Heat engines. zUse heat to do work in a cyclic process: same process is repeated over and over...
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Transcript of Heat engines. zUse heat to do work in a cyclic process: same process is repeated over and over...
![Page 1: Heat engines. zUse heat to do work in a cyclic process: same process is repeated over and over again.](https://reader036.fdocuments.in/reader036/viewer/2022082820/5697bf7b1a28abf838c83b0b/html5/thumbnails/1.jpg)
Heat engines
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Heat engines
Use heat to do work in a cyclic process: same process is repeated over and over again
![Page 3: Heat engines. zUse heat to do work in a cyclic process: same process is repeated over and over again.](https://reader036.fdocuments.in/reader036/viewer/2022082820/5697bf7b1a28abf838c83b0b/html5/thumbnails/3.jpg)
Reservoirs
Keep the same temperature
Supply the working substance with energy so that its temperature remains or becomes equal to Treservoir
All processes are “slow” and reversible from the reservoir’s point of view
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The Carnot cycle
Consists of four slow steps:
12: isothermal23: adiabatic34: isothermal41: adiabatic
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The first step
U=0 as T=0 (nearly true for non-ideal gas)
Stotal=0, hence this step is reversible
1
221 ln
VV
nRTWQ HH
1
221 ln
VV
nRTQ
SH
Hgas
H
Hreservoir
TQ
S 21
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Four questions
In the second step, Q, W, U, S are
a) positive
b) zero
c) negative
d) dependent on the
gas used
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Comments on the Carnot cycle
It is customary to define QC as positive even though heat flows out of the system
Heat loss to cold reservoir necessary to bring system back to original state
Scycle=0, hence
Shot reservoir+ Scold reservoir = 0 C
C
H
H
TQ
TQ
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Work and efficiency
Work done during cycle: W = QH – QC
This is independent of working substance!
HH
C
H
C
HCarnot T
TTT
QW
11
inheat outwork
efficiency
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Reversibility
There is no net change in entropy: the entropy of gas and environment are unchanged
Hence the entire Carnot cycle is reversible, and it can be run as a “fridge”
“Reversible” means: we can reverse the cycle and restore the original state without any residual changes outside the system
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Maximum efficiency
Imagine we have some heat engine converting heat into work, and a Carnot engine running backwards
The Carnot engine takes heat QC from the cold reservoir, work W is done on the gas and the engine gives QH to the hot reservoir
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Maximum efficiency II
other=Carnot: no net work done, no net heat transfer from either reservoir
other > Carnot: Say the other engine delivers QC to cold reservoir. Then it takes in QH + , which is used to do work W + . However:
This violates second law of thermodynamics
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Carnot engine is most efficient
If there was a more efficient engine, net work could be done by extracting heat from the hot reservoir without any other changes anywhere.
Thus disorderly thermal motion would be converted into orderly mechanical motion
Entropy would decrease, not allowed by 2nd law
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The Otto cycle
Four steps:
12: slow adiabatic23: fast isochoric34: slow adiabatic41: fast isochoric
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Temperature and heat
The temperature T2 (and T4) can be evaluated:
Heat added:41:23:
1
2
12
1 '
VV
TT
CTVCpV
H
)(
)(
2
4
CVC
HVH
TTnCQ
TTnCQ
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Entropy in the Otto cycle
Step 41: gas: fast process, calculate S as if it were slow reservoir: slow process, calculate as usual
HenceH
HV
H
Hreservoir
HV
T
TV
gas
TTT
nCTQ
S
TT
nCTT
nCSH
414
414 ln
d
4
01414 reservoirgas SS
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Entropy in the Otto cycle II
Likewise in step 23:
We conclude for the entire cycle
Since
The cycle is thus not reversible
03232 reservoirgas SS
0 reservoirgascycle SSS
0 ,0 reservoircyclegas SSS
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PS225 – Thermal Physics topicsThe atomic hypothesisHeat and heat transferKinetic theoryThe Boltzmann factorThe First Law of ThermodynamicsSpecific HeatEntropyHeat enginesPhase transitions