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.
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“The Wood experiment deals exactly with the underlying mechanism of the IPCC “greenhouse effect” (effect of trapped/back radiation) and demonstrates that it is bupkis.”
The Wood experiment is similar in some ways, but very different in others, which is the point hat Willis was making and which you apparently missed.
> The earth is surrounded by extreme cold (outer space), but Wood’s experiment is surrounded by materials at very similar temperatures to the experiment itself.
> The atmosphere of the earth is many KILOmeters in size, but Wood’s experiment is many CENTImeters in size.
These (and other rather fundamental differences) make Wood’s experiment a rather poor analogy to earth’s situation.
Let me quote Wood himself on the fact that is experiments are applicable to real greenhouses, but may not be applicable to otehr situations like the earth:
wayne says (February 7, 2013 at 9:35 am): “The shell as shown is outputting more energy, 470‹J/s/m2›, per time per area than is even available from the nuclear source in the first place,”
Assume the surface area of the “planet” and the outer surface area of the “shell” are both 1 square meter (or if you must, make the outer surface of the shell 1.003 square meter). Now do the calculation in watts and get back to us.
Hint: The inner surface area of the shell equals the outer surface area.
Experiment to test “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.”
We have two bodies on either side of a one-way mirror which is mostly transparent to light and IR in one direction and mostly opaque or reflective in the other. On the side of the mirror which is reflective, we have a black body # 1, and on the on the opposite side of the one way mirror, we have similar black #2. The black bodies are equidistant from the one-way mirror, and we will, for this experiment, stipulate the black body on the reflective side of the mirror begins at 30 degrees C, and the similar black body on the transparent side of the mirror begins at 20 degrees C. Since we are interested in the radiative energy flows, we have created an experiment so that conductive and convective flows are minimized. What happens? To exaggerate the above, we will place similar parobolic dishes behind each of the two black bodies with the black bodies at the focal points. What happens? Does the cooler black body “heat” the warmer black body. Is energy flowing from the cooler black body at a greater flux density than the other way round?
Wayne says: “Ever heard of the 1st law of thermodynamics? Don’t you see your problem?”
No, I don’t see a problem:
* the inner sphere gains 235 W/m^2 from the internal heater.
* the inner sphere gains 235 W/m^2 from the shell
* the inner sphere radiates 470 W/m^2 to the outer shell
** Net transfer = 0 W/m^2 → No temperature change
*** Obeys 1st Law
* the outer shell gains 470 W/m^2 from the inner sphere.
* the outer shell radiates 235 W/m^2 to the inner sphere
* the inner sphere radiates 235 W/m^2 to space
** Net transfer = 0 W/m^2 → No temperature change
*** Obeys 1st Law
I am not creating or destroying any joules of energy.
“It is like taking a sheet of polished aluminum foil and holding it up to a wall radiating by SB per T at 400‹W/m²› and 400‹W/m²› is reflected back to the wall by the foil at the temperature of the foil so now you have 800‹W/m²› at the wall so of course the wall’s T must rise”
To the extent that “polished aluminum” is shorthand for “a material that reflects back all thermal radiation that reaches it”, then this is exactly what will happen! This is why “space blankets” are so effective.
Basically you are saying that you keep pumping energy into a system (400 J into each m^2 of surface area each second) but don’t let any energy out (the full 400 W/m^2 is reflected back). So of course it will keep warming up!
(Of course, if there is convection or conduction that will limit things . And some radiation will be absorbed by the Al and then emitted to the outside, so it is not reflecting 100% and so that also will limit the temperature inside.)
“This is getting embarrassing for W.U.W.T.”
hmmmmm ….
Doug Allen says “We have two bodies on either side of a one-way mirror which is mostly transparent to light and IR in one direction and mostly opaque or reflective in the other.”
Can’t happen. This would be an example of “Maxwell’s Demon”. As such, the rest of your argument is wrong before you even start.
(“One way mirrors” are not “one way” at all. They are simply mostly reflective from both sides. But one of the rooms is dark and one is bright, which makes it very difficult to see from the bright room to the dark room. If you make the formerly bright room dark and the formerly dark room bright, then the “one way mirror” will work in the opposite direction. Watch any cop show on TV and you will notice that the “interrogation room” is always brightly lit, while the “observation room” is always dimly lit. http://en.wikipedia.org/wiki/One-way_mirror)
Respectfully, it seems to me that the anti-GHers have a fundamental issue that they need to address before anyone else will take them seriously.
