Guest Post by Willis Eschenbach.
For all of its faults, the IPCC (Intergovernmental Panel on Climate Change) lays out their idea of the climate paradigm pretty clearly. A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. Today I found myself contemplating the concept of radiative forcing, usually referred to just as “forcing”.
So … what is radiative forcing when it’s at home? Well, that gets a bit complex … in the history chapter of the Fourth Assessment Report (AR4), the IPCC says of the origination of the concept (emphasis mine):
The concept of radiative forcing (RF) as the radiative imbalance (W m–2) in the climate system at the top of the atmosphere caused by the addition of a greenhouse gas (or other change) was established at the time and summarised in Chapter 2 of the WGI FAR [First Assessment Report].
Figure 1. A graph of temperature versus altitude, showing how the tropopause is higher in the tropics and lower at the poles. The tropopause marks the boundary between the troposphere (the lowest atmospheric layer) and the stratosphere. SOURCE
The concept of radiative forcing was clearly stated in the Third Assessment Report (TAR), which defined radiative forcing as follows:
The radiative forcing of the surface-troposphere system due to the perturbation in or the introduction of an agent (say, a change in greenhouse gas concentrations) is the change in net (down minus up) irradiance (solar plus long-wave; in Wm-2) at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium, but with surface and tropospheric temperatures and state held fixed at the unperturbed values.
In the context of climate change, the term forcing is restricted to changes in the radiation balance of the surface-troposphere system imposed by external factors, with no changes in stratospheric dynamics, without any surface and tropospheric feedbacks in operation (i.e., no secondary effects induced because of changes in tropospheric motions or its thermodynamic state), and with no dynamically-induced changes in the amount and distribution of atmospheric water (vapour, liquid, and solid forms).
So what’s not to like about that definition of forcing?
Well, the main thing that I don’t like about the definition is that it is not a definition of a measurable physical quantity.
We can measure the average surface temperature, or at least estimate it in a consistent fashion from a number of measurements. But we can never measure the change in the radiation balance at the troposphere AFTER the stratosphere has readjusted, but with the surface and tropospheric temperatures held fixed. You can’t hold any part of the climate fixed. It simply can not be done. This means that the IPCC vision of radiative forcing is a purely imaginary value, forever incapable of experimental confirmation or measurement.
The problem is that the surface and tropospheric temperatures respond to changes in radiation with a time scale on the order of seconds. The instant that the sun hits the surface, it starts affecting the surface temperature. Even hourly measurements of radiative imbalances reflect the changing temperatures of the surface and the troposphere during that hour. There is no way that we can have the “surface and tropospheric temperatures and state held fixed at the unperturbed values” as is required by the IPCC formulation.
There is a second difficulty with the IPCC definition of radiative forcing, a practical problem. This is that the forcing is defined by the IPCC as being measured at the tropopause. The tropopause is the boundary between the troposphere (the lowest atmospheric layer, where weather occurs), and the stratosphere above it. Unfortunately, the tropopause varies in height from the tropics to the poles, from day to night, and from summer to winter. The tropopause is a most vaguely located, vagrant, and ill-mannered creature that is neither stratosphere nor troposphere. One authority defines it as:
The boundary between the troposphere and the stratosphere, where an abrupt change in lapse rate usually occurs. It is defined as the lowest level at which the lapse rate decreases to 2 °C/km or less, provided that the average lapse rate between this level and all higher levels within 2 km does not exceed 2 °C/km.
This is an interesting definition. It highlights that there can be two or more layers that look like the tropopause (little temperature change with altitude), and if there is more than one, this definition always chooses the one at the higher altitude.
In any case, the issue arises because under the IPCC definition the radiation balance is measured at the tropopause. But it is very difficult to measure the radiation, either upwelling or downwelling, at the tropopause. You can’t do it from the ground, and you can’t do it from a satellite. You have to do it from a balloon or an airplane, while taking continuous temperature measurements so you can identify the altitude of the tropopause at that particular place and time. As a result, we will never be able to measure it on a global basis.
So even if we were not already talking about an unmeasurable quantity (radiative change with stratosphere reacting and surface and tropospheric temperatures held fixed), because of practical difficulties we still wouldn’t be able to measure the radiation at the tropopause in any global, regional, or even local sense. All we have is scattered point measurements, far from enough to establish a global average.
