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…
Could someone (obviously way smarter than I) post a brief summary of what I learned reading thru all this, comparatively to what I would’ve learned (had I understood any of it) reading the paper that contains the ‘notorious Ternberth/Keihl energy balance schematic’. Perhaps I just missed it but what was the definitive answer to, ” . . . does this aspect of the global warming theory make any sense?”
MC had issues with ‘the numbers’ used in fig1 etc. and I was hoping among the learned posters here that some discussion/debate over those numbers would lead to an understanding of what numbers were most likely to be considered real or reasonable.
-Looking for knowledge in all the right places . . .
-Barn
Hello Izen
Quote “There is no way that the water surface acts like a one-way street for all incident radient energy. It is not Maxwell’s demon!”
But CO2 is a demonic gas, with its wavenumber of 666.666 (15 µm)!
Is it not Maxwell’s demons casting back all the IR radiation?
The blackbody temperature to emit a peak IR of 15 µm is according to Wiens law 193.18456 Kelvin.
This equates to -79.97 Centigrade (Please do not miss the MINUS sign, yes nearly minus 80C!)
How hot can this 15 µm -80 Centigrade IR heat anything?
NoIdea
Bomber_the_Cat says:
February 14, 2011 at 5:59 am
Thanks for the lesson on radiative physics theory – i’m quite aware if it. (The theory, that is)
Which is all that it is. A theory.
In theory, I receive energy from my surroundings, but they never surpass my basal metabolic rate of 58wm2. Similarly I transmit to cool night air energy at this basal rate, although only to the immediate few inches at most, and this energy soon thermalises. (disappears as the form of heat. Heat is not a permanent)
There’s little point in invoking the rather absurd SB equation, which is a thought experiment that leads to rather absurd conclusions.
The c02 blanket is a non existent phonomenon. Its a theoretical phenomenon – since it paints a human face on a climate system, just like Big Bang paints a human face on the unknown origins of time and the universe.
It was established well before AGW ideology that c02 delays the transit of 8% of total IR radiation, regardless of its quantity. As you know, as 14.77 microns the saturation window closes. 14.77 microns does broadly correspond to -28C. 10 microns corresponds to around 15C, the average global temperature. So the last place that c02 absorbs radiation is at the surface, which on average is above subzero. C02 is invisible to these temperatures.
In climatology the fix is made by saying that something is going up in the upper levels of the lower troposphere where indeed these subzero (C) temperatures occur and where c02 captures radiation at its peaks. Only it really isn’t, and what radiation is absorbed at subzero temperatures is quickly released.
There is no radiative mechanism by which -28C can increase to 15C or upwards from a base point of -28C. Unless, of course you increase the heat source (The sun). But then c02 becomes invisible again to SW radiation (It absorbs/delays at subzero).
if you want to prove that an ice cube will heat a cup of coffee -a tepid one, to give you the benefit of the doubt (which is the gist of your argument – that used by alarmists to prove the c02 conjecture) then we’d all be grateful for your video test results)
Dave Springer says:
February 14, 2011 at 7:01 am
only the problem with this theory is that the radiation goes in all directions equally, which does not change the amount of heat in the atmosphere.
NoIdea says:
February 14, 2011 at 7:29 am
“The blackbody temperature to emit a peak IR of 15 µm is according to Wiens law 193.18456 Kelvin.
This equates to -79.97 Centigrade (Please do not miss the MINUS sign, yes nearly minus 80C!)
How hot can this 15 µm -80 Centigrade IR heat anything?”
reply
oops. I’m way out on the temperature range at which c02 captures energy.
Bomber_the_Cat says:
February 14, 2011 at 5:59 am
Nice. It’s heartening to see some commenters who understand the physics. “School-boy” misunderstanding might be a bit harsh but I understand where you’re coming from. A good understanding of the four years of science taught in grades 9-12 covers almost everything you need to know about this subject in biology, chemistry, and physics from how the greenhouse gases work to what acid rain from sulfate emissions does to how these thing effect the biosphere. When PhDs get involved in the conversation they start such a pedantic bickering over small details in their fields of expertise that the big picture gets lost and laypersons then proverbially can’t see the forest because there are too many trees blocking the view.
@-P Wilson says:
February 13, 2011 at 11:23 pm
“heat leaves earth by convection. Clouds prevent convection. They don’t radiate energy”
Some energy leaves the SURFACE by convection, but it does not get very far.
It is not clouds that block it… strange idea!
Its the adiabatic lapse rate.
