Guest Post by Willis Eschenbach
Some days I learn a lot. Today was one of them. Let me start at the start. Back in 1987 in a paper entitled ‘The Role of Earth Radiation Budget Studies in Climate and General Circulation Research“, a prescient climate scientist yclept Veerabhadran Ramanathan pointed out that the poorly-named “greenhouse effect” can be measured as the amount of longwave energy radiated upwards at the surface minus the upwelling longwave radiation at the top of the atmosphere, viz:
The greenhouse effect. The estimates of the outgoing longwave radiation also lead to quantitative inferences about the atmospheric greenhouse effect. At a globally averaged temperature of 15°C the surface emits about 390 W m -2, while according to satellites, the long-wave radiation escaping to space is only 237 W m -2. Thus the absorption and emission of long-wave radiation by the intervening atmospheric gases and clouds cause a net reduction of about 150 W m -2 in the radiation emitted to space. This trapping effect of radiation, referred to as the greenhouse effect, plays a dominant role in governing the temperature of the planet.
And here is what Ramanathan was talking about:
Figure 1. All-sky (both cloudy and clear) greenhouse effect. In climate science, “upwelling” means headed for space, “downwelling” means headed for the surface, “forcing” means a change in downwelling radiation, “LW” is thermal longwave radiation, and “SW” is solar shortwave radiation.
The best modern information about this question comes from the CERES Energy Balanced and Filled (EBAF) dataset that I used to make Figure 1. It combines a number of satellite and other measurements into a single coherent group of individual datasets. Interestingly, Ramanathan’s estimate of the size of the greenhouse effect was “about 150 W/m2” and modern CERES data shows a number very close to that, 158 W/m2. Well done, that man!
Today, a chance comment got me thinking about top-of-atmosphere (TOA) downwelling longwave radiation versus what happens at the surface. A doubling of CO2 is supposed to lead to a 3.7 W/m2 increase in downwelling TOA longwave radiation … but what does that do to downwelling LW at the surface?
So what I did was to calculate on a monthly basis, the change in downwelling longwave radiation at the surface for each one W/m2 change in TOA greenhouse radiation. Figure 2 shows that result.
Figure 2. Change in downwelling radiation at the surface for each 1 W/m2 change in downwelling TOA radiation.
Now, this is curious. On average the change at the surface is a little less than half the TOA greenhouse effect change. So an increase of 3.7 W/m2 at the TOA from a doubling of CO2 becomes a 1.8 W/m2 increase at the surface. I would note that this value of 0.46 agrees in general with the published study of Feldman et al. in Nature magazine who found (from observations, not models) that surface forcing is 0.43 times the TOA forcing, quite near to the above figure.
Next, I got to wondering about something I’d never looked at—just how large an additional energy flux in watts per square metre of energy is needed to increase the surface temperature by 1°C. This is a simple calculation using the Stefan-Boltzmann equation, but I’d never done it for the entire globe. Figure 3 shows that result.
Figure 3. Increase in ongoing downwelling energy flux needed to increase the surface temperature by 1°C with everything else unchanged.
In Figure 3 you can see that as Stefan-Boltzmann says, it takes more energy to raise a hot surface by 1°C than to raise a cold surface by 1°C. And for the globe, the average is about 5.5 W/m2 per degree. That was a surprise to me, I didn’t expect it to be quite that large … but then as I said, I’d never calculated it.
So here’s the summary of today’s wanderings in CERESville.
• The long-accepted value for a doubling of CO2 gives a theoretical 3.7 W/m2 increase in downwelling TOA radiation. However, because of all of the factors that affect downwelling TOA radiation (changes in clouds, temperature, water vapor, eruptions, aerosols, etc.) and the fact that the log of CO2 is essentially a straight line, it’s not possible to determine that value experimentally. Here’s the problem:
Figure 4. Ramanathan’s greenhouse radiation, along with the change in CO2 radiation over the period. The CO2 radiation change has been set to the average of the greenhouse radiation for easy comparison.
Using that accepted 3.7 W/m2 figure for a doubling of CO2, that would give an increase in downwelling surface radiation of 1.8 W/m2.
• This doubling of CO2, in turn, would warm the surface by:
1.8 watts per square metre CO2 surface forcing / 5.5 watts per square metre per degree C ≈ 0.3°C …
By comparison, the IPCC says that a doubling of CO2 would increase the surface temperature by 1.5°C to 4.5°C. If we take the midrange value of 3°C, this would imply that there is some mysterious feedback increasing the CO2-caused surface temperature change by a factor of about ten …
The general view seems to be that this mysterious ten-fold increase is somehow the result of feedback from water vapor and clouds. The problem with that theory is that the CERES measurements I’ve used above include all of those feedbacks. That is to say, the GHE value includes the feedback effects of clouds and water vapor, and the surface downwelling radiation also includes those feedbacks.
Answers gladly accepted. Here on the northern California coast, despite the screaming about “PERPETUAL CALIFORNIA DROUGHT! CLIMATE EMERGENCY!” … it’s raining again, the trees are happy, and the cat is not.
My best regards to all,
w.
NOTES: For those unclear on the physics behind the poorly-named “greenhouse effect”, it works because a sphere only has one surface, and a shell has two surfaces, inside and outside. See “The Steel Greenhouse“, “People Living In Glass Planets“, and “The R. W. Wood Experiment” for further discussion.
MY USUAL REQUEST: When you comment please quote the exact words you are discussing, so we can all be clear on your exact subject.
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A comment about the climate sensitivity value:
Quote: “By comparison, the IPCC says that a doubling of CO2 would increase the surface temperature by 1.5°C to 4.5°C. If we take the midrange value of 3°C,…”
IPCC uses both ECS (Equilibrium Climate Sensitivity) and TCS (Transient Climate Sensitivity) concepts and summarizes the differences in AR5 [48], (p. 1110): “ECS determines the eventual warming in response to stabilization of atmospheric composition on multi-century time scales, while TCR determines the warming expected at a given time following any steady increase in forcing over a 50- to 100-year time scale.” IPCC has changed the TCS to TCR (Transient Climate Response). On page 1112 of AR5 [48], IPCC states that “TCR is a more informative indicator of future climate than ECS.” According to the IPCC science for the temperature changes during this century, the right CS is TCS, which is about 1.8 C on average according to the IPCC.
Quote: “So an increase of 3.7 W/m2 at the TOA from a doubling of CO2 becomes a 1.8 W/m2 increase at the surface.”.
Still I do not find any groundings, why Willis make this reduction and is it based on the IPCC science or on something else.
The K-T W/m^2 heat balance diagram and all its clones are thermodynamic garbage.
1.) Converting 289 K to 396 W/m^2 using S-B creates energy out of thin air, plus the surface cannot radiate BB.
2.) 333 W/m^2 continuously/perpetually looping/upwelling/downwelling/absorbing/”trapping” at 100% efficiency leaving nothing behind either place violates thermodynamics. YES IT DOES!!!
3.) 333 W/m^2 from the cold troposphere to the warm earth without added work violates thermodynamics. YES IT DOES!!!^2
4.) 288 K – 255 K = 33 C is rubbish. 288 K is pulled out of WMO’s butt. K-T uses 289 K. UCLA Diviner says 294 K. 255 K assumes the naked earth keeps the 0.3 albedo and that assumption is scientific if not criminal malfeasance.
It all reminds me of reading Karl Marx. First start with some arm waving, throw in some sophistry, mix it together to become a factoid, and then proceed with volumes and volumes of erroneous work based on that false premise.
