Guest Post by Willis Eschenbach
OK, a quick pop quiz. The average temperature of the planet is about 14°C (57°F). If the earth had no atmosphere, and if it were a blackbody at the same distance from the sun, how much cooler would it be than at present?
a) 33°C (59°F) cooler
b) 20°C (36°F) cooler
c) 8° C (15°F) cooler
The answer may come as a surprise. If the earth were a blackbody at its present distance from the sun, it would be only 8°C cooler than it is now. That is to say, the net gain from our entire complete system, including clouds, surface albedo, aerosols, evaporation losses, and all the rest, is only 8°C above blackbody no-atmosphere conditions.
Why is the temperature rise so small? Here’s a diagram of what is happening.
Figure 1. Global energy budget, adapted and expanded from Kiehl/Trenberth . Values are in Watts per square metre (W/m2). Note the top of atmosphere (TOA) emission of 147 W/m2. Tropopause is the altitude where temperature stops decreasing with altitude.
As you can see, the temperature doesn’t rise much because there are a variety of losses in the complete system. Some of the incoming solar radiation is absorbed by the atmosphere. Some is radiated into space through the “atmospheric window”. Some is lost through latent heat (evaporation/transpiration), and some is lost as sensible heat (conduction/convection). Finally, some of this loss is due to the surface albedo.
The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.15 (15% of the solar radiation hitting the ground is reflected by the surface back to space). So let’s take that into account. If the earth had no atmosphere and had an average albedo like the present earth of 0.15, it would be about 20°C cooler than it is at present.
This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.
Why is this important? Because it allows us to determine the overall net climate sensitivity of the entire system. Climate sensitivity is defined by the UN IPCC as “the climate system response to sustained radiative forcing.” It is measured as the change in temperature from a given change in TOA atmospheric forcing.
As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise 0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.
The UN IPCC Fourth Assessment Report gives a much higher value for climate sensitivity. They say it is from 2°C to 4.5°C for a CO2 doubling, or from four to nine times higher than what we see in the real climate system. Why is their number so much higher? Inter alia, the reasons are:
1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.
2. The climate models underestimate the increase in evaporation with temperature.
3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways .
4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.
5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.
6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind.
Note that the climate sensitivity figure of half a degree per W/m2 is an average. It is not the equilibrium sensitivity. The equilibrium sensitivity has to be lower, since losses increase faster than TOA radiation. This is because both parasitic losses and albedo are temperature dependent, and rise faster than the increase in temperature:
a) Evaporation increases roughly exponentially with temperature, and linearly with wind speed.
b) Tropical cumulus clouds increase rapidly with increasing temperature, cutting down the incoming radiation.
c) Tropical thunderstorms also increase rapidly with increasing temperature, cooling the earth.
d) Sensible heat losses increase with the surface temperature.
e) Radiation losses increases proportional to the fourth power of temperature. This means that each additional degree of warming requires more and more input energy to achieve. To warm the earth from 13°C to 14°C requires 20% more energy than to warm it from minus 6°C (the current temperature less 20°C) to minus 5°C.
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase. As a result, the equilibrium value of the climate sensitivity (as defined by the IPCC) is certain to be smaller, and likely to be much smaller, than the half a degree per CO2 doubling as calculated above.
It was not a Cooper’s hawk. it was an Eschenbach hawk!
Willis, you are only looking at the high frequency end of an amplifying process, there is definetely a frequency dependency in climate sensititivity noticable, although not as high on a century timespan as IPCC models make of it..
http://members.casema.nl/errenwijlens/co2/howmuch.htm
See also Pelletier for a temperature power spectrum
Jon D. Pelletier, 2002, Natural variability of atmospheric temperatures and geomagnetic intensity over a wide range of time scales, doi:10.1073/pnas.022582599 PNAS February 19, 2002 vol. 99 no. Suppl 1 2546-2553 Link
kevoka (00:24:44)
That would be the temperature of a blackbody earth if it only received the amount of energy that is not reflected by the earth’s albedo, or about 235 W/m2. But that assumes clouds and surface albedo, neither of which exist on a blackbody. On a blackbody earth the current distance from the sun, it would only be 8°C cooler.
w.
tallbloke (00:35:35)
Hey, tallbloke, always good to hear from you. If you look at the Kiehl/Trenberth paper cited above under Figure 1, he says 40 W/m2 for the radiation loss through the window.
