A detailed Qualitative Approach to the Cold Fusion Nuclear Reactions of H/Ni

By prof. Christos Stremmenos

After several years of apparent inaction, the theme of cold fusion has been recently revitalized thanks to, among others, the work and the scientific publications of Focardi and Rossi, which has been conducted in silence, amidst ironical disinterest, without any funding or support.  In fact, recently, practical and reliable results have been achieved based on a very promising apparatus invented by Andrea Rossi.  Therefore I want to examine the possibility of further development of this technology, which I deem really important for our planet.

Introduction
I will start with patent no./2009/125444, registered by Dr. Ing. Andrea Rossi. This invention and its performance have been tested and verified in collaboration with Prof. Sergio Focardi, as reported in their paper, published in February 2010 in the Journal of Nuclear Physics [1]. In that scientific paper they have reported on the performance of an apparatus, which has produced for two years substantial amounts of energy in a reliable and repeatable mode and they have also offered a theoretical analysis for the interpretation of the underlying physical mechanism.

In the history of Science, it is not the first time that a practical and reliable apparatus is working before its theoretical foundation has been completely understood! The photoelectric effect is the classic example in which the application has anticipated its full theoretical interpretation, developed by Einstein. Afterwards Einstein, Plank, Heisenberg, De Broglie, Schrödinger and others formulated the principles of Quantum Mechanics.  For the interactive Nickel/Hydrogen system it would be now opportune to compile, in a way easily understood by the non expert the relevant principles and concepts for the qualitative understanding of the phenomenon. Starting with the behavior of electrically charged particles in vacuum, it is known that particles with opposite electric charge attract themselves and “fuse” producing an electrically neutral particle, even though this does not always happen, as for instance in the case of a hydrogen atom, where a proton and a electron although attract each other they do not “fuse”, for reasons that will be explained later.   On the contrary, particles charged with electric charge of the same sign always repel each other, and their repulsion tends to infinity when their distance tends to zero, which implies that in this case fusion is not possible (classical physics).

On the contrary, according to Quantum mechanics, for a system with a great number of  particles of the same electric charge (polarity) it is possible that a few of them will fuse, as for instance, according to Focardi-Rossi, in the case of  Nickel nuclei in crystal structure and hydrogen nuclei (protons) diffused within it, Although of the same polarity,  a very small percentage of these nuclei manage to come so close to each other, at a distance of 10-14 m, where strong nuclear forces emerge and take over the Coulomb forces  and thus form the nucleus of a new element, either stable or unstable.

This mechanism, which is possible only in the atomic microcosm, is predictable by a quantum-mechanics model of a particle put in a closed box.  According to classical physics no one would expect to find a particle out of the box, but in quantum mechanics the probability of a particle to be found out of the box is not zero! This is the so called “tunneling effect”, which for systems with a very large number of particles, predicts that a small percentage of them lie outside the box, having penetrated the “impenetrable” walls and any other present barrier through the “tunnel”! In our case, the barrier is nothing else but the electrostatic repulsion, to which the couples of hydrogen and nickel nuclei (of the same polarity) are subjected and is called Coulomb barrier.

Diffusion mechanism of hydrogen in nickel: Nickel as a catalyst first decomposes the biatomic molecules of hydrogen to hydrogen atoms in contact with the nickel surface. Then these hydrogen atoms deposit their electrons to the conductivity band of the metal (Fermi band) and due to their greatly reduced volume, compared to that of their atom, the hydrogen nuclei readily diffuse into the crystalline structure of the nickel, including its defects. At this point, in order to understand the phenomenon it is necessary to briefly describe the structure both of the nickel atom and the nickel crystal lattice.

It is well known that the nickel atom is not so simple as the hydrogen atom, as its nucleus consists of dozens of protons and neutrons, thus it is much heavier and exerts a proportionally higher electrostatic repulsion than the nucleus of hydrogen, which consists of only one proton. In this case, the electrons, numerically equal to the protons, are ordered in various energy levels and cannot be easily removed from the atom to which they belong. Exception to this rule is the case of electrons of the chemical bonds, which along with the electrons of the hydrogen atoms form the metal conductivity band (electronic cloud), which moves quasi freely throughout the metal mass.

