# 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.

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 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.

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 0.03%,  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:

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 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
February 6, 2013 12:45 pm

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
February 6, 2013 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? Consider what happens as the size of the shell is increased to infinity?

Captain Dave
February 6, 2013 12:49 pm

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

Joe Public
February 6, 2013 12:50 pm

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

Sarge
February 6, 2013 12:53 pm

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
February 6, 2013 12:54 pm

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
February 6, 2013 12:56 pm

Ohh, Willis, you are inviting more posts from those who say that cooler bodies cannot transmit to warmer bodies (once again, yawn).
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.)
.

February 6, 2013 1:00 pm

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
February 6, 2013 1:01 pm

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.

February 6, 2013 1:02 pm

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
February 6, 2013 1:06 pm

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
February 6, 2013 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. 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
February 6, 2013 1:07 pm

aaaaaaaaaaaaaaaaaaaagh!
100 meters on a side.

Michael Moon
February 6, 2013 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.

Bill Thomson
February 6, 2013 1:13 pm

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.

February 6, 2013 1:15 pm

It may be worth a revisit to Perpetuum Mobile WUWT Eschenbach Jan 19, 2012.
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
February 6, 2013 1:15 pm

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
February 6, 2013 1:19 pm

@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
February 6, 2013 1:22 pm

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.

February 6, 2013 1:27 pm

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.

MarkB
February 6, 2013 1:33 pm

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

KevinM
February 6, 2013 1:35 pm

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
February 6, 2013 1:35 pm

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
February 6, 2013 1:37 pm

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
February 6, 2013 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.”

Don K
February 6, 2013 1:50 pm

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
February 6, 2013 1:52 pm

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

Noud Vermeulen
February 6, 2013 1:53 pm

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

Roger Clague
February 6, 2013 1:55 pm

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.

mkelly
February 6, 2013 1:57 pm

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
February 6, 2013 1:57 pm

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

Tamara
February 6, 2013 2:06 pm

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?

Allen B. Eltor
February 6, 2013 2:14 pm

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
February 6, 2013 2:17 pm

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
February 6, 2013 2:18 pm

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?

hmccard
February 6, 2013 2:22 pm

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.”

February 6, 2013 2:24 pm

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
February 6, 2013 2:26 pm

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
February 6, 2013 2:27 pm

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
February 6, 2013 2:28 pm

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
February 6, 2013 2:32 pm

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.

February 6, 2013 2:34 pm

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
February 6, 2013 2:34 pm

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
February 6, 2013 2:36 pm

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

February 6, 2013 2:36 pm

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.

KevinM
February 6, 2013 2:40 pm

Roger Clague says
“Watts are a unit of POWER, that is joules/ second.
ENERGY is measured in joules. There is no law of conservation of power.”
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.
Please remove the all caps. Do you talk to people that way in person?

davidmhoffer
February 6, 2013 2:40 pm

All these analogies are amusing but ignore Second Law.
>>>>>>>>>>>>>>>
You can’t pick and choose which laws of physics to use and when. They all exist at the same time. If the 2nd Law operates as you suggest, then it falsifies SB Law. You can have both, or neither, but you can’t have one and not the other.

Greg House
February 6, 2013 2:41 pm

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 guess I am the “commenter on another thread”. I do not “hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist”. Nor have I ever heard anyone claiming that. Nor did professor Wood hold that.
The Wood experiment demonstrates that “trapped/back radiation” has zero or negligible effect on the temperature of the source.
And this demonstrates that the underlying mechanism of the “greenhouse effect” as presented by the IPCC (http://www.ipcc.ch/publications_and_data/ar4/wg1/en/faq-1-3.html) does not work at all or is negligible.

bwdave
February 6, 2013 2:41 pm

With the same internal heat source, the outgoing W/m^2 simply decrease with the increased surface area, but the total heat escaping will remain the same as that supplied by the internal source. Everything else is irrelevant garbage.

February 6, 2013 2:42 pm

I’ve started using a IR non-contact thermometer to measure the temp of the sky on clear days, and so far on a 35F day, and a 28F day, it’s been lower than the temp the thermometer will read -40F. On a ~50F day, it read ~-35F.
My thermometer is measuring “back radiation”. Back radiation does add energy to anything it shines on, but the rate of transfer is very low, and while my 50F black driveway is getting radiated on, being at least 90F warmer, the rate of energy it’s radiating into space is much much higher than the other way around.
I plan to start logging the sky’s temp and logging it and air temp and humidity. I just have to wait for clear days, which I don’t get very often in NE Ohio.
From my work looking at the surface temperature record and night time cooling, I expect to see that humidity controls surface temps.

lou
February 6, 2013 2:45 pm

If the outer shell is radiating at 2x the flux of the inner core its temperature would have to be much higher than the inner core. How is that possible?

February 6, 2013 2:49 pm

MiCro says:
February 6, 2013 at 2:42 pm

my 50F back driveway

Is suppose to be “my 50F black driveway”
[Fixed. -w.]

Trond A
February 6, 2013 2:51 pm

Hi Willis!
Will much more of the energy be transferred as convection? Maybe, but this can be tested precisely with the example of the Wood experiment, the one with the rock-salt plate. Because the rock-salt plate is transparent to both short and long wave radiation, both the start value radiation of 65 W in my example, and the added long wave back radiation up to 390 W will enter this greenhouse box and none of the energy will slip away due to convection because no convection is allowed. This little box should experience a full greenhouse effect and reach a temperature of more than 300 C. Well, as Wood showed, it didn’t. None greenhouse effect of that caliber. And even in the desert the atmosphere has a cooling effect, and for the most equals out the excessive conditions. Both ways. There is probably a certain greenhouse effect contributing to the convection, and greenhouse gases most probably give the atmosphere a certain kind of heat capacity that speeds up the energy flow and equals out the temperature, but a heavy back radiation? Hmm..

Kev-in-Uk
February 6, 2013 2:52 pm

The basic physics is of course correct – but this hypothetical situation is not directly applicable to a planet warmed from an external source. For a start, the external radiation is both reflected and absorbed by the shell, and in a varying manner and in the case of a poorly conductive/convective and ‘reactive’ (or chaotic, if you prefer) atmosphere – the GHG properties are not constant. Once you have clouds, heat retaining liquid, surface solids, water vapour, aerosols, a ‘living’ Biosphere, etc, etc, etc – the situation is simply far too complicated to be realistically represented. Thus, when you realise just how complicated that it actually becomes in real life (as per our Earth) – it makes it even more highly suspicious to point to a SINGLE trace gas as a primary driver of GHG effect changes!
Further, when you then consider the actual carbon cycle and the natural CO2 present within the biosphere as a whole, and the potential natural variation of the ‘position/placement’ of CO2 within that biosphere (sinks and emitters!) – that adds even further complication.
Now, if someone wants to unravel that semi-chaotic non-linear mess and offer proof that CO2 based AGW is the proven real culprit, I’d be glad to hear it – as would millions of others!
Just sayin………..

lou
February 6, 2013 2:54 pm

What I meant was the outer shell is radiating over 2x the area of the inner shell at the same power so its temperature must be higher…

Editor
February 6, 2013 2:57 pm

Vacuum between shell and planet = planet gets a lot hotter.
Perfect conduction between shell and planet = planet doesn’t get any hotter.
Imperfect conduction between shell and planet = AGW.
As KevinM has said above, AGW is real. The main argument is about how large or small the effect is.

Greg House
February 6, 2013 2:58 pm

Guest Post by Willis Eschenbach: ” let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). … Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, …”
=============================================================
“Imagine”, I see.
I guess, it has never been proven experimentally that A warms B and then B warms A back, right? OK, this is a product of imagination, a fiction, and everyone has right to right a science-fictional story, no problem with that. But the readers need to be told clearly that this story is fictional, just to avoid confusion.

February 6, 2013 2:59 pm

“The Wood experiment demonstrates that “trapped/back radiation” has zero or negligible effect on the temperature of the source.”
Unfortunately it doesnt test how the “greenhouse effect actually works and can never test that.”
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.
back radiation from the silver lining of a thermos does not warm the coffee. It slows the rate at whch the coffee cools and keeps it warmer than it would be otherwise. If that radiation shield is “leaky” the coffee cools more rapidily.
Simple terms: Woods doesnt test the greenhouse hypothesis.
That hypothesis is.
1. Adding C02 RAISES the level of the ERL. woods experiment and any closed container experiment cannot test this.
2. Raising The ERL cause the source to cool less rapidly. The surface is not warmed by back radiation which is better understood as an effect of GHGs rather than the cause of warming.
back radiation in a thermos doesnt raise the temperature of the coffee, it slows the rate of energy loss.
Woods tested some other theory, some strawman version.

Trond A
February 6, 2013 3:01 pm

Correction: Because the rock-salt plate is transparent to both short and long wave radiation, both the start value radiation of 65 W in my example, and the added long wave back radiation up to 390 W will enter this greenhouse box and none of the energy will slip away due to convection because no convection is allowed.
Here it should be not 65 W but 1094 W, and instead of 390 W there should be 6562 W according to the example of the tropical desert. Sorry.

TomR,Worc,MA
February 6, 2013 3:02 pm

I think I am headed back to “Ilikebacon.com”, this thread makes my eyes bleed.
TR

davidmhoffer
February 6, 2013 3:03 pm

The Wood experiment demonstrates that “trapped/back radiation” has zero or negligible effect on the temperature of the source.
>>>>>>>>>>>>>>>
Yes, the joules of energy exist, they just don’t do anything. Let’s just change the definition of joule to suit our belief system and prove it using an apparatus that can’t possibly measure with enough accuracy to support such a conclusion. And let’s further propose that conclusion based on an experiment in 1906 in opposition to the findings of Wien, Planck, Einstein, Bohr, and Milliken’s Nobel prizes in 1911, 1918, 1921, 1922 and 1923 respectively.

Editor
February 6, 2013 3:03 pm

I agree with mkelly about figure 2 Willis. I can see what you were trying to say there but the arrows and amounts are confusing as presented.
I think it could be reworked to just key on what you were trying to say. Adding the shell makes the planet warmer without affecting the enery balance. At the steady state, planet produces 235, shell emits 235, but the produced 235 bounces around between the shell and the planet effectively warming it.
Still thinking about the Wood experiment.

Allen B. Eltor
February 6, 2013 3:09 pm

Don K says:
February 6, 2013 at 1:50 pm
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.
<<<<<<<<<<<<<<<<<
Woods set out to debunk the claim that specific gases, were slowing down the escape of a class of radiation. That radiation class was infrared light.
Woods let infrared in one box,
and stopped infrared from getting in another:
proving it wasn't the amount of infrared-class light getting in, OR out,
that assigned temperature in any measureable way in atmospheric air.
Claims of not being able to "see the point" are of those who don't WANT to see the point.
They have what's called a 'belief' system.
They BELIEVE in the effect, no matter how many experiments show them, it's erroneous fantasy.
Woods proved the claim of infrared-resonant gases, being in a fundamental way responsible for temperature assignment in atmospheric gas mixture, is utter falsehood.
When two boxes were put out in sun, and identical gases inside: atmospheric mix wherever he was –
the box which didn't let infrared light in, warmed identically in time,
with the box that let it in.
If the atmospheric gas mix was somehow holding infrared heat, creating detectable temperature readings, the box that let the infrared light in,
*would have warmed up faster. *
It didn't.
The one that had the infrared in, blocked,
*would have had to have waited for portions of the visible light coming in, to convert to heat, before temperature climb appoximated the other's.*
Anyone who says "they can't see what that proved," is simply trying to hang on to their self-prescribed "dignity"
and popularity.
Because we've got enough instruments from optical telescopes to infrared telescopes to
check on the stories about the magic gas.
And it's a bullshoot story.
Woods knew it,
anyone who sees Woods' experiment, who's honest, knows it,
and the only people who even still cling to it are those who staked their reputations as public figures on it's being real.
Nobody believes in that crap except what are called 'true believers.'

Editor
February 6, 2013 3:32 pm

Greg House – “I guess, it has never been proven experimentally that A warms B and then B warms A back, right? OK, this is a product of imagination, a fiction, and everyone has right to right a science-fictional story, no problem with that. But the readers need to be told clearly that this story is fictional, just to avoid confusion.“.
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.
To my mind, it would be a very good idea for someone to actually do the experiment, properly, and (if Roy Spencer is correct) lay to rest the argument that AGW violates the 2nd law of thermodynamics.

Kev-in-Uk
February 6, 2013 3:33 pm

Mike Jonas says:
February 6, 2013 at 2:57 pm
I don’t disagree in principle but as I observed earlier, in real life, we have a planet warmed from external radiation (with all the other associated effects) and also I take exception to using the term AGW instead of GHE (greenhouse effect) as that term is a false premise.

