Enrichment, or how I learned to stop worrying and love the SWU, part 3

OK, so in the last post we talked about the feed-to-product ratio and how it depended only on the levels of enrichment you chose for the product and the tails. Tonight we’ll begin to get into separative work.

When I first heard of separative work, I thought it was a very abstract and obscure concept. You read about it in a nuclear engineering text and they’ll talk about how it is the reduction of entropy in a mixture or something like that, and of course it is, but I think it is much more understandable than that.

Separative work is all around us. My first experience with separative work probably came about when I was a kid and I talked my mom into buying me some Lucky Charms. You know you’ve done it—I poured the cereal and immediately began to undertake some separative work on the delicious mixture of little marshmallows and less-exciting brown cereal bites. Separative work on Lucky Charms gave you happiness in the beginning, but you also know of the depressing moment when all you have left from your separative work are the soggy “tails” of the activity.

Later on, I realized that things would end a lot better if I ate the brown cereal at the beginning, thus “enriching” my Lucky Charms in little marshmallows. Then, after an appropriate increment of restraint, I could indulge myself in the culinary joy of highly-enriched Lucky Charms, and the sugar rush would send me running down the road to school.

Last night, I observed my little daughter, who is not quite two and has not yet been formally schooled in the principles of separative work and enrichment, nonetheless showing an effective grasp of the principles when I let her have a handful of Lucky Charms. She immediately set to work—separative work—extracting the marshmallows from the handful of cereal and casting the “tails” on the floor of the kitchen. So I have high hopes for her future as a nuclear engineer based on her success with separative work thus far.

The subtitle of this series has been “how I learned to quit worrying and love the SWU”, and SWU (pronounced “swoo”) is short for “separative work unit”, and it is a way to describe in a quantitative way this process of separative work. To separate one isotope from another in a mixture, we undertake separative work. Just how much separative work is required can be calculated using nothing but values of enrichment again, which is very handy. To calculate separative work, we use a function called the “value function”, which is based only on the value of enrichment.

 begin{equation} V(x) = (1 - 2 x) lnBigg(frac{1 - x}{x}Bigg) end{equation}

For those of you, who like me, occasionally require some mathematical refreshers, the statement “V(x)” is just mathematical shorthand for saying, “here is some value we call V, and it only depends on x, so whatever x is, just plug it in all the places you see an x on the right-hand side of the equation, and you’ll get an answer.”

So, if x was 0.03, for instance, we’d just plug in the value of 0.03 everywhere we see an x and calculate the answer:

 begin{equation} V(0.03) = (1 - 2(0.03)) lnBigg(frac{1 - 0.03}{0.03}Bigg) = 3.2675 end{equation}

The value function basically tells you the value of the mixture that you have. Surprisingly, if you have a mixture that has an enrichment of 50% (0.5), according to the value function, it has a value of zero! See, take a look:

 begin{equation} V(0.5) = (1 - 2(0.5)) lnBigg(frac{1 - 0.5}{0.5}Bigg) = 0 times ln(1) = 0 end{equation}

One might protest and say that a 50% enriched mixture of uranium is QUITE valuable, and in the real world one would be correct, but from the perspective of separative work, the least valuable mixture of all is the one where there is neither more or less of one component or another. And this makes sense if you stop and think about it for a moment. To have more or less of one component means that SOME separation has happened, therefore that mixture must have some greater value than a mixture of equal parts. So when we plot the curve of value function versus enrichment level, it looks like a deep valley, with the very bottom being at an enrichment level of 50%.

In order to calculate how much separative work it takes to get from one level of enrichment to another, we need to know the value of the value function for each of the three streams in the enrichment problem: feed, product, and tails. Now that we know the value function, calculating these values for each of the three streams is fairly straightforward:

 begin{equation} V(x_f) = (1 - 2 x_f) lnBigg(frac{1 - x_f}{x_f}Bigg) end{equation}

 begin{equation} V(x_p) = (1 - 2 x_p) lnBigg(frac{1 - x_p}{x_p}Bigg) end{equation}

 begin{equation} V(x_t) = (1 - 2 x_t) lnBigg(frac{1 - x_t}{x_t}Bigg) end{equation}

Where xf, xp, and xt are the enrichment values of the feed, product, and tails respectively. Remember when using enrichment values in the value function, do not input them as percentages! You will get the wrong answer! You need to input them into the value function as decimal values, not percentages.

Given that you have calculated the value function for each of the three streams, you can then calculate the separative work by a knowledge of the mass in each stream:

 begin{equation} text{separative work unit} = SWU = m_p V(x_p) + m_t V(x_t) - m_f V(x_f) end{equation}

The separative work is simply the mass of the product (mp) multiplied by its value function, plus the mass of the tails (mt) multiplied by its value function, minus the mass of the feed (mf) multiplied by its value function.

But I had said earlier that I could show that separative work was only a function of enrichment levels, right? To do so just takes a few more steps.

