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COP efficiency

  1. Jun 21, 2011 #1
    1. The problem statement, all variables and given/known data

    A reversible refrigerator has a coefficient of performance equal to 9.55

    What is its efficiency?

    2. Relevant equations

    COP = Qh/W

    e = W/Qh

    3. The attempt at a solution

    I thought e would simply be 1/9.55 = 0.1047 =0.105 (Sig. fig)

    This answer was incorrect. Can somebody please help?
     
  2. jcsd
  3. Jun 21, 2011 #2

    rock.freak667

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    Re: COP/efficiency

    For one I am not sure how you are calculating efficiency since the refrigerator is producing no work output but using work input to move heat from a region of low temperature to a region of high temperature.
     
  4. Jun 21, 2011 #3
    Re: COP/efficiency

    I don't know. That's the question in its entirety, and the answer is not 0, so there must be a way of figuring it out.
     
  5. Jun 21, 2011 #4

    rock.freak667

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    Re: COP/efficiency

    The problem will come in that your efficiency from a COP calculation can turn out to be more than 100%.

    COP = Qc/W

    Applying simple heat balance will give Qc=Qh+W

    But W is your work input and Qh is your heat output.
     
  6. Jun 21, 2011 #5

    Andrew Mason

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    Re: COP/efficiency

    For a refrigerator, output is considered to be the heat removed from the inside. So COP (output/input) is: Qc/W. For a heat pump, output is the heat delivered, Qh, so COP is Qh/W.

    So if they say the COP of the refrigerator is 9.55, this means Qc/W = 9.55

    Like rock.freak I am not clear on what efficiency means for a refrigerator. Does the question provide any further information? Try W/Qh using W+Qc = Qh

    AM
     
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