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288) CMNS researchers discuss chemistry


Ludwik Kowalski; 3/29/2006
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043



1) This evening I posted this message: “Dear Francesco [Celani]: Suppose a Pt cathode produces the COP=1.00 while a W cathode produces the COP=1.30, in a Muzino type cell, at the same voltage. Would this not be a sufficient proof that the excess energy is real when a W cathode is used? I think that this would be a valid argument against your suspicion that the inductance of the capacitor might be causing an error in the COP measurements.

It seems that the chemical origin of excess heat, in Mizuno type cells, becomes the central issue. I hope somebody can summarize what has been posted about that topic so far. The K2CO3 disappears when the salt is dissolved in water; it is replaced by + and - ions. An experiment begins when the electrolyte temperature reaches a constant temperature ~100 C. The K2CO3 no longer exists; it cannot be a fuel. If there is a fuel then it consists of ions. Which reactions are dominate and how many joules of energy can these reactions possibly deliver, per gram of the initially dissolved potassium carbonate? Significance of Paris-1 and Colorado-2 results depends on how is this question answered by chemists.”

Fauvarque is an electrochemist and I expect that he will write an appendix to our anticipated publication (see unit #271). Pierre, who works in his lab, took back a bottle with the electrolyte we used in the Colorado-2 experiment. Let me gather messages that people wrote about the possibility of chemical origin of excess heat. Even Jed Rothwell admitted that the excess heat in Mizuno type experiments (plasma electrolysis) might have chemical origin. Several days ago he wrote: “. . . but this open cell glow discharge is quite different, and one cannot as easily rule out conventional explanations.” In other words it is possible that some chemical reactions are responsible for what was interpreted, prematurely, (by Mizuno, Naudin, Fauvarque and Iorio as liberation of nuclear energy. High voltage plasma electrolysis is very different from the low voltage electrolysis used by most CMNS researchers.

2) On Michel Jullian asked:

“The mass loss corresponding to this large quantity of liberated O2 and H2 is attributed wrongly to evaporation isn't it? Has anybody worked out how significant this effect is, and which way (under or overestimation?) it affects output energy estimation? Responses would be appreciated.”
P.S. I forgot to include an earlier important observation made by X1. He wrote: I suggest that exothermic chemical reactions involving the electrolyte salt or its ions could be releasing heat in glow discharge experiments, on the bases that it is the only possible chemical reactant present in sufficient quantity (tens of grams) to explain observed excess heat, and that the color of the plasma "flame" depends on the nature of salt used.

3) The immediate reply from X2 was:
“First, the report of 80 times Faradaic efficiency for dissociation of H2O comes, I believe, only from Mizuno et al. The literature on Contact Glow Discharge Electrolysis (i.e. a section in the big multi-volume electrochemistry compendium edited by Bockris) says that up to 8 times has been reported. In our own experiments ( http://www.earthtech.org/experiments/Inc-W/2ndtry/run6.html ) we have observed only 2.5 times the Faradaic efficiency. I will use our observation in the calculations below.

Ignoring the fact that some mass is lost due to dissociation of H2O into H2 and O2 results in an underestimation of the heat energy produced by the cell. The error is around 1% for the cells I have run. Consider a cell running at 300 volts and 1 amp. Pin = 300 watts. The fraction of this power that goes into electrolysis is 1.48 volts (the thermoneutral dissociation potential) * 1 amp * 2.5, the latter factor from our own observations. So Pdissoc = 3.7 watts. That leaves 296.3 watts for boiling the water. Assuming no excess heat, if you measured the mass loss correctly and came up with 296.3 watts of heat power, then compared it to the input power of 300 watts, you would conclude that the process is only 98.8% efficient.

If the Faradaic efficiency did go as high as 80, things would be very different. Then Pdissoc would be 118.4 watts, leaving only 181.6 watts for boiling. Using mass-loss calorimetry the process would appear only 60.5% efficient.”
Let my mention that, according to Mizuno, the amount of H2 produced in plasma electrolysis is often much higher than what is expected by the Faraday’s law of electrolysis, for example, 80 times higher. Hydrogen is fuel and its excess must be counted as excess energy.

4) Ludwik Kowalski wrote:
Here is a little rough calculation worth sharing. In our typical 5 minutes test (input electric energy = 400 W * 0.0833 h = 33.3 Wh) the amount of excess heat is about 6.5 w*h (if the COP=1.2). This translates into 1.46*10^23 eV. The estimated mass of tungsten lost during the same time is about 0.5 grams. That translates into 1.63*10^21 atoms. Thus the excess heat per atom of tungsten seems to be close to 100 eV. That is about two orders of magnitude larger than what I associate with common fuels.

What happens to our spent tungsten? Part of it probably melts and then solidifies in the electrolyte. If so then pure tungsten must be found in the cell (in the deposits we see at the bottom, after some waiting). Suppose that 50 % melts and the rest combines with other atoms or ions. Not being a chemist I might be allowed to postulate that all possible chemical reactions are equivalent to one representative exothermic reaction. Suppose that our excess heat is due to that reaction. What would the Q value of the representative reaction be? It would be 178 eV. That is a lot !

The Q value would be two times larger if only 25 % of tungsten was used in chemical reactions while the rest was simply melted. This shows that Q increases when more tungsten is melted. At the other extreme the Q value would be 89 eV if nothing was melted (all consumed in exothermic chemical reactions). Even this number is unusually high. Did I make a wrong assumption, or a numerical mistake, somewhere? . . . Tungsten electrodes should be weighted before and after individual tests. Fortunately, a possible 50% error (in the estimated mass lost) would still support the conclusion -- too much energy per atom of tungsten.

I wish our real chemists, such as R. O. or Mike McKubre. and his team, were performing Mizuno type experiments. Confirmations made by them would be much more significant than by amateurs, like myself. I would be in a much better position to argue about nuclear particles and nuclear detectors than I am in arguing about chemical reactions. That is because I was trained as a nuclear physicist.

5) M. Jullian asked:
“ Dissolving solid K2CO3 in water is an endothermic reaction isn't it? (absorbs heat)

6) X2 responded:
My old 1948 "rubber bible" says it is exothermic. ... 6.5 kcal is released when 1 mole of K2CO3 is dissolved in 400 moles of water.

7) X6 wrote:
Whilst it is perfectly true that atomic hydrogen can give off lots of heat on reforming H2 (precisely 4.5 eV per molecule) this same energy is required to form it in the first place. Consequently any such reaction would produce no net energy.

8) M. Jullian wrote:
[Somebody asked] “As for hypothetical reactions with the electrolyte, what could they be?” I don't know. Whatever the high temperatures and the active species in the plasma can do to it. If you bring material past a plasma, like the electrolyte is brought past the plasma sheath around the cathode in a Mizuno cell, it will be processed happily and non reversibly. This phenomenon is called plasma processing, used to e.g. oxidize the material to be processed, exhaust gases, plastic film etc.

In any case it is beyond doubt that the electrolyte intervenes in the plasma reactions, have a look at all the beautiful different plasma colors one can get by trying different electrolytes

<:http://www.geocities.com/spfaile/plasma/Plasma.html>

Note that the above excellent experiment report (couldn't find authors names in LENR-CANR library) also confirms that initiating a plasma is a matter of electric field, not of voltage. They managed to generate plasmas with only 50V, using fine enough cathodes (thin wires, or sword shaped sharp plates). It was pointed out by Jed recently that Mizuno obtained the right kind of plasma at lower voltages than the 200-400V used in recent experiments, I submit that is because the lower radii of curvature of Mizuno's thin W plates' edges induced the required electric field for a lower voltage than with the relatively thick welding rods used in recent experiments.

Back to chemistry, my suspicion as a non-chemist familiar with only the _physical_ aspects of plasmas, is that the field effect ionization effect giving birth to the plasma, combined with the >3500°C temperature of melting tungsten, may be able to break the electrolyte into more active potassium species able to react exothermically with oxygen and/or water with a positive net energy balance.

But there is no need to know exactly what such reactions could be for now or if they are plausible, let's first check the hypothesis experimentally as discussed with Ludwik: is the electrolyte consumed at all in a _successful_ run? If not we can happily eliminate the electrolyte as a possible chemical fuel. Will remain your solvent hypothesis :)

9) Mike McKubre responded:
I have seen this thread bouncing around and hoped that there was a subtle meaning that I was not getting.  I may have missed the point but let me address the obvious. Potassium in the electrolyte is in the oxidized state  (K+) while oxygen is in its reduced state (O--).  If by some means potassium becomes reduced to K and oxygen oxidized to O2, and they somehow are allowed to react, the results is oxidized K and reduced O --- exactly as we started with.  Thermodynamics has the wonderful virtue of being promiscuous --- it is completely insensitive to pathway.  If the initial and final states are the same --- then no energy is absorbed or released. I am hoping that this is not the theme of this thread.  If it is, then let this answer be an end.

10) John Coviello wrote:
. . . Of course I agree with your obvious assessment but would like to include consideration of the possibility that the cell ingredients would undergo reactions that would result in different final compounds. For example, the plasma might stimulate the reaction

K2CO3 + H2O -> 2KOH + CO2

The CO2 would leave the cell as a gas. Adding up the heats of formation and solution of these compounds, this looks like a slightly endothermic reaction .... about 25 kcal per mole of K2CO3 converted. Compare this to 275 kcal/mole for the formation of K2CO3 from its elements. I suppose it is safe to assume that any "rearranging" reaction, i.e. one that leaves the metals oxidized and the oxidizers reduced will not have a great energy production or consumption.

Regarding the W cathode, I have examined this reaction:

W + 4H2O -> H2WO4 + 3H2

for which we presented evidence in this report:

http://www.earthtech.org/experiments/Inc-W/2ndtry/run6.html

and found it to be almost non-thermic. The energy required to dissociate 4 moles of H2O is about 274 kcal and the energy released when the H2WO4 is formed is 280 kcal. Plus, the total mass of W consumed in a typical run is quite small. In conclusion, it appears that there is no possibility that chemical energy could be the source of the apparent excess heat from the W-K2CO3-H2O system.

11) My own reply to Mike McKubre
Thanks, your comments, concerning the ongoing Mizuno type experiments, are highly desirable. Most of us performing, or planing to perform, these experiments, are not even chemists. The only way to compensate for this handicap would be to receive critical input, and help, from real electrochemists, like yourself, E. Storms, R. Oriani, S. Szpak, etc. Please help us to confirm reality of excess heat from plasma electrolysis cells.

12) Michael McKubre responding to X2:
The first reaction is endothermic as you state. In fact it is the bane of electrochemists running KOH electrolytes as it spontaneously reacts with CO2 from the air to form carbonate. I did not look up the numbers but your assessment of the W + H2O reaction seems right, and I agree with your general conclusion. For non chemists the heats of formation of all these materials are easily looked up and added up. You can answer these questions yourself, easily. In fact, thermodynamics is trivial - like nuclear physics. A few simple rules and a lookup table.

13) M. Jullian wrote:
Thanks M.McKubre Assuming the first reaction proposed by X2 occurs in a significant manner, we will have a significant mass loss via CO2 won't we? Since this mass loss will be wrongly attributed to evaporation, we will have to make sure each gram lost this way takes away as much heat, or more heat, than a gram of evaporated water. Haven't time to do the computation right now, maybe someone will beat me to it.

Ok so X2’s first reaction K2CO3 + H2O + heat -> 2KOH + CO2 (well
spotted X2 BTW, you made me look less foolish with my speculation of electrolyte chemistry ;) is endothermic, requiring about 25 kcal per mole of K2CO3. That's 25 kcal per mole of escaped CO2, i.e.25000*4.186 = 105000 J for 12+16*2 = 44 g of lost mass.

This works out as 2380 J/g, which seems to be slightly higher than H2O evaporation heat 2260J/g, but it would be worth double-checking the exact figure for the reaction heat (anywhere online where the formation heats you mentioned can be looked up?)

Pending confirmation of the figure, as far as this reaction is concerned we can happily keep counting all lost mass as evaporated water as this amounts to underestimating the output energy and the COP (it would have been bad if it had overestimated it!). But this result needed to be justified didn't it?

Since the outcome of the calculation had no particular reason to be favorable, I still think it would be a good idea to analyze the electrolyte after a successful run, to check that no other unexpected chemical reaction involving the electrolyte has occurred to mess up our results one way or another.

Note such analysis can be tricky though, as for example the above reaction could not be detected if the electrolyte was left to cool down in ambient air long enough: as Michael taught us the KOH would reabsorb the CO2 from the atmosphere and return to its initial K2CO3 composition. . . . .

Er... group, how about this new silly speculation about chemical heat from the electrolyte? (very basic this time, would probably also escape post-run chemical analysis depending on how the electrolyte is collected) Dissolving solid K2CO3 in water is an endothermic reaction isn't it? (absorbs heat)

So some electrolyte drying up on the cell walls after some splashing, the reverse reaction, would generate heat wouldn't it? Real heat this time, not an artifact of the measurement method. This effect could be far from negligible but it's getting quite late over here so I'll go to bed and ask a charitable soul to do the homework this time: - How many joules of excess heat per gram of dried-up salt on the cell walls? - How many grams end up this way, experimenters? - Couldn't this also affect P&F cells to a lesser extent?

14) X2 wrote:
You asked “Dissolving solid K2CO3 in water is an endothermic reaction isn't it?” My old 1948 "rubber bible" says it is exothermic. ... 6.5 kcal is released when 1 mole of K2CO3 is dissolved in 400 moles of water.

15) M. Jullian wrote:
Thanks X2, my mistake, and a very good news: again the outcome is favorable! I submit that a thorough paper on the subject would mention all the chemical reactions that have been thought of, and why they can only underestimate the COP or make it much lower than measured: - Faraday and thermal dissociation of water (underestimates COP) - reactions involving the tungsten (100 times not enough mass to explain the COP) - KOH and CO2 formation from K2CO3 (underestimates COP) - dried up electrolyte (underestimates COP) - ???

Other reactions with the electrolyte might be possible, is there any way at all to predict all possible chemical reactions that can occur in a given system? Analyzing the electrolyte before and after a successful run may reveal reactions we might have missed, and hopefully will give more consistence to the far from obvious claim that excess heat is not chemical in Mizuno experiments.

Such chemical analysis would also quantify the lost droplets/mist phenomenon as has been discussed: just saying it is negligible because some filtering means is used, or because no mist is seen at the surface of the covered cell on the photo (if we "see" water vapor in a covered cell it is not water vapor but means there is some nebulization occurring) may not be sufficient proof.

In any case the dream I had of a simple cell demonstrating reality of cold fusion to any skeptic in a few minutes is getting more remote every time a new artifact is found and ruled out. There is stronger and stronger hope that a simple cell (insulated and covered Fauvarque) could be sufficient proof, but it will also take a lot of reading!!! Unless of course it produces significant excess heat for days and months, but we are not there yet.

