A guest post by Ken Coffman and Mikael Cronholm
In clicking around on the Internet, I found an outstanding paper called Thermodynamics of Furnace Tubes – Killing Popular Myths about Furnace Tube Temperature Measurement written by Mikael Cronholm. The paper was clever and wise…and made a lot of sense. Clearly Mikael knows a lot about infrared radiation and I’m a guy with questions. A match made in heaven?
We exchanged e-mails. I want to be clear about this…Mikael corrected some of my wrong ideas about IR. I’ll repeat that for the slow-witted. Some of my ideas about infrared radiation were wrong. I am considered a hard-headed, stubborn old guy and that’s completely true. However, I want to learn and I can be taught, but not by knuckleheads spewing nonsense and not by authoritarians who sit on thrones and toss out insults and edicts.
Ken Coffman (KLC) is the publisher of Stairway Press (www.stairwaypress.com) and the author of novels that include Hartz String Theory and Endangered Species.
Mikael Cronholm (MC) is an industry expert on infrared radiation, a licensed, level III Infrared Training Center Instructor and holds two Bachelor of Science degrees (Economics and Business Administration).
The following is a summary of our conversation.
KLC: Hello Mikael. I found your paper called Thermodynamics of Furnace Tubes and I found it very informative, practical and interesting. I hope you’ll bear with me while I ask a couple of dumb questions. I am an electrical engineer, so I have some knowledge about thermodynamics of conduction and convection, but not so much about IR radiation. In return for your time, I would be happy to make a donation to the charity of your choice.
If I take an inexpensive IR thermometer outside, point it at the sky and get a temperature reading of minus 25°C, what am I actually measuring? Is there anything valid about doing this?
MC: Just as a matter of curiosity, how did you find my paper? I checked your website and I guess this has to do with the Dragon, no? If you want to make a donation I would be happy to receive that book. If you can, my postal address is at the bottom. I don’t follow the debate more than casually, but I am a bit skeptical to all the research that is done on climate change…it seems that the models are continuously adjusted to fit the inputs, so that you get the wanted output…and they argue “so many scientists agree with this and that”…well, science is not a democracy…anyway…
About radiation, then. There is more to this than meets the eye. Literally!
Looking at the sky with an infrared radiometer you would read what is termed “apparent temperature” (if the instrument is set to emissivity 1 and the distance setting is zero, provided the instrument has any compensation). Your instrument is then receiving the same radiation as a blackbody would do if it had a temperature of -25°C, if that is what you measure. It is a quasi-temperature of sorts, because you don’t really measure on a particular object in any particular place, but a combination of radiation, where that from outer space is the lowest, close to absolute zero, and the immediate atmosphere closest to you is the warmest. (I have once measured -96°C on the sky at 0°C ground temperature.) What we have to realize though, is that temperature can never be directly measured. We measure the height of a liquid in a common thermometer, a voltage in a thermocouple, etc, and then it is calibrated using the zeroth law of thermodynamics and assuming equilibrium with the device and the reference.
KLC: Global warming (greenhouse gas) theory depends on atmospheric CO2 molecules absorbing IR radiation and “back radiating” this energy back toward the earth. If you look at the notorious Ternberth/Keihl energy balance schematic (as shown in Figure 1 of this paper: http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf ), you see the back radiation is determined to be very significant…more than 300W/m2. From your point of view as an IR expert, does this aspect of the global warming theory make any sense?
MC: The paper you sent me mentions Stefan-Boltzman’s law, but it does not talk about Planck’s law, which is necessary to understand what is happening spectrally. I suggest you read up on Planck and Stefan-Boltzman at Wikipedia or something. Wien’s law would be beneficial as well—they are all connected.
Planck’s law describes the distribution of radiated power from a blackbody over wavelength. You end up with a curve for each blackbody temperature. The sun is almost a blackbody, so it follows Planck quite well, and it has a peak at about 480nm, right in the middle of visual (Wien’s law determines that). The solar spectrum is slightly modified as it passes through the atmosphere, but still pretty close to Planckian. When the radiation hits the ground, the absorbed part heats it. The re-radiated power is going to have a different spectral distribution, with a peak around 10um (micrometer). Assuming blackbody radiation it would also follow Planck’s law.
S-B’s law is in principle the integral of Planck from zero to infinity wavelength. Instruments do not have equal response from zero to infinity, but they are calibrated against blackbodies, and whatever signal they output is considered to mean the temperature of the blackbody. And so on for a number of blackbodies until you have a calibration curve that can be fitted for conversion in the instrument.
That means that the instrument can only measure correctly on targets that are either blackbodies, or greybodies with a spectral distribution looking like a Planck curve, but at a known offset. That offset is emissivity, the epsilon in your S-B equation in that paper. It is defined as the ratio of the radiation from the greybody to that of the blackbody, both at the same temperature (and wave length, and angle…). Some targets will not be Planckian, but have a spectral distribution that is different. If you want to measure temperature of those, you need to measure the emissivity with the same instrument and at a temperature reasonably close to the one you will measure on the target later.