Nick, have heart. It is a scientific model, used to scare the hell out of children. Also to put billions in pockets of properly connected people. It is not meant to be perfect, except for these listed purposes.
1) no it doesn’t, and it doesn’t make sense that you think that.
2) you are confusing thermodynamic heat flow from hot to cold, with electromagnetic components in the SB equation
3) no it doesn’t, see 2)
4) thats the number that demonstrates the radiative effect using albedo of .3 to illustrate the effect to high school students. Everybody knows it’s more complicated than that.
DM,
“2) you are confusing thermodynamic heat flow from hot to cold, with electromagnetic components in the SB equation”
Two bodies *do* exchange heat, i.e. a heat flow, via electromagnetic components. If the earth is considered to be a separate entity from the atmosphere then radiative heat flow must be considered. Please note that this heat flow *is* an exchange. The colder body does radiate toward the hotter body as well as the other way. Energy can’t be destroyed, only transformed. If no energy is left behind then where did it go?
“3) no it doesn’t, see 2)”
Heat flow from a hotter body to a colder body via radiation increases the kinetic energy of the colder body. How do you increase kinetic energy without performing any work?
“The colder body does radiate toward the hotter body as well as the other way.”
Does not happen.
If it did there would be refrigerators without power cords.
I haven’t seem any.
You?
This hopeless but anyway. How does the colder body know that it should not emit radiation according to Planck’s law because there is a warmer body somewhere in the vicinity. If you can give a scientific answer to this, you will win a Nobel prize.
He is to stupid to get that problem.
Apparently the photon leaving the sun knows ahead of time that earth is cooler and the photon leaving earth knows ahead of time that the sun is hotter 🙂
Hotter lose KE.
Colder gains KE.
Until equal temperatures and energy transfer stops.
Nick,
Hotter lose KE.
Colder loses KE.
Colder loses KE by radiating energy until it has Virtually none left.
All bodies not at absolute zero are radiating heat.
That radiated heat adds to the energy load of any other bodies it encounters warming them a little or slowing their rate of heat loss.
Not a Stokes puppet are you?
His sort of logic.
Nick is busy inventing his own physics because he isn’t using any physics that the rest of us know.
The interesting question to ask Nick would be what is the temperature of a laser beam?
I will have one shot at this and see if you can understand.
A photon has no temperature it a quantum entity all it has that you would recognize is energy. A laser beam is neither hot nor cold because temperature is not a fundamental property it is a classical physics construct. Your thermal emission is just a bunch of these photons it is called “thermal emission” because of the source but it actually has no temperature.
You have probably run across that fact when as a kid you used a magnifying glass and the sun to burn things … Concentrating the sunlight changed the apparent temperature at the focus point. What you are doing in concentrating the energy which becomes temperature in your classical world again. A photon in 10um infra-red (which is about the emission of a 300K earth) when concentrated forms the basis of CO2 metal cutting laser which reaches many thousands of degree at it’s focal point.
That is why your statement about cold not transmitting to hot fails …. QM doesn’t know what the temperature is it just transfers energy. So yes photons from cold sources do enter and get absorbed by hot sources because QM doesn’t know what temperature is.
Imagine a metal plate heated to say 400oC dangling in a vacuum. It’ll lose heat via radiation, and its core temperature can be plotted over time. Call this temperature profile A. Now repeat, but this time dangle metal plates heated to 300oC either side of the first plate, and plot the core temperature of the first plate over time again, calling it profile B. Will profile B (i) be identical to profile A; or (ii) trend above profile A (demonstrating lower rate of heat loss)? If you are saying that (i) is the correct answer, where is the radiative heat emitted from the two cooler plates in the direction of the hotter plate going?
“Energy can’t be destroyed, only transformed.”
How about sent back where it came from until they reach equilibrium.
Except they never do of course, a. because the CO2 also radiates to space, thus cooling everything and b. the Sun comes up next day to do it all again.
I don’t get B the sun is always shinning on Earth it’s just on the other side 🙂
Antero,
“This hopeless but anyway. How does the colder body know that it should not emit radiation according to Planck’s law because there is a warmer body somewhere in the vicinity. If you can give a scientific answer to this, you will win a Nobel prize.”
You nailed it. The heat transferred is dependent upon the total flow. (flow hot-to-cold) – flow cold-to-hot) determines total flow. Flow-cool -to-hot is *not* zero.
Wow, none of you guys get it. If there were heat, also known as energy, flowing from the cold body to the hot body, the hot body would get HOTTER. It does NOT, NEVER EVER!
Michael Moon January 21, 2020 at 4:26 pm
Suppose you have two stars far apart, one at say 4000°C and one at 6,000°C. You move them until they’re very close to each other.
BOTH OF THEM GET HOTTER.
Why? Because compared to when they were separated, both of them are receiving extra energy from the other star.
Inter alia, your problem is that you think that heat is “also known as energy”. It is not. Heat is a spontaneous flow of energy from a warm object to a cool object. Energy is just energy, and radiant energy flows both directions as in the example with the stars.
There is a NET spontaneous flow of heat from the hotter to the cooler star … but that is comprised of two separate independent heat flows, from star A to star B, and from star B to star A. And both of them end up hotter in the process
w.
Please read my explanation above … you start with the basic question how hot is a photon (which can be asked as how hot is a laser beam). You should quickly work out that a photon has no temperature it only has energy. That is why your statement quickly becomes nonsense because classical physics breaks down and you can’t apply it to the situation.
You were taught classical physics at school and there was a emphasis on temperature but there is no such fundamental property as temperature. It is an observational construct much like a rainbow of some deeper physics.
Michael Moon January 21, 2020 at 4:26 pm
“Wow, none of you guys get it. If there were heat, also known as energy, flowing from the cold body to the hot body, the hot body would get HOTTER. It does NOT, NEVER EVER!”
You didn’t bother to think about what I wrote, did you?
I said: “The heat transferred is dependent upon the total flow. (flow hot-to-cold) – flow cold-to-hot) determines total flow. Flow-cool -to-hot is *not* zero.”
If the energy received by the warmer body from the cooler body is less that the energy the warmer body emits toward the cooler body then the warmer body will *not* get hotter. It will see a net loss of energy and become “cooler”.
If the cooler body receives more energy from the warmer body than it emits toward the warmer body then the cooler body will see a net increase in energy and become “warmer”.
Temperature is the average kinetic ENERGY of the molecules of a substance. There are many binary stars in elliptical orbits, and the temp of the hotter one never increases as the cooler one approaches perihelion. Look it up. What you do not know fills all the textbooks about thermodynamics and heat transfer, actually basic physics. Energy is the capacity to do work. Heat is the average kinetic energy of a mass.
Cooler things never heat warmer things, as when something is heated, its TEMP goes UP. This has nothing to do with the way things cool down and your example with a plate between a heater and a human is sense-free.
Michael Moon January 21, 2020 at 4:26 pm
Wow, none of you guys get it. If there were heat, also known as energy, flowing from the cold body to the hot body, the hot body would get HOTTER. It does NOT, NEVER EVER!
Amazing, over 50 years of experience using radiation shields around thermocouples to get more accurate flame temperatures didn’t happen!
Detailed analysis of shielded thermocouples so old that they’re in NACA papers:
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19930084456.pdf
Michael Moon January 22, 2020 at 7:14 am
That’s true, but not for the reason you think. In such a system, when the stars are closer together, they move faster. And when they are further apart, they move slower. As a result, the total energy going in each direction (flow * time) stays the same.
This is true for the earth as well. You’d think we’d get hotter when we’re nearer the sun. But we spend less time there, and the two effects cancel each other out.
As you say … look it up …
w.
