8-1 lesson 8 objectives recap and iteration practice recap and iteration practice reduction of b.e....

8-1

Lesson 8 ObjectivesLesson 8 ObjectivesLesson 8 ObjectivesLesson 8 Objectives

• Recap and iteration practiceRecap and iteration practice• Reduction of B.E. to 1DReduction of B.E. to 1D• 1D Quadratures1D Quadratures

8-2

Recap of where we areRecap of where we areRecap of where we areRecap of where we are

1.1. The forward Boltzmann equation in space, energy, The forward Boltzmann equation in space, energy, angle equation was derived to be:angle equation was derived to be:

• The left side held the flux-dependent terms, the The left side held the flux-dependent terms, the right side held the non-flux-dependent term.right side held the non-flux-dependent term.

• We also derived the adjoint form of the equation We also derived the adjoint form of the equation (which we will ignore the rest of the course)(which we will ignore the rest of the course)

),ˆ,(

),ˆ,(),(

),ˆ,(),ˆˆ,(

),ˆ,(,),ˆ,(ˆ

40

40

Erq

ErErdEdE

ErEErdEd

ErErEr

ex

f

s

t

8-3

Recap of where we are (2)Recap of where we are (2)Recap of where we are (2)Recap of where we are (2)

2.2. In the absence of an external source, we found that In the absence of an external source, we found that we had to add an eigenvalue to the equation. We we had to add an eigenvalue to the equation. We learned four variations on this, the most common learned four variations on this, the most common being the k-effective (or lambda) eigenvalue:being the k-effective (or lambda) eigenvalue:

• The fission neutron source term became our new The fission neutron source term became our new “source”, so we moved it to the right hand side.“source”, so we moved it to the right hand side.

0 4

0 4

ˆ ˆ ˆ( , , ) , ( , , )

ˆ ˆ ˆ( , , ) ( , , )

1 ˆ( , ) ( , , )

1 ˆ( ) ( , , )

t

s

f

r E r E r E

dE d r E E r E

E dE d r E r E

E F r S r E

8-4


3.3. We then began attacking the equation, beginning with energy. We then began attacking the equation, beginning with energy. We approximated the energy dependence of the flux and We approximated the energy dependence of the flux and cross sections using the MULTIGROUP equation with (after cross sections using the MULTIGROUP equation with (after proper approximation of the group parameters), gave us (for proper approximation of the group parameters), gave us (for each group):each group):

• This “theme” of throwing more and more over to the “source” This “theme” of throwing more and more over to the “source” term will continue as we go!term will continue as we go!

1 4

,

1 4

ˆ ˆ ˆ( , ) ( , )

ˆ ˆ ˆ ˆ( , ) ( , ) ( , )

ˆ( , )ˆ( , )

1 ˆ( ) ( , )

g tg g

Gg gs g g

g

g ex

g

G

g fg gg

r r r

d r r S r

q r (Source problems)

S r or

(Fission problems)d r r

8-5


4.4. Since the only non-g term of the left hand side was the Since the only non-g term of the left hand side was the scattering from other groups, we then (what else) relegated scattering from other groups, we then (what else) relegated all of the non-g scattering to the right-hand-side to get:all of the non-g scattering to the right-hand-side to get:

• We label the new source “out” because it becomes the We label the new source “out” because it becomes the source for each group in an OUTER iteration, where each of source for each group in an OUTER iteration, where each of the energy groups are treated one-at-a-time.the energy groups are treated one-at-a-time.

• We learned two ways of computing the source, depending on We learned two ways of computing the source, depending on whether the inscattering sources from groups that have already whether the inscattering sources from groups that have already been handled in the current outer iteration use the OLD iteration been handled in the current outer iteration use the OLD iteration fluxes (“Jacobi”) or the NEW iteration fluxes (“Gauss-Seidel”)fluxes (“Jacobi”) or the NEW iteration fluxes (“Gauss-Seidel”)

4

1 4

ˆ ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , )

ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , )

g gg tg g s g

Gg g out

g s g ggg g

r r r d r r

S r d r r S r

8-6


• Adding in the outer iteration “counters”, , makes this:Adding in the outer iteration “counters”, , makes this:

