Mizuno reports increased excess heat

    Quote from Paradigmnoia

    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.


    That seems impossible. It would allow a perpetual motion machine. In any case, the steady state terminal air temperature in a flow calorimeter for any given power level must be the same, or the energy is either disappearing into the void, or appearing out of nowhere. With a flow calorimeter, when a body produces a given power level such as 100 W, the Delta T air temperature is always the same. With Mizuno's reactor, the duct is 66 mm, the air flow rate is 3.957 m/s, so the calibration constant is 0.0552 K/W. So, 100 W produces a 5.52 deg C Delta T. It makes no difference what color the sample is, what the emissivity is, or whether the sample is so small it is incandescent or so large you can touch it with your hand. None of that matters. The calorimeter cannot distinguish between those states. It always produces a 5.52 deg C temperature difference. The equation couldn't be simpler:


    Energy (kJ/s) = Mass of air (kg/s) * Heat capacity of air (kJ/kg) * Temp difference (K)

    Quote from anonymous

    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.


    I suggest you work with a flow calorimeter, if you have not done that before. You will see that you are wrong. What you propose violates the conservation of energy.


    I have spent a lot of time working with flow calorimeters, and I have made every possible mistake with them, including many you probably would not make, and others you haven't thought of yet. The one you did think of here is not possible. "An expert is a man who has made all the mistakes which can be made, in a narrow field." - Niels Bohr

    Quote from anonymous

    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.


    At 80 to 250 W? No way! More to the point, you couldn't not see this is happening. The pressure gauge would show it the reactor has gone up to 101,000 Pa. The mass spec would show massive amounts of water and other contamination. The D2 tank pressure would drop and then the tank would empty. It would stop after a few hours or days when the D2 ran out (depending on how full it is at the start). It would not continue for 111 days. There are so many reasons this cannot happen that I think you and THH contribute nothing to this discussion by harping on this. It cannot happen. It did not happen. Do not spend your time or add words to this discussion about hypotheses that are ruled out.

    Quote from JedRothwell

    That seems impossible. It would allow a perpetual motion machine. In any case, the steady state terminal air temperature in a flow calorimeter for any given power level must be the same, or the energy is either disappearing into the void, or appearing out of nowhere. With a flow calorimeter, when a body produces a given power level such as 100 W, the Delta T air temperature is always the same. With Mizuno's reactor, the duct is 66 mm, the air flow rate is 3.957 m/s, so the calibration constant is 0.0552 K/W. So, 100 W produces a 5.52 deg C Delta T. It makes no difference what color the sample is, what the emissivity is, or whether the sample is so small it is incandescent or so large you can touch it with your hand. None of that matters. The calorimeter cannot distinguish between those states. It always produces a 5.52 deg C temperature difference. The equation couldn't be simpler:


    Energy (kJ/s) = Mass of air (kg/s) * Heat capacity of air (kJ/kg) * Temp difference (K)



    Jed, you need to multiply that by the efficiency. From your figures, this varies between 100% (nearly) and 77% (sorry don't remember exact figure).


    Then, the efficiency depends on a number of factors that can vary from control to active, including (most importantly) the reactor case temperature.


    I agree that what happens inside the reactor is irrelevant except the total power, unless it alters the temperature distribution of the case which I doubt it does much. There is perhaps an issue about the temperature of the heater leads if these are not thermally connected to the reactor case. But I don't expect that to be significant.

    There are so many reasons this cannot happen that I think you and THH contribute nothing to this discussion by harping on this. It cannot happen. It did not happen. Do not spend your time or add words to this discussion about hypotheses that are ruled out.


    Jed, anon and I both have accepted, did some time ago, that the reactor is a closed system with v small D2 added, if any, and therefore this is not an issue.


    You are arguing it is not an issue anyway - but that is irrelevant and the fact that convinces us is the simplest "everyone will believe" reason.

    Both anon and I want to leave this matter, as you say it did not happen: arguing about under what circumstances it might happen is a waste of time.

    • Official Post

    Having spent time pondering the 'emissivity affects the airflow calorimetry hypothesis' I am sure that in this case where we are dealing with a reactor completely enclosed within a calorimeter chamber it makes no difference, Imagine putting a 500W chrome-plated heater coil into the calorimeter chamber and calibrating it to 500W. Then paint the heater black (with something!) and run the trest again. The skin temperature of the heater might be different, but all that happens is that the share of heat between radiation and convection changes. It's still putting 500W into the calorimeter, and you should not detect any significant difference despite the huge variation in emissivity.

