On Wed, Oct 8, 2014 at 7:21 PM, <mix...@bigpond.com> 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
