Mizuno reports increased excess heat

  • Dear all, I do enjoy these calcs. But not the company!

    In a peer review, even those doing the review don't get a pass! Part of the process. But you are doing a great job explaining yourself, and the dynamics involved, so please keep it up. Tough I know being a skeptic at LENR central, but someone has to do it.


    That said, I think the Mizuno side is winning this debate quite handily. Not that I am biased, but even to my untrained eye, it looks like M and R built a tight calorimeter, and accounted for all the input vs output.


    So far they have provided the superior answers to the probing questions. I have yet to see any weakness/error in their defense.


    I don't say that to discourage, but to encourage more poking around. Important this thing pass the most rigorous examination, before ICCF.

  • Do you think most of the calorimeter heat loss is conductive? Maybe you don't have as good insulation on the base?


    Do you mean between the reactor and the table? This is shown in the ICCF21 paper, Fig. 5. Not clearly, but you can see the reactor is sitting on a large yellow pad of heavy insulation. It is much better than the reflective padded bubble insulation in the box. The R-value is higher. I don't recall what it is.


    As I see it, the easy way to estimate this is to say the bottom has perfect insulation, the acrylic in the box does not insulate at all, and the reflective padded bubble insulation has an R-value of ~11. If you believe the manufacturer. I ran it the other way. I assumed the losses as measured by Mizuno are correct, and I computed the R-value. It came out roughly 11. Close enough for government work! Mizuno did a first principle estimate based on the properties of aluminum foil and the size of those bubbles with air in them.


    Try it yourself.


    I am not a stickler for accuracy or precision when I am writing on a piece of paper. Only when I use a computer program. Face it, I'm lazy. And careless . . . I tend to drop orders of magnitude. Something my late mother would frown upon. As she said, "you should learn to use a slide rule, because it forces you to keep track of the zeros." I think she wanted me to treat those numbers with respect.

  • Every joule of heat coming out of the reactor must end up heating the air. It is not possible for any heat to leave the reactor and not heat the air, because the chamber is well insulated with reflective insulation (see the ICCF21 paper).


    Jed, while I agree with your statement, I think what anon is contending (he can confirm my interpretation of his position) is that radiative heat causes the air circulating within the calorimeter to be heated more slowly than convective heat, and therefore the difference in the kind of heat transfer matters for air flow calorimetry. Can you comment to this specific critique?


    I suppose my response would be that the inside of the calorimeter soon reaches a steady-state temperature, and therefore any difference in the rate of heating of air between radiative/convective would be insignificant to the calorimeter measurements. And of course for R20, the COP is so large, such a difference wouldn't matter anyway.

  • To everyone: Please stay on topic and stop posting ill-intended distractions like that.


    Moderators: Time to up your game. Unfortunately, forums like these become unbearable if you don’t force people to stay on topic.

    We used to be stricter about that, and ran people off. Then at other times, we were too lenient, and the threads meandered so much they lost focus...which ran people off.


    Then we settled in the middle, and since then things seem to be going well. The feedback we get from our members guides us, so thank you.

  • It seems that between the ICCF21 paper and the current ICCF22 paper, there were some changes in method that I want to ask about.


    In the ICCF21 paper, page 11 and 12, the meshes are: 1) cleaned, 2) sanded, 3) placed in the reactor, 4) pressure brought down to .01Pa (room temp) for 2h, 5) heated at 110C for 5-20h. Then the meshes are 6) removed and 7) Pd is rubbed on, then 8) heated for 1-2h (to 110C?), and 9) cooled for 1-2h. After this, when the reactor is heated (to 100C?), it should 10) produce excess heat. I assume the temperatures are from the thermocouple on the surface of the reactor.


    In the current Iccf22 paper, page 17, it states, "After a) cleaning [and b) sanding] apply the palladium with ... c) rubbing..." Then d) place meshes in the reactor at 1-2Pa for 2h, e) heated to 110C for 5-20h, f) pressure to 0.01Pa (unspecified length of time), g) heat to 200C for 1-2h, h) cool 1-2h, i) introduce deuterium gas so that pressure becomes 100-300Pa, j) heat to 100C (using 50W of input power?) via sheath heater, then the reactor should k) produce excess heat.


    Questions:


    When/how is the deuterium introduced in the ICCF21 paper?


    Why did the step order of Pd rubbing change from ICCF21 paper (step 7) to the ICCF22 paper (step c)?


    Why did the 0.01Pa step change order (from step 4 to step f)?


