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I Isentropic not adiabatic

  1. Jan 14, 2016 #1
    Why do we always see the claim that an isentropic process for a system is adiabatic and reversible? The change in entropy for a process is the sum of the entropy transfer accompanying heat and the entropy production. The entropy production term is always at least zero, and the transfer term follows the direction of heat flow. Thus it seems reasonable that we could have an irreversible process where heat is transferred from the system (an entropy decrease for the system) in the right proportion to match entropy produced within the system.

    Thus the two processes would cancel and the entire process would be isentropic and yet not adiabatic.
     
  2. jcsd
  3. Jan 14, 2016 #2
    Please provide a single example of an irreverisble adiabatic process that is isentropic. Or, equivalently, provide a single example of any isentropic change between two thermodynamic equilibrium states, such that you can identify an adiabatic irreversible path between the exact same pair of thermodynamic equilibrium states (i.e., no heat transfer allowed).
     
  4. Jan 14, 2016 #3
    Hi ChesterMiller,

    Why would I need to provide an example when I am asking a theoretical question?
    Can you instead point out the error in my reasoning? For a certain system undergoing a process
    from state 1 to state 2 the change in entropy is:

    S2-S1 = Q/T + Sp

    Sp is the always positive (or zero) entropy production term.
    Q/T is the entropy transfer term. The sign follows the direction of heat transfer.

    Now for no change in entropy, S2-S1=0 so we are left with:

    Q/T = -Sp

    If the process transfers heat from the system it is not adiabatic.
    Why is the final equation impossible? In other words, why can the entropy production term not precisely balance the entropy transfer term?

    Thanks.
     
  5. Jan 14, 2016 #4
    If you carry out any adiabatic irreversible process, the change in entropy will be positive, and equal to Sg. If you try to carry out a reversible process between the same two thermodynamic equilibrium states, you won't be able to do it without transferring heat, since $$ΔS=\int{\frac{dq_{rev}}{T}}=S_g$$.
     
  6. Jan 14, 2016 #5
    "If you carry out any adiabatic irreversible process, the change in entropy will be positive, and equal to Sg."

    True. I agree with that if by Sg you mean what I called Sp, the entropy production term..

    "If you try to carry out a reversible process between the same two thermodynamic equilibrium states, you won't be able to do it without transferring heat"

    True, again. I agree with this statement too.

    But this doesn't address my question in any way that I can see.
    You addressed both adiabatic irreversible and nonadiabatic reversible.
    I am wondering about an irreversible and nonadiabatic process as I tried to show in the equations.

    Thanks again, ChesterMiller

    Anyone else have any ideas where my reasoning has gone wrong?
     
  7. Jan 14, 2016 #6
    Yes, you're right. I was confused. The question is "Can you have an non-adiabatic irreversible process that gives you zero change in entropy?" Is that a correct interpretation of your question?

    Chet
     
  8. Jan 14, 2016 #7
    "Can you have an non-adiabatic irreversible process that gives you zero change in entropy?"

    Yes, that is it. I always hear that an isentropic process is one that is both adiabatic and reversible. But if the change in entropy for a system undergoing a process is given by the sum of two terms (entropy transfer and entropy production), I don't see why they cannot cancel instead of both being zero. The only limit is that the production term cannot be the negative term.
     
  9. Jan 14, 2016 #8
    The answer to this question is YES. An isentropic process is certainly not exclusively something that is adiabatic and reversible, but an adiabatic reversible process is certainly isentropic. So adiabatic reversible processes are a subset of isentropic processes.

    Your analysis of the situation is absolutely correct. Tomorrow, I'll be back with a specific (simple) example that illustrates your conjecture. Sorry for my confusion up to now.

    Chet
     
  10. Jan 14, 2016 #9
    Thank you. I look forward to an example. :woot:
     
  11. Jan 15, 2016 #10
    I have 3 identical solid cubes of material (A, B, and C), each of mass m and heat capacity C. Cubes A, B, and C are initially at temperatures 100 C, 50 C, and 25 C, respectively. I carry out a two step process:

    Step 1:
    A and B are brought into thermal contact at one of their faces (full face) and allowed to irreversibly equilibrate. Their final temperature will be 75 C, and the temperature change of each will be 25 C. The amount of heat transferred from A to B will be 25mC. During this transient heat conduction operation, the temperature at the interface between A and B will jump initially to the average temperature of 75 C and stay at that value throughout the equilibration, while the other parts of cube A further from the interface drop more gradually to 75 C, and the other parts of cube B further from the interface rise more gradually to 75 C. So all the heat transfer between cubes A and B takes place at 75 C.