    I seem to remember an earlier post said that this Mizuno device was replicated already. Is anyone aware of who replicated and if they might publish? I assume is not one of Rossi's "highest echelon" customers. (just joking, no undies bundling please). Are the people associated with the Google funded experiments working on it now or they at least planning to attempt a replication? How about Hagelstein?

    Quote from Alan Smith

    but all that happens is that the share of heat between radiation and convection changes


    The main effect of putting low emissivity 0.04 foil on the calorimeter box is to

    increase the share of passive convection heat loss and decrease the share of the radiation loss.


    If one assumes 2 m2 SA for the box and an overall very crude HTC of 4 for passive laminar convection

    the heat loss rate is 160W for a surface temp of 20+20 = 40C where 20 is ambient,

    https://www.electronics-coolin…fficient-on-a-flat-plate/


    The radiation loss is 10W for emissivity of 0.04

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


    The reason for doing this is to more linearize the heat loss with respect to surface temp

    of the box rather than having it vary more greatly

    Convective loss is more linear(except for increases in HTC wrt to T)

    than radiation loss, which varies according to Stefan's 4th power.


    These calculations are indicative only.

    I ask experts help here concerning this "dark" notion of emissivity :)

    if a transmitter emits an IR spectrum on a source that has an almost full emissivity, it should return that spectrum as it is, i think.

    Now, If this receiver only has an 0,5 emissivity , what will become about original IR spectrum ?

    Will its tip be cut or will it be reduced in width, or both at once?


    Even if you are an expert with a hidden pseudo or a new one, you could help too :)


    Quote from Alan Smith

    Having spent time pondering the 'emissivity affects the airflow calorimetry hypothesis' I am sure that in this case where we are dealing with a reactor completely enclosed within a calorimeter chamber it makes no difference, Imagine putting a 500W chrome-plated heater coil into the calorimeter chamber and calibrating it to 500W. Then paint the heater black (with something!) and run the trest again. The skin temperature of the heater might be different, but all that happens is that the share of heat between radiation and convection changes. It's still putting 500W into the calorimeter, and you should not detect any significant difference despite the huge variation in emissivity.

    Quote from THHuxleynew

    Jed, you need to multiply that by the efficiency. From your figures, this varies between 100% (nearly) and 77% (sorry don't remember exact figure).


    I believe you mean the heat recovery rate, shown in Fig. 2. This depends on the difference in temperature between the air in the calorimeter box, and the ambient air. It is a function of the efficiency of the insulation. (The R-value, see the ICCF21 paper p. 8.)



    Quote from THHuxleynew

    Then, the efficiency depends on a number of factors that can vary from control to active, including (most importantly) the reactor case temperature.


    No, it does not depend on a number of factors. It depends on only factor: the R-value of the insulation. It never changes. As shown in Fig. 2, it is always the same at a given reactor case temperature, and it is linear.


    If you removed the insulation and put in a different type, the recovery rate would change, of course. You always have to calibrate before and after a test to make sure it has not changed. But that is just a precaution. In fact it does not ever change measurably. I suppose it might change if the insulation got loose, or got wet, or something like that.


    The recovery rate is completely independent of the temperature, size, emissivity and other characteristics of the reactor. Although, to be careful, Mizuno always calibrates with another control reactor that is the same, or very similar. It is always placed equidistant from the walls and the active reactor. Actually, calibration would work just as well with a different shaped object. We know that because when he does smoke tests with no insulation (so you can see inside the box), he sees that the air is swirling around and reaching every part of the box and all around the sample. Also because the sample is far away from the two outlet RTDs, so it cannot affect them directly. However, Mizuno always errs on the side of caution, so he calibrates with an identically shaped device, as shown in the ICCF21 paper.


    If the sample were affecting the inlet RTD we would know because the inlet RTD would register a different temperature than ambient. That is why he only needs one inlet RTD but it is safer to use 2 in the outlet.


    Quote from THHuxleynew

    I agree that what happens inside the reactor is irrelevant except the total power, unless it alters the temperature distribution of the case which I doubt it does much.