‘In climate science, “upwelling” means headed for space, “downwelling” means headed for the surface, “forcing” means a change in downwelling radiation, “LW” is thermal longwave radiation, and “SW” is solar shortwave radiation.’
Is there a formal definition of the boundary between “LW” and “SW”?
This is important because solar radiation is spectral, with units of watts per square meter per wavelength unit (um or nm). A total irradiance in W/m2 is calculated by integrating over a specific wavelength region. When total irradiance is measured with a radiometer, the instrument implicitly performs the integration (and the answer is always “colored” by its spectral transmittance).
So stating there is “156 W/m2 of LW radiation” is nearly meaningless without also stating the wavelength bounds that LW refers to.
Thermopile radiometers such as global pyranometers typically have uncertainties on the order of +/-5% at best.
The Instruments & Measurements
But wait, you say, upwelling LWIR power flux is actually measured.
Well, no it’s not.
IR instruments, e.g. pyrheliometers, radiometers, etc. don’t directly measure power flux. They measure a relative temperature compared to heated/chilled/calibration/reference thermistors or thermopiles and INFER a power flux using that comparative temperature and ASSUMING an emissivity of 1.0. The Apogee instrument instruction book actually warns the owner/operator about this potential error noting that ground/surface emissivity can be less than 1.0.
That this warning went unheeded explains why SURFRAD upwelling LWIR with an assumed and uncorrected emissivity of 1.0 measures TWICE as much upwelling LWIR as incoming ISR, a rather egregious breach of energy conservation.
This also explains why USCRN data shows that the IR (SUR_TEMP) parallels the 1.5 m air temperature, (T_HR_AVG) and not the actual ground (SOIL_TEMP_5). The actual ground is warmer than the air temperature with few exceptions, contradicting the RGHE notion that the air warms the ground.
I think that I have seen this comment before that there is no way to measure upwelling LW radiation emitted by the surface or even the LW radiation into space. These kinds of stories are part of the conspiracy stories. USA has a government that has money to invest in several satellites measuring LW radiation emitted by the Earth and the whole thing is a fairy tale.
I have analyzed the global results of upwelling LW radiation observations originating from GEWEX project. If somebody is interested in it, he/she can find this information just by googling.
My (poorly expressed) point should have been this: taking arithmetic sums or differences of these total irradiances (however measured) is simplistic because of the spectral nature of the quantities, the numeric definite integrations can’t be avoided. In other words, adding and subtracting irradiance is only valid at single wavelengths.
And I really would like to understand the spectral difference between what the climate modelers refer to as SW and LW.
For example, is it defined by specific instruments, such as pyranometers and pyrgeometers?
https://www.kippzonen.com/ProductGroup/3/Pyranometers
https://www.kippzonen.com/ProductGroup/4/Pyrgeometers
Both are thermopile instruments (with calibration constants of uV/W/m2). Their spectral bands are defined by the transmittance of the domes/windows: pyranometers typically have quartz or silica domes that have long-wavelength cutoffs of 2.5-4 um, pyrgeometer domes pass 4.5-42 um (I don’t what these materials are).
Based on these two instruments, the division between SW and LW would be about 4 um, is this correct?
And, oh, BTW,
5.) Dividing the discular 1,368 W/m^2 ISR by 4 to get an averaged 342 W/m^2 spherical ToA ISR is simplistic and dumb and and doesn’t even remotely resemble how the earth heats and cools.
6) The area of Earth used in the S-B equation for both absorption and emission is approximately 196,900,000 sq mi. This is an abstract number and not a real number that should concern scientists. In reality, the absorbing surface is much higher and the emitting surface is much much higher, because the Earth is not a polished marble but rather is covered by plants, geomorphic features, and rough surface waters.
If you think that’s bad, you should see Venus.
Meanwhile, back on mostly-ocean Earth, the sea is rarely smooth anywhere for long, resulting in constantly fluctuating surface area and zenith angles. Add in humidity and wind mass transport, and good luck modeling that. Using historical averages is a fool’s errand.
Your number is for the spherical area.
Only the lit hemisphere receives.
The entire sphere loses energy 24/7.
Because the lit side is hotter a large percentage leaves as radiation.
Because the dark side and poles are cold a rather small percentage leaves as radiation.
That’s why dividing by 4 and using averages for all flows is useless.
https://www.linkedin.com/posts/nicholas-schroeder-55934820_greenhouse-climatechange-activity-6503085690262216704-GDKH
5.5 W/m² per degree.
And the latest Trenberth heat budget says “net absorbed 0.9 W/M²” So that’s what, a little north of 0.15 deg C of warming?
5.5 w/m2 per degree C is surface warming without feedbacks. If Trenberth’s figure refers to the TOA then surface forcing will be ~1.5 w/m2
Also, since we’ve had over 0.5 degree warming since 1980 alone perhaps the alarmists are right and there is a strong positive feedback.
Feldman et al. referred to by Willis says the following:
“The climate perturbation from this surface forcing will be larger than the observed effect, since it has been found that the water-vapour feedback enhances greenhouse gas forcing at the surface by a factor of three and will increase, largely owing to thermodynamic constraints”.
Thanks, Phil. The data that I’m using includes all feedbacks. The data Feldman was working with didn’t. The citation for the factor-of-three increase clearly states that the changes occur within the same month as the temperature changes, so this is included in the CERES monthly data.
w.
Phil and Willis,
Thanks.
That answers my somewhat obtuse question in this thread about Paltridge et al (2009) as against Dessler and Davis (2010)regarding water vapour feedback.
If all the comments about CO2’s ability were true then two things must happen. 1. The specific heat of dry air must be revalued. 2. An addendum to specific heat tables must be published saying something like “If IR involved in energy supplied to increase temperature of either dry air or CO2 then the forcing equation must be included.”
As of now climate science is in conflict with thermodynamics area of specific heat. Thermodynamics says the energy can be of any form. Climate science has special case for IR and CO2. Anthony’s experiment demonstrated that increased CO2 does not cause increased temperature.
I’m just wondering how everybody deals with average temperature, while in the Stefan-Boltzmann’s law the temperature is in fourth power.
In other words if you have 1 square meter of 50 deg Celsius and 1 square meter of 0 deg. Celsius, their emission won’t be equal to 2 square meters at 25 deg Celsus, will it?
Luchezar Jackov January 21, 2020 at 9:28 am
Phrase that in Kelvin and see if you get something different.
It’s still in the 4-th power.
1 sq.m. of 50C = 323.15^4 = 10904773289.97600625
1 sq.m. of 0C = 273.15^4 = 5566789756.30100625
Total: 16471563046.2770125
2 sq.m. of 25C = 15804081127.5270125
Difference: 16471563046.2770125/15804081127.5270125 = 1.0422347818493162903409232289584
i.e. 4.22% difference
In other words, it is incorrect to calculate the outgoing radiation by averaging the temperature, be it day/night temperature or equator/pole temperature.
I don’t understand the focus on this question, since nowhere have I averaged temperatures. What are you all on about?
This is why I ask people to QUOTE THE EXACT WORDS you are discussing …
w.
I’d rather name my comment a food for thought. Looking at Figure 3 and the conclusion that it takes more energy to keep a hot surface one degree hotter than a cold surface provoked it.
Let’s take the tropics for example. The energy budget there is quite different from 1362/4 W/m2. It’s close to 1362/pi W/m2 (opposed to the poles where it is much less). The land surfaces heat up to more than 50C in the day (even though the air doesn’t get that hot) and when heated, they radiate. In the night they radiate, effectively cooling probably under the air temperature. Taking the average of the air day and night temperatures and assuming this is the temperature of emitting is not correct, because of the 4-th power in Stefan-Boltzmann’s law.
