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Unformatted text preview: ∂⎞
1 ∂⎛ 2 ∂⎞
1
⎜ sin(θ )
⎟+ 2
⎜r
⎟+ 2
∂θ ⎠ r sin 2 (θ ) ∂ϕ 2
r 2 ∂r ⎝ ∂r ⎠ r sin(θ ) ∂θ ⎝ (31) In case of symmetry around the angular variables θ and ϕ , these equations simplify in the one
dimensional case into: ∇2 ≡
Where: 1 d⎛ α d⎞
⎜r
⎟
r α dr ⎝ dr ⎠ α = 0, for the cartesian coordinates,
α = 1, for the cylindrical coordinates,
α = 2, for the spherical coordinates, and the partial derivative has been replaced by the total derivative.
When the flux is not a function of time, we use the steady state diffusion equation, or the
scalar Helmholtz equation; D∇2φ − Σa φ + S = 0 (32) Which is a partial differential equation of the elliptic type.
The Helmholtz equation can be written as: ∇2φ ( r ) −
Where L2 = 1
S (r)
φ (r) = −
2
L
D D
, L is the diffusion length.
Σa (33) If on the other hand, the neutron density and flux are independent of position, we can write
from Eqs. 28, considering div J = 0 : 1 d φ (t )
= S (t ) − Σ aφ (t )
v dt (34) which is a time dependent equation in the flux. 7. BOUNDARY CONDITIONS FOR THE STEADYSTATE DIFFUSION
EQUATION
Mathematically, for the Helmholtz equation the following boundary conditions are needed:
∂φ
, or a linear combination of the two must be specified.
either φ or the normal derivative
∂n
∂φ
cannot be specified independently. Normally, boundary conditions are
Both φ and
∂n
specified based on physical arguments that do not violate this condition.
1. Vacuum Boundary Conditions:
The mean free path of neutron in air is much larger than in the reactor, so that it is possible
to treat it as a vacuum in reactor calculations. If we consider no neutrons reflected from the
vacuum back to the reactor core (Fig. 3), then we can write the equivalent to Eq. 17:
J − ( x) = 1 Σs
1 Σ s ∂φ
φ0 +
=0
4 Σt
6 Σ t 2 ∂x 0 (35) Recalling that from the Fick’s law derivation:
D= Σs
3Σ t We can write:
J − ( x) = From which: 2 1 Σs
D ∂φ
φ0 +
=0
4 Σt
2 ∂x 0 1 ∂φ
1 Σs
=−
φ0 ∂x 0
2 D Σt (36) (37) If at the boundary the material is mostly a scatterer (Σ a << Σ t ) , then
substitute for D = λtr
3 Σs Σt ≈ 1 , and we can : 1 ∂φ
1
31
≈−
≈−
φ0 ∂x 0
2D
2 λtr (38) However from the geometry of Fig. 3, if the diffusion theory is linearly extrapolated, we can
write: φ0
d = tan(θ ) = − ∂φ
∂x (39)
0 where d is the “extrapolated length”.
Comparing Eqs. 38 and 39, we get:
1 ∂φ
31
1
≈−
=−
φ0 ...
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This note was uploaded on 02/28/2013 for the course NUC 150 taught by Professor N/a during the Spring '13 term at Berkeley.
 Spring '13
 N/A

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