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Homework Help: Air conditioner homework

  1. Nov 22, 2006 #1
    A home is kept cool by an air conditioner. The outside temperature is 311.75K and the interior of the home is 288.55K. If 127kJ/h of heat is removed from the house, what is the minimum power that must be provided to the air-conditioner? answer in kJ/h
    my work

    COP = Th / (Th - Tc)
    = 311.75 / (311.75 - 288.55)
    = 13.44

    W = 127kJ/h / 13.44
    = 9.45 kJ/h this is incorrect according to the homework server

    Anybody know where i went wrong on this one???
    any help is appreciated!:smile:

    Sergio
     
  2. jcsd
  3. Nov 23, 2006 #2

    marcusl

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    There's no way the electrical input can be less than the power removed! You need to multiply, not divide.
     
  4. Nov 23, 2006 #3
    Ok we went about this a different way and got the right answer...for those of you who may have the same problem...

    Th, Tc and Qc are given in the problem...isolate Qh using Qh/Th = Qc/Tc
    to find work
    W = Qh-Qc

    thank you for your help marcusl i realised i was going about it the wrong way :)
     
  5. Nov 24, 2006 #4

    OlderDan

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    Homework Helper

    I think that is not true. If the temperature difference goes to zero, it should ideally take no work to move the heat from one resevoir to another. I think the only problem with the original calculation is that the OP was using the COP for heating instead of the COP for cooling.

    http://en.wikipedia.org/wiki/Coefficient_of_performance

    Qh/Th = Qc/Tc
    Qh = Qc(Th/Tc)
    W = Qh - Qc = Qc(Th/Tc) - Qc = Qc[(Th/Tc) - 1] = Qc(Th- Tc)/Tc = Qc/COP_cooling

    COP_cooling = Tc/(Th- Tc) = 288.55/(311.75 - 288.55) = 12.44

    W = 127kJ/h/12.44 = 10.21kJ/h
     
  6. Nov 24, 2006 #5
    ahhh ok i see what i did wrong!

    Perfect thank you very much for your time!!!

    Sergio :D
     
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