A new energy – Exp.results

Experimental results

In this paper we report the results obtained with a process and apparatus not described here in detail and protected by patent in 90 countries, consisting of a system whose heat output is up to hundred times the electric energy input. As a consequence, the principle of the conservation of energy ensures that processes involving other energy forms are occurring in our apparatus.

The system on which we operate consists of Ni, in H atmosphere and in the presence of additives placed in a sealed container and heated by a current passing through a resistor. The maximum temperature value can be set to a wide range of values and an external meter allows us to measure the electric energy input. The container is in thermal contact with an external tank full of water and thermally insulated in order to minimize outside heat exchanges. As consequence of the energy production of the system, water boils and the water pipe is under pressure. The steam pressure cannot exceed a limit, whose value can be changed in the range 3-6 bar, because of the opening of a valve. When the valve opens, new water, whose amount is measured by a meter, enters from the supply. These data allow us to calculate the power produced by our system.

In stationary conditions the power output turns out to be much greater than the input (measured with an electric power meter). Some examples of the results obtained with this system (method A) in brief periods (~1-1,5 hours) are reported in lines 1-3 of the Table 1. The ratio between output and input power depends on changes occurring in the Ni-H system and on the time interval elapsed between the starting of the experiments and the measuring moments.

We have subsequently achieved a forced warm water movement through some radiators connected in series. In this case, the energy produced has been evaluated by measuring the power needed to obtain the same radiator temperature with a normal heating system (method B). In Table 1, lines 4 and 5, the results of these measurements are also reported. The patented apparatus is able of producing a constant and reliable amount of energy for a period of months.

A third method (method C) based on a closed circuit in which water is forced to circulate by means of a pump was used in order to measure the power generated: a section of the circuit contains the energy amplifier opportunely insulated in order to minimize thermal exchanges with outside.

Two thermocouples placed before and after the energy amplifier allow to detect continuously the water temperatures which are recorded on a computer. As a consequence the measured temperature difference allows to calculate the thermal energy transferred from the energy amplifier to the water. The electric input energy is measured by means an electric power meter.

In all cases the energy production is too high for any chemical process.

Table 1: Input and output energies, expressed in kWh, in some experiments.
(*) The anomaly in this experiment is due to contamination of the fuel.

In fact, assuming that each Ni atom in sample can realise, in optimal conditions, a typical chemical energy of some eV, the amount of energy emitted in the long lasting experiments would required at least 10ˆ28 atoms. That is something like a million of grams, a quantity enormously larger than the sample we have employed. For such a reason, we believe that form of energy involved is nuclear, and more specifically, due to fusion processes between protons and Nickel nuclei. They are exothermic with an energy release in the range 3-7,5 MeV, depending on the Nickel isotope involved.

It is remarkable that similar results have been obtained in the factory of EON in Bondeno (Ferrara, Italy) in a test performed with ENEL spa on June, 25th 2009 and in another sery of tests made in Bedford, New Hampshire (USA) in a lab of LTI with the presence of the DOE (November 19 2009) and of the DOD (November 20 2009).

The proton capture process performed by a Nickel nucleus produces a Copper nucleus according to the scheme


Copper nuclei, with the exception of the stable isotopes Cu63 and Cu65, decay with positron (e+) and neutrino (ν) emission in Ni nuclei according to the scheme


A process alternative to (4), electron capture, in abbreviated form indicated as EC, consists in the nuclear capture of an orbital electron which gives rise to the process


As a consequence, in this case, the reaction (4) must be replaced by


with emission of an antineutrino.

Table 2: Energy (in MeV) released by Ni->Cu and Cu->Ni transformations for different Ni isotopes.

The two decay processes (positron emission and EC) are alternative: their relative frequencies for the various copper isotopes are generally unknown with the only exception of Cu64 for which EC decay (6) is about twice as frequent as positron decay [4].

The capture rate of protons by Nickel nuclei cannot depend on the mass values of different isotopes: in fact they possess the same nuclear charge and the same distribution of electrons in the various atomic shells. In practice, starting from Ni58 which is the more abundant isotope, we can obtain as described in the two above processes, Copper formation and its successive decay in Nickel, producing Ni59, Ni60, Ni61 and Ni62. Because Cu63, which can be formed starting by Ni62 is stable and does not decay in Ni63, the chain stops at Ni62. In Table 2, for every Nickel isotope, we report, expressed in MeV, the energies obtained from the process Ni(A)+p -> Cu(A+1) (column 2), those obtained from the process Cu(A+1) -> Ni(A+1) (column 3) and their total for the complete transformation Ni(A) -> Ni(A+1) (column 4). Data reported in columns 2 and 3, are obtained as differences between the mass values of the initial and final state: that reported in column 3 contain also the neutrino (or antineutrino) energy, particles which interact weakly with the matter and does not hand their energy locally.

