@Thomas Clarke ,
From the plot I posted above, one could argue that a normal emissivity as low as 0.8 is feasible in the Optris range.
0.8 normal emissivity actually fits the dummy fairly well for power, in my cherry picking exercise, when the hemispherical emissivity is also fixed up in the low temperature range.
Data for total hemispherical emissivity of alumina was not very consistent at around 500K, however.
To discus the 'science' behind the dispute between Rossi and Industrial Heat
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When you integrate the bolometer response (which is lower at the high freq end where al203 emissivity crashes) with the al2o3 response you get 0.9. You can look at the numbers in my code - I took both responses from Bob's paper - he did a pretty good job of identifying likely response but of course there is quite a bit of variation possible.
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@Thomas Clarke ,
The response of the microbolometer array is affected by leveling corrections internal to the Optris.
I know that some work was done on it, but how the unit levels and how Bob or others have leveled the response compares is an open question.
I am just eyeballing the curve, and mentally weighting an average response.
Between 0.8 and 0.9 is something I can agree on, without buying an Optris.Maybe we can suggest that the MFMP use the Optris on the Glowstick cover tubes at 235°C and 450°C and fiddle with the emissivity setting at the end of the present experiment for maybe half an hour.
Edit: Image attached
Edit2: Notice that in the Manara et al plot, the big alumina spectral ε peak actually broadens with increasing temperature, especially in the far long wave side of the Optris view window, so that the Optris normal ε increases (a bit) with temperature for alumina. -
Hello boys !A very quick answer between two meetings. Paradigmoia, nice that you have found that reference but I presume that you have not understood it very well.Emittance and Emissivity are two different concepts. Emittance is an extensive property of the material that depends also on the thickness of the layer considered while Evissivity is intensive property of surface.When trying to measure the temperature of a surface we must consider Emissivity that, in simple terms, indicates how much power is radiated at a certain temperature from the surface respect the radiated power of a Black Body.Quantum Mechanics states that at a certain temperature nothing can radiate more power then a Black Body (also known as perfect absorber) at any wave length. So emissivity is defined by the ratio of the integrated power spectrum radiated by a body and the integrated power spectrum of the Black Body and have values always <1 and is a non dimensional quantity. Factory calibration of the infrared camera, optris provide calibration files for each single camera and IR optics, ensures that Is the emissivity value that must be used in the setup. Not surely the emittance. As a proof look first at your graph, noting that there is not much variation from 300K to 1050k, so if we accept your statesment the group shuold have used a 0.9 value even for the rods, BUT from the IR images we note that the reference dot (which has an emissivity 0.95) looks as bright spot . This meas that the alumina pipe has an emissivity much lower then the dot !Nevertheless the authors in fact have forgotten a detail, that doesn't create a real problem but should be noted. In the case of the reactor Alumina is put on the Kanthal wires and becoming partially transparent to IR at high temperatures part of the radiation to the camera comes directly from the wire. Because the actual material of the wire was unknown so also the alumina thickness it was impossible to make a correct calculus of the actual emissivity that was a convolution of three functions on T: wire emissivity, alumina transparency, alumina emissivity.Data of emissivity Alumina on Inconel is commonly available for IR measurements:http://www.scigiene.com/pdfs/4…erEmissivitytablesrev.pdf
As you see from the table the range given .69 at 427 °C and .45 at 1093 °C match quite well the values actually used by the Authors.Finally even if we suppose that a slightly higher figure for the emissivity parameter was a better choice for the reactor body during the run we should note that the actual measured emitted Energy has a very weak dependency on that parameter.That because IR camera use the emissivity in the inverse of the Stefen Boltzmann law to obtain the temperature and when we recalculate the energy the two factors cancel.The reported result seems correct and we estimated that in the most pessimistic and unrealistic scenario they should be 30% higher then real tacking the COP to 2.57. The rest of the analysis remains valid and due to the long amount of time of the run the energy produced seems far beyond any chemical source.This was a quick answer..... CU next time.... pardon me if there was any Typo. -
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"A pump can perfectly have a flow rate 60% more than its nominal flow rate if it is taking the water from a presurized pipe, I dont know if the exact configuration of the test is known, but as it is not provided by T Clark, he can't call Levi a fraud without taking this fact in account".(1) If you trawl the comments on mats site I personally raised this very point before agreeing that Ascoli's analysis was bulletproof. The water in this case was taken from a hose in a container - no mains water available. in any case the pump is dosimetric and therefore would likely break before giving wrong reading - but I did not buy into ascoli's analysis till the hose in water thing was clarified. My uncertainty over this, challenging ascoli, is all documented on that thread.
(2) I'm not calling Levi a fraud. Where have I done that? I'm just saying that based on his wrong calculation of the pump flowrate he is not a reliable witness for these tests.
QuoteDoes this bring into question your critique of Lugano?
