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Carnot Engine with finite reservoirs

  1. Feb 16, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    A Carnot engine operates between two finite heat reservoirs of total heat capacity CtH and CtC.
    a) Develop an expression relating TH to TC at any time.
    b) Determine an expression for the work obtained as a function of CtH ,CtC , TH and
    initial temperatures TH0 and TC0.
    c) What is the maximum work obtainable? This corresponds to infinite time, when
    the reservoirs obtain the same temperature.
    In approaching this problem, use the differential form of Carnot s equation,
    dQh/dQc=-TH/TC

    and a differential energy balance for the engine,
    dW - dQH - dQc = 0
    Here QH and QC refer to the reservoirs.


    2. Relevant equations



    3. The attempt at a solution
    I am having trouble with this problem really even getting started. I can't figure out how to use the hints to find T_H related to T_c through time. I gave it an attempt, but I am pretty sure my solution is wrong.
     

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    Last edited: Feb 16, 2014
  2. jcsd
  3. Feb 18, 2014 #2

    rude man

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    Well, since no one else has responded thus far:

    I guess this is to be viewed as a "differential Carnot engine" meaning changes in p and V are differentially small. In other words, on a p-V diagram it would be a teensy-tiny cycle, run an infinite number of times (except the reservoirs aren't constant-temperature reservoirs) ... until the two reservoirs have the same temperature.

    So write the differential forms given as hints.
    I haven't worked this out myself; will see if there is any response.
     
  4. Feb 18, 2014 #3

    Maylis

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    Yes, I wrote my differential form, but not sure if it's correct.
     
  5. Feb 18, 2014 #4

    rude man

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    For part (a), use the Carnot engine equation. Cross-multiply & you have an easy-to-solve diff. eq. Solve for T2(T1). I use T1 for Thot and T2 for Tcold. Initial reservoir temperatures are T10 and T20.

    for (b), use the 1st law equation combined with the result of part a. Integrate dW to get W.

    part c is really just part b integrated until the two reservoir temperatures are the same. Get this final temp. from part a again.
     
  6. Feb 18, 2014 #5

    rude man

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    OP, still out there? Don't give up, I think I have it figured out.
     
  7. Feb 18, 2014 #6

    Maylis

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    Yes. Here is my new solution. Let's compare. The algebra is ugly
     

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    Last edited: Feb 18, 2014
  8. Feb 19, 2014 #7

    rude man

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    OK, it's late my time, I have verified your part (a). Of course, that's the easy part.

    I'll finish up tomorrow.

    Looks like you're on very solid ground.
     
  9. Feb 19, 2014 #8

    rude man

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    Just got up, but let me start with:

    I have dW = -dQh - dQc = -Ch dTh - Cc dTc
    where remember dQh and dTh < 0 and dQc and dTc > 0.

    So just integrate each of the right-hand terms in their own variable:
    W = -ChTh0ThdTh - CcTc0TcdTc,

    keeping the T's separatet until after doing the trivial integrations w/r/t/ the respective T;
    then substituting Tc(Th) from part (a) for the second term on the right.

    For part (c) just substitute Tf for Th and Tc. You get Tf from part (a). Tf is the final temperature of the two "reservoirs".

    Main thing here is to keep track of the signs!
    How's that sound?
    .
     
  10. Feb 19, 2014 #9

    Maylis

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    I was told for part C that you need everything in terms of the initial hot and cold temperatures.
     
  11. Feb 19, 2014 #10

    rude man

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    That's what you get. You previously found Tc = Tc0(Th/Th0)(-Ch/Cc).

    So let Th = Tc = Tf and solve for Tf(Th0, Tc0).
     
  12. Feb 20, 2014 #11

    rude man

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    Just in case anyone drops in late here, my description of the p-V diagram was incorrect. It would look like a vertical sliver, with differentially small excursions along the isothermals but of course the adiabatics would have to go from T1 to T2 & back. The isothermals would gradually approach each other and finally merge, at which point the diagram is just a point on isotherm Tf.
     
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