16) M. Jullian also responding to what I wrote in (1):
X has just responded (he may be right about ceramic being better than plastic). My own rantings are not about the reality of electric power measurements either (like all or most people here I now have a high level of confidence that they are valid) but about their indisputability from the point of view of a reviewer.

About the COP depending on the cathode material, it would be an interesting result in itself and would certainly increase the level of confidence of the measurement, but would not yield indisputability, why should it?

As I have been stressing, no electrical measurement made in a high EMI environment will ever be indisputable. Please let us never forget that skeptic or even neutral reviewers will need EXTRAORDINARY PROOF for our extraordinary claims. Any reason to doubt, especially a good one such as abundant EMI, will be jumped upon to dismiss the whole stuff, and HAS BEEN actually. Simplicity, transparency, indisputability, should be maximized if we want to stop being considered as self deluding wishful thinkers, agreed?

Talking about EMI, it occurred to me that it might be a good idea if the anode itself, which could easily be made to surround entirely the discharge area (including top and bottom), could be used as a Faraday cage. This would mean connecting it to ground and having the cathode float below ground, Scott Francesco and all what is your opinion on such an arrangement?

P.S. About the chemical effects discussion, I would find it outrageous if the real chemists here did not get more actively involved in this, leaving amateurs such as Scott or myself do their homework as they have effectively done up to now:

1/ What are the possible reactions apart from the ones discovered (by chance more or less) by Scott and by myself?

2/ What can happen or not happen chemically speaking in the plasma at temperatures ranging at least from 100°C (outside of plasma sheath) to melting tungsten temperature (inside), and at high electric fields? Any missing parameters to answer this question? Why does the plasma color depend on the electrolyte? Real chemists please ENLIGHTEN US (if you have any clue that is).

17) Peter Gluck, who is a chemist, wrote:
. . . From the point of view of engineering, we need first of all, complete ENERGY and MATERIAL BALANCES. Has somebody a complete material balance for all the phases? Otherwise unknown data plus hidden parameters put any positive results in doubt; there more unknowns than equations of balance.
 
SOLID- e.g. W electrode at start  + ? --> electrode at finish + solid precipitate (composition?) I have told that my guess is that W is destroyed by spark erosion (as does Zr in the Cincinnati Cell) i.e. it does not react. This need a bit of analytical chemistry, however a simple balance: initial weight of the electrode vs final weight of electrode plus the washed and dried precipitate can tell something about the fate of W- is it broken, is it eroded, is it reduced, is it burn?
 
LIQUID- initial K2CO3 vs final K2CO3 plus KOH, K balance, CO3 minus 2 balance alternatively dissolved carbon can be measured
 
GAS- the most difficult part and the most critical, I dare to say!  Energy is consumed in order to get gases; Gases are coming out the cell and add to the weight loss;
Extremely hot gases are bubbling through the liquid that is already close to the boiling point and get saturated (supersaturated?) with water vapors. Was the volume/weight & composition of gases plus the weight of the water after condensation and separation measured?
 
By the way is there a connection between the depth of immersion of the electrodes and COP?  This is only an ad-hoc sketch of the chemical aspects of the material balance in these experiment.

18) John Coviello added:
. . . I don't think anyone would find anything exciting about what is going on in these cells if the heat measurements fell within the ranges of chemical reactions.

19) Ludwik Kowalski observed:
I am surprised that neither Mizuno nor Fauvarque, who are electrochemists, presented a complete discussion outlined by Peter Gluck. Why didn’t they do this? I am less surprised by the absence of such discussion on Naudin’s website; as far as I know he is not a chemist. Naming his device CFR (cold fusion reactor) he showed what he believes in nuclear origin of excess heat in Mizuno type cells. We have no evidence that the amount of “nuclear ashes,” discovered by Mizuno, matches the amount of excess energy, as it seems to the case in some Fleischmann type cells. Fauvarque et al, by the way, also refer to their setup as CFR.

20) M. Jullian responded:
Ludwik: the possibility of such chemical reactions may have escaped Mizuno or Fauvarque. X4 you wrote: “Isn't that exactly what Ed Storms and Mike McKubre have been doing on other threads regarding the Colorado replications regarding questions raised about the amount of heat produced via chemical reactions?” The short answer is no. Mike first implied that no chemical reaction with the K2CO3 electrolyte can affect excess heat. Then two effects with the K2CO3 which do affect excess heat or it's measurement by evaporated mass were found by non-chemists, one by X2 and one by myself, fortunately it was worked out that they can only lead to underestimating excess heat, and only slightly, which is not a problem.

My questions were about the possible reactions . . . Of course Mizuno excess heat would not be exciting if it could have a chemical origin, and the fact that two hitherto unthought of reactions REDUCING apparent excess heat have suddenly sprung up suggests that there might be more waiting to be discovered, and there is no guarantee they won't CONTRIBUTE to excess heat. It's awfully hot in the plasma, and it contains hydrogen, oxygen, carbon, and potassium, in various combinations, probably all ionized by high electric fields and temperature and in highly reactive forms. Can we be sure no reaction affecting excess heat can occur in this furnace, other than the ones mentioned? We count on chemists to suggest a detailed protocol to ascertain this.

21) J. Coviello wrote:
I agree with the spirit of your inquiry. There is no reason why we should just accept the excess heat measurements without considering all the possible causes for the observed excess heat and the potential that the excess heat measurements are the result of error. It is correct to explore alternative explanations and only when those have been exhausted to entertain the idea that the excess heat is of nuclear origin.

People like Mike McKubre and Ed Storms have been working on cold fusion for over a decade (or longer) and they have certainly considered at times whether the heat is of chemical origin or not. The arguments I've read against the heat being of chemical origin is the simple fact that the observed excess heat is just too high to be the result of chemical reactions. If a chemical reaction has the potential to cause a 5 milliwatt increase in heat in a cell and the observations are in the 20 to 50 milliwatt range for excess heat, then the chemical explanation quickly falls by the wayside. I believe you are reinventing the wheel here, as the chemical explanation has been considered and dismissed many times over.

The one thing cold fusion research really needs at this point is a larger experimental setup, one capable of outputting heat in the watt range consistently. That would put chemical explanations and error questions to rest forever. I'm not sure what exactly is preventing scaling up cold fusion cells, besides perhaps simple economics, there isn't all that much money for cold fusion research. Whatever money has been available, such as the SRI work funded by EPRI that Mr. McKubre performed, was spent on basic research just to determine if the cold fusion phenomenon is real and is worth pursuing. Now that it has been established as real, we need massive funding for research efforts that determine the optimum operating environment to produce the cold fusion reaction and to scale the reaction up to levels that might be useful for energy production.

22) Ludwik Kowalski wrote:
X4 wrote: “People like Mike McKubre and Ed Storms have been working on cold fusion
for over a decade (or longer) and they have certainly considered at times whether the heat is of chemical origin or not.”
That is true. But I do not remember them analyzing chemical reactions in Mizuno type cells (along the lines suggested by Peter Gluck). These cells are very different from Fleischmann type cells that were used by most CF researchers. But Mike, Ed, P. Boss, S. Szpak and R. Oriani are electrochemists; their opinion would mean much more than opinion of numerous amateurs. I suspect that our experts have had reservations about Mizuno type cells but preferred to remain salient. Was it because the chemical environment is too complex or because they decided not to create a conflict in our own "army?"

23) P. Gluck wrote:
Dear X1, Excuse me please for adding that some physical effects have to be also considered. An example: The solution has a concentration of 0.2m K2CO3 i.e. also 0.2 m in CO2. If the CO2 is released in the very hot reaction zone as CO2 gas, that is 4.5 liters in normal conditions (1 mol is 22.41 liters). The conditions are far from normal, it is very hot there! If CO2 is released as a very hot gas, it mixes with the also hot hydrogen and oxygen, the gases are bubbling through hot water and get charged (saturated or not, it is a problem of kinetics) with water vapor -- and this leads to extra weight losses. Therefore such problems as: -- is K2CO3 decarbonated during the tests? -- what is the total volume of gases during an experiment and can it be correlated with COP? have to be considered too.

24) Richard Slaughter wrote:
The reason I decided to try the Glow Discharge experiment was that it did produce excess heat in the 10 to 100 watt range or 10 to 30% of input power. [That is much more than fractions of watt, as in many experiments involving the Fleischmann type cells.] I'm not a chemist but it seems to me that we should be trying to figure out what remains chemical after a successful experiment.

I can think of several test we should be doing. I'm sure the chemist can add several more. Find out if there has been a change of mass in the system. If all we are doing is evaporating H20 the total mass should be constant. Find out what amount of K2C03 still remains. Find out how much pure W is still the beaker and how much has gone into solution. Find out what compound is holding the W in solution

If we answer these question don't we have a complete picture of the energy balance of the system? Do we really need to know the complex multistage relations that might have happen to reach this point? With this data can't we now calculate energy production per/atom and show that this production is outside normal chemical reactions and a potential energy source? Chemist, is there an inexpensive way to answer these questions?

25) Ludwik Kowalski wrote:
I think that this is good way to begin a discussion of "Mizuno cell chemistry for dummies." Suppose the water of used electrolyte is totally evaporated after an experiment. Then the recreated K2CO3 should be extracted from the deposit. No, I am not going to this; I do not know how to separate the K2CO3 from the remaining solid deposit. But a chemist can do this; I am sure Richard will be happy to send used electrolyte to a chemist willing to help us. Suppose that nearly 100% of the initial K2CO3 is recovered. This would show that no mass was lost in the form of the escaping CO2. The other extreme would be total absence of K2CO3 in the deposit. That would mean that concentration of the electrolyte was not constant in our experiments. Does this suggestion makes sense? I do not know which ions are formed when pure K2CO3 is initially dissolved in distilled water.

26) M. Jullian responded:
You wrote “no mass was lost in the form of the escaping CO2.” Not necessarily, as I explained in a recent post: “Note such analysis can be tricky though, as for example the above reaction could not be detected if the electrolyte was left to cool down in ambient air long enough: as Michael taught us the KOH would reabsorb the CO2 from the atmosphere and return to its initial K2CO3 composition."

27) Jed Rothwell wrote:
Ohmori and Mizuno are electrochemists, and they have done very thorough investigations of the cell chemistry.

28) Ludwik Kowalski wrote:
I do not recall anything of that kind in their papers downloaded from our <www.lenr-canr.org> library. It would be very useful if somebody who has access to Japanese papers, and who knows Japanese, could translate for us their analysis of the bulk chemistry in the hot plasma environment. In the context of this debate the bulk chemistry seems to more important than chemical impurities that might be confused with products of transmutations.

27) Mike McKubre wrote:
. . . My position is  that the magnitude of the excess energy reported in this thread (and elsewhere) from this class of experiment (as others) is far too large to be chemical in origin.  Actually I have been saying this for nearly 17 years.  This is a matter of eV vs. MeV.  It is a matter of quantification - something I see too little of in this somewhat airy discussion. . . . The Mizuno effect is of interest  to me, but not one of my primary foci.  In my view there are also far too many posts.  I am a little disturbed by the new tone of Chemist/Physicist specialization.  If you are going to do experiments or discuss them in this field you need to develop competence in both areas.  They are not so different, or difficult. Do not retreat behind your ignorance or ask others to train you or do your work for you. . .

Appended on 4/1/06

28) Peter Gluck wrote:

I think that material and energy balances based on measurements, chemical analyses and search for cause-effect are necessary in order to distinguish between successful (say, COP>1.20) and failed experiments. Without balances for the solid, liquid and gas phases we will not be able to understand what happens and to know if the effect is real - and adequate for intensification and scale-up. The problem is that we have a metastable system -- plasma, extremely hot gases, boiling water, great temperature gradients etc.,  and in these systems very small changes can have disproportionately great effects -e.g. fast evaporation, loss of mass. We have to establish if the difference between the successful and failed experiments is: - thermodynamic i.e. a real heat effect/excess - kinetic - in this case we need further investigation to decide if the effect is real.

29) Peter van Noorden wrote:
I want to stress that it is very important to look for the amount of non evaporated water/ evaporated water during the contact glow discharge experiments, because I think this can have a big influence the energy calculations of the cell. As I earlier mentioned I did a lot of experiments with a system which has a long vertical condensor (about 1 meter ). I used thallium chloride as an electrolyte to study remediation of radioactivity. I did not expect to see any electrolyte in the condensor flask but when I used radioactive thallium 201 I measured about 15-20% of the original activity in the condensor solution. This was very surprising so I assumed that 15-20% of the solution is carried away by ( non evaporated) droplets. Further I noticed that during gentle electrolysis the thallium concentrated on the kathode surface ( which is ofcourse normal), but as soon as the glow discharge started ,the thallium was rapidly expelled from the electrode and could be found at the bottom of the cell attached to the W debris. Probably this was the result of fast corrosion of the W electrode.

30) Ludwik Kowalski (not posted):
That is a very significant contribution. The COP close to 1.2, that we measured in Colorado-2 experiment would have to be reduced to 1.0 if only 80% of the water were known to be lost via vaporization and 20% in the form of tiny drops.

31) Johnny Coviello wrote:
I came across this critique [by Alexi Tekhasski] of the Mizuno evaporation measurements (see below) . The writer is saying that Mizuno and others who have found positive excess heat results from his experimental design are not accounting for liquid that is lost by the glow discharge cell due to mist created during the boiling process when bubbles violently break the surface of the liquid. The critic is saying that Mizuno and others are assuming that all the missing water has evaporated, and they are not accounting for mist created during the boiling process, which would reduce the amount actually evaporated during the experiment. I assume that Mizuno and others took this variable into consideration in their calculations, but this person says otherwise.

32) M. Jullian wrote:
Indeed. Mist becomes a central question obviously, it will have to be
elucidated in a very definitive way, and I don't think an open cell will
allow solving the controversy. . . .

33) Peter van Noorden:
This contact glow discharge reaction is very violent at 350V so you have a lot of mist which escapes from the electrolyte surface and is forced through the condensor. . . . The fact that the radioactivity is transported through the tubing (1 m high) and into the condensor vessel made me conclude that the mist above the electrolyte must be a combination of vapour and droplets

34) M. Jullian wrote:
Yes you are righ, it must be a combination. Indeed there is no way to tell condensing vapor from mist as they are the very same thing: liquid droplets. . . .

35) Peter Gluck wrote:

. . . The Peter’s experiment shows you that the very hot gases are very fast too and they carry microdoplets of solution at relatively great distances in fragments of a second. . . . I made Mizuno type experiments - very preliminary, with stainless steel electrodes ( in Gene Mallove's lab in 1998) and have observed that above the plasma zone there is splashing and it is very hot. But have not measured HOW hot and this is an important
unknown parameter. We also do not have the slightest idea now, if potassium carbonate is converted in hydroxide and in which extent.That is we have CO2 too.

I am an engineer (chemical, bien sur) and I am not very fond of experiments in which the electrodes are destroyed so fast. (the Cinci Cell has the same fatal flaw), therefore I have not continued this work. The system is not so simple as it seems and therefore the new experiments have to be planned carefully. In case the group needs my know-how, I am glad to help. Please read what I wrote re. the material balance.

