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Energy required to heat a house |
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| Feb28-11, 05:54 PM | #1 |
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Energy required to heat a house
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. |
| Feb28-11, 08:54 PM | #2 |
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| Feb28-11, 09:59 PM | #3 |
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Cop = 296.55 k / (296.55 k - 266.55 k)
cop = 9.89 (5440 w * 9.89*.3) = 16140.48w |
| Mar1-11, 06:36 AM | #4 |
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Energy required to heat a houseAM |
| Mar1-11, 06:09 PM | #5 |
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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
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| Mar1-11, 07:44 PM | #6 |
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| Mar1-11, 10:55 PM | #7 |
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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 |
| Mar2-11, 03:00 PM | #8 |
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| Mar2-11, 09:45 PM | #9 |
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Thanks for bearing with me and helping me figure it out :)
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| carnot engine, cop, thermodynamics |
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