Don't worry Abd, I'm am going somewhere with all this.
Edit: to speed things along while I eat, this is a good time to look at how a microbolometer functions.
Some Points Regarding a Recent Presentation at ICCF20 on the ‘Lugano Report’ (Rainer Rander)
-
-
And, along we go...
The microbolometer gives a varying electrical signal based on infrared power actually received by the little bolometer. This is attenuated and tuned by lenses and various doping agents, and is material dependent.
Long wave IR cameras use detector materials that strongly absorb IR energy in the 7 to 14 μm IR band, while using lenses that are very IR transmissive in this same range. The detectors are made into an array, so that many detectors can be used to generate an image by acting as one pixel each. Software and calibration adjust the electrical signal received (and unequal spectral band sensitivity to some degree) to generate a flat response to incoming IR energy, as best as possible, within the camera IR operating range. This levels the response of the camera to black body radiation received within the operating range of the device. High temperature IR cameras may concentrate on more careful conditioning of the signal generated by the shorter wavelengths of the camera spectral sensitivity band, due to the IR radiant energy peak moving towards the shorter wavelengths at high temperature (Wien's Displacement Law). An IR spot thermometer works on the same principle as the IR camera, but the spot thermometer is effectively only one pixel.
A black body by definition emits the maximum amount of power over the entire IR band at any given temperature. A grey body by definition emits less power than a black body across the entire IR band at any given temperature than a black body, but by the same factor in all wavelengths. A selective emitter, sometimes known as a real body, has an unequal distribution of IR energy emittance across the entire IR band, and so radiant power relative to the maximum possible in each wavelength is variable.
A grey body cannot emit the same radiant IR power that a black body can at the same temperature. This means that an IR camera will receive less energy in the spectral sensitivity range of its detectors, and report a lower temperature. A grey body will be hotter than a black body at the same level of radiant power. This is because it is not as effective at getting rid of heat by radiation as a black body at the same level of power. The IR camera measures temperature by receiving the heat radiated from the object. If the object does not radiate heat away well, it will not heat the camera receiver bolometers well, and the camera will not report as high a temperature for the grey body as a black body at the same temperature would.
A selective emitter may be hotter than either a grey body or a black body object of the same geometry and total radiant power. But the IR camera only needs to know how hot the bolometer can be made to be to determine temperature. And that depends only on the radiant power of the object within the effective bolometer detection range. Not the entire IR band, and not the total object radiant power.
The band(s) where the maximum emittance occurs is of much importance to the detector of an IR camera. The IR camera is measuring the radiant power within a restricted range of the IR band, and generating an electrical signal based on this input energy. If the object radiant power measured is low, (even if the temperature is very high) due to low emittance of the object in the camera IR detection range (spectral sensitivity), the camera will convert the low signal to a low temperature result. If the object radiant power measured is high, due to high emittance of the object in the camera IR detection range, the camera will convert the high signal to a high temperature result. An IR camera is considered to be ideally matched to the object being measured when both the camera IR detection range and the object peak IR emission ranges are as closely matched as possible. The more closely matched the object emittance and detector IR absorption ranges are, the less post-processing of the signal generated by the bolometer array is required to give the correct temperature based on received radiant power.
What is being matched is the effective emissivity of the microbolmeter to the emissivity of the object. By design the calibrated effective emissivity of the microbolometer array is 1.0 in the detector spectral sensitivity range. (A perfect emitter is also a perfect absorber - Kirchhoff's law of thermal radiation). For the Optris PI160 camera, this is 7.5 to 13 μm band of IR radiation.
-
To summarize my above comment:
The radiant power of an object being viewed by the Optris IR camera is important. But only the radiant power within the spectral sensitivity range of the camera, which is the only band of IR in which the camera is capable of determining the radiant power of an object.The camera can only compare this measured radiant power to the power expected of a black body object at that level of measured power, when viewed through a spectral window in the IR band, which is specific to the camera technology.
From this spectrally limited measurement, the camera reports a specific temperature that corresponds with the same level of radiant power of a black body as viewed through the same IR window.
