"Heinz Sause"... aha
This seems to be his brand new alter ego....
This probably means Heinz Ketchup because there are 57 Varieties yet they all taste the same.
JedRothwell
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Posts by JedRothwell
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Nope. It is measured with the anemometer. It does seem impossibly close, so I zoomed in some more to the individual data points. You see varying separation. Then I subtracted watts from air speed for each 5 s interval. That is, of course, a meaningless number. But if one number was derived from the other, the difference would be the same in all cases, I think. The differences were as large as 1 part in 10, and randomly distributed. Then I looked at what percent watts were of air speed. It varied, randomly. Watts were not multiplied by any single number to equal air speed.
It is possible I am confused about this. I intend to ask Mizuno when I get a chance. But as I said, I searched for a function that converts watts into air speed, and I cannot find one. There are small, random differences between them.
As I said, I suspect the HP gadget is homogenizing the data.
It is possible he massaged the air flow with data from the motor plus the anemometer, but I don't see anything about that in the notes. I have extensive notes on every field, which will be published in an upcoming paper. (Who knows when.) Some of the computed fields include complicated functions with multiple variables and physical constants. Unfortunately, in the transition from the old Japanese software to U.S. Windows 10, the functions are lost and only the computed results appear in the spreadsheet. So the spreadsheet is filled with copies of equations and polyglot notes saying, for example, "3 ohm resister here, so multiply by 0.3333." It is messy. Eventually I plan to convert all the fields back into computed values and upload some sample spreadsheets.
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This seems to imply that Mizuno has tried other nickels and had them turn out to be "ordinary nickel" and therefore "not produce excess heat."
Is the difference in the nickel how he scores it with sandpaper and burnished with a palladium rod or are we to believe that his source of nickel is superior?
I do not know if he has tried other sources. But he thinks the treatment makes it permeable. He thinks the gas may be is going into cracks and voids in the sandwich between the Ni and Pd, which is not the same as being absorbed. Maybe it is not as absorbent as it seems? It is difficult to tell. With high tech instruments you could find out.
As Alan Smith pointed out, repeated loading and deloading will improve permeability. He also wrote: "Measuring the actual amount absorbed on each of the cycles though requires some very precise measurement of pressure and temperature." You can see from the graphs that the measurements are somewhat crude in this case. The values are approximate, even with a very precise measurement of pressure. And a not-so-precise measure of temperature.
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I had the impression from the earlier answer you gave to me (about a week ago in the other thread) that the D2 valve and vacuum valve (separate valves) are shut "OFF" (closed) during a run, but that the pipes are still connected so that Mizuno can do the next run at a different pressure in the R19 run sequence by opening one valve or the other at the completing of a day's run.
The pressure is recorded every 5 seconds, so that has to stay open. I suppose the connection to the pump and the mass spec is closed with the Swagelok valves, until needed.
I do not think he pumped out or added gas every day. You can see from the table that he did from time to time.
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The anemometer info is here, by the way:
https://www.kk-custom.co.jp/emp/CW-60.html
You can Google translate it.
but in practice, these are averaged by the inertia (and diameter) of the anemometer's fan blade.
It is a hot wire type. But the same principle applies.
The HP data collection gadget does some averaging. I am not sure I recall how it is set up, but I think it takes 20,000 readings over a second, averages, writes one value, and then repeats after 5 seconds. That will smooth things out. I suppose it will diminish the lag between the blower motor speeding up and the air speed increasing.
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That diagram reveals that the air speed is clearly derived from the blower power throughout a mathematical formula.
Nope. It is measured with the anemometer. It does seem impossibly close, so I zoomed in some more to the individual data points. You see varying separation. Then I subtracted watts from air speed for each 5 s interval. That is, of course, a meaningless number. But if one number was derived from the other, the difference would be the same in all cases, I think. The differences were as large as 1 part in 10, and randomly distributed. Then I looked at what percent watts were of air speed. It varied, randomly. Watts were not multiplied by any single number to equal air speed.
I do not know what other derivation there might be, or how you would detect it.
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No, this is the wrong calculation. The heat from the cylinder is much lower where it is surrounded by insulated foil that reflect back a good proportion of the heat. You might perhaps get a 20X amplification factor (taking extreme values) due to emissivity difference and insulation if that was very high. But that is still ball park only 5W.
