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Calculating work heat, and efficiency given a TS diagram

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Is it possible to calculate the work done heat transfer, and efficiency of an object in a thermodynamic system, given the Temperature vs Entropy graph?

    Example:
    b7FPPj6.png


    2. Relevant equations

    ∆U=Q-W
    ∆S=dQ/T
    e=1-(Qc/Qh)
    e=W/Qh


    3. The attempt at a solution

    Since this is a T vs S diagram,i can understand how the product(area) will be heat.
    Since it's a cyclic process, I can also understand how Q=W , since ∆U=0
    Currently I'm thinking that the area inside the triangle will be W, which also corresponds to Q. However, I could barely find any documentation on this, which is why I wanted to confirm it.

    As for efficiency, we are given two equations:
    1-(Qc/Qh)
    and
    W/Qh
    Would both of them be correct here, and would Qh and Qc be T2 and T1 respectively?

    Thanks!
     
    Last edited: Nov 4, 2015
  2. jcsd
  3. Nov 4, 2015 #2

    SteamKing

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    Is a T-S diagram a plot of Temperature vs. Pressure?
     
  4. Nov 4, 2015 #3
    I'm so sorry, I meant entropy. I will edit it! :D
     
  5. Nov 4, 2015 #4

    rude man

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    Certainly, since these are both statements of the 1st law.
    For a Carnot cyccle, yes. A T-S diagram is valid for reversible processes only since entropy is defined for equilibrium states only.
    T-S diagrams are ideal for reading heats in a closed cycle:
    Qh = positive area under curve (going from left to right)
    Qc = negative area under curve (going from right to left)
    So Qh - Qc per cycle = area within the closed curve, as you correctly state, and = W also. And e = 1 - (Qc/Qh).
     
  6. Nov 4, 2015 #5
    Thanks for confirming!
     
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