The specific temperature corresponding to a black body that the camera reports can be modified (made higher only) by changing the camera user emissivity function, which tells the camera that it is not measuring as much radiated power as a black body would radiate in the spectrally sensitive range of the camera detection technology, and therefore the object is hotter than it appears. (Hotter than a black body would be at the same measured detector irradiance).
The IR camera only needs to detect power in the small segment of the entire IR band that the camera is capable of measuring, and compare that segment of radiant power to that which a black body produces in the same small segment of the entire IR band, in order to calculate a temperature.
I will expand upon this further in another chapter.
-
Paradigmnoia and I are obviously doing an psychic avatar swap. He is abs right but this time I'll do the short and pithy answer to BA and RB0.
Quote from BAFor being precise the camera is nit sensitive to temperature but to radiated energy. For making a measurement you must know the ratio of the total emitted power from the body surface and that that would be emitted by a Black Body. So all your reasoning means that total emissivity is an important factor to know during a measure. Total emissivity and not band emissivity
For such precision, you need radiated power, far different from radiated energy (which the camera is not sensitive to). But I know what you mean.
I've highlighted the word total used twice in the above which is total bunk. You don't need a physics degree to know that camera sensors are sensitive to a limited range of frequency (IR band 8-13u for the Optris). So how could total power (summed over all frequencies) be relevant? Similarly how could total emissivity be relevant?
I like serious techie arguments, and am happy to treat seriously anyone whatever their initial level of knowledge. We all learn stuff. But you and RB0 show on this topic an unusual and troll-like lack of ability to learn.
As polemic and rhetoric, your repetition probably works. As science it is a total (to reuse the word) fail.
Quotebecause, as you demonstrated, that for many bodies the energy emitted in a certain band could be identical.
I'm not quite sure how to parse this, but I think you are aluding to the fact that many surfaces are a good approximation to grey bodies when viewed with typical sensors. For grey bodies no care is needed and your cavalier statements are correct.
Paradigmnoia has posted many times the cliff-like graph proving alumina is very far from a grey body.
-
Hello again everybody,
Important for what?
Dear fellow, what game you are playing ? I will repeat some important points we agreed on.
Starting from first principles without considering any camera or detector we concluded that a Black Body Dog Bone would be much cooler then an Alumina Dog Bone, because the second has a lower total emissivity.Paradigmoia has also demonstrated that 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. So to calculate the temperature from the integrated energy the total emissivity of the material is necessary if that spectral range the same of the detector.
within the camera spectral sensitivity range only.
No Paradigmoia, we can discuss this privately, but because of the reasoning we have done is interdependent from any detector (so valid for all detectors) you can treat gray and selective emitters in the same way.
What is being matched is the effective emissivity of the microbolmeter to the emissivity of the object
This is a good point. You here demonstrated that Alumina is in fact one of the best materials to be used with the camera because have a good emissivity in rhe camera sensitivity range.
Because of the first point you stressed (with gray and BB) total emissivity is necessary to calculate the temperature.compare that segment of radiant power to that which a black body produces in the same small segment of the entire IR band, in order to calculate a temperature.
No Paradigmoia. the comparison is made on the total integrated spectrum. Otherwise even a simple measure for a gray body would impossible as you have shown.
So how could total power (summed over all frequencies) be relevant? Similarly how could total emissivity be relevant?
Total power is relevant as well as total emissivity for the reasoning explained.
Remenber that we started supposing a normal resistor in vacuum in equilibrium condition so that the radiated power must be equal to the input power- -
This is a good point. You here demonstrated that Alumina is in fact one of the best materials to be used with the camera because have a good emissivity in rhe camera sensitivity range.
Within 7.5-13 µm Aluminum is radiating very close to a blackbody, thus the Optris camera needs only a slight correction (0.85 - 0.90) of the emissivity to get the right temperature.
If we would measure between 4-7 µm there Aluminum would be a bad emitter and we must use a lower emissivity to get the right maximum temperature !Here the Optris extended instruction manual:
One thing Rb0 doesn't grasp because she obviously doesn't read manuals:
1) To get the right T measurement at 7.5-13 µm band you need to give the Optris the high (0.85 - 0.90) emissivity. (see formula on Page 9)
2) Then at the point where you calculate the Energy output of the reactor at the measured T you have to use the low total emissivity!!!!!!
