10-1 lesson 10 objectives chapter 4 [1,2,3,6]: multidimensional discrete ordinates chapter 4...

10-1

Lesson 10 ObjectivesLesson 10 ObjectivesLesson 10 ObjectivesLesson 10 Objectives

• Chapter 4 [1,2,3,6]: Multidimensional Chapter 4 [1,2,3,6]: Multidimensional discrete ordinatesdiscrete ordinates• Multidimensional quadraturesMultidimensional quadratures• Cartesian (x-y-z) difference equations and Cartesian (x-y-z) difference equations and

solution strategiessolution strategies• Ray effectsRay effects

10-2

MultiD quadraturesMultiD quadraturesMultiD quadraturesMultiD quadratures

• In contrast to the one direction cosine we considered In contrast to the one direction cosine we considered in Chapter 3, multidimensional D.O. use all 3:in Chapter 3, multidimensional D.O. use all 3:

• In terms of the eight octants of unit directional In terms of the eight octants of unit directional sphere:sphere:

1

;;222

zyx

OctantOctant

11 ++ ++ ++

22 -- ++ ++

33 -- -- ++

44 ++ -- ++

55 ++ ++ --

66 -- ++ --

77 -- -- --

88 ++ -- --

10-3

MultiD quadratures (2)MultiD quadratures (2)MultiD quadratures (2)MultiD quadratures (2)

• We have the following relationships between We have the following relationships between geometry type and octants that must be geometry type and octants that must be calculated:calculated:

GeometryGeometry Unique octantsUnique octants ReasonReason

1D slab/spherical1D slab/spherical 1,2 (1,2 ( only) only)

1D cylindrical1D cylindrical 1,2 (1,2 ( & & )) (same)(same)

2D geometries2D geometries 1,2,3,4(1,2,3,4( & & ))

3D geometries3D geometries All 8 (All 8 (, , , , )) No symmetryNo symmetry

,,,,

,,,,

10-4


• Note that 1D cylindrical has an angular dependence Note that 1D cylindrical has an angular dependence like 2D because there is no rotational symmetry like 2D because there is no rotational symmetry about the (one) spatial axisabout the (one) spatial axis

• Again, we will be handling the directional flux Again, we will be handling the directional flux dependence with a QUADRATURE: Calculating the dependence with a QUADRATURE: Calculating the angular flux only in particular directions and find the angular flux only in particular directions and find the flux moments as weighted sums of these “discrete flux moments as weighted sums of these “discrete ordinates”:ordinates”:

)()ˆ,()(

),,(,ˆ

)ˆ,()(

14

)()()(

rwrdr

rr

n

N

nn

nnnn

nn

10-5


• Let’s see how quadratures are laid out. First of Let’s see how quadratures are laid out. First of all, we will assume the quadrature has all, we will assume the quadrature has reflectionreflection symmetry symmetry

• If Octant 1 has a direction , then:If Octant 1 has a direction , then:• Octant 2 has the direction Octant 2 has the direction • Octant 3 has the direction , etc.Octant 3 has the direction , etc.

• This is NOT required, but it gives us all the This is NOT required, but it gives us all the angles needed for reflection B.C.’s and keeps angles needed for reflection B.C.’s and keeps things balancedthings balanced

• It is COMMON for shielding problems to use It is COMMON for shielding problems to use biased quadratures that have more quadrature biased quadratures that have more quadrature directions pointing in a direction of interestdirections pointing in a direction of interest

,,

,,

,,

10-6


3D quadrature—S?3D quadrature—S?

10-7

• The normal way of laying out quadratures is The normal way of laying out quadratures is the Level Symmetric Quadratures:the Level Symmetric Quadratures:

• Note that the Note that the -directions are not drawn, but -directions are not drawn, but you should be able to “see” themyou should be able to “see” them

• Be sure to keep the confusing subscripts Be sure to keep the confusing subscripts straight (I started over with 1 even though we straight (I started over with 1 even though we would normally call it would normally call it 55).).


Note the Note the and and “levels” “levels”(the (the -levels are there, too)-levels are there, too)

SS88 quadrature quadrature

10-8


• For Level Symmetric Quadratures (LSQ), the For Level Symmetric Quadratures (LSQ), the symmetry is extended to include 90 degree rotations symmetry is extended to include 90 degree rotations as well:as well:

• Somewhat surprisingly, there is only ONE degree of Somewhat surprisingly, there is only ONE degree of freedom in determining a LSQ: Once you choose the freedom in determining a LSQ: Once you choose the value of the first level, value of the first level, 11, the rest of the quadrature , the rest of the quadrature falls out automatically. Let’s see why.falls out automatically. Let’s see why.

