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Thermodynamics: Efficiency for Stirling engine

by armolinasf
Tags: efficiency, engine, stirling, thermodynamics
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armolinasf
#1
Nov16-11, 09:18 PM
P: 196
1. The problem statement, all variables and given/known data

I'm trying to find an expression for the efficiency of a stirling engine operating with an ideal diatomic gas, and cycling through a volume V and a multiple of its compression ratio, r, Vr.

2. Relevant equations

processes:

1-2 isothermal expansion
2-3 isochoric cooling
3-4 isothermal compression
4-1 isochoric heating

r=compression ratio
Th=high temperature
Tl=low temperature

Work=W1 proc. 1-2 (nRTh)ln(r)
Work=W2 proc. 3-4 (nRTl)ln(1/r)
Work Net= W1-W2= nRln(r)(Th-Tl) since ln=-ln(1/r)

Heat Input=Qh=nCv(Th-Tl)=(5/2)R(Th-Tl)

Efficiency=e=W Net/Heat Input=[nRln(r)(Th-Tl)]/[(5/2)nR(Th-Tl)

Canceling:e=(5/2)ln(r)

This does not Make sense since efficiency for an engine with an equal compression ration of say r=10 operating at a Temp high of 300k and low of 200k would have a carnot efficiency of (1/3) while with the above equation e=.92 which is impossible.
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Andrew Mason
#2
Nov16-11, 10:04 PM
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Quote Quote by armolinasf View Post

Heat Input=Qh=nCv(Th-Tl)=(5/2)R(Th-Tl)
You are assuming that heat flow into the gas occurs only in the 4-1 constant volume part. Apply the first law to the isothermal expansion (1-2): ΔQ = ΔU + W;

AM
armolinasf
#3
Nov17-11, 12:51 AM
P: 196
Yes, but the heat flow occurring in 2-3 is an out flow so it wouldn't be included in the efficiency calculation which is based on only on the heat input, right?

RTW69
#4
Nov17-11, 03:05 AM
P: 374
Thermodynamics: Efficiency for Stirling engine

I believe Qh=5/2*R*n*(Th-Tl)+R*n*Th*ln(r) in the denominator
armolinasf
#5
Nov17-11, 03:29 AM
P: 196
why is that? isn't nRThln(r) the work done from 1-2?
RTW69
#6
Nov17-11, 03:51 AM
P: 374
http://www.pha.jhu.edu/~broholm/l39/node5.html

This site might be helpful
Andrew Mason
#7
Nov17-11, 11:48 AM
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Quote Quote by armolinasf View Post
why is that? isn't nRThln(r) the work done from 1-2?
Exactly. Since it is isothermal, ΔU = 0. So, by the first law, ΔQ1-2 = W1-2 (where W = the work done BY the gas). You can see from the first law that heat flow into the gas occurs from 4-1 AND from 1-2.

AM
Andrew Mason
#8
Nov17-11, 11:52 AM
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P: 6,679
Quote Quote by armolinasf View Post
Yes, but the heat flow occurring in 2-3 is an out flow so it wouldn't be included in the efficiency calculation which is based on only on the heat input, right?
I did not say 2-3. I said 1-2. Apply the first law. You will see that there is positive heat flow into the gas from 1-2.

AM


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