Guest Post by Willis Eschenbach
Once again, the crazy idea that downwelling longwave radiation (DLR, also called infra-red or IR, or “greenhouse radiation”) can’t heat the ocean has raised its ugly head on one of my threads.
Figure 1. The question in question.
There are lots of good arguments against the AGW consensus, but this one is just silly. Here are four entirely separate and distinct lines of reasoning showing that DLR does in fact heat the oceans.
Argument 1. People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land. DLR is absorbed in the first mm of rock or soil. Yet the same people who claim that DLR can’t heat the ocean (because it’s absorbed in the first mm) still believe that DLR can heat the land (despite the fact that it’s absorbed in the first mm).
And this is in spite of the fact that the ocean can circulate the heat downwards through turbulence, while there is no such circulation in the land … but still people claim the ocean can’t heat from DLR but the land can. Logical contradiction, no cookies.
Argument 2. If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.
Nor can it be going to evaporation as many claim, because the numbers are way too large. Evaporation is known to be on the order of 70 w/m2, while average downwelling longwave radiation is more than four times that amount … and some of the evaporation is surely coming from the heating from the visible light.
So if the DLR is not heating the ocean, and we know that a maximum of less than a quarter of the energy of the DLR might be going into evaporation, and the DLR is not heating the air … then where is it going?
Rumor has it that energy can’t be created or destroyed, so where is the energy from the DLR going after it is absorbed by the ocean, and what is it heating?
Argument 3. The claim is often made that warming the top millimetre can’t affect the heat of the bulk ocean. But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers, heating the surface affects the entire upper bulk temperature of the ocean every night when the ocean is overturning. At night the top layer of the ocean naturally overturns, driven by the temperature differences between surface and deeper waters (see the diagrams here). DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer, and it also slows the speed of the overturning once it is established. This reduces the heat flow from the body of the upper ocean, and leaves the entire mass warmer than it would have been had the DLR not slowed the overturning.
Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. The DLR is about two-thirds of the total downwelling radiation (solar plus DLR). Given the known heat losses of the ocean, it would be an ice-cube if it weren’t being warmed by the DLR. We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?
Note that each of these arguments against the idea that DLR can’t warm the ocean stands on its own. None of them depends on any of the others to be valid. So if you still think DLR can’t warm the ocean, you have to refute not one, but all four of those arguments.
Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …
w.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
Hi Bob,
since energy is finally leaving Earth at around the same rate it’s arriving, and since as you correctly state the horizontal flux doesn’t do anything much in terms of heat transfer, I’m led to ask;
So what?
Seems to me that if (directional) IR meters are measuring 390 up from or near the surface and 320 down, then all the horizontally emitted radiation that builds up is just buzzing around doing no work until it just happens to get re-emitted up or down and become part of the up/down flux.
Or have I misunderstood something you are saying? (quite possible).
Tim Folkerts says:
August 30, 2011 at 3:39 pm
And, yes, changes in “DLR” also influence the climate.
Leaving aside the question of whether “changes in “DLR” also influence the climate” for now, can you point me to any empirical evidence that the DLR has changed? If Miscolzi is correct that Tau remains constant according to the radiosonde data and his line by line study, then it hasn’t, and we can all stop worrying.
So where is the empirical data to show he is wrong?
Thanks.
Tallbloke @ur momisugly August 31, 2011 at 1:14 am
Quickly: I don’t know if there is any good IR instrumental data that can give a global average result, but my understanding is that the Trenberth cartoon derives the 390 from an S-B calculation based on nominal global average T. To me there appears to be a paradox in that the 390 is radiated hemispherically from the surface and that a good part of that 390 is radiated horizontally (or close to), such that in terms of nominal global average T, it is continuous, and not in itself part of the vertical flux.
It seems that Trenberth and others claim that the 390 is polarised normal to the surface, and that is what I’m having difficulty with.
Belmiloud, Djedjiga, Roland Schermaul, Kevin M. Smith, Nikolai F. Zobov, James W. Brault, Richard C. M. Learner, David A. Newnham, and Jonathan Tennyson, 2000. New Studies of the Visible and Near-Infrared Absorption by Water Vapour and Some Problems with the HITRAN Database. Geophysical Research Letters, Vol. 27, No 22, pp. 3703-3706, November 15, 2000
http://www.tampa.phys.ucl.ac.uk/djedjiga/GL11096W01.pdf
Abstract. New laboratory measurements and theoretical
calculations of integrated line intensities for water vapour
bands in the near-infrared and visible (8500-15800 cm−1) are
summarised. Band intensities derived from the new measured
data show a systematic 6 to 26% increase compared
to calculations using the HITRAN-96 database. The recent
corrections to the HITRAN database [Giver et al., J. Quant.
