10KW LENR Demonstrator?

[D Tibbets]

Here is a post from the nextbigfuture that shows a positive energy production from NI58 to CU63 When you go through the calculation, you get positive energy output.

Again, the mass defect represents just that, the mass equivalent of a nucleus that is not measured on a scale. That is because it is in the form of energy. The total binding energy in fact.

Right, which is the total energy RELEASED by the nucleus during the reaction.

Why people believe that this energy represents some released energy is beyond me.

Becasue that is its definition. Why you can't understand that is beyond me.

It is a part of the nucleus. The nucleus cannot exist without this incorporated energy- it is the binding energy - a part of the nucleus.

No, the binding energy is the energy RELEASED. The remainder up to the ~14MeV is the part that stays with the nucleus as part of it's mass.

A lead rock sitting on the ground doesn't release any more energy than a lithium rock sitting on the ground. The lead rock has a lot more mass in the form of neutrons and protons and electrons, and binding energy, but that means nothing in this static situation. If you apply kinetic energy (an equal shove on both) it will not have any more energy than the light rock. If you divide up the mass in the lead into that represented by neutrons and protons, and mass deficit or binding energy, the total does not change. The binding energy may have increased, but this is a part of the nucleus, not the energy released in the formation. The total binding energy is simply the missing mass or the change in the balance between the defined masses of the particles in isolation and their mass within a nucleus. The mass- energy total does not change, only the ratio.

Dan, I begin to believe you are hopeless.
Go back to the empirical binding energy equation you linked to before. Notice that the FIRST term is a positive term of about 14MeV and the rest (except the final) are negative. Those NEGATIVE terms are what you are calling (inaccurately) the binding energy. Those terms relate to the amount of mass that DOESN'T go deficit upon a reaction. If no energy was absorbed by stretching bonds and going into surface tension, then each and every nucleon added would release 14MeV. But they don't. They release somewhat less, but still positive. Each and every time you add a nucleon.

The binding energy per nucleon is a different matter.

The binding energy per nucleon is by definition the amount of matter lost in the sum total of all reactions that make up the nucleus divided by the number of electrons. It is by Einstein's equation, the amount of energy lost (released) by the sum total of reactions divided by the number of nucleons.
DAN LISTEN, DON'T TRY TO MAKE IT ANY HARDER!!! It is just that simple. E=Mc². Mass deficit = energy released. This is called "binding energy". All those other terms (surface tension, couloum repulsion, symmetry) REDUCE the binding energy.

This experimentally derived table/ graph is less confusing (for me) if you consider it as the packing fraction, or nuclear density. Ni62 is the most tightly packed due to the interactions between the strong but short range Strong nuclear force, and the relatively weak but long range Electromagnetic repulsive force.

Seems it just makes it more confusing to you since you just refuse to get it.

Nuclear physic sometimes uses a liquid model (or gas) to describe things. As you compact/ increase the density of a fluid, the temperature increases (energy is released), as the fliud/ gas expands heat is absorbed. This is the same for nickel62. It is the most dense collection of nucleons possible for this consideration (neutron stars is different physics).

Dan, if you start condsing something, it may be the mopst compact form, but you continue to release energy from the least compact for if you condense it onto the unit. Same with nuclei. 62Ni may be the most dense, but it plus a proton are MUCH less dense that 63Cu. And it is the PROTON (or neutron) that releases the energy, NOT the nickel.

KitemanSA, you obvously do not consider my arguments as reasonable, so I will not present new ones, instead I will only give you referrals that you can persue if you truely wish to understand the issue.

