
#1
Jan1813, 04:27 AM

P: 205

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)




#2
Jan1813, 06:07 AM

P: 5,462

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? 



#3
Jan1813, 04:45 PM

P: 205

I think I do, and I do unerstand the demostration of the inequality derived fron the 2nd law of thermodynamics




#4
Jan1913, 01:40 PM

P: 5,462

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 Q_{H} units of heat from H and rejects Q_{c} units to C, performing W_{A} units of external work per cycle. So the efficiency, e_{A}, = W_{A} / Q_{H}, by definition. If A, being reversible, is now reversed then an input of Q_{C} units of from C and W_{A} units will supply Q_{H} units to H. Now consider the proposal that B is more efficient than A and let B extract Q_{H} from H, performing W_{B} units of external work per cycle. So the efficiency, e_{B}, = W_{B} / Q_{H}, by definition. Then we have the condition that e_{B} > e_{A} [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 W_{B} > W_{A} This implies that we can drive the reversed A engine from B and also use the additional external (W_{B}  W_{A}) 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 e_{B} < e_{A} Does this help? 



#5
Jan2013, 06:34 AM

P: 205





#6
Jan2013, 06:57 AM

P: 5,462

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? 



#7
Jan2013, 12:07 PM

P: 205

I'm sorry if it is simple but I yust don't see it 



#8
Jan2013, 12:17 PM

P: 5,462

OK the enxt step is to interchange the functions of A and B.
This then leads to the conclusion that e_{A} ≤ e_{B} But we have already shown that e_{B} ≤ e_{A} The only way they can both be true is if e_{B} = e_{A} Thus all reversible engines have the same efficiency. Thus those engines which are not reversible have a lesser efficiency. 



#9
Jan2113, 02:43 AM

P: 205

ok, thanks




#10
Jan2113, 05:28 AM

P: 5,462

"Thus all reversible engines have the same efficiency, working between the same heat reservoirs (or temperatures)", but I'm sure you know that. 



#11
Jan2113, 06:22 AM

P: 205

That's right



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