Guest post by Willis Eschenbach
There is a lot of misinformation floating around the web about how the greenhouse effect works. It is variously described as a “blanket” that keeps the Earth warm, a “mirror” that reflects part of the heat back to Earth, or “a pane of glass” that somehow keeps energy from escaping. It is none of these things.
A planetary “greenhouse” is a curiosity, a trick of nature. It works solely because although a sphere only has one side, a shell has two sides. The trick has nothing to do with greenhouse gases. It does not require an atmosphere. In fact, a planetary greenhouse can be built entirely of steel. A thought experiment shows how a steel greenhouse would work.
Before we start, however, a digression regarding temperature. The radiation emitted by a blackbody varies with the fourth power of the temperature. As a result, for a blackbody, we can measure the temperature in units of radiation, which are watts per square meter (W/m2). For objects with temperatures found on the Earth, this radiation is in the range called “longwave” or “infrared” radiation. See the Appendix for the formula that relates temperature to radiation.
This means that we can denote the temperature of a blackbody using W/m2 as well as the traditional measures (Fahrenheit, Celsius, Kelvin). The advantage is that while temperature (degrees) is not conserved, energy (W/m2) is conserved. So we can check to see if energy lost is equal to energy gained, since energy is neither being created nor destroyed by the climate.
For our thought experiment, imagine a planet the size of the Earth, a perfect blackbody, heated from the interior at 235 watts per square meter of surface area. How is it heated from the interior? Doesn’t matter, we’ll say “radioactive elements”, that sounds scientific.
The planet is in interstellar space, with no atmosphere and no nearby stars. The equilibrium surface temperature of this planet is, of course, 235 W/m2. To maintain the equilibrium temperature, it constantly radiates this amount of energy out to space. Coincidentally, this is the amount of solar radiation that makes it past the clouds to warm the Earth. If we convert 235 W/m2 to one of our normal temperature scales, it is -19 Celsius (C), or -3° Fahrenheit (F), or 254 Kelvins (K). It’s a possible temperature that the Earth might have if there were no greenhouse effect. That is to say … cold.
Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand meters above the surface, as shown in a cutaway view in the picture above, and in Figure 1 below. What happens to the surface temperature of the planet? (To simplify calculations, we can assume the shell has the same outer surface area as the surface. For an earth-sized planet with a shell two kilometers above the surface everywhere, the difference in area is only six-hundredths of one percent. This assumption makes no difference to the argument presented.)
In order to maintain its thermal equilibrium, including the new shell, the whole system must still radiate 235 W/m2 out to space. To do this, the steel shell must warm until it is radiating at 235 watts per square meter. Of course, since a shell has an inside and an outside, it will also radiate 235 watts inward to the planet. The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell. This will warm the planetary surface until it reaches a temperature where it radiates at 470 watts per square meter. In vacuum conditions as described, this would be a perfect greenhouse, with no losses of any kind. Figure 1 shows how it works.
Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K). [Clarification added] Note that the distance from the shell to the planet is greatly exaggerated. In reality, it is close enough to the planet that the difference in the areas can be neglected in practice.
The trick can be repeated again, by surrounding the planet and the shell with a second outer shell. In this two shell case, the planetary surface temperature (in W/m2) will be three times the initial surface temperature.
In nature, planets have atmospheric shells, of course, rather than steel shells. The trick works the same way, however. Rather than being warmed from the inside, the Earth receives 235 W/m2 from the sun. Solar radiation passes through the atmosphere. Outgoing longwave radiation is absorbed by the atmosphere, just as it is absorbed by the steel shell shown in Fig. 1.
So that’s the trick of the greenhouse. It has nothing to do with blankets, mirrors, or greenhouse gases. It works even when it is built out of steel. It depends on the fact that a shell radiates in two directions, inwards and outwards. This radiation keeps the planet warmer than it would be without the shell.
Now, it is tempting to think that we could model the Earth in this manner, as a sphere surrounded by a single shell. This is called a “two-layer” model, with the two layers being the surface and the atmospheric shell. In fact, a number of simplified climate models have been built in this way. Unnoticed by their programmers, however, is that is not physically possible to model the Earth as a two-layer system. Figure 2 shows why. Note that in all cases, the system has to remain in thermal equilibrium. This means that the surface must radiate as much as it absorbs, and the shell must also radiate as much as it absorbs. In addition, radiation from the shell to space (upwelling longwave radiation or ULR) must equal radiation from the shell to the Earth (downwelling longwave radiation, or DLR)
Figure 2. Single-shell (“two-layer”) greenhouse system, including various losses. S is the sun, E is the Earth, and G is the atmospheric greenhouse shell around the Earth. The height of the shell is greatly exaggerated; in reality the shell is so close to the Earth that they have about the same area, and thus the small difference in area can be neglected. Fig. 2(a) shows a perfect greenhouse. W is the total watts/m2 available to the greenhouse system after albedo. Fig. 2(b) is the same as Fig. 2(a) plus radiation losses Lr which pass through the atmosphere. Fig. 2(c) is the same as Fig. 2(b), plus the effect of absorption losses La. Fig. 2(d) is the same as Fig. 2(c), plus the effect of thermal losses Lt.
