Has this been discussed, or anyone with thoughts on the matter?
One would Expect that LENR have a different energy emission spectrum than electrical heated alumina.
Alumina transmittance and LENR energy spectrum is unkown parameters in the Lugano test.
I have reason to believe that the character of light produced by LENR is blackbody radiation associated with the temperature of the reactor core. In the case of the Lugano reactor that blackbody temperature would be between 1450C and 1500C.
Is this a known fact, a good supposition or a guess?
It is based on the exploded diagram of the Lugano device in the patent application.
I have fiddled with hot tubes a bit, and this experience leads me to believe that the "air" space gets rarefied due to the high heat, if open to the atmosphere even slightly (like a hot air balloon). If the outer void is completely sealed, or vented to the interior tube, etc., then other unforeseen effects might occur.
Also, embedding heater coils in alumina has been shown in numerous tests to result in early coil failure or tube breakage compared to those that are at least partially exposed to the atmosphere, and free to expand and contract.
Based on 15 Ga coil wire, I would expect the air gap to be not much more than 2.5 mm maximum, and no less than 1.5 mm minimum.
This in turn gives an idea of the outer diameter of the inner tube, which was supposedly plugged with a 4 mm diameter bushing containing a thermocouple. But the plug in the patent application image scales up from the photo to something closer to 7 mm.
Perhaps there was an additional bushing (already inserted), around the thermocouple probe orifice, so that the TC hole was 4 mm... and the bare hole was around 7 mm. The thermocouple entry is not depicted in the patent application exploded diagram.
Alumina spectral emittance in the visible
Based on the measurements from http://web.archive.org/web/200…logies/pdf/1999crd128.pdf after fitting their suggested polynomial model, I get:
epsilon(lamda,temp) = -0.00001864899582131242*lambda + 1.510506854256884e-7*lambda**2 - 3.1050084175084967e-11*lambda**3 + 0.0006781392396575488*temp - - 1.6184523809523763e-7*lambda*temp - 9.169987667661622e-7*temp**2 + 3.610740627829621e-10*temp**3
Not sure if extrapolating this to smaller wavelengths or different temperatures is valid.
Here's a plot showing the extrapolation:
Next step will be to integrate the spectral emissivity into the camera response calculations. (I am trying to match the MFMP picture, on which BTW the higher-temperature naked coils are saturated.)
Also, embedding heater coils in alumina has been shown in numerous tests to result in early coil failure or tube breakage compared to those that are at least partially exposed to the atmosphere, and free to expand and contract.
That makes sense.
But which patent were you referring to? The latest US one doesn't have any kind of tube drawing. The Italian one is vague IMHO and I don't see any dimensions.
Based on 15 Ga coil wire, I would expect the air gap to be not much more than 2.5 mm maximum, and no less than 1.5 mm minimum.
Isn't the coil wire braided? 1.5 mm seems too thin, the glowing wires from the pictures look thicker to me, I would have guessed 3 mm.
Patent application WO2015127263 (pamphlet 131) referencing PCT/US2015/016897 has the exploded diagram (photos) of the Lugano device. Newer line drawings have replaced the grainy B&W photos, but the original version should be still in the image file somewhere.
The wires are described as being braided in the Lugano report, but as 15 Ga in the patent application. My guess is that the braided wire is effectively/roughly 15 Ga, since 3 mm diameter wires would be very hard to make with acceptable resistance characteristics. 15 Ga wire is difficult to imagine coming in the resistance specified, even as braided smaller diameter wires, so there is some uncertainty in the actual wire size used. The wire does seem to be fairly substantial compared to what most replicators have been using.
I can't tell if the image is just too pixelated or if the braids are actually visible in the image.
@Antoine10FF Looks very interesting. I'll read it thoroughly later.
Regarding the diagram, if the outer caps are 4x4 cm, then the inner tube is about 1 cm in outside diameter.
I have "assembled" the reactor using the patent image pieces, traced from the image and maintaining their original relative scale, and they all nest together very nicely.
You can use clearer diagrams available here:
https://drive.google.com/file/…9g5EnNHBQVXVZaE5LY28/view
https://drive.google.com/file/…9g5EnNHBQVXVZaE5LY28/view
Look at Figure 2, item 16, the three entangled wires.
Two of the wires are spiralling around each other like threads, but the third is zigzagging across and through the double helix like in some drawing by Escher. Maybe there is a fourth dimension involved here? This may be needed in order to fit it onto the bobbin, item 18.
...
That must be the braiding they are talking about. 
(Deleted comment -- your PDF is not a full paper.)
I have almost solved the wire diameter/braid/resistance problem in order to arrive at the specified resistance per foot, as specified in the patent application, and arrive at the mathematically calculated resistance per phase/coil using inconel with the required number of wraps around the depicted core tube. The problem remaining is that this seems to lead to a rather cold-running heater coil system (good luck getting close to 750 C at 930 W). Possibly not even very effective at 3000 W....
@Antoine10FF, I may have some inconel emissivity data stashed away in my data files. I'll post what I have once I scrounge it up.
@Antoine10FF, I may have some inconel emissivity data stashed away in my data files. I'll post what I have once I scrounge it up.
Meanwhile, I extracted some Inconel emissivity data from "Thermal Radiation Heat Transfer", 5th ed, Howell & al. As usual, the emissivity depends a lot on the material condition, but in both cases it decreases with wavelength. The curve I picked goes from 0.9 at 340 to 0.7 nm at 800 nm.
