Flux Shape in Various Reactor Geometries in One Energy Group

1. Flux Shape in Various Reactor Geometries in One Energy Group B. Rouben McMaster University Course EP 4D03/6D03 Nuclear Reactor Analysis (Reactor Physics) 2013 Sept.-Dec.

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

3. Diffusion Equation • We derived the time-independent neutron-balance equation in 1 energy group for a finite, homogeneous reactor: • We showed that we could introduce the concept of geometrical buckling B2 (the negative of the flux curvature), and rewrite the equation as • where B2 has to satisfy the criticality condition

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 the diffusion 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 boundary condition makes Eq. (2) an eigenvalue problem, i.e., a situation where only distinct, separated solutions exist.

5. Solving the Flux-Shape Equation • We will now apply the eigenvalue equation to the infinite-slab geometry and the infinite-cylinder geometry, and solve for the geometrical buckling and the flux shape. • As concluded before, we will find that the curvature of the 1-group flux in a homogeneous reactor is always negative.

6. Exercise • Given that the curvature of the 1-group flux in a homogeneous reactor is negative, where do you thinkthe maximum flux would have to be in a regular-shaped reactor? • Explain.

7. Maximum Flux • In a 1-group homogeneous reactor of regular shape, the maximum flux must necessarily be at the centre of the reactor. • We can explain that by “reductio ad absurdum”. Suppose the flux is a maximum not at the centre of the reactor; if we draw a straight line from the maximum point to the centre of the reactor, then by symmetry, there would have to be another “maximum” on the same line on the other side of the centre. Therefore the centre of the reactor would be at a “local minimum” along that straight line, which would imply that the flux does not have negative curvature along that straight line!

8. Infinite-Slab Case • Let us study the simple case of a slab reactor of width a in the x direction, and infinite in the y and z directions. • Eq. (2) then reduces to its 1-dimensional version, in the x direction • We can without loss of generality place the slab symmetrically about x = 0, in the interval [-a/2, a/2]. cont’d

9. Infinite-Slab Reactor Geometry x = -a/2 x = a/2 x = aex/2 x = 0

10. Infinite Slab (cont’d) • 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  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) cont’d

11. Infinite Slab (cont’d) • However, symmetry about x = 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 of /2. Therefore B must satisfy: (8) cont’d

12. Infinite Slab (cont’d) • It looks as if there is an infinite number of values of B. • While that is true mathematically, the only physically possible value for the flux in the critical reactor is the one with the lowest value of B, i.e. for n =1: • [Why? What do the other solutions look like?] • See sketch in next slide cont’d

13. Infinite Slab (cont’d)

14. Infinite Slab (cont’d) • We can conclude that Eq. (9) is the only physical value of B, from the fact that the solutions with n = 3, 5, 7, … all feature regions of negative  in the reactor, and that is not physical. • Also to be noted from Eq. (9) is that the buckling increases as the dimensions of the reactor (here aex) decrease [as had also been concluded earlier] – the curvature needs to be greater to force the flux to 0 at a closer boundary! cont’d

15. Infinite-Slab Case (cont’d) • The 1-group flux in the infinite-slab reactor can then be written • The absolute value of the flux, which is related to the constant A1, is undetermined at this point. • This is because Eq. (3) is homogeneous – therefore any multiple of a solution is itself a solution. The physical significance of this is that the reactor can function at any power level. • Therefore, to determine A1, we must tie the flux down to some quantitative given data – e.g., the desired total reactor power. cont’d

16. Infinite-Slab Case (cont’d) • Because the slab is infinite, so is the total power. But we can use the power generated per unit area of the slab as normalization. Let this be P W/cm2. • If we call Ef the 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

17. Additional Notes • In solving this problem, we found that the reactor equation had very specific solutions, with only specific, distinct values possible for B. • 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 of eigenvalue problems, which have distinct solutions [eigenfunctions] - here the flux distribution  -, with corresponding distinct eigenvalues (here the buckling B2). • Although in this problem we found only 1 physically possible eigenfunction 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 of B) cannot singly represent the true flux in the reactor, they can exist as incremental time-dependent perturbations to the fundamental flux, perturbations which will die away in time as the flux settles into its “fundamental” solution.

18. Case of Infinite-Cylinder Reactor Infinite height Radius R r

19. Infinite-Cylinder Reactor • For a homogeneous bare infinite cylinder, the flux is a function of the radial dimension r only. All axial and azimuthal positions are equivalent, by symmetry. • We write the eigenvalue equation in cylindrical co-ordinates, but in the variable r only, in which the divergence is • The 1-group diffusion equation then becomes • By evaluating the derivative explicitly, we can rewrite Eq. (16) as cont’d

20. Infinite-Cylinder Reactor (cont’d) • We may be completely stumped by Eq. (17), but luckily our mathematician friend recognized it as a special case of an equation well known to mathematicians, Bessel’s equation (m is a constant): • Eq. (17) corresponds to m = 0, for which this equation has 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, isn’t it, even if Nobel didn’t like them!] cont’d

21. Infinite-Cylinder Reactor (cont’d) • I sketch the functions J0(x) and Y0(x) below: Although, mathematically speaking, the general solution of Eq. (17) is a combination of J0(x) and Y0(x), Y0(x) tends to - as x tends to 0 and is therefore not physically acceptable for a flux.cont’d

22. Infinite-Cylinder Reactor (cont’d) • The only acceptable solution for the flux in a bare, homogeneous infinite 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 negative values, B for the infinite cylinder can be given only by • Therefore the buckling for the infinite cylinder is

23. Infinite-Cylinder Reactor (cont’d) • The 1-group flux shape in the infinite homogeneous cylindrical reactor is then • As before, the absolute magnitude of the flux (i.e., the constant A) 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 density P, and the energy released in fission, we can write:

24. Infinite-Cylinder Reactor (cont’d) • The integral on the Bessel function may look forbidding, but it can be evaluated from known relationships between various Bessel functions. • I’ll just give the final result here without derivation. • If we ignore the extrapolation distance, • which gives for the 1-group flux the final equation

25. Exercise/Assignment: Apply to Other Shapes • Exercise: Apply the eigenvalue Eq. (3) to the following 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 appropriate dimensions, each with its own directional bucklings, which add to the total buckling.