• Assuming that we have a direction that comprises the Assuming that we have a direction that comprises the three levels i, j, and k, we must have:three levels i, j, and k, we must have:

2/,...,2,1 Nnnnn ,

1222 kji

10-9


• The key thing to “see” is that if we stay on a given The key thing to “see” is that if we stay on a given --level symmetry and increase the mu-level by one, then level symmetry and increase the mu-level by one, then the eta-level has to decrease by one:the eta-level has to decrease by one:

• If we subtract:If we subtract:

• we get:we get:

• But, since the mu, eta, and xsi values are all the same:But, since the mu, eta, and xsi values are all the same:

• This means that the differences of the squares of This means that the differences of the squares of consecutive levels must be some constant, C, that is consecutive levels must be some constant, C, that is constant for the quadratureconstant for the quadrature

2 2 2 21 1i i j j

1221

21 kji

1222 kji

2 2 2 21 1i i j j C

10-10


• So, if you know So, if you know 11 and C, you know the others: and C, you know the others:

• Using the one real point we can count on: That the Using the one real point we can count on: That the direction closest to the North Pole in our graph HAS direction closest to the North Pole in our graph HAS to have the lowest to have the lowest , the lowest , the lowest , and the HIGHEST , and the HIGHEST , therefore:, therefore:

Cj

CC

C

j 1

2

21

2

21

22

23

21

22

2 2 21 1 /2 1N

10-11


• which can be translated into:which can be translated into:

• So, after picking a So, after picking a 11, we can find C from the , we can find C from the above equation, then build the N/2 level values.above equation, then build the N/2 level values.

• With the level values, the actual quadrature With the level values, the actual quadrature comprises all combinations of i, j, and k such thatcomprises all combinations of i, j, and k such that

2 21 1 /2

2 2 21 1 1

2211

1

( 1) 12

2 1 31 3

212

N

NC

CN N

1222 kji

10-12


• The only quadrature this does not work for is SThe only quadrature this does not work for is S22, for which C is , for which C is

undefined.undefined.• For this special case (one direction per quadrature), symmetry For this special case (one direction per quadrature), symmetry

demands that:demands that:

• From the required relation:From the required relation:

we can easily see that:we can easily see that:

• Interestingly, this is the LARGEST that Interestingly, this is the LARGEST that 1 1 can be since the can be since the formula for C is only non-negative if:formula for C is only non-negative if:

3/11

111

121

21

21

57735.03/1111

10-13


• Returning to the original octant drawing, we see that the upshot of Returning to the original octant drawing, we see that the upshot of all this is that the number of all this is that the number of values in each “eta-level” decreases values in each “eta-level” decreases by 1 for each step we make:by 1 for each step we make:

• This also works vertically (for xsi-levels), although you cannot see it This also works vertically (for xsi-levels), although you cannot see it in the graph. (In looking at the graph, you should imagine that you in the graph. (In looking at the graph, you should imagine that you looking down on point drawn on the first octant of a basketball, e.g., looking down on point drawn on the first octant of a basketball, e.g., the lower left one is the HIGHEST one. Look at 10-6 again.)the lower left one is the HIGHEST one. Look at 10-6 again.)

• So, how many directions ARE there, total?So, how many directions ARE there, total?

10-14


• Let’s count them:Let’s count them:1.1. There are N/2 directions for 1There are N/2 directions for 1stst “eta-level” “eta-level”

2.2. There are (N-1)/2 directions for the 2There are (N-1)/2 directions for the 2ndnd eta-level. eta-level.

3.3. ……

4.4. There is 1 direction for the (N/2)th eta-level.There is 1 direction for the (N/2)th eta-level.

• For a 3D geometry (the whole basketball), there are 8 octants, soFor a 3D geometry (the whole basketball), there are 8 octants, so

• For 2D (x,y) geometry, we have symmetry up and down, so there For 2D (x,y) geometry, we have symmetry up and down, so there is no use calculating the “down” directions:is no use calculating the “down” directions:

/ 2 ( -1) / 2 ... 1

/ 2( / 2 1) ( 2)

2 8

# of levels per octant N N

N N N N

( 2)8 ( 2)

8

N N# of levels for 3D N N

( 2) ( 2)4

8 2

N N N N# of levels for 2D

10-15


Once the directions are found, the weights are found with a 3 Once the directions are found, the weights are found with a 3 step process:step process:

• All quadrant reflections are restrained to have the same All quadrant reflections are restrained to have the same weight:weight:

This means we only have to worry about the N(N+2)/8 values This means we only have to worry about the N(N+2)/8 values in the first quadrant.in the first quadrant.

