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Diffusion equation

  1. May 27, 2016 #1
    Hi

    In diffusion equation ,if we have a infinite slab of moderator with thickness ±a and the sources is

    s(x)= cos(x)

    i think the first boundary

    flux( ±a ) = 0

    what will be the second boundary condition ??
     
  2. jcsd
  3. May 27, 2016 #2
    I would say that this is the only boundary condition that you have. The symmetry around the center of the source and thus the flux, will lead you to have an even solution of the differential diffusion equation, which would be a Cosine..or a decaying exponential. So, you are left with two more constants to be determined; the constant in the exponential/cosine and the coefficient of the exponential/cosine function (the amplitude).

    To get the first one, you will have to apply the boundary condition of the total flux decay at the boundary; which you have already mentioned in your question. This will lead you to numerous possible solutions/harmonics.

    To get the second one, you have to know the power of the assumed reactor.
     
  4. May 27, 2016 #3
    this is for reactor but in my case it is a moderator
     
  5. May 27, 2016 #4
    To design a thermal reactor, you should have a moderator to slow down neutrons to the thermal region where fission cross section for U235 is high enough. So, a reactor, is a source(which is your case has a cosine distribution) and a moderator, at least.
     
  6. May 27, 2016 #5
    i do not have the power, i am thinking to use laplace transformtion
    if the boundary condition
    flux( ±a ) = 0
    Derivative flux(0)=0
     
  7. May 27, 2016 #6
    Although the flux vanishes at the boundaries, the derivative of the flux does not; this still annoys my intuition but this is what you will get when you follow the diffusion equation and the definition of the neutrons' current. How are you planning to use LT here, then? I think it is not applicable to such a problem - I would be happy to know that it is applicable, though!

    However, my first reply was the general scheme for the solution of the problem, but seems I underestimated your example as it seems that it will get more complicated with this cosine-shaped source.

    As far as I know the general scheme for your solution will be as follows..

    Solve using the diffusion equation using eigen functions solution (the way I mentioned in my first reply) after you apply your boundary condition (most cases, you can just expect the shape of the flux instead of solving the differential equation)..then take advantage of the orthogonality of the eigen set of functions that you have by putting your source term in an eigen functions summation as well as your flux (that you just got assuming no Source) term, and insert both of these terms in the diffusion equation. Now you are left only with the coefficient to be calculated (the constant that you were to get it by knowing the power), and you can get it. This is the over view of the solution, and obviously it may be not so helpful to just read it, so I advise you to read section 5-10 (especially subsection: distributed sources in a finite medium) from John R. Lamarsh, Introduction to Nuclear Reactor Theory.
     
  8. May 27, 2016 #7
    It is worth mentioning that this may not be the only way to approach the solution, but this is the way that I know and that way followed by Lamarsh in his solutions to the diffusion equation. May be other authors use a different/easier approach...I would be glad to know it!
     
  9. May 27, 2016 #8
    i read the lamarsh(intro to nuclear reactor ) book but in lamarsh(intro to nuclear enginnering ) problem 34 he use laplace transformtion

    that why i am confused now ,,anyway thank for helping
     
  10. May 27, 2016 #9
    Problem 34, which chapter?
     
  11. May 27, 2016 #10
    5
     
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