10KW LENR demonstrator (new thread)

[ladajo]

After all, anyone would think that a 1MW reactor is pretty easy to test, it works or it does not. However in Rossi land things can be more complex than that.

I prefer to think of it as Rossiworld and to paraphrase an Italian band (Ligabue) that I really like, Sta ballando sul mondo.

[Luzr]

If it does work, 1 megawatt will be difficult to explain away.

Much easier. Device with total input of 120Kw electricity producing steam.

The whole 1MW thing is total nonsense, waste of time. All that is needed for Rossi to become one of richest men in the world, if e-cat works, is proper test of single unit. The test without Rossi around.

[Skipjack]

As I said before, 1MW/h steam coming out of a 2" pipeline is not going to be healthy for any one near that machine.

This is indeed an interesting question. From what I understand, the pipes that we have seen are for the first loop?
If so, then this could be an explanation:

Andrea Rossi
October 15th, 2011 at 2:26 PM
Dear Bill Nichols:
1- very wide, we will use diathermic oil in the primary, toallow a wide range of choices
2- yes
3- up to 450 Celsius so far
Warm Regards,
A.R.

Otherwise, I dont think this is going to work, or rather, it would be quite catastrophic, if it worked as (allegedly) designed.

[ladajo]

Yes but even a 2 inch line in the primary means a high flow for the heat loads. And high flow rates introduce their own design problems, especially in a high temp potentially high pressure system.

Infinite Improbability Drive keeps coming to mind for some reason.

[tomclarke]

If it does work, 1 megawatt will be difficult to explain away.

Much easier. Device with total input of 120Kw electricity producing steam.

The whole 1MW thing is total nonsense, waste of time. All that is needed for Rossi to become one of richest men in the world, if e-cat works, is proper test of single unit. The test without Rossi around.

The 1MW thing without working low power prototype is one of the many obvious red flags here.

[D Tibbets]

I have not been following this thread much, at least partially because the enthusiasts are so entrenched. So, some of my comments may be inappropriate. First off adding a proton or a neutron to 62Ni will lose kinetic energy and increase the potential energy of the isotope. Several months ago I repeatedly tried to explain this.

And you repeatedly failed because you were repeatedly wrong.

THINK HYDROGEN

Dan, find ANYONE who agrees with you. Please. If you can't, please consider that you are just plain wrong. If you do find someone, introduce us, bring him (or her) into the discussion, and (s)he probably won't agree with you for long.

I have thought of hydrogen. You are the confused one.
A simple reaction describing the process of Binding energy per nucleon.

62 Ni + ~ 8.9 MeV (energy needed to pull one nucleon (proton in this case)off of the parent nuclei) --yields -- 61 Co + Proton (hydrogen nucleus.
The reverse reaction is 61 Co + P --yields- 62Ni + ~8.8 MeV
Note that the energy balance is not equal . It is this energy difference that allows for exothermic energy yielding nuclear reaction.. The energy yield is the difference between these two values. Whether it is a positive or negative KE yield depends on more stable ( the species with the least potential energy (or the most negative potential energy if you use that convention. This becomes obvious if you consider which species requires the most binding energy per nucleon to convert to it's neighbor. This is because the nucleus does not have as much potential energy to contribute to the reaction. This is of course the 62Ni nuclei.
Note that the energy on one side or the other of the equation will be shared between the products (if more than one). It is this energy (KE) that is the measure of the resultant velocity of the particle/s. If they slow down it is endothermic, and if they speed up it is exothermic. This is a simplification as it ignores gamma rays but irregardless of the specific particles emmited or absorbed, they have to share this energy.

The Reactant has a certain potential energy and the product has a certain potential energy. It is the difference between these potential energies as represented by the opposing strong force and electromagnetic force sum, along with some other miner interactions like the Pauli exclusion principle.
The proton in this case whether added or removed from the nucleus may share some of the KE of the reaction. But they contribute no potential energy to the reaction. After all the proton has zero binding energy.
This is why some of the links I provided use potential energy terms, that hopefully would avoid this confusion.
In a reaction whether chemical, gravitational, or nuclear. The only way you can get energy out (exothermic is if the potential energy of the products is less than the reactants. It is important to remember the potential. The negative value convention for attractive potential energy helps to keep the acounting straight. Rember that this potential energy is a balence between two forces. The strong force is the origin of KE yield due to clumping nucleons together. The electromagnetic or Coulomb force actually impeads this. That the reactions proceed at all is due to the greater strength of the strong force attractive interaction. The reason the interaction peaks and then reverses is due to the very short range of the strong force compared to the Coulomb repulsion force. This occurs at 62Ni. If there was not a reversal the universe would not exist as wee know it and it would not be possible to get any energy out of fission.
Ignoring complexities of excited nuclear isomers, this is why light elements "decay" towards iron/ nickel if the energy barrior -Coulomb barrior for nuclear reactions (Gibbs free energy for chemical reactions, the hill in front of the deep canyon in a gravitational reaction) can be overcome. The products have a lower potential energy than the reactants. For "decay" of heavier elements the reaction direction goes in the opposite direction, sheding KE as the become more stable, reach lower potential energy conditions , again ending at 6Ni. There are various permutations, restraints that effect the likely hood of these reactions , but it is o the potential enery difference that drives the reaction.

