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…
The solar day on Venus is about 582 days.
So it has long nights of 291 days.
It has a suface temp of 735 K, 460 C
Doesn’t that mean it should be emitting 36 times more W/m^2 than the
Earth at 300K? (735^4)/(300^4)
How is it, then, that Venus remains so hot, even on its night side?
It is closer to the Sun, but shouldn’t that mean it is only getting a little over 2.2 times the energy per unit area? Venus also has a high albeto — that ought to help keep it cooler.
Venus has almost 300,000 times more partial pressure of CO2 than does earth. It has almost no water. If the IR saturated spectra argument is valid, would 10% CO2 be just a bad as 99.5% CO2?
Surface pressure 93 bar (9.3 MPa)
Composition ~96.5% Carbon dioxide
~3.5% Nitrogen
0.015% Sulfur dioxide
0.007% Argon
0.002% Water vapor
0.001 7% Carbon monoxide
0.001 2% Helium
0.000 7% Neon
trace Carbonyl sulfide
trace Hydrogen chloride
trace Hydrogen fluorid
http://en.wikipedia.org/wiki/Venus
Is it something as simple as PV=nRT ?
with the high altitude, low pressure, low temperature stratosphere being the governor for heat loss rate?
I realize that many of you are not fond of the website realclimate.org. However, there are useful posts on the above subject (CO2 radiation exchange in upper atmosphere) at
http://www.realclimate.org/index.php/archives/2007/06/a-saturated-gassy-argument/
and more importantly at
http://www.realclimate.org/index.php/archives/2007/06/a-saturated-gassy-argument-part-ii/
Also an important data source for the absorptivity of CO2 is given in the data found in HITRAB at http://www.cfa.harvard.edu/hitran//
This software uses data from HITEMP https://kb.osu.edu/dspace/handle/1811/13476
“The HITRAN database has been recognized for more than 20 years as the international standard compilation of spectroscopic absorption parameters for atmospheric gases.”
The following paper has some of the CO2 data (available on line):
http://faculty.uml.edu/robert_gamache/papers/Rothman_et_al_Preprint.pdf
From what I have read the emissivity of CO2 does vary as a function of both pressure and wavelength. But I am not an expert on this subject.
Mikael, I am sitting in front of my wood burning heat-o-lator. When the door is closed and the fan running, the room heats up quite quickly, but my body doesn’t. If i want a quick kick of warmth, I turn off the fan, open the door, re-arrange the wood so that the ember side is now facing out, and Whammo! instant warming of the body.
Also reference the campfire effect. Side facing campfire feels warmth, as the body is cooler then the flames/embers. side away from campfire feels cool/cold, as body on that side is radiating heat to a lower temp atmosphere.
Jim D says:
February 13, 2011 at 5:48 pm
The measured IR from the sky is pretty much all emitted by CO2 and H2O molecules …
No Jim. If that were true then the warmth of the sun on my face would be coming in from all directions equally. But even a blind man …
Or do you mean at night?
And do you really mean to tell us that nitrogen can’t radiate IR?
Mikael Cronholm says:
February 13, 2011 at 9:18 pm
“@chico sajovic, 8.20. I think you talk about the furnace paper then, and in a furnace radiation is much stronger than any other mode. In the radiation section at the bottom, where the flames are, radiation is completely dominant. Up in the convection section, the tubes draw out the remaining energy they can from the exhaust gases, but that is much less than what is added to the feedstock in the radiation section. The clue is that radiation increases with the temperature to the power of 4, according to Stefan-Boltzman’s law, while conduction and convection don’t. Conduction is linear, convection probably less strong than linear, with temperature (logic: just because a surface is 1000C it does not create a storm around it by convection).”
Be careful. Just because radiation goes as the 4th power of temperature doesn’t mean it’s always the dominant form of heat transfer. It also has an extremely low coefficient compared to that of conduction or convection of most materials. Also, both conduction and convection are linear with temperature but their respective ratios are dependent on the particular fluid’s properties. Finally, that 1000C surface may not create a storm, but it most certainly will drive noticeable amounts of air or any other surrounding fluid flow. Of course, if you’re using your hand to measure that then you’re probably more worried about the burns you’re getting by holding your hand too close to a very hot object.
Ian W says:
February 13, 2011 at 12:35 pm
When water vapor in the atmosphere condenses into liquid water and then changes state again and becomes ice, it gives off latent heat for both state changes.
Does that latent heat release follow Stefan-Boltzmann’s radiative equation ?
