The Steel Greenhouse

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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David Ermer
November 17, 2009 3:05 pm

Figure 2?
Nice article, Thanks!
REPLY: WordPress didn’t like the Macintosh generated graphics – fixed now -A

November 17, 2009 3:14 pm

Might this have some application to the coronal heating problem?

November 17, 2009 3:30 pm

… but if, as K&T and you both agree, the surface radiation is ~390 W/m2, equivalent to 288 K (today’s mean global surface temperature) by the SB equation assuming emissivity = 1 for longwave radiation, then the first differential – i.e. temperature change per unit change in radiative flux – is just 0.185 K/W/m2, while the UN gives 0.313 K/W/m2, and also assumes temperature feedbacks on top, giving a final climate sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.

Jim
November 17, 2009 3:32 pm

Cool! Errr … warm!!

Jack Wedel
November 17, 2009 3:41 pm

I know this is OT, but I just couldn’t resist this release from NASA.
The image is from November 17, 2009, days after a large snow storm swept over China, covering much of the country in white. This image shows part of the North China Plain near the city of Shijiazhuang. On the plain, the snow is white where it fell on fields or natural landscapes. Artificial surfaces in cities, towns, and roads are gray. While the larger cities and towns would be visible on a day without snow or clouds, many of the smaller towns and roads would be difficult to distinguish from the surrounding landscape. In the snow, however, small towns and the roads that connect them stand out clearly.
The storm that brought the snow came unusually early in the winter. The snowfall, the heaviest in decades, killed at least 32 people, destroyed 300,000 hectares of crops, and caused more than 15,000 buildings to collapse, reported Xinhua, the Chinese news agency. The storm closed roads and curtailed train and air travel. Airports in many large cities, including Shijiazhuang, closed.

Tom in Florida
November 17, 2009 3:42 pm

You stipulate that the Earth acts like a perfect blackbody. Didn’t I read here on this blog that it doesn’t?

Steve in SC
November 17, 2009 3:43 pm

I see a giant capacitor.

Paul Linsay
November 17, 2009 3:46 pm

The more I look at the K/T diagram, the more I think it’s wrong. Without convection and the multiple phases of water vapor, the earth’s surface would be close to or above the boiling point of water. Radiation is not what keeps the temperature at roughly 20 C. A correct understanding of the atmosphere would derive the K/T diagram, not use it as a starting point.

Ian Schumacher
November 17, 2009 3:49 pm

One possible problem with your thought model of a greenhouse is that you have the energy coming from the Earth. This a completely unlike a greenhouse where energy first comes from outside in that anything you put to absorb and re-radiate heat inward will also stop heat from getting to the surface in the first place. This is the problem with the greenhouse theory applied to Venus (for example). The atmosphere of Venus is so thick that the light can’t even reach the surface to begin with. Venus hot temperature must be internally generated. If this wasn’t the case then our oceans should be boiling (infinitely thick absorber of IR and below, so thick light never reaches the bottom, just like Venus!)
Imagine your thought scenario with infinite shells (solid steel miles thick as an approximation). The temperatures on the surface would be infinite. This must be wrong, but why? It’s because of the idea of energy coming from inside at constant rate no matter what is not physical.
Maybe my thinking is muddled somewhere, but that’s how I see it at the moment. For an alternate though experiment how about this. http://www.ianschumacher.com/maximum_temperature.html

Robert Wood of Canada
November 17, 2009 3:53 pm

Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings). Now, the planet’s surface is around 270K while the outside (space) is around 4K.
Clearly. most energy is radiated to space.
2. The Earth’s surface temperature is not a function of greenhouse gas warming, but of thermodynamics. It is largely a function of three things: mass of planet; mass of atmosphere and distance from the Sun.

Editor
November 17, 2009 3:59 pm

Ok, mathematical quibble here:
The area of the surface of the earth, having a given radius, is going to always be less than the inner surface of the atmospheric shell, which is also going to be less than the outer surface of the atmospheric shell.
For this reason, the shell will receive less W/sqm from Earth radiation than is emitted by the surface, because the radiation will spread out by the inverse square law just as sunlight gets less intense the further from the sun you get, and the shell will radiate more to space than back to Earth.

morganovich
November 17, 2009 4:10 pm

ian-
i’m just sort of taking a stab here, but it occurs to me that one should be careful comparing water to gasses. our oceans behave nothing like the venusian atmosphere. venus has 90 atmospheres of pressure. that’s enough to make a gas or liquid very, very hot.
the exception is water. waters is weird, weird stuff. the fact that is in non-compressible is what allows deep oceans to be cold despite the pressure. the earth itself, by contrast, gets very hot as you go deeper. even stone and iron are more compressible than water.
further, the state changes of water require massive energy and are a big part of our climate cycle. raising one CC of water one degree C takes one calorie. but to change that same amount of ice to water takes about 80 calories. the state change to gas requires even more energy. this allows the ocean’s evaporative process to shed massive amounts of surface heat in a way that the venusian atmosphere could never do.

C Colenaty
November 17, 2009 4:12 pm

Would it be reasonable to speculate that the overall framework of this model might provide an explanation of ice ages as resulting from some sort of periodic fluctuation in the atmosphere?

kurt
November 17, 2009 4:12 pm

“Robert Wood of Canada (15:53:58) :
Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings).”
This is incorrect. Emitted radiation is proportional only to the temperature of the emitting body. Convection and conduction are the modes of heat transport that are proportional to the temperature differential across a boundary.

