Suppose a power company converts 40 percent of the heatingvalue of the oil it us

Question

Suppose a power company converts 40 percent of the heatingvalue of the oil it uses into electric power. Suppose also thahomeowners can buy heat pumps that are 30 percent as effective asan ideal Carnot machine. If a homeowner maintains the inside of herhouse at 68 degrees F, at what outside temperature would the heatpump be competitive with an oil furnace that recovered 80 percentof the heating value of the fuel? That is to say, at what outsidetemperature would the heat delivered by the heat pump per gallon ofoil burned at the power company equal the heat delivered by thefurnace per gallon of oil? The text provides the answer as-10.8F. I tried to use the ideal Carnot equation with .40, .30 and 68Fand then multiplying or dividing that answer with .80 but it is notworking and I can't seem to figure out another way of doing thisproblem. Please help. Thanks so much! Suppose a power company converts 40 percent of the heatingvalue of the oil it uses into electric power. Suppose also thahomeowners can buy heat pumps that are 30 percent as effective asan ideal Carnot machine. If a homeowner maintains the inside of herhouse at 68 degrees F, at what outside temperature would the heatpump be competitive with an oil furnace that recovered 80 percentof the heating value of the fuel? That is to say, at what outsidetemperature would the heat delivered by the heat pump per gallon ofoil burned at the power company equal the heat delivered by thefurnace per gallon of oil? The text provides the answer as-10.8F. I tried to use the ideal Carnot equation with .40, .30 and 68Fand then multiplying or dividing that answer with .80 but it is notworking and I can't seem to figure out another way of doing thisproblem. Please help. Thanks so much!

Explanation / Answer

68 F = 293 K Here it is given that input energy is same for both , So power output of furnace = 80% of fuel energy = 0.8 Qin So efficiency = Qout / Qin = 0.8 Also let the outside temperature be T , Then efficiency of ideal carnot heat pump = 293 / (293-T) Also efficiency of given heat pump = 0.4*0.3*293 / (293-T ) Equating the two , 0.12*293/(293-T) = 0.8 On solving we get , T = 249 K So 249 K = -11 F

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