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by
Valeriy Y.Tarasov
E-mail: vytarasov@yahoo.com
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Read the whole article
Download the ZIP file
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Abstract
The h-space theory is a variant of unified physical theory – a theory of everything.
This theory was built de novo, as the existing physical theories are incompatible and so unsuitable for unification.
A new approach is needed, and has been developed by re-evaluating the definitions of primary physical concepts.
The starting point for the re-evaluation was the following equation – Et = mvL, where energy – E, time – t, length – L, mass – m, velocity – v.
Analysis of these physical concepts resulted in the construction of a unique equation of the primary concepts such as space, length, energy and velocity.
From this, models could be developed that explain all well-known physical phenomena.
In addition, h-space theory predicts phenomena rejected by the current mainstream theories, such as limits to gravitational and electrostatic interactions, and the possibility of cold fusion (as a consequence of the electric charge definition, a modification of Coulomb’s law and the definitions of elementary particles in h-space theory).
The final section of this article describes a number of experimental tests that could be used to verify the h-space theory.
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Read the whole article
Download the ZIP file
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Dear Andrea Rossi,
Some of the differences between the eCat reactor (The Lugano experiment) and the experiment performed by Prof. Parkhomov (the Russian experiment) are:
a. the Lugano eCat had the dimensions of 200 mm in length, an outer diameter of 20mm and an estimated (by me) inner diameter of 12.7mm or 1/2″ while the Russian experiment had a length of 120mm, an outer diameter of 10mm and an inner diameter of 5 mm. So the ratio of the eCat to the Russian interior volumes is different by a factor of 10.
b. The Lugano eCat had 1 gram of fuel which I estimated to contain 0.55 grams of nickel, up to .389 grams of iron, and 0.061 grams of LiAlH4. The Russian experiment had 1 gram of nickel and 0.1 grams of LiAlH4. So the Russian experiment had 1.64 times the amount of LiAlH4 than the Lugano experiment.
c. The Lugano experiment used IR measurements of the surface temperature of the reactor while the Russian experiment used a phase transition of water (i.e., boiling) and loss of mass to estimate heat generation.
The Russian report estimated the interior pressure of the reactor at 100 atm. Given the ratio of the volumes and the mass of the LiAlH4, one would estimate the interior pressure of the Lugano experiment at around 6 – 10 atm – a valve previously reported to be in the correct range for eCat operation.
My conclusion is that the Russian experiment lacked the “catalyst” that enables LENR+ reactions are lower pressures. So this is a partial replication of the Lugano experiment.
Dear Dr Andrea Rossi:
How will you spend the New Year first day?
JC Renoir
The test made from the Russian scientist Alexander Parkhomov in his university laboratory seems to me very important
Dr Andrea Rossi:
Prof. Alexander Parkhomov has replicated the Rossi Effect: from Russia, with love.
You should be very welcome here.
Happy New Year,
D.T.
BroKeeper:
We are working to increase the maintainance time.
Warm Regards,
A.R.
http://www.ehs.wisc.edu/chem/LessonsLearned-LabExplosion.pdf
Here is one of the reports I have found of accidents with lithium aluminum hydride in an academic setting.
It shows the compound is indeed very dangerous – even in the hands of chemists.
Dear Andrea,
Are you still planning to refuel the customer’s E-Cats after six months or have you considered increasing the maintenance times? If so, how much have hoped to extend it?
Respectfully,
BroKeeper
Steven N. Karels:
I do not know the particulars of the replication made by the Russian scientist and I do not know how simple it is. The E-Cat is an extremely complex apparatus, though. Independent scientists can do all the experiments and analysis they deem opportune, without us having any voice on their laboratory work, whatever it is.
Warm Regards,
A.R.
Hank Mills:
Professionals know how to handle those materials in a laboratory. With non professionals I cannot get the liability to teach how to handle dangerous materials in the context of very dangerous experiments. The sole good sense based suggestion I can give to non professionals is to avoid to make this kind of experiments, unless directed by expert professionals, perfectly aware of all the safety issues implied by this kind of work.
Warm Regards,
A.R.
Wladimir,
1. Since background neutrons continually hit the nuclei of atoms, how does QRT explain the persistence of 1H1 in the Universe in a very large proportion? Should not 1H1 turn into 1H2, 1H3, etc, with all of these isotopes later decaying into 1H2, 1H3, or 2He3, but never back into 1H1?
2. How does QRT explain the following phenomenon:
For the lighter nuclei, continually adding a neutron will create a continual decrease in the half-life of the isotopes.
