10KW LENR Demonstrator?

[D Tibbets]

ANY REACTION TOWARDS Ni62 RELEASES ENERGY, ANY REACTION AWAY FROM Ni62 ABSORBS ENERGY.

And H+Ni=Cu is MUCH more TOWARD Ni62 than away.

Do the numbers, "think Hydrogen".

Your long pontifications don't hold any value.

Do the numbers, "think Hydrogen".

Do the numbers, "think Hydrogen".

To determine if KitemanSA's championed P (hydrogen nuclues) +62Ni --> Cu63 would release energy I calculated the numbers. A negative result implies an exothermic reaction. A positive result implies an endothermic reaction.

Using the power balance calculation as illustrated in this video:

http://www.youtube.com/watch?v=yTkojROg-t8


And the binding energy per nucleon (A) from this table:

http://www.nndc.bnl.gov/amdc/masstables ... mass.mas03

P + 62Ni ---> 63Cu
The binding energies per nucleon are P (or 1H if you prefer) = 0 MeV
62Ni= 8.794 MeV/A
63CU= 8.792 MeV/A
P= 0 MeV/A

These numbers multiplied by the number of nucleons in the respective nuclei :
62Ni = 545.2 MeV binding energy
63Cu = 553.8 MeV binding energy
P = 1*0 = 0 MeV

Reactants = 545.2 MeV
Product = 553.8 MeV
Reactants - product = +8.6
This is a positive result which implies an endothermic reaction.

58Ni + P ---> 59Cu
8.7320 MeV/A + 0 MeV/A ---> 8.6419 MeV/A
506.4 MeV + 0 MeV ---> 509.6MeV
506.4 MeV - 509.6 MeV = - 3.2 MeV
This is a negative result and would be an exothermic reaction. Of course you then have to deal with what happens to the unstable 59Cu..........


Comparative examples using other reactions:


[edited to correct- forgot to add the neutron to the Uranium 235 to start the fission- this will change the intermediate number for the transient U236, but not by much. ]

235Ur + neutron ---> 144Ba + 89Kr + several neutrons (Fission of heavy element)
7.509MeV/A + 0 MeV/a---> 8.2654 MeV/A + 8.6169 MeV/A + 0 NeV/A
1783.8 MeV ---> 1190.2 MeV + 766.9 MeV + 0 MeV
1783.8 - ( 1190.2 + 766.9 MeV) = - 173.9 MeV This is an example given in the video .


[EDIT to correct the binding energy of deuterium. 2 nucleons * 0.00231 MeV/A = 0.00462 MeV. This effects the results an extreamly trivial amount]

[Edited to correct wrong Deuterium binding energy / A initially used]
D + T --> 4He + Neutron
1.112 MeV/A + 2..827 MeV/A --> 7.073 MeV/A + 0 MeV/A
2.224 MeV + 8.481 MeV ---> 28.2956 MeV +0 MeV
10.705 MeV - 28.295 MeV = - 18.410 MeV This is the other example given in the video. Note that both are exothermic reactions as the result is a negative number


[EDIT-corrected BE/A for 11B]
11B + P --> 12C* --> 4He + 8Be --> 4He + 4He +4He
6.9297/a MeV/A + 0 MeV/A --> ---> 7.0739MeV/A + 7.0739MeV/A + 7.0739MeV/A
76.20 MeV + 0 MeV ---> 28.29 MeV + 28.29 MeV + 28.29 MeV
76.20 MeV - 84.88 = - 8.68 MeV. Note that this is a negative result (exothermic) and that the number is not real close to the energy output of the published P-11B output of ~ 8.6 MeV. I suppose this discrepancy is due to the intermediates. The 12 C that is initially formed is a higher energy isomer (meaning some energy is retained (endothermic in this regard when compared to the baseline stable 12C). The beryllium intermediate may also be complicating things.

