Energy required to heat a house


by Runaway
Tags: carnot engine, cop, thermodynamics
Runaway
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#1
Feb28-11, 05:54 PM
P: 48
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.
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Andrew Mason
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#2
Feb28-11, 08:54 PM
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Quote Quote by Runaway View Post
COP = Qh / W
....
I'm not sure how to set up an equation for part B.
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
Runaway
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#3
Feb28-11, 09:59 PM
P: 48
Cop = 296.55 k / (296.55 k - 266.55 k)
cop = 9.89
(5440 w * 9.89*.3) = 16140.48w

Andrew Mason
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#4
Mar1-11, 06:36 AM
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Energy required to heat a house


Quote Quote by Runaway View Post
Cop = 296.55 k / (296.55 k - 266.55 k)
cop = 9.89
(5440 w * 9.89*.3) = 16140.48w
I thought you said COP = Qh/W. What is this 16140 W supposed to be?

AM
Runaway
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#5
Mar1-11, 06:09 PM
P: 48
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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#6
Mar1-11, 07:44 PM
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Quote Quote by Runaway View Post
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
Yes. I see that. My question is why did you multiply COP by Qh? I thought you were supposed to find W.

AM
Runaway
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#7
Mar1-11, 10:55 PM
P: 48
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
Andrew Mason
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#8
Mar2-11, 03:00 PM
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Quote Quote by Runaway View Post
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
Correct. Technically W is really power and Qh is really rate of heat flow: dW/dt = (dQh/dt)/COP

AM
Runaway
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#9
Mar2-11, 09:45 PM
P: 48
Thanks for bearing with me and helping me figure it out :)


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