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.
Dave Springer @ur momisugly August 15, 2011 at 7:28 pm
And, attention Willis
I think you described that viscosity is important in the nano-skin of the ocean where most DLR is momentarily absorbed at the quantum level.
(I would add; that is before at astonishing rapidity relative to ocean dynamics, it is reemitted or causes higher energy H20 molecules to fly off, {evaporate}, partly via collisions).
Did you mean the ‘surface tension layer’ perchance? The surface tension layer is also maintained in most of the slow earthly dynamics of the ocean, and can surely only result in stability of the upper surface layer.
It seems to me that there are a lot of unknowns about surface tension, apart from the many observed effects such as meniscus. The following two links are interesting, but simply pose more questions to me.
http://www.engineeringtoolbox.com/surface-tension-d_962.html
http://en.wikipedia.org/wiki/Surface_tension
I see that this thread is still alive, so I’ll add a statistical perspective.
The temperature of a region is proportional to the mean kinetic energy of the molecules in that region. As always, there is a distribution of actual energies, and within a region some molecules are reasonably close to the region mean, some are way above the region mean, and some are way below the region mean. For the topic of the thread, the above average molecules in the air radiate energy down to the earth; in the earth’s top layer (solid or liquid) the radiated energy is absorbed by the molecules with below average energy. The above average molecules in the earth radiate to the air, where the energy is absorbed by the below average molecules. in the regions of high temperature, the distribution of the energies in the molecules is shifted upwards compared to the distribution of the energies in the regions of low temperature.
From region to region, there can be no net transfer of heat from the lower temperature regions to the higher temperature regions, but the back radiation from the higher energy molecules in the air to the lower energy molecules in the earth surface can for sure slow the cooling of the earth surface. There are in addition advective and convective processes, but their existence does not contradict the simple scheme I just described. The first major complication is the transfer of energy from the high energy molecules of the air to the low energy molecules of the air by collision. With a few modifications to the details of the total mechanism, the more GHGs there are in the air, other things being equal (which I doubt they are), the slower should be the cooling of the earth surface temperatures from their daytime maxima, hence a gradual increase of those maxima (seasonally adjusted) across time.
That’s the basic narration of radiative heat transfer. If the effect of the increased surface temperature is to increase cloud cover an thereby to reduce surface insolation, that is a major complication to the narration.
Willis Eschenbach @ur momisugly August 19, 2011 at 10:34 pm
(Sorry JAE if you think I interfere but but but Grhhhh!)
Willis, really! You said; I can feel the warmth from the cloud?
Some years ago there was a learned professor that made a claim in the McGraw-Hill scientific encyclopaedia that the moment that cirrus clouds formed (they comprising ice crystals at great very cold altitude) that they rapidly heated the Earth’s surface. I emailed them and that article was subsequently removed. There were similar bizarre claims in Encyclopaedia Britannica, that were similarly addressed.
Willis, your fertile imagination is often very interesting, promoting good debate, but you should not be so assertive about your hypothesises, and listen to what others say to you.
“You have an interesting and plausible hypothesis (energy absorbed from IR photons goes directly to evaporating the molecules, not to heating the surface). So do I (energy absorbed by IR photons heats the surface layer, which then fuels an increase in upward IR and evaporation).”
I may as well wade into this discussion with my own view. The energy found at the very surface is made up of energy from the DLR plus energy from the bulk that has conducted through the skin.
I dont believe you can say that the energy used in evaporation exclusively came from the DLR because energy from the bulk is there too. But you cant say its from warmed molecules from the DLR either because on average all the molecules are cooling not warming despite the DLR.
You might be able to look at individual molecules and look at the energy that kicked them into evaporation and I’m sure the DLR will do that often and the energy convected through the skin will also do it.
I do think you can probably say that the energy that makes it to the surface and becomes ULR and subsequently becomes DLR then essentially has two options. Its either going to be radiated again at the surface or its going to be used for evaporation. If its radiated again then its got two options its going to become DLR again or its going to escape to space.
I’m ignoring the exotic options like mixed down by a wave and residence time in the atmosphere…
So in a sense I guess all DLR is used for evaporation or subsequently lost to space.
