2
$\begingroup$

I am calculating the critical mass (radius) of $U^{235}$ sphere. I want to calculate the mass for three different cases:
1. air/vacuum surrounds the sphere (diffusion coefficient is infinite)
2. ideal reflector surrounds the sphere(diffusion coefficient is zero)
3. diffusion coefficient is the same in the surrounding matter as it is in the sphere.

There is no absorption or fission outside of $U^{235}$ sphere.

The equation i am using is : $$-D\nabla^2\phi + \Sigma_a\phi = \frac{1}{k}\nu\Sigma_f\phi.$$ (D is diffusion coefficient, $\phi$ is neutron flux, $\Sigma$ is macroscopic cross section, k is multiplication factor, $\nu$ is average number of secondary neutrons per fission )
This equation can also be written as $$\nabla^2\phi + B^2\phi = 0, $$ where $B^2 = \frac{1}{L^2}(\frac{k_{\infty}}{k} -1)$, $k = \frac{\nu\Sigma_f}{\Sigma_a}$ and $L^2 = \frac{\Sigma_a}{D}$.
Solution to this equation is: $$\phi(r) = \frac{a*\sin(B*r)}{r}.$$ This solution describes neutron flux in the $U^{235}$ sphere.


My question is about boundary conditions.
Is it OK if I say that $\phi(R)=0$ in the first case and $\frac{\partial\phi}{\partial r}|_{r=R}=0$ in the second (R is radius of the sphere)? In the third case I think i should calculate $-D\nabla^2\phi=0$ for the area outside of a sphere and then use: $$\phi_s|_{r=R} = \phi_o|_{r=R}$$ and $$j_s|_{r=R} = j_o|_{r=R}$$ (s refers to sphere and o to outside, j is neutron flux current). Are these boundary conditions correct for these cases?

$\endgroup$
3
  • $\begingroup$ have you read Serber's book "The Los Alamos Primer"? $\endgroup$ Aug 5, 2018 at 17:34
  • $\begingroup$ @niels I have not. $\endgroup$
    – Polihistor
    Aug 5, 2018 at 19:18
  • $\begingroup$ it may answer many of your questions. $\endgroup$ Aug 5, 2018 at 19:33

2 Answers 2

1
$\begingroup$

The equation ( sin B.r)/r leads to a critical radius of 3.14 x diffusion length and far too large a figure i.e. giving critical mass 200 Kg as opposed to 50 Kg. The reason for this is that the premiss on which the Helmholtz equation was formulated is that the fission material is of infinite length and a small cross-section examined. What the first equation is saying is that the material is just about 3 diffusion lengths and with an abrupt change to nothing. The diffusion equation as it stands cannot describe this situation..ss

$\endgroup$
0
$\begingroup$

Diffusion theory is not accurate for this situation. A rigorous solution requires use of the neutron transport equation. As others have commented, this is addressed in the Los Alamos Primer by Serber in section 10, which you can find online.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.