10KW LENR Demonstrator?

Point out news stories, on the net or in mainstream media, related to polywell fusion.

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KitemanSA
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Post by KitemanSA »

KitemanSA wrote:
Then he wrote: Have you watched the video?
Not yet. Don't have access at work and have limited time at home.
Just watched it. And other than a common mistake of calling the AVERAGE binding energy curve a binding energy curve, it provides the same infomation as I have been all along. The upshot is: to determine the energy release, subtract the TOTAL binding energy of the reactants from the TOTAL binding energy of the products. With 2H + 3H > 4He + n the equation is:
( 28 + 0 ) - ( 2 + 8 ) = 18MeV.
Do the same thing with p+Ni> Cu and you get ~8.7MeV. Same sign. Energy released. Dan, every link you have EVERY listed has supported my statements except when you read them wrongly or maybe just selectively.

Dan, I am tired of your wrong-headedness. Learn or don't, it is up to you. I am done trying to teach the unwilling. But I will correct you to others any time I see you spew this nonsense in the future.

stefanbanev
Posts: 183
Joined: Tue Jul 12, 2011 3:12 am

Post by stefanbanev »

KitemanSA wrote:
stefanbanev wrote:Dan Tibbets> You could cycle between say nickel and uranium,
Dan Tibbets> gaining energy from both pathways.

Yes, apparently such E+ cycle may be drafted:

i) Ni+p...Cu+p ...Pb+p..........-> U235 (fission-bang) go to (i)

Every step decreases the mass according M -= Ei/CC

Note(*): the fission products should have enough light enough atoms to ensure exothermic path.

Dan Tibbets> You now have an inexhaustible free energy system.

Absolutely not, once mass == 0 thee is no E.

Needless to say that such cycle is highly hypothetical nevertheless, it shows that 100% mass->energy conversion by means of nuclear reaction may be possible if Note(*) can be provided.

Stefan
As I pointed out in another post, the supply of energy is only as "inexhaustible" as the supply of protons and free neutrons. Once you no longer have either of those, the process stops. Not sure why he refuses to get that point.
Well, "protons and free neutrons" are the most "valuable" but the soup of fission products may have others less exothermic paths. Besides, recycling heavy atoms + external H supply indeed may improve mass->energy conversion efficiency. This is interesting to asses the limits of "nuclear" based space propulsion to get relativistic velocities (without antimatter involved).

D Tibbets
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Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

KitemanSA wrote:
stefanbanev wrote:Dan Tibbets> You could cycle between say nickel and uranium,
Dan Tibbets> gaining energy from both pathways.

Yes, apparently such E+ cycle may be drafted:

i) Ni+p...Cu+p ...Pb+p..........-> U235 (fission-bang) go to (i)

Every step decreases the mass according M -= Ei/CC

Note(*): the fission products should have enough light enough atoms to ensure exothermic path.

Dan Tibbets> You now have an inexhaustible free energy system.

Absolutely not, once mass == 0 thee is no E.

Needless to say that such cycle is highly hypothetical nevertheless, it shows that 100% mass->energy conversion by means of nuclear reaction may be possible if Note(*) can be provided.

Stefan
As I pointed out in another post, the supply of energy is only as "inexhaustible" as the supply of protons and free neutrons. Once you no longer have either of those, the process stops. Not sure why he refuses to get that point.
Actually for nucleosythesis of intermediate and heavy nuclei, alpha particals are the best ingredient from a stability standpoint.

You realize that you are ignoring many links to physics and stellar sites and texts that state repeatedly and unequivocally that the binding energy per nucleon determines the energy flow direction. It is not the ingredients so much as the end product . Nickel is the end of the line. The the holy grail, the most stable nucleus, the most tightly packed nucleus, the nucleus with the highest binding energy per nucleon.
Keep in mind that the proton has no binding energy. Any number that might be quoted is based on the weight of the proton in the stable carbon 12 nucleus. This is by convention, not on any underlieing property of the free proton. . Yet you say that the protons binding energy is added to the product nucleus. Where does this energy come from? It doesn't come from anywhere. You are only shifting the mass energy proportions within the product nucleus. Otherwise you would be creating energy out of nothing- a free energy, perpetual motion machine. Again, and again, and yet again, it is the density/ packing of the nucleons, specifically the average nucleon that is important. It is not a uniform homogenous environment within the nucleus. This is the property that determines the endothermic vs exothermic reaction.

You say that so long as there is hydrogen (protons) present, atoms can be built to unlimited size. You have no other limit in you system. This despite obvious consequences that is counter to the limits on how heavy a nucleus can become, and what drives radioactive decay and fission. I have pointed you to numerous sites that clearly defines the pertinent predictors and the reasoning behind them.

