Post on 20-Dec-2015
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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