36) Jed Rothwell wrote:
Look up "wet steam" and "dry steam" and you will see that engineers have been dealing with the enthalpy of steam and the issue of moisture in steam since the late 18th century. Measuring enthalpy by weight loss alone is a crude method, as I said here several times. Perhaps it proves there is excess heat but I would not trust it +/- 10%, so the significance of 20% excess is low. The other methods used by Mizuno -- such as bomb calorimetry and flow calorimetry combined with mass spectroscopy -- are much more reliable and convincing. Since there is no jet of steam leaving the cell it is not likely that droplets will be carried a long distance, but it is still better to condense the steam and let the hot water drip back into the cell, the way Mizuno has done. This eliminates the issue altogether.

37) Ludwik Kowalski wrote:
Yes, the COP=1.2 is probably the smallest one can trust by using a simple technique. But, according to Pierre, the COP close to 2.0 are constantly reproducible in Naudin's lab. Was it an exaggeration, Jean Louis? Please describe your most efficient excess heat generator for us. What metals are used as electrodes? Are you also using the 0.2 M K2CO3? What is your average current? At what voltage are you operating? Numerous reports about reproducible experiments with the COP>1.3, coming from students, teachers and amateurs, would have an effect on mainstream scientists. Simple and inexpensive techniques should be encouraged. More sophisticate techniques are for those who can afford them.

38) M. Jullian wrote:
. . . indeed there is no reason to expect the mist's droplets to have a given minimum size, most of it's mass could well be invisible, and it would have no particular reason to be entirely caught by a screen of any arbitrary fineness. . .

39) J. Rothwell wrote:
[The simple evaporated mass method] is wonderfully nice and simple, but not precise, and not very reliable. To make it precise you would have to bring in all kinds of complex instruments and measurement techniques. The same goes for any method of calorimetry, except the Seebeck method. Flow calorimetry is conceptually nice and simple, but Mike McKubre will tell you, beyond the conceptual stage it becomes messy and complicated in real life. The same is true of the evaporated mass method. If you want to make it better than 10% accurate you will have to spend a few thousand bucks and a few months I suppose. Why bother?

There is a reason why I kept saying this "wet steam" issue makes me nervous, and why I have not expressed overwhelming confidence in the results reported here. As I said, if you improve your calorimetry you may well find your excess heat results vanish. There is also a reason why Mizuno stopped using this technique many years ago.

40) Ed Storms wrote:
The plasma method dissipates generated heat mainly by boiling water. In addition, this is how the excess energy will eventually be used, i.e. by making steam. Rather than fighting this condition, why not use it to measure the heat produced? This can be done by surrounding the plasma cell by good insulation (no calorimeter), have water of known temperature delivered to the cell at a rate required to keep the level constant, and allow the steam to exit through another tube. Place a plug of glass wool in the exit tube, within the plasma cell, to trap any droplets. Measure the temperature of the steam as it exits and weigh the condensed water. You can measure the pH of the condensed water to determine if any droplets get through the filter. Of course, some H2, O2 and CO2 will be in the exit gas. The amounts of these gases are easy to determine. The H2 and O2 can be combined back to H2O using a catalyst and the resulting water weighed. Once steady state has been achieved, you have an easy measurement of produced power. In addition, the apparatus can run as long as you wish to show that the output is stable. If significant over unity is indicated, the arrangement would be easy to scale up. Almost all errors will result in a lower value for produced power, hence believable OU if it is claimed. The main challenge would be to make the size of the plasma and the exit pipe compatible to prevent too much pressure buildup in the cell. Eventually, this pressure can be increased to give steam at a higher temperature. Why is this approach not used?

41) Richard Slaughter asked:
Ed, I like your proposed approach. I know there are chemical catalyst to recombine the H2 and O2. I have the impression that these can be dangeroue if the system is not running at the right temperature and there is no visable way to know the catalyst is working. Would it be ill advised to use a spark generator to recombine the H2 and 02? I like the idea of a spark generator because I can tell instantly if it is working. You are proposing a semi-closed cell so I'm concerned about gas build up.

42) M. Jullian wrtote:
In terms of indisputability by a careful reviewer, even if Pt was cheap, a unity COP with Pt as Ed and you suggest wouldn't help in the least because conditions would be too different. Plasma temperature would be different for one thing (different melting temperature). As I suggested before, even a unity COP obtained with a _tungsten_ cathod in a negative run isn't proof that a 1.3 COP in a positive run isn't an artefact (it could be due e.g. to the right regime of discharge nebulizing a particularly large amount of mist).

Group, it occurred to me that the simplest and least expensive method to _quantify accurately_ the mist phenomenon would be to condense each run's exhaust gases as has been suggested, _measure the total salt content_ of the condensates(*), and if not zero use it to correct the run's COP precisely (the maths are easy). This would eliminate altogether the need for a mist filter and would be skeptic-proof I would think. Your opinions welcome.

43) Ed Storms wrote:
a) Wet or dry, the steam leaves as a gas and the droplets join the liquid
phase that slowly drips back into the cell. Droplets do not result from
the steam or form within the steam, but form as bubbles burst on the
electrolyte surface.

b) You wrote: “a way to renew the cathode continuously must be found for this.” This is easy to do by feeding a W rod through a glass tube into the cell. You might need to drain the cell periodically to remove the tungsten, tungsten oxide, and transmutation particles that would accumulate.

c) You asked about the mist filter. As I described above, the steam would push through any liquid in the plug. Because the plug is in the cell, the hot water returns to the cell where it adds its energy to the generation of more steam. Of course, the plug needs to be designed so that steam can leave at a slow rate, to avoid creating additional bubbles, and to allow the water to drain away. Such a design is easy to create.

44) L. Kowalski wrote:

a) Condensing escaping gasses (if practically possible) and vapors, makes sense to me. Please elaborate. (see the P.S. comment below).
b) Using a bomb calorimeter, as suggested by Jed, also seems appropriate when excess heat is generated at the rates of tens of watts, or more.
c) I wish to see more people performing experiments and reporting results.
d) Exploring the same type of phenomenon, and discussing the results on this list, could be very effective.

P.S. Suppose the mass of the water escaping from the condenser is measured. How do we determin what fraction of that mass is from the vaper and not from tiny drops escaping with it? The fraction must be known to calculate the COP correctly.

45) Ed Storms responded:
I don't understand why this is an issue. Removing the droplets is trivial. A resistively measurement of the condensate would give a good analysis for the amount of electrolyte that was not removed.

46) L.K. (not posted):
I am confused. If tiny droops are removed somehow (sparging ?) then what condenses should be pure water. Ed is referring to the study of the resistively of removed droplets. Obviousely, I did not understand some messages.

47) M. Jullian wrote:
. . . Ludwik I disagree with your objection that the condensate salt contents measurement method I suggested cannot measure escaped mist. Let me explain by using an example: if we find say 5g of salt in the condensate, along with _any_ amount of water, and if the electrolyte concentration is 50g per liter, then we can conclude 0.1 liter of electrolyte has escaped as mist. So 100g should be subtracted from the mass loss to determine evaporated mass, agreed? You also said the scale would not be needed anymore, why is this?

48) Jed Rothwell wrote:
. . . Conventional instruments and experimental techniques work fine. People have doing calorimetry and dealing with wet steam for 220 years. Find out how it is done and stop fretting about it. There are plenty of things that actually go wrong with this experiment, so stop inventing imaginary problems.

49) Ludwik Kowalski responded:
1) I disagree; it is better to anticipate what honest referees might say and to address the issues before submitting a paper. This debate was highly instructive to me. It showed that the issue of tiny droplets should be taken very seriously.

2) My last message showed that I did not understand the essence of Michel's suggestion. Only now do I do understand what he, and Storms, had in mind. It is a very clever method of determining the fraction of water escaping in the form of tiny drops. Yes, the scale would be needed, as in Paris-1 and Colorado-2. Am I the only one who did not understand the original proposal? Probably not. The Marseilles-1 experiment will probably produce more convincing results on reality of excess heat than earlier experiments.

50) M.Jullian wrote:
. . . As a lower cost alternative to a condenser, the mist collector could be an _external closed paper filter_ with a _large surface area_ (for low back pressure), airtightly connected to the cell's vent tube (a large vacuum cleaner bag with it's rubber gasket would do the trick), one numbered filter per run. The filter would be weighed once dried, and its new mass minus initial mass would yield the run's escaped salt mass (a more precise version of Mizuno's paper towels disposed around the cell as described by Jed).

Note the filter once dried and weighed would be reusable e.g. the following day ; successive dry weights, increasing with salt accumulation, could be handwritten on the filter itself vs the run numbers, e.g.:

0 -> 28.1g (initial mass)
1 -> 33.2g (+5.1) (run 1 has expelled 33.2-28.1=5.1g of salt)
19 -> 37.5g (+4.3) (run 19 has expelled 37.5-33.2=4.3g of salt)

IMPORTANT: the mist collector must exert no force on the cell: we mustn't count its weight with the monitored cell weight because at least some of the vapor will condense in the mist collector, which would make the evaporated mass erroneous. If we use a vacuum cleaner bag it must be independently suspended, with its opening on top, and the cell's rigid vent tube bent twice to connect airtightly but with zero vertical force with the bag's soft rubber gasket.

51) Ed Storms commented:

a) I get the impression that you have never seen glass wool and think it looks and acts like cotton. If it acted like cotton, it would not work as a filter because, as you say, it would block steam flow. Glass wool has a very open structure and it does not mat when it gets wet. Before you reject this method, I suggest you look at glass wool. I suggest a tube 2cm in diameter lightly packed, which is connected to a tube 0.5 cm in diameter that exits the cell.

b) Also, the paper filter method will not work. You need to have the filter where it remains hot so that steam will not condense. Otherwise, you will get a lot of water that did not come from droplets.

51) M.Jullian replied:
a) I personally trust your word that it would block all the mist without any
back pressure, but this claim would be hard to prove indisputably,
especially if the reviewer has never seen glass wool as you say. The only
way actually would be to have external capture AS WELL, and then there would
be no point in having an internal filter at all. So my point is that
external mist collection is all that is needed, and we have plenty of room
to do it.

b) You will indeed, but it doesn't matter as only the dry weight matters! As I
wrote earlier, and as Ludwik may be the only one to have understood in fact,
"if we find say 5g of salt in the condensate, along with
_any_ amount of water, and if the electrolyte concentration is 50g per
liter, then we can conclude 0.1 liter of electrolyte has escaped as mist. So
100g should be subtracted from the mass loss to determine evaporated mass"

52) L. Kowalski
I posted a long message (see in blue below) to which M. Jullian responded. He quotes my entire message and comments in different places.

53) M. Jullian:
You see I too think that mist is not really a problem, I don't even think chemical heat is a problem, but again what we think doesn't matter in the least, we need hard facts to put to rest any doubt a honest reviewer could have. Mist is one of those doubts, and chemical heat is another.

Hi Ludwik, my comments inserted into your blue text.

Let me try to review the situation, as i see it. Please correct me, if necessary. The goal is to reach an agreement on what is the best way to proceed. The bomb calorimeter, suggested by Jed, seems to be a good approach. But like in everything else, the devil is in details. I hope somebody will use a bomb calorimeter to confirm reality of OU ("over unity" COPs). Here I want to focus on the boiloff cell as used in Paris-1 and Colorado-2.

1) Use of the thermos is desirable but not essential because a large fraction of heat seems to be lost via convection above the open cell.
2) Layers of foam outside the cell walls are desirable (to keep the nonevaporative losses low).

I would think 1/ is better than 2/, both as a thermal insulator, and in terms of optical transparency (I suggest we use a transparent Dewar flask).

3) The cell should not be too large. In that way the electrolyte will boil and its temperature will be uniformly ~100 C, even when the wattage is low.

Agreed. It just has to be large enough to hold the electrodes and the heater.

4) A small ohmic heater (say 200 W) should be inside the electrolyte all the time. It should have its own W*h meter. In that way one should be able to turn it on and off at any time (to keep the electrolyte boiling, when plasma current is too low for this.

Agreed.

5) The COP is calculated as:

COP == Et / Ee (thermal energy released/electric energy received)

The electric energy is the sum of W*h (or joules) recorded by electric instruments (including the ohmic heater W*h measuring device) during a test. There are nuances in measuring the Ee, for randomly oscillating currents, but this is not our main problem. The thermal energy is:

Et = Ev + Ed + Ec + Er = Ev + En

It is the energy used to evaporate water (v) plus the energy escaping by conduction (d), escaping by convection (c) and escaping by radiation (r). The nonevaporative loss En, is simply the sum (Ed + Ec + Er). The method of measuring En with an ohmic heater, used in Paris-1, Texas-1 and Colorado-2 experiments is highly appropriate. Next we come to the main problem. The Ev was calculated as 2260*dm1, where 2260 J/g is the latent heat of evaporation while dm1 is the mass of water evaporated during an experiment. Unfortunately, what we measure with a scale is not the dm1 but the dm=dm1+dm2, where dm2 is the mass lost in the form of tiny droplets. Up to now we assumed that dm2<<dm1. But, as indicated in the discussion, this assumption must be experimentally validated. If the assumption is not valid then Ev must be calculated as 2260*(dm-dm2), where dm is that we measure with a scale and dm2 is the amount of water escaping in the form of tiny drops. Ignoring the dm2 we overestimate the COP.

Ejection of large (visible) drops can be practically eliminated by covering the cell with a fine wire mesh. Large drops are stopped be the mesh and fall back into the cell. Under favorable plasma conditions large drops are rare and they contribute to a small fraction of En (which we measured). The main problem is how to measure the dm2, or how to show that it is indeed negligible in comparison with the dm1?

Michel first solution was to collect all water escaping from the cell and to condense it. Suppose that the scale tells us that dm=10 grams while the mass of condensed water is 7 grams. It means that ~30 % was not condensed. It is important to know the percentage. Michel suggested that condensed water must be evaporated in order to collect solid deposits. The water that was evaporated did not contribute to that deposit. Only the tiny drops bring the dissolved chemicals with them.

We know how much deposit is produced when we evaporate one cubic centimeter of the electrolyte (taken from the cell). Suppose it is 0.3 grams. If the amount of deposit per cc of condensed water is 0.003 grams we can say that dm2 is indeed negligible (when the accuracy of 5% is sufficient). But what would we say if the measured amount of the solid deposit, per cubic centimeter of condensed water, turned out to be 0.1 grams. We would say that about one half of water lost was due to the dm2. Why one half? Because not 100% of lost water was condensed. This shows that the value of dm2 can be determined. Once determined it could be used in the calculation of COP.

Condensers, cooled by tap water, are routinely used by chemists. My guess is that the cost could be about $250. Please correct me if you can; I do not have any catalogues at home. What is not clear to me is a method to catch the escaping water (dm1 and dm2) into the condenser.

