Some Points Regarding a Recent Presentation at ICCF20 on the ‘Lugano Report’ (Rainer Rander)

Quote from Alan

Now who is trolling- smiley face or not? Mats went to the source as I did, and got a dusty answer. As I did.

Alan, I don't want to be tiresome about this - its not worth it - but I think we must be misunderstanding each other?

My point was that Mats said he would get independent (of TC, MFMP, Lugano authors etc) expert opinion on the Lugano themography and the TC etc correction thereof to see who was right. That is a definite possible thing to do. And it has a definite black or white scientific answer. Obviously, it would not require cooperation from the Lugano authors!

I tried to find the entry in Mats blog (not a comment, but a real entry) where he stated that he did not believe TC, because he was biased, but would get independent expert views on the TC etc critiques vs the original Lugano report. It seems to have disappeared, but perhaps that is just because I can't navigate that stuff easily - it is easy to lose things!

Anyway many people were waiting with baited breath for this independent view. I'm sure there are people in the LENR area with enough knowledge of the relevant physics to reach a definite answer, and therefore agree with TC. But as I said above it is understandable if Mats now just does not want to pursue that line of investigation since the results from his POV will be negative.

Regards, THH

Yes, Lugano has been flogged to a pulp.

Lets talk about water meters and DN40 pipes and supersonic vapor-steam.

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If this is a matter of so much concern, why not organise an investigation yourself?

I see no more need to investigate since I understand the Lugano report, what was wrong, and how to correct it. I don't need anyone else to tell me that though the fact than many others agree with the details, and that empirical data backs this up, makes me more confident I've not made a stupid mistake. It is always good to check.

Mats however does not have that, and he must judge based on what others say.

I have a rough idea now of how the hot tubes produce excess temperature externally. I think it is consistent with the roughly 'COP 1.1' experiments, and can probably go higher, but probably not to 'COP 2'. Note that I said excess temperature, not excess heat.
I will be able to test this soon, I hope. My shop is nearly put back together

Quote from Paradigmnoia

Again, the alumina on inconel emissivity was the total hemispherical emissivity in the Wade report.

And that is exactly the the value to use and that why is reported in emissity tables. Eventually, as the Lugano group has done, using the right emissivity vs temperature function.
I see that you select facts and report only what is not against your opinions. Should I remind you that the Lugano group had measured the emissivity of the Alumina pipes and found a value compatible with the literature ?

Quote from Paradigmnoia

The thickness of the coating of alumina and zirconia on the Inconel should also be taken into consideration

As you cite this is true just for very thin layers. Due that most likely (read the Lugano report) the thickness of the Alumina was more then 0.4 mm just the Alumina emissivity should be taken in account.

Quote from Paradigmnoia

The iteration method requires multiple IR intensities or multiple IR bands

No Paradigmoia this means that you are not understanding the math. Iterative methods and fixed point iteration methods are quite general and were successfully used by the Lugano group. Please read Wikipedia for more reference.
The methods you are citing are used for example for measure terrain emissivity from satellite data. We do that normally in climate and environmental research. Is not the Lugano case.

Good morning randombit0,

The total normal or total hemispherical emissivity is not the correct emissivity to insert for the camera, which cannot make calculations based on radiance it cannot see, ie: those outside of the camera spectral sensitivity.

The camera only reports a temperature based on the radiant IR band that it can see. It does not need to see the whole IR band to do this if the emissivity that is correct for the partial IR band that the camera can see is used.

Total hemispherical emissivity is correct for calculating the total radiant power of a hot object, using the correct temperature.

The Inconel does not contribute to the IR spectra of the alumina covered object once the alumina is at least 0.5 mm thick or so, therefore the alumina on Inconel emissvity value from the chart you suggested is not appropriate.

At about 200 C, the spectral emissivity near 7 to 14 um and total emissivity are very similar. The emissivity found at that time would have been close to the emissivity that should have been used for the rest of the test, for the camera. Instead the emissivity was changed constantly, and severely for high temperatures. The whole point was to find the correct emissivity for the camera. The emissivity in the LW IR does not change significantly over the temperature range from 200 C to 1200C, and maybe even higher.