February 6, 2013 3:36 pm

The top pain of glass was to make sure the IR that was reaching the boxes were equal in wavelength. Then the transfer was measured as longer wavelengths deduced by the temperature reading. Is this right?

Bob Roberts
February 6, 2013 3:43 pm

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.

Richard G
February 6, 2013 3:46 pm

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.

February 6, 2013 3:47 pm

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.

Mike M
February 6, 2013 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.
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…

Greg House
February 6, 2013 3:52 pm

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

Roger Clague
February 6, 2013 3:54 pm

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.

KevinK
February 6, 2013 4:04 pm

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.

KevinK
February 6, 2013 4:13 pm

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.

lou
February 6, 2013 4:18 pm

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.

Robert of Ottawa
February 6, 2013 4:20 pm

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.

Steve from Rockwood
February 6, 2013 4:23 pm

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?

Roger Clague
February 6, 2013 4:24 pm

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.

February 6, 2013 4:28 pm

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.

cd
February 6, 2013 4:33 pm

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.

Greg House
February 6, 2013 4:37 pm

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?

michael hammer
February 6, 2013 4:41 pm

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

Gary Pearse
February 6, 2013 4:45 pm

“..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?

February 6, 2013 4:49 pm

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

KevinK
February 6, 2013 4:50 pm

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.

Michael Moon
February 6, 2013 4:52 pm

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?

davidmhoffer
February 6, 2013 4:56 pm

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.

davidmhoffer
February 6, 2013 5:00 pm

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.

Anthony Zeeman
February 6, 2013 5:04 pm

What a silly example. Prior to the metal shell being put around the sphere, the temperature of the sphere would be just sufficient to radiate the wattage of the internal heat source to the vacuum of space. If the temperature was too low, the sphere would continue to heat until the temperature was just high enough to radiate away the energy. Adding the additional thermal resistance of a steel shell in series with the thermal resistance to the vacuum of space will actually reduce the temperature of the sphere as the emitting area is increased. Leaving a space, presumably containing a vacuum between the sphere and the shell would do nothing since this would have the same thermal resistance as from the sphere to space.
This would be the same as adding another heatsink on top of the existing heatsink on your cpu. The thermal resistance between the heatsink and the atmosphere is so high that adding additional metal actually helps cooling by increasing the surface area for emitting heat while adding only very slightly to the series thermal resistance between the chip and atmosphere.
For another example, adding a 1″ length of 20′ diameter pipe to a mile long length of 3/4″ hose will not appreciably increase the flow through the hose.
Also, note that your soup doesn’t get hotter when you put it into a Thermos in spite of the reflected IR radiation from the silvered sides of the Dewar flask.

davidmhoffer
February 6, 2013 5:05 pm

Problem #4
The radiative ghe changes the observed temperature of the earth as seen from space by exactly zero. With no ghe the earth as seen from space would be 255K and the temperature at surface would be 255K. As seen from space with a ghe, the temperature of the earth IS 255K and the temperature of the surface IS 288K, However, the average height at which 255K occurs is 14 km instead of at surface.
As this experiment only measured equilibrium temperature, and only at one elevation in each box, it does not and cannot be used as a simulation of the radiative ghe.

davidmhoffer
February 6, 2013 5:07 pm

Problem #5
The radiative ghe is distributed from earth surface to TOA. This experiment puts a sheet of glass in to represent the ghe of the atmosphere, but does so at what would be the TOA, but with conductance and convection still active, which doesn’t exist at the TOA.

davidmhoffer
February 6, 2013 5:22 pm

Greg, of course it has been proven, many times over. It’s in every college thermo textbook.
>>>>>>>>>>>>>>>>
Not to mention that verifying the thermo laws via experimentation is part of the curriculum in many universities. A quick search for SB Law alone gets:
http://uregina.ca/~szymanss/uglabs/p242/Experiments/EXPT1.pdf
http://www.fiziks.net/lifesciences/exp54.htm
http://www3.wooster.edu/physics/jris/Files/Carter.pdf
http://personal.tcu.edu/zerda/manual/lab22.htm
http://wanda.fiu.edu/teaching/courses/Modern_lab_manual/stefan_boltzmann.html
http://sky.campus.mcgill.ca/Exp/Manuals/en0049.pdf
Is that enough or should I copy and paste the entire 177,000 results that searching “stefan-boltzmann law experiment” produced?
Not to mention that when a bunch of engineers design, oh, I don’t know, let’s go with a nuclear reactor…they figure out exactly how much power the fission process will produce, how much of it they can capture and use to run generators, how much will turn into waste heat, how much that waste heat will raise the temperature of everything from the boilers to the turbines, exactly how much water at a given temperature and flow rate will be required to keep the system cooled to a given operating temperature, and they nail it to fractions of a percent before the construction even starts, and they do so using the precise same laws of physics that we’re discussing right now.
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?

cd
February 6, 2013 5:28 pm

Michael Moon
I’m not saying Willis is right but I don’t think he’s wrong because of what you say.
“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 by 2nd law you mean the movement of a system to maximum entropy this will not invalidate the above model. Remember this is a conceptual model? Why would it not radiate back to the planet, if the shell is cooler then the net transfer would be outward from the planet but there would be a some transfer of energy to the planet from the shell (radiative transfer is emitted from and toward bodies irrespective of current state (it doesn’t care about energy state); maximum entropy only requires that the net flow will be toward the cooler body). Eventually, if the planet were to run out of its internal heat then both would eventually become the same temperature (all other things being equal).
Your point about the steel being cooler. Yes to begin with, but with constant supply of energy it will warm, and yes it can’t get warmer than the emitter but it can get warmer and will return some energy back to the surface with the result that the surface of the planet will warm.

KevinK
February 6, 2013 5:31 pm

Willis wrote (re multiple light bulbs in close proximity);
”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.”
Please note that in this example you have TWO energy SOURCES.
Please replicate this thought experiment with two steel shells in close proximity (sans additional heat sources), do you speculate that steel shell A heats steel shell B (and vice versa) and the temperature of both rise ?
If that’s the case I could just replace my electric baseboard heaters with steel slabs, boy that would sure be a lot cheaper.
Cheers, Kevin.

Bob Roberts
February 6, 2013 5:34 pm

@ Kev-in-Uk: Actually “Greenhouse Effect” is also an incorrect term/expression, as greenhouses rely on physical barriers, not just varying amounts of gasses, to do what they do. You are correct that while the planet does NATURALLY warm and cool, anyone who suggests there is any measurable ANTHROPOGENIC warming is only fooling themselves.

February 6, 2013 5:50 pm

Willis Eschenbach says:
February 6, 2013 at 5:01 pm
“…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?”
The temperature will not exceed 160 watts.

Gino
February 6, 2013 5:59 pm

cd says:
“Your point about the steel being cooler. Yes to begin with, but with constant supply of energy it will warm, and yes it can’t get warmer than the emitter but it can get warmer and will return some energy back to the surface with the result that the surface of the planet will warm.”
Not true. Thermodynamic entropy is defined in units of energy per degree of temperature. Once temperatures are matched (delta T = 0), then no energy is transferred between bodies.

jae
February 6, 2013 6:17 pm

Willis:
I don’t know how many times I’ve seen your steel shell-game, but I still ain’t buyin’ it. You are creating energy from nothing there, IMHO. I’m certainly not spending any time disputing this nonsense.
But my main interest is in the Wood experiments, and despite reading your post several times, I still haven’t figured out just how the shell-game is relevant. I just want to know why the IR backradiation from the glass does not make the inside of the glass greenhouse any warmer than the air in the “salt” greenhouse, which cannot have any such backradiation. And how the atmosphere can cause warming (or slow down cooling), if even the glass cannot do it.
Then you end the post with a puzzle for us:
“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.”
After all those words, you are leaving it as an exercise to the reader….?? WOW, this seems like a cop-out! Aren’t YOU the one who is supposed to be demonstrating why this experiment does not tell us anything about how backradiation helps heat (or prevents heat loss, or whatever)??

davidmhoffer
February 6, 2013 6:19 pm

KevinK
Please replicate this thought experiment with two steel shells in close proximity (sans additional heat sources)
>>>>>>>>>>>
You’ve constructed a thought experiment which is in thermal equilibrium. By definition, net energy flux is zero. This tells you precisely nothing about a system which is not in thermal equilibrium (such as the baseboard heaters in your house).

Gino
February 6, 2013 6:21 pm

Sparks says:
“The temperature will not exceed 160 watts.”
Temperature is not measured in watts. You have the right idea though. The temperature measured in any body will be the equivalent of having absorbed 160Watts of energy, which is defined by the heat capacity of the body defined in terms such as J / (kg K).

davidmhoffer
February 6, 2013 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?

Editor
February 6, 2013 6:23 pm

davidmhoffer says: “Greg, of course it [that A warms B and then B warms A back] has been proven, many times over. It’s in every college thermo textbook.“.
Unfortunately, as I read them, all the links you provided were only tests of Stefan Boltzmann Law, not of “A warms B and then B warms A back”.
I don’t doubt that “A warms B and then B warms A back” but it seems that a test of that specific hypothesis is needed to settle the matter. Roy Spencer’s set-up is definitely testable.

February 6, 2013 6:32 pm

Willis, pretty picture but,
Let me fix that for ya.

February 6, 2013 6:32 pm

KevinK @ 4:50 – During the winter there is generally less humidity, so that would cause the air to feel cooler than in the summer with the same temperature; but, in line with your point, the folks around here with brick houses require less wood to keep their homes heated once the bricks absorb the heat. I place a pot of water on my wood stove to increase the percentage of water vapor in the house to help with the sensation of heat.

Gino
February 6, 2013 6:41 pm

the reason that houses with warm bricks require less ‘heat’ to maintain temperature is that the bricks are no longer absorbing heat, and their emissivity/conductivity is such that the rate of heat transfer from the exterior of the wall is very low. Hence the heating system only needs to keep up with the rate the bricks carry heat to the outside world .

Graham Green
February 6, 2013 6:41 pm

All the units and quantities are irrelevant because the implication here is that a photon leaves the planet surface, interacts with an electron in the shell then somehow one photon is emitted back and another one into space.
Doesn’t seem right.

davidmhoffer
February 6, 2013 6:47 pm

Mike Jonas;
Unfortunately, as I read them, all the links you provided were only tests of Stefan Boltzmann Law, not of “A warms B and then B warms A back”.
>>>>>>>>>>>>>>>>
My point was that the laws of thermodynamics are found in every university level text book and are validated through experimentation, of which SB Law is a single example. You really want me to go through the list of all the laws and post links to experiments validating each and every one of them?
On the other hand, your complaint that SB Law has nothing to do with a warms b suggests that you don’t have a clue what SB Law is or how to apply it. Otherwise you wouldn’t have said such a thing.

February 6, 2013 6:47 pm

Gino says:
February 6, 2013 at 6:21 pm
Sparks says:
“The temperature will not exceed 160 watts.”
Temperature is not measured in watts. You have the right idea though. The temperature measured in any body will be the equivalent of having absorbed 160Watts of energy, which is defined by the heat capacity of the body defined in terms such as J / (kg K).
The temperature will not exceed 160 watts, convert it to joules to watts and watts to temperature. My point is valid.

February 6, 2013 6:49 pm

Doing a simple calculation with the Stephan-Boltzman equation suggests that with black body radiation within a vacuum, the ratio of the planet surface temperature to the steel shell temperature is 2^.25 or (1.189). So if the surface of the planet was 100 degrees C, shell would have to be 40 degrees C.

wayne
February 6, 2013 6:56 pm

joletaxi in English so all can understand the exchange:

I have a problem with your figure 2
the ‘core’ issues 235 W corresponding to a t* mission, say TA and heats the steel sphere with t ° (forget surfaces) at a temperature corresponding to this received energy, or TA
this sphere emits both outward and inward, (according to you) but to the surface emission to thus doubled!(outer surface over inner surface)
and as the received energy remains 235 w, and to stay in balance, the steel sphere shall issue 235 w, we see immediately that it was unable to reissue inward
Indeed, in your configuration, there is no reason that the t° of the Interior of the sphere is different from the outside temperature.
or to stay in balance, need us TA to the outer surface, and also, TA indoors, with the source shall emit 2 times more energy.
but we do have 235 w
Now, imagine that the surface of the sphere fits perfectly the core and conduction to be perfect,
do you believe that under a film of steel the t* core will double?
Mr. Spencer had us already well occupied with this little paradox, (Yes Virginia…) which my friends and I have never been able to find a formulation vaiable (discussed frequently on the following site
http://www.skyfall.fr/
Whatever it is, thanks for your stories, I am a sailing

Willis responds:

If I understand you, I think that the problem is in the units. You use watts, and I use watts per square metre. The shell receives 470 W m-2 of the surface, and because it has twice the area surface, it radiates 235 W m-2 both inwards and outwards.
I am pleased to hear that you’re a sailor, for me, there is always more to learn about the sea…
Good luck, take care, because the ocean cares not at all…
w.

