Visualizing the "Greenhouse Effect" – Light and Heat

Guest Post by Ira Glickstein

Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).

My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)

Solar "light" energy in is equal to Earth "heat" energy out.
[Click on image for larger version]

As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.

The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)

I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).

Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?

The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.

Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.

The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.

DETAILED EXPLANATION

Left: Actual Solar radiation spectrum observed at top of Atmosphere, compared to black body model. Right: Black body Earth System radiation spectrum out to Space.

The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.

If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.

However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:

  • The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
  • The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.

After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.

The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.

Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)

The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.

However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.

NOTE ABOUT THE ABOVE ILLUSTRATION

WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?

Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:

  • The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
  • This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
  • The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
  • So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
  • Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.

But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)

ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE

First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.

So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.

Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.

Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.

Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.

Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.

Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.

Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.

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jae
May 26, 2011 7:18 am

Jim”
“You have obviously never sat out on a car hood till the late hours (with a member of the opposite sex) have you? And witnessed TRUE radiative cooling of the car’s hood, roof, until they eventually ‘dew up’ … the reason why FOG FORMS (surface radiative cooling).”
Oh, yes I have! And nothing you say is wrong. But I don’t see what that has to do with the subjects at hand.
Why do so many folks here think that I somehow deny the existence of radiation? Hell, I’ve used IR innumerable times to identify chemical compounds!
I just don’t think the GHE, however you want to define it, is caused by a radiative effect. It’ s essentially a simple heat storage phenomenon and is explained well by Siddons and some other links provided above. The radiation FOLLOWS from all this; it doesn’t CAUSE it.
You simply cannot have the sun furnishing an average of 342 wm-2 into the Earth system and then say that the system somehow “magnifies” that amount of energy to create a surface that emits at at an average of 390 wm-2. And it seems to me that that is exactly what is being proposed.
Folks, like Ira, keep saying that the GHE is just a result of “slowing down” the loss of radiation. If that was all that was being proposed, I would have no objection (although I would again say that radiation is only part of the mechanism). But that is NOT all that is being proposed!

May 26, 2011 7:31 am

So far, none of our experts out here in WUWT land – or anywhere else I have looked have made what I can call “a good job of explaining” as to how it is possible for energy that is transported away from the surface and back again via GHGs manages to warm the surface. (It can warm GHGs, yes but the atmosphere does not warm the surface, it cools it.)
Energy can neither be destroyed nor created and that’s the natural law that matters most in this case. – In other words if the total “wattage” exchange at the top of the Atmosphere (TOA) say 340 W/m² incoming radiation matches any outgoing W/m² radiation + reflection, as it does in K&T’s plan, then I cannot see why we must not (all of us) accept that there is neither warming nor cooling. The only thing that any radiation therefore can do is similar to what conduction, convection and advection do.
The difference is that radiation transports energy (that can facilitate work which creates motion which can create friction which does create heat or heat-energy, as it is sometime called, the strength of which can be measured by its temperature.

I know Ira has “disconnected” because I once said I do not “believe” that W/m² IR radiation contains heat, but even so he did a good job of explaining what “Kinetic energy” is. And I still do not believe “molecules in motion” can be transported by radiation.
Oh, – and Jim did you ever try to drive your “passion wagon” with the cooling water drained from the engine? If you believe LWIR radiation is anything but a minor player inside the earth system then, go ahead; drain away any cooling-fluid and let radiation do the job. – Or maybe perhaps rethink your “radiating away from the hood” idea.

Robert Stevenson
May 26, 2011 7:49 am

Phil says:
‘Where do you get your value of 0.2 emissivity from, ‘
Further to our recent posts concerning the above and Hottel; 0.2 (or 0.193) is the emissivity/absorptivity of CO2 in the atmosphere to the extinction point of 3600m for dry air.
Phil, what do your calculations give for the distance in which the first-generation photons from the Surface up into the atmosphere are absorbed to extinction in CO2’s case?

Robert Stevenson
May 26, 2011 9:34 am

Joel says:
‘Good God, are you guys blind? How many times do we have to debunk Postma’s nonsense?
http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat/#comment-655686
Powerful arguments Joel, but the GHE involving back radiation is just a theory right and I’m sure you could demolish it if you put your mind to it. Incidentally what is the magnitude of the emissive power of the atmosphere at 5km altitude?

Bryan
May 26, 2011 10:45 am

Joel says:
‘Good God, are you guys blind? How many times do we have to debunk Postma’s nonsense?
Now pay attention!
For Joel is an expert in nonsense.
He is a co author of an embarrassing collection of nonsense.
Read some earlier posts.
Halpern et al sets a new low standard by which all previous and subsequent nonsense will be judged.

