A heat pump refrigeration cycle is used to heat a building that is to be maintai

Question

A heat pump refrigeration cycle is used to heat a building that is to be maintained at 25 degree C. Heat is lost through the building's walls, windows, ceiling, etc. at a rate of 60,000 kj/hr. The minimum amount of power that is required by this heat pump is 2 kW. Calculate the maximum possible Coefficient of Performance for this heat pump and determine the outside air temperature. If the actual amount of heat added to the cycle is 50,000 kj/hr, determine the actual amount of power [kW] required by the heat pump.

Explanation / Answer

part (A)

for the room to be maintained at 25 degrees celsius; rate of heat lost by room= rate of heat supplied by heat pump

so rate of heat supplied by het pump=60,000Kj/Hr=16.66KW(since watt=joule/sec,divide 60,000 by 3600)

(1)maximum coefficient of performance = (rate of heat supplied by heat pump)/(minimum work input)

so,cop=16.66/2=8.33

(2)coefficient of performance=(T room)/(T room- T outside)

so, 8.33=(298)/(298-T)

gives you T outside=262.226 kelvin

part (B)

according to carnot principles

when heat pump operating between same temperaturescoefficient of performance will be same

heat added=50,000KJ/Hr=13.88KW

so by applying (1)

8.33=(13.88)/(work input)

so, work input =1.66 KW

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