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MODIFIED NUCLEAR ALGORITHM 1-A

In order to achieve some number, I built this algorithm. It's working like this:

- Base on the criticality of a Plutonium mass, about 5-11 kg
- You find the rayon, base on density of the correct phase (temperature related)
- The barn surface represent the probability of fission, with some deflector, it “increase”
- You find the barn surface of the critical mass: to get probability of 50 ±0.01, or more precise :)
- You repeat task 2-4, for the range where the reactor is going to operate, by keeping barn surface constant. Don't forget, rayon is modify by the density that is modify by temperature.
- Find the delta probability of the more dense phase minus the less dense phase
- For being able to produce heat we need 50+%
- Saturate the high probability with some more contamination, until that point
- Operate some calculation, to know the duration of reactor with that delta level of contamination versus the power of it: see sections “Design” and “Duration”

THE CODE

Written in C#, for simplicity, that way the form is simple to build. Watch out for your eyes, it's almost all red :) To copy the text: Shift at the beginning and another shift at the end!

THE DURATION (OLD DATA SET)

All this computation are relative to the criticality of an atomic Plutonium bomb at a diameter of 9 cm, with the neutron deflector. At that size, we know that the probability of producing 2 neutron by the occurrence of one neutron is close to 50+%. This fact will allow the reactor to produced self-sustain heat. Data:

Rayon: 0.045 m

Density of Sigma Phase: 15.85 g/cm³

Atomic weight: 239 g/mol

Contamination (purity): 5%

By the simulation software this configuration gives a barn surface for a probability of 50.079% of about: 2.895 b

Rayon: 0.04396 m

Density of Sigma Phase: 17 g/cm³

Atomic weight: 239 g/mol

Contamination (purity): 5%

barn surface: 2.895 b

By the simulation software, this configuration give a probability of 52.372%. The delta % of two phase = 2.293%, that amount could be lost in further contamination, with a safety margin of 0.293%, it gives 2% net lost.

0.293% gives a radius delta of about 50 micron. The subsequent contamination will need to be in order of 4.25% to fill the 2% probability left. This 4.25% of contamination could gives this amount of energy:

83.61E12 J/kg of fissile Pu-239

18 000 W of power for the reactor

2.153E-7 g/s of Pu-239

1 Years = 365 * 24 * 3600 = 3.15E7 s

The 4.25% give: 0.0425 * 6 kg(Pu-239) = 255 g

This 255 g could be constituted of 1/3 neutron depletion cycle

2.153E-7 g/s * 3.15E7 s = 6.79 g

255 g * (1/3) / 6.79 g = 10 years

1% gives a radius delta of about 0.17 mm. The subsequent contamination will need to be in order of 2.75% to fill the 1.293% probability left. This 2.75% of contamination could gives this amount of energy:

83.61E12 J/kg of fissile Pu-239

18 000 W of power for the reactor, thermal that gives 15 000 W electric

2.153E-7 g/s of Pu-239

1 Years = 365 * 24 * 3600 = 3.15E7 s

The 2.75% give: 0.0275 * 6 kg(Pu-239) = 165 g

This 165 g could be constituted of 1/3 neutron depletion cycle

2.153E-7 g/s * 3.15E7 s = 6.79 g165 g *(1/3) / 6.79 g = 8 years