Falsifying the Atmospheric CO2 Greenhouse Effect: G. Gerlich & R.D. Tscheuschner

[adb]

Instead of a sunbather, suppose we have a box made of thin IR transparent material. The box is evacuated and has a thick insulated base. It is placed so that the top of the base can only view the sky. A shade is positioned some distance away to block the sun. Only 400 watts of back radiation should be entering the box. We place a small beaker of water in the box and measure the time it takes to heat via back radiation.

Unfortunately the walls of the box will be at ambient temperature and will radiate to the beaker.

Somehow this setup seems similar to a pyranometer measuring the (low) temperature of the sky. If so, the water would cool rather than heat.

Is there any way that the 400 watts back radiation can be verified experimentally ?

 

[sylas]

Is there any way that the 400 watts back radiation can be verified experimentally ?

Yes, there is. It's quite routine, and was first done over 50 years ago. I previously cited Stern, S.C., and F. Schwartzmann, 1954: An Infrared Detector For Measurement Of The Back Radiation From The Sky. J. Atmos. Sci., 11, 121–129. (online). I commented further in [post=2128781]msg #73[/post] in the other thread.

What the instrument actually detects directly is the difference between forward and backradiation. But because we know the forward radiation (using the blackbody relation), the backradiation can be obtained directly. I don't know much at first hand about the workings of these instruments, but here are the wikipedia links. A pyranometer is for measuring solar irradiance, and a pyrgeometer is for measuring the IR backradiation, by measuring the difference between backradiation and the warmer surface temperature.

Cheers -- Sylas

 

[adb]

Instruments such as pyrometers bring the discussion full circle. They focus radiation from an object onto a thermocouple or an FPA array to determine its temperature by comparison to some standard, based on the amount of radiation it is emitting or absorbing. We point it at the sky and measure a temperature of say -40, we point it at the ground and measure a temperature of 20. From these values we can calculate the heat flow from the hot to the cold body taking into account radiation, view factors, transparency etc; convection; conduction; and latent heat.

 

[sylas]

I made a dumb mistake in giving numbers for daytime energy balance. It has pretty much no effect on all the discussions of temperature and radiation; but I did not need to worry about the heat capacity term... and I'd like to get the correction on record for posterity. Here's what I said previously:

We're looking at transfers at the beach. [...]

What's left? 850 in from the sun, 400 in from the atmosphere, and 500 out from the surface; we have 750 W/m² unaccounted for. This will be divided between absorbed energy heating up the surface (which will be given back again at night time, so it doesn't show up in the diagram) plus convection and latent heat from the surface (which is 102, on average, in the diagram).

Grabbing the back of an envelope: air has heat capacity of about 1000 J/kg/K. At night, you can get an "inversion", or reversal of the atmospheric temperature gradient, up to about 500m or so. [...]
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Now actually, if I am just considering energy flow at the beach, energy going to heat up the lower part of the atmosphere is simply a part of the upwards convection and latent heat. There's a little bit heating the ground that I could consider, I think, but it's small and can be ignored at this approximate level.

Hence what I should have said is simply that the upwards convection and latent heat processes would be about 750 W/m² at midday.

The average upwards convection and latent heat is given in the diagram, but since we are considering the sun is overhead, this is effectively a tropical summer, and so the geometric ratio for mean input solar radiation should be pi, rather than 4, at this latitude. The average convection is hence probably closer to 160 than the 100 shown for a global average. If the daytime peak is 750, then the nighttime minimum under the night inversion is probably about -430; or a net flow of special heat from the atmosphere back to the surface.

I'm still curious to know if my estimated flow of special heat fits any published estimates.

The radiant energy flows I am much more confident about. So my revised estimates for an example midday sunbather, all in W/m².

• 1000 downwards as sunlight.
• 400 downwards as infrared backradiation.
• 150 upwards as reflected sunlight.
• 500 upwards as infrared surface emission.
• 750 upwards as convection and special heat.

The paper I cited for measurement of backradiation has numbers which I previously quoted in [post=2130349]msg #34[/post], measured from Frederick, Maryland. Night time backradiation ranged from 206 to 312; daytime ranged from 314 to 405. Interpreted as a temperature of the sky, as seen at the surface, this would be a night time temperature from -28C to -1C, and a daytime temperature from -0.5 to 17.5. Even the -28C is surprising to me. It may be an outlier. A sky temperature of -40C would be really hard to understand, I think.


Cheers -- Sylas

 

[Phrak]

http://www.climateprediction.net/images/sci_images/ipcc_fig1-2.gif

(Source: http://www.climateprediction.net/content/basic-climate-science at climateprediction.net.)​

That's a very nice diagram, sylas.