On the other hand we have to consider the energy equivalent of the electron rest mass due to the positron annihilation. Cu64 also decays in Zn64 with negative electron emission; the energies relative to both decays are reported in Table 2 (third column); the value carried in column four takes into account the relative frequencies of both Cu64 decay modes. The two isotopes Ni59 and Ni63 are unstable, but because their long lifetime (8×10ˆ4 years and 92 years respectively for Ni59 and Ni63) can be considered as stable in the times of our experiments.

Ni64, coming from the decay of Cu64, decays with electron emission, releasing 2,14 MeV: such a value must be added to 8,22 MeV reported in Table 2 (line 6, column 4).The two isotopes Ni59 and Ni63 are unstable, but because of their long lifetime (8×104 years and 92 years respectively for Ni59 and Ni63) can be considered as stable in the times of our experiments.

For every nucleus in the mass range 58 – 64 amu, we have built Table 3 which contains:

  • the mass value expressed in amu (column 1)
  • the total energy obtainable from all transformations (column 2)
  • the percentage in natural composition (column 3)
  • the product of columns 2 and 3

The sum of the energy releases in the last column gives ≈ 35 MeV, which represents the mean energy value obtainable for every Ni nucleus (in the hypothesis that all nuclei give rise to the whole sequence of events).

Such a figure must be compared with E ≈ 200 MeV for every U235 fission in a nuclear reactor [5] and ≈ 18 MeV for every reaction between deuterium and tritium in not still existing fusion reactor.

For the same number of nuclei, the ratio between Ni and U masses is 0,25 and the ratio between the energies that can be obtained is ≈ 0,2. Taking into account the world reserves of these elements, their extraction costs and the great investments needed for the building and maintenance of a nuclear reactor, the nuclear processes (based on Nickel) appear on the economical point very interesting.

During experimental tests, continuous controls on the radioactivity levels in close proximity to the apparatus suitably lead shielded, were performed by using a gamma ray detector [6] and three passive neutron bubble detectors BTbubble [7], one of which for thermal neutrons: no radiation was observed at levels greater than natural radiation background. No radioactivity has been found also in the Nickel residual from the process.

The 10th of march 2009, during the run whose data are reported in Table 1, line five, measurements were performed, around the running Energy Amplifier, by the Bologna University Health Physics Unit which verified that emissions around the Energy Amplifier are not significantly different from the natural background. The water drawn from the Energy Amplifier has resulted to have the same concentration of natural radioisotopes of the tap water: therefore there is no difference between the tap water and the water from the Energy Amplifier.

Two different samples of material used in the experiments labelled in table 1as method A (288 kWh produced) and method B (4774 kWh produced) were analysed at Padova University SIMS. In the long period sample, the mass analysis showed the presence of three peaks in the mass region 63-65 a.m.u. which correspond respectively to Cu63, elements (Ni64 and Zn65) deriving from Cu64 decay and Cu65.

Table 3: Energy obtained by every Ni isotope due to all successive transformations

These allowed us the determination of the ratio Cu63/Cu65=1,6 different from the value (2,24) relative to the copper isotopic natural composition. The Zn64 derives from the β‾ Cu64 decay: as it.s shown in Table 3, formation of Cu64 requires the existence of Ni63 which, absent in natural Ni composition, must have been in precedence produced starting by more light nickel isotopes. More details on this analysis will be given in a successive paper [8].

Sergio Focardi
Andrea Rossi

110 comments to A new energy – Exp.results

  • Andrea Rossi

    Dear Mr. Rolando,
    Thank you for your appreciated attention.
    I don’t think the reaction is possible under the earth’s crust by spontaneous reactions, because the odds of the presence of the right catalyzers are one out of some billion. To get the reaction is not enough put together Ni and H at high temperature and/or pressure, you need also a proper control upon T and P and, mainly, a complex system of catalyzers.
    Warm regards,
    Andrea Rossi

  • Rolando

    Dear Mr. Rossi, I’m geologist in Italy. My second language is french, then I beg your pardon for my mistakes in english.
    A question : if the reaction involve Nickel and Hydrogen at hight pressure, is it possible that this reaction happens in nature, at conditions on deep crustal ambient ? If we suppose thermolisis of water, hydrogen may be present, and nickel is a component of earth’s nucleo. What do you think about ?
    Thank very very much for your job.

