Thermodynamics: Efficiency for Stirling engine

AI Thread Summary
The discussion focuses on deriving the efficiency of a Stirling engine using an ideal diatomic gas, analyzing the work done during isothermal and isochoric processes. The initial equation for efficiency, e = (5/2)ln(r), is questioned due to its unrealistic output compared to the Carnot efficiency for given temperatures. Participants clarify that the heat flow into the gas occurs during both the isothermal expansion (1-2) and isochoric heating (4-1), which must be considered in the efficiency calculation. The importance of applying the first law of thermodynamics to accurately account for heat flow and work done is emphasized. The conversation highlights the complexities of calculating efficiency in thermodynamic cycles.
armolinasf
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Homework Statement



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.

Homework 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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armolinasf said:
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
 
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 believe Qh=5/2*R*n*(Th-Tl)+R*n*Th*ln(r) in the denominator
 
why is that? isn't nRThln(r) the work done from 1-2?
 
http://www.pha.jhu.edu/~broholm/l39/node5.html

This site might be helpful
 
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armolinasf said:
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
 
armolinasf said:
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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