Guest post by Bob Fernley-Jones by Bob Fernley-Jones AKA Bob_FJ
CAUTION: This is written in Anglo-Oz English.
Here is the diagram as extracted from their 2009 paper, it being an update of that in the IPCC report of 2007 (& also 2001):
The unusual aspect of this diagram is that instead of directly showing radiative Heat Transfer from the surface, it gives their depiction of the greenhouse effect in terms of radiation flux or Electro-Magnetic Radiation, (AKA; EMR and a number of other descriptions of conflict between applied scientists and physicists). EMR is a form of energy that is sometimes confused with HEAT. It will be explained later, that the 396 W/m^2 surface radiation depicted above has very different behaviour to HEAT. Furthermore, temperature change in matter can only take place when there is a HEAT transfer, regardless of how much EMR is whizzing around in the atmosphere.
A more popular schematic from various divisions around NASA and Wikipedia etc, is next, and it avoids the issue above:
- Figure 2 NASA
Returning to the Trenberth et al paper, (link is in line 1 above), they give that the 396 W/m2 of EMR emitted from the surface in Fig.1 is calculated primarily by using the Stefan–Boltzmann law, and global year average conditions. Putting aside a few lesser but rather significant issues therein, it is useful to know that:
1) The Stefan-Boltzmann law (S-B) describes the total emission from a flat surface that is equally radiated in all directions, (is isotropic/hemispherical). Stefan found this via experimental measurement, and later his student Boltzmann derived it mathematically.
2) The validity of equally distributed hemispherical EMR is demonstrated quite well by observing the Sun. (with eye protection). It appears to be a flat disc of uniform brightness, but of course it is a sphere, and at its outer edge, the radiation towards Earth is tangential from its apparent surface, not vertical. It is not a perfect demonstration because of a phenomenon called limb darkening, due to the Sun not having a definable surface, but actually plasma with opacity effects. However, it is generally not apparent to the eye and the normally observed (shielded) eyeball observation is arguably adequate for purpose here.
3) Whilst reportedly the original Stefan lab test was for a small flat body radiating into a hemisphere, its conclusions can be extended to larger areas by simple addition of many small flat bodies of collectively flat configuration, because of the ability of EMR waves to pass through each other. This can be demonstrated by car driving at night, when approaching headlights do not change in brightness as a consequence of your own headlights opposing them. (not to be confused with any dazzling effects and fringe illumination)
4) My sketch below demonstrates how radiation is at its greatest concentration in the lateral directions. It applies to both the initial S-B hemispherical surface radiation and to subsequent spherical radiation from the atmosphere itself.
5) Expanding on the text in Figure 3: Air temperature decreases with altitude, (with lapse rate), but if we take any thin layer of air over a small region, and time interval, and with little turbulence, the temperature in the layer can be treated as constant. Yet, the most concentrated radiation within the layer is horizontal in all directions, but with a net heat transfer of zero. Where the radiation is not perfectly horizontal, adjacent layers will provide interception of it.
A more concise way of looking at it is with vectors, which put simply is a mathematical method for analysing parameters that 
possess directional information. Figure 4, takes a random ray of EMR (C) at a modestly shallow angle, and analyses its vertical and horizontal vector components. The length of each vector is proportional to the power of the ray, in that direction, such that A + B = C. Of course this figure is only in 2D, and there are countless multi-directional rays in 3D, with the majority approaching the horizontal, through 360 planar degrees, where the vertical components also approach zero.
6) Trenberth’s figure 1 gives that 65% of the HEAT loss from the surface is via thermals and evapo-transpiration. What is not elaborated is that as a consequence of this upward HEAT transfer, additional infrared radiation takes place in the air column by virtue of it being warmed. This initially starts as spherical emission and absorption, but as the air progressively thins upwards, absorption slows, and that radiation ultimately escapes directly to space. Thus, the infrared radiation observable from space has complex sources from various altitudes, but has no labels to say where it came from, making some of the attributions “difficult”.
DISCUSSION; So what to make of this?
The initial isotropic S-B surface emission, (Trenberth’s global 396 W/m2), would largely be absorbed by the greenhouse gases instantaneously near the surface. (ignoring some escaping directly to space through the so-called “atmospheric window”). However, a large proportion of the initial S-B 396 surface emission would be continuously lateral, at the Trenberth imposed constant conditions, without any heat transfer, and its horizontal vectors CANNOT be part of the alleged 396 vertical flux, because they are outside of the vertical field of view.
