Solving Square Box Reactor Question: Flux General Equation

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The flux general equation for a square box is defined as flux = A cos(πx/a) cos(πy/a) cos(πz/a), where 'a' represents the side length of the box. This equation is valid within the boundaries -a/2 < x < a/2, -a/2 < y < a/2, and -a/2 < z < a/2, assuming zero flux at the edges. The solution also incorporates the neutron diffusion equation, leading to flux(x,y,z) = S / (D[3π²/a² + 1/L²]) under steady-state conditions without a source. The general solution for steady-state conditions is expressed as flux(x,y,z) = A exp(-x/a) + B exp(x/a).

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Iam trying to work out question related to sequare box, my question is what the flux general equation for the sequare box?

I work it out and found it is

flux=Acos(pi*x/a)cos(pi*y/a)cos(pi*z/a)

where a is side lenght, am i right
 
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matt222 said:
Iam trying to work out question related to sequare box, my question is what the flux general equation for the sequare box?

I work it out and found it is

flux=Acos(pi*x/a)cos(pi*y/a)cos(pi*z/a)

where a is side length, am i right
Yes - with -a/2 < x < a/2, and similarly for y and z.

This assumes zero flux boundary condition at the edges (flux(-a/2) = flux(a/2) = 0), and max flux at flux(x,y,z) = flux(0,0,0) = A and even symmetry.

Flux is then described by 3 independent functions X, Y, Z which are described by X'' + (pi/a)2X = 0 and X(0) = Y(0) = Z(0) = A1/3
 
do you agree with me when we use neutron diffusion equation, that the general solution will be

flux(x,y,z)=S/(D[3pi^2/a^2+1/L^2])


and what about if we have no source and steady state? How the general equation going to be, should we take it as

flux(x,y,z)=Aexp(-x/a)+Bexp(x/a)
 

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