Jump to content

Need vaues for critical conditions of Uranium 235


Recommended Posts

Hello, I am working on theoretical research for what is turning into somewhat complicated nuclear physics. I am not a nuclear physicists, so I would appreciate help with finding the information I need. First and foremost piece of information I need to solve my system of equations is the critical density of Uranium 235.

 

I cannot find this information directly in books or the internet. However, online articles suggested statistics from bare spheres. It was suggested by Wikipedia that the critical conditions for an unspecified grade of Uranium 235 at an unspecific temperature (but I am assuming room temperature) and an unspecific pressure (but I am assuming 1 atmosphere) was 52000 grams in a sphere 17 centimeters large. But to me, this information seems inaccurate.

 

I calculated the natural density of pure Uranium 235 to be approximately 18.075g/cm^3. The density of elemental Uranium with mixes of different isotopes is 18.95-19.01 g/cm^3 with Uranium 238 being the most common isotope. So, it makes sense that the natural density of pure U235 is around 18g/cm^3.

 

With that said, it seems illogical that the critical density solid Uranium is 52000g/(4/3*pi*(17cm)^3)=2.526g/cm^3. The bare spheres were not in environments of exceedingly high temperatures in a vacuum, so I cannot figure out how the density of solid Uranium in spherical shape could be that dramatically lower in only that specific experiment.

 

Does anyone know of credible resources where I can find the information I am looking for and explain this discrepancy?

Link to comment
Share on other sites

excuse an ignoramus - but what is "the critical density"? I have heard this used in thermodynamics (ie where two phases can exist together) - I also know the usage in cosmology (will the universe continue expanding or contract?) - but I do not know of its use in nuclear physics.

 

There is of course critical mass - but this is measured in kilograms. The density will affect the mass - higher density means a lower mass will become critical. The mass required is proportional to the inverse square of the density. It is this relationship that was used to reach critical mass in the WW2 atomic weapons.

 

But I have not heard of a critical density - I realise you are also looking for info, but perhaps if you could point us towards some descriptions of your terms we can make progress

Link to comment
Share on other sites

Critical density is the ratio of mass to volume of a fissile material necessary to create a chain reaction in that fissile material. The critical density itself is related to the critical mass. The reason why density matters is because of the grade of fissile material and its environment. If only say, 20% of the isotopes of uranium are 235, you will need to compress a smaller mass of the uranium to make it critical or use neutron reflectors, and as the temperature of fissile material increases which it often does when trying to sustain a chain reaction, its density also decreases which makes a chain reaction less probable. But I am not an expert myself, that's just what I gather from research.

 

Some of the information I cannot find. Even if an article gives me information about a density relationship like the one you stated which I also found in Wikipedia, it won't specify the grade of Uranium or environmental conditions or sometimes not even the isotope.

Edited by MWresearch
Link to comment
Share on other sites

The critical density does deal with the actual number of uranium atoms, it is completely possible the sphere parameters are affected by the grade and neutron reflected but that information was not stated as being directly related to the statistics I saw.


It says on Swansont's profile that he or she specializes in atomic physics, maybe someone could get him or her to comment.

Link to comment
Share on other sites

I don't think you'll find a fixed critical density figure because it probably depends on the size of the original subcritical sample. The smaller the subcritical mass, the greater the density required to get a chain reaction going. I imagine the actual quantitive relationship is probably classified if you can't find it easily.

Edited by StringJunky
Link to comment
Share on other sites

It does depend on other variables, but I still can't find a clear relationship. And, why would it be classified when you can google how to make an atomic bomb with descriptions on the refining process? Making an atomic bomb isn't the hard part, its getting concentrated U235 at all. Luckily, I don't need any physical material in the realm of the imaginary and theoretical.

Link to comment
Share on other sites

It does depend on other variables, but I still can't find a clear relationship. And, why would it be classified when you can google how to make an atomic bomb with descriptions on the refining process? Making an atomic bomb isn't the hard part, its getting concentrated U235 at all. Luckily, I don't need any physical material in the realm of the imaginary and theoretical.

Making a dirty bomb is easy. Making a gun-type bomb is easy but it requires a supercritical quantity of fissile material. Making an implosion weapon with a subcritical quantity is not easy because it is an extremely maths-dense exercise which has to be accurate to nanoseconds to work correctly and I think what you want to know is part of that, which will be secret, because knowing this you can have portability and minimal amounts of fissile material to make a weapon. This and the difficulty of acquiring suitable material is what keeps it safe (so far) from causing harm by certain groups.