As in all transition metals, the nickel atoms in the solid state, and more specifically their nuclei, are located at the vertices and at the centre of the six faces of the cubic cell of the metal, leaving a free internal octahedral space within the cell, which, on account of the quasi negligible volume of the nuclei, is practically filled with electrons of the nickel atoms, as well as with conductivity electrons.

It would be really interesting to know the electrons’ specific density (number of electrons per unit volume) and its spatial distribution inside this octahedral space of the crystal lattice as a function of temperature.

Dynamics of the lattice vibration states
Another important aspect to take into consideration in this system is the dynamics of the lattice vibration states, in other words, the periodic three dimensional normal oscillations of the crystal lattice (phonons) of the nickel, which hosts hydrogen nuclei or nuclei of hydrogen isotopes (deuterium or tritium) that have entered into the above mentioned free space of the crystal cell.

It could be argued that the electrons’ specific density and its spatial distribution in the internal space of the crystal structure should be coherent with the natural frequencies of the lattice oscillations. This means that the periodicity of the electronic cloud within the octahedral space of the elementary crystal cell of Nickel generates an oscillating strengthening of shielding of the diffused nuclei of hydrogen or deuterium which also populate this space.

I believe that these considerations can form the basis for a qualitative analysis of this “NEW SOURCE OF ENERGY” and the phenomenology related to cold fusion, including energy production in much smaller quantities and various reaction products.

Shielding of protons by electrons
In the Focardi-Rossi paper the shielding of protons provided by electrons is suspected to be one of the main reasons of the effect, helping the capture of protons by the Ni nucleus, therefore  generating energy by fusion of protons in Nickel and a series of exothermic nuclear reactions, leaving as by-product isotopes different from the original Ni (transmutations). Such shielding is one of the elements contributing to the energetic efficiency of the system.  From this derives the opportunity, I think, to focus upon this shielding, both to increase its efficiency and to verify the hypothesis contained in the paper of Focardi-Rossi.  Of course, what we are talking of here is a theoretical verification, because the practical verification is made by monitoring the performance of the apparatus invented and patented by Andrea Rossi, presently under rigorous verification by many independent university researchers.

In my opinion, the characteristics of the shielding of the proton from the electrons should be defined, as well as the “radiometric” behavior of the system.

In other words, the following two questions should be answered:

  1. Which is the supposed mechanism that overcomes the powerful electrostatic repulse (Coulomb barrier) between the “shielded proton” and the Nickel nucleus?
  2. For what reason there is almost no radiation of any kind (experimental observation), while according to the Focardi and Rossi’s hypothesis there should have been some γ radiation (511 KeV) produced by the predicted annihilation of the β+ and β- particles that are being created during the Fusion?

I believe that some thoughts based on general and elementary structures, data and principles of universal scientific acceptance, might shed some light to this exciting phenomenon.  More specific, I refer to Bohr’s hydrogen atom, the speed of nuclear reactions (10-20 sec) and the Uncertainty Principle of Heisenberg.

I will take Bohr’s hydrogen atom as a starting point (figure 1a), which stays at its fundamental state forever in the absence of external perturbations, due to De Broglie’s wave, accompanying the sole electron.

As stated before, in contact with the metal, these atoms lose their fundamental state, as their electrons are being transmitted to the conductivity band.  These electrons, together with the “naked nuclei” of hydrogen (protons), form a freely moving cloud of charges (plasma at a degenerate state) inside the crystalline lattice. That cloud is being defused through the surface to the polycrystallic mass of the metal, covering empty spaces of the non-canonical structure of the crystalline lattice, as well as the tetrahedral and octahedral spaces between the molecules. As a consequence, the crystalline structure is covered by “delocalized plasma” (degenerate state), which is consisted by protons, electrons produced by the “absorbed atoms” of hydrogen, as well as by the electrons of the chemical valence of Nickel of the lattice, at different energy states (Fermi’s band). (Fig. 2)

Fig.1b

In this system, if one considers the probability of the creation inside the crystalline lattice of temporary (not at the fundamental state) “pseudo-atoms” of hydrogen with neutral charge, for example at a time of the order of 10ˆ-17 sec, then that possibility is not completely ill-founded. (Fig 1b)

