10KW LENR Demonstrator?

[DancingFool]

Sometime ago, Rossi said that he did not use zinc in his powder. There was an old post back in these pages about it.

Right. And the patent application specifically says that zinc is a byproduct of the reaction. Of course, the "ash" analyzed by Kullander doesn't have any zinc. So where did it go?

Ah. Rossi must have discovered _another_ reaction, and that's what he uses now. What a genius that Rossi is!

[Axil]

Sometime ago, Rossi said that he did not use zinc in his powder. There was an old post back in these pages about it.

Right. And the patent application specifically says that zinc is a byproduct of the reaction. Of course, the "ash" analyzed by Kullander doesn't have any zinc. So where did it go?

Ah. Rossi must have discovered _another_ reaction, and that's what he uses now. What a genius that Rossi is!

Correct me if I am wrong, as I understand the Kullander ash revelation, it was only a partial list of the element components of the ash.

[Giorgio]

Rossi contradicts himself on occasion. This makes Rossi reading very hard.

He contradicts himself on occasions but never when he claims extra heat?

This IMHO is the exact reason why you cannot trust one word of miraculous discoveries without independent verification of results.

[Axil]

Rossi contradicts himself on occasion. This makes Rossi reading very hard.

He contradicts himself on occasions but never when he claims extra heat?

This IMHO is the exact reason why you cannot trust one word of miraculous discoveries without independent verification of results.

This is the reason why it is great to have you as an unprejudiced undercover reporter of Rossi reaction test results from the U of Begonia test comment on this forum.

Reliable…totally honest…unopinionated…objective…just the facts

[Giorgio]

This is the reason why it is great to have you as an unprejudiced undercover reporter of Rossi reaction test results from the U of Begonia test comment on this forum.

Reliable…totally honest…unopinionated…objective…just the facts

There is nothing to report. There is no test coming. All this UoB story has been till now (and is looking more and more) a big waste of time.

If (and I make it a big IF) the UoB will finally end up in testing the E-cat than there will be no discussions on their results and no need to have additional undercover info.
Apart from Rossi's friends all the other professors involved in the test are people that are used to make experiment in a professional way and are going to go straight to the point.

[DancingFool]

Correct me if I am wrong, as I understand the Kullander ash revelation, it was only a partial list of the element components of the ash.

Hmmm. You may be right. It's a judgement call, I guess. If Kullander reported anomalous iron levels, it seems likely to me that he would have also reported anomalous zinc levels. Particularly levels anywhere near what you'd expect from the patent application.

[TallDave]

Rossi, BLP, Brillouin... what is it about nickel?

[Axil]

Rossi, BLP, Brillouin... what is it about nickel?

I believe that three basic factors are required to make a working LENR system.

1-Quantum mechanical coherence (QMC). This is needed to convert nuclear radiation to thermal power and to amplify the effects of the Yukawa potential (also called a screened Coulomb potential). When QMC has not yet been established, the nuclear radiation that is being produced in nuclear reactions at active nuclear sites will escape as gamma and x-ray radiation before it can be transformed into heat energy.

Nuclear reactions affect atoms in a large assemblage of coherent and entangled atoms by sharing the energy produced in any one member of the QMC assemblage among all its members.

In such a coherent collection, what happens to one member of such a coherent collection happens to them all. An averaging effect takes place where the nuclear energy output of one atom is averaged over a hundred or more atoms in the coherent collection.

In the Rossi reactor, at startup, a large amount of gamma radiation appears before QMC has established itself because the temperature of the nickel has not gotten to the relatively low Curie temperature (nickel has the Curie temperature of 631 K (~358 C)). Formation of QMC is magnetic in nature.

When nickel is ferromagnetic it cannot join the QMC assemblages. Nickel must first be made paramagnetic by heat to join the QMC assemblage. The QMC amplifies both the nuclear power produced and the heat output conversion of the cold fusion reaction.

Rydberg matter produces the QMC effect in nickel as Coherent hydrogen molecules are ionically attracted to the surface of the nickel. The strong dipole field of the Rydberg matter produced by a large coherent circulation of electrons with very large dipole moments induces a powerful paramagnetic field in the nickel within a quantum mechanical “blockade” distance of up to a micron. This field of coherent influence is directly proportional to the size of the Rydberg molecule which in turn is proportional to the temperature and pressure of the hydrogen envelope.

As the hydrogen envelope grows hot, more nuclear reactions take place in the coherent nickel on the surface of the nickel and more efficient gamma radiation conversion to heat occurs.

This is why nuclear ash transmutation products are found imbedded in the surface of the nickel.



