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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Le Sage's theory
November 17, 2009 10:39 pm

The difficulty with Greenhouse theories comes from a basic misunderstanding of how atmospheric temperature is achieved. It’s mainly the result of cosmic presure on the Earths atmosphere.
“The theory posits that the force of gravity is the result of tiny particles (corpuscles) moving at high speed in all directions, throughout the universe. The intensity of the flux of particles is assumed to be the same in all directions, so an isolated object A is struck equally from all sides, resulting in only an inward-directed pressure but no net directional force (P1).
P2: Two bodies “attract” each otherWith a second object B present, however, a fraction of the particles that would otherwise have struck A from the direction of B is intercepted, so B works as a shield,…”
http://en.wikipedia.org/wiki/Le_Sage's_theory_of_gravitation
Temperature is proportional to pressure, therefor Earth atmospheric temperature is formed by background radiation pressure, and distance from the shielding effect of the Sun
http://www.chm.davidson.edu/vce/KineticMolecularTheory/PT.html
CO2 plays no part in this.

November 17, 2009 10:54 pm

carlbrannen,
I’ve often thought that temperature itself is such bad concept. Its a state variable. It isn’t conserved. It depends on properties of the materials. Far better would be something like Energy density J/m^3. Energy density can be roughly translated into temperature when needed (for ideal gas, etc), we can integrate energy density to get energy which IS conserved. Various concepts come out of it naturally (energy flux energy density). Certain things come out of it easily, such as the concept of black body radiation and balance of energy flux with energy density in a cavity (for example).
Right away we could also see that the problem with Willis model is that he is starting off with an infinite internal energy source and therefore energy density inside the spheres is infinite to. How do we apply this internal energy source to the Earth with an external energy source. We can’t. Willis has put the Sun inside the Earth.
In a real model with external energy source, anything that we ‘put’ in place to restrict heat flux out will also restrict heat flux in. It is fairly obvious that putting a metal shell around the Earth would make it a whole lot colder, not warmer.
The key with global warming is conversion of wavelengths. Any global warming theory must address that. High energy short wavelengths come in, they heat the surface, lower energy longer wavelengths go out. This is what makes the Earth look more like a black body (absorbing more radiation then it otherwise would). Black bodies are often modeled as a small hole in hollow sphere. The greenhouse effect is a small hole in frequency spectrum, rather than space. The greenhouse effect makes the Earth more like a black body. But that is the limit, it can’t be MORE than a black body (hotter than a black body).

Willis Eschenbach
November 17, 2009 11:31 pm

Anton Eagle (22:12:14) :

Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.
As requested by others above… please remove this article.

The shell does absorb 470 W/m2. The point you seem to miss is that the surface area of the shell is twice the surface area of the planet.
Let’s take the surface area of the planet as X square metres. It is radiating a total of 470 W/m2 * X m2 = 470 X watts.
Now consider the shell. The surface area of the shell is 2X square metres. It is radiating 235 W/m2 * 2X = 470X watts … which is the same amount that it is absorbing. Energy in = energy out, and the system is at balance. That is what makes the system work, that a shell has two sides, so it has twice the surface area of the sphere.
So no, I see nothing in your claim that would require that I remove the article. Your “simple logical analysis” is simply incorrect.
Could we please hold off on requests to remove the article until we have at least discussed the objections to my article? This request, for action before consideration, is unseemly in a scientific discussion.

Willis Eschenbach
November 17, 2009 11:34 pm

yonason (21:37:01) :

Willis Eschenbach (20:36:37) :
“In a perfect greenhouse, one shell gives a surface temperature of two times the incoming radiation, two shells give us a surface temperature of three times the incoming radiation, three shells gives four times, and so on.”
So, how many shells will it take to get the core to 2,000,000 degrees?

Haven’t a clue, but the inner shells will melt long before that … not sure what your point is here.

Willis Eschenbach
November 17, 2009 11:37 pm

Thomas (22:14:36) :

Wait what? You add a shell, it obtains thermal equilibrium with the earth such that it radiates the same magnitude flux. Right: so you’ve got:
F(earth) = Flux(shell) (1)
but then you say:
Flux(shell in) = Flux(earth) and also Flux(shell out) = Flux(earth)
This is obviously wrong because
Flux(shell) = Flux(shell in) + Flux(shell out)
which is simply in violation of (1)
So really you should have
Flux(shell in) + Flux(shell out) = Flux(shell) = Flux(earth)
If you want, you can assume that the shell radiates equally in and out and that would make each 1/2 of the earth’s flux.

