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Cern - Nuclear data analysis framework

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We have just compiled & upgraded to the latest ROOT version. ROOT is a data analysis framework used mostly, but not limited to, high energy physics. It is used by the techs at CERN to explore the benefits of object oriented programming in physics data analysis. It consists of libraries, CINT or cling interpreter for C/C++, plotting interface and powerful parellel processing support. It is the preferred choice in many laboratories.

CERN - Conseil Europeen pour la Recherche Nucleaire | ECNR - European Council for Nuclear Research

What is Nuclear Energy? What is Nuclear Waste? ... and what to do with it ....

Nuclear waste is called Uranium Oxide ! Uranium Oxide Solar Panels, part 3.

Did you know that uranium dioxide is a semiconductor? In fact, it’s a really good one! That means that you could make solar panels, LEDs, and computer chips from it…can you imagine uranium oxide solar panels?

Example How Nuclear Waste can be useful Solar Panels !!!!

How a Nuclear Power Plant Works

Nuclear power plants run on uranium fuel. In the reactor, uranium atoms are split through a process known as fission(see sub menu item -Iovu Nuclear fission). When atoms are spilt, they produce a large amount of energy that is then converted to heat. The heat boils water, creating steam that is used to turn turbines, which spins the shaft of a generator. Inside the generator, coils of wire spin in a magnetic field and electricity is produced. Nuclear power plants in the United States use two types of reactors to achieve this process: pressurized water reactors(see animated PWR - just below) and boiling water reactors(see animated BWR - bottom of page).

The Pressurized Water Reactor (PWR)

The Pressurized Water Reactor (PWR)

The Pressurized Water Reactor (PWR)

Pressurized Water Reactors (PWR) keep water under pressure, so the water heats but does not boil. The heated pressurized water is run through pipes, which heat a separate water line to create steam. The water to generate steam is never mixed with the pressurized water used to heat it.

Boiling Water Reactor (BWR)

The Boiling Water Reactor (BWR)

The Boiling Water Reactor (BWR)

Boiling Water Reactors (BWR) heat water by generating heat from fission in the reactor vessel to boil water and create steam, which turns the generator. In both types of plants, the steam is turned back into water and can be used again in the process.

Courtesy of the Nuclear Regulatory Commission (NRC)


Lets get inside the BWR ... .



Left is the side view and under is the top view(for labels 1, 2 and 3)


1. Emergency stop bar - barre d'arrêt d'urgence


2. Control Bar - barre de contrôle


3. Fuel assembly - assemblage combustible


4. Biological protection - protection biologique


5. Steam outlet - sortie de vapeur


6. Water inlet - entrée de l'eau


7. Thermal protection - protection thermique







Introduction Enrichment.

First of all lets look at two way's uranium enrichment: the gas centrifuge, which promises vast efficiency improvements and far less power consumption than the previous way (gas diffusion) enrichment.

CANDU Nuclear Power Plant at Qinshan, China

Enrichment, part 1.

“Enrichment” or more particularly “uranium enrichment” is probably one of those phrases that the average person hears on television or reads on the Internet and has only the vaguest concept of what it means. They likely think “it’s bad” and “it has something to do with uranium, which I think is bad, for some reason”. Commentators and journalists decry the idea that Iran *gasp* might have access to uranium enrichment. Our leaders both domestic and international, loudly thunder that lesser nations shall not be allowed to develop enrichment.

Iranian enrichment President Iranian, Mahmoud Ahmadinejad

The world of the nucleus is a secret world, shielded from everything else by the cloud of electrons surrounding it. The electrons do all of the interacting, the nucleons do nearly nothing. The only hint they give to their existence is their charge, which is balanced by the electrons, and their mass. Practically all of the mass of the atom is in the nucleus.

Different isotopes of elements have different masses. They have the same number of protons, and hence the same charge and the same chemical nature, but different masses. From a chemical perspective, there’s no difference between most isotopes (with a few exceptions, like normal hydrogen and deuterium). But from a nuclear perspective there can be a huge difference between two naturally occurring isotopes. Lithium is a good example. Natural lithium is 90% lithium-7 and about 10% lithium-6. Chemically they’re the same. You can make batteries from lithium-6 just as well as you can make batteries from lithium-7. But from a nuclear perspective, they’re really, really different. Lithium-7 has almost no appetite to absorb neutrons, whereas lithium-6 has a HUGE appetite to absorb neutrons.

