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Postby Starside » Mon Dec 17, 2007 9:33 pm UTC

I am taking a calc based physics course, which is going well but thermodynamics is giving me a bit of a headache. I believe I understand a lot of it, including how to find the change in entropy of stuff, including in cyclic engines (which is 0). There is one inequality that was used a lot in recitation that we used to determine if a process, usually an engine/refrigeration cycle is possible. It is ds >= sum(Q/T). I am not sure why this is true. In some examples, which I cannot remember the exact numbers sum(Q/T) is negative and still possible. Which works for the inequality, I just cannot find a derivation in my text or lecture notes on why it is true, and it bugs the crap out of me when I do not know why some law is true. Does anyone know why ds >= sum(Q/T) is true? Or alternatly, ds >= Integrate(dQ/T) if you like more general stuff.

Part of my confusion I think also comes from the similarity to the equation ds=sum(Q/T). Like in a carnot engine where everything is isothermal and adiabatic, that should calculate ds, which should be 0 if the engine is possible and cyclic. But in some cases I believe I came up with negative values, which according to the inequality allow for a possible engine, but that makes no sense to me because ds has to be 0 for a cyclic process, and the inequality is identical to the equation in this case.

So can anyone straighten me out?

(This is not a homework problem, but I am trying to understand thermo for the final in 2 days)

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Re: Entropy

Postby Svalbard » Tue Dec 18, 2007 5:16 am UTC

Would you mind giving a quick example of what types of problems you're working with? I took thermodynamics a year ago, but if we ever wanted entropy values during cycle analysis we just looked everything up in a table in the back of the book and kept going.

Sorry for being of little help here :|
ne porvivajo nur mortigi tempo

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Re: Entropy

Postby iop » Tue Dec 18, 2007 3:17 pm UTC

Starside wrote:Does anyone know why ds >= sum(Q/T) is true?

Boltzmann knew it, but then he killed himself. Seriously, the "why" in this case is really complicated. I unfortunately don't understand it well enough to explain it.

The practical observation that goes with it is that whatever you do, there is friction, so some part of the energy you put into your process will be lost as heat, so Q ends up being positive.

In a closed system, the best you can get for ds is 0, but in reality, you don't even get that (the equal sign is only for ideal processes, and it is there to mock the engineers). In an open system you can get anything, of course, because you can neglect all positive Q.

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Re: Entropy

Postby TemperedMartensite » Wed Dec 19, 2007 3:38 am UTC

Cool I have a final in thermodynamics (for engineers so the >= is just there for relative efficiency as the above poster mentioned) tomorrow.

Anyways, here is the lowdown as I see it. The inequality, at least in my course, reads Sgeneration>=0. Perhaps viewing the inequality of Sgen will shed more light on this. Since Sgen = delta(Ssystem) + delta(Ssurroundings) >= 0 as I have it defined in my notes. So for a cycle, if your (cyclic intregal)ds = (cyclic integral)dQ/T(K), you are referring only to the entropy from heat transferred to say...the surroundings. In general, if you assume that your system is steady state (even flow etc) and causing delta(S) = 0, you have to assume that the entropy generation is equal to the negative of (cyclic integral)dQ/T(K).

Basically through that equation you are saying that the entropy generated is equal to the entropy leaving if you have a negative vector.

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