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

[sylas]

She's already in the shade. We are now blocking out the major source of radiation as the next step ... from the atmosphere. Use a sheet of glass as you suggest if you wish.

She felt an instant (instant because radiation is being blocked) chill when the sun was blocked; but nothing when the radiation from the sky was blocked as well !

Anything used to block radiation from the sky is going to reflect radiation from the surface. (Consequence of the second law.) She is in a bath of infrared radiation coming from all directions. There's a bit more coming up from the surface than back from the sky. It's thermal infrared radiation; and she's emitting it herself as well.

By the laws of thermodynamics, you are not going to be able to construct a barrier that will stop radiation getting in without also trapping inside the radiation that's already there. If you could do such a thing, you could use it to break the second law (think about it). Off hand I cannot think of any way to simply remove the infrared radiation. The glass room won't do it.

Cheers -- Sylas

 

[Borek]

So now you are claiming a perpetual energy as the atmosphere doubles the energy it receives from the surface or quadruples it and returns half back to the surface.
If this is true why are we useing hydrocarbons to provide energy. According to your theory we only need to capture all the extra energy being produced by the atmosphere.
I think that is called perpetual motion.

Mike, I don't know which one of you is right. What I know is that your "hit and run" strategy, compared with sylas efforts to explain in details his point of view, looks like if you were interested just in leaving impression that there is something wrong with sylas's explanation, no matter what the reality is.

 

[Mike Davis]

Borek:
Either I am reading what he is saying incorrectly (Which is possible). What he is saying is not what I have learned and I will leave it at that.

 

[Borek]

Borek:
Either I am reading what he is saying incorrectly (Which is possible). What he is saying is not what I have learned and I will leave it at that.

Can't you try to pinpoint why it is wrong in some more detailed manner?

 

[adb]

Anything used to block radiation from the sky is going to reflect radiation from the surface. (Consequence of the second law.) She is in a bath of infrared radiation coming from all directions. There's a bit more coming up from the surface than back from the sky. It's thermal infrared radiation; and she's emitting it herself as well.

It is not a "bath" of IR radiation. According to greenhouse, there is a hemispherical emitter (the sky) providing as much radiation to the top of the sunbather, as the sun was, before she was put in the shade.

Radiation shields are used everywhere. Your car exhaust is fitted with radiation shields. A tent acts as a radiation shield to prevent radiant heat loss to cold night skies for mountain climbers. Sit in front of an electric radiator ... feel the heat on your face ... then hold a sheet of paper in front of your face ... your face instantly feels cool because IR radiation is being blocked by the paper radiation shield.

By the laws of thermodynamics, you are not going to be able to construct a barrier that will stop radiation getting in without also trapping inside the radiation that's already there. If you could do such a thing, you could use it to break the second law (think about it). Off hand I cannot think of any way to simply remove the infrared radiation. The glass room won't do it.

The above examples show it is easy to block IR radiation. Adding additional isolated shields will further reduce radiation transmission between the radiation source and the cold body.

Of course all the above examples have IR radiation flowing from hot bodies to colder ones. I'm still mystified by how radiation shields are supposed to work (or not work) when you claim IR radiation traveling from a cold body (the sky) to our warm sunbather (now in the shade).

 

[sylas]

Borek:
Either I am reading what he is saying incorrectly (Which is possible). What he is saying is not what I have learned and I will leave it at that.

There's another possibility, which is that our differences come about in actually applying what we have both learned to a couple of specific situations.

We've both learned some basic physics, obviously. We both agree on laws like conservation of energy, and the second law of thermodynamics, and so on. We both know how to calculate the radiant energy coming from a blackbody. We both agree on the solar constant, and temperatures on the Moon, and so on. We both agree that an atmospheric greenhouse effect is different from a glass greenhouse effect. We both agree that the atmosphere helps to equalize temperatures around the Earth's surface.

In fact, when you really look at it, I think we agree on a lot more of the fundamentals than we disagree!

Is it going to possible to make a simple statement of any points of difference? I think if you make a deliberate effort to single out the points of real difference, you'll be able to resolve them, and you'll even find that you are not in fact "unlearning" much at all! All you really need is to apply what you've learned already just a little bit more accurately.

I'll propose two points where I think rapid progress should be possible.

(A) The association between greenhouse and equalization of temperatures

You have said that a greenhouse works by equalizing temperatures. Is that right? But were you ever actually taught that anywhere?

I think you may be mixing up two different but real effects, and simply labeling them the wrong way around.

