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Tc
Th
heat pump
Tc
Th
heat engine
Carnot’s Theorem
We introduced already the Carnot cycle with an ideal gas
Now we show:
1 Energy efficiency of the Carnot cycle is independent of the working substance
2 Any cyclic process that absorbs heat at one temperature, and rejects heat at one other temperature, and is reversible has the energy efficiency of a Carnot cycle
W
hQ
cQ
W
WQQ ch
cQ
W
QhP
reversibleRemark:
Note:P>1Textbook:coefficient of performance
Tc
Th
heat engine X
XW
XhQ
XcQ
Let’s combine a fictitious heat engine X with CX with a heat pump
realized by a reversed Carnot cycle
Tc
Th
heat pumpChQ
CcQ
X CCW
CX WWW
We can design the engine X such that Cc
Xc QQ
Let’s calculate XW XhQ X
cQ with Xh
XX
Q
W
X
XXh
WQ
XWXcQ
X
XW
XW
Xc
X
X Q1
If X would be a Carnot engine it would produce the work CWXc
C
C Q1
with CX XWXc
X
X Q1
> CWXc
C
C Q1
T
c
Th
heat engine X
XW
XhQ
XcQ T
c
Th
heat pump
ChQ
CcQ
X CCW
CX WWW
We can design the engine X such that Cc
Xc QQ
However:
CX WWW >0
0.0 0.2 0.4 0.6 0.8 1.00
2
4
6
8
10
21
01 1
d
d
X/(1-X)
/(1
-)
C X<
C/(1-C)
CX FalseLet X be the heat pump and the Carnot cycle operate like an engine
XC False
CX
1 Energy efficiency of the Carnot cycle is independent of the working substance.
2 Any cyclic process that absorbs heat at one temperature,and rejects heat at one other temperature, and is reversible has the energy efficiency of a Carnot cycle.
Why
Because: X can be a Carnot engine with arbitrary working substance
Carnot’s theorem: No engine operating between two heat reservoirs ismore efficient than a Carnot engine.
Proof uses similar idea as before:
We can design the engine X such that
Again we create a composite device
Tc
Th
heat engine X
W
XhQ
XcQ Tc
Th
heat pumpChQ
CcQ
X C
W operates the Carnot refrigerator
My statementholds man
Let’s assume that CX
Note: this time engine X can be also work irreversible like a real engine does
CX Xh
XQ
W >
ChQ
W
Ch
Xh Q
1
Q
1
0QQ Xh
Ch
Heat transferred from the cooler to the hotter reservoir without doing work on the surrounding
Violation of the Clausius statement CX Rudolf Clausius (2.1.1822 -24.8.1888)
Applications of Carnot Cycles
Any cyclic process that absorbs heat at one temperature, and rejects heat at one other temperature, and is reversible has the energy efficiency of a Carnot cycle.
We stated:
Why did we calculate energy efficiencies for
- gas turbine
- Otto cycle
Because: they are not 2-temperature devices, but accept and reject heat at a range of temperatures
Energy efficiency not given by the Carnot formula
But: It is interesting to compare the maximum possible efficiency of a Carnot cyclewith the efficiency of engineering cycles with the same maximum and minimum temperatures
Consider the gas turbine again
0 2 40
2
4
6
P
V
2 3
41
adiabates
(Brayton or Joule cycle)
1
h
l
P
P1
Efficiency
Ph
Pl
Maximum temperature:
@ 3
Minimum temperature:
@ 1
2 3Heating the gas (by burning the fuel)
4 1 cooling
: T3
: T1
1
h
l
P
P1 with
/)1(
l
1/)1(
h
2
P
T
P
T
2
1/)1(
h
/)1(l
T
T
P
P
2
1
T
T1
Efficiency of corresponding Carnot Cycle 3
1C T
T1
With 22,33 TTT 22,3
1C TT
T1
2
1turbinegas T
T1
22,3
1cycleCarnot TT
T1
0T 2,3
Unfortunately: Gas turbine useless in the limit 0T 2,3
Because: Heat taken per cycle 0
Work done per cycle 0
Absolute Temperature
We showed: Energy efficiency of the Carnot cycle isindependent of the working substance.
Definition of temperature independent of any material property
A temperature scale is an absolute temperature scale if and only if
C2
C1
2
1
Q
Q
T
T
where
,
C1Q and C
2Q are the heats exchanged by a Carnot cycle
operating between reservoirs at temperatures T1 and T2.
Measurement of 2
1c2
C1
C T
T1
Q
Q1
Temperature ratio2
1
T
TT1
T2
W
C2Q
C1Q
As discussed earlier, unique temperature scale requires fixed point
fixC2
C1T
Q
QT or fixC T1T
Kelvin-scale: Tfix =Ttripel=273.16K
It turns out:
proportional to thermodynamic Temperature T
empirical gas temperatureV303P
3g P
Plim
Why
Because: Calculation of efficiency of Carnot cycle based on nRPV
yields2
1C 1
Ta
With fixfix T a=1 T
C2
C1
2
1
Q
Q
T
TFrom definition of thermodynamic temperature
If any absolute temperature is positive all other absolute temperatures are positive
there is an absolute zero of thermodynamic temperature
when the rejected heat C1Q 0
T=0 can never be reached, because this would violate the Kelvin statement
however