• All permutations of the level subscript numbers have the All permutations of the level subscript numbers have the same weight.same weight.

• The unique weights that remain are found by either:The unique weights that remain are found by either:1.1. Correctly integrating as many 1D moments as possible (works up Correctly integrating as many 1D moments as possible (works up

to about Sto about S2020); or); or2.2. Associating an area of the unit sphere with each direction Associating an area of the unit sphere with each direction

(maintaining symmetry) and using (maintaining symmetry) and using as the weight. as the weight.

),,(),,( kjikji ww

etc. ,),,(),,(),,( kijjkikji www

10-16

Find the equal weightsFind the equal weightsFind the equal weightsFind the equal weights• Take this quadrature (what is its order?) and label with the Take this quadrature (what is its order?) and label with the

indices:indices:

• Now mark out the directions with the same weightNow mark out the directions with the same weight

10-17

Find the equal weights (2)Find the equal weights (2)Find the equal weights (2)Find the equal weights (2)• The unique weights that remain are found by: The unique weights that remain are found by:

1.1. Finding the weights for each of the N/2 “levels” to correctly Finding the weights for each of the N/2 “levels” to correctly integrate the EVEN 1D moments to order 2(N-1).integrate the EVEN 1D moments to order 2(N-1).

2.2. Solving the linear algebra problem to find the unique Solving the linear algebra problem to find the unique weights for the multidimensional directions—which will weights for the multidimensional directions—which will only have a solution for a particular value of only have a solution for a particular value of 11..

• They say this works up to about SThey say this works up to about S2020

• Book only goes to SBook only goes to S1616

• See Table 4-1 in text.See Table 4-1 in text.

1 /2

10

1 , 0, 2,..., 2

2

N

n nn

d w N

10-18

Cartesian difference equationsCartesian difference equationsCartesian difference equationsCartesian difference equations

• The spatial differencing for (x,y) geometry is a The spatial differencing for (x,y) geometry is a straight-forward extension of what we did in straight-forward extension of what we did in slab.slab.

• Integration of the continuous space equation Integration of the continuous space equation over a rectangular cell:over a rectangular cell:

gives us:gives us:

Cell ijCell ij

2/1ix 2/1ix

2/1jy

2/1jy

10-19

Cartesian difference equations (2)Cartesian difference equations (2)Cartesian difference equations (2)Cartesian difference equations (2)

• Again invoking the cell edge fluxes (this time 4 of them):Again invoking the cell edge fluxes (this time 4 of them):

1/21/2

1/2 1/2

1/21/2

1/2 1/2

( , ) ( , )( ) ( )

( , )

yxn n

n n t n

x y

yx

n

x y

x y x yz dx dy x x

x x

z dx dy S x y

2/1,, jin

jin ,2/1, jin ,2/1,

2/1,, jin

nij

10-20


iin

y

y

x

x

nij

iin

y

y

x

x

nij

in

x

x

jin

in

y

y

jin

nijnijtijjinjini

njinjin

i

n

yxyxSdydxS

yxyxdydx

xyxdx

yyxdy

Syx

/),(

/),(

/)2/1,(

/),2/1(

2/1

2/1

2/1

2/1

2/1

2/1

2/1

2/1

2/1

2/1

2/1,,

2/1

2/1

,2/1,

2/1,,2/1,,,2/1,,2/1,

:where

• Dividing by the cell volume Dividing by the cell volume xxyyz gives:z gives:

10-21


• This gives us 5 unknown fluxes with only 1 This gives us 5 unknown fluxes with only 1 equation, which we attack as we did for the equation, which we attack as we did for the slab equation:slab equation:

1.1. Two of the fluxes are known from previous cells Two of the fluxes are known from previous cells that have been calculated; andthat have been calculated; and

2.2. Two of the fluxes are found from the introduction Two of the fluxes are found from the introduction of auxiliary equations.of auxiliary equations.

