Lesson 1 Objectives
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Objectives of Course
Go over syllabus
Go over course
Overview of Course
The Transport Equation
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Assumptions
Definition of basic elements
Scattering cross sections
Use of Legendre expansions of angular
distribution
Fission neutron distribution
1-1
Objectives of Course
User setup should be:
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Materials and geometry
1.
2.
3.
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Source description
4.
5.
6.
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Source particles
Source energy distribution
Source spatial distribution
“Detector” response
7.
8.
9.
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Material makeup (isotopics)
Material energy interactions with particles
Material spatial distribution
Detector particle sensitivity
Detector energy sensitivity (response function)
Detector spatial location
Why are any other questions asked?
1-2
Objectives of Course (2)
Answer: Boltzmann gave us an exact equation,
but we cannot solve it.
We must simplify the equation:
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Space: Replace continuous space with
homogeneous blocks (“cells”) of material
Energy: Replace continuous energy with energy
“groups”
Direction: Constrain particles to only travel in
certain directions
Result: The deterministic discrete ordinates
equation, discretized for computer solution.
1-3
Objectives of Course (3)
“Deterministic codes give you exact solutions
to approximate models. Monte Carlo codes
give you approximate solutions to exact
models.”
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This situation puts an extra burden on you, the
user.
You are required to supply computer code
input that is NOT related to the description of
your problem, but is related to this simpler
model (which is the only one the computer
can solve).
My goal: Help you understand what is being
asked of you
1-4
Overview of Course
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First four chapters of:
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Lewis, E. E., and Miller, W. F., Jr.; Computational
Methods of Neutron Transport, American Nuclear
Society, La Grange Park, IL, 1993.
General flow of the course will be:
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Derivation of the continuous-energy
Boltzmann Equation (L&M, 1)
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Derivation of the forward equation
Differences in approach for source vs. eigenvalue
problems
Derivation and use of adjoint form of equation
1-5
Overview of Course (2)
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General flow of the course (cont’d):
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Energy and time discretization (L&M, 2)
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Multigroup approximation in energy
Fixed source solution strategies in energy
Eigenvalue problem solution strategies in energy
Time-dependent considerations
1D discrete ordinates methods (L&M, 3)
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Angular approximation
Spatial differencing
Curvilinear coordinates
Acceleration techniques
1-6
Overview of Course (3)
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General flow of the course (cont’d):
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2D and 3D discrete ordinates (L&M, 4)
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Angular quadrature
Cartesian treatments
Curvilinear treatments
Ray effects
Integral transport theory (L&M, 5)
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If time permits
1-7
The Transport Equation
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Introduction
Particle Interaction
Particle Streaming
Transport with Secondary Particles
The Time-Independent Transport
equation
The Adjoint Transport Equation
1-8
The basic physical assumptions
1.
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5.
6.
7.
Particles are points
Particles travel in straight lines, unaccelerated
until they interact
Particles don’t hit other particles
Collisions are resolved instantaneously
Material properties are isotropic in direction
Composition, configuration, and material
properties are known and constant
Only the expected (mean) values of reaction
rates are needed
You will think about these more deeply in HW problem 1-1.
1-9
Definition of basic elements
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Material cross sections: Particle/matter
interaction probabilities
We will use small sigma, s, for for
microscopic AND macroscopic cross
sections:
N

 ~i
s x r , E    ni r s x E 
i 1
=Probability of an interaction of type x per
unit path length
1-10
Definition of basic elements (2)
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where x=
‘c’ for capture=particle loss
‘f’ for fission
‘a’ for absorption=fission + capture
‘s’ for scattering=particle change of energy and direction
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For neutrons, the primary scattering
mechanisms are elastic scattering, inelastic
scattering, and (n,2n)
For photons, the scattering mechanisms are
Compton scattering and pair production
For coupled neutron/gamma problems, neutron
reactions that produce gammas are “scatter”
Unit of macroscopic cross section is cm-1
1-11
Definition of basic elements (3)
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Denoting the intensity of the flow of a beam of
particles as I(x), we have:
dI ( x)
 s t I ( x )
dx
 I ( x)  I (0)e st x
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This is the familiar exponential attenuation
In the book, the total cross section is sometimes
denoted by s (no subscript)
1-12
Definition of basic elements (4)
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Background: A “weighted average” of a function of x is defined as:

Average 
 w( x) f ( x)dx


 w( x)dx

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The most common variations we see in NE are:
• Unweighted average: w(x)=1 (over finite domain of x)
• Mean (or expected) value of x: w(x)=Pr(x) (probability of x
being chosen)
• Nth moment of x: w(x)=xn , 0<x<infinity
• Nth Legendre moment: w(x)=1/2 Pn(x), -1<x<1
1-13
Definition of basic elements (4a)
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The mean free path, l, is defined as the
average distance traveled before a collision:



xst I ( x)dx

xe st x dx
1
2
s
1
0
0
t
l





1
st
st I ( x)dx
e st x dx
st
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

