Experimental Evidence on Rossi Devices

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There were plenty of other elements in the experiment.

In order to save the idea that Rossi's experiment might be real, you are proposing reactions between contaminants from the ash holder (Si etc) and the Ni? I guess you could propose a reaction with the Al2O3 reactor housing. such a flight of fancy seems silly, and not enough to argue that my comment, contingent as it was, is ill-judged.

Eric,

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These photons are generally of an energy that will readily be stopped by the walls of the reactor.

You are I think in "react not think" mode here.

If stopped by the reactor walls then the energy does not escape, and the argument remains.

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It's not irrefutable, and I've refuted it. I acknowledge the error about assuming you lacked an understanding of balancing nuclear equations. But that does not negate most of the rest of what I've said. I believe your analysis to have been flawed, and I believe I've shown why it's flawed.

In this case I think it is "react not read". When I said irrefutable, and used such strong language, it was about the excess heat correction I made - note the context from which that must be clear. Whatever other uncertainties, the excess heat correction was the main (not second order) quantitative issue and my calculation is correct, theirs is wrong. Therefore my result is more correct than theirs. You have never refuted this. However you have stated in unequivocable terms that this is not true. Your repeated statement is false.

To be fair, I'm not confident about my error margins. They are "guestimates". I have not spent a lot of time trying to be precise about how large are the systematic errors here. If I were to do that the first thing I would do is get a better handle on the translucency error - which is the largest potential error I can see. There are two ways to do that: estimate it from the COP acceleration figures, or estimate it from known opacity of al2o3 at different frequencies. Neither looks very promising.

Sorry Colwyn,

As you will realise I got distracted by comments from Eric.

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...What I'm interested in answering is: If you set the alumina temperature to read, say, 1525C when Optris emissivity is set at 0.4, what does your code estimate the true temperature to be, assuming the correct emissivity setting is 0.95?

You'd need to use the code in the paper, or a web Planck calculator (easier). BTW the code should run fine on Python 2.7 but maybe it needs a bit of scipy or something - I can't remember to tell the truth - it should say in comments at the top, or look at the import statements.

Use a plank radiation calculator that gives you band radiation output, set the band to 7um - 13um (exact limits not so important).

Call the BB radiance you get from this for a BB of temp T: L(T). (You are measuring T in C here).
So L is the function you can calculate from T using the BB calculator. The units of L don't matter, as we will see. BTW my code does this by numerical integration over the band of the Planck curve.

Because we input 0.4 into the camera it will assume the real radiance is 2.5X larger than that measured, which will be 0.95L(Tx) where Tx is the temperature you had to turn the tube to. It is saying it thinks the real temperature is 1525C, so we have:
(0.95/0.4)*L(Tx) = L(1525)

You can see why the units for L don't matter though it must be proportional to optical power (rather than, say, the square root of this). Radiance does fine.

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Also, what would the temperature appear to be, according to your code, if the Optris was instead set to 0.7, in the situation described above?

As before but (0.95/0.7) factor.

If you don't want to use a BB calculator for L you can get an estimate - around these temperatures the band radiance varies as roughly T^2 or in your case (T+273)^2. (this is approximate, from memory, maybe 1.9 is more exact!).
So:
1800/(Tx+273) = 1.54 or 1.16 =>
T = 895 or T=1280 (all numbers approximate because I have not bothered to do exact band radiance calc).

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Does your model estimate radiative and convective transfer separately?

Yes but I cheat. There are actually many different parts of the reactor at different temps. I ignore this and use the scaling for convection from their equation, and assume the reactor body is the main radiative component. Obviously the lower temp parts have radiative output scaled differently. I think I actually separate body and rods radiation, with a separate average temperature for each. Not very precise but these really are second order corrections. luckily the convective components are small at the active test temperatures, and also the reactor body far predominates over anything else.

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What were the values you used to determine your COP figure? And at what alumina temperature?

I used the temperatures given in the report (average reactor body). You see, the report gives calculated temperatures, and all I have to do is work out the scaling factor and apply that to those.

Having got the corrected temperatures I then generate corrected radiant output powers from report radiant out powers by multiplying:
Pcorr = (epsiloncorr/epsilonrep)*Prep *(Tcorr/Trep)^4
Epsilonrep is their book value for Trep. Epsiloncorr is the same thing for Tcorr. The book values are as good as any, and variation in epsilon from this has only a linear affect on power out and hence COP.

And finally, are your integration calculations based on a derived-from-equation Rayleigh Jeans (or Plank) curve, or some empirical results from an alumina sample?

The band emisivity comes from Bob Higgins spectral emissivity values, which are empirical and his best estimate. But since at these frequencies epsilon is always pretty near to 1 I don't expect a big difference. The only other value relevant is epsiloncorr and that is (total radiance) from the Lugano book values. At the lower corrected temp this is higher and the book graph (itself empirical) seems OK. Of course there will be variation with different alumina microcrystalline structure etc.

If you wanted error bounds I'd say that epsiloncorr is the main issue, affected both by alumina properties and by the fact that the tube is translucent at higher frequencies so for part of the power the real emissivity and size are different.

PS - you could read the paper where this is all discussed, and the code for the details of the calculation.