The Front End of the Fuel Cycle:Enrichment-Gaseous Diffusion


In a gaseous diffusion enrichment plant, the
solid uranium hexafluoride from the conversion process is heated in its container until it
becomes a liquid. The container becomes pressurized as the solid
melts, and UF6 gas fills the top if the container. The UF6 gas is slowly fed into the plant’s
pipelines, where it is pumped through special filters called barriers, or porous membranes. The holes in these barriers are so small that
there is barely enough room for the UF6 gas molecules to pass through. The isotope enrichment occurs because the
lighter UF6 gas molecules made with U-235 atoms tend to diffuse faster through the barriers
than the heavier UF6 gas molecules made with uranium-238. One barrier certainly isn’t enough, though. It takes many hundreds of barriers, one after
the other, before the UF6 gas contains enough uranium-235 to be used in reactors. At the end of the process, the enriched UF6
gas is withdrawn from the pipelines and condensed back to a liquid that is poured into containers. This UF6 is then cooled and solidified before
it is transported to fuel fabrication facilities, where it is turned into fuel assemblies for
nuclear power reactors. All of this sounds very nice, but I want you
to get an idea of how freaking big and complicated these facilities are. Let me take you through a series of steps,
showing how the diagrams of the enrichment plant become more and more abstract until
only the details we want to think about are included. This is what a gaseous diffusion unit really
looks like, much like a great big stainless steel drum. This isn’t particularly useful. If we could look inside it, it would look
something like this. The feed stream brings UF6 to one side of
the barrier, and some fraction of it diffuses through the barrier, and the rest exits and
becomes the depleted stream. The gas that diffuses through the barrier
is slightly enriched in U-235 and becomes the enriched stream. These stages are put together, one after another,
to form what is called a cascade. Our next step is to draw the stages as boxes
with a feed stream and two exit streams. These exit streams become the feed streams
for the other stages in the cascade. Now let’s look at stage n in the figure
showing the cascade, and let’s begin to put some math to things. The enrichment factor, or the stage separation
factor, is called alpha, and it’s defined as follows. Where alpha is the ratio of the relative concentration
of U-235 in the enriched stream divided by the relative concentration of U-235 in the
depleted stream. Let’s look at this carefully, because I
want you to understand what all of the little symbols mean here. See Stage n in the figure above. Y sub n are the atoms or moles or molecular
concentration of U-235 of the enrichment side of the stage. That means that one minus Y sub n is the atoms
or moles or molecular concentration of U-238. Likewise, x sub n are the atoms or moles or
molecular concentration of U-235 on the depleted side of the stage. In words, this enrichment factor, alpha, is
the ratio of the relative molecular concentration, U-235 divided by U-238, in the enrichment
stream leaving the stage divided by the relative concentration of the depleted stream leaving
the stage. Now let me teach you to think as an engineer
thinks. We’re going to do something called material
balances. If we draw a huge ring around the whole cascade
system and look at the goes-in-a-s and the goes-out-a-s, we know that at steady state
these must balance. Looking at the number of moles in the feed,
the number of moles in the product, and the number of moles in the waste, we see that
the feed must equal the waste plus the product. We can also look at the feed product and waste
streams and do a balance on the uranium-235. The equations for these two material balances
are F equals P plus W, product plus waste equals feed. The feed times F sub F is equal to the product
times X sub P plus the waste times X sub W. Where F is the number of moles fed into the
cascade, P is the number of product moles, and W is the number of waste moles. X sub F, X sub P, and X sub W, are the mole
fractions of the individual streams. Let us suppose, as is usually the case, that
the feed is natural uranium, containing 0.711 atom percent of uranium-235. Let us suppose that the waste stream contains
0.3 percent of u-235, and the product contains 70 atom percent of u-235. For each mole of product, how much feed is
needed? Can you work this out? Hit the pause button, and go work it out. Hey, it’s easy. Two equations, two unknowns. Okay, I get 169.6 moles of feed for each mole
of product. What did you get? There’s a lot more to learn about cascades
than these two simple balance equations. That’s an understatement. Let’s take a balance across A-A prime in
the figure and find the net flow between stages in the cascade. Looking at the figure, we can write, “L
times Y sub N is equal to L minus P times X sub N plus 1 plus P times X sub P.” If
we further suppose there’s no mixing of streams that have unequal concentrations of
uranium-235, we have something called an “ideal cascade.” In symbols, this last statement is translated
as “Y sub N is equal to X sub N plus 2,” and, from the definition of alpha, we get
this equation. I could spend an hour and help you derive
all of these equations, but let’s not. I’m just going to tell you that, by taking
these equations and making some simple assumptions, we can show that the number of stages needed
to get a particular enrichment is given by the equation here for N ideal. We can also show that the inner stage flow
rates vary at each stage. That’s going to make things complicated,
right? And they are given by L ideal is equal to
this formula for the enriching stages of the cascade and equal to this formula for the
stripping part of the cascades. The equations we’ve looked at so far are
good for any cascade process. Now let’s look at some real examples to
give you an idea of how big a project it is to enrich uranium, particularly for gaseous
diffusion plants. For gaseous diffusion plants, the enrichment
factor can be assumed to be near 1.0043. Why this odd number? From molecular mechanics, we know that the
ratio of the speeds between two gas molecules at the same temperature is S one over S two
is equal to the square root of the masses of the inverse ratio. For U-235 and U-238 hexafluoride, the ratio
of their respective molecular weights are 238 plus six times nineteen divided by 235
plus six times nineteen, where the six times nineteen is the six fluorine atoms times their
molecular weight of nineteen. This ratio is equal to 1.0086. Therefore, the ratio of speeds is the square
root of this, or our famous 1.0043. To understand why this sort of works, if we
could arrange a very small tube so that 235 and 238 would race down this tube, the 235
would win, and this is basically what our barrier, pierced by these very small holes
does, and the enrichment across the barrier is 1.0043, theoretically. In practice, it is smaller than this, but
let’s use this number for our examples, and let’s make sure you understand the equations
I gave you above. Let’s let P, our product stream, be one
mole, and let the final enrichment be 80%. Let the final waste stream concentration,
otherwise known as the tails, be half of the input feed enrichment, or 0.00355%. Now, find the feed and tails needed to produce
one mole. Okay, next, how many enrichment stages do
we need, and how many stripping stages? Let me help you and say that the negative
number that you get for the stripping stages is correct. In this formula, negative means it’s part
of the stripping section. Finally, and here’s the surprising one,
what is the inner-stage flow rate at the feed stage? For an ideal cascade, that’s kind of a surprising
number, isn’t it? And it is even more surprising if I sum all
of the inner-stage flows over the whole cascade. There are over 34 million moles of gas pumped
around for each mole of product. Good grief! How these plants are laid out is roughly defined
by the inner-stage flow curve above. If the final stage is one diffusion unit,
then the feed stage is 51,805 units hooked in parallel, and, yes, these plants are that
big. They are freaking monsters! The one operating plant in the United States
is at Paducah, Kentucky, and sits on a 3, 425 acre site, containing 750 acres inside
the security zone. One process building covers 74 acres, and
there are four process buildings. The peak power capacity is 3,040 megawatts,
and the largest process motor is 3,300 horsepower. To cool the processes, the site uses 26 million
gallons of water a day and has approximately 400 miles of process piping. There are 19 miles of roadway in the site
and nine miles of railroad. It’s a big place!

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