Thomas, it looks like you are stuck on the same point that Anton Eagle didn’t consider, which is that the surface area of the shell is twice that of the earth. As a result, it radiates half of the received energy inwards, and half outwards. Go back to Figure 1, which shows the relationships. The shell receives 470 W/m2. It radiates it over twice the area, which gives us 235 W/m2 in each direction.

Willis Eschenbach
November 17, 2009 11:45 pm

Ron House (22:29:46) :

Stephen Goldstein (20:05:22) :
I am trying to get used to using W/M^2 as equivalent to temperature”
It isn’t. Firstly, radiation depends on a body’s resemblance to a black body. There is no reason a shell as postulated could not have different reflectance on the two sides (or even on parts of one side, as with the Earth and ocean/land/forest/bitumen etc.). Secondly, there is not a linear relation between WM^-2 and temperature, since radiation is proportional to T^4.
And I have my doubts that there is any reason to equate the reflectances of the inner and outer sides, since the rising temp on the inside will certainly give a different apparent reflectance.
As for the article, it starts out determined to be misunderstood by showing equal flux density at two different radii, which is just plain false. Sorry, Willis, but you will be misunderstood if you write fallacies and expect the reader to pick up on all the approximations. And there is, of course, no approximation with the Earth’s core and the surface, these radii are wildly different. If you want to promote this further, it should be rewritten in terms of total radiation and the approximation introduced only where appropriate.

First, as I said, this is a thought experiment with a blackbody planet and a blackbody shell. So your objections about reflectance, which are correct and which apply to the real world, do not apply here.
Next, as I showed above (Willis Eschenbach (20:32:25), the difference in the radii is less that 0.1%. You are right, I should have stated that in the head post, but I remedied that in the response above, so I’m not sure why you are bringing this up again.
The same is true about the difference between the core and the surface radii. I addressed this above, at Willis Eschenbach (20:36:37).
Finally, I wish people would let up on the talk of “writing fallacies” and the like. I have written this as clearly as I can, and I am clarifying what is not clear as we go. I am not “writing fallacies”, I’m doing the best I can. If you have questions or objections, that’s great, that’s why I put this out … but there is no need to insult me in the process.

Willis Eschenbach
November 17, 2009 11:53 pm

Le Sage’s theory (22:39:15) :

The difficulty with Greenhouse theories comes from a basic misunderstanding of how atmospheric temperature is achieved. It’s mainly the result of cosmic presure on the Earths atmosphere.
“The theory posits that the force of gravity is the result of tiny particles (corpuscles) moving at high speed in all directions, throughout the universe. The intensity of the flux of particles is assumed to be the same in all directions, so an isolated object A is struck equally from all sides, resulting in only an inward-directed pressure but no net directional force (P1).
P2: Two bodies “attract” each otherWith a second object B present, however, a fraction of the particles that would otherwise have struck A from the direction of B is intercepted, so B works as a shield,…”
http://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation
Temperature is proportional to pressure, therefor Earth atmospheric temperature is formed by background radiation pressure, and distance from the shielding effect of the Sun
http://www.chm.davidson.edu/vce/KineticMolecularTheory/PT.html
CO2 plays no part in this.

While it is good, albeit surprising, to get a message espousing the physics theories of the 17th Century, here we are discussing the Steel Greenhouse. I have fixed your Wikipedia link in what I quoted above so people can read your theory at their leisure, and I invite you to discuss it on another thread.
Thanks,
w.

Willis Eschenbach
November 18, 2009 12:05 am

carlbrannen (21:17:46) :

This post is wrong for a rather large number of reasons. It doesn’t take into account the radius and W/m^2 correctly, but most importantly, W/m^2 is not, nor ever has been, equivalent to temperature. It’s just a bad idea.

Please read up on the Stefan-Boltzmann equation that I list in the appendix. It allows us to calculation the amount of radiation which is emitted by a body at a certain temperature, or the temperature which a body needs to emit a certain radiation.

A better analogy to the greenhouse effect is “insulation”. You can get that by steel spheres and air gaps, but it’s better to analyze it as insulation. With different thickness of insulation, a fixed amount of heat flow results in a temperature difference proportional to the thickness.

Unfortunately, a “blanket” or “insulation” is a very poor model for the greenhouse effect. This is the difference between an insulated bottle and a Dewar flask (Thermos bottle). They operate on entirely different principles.
The Thermos bottle operates exactly like the steel greenhouse in my thought experiment, using a shell surrounding the inner chamber and separated from the chamber by a vacuum. This is why the Thermos is so much more efficient at keeping your coffee from losing heat than an insulated bottle. Half of the heat that is lost from the inner chamber to the shell is radiated back inward, keeping the contents warm and slowing heat loss.

