
#1
Jun210, 12:06 PM

P: 45

Hi Everyone, on page 238 of lamarsh, section 5.5, first paragraph, it says "flux must also be finite". What does it mean?




#2
Jun210, 01:37 PM

P: 18

Haven't got lamarsh but common sense says that its very hard to do numerical calculations involving infinite quantities!




#3
Jun210, 06:36 PM

Admin
P: 21,628

If one has a function Ae^{x} + Be^{x}, and as x > infty, Ae^{x} would > infty. To have a finite flux, the function describing the spatial distribution of the flux muxt be finite. Also, if the flux is described by a function proportional to 1/r, then as r> 0, it would go to infinity, so 1/r cannot be used to describe a flux at r=0. The flux, which is the number of neutrons passing through some unit area per unit time, usually neutrons/cm^{2}s, is finite because the number of neutrons and atoms is finite. 



#4
Jun1910, 10:54 AM

P: 9

Diffusion Equation 



#5
Aug2110, 11:07 PM

P: 30

I actually happen to have Lamarsh right next to me. For the clarity of everyone else, what the asker is referring to is Section 5.5, entitled "boundary conditions," and it is a general discussion on boundary conditions to the onespeed diffusion equation.
Some more of the excerpt: "Since the diffusion equation is a partial differential equation, it is necessary to specify certain boundary conditions that must be satisfied by the solution...For example, since a negative or imaginary flux has no meaning, it follows that [tex]\phi[/tex] must be a real, nonnegative function. The flux must also be finite, except perhaps at artificial singular points of a source distribution." That being stated, Astronuc is certainly spot on. But, one should be careful to not extend this in the following way: It is true that the flux being proportional to [tex]1/r[/tex], itself, if [tex]r =0[/tex] is a point in the solution domain, but the caution I wanted to spread was that do not dismiss all functions that simply have this term in it, one must consider the entire term. For instance, in a spherical, bare, reactor one finds the flux is a solution to the diffusion equation in steadystate as, [tex]\phi (r) = A\frac{\sin (r/L)}{r} + B\frac{\cos (r/L)}{r}[/tex] where [tex]L[/tex] is the optical (or diffusion) length, and [tex]A[/tex] and [tex]B[/tex] are constants. A suitable boundary condition is that the flux [tex]\phi < \infty [/tex] as [tex]r\rightarrow 0[/tex], enforcing this limit (although both terms involve [tex]1/r[/tex]), it is noted that the term: [tex]\lim_{r\rightarrow 0}\frac{\sin (r/L)}{r} \rightarrow \frac{1}{L}[/tex] while the cosine term becomes infinite (implying we require [tex]B=0[/tex]). (One can show that the sine term has a removable singularity by standard methods (e.g. expansion), or you may enforce the symmetry condition that [tex]\lim_{r\rightarrow 0} 4\pi r^2 J(r) = 0[/tex] to retrieve the same result). The moral I wanted to relate was just that do not think that any [tex]1/r[/tex] term is no good, one must take things as a whole. I realize this was posted awhile back, but perhaps it can help future visitors. 


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