The neutron diffusion coefficient

  • #1
I know that the neutron diffusion coefficient D= (The transport mean free path/3), but in some documents i read that:( D=The transport mean free path*v/3) where v=The neutron number.

How is that?

Answers and Replies

  • #2
Staff Emeritus
Science Advisor
Does one have an example of the latter statement?

[tex] D \thickapprox \frac{\lambda_t}{3} \thickapprox \frac{1}{\Sigma_t}[/tex].

I don't see where the number comes in. How is D used when related to neutron number?
  • #3
The lectures of R.Serber, page6.
  • #4
Staff Emeritus
Science Advisor
The lectures of R.Serber, page6.
Is this the The Los Alamos primer: the first lectures on how to build an atomic bomb By Robert Serber, Richard Rhodes? Unfortunately, I don't have a copy of that book.
  • #6
Duderstadt & Hamilton (D&H), as well as Lamarsh I believe, routinely use the approximation:

[tex]D \simeq \frac{\lambda_{tr}}{3} = \frac{1}{3\Sigma_{tr}} = \frac{1}{3(\Sigma_t - \bar{\mu}_0\Sigma_s)}[/tex]

The diffusion coefficient [tex]D[/tex] has units of length ([tex]cm[/tex]) in those texts. It is also not uncommon to enforce, perhaps more customary units on the diffusion coefficient, [tex]D = [cm^2/s][/tex] in other scenarios. Are you certain that 'v' in that image does not mean velocity? I notice on that document, that whenever they wish to typeset [tex]\nu[/tex] they handwrite it, yet in the case where they quote the value of the diffusion coefficient [tex]D[/tex], they type a letter v, leading me to believe they are distinct. I am guessing it is just the velocity (which would give units of the diffusion coefficient in [tex][cm^2/s][/tex].

This follows from the certain way that the authors choose to represent the neutron diffusion equation, in D&H, it is written:

[tex]\frac{1}{v}\frac{\partial\phi (\vec{r},E,t)}{\partial t} = -\nabla\cdot\vec{J} (\vec{r},E,t) - \Sigma_a (E)\phi (\vec{r},E,t) + S(\vec{r},E,t)[/tex]

Here, one can see that the equation is written for each term as having units of [tex]cm^{-3}s^{-1}[/tex]. The [tex]P_1[/tex] approximation is made in texts (e.g. D&H) such that:

[tex]J \simeq -D\nabla\phi[/tex], so that the above may be written:

[tex]\frac{1}{v}\frac{\partial\phi (\vec{r},E,t)}{\partial t} = +\nabla\cdot D\nabla\phi (\vec{r},E,t)} - \Sigma_a(E) \phi (\vec{r},E,t) + S(\vec{r},E,t)[/tex]

This demands the diffusion coefficient [tex]D[/tex] must be in units of [tex]cm[/tex] in this form, where it is noted that [tex]\phi = [cm^{-2}s^{-1}][/tex], so that each term has units of per unit volume per unit time. What I suspect is that that document uses a different definition of [tex]\phi = [cm^{-3}][/tex], as per the following:'s_laws_of_diffusion . In this case, it is identical to the neutron diffusion equation with the conventional definition (also in D&H) of the flux [tex]\phi [/tex] furnished by definition of the neutron number density [tex]n[/tex] and the velocity [tex]v[/tex] ([tex]\phi = nv[/tex]). With this definition, the neutron diffusion equation would be modified according to:

[tex]\frac{\partial n (\vec{r},v,t)}{\partial t} = +\nabla\cdot D \nabla n (\vec{r},v,t)} - v\Sigma_a (v) n (\vec{r},v,t) + S(\vec{r},v,t)[/tex]

I do not have access to a fuller document than the photo you have attached (the link is not able to be opened for some reason). But, maybe you could verify, if you find this idea promising, the units of each term defined in the neutron diffusion equation used in that text. Make sure that [tex]\phi[/tex] is really defined as it appears. Also, I hope I made sense with the arguments of each term, please feel to correct me if I am offbase.
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  • #7
Staff Emeritus
Science Advisor
I too was thinking along the lines of [tex]\phi = nv[/tex]

and the author has decoupled n and v, so apparently assumes some average v and the transport/diffusion equation is written in neutron density. For a thermal flux where v is not spatially dependent this is OK.

But in the Primer is looks like N is the neutron density and n is the atomic/nuclear density, [tex]\phi = Nv[/tex]. Also be careful not to confuse the neutron speed v with [itex]\nu[/itex] (nu) = the number of neutrons produce per fission.

I should have written the formula with {tr} vs {t} where tr indicates transport cross section rather than total cross section
[tex] D \thickapprox \frac{\lambda_{tr}}{3} \thickapprox \frac{1}{\Sigma_{tr}}[/tex]

In the primer, critical occurs for [itex]\nu'[/itex] = 0. [itex]\nu[/itex] is just the number neutrons produced per fission.

The term [tex]\frac{\pi^2D\tau}{R^2}[/tex] is just the leakage from the sphere in this example. As long as [itex]\nu - 1[/itex] = leakage, or [itex]\nu'[/itex] = 0, the system is critical.

The mean free path for any reaction [strike]1[/strike] or [itex]\lambda[/itex] = [tex]\frac{1}{\Sigma}[/tex], where [tex]\Sigma = n \sigma[/tex], and n = the nuclide density and [itex]\simga[/itex] is the microscopic cross section for the particular reaction.
  • #8
Thanx Astronuc and LawlQuals
It is the velocity.