The Carnot Cycle
Idealized thermodynamic cycle consisting of four reversible processes (any
substance):
! Reversible isothermal expansion (1-2, TH=constant)
! Reversible adiabatic expansion (2-3, Q=0, TH"TL)
! Reversible isothermal compression (3-4, TL=constant)
! Reversible adiabatic compression (4-1, Q=0, TL"TH)
1-2 2-3 3-4 4-1
The Carnot Cycle-2
Work done by gas = PdV, area under the
process curve 1-2-3.
1 dV>0 from 1-2-3
PdV>0
2
Work done on gas = PdV, area under the
process curve 3-4-1
subtract
1
Net work 1 Since dV<0
2 PdV<0
2
4 3 3
The Carnot Principles
The efficiency of an irreversible heat engine is always less than the efficiency of
a reversible one operating between the same two reservoirs. th, irrev < th, rev
The efficiencies of all reversible heat engines operating between the same two
reservoirs are the same. (th, rev)A= (th, rev)B
Both Can be demonstrated using the second law (K-P statement and C-
statement). Therefore, the Carnot heat engine defines the maximum efficiency
any practical heat engine can reach up to.
Thermal efficiency th=Wnet/QH=1-(QL/QH)=f(TL,TH) and it can be shown that
th=1-(QL/QH)=1-(TL/TH). This is called the Carnot efficiency.
For a typical steam power plant operating between TH=800 K (boiler) and
TL=300 K(cooling tower), the maximum achievable efficiency is 62.5%.
Example
Let us analyze an ideal gas undergoing a Carnot cycle between two
temperatures TH and TL.
! 1 to 2, isothermal expansion, U12 = 0
QH = Q12 = W12 = PdV = mRTHln(V2/V1)
! 2 to 3, adiabatic expansion, Q23 = 0
(TL/TH) = (V2/V3)k-1 " (1)
! 3 to 4, isothermal compression, U34 = 0
QL = Q34 = W34 = - mRTLln(V4/V3)
! 4 to 1, adiabatic compression, Q41 = 0
(TL/TH) = (V1/V4)k-1 " (2)
From (1) & (2), (V2/V3) = (V1/V4) and (V2/V1) = (V3/V4)
th = 1-(QL/QH )= 1-(TL/TH) since ln(V2/V1) = ln(V4/V3)
It has been proven that th = 1-(QL/QH )= 1-(TL/TH) for all Carnot engines since
the Carnot efficiency is independent of the working substance.
Carnot Efficiency
A Carnot heat engine operating between a high-temperature source at 900 K
and reject heat to a low-temperature reservoir at 300 K. (a) Determine the
thermal efficiency of the engine. (b) If the temperature of the high-
temperature source is decreased incrementally, how is the thermal efficiency
changes with the temperature. 1
0.8 Lower TH
T 300
th = 1 L = 1
Efficiency
= 0.667 = 66.7% 0.6
TH 900 Th( T )
0.4
Fixed T = 300( K ) and lowering T
L H 0.2
300 0
(T ) = 1
th H
200 400 600 800 1000
TH
T
Temperature (TH)
The higher the temperature, the higher the "quality" 1
of the energy: More work can be done 0.8 Increase TL
Efficiency 0.6
TH( TL )
0.4
Fixed T = 900( K ) and increasing T
H L
0.2
T
(T ) = 1
th H
L 0
200 400 600 800 1000
900 TL
Temperature (TL)
Carnot Efficiency
Similarly, the higher the temperature of the low-temperature sink, the more
difficult for a heat engine to transfer heat into it, thus, lower thermal efficiency
also. That is why low-temperature reservoirs such as rivers and lakes are popular
for this reason.
To increase the thermal efficiency of a gas power turbine, one would like to
increase the temperature of the combustion chamber. However, that sometimes
conflict with other design requirements. Example: turbine blades can not
withstand the high temperature gas, thus leads to early fatigue. Solutions: better
material research and/or innovative cooling design.
Work is in general more valuable compared to heat since the work can convert
to heat almost 100% but not the other way around. Heat becomes useless when it
is transferred to a low-temperature source because the thermal efficiency will be
very low according to th=1-(TL/TH). This is why there is little incentive to
extract the massive thermal energy stored in the oceans and lakes.