2 - Yukawa potential – to be supplied

3 – Rydberg matter – to be supplied

[D Tibbets]

* I have repeatedly struggled to explain this (often very poorly), but haven't succeeded, at least with some.
If you take a collection of ~ 63 free nucleons and combine them, then of course a lot of energy will come out. But if you add one nucleon, especially a proton, to Ni62 the energy balance is negative. Ni62 is the high water mark. You cannot pack a nucleus tighter, and that is what the thermal/ kinetic energy comes from. There are possibly combinations of smaller nuclei that could still release energy with a product past Ni62, but these are rare reactions and contrary to the Rossi's claims.

Dan Tibbets

Dude. That's what makes Rossi's work so, you know, revolutionary.

It is also what makes DT so, you know, wrong. Add a proton, binding energy zero, to ANYTHING (well, any nucleus) and you release energy. DT can't seem to understnad that. Don't know why that is.

Sigh, true, I indeed have not described it well, but the conclusion is not mine. It is that of several Nuclear fusion texts and physics web sites, which I have repeatedly quoted, and linked to.

Yet another probably confusing analogy. Hydrogen reacts with a lot of other elements to release chemcal energy (like with nickle). Consider hydrogen reacting with nitrogen. One hydrogen added to one nitrogen yields NH, and some exothermic energy/ heat. Add another hydrogen and you get NH2 and more heat. Add another hydrogen and you get NH3 and some more heat. Add another hydrogen and you get NH4, but now heat is absorbed, its is endothermic. Add some other combinations and you get additional heat absorption, the molecule is unstably and can fission , releasing energy in the process- an explosive ( comparable to the heavier elements that can undergo nuclear fission to release energy) .

As I have mentioned repeatedly, You cannot only consider a single combination of nucleons, but the mechanisms by which they combine- which is the derivation of the nuclear packing fraction, or nuclear binding energy per nucleon - This IS THE DETERMINING CONSIDERATION. The total binding energy per nucleus, is not the determinate of the energy release. It is the energy contained within the nucleus , not the energy released by the nucleus as it forms.
If you add a proton to Cl35, Fe 56, Ni 58, Ni62, Cu63, Ur 235, etc you are indeed adding energy to the product nucleus. Ignoring stability and probability issues, this is irrelevant to the energy output of the reaction. They are two completely separate issues. The total binding energy of the nucleus can be considered as the stored or potential energy of the nucleus. The nuclear binding energy per nucleon, can be considered as the surplus energy briefly stored by the formed nucleus which must be shed as heat/ kinetic energy through several possible pathways- gamma, beta, alpha, daughter product KE, etc.

So, why does the hydrogen (proton) not shed energy (exothermic) by itself. It is because I doens't have any energy to shed. The glueons hold the quarks together so well, that it is an entirely different ball game. Even then, some have argued that protons decay to a lower energy state, though they have failed to detect any experimental evidence of this.
The proton is the lowest energy state when considering quarks. Nickel is the lowest energy state when considering nucleons (protons and neutrons in combination). It is the tightest, most condensed state achievable for a nucleus. As you approach this state, whether from the low side or the high side, there is excess energy liberated- heat. As you move away from this state, you are adding energy to the system (note that I said system, not the nucleus alone), thus heat is absorbed.
Another analogy. A positive ion gains potential energy and loses kinetic energy as it moves away from a cathode. The energy gained or lost depends on the charge of the ion, not it's mass. A pos hydrogen ion or a 1 + gold ion gains or loses the same amount of kinetic energy despite the fact that the gold ion has a lot more mass/ energy stored in its nucleus. The velocity is different but that is irrelevant. If the velocity was the determinate of the energy exchange the total binding energy would be important, but the velocity is only a portion of what determines the kinetic energy of an object. It and mass essentially cancel out so that the interacting charges is what determines the acceleration. The nuclear binding energy per nucleon is comparable to this charge. It is maximum at Ni62 and represents the greatest delta energy that can be imparted- kinetic energy as it is approached, and potential energy as it is leaving.
It is the kinetic energy that you are interested in if you want an exothermic result for the system.

Dan Tibbets

[DancingFool]

It is also what makes DT so, you know, wrong. Add a proton, binding energy zero, to ANYTHING (well, any nucleus) and you release energy. DT can't seem to understand that. Don't know why that is.

Maybe because it isn't true. Look up the curve of binding energy. No, not the the book about Ted Taylor (although that's worth theprice of admission). Rather, for instance, http://hyperphysics.phy-astr.gsu.edu/hb ... ucbin.html , or the reliable standby, http://en.wikipedia.org/wiki/Nuclear_binding_energy
At a mass number of 56 - 62, binding energy peaks (the peak is shallow, so a precise number is largely irrelevant). The only way to get energy out of a nucleus is by driving it toward the peak, so once you're AT the peak, movement in either direction (fission or fusion) drives you away from the peak, and is endothermic.
And, just as a bonus, the steepness of the curve near hydrogen as opposed to uranium goes a long way toward explaining why fusion bombs are badder than fission.