Spectrosc. Radiat. Transfer, 66, 101-105, 2000] do not remove
these discrepancies and the differences change to 6 to
38 %. The new data is expected to substantially increase the
calculated absorption of solar energy due to water vapour in
climate models based on the HITRAN database.
Conclusions
Table 1 (Final column) also sets out values for the comparison
of our “best” total intensities of the water polyads
with those given in HITRAN-COR. It should be stressed
that the line intensities of our observations and the HITRAN
database differ from line to line and that the given values
are only valid for room temperature. The measurements
at 252K yielded slightly different ratios, but are omitted
here for clarity. For a major fraction of the lines, the principal
part of the change takes the form of a re-scaling of
the data on a polyad-by-polyad basis, we recommend the
use of the factors set out in Table 1 as an interim solution.
Other databases, such as GEISA [Jacquinet-Husson et
al., 1999], are based on the same laboratory data and will
therefore require the same correction. Detailed line-by-line
data including both experiment and theory will be published
[Schermaul et al., 2000].
There is another lesson to be learned. Making sure the
database is valid is necessary foundation for all modelling
of atmospheric radiation transfer, especially so when theory
and observation fail to agree.
tallbloke says:
August 31, 2011 at 3:50 am
It is apparent to me that there is a natural fluctuation to the oceans between SWR and DLWR. In other words same tropical area, less water vapor = more sun hitting the surface (in the near-infrared and visible)= oceans warm and release more water vapor which equates to greater spectral modification of TSI (in the near-infrared and visible) which of course then cools the oceans. So there is a natural flux whereby TSI is first absorbed more in the oceans, then due to water vapor spectral modification, (noted in your comment that it is greater then the models show) TSI is then absorbed more in the atmosphere. The oceans are alternately heated and then cooled by this natural process of changes in spectral modification of TSI.
Of course there are many other processes going on here which explains why none of them have a perfect correlaton to temperature, but this cycle of spectral modification is happening within them.
David A:
Yes, there are negative feedbacks operating at all levels and timescales of climate activity.
Folkerts:
You initially SAID the following, remember:?
““Do note that Phoenix cools down significantly more at night than Atlanta. Hmmm — now there indeed is an observed effect of the extra GHGs (water vapor) around Atlantla, providing DLR to limit the cooling of Atlanta overnight”.
Now you are changing your story to discuss diurnal variation, instead of “cooling down more?” The data show very clearly that Phoenix is much hotter than Atlanta, EVEN AT NIGHT.
Then you say:
“I would not necessarily assume that. How much difference do you think the difference in conditions from Alt to Pho will make on the DLR? Will it be more than 60 W/m^2 to make the temperatures similar in the two locations? Oh, and it has to make up for the difference in evaporative cooling edge Atlanta has, too.”
Well…if a few ppm CO2 can make a measureable difference, then a factor of three for GHGs should have way more effect than a mere 60wm-2 over 24 hours.
But let’s look at some other evidence that suggests no GHE is needed to explain Earth’s temperature:
http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
Don’t get hung up on the first part of the article. Juest look at the graph at the end. It seems that atmospheric temperatures on other planetoids have NO dependence upon the molecular makeup of those atmospheres. WTF??
jae says:
August 31, 2011 at 10:18 am
>>Folkerts: note that Phoenix cools down significantly more at night than Atlanta
>JAE: Now you are changing your story to discuss diurnal variation,
No, I said what I meant — you need to read more carefully. Phoenix cools down more at night (~ 15 C) than Atlanta cools down at night (~ 11 C). I did not say “Phoenix gets cooler than Atlanta …” It was always about diurnal variation.
>Well…if a few ppm CO2 can make a measureable difference,
>then a factor of three for GHGs should have way more effect
>than a mere 60wm-2 over 24 hours.
The typical estimate is that doubling CO2 leads to an increase of ~ 3.7 W/m^2. So tripling CO2 would be ~ 6 W/m^2. I have also seen estimates that H2O contributes ~ 5 times as much to the GHE than CO2. From this estimate, the increases water vapor will only add ~ 30 W/m^2, or half of difference from sunlight. So I conclude water vapor changes are less important than sunlight changes. Even if I am off a bit, the effect of water is still not “way more”.