Anyway, as said a million times, .. Optris does not recommend to use the 7.5-13 µm band for high temperature measurements.. except if you make an additional calibration!
-
Quote from RB0
1. Paradigmoia has also demonstrated that 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.
2. So to calculate the temperature from the integrated energy the total emissivity of the material is necessary if that spectral range the same of the detector.I've split your comment up into separate points so we can see it for what it is. Point 1. is tautologous. Since for a grey body of emissivity e radiation in any range is defined to be e*Prad where Prad is the corresponding BB radiation. You have just defined a grey body. Al203 is not a grey body.
Point 2. does not in any way follow from point 1. It is a complete non sequitur, unless the emitter is a grey body. In that case you can correctly infer that emissivity over the detector range is the same as total emissivity.
Quote from RB0No Paradigmoia, we can discuss this privately, but because of the reasoning we have done is interdependent from any detector (so valid for all detectors) you can treat gray and selective emitters in the same way.
I'd be interested for you to provide your reasoning. As you can see above what you state here is not reasoning. If, however, you feel unable to do this, perhaps a private discussion with Paradigmnoia will help you.
-
@randombit0,
You are getting close to the true solution.What you need now to consider is, when one looks at a black body, the temperature is always the same for a black body with the same total power, every time it is checked. The Planck curve will always be identical for the same black body temperature. This means that every point on the Planck curve will always have the same value for same black body temperature. The power for each wavelength will always be the same for the same temperature black body.
If each point on a black body Planck curve is always the same at the same temperature, then for any given temperature, there will always be only one value, a relative constant, that can be compared to. This one constant value can be used to determine whether measured power at any given wavelength is equal to, or less than that of a black body. The relationship between a blackbody temperature, the power of a blackbody at any given wavelength, and recieved power at any given wavelength, is sufficient to determine a temperature of an object by measuring radiated power in any given wavelength. All that needs to be known to compute a final temperature is if the emissivity in that wavelength is less than or equal to a blackbody, and by how much if it less than a black body.
No other wavelengths need to provide any other information.
Technology limits mean that single wavelengths are generally not used. So groups of wavelengths are used. Only the wavelengths in the group are used, as a slightly more complicated case of that described above. -
2) Then at the point where you calculate the Energy output of the reactor at the measured T you have to use the low total emissivity!!!!!!
Oh Dear are you crazy ? When calculating temperature (from energy) you use one vale of emissivity and when calculating the energy back you use another ?
You would have no energy conservation in this way !
Of course we know very well the manual of Optris and even page 9 on which there are many formulas, not only one, all of them only with one value of emissivity !I'd be interested for you to provide your reasoning.
Oh I see you have lost the short term memory...... I will repeat it for you even if I posted just few days ago.
Consider a normal resistor with at a fixed power in equilibrium condition in vacuum.
Then the input power and radiated power must be the same.
Consider first a Black Body.
If you integrate the Plank curve then you have the power density radiated from the surface of the resistor IBB , and multiplying than for the body area A you obtain the input power IBB*A .
Because Quantum Mechanichs impose that no material can have an emissivity higher then that of a Black Body if you have a gray or selective emitter body with identical geometry then to maintain Ib*A=IBB*A constant the body must be hotter then the Black Body in order to have Ib=IBB.
This simple considerations are independent from any detector. The ratio Ib/IBB is called total emissivity. -
Temperature isn't power or energy.
This is the Achilles heel of a large number CF-LENR experiments and conclusions. -
Consider a normal resistor with at a fixed power in equilibrium condition in vacuum.
Then the input power and radiated power must be the same.
Consider first a Black Body.
If you integrate the Plank curve then you have the power density radiated from the surface of the resistor IBB , and multiplying than for the body area A you obtain the input power IBB*A .
Because Quantum Mechanichs impose that no material can have an emissivity higher then that of a Black Body if you have a gray or selective emitter body with identical geometry then to maintain Ib*A=IBB*A constant the body must be hotter then the Black Body in order to have Ib=IBB.
This simple considerations are independent from any detector. The ratio Ib/IBB is called total emissivity.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.
