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Energy required to heat a house

  1. Feb 28, 2011 #1
    1. The problem statement, all variables and given/known data
    A) A house loses thermal energy through the exterior walls and roof at a rate of 5440 W when the interior temperature is 23.4 C and the outside temperature is −6.6 C.
    Calculate the electric power required to maintain the interior temperature at Ti if the
    electric power is used in electric resistance heaters (which convert all of the electricity
    supplied to thermal energy).
    Answer in units of W.

    B) Find the electric power required to maintain the interior temperature at Ti if the electric
    power is used to operate the compressor of a heat pump with a coefficient of performance
    equal to 0.3 times the Carnot cycle value.
    Answer in units of W.


    2. Relevant equations

    COP = Qh / W

    3. The attempt at a solution
    Part A is 5440 W because for the temperature to stay constant the energy leaving the house must be replaced at the same rate.

    I'm not sure how to set up an equation for part B.
     
  2. jcsd
  3. Feb 28, 2011 #2

    Andrew Mason

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    What is the COP for a Carnot heat pump operating between these temperatures? What is .3 times that (actual COP)? Use that actual COP of the heat engine and the above equation to determine the amount of work (per second) required to deliver that same 5440 J of heat (per second).

    AM
     
  4. Feb 28, 2011 #3
    Cop = 296.55 k / (296.55 k - 266.55 k)
    cop = 9.89
    (5440 w * 9.89*.3) = 16140.48w
     
  5. Mar 1, 2011 #4

    Andrew Mason

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    I thought you said COP = Qh/W. What is this 16140 W supposed to be?

    AM
     
  6. Mar 1, 2011 #5
    I used COP = Qh/(Qh-Qc) = Th/(Th-Tc) to find the COP then I took that and multiplied it by the output * .3 and got 16140 J/S, or in other words, 16140 Watts
     
  7. Mar 1, 2011 #6

    Andrew Mason

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    Yes. I see that. My question is why did you multiply COP by Qh? I thought you were supposed to find W.

    AM
     
  8. Mar 1, 2011 #7
    So if I follow you, the answer would be
    COP = Qh/W
    COP * W = Qh
    W=Qh/COP
    W= 5440 J/s/(9.89*.3) = 1833.5 J/s
     
  9. Mar 2, 2011 #8

    Andrew Mason

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    Correct. Technically W is really power and Qh is really rate of heat flow: dW/dt = (dQh/dt)/COP

    AM
     
  10. Mar 2, 2011 #9
    Thanks for bearing with me and helping me figure it out :)
     
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