Guest Post by Ira Glickstein
Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy) and energy/mass is conserved (cannot be created nor destroyed).
My Visualizing series [Physical Analogy, Atmospheric Windows, Emission Spectra, and Molecules/Photons] has garnered almost 2000 comments, mostly positive. I’ve learned a lot from WUWT readers who know more than I do. However, some commenters seem to have been taken in by scientific-sounding objections to the basic science behind the Atmospheric “Greenhouse Effect”. Their objections seemed to add more heat than light to the discussion. This posting is designed to get back to basics and perhaps transform our heated arguments into more enlightened understanding :^)
As I’ve mentioned before, during my long career as a system engineer I’ve worked with many talented mathematical analysts who always provided precise results, mostly correct, but some precisely wrong, usually due to mistaken assumptions. I got into the habit of doing a “back of the envelope” calculation of my own as a “sanity check” on their results. If their results matched within reasonable limits, I accepted them. If not, I investigated further. In those days my analysis was really done using a slide rule and scrap paper, but I now use spreadsheets.
The graphic above is based on an excellent spreadsheet from http://serc.carleton.edu/files/introgeo/models/mathematical/examples/XLPlanck.xls. It uses Planck’s Law to calculate the black body radiation spectrum from the Sun, as observed at the top of the Earth’s Atmosphere. It also may be used to calculate the radiation spectrum from the Earth System (Atmosphere and Surface, see below for explanation) at any assumed temperature. (I will refer to this spreadsheet as “Carleton” in this posting.)
I modified the Carleton spreadsheet to compute the mean Solar radiation per square meter absorbed by the Earth System, which turns out to be 240 Watts/m^2. I then used the spreadsheet to determine the effective mean temperature of the Earth System that would emit an equal amount of energy to Space, and that turned out to be 255 Kelvins (-18ºC which is 1ºF).
Since the mean temperature at the surface of the Earth is 288 Kelvins (+15ºC which is 59ºF), that leaves 33 Kelvins (33ºC which is 58ºF) to be accounted for. Guess how we acount for it?
The yellow curve (above left) shows that Solar radiation is in a tall, narrow “shortwave” range, from about 0.1μm (microns, or millionths of a meter) to about 4μm, which we call ultra-violet, visual, and near-infrared. The vertical axis is Intensity of the radiation, measured in Watts/m^2/μm, and the horizontal axis is Wavelength, measured in μm. If you divide the area under the yellow curve into vertical strips, and add up the total area, you get 240 Watts/m^2.
Since we humans sense the visual portion of this radiation as “light”, that is the name we give it, and that has led to the false assumption that it contains no “heat” (or “thermal”) energy.
The violet curve (above right) shows that, assuming a mean temperature of 255 K, Earth System radiation to Space is in a squat, wide “longwave” range, from about 5μm to beyond 40μm, which we call mid- and far-infrared. If you divide the area under the violet curve into vertical strips, and add up the total area, you get the same 240 Watts/m^2 as is under the yellow curve.
DETAILED EXPLANATION
The graph on the left shows the actual observed Solar radiation spectrum (in red) as measured at the top of the Atmosphere. It is superimposed on a black body model (in blue) showing very good correlation. Thus, while the Sun is not exactly a black body, it is OK to assume it is for this type of “sanity check” exercise.
If you calculate the area under the curve you get about 1366 Watts/m^2. That means that a square meter of perfect black body material, held perpendicular to the Sun, would absorb 1366 Watts.
However, the Earth is not a perfect black body, neither is it a flat surface perpendicular to the Sun! So, to plot the yellow curve at the top of this posting, I had to adjust that value accordingly. There are two adjustments:
- The Earth may be approximated as a sphere, with the Sun shining on only half of it at any given time. The adjustment factor for this correction is 0.25.
- The albedo (reflectiveness) of the Earth system, primarily clouds and light-colored areas on the Surface such as ice, causes some of the Solar radiation to be reflected back out to Space without contributing any energy to the Earth System. The adjustment factor for this correction is 0.7.