20Mg + 20Mg ---> 40Ca Note that this is an arbitrary selection. I don't know what the energy state of the 20Mg or 40Ca is. One or both bay be unstable higher energy isotopes so the large MeV yield may be misleading if the reaction has any possibility of occurring at all without adjustments.
6.7234 MeV/A + 6.7234 MeV/A ---> 8.5513 MeV/A.
134.4 MeV + 134.4 MeV ---> 342.0 MeV
268.8 MeV - 342.0 MeV = - 73.2 MeV

This last example will be done twice. The Aluminum and potassium isotope were both checked and they are stable isotopes. I doubt that the 66 Germanium is stable (formed by adding the 13, and 19 protons in the aluminum and potassium, along with the 14 an 20 neutrons). Because the Germanium is probably unstable I will then add neutrons to make 70Ge which is stable.

27 Al + 39K ---> 66Ge
8.3315 MeV/A + 8.5570 MeV/A ---> 8.67257 MeV/A
224.9 MeV + 333.7 MeV ---> 572.3 MeV
558.6 MeV - 572.3 MeV -= - 13.6 MeV

Same reaction except 4 neutrons added to reach a stable isotope of Ge. An alternative reaction might be beta decay to produce smaller element, but then I would have to worry about how much energy was carried away by the beta particle.

27Al + 39K + four Neutrons ---> 70Ge
8.3315 MeV/A + 8.5570 MeV/A + 4 Neutrons (0 MeV/A) ---> 8.7217 MeV/A
224.9 MeV + 333.7 MeV + 0 MeV ---> 610.5 MeV
558.6 MeV - 610.5 MeV = - 51.9 MeV

These examples shows exothermic reactions and one endothermic reaction. The endothermic one is the proton plus Ni62 reaction that is claimed by Rossi. He would have been better served by claiming that the heat producing reaction in his device was 58Ni + proton. It is exothermic. Of course, he would then have to explain the energy balance as the 59Cu isotope decays. Would the energy balance still be positive- I suspect so. The reason Rossi may not have chosen this reaction was through ignorance, or possibly that he would have a harder time explaining the nickel isotope distribution in the ash. It is much more convenient to claim reactions that result in natural stable isotopes. The problem with this is that despite high claimed output for long times, there was no consumption of the proportion of Ni62 that you would expect in a natural sample of nickel, which I believe someone mentioned testing. Rossi's excuse for this may be that he first enriched the 62 isotope content in the nickel, and the subsequent consumption reduced it to ~ natural levels. Now, how does he explain away that the Cu63 was also at natural levels (contamination) of at least some nickel samples?

The outputs of the exothermic reactions are ~ similar to what has actually been measured (D-T fusion, Ur fission), . This validates that this approach to determining output is reasonable. There is some variation, but this is a gross estimate based on the binding energy alone. It ignores the weak force, and various stabilities of differing isotopes (the neutron portion), excited isomer states, etc.

Dan Tibbets

[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.

Dan, it is not you and Kite. It is that you have created your own separate universe. I guess nobody else here, sceptic or not, would thinkg that the reaction being endotermic.

Nor anybody claims something like that in the whole blogosphere.

Everybody understands that such reaction, if possible, is exo. The only problem is that such reaction is unlikely given current physics....

Exactly, except for the "unlikely given current physics" part. I will accept "unlikely in the physical conditions within stars".
But even Dan points out that during a supernova the conditions are sufficient to synthesize higher elements by p+ reactions. Just hoiw much of a supernova output is bolstered by those reactions is not known to me.

Percisely, P reactions, or s-or r processes can build nuclei heavier than Ni, this is well acepted physics. BUT, that does not mean these are exothemic reactions past Nickel 62. There is plenty of thermal energyaviable in a aging gient star to drive these reactions. And during the final stage, if a supernova occurs this excess energy is profound for a few seconds.

And, as far as the entire blogosphere, is the links attached to this site -

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

denieing that the nuclear binding energy per nucleon graphs that show 62Ni as the critical point immaginary? Not to mention the hyperphysics links provided?


b

Dan Tibbets


Oh, and thank you by the way!

