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

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Homework Help Overview

The discussion revolves around the cost of heating a home using a heat pump with a specified coefficient of performance (CoP) in comparison to ideal electric heaters. The problem involves understanding the implications of thermodynamic principles on heating costs, particularly focusing on the CoP values and their calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the actual CoP and its application in determining heating costs. There is confusion regarding the use of the ideal CoP value in subsequent calculations. Questions arise about the assumptions made regarding heat transfer and the relationship between Qh and Win.

Discussion Status

Some participants have provided guidance on the correct application of the CoP in calculations, suggesting that the initial ideal CoP should not be reused. Multiple interpretations of the problem are being explored, particularly regarding the implications of the CoP values on the heating costs.

Contextual Notes

There is mention of potentially misleading information in the problem statement regarding the CoP of ideal heaters, which may affect participants' understanding and calculations. The discussion also highlights the need to clarify the relationship between the actual and ideal performance of the heat pump.

ewnair
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Homework Statement



Assume that you heat your home with a heat pump whose heat exchanger is at Tc = 2degrees celsius 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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