WTF? You have lost me here. The Lugano test had no water pump. -
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So emissivity is defined by the ratio of the integrated power spectrum radiated by a body and the integrated power spectrum of the Black Body and have values always <1 and is a non dimensional quantity. Factory calibration of the infrared camera, optris provide calibration files for each single camera and IR optics, ensures that Is the emissivity value that must be used in the setup.
you are making here the same error that Levi made, and the report authors. You see "emissivity" as a single parameter which you define as "total emissivity" should indeed be defined. But that is wrong; there are two values, quite different, both of which enter into the calculation.
The optris camera requires you to input the correct "emissivity" but what it does not explicitly tell you in the manual is that this is NOT the total emissivity. It is the band or "integrated spectral emissivity" over the 7-13um band. Whereas total emissivity integrates over the whole Plank function frequency curve, band emissivity integrates only over the Optris sensor bandwidth. I do this numerically using typical bolometer spectral response in my report, but you can easily do an eyeball calculation.
So for example at 1000C the typical alumina band emissivity is 0.92 and total emissivity is 0.45. Furthermore, it is total emissivity that is relevant for power emitted from a grey body, and band emissivty that is relevant for initially calculating the temperature of the grey body from the Optris instrument. You can see that mixing up the two parameters leads to a very serious error, which was the error made in the Report.
See http://lenr-canr.org/acrobat/ClarkeTcommentont.pdf
You will find references there to two independent groups who have come to similar conclusions (though not followed them through) as well as my analysis of this issue.perhaps you can reasure me that you are not sharing this mistake with the Lugano Report authors? Because without the magic words "total" and "band" or "optris effective" in front of emissivity I must conclude that you are...
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As a proof look first at your graph, noting that there is not much variation from 300K to 1050k, so if we accept your statesment the group shuold have used a 0.9 value even for the rods, BUT from the IR images we note that the reference dot (which has an emissivity 0.95) looks as bright spot . This meas that the alumina pipe has an emissivity much lower then the dot !Nevertheless the authors in fact have forgotten a detail, that doesn't create a real problem but should be noted. In the case of the reactor Alumina is put on the Kanthal wires and becoming partially transparent to IR at high temperatures part of the radiation to the camera comes directly from the wire. Because the actual material of the wire was unknown so also the alumina thickness it was impossible to make a correct calculus of the actual emissivity that was a convolution of three functions on T: wire emissivity, alumina transparency, alumina emissivity.
Well done! You correctly identify a possible problem caused by alumina transparency which adds a large and unknowable error to the results (see my comment above for various problems in this experiment - it is unknowable because it depends on the emissivity of the underlying surface, probably Kanthal wire (and BTW, though not relevant to the thermography issues we discuss now, if this is Kanthal wire it does not have a v large NTC and therefore you need to consider why the resistance changes by a factor of 3 between dummy and active tests).
However, you still have a badly incomplete analysis because you do not distinguish between band emissivity and total emissivity. That makes a big difference.
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The reactor used in the Lugano test was equipped with 3 heater coils in an AC 3-phase balanced delta configuration (see Figure 4 of the Lugano test report (LTR) "Observation of abundant heat production from a reactor device and of isotopic changes in the fuel").
There had been discussions, wether the power consumption values in the LTR are plausible using usual wires for the heater coils or if these values would require a miracle material.
Some calculations (published by discussion members) showed large differences between the single coil resistance of the dummy reactor, when compared to values calculated for the active E-Cat.
The lack of information about voltages, resistances etc. in the LTR, makes it not possible to calculate exact values for the active E-Cat. But it is possible to probe, wether the values given in the LTR might be plausible.
This task requires some knowledge of the conditions in 3-phase AC balanced delta configurations. So, there is a shot summery:
For 3 Phase AC and a balanced Delta Configuration the total power equals 3 times the power of a single phase.
The total current is sqrt(3) times the current of a single phase.
The total voltage is sqrt(3) times the voltage of a single phase (e.g. 400V total gives 230V single phase) and the total resistance is the resistance of a single phase divided by sqrt(3).Calculations of heater coil resistances
Dummy reactor at 450°C:
From the LTR one can take the values of 486 W input - 7 W joule heating (line-losses subtracted) = 479 W with 19.7 A current for a single phase.
An exact calculation of the single coil resistance is possible:
19.7A * sqrt(3) = 34.12 A (total current = single phase current * sqrt(3))
479 W / 34.12 A = 14.04 V (total voltage = total power / total current)
(The 14.04 V are for example comparable to the 400V line voltage. The voltage of a single line 230 V e.g. is calculated 400 V devided by sqrt(3)).