So, of course, the whole principle behind the greenhouse effect is that shorter wavelengths from the sun penetrates the atmosphere easily, whereas the re-radiated power—being at a longer wavelength—is reflected back at a higher degree. I have no dispute about that fact. It is reasonable. So I think the Figure 1 you refer to is correct in principle. My immediate question is raised regarding the numbers in there though. The remaining 0.9 W/m2 seems awfully close to what I would assume to be the inaccuracies in the numbers input to calculate it. You are balancing on a very thin knifes edge with such big numbers as inputs for reaching such a small one. An error of +/- 0.5% on each measurement would potentially throw it off quite a bit, in the worst case. But I don’t know what they use to measure this, only that all the instruments I use have much less accuracy than that. But with long integration times…well, maybe…but there may be an issue there.
KLC: I am interested in some rather expensive thermopile-based radiation detectors called pyrgeometers (an example is the KippZonen CGR 3 instrument http://www.kippzonen.com/?product/16132/CGR+3.aspx).
If a piece of equipment like this is pointed into the nighttime sky and reads something like 300W/m2 of downwelling IR radiation, what is it actually measuring? If I built a test rig from IR-emitting lightbulbs calibrated to emit 300W/m2 and placed this over the pyrgeometers, would I get the same reading?
MC: “What is it actually measuring?” Well, probably a voltage from those thermopiles…and that signal has to be calibrated to a bunch of blackbody reference sources to covert it either to temperature or blackbody equivalent radiation.
Your experiment will fail, though! If you want to do something like that, you have to look at a target emitting a blackbody equivalent spectrum, which is what the instrument should be calibrated to. IR light bulbs emitting 300W/m2 is simply impossible, because 300W/m2 corresponds to a very low temperature! Use S-B’s law and try it yourself. Like this: room temp, 20°C = 293K. The radiated power from that is 293K raised to the power of 4. Then multiply with sigma, the constant in S-B’s law, which is 5.67*10-8, and you get 419 W/m2 or something like that, it varies with how many decimals you use for absolute zero when you convert to Kelvin. For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it (yes, minus!). Pretty cool light bulb.
I don’t know what your point is with that experiment, but if it is to check their calibration you need a lot more sophisticated blackbody reference sources if you want to do it at that temperature. But you could do a test at room temperature though. Just build a spherical object with the inside painted with flat black paint, make a small hole in it, just big enough for your sensor, and measure the temperature inside that sphere with a thermocouple, on the surface. Keep it in a stable room temperature at a steady state as well as you can and convert the temperature to radiation using S-B’s law. You should get the same as the instrument. Any difference will be attributable to inaccuracy in the thermocouple you use and/or the tested instrument. Remember that raising to the power of 4 exaggerates errors in the input a lot!
I hope I have been able to clarify things a little bit, or at least caused some creative confusion. When I teach thermography I find that the more you learn the more confused you get, but on a higher level. Every question answered raises a few more, which grows the confusion exponentially. It makes the subject interesting, though.
Let me know if you need any more help with your project!
KLC: I found your paper because one of the FLIR divisions is local and I was searching their site for reference information about IR radiation. I know what a 100W IR lamp feels like because I have one in my bathroom. If someone tells me there is 300W/m2 of IR power coming from space, and I hold out my hand…I expect to feel it. What am I missing?
MC: Yeah, you put your hand in front of a 100W bulb, but how big is your hand…not a square meter, I’m sure. It is per area unit, that is one thing you are missing. The 100W of the bulb is the electrical power consumption, not the emitted power of the visual light from it. That’s why florescent energy-saving lamps as opposed to incandescent bulbs give much more visual light per electrical Watt, because they limit the radiation to the visual part of the spectrum and lose less in the IR, which we cannot see anyway. The body absorbs both IR and visual, but a little less visual.
And, here is the other clue. Your light bulb radiation in your bathroom is added to that of the room itself, which is 419 W/m2, if the room is 20°C. Your 300 W/m2 from space is only that. You will feel those 300 W/m2, sure. It will feel like -25°C radiating towards your hand. But you don’t feel that cold because your hand is in warmer air, receiving heat (or losing less) from there too.
Actually, we cannot really feel temperature—that is a misconception. Our bodies feel heat flow rate and adjust the temperature accordingly. It is only the hypothalamus inside the brain that really has constant temperature. If you are standing nude in your bathroom, your body will radiate approximately 648 W/m2 and the room 419 W/m2, so you lose 229 W/m2. That is what you feel as being cooled by the room, from radiation only. Conduction and convection should be added of course. The earth works the same way—lose some, gain some. It is that balance that is being argued in the whole global warming debate.
KLC: I still feel like I’m missing something. IR heat lamps are pretty efficient, maybe 90%? Let’s pick a distance of 1 meter and I want to create a one-square meter flooded with an additional 300W/m2. It must be additional irradiation, doesn’t it? That’s going to take a good bunch of lamps and I would feel this heat. However, I go outside and hold out my hand. It’s cold. There’s no equivalent of 300W/m2 heater in addition to whatever has heated the ambient air.
