A Carnot heat pump is to be used to heat a house and maintain it at 20°C in winter. On a day when the average outdoor temperature remains at about 2°C, the house is estimated to lose heat at a rate of 82,000 kJ/h. If the heat pump consumes 8 kW of power while operating, determine ...
a.) how long the heat pump ran on that day
b.) the total heating costs, assuming an average price of 8.5¢/kWh for electricity
c.) the heating cost for the same day if resistance heating is used instead of a heat pump
9 comments:
should the answer to part b be similar to part c?
Anon 11:34 PM
No. The resistance heater costs more than 16 times as much as our HP !
I'm confused on how you'd start to figure out the time necessary
Anonymous 6:12 PM:
The heat pump does not run ALL the time. Youneed to determine how many hours it must run each day in order to replenish the heat lost to the surroundings (82,000 kJ/h). The heat is LOST in all 24 hrs in a day. So determine how much energy is lost each day. This is Qh. Use your knowledge of heat pumps to determine the amount of work input to the heat pump required each day, in kJ. Then, divide this amount of work by the power of the heat pump, 8 kJ/s. The result is the time in seconds that the heat pump must operate in each 24 hour day.
I hope this helps. Where are you located ?
thanks! I got the Qh=1968000 kJ which gives the work needed to be 1.399 kJ. In the end this gave .175s...does this make sense?
Anonymous:
Your Qh is correct, but the other values you posted are wrong. I am just guessing that you used T in degC instead of T in Kelvin when you computed the COP. Try again !
hmm I did use kelvin so I have no idea..when doing the COP=Th/(Th+Tc) correct?
alright i got the COP=.5 which gives the work=949248.1017 and the time to be 118656.0127 seconds?
Anonymous 12:33 PM.
Your COP is not correct. The COP is 16.3 using Th = 293 K and Tc - 275 K.
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