Separative work, Enrichment of elements containing multiple isotopes

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Discussion Overview

The discussion centers on the separative work units (SWU) required for the isotopic enrichment of tungsten, which has multiple isotopes. Participants explore the applicability of the SWU equation typically used for binary systems to a multi-isotope scenario, examining both theoretical and practical aspects of enrichment processes.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions whether the SWU equation, designed for two isotopes, can be applied to tungsten with five isotopes.
  • Another participant suggests that while the equation may not directly apply, one could estimate upper and lower bounds for SWU by treating other isotopes as either the worst or best case.
  • A different viewpoint proposes that multiple isotopes can be managed through cascade separations, depending on the desired fractions of the product.
  • One participant provides a specific example of enriching tungsten-183 from a natural level to a higher concentration, calculating SWU based on given values.
  • Another participant reinforces that the equations are primarily for binary systems, citing the centrifuge process used in uranium enrichment as a relevant analogy.
  • Concerns are raised about the complexities of enriching isotopes that are close in mass, indicating that significant effort is needed for such separations.
  • Laser isotopic separation is mentioned as a potentially more effective method for targeting specific isotopes in the middle of the enrichment chain.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the SWU equation to multiple isotopes, with no consensus reached on a definitive method for enrichment. The discussion remains unresolved regarding the best approach for tungsten enrichment.

Contextual Notes

Participants highlight limitations in the SWU equation's applicability to multi-isotope systems and the complexities involved in separating isotopes that are close in mass. There are also unresolved assumptions regarding the efficiency of different enrichment techniques.

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Dear all

I'm trying to find the separative work units (SWU) required to enrich isotopically Tungsten.

I've come across the SWU equation (http://www.fas.org/programs/ssp/nukes/effects/swu.html)

SWU = P·V(Np) + W·V(Nw) – F·V(Nf)

where

V(x)=(2x-1)ln(x/(1-x))

and x is the concentration

My question is does this equation apply to elements with just two isotopes or will it work for tungsten which has 5 isotopes?

Thank you very much
 
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I don't even see a way to plug in more than two isotopes in this equation. It is designed for two isotopes.

If you want to enrich (or get rid of) the heaviest or the lightest isotope, treating all other isotopes as the worst case (the isotope closest to it) gives an upper bound, and treating them as the best case gives a lower bound for SWU.
I guess separating an isotope in the middle will need significantly more work.
 
It could be applied to multiple isotopes depending on the desired fractions of the product, but if one wants one out of five, one will have to have cascade separations. For example, of one wanted 1 of 5, then one might have to get 2 or 3 of 5, then 1 or 2 of the first stage product.

The trend these days is to use laser isotopic separation, which focuses on a specific isotope.
 
Thanks very much for the answers.

I just want to enrich one isotope at a time.

So if I've understood correctly is my following example correct?

For 1kg of P (product) enriched from a natural enrichment level of W183 = 14.31% up to 80%
with a F (feed) of 50kg at the natural enrichment (14.31%) and therefore 49kg of waste depleted to 6.4% is made.V(Np)=(2*0.8-1)*ln(0.8/(1-0.8)) =0.832
V(Nw)=(2*0.064-1)*ln(0.064/(1-0.064)) = 2.339
V(Nf)=(2*0.1431-1)*ln(0.1431/(1-0.1431)) = 1.278

SWU = 1 * 0.832 + 49 * 2.339 - 50 * 1.278 = 51.58
 
I believe that mfb is correct concerning the equations given in the OP are for a binary system, i.e., one heavy and one lighter isotope, and it is based on the centrifuge process. In the case of natural uranium, one separates U-235 from U-238, the two predominant isotopes, with the goal of extracting the rarer U-235 (natural abundance ~ 0.72%).

In the centrifuge process, the heavier isotope will concentrate to the outer region of the centriuge. In the case of U, it is in the form of UF6, a gas at the conditions in the centrifuge.

In the case of W, one can ignore W-180 (abund. ~0.12%) and focus on W-182, -183, -184 and -186, with abundances of 26.5%, 14.3%, 30.6% and 28.4%, respectively. It would be easy to separate W-186 from the lighter fractions ()in a centrifuge system), but then one would have to remove W-184 from W-183 + W-182, then separate W-183 from W-182, which is more difficult because the masses are so close (i.e., 1 amu).
 
It gets even worse - you can get a sample of highly enriched 186, but the remaining part will still have some 186 in it. If you try to enrich 184 there, this fraction increases again. I guess you can still use the same technique as for uranium, but every isotope will have a different distribution along the "enrichment chain" - to really separate them in the middle you need a lot of work.

Laser isotopic separation looks more promising, as it can select isotopes in the middle.