Mizuno reports increased excess heat

  • Situation 1: (Emissivity 85% calibration unit) -- If the heat is radiated out of the surface of the stainless steel cylinder (reactor or control) more of it makes its way _without being absorbed_ through the air in the convection chamber to the walls, thereby bypassing the air.

    The manufacturers claim it is the best reflecting insulation money can buy. Anyway, if some of the heat escaped without being reflected back, that would mean we underestimate the excess heat. It would mean there is more excess heat than we think, not less. Who cares? We have established that it produces at least 5 times more than a resistance heater calibration at the same power level. Calibrating with the resistance heater produces a balance of zero, so we know we are capturing all of heat from resistance heating.


    Do you suppose cold fusion heat has some special quality that allows it to escape from the reactor in a way that resistance heating does not? That seems kind of far-fetched. Why would the ratios of conduction to radiation to convection be different with cold fusion, after the heat passes through a steel wall?


    If the cold fusion heat has some means of going through the steel wall and bypassing the air in the calorimeter box, I suppose it would make the outside of the box walls hotter. That is not observed. Do you think the heat keeps moving, through the insulation and right out of the room, like a gamma ray? (By the way, gamma rays are not detected.) If the steel walls of the reactor and the air in the box do not stop the heat, and it does not reflect back from the bubble insulation, why would it stop outside in the room air? I suppose it would keep going right out of the solar system.


    The insulation on the inside of the calorimeter chamber is not a perfect reflector and is likely more absorbent than the optically transparent air at IR wavelengths.

    It is the best reflecting insulation you can buy at Home Depot. It is shiny, like a mirror.

  • If Mizuno says he valved off the D2 as an ongoing source of chemical energy, I believe him. Can we move on. This will be proven by the replicators.


    (Please note that for the future publication I would suggest making this fact explicit in the published paper.

    It is right there in Table 1! It shows the pressure. Do you see anything like 101,000 Pa? How could he keep it at 1000 Pa if he did not valve off the D2?? Leave the needle valve open and it will fill to whatever the pressure in the tank is. Of course you have to shut the valves to keep it at 1% of atmospheric pressure.


    What a weird comment.

  • It is right there in Table 1! It shows the pressure. Do you see anything like 101,000 Pa? How could he keep it at 1000 Pa if he did not valve off the D2?? Leave the needle valve open and it will fill to whatever the pressure in the tank is. Of course you have to shut the valves to keep it at 1% of atmospheric pressure.


    What a weird comment.


    Jed, I understand that you are frustrated with answering what you think are ridiculous questions, but table 1 only shows a set of pressures on each day. It doesn't say anything about the intraday history of the pressure after the day's test began. It doesn't say when in the day the pressure was measured. It doesn't say the procedure was to valve off the tank after the reactor cylinder was pressurized at the appropriate pressure (i.e. 300 Pa) . There is no schematic of the D2 and vacuum system. I am "skilled in the art". I did not mean to upset you with another ridiculous question. I am not a skeptic. I am trying to prove self-consistency of the data from your paper. My initial hypothesis was that the rig had some kind of D2 pressure+vacuum regulator that would allow you to maintain the stated low pressure. You then told me me that that was incorrect, and that the tank was simply valved off. I have used high vacuum systems, and their valves, and I know from experience that if the valve is functioning, it does indeed valve off the supply of gas (or vacuum). So your answer (that the reactor was valved off from the D2) corroborated the remainder of the paper and was good enough for me. But it was in no way obvious to me before you answered my question. I am a mature adult like you, and one skilled in the art of vacuum systems. It is also not in my opinion obvious to third party readers who will read your paper without the benefit of reading all 850 comments on this thread.


