subject:"RE\: \[Vo\]\:A bombshell of a different type\?"

In reply to  Axil Axil's message of Mon, 5 Jan 2015 22:07:23 -0500:
Hi,
[snip]
In the nuclear industry, there is a reactor type called the pebble bed
reactor. That reactor uses a uranium and plutonium nuclear fuel enclosed in
a graphite and Silicon carbide coating called TRISO fuel.

http://www.intechopen.com/books/metal-ceramic-and-polymeric-composites-for-various-uses/composite-materials-under-extreme-radiation-and-temperature-environments-of-the-next-generation-nucl

That pebble bed fuel has been tested to keep all the products of
fission sequestered  for years at a 100% reliability rate.

Hydrogen isn't really a fission product, though a little might be produced
through proton spallation.
(In the Fukushima disaster I think the hydrogen was produced by the chemical
reaction of the Zirconium cladding with water.)
However what I had in mind was actually a mini-molecule of Li bound to Hydrino
Hydride ions, and this mini molecule should pass through the interstitial gaps
in any lattice, so I doubt any sort of cladding would hold it for long (unless
it carries a net charge). The cladding might just have to be made thick to be
somewhat useful.
[snip]
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

http://luratia.com/graphene/graphene-is-impermeable

graphene is impermeable

Graphene is the most impermeable material Earth scientists have yet
discovered. It’s so impermeable the smallest element hydrogen can’t pass
through.

Real-life applications:

Graphene-lined hydrogen fuel tanks. Hydrogen is the smallest of all the
elements, making its containment very difficult. The impermeable nature of
graphene will enable leak-proof storage of hydrogen.

Graphene is one of the wonders of the science world, with the potential to
create foldaway mobile phones, wallpaper-thin lighting panels and the next
generation of aircraft. The new finding at the University of Manchester
gives graphene's potential a most surprising dimension – graphene can also
be used for distilling alcohol.

In a report published in Science, a team led by Professor Sir Andre Geim
shows that graphene-based membranes are impermeable to all gases and
liquids (vacuum-tight). However, water evaporates through them as quickly
as if the membranes were not there at all.

This newly-found property can now be added to the already long list of
superlatives describing graphene. It is the thinnest known material in the
universe and the strongest ever measured. It conducts electricity and heat
better than any other material. It is the stiffest one too and, at the same
time, it is the most ductile. Demonstrating its remarkable properties won
University of Manchester academics the Nobel Prize in Physics in 2010.

Now the University of Manchester scientists have studied membranes from a
chemical derivative of graphene called graphene oxide. Graphene oxide is
the same graphene sheet but it is randomly covered with other molecules
such as hydroxyl groups OH-. Graphene oxide sheets stack on top of each
other and form a laminate.

The researchers prepared such laminates that were hundreds times thinner
than a human hair but remained strong, flexible and were easy to handle.

*When a metal container was sealed with such a film, even the most
sensitive equipment was unable to detect air or any other gas, including
helium, to leak through.*

It came as a complete surprise that, when the researchers tried the same
with ordinary water, they found that it evaporates without noticing the
graphene seal. Water molecules diffused through the graphene-oxide
membranes with such a great speed that the evaporation rate was the same
independently whether the container was sealed or completely open.

Dr Rahul Nair, who was leading the experimental work, offers the following
explanation: Graphene oxide sheets arrange in such a way that between them
there is room for exactly one layer of water molecules. They arrange
themselves in one molecule thick sheets of ice which slide along the
graphene surface with practically no friction.

If another atom or molecule tries the same trick, it finds that graphene
capillaries either shrink in low humidity or get clogged with water
molecules.

*Helium gas is hard to stop. It slowly leaks even through a millimetre
-thick window glass but our ultra-thin films completely block it. At the
same time, water evaporates through them unimpeded. Materials cannot behave
any stranger, comments Professor Geim. You cannot help wondering what
else graphene has in store for us.*

This unique property can be used in situations where one needs to remove
water from a mixture or a container, while keeping in all the other
ingredients, says Dr Irina Grigorieva who also participated in the
research.

Just for a laugh, we sealed a bottle of vodka with our membranes and found
that the distilled solution became stronger and stronger with time. Neither
of us drinks vodka but it was great fun to do the experiment, adds Dr Nair.

The Manchester researchers report this experiment in their Science paper,
too, but they say they do not envisage use of graphene in distilleries, nor
offer any immediate ideas for applications.

However, Professor Geim adds 'The properties are so unusual that it is hard
to imagine that they cannot find some use in the design of filtration,
separation or barrier membranes and for selective removal of water

Read more at:
http://phys.org/news/2012-01-graphene-supermaterial-superpermeable.html#jCp

On Tue, Jan 6, 2015 at 1:55 AM, mix...@bigpond.com wrote:

In reply to Axil Axil's message of Mon, 5 Jan 2015 22:07:23 -0500:
Hi,
[snip]
In the nuclear industry, there is a reactor type called the pebble bed
reactor. That reactor uses a uranium and plutonium nuclear fuel enclosed
in
a graphite and Silicon carbide coating called TRISO fuel.

http://www.intechopen.com/books/metal-ceramic-and-polymeric-composites-for-various-uses/composite-materials-under-extreme-radiation-and-temperature-environments-of-the-next-generation-nucl

That pebble bed fuel has been tested to keep all the products of
fission sequestered for years at a 100% reliability rate.

Hydrogen isn't really a fission product, though a little might be

Re: [Vo]:A bombshell of a different type?

In reply to  Axil Axil's message of Mon, 5 Jan 2015 23:37:08 -0500:
Hi,
[snip]
The last MFMP test showed that stainless steal leaked hydrogen badly at
high temperatures.
Actually that wouldn't surprise me. Iron like Pt, Pd, Ti, and Ni, will absorb
some H. A process which I imagine would go faster at higher temperatures.
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

What if a coat of graphite was applied to the outside of the HotCat as a
hydrogen barrier during its fabrication and then a final thin veneer coat
of alumina cement completes the fabrication by covering the graphite and
forming the heat radiating fin structure.

The hydrogen could permeate throughout the alumina body of the remote not
being confined until the hydrogen hit the graphite coat on the outside of
the HotCat.

This method of fabrication would allow hydrogen to get into all of the
porous alumina structure throughout the entire HotCat reactor.

This would allow much more Oxygen 17 by many orders of magnitude to be made
available to the nuclear reaction under discussion.

On Mon, Jan 5, 2015 at 5:08 PM, mix...@bigpond.com wrote:

 In reply to  Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
 Hi,
 [snip]
 Have I missed something important?
 
 Eric

 Something else I just thought of:

 17O+6Li = 16O + 7Li + 3.107 MeV

 This reaction would provide a path for Li7 to be regenerated from O17 in
 the
 Al2O3.

 The same mechanism that enabled the transfer of a neutron from Li to Ni
 could
 also enable this regeneration transfer.

 0.037% of O is O17, so 450 gm of Al2O3 would contain about 3E21 O17 atoms
 allowing for the regeneration of another 3E21 Li7 atoms.

 This process would, optimistically, quadruple the amount of Li7 available,
 and
 also add considerable energy to the process.

 Regards,

 Robin van Spaandonk

 http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

In reply to  Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
Hi,
[snip]
Have I missed something important?

Eric

Something else I just thought of:

17O+6Li = 16O + 7Li + 3.107 MeV

This reaction would provide a path for Li7 to be regenerated from O17 in the
Al2O3.

