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.
Greg House says:
February 6, 2013 at 2:41 pm
In this case, Greg, actually it’s not all about you, your guess is wrong … sorry, but there it is.
I quoted what Wood said. You can find examples of people claiming it means the greenhouse effect doesn’t exist all over the web. See the misunderstandings on this thread as one of far too many examples.
w.
I believe you touched on a point I’ve made that I’m surprised isn’t much more widely appreciated and distributed. The Catastrophic Anthropogenic Climate Change Alarmists state, as “proof” of their theories humans are destroying the planet through Dangerous Anthropogenic Global Warming, the apparent ESTIMATES that one amount of energy is coming to the Earth from the sun and a different, lower amount of energy is escaping from the Earth back to space. Where is the missing energy, they shout. Why Anthropogenic carbon dioxide in the atmosphere must be trapping it! Only we’re not seeing the necessary changes that would cause.
So where does the missing energy go? Anyone? Surely you know, it was mentioned (to some extent) in this article.
It goes to take part in various physical, chemical and life processes here on Earth, that’s where. It drives the winds, convection in the atmosphere. It moves huge quantities of ocean waters. It evaporates water from any and all places it is present on the surface and subject to being evaporated. It powers many endothermic chemical reactions. It allows plants to grow and creates “free” energy through the “magic” of the photoelectric effect. And, if after estimating and accounting for ALL THAT ENERGY, there is still a discrepancy, I have an answer for that, too.
Check your math, you didn’t carry or borrow correctly somewhere.
TomR,Worc,MA says:
I think I am headed back to “Ilikebacon.com”, this thread makes my eyes bleed.
________
As with most things, Science tastes better with bacon.
How is it possible for the shell to be emitting twice as much energy as the source? I think Figure 2 should be showing half being returned to earth and half going out to space. The temperature of the shell would then be lower than the earth based on Stephan-Boltzman.
Roger Clague says: “There is no law of conservation of power.”
Power is merely the flow of energy which is the topic here – heat.
Willis: “I tried to head this incorrect argument off at the pass, but I was not emphatic enough. What I said was: ….”
No need – Sorry, I missed the caption because I was mesmerized by the pretty graphic…
Steven Mosher says, February 6, 2013 at 2:59 pm: “The green house effect operates by raising the ERL. A raised ERL means a earth that radiates from a higher colder region. That means a slower rate of energy release to space and the surface cools less rapidly in response. back radiation is an EFFECT of the greenhouse effect not a cause. The theory is not that back radiation warms the source. It does not. The rate at which the source cools is slowed.
===========================================================
First the less important part about “warming vs slowing down cooling”. Normally, people would call a slowing down cooling effect a warming effect, too. I nevertheless prefer saying “affect the temperature” or “have an effect on temperature”. And, again, the Wood experiment demonstrates that trapped/back radiation has a zero or negligible effect on the temperature of the source. By the way, do you have any problem with “global warming” being called “global warming” and not “reduced global rate of cooling” (LOL)?
Second, your “ERL” etc is absurd (http://wattsupwiththat.com/2012/08/30/important-paper-strongly-suggests-man-made-co2-is-not-the-driver-of-global-warming/#comment-1068226) and, more important, not the politically relevant “greenhouse effect” as presented by the IPCC. You and anyone else is absolutely entitled to come up with whatever hypothesis on anything, but these private “greenhouse effect” hypotheses have zero political relevance. The official concept of the IPCC is different. And they mean exactly that very old concept of the effect of trapped/back radiation professor Wood so easily debunked back in 1909.
Here is the official version of the “greenhouse effect” as presented by the IPCC: “The Sun powers Earth’s climate, radiating energy at very short wavelengths, predominately in the visible or near-visible (e.g., ultraviolet) part of the spectrum. Roughly one-third of the solar energy that reaches the top of Earth’s atmosphere is reflected directly back to space. The remaining two-thirds is absorbed by the surface and, to a lesser extent, by the atmosphere. To balance the absorbed incoming energy, the Earth must, on average, radiate the same amount of energy back to space. Because the Earth is much colder than the Sun, it radiates at much longer wavelengths, primarily in the infrared part of the spectrum (see Figure 1). Much of this thermal radiation emitted by the land and ocean is absorbed by the atmosphere, including clouds, and reradiated back to Earth. This is called the greenhouse effect.”
http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html
KevinM says:
February 6, 2013 at 2:40 pm
“No, but thermal equilibrium is the condition described by zero net energy trasfer, or net zero joules per second, right? Its not a law, its a condition.”