Perhaps I’m puzzled by something that is more like a flux…something that just is as a side-effect of a temperature difference and not really something that is capable of doing any work or as a vehicle for transporting heat energy.
It’s a canard of climate science that increasing atmospheric CO2 from 390PPM to 780PPM will raise the earth’s surface temperature by about 1°C (expanded to 3°C by positive feedbacks). From my way of thinking, the only thing CO2 can do is increase coupling to space…it certainly can’t store or trap energy or increase the earth’s peak or 24-hour average temperature.
Any comments are welcome.
MC: Efficiency of a lamp depends on what you want, if heat is what want then they are 100% efficient, because all electrical energy will be converted to heat, the visible light as well, when it is absorbed by the surrounding room. If visible light is required, a light bulb loses a lot of heat compared to an energy saving lamp. Energy cannot be created or destroyed—first law of thermodynamics.
When you say W/m2 you ARE in fact talking about a flux (heat flow is what will be in W). If you have two objects radiating towards each other, the heat flow direction will be from the hotter one, radiating (emitting) more and absorbing less, to the cooler one, which radiates less and absorbs more (second law of thermodynamics). The amount of radiation emitted from each of them depends on two things ONLY, the temperature of the object and its emissivity. So radiation is not a side effect to temperature, it is THE EFFECT. Anything with a temperature will radiate according to it, and emissivity. (If something is hotter than 500°C we get incandescence, emission of visible light.) Assuming an emissivity of unity, which is what everyone seems to do in this debate, the radiation (flux. integrated from zero to infinity) will be equal to what can be calculated by Stefan-Boltzmann’s law, which is temperature in Kelvin, raised to the fourth power, multiplied by that constant sigma. It’s that simple!
With regard to your thought experiment, it is always easier to calculate what an object emits than what it absorbs, because emission will be spreading diffusely from an object, so exactly where it ends up is difficult to predict. I am not sure where you are aiming with that idea, but it does not seem to be an easy experiment to do in real life, at least not with limited resources.
CO2 is a pretty powerful absorber of radiated energy, that fact is well known. Water vapor is an even stronger absorber. In the climate debate it is also considered a reflector, which probably also true, because that is universal. Everything absorbs and reflects to a degree. So I guess that the feedback you mention has to do with the fact that increasing temperature increases the amount of water vapor, which increases absorption, and so on. But my knowledge is pretty much limited to what happens down here on earth, because that is what matters when we measure temperature using infrared radiation. However, it is important to remember, again, that we talk about different spectral bands, the influx is concentrated around a peak in the visual band and the outgoing flux is around 10 micrometer in the infrared band, and the absorption may not be the same.
With so many scientists arguing about the effects of CO2 I am not the one to think I have the answers. I really don’t know what the truth is. And the problem that all these scientists have is that they will never be able to test if their theories are correct, because the time spans are too long. For a theory to be scientifically proven, it has to be stipulated and tested, and the test must be repeatable and give the same results in successive tests for the theory to be proven.
If not, it is not science, it is guessing.
More like a horoscope…
I appreciate the efforts of all those providing information &/or opinion that has been presented here, very much.
From what I’ve read here I’m finding even harder to believe that ‘heat’ &/or ‘heat energy’ or even just ‘energy’ could be transferred to the deep ocean depths >900m. Trenberth has lately claimed that that’s where he’ll find the ‘missing heat’ re: energy balance. This transfer was apparently undetected during the transfer and remains undetectable. From what I understand Trenberth believes ALL current and previous temp. data are wrong, as well as all interpretations of the available data.
Is it possible for any IR energy to penetrate >900m? Further could that happen without being detected with current technology?
Again, thank you for the time and effort of teaching . . .
-Barn
@ur momisugly Will. I have to be restrictive with my time here, so I will just ask you this, why do you think “Not black as they would do if they were opaque to IR.”? Why would a window look black if it is opaque? How many times have you actually held an infrared camera in your hand?
@ur momisugly barn E. rubble. “Is it possible for any IR energy to penetrate >900m?” Within the wavelengths of IR cameras, 2-14 um: Definitely NO! And unless Myrrh and Will can tell you one of their magic IR wavelength bands that penetrates everything, I seriously doubt that there is much radiation in any E/M band that reaches down there. As a SCUBA diver I know that the longer wavelengths of visual begin to disappear pretty fast, even at the depths of amateur SCUBA, red start to disappear and things look more blue. I have heard of fish that live at extreme depths having red color as camouflage. There is no red light there anymore, so nothing will reflect off them, and they “disappear”.
I don’t remember the exact numbers now, but I have read some time that the first 10 m or so of the oceans contain the majority of the heat stored in the oceans.