After applying these adjustments, the net Solar energy absorbed by the Earth System is 240 Watts/m^2.
The graph on the right shows the black body model for an Earth System at a mean temperature of 255 K, a temperature that results in the same 240 Watts/m^2 being emitted out to Space.
Of course, the Earth System is not a perfect black body, as shown by the graph in the upper panel of the illustration below, which plots actual observations from 20 km looking down. (Adapted from Grant Petty, A First Course in Atmospheric Radiation, Figure 8.2, http://www.sundogpublishing.com/AtmosRad/Excerpts/index.html.)
The actual measured radiation is the dark squiggly curve. Note that it jigs and jags up and down between the topmost dashed curve, which is the black body spectrum for a temperature of 270 K and a lower dashed curve which is the black body spectrum for 230 K. This data was taken over the Arctic, most likely during the daytime. The Petty book also has a graph looking down from over the Tropical Pacific which ranges from 300 K down to 210 K. Observations will vary by tens of degrees from day to night, summer to winter, and Tropical to Polar.
However, it is clear that my result, based on matching 240 Watts/m^2, is within a reasonable range of the true mean temperature of the Earth System as viewed from Space.
NOTE ABOUT THE ABOVE ILLUSTRATION
WUWT readers will notice some apparent inconsistencies in the graphs above. The top and bottom panels, from Petty, peak at 15μm to 20μm, while the purple, blue, and black curves in the middle panel, and the Earth System curves from the Carleton spreadsheet I used (see above) peak in the 9μm to 11μm range. Also, the Petty black body curves peak at a “Radiance” around 100 mW/m^2/sr cm^-1 while the black body curves from Carleton peak at an “Intensity” of around 14 W/m^2/μm. Furthermore, if you look closely at the Petty curves, the labels on the black body curves are mirror image! What is going on?
Well, I know some of the reasons, but not all. (I hope commenters who are more fluent in this than I am will confirm my explanations and provide more information about the differences between “Radiance” and “Intensity”.) I have Googled and Wikied the Internet and am still somewhat confused. Here is what I know:
- The horizontal axis in Petty’s plots are what he calls “Wavenumber”, increasing from left to right, which is the number of waves that fit into a cm (centimeter, one hundredth of a meter).
- This is proportional to the frequency of the radiation, and the frequency is the inverse of the wavelength. Thus, his plots are the mirror image of plots based on wavelength increasing from left to right.
- The spreadsheet I used, and my previous experience with visual, and near-, mid-, and far-IR as used in military systems, always uses wavelength increasing from left to right.
- So, when I constructed the above illustration, I reversed Petty’s curves, which explains why the labels on the black body curves are mirror image.
- Fortunately, Petty also included a wavelength legend, which I faithfully reproduced, in non-mirror image, at the top of each plot.
But, that still does not explain why the Petty black body curves peak at a longer wavelength than the Carleton spreadsheet and other graphics on the Internet. I tried to reproduce Petty’s blackbody curves by multiplying the Carleton values by the wavelength (μm) and that did not move the peak to the right enough. So, I multiplied by the wavelength again (μm^2) and, voila, the peaks agreed! (I hope some WUWT reader will explain why the Petty graphs have this perverse effect. advTHANKSance!)
ANSWERING THE OBJECTIONS TO BASIC ATMOSPHERIC “GREENHOUSE EFFECT” SCIENCE
First of all, let me be clear where I am coming from. I’m a Lukewarmer-Skeptic who accepts that H2O, CO2 and other so-called “greenhouse gases” in the Atmosphere do cause the mean temperature of the Earth Surface and Atmosphere to be higher than they would be if everything was the same (Solar radiation, Earth System Albedo, …) but the Atmosphere was pure nitrogen. The main scientific question for me, is how much does the increase in human-caused CO2 and human-caused albedo reduction increase the mean temperature above what it would be with natural cycles and processes? My answer is “not much”, because perhaps 0.1ºC to 0.2ºC of the supposed 0.8ºC increase since 1880 is due to human activities. The rest is due to natural cycles and processes over which we humans have no control. The main public policy question for me, is how much should we (society) do about it? Again, my answer is “not much”, because the effect is small and a limited increase in temperatures and CO2 may turn out to have a net benefit.
