metalresearcher
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How are extremely high temperatues measured ?
Looking at boiling point data of Osmium (5027oC), iridium (4827oC) or other even lower boiling transition metals (iron 2800oC) , how do they determine
such high temps and even at nearly 1 degree accuracy ?
A thermocouple is the most accurate device but no substances remain solid @ 4800oC and infrared radiation pyrometers have a far lower accuracy.
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Neil
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My understanding is that they use emissions and extrapolate from the collected wavelengths.
This site has a bit on high heat but not that high of heat
http://www.pyrometry.com/ultrahigh.php
I'd love to know how they measured the boiling point of graphite, even if it is an estimation.
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simba
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IR emission.
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watson.fawkes
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Quote: Originally posted by metalresearcher ![](images/xpblue/lastpost.gif) | Looking at boiling point data of Osmium (5027oC), iridium (4827oC) or other even lower boiling transition metals (iron 2800oC) , how do they determine
such high temps and even at nearly 1 degree accuracy ?
| What you do is to look at the black-body radiation spectrum. You do this by taking a large number of
samples from a spectrometer, not just a few easy-to-measure frequencies. Real matter always has added peaks from electronic excitation states, so you
throw away those and match the black-body spectrum to the remaining curve.
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annaandherdad
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Quote: Originally posted by watson.fawkes ![](images/xpblue/lastpost.gif) | Quote: Originally posted by metalresearcher ![](images/xpblue/lastpost.gif) | Looking at boiling point data of Osmium (5027oC), iridium (4827oC) or other even lower boiling transition metals (iron 2800oC) , how do they determine
such high temps and even at nearly 1 degree accuracy ?
| What you do is to look at the black-body radiation spectrum. You do this by taking a large number of
samples from a spectrometer, not just a few easy-to-measure frequencies. Real matter always has added peaks from electronic excitation states, so you
throw away those and match the black-body spectrum to the remaining curve. |
Can you give more detail? The link above on pyrometry talks about measuring the brightness at more than one frequency, and how you can get the
temperature from that if you assume the emissivity is the same at both frequencies. It also talks about how in practice the emissivity is not always
the same. I don't see how you can get really accurate measurements, no matter how many frequencies you sample, unless you have information about the
emissivity.
As for the peaks at the frequencies of electronic excitations, are you talking about a gas or a solid or both? And why are electronic excitations any
different than any other? As for a gas, if it's optically thick, then you should get the Planck spectrum, with no peaks (that's how it is for the
sun).
There is a historical element in this, too. Planck got his final (correct) formula for the spectrum of bb radiation toward the end of 1899, after
receiving experimental data by Rubens and other experimenters. How did those guys make their measurements? Did the use a genuine black body (a
cavity)? How did they know the temperature? What kind of instruments (bolometers?) did they use? I've always wondered.
Any other SF Bay chemists?
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froot
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Off topic but, how would you contain boiling Osmium, Iridium, Tungsten, etc? Always wondered about this.
We salute the improvement of the human genome by honoring those who remove themselves from it.
Of necessity, this honor is generally bestowed posthumously. - www.darwinawards.com
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metalresearcher
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Quote: Originally posted by froot ![](images/xpblue/lastpost.gif) | Off topic but, how would you contain boiling Osmium, Iridium, Tungsten, etc? Always wondered about this. |
Maybe levitating in a magnetic field as there is no crucible material at those temps.
Another way can be extrapolated vapor pressure values but that is VERY inaccurate.
The surface temperature of the Sun and other stars is also determined by blackbody radiation, so how (in)accurate is that ??
Stars like Aldebaran, Betelgeuse appear reddish to us but the surface temp is around 3000oC, but terrestial objects so hot ppear blindingly white hot
to us. That has an explanation as when putting a halogen lamp (3000oC) at a few km distance it appears as a reddish star as well.
So this temperature measurement is rather correct.