A thin rubber diaphragm connecting the cell output tube to the condenser input as I suggested should do the trick (think of the vacuum bag gasket if you have ever seen one of those ;)

Vertical leads to the anode and the cathode, in our cell, would interfere with efficient capturing of the escaping water into the condenser.

We need airtight seals for the leads and everything that goes across the lid. The lid itself must rest on top of the Dewar via a rubber gasket. The inverted U shaped output tube should be sealed to the lid.

So much for a condenser idea, as I understood it yesterday. Simple in principle but not easy to implement in practice, unless one is well equipped for custom-made devices.

6) I am not sure I understand your glass wool filtering method, Ed. How does it help us to determine the dm2? I can visualize a layer of glass wool above the cell, for example between two plastic screens (used on window). The screens have holes to insert electrodes. Suppose the scale tells that 10 grams of water was lost during an experiment. What else do you have to measure to determine the dm2?

7) Michel's method is conceptually clear to me. But it might indeed lead to a large underestimation of Ev (energy used to evaporate water), as indicated by Ed.

No! Wrong, as I have explained to Ed, and I thought you had understood! Dry mass of condensate is all that matters. Again: "if we find say 5g of salt in the condensate, along with _any_ amount of water, and if the electrolyte concentration is 50g per liter, then we can conclude 0.1 liter of electrolyte has escaped as mist. . . .

Underestimation, however, is not as bad as overestimation; it would mean, for example, that what is calculated as COP=1.2 might actually be 1.5 or more. But we want accurate determinations of COPs (to get the mean value and the standard deviation). I do not understand Ed's objection that the ionic composition of the electrolyte must be known.

Ed's objection could be valid maybe if a large mass of escaped salt is found, which we hope won't be the case.

A filter can always be calibrated by using the electrolyte taken from the cell. We can place 1 cc of the electrolyte on the filter, allow the water to evaporate and measure the amount of the remaining solid deposit. Then determinations of dm2 can be made, at least approximately, in consecutive tests. It would be important to often add 100 C water to the cell to keep the concentration of the electrolyte constant, more or less.

Yes.

My hope is that dm2 will be found negligible. A submitted paper stating that this was indeed the case would be much more difficult to reject than the paper in which the negligibility of dm2 is assumed.

Yes!

P.S. (next day): Now I realize I missed your point (and Ed's). You referred to the external paper filter solution, and assumed that the wet filter was on the scale along with the cell hence affecting it's weight, which indeed would lead to a large underestimation of Ev due to re-condensation of some of the vapor in the filter.

The answer is still "No! Wrong", and the explanation was in my proposal:
-----------
IMPORTANT: the mist collector must exert no force on the cell: we mustn't count its weight with the monitored cell weight because at least some of the vapor will condense in the mist collector, which would make the evaporated mass erroneous. If we use a vacuum cleaner bag it must be independently suspended, with its opening on top, and the cell's rigid vent tube bent twice to connect airtightly but with zero vertical force with the bag's soft rubber gasket.
-----------
In case your vacuum cleaner bags don't look the same as mine, mine feature a thin soft rubber membrane with a circular hole in them, which connects airtightly to a slightly conical rigid tube at the end of the cleaner hose by a push-on action when installing the bag in the cleaner. This kind of airtight coupling does not exert any significant force perpendicularly to the membrane plane, so if this plane is horizontal there is no vertical force which could affect the cell's weight. So there would be no underestimation of Ev at all. Same kind of no-vertical-force coupling should be used in the condenser solution IMHO. A bellows type airtight coupling would work too (probably better actually).

The corollary of this absence of vertical force between the cell and the output capture device is that the latter must be suspended independently of the cell of course.

Let me know if the above was still unclear, as this point is important: it's ok to underestimate the COP, but not grossly!

54) L. Kowalski again:
J.M. wrote “Mist is one of those doubts, and chemical heat is another.” According to private messages, Paris-2 results will allow us to rule out the anticipated "chemical heat" objection. I hope that new information will soon be published here. Then we can discuss significance of numerical data.

We want to know much water is lost as vapor (dm1), and how much water is lost as tiny droplets, (dm2). A separate experiment, not the one whose purpose is to measure the COP, can be conducted. Michel's idea is to capture a sample of what is escaping. This can be done by using glass wool (also used as thermal insulation) or by using a paper-like filter (also used in common vacuum cleaners). In each case the sample is weighted twice, first as wet and second as dried (after water is allowed to evaporate. The difference was tell us how much water was in the sample, the second mass will tell us how much deposits. A lot of deposit would indicated that dm2 was large, very little deposits would indicate that dm2 was very little.

Knowing the volume of water in the sample, and the amount of salt in it, we are able to calculate the salt concentration (for example, in mg/cc = grams/liter). Suppose the sample had 0.27 mg of solid deposit per one cubic centimeter of evaporated water. We then take some electrolyte from the cell and repeat the procedure. Suppose we find 27 mg of solid per cubic centimeter. Would it be appropriate to say that dm2, mass lost in the form of tiny drops, is 1% of the total mass lost? I think so. Note that the vapor escaping from the cell consist of pure water while the droplets escaping from the cell contain the salt. Consider the extremes. If 100% of the liquid was lost in the form of tiny drops then our sample would have the same concentration of salt as in the electrolyte taken the cell. And if 100% of the electrolyte was lost in the form of vapor then the concentration in the sample would be zero.

Did I describe your suggestions correctly, Michel (paper filter) and Ed (glass wool)?

Here is another way to accomplish the same thing. Suppose a large funnel is placed above an open plasma electrolysis cell. The leads to the electrodes, in this control experiment, can be on the side of an open plastic cell. The narrow exit tube of the funnel is plugged in and sticks up, like a blocked chimney. The wide end of the funnel is only several centimeters above the cell. A bag with ice can be used to keep the funnel cold. The diameter of the funnel should be larger than the diameter of the cell. Under such conditions what comes out of the cell will condense on the inner surface of the funnel and drip down into a large plate (on which the cell is standing). We can collect the liquid, measure its volume, and remove the water by evaporation. This again will give us concentration of the salt in the sample, for example, 0.27 mg/cc, as above.

I suggest that an estimation of dm2 should be made in Paris-2. Any one of these three approaches is likely to tell us the fraction of the liquid lost in the form of tiny droplets. If dm2<<dm1, for example, 1%, then the reported COP are fine. Otherwise, the COPs should be recalculated by taking the dm2 into account. Note that in Colorado2 COPs were calculated under the assumption that the dm2 is negligible. Will the Paris-2 data justify that assumption? That remains to be seen.
P.S.
Actually three mass measurements must be performed with each sample, not two. First before catching the liquid, second when the catcher is wet, and third after the catcher and remaining solid in it, are dry.

P.P.S.
On Apr 2, 2006, at 3:34 PM, Michel Jullian wrote:

You see I too think that mist is not really a problem, I don't even think
chemical heat is a problem, but again what we think doesn't matter in the
least, we need hard facts to put to rest any doubt a honest reviewer could
have. Mist is one of those doubts, and chemical heat is another.

1) According to private messages, Paris-2 results will allow us to rule out the anticipated "chemical heat" objection. I hope that new information will soon be published here. Then we can discuss significance of numerical data.

2) Go to Google type "distillation," in the search box, and click SEARCH. Then type on "images," above the search box. You will see standard glassware used in laboratories. Before deciding what is the best we must know what we want. How much water is lost as vapor dm1, and how much water is lost as tiny droplets, dm2. A separate experiment, not the one whose purpose is to measure the COP, can be conducted. Michel's idea is to capture a sample of what is escaping. This can be done by using glass wool (also used as thermal insulation) or by using a paper-like filter (also used in common vacuum cleaners). In each case the sample is weighted twice, first as wet and second as dried (after water is allowed to evaporate. The difference was tell us how much water was in the sample, the second mass will tell us how much deposits. A lot of deposit would indicated that dm2 was large, very little deposits would indicate that dm2 was very little.

55) J. Rothwel:
The ice would probably cool down the cell and interfere with calorimetry. Keep ice away from these experiments. I have seen some really dreadful experiments involving ice and calorimeters (and one impressive one). Mizuno used a much better condenser. The flow calorimetry cooling water condensed the steam and the water returned to the cell. See: <http://lenr-canr.org/acrobat/MizunoTconfirmatib.pdf>.

I do not understand why you want to use such complex and unusual methods, when simple, direct, foolproof methods are available. "Wet steam" is only a problem when you do take steps to prevent droplets from leaving the cell or to measure the total enthalpy of the effluent gas. Such steps are trivial, as Storms pointed out, and once you take them you need not worry about the problem again. You do not need to invent six different ways to address this. Simply select the most convenient method from the ones already suggested by Mizuno, Storms and me. (I suggest sparging and measuring the temperature rise of the water column.) It does not matter which method you use.

56) L. Kowalski:
Yes, ice should be kept away while measuring the COP. My suggestion is to conduct a control experiment whose only purpose is to determine the fraction of mass escaping in the form of tiny drops. Once know, for a given electric power, and given geometry, it can be used in experiments whose purpose is to measure the COPs.

Any method is OK for me. The funnel method was just an improvisation. The glass wool catcher seems to be the easier. But to avoid possible consequences associated with glass-wool needles I would use paper towels, napkins, or toilet paper instead. Keeping such collectors above the running plasma cell would accumulate escaping liquids at the rates of grams per minute (?). What can be more simple than this? Measuring the percentage of the liquid lost as droplets, in a separate control experiment, should be done by anybody performing Mizuno type experiments. Perhaps Richard will do this. Then we will know by how much our Colorado-2 COPs must be corrected.

57) M. Jullian:
I submit on the contrary that mist measurement, or any control measurement, must be done on the very runs for which we measure the COP. Otherwise we prove nothing at all, because the ODD overunity runs could be thought by an honest reviewer to be due to the PECULIAR, hard to get, right kind of discharge generating equally PECULIAR ARTEFACTS, like for example: - nebulizing a particularly large amount of mist (e.g. by this discharge generating ultrasounds of the same frequency as used in an ultrasound nebulizer). (a)- plasma reacting exothermically with the salt. (b) - generating an insulating foam layer (c) - or anything which may give the false appearance of excess heat and hasn't been thought about yet.

58) Jed Rothwell:
Michel Jullian wrote: “Hi Ludwik! I submit on the contrary that mist measurement, or any control measurement, must be done on the very runs for which we measure the COP.” Yes. This is essential. Also you should use the simplest and most conventional method possible. Never invent a new or unusual technique if there is already a standard, widely accepted one.

59) L. Kowalski:
Then the ohmic heater should not be used; the nonevaporative losses of thermal energy should be made without it. How can this be accomplished? Let the perfect not be the enemy of the good. I would also prefer a setup in which all measurements are made during one operation. What I am suggesting is a simple costs-nothing test that should make the overunity COPs more credible.

60) Chemistry facts (not posted):
a) What is formed when the K2CO3 is dissolved in H2O? It is K+, CO--, HCO3- and some H2CO3. How much of each depends on the temperature. The CO2 is lost to air because the H2CO3 breaks into H2O+CO2. The amount of CO2 going into air increases with temperature. After the electrolyte cool the CO2 may be taken back from the air but this is a slow process (hours or days) of recreating the H2CO3.

b) Can the K2CO3 be restored, for example, when the electrolyte was not used for many days and then allowed to evaporate, forming a solid residual deposit? That would create an illusion that no potasium carbonate was spent during the plasma electrolysis. The fact is that all K2CO3 disappears quickly when the electrolyte is made.

HCO3- + OH- --> <-- CO3-- + H2O

Some KOH might be formed in the electrolyte (from K+ and OH-). It is a complex, temperature-dependent kinetics.

61) Peter Gluck, who is a chemist, wrote:
Perhaps it would be useful- for understanding how the Mizuno system works- to perform some qualitative tests, before deciding re. the new experimental setup. It seems we have a system in which a solution of a K salt in water at ~ 100 C is in contact with a plasma flame at (?) 3000 C. A huge temperature gradient. Unnatural combination. It would useful to:

(a) measure the temperature above the plasma zone - a bit above the water level;

(b) to put a baffle - e.g. a sheet of shiny stainless steel in the same zone in order to collect the microdroplets and to see if what's released from the cell is K2CO3 or KOH (pH will tell, or you put a bit of acid and see if gas-CO2 is released or not)

(c) if it is KOH mainly, then to establish if the initial K2CO3 is transformed in KOH- that means CO2 is added to the gases, the material balance has to be made for the ion K+

(d) if this is the case, to see what happens - in the actual system if the carbonate is replaced with KOH from the start- is plasma formed? the COP is O/U? etc.

(e) to see if the W precipitate has the the structure I expect i.e. a core of salt covered by molten and frozen metal (formed by spark erosion) - a simple microscope is needed.

62) L. Kowalski:
Jed wrote: “I submit on the contrary that mist measurement, or any control measurement, must be done on the very runs for which we measure the COP. ”The paper-towel test (PTT), suggested yesterday, can be performed in that way. A typical Colorado-2 experiment lasted 12 minutes; after that we had white flashes indicating that the sticking end of the cathode was essentially consumed. Suppose the time from t~0 to t~2 min is used to perform one PTT and the time from t~8 to t~10 min is used to perform another PTT test. The COP can then be measured in 8 or 9 minutes between these two tests.

63) Michel Jullian:

Your paper towel test proposal consists in covering the top opening of the cell with paper towels to collect escaped salt. I don't see how you could perform a COP measurement while doing this test, as some amount of evaporated water will recondenses in the towels, wronging the evaporated mass measurement . . .

64) L. Kowalski:
I believe that the paper towel test is the same test you suggested (keeping the vacuum cleaner bag above the open cell). The only difference is that your bag is above the cell the entire time interval during which the plasma electrolysis is going on, for example 11 minutes while I would collect two small samples of the escaping liquid, one during the first minute and another near the end. The COP is measured between the sampling. Nothing is turned off or changed. In that way the COP measurement is performed under the same conditions as when the two samples are collected.

Yes, you can question "the same conditions" assumption but, as stated by Jed, that would not be reasonable, especially if the two samples give about the same % of mass lost in the form of droplets. Neither time nor mass lost by the cell have to be monitored when a paper towel (or toilet paper, or a handkerchief) is kept (by hands) about 5 cm above the open cell for about one minutes. That test might not answer all the questions but it is so simple that one must perform it. I hope that Pierre and Richard will perform such tests and share the results with us. We do expect the percentage to be small. That results should be reported in our anticipated paper.

65) J. Rothwell:
I doubt this test would prove anything. First of all, the method is crude. Some water vapor would condense on the towel (false positive). On the other hand if there are droplets in wet steam, many would miss the towel and drift away in the air (false negative). It is impossible to tell which is more likely. Second, there would only be 5 or 6 g of droplets over 1 minute, and you have no idea what fraction of this you can capture with a towel. In your message of 3/15/2006, you said that over a 20 minute run, the weight of the cell drops from 3052 down to 2801 g. That is 13 g per minute. You said the apparent COP was ~1.4, and a null run was around ~0.8. So about 4/10 of the lost water would be in droplets to explain away this excess heat.