Figure 2(a) shows the same situation as Figure 1(B), which is a perfect planetary greenhouse. In this case, however, it is heated by an amount of energy “W”, which is coming from the sun. The planet receives solar radiation in the amount of “W” from the sun, and longwave radiation “W” from the atmospheric shell. The surface temperature is thus 2W. All energy flows are in Watts/square metre (W/m2).
Figure 2(b) adds two losses. The first is the reflection from the Earth’s albedo (Wo – W). This is energy which never enters the system and is reflected back into space. We are still getting the same energy entering the system (W). The second loss Lr is from radiation which goes from the surface to space through the “atmospheric window”. Because of the second loss, the surface of the Earth does not get as warm as in a perfect system. In a perfect system, the temperature of the surface is 2W. But including the radiation loss Lr, the surface temperature drops to 2W – 2Lr.
Figure 2(c) adds another loss La, which is the solar radiation which is absorbed by the shell. This cuts the radiation hitting the surface down to W – La. Including this loss, the surface temperature is 2W – 2Lr – La.
Figure 2(d) adds the final loss. This is Lt, the thermal loss. It is the sensible energy (energy you can feel) and latent energy (evaporation) which is transported from the surface to the shell by convection. Including all of these losses, the surface temperature of the Earth is 2W – 2Lr – La – Lt.
Now, why can’t the Earth be modeled in this manner? A look at the size of the various losses shows why. Here is the canonical global energy budget, called the “Kiehl/Trenberth” budget, or the K/T budget.
Figure 3. The Kiehl/Trenberth Global Energy Budget. This is a “two layer” representation, with the surface and the atmosphere being the two layers. Lr, the radiation loss, is the 40 W/m2 labeled “Atmospheric Window”. La, the absorption loss, is the 67 W/m2 labelled “Absorbed by Atmosphere”. Lt, the thermal loss, is 102 W/m2. This is the sum of the 24 W/m2 labelled “Thermals” and the 78 W/m2 labelled “Evapo-transpiration”. W, the energy from the sun, is the incoming solar radiation of 342 W/m2 less the 107 W/m2 that is reflected by the surface and the clouds. This means that W is 235 W/m2. SOURCE
What’s wrong with this picture? Note that the temperature of the Earth is 390 W/m2, labeled as “Surface Radiation”. This is 15 C, or 59°F. But from Fig. 2(d), we know that the surface temperature of a greenhouse system with a single shell, as shown in the drawing, is 2W (470) – 2Lr (80) – La (67) – Lt (102) = 221 W/m2. This is far below the known surface temperature of the Earth. In other words, a single shell greenhouse system simply isn’t efficient enough to give a surface temperature which is warm enough to allow for the known losses.
So where is the problem with the K/T budget diagram? The hidden fault is that the upward radiation from the atmospheric layer does not equal the downward radiation. There is 195 W/m2 going to space from the atmospheric shell, and 324 W/m2 going down to the surface.
In order to get enough energy to allow for the known losses, the simplest model requires two atmospheric shells. A perfect greenhouse with two shells would give a surface temperature of 3W, or 705 W/m2. This is enough to allow for the known losses, and still give a surface temperature which matches that of the Earth. Figure 4 shows one such possible system.
Figure 4. Simplest greenhouse global energy budget capable of representing the Earth. Note that all major flows in the K/T energy budget have been maintained. There are two shells, which must be physically separate. These are the lower troposphere and the lowest stratosphere. They are separated by the tropopause.
This budget fulfills all of the requirements for thermal equilibrium. The same amount is radiated upwards and downwards by each layer. The amount absorbed by each part of the system equals the amount radiated. Further verification of this energy budget is the 147 W/m2 emission from just above the tropopause. This converts to a Celsius temperature of about -50 C, which is a typical temperature for the lowest part of the stratosphere.
I have written a simplified Excel radiation/evaporation/convection model which encapsulates the above system. It is available here. Click on the “File” menu on the webpage and select “Download”.