So, my motivation for this Visualizing series is not to add to the Alarmist “the sky is falling” panic, but rather to help my fellow Skeptics avoid the natural temptation to fall into an “equal and opposite” falsehood, which some of those on my side, who I call “Disbelievers”, do when they fail to acknowledge the basic facts of the role of H2O and CO2 and other gases in helping to keep temperatures in a livable range.
Objection #1: Visual and near-visual radiation is merely “light” which lacks the “quality” or “oomph” to impart warmth to objects upon which it happens to fall.
Answer #1: A NASA webpage targeted at children is sometimes cited because they say the near-IR beam from a TV remote control is not warm to the touch. Of course, that is not because it is near-visual radiation, but rather because it is very low power. All energy is fungible, and can be changed from one form to another. Thus, the 240 Watts/m^2 of visible and near-visible Solar energy that reaches and is absorbed by the Earth System, has the effect of warming the Earth System exactly as much as an equal number of Watts/m^2 of “thermal” mid- and far-IR radiation.
Objection #2: The Atmosphere, which is cooler than the Earth Surface, cannot warm the Earth Surface.
Answer #2: The Second law of Thermodynamics is often cited as the source of this falsehood. The correct interpretation is that the Second Law refers to net warming, which can only pass from the warmer to the cooler object. The back-radiation from the Atmosphere to the Earth Surface has been measured (see lower panel in the above illustration). All matter above absolute zero emits radiation and, once emitted, that radiation does not know if it is travelling from a warmer to a cooler surface or vice-versa. Once it arrives it will either be reflected or absorbed, according to its wavelength and the characteristics of the material it happens to impact.
Objection #3: The Atmospheric “Greenhouse Effect” is fictional. A glass greenhouse works mainly by preventing or reducing convection and the Atmosphere does not work that way at all.
Answer #3: I always try to put “scare quotes” around the word “greenhouse” unless referring to the glass variety because the term is misleading. Yes, a glass greenhouse works by restricting convection, and the fact that glass passes shortwave radiation and not longwave makes only a minor contribution. Thus, I agree it is unfortunate that the established term for the Atmospheric warming effect is a bit of a misnomer. However, we are stuck with it. But, enough of semantics. Notice that the Earth System mean temperature I had to use to provide 240 Watts/m^2 of radiation to Space to balance the input absorbed from by the Earth System from the Sun was 255 K. However, the actual mean temperature at the Surface is closer to 288 K. How to explain the extra 33 K (33ºC or 58ºF)? The only rational explanation is the back-radiation from the Atmosphere to the Surface.
Speaking of caveats I owe Phil an explanation on atmospheric nitrogen thermal radiation. It must remain true that nitrogen radiates. ALL matter above absolute zero has thermal radiation which is a consequence of atoms being composed of oppositely charged particle in motion relative to each other. However, nitrogen is a poor absorber of infrared so it is also a poor emitter of infrared. So while it must have a thermal spectrum characteristic of its temperature the intensity of that spectrum is down in the dirt so it contributes nothing by itself, per se, to the infrared glow of the atmosphere. However it can still itself be thermalized kinetically and it can transfer kinetic energy to good IR emitters like water vapor and CO2 so it can’t just be ignored like it isn’t there at all.
R Stevenson says:
May 12, 2011 at 2:54 am
My heat transfer book has a rough mean beam length formula of Le=3.6(V/A) where V is volume of gas and A is area.
