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I Energy change of reservoir in reversible weight process

  1. Sep 11, 2016 #1
    In what follows I refer to the ideas of "Thermodynamic: Foundations and Applications" by Gyftoploulos and Beretta. The abbreviated form of my question is: In a reversible weight process,
    Ω1R2R = E1 - E2 (see eqn. 6.18, p. 99) is transferred out of the composite of a system A and a reservoir R to the weight, where ΩR is the available energy of a system A with respect to reservoir R and E is the energy of system A.
    When the above equation is true, why does the book occasionally (problem 6.4, example 6.4 on p. 94) calculates the energy change of the reservoir as E1R - E2R ≠ 0 for the case of reversible weight processes?
    Because, according to the equation above, E1R = E2R must be true (see below). Also, (DE12R)revA should always be 0, p. 108/9. However, this quantity is used extensively in the discussion of entropy as if it was non-zero.



    My proof of E1R = E2R:
    Given two states, A1 and A2, of a system A and a reservoir R. Consider two reversible weight processes, α und β, consisting of two sub-processes, where A0R0 is the state where A and R are in mutual stable equilibrium. The only difference is the final state of the reservoir (R1 in process β instead of R2 in process α).

    α: A1R1 → A0R0 → A2R2
    β: A1R1 → A0R0 → A2R1

    The first sub-process in each process transfers energy of the amount of available energy to the weight: Ω1R. The second sub-process in α and the second sub-process in β both transfer the available energy Ω2R to the weight, because the stable equilibrium state is unique (see proofs on adiabatic availability on p. 76 and regard the adiabatic availability of A+R as the available energy von A) and Ω2R is independent of the initial (or end state in this case) of the reservoir (statement 3 on p. 89). The energy balances for both processes are:

    α: Ω1R2R = E1 - E2 + E1R - E2R
    β: Ω1R2R = E1 - E2

    Comparing both results, one gets that all states of the reservoir have the same energy: E1R = E2R
    So: Ω1R2R = E1 - E2
    The last equation corresponds to equation 6.18, p. 99.
     
  2. jcsd
  3. Sep 16, 2016 #2
    Maybe I should rephrase the problem above. The short version is:

    In a reversible weight process, in which system A goes from state A1 to A2 and reservoir R from R1 to R2, the energy transferred to the weight is
    (W12AR→)rev = E1 - E2 + E1R - E2R = Ω1R - Ω2R
    This is equation 6.5 in section 6.6.4.

    On the other hand, in section 6.6 the authors say:
    "[A] weight process for system A from A1 to A2 is [...] reversible only if"
    Ω1R - Ω2R = E1 - E2
    (eqn. 6.7) This would mean that the reservoir does not exchange energy, as E1R - E2R. However, in example 6.4 they get E1R - E2R = -50 kJ ≠ 0.

    Why do both equations contradict?
     
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