    So my "ask" is simple: do other readers the favor of explaining this with a simple few sentences in the next revision of the paper:


    "The reactor pressure was set to the desired pressure shown in Table 1 by letting in a small amount of D2 from the tank via a valve. If the pressure was to be lowered, by evacuating the system with the vacuum pump, the pressure was lowered via a different valve connected to the vacuum pump. Once the desired test pressure was obtained, the D2 tank and the vacuum were both valved off from the reactor. The pressure in the reactor was monitored with [type of vacuum measurement equipment, e.g. Pirani Gauge] and did not rise more than [xx%] for the 24 hour period of each test, thereby proving that there was no significant leaks or outgassing during each run that would effect the result."


    Thank you Jed. I appreciate your contribution to LENR now and over the years. Please continue -- you're achieving success with your work.


    (P.S. I am your anonymous supporter here, but when you make fun of me for asking a question, it makes it more difficult. Fortunately I can check my ego at the door.)

  • Question 1) Do you suppose cold fusion heat has some special quality that allows it to escape from the reactor in a way that resistance heating does not? That seems kind of far-fetched.


    Question 2) Why would the ratios of conduction to radiation to convection be different with cold fusion, after the heat passes through a steel wall?

    Answer 1): [Cold fusion heat has a special quality] No not at all. That is NOT my point. But perhaps that is a rhetorical device on your part. Of course it is far fetched -- I agree!


    Answer 2): Because you are doing the cold fusion in a reactor that is of different emissivity. The reactor steel cylinder which is shiny has a different emissivity (about 0.075) than the calibration steel cylinder (about 0.85), which is flat grey. It's not the "cold fusion" heat that is different, it's the cylinder emissivity that is radiating the heat to the walls of of the calorimeter.


    You have said here that some of the heat is lost thru the external ("Home Depo") reflecting insulation. We agree there. The HD insulation's inner surface emissivity is likely around 0.05% (i.e. it's very shiny), and it is likely backed with additional insulation, but heat radiation still gets through it. You know this because at 360 C only 77% of the heat is captured in the calorimeter by the mass flow, raising the exit air temperature. Those are quoted directly in the paper. 77% means that 23% of the heat is escaping, most likely mostly through this insulation. Now you lower the amount of thermal radiation that gets to the reflecting insulation by a factor or 7.5/85 = 1/11.3x at a given temperature. Making the reactor shiny is like putting more reflective insulation inside the calorimeter. So less heat now gets out of the calorimeter wall insulation, because less radiation reaches it at a given reactor surface temperature by factor 11.3x. That's a BIG difference. Where is the heat going to go? Out thru the airflow. The calorimeter now goes from 77% efficiency to what, 90%? We don't know how much it goes up because we didn't measure (calibrate) the calorimeter+reactor with an identical shiny reactor cylinder.


    Am I not clear? Is there something I missed.


    Respectfully,


    Anonymous


    P.S. This doesn't change the basic conclusion that the R20 reactor is making at least 225 watts out for 50 watts in, or COP of 4.5x. That is still a revolutionary improvement.

  • Some will perhaps think that the extra difficulty making the results stronger is unnecessary because these results are strong enough anyway. All that is needed is similar level, replicated independently. That would have some value, but IMHO not much. And the extra difficulty is not that large, so why not strive to do it?


    This is nonsense. The likes of THH will not be satisfied with a cold fusion experimental setup--ever. It would be a game of never-ending moving goal posts. The replications need to be done, and will be done, and those without conflicts of interest will then move the ball forward. Those whose interests conflict with a world filled with low-cost LENR devices will hop from one false trope to the next, and from one disingenuous attack to the next. I don't know whether THH has such conflicts, but I do know that he behaves as one who might.


    I'd contrast this with anonymous' critiques, which seem more genuine to me. Running the calibration with the same shiny cylinder as used for the test runs seems like a good suggestion. Anonymous is also careful to point out that the COP would still be 4.5x even taking the difference of emissivity into account. This is constructive. We need more of this kind of "skeptical" input and less of the "hand waiving pseudo-skeptical" flavor. I've been around long enough to spot the difference.

  • The reactor steel cylinder which is shiny has a different emissivity (about 0.075) than the calibration steel cylinder (about 0.85), which is flat grey.



