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:
The diffusion equation is mostly solved in media with high densities, such as neutron moderators (H2O, D2O, or graphite). The problem is usually bounded by air. The mean free path of the neutron in the air is much larger than in the moderator so that it is possible to treat it as a vacuum in neutron flux distribution calculations. The vacuum boundary condition supposes that no neutrons are entering a surface.
If we consider that no neutrons are reflected from the vacuum back to the volume, the following condition can be derived from Fick’s law:
Where d ≈ ⅔ λtr is known as the extrapolated length, for homogeneous, weakly absorbing media, an exact solution of the mono-energetic transport equation in this case yields d ≈ 0.7104 λtr. The geometric interpretation of the previous equation is that the relative neutron flux near the boundary has 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 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.
It may seem the flux goes to 0 at an extrapolated length beyond the boundary. This interpretation is not correct. The flux cannot go to zero in a vacuum because there are no absorbers to absorb the neutrons. The flux only appears to be heading to the zero value at the extrapolation point.
Note that the equation d ≈ 0.7104 λtr is applicable to plane boundaries only. The formulas for curved boundaries can differ slightly. However, the difference is small unless the radius of curvature of the boundary is of the same order of magnitude as the extrapolated length.
Typical values of the extrapolated length:
The most common moderators have following diffusion coefficients (for thermal neutrons):
D(H2O) = 0.142 cm
D(D2O) = 0.84 cm
D(Be) = 0.416 cm
D(C) = 0.916 cm
The thermal neutron extrapolated lengths are given by:
d ≈ 0.7104 λtr = 0.7104 x 3 x D
H2O: d ≈ 0.30 cm
D2O: d ≈ 1.79 cm
Be: d ≈ 0.88 cm
C: d ≈ 1.95 cm
As can be seen, this approximation is valid when the dimension L of the diffusing medium is much larger than the extrapolated length, L >> d.
These conditions are often used to eliminate unnecessary functions from solutions.
It must be added, as J must be continuous, the flux gradient will show a jump if the diffusion coefficients in both media differ from each other.
On this special interface, we shall apply an albedo boundary condition to represent the neutron reflector. Albedo, the latin word for “whiteness”, was defined by Lambert as the fraction of the incident light reflected diffusely by a surface.
In reactor engineering, albedo, or the reflection coefficient, is defined as the ratio of exiting to entering neutrons, and we can express it in terms of neutron currents as:
For sufficiently thick reflectors, it can be derived that albedo becomes
where Drefl is the diffusion coefficient in the reflector, and the Lrefl is the reflection’s diffusion length.
If we are not interested in the neutron flux distribution in the reflector (let say in the slab B) but only in the effect of the reflector on the neutron flux distribution in the medium (let say in slab A), the albedo of the reflector can be used as a boundary condition for the diffusion equation solution. This boundary condition is similar to the vacuum boundary condition, i.e., Φ(Ralbedo) = 0, where Ralbedo = R + de and
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