Generalized Theory of Bose-Einstein Condensation Nuclear Fusion for Hydrogen-Metal System

by Yeong E. Kim Department of Physics, Purdue University West Lafayette, Indiana 47907, USA

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ABSTRACT
Generalized theory of Bose-Einstein condensation nuclear fusion (BECNF) is used to carry out theoretical analyses of recent experimental results of Rossi et al. for hydrogen-nickel system.  Based on incomplete experimental information currently available, preliminary theoretical explanations of the experimental results are presented in terms of the generalized BECNF theory. Additional accurate experimental data are needed for obtaining more complete theoretical descriptions and predictions, which can be tested by further experiments.

I.  Introduction
Over the last two decades, there have been many publications reporting experimental observations of excess heat generation and anomalous nuclear reactions occurring in metals at ultra-low energies, now known as „low-energy nuclear reactions‟ (LENR).  Theoretical explanations of the LENR phenomena have been described based on the theory of Bose-Einstein condensation nuclear fusion (BECNF) in micro/nano-scale metal particles [1-3].  The BECNF theory is based on a single basic assumption capable of explaining the observed LENR phenomena; deuterons in metals undergo Bose-Einstein condensation.  While the BECNF theory is able to make general qualitative predictions concerning LENR phenomena it is also a quantitative predictive physical theory.  Some of the theoretical predictions have been confirmed by experiments reported recently.  The BECNF theory was generalized for the case of two species of Bosons [4].

Recently, there were two positive demonstrations (January and March, 2011) of a heat generating device called “Energy Catalyzer” [5]. The Energy Catalyzer is an apparatus built by inventor Andrea Rossi, Italy. The patent application [5] states that the device transforms energy stored in its fuel (hydrogen and nickel) into heat by means of nuclear reaction of the two fuel components, with a consequent observed production of copper [5,6]. According to Rossi‟s patent application [5], heating of the sample is accomplished by an electric resistance heater.  Details of March 2011 demonstration were reported by Essen and Kullander [7]. The report [7] also contains references to January 2011 demonstration. In the following, we describe hydrogen-nickel reactions in section II. Other possible reactions are discussed in section III.  Conclusions are given in section IV.

