Lesson 2 Objectives

2-1

Lesson 2 ObjectivesLesson 2 Objectives

• The Transport Equation (cont’d)The Transport Equation (cont’d)• Particle distributionsParticle distributions

• Interaction ratesInteraction rates• Boundary crossingsBoundary crossings

• Derivation of the Boltzmann EquationDerivation of the Boltzmann Equation• Initial conditions and boundary conditionsInitial conditions and boundary conditions

1-2

Definition of basic elementsDefinition of basic elements

• Material cross sections: Particle/matter Material cross sections: Particle/matter interaction probabilitiesinteraction probabilities

• We will use small sigma, We will use small sigma, , for for , for for microscopic AND macroscopic cross microscopic AND macroscopic cross sections:sections:

ErnEr ixi

N

ix

~,1

=Probability of an interaction of type x per =Probability of an interaction of type x per unit path lengthunit path length

1-3

Scattering cross sectionsScattering cross sections

• For scattering reactions, we must consider For scattering reactions, we must consider the post-collision properties as well as the the post-collision properties as well as the probability of interaction:probability of interaction:

)ˆˆ,()()ˆˆ,( EEfEEE ss

where:where:

scattering for section cross )(Es

particles emitted for function onDistributi )ˆˆ,( EEf

1-4

Fission neutron distributionFission neutron distribution

• Two data variables you need to know are:Two data variables you need to know are:

mean # of neutrons released from

fission caused by energy neutronsenergy distribution of fission neutrons,

(# emitted/unit energy)

E

EE

• The first is a function; the second is a The first is a function; the second is a distributiondistribution

2-5

Particle distributionsParticle distributions

• A basic concept we will use is the particle A basic concept we will use is the particle distribution:distribution:

ˆ( , , , ) the expected number of particles in a volume dV about traveling in a

ˆ direction about in the energy range dE about E at time t

N r E t dV dEdr

d

• We generally prefer the angular flux:We generally prefer the angular flux:

velocity particle

vtErvNtEr ),,ˆ,(),,ˆ,(

2-6

Particle distributions (2)Particle distributions (2)

• Angular flux is more useful for two types of Angular flux is more useful for two types of events:events:

1.1. Interaction ratesInteraction rates2.2. Boundary crossingsBoundary crossings

• Since the individual terms of the B.E. Since the individual terms of the B.E. involve these two types of events, the involve these two types of events, the angular flux is the primary unknown of the angular flux is the primary unknown of the equation. equation.

2-7

Interaction ratesInteraction rates

dtdEdVdtErEr

dtdEdVdtErvNEr

ErdtvdEdVdtErN

t

t

t

),,ˆ,(,

),,ˆ,(,

,),,ˆ,(

length) path unit per ninteractio ofty (Probabili particles)by traveled distance (Total

dt time in nsinteractio of #

2-8

Boundary crossingsBoundary crossings

• Consider an element of surface area dA Consider an element of surface area dA notnot perpendicular to : perpendicular to :

• Create a volume element by projecting Create a volume element by projecting dAdA backwards along the direction a distance backwards along the direction a distance vdtvdt

dtv

n

dA

2-9

Boundary crossings (2)Boundary crossings (2)

dtdAdEdErvNn

vdtdAndEdErN

dtdA

),ˆ,(ˆˆ

ˆˆ),ˆ,(

(Length) Area)(Projecteddensity) (Particle

(Volume)density) (Particle element volume the inside start that particles of #

time in crossing particles of #

dtv

n

dA

2-10


vector" current" where

),,ˆ,(ˆ),,(

),,(ˆ

),,ˆ,(ˆˆ

),,ˆ,(ˆˆ),,(

4

4

4

tErdtErJ

dtdAdEtErJn

dtdAdEtErdn

dtdAdEtErvNnddtdAdEtErJn

• Note that the total crossing rate (regardless Note that the total crossing rate (regardless of direction) can be found by integrating of direction) can be found by integrating over :over :

2-11


• This gives us the “net current”This gives us the “net current”• We are also sometimes interested in “partial We are also sometimes interested in “partial

currents”, which count particles crossing the currents”, which count particles crossing the surface in positive and negative directions surface in positive and negative directions (where positive is defined by YOU by which (where positive is defined by YOU by which way points):way points):n

),,(),,(),,(

),,ˆ,(ˆˆ),,(

),,ˆ,(ˆˆ),,(

0ˆˆ

0ˆˆ

tErJtErJtErJ

tErndtErJ

tErndtErJ

nnn

nn

nn

2-12

Derivation of Boltzmann Equation (Eulerian)Derivation of Boltzmann Equation (Eulerian)• Particle balance on the subset of particles occupying a Particle balance on the subset of particles occupying a

fixedfixed dEd dEddxdydzdt:dxdydzdt:

Term - Term - TermTerm

orcollision) to due element leaving particles of (Number-

element) of out streaming particles of number (Net-element) in born"" particles of (Number

element) space phase in particles of (Increase

4321

2-13

Term#1: Increase of particlesTerm#1: Increase of particles

• For this, we will use a Cartesian (dx,dy,dz) volume For this, we will use a Cartesian (dx,dy,dz) volume element:element:

• Obviously:Obviously: differs for the 6 different facesdiffers for the 6 different facesnˆ

dzdydxdV Volume

dxdy

dzinx ˆ

i

jny ˆ

j

knz ˆ

k

2-14

Term#1: Increase of particles (2)Term#1: Increase of particles (2)

• And the increase in the number of particles in the phase And the increase in the number of particles in the phase space element in the time interval is:space element in the time interval is:

dEdxdydzdtEzyxN

dttEzyxN

dEdxdydzdtEzyxN

dEdxdydzddttEzyxN

),,ˆ,,,(

),,ˆ,,,(

interval of beginningat # - interval of endat # ),,ˆ,,,(

),,ˆ,,,(

2-15

Term#2: Particles “born”Term#2: Particles “born”

• For now, we will combine all sources (fixed source, For now, we will combine all sources (fixed source, scattering, fission) into one term:scattering, fission) into one term:

dtdEddxdydztEzyxq ),,ˆ,,,(

2-16

Term#3: Particles streaming outTerm#3: Particles streaming out

• Each of the dimensions (x,y,z) has a Each of the dimensions (x,y,z) has a positive and a negative facepositive and a negative face

• Using the boundary crossing rates Using the boundary crossing rates from before, this is:from before, this is:

Net out-leakage

ˆ ˆ ˆ ˆ( , , , , , ) ( , , , , , )2 2

ˆ ˆ ˆ ˆ( , , , , , ) ( , , , , , )2 2

ˆ ˆ( , , , , , ) ( , , , , ,2 2

dx dxvN x y z E t vN x y z E t i d dE dydz dt

dy dyvN x y z E t vN x y z E t j d dE dxdz dt

dz dzvN x y z E t vN x y z E

ˆˆ)t k d dE dxdy dt

2-17

Term#4: Particle collidingTerm#4: Particle colliding

• Using the interaction rates from before, this is simply:Using the interaction rates from before, this is simply:

dtdxdydzdEdtEzyxvNEzyxt

),,ˆ,,,(,,,

ginteractin particles ofNumber

2-18

Putting it togetherPutting it together• Combining all the terms and dividing by Combining all the terms and dividing by dxdydzdtddxdydzdtddEdE gives gives

us:us:ˆ ˆ( , , , , , ) ( , , , , , )

ˆ ˆ( , , , , , ) ( , , , , , )2 2ˆ ˆ

ˆ ˆ( , , , , , ) ( , , , , , )2 2ˆ ˆ

ˆ ˆ( , , , , , ) ( , , , , , )2 2ˆˆ

, ,t

N x y z E t dt N x y z E tdt

dx dxvN x y z E t vN x y z E ti

dxdy dyvN x y z E t vN x y z E t

jdy

dz dzvN x y z E t vN x y z E tk

dzx y

ˆ, ( , , , , , )ˆ ( , , , , , )

z E vN x y z E t

q x y z E t

2-19

Putting it together (2)Putting it together (2)

• Taking the limits as and Taking the limits as and and converting to angular flux (= and converting to angular flux (=vNvN) ) gives us: gives us:

0,,, dtdzdydx

),,ˆ,,,( ),,ˆ,,,(,,,

),,ˆ,,,(ˆˆ),,ˆ,,,(ˆˆ

),,ˆ,,,(ˆˆ),,ˆ,,,(1

tEzyxqtEzyxEzyx

ztEzyxk

ytEzyxj

xtEzyxi

ttEzyx

v

t

2-20


• We can simplify (i.e., obscure) this by recalling We can simplify (i.e., obscure) this by recalling that the gradient operator, is defined as:that the gradient operator, is defined as:

which further allows us to write:which further allows us to write:

zrk

yrj

xrir

)(ˆ)(ˆ)(ˆ)(

zrk

yrj

xrir

)(ˆˆ)(ˆˆ)(ˆˆ)(ˆ

2-21


• And our final form of the Boltzmann Eqn. comes And our final form of the Boltzmann Eqn. comes from substituting this to give us:from substituting this to give us:

where I have simplified the notation by using:where I have simplified the notation by using:

),,ˆ,(

),,ˆ,(,

),,ˆ,(ˆ),,ˆ,( 1

tErq

tErEr

tErt

tErv

t

zyxr ,,

2-22

Derivation of Boltzmann Equation Derivation of Boltzmann Equation (Lagrangian)(Lagrangian)