Convection can only move air until the increased bouancy from the lower number of molecules per cubic metre is offset by the lower density with increasing altitude. It is the lower density of the atmosphere with altitude that blocks convection.
The low altitude of most clouds indicates that the temperature falls below freezing just a few Km above ground level and the atmosphere regains that latent heat of evaporation well below the tropopause. Above the main low cloud layer water vapor is much less important as an atmospheric absorber/emitter.
P Wilson says:
February 14, 2011 at 7:30 am
” As you know, as 14.77 microns the saturation window closes. 14.77 microns does broadly correspond to -28C. 10 microns corresponds to around 15C, the average global temperature. So the last place that c02 absorbs radiation is at the surface, which on average is above subzero. C02 is invisible to these temperatures.”
This is laughably wrong! PEAK emission frequency corresponds to those temperatures. A blackbody emits a CONTINUOUS spectrum with broad shoulders that fall off slowly on either side of the peak frequency. There’s plenty of 15um energy coming from a blackbody with 10um peak frequency. You obviously have no idea what a continuous blackbody spectrum looks like. What you’re saying is like saying no colors of visible light come from sun except for yellow because that’s the peak emission frequency of a 5200K blackbody source. Ridiculous misunderstanding.
Hello NoIdea
as usual you ask a deep question.
Any blackbody emitter above -80 Centigraqde will ALSO emit more energy in the 15 µm band, as well as more at shorter wavelengths.
But CO2 is NOT a blackbody emitter. Like the other atmospheric gases, including water vapor it is a very poor emitter over most of the spectrum. It just has strong bands of emission/absorption like water vapor at wavelengths related to its molecular vibrational modes.
Very hot CO2 will still have a peak emssion spectra in the 15 µm band becuase that is its surface ‘colour’, it does not have a blackbody emission/absorption spectra but one vastly biased to the energy level of the molecular vibrational modes.
In the Infra-red longwave spectra CO2 has a surface ‘colour’ that modifies its emission spectra; think of it as a ‘Deep Purple…..’
-grin-
Myrrh says:
February 14, 2011 at 4:06 am
Cal says Feb 14, 12:49 am, Slacko says, Jim says
Sort it out, will you.
http://www.wisegeek.com/how-does-infrared-heat-work.htm
“To observe how infrared heat works, the heating and cooling of the earth is probably the most helpful example. Apart from driving all weather events and patterns, as well as ocean currents, infrared heat from the sun is what the Earth during the day, and it is what the ground gives off at night after the sun sets. It is interesting to note that while the sun does emit a huge amount of infrared heat, only about half of the heat we feel on the ground is from direct infrared radiation from the sun. The other half comes from energy from visible light that is absorbed by objects on earth, and then emitted later as infrared heat.”
So which is it?
I accept that I should have said that the sun radiates at all wavelengths from UV to far infrared, but that the peak is in the visible region. Moreover a lot of the incoming energy in the infra red region is absorbed by CO2 and H2O in the atmosphere and is then re-radiated downwards (and upwards ultimately to space) in exactly the same way as the energy radiated by the surface. UV is also absorbed (particularly by ozone) but the peak energy at the surface is still in the visible region of the spectrum. So at night all the infrared radiaton is from CO2 and H2O and during the day the majority is. However the main point I was making was about infra red radiation at night.
Tsk Tsk says:
February 13, 2011 at 9:02 pm
Not sure I like that. The form of the equation shouldn’t change for different numbers unless there’s a discontinuity. It has to be logarithmic over the entire range of positive real numbers and obviously isn’t physical for negative numbers. I think you mean that for low concentrations of Co2 the dependence diverges only slightly from a simple linear dependence and the divergence grows as the concentration grows.
No need for a discontinuity, if you expand the terms in the equation for small [CO2] you get a linear dependence, for medium values you get √ln([CO2]) and for large values √[CO2]. For the range of values in the atmosphere ln[CO2] is a good fit.
Izen.
Thanks for the reply.
Clouds keep the earth warmer (or a given portion of the earth) by preventing heat from escaping. In fact, convectional currents can rise a long way into the atmosphere before they lose their buoyancy. Something like a hot air balloon does.
This uplift, and cooling of air is what causes most weather. Cumulonimbus clouds are often 10,000 metres in height from surface.
actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect.
eadler says
“If the mean error is know, the readings can be adjusted. If the mean error remains constant, than the temperature anomaly, which is what is being sought will not suffer in accuracy.”
And those “if’s” can be significant. Much of the mean error “adjustment” ability depends on the sensor, electronics, environment, calibration protocols, and how well these protocols are ACCUALLY followed.