Willis,
It disturbs me that you still do not acknowledge advice that it does not matter how much EMR* is whizzing around, unless there is a potential difference (PD) between two sources of it. Otherwise, it amounts to nothing in terms of HEAT transfer. Furthermore, unless there is a change in HEAT level in matter, there is by definition NO change in temperature.
The following illustration shows that by far the most intense EMR fields are towards the horizontal because initially from the surface, radiation is hemispherically equal in all directions, yet mostly nothing happens. Secondarily radiation above the surface is spherical in all directions. The vertical PD is relatively trivial.
http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1
There are other implications but let’s see if you can take-in the first step.
* (electromagnetic radiation; including your topical DLR and ULR)
Willis Eschenbach says:
August 19, 2011 at 10:13 pm
Now go convince tallbloke that the ocean absorbs DLR
tallbloke says:
August 15, 2011 at 1:46 pm
Hi Willis,
Argument one asks what the difference is between rock and water. Warm water molecules rise to the top. Warm rock molecules conduct heat to their neighbours, which can’t go anywhere.
Argument two asks where the energy goes. The answer is:
space.
Argument three is not an argument that DLR can warm the ocean, it’s an argument that it can slow its rate of cooling.
Argument four is a numerical misunderstanding. The ocean surface very efficiently absorbs 95% of DLR
Stop trying to misrepresent me and instead engage the substantive arguments.
Willis Eschenbach says:
August 19, 2011 at 10:34 pm
“When the cloud goes away, the clear sky just sucks the warmth out, but when the cloud is over me, I can feel the warmth from the cloud.”
And also said in the same post
“Basically, putting in a paragraph of that kind is like waving a big sign saying “I AM A NUTTER! IGNORE ME!”
Were you being serious in the first sentence. Or do you want to withdraw this comment?
Willis Eschenbach says:
August 19, 2011 at 10:34 pm
“There’s an entire industry out there making instruments that are specifically designed to measure the DLR. There are scientists who make a living measuring the DLR. There are a host of papers about the measurements of DLR”
This point has been discussed before on this site. And also on a Dr. Judith Curry thread
http://www.slayingtheskydragon.com/en/blog/102-climate-follies-encore
The most interesting involved the Holy Grail of Warmist worship, the phantom ‘back-radiation’ force. As the aforementioned “20 Milliseconds” explains there are Warmists, Luke Warmists, and Deniers. This phantom is the defining element of this fable and refuses to fit into classical Physics. The following is the actual exchange, playing the part of the English Lord will be Lord Monckton. The Luke Warmist chorus is played by the who’s who on the CC list and the Slayer is played by your narrator and co-author of Slaying the Sky Dragon.
Lord: “Back radiation can be simply demonstrated by pointing a simple infrared detector at the underside of a cloud. Try it.”
Chorus: “My IR detector only cost $60! Simple! Agreed! Agreed!”
Slayer: “Clouds do not absorb and re-radiate heat back to Earth. Clouds add THERMAL MASS which takes longer to heat and cool. Warmists ‘support’ this false hypothesis with IR thermometer readings, but the IR readings of a hot Barbie is the same from any distance; ENERGY is not. Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”
Jae
“I think I asked, twice, a very important question that goes to the heart of this (nonsense?)”
No need for the question mark
1- anything which absorbs energy easily is a conductor, not an insulator.
2- additional heat capacity improves the efficiency of a cooling system
3- co2 is not an IR mirror.
4- ‘black body’ is an idealization – there is no such thing IRL.
5- assumptions that an IR detector are measuring temperature of a surface through miles of atmosphere are absolutely false.
6- degrees do not measure joules. temperature is not heat energy
7- attribution of motives to a speaker is always done for the purpose of changing the topic.
8- misattribution of motives is a thinly disguised ad hominem argument.
9- third party dippers who do #8 need to let their sycopants out a notch, smokey.
10- if the numbers don’t match the facts, then the numbers are wrong; not the facts.
11- the facts don’t join anybody’s club and don’t require anybody’s approval.
