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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Anders L. (01:26:20) :
Hansen’s paper uses a cherry picked time frame and the usual Playstation™ models to get that number. I discussed this at the time, hang on … OK, thanks for waiting, it’s here.
In short, there is no way to determine whether the earth is in or out of balance, except that at any given instant we can pretty much guarantee that it is out of balance.
julien (01:27:30) :
See Fig. 2(a), which is the same as Fig. 1(b) except with external radiation flow.
yonason (03:43:14) :
yonason, I can’t make any sense of this. Please boil it down to the essentials and resubmit it. Brevity is clarity.
Joel Shore (07:21:07), thanks for the “Bad Greenhouse” link. I like the way it compares the Earth’s atmosphere to the sun’s photosphere. In a sense, I suppose one can think of the sun as the “steel greenhouse” model on steroids.
It also makes life here on our cold earth seem a bit more “sunny”. 🙂
TomVonk (06:30:56) :
My apologies, Tom, but I find your writing impenetrable. Please clarify and simplify what your objections are, and I’ll be happy to answer.
w.
Joel Shore, thanks for your assistance with the tsunami … much appreciated.
Anton Eagle (10:12:57) :
Sir, you are a gentleman. This is how science works. We throw our ideas out and let people try to find flaws with them. If they find flaws, we nod our heads, say “Dang … wrong again. Not the first time,” and move on. I can’t count how many times I’ve been wrong, that’s how my own scientific understanding moves forwards.
One caveat. You say:
One is not a consequence of the other. Both are just different ways of measuring the same thing, which is the average speed of the molecules involved. We could also measure it in meters/second. None of these are “a consequence” of the other. They are different ways of expressing the same thing.
My thanks to you for your honesty and style. Well done.
Willis – Nice article!
I’ll admit that as I read the article, I had similar reservations regarding a seeming confusion between energy balance vs. energy flux balance and the radii impact. I think you sufficiently explained your reasoning in the followup comments, but I do wonder if it is prudent not to generalize on assumed small radii differences.
Most readings I have come across use W/m2 to illustrate earth’s energy budget analysis, similar to your Figure 4. But a radii difference within atmospheric altitudes can give relevant W/m2 deviations.
As an example (and please doublecheck calculations for yourself) I refer to a statement by Anders L. (01:26:20) that quotes a max 1 W/m2 imbalance from Hanson et al. While I believe your explanation is very reasonable, I would like to note that 1W/m2 out of 492W/m2 surface flux (Fig4) can be offset by an altitude of approximately 20K ft. So, when I see energy balance charts such as fig4 given in units of W/m2 and attempting to tally in/out numbers from the surface, clouds, and space, I cannot help but question how anyone can assert an imbalance of 1 W/m2.
Juraj V. (11:16:41) :
Most scientists have a pretty good handle on it. I published the “Steel Greenhouse” to clarify the mechanism, and to point out that the usual “single shell” model used in innumerable places is inadequate to model the earth.
I have always wondered why we can’t concentrate LW the way we do with solar SW. Is it because the surface can’t be made smooth enough or what? I thought the main thing with a Thermos was just that, reflecting the (LW) radiation back inside.
SAM (12:56:40) :
The paper by Hansen asserting a 0.85 W/m2 imbalance is a joke, in my opinion. Even with satellites, we cannot determine either incoming or outgoing radiation to anywhere near that accuracy. Don’t know if you took a look at my citation above, but it discusses this question as regards Hansen’s paper.
Climate models are of no help in this regard. Their errors are legion, and large. Here’s an example, from the people who wrote the model:
Hansen et. al, Climate Simulations 1880-2003
Given Hansen’s own admission of 50 W/m2 errors, the idea that his model can detect a 0.85 W/m2 imbalance is … well … I’ll call it “extremely improbable” and leave it at that.
[A lot is happening behind the scenes. It is not being ignored. Much is being coordinated among major players and the media. Thank you very much. You will notice the beginnings of activity on other sites now. Here soon to follow. ~ ctm]
Willis Eschenbach (11:24:58) :
It’s clear enough. At “thermal equilibrium” the temperatures are all the same (see link I provided to definition).
The temperature of the shell is a “boundary condition” that you are choosing to solve your equation. That and the term “isothermal” are arbitrary because, while necessary to get the result you want, they are not consisstent with the initial conditions you’ve already given, or the final state that will evolve from those initial conditions after the addition of the shell. You already have enough information to solve for the temp of the shell, which will not come to the same as the earth’s initial temp., unless more heat is added. In fact, even then they will not both be the same, so it will never be isothermal.
The earth has, like P Wilson’s analogy of the lightbulb, a fixed output as postulated by your initial conditions. It cannot heat the iron shell to the same temp., without adding even more heat, but that violates your initial condition of constant output. It will also drastically raise the temp., of earth, thereby violating your condition that the system be isothermal.
While you are correct that the rate of heat emitted per unit area is the same for all perfect blackbodies at the same temperature, a larger surface area will radiate more total energy. It has to come from somewhere (conservation of energy). That’s why I looked at the Net Power, which is an absolute measure of available energy. Keep it constant, and the shell never warms to the same as the initial temp., of the earth, and the earth warms only slightly because the presence of the shell slows down the efflux of heat (it’s an insulator).
You are looking at emissivity, which is not a measure of total output, which is why your accounting doesn’t balance. Naturally the earth’s temp., will go up drastically if you add as much energy as needed to bring the shell to the same initial temp., of the earth. But it doesn’t come from the earth, which may be why John A made his comment about Maxwell’s demon?
Reconsider the shell and do the following thought experiment. Keep increasing it’s distance from the earth, and ask yourself what will need to happen in order to maintain it at a constant temp., and then ask where that energy comes from.