First read this reference carefully, I think it may be the most enlightening discussion of the terms and significance that I have found (Warning- you can read the text, but if you run the associated demonstrations, you may be required to register.)

http://www.furryelephant.com/content/ra ... ss-defect/


A whole bunch of binding energy/ nucleon graphs with links to the articles that discuss them. Click on the graphs to go to the links. Looking through these imparts a lot of knowledge.

http://www.google.com/search?q=nuclear+ ... 16&bih=591


Some other links, some of which has been referenced before

http://hyperphysics.phy-astr.gsu.edu/hb ... ucbin.html

Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship

The iron limit:
The buildup of heavier elements in the nuclear fusion processes in stars is limited to elements below iron, since the fusion of iron would subtract energy rather than provide it. Iron-56 is abundant in stellar processes, and with a binding energy per nucleon of 8.8 MeV, it is the third most tightly bound of the nuclides. Its average binding energy per nucleon is exceeded only by 58Fe and 62Ni, the nickel isotope being the most tightly bound of the nuclides.

The binding energy curve is obtained by dividing the total nuclear binding energy by the number of nucleons. The fact that there is a peak in the binding energy curve in the region of stability near iron means that either the breakup of heavier nuclei (fission) or the combining of lighter nuclei (fusion) will yield nuclei which are more tightly bound (less mass per nucleon).

The binding energies of nucleons are in the range of millions of electron volts compared to tens of eV for atomic electrons. Whereas an atomic transition might emit a photon in the range of a few electron volts, perhaps in the visible light region, nuclear transitions can emit gamma-rays with quantum energies in the MeV range.
The iron limit:
The buildup of heavier elements in the nuclear fusion processes in stars is limited to elements below iron, since the fusion of iron would subtract energy rather than provide it. Iron-56 is abundant in stellar processes, and with a binding energy per nucleon of 8.8 MeV, it is the third most tightly bound of the nuclides. Its average binding energy per nucleon is exceeded only by 58Fe and 62Ni, the nickel isotope being the most tightly bound of the nuclides.

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

As we go on to elements heavier than oxygen, the energy which can be gained by assembling them from lighter elements decreases, up to iron. For nuclei heavier than iron, one actually gains energy by breaking them up into 2 fragments. That, of course, is how energy is extracted by breaking up uranium nuclei in nuclear power reactors.

The reason the trend reverses after iron is the growing positive charge of the nuclei. The electric force may be weaker than the nuclear force, but its range is greater: in an iron nucleus, each proton repels 25 other protons, while (one may argue) the nuclear force only binds close neighbors.

As nuclei grow bigger still, this disruptive effect becomes steadily more significant. By the time plutonium is reached (94 protons), nuclei can no longer accommodate their large positive charge, but emit their excess protons quite rapidly in the process of alpha radioactivity—the emission of helium nuclei, each containing two protons and two neutrons. (Helium nuclei are an especially stable combination.) This process becomes so rapid that still heavier nuclei are not found naturally on Earth.

http://www-istp.gsfc.nasa.gov/stargaze/SnucEnerA-2.htm

The net binding energy of a nucleus is that of the nuclear attraction, minus the disruptive energy of the electric force. As nuclei get heavier than helium, their net binding energy per nucleon (deduced from the difference in mass between the nucleus and the sum of masses of component nucleons) grows more and more slowly, reaching its peak at iron. As nucleons are added, the total nuclear binding energy always increases--but the total disruptive energy of electric forces (positive protons repelling other protons) also increases, and past iron, the second increase outweighs the first. One may say 56 Fe is the most efficiently bound nucleus. (see reference #10b)

http://www.scienceinschool.org/print/257

Iron-56 has the most stable nucleus because it has the maximum nuclear binding energy (see box and diagram below). Nature cherishes stable configurations and therefore the fusion process described in our last article, which brings us from hydrogen up to heavier, more stable nuclei, will not continue beyond iron-56. So, where do heavier elements such as lead, silver, gold and uranium come from? There is no magic: the Universe provides other fascinating ways to produce all the heavy elements. In the high temperature and pressure of a star, fusion is as spontaneous as rolling down a hill (a process that releases energy). However, these new mechanisms are more laborious, like climbing a hill (a process that needs energy). Furthermore the next stages of nucleosynthesis are quite hectic, as they involve captures and explosions. Three types of capture are involved, two dealing with the capture of neutrons (the s- and r-processes) and one with the capture of protons (the p-process).