But for the heavier nuclei, starting from about 10Ne20, there is an alternating decrease and increase in the half-life of the isotopes.
All the best,
Joe
Dear Andrea,
For professionals in a laboratory – not amateurs dangerously playing around – what is the best reference you have found for the safe handling of lithium aluminum hydride?
Dear Andrea Rossi,
With the independent replication of the Rossi effect by a Russian scientist and with him using a rather simple experiment, it seems to me that the Rossi effect could be duplicated by the “masses”. Specifically, with an Alumina tubing, some alumina cement mixture, a few grams of nickel powder, some LiAlH4 and some heater wire, one could run the “non-professional eCat” for an extended period of time (say one year) and then analyze the “ash” products. Thoughts?
ing. Michelangelo De Meo:
Thank you for the information regarding the replication of the Rossi Effect made by an independent scientist.
Quite interesting indeed.
Warm Regards
A.R.
dear Dr. Rossi, the Russian researcher of Lomonosov Moscow State University, who says he has replicated the ECAT has different skills in different and interesting publications.
SKILLS AND EXPERTISE (11)
Radioactivity, Radiation Detection, Radiation, Radiation Physics,Ionizing Radiation
Radiation Protection, Nuclear Astrophysics, Radiation Measurements, Radiation, Dosimetry, Gamma Spectrometry,Dark Matter
http://www.researchgate.net/profile/A_Parkhomov/publications
Hello Dr. Rossi, I am sending you a report of a Russian researcher who says he REPLICATED end the Cat and confirm the “effect Rossi”. This is FANTASTIC !!!
Test of the replication design of High-temperature E-cat reactor of Rossi
Aleksandr Georgievich Parkhomov
Report at the workshop on Cold Fusion and Light Ball at the People’s Friendship University of Russia [1]
(Translation from Russian language by Stoyan Sarg/Bob Greenyer)
Conclusion (by A. Parkhomov): The replication test of the High-temperature E-cat reactor of Rossi loaded with a mixture of nickel and Lithium-aluminum-hydride shows that at temperature of 1000°C and greater this device really produces greater output energy than the input one.
https://docs.google.com/document/d/1Y3Bxr_aE2iosEKpGFUZiQgAcuT8AFN78RFCAlR-JqNw/edit?pli=1
Robert Curto:
Neil De Grasse: “Carneade, chi era costui?”
Warm Regards,
A.R.
Robert Curto wrote in December 28th, 2014 at 9:25 AM
I do not know him personally, but I pray for him to come here to answer you, and to give to the Readers of this blog the counterpoint from the point of view of the Standard Model, that I think is the right way ti go.
———————————————
Dear Robert,
in spite of to claim that a sphere has an elipsoidal shape is a masterpiece among the nonsenses said by Dr. JR, however the best among the masterpieces is his assumption saying that neutrons cannot be expelled from a nucleus due to the centrifugal force, because the centrifugal force is a ficticous force, and so it cannot expell the neutron out of the nucleus.
Robert,
as you are sure that Dr. JR is an infallible expert in Physics, I invite you to make the following experiment:
1) Take your car, and put it moving with 200km/h.
2) With the car moving with 200km/h, you can drive the car with no problem in a curve where the maximum speed allowed is 40km/h.
Dr. JR is sure you can make the turn without flipping over the car, because the centrifugal force is a ficticious force, and therefore the centrifugal force cannot expel your car out of the road.
Please, do it, and prove us that Dr. JR is right.
We will be waiting you come back here to tell us about the comprovation of the Dr. JR assumption.
If you survive… of course.
Good luck
regards
wlad
Joe wrote in December 28th, 2014 at 1:41 AM
Wladimir,
How does QRT explain the instability of 4Be10?
Since 4Be9 is stable due to the spin-interaction between a neutron and a deuteron, why can not the remaining deuteron have spin-interaction with another neutron?
———————————————
Joe,
first of all,
I would like Dr. JR come here to explain why 4Be10 is not stable by considering the nuclear models based on the Standard Model.
After all,
since 4Be9 is stable due to the spin-interaction between a neutron and a deuteron, why can not the remaining deuteron have spin-interaction with another neutron, by considering the Standard Model?
