## Enrichment, or how I learned to stop worrying and love the SWU, part 2

Okay, before we get into the techniques of enrichment, let’s spend a little time with the fun part—the numbers! How much does it take and how much does it cost? Here’s where I’ve really enjoyed learning more lately.

The basic equations that describe enrichment aren’t that hard to derive. Even I can do it! So here goes:

You know that you’re going to start out with some amount of input material—the “feed” as it is called. You know that after you’re done you will have split the feed into two: the “product” and the waste, which in this case is called the “tails”. Assuming that you didn’t mess up and lose lots of material along the way, you could add up the product and the tails and that would be the same amount of material that you started with. So we’ll call the mass of the feed “mf”, we’ll call the mass of the product “mp”, and we’ll call the mass of the tails “mt”. And we’ll write a really simple equation that relates all three.

[latex]

begin{equation}

m_f = m_p + m_t

end{equation}

[/latex]

Next, let’s assume that we have a mixture that has only two components. They might be U-235 and U-238, or Li-6 and Li-7, or Cl-35 and Cl-37. This derivation works for any two-component mixture. Let’s say that the fraction of the less abundant component (U-235 in the case of uranium) is given by a non-dimensional variable called “x”. And let’s say that that value of x is different for the feed, product, and tails. For instance, in the case of uranium, if the feed is natural uranium, 0.0071 of it is U-235. So the value for x for the feed would be 0.0071. xf = 0.0071. And let’s say that we want to make fuel for a light-water reactor out of this, and it needs the uranium enriched up to 3%, or 0.03. So the value of x for the “product” would be 0.03. xp = 0.03. The last thing to figure out would be how much U-235 we will tolerate in our “tails”, the part that we’re going to throw away.

At first blush, we might think, “hey, I don’t want to throw away any U-235, I want it to all go into my product. So I’ll set the tails to 0% and everything will be great.”

Hah—that seems like a good idea but it isn’t. Because the amount of effort it will take to strip every single last atom of U-235 from the mixture is infinite. So in reality don’t do that. So then you might think, “OK, that might be too hard, maybe I should set the tails enrichment higher so I don’t have to do as much work. I’ll set it to just a little bit less than the natural level of enrichment, something like 0.006%.”

Another bad idea. If you set the tails enrichment too low, you expend too much effort trying to get some small amount of stuff. If you set it too high, you end up throwing away most of the stuff you’re trying to get in the first place. So you have to be kind of careful about how you set your tails enrichment. Which will have everything to do with the economics of enrichment, as we shall see.

So I’ve spent this time telling you that you know the enrichment of the feed (easy, it’s the natural enrichment) and the enrichment of the product (easy, it’s what your customer wants) but the real question is what is the enrichment of your tails. That’s going to have to be a choice based on economics, and as economics change you might find yourself revisiting that decision.

OK, so let’s say we’ll set the tails enrichment at 0.003. xt = 0.003. That’s means about half of the U-235 in the original uranium is going to end up in your product and about half is going to end up in the tails. If we assume that the total amount of U-235 is constant, then we can write another equation:

[latex]

begin{equation}

x_f m_f = x_p m_p + x_t m_t

end{equation}

[/latex]

This is just another way of saying that all of the U-235 ends up in either the product or the tails, and if you added it all up it would be the same as the amount that you had in the feed. Now we can combine these two equations to start figuring things out. We can use some of those tricks we learned in eighth-grade algebra to solve for things when we have two equations. We can rewrite the first equation to equal the tails rather than the feed:

[latex]

begin{equation}

m_t = m_f – m_p

end{equation}

[/latex]

And we can substitute that definition into the second equation:

[latex]

begin{equation}

x_f m_f = x_p m_p + x_t (m_f – m_p)

end{equation}

[/latex]

Then, again using our eighth-grade algebra skills we can expand the equation:

[latex]

begin{equation}

x_f m_f = x_p m_p + x_t m_f – x_t m_p

end{equation}

[/latex]

and we can group the terms relating to the mass of the feed and the product:

[latex]

begin{equation}

x_f m_f – x_t m_f = x_p m_p – x_t m_p

end{equation}

[/latex]

We pull out the common factors…

[latex]

begin{equation}

m_f (x_f – x_t) = m_p (x_p – x_t)

end{equation}

[/latex]

