RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Jed Rothwell On Wed, 12 May 2010 13:33 Jed Rothwell said Abd ul-Rahman Lomax wrote: The entire palladium lattice can be considered a collection of cavities. No, it is a lattice. A lattice is not the same as a cavity. A cavity is a break in the lattice, in which D2 molecules can form. Deuterons cannot come together to form molecules in a lattice. They might be able to come together to form helium atoms. That's the subject of debate, but no one asserts they can form D2. Jed, Abd may have a point, if we consider the cathode a stack of Casimir plates that have all been pulled together by Casimir force to form a solid, atoms in small numbers form covalent bonds then clusters before they start to form metallic bonds - perhaps metallic bonds are a function of Casimir effect - the almost free electron generated by this squeezing together of the lattice. You also say deuterons cannot come together to form D2 in the lattice but I posit that fractional d2 formed in the cavities which act like our pump to fractionalize atoms then tie them together with a diatomic bond. These fractional d2 molecules would then be stuck in the lattice just like normal d2 is stuck outside of a Pd membrane. Where do Miley and Arata claim they observe the f/h or f/d? Regards Fran
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
OK, we are agreed that P1V1/T1 =P2V2/T2 even applies to state changes of gas atoms. Normally Anything that effects the volume of the overall population but here is where COE meets Casimir effect Which allows monatomic gas to translate Freely with very little opposition to fractional states, We can ignore the argument about this being relativistic or not and just treat this as a property of a Pd membrane that allows monatomic gas to pass but is a barrier to diatomic gas. My posit is that when f/h1 becomes f/h2 it finds itself surrounded by barriers where it must remain locked until it is again disassociated by random thermal energy. Unlike normal h2 this f/h2 does have one additional energy gradient that it can tap to help it translate back to normal h2. The sum of vacuum fluctuation energy that determined it's fractional value is decreasing as it leaves the cavity boundaries and enters the lattice, that differential wants to reshape the orbitals of the f/h2 but is prevented from doing so by the diatomic bond. I am humbly suggesting these f/h2 or f/d2 can take up positions normally occupied by a single d1 and that defects and cavities can be considered the pump houses where fractional diatomic bonds are utilized as containment vessels. Regards Fran Jones Beene Wed, 12 May 2010 20:20:58 -0700 -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. 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. 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. 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? 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 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. Jones
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
I am starting to feel a new animation coming on where the fractional diatoms squeeze very slowly into the lattice but every time the diatomic bond pops on one of them it causes a small chain reaction and further bursting of any other nearby fractional diatoms nearing their breaking threshold. Since the electrolysis or thermal energy creates monatomic gas which loads quickly into the cavities has been removed the de-loading slows to a crawl. the atoms cool and form diatoms which are confined by Pd's membrane property Initially the monatoms in the lattice will de-load quickly but their loss draws the fractional diatoms out of the cavity in a hard to predict population of fractional gas cycling between f/d1 and f/d2. I posit the diatom releases energy every time it forms but is restored to atomic energy level by the delta in vacuum energy giving the normally chaotic random motion of gas law a favored vector towards geometry where Casimir force is reduced. I am not saying COE is violated but that it normally ignores ZPE which always sums to zero inside a common inertial frame -I am saying this normal cancellation is avoided if you can build an energy capture system that cages the energy in one frame but releases it in another. A fractional diatomic bond may represent such a trap. Regards Fran
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
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
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
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.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Regarding this topic, Ed Storms asked me to post the following: A common mistake is being made here. In electrochemistry, when deuterium is generated at the cathode, it produces gas at ambient pressure and an activity of D in the Pd. In chemistry, activity is made equal to the IDEAL pressure when making calculations of Gibbs energy. However, in the real world activity and pressure are NOT the same phenomenon and are frequently not equal. For example, although the activity of D in Pd can be calculated and presented as 10e47 atm, this is NOT the pressure in the material. There is no gas pressure in the Pd because D2 does not exist there as gas. If a cavity exists and gas is generated to fill the cavity, the pressure will rise in the cavity. However, it will not rise to a value equal to the activity because H2 becomes nonideal at high pressure. As a result, the measured pressure will be much smaller than the activity.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 11:05 AM 5/13/2010, Jones Beene wrote: 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? I had referred to Arata, you then referred to Arata and Kitamura. So I looked for Arata and Kitamura together and then at the most recent work. I think this paper above was older, though it's not dated. Kitamura et al report: For Pd.Zr oxide nano-powders, anomalously large energies of hydrogen isotope absorption, 2.4 +/- 0.4 eV/D-atom and 1.8 +/- 0.4 eV/H-atom, as well as large loading ratio of D/Pd =1.1 +/- 0.0 and H/Pd = 1.1 +/- 0.3, respectively, were observed in the phase of deuteride/hydride formation. I know only a tiny amount about the heat of formation of palladium hydride. Kitamura does not state how large the anomaly is. I do have an impression, from my scans of various papers, that the heat of hydride formation varies with the exposure of the surface. What this means and whether or not it could explain the anomaly is beyond me. Life is full of anomalies that I will never explain.