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Efficiency of an ideal-gas cycle.

  1. Oct 10, 2010 #1
    1. The problem statement, all variables and given/known data
    A possible ideal-gas cycle operates as follows:
    (i) from an initial state (p1,V1) the gas is cooled at constant pressure to (p1,V2);
    (ii) the gas is heated at constant volume to (p2,V2);
    (iii) the gas expands adiabatically back to (p1,V1).

    Assuming constant heat capacities, show that the thermal eficiency is:

    1-ɣ[((V1/V2) - 1) / ((p2/p1) -1)]

    2. Relevant equations

    I used first law of thermal dynamics, n = W/Qh, differential forms of Cv and Cp.

    change in energy for the complete cycle = 0, therefor W = Qh + Ql.

    3. The attempt at a solution

    Using efficiency = W / heat absorbed.

    I attempted to find the W and Q of the 3 processes separately.
    I was using for Isobaric: dW = -Pdv but im pretty sure this is incorrect because no where does it state it is a reversible process. But the book for the class only shows reversible processes for examples and does not actually show a isobaric or isochoric process in any example.

    I tried using the differential form of Cv and Cp and rearranging them to find dQ, but failed.

    I have used 4 or 5 sheets of paper, trying every way i could think of to work this problem out. I know im missing something and if someone could just point me in the right direction I would gladly appreciate it and work out the problem on my own.

  2. jcsd
  3. Oct 10, 2010 #2


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    Welcome to Physics Forums.

    Actually, we would assume a reversible process here, so dW=-PdV as you said.

    You'll need W for the complete cycle, the tricky part is the adiabatic leg. Can you set up the integral for that part and show what you get? Here is an integral sign you can copy-and-paste: ∫

    You'll also need Q for one leg of the cycle (do you know which leg that is?).
  4. Oct 10, 2010 #3
    Using Qh for heat absorbed and Ql for heat lost I was using the equation as:

    n = w/Qh = (Qh +Ql)/Qh = 1 + Ql/Qh.

    Ql being from (i) and Qh being from (ii). Q = 0 from (iii).

    Qh = ∫Cv dT
    Ql = ∫Cp dt + ∫pdV

    For adiabatic ΔQ = 0 so ΔU = W.

    Since, you are ending where you started ΔU for the complete cycle should = 0 correct?

    so ΔU for adiabatic should equal ΔU for isochoric + isobaric.

    Isobaric: ΔU = Ql + pdV
    Isochoric: ΔU = Qh

    ΔU for adiabatic = Ql + Qh - ∫pdV.

    ΔU = W so W = Ql + Qh - ∫pdV

    and now i lost myself. Maybe i just need some sleep.

    Thanks for trying.
  5. Oct 11, 2010 #4
    Ok, so I tried 3 different ways and started to post my questions on here, and while posting one of my questions, I think I answered it myself.

    I stated that I wasnt sure how to relate the change in T to p and V, but I forgot about the ideal gas equation.

    So I just solved for Q's like before and put in pV/R in place of T. Doing that I ended up with

    n = Qin - Qout / Qin

    dQin = Cv/R(Vdp + pdV) and pdV = 0
    dQout = Cp/R(Vdp + pdV) and Vdp = 0

    n = Cv(V2(p2-p1)) - Cp(p1(V2-V1)) / Cv(V2(p2-p1)) because R's cancel out.

    n = 1 + (-Cp/Cv) *((V2-V1)/V2) * (p1(p2-p1)) sub in ɣ for Cp/Cv and clean up the rest

    and that shows that n = 1+ɣ[((V1/V2) - 1) / ((p2/p1) -1)]

    It is very close, only missing a - sign. I am not sure If i did a calculation wrong or that it is purely coincidental that I got this close.

    Now that part that scares me and why I didnt do this in the beginning is that I complete ignore the adiabatic leg, and all work done. I just simple used heat.

    It is close to the right answer but I am not convinced I did it correctly.

    Thanks for your help so far.
    Last edited: Oct 11, 2010
  6. Oct 11, 2010 #5

    Andrew Mason

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    The work done is the difference in heat in and heat out. It is also the work done in the adiabatic expansion less the work done in the isobaric compression. So you can approach it either by calculating 1) heat in and heat out (as you have done) or 2) heat in and work.

    If you are using 1 - Qout/Qin for efficiency you have to take the absolute value of Qout:

    [tex]|Q_{out}| = C_pP_1(V_1 - V_2)[/tex]

    That is where you got the signs mixed up.

  7. Oct 11, 2010 #6
    thanks, that makes me feel better.

    can you explain to me exactly why we take the absolute value? Just so i can understand it and apply it to future problems.

    thanks again.

    I think I understand. Work is actually equal to the sum of the heat. By me stating Qin - Qout I am taking the negative out of Qout already. To do it the way I did I should have used
    n = (Qin + Qout)/ Qin
    Last edited: Oct 11, 2010
  8. Oct 11, 2010 #7

    Andrew Mason

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    Right. If you define the net heat flow (=work) being the sum of all heat flows then you have to stick with the signs (+ being into and - being out of the system). If you define it as heat in MINUS heat out, you are implicitly using positive quantities for all heat flows.

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