There are some ifs and buts associated with this subject. It has been known for over a hundred years how that hydrogen will defuse through a hot metal enclosure.
The rate of diffusion is subject to the temperature and pressure of the hydrogen, together with the exact kind, thickness, and temperature of the metal. These are all variables in the calculation of the diffusion rate. Furthermore, the presence of oxides and/or carbides on the surface of the metal can reduce the rate of diffusion of hydrogen by up to 5 orders of magnitude. We don’t know for sure what the accurate values of some of these variables are and additionally they would vary widely within an operational range throughout the operational lifetime of the E-Cat. However, since hydrogen is very slippery and notoriously hard to contain, a good guess can be made that most of the hydrogen consumed by the Rossi reactor would be lost through diffusion through the hot walls of the stainless steel reaction vessel. Because of all these large uncertainties, calculation of the nuclear reaction rates as a function of hydrogen consumption implying a clue to the nuclear processes going on inside the E-Cat reaction vessel cannot be made in my opinion. With best regards, Axil On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner <hheff...@mtaonline.net>wrote: > From: > > http://www.rossilivecat.com/ > > Quote: > - - - - - - - - - - - - - - - - - - - - > Andrea Rossi > October 25th, 2011 at 4:59 PM > Dear Thomas Blakeslee: > Grams/Power for a 180 days charge > Hydrogen: 18000 g > Nickel: 10000 g > Warm Regards, > A.R. > - - - - - - - - - - - - - - - - - - - - > End quote. > > At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an atoommic > weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This means 10.48 atoms of > H need be provided per 1 atom of Ni. > > Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms of H is > consumed, thus 170.4 mols of H and a170.4 mols of Ni are consumed, maximum. > This involves the obviously wrong assumption that all the Ni atoms are > transmuted, not a more realistic 3 percent. There is also an outside > possibility the H reacts with daughter products, giving the possibility of > 10 subsequent daughter reactions per primary Ni+H reaction. Three such > reactions is an outside possibility. > > One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is (6.241x10^24 > eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms /mol) = 9.464x10^5 > eV/(Ni atom). If there is one reaction per atom and all Ni is consumed by > single reactions than that is 0.9464 MeV per Ni-H event. The gammas from > this would be lethal at short range, even through 2 cm of lead. If it is > assumed that 3% of the Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 > MeV per reaction. If there are an average of 3 daughter reactions per > primary reactions that is about 10 Mev per reaction. > > If 10 MeV gammas are produced then 5 cm of lead shielding will be of no use > in protecting the operators. If near 1 MeV gammas are produced the lead > shielding is inadequate. > > One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas, 6.24x10^18 > gammas per second. using: > > I = I0 * exp(-mu * rho * L) > > where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/cm^3, we > have for 5 cm of lead: > > > I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5 cm)) > > I = 2x10^18 free gammas per second. > > About half that, or 10^18 gammas/s would be directed toward the interior of > the container housing the E-cats, and most of the 2x10^18 gammas per second > would end up escaping the container. This is an approximate calculation. > Even if it is off by an order of magnitude, this kind of 1 MeV gamma flux, > even 1/32 of it from one E-cat, would be readily detected by a geiger > counter at significant range. > > It does not seem credible the energy from a Ni-H reaction, at least in the > form of one gamma per reaction, provides any explanation for 1 MW of heat, > if that thermal power is in fact achieved. > > Best regards, > > Horace Heffner > http://www.mtaonline.net/~**hheffner/<http://www.mtaonline.net/~hheffner/> > > > > >
