Diffusion equation and neutron diffusion theory

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Astronuc
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Basically the steady-state diffusion equation can be written in a form

[itex] \nabla^2\phi\,+\,k^2\phi\,=\,S[/itex]

When S = 0, this is just the Helmholtz equation - http://mathworld.wolfram.com/HelmholtzDifferentialEquation.html

See also - http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node107.html [Broken]

To solve it, like any differential equation, one simply applies the boundary conditions.

I will add more later.
 
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  • #2
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What about the 2 group diffusion equation? If you are familiar, is said equation only solvable by computer program?
 
  • #3
Astronuc
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In special cases (e.g. 1-D), there can be an analytical solution for two group diffusion theory, basically solving two simultaneous linear differential equations.

However, for most practical (real-world) cases, the two group diffusion theory requires a numerical solution. Most modern nuclear design codes use a modified 2 group approach. There has been some effort at employing transport theory, but it has proved difficult.
 
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Astronuc
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A brief overview of transport and diffusion theory with respect to neutron propagation.

http://lpsc.in2p3.fr/gpr/PPNPport/node28.html [Broken]

This is not comprehensive - just a summary.

The diffusion equation (second Boltzmann equation) is a special case of the transport equation.
 
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Just in time for my final today! :smile:
 
  • #6
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Weird - I've been solving this all day in cylindrical polars... :smile:

...subject to some nasty conditions :frown:
 
  • #7
Astronuc
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Yep - Bessel's functions are the solutions to Bessel's equation, which is the form of the diffusion equation in cylindrical or polar coordinates for radial dependence.
 
  • #8
Hi, I just started graduate level transport theory (sigh, the transport equation) and I'm having a heck of a time with it. Can anyone recommend any fantastic texts or websites? (I very obviously didn't do my undergrad in NE) Thanks!
 
  • #9
Clausius2
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Astronuc said:
Basically the steady-state diffusion equation can be written in a form

[itex] \nabla^2\phi\,+\,k^2\phi\,=\,S[/itex]

When S = 0, this is just the Helmholtz equation - http://mathworld.wolfram.com/HelmholtzDifferentialEquation.html

See also - http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node107.html [Broken]

To solve it, like any differential equation, one simply applies the boundary conditions.

I will add more later.
Can you explain a little bit more what "k" and "S" mean in the context of neutronic diffusion?.
 
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I think "k" is [tex]\frac {1} {L^2}[/tex] where [tex]L^2[/tex] is diffusion area. I think that makes "S" the source term.

Edit: I am having some problems with latex. Does the above look alright to everyone?
 
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theCandyman said:
I think "k" is [tex]\frac {1} {L^2}[/tex] where [tex]L^2[/tex] is diffusion area. I think that makes "S" the source term.

Edit: I am having some problems with latex. Does the above look alright to everyone?

If you mean the way it displays on white, that is a feature of our installation that showed up the last time we did an upgrade to our main software. It's on the list of problems to be looked at, but other problems have higher priority; people can read this just fine, after all.
 
  • #12
Astronuc
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The diffusion equation is derived from transport theory with several assumptions.

Basically the steady-state neutron diffusion equation can be written as:

[tex]D\,\nabla^2\phi\,-\,\Sigma_a\,\phi\,+\,S\,=\,0[/tex], where

D is the diffusion coefficient, here spatially independent, i.e. constant, and

[itex]\Sigma_a[/itex] is the macroscopic absorption coefficient, and

[itex]\phi[/itex] is the flux.

Buried in here is an assumption that the neutrons are more or less the same energy. If not the case, then one must account for different energy groups and the absorption coefficient becomes a removal coefficient which includes absorption and scattering out of the energy group.

Anyway, the above equation becomes,

[tex]D\,\nabla^2\phi\,-\,\Sigma_a\,\phi\,=\,-S[/tex], and dividing by D

[tex]\nabla^2\phi\,-\,\frac{\Sigma_a}{D}\,\phi\,=\,-\frac{S}{D}[/tex], or

[tex]\nabla^2\phi\,-\,\frac{1}{L^2}\,\phi\,=\,-\frac{S}{D}[/tex],

where [tex]L^2\,=\,\frac{D}{\Sigma_a}[/tex]

Then there is the case where S = a function of the flux [itex]\phi[/itex].
 
  • #13
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Two group slab reactor

In special cases (e.g. 1-D), there can be an analytical solution for two group diffusion theory, basically solving two simultaneous linear differential equations.

