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Neutron Flux Profile in a Spherical Moderator

  1. Nov 18, 2013 #1
    Hello People
    I need help with the following assignment:
    It states:
    Consider an ideal moderator with zero absorption cross section, Ʃa = 0, and a diffusion coefficient, D, which has a spherical shape with an extrapolated radius, R. If neutron sources emitting S neutrons/cm3sec are distributed uniformly throughout the moderator, the steady neutron diffusion equation is given by,
    D∇2[itex]\phi[/itex] -Ʃa[itex]\phi[/itex]=-S

    a) Simplify the above neutron diffusion equation for this moderator in spherical coordinates and state the appropriate boundary conditions.

    By solving the simplified diffusion equation, obtain the neutron flux profile, [itex]\phi[/itex](r).

    I know I need to divided the neutron diffusion equation and cancel out the absorption cross section and end up with something like:
    2[itex]\phi[/itex] = -S/D
    and the particular solution would be something like S/Ʃa
    but what's the general solution to:
    D∇2[itex]\phi[/itex] =0
    in spherical coordinates?
     
  2. jcsd
  3. Nov 19, 2013 #2
    Israakaizzy,

    I think it should be something like Asinh(λ.r)/r + Bcosh(λ.r)/r, applying the border conditions B=0.

    hope it helps,

    Hernán
     
  4. Nov 20, 2013 #3
    ok
    Just explain me what is [itex]\lambda[/itex] equal to? Is it 1/L ?
     
  5. Nov 21, 2013 #4
    Israakaizzy,

    You are right, it should be 1/L if Ʃa were different than 0.

    I did the maths for the homogeneous part:
    ∇[itex]^{2}[/itex][itex]\phi[/itex]=0

    saying that:

    [itex]\phi[/itex]=[itex]\frac{\widehat{\phi}}{r}[/itex]

    The Lapplacian inspherical coordintates turns:

    ∇[itex]^{2}[/itex][itex]\phi[/itex]=[itex]\frac{∂^{2}\widehat{\phi}}{∂r^{2}}[/itex] + [itex]\frac{2}{r}[/itex][itex]\frac{∂\widehat{\phi}}{∂r}[/itex]

    proposing an exponential solution:

    λ[itex]^{2}[/itex]e[itex]^{λr}[/itex] + [itex]\frac{2}{r}[/itex]λe[itex]^{λr}[/itex] = 0

    So:

    λ= -[itex]\frac{2}{r}[/itex]

    and finally:

    [itex]\phi[/itex]= [itex]\frac{A}{r}[/itex] + B

    Don´t forget that for the inhomogeneous part you have to use the Lapplacian in sphericals.

    Regards,

    Hernán
     
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