Tim Folkerts said ….”The clouds do indeed provide part of the 324 W/m “back radiation”, with the rest provided by the GHGs. But that 324 W/m^2 of thermal IR comes from the energy OF THE CLOUDS (and air molecules) at a rate determined by the temperature OF THE CLOUDS (and air molecules).”….
The natural meaning of Backradiation is radiation that has come “back” from somewhere.
The only place “back” can refer to in this context is the planet surface.
Since it plainly does not all originate from the planet surface the term should be avoided.
Downwelling long wavelength radiation is a much better title.
If we were to accept your definition then from KT97 we have 350W/m2 up and 324 W/m2 back.
The readily available spectra graphs show that this is wrong.
Bryan says:
May 12, 2011 at 12:57 am
Ira Glickstein might care to comment
Tim Folkerts says …..”I agree that the energies involved in the MB distribution for room temperature gases would put the IR energies out toward the tail.”……
Yes its all about probabilities for air at 250K.
1. From temperature considerations 4% CO2 molecules in the active vibrational state.
2. Now add a flux 15um radiation from Earth surface to further enhance the active numbers.
3. The relaxation time of vibrational state is longer than the chance of collision deactivation
4. The collision route will share the vibrational energy between CO2 and O2 molecules leading to local temperature increase.
5. The radiative route takes two paths
a. Collision activation of longer wave IR emissions from H2O is favoured for two reasons
(i) Nearly 40 times more H2O molecules than CO2
True near the surface, becomes less so as you increase altitude or go near the poles or to deserts. So near the surface the energy absorbed by CO2 is likely to be shuttled through N2/O2 to be emitted at longer wavelength by H2O. It’s like how a Helium/Neon laser works, the He is excited by electric discharge collides with Ne thereby exciting it (3s level) and then Ne emits (3s->2p) the familiar red beam. As seen from the ‘down’ spectra though some CO2 is still emitted.
(ii) Plenty H2O bands at wavelengths > 15um => much more probable.
Again depends where you are in the atmosphere. Also depends on the relative radiation lifetime of the excited states of H2O, if shorter than the CO2 lifetimes or not. From what I recall from shocktube studies, H2O has a slightly shorter lifetime but that was at wavelengths 14 micron and below.
mkelly says:
May 11, 2011 at 12:12 pm
“Pressure, Joel, pressure. There I have successfully told you. Whether you accept is a different story.”
So if I compress air into tank until the compressional heating raises it above the boiling point of water then I can use it to run a steam engine forever, right? Because it will never cool off until I let the pressure escape.
Bryan says: May 12, 2011 at 8:26 am
“Downwelling long wavelength radiation is a much better title.”
I think everyone agrees this is a much more descriptive and accurate name. So every time you hear “back radiation” just substitute in your mind “Downwelling long wavelength radiation”. The terminology is secondary, as long as everyone who really cares understands what is meant.
“The readily available spectra graphs show that this is wrong.”
Could you provide a link to specific spectra that you are referring to? What is wrong with them?
Phil thanks for your input and your qualification of particulars in my post are well founded.
Dave Springer says:
May 12, 2011 at 8:51 am
Sir, if you disgree that pressure causes an increase in temperature if volume is held steady that is well odd. PV= nRT discribes what is termed STP (Standard temperture and pressure). At 1 atmosphere the temperature of air is 0 C. On earth that accounts for 18 degrees of the so called 33 degrees of warming by GHG.
Is it hotter in Death Valley or the top of Everest on the same day of the year?
Further, I repeat that none of the heat transfer equations have an input for back radiation. So talking of something that does not exist in formulated heat transfer equations is strange. Please show me an equation that has an input for back radiation and is called back radiation in the formula. If you do I will speak no more of it.
mkelly says: May 12, 2011 at 9:42 am
“At 1 atmosphere the temperature of air is 0 C.”