What calculations support your claim of “way more”?
>But let’s look at some other evidence that suggests no GHE is needed
>to explain Earth’s temperature:
> http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
Actually, as I read the article, they discuss that (quite rightly) that other effects besides GHGs do indeed affect temperatures, which is not a surprise to anyone. But concluding that other factors are important does not imply (as you and they conclude) “no GHE is needed”. (Ironically, the article they quote to get the figure is all about how well radiative transfer & GHE predicts the profiles of atmospheres.)
Do you really think it is a surprise that heat capacity affects temperatures? Do you really think that people working in this field don’t know about lapse rates?
Its logically the same as saying “I know two people who eat the same amount of sugar each day but only one exercises; one is thin and the other is fat. Therefore sugar is not important, only exercise.” This is not proof that sugar is not important, any more than showing that heat capacity is important negates that GHE might be important. To positively declare that GHE is not needed, you have to not only show that other things are important, but that they can explain all the values without the GHE.
tallbloke says:
August 30, 2011 at 10:29 am
Myrrh says:
August 30, 2011 at 4:15 am
“It’s pretty well meaningless when water is transparent to visible and doesn’t absorb it.”
Empirically determined coefficients of absorption are strong evidence that water does indeed absorb visible light. Better evidence than anything you’ve come up with to say it doesn’t.
And how does this compare with thermal infrared absorption?
“That the atmosphere is not transparent to visible light is an object scientific fact.”
Yeah? How come I can see for 50 miles or more from a mountain top on a clear day, and yet it’s completely dark a lot less than a mile under the surface of the sea?
??
OK, I’m convinced you’re not able to see the disjunct, in what amused me about these claims. However, I am in awe of you being able to see the whole of the starry vastness around us through our transparent atmosphere during the day..
http://www.madsci.org/posts/archives/2000-03/953560966.Ph.r.html
This AGWScience fiction version of reality claims that the atmosphere is transparent to visible light, clearly, pun intended, it ain’t; we can see immediately for ourselves that on a clear day the sky is blue, it is not transparent, and on investigation we can discover that the sky is blue from light being scattered by the molecules of nitrogen and oxygen in electronic transitions, which means the light is absorbed by the electrons.
So, how much does this absorption visible light heat the atmosphere? Where is this heat reflected in the numbers of this fictional ‘energy budget’ which claims that absorption always creates heat?
Tim:
“Do you really think it is a surprise that heat capacity affects temperatures? Do you really think that people working in this field don’t know about lapse rates?”
Are you somehow “reading betweeen the lines?” Or something?
You don’t seem to comprehend the point of that graph. Heat capacity, lapse rate, and atmospheric pressure explain EVERYTHING! NO GHGs need be added!
Bob,
You ask some good question about radiation, but unfortunately I think this is a poor medium for trying to converse about such advanced topics. Without being able to sit and chat with a piece of paper between us, this medium of typed words in a blog (about a somewhat different topic to begin with) is just to unwieldy for discussions of surface integrals and radiation and idealizations and real-world conditions. So I think I will have to drop out of this conversation. I do hope that somewhere, somehow you succeed in your quest for knowledge about SB radiation and the paradox you think is there.
Myrrh says:
August 31, 2011 at 2:19 pm
tallbloke says:
August 30, 2011 at 10:29 am
Empirically determined coefficients of absorption are strong evidence that water does indeed absorb visible light. Better evidence than anything you’ve come up with to say it doesn’t.
And how does this compare with thermal infrared absorption?
So far as heat generated in the ocean is concerned, about 50/50 on average.
on investigation we can discover that the sky is blue from light being scattered by the molecules of nitrogen and oxygen in electronic transitions, which means the light is absorbed by the electrons.
Very little visible light is absorbed in the atmosphere. Which is why I can see 50 miles from mountain tops. Unlike in the ocean, where it gets pretty dark pretty fast as you go down.
Are you ready to concede this?
Tim:
You say: “The typical estimate is that doubling CO2 leads to an increase of ~ 3.7 W/m^2. So tripling CO2 would be ~ 6 W/m^2. I have also seen estimates that H2O contributes ~ 5 times as much to the GHE than CO2. From this estimate, the increases water vapor will only add ~ 30 W/m^2, or half of difference from sunlight. So I conclude water vapor changes are less important than sunlight changes. Even if I am off a bit, the effect of water is still not “way more”.