• The most common auxiliary equations are The most common auxiliary equations are the 2D equivalents of step, diamond-the 2D equivalents of step, diamond-difference, weighted diamond difference, difference, weighted diamond difference, and characteristicand characteristic

• We will work with diamond difference to We will work with diamond difference to show you how the math goesshow you how the math goes

10-22


• Again, average flux from incoming and outgoing. Again, average flux from incoming and outgoing. • This time we can do it in both dimensions:This time we can do it in both dimensions:

which can be rearranged to give:which can be rearranged to give:

• Substitution gives us:Substitution gives us:

222/1,,2/1,,,2/1,,2/1,

jinjinjinjinnij

2/1,,2/1,,

,2/1,,2/1,

2

2

jinnijjin

jinnijjin

i

n

i

ntij

jini

njin

i

nnij

nij

yx

yxS

22

222/1,,,2/1,

10-23

Cartesian solution strategiesCartesian solution strategiesCartesian solution strategiesCartesian solution strategies

• As for 1D, the solution “sweep” strategy consists of As for 1D, the solution “sweep” strategy consists of beginning at a known boundary and following the beginning at a known boundary and following the particles across the cellparticles across the cell

• This time there are 4 variations (one for each corner):This time there are 4 variations (one for each corner):

0,0 0,0

0,0 0,0

10-24

Non-Cartesian Equations/Solution StrategiesNon-Cartesian Equations/Solution StrategiesNon-Cartesian Equations/Solution StrategiesNon-Cartesian Equations/Solution Strategies

• For curvilinear geometries, we can develop a set of For curvilinear geometries, we can develop a set of equations that is discretized in direction and space, equations that is discretized in direction and space, like we saw in 1D spherical. The resulting like we saw in 1D spherical. The resulting relationships are too esoteric for me to force you relationships are too esoteric for me to force you through.through.

• Basic characteristics you should know are:Basic characteristics you should know are:1.1. The flux in any direction has an additional “linkage” to fluxes in The flux in any direction has an additional “linkage” to fluxes in

“neighboring” directions.“neighboring” directions.2.2. The only exception to the above is a direction (1 per “eta level”) The only exception to the above is a direction (1 per “eta level”)

that has that has =0. (See 10-6 again; they are plotted as 1, 4, and 9.)=0. (See 10-6 again; they are plotted as 1, 4, and 9.)3.3. Therefore, this “backward” direction must be added to the Therefore, this “backward” direction must be added to the

quadrature (with weight=0) and calculated FIRST to “jump-quadrature (with weight=0) and calculated FIRST to “jump-start” the processstart” the process

4.4. Many general SMany general SNN quadratures keep these “zero-weight” angles quadratures keep these “zero-weight” angles and use them even for Cartesian geometries.and use them even for Cartesian geometries.

10-25

Ray effectsRay effectsRay effectsRay effects

• The principal shortcoming of multiD discrete The principal shortcoming of multiD discrete ordinates is the existence of ray effects in ordinates is the existence of ray effects in some problem flux solutions.some problem flux solutions.

• Ray effects show up as non-physical Ray effects show up as non-physical oscillation “waves” in fluxes for regions far oscillation “waves” in fluxes for regions far away from source regions:away from source regions:

““Source in a corner”Source in a corner”

10-26

Ray effects (2)Ray effects (2)Ray effects (2)Ray effects (2)

• The problem arises, NOT from numerical errors, but from the The problem arises, NOT from numerical errors, but from the basic discrete ordinates representation of angular fluxes basic discrete ordinates representation of angular fluxes discretely.discretely.

• Ray effect “solutions” generally fall into three categories:Ray effect “solutions” generally fall into three categories:1.1. Increase the angular representation in problem directions. This Increase the angular representation in problem directions. This

either takes the form of increasing the number of directions or by either takes the form of increasing the number of directions or by “biasing” the directions in toward the problem directions.“biasing” the directions in toward the problem directions.

2.2. Get away from discrete ordinates by using a continuous Get away from discrete ordinates by using a continuous representation of direction (Spherical Harmonics methods)representation of direction (Spherical Harmonics methods)

3.3. Couple the problem with a first collision source estimator.Couple the problem with a first collision source estimator.

• The last solution consists of precalculating (i.e., before the The last solution consists of precalculating (i.e., before the DO solution) the uncollided contribution to the response and DO solution) the uncollided contribution to the response and the spatial, energy, and angle distribution of particles the spatial, energy, and angle distribution of particles emerging from their FIRST collision.emerging from their FIRST collision.

• Then the DO solution uses the (more evenly distributed) first Then the DO solution uses the (more evenly distributed) first collision source as its external sourcecollision source as its external source

9-27

Homework 10-1Homework 10-1Homework 10-1Homework 10-1

Define the SDefine the S1010 level symmetric quadrature by giving the level symmetric quadrature by giving the

direction cosines for the 15 directions in the first octant. direction cosines for the 15 directions in the first octant. Let Let 11 be 0.14. be 0.14.

9-28

Review for Test#2Review for Test#2Review for Test#2Review for Test#2

9-29

Review for Test#2Review for Test#2Review for Test#2Review for Test#2

10-1 lesson 10 objectives chapter 4 [1,2,3,6]: multidimensional discrete ordinates chapter 4...

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