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Click to see the list of links 303) Well known reactions or something else? Unit 302 was devoted to possible interpretational error in calculations of excess heat. But that is only one of our problems. Suppose that all the data
used to calculate the percentage of tiny droplets in wet steam are found to be reliable and that the COP (coefficient of performance) was indeed equal to
1.24, as described in unit #300. I will now assume that readers are already convinced that no experimental or interpretational errors were made when excess
heat was determined. What is next? We must also convince readers that what was measured (excess heat generated at rates of nearly 100 W, and equal to 24%
of the input energy) cannot be attributed to well known chemical reactions. Need for producing convincing arguments has been emphasized before. But no
progress was made, I my opinion.
Ludwik Kowalski (L.K.) responded: Why should we exclude a possibility that what is escaping in tiny droplets of the liquid mixed with dry steam is [colloidal?] tungsten, or something else different from K? The method used by Scott (measuring masses) seems to be more universal. And it is very simple. Most laboratories have scales which are sensitive enough to detect solid remains. Confirming or refuting Scott's finding is important to all of us. One does not have to measure the COP to demonstrate that a sizable fraction of the lost mass is not pure water. A “solid remains” test, by the way, can be performed by using stainless steel electrodes, as in many Naudin's experiments. = = = = = = = = = = = = = = = = = = = = P.G.: Tungsten - and what happens with with it - c'est une autre chose- a different problem. The electrode is eroded and a precipitate forms. What is this? Based on my experience with the Cincinnati Cell I think the precipitate is mainly molten and frozen W droplets that capture inside a quantity of K. The droplets are formed by a known process that's called spark erosion. In the Cinci Cell, all the salts present are fixed physically in the molten metal. being removed from the solution, in the precipitate. The aqueous phase ceases to be electrically conductive. The Cinci Cell has Zr electrodes - and Zr has a melting point of 2125 C, compared to 3680 C for W, but the mechanism is similar. However - to demonstrate this the precipitate has to be thoroughly analyzed. Anyway, it seems to be improbable that a part of the W electrode is flying with the droplets. Do we have soluble W present? And what's : "something else different from K?" So what is Scott weighing? We know that a part of CO2 is also lost - carbonate, bicarbonate, hydroxide all can be present - K is the only thing measurable equivocally. In this case- the mode of thinking- including a hypothesis how the system works is as important as the execution of the experiment. L.K.: That was my way of saying that "showing that amount of escaping K is negligible” is not the same as saying "what is escaping is pure water." The COPs were calculated by assuming that only pure water escapes. Now this assumption is being questioned. P.G. : Ok, I understand what you say, but flying W could be easily detected. However we are able to know what happens in the system and we can decide - with certainty: there is excess heat, when, when not and perhaps later - why. The system is not awfully complex - the unique strange thing is the huge temperature gradient - molten W in contact with boiling water. How pyrolysis of water interacts with its electrolysis???? Is some synergy that causes the apparent excess heat? L.K. : Yes, well known things and "strange" things do take place. The task is to summarize the well known reactions and to show that amount of heat they can possibly generate is orders of magnitude below what is measured. The system is not too complicated. And no additional experiments are needed to estimate chemical heat from known reactions. Perhaps I am wrong, but my guess is that for people with your background this should be trivial. It is only a matter of making reasonable assumptions about what is known and using the enthalpy table. Am I wrong, Peter? Can you perform such numerical calculations for us? Referees would certainly ask for convincing arguments that excess heat cannot be attributed to well known reactions. P.S. How much excess heat is generated in 5 minutes (a typical duration in Paris1 and Colorado2 test)? The answer is 100*5*60=30,000 J. How much gasoline must be burned to generate 30,000 J of heat? The answer is 30,000/50,000=0.6 grams. The heat of combustion is ~50 kJ per gram. Naturally, a Mizuno-type cell has no gasoline in it. But what about other potential