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Neutron Diffusion Equation/Spherical Geometry Source Problem

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve for the flux distribution using the 1D neutron diffusion equation in a finite sphere for a uniformly distributed source emitting S0 neutrons/cc-sec.

    My problem right now is that I can't figure out the boundary conditions for this problem. We usually work with point sources in infinite domains when working with spherical geometry, so I am unfamiliar with setting up boundary conditions for finite spheres.


    2. Relevant equations
    The governing differential equation is:
    [itex]
    \frac{1}{r^2} \frac{d}{dr} r^2 \frac{d\phi}{dr} - \frac{1}{L^2}\phi(r) = \frac{-S_0}{D}
    [/itex]

    with a boundary condition:
    [itex]
    \phi(r=R) = 0
    [/itex]
    This problem is going to be turned in to a computer code, so we were told not to use extrapolated boundary conditions. I just need help finding the boundary condition for r = 0

    As I mentioned above, we've never worked with finite spheres before so I am not 100% certain that the solution involves sin and cos. Maybe sinh and cosh?
    [itex]
    \phi(r) = \frac{C_1}{r}sin(\frac{r}{L}) + \frac{C_2}{r}cos(\frac{r}{L}) + \frac{S_0L^2}{D}
    [/itex]


    3. The attempt at a solution
    Physically, I know that the flux profile will be flat at the center of the sphere. I can't impose that the derivative equals zero because that would lead to a zero as the denominator in the coefficient terms. I've considered using a limit as r → 0, but that would cause the coefficient terms to shoot up to infinity. I don't think I can say that the neutron current at the center is equal to some value because there is a distributed source.

    I am completely lost in finding this boundary condition. I think what I have written so far for the equations is right. If someone could point me in the right direction, or enlighten me of a mistake I have made I would be forever grateful.
     
  2. jcsd
  3. Apr 30, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Why do you need a boundary condition for r = 0? Isn't r = 0 within the sphere? Why isn't the BC Phi = 0 at r = R sufficient?

    Obviously, you can substitute your trial function Phi(r) into the differential equation and see if the DE is satisfied. Have you tried this?
     
  4. Apr 30, 2013 #3
    I did check my trial function by plugging it in to the original ODE and it didn't take me long to figure out that it wasn't going to satisfy it. So I dusted off the old Differential equations textbook and found out what I was doing wrong. The actual solution is:

    [itex]
    \phi(r) = \frac{C_1}{r}sinh(\frac{r}{L}) + \frac{C_2}{r}cosh(\frac{r}{L}) + \frac{S_0L^2}{D}
    [/itex]

    with the boundary conditions [itex] \phi(r=R) = 0 [/itex] and [itex] \lim_{r \to 0} \phi(r) = finite [/itex]

    I reconsidered using the limit boundary condition after re-remembering L'hopital's rule. So my flux (with coefficients solved for) is:

    [itex]
    \phi(r) = \frac{-S_0L^2R}{rDsinh(\frac{R}{L})}sinh(\frac{r}{L}) + \frac{S_0L^2}{D}
    [/itex]

    Thank you for suggesting that I check my solution, I often forget to do that when I'm solving differential equations.
     
    Last edited: Apr 30, 2013
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