I’ve never considered the question of the atmospheric window and thunderstorms, not sure what the answer is. Thunderstorms definitely put more moisture into the high troposphere over the tropics.
Always more to learn, and more to think about.
w.
Brian W (00:44:44) :
It’s easy.
Incoming sunlight is 99% less than 4um in wavelength. Climate science calls this conventionally “shortwave”.
Radiation from the surface of the earth is 99.9% greater than 4um in wavelength. Climate science calls this conventionally “longwave”.
Which means solar and terrestrial radiation can be easily distinguished and lots can be learnt even without “fancy algorithms with built in beliefs.”
You can see some explanation about longwave and shortwave
Surprisingly, new NASA diagram shows no back-radiation..
http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/components2.gif
If Kiehl-Trenberth diagram is correct and backradiation of 321W/m2 is responsible for the hypothetical +33K effect, it means additional 3.7W/m2 will cause 0.38°C net warming.
kevoka (00:24:44) :
Everywhere I have read the calculations result in a blackbody temp of 255K.
This calculation is wrong, since it is calculated with present albedo of 0.3. 2/3 of this albedo is made by clouds, which should actually not be there if there are no GH gases or atmosphere.
More, much colder Earth with (frozen) oceans would turn partially to white snowball and its albedo would be mightily increased.
Re: Willis Eschenbach (00:44:35) :
“Regarding the moon, I’m not the one who brought it up”.
I know. I have used the moon as an example only because it is easy to see the implications as the temperaure swings are so huge. But the same idea could be applied to the Earth. The energy radiated by the Earth is not the same as the energy radiated by a black body of the same average but homogeneous temperature. It is quite bigger, because also the Earth has noticeable swings in the temperature between night and day (especially in the deserts), or the poles and the ecuator, or the summer and the winter hemispheres. The hottest parts radiate sooo much more.
interesting:::
I have a problem with your black body figures for the earth
http://en.wikipedia.org/wiki/Black_body
[edit] Temperature of Earth
If we substitute in the measured values for the Sun and Earth:
If we set the average emissivity to unity, we calculate the “effective temperature” of the Earth to be:
TE = 254.356 K or -18.8 C.
This is the temperature that the Earth would be at if it radiated as a perfect black body in the infrared, ignoring greenhouse effects, and assuming an unchanging albedo. The Earth in fact radiates almost as a perfect black body in the infrared which will raise the estimated temperature a few degrees above the effective temperature. If we wish to estimate what the temperature of the Earth would be if it had no atmosphere, then we could take the albedo and emissivity of the moon as a good estimate. The albedo and emissivity of the moon are about 0.1054[20] and 0.95[21] respectively, yielding an estimated temperature of about 1.36 C.
Estimates of the Earth’s average albedo vary in the range 0.3–0.4, resulting in different estimated effective temperatures. Estimates are often based on the solar constant (total insolation power density) rather than the temperature, size, and distance of the sun. For example, using 0.4 for albedo, and an insolation of 1400 W m−2), one obtains an effective temperature of about 245 K.[22] Similarly using albedo 0.3 and solar constant of 1372 W m−2), one obtains an effective temperature of 255 K.[23][24]
this is substantially different from your estimates is the author of the wiki misleading us?
the earth atmosphere albido is in the order of 30% based on satelliteMODIS values.
so the greengouse effect plus .3 albido gives an average temp of 14 deg c , a difference of 32deg c from the theoretical black body temp.
I’m scratching my ancient head a bit ! applying the albido of 30% should give the planet a temp of 279 deg k = 5.85 deg C meaning the greenhouse effect adds 8 deg c.
confusion abounds!
Steve Goddard (23:47:55)
The shell of your airplane heated due to air friction! Blankets indeed!
Science of Doom (23:43:15)
“We can measure downward longwave radiation (greater than 4um) at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation less than 4um.”
The sun is a “broadband radiator” it radiates at ALL frequencies. Since everyone likes w/m2 here is some watts for you. Bright sunlight provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is INFRARED light, 445 watts is visible light, and 32 watts is ultraviolet light(wiki). And these are only three tiny bands of frequencies in a much larger spectrum. So good luck with your pseudoscience of doom. You are just another warmist site promulgating calculated nonsense.
scienceofdoom (23:43:15) :
“Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.”