My attention is drawn to the gray layer called Greenhouse Gases with some sort of unlabeled cloud in it. There must be 7 items running in and out of it. If you were to take a wild guess, how much of this grayed-in area is due to the greenhouse gas, water vapor?

 

[Borek]

What are 165/30 emitted by atmosphere? Why split?

 

[sylas]

That's a very nice diagram, sylas.

My attention is drawn to the gray layer called Greenhouse Gases with some sort of unlabeled cloud in it. There must be 7 items running in and out of it. If you were to take a wild guess, how much of this grayed-in area is due to water vapor?

Including cloud as water vapour, I would guess that something from 65% to 85% of the absorption is to water vapour. Water is the most important part of the net greenhouse effect on Earth.

Just as a caution, however, it's not the the percentage absorption that really matters. This gets rapidly very technical, but basically, you can think of the consequences for temperature as following from the altitude at which radiant heat can escape to space, rather than simply the fraction of radiation absorbed.

And Borek, I believe the split is between thermal emissions from the atmosphere, and thermal emissions from cloud. I think.

Cheers -- Sylas

PS. Credit goes to adp for this diagram. He was the one who found it for us and posted the links to the thread. I just supplied some img tags to his link.

 

[Phrak]

Including cloud as water vapour, I would guess that something from 65% to 85% of the absorption is to water vapour. Water is the most important part of the net greenhouse effect on Earth.

With 20 percent of the solar insolation absorbed by greenhouse gases, water vapor contributes a lot. It must vary -99 to +200% or so, from place to place, day to day, year to year, and century to century. I imagine it's as predictable as next week's weather.

Any toy model of global weather prediction would be useless if the propagated error of the input data should become larger than the predicted change, wouldn't it?

Just as a caution, however, it's not the the percentage absorption that really matters. This gets rapidly very technical, but basically, you can think of the consequences for temperature as following from the altitude at which radiant heat can escape to space, rather than simply the fraction of radiation absorbed.

Go ahead; get technical. This is a physics forum.

 

[Count Iblis]

It must vary -99 to +200% or so, from place to place, day to day, year to year, and century to century. I imagine it's as predictable as next week's weather.

The average effect does not fluctuate much, otherwise the Earth's climate would be very unstable.

 

[Phrak]

The average effect does not fluctuate much, otherwise the Earth's climate would be very unstable.

How long is the coast line of California? I'm sorry, but that's not a meaningful statement. 'Average' means nothing without a timescale.

 

[sylas]

How long is the coast line of California? I'm sorry, but that's not a meaningful statement. 'Average' means nothing without a timescale.

... which is why you need to explain what you mean by -99% to 200%. I truly have no idea.

The general idea, of course, is right. Conditions vary a lot. We can quantify that also, in various ways. I just don't know what quantity you are thinking of with those particular numbers.

Although weather is not predictable, in the sense of knowing the particular insolation, humidity, pressure, temperature, precipitation, wind, etc on a given future date, physics does give a good basis for constraining the distribution of conditions for a given location, time of day and season of year. Hence, for example, it's not actually a mystery that Alice Springs has higher mean temperatures and bigger temperature swings than Sydney, even though you can't get an accurate prediction for conditions on April 30 until closer to the day.

That's not a bad way to think of the difference between weather and climate. Climate tells you a distribution of weather for a given point, season and time of day. Weather tells you conditions on a specific day.

I'm currently working on a reply to help explain why the total percentage thermal absorption is not a good way to describe the magitude of a greenhouse effect.

Cheers -- Sylas

 

[adb]

• 1000 downwards as sunlight.
• 400 downwards as infrared backradiation.
• 150 upwards as reflected sunlight.
• 500 upwards as infrared surface emission.
• 750 upwards as convection and special heat.

From an engineering perspective this would be expressed as:

• 1000 downwards as sunlight.
• 150 upwards as reflected sunlight.
• 100 upwards as infrared surface emission.
• 750 upwards as convection and special heat.

Another question:

If the atmosphere was 100% "greenhouse" gas, that is a strong IR absorber of the same specific gravity as a non greenhouse atmosphere, exactly what would be the difference in the troposphere lapse rate, if any ?

 

[sylas]

From an engineering perspective this would be expressed as:

• 1000 downwards as sunlight.
• 150 upwards as reflected sunlight.
• 100 upwards as infrared surface emission.
• 750 upwards as convection and special heat.

You are quoting the difference between thermal flux up and thermal flux down.

Whether an engineer uses this simple difference, or else goes into more detail on the flux in each direction, depends on what problem they are solving; not on whether they are an engineer.