After the initial atmospheric absorptions, the S-B law, which applied initially to the surface, no longer applies to the air above. (although some clouds are sometimes considered to be not far-off from a black body). Most of the air’s initial absorption/emission is close to the surface, but the vertical distribution range is large, because of considerable variation in the photon free path lengths. These vary with many factors, a big one being the regional and more powerful GHG water vapour level range which varies globally between around ~0 to ~4%. (compared with CO2 at a somewhat constant ~0.04%). The total complexities in attempting to model/calculate what may be happening are huge and beyond the scope of this here, but the point is that every layer of air at ascending altitudes continuously possesses a great deal of lateral radiation that is partly driven by the S-B hemispherical 396, but cannot therefore be part of the vertical 396 claimed in Figure 1.
CONCLUSIONS:
The vertical radiative flux portrayed by Trenberth et al of 396 W/m^2 ascending from the surface to a high cloud level is not supported by first principle considerations. The S-B 396 W/m^2 is by definition isotropic as also is its ascending progeny, with always prevailing horizontal vector components that are not in the field of view of the vertical. The remaining vertical components of EMR from that source are thus less than 396 W/m^2.
It is apparent that HEAT loss from the surface via convective/evaporative processes must add to the real vertical EMR loss from the surface, and as observed from space. It may be that there is a resultant of similar order to 396 W/m^2, but that is NOT the S-B radiative process described by Trenberth.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
ADDENDUM FOR AFICIONADOS
I Seek your advice
In figure 5 below, note that the NIMBUS 4 satellite data on the left must be for ALL sources of radiation as seen from space, in this case, at some point over the tropical Pacific. The total emissions, amount to the integrated area under the curve, which unfortunately is not given. However, for comparison purposes, a MODTRAN calculator, looking down from 100 Km gives some interesting information for the figure, which is further elaborated in the tables below. Unfortunately the calculator does not give global data or average cloud/sky conditions, so we have apples and pears to compare, not only with Nimbus, but also with Trenberth. However, they all seem to be of somewhat similar order, and see the additional tabulations.
| Compare MODTRAN & “Trenberth”, looking down from 2 altitudes, plus Surface Temperature | ||||
| Location | Kelvin | 10 metres | 100 Km. | (Centigrade) |
| Tropical Atmosphere | 300K | 419 W/m^2 | 288 W/m^2 | (27C) |
| Mid-latitude Summer | 294K | 391 W/m^2 | 280 W/m^2 | (21C) |
| Mid-latitude Winter | 272K | 291 W/m^2 | 228 W/m^2 | (-1C) |
| Sub-Arctic Winter | 257K | 235 W/m^2 | 196 W/m^2 | (-16C) |
| Trenberth Global | 288K ? | 396 W/m^2 | 239 W/m^2 | (15C ?) |
| Compare MODTRAN & “Trenberth”, looking UP from 4 altitudes: W/m^2 | ||||
| Location | From 10 m | From 2 Km | From 4Km | From 6Km |
| Tropical Atmosphere | 348 | 252 | 181 | 125 |
| Mid-latitude Summer | 310 | 232 | 168 | 118 |
| Mid-latitude Winter | 206 | 161 | 115 | 75 |
| Sub-Arctic Winter | 162 | 132 | 94 | 58 |
| Trenberth Global | 333 Shown as coming from high cloud area (= BS according to MODTRAN) | |||
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
Bob, I comprehend. It does seem you are still hanging on S-B calculations, mixing relative figures with net figures, that can so easily mess you up if all factors are not applied correct. It’s a well respected equation that with two temperatures and emissivities can hand you the maximum net transfer power across vacuum between but this ignores such factors as the absorption coefficients of any matter components lying between the two surfaces, like part of an atmosphere, and you get stuck again for the S-B’s answer is not correct unless the distance is so small that such factors are irrelevant. And then there is the geometric aspect. I still say net is correct as your article stated.
