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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Willis Eschenbach
November 20, 2009 1:31 pm

par5 (04:19:34) :

Cantobtainium? Are you no longer patient? Thought experiments have no ramifications?
“The planet is in interstellar space, with no atmosphere and no nearby stars.”
I’m pretty sure this was before you built your shell, yes? Perhaps it doesn’t matter?

A “thought experiment” often posits conditions which are not physically possible, in order to clarify some thorny question. Einstein used them frequently:

Here, and throughout his lifetime, he made up a thought experiment (Gedanken experiment). Gedanken expermiments cannot actually be performed, but nevertheless are a useful tool for testing a theory. Einstein imagined a person travelling at the speed of light and asked how that person would perceive the light wave.

Now, you could object to Einstein’s thought experiment as you have objected to mine, for a host of valid physical reasons … but both are useful nonetheless.
“Cantobtainium” is my humorous name for a mythical element which, as the name suggests, we can’t obtain. It is very useful in thought experiments, as it has no known properties. As Foghorn Leghorn used to say, “That’s a joke… I say, that’s a joke, son.” However, I could have just said “steel” instead, neither one vaporizes at 470 W/m2.

P Wilson
November 20, 2009 1:36 pm

Its a thought experiment. Well fine. A lot of physics contains thought experiments not yet proven when applied to areas they don’t belong. Would you use Boyles law to measure the area of a room, or would you use a standard measuring tool calibrated in centimetres?
The SB is pulled into climatology to justift some quite preposterous figures, applied to gases and liquids equally, irrespective of their molecular properties.

Willis Eschenbach
November 20, 2009 1:38 pm

TomVonk (04:22:04) :

The particular case used here is the case where the Planck’s law is integrated for solid angles in a “half sphere” e.g the plane surface sees only a “half space” .
That means concretely that the SB constant is not an universal constant but is only valid for plane surfaces emitting in a half space . For instance a ponctual source or a double faced shell emit in the whole space and the SB law for them has an SB constant which is DOUBLE of the one you used .
I went farther by saying that the constant you are using is only valid for convex bodies . No problem for your model because you have indeed chosen convex bodies but it would be wrong to think that the SB constant you use is valid for any body . Especially bodies with cavities are tricky .
2)
W/m² (or irradiances) generally do NOT conserve .
For spheres they definitely don’t (they do for INFINITE parallel plans) .
The irradiance of your shell is P.r²/R² where P is the irradiance of the inner sphere . As P.r²/R² < P follows that :
a) the shell has a lower temperature than the inner sphere
b) the shell and the inner sphere are not in thermal equilibrium (e.g there is a net flux going from the inner sphere to the shell)
c) The difference of temperature between the shell and the inner sphere (so the net flux) increases as R increases .

First let me say that you are correct. Second let me say that the differences between what you say and the results I show above are trivial, on the order of 0.2%. This is because for all practical purposes, there is no difference between the situation I describe above and two infinite parallel planes.
This is because the radii of the sphere and the shell are equal to within less than a tenth of a percent. As is common in analysing this particular situation, this tiny error is ignored, and they are treated as though they were infinite parallel planes. See e.g. the Kiehl/Trenberth diagram I show in Fig. 3.
I think that answers your questions, if not, let me know.

Willis Eschenbach
November 20, 2009 1:44 pm

P Wilson (05:29:08) :

lgl (10:37:03)
it still makes little sense. if 256 re-radiate back to the source and 256 exits the barrier, its still 256 one way and 256 the other that can’t be added together. (Addidion is the unified magnitude of several units) . translating it to temperatures. if an object radiates to an energy measured at 20C to a shell for a sufficent time to heat it to that temperature, then the shell isn’t going to be 40C because radiation leaves it bi-directionally, 20C each way. Temperatures don’t add up mathematically like that. If they did you’d have a litre of water at 10C and a litre at 30C combined to give 2 litres at 40C
blow up a balloon

I’ll repeat an explanation I gave upstream. Assume the surface area of the planet in Fig. 1(b) is “X” square metres. Total radiation from the planet is 470 W/m2 times X square metres, which is 470X watts.
The surface area of the shell is twice that of the planet, or 2X. Total radiation from the shell is 270 W/m2 times 2X square metres, which is 470 watts.
Finally, the reason for using W/m2 is that you can add them together, but you can’t add temperatures together.