• The “?” is there because the iteration counter depends on The “?” is there because the iteration counter depends on whether you are using Jacobi or Gauss-Seidel (and, if G-S, whether you are using Jacobi or Gauss-Seidel (and, if G-S, whether g’ is a group that has already been calculated in the whether g’ is a group that has already been calculated in the current outer iteration)current outer iteration)

• I gave you some practice using this in HW#6I gave you some practice using this in HW#6

1 1 1

4

? , 1

1 4

ˆ ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , )

ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , )g

g gg tg g s g

Gg g out

g s ggg g

r r r d r r

S r d r r S r

8-7


5.5. Just for a little practice in the numerical treatment of outer Just for a little practice in the numerical treatment of outer iterations, we reduced this to the infinite-medium-form, where iterations, we reduced this to the infinite-medium-form, where the flux does not depend on either space or direction, giving the flux does not depend on either space or direction, giving us:us:

• Following the traditional notation, this equation was written in Following the traditional notation, this equation was written in terms of the SCALAR flux (the direction-integrated ANGULAR terms of the SCALAR flux (the direction-integrated ANGULAR flux) because, for isotropic infinite-medium fluxes, they are flux) because, for isotropic infinite-medium fluxes, they are the same value (as long as you use unit solid angle):the same value (as long as you use unit solid angle):

1 1 ? , 1

1

, 1

g

Gg g g g out

tg g s g g s ggg g

outrg g g

g grg tg s

S S

S

where

("removal" cross section)

4 4

ˆ( ) ( , ) ( ) ( )g g g gr d r r d r

8-8


• Whenever you use an iterative method, if the numerical Whenever you use an iterative method, if the numerical method is convergent, the fluxes will change less and less method is convergent, the fluxes will change less and less with each iteration. The user has to decide when close with each iteration. The user has to decide when close enough is good enough.enough is good enough.

• Usually this is done with a convergence criterion of the Usually this is done with a convergence criterion of the MAXIMUM FRACTIONAL CHANGE of any iterating variable.MAXIMUM FRACTIONAL CHANGE of any iterating variable.

• After each iteration, you loop over each flux value and determine After each iteration, you loop over each flux value and determine the fractional amount it has changed during this iteration. You the fractional amount it has changed during this iteration. You find the maximum ABSOLUTE VALUE of any change and that find the maximum ABSOLUTE VALUE of any change and that becomes your “error” value for the iteration.becomes your “error” value for the iteration.

• You continue iterating until either:You continue iterating until either:1.1. The absolute value of your iteration error is less than some preset The absolute value of your iteration error is less than some preset

CONVERGENCE CRITERION; orCONVERGENCE CRITERION; or

2.2. You have exceeded the maximum number of iterations you will allow You have exceeded the maximum number of iterations you will allow (necessary so that a divergent problem doesn’t run forever)(necessary so that a divergent problem doesn’t run forever)

• In setting the convergence criterion, you must be careful not to In setting the convergence criterion, you must be careful not to demand more digits than the computer uses for a variable! (This demand more digits than the computer uses for a variable! (This is why FRACTIONAL change is used.) I like 1.0e-06is why FRACTIONAL change is used.) I like 1.0e-06

8-9


6.6. Concentrating our attention on the solution within each group Concentrating our attention on the solution within each group (dropping the outer iteration counter since they are all ):(dropping the outer iteration counter since they are all ):

we again simplified by throwing the more complicated within-we again simplified by throwing the more complicated within-group scattering term to the right-hand-side to get:group scattering term to the right-hand-side to get:

• Using an INNER iteration counter of Using an INNER iteration counter of kk, this becomes:, this becomes:

4

ˆ ˆ ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , ) ( , )g g outg tg g s g gr r r d r r S r

4

ˆ ˆ ˆ( , ) ( , )

ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , )

g tg g

out g g ing s g g

r r r

S r d r r S r

1

1 1

, 1

4

ˆ ˆ ˆ( , ) ( , )