Gino
February 6, 2013 7:00 pm

Sparks:
Watts do not convert to temperature without the heat capacity of the material. The point being that two items can have the same temperature but vastly different heat quantities.

KevinK
February 6, 2013 7:01 pm

Willis wrote;
“Your argument, as I understand it, is that a cooler object cannot radiate energy to a warmer object.”
NO, I have never said that. Energy flows back and forth between cooler objects and warmer objects all the time.
My argument is that you are mistaking these ENERGY FLOWS (fluxes) with ENERGY SOURCES.
An energy source ADDS energy to a system, an energy flow DIRECTS energy through a system. Think of cars leaving a parking lot, the number of people parking is like an energy source, the traffic light at the exit of the parking lot DIRECTS the flow of the energy (cars). Two totally different mechanisms.
If you go to a “Big City” Airport sometimes you miss the “exit here sign” and you have to loop back through the airport roads one more time to get on your way. You are not an “Extra” airport visitor; you just went through the ramp system twice before leaving. Same as the “greenhouse effect”, no more cars, they just looped through the airport roadways twice……….
Cheers, Kevin.

KevinK
February 6, 2013 7:03 pm

Greg House wrote (w.r.t. Roy Spencer’s”Yes Virginia there is a greenhouse effect”)
“This is a wonderful idea real experimental testing, Mike, thank you. Has Roy Spencer done it yet?”
As far as I know Dr. Spencer has never implemented this empirical experiment.
Please note that his description uses a “power supply” (a device that converts AC power from the electric company to DC current).
If instead you use an “energy supply” (i.e. a battery with a fixed supply of energy) you get a different result;
With a power supply you can indeed cause two metal slabs in a vacuum to achieve a higher temperature versus a single slab.
HOWEVER with an energy supply you simply cause a higher temperature for a short time and then the energy supply is exhausted and things cool back down.
With an energy supply you get a bit higher temperature for a time, but then things cool off and the average temperature is NEVER higher (although with proper techniques you might just “break even” although doubtful without “”perfect” (i.e. no efficiency loss) batteries).
Power supply, Energy Supply, two different things (apples vs oranges) as anybody that wished for a power supply to light up their flashlight when their energy supply (batteries) went dead should appreciate.
Cheers, Kevin.

Editor
February 6, 2013 7:11 pm

Kev-in-Uk says:”I take exception to using the term AGW instead of GHE (greenhouse effect) as that term is a false premise.“.
Point taken. Much more is needed to establish AGW. I think it’s there but negligible, but again it needs proper testing (not easy, because I can’t see the testing being able to be done in a lab).

Editor
February 6, 2013 7:15 pm

fhhaynie – If my understanding is correct, you need to use deg K not deg C. Also, the planet temperature you quote is the pre-shell temperature. Post-shell, the planet will be hotter.

February 6, 2013 7:27 pm

Mike,
I used Kelvin in calculating the ratio and then converted to C with 100 degrees as an example. The point is at equilibrium, the radiation from the planet will always be twice as much as the outbound radiation from the shell.

Editor
February 6, 2013 7:22 pm

davidmhoffer – It’s not me that has the problem. I’m suggesting a simple test to establish beyond doubt whether this interpretation of the law is correct – after all, science is supposed to be about testing. It is such a basic test, surely it is in a basic textbook somewhere, but no-one seems to have found it yet. If it’s not in the textbooks, then let’s encourage someone to do it.

February 6, 2013 7:23 pm

I have a historical excerpt from the 1980’s, explaining the poorly named “greenhouse effect”.
5 March 1984,
UK: Scientist warn of the Green House Effect; Concern is growing that carbon dioxide, produced by burning fossil fuels, will affect the climate. Carbon dioxide acts like the glass of a green house, trapping the suns heat, and the amount in the atmosphere is growing. The latest pointer to its effects comes from scientists at the University of East Anglia who have found that 1981 and last year were among the warmest on record. A warmer climate could damage agriculture and cause flooding by melting the polar ice caps.

February 6, 2013 7:27 pm

I may be a bit slow so correct me if I’m wrong but the take home message is that the experiment does *not* prove that the greenhouse effect (as described by radiation rather than convection or conduction) doesn’t exist but only shows that the greenhouse effect has virtually nothing to do with what keeps greenhouses warm. Have I got it or do I have to go sing about the rain in Spain while holding marbles in my mouth again?

davidmhoffer
February 6, 2013 7:29 pm

Mike Jonas says:
February 6, 2013 at 7:22 pm
davidmhoffer – It’s not me that has the problem. I’m suggesting a simple test to establish beyond doubt whether this interpretation of the law is correct – after all, science is supposed to be about testing. It is such a basic test, surely it is in a basic textbook somewhere, but no-one seems to have found it yet. If it’s not in the textbooks, then let’s encourage someone to do it.
>>>>>>>>>>>>>>>>>>
Once again you demonstrate that you have no idea what is in the text books or how to apply it. Shall I add you to the list of people who think nuclear reactors get designed by accident?

DR
February 6, 2013 7:31 pm

Why isn’t the greatest experiment of all, nature, being discussed in all this?

What we found consistently is that in a great part of the planet , the bulk of the atmosphere is not warming as much as we see at the surface in this region, and that’s a real head scratcher for us because the theory is pretty straight forward, and the theory says as the surface warms the upper atmosphere should warm rapidly. The rise in temperature of that part of the atmosphere is not very dramatic at all and really does not match the theory that climate models are expressing at this point. – Dr. John Christy

There is no “hot spot” and the stratosphere is not cooling; the climate models are wrong. So why all the kerfuffle with pictures of spheres and such? Is the atmosphere behaving according to GHE theory or not?

Joe Sixpack
February 6, 2013 7:42 pm

Power is not conserved. Energy is.
Go back and learn from someone who knows engineering and physics and then get back to us. Your amateurish analysis is laughable. Like flux is conseved! Figure out how much energy the “blackbody” absorbed and what the energy loss is. Also, steel does not transmit light in, like say, the atmosphere does. Your oversimplification is a tell that you have no clue what you’re talking about.
This is from someone who agrees that man made “global climate change” is trivial at best.
I’ve taught 1st year Mech Eng. students who have a better grasp of heat transfer than you.

LazyTeenager
February 6, 2013 7:50 pm

Thanks Willis. Agree totally with your article.
I quite like these discussions. They tend to be very amusing. The people who claim the people who get it right are idiots are a hoot. My absolute favorites are the guys who announce they are engineers. Fukushima anyone?

MattS
February 6, 2013 7:50 pm

DR,
You are confusing two very different theories.
GHE theory is that the atmosphere makes the Earth warmer than it would be without the atmosphere. While there are arguments about whether this results from radiative effects or some other effect, GHE is easy to prove. The Earth and the Moon recieve about the same amount of incomming radiation per m2 from the sun, but the Earth which has an atmosphere has a higher average surface temp than the Moon.
AGW is a theory that human actions (burning fossil fuels) is increasing the strength of the GHE which will result in a catostrophic increase in the Earth’s Average surface temp.
The “hot spot”, and stratospheric cooling are products of climate models that allege to prove AGW.
AGW is dependent on GHE but GHE by itself does not imply any particular temp trend.

jim bishop
February 6, 2013 7:52 pm

A thought experiment cannot refute a real experiment, ever.

February 6, 2013 8:04 pm

Gino says:
February 6, 2013 at 7:00 pm
Sparks:
Watts do not convert to temperature without the heat capacity of the material. The point being that two items can have the same temperature but vastly different heat quantities.
a 60w bulb and a 100w bulb equals 160w, put the two bulbs together while live, their temperature will never exceed 160w. They do NOT exceed 160w, no mater how much you mess around with ‘energy budgets’ and the material capacity, 160w in 160w out.

Jeff Carlson
February 6, 2013 8:04 pm

this post may be alot of things but a good example of physics it most certainly is not …
at the start of your experiment the steel shell is radiating no energy … (we are assuming the hot planet is the only energy source) therefore when it does begin absorbing energy it will then radiate that energy in all directions when it drops back down to equilibrium … if the planet radiates X amount of energy the shell can only absorb X but it cannot radiate X back at the planet … it can only radiate at best .5 X … and since that radiation is essentially a spherical projection from each molecule in the steel shell we know that quite a bit less than .5 X is radiated back to strike and supposedly be absorbed by the planet …
so no, the planet would never double its radiated output … ever …
2 metal balls … 1 foot apart in a vacuum … one very hot, one not so hot …
according to the 2nd law the hotter ball cannot he heated by the cooler ball yet we know the cooler ball is radiating heat … what happens to that radiated energy when it strikes the hotter ball ? Is it absorbed or reflected ?

KevinK
February 6, 2013 8:07 pm

Snake Oil Baron wrote;
“I may be a bit slow so correct me if I’m wrong but the take home message is that the experiment does *not* prove that the greenhouse effect (as described by radiation rather than convection or conduction) doesn’t exist but only shows that the greenhouse effect has virtually nothing to do with what keeps greenhouses warm.”
Yes that is about right, the “effect” exists, but it has nothing to do with the average temperature of the Earth, that’s determined primarily by the massive thermal capacity of the oceans.
The ”effect” simply delays the flow of energy from the Sun to the Earth and onwards towards space by a few tens of milliseconds. Since there are about 86 million milliseconds in a day the “effect” is so small we probably could never afford to measure it.
Cheers, Kevin.

Greg House
February 6, 2013 8:11 pm

MattS says, February 6, 2013 at 7:50 pm: “GHE theory is that the atmosphere makes the Earth warmer than it would be without the atmosphere.”
==========================================================
No, this is not true, anyway the “greenhouse effect” as presented by the IPCC is about particular “greenhouse gases” making the Earth warmer specifically by by returning back some of the Earth’s outgoing IR radiation, it is not about the whole atmosphere. They do not tell people, however, that this 150 years old hypothesis was debunked 100 years ago by the Wood experiment.
Note that the Wood experiment deals with the underlying mechanism of the “IPCC greenhouse effect”, namely the alleged effect of trapped/back radiation on the temperature of the source.

davidmhoffer
February 6, 2013 8:12 pm

2nd Law of thermo
Net energy flux between two bodies will flow from hotter to colder.
SB Law
Calculates the energy flux from a body at a given temperature.
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
Any thermo text, any physics curriculum, and yes, this does explain that A heats B and B heats A. The net flux is C. Proven over and over again by experimentation. Used to design nuclear reactors, kitchen ovens, air conditioners, automobile cooling systems, and more on a daily basis
Or you can subscribe to the theory that these laws of physics are not in the thermo texts, haven’t been proven by experimentation, and that engineers design nuclear reactors, kitchen ovens, air conditioners and automobile cooling systems that functions as designed in advance of being built by a complete fluke.

davidmhoffer
February 6, 2013 8:32 pm

Graham Green says:
February 6, 2013 at 6:41 pm
All the units and quantities are irrelevant because the implication here is that a photon leaves the planet surface, interacts with an electron in the shell then somehow one photon is emitted back and another one into space.
Doesn’t seem right.
>>>>>>>>>>>>>>>>>>
Because it isn’t. The theory is that a single photon gets absorbed and a single photon gets emitted in a random direction. For a large number of photons emitted in a random direction, about 1/2 will be generally upward and about 1/2 will be generally downward, but the same number get emitted as absorbed.

Richard G
February 6, 2013 8:33 pm

davidmhoffer says:
February 6, 2013 at 7:29 pm
…” people who think nuclear reactors get designed by accident?”
__________
Perhaps you mean by trial and error?
Because safety is no accident. *rimshot*

Greg House
February 6, 2013 8:34 pm

davidmhoffer says, February 6, 2013 at 8:12 pm: “2nd Law of thermo
Net energy flux between two bodies will flow from hotter to colder.”

=========================================================
No, the known historical statements do not contain the word “net” nor do they imply it, nor is there apparently any real scientific experiment confirming this notion.
I guess, that “net” thing is a trick of “climate scientists” to justify their “greenhouse effect”.

davidmhoffer
February 6, 2013 8:35 pm

greg house;
Note that the Wood experiment deals with the underlying mechanism of the “IPCC greenhouse effect”, namely the alleged effect of trapped/back radiation on the temperature of the source.
>>>>>>>>>>>>>>>
Is Lewandowsky paying you to blog?