JAE
May 26, 2011 1:16 pm

Joel waves arms again:
“Good God, are you guys blind? How many times do we have to debunk Postma’s nonsense? ”
And then provides links to….more armwaving… and more armwaving….and….
ROFLMAO.
Joel, you not only didn’t “debunk” anything, it appears you simply do not EVEN understand the issue. All you did is mumble about what if the planet had no ghgs, or something. I could not even understand what your point was.
OF COURSE, the radiation at 5 km is from what you are calling “greenhouse gases” (i.e. IR-active gases). So what?? The so called “black body temperature of the earth” (-18 C) is actually the temperature at 5 km (both measured and calculated). Because of the LAPSE RATE (NOT GHGs!), that means that the surface HAS to be about 15 C. The lapse rate has nothing to do with GHGs, depending only on heat capacity of the air and gravity. That is about as basic as physics gets.
Good God, did you say somewhere that you are a physicist?

JAE
May 26, 2011 1:24 pm

Perhaps another way to state the obvious is: Planet Earth stores just enough energy, as measured by an average temperature at the surface of 15 C, to support a column of air about 30 km deep.

Joel Shore
May 26, 2011 4:06 pm

Jae says:

OF COURSE, the radiation at 5 km is from what you are calling “greenhouse gases” (i.e. IR-active gases). So what??

So, the fact that the greenhouse gases raise the effective radiating level from 0 to 5km, together with the lapse rate in the troposphere, is what gives the greenhouse effect.
Do you learn in getting a PhD in chemistry how the equation for a line involves both a slope and a constant “intercept” term and how both are needed to determine y for a given x?

That is about as basic as physics gets.
Good God, did you say somewhere that you are a physicist?

Yes…which is apparently why I understand this stuff and you seem incapable of doing so (although most other chemists are intelligent enough to too). And, it is why all but a few crackpots or active deceivers in the physics community accept the greenhouse effect as reality. This includes even hard-core skeptics like Fred Singer and Will Happer.

Joel Shore
May 26, 2011 5:21 pm

Just a point of clarification on the last post: It is ALL of the IR-absorbing atmospheric constituents, i.e., greenhouse gases and clouds, that raise the effective radiating level from 0 to 5km.

wayne
May 26, 2011 7:30 pm

I just learned that even though the 390 Wm-2 shown on K&T graphics may be correct in a one-dimensional world, in an actual three-dimensional world, that is really only 130 Wm-2 radiating upward in the purest sense. The other 260 Wm-2 comprises of the horizontal two dimensions. When you integrate the three components across the hemisphere and account for random radiation in all directions from a flat surface like an ocean what you end up with is just that amount. In addition, due to horizontal homogeneity and hungry absorbers present in our atmosphere, all horizontal energy transfer cancels leaving the 130 Wm-2 upward. In this light the downward “back-radiation” would be 324-260 or 64 Wm-2 for the same reason (think cubic) leaving the NET upward to be 130-64 or 66 Wm-2.
The cancellation is the same thing that occurs between two opposing walls right there in the room you are sitting. Each wall has the power to transfer a great amount of energy if the other wall was 3 K, but that power to transfer never manifests due to the equal temperature and symmetry.
Of course, if the atmosphere could not absorb any radiation at all as a nitrogen atmosphere, the horizontal radiation would proceed right through the entire atmosphere unimpeded to exit the atmosphere in all directions. There would be no cancelling of transfer for there would be no absorption.
Finally, it is all falling in place and the now corrected numbers are actually making some sense.
We keep ignoring the other two dimensions in radiative transfer, nary a mention of them!
BTW: Does anyone know roughly how much energy can be radiated by 1.225 kg of air at 288 K as a cubic meter at sea level and at average relative humidity to empty space? If you have that, how much if much higher in altitude (like ~100 grams in the cubic meter as it would be near the 65-mbar level at 217 K, little humidity). What I’m getting at is it takes mass to radieate so how does the mass itself affect radiation rates.

Jim D
May 26, 2011 8:31 pm

JAE seems to agree that without greenhouse gases the surface would be -18 C, and with them it is +15 C. This, by definition, is the greenhouse effect, however you arrived at it. Postma, too.
wayne, your cube analogy is not correct. 390 W/m2 is the average outward component from a spherical surface. There is no sideways component to this number. It is equivalent to a surface radiative temperature of +15 C, which as you see from above is completely consistent with the greenhouse effect that JAE and Joel explain.

May 27, 2011 6:27 am

Robert Stevenson says:
May 26, 2011 at 7:49 am
Phil says:
‘Where do you get your value of 0.2 emissivity from, ‘
Further to our recent posts concerning the above and Hottel; 0.2 (or 0.193) is the emissivity/absorptivity of CO2 in the atmosphere to the extinction point of 3600m for dry air.
Phil, what do your calculations give for the distance in which the first-generation photons from the Surface up into the atmosphere are absorbed to extinction in CO2′s case?