I think you should stop dreaming up crude and unusual ways to address this problem, and simply use industry standard methods. Also, by the way, if you do this experiment again, I strongly recommend you use a computer, multiple thermocouples, digital cameras, and other modern instruments to record the event in detail, and a magnetic stirrer and other gadgets to improve accuracy. There is no point to going to all this trouble to do the experiment if you are going to use primitive 19th-century instruments.

66) Michel Jullian:
Ludwik I hadn't realized you meant to hold the paper towel _by hand_! But in this case as Jed just pointed out the steam and mist won't be forced through the paper as was the case with my airtightly connected vacuum bag suggestion, so you will only capture a tiny fraction of the mist because it will just go round the obstacle, so you'll overestimate the COP. It's a very sloppy solution, and it just won't work.

67) L.Kowalski:
Suspecting that the debate about a reliable estimate of the % of the liquid lost in the form of tiny droplets reached the point at which most people are no longer interested, I suggest that Michel, Jed and myself discuss this topic in private, at least for a while. My reply to the last message from Michel will go to him and to Jed only. But I will be happy to add names of those who might also be willing to participate. Write to at <kowalskil@mail.montclair.edu>.

68) L. Kowalski:
PRIVATE
1) Michel, please write again the latest version of your method. I suspect I am missing something important. Please attached a crude picture. Then I will comment.

2) Jed, you wrote: "First of all, the method is crude. Some water vapor would condense on the towel (false positive). On the other hand if there are droplets in wet steam, many would miss the towel and drift away in the air (false negative). It is impossible to tell which is more likely." I agree, that is a serious issue. I do not want to catch all that comes out, a large enough sample is sufficient. But it is true that the percentage of tiny droplets I catch might decrease when the paper towel is kept higher. Perhaps Michel's method will not suffer from this. I will wait for his picture plus detailed description.

69) Michel Julian:
Hi Ludwik, the vacuum cleaner bag solution was an earlier more complicated proposal, forget about it. As I said no accessory is needed in the current proposal. So I guess it's now time to address the more refined proposal, which allows for CO2 emission.

It is based on potassium mass loss (which can only occur through mist emission), rather than on total salt mass loss (which is affected not only by mist emission but also by gaseous CO2 emission as Peter Gluck pointed out). So basically we just replace "salt" by "potassium" in the method:

1/ weigh the cell's potassium before the run (when preparing the fresh electrolyte ; mass of potassium can be deduced easily from salt mass and salt formula) 2/ weigh the cell's potassium after the run (e.g. measure by precise chemical analysis K+ concentration in g/l, which any medical lab can do for you, and multiply by precisely measured volume) 3/ subtract to get escaped potassium mass
4/ infer escaped droplets mass from escaped potassium mass, based on
salt concentration and formula. Does it still make sense?

70) Jed Rothwell wrote:
We beat that horse to death days ago. Ed & I suggested several ways to solve this problem. If you do not like our suggestions, do it your own way. I am fed up with repeating these suggestions and I will not say another word about it. In fact, I am fed up with this whole discussion!

71) L. Kowalski wrote:
Too bad that you decided not to share me with what you know about old ways of measuring the percentage of droplets escaping with steam. I do not think our university library has book about this. But I will look for them. Perhaps measuring the temperature gradient in a pipe above the cell, as you hinted, will become clear to me. I know how to measure the gradient of temperature but I do not how to use this information to get what we need. Do not give up on me; I would be glad to endorse your recommendation after understanding it, and after playing with some realistic numbers.

72) J. Rothwell wrote:
Don't be tiresome Ludwik. Review the message archive here and look at the literature. I spelled it out several times and so did Ed Storms. I will reiterate the methods one last time.

Pons: reflux calorimeter. For more, enter the search term "reflux" into the Google search box at LENR-CANR.org.

Mizuno: flow calorimetry, where the cooling fluid goes through a condenser first. http://www.lenr-canr.org/acrobat/MizunoTconfirmatib.pdf

Storms: glass wool in effluent gas tube.

Rothwell: sparge effluent gas tube through a water column. I suggest you insulate the water column and measure the temperature rise. Storms suggests you measure the pH of the water column, or condense the steam by some other method and measure the pH of the condensate. Storms: use a large, closed Seebeck calorimeter. (I agree this method is the best by far.)

73) J. Rothwell wrote:
Here is my suggested method of doing the Ohmori-Mizuno glow discharge experiment while accurately accounting for wet steam. This is based upon many observations I made of both Ohmori and Mizuno at work in their labs, and it is also based upon suggestions made by Storms in this forum recently.

If the run is going to be short, 10 or 15 minutes, I suggest you use a well-insulated cell. (What I have designated "bomb calorimetry.") Seebeck calorimetry is preferred but it is more expensive.
 
Cover the cell with an airtight lid with various orifices in it to accommodate the anode and cathode lead wires, thermocouples and so on, plus an exhaust tube to capture all effluent gas. You might want to include an emergency valve in the lid. This can be as simple as a drinking straw bent into a "U" shape and stuffed into two holes.
 
Stuff glass wool into the end of the exhaust tube that goes into the cell. This will stop most of the moisture from leaving the cell.

1) WHAT IS MOISTURE? PROBABLY DROPS OF WATER, NOT VAPOR. THE INTERCEPTED DROPS WILL DRIP DOWN TO THE CELL, AS DESIRED. RIGHT?

Prepare a condensation water column. This is a well-insulated, tall beaker. Weigh it, fill it with ~200 mL of distilled water, and weigh it again to determine the exact mass of water. Measure the pH of the water.

--> Right. Liquid water entrained in the steam will drip back into the cell.

Put the other end of the exhaust tube at the bottom of the water column so that all effluent gas bubbles out through the column and the steam condenses. Also insert a thermocouple into the water and record the temperature.

2) THAT PART IS NOT CLEAR TO ME. HOW IS THE TOP OF THE EXHAUST TUBE (ABOVE THE GLASS WOOL) CONNECTED TO THE WATER COLUMN ABOVE? ON THE PICTURE I AM MAKING THE WATER IN YOUR CONDENSATION COLUMN WOULD RUN DOWN INTO THE CELL (THROUGH THE GLASS WOOL). WHAT AM I MISSING.

--> It is very simple, [see the drawing below]: That drawing shows that the water column is not above the cell, as i assumed. It is a beaker standing on the same table that the scale supporting the cell. The U-shaped tube containing the glass wool bends twice above the lead of the cell. It provides the only possible escape from the cell. In other words, the lid above the cell must be tight.

The part of the escape tube containing the glass wool is vertical. Thus the intercepted drops can fall down by gravity. Then the tube bends to become horizontal. After that it bends again and becomes nearly vertical. That is how it enters the beaker with water where the vapor is expected to condense. The right-side ending of the tube is situated near the bottom of the that beaker, deep below the surface of water, about as low as the glowing cathode in the cell. The O2 and H2 bubbles come out from the escape tube, rise in the condenser and disappear in the air above the open beaker. Yes, a simple picture is better than many words.


The cell should be equipped with at least three thermocouples separated vertically and horizontally. It should be mounted on a magnetic stirrer. The vibrations from the stirrer make it impossible to measure the weight of the cell during the experiment accurately, but mixing the fluid is more important than measuring the weight precisely. This is not important; you can measure the weight precisely before you turn on the stirrer, and after the experiment when the stirrer is turned off.
 
Record the appearance of the plasma with a video camera, and a digital still camera. You can probably set the digital still camera to take pictures automatically every 15 seconds or so, by connecting it to a computer. Be sure you turn on the video camera and still-camera date-time-stamp, but keep the date-time-stamp away from the important parts of the picture, such as the cathode. This will help you performed the experiment from a safe distance.
 
During the run, record all temperatures frequently, such as once per second, rather than every 10 seconds. If the cell explodes you will want a record of this -- assuming you survive. (If you think I am joking about this, you have not been paying attention and you should not be doing this experiment.) The video record may also be helpful in the event of an explosion. You should also record the water column temperature, ambient temperature, input power, weight scale reading, and all other relevant parameters in the same computer, in simultaneous time stamped records. This should be obvious but I suppose some people might record them in separate devices and then try to reconcile them manually.
 
Immediately after the run, momentarily turn off the magnetic stirrer to record the exact mass of the cell. Then turned back on and let it run until the cell cools to room temperature, which should take about an hour. Continue recording all parameters during this period.
 
Also immediately after the run, stir the water column (unless you also have a magnetic mixer for it), to be sure you have recorded the correct temperature. If the temperature has risen less than 1°C, you started with too much water in the column. If it rises more than 10°C you need to increase the starting volume of water. Measure the pH of the water column to estimate how much of the unboiled water has passed through the glass wool trap.

3) MY UNDERSTANDING WAS THAT ONLY THE VAPOR PASSES THE TRAP? THAT IS NOT THE UNBOILED WATER.

--> Mostly vapor, but a small amount of liquid water is bound to get through. Probably a tiny amount that does not matter.

4) I WOULD NOT KNOW HOW TO CALCULATE THE AMOUNT OF UNBOILED WATER PASSING THROUGH THE TRAP ON THE BASIS OF PH . . . A LITTLE NUMERICAL ILLUSTRATION WOULD HELP.

--> Neither would I but I am sure it would be easy to find out.


When the cell cools down, turn off the magnetic stirrer and record the exact mass one more time to account for any additional losses to evaporation, and measure the mass of the water column beaker.
 
That should do it. This should give you enough data to accurately account for heat lost to evaporation and radiation from the cell.

5) I SEE THAT THE OHMIC HEATER IS NOT MENTIONED. HOW IS IT USED?

--> It should be included to calibrate. I neglected to mention it, because it is obvious.

It should greatly reduce wet steam, and even if there is any wet steam, all of the enthalpy from it will be captured in the water column so it will not matter.

6) BY WET STEAM YOU PROBABLY MEAN VAPOR MIXED WITH TINY DROPLETS. IS THIS CORRECT?

--> Yes

Note that mass lost from the cell and not accounted for in the water column is mainly free hydrogen and oxygen. You can estimate whether this exceeds normal Faradaic levels, which would indicate that some hydrogen and oxygen was generated by pyrolysis.

7) THE ABOVE TWO SENTENCES ARE NOT CLEAR TO ME. MY UNDERSTANDING WAS THAT THE WATER COLUMN IS OPEN FROM ABOVE. UP TO THIS POINT I THOUGHT THAT THE COLUMN WAS OPEN, ALLOWING THE H2 AND O2 TO ESCAPE INTO THE AIR. WHAT DID I MISS?

--> You missed nothing. The mass lost from the cell is in three phases:

Liquid -- collected in water column
Vapor -- condensed in water column
Gas H2 and O2 -- escapes from column


You can estimate very roughly how much gas escapes by change in mass in the cell minus change in mass of column.

I leave the details of the heat balance computation up to you.
 
General advice: Do not stint on instruments or materials. Do not use haphazard methods, unusual methods, half-measures or unreliable equipment. If you are going to devote several weeks or months of your life to this work -- and perhaps even risk your life and health to accident -- you should either do this right or do not do it at all. I cannot over emphasize that. As Storms pointed out, this field has seen too many half-baked experiments already.

If the results show a positive heat balance I recommend you publish a paper along with every scrap of data and every single still photo on a web page, such as LENR-CANR.org. Do not hold back or summarize data; let those who would critique your results see every detail.

General advice to all researchers, everywhere: Read the last message written by Wilbur Wright, a few weeks before his untimely death. Wright was one of the most gifted experimental researchers who ever lived, and he survived hundreds of tests far more dangerous than a glow discharge experiment. His message is on page 11 of this document:

http://lenr-canr.org/acrobat/RothwellJthewrightb.pdf

Every word of Wright's message applies to cold fusion, in spades. Most research has failed because people ignored the precepts he outlines. I would post his text here, but I encourage you to read the whole document.

P.S.
I wrote:

You might want to include an emergency valve in the lid. This can be as simple as a drinking straw bent into a "U" shape and stuffed into two holes.

Cancel that. The exhaust tube is good enough. If there a buildup of steam too quick for that tube to vent, the drinking straw valve will also fail. The top will fly off and the cell will shatter. This happened to X. Zhang et al. three times, as I noted in a revised version of chapter 12 in my book, uploaded yesterday. See:

Zhang, X., et al. On the Explosion in a Deuterium/Palladium Electrolytic System. in Third International Conference on Cold Fusion, "Frontiers of Cold Fusion". 1992. Nagoya Japan: Universal Academy Press, Inc., Tokyo, Japan.

Message: Do not stand too close to these cells. Do not sit there with your eye a few centimeters away from the cathode the whole time, watching the glow discharge. Use safety goggles. For Goodness Sake. Have some common sense.

74) L. Kowalski wrote:
In trying something new Richard wants to use a closed cell (with a recombiner). Those who did this used a flow calorimeter. One must learn how to avoid problems with the recombiner, and how to build a reliable flow calorimeter. A bomb calorimeter, mentioned by Jed, probably offers a way to avoid the flow calorimeter.

(a) Considering the high rate at which excess energy is generated in a plasma electrolysis cell, for example 50 W or 100W, the calorimeter can be very simple. In fact, it can be a carton box surrounded by a layer of white packing styrofoam. A small air circulation fan is likely to be neceassary to keep the inner temperature uniform. No scale would be used, as in the water-evaporation calorimeter.

(b) One should be able to open the "door" of the calorimeter and put the working cell inside rapidly. Then it becomes a matter of recording the rising temperature for the duration of a test. Suppose the door is closed at t=0 and the glow discharge current is turned off at t=9 min. Suppose the change of temperature, in the last 5 minutes, is 24 C. How much thermal energy was generated during the test?

(c) The heat capacity of the calorimeter (K in joules per degree) must be known to answer this question. It can be determined by using an ohmic heater, for example, a 60 W light bulb. Suppose the temperature changed by 9 C when the light bulb was on for 10 minutes = 0.1667 hour. The electric energy received would be close to 60*0.1667=10 w*h. I would not trust the 60 W written on the ball; I would measure the w*h with an instrument. The measured 10 w*h translates into 36000 J. Thus K=36000/9=4000 J/C. That is the calibration of my bomb calorimeter. The value of K must be determined when the cell with the electrolyte is inside the calorimeter. This is important because by removing the cell, or by reducing the amount of the electrolyte, one would decrease the value of K.

(d) Now the question asked in (c) can be answered. The formula is Ethermal=K*dT=4000*24=96000 J. That is it. (It terms of power Pth = 96000J / 300s = 320 W).

