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Efficiency of Carnot engine

  1. Feb 28, 2013 #1
    The problem:
    An engine puts an ideal monatomic gas through a clockwise rectangular cycle on a PV diagram with horizontal and vertical sides. The lower left point has a pressure of 1 atm and a volume of 1m^3 and the upper right point has pressure and volume three times greater. Calculate the efficiency of a Carnot engine operating between the highest and lowest temperatures.

    Solution (so far):
    I know that for a Carnot engine e=1-T_c/T_h, but without being given the temperature differences I'm not exactly sure how you'd begin. I calculated the efficiency of the engine itself to be 22.2%, the work done in the cycle to be 4.04x10^7 J, and the heat absorbed in the cycle to be 434 kcal; if any of those quantities can be related to T_c/T_h. The answer comes to be 88.9%. Thanks in advance for any help.
     
    Last edited: Feb 28, 2013
  2. jcsd
  3. Feb 28, 2013 #2

    phyzguy

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    Try using the ideal gas law. Since you are given the ratios of pressures and volumes, you should be able to calculate the ratio of temperatures.
     
  4. Feb 28, 2013 #3
    Since there are 2 unknowns, namely the temperature difference and the number of moles, I don't see an obvious way that I could use the ideal gas equation to solve this problem.
     
  5. Feb 28, 2013 #4

    phyzguy

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    I promise you it will work. Can you write the ideal gas equation at Tc and Th?
     
  6. Feb 28, 2013 #5
    Oh nevermind,

    (P_3*V_3)/(P_1*V_1)=T_3/T_1
    (3*P*3*V)/(PV)=T_3/T_1
    9=T_3/T_1=T_h/T_c

    Therefore,

    T_c/T_h=1/9

    So,

    e=1-1/9=8/9=88.9%

    Thanks! Sorry, for posting such an obvious question; just a little low on sleep.
     
  7. Feb 28, 2013 #6

    phyzguy

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    You got it. Glad to help.
     
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