Link to comment
Share on other sites

It doesn't need to be nanosecond accurate, you only need to bring two sub-critical pieces together at a high velocity for a millionth of a second which total critical mass or more once together. You could easily assume the height from 30,000 feet is enough from data you can google and looking at past experients. Putting all of that under precise control to generate electricity in a nuclear reactor so you don't blow up a city full of civilians is what requires a lot of maths.

 

It might be more of an industry secret than a national secret.

 

In any case, I will simply have to wait for someone qualified to provide accurate answers.

Edited by MWresearch
Link to comment
Share on other sites

Ok. Well what made you think I was talking about an implosion weapon then? I'm just talking about a regular critical mass chain reaction, but the density does play a factor in that in terms of the number of U235 isotopes in a given volume. That's why the gun situation works, the velocity compresses the pieces together.

Link to comment
Share on other sites

Ok. Well what made you think I was talking about an implosion weapon then? I'm just talking about a regular critical mass chain reaction, but the density does play a factor in that in terms of the number of U235 isotopes in a given volume. That's why the gun situation works, the velocity compresses the pieces together.

I'm no physicist, so I'm very open to being corrected, but the function of the explosive is just to propel the U235 'bullet' to the U235 'target'. There will be compression by the enclosing tamper to enhance the yield as fission occurs but I don't think the conventional explosive contributes much in this configuration to the final yield; the conventional explosive forces are asymmetric - unlike in an implosion weapon - so I don't think you are going get a significant increase in the density of the fissile material.

Edited by StringJunky
Link to comment
Share on other sites

Hmm. Well from my research there is usually only such an enclose to reflect neutrons in order to increase the probability of a chain reaction with a smaller mass, as well as an additional casing to increase the thermonuclear yield with fusion such as with hydrogen. Compression or at least density changes can occur due to velocity as well as a decrease in temperature.

Edited by MWresearch
Link to comment
Share on other sites

"Atomic physics" does not mean "nuclear physics", and nuclear physics encompasses many fields, most of which don't relate with bombs nor chain reactions. So you don't need to kidnap Swansont for your project, we'd happily keep him here...

 

Since

  • uranium enrichment becomes easier over time, and fissile material can be acquired
  • the design of a bomb must be less obvious than the available descriptions
  • an ineffective bomb is already worrying enough
  • many different people read forums, not only the participants
  • several information requests on an other forum matched closely chemical weapons later used by the Assad regime

maybe some members here find unreasonable to add information on the Internet.

Link to comment
Share on other sites

 

It says on Swansont's profile that he or she specializes in atomic physics, maybe someone could get him or her to comment.

 

Which is not nuclear physics. But I taught reactor physics in the navy, so here goes. (edit: xpost with Enthalpy)

 

There is no critical density, per se. If you had a tiny ball of uranium, it will not go critical because there's a lot of surface area for neutrons to leave, compared to the volume. A larger ball at the same density could be critical, because surface/volume has decreased, and a larger fraction of the neutrons remain in the sphere. There is a minimum mass you must have, as imatfaal has mentioned, which can be made critical under optimal conditions. Wikipedia claims the mass is around 7 kg

 

The discrepancy in your numbers could be because you wouldn't have chemically pure uranium, but some compound, such as uranium dioxide. So your sphere would naturally be bigger than what you'd expect from pure uranium, since the oxygen takes up space. or maybe ~17 cm was the diameter. (I get 20g/cm3 using 8.5 cm for the radius, so it only needs to be a little bigger than that)

Ok. Well what made you think I was talking about an implosion weapon then? I'm just talking about a regular critical mass chain reaction, but the density does play a factor in that in terms of the number of U235 isotopes in a given volume. That's why the gun situation works, the velocity compresses the pieces together.

 

I think the explosive was just to make sure the pieces got together quickly and weren't prematurely forced apart by the reaction.

Link to comment
Share on other sites

I just had a ridiculously huge glitch when trying to insert a link and I had to erase everything I said be refreshing the page because the website wouldn't let me click on anything else.

 

Thanks for clearing up the discrepancy then, that makes more sense.