If you never got the chance to read Hans (Hans says: October 27, 2011 at 6:12 pm) comment and the small blue link “here” in his last sentence, you should. Somewhat deep depending on your level in physics but if you can just skim off the essence it will help you answer such questions and really understand why and why not. It’s on the spontaneous transfer of any energy, microstates, and it has everything to do with radiation and entropy. See if that might help fill in some of the holes.
Wayne,
I agree, very very very tiny. Then again, a .22 with a cyclical rate of fire of 1200rpm can shoot through steel, concrete, bullet proof vests… That extremely tiny amount has a rather large multiplier.
Wayne,
Hans post does not appear to be there. Do you have the link?
Wayne,
ignore that last post, found it.
Bob,
I agree that there are many places where the vector nature of photons is important. The direction that light is scattered is one of those cases.
For radiation emitted by a chunk of atmosphere:
* 1/2 of the energy goes up and 1/2 goes down
* 1/2 of the energy goes east and 1/2 goes west
* 1/2 of the energy goes north and 1/2 goes south
NOT 1/6 goes in each direction. In that sense I feel Figure 3 is misleading. It seem to be saying that (combined with Fig 4 showing vector components) a photon heading 10 degrees above horizontal toward the left should only be considered as carrying sin(10) = 17.4% of its energy upward and cos(10) = 98.5% of its energy leftward. Or maybe it is being treated as simply additive — if 10% is upward, then 90% is sideways. I’m really not sure what you intended there. In fact, it is carrying 100% of its energy up and 100 % of its energy to the left.
On the other hand, momentum IS a vector. The component of the photon’s momentum upward would be 17.45 % of the total momentum.
(And to be just a little more precise about energy, on average 1/2 + delta goes down and 1/2 – delta goes up, because of the average temperature gradient in the atmosphere. )
Bob Fernley-Jones says:
“…So ask yourself, if part of the energy moving from the surface is due to evaporation (80 Wm-2) and part is carried upward from thermals (17 Wm-2) then can the surface ALSO radiate at 396 Wm-2 upward by radiation even if the emissivity is one? …”
Yes, I know what you mean, and this had me puzzled too for a long time. I think that the answer is that the surface temperature is a consequence of ALL of the various heat transfers. Thus it may be valid to do an S-B calculation from that resultant surface temperature. (even though we do not have good T data for the actual surface! BTW).
To elaborate; without an atmosphere, it seems perhaps non-intuitively, that without the alleged 65% “convective” cooling, the surface would be significantly hotter, with thus a different S-B result. Thus whilst at first sight there seems to be additional energy over the top of S-B, I don’t think that this is actually the case.
Bob (and Wayne),
Remember, to the extent that non-radiative flux (latent heat and thermals) is leaving the surface, it’s largely being returned somewhere else in equal and opposite amounts (as the temperature component of precipitation, wind, weather, etc). Remember also, the surface upward radiative flux of 396 W/m^2 is also just the net energy flux entering the surface from the atmosphere. So what happens if more non-radiative flux leaves the surface than is returned to the surface on average (i.e. a net convective energy loss from the surface to the atmosphere)? The surface will lose energy and subsequently cool down and radiate less, right? Since energy cannot be convected to space – only radiated, in this particular example, non-radiative flux at the surface is just being traded off for radiative flux at the surface, requiring the surface to emit less to achieve equilibrium output power at the TOA.
Wayne,
“So back to your question. The 396 and 333, toss them. Change the 333 to zero for NET energy, never flows downward on an averaged Earth. The 396 is actually about 63 as you said in your article above. (We are converting these to NET transfer now). Of the ~63, ~40 Wm-2 is window frequencies that actually zip at the speed of light directly to space. So, that leaves us with a NET of ~23 Wm-2 instead of 356 that is attributed to IR radiation warming the atmosphere for this is NET absorbed from IR radiation from the surface above what it would be if it was never absorbed within the atmosphere.
That is my take after two years of deciphering this subject.
So, 161 Wm-2 absorbed = 80 + 17 + 63 + 1 (NET). Check!
And 239 Wm-2 at TOA = 78 + 80 + 17 + 63 + 1 (NET). Check!
Other papers show variances in these figures but all are closely clustered. It is 23 Wm-2 (63-40) we are talking about in relation to warming by GHGs, not 396 Wm-2.”
