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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Hank Henry
November 18, 2009 9:17 am

I’m not sure I follow. We have a surface and a shell both at 235 w per meter but then the radiation from the shell, by radiating inward, increases the inner 235 w/m so the shell and the surface were really never both 235? It seems like we begin saying that ” the steel shell must warm until it is radiating at 235 watts per square metre” but then we continue and argue the shell “will warm the planetary surface until it reaches a temperature of 470 watts per square metre” So why wouldn’t we have to go back and say again that the steel of the shell must warm until it is radiating at 470 w/m? I think we are talking of temperature in a different way than I understand it.
To me the lesson of hot greenhouses, hot attics or hot metal sheds is that if air is confined and not allowed to convect upward to the sky it will heat up mostly because the cooling of convection has been stifled and less so because of any differential transmission of radiation by glass. In other words, sure it’s true greenhouses get hot, but once the roof of a shed or attic heats up and the air inside remains confined they get hotter than the outdoor temperature also.

November 18, 2009 9:28 am

For Ralph and other people who struggle with physics, believing in their intuition and think its physicists’ fault that they don’t understand “analogies”, let me ask a hypothetical question and see if anyone grasps the concept.
Suppose somehow that the glowing disc called the Sun is smeared out into a larger and larger disc. As it gets larger suppose the intensity of each little part gets less so that the total energy we get from the whole disc remains the same as its size grows. Now suppose the disc were smeared out all over the sky so that there is no distinction between night and day. We still receive the same amount of energy that we received before the smear out even though it would be received uniformly over all 24 hours instead of just during daylight hours. If this were to happen then the Earth’s average temperature would
a) increase
b) decrease
c) stay the same

November 18, 2009 9:40 am

It’s beginning to sound like man made a tragic error when he left the cave and decided to live on the fickle surface; does anyone know the internal temperature of our moon?

SteveBrooklineMA
November 18, 2009 9:48 am

I think Lubos Motl is too kind in his comment above. This analysis, which implies an error with K/T, errs in adding model assumptions to K/T that are not required in K/T’s approach. K/T is not a “single shell model” it is simply an energy transfer accounting. There is nothing that requires the K/T accounting to equate the downward radiated flux from the atmosphere to the ground with the radiated flux from the atmosphere to space. As Lubos points out, the atmosphere is hotter near the ground than in the high atmosphere. This is very different from a thin shell, which has only one temperature. You can argue with the values that K/T come up with, but this post does not somehow invalidate method of the K/T accounting.

Willis Eschenbach
November 18, 2009 9:58 am

Anton Eagle (00:15:58) :

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, please read all of the comments. The shell is only slightly larger than the planet. The difference between the outer area of the shell and the outer area of the planet is less than a tenth of a percent. So (as is typical for this type of discussion), I have ignored this trivial difference.
I stated this clearly above. So yes, the total area of the shell is twice the surface area of the planet, to within two tenths of a percent. I’m not sure why you continue to argue this point.

Charlie
November 18, 2009 10:09 am

Willis Eschenbach — the same misconceptions keep showing up in the comments.
You should update the head post with an addendum that addresses these common misconceptions.
1. Difference in size of shell vs earth.
Although you say in the text that the steel shell is ” few thousand meters” above the surface of the earth, the diagram exaggerates the difference in diameters. Your addendum or FAQ should explicitly state that “the steel shell is assumed to be close enough to the surface that for this simple model the outer surface of the shell and the surface of the earth are considered equal in area”.
2. The effect of albedo.
“For simplicity, this model treats both the earth and the shell as perfect blackbodies. Assuming an albedo other than zero will change the temperatures, but will not change the general conclusions on how the steel greenhouse works.”.
3. Confusion about blackbody radiation proportional strictly to the blackbody temperature, vs the NET radiation which is the difference between incoming and outgoing fluxes. The addendum/FAQ should emphasize that the inner surface of the shell emits downward at 235W/m^2 while receiving incoming radiation at 470W/m^2. There should also be a short discussion on the energy balance in the shell: 470W/m^2 received from the surface. 235W/m^2 radiated back down from the inner surface. This leaves and excess of 235 W/m^2 which in conducted through the very thin shell and radiated outward at 235W/m^2.
4. The shell has two sides. A blackbody at a given temperature radiates from all surfaces at a given flux. This ties back to point #3, above.
5. At least one post tried to point out a logical fallacy by saying that if this article was correct, then hot liquid in a thermos could be raised to an arbitrarily high temperature simply by adding more shells around it. That comment showed that they didn’t understand that the model had a constant energy source in the core, not a constant temperature mass. The real equivalent with a thermos bottle would be if one had an electric heater inside the thermos. With better insulation, the heater will get hotter, with no limit as the insulation approached perfection.
—————————————–
It’s an excellent article. Good enough that IMO it is worth responding to the common misperceptions by addressing them directly in the article.

