I don't think the tungsten is vaporizing - at least there's no evidence
that supports that contention. In the example video, the surface of the
rod appears to be melted, so the flame must be at least 3500 C, higher
than produced by H2+O combustion. There's enough velocity in the flame
to blow the melted surface layer off the rod in small droplets, which
might oxidise when they leave the reducing part of the flame. This
process progressively continues to thin the rod until it breaks.
So the real question raised here is how monatomic hydrogen is created in
the system. It's called "HHO" but the simple electrolysis process as
used in typical "HHO" generators would yield 2H2O -> 2H2 + O2. So how
does it instead give 2H2O -> 4H1 +O2 ?
It doesn't seem likely a H2+O flame by itself could disassociate
unburned H2, because H2 splitting is highly endothermic. In the Langmuir
process the needed energy comes from the electric arc. I suppose there
could be plasma fusion somewhere in the flame. It would be nice to check
one of these torches for gammas.
AlanG
On 3/18/2014 8:39 AM, Axil Axil wrote:
In HHO welding, there is no electric current employed. HHO welding is
just the burning of hydrogen in oxygen.
But how does a hydrogen combustion process that produces only 2,660 °C
in heat vaporize tungsten at (5930 °C, 10706 °F).
This does not add up unless there is LENR involved.
On Tue, Mar 18, 2014 at 11:21 AM, Roarty, Francis X
<francis.x.roa...@lmco.com <mailto:francis.x.roa...@lmco.com>> wrote:
Axil, Langmuir was aware of this anomaly and advised not to pursue
it when he developed atomic welding with tungsten electrodes..
some will insist it is the energy of recombination but if so then
welding would not be a constant flow and one would have to
continually stop, build up a reservoir of atomic hydrogen [which
opposes retaining that state] and then weld a little bit to
exhaust the recombination energy in a very short burst to exploit
the stored energy enough to melt tungsten. Since atomic welding is
a smooth process and the electrical energy employed by the arc is
not to my knowledge significant enough to account for the melting
capability then yes.. your point is well taken.
Fran
*From:*Axil Axil [mailto:janap...@gmail.com
<mailto:janap...@gmail.com>]
*Sent:* Monday, March 17, 2014 11:11 PM
*To:* vortex-l
*Subject:* EXTERNAL: [Vo]:HHO welding is LENR
Why is a HHO flame able to vaporize tungsten and yet will not burn
the skin of your hand.
http://www.youtube.com/watch?v=Ax4sW3bo_dM
The HHO gas stream contains solid crystals of water. These
crystals act like nano lenses that concentrate infrared light in
the boundary layer between a shiny metal surface and a dielectric
gas like hydrogen or oxygen. The science that studies this effect
is called nanoplasmonics.
The heat energy is confined to the metal surface and locked in(AKA
dark mode) and concentrated their like in a EMF black hole.
The metal surface is said to have a negative coefficient of
reflectivity. This keeps the heat from leaving the metal surface.
In this way the heat energy builds up to huge temperatures to the
point where it will vaporize tungsten.
The skin on your hand has a positive index of reflectivity; it is
not shiny. The heat from hydrogen combustion is not confined to
the surface of your skin and can escape to the surrounding air. So
you will not be readily burned by the HHO flame.
This is a basic LENR effect (aka evanescent wave -
http://en.wikipedia.org/wiki/Evanescent_wave) of energy
concentration and focusing. This indicates that the upper
temperature limit of the LENR effect is beyond the temperature
required to vaporize tungsten (5930 °C, 10706 °F)
On the other hand, the combustion temperature of hydrogen is only
2,660 °C with oxygen. Do I need to spell this out any further?
http://www.youtube.com/watch?v=-ceOL83PM24
On the downside, spark ignition of HHO does not use the LENR
effect of the evanescent wave.
So burning hydrogen in oxygen is only combustion and not LENR.