The NI to Co reaction mentioned above is the same as the Ni to Cu reaction. but, note that the energy flow is reversed. In going from Co to Ni the energy difference as represented by the binding energy per nucleon of each nucleus. Again note the binding energy of the proton is Zero, so it does not contribute or take away any of the potential energy. It is the difference in the binding energy per nucleon of the Bound nucleus that is contributing to the energy balance. Before you get too excited by P-P fusion, it is the binding energy of the deuterium minus the binding energy of two protons (which is zero) that gives the yield.


At first glance both the Co-Ni and the Ni to Cu reaction would seam to yield ~ 0.1 MeV of energy ( the difference between the reactant and products. But consider the potential energy change. In the Co to Ni reaction the potential energy has decreased ( remember a negative value is assigned to the Potential energy derived from the strong force). The potential added to the nucleus as each nucleon is added due to the repulsive electromagnetic force is assigned a positive value. Due to the short range of the strong force, as the nuclear diameter increases the attractive force saturates- the attraction force growth will always increase, but the change will become smaller and smaller. Meanwhile the repulsive force will always increase in a more linear fashion, so at some point it will exceed the rate change of the nuclear force, which is at 62Ni, and eventually it will exceed the total accumulated strong force attraction, at which point the nucleus falls apart rapidly. This occurs at ~ above ~ 206 atomic mass units. Beyond ~ 250 AMU ? the nucleus cannot withstand the repulsive force at all so the decay halflives become vanishingly short.

Back to the Co- Ni- Cu reactions. When a nucleon is added to the Co nucleus the strong force attractive interaction increases an arbitrary amount of ~ -1.05 MeV. At the same time the electromagnetic force repulsion interaction increases ~+0.95 MeV. The net result is that the negative potential energy increases by ~ 0.1 MeV . Since the potential energy becomes more negative (smaller) the reaction releases ~ 0.1 MeV as KE. In the Ni to Cu reaction ~ -0.95 MeV of strong force attraction is added, while ~ +1.05 MeV of electromagnetic repulsion is added. This results in the net potential energy increasing in a positive direction by ~ 0.1 MeV. Therefor KE energy must be provided for the reaction to proceed. This is clearly evident from the shape of the Binding Energy per Nucleon graphs, and is why multiple links spell this energy flow on the graphs.

Again, the protons, and neutrons are along for the ride. As they do not have any usefull potential energy by them selves they can be ignored from an energy balance perspective.

Note that I used arbitrary values for the binding energy per nucleon. They are ~ accurate. All surely can agree that the binding energy per nucleon values for both 61Co and 63CU are a little less than for 62Ni, I didn't look up the specific values as they add nothing to this trend comparison.
Also note that this discussion relates to comparison between neighboring elements. This relates the relative contributions of the strong force and the electromagnetic force between the neighbors and thus the energy difference between two atoms. Obviously if you instantly strip all of the nucleons off of a nucleus, the total binding energy would represent the missing mass. But things don't work that way. Binding energy per nucleon changes as you go from step to step thus the results are adjusted for each step. And that adjustment involves the relationship of the two major contributors to the missing mass.
A simplified, but I suspect still problematic, comparison is the weight comparison. Comparing 63 grams of copper (one mole of 63Cu) to 62 grams of Ni is a handicap that you are ignoring. This may not make up the entire difference if you insist on using the Binding energy per Nucleon * number of nucleons in a mole) you have to accommodate the nature of the binding energy, not just the sum of the absolute values that gives the mass deficit. The electromagnetic and strong force contributions have opposite signs. If you wish to only use positive values for both you must adjust your formulas accordingly..

Finally, in short, the mass deficit does not describe the energy flow balance. The interaction of the components that contribute to the mass deficit does.

Dan Tibbets

[Giorgio]

After all, anyone would think that a 1MW reactor is pretty easy to test, it works or it does not. However in Rossi land things can be more complex than that.

I prefer to think of it as Rossiworld and to paraphrase an Italian band (Ligabue) that I really like, Sta ballando sul mondo.

Ehehe, nice one.