Yesterday’s post didn’t seem to make it so I’ll try again.
It depends on the wavelength, for the core of the 15 micron band less than 1000m but for the outer bands clearly more. The Q branch less than 100m.
http://i302.photobucket.com/albums/nn107/Sprintstar400/CO2spectra-1.gif
[As Phil demonstrates, it is best to always save a copy of your comments until they appear. WordPress is not always 100% reliable. On occasion I have also lost comments. ~dbs, mod.]

JAE
May 27, 2011 8:59 am

Well, Joel:
” And, it is why all but a few crackpots or active deceivers in the physics community accept the greenhouse effect as reality.”
IMHO, the jury’s still out regarding just who the crackpots are. You would have made a good pope back in Copernicus’ time.
But, to move beyond your insults, arm-waving, and fiats. There must be some other factors besides GHGs which determine the “effective radiating level,” like maybe convection. First, it is true that all radiation would have to come from the surface if there were no ghgs, but it is also likely that the atmosphere would be extremely hot, due to conduction/convection. The only way the gases could get rid of heat would be to conduct it back to the surface, but convection would try to prevent that. It would certainly be weird.
Second, there must be something else that affects the “effective radiating level” besides the types and quantities of ghgs. Maybe it’s density. As Alan Siddons has demonstrated in his paper, forget exact title, but something about moon ghe, almost all planets with atmospheres have an “effective radiating level” at about 0.1 bar, irregardless of the type(s) and amounts of ghgs present. The Earth’s level is lower, probably because of the unique properties of water.
Despite your convienient theory, you have no proof for your conjectures about the ghe. And the empirical evidence regarding increasing CO2 levels is not helping you any, either (no warming).
BTW, you didn’t respond to my reasoning for why the “positive feedback” part of the CAGW nonsense doesn’t make any sense.

JAE
May 27, 2011 9:02 am

Jim D:
“JAE seems to agree that without greenhouse gases the surface would be -18 C, and with them it is +15 C. This, by definition, is the greenhouse effect, however you arrived at it. Postma, too.”
Yes, but it’s still fun (for me, at least) to try to figure out just how it works. It’s unfortunate that I seem to be raising the blood pressure for certain folks, though. 🙂

Robert Clemenzi
May 27, 2011 9:13 am

The Hottel analysis uses an average “grey gas” emissivity. That is, it ignores the spectral lines and simply assumes that it absorbs (or emits) the same at all frequencies. As a result, it should not be used for more than 100 meters or so. Using the Hottel graph to try and find a distance where *all* the radiation is removed will give the wrong result because, even at large values, some of the radiation won’t be absorbed. This is because the radiation is completely absorbed at some frequencies, while CO2 remains mostly transparent at others.
To be clear, the Hottel average is across the entire blackbody emission for that temperature. As a result, trying to compute the power emitted (or absorbed) for a narrow band of frequencies will give wrong results.

May 27, 2011 9:28 am

Joel and others here is a paper on atmospheric radiative heat transfer. Please note his statements concerning CO2 on page 23.
http://www.scribd.com/doc/34962513/Elsasser1942

May 27, 2011 9:31 am

JAE says:
May 27, 2011 at 8:59 am
But, to move beyond your insults, arm-waving, and fiats. There must be some other factors besides GHGs which determine the “effective radiating level,” like maybe convection.

No, it’s related to the opacity of the atmosphere at altitude and is a function of wavelength depending on the atmospheric composition.
First, it is true that all radiation would have to come from the surface if there were no ghgs, but it is also likely that the atmosphere would be extremely hot, due to conduction/convection.
No the surface would be cold, say -18ºC, and the incoming solar radiation would be balanced by reflection and IR emissions, the atmosphere would be colder. Heat transfer would be slow between surface and atmosphere (mainly conduction), any convection would give rise to a profile determined by the lapse rate (dry).
The only way the gases could get rid of heat would be to conduct it back to the surface, but convection would try to prevent that. It would certainly be weird.
Convection wouldn’t prevent conduction, the highest temperature would be next to the surface (lapse rate).
Second, there must be something else that affects the “effective radiating level” besides the types and quantities of ghgs. Maybe it’s density.
It is related to density, it’s the altitude where thermalization via collision starts to not be as effective a means for the excited molecules to lose energy compared with radiation, it varies with altitude depending on the gas.