1. Per the Stefan-Boltzman law the *surface of the Earth* emits ~400W/m2
2. The Earth as a whole can only emit ~235W/m2 to *space*
As such, something must happen in the atmosphere (ie that which is between the surface of the Earth and space) to make up the difference. Given the complexity of the Earth-atmosphere system, it is easy to construct hypotheticals that *appear* to falsify the GH effect but much more difficult to do so and take into account 1&2.
Cheers, 🙂
Shawnet says
“1. Per the Stefan-Boltzman law the *surface of the Earth* emits ~400W/m2
2. The Earth as a whole can only emit ~235W/m2 to *space*
As such, something must happen in the atmosphere (ie that which is between the surface of the Earth and space) to make up the difference.”
You could also add that the Earth absorbs only 160W/m2 to make the apparent paradox more baffling!
However there are other examples in physics of a similar kind.
A child on a swing needs a very small input of energy to maintain much larger KE to PE interchange.
Energy can be stored in the system.
For our climate, the IPCC advocates say its all due to the GHE.
More rational people think that Latent Heat (both evaporation and fusion) and photochemistry are just two of the very many ways to store and release energy in the Earth system.
The bogus GHE theory (or theories because there are many variants) does not withstand rational scrutiny.
Ryan says: “You cannot, however, significantly warm the shed by covering it with tin foil (although many a claim has been made that you can….)”
ummm … yes you can.
http://littleshop.physics.colostate.edu/activities/atmos1/ColorAndCooling.pdf
Ryan says:
February 7, 2013 at 9:57 am
Ryan your comment about the shed made me think back to my SERE(survival evasion resistance escape) training in the Navy. One of the things they pounded into us was never use a downed aircraft body for shelter. Not in cold nor in hot areas. In cold the metal shell will suck the heat out of you and you freeze and in hot the metal shell will heat to oven like temps and bake you. small hyperbole.
I am sure if they had known that cold would warm us up we would not have been told that. sarc/
Shawnhet says:
February 7, 2013 at 10:35 am
Can you point to where this is defined and hopefully actually measured?
I’ll also note that if you look at nightly cooling since the 50’s (all 120 million NCDC records worth), there’s been no loss of cooling at night in the surface temperature record.
@davidmhoffer says:
February 7, 2013 at 5:34 am
So you are suggesting then that the IR from the snow kept you warm?
In Willis’ example, he mentions how you feel colder in the winter than in the summer within a house that is the same ambient temperature, incoherently attributing the extra chill you feel is from the colder walls. He may be correct about the walls, but in reverse. I would posit that you would feel cooler because the walls are absorbing more of the energy than you are, because they are cold, and you are NOT able to absorb energy from the cooler walls, because, well, errrrr, they are COOLER. Conversely, you may feel warmer in the summer because YOU are cooler than the walls and are absorbing more of the energy than the walls are, especially if the walls are warmer than the ambient temperature, in which case the walls are absorbing exactly ZERO energy from the ambient room (or YOU).
BTW, you stayed warmer in your house because you used the snow as INSULATION. An igloo comes to mind. When I was a scout, each year we would head up to the Blue Mountain in Washington for a week long camping trip. We did not stay in tents, we built igloos, because the snow provided such a good INSULATOR, not a back-radiation radiator…. sheeesh.. The only reason why it stayed warmer was because WE provided the heat. Without the heat from our bodies, the inside of the igloo was no warmer than the outside air, in fact, EXACTLY the same temperature!
Your house/snow/insulator example is a perfect test for Willis, and could be done this winter with the igloo example I just gave. This should be an easy, no brainer exercise for anyone to do. Make an igloo, does it get warmer inside when left to its own device? … I shall await your results.
Bryan:”More rational people think that Latent Heat (both evaporation and fusion) and photochemistry are just two of the very many ways to store and release energy in the Earth system.
The bogus GHE theory (or theories because there are many variants) does not withstand rational scrutiny.”
The problem with using latent heat as a proxy is that it doesn’t explain the difference. When latent heat is captured at the surface and released in the atmosphere, it acts to cool the surface (or bring the surface closer to the 235 w/m2 number). It doesn’t explain why the surface of the earth can remain warmer than the Earth as a whole). If there were no latent heat transfer at all (and nothing else changed) the surface of the Earth would emit 480W/m2 instead of 400W (IOW the GHE would be more intense *not* less).
