Friday, April 14, 2006
HW #7 - P #4 - Thermal Efficiency of an Internally Reversible HE with Multiple Heat Transfers - 4 pts
A system executes a power cycle while receiving 750 kJ by heat transfer at its boundary where the temperature is 1500 K and discharges 100 kJ by heat transfer at another portion of its boundary where its temperature is 500 K. Another heat transfer from the system occurs at a portion of the system boundary where the temperature of the system is 1000 K. No other heat transfer crosses the boundary of the system. If no internal irreversibilities are present, determine the thermal efficiency of the power cycle.
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7 comments:
Anon email
This reversible HE interacts with THREE thermal reservoirs instead of the usual two. The 1st Law always applies and the entropy generation is zero because the cycle is reversible. Because the cycle is reversible and it is interacting with thermal reservoirs (each at a constant T), the T inside the cycle at which the heat is absorbed or rejected must also be constant. This makes it relatively easy to evaluate the integral of dQ/T in the 2nd Law. Two equations in 2 unknowns (Q2 and Wcycle). Solve them and answer the question !
I'm still having trouble. The first law comes out as
Q1+Q2=Qout+Wcycle.
The second law integration of dq/T ends up being deltaS=Q1/T1+Q2/T2-Qout/T3. I don't know how to find deltaS though in order to get two equations with 2 unknowns.
washburn 6:28 PM
What is deltaS for any and all cycles ? Zero.
Your 2nd Law has a problem because Q2 is transferred OUT of the cycle.
Where did these equations even come from? I don't really understand what I am plugging into these equations and I only hope that the result is W/Qh = n
Katie 7:48 PM
The two equations are the 1st Law applied to the HE and the 2nd Law applied to the HE. The HE is a cycle, so deltaS = 0. The HE is reversible, so Sgen = 0. The HE receives and rejects heat reversibly to themal reservoirs. This means the T inside the cycle where the heat transfer occurs must be the same as the temperature of the reservoir with which heat is exchanged. Remember that reversible heat transfer only occurs between objects at the same temperature. Because the T of the system (HE) is constant during the heat transfer, the T pops out of the Integral{dQ/T} in the 2nd Law and becomes Q/T.
You are left with 2 equations (1st and 2nd Laws) in two unknowns: Q2 (to the reservoir at 1000 K) and Wcycle. Solve them simultaneously. Then, calculate the thermal efficiency of the cycle using the equation you provided in your comment.
Is it wrong to simply sove 0=DeltaS=Qh/Th-Q1/T1-Q2/T2 for Q2 and then plug that in to Eta=1-Qc/Qh where Qc= Q1+Q2 ???
I dont have any simultanious equations here, but i got the right answer...
nikfoo 8:37 PM
What you did is OK and correct, but you might understand it better if you used Eta = W / Qh and then used the 1st Law : Qh-Q1-Q2-W=0 to determine W. It all comes out the same, but my way you can see the two equations I am talking about and you don't have to make the small leap of faith that Qc= Q1+Q2.
Good work.
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