J. Nuclear Energy. 1956, Vol. 2. pp. 193 to 201. Pergamon Press Ltd., London

MINIMUM CRITICAL MASS AND FLAT FLUX*

GERALD GoERTzEL Nuclear Development Corporation of America

(Received 18 August 1955)

Abstract-The problem of distributing fuel in a homogeneous moderator region so as to obtain a reactor of minimum critical mass is set up, and a solution is obtained for thermal reactors. The condition for minimum critical mass is found to be constant thermal flux in the region where there is fuel (the core), with the thermal flux never greater than this constant value where there is no fuel (the reflector). It is necessary to solve an integral equation to find the fuel distribution. The size and shape of the core region is determined from this integral equation. An explicit solution is given for the multigroup reactor model.

The minimum problem is also considered with certain restrictions, such as confining the core to a given region. Further, the minimum critical mass for U*” in light water is computed (with infinite reflector), and found to be 30 per cent less than the corresponding best value for a uniform fuel distribution in the core.

INTRODUCTION

A FUNDAMENTAL problem in reactor theory is how to design a reactor with the least amount of fuel. This article treats the problem under the following assumptions: (1) the reactor is thermal, (2) the moderating and thermal transport properties are uniform throughout the core and reflector, and (3), when the reflector is finite, the neutron flux is assumed to vanish at an extrapolated outer boundary which is the same for all energies.

The problem considered, with the above assumptions, is this: Given a homo- geneous moderating material of definite size and shape and a fuel with definite properties, how shall the fuel be distributed in the moderator to achieve, with a minimum amount of fuel, a critical assembly? Note that, according to assumption 2, the presence or absence of fuel is assumed to have a negligible effect on the properties of the moderator.

The answer is that the fuel is to be distributed in an internal core region of the moderator in such manner as to make the thermal-neutron flux constant throughout that region. The remaining, unfuelled portion of the moderator constitutes the reflector. The thermal flux is variable in.the reflector, but nowhere exceeds its con- stant value in the core. Note that the flat-core flux makes all the fuel operate at the <ame specific power;? this leads to some interesting engineering considerations,

FORMULATION OF CRITICALITY CONDITION

In order to avoid, for the present, a special model for the description of the reactor, and to maintain as much generality as is left after the assumptions of the previous se&ion, it is desirable to express the condition that the reactor is critical in terms of the appropriate integral equations. These equations will now be presented.

* Research carried out under kontract with the U.S. Atomic Energy Commission. t R. P. FEYNMAN studied this problem in 1941 and obtained similar conclusions.

193

194 GERALD GOERTZEL

Let F(X) be the number of fission neutrons born per unit volume and time, and let q(x) be the number of neutrons becoming thermal per unit volume and time. There then clearly exists some linear operator K;such that

q(x) = V(x)

In general, K, may be expressed as an integral operator

K,F( 3) = J k,( x, X’) F( X’) dx’

The integration is carried out over the volume V of the moderator region, and dx’ is the volume element. According to assumption 2, above, K, (and k,) is independent of the fuel distribution.

The next task is to relate the thermal-neutron source q(x) with the thermal- neutron sink. Let I#(*) be the number of thermal neutrons absorbed by the moderator per unit volume and time, and let y(x)+(x) be the number of thermal neutrons absorbed by the fuel per unit volume and time. Clearly w(x) is proportional to the fuel density at x. Then the following relation holds:

4(x) = KMX) - Y(M41

This particular form of relation is chosen so that K, will be independent of the fuel distribution p. In fact, q is the source of neutrons, and y# is the extra sink due to the fuel; therefore q - y$ is the effective source for neutrons destined to be absorbed in moderator. Thus K, is the operator which characterizes the relation between source and sink for pure moderator.

If r] is the number of fission neutrons born for each neutron absorbed in fuel, then

4%) = VP(xM(x)

Thus the criticality condition is that the equations

4(x) = K&(x) - w(x)+(x)1

q(x) = rK,~(xW(x)

have a non-trivial solution. [A non-trivial solution is one other than q =d = 0. In addition, of course, physically relevant solutions have b(x) > 0 and q(x) > 0.1 These two equations may be combined to give the equation

4(x) = (~4, - KMxM(4 where K,, = K,K,

If we further define H = 11 Kf,9 - KS

then the criticality equation becomes deceptively simple in form

4 =H!@

This form, which is the usual criticality condition in integral formulation, will be used below.