    We don't know how much it goes up because we didn't measure (calibrate) the calorimeter+reactor with an identical shiny reactor cylinder.

    The calibration , I am pretty sure , is done on an equally shiny but inactive reactor

    with the same power electrical input as the active reactor.. For example both at 50W.

    This is the calibration talked about in point 1 (method 1)on page 5.

    This alone should be enough to demonstrate COP >>2

    .

    It does not take into the extra heat loss due to the higher temperature of the active reactor.

    and underestimates the COP. Another Method does but it is not used routinely.

    For the details of that (method 3) its probably best to talk with Mizuno.


    Both reactors have the same arrangement of shiny foil and or other insulation.

    this is the same situation as in 2017.

    "

    The same type of reactor is used in the calibration, and is installed as a control for calibration of the heat balance in the

    enclosure described below. The design, size, weight, and shape of this calibration reactor are exactly the same as the

    reactor used for testing. The internal reactants are the same nickel, of the same weight, size, dimensions and position.

    Both are washed and wound the same way. However, excess heat is not produced by the calibration electrode even

    though deuterium gas is added to the cell, because the nickel material is not processed as described in Section 2.5.


    It is possible to use the same reactor for the inactive and active tests.

    rather than two separate reactors.

    Probably do first the inactive test.=nickel foil without palladium??

    Then do the active test.= same amount of nickel foil + burnished with 50 mg palladium.


    For a COP of 5 the inactive/active reactor surface temperature difference should be>50C.

    but to estimate more accurately the energy output ... an air calorimeter is needed.


    This is based on

    https://www.engineeringtoolbox…-heat-transfer-d_431.html

    S.A.=0.24m2, emissivity =0.1 roomtemp 30C



  • Thx for this anonymous. You are making all the points I would make; but with greater tact!


    Re R20 I fully agree with you. If those sample R20 results are as reported this is far beyond all these issues of calorimetry errors. My reservation there is from the paper, where they are billed as sample results. However, I am sure that with such a working device M will in due course be able to provide more definitive data. My additional reservation (about stability) is really neither here nor there. If these results exist they will be easy to demonstrate, and why the reactor is apparently stable will then be interesting and to be understood.


    Re the R19 results I'd just point out that there are some uncertainties in the data (as I summarised above) which I do not think have all been dealt with. Thus the 23% heat loss from the reactor (which would be possibly different between calibration and control) is not necessarily fixed and so the potential calibration versus control error from this source might be higher than that. I'm not saying it is, just that it is unwise to assume that without more careful checking. Another thing - variation in air flow between the two reactors, if it existed, would have a similar affect, by allowing one to be cooled more effectively than the other and therefore have a lower case temperature. As would variation in the thermal resistance of the block on which the reactor is placed (possibly we are told those blocks are identical).

  • \ Another thing - variation in air flow between the two reactors, if it existed, would have a similar affect, by allowing one to be cooled more effectively than the other and therefore have a lower case temperature. As would variation in the thermal resistance of the block on which the reactor is placed (possibly we are told those blocks are identical).


    I skimmed your discussion on airflow. I did some of my own research and I decided that air mass flow is proportional to fan power input. If he is running at 6.5 watts the fan, then I believe that the mass flow will be about the same regardless of the accoutrements in the calorimeter. I also decided mass flow is independent of temperature. I have less concern about this part of the experiment. I see no need to measure the airflow at different spots, i.e. this is "good enough" for his proof.

  • Jed,


    I know this question wasn't asked of me, but I'd like to respond.


    Quote

    Question 1) Do you suppose cold fusion heat has some special quality that allows it to escape from the reactor in a way that resistance heating does not? That seems kind of far-fetched.



