Theoretical feasibility of cold fusion according to the BSM

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by
Stoyan Sarg Sargoytchev
York University, Toronto, Canada

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Abstract
Advances in the field of cold fusion and the recent success of the nickel and hydrogen exothermal reaction, in which the energy release cannot be explained by a chemical process, need a deeper understanding of the nuclear reactions and, more particularly, the possibility for modification of the Coulomb barrier.

The current theoretical understanding does not offer an explanation for cold fusion or LENR. The treatise “Basic Structures of Matter – Supergravitation Unified Theory”, based on an alternative concept of the physical vacuum, provides an explanation from a new point of view by using derived three-dimensional structures of the atomic nuclei.

For explanation of the nuclear energy, a hypothesis of a field micro-curvature around the superdense nucleus is suggested.
Analysis of some successful cold fusion experiments resulted in practical considerations for modification of the Coulomb barrier.
The analysis also predicts the possibility of another cold fusion reaction based on similarities between the nuclear structures of Ni and Cr.

617 comments to Theoretical feasibility of cold fusion according to the BSM

  • Steven N. Karels

    Andrea Calaon,

    I see that silicon is present in the ash. Could the following reaction be possible?

    27Al + 7Li + e -> 28Si + 6Li

    This would effectively remove the aluminum from the fuel and leave the silicon in the ash. Note the 28Si is major isotope in naturally occurring silicon.

    If the above reaction is possible, then the reaction within the eCat could be explained as well as the total energy output for one gram of fuel estimated and the percentage of fuel expended computed.

  • Herb Gillis

    Andrea Calaon:
    In your response to my last question you said:

    “The LENR do not take place between ANY of the nuclei present in a reacting powder/surface/whatever. The access to the reaction is controlled by chemical properties, not nuclear properties. So even admitting that any neutron rich isotope (apart from Li7) could work as a donor, then you would face the problem of having it react in the LENR.”

    1) Do you have any ideas on what the relevant chemical properties are (for accessing LENR)?

    2) If it turns out that only Nickel [or Ni/Li] has these chemical properties; then do you think it might be possible to use Ni [or Ni/Li] as a matrix alloy (ie. solid state “solvent”) for promoting neutron transfer reactions between other combinations of nuclei?

    Thank you for your insightful remarks.
    Regards; HRG.

  • Andrea Calaon

    Dear Joe,
    electron, proton and neutron are not points, they have intrinsic sizes.
    The wave function of all s orbitals overlaps significantly with the nucleus. But electrons do not fall into the nuclei of atoms. Fortunately 🙂
    If you would like to visualize an electron (I am not offering you the perfectly canonical description of the electron … ) imagine a point charge that runs along a helical trajectory at the speed of light. The diameter of the trajectory is fixed: 386 [fm]. And the frequency of the circular component of the motion is fixed as well: about 2.47E20 turns per second. Very quick indeed! The nature of the particle has to do with these fixed parameters. So that you can not have an electron without them. Now the radius of the hydrogen atom (as the most probable distance between the proton and electron in a hydrogen atom in its ground state) is 52.9 [pm]. Therefore the size of the electron is about 0.36% that size. Not a point that would fall onto the nucleus, nor something as big as the orbital.
    The best equations we have for the electron describe how the plane of the intrinsic rotation evolves (for dynamical conditions) or how is distributed on average in stationary conditions (like an atomic orbital).
    The problem of the precise size of the proton arose for an experiment where the size of the proton is estimated thanks to the interaction between an orbital and the nucleus. The experiment uses muons instead of electrons only because they, having a mass 207 times that of the electron, form orbitals that are 207 times more tight around the nucleus than an electron does. And the ratio between the size of the orbital and that of the nucleus is smaller.
    See for example: http://phys.org/news/2013-01-physicists-surprisingly-small-proton-radius.html

    Best Regards
    Andrea Calaon

  • Robert Curto

    Dr. Rossi, GENeco is a company in the UK that has a Plant that can convert food waste, and human waste, to provide fuel to power 8,500 homes, as well as to provide fuel for a Bio-Bus.
    With one tank of fuel the Bio-Bus can travel 200 miles, and emit 30% fewer emissions then a Diesel Bus.
    Google:
    GENeco
    Click on:
    GENeco
    Robert Curto
    Ft. Lauderdale Florida
    USA

  • Andrea Calaon

    Dear Herb Gillis,
    The only energetically possible neutron swap reaction with Ni64 acting as a donor is this:
    Ni64 (+ e) + Ni61 -> Ni63 (+ e) + Ni62 + 0.94 [MeV].
    I think however that Li7 acts as a donor in the LENR because it has very special nuclear properties, not found elsewhere.