fuels? That is my question. How much can they possibly contribute to 30,000 J from "strange" things. My hope is that it will be a small fraction of one percent. But we need numbers from a knowledgeable chemist. Wishful thinking is not enough. Yes, I know that you will agree with this. P.G.: No problem with the calculations. The problem is not with chemical phenomena. It is with a physical one, entrainment of droplets or mist by the very hot gases and this cannort be calculated yet. Many experimental data actually missing- composition of solution, of the precipitate, of the gases leaving the cell, we don't know a lot of things, sorry. No basis for a realistic calculation exists. = = = = = = = = = = = = = = = = = = = = L.K. : That is not what I asked you for; sorry for not being clear. The incandescent temperature of the cathode belongs to "strange" things, not to well known and understood things. Here is my request to you again. Assume we are both 20 years younger and CF remains to be discovered. A cell contains one liter of the 0.2 M K2CO3 electrolyte. The cathode is tungsten and the anode is platinum. The temperature is already 100 C. How much heat is generated when a constant potential of 30 V is applied for 100 seconds? Assume that a constant current of two amperes is flowing through the cell. Part of the answer is obvious, the thermal energy is 30*2*100 + X = 6,000 + X joules, where X is the energy released (positive or negative) via anticipated chemical reactions. Show, by making reasonable assumptions, that X cannot possibly exceed +60 J. That is what I am not able to do. How would you argue that chemical heat in only a negligible fraction of what is expected, Peter? P.G.: What are the expected or possible chemical reactions? Participants are H2O, K2CO3, W - ignore glass-it is inert. Loci for reactions: cathode, anode, hot surface of W, plasma. Nature of reactions electrolysis and pyrolysis. What is unknown and I have not found in the electrochemistry literature - what happens at the anode- in principle carbonate, hydroxyl ions ions are going there but only (?) oxygen is formed. I have asked an electrochemist friend about anodic reactions at carbonate electrolysis but it seems they are not well known- at least in this special form of electrolysis. However Mizuno has said he found some CO2 in the gases released from the cell too. How is this formed? Pyrolysis of K2CO3 is also a possibility. Carbonate captured by molten W is probably losing all its CO2. Anyway, I have hoped to have more data, systematic and complete after this series of experiments. A complete material balance could reveal what actually happens in this system. We still do not have such a balance. The positive fact is that these reactions are all endothermic- with high probability endothermic. But unfortunately a physical phenomenon-entrainement of droplets of solution is a possible explanation of the calculated excess heat and we don't know its quantitative contribution to the results. P.G. : I have already answered- we don't know exactly what happens at the anode, no pyrolysis is possible and in this case no droplets of solution are lost. Solution not boiling. (?) So what "anticipated chemical reactions"? This is a case of logical error - replacing a case with a different one- not justified. L.K. : 1) Why trying to find out what happens in a simple case (30 V) before addressing a more complicated case (300 V with GDPE, pyrolysis and CMNS) is called a logical error? One learns by progressing from what is known to what is unknown. I am sure you know this. 2) I agree with "no droplets." In my illustration, the solution was boiling because the outside temperature was assumed to be 100 C. That was a simplification made to avoid changes of temperature inside the cell. Thus, what comes out is the H2O vapor mixed with H2 and O2. What is the upper limit of X, Peter? 3) In your earlier message I see "Participants are H2O, Na2CO3, W - ignore glass . . . " Na was probably a typing error. Right? But then I see " Pyrolysis of CO3Na2 is also a possibility." In any case, I asked you to ignore the 300 V case, in the first round. 4) I want to know if theoretically predicting the upper limit of X, in the case of 30 V, is possible. What is the answer? P.G. : I am not able to find any exothermic reaction taking place in this system. In your opinion, what reactions have to be taken in account and used for the calculation? L.K. : 1) In other words, X must be negative. Right? That is an important point. I was not certain of this. 2) My opinion is that only dominant reactions should be taken in account. But I do not know reactions that accompany decomposition of water. And I do not know how to estimate their rates. That is why am asking you, a chemist. 