What calculations support your claim of “way more”? ”
There is something really, really wrong with your calculations and logic! How are you gonna get to even the 324 W/m2 for the “global average downwelling radiation,” using that kind of analysis? And we are not talking about radiation at some 15 C average global average here! The water vapor levels in Phoenix are typically in the neighborhood of 5 g/m^3 in Phoenix and around 20 g/m^3 in Atlanta. How can the difference be only 30 wm-2?
Please read the link I posted.
All the energy that is ‘reflected out’ in the fiction budget is being absorbed.
Kuhnkat @ur momisugly August 29, 2011 at 9:45 pm
Sorry, but I’ve only just got around to looking at your suggestion. I don’t think that Tim has suggested that, but what he does claim is that the vertical integration of the surface emission (altitude = 0) a la Trenberth average global T, is 390 W/m^2 normal to the surface, and that it remains normal regardless of altitude. Presumably he is ignoring other effects as inferred in the MODTRAN calculator.
If the horizontal components increase as you claim between 0.01 and 0.1 altitude, and at the same time the normal does also, isn’t the total energy greater? Why would that be so?
Tim Folkerts @ur momisugly August 31, 2011 at 3:25 pm
Thank you for your patience and good manners in the S-B debate. I agree that to continue it would be a bit like sawing sawdust. (I only like simple analogies)
Very little visible light is absorbed in the atmosphere. Which is why I can see 50 miles from mountain tops. Unlike in the ocean, where it gets pretty dark pretty fast as you go down.
Are you ready to concede this?
Just in case.. The point I was making is that you cannot see the stars in sunlight, even from the top of your mountain, therefore…
The ocean is around 800 times denser than our atmosphere, iirc, maybe someone has the maths to work it out, the relationship between the attenuation of light in the oceans and in the atmosphere, to reach the dark – I think that would have to be calculated by reflection of light back into atmosphere from the Earth to be comparable? Thinking of the thin blue line boundary as seen from space.
tallbloke says:
August 31, 2011 at 3:50 am
Belmiloud, Djedjiga, Roland Schermaul, Kevin M. Smith, Nikolai F. Zobov, James W. Brault, Richard C. M. Learner, David A. Newnham, and Jonathan Tennyson, 2000. New Studies of the Visible and Near-Infrared Absorption by Water Vapour and Some Problems with the HITRAN Database. Geophysical Research Letters, Vol. 27, No 22, pp. 3703-3706, November 15, 2000
http://www.tampa.phys.ucl.ac.uk/djedjiga/GL11096W01.pdf
Abstract. New laboratory measurements and theoretical
calculations of integrated line intensities for water vapour
bands in the near-infrared and visible (8500-15800 cm−1) are
summarised. Band intensities derived from the new measured
data show a systematic 6 to 26% increase compared
to calculations using the HITRAN-96 database. The recent
corrections to the HITRAN database [Giver et al., J. Quant.
Spectrosc. Radiat. Transfer, 66, 101-105, 2000] do not remove
these discrepancies and the differences change to 6 to
38 %. The new data is expected to substantially increase the
calculated absorption of solar energy due to water vapour in
climate models based on the HITRAN database.
========================================================
This might be a dumb question, but it won’t be the first time I’ve asked one of those. I think it’s related to some posts further up this humongous thread, or it may be on another. Could water vapor absorption band broadening be increased even further in the horizontal and horizontal-ish directions by motion due to wind ?? I’m not sure it would make much of a difference to boatloads of water vapor plus 280 ppm CO2 vs. boatloads of water vapor plus 390 ppm CO2, but just wondering.
One quick comment — when discussion the principles of radiation, I was assuming a transparent atmosphere. In MODTRAN, this would mean setting CO2 = CH4 = O3 = H2O to zero. With these settings, the radiation DOES remain pretty constant from the ground up, exactly as I predicted. There are some effects from 1.5 km to 20 km, bit these I suspect are other parameters in the model, like aerosols.
Once the principles of radiation are understood, then the details of absorption by GHGs can start to be added.
(I *am* a little surprised the output of the model does no change from 20 km to 100 km. Over this range, the curvature of the earth should definitely be noticeable, reducing the radiation by a few percent)
Re. this stuff about the Stefan-Boltzmann equation and hemispherical sections, would it help if it were pointed out that the derivation of said equation includes an integration over solid angles so that the quantity computed is the emission normal to the surface?
