Unformatted text preview: se of this law in reactor theory leads to the diffusion approximation. Let us make the
1. We consider an infinite medium.
2. The cross sections are constants, independent of position, implying a uniform medium.
3. Scattering is isotropic in the Laboratory (LAB) system.
4. The neutron flux is a slowly varying function of the position.
5. We use a one speed system where the neutron density is not a function of time. 6. A steady state system where the neutron density is not a function of time.
7. No fission source in the system.
Some of these assumptions will be later relaxed. For instance, the diffusing medium will be
taken as finite in size rather than infinite.
We shall attempt to calculate the current density at the center of the coordinate system of
Fig. 2. The vector J is given by:
J = J xi + J y ˆ + J z k
j , (8) so that we must evaluate the components J x , J y , J z . Fig. 2: Geometry for the derivation of the neutron current and Fick’s Law.
These net current components can be written in terms of the partial axial currents as: J x = J x+ − J x−
Jy = Jy − Jy J z = J z+ − J z− (9) Let us concentrate on the estimation of one single component: J z crossing the element of
area dS z at the origin of the coordinate system in the negative z direction, as shown in Fig. 2
Every neutrons passing through dS z in the x-y plane comes from a scattering collision.
Neutron scattering above the x-y plane will thus flow downward through dS z .
Consider the volume element:
dV = r 2 sin (θ )drdθ dϕ (10) The number of scattering collisions occurring per unit time in the volume element dV is: Σ sφ ( r )dV = Σ sφ ( r )r 2 sin(θ )drdθ dϕ (11) φ ( r ) is the particle flux,
Σ s is the macroscopic scattering cross section.
Since scattering is isotropic in the LAB system, the fraction arriving to dS z is that subtended
by the solid angle dΩ , given by:
d Ω r 2 dS z cos(θ )
4π r 2 (12) Thus the number of neutrons scattered per unit time in dV reaching dS z after being
attenuated in the medium by the exponential factor e − Σt r is: dN = e −Σt rφ ( r )Σ s r 2 sin...
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