Cheers, 🙂
MiCro:Can you point to where this is defined and hopefully actually measured?”
My number 1 is pulled directly from the Stefan-Boltzman law and the fact that the average temperature of the EArth is ~16C. Number 2 can be calculated from the TSI after accounting for the geometry of the Earth and its albedo.
HEre is a link to some good data for TSI IMO.
http://lasp.colorado.edu/sorce/data/data_product_summary.htm
Cheers, 🙂
tjfolkerts says:
February 7, 2013 at 10:27 am
Doug Allen says “We have two bodies on either side of a one-way mirror which is mostly transparent to light and IR in one direction and mostly opaque or reflective in the other.”
Can’t happen. This would be an example of “Maxwell’s Demon”. As such, the rest of your argument is wrong before you even start.
Certainly can, they’re called dichroic mirrors and are readily available, the type described is known as a ‘hot mirror’ whereby visible and IR below the cut-off wavelength is transmitted and IR above the cut-off wavelength is reflected. The Earth’s atmosphere acts as a dichroic although the mechanism is not reflection.
george e. smith says (February 7, 2013 at 9:32 am): “Black body radiation itself is fiction, in fact it is fictionally fictional. It involves the physical properties of no physical material whatsoever. No electronic energy states, or any other trappings of quantum mechanics of real materials is involved in the derivation of the black body radiation formulae.”
And don’t even get him started on the Ideal Gas Law! 🙂
Wayne says:
“The shell as shown is outputting more energy, 470‹J/s/m2›, per time per area than is even available from the nuclear source in the first place, there is only one energy source in Fig. 2 and it’s maximum output through the sphere’s surface is 235‹J/s/m2›. ”
NO! This is not what is happening. Go and look at Fig 2 again. The shell is radiating 235 w/m2 into space. In other words, the amount of energy leaving the system after the shell is the same as before the shell. All that happens is that the ball inside the shell will get a little warmer (the fourth root of 2 times warrmer to be exact).
It is like putting a coat on when you are cold. Your skin starts to warm up, but you have not created any energy from nothing.
Thanks for the interesting thought experiment, Willis. It kept me up mulling it over. For me, the “residue of nuclear material reacting in the core” is an unnecessary complication. If we get rid of it, I think I can see more clearly what back radiation from a blackbody shell does to the temperature of P in a dynamic rather than static case. The Shell doesn’t raise the temperature of an unpowered P, it slows its cooling. So if P is powered, then P will reach higher temperatures with a shell than without a shell.
Lets take a planet P, of uniform blackbody steel with finite specific heat and high thermal conductivity, in a void. Via some method, nuclear or otherwise, P is raised to an arbitrarily high temperature above Tp= 255 deg K. Then the heat source is turned OFF.
The planet P in a vacuum will radiate according to SB into infinite background at 4 deg K. P will start to lose energy and cool off. The surface of P will be cooler than the interior, but we have specified that it has high thermal conductivity, so P will be nearly isothermal.
When the surface of P cools past Tp=253.75 deg K, the surface is radiating at Fp = 235 W/m^2 in Willis’ example. SUDDENLY (at T=0) two very thin blackbody steel foil hemispheres, who’s radii is 1.1x the radius of planet P fly in from the 4 deg K void, and Pac-Man-like snap shut around Planet P as shell S. At this moment, the shell is at Ts = 4 deg K, but is being warmed up by radiation from P at Tp = 253.75 deg K.
For simplicity, let’s assume that P and S are made of the same stuff and that the mass of S is 10^-6 of P and specific heat is constant from T = 4 to above 255 deg K. Because the S has much less mass than P, we could raise Ts 250 deg K by thermal conduction and Tp would drop less than 0.001 deg K. Finally assume the area of P, Ap=10^8 m^2, giving radius Rp = 2821 m. So radius of shell Rs= 1.1*Rp = 3103 m and Area of Shell As = 1.21 *10^8 m^2. When Tp = 253.75 deg K, Fp = 235 W/m^2 and Planet P is radiating 23.5 GW = Fp*Ap
There will be a Time A>0, where the temp of the Shell reaches 50 deg K. and radiates 0.35 W/m^2 from both the inside and the outside surface. Tp will drop by (50-4)/1000000 deg K. The shell hardly matters at this point.
