Guest Post by Willis Eschenbach
Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.
To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.
Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation
This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.
Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.
Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)
[UPDATE: Misunderstandings revealed in the comments demonstrated that I lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%, I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]
Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.
And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.
Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.
Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.
Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.
Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.
Note on the Theory of the Greenhouse
By Professor R. W. Wood (Communicated by the Author)
THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.
I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.
To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.
There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.
Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.
I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.
Here would be my interpretation of his experimental setup:
Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.
Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.
Seems to me like with a few small changes it could indeed be a valid test, however.
Best regards,
w.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
My simple explanation of the Greenhouse effect.
Earth Shell Space
e1< e1< e1 e2> e2>
e3> e3/2>
<e3/2
e1 = energy recieved from the sun
e2 = energy directly radiated to space
e3 = energy indirectly radiated to space
Long term
e1 = e2 + e3/
Sorry. Should be
Long term
e1 = e2 + e3/2
Willis,
IR radiation has wave/partical duality properties as such it can be interferred with. Given your source in the thought experiment is the same the likelyhood is that the IR is coherent from both the steel plate and the core. This give an effective 0 W/M^2 between core and shell (much as there is no energy difference between core and slightly less core) as they are in equilibrium. No work is being done ergo no extra energy is needed for balance. If you place a thermometer between core and shell it will reach an equilibrium temperature faster than if no shell was there it will not be a higher temperature. As with all energy fields – it is not the amount that is important but the difference – in equilibrium there is no difference.
I really don’t get the quibbling by many here. Surely the point of R. Wood’s experiment is this . .
All things being the same, except that in one box infra red is retained by the glass and in the other it is released by the rock salt, it seems that the contribution to the temperature in the box made by reflected infra red is trivial at best.
The rest of the argy bargy is fine in that it would refine the experiment in various ways but it won’t change the observed fact by Wood that the greenhouse effect is not enhanced by the trapping of infra red radiation.
Ryan says: February 7, 2013 at 2:13 am
Ryan, Bad Clue! It will just get hotter and hotter. Or do you have some maths to show us why it won’t?
You have a constant heat source putting energy IN. You postulate a perfect reflecting surface that stops any energy getting out. The energy builds up ad infinitum. It will just get hotter and hotter.
Ryan says:
February 7, 2013 at 2:13 am
What part of “steel shell” leads you to think it is 100% reflective of IR? Steel has an IR reflectivity of about 5% or so …
Since nothing is 100% reflective, I fear I can’t answer you. It’s like the question about what happens when an irresistible force meets an immovable object. We can’t say because they are not possible even in theory, like a 100% reflective surface.
w.
Richard LH says:
February 7, 2013 at 2:17 am
That is so simple it is totally cryptic, I haven’t a clue what you’ve just said. What does e1 less than e1 less than e1 mean? (e1<e1<e1 in your explanation above)
w.
Keitho says:
February 7, 2013 at 2:37 am
Since the Wood experiment is totally unlike and has nothing to do with the greenhouse effect, it couldn’t reveal a dang thing about the greenhouse effect … read the post again. For the greenhouse effect to work, the shell has to/must/is required to be thermally isolated from the surface so it can take up a different temperature. In the Wood experiment the shell will always be at about the same temperature as the inside of the box, due to both convection and conduction.
So the two setups, the Wood experiment and the planetary greenhouse, have nothing to do with each other, and the experiment can’t show anything either way about the greenhouse.
w.
Thanks Willis, I was merely reflecting on the actual greenhouse that Wood was running and the fact that preventing the trapping of the IR made no measurable difference to the temperature in the boxes. The real cause of warming was the prevention of convection from cooling the box. The “reflected” IR is no big deal so surely the same thought could be applied to CO2, particularly in the presence of convection and conduction.
Your thought experiment at a global level is quite different and rather obvious assuming a vacuum between the planet and the steel shell where there is no convection or conduction, only radiation. Same difference really as the same point seems to be made.
Tim says:
February 7, 2013 at 2:32 am
My friend, if you can get coherent radiation from a chunk of steel, the laser scientists would like to talk to you … your whole claim is interesting, fanciful, and unfortunately, not possible. No coherent radiation comes from a planet or a steel shell, doesn’t happen.
w.
Willis Eschenbach says: February 6, 2013 at 2:33 pm
If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle.
____________________________________
Ahh, Willis, I did warn you that all the trolls would come out of the dark, claiming that a cool body cannot radiate to a warm body. (Cooler bodies have eyes, that can see a warmer body coming, and so they shut off their radiation immediately….)
But I must say that your analogy of a candle warming the Sun is much more elegant than my two radiator scenario. I like that idea that I can hold a candle up to the Sun and make it warmer. Anyone bold enough to say by how much?
.
“To compare with Willis’s example, you need to add a constant heat source to the inside of the thermos. Now what happens?”
