The R. W. Wood Experiment

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.

steel greenhouse 1Figure 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 energy 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.

steel greenhouse 2Figure 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 0.03%,  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%? 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 energy 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 energy 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:

r w wood experiment 2Figure 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 energy 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

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.
One of the (many) problems with this experiment is the ambiguity of the statement above. Passed through how? If glass directly atop the salt, it would render the two boxes effectively identical. Lots and lots of problems with the experiment itself, but how does one discuss it without knowing what the exact setup was?

Mike M

I’m confused by the use of units of radiated flux. Conservation of energy would be measured as watts versus watts not watts/m^2 versus watts/m^2? Consider what happens as the size of the shell is increased to infinity?

Captain Dave

Figure 2 text probably should be
“the planet and the shell have nearly the same surface area”

Joe Public

But in Diag 2, the steel shell has a much greater surface area than the core.

Sarge

I hate to have to point this out, but your math on the 2nd diagram does not work.
Any possible shell around any possible sphere is going to have a surface area larger than the sphere it contains, so you cannot just split the “w/m^2” as seen on the (smaller) planet’s surface in half, and assume the same sum. The outer shell is larger, so the watts per METER SQUARED must be proportionately lower to get the total energy sums correct.
I may just be an engineer, not a scientist… but I can add.

Joe Public

i.e. if the core emits 235 W/m2 and the shell is say 10% greater area, its outward emission will be 211.5W/m2.

Silver Ralph

Ohh, Willis, you are inviting more posts from those who say that cooler bodies cannot transmit to warmer bodies (once again, yawn).
For those about to post such comments consider this:
There is a coolish radiator, emitting a small amount of infrared heat into a cold room. Then, another warmer radiator is placed in the room, a meter or so from the cooler radiator. Does the cooler radiator suddenly stop emitting infrared? And why would it do so?
(The cooler radiator does, of course, continue emitting just as before.)
.

Willis: This is a classic case of – “People’s ability to extrapolate FAR beyond any justifiable level” from one set of observations to another. I personally have NEVER considered Dr. Wood’s experiment to indicate ANYTHING OTHER than the conclusion that the “mechanism of action” of an actual GREENHOUSE is that of a “convective boundary”. It is interesting to note that YOUR theory of the importance of the virtually “unlimited” convection mechanisms in the REAL atmosphere leads to the “thunderstorm thermostat” conclusion.
Just as an aside on Dr. Wood’s GENERAL GENIUS is to point out his “rotating table liquid mercury lens” experiment, which allowed him to take photos of GALAXIES well before Eddington identified them in the 1930’s. (The disadvantage was that there was about a 5 degree (steradian) arc which could be covered by any apparatus set up anywhere in the world.) It cost MARKEDLY less than the Mt. Polomar telescope, however!
Max

Joe Sixpack

This is stupid on stilts:
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.
Yes. Just like a real greenhouse with atmosphere and everything.
That experiment doesn’t tell you much because you’re not capable of interpreting it correctly.

The experiment is done completely wrong anyway. To properly illustrate the “greenhouse” effect, Wood SHOULD have placed a brick inside each box with a temperature measurement device at the center of the brick. Shine the light until the center of both bricks stabilizes in temperature. Then turn off the lights expose both boxes to a clear cold sky, and record the rate at which the temperature drops in the center of each brick. Greenhouse effect is mainly a nighttime effect, not a daytime.

davidmhoffer

Problem #1
The troposphere ranges in thickness from 8km to 16km, Using 14km as an average, and a putative ghe of 33 degrees, that’s about 0.0023 degrees per meter. The apparatus used glass thermometers from 1906, suggesting an accuracy of perhaps 0.2 degrees at best. The boxes would have to have been about 1,000 meters on a side just to be big enough to produce enough ghe that it could even be measured by the apparatus used.

mikerossander

I am struggling with your thought-experiment because it seems to imply that you could raise the temperature of the planet to any arbitrarily high level by wrapping it in additional shells of steel. That is, if instead of wrapping your hypothetical planet in one steel shell, you used two nested shells, the outer shell would still stabilize at 235, the inner shell would stabilize at 470 and the planet would stabilize at 705. Keep adding shells until the inner-most steel melts. Yet the energy source, 235 W/m2, never changes. Am I understanding that correctly?
That would be a simple experiment to test in any good vacumn system. In fact, if true you’d expect the effect to be a significant source of difficulty in evacuated experiments. It’s been a long time since I did any experiments in a vacumn lab but I don’t remember having to make any such adjustments.

davidmhoffer

aaaaaaaaaaaaaaaaaaaagh!
100 meters on a side.