No Ian, latent heat is not really related to S-B. Latent heat means that when you add or remove heat, energy, from a substance it will usually change its temperature, except when there is a phase change. At a phase change, for example melting or evaporation, you will be able to add or remove heat without a change in temperature taking place, the change of heat in the substance causes the phase change to take place instead. In the case of freezing water for example, there is molecular kinetic energy stored in translational movement, i.e. the molecules move around relative to each other (they have a mass, they move=energy). When the liquid turns to a solid crystalline structure, that energy of movement must be removed before the molecules can stand still in relation to each other. That is the latent heat that is released from the water when it freezes. Fruit orchard are actually sprinkled with water to prevent the plants from getting damaged on cold clear night in the spring. The water freezes and give off heat to the plant, preventing it from getting too cold.
So, as far as S-B is concerned, this process will only influence the input T in that equation, insofar as the cooling or heating of the substance reaches a plateau when latent heat takes effect. The radiation will still always depend on the temperature and emissivity of the substance.
To Dave Springer
In actuality, to my knowledge, the true greenhouse effect of any gases (including Nitrogen) has never been measured properly. They are calculated based on a lot of basically untested assumptions. (To explain all the assumptions currently used would require a very detailed and long technical paper.)
To do it correctly, you should use a radiative source at room temp to represent the typical earth radiation temperatures instead of a high heat source used historically in gas detectors. Then the detector itself should be near absolute zero to simulate outer space at night time rather than the room temperature state it normally is in a gas detector.
In other words, this proposed setup should be an exact simulation of night time conditions. The gases you put inside the tube can represent any kind of atmosphere conditions, or gases, that you wish.
THEN you DIRECTLY measure the longwave radiation transmission, to a great extent, etc of the various gases in the atmosphere.
For those of you with IR guns or devices, you can see true greenhouse effects directly at night by aiming your IR device at the center of the sky on a cloudless night, and then comparing that reading to another night by aiming at the center of the sky when the sky is full of clouds. That is a TRUE greenhouse effect differential measurement.
hotrod ( Larry L ) says:
February 13, 2011 at 5:49 pm
It is perhaps a subtle distinction for some folks, but the CO2 and water vapor “back radiation” does not “warm the ground”
Why not? Does it have to do with the 10 micron wavelength?
or the ocean surface,
Why not? Those photons have to go somewhere. Why does this look like a “missing heat” problem?
it only “reduces the rate at which it is cooling by IR radiation” to the apparent surface of the very very cold sky.
So “back radiation … only reduces … cooling.”
Huh??
@ur momisugly Tsk Tsk. No, but at high temperature it is definitely dominant, and my paper deals with furnaces where temperatures are very high.
Another point to make about that is that radiation and conduction are the most predictable and easy to calculate of the three, at least if conduction is in a solid. Convection is a nightmare!!!
chico sajovic says:
February 13, 2011 at 7:45 pm
“Back Radiation” is such a horrible analogy, description or way to think about heat transfer in the atmosphere, that I wish it would just go away. What we are concerned with is the flow of heat from the surface of the earth to space. Its not that the surface warms the co2 in the air then the co2 “back radiates” re-warming the surface, its that when the co2 in the air is warmed the flow of heat from the surface to the warmed co2 is reduced, thus increasing the temperature of the surface until the flow of heat returns to equilibrium.
——————————————-
I don’t quite get the preoccupation with “the surface” when it’s actually the air temperature that we measure, not the ground.
GHG’s don’t make the ground warmer in the daytime than insolation is able to do and at night, the paltry amount of energy accumulated in the surface doesn’t provide much residual warmth to the air, especially since it’s only a fraction of outgoing LWR that is absorbed.
The heat capacity of the air is not enormous but radiative heat loss is not really rapid at these temperatures, either.
Something like that?
in answer to the question up there, the SB equation doesn’t work with dimensional gases.
Secondly, it is a theoretical equation that doesn’t work with climate generally,
Practical demo: The basal human metabolic rate is around 58wm2. An average human is around 2m2, so the average energy a human generates is around 110wm2
This creates more heat than does the atmosphere or the ground at night under the view of spectroscopes.
It is therefore safe to assume that there is much less than even 50w m2 re-radiation.
true, there may be air currents rising, but these are invisible to c02 which absorbs (well, it doesn’t really absorb but delays by a billionth of a second) energy at around 15 microns, which corresponds to subzero temperatures.
