From: Bob Cook * It seems to me that there must be a separate cooling mechanism happening to remove or redistribute the heat within the reactor.
Well, if you are assuming that the net gain derives from 1/10th gram of hydride fuel, which starts off as a powder, then it is possible that the powder is first liquefied - and in a few hours, before the H2 has dissipated, is trapped on the inner surface, due to a liquid film layer deposited by the vapor pressure of Li-Al, and this alloy would be uniformly coating the wall of the reactor. The coating would “redistribute the heat”, no? It would also provide a perfect surface layer for SPP formation on the interface with the alumina. On another forum, it was suggested that the number (6.7e21) of lithium and hydrogen atoms, if giving up only 4 eV per atom which is the chemical maximum - would be 4.2 kilojoules, while the observed excess heat during 8 min “heat after death” was about 380 kilojoules - and during the whole experiment much more. This is one of several details indicating that the reaction could also be related to fractional hydrogen f/H, dark matter, or Mills’ hydrino, which would release the amount of energy which is seen. This level of energy is much less than nuclear but much more than chemical. In the famous Thermacore paper, for instance - the excess energy is shown to be ~54.4 eV per active proton, which is similar to what is seen here. A longer run is possible since the f/H is not depleted in the first few levels of redundancy. In fact, if the ash in the Russian experiment does not turn up the isotopes which Rossi claims, and it probably will not IMHOI, then the underlying reaction could be related to hydrogen “shrinkage” (ground state redundancy) with the energy coming from electron angular momentum instead of the nucleus. Jones
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