8: Source-Sink Problems in 1 Energy Group

Transcription

1 8: Source-Sink Problems in 1 Energy Group B. Rouben McMaster University Course EP 4D03/6D03 Nuclear Reactor Analysis (Reactor Physics) 015 Sept.-Dec. 015 September 1

2 Contents Solving the 1-group diffusion equation in the presence of external sources. 015 September

3 Source-Sink Problems Before studying reactor problems, we will study source-sink problems in diffusion theory. Source-sink problems are situations where the only neutron sources present are independent of the neutron flux, i.e., the environment is not really multiplying, it has only scattering and capture cross sections. We will study simple source-sink problems in 1-group diffusion theory. 015 September 3

4 Source-Sink Problems Because source-sink problems are situations which are driven by an external source (or sources), there is always a solution for the flux distribution, even if the solution cannot be easily obtained, for instance where the problem configuration is complex. i.e., the neutrons are emitted by the source, and they must go somewhere, so there is necessarily a flux distribution around the source(s), depending on the medium outside the source(s). We will see later that, in contrast, in critical systems there is not always a time-independent solution. 015 September 4

5 The Diffusion Equation The starting diffusion equation in 1 energy group was derived in a previous learning module: Sr r r r r Dr r 0 (1) f a In source-sink problems, f = 0. We will study the flux distribution: First, for a simple source extending throughout space And, secondly, for situations where the source has no extent in at least one spatial dimension. 015 September 5

6 Uniform Source in Infinite, Uniform Medium Assume a uniform source S(r) = S, constant independent of location r. If the medium is uniform, the nuclear properties are constant, i.e., a (r) = and D(r) = D, independent of location. In this completely uniform space, every point is equivalent to every other point. Therefore the leakage term (last term in Eq. 1) into or out of any point can only be 0! Also, the f term is 0. Eq. (1) becomes S r, and the solution is a spatially uniform flux a 0 r a 015 September 6 S

7 The Diffusion Equation Now consider cases where the source has no extent in at least one dimension. We still have the equation : S r r r r r Dr r 0 (1) f a In source-sink problems, f = 0. We will study the flux distribution outside the source, i.e., in regions where S(r) = 0. Let us also assume that the medium outside the source is homogeneous (uniform), i.e., the nuclear properties D and a are independent of r. The diffusion equation outside the source then becomes: D r r 0, or D r r 0 () a a 015 September 7

8 Exercise What is the operator form of Eq. ()? 015 September 8

9 Operator Form of Equation () The operator form of Eq. () (outside the source) is M r r Φ r where Φr M D r 0 and a r 015 September 9

10 Introduce the Diffusion Length and Area Define the quantity L diffusion length by L D a, i. e. L D a (3) L is called the diffusion area. and the diffusion equation can be rewritten in the form 1 r r 0 (4) L 015 September 10

11 Plane Source in Infinite Medium Let study the case of an infinite plane source in an infinite medium. Take the plane source to be the y-z axis located at x = 0. Everything is then a function of x only. If S is the source strength per cm, then the neutron current in either direction off the plane is S/, i.e. lim J x 0 x S : this becomes the " boundary" condition at x 0 (5) d In one dimension reduces to dx (6) 015 September 11

12 Flux from Plane Source in Infinite Medium The diffusion equation for the plane source becomes: x The x x d 1 x 0 (7) dx L The basic solutions to this equation are e The 1 flux (the appropriate one) By symmetry we must also have the same C on each side. : J most C e Ce x 0 general x / L cannot i. e., the general x / L C (9) solution x / L become solution The current at, say, the e is (8) can survive on each must right x d lim D x0 dx then be edge the source. CD S SL Comparing this to Eq.(5) gives, i. e., C (11) L D SL x / L The flux due to the plane source is x e (1) 015 September D 1 x0 of the CD L x / L infinite, only one of lim side e x / L and of e x / L the plane source is CD L. exponentials (10)

13 Current from Infinite Plane Source The current in an infinite uniform medium from an infinite plane source on the y-z plane is J x / L ˆ x / L x D x D e i e iˆ where and and [ The the source, iˆ is for Similarly the sign x for direction away unit in 0. front other of vector the of from the d dx iˆ SL D in locations must current the is plane.] of positive be the taken inf x inite S as perpendicular direction for 0 plane source. to x the plane 015 September 13

14 Exercise What happens if the medium is not infinite? How would our solution strategy change, if at all? What kind of boundary conditions would we apply? 015 September 14

15 If Medium Is Finite If the medium is not infinite, then the positive exponential is a perfectly good solution. We cannot argue to throw it out. The general solution would then indeed be a sum of the positive and negative exponentials. We would have to determine the coefficients of the two exponentials by applying boundary conditions at the source and at the outer boundary of the medium (most likely a vacuum boundary condition at the outer edges of the medium). 015 September 15

16 Point Source in Infinite Medium Let us now study the case of a point source in an infinite medium. Take the source to be located at r = 0. We assume the source is isotropic. Everything is then a function of r only. If S p is the source strength (neutrons emitted per s), then the neutron current away from the source in any direction must be S p /4, i.e., lim 4 r J r 0 r S (13) p cont d 015 September 16

