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,
You are on the right track but it’s horrendously difficult to create a description that avoids misinterpretation by others.
Your analogy is not perfect (in my humble opinion) but it helps to illustrate some important considerations that AGW theory ignores altogether.
You are dealing well with the naysayers who seem to lack specificity in proportion to the vigour of their opinions.
I consider the role of the stratosphere to be critical. It represents an area of transition between two separate energy transfer regimes (shells, if you will).
The troposphere works mostly via convection and the speed of the hydrological cycle. The upper atmosphere works mainly via direct radiative transfer.
The stratosphere is the buffer between the two and in a few days I will be releasing a fuller description of the significance of that (for climate changes) at climaterealists.com
This is an interesting post. Much of what it says I agree with but there are to me some glaring errors. To start, I totally agree with the steel shell model both the reasoning and the numerical conclusions. Where I start having problems is with the Kiehl and Trenberth model and the analysis flowing from that.
You see there is a second relationship which must also be met and that is the temperature of the radiating surface. Objects do not radiate because they receive radiation, they radiate because of their temperature. So one also has to check that the temperature data is correct.
Consider just the CO2 GHG component. The total absorbance of the CO2 in the atmospheric column at 280 ppm is 2000 absorbance. (For those who are not familiar with the term absorbance it is the amount of an absorber that absorbs 90% of the incident energy at the absorbing wavelength). Thus its not one shell or two shells but closer to 2000 shells. As the steel sphere correctly shows energy is only radiated from the outermost shell since energy radiated from innner shells is reabsorbed by shells above it. This means that energy is only radiated from the top of the atmospheric column containing the GHG.
For earth the most prominent green house gas is water vapour and the top of the water vapour column is the tropopause. For CO2 it could be somewhat higher. 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. So, maybe the tropopause is is not the radiating layer after all Kiehl and Trenberth talk about energy radiated to space from within the troposphere and even talk about an equivalent radiation altitude? Well there is simple proof it has to be, you see the tropopause is colder both than the atmosphere below it and the atmosphere above it. How does the tropopause stay cold when it is completely surrounded by warmer atmosphere? There is only one way and that is that it is radiating energy to a heat sink and the only heat sink colder than it is outer space. So the tropopause has to be radiating directly to space. But could not the lower atmopshere also be radiating? Well no because the ability to radiate and the ability to absorb are both governed by the same factor – the emissivity. If the tropopasue is radiating strongly to space it will also be absorbing strongly any radiation received from below, so it would be blocking radiation emanating from lower in the troposphere. This means the tropopause has to be the effective top of the atmosphere from a GHG point of view.
Now to come back to the steel shell analogy – if there really are 2000 shells the energy radiated back to the surface would be astonomical or put another way, almost none of the energy from the surface would be escaping to space. But that would mean the surface of the earth would be more like venus and it clearly isn’t so there has to be a flaw. Indeed there is. CO2 is not like a steel shell because it only absorbs and radiates over a small range of wavelengths rather than at all wavelengths as a steel shell would. with 2000 odd shells, indeed very little energy escapes from earth at the wavelengthes at which CO2 absorbs. However energy can escape freely from wavelengths within the atmospheric window where there is little if any atmospheric absorption. Thus Kiehl and Trenberth underestimates the emission to space in the atmospheric window and overestimates the energy absorbed by GHG in the atmosphere. I did a numerical analysis of this in some depth in my first paper at jennifermarohasy.com/blog.
The suggestion of two shells, one being the troposphere and the other the stratosphere cannot be correct because the outermost shell has to be cooler than the next inner shell and in this case the tropopause is colder than the stratosphere.
Interesting article and debate.
I’m missing a step here though, the power source supplies 235watts/m2 for a surface area of ‘a’, if the shell has twice the area, 2a, shouldn’t it emit half the energy; 117.5 W/m2.
Per m2 I cant see how the shell emits twice the energy it receives from the power source, the shell would be a power multiplier, Eout = 2Ein, which is impossible – no ?
Willis: I thought I would pipe up, since I had such vehement objections to your prior article on the Shindell paper, just to say that I thought this was a good post.
-Marcus
Willis,
excellent article, but I there is one thing that doesn’t make sense to me. If you start with a ball radiating at 235 w/m2, then if you add the 235 w/m2 radiated back from the steel shell aren’t you creating energy out of nothing? Isn’t this an example of perpetuum mobile of the third kind? Sorry if I’m being a bit thick but if we start with 235 joules worth of energy at the surface, then the other 235 joules that were transmitted downwards came from the original 235 to begin with, so then we have created energy out of nothing.