(e) There is no need to wary about splashing or about tiny droplets being mixed with the vapor. The electric energy, on the other hand, delivered to the cell during the last 5 min, would be measured in the same way as for an open cell. Suppose the Pel turns out to be 240 W (leading to Eel=240*300=72000 J. Then one would say that the COP=96000/72000=1.33. But one thing was neglected. The recombiner has to be heated and this must be taken under consideration before calculating the COP. Suppose the power needed for this is 40 W (?). This leads to 40*300/3600=3.333 w*h = 12000 J. Thus Tth =(96000-12000) = 84000 J and Pel=240-40=200 W. Thus Eel=200*300=60000 J and the COP=84000/60000=1.40. The energy delivered by the electric fan, for example 20 W (?) could also be accounted for.

(e) The nonevaporative losses are likely to be negligible, as I assumed, if the styrofoam is thick and and fits the box tightly. A sticky tape along the rims would help to reduce convectional losses. If losses are not negligible then they can be measured (see item i below).

(f) Unfortunately, the second big issue -- chemical origin of excess heat -- is not being addressed by switching from one type of calorimeter to another. That is where help from a good chemist would be essential.

g) Would this approach also work for an open cell? Probably yes, if one was willing the dry the calorimeter after each test. I would not try this.

h) I never used a bomb calorimeter. That is why the above description is probably too naive. What else should be taken under consideration?

i) Oh yes, the box should have a small mass, to make the calorimeter more sensitive (small K). Thermal inertia will probably be a factor. I suppose one would have to wait several minutes, after the current is turned off, till the temperature stops rising. Waiting much longer, perhaps several hours, one would see the temperature going down. The rate at which the temperature goes down could be used to calculate thermal losses due to conduction and convection. What else should be anticipated?

75) R. Slaughter wrote:
I'm confused...nothing new... I have two designs in mind. the first is a closed cell with a vent tube as describe by Michel and others in earlier post. We can measure the non-evaporated heat lost of this cell the same way we did with the open breaker cell. Instead of sparging through a condenser column we sparge the vent tube steam through water in a second beaker. The steam condense and adds water to the second beaker. We can even put the second beaker in an ice bath to make sure we condense all the steam. The weight gain of the second beaker is P-evaporated. We calculate COP the same as before. I'm counting on the small size of the vent tube to greatly decrease the possibility of mist and prevent expulsion of K2C03? We can also put glass wool as suggested by others in the vent tube to filter the steam. The liquid in the second beaker should just be distilled water if everything is correct. We can now check the condensed liquid for other chemicals and residues.

The second design is a truly closed cell with a condenser loop. The steam that escapes the vent tube sent to an open air condenser a small pump returns the liquid back to the cell. The trick to this design will be getting a steady liquid flow through the condenser. We measure the input and output temperatures of the condenser and with flow rate we have P-evaporated. We run the same setup using the ohmic heater and we can measure the non-evaporated loses in the cell. We again calculate COP as before.
P.S.
I should have added that both of these approaches would work best if the reaction can be keep going long enough for the system to reach equilibrium. I see a need to be able to feed the cathode to keep the system running for perhaps two hours. In the first design we would need a water reserve to keep the electrolyte level constant so the system could run for hours.

76) L. Kowalski (not posted):
I find

77) M. Julian:
Ok the proposals so far are (BOMs only for now please),

Michel J-1.0
- insulated cell (e.g. transparent Dewar)
- insulating lid, not airtight, vented, sitting on top of the cell ready to
pop off (holds the electrodes and heater)

Richard S-1.1
- Lid should be tight enough so that steam and gas can not escape around the edge. All steam and gas must exit the vent tube.
- Vent tube should be stainless steel. In the shape of a "U".
- Suggestion on the best way to recombine the H2 O2, for safety
concerns!
perhaps a spark plug or glow plug.
- Suggestion on best way to capture and condense the steam? My thoughts
are to run an insulated plastic tube from the vent to a second beaker of water on the balance and sparge the steam through water in the second beaker.

Michel J-1.0 being the simplest so far, let's make it the PUF (proposal under fire)

Michel J-1.0 method to determine mass loss due to mist (inspired by Peter Gluck's comment on potassium not being consumed):

Escaped mist mass = (Potassium mass loss)*K, where K is an easy to determine constant factor (65.71 for 0.2M K2CO3), and potassium mass loss is determined by chemical analysis of final electrolyte, initial potassium mass being known.

I initially thought this required a fresh electrolyte for each run, on second thought that's not necessary, initial potassium of one run being final potassium of the previous one. Small samples of the cell's water could be taken after each run for batch-analysing at the end of the series of runs. If no potassium has disappeared the COPs don't need correcting for mist, otherwise we have an accurate correction. Any potassium loss due to transmutations would be negligible of course.

Sample taking and analyzing is required for chemical heat assessment anyway, so added cost of the method is null.

Questions/comments/critics on this proposal please, let's quickly find it's flaws so we can amend it and move on to another proposal.

78) Peter Gluck:
Strong polycarbonate shield, . . . fire extinguisher, . . .goggles. [Also ] an analytical chemist friend who will tell you what on the earth happens with the potassium carbonate and all the other chemical, physical and nuclear (?) players involved in the game. [Also] an open mind - to decide to go along the line: "first understand, then measure again"

79) M. Julian:
Remembering unpublished information I have received from Pierre on preliminary Paris-2 results with their standard (silvered) Dewar, without changing the method for tracking mist and chemical heat I proposed, I would like to further simplify my proposal's BOM as follows:

MJ-1.1
- open insulated cell (e.g. narrow transparent cylindrical Dewar cf attached picture)

(insulating lid is not necessary if the insulated cell has a small enough open area for non-evap losses to be sizeably smaller than observed excess heat, which is the case with the Paris-2 narrow Dewar, where non-evap losses around 30W are only one third of excess heat!)

Electrodes and heater (the latter would probably only be needed for calibration) could be held as in Paris-1 and 2 experiments by a tripod weighed along with the cell, cf photo in http://www.lenr-canr.org/acrobat/FauvarqueJabnormalex.pdf

Periodical replenishing to approximately constant volume (level mark on the container) could be done by adding external boiling water from e.g. a saucepan on slow boil in counted constant doses of e.g. 20ml, or as in Fauvarque setup from an ambient temperature reservoir weighed along with the
cell.

What would then distinguish the MJ-1.1 proposal from Paris-2 Dewar experiment, apart from optional transparency of the Dewar (which allows for comfortable visual adjustment of the discharge and should not increase non-evap losses by more than a few Watts I would think), would mainly consist in the added method for tracking mist and chemical heat.

80) Ludwik Kowalski wrote:
Even two or three determinations of K1/K2 (see below), under typical conditions, would be sufficient for us at this stage. Do not forget what triggered this discussion. We want to convince potential referees, of our anticipated paper, that the reported values of the COP are not due to the unaccounted for droplets. I am expecting K1/K2 to be close to 1.00. We will not claim a confirmation of the Mizuno-type excess heat if K1/K2 turns out to be significantly smaller. Do you agree, Richard and Pierre? I am assuming that Texas-2, Paris-2 and Colorado-3 results will be similar to those from Paris-1 and Colorado-2 experiments. Otherwise the paper would have to wait till we understand the situation.

Appendix ?
a) Basic assumption:
The K2CO3 dissolved in water turns into ions of different kind. One kind is K+. In this analysis we will assume that what is escaping from the open cell, during a test, is only a mixture of water vapor and tiny droplets of electrolyte. That is an approximation; the mass of the H2 and O2, due to electrolysis, and possibly of other gasses, such as CO2 and CH4, is assumed to be a negligible fraction of the water mass loss. Note potassium ions are removed from the cell by the escaping droplets of the electrolyte; the vapor escaping from the cell does not cary the ions.

b) Conducting the test:
A small amount of the electrolyte is removed from the cell with a pipette, before a test in which the COP is to be measured. That is sample #1. A comparable amount of the electrolyte is removed from the cell after that test. That is sample #2. Both samples are sent to an analytical lab to determine concentrations of K+. Note that professional laboratories perform such measurements routinely on blood and urine. Units of ion concentration can be mg/cc or ions/cc.

c) Discussion:
Suppose that sample #1 shows 100 units of K+ while sample #2 shows only 99.0 such units. That would mean that 1% of the initial K+ was lost. Any arbitrary units can be used, for example, number of ions, micrograms, milligrams, etc.

If only the water vapor is escaping then the concentration of K+ in sample #2 should be practically the same as in sample 1. In that case K2/K1 would be close to 1.00. In a mixture of vapor and drops, however, the concentration of K+ in the sample #2 would be smaller than in the sample #1. The lowest possible value of K2/K1 corresponds to a situation when nothing but liquid droplets contribute to the mass escaping from the cell. In that extreme case (only drops are escaping) the value of the K2/K1 would be (M-dM)/M. That measurable ratio will be represented by the symbol W. If M=1000 g and dM=40 g then W=0.96. It is the lowest possible values of K2/K1 for a given experiment. .

In a mixture of vapor and droplets the K2/K1 ratio is a number between 1 and W, depending of the composition of the escaping mixture. We want to determine the composition of the escaping mixture on the basis of the actually measured W and K2/K1. Let R be a measure of the mixture composition; it is defined as the mass of tiny droplets divided by the total mass of the escaping mixture. To calculate the COP correctly we need to know R. The COP=1.2, from our Colorado-2 experiment, would be an illusion if R were close to 0.2. Assuming that the relation between R and K2/K1, for any given W, is linear one has:

R= (1 - K2/K1) / (1 - W)

As expected, this formula gives R=0 when K2=K1 (only vapor escapes) and R=1 when K2/K1=W (only droplets escape). If K2/K1 happens to be 0.988 (for W=0.96) then R is 0.3. That would indicate that 30% of the mixture consists of tiny droplets. The table below gives the expected values K2/K1 for different values of R, and vice versa. Note that the assumed W=0.96 is realistic; a typical test lasts about ten minutes and the mass lost during that interval is close to 40 grams.

R 0.010 0.040 0.070 0.100 0.200 0.300 0.500

K2/K1 0.9996 0.9984 0.9972 0.9960 0.9920 0.9880 0.9800

One thing becomes clear, to distinguish small differences in R, for example, between
4% and 7%, the K2/K1 determinations must be very accurate.

Note that the accuracy requirement becomes less severe (only 3 significant digits instead of four, as above) when W is reduced. Suppose that ten short tests, performed one after another. After each test about 40 grams of boiling water is added to the cell to keep the initial M constant. Then all ten tests can be treated as one test during which dM turns out to be close to 400. For that long test W=(1000-400)/1000 = 0.6. The relation between the R and K2/K1, for W=0.6, is tabulated below.

R 0.010 0.040 0.070 0.100 0.200 0.300 0.500

K2/K1 0.996 0.984 0.972 0.960 0.920 0.880 0.800


suppose the K2/K1 happens to be 0.984. That would indicate that R=0.04 -- 4% of the escaping mixture consists of tiny drops of the electrolyte.

d) Plan of action:
The mass of of escaping droplets must be taken under consideration when COPs are calculated. The mean COP=1.24, was calculated (after Colorado-2 experiments) under the assumption that R is very close to zero. That assumption must now be tested experimentally. Will the K2/K1 be very close to 1? I hope so. But that remains to be seen. Honest referees will probably take our claim seriously if the K2/K1 data are used to validate it.

81) Michel Jullian (posting even more simple description, publicly):
Ok so here is a revised version of the maths for the zero-hardware mist
evaluation method I propose.

NOTATIONS
---------
V Volume of solution in cell
CK concentration of potassium in cell solution
mK mass of potassium in cell
dmK mass of potassium lost (presumably via droplets)
Vd Volume of lost droplets
md mass of lost droplets (correction for droplets
to be subtracted from total mass loss)
StoK Ratio of salt mass to potassium mass
(1.77 for K2CO3 and 1.44 for KOH,
we will use 1.77 for conservatism)

RUN DATA
--------
1- BEFORE RUN
V1 (L) (will be roughly maintained during the run by adding pure water)
CK1 (g/L) (the same concentration will be assumed in the droplets)

2- AFTER RUN, RINSING OF CELL WALLS ETC WITH PURE WATER, AND MIXING
V2 (L) (a bit larger than V1)
CK2 (g/L)

CALCULATIONS
------------
mK1 = CK1*V1 (g)
mK2 = CK2*V2 (g)
dmK = mK1-mK2 (g)
Vd = dmK/CK1 (L)
md = Vd*1000+dmK*StoK (g)

Comments/corrections/critics welcome.

82) Ludwik Kowalski (after waiting the whole day for a reply to Michel):
1) Michel's method of studying wet steam -- a mixture of water vapor and droplets of the electrolyte escaping from an open cell -- seems to be good, at least in principle. It should be an essential part of our COP determinations. Fortunately, it is not too late to make the Paris-2 and the Colorado-2 claims more trustworthy. I hope that Richard and Pierre will confirm that the mass of lost droplets is only a negligible fraction of the liquid lost in a typical COP test. The percentage of the electrolyte droplets in wet steam must be reported in our anticipated paper.

2) My first comment is that the last formula ( dm= ...) is not needed. Knowing Vd (volume of escaping droplets) is sufficient. I am assuming that one gram corresponds to nearly one cc for the liquid water or the electrolyte. I am assuming that what is lost during a COP test is essentially pure water vapor and droplets of the electrolyte containing K+ ions. The combined mass of H2, O2, CO2, etc. is presumably negligible. For example, if 40 grams of the liquid was lost during a test (or 40 grams of water added to the cell to keep the same amount of liquid in the cell) then the Vd= 10 grams would indicate that droplets contribute 25% to the escaping mixture. In such case the mean COP=1.24, claimed in Colorado-2, would be an illusion. I hope that the Vd for such tests will be less than 1 gram; that would validate the tentative claim that excess heat was indeed close to 24% of the inputted energy.

3) Suppose we want to distinguish the ideal 0% "mixture" from a mixture containing 2% droplets. How accurate should the determinations of the CK1 and CK2 be? Suppose CK1=4.0000*10^19 ions of K+ per cubic centimeter. This corresponds to 27 grams of the K2CO3 salt dissolved in one liter of water, as in fresh 0.2 M electrolyte. Consider a typical test, lasting 10 minutes, in which 40 grams of mass escaped from the open cell. If the escaping mixture has 2% of tiny drops then the mass of droplets is 0.8 grams (the volume is close to 0.8 cc). This allows us to calculate the CK2. Unfortunately, the CK2 is only slightly smaller than the CK1. Replacing 0.8 grams, of the escaping electrolyte, by fresh water, changes the concentration of K+ ions by (0.8/1000)*100% = 0.08%. In other words the CK2 becomes 3.9968*10^19 ions per cubic centimeter. Measuring the CK1 and CK2 at the accuracy level of five digits is probably possible in a specialized lab. Very accurate determinations of CK2/CK1 (relative concentrations from two samples, one extracted at t=0 and another at t=10 minutes) are likely to be easier than very accurate determinations of individual CK1 and CK2.

