Thanks Peter. Lets revise the math. We still have approx. 1000J out, and lets presume 204 eV for each conversion of H to 1/4 hydrino state. 204 eV is 3.3E-17 joule, so 1000 J is 3.1E19 transitions. 3.1E19 / 6.02E23 = 5.1E-5 mole. One mole of H is 1 gram, so the number of H atoms converted is 51 micrograms.
Since the popper was started with 10 micoliters of water, this is 10 mg of water, there is 1.11 milligrams of H present. This would mean that 4.6% of the H atoms were converted to 1/4 hydrino state. This looks like a more reasonable percentage. As for the conversion of this back into electricity, with only a 2X gain to thermal, he would need better than 50% conversion efficiency to get to a self-sustaining mode - not counting the power drain of the supporting hardware (cryo-refrigerator, vacuum pumps, gear driving motors, cooling system, and catalyst recycler). Based on the fluid dynamic argument I previously posted, and presuming he does correct the direction of the B field, I don't think he will see more than a few percent conversion efficiency with the present design. Poor MHD fluid dynamics and poor conversion efficiency could be overlooked if the thermal gain were 10 or more; but at a gain of 2, he will be hard pressed to ever become self-sustaining. Maybe the thermal energy is under-estimated. He has these huge copper conductors that carry current to the reaction site. How he accounts for the thermal loss through all of this copper can make a big difference in the computed thermal energy. I would like to see how he has accounted for the thermal loads in his calorimetry. I would also like to see these pops monitored for gamma. However, I understand why Mills would not like to discover any gamma present - it could invalidate some of what he is describing as the heat source in the device, and potentially invalidate some patents. Still, if the calorimetry is remotely correct, it does appear to be an "over-unity" device. Let's not lose sight of that. Bob On Tue, Feb 4, 2014 at 11:15 AM, <[email protected]> wrote: > Hello Bob > > In the experiment the amount of fuel was according to R.Mills : 10 > microliter. > The amount of energy liberated from the transition of H to H1/4 is 204 eV > atom > > Peter > > > *From:* Bob Higgins <[email protected]> > *Sent:* Tuesday, February 04, 2014 5:04 AM > *To:* [email protected] > *Cc:* Bob Higgins <[email protected]> > *Subject:* Re: [Vo]:BLP video is out > > It is interesting to do a little math around this experiment. Presume > that the popper is operating with a fuel of 1 microliter of water and > produces a net excess energy of 1000 joules. Presume Mills to be correct > in assigning most of the reaction is conversion to 1/4 hydrino state that > is liberating ~54 eV per atom (8.65E-18 J). Then to get 1000 J of excess > heat, would require the transition to 1/4 hydrino state of 1.16E20 H > atoms. A mole is 6.02E23, so getting the 1000 J would take (1.16E20 H > atoms)/(6.02E23 atoms per mole) = 1.9E-4 mole. For H, one mole is 1 gram, > so getting 1000 J consumed 190 micrograms of H. > > In H2O, Hydrogen is 2/18 or 1/9 of the molecule. So, 1 microliter of > water is 1 mg of water and has 1/9 of 1 mg of H = .111 mg of H = 111 > micrograms of H. > > Hmmm. So 190 micrograms of H was converted to 1/4 Hydrino state, but only > 111 micrograms of H was present to start. So, if the reaction was 100% > efficient, it would require almost 2 microliters of water to begin. Or, > the H atoms would have to be sent to a smaller fraction hydrino state > liberating more energy per atom of H converted. > > This seems too efficient in conversion of H to hydrino, or much more > energy is being liberated per atom of H. > > Is my math correct? > > Bob > > > On Mon, Feb 3, 2014 at 2:14 PM, Nigel Dyer <[email protected]> wrote: > >> >> http://www.blacklightpower.com/whats-new/ >> >> >