Finally. The total binding energy goes up indefinatly based on the graph you insist on using, so if you use this as the only factor there is absolutely no reason why the fusion of any combination of nuclei will not release energy and become more stable, after all- the total binding energy is you predictor. You could merge two uranium nuclei , get energy out, and have a more stable nucleus. A star would condense into one single huge nucleus, before it ran out of energy to resist gravitational collapse (I an talking about nuclear fusion here, not other effects like overcoming the Pauli exclusion principle) or matter- antimatter annihilation. Using the excuse that free protons are needed is immaterial to this prediction, alphas, iron nuclei, Californium nuclei, etc would all work, the end product would always have more total binding energy. All of the physicists and astronomers are obviously stupid for presenting the binding energy per nucleon curve, with arrows and discussions on how Ni62 is the peak (or minimum if the reciprocal of the graph is used as some prefer), and turn around point . Your interpretation also has no room for exothermic heavy element fission reactions.

IF both both fusion and fission can produce- release excess energy, there has to be a tipping point or minimal energy (most stable) state between them, otherwise, again, it is a perpetual motion machine.

Basically you are confused about the energy contained within a nucleus and the energy that is released or absorbed in the formation of a nucleus through fusion/ fission mechanisms. Your arguments would be valid if you were talking about the total annihilation of a nucleus , but these nuclear processes are dealing with only a tiny and variable fraction of this total energy (~0.5% of the annihilation energy), whether it is in the form of binding energy, mass deficit, rest mass of the particles, etc.

Dan Tibbets
To error is human... and I'm very human.

Joseph Chikva
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Joined: Sat Apr 02, 2011 4:30 am

Post by Joseph Chikva »

KitemanSA wrote:Thank you, I appreciate that someone understands this!
Hehe
He appreciates
My little friend:
Two reactants with nuclei masses M1 and M2 respectively fuses.
M1+M2 = X
In result of fusion you get or perhaps get nucleus with mass M
If X>M exothermic
If X<M endo
Understand? What is a problem?
Or do you think that many words means a big sense?

D Tibbets
Posts: 2775
Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

Joseph Chikva wrote:
KitemanSA wrote:Thank you, I appreciate that someone understands this!
Hehe
He appreciates
My little friend:
Two reactants with nuclei masses M1 and M2 respectively fuses.
M1+M2 = X
In result of fusion you get or perhaps get nucleus with mass M
If X>M exothermic
If X<M endo
Understand? What is a problem?
Or do you think that many words means a big sense?

As I read it, this is not quite correct. It is not whether a reaction less than or greater than M (if you are saying that M= Ni62) it is the direction of the reaction. If two or more light isotopes fuse and the product binding energy per nucleon of the product is greater than the sum of the reactants, then it is exothermic. If one heavy nucleus fissions into two products, and the sum of the two products binding energy per nucleon is higher than the original heavy isotope's nuclear binding energy, then it is exothermic. but it shows the direction of the arrow. The video shows the comparisons well. You have to match the reactants and the products to their position on the nuclear binding energy per nucleon graph, not , as KitemanSA insists , on the total binding energy graph. If you did this with a fission reaction, it would always be endothermic, yet we manage to get energy out of fissioning uranium and plutonium. Radioactive decay obeys the same rules, though things can get more complicated. The preferred direction of a reaction is always towards the most stable end product (s). And the most stable = the lowest energy state. The lowest energy state is that which has given up the most energy. This has been found to be Ni62, thus the exothermic arrows always point towards this end product. It doesn't matter whether you reach this in many small steps, or one big step. Also, you can possibly overshot and still get energy out, provided the results have more binding energy per nucleon than the reactants. This obviously would be limited to the area near Ni62, though there is more overshoot room on the high side of Ni62 than the low side due to the steepness of the slopes.
Without looking up the numbers , I suspect that Fe59 + alpha --> Cu63 might release a little energy, because Cu63 is a little closer to Ni62.
I do not think Ni62+ P --> Cu 63 is exothermic Using ~ binding energy per nucleon numbers the reaction would be 8.9 MeV + 0 MeV --> 8.8 MeV. The number is lower, the nucleus is less stable. This is where the reciprocal of the graph (1/ Nuclear binding energy per nucleon) is more clear. It shows Ni62 as the lowest energy, most stable isotope The Cu63 is higher up this potential energy curve, and you had to add energy to get there.

For Rossi's purposes, he would have been wiser to claim that Ni56, or 58 were the reacting species that combined with a proton. These at least would reasonably produce a relatively small amount of exothermic energy ( the product is closer to Ni62 than the reactants.

Given enough time, all of the atoms in the universe would eventually combine in such ways that only nickel 62 would exist, because it is the most stable, lowest energy state possible (without going to degenerate matter).

Dan Tibbets
To error is human... and I'm very human.