Engineering involves the practical application of physics to real world problems. There's a "bath" of infrared radiation at the surface, coming from all directions. That's a fact. There's a bit more coming from below than above, and in any application where an engineer needs to keep something very cold, they'll certainly deal with IR coming from above as well as from below.

If the atmosphere was 100% "greenhouse" gas, that is a strong IR absorber of the same specific gravity as a non greenhouse atmosphere, exactly what would be the difference in the troposphere lapse rate, if any ?

It's a rather odd way of phrasing the question. Greenhouse gases have different strengths. You can speak of the fraction of IR absorbed in the gas per unit distance. This is the absorbtivity co-efficient. It's not a percentage, and it has units 1/distance. (For a given density.) The greenhouse effect results not so much from the fraction of photons being absorbed, but from the mean distance they can travel before being absorbed, or the "optical depth". It's not a percentage.

The question can be made more meaningful by simply asking what would happen if the atmosphere was a much stronger absorber of IR.

The answer is still the same as I gave for you last time. The greenhouse effect doesn't alter the lapse rate. For example, the atmosphere on Venus is massively more efficient at absorbing infrared. But it has about the same lapse rate as the dry adiabat on Earth.

Humidity has a significant effect on lapse rate; but that is mainly because of the effects of latent heat. It's got nothing particularly to do with greenhouse or IR absorption.

Cheers -- Sylas

 

[adb]

Engineering takes a pragmatic view. Given 2 bodies at different temperatures, engineers calculate the heat flow between them.

It the temperature of the atmosphere is the same at a given altitude, with or without greenhouse gases, I assume that this implies that if IR absorption becomes greater at a given altitude, the result is an increase the rate of vertical air movement ?

If it's temperature doesn't change, radiation from it can't change.

 

[sylas]

Engineering takes a pragmatic view. Given 2 bodies at different temperatures, engineers calculate the heat flow between them.

Engineers and physicists both calculate what they need for a given problem.

If you are an engineer dealing with cooling for a building, you're going to take atmospheric back radiation into account, even if you don't know enough physics to realize it. But a good engineer knows the physics relating to their problem.

They might, for example, use the reference book Passive Low Energy Cooling of Buildings, by Baruch Givoni (1994). It's specifically for engineers. There's a discussion of atmospheric back radiation there, because it's a real part of the physical world, and it is important for the efficiency of your cooling systems. It's in chapter 4, on "radiant cooling".

It the temperature of the atmosphere is the same at a given altitude, with or without greenhouse gases, I assume that this implies that if IR absorption becomes greater at a given altitude, the result is an increase the rate of vertical air movement ?

If it's temperature doesn't change, radiation from it can't change.

What two situations are you comparing here? It's hard to tell. By assuming the same temperature at given altitude, you are effectively also assuming the same surface temperature, since the lapse rate is fixed.

If you want to know the impact of IR absorption, then think of two planets A, and B, with the same insolation and the same lapse rate but A has more atmospheric IR absorption than B. Is that what you mean?

Then A will have a higher surface temperature, and ALSO a higher atmospheric temperature at a given altitude.

Cheers -- Sylas

 

[Xnn]

Be careful with altitude measurments.

Sometimes it is measured in pressure (millibars) while at other times it is a distance above a reference point. There is a slight difference depending on the convention used.

 

[adb]

None of the texts I have used or engineering collegues I have spoken to, make use of assumptions other than those based on the 2nd Law, in calculating heat flows an any situation we have encountered.

What two situations are you comparing here? It's hard to tell.

The same as in a couple of posts back where you stated :

The answer is still the same as I gave for you last time. The greenhouse effect doesn't alter the lapse rate.

 

[sylas]

None of the texts I have used or engineering collegues I have spoken to, make use of assumptions other than those based on the 2nd Law, in calculating heat flows an any situation we have encountered.

No offense intended, but I don't believe that can be true. The second law is not sufficient to calculate heat flows. If you are actually calculating heat flows, you're using more than the second law.

It's entirely possible that many engineers are not expert in all the relevant physics. As the book I cited for you previously demonstrates, there are other engineers who do know and use more of the relevant physics, including details of backradiation. I guess it will depend on the kinds of problems you work with.

For example, that book explains why you get better cooling with a radiator that has high emissivity over wavelengths about 8 to 13 microns, but is reflective in other parts of the spectrum. It's because atmospheric backradiation is very weak in that band. You see, for calculating heat flows, you sometimes need to distinguish different wavelengths of light. Identifying the best material to use for coating a rooftop radiator turns out to be such a case.

