Slab reactor

The diffusion equation reduces to a one dimensional equation

$$\frac{d\varphi(x,t)}{v\partial t}=D\frac{d^{2}}{dx^{2}}\varph... ...+\varphi (x,t)\Sigma_{a}\left( k_{\infty}-1\right) +S\delta(x)$$ (4.31)

Where we have used a single absorption cross section $\Sigma_{a}$, independent of x, and a plane neutron source at position x=0. At the boundaries $x=\pm\frac{a}{2}$, we require $\varphi(x=\pm\frac{a}{2},t)=0$. It is, therefore, convenient to use a Fourier development of $\varphi$ and $\delta:$
$$\begin{align*}\varphi(x,t) & =\underset{n~odd}{\sum} A_{n}(t)\cos B_{n}x\\ \delta(x) & =\underset{n~odd}{\frac{2}{a}\sum}\cos B_{n}x \end{align*}$$
With $B_{n}=\frac{n\pi}{a}~(n=1,3...)$. The coefficients A_n(t), are obtained by solving the equations:

$$\frac{dA_{n}(t)}{vdt}=\left( -DB_{n}^{2}+\Sigma_{a}\left( k_{\infty }-1\right) \right) A_{n}(t)+2\frac{S}{a}$$ (4.32)

If S=0, the solution is

$$A_{n}(t)=A_{n}(0)e^{\left( k_{\infty}-1-B_{n}^{2}\frac{D}{\Sigma_{a}}\right) \Sigma_{a}vt}$$

For $k_{\infty}<1+B_{1}^{2}\frac{D}{\Sigma_{a}}=1+\frac{\pi^{2}D}{a^{2} \Sigma_{a}}$ all terms vanish exponentially. For $k_{\infty}>1+\frac{\pi^{2} D}{a^{2}\Sigma_{a}},$ the first term, and possibly some other low order ones increase exponentially. The reactor becomes critical for $k_{\infty} =1+\frac{\pi^{2}D}{a^{2}\Sigma_{a}}$; in this case A₁(t) becomes time independent, while higher order terms decrease exponentially. Therefore, the neutron flux distribution becomes time-independent and is a solution of the time independent diffusion equation

$$D\frac{d^{2}}{dx^{2}}\varphi(x,t)+\varphi(x,t)\Sigma_{a}\left... ...^{2}}{dx^{2}}\varphi(x,t)+\frac{\pi^{2}}{a^{2}} \varphi(x,t)=0$$ (4.33)

which has the form

$$\varphi(x)=A_{1}\cos\frac{\pi x}{a}$$

Simple solutions are, also, obtained for spherical and cylindrical reactors