This is very unfortunate. It means that “radiative forcing” as defined by the IPCC is not measurable for two separate reasons, one practical, the other that the definition involves an imaginary and physically impossible situation.
In my experience, this is unusual in theories of physical phenomena. I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable. Climate science is already strange enough, because it studies averages rather than observations. But this definition of forcing pushes the field into unreality.
Here is the main problem. Under the IPCC’s definition, radiative forcing cannot ever be measured. This makes it impossible to falsify the central idea that the change in surface temperature is a linear function of the change in forcing. Since we cannot measure the forcing, how can that be falsified (or proven)?
It is for this reason that I use a slightly different definition of the forcing. This is the net radiative change, not at the troposphere, but at the TOA (top of atmosphere, often taken to mean 20 km for practical purposes).
And rather than some imaginary measurement after some but not all parts of the climate have reacted, I use the forcing AFTER all parts of the climate have readjusted to the change. Any measurement we can take already must include whatever readjustments of the surface and tropospheric temperatures that have taken place since the last measurement. It is this definition of “radiative forcing” that I used in my recent post, An Interim Look at Intermediate Sensitivity.
I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.
w.
The comments made by my fellow engineer have really got me thinking about this subject in a clear way.
What we have is a system where the tropopause as shown in the diagram is at -60Celsius. It is about 10,000m high. The ocean might be at say +25Celsius. Between the two we therefore have a temperature of 8%Celsius over a distance of 10,000m which is a very low temperature gradient indeed!
Now normally we think of air as being quite thermally conductive, because we use it as a coolant, but if you but enough of it in series you end up with a very good insulator. This is what is happening here. It is analogous to the electrical domain – copper is a very good conductor but if you put enough of it in series you end up with a very high resistance (a 7km loop of telephone wire has a resistance of about 2000ohms).
Due to the fact we have so much atmosphere between us and the tropopause the atmosphere itself at any two points say a metre apart are very nearly at thermal equilibrium – this means there can be almost NO cooling of the Earth’s surface either by radiation or by conduction through the atmosphere within the troposphere.
Above the tropopause the atmosphere actually starts to get warmer due to direct heating by incoming UV that reaches the ground. This tells us several things: The tropopause can only be cooling by being in direct thermal contact with space – i.e. it is radiating into space. It also tells us that the tropopause is still warmer than space largely because it finds it difficult to radiate the heat through the relatively warm layer of the upper stratosphere.
The stratosphere was not discovered (or expected) until 1900. Arrhenius proposed the greenhouse effect in 1876. Arrhenius had now way of knowing that the Earth’s atmosphere is actually a lot warmer at the top than at the bottom, so the actual physical model he proposed and which stands to this day was originally based on a lack of knowledge of the complexity of the Earth’s atmosphere.
Given that the troposphere has a very low temperature gradient I would suggest that very little of the Earth’s incoming solar energy can be dissipated through this route. I suggest it can only be dissipated via direct re-radiation to space. Measurements suggest this is not the case however. I propose that the measurements are wrong. That might seem like a bold claim but we have to remember that the Earth would be -18Celsius without an atmosphere and is +14Celsius with it – that’s about a 10% difference compared to absolute zero (-273Celsius). So you have to get those measurements spot on to be sure you have modelled the causes of the small impact of the atmosphere correctly. Considering that the heating of the Earth is likely to be greatest over particular parts of the land near the equator at noon, we must also consider that this is also the maximum time of cooling. Therefore you need to measure the cooling at the same time of maximum heating – a difficult task to accomplish accurately. Note also that using some kind of average won’t wash – you need to break up the surface of the Earth into small areas and make the measurements accurately for each one to be sure you understand the contribution that each makes to the heating effect at ground level and any subsequent cooling. I think that is this was done properly it would show that warming of the upper stratosphere by UV is what causes the Earth to be warmer than expected as it prevents significant cooling through the atmosphere. If this is the case then CO2 can be expected to have little impact even if raised by 2 orders of magnitude since the thermal gradient would only go from very low to even lower.