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 01:38 PM 5/10/2010, Jed Rothwell wrote: Jones Beene wrote: Can you cite the reference for this kind of bursting tube, due to internal pressurization, having being actually performed? See McKubre's replication of the Arata experiment. I didn't see mention of it there. Regarding the pressure from electrolysis, it far exceeds anything that can be accomplished with mechanical means such as pumps. See Mizuno's book, chapter 7. I think the fact that gas loading works means that high pressure is not required for a cold fusion effect. I guess it means that high pressure is not needed for high loading; there are other ways to achieve that, rather than brute force. Mizuno's book isn't available to me. When we are talking about a local effect, something that only involves perhaps, primarily, say two molecules of deuterium gas and their relationship in confinement, the concept of pressure is trickier. I'm thinking about how the Takahashi's TSC might come to exist. If I'm correct -- always a huge if -- that configuration requires energy to form, the molecules would have to be pushed together in some way. The pushing could merely be their momentum in a collision, but, again if I'm correct, a collision that would otherwise bring the four deuterons involved into the tetrahedral symmetric configuration would cause the molecules to dissociate. So there must be some resistance to dissociation, which could be supplied by the lattice. Deuterium gas doesn't ordinarily exist within the lattice, the lattice itself causes dissociation. I don't think anyone knows enough about the behavior of deuterium at the interface to calculate the frequency of TSC formation. It would be very low, but the incidence of fusion is very low. Meanwhile, I'm generally interested in this phenomenon of high pressure build-up within a hollow palladium tube, caused by electrolytic loading. That loading must create a barrier that the deuterium cannot cross, and while the barrier is low in voltage, the pressure equivalent must be high. It was said here that this could be used to generate free energy (and was therefore doubtful), but this assumed that the rate of flow was high enough to make up for the current flow necessary to maintain that barrier. You can get very high pressure, but not high flow rate, equivalent to, with charge, getting very high voltage but with only very low current available, as with a Van de Graaf generator. We already know that, even without electrolysis, loading ratios of 1:1 or higher can be obtained by gas loading. That's close to equivalent to hydrogen metal at room temperature, which is clearly a high-pressure equivalent. But that hydrogen cannot flow rapidly, unless the lattice is destroyed. Do not toss fully loaded palladium into a furnace, you might be quickly less one furnace, just from the release of pressure. Simply heating the palladium to a few hundred degrees doesn't cause extremely rapid release, or those cigarette lighters would have been really, really dangerous.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
ON Wed, 12 May 2010 07:49 Abd ul-Rahman Lomax said [quote] Meanwhile, I'm generally interested in this phenomenon of high pressure build-up within a hollow palladium tube, caused by electrolytic loading. That loading must create a barrier that the deuterium cannot cross, and while the barrier is low in voltage, the pressure equivalent must be high. It was said here that this could be used to generate free energy (and was therefore doubtful), but this assumed that the rate of flow was high enough to make up for the current flow necessary to maintain that barrier. You can get very high pressure, but not high flow rate, equivalent to, with charge, getting very high voltage but with only very low current available, as with a Van de Graaf generator. [/quote] Abd, Obviously Pd creates a barrier that d2 cannot cross, The controversy is over how the pressure is created. The deuterium ice, deuterium clusters, fractional hydrogen and other monikers all describe an unusual state of hydrogen Associated with this phenomena. The state is always associated with a catalyst and we must therefore assume the energy behind this effect is supplied by the cavity... Is this assuming too much? Regards Fran
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 01:20 PM 5/12/2010, Roarty, Francis X wrote: Obviously Pd creates a barrier that d2 cannot cross, The controversy is over how the pressure is created. No. This pressure is created by both hydrogen and deuterium. The high pressure mentioned is created by electrolysis with hydrogen or deuterium generated at the surface of a hollow palladium rod. Shanahan's cigarette-lighter explanation does not factor for the lack of excess heat when doing this with hydrogen. The deuterium ice, deuterium clusters, fractional hydrogen and other monikers all describe an unusual state of hydrogen Associated with this phenomena. It is an unusual state in palladium deuteride, almost like solid hydrogen (i.e., hydrogen metal) that will sublimate slowly. The state is always associated with a catalyst and we must therefore assume the energy behind this effect is supplied by the cavity Is this assuming too much? I think so. There are cavities involved, likely. However, they are not supplying any energy, apparently, rather they *configure* the reacting ingredient or ingredients. We know that the reaction rate increases with temperature. I suspect that the energy required -- there must be energy required, but energy is not the only ingredient -- is generally supplied by ordinary heat.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Abd ul-Rahman Lomax wrote: No. This pressure is created by both hydrogen and deuterium. The high pressure mentioned is created by electrolysis with hydrogen or deuterium generated at the surface of a hollow palladium rod. What experiment does this reference? Arata's double-structured cathode? The pressure is not all that high in that one, although it was high enough to rupture the hollow cylinder at SRI. Extremely high pressure is achieved by electrolysis at the surface of an electrode, in a microscopic domain. According to some sources it is 10E47 atmospheres (Mizuno, p. 101). I believe the maximum pressure in a cylinder can be achieved as easily with a pump hydraulically as with electrolysis. The limiting factor is the strength of the cylinder. Arata's double structured cathode results do not depend on brute force loading, nor do his more recent gas loading results. The state is always associated with a catalyst and we must therefore assume the energy behind this effect is supplied by the cavity Is this assuming too much? I think so. There are cavities involved, likely. However, they are not supplying any energy, apparently, rather they *configure* the reacting ingredient or ingredients. There is no significant volume of cavities in a successful cold fusion cathode. If there are cavities, the cathode will not produce the effect. It has to be an intact lattice. A sample with cavities will bend or expand much more than an intact lattice; see: http://lenr-canr.org/acrobat/StormsEhowtoprodu.pdf In a successful cathode, the deuterium is in solution with the metal, with deuterons interspersed in the metal lattice. The exact configuration -- with how many deuterons at each tetrahedral and octahedral site -- is disputed. - Jed
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
