Rossi's E-Cat (Cold Fusion).
In April, 2011, the Italian website Panorama has published an interview with Sergio Focardi, University of Bologna professor and colleague of Andrea Rossi who collaborated on the development of the E-Cat.
In the interview Focardi concludes by saying, “though I should not be the one to say, this is the greatest scientific discovery of all human history”.
Here follows a short the history of my intervention in Journal of Nuclear Physics.
In March, 2011, having read the basic on E-Cat in Journal of Nuclear Physics I wrote a short paper.
My intention was to publish an article on Journal of Nuclear Physics.
To do this, I read the rules for sending papers.
I read the following instructions:
___________________________________________________________
To publish your article on the Journal Of Nuclear Physics you have to mail it as an attachment in Word to:
[email protected].
Your article will be checked by an attorney to verify that the content is legal and will be peer reviewed by a physics professor. The peer reviewing will take from one to three months, depending on the complexity. If it will be accepted, it will be published for free, as all the papers of our Journal.
Warm Regards,
The Board Of Advisers
___________________________________________________________
So I sent the article,and it was then published into the blog.
In the following weeks several comments are followed.
I quote here the final edition.
Note that the version published in JONP is slightly different from this one for some corrections on the numbers. The corrections are listed there in the "Comments".
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Title
How can 30% of nickel in Rossi’s reactor be transmuted into copper?
Introduction
The interest on Andrea Rossi's Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that “this 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 produces a Copper nucleus according to the reaction
Ni58 + p → Cu59
Copper nucleus Cu59 decays with positron (e+) and neutrino (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 isotope, 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 starting by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each process means some MeV.
Of course how can a proton p gets captured by the Ni58 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.
In the interview Focardi concludes by saying, “though I should not be the one to say, this is the greatest scientific discovery of all human history”.
Here follows a short the history of my intervention in Journal of Nuclear Physics.
In March, 2011, having read the basic on E-Cat in Journal of Nuclear Physics I wrote a short paper.
My intention was to publish an article on Journal of Nuclear Physics.
To do this, I read the rules for sending papers.
I read the following instructions:
___________________________________________________________
To publish your article on the Journal Of Nuclear Physics you have to mail it as an attachment in Word to:
[email protected].
Your article will be checked by an attorney to verify that the content is legal and will be peer reviewed by a physics professor. The peer reviewing will take from one to three months, depending on the complexity. If it will be accepted, it will be published for free, as all the papers of our Journal.
Warm Regards,
The Board Of Advisers
___________________________________________________________
So I sent the article,and it was then published into the blog.
In the following weeks several comments are followed.
I quote here the final edition.
Note that the version published in JONP is slightly different from this one for some corrections on the numbers. The corrections are listed there in the "Comments".
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Title
How can 30% of nickel in Rossi’s reactor be transmuted into copper?
Introduction
The interest on Andrea Rossi's Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that “this 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 produces a Copper nucleus according to the reaction
Ni58 + p → Cu59
Copper nucleus Cu59 decays with positron (e+) and neutrino (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 isotope, 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 starting by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each process means some MeV.
Of course how can a proton p gets captured by the Ni58 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 is Ni59.
In order to have Cu59 (increase of atomic number from 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked proton becomes a proton.
The same process holds for all the subsequent transformations, until Cu63.
It remains to be understood the issue of the gamma radiation 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 possible, 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 decay in Nickel, producing Ni59, Ni60, and Ni62. The chain stops at Cu63 stable.
For simplicity I assume all the Nickel in the reactor 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 57.9353429 amu
The mass of a copper-Cu63 is 62.9295975 amu
Mass of Ni58 plus 5 nucleons is 57.9353429 +5=62.9353429 amu
Mass defect is 62.9353429 - 62.9295975 = 0.0057454 amu
1 amu = 931 MeV is used as a standard conversion
0.0057454 x931 MeV= 5.35 MeV
So each transformation of Ni58 into Cu63 releases 5.35 MeV of nuclear energy.
Nickel consumption
According to many blogs in the Internet “One hundred 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 operation, in order to generate 10 kW?
I follow a procedure outlined in [2].
10 kW is thermal or electrical (?) power. The nuclear 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 5.35 MeV of nuclear energy.
The number of Ni58 transformations should thus be equal to (1.25 x 10**17)/5.35 = 23.364 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 36.2 x 10**22.
This means 35 grams.
The order of magnitude seems to be plausible. This means also 35 grams of Nickel in Rossi’s reactor transmuted into (stable) 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 Fiction?”, 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 probable mechanism”, www.journal-of-nuclear-physics.com/?p=338&cpage, March 2011
In order to have Cu59 (increase of atomic number from 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked proton becomes a proton.
The same process holds for all the subsequent transformations, until Cu63.
It remains to be understood the issue of the gamma radiation 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 possible, 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 decay in Nickel, producing Ni59, Ni60, and Ni62. The chain stops at Cu63 stable.
For simplicity I assume all the Nickel in the reactor 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 57.9353429 amu
The mass of a copper-Cu63 is 62.9295975 amu
Mass of Ni58 plus 5 nucleons is 57.9353429 +5=62.9353429 amu
Mass defect is 62.9353429 - 62.9295975 = 0.0057454 amu
1 amu = 931 MeV is used as a standard conversion
0.0057454 x931 MeV= 5.35 MeV
So each transformation of Ni58 into Cu63 releases 5.35 MeV of nuclear energy.
Nickel consumption
According to many blogs in the Internet “One hundred 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 operation, in order to generate 10 kW?
I follow a procedure outlined in [2].
10 kW is thermal or electrical (?) power. The nuclear 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 5.35 MeV of nuclear energy.
The number of Ni58 transformations should thus be equal to (1.25 x 10**17)/5.35 = 23.364 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 36.2 x 10**22.
This means 35 grams.
The order of magnitude seems to be plausible. This means also 35 grams of Nickel in Rossi’s reactor transmuted into (stable) 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 Fiction?”, 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 probable mechanism”, www.journal-of-nuclear-physics.com/?p=338&cpage, March 2011