I understand that if you just have enough mass then the reaction goes critical, but whenever you add mass without compressing the material, you must also be increasing the volume as well, and multiple articles bring up the issue with density, compression, temperature and surface tension. So, density must have some role, at least in terms of the number of atoms in a volume or a cross section and there must be a way to put the final equation into terms of a relationship into mass and volume, whatever that final equation may be.

 

The Wikipedia article that scienceforums.net wouldn't let me link to references a physicist who worked on the Manhattan project and states "An ideal mass will become subcritical if allowed to expand or conversely the same mass will become supercritical if compressed."

Link to comment
Share on other sites

I understand that if you just have enough mass then the reaction goes critical, but whenever you add mass without compressing the material, you must also be increasing the volume as well, and multiple articles bring up the issue with density, compression, temperature and surface tension. So, density must have some role, at least in terms of the number of atoms in a volume or a cross section and there must be a way to put the final equation into terms of a relationship into mass and volume, whatever that final equation may be.

 

The Wikipedia article that scienceforums.net wouldn't let me link to references a physicist who worked on the Manhattan project and states "An ideal mass will become subcritical if allowed to expand or conversely the same mass will become supercritical if compressed."

Its' also true that a critical mass in an ideal geometry will become subcritical if you change the shape with no change in the density. Multiple factors affect (and effect) criticality.

Link to comment
Share on other sites

Hello, I am working on theoretical research for what is turning into somewhat complicated nuclear physics. I am not a nuclear physicists, so I would appreciate help with finding the information I need. First and foremost piece of information I need to solve my system of equations is the critical density of Uranium 235.

 

I cannot find this information directly in books or the internet. However, online articles suggested statistics from bare spheres. It was suggested by Wikipedia that the critical conditions for an unspecified grade of Uranium 235 at an unspecific temperature (but I am assuming room temperature) and an unspecific pressure (but I am assuming 1 atmosphere) was 52000 grams in a sphere 17 centimeters large. But to me, this information seems inaccurate.

 

I calculated the natural density of pure Uranium 235 to be approximately 18.075g/cm^3. The density of elemental Uranium with mixes of different isotopes is 18.95-19.01 g/cm^3 with Uranium 238 being the most common isotope. So, it makes sense that the natural density of pure U235 is around 18g/cm^3.

 

With that said, it seems illogical that the critical density solid Uranium is 52000g/(4/3*pi*(17cm)^3)=2.526g/cm^3. The bare spheres were not in environments of exceedingly high temperatures in a vacuum, so I cannot figure out how the density of solid Uranium in spherical shape could be that dramatically lower in only that specific experiment.

 

Does anyone know of credible resources where I can find the information I am looking for and explain this discrepancy?

The diameter of a sphere is not the same as the radius.

That makes the calculated density 8 times less than it should be.

2.526*8 = 20.2

That's close enough to the real value of 18 or so g/cc

Link to comment
Share on other sites

You're right, I completely missed that. That makes a lot more sense now.


That's why we need peer review :)


Although, even at that density, it still doesn't specify the specific compound, whether it's pure or not, what atmosphere it's at or what temperature, I'm still at a loss. Perhaps it can be confirmed by some complicated equation involving all of those variables which is still unknown to me.

Link to comment
Share on other sites

I don't suppose it's any military secret to say that they people who make bombs generally want them as small and light as possible.

Using uranium metal avoids "diluting" the reactive material with anything else and so it gives the smallest critical mass and critical volume.

Link to comment
Share on other sites

I wouldn't really care about critical mass if I knew the equations I am suppose to use and how to use them. When I was doing research an article said the number of uranium atoms that break into materials followed an equation of something like e^(k(t-1))... or something along those lines, and "k" is some criticality number, and if the mass is exactly critical it made k=1 which made the equation simpler and I also don't really know how I'd calculate a given *super*critical mass or density, let alone just regular critical mass.

Link to comment
Share on other sites

Another thing I'm confused about is the mechanism, because I see different examples like in a nuclear reactor and theoretical or in a bomb and it seems like they have different mechanisms for triggering fission. It seems like for the sake of a theoretical situation where ideal conditions are assumed, I don't need to calculate more math about a neutron gun right? If I have critical mass at critical volume, that's it, the decay from the high specific activity optizes the probability of the reaction triggers on its own right? Or no?

Edited by MWresearch
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.