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Calculate the ideal (Carnot) efficiency of a heat engine

  1. Dec 11, 2005 #1
    Calculate the ideal (Carnot) efficiency of a heat engine operating between 23.0 degrees and 515 degrees. How much heat would be rejected by the engine if 1.00 x 10^6 calories were taken from the high temperature resivoir.

    Would you just use e=1-(T(c)/T(H))?
    And for the second part i don't have any clue how to get started.
     
  2. jcsd
  3. Dec 11, 2005 #2

    lightgrav

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    You have to use Kelvin ... .

    efficiency is not "1" if your Energy input is split between 2 (or more) outputs. You have a million cal input, and only "e %" of them will be used for what you had intended. Once you get Tc/Th , you know what fraction are wasted.
     
  4. Dec 11, 2005 #3

    Andrew Mason

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    The efficiency is defined as the ratio of work output to heat input:

    Efficiency[tex] = W/Q_H = (Q_H-Q_C)/Q_H = 1 - Q_C/Q_H[/tex]

    It can be shown that for a Carnot engine (no entropy change), [itex]Q_C/Q_H[/itex] is equal to [itex]T_C/T_H[/itex].

    That should enable you to answer the question.

    AM
     
    Last edited: Dec 11, 2005
  5. Dec 11, 2005 #4
    So for efficiency e =1-T(c)/T(h)=1-(298.18/788.15)=62.171%
    And for the heat rejected .62171=x/1.00*10^6=6.22*10^5 cal.

    Is this right
     
  6. Dec 11, 2005 #5
    Or would you have to include the 1 in the efficiency equation again?
     
  7. Dec 12, 2005 #6
    No Body knows?!?
     
  8. Dec 13, 2005 #7

    Andrew Mason

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    You are trying to find the heat rejected. You have found the work done: W = 6.22 x 10^5 cal. The heat rejected (heat flow out to cold reservoir) plus the work done has to be equal to the heat flow into the engine.

    [tex]Q_C = Q_H - W = 1 \times 10^6 - 6.22 \times 10^5 = 3.78 \times 10^5 cal.[/tex]

    AM
     
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