Fig.2

According to the Uncertainty Principle of Heisenberg, the temporary atoms of hydrogen will cover during that small time interval Δt, a wide range of energies ΔΕ, which means also a wide range of atomic diameters of temporary atoms, satisfying the De Broglie’s condition.  A percentage of them (at fist a very small one) might have diameters smaller than 10ˆ-14 m, which is the maximum active radius of nuclear reactions. In that case, the chargeless temporary atoms, or mini-atoms, of hydrogen together with high energy but short lived electrons, are being statistically trapped by the Nickel nuclei at a time of 10ˆ-20 sec. In other words, the high speed of nuclear reactions permits the fusion of short lived but neutral mini-atoms of hydrogen with the Nickel nuclei of the crystalline lattice, as during that short time interval the Coulomb barrier (of the specific hydrogen mini-atom) does not exist.

Afterwards, it follows a procedure similar to the one described by Focardi and Rossi, but instead of considering the capture of a shielded proton by the Ni58 nucleus, we adopt the hypothesis of trapping a neutral temporary atom, or a mini atom, of hydrogen (with a diameter less than 10ˆ-14 m) which transforms the Ni58 nucleus into Cu59 (copper/59, short lived isotope*).

It follows the predicted “β decay” of the nuclei of the short lived isotope of copper, accompanied by the emission of β+ (positrons) and β- (perhaps the electrons of the mini atoms trapped inside that nucleus during the fusion). These particles are being annihilated with an emission of γ radiation (two photons of γ of energy 511 KeV each, for every couple of β+ and β-).

In other words, whoever has experimented with this system should have suffered the not-so-harmless influence of those radiations, but that never happened.  The radioactivity measured at the experiments is almost zero and easily shielded.

In any case, a rigorous, in my opinion, theoretical approach for the interpretation of that phenomenon with quantum mechanical terms, would give clear quantitative answers to the above stated models. With my Colleges of theoretical chemistry, we are already planning to face the problem using the time-depended quantum mechanical perturbation theory, bearing in mind the following:

  1. The total wave function (of the nucleus and the electrons) of temporarily, non-stable states.
  2. The total time-depended Hamiltonian, for temporarily states.
  3. Searching for the resonance conditions at that system.

Such an approach had a successful outcome at a similar problem of theoretical chemistry and we hope that it will be valid in this case as well.

Let’s go back to the intuitive, with ideal models, approach, in order to give a qualitative explanation for the (almost) absent radiations of the system, by using:

  • First of all the Boltzmann’s distribution (especially at the asymptotic area of high energies).
  • The photoelectric effect
  • The Compton effect
  • The Mössbauer effect

We have already mentioned that from the temporary mini atoms of hydrogen, the ones with diameter less than 10ˆ-14 m, have a larger probability of fusion. But, in order for them to be created, high energy bond electrons should exist at the “delocalized plasma” of the crystalline lattice.

1. Boltzmann’s statistics:
There are reasons to believe that the H/Ni system, at first at temperatures of about 400-500oC, contains a very small percentage of electrons in the “delocalized plasma” with enough energy to create (together with the diffused protons), according to the wave-particle duality principle, the first temporary mini atoms of hydrogen, that will trigger the fusion with the nickel nuclei and the production of high energy γ photons (511 KeV).

2. Photoelectric Effect:
It is not possible, the HUGE amount of energy (in kW/h), that the Rossi/Focardi reactor produces, as measured by unrelated scientists in repeated demonstrations (at one of them by the writer and his colleagues, Fig 3), to be created due to the thermalization of the insignificant number of  γ photons at the beginning of the reaction.

Fig.3

I believe that, as stated above, these photons are the trigger of fusion at a multiplicative series, based on the photoelectric effect inside the crystalline structure.

The two γ photons can export symmetrically (180°) two electrons from the nearest Nickel atoms. The stimulation, due to the high energy of γ, concerns electrons of internal bands of two different atoms of the lattice and has as a prerequisite the absorption of all the energy of the photon.  A small part of that energy is being consumed for the export of the electron from the atom and the rest is being transformed into kinetic energy of the electron (thermal energy).