@ur momisugly Will. On the site you refer to with your link they say, under “Infrared Windows”: “An IR transparent weatherproof shield or window can be made of polyethylene which is transparent to IR radiation in the 5 ~ 15 micrometer wavelength range. ” They suggest PE, because glass is opaque. The lenses they use are plastic, because glass is opaque.
And here something for you (and Myrrh):
“An example of a material whose emissivity characteristics change radically with wavelength is glass. Soda-lime glass is an example of a material which drastically changes its emissivity characteristics with wavelength (Figure 2-5). At wavelengths below about 2.6 microns, the glass is highly transparent and the emissivity is nearly zero. Beyond 2.6 microns, the glass becomes increasingly more opaque. Beyond 4 microns, the glass is completely opaque and the emissivity is above 0.97.”
And here you can find that quote, and a nice spectral graph (Figure 2-5 above) that describes how the transmission changes with wavelength. It even shows different thicknesses and how that changes the transmission.
http://www.omega.com/literature/transactions/volume1/theoretical3.html
And with that, I am done with the discussion about the IR transmission of glass. I could show you a live demo here at my office, but you would have to pay the ticket to Thailand yourself.
Mikael,
A substance which exhibits “hardly any transmissivity AT ALL!!!!!!” in the IR will obviously appear black in an IR image because it is not transmitting any IR, obviously.
I have never actually counted of how many times I have used my infra-red camera.
To Will
You are not reading what Mikael wrote very CAREFULLY. Mikael wrote that common glass starts decreasing transmission at 2 microns, and at 2.3 microns stops nearly all further transmissions.
So, yes indeed, glass does transmit ‘some’ IR, BUT ONLY A TINY AMOUNT compared to the entire infrared band!
Infrared originally simply meant ‘beyond the color red” because of Herschel’s discovery in 1800. However, later it was defined as the region of 0.7 to 1000 micron wavelengths.
re the pyroelectric motion detector
First of all, those are the cheapest IR systems you can buy. Any specs they give should be treated as ‘suggestions’ rather than ‘specs’. Most are built in China. The quality of the optical coating on the internal silicon window they use is the cheapest possible.
If you explain to me EXACTLY what you did with that glass and al foil in your test of that pyroelectric security sensor. I will tell you EXACTLY where your error in understanding is.
You cannot compare that cheap pyroelectric device with the equipment that Mikael uses. Mikael’s equipment is much more sensitive and precise, but it also has its limitations.
re Window coatings
You do not yet understand the concepts of emissivity and reflectivity. If you did, you would see where your reasoning has failed.
re O2 and N2
That part, I agree with you strongly. They have not been characterized fully for absorption, reflection, transmission, and emission over the ENTIRE infrared band 0.7 to 1000 microns. That is one the major flaws in the way the warmists have been defining greenhouse effect. I have been pointing that out in multiple posts here.
Response to George E. Smith
There are several frequency bands where the atmosphere really is IR opaque. Even though radiation is emitted in all directions, it is reabsorbed within 10 meters or so. Therefore, the fact that the line widths change with temperature and pressure does not have an effect. As a direct result, more energy is returned to the surface from the warm lower layers than released to space from the cold upper layers.
As for clouds, these are blackbody emitters and their energy passes through the windows where greenhouse gases don’t absorb. Except for a very small amount of energy from ozone, no IR energy from the upper atmosphere reaches the surface. None.
Your claim that “the escape path to space [should be] favored over the downward path” would be true if the atmosphere was only partly absorbing at those frequencies. To be clear, the atmosphere is IR opaque in most of those frequencies where water vapor and CO2 absorb, and IR transparent in the “windows”.
Sorry, I don’t understand your “drinking water bottle” comment. Could you please clarify that?
RE: Myrrh says:
February 17, 2011 at 5:28 am
“I’m not looking for a teacher.. ”
I am. And you’d be one of them. Having found most of what’s been posted here to be a tuff skate; it has been most informative. Altho I didn’t get the impression that much, if any, was in support of AGW. Do I understand correctly (from what I read from Myrrh) that CO2 levels, however increased, does not in fact mix or average out at increased levels globally but would accumulate locally, ie: where it is being produced and cycle about unless moved by other forces and dissipate sooner rather than later? Further, the GH effect of Co2 as per warming the surface temp by re-emitting IR/heat energy towards the surface is negligible? Is there no peer reviewed papers (for what that’s arguably worth) on the natural separation of in our atmosphere that Myrrh mentions? Growing ever so skeptical, it would appear the main plank in supporting AGW; well established physics, RE: CO2, seems to be . . . well, less so? Leaving the only other plank; consensus, somewhat shaky?
Excuse me for taking the discussion back a grades for just a moment. My understanding of AGW simply put: Energy from the sun comes in and hits the surface of the earth warming it. Energy is emitted back from the earth where CO2 (among others) blocks/absorbs and re-emits energy back to the surface. Increasing CO2 levels will enhance this by blocking/absorbing more energy and thereby re-emitting more energy/heat back to the surface. And so on, until the dreaded tipping point. Apparently that means our atmosphere must surely be getting bigger as well, like blowing up a balloon, if that is, there isn’t an equal (and natural) release somewhere. Does there not have to be a volume balance to deal with and not just the energy balance for AGW to work? Considering the panic of increased levels of one component should there not be an equal panic (at least a question or two?) about what components are being displaced?