In other words the hotter surfaces of the Earth dissipate much more energy than the cooler surfaces, and averaging the temperature and then calculating the dissipation based on the average temperature is incorrect.
It will be different, but still not correct.
The average of 0C, 273.15^4 and 50C, 323.15^ is 8235781523
Whereas 25C, 298.15^4 is 7902040564
So an error of 333740959
Near enough for Government Climate work of course.
This illustrates just one of the many problems with climate science. The surface of the Earth is not a cavity radiator. Thus, the actual LWIR characteristics of surface materials matters and one cannot just use the Stefan-Boltzmann formula without some adjustment. The average emissivity of snow, vegetation, soils, water, and so forth is probably in the neighborhood of 0.97. So at 288K (i.e. about 15C). The actual emission is more like
. The difference is already nearly four times bigger than the effect of doubling CO2 in the tropics (MODTRAN calculation). The whole science is filled with big numbers multiplied by fractional quantities that are poorly known then subtracted from other large, but error riddled numbers, and so on. Given enough imprecision and inaccuracy almost anything seems possible.
Kevin,
+10
Mr kilty, what is the difference between emissivity and reflectance, which during the winter in the NH is very high due to the angle of incidence.
Which means that the full range of Solar Radiation gets reflected rather than LWIR getting emitted.
Emissivity is a material property. It is a dimensionless quantity, and one would calculate it as the ratio of the measured power emitted from a surface by virtue of its temperature divided by the power a perfect blackbody would emit at the same temperature. Since real materials always show some angular and frequency dependence to their emitted power, we can say that “emissivity” is only an approximation to the complexity of emitted power from a surface that is not within a cavity. Emissivity is based on emitted power into a hemisphere above a flat surface.
Reflectance is the measured ratio of redirected to incident light intensity at a material interface. This definition is per the “Encyclopedia of Physics”. It is based on inteensity which is radiance energy in a pencil of radiation at some angle to a surface. You can see in this thread some disagreement over LWIR and solar radiation, and as the above definition uses the term “light” one might wonder if reflectance applies to LW radiation in the same way it does to visible or solar radiation or not. You mention that reflectance increases at low solar angle in the northern hemisphere winter, which it does, but a rough sea state probably also shows an enhanced reflectance as well.
In the context of climate science light would likely refer to solar radiation including radiation predominantly from about 2.8 um and shorter wavelengths (an Eppley PSP for instance uses this definition); while LWIR is 4 um and longer (the definition as per an Eppley PIR), leaving a no-man’s land of radiation between the two for people to fight over.
As I explained in a blog some months ago, total power travelling away from a surface is a combination of reflected and emitted power which is known as radiosity.
A C Osborn January 21, 2020 at 10:35 am
Emissivity refers to the absorption/emission of thermal infrared. By Kirschoff’s law, emissivity = absorptivity. Emissivity of fresh snow is about 0.986. This means that about 1.4% of impinging longwave is reflected, and 98.6% of impinging longwave is absorbed by snow.
Reflectance is also called “albedo” when referring to visible light. It generally refers to the amount of impinging radiation reflected from an object. There is no corresponding “emissivity” for visible light, objects in general don’t emit light. The albedo of fresh snow is about 0.95, meaning that about 95% of impinging sunlight is reflected, and 5% of impinging sunlight is absorbed by snow.
Note that the numbers are nearly reversed for visible and LW infrared … almost all visible is reflected by snow, and at the same time, almost all LWIR is absorbed by snow.
Go figure …
Also … what Kevin said.
w.
Kevin kilty…
The whole science is filled with big numbers multiplied by fractional quantities that are poorly known then subtracted from other large, but error riddled numbers, and so on. Given enough imprecision and inaccuracy almost anything seems possible.
My favorite is from the IPCC’s AR4 Chapter 5 Executive summary page 387 where they say:
Over the period 1961 to 2003, global ocean temperature has risen by 0.10°C from the surface to a depth of 700 m.
Of course Trenberth’s Heat Budget is right in there:
The noon-day sun puts out nearly 1370 wm² and these guys are claiming they’ve added up all the chaotic movements of heat over the entire planet and have determined an imbalance of 0.9 Wm². That’s an accuracy to five places. No plus or minus error bars or anything.
What it means is, all of the components
Reflected by clouds
Reflected by aerosols
Reflected by atmospheric gases
Reflected by surface
Absorbed by the surface
Absorbed by the atmosphere
Thermals
Evaporation
Transpiration
Latent heat
Emitted by clouds
Emitted by atmosphere
Atmospheric Window
AND
Back radiation
need to have an accuracy to those five places or better for the 0.9 Wm² to be true.
“Back radiation”
No such thang!!
Nick Schroeder January 21, 2020 at 12:41 pm
“Back radiation”
No such thang!!
How can you say that, it’s right there on Trenberth’s chart,
Global Energy Flows W m²Back Radiation 333
http://www.cgd.ucar.edu/staff/trenbert/trenberth.papers/BAMSmarTrenberth.pdf
Kevin, you’re correct. But in general, it’s a difference that makes no difference. This is because we’re usually looking at differences in temperature and differences in radiation.
Let’s say a surface warms by 1°C, from 15°C to 16°C
Using an emissivity of 1, that’s a change of 5.45 W/m2.
Using an emissivity of 0.97, that’s a change of 5.29 W/m2, a difference of a bit more than a tenth of a W/m2.
As a result, using emissivity =1 is the common practice.
w.
I understand your point, Willis, but my complaint is that the idea of radiation balance involves some numbers that make use of an effective emissivity, which we can’t possibly know better than
and others make use of an albedo which we surely don’t know better than 
and then we mix them numerically. I understand that we can juggle the numbers to obtain a balance, and then work with differences from there, but even so, how well do we know the resulting balance, or its trend? I am always skeptical of just-so arguments.
As an ocean science sort of fellow you can appreciate an analogy that drove me bonkers for years. The oceanography textbooks, and many articles (in the old Scientific American for instance) would always explain anomalously large tides, like at Truro, as a resonance phenomena; and then they would use a rectangular basin of water to explain the resonance — they would fudge the dimensions of this basin to get the answer they wanted. But any place having anomalously large tides is never on the shore of a rectangular basin. They are always an inlet in a channel or estuary having dimensions that diminish toward the port. The criteria for resonance in such an instance isn’t anything like the fudged rectangular dimensions; and the anomalously large tides are not a resonance, but are what a person would expect from jamming water into a restricted region.
Resonance requires a very fine set of dimensions balanced against a tight specification of driving frequency. It would occur in fact almost nowhere by coincidence, but by fudging dimensions the authors of textbooks and articles would have it happen everywhere. Fudge small differences all over and get an impossible result that looks reasonable.
Well anyway, as usual, you provoke a lot of thought.
Kevin kilty January 21, 2020 at 1:39 pm
Thanks, Kevin. Sorry, can’t deal with the theoretical. If you’d provide an exact quote and a link to someone actually doing what you’re describing that would be helpful.
Regards,
w.
A most interesting post, and as or more interesting comments following it–all indicating that this science is far from settled. But here’s one more monkey-wrench:
This planet is not 15C; far less than 1% of it is 15C. The average temperature is closer to 3,000C, and debating a few watts per square meter controlled by 400 ppm CO2 in the air at the surface of a 3,000 degree molten ball sure is a puzzling pursuit to a geologist. It seems like me trying to do careful calculations to explain the effect of CO2 on the surface temperature of my wood stove while ignoring that there’s a fire burning inside it. This planet is molten, and only barely skinned over.