[Giorgio]

As I said before, 1MW/h steam coming out of a 2" pipeline is not going to be healthy for any one near that machine.

This is indeed an interesting question. From what I understand, the pipes that we have seen are for the first loop?
If so, then this could be an explanation:

Andrea Rossi
October 15th, 2011 at 2:26 PM
Dear Bill Nichols:
1- very wide, we will use diathermic oil in the primary, toallow a wide range of choices
2- yes
3- up to 450 Celsius so far
Warm Regards,
A.R.

Otherwise, I dont think this is going to work, or rather, it would be quite catastrophic, if it worked as (allegedly) designed.

Well, if that's true is a good change in their test plans, at least is not going to blow up like a firework but more like a very hot spring fountain

[Giorgio]

Yes but even a 2 inch line in the primary means a high flow for the heat loads. And high flow rates introduce their own design problems, especially in a high temp potentially high pressure system.

Especially considering that a good thermal oil has generally only half of the heat capacity of water.
To get a meaningful heat removal capacity they should pump it out at least at 300'C, but I haven't seen any thermal joint or any other necessary ancillary equipment to compensate such a temperature load in their container.
I really hope they will broadcast the experiment on webcam as I have been reported, as this is going to be quite valuable material to show to the next generation of mechanical Engineers.

[KitemanSA]

I have thought of hydrogen. You are the confused one.
A simple reaction describing the process of Binding energy per nucleon.

62 Ni + ~ 8.9 MeV (energy needed to pull one nucleon (proton in this case)off of the parent nuclei) --yields -- 61 Co + Proton (hydrogen nucleus.
The reverse reaction is 61 Co + P --yields- 62Ni + ~8.8 MeV
Note that the energy balance is not equal . It is this energy difference that allows for exothermic energy yielding nuclear reaction..

Absurd! The binding energy needed to get from 62Ni to 61Co = 11MeV each way. (62*8.7945)-(61*8.7561)=11.1369
PLEASE learn this stuff and stop making a fool of yourself!

[parallel]

So far so good. It seems that Rossi will be using water in the 1 MW unit.
As I wrote before, this will enable the purchaser to do accurate heat flow measurements.

Andrea Rossi
October 16th, 2011 at 2:19 AM

Dear Italo:
1- The steam will be condensed in dissipators and recycled to the plant
2- see answer 1
3- Situation much more complex, will see the report
4- Exactly
Warm Regards,
A.R.

Andrea Rossi
October 16th, 2011 at 4:10 PM

Dear Enzo Amato:
I confirm: on the 28th so far there are no obstacles to run the test. Of course, if in the preliminar tests we are running something will go wrong, we will have to delay. But so far, so good.
Warm Regards,
A.R.

[seedload]

These are words of a prophet that his warming the crowd to keep it loyal in the eve of the coming failure.

That

[stefanbanev]

I have thought of hydrogen. You are the confused one.
A simple reaction describing the process of Binding energy per nucleon.

62 Ni + ~ 8.9 MeV (energy needed to pull one nucleon (proton in this case)off of the parent nuclei) --yields -- 61 Co + Proton (hydrogen nucleus.
The reverse reaction is 61 Co + P --yields- 62Ni + ~8.8 MeV
Note that the energy balance is not equal . It is this energy difference that allows for exothermic energy yielding nuclear reaction..

Absurd! The binding energy needed to get from 62Ni to 61Co = 11MeV each way. (62*8.7945)-(61*8.7561)=11.1369
PLEASE learn this stuff and stop making a fool of yourself!

Apparently he likes it...

[parallel]

Back of the envelope check on the exit pipe some have speculated at 2" diameter.

steam flow 10.27lb/min through 2" pipe - 240 diameters for 1 psi gauge pressure drop.
http://www.gutenberg.org/files/22657/22 ... /flow.html

Say 1150 btu/lb to boil water
1 BTU/minute equals 0.01757 kilowatt. http://www.boilerroomservices.com/Facts/SteamTables.pdf

= 11,500 btu/min = 201kW/min = 1275 kW/hr

So a 2" pipe is in the right ball park.

[stefanbanev]

Back of the envelope check on the exit pipe some have speculated at 2" diameter.

steam flow 10.27lb/min through 2" pipe - 240 diameters for 1 psi gauge pressure drop.
http://www.gutenberg.org/files/22657/22 ... /flow.html

Say 1150 btu/lb to boil water
1 BTU/minute equals 0.01757 kilowatt. http://www.boilerroomservices.com/Facts/SteamTables.pdf

= 11,500 btu/min = 201kW/min = 1275 kW/hr

So a 2" pipe is in the right ball park.

Anyway, once the pressure was not specified 2" diameter was a totally pointless point to make...