  • Andrea Rossi

    Dear Sir,
    Our measurements have always been done starting cold. Besides,I remember you that I am talking of a device which has not worked for some hours, but of a device which is working from years now. The effect of the starting period upon the energy balance is of no relevance anyway.
    I want also to say that I am building a 1 MW generator, based on my invention, in a very important site in the USA, and my presentation, with all the necessary informations, woill be made at that point.
    Warm Regards,
    Andrea Rossi

  • John Fisher

    Dear Prof. Rossi,
    Flow of heat in nickel powder permeated by hydrogen is carried from particle to particle largely by hydrogen molecules across the gaps between particles. Direct flow via metal-to-metal contact is small owing to the small areas of contact. If the space between particles is shorter than the hydrogen mean free path, thermal conductivity of the powder is roughly proportional to pressure. When pressure is increased thermal conductivity is increased, and energy flows more rapidly out of the nickel powder to the calorimeter. By continually adjusting the pressure, the power to the calorimeter can be stabilized and held constant for a period of time until the energy initially present in the powder has been depleted.
    It is understandable that you could come to believe that hydrogen pressure was responsible for increasing the rate of a nuclear reaction; that output energy to the calorimeter was equal in magnitude to the nuclear energy being generated; and that in consequence startup energy could be neglected when making an energy balance. However I believe that you are mistaken in this belief, and that output energy is simply withdrawal of startup energy.
    The question can be settled by a single experimental run in which total input energy, including startup energy and machine operation energy, is measured from a cold start.
    Because others are considering efforts to duplicate or extend your work, I encourage you to perform this experiment and to make the result public. If output energy exceeds input energy I will be embarrassed, but all interested parties will be encouraged to accept what you have done and to build on it. If output energy is less than startup energy the interested parties will take that into consideration in their research plans. Either way you will have performed a useful service.
    I am sharing this note with members of the CMNS community.
    John Fisher

  • Pierre Clauzon

    Dear Andrea,
    No news, good news ?

    PS: your answer on the gammas rays which are the origin of the power seems very correct but is quite questionnable around me. Could you give us more details on that point without of course giving confidential matters?

  • Jean de Lagarde

    Thank you again. At what temperature, is your one megawatt unit supposed to be working ? Can this temperature be raised in the future for other units ? How is the heat extracted from the shield ?

  • Andrea Rossi

    It is not mandatory, but for many reasons I can’t explain, for the safety of the operation is better that the electric flow goes on.
    Warm Regards,
    Andrea Rossi

  • Jean de Lagarde

    Thank you very much Mr Rossi for your kind answers. As I understand, in normal functionning there is a steady flow of electric energy going in and a steady flow of thermal energy, about a hundred times (or more) greater, going out. Is the electric flow mandatory ? In other words, what happens if you switch off the current ?

  • Andrea Rossi

    The shielding is necessary to allow the radiations to be thermalyzed: it’s from the thermalization of the gamma rays that comes the energy. The system is radiation free outside the reactor, and doesn’t produce radioactive wastes.

  • Hello I red a bit of your site which I found absolutely by mistake while doing a bit of online research for some of my projects. Please write lots more as it is rare that somebody has something appealing to say about this. I will be watching for more!

  • Ludwik Kowalski

    1) I suspect Andrea is away. Otherwise, the above question would have been answered by now. My guess is that the answer is either 10 or 9. Otherwise the apparatus would not be called reliable. But a guess is less useful than what is based on personal experience of the designer.

    2) Numbers in Table 2 (see Rossi and Focardi paper) are easy to verify by using an atomic mass table. But this is not true for numbers in Column 2 of their Table 3. The description states that they are “total energies obtainable from all transformations.” What does this mean? For example, why is the total energy 41.79 MeV  for A=58, and why is it  10.61 MeV for A=59?  Please list all transformations to be considered for A=58 and for A=59.

  • Ludwik Kowalski


    Suppose experiments of type A, for example, are repeated ten times. And suppose the “success” is defined as “at least 24 kWh in 24 hours.” This definition of success is deliberately low; your reported results, for three experiments of type A, are 83 kWh, 165 kWh, and 40 kWh.

    You say that the apparatus is reliable now. Does this mean that you would expect each of these ten experiments be successful? Yes, I know that “to expect” does NOT mean “to guarantee.” You have many uncertain factors to consider. My question refers to the ideal world. I am assuming that the remaining problems to solve are technical, not financial.

  • Jean de Lagarde

    Someone spoke of lead shielding. Is it the case or is the apparatus radiation free ?