II.  Hydrogen-Nickel Reactions
The generalized BECNF theory [4] can be applied to the case of hydrogen-nickel fusion reactions observed in Rossi‟s device (the energy catalyzer) [5] under the following two conditions: (1) additives used (not disclosed in the patent application) form Ni alloy and/or Ni metal/alloy oxide in the surface regions of nickel nano-scale particles, so that Ni atoms/nuclei become mobile with a sufficiently large diffusion coefficient and (2) local magnetic field is very weak in the surface regions, providing a suitable environment in which two neighboring protons can couple their spins anti-parallel to form spin-zero singlet state (S=0).  Relatively low Curie temperature (nickel has the Curie temperature of 631 oK (~358 oC)) is expected to help to maintain the weak magnetic field in the surface regions. If Rossi‟s device is operated at temperatures greater than the Curie temperature ~358 oC and with hydrogen pressures of up to ~22 bars, the conditions (1) and (2) may have been achieved in Rossi‟s device. The mobility of Ni atoms/nuclei (condition (1)) is enhanced by the use of an electric resistance heater to maintain higher temperatures. This may provide a suitable environment in which more of both Ni atoms/nuclei and protons become mobile, thus creating a favorable environment for the case of two species of Bosons (Ni nuclei and composite Bosons of paired two protons). If the velocities of mobile Ni atoms/nuclei under the condition (1) are sufficiently slow, their de-Broglie wavelengths become sufficiently large and may overlap with neighboring two-proton composite Bosons which are also mobile, thus creating Bose-Einstein condensation of two species of Bosons. The generalized BECNF theory can now be applied to these two-species of Bosons and provides a mechanism for the suppression/cancellation of the Coulomb barrier, as shown in [4]. Once the Coulomb barrier is overcome in the entrance reaction channel, many possible allowed exit reaction channels may become open such as reactions (i) ANi(2p(S=0), p)ˆA+1 Cu, with even A=58, 60, 62 and 64. These reactions will produce radioactive isotopes 59Cu and 61Cu with A = 58 and 60, respectively. 59Cu has a half-life of 81.5 seconds and decays by the electron capture to the 59Ni ground state (58.1%) which has a half-life of 7.6 x 10ˆ4 years and to the 59Ni excited states (41.9%) which in turn decay to the 59Ni ground state by emitting gamma-rays with energies ranging from 310.9 keV to 2682.0 keV [8]. 61Cu has a half-life of 3.333 hours and decays by the electron capture to the stable 61Ni ground state (67%) and to the 61Ni excited states (33%) which in turn decay to the 61Ni ground state by emitting gamma-rays with energies ranging from 67.412 keV to 2123.93 keV [8]. Gamma-rays (and neutrons) have not been observed outside the reactor chamber during the experiment [6]. These gamma-rays may have been present inside the reaction chamber. If no radiations are observed, reactions (i) are ruled out. Focardi and Rossi [6] reported that the experimental results of Rossi et al. indicate the production of  stable isotopes 63Cu and 65Cu with an isotopic ratio of 63Cu /65Cu ~ 1.6 (natural abundance is 63Cu/ 65Cu = 2.24). This production of Cu may be due to reactions (i). The production of 63Cu and 65Cu with isotopic ratio of 63Cu /65Cu different from the natural isotopic ratio is expected and can be explained by estimating the reaction rates for 62Ni(2p(S=0), p)63Cu and 64Ni(2p(S=0), p)65Cu.  Reaction rates estimates based on transmission probability calculated from a barrier tunneling model similar to the alpha-decay theory indicate that the reaction rates for stable Cu productions, 62Ni(2p(S=0), p)63Cu and 64Ni(2p(S=0), p)65Cu, are expected to be much larger than the reaction rates for production of radioactive Cu, 58Ni(2p(S=0), p)59Cu and 60Ni(2p(S=0), p)61Cu. This leads to the prediction that intensities of the gamma-rays from the decays of 59Cu and 61Cu are expected to be weak and do not commensurate with the observed heat production, which is mostly from stable Cu production  reactions 62Ni(2p(S=0), p)63Cu and 64Ni(2p(S=0), p)65Cu. There are other exit reaction channels which are (nearly) radiation-less, such as reactions (ii) ANi(2p(S=0), α)ˆA-2Ni, (even A=58, 60, 62, and 64) [9]. For this case, we expect that the natural isotopic ratio of Ni isotopes will be changed in a particular way, which can be checked from the  sample after each experiment.  Even though reactions (ii) produce radioactive isotope 56Ni, it can be shown using the alpha-decay theory that its reaction rate is much slower (by many order of magnitudes) than those of other reactions. Other exit reaction channels, ANi(2p(S=0), d)ACu, ANi(2p(S=0), 3HeA-1Ni, and ANi(2p(S=0), t)ˆA-1Cu (all with even A=58, 60, 62, and 64) are ruled out since these reactions all have negative Q-values.  There are possibilities of neutron-emission exit reaction channels, such as reactions (iii) ANi(2p(S=0), n)ˆA+1Zn, (even A= 62, and 64; Q is negative for A = 58 and 60).  However, reaction rates for reactions (iii) are expected be substantially smaller than those for reaction (i).  Reactions (iii) involve emission of a tightly bound neutron (62Ni -> 61Ni + n, Q = -10.597MeV or  64Ni -> 63Ni + n, Q = -9.657MeV) while reactions (i) involve emission of a loosely bound proton from an excited compound nuclear state consisting of ANi (even A) and 2p(S=0). Therefore, the transmission probability of a neutron tunneling through the centrifugal barrier in reactions (iii) is expected to be substantially smaller than that of a proton tunneling through the centrifugal barrier in reactions (i). The branching ratios of reactions (i) and (ii) need to be determined by measurements of gamma-ray energies and changes in isotopic ratios from future Ross-type experiments.  Theoretically, the branching ratios can be estimated by calculating transmission probability of an emitted charged particle tunneling through both Coulomb and centrifugal barriers in the exit reaction channel, as done in the alpha-decay theory.

