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Neutron flux in finite medium.

  1. Mar 4, 2006 #1
    In the text I use for class, the examples and derivations for functions showing the neutron flux at some point, are all about sources within infinite mediums. Now I have a probelm where I must show that neutron flux, for a point source within a finite sphere, is found by the following equation.

    \phi (r)= \frac{S}{4 \pi D sinh \frac{R + d}{L}} \frac{sinh( \frac{R + d - r}{L} ) }{r}

    If anyone can tell me where I could learn about flux in finite mediums, I would appreciate it.
    Last edited: Mar 5, 2006
  2. jcsd
  3. Mar 4, 2006 #2
    I believe I have solved this on my own. I found an example dealing with a infinite plane source within a slab.

    For reference for anyone else who may have the same question in the future, I will try to compile the answer into this thread later - it is a bit lengthy.

    Edit: I am a bit unsure of the hyperbolic sine in the numerator. In the book it is written like this:

    [tex]sinh \frac{1}{L} (R+d-r)[/tex]

    If that (R+d-r) term is not in the function, I have done the problem icorrectly. Does anyone know which is the correct formula?
    Last edited: Mar 5, 2006
  4. Mar 5, 2006 #3


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    What text are you using?

    Two classic texts are those by John Lamarsh, "Introduction to Nuclear Engineering" and "Nuclear Reactor Theory".

    Sinh is appropriate for an infinite planar source in an infinit slab, and I am trying to remember if A sinh (kr)/r or A sin (kr)/r is appropriate for a sphere. Basically, at r=0, the solution must be finite, and the flux is taken as zero at the extrapolated boundary.

    Basically, one is solving the diffusion equation in one, two or 3D in a finite system. The complete solution is determined by the boundary conditions. Unlike the inifinite system, where the only loss of neutrons is due to absorption, the finite system must deal with 'leakeage' of neutrons across physical boundaries.

    If you write the form of the diffusion equation and the boundary conditions you used, then we can discuss it.
    Last edited: Mar 5, 2006
  5. Mar 5, 2006 #4
    I am using the third edition of Introduction to Nuclear Engineering by Lamarsh.

    Here's my work:

    [tex]\phi = A \frac{e^{-\frac{r}{L}}}{r} + C \frac{e^{-\frac{r}{L}}}{r}[/tex]
    The general solution of point source within a medium flux, A and C are constants.

    [tex]\phi (R+d) = A \frac{e^{-\frac{R+d}{L}}}{R+d} + C \frac{e^{-\frac{R+d}{L}}}{R+d} = 0[/tex]
    Boundary equation.

    [tex]C = -Ae^{-\frac{2(R+d)}{L}}[/tex]
    Solving for the constant C.

    [tex]\phi = \frac{A}{r}(e^{-\frac{r}{L}} - e^{(\frac{r}{L}-\frac{2(R+d)}{L})})[/tex]
    Substituting in the value for C.

    [tex]J = -D \frac{d\phi}{dr}[/tex]
    [tex]r^2 J(r) = \frac{DA}{ L}(1 - e^{-\frac{2(R+d)}{L}})[/tex]
    Note: I took the limit as r goes to zero, so e^-(r/L) goes to 1.

    [tex]\lim_{r\rightarrow 0}r^2 J(r) = \frac{S}{4\pi}[/tex]
    This is the limit for the left hand side of the above.

    [tex]A = \frac{SL}{4\pi D}(1 - e^{-\frac{2(R+d)}{L}})^{-1}[/tex]
    Solving for constant A.

    [tex]\phi = \frac{SL}{4\pi D r} \frac{e^{-\frac{r}{L}} - e^{(\frac{r}{L}-\frac{2(R+d)}{L}})}{1 - e^{-\frac{2(R+d)}{L}}}[/tex]
    Equation after substituting in A.

    [tex]\phi = \frac{SL}{4\pi D r} \frac{e^\frac{R+d-r}{L} - e^{-\frac{R+d-r}{L}}}{e^{\frac{R+d}{L}} - e^{-\frac{R+d}{L}}}[/tex]
    Simplifying by multiplying numerator and denomator by e^[(R+d)/L]

    [tex]\phi = \frac{SL}{4 \pi D r} \frac{sinh\frac{R+d-r}{L}}{sinh\frac{R+d}{L}}[/tex]

    That is quite a bit of Latex coding, I will be fixing it if I there is a problem.
    Last edited: Mar 5, 2006
  6. Mar 5, 2006 #5


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    That looks right.

    Importantly sinh (kr)/r -> 1 as r -> 0.
  7. Nov 9, 2009 #6
    I have same exercise, but for the line source in a finite medium?
  8. Nov 9, 2009 #7
    I'm assuming you mean an infinite line source and aren't interested in the ends. It will be a very similar derivation, but you will be dealing with polar instead of spherical coordinates.
  9. Mar 15, 2010 #8
    theCandyman...you have an extra L in your solution...
  10. Apr 5, 2011 #9
    This is a more complete solution:

    Attached Files:

  11. Apr 1, 2012 #10
    So how would the complete solution change if a second finite moderating sphere surrounds the existing sphere and point source? Thanks in advance.
  12. Apr 3, 2012 #11
    Dunderstadt & Hamilton's Nuclear Reactor Analysis Chapter 5 discusses finite reactors. It gives the Geometric Bucking term for a bare sphere as [itex]\left(\frac{\pi}{\widetilde{R}}\right)^2[/itex] and the flux profile as [itex]r^{-1}sin\left(\frac{\pi r}{\widetilde{R}}\right)[/itex]
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