Back to IR:
A small aside here from examples used earlier on. There is no such thing as an IR sauna. Full stop. Perhaps just a misnomer but the literal translation of ‘sauna’ is steam bath. You can no more have a ‘dry’ sauna than you can have a ‘dry’ shower or bubble bath. Call it a warm room or sweat closet but a sauna it is not (pronounced: sow-na) it irritates your Fin friends, believe me.
Always sweating the small stuff . . .
-barn
@ur momisugly Robert Clemenzi. I think I can explain the “drinking water bottle” comment. Find the name “Will” somewhere in this thread and click on it… scroll down… very funny!
“”””” Mikael Cronholm says:
February 17, 2011 at 7:28 am
@ur momisugly barn E. rubble. “Is it possible for any IR energy to penetrate >900m?” Within the wavelengths of IR cameras, 2-14 um: Definitely NO! And unless Myrrh and Will can tell you one of their magic IR wavelength bands that penetrates everything, I seriously doubt that there is much radiation in any E/M band that reaches down there. “””””
Sea Water at it’s most transparent point; which occurs at about 460-70 nm wavelenght (blue), has an absorption coefficient of between 1 and 2 x 10^-4 cm^-1. So that means that that wavelenght will be attenuated down to 1/e (37%) in a dept of 10^4 cm max, or 50-100 metres for thatrange (I can’t read the logarithmic scale on the graph any more accurately). So five absorption lengths will reduce the residual down to 1%, which is 250 -500 metres. For 5% remaining you go three times, so 150-300 metres; and that is for the MOST penetrating wavelength. At the UV end of the visible spectrum, (380 nm) the absorption is 10 times as much; 0.001 cm^-1, so 99% extinction in 50 metres depth.
For the red end (780nm) the absorption is another ten times higher , or 0.01 cm^-1, so 99% is gone in 5 metres of good clean clear sea water, uncluttered with plankton etc.
At 1.5 microns wavelenght in the near IR, you run into a significant water absorption band, where alpha is 30 cm^-1, so the 1/e depth is 333 microns, or 1.6 mm for 99% loss. There’s a slightlyhigher peak at 2.0 microns, and then the highest of all at 3.0 microns where alpha is8-9,000 cm^-1 or 1.1 to 1.25 micron absorption depth. Longer than that, there’s some dips, but not below 100 cm^-1 for alpha, and after 7.0 microns, it settles down to around 1000 cm^-1, at least out to 100 microns wavelength which is about the limit of our interest for climate issues. Then water slowly gets more transparent about as the square root of the wavelength (or frequency) all the way into the radio spectrum. It finally gets back to around 1 cm^-1 at around three metres wavelength (100 MHz).
So no; there is no part of the EM spectrum, of concern for climate issues, that can penetrate to 900 metres depth in the ocean.
Those data are from two graphs in the same paper.
G.C. Ewing Oceanography from Space, Woods Hole Oceanogrqaphic Institution, Woods Hole Mass WHOI ref.No 65-10 April 1965
Domenic,
Your ad hom attacks on my understanding of these subjects will not gain any traction here.
These tactics are what we look for to discover the weakness in the arguments of self proclaimed experts. You are displaying the typical traits people like myself have come to expect.
It is interesting that as is a usual feature on these type of threads, there is a sort of tag team effort ongoing between yourself and Mikael.
I have pointed out once before that I’m not actually talking to you, yet you seem to want to answer all the points that Mikael appears to be struggling with.
There is a specific name for that technique. It’s called the Delphi technique. Have you heard of it?
That question is to you Domenic.
Mikael,
Before we leave the subject of glass I have a couple more questions.
If a substance has an emissivity of 0.97, with 1.00 being a perfect black-body emitter, what would be the absorptivity of that substance? And how could this be considered as a barrier to trap IR as in opaque?
Finally, are we going to continue to resort to semantics?
At the end of the day, the salient point is really in those last 3 paragraphs.
“If not, it isn’t science, it’s guessing.”
Kind of obvious really, but elegant in the way such quotes often are.
to Will
You have not given me the details of the test you performed on the pyroelectric sensor with window glass and al foil. There are multiple ways for you to introduce error in your ‘test’. How can I possibly address what you observed it you do not give me the all the details of your ‘experiment’? Do you think I am going to sit here and play a guessing game with you? That is not an ad hom attack.
The absorption would also be 0.97, but only for THOSE SAME WAVELENGTHS IT EMITS.
If an object has an emissivity of 0.97, then 0.03 is coming from either (1) reflection of wavelengths from materials around it or (2) transmission of wavelengths from materials going through it. If you can guarantee it is not one of those, then you are left with the other.
