I don't think that the 2mg are a problem , that is more the number of grains analysed.
it is like poling. the size of the sample have to be bigger if there is tiny minorities. few samples are ok to measure the majority. the problem like on polling is more about bias on the sampling, like taking more tiny or big dust, the one from one part or another from the tube. a thousand grains well mixed from the dust should be enough to be representative. we could do some compensation based on morphology like on poling. on a thousand grains, make statistics on morphologies, and analyse grains depending on morphologies, and then make a weighted average. I'm not an expert in polling, but this seems to apply well here. 2015-01-03 8:36 GMT+01:00 Eric Walker <[email protected]>: > On Wed, Oct 8, 2014 at 7:21 PM, <[email protected]> wrote: > > In reply to Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700: >> Hi, >> [snip] >> >> Li7 + Ni58 => Ni59 + Li6 + 1.75 MeV >> Li7 + Ni59 => Ni60 + Li6 + 4.14 MeV >> Li7 + Ni60 => Ni61 + Li6 + 0.57 MeV >> Li7 + Ni61 => Ni62 + Li6 + 3.34 MeV >> Li7 + Ni62 => Ni63 + Li6 - 0.41 MeV (Endothermic!) >> >> This series stops at Ni62, hence all isotopes of Ni less than 62 are >> depleted >> and Ni62 is strongly enriched. >> > > The authors of the Lugano report mention a total energy balance of 1.5 MWh > excess heat to be accounted for (p. 29). Translating that value, we > get 3.3e22 MeV. If the average reaction is 3.5 MeV (just to choose an > optimistic number), that means there were 9.4e21 reactions, and presumably > 9.4e21 7Li atoms to be consumed in the process. > > The authors mention that in the sample of the fuel they looked at, there > was 1.17 percent lithium (p. 53). If we extrapolate out from the 2 mg > sample they obtained to the 1 g of fuel from which it was taken (not > necessarily wise), there would have been 0.0117 g * 1 mole / 6.94 g * > 6.022e23 / mole = 1.0e21 atoms lithium in the total charge. If we assume > that that was 100 percent 7Li to be optimistic, that would mean there were > about 1/10th the number of 7Li atoms needed to account for the 1.5 MWh that > were produced. > > Judging from the fact that these calculations go back to the isotope > ratios found in a single 2 mg sample of fuel, there's a lot of room for > uncertainty. But in this instance we've been optimistic about the average > energy per reaction (3.5 MeV), about there being 100 percent lithium, and > about all of the 7Li being consumed. The actual heat balance is another > variable that can be adjusted to within one's sense of uncertainty. But it > would have to be pretty far off for the reaction to consist entirely of 7Li > neutron stripping reactions. > > Have I missed something important? > > Eric > >