Willis Eschenbach
November 20, 2009 1:46 pm

SteveBrooklineMA (07:53:29) :

Willis-
Here is a web-based calculator for your two-layer model:
http://home.comcast.net/~stevehaker/GlobalWarmingCalculator-II/
I put your figure in there for reference. I hope you don’t mind. I will take it out if you prefer.
I had fun making it. Playing around with the numbers is interesting.

Steve, that’s awesome. many thanks. Give me some time to play with it.
w.

Willis Eschenbach
November 20, 2009 1:51 pm

P Wilson (08:09:46) :

The SB may give total power emitted by surface, although other calibation equipment gives better measurements of power emitted form a surface. at 100w/m2 uniformly over a flat surface, is enough energy to produce 24C. 300-500w/m2 is enough energy to cook food from frozen in oil (see citiations above)
Why is this equation used to describe anything but incandescent/hot or superheated metals? If a black box were the absorbing heat at maximum potential, then it theoretically emits as much as it receives, but only if it is out of equilibrium. If it is in equilibrium then the rate for this theortical plane is 235w/m2 no more, no less between the two spheres. (if you thing the SB constant is a valid measure of energy).

You seem to be operating under a misconception. Maximum tropical sunlight on a flat surface at midday is on the order of a kilowatt per square metre. Call it two thirds of that in the US. But if you think that you can cook dinner by just laying it out on a wooden plank at noon, you are very mistaken, tropics or not.
I did a quick search for your “citation above” on google but found nothing. Please supply a URL.

Willis Eschenbach
November 20, 2009 1:54 pm

P Wilson (13:07:07) :

carrot eater (10:03:18)
The SB works well for IR radiation such as the sun, and heated metals in engines or but not normal temperature, or LW surfaces in equilibrium. If a black box in your room was next to a white box, according to SB they should be different temperatures and the black should be emitting more radiation than the white box, meaning they ought to have different temperatures. However, if they are both in thermal equilibrium at that room temperature then they’re the same temperature. Normal temperature matter just doesn’t don’t give off so much radiation that the constant gives it. (if its in equilibrium)

I fear I’m going to have to stop responding to you, P Wilson. The SB equation applies from the temperature of the interstellar microwave background to the temperature of the sun, and all temperatures in between. Please read up on it, as this is not the thread for disputing well known basic physics.

Willis Eschenbach
November 20, 2009 2:03 pm

BillG (13:11:21) :


The gravitational field is important in that as mass is convected to a higher altitude via expansion and work on its surroundings, thermal energy is traded off for potential energy. As a result, the temperature of the mass parcel decreases due to adiabatic expansion and the dry adiabatic lapse rate is created. But the total energy content remains constant. At higher altitudes radiation heat transfer becomes more significant and OLR is emitted and the parcel cools. This is one reason why the greybody temperature of the earth is lower than the surface temperature. Subsidence then occurs and the mass parcel descends and trades off potential energy for thermal energy and the parcel increases in temperature due to adiabatic compression. This is why the surface temperature is higher than the greybody temperature. This subsidence conversion of potential energy to thermal energy represents the bulk of the 324 W/m2 down arrow in the K&T cartoon. Very little is due to “back radiation”.

I was with you until I got to this paragraph. The downwelling longwave radiation from the atmosphere is not some theory that you can wave off with your own theory. It is a physical phenomenon which is measured routinely by scientists, and has been measured for decades all over the world. Do a google search on ‘surface downwelling longwave measurements’ to see thousands of examples.

Willis Eschenbach
November 20, 2009 2:08 pm

P Wilson (13:26:41) :

If you take the human basal metabolic rate as a standard of comparison – it is the calories required against calories burned, converted into w/m2, which for most people is 58w/m2, increasing on average to 100w/m2. In the following experiment radiation is measured fromleaving individuals.
in google enter:
Description of a human direct calorimeter
and it takes straight to the cambridge journals which conducted the experiment.
It stands to reason that if a human optimum emites 100w/m2 of radiation through a duration, that earth is emitting much less per square metre.

Sorry, missed this one. You have misread the reference. The values that it is giving are in watts, not watts per square metre.

