The problem is that deuterium is already .00016% of natural hydrogen, you would need to separate out the created deuterium 'signal' from natural deuterium in an experiment.
Interesting that the energies released in H-H fusion are relatively tiny - 0.42MeV for the fusion with 2 x 0.51 MeV gammas from the released positron annihilation = 1.44 MeV total (about 19145 kWh/g H2). No neutrons, no hard gammas. That would align with lack of observed radiation from NiH and low 0.4-0.5 MeV gamma signals I think were reported from one of Rossi or Piantelli's experiments. http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction Given about .04grams of hydrogen in a 50mL e-cat operating at 20bar and 300°C you would expect to see a doubling of the deuterium ratio (ie 0.0000065g extra Deuterium) with just 0.12kWh produced. Perhaps 10-100x in a typical 10(ish) kWh per ecat Rossi demo such as we have observed this year. It would be pretty straight-forward to check for anyone who had access to even a black box test (just sample the hydrogen released). It might also provide a reason why Rossi has not run his demo's for longer - the Hydrogen gets polluted by Deuterium and needs replacing, something he cannot do if the hydrogen bottle is disconnected from the units unless he has a large reservoir of hydrogen inside or attached to the Ecats. It would also explain why he states such a high rate of hydrogen consumption (about 1gram per e-cat per day) - simply to retain relative hydrogen purity. Those 511keV gammas have a half stopping thickness in lead of 4.1mm, 34mm for concrete, so that is still a lot of shielding required to cut down to safe/background levels (25mm would catch about 90%) if it were a straight hot fusion type process, but we know from numerous Pd-D systems that nasty gammas are not being released in appreciable numbers, so there must indeed be some sort of lattice mechanism to capture or limit gamma release. On 25 November 2011 15:57, Edmund Storms <stor...@ix.netcom.com> wrote: > Good thinking, Robert. I agree, deuterium is the natural nuclear product > when H is used. In addition, deuterium is easy to detect because it would > produce the DH molecule that has mass 3. Mass 3 has a very small peak in a > mass spectrum. > > Ed > > On Nov 25, 2011, at 8:50 AM, Robert Lynn wrote: > > Most interesting is the clear evidence for D-D fusion to make He4 in >> Pd systems, none of this Nickel transmutation business. >> >> Does that mean that it is more likely to be H-H fusion making >> deuterium in Ni-H systems? The deuterium would be very hard to detect >> as a reaction product. >> >> On 25 November 2011 02:44, Aussie Guy E-Cat <aussieguy.e...@gmail.com> >> wrote: >> >>> The Q & A session at the end is interesting. >>> >>> AG >>> >>> >>> On 11/25/2011 1:07 PM, Terry Blanton wrote: >>> >>>> >>>> On Thu, Nov 24, 2011 at 8:19 PM, Jouni Valkonen<jounivalkonen@gmail.** >>>> com <jounivalko...@gmail.com>> >>>> wrote: >>>> >>>> >>>> Quite nice lecture. Here are each video segments with direct links and >>>>> titles: >>>>> >>>> >>>> If you start with the first it will automatically play all 8 using >>>> autoplay. >>>> >>>> T >>>> >>>> >>>> >>> >>> >> >