And any level of black body radiant power is known not only for the total power, but also for the power radiated at each and every wavelength, which is specific to the one temperature belonging to a black body at that level of power.Which means that radiant power from any wavelength from the Planck curve can be measured from a hot object and normalized to the black body temperature equivalent power for that wavelength.
This means that there is a specific temperature for a black body at that level of power in that wavelength.
Only difference between the emissivity at that wavelength and a black body (1.0) at that same wavelength is needed to determine the true temperature, if the radiated heat energy at that wavelength has been normalized to the black body equivalent for which there is a known specific temperature. -
Quote from RB0
Oh I see you have lost the short term memory...... I will repeat it for you even if I posted just few days ago.
I try to answer what appear to be substantive comments when they are made. I pointed out exactly how that was wrong, at some length, soon after you posted it. You did not dispute my comments. Follow the link in case you did not read this first time.
So we need some correct argument now, not a repetition of what has been refuted.
-
Which brings us back to these images (below)...
Each of the segments has the same amount of total radiant power in that segment. The total power belonging to each spectral segment color is much different for each of them, when the entire IR band is considered (second image), as is the temperature for each. (Blue is ε = 1 in both images, Yellow is ε = 0.5 in both images, and Red is ε =0.1 in both images)
The Optris simply measures the power in the spectral segment, which is all of the IR range it can detect, and then the camera determines what the black body (emissivity 1.0) equivalent is.
The images show the spectral power between 7.5 and 13 μm for three different temperatures being all the same, but the spectral segment power at any level can be converted to a black body equivalent, from which the temperature of the black body equivalent can be determined, and then from that the temperature change due to emissivity. What happens is that the camera sees the 0.1 emissivity object, which has a very high temperature (7190 K here), then determines the black body temperature (1273 K). The camera only needs to know the true emissivity of the object in the spectral segment (0.1 here) and the correct temperature simply "falls out" of the equations. The full power curve is outside of the camera detection range, but is generated in entirety by the math knowing only the segment characteristics. A selective emitter will be converted to grey body equivalent also, by only measuring the spectral segment the camera can see, converting that power level into the black body equivalent, then generating the grey body equivalent if the emissivity is not a black body (ε = 1.0) in that segment.
If the emissivity is actually lower in that segment (0.5, 0.1 in the images) then a higher temperature can be attributed to the segment. The total power would be greater, but that does not change what the camera sees in its small view of the larger IR broadband spectrum. The spectral power in that segment does not change.
These methods are only useful for determining the object temperature, not the object total power, because these methods only deal with the radiant power and black or grey body equivalents based only on the wavelengths occurring within the spectral segment measured by the camera.
The broadband IR extrapolations (image 2) are for illustrative purposes only, and are only applicable to true grey body objects. Selective emitters, such as alumina, may deviate considerably from the extrapolations based on spectral measurements outside of the spectral range measured by the camera, greatly altering the total power output compared to a grey or black body. This does not affect calculations and equivalents of radiated power within the spectral sensitivity range of the camera. It simply means that the total power of selective emitter bodies cannot be extrapolated from spectral power, even though the correct temperature can be calculated. This also means that the spectral power cannot be extrapolated from the total power of a selective emitter, nor can the total emissivity of a selective emitter be used to determine spectral emissivity (and therefore temperature), because a selective emitter by definition does not have the same emissivity at all wavelengths.
The difference is that the IR camera actually measures the power emitted in a spectral segment. The result of that measurement can be mathematically manipulated to an equivalent power level within that segment to derive a known temperature, belonging to that of a black body. The grey body equivalent is a modification of the black body equivalent. The only extrapolations that can effectively be made outside of the measurement range are those that have the same emissive power function for each wavelength outside of that range, which are by definition either a grey body or a black body. One cannot expect to effectively extrapolate over the entire IR band a material-specific, wavelength specific, highly variable emissivity function based on a small segment of measured emissive power.
-
Abd Ul-Rahman Lomax wrote:
Dear fellow, what game you are playing ?
Truth or Consequences. And you?My question was a request to clarify what was affected by what RB0 asserted. He left the context out, making the question completely meaningless so that he could then question it as, perhaps, silly or an attempt to divert attention. His comment was:
QuoteThat's why the total emissivity parameter is so important !