III.  Other Possible Reactions
In addition to the above reactions described in II, there are possibilities of reactions involving additives used (not disclosed so far). For an example, if lithium is added as an additive, reaction (iv) 6Li(2p(S=0), p 3He)4He may be possible. As in cases of reactions (i) and (ii), Ni nano-particles would be still playing an important role of providing two-proton singlet composite Bosons for reaction (iv). Reaction (iv) would not change the isotopic ratios of Ni.

VI.  Conclusions
In order to explore validity and to test predictions of the generalized BECNF theory for the hydrogen-metal system, it is very important to carry out Rossi-type experiments independently in order to establish what are exact inputs and outputs of each experiment.  If the entrance and exit reaction channels are established experimentally, we can investigate selection rules as well as estimates of the reaction rates for different exit reaction channels, based on the generalized BECNF theory [1-4]. Once these experimental results are established, further application of the generalized BECNF theory can be made for the purpose of confirming the theoretical mechanism and making theoretical predictions, which can then be tested experimentally. Basic description of the above theoretical concepts for BECNF in the hydrogen-metal system will be included in an invited talk at a forthcoming nuclear physics conference [10], and will be published in the conference proceedings [10].

References

  1. Y. E. Kim, “Theory of Bose-Einstein Condensation Mechanism for Deuteron-Induced Nuclear Reactions in Micro/Nano-Scale Metal Grains and Particles”, Naturwissenschaften 96, 803 (2009) and references therein.
  2. Y. E. Kim, “Bose-Einstein Condensate Theory of Deuteron Fusion in Metal”, J. Condensed Matter Nucl. Sci. 4, 188 (2010), Proceedings of Symposium on New Energy Technologies, the 239th National Meeting of American Chemical Society, San Francisco, March 21-26, 2010.
  3. Y. E. Kim, “Theoretical interpretation of anomalous tritium and neutron productions during  Pd/D co-deposition experiments”, Eur. Phys. J. Appl. Phys.  52, 31101 (2010).
  4. Y. E. Kim and A. L. Zubarev, “Mixtures of Charged Bosons Confined in Harmonic Traps and Bose-Einstein Condensation Mechanism for Low Energy Nuclear Reactions and Transmutation Processes in Condensed Matter”, Condensed Matter Nuclear Science, Proceedings of the 11th International conference on Cold Fusion, Marseilles, France, 31 October – 5 November, 2006, World Scientific Publishing Co., pp. 711-717.
  5. Andrea Rossi, “METHOD AND APPARATUS FOR CARRYING OUT NICKEL AND HYDROGEN EXOTHERMAL REACTION”, United States Patent Application Publication (Pub. No.: US 2011/0005506 A1, Pub. Date: Jan. 13, 2011); http://www.wipo.int/patentscope/search/ja/WO2009125444.
  6. S. Focardi and A. Rossi, “A new energy source from nuclear fusion”, March 22, 2010. http://www.nyteknik.se/incoming/article3080659.ece/BINARY/Rossi-Forcardi_paper.pdf
    http://www.journal-of-nuclear-physics.com/?p=66 ,  February 2010
  7. H. Essen and S. Kullander, “Experimental test of a mini-Rossi device at the Leonardocorp, Bologna, 29 March 2011”, a travel report, April 3, 2011; http://www.nyteknik.se/nyheter/energi_miljo/energi/article3144827.ece
  8. Table of Isotopes, 8th Edition, Volume I: A = 1-150, edited by R. B. Firestone et al., published by John Wiley and Sons, Inc. (1999), pages 270 and 284.
  9. Reactions (ii) were suggested by T.  E. Ward, private communication, May 11, 2011.
  10. Y. E. Kim, “Deuteron Fusion in Micro/Nano-Scale Metal Particles”, an invited talk to be presented at the Fifth Asia Pacific Conference on Few-Body Problems in Physics 2011(APFB2011), August 22-26, 2011, Seoul, Korea. (http://www.apctp.org/conferences/2011/APFB2011/)

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859 comments to Generalized Theory of Bose-Einstein Condensation Nuclear Fusion for Hydrogen-Metal System

  • Guru Gurovic

    Dear Mr. Rossi,

    I read Your note about too inmature turbine systems.
    I have for You one hint:

    Kockums (Marine Propulsions) AB, Sweden already manufacture Stirling Engines for submarines. They also developed very compact Stirling Engines for Solar Parabolic Mirror systems. This compact Stirling Engine with generator has output 25kW and supposedly efficiency 30%. Ideally suitable for block of 20-25 of Your E-Cat units. Or one big Stirling submarine Engine for 150 your devices.

