PHYSICS 231
Engines and fridges
Heat Engines
Remco Zegers
PHY 231
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First Law of thermodynamics
∆U=Uf-Ui=Q+W
∆U=change in internal energy
Q=energy transfer through heat (+ if heat is
transferred to the system)
W=energy transfer through work (+ if work is
done on the system)
if P: constant then W=-P∆V (area under
P-V diagram
This law is a general rule for conservation of energy
PHY 231
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Types of processes
A: Isovolumetric ∆V=0
B: Adiabatic Q=0
C: Isothermal ∆T=0
D: Isobaric ∆P=0
PV/T=constant
First law of thermoDynamics:
∆U=Q+W
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Isovolumetric processes (line A)
p
∆V=0
W=0 (area under the curve is zero)
∆U=Q (Use ∆U=W+Q, with W=0)
In case of ideal gas:
∆U=3/2nR∆T
•if P↓ then T↓ (PV/T=constant)
so ∆U=negative Q=negative
(Heat is extracted from the gas)
•if P↑ then T↑ (PV/T=constant)
v
so ∆U=positive Q=positive
(Heat is added to the gas)
Cv=(3/2)R so ∆U=Cvn ∆T
Cv: molar specific heat at const. vol.
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isobaric process
p
∆P=0
v
Use ∆U=W+Q
In case of ideal gas:
W=-P∆V & ∆U=3/2nR∆T
•if V↓ then T↓ (PV/T=constant)
W: positive (work done on gas)
∆U: negative
Q: negative (heat extracted)
•if V↑ then T↑ (PV/T=constant)
W: negative (work done by gas)
∆U: positive
Q: positive (heat added)
Q=∆U-W=3/2nR∆T+ ∆(PV) = 3/2nR∆T+ nR∆T= 5/2nR∆T
(used ideal gas law) Q=nCp∆T with Cp=(5/2)R
Cp=molar heat capacity at constant
pressure
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p
isothermal processes
∆T=0
v
The temperature is not changed
Q=-W (Use ∆U=W+Q, with ∆U=0)
•if V↓
W=positive Q=negative
(Work is done on the gas
and energy extracted)
•if V↑
W=negative Q=positive
(Work is done by the gas
and energy added)
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p
Adiabatic process (line B)
Q=0
v
No heat is added/extracted from the
system.
∆U=W (Use ∆U=W+Q, with Q=0)
In case of ideal gas:
∆U=3/2nR∆T
•if T↓
∆U=negative W=negative
(The gas has done work)
•if T↑
∆U=positive W=positive
(Work is done on the gas)
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Cyclic processes
The system returns to its
original state. Therefore,
the internal energy must
be the same after completion
of the cycle (∆U=0)
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Cyclic Process, step by step 1
Process A-B.
Negative work is done on the gas:
(the gas is doing positive work).
W=-Area under P-V diagram
=-[(50-10)*10-3]*[(1.0-0.0)*105]
-½[(50-10)*10-3]*[(5.0-1.0)]*105=
=4000+8000 (work done by gas)
W=-12000 J (work done on gas)
∆U=3/2nR∆T=3/2(PBVB-PAVA)=
= 1.5*[(1E+5)(50E-03)-(5E+5)(10E-03)]=0
The internal energy has not changed
∆U=Q+W so Q=∆U-W=12000 J: Heat that was added to the
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system was used to do thePHY
work!
231
Cyclic process, step by step 2
Process B-C
W=-Area under P-V diagram
=-[(10-50)*10-3*(1.0-0.0)*105]=
W=4000 J
Work was done on the gas
∆U=3/2nR∆T=3/2(PcVc-PbVb)=
=1.5[(1E+5)(10E-3)-(1E+5)(50E-3)]=-6000 J
The internal energy has decreased by 6000 J
∆U=Q+W so Q=∆U-W=-6000-4000 J=-10000 J
10000 J of energy has been transferred out of the system.
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Cyclic process, step by step 3
Process C-A
W=-Area under P-V diagram
W=0 J
No work was done on/by the gas.
∆U=3/2nR∆T=3/2(PaVa-PcVc)=
=1.5[(5E+5)(10E-3)-(1E+5)(10E-3)]=+6000 J
The internal energy has increased by 6000 J
∆U=Q+W so Q=∆U-W=6000-0 J=6000 J
6000 J of energy has been transferred into the system.
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Summary of the process
-AREA
A-B
Quantity Work(W) Heat(Q)
Process
∆U
A-B
-12000 J 12000 J
0
B-C
4000 J
-10000 J -6000
C-A
0J
6000 J
6000
SUM
-8000 J
8000 J
0
B-C
C-A
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What did we do?
The gas performed net work (8000 J)
while heat was supplied (8000 J):
We have built an engine!
What if the process was done in
the reverse way?
Net work was performed on the
gas and heat extracted from the gas.
We have built a heat pump!
(A fridge)
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example
A gas goes from initial state I to
final state F, given the parameters
in the figure. What is the work done
on the gas and the net energy transfer
by heat to the gas for:
a) path IBF b) path IF c) path IAF
(Ui=91 J Uf=182 J)
a) work done: area under graph:
W=-(0.8-0.3)10-3*2.0*105=-100 J
∆U=W+Q 91=-100+Q so Q=191 J
b) W=-[(0.8-0.3)10-3*1.5*105 + ½(0.8-0.3)10-3*0.5*105]=-87.5 J
∆U=W+Q 91=-87.5+Q so Q=178.5 J
c) W=-[(0.8-0.3)10-3*1.5*105]=-75 J
∆U=W+Q 91=-75+Q so Q=166 J
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PHY 231
question
P (Pa)
3x105
Consider this cyclic
process. Which of the following
is true?