And we have as yet no idea how much internal heat is moved to the surface; black smokers were only discovered in the 70’s, explosive eruptions were impossible at Arctic Ocean depths until they were detected from a research vessel a couple of years ago, and we still have no idea how much submarine volcanism exists. We also don’t know how episodic that volcanism is; observations of volcanoes that we can see suggests that such heat delivery is not constant.
Doing detailed calculations of what a few ppm of CO2 in the atmosphere are or are not doing to temperatures at the earth’s surface seems quite premature–but I fully admit to having the biases of an earth scientist, not an atmospheric scientist. I guess only time and more research will reveal who was more down to earth.
I have not calculated this but other scientists have that the Earth’s surface receives 99.97 % of its energy from the Sun. On the other hand, other scientists say that below our feet is the core of the Earth having about the same temperature as the Sun – close to 6000 C. But who cares, if the deep sea has a temperature of 4 C. If that temperature would be about 15 C, then we should think otherwise.
Antero Ollila January 21, 2020 at 11:09 am
4 C is ~277K, some 22K ABOVE the infamous 255K, the maximum temperature the sun is supposedly able to provide.
How did these oceans get so hot? From the enormous amounts of heat inside the Earth, of from the downwelling LW from the atmosphere?
You can measure them
Heat flows of continental crust is 0.071 Watts per square meter
Heat flow of oceanic crusts is 0.105 Watts per square meter
The average is 0.090 Watts per square meter adjusted for land/sea ratio
So yeah 99.97 looks about right I didn’t do the maths.
Antero Ollila January 21, 2020 at 11:09 am
No answer to a simple question?
Len Werner January 21, 2020 at 10:11 am
Yes, the core of the earth is hot … but nobody has claimed different. We’re talking about the SURFACE TEMPERATURE.
Next, yes, there are underwater black smokers (hot springs) and there are underwater volcanoes. And there are abovewater hot springs and volcanoes.
So when you say “And we have as yet no idea how much internal heat is moved to the surface”, that’s absolutely not true. We know it is very small. If it wasn’t, snow would never stay on the ground. Wherever it has been measured over any large area, it’s less than a tenth of a watt/m2.
There’s an interesting scientific study of underwater heat flux here here. And a Google Scholar search for “geothermal heat flux” reveals some 126,000 scientific pages on the subject … your claim that we have “no idea” about it doesn’t pass the laugh test.
w.
Len Werner January 21, 2020 at 10:11 am
You’ll find that a lot of people believe that our cold, low density atmosphere is able to warm just about anything.
The Geothermal Flux through continental crust is on average just ~65 mW/m^2 and can supposedly be neglected.
Difficult to envision how the atmosphere can cause the increasing temperatures when moving deeper into the Earth.
Thanks for all the responses to the suggestion that our molted earth may have different thermodynamics than present climate science suggests; I also do know the numbers quoted. I spent a good part of a career in geothermal exploration so am conversant with crustal heat flows–that we know about. Please appreciate that having measured temperatures in a lot of drill holes, that I may have a different perspective on geothermal energy than someone who has not. When you tell me what the crustal heat-flow values are–I measured a lot of them.
However, most of that career pointed out how much we don’t know. Many of these numbers I feel are subject to revision as we learn more. For example, continental crustal heat flow is certainly insignificant compared to what is being delivered to the base of oceans. From ocean bottom it convects away, so ocean bottom temperatures may not be indicative of how much heat is delivered. We don’t have a saturation of sampling of ocean bottom temperatures; if we did we would know how much volcanic activity is down there. We are only beginning that discovery. That snow doesn’t melt immediately when it falls on land is irrelevant for the majority of the planet’s surface.
Willis, with all due respect I don’t think that today’s understanding of planetary and atmospheric thermodynamics is at all complete; in 2100, when all the present postulated climate catastrophes are to have happened and will be in the rear-view mirror I’m betting we will have a far different picture of how these systems work, and the exponential increase in knowledge will have led us in as different directions as we are now thinking compared to a century ago. I don’t laugh at you, and it may be unwise to subject me to a laugh test. After all, Michael Mann thinks he knows all there is to know too, and we don’t agree. The discussion should be kept civil.
Len Werner January 21, 2020 at 7:50 pm
Len, a question I’d like to have an answer for:
How long does it take for the Geothermal Gradient to completely re-adjust to a 1K sudden warming at the surface? So 1k higher average surface temperature for continental crust, and 1K warmer ocean floor water.
Geothermal heat is disregarded since the flux is so low. Yet the entire heat CONTENT of the crust is from geothermal origin, except for the upper 10-20 m or so of continental crust.
Same for the oceans, below a certain depth the (500m 1000m ?) the influence of solar energy is zero.
Although small, the ~100 mW/m^2 geothermal flux into the oceans alone is capable of bringing the oceans from freezing to boiling in just ~500.000 years.
Every liter of water warmed at the ocean floor has to be physically transported to (mostly) Antarctica before it can surface and “vent” this energy to the atmosphere and space.
A very slow and inefficient process.
There are the same old stories about the radiation. For example, the cold atmosphere cannot heat up the warm surface. Here are some facts. The Earth’s surface emits LW radiation according to Plank’s law and it is about 395 W/m2 corresponding to the temperature about 16 °C. The Earth receives SW radiation 165 W/m2 and absorbs it. There is an error in energy balance of 395-165 =230 W/m^2.
What about this balance: The surface receives the SW radiation 165 and LW radiation 345, totally 510 W/m2. The following energy fluxes leave the surface: LW radiation 395, latent heating 91, and 24 sensible heating, totally 510 W/m2. No error in energy balance.
If you say that there is no energy flow from a lower temperature to the higher temperature, you are wrong. The net energy flux is from higher to lower temperature, but the lower temperature surface radiates as well according to its temperature as stated by Max Planck’s law. My cheap infrared measuring device can measure the temperatures of my room surface and it is real. They radiate, even though I am standing there with my 35 °C skin temperature. I cannot stop the radiation of the surfaces of the room even they have lower temperatures. And neither can you. The radiates about 1360 W/m2 to the Earth and the Earth radiates back 240 W/m2 with the 4 times greater surface. The sun cannot stop it.
Antero, you say:
Compared to what? If we take away the atmosphere, the ground would be exposed to ~3 W/m2 background radiation of outer space. Instead, it’s getting ~320 W/m2 from the cold atmosphere.
Which one ends up with a warmer earth? Obviously, the one with the atmosphere …
See my post “Can A Cold Object Warm a Hot Object” for a full discussion. The key is “compared to what”?
Or you could look at it the other way. Can a hot object cool a cold object? Sounds crazy, right, but again, “compared to what”?
Suppose you have a stove at say 400°C. You’re getting too hot, so you put a metal plate between you and the stove. It warms up to say 100°C. What happens to you?
Obviously, you end up cooler … so a hot object (the metal plate at 100°C) has made you cooler.
Always remember, “compared to what”? The cool atmosphere leaves the surface warmer than it would be if it were exposed to outer space.
Best regards,
w.
I have to say that I did not understand what you tried to formulate. The basic question was if the LW radiation emitted by the atmosphere (345 or 320 no matter) adds energy to the surface or not. I say the same as all the researchers who have published energy balances of the Earth: Yes, it does.
I understood that you say in the way. If you do not say so, please correct me.
Thanks, Antero. My misunderstanding.
w.
”IR instruments, e.g. pyrheliometers, radiometers, etc. don’t directly measure power flux. They measure a relative temperature compared to heated/chilled/calibration/reference thermistors or thermopiles and INFER a power flux using that comparative temperature and ASSUMING an emissivity of 1.0.”