If your only point is that all this physics is new to you, that's no problem. Much of it is new to me too, as I keep trying to learn more about the details of atmospheric thermodynamics. I'm still sorting out details of that infrared window of transparency at 8-13 microns myself.

I guess the point is that different people have different areas of expertise. A civil engineer is not usually going to be as helpful as an atmospheric physicist when it comes to sorting out details of how the greenhouse effect works.

The same as in a couple of posts back where you stated :

The answer is still the same as I gave for you last time. The greenhouse effect doesn't alter the lapse rate.

That's what I thought; in which case you should not have have added the bit about temperatures in the atmosphere remaining the same at a given altitude. The greenhouse effect does result in a higher atmospheric temperature for a given altitude, within those parts of the atmosphere where you get circulation and the lapse rate. (The troposphere.)

The particular case we are considering is two otherwise identical planets, differing in the amount of IR absorption in the atmosphere. The planet with the greater IR absorption will have higher temperatures at the surface and within the troposphere. The lapse rate is not affected. The tropopause (where circulation ends and where the lapse rate comes back to zero) will be at a higher altitude.

Cheers -- Sylas

 

[adb]

Given :

Th = temperature of a hot body
Tc = temperature of a cold body

... what formula are you using for "back" radiative heat transfer from the cold body to the hot body ?

 

[sylas]

Given :

Th = temperature of a hot body
Tc = temperature of a cold body

... what formula are you using for "back" radiative heat transfer from the cold body to the hot body ?

That's not quite enough information in general; but if we add that the two bodies are facing one another flat on, with nothing absorbing the radiation between them and with each body being "black" (that is, absorbing all incoming radiation), then we use the Stefan-Boltzman radiative law. A black body surface at a given temperature T radiates σT⁴ Watts per square meter. The constant σ is 5.67e-8 W/m^2/K^4.

Here's an example. You have two flat black walls, facing each other, with vacuum between them. Behind one wall is boiling water; behind the other is cool water, at about 15C. Each wall is in thermal equilibrium with the water behind it.

The boiling water is at 373K, and radiates σT⁴ = about 1100 Watts per square meter. The cool water is at 288K, and radiates σT⁴ = about 391 Watts per square meter.

Thus the boiling hot wall is receiving about 391 Watts per square meter from the cool wall. There's no violation of the second law, because the cool wall is receiving 1100 Watts per square meter from the hot one. The net transfer of energy is 709 Watts per square meter from the boiling hot wall to the cool wall. To calculate the net flow of heat, you have to calculate the radiation in each direction, and take a difference.

There's a generalized form of the law that includes an "emissivity" factor for walls that are less than perfect at absorbing radiation. This factor is also a measure of how effective they are at emitting radiation, and you can prove this from the second law. In full generality the emissivity can be frequency dependent.

Cheers -- Sylas

 

[Borek]

adb, have you heard about kinetic approach to the equilibrium? I know this concept as used in chemistry. While details are different, that's very similar mechanism to the one present here. There is a constant flow of energy (mass in chemistry) in both directions. When part of the system has higher potential (be it chemical potential or higher temperature) net transfer is from the higher potential part of the system to the other part. Once potentials get even, net transfer becomes zero, but there is still transfer occurring - just the speed of the transfer in both directions is identical. System is in equilibrium.

If I understand the situation correctly if both bodies are perfectly black you don't need temperature of the hot body to calculate speed of heat transfer (ie power) from the cold body to the hot body - temperature of the cold body is enough. You need temperatures of both bodies to calculate NET transfer of the heat (which is what you observe in the real world).

ducks behind the chair and removes the green wig to be more difficult to spot and recognize

 

[sylas]

If I understand the situation correctly if both bodies are perfectly black you don't need temperature of the hot body to calculate speed of heat transfer (ie power) from the cold body to the hot body - temperature of the cold body is enough. You need temperatures of both bodies to calculate NET transfer of the heat (which is what you observe in the real world).

Exactly.

... with the proviso that the radiation in each direction is fully real, and fully observable.

In this case, the easiest thing to measure will probably be temperature of the cool water as it warms up, or the amount of energy you have to keep adding to the hot water to keep it boiling, etc; and those things will relate to net transfers of heat.

You can also use probes in the cavity between the walls, and measure the radiation going in each direction separately. It's all real.

An infrared probe has to be colder than the radiation it is observing so that you don't mess up readings with the probe's own thermal radiation. This principle is used in an infrared telescope, which needs to be kept very cold to work well.

Cheers -- Sylas

 

[adb]

In other words, you're taking half of the standard engineering equation for radiative transfer.

 

[sylas]

In other words, you're taking half of the standard engineering equation for radiative transfer.

I've tried to answer your questions by actually giving the formula used and showing a worked example.