    Ni64 is only 0.9% of all natural Nickel atoms. So in any case its role can only be minor both energetically and isotopically.

    The LENR do not take place between ANY of the nuclei present in a reacting powder/surface/whatever. The access to the reaction is controlled by chemical properties, not nuclear properties. So even admitting that any neutron rich isotope (apart from Li7) could work as a donor, then you would face the problem of having it react in the LENR.
    As far as I know the Hot Cat is the first device that seem to be based on a neutron swap mechanism activated by the LENR.
    Regards,
    Andrea Calaon

  • Andrea Calaon

    Dear Steven N. Karels,
    I agree with you, probably some grains or some other parts of the powder reacted fully, some others, not measured, much less.
    If 6Li can be turned into 7Li:
    Li6 + e + p -> Li7 + neutrino + (max) 6.47 [MeV]
    then hydrogen has a role, and turns into a neutron (together with one electron) first in this reaction. Then the neutron is transferred to xxNi.
    Therefore possibly it is not necessary to have all 7Li in the fuel powder at the beginning.
    Best Regards
    Andrea Calaon

  • Andrea Rossi

    Giuliano Bettini:
    As you know, in our laboratory we have analysed all the claims of the competitors and reproduced their apparatuses. We found one that works. I already spoke of it, but it is not correct that I speak on his behalf.
    I suppose publications will follow. For now, I just have to say, honestly, that this competitor of us has made a good job.
    Warm Regards,
    A.R.

  • Andrea Rossi

    JYD:
    Thank you for this interesting information.
    Warm Regards,
    A.R.

  • Giuliano Bettini

    Dear Andrea,
    extremely interesting the news that some lab was able to replicate the Rossi Effect (even in a minimal part) .
    A question, if I may:
    1. which lab?;
    2. is it “excess heat” (generally speaking)? or
    3. specifically what you call “Rossi Effect”?
    Thanks, kind regards,
    Giuliano Bettini

  • JYD

    Dear Andrea

    It could be the best friend for a Spatial HOT-Cat
    http://www.techno-science.net/?onglet=news&news=13372

    Futuristic regards
    JYD

  • Andrea Rossi

    Frank Acland:
    Of course I know also this paper that I received last week from Oscar Gullstroem (I write Gullstroem because I have not the dieresis to put on the “o”). I am studying it since I received it. It is worthwhile the time to be studied carefully.
    Warm Regards,
    A.R.

  • Steven N. Karels

    Andrea Calaon,

    I have not had a chance to go over your numbers in detail. Given the problem of producing too much energy than the measured test energy is a better scenario than the situation of not being able to produce the measured amount of energy with the measured or estimated components in the fuel.

    Perhaps Dr. Storms concept of a Nuclear Active Environment (NAE) is applicable and the ash was from such an environment and all the nickel at that site was converted.

    My original posting that you responded to asked whether the produced and naturally occurring 6Li could be transformed into 7Li. I understand you said you think it could be so transformed. If this is correct, then the supply of 7Li is only limited by the amount of hydrogen present.

    The Laguno report did not say all the fuel was consumed nor give any indication that the reactor was nearing fuel exhaustion. So the measurement that the ash was fully transformed to Ni62 only tells us what happened at the local site where the ash was produced.

  • Frank Acland

    Dear Andrea,

    In case you are not aware, there is a new paper published by Carl-Oscar Gullstrom titled “Collective Neutron Reduction Model for Neutron Transfer Reaction”.

    He writes by way of introduction:

    “So I have improved the neutron transfer theory. In my first attempt the radiation was still a bit high but it is solved now. The trick is to not have high energy protons to drag out the neutrons but instead neutrons that are so low in energy that they can’t enter the nucleon but at the same time they could drag out more neutrons. If it is of interest I attached a document with some simple calculations.”