http://www4.nau.edu/meteorite/meteorite ... saryb.html

BINDING ENERGY1 - Amount of energy released at the creation of a particular isotope. Protons and neutron are held together by the "strong force". The strong force only acts over very small distances but is able to overcome the electrostatic repulsion between protons. The magnitude of the bonding is measured by the binding energy per nucleon where "nucleon" is a collective name for neutrons and protons (sometimes called the "mass defect per nucleon"). The mass defect reflects the fact that the total mass of the nucleus is less than the sum of the mass of the individual neutrons and protons that formed it. The difference in mass is equivalent to the energy released in forming the nucleus. The general decrease in binding energy beyond iron is caused by the fact that, as the nucleus gets bigger, the ability of the strong force to counteract the electrostatic force between the protons becomes weaker. The most tightly bound isotopes are 62Ni, 58Fe, and 56Fe, which have binding energies of 8.8 MeV per nucleon. Elements heavier than these isotopes can yield energy by nuclear fission; lighter isotopes can yield energy by fusion

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Dan Tibbets

[Joseph Chikva]

This negative hydrogen ion is an electron quasiparticle;

May you please provide link? That hydrogen's negative ion can play electron's role in Nickel lattice. And how that forms?

One link about the matter with number density of 10^35 m^-3 I have already seen.
Can Rydberg matter be at lower number density?
For note: the real number density in Nickel lattice should be at order 10^31-10^32 or 1000-10000 times lower than 10^35.

Pardon, I have mistaken: number density of Nicken lattice not 10^28-10^29 m^-3 and so, 1,000,000-10,000,000 times lower than 10^35.

[cg66]

http://institute.lanl.gov/ei/LADSS/lect ... laytor.pdf

Claytor is an established resercher. But very far out of his area doing this stuff. Of course, he may be an expert on calorimetry & detecting very small quantities of tritium. Or he may, outside his normal area, overlook trivial contaminants like sources of natural tritium...

Given the low power outputs claimed, and short run times, the T produced as reaction product would have to be very low concentration. Somone else can do the calculations.

Still 2nd/3rd hand but interesting none the less - Tom Claytor reliably getting 5-16% excess energy.

http://www.mail-archive.com/vortex-l%40 ... 49051.html

[KitemanSA]

KitemanSA

And if Kim is correct in tses calculations, the transition temperature of polaritonic BECs could be quite high, since the Tc of a BEC is proportional to the particle density (to the 2/3) and INVERSELY proportional to its mass.

I am only partly familiar with polaritons in the form of SPPs. In the case of SPPs, they decay so fast that unless the photon flux is extremely high individual SPPs will never interact.

What form of polariton will form your postulated BEC?

I am not an expert on this so find data where I can, wikipedia in this case. My Konjecture was based on an assumed SPP as you stated, and for that I postulated that the "heater" in the Rossi device may actually be a UV laser which my unchecked calculation suggested was the frequency of photon needed to form the polariton. But I think I read where exciton polaritons might also exist in H:Ni lattices so they might also be the type. With excitons, the frequency needed is IR, so just heating the lattice up may suffice.

If a small number of polaritons "condense", would they still decay? Would they provide a "nuclation" site for the condensation and long life of other polaritons?

In either case, the available particle densities seem quite high, WAY higher than the cold plasmas I think are usually associated with BEC work, and the polariton masses should be way lower than the nuclear bosons usually used. Based on the equation I found, this suggests a very high Tc. In other words, they might just exist in the lattice without doing ANYTHING to cause them except to form the polaritons. They may be an undiscovered fact of lattice life! Just a konjecture!

[Joseph Chikva]

I am not an expert on this so find data where I can, wikipedia in this case. My Konjecture was based on an assumed SPP as you stated, and for that I postulated that the "heater" in the Rossi device may actually be a UV laser which my unchecked calculation suggested was the frequency of photon needed to form the polariton.

