Mizuno reports increased excess heat

  • I agree,

    especially when instant experts cannot do heat transfer calculations

    even halfway right.


    Even if they could, it is a complex system with multiple components and multiple heat transfer paths, i.e. a real time waster, i.e.


    P12 = k12c*(T1 - T2) + k12r*(T1^4 - T2^4) + f_naturalConvection(T1,T2)

    P23 = k23c*(T2 - T3) + k23r*(T2^4 - T3^4) + f_naturalConvection(T2,T3)

    P13 = k13c*(T1 - T3) + k13r*(T1^4 - T3^4) + f_naturalConvection(T1,T3)

    P14 = ...

    P = P12 + P13 + P14 ...


    As many elements as parts in the reactor can be modelled. Big time to setup and solve. It is possible that Jed is correct and emissivity doesn't matter, but a definitive solution would take time.


    R20 is the immediate goal. My preference is to change as little as possible between experiment and control, and that includes emissivity.

  • From the R19 paper: ”As the reactant temperature rises, the amount of deuterium in the reactant decreases. This causes the exothermic reaction to decrease and the reactant temperature to fall. When the temperature decreases, more deuterium enters the reactant again, and excess heat increases.”


    a) If this is correct and the reaction is driven by temperature, the reaction should stabilize at some optimal temperature level. Heating (slowly) above that temperature will cause COP to decrease. We will have a Peak COP Temperature (PCOPT).


    b) Reducing the heat loss high enough will, for any input power within a broad range, make the reaction temperature exceed this PCOPT. If then the power is switched off and COP is high enough, the reaction should become SELF-SUSTAINED and stabilize at PCOPT.



    Excess energy vs input power (current) seems to be a close to linear relation (table 1 and fig. 4, when dotting peak power out vs power in, in the R19 paper). Additionally, the 300 W input and up to 3 kW output does not produce particularly higher COP than 50 W in and 300 W out (R20). Ordinary exponential reaction temperature dependency (Arrhenius equation) does not seem to apply in this case. This indicates that temperature is NOT the main driving force for the reaction. But, if magnetic stimulation is, this might explain some other anomalies the presentations expose:

    1) When the heating coil is moved from the outside to the inside, COP raises 10-fold. This shift of position means that the electromagnetic shielding of the tube wall is bypassed and that the nickel mesh is subject to a substantially higher (even though relatively weak) magnetic influence.

    2) The cooling coils' killer effect on the reaction might possibly be explained by that the coils interact with and weakens the magnetic field created by the heater coil (positioned inside the cooling coils) on the inside of the reactor wall.


    3) This secrecy about the design of the heater in the reaction chamber. If the coil geometry is essential, and possibly other factors such as current pulses etc, to reach COP>5, the secrecy is more understandable – but it is not in the benefit of highly successful replications.


    4) A self-sustained reaction has not been demonstrated. Switching of the power seems to stop the reaction.

  • This secrecy about the design of the heater in the reaction chamber


    There is no secrecy.

    The heater runs along the central axis.

    Some people want it to be a magic curly wand to conform to their pet theories.


    The magic is in the combined effect of temperature and deuterium on the thirty or so nuclear isotopes in the wall and the mesh

  • As I understand it two reactors are used so that calorimeter temperatures of the fueled reactor can be compared to those of an un-fueled reactor with known power input. This is leading to some people to claim differences in emissivity may account for the results.


    For that to be the case some of the power from the un-fueled reactor must be escaping the calorimeter so that that 1kW in the un-fueled reactor gives similar temperature readings to a few hundred Watts in the fueled reactor. That seems possible but unlikely.


    Can't this easily be addressed by running just one reactor twice, first with untreated mesh, then with treated mesh? Has that not been done?

  • There is no secrecy.

    The heater runs along the central axis.


    The heater Jed stated that Mizuno uses is 2 meters long. We don't know how it is mounted (e.g. is the whole length inside the reactor...? any turns? etc).

