1 / 25

R6: Other Reactor Geometries

2010 January. 2. What Have We Learned Up to Now?. We derived the time-independent neutron-balance equation in 1 energy group for a finite, homogeneous reactor:We rewrote it as an eigenvalue equation:B2 is the

Mercy
Télécharger la présentation

R6: Other Reactor Geometries

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


    1. 2010 January 1 R6: Other Reactor Geometries B. Rouben UNENE Course UN 0802 2010 Jan-Mar

    2. 2010 January 2

    3. 2010 January 3 Apply to Other Shapes We will now apply the eigenvalue equation to some different geometries, and solve for the geometrical buckling and the flux shape. We will look at the following reactor shapes: Parallelepiped Infinite cylinder Finite cylinder Sphere

    4. 2010 January 4 Parallelepiped Reactor

    5. 2010 January 5 Parallelepiped Reactor For a parallelepiped, the eigenvalue equation becomes (1) To solve this, we get the brilliant idea of trying a separable form for ?, i.e. (2) Substituting this form into Eq. (1) gives (3) and if we divide both sides by f(x)g(y)h(z): (4) contd

    6. 2010 January 6 Parallelepiped Reactor (contd) 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: (5) and the 3 partial bucklings must add to the total buckling: (6) We solve each part of Eq. (5) in the same way as for the infinite slab, and, as in that case, we find the solution to be a pure cosine function: contd

    7. 2010 January 7 Parallelepiped Reactor (contd) If the parallelepiped has sides a, b, c in the x, y, and z directions, then (7) with the partial Bs as for the infinite slab: (8) 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.] contd

    8. 2010 January 8 Parallelepiped Reactor (contd)

    9. 2010 January 9 Infinite-Cylinder Reactor Infinite height Radius R

    10. 2010 January 10 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 (11) The 1-group diffusion equation then becomes (12) By evaluating the derivative explicitly, we can rewrite Eq. (12) as (13) contd

    11. 2010 January 11 Infinite-Cylinder Reactor (contd) We may be completely stumped by Eq. (13), but luckily our mathematician friend recognizes it as a special case of an equation well known to mathematicians, Bessels equation (m is a constant): (14) With m =0, this equation has 2 solutions, the ordinary Bessel functions of the 1st and 2nd kind, J0(Br) and Y0(Br). These functions are well known to mathematicians (see next slide)! [It sure helps having mathematicians as friends, isnt it, even if Nobel didnt like them!] contd

    12. 2010 January 12 Infinite-Cylinder Reactor (contd) Plots of the functions J0(x) and Y0(x):

    13. 2010 January 13 Infinite-Cylinder Reactor (contd) The only acceptable solution for the flux in a bare, homogeneous infinite cylinder is then (15) The flux must go to 0 at the extrapolated radial boundary . Therefore we must have (16) The figure in the previous slide shows that J0(x) has several zeroes, labelled xn [the 1st is x1 = 2.405, the 2nd is x2 ? 5.6] But because, physically, the flux cannot have regions of negative values, B for the infinite cylinder can be given only by (17) contd

    14. 2010 January 14 Infinite-Cylinder Reactor (contd) The 1-group flux shape in the infinite homogeneous cylindrical reactor is then (18) The flux constant A can be determined from the power per unit axial dimension of the cylinder, denoted P. We get: (19) We will not show this here, but the integral over the Bessel function in Eq. (19) can be evaluated without too much difficulty, using certain mathematically-known relationships. When the difference between the physical and extrapolated boundaries is neglected, the final result for the flux is (20)

    15. 2010 January 15 Finite-Cylinder Reactor Finite height Radius R

    16. 2010 January 16 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 (21) The 1-group diffusion equation then becomes (22) contd

    17. 2010 January 17 Finite-Cylinder Reactor Just as we did for the parallelepiped reactor, we try a solution in separable form: (23) Substituting this into Eq. (22) gives (24) and if we divide both sides by R(r)Z(z): (25) Because the terms in R and Z are separated, and their sum is a constant, they must each be equal to a constant: (26) contd

    18. 2010 January 18 Finite-Cylinder Reactor (contd) where the directional bucklings must add: (27) The separate equations are: (28) and (29) Eq. (28) 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. (29) gives cos(z) axially, as for the parallelepiped. So, in all (30)

    19. 2010 January 19 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 dont enter; the divergence is then (31) The 1-group diffusion equation then becomes (32) contd

    20. 2010 January 20 Spherical Reactor (contd) To solve Eq. (32), we hit upon the terrific idea of trying for ?(r) a form such as (33) where R(r) is an (as yet) unknown function, to be determined. When we substitute the form (33) into Eq. (32), we get (34) This reduces magically to (35) contd

    21. 2010 January 21 Spherical Reactor (contd) And we know the general solution of this equation! (36) from which we then get (37) But we can rule out the cos term, because the flux must be finite everywhere in the reactor, and (38) The sin term is O.K., because it remains finite at the origin: By LHpitals rule, (39) contd

    22. 2010 January 22 Spherical Reactor (contd) Therefore the flux shape in the bare homogeneous spherical reactor can finally be written (40) where, for the same reasons as in the other geometries, B must take the lowest value allowed, (41) to guarantee that there will not be regions of negative flux in the reactor.

    23. 2010 January 23 Flux Amplitude for a Spherical Reactor We can integrate Eq. (40) to evaluate the flux amplitude A for a given total reactor power P. Neglecting the extrapolation distance:

    24. 2010 January 24 Summary We have obtained the solution for the 1-group flux shape in bare homogeneous reactors of various geometries. In each case we determined also the directional bucklings, and 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.

    25. 2010 January 25 END

More Related