Diffusion equation

The first term of the r.h.s. of equation 3.22reads: $-div(\overrightarrow{J} (\overrightarrow{r} ,v,t))=D\nabla^{2}\varphi (\overrightarrow{r} ,v,t)$. The diffusion equation is obtained from the Boltzmann equation when neutrons are assumed to be monocinetic, or, in other words, to belong to a single group. This allows to drop the integration^4.9 over velocities in 3.22, thus

$$\frac{\partial\varphi(\overrightarrow{r} ,t)}{v\partial t}=D\... ...a}^{(j)}(\overrightarrow{r} )\right) +S(\overrightarrow{r} ,t)$$ (4.25)

Note that we have replaced $\Sigma_{T}$ in 3.22 by $\Sigma_{a}$, since diffusion has no effect on the flux in a one group formalism. Using relation 3.17equation 3.27 can be writen:

$$\frac{\partial\varphi(\overrightarrow{r} ,t)}{v\partial t}=D\... ...arrow{r} )\left( k_{\infty}-1\right) +S(\overrightarrow{r} ,t)$$ (4.26)

Equation 3.28 allows to make a few interesting remarks. In an infinite and homogeneous medium, with an evenly distributed neutron source, the equation should not include derivatives of $\varphi(\overrightarrow{r} ,t)$ , since $\varphi(\overrightarrow{r} ,t)$ should be independent of $\overrightarrow{r}$. Thus equation 3.28 simplifies to

$$\frac{\partial\varphi(t)}{v\partial t}=\varphi(t)\underset{j}{\sum}\Sigma _{a}^{(j)}\left( k_{\infty}-1\right) +S(t)$$ (4.27)

Consider first the case that for t>0, S(t)=0 and $\varphi(0)$ is finite. Then equation 3.29 has the solution:

$$\varphi(t)=\varphi(0)e^{v\underset{}{\left( k_{\infty}-1\right) t\underset{j}{\sum}}\Sigma_{a}^{(j)}}$$ (4.28)

which shows that if $k_{\infty}>1$ the flux diverges while it decreases to 0 for $k_{\infty}<1$. It is time independent only if $k_{\infty}=1.$ This condition can never be realized in reality. Rather, in critical reactors, one uses a time dependance of the absorption cross-sections, so that $k_{\infty }$ fluctuates around 1.

Consider, now the case when $k_{\infty}<1$, and S(t)=S₀ is time independent, but positive. The solution of 3.29 for stationary states reads:

$$\varphi=\frac{S_{0}}{(1-k_{\infty})\underset{j}{\sum}\Sigma_{a}^{(j)}}$$ (4.29)

The number of absorption reactions per seconds is, then

$$n_{reac}=\underset{j}{\sum}\Sigma_{a}^{(j)}\frac{S_{0}}{(1-k_... ...nderset{j}{\sum}\Sigma_{a}^{(i)}}=\frac{S_{0}}{(1-k_{\infty})}$$ (4.30)

which agrees with equation 3.13.