Ah yes, but there is yet another flaw in Willis’ over simple thought experiment. In his experiment you have an infinite energy source of unknown temperature. This is not the case in the Earth/Sun system. In the Earth/Sun system we know how hot the Sun is, and we also know how hot it could theoretically make the surface of a body heated by the IR emitted by the Sun (which is rather smeared out by the time it reaches Earth).
Willis’ thought experiment would indeed by more honest if you put a body of known temperature in the center of the sphere and asked if it will get any hotter if you but a metal shell around the whole thing. Only it can’t be a metal shell in willis’ experiment because an ideal metal shell won’t absorb any heat energy and get warmer – it needs to be a “black body” shell to absorb the heat. But it can’t be a black body as such because a black body only radiates otuward and not out and in at the same time. And you can’t measure the temperature of the atmosphere is Willis’ experiment because there is no atmosphere so no sensible heat in the atmosphere. And if there was an atmosphere his experiment wouldn’t work at all.
Time to move on from this isn’t it?
I do wish Willis would wake up and realise that IR is not “heat” nor does it have a “temperature”. IR is a form of electro-magnetic energy that can be CONVERTED to heat by an ideal black body. If you talk about IR then you need to discuss the laws of thermodynamics at the quantum mechanical level. If you don’t want to discuss quantum mechanics then you must ignore IR and stick to discussing simple heat energy transfers using the laws of thermo-dynamics alone. Willis gets himself in knots by mixing IR into simple models of heat energy transfer that were developed before IR was known or understood and which don’t need a knowledge of IR to work.
Ryan says:
February 7, 2013 at 2:03 am (Edit)
The emissivity of various materials is given here. Reflectivity is equal to [ 1 – emissivity ]. The reflectivity of steel is about 10%-15%. So what you see in the steel shell case is indeed back radiation. I should have clarified that the shell and planet were blackbodies … I’ve added a note to the head post, thanks.
w.
This was your thought experiment – so if we take multiple IR wavelengths you still get to the problem that in equilibrium there is no energy difference between core and plate (or it is not equilibrium) . The “vaccuum” merely acts as a conduction/convection filter from core to plate – energy is a flow not a substance (no phlogiston in them thar hills) as such when the plate has expanded (well it is steel) and in equilibrium there is no more work being done on the plate – the plate becomes the core for IR transmission purposes. The flow is one way, net IR transmission at the inner surface is still zero.
davidmhoffer says:
February 6, 2013 at 8:12 pm
Or you can subscribe to the theory that these laws of physics are not in the thermo texts, haven’t been proven by experimentation, and that engineers design nuclear reactors, kitchen ovens, air conditioners and automobile cooling systems that functions as designed in advance of being built by a complete fluke.
I call that BULL SHIT.
Which of those systems rely on a Vacuum.
All of them rely on being IN CONTACT with the Heat Source or the Air being heated by the Heat Source or Air to radiate the heat away.
consider the implications of your thought experiment. with a finite amount of input energy you can heat an object to infinite temperature. you should patent it.
— RADIATION DEPEND ONLY ON TEMPERATURE AND EMISSIVITY
Period. No exceptions. —
Actually there are some rather big exceptions. That’s because you forgot to mention when one can apply the SB-law.
“Sei ein absolut leerer Raum rings von für Wärmestrahlung undurchlässigen Wänden von
der absoluten Temperatur t umgeben;” – Bolzmann, 1879
You can only apply the law to a real blackbody in vacuum that doesn’t conduct heat to the other side. Steel is not a blackbody and you assume that it can conduct heat to the outside. So, you are not talking about the SB-law, but some madeup imaginary law.
By the way, your radiator also has to be fixed in space and it must have a defined surface. That’s why you cannot apply it to any gases. At least that’s what we learned in University here in Germany.
Whenever you use a ‘greybody’ you should first show that you are allowed to use some simplifications, and that this simplifications won’t change the essence of the law.
Sorry – bad ascii text formating.
My simple explanation of the ‘Greenhouse effect’.
Earth Atmosphere Space e1a> e1b> e1b> e2> e3> e3/m> e3/m> <e3/n <e3/ne1 = energy recieved from the sun
e1a = energy reflected by atmosphere
e1b = energy reflected by surface
e1c = energy absorbed by surface
e2 = energy directly radiated from surface to space
e3 = energy indirectly radiated from surface to space
e3 = e3/n + e3/m
n and m dermine the ratio of reflected energy to passthough.
Long term
e1c = e1 – e1a – e1b
e1c + e3/n = e2 + e3
e1 = e1a + e1b + e2 + e3/m
e3/n is the ‘Greenhouse effect’ as seen at the surface.
let’ s try another hypothesis
if I put a second shell around the first, and I make the same calculation like the mechanism described, this time, the core t° is 4 times higher?
if I repeat this process n times, can we say that the t ° of the core can become infinite?
weird is it not?
