They plan to use 2D + 3He -> 4He + 1H, so no neutrons https://en.wikipedia.org/wiki/Aneutronic_fusion#Candidate_re...

(I'm still not convinced of their explanations, but a fast proton may be easy to catch by the magnetic field and create the effect they want.)

Nitpick: You’re always going to get some parasitic D+D reactions because it’s reaction cross section is appreciably higher than D + 3He below 50 keV, and non-negligible even after the crossover point.

Doesn't D+D just yield 4He (and no neutrons) though?

It yields 3He plus a 2.45 MeV neutron at least part of the time.

See the diagram here: https://undsci.berkeley.edu/teach-resources/products-of-deut...

You could always add the liquid metal blanket if you want to eke out the extra 10% (or if you want to generate tritium). But it's not worth the complication in an early prototype.

Helion's fuel mix produces just 6% of its energy as neutron radiation, and if you harvest it you'll lose a third of that. As long as you have enough net energy, collecting that 4% might not make financial sense.

With fuel costs insignificant, your cost per kWh is mainly capital cost. Let's say it's all capital just to keep it simple. I don't know how much the input energy will be but if your choice is between, say, generating net energy of 50MW without a turbine or 54MW with a turbine, then you would skip the turbine if it adds more than 8% to the capital cost. I suspect Helion has done this calculation in detail.

It does seem to hinge on the cost of fuel, I have some doubts about whether they can secure a fuel supply so cheap as to skip out on that extra 4 MW, even after factoring in the cost of a small steam turbine installation.

Deuterium costs several thousand dollars/kilogram. But even though it is just one part in several thousand of the hydrogen in water, there's enough deuterium in your morning shower to provide all your energy needs for a year.[1] Cost of deuterium is definitely insignificant.

Helion's other fuel is helium-3 which they'll make themselves by fusing deuterium. So the helium-3 cost will directly depend on the capital cost of the reactor producing it.

[1] https://dothemath.ucsd.edu/2012/01/nuclear-fusion/