From: Arnaud Kodeck ➢ You [Jed} are assuming that D + D gives He4. In the Mizuno reactor, we still don’t know exactly what is the reaction taking place there. It could be Ni + D -> Cu or Pd + D -> Ag. Let’s hope that that the Pd is not consumed in the Mizuno reactor otherwise all you plans in the cost for fuels felt apart.
This is perceptive. Mills has done so much work with nickel and at the same time - using the low pressure hydrogen regime - that we can be almost certain that there is NO nuclear fusion going on with the nickel. That narrows the possibilities considerably. If palladium is being consumed then the economics are much less favorable – even when the correct price is used… <g> The best of all worlds and actually the most likely scenario is that palladium is not consumed or consumed very slowly. Deuterium could be involved in some kind of BEC reaction, very much like the scenario of Miley and Hora involving a Coulomb explosion, or else Holmlid’s muons. They have the photon signature for this – and If they are correct, the deuterium is hardly consumed (possibly less than in fusion). It would be easy for Mizuno to look for this signature. Jones