12- claques are for confidence hustlers.
tallbloke- plz don’t change your ways. you’re about the last one i can still admire – not that you should care; but i do.
if the question is to be answered, it won’t be by speculation, now, will it?
why not help design an experiment that can take this discussion out of the realm of opinion and provide indisputable facts?
the experiment would be to show what effect IR radiation has on the temperature of a body of water, right?
suppose i have a 30W CO2 laser… (which i do) so there should be no question about it putting out the pertinent LR wavelengths, agreed?
suppose i have an IR ‘thermometer’… (which i do)
suppose i have a cup or water… (which i do)
suppose i point the laser at the surface from a perpendicular direction…
now, where should i put thermometers and where should i take readings with the IR detector to satisfy the requirements of the experiment? what OTHER variables must be controlled for the results to be meaningful?
or would descent from the lofty realms of ‘what i think’ to the level of ‘what really happens’ make this the most boring blog on earth? (gratuitous dig… i acknowledge it without shame)
btw – when a pyrgeometer is pointed out toward space, it does not show the temperature of space, nor does it show the temperature of the earth when pointed the other way.
and a transparent substance is not in any way properly characterized as a ‘black body’
Willis
What I am about to say is not intended as a personal criticism of you. I should perhaps observe that I usually very much enjoy your articles and find them stimulating and insightful. Frequently, I find myself agreeing with much of what you have to say. However, this present article, falls well below your usual standard.
The article got off to a very bad start by suggesting that the arguments rejecting the contention that DLR does not HEAT the oceans is “just silly”. Given that the contention that DLR can heat (or slow down the cooling of) the oceans is fundamental to the AGW position, any argument contesting this presumption deserves due consideration. Given that the oceans cover approximately 70% of the globe and they account for 99% of the heat capacity of the earth system (ignoring that in the core etc) and given that the climate and weather is driven by the oceans and ocean conveyor belt and air currents that arise consequential thereto, I would suggest that there is no legs in the AGW conjecture if manmade changes to GHGs cannot result in a warming of the oceans. My point is that the issue raised by your Article is really really fundamental and therefore these issues require addressing in detail not by flimsy rebuttal.
Of the 4 arguments you raise against the validity of the contention that DLR does not heat the oceans, the first is patently bad in that it does not take account of the different nature and properties of matter (ie., the differences between solid land and liquid water) and in particular takes no account of evaporation (and the processes involved in that and the energy involved). The third argument, is also plainly wrong in that it confuses heating with a reduction in the rate of cooling. Thus, the entire thrust of your article is off to a shaky start.
Early on, I picked you up on the point that DLR cannot HEAT. You accepted that this was so and sought to justify your sloppy (my wording not yours) use of language on what is understood in common parlance by the expression to heat. I reverted on that saying that in a scientific article, one has to be precise especially when there is a fundamental difference between the concepts and principles involved since to do otherwise only leads to confusion and promotes misunderstandings. At the time you appeared to accept that.
Now look at your latest response with clouds. You suggest that “when the cloud goes away, the clear sky just sucks the warmth out” as if there was some giant vacumn cleaner up in the sky sucking warmth. There is no such process. This is not a scientific response. You state “I can feel the warmth from clouds”. It is extremely rare for clouds to be warmer than the ground still less warmer than the human body. You cannot feel the warmth from clouds. Again, your terminology is incorrect, and this is really very unfortunate since you were picked up on the incorrect use of these principles at an early stage, and you appeared to accept that you had not been scientifically accurate.
Turning now to the DLR measuring equipment. I have lost count of the number of times that I have had arguments with people who claim because there is DLR measuring equipment on the market DLR is real. There is a fundamental difference between signal and energy. For sure, DLR has a signal but that does not mean that it has sensible energy capable of doing sensible work for example capable of heating something.
If I stand on Mercury and point my DLR meter at the bright burning ball in the sky, I see that I am being bombarded with 5800K photons and I think to myself, no wonder I am ‘bloody’ hot. I then get into my space rocket and fly to Pluto. I get out and I feel rather cold. I point my DLR meter at the burning ball in the sky and observe that I am being bombarded with 5800K photons and wonder why it is so cold that it freezes ‘the balls’ off a brass monkey.