We know that feed, product, and tails are all related.

 begin{equation} m_f = m_p + m_t end{equation}

So it’s not hard to solve for one of these and incorporate it into the separative work calculations. Let’s solve for tails and substitute it into the separative work expression:

 begin{equation} m_t = m_f - m_p end{equation}

 begin{equation} SWU = m_p V(x_p) + (m_f - m_p) V(x_t) - m_f V(x_f) end{equation}

 begin{equation} SWU = m_p V(x_p) + m_f V(x_t) - m_p V(x_t) - m_f V(x_f) end{equation}

Now we can collect the terms that have to do with the mass of the product and the mass of the feed:

 begin{equation} SWU = m_p (V(x_p) - V(x_t)) + m_f (V(x_t) - V(x_f)) end{equation}

But recall from our last discussion that the mass of the feed and the mass of the product can be related by the feed-to-product ratio, which depends only on enrichment.

 begin{equation} FPR = frac{m_f}{m_p} = frac{x_p - x_t}{x_f - x_t} end{equation}

Let’s call this ratio “FPR” for feed-to-product ratio. mp*FPR = mf, right? So let’s substitute that into the expression for the mass of the feed:

 begin{equation} SWU = m_p (V(x_p) - V(x_t)) + m_p frac{x_p - x_t}{x_f - x_t} (V(x_t) - V(x_f)) end{equation}

Now we can simply divide through by the mass of the product, and get the amount of separative work required per unit of product:

 begin{equation} frac{SWU}{m_p} = (V(x_p) - V(x_t)) + Bigg(frac{x_p - x_t}{x_f - x_t}Bigg) (V(x_t) - V(x_f)) end{equation}

 begin{equation} frac{SWU}{m_p} = (V(x_p) - V(x_t)) + {FPR} (V(x_t) - V(x_f)) end{equation}

And that expression depends only on the enrichment levels of the feed, product, and tails.

An alternative would have been to solve for the enrichment in terms of the feed rather than the product. To do this, we would have again used the mass balance to remove one of the mass terms from the equation for separative work:

 begin{equation} m_p = m_f - m_t end{equation}

 begin{equation} SWU = (m_f - m_t) V(x_p) + m_t V(x_t) - m_f V(x_f) end{equation}

 begin{equation} SWU = m_f V(x_p) - m_t V(x_p) + m_t V(x_t) - m_f V(x_f) end{equation}

 begin{equation} SWU = m_f (V(x_p) - V(x_f)) + m_t (V(x_t) - V(x_p)) end{equation}

We employ the tails-to-feed ratio to complete the simplification:

 begin{equation} frac{m_t}{m_f} = frac{x_p - x_f}{x_p - x_t} end{equation}

 begin{equation} SWU = m_f (V(x_p) - V(x_f)) + m_f Bigg(frac{x_p - x_f}{x_p - x_t}Bigg) (V(x_t) - V(x_p)) end{equation}

 begin{equation} frac{SWU}{m_f} = (V(x_p) - V(x_f)) + Bigg(frac{x_p - x_f}{x_p - x_t}Bigg) (V(x_t) - V(x_p)) end{equation}

In the next post we’ll get into some specific cases where we use these expressions in real-world scenarios.

4 thoughts on “Enrichment, or how I learned to stop worrying and love the SWU, part 3

  1. What are the units of SWU? From the equations, they appear to be dimensionless (i.e., V() is dimensionless, FPR is a ratio, and therefore dimensionless, and there's literally nothing else). If I were to multiply product mass by the resultant number, I'd get a mass value.

    e.g., if I were to calculate the mass-independent SWU for enriching U-235 to 5% from natural(0.72%), with tails at 0.3%, I'd get ~1.619 (no unit). I don't understand what this number represents. "Separative work"? In Joules?

  2. I guess that the proper units for the answer would be kg*SWU/kg, which of course is just a SWU if you divide it out, which isn't exactly "dimensionless"…SWU itself is a dimension. A description of the reduction of entropy accomplished by a unit of separative work.

    Here were my calculation results for the problem you proposed:

    feed value: V(0.0072) = 4.856

    product value: V(0.0500) = 2.650

    tails value: V(0.0030) = 5.771

    feed-to-product ratio = (xp – xt)/(xf – xt)

    feed-to-product ratio = (0.05 – 0.003)/(0.0072 – 0.003)

    feed-to-product ratio = 11.19

    SWU/product = (V(xp) – V(xt)) + FPR*(V(xt) – V(xf))

    SWU/product = (2.650 – 5.771) + 11.19*(5.771 – 4.856)

    SWU/product = 7.13 kg*SWU/kg

  3. The unit for separative work is kg SWU. More often written as tons SWU (or MTSWU), since a kg SWU isn't much.

    Don't try to compare it with physical work as in joules. Separative work lives in its own little world. The amount of actual work you need to accomplish an amount of separative work varies greatly based on the process.

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