The question of the emissivity seems to keep rearing its head. One thing is certain and that is that the device looks very black within the visible spectrum at low temperatures. I assume that this suggests that it approaches a black body within that range, but I suppose that this may not be the case in the IR region.
Is there evidence that the emissivity changes with temperature? I have not heard of this behavior before, but some paints might have a problem. Has anyone found a reference to the actual paint used by Rossi for this test to determine how it functions as a emitter? Is it possible to scan the surface of the ECAT with some instrument to actually measure the emissivity just prior to the next test if is is to be tested in the same manner? What can the future testers do to enhance their ability to get accurate results? Joshua, if you were going to be a member of the test gang, what would you do to keep the skeptics at bay? Put yourself in the tester's shoes for a moment instead of casting stones. How much confidence should be placed in the white emission dots? Apparently they correlate well with a thermal probe place upon the test unit. What is the chance that Rossi would allow the test scientists an opportunity to spray some of their own paint upon a portion of the device for comparisons? This might only require paint over a small area. Does anyone offer additional suggestions to improve the acceptance of the future test data? Dave -----Original Message----- From: Joshua Cude <[email protected]> To: vortex-l <[email protected]> Sent: Tue, May 28, 2013 12:00 am Subject: Re: [Vo]:Ekstrom critique of Levi et al. On Mon, May 27, 2013 at 7:41 PM, Jed Rothwell <[email protected]> wrote: Eric Walker <[email protected]> wrote: Joshua's position is that in the present measurements, the emissivity is implicitly taken into account twice when using an IR camera, and that in assuming that a high epsilon is conservative (in the first calculation), people are neglecting to see what effect it has on the calculated power in the second calculation. For the fifth time, the authors addressed this! It is shown right there in Fig. 7. The camera software computes higher temperatures. The higher the temperature, the higher the power (all else being equal, which of course it is, since we are only changing one parameter). It's not all else equal. You're simply not paying attention. The emissivity is lower. So the higher temperature contributes to higher power, but the lower emissivity to lower power. The two effects work in opposite directions. Which one is bigger depends on the particular wavelength and temperature. It could not be shown more clearly! With this camera, when you lower the emissivity parameter, the computed temperature goes up. Right. No one disagrees with that. Cude asserts that if they lowered it all the way to 0.2 the temperature might be computed lower. No. You simply don't understand what I assert. You're not thinking about it carefully enough. Get Fletcher to help. I'm saying that while the temperature goes up, the effect on the power *including the lower emissivity* may be computed lower, depending on the effective exponent used to compute the temperature. Here's a simplified version of the math, ignoring ambient temperature and the temperature of the camera, since the point at issue is independent of those: The power measured by the camera is assumed to be given by P = C*e*T^n, where C is a constant, P is measured power within a range of wavelengths and n is the effective exponent determined by this range of wavelengths (which presumably depends on temperature). If the frequency range is the entire spectrum n = 4, as in the S-B equation. Solving for T gives T = (P/Ce)^(1/n) Now when you calculate power, you use Pcalc = C*e*T^4. You can see that if n = 4, Pcalc = P for any value of e. But if n is not equal to 4 as is the case in reality, to correct for the finite wavelength range, then Pcalc can differ from P, as it does in the 2 examples used in the paper. So it depends on what value of n gets used, and it may be very different when e=0.2. It should be possible to figure this out from the Planck law, but there is no mention of this in the paper, and no test to see what temperature gets computed for a lower emissivity. In any case, the correction only applies to grey bodies, where the emissivity is constant and there is no telling what the temperature means if its not. I don't know (and frankly doubt) that this is the problem, but all I'm saying is that it's not as simple as you or the authors have argued, and in the case of the authors, that's sloppy. And it's frustrating to see people trying to argue that it is blatantly obvious that e = 1 is conservative in all circumstances. It's not at all. It's quite subtle. And the failure to understand this is symptomatic of shoddy work. I am sure this is nonsense, but even if it were true it is irrelevant. There is not a shred of evidence the actual emissivity of this reactor is anything close to 0.2. It is 0.7 to 0.9. It makes no sense to talk about 0.2 anything. There is unfortunately no evidence at all what the emissivity is in the December run. I would be interested in a second opinion from someone familiar with IR cameras. In Fig. 7, the IR camera itself tells you the answer! That is the most authoritative answer you can get. Not for lower emissivities and not for non-grey bodies.
