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Found 2 results

  1. This is how the lossy image compression in a JPEG works.
  2. So You Want to Build a Fission Bomb? There are many reasons why one may want to build a fission bomb. Killing communists, for example, or sending a spacecraft to one of the outer planets. Building a bomb is not easy, but it can be done (see also Project, Manhattan). Even with publicly available information. Obviously, I’m not going to detail every little bit of our hypothetical bomb down to the last millimeter of wiring. First, I don’t know all that. If I did know it, posting it here might earn me a very long vacation to ADX Florence. The stuff here is just some equations and such to give you a general impression of how the design looks. The core of our hypothetical bomb is a sphere of highly enriched uranium. We want it to be subcritical (keff<1), but not by much. The more subcritical it is, the more we have to compress it to make it critical. In a real bomb, the core is usually surrounded by a layer of dense material such as tungsten or depleted uranium called the tamper. This helps keep the core together longer, and if it’s made of U-238, you can get some extra yield from the tamper fast fissioning. To simplify our analysis, our bomb won’t have a tamper. Then, you have a bunch of chemical explosives on the outside. This is what compresses the core, and takes it from subcritical to a super prompt critical state. When the core is super prompt critical, it’s going to heat up very quickly. Within milliseconds, the uranium at the center is going to become hot enough to be a gas (at very high pressure). At the edge of the core, you’re going to have very high pressure uranium gas next to an area of very low pressure. This is going to result in the uranium gas blowing off very quickly. This results in a “rarefaction wave” forming, as the core progressively evaporates away. This rarefaction wave proceeds inward at the speed of sound, and once it gets far enough in, the core becomes subcritical, and the reaction stops. Now, I’m going to make a few assumptions. These will result in some inaccuracy in our calculations, but the results will be close enough (also, it makes everything much simpler). Here they are; 1. The super prompt critical condition of the core will terminate once the rarefaction wave reaches the critical radius (rc). 2. The super prompt critical reactivity will remain constant until the core is subcritical. 3. The core is spherical with no tamper. 4. The temperature of the core is high enough that it can be treated as a photon gas (radiation pressure is the dominant force.) 5. No energy is lost to the surroundings during the process (adiabatic). 6. Our core is made of pure U-235. Since the rarefaction wave proceeds inward at the speed of sound, the device is critical for the following period of time; Where rmin is the radius of the core at maximum compression, and a is the speed of sound. We’ll also assume the gaseous core has a specific heat ratio of 4/3, so . Since the process is adiabatic, we know the following; Where Ecore is internal energy of the core at the end of the period of prompt criticality (this is the amount of energy released in the detonation). Substitute that into the speed of sound equation, and we get Putting that aside for a moment, let’s take a look at the point kinetics equation, which describes how power increases in a reactor following a sudden increase in reactivity (our bomb is essentially a reactor that’s undergone a massive increase in reactivity); (In this case, ρ represents reactivity, instead of density. β is the fraction of fission decay products which decay through neutron emission, and Λ is the average prompt neutron lifetime.) The second term in that equation gives us the power contribution from delayed neutrons, so we can ignore in this case (the bomb will have long since detonated by the time they become a factor). Also, in the case of super prompt criticality, ρ >> β. So our equation reduces to So to get the total amount of energy produced in the core during super prompt criticality, we need to integrate the power equation over the amount of time the core is super prompt critical. If we call that time tc, we get the following expression; Where E1 is the amount of energy produced by one fission event (202.5 MeV). Substituting that into our first equation and the speed of sound expression, and then doing a bit of algebra (which I’ll leave out for the sake of brevity), we end up with this; Solving for the Ecore expression, and defining Δr as the difference between criticality radius and the radius at maximum compression; Which gives us the total amount of energy released by the detonation. The main unknowns here are the reactivity (ρ) and critical radius (rc). Fortunately, both of these are fairly easy to determine. The critical radius is the radius at which a sphere of material has a keff (ratio of neutron production to neutron absorption) of 1. ν is the average amount of neutrons produced per fission event (~2.5), Σf is the fission cross section (σf = ~1 barn for fast neutrons), D is the thermal neutron diffusion distance (.00434m for U-235), Bg is the ‘geometric buckling’, and Σa is the absorption cross section (σa=~.09 barns for fast neutrons). Convert from σf to Σf using the following formulas; Bg for a sphere can be calculated using the following formula; Setting keff to 1 and solving for r, we find that the critical radius rc is roughly 5.2cm. A sphere of U-235 of this size will have a mass of about 11.25kg. Now that we have the keff equation, determining ρ is fairly simple. Since keff is going to be higher the more you compress the core, you obviously want to compress it as much as possible. The following equation gives the amount of explosive needed to compress the core by a given amount; Escfc is the amount of energy needed to compress the core by a given amount. η is the amount of energy contained in each unit of chemical explosive (4184kJ/kg for TNT), and ε is the efficiency of the implosion process. ε is about 30% in well-designed nuclear weapons, crude designs are probably closer to 5-10%. Congratulations! Now you have (a non-trivial portion of) the knowledge you need to build a working fission device! Edit: Updated 4/24 with corrected cross sections
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