@ur momisugly Will. Emissivity and absorptivity is essentially the total opposite of each other, emissivity tells you how well something ‘gives off’ radiation and absorptivity how well it ‘takes up and retains’ radiation. But here is the clue, and it is a very fundamental thing and basic in radiation science. At any given wavelength and angle of incidence the absorptivity and emissivity of a surface will ALWAYS be the SAME VALUE. This is called Kirchhoff’s Law, but if you google it make sure don’t end up with the one that deals with electrical circuits.
That law is fundamental, because without it we could not achieve equilibrium in a system.
So, if something has an emissivity of 0.97, the absorptivity is exactly the same.
Radiation equations:
a + r + t = 1
e = a
e + r + t = 1
For most objects we can assume t = 0, so
a + r = 1
e + r = 1
And I would still like to know why a window would look black in the IR camera because it is opaque?
@ur momisugly Will. Sorry, I see now that you have answered that already. So with your infrared camera, have you ever looked at windows or glass with different temperatures? And, just a question, what wavelength is your camera and what is normally your purpose for using it?
“”””” Robert Clemenzi says:
February 17, 2011 at 8:54 am
Response to George E. Smith
There are several frequency bands where the atmosphere really is IR opaque. Even though radiation is emitted in all directions, it is reabsorbed within 10 meters or so. Therefore, the fact that the line widths change with temperature and pressure does not have an effect. As a direct result, more energy is returned to the surface from the warm lower layers than released to space from the cold upper layers.
As for clouds, these are blackbody emitters and their energy passes through the windows where greenhouse gases don’t absorb. Except for a very small amount of energy from ozone, no IR energy from the upper atmosphere reaches the surface. None.
Your claim that “the escape path to space [should be] favored over the downward path” would be true if the atmosphere was only partly absorbing at those frequencies. To be clear, the atmosphere is IR opaque in most of those frequencies where water vapor and CO2 absorb, and IR transparent in the “windows”.
Sorry, I don’t understand your “drinking water bottle” comment. Could you please clarify that? “””””
What’s to understand? The mean Temperature of the earth (surface or Lower Troposphere) is purported to be 59 deg F or 15 deg C; 288 Kelvins. So a Black Body at that Temperature radiates a spectrum that peaks at 10.1 microns, and contains 98% of its total elergy emissions between 5.0 microns and 80 microns, with only 1% beyond each end. Now Temperature extremes range from as low as 183 K around Vostok to as high as 333 K in the tropical deserts (surface). So that will extend the spectrum somewhat.
But we should note that NO Black body emits MORE energy at ANY wavelength, than is emitted from a black body at a HIGHER temperature. So we can completely discount any areas that are substantially colder than the global average; their emissions don’t show up as any perceptible increase in the global average.
The high Temperature limit (333 K) will drop the spectral peak down to around 8.8 microns, so extending the low wavelength limit to say 4.4 microns.
So we can say that the LWIR spectrum of the earth covers perhaps 4.0 to 80 microns wavelength range.
Of that, CO2 carves out a notch from about 13.5 to 16.5 microns. At least half of the entire spectrum energy is emitted entirely below that 13.5 micron band edge of CO2; and at least 35%, maybe as much as 40% is entirely above the 16.5 micron upper edge of the CO2 band. And those higher Temperature tropical desert areas are emitting as much as 1.8 to 2.0 times the global mean radiant emittance; and that into a nearly water free atmosphere, and at wavelengths even further removed from the CO2 band; so even more escapes.
That leaves at MOST, 10-15% of the entire earth LWIR Radiant energy emissions subject to CO2 absorption.
The rest of that energy, with dry air, is pretty much free to leave.
Please don’t insult us here at WUWT by claiming that H2O is a greenhouse gas; the AGW folks are most strident in their assertions that H2O is merely a feedback amplifier for CO2 the Prince of green house gases.
Absent H2O, the atmosphere isn’t even vaguely IR opaque.
But back to our water bottle. As I said, the earth LWIR spectrum is reasonably from about 4.0 to 80 microns with a peak near 10 microns; and corresponds to a black body thermal radiation source (roughly) at a Temperature of around 288 K; 300 K or whatever; that is the Temperature of a source for the radiation that is involved in the greenhouse effect (which most WUWT readers readily acknowledge).
Now water has a reflection coefficient of about 3% tops, so a bottle of water at about room temperature is perfectly good laboratory source for the kind of LWIR radiation, that is absorbed by COt and causes the greenhouse effect.
!00 Watt light bulbs are NOT a good source for earth like LWIR thermal radiation. The wavelength is 10 times too short, and the Brightness is 10,000 times too large, and the Temperature is 10 times too high.
“”””” There are several frequency bands where the atmosphere really is IR opaque. Even though radiation is emitted in all directions, it is reabsorbed within 10 meters or so. Therefore, the fact that the line widths change with temperature and pressure does not have an effect. As a direct result, more energy is returned to the surface from the warm lower layers than released to space from the cold upper layers. “””””
When you say it that fast; you can almost get people to believe that. But that assertion is based on a popular fallacy; the notion that gases do NOT radiate a thermal spectrum based on their Temperature, so the “re-emission”, is the same 13.5 to 16.5 micron CO2 absorption spectrum. And that is clearly not the case.
