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Separative work, Enrichment of elements containing multiple isotopes

  1. Mar 3, 2014 #1
    Dear all

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

    I've come across the SWU equation (further details of terms)

    SWU = P·V(Np) + W·V(Nw) – F·V(Nf)



    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
  2. jcsd
  3. Mar 3, 2014 #2


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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.
  4. Mar 3, 2014 #3


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    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.
  5. Mar 4, 2014 #4
    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
  6. Mar 4, 2014 #5


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    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).
  7. Mar 5, 2014 #6


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    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.
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