    Link:

    http://www.scribd.com/doc/247067779/Collective-Neutron-Reduction-Model-for-Neutron-Transfer-Reaction

    Kind regards,

    Frank Acland

  • Joe

    Andrea Calaon,

    How can an electron get so close to a nucleus, in order to form a pseudo-particle, without the electron being forced to enter the nucleus due to electrostatic attraction?

    All the best,
    Joe

  • Wladimir Guglinski

    orsobubu wrote in November 20th, 2014 at 8:59 AM

    Wladimir, is this interesting, about strong force?

    http://press.web.cern.ch/press-releases/2014/11/lhcb-experiment-observes-two-new-baryon-particles-never-seen
    ——————————————–

    Dear orsobubu
    many new unstable particles can be created.

    However, they represent NOTHING for the working of the universe.
    By using the properties of the particles (baryon number, lepton number, parity, strangeness, etc), it is possible to predict new particles, because those properties of the particles is decurrent from the laws of intereaction for the formation of new particles, composed by the agglutination of the elementary particles of the aether (electricitons and magnetons).

    Strong force must be actually a kind of dynamic gravity (the magnitude of the strong force interactions depends on the velocity of the particles which are interacting).

    In spite of the strong force (dynamic gravity) can be responsible for the agglutination of the quarks in order to form the proton and the electron, it does not means that the nuclei are bound via the strong force.

    regards
    wlad

  • Herb Gillis

    Andrea Calaon:
    Thanks for responding to my question in such detail.
    As a possible alternative explanation: Do you think it possible that the Ni64 may be acting instead as a neutron donor to one of the lighter Ni isotopes (ie. 58, 60, 61) via the same mechanism as Li7? If this is true then perhaps LENR reactions can be achieved between any pairing of a relatively neutron poor nucleous and a relatively neutron rich nucleous?
    Regards; HRG.

  • Andrea Calaon

    Dear Steven N. Karels,
    the reaction
    6Li + e + p -> 7Li
    is possible, for what I know. And, given that Lithium7 is able to couple with the electron in the stimulated Hot-Cat powder, because we know it most probably reacts with Nixx, I would guess that Li6 should react as well. The magnetic dipole moment of Li6 is +0.8220.. [muN] whereas that of Li7 is 3.2564… [muN], therefore my theory would suggest that in the same conditions, Li6 should react significantly slower than Li7. And Li7 should do something like:
    7Li (+ e) + p -> 2He4 (+ e) + 16.84 [MeV]
    Another point is the abundance of protons in the Nickel metal structure. Is their number high enough to make this reaction “visible” among the neutron swap?
    The experimental results seem to suggest that Li can play the role of an interstitial like the proton. A LiH substructure in the Nickel? I really do not know.

    Checking today the data of my “energy analysis” of yesterday, I noticed a mistake in summing the number of atoms of Ni in the isotopic shift chain.
    I will not repeat the whole thing, but just give the (hopefully) right and important numbers:
    As a reference one gram of natural Nickel contains:
    6.985E21 nuclei of Ni58
    2.691E21 nuclei of Ni60
    1.170E20 nuclei of Ni61
    The total number of single one neutron shifts for a complete forward shift to Ni62 in one gram of Ni is 6.1377E22.
    The experimental average energy of a unitary Nickel forward shift reaction, would be around 1 [MeV]. Far too low.
    These data, together with the energies of the Ni isotopic shifts obtained via neutron swap with Li7 given yesterday, say that:
    A complete isotopic forward shift of Ni58, 60 and 61 to Ni62 of 0.55 [g] of Nickel would liberate 3.757 [MWh]. It is 2.5 times the energy measured during the test.
    For a 1.5 [MWh] are enough 0.22 [g] of natural Nickel, plus 0.17 [g] of natural Lithium.
    The minimum ratio between the weight of Lithium and the weight of Nickel in the powder for guarantying a complete isotopic shift of Ni is 77.4%.
    These corrected data say that the discrepancy between the energy measured and the isotopic and weight analysis is even wider than guessed yesterday. Possibilities:
    The shifts are due to a different reactions with an even lower energy. Does it exists?
    The estimated quantity of Ni in the charge was wrong. Ni was slightly more that 0.22 [g] and it underwent an almost complete isotopic shift.
    The sample showed a complete isotopic shift, but the value was not representative for the whole ash. In reality only 40% of the Nickel particles shifted completely while the others did not react. The non-reacted part was not present in the analyzed grains.
    What do you think?