First of all, the diagram is depicting a direct surface to space transmittance or ‘window’ transmittance of 70 W/m^2 (40 W/m^2 through the clear sky and 30 W/m^2 through the clouds or cloudy sky).
The surface emitted LW flux is solely due to the surface temperature and nothing else, because temperature is slaved to emitted radiative power via the Stefan-Boltzman law. Now I agree the emissivity of the surface is not 1, but it’s very close to 1 – probably 0.98-0.99+ on average. Taking this into account would only reduce the surface LW flux by about 4-8 W/m^2, which is negligible.
If the surface is emitting 396 W/m^2 in the LW infrared and 70 W/m^2 of this goes straight to space, that leaves 326 W/m^2 of it being absorbed or captured by GHGs and clouds in the atmosphere. Of this, 169 W/m^2 is emitted to space as part of the 239 W/m^2 flux leaving, and by deduction, 157 W/m^2 is returned or re-circulated back to the surface. So the net warming from GHGs is +157 W/m^2 into the surface (not 23 W/m^2). 157 W/m^2 from GHG re-direction + 239 W/m^2 from the Sun = 396 W/m^2 (the net energy flux entering the surface from the atmosphere).
I think you may not be realizing that the non-radiative fluxes from the surface to the atmosphere, from the atmosphere to other parts of the atmosphere, and from the atmosphere back to the surface are just moving energy around within the thermal mass so as to maintain the planet’s energy balance of about a net 396 W/m^2 flux into the surface and about 326 W/m^2 of atmospheric absorption from GHGs and clouds.
RW,
unless you are saying Trenberth is loony the flux entering the surface is simply 161 SW and 333 DLR for a total of 494 absorbed. Convection and evaporation is NOT returned as the heat is lost in the mid to upper trop and mostly radiated directly to space from there.
“First of all, the diagram is depicting a direct surface to space transmittance or ‘window’ transmittance of 70 W/m^2 (40 W/m^2 through the clear sky and 30 W/m^2 through the clouds or cloudy sky).”
No, that is 40 direct from ground + 169 from all atmosphere + 30 from clouds = 239. There is no 70 direct from ground to space.
The 169 would include the portions of the evapotranspiration and thermals as would the 30 from the clouds.
The emissivity of water is .98-.99. The ground ranges from about .93-.98 which includes plants. This is a mistake by Trenberth of a significant amount.
“Taking this into account would only reduce the surface LW flux by about 4-8 W/m^2, which is negligible.”
Trenberth’s missing heat is only .9 w/m2.
“If the surface is emitting 396 W/m^2 in the LW infrared and 70 W/m^2 of this goes straight to space, that leaves 326 W/m^2 of it being absorbed or captured by GHGs and clouds in the atmosphere.”
Again, there is only 40 going from ground to space. The other 30 is being emitted by the clouds. The diagram says what it says. You are overanalyzing something relatively straight forward.
17 thermal + 80 evapot + 396 + 1 = 161 + 333 = 494 (don’t ask me what the 1 is about!!)
The only involved issue is that it takes a number of cycles before the ground is emitting the full amount including the DLR coming back to it. That is, 396 and 333 include amounts that have been counted one or more times already.
A simplified way of seeing it is that when the OLR goes through the troposphere, tropopause, stratosphere… it is not counted when it goes through each layer, it is only counted once for the whole transit. In this diagram, because the flow reverses they count the same energy multiple times instead of taking a net figure. While a meter would “SEE” a flow there is no work being done or there wouldn’t be that large of a flux. Basically the atmosphere has EPICYCLES OF ENERGY!!! HAHAHAHAHAHAHAHAHAHAHA
Tim Folkerts @ur momisugly November 4, at 3:09 pm
Tim,
You wrote in part (a repeat from a few days ago):
So, you are claiming that ½ + ½ + ½ + ½ + ½ + ½ = 1 ?
The rest of what you wrote seems to be claiming that the vectorial treatment of a photon stream as discussed in the Lambertian optics article I cited, is not applicable to a photon stream originating from the Earth’s surface or within the atmosphere. You don’t adequately explain why, but just reject it. Would you please be kind enough to elaborate why you think this?
Bob F-J,
“So, you are claiming that ½ + ½ + ½ + ½ + ½ + ½ = 1 ?”