3) Am I wrong in assuming that what is already known to chemists is sufficient to estimate X numerically? We will be nowhere without numbers. What is the value of X? P.G. : I cannot speak for all my colleagues- by the way I am a chemical engineer, with Ph.D. in polymer technology. Now, for the real case I need the experimental results to know what reactions take place for all phases- solid, liquid, gases with analyses done correctly and professionally. L.K. : OK, perhaps someone else will estimate the value of X for me. Knowing that it will negative is important but that is not a numerical answer. Please help ! P.G. : Who has tried till now? And what were the obstacles- I have explained that there more unknowns than equations. L.K. : I was referring to an open cell with one liter of the 0.2 M K2CO3 electrolyte. The electrodes were W and Pt, potential 30 V, current 2 A and temperature 100 C. Was it wrong to assume that the bulk of what happens under such conditions is well known to chemists? P.G. : In this case, very different from the case of plasma electrolysis, the unique unknown factor is the fate of the carbonate ion at the anode, but there are no exothermic reactions conceivable. X is smaller than zero. L.K. : Yes, you already wrote this, Peter. Can someone estimate the value of the negative X on the basis of what is known? If X is also negative when the COP is measured, then the apparent COP must be increased on the basis of the value of X. But how can we start talking about X at 300 V when we are not able to get it at 30 V? P.G. : Let's take it stepwise - the main reaction is water electrolysis; its energy consumption depends on the temperature, current density, secondary reactions. How much of the electrical energy introduced in the system is used for water splitting? Theoretically! L.K. : [Hmm, Socratic dialog?] Yes, this should be part of X. What is your answer? And what are contributions to X from other dominant reactions, based on anticipated rates? P.G.: I think we are now near to the answer you wish. Normally electrolysis will take 1.48 V (approx.) from 30 V. The problem is with the "other dominant reactions"- I don't know what are they, where they take place and in which extent. L.K. : Thanks Peter. You probably meant 1.48 eV, not V. Is this per atom of hydrogen produced or per H2 molecule produced? Assuming this is per molecule I concluded that X, for that reaction only, is -295 J. Do you agree? a) 1.48 eV=1.48*1.6*10^-19 J = 2.36*10^-19 J Therefore X=2.36*10^-19 * number of H2 molecules. b) How many molecules are produced when I=2 A and t=100 s? The charge is 2*100=200 C = 0.00208 faradays c) IF (???) one Faraday produces one mole of H2 then the number of molecules is 0.00208*6*10^23 = 1.25*10^21 d) In that case, |X| = 2.36*10^-19*1.25*10^21 = 295 J. e) That is nearly 5% of the electric energy supplied ! ! ! How significant can this be? Suppose a claim is made, NOT REALISTICALLY FOR 30 V, That the electric energy = 6000 while the thermal energy =7000 J. Dividing 7000 by 6000 one could say that the COP=1.17. But that would be wrong. The correctly calculated COP = 7000/(6000-295) = 1.23. f) And that is only from one endothermic reaction, decomposition of water. We must also correct for other dominant reactions. I hope someone will calculate the value of X that accounts for all dominant reactions. Is it really so difficult for a chemist? P.G. : No..., splitting of water needs~1.48 volts minimum, in practice 1.6-2.0 volts are used. The difference to 30 V is heating the solution. Approx minus 5-6% (1.5- 1.6 of 30) So much of the energy introduced is consumed for the main endothermic reaction. I don't know any exothermic reaction possible in the system. Do you agree? L.K. : What is your value of X in joules? Volts are not units of energy. How many joules of energy were used in 100 seconds to produce hydrogen? I think that the answer has nothing to do with 30 V. It would be the same for 15 V or for 45 V. Only the total electric charge counts. Do you agree, Peter? Mel Miles (M.M.)wrote (7/1/06): The minimum voltage thermodynamically possible for splitting water must be based on "Delta G" and is 1.23 volts at standard conditions. For enthalpy calculations, "Delta H" is used and is 1.48 volts at standard conditions. L.K. : Does this mean that 1.23 eV is the binding energy of the H2O molecule with respect to its three separated atoms? I am thinking about this as work that has to be done to overcome attractive molecular forces. P.G. : [Incorporated into my message below, in blue] L.K. : On Jul 2, 2006, at 2:22 AM, Peter Gluck wrote: What's the amperage? 