So sorry.
Obviously a Mosher clone.
Willis
“Compared to what? If we take away the atmosphere, the ground would be exposed to ~3 W/m2 background radiation of outer space. Instead, it’s getting ~320 W/m2 from the cold atmosphere.”
–
Not fair.
Both get the background radiation 3.
The atmosphere is insulating the surface from the sun remember?
“ put a metal plate at 100 C Source at 400 C A hotter object cools you?”
Your quote.
The atmosphere actually stops the surface getting as warm as it could if exposed to the bare sun. Surface of the moon is hotter under the sun than on earth .
Colder on the other side and most importantly has the same TOA out as the earth.
Because they both get the same in just the TOA is virtually the moon surface not 100 km radius further out and so they both radiate the same amount out.
Angtech, I’m gonna pass on this subject. I can explain it to you. I can refer you to a long discussion on how it works.
But I can’t understand it for you.
w.
It also depends on what is considered “top of atmosphere”–between what two altitudes? If IR radiation is emitted from the surface of the earth, most of it will be absorbed and converted to heat (kinetic energy of molecules) at low altitudes (higher atmospheric pressures), where the number of molecules of water vapor and CO2 per unit volume are the highest. With increasing altitude, atmospheric pressure decreases, as does the number of molecules of water vapor and CO2 per unit volume, so the intensity of IR radiation absorbed and converted to heat also decreases. The thinner atmosphere may not affect the temperature change, since the specific heat per unit volume also decreases with decreasing pressure, but most of the energy is absorbed in the lower atmosphere.
Also, the calculation of the IR radiation rate varies not only with temperature, but also the type of surface. On dry land, the emission rate would depend primarily on albedo and temperature, but over oceans, some of the radiant heat received from the sun is converted into latent heat to evaporate ocean water, which can have a huge effect on ocean temperature during daylight hours, and the Stefan-Boltzmann equation would not take this into account.
“By comparison, the IPCC says that a doubling of CO2 would increase the surface temperature by 1.5°C to 4.5°C. If we take the midrange value of 3°C, this would imply that there is some mysterious feedback increasing the CO2-caused surface temperature change by a factor of about ten …”
There are probably many variables that both Willis Eschenbach (in this article) and the IPCC have not taken into account.
However, it is within the realm of possibility that the IPCC may be wrong!
Molecules disappear at 32 km. That’s my ToA. All radiation after that.
“With increasing altitude, atmospheric pressure decreases, as does the number of molecules of water vapor and CO2 per unit volume, so the intensity of IR radiation absorbed and converted to heat also decreases”
I cannot see much of a change in the greenhouse effect if the mean path to thermalization drops from say 5 meters to 3 meters above the surface with a doubling of CO2. The latency is still in milliseconds.
The average radiating temperature is warmer at 3 meters than at 5 meters. (abt 0.015C)
Thus the surface effect is extra heating from the average DWLR being up the Plank distribution to a more energetic level.
Seems trivially small to me.
This is the same answer I keep getting when looking at OHC too, and also proves strong negative feedback. If the “direct effect” of CO2 doubling is 3.7W/m^2 and this direct effect is supposed to be 1.2°C, my answer generally comes in at 1/4 that (0.3°C), not 3x that (3.6°C after + feedback).
This means the feedback is negative.
Twice now, I’ve seen articles about how the energy of N gazillion Hiroshima bombs have been input into the oceans, yet if you take the “area under the curve” of the CO2 forcing since, say, 1950, we “should have” seen MUCH more than we see, by the direct effect alone, using “top layer” of the ocean plus some mixing. Since oceans are the only true reservoir of energy of any substance on this rock, and this is a direct measurement of ocean heat (despite its troubles as a measurement), the amount of heat accumulating is nowhere near the “expected” amount. Each time it comes out to about 1/4 the “expected” amount. This agrees with your work shown here, and effectively destroys any argument about extreme ill effects or any emergencies.
So I always view these studies showing huge energy increases with great relief.
“Since oceans are the only true reservoir of energy of any substance on this rock….”
A pertinent reminder of the most appropriate measure of planetary climate.
In June 2009 Professor Robert Carter and 3 other scientists commented on the response of the Australian Chief Scientist and Will Steffen to the Wong- Fielding exchange ( Full Assessment available on line) as follows-
1. What is the most appropriate measure of planetary climate?
1.1 The government’s reply says “ When climate change scientists talk about global warming they mean warming of the climate system as a whole, which includes the atmosphere, the oceans , and the Cryosphere”, and then adds, “ in terms of a single indicator of global warming, change in ocean heat content is most appropriate.”
1.2 “
We agree that in an ideal academic discussion, and were accurate historical data available, ocean heat content might be a better criterion by which to judge global warming than would be atmospheric temperature. Use of this indicator was first pressed by Pielke (2007,2009) as a test of the dangerous warming hypothesis, but it has not been widely publicised by the IPCC.
1.3. In any case , Senator Fielding’s question was predicated upon the history of IPCC’s public advice which has consistently used the UK Hadley Centre near-surface air temperature record as the one that dominates in IPCC and government policy papers and discussion and is the criterion of judgement that both politicians and the public are familiar with.”
2. Natural variability in Air Temperature.
2.1 The government asserts that “at time scales of around a decade, natural variability can mask the atmospheric warming trend caused by the increasing concentration of greenhouse gases.”
2.2 It is widely agreed there is a considerable natural variability in air temperature on decadal timescales and longer. It is the IPCC who have previously denied the effect of natural variability.
For example the 2001 Summary for Policymakers, based on computer model simulations, that the climate system has only a limited variability. In turn, this claim was and is used to underpin the argument that carbon dioxide is the only plausible explanation for the late 20th century warming trend.
For the government to now invoke natural variability as an explanation for the elapsed temperature curve is to destroy the credibility of their previous arguments for carbon dioxide forcing.”
2.3 The government also claims that” in terms of the climate system as a whole, only about 5% of the warming since 1960 has taken place in the air”.
2.4 Using the Hadley CRU temperature record , the rise in air temperature since 1960 has been about 0,5 degrees C. Translating the 15×10 22 J of additional heat in the upper 700 metres of ocean since 1960 into a temperature rise, we find that this corresponds to an increase in upper ocean temperature of only 0.25C.
Thus, using these metrics air temperature increase since 1960 has been more than three times greater than ocean temperature increase.”
Willis Eschenbach
Excellent evidence and observations.
Your first figure of the difference in Long Wave Greenhouse Radiation vs TOA radiation appears related to regional rainfall and clouds. See especially the reds in the western Pacific. Comparing those may further your thermostat model.
Best wishes
David
To Willis Eschenbach
To Willis Eschenbach January 21, 2020 at 12:24 am
Willis, in the Stefan-Boltzmann equation the term including albedo is missing: E(1-A) ε = σ T4 . The Earth’s surface temperature 255 K (according to IPCC) can be obtained when E = 1368/4 = 341 W/m^2, albedo A=0.3, emissivity ε =1. However, albedo varies significantly: for water it is 0.06 – 0.1, for snow and ice 0.7 – 0.9, for sand 0.35, forests 0.07 – 0.18.
https://www.e-education.psu.edu/earth103/node/1002
This invalidates the accuracy of calculating both temperature and energy using the Stefan-Boltzmann equation.
Aleks, I don’t understand. Your claim is that somehow the Stefan-Boltzmann involves the albedo. It doesn’t. The equation is
Radiation = 5.67E-8 * emissivity * temperatureK^4
No albedo.
w.