[“8% Celsius rise” “85 (degrees) Celsius” rise ? Mod]
Willis: once again be careful about your tendency to be overly assertive. Gnomish is essentially correct that there is no direct relation ship between watts of radiated power and temperature in any given real-world case. This is because the radiated power of IR, for instance, is an electro-magnetic wave, it is not a beam of “hot” (in that sense it is quite different from conduction/convection which are indeed flows of “hotness” with no need for conversion between energy types). The electro-magnetic wave needs to be converted to heat energy to get an increase in temperature.
Stefan/Boltzmann theorised a perfect absorbtive object that converts the electro-magnetic energy to heat with 100% efficiency and therefore were able to develop a relationship between the electro-magnetic power reaching the object and the temperature. Any real-world application needs to take into account that the process is likely to be nowhere near 100% efficient. The process is nowhere near 100% efficient for the surface of the Earth.
Willis Eschenbach says:
December 20, 2012 at 12:19 am
….So, just what is your contention here? That DLR isn’t measured? I discussed the DLR measurements at the TAO buoys here.…..
>>>>>>>>>>>>>>>>>>>>>
Willis, I Just took a look at that information and I have a question.
First clouds are composed of tiny water droplets or ice crystals that are suspended in the air. Those clouds would have a temperature T, and would radiate as a gray body. This radiation would be a percentage of the theoretical black body radiation which is proportional to the fourth power of the absolute temperature as expressed with Stefan-Boltzmann Law. Would this not be confounding the whole idea of DLR from greenhouse gases. Heck would not the entire atmosphere have DLR because it is a gray body radiator and therefore the DLR would again have nothing to do with the Greenhouse gas theory except as a confounding contributor?
One site said the emissivity of the atmosphere is about 0.83 based on the gases but not the diatomic gases (Link will not connect so that is off the google blurb)
For clouds (liquid and solid) the emissivity of water and of frost crystals is 0.98 link
Now I am intrigued so I did more poking around and I found this which I think you might be very interested in, Willis.
This seems to indicate that at least at an altitude of 2500 m the DLR doubles on cloudy days. The statement “…a great facility for reaching the theoretical maximum value of 1 with cloudy skies.” seems to indicate the emissivity of water and of frost crystals is 0.98 holds true for clouds.
The bit of poking around that Sleepalot (comment) and I did (comment 1) and (comment 2) also seems to indicate the emissivity? GHG effect? of water vapor is much higher than that of dry air/CO2 even at sea level.
What was interesting about the data from the Barcelos, Brazil was that the average humidity was 90% in the month of May. The average humidity was 80% for the ten clear days and the temperature was nearly the same day or night whether it was clear or cloudy for that month.
Given the obvious effects of clouds, water vapor and the oceans I can not believe the IPCC pretty much ignores their effects on the earth’s climate. The radiative forcing attributed to water vapor in the IPCC Components of Radiative Forcing Chart makes absolutely no sense given the data from Barcelos, Brazil vs Adrar, Algeria. Heck they only list H2O in the statosphere! Where the heck is the rest of the radiative forcing from water ??? Dumped in with CO2 as a multiplier effect! (Grumble)
So now that I have finished my ramble, I will clarify my question. Are we looking at gray body radiation from the atmosphere because the sun warms the atmosphere/water vapor/clouds or are we looking at IR emitted from the earth and captured and re-emitted by greenhouse gases?
Now normally we think of air as being quite thermally conductive, because we use it as a coolant, but if you but enough of it in series you end up with a very good insulator.
–Ryan
Whenever talking about the conductive qualities of air, remember: whatever the atmosphere does, it is in series with the vacuum of space…which is the only reason we spend so much time considering the slippery effects of radiation. In the electrical analogy, the “resistance” of space is like a billion ohms, while the atmosphere would be much less and from radiative terms is mostly irrelevant…though the atmosphere can “smear” peak temperatures which has a great effect on peak radiation (keep in mind the t^4 relation). I think the real key to any comfort we enjoy on Earth is simply related to how ineffective radiation is at heating and cooling. Our comfort comes from the asymmetry of the hot ball of fire radiating very intensely and our cool ball of integrating/storing water radiating very weakly.