The same mechanism that enabled the transfer of a neutron from Li to Ni could
also enable this regeneration transfer.

0.037% of O is O17, so 450 gm of Al2O3 would contain about 3E21 O17 atoms
allowing for the regeneration of another 3E21 Li7 atoms.

This process would, optimistically, quadruple the amount of Li7 available, and
also add considerable energy to the process.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

In reply to  Axil Axil's message of Mon, 5 Jan 2015 17:58:58 -0500:
Hi,
[snip]

Providing that a graphite coat would actually perform this function, it could be
a good idea.

What if a coat of graphite was applied to the outside of the HotCat as a
hydrogen barrier during its fabrication and then a final thin veneer coat
of alumina cement completes the fabrication by covering the graphite and
forming the heat radiating fin structure.

The hydrogen could permeate throughout the alumina body of the remote not
being confined until the hydrogen hit the graphite coat on the outside of
the HotCat.

This method of fabrication would allow hydrogen to get into all of the
porous alumina structure throughout the entire HotCat reactor.

This would allow much more Oxygen 17 by many orders of magnitude to be made
available to the nuclear reaction under discussion.

Note that in my calculations here below, I already assumed that all of the O17
in the Alumina was used. That's why I said it was optimistic.


On Mon, Jan 5, 2015 at 5:08 PM, mix...@bigpond.com wrote:

 In reply to  Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
 Hi,
 [snip]
 Have I missed something important?
 
 Eric

 Something else I just thought of:

 17O+6Li = 16O + 7Li + 3.107 MeV

 This reaction would provide a path for Li7 to be regenerated from O17 in
 the
 Al2O3.

 The same mechanism that enabled the transfer of a neutron from Li to Ni
 could
 also enable this regeneration transfer.

 0.037% of O is O17, so 450 gm of Al2O3 would contain about 3E21 O17 atoms
 allowing for the regeneration of another 3E21 Li7 atoms.

 This process would, optimistically, quadruple the amount of Li7 available,
 and
 also add considerable energy to the process.

 Regards,

 Robin van Spaandonk

 http://rvanspaa.freehostia.com/project.html


Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

2015-01-05 Thread Bob Cook

I just heard Sunday that one of the first Thermo nuclear devices exploded in 
the Pacific yielded 15 Megatons of TNT vs the predicted 5 Mega tons.  This 
estimate or prediction was off by a factor  of 3.  

Maybe some of the extra energy came from reaction of Li-6 with O-17 in a 
reaction like that suggested by Axil, only with the coupling agent being 
kinetic energy of the reactants.  This of course assumes the presence of Li in 
the device to start with or its generation during the reaction. 

Bob  
  - Original Message - 
  From: Axil Axil 
  To: vortex-l 
  Sent: Monday, January 05, 2015 2:58 PM
  Subject: Re: [Vo]:A bombshell of a different type?


  What if a coat of graphite was applied to the outside of the HotCat as a 
hydrogen barrier during its fabrication and then a final thin veneer coat of 
alumina cement completes the fabrication by covering the graphite and forming 
the heat radiating fin structure. 


  The hydrogen could permeate throughout the alumina body of the remote not 
being confined until the hydrogen hit the graphite coat on the outside of the 
HotCat.


  This method of fabrication would allow hydrogen to get into all of the porous 
alumina structure throughout the entire HotCat reactor.


  This would allow much more Oxygen 17 by many orders of magnitude to be made 
available to the nuclear reaction under discussion. 


  On Mon, Jan 5, 2015 at 5:08 PM, mix...@bigpond.com wrote:

In reply to  Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
Hi,
[snip]
Have I missed something important?

Eric

Something else I just thought of:

17O+6Li = 16O + 7Li + 3.107 MeV

This reaction would provide a path for Li7 to be regenerated from O17 in the
Al2O3.

The same mechanism that enabled the transfer of a neutron from Li to Ni 
could
also enable this regeneration transfer.

0.037% of O is O17, so 450 gm of Al2O3 would contain about 3E21 O17 atoms
allowing for the regeneration of another 3E21 Li7 atoms.

This process would, optimistically, quadruple the amount of Li7 available, 
and
also add considerable energy to the process.


Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

2015-01-05 Thread Bob Cook

Axil--

That may be nice but it is probably a tall order to get the coating you
suggest--SiC/graphite to bond well to the alumina over the entire outer surface
and not crack with thermal cycling. I would think the distribution of H within
the Alumina could be accomplished better with a slow release of H as one would
have with LiH or AlLiH4. The controlled heating of the AlLiH4 could be
accomplished with the heating coils. I would think a better seal would be a
thin stainless steel or other high temperature ductile metal outer barrier to
contain the H, if that is desired.

The real advantage of the graphite outer shell in the pebble reactor (which is
not new) was to act as a neutron moderator as well as a long lived barrier to
corrosion and loss of fission products in the long-term management of the
depleted fuel. As I recall the old pebble reactors were to be cooled with He
gas. The pebbles were small spherical fuel cells whose diameter was
established to limit the internal temperature of the fuel at full power.

The item you reference discusses SiC and C/C fibers for structural strength at
high temperatures, NOT SEALING FOR H.

The conceptual design proposed in the reference is just pie in the sky
IMHO--not unlike the hot fusion concepts that have soaked up $B's over the
years and have not worked yet. I think the fast reactor ideas are non-sense.
The item you reference sounds like an add for research at Oak Ridge into
materials that can withstand high energy particle damage like one may get at
CERN.

Thankfully, the LENR coming revolution with put all this stuff to bed,
although way too slowly.

In summary, your TRISO LENR reactor would not compete with Rossi's drug store
variety IMO.

Bob
- Original Message -
From: Axil Axil
To: vortex-l
Sent: Monday, January 05, 2015 7:07 PM
Subject: Re: [Vo]:A bombshell of a different type?

In the nuclear industry, there is a reactor type called the pebble bed
reactor. That reactor uses a uranium and plutonium nuclear fuel enclosed in a
graphite and Silicon carbide coating called TRISO fuel.

http://www.intechopen.com/books/metal-ceramic-and-polymeric-composites-for-various-uses/composite-materials-under-extreme-radiation-and-temperature-environments-of-the-next-generation-nucl

That pebble bed fuel has been tested to keep all the products of fission
sequestered for years at a 100% reliability rate.

The same type of barrier element sequestering system could be used to keep
the Hot Cat type reactor element tight. The down side is that carbon has been
shown to be a poison to LENR in some LENR reactors.

Care in the design of the TRISO LENR reactor who be required.

From the analysis of the fuel mix in the Rossi reactor, carbon was found in a
high concentration. I take this as an indicator that carbon is being used in
the Hot Cat as a hydrogen barrier material.

I would put the graphite and SIC element barrier on the outermost serface of
the Hot Cat in my own reactor design.

On Mon, Jan 5, 2015 at 9:00 PM, mix...@bigpond.com wrote:

In reply to Axil Axil's message of Mon, 5 Jan 2015 17:58:58 -0500:
Hi,
[snip]

Providing that a graphite coat would actually perform this function, it
could be
a good idea.

What if a coat of graphite was applied to the outside of the HotCat as a
hydrogen barrier during its fabrication and then a final thin veneer coat
of alumina cement completes the fabrication by covering the graphite and
forming the heat radiating fin structure.

The hydrogen could permeate throughout the alumina body of the remote not
being confined until the hydrogen hit the graphite coat on the outside of
the HotCat.