And that doesn’t include the energy transferred through collision to non-GHGs!!!
kuhnkat,
I’m pretty sure it’s actually 30 W/m^2 from the surface directly through the clouds to space. The actual ‘window’ transmittance from the cloud tops directly space would be far greater than 30 W/m^2, as there is very little water vapor in between the cloud tops and space (and no clouds either).
“The 169 would include the portions of the evapotranspiration and thermals”
Yes it likely would, but those non-radiative fluxes originated from the surface and were in addition to the surface LW flux of 396 W/m^2. So if more non-radiative flux leaves the surface than returns to the surface on average, the surface will lose energy, cool down and subsequently radiate less. Because the diagram is assuming a steady-state condition, any non-radiative for radiative trade offs effects like this already are already embodied in the surface LW flux of 396 W/m^2.
BTW, I’m not defending the diagram. I think it’s a mess and really not an accurate depiction at all. I’m also aware of the counting of energy more than once, which make it look like there is more energy than there really is.
The bottom line is (ignoring Trenberth’s phony extra watt) there is 239 W/m^2 entering and leaving at the TOA and the surface is receiving a net flux of 396 W/m^2. If you really doubt if my depiction of the numbers is accurate, and you agree the atmosphere cannot create any energy of its own, what then is the origin of the +157 W/m^2 into the surface if not from GHG absorption and re-direction back towards the surface?
RW,
the lower atmosphere is very dense. Only 40 makes it past the first 10 meters (on average of course!!)
If you agree there is double counting why bother with:
“the surface is receiving a net flux of 396 W/m^2. ” The net surface flux is very small as it consists of the energy that is still in the surface each morning when the sun comes up.
According to Trenberth the surface receives 161 + 333 = 494 on average.
17 thermals + 80 evapo + 396 OLR + 1 = 494
Remember that .9 down at the bottom is the NET surface flux that is absorbed in the oceans and the ground and is why the outgoing needs the 1 added.
The 30 confusing you is direct from the surface into the clouds and then radiated by the clouds. The 30 is from the 356 that is going into the atmosphere.
The only values that are measured are the TOA in and out. The rest are estimated (made up).
Moderators,
I forgot to close the italics on the quote ““The 169 would include the portions of the evapotranspiration and thermals” in my last post. Can you fix? Thank you.
RW @ur momisugly November 1, at 9:30 pm & 9:44 pm, and November 4, at 5:29 pm
Sorry to be so tardy RW; but just a couple of quickies:
• I don’t see 70 leaving the surface direct to space, but only 40. The additional 30 you mention is shown as coming from the high clouds.
• The latent heat is released in clouds when the water vapour condenses into water particles, thus resulting in increased temperature related radiation. It is not returned to the surface via precipitation, whilst there is a complication that some precipitation does not even reach the surface because of intervening evaporation
Hopefully I can find more time to study your comments in more detail soon. Thanks for your big interest
Bob,
“The latent heat is released in clouds when the water vapour condenses into water particles, thus resulting in increased temperature related radiation. It is not returned to the surface via precipitation, whilst there is a complication that some precipitation does not even reach the surface because of intervening evaporation.”
While yes, some of the latent heat that condenses to form clouds ends up radiated into the atmosphere and ultimately radiated out to space. But a lot of it stays with the water and returns to the surface as the temperature component of precipitation. Some of it is also ends up radiated back down to the surface. Some of it can also be replenished by post albedo energy absorbed in the atmosphere from the Sun.
If you doubt this, might I ask you then what is the primary source of energy in the temperature component of precipitation, if not from the latent heat of evaporation?
“I don’t see 70 leaving the surface direct to space, but only 40. The additional 30 you mention is shown as coming from the high clouds.”
You’d have to double check with Trenberth himself on this, but I’m pretty sure the diagram is depicting the 30 W/m^2 flux from the surface directly through the clouds to space. As I said earlier, there is no way there is only a 30 W/m^2 direct transmittance from the cloud tops to space. That is just way too low. But again, neither the 30 W/m^2 or 40 W/m^2 values are really referenced in the paper and appear to be arbitrary estimates or just guesses.
There are really so many things wrong with this diagram, it’s hard to discuss it without their being a great deal of confusion and misinterpretation.
