Neutron Transport Theory: The diffusion approximation

Jonathan Evans

Department of Mathematical SciencesUniversity of Bath, UK

March 26 2015

Web pages: people.bath.ac.uk/masjde/teaching people.bath.ac.uk/masigg/ITT2

Outline

1. Modelling Process

2. Neutron Transport

1. Problem Formulation

2. Non-dimensionalisation

3. Asymptotic Analysis - (limit of small mean free path)

Modelling Process: Overview

Physical Description Mathematical Description

Mathematical ModelMathematical Problem

Statement

Physical ModelIdentification & Statement

of physical problem/process

Answer Original Questions of Physical Problem

Mathematical Results

Interpretation

Interpretation

AnalysisRefinement

Modelling Process: Analysis

Mathematical Statementof Problem

Dimensionless Model +

Dimensionless Parameters

Perturbation Analysis

Numerical Analysis on Full Problem

ModelRefinement

Numerical Analysison Reduced Problem

ProblemSolution

Modelling Process: Research Problems

Research Problems

Type 1 Novel Mathematics

New Techniques

TechnologyTransfer

Type 2Known Mathematics

Novel Aspects

TechnologyTransfer

MathematicalModel

Application

Area 1

Application

Area 2Application

Area N

Neutron Transport - Introduction

• References:

G.J. Habetler & B.J. Matkowsky, Uniform asymptotic expansions in tranport theory with small mean free paths and the diffustion approximation, J. Math. Phys. 16 (1975) 846-854.

K.M. Case and P.F. Zweifel, Linear Transport Theory, Addison-Wesley, (1967)

• Neutron Motion - Boltzmann Transport Equation - Linear Integro-differential Equation

• Diffusion Theory - Common Approximation - Neutron Flux = Fick’s law - Good working results

• Objective: Derive Diffusion Equation + B.C.s

Neutron Transport - Problem Formulation

• Neutron conservation statement: @N

@t+r ·H = q

1

4⇡

Z

S2v�s(r,⌦ ·⌦0)N(r,⌦0, t) dS(⌦0)+

⌫(r)

4⇡v�f (r)

Z

S2N(r,⌦0, t) dS(⌦0) Q(r, t)++

N(r,⌦, t) (angular) neutron density

(number per unit volume)

•

H = Nv v = v⌦ v speed (constant)

⌦ direction

Flux

r ·H = v⌦ ·rN

•

�(r) = �f (r) + �c(r) +1

4⇡

Z

S2�s(r,⌦ ·⌦0) dS(⌦0)

q = �v�(r)N(r,⌦, t)Source•

Total macroscopic cross-section

Neutron Transport - Problem Formulation

c(r) =�s(r) + ⌫�f (r)

�(r)

Isotropic scattering �s(r,⌦.⌦0) = �s(r)

mean number secondary neutrons

1

v

@ (r,⌦, t)

@t+⌦ ·r (r,⌦, t) + �(r) (r,⌦, t)

=�(r)c(r)

4⇡

Z

S2 (r,⌦0, t) dS(⌦0) Q(r, t)+

•

•

•

Neutron Transport - Problem Formulation

•

•

• 1-D problem

1 space coordinate x

1 direction coordinate

µ = ⌦ · i = cos ✓

= (x, µ, t)✓ 2 [0,⇡]

1

v

@ (x, µ, t)

@t

+ µ

@ (x, µ, t)

@x

+ �(x) (x, µ, t)

�(x)c(x)

2

Z 1

�1 (x, µ0

, t) dµ0 +Q(x, t)=

0 < x < d

�1 µ 1

slab geometry

x = 0

x = d

= f1(µ, t)

= f2(µ, t)

µ > 0

µ < 0

at t = 0 = g(x, µ) for 0 x d,�1 µ 1

Neutron Transport - Problem Formulation

x

µµ = 1

µ = �1x = d

= f1(µ, t)

= f2(µ, t)

Neutron Transport - Non-dimensionalisation

•

•

•

x = dx

t =d

vt = 0 Q = 0�(x)Q

f1 = 0f1 f2 = 0f2 g = 0g

� = �0�(x) c = c(x)

Non-dimensionalise:

Neutron Transport - Non-dimensionalisation

•

•

•

•

✏

�(x)

@

@ t

+✏µ

�(x)

@ (x, µ, t)

@x

+ (x, µ, t)

c(x)

2

Z 1

�1 (x, µ0

, t) dµ0 + Q(x, t)=

0 < x < 1,�1 µ 1,

on x = 0

= f1(µ, t)

= f2(µ, t)

= g(x, µ)

on x = 1

at t = 0

µ > 0

µ < 0

for 0 x 1,�1 µ 1

(d, v, 0,�0)

4 dimensional

parameters✏

1 dimensionless

parameter

Dimensionless Problem:

Dimensionless Parameter

✏ =1

�0d⇠ 10�9 � 10�3

Mean free path

Neutron Transport - Matched Asymptotics

•

• Matched Asymptotic Expansions ✏ ! 0

=

x

µ µ = 1

µ = �1

= f1(µ, t)

= f2(µ, t)

x = 1

Outer0 < x < 1

Inner InnerO(✏) O(✏)

c(x)

2

Z 1

�1 (x, µ0

, t) dµ0 +Q(x, t)

✏

�(x)

@

@t

+✏µ

�(x)

@ (x, µ, t)

@x

+ (x, µ, t)

Neutron Transport - Outer Expansion

•

•

• Outer region 0 < x < 1

= 0(x, µ, t) + ✏ 1(x, µ, t) + ✏

2 2(x, µ, t) + . . .

as ✏ ! 0

Pose

c(x) = c0(x) + ✏c1(x) + ✏

2c2(x) + . . .

Q(x, t) = Q0(x, t) + ✏Q1(x, t) + ✏

2Q2(x, t) + . . .

t =✏

�⌧ � = K0 +K1✏+K2✏

2 + . . .

•

Neutron Transport - Outer Expansion

•

•

K0

�(x)

@ 0

@⌧

+ 0 =c0(x)

2

Z 1

�1 0(x, µ

0, t)dµ0 +Q0(x, t)

• 0 = 0(x, t)K0

�(x)

@ 0

@⌧

+ (1� c0(x)) 0 = Q0(x, ⌧)

K0 = 0 c0 = 1 Q0 = 0

At O(✏0):

Neutron Transport - Outer Expansion

•

•

•

K1

�(x)

@ 0

@⌧

+µ

�(x)

@ 0

@x

+ 1 = c0(x)

2

Z 1

�1 1(x, µ

0, ⌧)dµ0

+c1(x)

2

Z 1

�1 0(x, ⌧)dµ

0 +Q1(x, ⌧)

1 = 10(x, ⌧) + µ 11(x, ⌧)

• K1 = 0 c1 = 0 Q1 = 0

1

�(x)

@ 0

@x

+ 11 = 0

K1

�(x)

@ 0

@⌧

= c1(x) 0 +Q1(x, ⌧)

At O(✏):

Neutron Transport - Outer Expansion

•

•

•

At O(✏2): K2

�(x)

@ 0

@⌧

+µ

�(x)

@ 1

@x

+ 2 = 1

2

Z 1

�1 2(x, µ

0, ⌧)dµ0

+c2(x)

2

Z 1

�1 0(x, ⌧)dµ

0 +Q2(x, ⌧)

2 = 20(x, ⌧) + µ 21(x, ⌧) + µ

2 22(x, ⌧)

K2

�(x)

@ 0

@⌧

=1

3 22 + c2(x) 0 +Q2(x, ⌧)

1

�(x)

@ 10

@x

+ 21 = 0

1

�(x)

@ 11

@x

+ 22 = 0

K2

�(x)

@ 0

@⌧

=1

3�(x)

@

@x

✓1

�(x)

@ 0

@x

◆+ c2(x) 0 +Q2

• Q(x, t) = ✏

2Q2(x, ⌧)t =

⌧

✏c(x) = 1 + ✏

2c2(x)

Neutron Transport - Inner Expansion

•

•

•

•

•

•

x = ✏y =

c(✏y) = 1 + ✏2c2(0) +O(✏3)

�(✏y) = �(0) Q(✏y, t) = ✏2Q2(0, ⌧)