    Answer: There seems to be an effect with some exotic energy systems in which odd particle emissions - often termed strange radiation - can penetrate the immediate walls of an inner reactor core (at least under certain conditions if they are not too thick) and then induce nuclear reactions with an outer layer of fuel. I don't think this has a great deal of relevance to Mizuno's current system. However, it is possible that a portion of the heat energy, even if only a few percent or less, is generated somewhere within the reactor wall rather than only on the nickel mesh.

  • I skimmed your discussion on airflow. I did some of my own research and I decided that air mass flow is proportional to fan power input. If he is running at 6.5 watts the fan, then I believe that the mass flow will be about the same regardless of the accoutrements in the calorimeter. I also decided mass flow is independent of temperature. I have less concern about this part of the experiment. I see no need to measure the airflow at different spots, i.e. this is "good enough" for his proof.


    Hi anonymous,


    I actually agree with all of that, though it took me some time to be sure. Except that people were at start talking about publishing a paper and if that is the case these details must be properly presented (or not claimed) so there is no inconsistency and the claims made are all valid. In this case the substantive is that flow variability across the pipe represents a likely 20% error (over-reading). The factor of 20% is not that variable with stuff as long as flow stays at Re = 5000 - 20,000, but to be safe we could say it is 15% - 25% which is significant in terms of absolute results, but not for control vs active ones.


    If the question was: "is extraordinary excess heat proven from the R19 results" this 20% would need to be fed into all the other analysis to check consistency of all assumptions, even though itself it is not large enough to invalidate the large results.


    As many have pointed out if the R20 results are correct then this is way beyond such issues, and any factor of 20% can be ignored, as can a load of other issues.


    I've noticed a few people on this thread view my interest in details as unhelpful. I'd rather argue it is helpful. I'm not trying to conclude "this proves LENR" or "this does not show LENR" because these results can do neither. I am interested in having these results as well understood as possible, and having anyone else replicating this in the best position - if they have the time and will - to generate the most robust results.


    One thing that modulates this is R20 - if this is real no-one needs a calorimeter, it will be blindingly obvious that there is some large unexplained source of heat. Much less care will be needed.


    THH

  • https://en.wikipedia.org/wiki/…stic_personality_disorder


  • Jed, I understand that you are frustrated with answering what you think are ridiculous questions, but table 1 only shows a set of pressures on each day. It doesn't say anything about the intraday history of the pressure after the day's test began. It doesn't say when in the day the pressure was measured.


    I believe that is the average. But you are missing the point. The paper says, in several places, the pressure should be low. Around 1% of atmospheric pressure. You and THH were asking whether the heat might come from burning deuterium and air getting into the cell. If that were the case, the supply of D2 and air would have to come in all day long, 24 hours a day, in large amounts. So the pressure would be 1 atm all the time. Neither the intraday nor the average would be ~1000 Pa.


    Also, when Mizuno says "keep the pressure low" and "be sure to exclude water and contamination" surely you understand that opening up the cell to a large flow of air and D2 gas would defeat that! Opening it on purpose, or by accident, would instantly add a lot of water and contamination. Not to mention that the flames would destroy the heater and contaminate the cell. Surely you realize that one glance at the instruments or the flames would tell Mizuno the experiment has failed drastically.



    It doesn't say the procedure was to valve off the tank after the reactor cylinder was pressurized at the appropriate pressure (i.e. 300 Pa) .


    It does not have to say that. I suppose anyone would know that from the many things the paper says.

  • Answer 1): [Cold fusion heat has a special quality] No not at all. That is NOT my point. But perhaps that is a rhetorical device on your part. Of course it is far fetched -- I agree!


    Actually, I think it is impossible, not just far-fetched. However, if it is true, it just means we have underestimated the excess heat. At this stage in the project I don't see why that would matter.



    Answer 2): Because you are doing the cold fusion in a reactor that is of different emissivity. The reactor steel cylinder which is shiny has a different emissivity (about 0.075) than the calibration steel cylinder (about 0.85), which is flat grey. It's not the "cold fusion" heat that is different, it's the cylinder emissivity that is radiating the heat to the walls of of the calorimeter.