HELP!
Quote:
John A (01:36:08) :
Let me demonstrate with a simple example:
Say you have a Dewar (vacuum flask) and you fill it with lukewarm soup. 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.
Its a basic mistake Willis, but you’re going to have to think about it first.
.
Will people please think before posting.
The soup analogy is wrong, because the soup does not have an internal energy source. The Earth system does (nuclear in this thought experiment, solar short-wave on the real Earth).
The soup analogy would only work if the soup had a small heating element inside it (nuclear powered?) giving of a constant 10w. Then it would heat up and up, depending on the number of vacuum layers.
.
You know what needs to be removed from this page?
Any comment that uses the term “re-radiate”.
This is a pet peeve of mine …. it demonstrates a complete lack of understanding of the concept of “radiation”.
There is quite a difference between the Stefan-Boltzmann equations (the fundamental equations governing radiation physics and temperature) and the climate models.
The surface radiates at 390 watts/metre^2 at a temperature of 15.0C.
For the surface to go up 3.0C to 18.0C, another 16.5 watts/metre^2 would have to be added to the surface.
I don’t know how the climate models can reconcile their prediction of a 4.2 watt/metre^2 increase in the tropopause radiation results in a 3.0C increase in surface temperatures (and a new surface radiation level of 406.5 watts/metre^2).
Their mistake is by converting everything into “linear equations”. Hansen took a shortcut in the early 1980s using his early climate model results (for the last glacial maximum for example) where his climate model came up with -6.6 watts change. Temperatures declined by 5.0C so therefore, the sensitivity is 0.75C per watt.
http://europe.theoildrum.com/uploads/465/cv_hansen_fig3.png
Between this shortcut/mistake (which violates the Stephan-Boltzmann equations and was copied by all the following climate scientists) and through the climate model’s assumption of a constant linear lapse rate of 6C/kilometre when it is probably not constant), they have changed all the logarithmic radiation equations into linear ones. And they are not.
http://img524.imageshack.us/img524/6840/sbearthsurfacetemp.png
http://img43.imageshack.us/img43/2608/sbtempcperwatt.png
Shortcuts and mistakes that were not corrected and here we are.
Willis Eschenbach (23:34:12)
yonason (21:37:01)
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.
It was a jocular reference, I am sure, to Al Gore’s recent claim, in the context of geothermal energy, that the temperature of the earth’s core is “several million degrees.” The WUWT thread on Gore’s TV appearance and this claim immediately precedes this thread.
Alan D. McIntire (21:13:02) :
The wavelength for CO2 absorption is ~14 microns, not ~14 nm. 14 nm is close to the soft x-ray region.
Re: Anton Eagle (00:15:58) :
Anton, you seem to be making the same mistakes that myself and, judging from the comments, many others made. There are a couple things that I missed when reading the article and the comments.
First this isn’t intended to be a climate model. It is in fact a model to show that our atmosphere does not act exactly like a greenhouse if I am understanding it correctly.
Second, let us look at the example of a ping-pong ball and a golf ball. We will assume that the golf ball has no dimples and is the same radius as the ping-pong ball, say 1cm radius. The surface area of the golf ball will be 4pi*r^2 or approximately 12.6 cm^2 as there is only an outside surface. The surface area of the ping-pong ball will be twice that, or about 25.2 cm^2 as there is an inside and outside surface.
Once I got my head around those items, the rest of the logic makes sense. Since the shell is radiating 235 W/m^2 from the side towards space (the net energy flow), it must also be radiating 235 W/m^2 from the other side towards the planet. Since the shell and planet must be at equilibrium, the energy radiated by the planet to the shell must raise to 470 W/m^2, where it will remain as the system is in “equilibrium” radiating a total of 235 W/m^2 towards space as it was before the steel greenhouse was added.
Also Re: Willis Eschenbach (23:34:12) :
I believe the comment by yonason was a very subtle dig at AlGore. If I understand that correctly he is referring to Mr. Gore’s comment that only a few kilometers below the crust the earth is a couple million degrees.
Charlie
Thanks Anthony for what is apparently a very rich article. I’m gonna bookmark this jewel.