As an engineer trying to cool hot CPUs, sometimes thermal radiation works in my favor and sometimes it works against me, but the bottom line is, compared to conduction and convection, radiation doesn’t do much…and its effects can often be ignored with little error. Radiation from weak radiators is a steady process, but it’s weak and slow…to do anything useful with radiation (like increasing the average temperature of something by 33C), the energy source must be very intense compared to the energy represented by the +33C I’m trying to get.
Ken, generally I agree with you, but I would say that space (being absorbtive and at absolute zero ) is the electrical analogue of ground (0V). The sun is the electrical analog of a voltage source at 5500V (!) although we probably need to split it into three supplies at different frequencies for UV, IR and visible. The stratosphere has a high impedance to UV, the earth has a high impedance to all three and the troposphere has a low impedance to all three (but there is an awful lot of it). Seems to me you could do a mesh analysis like you would do for electrical currents and “solve” for the atmosphere!
I don’t suppose climatologists are used to doing a mesh analysis. Maybe that is the crux of the problem. The science of multiple sources at different frequencies and multiple energy flows has been known for years in electrical/electronic engineering but is unknown outside this field.
exhibiting melodramatic monologue, willis said:
“You have your fantasy, and by God, neither Stefan nor Boltzmann are gonna get in your way.
Now, you come back with the meaningless statement that since emissivity isn’t a function of temperature, there is no “direct conversion” of watts to temperature … what on earth are you on about? For a given object,
W = C T4”
———————————————————————————————————————-
*sigh*… i reply:
just as there is a relationship between volume and mass, the properties are not convertible without the additional factor of density.
likewise, temperature and watts are not convertible without an additional factor.
you know this, right? you also understand the meaning of ‘directly converted’, right?
you removed that variable from the equation, thus misstating it. why so tricksy?
i may be about to learn something. i asked you please to stand aside and not interfere.
but nooo- you are desperate for attention. i can oblige for i am a connoisseur of schadenfreude and you are offering a feast.
pay close attention – there will be a test:
trolling your own thread doesn’t save WUWT from obscurity as you once claimed in a fever of unjustifiable self importance.
you are not well skilled in the art as you clearly imagine yourself to be.
whining about non-existent whining is an old trick of feeble old trolls that can be wrapped around their necks till they shriek.
i wonder if you even know how to keep score, because you’re losing badly right from the starting line.
Categorize this under: “The Truth is Out There”…
The good news about natural convection at high altitude is that the heat transfer doesn’t degrade as badly as it does in forced convection. Because the flow is generated by the temperature difference, it tends to be self-correcting, in a manner of speaking. With a fan, you a stuck with a fixed volumetric flow, so the mass flow rate, and the heat transfer, goes down linearly with density.
But in natural convection, if the density goes down, the mass flow goes down, and the temperature difference goes up. But that forces the flow rate to increase, and partially compensates for the loss of mass flow.
Tony Kordyban
http://www.coolingzone.com/library.php?read=515
Ryan says:
December 20, 2012 at 6:17 am
Thanks, Ryan, but if you and Gnomish were “essentially correct” that there is “no direct relationship between watts of radiated power and temperature”, then there couldn’t possibly be such a creature as an infrared thermometer.
But there are infrared thermometers, you can buy them at Home Depot. You point them at an object, they measure the IR of the object, and they tell you the temperature of the object.
So … if there is no direct relationship between IR and temperature as you and Gnomish foolishly claim … how does an IR thermometer work?
You also say:
How far it deviates from the theoretical perfect given blackbody is calculated using the “emittance”. Your claim that the process is “nowhere near 100 percent efficient” indicates that you don’t understand what is going on, because the emittance of the earth in the IR region is generally greater than 95%, which is not far from 100% in my book … and it’s not “efficiency” that you are measuring in any case, it is emittance.
Since all of the points that you made are incorrect, I’d have to say, Ryan … that once again you should be careful about your tendency to be overly assertive.
Sorry, couldn’t resist …
w.
gnomish says:
December 20, 2012 at 11:14 am (Edit)
Oh, OK, I see what you are on about. It seems that you think that something like
W = T4
is a “direct relationship” but
W = C T4
where “C” is a constant for any given object is not a “direct relationship”.
