Source-Sink Problems in One Energy Group

1. Source-Sink Problems in One Energy Group B. Rouben McMaster University EP 4D03/6D03 Nuclear Reactor Analysis 2008 Sept-Dec

2. Contents • Solving the 1-group diffusion equation in the presence of external sources. • Reference: • Pages 157-162 in Duderstadt & Hamilton • Also Appendix B in Duderstadt & Hamilton, for the various forms of the differential operators

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.

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.

5. The Diffusion Equation • The starting diffusion equation in 1 energy group was derived in a previous learning module: • In our source-sink problem, 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:

6. Interactive Discussion/Exercise • What is the operator form of Eq. (2)? • Do not turn the page until you have attempted/done this discussion/exercise.

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8. Operator Form of Equation (2) • The operator form of Eq. (2) (outside the source) is

9. Introduce the Diffusion Length and Area • If we define the quantity L “diffusion length” by • (L2 is called the “diffusion area”) • then the diffusion equation can be rewritten in the form

10. Point Source in Infinite Medium • Let us study the case of a point source in an infinite homogeneous 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 Sp is the source strength (neutrons emitted per s), then the neutron current away from the source in any direction must be Sp/4, i.e., cont’d

11. 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 ): • The diffusion equation outside the point source then becomes • This looks complicated, but someone got the brilliant idea of attacking this by writing cont’d

12. Interactive Discussion/Exercise • Where does the brilliant idea of changing variables according to Eq. (8) take us? • Derive the new form of Eq. (7) in terms of the new variable . • Can you identify particular solutions of this equation? • Do not turn the page until you have attempted/done this discussion/exercise.

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14. Point Source in Infinite Medium • In terms of the new function , Eq. (7) becomes • er/L and e-r/L are the solutions of this equation • The general mathematical solution of this is then a linear combination of these 2 functions:

15. Interactive Discussion/Exercise • A physical argument can be made to simplify the solution here from the most general mathematical form. • What is this physical argument and where does it lead? • Do not turn the page until you have attempted/done this discussion/exercise.

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17. Point Source in Infinite Medium (cont.) • The infinitely growing positive exponential is not a physical flux. • We are then left with: • The neutron flux as a function of r is then • Now we must determine the magnitude C1 from the “boundary” condition at the origin, Eq. (5). • From Eq. (12)

18. Point Source in Infinite Medium (cont.) • Applying then the limit at r = 0, Eq. (5), we get • 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.

19. Interactive Discussion/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? • Do not turn the page until you have attempted/done this discussion/exercise.

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21. 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).

22. Plane Source in Infinite Medium • Now study the case of an infinite plane source in an infinite homogeneous medium. • Take the plane source to be located at x = 0. • Everything is then a function of x only. • If S is the source strength per cm2, then the neutron current in either direction off the plane is S/2, i.e.

23. Interactive Discussion/Exercise • Solve the diffusion equation for the case of the plane source in an infinite medium. • Do not turn the page until you have attempted/done this discussion/exercise.

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25. Flux from Plane Source in Infinite Medium • By doing the exercise, you have shown that the flux distribution outside a plane source in an infinite medium is:

26. 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 travelled by a neutron from the source (r = 0) to the point of absorption. • The absorption rate at any point is a. Therefore the mean square distance to absorption

27. Interactive Discussion/Exercise • Using the flux shape for the case of the point source, evaluate the integral in Eq. (19) in terms of L, to get a physical interpretation of L • Do not turn the page until you have attempted/done this discussion/exercise.

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29. Meaning of Diffusion Length (cont.) Evaluating the integral in Eq. (19) has given you the following result:

30. Flux Curvature • Curvature is the mathematical term for the second derivative of a function. • What is the 2nd derivative of the 1-group flux shape from a plane source? From Eq. (18), for x > 0: • The same result is obtained for x < 0 • i.e., the curvature of the flux is positive everywhere

31. Interactive Discussion/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. • Do not turn the page until you have attempted/done this discussion/exercise.

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33. Positive Flux Curvature • Let’s take the 2nd derivative of the flux around a point source, from Eq. (15): • And generally, from the 1-group diffusion equation outside a source, i.e. Eq. (2): again showing a positive flux curvature • We will see that this will be different in critical systems!

34. 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 2nd derivative is positive. •  Net in-leakage of neutrons is associated with positive flux curvature.

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