QuoteUnfortunately, they reference their own graph, get the emissivity that matches the Optris temperature, and feedback the new emissivity value into the Optris or equation, for another iteration.
There is nothing wrong with the method - just the graph they used which was total emissivity and not band emissivity. Had they used band emissivity the variation with temperature would have been so small as to make regression hardly necessary.
@Thomas Clarke
The method was garbage.
It is a self-regulating feedback loop with one variable, which is detected radiance.
Try my experiment with invented T vs epsilon.
Make a fairly smooth curve, or you will be up all night iterating. A flat 0.5 is probably simpler.
Well, it obviously will not converge in some cases if the emissivity vs temperature curve is not smooth. There are, in principle, multiple solutions.
If you have an accurate curve for band emissivity vs temperature of surface BE(T), and an Optris camera, clearly the possible temperatures are those which solve:
e0 = BE(T0)
Optris reads T0 when given e0 as emissivity and pointed at the surface.
The regression will normally converge to such a solution, and with normal smooth emissivity curves there is only one such solution. The Lugano graph had a glitch in it that could possibly lead to regression instability but not in the region of interest. Of course that graph was total emissivity and not band emissivity (which is more or less flat at 0.9).
In case someone actually used random values, not merely invented values.... The epsilon value can move around quite a bit, but in gradual steps, and no weirdness will happen.
Trace a smooth line over a scaled plot of the average monthly rainfall of Switzerland onto a emissivity vs temperature plot. Then you can prove that the alumina normal emissivity versus temperature is directly proportional to the seasonal rainfall variation of Switzerland....
The glitch is not the Lugano plot specifically, but the self-satisfying feedback loop. They collect the predetermined epsilon value for the temperature from the plot, and keep feeding that in until the camera agrees with the plot.
This does not mean that they can show that the correct emissivity is being used.
This does mean that they can make the camera use any epsilon they put in, and make it look justified. When the justification is a farce.
Garbage in, garbage out.
How did they do the regression? Input emissivity to Optris instrument - measure temperature - calculate emissivity from curve - iterate by inputting new emisivity to instrument.
Since the temperature is variable zone by zone, the emissivity should be updated zone by zone...
Best way would be to contact the manufacture and use their own model.
Maybe the provided software can do that from recorded images?
There should be more data on that process.
We can do that with Planck curve, but it should be documented, at least in an appendix.
It can suffer artifact, especially about the sensitivity curve (even if my computation show window shape is not so important).
If it was done by software, it should be documented.
Zone by zone changes can be made by the Optris. See the old MFMP video where they changed the emissivity of only one of several zones.
The Optris has been factory optimized to level the unit specific IR reading bandwidth and response of the microbolometer array to an ideal greybody equivalent over a very large number of conditions. This means that standard equations will work very similarly (if not identically) to the Optris for temperature output, if the emissivity values and radiance values are the same.
A startling result, based on one fairly sloppy test with one of the IR heater tubes from the old toaster oven.
68% of the total radiant power apparently passes transparently through the quartz glass without leaving a heat signature in the glass at 516°C.
Total radiant power was calculated by subtracting the convective heat loss watts from the total electrical input watts.The epsilon required to make up the shortfall in the calculated total hemispherical radiant power, based on the correct temperature is 1.77 (LOL)
(thermocouple and IR probe in agreement; epsilon of 0.75 required to match IR T to TC at 516°C)
Edit: messed that crossed through part up. That was to make total power but forgetting convection.
The ε for making enough radiant power by total hemispherical emissivity is 0.968 at 516°C, but ε "should" be about 0.31
In other words, if I stick the tube into a blackbody sleeve at the same input power, it would be way hotter.
QuoteIn other words, if I stick the tube into a blackbody sleeve at the same input power, it would be way hotter.
Interestingly, that isn't what the math seems to show about this assumption.
Emitting as a blackbody, it seems to get cooler, based on preliminary calculations. Selective emitters seem to be trickier than I even suspected.
This needs to be tested.
Edit: 5 hours later...
Painted the same IR tube with BBQ black. Let dry a bit less than recommended. Got normal emissivity up to 0.86
T goes up to 559.2 C at same power. So up, but not lots, and not colder.
Have a better handle on hot R also, so will re-do earlier W and check.
Edit 2: I caught a math error, and the previous "68 % transparent heat loss" rate disappears, more or less.
Makes sense: how is a perfect absorber-emitter any less transparent than a transparent body? This kept me awake wondering last night. The only difference is that the emission spectrum is evened out.
However, the blackened version now convects about 6 W more heat, and is (because it is) definitely hotter, and the radiant portion is actually reduced by about 6W (according to the math). This somewhat suggests that the blackness is accumulating more temperature than a transparent body. A sort of delay before emitting, but becomes the equilibrium. Note that the temperature is 43°C higher, but total input power is exactly the same.
Measuring temperature only, it could look like a COP increase, about 10%, just from painting the tube black. It is actually radiating about 8% less heat.
And it isn't even fully black yet.
Calculating the required total hemispherical ε to make the power balance, the lower temperature selective emitter was an ε of 0.3225, and the blackened hotter version was 0.2394 at the same input power.
