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Easy thermodynamics problem

  1. Jun 16, 2005 #1
    This problem is supposedly easy but I just can't figure out how to do it.

    A heat pump is used to heat a house and maintain it at 24°C. On a winter day when the outdoor temperature is -5°C, the house is estimated to lose heat at a rate of 80,000 kJ/h. Determine the power required to operate this heat pump

    Only two equations I know are: COP = Qh/(Qh-Ql) = Qh/W
    where W is the net work done (what I'm looking for).

    In the book, they have an example just like it except the COP is given so all I gotta do is plug numbers in the equation and get the answer but I don't know how to do it when the temperatures are given instead of the COP.
  2. jcsd
  3. Jun 16, 2005 #2

    Andrew Mason

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    Does the question also say that the heat pump operates on an ideal Carnot cycle? If so:

    [tex]\Delta S = Q_H/T_H - Q_C/T_C = 0[/tex] so:

    [tex]Q_H/T_H = Q_C/T_C \rightarrow Q_H/Q_C = T_H/T_C[/tex]

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

  4. Jun 16, 2005 #3
    No, if it were a carnot I'd know how to do it but it doesn't say anything about that in the problem so I don't think I can just assume carnot cycle.
  5. Jun 16, 2005 #4
    I think I'm wasting too much time on this problem and I have a test tomorrow. Anyone know if there's a way to solve this problem if it were not a carnot cycle?

  6. Jun 16, 2005 #5

    Andrew Mason

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    Not possible. It would depend on the details of the thermodynamic cycle used. Without knowing that, you cannot answer the question. I would suggest that you assume a Carnot cycle.

  7. Jun 16, 2005 #6
    Alright. So, assuming it's a carnot cycle, this is how it should be solved:

    80,000 KJ/h = 2.22 KW

    COP = Th/(Th-Tc) = 297 k/(297 k-268 k) = 10.24

    COP = Qh/W ---> W = Qh/COP

    W = 2.22 KW/10.24 = 2.17 KW which should be equal to the power needed to operate the pump.

    Did I do anything wrong?
  8. Jun 16, 2005 #7

    Andrew Mason

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    Just that Qh = 22.2 kW but your answer is right.

  9. Jun 16, 2005 #8
    Yes, that is right except as Andrew explained.
  10. Jun 16, 2005 #9
    Aight thanks. You guys said that Qh isn't equal to 22.2 KW because that would be Qh/time. The units for Qh should be in KJ or J's I guess right?
  11. Jun 16, 2005 #10
    No, Qh is indeed 22.2kW but you said 2.22kW
  12. Jun 16, 2005 #11
    oh...I should probably go to bed now because my internal energy is going up, info in my head is starting to vaporize, my brain is becoming superheated, volume not changing since my skull is rigid, pressure increasing, enthalpy going up, aaaand I need to get some rest to cool down and condense the info again so it's usable by the time of the test. :zzz:
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