No, at 1 Atm the temperature of 1 MOLE OF GAS occupying 22.4 LITERS is 0 C. Go around the global at sea level and you will be right around 1 Atm the whole time, but I will guarantee the temperature will not be 0 C simply because the pressure is 1 Atm.
“STP” is simply an ARBITRARY definition for uniform comparisons. It is in no way related to what temperature the air “should” be on earth or Jupiter or Pluto. Choosing a standard is only done to allow — well — standard measurements that can be compared easily.
For example, in the US you buy natural gas by the cubic foot. But, of course, you would get a different number of moles in each cu ft depending on the temperature and pressure. The standard is 60 °F and 14.73 psi. This in no way says that natural gas “should” be 60 °F. It simply says that if you buy the natural gas at high pressure, you get charged more for the actual cu ft of gas because it would have more ‘standard cu ft’.
mkelly says:
Nobody is disagreeing with that. However, the volume is not “held constant” in the atmosphere.
You’ve had this all explained to you a million times: Considerations of the adiabatic lapse rate tell us the maximum rate at which temperature can decrease with height. However,
(1) It does not tell us that the temperature must decrease with height at this rate, and in fact it doesn’t once you get beyond the troposphere. The actual temperature distribution depends on how the atmosphere is heated, with the adiabatic lapse rate just providing a stability limit…i.e., an upper bound on the allowed lapse rate. Since the bottom of the atmosphere (the troposphere) is heated strongly from below, it turns out that the actual lapse rate in this part of the atmosphere is usually close to the (appropriate dry or moist) adiabatic lapse rate.
(2) Even once you know the lapse rate, that doesn’t determine the temperature at the surface alone. You must have an additional criterion, the temperature at some level in the troposphere, in order to determine the surface temperature. In particular, this criterion turns out to be that the average temperature at the effective radiating level in the atmosphere has to be equal to the “blackbody temperature” so that the earth system emits back out into space the same amount of energy it absorbs from the sun. This effective radiating level depends on atmospheric composition and, in particular, on those constituents that absorb terrestrial radiation.
If you repeat it a thousand times, it will be no less false than it was the first time. We have explained this to you. What you call it is totally irrelevant. If you don’t like the term, don’t use it. The greenhouse effect does not depend on calling the radiation that is downwelling from the atmosphere “back-radiation”. You can call it “Magical mystery radiation from the planet Zircon” and as long as you use the same equations, you will get the same result: namely, the greenhouse effect.
[Perhaps this is obvious, but since nothing seems to obvious to state explicitly here, when I say “you use the same equations”, I mean the standard equations used in all radiative transfer calculations, i.e., those used by scientists and engineers in practical calculations every day (as David M Hoffer has pointed out).]
Joel Shore says:
May 9, 2011 at 2:09 pm
“Any text on thermodynamics that treats it from a statistical physics perspective ought to do fine. In the context of radiative heat transfer specifically, any text that talks about the exchange of radiation between two objects or any object and its surroundings ought to give you the basic radiative transfer equations that are used in computing the greenhouse effect.”
Thank you Joel, but I’m afraid, not being as widely read as yourself, that I can’t find a textbook that supports your theory. Could you perhaps cite the document in question, and even better, quote the relevant passage(s)? I am familiar with blackbody and greybody (to their surroundings) radiation equations and I confess that I find no evidence of backradiation in the equations. Perhaps you could enlighten me?
Joel Shore says:
May 9, 2011 at 2:09 pm
“Yes…So, heat (which is a macroscopic concept by its definition) always flows from hotter to colder (in the absence of work). However, note that the interpretation that the radiative energy flows in both directions, while having an abundance of empirical support, is not necessary to show there is a greenhouse effect. All that is really necessary to say is that the heat flow between two objects depends on the temperature of both objects and not just on the temperature of the hotter object.