This method of fabrication would allow hydrogen to get into all of the
porous alumina structure throughout the entire HotCat reactor.

This would allow much more Oxygen 17 by many orders of magnitude to be made
available to the nuclear reaction under discussion.

Note that in my calculations here below, I already assumed that all of the
O17
in the Alumina was used. That's why I said it was optimistic.

On Mon, Jan 5, 2015 at 5:08 PM, mix...@bigpond.com wrote:

In reply to Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
Hi,
[snip]
Have I missed something important?

Eric

Something else I just thought of:

17O+6Li = 16O + 7Li + 3.107 MeV

This reaction would provide a path for Li7 to be regenerated from O17 in
the
Al2O3.

The same mechanism that enabled the transfer of a neutron from Li to Ni
could
also enable this regeneration transfer.

0.037% of O is O17, so 450 gm of Al2O3 would contain about 3E21 O17 atoms
allowing for the regeneration of another 3E21 Li7 atoms.

This process would, optimistically, quadruple the amount of Li7
available

Re: [Vo]:A bombshell of a different type?

The coefficients of thermal expansion is almost equal for Alumina, SiC and
Carbon, so cracking will not occur. These TRISO fuel pebbles were tested
for 24 months in a high fluence nuclear reactor and passed the test.

This TRISO fuel does leak silver but not hydrogen.

The last MFMP test showed that stainless steal leaked hydrogen badly at
high temperatures.

On Mon, Jan 5, 2015 at 11:05 PM, Bob Cook frobertc...@hotmail.com wrote:

Axil--

That may be nice but it is probably a tall order to get the coating you
suggest--SiC/graphite to bond well to the alumina over the entire outer
surface and not crack with thermal cycling. I would think the distribution
of H within the Alumina could be accomplished better with a slow release of
H as one would have with LiH or AlLiH4. The controlled heating of the
AlLiH4 could be accomplished with the heating coils. I would think a
better seal would be a thin stainless steel or other high temperature
ductile metal outer barrier to contain the H, if that is desired.

The real advantage of the graphite outer shell in the pebble reactor
(which is not new) was to act as a neutron moderator as well as a long
lived barrier to corrosion and loss of fission products in the
long-term management of the depleted fuel. As I recall the old pebble
reactors were to be cooled with He gas. The pebbles were small spherical
fuel cells whose diameter was established to limit the internal temperature
of the fuel at full power.

The item you reference discusses SiC and C/C fibers for structural
strength at high temperatures, NOT SEALING FOR H.

The conceptual design proposed in the reference is just pie in the sky
IMHO--not unlike the hot fusion concepts that have soaked up $B's over the
years and have not worked yet. I think the fast reactor ideas are
non-sense. The item you reference sounds like an add for research at Oak
Ridge into materials that can withstand high energy particle damage like
one may get at CERN.

Thankfully, the LENR coming revolution with put all this stuff to bed,
although way too slowly.

In summary, your TRISO LENR reactor would not compete with Rossi's drug
store variety IMO.

Bob

- Original Message -
*From:* Axil Axil janap...@gmail.com
*To:* vortex-l vortex-l@eskimo.com
*Sent:* Monday, January 05, 2015 7:07 PM
*Subject:* Re: [Vo]:A bombshell of a different type?

In the nuclear industry, there is a reactor type called the pebble bed
reactor. That reactor uses a uranium and plutonium nuclear fuel enclosed in
a graphite and Silicon carbide coating called TRISO fuel.

http://www.intechopen.com/books/metal-ceramic-and-polymeric-composites-for-various-uses/composite-materials-under-extreme-radiation-and-temperature-environments-of-the-next-generation-nucl

That pebble bed fuel has been tested to keep all the products of
fission sequestered for years at a 100% reliability rate.

The same type of barrier element sequestering system could be used to keep
the Hot Cat type reactor element tight. The down side is that carbon has
been shown to be a poison to LENR in some LENR reactors.

Care in the design of the TRISO LENR reactor who be required.

From the analysis of the fuel mix in the Rossi reactor, carbon was found
in a high concentration. I take this as an indicator that carbon is being
used in the Hot Cat as a hydrogen barrier material.

I would put the graphite and SIC element barrier on the outermost serface
of the Hot Cat in my own reactor design.

On Mon, Jan 5, 2015 at 9:00 PM, mix...@bigpond.com wrote:

In reply to Axil Axil's message of Mon, 5 Jan 2015 17:58:58 -0500:
Hi,
[snip]

Providing that a graphite coat would actually perform this function, it
could be
a good idea.

The hydrogen could permeate throughout the alumina body of the remote not
being confined until the hydrogen hit the graphite coat on the outside of
the HotCat.

This method of fabrication would allow hydrogen to get into all of the
porous alumina structure throughout the entire HotCat reactor.

This would allow much more Oxygen 17 by many orders of magnitude to be
made
available to the nuclear reaction under discussion.

Note that in my calculations here below, I already assumed that all of
the O17
in the Alumina was used. That's why I said it was optimistic.

On Mon, Jan 5, 2015 at 5:08 PM, mix...@bigpond.com wrote:

In reply to Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
Hi,
[snip]
Have I missed something important?

Eric

Something else I just thought of:

17O+6Li = 16O + 7Li + 3.107 MeV

This reaction would provide a path for Li7 to be regenerated from O17
in
the
Al2O3.

The same mechanism

Re: [Vo]:A bombshell of a different type?

2015-01-05 Thread Terry Blanton

On Mon, Jan 5, 2015 at 9:08 PM, Bob Cook frobertc...@hotmail.com wrote:

  I just heard Sunday that one of the first Thermo nuclear devices
 exploded in the Pacific yielded 15 Megatons of TNT vs the predicted 5 Mega
 tons.  This estimate or prediction was off by a factor  of 3.


Teller-Ulam.

Re: [Vo]:A bombshell of a different type?

In the nuclear industry, there is a reactor type called the pebble bed
reactor. That reactor uses a uranium and plutonium nuclear fuel enclosed in
a graphite and Silicon carbide coating called TRISO fuel.

http://www.intechopen.com/books/metal-ceramic-and-polymeric-composites-for-various-uses/composite-materials-under-extreme-radiation-and-temperature-environments-of-the-next-generation-nucl

That pebble bed fuel has been tested to keep all the products of
fission sequestered for years at a 100% reliability rate.

The same type of barrier element sequestering system could be used to keep
the Hot Cat type reactor element tight. The down side is that carbon has
been shown to be a poison to LENR in some LENR reactors.

Care in the design of the TRISO LENR reactor who be required.

From the analysis of the fuel mix in the Rossi reactor, carbon was found in
a high concentration. I take this as an indicator that carbon is being used
in the Hot Cat as a hydrogen barrier material.

I would put the graphite and SIC element barrier on the outermost serface
of the Hot Cat in my own reactor design.

On Mon, Jan 5, 2015 at 9:00 PM, mix...@bigpond.com wrote:

In reply to Axil Axil's message of Mon, 5 Jan 2015 17:58:58 -0500:
Hi,
[snip]

Providing that a graphite coat would actually perform this function, it
could be
a good idea.

The hydrogen could permeate throughout the alumina body of the remote not
being confined until the hydrogen hit the graphite coat on the outside of
the HotCat.

This method of fabrication would allow hydrogen to get into all of the
porous alumina structure throughout the entire HotCat reactor.

This would allow much more Oxygen 17 by many orders of magnitude to be
made
available to the nuclear reaction under discussion.