    I think you are completely wrong. No flow calorimeter can detect a difference in emissivity from the reactor inside it. It would never collect less heat for these reasons. The heat will not escape from the calorimeter chamber no matter what color or type of steel you use or whether you substitute ceramic or some other material.

  • I think you are completely wrong. No flow calorimeter can detect a difference in emissivity from the reactor inside it. It would never collect less heat for these reasons. The heat will not escape from the calorimeter chamber no matter what color or type of steel you use or whether you substitute ceramic or some other material.

    Actually the difference (a big difference) in emissivity can have an effect, and it is a bit insidious and counter intuitive. The high emissivity material will radiate heat more effectively, and therefore run cooler than the low emissivity material. The steady state temperature for both will be different at the same input power. This is more important once the reactor radiation heat loss exceeds convection heat loss, which occurs at high temperatures.

  • In response to Jed:


    "anonymous wrote: It doesn't say the procedure was to valve off the tank after the reactor cylinder was pressurized at the appropriate pressure (i.e. 300 Pa) ."

    Jed's response: "It does not have to say that. I suppose anyone would know that from the many things the paper says."


    Well -- I didn't know that (and I am an anyone) but you clarified that the tank was valved off here in your previous comments. If you want to clarify it in the paper that is up to you.


    Jed wrote:


    "I believe that is the average. But you are missing the point. The paper says, in several places, the pressure should be low. Around 1% of atmospheric pressure. You and THH were asking whether the heat might come from burning deuterium and air getting into the cell. If that were the case, the supply of D2 and air would have to come in all day long, 24 hours a day, in large amounts. So the pressure would be 1 atm all the time. Neither the intraday nor the average would be ~1000 Pa. Also, when Mizuno says "keep the pressure low" and "be sure to exclude water and contamination" surely you understand that opening up the cell to a large flow of air and D2 gas ... flames would destroy the heater"


    1) Now that I know the unit is sealed (rather than running under continuous regulated low pressure D2 supply) it is clear that the excess heat cannot be coming from the D2 that was already in the reactor. The possibility of making a low pressure D2 regulator that maintained the system at set pressure (i.e. 300 Pa) that included excess vacuum regulation capacity to remove any leaked air or heavy water vapor product was a possibility before your clarification. But you have clearly stated that this is not the case.


    2) Burning and flames ... would destroy the heater -- disagree. Stainless steel or platinum when heated act as a catalyst allowing chemical recombination of hydrogen (or deuterium) and oxygen to occur without flame and at low pressures and hydrogen concentrations. These are used "autocatalytically" (generating their own heat once started) as a safety mechanism in nuclear power plants to prevent hydrogen explosions, to combine the hydrogen released from hot zirconium fuel rods reacting with steam to be safely converted into water vapor without flame. These failed at Fukushima, resulting in the top of the reactor being blown off in a hydrogen/air explosion. This paper shows that the temperature of the catalytic plates can be as low as 450K or 175C. No flames, just heat. See https://www.degruyter.com/down…-0042/aoter-2013-0042.pdf I will not get into the technical calculations to prove or disprove whether catalytic recombination could have occurred at these pressures (i.e. 300 Pa) in the unit, but the possibility did exist that this could have occurred with no flames, just excess heat from the recombination. When you ruled out the resupply of D2 to the reactor in your comments, this ruled out this effect being significant in any way on the small and thus limited (finite) amount of D2 in the reactor when the test is run.


    Jed: "I think you are completely wrong. No flow calorimeter can detect a difference in emissivity from the reactor inside it. It would never collect less heat for these reasons. The heat will not escape from the calorimeter chamber no matter what color or type of steel you use or whether you substitute ceramic or some other material."


    We disagree technically with each other. My suggestion remains to use a reactor calibration cylinder of the same thermal (conduction, radiation, convection) characteristics. I respectfully agree to disagree and suggest we both move on.


    Thank you Jed,


    Anonymous