Thermodynamics: Efficiency for Stirling engine

  1. 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.
     
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,895
    Science Advisor
    Homework Helper

    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
     
  4. 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?
     
  5. I believe Qh=5/2*R*n*(Th-Tl)+R*n*Th*ln(r) in the denominator
     
  6. why is that? isn't nRThln(r) the work done from 1-2?
     
  7. Andrew Mason

    Andrew Mason 6,895
    Science Advisor
    Homework Helper

    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
     
  8. Andrew Mason

    Andrew Mason 6,895
    Science Advisor
    Homework Helper

    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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