This Eshenbach model and Trenberth’s assume Watts/m2, that is Joules/sec/m2 in and out at the surface and in and out at the shell are equal at each place.
They apply the Law of Conservation of Energy and average horizontally over time and space. They add 2 extra dimensions . It is not a condition it is an incorrect application of a law.
The Law of Conservation of Energy can be correctly applied to a small vertical column of air. The column does not vary with time . Troposphere height is constant from day to night.
mgh = mcT
m = mass of a small volume of air
h = height above surface ( variable)
T = temperature ( variable )
c = heat capacity ( almost constant for small molecules )
g = acceleration due to gravity ( constant )
cancelling m and rearranging
h/T ( lapse rate ) = c/g
This shows that the temperature of the troposphere at any height ( including at the bottom, the earth’s surface ) does not depend on its composition, such as the % of CO2 in it.
Willis, nice, but not even wrong.
The total surface of your shell is not off by 0.3% from surface of your sphere, it is off by ~200%, the sphere has both an interior surface and and exterior surface. Assuming the shell is directly above the sphere and is infinitely thin then the error becomes exactly 200%.
Also, to do a proper energy budget you need to keep real energy sources separate from redirected energy flows. The energy from your sphere is an energy source (real energy input to the system supplied by breaking chemical bonds). The energy returning to the surface from the shell is just a redirected energy flux. Sure you can add them, but not if you want a useful answer. It’s like getting two tens as change for a five dollar bill.
As an empirical example; after space going satellites are assembled on the ground they are tested inside large vacuum chambers. So start with a satellite with no energy supplied (all electrical circuits turned OFF) at room temperature (a source of stored thermal energy), roll it into a steel vacuum chamber (also at room temperature). Close the door and evacuate the air (also at room temperature). Per your example the satellite (radiating it’s stored heat) would heat the vacuum chamber walls and the temperature of the satellite would rise. EXCEPT when you do this NOTHING MUCH happens to the temperature of the satellite. Assuming whatever structure is supporting the satellite (the floor of the chamber for example) remains at room temperature the satellite and vacuum chamber walls do not change temperature. I have witnessed this empirical experiment many times, what you describe DOES NOT HAPPEN, not even once, since satellites are fairly expensive and surprisingly delicate we would sure as heck notice if it started heating up. And it’s not because we don’t watch the temperature of the satellite, they are covered with tens of temperature sensors (telemetry sensors they are called in the trade).
Of course if we turn on electrical systems on the satellite the waste heat will cause the satellites temperature to rise, but it would also rise if you wrapped it with normal thermal insulation.
What really happens in your steel shell example (assuming an infinitely thin shell with infinitely high velocity of heat) is that at time = zero the shell has no thermal energy stored inside it. As the sphere radiates the shell “backradiates” zero. As the shell heats up it eventually “backradiates” exactly an amount equal to the radiation from the sphere and they reach the same temperature. The outside of the shell is now re-radiating energy from the real energy source (the sphere). Once one little tiny bit of energy is radiated by the shell it is immediately replaced by a little tiny bit of energy radiated by the sphere and the temperature remains at equilibrium. Of course since nothing useful is actually infinitely thin or has an infinitely high speed of heat delays are involved and no real system exhibits “thermal equilibrium”.
Thermal equilibrium is strictly a textbook creature, nobody has ever bagged an example in the wild.
Cheers, Kevin.
Whoops, my surface area error figure should have been 100% (2x shell surface area, 1x sphere surface area, thus (2-1)/1 = 100% error).
Cheers, Kevin.