    Contacts here:
    http://www.kockums.se/en/products-services/submarines/stirling-aip-system/the-stirling-engine/

  • Andrea Rossi

    Dear Mr Martin:
    Thanbk you!
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Joseph,
    I really need help on this issue: I have contacted all the exotic turbines manufacturer I have been made aware of, but always found very, very “green” ( meaning not mature) technologies, not ready to fight in the battlefield of the market. Should I find a turbine fit for my E-Cats I would immediately buy one to test it, and if it goes I could buy thousands of pieces.
    We are working with some manufacturer, but I didn,t still found a solution to sell to our Customers.
    I need real proposals, not enlighting ideas.
    Thank you for real proposals coming: I am a customer ready to buy, preferably from the USA, where our reactors are manufactured.
    Warm Regards,
    A.R.

  • Cesar BP

    Dear Mr. Rossi.
    Congratulations, I am following the news about your technology and I think your discovery will radically change the world. Only a comment, probably hot air airships will be feasible using some kind of e-cat prepared to warm the air.
    Good luck.

  • Joseph Fine

    Dr. Rossi, Rick Meisinger & Friends,

    The best way to increase the generating efficiency of the steam turbine is not to use steam in the first place. Better to use Super-critical Carbon Dioxide or S-CO2. I hope we will all learn a lot more about turbines in the coming year. By the way, S-CO2 turbines are rather small.

    Super-critical regards, (with regards to turbines)

    Joseph Fine

  • Martin

    Haha mr Rossi,

    You surely are the number one in fast answering questions.
    I am certain you also will be the number one in resolving our energy problem.

    Thanks for your information.

    Best regards

    Martin

  • claudio puosi

    Buonasera Ing.Rossi le ho appena fornito sulla sua email info@leonardocorp…… i miei recapiti
    aspettando un suo contatto la saluto cordialmente

    Claudio Puosi

  • Andrea Rossi

    Dear Martin:
    1- We have no batteries
    2- the reactor is self maintaining with part of the energy produced by itself.
    Warm regards,
    A.R.

  • Martin

    Dear mr Rossi,

    I try to understand how the self sustaining of your ecat works.
    Do you use a battery? If not means self sustaining just start the reactor
    and the proces is beginning? I think the proces will last longer because the temperature in the reactor will rise slower. (is this assumption true? and if so how much longer will this last?)
    I am very courious about your information about this topic.

    Best regards!

    Martin

  • Andrea Rossi

    Dear Rick Meisinger:
    I am not an expert of turbines.
    Warm regards,
    A.R.

  • Rick Meisinger

    Dear Andrea Rossi;

    Is it practical to increase the electrical generation efficiency of the steam turbines by redirecting (recycling) the turbine exhaust?

    With much support and anticipation;
    Rick Meisinger

  • Andrea Rossi

    Dear Mattias Andersen:
    Yes, we are testing also this.
    Warm regards,
    A.R.

  • Mattias Andersson

    Dear Mr. Rossi,

    Have you been experimenting with other solubles that have a higher boiling point than water (such as glycol or glycerol)? I suppose that if you connect a lot of modules in parallel then it will also be desirable to raise the boiling point.

    Kind Regards,
    Mattias

  • Andrea Rossi

    Dear Wade:
    Few grams: your considerations are correct.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Marcia Pires:
    We will run the reactors without input energy so long the operation is safe and stable.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Paul Swanson:
    Yes,
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Roger:
    1- Enough
    2- It will be sent in October
    3- For safety reasons I can’t give information about the logistics
    Warm Regards,
    A.R.

  • georgehants

    Mr. Rossi a link that may interest you and all followers of the scientific disgrace associated with Cold Fusion research.

