A heat pump supplies heat to a house at the rate of 150 000 kJ/h when the house

Question

A heat pump supplies heat to a house at the rate of 150 000 kJ/h when the house is maintained at 25?C. Over the period of one month, the heat pump operates for 120 hours to transfer energy from a heat source outside the house to the inside the house. Consider two different outside sources: the first is the outside air at -3?C and the second is a lake at 7?. If electricity costs $0.06/kWh, determine the maximum money saved by using the lake water rather than the outside air as the outside energy source.

Explanation / Answer

1 kW-h = 1 kJ/s*3600 s = 3600 kJ

Heat transfer Q = 150 000 *120 = 18*10^6 kJ

For air:

Carnot COP = (25+273)/(25 - (-3)) = 10.643

10.643 = 18*10^6 / W

W = 1.6913*10^6 kJ

Cost = 1.6913*10^6 *0.06 / 3600 = $28.19

For lake water:

Carnot COP = (25+273) / (25-7) = 16.56

16.56 = 18*10^6 / W

W = 1.087*10^6 kJ

Cost = 1.087*10^6 *0.06 / 3600 = $18.12

Max. money saved = 28.19 - 18.12 = $10.07

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