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Adiabatic Engine Ratios

  1. Mar 16, 2005 #1
    This one had three bullets by it so it's a pretty tough one.

    Work and rambling thoughts so far... please tell me if I'm wrong.
    This would be a change in entropy problem for the first couple of ratios right?
    (delta)S=Sfinal-Sinitial=nRln(Vf/Vi)+nCvln(Tf/Ti)
    yet it is an adiabatic operation so
    pV^(1.3)=a constant during the adiabatic process
    and since pV=nRT
    TV^(1.3-1)= a constant
    or TiVi^(1.3-1)=TfVf^(1.3-1)

    now in the picture there are two adiabatic processes one when the gas is expanding 2 to 3 so temperature is decreasing and one from 4 to 1 where the gas is being compressed and temperature is increasing so the ratio of
    a) T2/T1 since there is no volume change= 3.00 (duh)


    other ratio's I can see T3/T4=3.00,

    P1V1=T1 3P1V1=T2 T1=1/3(T2) Still just not seeing how they get the other ratios

    Question: What formula do I use to find out b,c,d, and e?
    then I can do f.) on my own.

    Numerical answers b)1.98 c)0.660 d)0.495 e)0.165 f)34%
     

    Attached Files:

  2. jcsd
  3. Mar 17, 2005 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    I am assuming that P2 = 3.00P1, although your graph is a little confusing, So of course T2/T1 = 3.00

    For the adiabatic expansion from 2 to 3,

    [tex]P_2V_2^\gamma = P_3V_3^\gamma[/tex]

    [tex]P_3 = \frac{P_2V_2^\gamma}{V_3^\gamma}[/tex]

    Since V3 = 4V2 = 4V1 and P2 = 3P1, you can work out P3 in terms of P1. And the ideal gas equation will give you T3.

    Getting P4 is similar to finding P3.

    [tex]P_4 = \frac{P_1V_1^\gamma}{(4V_1)^\gamma}[/tex]

    The efficiency part is a little complicated, I think.

    Edit: To find the efficiency, I thought one would have to determine the work done in the cycle by integrating under the curves.
    [itex]\eta = \frac{W}{Q_H} = \frac{Q_H-Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H}[/tex]

    But looking at this again, I see that is not necessary since there is no heat change from 2 to 3 or 4 to 1 (adiabatic). The heat is gained and lost at constant volume from 1 to 2 and 3 to 4. So:

    [tex]Q_C = C_v\Delta T_{3-4}[/tex] and [tex]Q_H = C_v\Delta T_{1-2}[/tex]

    the efficiency is easily determined from the temperatures at 1, 2, 3 and 4.

    AM
     
    Last edited: Mar 17, 2005
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