Neutron Diffusion & Moderation Chapter 5

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Neutron Diffusion & Moderation
Chapter 5
Neutron Diffusion and Moderation
How are neutrons distributed throughout the reactor
core?
This is the fundamental question explaining the design
and subsequent operation of the reactor.
Unfortunately, determining the neutron distribution is a
difficult problem. An approximation can be arrived at
by solving the diffusion equation. This differential
equation describes how high concentration solute
diffuses to regions of lower concentration. This is
similar to how neutrons behave in a reactor.
Neutron Diffusion and Moderation
The diffusion equation is found as the combination of
two laws of physics;
1. Fick’s law of diffusion
2. Equation of continuity
Fick’s Law
Neutron Diffusion and Moderation
= − ∅
J x – The neutron current density is the net # of neutrons that
pass per unit time through a unit area perpendicular to the x
direction. J x has the same units as flux φ. {
}
D – the diffusion coefficient in units of m or cm.
∅()
x
If there is a negative
flux gradient, then
there is a net flow of
neutrons along the
positive x direction.
Neutron Diffusion and Moderation
Fick’s Law
Consider neutrons passing through the plane at x=0
from left to right as the result of collisions to the left of
the plane. Since the concentration of neutrons and the
flux is larger for negative values of x, there are more
collisions per cm 3 on the left. Therefore more neutrons
are scattered from left to right, then the other way
around. Thus the neutrons naturally diffuse toward the
right.
In general the flux and neutron current density are a
function of 3 spatial variables so that the law is
= −∅ = −∅
Fick’s Law
Neutron Diffusion and Moderation
= −∅ = −∅
Recall = =
, , =
+ + ̂
The grad of a scalar f(x, y, z) is a vector, whose
components are equal to the rates of change of f
along the component axes.
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Neutron Diffusion and Moderation
Ex… The flux at a distance r from a point source
emitting S n/sec is given by ∅ = /
where D
and L are constants.
a) Find expression for neutron current density vector J
b) Find expression for # of neutrons flowing out of a
sphere of radius r surrounding the source.
-----------------------------------------------------------
a) Because of the geometry, the current density vector
must point outward in the r radial direction, thus the r
component of the gradient is in spherical
Ex (cont.)
Neutron Diffusion and Moderation
So = −
=
= ∙
(
=
)/
+
b) Area of sphere = 4
so,
⁄
= ∙ = (1 + ⁄
) ⁄
coordinates is =
Neutron Diffusion and Moderation
It must be emphasized that Fick’s law is an
approximation and is not valid under the following
conditions;
1. In a medium that strongly absorbs neutrons.
2. Within ~3 mean free paths of either a neutron
source or a surface of a medium boundary.
3. When the scattering of neutrons is strongly
anisotropic.
To some extent, these limitations are valid in every
practical reactor. Never the less Fick’s law gives a
reasonable approximation. For more detailed
calculations, higher order methods are available.
Neutron Diffusion and Moderation
Equation of Continuity
Is the mathematical statement that the time rate of
change of neutrons in a volume V must be accounted
for by means of absorption and leakage.
[rate of change of n in V] =
[rate of production of n’s in V]
- [rate of absorption of n’s in V]
- [rate of leakage of n’s in V]
Neutron Diffusion and Moderation
Rate of change of neutrons
If n = neutron density, then is the total
number of neutrons in the volume V. The rate of
change is
(,,,)
.
, which can be written
Absorption rate
The rate (per second) at which neutrons are lost by
absorption per ∑ ∅ {units =
}.
= ∑ ∅
∑
Neutron Diffusion and Moderation
Leakage rate
Since the dot product of the current density with the
unit normal (u) to the surface { ∙ u } is the net number
of neutrons passing outward through the surface.
= ∙
This surface integral can be changed to a volume
integral by use of the divergence theorem.
∙
=
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Neutron Diffusion and Moderation
Divergence of a vector field
Given a vector F (x,y,z)
then
= ∙ =
+
+
.
Which is a scalar valued function.
Neutron Diffusion and Moderation
The equation of continuity thus becomes
=
− Σ ∅ −
Since all of the integrals are carried out over the same
volume, their integrands must be equal so,
or
= { − Σ ∅ − }
= S − Σ ∅ −
In steady state, when n is not a function of time,
S − Σ ∅ − = 0
Neutron Diffusion and Moderation
Diffusion Equation
Ficks Law: J = −D∅ . 1
.
