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The Carnot cycle, The Carnot principles, Carnot efficiency, and quality of energy
1.
The Carnot Cycle (YAC 5-7 to 5-11)
• Idealized thermodynamic cycle consisting of four reversible processes
(working fluid can be any substance):
• The four steps for a Carnot Heat Engine are:
Reversible isothermal expansion (1-2, TH= constant)
Reversible adiabatic expansion (2-3, Q = 0, THTL)
Reversible isothermal compression (3-4, TL=constant)
Reversible adiabatic compression (4-1, Q=0, TLTH)
1-2 2-3 3-4 4-1 Carnot cylce.ppt
Modified 10/9/02
2.
The Carnot Cycle (cont’d)
Work done by the gas = PdV, i.e. area
under the process curve 1-2-3.
1 dV>0 from 1-2-3
PdV>0
2
TL = co 3
nst.
Work done on gas = PdV, area under the
process curve 3-4-1
subtract
Net work 1 Since dV<0
2 PdV<0
2
4 3 3
3.
The Carnot Principles/Corollaries
1. The efficiency of an irreversible, i.e. a real, heat engine is always less than the
efficiency of a reversible one operating between the same two reservoirs. th, irrev
< th, rev
2. The efficiencies of all reversible heat engines operating between the same two
thermal reservoirs are the same. (th, rev)A= (th, rev)B
• Both of the above statements 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 (hope to) achieve. (see
YAC: 5.8, for proof)
• Thermal efficiency th=Wnet/QH=1-(QL/QH) = f(TL,TH)
• In the next slide we will show that th=1-(QL/QH)=1-(TL/TH).
• This relationship is often called the Carnot efficiency since it is usually
defined in terms of a Carnot Heat Engine .
4.
Carnot Efficiency
Consider an ideal gas undergoing a Carnot cycle between two temperatures T H and TL.
1 to 2, isothermal expansion, U12 = 0
QH = Q12 = W12 = PdV = mRT Hln(V2/V1) (1)
2 to 3, adiabatic expansion, Q23 = 0
(TL/TH) = (V2/V3)k-1 (2)
3 to 4, isothermal compression, U34 = 0 TL = co
nst.
QL = Q34 = W34 = - mRT Lln(V4/V3) (3)
4 to 1, adiabatic compression, Q41 = 0
(TL/TH) = (V1/V4)k-1 (4)
From (2) & (4): (V2/V3) = (V1/V4) (V2/V1) = (V3/V4)
Since ln(V2/V1) = - ln(V4/V3); substituting for ln(V4/V3) in (1)
(QL/QH )= (TL/TH)
Hence: th = 1-(QL/QH )= 1-(TL/TH)
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.
Example: A typical steam power plant operates between T H=800 K (boiler) and TL=300
K(cooling tower). For this plant, the maximum achievable efficiency is 62.5%.
5.
Factors which affect Carnot Efficiency
Example: Consider a Carnot heat engine operating between a high-temperature source
at 900 K and rejecting heat to a low-temperature reservoir at 300 K. (a) Determine the
thermal efficiency of the engine; (b) Show how the thermal efficiency changes as the
temperature of the high-temperature source is decreased; (b) Determine the change in
thermal efficiency as the temperature of the low-temperature sink is decreased
1
T 300 0.8 Lower TH
1 L
1 0.667 66.7%
T 900
th
Efficiency
H 0.6
Th( T )
Fixed T 300( K ) and lowering T
L H
0.4
300 0.2
(T ) 1
T
th H
0
H 200 400 600 800 1000
The higher the temperature, the higher the "quality" T
Temperature (TH)
1
of the energy: More work can be done
0.8 Increase TL
Efficiency 0.6
Fixed T 900( K ) and increasing T
H L
TH( TL )
0.4
T
(T ) 1 L
0.2
900
th H
0
200 400 600 800 1000
TL
Temperature (TL)
6.
Carnot Efficiency & Quality of Energy
• The previous example illustrates that higher the temperature of
the low-temperature sink, more difficult it becomes for a heat
engine to reject/transfer heat into it.
• This results in a lower thermal efficiency
• One reason why low-temperature reservoirs such as rivers, lakes and
atmosphere are popular for heat rejection from power plants.
• Similarly, the thermal efficiency of an engine, e.g a gas turbine
engine, can be increased by increasing the temperature of the
combustion chamber.
•This may sometimes conflict with other design requirements. Example:
turbine blades can not withstand high temperature (and pressure) gases,
which can leads to early fatigue. A Solution: better materials and/or
innovative cooling design.
7.
Quality of Energy cont’d
•This illustrates that the quality of energy is an important factor in
determining the efficiencies of systems. E.g. for the same amount
(quantity) of total energy, it is easier – more efficient – to produce
work from a high temperature reservoir than a low temperature
reservoir. Consequently, extracting energy from low-temperature
reservoirs such as rivers and lakes is not very efficient. E.g. solar
pond/lake have typical efficiencies of around 5%
•Also, work is in general more valuable – of a higher quality -
relative to heat, since work can convert to heat almost with almost
100% efficiency but not the other way around. Energy becomes
less useful when it is transferred to and stored in a low-temperature
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