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How can 30% of nickel in Rossi’s reactor be transmuted into copper?

by Dott. Giuliano Bettini
Retired. Earlier: Selenia SpA, Rome and IDS SpA, Pisa
Also Adjunct Professor at the University of Pisa
Adjunct Professor at Naval Academy, Leghorn (Italian Navy)

Abstract
In the present article I would like to answer a question posed by L. Kowalsky in a recent paper: 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 apologizes if I am too simplistic here.

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 calculated to be 57.95380± 15 amu
The actual mass of a copper-Cu63 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
0.04013×931 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 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 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 (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

 

842 comments to How can 30% of nickel in Rossi’s reactor be transmuted into copper?

  • The following information is a theory of LENR and could answer how 30% of nickel could be transmuted into copper. The theory comes about by my own understanding of atomic physics and unified fields. I refer to this technology as Unifying Field Oscillation technology that has,incidentally, been embodied but this is not what this information is about and I have to make perfectly clear that LENR is not my specialized field but like may others I find the subject fascinating. Anyway to begin with:-

    1. A four vibrational systemic reaction of nickel isotopes activated by positivity and negativity of the hydrogen atom. Resulting in 25% of the nickel transmutated into copper due to the most positive isotope transmuting and that equates as 30% of nickel content within the reactor upon completion of the reaction. My guess is that the catalyst is the heat to provide the initial heat energy and thereby pressure.

    What does this mean and does it draw parallels with the world around us. For me the answer is yes. I shall now attempt to explain why I have made such a statement. Every energy unit is comprised of a minimum of four mobile states of energy, each state occupies a 90 degree angle of the whole i.e. 360 degrees of a whole unit. Each state resonates to its own octave as it traverses/cycles the fission/fusion dimension that is of a static state (static with regards the life cycle with regards specifics i.e. quality of the unit). Consequently, four mobile states and two static states make for a six dimensional unit of energy. This can be referred to as a cubic neutral of energy. This configuration would comprise of an Absolute Positive static state, a positive positive mobile state, a positive negative neutral mobile state, a negative positive neutral mobile state, a negative negative mobile state and an Absolute negative static state. This configuration can either be recognised as a chain system being a link of a chain whereby the fission fusion static dimensions are back to back, as it were, or it can be recognised as a single unit being one of the four mobiles of a link. References with regards volume energy and size energy are a necessity. Volume energy encapsulates size energy. Mobile size energy has an affinity to the diametrical dimension of a unit whereas volume energy has an affinity to the circumferential dimension of a unit. What this is referring to is that within every structure/volume energy there has to be a minimum of four vibrations from a long wave energy to a short wave energy. These size units of energy constitutes the mass of a link controlled by a positive diametrical gravitational force of the negative fusion dimension that pyramids through the volume with its apex at the positive fission dimension and within every atom this same pattern will be found. These size units of energy rotate, oscillate upon the radial dimension as they ascend the pyramid to its apex whereupon by an act of fission they descend on the circumferential dimension only to re-cycle their path. It is when positive charges escape the gravitational field of the volume that transmutation takes place. Transmutation is connected with evolution and the economy flow system within structure. It allows for structure to exist in a more positive environment. Consequently, the ‘law of gravity’ can be considered as a ‘law of economy’ whereby the more gravity the more economy and the better quality with regards a structure. To bring about a systemic reaction you need a minimum of four harmonious vibrations in the volume of a unified positive/negative active field.. As I have stated I have not studied isotopes but I would suspect they are harmonious with regards their vibrations and to put these in the presence of an active expanded hydrogen atom to provide the two necessary static states of potential I can see no reason why a LENR would not occur. If indeed I am correct I suspect Mr Rossi is aware also with his reference ‘as to what goes on in a dwarf star.

  • David McCloud

    Hi Dr. Rossi,

    This is a great contribution to humanity that you can make. Do you use magnesium nano-particles mixed with the nickel micro particles?

    Thank you in advance.

  • Andrea Rossi

    Dear Eric Ashworth:
    Your explications and insights will be very welcome, indipendently from me. Do not worry, what you say is interesting anyway.
    Warm Regards,
    A.R.

  • Dear Mr Rossi, I have, like many people, been following your progress with regards your LENR technology and I am aware that for obvious reasons you will not disclose what is in your E-Cat other than to state nickle and hydrogen in the presence of a catalyst. However, in the article the question is asked “how can 30% of nickle in Rossi’s reactor be transmuted into copper?”. You must be aware that if the correct answer is given your present secret will be revealed and this is something you would not want. My question is, do you have any control over the published comments. This is because of my own understanding of transmutation with regards chain reactions and the duration of time with regards energy interactions that would in effect account for the the 30% figure to be mentioned. If you do have control over the published comments I would like to provide you with my explanation as understood from my own reasoning with regards internal atomic dimensions and there association with the two overall static states (static with regards the life of the unit) These overall states have been mechanically replicated by applying both geometry and maths to prove a theory. I would also like to take this opportunity to wish you all the best with regards your 1mw project. Regards Eric Ashworth.

  • Andrea Rossi

    Dear Bertil Nilsson,
    Delicious!
    Warm regards,
    A.R.

  • Bertil Nilsson

    Andrea Rossi, I hope you will enjoy and perhaps be a bit encouraged by this nice little song written and performed by some Swedish fan.
    http://www.youtube.com/watch?v=MNwOpSHNIQ4&hd=1

  • Andrea Rossi

    Dear Brian Cullen:
    1- Scheduling respected, no extra difficulties
    2-no
    3- now I am focused on my 1 MW plant
    4- I think so, as everything
    5-no
    6- exhausted from the attempts to stop the 1 MW plant operation, satisfied to have overcome the foes.
    Warm regards,
    A.R.