    No, that would not matter, unless it directly affected the 2 outlet RTDs. It would have to affect them both exactly the same way, which is impossible. If it heated one but not the other as much, causing a temperature difference between them larger than 0.1°C, Mizuno would see the error. (And I would see it from afar!)


    The temperature distribution in the case is uneven. It has to be, in any flow calorimeter. The fluid toward the outlet has to be warmer than the inlet. In this case the air is being blown all around the enclosure and the samples. It is swirling around, as revealed by smoke tests. So the air is well mixed and there is little difference from one part to another. But you can be sure that the air close to the inlet is cooler than the rest.


    If the air flowing through the box periodically became stagnant in places, and did not rapidly move out, the outlet temperature would fluctuate. Each five second interval would report a different amount of heat. That would be a real difference, not an instrument artifact. It would also be a lousy design for a flow calorimeter. You could measure total heat output by averaging, but you would have difficulty seeing minor variations in heat output from the reactor, from one minute to the next. It works much better when you have well mixed air, which is why Mizuno ensures the air is well mixed, and tests for that.


    The volume of the box is 210 L. The airflow rate is 13.4 L. So the air is replaced every 16 seconds. Actually it takes a little longer, as shown in smoke tests. Some of the smoke swirls around and lingers while new air comes in behind. But I think it is all replaced after ~20 seconds.

    Quote from JedRothwell

    The recovery rate is completely independent of the temperature, size, emissivity and other characteristics of the reactor

    because of the box insulation low emissivity.... radiation loss is minimised

    Passive convective loss is the major box heat loss mode.

    The recovery rate is mostly dependent on the calorimeter box surface temperature , dimensions and emissivity.

    As the temperature of the reactor increases the box surface temperature will increase but the effect on the HTC will be small.

    in the range 30-40.


    The important thing is that the heat recovery can be checked

    by for example placing any kind of heater in the box that raises the surface temp of the box

    In a similar fashion e,g a small quartz bar heater.

    https://www.electronics-coolin…fficient-on-a-flat-plate/

    Quote from Alan Smith

    Having spent time pondering the 'emissivity affects the airflow calorimetry hypothesis' I am sure that in this case where we are dealing with a reactor completely enclosed within a calorimeter chamber it makes no difference, Imagine putting a 500W chrome-plated heater coil into the calorimeter chamber and calibrating it to 500W. Then paint the heater black (with something!) and run the trest again. The skin temperature of the heater might be different, but all that happens is that the share of heat between radiation and convection changes. It's still putting 500W into the calorimeter, and you should not detect any significant difference despite the huge variation in emissivity.

    This intuitively seems correct to me also.


    However, in the example of a high emissivity null and a low emissivity active run, when the temperature is high for the active, the difference in temperature should result in a difference in the outlet heat calculations compared to a high emissivity null at the same input power since the heat transfer to the air should be greater if the object is hotter and the airflow the same.


    In order for the Heat In to balance Heat Out, this means that the surface temperature actually does not get hotter with decreases in emissivity while in the calorimeter, all else being equal. The increase in proportion of the air transfer compared to radiant heat keeps the temperature from increasing? I don’t see how this works, but it could be true.


    To be clear, I don’t at the moment know if this is relevant to these Mizuno experiments and don’t wish to insinuate or appear to insinuate that it does. I’ll see if I can model a simple cylinder with different emissivities and see what it does, to satisfy my curiosity.

    Quote

    I agree that what happens inside the reactor is irrelevant except the total power, unless it alters the temperature distribution of the case which I doubt it does much. There is perhaps an issue about the temperature of the heater leads if these are not thermally connected to the reactor case. But I don't expect that to be significant

    Look, the power level and power gain are so huge, it would take a mammoth error to account for them if in fact they aren't real. Having said that, what's the sense in having a control reactor which is significantly different from the experimental one? And I thought Mizuno frequently interchanged the active and the blank in the course of his experiments. Doesn't he? Or, being in a rush, did I miss something here? What I understood is that the control reactor has a dull finish while the experimental one is shiny (or maybe the other way around). Maybe a quick clarify would help in case other people also aren't getting this. It's a long string to reread in entirety even just for this issue.


    ETA: maybe that is what JedRothwell said happens in msg #874 (interchanging control reactors) but pls. clarify anyway for us nonreaders.

    Thanks Robert,

    why these questions ?

    let's take Mizuno's device as example.

    If we need Far IR (around Thz frequencies) then if our special japanese heater could do that, it's nice.

