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A Carnot engine using methane

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data
    A Carnot engine operates with 1 kg of methane, which we will consider an ideal gas.

    The ratio of specific heat capacities [itex]\gamma[/itex] is 1.35.

    The ratio of maximum volume to minimum volume is 4 and the cycle efficiency is 25%.

    Find the entropy increase during the isothermal expansion.

    2. Relevant equations

    The temperature is constant if the expansion. I wasn't sure which equations to start with, though, because the equations I saw int he text all seem to be predicated on the temperature being different.

    Now, I could put T in terms of P for the ideal gas, so if [itex]PV = nRT[/itex] and I keep T constant, [itex]T = \frac{PV}{nR}[/itex] and then use the following for entropy, since this is a Carnot engine and it's a reversible process:

    $$S_b - S_a = \int^a_b \frac{d'Q_T}{T} = \int^a_b \frac{d'Q_T nR}{PV} = nR \int^a_b \frac{d'Q_T}{PV}$$

    But I have no idea if I am even on the right track with this. So really I am hoping someone can tell me if I am going in the right direction. Because I am really not sure.

    Thanks in advance.
     
  2. jcsd
  3. Mar 17, 2014 #2
    Start by drawing a picture of the process.

    In isothermal expansion, ΔU=0.

    Now, [tex]W=\int_b^c PdV = \int_b^c \frac{nRT}{V}dV = nRT\ln(\frac{V_c}{V_b})[/tex]

    Do you see what you have to do from here?


    Also, I don't think this belongs in "advanced physics".
     
    Last edited: Mar 17, 2014
  4. Mar 17, 2014 #3
    I am not sure. I can use the work function and say it is Q2-Q1, but all of the functions I see that relate S to Q depend on temperature. Unless I can just plug in what T is in terms of P and go from there.

    I put it in advanced because it said upper level undergrad and a lot of Carnot engine problems here. So I went with that. If it belongs in beginner phys then so be it.
     
  5. Mar 17, 2014 #4
    How does efficiency relate to Q and T?
     
  6. Mar 17, 2014 #5
    well let's see. In a Carnot with two different temperatures thermal efficiency [itex]\eta = \frac{W}{Q}[/itex] but that's the thing that gets me here, because [itex]Q = c_P(T_2-T_1)[/itex]. But the process is isothermal. I could plug in pressure instead, assuming T constant as I did at the beginning. But that's where this whole problem sort of ceases making sense to me because the whole way a Carnot works is moving heat between reservoirs of two different temperatures and a free expansion of gas does no work at all...
     
  7. Mar 17, 2014 #6
    [tex]\eta = 1-\frac{T_1}{T_2}[/tex]
    [tex]\eta = 1+\frac{Q_1}{Q_2}[/tex]

    You can use this along with the adiabatic expansion equation to get Vc/Vb.


    Draw a picture of the process.
     
  8. Mar 17, 2014 #7
    Drawing a picture should be on a P-V diagram, yes? And I should have a diagram in which the let side is Vb and the right side is Vc, and the top is P2 and the bottom p1, yes?

    Also, in the second expression with eta, Q1 and Q2 are both dependent on T are they not? And if the process is isothermal T2-T1 is zero!!!! That means η=1. Well ok then, but the original problem says it isn't!

    What am I missing here?
     
  9. Mar 18, 2014 #8
    If you set them equal to each other you'll see that Q1/Q2=-T1/T2=-3/4, since the efficiency is .25. With this relationship and the adibatic expansion equation [itex] TV^{\gamma -1} \implies T_{2}V_{c}^{\gamma -1}=T_{1}V_{b}^{\gamma -1}[/itex]
    Now rearrange so that you have Vc/Vb
     
  10. Mar 18, 2014 #9
    thanks a lot - i actually figured it out once I looked at my diagram and realized what you were getting at with the b and c subscripts... stupid of me.
     
  11. Mar 18, 2014 #10
    Good job :)
     
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