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Heat Cycles, Work, and Efficiency

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    New Doc 13_1.jpg

    2. Relevant equations
    W=-∫PdV
    ε=W/Q
    PV=nRT
    COP=Q/W


    3. The attempt at a solution
    Using the first equation for work, I got W=-nrT*ln(3V1/V1) but I don't think that is right. Is this the wrong formula to use for this problem? My answer should be in terms of P1 and V1.
     
  2. jcsd
  3. Feb 23, 2016 #2
    That is not the correct integration of PdV for the process shown in the figure.
     
  4. Feb 23, 2016 #3
    What equation can I use to find the work then?
     
  5. Feb 23, 2016 #4
    You can use the integral of PdV. Just do it in a way that is consistent with what you see in the diagram.
     
  6. Feb 24, 2016 #5
    I just took the area of the rectangle. So W=2P1V1. I got an efficiency of 17% with that.
     
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