Greg House
February 6, 2013 8:39 pm

Willis Eschenbach says, February 6, 2013 at 8:17 pm: “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?”
=======================================================
I guess, every warmist would indeed feel warmer and insist on that even after getting acquainted with the Wood experiment.

Greg House
February 6, 2013 8:45 pm

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.

Gino
February 6, 2013 8:54 pm

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.

KevinK
February 6, 2013 8:57 pm

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.

February 6, 2013 8:59 pm

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.

jae
February 6, 2013 9:03 pm

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…

davidmhoffer
February 6, 2013 9:03 pm

Richard G says:
February 6, 2013 at 8:33 pm
>>>>>>>>>>>>>>>
LOL

davidmhoffer
February 6, 2013 9:05 pm

There is no basis in real science for your “C”, so simple is that.
>>>>>>>>>>>>>>>>>>>.
ROFLMAO

davidmhoffer
February 6, 2013 9:25 pm

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?

davidmhoffer
February 6, 2013 9:28 pm

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)

Michael Moon
February 6, 2013 9:37 pm

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….

Gino
February 6, 2013 9:41 pm

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.

davidmhoffer
February 6, 2013 9:47 pm

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.

jae
February 6, 2013 9:48 pm

[snip – take a 24 hour time out – Anthony]

Gino
February 6, 2013 9:52 pm

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.

gbaikie
February 6, 2013 9:52 pm

“[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.

Greg House
February 6, 2013 9:57 pm

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.

u.k.(us)
February 6, 2013 10:03 pm

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.

Gino
February 6, 2013 10:09 pm

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.

Björn
February 6, 2013 10:25 pm

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.

February 6, 2013 10:34 pm

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. 🙂

Jim G
February 6, 2013 10:50 pm

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

Nigel S
February 6, 2013 10:54 pm

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.)

donald penman
February 6, 2013 10:58 pm

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.

February 6, 2013 11:01 pm

Greg House says (February 6, 2013 at 4:37 pm): “This is a wonderful idea real experimental testing, Mike, thank you. Has Roy Spencer done it yet?”
As I’ve pointed out in other threads, Dr. Spencer has no incentive to perform the experiment, as it would only confirm what he and just about everybody else already knows.
The real question is why his critics haven’t performed it. If they really believe it would overturn established physics, it would be worth a Nobel Prize at least, and would earn the undying gratitude ot a world saved from the imaginary threat of Thermageddon. That they haven’t performed such a definitive experiment and claimed their laurels speaks volumes.

wayne
February 6, 2013 11:06 pm

Gino:
“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.”
Thanks Gino, Willis has always had this problem with power and flux.

wikeroy
February 6, 2013 11:40 pm

Gino says:
February 6, 2013 at 9:41 pm
“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.”
I agree with Dino. It is my opinion that Dino’s way of thinking is the way one applies physics.
And the word “Backradiation” was never mentioned in radiation physics classes back then. Maybe it is “Newspeak”?

February 6, 2013 11:43 pm

Anthony Zeeman says (February 6, 2013 at 5:04 pm): “Leaving a space, presumably containing a vacuum between the sphere and the shell would do nothing since this would have the same thermal resistance as from the sphere to space.”
You’re forgetting that the shell radiates both out and in. Very different from the one-way radiation from the surface only. Think of the shell as giving the sphere a “rebate” on its radiated energy. 🙂
“This would be the same as adding another heatsink on top of the existing heatsink on your cpu.”
No. The heat sink is in contact with the microprocessor. Willis’s shell is about the same diameter as the “planet”, is separated from it by a vacuum, and the whole shebang is surrounded by vacuum. For obvious reasons, my laptop is designed to work in air. It even has a fan to convectively cool the heat sink, which cools the cpu by conduction. As with Wood’s model greenhouses, radiative cooling is a minor factor. If I were Greg House, I’d say my laptop disproves the so-called “greenhouse” effect., but I’m not. 🙂
“Also, note that your soup doesn’t get hotter when you put it into a Thermos in spite of the reflected IR radiation from the silvered sides of the Dewar flask.”
To compare with Willis’s example, you need to add a constant heat source to the inside of the thermos. Now what happens?

paulinuk
February 6, 2013 11:48 pm

So what happens if the steel inner ball and the outer shell is replaced by nitrogen gas. Does the nitrogen gas temperature approach infinity as they say it can’t radiate heat away?

Bart
February 6, 2013 11:54 pm

wayne says:
February 6, 2013 at 11:06 pm
But, you and Gino are begging the question, by implicitly asserting that equilibrium has been achieved and that the equilibrium with the steel shell is the same as that without it. Allow me to interject some math.
Basically, without the shell, the planet is receiving heat at a rate which I will call P_core, the power from the nuclear core. It is radiating it at
P_planet_outgoing = sigma*T_planet^4*(4*pi*R_planet^2)
where sigma is the SB constant, T_planet is the temperature of the planet, and the last term in parentheses is the the surface area with R_planet being the radius of the planet. Thus, the temperature of the planet at equilibrium is
T_planet = (P_core/(sigma*(4*pi*R_planet^2)))^0.25
Now, the shell has an inner radius, which I will call R_inner, and an outer radius, which I will call R_outer. It has incoming and outgoing power at
P_shell_incoming = sigma*T_planet^4*(4*pi*R_planet^2)
P_shell_outgoing = sigma*T_shell^4*(4*pi*(R_inner^2+R_outer^2)
This is the nub of the issue: The shell has at least twice the surface area of the planet over which it radiates. At equilibrium, we have
T_shell = T_planet * (R_planet^2/(R_inner^2+R_outer^2))^0.25
[Tshell is at most 2^0.25 := 20% lower in temperature than T_planet] but, these being absolute temperatures, 20% can be quite a lot.
What is T_planet now, though? Is it higher than it was before? Yes. The planet power balance is now
P_planet_incoming = P_core + sigma*T_shell^4*(4*pi*R_planet^2)
P_planet_outgoing = sigma*T_planet^4*(4*pi*R_planet^2)
At equilibrium, these are equal, so
sigma*T_planet^4*(4*pi*R_planet^2) = P_core + sigma*T_shell^4*(4*pi*R_planet^2)
= P_core + sigma*T_planet^4*(R_planet^2/(R_inner^2+R_outer^2))*(4*pi*R_planet^2)
sigma*T_planet^4*(4*pi*R_planet^2)*(1 – R_planet^2/(R_inner^2+R_outer^2) ) = P_core
so, with the shell
T_planet_with_shell = T_Planet_without_shell / (1 – R_planet^2/(R_inner^2+R_outer^2) )^0.25
Again, the maximum is about 20% higher.
Things are slightly different for the atmosphere, in that gaseous particles can radiate in any direction, and not all those directions will intersect with the Earth’s surface. But, over 45% of them will at the top of the troposphere, instead of the ~50% from a thin solid shell, so this is not such a big deal.
A bigger deal is that, the atmosphere is in contact with the surface, and can exchange heat through conduction and convection. On that score, the steel shell thought experiment appears to become a bit of an academic exercise which may not be relevant for a planet with a gaseous atmosphere.

Bart
February 6, 2013 11:58 pm

Bart says:
February 6, 2013 at 11:54 pm
Erratum: “Tshell is at most 2^0.25 := 20% lower in temperature than T_planet…”

wayne
February 7, 2013 12:09 am

“To compare with Willis’s example, you need to add a constant heat source to the inside of the thermos. Now what happens?”
I’ll have them on the market in a blink! ☺
Don’t you just hate cold coffee. Think 5‹W/m²› would do it?

Kev-in-Uk
February 7, 2013 12:28 am

Bob Roberts says:
February 6, 2013 at 5:34 pm
I am fully aware of the physics of the way a greenhouse works, thanks, Bob – and I am thus also fully aware of the ‘misnomer’ status of the GHE. Unfortunately, until someone comes up with an alternative well accepted description for the ‘warming effect’ of the atmosphere, I think the consensus use of GHE is the only one we can realistically apply at the moment! (If someone has created a better term that can be widely used, perhaps hey can let us know?)
regards
Kev

February 7, 2013 12:48 am

I still haven’t heard an explanation, why 6,000 ppm of CO2 in Martian atmosphere (which is exact equivalent of 400 ppm CO2 and thousands ppm of water vapor here) has no visible effect on its surface temperature, e.g. its black body and average temperature is the same 210K.
http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
Second, almost nobody considers one simple fact, that is, simple thermal insulation and heat retention of the bulk atmosphere – nitrogen and oxygen – does the job, which some are still trying to stretch on radiation diagram arrows. Moon night temperature is -150°C. Earth night is +10°C. Which kind of invisible, IR-based dragons breath warms our night by incredible 160°C? Photons bounced/radiated from tiny part of air molecules, really?

Don
February 7, 2013 1:05 am

On my first day as a grammar school student (long, long ago but I recall it clearly) I became very upset when, after I gave my teacher the exact 50 cents required for the week’s “milk money”, she then gave “my” 50 cents to someone who had paid her a dollar. My mom eventually helped me to understand that the net result was correct and that it didn’t matter that my exact physical two quarters were no longer in the teacher’s possession; my milk was well and truly paid for.
It astonishes and dismays me that several rather didactic and even patronizing commentors are making similar conceptualization errors regarding basic radiative heat transfer as well as largely missing the point of Willis’ post.
A very wise man once wrote, “If any one thinks he is wise, let him become a fool, that he may become wise.” In other words, a humble, open mind is a prerequisite to learning.
Thank you for the clarifying and thought-provoking post, Willis. Well done.

cd
February 7, 2013 1:10 am

Gino says:
February 6, 2013 at 5:59 pm
Not true. Thermodynamic entropy is defined in units of energy per degree of temperature. Once temperatures are matched (delta T = 0), then no energy is transferred between bodiesI
You didn’t read the post properly. Not the reference to NET transfer. If both bodies are at the same temperature there will be no net transfer of energy but both will continue to radiate to each other (balance therefore no change in energy state of either body). The vector defining the path of a IR photon will not stop because the surface it comes across is at the same temperature as its source. There will still be radiative transfer but NET transfer = 0. For your point to be true, that is no transfer, then the bodies would have to reach a state where they no longer emitted radiation.

cd
February 7, 2013 1:12 am

Gino last point should have read:
Gino says:
February 6, 2013 at 5:59 pm
Not true. Thermodynamic entropy is defined in units of energy per degree of temperature. Once temperatures are matched (delta T = 0), then no energy is transferred between bodies.
You didn’t read the post properly. Note the reference to NET transfer. If both bodies are at the same temperature there will be no net transfer of energy, but both will continue to radiate to each other (balance: therefore no change in energy state of either body). The vector defining the path of a IR photon will not stop because the surface it comes across is at the same temperature as its source. There will still be radiative transfer but the net transfer = 0. For your point to be true, that is no transfer, then the bodies would have to reach a state where they no longer emitted radiation.

joletaxi
February 7, 2013 1:20 am

Mr. Eschenbach
thanks for attention
I remain convinced that whatever the conversion watt watt/m2, it does not change the fact that in Figure 2, you indicate that the core issue 235 w/m2 and, depending off the amount off energy in the core, and the surface (let suppose, it is an blackbody)and thus the surface of the sphere emits 235 w / m2 to stay équilibre.iff you forget a differences off the surface off the core, and the sphere.
Now you certainly indicate that, because You are aware that a diffrence in the surfaces matter.
and because you are in agreement that if the surface of the sphere is larger, the radiation for each squre m will decrease.
You then specify the sphere also emits 235 w/m2 inward.
But again, , you double the irradiation area!
By each square m off the sphere, You have 235 w émitted outwards, and also inwards, so the flux off energy is doubled?
Now, something else,
you indicate that the emission flux of 235 w/m2 of the core induce on the surface of the sphere equal flux irradiation outwards , but also inward.
I assume that if this time the core issues as indicated 470 w/m2, there is no reason that this time it would not be divided into two streams,
there should be 470 w/m2 inward as to the outside, if you follow your first pattern?
it is an circular patern?

February 7, 2013 1:20 am

The steel hollow sphere is in deep space (-273C) with an initial internal vacuum.
Its heated by an external variable power supply carefully monitored to maintain the temperature at 50C.
The voltage and current are noted, say as V1 and I1, to get power P1
Now suddenly insert an object inside the sphere at -40C ( i can because its a thought experiment! ).
What happens next?
Does the real radiation from the colder object warm the steel sphere when absorbed?
I think not.
To restore the steel sphere to 50C the supplied power (P2) would have to increase for a time.
P2 > P1
Point being that in these apparent two object problems there is always a third (often ignored )temperature …..the surroundings.
In the example above the cold object at -40C is separated from the surrounding deep space by the steel sphere.
So no Virginia, colder objects don’t always ‘warm’ objects at a higher temperature.

