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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

 

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

  • Andrea Rossi

    Jean Pierre:
    He,he,he…thanks for your attention. As a prize, here is the update as of 8 a.m. of Saturday July 18th: 1MW E-Cat stable; Hot Cat: I shut it down, because new inventions have to be applied to the apparatus, after the meditations made observing it in the last months; within a week a new version will be restarted.
    Warm Regards,
    A.R.

  • Thanks so much for the clarification, Andrea !
    I am a happy bunny (rabbit) now.

    Jean Pierre

  • Andrea Rossi

    Jean Pierre:

    1- the 1 MW E-Cat IS made entirely with LT (= low temperature) modules
    3- the 250 kW E-Cats ARE low temperature modules too
    Warm Regards,
    A.R.

  • Hi Andrea again.
    Sorry to query your reply to me.

    You said ‘yes’ to both items 1 and 3.

    It is not possible for the current 1MW plant in the customer’s factory to be made up from

    1) entirely (100%) LT e-cats and also

    3) a mixture of LT cats and 250kW cats

    It is one or the other. Please clarify for me .

    Thanks and apologies for disturbing you.

    Best wishes. Jean Pierre

  • Andrea Rossi

    Jean Pierre:
    1- yes
    2- no
    3- yes
    Thank you for your kind words,
    Warm Regards,
    A.R.

  • Dear Andrea.
    Please forgive me if the following question seems somewhat naif.

    With regards to the 1MW plant being tested this very day in the factory of your customer, would you please tell me the current configuration of the plant? Is it made up;

    1) entirely from LT cats?

    2) entirely from 250kW cats?

    3) from a mix of the two types of cat?

    Thank you for taking your valuable time to answer.
    Keep well and keep going, my friend. I can see that you are giving 100% to this very important project in order to make a breakthrough for the world. I don’t think that you can work harder.

    Warmest regards. Jean Pierre

  • Andrea Rossi

    Jean Pierre:
    1- yes
    2- not yet utilized, still unerd study
    3- deal!
    Warm Regards,
    A.R.

  • Hi Andrea,

    1) Are there any encouraging signs in the development of
    G-Cat (G=Gas) by the other half of your team ?

    2) Is the jet engine, that you bought, being utilised or
    has it been mothballed for the time being?

    3) I am trying to lose weight, and you need to gain some.
    How about a trade?

    Keep well and continue to enjoy your very valuable work.

    Jean Pierre

  • Andrea Rossi

    Jean Pierre:
    Thank you for the wishes and for your insight,
    Warm Regards,
    A.R.

  • Hi Andrea,
    A belated Happy Birthday! I have been away; sorry for my omission.

    The idea of augmenting a fossil- fuelled power station with a successful 1 MW LENR plant overcomes the need for a separate electrical storage device as a ‘direct- feed- back’ buffer to run the ecat, since a small part of the station’s electrical output can be fed back to the 1MW input in a direct and controlled amount; see next paragraph for further explanation. One feedback line could supply the necessary heating while a second would be dedicated to the electronic control system.

    Now for the explanation mentioned above. There should then be no concern for thermal runaway and safety because the ecat will not be feeding itself directly but borrowing from the station’s total output, the vast majority of which comes via the fossil fuel. One could imagine this station’s fossil fuel output portion to be a sort of storage device in itself—but one like a non- static electricity reservoir; an in-built, DYNAMIC storage device in fact.

    In the past, you seemed to be against feeding back part of the ecat’s output energy to its own input directly without some kind of storage buffer. Are you still thinking the same way now or will your control system be effective enough for all undesirable and unexpected conditions when you manage to generate ecat electrical energy and use part of it to run the system?

    I visualise that in the future the LENR contribution to the station’s input could be increased in steps, thus gradually reducing the huge amounts of fossil fuel currently needed.

    In the power station example above, if the 1MW had a mind of its own and was unaware of all the other machinery it would conclude that it was entirely self-sufficient, including its own start up.

    Happy birthday again. Jean Pierre.

  • Andrea Rossi

    Keriusene:
    Today we had a problem, due to series of gaskets that turned out to be defectous; we had to resolve the problem without turning out the plant. Got some trouble. Now is fixed.
    Great Team!
    Warm Regards,
    A.R.

  • keriusene

    Dr Rossi:
    Can you give an update ? How id going the 1MW E-Cat today?

  • Andrea Rossi

    Keriusene:
    Are you a ghostbuster?
    Warm Regards,
    A.R.

  • keriusene

    Dr Andrea Rossi:
    Voices are around about the fact that other very, very important replications have been made of the Lugano Report in Europe, after the replicas of Dr Parkhomov. Voices say that anomalous excess of heat has been measured independently from a group of scientists, well known, who have built a reactor copied from the Lugano report, as well as the charge, and they got an excess beyond any doubt! Do you have comments?

  • Andrea Rossi

    Keriusene:
    Sorry, I cannot give this information.
    Warm Regards,
    A.R.

  • keriusene

    Dear A.R.:
    How did you resolve the problem of dendrites in your Li based charges?

  • Andrea Rossi

    Keriusene:
    I will finish to write it within a couple of weeks ( I am using the nights inside the E-Cat, so I hope not to have troubles), then it will be peer reviewed: I hope to have it published in a month or two.
    Warm Regards,
    A.R.

  • keriusene

    Dr Rossi:
    You said you are writing a paper with calculations regarding the reverse Mossbauer Effect to explain the Rossi Effect as it has been observed in the measurements after the Lugano test of the ITP: when will it be published?

  • Andrea Rossi

    Keriusene:
    Here is the link related to the 3D printing of a jet engine:
    http://monash.edu/news/.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Keriusene:
    Yes, I read about it on “NASA Tech Briefs” of April 2015. Very interesting, that is a 3D tech hat can really interest us for the manufacturing of the E-Cat, but it is a prototype made by the Monash University: in this case 3D printing builds up layer after layer of metal powder, and this is a big step forward after the manufacturing of things made by caedboard or plastic.
    We are interested to these developments.
    Warm Regards,
    A.R.

  • keriusene

    Dr Rossi:
    Have you seen the 3D jet engine made in Australia by Prof Xinhua Wu?

  • Andrea Rossi

    Melanie Lisk:
    The safety certification for the domestic units is on course.
    Warm Regards,
    A.R.

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