    By Dr.Stoyan Sarg

    http://www.foreignpolicyjournal.com/2011/07/20/nuclear-safety-and-hazardous-radioactive-waste/

  • raul heining

    Dear Eernie,
    I liked your text on nuclear reactions. It is very clear and simple for people who do not
    have enough technical preparation.
    regards
    raul

  • Roger

    Dear Mr Rossi,
    Congratulations to your great discovery.

    I have three questions:
    1) How many people are working with the 1MW plant at your factory/factories in Florida?
    2) How long will the 1MW plant stay in Bologna before it goes to Greece?
    3) Will the 1MW plant be sent by boat or by plane to Europe?

    Warm regards,
    Roger

  • Paul Swanson

    Dear Andrea Rossi,

    Has the customer for the 1MW plant specified an input water rate and an output temperature?

    Warm Regards,
    Paul Swanson

  • Marcia Pires

    Dear Andrea Rossi,

    I made a research on wikipedia and I found what I wanted to know. What is the energy gain factor of an e-cat system to make it work without external input?

    http://en.wikipedia.org/wiki/Fusion_energy_gain_factor

    The e-cat alone with external power is stable for years, but why it operates for only a few hours without external input? I thought that a self feeding system wouldn’t make any difference to the reactor.

    Best wishes,

    Marcia Pires.

  • Wade

    Ok, well, how much Nickel and Hydrogen is in 1 charge for the 4kw reactor?

    Previously, I thought it was a “few grams” nickel, which was around maybe a mole or so.

    1 mole hydrogen per mole nickel, or 5 mole hydrogen per mole nickel to theoretically go through the entire proposed Nickel-Copper chain…that’s not much fuel at all.

    A mole of Nickel is currently only about $1.38 U.S…

    Assuming steam generator efficiencies, this would allow producing residential electricity for about 500 times cheaper than existing grid technology per KWh, neglecting the price of the reactor, which should totally pay for itself within a few years.

  • Andrea Rossi

    Dear Wade:
    1- We consume LESS than 1/6th of the output.
    2- yes
    3- few watts per charge and a charge works for 6 months in a 4 kW reactor
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Alessandro Casali:
    You are correct, the 1 MW plant will be made by the moduiles of the past test.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear H. Hansson,
    Here is a misunderstanding. Few reactors per turn are put in maintainance, so that the power remains constant at 1 MW.
    Warm regards,
    A.R.

  • H. Hansson

    Dear Mr. Rossi,

    In one of your replies you indicates that the 1 MV plant actually have a 1.200.000 W capacity (300 x 4 kw = 1,2 MV). But that approx 20% of the 300 units is down for maintenance or kept as a backup.

    How many maintenance hours does one (1) 1 MW unit requires on a yearly basis for maintenance and reloading??

  • Alessandro Casali

    Dear Dr. Rossi,

    Thank you for your previous reply.

    I have another question that i’d like to ask you:

    We all know that Defkalion Hyperions will be quite different from the modules we have seen in your pubblic tests with which they will only share the core (the nut).

    So is the first 1MW plant (the one you are building up now) made out of Hyperions modules or is it made out of the modules we have seen in the tests?

    I believe it is made out of the modules we have seen in the tests because, from what we know, Defkalion has not started production yet, therefore i think we can consider the first plant a prototype plant (obviously a working prototype), am i correct?

    Thanks again for your kindness.

    AC

  • claudio puosi

    Grazie Ing. Rossi , le siamo veramente grati.Ci possiamo sentire nei prossimi giorni per organizzare la diretta.

    Un saluto sincero Claudio Puosi

  • Wade

    Dear Dr. Rossi,

    How much input power you are feeding to the 1MW system?

    That’s the question.

    For the output, you used the simple linear of 300*4kw = 1200kw.

    I thought you said around 20% was used to maintain, so that seems to have a gain of about 6, which is on the low end of what is stated on the Defkalion website.

    Next question:

    If you are able to remove input for extended time and have self-sustaining reaction, then wouldn’t you have input of nearly zero?

    Finally, about how much energy is required for processing your fuel and catalysts for these reactors?

    I’m interested in that because I would like to try to figure out what the total gain is through the entire cycle of the production and operation lifecycle for any given amount of fuel.