I like your “Tinkertoy” model of a world with a steel shell ‘greenhouse’. It reminds me of the hours of fun I had back in the ’80’s playing around with a Daisy World simulator I’d programmed in Basic on an old Dragon PC. Happy days. I’m going to have some fun playing with your toy :-))
I think much can be learned from messing around with these sorts of ‘toys’, although as pointed out by many of the earlier comments, models of our climate system have little or no correlation to what goes on in the real world. Earth’s climate system is an immensely complex dynamic chaotic system, with much turbulence and boundary effects involved in the physics, chemistry and biology which go into make the total system. Even the best of the non-linear models being used by the IPPC do not provide results which can be used to predict the direction of our climate, although as with simpler models they are useful learning tools.
I think all models fall down because the many of the initial assumptions are wrong and there is poor quantitative understanding of how each non-linear process works. There is too little accurate data and the granularity of that date is insufficient by several orders of magnitude.
An example of a possible wrong initial assumption of real world models is the fact that Stafan’s radiation law is only valid only for in vacuum black body radiation, where all wavelengths are present. It is particularly important to understand this if the effects of GHG’s on climate are to be understood. Couple of links to this issue here:-
http://www.colin-baxter.com/tips_and_ideas/bb_radiation_gh/index.html
http://www.worldscinet.com/ijmpb/23/2303/S021797920904984X.html
Thanks Willis for posting such an enjoyable and thought provoking thread – this is what keeps me coming back to WUWT!
Please scratch the above . I did a mistake . This one is the right version :
=========================================
Willis I think that you made a mistake when writing :
“R = sigma * epsilon * T^4 where R = radiation (W/m2)”
.
The units in the Stefan Boltzmann formula are W/m²/sr .
When the solid angle is omitted , it implies that the integration has already been done (generally along a plane surface) .
This is a frequent nevertheless particular case .
Concentrical spheres are not part of this particular case .
What would it mean to your example ?
.
The internal sphere of radius r emits some radiation P
(units W/m²/sr) .
The total power (units W) emitted is P.4.Pi.r².2.Pi
The solid angle here is 2.Pi .
OK .
Now this power (units W) arrives at the big sphere of radius R and will be absorbed and emitted . As both are equal we have :
Emitted power (units W) = P.4.Pi.r².2.Pi .
This power (units W) will be divided in 2 halves , one emitted by each face e.g P.4.Pi.r².Pi per face .
However as this big sphere has 2 faces it only sees a half space from each face . So the solid angle for each face is 2.Pi .
And the power per sterradian (units W/sr) for each face is :
(P.4.Pi.r².Pi) / 2.Pi = P.4.Pi.r²/2
As for the specific power (units W/m²/sr) it is : P.4.Pi.r²/2.4.Pi.R² = P/2 . r²/R² . This is for each face . For the whole outer sphere it is P.r²/R² .
.
What follows ?
1) The outer sphere has not the same temperature as the inner sphere in equilibrium because P.r²/R² is not P .
.
2) Specific powers (units W/m²/sr) do not conserve because of those m² and sr that stand in the way . Total powers do .
.
3) If r~R then both spheres have approximately the same temperatures and the specific power (units W/m²/sr) of the outer sphere is half of the inner sphere for each face (P/2) .
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4) The absorbed part of the radiation returning to the inner sphere will decrease geometrically with reason r²/R² .
Thanks God for that because that will prevent the experiment of Michel going nuclear 🙂
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I do not chime in with some others that the post should be removed because it is an interesting exercice of radiative interaction but it must be corrected .
Clearly it can’t be left standing that the Stefan Boltzmann law has W/m² for units when the right units are W/m²/sr .
Neither can be left standing that specific power (units W/m²/sr) conserves .
>>Vincent (04:22:19) :
>> If you start with a ball radiating at 235 w/m2, then if you
>>add the 235 w/m2 radiated back from the steel shell aren’t
>>you creating energy out of nothing?
No, its just that the diagram is poorly drawn, in my humble opinion.
a. There is 235 going from the Earth to the shell.
b. Then there is 235 radiating out from the shell into space (eventually).
But then there is another 235 bouncing around between shell and Earth and back again. This is not ‘created energy’, it is just bouncing around until it can finally escape as ‘b.’. So the radiation out from Earth is not instantaneous, as in the unprotected Earth, there is a time delay as the energy bounces around for a while before escaping. Draw the diagram like that, and see if it makes more sense.
.
“If you start with a ball radiating at 235 w/m2, then if you add the 235 w/m2 radiated back from the steel shell aren’t you creating energy out of nothing?” Vincent
This bothered me a bit too till I thought of a heat source surrounded by a perfect insulator. Then the temperature of the heat source would rise indefinitely. The increase in temperature would (to my mind) simply be the accumulated energy from the heat source since it would be unable to dissipate any energy.
