Tuesday, April 24, 2007
HW #6, P3 - A Reversible HE Used to Drive a Reversible Heat Pump - 6 pts
A reversible power cycle receives QH from a reservoir at TH and rejects QC to a reservoir at TC. The work developed by the power cycle is used to drive a reversible heat pump that removes Q'C from a reservoir at T'C and rejects Q'H to a reservoir at T'H.
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9 comments:
i have tried qh=whe+qc and qh'=whe+qc' to get whe/qh+qc/qh=whe/qh'+qc'/qh' but cannot seem to get 4 temperature values for qh'/qh. am I approaching this the wrong way?
Anon:
You are doing fine. You have successfully applied the 1st Law. Now, make use of the fact that each cycle is reversible and therefore the Kelvin relationship applies.
using both whe=qh-qc and we=qh'-qc' but am ending up with 2 terms, qh'/qh on one side and qh/qh' on the other. sticking with only one whe term I am getting an expression in terms of only 3 temperatures. what am i doing wrong?
will Q'h / Qh have the Qc and Q'c terms in it since Qh / Qc = Th / Tc and you can't really get rid of the Qc's
or i can do it another way....nevermind
confused:
The two work terms are equal. Does that help ? Make use of the fact that each cycle is reversible and therefore the Kelvin relationship applies. This will let you eliminate Qc and Q'c from your 1st Law equations.
ernie:
Make use of the fact that each cycle is reversible and therefore the Kelvin relationship applies. This will let you eliminate the Qc and Q'c terms.
does W hp = W he or does W hp = -W he
bert:
In terms of the 2st Law, Whe = Whp. In terms of the tie fighter diagrams, Whe = Whp. It just depends on whether you are using the sign convention or not.
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