Complete the following involving reversible and irreversible cycles.
a.) Reversible and irreversible power cycles each discharge QC to a cold reservoir at TC and receive energy QH from hot reservoirs at TH and T'H, respectively. There are no other heat transfers involved. Show that T'H> TH.
b.) Reversible and irreversible heat pumpcycles each receive QC from a cold reservoir at TC and discharge QH to hot reservoirs at TH and T'H, respectively. There are no other heat transfers involved. Show that T'H < TH.
8 comments:
I just want to make sure I am reading this problem correctly. The Reversible cycle exchanges heat to a hot reservoir at T and the irreversible cycle exchanges heat to a different hot reservoir at T'. The cold reservoirs are at the same temp, Tc.
Jo 4:39 PM
I think you are reading the problem correctly. I would just insert some H's.
The reversible cycle gets Qh from a hot reservoir at Th and the irreversible cycle gets Qh from a different hot reservoir at Th'.
They both reject Qc to cold reservoirs that are at Tc.
I don't understand why I would get the complete opposite result with Kelvin's relationship. I assumed that Erev > Eirev. Then Qc/Qh is the same as Tc/Th. I know this should hold true because it's the material from Chapter 6. But why aren't they working in here? Thanks
This problem isn't working for me. I can prove both a and b. using equations from Chapter 6 but the algebra isn't working out using the enthalpy equations.... can you give me somewhere to start?
Kaifu 10:27 PM
I don't know what you did with Kelvin's relationship. I hope you did not apply it to the irreversible cycle ! Qc/Qh = Tc/Th, but this doesn't work for the irreversible cycle.
HairyPotter 6:22 PM
Start with deltaS = INT{dQ/T} + Sgen
For any cycle, deltaS = 0 because S is a state variable or property.
Apply this equation to the reversible cycle and then to the irreversible cycle. What do you know about Sgen for reversible cycles ? What about for irreversible cycles ? Do som algebra on these two equations and voila (hopefully) you've got it !
Dr. B - I think what kaifu did was take the fact that nrev = 1 - Tc/Th and nirrec < 1 - Tc/Th'
and because when rev Tc/Th = Qc/Qh
then with this and because Qc,Qh, Tc are constant the nin the relationship nirrev < 1 - Tc/Th' to be true Th' must be greater.
But how do you prove this using the entropy equations?
HairyPotter 6:49 PM
Yes, I think that is what Kaifu did.
I did my best to spell out how to use the fact that Sgen > 0 to show that Th' > Th. All that is left for you to do is some algebra.
At this point, I think you need to wait and see what the solution looks like.
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