Okay, so do the right calculation. Show us! You have 5 W left after the air removes most of the 50 W. Show us how that could all go out the hole, and fool two RTDs, a thermometer and two thermocouples.
Put a number on it. Do a quantitative analysis.
How the hell could it be 5 W reach a 5 cm hole 65 cm above the cylinder when 45 W are removed by the moving air? Does all of the heat bounce around, get reflected back, and then it all shoots right out that hole? Is the foil in the box in the shape of a parabola aimed at the hole? None of the heat radiates out from the wall? How would that work, anyway? Give us a model, instead of more hand waving, and baseless estimates of 5 W.
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Jed, my problem is, under what conditions was this done?
The conditions are described in the paper, in considerably more detail than most papers.
Once you have a significant effect which can be variable in unexpected ways (like this radiation issue) you need to be much more careful to ensure that all the other checks on which you rely are done under the conditions for which the active results are taken.
This is not significant. It is ~0.08 mW out of 300 W. It does not show up the calibrations, which produce a balance close to zero.
Much safer, and better, to design such effects out from the start.
"Such effects"??? This effect is imaginary. Or if it is real, it is 7 orders of magnitude below the excess heat level, and completely impossible to detect with this instrument. You might as well "design out" the effects of the gravity of Mars.
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I hate to watch two of our best minds argue over something that is hopefully immaterial. This is like arguing over your high school algebra grade after you graduated from university. Does it really matter that you could have gotten an A instead of a B 10 years later? THH has made his point. Jed has made his rebuttal.
It is indeed like arguing over high school algebra! You have hit the nail on the head. Geometry in this case, but that is exactly what we are arguing about. The thing is, THH is dead wrong and he is doing middle school and high school level geometry COMPLETELY WRONG. I hope you see that. It is important that readers understand that his geometry and physics would not pass a 7th grade test. It is all nonsense.
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Jed and THH -- I suggest that arguing over calorimetry details with each other is unnecessary here. THH's critique is noted and the validators can take this into account to produce a more accurate calibration.
That is incorrect. His critique is nonsense. It is physically impossible. If validators take it "into account" they will not produce a more accurate calibration. They will produce more physically impossible calculations. You might as well base the calorimetry on unicorn farts as THH's contrived, unscientific blather. I have given several indisputable reasons why he is wrong. He will not acknowledge them. He never acknowledges any mistake.
It is necessary to discuss this insofar as I do not wish to see Mizuno's results called into question for such nonsensical reasons. I do not wish to see it covered with a mountain of bullshit.
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So that is a separate set of data, not recorded in the paper. It would need to be documented and included for that to work.
It is in Fig. 2. The reactor temperature ranges from around 40 deg C to 360 deg C. Those are calibrations from 10 to 200 W, with 3 different reactors.
Plus I just documented it. I just told you, right here.
It is neither more nor less believable than it would be in a paper that I write and upload to LENR-CANR.org. If you don't believe what I write here, why would you believe what I wrote in a paper?
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I'm not here talking about the R20 results. I'm talking about R19, where the over-reading is much less.
It is 100 W. That's not much less than 250 W. It can measured with as much confidence as 250 W. It produces a large temperature difference, which is easily confirmed with multiple handheld sensors.
And it is then necessary to check that the RTD is not heated via conduction from a lead connected to a hot blower, etc.
As I said, that has been done, with hand-held thermometer and thermocouples. They confirm the air temperature is measured correctly by the two RTDs, so the motor heat has to be heating the air, not the RTDs directly. That's two RTDs, not one, and they are separated from the blower by a paper cylinder which has no metal leads in it. The blower cannot be much hotter than it is normally, and it cannot possibly be 100 to 250 W hotter, because it would burn up, and because there is nowhere near that much power going into the system.
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My point about the insulation relates to whether the insulation covers the blower inlet. If it does, then you are correct - and I asked this,
I misunderstood, as I said. It doesn't cover the hole. Now I suggest you estimate how much heat will reach that point. Start with 50 W. Remember, there is no anomalous heat; the effect you postulate is being caused by the instruments measuring 50 W incorrectly. Take into account the fact that 95% of the heat is removed by the moving air, as you see in Fig. 2. That leaves 2.5 W. Now estimate the solid angle based on the distance from the reactor to the top, and the size of the hole, 5 cm. What fraction of those 2.5 W reaches through that hole? I suggest you calculate this, although you seem strangely allergic to doing a quantitative analysis of your own hypothesis. So let me take a crack at it . . .