14.04 V / sqrt(3) = 8.11 V (single phase voltage = total voltage / sqrt(3))
8.11 V / 19.7 A = 0.41 Ohms (single coil resistance = single phase voltage / single phase current)
Probe:
8.11 V * 19.7 A = 159.77 W (single phase power = single phase voltage * single phase current)
159.77 W * 3 = 479.03 W (total power = single phase power * 3)
For a single heater coil at 450°C and a net input power of 479 W a resistance of 0.41 Ohms is calculated (the total resistance equals to 0.41 Ohms / sqrt(3) = 0.237 Ohms - further needed for verification using web-calculator).
Active E-Cat at 1260°C:
Consumption is 815.86 Watt subtracting the joule heating of 37.77 W one get a net consumption of the 3 heater coils in total of 778.09 W.
There is no exact data for the coil currents available. The report mentions 40 - 50 Amps for the total of the 3 coils.
But it is possible to prove whether the info in the report might be be plausible or not.
Let someone for example assume a total current of 43A.The current for a single heater coil equals to:
43 A / sqrt(3) = 24.83 A (single phase current = total current / sqrt(3))
The calculation steps as for the dummy reactor:
778.09 W / 43 A = 18.09 V (total voltage = total power / total current)
18.09 V / sqrt(3) = 10.45 V (single phase voltage = total voltage / sqrt(3))
10.45 V / 24.83 A = 0.42 Ohms (single coil resistance = single phase voltage / single phase current)
Probe:
10.45 V * 24.83 A = 259.47 W (single phase power = single phase voltage * single phase current)
259.47 W * 3 = 778.41 W (total power = single phase power * 3)
For a single heater coil at 1260°C and a net power input of 778.09W a resistance of 0.42 Ohms is calculated (the total resistance equals to 0.42 Ohms / sqrt(3) = 0.242 Ohms - further needed for verification using web-calculator).
Conclusions:
The information in the LTR is plausible, at least according to the input power consumption with respect to real existing and usual heater wire materials.
My calculations show a single coil resistance of 0.41 Ohms at 450°C and a single coil resistance of 0.42 Ohms at 1260°C. (The 0.42 Ohms for the active E-Cat at 1260° were calculated under the assumption, that the total input current is 43 A - other assumptions (40-50A) and values are possible).
Real existing and usual wire materials could have been used.With best wishes
Tom
P.S.
If you doubt my calculations, then you may check these using the 3-phase delta calculator at:https://www.watlow.com/Resourc…neering-Tools/Calculators
Be sure you have selected "3-Phase Delta (Balanced Load)" in the calculator selection box!
Then enter only two values:
Phase Current (lp)> 19.7
Watts (W)> 479
Leave the other fields empty and press calculate.The applet shows phase voltage 8.1049... and Ohms 0.2375... . The resistance in Ohms is that of the total circuit, for a single coil and multiplied by sqrt(3) that gives 0.411141... .
Press RESET
Enter only two example values for the active E-Cat:
Phase Current (lp)> 24.83
Watts (W)> 778.09
Leave other fields empty and press calculate.The applet shows phase voltage 10.4455... and Ohms 0.24288... . The resistance in Ohms is that of the total circuit, for a single coil and multiplied by sqrt(3) that gives 0.420668... .
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Tom,
The issue to explain is got from comparing the joule heating power (proportional to the square of the line current) with the total heater power (the same).
There is a X3 discrepancy in this ratio between dummy and active tests. So if the figures are correct we have:
(1) the connecting wire resistance changes by a factor of 3
(2) the line current changes by a factor of sqrt(3) for identical power
(3) the heater resistance changes by a factor of 3(1) and (3) look unlikely so we are then left with (2) - which would be true if the setup changed from Wye to Delta.
The solution is quite speculative, but the X3 discrepancy is fact and needs some explanation.
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Tom,
yes, there seems to be a problem with the calculations of the joule heating powers in the report.
But if we would assume the 46A single coil current, which Andrea S. calculated based on the joule heating power of the active E-Cat at 778 W, that would result to a phase voltage of only about 5.6 volts. Or about 9 V for 80A total ( not in the range mentioned in LTR ) compared to 14,4 Volt for 479 W of the dummy. The NTC and a very small phase angle would be a big challenge for the PID and the FUSION SCR.
Failure of a single phase (SCR channel) may also lead to higher currents in the cables (but not 3x).
P.S.
One miscalculation in the LTR is the multiplication of the single line loss by 3. Sqrt(3) would be correct. So the joule heating values are wrong and should have been re-calculated. -
See what you think of this:
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Tom,
yes, there seems to be a problem with the calculations of the joule heating powers in the report.
But if we would assume the 46A single coil current, which Andrea S. calculated based on the joule heating power of the active E-Cat at 778 W, that…
Yes, it could just be an error. But it is puzzling. Phase angle is unclear because (unless you know the controller better than me) we cannot know at what point in the mains sinusoid the triacs switch. There are different strategies possible.
A Wye to Delta configuration change does exactly account for the data without NTC?