Please re-read the reiterative method references. It does not matter if they are for measuring the temperature of planets, stars, trees, or whatever (even alumina tubes). The method needs two sources of measurements at least, or it will not work. As my rainfall in Ascona plot shows perfectly. The GIGO principle applies here.

Best Regards,
Paradigmnoia

Quote from Paradigmnoia

The total normal

.......

Quote from Paradigmnoia

The GIGO principle applies here.

My Goodness !
You are quite repetitive in your errors !
We have discussed that a long number of times ! and I presume we will discuss it for a long time more. All detectors are calibrated using Total Emissivity for the material. If that value is known that's the value to use.
I will send you a complete textbook on Thermography next time .
And also you are making a big confusion among a general math method and a procedure to measure emissivity if you are able to make two narrow band measures.
Lugano people used the general math method using the correct alumina curve.
Because the method is general if you use the wrong curve, i.e. the Ascona rainfall or any other curve, you will obtain a wrong number.
But at least I see that you agree for one important thing when you say:

Quote from Paradigmnoia

At about 200 C, the spectral emissivity near 7 to 14 um and total emissivity are very similar. The emissivity found at that time would have been close to the emissivity that should have been used for the rest of the test, for the camera.

then you say that you would use a value near 0.65 ( the value at 200 °C ). This is much less then the 0.95 claimed by others.
I really can't belive that you think that this value should not change with temperature when even MFMP has found that alumina became transparent (i.e. emitting less radiation !) at high temp.
Also is nice to find that you agree with me that the Inconel would not disturb the measure. So only the alumina curve should be used as the Lugano people has done.

@randombit0,
The temperature of an object can be determined using the Planck radiation formula using very limited wavelengths of IR radiation, and the correct emissivity for those wavelengths. Only one wavelength is required, if the correct emissivity for that one wavelength is used. Microbolometers are not that selective to use only one IR wavelength, except perhaps in very specialized equipment. The lens and microbolometers used in the Optris camera at Lugano use the 7.5 to 13 um IR band. So the emissivity value for 7.5 to 13 um IR is the correct one for the Optris camera.

The Optris is completely incapable of measuring radiance outside of its spectral sensitivity band. And it does not have to, in order to determine temperature. (See above).

The total emissivity then is unimportant to the Optris, and wrong to use for its emissivity function.

I said similar. Not exact. And it is not a wildly changing emissivity over a huge temperature range, for the camera, although it may be for the total emissivity, used to calculate radiant power in all IR bands. (Because although the camera sees only a small section of the total IR band, obviously the object is radiating heat in all IR wavelengths, even if the camera cannot see that).
I suggested this idea merely as a possible reason for conflating the emissivity types. At moderate temperatures (200 to 300 C for example), the temperature error caused by using the wrong emissivity value is somewhat numerically low, if the emissivity difference is not very large.
Perhaps I should have typed 300 C instead?

I see no sign of any adjustment to Plot 1 reflecting the changes made to represent the specific type of alumina, as reported in the Report. I have experimentally shown that the low temperature emissivity in the 8 to 14 um band is higher for alumina that the total emissivity value in Plot 1. I would say that the emissivity should be at least 0.8 from 20 C to at least 300 C, if not higher, in the 8 to 14 um IR region.

@randombit0,
I see that you disliked my excess temperature in hot tubes comment. Care to expand on your dislike?

The core of my idea is already shown in the recent data of Tom Conover.
Note that with about the same amount of power that the stainless steel rod blank is hotter than the alumina rod blank by about 100C, and the fueled tube is hotter than the stainless rod by about 100C. Power is a bit more complex for the fueled runs, since the temperature was limited by the electronics to 1200C.