…and then we can solve for the mass of the feed:

[latex]

begin{equation}

m_f = m_p Bigg(frac{x_p – x_t}{x_f – x_t}Bigg)

end{equation}

[/latex]

Or even better, we can figure out a ratio between the mass of the feed and the mass of the product:

[latex]

begin{equation}

frac{m_f}{m_p} = frac{x_p – x_t}{x_f – x_t}

end{equation}

[/latex]

Hooray! So why is this a big deal? Because now we have an answer that doesn’t really care how much actual mass of feed or product that we’re talking about. On the left-hand side of the equation is a ratio, and on the right-hand side of the equation are a bunch of enrichment values. Let me show you how useful this nifty little expression is.

Let’s say that you want to figure out how much natural uranium you will need to fuel a nuclear reactor that uses enriched uranium. This is pretty much the situation we are in in the United States. Let’s say that you know that a typical reactor takes about 35 tonnes of enriched uranium fuel per year, and that that fuel is enriched up to 3%. You already know that the enrichment level of natural uranium is 0.0071, and let’s say that you’re comfortable with your tails enrichment being 0.003. So then you run the numbers:

[latex]

begin{equation}

frac{m_f}{m_p} = frac{x_p – x_t}{x_f – x_t} = frac{0.030 – 0.003}{0.0071 – 0.003} = frac{0.027}{0.0041} = 6.58

end{equation}

[/latex]

See how the tails enrichment shows up in both the numerator and the denominator of the expression? In this case, the ratio is 6.58, which tells you that you’ll need almost seven times more uranium as a feed material than you’ll need for fuel. So if you need 35 tonnes, you multiply 35*6.58 to get 230 tonnes. You’ll need 230 tonnes of natural uranium to make 35 tonnes of enriched uranium. Where does the rest of the uranium go? Into the tails. There’s 195 tonnes of uranium tails in this example. One of the things that you find when you do uranium calculations is that you almost always make a lot more tails than you make product.

But let’s do another example, to broaden our perspective. Let’s say that we’re talking about enriching lithium now instead of uranium. We want to make lithium-7 for a LFTR, and we need it to be enriched in lithium-7 up to a point of about 0.9999 (the more nines the better). But natural lithium is only about 90% enriched in lithium-7. So how much lithium feed will we need to enrich to 0.9999? Again, it has everything to do with the enrichment we’ll tolerate in the tails. Let’s say (being foolish) that we set the tails enrichment to 0.80. Don’t want to work too hard, right? So let’s go run the numbers and find a feed-to-product ratio for our lithium enrichment.

[latex]

begin{equation}

frac{m_f}{m_p} = frac{x_p – x_t}{x_f – x_t} = frac{0.9999 – 0.8}{0.9 – 0.8} = frac{0.1999}{0.1} = 1.999

end{equation}

[/latex]

In this case, we need about twice as much lithium feed to get a particular level of product. Why is that? Because by setting the tails to 0.8 only about half of the lithium-7 ended up in the product. The other half if in the tails. So even in a case where we have something like lithium-7, where it is the dominant constituent of natural lithium, setting the tails level of enrichment has a lot to do with how much feed it will take.

So let’s go back and run it again, this time with a tails enrichment of 0.5. No sense letting all that valuable lithium-7 go to waste, right?

[latex]

begin{equation}

frac{m_f}{m_p} = frac{x_p – x_t}{x_f – x_t} = frac{0.9999 – 0.5}{0.9 – 0.5} = frac{0.4999}{0.4} = 1.25

end{equation}

[/latex]

Now we only a little bit more lithium feed (25%) than we expect as product—in other words, most of the lithium-7 ended up in our product stream rather than in the tails. Seems like a better way to go, right? Well, maybe. It still depends on a lot of other things, like how hard it actually is to do the separation.

And getting into that level of detail will be a subject of an upcoming post…

Lithium has a number of uses in which the isotopic composition is of no importance. This suggests that if the preperation of the Lithium for seperation is free than the tails would be very little depleted in Li 7 at least until LFTRs became common.

Note that seperation of isotopes of Lithium is an established procedure as purified Li 6 is used in production of Tritium for bombs. That means that the feed for the Li 7 needed in a LFTR should be the tails from the seperation of Li 6 rather than natural lithium.