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Temprature-Entropy cyclic process

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data
    There is a temperature-Entropy graph (T-S) (attachment),which illustrates a hypothetical cyclic process.
    a) Calculate the heat input or output along each of the paths.

    b) Find an expression for the efficiency η of the complete cycle in terms of T1 and T2 only.

    2. Relevant equations

    for b ,η=(Q1/Q2)-1=(T1/T2 )-1,HEAT INPUT: DH=DQ (ONLY in case of isobaric change)

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. May 6, 2010 #2


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    [PLAIN]http://img202.imageshack.us/img202/3227/temperatureentropygragh.jpg [Broken]
    I suppose the most complicated path is the linear process II-III (see attached image) in which both T and S change:
    [tex]T(S)= \alpha S + T_0[/tex]
    where [tex]\alpha, T_0 >0[/tex] are some constants.
    So, in this process we have
    [tex]\frac{dS}{dT} = \frac{1}{\alpha}[/tex]
    and hence
    [tex]\delta Q_{II-III} = TdS = \frac{1}{\alpha} T dT[/tex]

    When T changes from [tex]T_1[/tex] to [tex]T_2[/tex] you have
    [tex]Q_{II-III} = \frac{1}{\alpha}\int_{T_1}^{T_2} TdT = -\frac{1}{2\alpha} \left(T_1^2 - T_2^2 \right)[/tex]
    That is, the heat is extracted in this process.
    Now for the process III-I where the entropy is constant it is trivial that [tex]Q_{III-I}=0[/tex]. It is in fact an adiabatic process.

    Try to find the heat for isothermal process I-II. Note that the change of the entropy
    [tex]S_{II} - S_{I}[/tex]
    you can express in terms of [tex]T_2[/tex], [tex]T_1[/tex] and [tex]\alpha[/tex] just from the following equations for III and II points on the graph:
    [tex]T_2 = \alpha S_{I} +T_0[/tex]

    [tex]T_1 = \alpha S_{II} + T_0[/tex]
    Note also that [tex]S_{I}=S_{III}[/tex]
    Last edited by a moderator: May 4, 2017
  4. May 6, 2010 #3
    That's amazing.Thank you very much for both answers !
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