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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Don’t even try to make the claim you’ve got some secret telephone line to Entropy and that it calls you up and whispers “when it’s just not right” and you have to come here and proclaim migration of energy against principles that govern a heater.
You’ve already been shown by me to be capable of saying you never made utterances to particular effect, and shamelessly then, affecting pretense you, and I, had some other discussion than the one we HAVE had:
that I told you, there is not a single equation or mathematical inference ANYWHERE
that any energy in a system EVER flowed against the classical pressure-gradient mathematics that make it possible for a redneck with a ball cap to understand how his CB works.
I told you, not one single time in my entire 18 years PLUS three, has anyone EVER mentioned to me they believe in incalculable and unmeasurable “reverse entropic flow.”
Not once. Not ever.
Anything you’ve said here that implies that, is utter fabrication by a wannabe who never made past those “introductory texts” you claim you’re familiar with.
Radiative transfer mechanics is organized according to pressure-gradient concepts very much like pressurization of anything else, including an air tank. Not everything crosses exactly but
that’s why light energy is often seen referred to as “pressure”
and that’s why thermal energy is often referred to as “pressure”
and that’s why electronic flow is often referred to as “pressure.”
And you have been in here for MONTHS if not YEARS telling people *you saw a hose get connected to a PRESSURE and INJECT, using LOWER PRESSURE than the vessel that energy was injected into.
Bullshoot. Period. By a rank amateur who has no idea how few words it takes to chop that up and leave it on the deck, bait for GULLS.
Allen B. Eltor says:
February 8, 2013 at 9:00 pm
>>>>>>>>>>>>>>>>>>.
I refer you to Willis’ comment upthread:
Willis Eschenbach says:
February 7, 2013 at 9:11 pm
If you feel strongly about the matter, then I suggest you take it up with the authors of those text books and reference manuals in use at universities and engineering companies all over the world.
davidmhoffer says:
February 8, 2013 at 9:32 pm
If you feel strongly about the matter, then I suggest you take it up with the authors of those text books and reference manuals in use at universities and engineering companies all over the world.
<<<<<<<
I already did. They gave me a degree in electronic engineering, after thousands of hours face to face instruction,
and told me fakes on the internet would be making claims of magical "non pressure-gradient prescribed flow," that "no mathematics can show: it shows one way flow but actually, it's going both directions!"
and not to let them get my identification card and use it to get into the building.
Yes I agree with Willis that in his model of black bodies, they will work as described.
However, as BB are a theoretical construct, not able to be realised, give me and many others a problem when moving into the real world.
Perhaps this is the problem we should address, using a theoretical construct as though it is real?
We all know that a Parker warms you in the winter by increasing the thermal gradient between your body and the ambient air, no back radiation what so ever.
We all know that igloos are cold inside unless you are inside as well to warm them up, no back radiation what so ever.
Perhaps all of this confusion has been brought about by too enthusiastic application of an interesting theory into areas where is might be best to use others?
PS: we all know the earth is not a BB but some persist in using BB calculations as though they are a perfect fit.
Steve Richards
So you have planet and it’s radiating 235 watts per square meter. Or surface is about
-18 C [255 K].
You put steel shell around it and then planet surface radiates 470 watts per square meter.
What is surface temperature? Is about 26 C [300 K]?
Or what is the temperature on this surface which radiating 470 watts?
And could someone also describe this as heat from core [235 watts per square meter] is inhibited
from radiating enough energy as would be able to do if was radiating into the vacuum of space, therefore the ground temperature increases to 26 C, at which point the temperature is sufficient
to radiate the 235 watts per square generated from the core generated heat plus the increase heat of the surface which adding another 235 watts per square meter?
Steve Richards says (February 8, 2013 at 10:03 pm): “We all know that igloos are cold inside unless you are inside as well to warm them up, no back radiation what so ever.”
Funny you should mention igloos. According to Figure 18 on page 85 of this paper,
http://www.atmos.washington.edu/sootinsnow/PDF_Documents/Warren_Optical%20Properties%20of%20Snow.pdf
the infrared emissivity of snow varies from about 0.95 to very nearly 1.00, depending on temp, wavelength, and angle. In other words, snow is a very good emitter/absorber of infrared radiation, and the snow igloo is emitting it in all directions, inside and out. If you enter the igloo (or put a Willis-type nuclear heater in it), the inner surface of the igloo will absorb your IR radiation and send so-called “back radiation” in your direction.
Of course down here deep in our atmosphere, conduction and convection are the predominant methods of heat transfer, and the igloo “works” mainly by curtailing these processes. When you’re considering the heating and cooling of the Earth from and to outer space, however, radiation is the only game in town.
