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2 Stage Carnot Engine

  1. Oct 13, 2006 #1
    Consider a Carnot engine operating between temperatures T_h_ and T_c_, where T_c_ is above the ambient temperature T_o_. A second engine operates between the temperatures T_c_ and T_o_.

    Calculate (in terms of temperatures) the overall efficiency of the two-stage engine operating this way, and compare with the efficiency of a single Carnot engine operating between the high and low temperatures T_h_ and T_o_.
    Which has the higher efficiency, the two-stage or the single engine?

    e = 1 - (T_c_/T_h_)

    e_2stage_ = 1 - (T_c_/T_h_)(T_o_/T_c_) = 1 - (T_h_-T_o_)/(T_h_)

    If I set this up correctly, the 2tage carnot engine will be by far more efficent since it uses the discarded heat to create more work.

    I have a funny feeling as if this is violating some law of thermodynamics however..

    any advice and feedback is much appreciated.

  2. jcsd
  3. Oct 13, 2006 #2
    Maybe this image helps :)

    Attached Files:

  4. Oct 13, 2006 #3


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    What are the hot and cold temperatures for the single stage engine?
  5. Oct 13, 2006 #4
    hot temp = T_h_ and cold temp = T_c_ for single stage disregarding the second engine, which is T_o_ for the cold temp and T_c_ (above ambient temp) for the hot temp.
  6. Oct 13, 2006 #5


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    I think that what you are calling e_2stage is in fact the efficiency of a single stage operating between T_h and T_o. I can't see the equality between the two expressions you wrote. How did you come up with e_2stage?

    Which is more efficient, an engine operating between T_h and T_c or one operating between T_h and T_o?
    Last edited: Oct 13, 2006
  7. Oct 13, 2006 #6
    hmm, I see my mistake, let me try it this way

    e = W/Q
    e2stage = (W1 + W2)/(Q1+Q2)
    e2stage = W1/Q1 + W2/Q1 + W1/Q2 + W2/Q2
    e2stage = (Q1 - Q2)/Q1 + (Q1 - Q2)/Q2 + (Q2 - Q3)/Q1 + (Q2 - Q3)/Q2

    QH/QC = TH/TC

    e2stage = (T_h_ - T_c_)/T_h_ + (T_h_ - T_c_)/T_c_ + (T_c_ - T_o_)/T_h_ + (T_c_- T_o_)/T_c_

    is this it? I can't believe how far off I was earlier. Sorry for confusing you guys as much as I did.
    Last edited: Oct 13, 2006
  8. Oct 13, 2006 #7


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    The first two lines are good, but the third violates some basic rules of Algebra. The line after that I don't get what you are doing.

    The way I interpret your first two lines is this
    e is the efficiency of the single stage engine
    W is the work done by the single stage engine
    Q is the energy input to the single stage engine
    W1 and W2 are the work done by the individual stages of the two-stage engine, and Q1 and Q2 are the energy inputs to the two stages.

    To compare the efficiencies, e to e2stage, I would set the work outputs equal (as I did by adding the blue terms to yoru equation), and compare the energy input Q to the sum of the stage inputs Q1 + Q2. The known efficiencies for the carnot engine can be incorporated into the second equation to begin the analysis. The relative magnitude of Q and (Q1 + Q2) can be interpreted to tell which efficiency is greater. You need to state your conclusion, not just come up with an equation for e2stage
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