I have read wikipedia and then postulated.
Signed: Lev Landau

[KitemanSA]

KitemanSA, you obvously do not consider my arguments as reasonable, so I will not present new ones, instead I will only give you referrals that you can persue if you truely wish to understand the issue.

First read this reference carefully, I think it may be the most enlightening discussion of the terms and significance that I have found (Warning- you can read the text, but if you run the associated demonstrations, you may be required to register.)

http://www.furryelephant.com/content/ra ... ss-defect/

Thanks, I'll try to get to this tonight.

A whole bunch of binding energy/ nucleon graphs with links to the articles that discuss them. Click on the graphs to go to the links. Looking through these imparts a lot of knowledge.

http://www.google.com/search?q=nuclear+ ... 16&bih=591


Some other links, some of which has been referenced before

http://hyperphysics.phy-astr.gsu.edu/hb ... ucbin.html

These too, maybe.

Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship

This is exactly what I have been saying. Mass deficit IS binding energy which is the amount RELEASED by a reaction.

The iron limit:
The buildup of heavier elements in the nuclear fusion processes in stars is limited to elements below iron, since the fusion of iron would subtract energy rather than provide it.

Fusion of iron WITH WHAT??? I will bet you $1000 that the basic assumption here is iron with iron (or something up there on the graph). In which case, I agree. Iron with a proton? No true. Of course, in a star, there is probably no mechanism that will allow the reaction of Fe+p to shed the binding energy other than shedding the p! In a star. In plasmas. Not NECESSARILY in a lattice.

Iron-56 is abundant in stellar processes, and with a binding energy per nucleon of 8.8 MeV, it is the third most tightly bound of the nuclides. Its average binding energy per nucleon is exceeded only by 58Fe and 62Ni, the nickel isotope being the most tightly bound of the nuclides.

No argument here. The thing you seem unwilling to accept is that I MAY just know what I am talking about and I AGREE with all that you have posted but that YOU don't seem to want to hear the part about the PROTON being involve in the reaction. 62Ni + 62Ni, endo thermic, no doubt. 62Ni+ p, exothemic BECAUSE OF THE PROTON. No doubt.

[Joseph Chikva]

No argument here. The thing you seem unwilling to accept is that I MAY just know what I am talking about and I AGREE with all that you have posted but that YOU don't seem to want to hear the part about the PROTON being involve in the reaction. 62Ni + 62Ni, endo thermic, no doubt. 62Ni+ p, exothemic BECAUSE OF THE PROTON. No doubt.

2+2=4 No doubt.

[D Tibbets]

Nuclei are made up of protons and neutron, but the mass of a nucleus is always less than the sum of the individual masses of the protons and neutrons which constitute it. The difference is a measure of the nuclear binding energy which holds the nucleus together. This binding energy can be calculated from the Einstein relationship

This is exactly what I have been saying. Mass deficit IS binding energy which is the amount RELEASED by a reaction.

Please read the entire text, you are taking this out of context.

The iron limit:
The buildup of heavier elements in the nuclear fusion processes in stars is limited to elements below iron, since the fusion of iron would subtract energy rather than provide it.

[/quote]

Fusion of iron WITH WHAT??? I will bet you $1000 that the basic assumption here is iron with iron (or something up there on the graph). In which case, I agree. Iron with a proton? No true. Of course, in a star, there is probably no mechanism that will allow the reaction of Fe+p to shed the binding energy other than shedding the p! In a star. In plasmas. Not NECESSARILY in a lattice.

Iron-56 is abundant in stellar processes, and with a binding energy per nucleon of 8.8 MeV, it is the third most tightly bound of the nuclides. Its average binding energy per nucleon is exceeded only by 58Fe and 62Ni, the nickel isotope being the most tightly bound of the nuclides.

No argument here. The thing you seem unwilling to accept is that I MAY just know what I am talking about and I AGREE with all that you have posted but that YOU don't seem to want to hear the part about the PROTON being involve in the reaction. 62Ni + 62Ni, endo thermic, no doubt. 62Ni+ p, exothemic BECAUSE OF THE PROTON. No doubt.