Michael Moon

Silver Ralph,
You just flunked your first hourly in Thermo. The cooler radiator would be warmed by the warmer radiator, and begin radiating more. If you think this would warm the warmer radiator, then you will fail all your hourlies and never get through school.
Wilis, Joe Public has it exactly right. If you want to know what happens to the flux from a cooler source when it hits a warmer source, the answer is exactly nothing. It is not absorbed, but immediately re-emitted, transferring NO heat.
All these analogies are amusing but ignore Second Law.

Bill Thomson

To those who think the math in Figure 2 does not work: Please read the caption under Figure 2 as well as reading the picture.

It may be worth a revisit to Perpetuum Mobile WUWT Eschenbach Jan 19, 2012.
Rp = radius of Planet
Rs = radius of Shell.
Tsi, Tso = temp of Shell, inside and outside
Tp = temp of Planet.
By your setup, Rs > Re.
Tsi = Tso, since both are radiating the same energy flux.
By this, I conclude that the thermal conductivity of shell is very high and/or the shell is very thin.
Tp > Tsi (assuming black body).
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.
You would agree, wouldn’t you that if the planet had no radioactive core, then Tp would have to equal Tsi. Yet the addition of a tiny radioactive energy source is now able to raise Tp to a level where it has twice the radiant flux as Tsi.
Doesn’t Tsi have to be a function of (Rs/Rp)^1/4?

mikerossander

To Mike M, Joe Public and Sarge – Energy is conserved and yes, you could reframe the entire example in watts instead of watts per square meter. It is irrelevant to the thought-experiment, however, because you can make the shell arbitrarily small as long as it is infinitesimally separated from the planet. The effect would be identical if the shell were one millimeter out rather than the severely exaggerated separation shown in Figure 2. And while that one-millimeter increase in radius would increase the surface area very slightly, it’s WAY below the rounding error of the system.

Massimo PORZIO

@Silver Ralph
Your example is not pertinent, the two radiators are two different sources of energy in the room. Very different than asserts that the exterior sphere warmed by the inner one can warm more that last.
@Mike M
“Consider what happens as the size of the shell is increased to infinity?”
And consider what happens as the size of the shell is the one of the inner sphere plus just an atomic layer?
Uhmmm… Still skeptic

KevinM

A few typos, but correct to my understanding of heat transfer.
The difference in surface area of the actual atmosphere and the earth below it is not very big, and accounting for it would not change the argument. Integral over the surfaces of both would be equal number of Watts regardless of size.
So Co2 AGW is real. The point to argue is that it is benign, and probably swamped by larger natural variability over short and long time scales.

Bryan

The Greenhouse Effect goes missing on certain nights at Penn State University.
This is an interesting paper especially as it comes from a source with no “spin” on the AGW debate.
The way I read the paper is it gives strong support for the conclusions of the famous Woods experiment.
Basically the project was to find if it made any sense to add Infra Red absorbers to polyethylene plastic for use in agricultural plastic greenhouses.
Polyethylene is IR transparent like the Rocksalt used in Woods Experiment.
The addition of IR absorbers to the plastic made it equivalent to “glass”
The results of the study show that( Page2 )
…”IR blocking films may occasionally raise night temperatures” (by less than 1.5C) “the trend does not seem to be consistent over time”
Conclusion is that it makes almost no difference whether the material radiates or not.
http://www.hort.cornell.edu/hightunnel/about/research/general/penn_state_plastic_study.pdf

MarkB

There’s no sense trying to engage the sky dragon nutters on an intellectual basis.