Folkerts?? Your lack of reply is noticed. You morons have no science on your side and you know it.
Hey JAE, the whole world does not revolve around you — I have other things to do with my life. It gets old having you DEMAND that everyone answer your ill-posed questions, yet you provide no calculations yourself. So you fall back on ad hominem attacks..
Present with your OWN analysis and then I will be more impressed.
“How are you gonna get to even the 324 W/m2 for the “global average downwelling radiation,” using that kind of analysis?”
I don’t expect much, but here are two hints to get you started: 1) logarithms, 2) clouds. This will easily get you to the right rough answer using “that kind of analysis”.
PS. Using MODTRAN looking up,mid-latitude, summer settings,
* with the relative humidity = 0.25 –> I = 266.272 W/m^2 (~ phoenix)
* with the relative humidity = 0.75 –> I = 297.232 W/m^2 (~ Atlanta)
Frankly I am amazed my rough estimate of 30 W/m^2 extra due to the higher humidity using “that kind of analysis” was so close to MODTRAN’s result of 31 W/m^2.
@ur momisugly Folkerts:
“Even if I am off a bit, the effect of water is still not “way more”.
What calculations support your claim of “way more”?”
http://www.engineeringtoolbox.com/spesific-heat-capacity-gases-d_159.html
Gas or Vapor kJ/kg
Air 0.287
Carbon dioxide 0.189
Water Vapor 0.462
Steam 1 psia.
120 – 600 oF
That’s what it takes to change the temperature 1 degree K.
When CO2 changes from 1 to -1 C, a change of 2 degrees C, it radiates 2(0.189 kJ/kg) = 0.378 kJ/kg.
http://en.wikipedia.org/wiki/Enthalpy_of_vaporization
When water vapor changes from 1 to -1 (and condenses) it radiates 2257 kj/kg + 2(0.462 kJ/kg) = 2257.853776 kJ/kg.
It does this every single time you see a cloud.
But CO2 has no phase change so it carries no heat – the numbers:
All gases at the same temperature have the same number of molecules per unit volume.
Water, being light, masses 18g/mole and CO2 masses 44 g/mole
Using 1 mole of air, just to make math easy:
We lowball the water in the atmosphere at 1% of the molecules
So, in a mole of atmosphere, we have 0.01 moles of water = 0.18g
now we highball the CO2 at 500ppm which is 0.0005, or 1/2000 of a mole of CO2.
1/2000 * 44g/mole = 0.000484 moles of CO2 = 0.021296g
So in our mole of air with but 1% H2O and a generous 500ppm CO2-
the water condensing radiates 0.18g * 2257.853776 kJ/kg = 406.41367968 J
while the CO2 radiates 0.021296g * 0.378 kJ/kg = 0.008049888 J
the ratio of 0.008049888/406.41367968 = .00001980712855516645290496438242332
or as much to say that water vapor in the example carries 50486.873814890343815963650674393 times more heat than the CO2 does.
And that’s just condensation to liquid phase. If it turns to snow- multiply by 5-6.
Conclusion – the CO2 is insignificant retainer of heat in our atmosphere. Water vapor does 50,000 times more work.
(And that was even done with shorting the estimate on water while boosting the estimate on CO2)
oh, yah- please note that when water gas changes to water liquid, there is NO temperature change. Your boltzman forumla calculator measures NO change, eh.
it only calculates temperature and temperature is not heat.
Bob_FJ,
“If the horizontal components increase as you claim between 0.01 and 0.1 altitude, and at the same time the normal does also, isn’t the total energy greater? Why would that be so?”
The earth is curved. On a small ball you will see that even if you had radiation on a plane from the point of emission you would see that it quickly diverges from the surface. On earth the curve is much slower so the effect is very small comparatively. At the surface you will only have horizontal energy from a very small area to the point that where the emission happens there is no other area emitting to that point. As you move up in altitude you will have energy from an increasingly large area that can emit through the same space.
Think of overlapping hemispheres on the surface of the globe. The center of each hemisphere will not be able to emit directly at the center of the next hemisphere. You will see a wedge that it also cannot emit through being blocked by the curvature. This averages out from there being an almost infinite number of hemispheres. The end result is that the further apart the centers are the lowest area they can both irradiate will be higher above the surface.
I have to figure out how to draw pictures and post them some day. Any suggestions?