Yes, Willis, albedo is included to S.-B. equation for Earth, because Earth is not a black body. I can cite one from many links: 3. New Science 8: Applying the Stefan-Boltzmann Law to Earth. http://joannenova.com.au/2015/10/new-science-8-applying-the-stefan-boltzmann-law-to-earth/
Regards, a
You can apply the S-B equation to the earth. But you made a different claim, viz:
My point was simple and inarguable-there is no albedo term in the S-B equation.
w.
“Since oceans are the only true reservoir of energy of any substance on this rock….”
A pertinent reminder of the most appropriate measure of planetary climate.
In June 2009 Professor Robert Carter and 3 other scientists commented on the response of the Australian Chief Scientist and Will Steffen to the Wong- Fielding exchange ( Full Assessment available on line) as follows-
1. What is the most appropriate measure of planetary climate?
1.1 The government’s reply says “ When climate change scientists talk about global warming they mean warming of the climate system as a whole, which includes the atmosphere, the oceans , and the Cryosphere”, and then adds, “ in terms of a single indicator of global warming, change in ocean heat content is most appropriate.”
1.2 “
We agree that in an ideal academic discussion, and were accurate historical data available, ocean heat content might be a better criterion by which to judge global warming than would be atmospheric temperature. Use of this indicator was first pressed by Pielke (2007,2009) as a test of the dangerous warming hypothesis, but it has not been widely publicised by the IPCC.
1.3. In any case , Senator Fielding’s question was predicated upon the history of IPCC’s public advice which has consistently used the UK Hadley Centre near-surface air temperature record as the one that dominates in IPCC and government policy papers and discussion and is the criterion of judgement that both politicians and the public are familiar with.”
2. Natural variability in Air Temperature.
2.1 The government asserts that “at time scales of around a decade, natural variability can mask the atmospheric warming trend caused by the increasing concentration of greenhouse gases.”
2.2 It is widely agreed there is a considerable natural variability in air temperature on decadal timescales and longer. It is the IPCC who have previously denied the effect of natural variability.
For example the 2001 Summary for Policymakers, based on computer model simulations, that the climate system has only a limited variability. In turn, this claim was and is used to underpin the argument that carbon dioxide is the only plausible explanation for the late 20th century warming trend.
For the government to now invoke natural variability as an explanation for the elapsed temperature curve is to destroy the credibility of their previous arguments for carbon dioxide forcing.”
2.3 The government also claims that” in terms of the climate system as a whole, only about 5% of the warming since 1960 has taken place in the air”.
2.4 Using the Hadley CRU temperature record , the rise in air temperature since 1960 has been about 0,5 degrees C. Translating the 15×10 22 J of additional heat in the upper 700 metres of ocean since 1960 into a temperature rise, we find that this corresponds to an increase in upper ocean temperature of only 0.25C.
Thus, using these metrics air temperature increase since 1960 has been more than three times greater than ocean temperature increase.”
(Rescued from spam bin) SUNMOD
Johanus, just a small nitpick to your informative comment. Your linked diagram says the area under the solar radiation curve is 63,000,000 Wm^2 compared to 250 Wm^2 for the earth radiance curve.
Feldman et al did NOT measure downwelling IR directly.
The AERI instruments were designed to be extremely sensitive and they do measure downwelling IR in a number of regions. Just not in the 15 micron band. This was a major failing per Gero prior to Feldman getting involved. Feldman created a simulated spectra and then compared it to the measured spectra which showed nothing. Viola! We now have downwelling IR appearing.
It just seems to be overly Mannomatic.
I think that on TOA the SW should balance LW and this value is not dependent of greenhouse , only Sun and albedo : so the equation becomes SW (Sun- albedo) = LW (window + green house * K)
if greenhouse is increasing, then the window part is reduced .
When greenhouse = 0, SW(sun-albedo ) = LW (window)
On surface we have LW(surface) = LW (window + greenhouse)
So if greenhouse is increasing by 3.7 Watts/m2 , the window part is reduced and the increase of the total LW(surface ) is not 3.7 watts/m2, but less because the window part is reduced .
I hope it is clear .
You are right. GHE has no role in the energy balance at the TOA.
“This doubling of CO2, in turn, would warm the surface by: 1.8 watts per square metre CO2 surface forcing / 5.5 watts per square metre per degree C ≈ 0.3°C …”
So Willis, you are saying that the ECS is 1/10th of the number the IPCC used. If so we are going to have to burn a lot more fossil fuel in order to to dodge the next Stadial and have any hope of ending the Quaternary period ice age.
“Now, this is curious. On average the change at the surface is a little less than half the TOA greenhouse effect change. So an increase of 3.7 W/m2 at the TOA from a doubling of CO2 becomes a 1.8 W/m2 increase at the surface.”
“The key is to realize that the atmosphere is not heated by just Ramanathan’s ~150 W/m2.”
–
Hate that diagram.
Trying to explain things
There is a TOA of 237 W/m2.
At this level 100 km above the earth the incoming energy that is not reflected exactly balances the outgoing energy 237 W/m2.
The surface of the earth is radiating at 392 W/m2.
This is amazingly higher than the 342 W/m2. from the total incoming reflected and incident solar radiation.
The GHG effect is basically to add 321 W/m2. of back radiation to the heating of the earth surface to the 169 W/m2. from the incident solar radiation that reaches the earth.
Basically the surface should be at a temperature commensurate with 490 W/m2. ie hotter than it is.
It emits however at 392 W/m2.giving I presume a temp of 15C, because the other 98 W/m2. is lost by sensible heat 10 W/m2. and latent heat 18 W/m2.
–
Now there is no Ramanathan 150 W/m2. being absorbed all the time. Some energy has to absorbed to raise the temperature of the air and surface but this is almost instantaneous and trivial when considering all those hydrogen bombs of energy going through the atmosphere every second. Air temperature changes very quickly night to day. Once it is warmed up there is no 150 W/m2. being drained into an atmospheric greenhouse battery all the time.
The energy in equals the energy out at the TOA.
–
Now why do we have a seeming TOA imbalance from the surface when there is not one at the TOA?
Because we are not comparing oranges with oranges.
The total energy absorbed at the surface is for a much smaller sphere.
Earth Surface area: 510.1 million km² Radius: 6,371 km energy emitted 392.
TOA surface area 526.2 million km² Radius: 6,471 km energy emitted 237.
Is this enough to make these 2 figures equal is what I would like someone to answer.
On the surface it does not look likely but?
–
One cannot take energy figures per square meter of a much larger sphere from energy figures for a much smaller sphere without doing a calibration for surface area and attenuation.
Hmm seems the outgoing IR is measured at the TOA 100 KM out so spread over a bigger sphere surface area but the energy going into the ground is measured at earth surface area a smaller sphere so the energy budget diagrams are technically out of whack.
Technically the two have to balance to have a TOA in the first place
OLR is a critical component of the Earth’s energy budget, and represents the total radiation going to space emitted by the atmosphere.[3] OLR contributes to the net all-wave radiation for a surface which is equal to the sum of shortwave and long-wave down-welling radiation minus the sum of shortwave and long-wave up-welling radiation.[4] The net all-wave radiation balance is dominated by long-wave radiation during the night and during most times of the year in the polar regions.[5] Earth’s radiation balance is quite closely achieved since the OLR very nearly equals the Shortwave Absorbed Radiation received at high energy from the sun. Thus, the Earth’s average temperature is very nearly stable
angech January 21, 2020 at 11:04 pm
Angtech, I would be shocked if this were not taken into consideration. Scientists are often foolish but rarely dumb. Hang on, let me run the numbers …
… OK, The surface area of a sphere varies as R^2. The CERES satellites are actually at an altitude of about 500 km., not 100. That means that the area of the sphere where the satellites orbit is about 16.3% larger than the earth’s surface. The idea that scientists wouldn’t bot notice and adjust for a potential error of 16% is simply not reasonable.
w.
angech January 21, 2020 at 10:14 pm Edit
Back up. Explain what it is that you hate about my diagram. It is a representation of the simplest possible layout of the energy flows. Just what is it that you “hate” about it?