17 Point Source in Infinite Medium To solve the diffusion equation for this problem, we need to write the divergence operator in terms of r (for situations where there is no dependence on the angular variables and ): r 1 d dr r d dr (14) The diffusion equation outside the point source then becomes d dr This looks complicated, but someone got the brilliant idea of attacking this by writing d dr r dr r r r dr r r L cont d 015 September 17 r 0 (16) d dr (15)

18 Exercise Where does the brilliant idea of changing variables according to Eq. (16) take us? Derive the new form of Eq. (15) in terms of the new variable. Can you identify particular solutions of this equation? 015 September 18

19 Point Source in Infinite Medium In terms of the new function, Eq. (15) becomes d dr r r L 0 (17) e r/l and e -r/l are basic the solutions of this equation The general mathematical solution of this is then a linear combination of these functions: r / L r / L r C e C e (18) September 19

20 Point Source in Infinite Medium (cont.) An infinitely growing positive exponential is not a physical flux. We are then left with: r The neutron flux as a function of r is then Now we must determine the magnitude C 1 from the boundary condition at the origin, Eq. (13). From Eq. (0) 4r J r 4r 4C 1 r / L C e (19) r / L r C (0) d D dr Dr e 1 1 r r / L e r 1 1 rl r 4C D1 (1) 015 September 0 1 r L

21 Point Source in Infinite Medium (cont.) Applying the limit at r = 0 and comparing to Eq. (13): 4C 1 D S p S p r (3) 4D r / L 4D () This then is the final formula for the 1-group flux distribution around an isotropic point source. Note that we used the boundary condition at one end of the range (r = ) to ensure a physical solution, and a boundary condition at the other end (r = 0) to fix the magnitude of the flux. This is a typical strategy. e C r 1 S p 015 September 1

22 Current from Point Source The current in an infinite uniform medium from a point source at the origin is r / L d S p e J r Dr D rˆ dr 4 D r where rˆ is the unit vector in the direction location Similarly of for the point other source locations to of the the point S p 4 from r. 1 L the point source. 1 r e r / L r rˆ 015 September

23 Meaning of Diffusion Length We can use the flux distribution from a point source to derive an interpretation of the diffusion length L. Let us calculate the mean square distance from the origin to the point where the neutron is absorbed. The absorption rate at any point is a (r). Therefore the mean square distance to absorption r r 0 0 ( r)4r a a ( r)4r dr dr (4) 015 September 3

24 Exercise Using the flux shape for the case of the point source, evaluate the integral in Eq. (4) in terms of L, to get a physical interpretation of L 015 September 4

25 Meaning of Diffusion Length (cont.) Evaluating the integral in Eq. (4) gives the following result: r r 6L, i. e. L (5), interpreting the meaning of L 6 as the square root of one sixth of the average square of distance from the point source tothepoint of absorption. the 015 September 5

26 Flux from a Number of Sources A very important property of the diffusion equation (Eq. 4) is that it is linear. This means that the total flux from a number of external sources can be determined by summing up the flux solutions from the various sources. 015 September 6

27 Flux Curvature Curvature is the mathematical term for the second derivative of a function. What is the nd derivative of the 1-group flux shape from a plane source? From Eq. (1), for x > 0: d dx d SL D x / L S e DL x / L The same result is obtained for x < 0 dx e (6) i.e., the curvature of the flux is positive everywhere September 7

28 Exercise Show also that in the case of the 1-group flux shape from a point source, the curvature is positive. Further, show how it is inherent, from the 1-group diffusion equation around an isolated source, that the flux shape decreases away from the source with a positive curvature. 015 September 8

29 Positive Flux Curvature Let s take the nd derivative of the flux around a point source, from Eq. (15): d dr i. e., the d dr 0 flux S p 4D r / S p e 4D rl has L e a r / L r e r S p 4D r / L L positive e r e r curvature And generally, from the 1-group diffusion equation outside a source, i.e. Eq. (): a r r D r ar 0 r 0 Dr again showing a positive flux curvature We will see that this will be different in critical systems! 015 September 9 d dr r / L L r / e rl L r / L 3 e r r / L

30 Positive Flux Curvature & In-Leakage A positive flux curvature, as in the source-sink problems we have studied (absence of a multiplying medium), is associated with net neutron in-leakage: Consider any little volume in the medium, away from the source. Neutrons are streaming out of the source and passing through that little volume. Some neutrons come in ( leak in ), and some neutrons fly out ( leak out ). But since the medium is absorptive, and is not multiplying, then some of the neutrons are absorbed within the little volume. Therefore fewer neutrons leak out than leak in. The neutron current out (derivative of the flux) must be smaller than the current in, i.e., the derivative of the flux diminishes with distance from the source, i.e., the nd derivative is positive. Net in-leakage of neutrons is associated with positive flux curvature. 015 September 30

31 END 015 September 31