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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davidmhoffer says, February 6, 2013 at 8:12 pm: “For two bodies in proximity to each other, suppose a net flux of C.
Via SB Law, calculate the energy flux from hotter = A
Via SB Law calculate the energy flux from colder = B
A – B = C”
============================================================
There is no basis in real science for your “C”, so simple is that.
Willis said:
Here’s another example. You are standing next to a block of ice. It is suddenly replaced by a block of metal about 5° below body temperature. Do you feel warmer? Why?
w.
——————————————————————
Your body is transmitting less heat to the block. In other words, the energy flux (watts/m2) drops, and the draw on your bodies energy supply is lower. Your body requires less energy to maintain the same temperature.
LazyTeenager wrote;
“Fukushima anyone?”
Reliable (24/7/365 +/- a few minutes) AC electricity anyone ?
Fresh Water anyone ?
Safe transportation anyone ?
Bridges and building that stay put when you walk on/in them anyone ?
Cheap airplane travel across the globe anyone ?
Cheap (really really cheap) computing power anyone ?
Safe/warm houses in cold climates anyone ?
Apollo, SST, a safe end to the cold war anyone ?
Or how about endless doom and gloom predictions that NEVER EVERY MATCH REALITY anyone ?
I for one will take RESULTS instead of PREDICTIONS/PROJECTIONS/WILD ASS GUESSES.
Snake Oil Baron,
Congrats on your hole in one, Willis says you passed. I have a greenhouse, and my plants and I needed to hear this. This all needs to be cleared up.
For some, though, the rain in Spain will remain much too plain.
davidmhoffer says:
February 6, 2013 at 6:22 pm
“Who out there thinks the laws of physics don’t work like the thermo texts say they do and the engineers just nailed their design by shear luck? A few hundred nuclear reactors in a row? C’mon, hands up. Greg and who else?
>>>>>>>>>>>>>>>>
OK two hands. Greg and jae. Anyone else?”
Very funny, huff-man. NOW, please put up some detailed facts. Specific stuff, like a named text and page number. You’re good at broad insults, but I cannot think of any specific references to your snarks. Now let’s see something concrete, fella…
Richard G says:
February 6, 2013 at 8:33 pm
>>>>>>>>>>>>>>>
LOL
There is no basis in real science for your “C”, so simple is that.
>>>>>>>>>>>>>>>>>>>.
ROFLMAO
jae;
You’re good at broad insults, but I cannot think of any specific references to your snarks.
>>>>>>>>>>>>>>>
Is there some part of “any university level thermodynamics text” that you failed to understand? Do you want me to read it for you as well so that you don’t have to?
Wait… I already did.
No, the known historical statements do not contain the word “net”
>>>>>>>>>>>>>>>>>
There have been nearly as many formulations of the second law as there have been discussions of it.
—Philosopher / Physicist P.W. Bridgman, (1941)
Gino says:
February 6, 2013 at 8:54 pm
The “draw” on my body’s energy? The energy flux that I am radiating drops?
My friend, you desperately need to read about the great discovery of Stefan-Boltzmann. They showed that
RADIATION DEPEND ONLY ON TEMPERATURE AND EMISSIVITY
Period. No exceptions. No special clause that says you are radiating less energy because you are standing next to an ice block. No way to “draw” on something to make it radiate more or radiate less. No way for the energy flux to “drop”. Since for a given object emissivity is fixed, radiation depends only on temperature. Radiation does NOT depend on what is absorbing its radiation. It does NOT depend on the mass of the object, or the thermal mass. For a given emissivity, radiation is only and solely and completely a function of temperature.
In other words, your explanation is … well … wrong. The difference has nothing to do with you. It has to do with the fact that the ice block radiates less energy than the warm metal block … and with the fact that your body absorbs radiation from the ice block just as it does from the metal block. With the ice, your body gets less total radiation, so it feels colder.
Note that the flow of HEAT, which is different from ENERGY, is always and forever in one direction, from me to the ice block or to the metal block, because my body is warmer than they are … but the energy flows in both directions.
w.
Willis,
You must consider something here. Energy, in the form of heating your house, or keeping your lights on with the electrical bill, has been the subject of, shall we say, INTENSE scrutiny, since
Edison in the 1870’s. When a large-gigantic-immense-overwhelming amount of money is spent on the subject of energy, the community of suppliers tend to research this subject, to learn how to satisfy their customers at minimal cost.
Well, we did.