Now, I drew that up about 20 years ago because of the problems with the Trenberth version, which has lots of handwaving. Mine, on the other hand, obeys the physical laws—energy is conserved at all levels, and radiation up = radiation down.
Now, the numbers are slightly out per CERES … but then two decades ago I didn’t have CERES data. But other than that … what’s wrong with it?
Finally, the top layer is not 500 km out, or a hundred KM out. The bottom layer of the stratosphere is the effective radiating layer. We know this from the brightness temperature of the radiation. It’s at about 10 km. This difference in altitude introduces an error of 0.3% in the simplified energy diagram … lost in the noise.
w.
Willis,
I notice that you’ve got, (in your diagram), 321 watts/sq.m of “backradiation” from the “greenhouse” gases coming down from the atmosphere and absorbed by the surface.
According to the diagram you only get 169 watts/sq.m impinging on the surface from the sun…the sun Willis,… in summer hot enough to melt tar on the roads.
I was wondering if you leave your bacon and eggs out on the porch overnight and have them cooked for you in the morning by that backradiation from the atmosphere.?
Mack January 22, 2020 at 2:47 am
Summer roads are not heated by the average radiation of 169 W/m2. They’re heated by something like a kilowatt per square metre or so of sunshine, plus thermal radiation from the atmosphere.
Next, it seems you think that the idea that the atmosphere emits thermal radiation to be somehow incredible or impossible. Not sure why. It’s been measured, not theorized but measured, thousands and thousands of times by scientists around the planet.
w.
Well, I thought those numbers would have pricked up your ears, Willis I would have thought that 321 watts/ sq.m. of “backradiation” belting down from the ATMOSPHERE 24/7, would have triggered some form of thought process in your head….. particularly since it’s nearly TWICE the amount of solar radiation impinging upon the surface.! Is there nothing about that which really unsettles you? Is there nothing about that which says…”hang on, there could be some mistake in these diagrams.” ?
Thanks, Mack. You clearly think downwelling longwave infrared radiation is imaginary.
Me, I know that it’s been measured all over the planet by scientists. It’s measured at all the SURFRAD sites. It’s measured by the TAO buoys. It’s measured at the ARM sites.
Do you truly think that those hundreds of scientists are just making it up?
Also, if the ≈ 169 w/m2 of sunlight was the only thing heating the surface, it would be at about -40°C or so … is there nothing about that which really unsettles you?
w.
Willis Eschenbach January 23, 2020 at 12:35 am
Why do you keep pushing this radiative balance temperature nonsense?
If the surface temperatures on Earth were in radiative balance with incoming solar we would see temperature swings from ~3K during the night to 365K or higher during the day.
Is not happening.
169 W/m^2 is ~14,6 MJ/m^2 over 24 hrs. This seems close to the world average as shown in these charts:
https://www.pveducation.org/pvcdrom/properties-of-sunlight/isoflux-contour-plots
14,6 MJ/m^2 between sunrise and sunset is enough energy to INCREASE the temperature of the upper 4 m of ocean water 1K.
Has nothing to do with RADIATIVE balance.
Backradiation does not warm the surface, it reduces the energy loss from the surface to the atmosphere. Otherwise we would see your 321 W/m^2 + ~1000 W/m^2 at noon giving temperatures of ~390K.
Ben Wouters January 23, 2020 at 4:03 am
Seriously?
Ben, if you don’t know the answer to that question there’s no point in discussing it further. Average energy budgets are a very useful tool for studying the nature and size of the energy flows of the climate system. Come back when you understand that and we can continue the discussion.
w.
Willis Eschenbach January 23, 2020 at 10:01 am
Absolutely.
Absolutely agree: average ENERGY budgets.
The moment you assign a SB temperature to these energy flows, you implicitly assume RADIATIVE balance.
On Earth we do NOT see radiative balance temperatures. Nigh time temperatures do not drop to ~3K, nor do daytime temperatures soar to ~364K, the RADIATIVE balance temperature for 1000 W/m^2 solar radiation.
The 169 W/m^2 average solar that actually heats the surface is enough energy to increase to temperature of ~3,5 m^3 ocean water 1K.
169 J/s/m^2 x 60 x 60 x 24 = 14,6 MJ/m^2/day.
On our moon we DO see ~radiaitve balance temperatures during the day.
Not so for the lunar night. Temperatures do not drop to ~3K as radiaitve balance would suggest, but to ~80K on average.
Difference between the two obviously our oceans vs lunar regolith.
The thing that unsettles me ,Willis, is that you think the ATMOSPHERE raises the temperature of the planet from that… ” -40deg C or so” to what we have in reality of about 15deg C.
“On our moon we DO see ~radiaitve balance temperatures during the day.”
No Ben, DIVINER radiometer results show brightness temperatures are always changing lunar day and night (diurnal) so no ~radiative balance. Or define your ~. Apollo derived results though show temperature unchanging diurnally about 0.3m deep in the lunar soil.
Mack January 24, 2020 at 1:16 pm
Compared to the moon it’s even more unsettling’.
On the moon ~90% of incoming solar is available for heating, resulting in avg temperature of ~197K. Increasing its rotation to ours should result in ~220K.
Adding an earth-like atmosphere reduces the amount of solar that actually reaches the surface to <50%. Yet this same atmosphere is supposed to heat the surface to ~290K.
Trick January 24, 2020 at 8:12 pm
See https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2011JE003987
(search for equilibrium)
Nighttime temperatures never go to 3K, the radiative balance temperature with the Cosmic Background radiation.
Noon temperature on the equator is even slightly above radiative balance temperature, reason probably the high temperature at sunrise (~100K or so) Reason for this some geothermal plus heat storage from the previous day.
Apologies,
no offence intended.
It was a comment on the complexity of the radiation flows and trying to work through them, not on the authorship of the diagram.
You see it everywhere. I did not know it was yours.
Love it.
Willis
Finally, the top layer is not 500 km out, or a hundred KM out. The bottom layer of the stratosphere is the effective radiating layer. We know this from the brightness temperature of the radiation. It’s at about 10 km. This difference in altitude introduces an error of 0.3% in the simplified energy diagram … lost in the noise.”
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Respectfully,
I may be wrong.
I thought the Top of Atmosphere is defined as where radiation in equals radiation out by convention.
For a sphere the average distance is said to be about 100 km.
In truth the TOA is very close to the poles and further away at the equator during the day.
The effective radiating layer may be the bottom of the stratosphere but the TOA is where the energy in balances the energy out by definition.
The satellites may be out 500 km but what they are measuring is the output of IR from those top layers and formulating an overall TOA value.
Incidentally they rely on ground stations for the IR out and in at surface level because the satellites have a lot of trouble trying to assess IR levels below 800 meters and cannot do it through clouds, apparently.
CERES apparently does it’s TOA from an amalgam of satellites and a computed input from GCMs with an inbuilt warming and it is extremely variable with large margins of error and does not do clouds well.
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“The idea that scientists wouldn’t notice and adjust for a potential error of is simply not reasonable.”
True.
Very hard to actually find out. Perhaps Dr Spencer could enlighten us if he is reading.