Comp. Physics: Finite time Carnot cycle

  1. Hi,

    For my Bachelor's thesis I've been working on a finite time Carnot cycle.
    I've finished my numerical analysis using the differential equations governing the time evolution.

    My next step should be a simulation.
    First I should stick to a 1 dimensional system.
    This system consists of a piston and a thermalising wall.
    This thermalising wall acts as a heat reservoir. Every particle colliding with this wall is absorbed.
    The wall than ejects a 'new' particle with a certain velocity.

    This velocity is governed by the stochastic distribution
    [itex]f(\vec{v},T_i)=C v \exp\left(-\frac{m v^2}{2kT_i}\right)[/itex] with [itex]i=h[/itex] while expanding and [itex]i=c[/itex] while compressing. C is the normalisation constant.

    Since I use a 1D system in this first approximating step, the Maxwell-Boltzmann distribution isn't necessary.

    The piston has a constant velocity u. This is chosen because of the fact that the article I base my calculations on is targeting a system that is easy to control.
    The article is "Molecular Kinetic analysis of a finite-time Carnot cycle" by Y. Izumida and K. Okuda published in september 2008.

    I reckon I have to use some sort of Monte-Carlo method because of the stochastic nature of the reservoir. I have however not a clue on how to start.

    But the paper talks about Molecular Dynamics. Am I thinking about it in a wrong way?

    My previous experience with computational physics is small.
    I've only worked with a driven pendulum using the GSL library and the Ising model using the metropolis algorithm.


    Last edited: Apr 2, 2013
  2. jcsd
  3. DrClaude

    DrClaude 2,403
    Science Advisor
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    Gold Member

    First of all, I don't think you'll be able to do it in 1D. One-dimension problems are peculiar because the particles can't overtake each other, and this can create strange effects. Since you have a purely classical system, I see no point in not doing it in 3D from the start.

    To find out how to do it, you have to look up molecular dynamics. A book about computational physics would be a good start, for instance, J. M. Thijssen, |i]Computational Physics[/i] (Cambridge University Press, 1999).
  4. Thanks for the quick response.
    I get there as well that 1D is not really useful.
    Further more I understood the teacher wrong. We decided to move the thermalising wall to make the problem in essence 1D. The change was that I could neglect collisions in the y-direction because they became perfectly elastic.

    Thanks for the book reference.
  5. DrClaude

    DrClaude 2,403
    Science Advisor
    Homework Helper
    Gold Member

    Just keep in mind that simulating this 1D problem is not equivalent to simulating one dimension of a 3D problem.
  6. You're absolutely right. Forgot that initially.
  7. Ok, I'm stuck again.

    The book was useful for grasping the basic ideas.
    I found how I could initialize the system. And understood the idea from appendix B.3 (box-muller transform).

    However I believe I have to find something similar for collisions with the thermalising wall.
    The distribution now (in the expanding stage) is
    [itex]f( \vec{v} ,T_h )=C v_x \exp\left( -\frac{m v^2}{2kT_h} \right) [/itex].

    If I try to use the box-muller transform I'd get following expression
    [itex]P(v_x,v_y)dv_xdv_y= C v_x\exp \left(\frac{-v^2}{2}\right) dv_xdv_y=C v^2\exp \left(\frac{-v^2}{2}\right) \cos (\phi )dvd\phi=P(v,\phi)dvd\phi[/itex]

    After that I would find (in analogy of the appendix mentioned before)
    [itex]\left( g^{-1}\right) ^\prime (v,\phi)=C v^2\exp \left(\frac{-v^2}{2}\right) \cos (\phi )[/itex] (1)
    I used g for to stress the difference with the original distribution

    I don't see how I can get similar expressions as for the Maxwell-Boltzmann distribution, to get random numbers distributed like [itex]f(\vec{v},T_h)[/itex].

    I suspect I need another way to do this. Box-Muller doesn't seem to work at first sight.

    Furthermore does the derivative of the inverse function in (1) even make sense?
    Does it mean [itex]g^\prime(x,y)=\frac{\partial g(x,y)}{\partial x}+\frac{\partial g(x,y)}{\partial y}[/itex]?
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