On Wed, 12 May 2010 Abd ul-Rahman Lomax said I think so. There are cavities involved, likely. However, they are not supplying any energy, apparently, rather they *configure* the reacting ingredient or ingredients. We know that the reaction rate increases with temperature. I suspect that the energy required -- there must be energy required, but energy is not the only ingredient -- is generally supplied by ordinary heat. Abd, I don't think the heat is ordinary or it would dissipate see quote from Moddel paper below, Regards Fran Quote from Professor Garret Moddel dated 30 October 2009 Assessment of proposed electromagnetic quantum vacuum energy extraction methods http://www.calphysics.org/articles/Moddel_VacExtrac.pdf There is a fundamental difference between the equilibrium state for heat and for ZPE. It is well understood that one cannot make use of thermal fluctuations under equilibrium conditions. To use the heat, there must be a temperature difference to promote a heat flow to obtain work, as reflected in the Carnot efficiency of Eq. (4). We cannot maintain a permanent temperature difference between a hot source and a cold sink in thermal contact with each other without expending energy, of course. Similarly, without differences in some characteristic of ZPE in one region as compared to another it is difficult to understand what could drive ZPE flow to allow its extraction. If the ZPE represented the universal ground state, we could not make use of ZPE differences to do work. But the entropy and energy of ZPE are geometry dependent.32 The vacuum state does not have a fixed energy value, but changes with boundary conditions.33 In this way ZPE fluctuations differ fundamentally from thermal fluctuations. Inside a Casimir cavity the ZPF density is different than outside. This is a constant difference that is established as a result of the different boundary conditions inside and out. A particular state of thermal or chemical equilibrium can be characterized by a temperature or chemical potential, respectively. For an ideal Casimir cavity having perfectly reflecting surfaces it is possible to define a characteristic temperature that describes the state of equilibrium for zero-point energy and which depends only on cavity spacing. In a real system, however, no such parameter exists because the state is determined by boundary conditions in addition to cavity spacing, such as the cavity reflectivity as a function of wavelength, spacing uniformity, and general shape.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Abd ul-Rahman Lomax wrote: We know that the reaction rate increases with temperature. I suspect that the energy required -- there must be energy required, but energy is not the only ingredient -- is generally supplied by ordinary heat. I assume this reaction refers to heating from the hydrogen or deuterium that emerges from highly loaded palladium in air. This reaction is what was used to make cigarette lighters with palladium hydride, in the 19th century. When you expose highly loaded palladium to air, the hydrogen emerges and the surface of the palladium catalyzes recombination with oxygen in air, so the cathode heats to incandescence, and can be used to light a cigarette. You have to have quite a lot of palladium to pull this off -- roughly an ounce I suppose (32 g) which is much more than most cold fusion cathodes. This never happens with heat after death because, as I mentioned, there is no air or oxygen in the headspace, only water vapor. It can only recombine in the atmosphere outside the cell. Ikegami's group at the Nat. Inst. for (plasma) Fusion Science did extensive testing of this. They got highly loaded cathodes to self-heat, and they also heated them in air with blow torches and other methods, trying to make them explode the way Fleischmann's cathode did. They never got it do anything other than the standard cigarette lighter effect in air. The gas will always come out of a saturated cathode. When the cathode is submerged in water, you can see fine gas bubbles appear on the cathode surface, looking like carbonation bubbles coming out of solution in soda-pop at 1 atm (except they are smaller). The only way to maintain high loading is to keep doing electrolysis, and then you have a steady stream of gas coming out of the cathode while other gas goes in. The gas emerges in two phases, corresponding to the alpha and beta phase loading, which refers to the sites in the lattice that are loaded. Assume it starts highly loaded to the beta phase. Lots of bubbles form and emerge. The rate gradually slows down -- which is readily apparent to the naked eye. It slows down and down until it reaches the alpha phase, and then suddenly starts up again, until the alpha phase deuterons are mostly gone from the cathode. It is very difficult to purge all of the deuterons from the lattice. There is always some left. Heating up the cathode will increase the rate at which it de-loads, but it always de-loads on its own, and as far as anyone knows (per Ikegami) there is no way to speed up the rate of degassing so much that the metal explodes. Of course there is not enough energy in the gas to make the metal melt or vaporize, the way some cathodes have done! (Note that it can take far less energy to make metal fracture or explode than it takes to melt it. Fracturing is function of the speed of energy release, and how confined the site of the release is.) As Fleischmann pointed out, the rate of de-loading is far too slow to account for cold fusion heat, and the total energy release is far too small. It was ~1,700 times too small when Morrison and others suggested this hypothesis back in 1990, and it must be millions of times too small by now. Shanahan and his editors must be innumerate. - Jed
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
-Original Message- From: Abd ul-Rahman Lomax Shanahan's cigarette-lighter explanation does not factor for the lack of excess heat when doing this with hydrogen. Well, at the risk of defending a repulsive argument, many experiments do show excess heat with hydrogen, including some of the Arata experiments. To get the repulsive punage, you will have to read all the way through this - sorry for the lack of an instant smiley. However, with H2 the heat generated is most likely NOT nuclear. With D2 some of the heat may not be nuclear, either. Excess heat with H2 or D2 could be related to Mills shrinkage or to ZPE. With D2, it is even possible that actual fusion may be a time reversed QM balancing act and only replaces energy lost in other methods. Here is a SciNews article, recently updated, which gives hope that Casimir cycling for gain is ultimately doable: http://www.sciencedaily.com/releases/2009/12/091210153657.htm ... indicating at least that some high-priced brain power at DoE and Argonne believe that Casimir attraction can be manipulated into repulsion. That would be the key to finding a useable asymmetry in the heat of adsorption, at this geometric level. The ultimate source of energy would be ZPE. It could be as simple as rapidly cycling an electric field in which the fully loaded nanoparticles are allowed to express whatever Casimir asymmetry can be engineered into the nanostructure ... As characteristic device dimensions shrink to the nanoscale, the effects of the attractive Casimir force becomes more pronounced making very difficult to control nano-devices. This is a technological challenge that need to be addressed before the full potential of NEMS devices can be demonstrated ... The goal is to not only to limit its [Casimir force] attractive properties, but also to make it repulsive. Not that Shanahan's ignorance of the repulsive Casimir makes his net ignorance of the entire field of LENR any less repulsive :) Jones