If there were a way to heat anything by putting a shell around it which would magically double the “heat input,” otherwise known as Flux, then it would have been seen and exploited many many years ago.
This is just a blog run by a weatherman, who is searching for something to post that he did not have to write, and you are a good writer who is supportive of our host. Why would you go here, with no fundamental understanding of the physics? You should stop.
You are not stupid, but you seem to be over-confident, posting science of which you know NOTHING. Please stop, otherwise people like GS and JH will shred you and destroy this blog. Anthony, I previously begged you to stop this.
If you would like me to edit posts concerning physics I could do it for a short time now….
Jeff Carlson said:
what happens to that radiated energy when it strikes the hotter ball ? Is it absorbed or reflected ?
=======================================
Higher temperatures emit more energy than lower temperatures so even though it “sees” the lower temperature object, the higher temperature object overpowers the lower on so to speak. That is the reason heat transfer calculations are based on temperature DIFFERENCE. Energy is transmitted at various wavelengths. Best explanation I’ve read involves wave theory. When two bodies at the same temp emit at the same wavelength, they form a standing wave. A standing wave does not transport energy. When they emit at different wavelengths, they ‘exchange’ energy as in absorb in one bandwidth and radiate in another with a net of zero to maintain a delta T of zero.
Cop; Do you know how fast you were going?
Heisenberg; No… but I know where I am.
Cop; You were going 80 in a 60 zone….
Heisenberg; Impossible. There was an experiment in 1906 proving that cars would never go faster than 40.
Cop; License and registration please…
[snip – take a 24 hour time out – Anthony]
Greg House says:
February 6, 2013 at 8:45 pm (Edit)
Greg, that is so precious, you actually think your unsupported opinion means something … you might try here for the relevant equations and a calculator.
There’s also an overview here … a text here …
And every one of them says the same thing … that C is the actual heat flow between the two bodies. If you have say 140 W/m2 of radiation going one way, and 290 W/m2 of radiation going the other way, the net heat flow is the difference “C”, of 150 W/m2.
Truly, my friend this is absolutely basic stuff, first year physics. You’re embarrassing yourself, or while you may not be, at least I’m embarrassed watching your ludicrous claims …
w.
Willis said:
Since for a given object emissivity is fixed, radiation depends only on temperature.
—————————–
Energy transmission however is dependent on temperature DIFFERENCE. Thank you for making my point. When the temperature difference drops (60F to say 5 f) then the energy flux drops as well. Your body maintains it’s surface temp with less energy flux. Your body is a generator that uses different amounts of energy to maintain a minimum core temp. It generates POWER from potential ENERGY. This manifests as TEMPERATURE, which is dependent on environment.
Willis Temperature is NOT energy, is NOT power. They are all different physical concepts and you seem to be having trouble separating them.
“[UPDATE: Note that because the difference in exterior surface area of the shell and the surface is only 0.03%, 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%?]”
Why does it have made of steel and and kilometers above earth?
Why not inch of dirt and inch of space separating it?
Why not do this 100 times. So it’s 200″ high.
Does prove anything other than air is pretty good insulator.
Now how much is earth being warmed per second?
Are suppose to think it’s somewhere around “235 watts per square meter”
heat coming from the surface?
Take a 10 meter by 10 meter floor space of a house- 100 square meter.
So 235 watts per square meter is 23500 watts of heat. Or using 23.5 kilowatts per
hour of electricity used to heat the house and the house is well insulated. And we suppose
or imagine it’s not going to get hot in the house?
Now, you can go into 100 square meter floor space room and measure the temperature of
the walls, floor and ceiling and convert this to watts per square, but if total area, this not how
much energy is being used to warm the room.
Say you planet and it’s geothermally heated so it’s uniform temperature of 15 C at it’s surface.
So no ice, and snow could not last very long, ocean will warm to 15 C at surface, and air will warmed to 15 C near the surface. So whenever surface of ocean cools below 15 C and air just above surface cools below 15 C, one added more heat. So we have a little Utopia, lack any exciting weather and somewhat cool- warmer for Russia and cooler for Tropics.
If mountains were considered surface, then you get no skiing. If you exclude mountains or certain mountains, then you can get surface and air temperature cold enough- and if high enough and made artificial snow- a mountain about 3000 meters gets cold enough for snow.
So for fun, let say in Utopia even though some might get injured, skiing is desired and allowed.
Now some genius decides putting a steel sphere around earth, to lower energy costs.
So we get the “Deathstar is too small project” and we going to put sphere of steel around the planet.
We have few option, we could try to make sphere strong enough, but that is
hopeless.
