Lugano performance recalculated - the baseline for replications

99% alumina has a melting point of 2016 degC.

Thomas, or anyone else, I would like an opinion on where this idea might be going wrong, if it is.

I have put the formula for radiative heat used in the Lugano report in Excel, as it was written, and tested it against Table 3 (page 17) to make sure it is giving the same answers in Watts. It does, when the temperatures supplied for the rows in the same Table are used. Sometimes there are very minor (second decimal place) differences. (Close enough, probably due to some third decimal place rounding in the report.)
I note that the example of the formula ([14], page 16) has an error in the paper, compared to Row 5 in the Table. (Uses 454.3°C for the formula example, not 454.03° as in the Table). ....So far no concerns....

Now using the temperatures from Table 6 inserted into the formula, and the Watts calculated in the report, and some manual iteration for the first ε in the formula, the ε values used by the professors, as an average over the ten subdivisions of the main heater tube section (multiply the area by ten again to obtain the area of the whole tube, not a subdivision) can be derived. They are low, like 0.44 etc, as expected from Figure 6 and Plot 1, page 9. (I left the ambient temperature part as it was). We know this ε (camera) value for the reactor body is wrong, as Thomas Clarke, Bog Higgins and others have worked out. However, by iterating until the report Watts for the reactor body is derived, the value originally used for the report can be worked out..... Still not too controversial.....

Now, the idea is to obtain the original temperature seen by the Optris camera. I did this (I think, within a fairly high degree of probability, I hope) by inserting a probable ε value (like 0.9) and changing the temperature value in as many iterations as required in order to get the newly calculated Watts to match the original (report) Watts for that row. Ignore that those Watts are more than probably wrong. The idea is to match them up. The calculations are incredibly sensitive in the vicinity of the correct temperature to get the Watts to match. It often takes three decimal places to lock it in. Going over in temperature creates a huge over-watts value in the new value. I also inserted a 1.0 for the ε value, and re-iterated T for a Watts match, and the difference is not often that much in temperature from that at 0.90 ε .... This part seems like it should work to get the approximate reactor body temperature. One could re-do the power using the correct ε for hemispherical emissivity... but I haven't gone that far until I have decided that this part above is reasonable. More precise values for the normal emissivty ε would be useful (for the spectral range of the camera), but since most agree it should be around 0.9 or so, and the results are not too sensitive to it, the temperatures the camera actually saw can be fairly closely estimated.

I have obtained 1119.663°C for an ε of 0.9 for Row 12 (Table 7) and 1083.42°C with an ε of 1.0 for Row 12. (FWIW, the average of these two values for T is 1101.5°C, essentially the same as what Higgins calculated.)
For Row 1, I have 1012.657°C for an ε of 0.9
For Row 5, I have 1000.68°C for an ε of 0.9

I manually raise and lower the values to arrive at a match in as many iterations as it takes. It isn't too time consuming once the values that are too high and low low are found. Just a matter of closing in on the value that works...

Basically I am running the formula in reverse, to obtain a T value that matches the incorrectly ( report) calculated watts. This mostly a check to see if the report math is right, but also finds the epsilon used for the report watt calculations. Then I satisfy the case where they could have calculated that particular wrong value for watts, but instead I insert the new, more likely epsilon when running the same formula forwards, and match the wrong watts they decided on when they used the wrong T. When the watts are in agreement, by adjusting the temperature value up and down, the camera view temperature at the new epsilon is found.

This will return the original camera view temperature without complex IR window calculations, if the correct improved epsilon is used. The only outstanding problem is the precise value for the epsilon. However, if the estimate is close, the formula is not terribly sensitive.

(ε ∙ T^4–εamb ∙ Tamb^4)∙σ∙ Area = Watts

That makes no difference. I am undoing the error made by the professors, by reversing their math.
To see that it is the method they used, the original ε they used can be derived by reversing the formula they presented and comparing it to their plot of alumina ε. They agree.
Once you know their calculation style, you can hold other portions of their calculations as constant, and re-solve for the original condition. This is essentially how we worked out the current for the device, by reverse calculation of the Joule Heat values.
We all know that radiant Watts are wrong. Because T was wrong, because ε was wrong. I am solving for T.

I think I get what you are saying. I will consider it.

I have the Optris software, and it has an emissivity calculator.
I'll see if I can "push" the required values in, or if it needs a real reading to make a calculation.
I don't have an Optris camera, however.

So, for sake of my knowledge:
if I set ε to 1.0, with my method (above) and get 1083°C (Row 12, max T), how does your method result in a temperature (779°C) that is 304°C lower? It seems like a huge variance. I actually expected the values to be closer.

(Probably coincidentally looks like the difference is approximately the room temperature in K, which started me looking at this in more detail in the first place; wondering if the professors accidentally inserted K for °C at some point in a spreadsheet - determined to be unlikely).

What I think you are saying is that the internal formula of the Optris for dealing with radiant flux causes this 300°C difference between our methods. The MFMP tests should be informative, since they did change ε on the fly.

OK, using limited MFMP values, it is apparent my method is not working well.
There is something goofy going on, that looks like the °C to K conversion is somehow done an extra time, possibly within the camera. Or at least a value close to 273 is tangled up in the equations somehow. I can't quite put my finger on what it is. I'll see if dropping that ambient term clears it up and try a few other things.