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 03:06 PM 5/12/2010, Jed Rothwell wrote: Abd ul-Rahman Lomax wrote: No. This pressure is created by both hydrogen and deuterium. The high pressure mentioned is created by electrolysis with hydrogen or deuterium generated at the surface of a hollow palladium rod. What experiment does this reference? Arata's double-structured cathode? The pressure is not all that high in that one, although it was high enough to rupture the hollow cylinder at SRI. Hey, I'm no expert on this, at all. I was skeptical about the *very* high pressures, but I thought you had confirmed it. Extremely high pressure is achieved by electrolysis at the surface of an electrode, in a microscopic domain. According to some sources it is 10E47 atmospheres (Mizuno, p. 101). And what does this mean? I believe the maximum pressure in a cylinder can be achieved as easily with a pump hydraulically as with electrolysis. The limiting factor is the strength of the cylinder. Arata's double structured cathode results do not depend on brute force loading, nor do his more recent gas loading results. The state is always associated with a catalyst and we must therefore assume the energy behind this effect is supplied by the cavity Is this assuming too much? I think so. There are cavities involved, likely. However, they are not supplying any energy, apparently, rather they *configure* the reacting ingredient or ingredients. There is no significant volume of cavities in a successful cold fusion cathode. The entire palladium lattice can be considered a collection of cavities. If there are cavities, the cathode will not produce the effect. It has to be an intact lattice. A sample with cavities will bend or expand much more than an intact lattice; see: http://lenr-canr.org/acrobat/StormsEhowtoprodu.pdf In a successful cathode, the deuterium is in solution with the metal, with deuterons interspersed in the metal lattice. The exact configuration -- with how many deuterons at each tetrahedral and octahedral site -- is disputed. Yes. My suspicion is that overloading is required, but this could be transient, i.e., a lattice might be loaded to 1:1, but then a shock wave or other phenomenon causes some level of overloading locally. Loading generally refers to the bulk; the nuclear effect is apparently at the surface. Subsurface conditions can affect surface conditions, which would be the connection.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 03:06 PM 5/12/2010, Jed Rothwell wrote: Abd ul-Rahman Lomax wrote: No. This pressure is created by both hydrogen and deuterium. The high pressure mentioned is created by electrolysis with hydrogen or deuterium generated at the surface of a hollow palladium rod. What experiment does this reference? Arata's double-structured cathode? The pressure is not all that high in that one, although it was high enough to rupture the hollow cylinder at SRI. Hey, I'm no expert on this, at all. I was skeptical about the *very* high pressures, but I thought you had confirmed it. Extremely high pressure is achieved by electrolysis at the surface of an electrode, in a microscopic domain. According to some sources it is 10E47 atmospheres (Mizuno, p. 101). And what does this mean? I believe the maximum pressure in a cylinder can be achieved as easily with a pump hydraulically as with electrolysis. The limiting factor is the strength of the cylinder. Arata's double structured cathode results do not depend on brute force loading, nor do his more recent gas loading results. The state is always associated with a catalyst and we must therefore assume the energy behind this effect is supplied by the cavity Is this assuming too much? I think so. There are cavities involved, likely. However, they are not supplying any energy, apparently, rather they *configure* the reacting ingredient or ingredients. There is no significant volume of cavities in a successful cold fusion cathode. The entire palladium lattice can be considered a collection of cavities. If there are cavities, the cathode will not produce the effect. It has to be an intact lattice. A sample with cavities will bend or expand much more than an intact lattice; see: http://lenr-canr.org/acrobat/StormsEhowtoprodu.pdf In a successful cathode, the deuterium is in solution with the metal, with deuterons interspersed in the metal lattice. The exact configuration -- with how many deuterons at each tetrahedral and octahedral site -- is disputed. Yes. My suspicion is that overloading is required, but this could be transient, i.e., a lattice might be loaded to 1:1, but then a shock wave or other phenomenon causes some level of overloading locally. Loading generally refers to the bulk; the nuclear effect is apparently at the surface. Subsurface conditions can affect surface conditions, which would be the connection.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Jones Beene wrote: Shanahan's cigarette-lighter explanation does not factor for the lack of excess heat when doing this with hydrogen. Well, at the risk of defending a repulsive argument, many experiments do show excess heat with hydrogen, including some of the Arata experiments. Which ones? I don't recall any Arata-style Pd nanoparticle experiments that produced heat with hydrogen. Only experiments with Ni have done that as far as I recall. - Jed
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 03:33 PM 5/12/2010, Roarty, Francis X wrote: On Wed, 12 May 2010 Abd ul-Rahman Lomax said I think so. There are cavities involved, likely. However, they are not supplying any energy, apparently, rather they *configure* the reacting ingredient or ingredients. We know that the reaction rate increases with temperature. I suspect that the energy required -- there must be energy required, but energy is not the only ingredient -- is generally supplied by ordinary heat. Abd, I don't think the heat is ordinary or it would dissipate see quote from Moddel paper below, No, by ordinary heat I mean the movement of molecules and atoms at the involved temperature. That is a distribution of energies, so a given molecule may have more or less than the mean energy which is present at a given temperature. The heat is not used to generate energy. Suppose that some reaction requires a certain energy to occur. That energy is present from the collision velocity of two reactants, if they are moving rapidly enough. Suppose that we put these reactants together at a temperature that is below the temperature necessary for the average molecule to have the necessary energy. What will the reaction rate be? Will it be zero? Let's suppose that the reaction is exothermic. However, a single molecular reaction will not raise the reaction rate enough to cause the reaction to self-propagate. An ignition source, though, could raise the reaction rate in an area in contact with it, causing the reaction to them propagate. A stoichiometric mixture of hydrogen and oxygen gas at room temperature the mixture is stable, the gases do not combine. However, given an ignition source, the mixture will rapidly combine (if it's dense enough). It explodes. Now, there is some level of combination that occurs in the un-ignited mixture, but the rate of this is low enough that the mixture doesn't spontaneously ignite. It may be (I don't know), that the formed water molecules are dissociated as rapidly as they form. In the case of cold nuclear fusion, the energy necessary to *catalyze* it need not be anything more than the kind of energy that would cause some level of combination of hydrogen and oxygen in a mixture as described. I don't think that ZPE is relevant to cold fusion. But since we don't know, with any level of certainty, what is going on in the CF cells, beyond some clear evidence that there is deuterium fusion (not necessarily d-d fusion!) taking place, nothing can really be ruled out. As I've described, one potential candidate for a fusion reaction is known, the Tetrahedral Symmetric Condensate, which is formed when four deuterons and their electrons (or, I think, three of the electrons works) form a tetrahedral configuration of a certain size. My sense of this is that this is a compressed form, a cluster like this would not stay together if unconstrained. To compress it takes energy, but since we are only talking about two molecules, the energy involved would be the relative velocities of the two molecules in collision. This kind of energy is available thermally. It is nowhere near enough to cause fusion, itself, it simply puts the nuclei and electrons into a position where they can collapse to a Bose-Einstein condensate and then, according to Takahashi's calculations, spntaneously fuse 100% to Be-8.