We define a scalar product as

(f,g) = s (j-4 g(s) dx

Minimum critical mass and flat flux 195

With this a symmetric operator L may be defined such that (f.Lg) = (g&f). It is shown in Appendix A that H is symmetric for many cases of interest. We assume symmetry of H in this discussion. In addition, His independent of w by construction.

FLAT-FLUX MINIMAL PROPERTY

As was seen in the previous section, a reactor is critical if the equation

has a non-trivial solution for 4. Clearly this requirement defines a class of functiops y. This class of functions is further restricted by the requirement, arising from the definition of y’, that Y(X) 2 0.

Also note that the mass of fuel is proportional to j’ w dx = m. Thus the problem is to find, of all the values of y > 0, the one which minimizes m. In order to simplify the solution, we change variables in such a way as to explicitly introduce the mass m of fuel. _

Let Y(X) =

mu(x) ju( x’)dx’

Then 4(x) J u(x’) dx’ = m&(x)$(x)

is an equation which for given u(x) will have a solution for only certain eigenvalues m. Our problem is now to find a function u(x) > 0 such that the lowest eigenvalue m has its minimum value.

The usual approach to this sort of problem is that of the calculus of variations. Changes in m which are due to variations of u about that u which minimizes m are studied. To eliminate the effect of changes in 4, consider

(u&#) j u dx’ - m(u$,Hu$) = I

and note* that I = 0 and 6Z= 0 if 4 satisfies

$judx’ =mHu# Now consider 6u and Sm

(u#,Hu+) 6m = (6u&#) J u dx’ + (u&4) j 6u dx’ - m(cYu~,Hu$) - m(u&H 6~4)

= (u+,+) J 6u dx’ - (6ucj,#) j u dx’ or

(Jqh$dxJudx’); = J Gu(x’)[J $ucj dx - d2(x’) j u dx] dx’

The coefficient of 6m is positive. Also, by hypothesis, u is such as to minimize m. Thus bm must be incapable of taking on negative values, whatever may be selected for du. If u > 0, 6u being arbitrary, it is necessary to have

+2Judx ==j++dx

or 4 = K, where K is some constant. On the other hand, if u 10, when 8u > 0, it is nkcessary to have 4 < K to ensure

that bm 3 0. * It is here and in the variations considered later in the article that use is made of the symmetry of H.

196 GERALD G~ERTZEL

Thus for minimum critical mass the thermal flux has a constant value wherever there is fuel, and is not greater than this constant value where there is no fuel. This is the fundamental result determined from this article.

To find the fuel distribution which will achieve minimum critical mass and there- fore will achieve uniform flux (4 constant) in the core (where u > d), it is necessary to solve the equation 4 = H+& in the core, where it becomes 1 = Hy. The solution of this equation for special forms of H is considered below. A typical solution is indicated in Fig. 1.

i

I

hj \2, -CORt~REFLECTOR----k

FIG. I.-Flux and fuel distribution for minimum critical-mass reactor.

FIG. 2.-Flux and fuel distribution for flat-flux reactor, core required to be smaller than that-for minimum critical mass. The crosshatched area is

concentrated at the core-reflector interface.

MINIMIZATION WITH RESTRICTIONS The conclusion presented in the previous section may be extended to the following

more general problem: Find the condition in which m is minimum for a y such that f-<Y<g. The solution may be stated as follows:

Wheref -=c y -=c g, #=K Where f = y, 4<K

Where w =g, 4>K That is, where w is free, 9 is ftxed with the value K; where y is bounded from below, 4 is bounded above; and, where w is bounded above, $ is bounded below. If w is fixed, C# is free, Our previous result was for f = 0 and g = co everywhere in the system.

A useful application of this extension arises. Suppose we require a reactor with specified core size. This corresponds to f = 0, with g = ~ZI in the core and g = 0 in the reflector. If the desired core size is smaller than would be achieved for minimum fuel, then the fuel tends to lump at the core-reflector interface, giving rise to a solution like that of Fig. 2. We note that since f = g = 0 in the reflector, no restriction applies to 4 in that region, i.e., 6~ = 0; thus the thermal flux in the reflector can now exceed the value in the core.