Concerning Stars, the iron group (Fe56, Ni58, Ni62) are the most tightly bound nuclei and they are the end of the road for heat generation. When this occurs there is no longer enough radiation pressure to support the star against further gravitational collapse. It is a huge unstable hot ( and getting hotter due to gravitational collapse) plasma. There are all sorts of nuclear reactions occurring. At this stage almost all are endothermic, neutrons are produced, etc. These can build heavier atoms easily, but you have to consider the cost of obtaining these free neutrons. Also any stray protons can form heavier elements, described as the rp- process. But again this is endothermic past Ni62. It is not that heavier elements cannot be made in stars, obviously this occurs in huge quantities, it is that these processes are not exothermic and that is why the star (if massive enough) explodes at this point. It is the sudden release of the gravitational energy that was previously resisted that drives the explosion, and provides some of the heat (along with residual heat from exothermic light element fusion that has just been exhausted)

As far as summing (fusing) parts to get a final particle, a proton + a Ni62 is roughly the same as a any other combination (with the exception as noted below) of 2 or more elements, if the sum is heavier than Ni 62, the energy out is not maximized. This does not mean you cannot get energy out, just, that it is less efficient. If Rossi said he was combining two isotopes, of 32 and 31 neucleons, he would get energy out, but less than isotopes of 32 and 30 combining. But Rossi claims he starts with Ni62, and adding or subtracting single or multiple nucleons to this lowest energy state (or highest energy state- depending on how you define the system- that is part of the problem, they are not consistent in the methods in which they approach the problem) starting point is a losing proposition.

So much for not arguing further!

The links I provided explains things better and repeatably. You need to read the entire descriptions without taking things out of context (they surely could have developed their definitions better- it must have been done by committee). If you are interested in the stellar perspective, Google stellar nuclear synthesis.

Dan Tibbets

[Axil]

This negative hydrogen ion is an electron quasiparticle;

May you please provide link? That hydrogen's negative ion can play electron's role in Nickel lattice. And how that forms?

One link about the matter with number density of 10^35 m^-3 I have already seen.
Can Rydberg matter be at lower number density?
For note: the real number density in Nickel lattice should be at order 10^31-10^32 or 1000-10000 times lower than 10^35.

Pardon, I have mistaken: number density of Nicken lattice not 10^28-10^29 m^-3 and so, 1,000,000-10,000,000 times lower than 10^35.

Heavy Rydberg states, E. Reinhold, W. Ubachs, Molecular Physics, Vol. 103, No. 10, 20 May 2005, 1329–1352

Piantelli in his 2010 patent

[Giorgio]

http://institute.lanl.gov/ei/LADSS/lect ... laytor.pdf

Claytor is an established resercher. But very far out of his area doing this stuff. Of course, he may be an expert on calorimetry & detecting very small quantities of tritium. Or he may, outside his normal area, overlook trivial contaminants like sources of natural tritium...

Given the low power outputs claimed, and short run times, the T produced as reaction product would have to be very low concentration. Somone else can do the calculations.

Still 2nd/3rd hand but interesting none the less - Tom Claytor reliably getting 5-16% excess energy.

http://www.mail-archive.com/vortex-l%40 ... 49051.html

So, tritium formation as intermediate step for the extra heat generation, with nickel acting as a catalyst...
I'll have to think this over a little bit.

[KitemanSA]

Concerning Stars, the iron group (Fe56, Ni58, Ni62) are the most tightly bound nuclei and they are the end of the road for heat generation. When this occurs there is no longer enough radiation pressure to support the star against further gravitational collapse. It is a huge unstable hot ( and getting hotter due to gravitational collapse) plasma. There are all sorts of nuclear reactions occurring. At this stage almost all are endothermic, neutrons are produced, etc. These can (endothermically) build heavier atoms easily. Also any stray protons can form heavier elements, described as the p- process. But again this is endothermic past Ni62.