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Abd ul-Rahman Lomax wrote: Extremely high pressure is achieved by electrolysis at the surface of an electrode, in a microscopic domain. According to some sources it is 10E47 atmospheres (Mizuno, p. 101). And what does this mean? Oops, that was described as a misinterpretation. The correct figure, from p. 103, is 10E23. Not quite as high but still -- if this is correct -- on the scale of a few atoms a Pd surface undergoing electrolysis can produce more pressure than the core of a neutron star. That would be an unexplored domain. There is no significant volume of cavities in a successful cold fusion cathode. The entire palladium lattice can be considered a collection of cavities. No, it is a lattice. A lattice is not the same as a cavity. A cavity is a break in the lattice, in which D2 molecules can form. Deuterons cannot come together to form molecules in a lattice. They might be able to come together to form helium atoms. That's the subject of debate, but no one asserts they can form D2. Cavities rapidly expand to form cracks, which can readily be seen on the surface of a submerged cathode. Lines of bubbles form on them and then detach and float up. These are much larger bubbles than the ones formed at the surface by deuterium coming out of solution with the lattice. - Jed
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 03:40 PM 5/12/2010, Jed Rothwell wrote: Abd ul-Rahman Lomax wrote: We know that the reaction rate increases with temperature. I suspect that the energy required -- there must be energy required, but energy is not the only ingredient -- is generally supplied by ordinary heat. I assume this reaction refers to heating from the hydrogen or deuterium that emerges from highly loaded palladium in air. No. I'm referring to the nuclear reaction. This reaction is what was used to make cigarette lighters with palladium hydride, in the 19th century. When you expose highly loaded palladium to air, the hydrogen emerges and the surface of the palladium catalyzes recombination with oxygen in air, so the cathode heats to incandescence, and can be used to light a cigarette. You have to have quite a lot of palladium to pull this off -- roughly an ounce I suppose (32 g) which is much more than most cold fusion cathodes. This never happens with heat after death because, as I mentioned, there is no air or oxygen in the headspace, only water vapor. It can only recombine in the atmosphere outside the cell. That's correct. Ikegami's group at the Nat. Inst. for (plasma) Fusion Science did extensive testing of this. They got highly loaded cathodes to self-heat, and they also heated them in air with blow torches and other methods, trying to make them explode the way Fleischmann's cathode did. They never got it do anything other than the standard cigarette lighter effect in air. Cool. Sounds like fun. The gas will always come out of a saturated cathode. When the cathode is submerged in water, you can see fine gas bubbles appear on the cathode surface, looking like carbonation bubbles coming out of solution in soda-pop at 1 atm (except they are smaller). The only way to maintain high loading is to keep doing electrolysis, and then you have a steady stream of gas coming out of the cathode while other gas goes in. The gas emerges in two phases, corresponding to the alpha and beta phase loading, which refers to the sites in the lattice that are loaded. Assume it starts highly loaded to the beta phase. Lots of bubbles form and emerge. The rate gradually slows down -- which is readily apparent to the naked eye. It slows down and down until it reaches the alpha phase, and then suddenly starts up again, until the alpha phase deuterons are mostly gone from the cathode. It is very difficult to purge all of the deuterons from the lattice. There is always some left. Heating up the cathode will increase the rate at which it de-loads, but it always de-loads on its own, and as far as anyone knows (per Ikegami) there is no way to speed up the rate of degassing so much that the metal explodes. Of course there is not enough energy in the gas to make the metal melt or vaporize, the way some cathodes have done! Right. (Note that it can take far less energy to make metal fracture or explode than it takes to melt it. Fracturing is function of the speed of energy release, and how confined the site of the release is.) As Fleischmann pointed out, the rate of de-loading is far too slow to account for cold fusion heat, and the total energy release is far too small. It was ~1,700 times too small when Morrison and others suggested this hypothesis back in 1990, and it must be millions of times too small by now. Shanahan and his editors must be innumerate. Nice word, that.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 03:50 PM 5/12/2010, Jones Beene wrote: -Original Message- From: Abd ul-Rahman Lomax Shanahan's cigarette-lighter explanation does not factor for the lack of excess heat when doing this with hydrogen. Well, at the risk of defending a repulsive argument, many experiments do show excess heat with hydrogen, including some of the Arata experiments. [...] No. Arata does not show excess heat with hydrogen. He shows the normal heat of formation of palladium hydride (same as with deuterium). He uses hydrogen as a control, showing the extra heat with deuterium. There are other experiments of other kinds that may show excess heat with hydrogen. My sense is that there are many kinds of possible reactions, under different conditions. Not just one. However, some low level of effect may be possible with hydrogen due to the presence of deuterium in it. That's a different story. However, with H2 the heat generated is most likely NOT nuclear. With D2 some of the heat may not be nuclear, either. Excess heat with H2 or D2 could be related to Mills shrinkage or to ZPE. With D2, it is even possible that actual fusion may be a time reversed QM balancing act and only replaces energy lost in other methods. The formation of palladium deuteride is exothermic. Thus, like most evaporations, the escape of deuterium from the lattice must cool it. Slowly. In the cigarette lighter, this effect is swamped by recombination with oxygen from the air.