[KitemanSA]

Using the power balance calculation as illustrated in this video:

http://www.youtube.com/watch?v=yTkojROg-t8


And the binding energy per nucleon (A) from this table:

http://www.nndc.bnl.gov/amdc/masstables ... mass.mas03

P + 62Ni ---> 63Cu
The binding energies per nucleon are P (or 1H if you prefer) = 0 MeV
62Ni= 8.794 MeV/A
63CU= 8.792 MeV/A
P= 0 MeV/A

These numbers multiplied by the number of nucleons in the respective nuclei :
62Ni = 545.2 MeV binding energy
63Cu = 553.8 MeV binding energy
P = 1*0 = 0 MeV

Reactants = 545.2 MeV
Product = 553.8 MeV
Reactants - product = +8.6

Actually, it is PRODUCTS-REACTANTS, but you did it backwards by YOUR terms and got it right.

This is a positive result which implies an endothermic reaction.

Got it all right except that a positive value indicates EXOthermic. Do D+T like they did in the video and you will see that their result was positive too. + is EXO.

58Ni + P ---> 59Cu
8.7320 MeV/A + 0 MeV/A ---> 8.6419 MeV/A
506.4 MweV + 0 MeV ---> 509.6MeV
506.4 MeV - 509.6 MeV = - 3.2 MeV
This is a negative result and would be an exothermic reaction. Of course you then have to deal with what happens to the unstable 59Cu..........

Sorry, you got this one backward. This is a +3.2 which is EXO too, but much less so. It is the "much less so" that makes it less stable.

Comparative examples using other reactions:

235Ur ---> 144Ba + 89Kr + several neutrons (Fission of heavy element)
7.509MeV/A ---> 8.2654 MeV/A + 8.6169 MeV/A + 0 NeV/A
1783.8 MeV ---> 1190.2 MeV + 766.9 MeV + 0 MeV
1783.8 - ( 1190.2 + 766.9 MeV) = - 173.9 MeV This is an example given in the video .

Again, backwards. PRODUCT-REACTANTS. ( 1190.2 + 766.9 MeV) - 1783.8 = +173.9... EXO.

D + T --> 4He + Neutron
0.00231 MeV/A + 2..827 MeV/A --> 7.0739 MeV/A + 0 MeV/A
0.00231 MeV + 5.654 Mev ---> 28.2956 MeV +0 MeV
5.656 MeV - 28.2956 MeV = - 22.6392 MeV This is the other example given in the video. Note that both are exothermic reactions as the result is a negative number

Wow, this one is ALL screwed up. D is 1.112MeV/A, not what you have.

I see you use a backward convention. Your convention is Rect-prod = BE and negative equal exothermic. Given this, your first example is BACKWARD by your convention and the result should (by your convention) have been negativce (exo). The other two prior examples were also exo. So far, by your convention, ALL have been negative and all have been exothermic, including the first one which you calculated backward by your convention.

11B + P --> 12C* --> 4He + 8Be --> 4He + 4He +4He
6.297/a MeV/A + 0 MeV/A --> ---> 7.0739MeV/A + 7.0739MeV/A + 7.0739MeV/A

Should be 6.927, not 6.297.

69.26 MeV + 0 MeV ---> 28.29 MeV + 28.29 MeV + 28.29 MeV
69.26 MeV - 84.88 = - 15.62 MeV. Note that this is a negative result (exothermic)

Only by your convention but I can work with that too.

and that the number is not real close to the energy output of the published P-11B output of ~ 8.6 MeV.

Correct the mistake in your BE/A for 11B and the answer will come out as it should. The correct value of course, changes the answer. Don't you check your math?

I suppose this discrepence is due to the intermediates.

No, it is due to your math error.

The 12 C that is initially formed is a higher energy isomer (meaning some energy is retained (endothermic in this regard when compared to the baseline stable 12C. The beryllium intermediate may also be complicating things.....

These examples shows exothermic reactions and one endothermic reaction.

Since you did the first calculation BACKWARD by your own convention, it turns out, by your convention, to be negative and EXOthermic too. Do you TRY to get your calculations wrong? You got your first calculation wrong in your other set of calcs in chrismb's total binding energy topic over in "General".

The endothermic is the proton plus Ni62 reaction that is claimed by Rossi.

It is only "endo" if you do your own convention backwards. Do you TRY to do the math wrong?

He would have been better served by claiming that the heat producing reaction in his device was 58Ni + proton. It is exothermic.

