(from Was something else, more chem data thoughts, CG):
Start with trying to find the rpm. Assume the diameter as .5m, with a radius of .25m, plugged into pi r^2 = pi (.25m^2) = 1.9634m the circumference. The circumference C travels L with the number of rotations L/C. 350m/1.934m = 178.262m/s, or 10695.72 rpm.
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I was wrong, since the circumference is 2(pi)r, not (pi)r^2. So redoing the estimated rpm goes like this: 2(pi)r = C, 2(pi)(.25m) = 1.5707 and L/C = 350ms/1.5707 = 222.8169m/s, or 13,369 rpm.
Any way, WDK wrote:
``I think you're interpreting SWUs (more exactly, SWU-kgs) incorrectly. SWUs are a measure of a certain amount of isotope separation, not the amount of input energy required for that given amount of separation by whatever process one uses to achieve that separation. For example, here:
http://www.globalsecurity.org/wmd/intro/u-centrifuge.htm ''
The above was a good article. Thanks.
I might be interpreting SWU's wrong. But think about it. SWU is separative work units, meaning the amount `work' done to separate a given quantity of isotopes. But notice there is no ratio of work units to time units.
``The unit is strictly: Kilogram Separative Work Unit, and it measures the quantity of separative work (indicative of energy used in enrichment) when feed and product quantities are expressed in kilograms...''
I read the above to say kilograms or tons relates not to the SWUs per se, but to the units used to measure the weight of the gas processed; that is why the notation puts kg in front of SUW rather than after it, ie. kg-SWU in the Urenco paper rather than SWU kg. SWUs depend on units of mass, but are not a unit of mass.
So my real point is that it looks to be difficult or maybe impossible to derive a unit of time out of SWUs. And to re-iterate, the current ramp up of US propaganda depends critically on guessing how long it would take Iran to get enough enriched uranium to build a nuclear weapon.
(According to wikipedia the bomb for Hiroshime took 60kg, of which only 0.7 went into fission.)
As you note above, kWh/SWUs. I take this to mean an empirically derived ratio of kilowatt hours to separative work units. These kinds of empirical factors are of a different quantitive type than an equation rearranged and solved as a function of time. We can argue, fudge, or dispute an empirical factor. As Russell noted about such empirical results, they become increasingly probable approaching certainty as the repetition of their measure approaches infinity...
The best part of the above article occurs just above your quote:
Although the separation factors obtainable from a centrifuge are large compared to gaseous diffusion [ranging from 1.01 to over 1.10], several cascade stages are still required to produce even LEU material. Furthermore, the throughput of a single centrifuge is usually small, which leads to rather small separative capacities for typical proliferator centrifuges. Separation factors depend on the absolute mass difference between isotopes (not the mass ratio) and the square of the peripheral speed. Separation factors for U-235/238 range from 1.026 for a 250 m/sec centrifuge to over 1.233 for a 600 m/sec centrifuge.
A single centrifuge might produce about 30 grams of HEU per year, about the equivalent of five Separative Work Unit (SWU). As as a general rule of thumb, a cascade of 850 to 1,000 centrifuges, each 1.5 meters long, operating continuously at 400 m/sec, would be able to produce about 20-25 kilograms of HEU in a year, enough for one weapon. One such bomb would require about 6,000 SWU...''
The mention of `separation factor' is probably the same equation given in the Urenco paper. It's an exponential function using the difference between masses U-235 and U-238 multiplied by centrifuge velocity squared, divided by temperature and the gas constant, all multiplied by the length of the centrifuge over the diffusion constant--all times sqrt of two. The separation factor depends on process gas properties and the dimensions and velocity of the centrifuge.
In other words, it seems to me calculating the separation factor of a centrifuge is a more realistic measure. That's why I was trying to assemble the necessary values. Notice the above example uses 1.5m L and 400m/s for v. That is about the same as the Iranian's, 1.8m going at 350m/s. The quote above figures it would take 850-1000 cascade to get 20-25kg U-235. Remember Hiroshima used 60k.
Oh well, back to the pencil and paper, waiting for the big afterglow...
CG