I invite people to play with the model. It is what I term a “Tinkertoy” model, which is what I call the simplest possible model that can represent a particular reality. One of the interesting results from the model is that there can be very little thermal loss between the inner and the outer atmospheric layers. If there is more than a trivial amount of leakage, the surface cools significantly. Among the other insights yielded by the model is that a change equivalent to a doubling of CO2 (an increase of 3.7 W/m2 downwelling radiation at the top of the atmosphere) can be canceled by a 1% increase in the upper and lower cloud reflections.
Experiment with the values, it is an interesting insight into the energy flows in the simplest possible climate model that can represent the Earth’s greenhouse system.
APPENDIX
The formula that relates the temperature to radiation is called the “Stefan-Bolzmann” equation:
R = sigma * epsilon * T^4
where r = radiation (W/m2)
sigma = the Stefan-Boltzmann constant, 5.67 * 10^-8
epsilon = the emissivity of the body, which for a blackbody = 1
T^4 = absolute temperature in Kelvins (which is Celsius plus 273.15) raised to the fourth power
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if an average man gives off 180 watts at night, though i record 228 watts sometimes, then how could earth be emitting 255w/m2?
To calculate energy at this level you’d need to divide human surface area by 148, so 1.75/148=85w/m2, or 1.75/228=130w/m2
P Wilson (06:26:15) : Well, if you trust your thermal imaging device at all, then you’re using the very physics that you think are incorrect. The only trick is knowing the emissivity of a surface (and whether the emissivity is constant over the wavelength range in question), but that doesn’t really seem to be your difficulty.
I can’t even figure out what you’re trying to say, otherwise.
Nick Stokes (00:45:31) : Thank you for your contribution. There are points here which got unfortunately totally lost in the discussion, because so many people were getting hung up on the basics of radiation. Namely, that either in the steel cartoon or with a real atmosphere, the final emission to space is about the same. What matters is the temperature where this emission takes place. Willis’s radiation-only no-atmosphere example cannot describe the actual lapse rate within our atmosphere; this has been touched on in other comments as well. But it becomes difficult to write a simple tinkertoy model that captures what actually happens in the atmosphere – the lapse rate, spectral absorption and emission, pressure effects, and so on. You quickly end up with a computer code of the sort that sceptics don’t like, because it’s complicated.
P Wilson
Your body is emitting about 930 W (if you are 2 m2 and wearing clothes)
but you are also receiving 830 W from the surroundings at 20 C.
http://en.wikipedia.org/wiki/Black_body
Nick,
This idea that the outer atmosphere represents the blackbody is a common thing I hear. However it seems to me I can imagine a new sphere just inside the outer atmosphere. I can image the outer atmosphere as the source now, and our new smaller sphere the new body. Now we realize that the outer part of our new smaller sphere must also absorb and radiate like a blackbody … and so on, all the way to the surface.
wavelength expressed in colour as it accords to temperature is normally the way such devices are calibrated.
I know all about the SB theory, though 7-14 microns is typical range of detection. At sunset- night, most outside objects are invisible to these wavelengths in a typical suburban environment -and humans are clearly visible across a quite a few wavelengths. A human emitting around at 8 microns average, corresponds to around 100w/m2, as the BMR is around 58w/m2 where radiation emitted is equalt to the BMR
It leads to whether we should accept the SB for *normal* temperature matter at a constant and whether it should be reformulated to a coefficient, divided by as much as 12, to produce 30w/m2 for outgoing longwave from earth on average, since it’s clear that earth isn’t attempting to reach absolute zero in order to reach equilibrium, and emit as much as it possible can to achieve it.
Lgl. I’m aware of the absurd figures produced from the SB constant. The only difficulty with absorbtion of radiation is that at 20C, matter is cooler than the human body, so its unlikely a human will absorb that much radiation. Experimental procedures show a recorded 85watts emitted at 27C.
Nick,
To address the second part. As in Willis model, the inside temperature can get to infinity with insulation and no holes. Holes make all the difference.
Image a steel container of air, completely closed off. Now imagine there is a source of new air creating new molecules inside the container. The container pressure can build up to infinity. That is Willis model with a totally reflecting steel shell (or infinite absorbing shells).
Now image a steel container that we fill up with air using an external source. We need a hole for the air to enter. Can we build up the pressure inside our steel container higher than the pressure of the source? No. This is the same with our energy flux from the sun. We can’t build up the energy density (a rough analogy to our air pressure) to be higher than the source.
correction: at 2m2 85/2=42.2w/m2 at 27C
“It leads to whether we should accept the SB for *normal* temperature matter at a constant and whether it should be reformulated to a coefficient”
Yeah, and think of all the fun you can have if you do the same with the speed of light.