    In step j) should 50W from the sheath heater result in 100C on the reactor outside wall measured via the thermocouple?


    If I understand what Jed has written above, between steps i) and j), the deuterium supply should be cut off. Forever? Did R20 produce excess heat for months with no deuterium supply?


    Could you give some clarification regarding the text on page 11 of ICCF22? More details on the one time procedure described to "confirm the material is more permeable than ordinary nickel".

  • What happens to a plastic floor which is in contact with a 300C ss reactor?

    There is no acrylic plastic on the bottom. The top box fits into the chunk of yellow insulation at the bottom.


    I think Mizuno referred to the stuff at the bottom as "bricks." Maybe they are bricks? High temperature insulation? I did not look closely. Anyway, the bottom is well insulated.

  • Jed, while I agree with your statement, I think what anon is contending (he can confirm my interpretation of his position) is that radiative heat causes the air circulating within the calorimeter to be heated more slowly than convective heat, and therefore the difference in the kind of heat transfer matters for air flow calorimetry. Can you comment to this specific critique?

    I do not know if it is true that the calorimeter air is heated more slowly with one path or another (radiation versus convection). If it is slower, the calorimeter will show it takes more time to heat up. Eventually, all the heat will emerge. As long as it emerges with a power level of ~5 W or more, this calorimeter will detect it. A calorimeter measures heat energy, not instantaneous power.


    Suppose you get an energy burst of 9,000 joules. For some reason it emerges very slowly, at a power level of ~1 W. This calorimeter will not see it. That's in the noise. The margin of error. But if it comes out at 5 W for 30 minutes or 10 W for 15 minutes, the total energy is the same, and the calorimeter will show pretty much the same 9,000 J. It becomes less accurate at low power, as it enters the noise. Then at high power it goes to pieces again. Any calorimeter has a best range of operating power.

  • Jed, while I agree with your statement, I think what anon is contending (he can confirm my interpretation of his position) is that radiative heat causes the air circulating within the calorimeter to be heated more slowly than convective heat, and therefore the difference in the kind of heat transfer matters for air flow calorimetry. Can you comment to this specific critique?


    I suppose my response would be that the inside of the calorimeter soon reaches a steady-state temperature, and therefore any difference in the rate of heating of air between radiative/convective would be insignificant to the calorimeter measurements. And of course for R20, the COP is so large, such a difference wouldn't matter anyway.



    No IH FB -- I am looking at steady state so that the speed or rate of transfer is not significant once it finally reaches equilibrium.


    I am concerned about a different amount of radiation leaving the reactor and striking the walls of the calorimeter (even with emissivity claimed of 0.04) and thereby more effectively heating up the walls of the calorimeter which in turn would radiate, conduct, or convect the heat to the outside without being measured by the air flow Delta-T.


    None of this matters for R20 with its "uncorrected" power output of 225 watts for 50 watts in, and at least some of our friends around the world will make their own replications using identical emissivity so that this will not be a factor.


    If I was in high school and needed a science fair project it could be "Does Emissivity of a heat source effect an Air Mass Flow calorimeter's Heat Capture". It's easy enough for a high schooler to build -- fans, acrylic boxes, home depo insulation. It also lends itself to enough high school algebra to model the results. I hear all the well thought out arguments here on both sides but in the end, it just doesn't matter for R20 so we should move on. If we can replicate an R20, this debate here is a time wasting footnote.

  • If I was in high school and needed a science fair project it could be "Does Emissivity of a heat source effect an Air Mass Flow calorimeter's Heat Capture". It's easy enough for a high schooler to build -- fans, acrylic boxes, home depo insulation.

    If emissivity does have an effect, I guarantee you would not see it with this instrument. It would be much too small. The margin of error is 1 or 2 W. The noise from ambient temperature change is huge, as you see in Fig. 7. If you put this instrument in a lab at SRI or MIT, where the ambient temperature is controlled to less than 1 deg C, those fluctuations would be 10 times smaller.


    There are many known sources of error with this instrument. You don't need to go looking for hypothetical tiny ones. The big ones are readily apparent! I expect the kind of thing you are talking about is in the milliwatt level, if it exists at all, and we can't even measure 1 W with confidence. At low power, the acrylic box and insulation radiate ~3 W. Emissivity would have to be a small fraction of that.


    A high school kid building one of these would learn a lot, but the calorimeter probably would not work well. There are many people here who would learn a lot, and I wish they would undertake the exercise.

