Some Points Regarding a Recent Presentation at ICCF20 on the ‘Lugano Report’ (Rainer Rander)

    Quote from Abd Ul-Rahman Lomax

    It is not necessary to follow the math here, but the integral of a constant times a rate is the constant times the integral of the rate, i.e., the total accumulation over the full range. The constant is the gray body emissivity.


    I think this a case in which RB0 is correctly summarizing a point Paradigmnoia made. The important detail in this case is that even though the integrated blackbody spectrum and the integrated graybody spectrum are constant multiples for the same total power (and temperature), if the temperatures of the two bodies are allowed to differ, the graybody spectrum for an object at one temperature can within a small window be made to look very similar to that of the blackbody spectrum at a different temperature. The key detail here is that the graybody spectrum will extend over a larger domain of wavelengths (thus leading to a larger integral). This is shown in the second image in Paradigmnoia's post.

    @Abd Ul-Rahman Lomax,
    I think what randombit0 meant is that a grey body of one temperature can emit a similar level of power within a spectral range as a black body of another temperature within the same spectral range. The individual wavelengths might not emit exactly the same power, wavelength to wavelength, black body to grey body, but if the temperatures are different, at one wavelength the two Planck curves can cross. There will be one point where the emitted power is the same in both curves. When the temperature increases but the emissivity does not change, the Planck curves can never cross.

    Quote from Eric Walker

    Abd Ul-Rahman Lomax wrote:


    I think this a case in which RB0 is correctly summarizing a point Paradigmnoia made. The important detail in this case is that even though the integrated blackbody spectrum and the integrated graybody spectrum are constant multiples for the same total power (and temperature), if the temperatures of the two bodies are allowed to differ, the graybody spectrum for an object at one temperature can within a small window be made to look very similar to that of the blackbody spectrum at a different temperature. The key detail here is that the graybody spectrum will extend over a larger domain of wavelengths (thus leading to a larger integral). This is shown in the second image in Paradigmnoia's post.


    The full quote:


    Quote

    Black Bodies and grey bodies could have a very similar emission in a selective spectral range even if the gray body is much more hotter than the Black Body.


    Mmmm.... if the black body has a certain emission, i.e, radiated power, within a spectral range, at a particular temperature, a gray body (emissivity less than one) would have to be at a higher temperature to have that same emission. This was true.


    I read "emission" differently, that was the problem. The spectrum actually does change, though, as shown by the plots. The slope differs.

    Quote from Paradigmnoia

    When the temperature increases but the emissivity does not change, the Planck curves can never cross.


    Just to complete a never ending discussion:



    Now we have three calculations and two experiments, that all give the same Lugano COP of 2 (+-15%):


    1) MFP experiment/calculation leads to a COP of about 2. ( Most trusted finding. )
    2) Using linear heat transport equations between caps/rods and Lugano measurement lead to a COP of 2. (reasonably trustable)
    3) Adjusting Lugano Emissivity from 0.42 (out of report table) to 0.85 (recommended) reduces the reported COP 4 by 4*(0.42/0.85) to a COP of 2! (highly trustable if no fraud is involved..)


    I hope that nobody is satisfied with this findings except the people who adhere to scientific standards...

    Quote from JedRothwell

    Please show the the MFMP experiment/calculation that the MFMP researhers themeselves believe shows a COP of two.


    Your objection is correct. Our analysis based on multiple calibrations of a physical replica with core Joule heating suggest a COP of ~1.08-1.25 in the Lugano test. Here's an excerpt from our pending paper in JCMNS vol.21:


    "Additional analyses of the MFMP replica data, as applied to the Lugano results, showed that if the single, hottest zone (8) of the MFMP replica was used as the basis for comparison, instead of the zones 5 to 9 average of Figure 4, less excess power is estimated. Hottest zone analysis estimates the lowest excess heats: File-4 excess heat of 63.3 W, and File-8 excess heat of 220.8 W."

    Quote from Paradigmnoia

    I am having trouble deciding if you are making a joke or not.


    As you might have noticed this was a wake-up call!


    If you give the optris a wrong (band) emissivity like .45 instead of .90 the temperature is off by 2 1/4 : Thus the effective measured temperature of e.g. 1570K goes down to 1320... Not to 900 as some believe...


    The measured energies wrong/correct are just the ratio of the two emissivities...

    Quote

    If you give the optris a wrong (band) emissivity like .45 instead of .90 the temperature is off by 2 1/4 : Thus the effective measured temperature of e.g. 1570K goes down to 1320... Not to 900 as some believe...


    Wyttenbach. I challenge you to prove that statement.


    I happen to know that the relationship between epsilon change and temperature is in this case is roughly 2^(1/2) not 2^(1/4).


    And I know why, both analytically (think Planck curve shape) and numerically. As have others. Would you care to take back this superficially plausible, but on examination obviously wrong, assertion? I've given you enough of a hint to rethink matters.


