Tuesday, April 11, 2006

HW #4 - P #11 - Coefficient of Performance of a Household Heat Pump

A heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h. If the COP of the heat pump is 2.8, determine the power the heat pump draws when running.

6 comments:

Anonymous said...

For this problem I am using the equation COPh=Qh/(Qh-Qc) and W=Qh-Qc. I am given COP and I put in 3*(rate the house is loosing heat) for Qc. I converted the KJ/hr to KJ/sec=KW and I am getting W=10.19 KW, which is wrong. Do you see where I am messing up?

Anonymous said...

I am using the same equation as greenpepper except that I assigned Qc to equal 22,000 kJ/hr. I then used the COP=Qh/(Qh-Qc) equation to solve for Qh (the value if the pump was running all the time). And so from the hint, I said that the actual Qh (which I will from this point on call Qh') would have to be three times the first Qh I found...that is Qhactual = Qh'= 3Qh. I've probably completely misinterpreted this problem. Help! What has a value of 22,000kJ/h? Can someone please further clarify what I'm supposed to do with the info. that the pump only works 1/3 of the time? Thanks for any help :)

Dr. B said...

greenpepper 3:52 PM
COP definition and 1st Law are correct and useful.

The KEY to this problem is to realize that if the temperature of the house is going to remain constant, the heat pump must push the same amount of heat into the house as it is losing to outdoors. Therefore QH = 3*Qlost, where Qlost=22,000 kJ/h.

We do not know QC in this problem. We could calculate it, but it isn't very important unless we are studying global warming. QC in a heat pump is the amount of thermal energy absorbed from the outside air by the heat pump.

Dr. B said...

yoda 5:23
We don't know QC. See the comment I just made to greenpepper. All those different QH's you have got me confused. The ONLY QH is the rate at which heat flows from the heat pump into the house WHILE it is running. QH = 3*22,000 kJ/h.

I hope this combined with my comment for greenpepper is helpful.

Anonymous said...

In the Thermo CD handbook it states that all quantities are positive, for the cycles, but you have the answer being negative work. I can see that it would be negative becuase according to the work sign conventions, we are doing work on the sustem which is negative, but I am confused as to why the handbook states that with emphasis.

Dr. B said...

Claire 2:32 PM
That is a good question. My answer on the HW is not specific enough. It might be best to say something like "The work INPUT is 6.6 kW". This does not leave the reader with any doubt about things like the sign convention.