At 09:10 AM 5/10/2010, Michel Jullian wrote:
No Abd, Shanahan may be wrong on many points but the "equivalent to
many atmospheres of hydrogen gas pressure exposure" assertion is
correct, it is even a gross understatement, in the P&F original paper
they computed something like 10^26 atm IIRC. That's electrolytic
compression: if you use a hollow Pd cathode, the pressure inside will
rise to tens of thousands of atmospheres, until the palladium envelope
bursts. It has been done.
Fascinating. So a very low voltage is equivalent
to a very high hydrogen pressure. The real issue
is how this information is being used. Shanahan
uses the high pressure to imply that this kind of
pressure is behind the outgassing, whereas the
affinity of palladium for deuterium inhibits the
outgassing. The cigarett lighter effect is
important, for sure. If oxidation, or anything,
at the surface of the cathode heats the cathode,
it will increase outgassing (once the electrolytic pressure is removed).
Palladium deuteride is a truly remarkable
material. It is, as it were, an alloy of
palladium and hydrogen *metal.* That is, the
density of hydrogen in it is close to that of
hydrogen as a metal, which nornally requires
tremendous pressure, partiuclarly at room temperature.
But a piece of hydrogen metal of the size of the
cathode is still just that much material to
oxidize. And it won't oxidize all at once, unless
you could melt (?) or boil the palladium while you were feeding oxygen to it.
The cell only contains a little oxygen.
2010/5/8 Abd ul-Rahman Lomax <[email protected]>:
> The kicker is this "equivalent to many
> atmospheres of hydrogen gas pressure exposure." No, the D2 pressure at the
> cathode is roughly one atmosphere through the whole experiment.
This is the original Shanahan statement:
K&M go on to discuss a specific type of
calorimetric result obtained by F&P commonly
known as eheat-after-deathf (HAD). In the HAD
experiment, a F&P electrolysis cell is
allowed to lose enough electrolyte via
evaporation, entrainment, and electrolysis that
electrical contact is broken and current flow
stops. Such an event is shown in K&Mfs
Figures 4 and 5, where an excess power signal is
observed for approximately 3 hours
after this point is reached.
It is instructive to review this result with the
prior discussion of the CCS in mind. Once
current stops, the driving force to load the Pd
electrode with hydrogen is removed, and
the system seeks to obtain equilibrium under the
new state by releasing gas, converting
SRNS-STI-2009-00825
the situation from an electrolysis cell to a gas
unloading experiment. At the point where
electrical contact was broken, the cell gas was
very nearly a stoichiometric mixture of
hydrogen and oxygen. Significant hydrogen
release will occur because the electrolytic
loading was equivalent to many atmospheres of
hydrogen gas pressure exposure, but the
hydrogen present will inhibit complete Pd
unloading. The equilibrium plateau pressure
of Pd-D at ~70-100 C is ~300-1000 mbar8, and
that the cell pressure is ~1000 mbar since
it is an open cell, with at least 2/3 of that
consisting of D2. Thus initial unloading to ~0.6
D/M units should occur, and not much more,
leaving plenty of hydrogen in the electrode
as hydride.
he's got the environment wrong. At the time the
current flow stops, most of the gas in the cell
is water vapor, not deuterium. The water has been
boiling, at the same time as deuterium/oxygen
evolution declines and then stops.
Yes, there is plenty left. And it takes a long time to diffuse out.
With the Pd and Pt electrodes exposed, a metal
surface is presented which will catalyze
the recombination of hydrogen and oxygen. That
process does three things; a) it reduces
the hydrogen pressure, causing the Pd to unload
further, and b) it reduces the overall
pressure, causing air to be drawn back into the
cell, resupplying the oxygen content of the
cell gas somewhat, and c) it produces heat,
which will presumably be detected by
temperature sensing devices in the cell.
So he has burning of hydrogen as a
pressure-relieving process. There goes the idea
of using hydrogen to power an internal-combustion
engine, except maybe by the vacuum formed as the
hydrogen and oxygen gases collapse to form water.
Clearly, the steady state is now radically
different and the system would have to be
recalibrated under the new steady state to
translate those temperatures into heats.
To do so accurately, yes. To do so in bulk, no.
However, K&M report that the energy detected
was gfar beyond the quantity of possible stored
chemical energyh. No recalibration was
reported in the 1993 F&P paper, so one wonders
how this was determined accurately.