The result of that procedure is to enrich the “delocalized plasma” with high energy electrons that will contribute multiplicatively (by a factor of two) at the progress of the cold fusion nuclear reactions of hydrogen and nickel and at the same time transform the hazardous γ radiation into useful thermal energy.

3. The Compton Scattering:
It gives the additional possibility of multiplication, this time due to secondary photons γ, in a wide range of frequencies, as a function of the angular deviation from the direction of the initial photon of 511 keV. That has as a result the increase of the export of electrons, due to the photoelectric phenomenon at the crystalline mass, in many energy/kinetic levels, which gives an additional possibility of converting the γ radiation into useful thermal energy.

4. The Mössbauer effect:
It gives another possible way of absorbing the γ radiation and transforming it into thermal energy. It is based on the principle of conservation of momentum at the regression of the new Cu59 nucleus/ from the emission of a γ photon. Relative calculations (Dufour) showed that this mechanism has an insignificant (1%) contribution.

It follows that, according to given data, the Photoelectric phenomenon and the Compton Effect, could explain the absence of radiations in the Focardi-Rossi system, which, from the amount of producing energy versus the consumption of Ni and H2, as well as from the experimental observation of element transformations,  lead undoubtedly to the acceptance of hydrogen cold fusion.

ACKNOWLEDGEMENTS: The author wishes to acknowledge Aris Chatzichristos for the contribution in formulating this paper in English

References:
(1)www. journal-of-nuclear-physics.com /Focardi Rossi/  (A new energy source from nuclear fusion)

* I believe that the phasmatometric tracing of copper is the most definitive sign of nuclear fusion: From the relative bibliography (HANDBOOK OF CHEMISTRY AND PHYSICS, 66TH edition), it follows that the stable non radioactive isotopes of nickel are the following five:

58, 60, 61, 62 and 64. These, when fused with a hydrogen nucleus, are being transmuted relatively to Cu-59, Cu-61, Cu-62, Cu-63 and Cu-65. From these isotopes of copper only the last two (Cu-63 and Cu-65) are not radioactive, i.e. they are stable. The other three Cu-59, Cu-61, Cu-62, are being transmuted again to Nickel, with an average life expectancy of some hours and the most unstable Cu-59 in 18 seconds.

By prof. Christos Stremmenos


840 comments to A detailed Qualitative Approach to the Cold Fusion Nuclear Reactions of H/Ni

  • Andrea Rossi

    Dear Lars:
    First of all, sorry if I cut your comment: it contained names in a context I do not accept to publish. I never comment the work of my competitors; besides, the interview you talked about is fake, the person you named didn’t release any interview; the statement, of course, was a stupidity and is not true.
    About the cleared part of your comment:
    Focardi referred to the efficiency of the conversion cycles. I think that the efficiency is less, I don’t think you can overcome an efficiency of 35% when you convert thermal energy into electric energy.
    The progression of my work, that as you know is focused on the manufacturing of the 1 MW plant, is very good and we are perfectly in time for the scheduled delivery in October. You bet. If some skeptic will tell you that for some reason I will not succeed, make a bet, do not worry: you will win. Today I made a wonderful test, running most of time with no energy at the input, again and again, during the stress tests of the modules. I am making a lot of modifications, by the way, learning a lot from many reactors working: evolution is exponential in these conditions. I remade all the reactors, because I discovered new important things and decided to make a new generation of E-Cats already in the first 1 MW plant. Big waste of components, but very big progress.
    Warm regards,
    A.R.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Lars:
    Ampenergo has made a tremendous work, and in October you will understand the tremendous importance of the agreement that Ampenergo has made. Yes, now we have to maintain a low profile, also because our policy is: first, make; second, talk of what you have made.
    Warm regards,
    A.R.

  • Lars

    Sorry I didnt finish the last question
    I dont hear much from Ampenergo but much from Defkalion. Is Ampenergo not active because the owners have so much else to do or is it just they keep a low profile for now?
    Thanks

  • Lars

    Dear Mr Rossi

    I heard Mr Focardi said in a video from the beginning of your experiments that when converting to electricity you have to divid the energy out put in 2 so only half is left. Is that still soo?

    Can you tell us anything about the progression of your work?