I hope this is correct and helpful; I am just a rusty engineer.
>>Myrddin Wyn (04:09:21) :
>>I’m missing a step here though, the power source
>>supplies 235watts/m2 for a surface area of ‘a’, if the
>>shell has twice the area, 2a, shouldn’t it emit half the
>>energy; 117.5 W/m2.
No, as per my post above.
The shell receives 470 units.
a. 235 from the Earth
b. 235 from the energy bouncing around in the Earth-shell void.
And the shell re-radiates 470 units.
c. 235 to space
d. 235 back to the Earth-shell void, where it bounces around for a while.
a + b = c + d, so the system is in equilibrium
And there is no ‘creation of energy’, as elements b. and d. are simply time-delayed manifestations of a.
There is only energy ‘a.’ (energy from the Earth). Energy a. (a 235-unit ‘package’) goes up to the shell, then down to Earth and back again a few times (bouncing around), and then finally escapes into space. But there is only one 235-unit energy ‘package’).
.
A couple of scientist wanted to know what the conman man thought was the greatest invention. So they went down to hills and pose this question to Bubba. He told them it was the thermos bottle. He said when I put something hot in it it stays hot and if I put something cold in it it stays cold. How do it know?
I agree withn John A. (20:11:22) this load of [snip] has no place on Wuwt.
Wuwt’s reputation has now been tarnished. This is not science but fantasy.
The thing is so sorely messed up that it doesn’t even qualify for print.
Has Wuwt fallen prey to a belief in pseudoscience? I am truly disappointed!
If I get started it will be 2 pages and space does not permit.
Warning: any serious consideration of this or any other AGW fantasies
will ruin your mind and lead to a belief in superstitions.
Back2Bat (07:10:21) :
“This bothered me a bit too till I thought of a heat source surrounded by a perfect insulator. Then the temperature of the heat source would rise indefinitely. The increase in temperature would (to my mind) simply be the accumulated energy from the heat source since it would be unable to dissipate any energy.
“I hope this is correct and helpful; I am just a rusty engineer.”
Exactly! Hence, in the context of nuclear power reactors, there is a need for post-shutdown core cooling to remove residual decay energy (think Three Mile Island and the consequence of a cooling system failure).
a simple diagnostic comparison that the SB constant is irrelevant to climatology:
100w/m2 is the optimum surface heat for floor based central heating. This creates 24C at floor level, cooling as hot air rises. 15C, or 59F is certainly cooler than 24C. It is improbable that the floor will re-absorb radiation it has already emitted.
I think John A and Brian have made the same mistake that I made. I interpreted the other 235 w/m2 of back radiation as energy added to the original energy – ie energy out of nothing. Both Ralph and Back2bat have answered by question – thanks to both you guys.
I can see that if you take the radiation going from the steel shell into space, it is no more than the original value W, not 2W. Therefore, no extra energy is being radiated into space, but the inside gets hotter.
Obviously, if there was no energy input to begin with, the ball would continue to cool to absolute zero, since there would be a net loss of energy.
These sorts of problems are certainly tricky, and have fooled even some physicists. G&T come to mind.
Admin :
Please remove my first post (06:42:40) . I made a typo which inverted some figures and made it wrong . Thanks .
.
For Willis : the right post is the second
For those with a short attention span that doesn’t allow to read longer posts : 99% of those who posted here are confusing specific power (unit W/m²/sr) and power (unit W) .
The former does NOT conserve and the latter does .
John A (01:36:08) :
Your example is simple, but so simple you forgot one simple little point, which I will explain.
Good so far.
Oops. You jumped to a conclusion that cannot be made based on his description – irrespective of the validity of his claim. Your system is not even close to the same as his system – his has a constant input power source, the sun, but I see none in your example.
Indeed, because there is no energy input to the system, the temperature will decay as a function of the reflected and radiated energy.
Perhaps you should take a dose of your own medicine.
Part of the problem I’m seeing here is that people don’t understand the basic mechanism of storage, which is a very simple feedback concept. While I understand that the concept of feedback is not part of the general curriculum of any typical education, people need to recognize this before making claims of psuedoscience or fantasy. To not do so is nothing more than arrogance.
Mark
Willis,
If the energy source is external, your spheres must have a hole in them to let light in. You now have a spherical cavity with a hole in it, i.e. an approximation to a black body. No matter what you do inside this sphere you can not make it hotter inside then a black body as you model allows.
In the case of the earth the hole is in wavelength, not a physical hole. The result is the same though.