Willis Eschenbach
November 18, 2009 10:16 am

Anton Eagle (00:30:43) :

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.

While it is true that a passive object cannot heat an energy source infinitely, it can certainly heat it. Look at the energy flows in the K/T budget in Fig. 3 above. The earth heats the atmosphere, but the atmosphere also heats the earth.
In this case the shell heats the planet, but not forever, just until a new equilibrium is reached. If it were as you say, putting a shell around a planet would have no effect on the planet’s temperature, which flies in the face of experience.

Mark T
November 18, 2009 10:28 am

Good points, Charlie.
John A, I’m not exactly sure what point you are trying to make in your “smeared” sun example. It is immaterial to anything Willis’ has claimed. Certainly an assumption of an input averaged out over the whole surface makes computation simpler, but that is its only purpose, i.e., the assumption is made only to remove integration.
Are you the same John A that missed the external source point in the thermos example, as both Charlie and I have noted? If so, really, try a different approach, e.g., an approach that actually addresses Willis’ argument.
Mark

November 18, 2009 10:32 am

Robert Wood, and JaneHM have it correct.
The steel shell analogy fails. The S-B law uses the difference in temperature between the radiating body and the absorbing body, each raised to the fourth power.
If a steel shell were used as posited in the article, with equal heat transmission from the outer surface and the inner surface, then there would be a substantial temperature gradient across the steel shell. The inner surface would be much hotter than the outer.
There are real-world applications of this principle in thousands of applications, in particular double-walled cryogenic liquid storage tanks. The annular space between the inner and outer walls is evacuated, thus heat transferred is almost entirely due to thermal radiation.

MartinGAtkins
November 18, 2009 10:39 am

Fom the original post.
The planet is now being heated by 235 W/m2 of energy from the interior, and 235 W/m2 from the shell.
The universe is an open system. Even with a shell an entity will reach equilibrium with the input of the energy supplied.
The paper is rubbish.

Roger Knights
November 18, 2009 10:41 am

Typo–replace “it” with “is” in:
“Un-noticed by their programmers, however, it that…”

Willis Eschenbach
November 18, 2009 10:41 am

Luboš Motl (00:32:39) :

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.

Thanks, Lubos, since you are a physicist your comments are much appreciated.
You are correct that the lapse rate explains the differences in the K/T budget upward and downward radiation. However, this doesn’t solve the underlying problem with the K/T budget — it doesn’t hold in enough heat. Unless there are two shells which are physically separated to prevent thermal and evaporative heat loss, the system will not heat the surface enough to explain the known surface temperature plus losses. There is a fixed physical limit on how much heat a single shell greenhouse system (with or without a lapse rate) can concentrate, and it’s not enough to explain the Earth’s temperature.
The objects are not black bodies, as you point out, which will affect the temperature of each object. Alternately, if the temperature is know, it will affect the amount of radiation.
However, as with the area of the shell and the area of the planet, this is not a large difference. For the shortwave (solar energy) part of the equation, it is explicitly accounted for by the surface albedo. For the longwave, most objects are very good at absorbing and emitting infrared radiation. Even snow and ice, which are blinding white at visual wavelengths, are almost black (excellent absorbers) in infrared. My well-thumbed copy of “Climate Near The Ground” gives the following emissivities for longwave radiation:
Water 0.98
Snow 0.986
Leaves 0.97
Grass 0.986
Coniferous Forest 0.97
Dry Peat 0.97
Dry sand 0.95
So if we assume 0.97 as the emissivity of the planet, it only makes a difference in the surface temperature of two degrees.
However, I must ask you to explain your statement that “The whole greenhouse effect would disappear if there were no lapse rate”. As my steel greenhouse shows, a greenhouse effect can exist without any atmosphere at all.

Roger Knights
November 18, 2009 10:42 am

PS–and delete the hyphen from “un-noticed”

Willis Eschenbach
November 18, 2009 10:45 am

Michael D Smith (00:57:13) :

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.

I could not agree more. I discuss this at length on another thread here at WUWT. The greenhouse effect exists, but it does not control the temperature of the planet. That honor goes to clouds and thunderstorms.

November 18, 2009 10:45 am

>>For Ralph and other people who struggle with physics,
>>let me ask a hypothetical question and see if anyone
>>grasps the concept.
>>>Smeared out, 24-hour Sun.
The Earth would stay at the same temperature. But you have to understand that the climate models assume that the Sun has very little direct impact on the atmosphere (on the steel shell). Thus there is no Solar warming of the steel shell.
Thus the thought experiment remains the same. The internal energy source represents the Solar flux passing though a ‘transparent’ steel shell, and warming the Earth, while the outgoing energy from the Earth is faced with a suddenly opaque steel shell, and is forced to warm it up (change in wavelength).
.