  • Andrea Rossi

    Dear Craig:
    Sorry, I do not like foot-ball. I prefer active sport: tennis, swimming, skiing, jogging. I do not like passive sports, as I do not like passive things.
    Warm Regards,
    A,R,

  • Hi Andrea,

    Been meaning to ask you this fundamentally important question for a while. As a native to Milan which side of the city do you cheer on – Inter or AC?

    Craig

  • Andrea Rossi

    Dear Luca T:
    Our mail target now is the production of thermal kWh. Of course we will also research for declinations.
    Warm Regards,
    A.R.

  • LucaT

    La NASA, con cui hai avuto dei colloqui, è evidentemente interessata all’applicazione della sua tecnologia in ambito spaziale.

    Le attuali prospettive per applicare la tecnologia Ecat sono rivolte:

    – ad utilizzare l’Ecat per laproduzione di energia elettrica con cui alimentare un sistema propulsivo di tipo ionico come ad esempio un vasimir, oppure

    – una nuova forma di applicazione della reazione Ni-H per la realizzazione di un nuovo tipo di propulsione a razzo.

    Distinti saluti

    L

  • Andrea Rossi

    Dear Pedro Ribeiro:
    You are oerfectly right !!!
    Warm Regards,
    A.R.

  • Pedro Ribeiro

    Please don’t receive more visits to prove or show your invent (we know what happens to the people that mennace the imposed system), just finish the energy plant in Greece, that will prove your invention.

    Kind Regards,
    P.R.

  • Andrea Rossi

    Dear Giovanni Pedrotti:
    When we will have production of electric power we will see. I suppose that we will either go directly to the loads of the Customers, or to the grid. To go to the grid takes substantial work for the interconnections and the authorizations. We know very well the issue, because we used to manufacture bio-masses fueled power plants, many of which have been connected with the grid.
    Warm Regards,
    A.R.

  • Dear Ing. Rossi
    which is your thought regarding the connection of a E-CAT to the grid? If not connected, everything is fine. No problems with the grid administrator (ENEL, ACEA, etc.). The generated kWh will be used exclusively by the site. In case of poor consumption, kWh will be lost. On the contrary, if grid connected, kWh produced in excess could be send to the grid (like a photovoltaic plant), and SOLD. The implication of this case is that the E-CAT should be compliant to the safety and technical rules concerning a power generator connected to the public grid. One example: if not powered by the grid, the apparatus must stop, preventing the apparatus itself to input electricity into the grid and hurt let’s say workers repairing the distribution line. Lot of other prescriptions exist.
    With my best regards and looking forward hearing from your plant in October
    Giovanni

  • Andrea Rossi

    Dear Georgehants:
    Maybe, who knows? In what centuries ago was named “Alchemy” there were the roots of the modern Chemistry and Physics, among a lot of magic bullshit which had nothing to do with the work of Scientists like Newton…like a baby in the dirt water.
    Warm Regards,
    A.R.

  • georgehants

    Dear Mr. Rossi,

    May I ask, is it possible, that one of the great true scientists of early years who did so much wonderful scientific work with Alchemy could have stumbled across your secret.
    Could the conditions necessary for the reaction have been achieved by Newton or any of the other pioneering seekers of truth.

  • Andrea Rossi

    Dear Lars:
    If this Cap works, I will wear it !
    Warm regards,
    A.R.

  • Lars

    Dear Mr Rossi,
    have you seen there is an e-cap for the e-cat?

    http://www.eisenmann.us.com/pdfs/brochures/e-cap_brochure.pdf

  • Andrea Rossi

    Dear Bernie Koppenhofer:
    No, muons are not invelved, we are far from the energy levels necessary to get them.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Snattin:
    As I said before, we are still searching a solution, let alone the patents.
    Warmest Regards,
    A.R.

  • Snattin

    Dear Mr Rossi,

    I think the turbines you mention in an earlier reply may be of great interest in many other applications due to their impressive performance (Carnot cycle efficiency). Are they patented? Will they be sold separately from the E-cat?

    Warm regards,

    Snattin

  • Bernie Koppenhofer

    Congratulation! Do you think Muons are in any way a part of the reaction taking place?

  • Andrea Rossi

    Dear Lars:
    We are designing with specialized companies the specific turbines for our reactors.
    The efficiency will be the same as usual for Carnot Cycle.
    Warm regards,
    A.R.

