Neutron Flux Profile in a Spherical Moderator

  • #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?
 

Answers and Replies

  • #2
15
4
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
 
  • #3
ok
Just explain me what is [itex]\lambda[/itex] equal to? Is it 1/L ?
 
  • #4
15
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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