Joe,
look at the Figure 14 at the page 18 of the paper Stability of Light Nuclei, published in the JoNP, where is shown the structure of 4Be8:
http://www.journal-of-nuclear-physics.com/files/Stability%20of%20light%20nuclei.pdf
The 4Be8 is not stable because due to the perfect symmetry of the nucleus the two deuterons D1 and D2 are captured by the central 2He4, because of the spin-interaction. So, the 4Be8 is desintegrated in two alpha particles 2He4.
In the case of 4Be10, each deuteron of the 4Be8 captures a neutron. However, with the capture of the neutron, because the mass of each 1H2-n is greater, they move away from the central 2He4 by the action of the centrifugal force, and the 4Be10 gets stability.
The 4Be10 gets stability along 1,39×10^6 years, which is its half-life.
After 1,4 million years the 4Be10 decays to 5B10, by beta decay.
regards
wlad
Dr. Rossi, what do you think of the tweet of the astrophysics Neil De Grasse,
who said that December 25 is known as the birthday of Isaac Newton ?
Robert Curto
Ft. Lauderdale Florida
USA
Wlad, Probably Dr. JR has not time to answer you again, after all the answers
he already gave you, and thinks to answer you further is worthless.
I do not know him personally, but I pray for him to come here to answer you, and to give to the Readers of this blog the counterpoint from the point of view of the Standard Model, that I think is the right way to go.
Robert Curto
Joe,
I dont understand why Dr. JR did not come here to explain why the nuclei do not capture neutrons from the background of neutrons.
After all, Dr. JR claims that it is possible to explain everything by considering the Standard Model.
I would be very glad if Dr. JR comes here to explain why the oceans and rivers are not formed by heavy water D2O.
Because if he does not come here, the readers may be stay with the wrong impression that Dr. JR does not know to explain the puzzle.
Dear Robert Curto,
may you please invite to Dr. JR to come here for explaining to us how this puzzle is solved by the Standard Model ?
Please, Robert,
convince Dr. JR to come here to give us the pleasure of hearing his wise explanation.
Many thanks
wlad
Wladimir Guglinski
December 27th, 2014 at 6:48 PM
Joe wrote in December 27th, 2014 at 4:27 PM
A neutron must have some specific energy in order to settle stably within a nucleus. That is why a shell model exists for the nucleus, just like one does for the electrosphere of the atom.
———————————————–
Joe,
and what about the proton?
There is not any shell around the proton, according to the Standard Model. Surrounding the proton there is only its positive electric field.
So, as the neutron has no charge, any low energy neutron would penetrate in the proton’s electric field, and they would form the deuterium, because the proton and the neutron have attraction through the strong force, according to the Standard Model.
Therefore there would not exist hydrogen in the universe, because all the protons would be converted to deuterium by the capture of a neutron from the background of neutrons existing in the enviroment.
And the oceans would be formed entirely by heavy water composed by D2O, instead of H20.
We could perform the Fleischmann-Pons with the water of the oceans, without the need of doing distillation for the obtainment of the heavy water.
regards
wlad
Wladimir,
How does QRT explain the instability of 4Be10?
Since 4Be9 is stable due to the spin-interaction between a neutron and a deuteron, why can not the remaining deuteron have spin-interaction with another neutron?
All the best,
Joe
Joe wrote in December 27th, 2014 at 4:27 PM
Wladimir,
You write,
“And because protons and neutrons have attraction by the strong force, when the neutron hits the nucleus it must be captured.”
It is true that neutrons are always hitting the nucleus, but they are not always captured since we do have isotopes which are stable. A neutron must have some specific energy in order to settle stably within a nucleus. That is why a shell model exists for the nucleus, just like one does for the electrosphere of the atom.
———————————————–
Joe,
the shell model is wrong, because it is unable to explain a lot of nuclear properties. If the shell model was correct, there would not need to propose other nuclear models, as the Collective Model, The Model of Layer, the Fermi’s model, the model of Clusters, etc.
Each one of these models are proposed because no one of them is able to explain all the nuclear properties.
Concerning the shell model, for instance there is no way to explain the levels of energy of the nuclei.
Also, the calculation of the binding energy of the nuclei by the shell model is possible only after the oxygen nucleus. The shell model does not work for the lightest nuclei.
Besides, light nuclei have no sufficient quantity of nucleons, in order of forming a shell model. A shell model needs to have the shape of a orange’s peel.
For instance, consider the 3Li6. It has only 6 nucleons. There is no way to form a shell shape. Then a neutron with any low energy could be captured by 3Li6, transmuting to 3Li7.
