Neutron Diffusion and Reactor Flux
Neutron Diffusion and Reactor Flux
Inside a nuclear reactor, neutrons are constantly being produced by fission, scattered by collisions with atomic nuclei, absorbed, and lost through leakage. Tracking the spatial distribution of these neutrons is essential for predicting where power is generated, designing fuel arrangements, and ensuring safe operation. The mathematical framework for this is neutron diffusion theory.
The central quantity is the neutron flux, denoted phi, which represents the number of neutrons passing through a unit area per unit time. Flux is directly proportional to the power density in a reactor — wherever the flux is highest, the most fission is occurring and the most heat is being generated.
In a simple one-dimensional model of a slab reactor, the steady-state neutron flux follows a cosine profile: phi of x equals phi-max times cosine of pi x over the extrapolated width. The flux is highest at the center and drops to zero at the boundaries. For a cylindrical reactor, the radial flux profile involves a Bessel function of the first kind, J-zero.
The diffusion equation that governs neutron behavior is remarkably similar to the heat equation in thermal engineering. The diffusion coefficient D relates to how far neutrons travel between scattering events, and the diffusion length L characterizes how far thermal neutrons travel before being absorbed. These parameters depend on the moderator material — water, graphite, or heavy water — and their optimization is a key part of reactor design.
$$D \nabla^2 \phi - \Sigma_a \phi + S = 0$$
where $D$ = diffusion coefficient (cm), $\Sigma_a$ = macroscopic absorption cross-section (cm$^{-1}$), $\phi$ = neutron flux (n/cm²·s), $S$ = source term.
$$L = \sqrt{\frac{D}{\Sigma_a}}$$
The diffusion length $L$ is the average distance a thermal neutron travels before being absorbed. For water: $L \approx 2.85$ cm. For graphite: $L \approx 50$ cm.
For a homogeneous slab reactor of width $a$ (with extrapolation):
$$\phi(x) = \phi_{\max} \cos\left(\frac{\pi x}{\tilde{a}}\right)$$
where $\tilde{a}$ is the extrapolated width, $x$ is measured from the center. The cosine shape means peak power at the center.
A reactor has $D = 0.84$ cm and $\Sigma_a = 0.12$ cm$^{-1}$. Find the diffusion length and interpret it.
Step 1: Apply the formula.
$$L = \sqrt{\frac{D}{\Sigma_a}} = \sqrt{\frac{0.84}{0.12}} = \sqrt{7.0} \approx 2.65 \text{ cm}$$
A thermal neutron travels on average about 2.65 cm before being absorbed. This is typical for a water-moderated reactor.
A slab reactor has extrapolated width $\tilde{a} = 300$ cm. If $\phi_{\max} = 10^{14}$ n/cm²·s, find the flux at $x = 75$ cm from center.
Step 1: Apply the cosine profile.
$$\phi(75) = 10^{14} \cos\left(\frac{\pi \times 75}{300}\right) = 10^{14} \cos(\pi/4)$$
Step 2: Evaluate.
$$\phi(75) = 10^{14} \times \frac{\sqrt{2}}{2} \approx 7.07 \times 10^{13} \text{ n/cm}^2\text{·s}$$
At 25% of the way from center to edge, the flux drops to about 71% of its maximum.
A point source emits $S_0 = 10^8$ neutrons/s in an infinite moderator with $L = 3$ cm and $D = 1$ cm. Find the flux at $r = 10$ cm.
Step 1: The flux from a point source in an infinite medium is:
$$\phi(r) = \frac{S_0}{4\pi D r} e^{-r/L}$$
Step 2: Substitute values.
$$\phi(10) = \frac{10^8}{4\pi(1)(10)} e^{-10/3} = \frac{10^8}{125.66} \times 0.03567$$
$$\phi(10) \approx 2.84 \times 10^4 \text{ n/cm}^2\text{·s}$$
Practice Problems
Show Answer Key
1. $L = \sqrt{0.84/0.000335} = \sqrt{2507} \approx 50.1$ cm.
2. $\cos(\pi x/200) = 0.5 \Rightarrow \pi x/200 = \pi/3 \Rightarrow x = 200/3 \approx 66.7$ cm.
3. $\phi(100) = 5\times10^{13} \cos(\pi \cdot 100/400) = 5\times10^{13} \cos(\pi/4) \approx 3.54\times10^{13}$.
4. Since neutrons that reach the boundary escape and don't return (vacuum boundary condition), the flux must vanish at the extrapolated boundary.
5. Mean free path $= 1/0.1 = 10$ cm.
6. Graphite has a much longer diffusion length, so neutrons travel farther before absorption. The reactor needs to be much larger to prevent excessive leakage (larger critical size).
7. $\bar{\phi} = \frac{1}{\tilde{a}}\int_{-\tilde{a}/2}^{\tilde{a}/2} \phi_{\max}\cos(\pi x/\tilde{a})\,dx = \phi_{\max} \cdot \frac{2}{\pi} \approx 0.637 \, \phi_{\max}$.
8. At the center, where flux is maximum.
9. $B^2 = (\pi/300)^2 = 1.097 \times 10^{-4}$ cm$^{-2}$.
10. A flatter profile means the fuel at the periphery contributes more power, increasing total power output for the same peak flux. This is achieved with enrichment zoning or reflectors.