0
0
Work this out (Prob. 1-2)
For reaction rate x, we have: l x ( E ) 
1
sx (E)
1-14
Scattering cross sections
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For scattering reactions, we must consider
the post-collision properties as well as the
probability of interaction:
ˆ 
ˆ )  s ( E) f ( E  E, 
ˆ 
ˆ )
ss ( E  E, 
s
where:
s s (E )  cross section for scattering
ˆ 
ˆ )  Distributi on function
f ( E  E , 
for emitted particles
1-15
Scattering cross sections (2)
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Based on Assumption #5, the angular
dependence is dependent on the deflection
angle between the two directions:
ˆ 
ˆ )  f ( E  E, 
ˆ 
ˆ )  f ( E  E,  )
f ( E  E, 
0
where:
ˆ and 
ˆ
0  cosineof the anglebetween 
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Note that there is no azimuthal angular
dependence
1-16
Scattering cross sections (3)
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The distribution function is normalized to
integrate to the number of particles that are
emitted by the reaction. For example, elastic
scattering has:

1
 dE  d
0
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0
f es ( E  E,  0 )  1
1
whereas for (n,2n) we have:

1
 dE  d
0
0
f n 2 n ( E  E,  0 )  2
1
1-17
Scattering cross sections (4)
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Note that since we combine cross sections
linearly, the relationship between the
macroscopic and microscopic distribution
functions is given by:
s s ( E  E ,  0 )  s s ( E ) f ( E  E ,  0 )
Isotopes


~ ( E ) f ( E  E ,  )
ni s
si
i
0
i 1
Isotopes
f ( E  E ,  0 ) 

~ ( E ) f ( E  E ,  )
ni s
si
i
0
i 1
Isotopes

i 1
~ (E)
ni s
si
1-18
Scattering cross sections (5)
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The most familiar distribution is the elastic
scattering distribution (from kinematics):
1

  0    , for  E  E   E

f i ( E  E , 0 )   (1   i ) E

0, otherwise

where:
A i -1

i 
,
2
 Ai  1
2
A i  atomic mass of isotope i (multiples of neutron mass)
1
E
E
   A i  1
  A i  1

2
E
E 
1-19
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Functional expansions
The general field of functional expansion
involves the approximation of a continuous
function as a linear combination of a basis
function set:
L
f ( x)   f  e  x 
 0
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Use of a “complete” basis functions set
means that as L goes to infinity, the
approximation approaches the function
The trick is finding the coefficients f  that
“best” fit the function
1-20
Functional expansions (2)
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Determining the “best” comes down to
finding the approximation (for a given L) that
is “closest” to the function according to some
“norm”
A “norm” is simply a measure of difference
between two values (or functions) that
satisfies two simple criteria:
1.
2.
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The value of the norm is always non-negative
The value is only zero if the two are identical
Example: Distance norms for points
1-21
Functional expansions (3)
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The norm that we will use is the least-square
norm, L2 , which is a member of the Ln series
defined by:
Ln 
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 f x   f x 
1
domain
2
n
dx
The “distance” from a function and its
approximate expansion is then:
2


Ln    f x    f  e x  dx
 0

domain
L
1-22
Functional expansions (4)
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We find the optimum coefficients by setting the
partial derivatives to zero:
L
Ln


  2  f  x    f e  x  ek  x  dx  0
f k domain 
0

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This gives us a set of linear (matrix) equations:
L
 f xe xdx   f  e xe xdx
k
domain
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 0


domain
k
Thus minimizing the least squares norm comes
down to preserving the moments of the expansion
functions themselves (Galerkin method).
1-23
Functional expansions (5)
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Solving for the coefficients becomes much easier if
the basis functions are orthogonal (which means
that the integral of the product of any two different
basis functions is zero):
 e xe x dx  0 if k  

domain
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k
This turns the previous equation into:
fk 
 f xe xdx
k
domain
 e xe xdx
k
domain
k
1-24
Use of Legendre expansions
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Using the cosine of the deflection angle, we
can represent the angular dependence of the
distribution in a Legendre expansion:
s s ( E  E ,  0 ) 

 2  1s s ( E  E ) P  0 
 0
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This allows us to represent the scattering
distribution by determining the Legendre
coefficients:
s s ( E  E ),   0, 1, ..., 
1-25
Use of Legendre expansions (2)
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Using the orthogonality of the Legendre
polynomials:
1
2
1 d 0 P  0  Pm  0   2  1 
m
we can operate on both sides of the
expansion (1st eqn. on previous slide) with:
1
 d
0
Pm  0 
1
1-26
Use of Legendre expansions (3)
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To get:
1
1
s sm ( E  E)   Pm  0  s s ( E  E , 0 ) d 0
2 1
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Work this out for yourself (Prob. 1-3)
1-27
Fission neutron distribution
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Two data variables you need to know are:
E   mean # of neutrons released from
fission caused by energy E neutrons
E  energy distributi on of fission neutrons,
(# emitted/un it energy)
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The first is a function; the second is a
distribution
1-28
Homework 1-1
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For each of the assumptions listed on
slide 1-9, give a physical situation (if
you can think of one) for which the
assumption may not be a good one.
1-29
Homework 1-2
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Use integration by parts and l’Hopital’s
rule to show that:



xst I ( x)dx

xe st x dx
1
2
s
1
0
0
t
l





1
st
 st x
st I ( x)dx
e
dx
st


0
0
1-30
Homework 1-3
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a. Show that if we expand:
s s ( E  E ,  0 ) 

 2  1s s ( E  E ) P  0 
 0
that the coefficients can be found from:
1
1
s sm ( E  E ) 
d 0 Pm  0 s s ( E  E ,  0 )
2

1
x


f
x

e
,1  x  1
b. Use this fact to expand
Expand it to enough terms so that the error is
less than 1% at all values of x.
1-31