Diffusion Equation


by candice_84
Tags: diffusion, equation
candice_84
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#1
Jun2-10, 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?
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statphys
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Jun2-10, 01:37 PM
P: 18
Haven't got lamarsh but common sense says that its very hard to do numerical calculations involving infinite quantities!
Astronuc
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Jun2-10, 06:36 PM
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Quote Quote by candice_84 View Post
Hi Everyone, on page 238 of lamarsh, section 5.5, first paragraph, it says "flux must also be finite". What does it mean?
What is the context.

If one has a function Aex + Be-x, and as x -> infty, Aex 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/cm2-s, is finite because the number of neutrons and atoms is finite.

friendboy
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#4
Jun19-10, 10:54 AM
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Diffusion Equation


Quote Quote by candice_84 View Post
Hi Everyone, on page 238 of lamarsh, section 5.5, first paragraph, it says "flux must also be finite". What does it mean?
The section is about determining boundary condition, and to do this, we have to consider of what is physically possible values of flux. By this logic we could deduct that the value of flux must be real, nonnegative and also finite.
LawlQuals
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#5
Aug21-10, 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 one-speed 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, non-negative 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 steady-state 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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