Another thing I learned is that blackbody colors below approx. 1500°C are outside of the sRGB gamut:
This may explain why the MFMP calibration colors are weird and purplish - what happens to colours outside the standard sRGB gamut may be very camera-dependent.
This is also why I paused working on the camera spectral response. I may resume when I find the little Olympus camera and analyze some fireplace pictures with it, to see what that particular camera does to non-sRGB colours.
Now I'm just working in the XYZ color space, which is linear. Previous estimates were using the sRGB space, which is highly nonlinear. Those previous estimates are therefore grossly incorrect.
Now, having looked at reasonable values for indoor illuminance levels, I have determined those to range from 50 lux (quite dim) to 2200 lux (equivalent to overcast daylight). Higher values would be plausible only under special conditions such as operating rooms, etc. — or under broad daylight.
I have estimated the reactor glow by substracting the XYZ value of the brightest reactor spot (within the "Reactor" zone below) from the average value of the "ReactorDim" region. The XYZ value of the wall is taken as the average of the "WallTop" region.
The glow from the reactor core is diffused by the alumina tube, thereby reducing its effective (visible!) emissivity. For the cylindrical geometry the effective emissivity should be equal to the ratio of the diameters for an ideal diffuser.
For any given luminance value assumed for the wall portion, and for any given effective emissivity, there is a single temperature value for the reactor core (whose radiating part I assume to be inconel steel) that matches the ratio measured from the images.
In the next plot, each curve corresponds to one effective emissivity ratio and gives the relationship between wall luminance and estimated reactor temperature.
As usual, all of this is rough, unverified, etc.
If we are to stay within the luminance values one would get from a wall having a reflectivity of 0.7 under illuminance levels of 50 to 2200 lux, we have to use an effective visible emissivity that is one fifth of the inconel emissivity I selected to get Bob Higgins' 1100°C estimate, and then we are stuch with a luminance of 280 cd/m2 which requires an illuminance of 1200 lux.
As the alumina housing has a diameter of 20 mm, this would imply that the core has a diameter of about 20/5 = 4 mm.
EDIT: Fix luminance range.
A few quick notes:
There is supposedly a skylight in the room in Lugano. We were told that it was opened to let heat out of the room.
The purple digital camera colour caused by IR can be easily seen by taking a picture of a TV remote or similar where the LEDs are visible. Active IR LEDs will appear violet in the photo, but unchanged from off by eye.
Do you mean 4 mm OD for the core?
.
A few quick notes:
There is supposedly a skylight in the room in Lugano. We were told that it was opened to let heat out of the room.
That is possible. There is some kind of diffuse, bluish light source that is apparent in some pictures. We could try to estimate the relative illuminance from that light source based on the contrast of the shadows from the point sources.
Quote
The purple digital camera colour caused by IR can be easily seen by taking a picture of a TV remote or similar where the LEDs are visible. Active IR LEDs will appear violet in the photo, but unchanged from off by eye.
I did look into the IR effect last week, and it's true for a lot of cameras, but some others have relatively good IR cut-off filters. Yet I still managed to get purple/pink colors using spectral responsivity curves that were zero in the IR; this is due to the way the channels saturate, and on how the camera firmware produces XYZ values from sensor readings.
This could explain why the naked wire appears orange in the Lugano images but pinkish in the MFMP images - different cameras, different color balance settings, etc.
QuoteDo you mean 4 mm OD for the core?
Yes, 4 mm OD — but of course there are large uncertainties and big assumptions in all the quantities.
@Antoine10FF,
We have a diagram and enough measurements to establish that the outer diameter of the inner tube is close to 1 cm. Using 10 mm instead of 4 mm for the OD, how does that affect the results of your analysis?
@Antoine10FF,
We have a diagram and enough measurements to establish that the outer diameter of the inner tube is close to 1 cm. Using 10 mm instead of 4 mm for the OD, how does that affect the results of your analysis?
I think the effective emissivity would then be 10/20=0.5 of the inner tube emissivity. The resulting curve would lie between the green and the blue curves in the last plot (which are for an emissivity scaling factor of 0.6 and 0.4). The temperature estimate would be between 825 and 1150°C if we discount skylight.
It is possible that fig. 12a is actually mostly daylight illuminated.
Tracing the shadows from the photos, it appears that one of the point light sources is pretty high, maybe 4 or 5 meters. Therefore the room has a high ceiling. Therefore it's not a "typical room." Maybe it's some kind of old work shop, or a converted barn, with high windows, or maybe skylights?
If the room has large windows, and if the picture was taken on a bright day, the illumination level might have been significantly higher than the 500 lux artificial lighting Iimit I selected. In that case temperatures of 1200°C are plausible.
Reactor core temperatures below 900°C do not seem compatible with this model.
Also there is one very important thing that is being overlooked, and that is the visible bulk transmittance, reflectance and thickness of the alumina. That is a major factor. One of the next things to do would be to use the assumed dimensions to estimate that.
These estimates for the wire temperature (unless you believe the strange "shadows" argument) seem perfectly plausible, and compatible with the known surface temperature of the alumina (780C).
The temperature gradient can be related to the heat flux precisely if the wire helix OD / reactor diameter is known. (The fins complicate things but can be well approximated, since the ratio applies inside a log and therefore does not matter much). However unfortunately the power out comes from both heat flux and radaition from alumina surface, and also direct radiated power out from the wire, so things remain unclear.
