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Enrichment: Visualizing the Value Function

Kirk Sorensenblog, Isotope Separation

(1)   \begin{equation*} V(x) = (1 - 2 x)\ln\left(\frac{1 - x}{x}\right) \end{equation*}

I’m a math minor, and an equation like that is not a very appealing way to begin a blog post. But I got the unpleasant part out of the way early because the good news that this equation (the value equation) is totally central towards understanding how much separative work we need to separate isotopes from one another, and the even better news is that the value function makes a lot more intuitive sense than we might initially think.

Let me show you what the value function looks like when you plot it on a graph. The input value “x” can only range from zero to one. No negative numbers or values greater than one allowed here!


You can see that the value function looks like a deep valley with very steep sides. Reading the values of the graph, you can see that if the input “x” is one-half (0.5) then the value function is zero. And it’s hard to see, but you can imagine that if the value function is zero or one, then it has an infinite value. So what does all this actually mean?

To help answer this question recently I bought a bunch of red and white marbles online (www.megaglass.com if you’re interested). Red and white marbles are going to represent two different nuclear isotopes in my explanation of the value function. They’re pretty good at doing this because they’re the same size and only differ in color.

Let’s say I only have white marbles in some kind of mixture. Well, then it isn’t much of a mixture! But according to the value function, when I have all white marbles in my mixture (and we’ll say that “x” is the fraction of red marbles in my mixture) then my value function would be:

    \begin{displaymath} V(0) = (1 - 2 (0))\ln\left(\frac{1 - 0}{0}\right) = 1*\ln(1/0) = \infty \end{displaymath}


Let’s imagine that I only have red marbles in my “mixture” (x = 1.0):

    \begin{displaymath} V(1) = (1 - 2 (1))\ln\left(\frac{1 - 1}{1}\right) = -1*\ln(0) = \infty \end{displaymath}


But let’s say I have a half-and-half mixture of red and white marbles (x = 0.5):

    \begin{displaymath} V(0.5) = (1 - 2 (0.5))\ln\left(\frac{1 - 0.5}{0.5}\right) = 0*\ln(1) = 0 \end{displaymath}

Then my mixture doesn’t have any value at all?

Here was my first clue that there was something intuitive going on with the value function. Why does a half-and-half mixture have no value at all? Well, think about it. If you have a half-and-half mixture of marbles, you’re no closer to having a mostly white mixture than you are to having a mostly red mixture. It all depends on which kind of marble you start picking out.

But if you have a mixture that’s mostly white, or a mixture that’s mostly red, then those mixtures already have some “value”, and it’s easier to increase their “value” by removing the minority component. When it comes to enriching lithium, that’s exactly what we do. We remove more and more lithium-6 until the mixture is nearly all lithium-7.

But with uranium, surprisingly, we move the other direction on the value function. We “decrease” the value of the mixture as we “enrich” it! Hard to believe, huh?

Consider natural uranium. It’s only got 0.71% uranium-235. Let’s imagine that in white and red marbles:

    \begin{displaymath} V(0.0071) = (1 - 2 (0.0071))\ln\left(\frac{1 - 0.0071}{0.0071}\right) = 4.870 \end{displaymath}

Now let’s imagine this “enriched” up to 3.5% red marbles, like enriching uranium up to 3.5%:

    \begin{displaymath} V(0.035) = (1 - 2 (0.035))\ln\left(\frac{1 - 0.035}{0.035}\right) = 3.085 \end{displaymath}

Wait a minute! Are you telling me that natural uranium with a value of 4.870 is “better” than enriched uranium with a value of 3.085?

Yup, because these equations don’t say anything about whether what you’re separating is good or bad or cheap or valuable. They just describe how “separated” it is. And natural uranium, with only 0.71% U-235, is more “separated” than enriched uranium at 3.5%.

Kinda wild, isn’t it?

What about the tails? What are they worth?

    \begin{displaymath} V(0.003) = (1 - 2 (0.003))\ln\left(\frac{1 - 0.003}{0.003}\right) = 5.771 \end{displaymath}

HAH! So the tails are worth the most of all, according to the value function! All of this seems kind of counter-intuitive, but if you think about it for a moment, the tails are much closer to being a pure mixture (in this case of U-238) than either the feed or the product is, so they have the highest value function.

Now of course, we would not consider 50% enriched uranium to have a value of zero (in fact we would consider it QUITE valuable) but that’s how the equations come out. And that’s because 50% enriched uranium would be neither more U-235 or more U-238, but rather a mixture that could go either way.

Isn’t it interesting that when you plug all this into the SWU equation you get something that makes sense? It still amazes me, and I wrote the dang program that shows it.

Well, that’s probably enough fun with marbles and value functions for one night–I hope you liked it! Let me know if you did…

7 thoughts on “Enrichment: Visualizing the Value Function”

  1. Hi Kirk

    Great work – you are making this compelx subject very understandable. keep it up!

    Do you have a source which explains the derivation of the value function?

    Reply

  2. Chapter 12 of "Nuclear Chemical Engineering" has some extensive derivations on aspects of separation, but I haven't found the particular one for the value function yet.

    Here's another good resource:

    Uranium Enrichment and Nuclear Weapon Proliferation

    You can download the whole book in PDF form for free on that site. Chapter 5 is particularly interesting.

    Reply

  3. Yep, the work of uranium enrichment is all about making those tailings, squeezing the drops of U-235 out of them into the richer U stream.

    Which is why there are some relatively "rich" tailings around, since diffusion is a relatively expensive/inefficient separative process. The more efficient processes (centrifuge and SILEX) can economically squeeze some more U-235 out of them that was too expensive to get with diffusion.

    Reply

  7. Unfortunately the LaTeX form of the equations seems to be out of date, and the equations are not readable. I have been looking for some sort of derivation of a generalized form for the value function for enrichment.

    Does one exist also for thorium breeding in a reactor of U233? Perhaps one that considers the value lost of not burning enriched U235 or U238 to become Pu239.

    Is it valid (and who proved) for a value function from gas diffusion to be applied to other enrichment processes such as centrifuge enrichment?

    Thanks

    Reply

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