• Now that we have successfully derived the Boltzmann equation, let’s do it again…..Now that we have successfully derived the Boltzmann equation, let’s do it again…..• This time let’s use a Lagrangian grid, that moves with the particlesThis time let’s use a Lagrangian grid, that moves with the particles• With this approach, instead of letting all of the variables (x,y,z,E,With this approach, instead of letting all of the variables (x,y,z,E,,t) define the state of the particle, we characterize the state in terms of particle incoming parameters at the ,t) define the state of the particle, we characterize the state in terms of particle incoming parameters at the

boundary plus a SINGLE parameter that takes care of changes that have occurred since that initial stateboundary plus a SINGLE parameter that takes care of changes that have occurred since that initial state• For this derivation, we will let DISTANCE, s, from the boundary (in the direction of travel of the particle) be the single parameter that we use.For this derivation, we will let DISTANCE, s, from the boundary (in the direction of travel of the particle) be the single parameter that we use.

2-23

Lagrangian state descriptionLagrangian state description

• In this alternative viewpoint, we have:In this alternative viewpoint, we have:

Problem boundary

0,,ˆ,,,, 0000000 stEzyx

vstsd

dsdttst

sddsds

EsddsdEEsE

skzsddsdzzsz

sjysddsdyysy

sixsddsdxxsx

stEzyxs

s

s

s

s

s

s

00

0

00

0

00

0

00

0

00

0

00

0

000000

ˆˆˆˆ

ˆˆ

ˆˆ

ˆˆ

,,ˆ,,,,

s

2-24

Lagrangian Derivation of BE (2)Lagrangian Derivation of BE (2)

• We have replaced one 7D PDE with seven We have replaced one 7D PDE with seven 1D ODEs1D ODEs

• The advantages of this form are:The advantages of this form are:1.1. The seven are easy to solve (I solved 6 of them The seven are easy to solve (I solved 6 of them

on the previous slide!)on the previous slide!)2.2. This is the form of the equation that integral This is the form of the equation that integral

transport methods and Monte Carlo methods transport methods and Monte Carlo methods begin withbegin with

3.3. It is much easier to understand curvilinear It is much easier to understand curvilinear geometries with this formgeometries with this form

4.4. It is much easier to derive the adjoint equation It is much easier to derive the adjoint equation with this formwith this form

2-25


• Now our question is how does the flux Now our question is how does the flux change as the particle moves from s to change as the particle moves from s to s+ds?s+ds?

Problem boundary

s s+ds

s dss

2-26


• Beams are particularly easy to solve Beams are particularly easy to solve • The particle flux is depleted by any interaction, with The particle flux is depleted by any interaction, with

the probability of interaction per unit path given by the probability of interaction per unit path given by the total cross sectionthe total cross section

• Therefore the total probability of interaction is Therefore the total probability of interaction is total cross section times dstotal cross section times ds

• The gains per unit path are given by the source term The gains per unit path are given by the source term times the unit path, so the balance equation is:times the unit path, so the balance equation is:

• or:or: dssqdssssdss t )()()(

sqssdssd

t )()(

2-27


• This equation will be our starting point for the This equation will be our starting point for the integral transport equation in Chapter 5.integral transport equation in Chapter 5.

• For now, except for the derivative term we can jump For now, except for the derivative term we can jump to the “deterministic” equation by simply substituting to the “deterministic” equation by simply substituting the dependencies:the dependencies:

• For the derivative term, we use the chain rule:For the derivative term, we use the chain rule:

tEzyxs ,ˆ,,,,

dsd

dsdE

Edsdz

z

dsdy

ydsdx

xdsdt

tyx

dssd

ˆˆ

,...,)(

2-28


• Plugging in the derivatives (the constants that were Plugging in the derivatives (the constants that were buried in the integrals of Slide 2-20) gets us to:buried in the integrals of Slide 2-20) gets us to:

• This is the equation from Slide 2-16, which leads to This is the equation from Slide 2-16, which leads to the same final form as before (Slide 2-18)the same final form as before (Slide 2-18)

),,ˆ,,,( ),,ˆ,,,(,,,

),,ˆ,,,(ˆˆ),,ˆ,,,(ˆˆ

),,ˆ,,,(ˆˆ),,ˆ,,,(1

tEzyxqtEzyxEzyx

ztEzyxk

ytEzyxj

xtEzyxi

ttEzyx

v

t

2-29

Homework ProblemsHomework Problems

(2-1) Repeat the Lagrangian derivation using time, (2-1) Repeat the Lagrangian derivation using time, t, as the parameter instead of distance, s, along t, as the parameter instead of distance, s, along the direction of travel.the direction of travel.

(2-2) How would gravity (in –z direction) change (2-2) How would gravity (in –z direction) change the equation?the equation?

(2-3)(2-3) How would the equation look for a charged How would the equation look for a charged particle with stopping power (i.e., energy loss per particle with stopping power (i.e., energy loss per unit distance) of unit distance) of S(x,y,z,E)?S(x,y,z,E)?

Lesson 2 Objectives

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