Question

A heat pump that supplies energy to a house at a rate of 8000 kJ/h for each kW of electric power it draws. a. Determine the COP. (Points 7) b. Determine the rate of energy absorption from the outdoor air (MJ/h). (Points 8)

Note: could you plz explain little about it rather than just solving it

Answer 1

COP is given by the following relation:

where Q is the useful heat supplied by the heat pump

W is the input work given

Thus in our case:

Q = 8000 kJ/h

W = 1 kW = 3600 kJ/h

Thus

COP = 8000/3600 = 2.23

For the second part, you need to find the efficiency of the heat pump

8000 kJ/h of heat is supplied to the house. All the heat is acquired from the air outside along with some losses. The losses is given by:

energy loss = 8000 x eff

Thus the rate of energy absorption from the outdoor air is foud as:

E = 8000 + 8000 x eff

= 8000(1 + eff)