7 Flux Shape in Various Reactor Geometries in 1 Energy Group

8/3/2019 7 Flux Shape in Various Reactor Geometries in 1 Energy Group

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2008 September 1

Flux Shape in Various Reactor Geometries

in One Energy Group

B. Rouben

McMaster University

EP 4D03/6D03

Nuclear Reactor Analysis

2008 Sept-Dec


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2008 September 2

Contents

We derive the 1-group flux shape in the critical

homogeneous infinite-slab reactor and infinite-

cylinder reactor + Exercises for other

geometries. Reference: Duderstadt & Hamilton Section 5 III


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2008 September 3

Diffusion Equation

We derived the time-independent neutron-balanceequation in 1 energy group for a finite, homogeneous

reactor:

We showed that we could introduce the concept ofgeometrical buckling B2 (the negative of the flux

curvature), and rewrite the equation as

where B2

has to satisfy the criticality condition

)1(02 !77 JRJJ faD

)2(022 ! JJ B

)3(112

2

L

k

DB af

!77! gR


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2008 September 4

Solving the Flux-Shape Equation

Eq. (2) is the equation to be solved for the flux shape. We will study solutions of this shape equation for various

geometries, and will start with the case of an infinite

cylindrical reactor.

The thing to remember is that the solution must satisfy thediffusion boundary condition, i.e. flux = 0 at the

extrapolated outer surface of the reactor.

While Eq. (2) can in general have a large multitude of

solutions, we will see that the addition of the boundarycondition makes Eq. (2) an eigenvalue problem, i.e., a

situation where only distinct, separated solutions exist.


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2008 September 5

Solving the Flux-Shape Equation

We will now apply the eigenvalue equation tothe infinite-slab geometry and the infinite-

cylinder geometry, and solve for the

geometrical buckling and the flux shape. As concluded before, we will always find that

the curvature of the 1-group flux in a

homogeneous reactor is always negative.


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2008 September 6

Interactive Discussion/Exercise

Given that the curvature of the 1-group flux in a

homogeneous reactor is negative, where do you think

the maximum flux would have to be in a regular-shaped

reactor?

Explain.

Do not turn the page until you have attempted/done this

discussion/exercise.


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2008 September 7

Intentionally Left Blank


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2008 September 8

Maximum Flux

In a 1-group homogeneous reactor of regular shape, themaximum flux must necessarily be at the centre of thereactor.

We can explain that by reductio adabsurdum.

Suppose the flux is a maximum not at the centre of thereactor; if we draw a straight line from the maximumpoint to the centre of the reactor, then by symmetry,there would have to be another maximum on thesame line on the other side of the centre. Therefore the

centre of the reactor would be at a local minimumalong that straight line, which would imply that the fluxdoes not have negative curvature along that straightline!


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2008 September 9

Infinite-Slab Case

Let us study the simple case of a slab reactor of width

a in thex direction, and infinite in they andz

directions.

Eq. (2) then reduces to its 1-dimensional version, in

thex direction

(4)

We can without loss of generality place the slabsymmetrically aboutx = 0, in the interval [-a/2, a/2].

contd

022

2

! JJ

Bdx

d


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2008 September 10

Infinite-Slab Reactor Geometry

x = 0

x = a/2 x =

aex/2

x = -a/2


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2008 September 11

Infinite Slab (contd)

We also know that:

The flux must be symmetric about x = 0, and

The flux must be 0 at the extrapolated boundaries, which

we can call s aex/2

Eq. (4) has the well-known solutions sin(Bx) and

cos(Bx).

Therefore the most general solution to Eq. (4) may

be written as(5)

contd

BxABxAx sincos 21 !J


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2008 September 12


However, symmetry aboutx = 0 rules out the sin(Bx)component.

Thus the reactor flux must be

(6)

We can determine B from the boundary condition at aex:

(7)

Now remember that the cos function has zeroes only at

odd multiples ofT/2. Therefore B must satisfy:(8)

contd

BxAx cos1!J

02

cos1 !

exBaA

oddnwitha

nBei

nBa

ex

ex !!!TT

.,.,22


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2008 September 13


It looks as if there is an infinite number of values ofB. While that is true mathematically, the only physically

possible value for the flux in the critical reactor is the onewith the lowest value ofB, i.e. forn =1:

(9)

We can conclude that Eq. (9) is the only physical solutionfrom the fact that the solutions with n = 3, 5, 7, all featureregions of negative J in the reactor, and that is not physical.