    Best Regards

    Andrea Calaon

  • Andrea Calaon

    Dear Herb Gillis,
    Ni64 can not be given a neutron from Li7 because the reaction
    Ni64 (+ e) + Li7 + 1.15 [MeV] -> Ni65 + Li6 (+ e)
    requires 1.15 [MeV] to take place and apparently there are no such energetic photons around.
    Neither it is possible for Ni64 to shift “back” and lose one neutron to a Li6:
    Ni64 (+ e) + Li6 + 2.41 [MeV] -> Ni63 + Li7 (+ e).

    Therefore the disappearance of this Nickel isotope must happen in a different way.

    Let me spend a few words to advertise my theory :).
    In the reactions above I explicitly added (+e) because I believe that the “secret” of the LENR is a coupling with the electron. Li7 in a uncommon “physical-chemistry” event in the metal matrix, couples with one electron becoming a sort of “new particle”: Li7e. Then this pseudo-particle, which is neutral already at picometric scales, can easily couple (through the same mechanism) with a Ni isotope: Li7eNixx. Li7 and Nixx become forced to travel inside the “circular” electron potential well. Soon they reach “nuclear contact” (at 2-3 [fm]) with very low excess kinetic energy, and can exchange the neutron because it is energetically convenient and probably Li7 offers it on the plane orthogonal to its magnetic moment, right where Nixx can easily “grab it”.

    Back to Ni64
    Remember that Andrea Rossi for a certain time (about 2011-2013) stressed that he had a way to enrich isotopically Ni in number 62 AND 64. Now we know something of what that process is. However that process in its present form seems to consume Ni64 as well.
    Since I believe the mechanism at the base of LENR is always the electron coupling, Ni64 couples with an electron that is itself coupled with another nucleus. In this case most probably the other nucleus is a proton, and the most likely result is:
    Ni64 (+ e) + p -> Cu65 (+ e) + 6.94 [MeV]
    In a small percentage of the reactions the electron crosses the two nuclei right while they are reacting and takes part in the nuclear reaction. This leads to the production of Ni65, which decays beta with a half life of about 2.5 [h]:
    Ni64 + e + p -> Ni65 + neutrino + 5.32 [MeV]
    Ni65 -> Cu65 + e + antineutrino + (max) 2.138 [MeV] (beta decay)

    With the tiny charge the presence of the second process and its beta decay should almost be undetectable (but I haven’t done the numbers).

    I will later answer to Steven N. Karels as well.
    Thank you all for these interactions

    Regards
    Andrea Calaon

  • Andrea Rossi

    Marco Serra:
    Yes.
    Warm Regards
    A.R.

  • Andrea Rossi

    Italo R.:
    Thank you for the information.
    Warm Regards
    A.R.

  • Steven N. Karels

    Andrea Calaon,

    An interesting analysis. Thank you. Please consider if the reaction
    6Li + e + p -> 7Li
    is possible.

    If this were to occur, would not the available amount of 7Li increase?

    Perhaps the rate equations favors the 7Li + Ni over the 6Li -> 7Li to establish the ash lithium isotropic ratio?

    So what would happen is the 7Li fuses with the Ni to become 6Li and the next Ni isotope. Then occasionally an hydrogen nucleus fuses with a 6Li to replenish the 7Li nuclei. And the reaction continues until the hydrogen is depleted. Since there are four times as many hydrogen atoms in LiAlH4 than lithium atoms, the “fuel” is larger than you assumed.

    Thoughts?

  • Marco Serra

    Dear Andrea,
    you said that “We cannot feed more information to our competition, which now is very powerful”. My question is: how can any competitor be powerful without knowing the core effect that drive the ECat ? Do you know of any lab that succeded in replication of the Rossi Effect even in a minimal part ?

    God bless you
    Marco

  • Andrea Rossi

    Dr Joseph Fine:
    Thank you: fantastic translation. Happy Thanksgiving (in advance) to you and all our American Readers.
    Warm Regards,
    A.R.

  • Joseph Fine

    Andrea Rossi, Silvio Caggia,

    The Navajo message translates as “Thanks, Warm Regards.”