Phil. says:
February 8, 2013 at 10:18 am
“Not correct there are indeed counter flowing flows of electromagnetic radiation, interference does not prevent the propagation of radiation in either direction, if it did a laser cavity wouldn’t work!”
Optical resonators form a standing wave oscillation. This is a singular condition, and there is no net flow of energy in either direction. So, here also, your interpretation of what is happening is a mental construct, a “useful fiction” to help understand the system, and what it will do when you perturb it from that condition.
Don’t you think it is ironic on these blogs that sceptics like myself have to spend all our time defending the greenhouse effect because so many commentators are unable to grasp even basic physics?
I would rather be debating how limited its future effect will be.
And now, in addition to the inappropriately named Greenhouse Effect we have inappropriately named ‘ThePhysicsGuy’
Sad, n’est pas.
There is an error in my equations above. If others had been interested in doing the math, instead of insisting on their individual infallible arrangement of words, it might have been detected earlier. I want to lay this out very carefully, so please check my math, any who are interested.
The total incoming power to the planet is
This is the power from the nuclear core, plus the reflected power from the inner shell. I had it incorrect previously because the reflected power must be integrated over the area of its source, not its sink.
The total outgoing is
Thus, at equilibrium when these two quantities balance, we have
(1)
For the shell, we have as before (I got it right in this one, for some reason)
When these balance, we have, as before
(2)
Plugging (2) into (1), we get
(3)
We know that without the shell
(4)
Hence,
(5)
Please, by all means, check my math and make constructive comments using math because this is a rather astounding result, the implications of which we can explore when others have verified the formula.
If others verify the formula. It is late, and I could be befogged. If this is right, it stands the whole discussion up on its head, so I am filled with self-doubt at the moment. It will not surprise me if I have made an error somewhere. But, if I haven’t…
Allen B. Eltor:
My post at February 8, 2013 at 2:58 pm genuinely tried to help you by suggesting your posts would be more understandable if they each provided one point clearly expressed.
Your post at February 8, 2013 at 6:01 pm has replied to that but, sadly, I am still failing to understand what you are trying to say. I read your words but I fail to see any clear point. There is no summarising statement and I fail to read any explanation, justification or exposition of it.
When reading your words I am reminded of the response by Eric Morecombe when answering complaint at his piano playing; i.e.
I regret this because you may be trying to tell me something which I would like to know.
Richard
Mr Hoffer has a serious problem distinguishing between thermal insulation and radiation.
This thread is strewn with examples where he continually confuses the two. Using his igloo analogy. The snow acts as a insulator. There will be “back-radiation” but no thermalisation can occur from it to anything above 273 K.
If I am wrong then ice would melt ice. Or rather no ice could ever form because everything would be continuously making everything else ever hotter and hotter.
As I said above, a colder sky cannot possibly heat a warmer surface, any more than a trillion 1 ton blocks of frozen CO2 radiating at 194.56 K could melt a 1 inch water ice cube radiating at 273.15 K.
In both cases it is forbidden by the laws of thermodynamics.
The colder substance/object radiates IR with a lower flux density than the warmer object. The IR from the colder substance/object cannot be thermalised by the warmer substance/object and therefore is just scattered/re-emitted at light speed.
No thermalisation occurs when IR travels from cold to hot.
Willis’ “shell game” generates a paradox. There is no mechanism preventing the system from getting ever hotter and hotter. You could not maintain the inner vacuum in reality, for a start. Then the system would simply return to radiative equilibrium with the core.
It’s a “thought experiment” for a very good reason.
Bart,
The radius of the shell doesn’t matter. For a thin shell it cancels out. You get the same result whatever the radius of the shell (see my first post).
By the way, the shell is not ‘reflecting’, it is absorbing and then emitting.
Sorry Willis this is complete sophistry.
Your model is nothing like reality and reality shows a different scenario. The planet is a revolving sphere with insolation at 1370W/m2. This is seen on only half the planet, since reality has night time, and is reduced by albedo and adsorption to 960W/m2, over half the planet. since this power, yes power since a Watt is a unit of power not energy, arrives as a disc and has to spread over a hemisphere, we divide by 2 since a hemisphere has twice the area if the subtended disc, giving 480W/m2 average over the hemisphere. True radiated energy is at 240W/m2 but this is for all the planet’s area. 480W/m2 gives a temperature of +33C.
