Question

A heat pump cycle operating at steady state receives energy by heat transfer from well water at 10oC and discharges energy by heat transfer to a building at the rate of 1.4x105 kJ/h. Over a period of 14 days, an electric meter records that 1620 kW·h of electricity is provided to the heat pump. These are the only energy transfers involved. Determine the total amount of energy transfer by work into the heat pump over the 14-day period, in GJ. Determine the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period, in GJ. Determine the heat pump’s coefficient of performance.

Answer 1

Answer: the total amount of energy transferred by work into the heat pump over the 14-day period is 5.8GJ

Explanation: since the meter reading for the 14 days period is 1620kW.h

Electrical power Ep is

Ep = 1620*10^3W.h

In 14 days there are 14*24 hrs = 336hrs

Ep = (1620*10^3)/336 = 4821.42W

This means 4821.42J of work is done per sec. Therefore for 14 days we have,

3600*24*14*4821.42J of enegy

= 5831989632J = 5.8GJ

Answer 2 = the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period is 41.2GJ

Explanation: rate of energy transfer as heat to building is 1.4*10^5 kJ/hr

That is 1.4*10^8 J/hr

In 14 days it becomes

24*14*1.4*10^8 = 47GJ

Energy received from well by heat pump =

Energy transfered - energy received

= 47 - 5.8 =41.2GJ

Answer 3. the heat pump’s coefficient of performance is 0.12

Explanation: COP = energy obtained/ work done

COP = 5.8/47 = 0.12