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10 July 2012

Back-Radiation and the Highly Fallacious Kiehl-Trenberth Energy Budget

The usual greenhouse gas calculation such as is offered up by the various Trenberth diagrams is highly dominated by the radiative transport of energy in the Earth’s atmosphere.  It leads to an unreasonable result for the situation I am going to describe here.

But first, my approach to the back-radiation problem was always to show that it was unrealistic on many levels.  One cannot make a net gain of heat energy in a surface by returning a portion of the energy lost from a surface to it.  This is especially true of any process that is said to lose half of the surface emitted energy to space immediately and to only re-absorb some part of half of the energy that was returned to the surface.  I went further and explained that only a small fraction of the energy emitted as radiation from the surface was re-emitted as radiation from an absorbing water or CO2 molecule.  With just these considerations alone, the upper limit on once radiated energy from the surface which is re-absorbed by the surface would be:

(0.5) f,

where the 0.5 is the half of any IR absorbing molecule radiation which was not radiated into space and is radiated toward the surface.  The fraction f is the fraction of the IR energy lost by the surface and absorbed by an IR absorbing molecule which is re-emitted before collisions have dissipated the energy absorbed.  The fraction f is much less than one because most often the IR absorbing molecule undergoes collisions with other molecules and transfers much of the IR absorbed energy to other molecules before re-emission occurs.  These other molecules are rarely water or CO2 or other IR emitting molecules, so that energy is not returned to the surface as IR radiation.  Thus much of the surface emitted IR radiation is dissipated to the 99.97% of the atmosphere molecules which are nitrogen, oxygen, or argon and are not IR emitters.

Let us estimate the value of f.  At sea level, the mean gas velocity is 459 m/s, the mean free path or distance between collisions is only 6.6 x 10^-8 m or 66 nm, and the collision frequency is 6.9 billion/s.  At an altitude of about 4000 m, the radiative transfer of energy competes about evenly with transfer by collisions.  At 4000 m altitude, the frequency of gas molecule collisions is about 4.4 billion/s.  We can use the equivalency of energy transfer by radiation and gas molecule collisions at the 4000 meter altitude to estimate the fraction of energy transfer by radiation of the total of energy transferred by radiation plus gas molecule collisions. At sea level, energy transfer by radiation is equivalent to about 4.4 x 109 collisions per second, so the fraction of energy transferred by radiation is about 4.4/(4.4 + 6.9) = 0.39 of the total by gas molecule collisions and radiation.  This suggests that about 1.5 times as much energy is transferred by gas collisions as by radiation at sea level.

So at this point, the upper limit on IR radiation emitted from the surface which can be returned to the surface and absorbed by it is about:

(0.5) (0.39) = 0.195

It also has to be remembered that this is the upper limit for that portion of the IR radiation from the surface which can be absorbed by an IR-absorbing molecule such as water or CO2.  Much of the surface-emitted IR radiation is of such wavelengths that no IR-absorbing molecule can absorb it in the first place.  If one takes this fact into account, the 0.195 upper limit is a hugely generous upper limit.

But, it does not follow that simply because this much back-radiated IR is incident upon the surface that it will be absorbed by the surface.  During the roughly 8 hours of a day when the surface is warming under increasing sunlight and with a 2-hour lag for warming the ground, water, or air in a vicinity, none of this radiation may be absorbed, except when a cloud casts a shadow on a part of the area or at such points under the shadow of a tree of some such object.  The absorbing ground has to be cooler than the ground from which the photon was emitted and subsequently absorbed by an IR-absorbing molecule in the atmosphere.  During the remaining cooling hours of the day, roughly 16 hours, the surface is more likely to absorb such back-radiated energy.  As a mean value for re-absorption of IR back-radiation over the daily cycle, the value of 0.95 is often used.  I believe that value is much too high.

Let us look at one of the Kiehl-Trenberth energy budgets for a moment:

According to this diagram, 67/ 342 = 0.196 of the incoming solar radiation is absorbed by the atmosphere.  77/342 = 0.225 is reflected by clouds and aerosols.  The surface reflects 30/(30 + 168) = 0.152 of the solar radiation incident upon the surface.  Thermals cooling the surface dissipate 24/168 = 0.143 of this incident radiation.  Let us note that the surface emits a flux of 390 W/m^2 of IR radiation and according to this diagram the atmosphere has absorbed a highly efficient 324 W/m^2 of this or 83.1% of this.  Yet, above I showed in very simple terms that 19.5% is more than the upper limit by far of the amount of IR radiation that the atmosphere can return, which is not from IR-absorbing gas molecules that are substantially cooler than the Earth's surface.

