Using the Kelvin-Planck statement of the 2nd Law, demonstrate the following corollaries.
a.) The coefficient of performance (COP) of an irreversible heat pump cycle is always less than the COP of a reversible heat pump when both heat pumps exchange heat with the same two thermal reservoirs.
b.) All reversible heat pump cycles exchanging heat with the same two thermal reservoirs have the same COP.
8 comments:
In the example on p. 129 today in class, why did you choose to make the Qh's of the irreversible and reversible HP/HE equal? Is it arbitrary whether the Qh's or Qc's are chosen?
Giro:
It is not arbitrary, but it is difficult to tell which Qs to make the same between the two cycles. I THINK one way to choose is to look at the measure of the performance of the cycle (efficiency or COP) and see which Q (QH or QC) appears in the equation that defines the measure of performance. Keep the Q that appears in the measure of performance the same for both cycle. There are only two possibilities: keep QH or QC the same. Either will work. It's just that one choice makes the solution a bit easier.
Professor or anyone who can help,
I don't really know how to start this problem, so I was wondering if you could give me any advice on this problem. I'm sorry if this problem was discussed in class on Wed, I had an appointment.
anon:
This problem is just like the two proofs we did in class on Tuesday. Reverse the reversible HP, keep QH the same for both HPs. Include the hot reservoir in a new system. This new system completely converts heat into work and that violates the KP statement of the 2nd Law. I must say asking about P#1 at 8:30 PM the night before it is due does not look too good. Best of luck anyway though !
don't worry professor, that's the only one i have left.
6E page 20-22 has the answer (I think) you just switch the HE into heat pumps by changing arrows/labels.
Anon:
Whew, I am glad you did not just start the HW last night !
Anon:
Yes, this problem is very similar to showing why the Carnot Principles are true.
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