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Another contradiction in thermodynamics?

  1. Feb 23, 2015 #1
    Consider an ideal gas operating in a quasi-static (very slow) cycle that is identical to the heat engine version of the carnot cycle in every aspect, except that friction is present. So even though the cycle is quasi-static, it is irreversible due to friction.

    Now the question is: How does the efficiency of this quasi-static cycle compare to ideal carnot efficiency? Let ηrev be the efficiency of the carnot cycle and ηirrev be the efficiency of the quasi-static cycle:
    Is ηirrevrev?
     
  2. jcsd
  3. Feb 23, 2015 #2

    mfb

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    Its efficiency is worse. You convert mechanical work to heat with a 1:1 ratio instead of the thermodynamic optimum.
     
  4. Feb 24, 2015 #3
    I'm guessing you're talking about work lost due to friction, correct?
    But how does this affect our equations? Take the adiabatic step for instance. Since the process is quasi-static, we could still use dW=-PdV with p referring to the system's pressure. dq=0 right? It's an ideal gas, so dU=CvdT. I'm taking the gas confined in the cylinder as my system. Where does friction between piston and cylinder come into the picture? If CvdT = -PdV, then we're gonna get the same PVgamma= Const equation.
     
  5. Feb 24, 2015 #4
    I can help you work this out. There are two ways of doing this, but the easiest way is to treat the piston and cylinder as a combined system. Assume that all the "frictional heat" goes into the gas. To get things started, let F be the frictional force acting on the piston by the cylinder wall. Do a free body diagram on the piston, and then write a force balance on the piston. Don't forget the supplemental force you need to apply to the piston to keep the conditions quasistatic. What do you obtain?

    Chet
     
  6. Feb 24, 2015 #5
    Thnx Chet. I attached a diagram with three forces/pressures acting on the piston: internal pressure of the gas P, external pressure Pext, and fk due to kinetic friction. I've assumed the gas is undergoing expansion. I'm not sure about the supplemental force that you've mentioned though. I'm also confused about your choice of system.
    Is it gas + piston, with the cylinder as surroundings? or is it piston+ cylinder?

    Why all of the heat?

    And what ratio are we trying to calculate here?
     

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  7. Feb 24, 2015 #6
    Yes. The diagram is correct. I see that you had the cylinder sitting in a bath. I was assuming you were going to be taking the system to be adiabatic when you wrote PVγ=const.

    The "supplemental" force I was referring to is what you called Pext. So the force balance on the piston is, as you indicated,

    ##PA - F = P_{ext}A##

    Sorry. I meant to say, treat the gas and piston as a combined system, and adiabatic cylinder as part of the surroundings.

    To do this problem, you have to decide where the frictional heat is going. If it all somehow goes into the surroundings, then the gas experiences exactly the same treatment as in an adiabatic reversible expansion, and your PVγ is correct. On the other hand, if part or all of the heat goes into the gas, it experiences a different treatment, equivalent to a non-adiabatic reversible expansion. I was assuming that you wanted to consider the case in which all the frictional heat goes into the gas.

    Chet
     
  8. Feb 24, 2015 #7
    Continuing: If we multiply your force balance by dx (the displacement of the piston) we obtain:
    $$PdV-\frac{F}{A}dV=P_{ext}dV=dW$$
    where dV = Adx and dW is the work done by the system on the surroundings. If you combine this equation with the differential version of the first law and assume that the cylinder is adiabatic, what do you get?

    Chet
     
  9. Feb 24, 2015 #8
    So what ratio do you have in mind as efficiency? Are you considering the Work done by Pext as the numerator? But what if we choose the gas as our system, then it should be work done by the gas (PdV), no?
     
  10. Feb 24, 2015 #9
    I understand you're writing the wok-energy theorem for the massless piston.
    PdV-PextdV-fkdx=dkpist=0

    dU=dW+dq+dW*= -PextdV-Fdx
     
  11. Feb 24, 2015 #10
    where F= (P-Pext)A
    Substituting for f yields:
    dU= -PextdV+ (PextdV-PdV)= -PdV
     
    Last edited: Feb 24, 2015
  12. Feb 24, 2015 #11
    Let's not worry about the efficiency yet. If we chose the gas as our system, then, yes, this would be the work. We'll do the problem that way after we do it this way, OK?
     
  13. Feb 24, 2015 #12
    I don't know exactly what you did here, but the result should be with a plus sign in front of the F and it should be P, not Pext:
    $$dU=\frac{F}{A}dV-PdV$$

    Now, the next step is to substitute the ideal gas relationships into this.

    Chet
     
  14. Feb 24, 2015 #13
    my equation is dU= -PextdV - Fdx. The first term on the right side of my equation equals the entire right side of your equation, i.e. -PextdV= F/AdV - PdV.
    The difference between my equation and yours is that I have added the term -Fdx, which is the work done by the cylinder on the piston. Considering that the cylinder is part of the surroundings and the piston is part of the system, therefore the force of friction, F, is external to our system, and since its point of application is moving, we must include a term for its work.

    I added -Fdx because I believe it's part of the boundary work for our choice of system.
     
  15. Feb 24, 2015 #14
    Remember that our system is working against TWO forces, not one. dW = PextdV is only PART of the the work done by the system on its surroundings. The other half is done against friction.
     
  16. Feb 24, 2015 #15
    How many times do you want to include it? We already included it once.

    The cylinder is stationary, so our system does no work on the cylinder. The cylinder is considered part of the surroundings, so our system does no work on that part of the surroundings.

    Chet
     
  17. Feb 24, 2015 #16
    No, PextdV is all the work done by our system on its surroundings.
     
  18. Feb 24, 2015 #17
    You are right, the piston does no work on the cylinder as the cylinder isn't moving. So PextdV is the only work done BY the system ON the surroundings. So I made a mistake here. But I don't think you should use the engineering convention of the first law of thermodynamics here. Start with dU=dW +dq, where dW is the work done by the surroundings on the system. This I believe is a generalization of conservation of mechanical energy, and is consistent with the conventions used there.
     
  19. Feb 24, 2015 #18
    In other words, in this particular problem, the work done on the surroundings by the system is not equal to the negative of the work done on the system by surroundings, and thus you cannot use dU= dq-dW where dW is the work done by the system as a valid form of the first law.
     
  20. Feb 24, 2015 #19
    Actually, the version of the first law that I am accustomed to is dU = dq - dW, where dW is the work done by the system on the surroundings. So, from that perspective, what I said was OK.

    So, it's a bit paradoxical, huh? But I am confident that I can resolve the paradox for you, so that we are both in agreement. But, for sure, handling the friction effect at the interface between the piston and the cylinder wall is a bit tricky.

    I am going to analyze the friction effect fusing a different approach. Are you familiar with viscous fluids and Newtonian viscosity?

    Please stay tuned. But right now, my wife an I are going out to dinner. So, see ya later.

    Chet
     
  21. Feb 24, 2015 #20
    Ok take your time thank you and I hope you have a great time.
     
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