Dyspeptic Curmudgeon
November 20, 2009 2:16 pm

Interesting thought experiment, but I think it is totally and completely wrong.
You have improperly deleted all reference to the energy inputed to the system from the sun and discussed only the energy produced by the interior of the earth. In reality, together those are in equilibrium. Put a steel greenhouse around the earth, and the surface will be heated by the radiation coming from the inside of the shell (instead of directly by the sun). To which the energy from the earth’s core will continue to be added.
Your experiment is fatally flawed in real terms. Moreover you have conflated cause and effect. The surface will not double in temperature because of the effect of the radiant energy of the shell. The surface and the shell will stabilize (at the same temperature since this is a ‘perfect’ experiment.) In your version, the energy source is the earth’s core giving a 235 W/m2 level. In reality, the OUTSIDE of the shell will be heated, and will radiate about 234 W/m2 inward (I find it hard to believe the the earth’s core produces much more than 1 W/m2 at the surface, although I am almost surely wrong about the exact number….it is probably whatever equates to 3 degrees C…which is the average deep oceanic water temperature.)
To claim that since the shell ‘must’ radiate both inwards and outwards, that the surface necessarily will be heated to
Interestingly I was recently pointed to a paper which discusses the Greenhouse effect and falsifies it *on a physical basis*. CO2cannot do the job.
http://arxiv.org/pdf/0707.1161v4 (link is to a pdf file at Cornell U).

Dyspeptic Curmudgeon
November 20, 2009 2:31 pm

Opps, missed something in the editing. Should be:
‘To claim that since the shell ‘must’ radiate inwards and outwards, that the surface necessarily will be heated to twice the 235 W level, since it is already at that temperature, and will also receive that amount of radiated energy is mindboggling. Yes you *can* add numbers representing similar units but that does not necessarily mean that you *can* add meaningfully add those items of those units and have the result mean something.
Short thought experiment: 2 small steel plates suspended closely in parallel in your kitchen at room temperature: say 20C. Remove one plate. Does the temperature (black body radiation) of the other change? NO. The temperature is the result of external forces. Replace the plate. Heat the back of one, with a heat lamp, until its temp is stable over the entire front surface.No direct light impinges on the other, and the room stays the same. Say the front of the plate is now 40C. What is the temperature of the front face of the other plate? In your thought experiment that plate would be 20C (in W/m2) plus what is radiated from the other 40C (in W/m2)…. Nah….Does not compute…and it does not happen like that. Agreed, the cool plate might climb a little but not much.
Even allowing for perfect conduction of energy between the plates, what you hypothesize DOES NOT HAPPEN.
An Inconvenient Truth perhaps, but like the goreacle you are wrong.

P Wilson
November 20, 2009 2:40 pm

Willis Eschenbach (14:08:45) :
whoops you’re right. A human can be anything from 0.75-2m2

P Wilson
November 20, 2009 2:52 pm

Its in a textbook:
Vijayan and Singh
“Advances in deep fat frying of food”
1997 CRC Press
Not sure if it can be found online.
However, we’re all agreed black tarmac surface will absorb more heat and transmit more heat than the equivalent area of grey or white matter. However, once equilibrium is as close as it can be then theoretically, re-radiation should occur at night, or at the point when the temperature goes down, and presumably this is where the greenhouse effect should be felt the most – yet matter emits heat very quickly – and quickly thermalise to new temperatures, so as not to give off that much radiation. That can be seen by thermal imaging equipment at night that traces 7-14 microns. they’re specifically used to trace humans

Gary Hladik
November 20, 2009 4:14 pm

Willis, I see you’re still here answering questions. That’s the mark of a true educator.
This all reminds me of the earlier discussion on whether CO2 could freeze out at earth’s south pole. As here, a number of misconceptions had to be addressed, and as here, I had to hit the references pretty hard to sort the whole thing out.
It’s a good thing you presented a simple model, otherwise you’d be here until some hacker exposed the whole CAGW fraud. 🙂
Oh, wait…

carrot eater
November 20, 2009 4:18 pm

P Wilson (13:07:07) :
P Wilson, you are continually confusing yourself into thinking things which aren’t true, and then blaming the Stefan-Boltzmann law for it.
If this matter so much to you, please go to the hardware store and buy an infrared thermometer. It’ll cost you about $50. Then aim it at different things around the room. You’ll find that the thermometer works. And it works because all those things around the room are emitting radiation, just as the SB law says they should.
It’ll be as direct a confirmation of the principles as I can offer you.

P Wilson
November 20, 2009 5:16 pm

of course they’re emitting some radiation. Just not as much energy as is inferred from an equation

carrot eater
November 20, 2009 6:00 pm

P Wilson: Buy an infrared thermometer. It uses Stefan-Boltzmann. It works. The only reason you think it doesn’t work is because you’ve confused yourself in applying it, and you’re saying things like “300-500w/m2 is enough energy to cook food from frozen in oil”, which has zero meaning.
It’ll only not work if you point it at some shiny metallic surface with a low emissivity.