We may also ask under what circumstances a reactor restricted to constant fuel density in the core will have the least critical mass. In this case, in the core u = 1 and in the reflector I( = 0. Since 6u = 0 except at the core reflector interface

(4 edge)e j u dx = f I$% dx i.e., for the optimum core size the flux at the interface has its root-mean-square average value.

Minimum critical mass and flat flux 197

SOLUTION FOR MULTI-GROUP KERNELS We limit our considerations here to one-, two-, or three-dimensional problems

with symmetry such that one co-ordinate suffices. Thus let v = 1, 2, or 3 and V2 =x -(“-l) (d/dx) x”-l (d/dx). Let x = a be the edge of the core, let c(x) = 1 when x < a and l(x) = 0 when x > a, and let x = A be the edge of the moderator region.

The multi-group formulation* corresponds to

H - q ii K, - K,, i=O

where L,K, = 1, Li = -ai2v 2 + 1, and Ki vanishes at x = A. The xi values, along with ‘7, are the physical parameters of the problem.

For x < a

1 - Hy = (q i K, - K,)y i=o

12 applying the operator Il L, to this equation

i=O

1 ‘?&iL# i-1

The solution of this differential equation is 1

- + 2 4 F,(x) !q - I o=l

where (;Z0k,2 + 1) F,(x) = 0 and F,(x) - F,(-x) = 0. We thus consider for y

y(i) = c(x) [ hl + 2 B, sd + aR 6(x - a) 0=1 0

(1)

where the term proportional to A enables us to consider the problem of Fig. 2 along with that of Fig. 1, A = 0 corresponds to the unrestricted minimum fuel condition. In order for v to satisfy its differential equation for x < a, the 1, values are roots of

7-k 9+1 =() ( 1 I., (2)

i=l

To determine the B, and A consider 1 = Hy when x < a. Now if

Cj =II 1

i+j I - (ai/Q2

then

H = 17 5 C,K, - K, = 5 D,K, i=O i=O

(See Appendix B.) Also

+ aL2FjW [G’La) Fe-,(a) - Gi(a) F’,(a)]} where, if K&x) = f k,(x,x’)f(x’) dx’ defines Qx,x’), then k,(x,x’) = F,(x) Gi(.u’) for x < x’ and F,(x’)G,(x) for x > x’.

* Use is made here of the fact that the Ki commute (see Appendix A); the notation otherwise is ambiguous.

198 GERALD GOERTZEL

Thus Hy = T DiKiw = 1, together with equation (2) yields the equations i=o

i: l O=I 1 + (a,/&)”

where i =O, 1,. . . n. The solution of equation (3) substituted in equation (1) yields y. We also need

j 7yx’-l dx = m But

s

a F,(x) ~“-1 dx = --&2

s ‘d -_x “-1 d

0 o dx dx F,,(x) dx = - > a”aF’,(a)

0 Thus

m 1 1 n 2, * _ =-- - a’ Y?7--1 c B 0 ‘+ A

UP1 H a (4)

The following is a summary of the solution in the multigroup case: 1. Solve equation (2) for the 1,. 2. Solve equation (3) for the B, and A or for a such that A = 0. 3. Use equation (4) to find m. 4. Use equation (1) to find y.

Table 1 gives F,, and G, for v = 1, 2, 3.

TABLE l.-THE FUNCTIONS F,, AND G,

Slab (Y = 1) F&r) = cos (x/lo)

G,(x) = sinh [(A - x)/at]

Cylinder. (Y = 2) F,(x) = J 0(x/&)

Zo(x/a3 &(x/aJ -_- Gi(x) = Zo(A/ai) K&4/a,)

Sphere (v = 3) sin(x/J.J

FAX) = - X

G,(x) =’ sinh [(A - x)/a,]

X

MINIMUM CRITICAL MASS OF IPs IN LIGHT WATER

A detailed calculation of the kind outlined in the last section was performed for a spherical light-water reactor with pure U 235 fuel and an infinite light-water reflector. The three group kernel for the slowing down distribution in water introduced by GREULING was used*.

A list of the parameters used in the calculation is presented below.