1) Let me post extracts from what I already wrote. This is mostly for people like Mike C, who missed the beginning. "A cell contains one liter of the 0.2 M K2CO3 electrolyte. The cathode is tungsten and the anode is platinum. The temperature is already 100 C. How much heat is generated when a constant potential of 30 V is applied for 100 seconds? Assume that a constant current of two amperes is flowing through the cell. Part of the answer is obvious, the thermal energy is 30*2*100 + X = 6,000 + X joules, where X is the energy released (positive or negative) via anticipated chemical reactions. Show, by making reasonable assumptions, that X cannot possibly exceed +60 J. That is what I am not able to do. How would you argue that chemical heat in only a negligible fraction of what is expected, Peter?" It is not established a priori - depends on the experimental conditions. 2) Yes. And my request for X (number of joules) was for a specific (assumed) experimental conditions. Such concrete-operational approach is often helpful. Excuse me- but I really think this hypothetical case is much too different from plasma electrolysis in order to be of any use. 3) I strongly disagree. Why did I ask you to calculate X (positive or negative contribution of each dominant chemical reaction to excess heat) for 30 V? Because this is a case of basic electrolysis. It is not complicated by glow discharge plasma electrolysis. How can we be confident that our analysis of a more complex situation is correct if we are not able to get the answer for a much simpler case? Mike Carrell has presented you a good idea that has to be taken in account when finally we will have a complete material balance of the Mizuno process. 4) Yes, we should not stop this discussion after the X is determined for a simple case. Recall what we decided to do this at the beginning. We agreed that "things" should be divided into two categories; well known and unknown. You used the term "strange" for pyrolysis and I placed the CMNS under the same category. The 30 V case was invented to deal with well known things only. Till then, good old Kotarbinski, the father of praxeology, will roll angrily in his grave. Let's be efficient and systematic, friends! That is right, First 30 V and then 300 V. The Polish philosopher would agree with this. I am really disappointed that other chemists on this list do not help us to estimate X. The only thing we were able to do, so far, was to find the contribution of one reaction (decomposition of H2O) to X. Was my answer, -295 J, correct, Peter? Is it reasonable to assume that contributions from other known reactions (at 30 V) are negligible in comparison with that number? Michel Jullian (M.J.) wrote: Ludwik: here is the explanation for the 1.48V (V, not eV) our friends keep throwing at you. For water dissociation the enthalpy tables (e.g. those in my calculator) give: H2O(l) -> 0.5 O2(g) + H2(g) - 285.83 kJ/mol (endothermic) The delta H 286 KJ per mole is of course Avogadro's number NA (6.02*10^23) times the required electric energy to dissociate one single molecule, which is therefore: E=(286 * (10^3)) / (6.02 * (10^23)) = 4.75*10-19 J per molecule We also know that we need to circulate 2 electrons per molecule, so if V is the part of the electrolysis voltage used to dissociate the molecule and e is the electron charge 1.6*10^-19 C, another expression for the required electric energy (work) for the molecule dissociation is: E=V*2e. Thus: V=E/2e = 4.75*10-19/3.2*10^-19 = 4.75/3.2 = 1.48V Let me know if it makes sense, and if in this light you agree that the short paragraph I suggested for the "water electrolysis" section of your paper is sufficient. L.K. : I do remember, vaguely, the dG and dH and T*dS. But how does this relate to the fact that 1.48 eV is the "electric energy to dissociate one single molecule?" Assuming that the 1.48 eV is the depth on the potential-energy-versus-distance plot what is the meaning of 1.23 eV mentioned by Mel yesterday? And what is the meaning of the 1.6 to 2.2 V range mentioned by Peter? P.G. : 1) [You asked] “and what is the meaning of the 1.6 to 2.2 V range?” That's because the efficiency of water electrolysis is only 70-50%. Working at more higher voltages does not help more. 2) [You asked about 295] In this very case 1.48 X 2X 100= 296 J and you are right. Endothermic reaction. As told I have no idea re possible exothermic reactions but my (our) worry is not reactions but physical phenomena as droplet mist entrainment. I acknowledge my ignorance or lack of imagination re exothermic reaction in the system at 30V. L.K. : 1) And what about endothermic reactions, Peter? My impression was that you gave up on calculating the X for them. Is it not true? 2) It appears that you are not the only chemist on this list who did not estimate contributions of well known chemical reactions to X. I was probably wrong in assuming that this should be a trivial problem for most chemists. 