Please show me a specific reference to this contention. Show me ANY text anywhere that says that p+Ni is endotermic. Any text please.

It is not that heavier elements cannot be made in stars, obviously this occurs in huge quantities, it is that these processes are not exothermic and that is why the star (if massive enough) explodes at this point. It is the sudden release of the gravitational energy that was previously resisted that drives the explosion, and provides some of the heat (along with residual heat from exothermic light element fusion that has just been exhausted)

Show me any equation in a stellar evolution text that shows p+ANYTHING being endo-thermic.

As far as summing parts to get a final particle, a proton + a Ni62 is roughly the same as a any other combination of 2 or more elements, if the sum is heavier than Ni 62, the energy out is not maximized.

Who said anything about "maximized"? I said p+Ni was exothermic. It may not be AS "exothermic" as p+something smaller than Ni, but it is still exothermic, if it sticks.

This does not mean you cannot get energy out, just, that it is less efficient.

Ok, now it seems you are agreeing that the reaction of a p with something heavier than Ni is still exothermic. Is that what you are saying here? Because that is what I have been saying since day one.

If Rossi said he was combining two isotopes, of 32 and 31 neucleons, he would get energy out, but less than isotopes of 32 and 30 combining. But Rossi claims he starts with Ni62, and adding or subtracting single or multiple nucleons to this lowest energy state (or highest energy state- depending on how you define the system- that is part of the problem, they are not consistent in the methods in which they approach the problem) starting point is a losing proposition.

But he is claiming to combine two particles, Ni (which by your jargon is at this lowest energy state) with a series of solitary nucleaons, each of which are at the HIGHEST possible energy state. It is THAT energy, from the proton or neutron, that causes the energy release. It is truly simple. Don't get stuck looking only at the middle of the curve when one of the reactants is at the very beginning.

[KitemanSA]

2+2=4 No doubt.

Do you doubt it?

[Joseph Chikva]

Heavy Rydberg states, E. Reinhold, W. Ubachs, Molecular Physics, Vol. 103, No. 10, 20 May 2005, 1329–1352

Piantelli in his 2010 patent

Unfortunately I have not access to first and by some reasons Piantelli patent is lees interesting.

I really do not know about Rydberg matter.
And first that I found was wikipedia in which I read the following:

The difficulty in the production of heavy Rydberg systems arises in finding an energetic pathway by which a molecule can be excited with just the right energy to form an ion pair, without sufficient internal energy to cause autodissociation (a process analogous to autoionization in atoms) or rapid dissociation due to collisions or local fields.

Currently production of heavy Rydberg systems relies on complex vacuum ultra-violet (so called because it is strongly absorbed in air and requires the entire system to be enclosed within a vacuum chamber) or multi-photon transitions (relying on absorption of multiple photons almost simultaneously), both of which are rather inefficient and result in systems with high internal energy.

And the applicability of Rydberg matter for producing the certain reaction for generation of net power is just another issue.

PS: so, 10^35 m^-3 number density is not obligatory?

[Joseph Chikva]

2+2=4 No doubt.

Do you doubt it?

I even have no doubt that you know how to calculate the reaction energy release by calculation of mass defect. Perhaps learn that yesterday. And today are very proud.

[KitemanSA]

KitemanSA, you obvously do not consider my arguments as reasonable, so I will not present new ones, instead I will only give you referrals that you can persue if you truely wish to understand the issue.

First read this reference carefully, I think it may be the most enlightening discussion of the terms and significance that I have found (Warning- you can read the text, but if you run the associated demonstrations, you may be required to register.)

http://www.furryelephant.com/content/ra ... ss-defect/

Thanks, I'll try to get to this tonight.

Just read it, and except for a few small faux pas and a remaining question on how you define fusion, the document said exactly what I have been saying all along.

So Dan,
I have asked several times and you have not done so, but now I think you must... DEFINE "FUSION". Until you do, we cannot have a meaningful conversation. I have given you MY definition several times. What is yours?