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Home heat pump

  1. Feb 10, 2006 #1
    In a home heat pump, heat is absorbed from the earth by a fluid circulating in buried pipes and delivered to the home interior. Suppose that the heat pump is 40% as efficient as a Carnot refrigerator. If the ground surrounding the pipes is at 5`C and the interior temperature of the house is kept to 20`, how many kilowatt-hours of heat would be supplied to the home interior for ever kilowatt-hour of electrical energy needed to operate the refrigerator?

    omega = .4
    [itex]\ Q_C=20 C [/itex]
    [itex]\ Q_H=5 C [/itex]
    [itex]\omega = Q_C/W [/itex]

    Well we can solve for the work but were looking for kilowatt-hours,

    E=Power X time.

    Can someone lead me in the right direction and provide the general equation suited for this problem?

    Last edited: Feb 10, 2006
  2. jcsd
  3. Feb 10, 2006 #2


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    What else can one say about the efficiency as it relates to the hot and cold temperatures, which should be in absolute temperature (K) not (C), although when one is looking at temperature difference, either unit is acceptable.
  4. Feb 10, 2006 #3
    So what formula is associated with this problem?
  5. Feb 10, 2006 #4

    Andrew Mason

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    As Astronuc says, work out the coefficient of performance of a Carnot cycle between these temperatures.

    Since the COP ([itex]\omega[/itex]) of this cycle is 40% that of the Carnot cycle, you can determine the input work (W) required using your formula [itex]\omega = Q_C/W [/itex]. [Note:[itex]Q_C[/itex] is the heat removed from the cold reservoir. The output heat [itex]Q_H[/itex] is [itex]Q_C + W[/itex]]

    Last edited: Feb 10, 2006
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