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Thermodynamics, two engine device

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose you build a two-engine device with the exhaust energy output from one heat engine supplying the input energy for a second heat engine. We say that the two engines are running in series. Let e1 and e2 represent the efficiencies of the two engines. (a) The overall efficiency of the two-engine device is defined as the total work output divided by the energy put into the first engine by heat. Show that the overall efficiency is given by:
    e = e1 + e2 - e1e2

    2. Relevant equations
    e = Weng / |Qh| = 1 - |Qc| / |Qh|
    e = 1 - Tc/Th

    3. The attempt at a solution
    Since the first engines Qc is supplying the Qh for the second engine, I said Qc1 = Qh2, and then I tried using that equality in the first formula there but I don't think thats working out for me
    Last edited: Mar 25, 2009
  2. jcsd
  3. Mar 25, 2009 #2


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    Write down the expression for e1 and e2.
    From these expression find the ratios of Qc1/Qh1 and Qc2/Qh2.
    If you multiply these ratios you, will get the overall efficiency.
  4. Mar 25, 2009 #3
    assuming that Qc1 = Qh2,

    e1 = 1 - Qc1/Qh1
    e2 = 1 - Qc2/Qc1

    Qc1/Qh1 = e1 - 1
    Qc2/Qc1 = e2 - 1

    multiplying them, i get e1e2 - e1 - e2, which is -1 times the answer im trying to show. why is this?
  5. Mar 25, 2009 #4


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    Your simplification is wrong.
    Qc1/Qh1 = e1 - 1
    Qc2/Qc1 = e2 - 1

    it should be
    Qc1/Qh1 = 1 -e1
    Qc2/Qh2 = 1 - e2
  6. Mar 26, 2009 #5
    thank you i got it now.
    one question though: how do you know to multiply the ratios? as far as i understand, it gives you the efficiency for Qc2 / Qh1 so that would be like the energy input from the first engine to the energy output of the second engine so it would be the efficiency for the entire cycle?
  7. Mar 26, 2009 #6


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    Work done in the first engine = Qh1 - Qc1 = Qh1*e1 = Qh1 - Qh2.......(1)
    Work done in the second engine = Qh2 - Qc2 = Qh2*e2..........(2)
    So the total work done = Qh1 - Qc2 = Qh1*e1 + Qh2*e2......(3)
    From the equation (1) Qh2 = Qh1 - Qh1e1
    Put this value in eq. (3) and find the efficiency.
  8. Mar 27, 2009 #7

    Andrew Mason

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    Overall efficiency is:
    [tex]\eta_T = 1 - \frac{Q_{c2}}{Q_{h1}}[/tex]


    [tex]\frac{Q_{c2}}{Q_{h1}} = \frac{Q_{c2}}{Q_{c1}} \frac{Q_{c1}}{Q_{h1}}[/tex]

    And since:

    [tex]\frac{Q_{c1}}{Q_{h1}} = 1 - \eta_1[/tex] and

    [tex]\frac{Q_{c2}}{Q_{c1}} = \frac{Q_{c2}}{Q_{h2}} = 1 - \eta_2[/tex]


    [tex]\eta_T = 1- (1-\eta_1)(1-\eta_2)[/tex]

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