ˆ ˆ ˆ ˆ ˆ( , ) ( , ) ( , ) ( , )

g g

g

k ktg

out g g k in kg s g

r r r

S r d r r S r

8-10


7.7. Just for a little practice in the numerical treatment of outer Just for a little practice in the numerical treatment of outer iterations and inner iterations at the same time, we reduced this iterations and inner iterations at the same time, we reduced this to the infinite-medium-form, where the flux does not depend on to the infinite-medium-form, where the flux does not depend on either space or direction, giving us:either space or direction, giving us:

• Notice that I did NOT use the removal cross section since the Notice that I did NOT use the removal cross section since the total cross section and within-group scattering are no longer on total cross section and within-group scattering are no longer on the left-hand-side of the equationthe left-hand-side of the equation

• YOU MUST BE SURE to start each inner iteration with the YOU MUST BE SURE to start each inner iteration with the initial fluxes set to the BEST AVAILABLE fluxes for that group initial fluxes set to the BEST AVAILABLE fluxes for that group (i.e., the results of the last time this group was calculated). DO (i.e., the results of the last time this group was calculated). DO NOT reset them to zero; they are only set to 0 the first time.NOT reset them to zero; they are only set to 0 the first time.

1, 1 , 1 1,

, 1 1

1 1g g

k out g g ktg g g s g

G Gout g g g gg g s s

g gg g g g

S

S S

8-11


8.8. Whenever you are dealing with BOTH inner and outer Whenever you are dealing with BOTH inner and outer iterations, you have to decide how finely to converge the inner iterations, you have to decide how finely to converge the inner iterations before moving on to the next group.iterations before moving on to the next group.

• Basically, this comes down to setting the maximum number of Basically, this comes down to setting the maximum number of inner iterations to run for each outer sweep through a group.inner iterations to run for each outer sweep through a group.

• The two limiting cases of this parameter are 1 and infinity, i.e.,The two limiting cases of this parameter are 1 and infinity, i.e.,• Running just one inner iteration per outer iteration; andRunning just one inner iteration per outer iteration; and• Converging the inner iterations FULLY before moving on, no Converging the inner iterations FULLY before moving on, no

matter how long it takes.matter how long it takes.

• Although this decision is problem dependent, usually a good Although this decision is problem dependent, usually a good value can be found. (At SRP, we used 4.)value can be found. (At SRP, we used 4.)

• FYI, it is frequently done that the INNER iteration FYI, it is frequently done that the INNER iteration convergence criterion is smaller than the OUTER iteration convergence criterion is smaller than the OUTER iteration convergence criterion (I like to use half). This helps keep the convergence criterion (I like to use half). This helps keep the number of outer iterations lower.number of outer iterations lower.

8-12

Treatment of Space and DirectionTreatment of Space and DirectionTreatment of Space and DirectionTreatment of Space and Direction

• Now we move on to solve the group equations in both Now we move on to solve the group equations in both direction and spacedirection and space

• Direction this weekDirection this week• Space next weekSpace next week

• You might THINK that we would treat one of them first and You might THINK that we would treat one of them first and then move on to the next one. BUT, since multidimensional then move on to the next one. BUT, since multidimensional problems are more difficult than one-dimensional problems, problems are more difficult than one-dimensional problems, we shall instead attack:we shall instead attack:

• Both space and energy in one-dimensional problemsBoth space and energy in one-dimensional problems and THENand THEN

• Both space and energy in multi-dimensional problemsBoth space and energy in multi-dimensional problems

• So, next is the treatment of one dimensional problems. We So, next is the treatment of one dimensional problems. We shall also break THIS up by difficulty, looking at:shall also break THIS up by difficulty, looking at:

• One dimensional SLAB geometries; and thenOne dimensional SLAB geometries; and then• One dimensional CURVED geometries (cylindrical and One dimensional CURVED geometries (cylindrical and

spherical).spherical).

8-13

Reduction to 1D SlabReduction to 1D SlabReduction to 1D SlabReduction to 1D Slab

• To concentrate our attention on our angular To concentrate our attention on our angular treatments, we will simplify space to 1D slab treatments, we will simplify space to 1D slab geometry:geometry:

• where the second relation reminds that the where the second relation reminds that the angular dependence of the source is given angular dependence of the source is given as Legendre source moments.as Legendre source moments.