Now let us consider a midday calculation of the surface and do so where there is no cloud cover and where it is so dry that there is no evapotranspiration.  What kind of surface temperature will we have.  Since there are no clouds and I think clouds are the better part of the summed cloud and aerosol effect, let us assume the aerosol effect alone is 0.08.  The midday radiation incident on the upper atmosphere is 1367 W/m^2.  Note that at midday, the incident radiation path length through the atmosphere is shorter than it is for the average daily values normally used, so losses in the atmosphere should be lower than these numbers.  I will use them nonetheless.

The radiation upon the ground is then:

(1 - 0.196) (1- 0.08) (1 - 0.152) (1 - 0.143) 1367 W/m^2 = 734.8 W/m^2

The ground temperature is then found from:

734.8 W/m^2 = ε σ T^4,  ε = 0.95

T = 341.8 K = 68.6̊C = 155.5̊F

But if you believe that the upper limit amount of back-radiation is 95% absorbed by the surface, then the incident radiation is:

(734.8 W/m^2) (1 + 0.195(0.95)) = 870.9 W/m^2

T = 356.6 K = 83.4̊C = 182.2̊F

Now this is clearly much too hot and implies that even the addition of this upper limit of back-radiation which is much smaller than the back-radiation in the Kiehl-Trenberth diagram and energy budget is not physical.  In reality, air convection or thermal effects are more important in cooling the surface than the Kiehl-Trenberth energy budget allows for.  The same is true of water evaporation, transport, and condensation effects.

Note that I did not follow the Kiehl-Trenberth diagram in subtracting the emitted radiation that corresponds to the Earth’s surface temperature.  They subtract 390 W/m^2 and add back in 324 W/m^2 of back-radiation for their mean daily calculation.  This would imply that if we had no atmosphere with IR absorbing gases in it, then there would be no back-radiation, so the total energy budget would look like:

(168 - 24 - 78 - 390) W/m^2 = -324 W/m^2,

which is nonsense.  Of course you might say that the thermals would be different and if I have no absorbing IR gases, then I certainly do not have water evaporation and movement.  This perhaps is a muddled situation.

So let us consider the equivalent calculation technique for an isolated black body radiator in space with incident energy flux of Ii and an emitted radiation of Ie.  But Ii = Ie, so then this approach would have us fallaciously conclude that

Ii - Ie = 0 = σ T^4 and T = 0 K.

The temperature of the isolated black (or gray) body is determined by the incident radiation on it.  The basic approach of the Kiehl-Trenberth diagram to radiation is nonsense.  The energy budget is a farce based on bad physics.  Indeed, when challenged on this issue, many proponents of man-made catastrophic global warming back down and say that the General Circulation Models are calculated primarily as air and water vapor circulation models and are not really consistent with the several variations of the Kiehl-Trenberth Energy Budget.  Yet it is such fallacious energy budgets that the public has been fed as the basis for the claim that there are substantial effects on the Earth's surface temperature due to man's emissions of CO2.  Government websites have been full of these energy budgets, as have college classes.  For their part, the range of results in the GCM computer models is too large to be consistent with the idea that climate science is settled and everyone agrees that it is understood.


MS said...

Thanks for this excellent post.

The APS explanation of the greenhouse effect

Eq 15 erroneously doubles the IR flux of the atmosphere

Eq 17 shows lower atmosphere DOWN emissivity = 1, generating imaginary energy. To assume reduced atmospheric emissivity is no solution because you must also reduce the earth’s emissivity AND the CAGW scare would end. The authors of this report maintained professional integrity by quietly pointing out this mistake.

To correct it you have to correct the IR physics. The excess energy [333.[1-0.76] = 80 W/m^2] is 50 times claimed AGW. The IPCC offsets it using other incorrect physics.

Anonymous said...