Isn’t this because the absorption of IR at specific wavelengths raises the internal (vibrational) energy of the molecule (of CO2, H2O), which then promptly re-emits energy at the same wavelength but in a random direction, hence the notches in the upwelling radiation? The absorption is an internal process, not a thermal process.
Hi Willis,
good post i think. I’ve been studying some time this kind of radiation balances, So some things can be added for further knowledge:
1. The graph is old in this authors (Trenberth et altea, 2008), the new one is clearly unbalanced to warm ( i couldn’t see clear what was based on)
2. I think they don’t have applied some basic laws of energy and thermodynamics like: internal energy of the system (U), all the kinds of energy that the earth also receipt, f. e., magnetic, gravitational, etc…. Also thety didnt take into account the effect of winds as cinetic energy, etc… this graph suffers from too much simplicity (only radiation).
Juraj V. (01:34:32)
Once again, people are confusing net radiation flows with the individual radiation flows.
Suppose I give you a hundred dollars, and you give me seventy-five dollars. There are two equally valid and accurate ways of representing this transaction:
Figure 3. Two different ways of correctly illustrating the same transaction. 3(a) shows the two individual transactions. 3(b) shows the net effect of the two transactions.
Now, if we look at Figure 3(b) above, does the lack of an arrow from you to me mean that you didn’t give me $75? No, of course not. But since your $75 is smaller than the amount I gave you, it does not show up in the net flow diagram.
What the new Nasa diagram is showing is net flows. What Figure 2.3 above is showing is net flows. But as my drawing above illustrates, that doesn’t mean that the individual flows don’t exist. Downwelling radiation is real. But since it is smaller than the upwelling radiation, it doesn’t show up in the net flow diagram. This does not mean that it doesn’t exist …
Not true by the UN IPCC definition. That definition relates warming to top of atmosphere (TOA) radiation, not to downwelling radiation from the bottom of the atmosphere.
Lindsay H (01:49:50)
Blackbody earth calculations:
Insolation: 345 W/m2
Stefan-Boltzmann Constant (S): 5.67 E -8
Equation: R = S T^4
where R is radiation (W/m2) and T is temperature (K)
Solving for T, we get:
T = (R/S) ^ (1/4)
= (345 / 5.67 E-8) ^ (1/4)
= 279K = 6°C
The Wikipedia article is using the amount of sun after albedo losses, which is about 235 W/m2. That gives the colder value you quote above.
But I’m interested in what happens without the albedo (or to be more accurate, without the cloud albedo). Calculating the temperature with only the surface albedo shows the change that is due to the atmosphere with all its components (GHGs, aerosols, clouds, dust, and all the rest).
What it shows is that the addition of an atmosphere gives us a TOA radiation of ~ 150 W/m2, and a temperature rise of ~ 20°C. This gives the average climate sensitivity shown above.
I thought incoming solar radiation was about 1400W/m^2 i.e. the TSI figure.
Why is it 342 W/m^2 in the diagram? Or is it a different physical quantity?
Thanks
Dermot
Willis
You might have fun adding a graph showing the Earth’s temperature profile up to 500km above surface – it might surprise quite a few that the temperature does not continue dropping the further from Earth but actually rises significantly into the thermosphere.
So which temperature would be representative of the Earth’s temperature, and does the temperature at the thermosphere change diurnally etc etc etc.
The climate people are waxing thermally over the bottom part of the plot – but presumably remain ignorant of what causes the rest of the profile up to 500 km etc.
http://www.windows.ucar.edu/tour/link=/earth/images/profile_jpg_image.html
Have fun folks – I am bush till Easter drilling holes in the ground testing scientific hypotheses in situ.
Steve Goddard:
But blankets don’t warm by suppressing radiation.
I think that in time the Kiehl/Trenberth diagram will be considered an even bigger joke than the Mann hockey stick.
Proof
The hockey stick had at least a possibility of being correct, if the predictions turned out to true.
Its a travesty for him that reality dealt a massive raspberry.
The Kiehl/Trenberth diagram is wrong from the start
Just two points
1. They seem at great pains to show that the radiation intensity is conserved.
There is no such conservation law in Physics.
2. The magnitude of so called back radiation from the trace gasses co2 and water vapour is shown as almost twice the total solar radiation must be physical nonsense.