  • Andrea Rossi

    It’s supposed to work continuously, until the charge is consumed, and this takes about 6 months.
    In a modular array there is always one module more than necessary, so when a recharge or a reparation is made there’s no decrease of production.

  • Andrea Rossi

    It’s OK.
    Thank you anyway for the correction

  • Jean de Lagarde

    Is your one megawatt unit supposed to work continuously or will it be necessary to stop it from time to time to restore it before restarting ?

  • kowalskil

    There was a typing error in my sentence above. The word NOT was omitted. Here is the corrected version:

    Yes, I know that “to expect” does NOT mean “to guarantee.” This was probably obvious to most of you. But I prefer to post the correction.

  • Andrea Rossi


  • Andrea Rossi

    Thank you for the attention. Yes, the power density is 1w/cc

  • Ludwik Kowalski

    Reading the answer to the Jean’s question, I see that the 1 MW plant will consist of 50 modules, 20 kW each. The volume of each module will be 20 liters (this is only 5 gallons). In other words 1 watt of power per cubic centimeter.

  • Ludwik Kowalski

    And I thank you for answering, Mr. Rossi.

    My question about reproducibility on demand was about the present, not about the past. So let me repeat it. Suppose experiments of type A, for example, are repeated ten times. And suppose the “success” is defined as “at least 24 kWh in 24 hours.” This definition of success is deliberately low; your reported results, for three experiments of type A, are 83 kWh, 165 kWh, and 40 kWh.

    1) You say that the apparatus is reliable now. Does this mean that you would expect each of these ten experiments be successful? Yes, I know that “to expect” does mean “to guarantee.” You have many uncertain factors to consider. My question refers to the ideal world. I am assuming that the remaining problems to solve are technical, not financial.

    2) You wrote: “Soon we’ll put in operation the first section of the 1MW plant, in the USA and when we will have everything well in operation we will communicate the data.” Does “everything” refer to “the entire plant” or does it refer to the “first section” only? What will the power of the first section be?

    Have a good day,

    = = = = =

  • Andrea Rossi

    Dear Mr Kowalsky:
    Thank you for your attention.
    Your calculation of the enrgy necessary to make a delta T of 80 Celsius in a liter of water, of course, is correct.
    Here are the answers to your questions of the points 1,2,4,5 of your comment:
    1- the question was which is the power of the resistances used to activate the reactors
    2- I deem this year: I am working very hard, now in the USA on the toy
    4- because there was a contamination of the fuel
    5- the results reported are paradigmatic of the relative campaigns; if you are asking if we had unsuccessful tests, the answer is yes. I struggled years before making real energy. How many times I worked for nothing? Thousands.
    The core of the novelty is that now the apparatus is reliable. Our modules have a fixed power of operation and just make their job.
    But you are right: to arrive here has been a Calvary.
    Andrea Rossi

  • Ludwik Kowalski

    1) Above this box I see this answer: “From 100 w through 1000 w of power.” What was the question?

    2) My first question has to do with how Andrea answered Jean. He wrote: “Soon we’ll put in operation the first section of the 1MW plant, in the USA and when we will have everything well in operation we will communicate the data.” That is very exciting. How should the word “soon be interpreted?”

    Does it mean “this spring?” Does it mean “this summer?” Does it mean “this decade?” What is your plan now?

    3) My second question has to do with the table in which results are reported (methods A, B, and C). Energies are expressed in kWh. Unless I made an arithmetic mistake, the amount of energy needed to increase the temperature of one liter of water from 20 C to 100 C (from the room temp to boiling) is close to 0.1 kWh. For one gallon the amount is of energy would be about 0.4 kWh.

    In the the second experiment B you produced 3750 kWh in two days (March and April). That translates into boiling

    3750/0.4=9375 gallons of water. That is indeed very impressive. Even in the experiment C, lasting one day,you would boil 8 gallons of water.

    Three experiments A, lasting one day each, seems to be reproducible, if success is defined as “unless 30 kWh of excess energy.”

    4) Why do you say that the third experiment A was an anomaly?

    5) How many experiments of type A were performed? This question has to do with reproducibility. I suspect that some experiments produced much less excess heat per day than 30 kWh and were not reported. Is this a good guess? The same question can be asked for the experiments of type B.

    Ludwik Kowalski

    How many experiments of type A did not produce at least 30 kWh in one day?

    Are you reporting only successful experiments or are you performing reproducible


  • Andrea Rossi

    From 100 w through 1000 w of power.
    Warm Regards,
    Andrea Rossi

  • John Fisher

    Your answers to questions have improved my understanding of the reactor. You are achieving remarkable results. I have another question: What is the amount of start-up energy employed to initiate nuclear reaction in the methods A, B, C of Table 1?

