Derivation of neutron flux equation

[Oxlade]

Homework Statement

reactor consists of nested spheres (sorry, this is my first time posting):

Sphere 1 --> Lead target with radius of R0 (Target-region)
Sphere 2 --> Actinide Fuel surrounding lead target; has outer radius of R1 (A-region)
Sphere 3 --> Reflector surrounding the fuel with outer radius of R2 (E-region)
Neutron flux vanishes at radius R2

Derive the neutron flux equation in A-region

Homework Equations

the neutron flux at A-region is given as:

Φ(r) = S/4πD [1/r - 1/R2]

The Attempt at a Solution

I've attempted this solution assuming the lead target as a point source so I get Φ(r) = B e^-r/L
where I determine the constant 'B' using neutron current going to zero at radius R2, but I always end up with a final answer containing exponentials...I'm really stuck even though this was supposed to be a very simple derivation we were to get to work on in the class hour (prof made it a take home aspect)

 

[nilesh_pat]

A:I'm not sure what you mean by 'a point source' but the equation you gave for the flux is correct.To derive it, start with the diffusion equation$$\nabla^2 \phi = - \frac{S}{D}$$where $\phi$ is the neutron flux and $S$ is the source term. Then, assume spherical symmetry so that the diffusion equation reduces to$$\frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial \phi}{\partial r} \right) = -\frac{S}{D}$$You can solve this equation using the method of separation of variables, which will yield a solution of the form$$ \phi(r) = C_1 + \frac{C_2}{r} + \frac{S}{4 \pi D} \ln r$$The constants $C_1$ and $C_2$ can be determined by applying boundary conditions. At $r=R_2$, the flux goes to zero, so we have$$ 0 = C_1 + \frac{C_2}{R_2} + \frac{S}{4 \pi D} \ln R_2$$At $r=R_1$, the flux is given by$$ \phi(R_1) = C_1 + \frac{C_2}{R_1} + \frac{S}{4 \pi D} \ln R_1$$Plugging these two equations into each other then gives$$ \phi(R_1) = \frac{S}{4 \pi D} \left[ \ln R_2 - \ln R_1 \right]$$Substituting this back into the original solution for the flux yields$$ \phi(r) = \frac{S}{4 \pi D} \left[ \frac{1}{r} - \frac{1}{R_2} \right]$$This is the solution you wrote down. Note that this assumes that $C_1=0$. If $C_1 \neq 0$, then the first term in the equation would be $C_1 + \ldots$.