=

✏2

�(0)

@

@⌧+

µ

�(0)

@ (y, µ, ⌧)

@y+ (y, µ, ⌧)

c(✏y)

2

Z 1

�1 (y, µ0, ⌧) dµ0 +✏2Q2(0, ⌧)

0 < y < 1,�1 µ 1Domain:

Inner region at x = 0

Neutron Transport - Inner Expansion

•

•

•

•

Pose:

= 0(y, µ, ⌧) + ✏ 1(y, µ, ⌧) + ✏2 2(y, µ, ⌧) + . . .

as ✏ ! 0

At O(✏0): µ

�(0)

@ 0

@y+ 0 =

1

2

Z 1

�1 0(y, µ

0, ⌧)dµ0

0(0, µ, ⌧) = f1(µ, ⌧)

�⌫(µ) =⌫

2P

1

⌫ � µ+ �(⌫)�(⌫ � µ) �(⌫) = 1� ⌫ tanh�1 ⌫

A0(⌫, ⌧) = 0 ⌫ < 0

General solution:

0(y, µ, ⌧) = a0(⌧) + b0(⌧)(�(0)y � µ)

+

Z 1

�1A0(⌫, ⌧)�⌫(µ)e

�y�(0)/⌫ d⌫

•

Neutron Transport - Inner Expansion

•

•

•

•

•

Z 1

0�⌫(µ) �(µ) dµ = 0

Z 1

0�⌫(µ)�⌫0(µ) �(µ) dµ =

�(⌫)

⌫N(⌫)�(⌫ � ⌫0)

�(µ) =3µ

2X(�µ)N(⌫) = ⌫

✓�(⌫)2 +

⇡2⌫2

4

◆

X(z) =1

1� zexp

✓1

⇡

Z 1

0

1

(µ0 � z)tan

�1

⇡µ0

2�(µ0)

�◆dµ0

Orthogonality Conditions:

Neutron Transport - Inner Expansion

•

•

• a0(⌧) =�1�0

b0(⌧) +1

�0

Z 1

0f1(µ, ⌧)�(µ)dµ

A0(⌫, ⌧) = � ⌫2�0b0(⌧)

2�(⌫)N(⌫)+

⌫

�(⌫)N(⌫)

Z 1

0f1(µ, ⌧)�⌫(µ) �(µ) dµ

b0(⌧) determined by matching to outer

�i =

Z 1

0µi�(µ)dµ for i = 0, 1

Neutron Transport - Matching

• •

•

•

✏ ⌧ x = ✏y ⌧ 1Overlap domain:

Outer in Inner variables:

= 0(✏y, µ, ⌧) + ✏ 1(✏y, µ, ⌧) + ✏2 2(✏y, µ, ⌧) + . . .

= 0(0, ⌧) + ✏

y

@ 0(0, µ, ⌧)

@x

+ 1(0, µ, ⌧)

�

+✏2y

2

2

@

2 0(0, µ, ⌧)

@x

2+ y

@ 1(0, µ, ⌧)

@x

+ 2(0, µ, ⌧)

�+ . . .

Outer limit of Inner:

= 0(y, µ, ⌧) + ✏ 1(y, µ, ⌧) + ✏2 2(y, µ, ⌧) + . . .

as y ! 1• At O(✏0): lim

y!1 0(y, µ, ⌧) = 0(0, ⌧)

Neutron Transport - Matching

•

• 0(y, µ, ⌧) = b0(⌧)�(0)y + (a0(⌧)� b0(⌧)µ) + o(1)

as y ! 1

b0(⌧) = 0 a0(⌧) = 0(0, ⌧)

Neutron Transport - Summary

•

•

•

K2

�(x)

@ 0

@⌧

=1

3�(x)

@

@x

✓1

�(x)

@ 0

@x

◆+ c2(x) 0 +Q2

in 0 < x < 1, ⌧ > 0

Boundary Conditions:

0(0, ⌧) =

Z 1

0f1(µ, ⌧)�(µ)dµ

0(1, ⌧) =

Z 1

0f2(µ, ⌧)�(µ)dµ

Initial Condition:

0(x, 0) = G(x) ??

Governing Equation:

Thank You !

The End