion)

ln ln ( )dC cd C C c d

c

c

d

V VQ W nRT nRT V V

V V

(cd: isothermal compression)


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Details on the Carnot Cycle

(by definition)

The adiabatic expansion (bc) and compression (da):

0adiabaticQ

From Section 19.8, we learned that1

for adiabatic processes.TV const

(bc: adiabatic expansion)1 1H b C cT V T V

(da: adiabatic compression)1 1H a C dT V T V

Dividing these two equations gives,

b c

a d

V V

V V


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Efficiency of the Carnot Cycle

From definition, we have 1 CH

QeQ

Using our results for QCand QHfrom the isothermalprocesses,

ln ln1 1

ln lnC c d C c d

H b a H b a

nRT V V T V V e

nRT V V T V V

From the adiabaticprocesses, we have

1 1C

H

C

H

Te

Q T

Q

b a c d

V V V V

ln1

lnc d

b a

V V

V V

(Carnot Cycle only)

(Tmust be in K)


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Efficiency of the Carnot Cycle

1 CcarnotH

TeT

(Carnot Cycle)

General Comments:

Higher efficiency if either TCis lower and/or THis higher. For any realistic thermal process, the cold reservoir is far

above absolute zero and TC> 0. Thus, a realistic e is strictly less than 1! (No 100%

efficient heat engine)Realistic heat engines must take in energy from the high T

reservoir for the work that it produces AND some heatenergy must be releasedback to the lower Treservoir.

(Kelvin-Plancks Statement)skip


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Internal Combustion EngineThe Otto Cycle

A fuel vapor can be compressed, then detonated to reboundthe cylinder, doing useful work.

animationhttp://auto.howstuffworks.com/engine1.htm


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The Otto Cycle

a b b, c c d

2 ( ) 0H V c bQ U nC T T

4 ( ) 0C V a d Q U nC T T

ris the compression ratio (8 to 13)

intake

exhaust

intake exhaust

For the two constant Vprocesses: 2 and 4,we can calculate,


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The Otto Cycle

Applying the definition of efficiency,

1 1 1V a dC a d

H V c b c b

nC T T Q T Te

Q nC T T T T

Now, we can utilize the two adiabatic processes: 1 and 3,

1 1a a b b

T V T V and

1 1c c d d

T V T V

1 1

a bT rV T V

11

c dT V T rV

1a b

T r T 1c dT T r


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The Otto Cycle

Substituting Tb and Tc into the efficiency equation, we have,

1 1 11 1 1a d a

d

a

c b da

d d

a

T T T T T T e

T T T r T r T T r

1

11e

r

Using a typical value for the compression ratio r = 8 and = 1.40 gives,

0.56 (or 56%)eNote: This is a theoretical value.Realistic gasoline engine typically

has e ~ 35%.


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The Diesel Cycle

Key difference: No fuel in cylinder at the beginning of the

compression stroke (process 1)

Fuel is injected only moments beforeignition in the power stroke

No fuel until the end of the adiabaticcompression can avoid pre-ignition

Compression ratio rvalue can be higher (15to 20)

Higher temperature can be reached duringthe adiabatic compression

Higher e and no need for spark plugs

Dsteam engine


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RefrigeratorsRefrigerators are basically heat engine running in reverse.

Heat from inside the refrigerator (cold Treservoir) is absorbed and releasedinto the room (high Treservoir) with the inputof mechanical work.


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Refrigerators

what you get

what you put inC C

H C

Q QK

W Q Q

Coefficient of Performance for a Refrigerator

From 1st

Law, 0cycle C H H C

U Q Q W Q Q W

(Note: we have put in the explicit signs according to our sign convention.)

|QC| (absorbed)positive |QH| (released) negative |W| (work is done on working substance by motor) negative

explicit signs


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A perfect Refrigerator

C C

H C

Q Q

K W Q Q

A perfect refrigerator means ( ).This means that

0H CQ Q or W

No mechanical work Wis needed to

transfer heat from the cold reservoirto the hot reservoir.