)()(12),(

),(),()(),(

0

PxSxS

xSxxxx

L

t

8-14

Quadrature integrationQuadrature integrationQuadrature integrationQuadrature integration

• In a minute, we are going to represent the angular dependence In a minute, we are going to represent the angular dependence of the angular flux as a of the angular flux as a quadraturequadrature..

• But, since this is a relatively obscure mathematical practice, let’s But, since this is a relatively obscure mathematical practice, let’s review what a quadrature review what a quadrature isis and the specific properties of a and the specific properties of a Gauss-Legendre quadrature.Gauss-Legendre quadrature.

• Whereas an expansion is used to approximate a Whereas an expansion is used to approximate a functionfunction at all at all points of the domain:points of the domain:

a quadrature is used to approximate the a quadrature is used to approximate the integralintegral of a function of a function between between aa and and bb by sampling specific points of the function and by sampling specific points of the function and using a weighted sum (with weights summing to 1):using a weighted sum (with weights summing to 1):

11),(12)(0

xxPfxfL

1

( ) ( ) ( )b N

n nna

f x dx b a w f x

8-15

Quadrature integration (2)Quadrature integration (2)Quadrature integration (2)Quadrature integration (2)

• The different quadratures differ in the number of points sampled, the prescription The different quadratures differ in the number of points sampled, the prescription of the values to be sampled, and how much weight each is given:of the values to be sampled, and how much weight each is given:

• The most common quadratures use equally spaced x values, but use specially The most common quadratures use equally spaced x values, but use specially chosen weights:chosen weights:

• Reimann sum: N cell-centered x’s, weights=1/NReimann sum: N cell-centered x’s, weights=1/N• Trapezoidal rule: N+1 cell-edged x’s, weights=(0.5,1,1,…,1,0.5)/NTrapezoidal rule: N+1 cell-edged x’s, weights=(0.5,1,1,…,1,0.5)/N• Simpson’s rule: N+1 cell-edged x’s, weights=(1,4,2,4,2,…,2,4,1)/(3N)Simpson’s rule: N+1 cell-edged x’s, weights=(1,4,2,4,2,…,2,4,1)/(3N)

• Example:Example:

• Reimann: 2(0.5^3+1.5^3)=3.5Reimann: 2(0.5^3+1.5^3)=3.5• Trapezoidal: 2(0.5*0+1*1+0.5*8)=5Trapezoidal: 2(0.5*0+1*1+0.5*8)=5• Simpson’s: 2(1*0+4*1+1*8)/3)=4 (Simpson’s rule is exact up to 3Simpson’s: 2(1*0+4*1+1*8)/3)=4 (Simpson’s rule is exact up to 3 rdrd order) order)

N ... 2, 1,n for nn xw ,

242

0

3 Nwithdxx

8-16

Gauss-Legendre QuadratureGauss-Legendre QuadratureGauss-Legendre QuadratureGauss-Legendre Quadrature

• An Nth-order Gauss-Legendre quadrature is An Nth-order Gauss-Legendre quadrature is RESTRICTED to integrals between -1 and 1 (although RESTRICTED to integrals between -1 and 1 (although they can be adapted for other domains).they can be adapted for other domains).

• This quadrature defines the wThis quadrature defines the wnn’s and the x’s and the xnn’s by the ’s by the following rules:following rules:• The xThe xnn’s are chosen to be the roots of the Nth order Legendre ’s are chosen to be the roots of the Nth order Legendre

polynomial:polynomial:

• The wThe wnn’s are chosen so that the Legendre moments up to order ’s are chosen so that the Legendre moments up to order N are satisfied exactly, i.e,N are satisfied exactly, i.e,

NnxP nN ,...,2,10)( for

1

11

2( ) ( ) 2 1

0

N

n nn

, if is evenP x dx w P x

, if is odd

3x

8-17

Gauss-Legendre Quadrature (2)Gauss-Legendre Quadrature (2)Gauss-Legendre Quadrature (2)Gauss-Legendre Quadrature (2)

• The power in a Gauss-Legendre quadrature is The power in a Gauss-Legendre quadrature is that an Nth-order Gauss-Legendre quadrature that an Nth-order Gauss-Legendre quadrature integrates a (2N-1)th order polynomial exactly.integrates a (2N-1)th order polynomial exactly.