MikeB
February 7, 2013 1:24 am

Even If the steel shell was much larger than the planet surface, let’s say twice the surface area, it makes absolutely no difference to the argument. It this case the outer surface of the shell would have to radiate 117.5 W/sq.m to balance the 235 W/sq.m at the planet’s surface. This accords with the shell having a temperature of 213K. Because the shell is at 213K it also radiates from its interior surface 117.5 W/sq.m (Stefan’s Law). This energy falls on the planet which, because it has only half the surface area of the shell receives 235 W per sq.m of planetary surface.
So, the same calculation ensues. The planet is now receiving 235 W/sq.m from its interior and a further 235 W/sq.m ‘back-radiation’ from the shell, that is 470W/sq.m altogether and so the planet becomes much warmer because of the presence of the surrounding shell.
Did you notice the provocative ‘back-radiation’ I slipped in there?
Unfortunately, most commentators don’t seem to have the ability to grasp this, but it is so simple and fundamental it separates the sheep from the goats.

February 7, 2013 1:29 am

The shell game is a foolish game to play. The shell game is how the AGW pseudo scientist came up with their ludicrous 33 degrees warmer because of “GHGs” meme. Once again let’s review the critical “Do Nots” of atmospheric modelling-
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”*
As you can see the shell game commits the following scientific atrocities 2,3,5,6 & 7. It’s so bad at modelling the atmosphere it cannot even commit 4.
Does the surface emit IR? Yes. Does the atmosphere absorb some IR? Yes. Does the atmosphere radiate some IR back to the surface slowing it’s cooling rate? Yes. Is there a radiative GHE? Yes. Is the NET effect of radiative gases in our atmosphere warming? NO!
Radiative gases are critical for the continued vertical convective circulation observed below the tropopause. Without radiative energy loss at altitude, convective circulation would stall, and our atmosphere would heat and go isothermal. Increased radiative cooling of the surface under a non radiative atmosphere is no substitute for radiative cooling at altitude. Avoid the “Do Nots” and observe the science Willis avoids and you will find –
– Due to gravity, pressure gradient and resultant IR opacity gradient, an atmospheric column emits more IR to space than to the surface. 2 & 6
– Most energy emitted as IR to space from the atmosphere was obtained by the release of latent heat of evaporation above the level of max atmospheric IR opacity. 2 & 6
– Radiative gases emit more energy to space than the atmosphere ever intercepted from NET surface IR flux 2, 5 & 6
– Radiative gases return far, far less energy to the surface than emitted from the surface. 7*
– The conductive transfer of energy and the transfer of latent heat to the atmosphere is a function of surface Tmax not Tav. 3, 4 & 5
– Surface cooling by conduction + convection is more effective than gas conduction alone. 5
– Under a non-radiative atmosphere, increased radiative surface cooling will not result in significantly increased atmospheric cooling. 5 & 6
– Deep vertical convective circulation cannot continue under the tropopause without radiative gases. 5 & 6
And the summary? AGW is tripe. Radiative gases cool our atmosphere at all concentrations above 0.0ppm. Adding radiative gases to the atmosphere will not reduce the atmospheres radiative cooling ability.
* Willis has yet to concede that DWIR has no effect on the cooling rate of liquid water that is free to evaporatively cool. I base my claim on the several empirical experiments I have conducted myself. To this date I am unaware of any empirical work Willis has conducted in this area, or indeed any empirical experiment he has conducted.

MikeB
February 7, 2013 1:31 am

Michael Moon says: February 6, 2013 at 9:37 pm

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.

That shell is called ’insulation’ Mike; I think heating engineers know about it. But note, in Willis’s article he says nothing about doubling the heat input. That stays the same. Why not go back and read it very carefully and see if you can understand it this time.
.

MikeB
February 7, 2013 1:46 am

There seems to be great confusion about fundamentals and the ‘imaginary’ 2nd Law of Thermodynamics.
Understanding electromagnetic radiation is a prerequisite to understanding the greenhouse effect. Radiation is a wave motion which transports energy through empty space at the speed of light by means of self-propagating electric and magnetic fields. It manifests itself in many forms, for example, as gamma rays, X-rays, light, infrared and radio waves. The only intrinsic difference between these various forms is that the waves have different wavelengths.
Radiation doesn’t ‘know’ if it is coming from a cold object or heading towards a hot object. How could it?
When radiation strikes the surface of a material one of three things may happen. It may be reflected, in which case it continues in a new direction with its energy conserved. It may pass through the material, like light through a window, radio waves through the walls of a house or like X-rays through your body. Or it may be absorbed by the material. Usually, a combination of two or three of these effects apply.
When radiation is absorbed the energy it conveys is also absorbed. Energy must be conserved and, most commonly, the absorption of radiation causes the absorbing material to heat up (the energy could, however, produce some alternative effects instead; photo-electric emission for example).
The Earth approximates to a black body in the long wave infrared region and so inevitably absorbs nearly all the down-welling radiation from the ‘cold’ atmosphere. And of course, this makes the surface warmer than it would otherwise be.

joletaxi
February 7, 2013 1:49 am

That shell is called ’insulation’ Mike
that something different no?
there is no matter off insulation here,

Mark Harvey aka imarcus
February 7, 2013 1:51 am

Willis
I have enjoyed your philosophy over the years, particularly your views on the safety valve action of thunderstorms, but your steel shell model here jars rather with my concept of the physics – bearing in mind that my degree is in archaeology, I might of course be missing a trick, but……
In my view your back radiation at 235 w.m-2 would only work if the steel shell is hotter than the core, which I don’t believe it is otherwise your model would be negated by the 2nd law of Thermodynamics. (Heat only flows from hot body to cold body, never in the other direction.) Otherwise the back radiation must just be deflected back to the steel shell, from whence it came and it wouldn’t heat that either, as it has just emanated from there and would eventually be re-radiated into space.
The same would apply, even if the core and shell were perfect black body radiators.
And as your analogy is, I assume, to ultimately represent in the end the earth and atmosphere, and particularly CO2 , the same principle applies – unless you assume that the CO2 molecules are hotter than the earth’s surface below the concept of ‘back radiation’ on which AGW seems to heavily revolve is not physically possible! Bit of a fatal flaw in the AGW concept.
Tx. Imarcus.

Ryan
February 7, 2013 1:53 am

Your first model is OK for greenhouse gas theory, BUT IT IS COMPLETELY WRONG FOR PLANET EARTH!
Why is that? Well because the TOP os Earths’s atmosphere is actually HOTTER than the bottom. This is because UV radiation is absorbed by the top of the stratosphere to such an extent that the top of the atmosphere is the hottest part.
You might like to consider how this might change your simple metallic atmosphere model, Willis.

Ryan
February 7, 2013 2:03 am

“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.”
This is untrue. What you see in the steel shell case is not “back-radiation” at all. It is merely reflection of the existing radiation from the planet – there is no change in wavelength. The steel shell will reflect the IR just as a mirrored surface will reflect visible light unchanged, and it will do so at the same angle of incidence too. CO2 actually re-radiates the IR but in all directions and at a different wavelength – quite different from your simplified model. Also, the CO2 only makes up 500ppm so it is nowhere near as dense as a steel shell at the photon level.
Furthermore, if the only heat source is actually outside the planet, how warm will the planet get then?
This steel shell model is typical of the over-simplified atmospheric models we get from Team-AGW to justify their claims. It has no business being considered here.

pochas
February 7, 2013 2:13 am

February 7, 2013 at 1:29 am
Your comment above contains many surprises, at least for me, but I find all of them plausible. You need to get published.

Ryan
February 7, 2013 2:13 am

“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.”
Are you sure it warms up? If it is 100% reflective of IR it won’t absorb any energy – so it won’t warm up. What will be the temperature of the planets surface then? (Here’s a clue – it won’t just get hotter and hotter….)

Richard LH
February 7, 2013 2:17 am

My simple explanation of the Greenhouse effect.
Earth Shell Space
e1< e1< e1 e2> e2>
e3> e3/2>
<e3/2
e1 = energy recieved from the sun
e2 = energy directly radiated to space
e3 = energy indirectly radiated to space
Long term
e1 = e2 + e3/

Richard LH
February 7, 2013 2:24 am

Sorry. Should be
Long term
e1 = e2 + e3/2

Tim
February 7, 2013 2:32 am

Willis,
IR radiation has wave/partical duality properties as such it can be interferred with. Given your source in the thought experiment is the same the likelyhood is that the IR is coherent from both the steel plate and the core. This give an effective 0 W/M^2 between core and shell (much as there is no energy difference between core and slightly less core) as they are in equilibrium. No work is being done ergo no extra energy is needed for balance. If you place a thermometer between core and shell it will reach an equilibrium temperature faster than if no shell was there it will not be a higher temperature. As with all energy fields – it is not the amount that is important but the difference – in equilibrium there is no difference.

Editor
February 7, 2013 2:37 am

I really don’t get the quibbling by many here. Surely the point of R. Wood’s experiment is this . .
All things being the same, except that in one box infra red is retained by the glass and in the other it is released by the rock salt, it seems that the contribution to the temperature in the box made by reflected infra red is trivial at best.
The rest of the argy bargy is fine in that it would refine the experiment in various ways but it won’t change the observed fact by Wood that the greenhouse effect is not enhanced by the trapping of infra red radiation.

MikeB
February 7, 2013 2:42 am

Ryan says: February 7, 2013 at 2:13 am

If [the shell] is 100% reflective of IR it won’t absorb any energy – so it won’t warm up. What will be the temperature of the planets [sic] surface [be] then? (Here’s a clue – it won’t just get hotter and hotter….)

Ryan, Bad Clue! It will just get hotter and hotter. Or do you have some maths to show us why it won’t?
You have a constant heat source putting energy IN. You postulate a perfect reflecting surface that stops any energy getting out. The energy builds up ad infinitum. It will just get hotter and hotter.

Editor
February 7, 2013 3:49 am

Thanks Willis, I was merely reflecting on the actual greenhouse that Wood was running and the fact that preventing the trapping of the IR made no measurable difference to the temperature in the boxes. The real cause of warming was the prevention of convection from cooling the box. The “reflected” IR is no big deal so surely the same thought could be applied to CO2, particularly in the presence of convection and conduction.
Your thought experiment at a global level is quite different and rather obvious assuming a vacuum between the planet and the steel shell where there is no convection or conduction, only radiation. Same difference really as the same point seems to be made.

Silver Ralph
February 7, 2013 3:07 am

Willis Eschenbach says: February 6, 2013 at 2:33 pm
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.
____________________________________
Ahh, Willis, I did warn you that all the trolls would come out of the dark, claiming that a cool body cannot radiate to a warm body. (Cooler bodies have eyes, that can see a warmer body coming, and so they shut off their radiation immediately….)
But I must say that your analogy of a candle warming the Sun is much more elegant than my two radiator scenario. I like that idea that I can hold a candle up to the Sun and make it warmer. Anyone bold enough to say by how much?
.

Ryan
February 7, 2013 3:10 am

“To compare with Willis’s example, you need to add a constant heat source to the inside of the thermos. Now what happens?”
Ah yes, but there is yet another flaw in Willis’ over simple thought experiment. In his experiment you have an infinite energy source of unknown temperature. This is not the case in the Earth/Sun system. In the Earth/Sun system we know how hot the Sun is, and we also know how hot it could theoretically make the surface of a body heated by the IR emitted by the Sun (which is rather smeared out by the time it reaches Earth).
Willis’ thought experiment would indeed by more honest if you put a body of known temperature in the center of the sphere and asked if it will get any hotter if you but a metal shell around the whole thing. Only it can’t be a metal shell in willis’ experiment because an ideal metal shell won’t absorb any heat energy and get warmer – it needs to be a “black body” shell to absorb the heat. But it can’t be a black body as such because a black body only radiates otuward and not out and in at the same time. And you can’t measure the temperature of the atmosphere is Willis’ experiment because there is no atmosphere so no sensible heat in the atmosphere. And if there was an atmosphere his experiment wouldn’t work at all.
Time to move on from this isn’t it?
I do wish Willis would wake up and realise that IR is not “heat” nor does it have a “temperature”. IR is a form of electro-magnetic energy that can be CONVERTED to heat by an ideal black body. If you talk about IR then you need to discuss the laws of thermodynamics at the quantum mechanical level. If you don’t want to discuss quantum mechanics then you must ignore IR and stick to discussing simple heat energy transfers using the laws of thermo-dynamics alone. Willis gets himself in knots by mixing IR into simple models of heat energy transfer that were developed before IR was known or understood and which don’t need a knowledge of IR to work.