  • Lars

    Dear Mr Rossi
    Can you tell anything about how it is going with the work converting to electricity?
    Do you have big losses of energy when converting?
    How many watts of electricity is produced by the 10 kW now?
    Thank you

  • Andrea Rossi

    Dear S. Woosnam:
    Good question. We put boron just for safety, as Prof. Focardi teached to me. Kind of just in case…
    Warm regards,
    A.R.

  • S Woosnam

    Dear Mr. Rossi,
    First may I say how refreshing it is for a scientist such as yourself to engage with the public in the way you have done here. I think it is commendable.
    Second could I ask you about a technical aspect of your invention? I know you do not subscribe to the Widom-Larsen theoretical explanation of your empirical results; I found their theory rather plausible save for the neutron capture gammas which one would expect but aren’t observed. If neutrons aren’t generated (whether or not in the way they propose), what is the purpose of the boron shield?

  • Andrea Rossi

    Dear Daniele Passerini:
    So, now we know that the Snake had the arrogance to insult a Nobel Prize: every comment is useless. I am nobody, he can insult me as he pleases, but a Nobel price must be respected.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Hamdi Ucar:
    Thank you for your interesting considerations.
    Warm Regards,
    A.R.

  • Hamdi Ucar

    Dear Mr. Rossi,

    I understand that scalability depends on stability. I am just considering some possibilities where stability can be improved by passive ways (i.e. a material/agent which react with gamma rays and increasingly radiation can turn this agent to slowdown the fusion, therefore providing a negative feedback to regulate the fusion rate.) and by active ways. I had experiences with electromagnetic waves in resonance, specifically resonances involving two or more frequencies. These resonances are not well known but I had opinion that they cause break-up of water molecules and ozone generation, ionization. It may be possible to control the atomic hydrogen submission with such a external magnetic/electromagnetic fields effectively, offering better stability and therefore scalability.
    On the business point of view, one may foresee countless of new designs and phenomena which may improve the Ni-H fusion, therefore exclusivity on this technology cannot be kept for long time. Fortunately, innovation and good research is not an easy thing, and who have it will be in at head of the race.

    Best Regards,
    Hamdi Ucar

  • Dear ing. Rossi,
    three years ago Nobel Prize Brian Josephson wrote WHAT DRIVES STIVEN KRIVIT, a “report” about SK and New Energy Times which seems written now! I advise all to read it. The link is in this post on my blog.

  • Andrea Rossi

    Dear Lars:
    Thank you!
    Warm regards,
    A.R.

  • Lars

    Dear Mr Rossi
    You are right. I checked the numbers better for electricity production. So your invention is better than I understood. Congratulations :)

  • Andrea Rossi

    Dear Carlo,
    Yes, the costs can be reduced up to 60% with strong economy scale.
    Warm Regards,
    A.R.

  • Carlo

    Dear Rossi
    in a recent interview you stated that the expected cost of the ecat will be between $1000 to $2000 per KW of power. I suppose you were referring to KW of thermal power.
    I see very few reasons why one should concentrate the ecats in large plants as we do for fission nuclear power, but just to the sake of feasibility, given your expected cost, to build a plant for 1GW electric (3GW thermal) it will cost from $3 to $ 6 billion. It is a competitive cost compared to a nuclear plant which now costs $7B and up without considering the decommissioning cost at the end of life and all associated environmental costs.
    Do you think for the ecat, from a cost per KWe prospective, we may expect big cost reductions in the next few years or not?
    I’m asking this because it can influence, together with other variables of course, the speed of adoption of the ecat technology.

    Best regards

  • Andrea Rossi

    Dear Riccardo T.:
    We always said that no more public test would have been made before the start up of the 1 MW plant in October. All the tests we are making are not to be published and there are no substantial differences from the results already obtained. We have been extremely disappointed from the disinformation made from snakes and clowns, able to distort any data we give, therefore we will publish from now on only the data we guarantee to our Customers, while R&D data will be strictly reserved to the Scientists we are working with in the USA and in Europe.
    By the way we are using presently a more sofisticated system to measure the steam quality and the results are the same, as we expected.
    Warm Regards,
    A.R.