Kuhnkat, you wrote in part:
Trenberth’s missing heat is only .9 w/m2.
Hey look; I have it from a confidential source that Trenberth has calculated the “missing heat” as greater at 0.931459265358979323846 W/m^2, but decided to round-it-off to avoid controversy. Apparently it has something to do with ye old ‘Pi’, a convenient mathematical constant.
Bob,
“I have it from a confidential source that Trenberth has calculated the “missing heat” as greater at 0.931459265358979323846 W/m^2, but decided to round-it-off to avoid controversy.”
You sure they carried the “1” correctly??
I calculated that the emissive power of the entire spectrum @ur momisugly 15 C is 392 W/m^2.
I calculated also that CO2 at 380ppm absorbs 79.8 W/m^2 of these ‘first generation’ photons to extinction, and H2O at 23000ppm absorbs 248.2 W/m^2 in relatively short distances.
I may be too high with the H2O calc, does anyone else have any figures?
kuhnkat,
“According to Trenberth the surface receives 161 + 333 = 494 on average.
17 thermals + 80 evapo + 396 OLR + 1 = 494”
Yes I know, but 97 W/m^2 (from latent heat and thermals) of the 494 W/m^2 originated from the surface in addition to the radiative flux, so the net flux received at the surface is 396 W/m^2 (494 – 97 – Trenberth’s phony extra watt = 396). Why is this so hard to see? Do you not understand the fundamental T^4 relationship between temperature and emitted power? Whatever temperature a body is radiating at (in this case the surface of the Earth), the amount of energy it is radiating has to be continually replaced or else the body will gain or loss energy and subsequently warm up or cool down. If the surface is emitting 396 W/m^2 in the steady-state, it has to be receiving a net of 396 W/m^2 from the atmosphere in order to sustain it.
“The 30 confusing you is direct from the surface into the clouds and then radiated by the clouds. The 30 is from the 356 that is going into the atmosphere.
Not really. The cloud tops radiate according to their temperature an emissivity. If the direct surface to space transmittance were actually 40 W/m^2 as you claim, the 356 W/m^2 is the amount absorbed by GHGs and clouds in the atmosphere, some of which (half) ends up radiated up out to space and some radiated down back to the surface.
RW,
“there is no way that the clear sky atmosphere is radiating 169 W/m^2 to space.”
You continue to confuse a very simple and straightfoward, although wrong, diagram. The 169 is an AVERAGE and comes from clear sky AND cloud areas, night/day/poles. The 30 from clouds is only the extra from clouds AVERAGED over the whole earth.. That is, remove the clouds and that 66% of the atmosphere would still be radiating depending on the ground temp pumping it. The atmosphere above the clouds would still be radiating unhindered by the clouds whether they are there under it or not. Some cloud areas emit little above average clear sky and others emit a huge amount more. Some clouds actually reflect IR which means they do not allow the atmosphere under them to radiate out. This reduces the net cloud output as the radiation is from the upper surface instead of a 3d opaque atmosphere.
These are GLOBAL AVERAGES that include clear sky and clouds and monsoons and hurricanes…
All the AVERAGING has already been done on these figures. There is no percent taking to be done unless you wish a more detailed analysis.
You are worried about the 169 number Trenberth made up. Try this,
incoming SW absorbed by atmosphere 78
incoming SW absorbed by earth 161
ignoring how it gets back into the atmosphere that is 239 available to make its way out on average 24 hours a day seven days a week… That is about what DOES make its way out. I think we can agree that Trenberth has a problem of distorting the details of reality. Quibbling over which pieces doesn’t get us too far unless we have hard data to show it is different. Notice that 239 doesn’t leave .9 to be absorbed anywhere!!! Trenberth can’t even add!! Of course he wouldn’t want to be too clear because he might have to admit some of that absorbed is in the ground instead of the ocean or photosynthesis or any number of other chemical processes driven by temp that doesn’t spontaneously reverse!!!
Yes, his figures may be off, but, it is a cartoon. If anyone tries to use this to create a GCM they deserve what they get. Willis or anyone else depending on these numbers are wasting everyone’s time. Depending on the imaginary 396 up and 333 down (or whatever it is today) are REALLY wasting our time.