February 7, 2013 3:21 am

This was your thought experiment – so if we take multiple IR wavelengths you still get to the problem that in equilibrium there is no energy difference between core and plate (or it is not equilibrium) . The “vaccuum” merely acts as a conduction/convection filter from core to plate – energy is a flow not a substance (no phlogiston in them thar hills) as such when the plate has expanded (well it is steel) and in equilibrium there is no more work being done on the plate – the plate becomes the core for IR transmission purposes. The flow is one way, net IR transmission at the inner surface is still zero.

A C Osborn
February 7, 2013 3:48 am

davidmhoffer says:
February 6, 2013 at 8:12 pm
Or you can subscribe to the theory that these laws of physics are not in the thermo texts, haven’t been proven by experimentation, and that engineers design nuclear reactors, kitchen ovens, air conditioners and automobile cooling systems that functions as designed in advance of being built by a complete fluke.
I call that BULL SHIT.
Which of those systems rely on a Vacuum.
All of them rely on being IN CONTACT with the Heat Source or the Air being heated by the Heat Source or Air to radiate the heat away.

lou
February 7, 2013 3:49 am

consider the implications of your thought experiment. with a finite amount of input energy you can heat an object to infinite temperature. you should patent it.

Joel Henrique
February 7, 2013 4:07 am

— RADIATION DEPEND ONLY ON TEMPERATURE AND EMISSIVITY
Period. No exceptions. —
Actually there are some rather big exceptions. That’s because you forgot to mention when one can apply the SB-law.
“Sei ein absolut leerer Raum rings von für Wärmestrahlung undurchlässigen Wänden von
der absoluten Temperatur t umgeben;” – Bolzmann, 1879
You can only apply the law to a real blackbody in vacuum that doesn’t conduct heat to the other side. Steel is not a blackbody and you assume that it can conduct heat to the outside. So, you are not talking about the SB-law, but some madeup imaginary law.
By the way, your radiator also has to be fixed in space and it must have a defined surface. That’s why you cannot apply it to any gases. At least that’s what we learned in University here in Germany.
Whenever you use a ‘greybody’ you should first show that you are allowed to use some simplifications, and that this simplifications won’t change the essence of the law.

Richard LH
February 7, 2013 4:15 am

Sorry – bad ascii text formating.
My simple explanation of the ‘Greenhouse effect’.

```Earth       Atmosphere        Space
e1a>
e1b>                          e1b>
e2>
e3>            e3/m>        e3/m>
<e3/n       <e3/n
```

e1 = energy recieved from the sun
e1a = energy reflected by atmosphere
e1b = energy reflected by surface
e1c = energy absorbed by surface
e2 = energy directly radiated from surface to space
e3 = energy indirectly radiated from surface to space
e3 = e3/n + e3/m
n and m dermine the ratio of reflected energy to passthough.
Long term
e1c = e1 – e1a – e1b
e1c + e3/n = e2 + e3
e1 = e1a + e1b + e2 + e3/m
e3/n is the ‘Greenhouse effect’ as seen at the surface.

joletaxi
February 7, 2013 4:40 am

let’ s try another hypothesis
if I put a second shell around the first, and I make the same calculation like the mechanism described, this time, the core t° is 4 times higher?
if I repeat this process n times, can we say that the t ° of the core can become infinite?
weird is it not?

EForster
February 7, 2013 4:56 am

Might I suggest that the energy radiated by the planet is absorbed by the interior blackbody shell and an equal amount of radiation is radiated outwards by the shell. There is no temperature increase at the surface of the planet.

squid2112
February 7, 2013 5:23 am

Wow, Willis, I have enjoyed so many of the things you have written here over the years, but this one, wow. Still clinging on to the “cold objects can make warmer objects warmer still” garbage? You “thought” experiment here is nothing more than a hodgepodge of card shuffling. I recently played this same game with my father over Dr. Spencer’s cold object/hot object claim. My father lost, and lost of one very simple reason, 2nd law. As indicated by others here, no matter how you slice it, twist it, dice it, a colder object cannot warm a warmer object, not in gas, not in steel, not in any way. The GHE either doesn’t exist, or is of such utterly small consequence that we will never be able to actually measure it. As for the IPCC form of the GHE theory, it is impossible.

davidmhoffer
February 7, 2013 5:26 am

A C Osborn says:
February 7, 2013 at 3:48 am
davidmhoffer says:
February 6, 2013 at 8:12 pm
Or you can subscribe to the theory that these laws of physics are not in the thermo texts, haven’t been proven by experimentation, and that engineers design nuclear reactors, kitchen ovens, air conditioners and automobile cooling systems that functions as designed in advance of being built by a complete fluke.
I call that BULL SHIT.
Which of those systems rely on a Vacuum.
All of them rely on being IN CONTACT with the Heat Source or the Air being heated by the Heat Source or Air to radiate the heat away.
>>>>>>>>>>>>>>>>>>>>>>.
Huh? Are you of the impression that systems being in contact suspend all other modes of energy transfer? It isn’t A or B. It is A and B. If the systems are in contact then they xfer energy via conduction and radiance. Makes the math even more complicated. But if you think the engineers can get it right by simply ignoring the laws of thermodynamics because the systems aren’t in a vacuum….I’ll add you to the list of people who think nuclear reactors work by shear fluke.

davidmhoffer
February 7, 2013 5:34 am

squid2112;
As indicated by others here, no matter how you slice it, twist it, dice it, a colder object cannot warm a warmer object, not in gas, not in steel, not in any way.
>>>>>>>>>>>>>>
When I was a kid we didn’t have enough money for heating fuel. We piled snow up against the side of the house to the point we could pretty much keep it warm with just the body heat of the people inside. Good to know that the snow must have been warmer than the house, because while shoveling it at -20C I could have sworn it was colder.

February 7, 2013 5:49 am

davidmhoffer says:
February 7, 2013 at 5:34 am
“When I was a kid we didn’t have enough money for heating fuel. We piled snow up against the side of the house to the point we could pretty much keep it warm with just the body heat of the people inside. Good to know that the snow must have been warmer than the house, because while shoveling it at -20C I could have sworn it was colder.”
What the snow did was increase the insulation factor for the house, it’s the same way igloo’s can be a lot warmer inside than out, while made of snow. Also note, snow has a lot of trapped air, which is a very good insulator.

Vince Causey
February 7, 2013 5:56 am

“Mark Harvey aka imarcus says:
February 7, 2013 at 1:51 am ”
In my view your back radiation at 235 w.m-2 would only work if the steel shell is hotter than the core, which I don’t believe it is otherwise your model would be negated by the 2nd law of Thermodynamics. (Heat only flows from hot body to cold body, never in the other direction.) ”
When people read “heat flow can only flow from a hot body to a cold body” they nearly always make the same mistake. They take this statement from the point of view of an omnidirectional flow of heat. They mistakenly believe that it means that there can never be any transfer of the energy from the cooler body to the warmer body. This is an incorrect understanding.
The 2nd law always refers to coupled systems. It says that if you have 2 bodies of different temperatures there cannot be a spontaneous transfer of heat from the cooler to the warmer. If heat did in fact transfer in such a way, the warmer body would grower warmer still, at the expense of the cooler body. But it doesn’t say that energy cannot flow from cooler to warmer. Energy does in fact flow from the cooler to the warmer body, but at the SAME TIME, even MORE energy flows from the warm to the cooler body.
Before you say this can’t happen – it would lead to pertual motion, etc – just consider this. It is impossible to devise any hypothetical situation where you can allow energy to radiate from the cooler to the warmer body WHILST simultaneously preventing the energy flowing the other way. Energy will ALWAYS flow to the cooler body at a greater flux density than the other way round. The 2nd law is not violated.

Alberta Slim
February 7, 2013 6:14 am

February 7, 2013 at 1:29 am
“The shell game is a foolish game to play……………….”
Konrad’s explanation wins the debate IMO.
Perfect.
Where’s Al Gore? I want to tell him the science is settled. ;^)

beng
February 7, 2013 6:14 am

Don’t think I’ve ever seen a thread where Mosher is right, and many other posters are wrong….

MattS
February 7, 2013 6:16 am

Greg House,
“No, this is not true, anyway the “greenhouse effect” as presented by the IPCC is about particular “greenhouse gases” making the Earth warmer specifically by by returning back some of the Earth’s outgoing IR radiation, it is not about the whole atmosphere. They do not tell people, however, that this 150 years old hypothesis was debunked 100 years ago by the Wood experiment.”
1 GHE theory in general predates the IPCC significantly.
2 Willis shows quite convincingly why the Wood experiment demonstrates/debunkes NOTHING.
3) IPCC claims the GHE is about particular gasses returning back some of the Earths’ outgoing IR radiation. Others here claim that the above applies to the whole atmosphere and others still claim it is a gravitational effect not radiative. Some here even claim simple convetive distribution of heat fully explains the difference.
I did note that the mechanism by which the atmosphere keeps the surface warmer than it would be without one was in dispute. However, a simple comparison with the Moon shows that whatever the mechanism is the atmosphere clearly affects the Earths average surface temp.

OzWizard
February 7, 2013 6:42 am

‘Thought experiments’ may fit comfortably into either Philosophy or Mathematics, but they are not a proper part of Physics. [Sorry, Einstein!]
When others try to repeat Willis’ ‘thought experiment’ and get different ‘results’, it is quite clear that there is confusion of mind somehere. I’m not the one to point my finger at anyone in particular, or waste my time trying to disentangle where the mistaken thoughts might be, because all (including me) are equally susceptible to mistakes in this regard.
One of the commonest mistakes in ‘thought experiments’ is introducing unspecified assumptions without realizing you are doing it. Merely stating that one “knows” what a certain hypothetical object will do in specified circumstances reeks of hubris.
2. Measure its equilibrium temperature;
3. Enclose it in a steel sphere of specific size (without contacting the ‘planet’);
4. Re-measure the temperatures of the two spheres (good luck with that!); and
5. Report the results.
All else is imaginary, with real numbers, i.e. Mathematical games. The gulf between ‘what we think will happen’ and ‘what actually happens in a real experiment’ is often surprising and should be educational.

Ryan
February 7, 2013 6:46 am

“The emissivity of various materials is given here. Reflectivity is equal to [ 1 – emissivity ]. The reflectivity of steel is about 10%-15%.”
Ummm, I think you need to go back and re-read that table again!
Anyway, I’m glad you admit the shell cannot be made of steel at least. Rusty steel perhaps (iron oxide will get you closer to being a black body) but not steel. So now you know we are talking about black bodies – definitely not steel, not anything like steel.
Lets consider what ACTUALLY happens with a CO2 molecule when a IR photon hits it. Remember that IR is not HEAT. It is an electromagnetic radiation with energy that can be converted to heat in certain circumstances. So a photon with energy in the IR spectrum reaches a CO2 molecule. What happens to it? Well if it is within the scope of the covalent bonds of the CO2 molecule it could be absorbed by the electrons in the bonds causing the electrons to be lifted to a higher energy level. But here is the crux of the matter – if the electron is already at that higher energy level then an incoming IR photon WILL NOT BE ABSORBED (at least not at the same frequency) – the photon will just travel straight through. But also bear in mind that lifting an electron to a higher energy level does not in itself cause heat to occur – the electron must drop back to its lower energy level and the assumption is that the electron then loses energy and that energy (probably) ends up at some point exhibiting itself as heat (just following the laws of entropy) but maybe not right away because the CO2 may just emit IR of a different wavelength (i.e. not heat as such) which may then be absorbed (or then again may not) by some other object which then gets a bit warmer.
Does any of this actual description of what happens to CO2 with an incoming photon strike you as being in any way related to ideal black bodies suspended in space over planet earth with no atmosphere when the Earth is being heated from the inside from an essentially infinite energy source of no known specific reaction temperature? Because it seems the model you are creating is in no way related to reality and therefore not useful, before we even begin to get into the details of whether your description of the maths is plausible.

February 7, 2013 6:49 am

davidmhoffer says:
After reading all of these, I think you’re one of the ones who have it right.
Willis, Working from a flux rate (watt’s/sq) alone makes the analogy harder to follow. S-B is based on a difference in Temps, with that an an emissivity, you can calculate a flux rate. In the case of this example the 100/40 temps (fhhaynie’s calculation) for the inner/outer shells(if the steel was a blackbody) sounds about right. The net flux from the surface should then match your 235w flux (net) from the inner to the outer, the back radiation is why the inner sphere is warmer than the outer.
Ryan says:
February 7, 2013 at 3:10 am
“I do wish Willis would wake up and realise that IR is not “heat” nor does it have a “temperature”.”
It has an equivalent temp. CO2’s main absorption bands are ~4u @ ~850F, and then 10u-15u 60F- -112F But the 10u-15u band is very weak, and much of it is overlapped by water vapor.
Greg House, Please, please go to the hardware store, and buy yourself a handheld non contact IR thermometer. Once you can explain/understand how that works, you’ll at least get that S-B is well tested, and how everything radiates IR, even if it’s cold. Every things temp is the net of incoming radiation and out going radiation. Just as described by S-B.
My Ryobi IR thermometer on a clear sky 35F day, reads the skies temp at it’s minimum scale -40F, or it’s maximum scale of 608F. The less than -40F is the back radiation, instead of being the absolute zero of space, this lowers the temp difference between the earth and space slowing the cooling of the surface. It’s this -40F that all of the hubbub is about. But after digging through 120 million surface temp records, it doesn’t reduce reduce the amount of night time cooling any. The 608F is unexpected, but I think it’s just a reflection of solar IR off the atm. If there is a temp difference from increasing CO2, IMO it’s from this. Since a CO2 molecule will only absorb, and emit a single photon at a time, more Co2 will reflect more IR.
Lastly Dr Spencer did attempt to do this experiment.