Also to be noted from Eq. (9) is that the buckling increases asthe dimensions of the reactor(here aex) decrease [as had alsobeen concluded earlier] the curvature needs to be greater toforce the flux to 0 at a closer boundary!

contd

exa

BT

!


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2008 September 14

Infinite-Slab Case (contd)

The 1-group flux in the infinite-slab reactor can then bewritten

The absolute value of the flux, which is related to theconstantA

1

, is undetermined at this point. This is because Eq. (3) is homogeneous therefore any

multiple of a solution is itself a solution. The physicalsignificance of this is that the reactor can function atany power level.

Therefore, to determineA1, we must tie the flux downto some quantitative given data e.g., the desired totalreactor power. contd

)10(cos1

!

exa

xAx

TJ


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2008 September 15

Infinite-Slab Case (contd)

Because the slab is infinite, so is the total power. But we can use the powergenerated per unit area of the slab as normalization. Let this be P W/cm2.

If we call Efthe recoverable energy per fission in joules (and we know that

this is ~200 MeV =3.2*10-11 J), then we can write

And doing the integration will allow us to find A1:

i.e.,

So that finally we can write the absolute flux as

)11()cos(2/

2/

2/

2/

1 dx

a

xAEdxxEP

a

a

a

a ex

ffff

7!7!T

J

)12(2

sin2

sin 1

2/

2/

1

7!

7!

exff

ex

a

aex

ffex

a

aAE

a

a

xAE

aP

ex

ex

T

T

T

T

)13(

2sin2

1

7!

ex

ffa

aaE

P

A T

T

)14(cos

2sin2

7

!ex

ex

ff

a

x

a

aaE

Px

T

T

TJ


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2008 September 16

Additional Notes

In solving this problem, we found that the reactor equation had veryspecific solutions, with only specific, distinct values possible forB. Actually, there is a general point to be made here: Equations such as

Eq. (3), holding over a certain space and with a boundary condition,i.e., the diffusion equation for the reactor, fall in the category ofeigenvalue problems, which have distinct solutions [eigenfunctions] -here the flux distribution J -, with corresponding distinct eigenvalues(here the buckling B2).

Although in this problem we found only 1 physically possibleeigenfunction for the steady-state reactor(the fundamental solution),the other eigenfunctions are perfectly good mathematical solutions,which do have meaning.

While these higher eigenfunctions (which have larger values ofB)

cannot singly represent the true flux in the reactor, they can exist asincremental time-dependent perturbations to the fundamental flux,

perturbations which will die away in time as the flux settles into its

fundamental solution.


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2008 September 17

Case of Infinite-Cylinder Reactor

Infinite height Radius Rr


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2008 September 18

Infinite-Cylinder Reactor

For a homogeneous bare infinite cylinder, the flux is a function ofthe radial dimension ronly. All axial and azimuthal positions are

equivalent, by symmetry.

We write the eigenvalue equation in cylindrical

co-ordinates, but in the variable ronly, in which the divergence is

The 1-group diffusion equation then becomes

By evaluating the derivative explicitly, we can rewrite Eq. (16) as

contd

022 ! JJ B

)15(12

!

dr

dr

dr

d

r

JJ

)16(0)(1 2 !

rB

dr

rdr

dr

d

rJ

J

)17(01 2

2

2

! JJJ

Bdr

d

rdr

d


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2008 September 19

Infinite-Cylinder Reactor(contd)

We may be completely stumped by Eq. (17), but luckilyour mathematician friend recognized it as a special caseof an equation well known to mathematicians, Besselsequation (m is a constant):

Eq. (17) corresponds to m = 0, for which this equationhas 2 solutions, the ordinary Bessel functions of the 1st

and 2nd

kind, J0(Br) and Y0(Br) respectively. These functions are well known to mathematicians (see

sketch on next slide)![It sure helps having mathematicians as friends, isnt it,even if Nobel didnt like them!] contd

)18(01

2

2

22

2

!

