Heat Engine
• A heat engine is a device that absorbs heat (Q) and uses it
to do useful work (W) on the surroundings when operating
in a cycle.
• Sources of heat include the combustion of coal, petroleum
or carbohydrates and nuclear reactions.
• Working substance: the matter inside the heat engine
that undergoes addition or rejection of heat and that does
work on the surroundings. Examples include air and water
vapour (steam).
• In a cycle, the working substance is in the same
thermodynamic state at the end as at the start.
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Heat Engine
Hot Body
(source of heat)
Q1
W
E
Q2
Cold Body
(absorbs heat)
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Example of a Heat Engine
Open system
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Internal Combustion Engine
d
a
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Comparison of Otto and Diesel Cycles
Work per cycle
= Area inside
combustion
Q=0
Q=0
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Nuclear Power Plant: A Very Large Heat Engine
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Heat Engine
Hot Body
(source of heat)
Q1
W
E
Q2
Cold Body
(absorbs heat)
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Efficiency of a Heat Engine
Efficiency, h = Work out/Heat in:
W
h
Q1
Apply First Law to the working substance:
DU = Q1 – Q2 – W
But in a cycle, DU = 0
Thus, W = Q1 – Q2.
Substituting:
W Q1  Q2
Q2
h 
 1
Q1
Q1
Q1
Lesson: h is maximum when Q2 is minimum.
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The Stirling Engine
•Closed system
See:
http://www.animatedengines.com/ltdstirling.shtml
•Operates between two bodies with (small) different temperatures.
• Can use “stray” heat
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The Stirling Cycle
TH >TC
isothermal
Heat in
(TH - TC ) is proportional
to the amount of work
that is done in a cycle.
= air temp
=hot water
2
isothermal
Heat out
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Carnot Cycle
Hot Reservoir
T1
Q1
W
C
Q2
Cold Reservoir
T2
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Carnot Cycle
Pressure
a
•
nRT1
P=
V
Q1
b
•
Q=0
nRT2
P=
V
T1
Q=0
•
d
Q2
P=
const.
V
•c T2
Volume
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Carnot Cycle
Pressure
a
•
nRT1
P=
V
Q1
b
W
Q=0
nRT2
P=
V
•
d
Q2
•
T1
Q=0
P=
const.
V
•c T2
Volume
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Carnot Cycle
From a to b: isothermal, so that DU = 0 and Q = - W
Thus, Q1 = +nRT1ln(Vb/Va)
(+ve quantity)
From b to c: adiabatic, Q = 0, so that TV-1 is constant.
 1
Thus, T1Vb-1 = T2Vc-1 or
T1  Vc 
  
T2  Vb 
Similarly, from c to d: isothermal, so that DU = 0 and Q = - W
Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve)
Similarly, d to a: adiabatic, Q = 0, so that TV-1 is constant.
Thus, T2Vd-1 = T1Va-1 or
 1
T1  Vd 
  
T2  Va 
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Carnot Cycle
We see that:
Which means that
Now also:
But as the volume
ratios are equal:
T1  Vc 
  
T2  Vb 
 1
V
  d
 Va



 1
Vc Vb

Vd Va
Q1 nRT1 ln(Vb / Va ) T1 ln(Vb / Va )


Q2 nRT2 ln(Vc / Vd ) T2 ln(Vc / Vd )
Q1 T1

Q2 T2
This is an important result. Temperature can be defined (on the
absolute (Kelvin) scale) in terms of the heat flows in a Carnot
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Cycle.
What’s Special about a Carnot Cycle?
(1) Heat is transferred to/from only two reservoirs at fixed
temperatures, T1 and T2 - not at a variety of temperatures.
(2) Heat transfer is the most efficient possible because the
temperature of the working substance equals the temperature
of the reservoirs. No heat is wasted in flowing from hot to cold.
(3) The cycle uses an adiabatic process to raise and lower the
temperature of the working substance. No heat is wasted in
heating up the working substance.
(4) Carnot cycles are reversible. Not all cycles are!
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What’s Special about a Carnot Cycle?
(5) The Carnot theorem states that the Carnot cycle (or any
reversible cycle) is the most efficient cycle possible. The Carnot
cycle defines an upper limit to the efficiency of a cycle.
• Recall that for any cycle, the efficiency of a heat engine is
given as:
W
Q2
hE = = 1
Q1
Q1
• For an engine using a Carnot cycle, the efficiency is also
equal to:
T2
hC = 1
T1
• Where T1 and T2 are the temperatures of the hot and cold
reservoirs, respectively, in degrees Kelvin.
 As T2 > 0, hc is always <1.
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Efficiency of a Stirling Engine
Question: What is the maximum possible efficiency of a
Stirling engine operating between room temperature (25 °C)
and boiling water (100 °C)?
Maximum efficiency would be achieved by a Carnot cycle
operating between reservoirs at T1 = 373 K and T2 = 298 K.
298
W
hc = 1
= 0.20 =
373
Q1
Question: What is the maximum possible efficiency of a
Stirling engine operating between room temperature (25 °C)
and ice (0 °C)?
273
W
hc = 1
= 0.08 =
298
Q1
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Kelvin-Planck Statement of the Second Law of
Thermodynamics
“It is impossible to construct a device that - operating in a
cycle - will produce no other effect than the extraction of
heat from a single body and the performance of an
equivalent amount of work”
Or…A cyclical engine cannot convert heat from a single
body completely into work. Some heat must be rejected
at a lower temperature. Thus, efficiency, h < 1!
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Heat Engine
Hot Body
(source of heat)
Q1
W= -Q1
E
Q2 = 0
Cold Body
(absorbs heat)
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Heat Engine
Hot Body
(source of heat)
Q1= 0
POSSIBLE!
W
E
Examples: friction
creating heat;
isothermal
compression of ideal
gas
Q2 = W
Cold Body
(absorbs heat)
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Refrigerator: A heat engine operating in reverse
Hot Body
Q1
Refrigerator Efficiency:
heatout Q2
hR 

work in W
W
E
Q2
Cold Body
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Refrigerator Efficiency
hR 
heatout Q2

work in W
First Law tells us that Q2 + W -Q1 = 0.
Thus, W = Q1 – Q2
Q2
hR 
Q1  Q2
For a Carnot refrigerator, the efficiency is:
1
hR c
hR
c

Q1  Q2 Q1
T
T T

1  1 1  1 2
Q2
Q2
T2
T2
T2

T1  T2
Efficiency is usually >1!
The smaller the T difference, the more efficient is the
refrigerator.
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Clausius Statement of the Second Law of
Thermodynamics
(applies to refrigerators)
“It is impossible to construct a device that - operating in a
cycle - will produce no other effect than heat transfer from
a colder body to hotter body.”
“Or…Heat cannot flow from a cold body to a hotter body
by itself. Work has to be done in the process.”
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Efficiency of a Heat Pump
The purpose of a heat pump is to extract heat from a cold body (such
as the River Thames) and “pump” it to a hot body (such as an office
building).
The efficiency is defined as the amount of heat pumped in to the hot
body per the amount of work done:
Q
h hp C  1
W
The First Law tells us that W = Q1-Q2 So, substituting, we find:
Q1
T1
1
hhp C 


Q1  Q2 T1  T2 1  T2 / T1
hhp is always > 1! For maximum h, T2 should be  T1 (just slightly less).
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