Note that in my calculations here below, I already assumed that all of the
O17
in the Alumina was used. That's why I said it was optimistic.

On Mon, Jan 5, 2015 at 5:08 PM, mix...@bigpond.com wrote:

In reply to Eric Walker's message of Fri, 2 Jan 2015 23:36:57 -0800:
Hi,
[snip]
Have I missed something important?

Eric

Something else I just thought of:

17O+6Li = 16O + 7Li + 3.107 MeV

This reaction would provide a path for Li7 to be regenerated from O17 in
the
Al2O3.

The same mechanism that enabled the transfer of a neutron from Li to Ni
could
also enable this regeneration transfer.

0.037% of O is O17, so 450 gm of Al2O3 would contain about 3E21 O17
atoms
allowing for the regeneration of another 3E21 Li7 atoms.

This process would, optimistically, quadruple the amount of Li7
available,
and
also add considerable energy to the process.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

2015-01-03 Thread Alain Sepeda

I don't think that the 2mg are a problem , that is more the number of
grains analysed.

it is like poling. the size of the sample have to be bigger if there is
tiny minorities. few samples are ok to measure the majority.

the problem like on polling is more about bias on the sampling, like taking
more tiny or big dust, the one from one part or another from the tube.

a thousand grains well mixed from the dust should be enough to be
representative.

we could do some compensation based on morphology like on poling.
on a thousand grains, make statistics on morphologies, and analyse grains
depending on morphologies, and then make a weighted average.

I'm not an expert in polling, but this seems to apply well here.

2015-01-03 8:36 GMT+01:00 Eric Walker eric.wal...@gmail.com:

 On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

 In reply to  Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700:
 Hi,
 [snip]

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.


 The authors of the Lugano report mention a total energy balance of 1.5 MWh
 excess heat to be accounted for (p. 29).  Translating that value, we
 get 3.3e22 MeV.  If the average reaction is 3.5 MeV (just to choose an
 optimistic number), that means there were 9.4e21 reactions, and presumably
 9.4e21 7Li atoms to be consumed in the process.

 The authors mention that in the sample of the fuel they looked at, there
 was 1.17 percent lithium (p. 53).  If we extrapolate out from the 2 mg
 sample they obtained to the 1 g of fuel from which it was taken (not
 necessarily wise), there would have been 0.0117 g * 1 mole / 6.94 g *
 6.022e23 / mole = 1.0e21 atoms lithium in the total charge.  If we assume
 that that was 100 percent 7Li to be optimistic, that would mean there were
 about 1/10th the number of 7Li atoms needed to account for the 1.5 MWh that
 were produced.

 Judging from the fact that these calculations go back to the isotope
 ratios found in a single 2 mg sample of fuel, there's a lot of room for
 uncertainty.  But in this instance we've been optimistic about the average
 energy per reaction (3.5 MeV), about there being 100 percent lithium, and
 about all of the 7Li being consumed.  The actual heat balance is another
 variable that can be adjusted to within one's sense of uncertainty.  But it
 would have to be pretty far off for the reaction to consist entirely of 7Li
 neutron stripping reactions.

 Have I missed something important?

 Eric

Re: [Vo]:A bombshell of a different type?

2015-01-02 Thread Eric Walker

On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

In reply to  Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700:
 Hi,
 [snip]

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.


The authors of the Lugano report mention a total energy balance of 1.5 MWh
excess heat to be accounted for (p. 29).  Translating that value, we
get 3.3e22 MeV.  If the average reaction is 3.5 MeV (just to choose an
optimistic number), that means there were 9.4e21 reactions, and presumably
9.4e21 7Li atoms to be consumed in the process.

The authors mention that in the sample of the fuel they looked at, there
was 1.17 percent lithium (p. 53).  If we extrapolate out from the 2 mg
sample they obtained to the 1 g of fuel from which it was taken (not
necessarily wise), there would have been 0.0117 g * 1 mole / 6.94 g *
6.022e23 / mole = 1.0e21 atoms lithium in the total charge.  If we assume
that that was 100 percent 7Li to be optimistic, that would mean there were
about 1/10th the number of 7Li atoms needed to account for the 1.5 MWh that
were produced.

Judging from the fact that these calculations go back to the isotope ratios
found in a single 2 mg sample of fuel, there's a lot of room for
uncertainty.  But in this instance we've been optimistic about the average
energy per reaction (3.5 MeV), about there being 100 percent lithium, and
about all of the 7Li being consumed.  The actual heat balance is another
variable that can be adjusted to within one's sense of uncertainty.  But it
would have to be pretty far off for the reaction to consist entirely of 7Li
neutron stripping reactions.

Have I missed something important?

Eric

Re: [Vo]:A bombshell of a different type?

On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.


This is very nice.  I've been too attached to deuterium.  In this
particular instance, deuterium reactions above 62Ni would be exothermic:

   - 62Ni + n → 63Ni + p + Q (5.1 MeV)
   - 63Ni + n → 64Ni + p + Q (7.9 MeV)

Since neither 62Ni nor 63Ni were seen in significant quantities in the ash,
I think we can rule deuterium out for this particular test.  Note that
while 64Ni(7Li,6Li)65Ni is also endothermic, 63Ni(7Li,6Li)64Ni is
exothermic.  Since 63Ni is not found in nature, however, and since it won't
be coming from the 62Ni(7Li,6Li)63Ni reaction, none will arise unless there
is deuterium in the mix.  It all feels a little precarious, because if you
get any 64Ni, you can get penetrating radiation from deexcitation gammas
from inelastic collisions.

To add to your thought about the kinetic energy of the daughter 6Li being
relatively low, for the maximum Q value in your list above, there would be
4.14 MeV / 6 nucleons = 690 keV per nucleon, which seems manageable.  I
will nominate you for the Vortex Nobel Prize for your insight about neutron
stripping from lithium.

Two questions I have:

   - Why use hydrogen at all if the reaction can be sustained with lithium?
   - What is the amount of force that would be needed to bring a 7Li to
   within a sufficient distance of a nickel nucleus for stripping to occur?
   It seems like it would be high.

Eric

Re: [Vo]:A bombshell of a different type?

2014-10-10 Thread Axil Axil

The isotopic shift observed is only a side effect of the real reaction that
are taking place. From others LENR experiments one can suspect that
hydrogen is the fuel and that Ni is just modified by whatever is in its
vicinity.

Do you remember all the internet ink was used to debate the copper ash in
the nickel powder; now all that is for naught.

The transmutation pattern is based on the geometry of the reactor. As that
geometry changes so does the transmutation patterns.

The analysis of transmutation was incomplete and much of the many reactions
were missed.

For example from page 53...

Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a very
fine powder. Besides the analyzed elements it has been found that the fuel
also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and these
are not found in the ash.

And there was transmutation of aluminum.



On Fri, Oct 10, 2014 at 2:06 AM, Eric Walker eric.wal...@gmail.com wrote:

 On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.


 This is very nice.  I've been too attached to deuterium.  In this
 particular instance, deuterium reactions above 62Ni would be exothermic:

- 62Ni + n → 63Ni + p + Q (5.1 MeV)
- 63Ni + n → 64Ni + p + Q (7.9 MeV)

 Since neither 62Ni nor 63Ni were seen in significant quantities in the
 ash, I think we can rule deuterium out for this particular test.  Note that
 while 64Ni(7Li,6Li)65Ni is also endothermic, 63Ni(7Li,6Li)64Ni is
 exothermic.  Since 63Ni is not found in nature, however, and since it won't
 be coming from the 62Ni(7Li,6Li)63Ni reaction, none will arise unless there
 is deuterium in the mix.  It all feels a little precarious, because if you
 get any 64Ni, you can get penetrating radiation from deexcitation gammas
 from inelastic collisions.