But also 3Li7 cannot have the shape of an orange’s peel, and so 3Li7 would have to capture a neutron from the background of neutrons, transmuting to 3Li8, which is no stable. 3Li8 transmutes to 4Be8, which decays in two alpha particles.
Therefore 3Li6 and 3Li7 could not exist in the nature.
4Be9 has only 9 nucleons. It also cannot have the shape of a orange’s peel. So 4Be9 would have to capture a neutron, and transmute to 4Be10, which is no stable and transmutes to 5B10.
As 4Be9 is the unique stable isotope of beryllium, then beryllium could not exist in the universe.
There is a lot of puzzles impossible to be explained via the nuclear models based on the Standard Nuclear Theory.
If you succeed to solve the puzzles of Nuclear Physics, you will win the Nobel Prize in Physics.
Good luck.
regards
wlad
Tom Conover:
Thank you for the link. Yes, it is quite interesting.
Warm regards,
A.R.
Wladimir,
You write,
“And because protons and neutrons have attraction by the strong force, when the neutron hits the nucleus it must be captured.”
It is true that neutrons are always hitting the nucleus, but they are not always captured since we do have isotopes which are stable. A neutron must have some specific energy in order to settle stably within a nucleus. That is why a shell model exists for the nucleus, just like one does for the electrosphere of the atom.
All the best,
Joe
Dear Dr Rossi,
Here is a link to a google translation of the Lugano replication for interested readers. It seems to be quite interesting.
Tom
https://www.scribd.com/doc/251130826/Parkhomov-Alexander-Rossi-Replication-Paper-2014-12-25
Ing. Michelangelo De Meo:
Thank you very much, very interesting and well done.
Warm Regards,
A.R.
Dear Dr. Rossi is now that the plant Hot Cat is put on the market to save the world from extinction.
Beautiful animation that explains the history of our civilization, how it has changed with the advent of oil, which has reached the limits and what we can do.
https://www.youtube.com/watch?v=uA2WYGIFwwo
Keith T.:
Excuse me, but why have I to pull a cart with square wheels?
Warm Regards,
A.R.
Dear Andrea,
For the current generation of thermoelectric generators of 5% to 8% efficiency, if these were driven by an Ecat you would require a COP of from 12 to 20 just to break even.
With your current success of achieving self sustain mode of up to 2 hours, with further research and development is there a possibility of achieving a sufficiently high COP for either the Ecat or Hot Cat to enable current commercial thermoelectric generators to be used for mall scale electricity production.
Kind regards,
Keith Thomson.
AlainCo:
Thank you for the important information.
I do not know the particulars, therefore cannot comment, but it is possible that the so called “Rossi Effect” is replicable after the data published in the Report of Lugano.
Warm Regards,
A.R.
Dear Mr Rossi,
you are probably aware of a successful replication tentative of your technology by a Russian scientists “Alexander G. Parkhomov” using mass calorimetry.
https://yadi.sk/i/sZBuzGRWdeE4c
http://www.e-catworld.com/2014/12/27/lugano-confirmed-replication-report-published-of-hot-cat-device-by-russian-researcher-alexander-g-parkhomov/
what is your opinion on that result.
do you think it would be possible for your company to organize a similar test of your E-cat, with Pr Parkhomov, or at least following the same protocol.
Steven N. Karels wrote in December 26th, 2014 at 3:33 PM
Wlad,
Remember the iron is my estimate based on The Report. It could be much lower. Perhaps the iron is added for magnetic reasons? Such as to increase the inductance?
—————————————————
Steven,
as Andrea Ross never responds any question concerning the interior of the eCat, I asked to God, and He told me the following:
In the last page of the Lugano Report is writen:
Besides the analyzed elements it has been found that the fuel also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and these are not found in the ash.
As the percentage of Fe is smaller than the percentage of Ni, then all the quantity of Fe is transmuted before the consumption of Ni.
But the eCat continues working after the total consumption of Fe, and therefore the iron does no play any fundamental role in the Ni-Li reactions, even concerning the inductance.
The iron only contributes producing heat because it has fusion with 3Li7. However the portion of the heat produced by Fe can be produced by Ni if iron is not used in the fuel.
regards
wlad
Joe wrote in December 26th, 2014 at 9:33 PM
Wladimir,
Since there exists a shell model of the nucleus in standard physics, would not therefore every nucleon, including neutrons, need a certain amount of energy to be able to place itself within the nucleus in a stable way?
————————————————–
why?