    As receiver our nickel mesh with a low emissivity, for example, we should find some hidden frequencies from heater in this case, so that would imply a necessary frequency adjustment between heater spectrum and mesh emissivity to find best results, i suggest :)


    Quote from RobertBryant

    I am no expert but this may be helpful


    https://www.google.com/url?sa=…Vaw0QmU8Ga07yCKaLBiB-VrXK


    Of course this may be the secret of dark energy:)

    Quote from Cydonia

    If we need Far IR (around Thz frequencies) then if our special japanese heater


    That is a refinement.. the special Japanese heater used by Mizuno

    has no nationality its not special or teriyaki Thz


    I believe what is special is the magnetoelectric interaction btw the ten twenty or so isotopes in the Ni/Pd foil

    and the 30 or so isotopes in the stainless steel wall

    and that this is not French or Japanese but a Dieu.

    Quote from seven_of_twenty

    Having said that, what's the sense in having a control reactor which is significantly different from the experimental one?


    It is not significantly different, as you see in the photos in the ICCF21 paper. However, even if it were different, the smoke tests and other tests show that would not affect the measurements. He has tested with a wide variety of reactor shapes, as you see in the ICCF21 paper. Big ones and small ones. Cross-shaped ones and cylinders. In every case, in every calibration, the response and the heat recovery rate (Fig. 2) was the same, to within the margin of error. The calibration constant is the same to many digits. * The shape of the reactor has no effect on the calorimeter performance. However, he errs on the side of caution and calibrates with closely similar or identical reactors. In the ICCF21 paper Fig. 6 the control reactor is a slightly different color. That cannot have any effect on the calorimeter performance. People who think it might need to get some experience using flow calorimeters.


    Various things can affect the calorimeter performance, especially changes in ambient temperature, which unfortunately are large in this laboratory. However, the size, shape and placement of the control reactor do not have any measurable effect.



    Quote from seven_of_twenty

    And I thought Mizuno frequently interchanged the active and the blank in the course of his experiments. Doesn't he?


    Not in the 111 days shown in Table 1. There are some days not accounted for, but on those days he wasn't in the lab. He was not running a control reactor.


    As you see in the ICCF21 paper, he did interchange them in that data set.





    * It is the same, but it is recomputed for each five second interval taking into account the atmospheric pressure in the lab, the instantaneous power consumption of the fan, and other factors. Which, as I recall, the HP gadget measures 20,000 times in one second, and then averages. It is remarkably accurate. The atmospheric pressure is recorded every 5 seconds along with everything else. So, for example, the weight of air in kilograms per second for a series of 5-second intervals is:


    0.017115
    0.017109
    0.017114
    0.017096
    0.017106
    0.017117
    0.017112
    0.017096
    0.017116
    0.017109
    0.017103
    0.017108
    0.017091


    As long as you have a computer, you might as well tell it to account for such small differences.

    Dear Robert we can be friend therefore don't share all ideas :)

    You share same ideas as Jurg Wyttenbach or Russ, i think.

    From my side, i never believed in magic powders.

    Electrons are needed to be slow enough, long parked too between 2 cores to fuse them.

    Also IRs could generate EM waves that annihilate together under certain circumstances and produce this effect.



    Quote from JedRothwell

    So, for example, the weight of air in kilograms per second for a series of 5-second intervals is:

    0.017115
    0.017109
    0.017114


    That's nifty. It is pleasing to a computer geek such as me. The problem is, "skeptics" may pounce on this and say: "Ha! That means if you do not record atmospheric pressure in the lab and you do not compute the weight of air to 6 digits, your calorimetry stinks! Your result is invalid!" They will not realize that Mizuno is going overboard and recording things with far more precision than he needs to. He does that out of force of habit. Because he can. Because, as I said, as long as you have a computer you might as well tell it to account for small differences. That's why God gave us computers.


    I like to do it this way with the spreadsheets and their long formulas. But I also like to compute the calibration constant from the textbook values for air at STP (the density and heat capacity). That comes out to 0.05522 K/W. Which is close enough for government work. You can apply that to any calibration Mizuno has done, with any shaped reactor, at any power level. Use a hand calculator. Throw in the recovery rate. (You have to use the recovery rate for the air temperature, which is less precise, since we are doing this for different reactors.) The answer will be smack on target every time. That proves the calorimeter is working right, and has been all along.

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