Willis, for the outer shell to raise the temperature of the inner core its temperature must be higher than the inner core right? How can you raise the temperature of the outer shell higher than inner core so that there is net energy flow between the outer shell and inner core (a necessary condition to raise the temperature of the inner core) ?
The system will not reach an equilibrium if you posit that the outer shell will heat the inner core. The temperature will just rise higher and higher.
Great explication, WIllis. I particularly liked
“if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell”
because, in fact, convection in the atmosphere – storms, clouds, winds, etc. are the huge thick pillars that reduce the temperature differential between Earth’s surface and the top of the atmsophere.
Haven’t read the comments but…if you put a steel shell around the earth aren’t you shielding the earth from the sun and wouldn’t that lead to immediate cooling?
Mike M says:
February 6, 2013 at 3:52 pm
Roger Clague says: “There is no law of conservation of power.”
Power is merely the flow of energy which is the topic here – heat.
The flow of energy to the earth is not constant. During daytime, energy flow in and out, but during
the night energy flow out but no energy flows in.
The Law of Conservation of Energy can only be applied vertically not horizontally and gives us the result
h/T = c/g.
As is confirmed by observation.
Steven Mosher says:
February 6, 2013 at 2:59 pm
“The green house effect operates by raising the ERL. A raised ERL means a earth that radiates from a higher colder region.”
———————————————————————————————————————-
No, running back to the ERL thing won’t work. The ERL game was only cooked up after it became impossible to ignore that most of the energy that radiative gases radiated to space was acquired through conduction and release of latent heat, not IR from the surface.
And of course we can see cloud tops radiating strongly in IR images from space. Far hotter than the surrounding air at their altitude. The altitude of radiative gases provably does not set the temperature of much of those gases at the time they are radiating the most IR. Try again.
Stephen Rasey
“As we reduce Rs to approach Rp in the limit, then Tp > Tsi and an infinite temperature gradiant which seems to be a logical impossibility.”
Firstly, without a conductive medium there is no gradient – was that not the reason why he stated no atmosphere.
Secondly, if Rs = Rp then the surface of the planet changes and will be first heated to a temperature at which it starts emitting at the equilibrium state; the surface energy state is doubled – but more importantly you’d also have conduction to deal with. The point of why the shell is off-surface is to avoid the point that Willis was trying to make: convection + conduction provide methods for thermal transfer from lower to upper atmosphere; however thermal conduction does not always translate to radiative transfer at some point and the conductive process obviously stops at the atmosphere-space interface, so it may just start doing different work such as expanding the volume of the total atmosphere rather than releasing IR to space (I don’t know). BTW a nominal distance will do for the purposes of the experiment.
Mike Jonas says, February 6, 2013 at 3:32 pm: “http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
The advantage of Roy Spencer’s example is that it can be set up and tested in a lab.
==========================================================
This is a wonderful idea real experimental testing, Mike, thank you. Has Roy Spencer done it yet?
Hello Willis;
If you have time please look at my post at 2:17. The glass covered box is far more interesting than it appears at first sight. Neither prevention of convection nor interception of thermal IR by the glass should make any difference at all if the glass is at the same temperature as the inside of the box. It will only work if the glass is colder than the inside of the box. The analogy to the steel shell is far closer than one might think at first sight.
cheers
Mike Hammer
“..Somehow I can’t see 235w radiating energy into 236w….”
There is a lot of this type of thinking by commenters. The radiation coming from a bright flashlight, trained on a light beam from a weaker flashlight doesn’t prevent the radiation from the weak flashlight (the batteries in the weak one rundown just as fast with or without its adversary). The light filaments in both glow and burn energy which is emitted as light. Ditto, the radiators. This is akin to the crystallization of salts in saturated solution: equilibrium is reached when dissolution from the salt crystals just equals the crystallization of solid salt – it doesn’t stop the action. Imagine turning on a weak flashlight just before turning on the strong one trained on it. What happens to the radiation that took off from the weak one just before it was “flooded” by the strong one? The fellow holding the strong flashlight in the dark can still see the light from the weak one. What about stars in the southern hemisphere shining “on” the stars from the northern hemisphere?
