Energy required to heat a house

  • Thread starter Thread starter Runaway
  • Start date Start date
  • Tags Tags
    Energy Heat
AI Thread Summary
The discussion revolves around calculating the electric power required to maintain a house's interior temperature using electric resistance heaters and a heat pump. For part A, it is established that 5440 W is needed to replace thermal energy lost through the house's exterior. In part B, participants explore the coefficient of performance (COP) of a heat pump, ultimately determining that the work required to maintain the same heat flow is 1833.5 J/s when considering a COP of 2.967 (0.3 times the Carnot value). The conversation emphasizes the importance of correctly applying the COP formula to find the required power for heating. Overall, the calculations illustrate the differences in energy requirements between direct electric heating and heat pump systems.
Runaway
Messages
48
Reaction score
0

Homework Statement


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.


Homework Equations



COP = Qh / W

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.
 
Physics news on Phys.org
Runaway said:
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
 
Cop = 296.55 k / (296.55 k - 266.55 k)
cop = 9.89
(5440 w * 9.89*.3) = 16140.48w
 
Runaway said:
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
 
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
 
Runaway said:
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
 
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
 
Runaway said:
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
 
Thanks for bearing with me and helping me figure it out :)
 
Back
Top