I suspect that the specific heat capacity changes with the change from solid powder to a gas, liquid and solid fuel combination is enough that the core temperature increases with the same power input, which ultimately leads to the exterior getting to a higher temperature as well. The amount of Heat in the system will be constant, however. The geometry of the system can accentuate the temperature change, and certainly the tube shape is very important in concentrating the heat gradient to the center to allow the effect to occur in its simplest embodiment. It is, I think, ultimately about the degrees of freedom of the "fuel".

Edit: Some comments and predictions based on this idea:
1) The temperature-based low "COP" experiments are explained by this idea. The molten metal (Al, Li)-gas(H) core tube will show a higher temperature than a solid core or empty core tube. This includes the recent the MFMP experiments, and the Conover-Wizkid experiments, and several others with elevated "COP" that are using the calculated power difference required to reach a certain temperature as a measuring stick for COP. (IOW, extrapolating power based on temperature as a proxy for measuring Heat compared when to a "null fuel" that does not have the same effective heat capacity)
2) A molten salt will show excess temperature compared to an empty tube blank, an alumina rod blank, and a solid stainless rod blank in hot tube experiments.
3) A molten salt should be a more effective blank for the excess temperature hot tube experiments. The ideal salt may take a bit of work, and the roles of H gas (or alternate) and particulate solid (Ni or proxy) may be important, but this has yet to be determined.
4) A good combination of tube design, molten material and gas (Maybe with some solid particulate material [like but not necessarily Ni]) can show strong excess temperature compared to solid fuel blanks, and especially air (empty cell) blanks.
5) Calorimeter-measured experiments will fail to show excess heat in hot tube experiments that showed excess temperature using benign molten +/- gas (and even the LAH-Li-Ni fuels) compared to solid core or air blanks.
6) If there is a LAH-Li-Ni reaction of unknown nature, the calorimeter experiments will show excess heat above a molten fuel blank, and this would be much more solid proof of the reaction. If the temperature-based COP experiments do not show any excess, then there is no point in moving up to a calorimeter to test further.
7) The hot tube may need to be in open air to generate the required thermal gradient to make an "excess temperature effect". This means that tubes insulated in bricks might be far less effective, but this needs to be tested.

Quote from randombit0

We have discussed that a long number of times ! and I presume we will discuss it for a long time more. All detectors are calibrated using Total Emissivity for the material. If that value is known that's the value to use.I will send you a complete textbook on Thermography next time .

This is trolling. The quality of explanation here (as you point out we've done it many times) has been exceptional. I can't believe you are for real.

Since different cameras have different detection frequency it is quite impossible for calibration for all to be based on total emissivity - unless all objects are assumed to be grey bodies.

If you show me a thermography textbook that makes such an assumption without explicitly stating the caveat I will write to the author and point out the omission, posting the no doubt apologetic reply. As many people have posted here, alumina is most certainly not a grey body.

Quote from Paradigmnoia

The geometry of the system can accentuate the temperature change, and certainly the tube shape is very important in concentrating the heat gradient to the center to allow the effect to occur in its simplest embodiment. It is, I think, ultimately about the degrees of freedom of the "fuel".

@P: Did you ever calculate the measurement spot distortion, when pointing the Optris toward a cylindric surface with a small radius?. This might significantely under estimate the measured temperature.

@Wyttenbach,
I haven't, but the camera is relatively close to the reactor. (less than a meter it seems)
The pixel to surface area ratio is pretty good, I think.
The reactor is well-defined in the images provided, and outlined with the measurement boxes quite clearly. The camera was able to select quite small spot sizes (Figure 7), and the body of the reactor, is rough, porous alumina, which should have quite isotropic radiant power properties.

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@P: Did you ever calculate the measurement spot distortion, when pointing the Optris toward a cylindric surface with a small radius?. This might significantly under estimate the measured temperature.

[Edit - maybe you are talking here about a different experiment - I'm talking about Lugano]

There is no measurement spot, just an array of photodiode detectors whose output is averaged. You might get aliassing effects though I'd think it highly unlikely they create systematic problems when averaged. A cylindrical surface (say the apex of the ridges) will radiate differently because the view factor correction will not completely apply (and it would lead to a slight temperature overestimate due to emissivity reduction). But this is a relatively small adjustment anyway, and only applies for a small number of pixels. Totally insignificant.