Hmmm … I’m afraid I can’t see the difference. Perhaps you could explain it to the class.
OK, gnomish, have it your way. For me a “direct relationship” means that one of the variables is directly PROPORTIONAL to the other variable in some sense. For example, I would say the nutrition of a child has a direct relationship with its health. Similarly, I would say that the amount of gunpowder in a cannon has a direct relationship with how far the cannonball goes.
Now, to make the actual calculation for the cannonball’s trajectory you need other details—air temperature, weight of the ball, air density, and the like.
But for me, none of that wipes away the direct relationship between gunpowder and distance.
Obviously, you have some very different meaning for a “direct relationship”, my bad for not reading your mind and knowing that.
w.
gnomish says:
December 20, 2012 at 11:14 am
Yes, and I pointed out that there is no possible way for me to interfere with your learning something here on the blog, whether I post or not.
Either you learn or you don’t, gnomish, and your claim that I am somehow standing in the way of you learning something is a risible and physically impossible excuse.
w.
Gail Combs says:
December 20, 2012 at 6:18 am
Thanks, Gail. The emissivity of clouds is generally taken as 1.0, they are so close to a theoretical black body.
Would this “confound the whole idea of DLR from GHGs?
Well, no. They are different phenomena. You can see it in records of downwelling longwave radiation (DLR). There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation.
Take a look at the graphic I posted above. The background DLR is about 300 W/m2, and you can see the variations in that value due (mainly) to clouds.
w.
Willis Eschenbach says, December 20, 2012 at 12:07 pm:
“Thanks, Ryan, but if you and Gnomish were “essentially correct” that there is “no direct relationship between watts of radiated power and temperature”, then there couldn’t possibly be such a creature as an infrared thermometer.
But there are infrared thermometers, you can buy them at Home Depot. You point them at an object, they measure the IR of the object, and they tell you the temperature of the object.
So … if there is no direct relationship between IR and temperature as you and Gnomish foolishly claim … how does an IR thermometer work?”
========================================================
Anyway, not the way you imagine it works, dear Willis. Generally, do not just believe your eyes, dig deeper.
From my humble knowledge, IR thermometer calculates temperature on the basis of assumed emissivity of the target. Guess, what happens, if emissivity of the target is significantly different from the assumed one?
Hence, Willis, if an IR thermometer was not fed with that assumption (by the manufacturer), it would be unable to calculate the temperature of the target at all.
As you can possibly see now, things are sometimes different from what they appear to be. I hope it will be sufficient reason for you to reduce a little bit the teaching connotation your comments sometimes probably unintentionally contain.
Greg House says:
December 20, 2012 at 1:54 pm
Say what? Is any of that supposed to be news?
Of course the IR thermometers have to deal with the emissivity, Greg. Stefan-Bolzmann’s law contains emissivity, how could they not have to deal with that? The good IR thermometers have an emissivity adjustment, viz:
I’m surprised you would assume I don’t know these things …
My issue is, I’m not at all sure what your point is here. We seem to have a semantics problem. I call a relationship like
W = C T4
where C is a constant for any given body a “direct relationship”.
You don’t.
Having established that, what is the underlying point that you are trying to make? That objects don’t radiate in the IR? That their radiation is not a function of temperature and emissivity? I’m not clear on what you think I’ve done wrong.
Let me see if I can explain this a different way:
If I have a lump of steel, in any condition, and I double its temperature, I will get sixteen times the wattage of radiation from it that I got at the original temperature.
If I have a lump of copper, and I double its temperature, I will get sixteen times the radiation.
The same is true for every object that you can name. If you double its temperature, you get sixteen times the radiation. The is because for any given object that you can name, radiation (watt/m2) is directly proportional to the fourth power of the temperature (K).
I call that a “direct relationship”. YMMV, you may wish to call it something else, but either way, what is the underlying error that you seem convinced that I am making?
Regards,
w.
One last comment on IR thermometers. While it is true that emissivity plays a part in the relationship between temperature and IR radiation, the difference is more apparent than real.
