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Heat engine: moles/temp unknown

  1. Mar 3, 2013 #1
    1. The problem statement, all variables and given/known data

    A heat engine uses the closed cycle shown in the diagram below. (Figure 1) The working substance is n moles of monatomic ideal gas. Find the efficiency of such a cycle. Use the values for pressure and volume shown in the diagram, and assume that the process between points 1 and 3 is isothermal.

    1065003.jpg

    2. Relevant equations

    pV=nRT
    efficiency = Wout/QH

    3. The attempt at a solution

    We know that T3 and T1 are on an isotherm, and so are both exactly 3 times T2. So, the absolute value of ΔT is 2 times T2. The reason I refer to the absolute value here is because the sign of ΔT for different processes will be different.

    Wout = nRT*ln(V1/V3) - pΔV

    So, Wout = nRT*ln(V1/V3) - nRΔT

    QH= nRT*ln(V1/V3) + nCvΔT

    So the efficiency is:

    3nRT2*ln(V1/V3) + 2nRT2 / 3nRT2*ln(V1/V3) + 2nCVT2

    Then the idea is to cancel out all the n's and T2's. All the rest are knowns. I get a value of 84.1%. There's definitely something very wrong about my approach.
     
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 3, 2013 #2

    mfb

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    Staff: Mentor

    Did you consider the heat involved in 1->2?

    I don't understand the last, long expression and I think there are brackets missing.
     
  4. Mar 3, 2013 #3
    The heat involved in 1 -> 2 is actually leaving the system. So, QH is just the heat we put in during the isothermal and isochoric legs. So, the heat associated with the isobaric process, nCPΔT, does not enter into the expression for QH.

    Yes, there were brackets missing. Let me clarify as to where this expression comes from.

    [3nRT2*ln(V1/V3) + 2nRT2] / [3nRT2*ln(V1/V3) + 2nCVT2]

    We know that T3 is three times greater than T2. This should help to explain the first term of the numerator and the first term of the denominator. The former is the heat intake during the isothermal process, and the latter is the work done during the isothermal process. These expressions are the same because in an isothermal process, W = QH.

    The expression begins as nRT*ln(V1/V3). Now T can refer to the temeperature at point 1 or point 3, because we are dealing with an isothermal process. And we said in the last paragraph that 3T2 = T3. So this is where the 3T2 comes from in both expressions.

    The second terms in the numerator and the denominator contain the factor 2T2. That's because ΔT = T3 - T2, so ΔT = 3T2 - T2.

    Note the sign change on the second term in the numerator. That occurs because we lose temperature moving from 1 -> 2, so ΔT is negative. ΔT is not negative in the other place it occurs, moving from 2 -> 3.

    EDIT: Let's not get trapped in the paradigm of my thinking. It hasn't gotten me anywhere with this problem.
     
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