How dos the He3 catalysed DD cycle work?
As I understand it.
Reactor parameters are chosen so the Fusion Reaction Cross-Section is much bigger for D+He3 than for D+D, so al He3 is fast consumed. No unnecessary D+D reactions.
But if so, there produce same amount of T as He3.
From this you have D+T reactions and the nasty 14Mev neutron or T as waste product.
Hope I got something wrong.
He3 catalysed DD cycle?
A little more than half way down this page is a list of fusion reactions with large cross section. It gets complicated fast if you really want to include all the probable reactions, and it will depend on the over all energy distribution. Most all computations assume thermal distributions, and the Polywell isn't thermal (although the center may be close enough). I think we can deal with a few neutrons, it's a major problem if the majority of reactions produce one.
From what I read at that place and from this http://en.wikipedia.org/wiki/Fusion_power
I can not see you can reduce the tritium production. You get more energy out of the fuel but not small part ass 14Mew neutrons.
I can not see you can reduce the tritium production. You get more energy out of the fuel but not small part ass 14Mew neutrons.
There are tricks that nuclear scientists/engeneers can use to increase the deuterium H3 reaction rate over deuterium side reactions. As shown in the referance by DrMike at 100,000 electron volts the D-H3 reaction cross section is 7-8 times that of D-D. And if if excess (10-20?, 10,000?) times more H3 is added than D then the chances of D-H3 collisions are increased over D-D. There would presumably be loss of efficiency due to increased nonproductive H3-H3 collisions, but so long as you have adequete 'Q' you can play with the ratios (and I'm sure other tricks) to meet some neutron tolarance goal. I beleive I have seen on the Wikipedia reference to aneutronic fusion that D-H3 is technically considered aneutronic as less than 1% of the energy is given off as neutrons- presumably with the above and other 'tricks'.
Dan Tibbets
Dan Tibbets
To error is human... and I'm very human.
Because the energy out per reaction is 1/10th that of a fission reaction, aneutronic only means that your neutron production is in the same range as a fission plant.
If you get a neutron production of 1E-4 per reaction or better you are getting somewhere. No shielding (distance, mass) requires something on the order of 1E-12 or better.
Sadly we do not know how to do that. Yet.
If you get a neutron production of 1E-4 per reaction or better you are getting somewhere. No shielding (distance, mass) requires something on the order of 1E-12 or better.
Sadly we do not know how to do that. Yet.
Engineering is the art of making what you want from what you can get at a profit.