The atmosphere clearly does radiate a normal Tempertaure based thermal Spectrum, and yes it does contain some absorption dips notably a narrow one aorund 9.6 microns, from the thin high level Ozone layer, and the 13.5 to 16.5 micron CO2 band; heck, take 13 to 17 microns if you like. The point is that the bulk of the thermal radiation from the atmospehre qand from the surface is NOT within the CO2 absorption band, so it can proceed to space through a dry atmosphere, with just a small recapture by CO2. what the CO2 doesn’t recapture; which is most of the spectrum, escapes; subject ONLY to the water conditions; and we know that water is not a GHG; just a CO2 helper.
My source for the atmosphere radiation data is:- H. Rose, et al., “The Handbook of Albedo, and Thermal Earthshine”; Environmental Research Institute of Michigan (ERIM), Ann Arbor, MI, Report No. 190201-1-T
Yes it is true that in the near vaccuum of the stratosphere, the mean free paths are long enough for CO2 to spntaneously re-radiate its 13.5 to 16.5 micron band; but in the lower levels (tou mentioned the 10 metres or so near the surface), the capured energy is completely thermalized through molecular collisions; so the source of the atmsopheric LWIR is ordinary thermal emission form the ordinary atmospehric gases, and that emission is entirely independent of any trace GHG content; and depends ONLY on the temperature of the Atmosphere.
And if you really want to permit H2O to take its proper place in the mechanics of the atmospheric energy (radiant) processes; then you will quickly discover how inconsequential CO2 really is.
Since H2O is the ONLY condensing GHG, it is the only one that can form clouds, which immediately introduce a huge net cooling effect. Nobody EVER observed it to warm up (go to a higher Temperature) in the shadow zone, when a cloud moves in front of the sun; it ALWAYS cools down.
“”””” …….. no IR energy from the upper atmosphere reaches the surface. None.
Your claim that “the escape path to space [should be] favored over the downward path” would be true ……….. “””””
Somewhat conflicting aren’t these notions ?
to Barn
The most pristine data sets to relate these IR discusssions to are those from Amundsen Scott AFB and Vostok bases in Antarctica.
The sun is a minor effect there. Temperatures are almost completely dominated by radiational heat transfer to outer space through the atmosphere.
Outer space is a constant. Nighttime sky does not vary as a blackbody target.
If CO2 was acting as a powerful greenhouse gas, it would show up in their data first, as CO2 increases in the atmosphere there would trap more heat radiationally.
But it is not.
http://icecap.us/images/uploads/VOSTOK.pdf
And the data would be more clear if someone went and picked out simply the lowest recorded temperature per year at those two locations. And then graph those.
You see, there is absolutely nothing on earth that can drive those temperatures lower.
Nothing.
Except the ‘greenhouse effect’ lessening.
Bill Illis says:
February 16, 2011 at 2:45 pm
Its just that you cannot find anywhere on the internet, someone that shows N2 and O2 and Argon absorb any IR at all. Everyone seems to believe/has been taught, that N2 and O2 do not absorb any IR at all, even blackbody-type radiation.
This was the only page I’d ever found, don’t know how good it is: http://www.spinonthat.com/CO2_files/02_N2_IR_absoption.html
He makes the same point you do, that both sides take this for granted as if confirmed scientific fact. And says:
“The evidence I have presented here conclusively shows that Oxygen and Nitrogen which constitute 99% of the Earth’s atmosphere, as one would expect, do in-fact absorb infrared radiation.
The claim that these two gases are transparent to IR is completely fallacious. Which goes a long way to explain why the “Greenhouse Effect” hypothesis is one of the longest standing unsubstantiated hypothesis in the history of science.
The so called “Greenhouse Effect” will always remain an hypothesis for the simple reason that it is based on pseudo science.”
“”””” Bill Illis says:
February 17, 2011 at 5:01 am
George E. Smith says:
February 16, 2011 at 2:45 pm
… thermal radiation; that is Electromagnetic Radiation following Maxwells equations are emitted from ALL materials at above absolute zero, including ALL gases; including N2 and O2 and even Ar; and he agrees that the fundamental Physical mechanism is simply accelerated electric charges following Maxwell’s Equations for the EM field.
You see the molecules/atoms in a gas are zipping this way and that, and colliding with each other; and everytime two molecules collide, …
—————————————————-
At sea level pressure, each atmospheric molecule collides with another molecule 6.7 billion times per second. A rate that is 33,000 times faster than the average emission time for CO2. So, an excited CO2 molecule (which is now effectively many times warmer than the surrounding air) crashes into another atmospheric molecule 33,000 times before it can emit. In the troposphere, the numbers are not much different, 10,000 times.