KevinM

The steel shell analogy does break down when you consider the atmosphere as a fluid instead of a solid.
Hot and cold air units circulate.Hot moves farther from the surface. The farther a unit of atmosphere gets from the surface, the less the earth obstructs its radiation into space. – just like your hand blocks more or less light as you move it closer or farther from your eyes.
Net result: Atmosphere is less reflective than a steel shell, and also because it circulates.

BigDon

It’s been 30 or so years since I studied heat transfer in engineering school, but I seem to recall radiation heat transfer being proportional to delta-T to the fourth power, hence the shell would be radiating much (much!) more IR out to (nearly) absolute zero space than to the warmer planet surface. And as for the radiators-in-a-room example, I do believe that there indeed would be no IR radiated from the cooler unit to the warmer unit, but just in that particular direction — exactly at the points normal to the warmer radiator. I believe it would continue to emit IR in all other directions, however.

eo

The biosphere is still in existing. I think the University of Arizona is operating it. Why not just carry out an experiment on the effect of various carbon dioxide concentration on the temperature inside the sphere. The experiment could be a super simplification but at least there will be some empirical data rather than all the computer models and assumptions.

TimTheToolMan

Your diagram appears to be missing a glass plate that covered both enclosures to stop IR_in being a confounding factor. The relevent quote is
“In order to eliminate this action the sunlight was first passed through a glass plate.”

Don K

I’m probably wrong, but wouldn’t In order to eliminate this action the sunlight was first passed through a glass plate. imply a second glass plate above both the glass and rock salt plates in order to condition in incoming spectrum to be the same for both boxes?
Please pardon me if that’s silly or stupid. I’m too old and fuzzy minded to work through stuff like this quickly. I think I may have been smarter/quicker about 5 decades ago. Or at least I thought I was smarter/quicker.

seine

Somehow I can’t see 235w radiating energy into 236w.

Noud Vermeulen

This: http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html is in my opinion a much simpler, more elegant explanation.
Noud

Roger Clague

Mike M says:
February 6, 2013 at 12:49 pm
“Conservation of energy would be measured as watts versus watts ”
Watts are a unit of POWER, that is joules/ second.
ENERGY is measured in joules. There is no law of conservation of power.
The earth get energy for only part of the day. At night half of the surface only loses energy.
Introducing the time dimension, that is averaging over 24 hours, as in this model and Trenberth’s, is not realistic.

Willis Eschenbach

Mike M says:
February 6, 2013 at 12:49 pm

I’m confused by the use of units of radiated flux. Conservation of energy would be measured as watts versus watts not watts/m^2 versus watts/m^2?

I tried to head this incorrect argument off at the pass, but I was not emphatic enough. What I said was:

… 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.)

It is an acceptable simplification for a first-order analysis such as this one because the exterior surface are of the shell is approximately equal to the surface of the sphere ( less than half a percent difference).

Consider what happens as the size of the shell is increased to infinity?

If the shell size increases to infinity, you’re no longer in this thought experiment, but in a very different one.
w.

mkelly

Silver Ralph says:
February 6, 2013 at 12:56 pm
There is a coolish radiator, emitting a small amount of infrared heat into a cold room. Then, another warmer radiator is placed in the room, a meter or so from the cooler radiator. Does the cooler radiator suddenly stop emitting infrared? And why would it do so?
Silver Ralph which radiator or combination of radiators determines the maximum temperature the room can attain? And why?
From a radiative heat transfer stand point picture two in wrong. As soon as Tsphere and Tshell are the same W/m^2 goes to zero. q/a= e SB (T1^4-T2^4) The surface area of the shell dictates that heat will only go in one direction from sphere to inner shell to outer shell to space.

Miket

Sarge, Joe Public,
If you read the whole piece, you will see that Willis did address that issue.

Willis Eschenbach

I’ve added the following update to the head post:

[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%?]

w.

Willis Eschenbach

TimTheToolMan says:
February 6, 2013 at 1:50 pm

Your diagram appears to be missing a glass plate that covered both enclosures to stop IR_in being a confounding factor. The relevent quote is
“In order to eliminate this action the sunlight was first passed through a glass plate.”

True, thanks, Tim. I’ve updated the drawing.
w.

Tamara

One atom of the planet transfers one unit of energy to one atom of the shell. The shell can radiate in any direction. How many units of energy does it have available to transfer to space and to the shell?