At a later Time B>A, the temp of the shell reaches 100 deg K, radiating 5.67 W/m2 from both the inside and outside. The heat needed to raise the shell from 4 to 100 deg K only drops P by 96/1000000 deg K, so P no warmer than 253.749908 deg K, but the shell is radiating 0.6 GW into space while absorbing 23.5 GW from P. It is warming quickly.
At Time C>B, the temp of the shell reaches 150 deg K and radiates 28.7 W/m2 from both the inside and outside = 3.47 GW from each surface. The shell is receiving 23.5 GW from the planet and losing 6.94 GW from its own radiation, so it gaining energy and still warming up. To heat the shell 146 deg, planet P has had to give up at least heat equal to a drop in temperature of (150-4)/1000000 = 0.000146 deg. So P is still radiating 235 W/m2. But now we have a not insignificant 28.7 w/m2 back radiation from the shell.
Do we add that 28.7 W/m2 to the 235 W/m2 from the planet?
No. If we do, we have 263.7 W/m2 and if treated as from a black body, we get a temperature Tp = 261.15 deg K. We would think the planet warmed without a source of energy.
Do we subtract 28.7 W/m2 because the NET flow from the planet is 206.3 W/m2?
No. If we do, then 206.3 W/m2 from a blackbody is Tp = 249.6 deg K when it really still above Tp=253.749 deg K.
The back radiation flux cannot be used to change the temperature of P directly, but it does change the net energy flow, so it reduces the rate of P cooling. The planet is still at Tp = 253.749+ deg K and radiating 235 W/m2 at the same time it is absorbing 28.7 w/m2 from the shell.
To recap, we have a blackbody planet heated above 255 K, let to cool down to a point T=0 where it emits 235 W/m2 at 253.75 deg K. We wrap it in a blackbody foil shell (one millionth the mass of the planet) at 4 deg K. Via radiation, the planet transfers heat to warm the shell, by huge difference in mass, the planet cools 1/10000 deg for the shell to warm 100 deg. As the shell warms past 150 deg K at T=C, it is now back-radiating to the planet 28.7 W/m2. The Planet must still radiate at Tp = 253.74985 deg, Fp = 235 W/m2, losing energy into the shell at 23.5 GW. But the Back radiation from the shell of 28.7 W/m2 amounts to 2.8 GW of energy returned to the planet for a net planet energy loss of 20.7 GW instead of 23.5 GW when there was no shell.
Wrapping P in shell S, does not cause the temperature of P to rise, only allow P to cool more slowly.
In this setup, there is no equilibrium. Ts will rise until the net energy it receives from P (less back radiation) equals what it loses to the void. That would be 11.75 GW, over As, so Fs = 97.1 W/m2, which would be 203.4 deg K. At which point, both P and S begin a coupled cooling though they are at different temperatures.
Turn on P’s power source (a 23.5 GW, high temp) at the core, and P will indeed find an equilibrium Tp above its 253.75 deg K equilibrium when exposed to the 4 deg K void instead of a Shell at greater than 200 deg K.
Tosh! Do you realise that is a completely meaningless statement? Equivalent temperatures are based on power emitted, not on any particular chemical molecule.
There are two reactions you may get when you point someone in the right direction. Either they go in that direction or they stubbornly refuse because what they were seeking was confirmation of their belief in the wrong direction.
In which case, if you don’t want to learn, I can do no more.
As said on WUWT earlier this week,
(Mark Twain)
There are over 200 comments on this thread. At the time of my last comment only about 5 contributors seemed to understand what they were talking about (although a couple more have weighed in since). The rest are totally confused.
But the steel shell model is a simple example. If you don’t understand that then try to get up to speed. You make sceptics look bad.
joletaxi says:
February 7, 2013 at 4:40 am
It’s crazy but true. I discussed this above. Your numbers are a bit off, in that 1 shell gives double the watts, and two shells give triple the watts. N shells will multiply the watts by N+1.
However, the temperature of core does not go “infinite”. It goes up by the fourth root of N+1. So merely to double the temperature, you need 15 shells …
Long before that, of course, it would melt. Imagine putting a light bulb inside a vacuum flask (a Dewar or a Thermos). Then put that inside another vacuum flask, and then put the whole thing in another …
You will melt the glass or the light bulb.