The main process by which a glass greenhouse works is that it prevents hot air from rising out through the roof. Sunlight comes in, and heats up the floor, which in turn heats up the air above the floor. Now normally, hot air would rise up away from the floor, and cooler air from up higher in the atmosphere would be circulated back down again. It's why we have thermals and air circulation out in the open. Basically, there is a convective flow of heat up from the surface into the atmosphere caused by moving air masses. But in the greenhouse, the hot air is trapped near the floor. The glass roof prevents the processes by which temperatures are equalized between air at different heights above the ground, and results in a larger temperature difference between air in the green house, and in the atmosphere above it.

The circulation of air horizontally is also important for equalizing temperatures. This does not have much to do with a greenhouse. A greenhouse with no walls will continue to work, despite loss of heat out the sides; but a greenhouse with no roof can't even get started.

Horizontal circulation is important for equalizing temperatures around the planet. Winds tend to blow from a hot high pressure area into a cooler low pressure area. You see this on daily weather maps. So we agree that the atmosphere helps equalize temperatures around the Earth. The thing is... that is not the greenhouse effect! I suspect you've just associated the wrong label with the phenomenon. The greenhouse (glass greenhouse) prevents circulation (mainly vertical circulation, and also horizontal if you have walls). The equalizing effect around the Earth is caused by circulation.

As for the atmospheric greenhouse effect, this is not the same as a glass greenhouse. What they have in common is that they both work by inhibiting movement of heat up from the surface. But they do it by very different processes. The atmospheric greenhouse works by blocking radiant heat flow from moving up into space. The glass greenhouse works by blocking convective heat from moving up into the atmosphere.

(B) Perpetual motion

You've suggested that the atmospheric greenhouse effect is described as a kind of perpetual motion machine.

In this case, we all agree that energy is conserved, and you can't get a continuous flow, or circulation, without a continuous source of energy from somewhere. Of course, the energy coming in from the sun drives all kinds of continuous cycles. The water cycle, with water circulating continuously up into the atmosphere and back down to the surface, through rivers to the sea, is basically "perpetual", for as long as the Sun shines on the ocean, and we can use this to drive, for example, a hydroelectric power station.

We've spoken of atmospheric "backradiation"; which is a continuous flow of radiant heat from the atmosphere back down to the surface. You've apparently taken this as a "perpetual motion" in violation of the first law, since it is being driven from the surface.

But in this case, the problem goes away as soon as we take into account the flow of heat in both direction; just like we explain why the apparently endless flow of water from rivers doesn't overflow the ocean. There's another flow of water in reverse to consider.

In the greenhouse effect, there is a larger flow of radiant heat up from the surface than there is backradiation from the atmosphere down to the surface. When you include also the energy exchange from the atmosphere out into space, it all balances. The atmosphere receives exactly as much energy as it emits. The flow of energy through the atmosphere proceeds mainly from the surface and out into space, with a balance between the total energy in and total energy out, and this flow will continue as long as the Sun shines.

I'm quite sure you've never actually been taught that there's a perpetual motion machine in the atmosphere. I think you have picked this up from incomplete descriptions, which don't attempt a complete account of all the energy flows involved. If you take a full description of the energy flows as given in energy balance diagrams for sunlight and the atmosphere, it's quite clear that there's no perpetual motion machine involved.

-------

If there are other points where you think I am saying something different from what you have learned, just try to spell it out as carefully as you can, identifying where the difference arises. I'm a pretty friendly chap, and I won't abuse you when I think I can see a point of error. I'll just point it out, and if you find real problems with my descriptions I'll take them on board gratefully. Getting things right is a win/win situation.

One difficulty is that discussion can be confused by use of poor quality reference material. If you use conventional reference books on Earth science, or climate, or thermodynamics, or astronomy, you are pretty safe for the most part. Sometimes a description can be incomplete or poorly expressed, but you won't get errors at the level of violating fundamental laws of physics. On the other hand, there are some webpages or articles that are completely wrong, even outright pseudoscience. Sometimes they come from individuals who ought to know better, and who have legitimate scientific credentials (usually exaggerated). Usually they'll mix in a bit of real physics with some completely counter factual claims or bad errors in analysis, in such a way that it can look superficially plausible. From your posts, I think you have enough physics background to engage a discussion of such articles on their own merits, and that you won't actually be "unlearning" anything in the process. Just clearing up how to apply what you know already.

Good luck with it. Don't give up! Cheers -- Sylas

 

[adb]

We've spoken of atmospheric "backradiation"; which is a continuous flow of radiant heat from the atmosphere back down to the surface.

We're both intelligent people and I'd like to tie things down. I don't work in the industry and have nothing to gain whether greenhouse exists or not.

No question that real greenhouses have no analogy with the atmosphere.