  • Riccardo T.

    Dear Dr. Rossi,

    Defkalion forum administratore informed us (curios people like me :)) that at the beginning of July there were plannet test for stability, efficiency and safety of Hyperion modules and that the result would be published on Defkalion Forum.

    Unfortunately the forum was closed before the end of the test. So we are left here hanging for results :)
    Do you have any news on the subject?

    Thank you so much

    Riccardo T.

  • Sebastian

    Dear Lars,
    your are extremely wrong…
    Just take a look at what is going to be the world’s most powerful nuclear power plant (Olkilouto in Finland): http://en.wikipedia.org/wiki/Olkiluoto .The new reactor has a power of 1600MW.
    If we had Terrawatt power plants, well then we would not be speaking about energy problems…

    Secondly, A. Goumy is also right, comparing electric with thermal energy results in bad numbers. So, in order to “replace” Olkiluoto with cold fusion plants, (the 3.45 MW is going the biggest reactor which is to be produced by Defkalion. They will also produce the 1MW reactors earlier! Read the other sources…) it will require approximately 1400 containers of a 3.45MW plant.
    That is not really much, taking into account container ships which can carry up to 15000 standard containers! So one ship could carry the power of 11 nuclear fission power plants 😉

    Best regards
    Sebastian

  • Andrea Rossi

    Dear A.Goumy:
    1- Yes, you are right
    2- Containers can contain more modules: we have to use standard containers, for many reasons.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Lars,
    I am very sorry, but I must say that you are wrong. A nuclear plant has a power in the order of thousands of MW, not millions of MW: for example, the plant of Fukushima is a plant with a power of 1,000 MW (one thousand MW).
    Warm Regards,
    A.R.

  • Lars

    Dear Mr Rossi
    A nuclear plant has aprox 3,5 TW so it is one million containers of 3,5 MW.
    3,5 million MW= 3,5 000 GW = 3,5 TW

  • A. Goumy

    Dear Mr Rossi,
    In the answer you made to Lars, your calculation takes into account the electric power of nuclear plants (~ 1GW), instead of its thermal power (~ 3 times more), so I think a better number for containers is around 900, and not 290. Am I right ?
    Secondly, you give a 3.45 MW figure for container power. As presently, the value given for Hyperions is 1 MW, where does the difference come from ? Will the planned containers contain more E-Cats, or will you increase from 6 to 20 the power ratio ?
    Best regards.
    A.G.

  • Andrea Rossi

    Dear Maryyugo:
    Why have I to? I do not understand why this issue has to be of your interest. Anyway: we have presently plants working in the USA and in other locations. What’s your problem about this?
    Warm regards,
    A.R.

  • maryyugo

    Dear Mr. Rossi,

    I am confused about where the 330 E-cats for the megawatt plant are being built? Is that Florida or Europe? Can you narrow the location a bit more?

    Thanks!