It then quickly thermalises with nitrogen and oxygen at its most active region – which is quite high in th eatmsphere where freezing temperatures correspond to c02 absorbtion. However, at this height (around -28C in the troposphere) it also competes with the peaks of nitrogen and oxygen absortion (they absorb radiation too).
It is questionable that c02 is even a greenhouse gas in real observable terms.
Slacko 10.33, perhaps I can explain. If you look at the beginning of Ken’s and my conversation I mention something we call “apparent temperature”. It is the blackbody equivalent temperature that something radiates. We use it to determine the reflected radiation in commercial IR measurements. That is also what would be measured in that IR image at the top if the camera was able to measure that low (and emissivity was set to 1 and distance to zero).
So consider the exchange of heat between earth and the sky. It depends on the balance between incoming and outgoing radiation, the net difference is the gain or loss. If you have a very clear sky, the sky will have a low apparent temperature, so the temperature difference between the earth and the sky is the greatest. (You will still have a big influence from the atmosphere, otherwise the apparent temperature of the sky would approach absolute zero. -273C.)
Clouds will have a higher apparent temperature, meaning that they will radiate more towards the earth than a clear sky would do, so the difference is smaller and hence the heat loss is also smaller, than with a clear sky.
So, yes, it is true that radiation from the clouds will prevent cooling of the earth, allowing the earth to keep its heat to a larger degree. But as the clouds will not be hotter than the earth, they will not reverse the heat flow from the earth to become a gain rather than a loss, they can just make it less of a loss. The sun will give the positive contribution.
Slacko says: (February 13, 2011 at 7:47 pm)
…that empty and useless trollbuttal.
On the web: No definitions were found for trollbuttal.
Anthony, Slacko may have coined a new, very expressive, and very useful word here on WUWT.
I hope you will note it with appropriate applause.
Mikael Cronholm says:
February 13, 2011 at 11:12 pm
heat leaves earth by convection. Clouds prevent convection. They don’t radiate energy
Falicoff — Further, your statement that is not possible to achieve a measurement less than the accuracy of an instrument is also not correct, as the accuracy can be increased by taking many measurements (up to a threshold). The resultant accuracy is approximately proportional to the square of the number of readings.
Apples and oranges, and utter nonsense.
What you describe is simple measurement repeatability with an assumed constant source and then averaging out many multiple readings. All modern measurement devices derive their accuracy spec (traceable to NIST etc) in this manner. Of course, I know this because I made measurement equipment for many years… The measurements in question are those of thermometers looking at a different condition each time, not a fixed, constant source, and the result is that the error is the stated resolution of the device.
If an 1880’s era thermometer was good to +/- 1 degree, that’s your error range. You can’t average out 220x 1880’s era thermometers and derive accuracy tighter than +/- 1 degree.
richard verney says:
February 13, 2011 at 4:11 pm
I believe this is a valid question. The response –
Dave Springer says:
February 13, 2011 at 8:14 pm
fails to address the question in ignoring the diurnal cycle of solar energy input for a given location on the Earth’s surface. For CO2 to cause CUMULATIVE warming, the near surface temperature just before dawn for a single point on the Earth’s surface would have to be greater for a local air mass with greater CO2 concentration than for one with less. A simple empirical experiment could clarify this. I would be interested if anyone could point me to the results of such an experiment.
richard verney
You asked what sort of time delay does CO2 impose on the re-emission of the sun’s energy back into space.
I do not know the answer, but suggest we consider what happens when the hot summer ends and chilly winter appears?
What happens when day follows night?
What happens to the measured temperature at individual locations for well over a century, when you are able to adjust for the rising level of UHI?
Not much – that’s my simple answer.
John S says:
February 13, 2011 at 4:05 pm
How can it be explained that ANY increase in temperature does not cause this “thermal runaway” ‘tipping point’? If the temperature as recently as the MWP was higher than today, which it certainly was in at least some areas and possibly all, why was this mythical tipping point not reached? As it was not, what makes you believe it will be during the present warm period? Is there any evidence for this? Models are NOT evidence, BTW, as they ONLY produce what they are programmed to produce.
chico sajovic says:
February 13, 2011 at 7:45 pm
When I am hot I should:
a) stand outside naked on a low humidity night
or
b) get soaked in water and stand in front of a fan
from what we’re told. you should do c) go to a special room that has 280ppm c02, regardless of its temperature
Slacko says:
February 13, 2011 at 10:33 pm
hotrod ( Larry L ) says:
February 13, 2011 at 5:49 pm
QUESTION
It is perhaps a subtle distinction for some folks, but the CO2 and water vapor “back radiation” does not “warm the ground”
Why not? Does it have to do with the 10 micron wavelength?