 To add to your thought about the kinetic energy of the daughter 6Li being
 relatively low, for the maximum Q value in your list above, there would be
 4.14 MeV / 6 nucleons = 690 keV per nucleon, which seems manageable.  I
 will nominate you for the Vortex Nobel Prize for your insight about neutron
 stripping from lithium.

 Two questions I have:

- Why use hydrogen at all if the reaction can be sustained with
lithium?
- What is the amount of force that would be needed to bring a 7Li to
within a sufficient distance of a nickel nucleus for stripping to occur?
It seems like it would be high.

 Eric

Re: [Vo]:A bombshell of a different type?

On Thu, Oct 9, 2014 at 11:33 PM, Axil Axil janap...@gmail.com wrote:

Do you remember all the internet ink was used to debate the copper ash in
 the nickel powder; now all that is for naught.


Yes, and thankfully so.  Ni(p,ɣ)Cu can go away and die a peaceful death.
 (Not to say that it doesn't happen in small quantities.)

Like you, I wonder about other reactions, though.

Eric

Re: [Vo]:A bombshell of a different type?

2014-10-10 Thread Teslaalset

Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a
very fine powder. Besides the analyzed elements it has been found that the
fuel also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and
these are not found in the ash.

This indicates that also virgin powder was analyzed. This was not
explicitly mentioned in TIP2, was it?


On Fri, Oct 10, 2014 at 8:33 AM, Axil Axil janap...@gmail.com wrote:

 The isotopic shift observed is only a side effect of the real reaction
 that are taking place. From others LENR experiments one can suspect that
 hydrogen is the fuel and that Ni is just modified by whatever is in its
 vicinity.

 Do you remember all the internet ink was used to debate the copper ash in
 the nickel powder; now all that is for naught.

 The transmutation pattern is based on the geometry of the reactor. As that
 geometry changes so does the transmutation patterns.

 The analysis of transmutation was incomplete and much of the many
 reactions were missed.

 For example from page 53...

 Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a very
 fine powder. Besides the analyzed elements it has been found that the fuel
 also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and these
 are not found in the ash.

 And there was transmutation of aluminum.



 On Fri, Oct 10, 2014 at 2:06 AM, Eric Walker eric.wal...@gmail.com
 wrote:

 On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.


 This is very nice.  I've been too attached to deuterium.  In this
 particular instance, deuterium reactions above 62Ni would be exothermic:

- 62Ni + n → 63Ni + p + Q (5.1 MeV)
- 63Ni + n → 64Ni + p + Q (7.9 MeV)

 Since neither 62Ni nor 63Ni were seen in significant quantities in the
 ash, I think we can rule deuterium out for this particular test.  Note that
 while 64Ni(7Li,6Li)65Ni is also endothermic, 63Ni(7Li,6Li)64Ni is
 exothermic.  Since 63Ni is not found in nature, however, and since it won't
 be coming from the 62Ni(7Li,6Li)63Ni reaction, none will arise unless there
 is deuterium in the mix.  It all feels a little precarious, because if you
 get any 64Ni, you can get penetrating radiation from deexcitation gammas
 from inelastic collisions.

 To add to your thought about the kinetic energy of the daughter 6Li being
 relatively low, for the maximum Q value in your list above, there would be
 4.14 MeV / 6 nucleons = 690 keV per nucleon, which seems manageable.  I
 will nominate you for the Vortex Nobel Prize for your insight about neutron
 stripping from lithium.

 Two questions I have:

- Why use hydrogen at all if the reaction can be sustained with
lithium?
- What is the amount of force that would be needed to bring a 7Li to
within a sufficient distance of a nickel nucleus for stripping to occur?
It seems like it would be high.

 Eric

Re: [Vo]:A bombshell of a different type?

2014-10-10 Thread frobertcook

Eric--

It may not be so high in the local mag. field if the Li7 is lined up properly 
with the Ni nucleus.  

The question remains, what is the mechanism that transfers mass energy to 
thermal energy at Mev levels without gammas?

The paper on neutron hopping that Jones identified may help answer this 
question .

I still think  spin coupling is involved because of the small differential 
energy states associated with spin quanta.

Bob Cook


Sent from my Verizon Wireless 4G LTE SmartphoneTeslaalset 
robbiehobbiesh...@gmail.com wrote:Sample 2 was the fuel used to charge the 
E-Cat. It’s in the form of a very fine powder. Besides the analyzed elements it 
has been found that the fuel also contains rather high concentrations of C, Ca, 
Cl, Fe, Mg, Mn and these are not found in the ash.

This indicates that also virgin powder was analyzed. This was not explicitly 
mentioned in TIP2, was it?


On Fri, Oct 10, 2014 at 8:33 AM, Axil Axil janap...@gmail.com wrote:
The isotopic shift observed is only a side effect of the real reaction that are 
taking place. From others LENR experiments one can suspect that hydrogen is the 
fuel and that Ni is just modified by whatever is in its vicinity.

Do you remember all the internet ink was used to debate the copper ash in the 
nickel powder; now all that is for naught.

The transmutation pattern is based on the geometry of the reactor. As that 
geometry changes so does the transmutation patterns.

The analysis of transmutation was incomplete and much of the many reactions 
were missed.

For example from page 53...

Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a very fine 
powder. Besides the analyzed elements it has been found that the fuel also 
contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and these are not 
found in the ash.

And there was transmutation of aluminum.




On Fri, Oct 10, 2014 at 2:06 AM, Eric Walker eric.wal...@gmail.com wrote:
On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

This series stops at Ni62, hence all isotopes of Ni less than 62 are depleted
and Ni62 is strongly enriched.

This is very nice.  I've been too attached to deuterium.  In this particular 
instance, deuterium reactions above 62Ni would be exothermic:
62Ni + n → 63Ni + p + Q (5.1 MeV)
63Ni + n → 64Ni + p + Q (7.9 MeV)
Since neither 62Ni nor 63Ni were seen in significant quantities in the ash, I 
think we can rule deuterium out for this particular test.  Note that while 
64Ni(7Li,6Li)65Ni is also endothermic, 63Ni(7Li,6Li)64Ni is exothermic.  Since 
63Ni is not found in nature, however, and since it won't be coming from the 
62Ni(7Li,6Li)63Ni reaction, none will arise unless there is deuterium in the 
mix.  It all feels a little precarious, because if you get any 64Ni, you can 
get penetrating radiation from deexcitation gammas from inelastic collisions.

To add to your thought about the kinetic energy of the daughter 6Li being 
relatively low, for the maximum Q value in your list above, there would be 4.14 
MeV / 6 nucleons = 690 keV per nucleon, which seems manageable.  I will 
nominate you for the Vortex Nobel Prize for your insight about neutron 
stripping from lithium.

Two questions I have:
Why use hydrogen at all if the reaction can be sustained with lithium?
What is the amount of force that would be needed to bring a 7Li to within a 
sufficient distance of a nickel nucleus for stripping to occur?  It seems like 
it would be high.
Eric

Re: [Vo]:A bombshell of a different type?

