A heat pump cycle operating at steady state receives energy by heat transfer fro

Question

A heat pump cycle operating at steady state receives
energy by heat transfer from well water at 108C and
discharges energy by heat transfer to a building at the rate
of 1.2 3 105 kJ/h. Over a period of 14 days, an electric meter
records that 1490 kW.h of electricity is provided to the
heat pump. These are the only energy transfers involved.
Determine (a) the amount of energy that the heat pump
receives over the 14-day period from the well water by heat
transfer, in kJ, and (b) the heat pump’s coefficient of
performance.

Explanation / Answer

Electric meter reading (or Work input, W) = 1490 kW.h = 1490 kW. 3600 s = 5364 MJ

Heat transfer to the building (Q2) = Heat transfer rate*number of hours in a day* no. of days

Heat transfer to the building (Q2) = 123105*24*14 = 41363.28 MJ

a)

From the first law

Q2 = W + Q1

Q1 = Q2-W

Q1 = 41363.28 MJ- 5364 MJ

Q1 = 35999.28 MJ

Q1 = 35999280 kJ

This is the amount of heat received by the HP from the well water

b)

COP = Q2/W = 7.711

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