[KitemanSA]

Dude. That's what makes Rossi's work so, you know, revolutionary.

It is also what makes DT so, you know, wrong. Add a proton, binding energy zero, to ANYTHING (well, any nucleus) and you release energy. DT can't seem to understnad that. Don't know why that is.

Sigh, true, I indeed have not described it well, but the conclusion is not mine. It is that of several Nuclear fusion texts and physics web sites, which I have repeatedly quoted, and linked to.

No, your incorrect conclusion is due to your assiduous mis-reading of such texts.

Yet another probably confusing analogy. Hydrogen reacts with a lot of other elements to release chemcal energy (like with nickle). Consider hydrogen reacting with nitrogen. One hydrogen added to one nitrogen yields NH, and some exothermic energy/ heat. Add another hydrogen and you get NH2 and more heat. Add another hydrogen and you get NH3 and some more heat. Add another hydrogen and you get NH4, but now heat is absorbed, its is endothermic. Add some other combinations and you get additional heat absorption, the molecule is unstably and can fission , releasing energy in the process- an explosive ( comparable to the heavier elements that can undergo nuclear fission to release energy) .

Totally beside the point. Not analogous at all.

As I have mentioned repeatedly, You cannot only consider a single combination of nucleons, but the mechanisms by which they combine- which is the derivation of the nuclear packing fraction, or nuclear binding energy per nucleon - This IS THE DETERMINING CONSIDERATION.

Absolutely true... for stability. Not for energy release.

The total binding energy per nucleus, is not the determinate of the energy release.

Yes, it is. Very simply and is the thing you refuse to understand.

It is the energy contained within the nucleus , not the energy released by the nucleus as it forms.

This statement is incorrect. The binding energy, per nucleon or total, it the energy released per nucleon or total. Learn that and you will finally get it.

If you add a proton to Cl35, Fe 56, Ni 58, Ni62, Cu63, Ur 235, etc you are indeed adding energy to the product nucleus.

Dan, you are adding ~14MeV per nucleon and absorbing ~5-7MeV into the nucleus, leaving ~7-9MeV to be released. The amount RELEASED is the binding energy.

Ignoring stability and probability issues, this is irrelevant to the energy output of the reaction. They are two completely separate issues. The total binding energy of the nucleus can be considered as the stored or potential energy of the nucleus.

No, the binding energy is the amount RELEASED. It is analogous to GRAVITATIONAL binding energy. Gravity binds things together RELEASING a lot of energy in the process.

The nuclear binding energy per nucleon, can be considered as the surplus energy briefly stored by the formed nucleus which must be shed as heat/ kinetic energy through several possible pathways- gamma, beta, alpha, daughter product KE, etc.

That is exactly what I said above. "Must be shed" = released. It is the energy RELEASED.

So, why does the hydrogen (proton) not shed energy (exothermic) by itself. It is because I doens't have any energy to shed.

Why doesn't a small asteroid, ~infinately far away from a black hole not shed any gracvitational binding energy by itself. Because BINDING is and act of bringing TOGETHER of two or more items. One single item can't BIND. Duhh!

The glueons hold the quarks together so well, that it is an entirely different ball game. Even then, some have argued that protons decay to a lower energy state, though they have failed to detect any experimental evidence of this.

An asteroid ~infinitely far from a black hole is at the lowest state too.

The proton is the lowest energy state when considering quarks. Nickel is the lowest energy state when considering nucleons (protons and neutrons in combination).

Lowest state of WHAT? SPECIFICALLY.

It is the tightest, most condensed state achievable for a nucleus. As you approach this state, whether from the low side or the high side, there is excess energy liberated- heat. As you move away from this state, you are adding energy to the system (note that I said system, not the nucleus alone), thus heat is absorbed.

Right, but whan you start with a proton, you are ALWAYS moving toward that state. Even if the other particle is UnUnNilium, adding a proton (or neutron) moves toward a MORE bound state.

Another analogy. A positive ion gains potential energy and loses kinetic energy as it moves away from a cathode. The energy gained or lost depends on the charge of the ion, not it's mass. A pos hydrogen ion or a 1 + gold ion gains or loses the same amount of kinetic energy despite the fact that the gold ion has a lot more mass/ energy stored in its nucleus. The velocity is different but that is irrelevant. If the velocity was the determinate of the energy exchange the total binding energy would be important, but the velocity is only a portion of what determines the kinetic energy of an object. It and mass essentially cancel out so that the interacting charges is what determines the acceleration. The nuclear binding energy per nucleon is comparable to this charge. It is maximum at Ni62 and represents the greatest delta energy that can be imparted- kinetic energy as it is approached, and potential energy as it is leaving.

