The main point Abd, ... which we seem to be digressing away from - is that you were/are a little rough on Shanahan, when in fact you missed the very thermodynamic issues that he may (or may not) have known.
This is not to say that he is correct, by any means, as he is probably babbling on blindly about what others have told him years ago - to wit: there could be anomalous heat due to the inversion temperature - which quite frankly, many researchers in LENR still do NOT understand! And in fact, IF the models being put forth by Roarty and others about ZPE (Casimir) contribution to heating are correct, or even partly correct, then get this: THE PALLADIUM CIGARETTE LIGHTER ITSELF IS OVERUNITY, but not from LENR - ... instead it overunity is from Casimir heating - which is almost as far out, if not more. Except this exact thing is now under investigation at Argonne and elsewhere, and cannot be ruled out (i.e. Casimir heating). That is precisely why I brought up Kitamura's version of the Arata experiments: http://www.lenr-canr.org/acrobat/KitamuraAanomalouse.pdf ... which apparently you mistook for another Kitamura paper. Kitamura also finds anomalous heat from hydrogen in a Phase 1 (non-nuclear) loading phase. This is above the heat of formation of the hydride. Please look at Table 1 to see what I am talking about. Do you disagree with that these findings are accurate? Do you have another explanation? There could well be more going on here than LERN ... but LENR researchers apparently do not want to hear it - but in fact the Casimir heating may itself be a PRECURSOR to later nuclear reactions (Phase II in the Kitamura paper). Of course anomalous heating without fusion could also be a Millsean precursor step as well, but dyed-in-the-wool LENR people do NOT want to hear that sentiment expressed either. Jones -----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 ±10Ž) at high >temperature (T „100Ž) 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
