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Engine Efficiency

  1. Jun 9, 2017 #1
    1. The problem statement, all variables and given/known data
    An ideal gas with Cv = 5 2R, and γ = 1.4 starts at a volume of 1.5m3 , a pressure of 2.0×105Pa ,and a temperature of 300K. It undergoes an isobaric expansion until the volume is V , then undergoes an adiabatic expansion until the volume is 6.0m3 , and finally undergoes an isothermal contraction until it reaches the original state.

    2. Relevant equations
    What is the efficiency of this heat engine?

    3. The attempt at a solution
    I am aware that the answer is 0.19, and that efficiency is ∆W/∆Qh but I am unsure of how to find ∆W and ∆Qh in this scenario.
     
  2. jcsd
  3. Jun 9, 2017 #2

    DrClaude

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    Staff: Mentor

    Make a PV diagram and calculate W and Q for each leg of the cycle.
     
  4. Jun 9, 2017 #3
    There is usually a formula for W, but what is the formula for Q?
     
  5. Jun 9, 2017 #4
    I did the calculations for the W and Q for each PV leg and got:
    W= 146000 for isobaric, W = 346524 for adiabatic, and W = -415888 for isothermic.
    Q = 511000 Q= 0 Q = W = -415888
    Adding these up I get 94636/95112 = 0.99. I know the answer is not that. Any ideas of what I could be doing wrong?
     
  6. Jun 9, 2017 #5
    What is the next step after I calculate the W and Q for each leg of the cycle?
     
  7. Jun 9, 2017 #6
    Please show the details of your work.
     
  8. Jun 9, 2017 #7
    Isobaric: W = P∆V = (2x10^5)(2.23-1.5) = 146000J Q = (C/R)(Vf-Vi)p = 3.5(2.23-1.5)2x10^5 = 511000J
    Adiabatic: W = (K (Vf1-gamma - Vi1-gamma) )/(1-gamma) where K = PV1-gamma = (2x10^5)2.231.4 = 614690
    W = 614690 ( 0.4884-0.72556) / -0.4 = 364524J Q = 0
    Isothermic: W = PiVi ln(vf/vi) = (6)(5x10^4) ln(1.5/6) = -415888 Q = -W

    Now to add the Ws, and the Qs. W = 146000 + 364524 + -415888 = 94636 Q = 511000 + 415888 = 926888
    e = W/Q = 94636/926888 = 0.10
    which is incorrect. Any ideas?
     
  9. Jun 9, 2017 #8

    TSny

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    Homework Helper
    Gold Member

    OK. For the isothermic you have that Q = -W. Should that be Q = +W?

    The definition of efficiency is Wcycle/Qin where Qin is the amount of heat that went into the engine during the cycle rather than the total Q for the entire cycle. So, identify the part or parts of the cylce where heat went into the engine.
     
  10. Jun 9, 2017 #9
    i found the answer, thanks everyone
     
  11. Jun 9, 2017 #10
    Actually, to do this, all you have to do is. to get the three heats:

    Q1 =511000

    Q2 = 0

    Q3 = -415888

    So, the work is W = Q1 + Q2 +Q3 = 95112

    So the efficiency is 95112/511000
     
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