Fuel input per gigawatt output: 250 tons raw uraniumI liked the article, but dumb unit-conversion oversights like this make me twitch a little. Just to put it in more familiar terms, this is like saying that a 150 horsepower engine requires 10 gallons of gas. You have a rate on one side, and an absolute quantity of fuel on the other. That doesn't compute. I suspect they mean GW*yr, or maybe GW*powerplant-lifetime, but you can't tell without additional information.
Also, since we're talking of Thorium, scientists have found the first stable ultra-heavy element, provisionally referred to as unbibium, lurking in garden-variety samples of terrestrial Thorium.
The magical thing about uranium is that it goes critical all by itself. Natural uranium, sitting in the ground, has some fraction that spontaneously fissions and makes a few free neutrons. All you have to do is put enough uranium together that the best place for the neutrons to go is onto another uranium nucleus. A uranium nucleus eating a neutron breaks up, freeing two or three more neutrons; they are already in your pile, and so the best place for them to go is onto another uranium. It used to be that you didn't even need people.posted by fantabulous timewaster at 3:53 AM on December 22, 2009 [1 favorite]
Reactor folks talk about the "neutron multiplication factor," usually called k or keffective, which is the number of fission neutrons that go on to produce another fission. A heavy element like uranium or thorium has half again as many neutrons as protons; two fragments from the middle of the periodic table only want to have about 20% more neutrons than protons; this is why you get a few free neutrons from a fission. An "average" fission in U-235 makes about 2.4 neutrons. It took one neutron to induce the fission, so if all the fission neutrons made other fissions you would have k = 1.4 and a runaway reaction. Your reactor must contain some material that eats 0.4 neutrons per fission, and it should be possible to add or remove this "neutron poison" as different parts of the fuel burn faster or slower.
To induce fission on Th-232 takes two neutrons: one to initiate the sequence Th-233 → Pa-233 → U-233, and a second to induce fission in U-233. If thorium reactors exist, U-233 must produce more than three neutrons per fission. But now there is a much bigger variation in the amount of control that's needed. Early in the fuel cycle you need a neutron source, which is usually enriched U-235. The protactinium takes weeks to decay to U-233; during this time you'd like it to be out of the neutron flux, since it apparently likes to send off an extra neutron and turn into the nastier U-232. Finally your U-233 has to be dense enough to sustain a reaction, and you need some neutron absorber to control its rate. If the protactinium decay were faster this would be a nice little system. As it is, it sounds pretty touchy.
Thorium reactors are probably doable in the long term. But sustaining a reaction is much more complicated than simply assembling a pile of enriched uranium, and politically the "simple" reactors are hard enough to build.
The fission of a single uranium-235 (or similar) nucleus is thus accompanied by the release of over 200 Mev of energy. This may be compared with about 4 ev which are released by the combustion of an atom of carbon-12 […] Alternatively, it may be stated that 1 pound of fissile material should be capable of producing the same amount of energy as 1400 tons of 13,000 Btu/lb of coal. […]The actual true fuel consumption of a nuclear reactor, even a large one, is fairly small, on the order of a few kilograms per fueling. The only way Wired is getting anything close to the number they are quoting is by counting as consumed the mass of the entire fuel assemblies, which include not only the actual (fissile) fuel but also a lot of cladding and then all the nonfissile 238U that's just along for the ride and occasional neutron.
A useful fact to remember is that the power production corresponding to the fission of 1 gram of material per day would be roughly 104 watts, i.e., 1 megawatt. The mass consumed is, however, greater than 1 gram because some of the fissile nuclei are lost as a result of non-fission capture reactions.
from Glasstone, S., Sesonske, A., Nuclear Reactor Engineering, 1967. §1.44 - 1.46.
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posted by aeschenkarnos at 6:26 PM on December 21, 2009 [20 favorites]