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Is every reversible process coincident with a quasi-static process?

  1. Sep 30, 2014 #1
    For a closed system:
    If we define a quasi-static locus (process) as an ordered and dense succession of equilibrium states in the thermodynamic configuration space. Then we define a reversible process as one in which no entropy is generated. Then it is clear that there are some quasi-static loci which are not necessarily reversible simply because there is entropy generation along these loci. However, is it necessary that every reversible process has to be coincident with a quasi-static process? I think another statement for the same question, is it possible to have no entropy generation during a non-equilibrium process?
     
  2. jcsd
  3. Sep 30, 2014 #2
    This statement means that no entropy is generated for the combination of system and surroundings. In a reversible process, the entropy of the system and the entropy of the surroundings can each change. But, the sum of these changes must be zero.

    Chet
     
  4. Oct 1, 2014 #3
    Chet, sure I agree to what you mentioned. But I'm trying to limit the discussion to a closed system first. Any suggestions to related the question I posed?
     
  5. Oct 1, 2014 #4
    A closed system is one in which no mass enters or leaves; is that your understanding? Or are you referring to an isolated system?

    Chet
     
  6. Oct 1, 2014 #5
    Good point to clarify, the closed system I mean here is one which does not exchange neither energy nor matter with the surrounding. I guess this is what you called isolated.
     
  7. Oct 1, 2014 #6
    Yes. So now I need to go back to your original post and substitute the word isolated for the word closed in my thinking. It's bed time here (midnight), so I'll get back to it tomorrow.

    Chet
     
  8. Oct 2, 2014 #7
    Can you please give a specific example of this so that we have a point for discussion?
    For an isolated system, the answer to this is a definite no. For a non-equilibrium process (i.e., irreversible process) in an isolated system, the change in entropy must be positive.

    Chet
     
  9. Oct 3, 2014 #8
    That what I was guessing but wanted to make sure that is correct. Thank you!

    The example is the quasi-static heat transfer between two subsystems who can exchange heat between each other but otherwise are isolated. If T1=T2 the heat transfer between them is reversible. If T1<T2, the heat transfer between then is irreversible.
     
  10. Oct 3, 2014 #9
    You're welcome.
    Nucleus, I was hoping this was the example you had in mind. You're thinking is great.

    The term quasi-static refers exclusively to the work W, not to the heat Q. In this example, even though the process within the enclosure is quasi-static (actually no work gets done at all), the process is still not reversible, because of the reason that you gave. So the entropy within the enclosure increases when the hotter body is equilibrated thermally with the colder body. Knowing the initial temperatures of the bodies, we can calculate precisely what the entropy change is.

    It is also possible to carry out the process of letting the hotter and the colder bodies reach the same temperature within the enclosure reversibly, but we would have to introduce some additional equipment. One way would be to introduce a continuous sequence of constant temperature reservoirs, with temperatures varying between the temperatures T1 and T2. Of course, if we did that, the entropy increase of the bodies would exactly equal the temperature decrease of the reservoirs. That way, the total change in entropy within the enclosure would be zero. When the final state of the system of bodies and reservoirs was reached, it would then be possible (conceptually) to reverse the process and return both the bodies and the reservoirs to their original states.

    Chet
     
  11. Oct 6, 2014 #10
    Chet,

    I'm a little confused by your definition of quasi-static.

    From the definition that Useful nucleus gave in the first post, it would seem that the "contact with a succession of thermal reservoirs" case would be a quasi-static process whereas the "contact with finite temperature difference" case would not. (In the intermediate states of the latter, the systems are not in perpetual thermal equilibrium.)

    If we define a quasi-static process to be one such that the system progresses through equilibrium states, where is it implied that this classification refers exclusively to work?

    Nayan
     
  12. Oct 7, 2014 #11
    Hi Nayan,
    Thanks very much for your post. It is extremely helpful to our discussion.

    I must confess that, over the years, I have been somewhat confused over the term quasi-static process. That being said, I will also mention that this has never limited my ability to get correct answers to thermodynamics problems.

    As a result of your post, I have researched the definition of the term quasi-static process. I have found that such a process is simply one that is carried out extremely slowly. So the term quasi-static doesn't only apply exclusively to work (as I previously believed), but also to heat transfer and mass transfer.

    The definition that nucleus gave in his original post is also an incorrect definition of a quasistatic process. In particular, it is overly restrictive. A process that conforms to his definition is a Reversible Process. Every Reverislbe Process is a quasi-static process, but not every quasi-static process is a reversible process. Thus, reversible processes represent a subset of quasi-static processes.

    An example of a quasi-static process that is not reversible is the transfer of heat from a hot body to a cold body through a barrier of very low thermal conductivity. This process takes place very slowly (so that it is quasi-static), but it is irreversible. On the other hand, the process I described in my previous post involving contact with a continuous succession of isothermal reservoirs is both quasi-static and reversible. Once the bodies and reservoirs arrived at their final states, they could be returned to their original states without incurring any significant change in either of the bodies by executing the process in reverse.

    I hope this clears up some of the confusion.