Willis Eschenbach
November 18, 2009 12:08 am

Ian Schumacher (22:54:19) :


In a real model with external energy source, anything that we ‘put’ in place to restrict heat flux out will also restrict heat flux in.

Curiously, in the Earth this isn’t true. The atmosphere is basically transparent to incoming solar radiation, but it absorbs outgoing longwave radiation. Heat flux out is restricted, but heat flux in is not restricted.

Anton Eagle
November 18, 2009 12:15 am

Willis, your reply to my comment is even less logical than your original article. You state… “The shell does absorb 470 W/m2. The point you seem to miss is that the surface area of the shell is twice the surface area of the planet”… blah blah…
No, your shell is not necessarily “twice the surface area of the planet”. In fact, in your model, the distance of the shell from the planet has no bearing at all on the model…

and thus the area of the shell has no bearing on the model (the further the shell, the larger the area), and thus your attempt to show the radiation of the shell being allocated to some larger area falls apart, since the area can be any value that is larger than the area of the planet. Its not any specific quantity.
Your comments don’t even hold up consistently against your own model. Just give it up. Your model doesn’t make any sense… it’s non-physical.

Anton Eagle
November 18, 2009 12:30 am

Look. Lets simplify this argument.
You are treating this shell as if its a source of energy. But, its not a source of energy. The only source of energy in your model is the planet.
So, you have an energy source that is heating up a passive object. That passive object cannot then turn around and heat up the energy source. If it could, you would end up in some kind of infinite loop that violates the laws of thermodynamics. This simply cannot happen. Period. End of discussion.
Another way of looking at your model…
Pretend your shell is in contact with the surface. Its easy to see then, that the shell would simply be the same temp as the planet, and the planet would be the same temp as it was without the shell. Raising the shell up off the ground does not change this situation. Again, the shell does not magically heat up the planet… anymore than the surface material of the planet heats up the material below the surface.
You are violating laws of thermodynamics by essentially having energy flowing up hill (so to speak)… from the passive shell to the energy source. This just doesn’t happen.
If it could, then you could just as easily argue that two space heaters placed next to each other could heat each other up to millions of degrees with the same kind of infinite loop. Again… just doesn’t happen.

November 18, 2009 12:32 am

Exactly, Willis, very nice and analytic in Nature.
At the beginning, I was afraid that you would argue that the difference between 195 and 324 would permanently invalidate any such model.
The two figures are different simply because 195 refers to the radiation of the (colder) upper layers of the atmosphere (up), while 324 is the radiation of the (warmer) lower layers of the atmosphere. They don’t have to be equal. The whole greenhouse effect would disappear if there were no lapse rate – i.e. cooling of the air with the altitude which is close to the adiabatic one in the real system (as opposed to isothermal).
With the two shells, you may cure any contradictions of this sort because the two shells may have different temperatures and any discrepancy can be covered by non-radiative heat transfers in between the two shells. Of course, such additional terms in the budget are not arbitrary: heat convection is proportional to gradients, and so forth.
However, I am still not quite certain that you have fully accounted for the objects’ not being black bodies.
By the way, your figure 1) would suffer from an entertaining new huge problem – which is much smaller in reality. Your figure has the outer sphere which has approximately a 3x bigger radius than the Earth. Consequently, its surface area is 9x bigger, and “per squared meter” data should be recalculated to a much bigger area, changing them by a factor of nine which is not quite negligible. 😉
In fact, if the spheres were that different, it would still be a priori natural for them to have the same temperature at equilibrium. However, the inner sphere would be getting most of the energy from the outer sphere, and a negligible one from the Sun.
In reality, the atmosphere’t thickness is something like 0.3% of the Earth’s radius, making the external surfaces 0.6% bigger or so. Neglecting that the Earth is not round ;-), one could therefore get errors as big as 0.6% of 342 W/m^2 which is something like 2 W/m^2 only.
All the numbers are global averages. The actual flows depend heavily on the place, especially on the latitude, and there are additional heat transfers in between the different latitudes.