Of course the energy flow between the two objects depends on the difference in temperature between the two objects. However that does not imply that energy flows from the cooler object to the warmer object. That merely accounts for the difference in energy states. The greater the difference in energy states, the more energy will flow from the higher state to the lower state. The potential energy of an object 5 meters above the ground is greater than that of an object 1 meter above the ground. The energy transfer from the higher object to the ground (via transformation to kinetic energy) will be greater than the energy transfer from the lower object to the ground. However the energy transfer does not require that the ground move towards either object. The energy transfer (current) in an electric circuit is unidirectional and proportional to the difference in the energy states between the source of the energy and its destination. Energy is energy, it doesn’t change nature or characteristics, whether it is potential, kinetic or radiative.
I don’t dispute that radiative energy can flow in both directions, I don’t think anyone does. Any object above 0K will radiate some energy, just because you put a hotter one next to a colder one doesn’t mean the colder one stops radiating – but neither does it mean that the colder one will warm the hotter one.
The SB law of the radiation energy of a black body
q = σ*T^4*A
states that the radiation energy of an object is proportional to the fourth power of its absolute temperature.
The grey body equation for the radiation loss rate of an object to its surroundings
q = ε*σ*(Th^4 – Tc^4) Ah
must correct for the temperature of the surroundings in order to correctly reflect the absolute temperature of the object, hence the term (Th^4-Tc^4). This however in no way implies that the surroundings (which are radiating energy, not being at 0K) transfer energy to the object.
Furthermore the equation has a term for the emissivity of the object and no term for the emissivity of the surroundings or the absorptiveness of the object, both of which would be required if the surroundings were transferring energy to the object.
If you put a mirror a few feet away from an infrared source, say a heat lamp, the mirror radiates energy as visible white light and LWIR, both radiative energy, but that won’t increase the temperature of the heat lamp any.
Joel Shore says:
May 9, 2011 at 2:09 pm
“I have no clue what you are trying to say in this last sentence. The term “heat pump” is sometimes used as a category that includes air conditioners and refrigerators. Other times, it is used to represent the subcategory of the “heat pump” in the above context that are actually used to heat a house rather than cool it. However, regardless of how the term is used, heat pumps used for heating and air conditioners operate by the same basic principle. Both use work to “pump” heat energy from cold to hot, i.e., in the opposite direction from which heat spontaneously “flows”.”
I was using the term “heat pump” in the colloquial sense, meaning of a machine which heats a house by extracting heat from the surroundings and moving it into the house. The reason I contrasted the two types of ‘heat pump’, heat pumps and air conditioners, was to illustrate the large amount of work (energy) required to circumvent the Second Law – ie to cool an object further by moving energy from it to a warmer one or warmer surroundings.
I don’t necessarily disagree at this point with the concept that GHGs trap heat, I disagree with the idea contention that they heat the earth through backradiation. They can’t, they don’t.
But just for the sake of argument, tell me in your theory what the temperatures of the earth and atmosphere are, the amount of energy radiated from the earth to the atmosphere and backradiated from the atmosphere to the earth. Please state all units in Kelvin if you don’t mind. I’ll run the calcs and see if the numbers prove me wrong.
Joel Shore says:
May 12, 2011 at 10:42 am
What you call it is totally irrelevant.
Sorry I totally disagree. What a thing is named is important. What the equations call temperaturee you seem to want to call back radiation. Temperture is temperature and radiation is radiation. They maybe intertwined but they are not interchangeable.
And since you did not supply an equation with a back radiation input I must declare I am correct.
Alleyne says:
May 12, 2011 at 11:29 am
q = ε*σ*(Th^4 – Tc^4) Ah
The above from an astute person shows clearly that if two objects of the same temperature are radiating at each other there is no (none, nada, zippo) energy exchange. No back radiation heating the other up as you claim back radiation heats the earth.
MKelly says:
My heat transfer book has a rough mean beam length formula of Le=3.6(V/A) where V is volume of gas and A is area.