The shape of the distribution of protons and neutrons in the nucleus has nothing with the penetration of a particle into the nucleus, because the repulsion on the proton happens in the electrosphere of the nucleus. The charge of electrosphere of the nucleus is the same no matter what nuclear model is considered (the charge of the electrosphere depends only on the quantity of protons of the nucleus).
The proton cannot penetrate into the nucleus because it has positive charge. And as the nucleus has positive charge, the proton is repelled by the positive electrosphere of the nucleus.
The neutron has no charge, and therefore it can penetrate the positive electrosphere, and hit the nucleus, no matter what is the kinetic energy of the neutron.
And because protons and neutrons have attraction by the strong force, when the neutron hits the nucleus it must be captured.
regards
wlad
Robert Curto:
Thank you for your information.
Warm Regards,
A.R.
Wladimir,
Since there exists a shell model of the nucleus in standard physics, would not therefore every nucleon, including neutrons, need a certain amount of energy to be able to place itself within the nucleus in a stable way?
All the best,
Joe
Joe,
by considering the nuclear models of the Standard Model, actually there would be no any stable element existing in the nature.
For instance, the stable isotopes of neon are 20Ne, 21Ne, 22Ne.
By capturing a neutron from the background of neutrons, 20Ne would transmute to 21Ne.
21Ne would capture a neutron and transmute to 22Ne.
22Ne would capture a neutron and transmute to 23Ne
But 23Ne is no stable, and transmutes to 23Na.
So, stable isotopes of neon would have do not exist in the nature.
.
Beryllium has only one stable isotope: 4Be9
But it would have to capture a neutron, and to transmute to 4Be10.
4Be10 is not stable, and decays to 5B10.
So, no one stable isotope of beryllium could exist.
.
Carbon has two stable isotopes, 12C and 13C.
By capturing a neutron, 12C would transmute to 13C.
By capturing a neutron, 13C would transmute to 14C.
But 14C is no stable, and decays to 14N.
So, no one stable isotope of carbon could exist.
.
By considering the nuclear models based on the Standard Model, we would be living in a universe without stable elements.
regards
wlad
Joe,
according to my Quantum Ring Theory,
in order to penetrate into a nucleus, a neutron must perforate the gravitational flux n(o) of a nucleus:
http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png
So, according to my QRT, only neutrons with a minimum energy can enter into a nucleus.
However,
the neutron has no charge, and so from the nuclear models of the Standard Nuclear Physics any low energy neutrons would have to succeed entering into a nucleus, and be captured. There would not be a minimum energy required for a neutron entering a nucleus.
Therefore many nuclei could not exist.
For instance, the stable 10Ne20 could not exist, because it would capture a neutron from the background of neutrons, and transmute to 10Ne21.
Also 10Ne21 could not exist, because it would capture a neutron and would transmute to 10Ne22.
I would like to invite some nuclear physicists, like Dr. JR, Dr. Wilfried Nörtershäuser, Dr. S.Lakshminarayana, Dr. Martin Freer, etc., so that they come here to explain how is possible the existence of stable isotopes like 20Ne and 21Ne (among a lot of other isotopes, as for instance 24Mg, 25Mg, 56FE, 57Fe, etc, etc, etc… all they could not exist).
However,
unfortunatelly it is a waste of time to invite the nuclear physicists to come here to explain some questions, when the questions defy the Standard Nuclear Physics.
Joe,
dont you think there is a lack of honesty among the nuclear physicists in general ?
Do you think is it possible to advance the Physicis from such lack of honesty?
regards
wlad
Dr. Rossi, Feed-in Tariff is in effect in 50 Countries.
It allows home owners, and small business owners, who generate their own electricity with wind, solar, etc. to sell the electricity they do not need back to the Grid !
I believe E-Cat will have an easy time to qualify for this benefit.
E-Cat will not need any help, it will supply more electricity then they need.
But this will be a bonus on top of a bonus.
I can see millions of E-Cat owners selling electricity to the Grid at a cost of less then their present cost.
What ever they pay the E-Cat owners is 100% profit.
I think they call this a win win situation.
For more information Google:
Feed-in Tariff
And read 10 or 20 websites.
Robert Curto
Ft. Lauderdale Florida
USA
Steven N. Karels wrote in December 26th, 2014 at 3:33 PM
Wlad,
Remember the iron is my estimate based on The Report. It could be much lower. Perhaps the iron is added for magnetic reasons? Such as to increase the inductance?
—————————————————
Dear Steven,
only two persons know it: Andrea Rossi, and God
regards
wlad
Joe wrote in December 26th, 2014 at 3:53 PM
Wladimir,
You write,
“It is the charge of the electricitons which give the electric charge of the proton.”