On Fri, Oct 10, 2014 at 12:58 AM, frobertcook frobertc...@hotmail.com
wrote:

The question remains, what is the mechanism that transfers mass energy to
 thermal energy at Mev levels without gammas?


As you allude, gammas were not seen in this test run.  That leads me to
adopt Robin's hypothesis, that it was lithium that yielded the heat this
time around.  When a neutron is stripped from 7Li and added to a nickel
lattice site, I believe the highest Q value of the various possible
reactions is ~ 4 MeV.  When divided among the number of nucleons in the
daughter 6Li, that comes out to about 0.6 MeV per nucleon in kinetic energy
of the daughter.  This number is not tiny, but I think you would not
necessarily see bremsstrahlung in the way that you would for a 10 MeV
proton.

The mechanism itself I'm guessing is acceleration via arcing between
electrically insulated grains, brought about through the current that
drives the resistance heating somehow.

I'm also guessing that Rossi and Industrial Heat are experimenting with
deuterium-based reactions in which fast protons are generated, which they
did not make use of for this demo.  Those reactions are likely to produce
ionizing radiation.  (This was a possibility that Robin raised in an
earlier thread.)

Eric

Re: [Vo]:A bombshell of a different type?

2014-10-10 Thread Axil Axil

On Fri, Oct 10, 2014 at 3:58 AM, frobertcook frobertc...@hotmail.com
wrote:

 Eric--


 The question remains, what is the mechanism that transfers mass energy to
 thermal energy at Mev levels without gammas?


Superabsorbsion of a polariton (SPP) boson condensate. SPP last about 50
picoseconds, When they decay, they release the XUV photon that then decays
via multiple cascading photon absorptions by multiple atoms.


 The paper on neutron hopping that Jones identified may help answer this
 question .


neutrons are not involved.


 I still think  spin coupling is involved because of the small differential
 energy states associated with spin quanta.


At 1400C all atoms are either ionized or in dipole vibrations.  There is an
electron plasma formed from which polaritons are then formed from electron
shielded infrared photons,




 Sent from my Verizon Wireless 4G LTE Smartphone

 Teslaalset robbiehobbiesh...@gmail.com wrote:
 Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a
 very fine powder. Besides the analyzed elements it has been found that the
 fuel also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and
 these are not found in the ash.

 This indicates that also virgin powder was analyzed. This was not
 explicitly mentioned in TIP2, was it?


 On Fri, Oct 10, 2014 at 8:33 AM, Axil Axil janap...@gmail.com wrote:

 The isotopic shift observed is only a side effect of the real reaction
 that are taking place. From others LENR experiments one can suspect that
 hydrogen is the fuel and that Ni is just modified by whatever is in its
 vicinity.

 Do you remember all the internet ink was used to debate the copper ash in
 the nickel powder; now all that is for naught.

 The transmutation pattern is based on the geometry of the reactor. As
 that geometry changes so does the transmutation patterns.

 The analysis of transmutation was incomplete and much of the many
 reactions were missed.

 For example from page 53...

 Sample 2 was the fuel used to charge the E-Cat. It’s in the form of a
 very fine powder. Besides the analyzed elements it has been found that the
 fuel also contains rather high concentrations of C, Ca, Cl, Fe, Mg, Mn and
 these are not found in the ash.

 And there was transmutation of aluminum.



 On Fri, Oct 10, 2014 at 2:06 AM, Eric Walker eric.wal...@gmail.com
 wrote:

 On Wed, Oct 8, 2014 at 7:21 PM, mix...@bigpond.com wrote:

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.


 This is very nice.  I've been too attached to deuterium.  In this
 particular instance, deuterium reactions above 62Ni would be exothermic:

- 62Ni + n → 63Ni + p + Q (5.1 MeV)
- 63Ni + n → 64Ni + p + Q (7.9 MeV)

 Since neither 62Ni nor 63Ni were seen in significant quantities in the
 ash, I think we can rule deuterium out for this particular test.  Note that
 while 64Ni(7Li,6Li)65Ni is also endothermic, 63Ni(7Li,6Li)64Ni is
 exothermic.  Since 63Ni is not found in nature, however, and since it won't
 be coming from the 62Ni(7Li,6Li)63Ni reaction, none will arise unless there
 is deuterium in the mix.  It all feels a little precarious, because if you
 get any 64Ni, you can get penetrating radiation from deexcitation gammas
 from inelastic collisions.

 To add to your thought about the kinetic energy of the daughter 6Li
 being relatively low, for the maximum Q value in your list above, there
 would be 4.14 MeV / 6 nucleons = 690 keV per nucleon, which seems
 manageable.  I will nominate you for the Vortex Nobel Prize for your
 insight about neutron stripping from lithium.

 Two questions I have:

- Why use hydrogen at all if the reaction can be sustained with
lithium?
- What is the amount of force that would be needed to bring a 7Li to
within a sufficient distance of a nickel nucleus for stripping to occur?
It seems like it would be high.

 Eric

Re: [Vo]:A bombshell of a different type?

On Fri, Oct 10, 2014 at 7:22 PM, Axil Axil janap...@gmail.com wrote:

At 1400C all atoms are either ionized or in dipole vibrations.  There is an
 electron plasma formed from which polaritons are then formed from electron
 shielded infrared photons,


Sometimes I wonder whether there's a correlation between the quality and
quantity of your teachings and the interesting and juicy tidbits that
others are sharing over this list.

Eric

Re: [Vo]:A bombshell of a different type?

2014-10-09 Thread frobertcook

I for one consider a hot  Li6 is inconsistent with no radiation.

The enerrgy  release must  be by a different mechanism.

Bob Cook


Sent from my Verizon Wireless 4G LTE SmartphoneAxil Axil janap...@gmail.com 
wrote:
I agree, you really can tell where that Li6 came from.

On Wed, Oct 8, 2014 at 10:21 PM, mix...@bigpond.com wrote:

 In reply to  Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700:
 Hi,
 [snip]

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.

 I have only briefly skimmed the report, but the basic reaction appears to
 be a
 neutron transfer reaction where a neutron tunnels from Li7 to a Nickel
 isotope.
 The excess energy of the reaction appears as kinetic energy of the two
 resultant
 nuclei (i.e. Li6  the new Ni isotope), rather than as gamma rays. Because
 there
 are two daughter nuclei, momentum can be conserved while dumping the
 energy as
 kinetic energy in a reaction that is much faster then gamma ray emission.
 Because both nuclei are heavy and slow moving, very little to no
 bremsstrahlung is produced. There is effectively no secondary gamma from
 Li6
 because the first excited state is too high. (I haven't checked Li7).
 There is
 unlikely to be anything significant from Ni because the high charge on the
 nucleus combined with the 3 from Lithium tend to keep them apart (minimum
 distance 31 fm).

 It would be nice to know if the total amounts of each of Li  Ni in the
 sample
 were conserved (I'll have to study the report more closely).
 Regards,

 Robin van Spaandonk

 http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

2014-10-08 Thread Craig Haynie


On 10/08/2014 12:22 PM, Jones Beene wrote:

You have to wonder - given the tiny amount of hydrogen at the start, and the
isotopic analysis at the end, if hydrogen was necessary for this reaction.
This looks like a lithium burner.



Captain, we're going to need more dilithium crystals!

Craig

RE: [Vo]:A bombshell of a different type?

-Original Message-
From: Craig Haynie 

 You have to wonder - given the tiny amount of hydrogen at the start, and the
 isotopic analysis at the end, if hydrogen was necessary for this reaction.
 This looks like a lithium burner.