Paul
November 17, 2009 4:16 pm

Something is puzzling me about the “greenhouse effect”
CO2 molecules are supposed to re-radiate IR Radiation Energy. Surely, if you double CO2 in the atmosphere, the amount of IR energy being re-radiated will be halved per CO2 molecule? From my understanding of Physics, energy can’t be created from nothing, so why would the CO2 theory mean increased re-radiation of IR energy?

Bill Illis
November 17, 2009 4:20 pm

Thanks very much Willis,
In your model, the orange cell (long-wave absorption by GHGs) is set to 95%, what levels do we set it at for changing GHG levels.
I am interested in simulating the last glacial maximum mainly right now – so CO2 would be around 185 ppm. Just changing the Albedo percentage to a number I believe it was (0.333) and then changing the orange cell to 89% simulates the surface temperature properly but the troposphere temperature drops to -11.9C, which is not the way I understand the climate models would simulate it. So how does the orange GHG cell change.

Charlie
November 17, 2009 4:34 pm

This is the clearest explanation of the greenhouse effect that I have seen anywhere.
Thank you!!!!
An oversimplified qualitative explanation works well for some things, but in this case, actually attaching some numbers to the various models really put the explanation in focus for me.

Alan D. McIntire
November 17, 2009 4:36 pm

That 390 watts is for a perfect blackbody. According to
Hartwig Volz
http://www.klimanotizen.de/2006.06.17_Sea_Water_Emissivity_Volz.pdf
the emissivity of sea water is somewhere between 0.92 and 0.965.
Splitting the difference, you get 0.9425.
You only need a surface flux of 390 * 0.9425 = 368 watts to
get 15 C.
In response to Mike Lorrey, the scale height of the atmosphere
is about 8 km. The radius of the earth is about 6400 KM,
so the difference in area between the two surfaces is
(6408/6400)^2 =1.0025. As Mike said, it’s a quibble.
I remember in high school we performed an experiment with water waves. We had a lever repeatedly strike the water at a specific point, and what we got was a series of concentric circles spreading out.
Stick a divider with a slit in the water, and the water passing
through the slit formed a new series of concentric circles, sort of
like this:
http://www.acoustics.salford.ac.uk/feschools/waves/diffract3.htm
The only reasonable explanation is that at each point, a new
concentric series of waves is generated, but most waves interfere
with each other, cancelling out. The same thing happens in
our atmosphere. At each point, a series of concentric waves is
generated, but most ot those cancel each other out. The net result is two virtually parallel patterns between the earth and atmosphere. Half the radiation from the shell will go to space, half will go back to earth into that insignificantly smaller shell,

Bird Stewart Lightfoot
November 17, 2009 4:46 pm

This article should be removed. The author obviously knows nothing about radiative heat transfer.
All of the numbers based on constant heat flux are meaningless because the surface area of the emitting surface is smaller than the surface area of the receiving shell. The actual radiative heat transfer is, ideally, proportional to the fourth power of the temperature, and the analysis must be based on this.
While total energy is conserved, energy flux is not.

ShrNfr
November 17, 2009 4:47 pm

Wood of Canada on your point 1. If the earth were in deep space with no energy input, what you say is true. However, there is 342 W/m^2 being pumped into the surface. This energy is partially absorbed and partially reflected. Depending on how “black” the steel is, and I think the article regards it as an ideal black body with zero albedo on the interior and exterior the fact that open space is 4 degrees K is irrelevant. Its the 342 W/M^2 input that has to be dispensed with.

DaveF
November 17, 2009 4:48 pm

I know that this is a little off-topic, but I thought I’d just mention that today’s “Independent” (UK) reports that Prof. Corrine Le Quere of the University of East Anglia (where else?) thinks the world is heading for a 6 deg C rise in temperature by the end of the century. Apparently this will mean the mass extinction of almost all life and that Copenhagen is the last chance to stop it. Head for the hills, folks, if this is true, or do I smell the whiff of desperation in the air?

michel
November 17, 2009 4:49 pm

“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.”
OK, I’ll bite. How much would 10 shells raise the surface temperature by? Or 100 shells?

JaneHM
November 17, 2009 4:58 pm

“kurt (16:12:26) :
“Robert Wood of Canada (15:53:58) :
Whoa there! I haven’t finished this article but I must make a couple of points:
1. The steel plate will not radiate equally in both directions. The radiation is in proportion to Ts^4 – Ta^4. (Ts = temperature of radiating surface; Ta = temperature of ambient surroundings).”
This is incorrect.”
Kurt,
no Robert is actually correct (although he didn’t express it precisely). The net ‘radiation’ is the emission (Ts^4) minus the absorption (Ta^4). Every blackbody above 0 K is both emitting and absorbing thermal radiation
JaneHM (University of Cambridge PhD in Theoretical Physics)

Clayton Hollowell
November 17, 2009 5:11 pm

Paul,
As you increase the quantity of CO2 molecules in the atmosphere, the fraction of IR radiation absorbed increases. The relationship is far from linear, however. Roughly speaking, the probability that any photon in the absorption band of CO2 passes the barrier is an exponential relation, but the absorption band changes with the CO2 pressure.

Steve
November 17, 2009 5:12 pm

You can’t say that the earth radiates 235 watts per square meter from the center and that this amount still ends up being 235 watts per square meter at the surface. As you go out in radius, the square meters of surface increases. For the total amount of energy radiated to be the same, the watts per square meter have to go down.
For example, if you measure the watts per square meter radiated by the sun at a distance of 1 AU from the sun’s surface, you aren’t going to get the same measurement of watts per square meter if you back away to 5 AU.
I haven’t done my geometry in a while, but I recall something about the surface area increase being proportional to the square of the increase in radius.

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