1x105
1
a)
b)
c)
d)
e)
3 V (m3)
This is a heat engine and the W done by the gas is +4x105 J
This is a fridge and the W done on the gas is +4x105 J
This is a heat engine and the W done by the gas is +6x105 J
This is a fridge and the W done on the gas is +6x105 J
This is a heat engine and the W done by the gas is –4x105J
clockwise: work done by the gas, so engine
5 J
W by gas=area enclosed=(3-1)x(3x105-1x105)=4x1015
PHY 231
More general engine
turns water to steam
W=|Qh|-|Qc|
efficiency: W/|Qh|
e=1-|Qc|/|Qh|
heat reservoir Th
Qh
the steam moves the piston
engine
Qc
work
W
the steam is condensed
cold reservoir Tc
work is done
The efficiency is determined
by how much of the heat you
supply to the engine is turned
into work instead of being lost
as waste.
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Reverse direction: the fridge
heat is expelled to outside
heat reservoir Th
Qh
a piston compresses the coolant work is done
engine
Qc
the fridge is cooled
work
W
cold reservoir Tc
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The 2nd law of thermodynamics
1st law: ∆U=Q+W
In a cyclic process (∆U=0) Q=W: we cannot do more work
than the amount of energy (heat) that we put inside
2nd law: It is impossible to construct an engine that,
operating in a cycle produces no other effect than the
absorption of energy from a reservoir and the performance
of an equal amount of work: we cannot get 100% efficiency
What is the most efficient engine we can make
given a heat and a cold reservoir?
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A→B isothermal expansion
B→C adiabatic expansion
Carnot engine
W-, Q+
Thot
W+, ∆T+
W-, ∆T-
W+, Q-
Tcold
D→A adiabatic compression
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C→D isothermal compression
PHY 231
inverse Carnot cycle
Carnot cycle
Work done by engine: Weng
Weng=Qhot-Qcold
efficiency: 1-Tcold/Thot
PHY 231
A heat engine or a fridge!
By doing work we can
transport heat
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example
The efficiency of a Carnot engine is 30%. The engine absorbs
800 J of energy per cycle by heat from a hot reservoir at
500 K. Determine a) the energy expelled per cycle and b)
the temperature of the cold reservoir. c) How much work
does the engine do per cycle?
a) Generally for an engine: efficiency: 1-|Qcold|/|Qhot|
0.3=1-|Qcold|/800, so |Qcold|=-(0.3-1)*800=560 J
b) for a Carnot engine: efficiency: 1-Tcold/Thot
0.3=1-Tcold/500, so Tcold=-(0.3-1)*500=350 K
c) W=|Qhot|-|Qcold|=800-560=240 J
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An engine is operated between a hot and a cold reservoir
with Qhot=400J and Qcold=300J. a) what is the efficiency
of the engine?
The engine is modified and becomes a carnot engine. As a
result the efficiency is doubled. b) what is the ratio
Tcold/Thot. c) what is the maximum efficiency of this engine?
a) efficiency=1-Qcold/Qhot=1-300/400=0.25
b) new efficiency: 0.5=1-Tcold/Thot Tcold/Thot=0.5
c) 0.5 (Carnot engine has maximum efficiency).
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A new powerplant
A new powerplant is designed that makes use of the
temperature difference between sea water at 0m (200) and
at 1-km depth (50). A) what would be the maximum efficiency
of such a plant? B) If the powerplant produces 75 MW, how
much energy is absorbed per hour? C) Is this a good idea?
a) maximum efficiency=carnot efficiency=1-Tcold/Thot=
1-278/293=0.051 efficiency=5.1%
b) P=75*106 J/s W=P*t=75*106*3600=2.7x1011 J
efficiency=1-|Qcold|/|Qhot|=(|Qhot|-|Qcold|)/|Qhot|=
W/|Qhot| so |Qhot|=W/efficiency=5.3x1012 J
c) Yes! Very Cheap!! but… |Qcold|= |Qhot|-W=5.0x1012 J
every hour 5E+12 J of waste heat is produced:
Q=cm∆T 5E+12=4186*m*1 m=1E+9 kg of water is heated
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by 10C…perhaps not!
PHY 231
Story of Hawaiian deep water project
•
•
•
Keahole sits at a point where
underwater land slopes sharply
down into the sea, it was a place
where warm water can be piped
from the surface of the sea and
cold water can be piped from
•
depths of about a half-mile.
A process called ocean thermal
energy conversion, or OTEC, used
the temperature difference
between hot and cold sea water
to produce 50 KW of electricity
at Keahole in 1993. The process
worked but it was uneconomical.
PHY 231
KAILUA, HAWAI'I — Koyo
USA Corp., a company selling
deep-sea water from Keahole
Hawai'i, is expanding its plant
and has applied to sell the
water in the United States
The company is producing
more than 200,000 bottles a
day and says it can't keep up
with demand in Japan, where
it sells 1.5 liter bottles of its
MaHaLo brand for $4 to $6
each.
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