Wood’s experiment had nothing to do with re-radiated heat but to confirm that despite the IR opaqueness of glass visible light gives heat as well. This was confirmed using a rock salt window and glass window in side by side vessels where temperatures were similar despite the glass being IR opaque.
See:- http://www.climateofsophistry.com
Will wrote “Willis’ “shell game” generates a paradox. There is no mechanism preventing the system from getting ever hotter and hotter.”
This is incorrect, the only source of heat involved is the nuclear core, and as the shell radiates heat from its outer surface. The rate of heat loss from the planet/shell system is proportional to the fourth power of the temperature of the outer shell, so as soon as the outer shell warms enough to radiate energy at the same rate it is generated by the nuclear core, the system will reach thermal equilibrium and will cease to get any hotter. There is no paradox.
Allen B. Eltor:
Perhaps my difficulty in grasping your words would be understandable by comparing these quotations of your words.
At February 8, 2013 at 9:18 pm you wrote
Well, your maths are wrong when your maths dispute reality.
There are several systems where some energy in a system flows against the pressure-gradient; e.g.
ater pumps which use some the energy of the flowing water to raise some of the water to above its pressure head,
refrigeration systems,
etc.
At February 8, 2013 at 9:00 pm you wrote
So what? I rely on an auto mechanic to “calibrate analyze & maintain” my car but I would not expect him to understand the combustion kinetics in the car’s cylinders.
The only way I can make sense of your rants in this thread is to assume you are one of the ‘new bread’ of scientists and engineers who graduate with a good understanding of how to ask a computer to provide an “answer”, and who think they understand what the computer does, but they don’t really understand because they only have a superficial theoretical knowledge.
At the moment, all I can see in your several posts on this thread is assertions that everybody except you is wrong because you are qualified to say they are but you have not shown their errors.
Perhaps you can try to provide a coherent and cogent post based on evidence and/or logic instead of claims to your own authority. That would alter my opinion of your posts which I have here explained.
Richard
Willis,
Over at Tallbloke’s Talkshop we are debating the same issue:
http://tallbloke.wordpress.com/2013/02/04/david-cosserat-atmospheric-thermal-enhancement-part-i-the-great-debate-begins
Coincidence? 🙂
We are currently trying to establish sensible ground rules there on a number of technical issues one of which is this vexed issue of so-called back radiation. I describe there a thought experiment which may help people understand that radiative energy flowing between bodies separated by a vacuum does of necessity flow in both directions. Like you, I have found it a tough task to persuade some commentators that two-way flow really does (and must) happen.
The difficulty, I think is psychological. Skeptics such as your ludicrous, bombastic Allen B. Eltor, who have a perfect right not to believe in CAGW or AGW, appear to think that they are at Custer’s Last Stand. So they try to defend the indefensible without thinking whether they need to do so in order to sustain their position.
In fact, as I have demonstrated at the above link, skeptics should welcome the concept of back radiation because it shows that the fully-thermalised lower atmosphere is behaving as it should, given its elevated temperature. It is hardly surprising that it radiates back to the surface nearly as much power (333Wm-2) as the surface radiates upwards (356Wm-2).
The net difference beween the up and down radiative flows is a piddling 23Wm-2. This net flow is upwards of course from warmer surface to slightly cooler lower atmosphere in full conformance with the 2nd Law. That 23Wm-2 is in any case almost immediately thermalised in the atmosphere and joins the 17Wm-2 of conduction/convection thermalisation and the 80Wm-2 of latent heat (later thermalised at cloud level) to provide an overall thermalised through-flow of ~200Wm-2 of energy from Sun-to-Earth-to-space.
People need to understand that the enhanced radiation of 333Wm-2 apparently ‘going round in circles’ between surface and atmosphere is is not the cause of their respective elevated temperatures. It is simply a consequence of them! It does no work. It is not dissipated. So it does not count in the overall (relatively modest) 199Wm-2 through-flow of energy from Sun-to-Earth-to-space that is needed to make up for energy losses, and which keeps the surface and atmosphere at the stable enhanced mean temperature levels that we humans enjoy.
Rather than call it the ‘Greenhouse Effect’ I prefer to use the term ‘Atmospheric Thermal Enhancement effect’ or ATE for short. This is a perfectly neutral term which allows us to go on to debate in a mature, and non-confrontational, way whether or not the thermal enhancement is due to GHGs. This we intend to discuss, rationally, in our upcoming Part II.
johnmarshall, you are being grossly unfair to Willis, who is doing a good job of explaining some very basic concepts. The shell model is obviously intended to be only the most basic representation of one aspect of the greenhouse effect (backradiation), only complex enough to reveal the problem with the Woods experiment. nothing more. Using such simplified models is common place in science, for good reason, and is hardly sophistry.
