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.
Solution for the Finite Cylindrical Reactor
Let assume a uniform reactor (multiplying system) in the shape of a cylinder of physical radius R and height H. This finite cylindrical reactor is situated in cylindrical geometry at the origin of coordinates. In order to solve the diffusion equation, we have to replace the Laplacian by its cylindrical form:
Since there is no dependence on angle Θ, we can replace the 3D Laplacian by its two-dimensional form, and we can solve the problem in radial and axial directions. Since the flux is a function of radius – r and height – z only (Φ(r,z)), the diffusion equation can be written as:
The solution of this diffusion equation is based on use of the separation-of-variables technique, therefore:
where R(r) and Z(z) are functions to be determined. Substituting this into the diffusion equation and dividing by R(r)Z(z), we obtain:
Because the first term depends only on r, and the second only on z, both terms must be constants for the equation to have a solution. Suppose we take the constants to be v2 and к2, sum of these constants must be equal to Bg2 = v2 + к2. Now we can separate variables, and the neutron flux must satisfy the following differential equations:
Solution for radial direction
The differential equation for radial direction is called the Bessel’s equation and it is well known to mathematicians. In this case, the solutions to the Bessel’s equation are called the Bessel functions of the first and second kind, Jα(x) and Yα(x) respectively.
For r > 0, this differential equation has two possible solutions J0(vr) and Y0(vr), the Bessel functions of order zero, which give a general solution:
From finite flux condition (0≤ Φ(r) < ∞), that required only reasonable values for the flux, it can be derived, that C must be equal to zero. The term Y0(vr) goes to -∞ as r ➝0 and therefore cannot be part of a physically acceptable solution. The physically acceptable solution for must then be:
R(r) = AJ0(vr)
The vacuum boundary condition requires the relative neutron flux near the boundary to have a slope of -1/d, i.e., the flux would extrapolate linearly to 0 at a distance d beyond the boundary. This zero flux boundary condition is more straightforward and is can be written mathematically as:
If d is not negligible, physical dimensions of the reactor are increased by d and extrapolated boundary is formulated with dimension Re = R + d and this condition can be written as Φ(R + d) = Φ(Re) = 0.
Therefore, the solution must be R(Re) = A J0(vRe) = 0. The function of J0(r) has several zeroes, the first is at r1 = 2.405, and the second at r2 = 5.6. However, because the neutron flux cannot have regions of negative values, the only physically acceptable value for v is 2.405/Re. The solution of the diffusion equation is:
Solution for axial direction
The solution for axial direction have been solved in previous sections (Infinite Slab Reactor) and therefore it has the same solution. The solution of in axial direction is:
Solution for radial and axial directions
The full solution for the neutron flux distribution in the finite cylindrical reactor is therefore:
where Bg2 is the total geometrical buckling.
It must be added the constants A and C cannot be obtained from the diffusion equation, because these constant give the absolute value of neutron flux. In fact the neutron flux can have any value and the critical reactor can operate at any power level. It should be noted this flux shape is only in theoretical case in a uniform homogeneous cylindrical reactor at low power levels (at “zero power criticality”).
In power reactor core, the neutron flux can reach, for example, about 3.11 x 1013 neutrons.cm-2.s-1, but this values depends significantly on many parameters (type of fuel, fuel burnup, fuel enrichment, position in fuel pattern, etc.).
The power level does not influence the criticality (keff) of a power reactor unless thermal reactivity feedbacks act (operation of a power reactor without reactivity feedbacks is between 10E-8% – 1% of rated power).