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Thermodynamics -- Carnot Machine

  1. Sep 15, 2014 #1
    Why can we say that a Carnot's machine efficiency does not depend on the fluid that you are heating? Is it because Carnot's machine is very ideal, a mere thought experiment? I have another question. I know that the efficiency of a heat engine depends on the quantity of heat that enter and leaves. I also understand that in a reversible machine the efficiency only depends on the temperature of the reservoirs, but if I had a reservoir that was at 20 C and another one at 10 C and my fluid was at 30 C, the fluid would contract, nonetheless, right? Now if we were talking about the number of cycles produced in a certain time, then the 10 C would be more "efficient" since you would transfer heat faster. So where am I wrong? Is it because the efficiency is the same whether you divide energy or power?( after all the times will cancel out.)

    Thank you!
     
  2. jcsd
  3. Sep 16, 2014 #2

    DrClaude

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    Staff: Mentor

    The fact that there is a maximal possible efficiency is based on the first and second law of thermodynamics only, resulting in
    $$
    e \leq 1 - \frac{T_\mathrm{c}}{T_\mathrm{h}}
    $$
    It then turns out that the Carnot cycle's efficiency is at the upper limit of efficiency (equality in the above equation). The Carnot cycle itself makes no assumption as to the working substance, so long as you can get the cycle of isotherm - adiabat - isotherm - adiabat.

    If the working substance is at a higher temperature than both reservoirs, then it just loses heat to both reservoirs and is not a cyclic engine anymore.
     
  4. Sep 16, 2014 #3
    Because in Carnot's cycle work is done only in isothermal process(addition or rejection), and for ideal gases isothermal process is independant of gas(or the adibatic index). W = nRTln(P2/P1). With the change in ideal gas only adiabatic expansion would affect, and therefore, this wouldn't make any difference.

    If the source is at 20C, sink at 10C, and gas(initially) at 30C then during isothermal expansion the gas at 30C would irreversibly transfer heat to source and then be at 20C. This first stroke would have higher efficiency because the gas was at higher temperature, but further cycles would have efficiency that of Carnot's.
     
  5. Sep 16, 2014 #4
    Also Carnot's cycle is not only the upper limit but the only efficiency that is available for a reversible engine at given working temperatures.
     
  6. Sep 16, 2014 #5

    DrClaude

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    Staff: Mentor

    Can you please clarify what you mean? Because there are other [STRIKE]reversible[/STRIKE] engines working at the same temperatures that have lower efficiency (such as the Otto cycle).

    [Edit: I was thinking of engines working on closed cycles, not reversible. AbhiNature is correct. Apologies.]
     
    Last edited: Sep 16, 2014
  7. Sep 16, 2014 #6
    Two different reversible cycles that have different efficiencies at the same operating temperatures when used in conjugation will create a PMM of second kind.
    One that is more efficient can be used to drive the second(now used as heat pump instead of engine). This will pump more heat from the sink than was rejected by the more efficient engine.

    Otto cycle, if reversible, will have same efficiency as Carnot's. Otto cycle's efficiency is a function of compression ratio. Greater the compression ratio lesser will be the allowed sink temperature, and hence greater the difference between source and sink temperatures, therefore greater efficiency. If compression ratio is converted equivalantly to operating temperatures and then these temperatures used to calculate Carnot's efficiency, then it should be same as that of Otto cycle's efficiency, if not, then Otto cycle is not reversible.
     
  8. Sep 16, 2014 #7
    Thank you all for your answers! What I meant with the 20 and 10 C reservoirs is that to me it wouldn't seem to matter if my low temperature reservoir was at 20 or 10, it would still work. It would matter if we measured the number of cycles it can output in a given time, but if we just calculated one cycle, either the 10 C or 20 C low temperature reservoir would work. Why is it then that the 10 C reservoir would make it more efficient? Is it because of the power it can otput, like I just mentioned? Or is there another thing I haven't realized?
     
  9. Sep 16, 2014 #8

    DrClaude

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    Staff: Mentor

    $$
    \Delta S = \frac{Q}{T}
    $$
    The lower the temperature, the lower of the amount of entropy "dumped" in to the cold reservoir, so the higher the efficiency, which depends on the amount of entropy created during the process.
     
  10. Sep 16, 2014 #9
    I'm sorry. I haven't actually read about entropy. I'm still in reversible processes and cyclic engines. Is there another way to explain it?
     
  11. Sep 16, 2014 #10
    Chacabucogod,

    For Carnot's cycle, efficiency is independant of energy or power.
    The gas contraction that you mentioned from 30C to 20C or 10C is not a part of Carnot's cycle. That is not a reversible process.

    Power from Carnot's engine would be same for any temperatures because isothermal process is infinitely slow, unless there are different infinities for different temperatures!
     
  12. Sep 16, 2014 #11
    Alright... Let me think about it. I haven't quite grasped it, but I'll let you know if another question comes out. Thank you both!
     
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