Willis Eschenbach

mikerossander says:
February 6, 2013 at 1:07 pm

I am struggling with your thought-experiment because it seems to imply that you could raise the temperature of the planet to any arbitrarily high level by wrapping it in additional shells of steel.

Indeed, you can, you just need vacuum. Crazy but true, do the math. For each additional layer, you get an additional multiplication, i.e. 1 shell doubles the watts, 2 shells triples the watts, and so on. Of course, this does not double or triple the temperature. If you double the watts, the temperature goes up by only 20%, and tripling the watts only raises it by 30%.
w.

Allen B. Eltor

The number of errors in just the fundamentals associated with this thread, make it beyond redemption into anything associated with science.
[perhaps you would like to point these errors for the benefit of all. thanks . . mod]

michael hammer

The example and analysis of the steel shell given above is completely correct. If one was to add a second shell around the first the effect would be still greater. In fact, there is a form of commercial insulation made up of many layers of aluminium foil stacked one on top of the other which is highly effective and works on exactly this principle. It is also a commonly performed science experiment at secondary school level.
The claim that if there was sufficient conduction (or convection) so that the surface and steel shell were at the same temperature the effect would disappear is also correct. However in that case the radiation from surface to shell and from shell to surface would be the same so that radiation would no longer play a part – in effect the surface and the shell would become all the one body.
The thing is that in our planetary system the condition in the above paragraph is NOT met. The equivalent of the steel shell is the tropopause and it is NOT at the same temperature as the surface, it is quite a bit colder. In fact there is a non zero lapse rate between the surface and the tropopause whcih maintains that temperature difference. Why do I say the tropopause is the equivalent of the steel shell? Because this is the effective top of the CO2 and H2O columns and it is these gases that are capable of absorbing and radiating energy (oxygen and nitrogen do not absorb or radiate in the thermal infrared range of wavelengths).
The “steel shell” is not in fact opaque at all wavelengths, only at the green house gas wavelengths. Without green house gases there would be no shell and the surface would radiate freely to space. If green house gases absorbed all wavelengths the shell would be the equivalent of a steel shell and the surface temperature would be raised by in effect the integrated lapse rate from surface to shell. That explains why the temperature on Venus is so high. The extremely high concentration of GHG has caused so much line broadening that the GHG intercept virtually all long wave radiation and the atmosphere is so thick that the integrated lapse rate is huge.
With regard to the box experiment, this is complicated by the fact that a box as described can lose energy by radiation, convection or conduction whereas a planet surrounded by vacuum can only lose energy by radiation. The popular explanation is that the covered box works by blocking convection but consider the following. If convection occurs from the hot surface of the box why would it not also occur from the hot surface of the glass? Further if we argue that the important effect is interception of long wave radiation we again have a problem. If the glass is an effective absorber of long wave radiation it is also an effective emitter of long wave radiation so why would the glass if at the same temperature not emit as much long wave radiation as it absorbs? In that case it would make no difference whether or not the glass absorbed long wave radiation or not. Thus if the glass and box were at the same temperature, conduction, convection and radiation would be unchanged and there should be no difference in temperature.
The only answer to this apparent paradox (since clearly the glass covering does work) is that the glass is colder than the surface of the box – like the steel shell analogy. In fact what I think will be found is that the dominant energy transfer mechanism between box and glass is convection and that this is far from infinite so that there is a substantial temperature difference between box and glass, which is much closer to the steel shell example than one might think.
There are many inferences from this that should be testable but this post is already long and I think its appropriate to let others comment on the above first
cheers
Mike Hammer