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 03:40 PM 5/12/2010, Jed Rothwell wrote: Abd ul-Rahman Lomax wrote: We know that the reaction rate increases with temperature. I suspect that the energy required -- there must be energy required, but energy is not the only ingredient -- is generally supplied by ordinary heat. I assume this reaction refers to heating from the hydrogen or deuterium that emerges from highly loaded palladium in air. No. I'm referring to the nuclear reaction. This reaction is what was used to make cigarette lighters with palladium hydride, in the 19th century. When you expose highly loaded palladium to air, the hydrogen emerges and the surface of the palladium catalyzes recombination with oxygen in air, so the cathode heats to incandescence, and can be used to light a cigarette. You have to have quite a lot of palladium to pull this off -- roughly an ounce I suppose (32 g) which is much more than most cold fusion cathodes. This never happens with heat after death because, as I mentioned, there is no air or oxygen in the headspace, only water vapor. It can only recombine in the atmosphere outside the cell. That's correct. Ikegami's group at the Nat. Inst. for (plasma) Fusion Science did extensive testing of this. They got highly loaded cathodes to self-heat, and they also heated them in air with blow torches and other methods, trying to make them explode the way Fleischmann's cathode did. They never got it do anything other than the standard cigarette lighter effect in air. Cool. Sounds like fun. The gas will always come out of a saturated cathode. When the cathode is submerged in water, you can see fine gas bubbles appear on the cathode surface, looking like carbonation bubbles coming out of solution in soda-pop at 1 atm (except they are smaller). The only way to maintain high loading is to keep doing electrolysis, and then you have a steady stream of gas coming out of the cathode while other gas goes in. The gas emerges in two phases, corresponding to the alpha and beta phase loading, which refers to the sites in the lattice that are loaded. Assume it starts highly loaded to the beta phase. Lots of bubbles form and emerge. The rate gradually slows down -- which is readily apparent to the naked eye. It slows down and down until it reaches the alpha phase, and then suddenly starts up again, until the alpha phase deuterons are mostly gone from the cathode. It is very difficult to purge all of the deuterons from the lattice. There is always some left. Heating up the cathode will increase the rate at which it de-loads, but it always de-loads on its own, and as far as anyone knows (per Ikegami) there is no way to speed up the rate of degassing so much that the metal explodes. Of course there is not enough energy in the gas to make the metal melt or vaporize, the way some cathodes have done! Right. (Note that it can take far less energy to make metal fracture or explode than it takes to melt it. Fracturing is function of the speed of energy release, and how confined the site of the release is.) As Fleischmann pointed out, the rate of de-loading is far too slow to account for cold fusion heat, and the total energy release is far too small. It was ~1,700 times too small when Morrison and others suggested this hypothesis back in 1990, and it must be millions of times too small by now. Shanahan and his editors must be innumerate. Nice word, that.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Abd ul-Rahman Lomax wrote: A stoichiometric mixture of hydrogen and oxygen gas at room temperature the mixture is stable, the gases do not combine. However, given an ignition source, the mixture will rapidly combine (if it's dense enough). It explodes. As I said, with ordinary palladium hydride exposed to air, the palladium acts as a catalyst and causes the emerging gas to ignite. It does not explode; it glows. To stop the reaction with the cigarette lighter, they simply covered it up. I assume the gas kept leaking out, under the cover, but there was no air so it did not ignite. Not sure I would want to carry around something like that, but people used to carry Lucifer strike-anywhere wooden matches that would occasionally rub together and ignite. That's not what you want rubbing together in your pants pocket. I think they formed the hydride in those cigarette lighters by exposing them to high pressure hydrogen gas, which would only achieve low loading. I have not seen a photo or detailed description of a German palladium cigarette lighters. I did a very rough estimate of the amount of palladium needed simply by making some assumptions: they wanted a lighter that had roughly as many lights as a book of matches; loading was low from gas only (maybe 30%); they wanted it small enough to fit in a pocket; and they did not want it to cost too much. Palladium has always been expensive. Anyway, I figure that takes something like 30 to 100 g of palladium. - Jed
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
-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 :) 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. 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. The inversion temperature of hydrogen is lower than ambient as the table below demonstrates: GasInversion Temp Deg K Helium 51 Hydrogen (H2) 205 Neon 242 Nitrogen (N2) 621 Argon 723 Krypton727 Oxygen (O2) 893 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) Jones
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
-Original Message- From: Abd ul-Rahman Lomax Shanahan's cigarette-lighter explanation does not factor for the lack of excess heat when doing this with hydrogen. Well, at the risk of defending a repulsive argument, many experiments do show excess heat with hydrogen, including some of the Arata experiments. OK - I should have said including Kitamura's version of the Arata experiments http://www.lenr-canr.org/acrobat/KitamuraAanomalouse.pdf At first read, I thought this was going to be an Arata replication, and little else.. Wrong. There is excess energy with both hydrogen and deuterium The important implications are in Table 1, the line entry begins with [H-PZ4#1]. In fact he gets more energy from hydrogen than from a few of the deuterium runs and ostensibly without LENR, without nuclear reactions, and without combustion. I suspect that it is even without Mills hydrino, but that is unclear. Although, I should add that the authors do not state the glaring implication of this in so many words. I do not think they could have missed this. That is why I called it an alternative energy paper instead of an LENR paper. The active modality for the first phase is surely ZPE. It also bolsters the view that deuterium gets most of its total gain in the second phase, but hydrogen gets none there. The second phase is nuclear. Jones
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
I electrolyzed a 1 oz. palladium wafer in D2O and it sure looked as if it made its own cavities -- it puffed up like a pillow with bubble like protrusions and it seemed it was hollow inside. -Original Message- From: Jed Rothwell [mailto:jedrothw...@gmail.com] Sent: Wednesday, May 12, 2010 12:06 PM To: vortex-L@eskimo.com Subject: RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis Abd ul-Rahman Lomax wrote: No. This pressure is created by both hydrogen and deuterium. The high pressure mentioned is created by electrolysis with hydrogen or deuterium generated at the surface of a hollow palladium rod...