[Edited on 2011-10-28 by metalresearcher]
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watson.fawkes
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Quote: Originally posted by annaandherdad ![](images/xpblue/lastpost.gif) | The link above on pyrometry talks about measuring the brightness at more than one frequency, and how you can get the temperature from that if you
assume the emissivity is the same at both frequencies. It also talks about how in practice the emissivity is not always the same. I don't see how
you can get really accurate measurements, no matter how many frequencies you sample, unless you have information about the emissivity.
| It's a curve-matching problem. The theory about how black-body radiation arises does not depend on the
particulars of atomic processes. Rather, it's arises from the equipartition of radiation in a closed system. See Planck's Law.
Of note is that the derivation only applies when you have equilibrium conditions in a closed environment. As soon as you point a pyrometer at it,
you're no longer in radiation equilibrium, because the instrument is at lower temperature than the black-body being measure. In this situation, you
get energy pushed into the electronic excitations of the materials present. The excitations will be atomic, molecular, or solid state depending upon
the state of matter, but there will always be something.
In the laboratory, you get more accurate measurements by minimizing the amount of radiation leaking out of the black body. Also, you maximize the
amount of insulation, so that the equilibrium condition is closer. Thus practical measurements in a foundry are more difficult, because you have
neither high insulation nor a small radiation aperture.
I don't know the historical details of the original measurements. I'm sure it's well-documented, though.
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D4RR3N
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You could do it by putting a thermocouple on the end of a bar of the material being heated, If you know the thermal conductivity of the bar you could
calculate the temperature at the heated tip.
Another way could be to use a wire of the material and send an electric current through it to heat it, there should be a correspondence between
measured electrical resistance and temperature.
Having said all that I believe in modern times it is done by a non contact process of analysing the spectra given off by the heated material.
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annaandherdad
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Quote: Originally posted by watson.fawkes ![](images/xpblue/lastpost.gif) | Quote: Originally posted by annaandherdad ![](images/xpblue/lastpost.gif) | The link above on pyrometry talks about measuring the brightness at more than one frequency, and how you can get the temperature from that if you
assume the emissivity is the same at both frequencies. It also talks about how in practice the emissivity is not always the same. I don't see how
you can get really accurate measurements, no matter how many frequencies you sample, unless you have information about the emissivity.
| It's a curve-matching problem. The theory about how black-body radiation arises does not depend on the
particulars of atomic processes. Rather, it's arises from the equipartition of radiation in a closed system. See Planck's Law.
It's not a curve matching problem if you don't know what the curve is. The Planck radiation law applies to a black body, but a surface radiating out
into space is not a black body. Instead the Planck formula for the intensity is multiplied by the emissivity, which in general is a function of
temperature, frequency, and the type of surface. So the actual curve, emissivity times Planck formula, is not known unless you know the emissivity.
That was my point.
Of note is that the derivation only applies when you have equilibrium conditions in a closed environment. As soon as you point a pyrometer at it,
you're no longer in radiation equilibrium, because the instrument is at lower temperature than the black-body being measure. In this situation, you
get energy pushed into the electronic excitations of the materials present. The excitations will be atomic, molecular, or solid state depending upon
the state of matter, but there will always be something.
It's not just pointing a pyrometer at it. It's the fact that even without the pyrometer the body does not radiate as a black body, as I just
explained. If you have closed cavity whose walls are at a constant temperature, then you do get black body radiation, but the radiation received,
coming from a part of the wall, is the sum of the radiated and reflected radiation. That *sum* follows the Planck law, not the radiated part by
itself.
In the laboratory, you get more accurate measurements by minimizing the amount of radiation leaking out of the black body. Also, you maximize the
amount of insulation, so that the equilibrium condition is closer. Thus practical measurements in a foundry are more difficult, because you have
neither high insulation nor a small radiation aperture.
I don't know the historical details of the original measurements. I'm sure it's well-documented, though. |
Any other SF Bay chemists?
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