The Diffusion Equation
In previous chapters we introduced two bases for the derivation of the diffusion equation:
which states that neutrons diffuses from high concentration (high flux) to low concentration.
which states, that rate of change of neutron density = production rate – absorption rate – leakage rate.
We return now to the neutron balance equation and substitute the neutron current density vector by J = -D∇Ф. Assuming that ∇.∇ = ∇2 = Δ (therefore div J = -D div (∇Ф) = -DΔФ) we obtain the diffusion equation.
See also: Diffusion Coefficient
See also: Neutron Cross-section
See also: Neutron Flux DensityThe derivation of diffusion equation is based on Fick’s law which is derived under many assumptions. The diffusion equation can, therefore, not be exact or valid at places with strongly differing diffusion coefficients or in strongly absorbing media. This implies that the diffusion theory may show deviations from a more accurate solution of the transport equation in the proximity of external neutron sinks, sources and media interfaces.
Physical Interpretation of Fick’s Law[xyz-ihs snippet=”diffusion”]
The physical interpretation is similar to fluxes of gases. The neutrons exhibit a net flow in the direction of least density. This is a natural consequence of greater collision densities at positions of greater neutron densities.
Consider neutrons passing through the plane at x=0 from left to right as the result of collisions to the left of the plane. Since the concentration of neutrons and the flux is larger for negative values of x, there are more collisions per cubic centimeter on the left. Therefore more neutrons are scattered from left to right, then the other way around. Thus the neutrons naturally diffuse toward the right.
To solve the diffusion equation, which is a second-order partial differential equation throughout the reactor volume, it is necessary to specify certain boundary conditions. It is very dependent on the complexity of certain problem. One-dimensional problems solutions of diffusion equation contain two arbitrary constants. Therefore, in order to solve one-dimensional one-group diffusion equation, we need two boundary conditions to determine these coefficients. The most convenient boundary conditions are summarized in following few points:
Solutions of the Diffusion Equation – Non-multiplying Systems
in various geometries that satisfy the boundary conditions discussed in the previous section.
We will start with simple systems and increase complexity gradually. The most important assumption is that all neutrons are lumped into a single energy group, they are emitted and diffuse at thermal energy (0.025 eV).
In the first section, we will deal with neutron diffusion in non-multiplying system, i.e., in system where fissile isotopes are missing and therefore the fission cross-section is zero. The neutrons are emitted by external neutron source. We will assume that the system is uniform outside the source, i.e. D and Σa are constants.
Solutions of the Diffusion Equation – Multiplying Systems
in various geometries that satisfy the boundary conditions. In this equation ν is number of neutrons emitted in fission and Σf is macroscopic cross-section of fission reaction. Ф denotes a reaction rate. For example a fission of 235U by thermal neutron yields 2.43 neutrons.
It must be noted that we will solve the diffusion equation without any external source. This is very important, because such equation is a linear homogeneous equation in the flux. Therefore if we find one solution of the equation, then any multiple is also a solution. This means that the absolute value of the neutron flux cannot possibly be deduced from the diffusion equation. This is totally different from problems with external sources, which determine the absolute value of the neutron flux.
We will start with simple systems (planar) and increase complexity gradually. The most important assumption is that all neutrons are lumped into a single energy group, they are emitted and diffuse at thermal energy (0.025 eV). Solutions of diffusion equations in this case provides an illustrative insights, how can be the neutron flux distributed in a reactor core.
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