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RE: starship-design: Staged Fusion Power



Timothy,



> The first stage D+D->He3+n+3.27MeV does not really add a lot of energy
> relative to 18.3MeV and as you mention it has a neutron as reaction
product
> which may disturb the second stage so much that the second stage reaction
> isn't as efficient anymore. (Maybe even more than 3.27MeV less efficient.)

Probably much less.

> I wonder why our tables don't show p+D->He3+5.5MeV it looks like an
> excellent candidate to be followed by a second stage (as do all single
> end-product reactions). If this is a valid reaction, it may be followed by
> my D+He3->He4+p+18.4MeV that would give a yield of 4.78MeV per nucleon.
> (Fusing upto Fe54 starting with H2 and H3 will give a yield of 7.43MeV per
> nucleon.)
> But again I'm not sure if a second stage is as "easy" as we hope, it still
> is very far away from the current fusion technology.


>>Too bad we can't use the gamma rays produced in the D+D reaction though...

> What would you use them for?

Well, I just hate to see 20 MeV go to waste!

>>As you said, your notation is a little different but it is understandable.
>>Where did you get the 18.4 MeV though?

> D+He3 -> He4 (3.6 MeV) + p (14.7 MeV)
> 3.6+14.7=18.3MeV  My 18.4MeV is actually 18.35MeV, so the difference is
> likely due to rounding errors.

Umm, I don't think the 3.6 MeV is usable, that is potential energy bound up
in the He4.

> p+B11 would be better since it doesn't have D+D reactions as D+Li6 has.
And
> a bonus is that both p and B11 are abundant in nature, while D and Li6 are
> harder to find. The only disadvantage is that it only yields 8.7MeV while
> its reaction  materials consist of already 12 nuclei in total. (0.725MeV
> per nucleon)
> I wonder if a second stage reaction could significantly increase this.

Li6 is common as mud on the back side of the moon. The D+D reactions are
very dependent upon temperature. If the D+Li6 reaction is hot enough there
are virtually no D+D reactions and hence very few neutrons.

Lee