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Wikipedia gives the following definition for critical mass.

A critical mass is the smallest amount of fissile material needed for a sustained nuclear chain reaction.

No mention is made of a neutron moderator in this definition. If critical mass is defined like this, then that should allow infinite moderator material to be used if doing so would lead to the minimum amount of fissile material.

My question: What is the universally minimum amount of fissile mass needed to achieve a critical configuration (allowing a neutron moderator and no restriction on the moderator) for something like Uranium-235 and what would its configuration be?

Uranium-235 has a bare sphere critical mass (BSCM) of around $52 kg$, which is the mass of a critical sphere containing only the fissile material where no neutrons are reflected back into the sphere after leaving the surface. Wikipedia has a good illustration of a bare sphere versus a sphere surrounded by a moderator.

BSCM and moderator blanket Image: First item is a BSCM illustration, 2nd item is a critical sphere that uses less fissile mass than the BSCM due to the introduction of a neutron moderator blanket

This example illustrates how a critical configuration can be made with U-235 that uses less than the BSCM, or $52 kg$. It is possible, although unlikely, that the above sphere surrounded by a moderator could be the configuration that answers my question. Alternatives would include a homogenous mixture of the moderator and the fissile material, a mixture of the two that varies radially, or something I have not thought of.

The hard part is showing that the particular fissile material/moderator mix can not be improved by any small change. It would also be necessary to show that it can not be improved by adding more than 1 type of moderator (I suspect this could be done with a reasonably short argument).

Technical mumbo junbo

Everything I write here is just a suggestion, answer however you want to or can

Both the moderator and fissile material have a certain density. I would make the simplifying assumption that the density of a homogenous mixture of the fuel and moderator would be a linear combination of their specific volumes, with the understanding that this is not true in real life. Changing the mix changes the macroscopic cross section of both. It might help to note that the BSCM almost exactly determines the macroscopic cross section of the pure fissile material, one over the macroscopic cross section is the path length, which is on the same order of magnitude of the radius. Generally fast and thermal scattering from U-235 is small compared to the fission cross section.

I would use 2 neutron energy groups, and assume either diffusion or immediate absorption after scattering to thermal energies. In fact, I would just assume immediate absorption. At that point, all you would need is the scatter to absorption ratio and microscopic cross section at both thermal and fast energies for the moderator and fissile material, in addition to the densities of course. Even then some type of calculus of variations may be necessary (if you're looking for a radially varying mixture) in addition to the fact that it would be hard to describe the fast group without a fairly complicated form of neutron transport. The BSCM is comparatively simple since it has a constant number density of the fissile material and even that is actually pretty complicated.

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I don't know enough about critical mass calculations to answer your question definitively, but the minimum critical mass is less than 780 g of U-235 (this example is a solution-type core). –  mmc Aug 9 '11 at 17:18
    
@mmc Your comment was exactly the type of answer I was looking for, thanks! I understand definitively answering it is a very BIG problem, but your answer knocks it down a whole lot. –  Alan Rominger Aug 9 '11 at 17:23
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Somewhat off topic, but I've seen speculation that if isotopes at the island of stability en.wikipedia.org/wiki/Island_of_stability had half-lives long enough to allow macroscopic samples to exist, then you could make a nuclear bomb the size of a pencil eraser. Luckily for world peace, these isotopes are both incredibly hard to produce and unlikely to have such long half-lives. –  Ben Crowell Aug 9 '11 at 17:52
    
@mmc could you make your comment an answer so I can select it? Thanks. –  Alan Rominger Aug 9 '11 at 18:53
    
@Zassounotsukushi Done. –  mmc Aug 9 '11 at 22:27

4 Answers 4

up vote 3 down vote accepted

I'm not an expert in critical mass calculation, but the minimum critical mass for U-235 must be less than (or equal to) 780 g as shown in "Mass Estimates of Very Small Reactor Cores Fueled by Uranium-235, U-233 and Cm-245" for a solution type core.

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to add to this answer, the naturally occurring uranium has very small percentage of U235, hence you need at least 100kg of nat. Uranium to achieve critical mass. –  Vineet Menon Jan 3 '12 at 4:53

I'm just a layman, but if a nuclear reactor depends on minimum sustained criticallity, then isn't it possible to "burn-out" nuclear waste by combining just enough of it to achieve "near critical mass" ? If the "unlined ditch" accident at Hanniford was a real event, (and not just a Stephen King invention), then it seems that this process would be possible in a very controlled environment. It would result in the half-life of that nuclear material being made much longer, and the nuclear material's radioactivity being reduced very quickly. Soon there would be no more nuclear waste, because in order to continue to achieve even the most minimal sustained criticality, all the nuclear waste of that isotope would have to be combined to achieve it.

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You might theoretically be able to get rid of all the fissile isotopes, but there are an enormous number of non-fissile highly radioactive fission reaction products. How do you plan to get rid of all the Cesium 137 and Strontium 90, for example? –  Peter Shor Jan 2 '12 at 23:17
    
Hi Mark - I've merged your new account into your old one. You should now be able to edit this answer. Please do that if you have more to add, instead of posting separate answers. –  David Z Jan 2 '12 at 23:27

To further add to your confusion, you might want to distinguish between "prompt critical" and "delayed" critical. Reactors rely on delayed neutrons, which is why they must be cooled even after they drop below critical. Bombs use prompt critical because they operate much faster. One rule of thumb is that an external relfector drops the spherical critical mass by about a factor of three. But obviously the lowest mass is achieved by homogeneous moderated systems (eg. U-235 hydride).

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While the factor of 3 is an attractive rule of thumb, I should add that it depends on the absorption and scattering cross section ratios which depends on the moderator material. –  Alan Rominger Jan 3 '12 at 6:18

You are correct in assuming that the presence of a moderator (or reflector) can change the mass required for a critical configuration. But there are too many other variables to consider to provide a concise answer to your question. All I can offer is a qualitative explanation:

The smallest critical mass would be for a high density spherical core of fissile fuel with a low neutron absorption cross section and high fission cross section surrounded by an infinite (or very thick) high density moderator with a high neutron scatter cross section and low neutron absorption cross section.

The critical mass will depend on the geometry and material properties of the fissile fuel and moderator. A spherical mass of pure fissile fuel surrounded by an infinite moderator will generally require the lowest critical mass. This is for 2 main reasons: 1) The spherical configuration is the most efficient when it comes to the use of neutrons in a simple homogenious reactor (spherical configurations give the highest 'non-leakage factor', which means that fewer neutrons escape the system without causing a fission). 2) The moderator surrounding the spherical fuel configuration will scatter some of the neutrons escaping from the fissioning core back into the core where they can cause more fissions, effectively acting a a reflector to improve the usage of neutrons in the system.

For a specific answer to your question, you need to know:

  1. Fissile fuel composition (pure U-235, 90%/10% U-235/U-238, etc.)
  2. Fuel configuration (geometry, density)
  3. Moderator/reflector material (water, U-238, etc.)
  4. Moderator/reflector configuration(geometry, density)

If you then had the mass of the fuel, you would be able to compute the 'neutron multiplication factor' (most often represented as 'k') for the system. Methods for computing k depend on the system, application, and assumptions made and can be quite complex, so I won't list a particular equation here. For a 'critical' system, k = 1 (sub-critical is k < 1, super-critical is k > 1). To compute the 'critical mass' you hold k = 1 and can then solve for mass.

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protected by David Z Aug 20 '13 at 5:03

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