So, if N2 and O2 and Argon are actually emitters and absorbers of IR (albeit at a far reduced level than the specific absorption and emission frequencies of CO2), and are crashing into CO2 at a far faster rate than CO2 emits at, the entire picture of atmospheric radiation changes. “””””
Bill, YOU get it I get it, some others get it. Why is it so hard for others to grasp.
The LWIR radiant energy that is specific to the GHG absorption bands (13.5-16.5 microns for CO2), is what the GHG molecules CAN and DO capture; from the entire 4.0-80 micron LWIR spectrum corresponding to earth Temperatures; and that capture energy is immediately thermalized (thanks for coming up with the numbers); and as of that point; the GHG molecule is totally removed fromt he picture (well to capture again).
And even thogh N2 and O2 or Ar, and other non dipolar molecules do not capture IR through resonance absorption bands, as do CO2 and H2O, those air molecules are still quite free to radiate a perfectly normal thermal emission spectrum, that depoends ONLY on the atmospheric Temperature, and is quite insensitive to the mechanism that heated the atmosphere in the first place.
That heating could have been incoming soalr heating, or ground contact conduction and convection; by whatever means the atmosphere can be warmed (including deposit of latent heat from water condensation (or freezing); it will then radiate according to thermal radiation laws; and MOST of that thermal radiation spectrum is NOT subject to further assault by CO2; more is by H2O though.
As Phil several times pointed out here, in the stratosphere mean free paths can be long enough for GHGs to spontaneously decay to the ground state.
Even thoguh the atmospheric gases are low thermal mass, and must therefore cool from radiating thermal emission; they are maintained at the ambient Temperature by immediate resupply from the molecular collision processes; so it is a continuous energy pumping process.
Mikael,
“@ur momisugly Will. Emissivity and absorptivity is essentially the total opposite of each other, emissivity tells you how well something ‘gives off’ radiation and absorptivity how well it ‘takes up and retains’ radiation. But here is the clue, and it is a very fundamental thing and basic in radiation science. At any given wavelength and angle of incidence the absorptivity and emissivity of a surface will ALWAYS be the SAME VALUE. This is called Kirchhoff’s Law, but if you google it make sure don’t end up with the one that deals with electrical circuits.
That law is fundamental, because without it we could not achieve equilibrium in a system.
So, if something has an emissivity of 0.97, the absorptivity is exactly the same.
Radiation equations: a + r + t = 1 e = a e + r + t = 1 For most objects we can assume t = 0, so a + r = 1 e + r = 1”
Yes of course I knew the answer to the question because I know all about Kirchhoff’s Law. And the last and most important part of my question, which you didn’t manage to answer was, “And how could this be considered as a barrier to trap IR as in opaque?”
So you have stated quote: “Glass will start to lose transmissivity around 2 um and nosedive completely at 2.3um. Over 3um there is hardly any transmissivity AT ALL!!!!!! How can I know for sure? Because glass is totally opaque to any IR camera I use, in addition to all the spectral charts that are available. Can’t see through it at all.”
Meaning that IR cannot pass through glass because it is “opaque” to IR and therefore traps IR as in the “greenhouse effect”.
Yet here you are tying yourself up in knots with your own semantics.
No further questions your honour.
If I were a fotong from the stellar entity Sol who had somehow been cast abruptly down to the Earth’s surface, I would be trying to get back to the mother ship as quickly as I could. Falling into the clutches of CO2 and having to ping-pong my way back to freedom through the troposphere, I would be delighted to be back-radiated to the ground ‘cos then I’d have another chance at a clear shot through the window.
Gettng my bearings after that might prove difficult but, if I did make it back to Sol, I would make my home “warmer than it otherwise would have been”.
to Will
you wrote: “Also infra-red detectors on security lights which are filtered to detect between 8-14 µm will still work perfectly well with a thick piece of glass covering the sensor even when the housing is encased in tinfoil. This is a test I have carried out many times.”
You are making claims that go counter to common knowledge to many infrared professionals.
In fact, I have performed those same tests you describe with many, many IR detectors. And gotten completely different results than you claim.
I asked you for specific details of how you performed those tests.
You have not supplied those details.
Why?
@ur momisugly Will. And no further answers to you! I find it a bit insulting that you ask me to explain things to you that you already claim to know. But still don’t seem to understand at all…
@ur momisugly Domenic. I think I know. Because Will is a dishonest trickster who is playing games here to make us waste our time. No more of that…
Will is obviously in the business of peddling misinformation. Look at the silly stuff with water bottles that he has on his website. Click his name. How many scientific errors can you find in that experiment? But to a layman it might seem like it would be true, although it does not prove anything that it claims to prove.
Mikael,
I thank you for allowing us to eavesdrop on your conversation with Ken. It was both enjoyable and informative. Additionally, you also joined us here to continue the discussion. That probably made this thread one of the better ones to read and digest on WUWT.
The only thing I may disagree with you is your definition of heat, as I learned the classical thermodynamic system boundary version. Old habits are hard to break.
Thanks again,
Jim