Something had to heat the cell.
At an even more subtle level is the
consideration of the rate of hydrogen release from the
Pd electrode, which would impact the amount of time heat would be produced by
recombination. It is well known that surface
contaminants affect the ability of Pd to
absorb and desorb hydrogen9,10, and after being
immersed in an electrolyte solution for
hundreds of hours, it is unlikely the Pd surface
is pristine, and in fact there are many
studies showing it is not (see below).
Furthermore, crack-free Pd is supposedly required
to allow high loading to be attained, and high
loading is supposedly required to obtain the
FPHE. This in turn implies a low
surface-to-volume ratio (SVR) is the desired condition.
Hydrogen desorption from Pd is a diffusional
process, and the lower the SVR, the slower
the desorption will proceed. To maintain, or
even slightly increase, the registered
temperature would only require supplying enough
heat to replace that lost from the cell,
which is now an unstirred gas cell instead of a well-stirred liquid cell.
The original cell design was well-insulated,
implying the natural heat loss rate would be
low. It seems eminently reasonable to assume
that that amount of heat lost would be
small per unit time. The Pd in the 1993
experiment likely evolved hydrogen slowly,
allowing for heat production over extended periods. Thus the HAD event was
completely predictable by conventional chemistry.
Not at the level of energy generation found. The
calorimetry will measure heat being lost from the
cell. If the cell contents become gaseous, the
heat transfer will slow, but as Fleischmann notes
in the reply to Morisson that Rothwell cited, the
critical impedance is the transfer of heat across
the vacuum in the silvered Dewar flask being used
for the experimental cell, and that has been
calibrated at all levels of electrolyte. The
transfer that slows is internal transfer,
presumably, between the cell contents and the
wall of the flask. So the heat loss being
measured by the calorimetry is *below* the actual
heat being generated, for a time, until steady state is achieved.
This is typical of Shanahan. "Calibration
Constant Shift" is asserted as if it were the
whole story, if your calibration is off, your
whole set of measurements must be tossed out the
door. In fact, with the calorimentric design, the
Shanahan effect would only cause energy transfer
to be time-inaccurate. The integral of heat
evolved over time would still represent accurate
measurement of energy, excepting only differences
in stored heat between the beginning and the end
of the measurement period. That's a limited error.
This author knows of no HAD event
reports that recalibrate due to the obvious CCS
and consider the above concerns. Thus
reported HAD events certainly do not
significantly advance the nuclear hypothesis. What
is required to fully understand HAD events is a)
recalibration (with supplied data/results
as above) and b) detailed surface examinations of the Pd coupled with gas
loading/unloading studies. With that information the above concerns might be
addressable.
From Fleischmann's response, the calorimeter has
indeed been calibrated at various levels of
electrolyte and at various temperatures.
Shanahan assumes plenty of oxygen, but he
realizes that the oxygen present in the cell
would have to be replaced, because it could
oxidize only a little hydrogen (deuterium). So he
postulates the preposterous, that the catalyzed
burning of hydrogen would reduce pressure,
causing air to be sucked in. In fact, if there
were a stoichiometric mixutre of hydrogen and
oxygen in the cell, sudden oxidation would cause
an explosion, not implosion. It is true that the
ratio of deuterium and oxygen in the cell is a
stoichiometric one at the point of "death." But
the percentage of water vapor is much higher at
that point, from the boiling of the electrolyte.
Indeed, the cathode would probably catalyze the
combustion of hydrogen at this point, but the
increase in temperature would increase pressure,
not decrease it, driving more of the deuterium
and oxygen out of the cell, not attracting it to the cathode.
I'd say that the cell should be designed to not
allow air to be sucked in, and conditions of
negative pressure should be noted to completely
rule out the influx of oxygen from the air. A simple flap valve should do it.
Air would not be sucked into the cell until the
cell began to cool; if the cell temperature
continued to increase, the flow would only be out.
I'd be very interested in the behavior of a
deuterium-loaded cathode, removed in that
condition from a cell. Morisson claims that the
evolution of deuterium from loaded palladium is
exothermic, itself, that's what heats the
cigarette lighter to cause catalytic combusion to
occur with the ambient air. Is this correct?
So many questions, so little time.