Michael D Smith
November 18, 2009 12:57 am

Monckton of Brenchley (15:30:32) : “…giving a final climate sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.”
Nice to see you here, damn fine job on Glenn Beck. Next time maybe he’ll let you do the talking for the full hour.
The non-radiative effects (convection) and phase changes carry more heat into the upper atmosphere where there is a greater chance for energy to be radiated directly to space, less chance of radiative interaction with molecules on the way out. A small localized change in surface temperature can cause a convection burst (thunderstorm) and a large increase in convection height, improving both reflection of incoming solar radiation, and conveying sensible heat to a higher altitude where it can then escape to space via radiative processes with far less interference. Convection at lower altitudes will punch through the thicker lower layers allowing a much more direct path for radiative effects to become efficient and unhindered in delivering the heat to space.
Clouds and condensation are the balancing outgoing delivery mechanism of heat on this planet, and overwhelm the radiative effect with convection, and as a bonus also block incoming radiation, especially in the tropics, leading to a natural, self regulating thermostat effect. This results in the lower climate sensitivity we are seeing in many recent studies like Lindzen’s. The notion of an H2O positive feedback (which probably is present on a clear day) is squashed by this process.
While warmer air can hold exponentially more water vapor, presumably increasing greenhouse effects (an process the IPCC hangs its collective hat on), it is also this exact same property that vastly improves the chances of convective and phase change heat transport by thunderstorms. Once triggered, the radiative effects of H2O are completely overwhelmed by the storms, resulting in a very strong localized negative feedback. Cumulus clouds will have the same effect, but more in balance with the positive effects, resulting in less negative net feedback, but with the same result, much lower climate sensitivity than the IPCC would have you believe.
I realize that climate sensitivity is not usually discussed as a local phenomenon, but it should be, since it is the integral of all local phenomena.
This should be fairly obvious. Suppose some factor, like CO2, or weather, it doesn’t matter, produces a warmer day. More water will evaporate. And then it is more likely that a storm will develop. If one does, negative feedback by non-radiative effects. If one doesn’t, positive feedback by radiative effects. More positive feedback also evaporates even more water, further increasing the chances of a storm developing. If the storm develops, you can be guaranteed that the excess heat will, without a doubt, be taken to space in an orders of magnitude more direct way. The radiative changes will be fairly small no matter what, but the non-radiative effects will be most extreme where needed the most (tropics and mid-latitudes), and very mild where temperature differentials are smaller (toward the poles).
An important study would be to nail down these localized effects using measurements in very small increments of time and space. Once the local effects are understood, it should be a much simpler matter to integrate them over time and space using satellite cloud measurements to a more generalized climate sensitivity over various bands / latitudes of the earth. It seems most researchers are using data over far too much time and distance to derive the feedbacks properly, resulting in a very smeared picture of what’s going on. The real differences between radiative and non-radiative are very extreme but short-lived. Capturing this signal and integrating it seems to me to be the only way possible to better understand the broader effects and how sensitivity changes over time and space.
When the non-radiative effects are orders of magnitude more effective than the radiative ones, and we see this clearly on the surface of the sun (or in a boiling tea kettle), does this not scream net-negative feedback in the climate system?

Cyril
November 18, 2009 1:05 am

Thin foil metal (solid) are very efficient to protect you from very hot source, but as stated by doctor Pierre-Marie Robitaille (Ohio State University) in “An Analysis of Universality in Blackbody Radiation”
http://arxiv.org/ftp/physics/papers/0507/0507007.pdf :
“The application of the laws of Planck, Stefan and Wien to non-solids is without both
experimental and theoretical justification”
Neither the hearth (mainly ocean) nor the atmosphere are solid. This model will not work.

Ed Zuiderwijk
November 18, 2009 1:08 am

I’m sorry, but this is not even wrong …..

MartinGAtkins
November 18, 2009 1:25 am

In order to maintain its thermal equilibrium, 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 metre. 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.
I haven’t had time to read other responses yet but.
I believe the model to be fundamentally wrong.
“To do this, the steel shell must warm until it is radiating at 235 watts per square metre.”
1. The entire surface of the outer shell would only have to radiate the total energy output of the core. Since the surface area of the outer shell has a greater area than the core then it would only radiate a fraction of the total energy at the outer surface area per unit that the expanded volume would dictate. Not 235 watts per square metre.
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.
2. The entire shell would radiate 235 watts in all directions so in a two dimensional world only 235/2 or 117.50 watts would be returned to the system.
If my obsevations are correct then post is misleading and of no help to those who wish to learn.

November 18, 2009 1:28 am

Re: Anton Eagle (22:12:14)
Logic. Science and logic should be inseperable.
This article is utter nonsense, and a simple logical analysis reveals this without even resorting to physics (and I am a physicist).
Using the author’s argument… the mythical planet radiates 235 W/m2. The steel shell absorbs this and re-radiates it all in both directions. The planet obligingly absorbs the re-radiated 235 coming off the shell, and re-radiates it, for a total radiation “budget” of 470 W/m2. The author conveniently stops at this point.
But wait… there’s more. If we are to follow the author’s logic, then the shell should absorb this 470 W/m2 and re-radiate it again… 470 out and 470 in. The planet should then absorb this 470… add it to the orignal 235 coming from the core, and now be radiating 705 W/m2… this would continue till everything anihilated at a gazillion degrees. Utter nonsense.

