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Homework Help: Efficiency of a cycle in TS diagram

  1. Jun 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A working substance goes through a cycle within which the absolute temperature varies n-fold, and the shape of the cycle is
    66w082.jpg ,
    where T is the absolute temperature, and S the entropy. Find the efficiency cycle.

    2. Relevant equations

    [tex]\frac{\delta Q_R}{T}=dS[/tex], [tex]\eta=1-\frac{|Q_{out}|}{Q_{in}}[/tex]


    3. The attempt at a solution

    The total heat is [tex]Q=\int T dS[/tex], that is the area of the surface in the picture. I could just say: it's a triangle so I'll use the formula for the triangle surface:
    [tex]P=\frac{1}{2}ab[/tex].

    The [tex]Q_{in}[/tex] is easy to calculate:[tex]Q_{in}=T_0\int_{S_0}^{S_1}dS=T_0(S_1-S_0)[/tex].

    But how do I get the [tex]Q_{out}[/tex]? The temperature changes. I have found in solution (without explanation) that the answer is:

    [tex]Q_{out}=\frac{1}{2}(T_0+T_1)(S_1-S_0)[/tex], but why [tex]T_0+T_1[/tex]? and where does that 1/2 comes from? The triangle area formula? :\
     
  2. jcsd
  3. Jun 21, 2010 #2
    No, it's trapezium's area.
    But why don't you look at it another way? The "total heat" you calculated is actually work done A! And for a cycle: [tex]A = Q_{in} - |Q_{out}|[/tex]. So instead of doing an integration or wondering it's triangle or something else, just calculate a simple sum.
     
  4. Jun 21, 2010 #3

    Ummm it's triangle :D

    I don't get it how to calculate the sum when I don't have the work :\ [tex]A=W=pdV[/tex] right? 1-2 is isotherm, 2-3 is isenthalp (no Q) and 3-1 is sth I don't know :D I could say that its a line with slope k, which is not that hard to calculate, but I don't have the y-intercept (T in my graph)...
     
  5. Jun 21, 2010 #4
    Eek, if you rotate your diagram 90 degrees, you will see a trapezium.
    And again, the "total heat" you got is the work A. It's a complete cycle, so [tex]\Delta U = 0[/tex].
     
  6. Jun 21, 2010 #5
    Maybe you cannot see it from my drawing, but it is a triangle, it's such in the original form. There's just no way to make it into trapezium. But that aside, I still don't see how to calculate work XD
     
  7. Jun 21, 2010 #6
    The trapezium corresponds to [tex]Q_{out}[/tex]. The formula you got for [tex]Q_{out}[/tex] is the trapezium's area. See the area you need to calculate? Rotate it 90 degrees.

    For a complete cycle: [tex]\Delta U = Q_{total} - A = 0[/tex]. You see?
     
  8. Jun 21, 2010 #7
    I get it now, the area UNDER 3-1 line is [tex]Q_{out}[/tex]! And that is trapezium. I didn't understand that XD Thnx anyhow :D
     
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