The above is the crux on which I am trying to understand your thinking. The amount of IR "back radiation" from the atmosphere is claimed by greenhouse theory to be comparable to the total radiation from the sun.

Radiation from the atmosphere or anywhere else travels in straight lines. Our sunbather will experience downward radiation from the sun as well claimed downward radiation from the hemispherical sky source. If the sunbather is shielded from the sun (ie placed in shadow), she will feel it immediately, as expected. However, there is no effect on subsequently shielding from the second source, the sky.

We can only conclude that any "back radiation" is insignificant.

This is as expected. Where temperature differences are small, heat transfer is primarily by convection and conduction. Radiation becomes more significant as temperature differences increase.

It is also interesting that you agree that greenhouse does not have a significant effect on atmospheric temperature profiles. This agrees with what Thieme suggests.

 

[sylas]

It is not a "bath" of IR radiation. According to greenhouse, there is a hemispherical emitter (the sky) providing as much radiation to the top of the sunbather, as the sun was, before she was put in the shade.

The notion of a "bath" of radiation is standard, and it applies here because there are large flows in both directions. Any IR detector is going to see radiation coming from all sides, and that is called a bath of radiation.

If you want to know what occurs according to the greenhouse, use the description of the greenhouse you've provided already. Here it is again, inserted visually.
http://www.climateprediction.net/images/sci_images/ipcc_fig1-2.gif

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

Remember, that is averaged over night and day, summer and winter, poles and equator, clear and cloudy. If you want to get concrete numbers for the sunbather, you'll have to adjust the numbers for some typical daylight sunbathing.

OK. Let's do some physics! I'll welcome corrections to my numbers; but they have to stay consistent with basic physics.

You can skip over this indented section at first reading. It's only here because I want to use more realistic daytime numbers. The real answer to the puzzle is simply that you can't block the backradiation without trapping the upwards radiation also.

We're looking at transfers at the beach. The 198 W/m² (168+30) sunlight coming down will be roughly 1000 W/m². This is a scaled up by four for the sun being overhead; plus a bit more for clear sky and reduced atmospheric reflection. This corresponds well to measurements of summer sunlight at midday on a clear day. The ground reflection scales the same way, up from 30 to 150.

The surface radiation upwards relates to ground temperature. 390, by Stefan-Boltzmann, corresponds to about 15C. But in the day, when sunbathing, it will be more like 30C, which would be radiation of about 480. To stick with rough figures, I'll use 500 W/m² upwards, which is a surface temperature of 33C (or 92F if you like Fahrenheit). This also has the benefit of keeping temperatures close to body skin temperature; because we want to ignore complications of body heat.

The radiation down from the atmosphere tends to remain roughly in proportion to surface air temperature. (See the paper I cited previously on measurements of backradiation.) So we can scale the 324 backradiation to roughly 400.

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. This is the air giving heat back to the surface at night. Density is about 1.2 kg/m^3. Hence the upper 500m of the atmosphere is about 6e5 J/K/m^2. Assuming a temperature difference of about 20 degrees from night to day, we get about 1.2e7 J/m^2 stored energy difference. Assuming a transfer of this over 8 hours, or about 2.9e4 seconds, I get a bit over 400 W/m² flux. That sounds plausible as about the flux of excess energy up into the daytime sky to heat the air, storing energy that is given back at night. So in the daytime, there's about 400 W/m² energy actually being soaked up in the air. The 350 still unaccounted for needs to show up as convection and latent heat… and this looks a credible value also, because the average over night and day is given in the diagram as 102. If anyone has actual measurements of daytime energy balance, I'd love to see them for comparison with this guesstimate for division of the remaining 750.

Final numbers I propose for the 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.
• 350 upwards as convection and special heat.
• 400 excess being absorbed to heat up the air.

[/size]​

OK. The real question is, can't we make the sunbather feel cold by shielding her from the 400 Wm² backradiation? The answer is… no, because there's no physical way to remove that without trapping the radiation she emits herself.

Radiation shields are used everywhere. Your car exhaust is fitted with radiation shields.

Right; and this is useful, because the exhaust pipe is so much hotter than the air outside. What you CAN'T do is make a shield that keeps all the thermal heat inside at the exhaust pipe, but still let's in heat from the other direction. If you could make such a shield, you could take a warm brick, which is cooler than the exhaust pipe, put it next to the radiation shield, and have heat from the brick flowing in through the barrier, while heat from the exhaust pipe was prevented from flowing the other way. You could use such a shield to heat up a hot object from a colder one, in violation of the second law.

As I said previously, we CAN block infrared radiation with glass. It does make a very good shield for IR radiation. So what happens if we put a glass sheet over the sunbather?