This is because most objects have an emissivity of 0.9 or better. Let me quote from a previous post:
As you can see, an IR thermometer with an assumed emissivity of 0.95 will only have an error of a couple of percent for everything from sand to glass to trees to mouse fur …
w.
a couple of percent Kelvin, you might point out, translates to @ur momisugly 10 degrees F.
that might be considered significant when the matter of global warming hinges on tenths and hundredths.
wrt your misdirection: i said ‘direct conversion’ not ‘direct relationship’. but you knew that.
presently, i hope R. Clemenzi may give a conclusive reply to my question.
his comments are more digestible because he has no need to assuage a deep insecurity by fighting with ignorant nobodies like myself.
Willis Eschenbach says, December 20, 2012 at 12:07 pm:
“So … if there is no direct relationship between IR and temperature as you and Gnomish foolishly claim … how does an IR thermometer work?”
Willis Eschenbach says, December 20, 2012 at 12:24 pm:
“For me a “direct relationship” means that one of the variables is directly PROPORTIONAL to the other variable in some sense.”
Willis Eschenbach says, December 20, 2012 at 2:33 pm:
“My issue is, I’m not at all sure what your point is here. We seem to have a semantics problem.”
===========================================================
No, Willis, it is that you present a (bad) semantic “solution” to a non-semantic problem.
Your “directly PROPORTIONAL to the other variable in some sense” is a contradiction in itself.
No IR thermometer can determine the temperature of the target from IR alone. Which means, it can receive more IR from an object having lower temperature and the same amount of IR from another object having higher temperature. This is not what people would call, in your words, “directly PROPORTIONAL”. It is just your semantic “solution” to your problem.
This is the 4th “semantic” solution you have presented to the public recently. Let me remind you of the other 3 modern warmism is based on:
1) The “semantic” solution to the problem with the 2nd Law of Thermodynamics: “climate science” invented the “NET” thing that is not to be found in the historical formulations of the 2nd Law, nor is it apparently proven experimentally.
2) The “semantic” solution to the experimentally unproven “warming via back radiation”: to somehow theoretically substantiate it they cut the solar power in half.
3) And last not least, the “semantic” solution to the problem with the calculations of so called “global temperature”: they just call what they calculate without any basis in science (“extracting” temperatures for large areas from single or few weather stations, “reconstructing” temperatures etc.) “global temperature”, apparently even without any scientifically established definition.
Well as a result of all this carry on here you won’t be blathering on any more about oceans freezing will you Willis. Here’s some Xmas holiday reading for you….. call it Xmas cheer. because the message should bring comfort and joy.
http://jennifermarohasy.com/author/nasif-s-nahle/
Greg House says:
December 20, 2012 at 3:36 pm
Greg, I don’t care about the semantics. I don’t care what you call these things. I just want to know what your issue is. You seem very emphatic about something, but what?
I assure you that statements like:
do not assist me in trying to find out what it is that your are trying to say, or what error of mine you are pointing out. What on earth does that statement of yours mean? On second thought, never mind, I don’t want to know. What I do what to know is, what is it that you are on about?
Please boil it down to as few and as clear words as possible, because I can’t make sense of the above.
w.
Willis Eschenbach says:
December 20, 2012 at 1:26 pm
…Well, no. They are different phenomena. You can see it in records of downwelling longwave radiation (DLR). There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation…..
>>>>>>>>>>>>>>>>>>>>>>>>>>>
I am still confused. Maybe it is just semantics. I interpret what you are saying to mean that without a Greenhouse Gas (GHG) Effect the atmosphere immediately after sunset would drop to absolute zero and therefore all DLR is due to the GHG and there is no DLR (IR) from the atmospheric gases being at a temperature above 0K.
1. I think we all agree there is IR radiation from all matter that is not at absolute zero, correct?
I did have a bit of difficulty finding what the gray body radiation for the atmosphere/gas was. (White screen of death problem) However see bottom of comment.
2. Gases do emit IR radiation and the amount is dependent on the partial pressure and the temperature. (see bottom)
3. Radiation from a gas not at absolute zero is due to temperature and has nothing whatever to do with the greenhouse gas effect if you are considering only gray/black body radiation.