Or we orbit bands of steel at varying orbital height. This allows us use very thin plate steel. There going to be orbital mechanic issues, but we put say 1000 mile wide band in a polar orbit, and need 20 of them [circumference at equator 40,000 km]. if 50 km orbital height difference it’s 1000 km in total. So starting 100 km, and longitude 360, and second 50 km higher is 180 longitude, etc.
The question is would this reduce the heating costs. Simple answer is, no.
It would help [at least show that you trying] if you put shiny stuff on interior surface of it, but that would not keep the bands very warm [assuming earth was only thing keeping it warm].
Another plan is you want to do is keep the outside part of band as cold as possible. And this done by making layers of material. So say 10 layer of shining material which separated and insulated against heat being conducted. That should make the outside of bands fairly cold- very heat passing thru it.
And this whole band thing is quite cold with warmest part being the reflective surface facing Earth.
Now, how much is the this reflective sphere going warm the earth- near zero.
You would save a lot more energy if you decided to keep all mountains above 2000 meters non heated.
And if did this, these mountains would receive the most measurable heating from the sphere- so instead being at 0 C it might be 1/2 degree warmer.
One test this, have insulated box, put glass which is transparent to IR, put on balloon fly it to 60,000′ point window box at ground night side surface and see how warm it gets. If get a greenhouse effect which warms the air [or black plate] over 10 C- I am wrong.
Or easier use same box in room with a high ceiling and have ceiling 10 C warmer than floor. Put box on floor. See if it warms up by 5 C.
And I would be really impressed if the box which is warmed from the ceiling is warmer than the ceiling.
Willis Eschenbach says, February 6, 2013 at 9:49 pm: “you might try here for the relevant equations and a calculator. There’s also an overview here … a text here …”
==========================================================
Willis, your links show only that you are not the only one who is practising this “radiation arithmetic”. There are also others, I get it, thank you.
What I mean is that this “radiation arithmetic” has apparently no basis in real science, please, note this point. Apparently there is no real scientific physical experiment confirming that this “radiation arithmetic” is correct. This is the point.
Willis Eschenbach says:
February 6, 2013 at 9:31 pm
RADIATION DEPEND ONLY ON TEMPERATURE AND EMISSIVITY
Period. No exceptions.
———-
And the Gods laughed, and angels wept, at the temerity.
Of things never imagined.
Willis, HEAT is the transfer of energy. The transfer of energy is work (Joules), the rate at which energy is transferred is power (Watts). If no energy is transferred, no work is performed, If no work is performed, no power is used. If two bodies are at the same temperature the may “see each other” through radiation (see my standing wave post), but no work is performed transferring energy between the two bodies, therefore no energy moves between them, hence there is no power consumed.
Willis, a minor point/nitpick but I think you overstate to the percentage diffrence of your planetary globe and the steel shell areas by a factor of 10, assuming the globe is an earth sized and the shell height is top ~ top of our atmosphere . The area of the steel shell would be equal to 4*pi*(R+h)² and the area of the planet 4*pi*R² , where R is the radius of the globe and h is the heigth of the shell above the globe surface so the diffrence D comes out as D = 4*pi*[(R++h)² – R²] = 4*pi*(2*R+h²) , now using R= 6200 km and h= 10 km , giving D = 4*pi*12500 km² ~ 1.57*10⁵ km² , and the globe surface would be somthing like 4.85 * 10⁸ giving the ratio of D/Globearea ~ 0.0003 or 0.03%
not 0.3% , so the diffrence is even slighter than you state.
Michael Moon says (February 6, 2013 at 9:37 pm): “If you would like me to edit posts concerning physics I could do it for a short time now….”
When I could finally pick myself up off the floor, it occurred to me that I should nominate this comment for February Funny. 🙂
Perhaps I’m missing something.
If you chose a planet whose radius is r, and a shell with a radius of 2r, would not the radiative heat flux reach at the shell be 1/4 (235W/m^2)? Surface area being A=4pi*r^2
Polyethylene is transparent to IR but polytunnels work fine for friut and veg farming. (In fact an IR absorbent grade is sometimes used to prevent scorching.)
The Earth is never in equilibrium with space because there is virtually nothing in space to be in equilibrium with.Despite the Sun getting brighter and radioactivity in the Earths core the Earth has got colder over time no amount of insulation is going to slow that rate of cooling to space.Space has a density of almost nothing,there is no analogy that we can use to understand how the Earth and Space interact ,analogy is a poor argument.