Yes it is and you calculated that by doing your own convention correctly. (backward though it is to the rest of the world).

Of course, he would then have to explain the energy balance as the 59Cu isotope decaye. Would the energy balance still be positive- I suspect so.

Since he correctly knows that the p+62Ni is exothermic, he needs no such circumlocutions. You are still batting ZERO.
The rest of your post was not worth commenting on.
I am going to change my signature, Do the numbers... correctly.

[KitemanSA]

Percisely, P reactions, or s-or r processes can build nuclei heavier than Ni, this is well acepted physics. BUT, that does not mean these are exothemic reactions past Nickel 62. There is plenty of thermal energyaviable in a aging gient star to drive these reactions. And during the final stage, if a supernova occurs this excess energy is profound for a few seconds.

And, as far as the entire blogosphere, is the links attached to this site -

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

denieing that the nuclear binding energy per nucleon graphs that show 62Ni as the critical point immaginary? Not to mention the hyperphysics links provided?


b

Dan Tibbets

Dan, in each and every example calculation you provided, even the one you did backward by your own convention, you FIRST converted BE/A into BE by multiplying each term by the appropriate A. So by your own calculations, it is the BE (total binding energy per nucleous), not the BE/A (binding energy per nucleon) that is what counts. Do you deny that?

[D Tibbets]

Using the Total binding energy graph as the final description, fusion always adds decreasing,

I went to considerable effort to plot out a 'total binding energy' graph for you, and you still don't recognise what the data is telling you.

Sorry, Dan, you are flat-out wrong on this. I said the same as you initially, then it became obvious that it wasn't true.

Think not what nikel can do for H, but what H can do for Ni. Picture, if you will, that the H is the one giving up its exothermic potential. H has the capacity to fuse to released energy - when it fuses it has release the exothermic potential, whilst the Ni, being at the top of the per nucleon binding energy, changes only a little. The Ni DOES, effectively, consume energy (2MeV or so) to become its part of the 53Cu, but the H part of that reaction releases more (8MeV). The result is 6MeV released. The Ni, in your parlance DOES consume energy endothermically, but what happens to the H is to release more energy than that, so the overall is exothermic.

Capiche?

No I don't.
Hydrogen doesn't give anything. It has zero binding energy. Nuclear binding energy only exists when two or more nucleons are bound together. You can consider any combination of two or more bound nucleons and the binding energy associated with them. The real starting point, or ending point or zero point if you want is the Ni62. This is why the reciprocal of the nuclear binding energy per nucleon graph is more reveling. The Ni62 is at the low point on the graph. If you add or subtract a single nucleon, like a hydrogen proton, you move upwards on the graph. As the Ni62 represents the minimal potential energy state, moving up the graph increases the potential energy. It doesn't matter whether you move by one nucleon or many. You are always inputing kinetic energy, there is no other way you could be raising the potential energy. . Your graph is the total binding energy. It would apply if you were totally dissasembling or totally assembiling a nucleus from/ to totally free single nucleons. This allows the average binding energy per nucleon to apply as all of them are involved. This is your graph- the total binding energy of the nucleus. You cannot simply subtract out nucleons from this and assume each reaction would result in the average energy change. But in any sequential small or moderate or large (not total) steps, the nucleons are being added or removed at different energies depending on the starting point. As you approach Ni62 from the low side, each added nucleon adds less energy than the previous step. This of course continues past 62Ni if you are talking about the total binding energy, but as repeatedly pointed out , this total energy is made up of primarily two opposing energies. The total goes up but the effect of this total energy on a particular nucleon (especially those on the surface where the cohesive strong force mediated attraction is waining rapidly, while the repulsive electromagnetic force is waxing. This surface nucleon experiances forces much different than a nuclon in the center of the nucleus. That is why using an average term cannot apply to the surface nucleon and a centrally located nucleon at the same time. You have to incorporate this nucleon specific effect in your graph, And you also have to consider these effects for the two actual forces involved (it is more complicated than this, see the 2nd link) So you have to recalculate the effects for all possible nuclei and nucleon combinations. This is why the Nuclear Binding Energy per Nucleon graph is used. It incorporates these changing and competing effects. That is why this graph is used to calculate the energy balance, not the total binding energy graph, as is obvious from all of these examples-