A PVR is essentially a DLR meter tuned to the wavelength of solar radiation being received from the sun. It can convert the solar radiation into energy which can be used to power something. By contrast the DLR meter cannot convert the DLR from the atmosphere into energy or electricity capable of doing work. IF it could we would have solved the world’s energy problems. The DLR being radiated has no sensible energy and one should not confuse signal with energy.
For example, I have a problem with my car. It will not start. I measure the battery and the volt meter suggests that everything is fine. It reads 12 volts. I turn on the ignition, a little red light comes on suggesting that the alternator is not yet charging. I turn the ignition further (to energise the starter) the engine does not turn and even the little red warning light goes out. The battery may be producing 12 volts but is has no power. I am not saying that this is a like for like analogy. I am merely using it as an illustration of the principle that although one can detect a signal, it does not in itself mean that there is any energy in that signal.
As far as I am aware, there is not a single sceptic who denies the existence of DLR. However, there are many who challenge precisely what DLR is capable of doing. Your argument that there are DLR measuring devices on the market does not address that issue. Why people raise it, I do not know.
Finally, I would observe that (in the case of this article) you frequently answer a question with another question or in some other way so as to side step the issue. Now I know that this is difficult, and I know human nature adds to this difficulty but it would be better to respond along the lines: ‘I do not know, but I consider that this is a more telling point etc’ Or, ‘the question you raise is not capable of a quick answer and I do not have the time to properly address it,. but I think that this or that is a more significant point/or you fail to appreciate this or that which is also (or more) relevant’ etc. I can see that many commentators have become frustrated by this side stepping and of course, this side stepping has hindered taking this debate further and has hindered getting to the root of the issues raised.
I emphasise that this comment is not meant as a personal criticism. I do not like to see personal criticisms of the authors of articles posted on this site, and I have on a number of occasions commented that this is disrespectful. Those who take the time to post an article deserve the respect of all those who visit this site. In this you have my respect although I find myself very much in disagreement with much of what you have said in this particular article and your subsequent comments. I consider that the issues raised by your article to be extremely complex ones, and of such complexity compounded by the lack of any empirical data and experiments, to be such that most of us are out of our depths. That said, the burden of proof is firmly upon those who contend that DLR heats the oceans, or slows down the cooling or that the oceans would be frozen but for DLR and to date this burden has not been discharged.
Dave Springer: (if you’re still reading)
I have had this thread open since it started, have gradually read the whole thing (in sequence), and have saved most of your comments to a separate file for further reading and rumination. For what it’s worth, I am impressed by your comprehensive grasp of this entire matter, and find your explanations both lucid and cogent. I’ll be plagiarizing/paraphrasing some of your observations next semester for the atmospherics part of my thermodynamics course.
My thanks and compliments,
/dr.bill
Sorry, Willis, I’ve read your exposition, Dr. Curry’s exposition, and ScienceOfDoom’s exposition — which are all basically the same — and I simply can’t buy it. No amount of armwaving about turbulence and waves can overcome the enormous differences in magnitudes here. The best picture we have of the ocean’s surface seems to be:
A couple of microns where IR is actually absorbed (and radiated);
About a millimeter of “skin” where the temperature is perhaps a third of a deg C lower than the next layer down, presumably due to evaporation; and
A couple of meters at the “real” ocean surface temperature.
There is some casual talk above confusing microns and millimeters. Please: that is, quite literally, like confusing millimeters and meters. So the claim amounts to:
Turbulent mixing of the topmost membrane has a thermal effect on the skin layer, which is cooled by evaporation and is ~500 time thicker than the membrane; then
Turbulent mixing of the skin layer has a thermal effect on the “real” layer, which is ~1000 times thicker than the skin layer.
This is nearly impossible to believe. As I said, the differences in magnitudes are just too great, no matter how stormy the sea or impressive the surf, and no amount of armwaving and sailing experience (which I also have) can get around that. My conclusion is that nobody really has the faintest idea of what’s actually going on at the water/air interface.
