Posts by JedRothwell

Quote from THHuxleynew

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.

Quote from THHuxleynew

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

Quote from THHuxleynew

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.

Quote from THHuxleynew

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.

Quote

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?

THH wrote:

Quote

The 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.

Quote

Umm... 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?

Quote from THHuxleynew

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.

Quote from THHuxleynew

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.

Quote from dartin

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.

Quote from Alan Smith

Why nit use a control with the only difference being no Pd on the mesh. 50 mg won't be missed.

I would just use a reactor with a heater only.

Quote from dartin

If the unit is heating a room as was demonstrated in figure 1 and the COP is 10, what would happen if one built a brick oven structure around the unit making it more difficult for the heat to escape?

Would it reach a higher temperature and a higher COP?

Will it stop working at some higher temperature caused by this procedure?

What would be the highest COP reached before it ceased operation?

Would it reach thermal runaway?

I have no idea.

Quote from dartin

Does that mean it is in self-sustaining mode or simply not working?

I think it was not operating when that photo was taken.

Quote from dartin

If the unit is working and is shut down to be moved to another location, will it simply require the power to be applied to the heater in order to start up again?

I strongly recommend you leave it in one place. Experiments attached to pressure gauges and pumps with Swagelok valves do not like to be moved or monkeyed with.

If you must move it, I urge you not to mail it. Put it in a car, and keep an eye on it. I think it may be possible for the thing to self-heat spontaneously.

Quote from dartin

I have unanswered questions. Will I have to provide D2 at above 1 atmosphere pressure?

Nope. You might wreck the reactants with that. Keep the pressure below 6,000 Pa.

Quote from Alan Smith

All the Ni mesh is now gone. I am ordering another 20 pieces -should be in my hands around the 20th july.

Did you also order Pd? And Kyukyutto Orange scented detergent?

Here is an example of how closely the blower power correlates with the air speed measured by the anemometer:

Air temperature and other factors have no visible impact on the air speed. No doubt they do have an effect, but the fluctuations in electric power are so large they swamp these effects. You would need much more sensitive instruments to detect them.

Quote from Shane D.

He says the power supply "is not connected in this photo":

Yup. I don't know why not, but that's what the man said.

Quote from THHuxleynew

What is the exact methodology of the tests. Is the reactor capped from the vacuum pump (with pipes disconnected).

It is connected as I said. That is clear from the paper and Table 1, which shows the pressure, and shows changes in the pressure. If it were disconnected you could not read the pressure, or pump out the gas, or add new gas.

Quote from THHuxleynew

I'd also like to point out that there is over this no difference between skeptics like me who expect that some artifact will be found,

If you "expect" some artifact then you are not a skeptic. You are a true believer. You have not found any reason to think there is artifact, and neither has anyone else. The results have a high signal to noise ratio. As shown in Fig. 5, the outlet air is 10 deg C hotter with excess heat than during a 50 W calibration. That is confirmed with other thermometers and thermocouples. It cannot be a mistake. It is not possible the fan is running that much slower; it would stop running completely, and burn. The flow calorimetry result is backed up by the fact that the reactor temperature during a 50 W calibration is 27 deg C, whereas with excess heat it is 380 deg C. So your "belief" is entirely based on faith, without a scrap of evidence to support it, and it is contrary to many physical laws. This is the opposite of skepticism.

If you were to say "I suppose there an artifact" I guess that would be skeptical. I would call it cynical or unreasonably pessimistic. But to "expect" an artifact where there no evidence for one, in a conventional instrument, using techniques that are employed by hundreds of thousands of HVAC engineers every day, is to "expect" a miracle.

Quote from Curbina

Also, much thanks for sharing the LENR-Aviation paralell!!! I took a quick read of it and found It was really informative and completely appropiate in many more ways than I thought before!!!

As Mark Twain probably did not say: "History doesn't repeat itself but it often rhymes."

As George Santayana definitely did say, "Those who cannot remember the past are condemned to repeat it."

Quote from Alan Smith

We decided not to use social media to promote our work, but to do it 'face to face' via personal contacts instead.

I suggest you write a scientific paper. That is not social media. You can publish it on your own website or present it at an ICCF conference.

Quote from Alan Smith

Because this is nobody's business but mine and Russ's.

You are saying it is secret! That's what I just said. We agree. So stop saying it isn't secret.

Quote from Alan Smith

Who visits your office?

No one, but everything I do that has any value ends up being uploaded to LENR-CANR.org, mostly in the JCMNS. Mostly in the form of other people's papers. I would never do all this tedious work if I did not mean to publish it. What would be the point? I do not edit papers or spend all day noodling with calibration spreadsheet data for the fun of it.

Quote from Alan Smith

Because LENR patents are already being extensively trolled, as would anything we filed, fun for lawyers but a distraction we don't need, priority we don't care a fig. As for keeping secrets, see my first answer.

Okay, so you are not planning to get intellectual property protection. Patents are the only way to do that with this kind of technology. Trade secrets will not work. If your claims are true, and they have any commercial value, they will be snatched away from you soon after you reveal them. That being the case, you might as well reveal them now, I suppose. You might as well get priority and scientific fame, because you not get any other reward without a patent.

If you enjoy working in obscurity, go ahead. Just don't expect credit, or priority, or money. I personally cannot see why anyone would want to do that, but I am not you.

It sounds more like a hobby than research.

Quote from Wyttenbach

The only thing I do not estimate is dirty tricking as in the Mizuno case.

There has been no "dirty tricking" as far as I know. Not by IH, and not by Mizuno. I know much more about the "Mizuno case" than Wyttenbach does. There were some differences of opinion, and some frustrating problems attempting to replicate. Such things always happen in experimental science. Always.