    So I agree there is no "secrecy" around the heater, but some more info around how it is mounted would be good to be able to replicate it.

  • Jed has passed on a little info about the heater - it is entirely within and centralised in to reactor and folded (in some way) so that it fits. Thar is I think, all we currently know.

     

    There aren't many practical ways to fold a 2 meter long heater in a less than 1 meter reactor. It's either folded in zig zag or rounded, I don't think it's shaped like a Moebius strip albeit it would be very nice if that's the way Dr. Mizuno folded it.

    I certainly Hope to see LENR helping humans to blossom, and I'm here to help it happen.

  • So to continue with the heat issue: which no-one has answered yet. just to be clear: I'm trying to understand the given figures from the paper. Always a good thing to try and do. I'm not trying to debunk the paper. I'm happy for others to propose explanations, and I have proposed some, so I am on the side of the angels here - if you think that angels believe M has working LENR in his R19 experiments.


    If you reckon the R20 measurements are definitive then why bother thinking about R19? R20 has such large heat emitted that you do not need a calorimeter to be clear about it. But in that case you should just ignore the R19 debate.


    (1) RB does not like my calculations for radiation vs convection. Key point - I'm not trying to do it exactly. I'm pointing out that RB's calculation at 40C (that radiation << convection) changes at 380C because the radiation becomes 10X larger relative to (linear to first approx) forced convection, and therefore RBs point that radiation can be ignored is wrong from his own figures and the power law.


    (2) RB asks me to do calculations. I have done an analytic calculation. Showing that the T^4 term, when used to determine radiation between two surfaces makes radiation much larger at higher temps. That is obvious from T^4. If RB questions this he can do a more accurate calculation but it is difficult to be very accurate due to unknowns.


    (3) If I am wrong about radiation, then there is as yet no explanation for the 60W lost from the 5/1 R19 data except the unconvincing ones I've also proposed - supposing the calorimeter calibration heat out/ heat in vs reactor temp graph is correct. However, if Jed is right and the inner foil of the insulation does not heat up significantly there is still no explanation (other than as before). 60W and the area and R value of the insulation force a large temperature difference: maybe I've got this wrong? I was taking the values from the paper.


    (4) Alternatively, the 60W lost (from 77% efficiency) could go somewhere else. But where? it is a mystery and I'd like Jed's comment since he has more experience here than others.


    (5) Whether the "radiation excuse" works for this anomaly is still not clear. If RB thinks I'm all wrong here then maybe it does not work and an anomaly remains. Does it matter? Not necessarily, but as with a possible iceberg it is always worth checking what lies beneath things that you are given that do not quite make sense. Personally i'm not sure that the radiation excuse works - I'm with RB in being skeptical of it - and I'd need to do much more work to know, but it cannot easily be dismissed. It has the desired characteristic of decreasing efficiency at higher reactor temperatures.


    (6) Where did the lost 60W go if not through the calorimeter box insulation? Jed - were there v hot metal pipes emerging from the box connected to the reactor?

  • The magic is in the combined effect of temperature and deuterium on the thirty or so nuclear isotopes in the wall and the mesh

    For sure, Robert, it's obvious this is the truth what you said, on the other hand, just a small remark, why R19 worked much less well than R20 despite it have had the same 30 nuclear isotopes on the wall ?

  • (6) Where did the lost 60W go if not through the calorimeter box insulation? Jed - were there v hot metal pipes emerging from the box connected to the reactor?

    What about the heater wires? Copper is an excellent thermal conductor. If one end of those leads is at 380C and the other end is at 20C, much of the total length will be warm. Jed, do you know what gauge wire was used? To do the calculation, you would also need to know the wire length inside and outside the calorimeter.

  • For sure, Robert, it's obvious this is the truth what you said, on the other hand, just a small remark, why R19 worked much less well than R20 despite it have had the same 30 nuclear isotopes on the wall ?