The reactor is a 12 cm diameter cylinder, 50 cm long, sitting at the bottom of a 70 cm tall box. The hole is a circle 5 cm in diameter, ~8 cm^2 area. Assume the 50 W is distributed evenly over the entire surface of the cylinder, but none comes out the end. Assume it is directly under the hole. The heat radiates in all directions equally. 5 cm is one-tenth of 50 cm length, so 1/10 of the heat that is not removed by the air radiates from a 5-cm segment. That's 0.25 W.
I am going to treat this 5-cm segment of the cylinder as a sphere, because my geometry is rudimentary. The diameter puts the center of the cylinder it 6 cm above the floor, so you can think of it as a sphere 64 cm in radius by the time it reaches the top of the box. The solid angle is . . . ummm 0.002 sr. See:
https://rechneronline.de/winkel/solid-angle.php
An entire sphere is 4 pi = 12 steradian. The imaginary sphere is 64 cm in radius, 51,472 cm^2 in surface area, or 25,736 cm^2 per hemisphere. 0.002 sr is 0.0003 hemispheres, or 8 cm^2 out of 25,736 cm^2, or 0.03%. To put it simply, we have 8 cm^2 piece of the surface of a sphere with 51,472 cm^2 surface area, which is 0.03%.
In other words, 0.03% of the 2.5 W of heat reaches the hole. Right? That's 0.08 mW.
You want to check my arithmetic? Please do.
Now explain how 0.08 mW can raise the temperature of the air by 10 deg C. That 10 deg C cannot be an error in the two RTDs on the top, because it is confirmed with other temperature sensors.
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Where does it say the reactors are equidistant from the blower?
Right here. I just told you.
Plus, I think it is apparent from p. 7 here:
https://www.lenr-canr.org/acrobat/MizunoTexcessheat.pdf
Another picture, Figure 6, shows reactors of different size and shape.
They are almost the same. The camera angle makes them look more different than they are. In any case, you could use a different shaped object, and you could put one directly under the fan, but it would still radiate in a sphere. The heat would not go exclusively directly up into the fan. You can estimate what fraction of it is carried off in the moving air. You have the dimensions of the box, the distance from the reactor to the hole, and the size of the hole. You can compute the steradian and estimate what fraction of 50 W reaches the hole. You will see it is insignificant. No matter what, it cannot add 250 W to the air when there is only 50 W coming into the reactor. It cannot make the reactor TC go haywire exactly in synch with the other temperature sensors. That TC does not go haywire during normal calibration.
I suggest you let go of this. Your hypothesis is physically impossible for a long list of reasons. It is not adding anything useful to the analysis here.
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Yes. the adjacent wall is insulated. The blower (entrance) is not (although if Jed says this is baffled by similar insulation
I misunderstood. It cannot be baffled because the blower is over the hole, as you say. The insulation is between the RTDs and the reactor, and between handheld thermometers and the reactor. However, you are overlooking many things that make your hypothesis impossible:
The outlet air is definitely 10 deg C hotter than it is in the 50 W calibration. This is confirmed with thermometers. So something is adding 250 W to the air. Even if the heat comes from the reactor and magically goes into the blower, it is still anomalous, and far more than input.
It is not possible for the reactor to pass heat to the blower only, in a narrow path through the moving air to the blower. The heat goes out from the reactor everywhere, spherically in all directions. Most of it is carried off in the moving air. If there is no anomalous heat, then only a fraction of 50 W will reach the blower. Not enough to measurably change the temperature of it, or to heat up the air by 10 deg C.
You have overlooked the reactor temperature. The thermocouple on it shows a temperature hundreds of degrees higher when there is anomalous heat compared to the calibration. Perhaps you think the thermocouple is malfunctioning. That is ruled out. It works correctly during calibration at low power and high power. When there is anomalous heat, it tracks the RTDs and thermometers, and agrees with them (based on the calibration). That cannot be a coincidence. It cannot be malfunctioning only when the other problems you postulate occur, exactly to the same extent.
