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Conceptual thermodynamics question |
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| Jan18-13, 04:27 AM | #1 |
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Conceptual thermodynamics question
According to Carnot's theorem η≥η' where η corresponds to a reversible machine and η' is not necessarily reversible, so far so good, the problem is that it is ussually assumed that when η' is irreversible the inequality must hold however a could not find a rigorous demostration of this within the context of classical phenomenological thermodynamics, can anyone tell if such a demostration exists and where I can find it?(obs: η is the machine efficiency)
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| Jan18-13, 06:07 AM | #2 |
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The proof you are seeking depends upon the fact that the machine works in a cyclic process.
Do you understand this as it is most important? |
| Jan18-13, 04:45 PM | #3 |
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I think I do, and I do unerstand the demostration of the inequality derived fron the 2nd law of thermodynamics
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| Jan19-13, 01:40 PM | #4 |
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Conceptual thermodynamics question
So we are equipped to compare the efficiencies of heat engines.
Consider two heat engines, A and B, working between reservoirs H and C. Let A be a reversible engine so that it takes in QH units of heat from H and rejects Qc units to C, performing WA units of external work per cycle. So the efficiency, eA, = WA / QH, by definition. If A, being reversible, is now reversed then an input of QC units of from C and WA units will supply QH units to H. Now consider the proposal that B is more efficient than A and let B extract QH from H, performing WB units of external work per cycle. So the efficiency, eB, = WB / QH, by definition. Then we have the condition that eB > eA [itex]\left( {\frac{{{W_B}}}{{{Q_H}}}} \right) > \left( {\frac{{{W_A}}}{{{Q_H}}}} \right)[/itex] : note we are going to prove this condition false WB > WA This implies that we can drive the reversed A engine from B and also use the additional external (WB - WA) to perform some additional task. Since we are extracting (via B) and returning (via A) the same amount of heat from the hot reservoir H there will be zero net extraction of heat from H and so it is unecessary. This is contrary to the experience that forms the Second Law and is based on the original statement of it (by Carnot) that Two heat engines working together as a single self acting machine are unable to perform external work without a separately maintained heat reservoir. This proves that at best the efficiency of B cannot be greater than that of A. This argument can be developed further to show that equality holds for B also reversible, but for B irreversible eB < eA Does this help? |
| Jan20-13, 06:34 AM | #5 |
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| Jan20-13, 06:57 AM | #6 |
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Carnot did not propound a theorem. He put into words what was technical experience at the time. Ref: Carnot : Reflexions sur la Puissance motrice du Feu. The whole point is that there is no more fundamental law or axiom of physics to appeal to in order to prove the laws of thermodynamics. They just make sound common sense and have never been observed to be broken. Carnot's monograph was the first to offer a perpetual motion machine of the second kind and to disbar it. Such a machine does not contravene the First Law. Since you were enquiring deeply I thought, in my last line, that you might like to try to extend the line of reasoning to irreversible machines to deduce your inequality for yourself, before doing all the work for you. Edit : Hint What are the consequences of B being irreversible? ie what does that mean about the work that B generates? |
| Jan20-13, 12:07 PM | #7 |
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I'm sorry if it is simple but I yust don't see it |
| Jan20-13, 12:17 PM | #8 |
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OK the enxt step is to interchange the functions of A and B.
This then leads to the conclusion that eA ≤ eB But we have already shown that eB ≤ eA The only way they can both be true is if eB = eA Thus all reversible engines have the same efficiency. Thus those engines which are not reversible have a lesser efficiency. |
| Jan21-13, 02:43 AM | #9 |
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ok, thanks
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| Jan21-13, 05:28 AM | #10 |
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"Thus all reversible engines have the same efficiency, working between the same heat reservoirs (or temperatures)", but I'm sure you know that. |
| Jan21-13, 06:22 AM | #11 |
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That's right
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