My first “outside” job, over 60 years ago, was taking the hi-lo temperatures, for a neighbor, on a old Taylor mechanical thermometer. It had mechanical slides that indicated the hi-lo points, and if I remember, 2 deg. graduations. The neighbor then sent the data, to some government agency. At $0.50 a week that was big money back then. A long way from 3/4 wire platinum RTD’s
Falicoff’s
comment about taking multiple readings may “enhance” the resolution, or quantization, but not necessarily the accuracy. This is also known as “resampling”, as denoted in p.334, “The Handbook of Astronomical Image Processing”, Berry & Burnell.
P Wilson says:
February 14, 2011 at 8:04 am
“How hot can this 15 µm -80 Centigrade IR heat anything?”
Because it sits between a 15C ocean surface and the -253C of the cosmic void. It doesn’t heat anything. It slows down the rate of cooling by interposing something warmer than the black of space between the surface and the black of space. This isn’t rocket science. It’s about as difficult to understand as is understanding why sitting a cup of hot coffee on a block of dry ice versus a block of water ice. In both cases the coffee won’t get any hotter but it will cool a lot faster sitting on the dry ice versus sitting on the regular ice. Or even better hot coffee in an insulated thermos vs. a non-insulated thermos. In both cases the coffee is going to get cooler but the insulated thermos will slow down the rate of cooling. To get an even closer analogy consider an insulated versus and uninsulated hot water heater where the heating element is turned on once a day for a set period of time. The insulated water heater will have a higher temperature because the amount of energy added to each one is equal but the rate of escape of that added energy is lower for the insulated vessel. The end result is the water in the insulated vessel will have a higher maximum and minimum daily temperatures. In the case of the ocean it’s the sun doing the heating on a daily basis and greenhouse gases are the insulators which slow down how fast the ocean cools when the sun isn’t heating it. As Ernest Rutherford said “if you can’t explain a theory in physics such that a bartender can understand it then the theory is probably wrong”. A bartender can understand the difference between an insulated and uninsulated thermos. Why can’t you?
Bomber_the_Cat says:
February 14, 2011 at 5:59 am :
“If that radiation impacts a warmer object then it is absorbed by the warmer object. The warmer object therefore receives energy that it wouldn’t receive if the cold object was not there – and so the warm object is kept warmer than it otherwise would be. Radiation is not somehow preferentially attracted only to colder objects. Thus the presence of cold objects can keep warm objects warmer!!! This does not infringe the 2nd law of thermodynamics, only a schoolboy misunderstanding of it.”
———————-
Hoping to have eluded the Groundhog Day Effect that your comment was caught in, I’ll remark that the warmer object will now be inclined to radiate a lot more energetically than it otherwise would have.
At lower troposphere densities it will be even more inclined to pass on the energy, through collisions, to non-absorbing species which then convect upwards.
When your feet are cold, it takes a painfully long time for them to warm up by merely putting on wool socks. If you change those socks out repeatedly for a colder pair, your toes will turn blue.
It’s not hard to understand that GHG’s provide the air with warmth, but it seems the ground is not really significant in the back-radiation scenario. In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.
“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”
Is that so? What if we had a month off for CO2 emissions as a globe, with CO2 output deliberately reduced by say 30%. This would create a point of inflexion in our CO2 output which in theory should then be followed by a point of inflexion in rising temperatures, thus proving AGW theory and demonstrating cause and effect. It would then allow us to go further and estimate the actual impact of given increases in CO2. Got to be worth doing just as a global experiment before making any solid commitments to changing our habits.
(Well in theory, in practice I guess it would just get lost in the noise of normal weather patterns which would only prove the futility of the whole AGW nonsense)
P Wilson says:
February 14, 2011 at 8:35 am
“actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect”
Water vapor is invisible. A cloud is composed of water droplets not water vapor. Water vapor coming out of a tea kettle is invisible until it has cooled enough to condense into water droplets. This is barely high-school level physical science. You need to go back to high school and relearn this stuff and that’s presuming you ever learned it in the first time through which at this point is a matter of grave doubt. You’d do miserably on the TV show “Are you smarter than a fifth grader?”
I didn’t mean to suggest that heating the surface more would cool the air!
Just that returning energy to the surface provides more of a direct avenue for escape than if it relied only on radiation from GHG’s.
Dave Springer says:
February 14, 2011 at 8:21 am
Largely irrelevant comments re c02 though correct ones regarding the sun. However, the argument is that c02 is largely irrelevant to the climate, since most heat escapes via convection and conduction, hence the quick cool down at night. this cooldown, and corresponding cool upper layers show just how quickly this takes place. Nothing to do with the SB equation. To do with observed phenomena
the main reason is that there is no heat being generated, or stored by greenhouse gases, and so back radiation doesn’t occur. Even if it did, the quantities would be almost infinitessimal.