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Hoyt A. Stearns Jr wrote: I electrolyzed a 1 oz. palladium wafer in D2O and it sure looked as if it made its own cavities -- it puffed up like a pillow with bubble like protrusions and it seemed it was hollow inside. Yes, that often happens, but Pd that does that will never produce the cold fusion effect. - Jed
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
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.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 06:04 PM 5/12/2010, Hoyt A. Stearns Jr wrote: I electrolyzed a 1 oz. palladium wafer in D2O and it sure looked as if it made its own cavities -- it puffed up like a pillow with bubble like protrusions and it seemed it was hollow inside. My condolences.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
-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. 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. 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. 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? 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 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. Jones
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
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 PF 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. Michel 2010/5/8 Abd ul-Rahman Lomax a...@lomaxdesign.com: 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.
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Michel, Can you cite the reference for this kind of bursting tube, due to internal pressurization, having being actually performed? I have heard this before but not been able to verify it. The reason that I think it would be unlikely is that it presents an easy avenue for demonstrating gain - via direct conversion of oscillating pressure into electricity (via a magnet/coil attached to a bellows tube). I've mentioned this before. Someone with a little inventive curiosity would have done it by now if enormous pressure results from applied current only, and should have at least reported the results (if null). A high internal pressurization would be easily oscillated by AC in a palladium bellows tube of course, since it would have to be reversible - even if not gainful. A putative gain would be the expected result of the added heat from fusion. Direct conversion alone (of oscillating pressure into electricity) would be an important advance in getting LENR to market on a small scale without the need for lossy systems like TEG (5-7% efficiency) or steam (hard to scale down). Well - not just an important advance but *huge*. If I had not mentioned this possibility several times in past years, to no avail - then I might be inclined to rush off to the patent office, but methinks the bursting tube is a myth. Jones -Original Message- From: Michel Jullian 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 PF 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. Michel 2010/5/8 Abd ul-Rahman Lomax a...@lomaxdesign.com: 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.
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
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 PF 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 a...@lomaxdesign.com: 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: KM go on to discuss a specific type of calorimetric result obtained by FP commonly known as eheat-after-deathf (HAD). In the HAD experiment, a FP 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 KMfs 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, KM report that the energy detected was gfar beyond the quantity of possible stored chemical energyh. No recalibration was reported in the 1993 FP 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
RE: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Jones Beene wrote: Can you cite the reference for this kind of bursting tube, due to internal pressurization, having being actually performed? See McKubre's replication of the Arata experiment. Regarding the pressure from electrolysis, it far exceeds anything that can be accomplished with mechanical means such as pumps. See Mizuno's book, chapter 7. I think the fact that gas loading works means that high pressure is not required for a cold fusion effect. I guess it means that high pressure is not needed for high loading; there are other ways to achieve that, rather than brute force. - Jed
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
[quote] 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.[/quote] Once power is removed disassociation drops off dramatically. Only atoms can translate freely in the lattice so the sudden majority of diatoms Will also oppose unloading. If these atoms were fractional then the diatoms are also fractional and as such store this transitional energy. Fran
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
Roarty, Francis X wrote: [quote] At the point where electrical contact was broken, the cell gas was very nearly a stoichiometric mixture of hydrogen and oxygen. . . . I think that is a quote from Shanahan but anyway it is completely wrong, as I said. The HAD cells described by Fleischmann and Pons were all open, and the oxygen was long gone when the hydrogen emerges. The only thing in the headspace after the boil-off is boiled off water vapor. It drives out the oxygen, and air. For a closed cell, there is a little orphaned oxygen in the headspace -- obviously just enough to recombine with the emerging hydrogen. But there is hardly any gas. FP used relatively large cathodes that held enough hydrogen to produce ~600 J maximum, which -- as Fleischmann pointed out -- would last a few seconds at this power level. Most cathodes nowadays are smaller. Shanahan either did not realize that his hypothesis is preposterous and innumerate, or he realized it and he plans to publish it anyway. Either way he is no scientist. Shanahan's arguments are pretty much the same as Morrison's circa 1990. There has been no change in the skeptical camp; no learning, and argument worth discussing. These people understand nothing about conventional electrochemistry or cold fusion. In a sane world, Shanahan would not be able to publish this rubbish anywhere but his own blog. - Jed
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 08:29 PM 5/10/2010, Jed Rothwell wrote: Shanahan's arguments are pretty much the same as Morrison's circa 1990. There has been no change in the skeptical camp; no learning, and argument worth discussing. These people understand nothing about conventional electrochemistry or cold fusion. In a sane world, Shanahan would not be able to publish this rubbish anywhere but his own blog. Well, we don't know that it's been accepted, and it might not be. It's pretty bad. It does appear, though, that Shanahan may have been funded to write this by the DoE. Now, there's a project for Krivit. What happened? Why Shanahan?
