Neutron flux in an infinite slab reactor

In this post we derived the Helmholtz equation

$$\begin{equation*} \begin{aligned} \nabla^2\phi + B^2\phi = 0 \end{aligned} \end{equation*}$$

We saw how this can be solved in 2-dimensions. However, in the simple case of a slab reactor of width $a$ in the $x$ direction, and infinite in the $y$ and $z$ directions the Helmholtz equation reduces to 1-dimension:

$$\begin{equation*} \begin{aligned} \frac{\text{d}^2 \phi}{\text{d} x^2} + B^2\phi = 0 \end{aligned} \end{equation*}$$

We can place the slab at $x = 0$ with its width in the interval $\left[-\frac{a}{2}, \frac{a}{2}\right]$. We know that the flux must be symmetric about $x = 0$, and that the flux must be 0 at the extrapolated boundaries, $\pm\frac{a_{ex}}{2}$. The boundary conditions are then

$$\begin{equation*} \begin{aligned} \phi\left(\frac{a_{ex}}{2}\right) = \phi\left(-\frac{a_{ex}}{2}\right) = 0 \end{aligned} \end{equation*}$$

Furthermore, because of the symmetrical geometry there is no net flow of neutrons at the center of the slab. Because the neutron current density is proportional to the derivative of $\phi$, at the center of the slab i.e. at $x = 0$,

$$\begin{equation*} \begin{aligned} \frac{\text{d} \phi}{\text{d} x} = 0 \end{aligned} \end{equation*}$$

Again, because the geometry of the slab is symmetric around $x = 0$ this is equivalent to the condition of $\phi$ being an even function

$$\begin{equation*} \begin{aligned} \phi(-x) = \phi(x) \end{aligned} \end{equation*}$$

The 1-dimensional form of the flux equation has the general solution

$$\begin{equation*} \begin{aligned} \phi(x) = A\cos(Bx) + C\sin(Bx), \end{aligned} \end{equation*}$$

where $A$ and $C$ are unknown constants. Taking the derivative and using the condition that the derivative is 0 at $x = 0$ we get that $C = 0$. The flux equation, therefore, becomes

$$\begin{equation*} \begin{aligned} \phi(x) = A\cos(Bx) \end{aligned} \end{equation*}$$

Applying the boundary condition stated above that the flux at the extrapolated boundaries is 0,

$$\begin{equation*} \begin{aligned} \phi\left(\frac{a_{ex}}{2}\right) = A\cos\left(\frac{Ba_{ex}}{2}\right) = 0, \end{aligned} \end{equation*}$$

gives $A = 0$ with the trivial solution $\phi(x) = 0$, or that

$$\begin{equation*} \begin{aligned} \cos\left(\frac{Ba_{ex}}{2}\right) = 0 \end{aligned} \end{equation*}$$

This solution leads to $B$ taking values $B_n$,

$$\begin{equation*} \begin{aligned} B_n = \frac{n \pi}{a_{ex}}, \end{aligned} \end{equation*}$$

for odd values of $n$. The square of the lowest value of $B$, i.e. $B_1^2$, is the buckling of the reactor, which represents the fuel material in an infinite medium. Under non-critical conditions, the reactor flux is the sum of the cosine functions for all values of $B_n$, i.e.

$$\begin{equation*} \begin{aligned} \phi(x) = \sum_{n \geq}^{} A\cos(B_n x) = \sum_{n \geq 0}^{} A\cos\left(\frac{n\pi x}{a_{ex}}\right) \end{aligned} \end{equation*}$$

However, the only physically possible case is that of a critical reactor where all the cosine functions except the first one die out in time. And so, the flux in a critical slab reactor is given by

$$\begin{equation*} \begin{aligned} \phi(x) = A\cos(B_1 x) = A\cos\left(\frac{\pi x}{a_{ex}}\right) \end{aligned} \end{equation*}$$

Note that the value of the constant $A$, which determines the magnitude of the flux, has not been determined. Because the Helmholtz equation we started with is homogeneous, multiplying the flux by any constant still gives a valid solution. Furthermore, the magnitude of the flux in the reactor is determined by the reactor power rather than its material properties. The recoverable energy per fission in a nuclear reactor, $E_f$, is $~200 \text{ MeV} = 3.2 \times 10^{-11} \text{ J}$. There are, therefore, $\Sigma_f \phi(x)$ fissions per cm$^3$s$^{-1}$ at the point $x$, where $\Sigma_f$ is the macroscopic cross section. And so, the power, $P$ (watts/cm$^2$), is

$$\begin{equation*} \begin{aligned} P = E_f\Sigma_f \int_{-\frac{a}{2}}^{\frac{a}{2}} \phi(x) \text{d} x \end{aligned} \end{equation*}$$

Substituting in and integrating the expression we found for the reactor flux we get

$$\begin{equation*} \begin{aligned} P = \frac{2a_{ex}E_f\Sigma_fA\sin\left(\frac{\pi a}{2a_{ex}}\right)}{\pi} \end{aligned} \end{equation*}$$

Rearranging this we get the constant $A$ as

$$\begin{equation*} \begin{aligned} A = \frac{\pi P}{2aE_f\Sigma_f\sin\left(\frac{\pi a}{2a_{ex}}\right)} \end{aligned} \end{equation*}$$

And so, the thermal flux in a slab reactor can be expressed as

$$\begin{equation*} \begin{aligned} \phi(x) = \frac{\pi P}{2a_{ex}E_f\Sigma_f\sin\left(\frac{\pi a}{2a_{ex}}\right)}\cos\left(\frac{\pi x}{a_{ex}}\right) \end{aligned} \end{equation*}$$

References:

Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.