In reply to David Roberson's message of Thu, 3 Sep 2015 11:04:18 -0400: Hi, [snip] >In order to operate it uses a mixture of deuterium and tritium fuel. I just >remember reading that the reaction process results in the regeneration of >additional tritium fuel but that the process is barely able to replace the >original quantity of fuel. Perhaps someone who understands how that fuel is >regenerated can set me straight concerning the excess production of tritium >you mention. > >When the Takamak reaction takes place, how many neutrons are released per >event and do you need to capture most of those in order to produce new >tritium? That is what I read about which suggested that this was not going to >be such an easy task. Perhaps a technique is now in place which offers >overkill for this requirement? > >Dave
The primary reaction is D + T => He4 + n (14 MeV) If I'm not mistaken the secondary reaction is Li6 + n (slow) => T + He4 (both fast) So you would need to capture every single neutron, which will be difficult, because they are very fast neutrons, with considerable penetrating power. However there are likely to be some other reactions that make the task a little easier, such as D + n (fast) => H + 2n (slower) Li7 + n (fast) => Li6 + 2n (slower) and He4 (fast) + D => H + n (slow) + He4 (slow) These three are neutron multiplying reactions and n(fast) + Li7 => T + He4 + n (slower) and n (slower) + Li7 => T + He4 + n (even slower) etc. until the neutron is either captured, escapes, or has insufficient energy. This reaction creates more T without consuming the neutron (just it's energy). Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