  • Andrea Rossi

    Dear Pierre:
    Thank you for your kind message. Please read the answer I gave to Dr Jean de Lagarde: it answers to part of your questions. For the remaining part, there are at the moment concerns about the proprietary parts of the know how:you are perfectly right saying that witout more substantial details you can’t reproduce the effect, but we have to decide, with our US Partners, the limit between possible divulgation and industrial technological property. It is not an easy issue, also considering the ethics involved. Anyway, after our unit in the USA will have been started, we will give more informations. Anyway, I understand you, perfectly and from your point of view you are right.
    Andrea Rossi

  • Andrea Rossi

    Yes, we are making a 1 MW power reactor constituted by 50 modules of 20 kw each: I prefer for safety reasons to add series and parallels with the cooling fluids, not making bigger reactors, to maintain small and well tested reactors which we learnt perfectly to control. Soon we’ll put in operation the first section of the 1MW plant, in the USA and when we will have everything well in operation we will communicate the data. We want not to make press conferences if we have not an industrial plant operated not by us, as it has been up to now, but by the very high level Institution which is going to test and use it in the USA. This to avoid the usual logosturbations which affected the scientists who worked in this field.
    Andrea Rossi

  • Andrea Rossi

    A module with a power of 20 kw has a volume of 20 liters and weights 30 kg. Bigger powers are made with more modules, because for safety reasons I prefer to add up series and parallels with the cooling fluids , not with the reactors, to maintain small energy reactors.
    Andrea Rossi

  • Pierre Clauzon

    Dear Andrea, very sincere congratulations for your success, for you and your team. But in the past, we were quite disappointed with the BLP achievements on Ni-H experiments since years and years… So , could you tell us when you will give us more details on your experiments, operation, control, lack of nuclear ashes, general views etc. Duplication in others labs would be also a guarantee of validity…

  • Jean de Lagarde

    Can you confirm that you are working in the megawatt range ?

  • John Fisher

    I have been thinking about how to use your nuclear machine, and it would help if I knew how big it is. Can you tell me its approximate weight?

  • Andrea Rossi

    Be sure, if the article respects the law and is scientifically sustainable, we’ll publish it: of course, the publication will be for free and you will remain free to publish also in other magazines.

  • John Fisher

    Thank you. You have been very helpful. Now I have a much better understanding of your new energy source.

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  • Andrea Rossi

    The data reported are related to one hour measurement of the method C. Eventually, other measurements made in 24 hours per day operation have maintained the same efficiency

  • John Fisher

    Sorry I did not ask my question clearly. I will try again. Method A energies were measured for brief time periods (1-1.5h). Method B energies were measured for long time periods (15 days, 53 days). Method C energies were measured on date 2009-6-24 but length of time period was not mentioned. My question: what was the time period for method C measurement?

  • Andrea Rossi

    Temperature are registered in a datalogger always connected with the reactor,s loop.

  • John Fisher

    I have one more question: how long was the period of time during which method C input and output energies were measured?

  • Andrea Rossi

    Yes, Sir.

  • John Fisher

    In your comment does 30′ mean 30 minutes?

  • Andrea Rossi

    No, the reactor doesn’t work that way. As I said it’s far more complex. I can say that: the measurements start after about 30′ from the start up, because at that point it is surely stabilized.

  • John Fisher

    Now I try to ask my first question again without raising confidential issues. Did you wait until nuclear reaction began before turning off start-up power and starting measurements (so you could measure nuclear operation without interference from start-up)?

  • Andrea Rossi

    Because we wanted to make a test with a lower energy to test a different configuration, destined to other kinds of employ.

  • John Fisher

    I have another question. According to Table 1 the average output power of the method A experiments is about 77kW. The average output power of the method B experiments is only about 3 kW. Why is the method B output power so small?

  • Cheers for giving that website link, but in my case at any rate the destination doesn’t work. Perhaps it’s just a problem here, and does anyone have an alternative or mirror?

  • Andrea Rossi

    Actually, is more complex. You are asking confidential issues. Sorry.

  • John Fisher

    As I understand it you can control the rate of energy production in the nickel by adjusting the hydrogen pressure, and this method was used to maintain constant output power during the periods of energy measurement. Is this correct?

  • Andrea Rossi


  • Andrea Rossi

    Thank you: we work hard and messages like yours make our job easier.
    We all of the
    Journal of Nuclear Physics

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