Summing up: Your drastic increases in earth’s temp., appear to be an artifact of improper boundary conditions and incorrect assumptions.
lgl (13:03:55) : “I have always wondered why we can’t concentrate LW the way we do with solar SW.”
“Concentrate” meaning what?
Willis Eschenbach (12:38:02) :
“None of these are “a consequence” of the other. They are different ways of expressing the same thing.”
Well, if it’s blackbody emission, yes. But if you don’t know the emissivity, then you can’t go from the total emitted flux to temperature.
TomVonk (06:54:18) :
You’re overly worried about the solid angle. Yes, the flux found from S-B is integrated over all directions. And this is perfectly fine: all of the energy emitted by the earth ends up at the first shell. All of the energy emitted by the inner surface of his first shell ends up at the earth. No need to worry about angles or directions or solid angles; the view factor is 1 in both directions.
Curt:
Very touching. Unfortunately for you I took the course and passed it. I would be asking the Dean why someone like you was arguing against results known and reproduced since the 19th Century, and why such a crackpot is on the faculty.
Duh. I did not make the statement that the cooler objects cannot be detected by warmer objects. I made it clear that all objects above absolute zero radiate.
Duh. See above.
Duh. Except that the laws of mathematics would have me point out that T1 must be greater than T2 otherwise the net radiative heat transfer is negative.
The heat flow is one way from hotter to colder, which is exactly the point I’ve continued to make, consistent with the laws of thermodynamics.
I make no such claim. I have made no such claim. I pointed out that cooler objects do not warm warmer objects because the net radiative heat transfer would be negative and because it would imply that the cooler object spontaneously loses entropy without doing any work.
Unfortunately for you, all power plants, power electronic systems blah blah blah obey the laws of thermodynamics.
SImple experiment: put a sample of a radioactive element (which has a measured temperature) inside of a Dewar flask and measure if the temperature of the radioactive element rises as a result of radiative reflection from the inner walls of the flask.
A Nobel Prize awaits if it does.
But I’m not holding my breath.
Hence or otherwise, grow up. Crack a book on introductory thermodynamics and read. You may be pleasantly surprised at how much you can learn from a book.
yonason (13:18:36) :
I’m sorry, but this is still not “clear enough” at all. You have asserted, not shown but asserted, that at equilibrium all the temperatures are identical.
You keep saying no, it doesn’t work out like that … but simply saying so doesn’t advance the discussion. For example, you say “That and the term ‘isothermal’ are arbitrary …”, but since I haven’t used the term “isothermal” in this entire thread, I haven’t a clue what you are talking about.
Try to boil your assertions down to a single paragraph where, instead of assertions, you pose questions that point to what you think I’ve done wrong. That might help.
Best regards,
w.
John A (13:35:21) :
You are still ignoring the difference between heat flow and net heat flow. The net heat flow is one way, as you correctly point out. But as the equation clearly shows, it is the net of two individual heat flows in two different directions. See Fig. 1 for an example. The net heat flow is from the planet to the shell. But that is made up of two individual heat flows, one from the planet to the shell, and one from the shell to the planet.
Of course it will rise. The temperature of anything will rise until the system loses heat at the same rate that heat is added. Since the radioactive element is constantly producing heat at a steady rate, the temperature inside the Dewar will increase until the loss from the outside of the Dewar is equal to the heat produced inside the Dewar. But since the Dewar slows the heat loss, the contents of the Dewar must be much warmer for the outside of the Dewar to radiate the heat produced.
Until then, the heat lost by the entire system will be less than the heat produced, so the temperature inside the Dewar must rise.
We are here to attack scientific ideas, not individuals. If you want to convince us that you are right, attacking the person you are discussing an idea with is bad tactics, as well as bad manners.
w.
carrot eater (13:23:57) :
“Concentrate” meaning what?
Hm.. perhaps focus works better in english, using a parabola for instance.
Carrot eater,
“I’m not sure where you’re going with energy density. You’re best off going back to the temperature of the two bodies. The sun is much hotter than the earth; thus NET radiation exchange will be from sun to earth, by a huge margin. No need to complicate things beyond that.”
The sun is way hotter yes, but it is also very far away. The energy flux from it has decreased significantly by the time it reaches us. Imagine there was an equivalent energy source just outside our atmosphere that generated heat identical to the heat from the Sun when it reaches our atmosphere. What would be the equivalent temperature of this source? That is the hottest we can get then, not the temperature of the sun 8 light minutes away.
That’s why the focus on energy density. Energy density just outside the atmosphere, just before the light enters the atmosphere is the upper limit of the average energy density on earth.
lgl,
Long wave radiation can be concentrated also. A microwave dish for receiving microwaves from satellites is an example.
John A: Pick out a single instance in the Figures above where net heat flux is going from a cold object to a warmer one. If you are unable to do so, then I suggest toning down the attitude.
lgl (13:03:55) :
Lenses can concentrate longwave, applications are common.
Ian Schumacher (14:05:39) :
But I can’t point a disk to the ground and focus the 400 W/m2 (avg. yes) emitted to boil water.
Willis-
I worked out steady-state formulas that allow for making calculations in your two-layer model without iteration. In case you are interested, they are in here:
http://home.comcast.net/~stevehaker/GlobalWarmingCalculator/Willis1.xls
I was thinking of using these formulas and your two-layer model to extend this single-layer “Global Warming Calculator” I made a couple of years ago:
http://home.comcast.net/~stevehaker/GlobalWarmingCalculator
In any case, thanks for your post. I am impressed at how many comments you have replied to!
Willis Eschenbach (14:16:08) :
So I can point a 3 m2 lense to the 15 C warm ground and get 1 kW in the focal point. Don’t think so.