Might I suggest that the energy radiated by the planet is absorbed by the interior blackbody shell and an equal amount of radiation is radiated outwards by the shell. There is no temperature increase at the surface of the planet.
Wow, Willis, I have enjoyed so many of the things you have written here over the years, but this one, wow. Still clinging on to the “cold objects can make warmer objects warmer still” garbage? You “thought” experiment here is nothing more than a hodgepodge of card shuffling. I recently played this same game with my father over Dr. Spencer’s cold object/hot object claim. My father lost, and lost of one very simple reason, 2nd law. As indicated by others here, no matter how you slice it, twist it, dice it, a colder object cannot warm a warmer object, not in gas, not in steel, not in any way. The GHE either doesn’t exist, or is of such utterly small consequence that we will never be able to actually measure it. As for the IPCC form of the GHE theory, it is impossible.
Please, go back to writing your “good” articles. I really enjoy reading a lot of your stuff, but this article just blew away about a years worth of reading from you. argh….
A C Osborn says:
February 7, 2013 at 3:48 am
davidmhoffer says:
February 6, 2013 at 8:12 pm
Or you can subscribe to the theory that these laws of physics are not in the thermo texts, haven’t been proven by experimentation, and that engineers design nuclear reactors, kitchen ovens, air conditioners and automobile cooling systems that functions as designed in advance of being built by a complete fluke.
I call that BULL SHIT.
Which of those systems rely on a Vacuum.
All of them rely on being IN CONTACT with the Heat Source or the Air being heated by the Heat Source or Air to radiate the heat away.
>>>>>>>>>>>>>>>>>>>>>>.
Huh? Are you of the impression that systems being in contact suspend all other modes of energy transfer? It isn’t A or B. It is A and B. If the systems are in contact then they xfer energy via conduction and radiance. Makes the math even more complicated. But if you think the engineers can get it right by simply ignoring the laws of thermodynamics because the systems aren’t in a vacuum….I’ll add you to the list of people who think nuclear reactors work by shear fluke.
squid2112;
As indicated by others here, no matter how you slice it, twist it, dice it, a colder object cannot warm a warmer object, not in gas, not in steel, not in any way.
>>>>>>>>>>>>>>
When I was a kid we didn’t have enough money for heating fuel. We piled snow up against the side of the house to the point we could pretty much keep it warm with just the body heat of the people inside. Good to know that the snow must have been warmer than the house, because while shoveling it at -20C I could have sworn it was colder.
davidmhoffer says:
February 7, 2013 at 5:34 am
“When I was a kid we didn’t have enough money for heating fuel. We piled snow up against the side of the house to the point we could pretty much keep it warm with just the body heat of the people inside. Good to know that the snow must have been warmer than the house, because while shoveling it at -20C I could have sworn it was colder.”
What the snow did was increase the insulation factor for the house, it’s the same way igloo’s can be a lot warmer inside than out, while made of snow. Also note, snow has a lot of trapped air, which is a very good insulator.
“Mark Harvey aka imarcus says:
February 7, 2013 at 1:51 am ”
In my view your back radiation at 235 w.m-2 would only work if the steel shell is hotter than the core, which I don’t believe it is otherwise your model would be negated by the 2nd law of Thermodynamics. (Heat only flows from hot body to cold body, never in the other direction.) ”
When people read “heat flow can only flow from a hot body to a cold body” they nearly always make the same mistake. They take this statement from the point of view of an omnidirectional flow of heat. They mistakenly believe that it means that there can never be any transfer of the energy from the cooler body to the warmer body. This is an incorrect understanding.
The 2nd law always refers to coupled systems. It says that if you have 2 bodies of different temperatures there cannot be a spontaneous transfer of heat from the cooler to the warmer. If heat did in fact transfer in such a way, the warmer body would grower warmer still, at the expense of the cooler body. But it doesn’t say that energy cannot flow from cooler to warmer. Energy does in fact flow from the cooler to the warmer body, but at the SAME TIME, even MORE energy flows from the warm to the cooler body.
Before you say this can’t happen – it would lead to pertual motion, etc – just consider this. It is impossible to devise any hypothetical situation where you can allow energy to radiate from the cooler to the warmer body WHILST simultaneously preventing the energy flowing the other way. Energy will ALWAYS flow to the cooler body at a greater flux density than the other way round. The 2nd law is not violated.
Konrad. says:
February 7, 2013 at 1:29 am
“The shell game is a foolish game to play……………….”
Konrad’s explanation wins the debate IMO.
Beautifully summarized Konrad.
Perfect.
Where’s Al Gore? I want to tell him the science is settled. ;^)
Don’t think I’ve ever seen a thread where Mosher is right, and many other posters are wrong….