  • Andrea Rossi

    Dear Wade:
    Sincerely,I did not understand your question. Can you rephrase, “idiot proof”?
    Warm regards,
    A.R.

  • eernie1

    Sorry about the interruption, I hit the wrong button.
    As I was writing,injecting the quasi proton into the Nickel nucleus will create a metastable state in the nucleus which must be balanced by some sort of rection.One such reaction may be the decay of a neutron into a proton creating a copper atom with the ejection of a beta particle enhanced with extra energy from the mass conversion of the decay processes.This beta particle then enters the electron cloud of the Nickel atom and interacts with the lattice electrons of the Nickel crystals. this creates IR energy which heats up the Nickel nodules.

  • Andrea Rossi

    Dear Daniel De Francia:
    I will try again to be understood: we produce thermal energy.
    When we will produce electric energy, of course we will be subject to the efficiency of the generators, e.g. the Carnot cycle.
    Our gain is on the production of the thermal energy, with which you eventuallt make what you want, paying the price of the efficiencies of the devices you are going to make use of.
    Warm Regards,
    A.R.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Eernie 1:
    Thank you for your insight,
    Warm Regards,
    A.R.

  • eernie1

    Dear blogers;
    I see many of you asking questions about the physics of the Rossi reaction.To the ones who are less sophisticated enquirers I can offer perhaps a simple analogy to explain the basic function.The physics of a common fission process can be used to illustrate.To cause the fission of a nucleus we must have a source of neutrons and unstable nuclei.Neutrons because they have no charge and can pass through the coulombic barrier of the positively charged nucleus.Both are provided by atoms such as Uranium or Plutonium isotopes.However the neutrons are too energetic(move too fast)to enter the nuclei.Think of the nucleus having a revolving door. If the neutron is moving too fast it will hit the door and rebound because the door will not open fast enough.If the neutron is moving at the speed of the door’s revolution the neutron will enter the nucleus,just like we enter through the door.So we must decrease the energy(speed)of the neutrons to a value consistant to the doors revolution. this is done by using what is called a modulator.Substances like carbon and heavy water are used and slow down the neutrons by collisions which absorbs the energy, creating what are called thermal neutrons reduced to energy levels of less than electron volts(ev). When the neutron enters the unstable nucleus it creates an energy imbalance which the nucleus must rebalance.It does this by emitting more than one neutron,excess energy and splitting the remainder of its substances into lower weight atoms.The multiple created neutrons then continue the process called a chain reaction.
    In Rossi’s configuration the place of the neutrons is taken by a hydrogen atom which consists of a proton and an electron revolving around it.The difference is the energy of the angular momentum of the electron keeping the electron from decaying into the proton.If you remove this energy you wind up with a quasi neutron.By using a medium such as a transition metal,and If you introduce the hydrogen atom into its electron cloud,the electrons of the metal can interact with the hydrogen electron and remove some of its momentum making the hydrogen atom more like a neutron.Transition elements are the best choice because their electrons are the most volatile of the common elements thereby more likely to interact with a passing electron in its cloud.If the atom then becomes less charge separated it can more easily pass through the coulombic charge of the transition metal nucleus and cause an imbalance which must be compensated for as in the fission process of Uranium.Although the nuclei of transition metals are relatively stable,it is possible to create what is called a metastable state by injecting the quasi proton

  • Andrea Rossi

    Dear Koen Vandevalle,
    Yes, you got it.
    Warm regards,
    A.R.

  • Daniel de França MTd2

    Dear Andrea Rossi,

    I will try again :). How much of the output energy is used to generate electricity? Because of the carnot cycle, it cannot be 100%, so, more than 700W of the output energy must be used for a 4-5KW e-cat.

  • Andrea Rossi

    Dear Claudio Puosi:
    Congratulations for your important meeting in Viareggio.
    Some day before it, please contact me to say at what time you want me to be available at phone (I will be in the USA), to salute you and all your Guests.
    Warm Regards,
    A.R.

  • Wade

    So, you’re feeding input power to each module individually?

    Or only some of them?

    I guess it really doesn’t matter either way, now that I think of it, since if you used the power to make a steam generator, you could feed some of the electricity back to maintain inputs…

    I guess I just assumed it worked like an exponential growth tree, but that doesn’t jive with what you just said.

