When Does a Heat Pump Become Ineffective Based on COP?

[rvaafrica]

1. You have a heat pump which has a coefficient of performance (C.O.P.) that is 17% of the maximum achievable. You want to heat your house to 20◦C. At what outside temperature does the heat pump become not useful?



2. C.O.P. = 1/(T1/T0 - 1)
T1= Temperature of the outside
T0 = Temperature of the house



3. Hi. Basically I am NOT looking for someone to give me an answer. I would just like some help on how to approach this problem. I know how to solve for T1 if they give the maximum C.O.P but I am stuck on how to complete this problem since the max. C.O.P is not given. Any help would be greatly appreciated.

 

[kreil]

The formula you gave can be rewritten,

$$COP_{heating} = \frac{T_0}{T_1-T_0}$$

Assuming the temperature outside is colder than that inside, this formula represents the coefficient of performance for heating at maximum theoretical efficiency.

Read more: http://en.wikipedia.org/wiki/Coefficient_of_performance#Derivation

 

[Andrew Mason]

1. You have a heat pump which has a coefficient of performance (C.O.P.) that is 17% of the maximum achievable. You want to heat your house to 20◦C. At what outside temperature does the heat pump become not useful?
2. C.O.P. = 1/(T1/T0 - 1)
T1= Temperature of the outside
T0 = Temperature of the house
3. Hi. Basically I am NOT looking for someone to give me an answer. I would just like some help on how to approach this problem. I know how to solve for T1 if they give the maximum C.O.P but I am stuck on how to complete this problem since the max. C.O.P is not given. Any help would be greatly appreciated.

The equation for COP gives you the maximum (Carnot) COP. (Note: the denominator should be (1 - T1/T0)). Just multiply the maximum COP by .17

What is the value of COP at which the heat pump stops being useful?

AM