Richard G

I recently showed a film during middle school science class entitled “Heat”. The film makers conducted an experiment as follows.
4 identically sized double walled flasks: one no vacuum with silver coating, one no vacuum no silvering, one with vacuum with silvering, one with vacuum no silvering. Each flask equipped with a digital temperature sensor and insulated stopper.
Equal measures of water heated to the same temperature were introduced into each flask. Heat loss was measured over time and compared between the 4 flasks.
Result: Heat loss from the two vacuum flasks was markedly less than the two non-vacuum flasks. (I don’t recall any of the numbers). Heat loss from the two silvered flasks compared to the counterpart non-silvered flasks was minimal.
Their conclusion: In this experiment the predominant mechanism for heat loss was convection and conduction. Radiative heat loss accounted for a tiny fraction of total heat loss.
The term Green House Gas is a misnomer. Greenhouses work by stopping convective heat loss with a mechanical barrier. The equivalent barrier to convection in the atmosphere is called the tropopause. Greenhouse operators introduce enriched CO2 atmospheres into greenhouses to enhance growth, not to reduce radiative heat loss. Does this mean that CO2 does not change the radiative balance? No, radiative loss only becomes dominant at and above the tropopause which is above 80% of the worlds atmosphere. In other words above 80% of the atmosphere’s CO2. So I guess my question is: Do the radiative model calculation outputs need to be reduced by 80% to bring them into conformance with the real world?
Just asking.

hmccard

Over at Climate, Etc., Pekka Pirilä says (February 5, 2013 at 5:27 am):
“It’s not totally clear what would happen for the atmosphere in total absence of all radiative gases, i.e. with exactly zero emissivity/absorptivity. I have been arguing for the mostly isothermal atmosphere but others have argued that the diurnal and latitudinal variability could still maintain circulation over an altitude range comparable to the present troposphere. As I haven’t heard of any credible analysis of this case i consider the case open.”

What about people who tried the standard scientific procedure of reproducible results? Apparently more “nutters”.
http://principia-scientific.org/supportnews/latest-news/34-the-famous-wood-s-experiment-fully-explained.html

michael hammer

Further to my previous post, from what I said there one would predict that two layers of glass over the box, separated from each other, would be more effective than one just as two concentric steel shells would be more effective than 1. In fact this is exactly the case and some glass houses are now made with two transparent layers (usually thin plastic not glass) for exactly that reason. Especially effective at retaining heat over night.

Tamara

I am struggling with this. I am not a Sky dragon, and I accept the GH theory. But your diagram implies that there is no reason for heat energy to flow from a high energy state to a lower energy state. For instance, when a CO2 molecule is struck by radiation from the sun, it assumes a higher energy state than its surroundings. If, somehow, it were the last low energy CO2 molecule in the atmosphere, it would not emit the radiation. A blackbody the size of the universe, would emit no radiation, because there would be no lower energy state for the energy to flow to.

Ray

Now, let’s say that instead of a steel shell or atmosphere you envelop the radiating planet with water. The water has a much greater heat capacity than air or vacuum…

davidmhoffer

I’ve updated the drawing.
>>>>>>>>>>>>>
Which makes the problem worse. How far above the two boxes is the extra layer of glass? Glass absorbs LW, conducts well, and also radiates. So, upward LW from the rock salt box hits the glass, is absorbed, conducted, and re-radiated toward both the rock salt covered box and the glass covered box. Then you’ve got the glass in the glass covered box heating up by conduction, causing it to radiate, and the LW that it radiates being absorbed by the higher level of glass which then conducts and re-radiates to both boxes as well. The LW radiated from the upper layer of glass heats both boxes, one directly and one by heating the glass shield which then heats the box below by both radiance and conduction.
All of which happens at an order of magnitude in which the direct effects are easily measured with a glass thermometer but which isn’t even close when comes to the accuracy required to measure LW absorption and re-radiance.

Willis Eschenbach

Michael Moon says:
February 6, 2013 at 1:11 pm

Silver Ralph,

You just flunked your first hourly in Thermo. The cooler radiator would be warmed by the warmer radiator, and begin radiating more. If you think this would warm the warmer radiator, then you will fail all your hourlies and never get through school.

Wilis, Joe Public has it exactly right. If you want to know what happens to the flux from a cooler source when it hits a warmer source, the answer is exactly nothing. It is not absorbed, but immediately re-emitted, transferring NO heat.
All these analogies are amusing but ignore Second Law.

