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Carnot Cycle and Reversibility

  1. Jul 16, 2012 #1
    I'm trying to resolve some of my conceptual sticking points with the Carnot cycle.

    For one thing, I'm reading Reiss's Methods of Thermodynamics, and he offers a proof that the Carnot cycle obtains maximum efficiency when conducted reversibly.

    How can the cycle be conducted reversibly? During the two isothermal expansions, heat is exchanged and temperature is nonzero. Therefore there is is a finite nonzero change in entropy. I was under the impression that processes for which ΔS≠0 are irreversible. Is this correct?

    Secondly, since an adiabatic process has no heat exchange by definition, what is the difference between an adiabatic and a reversible process (or is there none)?

    Thirdly, there are two isothermal legs of the Carnot cycle, one wherein the engine performs work on the environment, and one for the opposite. Hence dS must be positive in one direction, and negative in the other. Since dS [itex]\geq[/itex] 0 by the second law, am I right in thinking there must be some associated dS of the environment which balances things here? Otherwise, I do not see how one can have a process for which dS < 0.
  2. jcsd
  3. Jul 16, 2012 #2
    Good question, DreadyPhysics! Here's the catch with reversibility: a process can be reversible (ΔS = 0) so long as the process takes infinitely long to complete. We can see this most simply in our equation ∂S = ∂Q/T. If ∂Q → 0 at every point in the cycle, ∂S will also be zero. (This is why no process can truly be reversible: we can't reach absolute zero and we can't perform a process infinitely slowly, and even if we could, it would do us much good, would it?)

    In the case of the isothermal process, as you have already pointed out, there is steady heat transfer, but if that transfer is infinitesimal (∂Q → 0) then reversibility is saved.

    For the adiabatic case, let's look at the fundamental thermodynamic relation:
    dU = TdS - PdV
    In this case, dU = 0, so TdS = PdV. For P =/= 0, We achieve dS = 0 ONLY if dV → 0, which implies infinitely slow expansion or compression, much like the isothermal case. In fact, we can also use this relation to examine our isothermal case in the same way (it might be a better way to look at it, actually: I think the above formula is more general, I'm not entirely sure.)
  4. Jul 17, 2012 #3
    1) The heat is exchanged across two bodies that are at the same temperature. Thus the entropy increase in the receiving body is equal to the entropy decrease in the donor body.

    2) Adiabatic is reversible ONLY IF it is an "ideal" process which again means very slowly such that the system is always in equilibrium.

    3) Yes, see 1)
  5. Jul 17, 2012 #4
    Entropy is not created in a reversible cycle. However, entropy can be moved in any cycle, whether it is reversible or irreversible. Entropy is an extensive property. In any volume of space, there is a certain amount of entropy. Therefore, entropy always has a location associated with it.
    In the Carnot cycle, entropy is moved from the high temperature reservoir to the low temperature reservoir. No entropy is created in the ideal Carnot cycle. When the amount of entropy per cycle is calculated, the entropy that is determined is the entropy that moves.
    At the end of the cycle, the entropy that was originally at the high temperature reservoir is not in the low temperature reservoir. Work was done by the engine. However, no entropy was created. If work is done on the entropy, the same amount of entropy can move from the low temperature reservoir to the high temperature reservoir.
    Entropy can be thought of as an indestructible fluid. Entropy can be moved or it can be created, but it can never be destroyed. Reversible processes move entropy without creating it. Irreversible processes create entropy. The temperature is analogous to the pressure of this fluid called entropy.
    I find the difficulty is recognizing when entropy is created rather than moved. Frictional forces create entropy. So do many chemical reactions. However, there are many systems where the creation of entropy is negligible.
  6. Jul 17, 2012 #5
    Thanks guys, that actually helps a great deal.

    I'm still unclear on the distinction between adiabatic and reversible, though. It would seem to me that all adiabatic processes have to be reversible (if done slowly, I guess) in the sense that Δq = 0 by definition, so ΔS=0 by definition, and as long as ΔS=0 it should be reversible, right?

    But I know that there is some distinction, I'm just trying to understand it.
  7. Jul 17, 2012 #6
    You must have a deeper understanding of what entropy is in order to understand that fully. A good start is to look at Feynman's explanation, it's probably the best for 1st time learners (I wish I learned it that way at first, it would have saved me a lot of time and headache)
  8. Jul 17, 2012 #7
    An adiabatic process may or may not be reversible. Here is the way I think of it.
    Entropy can either be moved or created.
    The two processes are different. Moving entropy and creating entropy are two entirely different things. Frictional forces can create entropy. Heat conduction and convection can move entropy.
    1) The definition of adiabatic is that the entropy can't move. Entropy can be created in an adiabatic process.
    2) The definition of reversible is that entropy can't be created. Entropy can still move in a reversible process.