RW,
maybe y’all can answer a question I have. What size meter do all these figures fit. That is, does the 239 w/m2 out cover a meter of the sphere’s surface at orbit, a meter of the sphere’s surface at average radiation altitude, or a meter of the surface of the earth, which varies due to roughness and altitude!! It DOES make a substantial difference. Since they claim it is measured great. Is there a conversion factor or what??
kuhnkat,
Moreover, there is no way that the clear sky atmosphere is radiating 169 W/m^2 to space. According to the ISCCP data, clouds cover an average of about 2/3rds (66%) of the Earth’s surface, which means 1/3 (33%) is clear sky. 396 W/m^2 x 0.33 = 131 W/m^2 emitted to the clear sky, which is less than 169 W/m^2. And this is before even subtracting the ‘window’ of 40 W/m^2. The clear sky cannot emit more to space than is emitted to it from the surface.
One of the biggest problems with Trenberth is he does not clearly separate or distinguish the clear from the cloudy sky atmosphere, which only adds further to the confusion and ambiguity.
kuhnkat,
“You continue to confuse a very simple and straightfoward, although wrong, diagram. The 169 is an AVERAGE and comes from clear sky AND cloud areas, night/day/poles. The 30 from clouds is only the extra from clouds AVERAGED over the whole earth.. That is, remove the clouds and that 66% of the atmosphere would still be radiating depending on the ground temp pumping it. The atmosphere above the clouds would still be radiating unhindered by the clouds whether they are there under it or not. Some cloud areas emit little above average clear sky and others emit a huge amount more. Some clouds actually reflect IR which means they do not allow the atmosphere under them to radiate out. This reduces the net cloud output as the radiation is from the upper surface instead of a 3d opaque atmosphere.”
I’m having trouble understanding exactly what you’re referring to here. Yes I know the 169 W/m^2 depicted is from the whole atmosphere (cloudy and clear sky). I thought you were saying the 30 W/m^2 was the total amount emitted up from the cloud tops. If you’re saying that the 30 W/m^2 depicted is the ‘window’ transmittance from the cloud tops directly to space, I can understand this, although as I said, I dispute the number. I’m also well aware that the whole atmosphere radiates, both below and above the clouds, so I’m not sure what you are implying by this.
RW,
I am not claiming any of the numbers are right either, only that the TOA in and out are probably close.
What I am trying to get across is that Trenberth tried to simplify too much. I THINK he gives the 30 as the difference from what the average clear air and the average cloud is. There would be 33% of the area with 0 extra as there is no cloud. Maybe another 33% with a small extra contribution. So only about 33% of the area has a significant contribution to this 30 that is an average of the total cloudiness spread over 100% of the atmosphere.
Here is a newer cartoon from NASA:
http://mynasadata.larc.nasa.gov/Radiation_Explanation.html
kuhnkat,
“maybe y’all can answer a question I have. What size meter do all these figures fit. That is, does the 239 w/m2 out cover a meter of the sphere’s surface at orbit, a meter of the sphere’s surface at average radiation altitude, or a meter of the surface of the earth, which varies due to roughness and altitude!!”
Interesting question. I believe it means a path straight from the surface to the TOA, globally averaged. Or the shortest distance from the surface to the TOA, globally averaged. So it’s relative to a m^2 from the surface, I think is the answer to your question.
RW @ur momisugly November 4, at 11:31 pm
You raise an interesting question RW, with:
If you [Bob] doubt this, might I ask you then what is the primary source of energy in the temperature component of precipitation, if not from the latent heat of evaporation?
If I remember correctly, the evapotranspiration of 80 was derived from estimates of rainfall, which was rather simplistic, but I think possibly a little better than a wild guess, and in the too-hard category.
To your issue, I don’t think the precipitation would warm the surface, because it comes from a colder region. Thus, if anything, it might cause further cooling? There would be some warming from braking of gravity driven KE but I guess it would be slight. (friction from air drag and impact with surface)
So I reckon it is a small issue compared with other stuff in the diagram.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Incidentally, if the 80 is real, it might be that a small change in it would be more significant than the effect of a small change in the net surface radiative effects. I’ve had exchanges with Roy Spencer on this, and it seems that he and everyone are too busy competing on the radiative stuff to look at it. Strange; it could be a bigger negative feedback than in the area they are looking.
