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Thermodynamics, engine cycles , entropy, concepts.

  1. Jan 11, 2014 #1
    Okay, I am considering a cycle, where the working fluid is an ideal gas, with heat capacities Cv and Cp, the cycle consists of: isochoric increase in volume, adiabatic expansion back to initial pressure and a isobaric compression back to initial conditions.


    Questions:
    -
    q1) I am asked to draw an T-S diagram. The diagram indicates the entropy at the end returning to its initial entropy. I am not sure why this is:
    Thoughts
    - for there to be no change in entropy, a process must be reversible. I can see that we have returned to the initial conditions and I know this is the definition of reversible.
    - But am I then correct to say that, whilst the individual processes are not reversible, the cycle, considered as a whole, is reversible?*

    q2) For the adiabatic reaction, it is drawn at constant entropy. I know this equality only holds for a reversible process, and increases if not. I'm struggling to see how you justify that this is a reversible adiabatic reaction?

    q3) The expression for the efficiency includes ΔQ. (whilst I am able to attain the correct value by plugging in values attained for Qh and Ql - heat input and output respectively -(more information is specified than I have said at the start of the post), I am not able to justify why the following is flawed:
    - Any change in heat comes from the isocharic and isobaric processes.
    - We must have that ΔSp + ΔSv = 0 (for the cycle to be constant entropy consistent with the diagram), where Sp is the entropy of the isobaric process and Sv the isocharic
    - I then know that [itex]\int[/itex]dS=[itex]\int[/itex]dQ/T** for a reversible process only
    - Whilst I know that these individual processes are not reversible , st ** can not be applied, surely they can be considered reversible together? ( consistent with the diagram)
    - So could you not apply **to the processes together? (which yeilds efficiency=0 as ΔS=0 => ΔQ=0...complete nonsense)

    Many thanks to anyone who can help shed some light on this, greatly appreciated ! :)
     
  2. jcsd
  3. Jan 11, 2014 #2
    Doesn't isochoric mean at constant volume? How can you increase the volume at constant volume?
     
  4. Jan 12, 2014 #3
    Sorry. An isochoric increase in pressure
     
  5. Jan 12, 2014 #4

    rude man

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    A contradiction in terms.

    NM.
     
  6. Jan 12, 2014 #5
    post 3?
     
  7. Jan 12, 2014 #6

    Andrew Mason

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    The system returns to its initial state at the end of the cycle. Since entropy is a state variable, there can be no change in entropy of the system regardless of the process that it underwent.

    The process is reversible only if the total change in entropy (system + surroundings) is zero.

    AM
     
  8. Jan 13, 2014 #7
    Andrew Mason already addressed part of this question. But, in addition, who says that the individual segments of the process are not reversible? I think they meant to say in the problem statement that all the process segments are reversible. Then, certainly, the adiabatic segment would be at constant entropy for the system.

    Why are you saying that the adiabatic expansion can't be reversible? It most certainly can.
     
  9. Jan 13, 2014 #8
    Oh thanks for that. I did not realise it was the total entropy. So am I correct in thinking that whilst , if a process is reversible, it must follow that the total entropy change is 0, but that if the entropy change is 0 it is not at all necessarily true that the process is reversible? I.e that the contrapositive of the first statement does not hold?

    I am also wondering how it follows, that we are referring to the total entropy and not just the entropy of the system, from the derivation of the inequality ds=[itex]\int[/itex] [itex]\frac{dQ(rev)}{T}≤[/itex][itex]\int[/itex][itex]\frac{dQ}{T}[/itex], where dQrev is the change in heat for a reversible process.

    A standard textbook derivation is to consider a irreversible process - from state A to B- followed by a reversible process from B to A.
    It then uses clausius inequality that for any closed cycle: [itex]\oint[/itex][itex]\frac{dQ}{T}[/itex]≤0 , with equality holding for a reversible cycle,to deduce that:

    [itex]^{B}_{A}[/itex][itex]\int[/itex][itex]\frac{dQ}{T}[/itex]≤[itex]^{B}_{A}[/itex][itex]\int[/itex][itex]\frac{dQ(rev)}{T}[/itex]

    ∴ dS = [itex]\frac{dQ(rev)}{T}[/itex]≥[itex]\frac{dQ}{T}[/itex]

    - So my question is, how does it follow that dS is the total entropy , of both the surroundings and the systems?
    Does this mean that dQ, is the total heat change, of both the surroundings and the system in Clausius Ineqaulity?
     
  10. Jan 13, 2014 #9
    I have not said that it can't be.My point is that, I can see that it seems natural for it to be reversible, but that nothing in the question seems to allow for us to make this assumption. My issue was, that I am struggling to find a way to fully justify/ prove that the process is reversible, and equally as much, as implied by the question, that is is irreversible.

    Am I correct in thinking that there is not enough information in the question to make such a deduction. e.g , whilst we know ΔSp, for the system, we do not have any information to calculate the entropy change of the surroundings ? (If we did, we could simply check whether the two were equal?)
     
  11. Jan 13, 2014 #10
    You are correct in thinking that there is nothing in the problem statement to indicate that the process is reversible. However, if their answer sheet showed the entropy did not change along the adiabatic segment, it would appear that they meant for you to assume that the process was carried out reversibly. They just forgot to mention this in the problem statement. This caused you to expend much valuable time putting in lots of unnecessary work and thinking about the problem and agonizing over it. That's very unfortunate. But, your deliberations seem to indicate that you really know what you are doing. Just remember that, even if a cyclic process is irreversible, the changes in all the thermodynamic functions for the system over a cycle must be zero. This automatically means that the entropy of the surroundings must have increased, although we don't usually know how much.