    Here is how an adiabatic process can also be irreversible. Look at a piston with a cylinder that is a perfect insulator of heat. Entropy can't move from the piston.
    If the piston can slide in the cylinder without friction, entropy won't be created. In that case, the adiabatic process is also reversible. The absence of friction is one of the hypotheses of the Carnot cycle.
    Friction may slow down or stop the piston from sliding in the cylinder. The frictional force of the cylinder on will create entropy. Therefore, the presence of friction makes the adiabatic process irreversible.
    The main difficulty in understanding thermodynamics is distinguishing between physical processes that move entropy, and physical processes that create entropy. The equations for the two processes may look very similar. However, the application of these equations are very different.
    Some textbooks shows a calculation of a change in entropy during the Carnot cycle where ΔS>0. This quantity is the entropy that is moved from the hot reservoir to the cold reservoir. Other textbooks makes a statement that in the Carnot cycle ΔS=0. This is the amount of entropy that was created.
    These are two different types of ΔS. When reading about any heat engine, focus on distinguishing between entropy transport and entropy generation.
    Just as a historical note: Sadi Carnot was the one who came up with the Carnot cycle. However, he learned about friction from his daddy, Lazare Carnot. Lazare did an in depth study of friction that he was very proud of.
  9. Jul 17, 2012 #8
    I get that heat won't leave or enter the piston, thanks to its adiabatic walls. But will friction within the system generate heat (as it generates entropy?) I suppose it must- but then Δq must refer to heat entering or leaving the system; a heat flux if you will (as ΔS would be an entropy flux.)

    Is this right? I feel I'm starting to understand...
  10. Jul 18, 2012 #9
    You're using the relation ΔS = Δq/T to prove the adiabatic system MUST be reversible, but remember that that relation only works for reversible processes, so what you're saying is "it is reversible if it is reversible."

    Instead look at the fundamental thermodynamic relation:
    dU = TdS - PdV

    It applies to irreversible processes.

    You are correct that if done slowly enough, the adiabatic process MUST be reversible, and you obtain the first relation we had for entropy, but you can also show that ANY process done slowly enough will be reversible. The adiabatic process is not unique in this regard, so don't get too hung up about it. Remember: in reality, no process is reversible.
  11. Jul 18, 2012 #10
    That is correct. If there was friction in the walls, entropy would be created. However, in the ideal case there is no friction. So entropy is not created.
    The Carnot cycle is a hypothetical ideal engine. Therefore, there is no friction on the walls at anytime in the cycle. Therefore, entropy is never created in the Carnot cycle. If someone says ΔS>0 in the Carnot cycle, then he is referring to the entropy that is transferred.

    It would be useful to know what happens to the entropy at each step in the Carnot cycle. A certain quantity of entropy, ΔS, goes through the following steps:
    1) During the first isothermal process, ΔS is transferred from the hot reservoir to the chamber while the chamber expands.
    2) During the first adiabatic process, the amount of entropy in the chamber is unchanged while the chamber expands.
    3) During the second isothermal process, ΔS is transferred from the chamber to the cold reservoir while the chamber shrinks.
    4) During the second adiabatic process, the amount of entropy in the chamber is unchanged while the chamber shrinks.

    It would be useful to review what happens to the temperature in the chamber at each step of the Carnot cycle. In the chamber, ...
    1) the temperature doesn't change.
    2) the temperature decreases.
    3) the temperature doesn't change,
    4) the temperature increases.

    That is basically right. ΔS is proportional to the energy flux.
    I said proportional since the units are a bit wrong to be a flux. The word "flux" has a formal definition. Flux is usually defined in terms of rate of transfer versus area. So technically the entropy flux would be determined by,
    where F_S is the entropy flux, Δt is an increment in time, and ΔA is an increment of area normal to to the direction of the flux.
    In any case, you have the idea. The flux is "carrying" a certain amount of entropy, ΔS.
    In the case of an irreversible engine, there would be a second source of entropy that was created. However, entropy can never be destroyed according to the second law of thermodynamics.
    I think you understand.
  12. Jul 18, 2012 #11
    Point taken, "flux" was a poor choice of words. A further point in the thread that the second law only applies to reversible processes is also an important distinction.

    This thread has been very helpful, and I might return with more questions shortly.
  13. Jul 23, 2012 #12
    The second law applies to all processes. It doesn't matter if the process is reversible or irreversible.

    The second law states that entropy can never be destroyed by any process.
    A reversible process is a process where entropy is not created.
    An irreversible process is one where entropy is created.

    A consequence of the second law is that entropy can never spontaneously move from a low temperature to a high temperature by any process.
    An adiabatic process is a process where entropy does not move in either direction.
    An isothermal process is a process where entropy moves from one system to another system that has the same temperature.
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