At 05:29 PM 5/12/2010, Jones Beene wrote:
-----Original Message-----
From: Abd ul-Rahman Lomax
> Thus, like most evaporations, the escape of deuterium from the lattice
must cool it.
No. The inversion temperature of hydrogen is low, and there is no cooling on
expansion from the Joule-Thomson effect - as with an ideal gas.
In fact it heats. That is the very crux of the cigarette lighter hypothesis,
which apparently you missed :)
Hey, I'd miss my nose if it didn't itch occasionally.
No, the cigarette lighter does not generate heat from release of
hydrogen, it generates it by catalyzing the oxidation of hydrogen at
the surface of the palladium.
I don't think I made this up about the loading being exothermic, and
that's very clear in the Arata experiments, you can see the heat
released from the heat of formation of palladium deuteride. Now if it
releases heat when it is formed, wouldn't it absorb heat when the
deuterium evaporates?
If not for the inversion temperature heating effect, the Pd in the lighter
would never release much gas to begin with, and no cigarettes would get lit.
Note this was an actual lighter design at one time and addicts do not like
to get fooled.
Yes, I understood that. The heat, though, doesn't come from expansion
of hydrogen.
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). 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.
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. But it would not explain heat after
death, the opposite, one would expect some cooling as the deuterium
evaporates, though not much.
[...]
> No. Arata does not show excess heat with hydrogen.
Maybe it is Kitamura then, but one of them shows substantial heat from
hydrogen nanopowder too, with no added input other than the trigger heating
- but more from deuterium. Citation to follow (when I get a spare moment)
I don't think you have understood those experiments. They are
basically the same.
From the Sasaki/Kitamura/et al paper,
http://www.lenr-canr.org/acrobat/SasakiYdeuteriumg.pdf:
The results for the case of D2 and H2 absorption are
compared in Fig. 4. After the gas was introduced,
pressure did not begin to rise for a while. During this
first phase the Pd powder absorbed almost all of the D2
(H2) gas atoms as they flowed in, and heat was released
as a result of adsorption and formation of deuterides (or
hydrides). After about 30 minutes, the powder almost
stopped absorbing gas; the gas pressure began to rise,
and the heat release from deuteride (hydride) formation
subsided. This is the beginning of the second phase.
If you look at the time/temperature/pressure charts for these
experiments, you can see the substantial heat released from
hydride/deuteride formation. D and H are about the same. What is not
the same is a low but long-continued heat with deuterium. It's quite
striking in Arata's temperature plots. Kitamura was doing
calorimetry, so he's measuring watts, and it's noiser.
There is no "trigger heating."