• The resulting quadrature sets are given in text on The resulting quadrature sets are given in text on Table 3-1. Table 3-1.

• NOTE: Only the positive values of the xNOTE: Only the positive values of the xnn are given. are given. You must include the negative values as well (with the You must include the negative values as well (with the same weights.)same weights.)

• NOTE: The ORNL code convention is to have all NOTE: The ORNL code convention is to have all weights add up to 1. The text follows the convention weights add up to 1. The text follows the convention that the weights add up to the width of the domain of that the weights add up to the width of the domain of integration (i.e., 2 for our problems). In effect, they integration (i.e., 2 for our problems). In effect, they incorporate the (b-a) term the weights.incorporate the (b-a) term the weights.

8-18

Gauss-Legendre Quadrature (3)Gauss-Legendre Quadrature (3)Gauss-Legendre Quadrature (3)Gauss-Legendre Quadrature (3)

8-19

Application of SN quadrature to directionApplication of SN quadrature to directionApplication of SN quadrature to directionApplication of SN quadrature to direction

• We will now apply the Gaussian quadrature to our We will now apply the Gaussian quadrature to our direction variabledirection variable

• As engineers, we are seldom interested in the angular flux As engineers, we are seldom interested in the angular flux itself. We are much more interested in REACTION itself. We are much more interested in REACTION RATES, which are determined from the scalar flux (using RATES, which are determined from the scalar flux (using unit :unit :

1

1

1

( ) ( )

( ) ( , )

( , )

V

N

n nn

Reactions rates of type x x dx

x x d

w x

8-20

Applic. of SN quadrature to direct’n (2)Applic. of SN quadrature to direct’n (2)Applic. of SN quadrature to direct’n (2)Applic. of SN quadrature to direct’n (2)

• This approach also fits well with our definition of the This approach also fits well with our definition of the angular source, for which we use flux moments:angular source, for which we use flux moments:

'0

' 1

0

( ) ( ) ( ) ( ) ( )

( )

( )

Gg g

ext s gg

ext

S x S x F x x x

where S x external isotropic source

F x fission particle production

Kronecker delta needed because

fission and external source

'

1

11

( )

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

g gs

N

g g n g n nn

isotropic

x th coefficient of g' g scatter

x x P d w x P

8-21

Resulting equationResulting equationResulting equationResulting equation

• The final result is that we calculate the angular flux ONLY The final result is that we calculate the angular flux ONLY IN PARTICULAR DIRECTIONS (“discrete ordinates”):IN PARTICULAR DIRECTIONS (“discrete ordinates”):

• We will begin the spatial attack next weekWe will begin the spatial attack next week

0

1

( ) ( ) ( ) ( , )

( ) ( , )

( , ) 2 1 ( ) ( )

( ) ( ) ( )

n n t n n

n n

L

n n

N

n n nn

d x x x S x

dxwhere

x x

S x S x P

From which we get the flux moments for the current group:

x w x P

and the source moment

s from 1st equation on previous slide.

8-22

Homework 8-1Homework 8-1Homework 8-1Homework 8-1

• Using the information on slide 8-16, find the Legendre weights and values for N=3 and N=5

You may find the following recurrence relation to be of use:

0

1

1 2

( ) 1

( )

(2 1) ( ) ( 1) ( )( )

P x

P x x

xP x P xP x

8-23


• Solve for :

a.

b.

using the Legendre-based quadratures from Table 3-1 of the text. For each of them, use the 2, 4, 6, 8, 10, and 12 order quadrature parameters.

• Solve for by using

proportional x values (same weights)

1

1

f x dx

2( ) xf x e

, 1 0.2( )

1 , 0.2 1

0 for xf x

for x

2

1

f x dx2( ) xf x e

8-24


• Demonstrate (not prove) that use of the N=12 quadrature can exactly integrate powers of xl between -1 and 1 for up to l=2N-1=23. between.

• Since roundoff error is unavoidable, it is sufficient to show me results for l=0 to (say) 30 and show that very low errors become not-so-low errors after 23.

8-1 lesson 8 objectives recap and iteration practice recap and iteration practice reduction of b.e....

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