So perhaps it would be better to stick to something like “slows down the cooling” or the impossibly opaqe and convoluted lapse rate explanation. The direct heating by IR simply makes no sense.
I’m surprised you don’t get this, Willis. Here’s your definitive statement:
My argument is that the ocean is warmer, both at the surface and in the bulk of the mixed layer, than it would be if it were not absorbing DLR.
– Willis Eschenbach
This is only true if you cherry-pick the time where the temperature reading is taken. Yes, there are times when a small delay in outgoing radiation will create a temperature higher than it otherwise would have been. But it’s also true there are times when the delay causes the temperature to be lower. If you integrate over a whole cycle (like a day or a year). then a small delay in outgoing radiation getting to space contributes nothing to the integrated surface temperature or the integrated ocean temperature. It can’t. It doesn’t.
In addition, a small delay in outgoing radiation cannot lead to an increase in the peak temperature. It contributes nothing toward the warmist claims of higher and higher record temperatures. Re-radiated radiation does not make the emitting surface hotter than it started out…the only thing it can do is modulate (slightly) the rate of cooling. It does not add energy to the system. It can’t. it doesn’t.
I admire you and your work, Willis. I really do. But, in this case, with all due respect, your thinking cap is askew.
Any increase in water vapor (Tallblokes invisible layer which I am surprised Willis considers a fairy tale) is in and of itself a spectral modification of incoming TSI reducing SW radiation at the surface, (See A) which we all agree is the PRIMARY mechanism which heats the ocean below the surface. Any increase in cloud cover is an even greater spectral modification of SW TSI at the surface. The question needing to be quantified; Is this surface reduction of SWR entering the oceans, caused by additional GHG, adequate, over time, to reduce the subsurface temperature, despite the change in gradient between the subsurface and the skin? It is amazing, people are throwing around lots of hypothesis, with very little quantifiying of the numbers, this includes Willis. Numbers anyone?
(A) Looking at a solar spectrum chart; “ It show that about 98% of that energy lies between about 250 nm in the UV and 4.0 microns; with the remaining as 1% left over at each end. Such graphs often have superimposed on them the actual ground level (air Mass once) spectrum; that shows the amounts of that energy taken out by primarily O2, O3, and H2O, in the case of H2O which absorbs in the visible and near IR about 20% of the total solar energy is captured by water VAPOR, (CLEAR SKY) clouds are an additional loss over and above that.
Gnomish,
I love the idea of trying an experiment! I had contemplated this before, but lack of access to a CO2 laser made me forget the idea
However, I suspect that getting the details right will be devilishly difficult.
Thermometry: Since the temperature gradient between skin and bulk seems to be a key feature, it would be great to confirm that difference even exists, which would require two thermometers (the IR surface thermometer and the bulk thermometer) that could measure accurately to within ~ 0.02 C.
Insulation: We are talking about relatively small differences in energy flow rates. Minimizing unwanted energy flows would be a challenge.
Atmospheric conditions: small changes in temperature or humidity will have noticeable impacts.
With that said, here is one experiment I might propose. It is about as simple as I can imagine to approach the conditions needed. Others are welcome to suggest improvements.
Fill an insulated, uncovered container with water (like a styrofoam cooler). This will ensure that energy primarily enters and leave the system via the surface.
Put a small heater into the water and leave it on until the temperature reaches some equilibrium value. The value should be ~ 170 W per square meter of water surface to approximate the solar energy into the system (although being within a factor of 2 either way would almost certainly work). The water needs to be significantly warmer than the surroundings to ensure appropriate evaporation and to ensure that there is a net flow of IR from the water to the surroundings. Care should be taken not to set up large convection currents.
Aim the CO2 laser down at the water. Run the experiment several times, half the time with the laser on and half the time with the laser off. See if there is a statistically significant difference in water temperature in the two cases. (A single pair of runs may well not be enough, because of other factors like slightly variations in room temperature or humidity.)
For added fun (and rigor), you could intentionally adjust some of the parameters (like room temp, heater power, laser power, humidity). At a minimum, these parameters should be recorded.