It’s not the complication, at least not for me. A line-by-line atmospheric radiative transfer program may be complex, but it’s truly physics based and produces results that conform to observation provided the input data (lapse rate, humidity profile, etc.) are correct. It’s the oversimplification or parameterization of complex processes and other kludges in GCM’s necessary to get the programs to run without blowing up and in a finite amount of time that are the problem. GCM’s don’t use line-by-line or even band models for radiative transfer. They take too much computing time. We know that even fairly recently some models used incorrect parameterizations of radiation transfer. Model cloud physics produces results that are wrong in both extent and location of cloud cover. Aerosol forcings, particularly the aerosol indirect effect, don’t seem to be based on observation, but are used as a fudge factor to produce a more or less correct hindcast whether the climate sensitivity of the model to ghg forcing is high or low. The list goes on and on.
DeWitt Payne (11:01:16) :
Well, as you can see from this thread, you might be OK with radiative transfer code, but plenty of others are confused about radiation and the conservation of energy, in general. You can see how controversial this simple toy model was. As for how the radiation transfer is parameterised, I think that’s pretty well straightened out. Cloud parameterisations would be the topic of a whole other thread. Remember that no model is ever perfect – it just has to be good enough to be useful for whatever you are trying to learn.
Ian
Now we realize that the outer part of our new smaller sphere must also absorb and radiate like a blackbody … and so on, all the way to the surface.
Yes, you’re right. The top shell gets 470 W/m2 from below, and radiates 235W/m2 up and down. The next must radiate that 470 up, and so also 470 down; it gets 235 from above, and so must get 705 from below. And so it goes. Each shell successively has a nett flow through of 235 W/m2, as it must. And each one gets hotter, emitting in total 235 W/m2 more than the one above.
This works because each is fully opaque, conducting (so both faces are at the same temp) and insulated (by vacuum) from each other. Just like the Multi-layer_insulation of spacecraft (read the description – it’s a steel greenhouse). But the atmosphere is not like this. The “layers” are only partly opaque, not fully conducting, and not separated.
As to your point about the Sun – yes, the sun’s temp is another theoretical limit. In the air, the lapse rate is a much lower and more real limit. But OK, for Willis’s cartoon, your issue has some meaning. It’s true that for Willis’s postulate of a radioactive energy source, you could reach an arbitrarily high temp. But the analogy heat source he really means is where the “steel” layers, like GHG, allow sunlight in, but block outgoing IR through frequency selectivity, creating a 235 W/m2 heat source at the surface.
As the temperature rises, the thermal outgoing radiation rises in frequency, and would approach the solar spectrum if the Earth got that hot. In that case, the shells can no longer exercise that frequency selectivity. The outgoing radiation would pass through as easily as sunlight got in. And since it goes out to all the sky, that exit radiation is far greater than the incoming, so in fact the bottom surface temperature could not rise to anything like the Sun’s.
After 10 exchanges (approx 2 minutes) it would be 2350w^m2.
golly thats hot for a subzero source.
Nick,
Whether the layer is opaque or partially transparent doesn’t, we can still treat it as a source to the next layer.
I’m not sure if you read my article yet (http://www.ianschumacher.com/greenhouse_effect_maximum.html) but in there I get a result which I ‘think’ is what you are thinking. A balance that is hot, less than the sun, but greater than blackbody equivalent, but then I discuss why this would not be the case because it assumes the outgoing energy spectrum is static, when in fact the energy will continously redistribute (trying to maintain the blackbody spectrum shape) to pour out the visible window.
Anyways, that my thought.
Fun experiments to try with your IR thermometer (strictly speaking a pyrometer), available for ~$50 at auto parts and kitchen stores (no home should be without one):
1. Open your refrigerator and measure the temperature.
2. Open your freezer and measure the temperature.
Do you still think that colder objects do not emit IR radiation? Explain how the thermometer can get a reading if they don’t.
3. Go outside and point it at the sky and measure the temperature. Don’t include the sun or moon in the field of view. Last night with a clear sky I measured a temperature of -27 C. This morning with cloud cover it was -4.9 C.
Do you still think the atmosphere doesn’t radiate in the IR? Do you understand why it doesn’t feel as cold on a winter night under a cloudy sky compared to a clear sky? Now think how cold it would be if the atmosphere was completely transparent to IR so that the sky brightness temperature was -270 C.
