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HE2 Thermal Physics
HE2 Thermal Physics
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.
HE2 Thermal Physics
Heat Engine
E
Hot Body
(source of heat)
Q1
Cold Body
(absorbs heat)
Q2
W
HE2 Thermal Physics
Open system
Example of a Heat Engine
HE2 Thermal Physics
a
d
Internal Combustion Engine
HE2 Thermal Physics
Comparison of Otto and Diesel Cycles
combustion
Q=0
Q=0
Work per cycle
= Area inside
HE2 Thermal Physics
Nuclear Power Plant: A Very Large Heat Engine
HE2 Thermal Physics
Heat Engine
E
Hot Body
(source of heat)
Q1
Cold Body
(absorbs heat)
Q2
W
HE2 Thermal Physics
Efficiency of a Heat Engine
Efficiency, η = Work out/Heat in:
Apply First Law to the working substance:
∆U = Q1 – Q2 – W
But in a cycle, ∆U = 0
Thus, W = Q1 – Q2.
1
2
1
21
1
1
Q
Q
Q
QQ
Q
W
−=
−
==ηSubstituting:
1Q
W
=η
Lesson: η is maximum when Q2 is minimum.
HE2 Thermal Physics
The Stirling Engine
•Closed system
•Operates between two bodies with (small) different temperatures.
• Can use “stray” heat
See: http://www.animatedengines.com/ltdstirling.shtml
HE2 Thermal Physics
isothermal
isothermal
= air temp
=hot water
Heat in
Heat out
TH >TC
The Stirling Cycle
(TH - TC ) is proportional
to the amount of work
that is done in a cycle.
2
HE2 Thermal Physics
HE2 Thermal Physics
Carnot Cycle
Hot Reservoir
T1
Cold Reservoir
T2
C
Q1
Q2
W
HE2 Thermal Physics
Volume
Pressure
•
•
a
b
•
d
T1
Q1
Carnot Cycle
Q2
V
nRT
P 1
=
γ
V
const
P
.
=
T2
•c
Q=0
Q=0
V
nRT
P 2
=
HE2 Thermal Physics
Volume
Pressure
•
•
a
b
•
d
T1
Q1
Q2
V
nRT
P 1
=
γ
V
const
P
.
=
T2
•c
Q=0
Q=0
V
nRT
P 2
=
W
Carnot Cycle
HE2 Thermal Physics
From a to b: isothermal, so that ∆U = 0 and Q = - W
Thus, Q1 = +nRT1ln(Vb/Va) (+ve quantity)
Carnot Cycle
Similarly, from c to d: isothermal, so that ∆U = 0 and Q = - W
Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve)
From b to c: adiabatic, Q = 0, so that TVγ-1
is constant.
Thus, T1Vb
γ-1
= T2Vc
γ-1
or 1
2
1
−






=
γ
b
c
V
V
T
T
Similarly, d to a: adiabatic, Q = 0, so that TVγ-1
is constant.
Thus, T2Vd
γ-1
= T1Va
γ-1
or 1
2
1
−