Sure enough, we've blocked out the 400 W/m² atmospheric backradiation. But the sunbather, who is at a nice comfortable 33C, is radiating herself, upwards, at 500 W/m²! And we've trapped that inside! She gets HOTTER as a result. You simply cannot block out the 400 W/m² down without also blocking the 500 W/m² going up.

All your examples of cooling something by blocking radiation work because they are shielding you from something hotter than you are. But in the greenhouse effect, the surface (and the sunbather) is hotter than the atmosphere. If you put in a block, you're actually making it harder for the sunbather to shed the thermal radiation she needs to emit to keep cool.

It's the same problem in all these discussions. All the attempts to portray greenhouse, or backradiation, as some kind of violation of thermodynamics or perpetual motion are ignoring the energy flows in the other direction.

The paper by G&T is just the same. It may look superficially plausible, but it's not going to take in anyone who works with atmospheric thermodynamics for a second. If they were first year physics students, we'd simply say that they need to learn a bit more about basic physics. But these clowns have had their errors pointed out to them at length, for years. The paper came out ages ago in arxiv, and the errors were identified quickly and publicly. Recently, they actually managed to get their paper into a small mainstream journal, bypassing the normal technical review by appearing in an "invited" category. Whichever editors made the invitation have slipped up badly, and G&T are plainly pseudoscientific cranks – on this topic, at least. Neither one of them has any publication record in climate or basic thermodynamics, nor indeed do they have much of a publication record at all. They are very minor players in real science, with a sideline in gibberish. It's not unheard of, and there are plenty of other isolated examples in other fields of science. From time to time such individuals do manage to get some of their stuff into the scientific literature, because editors are not infallible.

------

And also, in response to your next post:

We're both intelligent people and I'd like to tie things down. I don't work in the industry and have nothing to gain whether greenhouse exists or not.

Thanks muchly; I feel the same way. I'll add that there's no animosity between us either, as far as I am concerned, and I'm sure you feel the same. We're just having an "energetic" discussion of points of physics. It's fun, and hopefully it's educational. I expect we'll be able to tie most of it down. The point about radiation shields working in both directions will be crucial.

… It is also interesting that you agree that greenhouse does not have a significant effect on atmospheric temperature profiles. This agrees with what Thieme suggests.

Sure. Thieme's errors – and his essay is full of them – are on other matters. His comments on convection are merely a distracting non-sequitur. He might as well point out that greenhouse has no effect on atmospheric scattering of light, and the consequent colour of the sky.

Cheers -- Sylas

 

[adb]

OK. The real question is, can't we make the sunbather feel cold by shielding her from the 400 Wm² backradiation? The answer is… no, because there's no physical way to remove that without trapping the radiation she emits herself.

How about reflective foil, blackened on the sunbather side ?

I'm an engineer, not a physicist ... I still find the concept of heat traveling from cold bodies to hot bodies difficult to get my head around.

 

[sylas]

How about reflective foil, blackened on the sunbather side ?

It makes no difference. If it is "black" then it is absorbing and emitting thermal radiation. Think of a thermodynamic "blackbody". There's still energy coming out. It's just the that all the energy is emitted as thermal radiation, with a spectrum depending on temperature. It can't just soak up all the thermal radiation and give nothing back, unless it is initially cooled to absolute zero.

If you use a super cold glass shield you could just soak up all the IR radiation, and emit nothing in return, and still let though the sunlight.

With a supercold shield, you would indeed cool down the sunbather. But it's not because of "cold" flowing from the shield to the sunbather.

Add in a second transparent vacuum shield to prevent any loss into the cold shield by conduction, or (equivalently) to trap a small layer of warm air around the sunbather. She's still going to feel significantly cooler as a result. It's not because she's actually touching anything cold; there's still a layer of warm air around her trapped by the vacuum shield.

Of course, this is far from equilibrium. The cold shield will soon heat up again, unless you apply refrigeration. But the immediate effect of moving the cold shield into place should have the effect of soaking up the downwards flux without reflecting any of the upwards flux, which is what we want. When I previously said this was impossible, I should have said it was impossible at equilibrium.

I'm an engineer, not a physicist ... I still find the concept of heat traveling from cold bodies to hot bodies difficult to get my head around.

Think in terms of a warm brick next to a hot brick. From above the two bricks, you can feel heat coming from each one. There's more coming from the hot one. They are both radiating in all directions. Drop a cold slab of insulating material between the bricks. Both sides will warm up; and the side facing the hotter brick will warm up more quickly.

Evidently, there's heat from the hot brick moving towards the warm brick, and also visa versa.

Cheers -- Sylas

 

[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. [...]
[/size]​

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.