KitemanSA
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Location: OlyPen WA

Post by KitemanSA »

D Tibbets wrote: You realize that you are ignoring many links to physics and stellar sites and texts that state repeatedly and unequivocally that the binding energy per nucleon determines the energy flow direction.
And do you realize that you are ignoring the fact that Hydrogen has a LONG way to go to get to the Ni stage of BE/A?

Dan, until you have personally and correctly done the numbers on p+10B, p+20Ne, p+31P, p+40Ca, p+62Ni, p+72Ge, and p+83Kr and are ready to discuss the meaning of the results, you are just spewing odeous hot air. Here are some of the values to get you started.
1-H 0
10-B 64750.7
11-C 73439.9
20-Ne 160644.85
21-Na 163076.15
You can get the rest, courtesy of chrismb here.
viewtopic.php?p=64649#64649
No more links that you refuse to understand. The only link from you is one that unambigously states that p+62Ni is endothermic. You won't find one except by some unknowing dufus.
Your mantra is "think Hydrogen"!

Do the numbers, "think Hydrogen".

Do the numbers, "think Hydrogen".

Do the numbers, "think Hydrogen".

ok? :)

Joseph Chikva
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Joined: Sat Apr 02, 2011 4:30 am

Post by Joseph Chikva »

D Tibbets wrote:As I read it, this is not quite correct. It is not whether a reaction less than or greater than M (if you are saying that M= Ni62) it is the direction of the reaction. If two or more light isotopes fuse and the product binding energy per nucleon of the product is greater than the sum of the reactants, then it is exothermic. If one heavy nucleus fissions into two products, and the sum of the two products binding energy per nucleon is higher than the original heavy isotope's nuclear binding energy, then it is exothermic.
No, that is correct. If you loss mass, you gain energy. If you have created nucleus with mass more than reactants' mass addition, you spent energy.

ScottL
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Joined: Thu Jun 02, 2011 11:26 pm

Post by ScottL »

I don't know much about binding energies, just started looking at the math today. From my math and looking at Joseph's endo vs. exothermic explanation, I'd say Ni62 + P -> Cu63 is exothermic although my numbers could be wrong.

I got something like (fudging sig figs):

Ni62(61.9283) + (P)1.0078 -> (Cu63)62.9295

62.9361 - 62.9295 = 0.0066

This would be exothermic by Joseph's explanation correct?

Figured I might as well learn while I read the arguments back and forth.

KitemanSA
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Location: OlyPen WA

Post by KitemanSA »

ScottL wrote:I don't know much about binding energies, just started looking at the math today. From my math and looking at Joseph's endo vs. exothermic explanation, I'd say Ni62 + P -> Cu63 is exothermic although my numbers could be wrong.

I got something like (fudging sig figs):

Ni62(61.9283) + (P)1.0078 -> (Cu63)62.9295

62.9361 - 62.9295 = 0.0066

This would be exothermic by Joseph's explanation correct?

Figured I might as well learn while I read the arguments back and forth.
Surprising to say that despite his normally juvenile writings, yes, he is correct in his explanation and YOU are correct in you calculation. It is exothermic. E=mc².
This is the point that Dan T just can't seem to get.

Dan,
Do the numbers, "think Hydrogen".

PS: ScottL, just as a nit-pick, but may keep things clearer in the future, a proton is designated either H, or p. P is phosphorus. :wink:

ScottL
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Post by ScottL »

PS: ScottL, just as a nit-pick, but may keep things clearer in the future, a proton is designated either H, or p. P is phosphorus.
I thought so, but Dan wrote it as P so I figured maybe I was wrong. :)

KitemanSA
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Post by KitemanSA »

ScottL wrote:
PS: ScottL, just as a nit-pick, but may keep things clearer in the future, a proton is designated either H, or p. P is phosphorus.
I thought so, but Dan wrote it as P so I figured maybe I was wrong. :)
We all get a bit sloppy at times, myself included. :o

Enginerd
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Joined: Sun Feb 07, 2010 5:29 am

Post by Enginerd »

ScottL wrote:I got something like (fudging sig figs):

Ni62(61.9283) + (P)1.0078 -> (Cu63)62.9295

62.9361 - 62.9295 = 0.0066

This would be exothermic by Joseph's explanation correct?
Correct. Of course, this bit of math has been shown about 100 times over the last 235 pages...

Code: Select all

62Ni   61.9283451
+
1H     1.0078250
-
63Cu  62.9295975
      ----------
0.0065726
The devil of course is in the details, as even if you win this particular battle you will still generally lose the war, i.e. even if you manage to get some fusion, you lose energy overall. i.e.