3) Do you agree that the Colorado2 claim, COP=1.24, would not be taken seriously unless it is accompany with a credible statement that "contribution of known chemical reactions to excess heat has been shown to be less than 1%," or something like this? P.G. : I am contented with electrolysis as a great entoderm contribution and do not think there are other such reactions. What should motivate me to search for nonexistent reactions when I see that existent physical phenomena make us Great trouble? I need that complete material balance to understand the system. And we are discussing hypotheses as 30 V electrolysis. Not effective and not efficient. Tell please explicitly what real reactions do you will see analyzed dissected and measured. The Colorado claim will not be taken seriously till we do not know what happens in the system, the real experimental system. We can exercise our fantasy beyond any limits but the paper has to be based on reality. My question is: How great is the contribution of the entrainment of droplets of solutions to the material balance? As soon I know it, plus we make the complete balance, our dialogue will continue. I am seeking for kairos not chronos at CMNS L.K. : [Peter wrote] I am contented with electrolysis as a great endothermic contribution and do not think there are other such reactions. 1) I am willing to accept this. --> The value of X is close to -295 J; contributions of other reactions to X are negligible. That means we can go to the next step. Suppose we start increasing the voltage. How does the 60 V situation, for example, differs from the 30 V situation? I am assuming that the current is still 2 A. The electrical energy delivered becomes 12000 J (instead of 6000 J in 100 seconds) but X remains essentially the same because the current is the same. Is this an acceptable expectation? I hope that other chemists on this list will agree. [Peter wrote] The Colorado claim will not be taken seriously till we do not know what happens in the system, the real experimental system. We can exercise our fantasy beyond any limits but the paper has to be based on reality. 2) I hope you are not referring to the draft of our paper at: http://blake.montclair.edu/~kowalskil/cf/300positive.html Do you see any "fantasy" in that draft? [Peter wrote] How great is the contribution of the entrainment of droplets of solutions to the material balance? 3) Scott Little described a method by which this question can be answered. His preliminary answer is that droplets of the electrolyte contributed about 7% to the total mass lost during an experiment. That remains to be independently confirmed. . . . M.M. : It is Delta G and not Delta H that determines if a reaction is thermodynamically possible at constant temperature and pressure (usual laboratory conditions). Consult any Physical Chemistry text for this result. The splitting of one mole of H2O requires an applied energy of Delta G = +237129 Joules(1 J= 1 V*A*s). Converting to eV per molecule yields 2.45766 eV. Using Delta G=-zFE gives E=-1.22884 V at standard conditions( 298.15 K temperature and 100000 Pascals pressure,slightly below one atmosphere). Note that 2.45766 eV and -1.22884 V differ by a factor of two but have different units and sign. The factor of two arises because the splitting of one molecule of water, i.e. H2O = H2 +0.5 O2, involves the transfer of two electrons for the electrode reactions at the anode and cathode. Note that these calculations are for splitting water into hydrogen and oxygen gases and not for the splitting into H and O atoms. To state in simple terms, multiplying the 1.23 V by 2 gives the minimum energy in eV(2.46 eV) required to split a single water molecule at standard conditions. M.M. : Peter, for the LiOH-H2O electrolyte, I believe the reaction at the anode would be 2 OH- = H2O +0.5 O2 + 2e- I cannot think of any other likely anodic reactions for the LiOH-H2O system. For carbonates, various electrode reactions are possible. This is especially true for the high voltages of plasma electrolysis. Reactions of the metal used as the anode should also be considered under the extreme conditions of plasma electrolysis. P.G. (A private message 7/13/06)): Please feel free to use my name -- for everything I wrote and (hopefully) will write at CMNS. . . . Think about temperatures in plasma, and in the gases bubbling through the water phase -- in the vicinity of the cathode and the anode. This can explain a lot. Click to see the list of links |