    Step 2:
    B and C are brought into thermal contact at one of their faces (full face) and allowed to irreversibly equilibrate. Their final temperature will be 50 C (cube B starts this step at 75 C), and the temperature change of each will be 25 C. The amount of heat transferred from B to C will be 25mC. During this transient heat conduction operation, the temperature at the interface between B and C will jump initially to the average temperature of 50 C and stay at that value throughout the equilibration, while the other parts of cube B further from the interface drop more gradually to 50 C, and the other parts of cube C further from the interface rise more gradually to 50 C. So all the heat transfer between cubes B and C takes place at 50 C.

    At the end of this two-step process, cube A is at 75 C, and cubes B and C are at 50 C.

    We are going to look at the changes in entropy of all three cubes, the amount of entropy entering each cube, the amount of entropy leaving each cube, and the amount of entropy generated within each cube.

    Cube A:
    Entropy entering = 0

    Entropy leaving = ##\frac{25mC}{(273+75)}=0.07184mC##

    ##\Delta S_A=mC\ln(348/373)=-0.0.06938mC##

    Entropy generated in cube A = ##G_A= -0.06938mC+0.07184mC=+0.00246mC##

    Cube B:
    Entropy entering = ##\frac{25mC}{(273+75)}=0.07184mC##

    Entropy leaving = ##\frac{25mC}{(273+50)}=0.07740mC##

    ##\Delta S_B = 0##

    Entropy generated in cube B = ##G_B = 0-0.07184mC+0.07740=+0.00556mC##

    Cube C:
    Entropy entering = ##\frac{25mC}{(273+50)}=0.07740mC##

    Entropy leaving = ##0##

    ##\Delta S_C = mC\ln\frac{323}{298}=0.0856mC##

    Entropy generated in cube C = ##G_C = 0.0856-0.07740=+0.00820mC##

    Note that there is entropy generation in all three cubes as a result of this irreversible process. The sum of the three entropy generations is equal to the sum of the three entropy changes. Since the three blocks constitute an isolated system, the entropy change for the combination is positive. Most importantly, if we regard block B as our system (and blocks A and C as its surroundings), even though it has undergone an irreversible non-adiabatic process (and entropy has been generated within this cube), its change in entropy is zero.

    Chet
     
    Last edited: Jan 15, 2016
  12. Jan 15, 2016 #11

    DrDu

    User Avatar
    Science Advisor

    Maybe the simplest example is that of stirring a viscous solution in a heat bath, so that its temperature is hold constant (its volume shall also be fixed). The work done on the system will be converted into heat passed to the heat bath, but as T and V of the system are constant S(T,V)=const. In fact, this corresponds to the experiment of Lord Rumford, who first showed the equivalence of work and heat (He used a drill and canon as a system, but the principle is the same): http://www4.ncsu.edu/~kimler/hi322/Rumford-boring.html
     
  13. Jan 15, 2016 #12
    Good one.
     
  14. Jan 15, 2016 #13
    spamanon:

    I don't know who is teaching you this stuff in this way, but I really like the approach. It is very intuitive and appealing. Is there a textbook that goes along with this?

    Chet
     
  15. Jan 16, 2016 #14
    Hello,

    The textbook I first learned from is Engineering Thermodynamics by Moran and Shapiro. It is a standard text for mechanical engineering and has been for a long time.
    Thanks for the examples, everyone. I do not know why it is so common to say an isentropic process is only adiabatic and reversible when there are clearly counterexamples.


    Here is the book: Engineering Thermodynamics
     
  16. May 30, 2016 #15
    Can you please elaborate how this is related to OP question
     
  17. May 30, 2016 #16

    DrDu

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    Science Advisor

    The OP was asking for a process which is isentropic but not adiabatic. I think my example is maybe one of the simplest realisations of such a process.
     
  18. May 30, 2016 #17
    I understood that it is not adiabatic and reversible also. But how it became isentropic.
     
  19. May 30, 2016 #18
    The entropy generated within the liquid is transferred to the reservoir, so the entropy of the liquid does not change. Entropy is a physical property of the liquid. In the final state, its temperature is equal to its initial temperature. So the initial and final states of the liquid are the same, and there is no change in its entropy. However, the entropy of the reservoir has increased, and the entropy of the combination of liquid and reservoir has increased.
     
  20. May 30, 2016 #19
    Ok got it sir
     
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