The original charts were for a hemispherical gas mass of radius L. Constants by which the characteristic dimensions of simple shapes are multiplied use 3.4(V/A) to obtain an equivalent mean hemispherical beam length L.
Ira, thank you very much for your reply to my comment on your article “IraGlickstein, PhD says May 11, 2011 at 2:23 pm: “——-“
I have not had a chance to reply until now. – Well, “until now” is not strictly correct as I finished an explanatory reply as to why you “get a disconnect”. However before I submitted my posting I read through some of the later comments and found that Tim Folkerts says it much more elegantly in his comment on May 11, 2011 at 3:12 pm, than I did in mine so I scrubbed my entry and recommend his instead.
Mkelly,
The above from an astute person shows clearly that if two objects of the same temperature are radiating at each other there is no (none, nada, zippo)
energy exchangeHEAT (ie no NET energy exchange). You even say specifically that they are radiating at each other; since radiation = traveling EM energy, you just admitted that energy is being exchanged in the very sentence where you say there is no energy exchange.As you just said, what you call things DOES matter — ESPECIALLY if you call them the wrong thing (ie saying that “q” is “energy” rather than “heat”)! There IS energy exchange when the temperature is the same — but the energy going each way is the same. There is no HEAT from one to the other.
“What the equations call temperature you seem to want to call back radiation. ”
This is just silly! Joel has always been way too careful to mistake temperature (measured in K) with radiation (in W in this case). Where does he do such an odd thing? Or is this another case where being careful to use the right words was not important?
mkelly says: “Sir, if you disgree that pressure causes an increase in temperature if volume is held steady that is well odd. PV= nRT discribes what is termed STP (Standard temperture and pressure).”
How are you applying pressure to your fixed volume? If I increase the pressure inside a scuba cylinder by pumping in more air, I get a proportional increase of n, not T, for my fixed volume.
Are you trying to imagine some other way to apply pressure to a fixed volume without compressing it (not lowering V) AND without adding more gas (not increasing n)? That would indeed be an odd theoretical device – the non-compressing presser that heats a fixed volume. If I was trying to increase the pressure of a fixed volume, I would work the other way around – heat it to induce a temperature increase, leading to a pressure increase.
In the real world, I do not think you would have “pressure causes an increase in temperature if volume (and moles!) is held steady”, I think you would have “temperature causes an increase in pressure if volume (and moles) is held steady.”
Robert Clemenzi says:
‘Using 1.0% water vapor, my program computes 282.68 W/m2 .. close to your value.
For power absorbed, my program incorrectly assumes 1 km at constant temperature and pressure. However, it also plots the distance to absorb x% of the available radiation. As long as those values are under 100 meters, I feel that assuming constant temperature and pressure causes only minor errors.
At any rate, I would like to investigate WHY the programs are giving different values to absorb the radiation.’
Robert,
For water vapour I use a partial pressure of 0.231 atm. I calculate the emissivity/absorptivity of water vapour again using Hottel’s emissivity v absolute T graphs.
This equation has shown up several times:
q = ε*σ*(Th^4 – Tc^4) Ah
Every equation applies in some specific set of circumstances (for example, d= vt only applies when v is constant). Looking at this equation, I would say it applies to radiation between two large, flat surfaces with area Ah with the same emissivity close to each other. It should work for a convex polyhedron of area Ah completely by the second object. It might well work for some other circumstances. (People are welcome to suggest other circumstances, or to suggest why I might be wrong, but that is not the point.)
The given equation most certainly does NOT work for two arbitrary objects at arbitrary distances with arbitrary orientations with different emissivities. Expecting this simplified equation to work in general is a losing proposition.
If you want an equation that applies to more general circumstances, you will need a more general equation. So the conclusion (furthermore the equation has a term for the emissivity of the object and no term for the emissivity of the surroundings or the absorptiveness of the object, both of which would be required if the surroundings were transferring energy to the object.) is more a statement about the limitation of the equation, not a statement about whether objects with different emissivities can transfer energy just one direction or both directions.