    Chet
     
  13. Oct 8, 2014 #12

    DrDu

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    I also wouldn't call a succession of equilibrium states in state space a quasi-static process, probably not even a process but simply a path. It is quite important to realize that you can calculate the change of thermodynamic quantities like entropy by integrating along such a path without claiming that the path is really the limit of some thermodynamic process.
    That doing a process quasi-statically does not always help to make it reversible can be seen in the case of friction. The heat generated per unit of time is proportional to v. If you perform a process more slowly, the velocity decreases but the process will also take longer so that in the end, the amount of heat generated remains the same.
     
  14. Oct 8, 2014 #13
    Chet,

    Of course it hasn’t limited your ability to get correct answers - we’re only discussing definitions and not some behavior of the universe after all!

    These terms seem to be very ambiguous, even in standard thermodynamic texts. Smith, Van Ness, and Abbott (which I know you are fond of) draws no distinction and only defines reversible processes. Some authors even define a third type of process, a quasi-equilibrium process, which is more in line with Useful nucleus’s definition.

    Here is my present understanding:

    Quasi-static process - a process carried out infinitely slowly

    Quasi-equilibrium process - a process carried out such that the generalized force assigned to the surroundings is at most ever differentially apart from the generalized force assigned to the system, such that the system can be considered in perpetual thermodynamic equilibrium (all properties are spacially homogeneous) throughout the process

    Reversible process - a process which generates no entropy in the universe
    Equivalently, a process taking a system from state I to state II is said to be reversible if when the system is brought back to state I (by whatever means necessary) the surroundings ALSO return to the initial state.

    So now the question is: where do these processes overlap?

    The general consensus, I think, is:
    1) All quasi-equilibrium processes must also be quasi-static.
    2) All reversible processes must also be quasi-equilibrium (and thus also quasi-static).

    The difficulty now is going backwards.

    You showed in your previous response that quasi-equilibrium and/or quasi-static processes need not be reversible when there is dissipation. That is one viewpoint. I saw a paper online which claimed that the “infinitely slow” nature of a quasi-static process directly implies no dissipation via friction, viscosity, hysteresis, etc. So for example, the slower you carry out the expansion of a gas in a piston-cylinder system, the less dissipation you get due to friction - to the limit where as time tends to infinity, dissipation tends to zero. Based off of this, the low thermal conductivity heat transfer would also be reversible - but only truly so if the thermal conductivity were infinitely low. Again, I’ve seen no consensus.

    Nayan
     
    Last edited: Oct 8, 2014
  15. Oct 8, 2014 #14
    Hi Nayan,
    I've never hear the term quasi-equilibrium process before, but it sounds awfully like a reversible process.
    Yes this is true for mechanical energy dissipation, but there can also be irreversibilities involving heat transfer, mass transfer, and chemical reaction.

    As you correctly point out, the low thermal conductivity heat transfer situation I discussed in my previous post would definitely not be reversible, even if the thermal conductivity were very very slow. As you said, for a process to be reversible, "if when the system is brought back to state I (by whatever means necessary) the surroundings ALSO return to the initial state." This could not be accomplished for the heat transfer process I described. Of course, in the limit of infinitely low thermal conductivity, there would be no heat transfer at all, so there would be no change from the initial equilibrium state.

    Chet
     
  16. Oct 8, 2014 #15
    Thanks DrDu. Your point about the effect of friction is a good one. This is definitely the situation in the typical case where the frictional force is assumed constant. But, if the frictional force were, say, to be proportional to the deformation rate (i.e., the piston speed), then as the process were run more slowly, the heat generated by friction would decrease and approach zero as quasi-static operation were approached (and the total amount of heat generated would approach zero). This is very analogous to what happens with viscous dissipation, in which the viscous stresses are proportional to the deformation rate.

    So, as you correctly indicate, the case with a constant frictional force is an exception to the "rule" that, under quasistatic operation, the dissipation of mechanical energy is negligible; the existence of the constant frictional force clearly prevents the process from being reversible. But, other functional relationships between the frictional force and the velocity can lead to different results.

    Chet
     
  17. Oct 8, 2014 #16

    DrDu

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    Oh Chet, I think yo got me. You are right that the power due to usual friction terms goes as #v^2# and not as v.
     
  18. Oct 8, 2014 #17
    Hi DrDu,

    Please understand that I wasn't trying to "get" you. I hold you in very high regard.

    I was just trying to elaborate on what you said, and make the point that, if the frictional force is constant (an assumption commonly used in many thermo analyses), the power is proportional to v, but, if the frictional force is proportional to v, the power goes as v2.

    Chet
     
  19. Oct 8, 2014 #18

    DrDu

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    Ok, but nevertheless I am not sure whether there are physical processes where the frictional force is independent of v. Spontaneously, I can't think of any.
     
  20. Oct 11, 2014 #19

    Philip Wood

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    Dare I chip in with the remark that the tangential force between one solid surface sliding over another is said to be independent of relative velocity? This is, I believe, one of the so-called laws of friction between solids. The laws are not obeyed very accurately.
     
  21. Oct 12, 2014 #20

    DrDu

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    Oh, yes,Philip, in deed! I was distracted by thinking of viscuos friction.
     
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