The error is more profound. It is assumed that all the radiation comes from the sun-heated surafce. But this is not true. The tempoeature of the surface is lower because part of the heat has been removed by convection and by evaporation of water, and is then radiated highere up. The temperature of the suraface cannot be calculated unless these form sof heat bremopval have been allowewd for.

All this has nothing to do with "back radiation" , which undoubtedly occurs, because the Stefan.Boltrzmann Law requires it.

Charles R. Anderson, Ph.D. said...


I agree and have stated the problem in similar terms in the past.

MostlyHarmless said...

"Thermals cooling the surface dissipate 24/168 = 0.143 of this incident radiation."

They do not - thermals remove heat from the surface.

"Let us note that the surface emits a flux of 390 W/m^2 of IR radiation and according to this diagram the atmosphere has absorbed a highly efficient 324 W/m^2 of this or 83.1% of this."

No, the back-radiation is emitted by both GHGs and clouds, The amount absorbed by clouds includes some incoming from the sun as well as outgoing, and the individual figures aren't shown on the diagram.

"where it is so dry that there is no evapotranspiration" - WHAT? Lack of water vapour PROMOTES evapotranspiration.

Your calculation of IR incident on the surface (not "ground", it includes ocean) is incorrect. The KT diagram simply divides the "sun overhead" figure by 4 to average over the surface of the globe. The figure is thus 4*168 = 672 W/m^2

You can't use SB to calculate temperature from incoming radiation. SB relates to emitted, therefore outgoing only.

Charles R. Anderson, Ph.D. said...

Mostly Harmless:

Of course the thermals remove heat from the surface. The supply of heat at the surface is the incident radiation that is absorbed by the surface, which is what the 168 W/m^2 is.

Back-radiation is not radiation absorbed in the atmosphere directly from the sun, but who knows what innovative and wild terminology may be used by some. However, it is true that some incoming solar radiation is absorbed in the atmosphere and a portion of that may be re-emitted toward the ground. In the K-T diagram the total flux absorbed by the atmosphere of the incoming radiation is 67 W/m^2. Some of what was absorbed has been reduced in energy by many collisions with cooler higher altitude nitrogen, oxygen, and argon molecules. If we assume the surface absorbs a generous 95% of half of the radiation incident on it, this source provides a flux less than 31.8 W/m^2.

Clouds and aerosols reflect 77/342 of the incoming solar radiation, which is 0.225 of it. The underside of clouds is sometimes less reflective then the upper surface, so it may be reasonable to assume this is an upper bound for the surface-emitted radiation reflected from clouds and aerosols. If 95% (a generous upper bound) of this is absorbed by the surface, we have (0.95)(0.225)(390 W/m^2)= 83.4 W/m^2 from back-reflection off clouds and aerosols. In reality, aerosols should play a smaller role here than for incoming solar radiation, so again this is a generous upper bound.

The IR-absorbing molecule upper limit on IR radiation returned to the ground which had been emitted by the ground is (0.95)(0.195)(390 W/m^2 = 72.2 W/m^2.

In this case, the sum of upper bounds on these three sources of radiation from the atmosphere is then (31.8 + 83.4 + 72.2) = 187.4 W/m^2. This is still much less than 324 W/m^2 claimed in the K-T diagram.

In reality, less of this radiation is actually incident on the surface than assumed in these upper bounds and much less is actually absorbed by the surface.

No, 4(168)= 672 W/m^2 assumes mean cloud cover and my calculation was for a spot on land (so it could be very dry) and without clouds. As a result, my case has a higher incident solar radiation on the ground.

SB incoming and outgoing radiation for the cooler body are equal when a hotter body (sun) radiates a cooler body (Earth), assuming that other cooling mechanisms or sources of heat do not exist. Here we have thermals and evapotranspiration in general, but they will have to be stronger effects to provide the necessary cooling than is the case in the K-T diagram.

Anonymous said...

The error is even more profound than that!

Most people know that hot air rises. Many know that when you heat a gas it expands.

So how much potential energy does the 10km and 10tonnes of air above every sq m of surface store when it is expands and rises up during the day?
How muchenergy is returned wehen it drops back down during the night?

Earths Energy Budget in reality is based on a transfer of thermal energy to gravitational potential energy during the day and the reverse transfer of potential energy to thermal energy at night.

The current climate models are fundamentally wrong.