November 20, 2009 7:35 pm

Steel shells, mirrors, blankets, they are all fine metaphors but I have always felt they failed at getting to the heart of how the greenhouse effect works i.e. letting light in at one frequency and rejecting it at others.
They don’t give us any idea of limits, if there are any, what they are and if they are important.
I had written up a document some time ago about ‘limits’ being those of a blackbody, but the document wasn’t that well put together and I don’t think convinced many. The resistance of most to even considering this concept seems strange to me. To accept, without thought, that we can create something that absorbs more energy than a blackbody … I still don’t understand that. Ca la vie.
I have taken yet another attempt at writing up a description of why there must be limits to the greenhouse effect and why this limit is almost certainly that of a blackbody. I hope this document is much clearer then the previous one. The document can be found here:
http://www.ianschumacher.com/greenhouse_effect_maximum.html
I accept again that most will reject it. That’s ok. The unquestioned wisdom that greenhouse effect can build up to infinity is strong. I accept that I’m probably wrong even. I make a lot of mistakes … But I would ask then that those who quickly dismiss this idea consider then, ‘what is the limit?’ Is it inifinity? Unlikely. Is it the temperature of the Sun. Maybe, but if so, then with the the right atmosphere, Pluto could become as warm as Mercury. Distance from the Sun wouldn’t matter. Somehow I doubt it. So in that case, what is the limit? Think about it …
Cheers.

Willis Eschenbach
November 20, 2009 7:36 pm

Dyspeptic Curmudgeon (14:16:36) :
Dyspeptic, please read the entire thread. Some things, such as your link to the arxiv paper, have already been discussed.

Interesting thought experiment, but I think it is totally and completely wrong.
You have improperly deleted all reference to the energy inputed to the system from the sun and discussed only the energy produced by the interior of the earth. In reality, together those are in equilibrium. Put a steel greenhouse around the earth, and the surface will be heated by the radiation coming from the inside of the shell (instead of directly by the sun). To which the energy from the earth’s core will continue to be added.
Your experiment is fatally flawed in real terms.

Also, please re-read what I wrote in the opening post. For example, I said “The planet is in interstellar space, with no atmosphere and no nearby stars.” No nearby stars means that there is no radiation from the sun.

Moreover you have conflated cause and effect. The surface will not double in temperature because of the effect of the radiant energy of the shell. The surface and the shell will stabilize (at the same temperature since this is a ‘perfect’ experiment.) In your version, the energy source is the earth’s core giving a 235 W/m2 level. In reality, the OUTSIDE of the shell will be heated, and will radiate about 234 W/m2 inward (I find it hard to believe the the earth’s core produces much more than 1 W/m2 at the surface, although I am almost surely wrong about the exact number….it is probably whatever equates to 3 degrees C…which is the average deep oceanic water temperature.)

Again, what is shown in Fig. 1 is a thought experiment, not a description of the earth..

To claim that since the shell ‘must’ radiate both inwards and outwards, that the surface necessarily will be heated to
Interestingly I was recently pointed to a paper which discusses the Greenhouse effect and falsifies it *on a physical basis*. CO2cannot do the job.
http://arxiv.org/pdf/0707.1161v4 (link is to a pdf file at Cornell U).

Willis Eschenbach
November 20, 2009 7:39 pm

Dyspeptic Curmudgeon (14:16:36) :

Interesting thought experiment, but I think it is totally and completely wrong. …

Dyspeptic, please re-read first the original post, then the entire thread. You keep talking about the sun heating the steel shell, when I said “The planet is in interstellar space, with no atmosphere and no nearby stars.” So there is no solar heating, because no star is nearby.
Your other objections have been dealt with in the thread, including your reference to the arxiv paper.

par5
November 20, 2009 8:58 pm

Willis, I think that I have solved your conundrum for you. I did not realise that the planet Cantobtainia was completely sealed with a thick layer of cantobtainium (which prevents outgassing). At first, I thought that my questions were valid, on topic and had merit. I was wrong and I do apologise. Anyway, I glanced at my atomic chart shower curtain today and realised that cantobtainium has to be some sort of radioactive isotope or trans-uranium element. Giggle. So you see, it is your encasement of cantobtainium that is producing your 235W/m^2.
It is a win/win situation for both of us. You get to keep your planetary greenhouse without an atmosphere, and I have the pleasure of knowing that your planet isn’t going to last much longer due to its high rate of decay.
Can we now shake hands, thank each other for the polite conversation and walk home happy and content?