    The critical difference (IMHO) is that the heater is inside on R20 and was (I think) outside on the others- heat flow from heater to reactor wall via the Ni/Pd/D loaded mesh

  • Yes ! Alan !

    Another interesting point from Birger below:


    .From the R19 paper: ”As the reactant temperature rises, the amount of deuterium in the reactant decreases. This causes the exothermic reaction to decrease and the reactant temperature to fall. When the temperature decreases, more deuterium enters the reactant again, and excess heat increases.”


    convoluted explanation i think therefore now, when you increase temperature, both you move IR spectrum to shorter wavelenghs both you forget too far IR then reaction stops.

    It's strange it seems to work by cycles like cat / mouse..HOULALA hell and damnation what i said ?


    The critical difference (IMHO) is that the heater is inside on R20 and was (I think) outside on the others- heat flow from heater to reactor wall via the Ni/Pd/D loaded mesh

  • What about the heater wires? Copper is an excellent thermal conductor. If one end of those leads is at 380C and the other end is at 20C, much of the total length will be warm. Jed, do you know what gauge wire was used? To do the calculation, you would also need to know the wire length inside and outside the calorimeter.


    That was my guess - heater wires or vacuum pipe. But they are both a bit surprising, because it is easy and sensible to keep all the hot parts inside, and they get force air cooled by the airflow. I'm hoping Jed can shed light on this

  • I have created a new document to supplement Mizuno's recent paper:


    Mizuno, T. and J. Rothwell, Supplemental Information on 'Increased Excess Heat from Palladium Deposited on Nickel' 2019: LENR-CANR.org


    https://www.lenr-canr.org/acrobat/MizunoTsupplement.pdf


    So far it includes three sections:



    Nickel Toxicity


    WARNING: The steps described in this paper can produce fine nickel powder, which can be toxic. The procedures in this paper should be performed in properly equipped laboratory, in a glove box or other enclosure. Disposable gloves and masks should be worn when handling these materials. They should also be worn to avoid contaminating the materials. Only people skilled in the art should attempt to replicate this experiment.



    Q&A Responses by Mizuno


    Uploaded here already



    Temperature Distribution Study


    An image and graph showing the surface temperature of the reactor.

  • What about the heater wires? Copper is an excellent thermal conductor. If one end of those leads is at 380C and the other end is at 20C, much of the total length will be warm.


    This would show up during calibrations at high power. It does not show up. The balance is zero at all power levels, after taking into account known losses from the calorimeter chamber walls.


    No doubt a little heat does escape by this path, but the power is too low to be measured with this calorimeter. It is in the noise.

  • If you reckon the R20 measurements are definitive then why bother thinking about R19?


    That's a strange question. Is there a rule you cannot study two reactors at the same time?


    If I am wrong about radiation, then there is as yet no explanation for the 60W lost from the 5/1 R19 data except the unconvincing ones I've also proposed


    What on God's Green Earth are you talking about?? There is no 60 W loss on 5/1, or any other date. Input + excess is ~300 W. The reactor temperature is the same as other days with 300 W. The calorimetry measures 300 W output normally.


    I must have missed this discussion about missing watts. Anyway, there are none. It is some sort of confusion. I am too busy to look back through the messages to find out who is confused about what. Just forget it. No calorimeter loses track of 60 W.

  • RB asks me to do calculations. I have done an analytic calculation

    who is confused about what

    THHnew

    You have not done heat transfer calculations.

    You have not done analytic calculations

    Analytic is the wrong word

    Please use the standard formulas for convective heat transfer and radiative het transfer

    State your HTCs and emissivities ,areas etc


    Until you do analytic

    as described here Mizuno reports increased excess heat

    what you say is nonanalytic blah. and will remain so even if you become THHold

    who is confused about what. Just forget it. No calorimeter loses track of 60 W.

    THHnew is confused nonanalytically... 60W of blah with no light bulb on