February 7, 2013 7:02 am

February 7, 2013 at 1:29 am
“Does the surface emit IR? Yes. Does the atmosphere absorb some IR? Yes. Does the atmosphere radiate some IR back to the surface slowing it’s cooling rate? Yes. Is there a radiative GHE? Yes. Is the NET effect of radiative gases in our atmosphere warming? NO!”
This is a nice summary, but IMO the last answer is a possible slight warming, I put a max of maybe 25% of the measured trend, a slight beneficial warming to go along with all of the plant food. Plus if the Sun is getting ready to go on vacation, we’ll want all the extra warming we can get.
As a side note, we’ve had snow on the ground all but 4 or 5 days since a few days before Christmas in NE Ohio, it’s been quite a while since that’s happened!

lou
February 7, 2013 7:05 am

Please explain how the low energy photons from the colder body gets absorbed by the atoms of the hot body already in higher excited states.

davidmhoffer
February 7, 2013 7:11 am

micro;
What the snow did was increase the insulation factor for the house, it’s the same way igloo’s can be a lot warmer inside than out, while made of snow. Also note, snow has a lot of trapped air, which is a very good insulator.
>>>>>>>>>>>>>>>>>>
Well gee, the air in the snow would be the same temperature as the snow, would it not? And how, exactly, do you think insulation works? It works by absorbing radiated energy and re-emitting some portion of it back.

davidmhoffer
February 7, 2013 7:16 am

Willis:
here’s how the Woods experiment was set up.
Whole Bunch of People Who Have Studied and Worked with Thermodynamics:
here’s about 20 different reasons why that experiment could not possibly demonstrate any meaningful data in regard to the radiative ghe
Whole Bunch of People Who Have NOT Studied or Worked with Thermodynamics:
you guys don’t know what you are talking about
I’ve scored the whole discussion and the winner is….
Lewandowsky

Ned
February 7, 2013 7:17 am

I’d like to make a quick comment about willis’ light bulb example and the assertion that a nice cool 60 watt bulb can increase the temperature of a hot 120 watt bulb.
Instead of light bulbs, imagine 2 batteries, one with 12 volts of electrostatic potential, one with 6. If you connect the two batteries in parallel (maybe through a resistance), the 6 volt battery doesn’t increase the potential of the 12 volt battery. That’s just obvious. Current only flows in one direction, from high to low. The potential of the 12 volt battery drops, and the potential of the 6 volt battery increases.
The exact same thing happens with a 120 and 60 watt bulb. The 60 watt bulb takes heat from the 120 watt bulb, cooling it.
Nevertheless, I completely agree with Willis, the 120 watt bulb is hotter because the 60 watt bulb is there. That’s because a 60 watt light bulb is a much poorer heat sink than the alternative, some nice cool air, which will absorb heat than waft away and be replaced by more cool air. The hot bulb replaces cool air and so the 120 watt bulb has a harder time getting rid of its heat and it runs hotter still.
So the 60 watt bulb is like a 6 volt battery and the air is like, say, a 1.5 volt battery. Less current flows from a 12 volt battery into a 6 volt battery than it would into a 1.5 volt battery, and as a result the voltage you’d measure at the junction of a 12 and 6 would be higher than at a 12 and 1.5. So in a sense, the 6 volt battery “increases” the potential of the 12 volt battery, but only in comparison to the alternative.
Now, before anyone jumps on me, let me state my awareness that electrical flow may not be a perfect analogy to heat flow. Electricity really really (pretty much really) only flows in one direction. What about heat? Does heat flow in one direction, or does it flow back and forth? Who Cares.
Quantum mechanics says that radiation scatters in all directions without preference. The presence of a hot light bulb doesn’t resist the flow of photons the way a positive electrical field would resist the flow of electrons, so the 60 watt bulb scatters in the direction of the 120 watt bulb exactly the same amount as it would in the direction of empty air or an ice cube or a vacuum. But thermodynamics says that heat only flows one direction, from hotter to cooler. That’s because in thermodynamics, which was come up with before corpuscles and turns out to be statistical law, you’re really talking about a net flow. In thermodynamics it’s ok to think of the 60 watt bulb as a heat sink, cooling the 120 watt bulb, even though the reason it’s a poorer heat sink, taking less heat away from the 120 than air would, is actually because it gives more back. Whether you think of heat as a flux flowing in only one direction, or as discrete bits of radiation bouncing between them back and forth, it really shouldn’t matter; as long as you just stay consistent about it you’ll get the same answer.
In other words, I think most, or at least a lot of, the arguing here is irrelevant and just people defending their initial semantic choices. But as for me and my own personal morality, I think it’s probably best when talking about “heat” to stick with thermodynamics convention and think of it as a flux and say it only flows from hot to cold, and if you’re talking about “radiation” it’s ok to say it goes everywhere.

MikeB
February 7, 2013 7:24 am

Micro.
You can’t measure back radiation with an IR thermometer. They are specially designed to work within the IR atmospheric window, which means that they won’t see radiation from the atmosphere.
CO2’s absorption band relevant to the greenhouse effect is around 15 microns. It is not weak!
In fact the small amount of CO2 in the atmosphere will absorb 95% of radiation at 15 microns within a distance of 1 metre

Vince Causey
February 7, 2013 7:24 am

lou says:
February 7, 2013 at 7:05 am
“Please explain how the low energy photons from the colder body gets absorbed by the atoms of the hot body already in higher excited states.”
Ok, I’ll have a go. Vibrating atoms of the “hot” body are not all moving at the same speed. Because they are colliding off each other, their movements are random. Some atoms will be moving at a speed higher than the temperature of the body would suggest, others moving slower. The slower moving atoms are able to absorb the photons from cooler body.
Though, having written that, I am not sure whether you are conflating two different things – the temperature of the body which consists of the average KE of the atoms, and whether the atoms exist in highly excited states. The latter means their electrons have moved to higher energy shells, and is not a function of their KE (temperature). Higher temperatures do not necessarily mean that their atoms exhibit highly excited (electronic) states – it merely refers to the average speed of motion of the atoms. This has nothing to do with whether or not the atom can absorb a photon, imo.

Andyj
February 7, 2013 7:24 am

@ Willis,
Figure 2 is WRONG!
There is no 470W/m^2.
.
Even IF the steel shell was a perfect reflector and insulator – not allowing any losses. The internal volume will still be 235W/m^2.
.
What this theory does not cover is the rate of loss against the rate of heating. The effect of shutting the oven (or fridge) door.
.
To comprehend more clearly with fewer issues please see across temperatures as a “potential difference” the same mindset to Voltage differences. Then consider the rate of heating and cooling as Amps. The outcome is, of course, in Watts.

February 7, 2013 7:50 am

davidmhoffer says:
February 7, 2013 at 7:11 am

Well gee, the air in the snow would be the same temperature as the snow, would it not? And how, exactly, do you think insulation works? It works by absorbing radiated energy and re-emitting some portion of it back.

Fair enough.

DR
February 7, 2013 8:00 am

@MattS
I quoted John Christy for a reason, as I think he knows a little bit about atmospheric physics hence it is “a real head scratcher” the atmosphere isn’t warming per the “theory”. Which theory was he speaking of? Atmospheric CO2 has been increasing. The upper atmosphere should be warming at a higher rate than the surface. The stratosphere should be cooling because we’re told the troposphere “traps” the heat. The opposite is true, so the idea of an “enhanced” GHE is wrong. If the basic physics according to Christy is the

theory says as the surface warms the upper atmosphere should warm rapidly.

then something is wrong with the “theory”.
We are constantly lectured that increasing CO2 levels WILL warm the atmosphere; it is all about basic high school physics we’re told. Well, I don’t think it is about basic physics. I think it is about how the climate system works apart from “basic physics”. The physics is not well understood in my view.
We’re also told OHC increases in the past 30 years is due to AGW. There is zero empirical evidence that even a doubling of atmospheric CO2 can warm the oceans to any measurable degree. Instead, we’re told there is mixing by ocean waves that magically transports the top few microns of “heated” water affected by “back radiation” from a few extra ppm’s of atmospheric CO2. Sure, everyone stirs their hot coffee to make it warmer. How can anyone try to pass off the idea that anything other than SW solar radiation can warm the oceans and expect thinking people to believe that is frankly, insulting.
We’re also told water vapor is a positive feedback. Ok, then explain why deserts with very little water vapor become much warmer than tropical regions with much higher water vapor content. There seems to be a limit to how much the earth can warm, and water in its various forms appears to be the mediator.
I understand radiation is the only method by which heat is transferred to space. However, it is convection that dominates during the day when the sun shines which cools the surface by transporting heat to the upper regions of the troposphere. The surface of the earth would be uninhabitable if not for convection. I question whether the “basic physics” properly account for these convective processes.
I am a bit disappointed Willis did not address Tim Ball’s reference to http://principia-scientific.org/supportnews/latest-news/34-the-famous-wood-s-experiment-fully-explained.html . I claim no scientific expertise, only that I have been following the “greenhouse effect” meme for the last 25+ years and it is not working as advertised. Obviously the physics are not all accounted for or that well understood.
If the Woods experiment did not demonstrate/disprove anything, then neither did Arrhenius.
Finally, we’re told the earth’s atmosphere is not like a real glass greenhouse, yet I have seen dozens of references by various government science agencies and universities describing the earth’s atmosphere being like a real glass greenhouse in which the troposphere “traps” the heat thereby cooling the stratosphere.

February 7, 2013 8:18 am

MikeB says:
February 7, 2013 at 7:24 am

Micro.
You can’t measure back radiation with an IR thermometer. They are specially designed to work within the IR atmospheric window, which means that they won’t see radiation from the atmosphere.

They have to measure IR in the same range the meter can measure temps in, 608F to -40F is the full scale range of the meter. Now the meter’s emissivity setting is set to .95, and the sky might not be .95, so the temp displayed could be wrong, but it has to measure IR in the 4.5u to 12.5u range. I would love to have an IR spectrometer, but it is out of my price range. And it obviously reads a IR signal from the sky. Besides NASA thinks it can be done.
CO2’s absorption band relevant to the greenhouse effect is around 15 microns. It is not weak!
In fact the small amount of CO2 in the atmosphere will absorb 95% of radiation at 15 microns within a distance of 1 metre
Here’s the spectrum of Co2 from NIST. 15u IR has an equivalent temp of -112F, that might keep Co2 from freezing, but it isn’t going to keep all the water on the planet from freezing.

February 7, 2013 8:18 am

Sorry meant to quote this

CO2’s absorption band relevant to the greenhouse effect is around 15 microns. It is not weak!
In fact the small amount of CO2 in the atmosphere will absorb 95% of radiation at 15 microns within a distance of 1 metre

Greg House
February 7, 2013 8:26 am

Willis Eschenbach says, February 7, 2013 at 2:57 am: “Since the Wood experiment is totally unlike and has nothing to do with the greenhouse effect, it couldn’t reveal a dang thing about the greenhouse effect …”
==========================================================
I provided a quote from the IPCC explanation of the “greenhouse effect” and the link to it on this thread, you must have overlooked it.
The Wood experiment deals exactly with the underlying mechanism of the IPCC “greenhouse effect” (effect of trapped/back radiation) and demonstrates that it is bupkis.

Gino
February 7, 2013 8:32 am

cd says:
February 7, 2013 at 1:12 am
CD you are correct. I was sloppy in my expression at that point. I addressed that topic in later posts.

Gino
February 7, 2013 8:49 am

Bart says:
February 6, 2013 at 11:54 pm
P_shell_outgoing = sigma*T_shell^4*(4*pi*(R_inner^2+R_outer^2)
This is incorrect. There is no net energy flow from the shell to the core. The shell is not a generator. With no net energy there is no power flow from the inner surface. Net energy transfer back is zero. At best what you add is thermal resistance (what the rest of the world call insulation). But you need to be careful here because there is something called a critical thickness. This is the thickness of insulation that actually increases heat transfer due to greater surface area.

