[KeithPickering]

dennisp wrote: Being that focus fusion is already a rocket, I’m wondering: what is the exhaust velocity?

Beyond that, has anyone figured specific impulse and thrust? Would it make sense to simply mix additional reaction mass into the exhaust stream for a higher-thrust application?

I compute it as follows:
1. The energy released in a single reaction (p+11B -> 3He) is 8.7 MeV, or 1.3939e-12 Joules.
2. I will assume here that ALL of this energy is kinetic energy carried away by the He ions. In actual fact, they tend to run into electrons in the plasma (which slows them down) and that energy is emitted by the electrons as x-rays. But let’s assume best-case scenario for now.
3. The mass of a 4He nucleus (2 protons, 2 neutrons) is 6.7e-27 kg, and three of those have a combined mass of 2.01e-26 kg.
4. Since e = 1/2 m v², we can compute v = √ (2e / m)
5. From this, the velocity of the three helium nuclei, in the best case scenario, would be 11781 km per second, about 4% of c.

Specific impulse and thrust depend on how much and how rapidly fuel is burned. I don’t believe we know the answers to those questions yet.

[KeithPickering]

HermannH wrote:

It may turn out that you can get to theoretical breakeven, but not quite to practical breakeven … or just barely over the line, which would mean that the excess salable energy would be a lot more expensive than assumed.

That is the worry I have as well. According to Eric’s simulation results in the Technical Paper 1 the maximum reasonably achievable ratio of (Xray + Beam) / Input is 1.57.

Thanks for that link, and the very interesting technical paper. Apparently EL’s zero-dimensional simulation has addressed this problem pretty well, and assumes that energy recovery is 80% for both ion beam and for x-ray. As the input energy increases, the recoverable energy also increases, but the ratio of x-ray to ion beam also increases … so more and more of the fraction must be recovered by the x-ray capture shell. Since EL invented it himself, we must assume that his 80% figure is roughly correct.

A couple of final results then: at a ratio of 1.57 (energy recovered to energy input), the salable output of a reactor using 5 kg of borax per year (see my previous post for computations) would be 1111 kW. To convert ALL electric production worldwide (2.5 million megawatts) to pB11 fuel would require 12,500 tonnes or borax per year, or about 8% of current worldwide production.

[KeithPickering]

Lerner wrote: A 5 MW reactor takes about 5 kg of fuel per year. For an experiment, we purchased decaborane for $5 per gram or $5,000/kg. At that price fuel costs would be $25,000/yr or 0.05 cents/kWh. Electricity now costs about 5 cents/kWh.

Isn’t this rather optimistic in terms of the efficiency of the energy capture and net energy gain?
The good news is, the current price of borax is $900 per short ton, or $1 per kg. But what’s the real net energy we can get from that?

Here’s my computation: Each pB11 reaction nets 8.7 MeV or 1.4e-12 J
Borax (B2O3) has a gram molecular weight of 69.619, meaning 7.18 mol in 5 kg.
So 5 kg of borax contains 8.65e25 boron atoms, of which 6.92e25 are B11 atoms.
Fusing all of those B11 atoms would release 9.64e7 MJ or 2.68e7 kWh.
Since there are 8760 hours in a year, running at a 5kg per year rate would produce at total of 3.06 MW.

But that’s assuming that 100% of the energy produced is captured, and that the device itself uses no energy.

It seems to me that the key economic question is not, Can we achieve net energy gain?, but rather, How Much net energy gain can we achieve? Because first, the device needs to PRODUCE more energy than it uses; second, the device needs to CAPTURE more energy than it uses; and third, only that excess fraction is actually usable and saleable.

Now I can see how the electrons and ions produced might be captured with nearly 100% efficiency (although this too needs to be demonstrated). But 40% of the energy produced is in the form of x-rays, which are supposed to be captured by the layered foil shell. How efficient is that shell? If it’s only 20% efficient (which is comparable to solar PV cells), the device itself will only capture 68% of produced energy. And that would mean that the effective breakeven point is 47% higher than the theoretical breakeven point.

It may turn out that you can get to theoretical breakeven, but not quite to practical breakeven … or just barely over the line, which would mean that the excess salable energy would be a lot more expensive than assumed.

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