However, for most practical (real-world) cases, the two group diffusion theory requires a numerical solution. Most modern nuclear design codes use a modified 2 group approach. There has been some effort at employing transport theory, but it has proved difficult.
Would you say finding the fluxes for a two group is possible analytically? I have to solve the two equations below:

[tex]-D_{2}d^{2}\phi_{1}[/tex]/dx[tex]^{2}[/tex]+[tex]\Sigma_{R1}\phi_{1}[/tex]=1/k(v[tex]_{1}[/tex][tex]\Sigma_{f1}\phi_{1}[/tex]+v[tex]_{2}[/tex][tex]\Sigma_{f2}\phi_{2}[/tex])

[tex]-D_{2}d^{2}\phi_{2}[/tex]/dx[tex]^{2}[/tex]+[tex]\Sigma_{a2}\phi_{2}[/tex]=[tex]\Sigma_{s12}\phi_{1}[/tex]

This is solving a two group diffusion in a slab reactor with plane source at the center of the subcritical slab. So analytical or numerical? Any guess on the method to solve for flux?

Sorry, but v1 and v2 are not squared. They are just v_1 and v_2. Some reason it puts as a square.
 
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  • #14
Astronuc
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Certainly the problem can be solved numerically (FD or FE), and I believe analytically, but I'd have to dig back in my archives for that.

one could write \nu_1 and \nu_2 in thex LaTeX expressions before \Sigma.

I think this is how the equations are supposed to look:

[tex]-{D_1}\frac{{d^2}\phi_1}{dx^2}\,+\,\Sigma_{R1}\phi_1\,=\,\frac{1}{k}({\nu_1}{\Sigma_{f1}\phi_1}\,+\,{\nu_2}{\Sigma_{f2}\phi_2})[/tex]


[tex]-{D_2}\frac{{d^2}\phi_2}{dx^2}\,+\,\Sigma_{a2}\phi_2\,=\,{\Sigma_{s12}}\phi_1[/tex]


Just looking these, one could collect coefficents and rewrite the equations as:

[itex]\phi_1[/itex]'' + A [itex]\phi_1[/itex] = B [itex]\phi_2[/itex]

[itex]\phi_2[/itex]'' + C [itex]\phi_2[/itex] = D [itex]\phi_1[/itex]
 
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  • #15
Astronuc
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[tex]-D_{2}d^{2}\phi_{1}[/tex]/dx[tex]^{2}[/tex]+[tex]\Sigma_{R1}\phi_{1}[/tex]=1/k(v[tex]_{1}[/tex][tex]\Sigma_{f1}\phi_{1}[/tex]+v[tex]_{2}[/tex][tex]\Sigma_{f2}\phi_{2}[/tex])

[tex]-D_{2}d^{2}\phi_{2}[/tex]/dx[tex]^{2}[/tex]+[tex]\Sigma_{a2}\phi_{2}[/tex]=[tex]\Sigma_{s12}\phi_{1}[/tex]

This is solving a two group diffusion in a slab reactor with plane source at the center of the subcritical slab. So analytical or numerical? Any guess on the method to solve for flux?
The equations as given, assuming that [tex]\phi_{1},\,\phi_{2}[/tex] are functions of x, represent a spatially distributed sources terms as opposed to a planar source, which would be a delta-function [itex]S\delta(x)[/itex] at x=0 (with x=0 being the center of the slab). S could be a function of [itex]{\nu_i}{\Sigma_{fi}}{\phi_i(x=0)}[/itex]
 
  • #16
this forum has helped with this challenge problem but I am still having a hard time getting started, the problem is:

"There is a waiting room on the opposite side of a very large wall adjacent to a proton therapy treatment room at the local hospital.

Compute the neutron flux [tex]\phi[/tex](x) into the waiting room using a diffusion theory approximation. Assume the neutrons are emitted from the wall surface via a uniform planar surface source emitting Snot neutrons/cm^2/s. ( At wall surface, x=0)

The diffusion equation is D*(d^2[tex]\phi[/tex]/dx^2)-[tex]\sigma_{a}[/tex]=0 for x not equal to 0.

Assuming a 1-dimensional flux approximation, and other dimensions of the room (y,z) relative to the wall (at x=0) are infinite, use the following conditions:

(i)Flux must remain finite as x---> infinity
(ii)The X-ray all current(Neutron coming out of the wall) has a limit as x-->0, where:
limx-->0 J(x)*i^---->limx-->0(-D(d[tex]\phi[/tex]/dx))=(Snot/2)
(iii) D, [tex]\sigma_{a}[/tex], Snot are all constants"

any help is appreciated!
 

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