Heat Pump Q: Cost of Winter Heat w/ 75% CoP

AI Thread Summary
The discussion revolves around calculating the cost of heating a home using a heat pump with a coefficient of performance (CoP) of 5.33, derived from a theoretical maximum CoP of 7.11. Participants clarify that the actual CoP should be used directly in the equation Qh/Win, rather than involving the ideal CoP again. The correct cost calculation indicates that heating the home would cost $187.50, which seems counterintuitive since the actual heat pump is more efficient than the ideal scenario. Misinterpretations of the CoP and its application in the equations led to initial incorrect calculations of $1330. The conversation emphasizes the importance of using the correct CoP value for accurate cost estimations.
ewnair
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Homework Statement



Assume that you heat your home with a heat pump whose heat exchanger is at Tc = 2degrees celcius and which maintains the baseboard radiators at Th = 47 degrees celcuiys. If it would cost $1000 to heat the house for one winter with ideal electric heaters (which have a coefficient of performance of 1), how much would it cost if the actual coefficient of performance of the heat pump were 75% of that allowed by thermodynamics?

Homework Equations



I understand that K = Th/(Th-Tc)

The Attempt at a Solution



I converted the temperature to kelvin and solved for the ideal coefficient which was 7.11. 75% of that is 5.33. Equated this to Qh/Win and assumed that Qh is the same for both actual and ideal pump. I multiplied 1000 with 7.11/5.33 and got an answer of $1330. However the answer is incorrect. Can anyone help please?
 
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Once you have figured out that the heat pump's actual COP is 5.33, the 7.11 figure should not enter into the calculation again.

... 75% of that is 5.33. Equated this to Qh/Win and assumed that Qh is the same for both actual and ideal pump.

Yes, absolutely. So what is Win, using this equation?
 
I got $1330 which isn't correct. i equated the 2 equations and plugged in Win = 1000 and K = 7.11 to find out Qh. Then i used this value of Qh and K= 5.33 to get Win which is $1330 but the answer is wrong.
 
K is not 7.11, it is 5.33.

ewnair said:
I converted the temperature to kelvin and solved for the ideal coefficient which was 7.11. 75% of that is 5.33. Equated this to Qh/Win ...

Yes, 5.33 should be equated with Qh/Win. Then solve the equation for Win.
 
Redbelly98 said:
K is not 7.11, it is 5.33.
Yes, 5.33 should be equated with Qh/Win. Then solve the equation for Win.

The Problem states that K=1 (which is kind of misleading). So if we have a actual engine with K=5.33, our work should be less.

So, 1=Qh/(1000) and
5.33=Qh/(dollars we pay)

you will get $187.5 as your answer (you might think that is weird because an "actual" engine is more efficient than a carnot engine)
 
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