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Carnot Air Conditioner

  1. Nov 28, 2004 #1
    Hi there,

    I have what should be a simple problem, but it's driving me insane.

    Here's the problem: A Carnot air conditioner maintains the temperature in a house at 290 K on a day when the temperature outside is 315 K. What is the coefficient of performance of the air conditioner?

    So I know to find the coefficient it would be Qc/W. I have neither of those numbers in the problem. I do have Th and Tc and even if I do substitutions I will ALWAYS have two unknowns. At least the way I am attempting this.

    My thinking was originally that I could substitute for Qc, since Oc = W + Qh but I still have two unknown variables... What am I forgetting??

    Any help will be greatly appreciated.
  2. jcsd
  3. Nov 28, 2004 #2

    Andrew Mason

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    They expect you to determine the coefficient of performance of an ideal Carnot cycle using:

    [tex]COP = \frac{T_C}{T_H-T_C}[/tex]

  4. Nov 29, 2004 #3
    Thanks for replying but how did you come to that point? In my book the equation for the coefficient of performance is Qc/W. No where in my book is your equation listed.

    Thanks for your help.
  5. Nov 29, 2004 #4

    Andrew Mason

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    For the refrigerator,
    [tex]Q_H = \Delta W + Q_C [/tex]
    This means that the heat flowing out of the cold reservoir + work added is equal to the heat flowing into the hot reservoir.

    For the reversible (ideal) Carnot cycle, [tex]\Delta S = 0[/tex] where S = Q/T
    [tex]Q_H/T_H - Q_C/T_C = 0[/tex]

    Combining these two equations:
    [tex]Q_H = Q_C(T_H/T_C) = \Delta W + Q_C[/tex]
    [tex]T_H/T_C = \Delta W/Q_C + 1[/tex]
    [tex]Q_C/\Delta W = 1/(T_H/T_C - 1)[/tex]

    Which simply reduces to:
    [tex]Q_C/\Delta W = T_C/(T_H-T_C)[/tex]

    Last edited: Nov 30, 2004
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