Robert Clemenzi
May 27, 2011 9:35 am

“effective radiating level” is nonsense. The atmosphere is about -18C at 5km, 45.5km, 56.5km, and another higher up. Which of the 4 is closer to the surface? Which is closer to space?
At 49km, the atmosphere is about -2.5C. Get it, above the “effective” 5km level, the atmosphere is warmer?
This is one of the reasons that I started questioning the consensus. When the consensus claims that the “effective radiating level” is below 2 levels of atmosphere that are much warmer, then I think that a good explanation is required.

Robert Clemenzi
May 27, 2011 10:34 am

Phil. says:
No the surface would be cold, say -18ºC,
No, that is simply the average temperature. The daily fluctuation would be from hot enough to boil water to extremely cold.
the atmosphere would be colder. Heat transfer would be slow between surface and atmosphere (mainly conduction), any convection would give rise to a profile determined by the lapse rate (dry).
No – The atmosphere would be the same temperature as the peak surface temperature. I agree that the main method of heating would be convection until the temperature profile matched the dry adiabatic lapse rate. After that, you are right that slow conduction would take over. However, at some point, the atmosphere would have a single constant temperature, from top to bottom, equal to the daily peak value.

Robert Stevenson
May 27, 2011 1:49 pm

Joel says:
‘Good God, are you guys blind? How many times do we have to debunk Postma’s nonsense?
I haven’t studied Joseph Postma’s paper in detail (all 32 pages), but I noted that there is no mention of the LWIR radiant heat which is supposed to be absorbed by water vapour, CO2 etc. Most contributors accept that first generation photons from the surface are absorbed to extinction by water vapour in the first 120m of traverse through the atmosphere. This means that 63.3% (248Wm-2) of the 391Wm-2 emitted as LWIR radiant energy from the surface is absorbed by water vapour in 120m of traverse. On page 3 Postma states that anthropogenic global warming means a general warming of the atmosphere theorized to be human emission of carbon dioxide (CO2 ), which is then theorized to cause a strengthening of the effect of the Greenhouse Theory, which actually causes said warming. It should be noted however that CO2 absorbs very little heat in the first 120m of traverse and as has been mentioned by many contributors before is a very weak GHG. So, AGW or ‘any global warming’ as it should be known is therefore down to water vapour and governments could save millions of $ expenditure by reducing water vapour emissions instead of CO2 emissions by merely paying lip service (10$) to it (water vapour that is).

Robert Stevenson
May 27, 2011 2:38 pm

Robert Clemenzi says:
The Hottel analysis uses an average “grey gas” emissivity. That is, it ignores the spectral lines and simply assumes that it absorbs (or emits) the same at all frequencies. As a result, it should not be used for more than 100 meters or so.
Have you read Hottel and Sarofim, “Radiative Transfer,” Chap 11, McGraw Hill, New York 1967. Data is presented on CO2 band emissions for use in calclations.

Robert Clemenzi
May 27, 2011 5:00 pm

Robert Stevenson says:
Have you read Hottel and Sarofim, “Radiative Transfer,” Chap 11, McGraw Hill, New York 1967. Data is presented on CO2 band emissions for use in calculations.
No. I am using “Principles of Heat Transfer” by Krieth, 1967, which provides only 5 pages on Hottel’s methods. Since the years are the same, and based on your comment, I assume that I am missing a more advanced analysis. Unfortunately, the book is $110 to $160.

wayne
May 27, 2011 5:32 pm

mkelly, thanks for the link to Elsasser (1942), he explains much of what I have harping on (even my 3d examples) above in the first few pages. He has a good way to explain the full (or even partial equalibrium with residual transfer) radiative equalibrium that most of our atmosphere is in at all times. Thanks, his words will help me explain myself better.

jae
May 27, 2011 5:56 pm

Phil:
I’m having trouble putting these two statements together:
“No, it’s related to the opacity of the atmosphere at altitude and is a function of wavelength depending on the atmospheric composition.”
and
“It is related to density, it’s the altitude where thermalization via collision starts to not be as effective a means for the excited molecules to lose energy compared with radiation, it varies with altitude depending on the gas.”
Please help.
Also,
“No the surface would be cold, say -18ºC,”
Robert is on the right track, I think. The surface on most of the planet would warm during the day to at least 90 C (probably higher), which would warm the “no-ghg-atmosphere” by conduction/convection. I don’t know (and you probably don’t, either) how much of that energy would be lost at night, especially since there would be an inversion. If not all the energy was re-irradiated at night, the next day would add heat. An equilibrium would be reached, and it just might be very warm. I don’t know how to determine what it would be, and neither do you, I suspect.
Maybe Joel knows….

Jim D
May 27, 2011 7:03 pm

The effective level concept is a simplified way of looking at all the wavelengths together. In reality, each wavelength of IR emission has a different altitude which is up in the stratosphere for some where GHGs have strong lines, and down at the surface for others in the “window” region. 5 km just happens to be the weighted average over the emitted wavelengths.