4) Michel suggested much longer tests. Suppose that ten consecutive tests are treated as one 100 min test. In such test 400 of the mixture would escape. This corresponds to 8 grams of the electrolyte (instead of 0.8 g). The final CK2 would be 0.8% smaller than the CK1 (3.967*10^19 ions per cc). Under such condition the accuracy of only four significant digits would be sufficient to distinguish the wet steam containing 2% of droplets from the pure vapor.

5) The earlier suggestion of Michel -- capturing the escaping mixture and condensing the two samples (to determine the CK2/CK1 ratio) -- seems to be more attractive. In that case the accuracy at the level of two or three significant digits would be sufficient to distinguish the 2% mixture from the ideal 0% mixture. Measuring the CK2/CK1, in an open Mizuno-type cell, can be decoupled from the COP measurements. I already wrote how I would do this by using a large funnel. Jed's objection to the use of ice (to keep the funnel cold) would be valid if the COP were to be measured. But it is not valid for an experiment whose purpose is evaluate a typical steam wetness. The wetness of a sample, R, is defined as the ratio of the mass of electrolyte droplets and the total mass of wet steam. That is what we need in order to calculate the COP correctly. The description of my cold-funnel condenser can be found in the message #54 above.

Michel wrote that medical laboratories routinely measure concentrations of potassium ions. How accurate are these measurements? In what units are medical concentrations expressed? Should a typical chemist be expected to know how to measure the CK2/CK1 ratios (at the level of ~3% accuracy) in an improvised lab?

82) Michel Jullian:
. . . The last formula md = Vd*1000+dmK*StoK (g) is needed actually, to convert Vd (in liters) to the mass of escaped droplets (grams) which is what we are after. My conversion accounts not only for the mass of the water but also for the mass of the salt, as it should (even if it doesn't make much difference we might as well use the exact formula since it is quite simple)

Since Vd=dmK/CK1 the final formula can also be written md=dmK*(1000/CK1+StoK), which e.g. for 0.2M K2CO3, i.e. 0.2 mole of K2CO3 per liter, i.e. 0.4 mole of K per liter, i.e. CK1=0.4*39.1 g/L = 15.64g/L, would boil down to: md=dmK*(1000/CK1+StoK)=dmK*(1000/15.64+1.77). As expected we find: md=65.71*dmK (66*dmK will be used to compute accuracy below)

[You asked] “How accurate are these measurements?” We can see by ourselves in our own blood tests. Potassium is an important blood electrolyte, and it's concentration in routine blood tests is given to the hundredth of a millimole per liter (e.g. potassium (mmol/L) 4.47). Since we have about one liter of solution in the cell, we would know potassium mass to the hundredth of a millimole, i.e. to 0.39 mg (molar mass of K being 39g or so). Even if the error on dmK was five times that, it would only be about 2mg, so the error on the total escaped droplets mass would be only about 66*0.002 ~ 0.1g which you will admit is quite accurate!

83) L. Kowalski:
My recent blood report shows 4.2 mmol/L of potassium. The normal range is 3.5 to 4.5. I do not know how accurate the 4.2 result is. Several results (for the blood) would be needed to estimate the standard deviation. My guess is that it could be as large as 0.3, for a routine medical test. That would be 7%. By the way, the unit of concentration should be mmol, not the mmol/L. Do you agree? It could also be grams per liter or number of molecules per liter, or mg/L, or mg/cc, etc.

Suppose that the first sample has no droplets at all, its potassium concentration, CK1, is 0.000 mmol. Suppose the second sample has 100% of droplets. What is the potassium ion concentration, CK2, in the second sample? It is the same as in the original electrolyte. That amounts to 0.4 mol = 400 mmol (27.6 grams of K2CO3 are dissolved in one liter of water to produce a 0.200 M solution of the salt). A wet steam containing 1% of droplets would have 100 times smaller molarity -- 4 mmpl. That is about the same as for my blood. For the third sample, containing 2% of droplets molarity would be 8 mmol, etc. A difference in the potassium concentration resulting from only 1% difference in the steam wetness seems to measurable with automated medical analyzers. That is good news. But such analyzers could not be used when samples of the electrolyte are collected from the cell (before and after an experiment), as illustrated yesterday.

Condensing several grams of wet steam, with a cold funnel test, should not take longer than two or three minutes. That would be sufficient for several samples needed to establish the accuracy at which potassium concentrations are measured. My advice to Peter, Richard and Scott is to perform the cold funnel test, and to interpret it in the context in which the test was originally suggested by Michel. He is now preoccupied by other nuances but this should not delay a simple test described above. Let the perfect not be the enemy of the good.

Appended on 4/13/06:
On Apr 12, 2006, Michel asked:

Do you agree now that 0.025 mmol/L accuracy on CK1 and CK2 yields
about 0.1g accuracy on escaped mist mass if volumes are exactly
known and of the order of 1L?


1) What are we discussing? It is a method of measuring wetness of the steam escaping from an open cell during a Mizuno-type experiment. I agree with Jed (and others) that this problem can be eliminated by using the closed cell method. But that was the method we used. Paris-1 and Colorado-2 experiments were performed with open cells and we want to know the composition of the lost liquid (a mixture of water vapor and droplets of the electrolyte). The mass of the H2, O2, CO2, etc. was considered to be negligible. That is how the problem was originally formulated by Michel. If droplets of the electrolyte represent ~20% of the escaping mixture then the reported COP~1.20 can be explained as an artifact. On the other hand, if the percentage of droplets is close to zero then the reported COP~1.20 should be taken more seriously.

Two methods of measuring the percentage of droplets have been proposed by Michel. Which of these two methods is more desirable, from the point of view of accuracy? That is the issue.

a) Method 1: compare the concentration of K+ in two samples of the electrolyte, one extracted from the cell before the plasma electrolysis experiment and another extracted after the experiment. If two concentrations are identical then only a negligible fraction of the escaping mixture consists of droplets.

b) Method 2: compare the concentration of K+ in two samples of condensed mixture. One when the ohmic heater is used and another when plasma electrolysis is in progress. If the two two concentrations are identical then only a negligible fraction of the escaping mixture consists of droplets.

2) The numerical example below will show why I am in favor of the method (b). To prepare a 0.2000 M solution of K2CO3 we dissolve 27.000 grams of the salt in 1000.0 grams of water. Five digits of accuracy will be used in my arithmetic. The original concentration of K+, based on known atomic weights of K, C and O, is:

27.000 * (2*39.098 / 138.19) = 15.278 grams of K+ in 1015,3 grams of the electrolyte.

In other words, the original concentration, CK1 is 15.048 units (the units are grams of K+ per 1000 grams of mixture). That is the value CK1 for the method (a). Suppose a test, lasting 10 minutes, was conducted and t50.000 grams of the mixture escaped. To compensate for this we add 50.000 grams of water to the cell and measure the CK2. What is the expected value of CK2 when 20% of the mixture consists of droplets of the electrolyte? The fraction of the K+ removed during the test is (50 / 1015.3) * 100% = 4.9246 %. That amounts to 0.75238 grams of K+. The new mass of K+ is (15.278-0.75238) = 14.526 grams of K+. It is dissolved per 1015.3 grams of the mixture. Thus the CK2= 14.307 units.

4) Now let me compare the CK1 and CK2 for the second method. The CK1=0.0000 because no K+ is expected to be found in condensed "mixture" when droplets represent 0%. And what is the value of CK2 when 20% of the mixture consists of droplets of the electrolyte? We now have 0.75238 grams of K+ in 50 grams of condensed mixture. That gives us CK2=0.015048. Distinguishing 0.0000 from 0.015048 is likely to be much easier than distinguishing 15.048 from 14.307.

5) To summarize let me show a table of expected values CK for different percentages of droplets in the mixture, first for the method (a) and then for the method (b). Units of concentration are g/1000g, as explained above.

drops --> 0% 1% 2% 3% 4% 20%
==========================================================================
CK (a) --> 15.048 15.011 14.974 14.937 14.900 14.307
CK (b) --> 0.0000 0.0015 0.0030 0.0045 0.0060 0.0300

The table shows that five digits of accuracy would be needed to measure concentrations from the method (a) and only three digits from the method (b). That is why I wrote that the method (b) is more desirable. It is easier to deal with situations in which what we want to know (% of droplets) is directly proportional to what we actually measure. Note that numbers in the last line double and triple with doubling and tripling of the corresponding numbers in the first line.

6) When Richard and Pierre send me samples (one from condensing the mixture during the ohmic heater boiling and one from condensing the mixture during the plasma electrolysis) then I will try to measure their concentrations on the basis of conductivities. To do this I will build a small cell with two Pt electrodes. The current flowing through the cell is likely to be directly proportional to CK, provided the voltage (such 0.5 V) and temperature (such as 20 C) remain constant. I will calibrate that cell by using several electrolytes of known concentrations. Should I anticipate problems with this simple approach?

Yes, I know that other ions, if present, would also contribute to the current. The conductivity test is not selective; the current will be proportional to the total concentration of all cations. But suppose the two currents are practically identical for my two samples. In that case it would be possible to say that the contribution of droplets was negligible during our experiments.

84) Michel Jullian wrote:
Another possible chemical fuel which could be supplied indefinitely by the atmosphere: nitrogen (80% of ambient air). As Bill Collis rightly mentioned in the Jules Vernes thread: “Ah but one of the benefits of burning in pure oxygen rather than
air is that you cannot form any polluting nitrogen oxides (NOx).”

My scenario for this other possible Mizuno artefact: (1) atmospheric N2 dissolves spontaneously in the electrolyte solution (SCUBA divers know this phenomenon quite well) (2) N2 in solution is convected towards the plasma area (3) N2 burns in the high temperature plasma where explosive O2+H2 recombination occurs, forming NOx (a good soul with one of those rubber bibles may tell us how much heat this produces per gram of burnt N2 (4) GOTO (1) I wonder how this artefact could manage not to happen if the cell is exposed to ambiant air BTW. Another good reason not to let atmosphere interfere with the experiment isn't it?

85) Ludwik Kowalski:
As shown in unit #290, several people think that Colorado 2 experiment was a waste of time. I have no idea how to estimate the amount of excess heat generated by N2 reactions. Why should these reactions produce 24% of excess heat and not, for example, 50% or 500 %?

Not being a chemist I will not be able to address every possible objection. Perhaps it was a mistake to go ahead without a good chemist. The goal was to perform an experiment as in Paris 1. We did this and we got similar results. But this will not be enough for referees. A paper without convincing evidence, against chemical origin of excess heat, presented by a reputable chemist, is likely to be rejected. I would not criticize the editor for this. It is too bad that original results of Mizuno, or results of numerous reports from Naudin, were not discussed, as far as i know, in terms of exothermic chemical reactions. It is too bad that nobody waned me about the issue of safety after units #252and #253 was posted.

86) Michel Jullian:
Group, I was quite glad to find out, thanks to a Google search, that one can do thermochemistry without one of those rubber covered chemistry handbooks and associated expertise and tedious calculations :)

I was able to check the hypothesis in my previous post, and more, with a trial version of CHEMIX ("lab" edition, available at http://home.c2i.net/astandne/ ), a very helpful tool whose thermochemistry page could be dubbed "thermochemistry for dummies", so simple that even it being in Norwegian language was not an obstacle to using it! (I must have missed the language option at download or install time).

Example of use, you type in the following unbalanced reaction (electrolysis of water):

H2O(l) > H2(g) + O2(g)

it balances it for you (works out the coefficients), works out the enthalpy change, and outputs:

2H2O(l) = 2H2(g) + O2(g) - 571.6kJ

meaning that the reaction is endoenergetic, 1 mole of the elementary reaction requiring 571.6kJ of energy (electrical or thermal or whatever) You can even get it to display the energy for a given mass of either a reactant or a product, e.g. you type: 1gH2O(l) > H2O(g) and it outputs:

H2O(l) = H2O(g) - 2.44237kJ

meaning vaporizing 1 g of liquid water requires 2442J (and not 2260J as we have been using, can anybody explain this discrepancy BTW?)

The various Nox (NO or NO2) forming reactions that I tried turned out to be globally endothermic (absorb heat) so we probably don't have to worry about those.

A more preoccupying reaction is the spontaneous and exothermic (produces heat) reaction between N2 and H2 forming ammonia gas:

N2(g) + 3H2(g) -> 2NH3(g) (+ 15.1795kJ per g of H2)

but fortunately this produces much less energy per g of H2 than was used to produce the same gram of H2 by dissociation of water (e.g. thermal dissociation):

2H2O(g) -> 2H2(g) + O2(g) (- 119.948kJ per g of H2)

so we don't have to worry about this reaction either. However the possibilities of reactions between the various substances in play being quasi infinite it would be very presomptuous to claim none of the reaction paths is globally exothermic.

Whereas preventing the atmosphere and it's unlimited supply of various substances from interfering with the experiment, and still getting over hours/days/months much more excess heat than could possibly be produced chemically by the precisely known limited amount of substances we will have deliberately put into the cell, would make the claim of non-chemical origin of the excess heat much stronger obviously.

About the method 3 I suggested below to achieve this (keeping the solution boiling at all times to maintain an outgoing flow of steam), of course the cell opening would have to sufficiently narrow, and the liquid level sufficiently low, to make sure the outgoing steam constitutes an efficent barrier against atmospheric gases contacting the liquid surface.

87) Michel Jullian:
Hi John. Yes, liquid and gas H2O are at same temperature (100°C). I have found that the discrepancy comes from formation enthalpies variations with temperature being significantly different for liquid and gas between standard temp 25°C and 100°C.

According to
<http://webbook.nist.gov/cgi/cbook.cgi?Formula=H2O&MatchIso=on&NoIon=on&Units=SI&cTG=on&cTC=on> the standard (25°C) formation enthalpies are:

DH0f H2O(g) -241.83 kJ/mol
DH0f H2O(l) -285.83 kJ/mol

So the standard enthalpy change is +44kJ/mol, which for water's 18.02g/mol is 44/18.02=2442J/g as given by CHEMIX.

But if you now take into account the temperature dependencies, also given in the NIST page, you find the actual enthalpy change at 100°C is only about +41kJ/mol (based on rough interpolations from their tables: gas DH0f increases by about 2.5kJ, whereas liquid DH0f increases by about 5.5kJ, so gas minus liquid decreases by about 3kJ), which is 41/18.02=2275J/g very close to the 2260J/g we have been using. So 2260J/g must be the correct value for 100°C evaporation heat of water indeed.

This shows that the usual approximation that enthalpies do not depend on T can lead to significant inaccuracies in some cases. What would be needed is a super-CHEMIX which wouldn't make such approximations.

P.S.
Powerful consequences of the proposed "permanent one way steaming" scheme with very long duration runs:

1/ After a while, the only gases that can possibly escape the cell in meaningful amounts are water vapor, hydrogen and oxygen. 2/ The long duration of the run renders the potassium mass loss significant if there is any significant salted mist escaping, which makes it easily measurable.

