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Homework Help: Efficiency of an engine performing the Joule ideal-gas cycle.

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The figure represents a simplified PV diagram of the Joule ideal-gas cycle. All processes are quasi-static, and [tex] C_P[/tex] is constant. Prove that the thermal efficiency of an engine performing this cycle is
    [tex] 1 - \left(\frac{P_1}{P_2}\right)^\frac{\gamma - 1}{\gamma} [/tex]
    http://img50.imageshack.us/img50/7734/thermo1ym3.jpg [Broken]
    http://g.imageshack.us/img50/thermo1ym3.jpg/1/ [Broken]

    2. Relevant equations
    [tex] PV = RT [/tex]
    [tex] \gamma = \frac{C_P}{C_V} [/tex]
    [tex] dE = dq + dw [/tex] (should have strokes through the d's on dq and dw, but I don't know how to latex inexact differentials)
    [tex] \eta = 1 - \frac{|q_{out}|}{|q_{in}|} [/tex]
    On adiabatic processes,
    [tex] TV^{\gamma - 1} = [/tex] constant
    [tex] PV^{\gamma} = [/tex] constant


    3. The attempt at a solution
    First of all, [tex] \eta = 1 - \frac{|q_{out}|}{|q_{in}|} [/tex]
    [tex]q_{in} [/tex] is only path 2->3 and [tex] q_{out} [/tex] is only path 4->1.

    2->3
    Since we have an ideal gas,
    [tex] dq = C_{P}dT [/tex]
    [tex] q_{in} = \int^{T_3}_{T_2} C_{P}dT [/tex]
    However,
    [tex] dT = \left(\frac{\partial T}{\partial P}\right)_{V} dP + \left(\frac{\partial T}{\partial V}\right)_{P} dV [/tex]
    since dP = 0 in the 2->3 process, we have:
    [tex] dT = \left(\frac{\partial T}{\partial V}\right)_{P} dV [/tex]
    Now, using the ideal-gas equation of state and solving for T:
    [tex] T = \frac{PV}{R} [/tex]
    Differentiating:
    [tex] \left(\frac{\partial T}{\partial V}\right)_{P} = \frac{P}{R} [/tex]
    Now, substituting into the above expression, we get:
    [tex] q_{in} = \frac{C_{P} P_{2}}{R} \int^{V_{3}}_{V_{2}}dV [/tex]
    So,
    [tex] |q_{in}| = \frac{C_{P} P_{2}}{R} (V_{3} - V_{2}) [/tex]

    Now, considering process 4->1

    4->1
    We have the same process as above, but with different pressures and volumes. Therefore,
    [tex] q_{out} = \frac{C_{P} P_{1}}{R} \int^{V_{1}}_{V_{4}} dV [/tex]

    Since q_{out} is negative, we switch signs,
    [tex] q_{out} = \frac{C_{P} P_{1}}{R} (V_{4} - V_{1}) [/tex]

    Plugging into the efficiency formula [tex]\frac{C_{P} P_{1}}{R} [/tex] cancels, and we get:
    [tex] \eta = 1 - \frac{P_{1} (V_{4} - V_{1})}{P_{2} (V_{3} - V_{2})} [/tex]

    Now, I'm pretty sure I have to use the identities:
    [tex] P_{1} V^{\gamma}_{4} = P_{2} V^{\gamma}_{3} [/tex]
    [tex] P_{1} V^{\gamma}_{1} = P_{2} V^{\gamma}_{2} [/tex]

    I have tried dividing these two equations so that all P's cancel.
    I have also tried subtracting the equations.

    I can't, for the life of me, get my efficiency in the form that the problem asks me to put it in.

    Any help would be greatly appreciated. Thanks.
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Sep 14, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi Jacobpm64,

    To cut down on the algebra, try writing the adiabatic relations as

    [tex] P_{1}^{1/\gamma}\ V_{4} = P_{2}^{1/\gamma}\ V_{3} [/tex]
    [tex] P_{1}^{1/\gamma}\ V_{1} = P_{2}^{1/\gamma}\ V_{2} [/tex]

    and use them to eliminate [itex]V_4[/itex] and [itex]V_1[/itex]. Do you get the answer?
     
  4. Sep 14, 2008 #3
    I sure did get the answer!

    You don't know how much you're helping me.

    This course is really hard to me, and I appreciate the help that I'm getting here.

    I have one question though.

    How did you come up with that adiabatic relation? (I didn't know that it was one).
     
  5. Sep 14, 2008 #4

    alphysicist

    User Avatar
    Homework Helper


    It's the same adiabatic relation as you had in your post. Just raise both sides to the [tex]1/\gamma[/tex] power to move the exponent onto the P instead of the V:

    [tex]
    \begin{align}
    P_{1} V^{\gamma}_{1} &= P_{2} V^{\gamma}_{2} \nonumber\\
    \left( P_{1} V^{\gamma}_{1}\right)^{1/\gamma} &= \left(P_{2} V^{\gamma}_{2} \right)^{1/\gamma}\nonumber\\
    P_{1}^{1/\gamma}\ V_{1} &= P_{2}^{1/\gamma}\ V_{2} \nonumber
    \end{align}
    [/tex]

    Since in your problem you were trying to cancel the V variables in your numerator and denominator, I think it gives easier algebra to use the adiabatic relation in a form that has V to the first power
     
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