It will help if you can do the same. What do you mean by "half the standard engineering equation"? Engineers don't use different equations from physicists, and they don't get different answers for pragmatic questions, such as heat flow.

How would you describe the particular example I have given, of the radiant heat flow between two walls with vacuum between them as the insulator?

I can do more complicated examples as well, where you consider heat flow by convection and conduction and heat of evaporation and radiation. The more complex examples will change the total flow of heat, but if as a pragmatic engineer you want to consider all the factors involved then you are going to include radiant transfers along with everything else. And for that, you'll need to include radiation moving in both directions.

It's real. It exists. It can be measured. Deal with it.

Cheers -- Sylas

 

[Phrak]

I though you were interested in addressing the variability, uncertainty and experimental error of the most predominant material effecting weather, sylas.

 

[sylas]

I though you were interested in addressing the variability, uncertainty and experimental error of the most predominant material effecting weather, sylas.

That remark is merely disruptive of the discussion. There's nothing in my posts which suggests I am trying to single out one particular "material". I'm also not all that interested in weather, as such.

Weather variation is chaotic. You cannot possibly predict the particular weather on a given day in the future, except in the very near future.

You CAN, however, make some perfectly sensible predictions based on simple physical principles of the range within which weather can be expected. This is climate, and THIS is where my primary interest lies. To take a really obvious example, winter tends to be colder than summer in mid-latitudes, but there's no such thing as summer and winter at the equator. Winter in Canada tends to be more harsh than in the same latitudes in Norway. We know why this occurs. It can be explained simply, and without direct reference to all the other processes giving rise to the unpredictable chaos of specific weather conditions on a given day. And it can certainly be explained without distracting asides about one supposedly most important material. (Air? Water? Sunlight? Radiation? It's silly to single out one as "predominant".)

Most of the discussion here is not at a level of full complexity. Here we are mostly still at the stage of sorting out elementary thermodynamics. The paper in the first post of this thread is an example. It's physical nonsense, and it is well worth while explaining why. You don't need all the full complexity of climate analysis for this. Getting these basics right is a solid basis for going on to deeper understanding.

If you want all the full details, don't look to me to write a textbook. I'll stick to addressing things at a more basic level, appropriate to this forum, dealing with particular points of confusion as they arise.

There are good textbooks for more detail. When I started learning about technical details of this myself, I got a lot of value from a long on-line text, called "Principles of Planetary Climate", by R.T. Pierrehumbert at the Uni of Chicago. It can be used for advanced undergraduate course work. It deals with physical basics that can be applied to any planet, and covers physics of lapse rate, radiative transfers, circulation, condensation, etc. It can take you to a level of understanding that will allow you to calculate bounds on the lapse rate and height of the troposphere, and show a generalized definition of "tropopause" or "stratosphere" that continues to work on a planet with a vastly different atmsopheric profile of temperature and pressure. It is available online, and look on that page for the 13.6 Mbytes pdf download of the current working draft. Not an easy read, but I've learned a heck of a lot from it so far. I am still struggling with the harder parts. Some readers might find it of interest.

In the meantime, let's stick with calculating a radiative transfer. What do you reckon would be the equilibrium temperature, approximately, of a small ball of black iron suspended above a large pan of liquid nitrogen, under a sunlit summer sky. Rough estimate will do.

It's a useful exercise, and the answer let's you get to grips with one of the important features of Earth's climate that some people find confusing: atmospheric backradiation.

Cheers -- Sylas

 

[Phrak]

It's difficult to dissrupt a dead thread.

 

[suibhne]

Silas
My interest in climate change was sparked by the East Anglia e-mail scandal.
I have a physics degree obtained in 1967 and have taught high school physics until I retired.
The G and T paper seemed to be a genuine contribution to the ongoing debate.
I followed the discussion in this forum with interest.
My take on the Moon/Earth difference is that the Moon does not rotate or have Oceans or an atmosphere.
It is true that the atmosphere stores energy in the form of translation KE of the gas molecules and latent heat and so on but the huge contribution that your diagram shops of back-radiated is not backed up with realistic calculations nor have you answered abb well made point that if the Suns EM radiation can be blocked or reflected this seems not to happen in the case of the atmospheres.
The G and T paper gets quite technical in places and I don't intend to revise my vague knowledge of non linear partial differential equations to nail every last point but it seems to me rather unlikely to say the least that they have made fundamental mistakes in basic physics.

 

[Evo]

Due to the inability to moderate discussions pertaining to Global Warming or Climate Chnge. Discussions on the subject have been temporarily suspended. It would appear that this thread slipped through the cracks, I apologize for any inconvenience.