Willis:
I was referring to a snide comment about “nutters” in an earlier post. Here is the link to the experiments for those who didn’t look last time.
http://principia-scientific.org/supportnews/latest-news/34-the-famous-wood-s-experiment-fully-explained.html
Willis wrote (re the Woods experiment);
“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 hold that “backradiation” does indeed exist; after all the walls of your house are “back radiating” towards you all the time. Did you ever notice (for folks in the Northern climes) that no matter how warm the air in your room is you still feel a little bit chillier in the winter than the spring or fall ? Why ? Because in the winter the inter walls of your house are just a few degrees cooler than in the spring. Thus they “back radiate” less and you feel just a bit chillier even though your furnace is holding (as well as it can) the air temperature constant.
The poorly named “greenhouse effect” is in fact equal to ZERO, not “trivally small”.
When you design a heating system for the interior of a house you consider the room size and select a heat source (say electric baseboard heaters) to warm that volume to a desired temperature. You add some overhead for really cold days and add a thermostat to regulate the average temperature. You DO NOT subtract the “back radiation” from the walls from the size of your heaters. The “back radiation” IS NOT an energy SOURCE, no way, no how. It is just re-directed energy and since it’s moving at the speed of heat it has no effect on the average temperature.
Now of course if you replaced the thermal insulation in your walls with steel you would have a very uncomfortable house. The speed of heat through the steel walls would mean that your furnace would never keep up and it would be damn chilly.
Sorry Willis, but R. W. Woods was correct, the temperature rise inside a greenhouse is ONLY caused by the restriction of convection. The opaque nature of some materials at some wavelengths only delays the flow of energy through the system. A real thermal insulator slows the velocity of heat flow, slowing and delaying (via multiple passes) are not interchangeable effects.
Cheers, Kevin.
Willis,
Please take some physics courses, or stop talking about it. You are embarassing yourself in front of thousands.
Your post is sense-free. 235 Watts per square meter would heat the steel, which would be cooler than the planet because it is 0.3% larger. The steel would radiate to space, but transfer ZERO heat back to the planet, as we all know that the Second Law is true. If you do not think the Second Law is true, well, I will read your fun stories about dolphins but nothing more concerning physics.
Absorbed and re-emitted, transferring NO heat, was I not clear? How about, absorbed and re-emitted in an infinitesimal? Does that help?
Do you know what heat is?
Problem #2
Since this is a closed system (unlike the earth) which is heated by SW that is absorbed into both systems in an identical manner, the question becomes, is this a model of what happens in the atmosphere? It is not.
In the atmosphere, 100% of all energy leaving the system does so via radiance. In this experiment, energy leaves the system by conductance/convection plus radiance. The ratio of conductance/convection to radiance is so large that in a system this small it could only be measured by instrumentation that did not exist in 1906.
Problem # 3
The ratio of conductance/convection to radiance is so large that in a system this small it could only be measured by instrumentation that did not exist in 1906.
Since both boxes heat up to about the same temperature, and both the salt rock and glass radiate the same amount at the same temperature, we’d need to be able to measure the energy radiated directly by each versus the amount of energy radiated by the salt rock plus the LW that passed through from the box itself. This would be also be a ratio to minute that instrumentation in 1906 could not possibly detect it.
Greg House says:
February 6, 2013 at 2:58 pm
Greg, of course it has been proven, many times over. It’s in every college thermo textbook. Go buy one. Read it. Come back. Participate intelligently, instead of just asserting your misconceptions.
Here’s an example to consider. Mount a light bulb socket on a table. Insert a 60-watt bulb. Measure the temperature, we’ll say it’s 150°C. Try a 100-watt bulb, it’s hotter, say 200°C.
Now, mount another socket right next to the first one. Insert a 100-watt bulb in one and a 60-watt bulb in the other. What will the final temperatures of the bulbs be?
I say both of them will run hotter in that situation, because A warms B, and B warms A. If you disagree, I urge you to buy a thermometer and bulbs and sockets and do the test. Me, I don’t need to do it, the physics of the situation is quite clear to me.
Regards,
w.
Steven Mosher says:
February 6, 2013 at 2:59 pm
What he said.
w.