[Vo]:Shanahan is proposing the cigarette lighter hypothesis
Shanahan is proposing the oldest and least credible prosaic explanation for cold fusion: 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 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. With Fleischmann's open cell, that is completely wrong, as Abd noted. The cell head space is filled with water vapor only, with no significant amount of 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. . . . This is nonsense. See Fleischmann's response to Morrison about this, p. 10 and 11: http://lenr-canr.org/acrobat/Fleischmanreplytothe.pdf To summarize, the amount of hydrogen in cathode is ~1,700 times too small to explain the heat after death, and heat after death occurs over 3 to 6 hours, whereas it would take 29 days for the hydrogen to emerge from the cathode. Quote: In the first place . . . the vapourisation of the D2O alone would have required ~1.1MJ of energy whereas the combustion of all the D in the palladium would at most have produced ~ 650J (assuming that the D/Pd ratio had reached ~1 in the cathode), a discrepancy of a factor of ~ 1700. In the second place, the timescale of the explanation is impossible: the diffusional relaxation time is ~ 29 days whereas the phenomenon took at most ~ 6 hours . . . Thirdly, Kreysa et al [8] confused the notion of power (Watts) with that of energy (Joules) which is again an error which has been promulgated by critics seeking Chemical Explanations of Cold Fusion. Thus Douglas Morrison reiterates the notion of heat flow, no doubt in order to seek an explanation of the high levels of excess enthalpy during Stage 4 of the experiments. We observe that at a heat flow of 144.5W (corresponding to the rate of excess enthalpy generation in the experiment discussed in our paper [2] the total combustion of all the D in the cathode would be completed in ~ 4.5s, not the 600s of the duration of this stage. Needless to say, the D in the lattice could not reach the surface in that time (the diffusional relaxation time is ~ 10^5s) while the rate of diffusion of oxygen through the boundary layer could lead at most to a rate of generation of excess enthalpy of ~ 5mW. . . . - Jed
Re: [Vo]:Shanahan is proposing the cigarette lighter hypothesis
At 08:42 AM 5/8/2010, Jed Rothwell wrote: Shanahan is proposing the oldest and least credible prosaic explanation for cold fusion: This is nonsense. See Fleischmann's response to Morrison about this, p. 10 and 11: http://lenr-canr.org/acrobat/Fleischmanreplytothe.pdfhttp://lenr-canr.org/acrobat/Fleischmanreplytothe.pdf To summarize, the amount of hydrogen in cathode is ~1,700 times too small to explain the heat after death, and heat after death occurs over 3 to 6 hours, whereas it would take 29 days for the hydrogen to emerge from the cathode. If Fleischmann had stopped there, it would not have been enough. The cigarette lighter effect is based on the fact that a loaded palladium rod, *exposed to air* will heat up from catalyzed combustion at the surface. Morrison correctly points out that such a rod can burn something it is placed upon. As the rod heats up, evolution of gas increases, so it would burn up the deuterium, probably much faster than 29 days. Or not, if Fleischmann considered the operating temperature. So he points out that if the combustion was taking place at the measured enthalpy, it would burn up the entire store of deuterium in 4.5 seconds. Further, Fleischmann points out that (as Morrison had acknowledged) the amount of palladium present was quite small compared to what was used in the cigarette lighters. To heat up the rod, oxygen must reach a rod which is supposedly outgassing deuterium at the same time. My own analysis of this suggestion of recombination through small oxygen bubbles was that the recombination would create water or water vapor, which would push oxygen away from the palladium; an oxygen bubble alighting on the palladium surface would, I'd expect, recombine, but this would be self-limiting. For an explosion to occur, it would have to be pre-mixed with deuterium. In the cells taken to boiling, water vapor is rapidly being generated, and that is, as Fleischmann noted, the principle outgas at that point. Small amounts of oxygen would be generated by the remaining electrolysis, but would be rapidly swept out of the cell (with generated deuterium as well). The concentration of deuterium and oxygen in the head gas would be relatively low, not an explosive mixture, my estimation. Morrison wanted to see closed cells, and that's understandable, but, as Fleischmann noted, irrelevant. Closed cells require recombination and are dangerous, because the gas inside can, indeed, be an explosive mixture. This is what occurred to me, reading Morrison and Fleischmann's response. The defenders of traditional fusion theory from the chemistry intruders took the position that it was not necessary for them to do any real work, any more. After all, some of them had tried in 1989, and failed to confirm, and they blamed their failures on Pons and Fleischmann. So, they thought, the burden of proof is on FP. And then they sat in judgment on every new experiment and piece of evidence, simply thinking up hypothetical explanations that might leave the phenomenon in the chemical realm, and it didn't matter if the explanation was preposterous or did not match the experimental conditions. They did not apply, at all, the same standards for cogency of theory, to their own theories, that they were demanding of Fleischmann. A basic principle was forgotten: preponderance of the evidence. They actually had *no* evidence that nuclear reactions were *impossible*, under the FP conditions; all the proofs depended upon assumptions that the reaction being considered was known, typically d-d fusion, and that it would therefore have known characteristics for that reaction. What accumulated over the years was evidence that there was an anomaly here. For decisions involving funding and allocation of research resources, the issue should not be proof, but rather the simplest explanations of what's being observed, and such explanations do not require comprehensive theoretical understanding. Fleischmann was quite careful to point out that he wasn't claiming nuclear fusion, and that wasn't merely a device to avoid approbation. He was claiming excess heat, at levels that were unexplainable by chemistry. That does leave some kind of nuclear reaction as likely. But excess heat is the fact being exposed, and there was and is no special burden on Fleischmann to explain it. This is what a real debunker would do, and it's what was done with N-rays and polywater. Fleischmann was claiming high reproducibility for heat after death, and a dramatic effect, so dramatic that the accuracy of the calorimetry, he claimed, became moot. A debunker would have reproduced the experiment, generating the kind of heat shown. And then would have attempted to show that this heat was coming from recombination, or that the calorimetry was in error by correcting the errors, or the like. But, of course, this would be risky politically. What if they couldn't find