= − Σ ∅ − . 2
Neutron Diffusion and Moderation
Diffusion Equation
Where
given by
= div grad is called the Laplacian and is
(, , ) =
+
+
Eq. 2
From Eq. 1
So
= − Σ ϕ −
= − ∅ = − ∅
∅ − Σ ∅ + =
Since we will consider only time-independent problems
The steady-state diffusion eqn. is given by:
∅ − Σ ∅ + = 0
Diffusion Equation
Dividing by D we get:
∅ − Σ ∅
+ / = 0
Define: =
= diffusion area.
And L = diffusion length in meters
L is a measure of the average distance traveled by a
Neutron prior to absorption. It is determined by
the geometry and physical properties of the core.
Ex…
∅ − ∅ + / = 0
5.3 Two point sources each emitting S {n/sec}, are
located a distance 2a cm apart in an isotropic medium
with diffusion constant D. Derive an expression for the
flux φ and current density J at the midpoint P.
-----------------------------------
Solution --- a --- P --- a ---
S1
S2
The flux φ is a scalar so, ∅ = ∅ 1 + ∅(2).
∅ = /
4 + /
4 = /
2
The current densisty J(r) is a vector, and because the
current vectors from each source are equal and
opposite at point P, J(P) = 0.
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Solution to the Diffusion Equation for a point source
Consider a small sphere point source emitting S
neutrons per second into an isotropic medium.
The spherical form of the Laplacian is
=
(
EQ 4.
=
Solution to the Diffusion Equation for a point source
So,
= 0 has solution
= / + /
)+ ….. (see App III of text)
thus
∅ = /
+ /
∅ = /
.
The diffusion equation becomes (for r ≠ 0 )
∅()
− ∅ = 0 EQ. 3
Let = ∅ , then (see aside notes) EQ. 3 becomes
Since physical conditions dictate that φ remain finite as
r increases, then = 0. So that
To find C1 we use the fact that the current density J(r)
through a small sphere (by simple geometry) must be
Solution to the Diffusion Equation for a point source
So, using Fick’s Law = − ∅
, substituting
Evaluating at r ->0 (where J(r) = S)and solving for S yields
= lim
→
4
thus =
4 =
+ 1 / = 4
, and so finally
1
+ 1 / ASIDE
∅() = /
4
Ex… 5.6 A point source emitting 10 7 {n/sec}, is located
in an infinite body of unit density water at room
temperature. What is the flux φ 15cm from the
source?
-----------------------------------
Solution
For a point source the flux φ is ∅ = /
Using table 5.2: L = 2.85 cm, and D = 0.16 cm.
∅ = 10 /.
4 0.16 15 =
1.7210 −
Solution to the Diffusion Equation for a Infinite Plane
Source.
We have an infinite planar source at x = 0, emitting S n/s.
There is no variation in ‘y’ or ‘z’ flux since plane is infinite
in those directions. There are no source except at x = 0.
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Solution to the Diffusion Equation for a Infinite Plane
Source.
The diffusion equation becomes
∅
− 1 ∅ = 0 ≠ 0
S (n/sec)
x = 0
Because of symmetry, we need only solve the equation
in one-half of the plane. Then, by an appropriate
transformation we can get the solution for the other
half.
Solution to the Diffusion Equation for a Infinite Plane
Source.
The general solution is
∅ = / + /
Since ∅′ = ⁄ / , ∅′′ = ⁄ /
So,
0 = ⁄ / + ⁄ /
Considering the right half plane x > 0, then C2 = 0, since
the φ can not increase without bound.
So,
∅ = /
Solution to the Diffusion Equation for a Infinite Plane
Source.
To find C1 we use the source condition, and Ficks Law,
= − ∅
=
/
Imagine a unit area box constructed at the plane, the
net flow of neutrons parallel to the plane source
through the box is 2 J(x). Then at the surface of the
plane, as the box shrinks to x = 0, the net flow of
neutrons out of the box must be equal to S n/s.
Thus,
lim
→
2 ∙ =
Solution to the Diffusion Equation for a Infinite Plane
Source.
So, lim = ⁄ = lim
→
Thus,
→
=
/ =
And, ∅ =
for x >0
and by symmetry,
So,
∅ − =
() for x