This was in reply to Paradigmnoia, itself quoted in a way that made the meaning utterly obscure, one would have to read back. If anyone wants to, here is RB0's post, which links to Paradigmnoia's comment.
Some Points Regarding a Recent Presentation at ICCF20 on the ‘Lugano Report’ (Rainer Rander)Context had been completely lost, hence the question, which RB0 does not answer.
I will therefore answer. Total emissivity is the fraction of total thermal power emission of a body compared to a black body at the same temperature. As a black body is the most efficient radiator of thermal energy, total emissivity is always a value between zero and one. In order to know how much power is radiated by, say, alumina at a certain temperature, one must know the total emissivity of the alumina at that temperature. (Emissivity also varies with temperature, generally.) So if we know the temperature of the alumina, and we know the total emissivity at that temperature, then total emissivity is important to determine total radiated power. It will be needed to calculate emitted power from temperature.
Total emissivity is not relevant to determining the temperature of the body from the readings of an IR camera with a certain band sensitivity. Rather what is needed for this determination is band emissivity, and since the sensitivity of the camera is rather complex, this is done with calibrations, either by the manufacturer or by one using it, in the latter case by comparing IR camera readings with temperature measured by other means. Proper calibrations would have resolved the Lugano problem. MFMP has, later, simulated this.
Generally, RB0 is confusing two distinct issues, the determination of temperature and the determination of power emission. This is not unclear or marginal or some ordinary disagreement. The facts of this are so clear that if RB0 is knowledgeable, then RB0 is likely deliberately deceptive, spreading a form of FUD.
QuoteI will repeat some important points we agreed on.
Possibly we agreed. RB0 is often saying things that are true, but also often radically misapplied.QuoteStarting from first principles without considering any camera or detector we concluded that a Black Body Dog Bone would be much cooler then an Alumina Dog Bone, because the second has a lower total emissivity.
The meaning of this, standing alone, is utterly unclear. What is the context, on what information is the conclusion about "cooler" based? Given the same temperature, a Black Body will emit more power than something else. RB0 must mean something else. Given the same radiant power, at equilibrium, the Black body will be cooler. How, then, is this applied? Why is this statement of importance?Does it have anything to do with the determination of temperature? No. It has to do with something that comes later, and that is much more complex, in reality, given that emissivity varies with temperature (both total and band emissivity. They are apparently not co-variant, as well.). As well, there is cooling by convection, another complication. Total radiant power varies as the fourth power of the absolute temperature, generally, but shifts in emissivity can lead to variation in this.
(Using rough figures, 800 C is 1073 absolute, and 1400 C is 1673 absolute, so the ratio would be about 6. I.e. if emissivity variation doesn't mess this up -- it could -- the error in calculating emitted power could be a factor of 6. The emissivity used to determine total radiated power from temperature would be total emissivity, which is well-known for alumina as a function of temperature. Levi's comment on this (about using emissivity of 1.0, betrayed that he still, years later, doesn't understand the issue. If band emissivity were 1, the temperature is something like 800 C. (I am not certain of this figure, it's approximate.) But then to calculate radiated power, total emissivity would be used, something like i.e., 40%. Not 1.0, which would here be band emissivity only. 1.0 would only give relative power radiated in that band.)
QuoteParadigmoia has also demonstrated that 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.
I knew that Paradigmnoia was setting up a problem by what he wrote. The restatement here is true, but massively confusing.In any given spectral range, the black body emission is always the maximum. A gray body is defined as a body that has the same emissivity in all bands, so with a gray body, so if we assume constant power dissipation -- again RB0 omits the condition -- the so-called gray body, with "very similar emission in a selective spectral range," or band, is "very similar' to a black body and would be close to the same temperature, not "much more hotter," even with the poor grammar. So, no, "Paradigmnoia" did not "demonstrate" that, because it is an oxymoron.
QuoteSo to calculate the temperature from the integrated energy the total emissivity of the material is necessary if that spectral range the same of the detector.
