At 11:19 PM 5/12/2010, Jones Beene wrote:
-----Original Message-----
From: Abd ul-Rahman Lomax
> Yes, I understood that. The heat, though, doesn't come from expansion
of hydrogen.
Wrong. Some of the heat does come from expansion of hydrogen. Of course much
more comes from combustion.
What hydrogen is "expanding" in the cigarette
lighter? I mean expanding more than a small
amount, a fraction of an atmposphere. If the
lighter is closed, how high does the pressure
build inside of it? I think the whole point is that this is low pressure.
>When air initially at a moderate temperature is expanded through a valve,
>its temperature decrease because it starts out below the inversion
>temperature of the constituent gases, and the expansion will cause a
>temperature reduction as the result of the Joule-Thomson effect. Any gas
>expanded at constant enthalpy will experience a temperature decrease ONLY
if
>is below the inversion temperature, however, and if above it will usually
>experience a temperature increase.
> I was not assuming endothermy from expansion, but from evaporation
(more like sublimation in this case).
There is no difference.
There certainly is a difference. There is what is
effectively solid hydrogen, an alloy with
palladium. However, this alloy has the property
that the hydrogen can move from site to site
within the lattice, with only a small energy hump
between each site. At the surface, the hydrogen
can escape, but it requires energy for it to
escape from a lattice position, roughly the same
as as released when it occupies the position. It
is not a gas, where the hydrogen can move
completely freely. The pressure at the surface is
ambient. The pressure does not change. There is
no "expansion." It is not simply like what is
involved with "transition temperature," a gas
escaping from pressure through some valve, where
it expands, pressure being reduced. The movement
of the hydrogen is more like diffusion, but constrained.
At the surface, hydrogen will escape, due to
thermal variations and impacts from the gas. Just
as capture of hydrogen by the lattice releases
energy, escape, the reverse process, must absorb
it. In equilibrium, the release and absorption
are equal, but if gas is moving into the lattice,
it heats. If it is escaping, it *must* cool, for
if it were exothermic, as Mr. Beene seems to be
claiming, we could then create a perpetual motion machine.
> It's simply the reversal of the heat released from absorption. If
hydride/deuteride formation is exothermic, and it is, then de-formation is
endothermic.
Wrong. You are missing the balance point. The balance is between the energy
used to pressurize the gas before loading and the net energy returned by
both hydride formation and hydride release.
Okay, we have a low-pressure cell (evacuated,
actually, I think) with nanoparticle palladium in
it. We have a high-pressure hydrogen gas. The
high pressure represents potential energy, yes,
stored in the gas by the compressor.
When we admit this gas to the cell it is rapidly
absorbed by the palladium, so cell pressure does
not rise immediately to the source pressure.
Because the temperature is above the inversion
temperature for hydrogen, and neglecting energy
lost in the valve, we would expect the gas to
heat as it is admitted to the cell. This is not
the heat of formation of palladium hydride, it is
Joule-Thompson heating. This would add to the
heat of formation of the hydride, which is
estimated in one source: http://www.lenr-canr.org/acrobat/TianJheatmeasur.pdf
When the laser beam passed through the glass
wall, there was a spot on the surface of Pd
sample. The volume of Pd being irradiated was
only about one fifth (9mm)of the whole wire. The
number of Pd atoms is about 1.0×1020. If they
all combined with hydrogen atoms to form
palladium hydride they would release heat of
formation of about 6.4 joule, which corresponds to
about 6.4×10-20 for each PdH. When the loading
ratio is 0.3, as in our experiment for example,
there should be 3.0×1019 hydrogen atoms
dissolved in the bulk of palladium lattice, that is to say
they would release 1.9 joule when these hydrogen
atoms combined with Pd. But we measured
232.4±0.3 joules heat released from the
experiment that had the most significant exothermic
behavior, which corresponds with 7.73×10-18
joule per atom Pd. Obviously the heat we get in
experiment is about 130 times more than that
released in the formation of a PdHx (x=0.3).
However, the of temperature of the Pd wire
varied greatly (more than ±10) at high
temperature (T 100) duo to the instability
of the laser power output. Also some H2 was
released from the Pd bulk when it was being
irradiated. Because of the large uncertainty in the
measurements, we will not make claims of excess heat in this paper.
As I understand the matter, if the valve and
piping to the cell are at room temperature, so
that any J-T heat is lost to the environment,
there will still be heating in the cell. Kitamura
et al ascribe the heating they see on the
admission of gas to the cell to the heat of
hydride (or deuteride) formation. That means to
me that they consider J-T heating to be unimportant in this effect.
If the energy needed to compress hydrogen for loading a tank is say 10
W-hrs, then that can be balanced exactly against 5 W-hrs of exotherm for
deuteride formation and another 5 W-hrs of exotherm for expansion. Get it?
Sure, though the 5 Wh don't necessarily end up in
the cell. I haven't seen actual calculations, but
my sense is from what I've cited in this mail and
my previous is that the J-T heating is low in
comparison to the heat of hydride formation.
> Note that failure to account for the heat of formation of palladium
deuteride could be a possible source of error in the calorimetric
analysis of CF experiments.
You finally got something right!
But ... that would only account for heating in
the early stages of the experiment, and goes to
zero as loading stops increasing, and becomes a
negative term when loading starts to decrease, as
it will when electrolysis stops (unless deloading
is inhibited by the cell being sealed?)
> But it would not explain heat after death.
Partly wrong. It can explain some of it, but usually there is much more heat
after death than can be explained by expansion above the inversion
temperature.
You already got the expansion heat by admitting
the gas to the cell... But, of course, as the
cell pressure increases, you then have gas under
pressure, with potential energy to be released, but only if the cell is opened.
In HAD experiments reported by Fleischmann, the
cell was open, so there was no J-T energy stored.
J-T energy is not stored in the palladium
hydride, it is not under (much) pressure. If a
piece of palladium hydride is placed in a closed
cell, how high will the hydrogen pressure rise?
That would be a measure of the effective pressure
of the hydrogen. I don't think it's high at all.
However, what is it? Surely this is known.
Just some links that might be useful:
http://iopscience.iop.org/0305-4608/14/6/012;jsessionid=6463902CDDD085E6B1B080CC70BAF4F8.c3
http://cat.inist.fr/?aModele=afficheN&cpsidt=18010160