Since protons and electrons are different sizes, how do you explain that they have the same charge?
————————————————-
Joe,
this is explained in my paper Aether Structure for unification between gravity and electromagnetism, submitted for publication in JoNP six months ago.
Be patient and wait the publication.
regards
wlad
Wladimir,
You write,
“It is the charge of the electricitons which give the electric charge of the proton.”
Since protons and electrons are different sizes, how do you explain that they have the same charge?
All the best,
Joe
Wlad,
Remember the iron is my estimate based on The Report. It could be much lower. Perhaps the iron is added for magnetic reasons? Such as to increase the inductance?
Giannino Ferro Casagrande:
We are working very hard also on the small modules, whose technology is becoming stronger by the day, but stoll is pending the safety certification .
We are working also on that…
Thank you for your attention and kindness.
Warm Regards
A.R.
Joe wrote in December 26th, 2014 at 2:25 AM
Wladimir,
Do you think it is possible that gravitational fluxes n(o) also exist in the electrosphere, and not just in the nucleus, of an atom? They would be much weaker in strength since their radius would be much greater than the radius within the nucleus. This might explain similarities between the organizational structures of the electrosphere and those of the nucleus, such as the presence of shells.
——————————————-
Joe,
the gravitational flux n(o) exists in the electrosphere. For instance, you can see it in the electrosphere of the proton, in the figure:
http://peswiki.com/index.php/Image:FIGURE_1-_3_fields_of_the_proton.png
Regarding the formation of the electric field of the proton, in the figure we have:
1- Electric field (yellow)
2- Flux n(o): the blue lines in the electric field
3- Electricitons captured by the flux n(o): the figure shows only 4 electricitons, but there is an imense amount of them captured by each flux n(o). It is the charge of the electricitons which give the electric charge of the proton.
4- The electricitons move with the speed of light along the flux n(o). They have the shape of a ring, and the flux n(o) crosses into the ring.
regards
wlad
Sono già da un paio d’anni che seguo con gran simpatia l’evolversi continuo di queste entusiasmanti ricerche sulle LENR ! Ho conosciuto Lei egregio signor Rossi di persona quando ha esposto alcuni risultati sula fusione fredda a Pordenone ! Sin dall’inizio ho sottoscritto il mio impegno ad acquistare un E.CAT di tipo casalingo , quindi sono uno tra i primi ! Ora questa mia proprio per augurare a Lei e a tutta la Vostra squadra al lavoro sul grosso dispositivo da 1 MEGA ,assieme a tutti i lettori ed interventisti del sito JONP i più fervidi auguri per le ricorrenze delle festività ed in calce Le chiedo quali possibilità abbiamo noi piccoli clienti usufruttuari di poter disporre di un E-CAT casalingo ! Giannino di Udin
To the Readers of the JoNP:
I am continuing to receive comments asking information regarding the fuel, the composition of the powders, etc, and I am very embarassed to have to continue to answer ” I cannot give information about this issue”. To avoid this, from now on all the comments asking from me information about fuel and operation inside the reactors, beside all the information (very substantial) already given, such comments will be automatically spammed, without answer.
On the contrary, assumptions and theoretical speculations made from the Readers are welcome, obviously on the base of the fact that the publication of such speculations, theories and assumptions are totally independent from my position.
Again wishes for a Great 2015 to all,
Andrea Rossi
Element % by Weight
Nickel 55.0
Iron 38.9
Aluminum 4.3
Lithium (total) 1.2
Hydrogen (no Deuterium) 0.6
Total 100.0
———————————————–
Steven
I dont understand why Rossi put so much iron in the fuel.
In spite of the iron also is transmuted by reaction with 3Li7, however the iron causes a litle decrease in the COP, because Ni is the best fuel.
I guess he put so much iron in the eCat sent to be tested in Lugano with the aim to suggest to everybody that iron plays some important role in the reactions.
I think Rossi does not put iron in the fuel of the eCat used for the 1Mw plant
regards
wlad
Steven,
perhaps 3Li7 is not formed.
Probably 3Li6 captures a proton, transmuting to 4Be7.
The newborn 4Be7 has only the electrons of the 3Li6 isotope: 1s1, 1s2, 2s1.
The electron 2s1 of 4Be7 is shared with one Ni atom, and the proton of 4Be7 is captured by the orbit of the electron 2s1.
regards
wlad