This is directly contradictory to instructions for using the camera, and is directly contradictory to common sense. The camera's microbolometer will measure thermal emission in the band which it absorbs, ignoring energy in all other bands. To relate that absorbed heat to the object's temperature, total emissivity is irrelevant, what is relevant is emissivity in the band. It's obvious: the camera cannot see anything else.The Lugano researchers made a mistake, which easily happens when one is not familiar with a process or technique and simply does what seems reasonable, based on incomplete understanding. RB0, "Rainer Rander," and the other fellow, are now defending that mistake with increasingly preposterous arguments. I'm not seeing a sign that anyone is fooled other than knee-jerk Rossi promoters, and not even many of them. RB0 and Rander, etc., might actually be Rossi, or might be someone close to Rossi, and it doesn't matter. Even if Rossi has the greatest heat engine since ever, the facts about how the camera work aren't going to change. An error was made, and it's completely obvious.
-
And just in case someone likes to test these ideas out, here is the website for the USGS-NASA radiance calculator,
and below are data from the MFMP Lugano Thermal Verification video plotted up on the calculator. (I'm not sure if the first one is supposed to have an ε of 1.0, but you can try 0.95 if you like). To make sure I haven't messed the numbers up, or I am not pulling the wool over your eyes, I suggest watching the MFMP video.(somewhere around 1:57 is a good spot, if you are in a hurry. 2:03 is good)External Content www.youtube.comContent embedded from external sources will not be displayed without your consent.Through the activation of external content, you agree that personal data may be transferred to third party platforms. We have provided more information on this in our privacy policy. -
Wyttenbach wrote:
Oh Dear are you crazy ? When calculating temperature (from energy) you use one vale of emissivity and when calculating the energy back you use another ?
You would have no energy conservation in this way !
Actually, I some time back I blocked Wyttenbach because of the low density of cogency, but his explanation here was right on, and to be able to upvote I had to unblock and I'm leaving him that way for now. (The software changed, I think, I used to be able to upvote and downvote blocked posts, though normally I ignored them.)The RB0 reaction is amusing, showing the major brain fault involved, if not deliberate deception. First of all, temperature is not being calculated from energy, using the camera, except in one way: the camera measures thermal energy that heats the microbolometer, which only absorbs in a relatively narrow band in the infra-red. From that energy, with the emissivity properly set for band emissivity of the material -- the percentage of black body radiation that it emits in that band -- the temperature can be calculated, and the camera will do this. However, the body itself will radiate in all bands, and that is what total emissivity expresses. All-band-radiation.
There is absolutely no violation of conservation of energy. The measurement of temperature only considers radiation in a narrow band. Total radiated power uses all bands. As the material has high emissivity in the IR band used, it follows, from the discussion, that it has lower emissivity elsewhere. Elsewhere in the discussion these values can be found, or in the Clarke paper or elsewhere.
The Lugano paper is at http://www.elforsk.se/Global/O…er/LuganoReportSubmit.pdf
The report discusses emissivity on page 8. Then Figure 6 on page 9 shows emissivity of alumina vs. temperature. Nowhere in the discussion on page 8 nor on page 9 is band emissivity mentioned. These are all about total emissivity. Plot 1, page 9, shows the same data as Figure 6. (The paper has many signs of poor editing, this is a small one.) You can see that at 1400 C, total emissivity is about 0.4. At 800 C., it's about 0.5.They do not actually state what emissivity values they used for the experimental reactor, but with the analysis of the dummy reactor, at much lower temperature, it's clear that they used total emissivity, which was an error. It may be, however, that the error was not so serious there. What would the correct emissivity to use be?
The Optris document quoted by Wyttenbach, repeated here: http://www.optris.com/search?keywords=irbasics&x=0&y=0&file=tl_files/pdf/Downloads/Zubehoer/IR-Basics.pdf
gives directions on page 12 for determining the emissivity to use. One of the fascinating ideas there is to drill a hole in the material, of certain specifications, which then becomes a black-body radiator at the temperature. Way cool! Or hot, as the case may be. Now why didn't I think of that? In fact, I've seen nobody mention it, most ideas are much more complicated!QuoteOf course we know very well the manual of Optris and even page 9 on which there are many formulas, not only one, all of them only with one value of emissivity !
