Engines
(Finn: 4.7, 4.8, 4.9, 4.10)
Carnot Efficiency
Thermodynamic temperature scale.
Carnot devices.
A (nearly) real engine.
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
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Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
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British Empire - built by thermodynamics
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
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Carnot Engine efficiency
Carnot efficiency is independent of
working substance.
Heat flows (Q1 and Q2 ) are determined
solely by the reservoir temperatures (T1
and T2 ).
Can define a temperature scale based
on thermodynamic principles.
Is it consistent with ideal gas scale?
... consider ideal gas as a working substance.
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
4 / 14
Heat flow and work done in a Ideal gas Carnot cycle
Isothermal expansion “ab”:
Ideal gas U = U(T ); First Law gives dU = 0 = d¯q − PdV =⇒ ,
Z
Vb
qab =
Z
Vb
PdV = nRT1
Va
Va
Vb
dV
= nRT1 ln( ), which is positive.
V
Va
Similarly, isothermal compression “cd”:
Z
Vd
qcd =
Z
Vd
PdV = nRT2
Vc
Vc
Vd
dV
= nRT2 ln( ), which is negative.
V
Vc
Heat flows using “engine” sign convention. Q1 = qab ; Q2 = −qcd .
Q2
nRT2 ln(Vc /Vd )
T2 ln(Vc /Vd )
=
=
Q1
nRT1 ln(Vb /Va )
T1 ln(Vb /Va )
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
5 / 14
Q2
nRT2 ln(Vc /Vd )
T2 ln(Vc /Vd )
=
=
Q1
nRT1 ln(Vb /Va )
T1 ln(Vb /Va )
Adiabats
T1 Vbγ−1 = T2 Vcγ−1
and
T1 Vaγ−1 = T2 Vdγ−1
So Vb /Va = Vc /Vd =⇒ ln(Vb /Va ) = ln(Vc /Vd ). Substitution of
this result in the formula above gives
T2
Q2
=
Q1
T1
Conclusion
Defining thermodynamic temperature Tth via
Tth2
Tth1
=
Q2
Q1
makes the thermodynamic temperature scale the same as the ideal
gas temperature scale. Henceforth we denote temperatures on both
scales by T , measured in kelvin.
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
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Efficiency for engines, refrigerators and heat pumps
Can define ideal efficiency via of the temperatures of the reservoirs.
Efficiency is always defined by (what you want out)/(what you put in). So
For an Engine you put heat in and get work out.
For a Refrigerator, you put work in, and take heat out (from the cold
region).
For a Heat Pump you put work in, and get heat out (into the warm
region).
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
7 / 14
Heat Engine
hot reservoir
T1
Refrigerator
hot reservoir
T1
?
Q
Q1
1
W
C
6
W
C ?
Q
Q
2
62
T2
cold reservoir
T2
cold reservoir
Engine, efficiency
Refrigerator, coefficient of performance
Heat pump efficiency:
Carnot only: ηC = 1 −
Graeme Ackland ()
T2
T1 ;
ηCHP =
Q2
η =1− Q
1
Q2
2
ηR = Q
=
W
Q1 −Q2
Q1
1
η HP = Q
=
W
Q1 −Q2
T1
T1 −T2
Lecture 7: FIRST LAW OF THERMODYNAMICS
ηCR =
T2
T1 −T2
September 20, 2015
8 / 14
infinitely efficient?
Heat Pump and Refrigerator are the same device, used for different
purpose.
Refrigerator, coefficient of performance η R :
ηR =
Q2
Q2
=
W
Q1 − Q2
ηCR =
T2
T1 − T2
Heat pump, heat pump efficiency η HP :
η HP =
Q1
Q1
=
W
Q1 − Q2
ηCHP =
T1
T1 − T2
QUESTION : Explain (using the engineering definition of the efficiency of
a heat pump) why heat pumps are best used to produce domestic
background heating.
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
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Steam Engine
Steam engine works with a liquid+vapour mixture, which combines big
volume expansion (steam) and easy to pump/heat (water)
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
10 / 14
The air standard Otto cycle: a “nearly real” engine
Simplified two-stroke petrol engine.
Assume single working substance with
“external” heating
Two adiabats and two isochores.
Heat exchange takes place in the isochoric
processes
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
11 / 14
1
a – b: reversible adiabatic compression (Piston
moves in)
Ta V1γ−1 = Tb V2γ−1
2
b – c: heat added (actually, combustion) at
constant volume.
Q1 = CV (Tc − Tb )
3
c – d: reversible adiabatic expansion (the
“power stroke”: piston moves out)
Td V1γ−1 = Tc V2γ−1
4
d – a: heat rejected (actually exhaust) at
constant volume.
Q2 = CV (Td − Ta )
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
12 / 14
Otto efficiency
The efficiency η for the engine has to be specified in terms of Q1 and Q2 .
From (4) and (2):
Q2
Td − Ta
η =1−
=1−
Q1
Tc − Tb
To get more insight into the factors controlling efficiency use (1) and (3)
to give:
γ−1
V2
1
η =1−
= 1 − γ−1
V1
rc
where rc = V1 /V2 is called the compression ratio. If rc is ∼ 5, η ∼ 50%.
Other considerations mean real engines are well below this.
Four-stroke engines have exhaust and intake stages between da and ab.
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
September 20, 2015
13 / 14
Graeme Ackland ()
Lecture 7: FIRST LAW OF THERMODYNAMICS
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