    If you'd bothered to read TC's paper you would not have made this mistake.


    Or do you want me to prove you both wrong and stupidly obstinate? :)


    regards, THH


    Two problems here: The first comment is apparently based on total radiated thermal power for a black body varying with the fourth power of the temperature. However, total radiated power depends on total emissivity, not band emissivity. Because band emissivity can shift with temperature, I am not happy with the simplified statement.


    Second, 1320 K is about 1050 C. Wyttenbach mixes Centigrade and Kelvin temperatures. Nobody "believes" 900 K.


    Then there is the calculation of energy from temperature. This is far more complex, and radiated power is not the only dissipation mode.


    Lugano failed in a very simple way: failure to calibrate. We may then try to recover data from the mess created, but it will never be as conclusive as calibration could have created.

    Quote from Paradigmnoia

    And an object hotter than a black body still has a mathematical equivalent to a black body at the same radiant power level which will have a temperature specific to that amount of power.


    Paradigmoia try to write in plain english ;) . What you are writing is pure non sense. I will try to rephrase it.
    "A gray body or a selective emitter must be hotter then a Black Body if it radiates the same total power." This is due Quantum Mechanics, a law of Nature.
    Does not regard any "mathematical description" but merely physics.



    Quote from Paradigmnoia

    The Optris simply measures the power in the spectral segmen


    Now to be clear. All you have written, the plots and the rest just demonstrate that without the correct epsilon, the total emissivity, the same factor that you are using to produce your plots.

    Quote from Wyttenbach

    @THH: Just consult the Optris manual. It's all written there on the famous page 9... Of course there are other constants and this is a first approximation only, but more close than way off as others...


    From that page 9, a formula for object temperature is given that has it vary with the nth root of the emissivity. Wyttenbach's claim assumed n was 4, but the guide goes on to point out the effect of the limitation of the IR thermometry to a band. It then states:


    Quote

    Since infrared thermometers do not cover the total wavelength range, the exponent n depends on the wavelength λ. At wavelengths ranging from 1 to 14 µm. n is between 17 and 2 (at long wavelengths between 2 and 3 and at short wavelengths between 15 and 17).


    Paradigmnoia stated it was more like 2, not 4, I think. I highly recommend taking deep breaths, slowing down, and noticing and setting aside "I'm right" reactivity. It can make for far better communication.

    Quote from Abd

    Two problems here: The first comment is apparently based on total radiated thermal power for a black body varying with the fourth power of the temperature. However, total radiated power depends on total emissivity, not band emissivity. Because band emissivity can shift with temperature, I am not happy with the simplified statement.Second, 1320 K is about 1050 C. Wyttenbach mixes Centigrade and Kelvin temperatures. Nobody "believes" 900 K. Then there is the calculation of energy from temperature. This is far more complex, and radiated power is not the only dissipation mode.


    If it were just that Wyttenbach would have a point (though your point about C vs K is significant - it alters the ratio a bit). But the real reason he is wrong is that while the area under the Planck curve scales as T^4 he mistakenly assumes the amplitude of the curve at a point well below the peak emissivity varies with T the same way.


    If you look at the various Planck approximations the long wavelength approximation (where the peak is at a much higher wavelength) varies as T^1 (T in K of course).


    This case is not as bad as that, but the power of T is much smaller than you would get from the Planck integral. Wyttenbach should have seen that you cannot generalise from the behavior wrt total power to the behavior wrt power in a narrow band far from the Planck curve peak.


    You can see for yourself using a Planck curve black body radiation calculator which does bands, giving it a 7-13u band, and comparing radiation at say 800C and 1400C.


    https://www.sensiac.org/extern…d_radiance_calculator.jsf


    1073K (800C) -> 0.204 W/(cm^2Sr)
    1673 (1400C) -> 0.427 W/(cm^2Sr)
    1673/1073=1.56
    0.427/0.207=2.06


    1.56^(1.6) = 2.06


    So the power law here is 1.6 << 4 (strictly, as W quoted it, we have 1/1.6 not 1/4) as Wyttenbach erroneously claimed. It is an easy mistake to make, except that TC wrote it up nicely, and its been discussed here many times. You have to be pretty full of yourself to ignore all that if you look at the technical stuff.


    There are a few adjustments to what you'd expect from this, but it is a good 1st pass approx.


    EDIT
    Wyttenbach. Please address the point above? It directly contradicts your previous post. But, following YOUR logic (which is correct) with the correct power law - 1/1.6 not 1/4 (which you can verify for yourself) shows TC et al right and you wrong.


    You should man up on this one or you'll look bad.

    Quote from RB0

    Also because I'm not alone.


    Well I'd expect you are now. Wyttenbach is as able as anyone to use a BB radiance web calculator, and unlike you (at various previous times on this thread) he admits mistakes!


    EDIT. I hope...

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