The formulas there are for gray bodies, so they use a single value. But at the end of the page is this:QuoteThe formula shows that the emissivity ε is essential, if you want to determine the temperature with radiation measurement. The emissivity measures the ratio of thermal radiation, which is generated by a gray and a black body of equal temperature. The maximum emissivity for the black body is 1. A gray body is an object, that has the same emissivity at all wavelengths and emits less infrared radiation than a black radiator (ε <1). Bodies with emissivities, which depend on the temperature as well as on the wavelength, are called nongray or selective bodies (e.g. metals).
Alumina is a nongray body. There is a table on page 35 of the manual, which gives a value for alumina (aluminum oxide), but this is for powder, not cast, and no temperature is given and two forms are shown, one for "activated," with an emissivity of 0.46, and then another for ordinary clean powder of 0.16. This is not helpful for the purpose here. Indeed, because manufacturing of the material may introduce variations, one would need to calibrate, as by one of the methods shown on page 12. There may be more data in the specific camera manual.
QuoteTHHuxley wrote:
Oh I see you have lost the short term memory...... I will repeat it for you even if I posted just few days ago.
Actually, as I recall, a few days ago RB0 said that he would provide the math. I didn't see that. But maybe I overlooked something.QuoteConsider a normal resistor with at a fixed power in equilibrium condition in vacuum.
Then the input power and radiated power must be the same.
Yes. Equilibrium requires this.QuoteConsider first a Black Body.
If you integrate the Plank curve then you have the power density radiated from the surface of the resistor IBB , and multiplying than for the body area A you obtain the input power IBB*A .
The "Plank [Planck -- this was pointed out by another already] curve is the black-body radiation distribution for the temperature. It is radiation per unit surface area plotted according to frequency, and, integrated and then multiplied by the surface area, this is simply total radiated power. This is a statement that the total radiation for the black body (the integrated Planck curve) will then be the same as the input power, once the body reaches equilibrium. Yes. Absolutely true for thermal radiation and "vacuum" was specified, great. No worries about conduction or convection.QuoteBecause Quantum Mechanichs [sic] impose that no material can have an emissivity higher then that of a Black Body
This explanation is far more complex than need be. It is not necessary to mention quantum mechanics, for example. There is no dispute here that a black body has maximum thermal emissivity, and that no other material, as to thermal emission, can have higher emissivity. A black body is a "perfect thermal emitter."Quoteif you have a gray or selective emitter body with identical geometry then to maintain Ib*A=IBB*A constant the body must be hotter then the Black Body in order to have Ib=IBB.
This is saying that if the resistor is not a black body, and holding input power constant, the resistor will be hotter than if it is a black body. Yes. Absolutely true, again, while stated in a confusing way.QuoteThis simple considerations are independent from any detector. The ratio Ib/IBB is called total emissivity.
Again, true. And also thoroughly understood by the people RB0 is so condescendingly attempting to correct. This has nothing to do with the problem at hand! It is all completely accepted and known. It is, indeed, independent of any detector.And it says nothing about how to determine the temperature of the body with a thermal camera with a limited detection range. That requires band emissivity to be known or determined from calibration. That calibration must be (that is, include) observations at the temperature being measured, and will vary with the material and sometimes with how the material has been processed, matters such as surface texture.
-
Temperature isn't power or energy.
This is the Achilles heel of a large number CF-LENR experiments and conclusions.
Well, there are two problems. There are experiments done by some with poor calorimetry. Then there is poor writing, sometimes, or we could also call it poor editing. Power and energy are often confused.Some use what amounts to isoperibolic calorimetry without understanding it and how to improve accuracy under dynamic conditions. This is common among newcomers. This could be a whole topic of its own. I would be careful, though, about characterizing the field that way. There are true experts involved, with comprehensive understanding.
Almost any calorimetry can be used if well-calibrated. When the heat characteristics of a body are known and calibrated, and if the body exists in a constant-temperature environment, there will be a definite relationship between power dissipation and temperature, *at equilibrium,* and if rate of change of temperature is considered, this can be tightened to be almost continuous. This was Fleischmann-Pons calorimetry, it was modified "isoperibolic," whereas the negative replicators used naive isoperibolic, thus creating artifacts from power shifts.
Storms claims that isoperibolic calorimetry might be as imprecise as +/- 250 mW, though he is looking at worst-case, pretty much. Fleischmann-Pons calorimetry appears to have had a precision of about 0.1 mW.