Brian W (01:59:09)
Brian, there is infrared and there is infrared. ScienceOfDoom points out that 99% of the sun’s radiation is at less than 4um wavelength. While (as you point out) this contains UV, visible, and infrared, it is not infrared in the frequency range of the downwelling longwave “greenhouse” infrared radiation, which is all greater than 4um wavelength.
So ScienceOfDoom is correct, they can easily be distinguished. It is also the reason that the infrared from the sun passes through the atmosphere, because the atmosphere is basically transparent to IR in the solar wavelengths, but absorbs in the “greenhouse” wavelengths.
Finally, a “warmist” site??? I think that downwelling IR exists because PEOPLE MEASURE IT ALL THE TIME, ALL OVER THE WORLD. That doesn’t make me a “warmist” … it makes me a realist.
So I ask again, if you want to argue that downwelling IR doesn’t exist, please do so elsewhere. This is not the site for that discussion. Here, we believe in things that can be measured scientifically, and downwelling IR is one of them. I have neither the time nor the inclination to argue with people who think that things that can be measured don’t exist.
Dermot O’Logical (02:29:29)
Excellent question. If the earth was a flat disk (area of pi R^2), one side of it would receive the 1372 W/m2 from the sun. But the area of the earth is much larger than that. It is 4 pi R^2. So we need to divide the solar radiation by the ratio of the surface areas, which gives us 1372/4 = ~ 345 W/m2.
Juraj V. (01:42:28) :
“More, much colder Earth with (frozen) oceans would turn partially to white snowball and its albedo would be mightily increased.”
There can’t be exposed ice if there is no atmosphere. Ice without an atmosphere vaporizes. That is why there is no exposed ice on Mars or the Moon.
Willis,
Another consideration of the energy budget is in figure 2 of http://physicsworld.com/cws/article/print/17402 . You will need to log in to physicsworld.com (IOP website)but it is free. As far as I know has discredited this assessment and the work of Ahilleas Maurellis.
In my assessment (from texts books on heat transfer and personal experience and measurement) if there are two molecules of CO2 near each other in the atmosphere and they are at the same temperature there will be no heat transfer them because both molecules will be saturated at the wavelengths of absorption and emission. If the temperature is different, heat flux will be from the higher to the lower temperature and never the reverse. The same thing applies to applies to the earth’s surface which is about 70% water with an emissivity of about 0.95 (almost a black body). If the surface temperature is above that of the atmosphere, which is the case due to due the the lapse rate and radiation from H2O and CO2 heat will only from the surface.
In my assessment the water vapor (and water and ice in clouds) and to a minor extent CO2 can be compared to a perforated wide spaced sieve. A large quantity of heat flows straight through while some is absorbed and re-radiated upwards reducing the overall heat flux (or heat flow rate). The heat flux will vary considerable around the globe and at different times of the various cycles which affect the earth.
I have no respect for the Kiehl and Trenberth paper. However, I do not dispute (more correctly agree) some of the other points you make such as about evaporation and forced (wind) convective heat transfer.
Dermot O’Logical (02:29:29) :
“I thought incoming solar radiation was about 1400W/m^2 i.e. the TSI figure.
Why is it 342 W/m^2 in the diagram? Or is it a different physical quantity?”
1400W/m2 is what you would get in space, in a surface perpendicular to the incident light. 342W/m2 represents an average in the surface of the Earth between night and day zones as well as the different angles of the incident light wrt the Earth’s surface. That is, if you multiply 342 times the area of the surface of the Earth you would get the total energy we receive from the Sun at any given time.
So, Willis, if a station is located where it is naturally near maximum for it’s latitude (due to phsyical geography and circulation), then UHI from any C02 increase will be negligible.
Give me a day or so, I am almost done with just such a station.
Willis Eschenbach (02:51:43)
Much obliged. Kind of obvious when you…. think… about…. it. I must learn to do that.
But why divide by the _whole_ surface area of the Earth? We are only getting sunlight on half of the surface area i.e. the daylight side. It seems invalid to average the incoming insolation over the whole surface area, as that would imply complete absorption of the energy, then being evenly distributed across the whole surface, _before_ being re-radiated away. Stefans law would suggest the daylight side radiates away significantly more energy that the night side, and that’s non-linear, at T^4.
(Please forgive amateur analysis of a physics graduate 20 years out of University – the day job doesn’t require me to keep up to speed with thermodynamics and statistics)
Dermot.