PS As I pointed out once before, a more general equation is in Wikipedia http://en.wikipedia.org/wiki/Thermal_radiation . This equation DOES include the areas of the two objects, the emissivities of the two objects, and the geometry of the two objects. This removes the previous objection.
Phil informs:
“It goes on to assert that any point in the Venusian atmosphere at the same pressure as the Earth’s would be at the same temperature with appropriate adjustment for the relative power received from the sun. This is said to be the 4th root of the ratio of the powers (∜1.91), and lo and behold they match!
The trouble is that ~30% the power received at the Earth’s TOA is reflected and ~90% of the power received at the Venusian TOA is reflected, so by his argument Venus receives less power than Earth!”
Can you please share with us the source of your 90% reflectance number?
Wayne,
Sorry I missed this reply of yours. Thank you for the kind words, you may of course use my words anytime you wish, for what they are worth 🙂
I too don’t remember the concept of “back radiation” occuring before AR4. As with many things the IPCC claims, I find the actions they attribute to it unscientific and a misjudgement of cause and effect, as with the effect of clouds. But I don’t want to open that can of worms at this time…
Alleyne says:
The -18 C “average temperature of the Earth” is the temperature where more radiation is leaving than is entering–about 5 km in the sky. As measured accurately by satellite. Because of the lapse rate (environmental lapse rate = -6.49 K/km), that means that the surface of the Earth has an average temperature that is about 33 C higher, or about 15C. We know that the lapse rate is not caused by a GHE, since it is simply a function of gravity and thermal capacity. I is caused by the relationship explained by the ideal gas law pV=RT. Therefore, it looks to me like there is no need to “explain” anything with a GHE.
Of course, other planets with an atmosphere should show the same relationships, and it appears that they do. They all show the same type of increase in temperature with a decrease in altitude, such that the “black body temperature” of the planetoid is at a level that coincides with about 100 mb pressure (Earth is somewhat different, probably due to the presence of water). Doesn’t seem to matter what the gaseous constituents are. See the figure towards the end of this article:
http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
All right, now, tigers. Tell me why all this is wrong.
Alleyne says:
Both a heat pump and an air conditioner use work to move heat from colder to hotter, so I don’t really see the distinction. The only substantive qualitative difference is that for the heat pump, the “hot reservoir” is your house and the “cold reservoir” is the outside and for the air conditioner, the “cold reservoir” is your house and the “hot reservoir” is the outside.
At some point, this comes down to terminology. It is most correct to say that because of back-radiation (in combination with the radiation from the sun), the earth is warmer than it would be if all the radiation from its surface escaped into space.
This is an ill-defined question. The average flows of energy radiated and back-radiated have been measured to fairly good accuracy and are summarized in the diagram of Trenberth and Kiehl (e.g., http://www.nar.ucar.edu/2008/ESSL/catalog/cgd/images/trenberth9.jpg ). However, the atmosphere is not just at one temperature (nor is it a blackbody emitter). To compute what happens quantitatively, one must solve the equations for radiative transfer absorption-line by absorption-line through the atmosphere. This is quite an undertaking (although I think there are codes available on the web).
To get a qualitative picture, you can use simpler blackbody (or graybody) shell models like the one we discuss in Section 2.3 of our comment on G&T: http://scienceblogs.com/stoat/upload/2010/05/halpern_etal_2010.pdf
mkelly says:
Perhaps the most fundamental thing to understand about that equation is that if Tc was zero then there would be more heat going from the hotter object to the colder object than when Tc is not zero. That, along with the fact that the surface temperature of the earth is determined by balancing the heat it receives from the sun with the heat it radiates away tells you there is a greenhouse effect. In other words: the temperature Th necessary to radiate away the 240 W/m^2 of radiant energy that the earth receives from the sun depends on the temperature Tc.
It is sort of amazing how such a simple concept seems to illusive for some people!