4. Pure nitrogen will warm to room temperature through conductivity and convection. Pure air (without GHGs) will also warm without the influence of GHG effect. Although this observational data seems to indicate sunlight has a heck of a lot to do with the warming of the atmosphere.
5. The links I made in the last post indicate the emissivity of air, at least in those mountains, was 0.5 or less without clouds. The below articles indicate CO2 has an emissivity of 0.0027. Since you indicated in another comment you consider all emissivities ~ 0.95 I think this may be the key concept.
6. I think we can at least agree that the emissivity of clouds is 0.98 to 1.0 and the emissivity of dry air is less, somewhere between 0.5 to 0.83.
7. There is also the Specific Heat Capacity of Water water has to absorb 4.184 Joules of heat for the temperature of one gram of water to increase 1 degree celsius (°C). Nitrogen has a specific heat at constant pressure of 1.04 @ur momisugly at 68F (20C) and 14.7 psia (1 atm) and for air ~ 1.00
Could the difference in DLR between clear and cloudy days be due mostly to the difference in the emissivity and heat capacity of air vs clouds without considering IR energy bounced from the earth to the clouds and back again?
That is emissivity for CO2 is 0.0027
gnomish says:
December 20, 2012 at 3:34 pm
I’m not fighting with you, gnomish, I’m trying to find out what you are so concerned about. Again, I have to ask, what is your point? Surely it is not that a couple of percent kelvin is around 6°C, we both knew that. But what is it that you find so important? It seems you think I’ve make some terrible error in saying that my desk radiates about 400 W/m2, or maybe it was something else that I said, but I haven’t a clue.
Is your point really that you said “direct conversion” and I said “direct relationship”? Are you serious when you claim that this was somehow to “misdirect” you?
Rather than spending your time accusing me of “deep insecurity” and “misdirection”, and complaining that I’m not as good as someone else at making “digestible” comments, how about you focus on the science instead? I freely confess that my failings are manifold, I was born yesterday, but my morals and motives have nothing to do with the science. Your continued insistence on attacking me is warping your thinking and causing you to lose focus on the science itself. It’s gotten so bad you think I’m attacking you. I’m not. I’m just saying, both I and a variety of knowledgeable people have commented upthread that you are badly mistaken in some of your views. You should at least consider the possibility that they might be right …
My regards to you,
w.
PS—The error from a preset emissivity of 0.95 being in error by a couple of percent is not ten degrees F in any case:
Calculated temperature, 390 W/m2, emissivity 0.95, 18.5°C
Calculated temperature, 390 W/m2, emissivity 0.97, 17.0°C
Difference, 1.5°C, or 2.7°F
So the 2% error in emissivity translates to about a 1.5°C (2.7°F) error, not a 10°F error as you state.
Willis:
Thanks for the links to the radiation charts. I was unaware of them. I’m intrigued by the fact that the downwelling radiation appears to be little affected by virtually ANYTHING during warm days when it is clear. Pick a day in July, August, or September and plot downward solar radiation, downward IR, and temperature–and then look at all the sites. You get about 400 Wm-2, regardless of altitude, latitude, temperature, time of day, and amount of GHGs (water vapor). I find this weird. Speechless for now…
Willis writes “There is a background DLR level, due to water vapor and CO2 and other GHGs. But when a cloud comes over, the DLR (generally) increases, because the clouds such excellent absorber-emitters of longwave radiation.”
This IS semantic but IMO important. The DLR increases because the flow of IR energy away is reduced, not inherantly because clouds are good absorber-emitters. That happens miles away. The DLR that increases, the stuff you might measure at the ground, isn’t DLR that was emitted by the cloud.
Its just the way you think about it, Willis and its cooling not warming.
jae says:
December 20, 2012 at 5:42 pm
Every time I look at some of those charts, I come away mystified by something.
Regarding variations in the DLR, what kind of variations were you expecting? All of them have diurnal variations on the order of 80-100 W/m2. It’s difficult to say what the average might be for any of them without downloading the data. I see some that vary maybe 320-420 W/m2, average maybe 370, and others that vary 290-370 or so on the same day, average maybe 330.
The S-B temperatures corresponding to those average radiation levels with emissivity = 0.95 are about 56°F (13°C) for the larger average and about 44°F (7°C) for the smaller average of 330 W/m2.
