## Solutions of the Diffusion Equation – Multiplying Systems

**non-multiplying**. In non-multiplying environment neutrons are emitted by a neutron source situated in the center of coordinate system and then they freely diffuse through media. We are now prepared to consider

**neutron diffusion**in

**multiplying system**, which contains fissionable nuclei (i.e.

**Σ**

_{f }**≠ 0**). In this multiplying system we will also study spatial distribution of neutrons, but in contrast to non-multiplying environment these neutrons can trigger

**nuclear fission reaction**. [/su_accordion]

**diffusion equation**

in various geometries that satisfy the **boundary conditions**. In this equation** ν** is number of neutrons emitted in fission and **Σ _{f}** is macroscopic cross-section of fission reaction.

**Ф**denotes a

**reaction rate**. For example a fission of

^{235}U 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 **v ^{2}** and

**к**, sum of these constants must be equal to

^{2}**B**. Now we can

_{g}^{2}= v^{2}+ к^{2}**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, **J_{α}(x) and Y_{α}(x) respectively.

For r > 0, this differential equation has two possible solutions **J _{0}(vr)** and

**Y**, the Bessel functions of order zero, which give a general solution:

_{0}(vr)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 **Y _{0}(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) = AJ _{0}(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 **R _{e} = R + d** and this condition can be written as

**Φ(R + d) = Φ(R**.

_{e}) = 0Therefore, the solution must be **R(R _{e}) = A J_{0}(vR_{e}) = 0**. The function of

**J**has several zeroes, the first is at

_{0}(r)**r**, and the second at r

_{1}= 2.405_{2}= 5.6. However, because the neutron flux cannot have regions of negative values, the only physically acceptable value for

**v**is

**2.405/R**. The solution of the diffusion equation is:

_{e}**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 **B _{g}^{2} **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 10**^{13 }**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 (k_{eff}) 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).