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Homework Help: Maximum work done by a Carnot engine

  1. Dec 31, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-12-31_11-56-44.png

    2. Relevant equations


    3. The attempt at a solution


    Taking the engine to be a Carnot engine,

    ## \eta = \frac { T_h – T_c } {T_h} = \frac { W} {Q_h} ## .....(1)

    ## Q_h = C(T_h – T_f) .....(2)

    \\ Q_c = C(T_f – T_c) .....(3)

    \\ W = Q_h – Q_c = C(T_h + T_c – 2 T_f) ## .....(4)


    Solving the above equations,

    ## T_f = \frac { 2T_c T_h }{T_h +T_c} ## .....(5)


    This solution does not lead me to any of the given option.

    Is this correct?
     
  2. jcsd
  3. Dec 31, 2017 #2

    TSny

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    Does ##\eta## remain constant during the process?
     
  4. Dec 31, 2017 #3
    Algebraically, in terms of ##T_h##, ##T_f##, and the heat capacity C, what is the entropy change of the hot system?
     
  5. Jan 3, 2018 #4
    As I understand: ## \eta ## is not defined during a process. ## \eta ## is defined for a process. For a Carnot engine working between temperatures ## T_h ## and ## T_f ##, ## \eta ## is defined as ## \frac W{Q_h} ##. This is what I have used.

    I am not getting the intention behind this question. Please give some more hint.
     
  6. Jan 3, 2018 #5
    The entropy change of the hotter system is equal and opposite to that of the colder as it is a reversible process.

    ## dS = C \frac { dT } T ## .....(1)

    ## \Delta S_h = - C \ln \frac { T_h}{T_f} = - \Delta S_c = - C \ln \frac { T_f}{T_c} ## .....(2)

    ## T_f = \sqrt{ T_h T_c} ## .....(3)

    ## W = Q_h – Q_c = C(T_h + T_c – 2 T_f) ## .....(4)

    ## W = C(\sqrt{ T_h } - \sqrt{ T_c})^2 ## .....(5}

    So, the answer is option (d).


    Why is using the efficiency equation not a correct step?

    The engine is working between the two temperatures through a reversible cycle. So, it is a Carnot engine.

    For calculating maximum work done, I have to take maximum efficiency. This is what I did in the OP.
     
  7. Jan 3, 2018 #6
    The temperatures of the two reservoirs are changing during the process. So you can't use the results for a single Carnot cycle with constant reservoir temperatures. You can't get from the initial state of this system to the final state reversibly with a single Carnot cycle. You need to use multiple Carnot cycles, each with slightly different pair of reservoir temperatures. And, for each of these multiple Carnot cycles, the efficiency is different.
     
  8. Jan 3, 2018 #7
    I didn't realize that the temperatures of the reservoirs are changing and the efficiency eqn is valid for constant reservoir temperatures even after solving the question.
    Thanks for it.
     
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