Captain, we're going to need more dilithium crystals!


Truth is often stranger than fiction...

...or are we talking about life imitating art ?

Re: [Vo]:A bombshell of a different type?

I have postulated for years that a alkali metal would be the secret sauce
based on the operating temperature of the reactor.

When the operating temperature is about 1200C, this makes lithium the best
fit to vaporize at about 1330C and at lower temperatures condense into
nano-particles in areas of the reactor that are below the lithium boiling
point.

Nano-particles of Lithium and lithium hydride form the dynamic nuclei
active environments that are central to the Rossi reactor/

Rossi had to move his secret sauce from cesium and potassium to lithium as
the operational temperature of his reactor increases to 1200C and above.

On Wed, Oct 8, 2014 at 12:22 PM, Jones Beene jone...@pacbell.net wrote:


 In one way, this report is shaping up as an amazing piece of oversight -
 which Levi and the Swedes may have failed to grasp, or at least failed to
 fully appreciated in its ultimate significance. There could be a shadow
 over
 this story which goes back to 1989.

 Moreover, do we even need hydrogen at all?

 You have to wonder - given the tiny amount of hydrogen at the start, and
 the
 isotopic analysis at the end, if hydrogen was necessary for this reaction.
 This looks like a lithium burner.

 Perhaps it is basically a new kind of lithium reaction… or maybe it is not
 so new.

 As mentioned in many prior posts here, Nickel-58 is extremely neutron
 deficient. Nickel 58 is the most abundant isotope of element 28, but is
 out-of-place in the periodic table, being lighter in amu than any stable
 cobalt isotope, the element to the left of nickel having one less proton;
 and it should be heavier (essentially all cobalt is Co-59). By itself, that
 factoid would be somewhat unique - in that it only happens in two other
 places in the entire periodic table, where elements routinely increase in
 average amu, in step with Z.

 So, we have Ni-58 which is is strongly neutron deficient, in the vicinity
 of
 gaseous Li7 which has an anomalous excess – even if the excess is a single
 weakly bound neutron, such that the nickel is acting in some ways like a
 “neutron sink” for a low energy transfer from Li-7.

 If hydrogen is necessary at all, its role could be limited to that of a
 transfer mechanism to facilitate the movement of the excess neutron from
 lithium to nickel.

 Unfortunately, the strong overtone here could relate to non-proliferation
 issues which reverberate back to 1989. After all, if helium is seen in any
 kind of lithium reaction, when nickel is not present – it could derive from
 Li7. At that time in history, PF using lithium, plus that other dreaded
 ingredient (heavy water) may have worried strategists who knew a few things
 about lithium which are still not in the public domain.

 This is probably not going to be the instant bombshell, or
 extremely well-prepared announcement from truly independent scientists that
 we had hoped for.

 Agreed. I don't think any of us should be pinning all our
 hopes on this overturning establishment beliefs, but I think it's a rather
 large/important piece of the puzzle, no?

RE: [Vo]:A bombshell of a different type?

One more interesting thing about this report. If lithium is the active fuel, 
and not hydrogen, which seems to be the case, then the ash which is lithium-6 
is as valuable for batteries as is the natural metal. Maybe more valuable.

Thus the fuel is essentially free, since the ash can be sold for cost (or even 
marked-up).

Not that cost is as big an issue for IH as it is for Tesla, but it could also 
be that Li-6 is safer. Safer is worth paying for.

Remember the Boeing lithium fires on their new plane?

They were NEVER able to locate the real problem. Perhaps it was lithium-7 and 
the problem will go away once they get an adequate supply of Li-6 to make 
batteries only from the lighter isotope ... which of course will be the ash of 
the Rossi/IH reactor and the others which follow.

I could see the necessity of Safety Laws being necessary, sometime in the 
future to demand that only Li-6 be used in batteries.

Why do I think Elon Musk is already onto this ?

Jones

Re: [Vo]:A bombshell of a different type?

What is missing from LENR theory is how any nuclear radiation types are
not detected in the Rossi reaction. I have put forward the Super-absorber
theory made possible by boson condensation. The testing and analysis group
are afraid to put their names onto a nuclear reaction mechanism that is
nuclear radiation free.

Yes, these testers are hopelessly conflicted. They state that the reaction
must be nuclear because of the large amounts of energy produced, but say
that nuclear reactions cannot occur without the detection of nuclear
radiation. They cannot believe the evidence of their own eyes. They are
afraid to stake their reputations on proclaiming the obvious conclusions of
this test



On Wed, Oct 8, 2014 at 2:08 PM, Jones Beene jone...@pacbell.net wrote:

 One more interesting thing about this report. If lithium is the active
 fuel, and not hydrogen, which seems to be the case, then the ash which is
 lithium-6 is as valuable for batteries as is the natural metal. Maybe more
 valuable.

 Thus the fuel is essentially free, since the ash can be sold for cost (or
 even marked-up).

 Not that cost is as big an issue for IH as it is for Tesla, but it could
 also be that Li-6 is safer. Safer is worth paying for.

 Remember the Boeing lithium fires on their new plane?

 They were NEVER able to locate the real problem. Perhaps it was lithium-7
 and the problem will go away once they get an adequate supply of Li-6 to
 make batteries only from the lighter isotope ... which of course will be
 the ash of the Rossi/IH reactor and the others which follow.

 I could see the necessity of Safety Laws being necessary, sometime in the
 future to demand that only Li-6 be used in batteries.

 Why do I think Elon Musk is already onto this ?

 Jones

Re: [Vo]:A bombshell of a different type?

I also see that the tubules on the a  fraction of the micro-particles are
melted by the high heat. This leads be to the conclusion that reaction is
also being carried by lithium based nano-particles produced by plasma
condensation. The testers are only looking at the nickel particles for
isotope shifted reaction products but this method of analysis will not
detect reaction products made by the lithium based nano-particles. They
will only find what they expect to find not what is actually happening.

On Wed, Oct 8, 2014 at 2:35 PM, Axil Axil janap...@gmail.com wrote:

 What is missing from LENR theory is how any nuclear radiation types are
 not detected in the Rossi reaction. I have put forward the Super-absorber
 theory made possible by boson condensation. The testing and analysis group
 are afraid to put their names onto a nuclear reaction mechanism that is
 nuclear radiation free.

 Yes, these testers are hopelessly conflicted. They state that the reaction
 must be nuclear because of the large amounts of energy produced, but say
 that nuclear reactions cannot occur without the detection of nuclear
 radiation. They cannot believe the evidence of their own eyes. They are
 afraid to stake their reputations on proclaiming the obvious conclusions of
 this test



 On Wed, Oct 8, 2014 at 2:08 PM, Jones Beene jone...@pacbell.net wrote:

 One more interesting thing about this report. If lithium is the active
 fuel, and not hydrogen, which seems to be the case, then the ash which is
 lithium-6 is as valuable for batteries as is the natural metal. Maybe more
 valuable.

 Thus the fuel is essentially free, since the ash can be sold for cost (or
 even marked-up).

 Not that cost is as big an issue for IH as it is for Tesla, but it could
 also be that Li-6 is safer. Safer is worth paying for.

 Remember the Boeing lithium fires on their new plane?

 They were NEVER able to locate the real problem. Perhaps it was lithium-7
 and the problem will go away once they get an adequate supply of Li-6 to
 make batteries only from the lighter isotope ... which of course will be
 the ash of the Rossi/IH reactor and the others which follow.

