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
Messages
3
Reaction score
0

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?
 
Physics news on Phys.org
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)
 
Thread 'Voltmeter readings for this circuit with switches'

Thread 'Voltmeter readings for this circuit with switches'

TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
View full post »
Thread 'Correct statement about a reservoir with an outlet pipe'

Thread 'Correct statement about a reservoir with an outlet pipe'

The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
View full post »

Similar threads

Heat Pump Problem: Max Heat Transfer/COP Calculation
Replies
8
Views
5K
Saving Electrical Energy with a Heat Pump: 75%
Replies
1
Views
2K
B Heat pump as a heat engine that operates in reverse
Replies
19
Views
4K
Calculating the Cost of Heating with a Less Efficient Heat Pump
Replies
1
Views
3K
B Thermal physics, COP upper bound for heat driven heat pumps
Replies
1
Views
3K
B Lorenz COP of a heat pump and the temperatures of the hot/cold reservoirs
Replies
3
Views
3K
Heat Pump Power and COP: How Does Temperature Affect Efficiency?
Replies
1
Views
2K
Heat pump problem with heat loss?
Replies
2
Views
2K
Geothermal heat pump or heat engine power requirements
Replies
8
Views
4K
What is the power input (in kW) needed to run the heat pump?
Replies
2
Views
5K

Hot Threads

Recent Insights

Back
Top