For those few apex pixels, ignoring view factor, a cylindrical surface would only radiate differently if the radiation was not isotropic - but it is to a very good approximation. The curvature has no relevance except in altering local heat balance - another story but again a second order effect and it averages out. Still there is enough nonlinearity around that checking the effect of the ridge temperature differences might lead to a useful small correction.

Lack of isotropy would be more relevant for the 45 degree ridge angle on the whole surface, something different again, but not expected.

Maybe there is something I'm not seeing here, but I suspect that you are getting confused about active vs passive IR measurements

Quote from Paradigmnoia

I haven't, but the camera is relatively close to the reactor. (less than a meter it seems)

The measurement dot (fig.7 Luganoreprot) covers 1/3 of the cylinder, Further on they hold it to the upper part of the cylinder. This could explain why they underestimated the dummy reactor temperature! And also partly why the rod's temperature is off!
You have to integrate the intensity with a decreasing cos of the cylinder angle you measure.

@THHuxley, @Wyttenbach,
IR "guns" often exhibit the sort of spot size problem referred to. Cheap ones especially, where the laser pointer often doesn't really point at the IR detection spot. Better quality IR guns generate a laser circle that closely matches the gun's restricted IR sensor view location and distance-related focal width.

Quote

The measurement dot (fig.7 Luganoreprot) covers 1/3 of the cylinder, Further on they hold it to the upper part of the cylinder. This could explain why they underestimated the dummy reactor temperature! And also partly why the rod's temperature is off!You have to integrate the intensity with a decreasing cos of the cylinder angle you measure.

That is not true, unless the radiation is non-isotropic. The normal (good) assumption is that radiation is isotropic.

But, if it were true, you are forgetting a more significant effect which is the ridges. These mean that the whole surface is at a 45 degree angle to the camera. All the time.

Finally we can get very little out of the dummy reactor measurements since the IR data was essentially replaced by TC data, through chnaging the emissivity graph. For the dummy that just means that the IR measurements were OK. For the real reactor there is no implication, since the graph is only changed at the dummy temperatures. We don't actually know quite how they did this, but they say they did it to make the dummy calibration correct independent of the (wrong, anyway) IR calculations, so I'd expect that.

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IR "guns" often exhibit the sort of spot size problem referred to. Cheap ones especially, where the laser pointer often doesn't really point at the IR detection spot.

Yes, as I said above, I think Wyttenbach is thinking of an active spot measurement device where you do indeed get angle dependence. (Do you? Yes because the surface angle affects the incident light power per unit of surface).

@Wyttenbach,
Regarding Figure 7, I suspect that the greatest potential for error is the left-right comparison of the two spots, where there is clearly a strong thermal gradient (logarithmic) towards the outside end of the rods.
I would expect that left-right thermal gradient to be greater than the middle to top gradient of the Rod, the same distance form the reactor (although there could be one there as well).
The measurement spot contains many pixels. I suppose it could be worked out how many, and the what the effect might be due to curvature. (They did calculate the pixel vs area in the Levi 2013 paper.) If the IR camera calculation dot ran right off the top of the Rod, that would be a problem, probably.

Quote from THHuxley

Yes, as I said above, I think Wyttenbach is thinking of an active spot measurement device where you do indeed get angle dependence. (Do you? Yes because the surface angle affects the incident light power per unit of surface).

Lamberts cosine law takes care of the incident angle problem in isotropically radiant bodies. As long as there is no weird reflective effects, prismatic action, or parallel ridges, Lambert's cosine law should hold. The view factor is reduced by the same amount as as the cosine of the emission angle. This is the same reason that total hemispheric emissivity is the same as total normal emissivity for most objects. The rough alumina surface and porous composition of the Lugano materials ensures an isotropic emission pattern. (According to Wikipedia, this is also the reason the (full) moon looks just as bright right to the edge.)