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@Thomas Clarke
I do get a slight drop in R from the dummy to the active runs, about 0.05 Ω.
Considering the potential measurement inaccuracies, that is probably insignificant.
I also did not consider any inductance in my analyses.
The resistance calculated does not increase relative to temperature in the way any type of calibrated resistance wire I could find normally does. -
<a href="https://www.lenr-forum.com/forum/index.php/User/603-Thomas-Clarke/">@Thomas Clarke</a>
I do get a slight drop in R from the dummy to the active runs, about 0.05 Ω.
Considering the potential measurement inaccuracies, that is probably insignificant.
I also did not consider any inductance in my analyses.
The resistance…Yes, various Inconel alloys have differing R vs T graphs and slight drop or increase are both possible...
But not X3, without SiC.
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but what it does not explicitly tell you in the manual
LOL Come ON !!!!! They forget this "detail"The FLIR IR camera makers also seem to have forgot this !
https://www.ivytools.com/v/fli…lir-t420-users-manual.pdfYou boy are really the ONLY expert in the world !
Well done! You correctly identify a possible problem
Thanks!
Note that the fact that the Alumina was partially transparent means that has low emissivity !
As usual in your answers you don't mention the rest and the evidences selecting ad hoc lines self referring to your own comment so not adding any information !
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The calculation of the joule heating values is wrong in the report. They used factor 3, but sqrt(3) would have been correct. So I think that it doesn't make sense to evaluate other configurations, based on these false values.
I will do a recalculation of the joule heating values, when i find some time to do so.
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What Optris says:
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See what you think of this:
Just checked your "Dummy Run" picture. The single phase voltage is too high (14.x V). Should be 8.x V.
Please refer to my calculation for the dummy reactor. -
Dear randombit0,
We are discussing specific technical matters with definite answers, so perhaps we can slowly and carefully deal with any possible disagreements one by one?
Quote from randombit0
Note that the fact that the Alumina was partially transparent means that has low emissivity !
Not quite, it means that it has low total emissivity. However, at longer wavelengths it remains opaque, and therefore has high spectral emissivity. Would you like to check a graph for the spectral emissivity of alumina?The key dimension of this problem which you are not mentioning (and which Levi clearly did not consider) is that alumina has a highly variable spectral emissivity - it is very far from a grey body. Most of the total emissivity change with temperature is simply the Planck curve integrated over this varying emissivity with frequency.
QuoteAs usual in your answers you don't mention the rest and the evidences selecting ad hoc lines self referring to your own comment so not adding any information !
That sounds unlike me. I'll address what I can think of here and you could ask me for further comment if I've left anything out. I agree that the translucence issue can generate serious errors in this experiment. I agree that if the reference dots look brighter than the adjacent alumina in the IR picture then they have higher band emissivity, but I do not know quantitatively how much higher that is - which would depend on how the photo was reproduced. If they look brighter in an optical picture is means their emissivity at optical wavelengths is higher (certainbly true) but not necessarily that the IR band emissivity is higher.
However, my self-reference to my comment was because you were not directly addressing the most obvious and large error in the Lugano experiment which was that they badly miscalculated temperature by thinking band and total emissivity are identical. I'm still waiting for you to acknowledge that this is indeed an error. Until you do this, you have a badly incomplete analysis. Of course you may be doing this but not saying it, in which case my insistence on this point is not needed.
QuoteLOL Come ON !!!!! They forget this "detail" The FLIR IR camera makers also seem to have forgot this ! ivytools.com/v/flir-manuals/t4…lir-t420-users-manual.pdfYou boy are really the ONLY expert in the world !
I really don't know here whether you are saying everyone knows the band vs total emissivity issue, and I don't need to raise it, or whether you are saying because it is not in the FLIR manual it does not exist? Certainly the Lugano authors got it wrong, and Levi got it wrong in his recent reply to Mats.
The FLIR manual, and other IR thermography references, do not need to say anything about this because they ALWAYS say you must determine emissivity experimentally. Therefore you are automatically getting band emissivity. Whether the surface is a true grey body (with total = band) or not is then irrelevant. For thermography "emissivity" is always "band emissivity" and "total emissivity" is not needed. Most materials have similar total and band emissivity which further confuses the issue, but alumina is a spectacular exception.
I don't consider myself an expert here at all. I spent a good 6 months reading the Lugano report and learning themography and the properties of alumina (my basic physics is ok, as is my maths, and I am pretty good at learning new stuff. For example, I greatly enjoyed working out how those ridges on the reactor affect the emissivity - some stuff called shape factor or view factor!). So, if you think all I say is obvious, then of course I'll agree. It took quite a while for me to work it all out but an expert would know immediately.
So: perhaps you now could do me the courtesy of stating whether of not your analysis addresses this issue of the band and total emissivity being very different in alumina at temperatures above 600C?
