• Corpus ID: 3175953

Cold Fusion. Andrea Rossi’s Method-bis

@inproceedings{Bettini2011ColdFA,
  title={Cold Fusion. Andrea Rossi’s Method-bis},
  author={Giuliano Bettini},
  year={2011}
}

• G. Bettini
• Published 2011
• Physics

The interest on Andrea Rossi's Nickel-Hydrogen Cold Fusion technology is accelerating. In the present article I would like to answer a question p sed by L. Kowalsky in Vixra: how can 30% of nickel in Rossi’s reactor be transmuted into copper ? "Everything should be made as simple as possible, but not simpler", says a guy. I apologize s if I am too simplistic here. ________________ * Retired. Earlier: Selenia SpA, Rome and IDS SpA, Pisa. Also Adjunct Professor at the University of Pisa, Adjunct Professor at Naval Academy, Leghor n (Italian Navy). E-mail: [e-mail] Introduction The interest on Andrea Rossi's Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was tu rned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that “t his seems to be impossible because the produced copper isotopes rapidly decay into Ni”. But how it works? How it works Following Focardi Rossi [3]. a Ni58 nucleus produce s a Copper nucleus according to the reaction Ni58 + p → Cu59 Copper nucleus Cu59 decays with positron (e+) and n eutrino (ν) emission in Ni59 nucleus according to Cu59 → Ni59 + ν + e+ Then (e+) annichilates with (e-) in two gamma-rays e+ e+ → γ + γ Starting [3] from Ni58 which is the more abundant i sotope, we can obtain as described in the two above processes Copper formation and its successive decay in Nickel, producing Ni59, Ni60, Ni61 and Ni62. Because Cu63, which can be formed startin g by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each pro cess means some MeV. Of course how can a proton p gets captured by the N i58 nucleus? (and subsequent Ni59, Ni60, Ni61 and Ni62). Following Stremmenos [4]. a neutron-like particle, an electron proton pair, a mini-atom, a proton masked as a neutron, gets captured by the Ni58. If the masked proton becomes a neutron the result i s Ni59. In order to have Cu59 (increase of atomic number fr om 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked p roton becomes a proton. The same process holds for all the subsequent trans fo mations, until Cu63. It remains to be understood the issue of the gamma radi tion in the MeV range. Numbers I am an electronic engineer, so I need easy numbers in order to understand. However "Everything should be made as simple as pos ible, but not simpler", says a guy. Maybe I am too simple here. Let’s calculate. MeV for each Ni transformation I read that starting from Ni58 we can obtain Copper formation and its successive d ecay in Nickel, producing Ni59, Ni60, and Ni62. The chain stops at Cu63 stable. For simplicity I assume all the Nickel in the react or in the form Ni58. For simplicity I suppose for each Ni58 the whole sequence of events from Ni58 to Cu63 and as a rough estimate I calculate the mass defect between (Ni58 plus 5 nucleons) and the final state Cu63. Ni58 mass is calculated to be 57.95380± 15 amu The actual mass of a copperCu63 nucleus is 62.91367 amu Mass of Ni58 plus 5 nucleons is 57.95380+5=62.95380 amu Mass defect is 62.95380-62.91367=0.04013 amu 1 amu = 931 MeV is used as a standard conversion 0.04013x931 MeV=37.36 MeV So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy. Nickel consumption According to many blogs in the Internet “One hundre d grams of nickel powder can power a 10 kW unit for a minimum of six months”. How much of Ni58 should be transformed, in six months of continuous peration, in order to generate 10 kW? I follow a procedure outlined in [2]. 10 kW is thermal or electrical (?) power. The nucle ar power must be larger. Assume a nuclear power twice: 20 kW = 20,000 J/s = 1.25 x 10**17 MeV/s. Each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy. The number of Ni58 transformations should thus be equal to (1.25 x 10 **17)/37.36 = 3.346 x 10**15 per second. Multiplying by the number of seconds in six months (1.55 x 10**7) the total number of transformed Ni58 nuclei is 5.186 x 10**22. This means 5 grams. The order of magnitude is not exactly the same but seems to be plausible. This means also 5 grams of Nickel in Rossi’s reactor transmuted into (stabl e) Copper after six months of continuous operation at the rate of 10 kW. Conclusions Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. At first glance this seems to agree with calculations based on simple assumptions. References [1] Link, Cold Fusion "Andrea Rossi" Method [2] L. Kowalski, “Rossi's Reactors Reality or Ficti on?”, http://vixra.org/abs/1103.0080 , March 2011 [3] A.Rossi, S. Focardi, http://www.journal-of-nuclear-physics.com [4] E. Stremmenos, “Hydrogen/Nickel cold fusion pro bable mechanism” , www.journal-of-nuclearphysics.com/?p=338&cpage , March 2011