  • Caro Rossi,
    era chiaro che non le avrebbero permesso di cambiare lo stato dell’energia nel Mondo e le manovre in atto lo stanno dimostrando. Forse la costringeranno con le buone o con le cattive a insabbiare tutto e dichiarare alla fine che era tutto un Bluff. Forse la su migliore assicurazione sarebbe quella di rivelare al mondo la sua formula , conquistare un nobel e rimanere il piu’ grande inventore del millenio e benemerito dell’umanita’.

  • Brian Cullen

    Dear Mr. Rossi,
    I have some non-tech related questions I was hoping you could answer.

    1. Is the 1MW facility on schedule for October? and was the journey more difficult than you had previously expected?
    2. Have you been surprised by the reception you have received from both skeptics and non-skeptics?
    3. How are you feeling about the future possibilities of your invention at this stage? – in terms of the range of applications it may be useful for in the future.
    4. Is there still some room for the Ecat to evolve with further R & D in the coming years? – can it become even more efficient?
    5. Do you have any surprises up your sleeve for October?
    6. What is your general mood at this late stage in development? – (my guess) exhausted but happy and excited to be working on a revolutionary product?

    I would be more than grateful if you could satisfy my (and hopefully others readers’) curiosity regarding these issues.
    Yours Sincerely – Brian Cullen

  • Andrea Rossi

    Dear Mr Domenico:
    The risk is zero, because we do not use radioactive material and we do not produce radioactive wastes. In all the cases you listed H is cut. If the temperature rises too high, Nickel melts, powder becomes ingot and the process is stopped: intrinsecally safe situation.
    Warm Regards,
    A.R.

  • Domenico

    Egregio ing. Rossi

    (If you already have answered this question previously in your blog, please apologise) – What are the risks of an E-Cat in operation being damaged/destroyed by a collapsing building (e.g. earthquake), by shieldbreaking firearm bullets or explosives (never underestimate human madness or stupidity), by floods/tsunamis or other destructive action, in terms of gamma ray exposure intensity and duration?

    Thank you – in your work lies hope of many people.

    Cordialmente

    Domenico

  • Ed

    hmmm. Yes, you have the bragging rights, but compare to the size of an average fist, symbol of power and revolution. A lemon has a bad connotation, my car is a lemon…

    Now, get back to work!! Billions of people are counting on you, no pressure!! ;o)

    Regards
    Ed

  • Andrea Rossi

    Dear Arnoldo Giordano:
    The strict E-cat, net of insulation and shielding , is as big as a lemon.
    Warm Regards,
    A.R.

  • Arnoldo Giordano

    I understand if you can not answer this, but what are the dimensions of the Ecat? (Cm, height, width, length). I am surprised something so small could produce so much energy. Thank you.

  • Ed

    Dear Mr. Rossi

    I think you should leave these inquiries to your publisher. You have far more important issues to deal with. Spies with “The Saint” alias: Thomas Moore, Emma Russel can chew on the wind for answers.

    Best of Luck.
    Ed

  • Thomas Moore

    Dear Ing. Rossi,

    Forgive me for speculating. I understand that some things have to stay undisclosed for the time being. By the way, I made a typo in my last post. I meant “… the half-life of Ni-65 is 2,5 hours.”

    Respectfully yours,
    Thomas Moore

  • Andrea Rossi

    Dear Thomas Moore:
    I cannot give info regarding what happens inside the reactor.
    Warm Regards,
    A.R.

  • Thomas Moore

    Dear Ing. Rossi,

    Reading your answers to other posts at this site I have learnt that the inherent process in the E-Cat that, as we have seen, generates a lot of heat is based on transmutation of nickel to copper, more specifically a mixture of Ni-62 and Ni-64 that turns into Cu-63 and Cu-65, both stable copper isotopes.

    This process is in agreement with compiled data stating that Ni-63 is a possible precursor for Cu-63 as is Ni-65 for Cu-65. Let us leave aside how to get from Ni-62 to Ni-63 and from Ni-64 to Ni-65. Apart from that I can see another problem: bringing an E-Cat up to full steam takes about a quarter of an hour, but the half-life of Ni-64 is 2,5 hours. So I can only assume that you have a catalyzer for that step too. Is that correct?

    Respectfully yours,
    Thomas Moore

  • Andrea Rossi

    Dear Alessandro Ghio:
    Interesting, thank you for the suggestion.
    Warm Regards,
    A.R.

  • Buongiorno Dottore, innanzitutto complimenti per gli sviluppi di questa nuova branca di studi e spero che faccia cambiare qualcosa per noi e per le generazioni future. Volevo chiederle se avete mai pensato di fare delle simulazioni chimico-fisiche-termodimnamiche con modelli 3d usando softwere in commercio, tipo COMSOL o MAGMA o simili (che saranno comunque da sviluppare..). Penso che potrebbe aiutare a capire cose in più ed ad avere una migliore performance per il vostro strumento. Io lavoro in Avio, propulsione aerospaziale, e di solito usiamo delle simulazioni per svariate applicazioni quindi ho pensato anche ad una qualche opportunità per voi.. Cordiali Saluti e buon lavoro!

  • Andrea Rossi

    Dear Emma Russel:
    You are correct.
    Warm Regards,
    A.R.

  • Emma Russel

    Dear Andrea Rossi,

    About my faulty calculations of July 13th, 2011 at 8:41 AM I think that I found the error. I was not aware that you enrich the Ni62 and Ni64. And I am thrilled to learn that you have invented a new cheap way of enriching nickel isotopes. That is truly wonderful and worth a Nobel prize on its own merits. Did you already apply for a patent on this invention?

    Kind regards, Emma Russel

  • H. Visscher

    Dear Mr. Rossi,

    Thank you for answering my questions. I thank you for being op approachable and taking the time for answering all our questions. I do realize that his is something you don’t have to do as there is no incentive for you. I applaud you for exposing yourself like you do!

    H. Visscher

  • Andrea Rossi

    Dear Emma Russel:
    You did not read carefully my former answer. Please read carefully and remake the maths.
    Warm Regards,
    A.R.

  • Emma Russel

    Dear Andrea Rossi,

    Many thanks for your answer regarding the isotopic ratios. But I have only taken a one semester course in nuclear physics, and I feel a little confused here.

    Perhaps you can explain somewhat better to me. We start out with nickel powder with natural isotopic ratios. (I forgot to say so, but the unused powder had natural isotopic ratios as well.) Then we burn nickel isotopes but only Ni62 and Ni64 according to what you have told us in an answer to another recent post.

    Since 100% natural nickel is made up from 68% Ni58, 26% Ni60, 1,1% Ni61, 3.6% Ni62 and 0.92% Ni64, reducing only the percentage of Ni62 and Ni64 would increase the other percentages. Thus the natural isotopic ratios of nickel would not be preserved, contrary to the ICP-MS results. Could you please explain this to me?

    Kind regards, Emma Russel

  • Andrea Rossi

    Dear H. Visscher:
    Of course I am sure that Professors of the Uppsala University ( and of the Stockolm University too) will participate to the work of the Bologna University, but this fact does not necessarily imply a contract, as well as it does not necessarily exclude it. It depends on the evolution. Nothing at all is changed respect what foreseen. I never said that a contract with Uppsala has been signed. What is sure is a free collaboration of scientific reciprocal interest. And I want to add one important statement: in the meetings with the Professors of Uppsala and Stockolm I learnt a lot.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Emma Russel:
    Yes, my theory is consistent with these results.
    Warm Regards,
    A.R.

  • Emma Russel

    Dear Andrea Rossi,

    Some time ago you sent two nickel powder samples to Professor Sven Kullander in Sweden for analysis. One of the nickel powder samples had been used for 2.5 months in an E-Cat, the other was unused nickel powder.

    Chemical analysis showed that the used powder contained 10% copper and 11% iron. Isotopic analysis using ICP-MS showed that within measuring errors the isotopic ratios in the nickel as well as in the copper were natural.

    The theory for the E-Cat nuclear process that you hopefully will make public in October, is it consistent with these results?

    Best regards Emma Russel

  • H. Visscher

    Dear Mr Rossi,

    Thank you for your quick answer! However, I do not quite understand. This is what you answered on May 29:
    ———
    Andrea Rossi
    May 29th, 2011 at 7:02 PM

    Dear Ms Rosie Andreasson:
    Prof. Sergio Focardi is Prof. Emeritus of the Bologna University and Prof. Levi, Prof. Villa and Dr Bianchini of the University of Bologna made the tesp of the Jan 14th reported on the Journal Of Nuclear Physics.
    Presently, we are making a contract with the University of Bologna to develope R&D for the technology related to my patent. We will ask to the University of Uppsala to participate: I will be in Uppsala for this reason in July.
    About the patent: It has been granted in Italy (20 years validity) and is still pending for theInternational Application.
    The Countries which respect the patents in the world are 90.
    Thank you for your kind attention,
    Warm Regards,
    A.R.
    ——–
    If I read it correctly it stated clearly that you will ask the University of Uppsala to participate in R&D. Has something changed from what you answered or did it not go through? May be you will have another meeting later this month?

    Thank you for clearing this up!

    H. Visscher

  • Andrea Rossi

    Dear H. Visscher:
    The object of the very important meeting I had in Uppsala with top level international Scientists is confidential. We did not sign any contract with the University of Uppsala:
    I never said or wrote we had. Beware false statements of spies.
    Warm Regards,
    A.R.

  • H. Visscher

    Dear Mr. Rossi,

    On May 29 you answered a question from Ms. Rosie Andreasson in which you said that you were in Uppsala in July in order to get the University of Uppsala to participate in R&D on E-Cat. Have you been successful in that? The reason I ask is that -according to “snake” – the university denies there is agreement or any preparation of an agreement in this matter.
    May be you can clarify this? Will this mean that the University of Bologna will work on his own?

    Thank you!

    H. Visscher

  • Dear Ing. Rossi
    I am reading all messages in your blogs about your extraordinary E-Cat and I am really happy to live in this time that is the beginning of a new era for humans.
    I have prepared a web site where I collect a part of all messages present in your blogs, and I am sharing these web pages in several News Group.
    The link is the following:
    http://www.favorscake.com/E-Cat/ecat-01.html

    Till now it only has some hundreds of messages, but it will grow fast during time.

    Tanti auguri di Buon Lavoro da un italiano orgoglioso di Lei!!
    I.R.

  • Guru Gurovic

    Is visible light a form of electromagnetic waves ?
    Yes.

    Is Gamma radiation inside E-Cat a form of electromagnetic waves ?
    Yes.

    It already exists twenty versions/technologies for capture visible light and convert this into electricity ?
    Yes.

    So soon there will exist materials for capturing Gamma waves directly inside E-Cats and convert this directly into electricity.

    Plus a lot of waste heat as premium.

  • Andrea Rossi

    Dear Stefano Cicchiello:
    Yes.
    Warm Regards,
    A.R.

  • stefano cicchiello

    Dear Sir Ing Andrea Rossi .
    Your kindness in answering so many people is amusing
    My Question is : there is any relation between the pressure of Hidrogen in the reactor vessul Thus the efficency and self-substaining process?

    Thank you very much Sir for reading this

  • Andrea Rossi

    Rossi Site Down:
    Yes, we are often attacked, our Informatic consultant fixed it.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Christian Scholl,
    I used to produce Hydrogen by means of electrolysis, but I prefer the new solution, which gave better results, for many reasons that I cannot disclose. Good point, anyway.
    Warm Regards,
    A.R.

  • Christian SCHOLL

    Dear Rossi,

    You commonly use pressurized hydrogen.
    Don’t you bother about leakage risk?
    Electrolysis is a surest way to release hydrogen.

    Best regards

    C.SCHOLL

  • [...] that the site is still accessible if you know a specific address. By jumping direct to a page (for instance) everything appears normal. Posted in Media/Blogs, Rossi | Tagged Kournal of Nuclear Physics, [...]

  • Sam Morton

    Dear Andrea Rossi,

    You may have information already about this Panasonic Thermoelectric Generator.
    It maybe perfect solution to turn E-CAT produced heat to electricity with out moving parts.

    http://www.green-energy-news.com/arch/nrgs2011/20110051.html

    Best Regards,
    Sam Morton

  • Monsieur Rossi,
    Note provisoire sur la fusion Ni2858 / Ni2859 —- > Cu2963 / Cu2965 du réacteur E-CAT de Rossi et Focardi.
    Question posée : Est-ce qu’à la lumière de vos travaux vous pouvez apporter des explications aux expériences de Rossi et Focardi sur la fusion « froide » catalysé. R. P.

    A._ Nucléons et éléments atomiques de base pris en compte.
    e électron : …. 0,000548652 _ 510 999 eV _ 9,11000 10-31 kg
    p proton : …. 1,007406568 _ 938 272 071 eV _ 1,672773 10-27 kg ….. p proton : noté aussi comme : 0,999451H
    11H Hydrogène : …. 1,00796 _ 938 783 071 eV _ 1,67364 10-27 kg masse/énergie : – 0,000544 e _ Soit : -0,00054 e par nucl. _ : – 0,000276 MeV/nucl.
    n neutron ou deuton : …. 1,0085 _ 939 294 071 _ 1,67455 10-27 kg masse/énergie : – 0,00898 e _ Soit : -0,0089 e par nucl. _ : – 0,00455 MeV/nucl.
    n neutron : noté aussi comme : 1,0005441Deuton
    11H + 11H = bi atome de Bohr (molécule d’hydrogène) ou bi atome de Perrin ~ (atome protonique et neutronique couplés en alternance : p / bohr / n / bohr) ;
    C’est de cette structure qu’émergent électron, proton et neutron des nucléons (atome / proton / neutron / biatome) liés dans les différents éléments de la nature.
    21H Deutérium : …. 2,014102 _ 1 877 566 700 eV _ 3,34729 10-27 kg masse/énergie : – 3,297 e _ Soit : -1,63 e par nucléon : – 0,836 MeV/nucl..
    42He Hélium : …. 4,0026 _ 3 755 133 399 eV _ 6,69458 10-27 kg masse/énergie : – 53,26 e _ Soit : -13,306 e par nucl. : – 6,799 MeV/nucl.
    5626 Fe Fer : …. 55,845 _ 52 573 912 702 eV _ 9,37277 10-26 kg masse/énergie : – 1098,5 e _ Soit : -19,67 e par nucl. _ : – 10,051 MeV/nucl.
    5828Ni Nickel : …. 57,935 _ 54 450 456 845 eV _ 9,70732 10-26 kg masse/énergie : – 961,481 e _ Soit : -16,59 e par nucl. _ : – 8,48 MeV/nucl.
    5928Ni Nickel : …. 58,6934 _ 55 889 751 473 eV _ 9,87477 10-26 kg masse/énergie : – 1417,33 e _ Soit : -24,14 e par nucl. _ : – 12,339 MeV/nucl.
    6329Cu Cuivre :…. 62,929 _ 59 145 907 429 eV _ 1,05444 10-25 kg masse/énergie : – 1047,92 e _ Soit : -16,67 e par nucl. _ : – 8,509 MeV/nucl.
    6429Cu Cuivre :…. 63,546 _ 60 086 224 613 eV _ 1,07119 10-25 kg masse/énergie : – 1761,49 e _ Soit : -27,71 e par nucl. _ : – 14,1649 MeV/nucl.
    6529Cu Cuivre :…. 64,927 _ 59 145 907 429 eV _ 1,08793 10-25 kg masse/énergie : – 1082,57 e _ Soit : -16,67 e par nucl. _ : – 8,52 MeV/nucl.
    …Pour mémoire :
    63Li Lithium : …. 6,08 _ 5 632 700 099 eV _ 1,00419 10-26 kg masse/énergie : + 58,99 e _ Soit : +9,7027 e par nucl. : + 4,958 MeV/nucl.
    105 B Bore : …. 10,115 _ 9 387 833 498 eV _ 1,67364 10-26 kg masse/énergie : + 212,238 e _ Soit : +20,815 e par nucl. _ : + 10,636 MeV/nucl.
    …Note :
    Dans les tables numériques donnant les masses/énergies des différents éléments naturels (42He, 5626 Fe, 6429Cu, etc.) et des différents isotopes de chacun d’eux, y compris pour eux-mêmes, règne une certaine ambiguïté sur la référence « uma » prise en compte. Ainsi les valeurs relevées dépendent des auteurs et de la précision attachée à chacune des valeurs. Ceci dit les tendances observées pour les écarts (masse/énergie ) de masse/énergie libérée lors de fusion ou à fournir lors de fission donnent de bonnes indications.
    On observera par ailleurs que toute libération d’électron (ou amas équivalent) du niveau orbital n1 par exemple, pour tout atome sous l’effet de la fission atomique : atome = > proton et électron éjecté, s’effectue sous une pression énergétique extérieure de 13,6 eV ; qu’elle soit photonique, thermique, électromagnétique, biologique, masse volumique, rayonnement gamma ou présence rapproché d’atomes voisins. Toute nouvelle pression énergétique équivalente tirera un autre électron prélevé du niveau : > à n1 du même atome, celui-ci pouvant être considéré dans son état : atome grave.
    Référence : Communication n°3 et Communication n°4 de 2003 et 2004 _ http://jcvillame.free.fr .
    Ce que certains autres auteurs évoquent à travers la théorie de l’ hydrino ou du vertex.
    Réciproquement à l’inverse et pour les produits de fusion utilisés dans l’E-CAT de Rossi et Focardi :
    toute libération d’énergie (radiatif, thermique, électrique…) d’un atome sous l’effet de la fusion (proton + électron = > atome de base ou, atome + électron = > neutron), à partir du niveau orbital n11,706 _ s’effectue par la capture d’un électron sur le niveau n1.(pour chaque atome du crayon de 50 gr de Ni). Ce premier électron s’enfonçant sous n1, dû par exemple à une très forte concentration atomique des éléments de Ni, conjugué à une élévation thermique du mélange gazeux H1 également sous pression et qui l’entoure, re – libère une quantité de matière/énergie équivalente, etc.
    Ceci peut participer à expliquer le phénomène observé dans l’E-CAT. Par contre si tel est le cas, il convient d’observer que l’état potentiel d’atome grave du Ni doit être préparé au préalable dès la fabrication du crayon de Ni. Ce potentiel se libérant avec la montée en température de H1. Alors que la seule différence d’écart de masse/énergie libérale dit improprement « masse manquante », fait partie intégrante du phénomène mis en œuvre dans l’E-CAT ; et le soutient théoriquement .

    B._ Application au cas de la fusion Ni58 — > Cu63 / Cu65 du réacteur E-CAT de Rossi et Focardi.

    Première évaluation comparative.
    Efusion du Ni2859 + 5 nucléons (4 deuton/neutron01,00054 et 1 proton10,99945) en…. Cu2964. Les éléments Ni2859 et Cu2964 étant les plus courants du Ni et du Cu.
    Relation de fusion de base : 5928Ni + 4 x 1,0005440Deuton + 0,999451H === > 6429Cu + Différentiel énergétique libéré : ? MeV.
    Ce qui donne numériquement comme différentiel énergétique libéré, (colonnes J et M_ de la table de référence donnée en Communication_jcv : n° 6 de 2006 ) — >
    — > – [[1417,333 él.] – [(4 x 0,00898) él.) + ~0,00él.] + [1761,49 él.] = ? MeV.
    Soit — >> [-1417,297 él. + 1761,49 él.] = 344,xx193 électrons ! Ou… : 175,8xx MeV pour la réaction nucléaire globale (2,74xx81 MeV par nucléon).
    Soit ~ 9,9 fois plus que le potentiel de l’hélium obtenu par fusion du deutérium et du tritium : 34,72 électrons ou 17,74 MeV (4,435 MeV par nucléon).
    Mais ~ 0,618 fois plus par nucléon, que le potentiel de l’hélium.

    Efusion du Ni2858 + 6 nucléons (5 deuton/neutron01,00054 et 1 proton10,99945) en…. Cu2964. L’élément Ni2858 étant communément le plus cité comme base de l‘E-CAT.
    Relation de fusion de base : 5828Ni + 5 x 1,0005441Deuton + 0,999451H === > 6429Cu + Différentiel énergétique libéré : ? MeV.
    Ce qui donne numériquement comme différentiel énergétique libéré, — >
    — > – [[961,481 électrons] – [(5 x 0,00898) électron + ~ 0,00 électron +]] + [1761,49 électrons] = ? MeV.
    Soit — >> [-961,436 él. + 1761,49 él.] = 800,054xx44 électrons ! Ou… : 408,78xx138 MeV pour la réaction nucléaire globale (6,38xx7934 MeV par nucléon).
    Soit ~ 11,77 fois plus que le potentiel de l’hélium obtenu par fusion du deutérium et du tritium : 34,72 électrons ou 17,74 MeV (4,435 MeV par nucléon).
    Ou encore ~ 1,44 fois plus par nucléon, que le potentiel de l’hélium.

    Deuxième évaluation complémentaire.
    Efusion du Ni2858 + 5 nucléons (4 deuton/neutron01,00054 et 1 proton10,99945) en…. Cu2963.
    Les éléments Cu2863 et Cu2965 étant les plus cités comme produits de la fusion dans l’E-CAT (Cu2863 / Cu2965 = 1,6).
    Relation de fusion de base : 5828Ni + 4 x 1,0005441Deuton + 0,999451H === > 6329Cu + Différentiel énergétique libéré : ? MeV.
    Ce qui donne numériquement comme différentiel énergétique libéré, (colonnes J et M_ de la table de référence donnée en Communication_jcv : n° 6 de 2006 )— >
    — > – [[961,481 él.] – [(4 x 0,00898) él.) + ~ 0,00 él.] + [1047,92 él.] = ? MeV.
    Soit — >> [-961,445 él. + 1047,92 él.] = 86,476 électrons ! Ou… : 44,18xx89 MeV pour la réaction nucléaire globale (0,xx7014 MeV par nucléon).
    Soit ~ 2,49 fois plus que le potentiel de l’hélium obtenu par fusion du deutérium et du tritium : 34,72 électrons ou 17,74 MeV (4,435 MeV par nucléon).
    Mais ~ 0,158 fois plus par nucléon, que le potentiel de l’hélium.
    1047,92 électrons
    Efusion du Ni2858 + 7 nucléons [6 deuton/neutron01,00054 et 1 proton10,99945) en…. Cu2965.
    Relation de fusion de base : 5828Ni + 6 x 1,0005441Deuton + 0,999451H === > 6529Cu + Différentiel énergétique libéré : ? MeV.
    Ce qui donne numériquement comme différentiel énergétique libéré, --- >
    --- > - [[961,481 él.] – [(6 x 0,00898) él.) + ~ 0,00 él.]] + [1082,57 él.] = ? MeV.
    Soit — >> [-961,42xx58 él. + 1047,92 él.] = 121,14xx42 électrons ! Ou… : 61,9xx04 MeV pour la réaction nucléaire globale (0,9xx5238 MeV par nucléon).
    Soit ~ 3,489 fois plus que le potentiel de l’hélium obtenu par fusion du deutérium et du tritium : 34,72 électrons ou 17,74 MeV (4,435 MeV par nucléon).
    Mais ~ 0,214 fois plus par nucléon, que le potentiel de l’hélium.

    Troisième évaluation comparative.

    Efusion de 2 Ni2858 + 3 nucléons (4 deuton/neutron01,00054 – 1 proton10,99945) en…. Cu2963 et… Fe2656. L’hypothèse formée par un auteur mais rejetée par les inventeurs de l’E-CAT. Théoriquement cette hypothèse s’exclut d’elle-même… ainsi que le confirme cette troisième évaluation.
    Relation de fusion de base : 2 x 5828Ni + 4 x 1,0005441Deuton – 0,999451H === > 6329Cu + 5626Cu + Différentiel énergétique libéré ou à fournir : ? MeV.
    Ce qui donne numériquement comme différentiel énergétique, — >
    — > – [[2 x 961,481 électrons] – [(4 x 0,00898) électron) - ~ 0,00 él.] + [1047,92 électrons libérés - 1098,52 électrons libérés ] = ? MeV.
    Soit — >> [-1922,968 él. + (2146,44) él.] = 223,47 à fournir !!! ! Ou… : 114,19 MeV à fournir !!!! Pour la réaction nucléaire globale (6,437 MeV par nucléon)… A fournir !!!. Ce qui ne peut valider l’hypothèse, compte tenu des résultats concrets de l’E-CAT.

    A._ Conclusion provisoire.

    La fusion thermonucléaire n’étant qu’une vision partielle du phénomène générique de la transmutation (fusion / fission) de la matière/énergie, il ne peut y avoir des a priori sur ce sujet générique.
    En fonction de mes analyses antérieures, les travaux et les résultats obtenus sur la machine E-CAT de Rossi et Focardi sont cohérents et patents.
    Mes thèses (Agrégation-gravitation électromagnétique – Atome grave – Echanges équilibrés « électrons / raies spectrales – Ecarts de masse/énergie équilibrés fusion/fission, etc.) permettent de les expliquer, tout au moins dans l’instant, et dans l’attente des précisions complémentaires concernant les composés mis en œuvre dans la fusion E-CAT ainsi que celles concernant les produits fusionnés.
    Les résultats actuels démontrent la faisabilité du procédé. Ce qui constitue une fabuleuse découverte tecnologique
    Pour définir plus avant la théorie globale du procédé, nous devons attendre les futures informations techniques que les auteurs prévoient livrer en octobre – novembre 2011.
    J’espère apporter par ces quelques notes provisoires une réponse positive à la question posée et un point de vu complémentaire au contenu des Questions / Réponses du site de Rossi et Focardi. Réf. : http://www.nyteknik.se/nyheter/energi_miljo/energi/article3144827.ece
    Jean-Claude Villame _ 4/7/2011.
    Extrait de la communication n°6, pour application : fusion Ni58 — > Cu63 / Cu64 du réacteur E-CAT de Rossi et Focardi.

    Premier extrait.
    II.6c-3 Fusion nucléaire : amélioration du rendement énergétique / choix des meilleurs éléments (lithium7, bore11 Vs H1, …)

    Quand elle est technologiquement possible, la fusion de deux éléments X1 et X2 [l’un de masse atomique rapportée à son nombre de nucléons : Mnucl. 1, l’autre de masse atomique rapportée à son nombre de nucléons : Mnucl. 2], aboutit à la création d’un élément X3 [de masse atomique rapportée à son nombre de nucléons : Mnucl. 3].
    L’opération de fusion s’effectue soit avec un gain de masse/énergie ou avec une perte de masse /énergie.
    La variation énergétique résultante (équivalente à la modification de masse atomique par unité nucléique) est proportionnelle à la différence des masses atomiques nucléiques de départ et d’arrivée.
    Ainsi, Efusion de (X1 + X2), en X3 = fonction de [n3Mnucl. 3 - (n1Mnucl. 1 + n2Mnucl. 2)]. Où ni est le nombre de nucléons des éléments Mnucl.i concernés.
    Le gain ou la perte d’énergie est directement fonction du choix des éléments X1, X2 fusionnés et de l’élément X3 résultant.
    Dans les centrales atomiques on recherche un gain d’énergie, avec par exemple la fusion des éléments deutérium ou tritium pour obtenir de l’hélium, qui est la réaction nucléaire la plus usitée. Généralement, les atomistes en donnent le bilan suivant : 21H + 31Ti === > 42He + 10Deuton + 17,58 MeV.

    Application et vérification, en fonction des valeurs issues des tables précédentes.
    Les masses atomiques nucléiques de ces éléments sont respectivement : 0,9991029 / 0,9974153 / 0,992752 et 1,00054 pour le deuton (résidu de la réaction). (Référence colonne V). On obtient l’identité suivante :
    Efusion du deutérium et du tritium en hélium = [4 x 0,992752 + 1 x 1,00054] – [2 x 0,9991029 + 3 x 0,9974153] = 0,0189037 x 938,783 MeV ;
    [3,971008 + 1.00054] – [1,9982058 + 2,9922459] = 0,018903 x 938,783 = 17,746472 MeV ou 0,018903 x 1837,15 = 34,72893 électrons
    [+ 4,971548 - 4,9904517] = 17,746472 MeV / 0,511 == > 34,7289 électrons
    Soit 17,746 MeV ou = 34,7289 équivalent électrons ! C.Q.F.D.
    On notera que : 938,783 MeV n’est autre que la masse/énergie de l’atome classique, équivalent à la masse/énergie de 1837,15 électrons de 0,511 MeV.

    Cette application chiffrée est mentionnée pour bien démontrer la validité des principes exposés et partant de la réalité des conclusions qui suivent qu’aucun physicien ne peut mettre en doute. Ainsi on peut directement utiliser les valeurs approchées des excès ou défauts de masse/énergie de chacun des éléments de la colonne J exprimée en équivalent électrons ou de la colonne M (équivalent électrons par nucléide).
    Soit : 21H + 31Ti === > 42He + 10Deuton + 17,58 MeV (ou 34,40 équivalent électrons).
    — > [53,261 él. + 0,00898 él.] – [3,297 él. + 15,247 él.] = 53,269 – 18,544 — > 34,725 électrons à 0,6 % près.
    Ceci étant établi et vérifié il devient très aisé de comparer l’efficacité de la réaction de fusion avec différents éléments naturels, avec cette dernière relation approchée en utilisant les valeurs établies dans la colonne J ou M. D’emblée, on constate que les transmutations atomiques des éléments lithium (63Li en 73Li) et bore (105B en 115B) sont particulièrement favorables. Comme le seraient, à moindre degré, des réactions de fusion à partir du néon20, du magnésium20 et du clore35 ou 37, vers du lithium7, du bore11, de l’hélium4 ou du carbone12 ou 13… Par exemple.
    L’exemple numérique réalisé avec le lithium ou le bore, illustre bien l’immense potentiel industriel, dont je fournis le fondement théorique inédit.
    Naturellement sous réserve de la faisabilité technique de la transmutation atomique.

    Efusion du lithium6 et de l’hydrogène1 en lithium7.
    Relation de fusion de base : 11H + 63Li === > 73Li + ? MeV. Ce qui donne numériquement comme différentiel énergétique, (Référence colonne J) :
    — > [210,037 él.] – [0,00054 él. + (- 58,9935) él.] = 269,0299 électrons ! Soit : 137,474 MeV pour la réaction nucléaire globale (19,639 MeV par nucléon).
    Relation de fusion : 11H + 63Li === > 73Li + 137,474 MeV.
    Soit ~ 7,749 fois plus que le potentiel de l’hélium obtenu par fusion du deutérium et du tritium : 34,72 électrons ou 17,74 MeV (4,435 MeV par nucléon).
    Ou encore ~ 4,428 fois plus par nucléon, que le potentiel de l’hélium
    Efusion du bore10 et de l’hydrogène1 en bore11.
    Relation de fusion de base : 11H + 105B === > 115B + ? MeV. Ce qui donne numériquement comme différentiel énergétique, (Référence colonne J) :
    — > [504,982 él.] – [0,00054 él. + (- 212,238) él.] = 717,219 électrons ! Soit : 366,499 MeV pour la réaction nucléaire globale (33,318 MeV par nucléon).
    Relation de fusion : 11H + 105B === > 115B + 366,499 MeV.
    Soit ~ 20,659 fois plus que le potentiel de l’hélium obtenu par fusion du deutérium et du tritium : 34,72 électrons ou 17,74 MeV (4,435 MeV par nucléon).
    Ou encore ~ 7,5125 fois plus par nucléon, que le potentiel de l’hélium

    Le rendement de la fission de l’uranium 23592U en, 9438 Sr et 14054Xe, n’étant que de : ~ 0,85 MeV (1,663 équivalent électron) par nucléon, pour une énergie globale libérée de : ~ 200 MeV (391,39 équivalent électrons).
    [Si on remplace l’hydrogène par du deutérium, du tritium ou de l’hélium, les produits secondaires seront légèrement différents. Le rendement énergétique sera évidemment moindre].

    Deuxième extrait. Pour référence théorique du phénomène.
    Voir texte du chapitre II.6 de la Communication n° 6 – deuxième partie du 15/11/2006. Réf. http://jcvillame.free.fr Page 14 du site.
    La découverte et la théorie de l’atome grave (Situation de l’atome pour tous les niveaux orbitaux sous le niveau de Bohr sont exposées dans les
    Communications n° 3 et n°4 : (pages 7 et 8 du site de l’auteur).

  • Andrea Rossi

    Dear Bob Dingman:
    Thank you.
    Andrea Rossi

  • Bob Dingman

    Dear Andrea Rossi:

    Well I thought I’d take a shot. Some of us have too much time on their hands. Yes, I read the Petroldragon story. Very few people have the fortitude to survive such an onslaught of organized, despicable, corruption. Words are not sufficient to convey the disgust I felt when I read the account.

    Obviously, the technology is not challenge enough. You have to battle vested interests, greed, egomaniacs, and entrenched scientific dogma. One can wonder if mankind will survive itself. Your latest achievement is evidence that there is much hope, in contrast to those that seek to serve the evil forces. unPatiently looking forward to November,

    Best Regards,

    Bob Dingman

  • Andrea Rossi

    Dear Bob Dingman:
    No, there are no connections at all between my patent to make energy from wastes (1978) and the eventual R&D on the LENR. With one exception: I learnt how to fight to defend a revolutionary technology. Today I am less vulnerable. See http://www.ingandrearossi.com to know what happened to me many years ago…
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Bob Dingman,
    Thank you for the translation, very useful.
    Warm Regards,
    A.R.

  • Bob Dingman

    I wanted to ask Signor Rossi if his previous work with Petroldragon had led him to his amazing discovery. I was trying to imagine how the reconversion of plastic refuse under heat and pressure into liquid hydrocarbon could inspire anyone into thinking about the nuclear transmutation of nickel into copper. It didn’t seem possible or feasible. And yet, it also seems to be more than mere coincidence. There has to be a connection! History has shown us numerous examples of the greatest discoveries of mankind have occurred quite by accident. I kept wondering what does scrap plastic have to do with nickel? I know plastic is often chrome plated, but where would the nickel come from? Then it finally dawned upon me. What is nickel plated? Fittings, tubing, pressure vessels? Is it just possible that the apparatus that was initially nickel plated came to take on a copper color?

    Best Regards,

    Bob Dingman

  • Rick Meisinger

    Dear Andrea Rossi:

    From the ashes shall come forth a shaking, inventions, and an innovative people that shall begin to search and say, ‘I know that there’s a mystery. I know that there is something that’s hidden that is for this generation and for us as a people’. We will not stop until we have found that, and that light that God has for our generation. Happy 4th of July!

    With much support and appreciation;
    Rick Meisinger – USA

  • Bob Dingman

    The following is the google translation of the post from Mario dell Pozzo, July 3, 10:11 AM.

    The writer is a former engineer, electrical engineer, plant, still active, nearly half a century of experience, including significant management Enel type where at least until the early 80′s things were set very seriously. Mine is the latest generation of graduate engineers with slide rules, generalist, accustomed as they were called to deal.
    I read your story. I’ve seen worse. Companies without fail even without administrative intervention, simply because larger companies have said the bank “or remove the exposure to that or do not work with you anymore.” Admittedly, her face showed a great reaction. Most people would be torn down. Congratulations. There is a note from old engineer, however, that I would do it. She gives too much importance to the title of engineer. If a young man had done the polytechnic would probably today the mental agility that led to the results obtained.
    I come to his Ecat I follow with interest. I have already set up projects, particularly in some institutes Salesians that often amount to medium businesses with regard to energy consumption, to try to understand the profitability dell’Ecat. Because it is obvious that any equipment is placed in a facility that is designed and the costs must be calculated. Of course not having the flexibility and response curves dell’Ecat I had to make many assumptions, and this experience helps a lot.
    Summarize some data, in pounds, so why do I express myself better. Today, July 2011, in an average company kWh electricity costs about 350 pounds (45% costoc energy, 12% for transport and distribution,
    18% alternative energy to fuel the fund, 25% taxes and duties).
    The transformation of the electric thermal kWh kWh, taking into account the yield, the cost of thermal kWh, kWh of electricity, is about 400 pounds (a). Instead, the thermal kWh of gas still costs 230 pounds recital performance (b). The flexibility of these energies is complete, there is no need for other facilities.
    I read that the Ecat would be sold at 1380 euros / kW + 2 refills / year at 70 euros each. Consider a 1 kW Ecat that lasts 10 years. If today I ask a loan of 1380 euros to 10 years pay 230 euros / year. That would cost the Ecat 230 + 2 × 70 = 370 euros / year. That is 716500 pounds. The thermal energy can be produced (yield 90%) 7500kWh/anno. That is, the thermal kWh costs about 100 pounds. The flexibility is zero. If I want to build a plant to exploit fully the energy that can produce the experience teaches that virtually doubles the cost of energy, that becomes 200 pounds, very close to (b). Just a heavy tax burden for even making the assumption is not convenient. In the past, machines like the TOTEM microturbines or gas from aircraft have not been successful because the lack of flexibility was falling productivity and increase the cost of energy to make it competitive in the end.
    Now a machine like the one you created that produces clean energy is in any way supported and diffused and engineering must be developed to its maximum.
    Still, I talked to some charge of Vocational Training Centres Salesians (popular in the world) just because things will be clearer and you can then develop an engineering system classes of young people are trained to work in this new field, and create good jobs work everywhere in the world is no small thing.
    You ever need a piece of information will allow me to bother you.
    Best wishes and congratulations again.
    mario pozzo

  • Andrea Rossi

    Dear John Cauchi:
    We will not make new public tests, we will make products that our Customers will put in operation: as you do when you phone to your mum from Kenya, and can look at her too: isn’t it a test more convincing than a public test made in some lab from the phone company?
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Adriano Zapparoli:
    Perhaps I gave you a wrong address?
    Please try again: stsalikoglou@gmail.com
    Copy to me: info@leonardocorp1996.com
    Warm Regards,
    A.R.

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