In reply to Mike Carrell's message of Fri, 24 Oct 2008 16:54:12 -0400: Hi, [snip] >To: >Robin van Spaandonk >Jones Beene >Ed Storms >Scott Little >[and lurkers] > >This has been a very useful discussion. If you have not done so, I recommend >downloading http://www.blacklightpower.com/papers/WFC102308WebS.pdf and >printing pages 10-14 and 48. Figure 7 on p48 is a scan of NaH using >Differential Scanning Clorimetry. It is most instructive. At 350 C there is >endothermic decompoisition of NaH. Beginning at 640 C is a very strong >exothermic reaction, which I think is conventionally unexpected. The NaH was >in 760 Torr He.
This is unfortunate given that He+ is also a catalyst. > >The reactions involved in the test cell are complex, and discussed on pp >10-12, equations 23-35. The next-to-bottom paragraph of p11 is specially >interesting. > >NaH apparently qulaifies as a catalyst because heating can intiate a >reaction resulting in H[1/3] which is a hydrino catalyst. That is secondary. The primary reason it qualifies as a catalyst is that the sum of the three components of the dissociation energy into the specified components adds to 54.35 eV, which is a close match for 54.4 eV. >It still is not >clear to me where the 54.35 eV for ionizing Na to catalyze H comes from. Mills has this weird way of writing his equations. Note that the Hydrino reaction itself on the right hand side of equation 23 actually produces 108.8 eV, half of which goes into the electron hole, and the other half of which is just direct free energy. Any one else would just have written eq. 23 with an excess of 54.45 eV on the right hand side, and nothing on the left. He writes it the way he does, in order to indicate that the energy release occurs in 2 phases, the first resonant energy dump into the "hole" (which in this case is 54.35 eV), and the second phase release, which is likely in the form of kinetic energy. However don't mistake the 54.35 eV on the left as external input to the reaction. It isn't. (it's just a quantity of -54.35 eV that Mills has transferred from the right hand side of the equation to the left hand side). What he should have done was: NaH -> Na++ + 2 e- -54.35 eV + H[1/3] + 108.8 eV (note that the net on the right hand side is 54.45 eV) This makes it obvious that 54.35 eV is needed to break up the molecule, while the shrinkage yields a total of 108.8 eV. After the Hydrino forming reaction is complete, there is still free Na++ in the environment, and when this reacquires its missing electrons and recombines with a free H atom, to form a new molecule of NaH, a total of 54.35 eV is released. So in total for the two reactions (23 & 24) we get 54.45 (from 23) and 54.35 (from 24) = 108.8, which is precisely the total released during Hydrino formation. To make a long story short, when the Hydrino forms, part of the energy released is stored in chemical form (Na++ etc.) and part is released directly to the environment. The part stored in chemical form is then shortly (and separately) also released to the environment as per equ. 24. [snip] Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