26. Parallelepiped Reactor y +b/2 -b/2 z c/2 -c/2 x -a/2 +a/2

27. Parallelepiped Reactor • For a parallelepiped, the eigenvalue equation becomes (27) • To solve this, we get the brilliant idea of trying a separable form for , i.e. (28) • Substituting this form into Eq. (27) gives (29) and if we divide both sides by f(x)g(y)h(z): (30) cont’d

28. Parallelepiped Reactor (cont’d) • Since the 3 terms on the left-hand side are functions of x only, y only, and z only respectively, and the sum is a constant, then each term must be a constant: (31) and the 3 partial bucklings must add to the total buckling: (32) • We solve each part of Eq. (31) in the same way as for the infinite slab, and, as in that case, we find the solution to be a pure cosine function: cont’d

29. Parallelepiped Reactor (cont’d) • If the parallelepiped has sides a, b, c in the x, y, and z directions, then (33) with the partial Bs as for the infinite slab: (34) since the flux must go to 0 at the extrapolated boundaries. [Any odd multiple of each of these values is mathematically possible as well, but as before, only the lowest value is physically possible, as the higher values give regions of negative flux.]

30. Parallelepiped Reactor (cont’d) We can finally write the complete solution for the flux shape by substituting Eq. (33) into Eq. (28), getting the form (35) As in the case of the infinite slab, the flux “amplitude” A can be determined only by “anchoring” the flux to some measured or desired quantity. Usually this is the total power P, or perhaps the power at some point in the reactor. If we use P, we can integrate Eq. (35) over the volume to find (neglecting the extrapolation distance) (36)

31. Parallelepiped Reactor • Directional bucklings: • Total buckling: • Flux shape: • If total power = P, and neglecting extrapolation distance:

32. Finite-Cylinder Reactor Finite height Radius R +H/2 -H/2 r

33. Finite-Cylinder Reactor • For a homogeneous bare finite cylinder, the flux is a function of r and also of the axial dimension z. All azimuthal positions are equivalent, by symmetry. • In the divergence we must therefore add the 2nd derivative in z (41) • The 1-group diffusion equation then becomes (42) cont’d

34. Finite-Cylinder Reactor • Just as we did for the parallelepiped reactor, we try a solution in separable form: [Function R(r) not to be confused with radius R] • Substituting this into Eq. (22) gives (44) and if we divide both sides by R(r)Z(z): (45) • Because the terms in R and Z are separated, and their sum is a constant, they must each be equal to a constant: (46) cont’d

35. Finite-Cylinder Reactor (cont’d) where the directional bucklings must add to B2, so that: (47) • The separate equations are: (48) and (49) • Eq. (48) is the Bessel equation of the first kind, order 0 (as before), and so the radial solution is J0(r) as for the infinite cylinder, and Eq. (49) gives cos(z) axially, as for the parallelepiped. So, in all (50)

36. Finite-Cylinder Reactor • Directional and total bucklings: • Flux shape:

37. Spherical Reactor • For a homogeneous bare sphere, the flux is a function of the radial dimension r only. All “latitudinal” and “longitudinal” (azimuthal) positions are equivalent, by symmetry. • We write the eigenvalue equation in spherical co-ordinates, but in the variable r only since the other dimensions don’t enter; the divergence is then (53) • The 1-group diffusion equation then becomes (54) cont’d

38. Spherical Reactor (cont’d) • To solve Eq. (54), we hit upon the terrific idea of trying for (r)a form such as (55) where R(r) is an (as yet) unknown function, to be determined (again, not to be confused with radius of sphere).. • When we substitute the form (55) into Eq. (54), we get (56) • This reduces magically to (57) cont’d

39. Spherical Reactor (cont’d) • And we know the general solution of this equation! (58) from which we then get (59) • But we can rule out the cos term, because the flux must be finite everywhere in the reactor, and (60) • The sin term is O.K., because it remains finite at the origin: • By L’Hôpital’s rule, (61) cont’d

40. Spherical Reactor (cont’d) • Therefore the flux shape in the bare homogeneous spherical reactor can finally be written (62) where, for the same reasons as in the other geometries, B must take the lowest value allowed, (63) to guarantee that there will not be regions of negative flux in the reactor.

41. Flux Amplitude for a Spherical Reactor We can integrate Eq. (66) to evaluate the flux amplitude A for a given total reactor power P. If ER is the energy released per fission, and neglecting the extrapolation distance: 41

42. Spherical Reactor • In summary then for a sphere of extrapolated radius Rex: • Buckling: where, for the same reasons as in the other geometries, B must take the lowest value allowed, to guarantee that there will not be regions of negative flux in the reactor. • and flux shape:

43. Summary • We can obtain the solution for the 1-group flux shape 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 lowest mathematical values allowed, to ensure that the flux solution is physical everywhere in the reactor.

44. END