PS. To me, the most fundamental way to start is looking at the top of the atmosphere. Here there is very little water vapor to worry about because it is so cold ! If more CO2 (or other GHGs are added to the atmosphere, the layer from which the atmosphere radiates will be higher. To maintain energy equilibrium, the temperature of the new, higher radiation layer must adjust to be the same as the older, lower layer had been.
Everything else being equal*, the lapse rate will stay the same, so by the time you get to the surface, it must be warmer than it had been. At this level of analysis, the details at the surface are mostly immaterial, since they are occurring within the system.
I don’t really want to start a new discussion here. There plenty of other places that have discussed this idea, and plenty of other places that will discuss it again.
PPS. Yes, I know that “everything else being equal” sweeps a lot of things under the rug. But Occam’s Razor suggests this as a starting point until some other explanation points otherwise.
Tim writes “What I am saying is that IF there is some extra energy to the skin (for example from extra downward thermal IR, then the balance will get upset. The temperature gradient will decrease (and not merely increase in evaporation rate as Dave Springer is arguing). Less of a gradient –> less conduction of energy from below –> the temperature of the bulk of the ocean must increase. ”
This is the argument that RC (and Peter Minnett) used to explain how the DLR warms the ocean. You are quite correct in suggesting the temperature gradient will change with a change in conditions (ie changed DLR or changed DSR) to either increase the rate of conduction through the skin or reduce it.
But the upshot of the change is to alter the “hook” temperature and consequently the actual SST. As the equilibrium is being re-established, the skin’s temperature gradient will assume a value that is optimal to keep the equilibrium and it wont be the same as the value it initially assumed to reduce (or increase) the conduction to get the SST to that value.
So a typical example might be that the ocean starts off in reasonable equilibrium with the DSR balanced by radiative losses to space and evaporation and then thick clouds come over. DSR drops and DLR increases. The immediate effect of DSR dropping is that the energy input into the ocean drops and the temperature profile due to the DSR vanishes as the residual warmth convects towards the surface.
Having the temperature profile drop means an immediate cooling effect throughout the bulk and although the convecting residual warmth will tend to keep the “hook” temperature up for a time, the “hook” is now tending to drop in temperature because its being radiated/evaporated away and not fully replaced by the dwindling warm convected waters from below. The SST itself can tend to remain at its temperature because the ocean’s requirement to radiate is now being “subsidised” by the increased DLR which is absorbed at the “very surface” and (for want of a better term) re-radiated.
So the “very surface” temperature has a tendency to remain while the “hook” temperature has a tendency to drop and that means the temperature gradient of the skin is reducing and therefore energy conducted through the skin is also reducing. The ocean is cooling less as a result of the effect of the DLR.
As the “hook” temperature drops (the energy is being radiated/evaporated away) it necessarily takes the SST with it though. Eventually the SST temperature will reduce to the point where the balance is again restored between the DSR and radiated loss plus evaporation and at this time the temperature gradient of the skin will have assumed the optimum value for the new level of radiation and evaporation.
Tim Folkerts-
say it was still water – not being mixed …
say no cross-wind – no foam or spray, unrestricted convection only ..
say water is at ambient …
say no plankton or other particulate matter…
(unlensed, laser is @ur momisugly 5mm diameter beam, delivers @ur momisugly 25-30W IR)
easy numbers to get:
bulk temp of water – it’s at ambient
alleged reading of ‘surface’ by IR meter, alleged to be a thermometer, with and without laser on…
…at various distances from the spot of the beam
(it will really be registering the heated cloud of water vapor immediately above and around the hot spot, not the surface of the water)
water does not even need to convect in order to rise once it becomes gas because its density is so much lower than the other gases.
is already done some experiment which determined the attenuation of IR as it passes through water vapor?
attenuation, wrt h2o^ (or co2), is synonymous with absorption, right?
what if –
2 identical cups with same mass of water-
measure the loss over a specific period with IR on one. then stir the water and take its temp.
and the loss over the same period without IR on the other. then stir the water and take its temp
easy math would give the joules absorbed from IR, yes?
it would also need for the heat sourcing beam to be dispersed to ‘dilute’ the power to actual insolation levels…
and also specify the surface area to be completely irradiated…
Longwave absorbs in the ocean surface just like microwaves absorb on the surface of the food you are heating. To heat the food evenly you have to stir it while heating it. The same applies to the surface layer of the ocean. To understand why the surface layer is not at a radically different temperature from the underlying layers, consider an initial state where the ocean, the surface layer and the atmosphere are all in local thermal equilibrium (LTE), which by definition means that they are at the same temperature. In this case there will be no net heat flux from the surface to the ocean sublayers.