    M. Y.

  • Andrea Rossi

    Dear Paul Story:
    Very funny: this clown, named Julian Brown, wrote me saying he was an officer of the Patent Office and that he wanted give me suggestions.
    I received him to get those suggestions, curious to know about what he had to suggest. I was working in my Bologna lab when I received him and he saw one E-Cat under test for no more that 30 seconds, after which I invited him to exit. He made no tests, he saw nothing, he just has taken a 30 seconds glance at a totally closed box. He saw nothing, I said nothing, also because he inspired me no trust, because said he is a Quantum Physicist, but said so much stupidities that not even a 13 years old student of middle school could say, so I understood he was an impostor.
    We agreed, after a short meeting, that he would have mailed to me a text of a patent that he thought would have had many probabilities to be accepted: I was very baffled, because I could not understand how an officer of the European Patent Office could behave like that: he asked me to be paid by my company’s shares in change of his help!
    After some day I received from this clown ( who until that moment spoke only positively of all what he saw) some text, simply ridiculous, for a new patent (I conserve the copies, of course) and in these texts there was written that my patent was to be based on an already granted patent made from one competitor of mine, obviously a “friend” of his. I made a research and discovered that the patent of my competitor he referred to had not been granted, but had been refused. Of course he has been sent from such competitor to spy and to try to mess up with patents. I wrote him a mail inviting him not to contact me again: he was clearly an impostor. At this point he made the comment on Ecatnews…
    Conclusion: he lies when he says he made a test, he lies when he says he has seen an E-Cat enough to say anything about it (just has taken a 30 seconds look), he lies if he says he has seen the steam, he saw nothing because the circuit was ermetically closed, he lies when he says that he is an officer of the Patent Office (if he is, he made a crime, because he asked me, in change of his help, shares of Leonardo Corporation), he lies when he says he is an expert of patents, he lies when he says that my competitor he works with has a patent on this matter granted in 1995.
    It is clear that somebody, desperate of the fact that a 1 MW plant is close to be started up from us, is trying to use all the methods of a snake to try to put clubs in the wheels. But all this is just clownery: my plants will give evidence in the real market of the validity of my effect.
    It is also clear that at this point I cannot allow any more info or courtesy visit before the start up of my 1 MW plant.
    I have evidence, registrations and witnesses of all what I wrote here: BY THE WAY, ALL MY LABS AND FACTORIES, FOR SECURITY ISSUES, ARE SUPPLIED BY HIDDEN CAMERAS AND MICROPHONES TO REGISTER EVERYTHING WHICH HAPPENS AND IS SAID INSIDE.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Lars:
    Dear Lars,
    I tink you must review your maths: if one nuclear plant makes , say, 1.000 MW, it takes 1.000/3.45 containers of 3.45 mw each.
    This makes 290 containers.
    I stick on the concept of small safe modules.
    Warmest Regards,
    A.R.

  • Lars

    Dear Mr Rossi
    Thank you for your excellent work and your kind answers.
    It takes one million 3,45 MW containers to produce as much electricity as one nuclear plant. Is there any plans to make bigger Hyperions. As it develops now it will take a very long time to make a difference. I understand you do as much as you can and work very hard, but I think the pressure from people when this reaches the big public will be enourmous. Dont you think people will demand this technology instead of risking nuclear accidents like the one on Fukochima? What is your thoughts about this?
    Best regards

  • Dear Dr Rossi,

    Someone calling himself Julian Brown from the European Patent Office said (on the eCatNews site) that you demonstrated the eCat to him personally on the 8th. His comment is curiously detailed and my first reaction is to doubt its authenticity. Can you confirm that this informal demo did or did not take place? I intend to make a post about it to demonstrate the type of attacks you are under if that is the case.

    The comment is here:

    http://ecatnews.com/?p=489#comment-70

    Good luck.

    Paul

  • Andrea Rossi

    Dear Marco De Leonardis:
    I didn’t receive your former comment, probably for some reason our robot has spammed it. I do not have the time to go through the spam (hundreds of comments per day), so if a comment goes there it gets lost.
    Anyway your question is clear: of course we respect the law, therefore we are making all the necessary authorizations. We have specialized attorneys working on this issue. Of course the fact that we do not use rafioactive materials and do not produce radioactive wastes helps a lot. Anyway our Customers have to go through the authorizations issues with the help of our attorneys.
    Thank you for your contribution,
    Warm Regards,
    A.R.

  • Marco De Leonardis

    Dr. Rossi,
    Il mio post sul rischio incombente relativo alla legislazione sulla radioprotezione sembra essere stato censurato.
    Sto cercando di contribuire positivamente a questo progetto, individuando rischi e problemi da mitigare/risolvere.
    Non mi sembra di aver usato un tono o degli argomenti degni di censura.
    Grazie
    Marco

  • Andrea Rossi

    Dear Lars Baudot:
    Please read the answer to Martin.
    Warm Regards,
    A.R.

  • Lars Baudot

    Sorry for this second message.
    Do you have anything that you can report to us about yesterdays meeting of Defkalion & Ampenergo with NASA? Is it allowed for you name the participating people?

    Regards,
    Lars

  • Andrea Rossi

    Dear Lars Baudot:
    I referred to the cost of the energy made by the reactor. The price of the auxiliary energy has not to be layered upon the cost of the energy from the reactor, because the thermal energy produced by the resistance goes anyway to the fluid, so you use it, you do not lose it in the reactor, once the system is in equilibrium.
    Warm Regards,
    A.R.

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