Frankly, your hypothesis are more like Just So stories than physics.
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This hypothesis cannot explain why the outside of the reactor box is measurably warmer when there is excess heat.
? Do you have data for this
Do you need data for this? There is a hot object inside the box. However good the insulation is, it is not perfect, so the box has to radiate some heat. It is not possible for the heat to go directly from the reactor, straight through one segment of the insulation, and into the fan only, without reaching the rest of the insulated walls. Figure 2 shows that at high power calibrations, when the reactor temperature reaches 360 deg C, 22% of heat escapes from the walls. Where else could it be escaping from, if not the walls?
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THH wrote:
QuoteThe control reactor is to one side, the active reactor directly underneath and close to the blower. This the active reactor surface radiation would heat up the blower via the exhaust hole, unless this was baffled.
As stated in the paper and shown in the photos, they are equidistant from the walls and the blower, which is in the center. But again, how can a 50-W calibration add 40 to 250 W to a blower? That violates the conservation of energy.
Furthermore, if you have any experience with insulation, you will know that all 50 W of heat cannot penetrate straight through the insulation and go into the fan only. It will go everywhere in the box. Even if there were no insulation, and the calibration reactor was placed directly under the fan, the heat from the calibration reactor would go in all directions. Only a tiny fraction of it would heat the fan. It could not magically transfer 50 W to the fan directly, which then heats the air by 10 deg C extra, indicating 250 W. The air is definitely heated. This is confirmed by two RTDs and handheld thermometers and thermocouples.
QuoteUmm... I don't assume there is anomalous heat
Then why would the calibration be any different at all from the apparent anomalous heat? Why would it affect the RTDs and handheld instruments differently? Why would it heat the air with far more net energy than the 50 W going into the system? Whether the 250 W of heat emerges from the reactor or the blower, where would it come from?
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Exit sensor after blower means that we have the indirect effect (from blower heating up) not the direct effect.
Again, you assume there is no anomalous heat. So the heat in both cases is 50 W coming from the resistance heater inside the reactor. Why would this same 50 W heat the blower in one instance, but not the other? It is same level of heat, coming from the same heater, in the same place.
If there is no excess heat, and both the calibration and the excess heat test are at 50 W, how can they heat the blower up to 40 to 250 W?
If the blower heated itself up by 40 W to 250 W, this would be recorded in the data, and it would burn up. There is no way it could produce this much heat itself. It consumes at most 5 W.
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For a reactor placed in front of the exit tube from the calorimeter, the reactor surface is visible from the exit. Therefore a RTD , measuring exit gas temperature will view the reactor surface. At high reactor surface temperatures the radiation from the visible part of the reactor
There are several reasons why this hypothesis is incorrect.
1. The reactor surface is not visible to the RTDs. As noted by Robert Bryant above, the outlet RTD is placed after the blower. (Actually, there are 2 RTDs, both after the blower.) There is a layer of reflective insulation between the RTDs and the reactor surface.
2. The temperature of the outlet air has been confirmed with thermometers and hand-held thermocouples. The reflective insulation also blocks the reactor from these instruments.
3. This hypothesis cannot explain why the outside of the reactor box is measurably warmer when there is excess heat.
4. Why would this happen with excess heat, but not with calibrations? If the excess heat is an artifact, as THHuxley postulates, it would be exactly the same 50 W coming from the same heater in both the excess heat tests and the calibration. It would be ordinary heat from a resistance heater inside the reactor. Why would it have a drastically different effect on the RTDs in the two tests?
5. Assuming there is anomalous heat (which makes the THH hypothesis wrong), again, why would it drastically change the performance of an RTD? Heat is heat. Once it passes through 1/8" of stainless steel, I cannot imagine it would look different to RTDs.
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Maybe next winter when it is heating his home, he can take a picture of the thermostat showing that it is off
The Japanese houses I have lived in do not have thermostats or central heating. They have room heaters, gas, electric or kerosene. Maybe houses in Sapporo have central heat. I wouldn't know. Actually, the houses I lived in did not have walls, strictly speaking. More like paper. In the morning the toothpaste is frozen. They were from the Meiji era. I am pretty sure Mizuno's house is more modern than that! He says it is well insulated.
Anyway, sooner or later I hope he will get around to testing this thing in a large calorimeter.
