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Heat Pumps

  1. May 22, 2005 #1
    A perfectly reversible heat pump heats a building at 20deg from the atmosphere at 5deg. If the heat pump is run by an electrical motor whose efficiency is 80%, what is the cost of 1KWhr of heat supplied? (1KWhr of electricity is 6p)

    I am confused by this qn as I thought the efficiency, e = 1-(T1/T2), which in this case would be 0.75 and not 0.8?!
    Could someone give me a clue on where I'm going wrong please!
  2. jcsd
  3. May 22, 2005 #2


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    Let's see. You have a perfect reversible heat pump of coefficent of operation [tex]COP[/tex] connected to an electric motor of efficiency [tex]\eta_e[/tex].

    The coefficient of operation of the heat pump is defined as:

    [tex]COP=\frac{Q_h}{W}=\frac{T_h}{T_h-T_c}[/tex] being [tex]W[/tex] the work exerted on the cooling fluid.

    So that you can work out the COP because you have all temperatures. The net COP of the whole would be:

    [tex]COP_n=\frac{Q_h}{W_e}=\frac{Q_h \eta_e }{W}=\eta_e COP[/tex]

    so this yield another number.

    for [tex]Q_h=1KWh[/tex] the cost of the electrical power which must be supplied is:

    [tex]C=\frac{1}{COP_n}c[/tex] being [tex]c[/tex] the cost of 1 KWh of electricity (6p)
  4. May 22, 2005 #3

    Andrew Mason

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    You are confusing motor efficiency with the efficiency or coefficient of performance of the heat pump. They are quite different and independent concepts. Also you have to use K not [itex]\degree C[/itex]

    The efficiency of the heat pump is given by the coefficient of performance:

    [tex]cop = \frac{Q_h}{W} = \frac{Q_h}{Q_h-Q_c} = \frac{1}{1 - Q_c/Q_h}[/tex]

    For a Carnot (reversible) cycle, [itex]\Delta S = Qh/Th - Qc/Tc = 0[/tex] so:

    [tex]Q_c/Q_h = T_c/T_h[/tex]

    [tex]cop = \frac{1}{1 - T_c/T_h}[/tex]

    Using K, work out the cop. From that you can determine the amount of heat delivered to the building for each joule of work supplied. Since the motor is 80% efficient, for each joule of electrical energy consumed it delivers .8 J of work to the heat pump.

  5. May 22, 2005 #4
    V Helpful, I can now get the correct answer!
    Thanks for your time guys.
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