    ” Jo, jo, jo. ”

    Happy Thanksgiving (in advance),

    Joseph Fine

  • Andrea Rossi

    Frank Acland:
    We offered to all our Licensees to buy back from them the licenses. Some of them have accepted the offer, and signed an Agreement that is under NDA, some preferred to hold the licenses: obviously, the licensees that preferred to hold the licenses have continued their Licensee status.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Bernie Koppenhofer:
    No, it is not true. The point is not the fuel.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Steven N. Karels:
    No, I cannot comment.
    Warm Regards,
    A.R.

  • Steven N. Karels

    Dear Andrea Rossi,

    Can you comment on the accuracy of The Report regarding the fuel composition?

    a. Was LiAlH4 used as the source of hydrogen?
    b. Was the nickel content in the fuel 55% by weight?
    c. Was the lithium content of the fuel higher than The Report estimated?

  • Wladimir Guglinski

    Andrea Calaon wrote in November 19th, 2014 at 10:35 AM

    I am convinced that these are the reactions Andrea Rossi is investigating (remember his excitement at the article speaking about nucleus tunneling between Lithium7 and Nickel?). Note that, if the nuclear binding energy is not related to the nuclear force (as suggested by many and as in my theory), these reactions do not even involve the Weak interaction: They are purely electromagnetic.
    Here are the reactions and the energies:
    Ni58+Li7 ->Ni59+Li6 + 1.749 [MeV]
    Ni59+Li7 ->Ni60+Li6 + 4.138 [MeV]
    Ni60+Li7 ->Ni61+Li6 + 0.570 [MeV]
    Ni61+Li7 ->Ni62+Li6 + 3.346 [MeV]

    =================================================

    Dear Andrea Calaon
    Actually we have to be astonished with the question: why did not the nuclear theorists realize 80 years ago that strong nuclear force cannot be responsible for the nuclear binding energy of the nuclei??? Because if the strong force was interaction which responsible for the attraction proton-neutron and neutron-neutron, then the dineutron would exist in the nature.
    Two neutrons linked by the strong force cannot be separated by the isospin proposed by Heisenberg, because only a force of repulsion would be able to win the attraction by the strong force between two neutrons, and an abstract mathematical concept as the isospin cannot create a force of repulsion.

    So, from a simple question of logic, the strong force must be discarded as the responsible for the attraction proton-neutron and neutron-neutron within the nuclei.

    But the question is not so easy as it seems.
    By considering the Coulomb repulsion in the distances of 1fm between protons and neutrons within the nuclei, the electromagnetic interaction is not able supply the necessary force for the agglutination of the stable nuclei. There is need an interaction 100 times stronger than that promoted by the electromagnetic interaction in a distance of 1fm, and this is the reason why the nuclear theorists had discarded 80 years ago the electromagnetism as the promoter of the nuclear binding energy, and they had adopted the strong nuclear force, which interaction is 100 times stronger than the electromagnetism in the distance of 1fm.

    And now finally the E-Cat is showing what the logic was suggesting to us, when we had faced the obvious: as two neutrons do not form the dineutron, then the strong nuclear force cannot promote the agglutination of the nuclei.
    And the consequence: the nuclear theorists will be obliged to accept this unavoidable fact.

    Nevertheless a problem arises: as the nucleus is not bound via the strong nuclear force, but in the distances of 1fm there is need a force 100 times stronger than that promoted by the electromagnetism, how can the nuclei be bound via the electromagnetism?

    Obviously an acceptable new nuclear model must be able to explain such paradox, and the nuclear theorist will discard the theories which do not solve the puzzle.

    You said: “as suggested by many and as in my theory”.
    However, how do you (and the many) explain it ?

    .

    In the nuclear model proposed in my Quantum Ring Theory the puzzle is solved as follows:

    1) The nuclei are surrounded by an electric field.
    See Figure 1:
    http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png

    2) Suppose a proton fuses with a nucleus. The fusion occurs as explained ahead.

    3) The proton must perforate the electric field of the nucleus, so that to be captured by the nucleus.

    4) When the proton perforates the electric field of the nucleus and they have fusion, the electric field of the nucleus has no repulsion with the electric field of the proton, because the two electric fields fuse by forming one unique electric field, surrounding the nucleus and the proton. By this way, the proton is not submitted to that Coulomb repulsion considered in the Standard Nuclear Model, in the order of 100 times stronger than the electromagnetic interaction.