J

JJr

mB

dr

d

rdr

d


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2008 September 20


I sketch the functions J0(x) and Y0(x)below:

Although, mathematically speaking, the general solution of Eq. (17)

is a combination ofJ0(x)andY0(x), Y0(x) tends to -g as x tends to 0and is therefore not physically acceptable for a flux. contd


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2008 September 21


The only acceptable solution for the flux in a bare, homogeneousinfinite cylinder is then

The flux must go to 0 at the extrapolated radial boundary .

Therefore we must have The figure in the previous slide shows that J0(x) has several

zeroes, labelled [the 1st is at x1 = 2.405, the 2nd at x2 } 5.6]

But because, physically, the flux cannot have regions of negativevalues, B for the infinite cylinder can be given only by

Therefore the buckling for the infinite cylinder is

)19(0 BrAJr !J

exR

)20(00 !exBRJ

)21(405.211exex RR

xB !!

)22(405.2

2

2

!

exR

B


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2008 September 22


The 1-group flux shape in the infinite homogeneouscylindrical reactor is then

As before, the absolute magnitude of the flux (i.e., the

constantA) can be determined only from some

quantitative information about the flux, for example

the power per unit axial dimension of the cylinder.

If we denote that power densityP

, and the energyreleased in fission, we can write:

)23(405.2

0

!

exR

rAJrJ

)24(220 0

00 7!7!7!R R

ffffff rdrBrJAErdrBrJAEdVEP TTJ


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2008 September 23


The integral on the Bessel function may lookforbidding, but it can be evaluated from known

relationships between various Bessel functions.

Ill just give the final result here without derivation.

If we ignore the extrapolation distance,

which gives for the 1-group flux the final equation

)26(405.2738.0

02

7

!R

rJ

RE

Pr

ff

J

)25(738.0

2RE

PA

ff7!


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2008 September 24

Exercise/Assignment: Apply to Other Shapes

Exercise: Apply the eigenvalue Eq. (3) to thefollowing geometries to find the geometrical

buckling and the flux shape: Parallelepiped

Finite cylinder Sphere

Note: in the cases of the parallelepiped and of the

finite cylinder, invoke separability, i.e., write the

solution as a product of functions in the appropriatedimensions, each with its own directional bucklings,

which add to the total buckling.

Do not turn the page until you have attempted/done

this discussion/exercise.


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2008 September 25

Intentionally Left Blank


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2008 September 26

Parallelepiped Reactor

-a/2 +a/2

+b/2

-b/2

c/2

-c/2x

y

z


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2008 September 27

Parallelepiped Reactor

Directional bucklings:

Total buckling:

Flux shape:

If total power = P, and neglecting extrapolation distance:

)27(,,

2

2

2

2

2

2

!

!

!

ex

z

ex

y

ex

xc

Bb

Ba

BTTT

)28(2222

zyx BBBB !

)29(coscoscos,,exexex c

z

b

y

a

xAzyx

TTTJ !

)30(8

3

ffabcE

PA

7!

T


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2008 September 28

Finite-Cylinder Reactor

Finite height Radius Rr

+H/2

-H/2


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2008 September 29

Finite-Cylinder Reactor

Directional and total bucklings:

Flux shape:

)31(,,405.2 222

2

2

2

2

zr

ex

z

ex

r BBBH

BR

B !

!

!

T

)32(cos405.2

, 0

!

exex H

z

R

rAJzr

T

J


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2008 September 30

Spherical Reactor

Sphere of radius R Buckling:

Flux shape:

)34(sinr

R

r

Ar ex

!

T

J

)33(

2

2

!

exR

BT


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2008 September 31

Summary

We can obtain the solution for the 1-group fluxshape in bare homogeneous reactors of various

geometries.

In each case we determine directional bucklings(if applicable) and the total buckling, in terms of

the dimensions of the reactor.

The buckling(s) must take the lowestmathematical values allowed, to ensure that the

flux solution is physical everywhere in the reactor.


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2008 September 32

END

7 Flux Shape in Various Reactor Geometries in 1 Energy Group

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