Solutions of the Diffusion Equation – Non-multiplying Systems

As was previously discussed the diffusion theory is widely used in core design of the current Pressurized Water Reactors (PWRs) or Boiling Water Reactors (BWRs). This section is not about such calculations, but provides an illustrative insights, how can be the neutron flux distributed in any diffusion medium. In this section we will solve diffusion equations:

solution of diffusion equation - equations

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.

Solution of diffusion equation

Neutron flux distribution in various geometries of non-multiplying system.

Neutron Diffusion - planar source-minLet assume the neutron source (with strength S0) as an infinite plane source (in y-z plane geometry). In this geometry the flux varies so slowly in y and z allowing us to eliminate the y and z derivatives from ∇2. The flux is then a function of x only, and therefore the Laplacian and diffusion equation (outside the source) can be written as (everywhere except x = 0):

solution of diffusion equation - equations2

For x > 0, this diffusion equation has two possible solutions exp(x/L) and exp(-x/L), which give a general solution:

Φ(x) = Aexp(x/L) + Cexp(-x/L)

Note that, B is not usually used as a constant, because B is reserved for a buckling parameter. To determine the coefficients A and C we must apply boundary conditions.

From finite flux condition (0≤ Φ(x) < ∞), that required only reasonable values for the flux, it can be derived, that A must be equal to zero. The term exp(x/L) goes to ∞ as x ➝∞  and therefore cannot be part of a physically acceptable solution for x > 0. The physically acceptable solution for x > 0 must then be:

Φ(x) =  Ce-x/L

where C is a constant that can be determined from source condition at x ➝0.

If S0 is the source strength per unit area of the plane, then the number of neutrons crossing outwards per unit area in the positive x-direction must tend to S0 /2 as x ➝0.

source condition - planar source

So that the solution may be written:
solution of diffusion equation - planar source

Neutron Diffusion - point source-minLet assume the neutron source (with strength S0) as an isotropic point source situated in spherical geometry. This point source is placed at the origin of coordinates. In order to solve the diffusion equation, we have to replace the Laplacian by its spherical form:

laplacian - spherical geometry

We can replace the 3D Laplacian by its one-dimensional spherical form, because there is no dependence on angle (whether polar or azimuthal). The source is assumed to be an isotropic source (there is the spherical symmetry).  The flux is then a function of radius – r only, and therefore the diffusion equation (outside the source) can be written as (everywhere except r = 0):

solution of diffusion equation - spherical geometry

If we make the substitution Φ(r) = 1/r ψ(r), the equation simplifies to:

substitution - diffusion

For r > 0, this differential equation has two possible solutions exp(r/L) and exp(-r/L), which give a general solution:

solution of diffusion equation - point source

Note that, B is not usually used as a constant, because B is reserved for a buckling parameter. To determine the coefficients A and C we must apply boundary conditions.

To find constants A and C we use the identical procedure as in the case of infinite planar source. From finite flux condition (0≤ Φ(r) < ∞), that required only reasonable values for the flux, it can be derived, that A must be equal to zero. The term exp(r/L)/r goes to ∞ as r ➝∞  and therefore cannot be part of a physically acceptable solution for r > 0. The physically acceptable solution for r > 0 must then be:

Φ(r) =  Ce-r/L/r

where C is a constant that can be determined from source condition at x ➝0.

If S0 is the source strength, then the number of neutrons crossing a sphere outwards in the positive r-direction must tend to S0 as r ➝0.

source condition - point source

So that the solution may be written:

solution of diffusion equation - point source2

Neutron Diffusion - line source-minLet assume the neutron source (with strength S0) as an isotropic line source situated in an infinite cylindrical geometry. This line source is placed at r = 0. In order to solve the diffusion equation, we have to replace the Laplacian by its cylindrical form:

laplacian - cylindrical geometry

Since there is no dependence on angle Θ and z-coordinate, we can replace the 3D Laplacian by its one-dimensional form, and we can solve the problem only in radial direction. The source is assumed to be an isotropic source. Since the flux is a function of radius – r only, the diffusion equation (outside the source) can be written as (everywhere except r = 0):

solution of diffusion equation - cylindrical geometry
modified bessel functions - first and second kindThis differential equation 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 modified Bessel functions (or occasionally the hyperbolic Bessel functions) of the first and second kind, Iα(x) and Kα(x) respectively.

For r > 0, this differential equation has two possible solutions I0(r/L) and K0(r/L), the modified Bessel functions of order zero, which give a general solution:

solution of diffusion equation - bessel functions

To find constants A and C we use the identical procedure as in the case of infinite planar source. From finite flux condition (0≤ Φ(r) < ∞), that required only reasonable values for the flux, it can be derived, that A must be equal to zero. The term I0(r/L) goes to ∞ as r ➝∞  and therefore cannot be part of a physically acceptable solution for r > 0. The physically acceptable solution for r > 0 must then be:

Φ(r) =  C.K0(r/L)

where C is a constant that can be determined from source condition at r ➝0.

If S0 is the source strength, then the number of neutrons crossing a cylinder outwards in the positive r-direction must tend to S0 as r ➝0.

So that the solution may be written:

solution of diffusion equation - line source

A diffusion environment may consist of various zones of different composition. The consequence of this is that the diffusion coefficient, absorption macroscopic cross-section, and therefore, the neutron flux distribution, will vary per zone. For the determination of the flux distribution in various zones, the diffusion equations in zone 1 and zone 2 need to be solved:

diffusion equation - two different media

where a is the real width of zone 1 and b the outer dimension of the diffusion environment including the extrapolated distance. With problems involving two different diffusion media, the interface boundary conditions play crucial role and must be satisfied:

At interfaces between two different diffusion media (such as between the reactor core and the neutron reflector), on physical grounds the neutron flux and the normal component of the neutron current must be continuous. In other words, φ and J are not allowed to show a jump.

1., 2. Interface Conditions

interface condition - equations

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. Since the solution of these two diffusion equations requires four boundary conditions, we have to use two boundary conditions more.

3. Finite Flux Condition

The solution must be finite in those regions where the equation is valid, except perhaps at artificial singular points of a source distribution.  This boundary condition can be written mathematically as:

finite flux condition - equation4. Source Condition

The presence of the neutron source can be used as a boundary condition, because it is necessary that all neutrons flowing through bounding area of the source must come from the neutron source. This boundary condition depends on the source geometry and for planar sourve can be written mathematically as:

source condition - planar source2

For x > 0, these diffusion equations have the following appropriate solutions:

Φ1(x) = A1exp(x/L1) + C1exp(-x/L1)

and

Φ2(x) = A2exp(x/L2) + C2exp(-x/L2)

where the four constants must be determined with use of the four boundary conditions. The typical neutron flux distribution in a simple two-region diffusion problem is shown at the picture below.

diffusion equation - two media

Solution of diffusion equation in two different non-multiplying diffusion media

Nuclear and Reactor Physics:

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Advanced Reactor Physics:

  1. K. O. Ott, W. A. Bezella, Introductory Nuclear Reactor Statics, American Nuclear Society, Revised edition (1989), 1989, ISBN: 0-894-48033-2.
  2. K. O. Ott, R. J. Neuhold, Introductory Nuclear Reactor Dynamics, American Nuclear Society, 1985, ISBN: 0-894-48029-4.
  3. D. L. Hetrick, Dynamics of Nuclear Reactors, American Nuclear Society, 1993, ISBN: 0-894-48453-2. 
  4. E. E. Lewis, W. F. Miller, Computational Methods of Neutron Transport, American Nuclear Society, 1993, ISBN: 0-894-48452-4.

See above: