Type your question here A heat pump receives heat from a lake that has an averag
ID: 1862058 • Letter: T
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Type your question here
A heat pump receives heat from a lake that has an average winter time temperature of and supplies heat into a house having an average temperature of 27^C. If the house loses heat to the atmosphere at the rate of 64 000 kJ/h, determine the minimum power supplied to the heat pump, in kW If the lake water pump requires 0.53 kW to circulate the water to the heat pump, determine the rate of heat absorbed for the lake. From part b, determine the actual COP of the heat pump. Comment on the reason(s) for the difference in COP.Explanation / Answer
We have
TL=6+273=279K, TH=27+273=300K,Qlost=QH=64000/3600=17.78kW
(a) For minimum power supplied to the heat pump we consider heat as an ideal heat pump, so
QH/Win=TH/(TH-TL)
17.78/Win=300/(300-279)
Win=1.25kW
COP=TH/(TH-TL)=300/21=14.28
(b) here
Net work input=Wnet=1.25+0.53=1.78kW
QL=QH-Wnet=17.78-0.53=17.25kW
(c) actual COP=QH/Wnet=17.78/1.78=9.98
(d) Here we have seen that the actual COP is less than the ideal COP because in actual case the value of work input has increased.
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