I think you are making a slight error with your assumptions above.
The shared reservoir, T, is Th for the heat engine and Tc for the heat pump (heat engine takes from a high temp reservoir and dumps into low).
Q,HE = W / (1 - (TL / T) )
Q,HP = W / (1 - (T / TH) )
They certainly aren't...
Why do you say that?
I thought that since they were reversible the efficiencies were as follows:
\eta HE = W / QH1 = 1 - (TL/T)
and COP HP = QH2 / W = 1 / (1 - (T/TH))
unless I am misstaken.
I do confirm that the diagram is correct.
Keep in mind that they DID specify the values of T, TL and TH, I just can't remember them.
Also remember that both the heat pump and the heat engine are reversible.
I talked to some other people and they got 5 as well so that makes me feel a...
Please find attached my drawing of the system, which was given on the exam.
What if you assume that all of the work generated by the heat engine is used to power the heat pump?
But then comes the question, is that a valid assumption?
Thanks for the response.
So I wrote my Thermodynamics final today and one of the questions were as follows:
A reversible heat engine supplies a reversible heat pump. Both draw off of a temperature reservoir T (can't remember the exact value, but there was one). The heat engine discharges heat into TL, and the heat...