If after all that, you still don’t believe that there is a greenhouse effect from IR active molecules like water and CO2 in the atmosphere, then I give up. I don’t have Willis’ patience.
carrot eater,
I’m appalled by what seems to be willful ignorance on both sides. It’s truly depressing that physicists with advanced degrees could think that the greenhouse effect somehow violates the Second Law. I also find statements by people who should know better that Antarctica would be the only habitable continent if GMT goes up 6 C, which is still nowhere near as high as the Eocene Optimum 55 Mya. That doesn’t mean that a 6 C rise wouldn’t cause immense problems, but it wouldn’t make most of the planet unihabitable for a species that can live in climes ranging from the Arctic to the Arabian desert.
DeWitt Payne (10:22:23) :
of course radiation emanates from the sun and heats earth surface, oceans and atmosphere, whilst parts of the atmosphere maintain whats known as a greenhouse effect.. Its the theoretical magnitude as opposed to the real measured magnitude that we’re fascinated by. Just that with certain constants taken from physics and applied to gases, which exagerrate their radiative effect by 12 times gives us a 2-6C rise.
those who believe the SB will need to accept the validity of AGW
DeWitt,
You’re appalled because a couple of crackpots (out of thousands of non-crackpots that read the article) post crackpot ideas on a public forum?
You don’t get out much do you 🙂
I assume most of the people here are laymen, not professional scientists. Really, assuming you are a real scientist, why are you here except to feed a superiority complex? Otherwise, gently correct, teach, help, and ignore. Simple.
DeWitt Payne (10:22:23)
Ok. Here’s how to proceed. Open your freezer and line it with black paper leaving a 5 inch perimeter between the paper and the wall. Set the temperature to -19C. (what the constant says is the relationship between 235w/m2 and the temperature). Put in a black matt box, which will radiate at a theoretical 235w/m2, and see if the temperature is 10 times warmer than -19C after 1 day… Or something along those lines.
Actually this thought experiment is replicable as an experiment even with black spheres and a black steel insulator
incidentally if it does what was said above somewhere – 470w/m2 – the SB constant puts that at 29C
Nick Stokes (02:22:27) : You beat me to it; it also occurred to me today that the theoretical limit in Willis’s cartoon would be the point where the Earth got so warm that it was emitting in wavelengths which would be transmitted by the ‘steel greenhouse’.
DeWitt Payne (10:28:30) : If somebody posing as having an advanced degree in physics is making simple mistakes like those seen above, I would assume that person is making up credentials. Maybe they had a relevant undergrad class or two.
What bothers me is the lack of humility and self-awareness. If you haven’t a clue what you’re talking about, you should realise this, instead of spewing attitude.
DeWitt Payne (10:28:30) :
There is ignorance on both sides, but I think it is honest ignorance, not wilful ignorance.
P Wilson (08:48:58) :
P Wilson, I think by now everyone knows that you think a basic and long-established law of physics, the Stefan-Bolzmann Law, is incorrect. I note that the page I cited give two different ways to derive the law from fundamental principles, so to overthrow the SB Law you’ll have to show that both derivations are wrong …
However, this is not the forum to argue that issue. This thread is for people who do believe the SB Law is correct, and are trying to understand the implications of that Law.
I’m sure there is a forum on the web somewhere for like-minded people where you can discuss your ideas. I wish you the best of luck in overturning the SB Law. However, I fear I must I ask you to let it go here and refrain from further posting on the SB law, as it is far, far removed from what we are discussing here.
Thanks,
w.
Willis Eschenbach (15:00:35) :
fine, as long as you can resolve the problems outlined aabove with the values taken from biology and the problem of the difference in temperature values from a black box at 235 increasing to 470w/m2 with no additional radiation input- along with quite a number of problems associated with its application rather than just censoring them.
you’d also have to accept that heat as an entity were a fixed permanent constant, like solid matter that doesn’t change its form – like the ideal titanium spanner, whether it is an internal energy pressure caused through friction and/or whether heat is reduced and disippated from autonomous cooling, as Kelvin considered it as “heat loss”.
Final comment prior to censorship: What happens to heat from a red hot poker from a hot furnace and exposed to normal temperature air? Does the heat disippate/disappear or does it have to be integrated into some other entity, or is it just a case that when objects cool and emit radiation they thermalise to the temperature set by the air, as the electrons and atoms lose their excited state?
P Wilson (18:17:03) :
I fear I can’t answer your questions, P Wilson, not because I don’t want to, not because I want to “censor” you, but simply because we have such radically different fundamental assumptions. As a result, we are talking past each other, rather than to each other, and anything I say in answer to your questions will be misunderstood. Or perhaps even “misunderstood” is wrong, but understood from a very different point of view, like a reflection in a fun-house mirror where everything is there but it is distorted in a host of ways that make discussion impossible.
I wish it weren’t so, but there it is.
Best regards,
w.