    But I do see that, given your corrected numbers on the power of each module, there isn’t too much room for practical improvement either way, since to do it the way I was thinking would only save about 3% of net power even “on paper,” in pure mathematics.

    I also see that you have revealed the 1MW plant doesn’t necessarily need permanent input power any more anyway, so the difference might not even matter.

    Thanks for responding.

  • Andrea Rossi

    Dear Georgehants:
    I do not know. ( Is it a siggestion to me to go to the moon to take away the disturb? He,he,he…)
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear wade,
    your question was good, not stupid.
    Warm regards,
    A.R.

  • Andrea Rossi

    Dear Claud:
    Actually, we have no noise problems at all.
    By the way, the pumps are powered from a different power source than the reactors.
    Warm Regards,
    A.R.

  • Andrea Rossi

    Dear Marcia:
    I am very sorry, but I did not understand what you said.
    Warm Regards,
    A.R.

  • Wade

    Sorry if I accidentally said something stupid or offensive, or if I accidentally stumbled onto something you didn’t want known publicly yet. Maybe I just misunderstand and my math was overly optimistic.

  • Andrea Rossi

    Dear Wade:
    1 module makes 4 kW
    300 modules sum up 1,200.00 kW.
    Take 20% of margin for maintainance turns, remains a power of 1 MW.
    Warm regards,
    A.R.

  • Wade

    Dear Dr. Rossi,

    I have a question about the way the individual modules are networked(right term?) to make the larger reactor.

    Is what I’m outlining below correct, roughly speaking?

    I’m positive you know these relationships anyway, as most should if they think about it a while…

    I noticed, using some round numbers based on previously published estimates, I derived a way which “probably” is similar to what you did to make 1 MW using 300 E-cats, using the round number of the assumption of a gain of 8:1 of output to input for individual modules.

    Now if they are linked in series and parallel, making a tree, this gives, as I posted elsewhere:

    With exponential growth of devices per tier, as 8^(n-1):

    Tier = total devices = net gain ratio = output watts

    1 tier = 0001 devices = 8:1 = 3840 watts
    2 tier = 0009 devices = 64:1 = 30720 watts
    3 tier = 0073 devices = 512:1 = 245760 watts
    4 tier = 0585 devices = 4096:1 = 1.966 megawatts*
    5 tier = 4681 devices = 32768:1 = 15.7 megawatts

    So to get 1 megawatt, you would need:

    1 + 8 + 64 + 227 ( 256 is over the number) = 300 units

    …with less than half of the 4th tier (8^(n-1)) used, and summing the units in each Tier or part of a tier from 1 to N, giving 300 units.

    In this system, if each Module has a gain of 8:1, then the total net gain of the system 2072:1, which suggests about half of the mathematical tier 4 I listed above…

    This gives a net power output of around 994,560 watts, which is within a half a percent of a megawatt…

    So it seems that while the details of my understanding are probably not entirely correct, the net power production sounds about right based on the assumption of exponential returns of reactors providing the input power for exponentially more other reactors, if each module individually has a gain of around 8:1…

    Sorry for being so long-winded, but is this roughly similar to what you are doing??

  • Marcia Pires

    I made some mistakes in the previous messages, sorry.

    Dear Andrea Rossi,

    I made a mistake in my question number 1. I will give you an example of what to see I you can understand. I will use approximate quantities

    Suppose the output is 4 KW and the input is 700W. But a part of the input, say X is used to generate energy for the input. If the efficiency of the energy generator is 40%, you will have to use for X the quantity 700*10/4= 1750W. So, the positive energy, the profit of energy is (5KW– 1750W (needed to input 700W))*4/10 = 2100W.

    In your case, what is the efficiency of the machine to generate electricity?

    Best wishes,

    Marcia.

  • Claud

    Dear Mr. Rossi, besides the peristaltic pump is any other significant source of noise in the e-cat system? In other words, could any noise problem arise in a household application such as nigthtime operation?
    Thank you.

    Claudio Rossi

  • georgehants

    Dear Mr, Rossi,

    Would your E-Cats work o.k. in a vacuum such as on the moon.

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