As someone pointed out above, high end vacuum flasks use this exact principle by putting multiple layers of aluminium foil inside the flask … note that they have to be in vacuum for this to work efficiently. Even inefficiently it work well, that’s how they make the so-called “Space Blankets”.
w.
mkelly says:
“In cold the metal shell will suck the heat out of you ….
Yes, metal is a very good conductor, so a metal frame at 0C will cool you much faster than a rock or a tree at 0C.
“ and in hot the metal shell will heat to oven like temps and bake you”
Yes, metal is a very good conductor, so a metal frame at 45C will heat you much faster than a rock or a tree at 45C.
And then a second thought occurred to me. A quick search shows that metals tend to be more reflective for IR than for visible. http://www.tvu.com/PNextGenTFwebFig3.jpg
This means metal emits IR poorly but absorbs visible light (relatively) well. Or put the other way around, metal has a hard time absorbing sunlight, but an even harder time radiating IR. This make metal (in some sense) a “natural greenhouse material” (sunlight gets in, but IR can’t get out so well). Sun shining on metal makes it very warm (as everyone knows). Compare this to a light colored rock. The rock will absorb a little more sunlight than the metal, but will emit IR much better, so it will not warm up as much.
This could make an interesting little experiment …
q = ε σ (Th4 – Tc4) Ac
A typical radiative heat transfer equation from Engineering Tool Box. Where in this formula can the radiation from the shell cause an increase in the temperture of the globe? Again it seems to me that when Th = Tc then q is zero.
I may have to throw out my heat transfer book it may be all wrong.
Keitho says:
February 7, 2013 at 2:37 am
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.
Wood’s experiment is fatally flawed as a description of the atmospheric GHE for several reasons. The fundamental one is that there is an atmosphere above the glass/salt plates, to be accurate it should be a vacuum and therefore only permit radiational heat transfer outwards. This could be achieved by putting the boxes inside a vacuum chamber with the top window being made from IR transparent material (preferably cooled to liq N2 temps). The other flaw which has been mentioned is that the size of the boxes is less than the convective length scale of the atmosphere contained in them, in order to get that right you’d either need a much bigger box or change the atmosphere inside to greatly change the length scale (Peclet No?)
I hear you Phil, and thank you for your reasoned response. I do wonder though how much more warming would be caused by the IR if the problems you outline were resolved. From my understanding it wouldn’t make a huge difference to the outcome, although it would greatly improve the experiment.
wayne:
In your post at February 7, 2013 at 9:35 am you say
Well, if by that you mean your post, then I agree.
For example, you ask about the hypothetical shell model
But nobody is suggesting that.
The shell inhibits the rate of heat loss from the planet by radiating some energy back towards the planet and, thus, reducing the net radiation from the planet. Thus, to obtain the same rate of heat loss in both cases the planet’s surface temperature is higher than it would be in the absence of the shell. But the shell does not ‘heat’ the planet.
Similarly, if you wrap yourself in a blanket on a very cold night then your surface temperature is higher than it would be if you were naked. The blanket inhibits the rate of heat loss from your skin by reducing some thermal conduction to the air. But the blanket does not ‘heat’ your skin.
The point you seem to be missing is that heat radiated to space from the hypothetical planet’s shell always equals the heat supplied to the surface from the hypothetical planet’s radioactive core. Similarly, when considering the Earth’s GHE, the heat radiated to space from the Earth’s surface and atmosphere always equals the heat supplied to the Earth from the Sun.
The effect of GHGs in the atmosphere is to inhibit the rate of heat loss from the Earth’s surface by radiating some energy back towards the surface. (This is like the shell of the hypothetical planet.) Calculating this GHE effect is not a simple application of the SB Law for the real Earth. Temperatures and concentrations of GHEs vary with time, geographical position and altitude over the Earth’s surface. It is difficult to conceive of an integration of SB which would enable direct calculation of the inhibition to radiative heat loss from the Earth’s surface.
Estimation of ‘back radiation’ is an attempt to quantify the magnitude of the inhibition to radiative heat loss from the Earth’s surface in the absence of the needed integration of the SB Law.
In my opinion, the Trenberth energy budget is very wrong. But it is not flawed in principle.