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



Compare it to the electron orbitals in an atom. By assigning an average energy to all of the electrons, and then using this to predict the chemical reaction possibilities- oxidation, reduction preferences, etc. is fruitless. The only exception would be totally ionizing the atom/ now an ion. You could then apply an average energy to all of the electrons and talk about the total energy involved and the energy carried by each 'average' electron. This is a special case only where the electrons are totally disassembled from the atom (completly ionized)and is equivalent to the total binding energy curve. The ionization energy of all of the electrons are not the same, or average. It takes only a few eV to pull off a surface electron from the outer shell, but it takes at least hundreds of eV to pull off the deeper electrons from a high Z atom. And this is with only one force involved. The complexity increases with two competing forces involved in a nucleus. The total binding energy per nucleon, reflects the differing energy status of the bound nucleons within different nuclei positions.


A little more information with some different language:
http://www.flickr.com/photos/mitopencou ... 773069704/

and then look up the semi-empirical Weizsäcker formula.

And, I'll ask you, by using the total binding energy graph to predict the energy balance, how do you explain the existance of exothermic Uranium fission, if the the slope of the graph is always positive? *

And, finally, look at the examples I worked a couple of posts back. I answered Kiteman'SA's challange and determined the energy balance for the hypothetical 62Ni + proton to 63Cu reaction along with a few other examples.

* The answer is of course the further refinement of the interacting physics that leads to the understanding reflected in the nuclear binding energy per nucleon graph.

Dan Tibbets

[Joseph Chikva]

No I don't.
Hydrogen doesn't give anything. It has zero binding energy. Nuclear binding energy only exists when two or more nucleons are bound together. You can consider any combination of two or more bound nucleons and the binding energy associated with them. The real starting point, or ending point or zero point if you want is the Ni62.

Dan,
Sorry. It is too difficult for me to read your texts. Those are too long for me.
So, I will answer on above quoted.
Yes, hydrogen has zero binding energy.
But there is not a big difference between:
• p+Ni62=>Ni63 (in case if that will occur)
• p+B11=>3He4
As in both cases for calculation of energy yield for a single event you simply should add the masses of both reactants and then compare that with mass (or addition of masses in the second case) of product/products.
That's all.

PS: pardon for mistake. I wrote:
p+Ni62=>Ni63 (in case if that will occur)
Not Ni63 but Cu or whatever follows Ni in Mendeleev tab.

[Giorgio]

Dan,
Sorry. It is too difficult for me to read your texts. Those are too long for me.

You find Dan posts too long and complicated to read, YET you claim to have read and understood all Polywell related papers that are orders of magnitude longer and complicated!

You are a funny guy indeed.

[Joseph Chikva]

You find Dan posts too long and complicated to read, YET you claim to have read and understood all Polywell related papers that are orders of magnitude longer and complicated!

You are a funny guy indeed.

During my life and I am 47 I have read a lot of things.
Some of them are much longer than even all that was written on Polywell.
And also much more complicated.
In Dan's post I saw the mistake for the beginning. Why I should go on?
As a rule many words do not make a big sense.

[Giorgio]

You find Dan posts too long and complicated to read, YET you claim to have read and understood all Polywell related papers that are orders of magnitude longer and complicated!

You are a funny guy indeed.

During my life and I am 47 I have read a lot of things.
Some of them are much longer than even all that was written on Polywell.
And also much more complicated.
In Dan's post I saw the mistake for the beginning. Why I should go on?
As a rule many words do not make a big sense.

This has nothing to do with Dan post, this has to do with your attitude.
You claim that the Polywell cannot work and yet you have not read nor understood the papers describing the Polywell idea.

You should really take a step back, stop behaving as if you know everything and have the humility to study the papers (or ask clarifications) about the the theoretical framework of Polywell.

Lastly, age does not make us better persons. Attitude does.

[Joseph Chikva]

This has nothing to do with Dan post, this has to do with your attitude.
You claim that the Polywell cannot work and yet you have not read nor understood the papers describing the Polywell idea.