If you want a more realistic model, including features such as rotation, then I suggest you try Raymond Pierrhumberts book, “Principles of Planetary Climate”, chapter 3 deals with radiation balance of planets.
For those who argue there’s no back radiation, no “Net”, notice SB equations have a T1-T2 term, if you look up the definition of Net, it say one definition is a difference. This difference does always flow from hot to cold, but to get the right answer you have to account for radiation being exchanged between both objects.
What I think we should discuss next is that the temp of the sky is lower that the earth( 75 F colder on a clear 35F day in NE Ohio). Much lower when there’s little water vapor in the atm. These facts explain why when the Sun goes down the temp drops, a lot in dry locations. And the difference in day night cycle controls whether It’s warm or cold, ie the seasons.
The average temp swing is about 18F when averaged over a year (follow the link in my name to see the graphs).
Yet, it can be as much higher in dry locations, and much lower in humid locations. This average has not really changed over the last 60+ years.
dikranmarsupial says:
February 9, 2013 at 3:30 am
“Will wrote “Willis’ “shell game” generates a paradox. There is no mechanism preventing the system from getting ever hotter and hotter.”
This is incorrect, the only source of heat involved is the nuclear core, and as the shell radiates heat from its outer surface. The rate of heat loss from the planet/shell system is proportional to the fourth power of the temperature of the outer shell, so as soon as the outer shell warms enough to radiate energy at the same rate it is generated by the nuclear core, the system will reach thermal equilibrium and will cease to get any hotter. There is no paradox.
You have only half quoted me and then simply repeated the rest of my comment in your own words and then claim that I am incorrect. This is an over used trick of semantics.
My quote in full:
“Willis’ “shell game” generates a paradox. There is no mechanism preventing the system from getting ever hotter and hotter. You could not maintain the inner vacuum in reality, for a start. Then the system would simply return to radiative equilibrium with the core.
Which is the same as yours, sans waffle. Neither of which are anywhere near Willis’ impossible thought experiment because he double counts energy in order to imply additional heating.
So in effect you are agreeing with me while implying that you support Willis.
Very tricksy.
David Socrates,
I took a look at Tallbloke’s “The Great Debate”, and the ground rules establish essentially the “alarmist” GHG model as developed by Trenberth as the starting point. So if you don’t agree with the basic concepts of the model as set forth, too bad, that is not up for debate. No thanks. Sounds more like an indoctrination session.
For you electronics guys who still struggle with this.
Picture 3 resistors in series, with each end connected to separate 12v power supplies, now calculate current, voltage and power for each node. Now change one supply to 6v, recalculate, now set it to 0v, recalculate again.
It’s not exactly the same as SB between two objects, but its similar.
Will says:
February 9, 2013 at 2:41 am
Mr Hoffer has a serious problem distinguishing between thermal insulation and radiation.
This thread is strewn with examples where he continually confuses the two. Using his igloo analogy. The snow acts as a insulator. There will be “back-radiation” but no thermalisation can occur from it to anything above 273 K.
>>>>>>>>>>>>>>>>>>>>>>..
Mr Hoffer did not propose an igloo analogy in this thread. Your statement that back radiation exists but no thermalisation occurs is a violation of the Law of Conservation of Energy, First Law of Thermodynamics and Stefan-Boltzmann Law and to be true would require that both the joule and watt as fundamental units of physics be re-defined. The rest of your rant suffers from the same errors as your first paragraph.
Bart says:
1) As mentioned before, the shell “emits” its own power, it doesn’t “reflect” power.
2) More importantly, you missed power transfer from the inner surface of the shell back to the inner surface of the shell. In the limit that the shell is the same radius as the planet, this term doesn’t matter. In the limit that the planet is small compared to the shell, then the thermal radiation emitted by the inner surface of the shell will simply hit some other part of the inner surface of the shell.
I started to write out all the equations, but it got too messy. Suffice it to say that not all the power emitted by the inner surface of the shell hits the planet. If you want to look at the “realistic” situation where the planet is some finite distance above the planet’s surface, then you will have to account for the photons the get emitted by the inner surface of the shell but “miss” the planet and instead land back on the shell. (For earth, this amounts to a fraction of a percent.)
P(in to the planet from the shell) = σ (T_shell)^4 (area of PLANET)
P(out from the shell) = σ (T_shell)^4 (area of SHELL)
That would mean
P(from shell back to shell) would be σ (T_shell)^4 (area of SHELL – area of PLANET)