So why isn't F-P calorimetry used more often? Well, it is mathematically complicated, and the field wanted to be simple in order to convince skeptics. At least that is my impression.
It was a bad idea. The math can be hidden under the hood, so to speak, in Labview setup, and then the entire system fully and precisely calibrated, which, then, will show how well it is working! And calibrations are easy to understand!
-
I do hope that Paradigmnoia's and Abd's comments help matters for Randombit0.
I have been replying to her incorrect assertions pointing out why they are incorrect. This is a lot easier than the general "show what is going on" and perhaps more useful for her.
It seems I missed one, which Abd quoted. It is very easily seen to be incorrect:
Quote from RB0Oh Dear are you crazy ? When calculating temperature (from energy) you use one vale of emissivity and when calculating the energy back you use another ?You would have no energy conservation in this way !
The issue is precisely encapsulated here. When calculating temperature from power you are not using all the power but only the small fraction of it in a wavelength for which the camera sensor is sensitive. When calculating power out you are using all the power, regardless of wavelength. If you think the two powers must be equal (and hence the same emissivity applies) you are most certainly violating conservation of energy!
I'd like you to comment on this RB0, and engage with it and admit you are wrong in the above statement.
PS - I've substituted power for energy here because although RB0 mixes up the two I cannot bear myself to say energy in a context where power is meant. However, in this case, that is orthogonal to RB0's error. If you replace energy by power (which is what she means) her statement is simply wrong as I explain above.
-
The issue is precisely encapsulated here. When calculating temperature from power you are not using all the power but only the small fraction of it in a wavelength for which the camera sensor is sensitive. When calculating power out you are using all the power, regardless of wavelength. If you think the two powers must be equal (and hence the same emissivity applies) you are most certainly violating conservation of energy!
Nice TTH. I think it was BH who long ago used "wavelength" to put it all in perspective. That always stuck in my mind afterwards while reading the much more complex explanations to follow. All that "color", and temp stuff is just another layer of complexity...or at least it is to us low tech guys.
My earlier question somewhere on LF: "so is Lugano dead?" was rhetorical BTW. Yes, it is dead. Sort of. Seems there is still some life left in it though as most, including you, leave open the possibility of overunity...however slight that may be be. Some such as MFMP, BH assign some significance to that chance, while you do the opposite.
-
It seems I missed one, which Abd quoted. It is very easily seen to be incorrect:
There is another. RB0 wrote:Some Points Regarding a Recent Presentation at ICCF20 on the ‘Lugano Report’ (Rainer Rander)
QuoteParadigmoia has also demonstrated that 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.
Paradigmnoia, of course, did not demonstrate that, because it is an oxymoron.A gray body, by definition, is a body that has the spectral distribution of a black body, merely at a lower level. I.e., the emissivity of a gray body is constant across the entire spectrum. If the gray body has a "very similar emission" in a selective spectral range, it has that everywhere, or it is not a gray body. If it has that everywhere, its total emission will also be "very similar," so, with the same dissipated power, the temperature will be "very similar."
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. If it is not a gray body, then the emissivity is not a constant as the measured or posited frequency varies.
I.e., not more than a little hotter. Not "much more hotter," or, for decent English, simply "much hotter." (We could also say, "much more hot." Both would mean the same thing, but "much more hotter" is redundant and poor grammar. As a copy editor I would definitely choose "much hotter.")
(Just like cooler means more cool, so we would never say "more cooler," likewise with hipper, smarter, and better. Children will say "more better," its an easy mistake. How about butter? More butter is what I always ask for since I was a kid. Have some bread with your butter.)
The description of this "gray body" is intrinsically contradictory.
Added after comments below.
I read "very similar emission in a selective spectral range" as referring to the emission behavior of the materials, i.e., the full behavior within that spectrum." I was over-narrow in that interpretation. If we take "similar emission" to mean "similar total energy," it is obvious that the condition described can exist. As well, the emission within band can be "similar" though unlikely to be the same. Confusing is that this thread was primarily about determining Lugano temperature from band emissivity vs total emissivity, and then, consequentially, about determining power dissipation from temperature. The statement, taken as now interpreted, was true but irrelevant.