 I could see the necessity of Safety Laws being necessary, sometime in the
 future to demand that only Li-6 be used in batteries.

 Why do I think Elon Musk is already onto this ?

 Jones

RE: [Vo]:A bombshell of a different type?

On the subject of energy balance and radiation

Mass of Ni58 =57.935 mass of Ni62 =61.928 difference 3.993 amu

Mass of neutron 1.0087 x 4 = 4.035 

Mass of Li7=7.016 Mass of Li6= 6.015 difference 1.001 x 4= 4.004

 

The problem is that on paper - lithium cannot give up a neutron easily as there 
is a mass deficit going to neutrons - and at the same time, nickel cannot add 4 
neutrons without shedding a large amount of energy. 

 

Curiously, if one looks at it from the perspective of 4 atoms of the initial 
lithium, 4.004 amu has been lost and only 3.993 gained, so the reaction is 
ostensibly lossy; but from the perspective of the nickel 4 neutrons have been 
added, and the gain is massive – in the range of 42 MeV net.

 

 

One more interesting thing about this report. If lithium is the active fuel, 
and not hydrogen, which seems to be the case, then the ash which is lithium-6 
is as valuable for batteries as is the natural metal. Maybe more valuable.

Thus the fuel is essentially free, since the ash can be sold for cost (or even 
marked-up).

Not that cost is as big an issue for IH as it is for Tesla, but it could also 
be that Li-6 is safer. Safer is worth paying for.

Remember the Boeing lithium fires on their new plane?

They were NEVER able to locate the real problem. Perhaps it was lithium-7 and 
the problem will go away once they get an adequate supply of Li-6 to make 
batteries only from the lighter isotope ... which of course will be the ash of 
the Rossi/IH reactor and the others which follow.

I could see the necessity of Safety Laws being necessary, sometime in the 
future to demand that only Li-6 be used in batteries.

Why do I think Elon Musk is already onto this ?

Jones

RE: [Vo]:A bombshell of a different type?

It may have been noted that lithium has a heat of vaporization of about 140 
kJ/mol and that the boiling point is close to the operating temp of this 
reactor on the inside 1350C.

 

This seems to work out to a whopping 20 MJ/kg which is high (did I get that 
wrong?) Consequently a lot thermal energy could be balanced, cycling around 
this phase change point – which is why they must keep the reactor hot. Phase 
change is one of the usual suspects in thermal anomalies.

 

If one were to be looking for an alternative energy source which was 
gamma-free, but might cause mass-to-energy conversions as a side effect then 
any asymmetry here would do it. 

 

I say that without much risk - since in looking at 5 authoritative source for 
lithium properties, all of them had different heat of evaporation values - 
indicating that no one has a clue !

 

 

On the subject of energy balance and radiation

Mass of Ni58 =57.935 mass of Ni62 =61.928 difference 3.993 amu

Mass of neutron 1.0087 x 4 = 4.035 

Mass of Li7=7.016 Mass of Li6= 6.015 difference 1.001 x 4= 4.004

 

The problem is that on paper - lithium cannot give up a neutron easily as there 
is a mass deficit going to neutrons - and at the same time, nickel cannot add 4 
neutrons without shedding a large amount of energy. 

 

Curiously, if one looks at it from the perspective of 4 atoms of the initial 
lithium, 4.004 amu has been lost and only 3.993 gained, so the reaction is 
ostensibly lossy; but from the perspective of the nickel 4 neutrons have been 
added, and the gain is massive – in the range of 42 MeV net.

 

 

One more interesting thing about this report. If lithium is the active fuel, 
and not hydrogen, which seems to be the case, then the ash which is lithium-6 
is as valuable for batteries as is the natural metal. Maybe more valuable.

Thus the fuel is essentially free, since the ash can be sold for cost (or even 
marked-up).

Not that cost is as big an issue for IH as it is for Tesla, but it could also 
be that Li-6 is safer. Safer is worth paying for.

Remember the Boeing lithium fires on their new plane?

They were NEVER able to locate the real problem. Perhaps it was lithium-7 and 
the problem will go away once they get an adequate supply of Li-6 to make 
batteries only from the lighter isotope ... which of course will be the ash of 
the Rossi/IH reactor and the others which follow.

I could see the necessity of Safety Laws being necessary, sometime in the 
future to demand that only Li-6 be used in batteries.

Why do I think Elon Musk is already onto this ?

Jones

Re: [Vo]:A bombshell of a different type?

2014-10-08 Thread mixent

In reply to  Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700:
Hi,
[snip]

Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

This series stops at Ni62, hence all isotopes of Ni less than 62 are depleted
and Ni62 is strongly enriched.

I have only briefly skimmed the report, but the basic reaction appears to be a
neutron transfer reaction where a neutron tunnels from Li7 to a Nickel isotope.
The excess energy of the reaction appears as kinetic energy of the two resultant
nuclei (i.e. Li6  the new Ni isotope), rather than as gamma rays. Because there
are two daughter nuclei, momentum can be conserved while dumping the energy as
kinetic energy in a reaction that is much faster then gamma ray emission.
Because both nuclei are heavy and slow moving, very little to no
bremsstrahlung is produced. There is effectively no secondary gamma from Li6
because the first excited state is too high. (I haven't checked Li7). There is
unlikely to be anything significant from Ni because the high charge on the
nucleus combined with the 3 from Lithium tend to keep them apart (minimum
distance 31 fm).

It would be nice to know if the total amounts of each of Li  Ni in the sample
were conserved (I'll have to study the report more closely).
Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?

I agree, you really can tell where that Li6 came from.

On Wed, Oct 8, 2014 at 10:21 PM, mix...@bigpond.com wrote:

 In reply to  Jones Beene's message of Wed, 8 Oct 2014 09:22:13 -0700:
 Hi,
 [snip]

 Li7 + Ni58 = Ni59 + Li6 + 1.75 MeV
 Li7 + Ni59 = Ni60 + Li6 + 4.14 MeV
 Li7 + Ni60 = Ni61 + Li6 + 0.57 MeV
 Li7 + Ni61 = Ni62 + Li6 + 3.34 MeV
 Li7 + Ni62 = Ni63 + Li6 - 0.41 MeV (Endothermic!)

 This series stops at Ni62, hence all isotopes of Ni less than 62 are
 depleted
 and Ni62 is strongly enriched.

 I have only briefly skimmed the report, but the basic reaction appears to
 be a
 neutron transfer reaction where a neutron tunnels from Li7 to a Nickel
 isotope.
 The excess energy of the reaction appears as kinetic energy of the two
 resultant
 nuclei (i.e. Li6  the new Ni isotope), rather than as gamma rays. Because
 there
 are two daughter nuclei, momentum can be conserved while dumping the
 energy as
 kinetic energy in a reaction that is much faster then gamma ray emission.
 Because both nuclei are heavy and slow moving, very little to no
 bremsstrahlung is produced. There is effectively no secondary gamma from
 Li6
 because the first excited state is too high. (I haven't checked Li7).
 There is
 unlikely to be anything significant from Ni because the high charge on the
 nucleus combined with the 3 from Lithium tend to keep them apart (minimum
 distance 31 fm).

 It would be nice to know if the total amounts of each of Li  Ni in the
 sample
 were conserved (I'll have to study the report more closely).
 Regards,

 Robin van Spaandonk

 http://rvanspaa.freehostia.com/project.html

Re: [Vo]:A bombshell of a different type?