Uranium is in an analogous situation. Uranium-235 and uranium-238 are the two naturally occurring isotopes of uranium. One of them is far more common than the other. Can you guess which? Yeah, it’s the one that we’re not so interested in. Kind of like how when your toast falls down it always seems to land on the buttered side. Nevertheless, we’re a lot more interested in uranium-235, which is only 71 out of every 10000 uranium atoms than we are about uranium-238, which is the other 9920 out of 10000 uranium atoms. (1 atom out of 10000 is uranium-234, but we won’t worry about that right now).

Enrichment is the process where you separate one kind of isotope from another, and the overwhelming reason to do it is because you intend to use your product in a nuclear process. Because if you want to do a chemical process, you usually don’t care too much about what kind of isotope mix you have. Sometimes, but very rarely.

So how do you conduct enrichment? Well, your toolbox is pretty sparse. All of the chemical tricks that are so commonly used to separate one thing from another don’t work when you’re dealing with one kind of element, because chemically it’s all the same. You have to try to separate one thing from another based on something that’s different between the two.

Almost always that difference is mass. Lithium-6 has six units of atomic mass, lithium-7 has seven. Uranium-235 has 235 units of atomic mass, uranium-238 has 238. So there’s a difference there. It’s more pronounced for lithium than uranium, but it’s there nevertheless.

Separating uranium is a big deal because U-235 is fissile and U-238 is not. The original motivation to separate uranium was to make a bomb during the Manhattan Project. The first nuclear bomb ever used in war, 65 years ago today, was based on separated uranium-235. And enrichment has had a bad rap ever since.

To separate U-235 from U-238, you need to put them in a chemical form where it’s easier for them to get away from one another. Thank goodness for fluorine! Uranium hexafluoride (UF6) is a gaseous form of uranium. It has one uranium atom in the middle of six fluorine atoms. It looks a lot like one of those little jacks my sister used to play with. There’s another aspect to UF6 that’s really important. If fluorine itself had a number of different isotopes then this whole idea wouldn’t work. For instance, if there were isotopes of fluorine that had mass numbers of 18, 19, and 20, then you could go build a UF6 molecule and one of its fluorine’s might be 18, and another would be 20 and another could by 19, and that mass difference from uranium could get “lost in the noise” from all of the different isotopes of fluorine.

The crystal structure of This is a simple UF6 in a glass Ampoule
uranium hexafluoride (UF6) mononuclear molecule(UF6)

But fortunately, that’s not how it works. Fluorine has only one natural isotope, number 19. So when you go add six fluorine's to uranium, you know that they’re all going to be 19. Any difference in mass from the molecule all comes from the uranium, not the fluorine. So we’ll do a little math. If you have a uranium atom that has a mass of 235, and you go and add six other atoms each with a mass of 19 to it, what do you get? (235 + 6*19) = 349. (I won’t lie, I used my calculator) And if you have a uranium atom with a mass of 238? (238 + 6*19) = 352.

So the whole trick in uranium enrichment is to separate molecules in a gas that have a mass of 349 from molecules that have a mass of 352. Folks, that’s not very much, and you can see why uranium enrichment is hard and expensive. But, thanks to the gaseous nature of uranium hexafluoride and the monoisotopic existence of fluorine, it’s possible in the first place, where otherwise it might not be.

Enrichment, part 2.

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

mf = mp + mt

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:

xf*mf = xp*mp + xt*mt

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(4eme College) algebra to solve for things when we have two equations. We can rewrite the first equation to equal the tails rather than the feed:

mt = mf – mp

And we can substitute that definition into the second equation:

xf*mf = xp*mp + xt*(mf – mp)

Then, again using our eighth-grade(4eme College) algebra skills we can expand the equation:

xf*mf = xp*mp + xt*mf – xt*mp

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

xf*mf – xt*mf = xp*mp – xt*mp

We pull out the common factors…

mf*(xf – xt) = mp*(xp – xt)

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

mf = mp*(xp – xt)/(xf – xt)

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

mf/mp = (xp – xt)/(xf – xt)

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 World. 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:

mf/mp = (xp – xt)/(xf – xt) = (0.03 – 0.003)/(0.0071 – 0.003) = 0.027/0.0041 = 6.58

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.


mf/mp = (xp – xt)/(xf – xt) = (0.9999 – 0.8)/(0.9 – 0.8) = 0.1999/0.1 = 1.999

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?


mf/mp = (xp – xt)/(xf – xt) = (0.9999 – 0.5)/(0.9 – 0.5) = 0.1999/0.1 = 1.25

Now we have 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.

Last Updated on Sunday, 08 October 2017 13:32  


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