A heat pump draws cool air from outside a building where the air temperature is
ID: 2180900 • Letter: A
Question
A heat pump draws cool air from outside a building where the air temperature is 5 degree C and delivers warm air to the inside of the building where the air temperature is 20 degree C, as shown. The heat pump has a coefficient of performance which is 60% that of the ideal Carnot value for a refrigerator. The motor driving the heat pump supplies 0.5 kW. Find the rate at which heat is transferred into the building by the pump. In summer the outside temperature changes to 30 degree C and it is desired to keep the interior of the building at 20 degree C. Explain how the heat pump can be used to cool the building and calculate the maximum rate at which heat can be extracted from the interior.Explanation / Answer
a)The Carnot COP is given by COP(max)=Tcool/(Thot-Tcool) (for a refrigerator) =278/(293-278) (in Kelvin) =18.533 So, COP(pump)=0.6*COP(max) =11.12 Supplied energy=0.5kW So , max rate of heat transfer = W*COP =0.5*11.12kW=5.56kW b)It can be used as a cooler by reversing the cycle In this case, COP(max)=(293)/(303-293)=29.3 Hence, COP(pump)=29.3*0.6=17.58 So, max rate of heat transfer=0.5*17.58=8.79kW