Edit: I dropped the ambient term to test that and it is only good for a few degrees difference. Might be way more critical around room temperature.
Edit2: Using the approximate difference of an average of MFMP figures in the same temperature range, I get about 820°C for Row 12.
So that puts me into Thomas' ballpark for a maximum temperature.
Which means my electrical resistance coefficient curve vs T looks really weird... So does the 2012 test R curve, BTW, which actually has V and I measured...

@Thomas Clarke
Yes, I fiddled with an on-line calculator a bit. Perhaps that one is better.

I mostly was attempting to squeeze a few more drops of info from the report, in a way that I hadn't seen done to this point.
The method I attempted seemed feasible, at a first pass. It was not so cut and dried that I felt it necessary to recalculate the whole power, COP, etc.
Thanks for your earlier constructive comments. I had an inkling that there was more hidden complexity in producing the ε change from one to another.

Edit: this one works nicely. (Still doesn't help with the camera internals)
http://astrogeology.usgs.gov/t…rmal-radiance-calculator/

@Thomas Clarke , I have thought some more about this.
I have tested a modification of my earlier idea against the MFMP emissivity test values and have managed a close agreement (~20°C higher in my results).
I will explain my reasoning below.

Going back to my earlier idea of Watt matching...

Consider for an approximation that in the long wave IR, the peak of the Planck curve is far away, so that the power curve in the long Wave IR is approximately flat. (The idea will fail at other frequency ranges closer to the peak. It might be better to say that camera has flattened the curve to some extent instead, or in addition).
Then we consider that the Watts calculated by the formula (ε ∙ T4–εamb ∙ Tamb4)∙σ∙ Area = W (as used in the report) gives total hemispherical power.

What we want is the normal power value in order to reverse the formula for a new ε of our choosing. Since the Planck curve is effectively flat at the camera spectral range and in this temperature range (I'll use Row 12 again, peak power and temp for the report) we can factor the total power by the result of the ratio of total ε to normal ε. For Row 12, the total hemispherical power ε was calculated (by me) by reversing the report formula, and the result is an average of 0.4313 for the length of the main hot tube. The "real" normal ε is estimated at 0.95 .
So we divide 0.4313 by 0.95 to get the power adjustment factor, which is 0.454 .
Multiplying the total hemispherical power by this result (2397.97 W), we arrive at a value of 1088.664 W. This value should be approximately the normal radiant power. (First approximation estimate).

Now, using my Watt matching idea, and iterating T until the Watts fit the newly calculated normal power value estimate, I arrived at a T of 855.215°C. This will of course be an approximation, with an unknown level of uncertainty. Compared to the MFMP values I plugged in to see how well this worked, which are similar, I would say that this estimate is about 20°C high. (This is probably due to the actual change in Planck curve angle in this spectral range and the camera's response using it's internal formula).

So I conclude that the real T measured by the Optris camera, using an ε of 0.95 would be about 835°C .

Does my hemispherical to normal power conversion idea above makes sense?
Of course it is a rough approximation. But is the overall idea sound?
(Clearly doing it the right way is better, and your result is probably more accurate).

Edit: Don't be concerned with the the high normal power value obtained for now compared to the report input power. I am looking for a constant to work with, at this stage.

Anyways, I critiqued myself with more data, and the above doesn't hold up to scrutiny very well. It seems like mostly coincidence that it worked out sort of close in a few cases.

Oh well, nothing ventured, nothing gained.

On the other hand, the NASA radiance calculator works very well for an illustrative look at emissivity changes.
http://astrogeology.usgs.gov/t…rmal-radiance-calculator/

What can be easily seen, without too much complex math (you need to extract the report ε, and add 273.3 to the °C values to get Kelvin values), one can easily input the emissivity and the temperature for a known value in the Optris range (7 to 13.5 microns), and simply add a second emissivity and fiddle with the temperature for the second emissivity until the lines are aligned as closely as possible to each other, crossing just above the center of each other. Zoom the scales in to get as close to the lines as reasonable. Convert the K value into °C and there is your answer.

And lo and behold, Thomas' 780°C (1053.3 K) at an ε of 0.95 fits extremely well to 1410°C (1683.3 K) at an ε of 0.4313 . I might think that possibly 0.96 could fit slightly better... but close enough.

The MFMP values also work out perfectly (as close as the eye can judge).

Handily, if you feel like some integration math, the calculator spits out the values for you use for a fairly precise Power value (considering that the T is gained by crossing the streams "just so").

Did you do the temperature adjustments to all the parts when you did your COP estimate? Including to the convection? I'll go over your paper again after a coffee,and then dig into Bob's paper.

I'd to see what Bob works out, if he incorporates your suggestions also. Maybe I can work that out myself....

For a double neologolization in a sentence, I am looking to preponderize the evidence towards a temperature range.

Do you think it is reasonable to say that the Optris software is designed to calculate the equivalent black body response (curve section) from the spectral sensitivity range used and the detected radiant signal? In other words, the Optris does the "levelling" operation internally to fix up the bolometer sensitivity variations for the various wavelengths it uses, considers the detected powers in the sensitive range, and integrates these to arrive at a consolidated bandwidth specific epsilon that is representative for the entire detected signal.
As far as I can tell, it is the only way the camera can be effective.
Therefore, what the camera sees, (ridges, valleys, and weird selective emissivity spectral patterns, etc.) gets converted to a pure Planck black body curve section for the camera detection range so that it can use that as a basis for temperature calculation.