    5) The equilibrium of the newborn nucleus formed by the proton+(original nucleus) is promoted via the equilibrium between the action of the centrifugal force trying to expel the protons and neutrons of the newborn nucleus and the magnetic force actuating on the protons.

    6) The stability of the light nuclei via equilibrium between magnetic force and centrifugal force is shown and calculated in my paper Stability of Light Nuclei, published in the JoNP:
    http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf

    .

    As the nuclear theorists are now accepting the reality of the cold fusion, thanks to the performance of the E-Cat, sure that they will realize that there is need a new nuclear model for explaining the results obtained by Andrea Rossi.

    And as the results obtained by the E-Cat are pointing out that the strong nuclear force cannot be the responsible for the agglutination of the nuclei, then facing the question on what will be the new nuclear model to be chosen (between the many new nuclear models proposing the electromagnetism as the cause of agglutination of the nuclei) of course they will consider only those models capable to explain the puzzle:
    how can the electromagnetism to promote the agglutination of the nuclei, since there is need a force 100 times stronger than that promoted by the electromagnetism?

    The best new nuclear model able to solve the puzzle will be chosen.

    Regards
    wlad

  • Andrea Calaon

    Dear All,
    if Andrea Rossi allows me I will shortly abuse of the JoNP for a personal message.
    Immediately after my last post, despite the grammatical and lexical mistakes, many peopled re-accessed the site where I have my theory written down.
    I would like to stress that the documents in my theory-site are not up-to-date with the latest “news and changes”. Still new things appear on a daily basis, therefore I will wait first the ideas to settle a bit and then I will write them.
    When a complete review of my theory will be ready I will communicate it in one of my messages to the JoNP.
    Best Regards,
    Thank you Andrea

  • Bernie Koppenhofer

    Dr. Rossi: It has been suggested safety certification will be a lot easier for the Gas Cat than for the Electric Cat. Is this true?

  • Frank Acland

    Dear Andrea,

    You mention Leonardo Corporation bought back some licensing contracts from licensees. Going forward, are there any licensees that will continue with their licensee status?

    Many thanks,

    Frank Acland

  • Herb Gillis

    Andrea Calaon:
    How does your proposed reaction mechanism explain the observed reduction in the concentration of Ni64?
    Regards; HRG.

  • Joseph Fine

    Andrea Rossi, Silvio Caggia.

    I have not translated the earlier comment made using the Navajo Codetalkers ‘dialect’.

    http://asecuritysite.com/challenges/nav

    Best regards,

    Joseph Fine

  • Andrea Rossi

    Andrea Calaon:
    I appreciate your efforts. Obviously I cannot comment.
    Warm Regards,
    A.R.

  • Andrea Calaon

    Dear Readers of the JoNP,
    Andrea Rossi a few days ago, commenting the recent exchange of ideas about reactions that can justify the large isotopic shifts in Nickel and Lithium, said:
    “The contradictions or errors possibly emerging from such kind of comments or articles cannot be commented by me”.
    In the “comments” of the Readers (like myself) for sure there are mistakes. But what is Andrea referring to with the word “articles”?
    My guess is that if he could he would have criticized the conclusions of the report regarding the total content of Li and Ni in the powder and in the ash (may be not only these …).
    The ICP-AES sample was 0.21% of the total powder and ash. Since the powder is a mixture of grains of different origin, the sample could well be not representative of the whole population. And so the real total content of Li7 could have been much more than 0.0117 [g] (1.17% of the charge powder), or conversely the Nickel content could have been much less than 0.55 [g].

    In the last few days I eventually extended the study of the reactions to the energy they produce, comparing it to the measured energy released in the experiment. Well, …. I am a bit late, now is one and a half month after the publication of the report. But better late than never. The conclusions I arrive at contradict some of my guesses so far. Fortunately they also strongly point towards interesting conclusions.
    Two starting points:
    – 1.5 [MWh] are equal to 3.37E22 [MeV],
    – The number of Nickel nuclei 58, 60 and 61 (all the forward shifting) present in the 0.55 [g] of Nickel (if the w% estimation is correct) are respectively: 3.84e21, 1.48e21, 6.433e19. (You have to actually to do some numbers from those in the report to get these). These values are obtained using the isotopic ratios.