ANSWER: Because the downward re-reradiated LW is MUCH more likely be be absorbed by water vapour or other absorbers and be re-radiated back up than it is to ever hit the ground. Double CO2 instantly, wait for a new equilibrium to be established, and you have the EXACT same amount of SW going in, and the EXACT same amount of LW coming out. The EFFECTIVE black body temperature of the earth is what the IPCC claims is going to rise by 1 degree, and this is at a point high up in the atmosphere (about 14,000 feet if I recall) not at earth surface.
QUESTION
or the ocean surface,
Why not? Those photons have to go somewhere. Why does this look like a “missing heat” problem?>>
QUESTION
LW radiatiance cannot penetrate more than a micron or so of water before being absorbed. The result being that any longwave that does strike water is absorbed in a layer so thin that it immediately evaporates taking the extra energy from the LW, plus any energy that was already in that water with it into the atmosphere. Any temperature changes in the ocean have to be attributed as a consequence to other factors such as fluctuationm of SW from the Sun, rainfall, runoff and so on. As for “missing heat” that’s a measurement problem.
QUESTION
it only “reduces the rate at which it is cooling by IR radiation” to the apparent surface of the very very cold sky.
So “back radiation … only reduces … cooling.”
Huh??
ANSWER – that is an imperfect way of thinking about it, but at day’s end it is a fair description. A given photon might, in theory, be radiated upward and go straight out to space. Or it might hit one CO2 molecules be absorbed, and then re-radiated in a random direction. Up, sideways, down, what ever. For rough figuring, call it 2/3 up or sideways and 1/3 down. So no matter how much CO2 you have, the end results is always more ups than downs, and the photon eventually escapes to space, but not have hundred, to millions of absorptions and re-radiations. So adding CO2 does not add a single additional photon, not one, to the equation. All it does is increases the average number of zig zags through the atmosphere before eventually escapting to space. So yes, extra CO2 would add no extra heat at all, it would just increase the amount of time it takes any given photon to escape, and “slow down the escape” could roughly be equated to “slow down the cooling”.
Slacko says:
February 13, 2011 at 9:45 pm
Jim D says:
February 13, 2011 at 5:48 pm
The measured IR from the sky is pretty much all emitted by CO2 and H2O molecules …
No Jim. If that were true then the warmth of the sun on my face would be coming in from all directions equally. But even a blind man …
Or do you mean at night?
And do you really mean to tell us that nitrogen can’t radiate IR?
Sorry Slacko but Jim D is right. The sun emits energy in the visible and UV part of the spectrum and a good job it does too. The atmosphere is almost transparent at these wavelengths so unless it gets reflected by clouds or the surface it will make it to the ground where it will be absorbed to warm the earth. The earth on the other hand emits at longer wavelengths with a peak at 10 micron (infra red starts at about 0.8 micron) and a lot of this does get absorbed by the CO2 and water vapour in the atmosphere before being finally radiated into space by the same molecules.
And no, nitrogen does not radiate (or absorb) in the infra red.
On a more general point. The picture at the top of this piece is a bit misleading. When you point an infra red camera at the sky at night you will only capture the infrared radiation from molecules (mainly CO2 and H2O ) in the atmosphere. There is no other significant source of infra red radiation in space to detect. Since these molecules are radiating in narrow bands and a lot of it in the far infrared that the camera may not detect the total energy received by the camera is quite small. As explained in the responses by MC the camera will assume it is looking at a grey or black body with energy distributed across the whole spectrum according to Planck’s law. So it will calculate what temperature a back or grey body would have to be to output that amount of energy. Because the energy in the narrow bands that are detected is then spread out over the whole spectrum that calculated temperature will appear much lower than it really is and the sky will be shown as black.
” 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…”
Love it!
I agree with Berényi Péter with respect to
“For 300 W/m2 radiation I get -23.4°C at 300 W/m2 when I calculate it”
My temperature conversion calculator correlates 300 W/m2 with -3.3°C (assuming an emissivity of one), and -23.4°C with either an emissivity of 73.4% or 220 W/m2. A few lines later
your body will radiate approximately 648 W/m2
implies that your skin temperature is at least 129°F (don’t think so).