I hope that helps to overcome your embarrassment.
Richard
Greg House says:
February 6, 2013 at 2:58 pm
Guest Post by Willis Eschenbach: ” 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). … 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, …”
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“Imagine”, I see.
I guess, it has never been proven experimentally that A warms B and then B warms A back, right? OK, this is a product of imagination, a fiction, and everyone has right to right a science-fictional story, no problem with that. But the readers need to be told clearly that this story is fictional, just to avoid confusion.
Actually the readers need to be told repeatedly that you’re wrong and that you don’t understand undergraduate level radiation heat transfer! Try reading H C Hottel’s text for example. Read about the application of radiation shields for thermocouples as a practical example of the phenomenon, add a radiation shield around the ThC which is colder than the ThC and the measured temperature goes up!
Willis says: “The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2.”
tjflokerts says:
No, I don’t see a problem:
* the inner sphere gains 235 W/m^2 from the internal heater.
* the inner sphere gains 235 W/m^2 from the shell
* the inner sphere radiates 470 W/m^2 to the outer shell
** Net transfer = 0 W/m^2 → No temperature change
Willis used the word “warms” which means the temperature increased. So there was a temperature change. Does this change your thinking?
DR says:
February 7, 2013 at 8:00 am
We’re also told OHC increases in the past 30 years is due to AGW. There is zero empirical evidence that even a doubling of atmospheric CO2 can warm the oceans to any measurable degree. Instead, we’re told there is mixing by ocean waves that magically transports the top few microns of “heated” water affected by “back radiation” from a few extra ppm’s of atmospheric CO2. Sure, everyone stirs their hot coffee to make it warmer. How can anyone try to pass off the idea that anything other than SW solar radiation can warm the oceans and expect thinking people to believe that is frankly, insulting.
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I have for years been pointing out to Willis, the problems with DWLWIR and the oceans. Water is essentially opaque to DWLWIR. The absorption characterics of LWIR in water is that approximately 50% is fully absorbed within just 4 microns and virtually everything within 10 microns.
The energy flux is upwards not downwards in the top micron layers of the ocean so absorbed energy in those micron layers cannot be conducted downwards (ie., against the direction of flux) and ocean overturning (even if it could wrap and overturn the very top microns) is a mechanical process measured in many many hours whereas the absorption of DWLWIR is close to a light speed event.
I have often asked Willis to explain what physical process carries the absorbed DWLWIR down to the deep ocean in the light of the above facts, and, despite many requests for an explanation of the physical process, he has never provided one; instead he opts for the cop out, the oceans would freeze if they did not absorb DWLWIR.without actually proving that the tropical ocean would freeze (and of course, the tropical ocean acts as a heat pump distributing its heat to the other oceans).
But the problems with oceans is even more difficult. Oceans are covered with a fine mist of windswept spray and spume. This is significant since it would act as a LWIR block preventing DWLWIR even reaching the top micron layer of the ocean.
The average global wind speed over the oceans is about 7m per second, ie., about BF force 4. The human eyesight does not have good resolution and hence why TV works acceptably well (pre the digital era, TV in the UK had 405 lines per inch later increased to 625 lines per inch when colour TV was introduced, and for many years in the States, they had even lower resolution – nonetheless the human eye did not resolve these individual lines, just as it does not resolve the individual pixels in digital pictures). Between force 2 and force 3, the human eye can just begin to detect water being skimmed off the surface of the ocean. By force 3 we see scattered whitecrested wavelets. Force 4 white crested wavelets abound. What we are seeing here is the removal of more than just a few microns of water from the top layer of the ocean.. At this stage, immediately above the oceans is a fine layer of mist much more than 6 microns thick.. This mist is absorbing DWLWIR from above and it is preventing DWLWIR even reaching the top micron layer of the ocean below. It is only in still conditions that the main body of the ocean could actually receive any DWLWIR since in most conditions there is a layer of windswept spray and spume lying parellel and above the ocean below that acts as a complete LWIR block just like a sunumbrella/parasol would do for humans, or sunblock cream.
It would appear that DWLWIR cannot practically heat the oceans since in most conditions it cannot even reach the top of the ocean still less penetrate, or in someway be over-turned down into the deep ocean below.
IF DWLWIR cannot effectively heat the oceans since it is the oceans that ultimately drive the climate, CAGW is a non starter.