The Clausiuss statement of the 2ndLaw does notallow this !

realisticK

K


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EntropyRecall from a Carnot Cycle, we have derived the following relationship:

C C

H H

Q T

Q T 0H C

H C

Q Q

T T

Formally, we can rewrite this as,

0cycle

Q

T

(We have absorbed the explicitsign back into the Q variable.)

where Q represents the heat absorbed/released along the isotherm at temp T.


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Entropy

Any reversible cycles can be approximated as a seriesof Carnot cycles !


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Entropy

This suggests that the following generalization to be true for any reversibleprocesses,

0r

cycle

dQ

T

where,dQr is the infinitesimal heat absorbed/released by the system at an

infinitesimal reversible step at temp T.

denotes the integration evaluated over one complete cycle.cycle


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Entropy

We have seen this property previously,

0cycle

dU Changes in the internal energy U over aclosed cycle is zero!

This is a consequence of the fact thatU

is astate variable

anddU

for anyprocesses depends on the initial and final states only.

Thus the result indicates that there is another state variable Ssuch that,

0r

cycle

dQ

T

and 0cycle

dSdQdST

This new state variable Sis called the entropy of the system.


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Entropy: DisorderRecall that the 2ndLaw of Thermodynamics is a statement on natures

preferential direction for systems to move toward the state of disorder.Let see how Entropy is a quantitative measure of disorder.

Recalling our previous derivation, was from the isothermal branch ofthe infinitesimal Carnot cycle. Let look at an isothermal expansion of an ideal

gas microscopically:Intuitively, as the gas expands into a biggervolume, the degree of randomness for thesystem increases since molecules now have

more choices in position for them to moveabout. One can associate the increase inrandomness to the ratio:

V dVor

V V

/d Q T

(Tstays the same avg. KEstays the same)


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Entropy: DisorderSince this is an isothermal process, we have the following relation from the 1st Law:

nRTdQ dW PdV dV

V

1 dQ dV

nR T V

So, the newly introduced macroscopic variable S(entropy),

dQdS

T

is proportion to the degree of disorder of the system.

f

i

dQS

T

/S J K

dSis an infinitesimal entropy change for a reversible process at temperature T.For any finite reversibleprocess, the total entropy change Sis,


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2ndLaw (Quantitative Form)

The total entropy (disorder) of an isolatedsystem in anyprocesses can never decrease.

0

( 0 ; 0 )

tot sys env

tot tot

S S S

S reversible S irreversible

Nature always tends toward the most probable macrostates

[states with the highest S(disorder)] in any processes.

system

environmentQ

W


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Entropy Changes for DifferentProcesses

1. GeneralReversible Processes:

V

i

f

P

f f

r

i i

dQS dS

T

NOTE: in most applications,it is the change in entropy Swhich one typically needs tocalculate and not Sitself.

Note: Sis a state variable, Sis the same for allprocesses(including irreversible ones) with the same initial and finalstates!


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Dividing Ton both sides and integrating,

f f

Vr

i i

nC dTdQ dV S nR

T T V

We have,

ln lnf fVi i

T VS nC nR

T V


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4. Calorimetric Changes:

f f

i i

dQ mcdT

dQ mcdT S

T T

If c is constant within temperature range,

ln f

i

TS mc

T


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5. During Phase Changes (or other isothermalProcesses):

1dQS dQ

T T

(Tstays constant during aphase change.)

Q mLS

T T


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6. Irreversible Processes:

Although for a given an irreversibleprocess, we cannot writedS = dQr/T, Sbetween a well definite initial state a and final stateb can still be calculated using a surrogate reversible processconnecting a and b. (Sis a state variable!)

Example 20.8: (free expansion of an ideal gas)

Initial State a: (V,T) Final State b: (2V,T)

Since Q=W=0, U=0.For an ideal gas, thismeans that T=0 also.

Although Q=0, but Sis not zero!


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Sin a Free Expansion

Important point: Since Sis a state variable, Sis the same forany processes connecting the same initiala and final b states.

In this case, since Tdoes not change, we can use an surrogate isothermalprocess to take the ideal gas from state a (V,T) to state b (2V,T) to calculate S.

Surrogate

Isothermal Expansion

Applying our general formula,

ln lnf fV

i i

T VS nC nR

T V

we have,

lnV

TS nC

T

2ln ln 2 5.76 /

VnR nR J K

V

(n=1)

Ch20 Young Freedman2

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