Bob,
“To your issue, I don’t think the precipitation would warm the surface, because it comes from a colder region. Thus, if anything, it might cause further cooling? There would be some warming from braking of gravity driven KE but I guess it would be slight. (friction from air drag and impact with surface)
So I reckon it is a small issue compared with other stuff in the diagram.”
I’m not claiming it warms the surface. Just that it’s a significant portion of the energy entering the surface from the atmosphere that is not depicted or accounted for in the diagram. If not primarily from the left over latent heat of evaporation that doesn’t get radiated out to space or back to the surface, then what is the source? Don’t forget too that some of the energy from the latent heat of evaporation condensed in clouds that is radiated into the atmosphere is also replenished from energy from the atmosphere radiated into the clouds, effectively sustaining much of the energy lost from the clouds via radiation.
I also agree that quite often precipitation is colder than the surface, but globally averaged at least, I wouldn’t think this in and of itself has a cooling effect the surface, especially if the energy primarily originated from the surface to begin with. Of course, since precipitation emanates from clouds and clouds reflect sunlight, there is a cooling connection associated with precipitation – just not a direct one.
Bob,
“Incidentally, if the 80 is real, it might be that a small change in it would be more significant than the effect of a small change in the net surface radiative effects. I’ve had exchanges with Roy Spencer on this, and it seems that he and everyone are too busy competing on the radiative stuff to look at it. Strange; it could be a bigger negative feedback than in the area they are looking.”
Agreed. I know Roy’s main focus is on the cloud feedback (i.e. changes in radiative flux at the TOA), but all of these components — the latent heat of evaporation, water vapor and clouds — are all interconnected to one another and constitute the whole of atmospheric the water cycle (ground state water -> evaporation -> water vapor -> clouds -> precipitation -> ground state water), which is the primary thing controlling the planet’s energy balance and ultimately the globally averaged surface temperature.
It’s true that warmer temperatures are associated with increased water vapor from evaporation, but increased water vapor from warming also increases the amount of energy removed from the surface as latent heat of evaporation, which has a cooling effect on the surface. And of course water vapor from evaporation is ultimately what drives cloud formation, and as the clouds form they reflect more and more sunlight, which further cools the surface. While the water vapor feedback by itself is likely positive, it’s ultimately the net combined feedback of all three of these things being positive that is nonsense and cannot be supported by any real science or data.
kuhnkat,
“What I am trying to get across is that Trenberth tried to simplify too much. I THINK he gives the 30 as the difference from what the average clear air and the average cloud is. There would be 33% of the area with 0 extra as there is no cloud. Maybe another 33% with a small extra contribution. So only about 33% of the area has a significant contribution to this 30 that is an average of the total cloudiness spread over 100% of the atmosphere.”
Forgive me, but I just cannot understand what you’re talking about here. Sorry.
The bottom line is the neither the value of 30 W/m^2 or 40 W/m^2 is referenced in the paper, and neither value is even in the right ball park. They are useless, as is most of the diagram.
As I mentioned before, if you really want an accurate depiction (and proof) of the net energy flux in and out of the system, I recommend this one here:
http://www.palisad.com/co2/div2/div2.html
RW @ur momisugly November 5, at 6:23 pm
Hi RW, I don’t have a significant issue with your thoughts. I agree that precipitation will be initially warmed by the release of latent heat to its ambience. However, there is no doubt that precipitation typically comes from much colder regions than the surface, and that includes ice-hail and snow. You correctly argue that there is less cooling of the surface than if there were not evapotranspiration. It then boils down to whether it is a significant consideration; a matter of scale.
Furthermore I think if Trenberth tried to include every detail in his cartoon, it would be too complicated to follow. Nevertheless, some aspects of his team’s draftsmanship and analysis are very sloppy.
For instance, they show 80 leaving the surface and being released as latent heat, (the same value), in the clouds. However, this cannot be correct. The heat loss from the surface has an un-shown element known as evaporative cooling. In the process of evaporation, the higher energy water molecules are more readily able to escape and become gas, with the consequence that a higher proportion of lower energy molecules are left behind, resulting in cooling of the skin of water etc.
I don’t know the relative scales of this, and probably no one does, but I’d rather drive a nail into the coffin with the S-B consideration, than with these lesser and controversial considerations.