So you agree that it is re-emitted. To be re-emitted, it had to be absorbed. You just say it happens really fast.
Actually, once energy is absorbed, there’s no way to distinguish it from any other energy, so there’s no way to tell when that particular energy was re-emitted.
Finally, you are correct that it transfers no heat, but not for the reason you think. Heat is NET energy transfer, which is different from radiative energy transfer from one body to another. It is the latter we are discussing, energy transfer.
Radiative energy doesn’t care where it is coming from or going go, or what the temperature is on either end. If I light a candle on the earth during the day, the sun ends up warmer than it would be if I didn’t light the candle. Of course the reverse is true as well, the candle ends up warmer than if there were no sun. Since NET heat flow is from the sun to the candle, no thermodynamic laws are broken … but that doesn’t mean that the light from the candle is not absorbed by the sun. It is definitely absorbed, and the sun ends up warmer because of that radiation.
Physics. Don’t leave home without it.
w.

Konrad

W – “Seems to me like with a few small changes it could indeed be a valid test, however.”
Been there, done that –
http://i49.tinypic.com/34hcoqd.jpg
Two insulated boxes with double glazed LDPE film windows. Circulation fans and thermometers shielded from incoming light and out going IR. Matt black aluminium target plates. Halogen lights. One box filled with CO2 the other air. Which box heats up faster? Which box cools slower when the lights are switched off?
Answer – Over a 20 degree temperature change there is no measurable difference between the boxes using an electronic dual probe thermometer with 0.1 degree resolution. The reason is that while CO2 can intercept some outgoing IR from the target plate, it can also radiate energy it has acquired conductively from the target plate.
Willis you were almost there when you wrote on another thread –
W – “Some gases most assuredly absorb and emit infra-red, some gases don’t. In a mixture of the two, those that do absorb infrared immediately (nanoseconds) pass that energy on via collisions to the other gases that do not absorb or emit infrared and thus warm the mass of air. The reverse is true when they emit infrared, within nanoseconds they absorb energy from the other gases and cool the mass of air.”
However in modelling the role of radiative gases in the atmosphere the shell game is the wrong approach. This is the failed model used by the AGW pseudo scientists. It involves most of the “Do Nots” of atmospheric modelling. Let’s review –
1. Do not model the “earth” as a combined land/ocean/gas “thingy”
2. Do not model the atmosphere as a single body or layer
3. Do not model the sun as a ¼ power constant source without diurnal cycle
4. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
5. Do not model a static atmosphere without moving gases
6. Do not model a moving atmosphere without Gravity
7. Do not model the surface as a combined land/ocean “thingy”
Avoid the “Do Nots” and you will fine the true role of radiative gases in the atmosphere. They cool at all concentrations above 0.0ppm. Without radiative gases full convective circulation below the tropopause will stall and the atmosphere will heat. Build this experiment to find out why convective circulation is critical to atmospheric temperatures and why radiative gases are critical to convective circulation –
 http://i48.tinypic.com/124fry8.jpg http://tinypic.com/r/zmghtu/6  http://i49.tinypic.com/2a106x.jpg
Where are almost all the radiative gases in the atmosphere? Below the tropopause.
Where does almost all the vertical convective circulation occur? Below the tropopause.
Adding radiative gases to the atmosphere will not reduce the radiative cooling ability of the atmosphere. Play the shell game with the AGW pseudo scientists and you will always get the wrong answer.
And for the record – No, I am not one of the “slayers”. I do accept that the atmosphere radiates IR back to the surface. There is no easy out there.

Curt

mikerossander says:
February 6, 2013 at 1:07 pm
I am struggling with your thought-experiment because it seems to imply that you could raise the temperature of the planet to any arbitrarily high level by wrapping it in additional shells of steel.
**************************************
It implies it because it is true. Examine the construction of a high-end vacumm flask. Of course, the vacuum between the outside and the inside is to eliminate the gas conductive/convective heat transfer mechanism. But the good flasks also have multiple thin silverized plastic sheets that act as radiation barriers (combination of reflective and absorbed re-admitted) with non-thermally-conductive sheets in between. This is more difficult and expensive to manufacture than a single radiation barrier, but it is much more effective at reducing heat transfer.
If there were a heat source in the inner chamber (e.g. an electrical resistance heater) adding energy at a constant rate, each additional layer of radiative barrier would result in a higher temperature in the inner chamber.

