Robin,
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.
Actually it is close to unity. The (n,2n) reaction is extremely
rare with most light elements - 7Li and Be - being the ONLY
exceptions. Fast neutrons simply do not breed many neutrons with
either heavy water or 6Li.
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.
Yes It is way off. Fast neutrons are very funny - and you simply
cannot base the situation on what thermal neturons do. Fast
neutrons only react certain nuclei. This has all been worked out
by careful experiment, and has been known for decades.
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.
There are always too few without a true multiplier - which is the
(n,2n) reaction. It just doesn't happen with light lithium or
heavy water. Sorry. You need to get hold of old issues of "Fusion
Technology" for the confirmation of this, but I am pretty certain
of the details
Since nearly all of these neutrons are going to react with Li6
to
create T, this would appear to actually be a T breeder.
This is based on an incorrect assumtpion. Fast neutrons do not
multiply with light lithium or heavy water, so there is always
less than one thermal for every incident 14 MeV high energy one
(since some few will go though a meter of heavy water without
interacting - even the scattering cross section for fast neturons
is low).
(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).
Yes. Most thermal neutrons will react with the 6Li - the problem
is that fewer are multiplied than the ones which escape, so the
result is too little tritium to sustain.
Taking into account the fact that there are going to be about
2.7
of these reactions for every fusion reaction,
That is not a fact. In fact it is a considerably large error,
because you are assuming (n,2n) multiplication in a situation
where there is none.
Jones