  • No IH FB -- I am looking at steady state so that the speed or rate of transfer is not significant once it finally reaches equilibrium.

    Oh okay, thanks for clarifying.


    Quote

    None of this matters for R20 with its "uncorrected" power output of 225 watts for 50 watts in, and at least some of our friends around the world will make their own replications using identical emissivity so that this will not be a factor.

    Agreed.


  • Shane,


    I am not somone who dislikes robust argument. In fact I welcome it, as all here know well. If you can find the robust argument - rather than rhetoric, repeating old matters that are not relevant and not in contention, and personalisation - it would make me happy.


    I think you totally misunderstand this issue. It is not about sides. Nor is understanding experimental results a contest. Nor has my stance on this matter ever been one of being on any side, except what M and any other decent scientist would want which is the side of making experimental write-ups as rigorous as possible. What so dissapoints me is that I am therefore seen as somone on a different side.


    No-one has to be a skeptic here: and I feel no obligation to fulfill this role. I actually enjoy all this stuff since it is a puzzle, and I also enjoy learning (where needed) stuff to do this analysis, or I would not do it.


  • That is very helpful Jed, and it is also what I expected (I really don't understand what RB expected - or why he thinks my wanting to check this is so extraordinary).


    In that case we have a problem with your contention that the inside of the box insulation stays relatively cool (not 320C) unless you can suggest some path for the power emitted other than the exhaust air?


    The paper shows at 380C reactor case 23% of input power is lost (not measured in output). Where does it go? I suggested heater leads but it does not seem likely? Maybe a whole load of vacuum piping comes out of the calorimeter at 300C or so, and heat is lost by conduction that way? The problem there is that we need a non-linear heat loss ramping up to 60W for reactor case = 380C, but proportionately a much smaller fraction of this for lower reactor temperatures to give the efficiency graph shown in the paper.


    My workings here are on the previous page where I asked for this answer?


    It is a mystery to me: however RB I hope from his posts has seen something here that is obvious and explains it: RB would you care to share?

  • I do not know if it is true that the calorimeter air is heated more slowly with one path or another (radiation versus convection). If it is slower, the calorimeter will show it takes more time to heat up. Eventually, all the heat will emerge. As long as it emerges with a power level of ~5 W or more, this calorimeter will detect it. A calorimeter measures heat energy, not instantaneous power.


    Suppose you get an energy burst of 9,000 joules. For some reason it emerges very slowly, at a power level of ~1 W. This calorimeter will not see it. That's in the noise. The margin of error. But if it comes out at 5 W for 30 minutes or 10 W for 15 minutes, the total energy is the same, and the calorimeter will show pretty much the same 9,000 J. It becomes less accurate at low power, as it enters the noise. Then at high power it goes to pieces again. Any calorimeter has a best range of operating power.


    The calorimeter exact operation is complex.


    The air is heated by convection.


    However the foil on the inside of the insulation is heated by both convection and radiation. Convection can't push its temperature higher than the exhaust air. Radiation can push its temperature higher than that. At least according to RB's rough figures above (1 or two pages back) modified by the analytic result of the T^4 factor, which amplifies radiation relative to convection 10X at 380C relative to 40C. My post on last page gives the working.

  • The calorimeter exact operation is complex.


    The air is heated by convection.


    If you want someone to diligently consider your calculation for

    the ht btw the reactor and the box..

    Mizuno reports increased excess heat

    do due diligence.

    Please use Stefans Law with areas and emissivities

    to calculate the heat emitted from both surfaces.


    Please use the standard convective ht equation

    The overall convective HT , "Qc Reactor" is in two parts

    not shown in the diagram

    1) btw the reactor and the moving air

    2) btw the moving air and the box,

    1) and 2) have very different values.


    Use appropriate areas for the box and for the reactor.. they are very different


    State what convective HTCs , h, you are using ... 4, 40 or 400...

    State how you justify the HTCs.

    Don't use facile shortcuts like "Thus while convective transfer stays roughly proportional to dT"




    Is this the THH equation for convective heat transfer?

    Then if you come up with missing energy write it down

    with units

    on a diagram such as this so its easy to understand,


  • Radiation is I believe needed to explain the calorimeter efficiency graph in the paper, unless the vacuum piping etc emerges from the calorimeter very hot. In that case nonlinear natural convection might perhaps do it, but it seems unlikely. Because we have a lot of info on the insulation, and the change in efficiency with reactor temperature, this is something that can be understood I think.


    Agreed about R20.