February 7, 2013 9:15 am

There’s waaaay too much here to respond to it all, but let me pick a few random bits to reply to.
“But here is the crux of the matter – if the electron is already at that higher energy level then an incoming IR photon WILL NOT BE ABSORBED (at least not at the same frequency) – the photon will just travel straight through.
NO! That is more or less what happens for visible light, but not for IR. IR absorption is due to quantized VIBRATIONS of the molecules, not EXCITATIONS of electrons. The molecule can have multiple units of vibrational energy, so even if a molecule is already vibrating, it can pick up additional vibrational energy by absorbing another photon.
“… because all (including me) are equally susceptible to mistakes in this regard.
NO! People with more education and experience and intelligence are less susceptible to making mistakes. Not everyone is created equal when it comes to analyzing science.
“Energy does in fact flow from the cooler to the warmer body, but at the SAME TIME, even MORE energy flows from the warm to the cooler body.
YES! As an additional example, consider heat conduction between two pieces of metal at different temperatures — which is really just the sum total of the energy transferred at a molecular level by collisions at the interface between the two blocks of metal. Some of the collisions will involve faster-than-average atoms in the cool block hitting slower-than-average atoms in the warm block. ENERGY can and will occasionally be transferred from the cool block to the warm block but such collisions, but more often the transfer will be from warm to cool. The NET transfer of energy (ie the “heat”) will always be from warm to cool.
“Still clinging on to the “cold objects can make warmer objects warmer still” garbage?
I am (unfortunately) not amazed that people still cling to this objection. It is very seductive to people to don’t really understand thermodynamics or the statistical nature of the 2nd Law.
> “heat” does not move from cool to warm.
> “energy” can and does move from cool to warm (just not as much as moves from warm to cool).
Others have given other examples, but try this one. Take a shed and put a 1,000 W heater inside. On a very cold day (say-20 C) the interior might reach 10 C. Now put the shed in contact with other cold air (say -10 C). Now the interior might reach 15 C.
You can slice and dice and discuss this anyway you want, but the simple fact is that the presence of “cold material around the heated object” (at -10C) was an integral part of “warming” the interior of the shed. We used merely “sort of cold surroundings” instead of “really cold surroundings” and the result was a warmer interior.
Just like using “sort of cold surroundings” (the atmosphere) can warm the surface compared to having “really cold surroundings” (2.7 K outerspace).
if I repeat this process n times, can we say that the t ° of the core can become infinite?
weird is it not?

Yes, it is weird, but it is not wrong! There is is a significant limitation to the thought experiment that would prohibit infinite temperatures (besides the mere fact that the whole thing would eventually melt). The heater itself must continue to provide the required power even when the surroundings are hot — ie the heater itself must be HOTTER than the surroundings so heat can still flow from the hotter heater to the cooler surroundings. Limitations on the actual heater used would eventually impose an upper limit. For example, if the sun (5780 K) is used as the heater, then the surface could never get above 5780 K no matter what sort of insulating layers (or mirrors or lenses) you tried to use.

mkelly
February 7, 2013 9:18 am

davidmhoffer says:
February 7, 2013 at 7:11 am
And how, exactly, do you think insulation works? It works by absorbing radiated energy and re-emitting some portion of it back.
As a general statement I disagree. Most insulation tries to prevent convection and conduction.

george e. smith
February 7, 2013 9:32 am

Regardless of the Woods experiment; or irregardless, as the case may be, The Steel Shell experiment Willis presents simply doesn’t work.
Willis asserts his nuclear powered planet (sans shell) emitting 235 W/m^2, is in equilibrium.
Not true, it is cooling down, and eventually be quite cold when the finite amount of radioactive materials has all decaysed. OK so it may be a long time.
So now add the steel shell, and wait for it to reach equilibrium, which Willis asserts it does.
Well now we know for sure that it is not in equilibrium, since it is not all isothermal. The shell is presumably cooler than the planet. Furthermore, if it WAS in equilibrium, then Kirchoff’s Law would apply, and the shell would be absorbing and receiving the exact same spectrum of whatever radiation was in the cavity between the planet and the steel shell.
So a fundamental assumption of the experiment cannot be met.
Whatever the radiation flying around this shell game, it certainly is not black body radiation, so SB law and Planck’s radiation law do not apply here.
The Planck derivation was for the radiation escaping from a small hole in a THERMALLY ISOLATED ISOTHERMAL cavity, from which NO energy can escape through the insulating walls.
Steel of course is not a black body radiator, nor is any other known material.
The electromagnetic radiation spectrum, extends from down to but not including zero frequency (DC) and up to but not including infinite frequency, and a black body must perfectly absorb ALL of that radiation completely.
NO physical material has zero reflectance for EM radiation; that would reuire it to have unity refractive index like the vaccuum, and no material does.
Black body radiation itself is fiction, in fact it is fictionally fictional. It involves the physical properties of no physical material whatsoever. No electronic energy states, or any other trappings of quantum mechanics of real materials is involved in the derivation of the black body radiation formulae.
It is a quite hypothetical radiation spectrum that does not and cannot exist, and the knowledge of no material properties of real physical materials, was used in its derivation.
Yes it is an extremely useful and improtant result, but Willis’s shell planet, is not emitting BB radiation.

wayne
February 7, 2013 9:35 am

@davidmhoffer: February 7, 2013 at 5:34 am
Well davidmhoffer, you seem so ripe for being sold one of these new inventions today, run out and talk to one of the I.P.C.C. salesmen, get yourself one! I know, not quite as nifty as when you were heating your home with the backradiation from snow or you were staying so toasty warm in an igloo with all of that parabolic backradiation, but close.
http://www.ilovemycarbondioxide.com/IPCC_oven.HTML

Seriously davidmhoffer, you and Willis have got some big problems here.
The shell as shown is outputting more energy, 470‹J/s/m2›, per time per area than is even available from the nuclear source in the first place, there is only one energy source in Fig. 2 and it’s maximum output through the sphere’s surface is 235‹J/s/m2›. Ever heard of the 1st law of thermodynamics? Don’t you see your problem? This is getting embarrassing for W.U.W.T.
Ever heard that an energy source can never heat itself?
I think now you too have mixed possible power with flux.
But this does shine a bright light on exactly where the Stefan-Boltzmann equation is being used and misapplied in climate science. Reflected energy from a source, in any manner, can never then be added back to the source’s original hypothesized flux by SB and then fed back recursively into the SB equation to hypothesize a new higher temperature of the source itself. This is where the I.P.C.C.’s Free Energy Oven fails.
This is also where Trenberth slips in that extra 23‹W/m²› in his “global” energy budget to boost the “backradiation”.
It is like taking a sheet of polished aluminum foil and holding it up to a wall radiating by SB per T at 400‹W/m²› and 400‹W/m²› is reflected back to the wall by the foil at the temperature of the foil so now you have 800‹W/m²› at the wall so of course the wall’s T must rise… if you can really see that this happening and being something physically real then, rather than trying to teach radiative energy science to all of the scientists and engineers coming here and to comment, maybe you should just stick to selling computer systems and Willis to construction.
However, David, if a cloud or the atmosphere gases absorb solar energy that came from an energy source external to the Earth’s climate system, the sun, then that energy can be deemed an original energy source in the system and that energy can be handled isotropically with down welling LW radiation that can raise the surface’s temperature if it ever increased.
I myself have been wrong in this respect too in the near past, to TFK09’s energy budget graphic, the only down welling source radiation is one-half of the 78, or 39, absorbed by the atmosphere and clouds plus the 161+1, or 162 absorbed by the surface. That gives the surface 200‹W/m²› of absorbed energy. From that the other one-half of the 78, or 39, goes to space plus the 98 sensible and evapotranspiration plus 102‹W/m²› net up radiated from the surface that gives 239‹W/m²› as expected, TFK’s one off. The net LW flux is the 102 from the surface, up, minus the one-half of the 78, or 39 down, giving the 63‹W/m²› net that you can see day in and day out of the net surface IR graphs at the ESRL.NOAA site for various locations. Finally even that now makes sense and balances and matches the data on the government climate and weather sites. I learn something new every day so I guess Willis’s so wrong post ends up with a good ending, for me anyway.

Ryan
February 7, 2013 9:57 am

“Others have given other examples, but try this one. Take a shed and put a 1,000 W heater inside. On a very cold day (say-20 C) the interior might reach 10 C. Now put the shed in contact with other cold air (say -10 C). Now the interior might reach 15 C.”
This has absolutely nothing to do with the cold air making the warm air warmer!!! Once again this is due to conduction. Heat inside the shed will be conducted to the outside. The rate of conduction is dependent on the thermal gradient. The thermal gradient is dependent on the outside air temperature. Make the outside air warmer and the rate of conduction will decrease so the interior will be warmer. You can also slow the rate of conduction by covering the shed in rockwool. You cannot, however, significantly warm the shed by covering it with tin foil (although many a claim has been made that you can….)

Alan S. Blue
February 7, 2013 9:58 am

I think it would be useful to back up one step and have -just- the argument about a hot plate and a not-so-hot plate. There are plenty of examples in textbooks to quote. Having the detailed discussion about how individual atoms in the “cold plate” can be hot and how individual atoms in the “hot plate” can be cold would be useful. Perhaps even a discussion over spectroscopy and how exceedingly hot things can -still- absorb IR photons – particularly if they’re molecules and happen to have vibrational modes as well.
And how many of the general thermodynamics rules are over NET energy transfers – while -statistical- thermo really lays out what temperature actually means atomically.
Until this particular discussion is firmly laid out, the entire “Steel Greenhouse” discussion inevitably latches onto misunderstandings.

February 7, 2013 9:59 am

“The Wood experiment deals exactly with the underlying mechanism of the IPCC “greenhouse effect” (effect of trapped/back radiation) and demonstrates that it is bupkis.
The Wood experiment is similar in some ways, but very different in others, which is the point hat Willis was making and which you apparently missed.
> The earth is surrounded by extreme cold (outer space), but Wood’s experiment is surrounded by materials at very similar temperatures to the experiment itself.
> The atmosphere of the earth is many KILOmeters in size, but Wood’s experiment is many CENTImeters in size.
These (and other rather fundamental differences) make Wood’s experiment a rather poor analogy to earth’s situation.
Let me quote Wood himself on the fact that is experiments are applicable to real greenhouses, but may not be applicable to otehr situations like the earth:

“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.”

February 7, 2013 10:06 am

wayne says (February 7, 2013 at 9:35 am): “The shell as shown is outputting more energy, 470‹J/s/m2›, per time per area than is even available from the nuclear source in the first place,”
Assume the surface area of the “planet” and the outer surface area of the “shell” are both 1 square meter (or if you must, make the outer surface of the shell 1.003 square meter). Now do the calculation in watts and get back to us.
Hint: The inner surface area of the shell equals the outer surface area.

February 7, 2013 10:10 am

Experiment to test “It is impossible to devise any hypothetical situation where you can allow energy to radiate from the cooler to the warmer body WHILST simultaneously preventing the energy flowing the other way. Energy will ALWAYS flow to the cooler body at a greater flux density than the other way round. The 2nd law is not violated.”
We have two bodies on either side of a one-way mirror which is mostly transparent to light and IR in one direction and mostly opaque or reflective in the other. On the side of the mirror which is reflective, we have a black body # 1, and on the on the opposite side of the one way mirror, we have similar black #2. The black bodies are equidistant from the one-way mirror, and we will, for this experiment, stipulate the black body on the reflective side of the mirror begins at 30 degrees C, and the similar black body on the transparent side of the mirror begins at 20 degrees C. Since we are interested in the radiative energy flows, we have created an experiment so that conductive and convective flows are minimized. What happens? To exaggerate the above, we will place similar parobolic dishes behind each of the two black bodies with the black bodies at the focal points. What happens? Does the cooler black body “heat” the warmer black body. Is energy flowing from the cooler black body at a greater flux density than the other way round?

February 7, 2013 10:19 am

Wayne says: “Ever heard of the 1st law of thermodynamics? Don’t you see your problem?
No, I don’t see a problem:
* the inner sphere gains 235 W/m^2 from the internal heater.
* the inner sphere gains 235 W/m^2 from the shell
* the inner sphere radiates 470 W/m^2 to the outer shell
** Net transfer = 0 W/m^2 → No temperature change
*** Obeys 1st Law
* the outer shell gains 470 W/m^2 from the inner sphere.
* the outer shell radiates 235 W/m^2 to the inner sphere
* the inner sphere radiates 235 W/m^2 to space
** Net transfer = 0 W/m^2 → No temperature change
*** Obeys 1st Law
I am not creating or destroying any joules of energy.
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