joletaxi

bonsoir Mr Eschenbach
j’ai un problème avec votre figure 2
le “core” émet 235 W correspondant à une t* d’mission, disons TA et chauffe la sphère d’acier dont la t° s’établit (oublions les surfaces) à une t° correspondant à cette énergie reçue,soit TA
cette sphère émet à la fois vers l’extérieur et vers l’intérieur,(selon vous) mais de ce fait la surface d’émission à ainsi doublé!(surface extérieure plus surface intérieure)
et comme l’énergie reçue reste 235 w, et que pour rester en équilibre, la sphère d’acier doit émettre 235 w, ,on voit immédiatement que celle-ci n’a pu réémettre vers l’intérieur
en effet , dans votre configuration, il n’y a aucune raison que la t° de l’intérieur de la sphère soit différente de la t° de l’extérieur.
or pour rester en équilibre, il nous faut TA à la surface extérieure, et aussi, TA à l’intérieur, et dont la source doit émettre 2 fois plus d’énergie.
mais nous ne disposons que de 235 w
Maintenant, imaginons, que la surface de la sphère épouse parfaitement le core, et que la conduction soit parfaite,
croyez vous que sous la pellicule d’acier la t* du core va doubler?
Mr. Spencer nous avait déjà bien occupé avec ce petit paradoxe,(Yes Virginia…) dont mes amis et moi, n’avons jamais pu trouver une formulation valable(nous en discutons fréquemment sur le site suivant
http://www.skyfall.fr/
Quoi qu’il en soit, merci pour vos récits , je suis un voileux

Konrad

W – “Seems to me like with a few small changes it could indeed be a valid test, however.”
Been there, done that –
http://i49.tinypic.com/34hcoqd.jpg
Two insulated boxes with double glazed LDPE film windows. Circulation fans and thermometers shielded from incoming light and out going IR. Matt black aluminium target plates. Halogen lights. One box filled with CO2 the other air. Which box heats up faster? Which box cools slower when the lights are switched off?
Answer – Over a 20 degree temperature change there is no measurable difference between the boxes using an electronic dual probe thermometer with 0.1 degree resolution. The reason is that while CO2 can intercept some outgoing IR from the target plate, it can also radiate energy it has acquired conductively from the target plate.
Willis you were almost there when you wrote on another thread –
W – “Some gases most assuredly absorb and emit infra-red, some gases don’t. In a mixture of the two, those that do absorb infrared immediately (nanoseconds) pass that energy on via collisions to the other gases that do not absorb or emit infrared and thus warm the mass of air. The reverse is true when they emit infrared, within nanoseconds they absorb energy from the other gases and cool the mass of air.”
However in modelling the role of radiative gases in the atmosphere the shell game is the wrong approach. This is the failed model used by the AGW pseudo scientists. It involves most of the “Do Nots” of atmospheric modelling. Let’s review –
1. Do not model the “earth” as a combined land/ocean/gas “thingy”
2. Do not model the atmosphere as a single body or layer
3. Do not model the sun as a ¼ power constant source without diurnal cycle
4. Do not model conductive flux to and from the surface and atmosphere based on surface Tav
5. Do not model a static atmosphere without moving gases
6. Do not model a moving atmosphere without Gravity
7. Do not model the surface as a combined land/ocean “thingy”
Avoid the “Do Nots” and you will fine the true role of radiative gases in the atmosphere. They cool at all concentrations above 0.0ppm. Without radiative gases full convective circulation below the tropopause will stall and the atmosphere will heat. Build this experiment to find out why convective circulation is critical to atmospheric temperatures and why radiative gases are critical to convective circulation –
 http://i48.tinypic.com/124fry8.jpg http://tinypic.com/r/zmghtu/6  http://i49.tinypic.com/2a106x.jpg
Where are almost all the radiative gases in the atmosphere? Below the tropopause.
Where does almost all the vertical convective circulation occur? Below the tropopause.
Adding radiative gases to the atmosphere will not reduce the radiative cooling ability of the atmosphere. Play the shell game with the AGW pseudo scientists and you will always get the wrong answer.
And for the record – No, I am not one of the “slayers”. There is no easy out there.