Fast slowing down lengths, cm aI =4*49 a, = 2.05 aa = 1.00

Thermal slowing down length, cm a0 = 2.88 Fission neutrons/thermal neutron absorbed in fuel ?I ==2*10 Absorption cross-section for thermal neutrons in water, cm2/g (T,, = 0.0207 Absorption cross-section for thermal neutrons in U235, cm2/g (T,, = 1,639

* Unpublished, see GLASSTONE and EDMUND, The E/twents of Ndear Rmrtor T/wor,v (1952). van Nosirand, pages 369-371.

Minimum critical mass and flat flux 199

Results of the calculation are exhibited in Figs. 3, 4, and 5. As shown in Fig. 3, the minimum critical-mass (MCM) reactor radius is 167 cm, and the corresponding uranium investment is 0.69 kg of U23S. The minimum radius for which a critical assembly is possible lies at 9.6 cm.

RADIUS - cm

FIG. 3.-Uranium investment vs. core radius for flat-flux, spherical U2a5, light-water reactor with infinite moderator region.

0 2 4 b 8 IO I2 14 lb

RADIUS- cm FIG. 4.-Fuel density as a function of position for a minimum critical-mass, light-water reactor

with infinite moderator region.

0 2 4 b 8 IO I2 14

RADIUS - cm

FIG. S.-Fuel density as a function of position for a flat-flux, spherical U2a5, light-water reactor

with infinite moderator region.

Fig. 4 exhibits the uranium density for the MCM situation. For comparison Fig. 5 shows the density for a 13-cm flat-flux reactor.

Comparison of the MCM reactor with the best critical situation achieved in a homogeneous reactor indicates a saving of fuel of approximately 30 per cent by the former.

200 GERALD GOERTZEL

Some of the parameters for the solutions of Figs. 4 and 5 are given in Table 2.

TABLE Z.--CALCULATED PARAMETERS

Roots of the secular equation

- = 0.037237, -0.962532, -0.362264 A” 1 - = 1tO.19297, +0*98109j, f0%0188j rz

A, cm B1 & & h 16.676 - 660303 -0.148612 +@358454 oaOO33 13 -9.68983 +0521485 -0.620213 0.358112

Acknowledgements-The writer is deeply indebted to many of his colleagues for their interest, encouragement, and assistance. To be mentioned especially are EUGENE

GREULING, A. M. WEINBERG, J. ERNEST WILKINS, JUN., and GALE YOUNG. The cal-

culations for light water were carried out by LEONARD SOLON.

APPE’NDIX A

Irfinite Regions SYMMETRY OF H

In an infinite homogeneous and isotropic moderator region H is a displacement

operator

Hf(x) = J h ( j x - x’ / )f(x’) dx’ and is per se symmetric.

Finite Regions, Multi-group Operators We note Ki is defined so that

(--aFV2 + 1) ki (x,x’) =6(x - x’)

and ki(x,x’) vanishes whenever x or x’ lies on the surface S of the moderator region. . If all solutions of

(V2 + P2W,W =o for 4,(x) = 0, whenever x lies on S, define a set of eigenvalues p and eigenfunctions +,, this is a complete orthogonal set. We may normalize the 4$+&J = S,,] and expand ki in terms of them. We take a set of real functions $r for convenience. Then

6(x - x’) = 2 ~~(~M~(~‘) P

so

Thus KS is symmetric.

Minimum critical mass and flat flux 201

Furthermore KiKj is symmetric, as the following argument shows,

j k,( X,X”) kj( x”,x’) dx” = 2 4,wP,<x’) ti (1 + p2ai2) (1 + p’aiz)

It is now clear that H is symmetric for the multi-group case in a finite region.

APPENDIX B

PRODUCTS OF THE OPERATORS Ki

We show that

KiKj = 1

1 - (Mi/CCj)2 Kj + 1 1

- (tcj/c(J’ Ki

if ai # M aj. From this the conclusion reached in the text follows.

j ki(x,x”) k&x”,x’) dx” = 1 b,(x) $ux’> p ( 1 + a?p21 (1 + atp2)

= Cd,W,W [ l l P 1 - (ai/ai)2 I + aj”p2

+ l 1

1 - (ai/aj)2 1 + aFp2 I = kj k

1 - (a,/aj)” + 1 - (aj/aJ2