Thank you Anton. It wasn’t just me who spotted that the logic is simply busted.
If the shell were a perfect reflector, the temperature of the atmosphere would simply rise to the temperature of the input and stay there. Regardless of how many shells were outside, the temperature would never rise beyond the input.
Willis, next time contact a physicist before doing something like this.
This isn’t about the greenhouse effect, because you’ve misunderstood what the greenhouse effect is.
Case closed.

November 18, 2009 1:31 am

>>In a sense, what Willis has done is manage to
>>make the same mistake as the people who think
>>that the Earth is in “radiative balance” and try to work
>>out a tedious budget of everything going in and out.
Ummm. The Earth HAS to be in “radiative balance” over long time intervals, otherwise it would be warming up or cooling down.
.

November 18, 2009 1:36 am

Willis,
Let me demonstrate with a simple example:
Say you have a Dewar (vacuum flask) and you fill it with lukewarm soup. The energy is radiated back from the outer reflective skin (conduction and convection being minimal.
According to your recipe, the temperature of the soup would rise. And if you put that Dewar flask inside of another, the temperature would rise still further… and so on ad infinitum.
Except that that isn’t what happens at all. The soup remains lukewarm and slowly cools.
Its a basic mistake Willis, but you’re going to have to think about it first.

Jon-Anders Grannes
November 18, 2009 1:45 am

If i measure IR temp from clouds above I find that its much colder(density of the air) than on the ground, so this is not where the IR is re-emitted to ground.
The IR from the ground will have to “absorbed” by the air just above it and i really have a problem imagening that this air is re-emitting this energy as IR?
I imagine the climate on Earth mostly as consequence of density of the air an geographic position.
If we depleted the atmosphere over Earth from todays 1013 hPa to 500 hPa we would have an average temperature over Equator at the surface as it is today at 18.000 feet, that is below freezing!
So for me “greenhouseeffect” is just that energy(light and IR etc) is trasmitted trough the atmosphere and that resistance in the atmosphere, density of the air, leads to some warming of the media(atmosphere) this energy is going trough.
?

November 18, 2009 1:47 am

>>But wait… there’s more. If we are to follow the author’s
>>logic, then the shell should absorb this 470 W/m2 and
>>re-radiate it again… 470 out and 470 in. The planet
>>should then absorb this 470… add it to the orignal 235
>>coming from the core, and now be radiating 705 W/m2…
>>this would continue till everything anihilated at a gazillion
>>degrees. Utter nonsense.
The outer shell DOES re-radiate 470 W/m2 — 235 inwards and 235 outwards. That makes 470 to me and the outer shell is in equilibrium (as is the Earth).
And you are a physicist??
This is something I find with scientists in general. Quite brilliant in their specialist field, but all at sea in the real world. The trouble with climate, of course, is that it is a hugely multi-disciplinary field, and there are not that many people who have both the depth and breadth of knowledge to be able to deal with it logically – as this article has demonstrated.
.

SNRAtio
November 18, 2009 1:50 am

“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.”
OK, so now the whole tradition of which Kiehl/Trenberth is a part is proven wrong.

November 18, 2009 1:55 am

>>Ron House (22:29:46) :
>>As for the article, it starts out determined to be
>>misunderstood by showing equal flux density at two
>>different radii, which is just plain false.
Hardly. The shell is 2 km above the surface of a planet that has a radius of 6370 km. Hardly a difference in radii for a simple thought experiment.
.

lgl
November 18, 2009 2:01 am

Positive feedback from wiki: “Ai = (output voltage/input voltage) = A/ (1 − Aβ). Here A is the gain of the feed-forward active part of the amplifier without feedback, and β is the gain of the feedback element”
In your ‘Steelhouse’ A=1 and β=0.5 giving Ai=2 so that works fine.
Your problem is: “It has nothing to do with blankets, or mirrors, or greenhouse gases”. It has everything to do with greenhouse gases. O2 and N2 do not radiate (almost) so without the GHGs you would not get much downward LW.

Robert Wood of Canada
November 18, 2009 2:02 am

kurt (16:12:26) :
I stand corrected. I meant that the total radiative flux is proportional to Ts^4-Ta^4. Thanks for clarifying me, JaneHM.

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