In reply to Jones Beene's message of Sun, 12 Mar 2006 08:26:54 -0800: Hi, [snip] >CUT TO THE CHASE: > >The BEST (by far) method of getting the needed tritium is to breed >most of it from a cheaper reactor thermal blanket, as Robin/Horace >suggest using heavy water and natural lithium hydroxide - this >gives you about 75% of the tritium you need to self-sustain. This >was what we were planning on doing in the 1960s, until it was >discovered that it came up short. [snip] Using only Li6 (in heavy water as Horace suggested), and running a few numbers I get a neutron breeding ratio of about 2.7 (from the deuterium), i.e. for every incident fast neutron about 1.7 extra slow neutrons. This is probably a bit off because it assumes that the ratio of the fission and elastic scattering cross sections is constant for any incident neutron energy above 2.2 MeV. Nevertheless it takes no account of extra neutrons flowing from secondary collisions, of energetic deuterium nuclei, which in turn result from elastic collisions with energetic neutrons. So it may not be too far off.
Since nearly all of these neutrons are going to react with Li6 to create T, this would appear to actually be a T breeder. (The reaction cross section for the reaction n + Li6 -> He4 + T + 4.78 MeV is about 940 b for thermal neutrons, which totally swamps the cross section for any other potential reaction). Taking into account the fact that there are going to be about 2.7 of these reactions for every fusion reaction, the net energy from this reaction increases to 12.9 MeV / fusion reaction, which is about 73% of the fusion energy. IOW the breeding step nearly doubles the overall energy output. Regards, Robin van Spaandonk http://users.bigpond.net.au/rvanspaa/ Competition provides the motivation, Cooperation provides the means.
