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Comp. Physics: Finite time Carnot cycle

by JorisL
Tags: carnot, comp, cycle, finite, physics, time
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Apr2-13, 06:59 AM
P: 46

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.


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Apr2-13, 07:17 AM
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P: 1,344
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).
Apr2-13, 08:16 AM
P: 46
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.

Apr2-13, 08:38 AM
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P: 1,344
Comp. Physics: Finite time Carnot cycle

Quote Quote by JorisL View Post
The change was that I could neglect collisions in the y-direction because they became perfectly elastic.
Just keep in mind that simulating this 1D problem is not equivalent to simulating one dimension of a 3D problem.
Apr2-13, 12:09 PM
P: 46
You're absolutely right. Forgot that initially.
Apr2-13, 03:23 PM
P: 46
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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