Code: Select all

Energy output = (0.0065726 * (number of fusion events)) - (energy loss due to thermodynamic cycle used for electrical conversion) - (energy loss due to escaping neutrons and escaping radiation) - (energy loss due to all other mechanisms) - (energy required to initiate the fusion reactions)
The energy required to initiate the fusion reactions generally speaking overwhelmingly dominates. Which is why when people propose a short cut to net positive fusion (i.e. the e-cat), with little to no input energy, people are understandably interested but also seriously skeptical.

KitemanSA
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Location: OlyPen WA

Post by KitemanSA »

Enginerd wrote: Correct. Of course, this bit of math has been shown about 100 times over the last 235 pages...

Code: Select all

   62Ni  61.9283451
+  1H     1.0078250
-  63Cu  62.9295975
         ----------
          0.0065726
The devil of course is in the details, as even if you win this particular battle you will still generally lose the war, i.e. even if you manage to get some fusion, you lose energy overall. i.e.

Code: Select all

Energy output = (0.0065726 * (number of fusion events)) - (energy loss due to thermodynamic cycle used for electrical conversion) - (energy loss due to escaping neutrons and escaping radiation) - (energy loss due to all other mechanisms) - (energy required to initiate the fusion reactions)
The energy required to initiate the fusion reactions generally speaking overwhelmingly dominates. Which is why when people propose a short cut to net positive fusion (i.e. the e-cat), with little to no input energy, people are understandably interested but also seriously skeptical.
{quote edited} This is indeed the question. My only intent was to debunk the erroneous statement that Rossi's machine can't work because the reaction is endotherminc. As shown, it is exothermic. The questions remaining are, do the reactions actually happen and can you get more out than in; which is just a restatement of your second "code".

D Tibbets
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Joined: Thu Jun 26, 2008 6:52 am

Post by D Tibbets »

So long as KitemanSA, etel continue to use the total binding energy graph/ tables as the indicator of the energy balance while ignoring the many links I have provided that clearly explain the proper way of using the binding energy per nucleon graph/ tables, he and I exist in separate universes- literally,, as stars would evolve in ways that would not allow us to exist in his universe.

The total binding energy is a part of the nucleus, it can be measured (as mass deficit). How can something that is an integral part of the nucleus somehow also release energy to the enviornment. Would not the nucleus have a resultant smaller mass/ energy weight if such was the case?

The totasl binding energy of a nucleus is defined as the amount of energy you would have to expend to tear ALL of the nucleons off of the nucleus so you only have the free nucleons (proptons and neutrons). Thus if you fissioned uranium 235 all of the way to free nucleons, the total binding energy graph would reviel how much energy was released. It tells you nothing about the intermediate possibilities.

The nucleons are bound by the strong nuclear force, and this has a range of ~ 2 to 2.5 nucleons. So the strong force assoiated with one bound nucleon can reach all other nucleons, and visa verse, to a nuclear radius of up to ~ 2 to 2.5 nucleon diameters. This would be a clump 4-5 nucleons wide, which cubed would result in ~ 60 to 70 nucleons . These are all tightly bound as they are interacting with all other nucleons. At the same time the electromagnetic repulsion is increasing with the number of bound protons. Once additional nucleons are added beyond this limit, they are more loosely bound as they cannot interact via the strong force with all other nucleons in the nucleus. . You can represent this with the binding energy per nucleon graph. It is known experimentally (not theoretically) that this maximally bound nucleus is Ni62. It is the most stable, and thus the lowest energy nucleus possible. . When ever a reaction releases energy it is going towards this lowest energy state. Those of you who have been exposed to chemistry classes know that is the case. The reaction releases energy as it progresses towards the most stable lowest energy state, and energy input is needed to push the reaction in the opposite direction. You can complicate the picture by talking about catalysts, Gibbs free energy, etc. but this basic tenet of chemical and nuclear reactions is absolute.
Please peruse the many binding energy graphs with linked articles that discusses this.

And speaking of KitemanSA's insistence that I present proofs, authoritative sources (of which I have presented many) where are his authorities (please do not take items out of context, such does allow for confusion due to the complexity and poor choice of labels).

Finally, I again supply a link to a source that is well respected. The graph labels and text clearly illustrates the accepted physics. please read the entire page, not just the first slide.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Dan Tibbets
To error is human... and I'm very human.

Enginerd
Posts: 191
Joined: Sun Feb 07, 2010 5:29 am

Post by Enginerd »

D Tibbets wrote:Finally, I again supply a link to a source that is well respected. The graph labels and text clearly illustrates the accepted physics. please read the entire page, not just the first slide.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Dan Tibbets
I presume you were intending to link to a specific article? i.e.

http://hyperphysics.phy-astr.gsu.edu/hb ... in.html#c1
http://hyperphysics.phy-astr.gsu.edu/hb ... in.html#c2
http://hyperphysics.phy-astr.gsu.edu/hb ... n2.html#c1
http://hyperphysics.phy-astr.gsu.edu/hb ... yn.html#c1

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