    Thus the average energy of a unitary Nickel forward shift reaction, FORGETTING LITHIUM, would be around 1.83 [MeV] (you have to consider that the number of reacting nuclei of Ni60 in a complete shift is actually equal to Ni58 plus all initial nuclei of Ni60, and that Ni58 reacts twice since it has to shift by two neutrons, …). If Lithium were added as a separate shift the average value would decrease. 1.83 [MeV] is a particularly low value for isotopic shifts.
    Let us now compare this average energy with the energies that would be released by Nickel shifting separately from Lithium, and starting from lowest masses: father nucleus plus proton plus electron. Here they are:

    Ni58+e+p ->Ni59+neutrino+ (max) 8.22 [MeV]
    Ni59+e+p ->Ni60+neutrino+ (max) 10.61 [MeV]
    Ni60+e+p ->Ni61+neutrino+ (max) 7.04 [MeV]
    Ni61+e+p ->Ni62+neutrino+ (max) 9.81 [MeV]

    Well, it is not necessary to check that the average (in the sense explained above) energy of all these reactions is far above the average experimental value. An this is only for the Nickel isotopic shift, than you should add the reactions that would shift Lithium. Conclusion: in the Hot Cat used in the first 32 days these reactions are not taking place.

    Then let us look at the reactions in which Li7 swaps a neutron with the Ni isotopes. These should be the reactions providing the lowest possible energy, since they realize the two isotopic shifts without further father particles (energy).
    I am convinced that these are the reactions Andrea Rossi is investigating (remember his excitement at the article speaking about nucleus tunneling between Lithium7 and Nickel?). Note that, if the nuclear binding energy is not related to the nuclear force (as suggested by many and as in my theory), these reactions do not even involve the Weak interaction: They are purely electromagnetic.
    Here are the reactions and the energies:

    Ni58+Li7 ->Ni59+Li6 + 1.749 [MeV]
    Ni59+Li7 ->Ni60+Li6 + 4.138 [MeV]
    Ni60+Li7 ->Ni61+Li6 + 0.570 [MeV]
    Ni61+Li7 ->Ni62+Li6 + 3.346 [MeV]

    All these reactions, in the case of a complete forward isotopic shift of 0.55 [g] of natural Nickel would have liberated a total energy 2.248 [MWh]. This value is about 1.5 times the measured energy (with an almost complete isotopic shift towards Ni62. But is as near as I can get.
    I think there are no realistic reactions with lower released energy.

    My conclusions are now these:
    – The reactions that take place in the Hot Cat are those in which the neutron is exchanged between Li7 and Nickel.
    – Hydrogen would seem unnecessary, apart from its possible lattice distortion effects.
    – The reaction of Ni61 to Ni62 should be more “efficient” than the others because experimentally there was absolutely no Ni61 left, whereas all other shifts lead to Ni61 before the final jump to Ni62. In fact the theory I am proposing says that this reaction should be quicker than the others, since Ni61 is the only Nickel isotope that has a magnetic dipole moment.
    – I think that the sampling of the powder for the ICP-AES was not lucky and led to numbers that can be misleading. The almost total isotopic shift of Ni and the analysis above tell me that most probably the charge powder contained a lower % of Ni: more or less 0.37 [g] for each gram. The corresponding Li (with natural isotopic ratio) for a complete Nickel shift would be 0.156[g] for each gram of fuel. Of which Li7 would be 0.144 [g] and Li6 0.0118 [g]. Is looks like the quantity that has the weight indicated in the report (0.0117 [g]) is precisely Li6, and not Li7. The other 47.4% in weight of the charge is made of the other nuclei as described in the Report.

    I am convinced that who is studying this phenomenon for commercial purposes did similar analyses immediately after the report arriving to similar conclusions, independently from the theory applied.

    Best Regards

    Andrea Calaon

  • Andrea Rossi

    Silvio Caggia:
    Ugh!
    A.R.

  • Silvio Caggia

    Dear Andrea Rossi,
    Than-zie Lin Wol-la-chee Nesh-chee Klizzie-yazzi Dibeh,
    Gloe-ih Wol-la-chee Gah Na-as-tso-si Gah Dzeh Klizzie Wol-la-chee Gah Be Dibeh!