    Chet
     
  12. Jan 14, 2014 #11

    Andrew Mason

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    The Clausius inequality deals with the system only. Essentially it says that since the system undergoing an irreversible process has no change in entropy after one cycle, the increase in entropy of the surroundings must be because more entropy left the system than entered it (so the total < 0). Mathematically:

    [tex]\Delta S_{tot} = \Delta S_{sys} + \Delta S_{surr} = \oint \frac{dQ_{rev-sys}}{T} + \oint \frac{dQ_{rev-surr}}{T} > 0[/tex]

    Since [itex]\oint \frac{dQ_{rev-sys}}{T} = 0[/itex] this means that [itex]\oint \frac{dQ_{rev-surr}}{T} > 0[/itex]

    But since the surroundings remain at the same temperature during the process, the reversible heat flow vis a vis the surroudings is just the actual heat flow to/from the surroundings, which is the negative of the heat flow to/from the system. So: [itex]\Delta S_{surr} = \oint \frac{dQ_{rev-surr}}{T} = -\oint dQ/T > 0[/itex]

    This last statement is equivalent to:

    [tex]\oint dQ/T < 0[/tex] where dQ is the heat flow to/from the system.

    AM
     
  13. Jan 14, 2014 #12
    Which processes is it that the temperature of surroundings remains constant? Is it for all of them? Sorry I think the issue is that I am not 100% sure how we define the surroundings and the system - are the reservoirs part of the surroundings and not the system?
     
  14. Jan 14, 2014 #13

    rude man

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    The system is the gas, for example. The surroundings (aka 'universe') includes everything outside the gas, so reservoirs are part of the surroundings.

    A reservoir by definition does not experience a change in temperature, no matter how much heat goes into or out of it. It therefore has a simple (reversible process) entropy change = Q/T where Q is the heat in (+) or out (-) and T is its constant temperature.
     
  15. Jan 14, 2014 #14

    Andrew Mason

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    There are probably many different nuances to the meaning of a thermodynamic system, but the distinction is made between the system and the surroundings. This may not be a perfect explanation, but...: The system is a defined, bounded region in which a cyclical thermodynamic process occurs that results in heat flow. The surroundings may be in thermal contact during all or part of the thermodynamic process that occurs in the system and the heat flows in one direction from one part of the surroundings to another as a result of the process. Generally, either a flow of heat is being used to do work or work is being done to make heat flow from one part of the surroundings to another.

    AM
     
  16. Jan 15, 2014 #15
    We specify what we choose to regard as the system. The surroundings are everything else. As far as the system is concerned, all exchange of heat and work takes place at the interface between the system and surroundings. We can specify what is happening to the system at this interface, but we often have less control (and/or concern) over what is happening throughout the surroundings. For more perspective on this, check out my Blog at my PF personal web page.https://www.physicsforums.com/member.php?u=345636

    Chet
     
  17. Jan 16, 2014 #16
    Is this considering the cycle from beginning to end - the reversible process entropy change? Whilst we know P,V,T , in a cycle, do not change for the system, how do we know that P,V do not change for the reservoirs?


    Surely it's entropy change for intermediate processes- isobaric and isocharic will not be given by Q/T?
     
  18. Jan 16, 2014 #17

    rude man

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    P and V are not considerations for a temperature reservoir. You're concerned with P and V for the systrem only.
    Yes it will. Always. For the reservoirs only, of course. But for the system, T is not constant so you have to integrate dQ/T to compute entropy changes along reversible paths.

    Entropy change is zero for any reversible adiabatic path.
     
    Last edited: Jan 16, 2014
  19. Jan 16, 2014 #18
    When we talk about an ideal constant temperature reservoir in thermo, what we we mean is a bath with a very high capacity to absorb or release heat without its temperature changing. We can approximate such a bath in practice by just having a huge amount of liquid in the reservoir, or by having a smaller reservoir in which a phase change is occurring, like, for example, an ice bath.

    Chet
     
  20. Jan 17, 2014 #19
    Oh okay, I was going by the fact that entropy is a state variable, so for there to be no change in entropy the state of the body must be unchanged. So the parameters we use to describe the state can change? I've only seen P,V,T and used, but for a reservoir we only have to use T?
     
  21. Jan 17, 2014 #20
    The nice thing about splitting everything into "system" and "surroundings" is that our focus can be placed on the system, and we don't need to concern ourselves with the thermodynamic states of the various parts of the surroundings. A reservoir is just a part of the surroundings that is used to transfer heat to the system at a constant interface temperature. So the P and V of the reservoir are not relevant to us if all the reservoir does to the system is transfer heat.

    Incidentally, "no change in the state of the body" is a sufficient (but not necessary) condition for the entropy of the body to be unchanged. The entropy can be unchanged between many different states of the body. Consider, for example, a single reversible adiabatic path. Or consider a path that includes two reversible isothermal segments (with an adiabatic reversible segment in between) in which the sum of Q/T for the two isothermal segments is zero (i.e., a portion of a Carnot cycle).

    Chet
     
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