Sorry, your analogy doesn't make any sense to me.

[KitemanSA]

It is also what makes DT so, you know, wrong. Add a proton, binding energy zero, to ANYTHING (well, any nucleus) and you release energy. DT can't seem to understand that. Don't know why that is.

Maybe because it isn't true. Look up the curve of binding energy. No, not the the book about Ted Taylor (although that's worth theprice of admission).
At a mass number of 56 - 62, binding energy peaks (the peak is shallow, so a precise number is largely irrelevant). The only way to get energy out of a nucleus is by driving it toward the peak, so once you're AT the peak, movement in either direction (fission or fusion) drives you away from the peak, and is endothermic.

Oh lord, another person who can't read.
If you start with two nuclei at or above that mass, I agree with you since they have already given up their binding energy. BUT, if you start with a proton (BE/A~13eV) or neutron (BE/A=ZERO!) and you add it to ANYTHNG, BE/A~8MeV, you will release ~8MeV. Smple math. Look back thru the several discussions, you may figure it out.

And, just as a bonus, the steepness of the curve near hydrogen as opposed to uranium goes a long way toward explaining why fusion bombs are badder than fission.

And the fact that OPs&Ns have ~ZERO BE/A explains why the release energy when they bind into something at about 8MeV BE/A.
You may want to dig thru this.
viewtopic.php?t=3137

[Joseph Chikva]

Rossi, BLP, Brillouin... what is it about nickel?

I believe that three basic factors are required to make a working LENR system.

1-Quantum mechanical coherence (QMC). This is needed to convert nuclear radiation to thermal power and to amplify the effects of the Yukawa potential (also called a screened Coulomb potential). When QMC has not yet been established, the nuclear radiation that is being produced in nuclear reactions at active nuclear sites will escape as gamma and x-ray radiation before it can be transformed into heat energy.

Nuclear reactions affect atoms in a large assemblage of coherent and entangled atoms by sharing the energy produced in any one member of the QMC assemblage among all its members.

In such a coherent collection, what happens to one member of such a coherent collection happens to them all. An averaging effect takes place where the nuclear energy output of one atom is averaged over a hundred or more atoms in the coherent collection.

In the Rossi reactor, at startup, a large amount of gamma radiation appears before QMC has established itself because the temperature of the nickel has not gotten to the relatively low Curie temperature (nickel has the Curie temperature of 631 K (~358 C)). Formation of QMC is magnetic in nature.

When nickel is ferromagnetic it cannot join the QMC assemblages. Nickel must first be made paramagnetic by heat to join the QMC assemblage. The QMC amplifies both the nuclear power produced and the heat output conversion of the cold fusion reaction.

Rydberg matter produces the QMC effect in nickel as Coherent hydrogen molecules are ionically attracted to the surface of the nickel. The strong dipole field of the Rydberg matter produced by a large coherent circulation of electrons with very large dipole moments induces a powerful paramagnetic field in the nickel within a quantum mechanical “blockade” distance of up to a micron. This field of coherent influence is directly proportional to the size of the Rydberg molecule which in turn is proportional to the temperature and pressure of the hydrogen envelope.

As the hydrogen envelope grows hot, more nuclear reactions take place in the coherent nickel on the surface of the nickel and more efficient gamma radiation conversion to heat occurs.

This is why nuclear ash transmutation products are found imbedded in the surface of the nickel.



2 - Yukawa potential – to be supplied

3 – Rydberg matter – to be supplied

May you please note me who is developer of this theory? Rossi, Levi and the third person?
As it weighs at least on Nobel Prize.
Paramagnetic matter stops neutrons and photons?
Thanks. This is very new for me.

May you please also note the energy of neutrons and gamma-photons before Nickel become paramagnetic?

[Ivy Matt]

Rossi, BLP, Brillouin... what is it about nickel?

It's a lot cheaper than palladium.

[Axil]

http://nextbigfuture.com/2011/04/how-ca ... actor.html

Here is a post from the nextbigfuture that shows a positive energy production from NI58 to CU63

So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy.

When you go through the calculation, you get positive energy output.

I ran the mass calculations with H = 1.007825032 and all the beta particle included but to no avail there math is correct I got 37.1778MeV the discrepancies being the difference in our mass tables. So despite what they tell you about the curve of binding energy, fusions for elements heavier then Iron can make energy (it surprised me anyway).