These were early summertime records I looked at, June 1st. The downwelling radiation represents the temperature of the atmosphere from which the radiation originated.
It is not widely appreciated that almost three-quarters (72%) of the downwelling longwave radiation at the surface comes from the lowest 87 metres (285 feet) of the atmosphere. Or to look at it another way, 90% of the downwelling radiation hitting the ground comes from the lowest half kilometer (a third of a mile) of the atmosphere. SOURCE: My bible, Geiger’s classic text “The Climate Near The Ground”, sixth edition (2003). An early version (1950) is a free download here. Modern version is on Amazon, new $125, used $45.
As a result, the intensity of the downwelling radiation is mostly representative of the air temperature of those lowest atmospheric layers. For these lowest levels, that range of 45-55°F in the summertime, seems reasonable. I usually figure that temperature drops about a degree C per 100 metres. At an altitude of 500 metres, that would be a drop of 5°C (9°F) from the surface air temp or so.
Finally, I would expect the variations in lower atmospheric temperatures over hundreds of miles to be far less than those observed at ground stations, with their host of microclimates. I’d expect the atmosphere to average a whole lot of that out, and over large areas.
… But of course with steep thermoclines where you least expect them, you gotta love Nature, she does pierced, patched, and plotted very well, but she doesn’t do gradual all that stunningly. Even in the atmosphere.
Anyhow, that’s my thoughts on the matter. I hardly pretend to understand the nuances (and in some cases the broad strokes) of what I find in the actual observations … gotta love it, always more to learn.
w.
Willis Eschenbach says, December 20, 2012 at 4:57 pm: “I assure you that statements like: “The “semantic” solution to the experimentally unproven “warming via back radiation”: to somehow theoretically substantiate it they cut the solar power in half.” do not assist me in trying to find out what it is that your are trying to say, or what error of mine you are pointing out. What on earth does that statement of yours mean?
==========================================================
OK, Willis, I will explain it again, because the point is very important. And it is not your error primarily, because it probably was not you who invented that bogus calculation about the -18C cold “Earth without atmosphere”. Let us do it point by point.
1. The basic NASA’s statement about total solar irradiance (TSI) is this: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”. There are at least 2 things in this statement indicating that TSI goes for the whole hemisphere and not for an imaginable circle/disk.
First, they say reaching the top of the atmosphere directly facing the Sun. It is obvious that the top of the atmosphere directly facing the Sun is not a flat disk, but a hemisphere, exactly like the half of the Earth directly facing the Sun.
Second, they call it “TOTAL solar irradiance”. The word TOTAL is another indication that the whole hemisphere is meant.
Conclusion: in your calculation the real solar power is cut in half. This is one error.
2. The other statement you quoted from the same NASA page contradicts this one, they replace in fact “average” and “total” with “maximum” for no scientific reason.
This replacement is what I call a semantic “solution” to a problem. The problem is, as I said earlier, that there is apparently no physical experimental proof of warming via back radiation, and that “solution” helps a little bit.
Another problem with you calculation is, even if we put aside that hemisphere/disk error, that your Earth does nor rotate at all. You consider only the half of the Earth to receive solar power. This is apparently another error, because the Earth does rotate. The WHOLE rotating Earth is receiving solar power and the part that temporarily is not, has not been proven to cool down to the absolute zero, before it faces the sun again. If it does not cool down to the absolute zero, the average temperature will be higher. In your calculation it is not taken into consideration, this is the second error.
The third error is your way of operating with averages, Ryan already said that on this thread (December 18, 2012 at 4:57 pm) and gave you an easy example. Unfortunately, you failed to comment on that, as well as on the “rotating Earth” argument of mine.
I hope this explanation will help.
TimTheToolMan says:
December 20, 2012 at 6:39 pm
You are correct, the downwelling radiation from upper layers (clouds or not) will be absorbed at lower levels. Those levels will radiate in turn, upwards and downwards. Some of that eventually reaches the ground.
When a cloud comes over, it intercepts 100% of the upwelling radiation. As you point out, this reduces the flow out to space, with half of the radiation being redirected back to the surface.
Not clear what this means.
w.