In your later analogy however, it is convection that is being vastly reduced in a thermal flask, which prevents quick cooling.
this convectional trapping process does not occur with outgoing IR radiation through c02 or any other ghg.
however, even at ground-10,000 metres, what c02 does is so infinitessimally small that it might as well be factored out of climate projections.
at 15microns peak, that makes not an iota of difference to atmospheric temperatures. Besides, gases are not blackbodies, and are 3 dimensional. Radiative equations can’t be applied to them (air has a very poor conductor), and so is irrelevant to climatology. With ghg’s, *blocking* operates in all directions, so doesn’t make any difference to the atmospheric heat content.
if you’re talking about the shoulders of c02, then at this level there is more nitrogen and oxygen per CO2 molecule in this area. Dilution reduces the temperature increase per unit of energy.
Dave Springer says:
February 14, 2011 at 8:54 am
i said
“actually, watervapour above this level acts as an absorber of radiation. the so called greenhouse effect”
Water vapour is indeed invisible.
the above comment was a contradiction of Izens, who said that water vapour higher than clouds have no effect. He said “Above the main low cloud layer water vapor is much less important as an atmospheric absorber/emitter.”
I think you’ve got it muddled somewhere, although if you wish to trace a source of the confusion between clouds and water vapour, I suggest you consult Izens first.
An interesting dialog; but perhaps KLC didn’t ask the right questions, so in turn MC didn’t really provide the needed answers; although what MC DID say is not incorrect.
One issue raised in the discussion was the 100 Watt light bulb, versus the 300 W/m^2 from the sky. the first one feels “hot”; while the second one doesn’t. MC says that the 300 W/m^2 corresponds to -23.4 deg C. That ties in well with Trenberth’s 390 W/m^2 corresponding to +15 deg C or 288 K.
But here’s what MC didn’t tell KLC; and we have all seen it with that 100 W light bulb “radiating” onto two samples of dry air; one containing more CO2 than the other. Well “The Science Guy” Bill Nye, has performed that quite fraudulent experiment in public.
Here’s what’s wrong with it and what MC forgot to tell KLC.
The 100 Watt light bulb, is a tungsten filament lamp (likely gas filled), and it will emit a roughly black body like spectrum (thermal spectrum); and MC explained how near BB and real BB sources can be related via an emissivity; and of course most practical sources, are not a singl4 Temperature source so they will be somewhat spectrally discombobulated as well.
But back to that 100 Watt lamp. Which is fairly typical of the so called “warm white”light source. It will radiate a spectrum that is something like a 28-2900 deg thermal spectrum; ie a somewhat BB like source of radiation with a color temperature of about 2900 K. THAT IS HALF THE TEMPERATURE OF THE SUN SURFACE.
The sun emits a thermal (BB like) spectrum at about 64 million Watts per square metre (at the sun’s surface). Our innocent 100 Watt lamp, at half the temperature is only emitting 4 million Watts per square metre, since that goes as T^4, per S-B.
More importantly, the 100 Watt “light bulb” is ten times the Temperature of the 288 K Trenberth average Temperature of the earth, and the presumed source of that 390 W/m^2.
So the incandescent light bulb is ten times the average earth Temperature, and is radiating 10,000 times as much energy per square metre as the average earth.
So now we kick in with the Wien Displacement Law that MC also mentioned.
At 288 K (+15 deg C or +59 deg F), the LWIR radiation spectrum, that is emitting a total 390 w/m^2, has a spectral peak wavelength at 10.1 microns. And assuming a single Temperature source, 98% of the LWIR radiation is contained between 1/2, and 8 times the peak wavelength so from 5.0 to 80 microns, is the spectral range of the average earth Temperature roughly BB spectrum. So colder places, will radiate even longer wavelengths, but being colder, they also are radiating even less, so they don’t really contribute much energy on the long wave end. For the highest surface desert Temperatures which can be above +60 deg C, the spectral peak could be as low as about 8.8 microns from the Wien Law, so that will radiate down to maybe 4.4 microns, and also the total S-B emission can be 1-8-2.0 times as high as Tenberth’s 390 W.m^2. So it is those hot dry desert regions that are the principle radiative coolers of the earth. The 8.8 micron peak is even further away from the 15 micron CO2 band, and the peak moved further into the “atmospheric window” where water vapor is somewhat benign. Well there’s little water vapor in those arid deserts anyway. But don’t forget that Ozone kicks in a dip at around 9.6 microns; but it is quite narrow, because of the height, and low density and Temperature of the ozone layer (less line broadening)
Now back at our phony heat lamp at 2880 K,; not only is it emitting 10,000 times the emittance of the average earth; but the spectral peak is not 10.1 microns either, but is now 1.0 microns.