  • Andrea Rossi

    Daniele Passerini (blogger of “22 Passi”)
    You asked me few days ago about why some of our commercial Licensees have cancelled their websites. The reason is that we decided to offer to all our commercial Licensees to buy back their licence at a price, obviously, superior to the price they paid for it. Some of our Licensees have accepted our proposal and sold us back their license.
    The details of the agreements are covered by NDA ( Non Disclosure Agreement).
    We maintained with our former Licensees a friendly and collaborative relationship, open to the possibility of future collaboration upon specific issues.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Buck:
    As I wrote on this blog one hour ago, now the competition is very serious. Thanks to the work of my Team, LENR, that 5 years ago were very “low”, not only in temperature, but also in global consideration, have gained momentum at high level. My Team merits recognition for this: our action and our fight have been the real game changer.
    Warm Regards,
    A.R.

  • Buck

    The news about CF has reached a critical point as far as the Oil Industry is concerned.

    Gulf News, the largest English language newspaper in the Gulf region (UAE, Dubai, Oman, Bahrain, Qatar, Kuwait, Yemen, and Saudi Arabia), with a daily circulation of about 110,000, has just reported that India is moving towards getting back into CF research. Your work, the China Nickel Energy connection in Baoding, and Bill Gates’ recent visit with Vittorio Violante were cited in a factual positive fashion.

    To me, the tone takes on an alarmist quality as it presents CF phenomena as fact and CF technology as imminent.

    LINK>> http://gulfnews.com/news/world/india/indian-government-urged-to-revive-cold-fusion-1.1413814

  • Andrea Rossi

    Silvio Caggia:
    Navajo dialect: between Navajo and Cherokees cultural exchanges were frequent. Please keep this a jealously guarded secret: must remain strictly between you and me.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Franco Sarbia:
    Gas fuel will substitute electric energy to activate the reactor and drive it; I cannot give more particulars until we will have a product ready for the market. We cannot feed more information to our competition, which now is very powerful. We need to reach extreme commercial competitivity before leaking more information. When we will have reached the necessary economy scale our prices will discourage any competition, but before that phase we must be aware of the fact that our Competitors are eating voraciously any single bit of information we are leaking.
    Warm Regards,
    A.R.

  • Franco Sarbia

    Dear Andrea Rossi
    1) In the gas fueled Hot Cat, the gas feeding performs a function of “starter” required for
    a) bring the system to operating temperature and make it independent from any network or external power supply?
    b) or also for operating at the regime?
    2) This means that, in both cases, the Hot Cat functioning at regime through its own generator can feed itself the reaction and its electronic control?
    3) The total autonomy in each stage would make safe and completely self-sufficient its cogeneration capacity for the production of electricity and heat even in hostile and remote environments, without any connection to wired networks, eliminating the supply of powerful, expensive, and dirty accumulators to start a Hot Cat System equipped with a battery of hundreds of modules, necessary to provide electricity and heat at homes and production activities of a small town or an urban neighborhood, both in developed countries and in the underdeveloped countries not yet equipped with wired electrical infrastructure. Can you confirm this strategic purpose?

  • silvio caggia

    Dear Andrea Rossi,
    I understand that you can use electric current in order to “communicate” with the e-cat reactor in a sort of morse-code like:
    ———- (Prepare)
    ………. (Activate)
    ———- (Deactivate)
    But with gas e-cat what will you use? “Smoke signals” like native Cherokees? 😀

  • Andrea Rossi

    Eric Ashworth:
    1- yes
    2- no
    3- among others, a, not b.
    Warm Regards,
    A.R.

  • Eric Ashworth

    Dear Andrea, Your reply November 17th to Steven N. Karels. You say if the temperature reaches the safety limit the reactors turn off by a law of nature, whatever the source of the heat that causes a rise of the temperature.

    I have several questions regarding your answer:-

    1. If the reactors turn off by a law of nature/response mechanism can you be absolutely positive that a limit of safety has actually been reached?.

    2. Would you consider this law of nature to be a hindrance to the performance of LENRs?.

    3. Your ongoing R&D with regards the e-cat, is this with regards (a) the possible applications of the e-cat?. (b) An attempt to overcome this law of nature?. or
    (c) Both?.

    Regards Eric Ashworth.

  • Wladimir Guglinski

    Errata:

    In my last comment,

    instead of the relative kinetic energy of the deuteron regarding the 3Li7 is E= 0,5.m.(V² – v²)

    actually the relative kinetic energy is E= 0,5.m(V-v)²

    regards
    wlad

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