Denny
November 20, 2009 11:22 pm

anna v, 22:10:40
Thank you for your input! I thought about that also, Wood doesn’t burn as hot and I liked what Willis stated about “more flame” produced by Wood than Coal. Coal being denser, it makes sense.
Anna, a wood/addon is a stove that has a box around it and a Phlenum on top. The heated air is sent around the Fire box by a Squirel Cage Blower, out the Phlenum and into the Duct system that carries the heated air to the Rooms. When I use “only” Wood for heat, it creates the specified sense of warmth I mentioned in my first post. The warm air surrounding the Living Space has this feeling. It’s false only in the way that the RH is very dry. Without Humidifing, my RH goes down to below 30%. Not good for your health I might add.
This feeling I’m talking about in the heated conditioned air going into the Living area, changes with a noticable difference in “confort”. Yes, Willis is also correct that the “smell” also changes and that’s because a little bit of Coal smoke in the House can go a long way in staying around odor wise. It’s a very “heavy Smoke”!
Willis stated that I can smell the Heat…Well, that’s true to a degree but it’s when I can actually “feel” the temperature, does it feel warmer than other Fuels? Yes is my answer. I’ll put in order a list of the “best” heat to the “worst” heat IMO.
1. Wood Heat- because you instantly feel warmth when it’s around you.
2. Hydronic or Baseboard Heat- this heat produces a warm heat because it uses convection and doesn’t dry the air out as much. Type of Fuel doesn’t make any difference. Also depends on the efficiency of insulation in the House.
3. Nat., Propane Gas, Fuel Oil – forced air to which I have Propane for warmer days. I live in the Rural area. Fuel Oil is slightly different but close.
4. Electric Baseboard – As long it’s convection to which I have but do not use. It’s expensive to operate.
5. Heat Pump- forced air obviously used with electric backup if the Heat Pump can’t keep up. Very dry heat..Hard to stay warm. Must Humidify.
6. Electric Forced Air- this heat is the worst. If I’m in this for more than 5 minutes, I nose goes crazy because the RH is getting to 20% which is very dry and unhealthy for obvious reasons. If you have a very well insulated structure Humdity levels can be maintained. Wind blowing on a structure wicks the RH out if it’s not insulated properly.
I hope this helps Anna…stay warm, dry and Healthy! 🙂

Nick Stokes
November 21, 2009 12:45 am

Ian Schumacher (19:35:21) :
The Greenhouse Earth doesn’t absorb more energy than a black body – in these models it is in equilibrium, with no nett absorption at all. I think your issue is, how high can the surface temperature go. There is a limit. The GHE happens because part of the outgoing 235 W/m2 is emitted from the high troposphere, at a temperature of about 225K. If it weren’t for the atmospheric window emitting at higher intensity, this emitting layer would reach the snowball earth temp of about 255K, at which the outgoing IR balances incoming sun light. It can’t go higher.
The surface could still get very hot if there were sufficient insulation – say several layers of “steel” plus vacuum. But the ability of the real atmosphere to insulate is sharply limited by the dry adiabatic lapse rate . This is about 9.8 K/km, and is a stability limit. If it were ever exceeded, the atmosphere would become very turbulent, and conduct heat upward very readily.
So no matter how much GHG there is, the tropopause will be limited to about 255K, and the surface to an excess determined by that maximum lapse rate. That’s still hot.

Willis Eschenbach
November 21, 2009 12:57 am

par5 (20:58:51) :

Willis, I think that I have solved your conundrum for you. I did not realise that the planet Cantobtainia was completely sealed with a thick layer of cantobtainium (which prevents outgassing). At first, I thought that my questions were valid, on topic and had merit. I was wrong and I do apologise. Anyway, I glanced at my atomic chart shower curtain today and realised that cantobtainium has to be some sort of radioactive isotope or trans-uranium element. Giggle. So you see, it is your encasement of cantobtainium that is producing your 235W/m^2.
It is a win/win situation for both of us. You get to keep your planetary greenhouse without an atmosphere, and I have the pleasure of knowing that your planet isn’t going to last much longer due to its high rate of decay.
Can we now shake hands, thank each other for the polite conversation and walk home happy and content?

Done. Thanks.

P Wilson
November 21, 2009 6:26 am

carrot eater (18:00:48) :
Thanks carrot eater but I regularly use IR thermal imaging devices outdoors