Well wouldn’t you know it; water (H2O) has several absorption bands near there at 0.94 microns, and also at 1.1; and supposedly humans are 95% H2O.
So the peak radiant emission from a 100 Watt light bulb is at the correct wavelength to cause strong absorption in human flesh; no wonder it feels warm.
The 288 K mean earth surface on the other hand is radiating at 10.1 microns, and the water absorption coefficient there is about 1000 cm^-1; which gives a 1/e transmission depth of 10 microns; or a 99% absorption depth of 50 microns; 2/1000 inches.
It doesn’t even make it through the surface of your skin; and even if it did, it wouldn’t register against your body Temperature of 98.6 deg F or 37 deg C.
So no wonder the 100 Watt lamp feels warm to your skin, and no wonder it warms the air samples. CO2 has absorption bands at around 2.7 microns, and I think also at 4.0 microns. The 2880 K lamp spectrum is going to put 98% of its energy in the spectral range from 0.5 Microns (green) to 8.0 microns, so both of those CO2 bands are going to be activated. And even the 15 micron band will absorb much more energy from the 2880 K source than from the 288K source.
One should also note that although the S-B equation says the total emittance goes as the 4th power of temperature; the spectral peak emittance actually goes as the fifth power of the Temperature (T^5).
I’ll leave it to the math geeks out there to figure out how much more spectral emittance you get at 15 microns, from a 2880 K sourcves compared to a 288 K source (Watt’s per m^2 per micron wavelength.
So like I said the 100 Watt lamp demonstration is a total fraud. A much better source fo 300-390 W/m^2 LWIR radiation that is properly spectrally peaked at about 10 microns, would be an ordinary 16 ounce bottle of water. I don’t recommend the French Evian water; it’s too expensive and doesn’t radiate any more than any other water; use tap water it’s cheaper.
“”””” P Wilson says:
February 14, 2011 at 9:10 am
Dave Springer says:
February 14, 2011 at 8:21 am
Largely irrelevant comments re c02 though correct ones regarding the sun. However, the argument is that c02 is largely irrelevant to the climate, since most heat escapes via convection and conduction, hence the quick cool down at night. this cooldown, and corresponding cool upper layers show just how quickly this takes place. Nothing to do with the SB equation. To do with observed phenomena
the main reason is that there is no heat being generated, or stored by greenhouse gases, and so back radiation doesn’t occur. Even if it did, the quantities would be almost infinitessimal.
In your later analogy however, it is convection that is being vastly reduced in a thermal flask, which prevents quick cooling.
this convectional trapping process does not occur with outgoing IR radiation through c02 or any other ghg.
however, even at ground-10,000 metres, what c02 does is so infinitessimally small that it might as well be factored out of climate projections.
at 15microns peak, that makes not an iota of difference to atmospheric temperatures. Besides, gases are not blackbodies, and are 3 dimensional. Radiative equations can’t be applied to them (air has a very poor conductor), and so is irrelevant to climatology. With ghg’s, *blocking* operates in all directions, so doesn’t make any difference to the atmospheric heat content.
if you’re talking about the shoulders of c02, then at this level there is more nitrogen and oxygen per CO2 molecule in this area. Dilution reduces the temperature increase per unit of energy. I don’t know about other readers; but I have absolutely no idea who said what, in this joint effort by P-Wilson, and Dave Springer.
Perhaps, P. Wilson, being the outermost poster, can find some simple methodology to separate what he is citing from Dave Springer (if anything) and what is the product of his own thoughts.
I often see people ascribe whole rafts of stuff to me; when I never said one word of what they posted.
Trying to italicize things doesn’t do a thing, since the browsers can make mincemeat out of font twiddling.
So I dunno P. Wilson; just what did you add if any to what Dave said if any ?
Don V says:
February 14, 2011 at 1:47 am
Thanks.
Oliver Ramsay says:
February 14, 2011 at 8:51 am “In fact the warmer the surface becomes the more the energy will be radiated to space without bouncing around in the atmosphere.”
WUWT?
So the surface has a direct connection to space then?
Can I ask all the Physics guys a simple question?
What happens when 2 photons travelling in opposite directions collide?