For some purposes, LTE is a good assumption, but the deep ocean is quite cold, and there is a slow conductive heat flow downward, enhanced by mixing. The top underlying layer is slowly losing heat to an even deeper layer. To keep things in “near” LTE the top underlying layer must slowly gain heat from the top, IR absorbing layer. The key word is slowly. Not much heat flow downward is necessary to maintain shallow sublayers in near-LTE with the surface, which is why shallow sublayers will be at nearly the same temperature as the surface.
TimTheToolMan says:
August 20, 2011 at 7:47 am
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Tim, it seems the surface and skin layer and will ‘adjust’ to maximise heat loss from ocean to air at all times.
Is this the constructal law which Willis likes at work? Or simply the entropy law at work?
Tim F: It’s fine to move the action back up 5km into the air, but you thereby concede the ocean surface. As I’ve been saying, the air temperature affects the rate the ocean will cool at, but since DLR isn’t directly heating the ocean bulk, and air temp follows ocean surface temp, it’s clear the effect from changes in DLR on air temp is small in comparison.
Excellent discussion from all by the way.
Gnomish says:
August 20, 2011 at 7:52 am
say it was still water…no cross-wind… no foam or spray ..
unrestricted convection only … no plankton or other particulate matter…
Those all sound like good conditions — keeping it simple is always a good starting point.
(unlensed, laser is @ur momisugly 5mm diameter beam, delivers @ur momisugly 25-30W IR)
This is problematic. You are delivering ~ 20 W to ~ 0.00002 m^2 = ~ 1 million W/m^2. To be comparable to real conditions, you would want to add only ~ 1-10 W/m^2 extra IR to the surface of the water, and you would want to add it uniformly over the surface. Highly concentrated power over a small bit of the surface would be sooooo far from the conditions we are simulating as to be useless.
easy numbers to get:
This reminds me of a humorous sign I saw once. Paraphrasing slightly … “We offer accurate, easy, inexpensive results. You can pick any two.”
I’m afraid that any “easy” answers will not be sufficient accurate. This is actually a subtle experiment, since we are only looking at small differences that will occur slowly.
Some back-of-the-envelope calculations …
Consider a tank of water 1 m x 1 m that is only 0.1 m deep = 100 kg of water. The net heat capacity is (4200 J/kg*K) x (100 kg) = 4.2E5 J/K. A 420 W heater (similar to daytime solar energy) will raise the temp only ~ 1/1000 K per second (assuming perfect insulation). So it will take ~ 15-20 min to warm 1 K. This heater will raise the temperature quite a bit above ambient temperature before equilibrating, so this might take several hours. (Alternately, we could start with warm water so it will equilibrate sooner).
We want to add just a few W/m^2 of IR to see how it affects the equilibrium temperature. First of all, 10 W/m^2 extra IR will only affect the temperature at a rate of 10 J/s / 4.2E5 = 2E-5 K/s. This would be 1 K in 12 hr. So the experiment will need to run most of the day before the temperature changes are easily noted. (This assumes I am right and there is indeed a change. Dave would presumably expect no change in temperature, only a change in evaporation rate.)
All the while we want to control IR into the water from the room, so the temperature of the walls & ceiling needs to be well maintained. At 300 K, a 1 K shift in the temperature of the walls will change the IR flux about 6 W/m^2. So we would want to control the wall temp within ~ 0.1 K for the control experiment, then raise the wall temperature ~ 2 K to increase the the IR by an appropriate amount.
But the air temperature and humidity must not change, because that would change the evaporation rate and confound the results!
This is not a “throw a thermometer into a cup of water” type of experiment!