To Jim D:
Yes, IR is reflected. The 324 W/m2 back radiation from Kiehl and Trenberth (1997) implies a temperature of 34°F if the emissivity is one, but 58°F (15°C) with an emissivity of 0.8 (-20% in my calculator). Since I don’t think that the atmosphere is able to be the same temperature as the surface (remember, the troposphere cools with increasing height), some of the energy must be reflected and not emitted. Some calculations indicate that even 80% is too high .. in the absence of clouds. On the other hand, since about 50% of the planet is covered with clouds, it sort of makes sense to assume that some of the energy is “reflected” by the cloud bottoms. In fact, on cloudy nights, the hand held radiometers indicate an apparent cloud temperature within 2°F of the surface temperature. In addition to the expected reflection by the droplets themselves, the heat from the surface evaporates droplets at the cloud bottom, which then causes the vapor rise a few inches, where they recondense. For both evaporation and condensation, the associated spectrum is nearly blackbody, with no spectral lines. Notice that this is similar to reflection since the energy emitted is not affected by the temperature of the cloud, but only by the temperature of the surface. (“Scattering” may be a better term.)
For Jim Masterson:
In Kiehl and Trenberth (1997), the downward flux of 324 W/m2 is specifically for cloudy days, it is 278 W/m2 for clear days.
To all:
At a single frequency, the change in absorption is logarithmic. However, when the entire IR band is considered, and over the range from 200 ppm to 500 ppm, the change in CO2 absorption is logarithmic with R^2=0.9988 and linear with R^2=0.9932. This is because as each frequency becomes more saturated, a new frequency starts to absorb.
P Wilson says:
February 13, 2011 at 11:23 pm
Mikael Cronholm says:
February 13, 2011 at 11:12 pm
heat leaves earth by convection. Clouds prevent convection. They don’t radiate energy
Reply: Sorry Mr Wilson, but I do not agree with what you say. (To other readers: I was not the one who made the statement above.)
Convection takes place in substances in the form of fluids and the atmosphere itself is a fluid, so convection can take place in it, including within the clouds themselves (they don’t “prevent” convection, they take part in it). It is true that heat transfers from the ground to the atmosphere, but it is still part of the heat that surrounds the planet. So assuming that we count the atmosphere as a part of our planet, as opposed to space around it, no heat will enter space by convection because there is no fluid that can circulate into space. Clouds do radiate, because everything with a temperature does.
Let us make this clear. The only way that Earth, and its atmosphere, can exchange heat with space is by radiation.
I have been fascinated by this posting – as well as all of the many cogent comments from the insightful experts that have been attempting to answer many of the questions that have been posed in the subsequent comments here.
If I may, I would like to add my two cents worth to clear up some things I have read here, and use this forum as a sounding board/perhaps clarify some misconceptions. Please correct me if I am wrong.
First, regarding reflection, vs absorption/re-radiation. Although, at a molecular level, it may seem they are the same, there is a difference. Corner cube reflection in a prism of glass or reflection that occurs inside a droplet of water generally 1)absorbs incoming light at the first material boundary, transmits through the second material, then reemits the incoming light back into the first material and maintains all spectral integrity and most of the intensity of the incoming light, (polarization may occur) 2)there is very little if any loss of energy EXCEPT at those wavelengths that the second material absorbs light and 3) reflection/refraction occurs where either a phase change, or significant density change is encountered. It isn’t widely understood but, gases can be seen to refract and change the apparent direction of incoming light and this could be considered “reflected” light, but it is nowhere near as dramatic an effect as what occurs in the reflection of nearly all visible light when it encounters many many tiny droplets of liquid water (clouds or rainbows). Two examples of air refracting/reflecting light are the Schlieren waves you see when you look at a distant scene across a hot parking lot or runway http://hiviz.org/hsi/ss/schlieren/index.htm
and the “mirage” you see of a distant city or the reflection of the sky to give the appearance of an oasis of water on a hot desert.