1/ and 2/ make the mist-corrected evaporated mass calorimetry method simpler, easier and more convincing than ever. For example it is not needed anymore to explain why mass or energy balance of other gases such as CO2 cannot overestimate the COP, since there are no other gases. All there is to explain is that our counting of H2 and O2 escaped mass as evaporated H2O mass only underestimates the COP since it takes much more energy to dissociate 1g of water than to evaporate it.

Of course for the runs to last hours/days the necessary cathode replacements, whether continuous or not, have to be part of the run. If we keep using the easily procured 2.4mm welding sticks we could consider forming them beforehand by operating them in short preliminary dummy runs (e.g. the day before) so they are already suitably pointed and cracked when they are inserted into the experiment, ready to produce the right kind of discharge right away.

Regarding my proposal of an immersed smoothing filter to make input power and energy measurement easier and more convincing by suppressing spiky i and v waveforms and associated EMI, it has occurred to me that only the most dissipative part of it (the resistor) really has to be immersed, as neglecting the comparatively small losses in the capacitor and inductor only underestimates the COP, and only slightly. This simplifies things by allowing the use of the already included ohmic heater as the filter resistor at electrolysis time, and suppressing the need for any additional submersed component: the cap and inductor just have to be located in an EMI-shielded area, e.g. a grounded Faraday cage made of screen which would also enclose the cell.

The pieces of the puzzle seem to all come together nicely, making my dream of a both simplistic and indisputable Mizuno experiment less unrealistic than it was.

Michel Jullian (private message):
I know our common objective is to prove to anyone interested -not just the converted few but also honest scientists and politics- that the Mizuno excess heat is (a) indisputable and (b) indisputably abnormal. This should have been done long ago by Mizuno himself or others but hasn't. To fully realize this objective, we both know that some changes are required wrt existing experiments.

I have proposed yesterday what is probably a minimal (let me know if you agree on this qualifier(*)) set of such changes wrt the nice and simple Fauvarque-ICCF12 version of the experiment run in Paris and Colorado, in my "making the puzzle fit together" post. Note the cell remains _physically open_ in this proposal so it can't turn into a bomb a la Mizuno ICCF12 (no open cell has ever exploded I don't think, correct me if I am wrong)

What this proposal needs now (if it works, and even more if it doesn't so we can quickly find it's flaws and move to a better one without regrets) is commenting and criticizing by the group so an unbiased and argumented consensus can be reached wrt it's fitness to achieve the objective.

88) Richard Slaughter (private):
Ludwik, Good morning hope all is going well. I'm still waiting on getting my boards back so I can't measure the input power correctly. I've designed a new cell that is looking very promising. The new cell is closed except for a 0.5 inch hole in the top to let out the H2 02 gas. I've put a (condenser) coil of tubing just below the lid of the cell. Between the lid and the coil are several layers of stainless steel screen. The screen is in contact with the tubing so it stays the temperature of the tubing.

I've a large ice chest with ice water and I'm running the ice water though the condenser. Using my 0.25 inch tungsten rod I'm able to run the cell for an hour before the tungsten burn away. The water in the ice chest is staying at 1C for duration of the test so I only need to measure the outlet temperature of the condenser and the flow rate to get my output power. The outlet water from the condenser circulates back into the ice chest. I spray it as a fine mist back into the ice chest so it looses a lot of heat to the air.

The temperature of the electrolyte is about the same as before 80C. When I run the test there is very little loss in mass. Looks like input power is going mostly into making steam and the steam is immediately condensing on coil of tubing. Haven't got any real data yet but I'm encouraged. If I can make this type of cell run for several hours at equilibrium with excess heat it should put to rest most of the objections were hearing on CMNS.

89) Michel Jullian:
The changes wrt Fauvarque-ICCF12 in the new proposal (let's call it BM_MJ-2.0, BM for boiloff Mizuno) are really few and simple, I'll summarize them here:


- insulated cell with a not-too-wide and not-too-low opening (e.g. same double walled reactor as in JLN's enhanced CFR, only with a closed airspace in the double wall) - a couple protocol changes to prevent ingress of known (CO2, N2) and unknown reactive atmospheric substances, make the runs long-duration (hours/days so any initial chemical fuel is exhausted), and correct evaporated mass for escaped mist (note long runs make the cell potassium concentration method more accurate)

- plus an _optional_ hardware change which boils down to adding an inductor and an EMI shielding screen and doing some rewiring to make electrical measurements totally convincing, even if done with a dumb AC mains watt-hour meter which anyone can use and read.

Note the cell remains _physically open_ in this proposal so it can't turn into a bomb a la Mizuno ICCF12.

What this "minimal change" proposal needs now (if it works, and even more if it doesn't so we can quickly find it's flaws and move to a better one without regrets) is commenting and criticizing by the group so an unbiased consensus can be reached wrt it's fitness to realize it's objective.

It's objective being what I assume to be a common objective of this group: proving to anyone interested -not just the converted few but also honest scientists and deciders- that the Mizuno excess heat, and hence cold fusion excess heat in general, is (a) indisputable and (b) indisputably abnormal. As I suggested before, simpler solutions should be discussed/eliminated first, because they are more likely to realize the above objective than more complicated solutions. Or am I missing something?

90) Richard Slaughter:
MIchel,
I'm working on a design for a new cell. I'm trying to incorporate as much as I can about what has been said here. I think to have an indisputable experiment we need to do the following. The cell has to be completely closed. A high speed analog measurement of input v(dt)*i(dt) something like the AD835 multipllier chip. The H2-02 needs to be recombine inside the cell. The cells need to run at equlibrium for at least 1 hour producing excess heat.

I think this can all be accomplished and still have as simple a system as used in the Fauvarque experiment. The key is to condense the steam inside the cell at the same rate it is being generated. I propose we put a liquid to air condencer inside the cell. The condencer only has to disapate about 300Watts to bring the system into equlibrium. A constant temperatue ice bath provide cooling water to the condenser. We only need to measure the output temperatue and flow through the condenser to calculate output power. I'm not sure of the best way to recombine the H2-02. I'm thinking of using a spark plug or glow plug. The whole cell can be well insulated so the only heat output is via the condenser.

If flow though the condenser is 0.5 liters/min and we have a temperature rise of 15C that's over 500 watts. This seems reasonable approach.

91) Michel Jullian:
Richard, I will be glad to discuss your new proposal (RS-2.0?). Shall we discuss mine next then?

92) Richard Slaughter:
Michel, Lets dicuss yours first. I concerned that as long as you are trying to do a boil off it will be difficult to keep the system running for hours or days? Replinishing of the lost water greatly adds to the complexity of calculated the heat loss. Can the critics say that fresh water is adding "fuel"? I tried to replinish the water for a while and found, it difficult to maintain the fresh water temperature because of the output steam. I'm sure this problem can be solved. The boil off approach is always going to be suseptable to changing conditions inside the lab such as Temperature, humidty and pressure. I don't see the system ever reaching an equlibrium where COP is constant. If we can devise an experiment where the COP is greater than 1 and constant for hours or days that is going to be very convincing.

You comments on the EMI sheild and computing input power are correct. I think this is just a case of selecting and building the correct hardware. The other problem with all system is how to keep the W rod active for hour or days? I see two approaches to this problem. One is to build a system that auto-feeds the electrode. The other is devise an experiment that can reaches equilibrium quickly and thus doesn''t need as much electrode. My Musing, Richard

93) Michel Jullan:
Richard wrote: Lets dicuss yours first. I concerned that as long as you are trying to do a boil off it will be difficult to keep the system running for hours or days? In any non-automated scheme, boiloff or otherwise, hours will be easy, days will require taking shifts like at the helm of a boat... or more appropriately like at the controls of a power plant :)

Replinishing of the lost water greatly adds to the complexity of calculated the heat loss. Easy, we add water from pre-weighed bottles and weigh the last unfinished
bottle to know how much pure water we have added.

Can the critics say that fresh water is adding "fuel"? No :) Unless we add gasoline or whisky to it to improve our COP ;) Using water as a fuel to produce heat would be quite an achievement anyway.

I tried to replinish the water for a while and found it difficult to maintain the fresh water temperature because of the output steam. I'm sure this problem can be solved. Let's keep the bottles away from the steam.

The boil off approach is always going to be suseptable to changing conditions inside the lab such as Temperature, humidty and pressure.
No, lab conditions have no influence whatsoever on this scheme since non-evap losses are small by design (<30W) and therefore can be neglected altogether if we can sustain a COP>1.1 at 300W. This prevents any objection based on bubbles or foam reducing those losses since we already assume they are null!

I don't see the system ever reaching an equlibrium where COP is constant.If we can devise an experiment where the COP is greater than 1 and constant for hours or days that is going to be very convincing. Indeed, it will be very much like a car going on for ever without ever refueling.

You comments on the EMI sheild and computing input power are correct.I think this is just a case of selecting and building the correct hardware.
Actually an off-the-shelf residential electricity meter (joule or watt-hour meter) working from 100-400VAC installed between the variac (or insulation transformer) and the rectifier will be perfect with a proper smoothing filter, which only requires adding an inductor to existing setups and doing some rewiring to use the ohmic heater as the filter's resistor at electrolysis time ; flipping a multipole switch to go from electrolysis mode to heating mode will put the heater in the right circuit.

The other problem with all system is how to keep the W rod active for hour or days? I see two approaches to this problem. One is to build a system that auto-feeds the electrode. The other is devise an experiment that can reaches equilibrium quickly and thus doesn''t needas much electrode. As I said in an earlier post we can also keep using the easily procured 2.4mm welding sticks. I suggested to pre-form them (by operating them in a "dummy" glow discharge before the actual run) so they can produce a high COP immediately when inserted into the system. While changing the cathode, which won't take long with a proper holder/connector, we switch to ohmic heating mode to keep the vapor barrier active, and then flip back to electrolysis mode.

More constructive comments/criticisms please, it seems we haven't succeeded
in ruling out this scheme yet and there are others awaiting discussion so we must validate it or debunk it fast.

94) Ludwik Kowalski:
Colorado2 experiments replicated the COP>1 claim of the Paris-2 team. They showed that by doing the same things we obtain similar results. That was an important step forward. To make Colorado-2 results publishable we must be able to convince honest referees that excess heat cannot be explained in terms of known chemical reactions. That is the main issue. Unfortunately, it has not been addressed by recognized authorities in chemical aspects of high voltage electrolysis. Did Mizuno, or some of his collaborators, publish something along these lines in Japanese?

My understanding was, and still is, that a complete analysis of potentially possible exothermic chemical reactions is now being conducted in Paris. Im I correct? Planning for new setups, ahead of such analysis, does not seem to be productive, at this stage. Writing a paper based on Colorado-2 results also seems to be premature, for the same reason.

But we should compare the conductivity of water (from which the electrolyte was made) with conductivity of condensed liquid escaping from the cells (when the COP was measured). In that way we can estimate the percentage of tiny droplets of the electrolyte in the escaping steam. Several people suggested that the COP>1 might be due to our inability to distinguish between water vapor and droplets of the electrolyte. Yes, this problem would be eliminated in a closed cell, as illustrated by Jed. But Paris-1 and Colorado-2 experiments were performed by using open cells. Our anticipated paper will describe these experiments, and results should be made trustworthy.

95) Richard Slaughter:
Michel, If the goal you've set for your experiment is to make the existence of excess heat "indisputable, and indisputable abnormal". I think that it is essential that your data colection system run continuous for hours or days. You must collect the input and output power measurments at least once a minute and ideally once a second. You must be able to add water to your system and adjust the electrode without disrupting the data stream and keeping the reaction going. In the French 1 and Colorado 2 experiments this was one of the shortcomming. We could only collect continuous data for 5 to 20 minutes so we could only say we had excess heat for a 5 to 10 minutes

I think from an engineering standpoint it will be very difficult in the boil off experiment to maintain the reaction and the continuous data stream. Any disruption in the system seems capable of stopping the production of excess heat. In Colorado we would add hot water to replinish the lost water so our down time was a minimum. Still adding the hot water could stop the reaction. Adding cold water always stops the reaction. Adjusting the electrode required turinng off the input power and killing the reation.

The continuous data stream is essential to the primary goal of being indisputable. If you can show that excess heat continues for hours or days it's hard to argure that it is chemical in nature. It is this practical problem of keeping the reacting going for hours or days that is the biggest downside I see to the boil off experiment. If you have a engineering solution to add water and adjust the electrode so the reaction continues that's great. I don't see a way and keep the experiment simple. That why I'm leaning toward the closed cell approach. Richard

96) Michel Jullian:
Hi Richard, my comments in your text below, feel free to do the same.

If the goal you've set for your experiment is to make the existence of excess heat "indisputable, and indisputable abnormal". I think that it is essential that your data colection system run continuous for hours or days.

Long runs would be nice, highly desirable even, but are not essential in my proposal. Short runs as you have been doing would be still ok in this scheme: provided the vapor barrier is not interrupted in-between, there can be no ingress of potential chemical fuels.

You must collect the input and output power measurments at least once a minute and ideally once a second.

I was thinking of collecting them permanently by integration: measuring cumulated input energy (watt-hours from a simple electricity meter working on the digital multiplier principle, that's how recent meters work it seems) and cumulated output energy (evaporated mass), and reading them as often as we can so we get a smooth curve, but without this being an obligation: if there is no reading for 2mn it doesn't affect the run's global COP.

You must be able to add water to your system and adjust the electrode without disrupting the data stream and keeping the reaction going. In the French 1 and Colorado 2 experiments this was one of the shortcomming. We could only collect continuous data for 5 to 20 minutes so we could only say we had excess heat for a 5 to 10 minutes

I think from an engineering standpoint it will be very difficult in the boil off experiment to maintain the reaction and the continuous data stream. Any disruption in the system seems capable of stopping the production of excess heat. In Colorado we would add hot water to replinish the lost water so our down time was a minimum. Still adding the hot water could stop the reaction. Adding cold water always stops the reaction. Adjusting the electrode required turinng off the input power and killing the reation.

All this is useful information Richard. Why did you have to turn off the input power BTW? The cathode was grounded wasn't it?

The continuous data stream is essential to the primary goal of being indisputable. If you can show that excess heat continues for hours or days it's hard to argure that it is chemical in nature. It is this practical problem of keeping the reacting going for hours or days that is the biggest downside I see to the boil off experiment. If you have a engineering solution to add water and adjust the electrode so the reaction continues that's great.

Replenishing with boiling water is one thing. Weighing it would be trickier but we could use the magnitude of the steps on the scale readings to know precisely how much we have added couldn't we? To adjust the electrode, how about holding/connecting it in a sliding arrangement so it could be tapped gently with some tool to push it down a few mm at a time?

I don't see a way and keep the experiment simple. That why I'm leaning toward the closed cell approach.

I don't follow your reasoning. A closed cell would make things even more difficult wrt the above engineering questions than an open one it seems to me. Also what would be the advantages of the closed cell that the one-way scheme of the vapor barrier doesn't offer? They would have to be worth the risk of explosion...Michel

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