How many stages?

A container divided by porous membrane, contains


uranium hexaftoride gas UF6 in the left half and vacuum


in the right half.





The gas slowly diffusing through the membrane is extracted,


and stored for future enrichment.





How many stages of enrichment are needed, if natural


concentration of U235 is 伪=0.7%, and desired final


concentration is 蠅=90% ?|||Okay, first a little probability thought. Let's say that I have an urn full of black and white marbles, in the ratio R/1-R where 0%26lt;R%26lt;1. Let's say that if I pick a black marble, the odds that I will decide to toss it aside is p, and for white, it's q. Then if I'm picking marbles at random from the urn, the expected ratio of blacks and whites in the tossed pile is Rp/(1-R)q, or (R/(1-R))k, where k = p/q. Letting





r/(1-r) = R/(1-R))k





and solving for r, we see that doing this recursively, we get:





r = (R k^n) / (1+R (k^n-1))





where R is the original concentration 0.007, r is to be 0.900, n is the number of stages to be determined, and k is the magic ratio. What shall k be? Well molecular diffusion occurs at a rate proportional to the inverse square root of atomic mass, so since we know that the atomic mass of F is roughly 19, we can compute the ratio k to be 1.00429 theoretically. For this value of k, n will have to be about 1671 stages for r to be over 0.900. However, in practice the figure of k is closer to 1.0014, so that n will have to be about 5113 stages.|||Long ago, I worked in Uranium diffusion for about a year and if my memory serves me right, we had 5,000 centrifuges pushing the hexafluoride through 88 miles of 10" aluminium piping at very high vacuum with a membrane after each stage. Out of 2 tons of gaseous feed we got about a pint of solid 235.


When performing cell maintenance, the cell was isolated and the hexafluoride was recovered before opening up the cell, by pulling it out through a liquid nitrogen cooled vessel.