Willis Eschenbach
November 18, 2009 10:50 am

Cyril (01:05:22) :

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.

Relying on a single unpublished paper in arxiv which has no other support is generally not a good plan. The applicability of the Stefan-Boltzmann equation to liquids and solids has extensive support, both theoretical and experimental. It forms the basis of such things as infrared cameras, which show the temperature of water as easily as that of solids. It is also used in satellites to measure air temperature. I fear your basic thesis is flawed, but this is not the place to debate it.
Thanks,
w.

Willis Eschenbach
November 18, 2009 10:53 am

MartinGAtkins (01:25:06) :


I haven’t had time to read other responses yet but.

Dear friends, first rule of long threads. Read the other responses first, as many mysteries are solved therein. If you still have idea or questions or objections after reading the thread, then add them to the thread.
Thanks,
w.

November 18, 2009 10:59 am

>>>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.
Of course it can!!!!
Stand outside on a cloudy night, and you will invariably find the temperature quite balmy.
Then stand outside on a cloudless night (preferably the next day), and you will invariably find the temperature is quite cold in comparison.
The difference is that the clouds (your so-called passive objects) are heating up the surface of the Earth. (That term re-radiating, that someone does not like. Ok, absorption and radiation, so what.)
.
What this article has demonstrated, in an all-too-clear manner, is how easy it must be to dupe gullible politicians into believing the Global Warming hoax. Here we have a good cross-section of the educated public, and about half are totally stumped by a simple thought experiment involving a energy source and a surrounding barrier.
What we must have, therefore, is a degree course for politicians – a compulsory qualification that all budding politicians must acquire before standing for office.
.

Willis Eschenbach
November 18, 2009 11:03 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.

Second rule of long threads. Any time you see “Period. or “Case closed.” or “Discussion over.” or the like … it isn’t.

For example, I said in the head post:

The planet is in interstellar space, with no atmosphere

You reply

If the shell were a perfect reflector, the temperature of the atmosphere would simply …

As you can see, with that kind of misunderstanding, the case can’t possibly be closed. You go on to say:

If the shell were a perfect reflector, the temperature of the atmosphere would simply rise to the temperature of the input and stay there.

Since energy is constantly being added from the decay of the radioactive elements in the planet’s core, the temperature inside a perfect reflector would rise until the reflector melted.
So no, the case is not closed. Nor is that kind of false certainty an asset to you. Yes, I think I’m right, but I might be wrong … I have been before. I suggest that if you want to have a discussion, you adopt a like attitude.

Gary
November 18, 2009 11:05 am

Willis,
Good analogy to radiation through an absorbing medium. However just on point. The comparison of the temperature with and without an atmosphere is theoretical only. If the earth did not have an atmosphere the solar insolation would not be 235 but around 430 W/m2. This is because there would be no reflection or absorption of incoming solar radiation. So the atmosphereic effect includes more than just greenhouse. The difference becomes much reduced.

Willis Eschenbach
November 18, 2009 11:05 am

Sorry, that last one should have been addressed to John A (01:28:22), who was quoting Anton Eagle.

Willis Eschenbach
November 18, 2009 11:08 am

SNRAtio (01:50:42) :

“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.

Slow down there, my friend. There is a tendency of AGW skeptics to say “If X is wrong then the whole edifice crumbles”. Generally, this is not true. Often if X is wrong it just means that X and perhaps a few other things are wrong.

Willis Eschenbach
November 18, 2009 11:11 am

lgl (02:01:11) :

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.

Since I can build a steel greenhouse, clearly the greenhouse effect itself has nothing to do with greenhouse gases.
On Earth, as you point out, the particular greenhouse system found on the planet requires GHGs.

Willis Eschenbach
November 18, 2009 11:14 am

michael hamnmer (03:15:37) :

But herein lies a problem, the tropopause is too cold to radiate the 165 watts/sqM the Kiehl and Trenberth model say is radiated by the atmosphere.

As I note in the head post, in my two-shell model, the temperature of the tropopause is in agreement with measurements.

Jim
November 18, 2009 11:19 am

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John A (20:11:22) :
I’m sorry but I thought the article was tremendous nonsense, and will cause immense damage to the reputation of this website.
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I. Newton was wrong, but he didn’t take “immense damage.” Discussion and argument is good, that is a mainstay of science. Kick back and enjoy it 🙂

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