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Homework Help: Efficiency of a Heat Engine

  1. Jul 9, 2008 #1
    1. The problem statement, all variables and given/known data
    An ideal gas engine that works according to the following cycle.
    2652160531_9135b3c55b_m.jpg

    Find the efficiency of this engine assuming that the heat capacities of the gas may be taken to be constant. Recall that the efficiency may be defined as:
    [tex] \eta = \frac{Net work done over full cycle}{Heat absorbed along an isotherm} [/tex]

    Express your answer in terms of the volumes [tex]V_1[/tex] and [tex]V_2[/tex] and the pressures [tex]P_1[/tex] and [tex]P_2[/tex] and the heat capacities at constant pressure and volume [tex]C_p[/tex] and [tex]C_v[/tex].

    2. Relevant equations
    [tex]W = \int P dV [/tex]
    [tex]P V = N kB T [/tex]
    [tex]C_v = 3N k_b /2 [/tex]
    [tex]C_p = 5N k_b /2 [/tex]


    3. The attempt at a solution

    [tex] \eta = \frac{W}{Q} = \frac {W_a + W_b + W_c}{Q_c} [/tex]
    [tex] W_a = P_2 (V_2 - V_1) [/tex]
    [tex] W_b = 0 [/tex]
    [tex] W_c = \int^{V_1}_{V_2} P dV = \int^{V_1}_{V_2} \frac{N k_b T}{V} dV = N k_b T ln (V_1 / V_2)[/tex]
    To find [tex]Q_c[/tex], recall [tex]U = Q + W[/tex] and [tex]U=(3/2)N k_b T[/tex] which is constant on an isotherm. So [tex]Q_c=-W_c= -N k_b T ln (V_1 / V_2) [/tex]

    Thus, by the equation of the efficiency,

    [tex]\eta = \frac{N k_b T ln (V_1 / V_2)+P_2 (V_2 - V_1)}{-P_2 (V_2 - V_1)}[/tex]

    However, I'm not sure if all my assumptions above are valid, and I don't see how the above could be put in terms of [tex]C_v[/tex] and [tex]C_p[/tex], since there's no relation between them and the temperature.

    Any help is appreciated.

    (So you know, I'm studying for a qualification exam this fall in graduate school. These problems are from past exams that they've given us to help study. I've been out of the physics world for a year and a half, so this stuff comes back slowly sometimes. I appreciate everyone who has responded so far to my questions.)
     
  2. jcsd
  3. Jul 9, 2008 #2

    Andrew Mason

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    Where did you get this definition from? Efficiency is:

    [tex]\eta = \frac{W}{Q_h}[/tex]

    where W is the net work done per cycle and [itex]Q_h[/itex] is the heat absorbed from the hot reservoir. Since [itex]W = Q_h - Q_c[/itex] as there is no change in U in one complete cycle, you can replace the denominator with [itex]W + Q_c[/itex]. In this case, heat is absorbed in parts b and c of the cycle and released to the cold reservoir in a.

    This should be:

    [tex] \eta = \frac{W}{Q_h} = \frac {|W_b| + |W_c| - |W_a| }{Q_h} [/tex]

    This is a negative value since work is being done ON the system (ie should be:

    [tex] W_a = P_2 (V_1 - V_2) [/tex]

    Correct.

    To work out the heat flow into the cold reservoir (part b) you should recognize that heat flows at constant pressure so:

    [tex]Q_c = nC_p\Delta T[/tex]

    You will have to work out the temperature change, which is simple to do.

    AM
     
  4. Jul 9, 2008 #3
    Good question Andrew. Interestingly enough, it's verbatim on the graduate entrance exam for a state university. I was wondering about it myself, but that's why I asked here. Thanks, I think with your help I can solve the problem.
     
  5. Jul 9, 2008 #4

    Andrew Mason

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    Someone may be confusing coefficient of performance of a refrigerator with efficiency of a heat engine. They are quite different. COP of a refrigerator is W(input)/Qc

    Incidentally, I had the volumes backward in my mind so ignore my comment that Wa = P2(V1-V2). You had it right.

    AM
     
  6. Jul 9, 2008 #5
    I had to double check my work, but I had convinced myself that you were mistaken. Thanks for letting me know though.

    Here's the solution I worked out.

    [tex] \eta = \frac{W_{total}}{Q_{hot}} = \frac{W_a+W_b + W_c}{W_{total}+Q_{cold}} = \frac{-P_2 (V_1 -V_2)+ 0 + nRT ln(V_1 / V_2)}{-P_2 (V_1 -V_2)+ 0 + nRT ln(V_1 / V_2)+Q_{a}} [/tex]

    where [tex] Q_a = n C_p \Delta T = n C_p (T-P_2 V_2 /(nR)) = C_p (n T - P_2 V_2 /R) [/tex]

    so [tex] \eta = \frac{-P_2 (V_1 -V_2) + nRT ln(V_1 / V_2)}{-P_2 (V_1 -V_2) + nRT ln(V_1 / V_2)+C_p (n T - P_2 V_2 /R)} [/tex].

    Which can be re-written

    [tex] \eta = \frac{- (V_1 -V_2) + ( V_1) ln(V_1 / V_2)}{( V_1) ln(V_1 / V_2)+(C_p/R - 1) (V_1- V_2 )} [/tex]

    Howabout that.... no explicit P2 dependence. That answer, though, doesn't fit well with what the problem asked for. It may just be a poorly asked question (as the [tex]\eta[/tex] definition they gave already showed).
     
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