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One of the counterpoints to the Thorium fuel cycle is that Protactinium, which is generated in this cycle degrades reactor efficiency and thus needs to be removed, at least from the liquid fluoride or molten salt reactors. However, as far as I can tell, Protactinium was not removed during the functioning of the first solid-fuel Thorium reactor, which was the third core used at Shippingport; or at least I can find no mention of Protactinium being removed (during operation) in the official fuel report.

So, question(s):

  • Quantitatively, how bad can reactor efficiency be degraded by Protactinium not being removed?
  • To what degree does this degradation depend on the type and other parameters (geometry etc.) of the reactor?
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    $\begingroup$ This is one of the more interesting questions I've seen on this site. $\endgroup$ – Fred Feb 11 '15 at 1:19
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This is a rather complex question as there are multiple variables and many proposed designs for thorium fuel cycles to consider as well; but it seems your primary interest is in whether or not leaving the Pa-233 in solution will adversely affect the thorium nuclear cycle to the point where it makes more sense to remove this isotope to be reintroduced after it beta decays to our much needed U-233 fuel.

To answer this question briefly let's first assume a thermal reactor (as in the neutrons are well moderated and have energies ideal for fission of U-233). Next let's make an assumption about composition with 98% Th-232, 1% Pa-233 and 1% U-233.

The cross sections of each of these isotopes (how 'big' they are to a thermal neutron) are approximately: Th-232, 7.37 barns for absorption; Pa-233, 40 barns for absorption; U-233, 529 barns for fission. If you do not know what a 'barn' is, basically it is nothing more than describing the 2D size of the target nuclei as far as having an interaction with the incoming neutron. 1 barn = 10-24 cm2 and was named such because on atomic scales, as the old saying goes, "...is as big as a barn."

This information can be used to derive the average distance a neutron will travel before it has a 'collision/interaction' with one of these atoms (also known as the transport mean free path). The function is as follows:

$$l=\frac{1}{\sigma N-\frac{2}{3A}}$$

Where:

Since they are all very similar in number of protons and neutrons we can eliminate the $\frac{2}{3A}$ term. Also, this function is primarily used for scattering and calculating energy loss of a neutron through a given depth of material but it works just as well for absorption leaving us with:

$$l \sim \frac{1}{\sigma N}$$

This formula gives the average (ish) distance a neutron will travel through a material before having an interaction with an atom (absorption, fission, scattering, etc.).

With some quick number crunching (skipping the exact number densities and going with the % of compositions) we can easily see the average distance traveled by the neutron is over an order of magnitude shorter for the U-233 and Th-232 vs. the Pa-233 isotope so its effects on the 'efficiency' of this reactor would be negligible.

To answer your questions:

  • Does Pa-233 formation affect the reactor efficiency? Yes.
  • Is it critical to remove Pa-233 to have a viable thorium fuel cycle? No.
  • Does the geometry of the reactor effect efficiency? Yes, but that is a whole other question. ;)

Hope this helps!

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  • $\begingroup$ Please double-check the correctness of your equations after having been formatted. I'm not sure what property/quantity "# of these isotopes" referred to, so used a generic N as the symbol. $\endgroup$ – Air Jan 4 '16 at 17:05
  • $\begingroup$ Nicely edited Air. The, "# of these isotopes" is a reference to atomic number density which happens to use the capital 'N' so well done on all counts! My only misgiving is it is apparent I need to work on my Latex skills... $\endgroup$ – eatscrayons Jan 4 '16 at 21:47
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The protactinium separation is a nice benefit of a liquid fluoride thorium reactors, made possible by the fact that the fuel (and the protactinium) are in liquid form. That it easy to pump around and do chemistry-stuff with.

The Shippingport reactor was a solid fueled (thorium oxide) reactor with water as a coolant and moderator. So the protactinium would have been stuck in the fuel elements.

Other fuel cycles (e.g. U-235) generate reactor poisons as well. These actually render solid fuel elements useless before all of the fuel has been consumed. It is possible to melt the fuel down and recover the useful fissionable material. This process has not enjoyed the level of adoption that it might otherwise due to politics, bureaucracy etc. Often, spent fuel is simply disposed of without reprocessing.

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    $\begingroup$ The OP asked specifically about the quantitative effect on efficiency, but you don't seem to have covered that at all. Are there are any real-world implementations of the technique in your first paragraph, that you could use to illustrate your answer? $\endgroup$ – EnergyNumbers Feb 11 '15 at 7:12
  • $\begingroup$ One of the more recent articles on rapid Protactinium extraction: dx.doi.org/10.1080/19443994.2012.664263 $\endgroup$ – Deer Hunter Feb 11 '15 at 8:10
  • $\begingroup$ @EnergyNumbers, I'd say you're quite right. I'll keep looking but the books that I have on hand only talk about poison the the U-235 cycle (mostly by Xe). As it stands my answer is pretty weak. $\endgroup$ – Dan Feb 12 '15 at 2:01
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The answer (believe it or not) is no. Why? Because Protactinium is now being discussed as a "additive" to nuclear reactors to improve fuel burn. The cost of removing Protactinium is not necessary at all.

$$\begin{gather} 238\text{Th} + 1\text{n}= 233\text{Pa} = 233\text{U} \\ 233\text{Pa}+1\text{n}=\text{U}234+1\text{n}=\text{U}235 \end{gather}$$

Both are fissionable. So, short answer is no.

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