=
γ
a
d
V
V
T
T
HE2 Thermal Physics
Carnot Cycle
We see that:
11
2
1
−−






=





=
γγ
a
d
b
c
V
V
V
V
T
T
)/ln(
)/ln(
)/ln(
)/ln(
2
1
2
1
2
1
dc
ab
dc
ab
VVT
VVT
VVnRT
VVnRT
Q
Q
==
a
b
d
c
V
V
V
V
=
Which means that
Now also:
This is an important result. Temperature can be defined (on the
absolute (Kelvin) scale) in terms of the heat flows in a Carnot
Cycle.
But as the volume
ratios are equal: 2
1
2
1
T
T
Q
Q
=
HE2 Thermal Physics
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!
HE2 Thermal Physics
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.
• Where T1 and T2 are the temperatures of the hot and cold
reservoirs, respectively, in degrees Kelvin.
⇒ As T2 > 0, ηc is always <1.
• Recall that for any cycle, the efficiency of a heat engine is
given as:
1
2
1
1==
Q
Q
Q
W
Eη
• For an engine using a Carnot cycle, the efficiency is also
equal to:
1
2
1=
T
T
Cη
HE2 Thermal Physics
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)?
Question: What is the maximum possible efficiency of a
Stirling engine operating between room temperature (25 °C)
and ice (0 °C)?
Maximum efficiency would be achieved by a Carnot cycle
operating between reservoirs at T1 = 373 K and T2 = 298 K.
1
=20.0=
373
298
1=
Q
W
cη
1
=08.0=
298
273
1=
Q
W
cη
HE2 Thermal Physics
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, η < 1!
HE2 Thermal Physics
Heat Engine
E
Hot Body
(source of heat)
Q1
Cold Body
(absorbs heat)
Q2 = 0
W= -Q1
HE2 Thermal Physics
Heat Engine
E
Hot Body
(source of heat)
Q1= 0
Cold Body
(absorbs heat)
Q2 = W
WPOSSIBLE!
Examples: friction
creating heat;
isothermal
compression of ideal
gas
HE2 Thermal Physics
Refrigerator: A heat engine operating in reverse
E
Hot Body
Q1
Cold Body
Q2
W
W
Q
work
heat
in
out
R
2
==η
Refrigerator Efficiency:
HE2 Thermal Physics
Refrigerator Efficiency
W
Q
work
heat
in
out
R
2
==η
21
2
QQ
Q
R
−
=η
21
2
TT
Tc
R
−
=η
First Law tells us that Q2 + W -Q1 = 0.
Thus, W = Q1 – Q2
2
21
2
1
2
1
2
21
11
1
T
TT
T
T
Q
Q
Q
QQ
c
R
−
=−=−=
−
=
η
For a Carnot refrigerator, the efficiency is:
Efficiency is usually >1!
The smaller the T difference, the more efficient is the
refrigerator.
HE2 Thermal Physics
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.”
HE2 Thermal Physics
HE2 Thermal Physics
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 First Law tells us that W = Q1-Q2 So, substituting, we find:
1221
1
21
1
/1
1
TTTT
T
QQ
QC
hp
−
=
−
=
−
=η
ηhp is always > 1! For maximum η, T2 should be ≅ T1 (just slightly less).
W
QC
hp
1
=η
The efficiency is defined as the amount of heat pumped in to the hot
body per the amount of work done:

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Heat engine-introduction

  • 2. HE2 Thermal Physics 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.
  • 3. HE2 Thermal Physics Heat Engine E Hot Body (source of heat) Q1 Cold Body (absorbs heat) Q2 W
  • 4. HE2 Thermal Physics Open system Example of a Heat Engine
  • 6. HE2 Thermal Physics Comparison of Otto and Diesel Cycles combustion Q=0 Q=0 Work per cycle = Area inside
  • 7. HE2 Thermal Physics Nuclear Power Plant: A Very Large Heat Engine
  • 8. HE2 Thermal Physics Heat Engine E Hot Body (source of heat) Q1 Cold Body (absorbs heat) Q2 W
  • 9. HE2 Thermal Physics Efficiency of a Heat Engine Efficiency, η = Work out/Heat in: Apply First Law to the working substance: ∆U = Q1 – Q2 – W But in a cycle, ∆U = 0 Thus, W = Q1 – Q2. 1 2 1 21 1 1 Q Q Q QQ Q W −= − ==ηSubstituting: 1Q W =η Lesson: η is maximum when Q2 is minimum.
  • 10. HE2 Thermal Physics The Stirling Engine •Closed system •Operates between two bodies with (small) different temperatures. • Can use “stray” heat See: http://www.animatedengines.com/ltdstirling.shtml
  • 11. HE2 Thermal Physics isothermal isothermal = air temp =hot water Heat in Heat out TH >TC The Stirling Cycle (TH - TC ) is proportional to the amount of work that is done in a cycle. 2
  • 13. HE2 Thermal Physics Carnot Cycle Hot Reservoir T1 Cold Reservoir T2 C Q1 Q2 W
  • 14. HE2 Thermal Physics Volume Pressure • • a b • d T1 Q1 Carnot Cycle Q2 V nRT P 1 = γ V const P . = T2 •c Q=0 Q=0 V nRT P 2 =
  • 15. HE2 Thermal Physics Volume Pressure • • a b • d T1 Q1 Q2 V nRT P 1 = γ V const P . = T2 •c Q=0 Q=0 V nRT P 2 = W Carnot Cycle
  • 16. HE2 Thermal Physics From a to b: isothermal, so that ∆U = 0 and Q = - W Thus, Q1 = +nRT1ln(Vb/Va) (+ve quantity) Carnot Cycle Similarly, from c to d: isothermal, so that ∆U = 0 and Q = - W Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve) From b to c: adiabatic, Q = 0, so that TVγ-1 is constant. Thus, T1Vb γ-1 = T2Vc γ-1 or 1 2 1 −       = γ b c V V T T Similarly, d to a: adiabatic, Q = 0, so that TVγ-1 is constant. Thus, T2Vd γ-1 = T1Va γ-1 or 1 2 1 −       = γ a d V V T T
  • 17. HE2 Thermal Physics Carnot Cycle We see that: 11 2 1 −−       =      = γγ a d b c V V V V T T )/ln( )/ln( )/ln( )/ln( 2 1 2 1 2 1 dc ab dc ab VVT VVT VVnRT VVnRT Q Q == a b d c V V V V = Which means that Now also: This is an important result. Temperature can be defined (on the absolute (Kelvin) scale) in terms of the heat flows in a Carnot Cycle. But as the volume ratios are equal: 2 1 2 1 T T Q Q =
  • 18. HE2 Thermal Physics 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!
  • 19. HE2 Thermal Physics 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. • Where T1 and T2 are the temperatures of the hot and cold reservoirs, respectively, in degrees Kelvin. ⇒ As T2 > 0, ηc is always <1. • Recall that for any cycle, the efficiency of a heat engine is given as: 1 2 1 1== Q Q Q W Eη • For an engine using a Carnot cycle, the efficiency is also equal to: 1 2 1= T T Cη
  • 20. HE2 Thermal Physics 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)? Question: What is the maximum possible efficiency of a Stirling engine operating between room temperature (25 °C) and ice (0 °C)? Maximum efficiency would be achieved by a Carnot cycle operating between reservoirs at T1 = 373 K and T2 = 298 K. 1 =20.0= 373 298 1= Q W cη 1 =08.0= 298 273 1= Q W cη
  • 21. HE2 Thermal Physics 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, η < 1!
  • 22. HE2 Thermal Physics Heat Engine E Hot Body (source of heat) Q1 Cold Body (absorbs heat) Q2 = 0 W= -Q1
  • 23. HE2 Thermal Physics Heat Engine E Hot Body (source of heat) Q1= 0 Cold Body (absorbs heat) Q2 = W WPOSSIBLE! Examples: friction creating heat; isothermal compression of ideal gas
  • 24. HE2 Thermal Physics Refrigerator: A heat engine operating in reverse E Hot Body Q1 Cold Body Q2 W W Q work heat in out R 2 ==η Refrigerator Efficiency:
  • 25. HE2 Thermal Physics Refrigerator Efficiency W Q work heat in out R 2 ==η 21 2 QQ Q R − =η 21 2 TT Tc R − =η First Law tells us that Q2 + W -Q1 = 0. Thus, W = Q1 – Q2 2 21 2 1 2 1 2 21 11 1 T TT T T Q Q Q QQ c R − =−=−= − = η For a Carnot refrigerator, the efficiency is: Efficiency is usually >1! The smaller the T difference, the more efficient is the refrigerator.
  • 26. HE2 Thermal Physics 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.”
  • 28. HE2 Thermal Physics 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 First Law tells us that W = Q1-Q2 So, substituting, we find: 1221 1 21 1 /1 1 TTTT T QQ QC hp − = − = − =η ηhp is always > 1! For maximum η, T2 should be ≅ T1 (just slightly less). W QC hp 1 =η The efficiency is defined as the amount of heat pumped in to the hot body per the amount of work done: