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

 

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

  • Andrea Rossi

    Dear Mr Joshs:
    I cannot give information about the operation of the reactor.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Mr Claud:
    Thank you! (How do you know?)
    If you know that today is my birthday you also know my direct line. To help me, is necessary I know who you are.
    Warm regards,
    A.R.

  • Claud

    Dear Mr. Rossi, my best wishes for your project and “happy birtday”.
    Should you ever need a business plan specialist I’d like to offer my experience (for free).
    C.R.

  • Joshs

    Dear Mr Rossi,

    Am I correct in concluding that pressure along with micro fissures causes a leverage effect that thus overcomes the coulomb forces? If so is there a correlation with pressure and reaction rate?

    Further if so then is not pressure regulation the simplest form of control, ie to shut down the reaction simply release the pressure.

    Warmest Regards

    Joshs

  • Andrea Rossi

    Dear Ms Maryyugo:
    Not yet. Soon, yes.
    Warm regards,
    A.R.

  • maryyugo

    Dear Mr. Rossi,

    Than you for your answer however I was not asking about public conferences. What I was hoping to determine was simply whether or not the University of Upsala (or any other university) has in their physical possession at this present time an E-Cat to test. Or whether such tests are for the future. I was not requesting that you reveal any results prematurely. Only whether or not the University has an E-Cat to test. Sorry if my question was not clear. Best regards, M. Y.

  • Andrea Rossi

    Dear Ms Maryyugo:
    Before October there will be no public conferences regarding our E-cats. Until then we will make R&D, closed doors, with Bologna and Uppsala Universities.
    Warm regards,
    A.R.

  • maryyugo

    Dear Mr. Rossi,

    Thank you again for your patience in responding to questions. There are people who are saying that two E-Cats are being tested independently at this time by universities and that one of those is University of Upsala. This information appears, for example, at peswiki.com. I had thought you said this would not happen until October. Which is correct? And if you are waiting until October, perhaps you would be kind enough to explain why, considering that you have many E-Cats already constructed. Thank you. M.Y.

  • Andrea Rossi

    Dear Dr M.S. Meyers:
    The 58-Ni doesn’t work.
    Warm Regards,
    A.R.

  • M S Meyers

    I recalculated this articles numbers using Ni-58 and Cu-63 masses and come up with a number of ~35% conversion to copper, which matches the claims. This assumes starting with 100% Ni-58, a simplification. I’ve seen several references to Fe and Zn as by products which, if true, means other reactions must be occurring. My only reservation is that FE-58 to Cu-63 requires 5 reactions and this will be intriguing to explain.

  • Andrea Rossi

    Dear Mr Sam:
    For commercial issues, please contaqct starting November info@leonardocorp1996.com
    Leonardo Corp will forward the requests to the proper persons.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Mr Thon Brocket:
    We will make thermal MWh.
    I do not know if it will be in Xanthi or somewhere else, the Customer knows.
    Warm regards,
    A.R.

  • Andrea Rossi

    Gent. Sig. Guido B.:
    1- The isotopoical concentration is not equal, and the equations after the enrichment of Ni are not that simple
    2- The R&D program with UNIBO is a private, not a public, issue
    3- Yes
    Warm Regards,
    A.R.

  • Guido B.

    Gentile Dott. Rossi,
    cerco di seguire il più possibile la vicenda e-cat,
    spero francamente che tutto vada per il meglio.
    Avrei qualche domanda alle quali spero che Lei possa rispondere:
    1) Avete sostenuto più volte che dopo diversi mesi di funzionamento nel reattore il rame prodotto dalla fusione non aveva concentrazione isotopica naturale, i fisici svedesi Essen e Kulander sostengono di aver ricevuto da Lei delle scorie di un reattore che aveva funzionato diversi mesi e che le scorie stesse hanno un rapporto isotopico del rame in concentrazioni naturali. C’è qualche spiegazione a questa apparente contraddizione ?
    2)Ci sono novità per il programma di sperimentazione all’ Unibo ? Sarà operativo entro giugno ?
    3)La commercializzazione dell’ecat partirà in ogni caso entro fine anno ?

    Grazie

    Cordiali saluti e buon lavoro

  • Thon Brocket

    Mr Rossi

    What is the configuration of your 1MW plant in Xanthi? Would I be correct to assume that it is a conventional steam turbine and generating set, perhaps of the sort you might find in a geothermal power plant?

    Best regards, and wishing you all success in your venture.

    Thon Brocket

  • Sam

    Hi Andrea Rossi ,

    For Commercial issue , Can you kinldy advice your email address .

  • Andrea Rossi

    Dear Mr Adriano Z.:
    Please contact us in November for commercial issues.
    The visits to the 1 MW plant will be reserved to a restrict number of persons.
    Warm Regards,
    A.R.

  • adriano z.

    Preg.mo Ing Rossi,
    già le scrissi relativamente allle possibilità di commercializzazione del suo Ecat in indonesia ed altri.
    Le chiedevo per maggior chiarimenti quali sono
    le varie tipologie di impianto che nasceranno
    dopo l’avvio del protoripo greco, sicuramente
    verranno assemblati impianti diversi ma di
    quanti kwh?
    impianti come il prototipo saranno solo un esperimento oppure possono essere riproducibili anche di MGW superiori.
    Lei pensa che potranno essere utilizzabili
    anke in piccole e medie aziende?
    L’eventuale allacciamento in rete elettrica
    potrà consentire degli stessi privilegi del fotovoltaico o altre rinnovabili?
    Sarà possibile assistere all’avvio del suo
    primo impianto in Grecia? il mio socio abita a Cipro con la sua societa di trading.
    Io invece abito a 20 Km da Bondeno:
    La ringrazio per la sua risposta , naturalmente fin dove lei può?
    Saluti
    adriano z.

  • Peter G

    Dear Mr. Rossi

    I have expertise in remote predictive equipment diagnostics where equipment worldwide can be monitored; data collected and trended for the purpose of predicted maintenance, all from a single location. I am not associated with any company. If you have any interest in this please contact me by my private e-mail.
    Peter

  • Andrea Rossi

    Dear Mr Luke Mortensen:
    1- NDA are useless, totally. We maintain the industrial secret. With our Customers, we respect the limits fixed in contracts.
    2- Every day happens something exciting: I have the luck to get fun from the work I do.
    3- Not yet.
    Warm regards,
    A.R.

  • Luke Mortensen

    Mr Rossi,

    Had a couple questions:

    1. Could you tell us about the nature of the NDA agreements you have in force?
    I understand that the reactor design, the fuel composition, the catalyst, the output particles, and your operational discoveries are all confidential. How much of this is a contractual obligation, and how much is your discretion?

    2. Anything exciting happening in your workshop this week?

    3. What is on the agenda for your trip to Upsala in July? Bringing any ecat toys with you?

    Thanks,
    -Luke

  • Andrea Rossi

    Dear Mr Craig:
    I prefer not to talk about these issues before the 1 MW operational start up.
    Warm Regards,
    A.R.

  • Dear Mr Rossi,

    As you may have heard today the big news is that the German government have now announced an end to nuclear power generation by 2022 in the wake of the Fukoshima Nuclear Disaster. There is no mention of how they will replace 22% of their energy needs in the next 11 years. Other countries look sure to follow down this path, so they too will need vast amounts of energy to be replaced by other means. Perhaps your e-catalyzer can assist them?

    Have you had any inquiries to date from foreign governments regarding the use of your new technology for power generation and can you comment on this announcement by the German government. Is a nuclear free world possible?

    Craig

  • Andrea Rossi

    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.

  • Rosie

    Dear Mr Andrea Rossi

    I read your and prof Focardis report dated march 22 2010. It states that the Physics Department Bologna University is behind the report. Am I correct?

    It also states that the invention is patented in 90 countries. Is that still correct, or is it now patented in more countries?

    Sincerely
    Rosie Andreasson
    Sweden

    http://www.nyteknik.se/incoming/article3080659.ece/BINARY/Rossi-Focardi_paper.pdf

  • Joseph Fine

    I exchanged a few messages with Dott. Paolo Savaris. He has some very interesting results, but he’s busy now. I will let him comment on his own work – when he is ready. I suggested that when diatomic Hydrogen breaks up into its monatomic form, in addition to getting (mostly) neutral hydrogen atoms – some of the (diatomic) H2 molecules could break up into a proton (a positive ion – of course) and a negative Hydride ion. There has been a lot of discussion of the proton here (of course), but almost nothing mentioned about the hydride ion. A hydride ion is a proton that has two electrons and a net negative charge. I don’t know how long a hydride ion remains stable, but a possible answer may be – long enough for practical purposes.

    The hydride ion would be able to get closer to the nucleus than the proton. I don’t want to write a long discussion here since it is late. But in view of the electric attraction, it may be a worthwhile consideration.

    J.F.

  • Andrea Rossi

    Dear Mr Rick Meisinger:
    I am sure that the E-Cats will evolve, like the Ford T evolved in a 2011 NASCAR Ford.
    Thank you for your kind attention,
    Warm regards,
    A.R.

  • Rick Meisinger

    Dear Andrea Rossi:

    Do you think that the current E-Cats are obtaining the peak efficiencies of this technology or do you believe that further refinements will bring even higher performance in the future? I understand if you decide not to answer a speculative question.

    With much support;

    Rick Meisinger – USA

  • Alessandro Casali

    thank you mr. Rossi for the explanation,

    from simple calculation i assume there will be roughly 400 running e-cats and i guess there will also be another undefined number of back-up e-cats, am i correct?

    warm regards,

    AC.

  • Andrea Rossi

    Dear Mr Alessandro Casali:
    It seems to me that there is no contrast; I repeat: 300 was an approx number, the reactors will be more than this number.
    Warm Regards,
    A.R.

  • Alessandro Casali

    sorry Mr. Rossi but the following statements are in open contrast, i want to trust you but i think we all need some clarification on this matter.

    Andrea Rossi
    May 4th, 2011 at 5:06 AM
    Dear Luke Mortensen:
    1- up to now we have in operation 170 modules of the 300 that will compound the 1 MW plant.
    2- Thank you: You cannot imagine how much in this moment I need moral sustain.
    Warm regards,
    A.R.

    Andrea Rossi
    May 25th, 2011 at 9:07 AM
    Dear Mr Alessandre Casali: you are right, 300 is not the exact number, is an order of magnitude. Actually, the configuration of the 1 MW plant is complex, and confidential. I can’t give details: of course you are right: 2,5 x 300 makes 750. We will produce 1,000 and the Ecats inside are much more than the necessary number…
    Warmest Regards,
    A.R.

    Best Regards,
    AC

  • Andrea Rossi

    Dear Mr Johan Gustafson:
    Not yet. We are stress testing all the modules that will compound the 1 MW plant we are manufacturing. In past we used a module to heat our offices ( please see the Focardi-Rossi paper on the Journal of Nuclear Physics): we made it to make R&D, but also saved thousands of euros to heat our offices.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Mr Raffaele:
    No, in the worst case you have to supply 167 kWh/h to produce 1 MWh/h.
    The efficiency of the Carnot cycle or of any other cycle is the same of any other system, it does not depend from us.
    Warm Regards,
    A.R.

  • Raffaele

    Quindi nel caso peggiore dovrei immettere 1,79 MWh elettrici. Ora se volessimo convertire in elettrici i 10,75 MWh termici, con un rendimento ipotizziamo del 50%, otterrei 5,37 MWh elettrici. Sarà questo lo scenario dell’e-cat in versione “elettrica” ? Oppure il rendimento sarà diverso? Grazie

    So in the worst case I should enter 1.79 MWh of electricity. Now if we convert 10.75 MWh to electricity, assuming an efficiency of 50%, we would get 5.37 MWh of electricity. Is this the scenario of the “electric” e-cat version? Or the efficiency will be different? Thanks

  • Johan Gustafson

    Dear Mr. Rossi
    It seems that one can easily control the heat and a heating system, in a house, with an e-cat.
    Is there currently buildings whose heating system powered by an e-cat? And if so, how many such test facilities are connected and run against a heating system today?

    varm regards
    J Gustafson

  • Andrea Rossi

    Dear Mr Raffaele:
    We guarantee a factor K+6 as a minimum limit.
    Warm regards,
    A.R.

  • Raffaele

    Caro ing. Rossi, complimenti!
    Dato che in 4300 ore (6 mesi) l’e-cat da 2,5 kWh produce 10,75 MWh termici,
    vorrei sapere quanti kWh elettrici è necessario fornire in ingresso durante queste 4300 ore.
    Grazie e complimenti di nuovo!

    Dear Mr. Rossi, congratulations!
    We know that in 4300 hours (6 months) the 2.5 kWh e-cat produces 10.75 MWh (thermal),
    I would like to know how many electrical kWh is necessary to input during those 4300 hours.
    Thanks and congratulations again!

  • Richard Brown

    Mr. Rossi,

    An even better way to convert thermal to electric energy can be found here –
    http://esciencenews.com/articles/2011/05/16/new.solar.product.captures.95.percent.light.energy

    Best Wishes.

    Richard

  • Andrea Rossi

    Deaqr Mr Italo A. Albanese:
    Interesting. This is a very interesting line of research, I always believed in it, also if now I got no time for this research. But as soon as a product will be ready, I will test it coupled with my toy.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Mr Alessandre Casali: you are right, 300 is not the exact number, is an order of magnitude. Actually, the configuration of the 1 MW plant is complex, and confidential. I can’t give details: of course you are right: 2,5 x 300 makes 750. We will produce 1,000 and the Ecats inside are much more than the necessary number…
    Warmest Regards,
    A.R.

  • Alessandro Casali

    Dear Mr. Rossi,

    Thanks in advance for your patience, i have a simple question that i’m sure you can clarify.

    You have recently stated that each signle e-cat (in the 1 MW Greek plant) will produce 2.5 kW so how can you reach 1 MW with only 300 e-cats? the total energy whould be 750 kW.

  • Italo A. Albanese

    About thermoelectric generators, I fond an interesting article on technology review. http://www.technologyreview.com/energy/37621/?p1=A3

  • David Roberson

    Dear Mr. Rossi,

    Congratulations! Thank you for the prompt answer and I wish you good fortune as you proceed in your endeavor. It is exciting to bear witness to a truly world changing event.

    Warm Regards,

    D.R.

  • Andrea Rossi

    Dear Mr David Robertson:
    Our E-Cats are perfectly reproducible and their performance is standard. Out of 100 we guarantee that 100 respect the performance data.
    Should not be so we could not go deliver in Greece a 1 MW plant, made by 300 E-Cats.
    Warm Regards,
    A.R.

  • David Roberson

    Dear Mr. Rossi,

    I appreciate the timely response you submitted to my last question. It appears that you have achieved fantastic results with your invention and I expect to see the skeptics run quickly toward the exit when you complete the large project(1 Megawatt) for your first customer.

    One question which has always been a serious issue with LENR may be answered by you at this point. How consistant is your ability to create large amounts of excess power? For example, if you made 100 E-Cat devices and began testing, how many would achieve the desired ratios of output to input power? My gut feeling is that your answer would be greater than 90, which would be wonderful. As you know, the original work in LENR was shot down because of the lack of abilitiy to reproduce results. I am sure many other science minded followers like myself would appreciate your kind response.

    Thank you for your consideration,
    D.R.

  • Guy Ben-Zvi

    Thanks to all those who sent me the paper!
    I checked and the downloading problem happens in Internet Explorer 8. This problem is not only for me but for others also.
    The paper downloads fine in Chrome and Safari.
    Guy

  • Joseph Fine

    Guy Ben-Zvi

    I believe this is the link to the report you are seeking.

    The Villa paper is embedded inside of this document.

    Sorry to post it here, but others might not be able to get it for some unknown reason.

    http://www.journal-of-nuclear-physics.com/files/Levi,%20Bianchini%20and%20Villa%20Reports.pdf

    Best regards,

    Joseph Fine

  • @ Guy Ben-Zvi
    I have sent you a copy of the report.
    Anyway, there is a problem with Firefox 4 I noted. You can not open the report with FF4. You can save it and open it with Adobe Acrobat Reader.

  • Andrea Rossi

    Dear Mr Guy Ben-Zvi:
    We have controlled, our site works regularly. Our infiormatic assistant told us that you probably have a browser problem.
    Sorry for this,
    Warm regards,
    A.R.

Leave a Reply

You can use these HTML tags

<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>