But that is NOT what happens when a molecule like water or CO2 ABSORBS light at their respective infrared bands. The definition of absorbance is A = 1 – R – T where R is the reflectance and T is the transmittance. Absorbance is the loss of light at specific wavelengths of light, and these unique wavelenghts for both CO2 and water correspond to electronic, vibrational, and rotational energy modes that are unique to each molecule. They happen at unique wavelengths because only quanta of energy that match the exact electronic, vibrational and rotational energies for each of these molecules can cause them to absorb and become excited at these quanta of energy. Generally, if a molecule is excited at a specific wavelength, but it immediately gives up that energy and decays back to an unexcited state and thus reemits that photon, it is no different than transmitted light. The resulting photon seems to pass right through the excited->unexcited molecule. Outside of absorbance bands this is usually what happens to both visible and infrared light when it interacts with gases like CO2 and water. However, at very short UV wavelengths light absorbance/scatter has the effect of making the absorbing gases appear to “glow” in the visible spectrum – Raleigh scattering and Mie scattering – which is why the sky appears sky blue, but at sunrise or sunset appears to have a rainbow of colors, and why the sun appears yellow.
At absorbance bands, however, scatter, transmittance and reflectance don’t often happen. The incoming photons with quanta of energy that match absorbance bands of any given gas molecule are absorbed by that gas molecule and it is excited up into a higher electronic or vibrational, or rotational energy state, and if or when it reemits that energy it necessarily must experience some LOSS – entropy gets it’s cut. Because the molecule has an “affinity” for quanta of energy that correspond to its natural “frequencies”, it is also very rare that light that is “absorbed” at a given wavelength is reemitted at that wavelength. If it were you wouldn’t see the loss of light – absorbance – at that wavelength. Instead, energy that is absorbed at one quanta is usually released only when 1) even more energy is pumped into the molecule at the same quanta (rare), or 2) it reemits that light at a longer wavelength (fluorescence – does CO2 fluoresce in the infrared? I’m not sure but I doubt it.), or 3) the molecule collides with another molecule that has less energy (the usual case). When collisions occur, (very frequently) the absorbed quanta of energy is transferred between the two molecules so that they balance out – generally both of them just pick up more translational energy – they move a little faster.
In general devices that “measure” the temperature of a gas or a liquid actually transfer (come into equilibrium with) the translational energy of that gas or liquid to a visual media – indicator liquid that expands in a fixed volume – or transfer the translational energy to an equivalent vibrational energy in a solid material which is used to derive an electronic (thermistor, thermocouple etc.) or visual signal (bimetal). So an observed “temperature” of a gas is an average of all of the molecule’s translational energies in a given volume that interact with the surface of the measuring device.
Now having said that, I would like to make the following ascertions:
1) As stated by several previous posters, water has many more, and much broader absorbance bands in the infrared than CO2,
2) Water vapor, and water droplets are much, much more abundant in the lower atmosphere than CO2,
3) Even at the now greater concentration of CO2 in the atmosphere the relatively small amount of IR energy it might absorb is quite quickly transferred during collisions to the much greater abundance of water liquid, water vapor, N2, or O2 molecules,
4) Water vapor, water liquid, and ice all have greater heat capacity than any of the other gas molecules in air. Liquid water’s heat capacity is significantly greater!
5) Because water experiences phase changes and significant density changes, it is redistributed throughout the atmosphere in ways that have a significantly greater impact on the transport of energy from the lower atmosphere and planet’s surface to outer space than any of the other gases that make up our air,
5) Since water vapor regularly experiences both gas to liquid, liquid to solid, and even gas to solid transitions in the atmosphere:
a) At any given instance the loss of incoming solar radiation energy (on the day sid of the planet) by direct reflection, refraction and back scatter, by clouds, snow, rain and the oceans are significantly greater in magnitude than all of the IR absorbed by CO2 combined.
b) At any given instance the total amount of incoming solar radiation energy that results in phase change of water liquid to water vapor, greatly and significantly exceeds all of the IR energy absorbed by CO2 combined.
c) At any given instance the total amount of energy contained in water vapor, water liquid droplets (clouds) and water ice particles (high clouds) in the planet’s atmosphere by a huge amount dwarfs the total amount of energy contained in IR excited CO2 gas molecules, so much so that I would think (I have no proof) that most of CO2’s translational energy content is created more by collisions with water molecules than IR photons by a very large ratio.
d) And finally and most imortantly, changes in the concentration of CO2 that have been observed have gradually been increasing over DECADES. But in any given DAY the concentration of water vapor, water liquid and ice particles in the atmosphere change by orders of magnitude greater concentration than the few hundred parts per million that CO2 has changed. These significant concentration changes even at the poles have a far greater effect on both immediate weather and of course long term climate. Alarmists would have us believe that water cycle does not now have the capacity to mitigate and regulate the small change in CO2 concentration that have occurred after centuries and centuries of significantly higher CO2 concentrations and significantly LOWER CO2 concentrations in the past.