You should really take a step back, stop behaving as if you know everything and have the humility to study the papers (or ask clarifications) about the the theoretical framework of Polywell.

Lastly, age does not make us better persons. Attitude does.

I have read regarding Polywell enough for undarstanding how that should work. And have my own opinion.
I have not any necessity to look good in your eyes.
I thought that the forum is an exchange of opinions. But opinions of the people understanding the discussing issues at least at minimal level. But here basically I see only excited amateurs.

[MSimon]

This has nothing to do with Dan post, this has to do with your attitude.
You claim that the Polywell cannot work and yet you have not read nor understood the papers describing the Polywell idea.

You should really take a step back, stop behaving as if you know everything and have the humility to study the papers (or ask clarifications) about the the theoretical framework of Polywell.

Lastly, age does not make us better persons. Attitude does.

I have read regarding Polywell enough for undarstanding how that should work. And have my own opinion.
I have not any necessity to look good in your eyes.
I thought that the forum is an exchange of opinions. But opinions of the people understanding the discussing issues at least at minimal level. But here basically I see only excited amateurs.

Joe,

My advice? Quit hanging with the idiots and go to a blog where the really smart people hang out. And don't tell anyone. Because before you know it there will be users and idiots attracted like moths to a flame for the merest hope that some of the shining brain power would reflect well on them.

Your talent is wasted here.

[Joseph Chikva]

Joe,

My advice? Quit hanging with the idiots and go to a blog where the really smart people hang out. And don't tell anyone. Because before you know it there will be users and idiots attracted like moths to a flame for the merest hope that some of the shining brain power would reflect well on them.

Your talent is wasted here.

Ok, thanks.

[KitemanSA]

Using the Total binding energy graph as the final description, fusion always adds decreasing,

I went to considerable effort to plot out a 'total binding energy' graph for you, and you still don't recognise what the data is telling you.

Sorry, Dan, you are flat-out wrong on this. I said the same as you initially, then it became obvious that it wasn't true.

Think not what nikel can do for H, but what H can do for Ni. Picture, if you will, that the H is the one giving up its exothermic potential. H has the capacity to fuse to released energy - when it fuses it has release the exothermic potential, whilst the Ni, being at the top of the per nucleon binding energy, changes only a little. The Ni DOES, effectively, consume energy (2MeV or so) to become its part of the 53Cu, but the H part of that reaction releases more (8MeV). The result is 6MeV released. The Ni, in your parlance DOES consume energy endothermically, but what happens to the H is to release more energy than that, so the overall is exothermic.

Capiche?

No I don't.
Hydrogen doesn't give anything. It has zero binding energy. Nuclear binding energy only exists when two or more nucleons are bound together.

Dan
Perhaps this is the source of your problem.
While it is true that H has no binding energy, it has the POTENTIAL energy (mass excess) that is released when binding. That RELEASED energy, part of its total potential (total mass), is the binding energy.

Someone mentioned the "whole blogosphere" being against you. While I don't know how to poll the "whole blogosphere", you might want to poll this forum. If I do it, I'm beating you up. If you do it, you are seeking knowledge. Simple question.
Given a reaction p+62Ni=63Cu; is the reaction endothermic or exothermic? Then provide those two answers in the poll.

Would be interesting.

[Luzr]

Hydrogen doesn't give anything. It has zero binding energy. Nuclear binding energy only exists when two or more nucleons are bound together.

Ah, I see. Means H+H -> He is endotermic. Must be quite dark in your personal universe

[Tech]

This was posted on his blog 4 days ago, it sounds interesting:

Martin
July 14th, 2011 at 1:57 PM
Dear mr Rossi,
Today there was a meeting with NASA about your invention. Is it possible to give some information about this meeting? (omissis)

Andrea Rossi
July 15th, 2011 at 7:24 AM
Dear Martin:
I am not authorized to give this information.
I can only say that there is really to learn.
I met extremely high level scientists. I have been really surprised and honoured to discover that they have got indipendently throughly information about this technology. All I can say, so far, is that we will work together.
Warm Regards,
A.R.
Andrea Rossi