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.
Thanks for addressing my comment, and I understand your claim here.
Can you answer this question that I posted?
Isn’t it possible that the DLR fueled evaporation process displaces a mechanism that would materially cool the ocean more than the ocean is currently cooled?
kuhnkat says:
August 16, 2011 at 6:49 pm
Willis, you can say it as many times as you want and it still is only a half truth.
I get about 275K for the average with the 170 in. Of course, this close averages may hide stuff, but, 275k isn’t frozen. Also of course, I may have screwed up trying stuff I barely understand even with the BB Calculator
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Kuhnkat
Please will you post your calculation.
Your result reflects reasonably well the average temperature of the ocean
From Ken Coffman on August 17, 2011 at 4:12 am:
Sure you can. Just be the sole owner of a bank.
1. Deposit the $100 in your bank.
2. Give yourself a $100 no-interest no-fee loan.
3. The loan is an asset on the bank books, so your bank has $100 from that.
4. You have the $100 in your wallet.
5. Add what’s in your wallet and your bank assets together, you have $200.
Alternate 5: $100 in your wallet, $100 deposited in the bank, you have $200.
5 with alternate 5: Actually you have $300.
You should study governmental accounting methods. For example, the US federal government has generated fantastic sums of money over the years by issuing itself US Treasury notes for Social Security tax surpluses so it can direct that money to the general budget. These notes are then easily repaid by the US Treasury printing itself many pretty pieces of paper, which can cost nothing as it only needs to be done “on paper.”
Alexander Duranko, richard verney, et al:
Thanks for pointing out that climate science incorrectly assumes gross radiative transfers from cold to hot and vice versa can be assumed in isolation, rather than the correct calculation using only net FLOW of heat.
This is a point repeatedly made by Claes Johnson, who explains very simply how this error came about here:
http://claesjohnson.blogspot.com/2011/08/how-to-fool-world-by-measuring-masive.html
http://claesjohnson.blogspot.com/2011/08/how-to-fool-yourself-with-pyrgeometer.html
richard verney says:
August 17, 2011 at 8:04 am
No, I hadn’t seen the question. Nor do I understand it. Is it your clam that the ocean should freeze everywhere at the same latitude? Why would that be?
And having not understood it, I understand even less what it has to do with the subject under discussion. I don’t have time to read every unrelated post. If you start out with something that doesn’t make sense, or that doesn’t catch my interest, I’m likely to just go “Next!” and move on. I said before, to get someone’s attention you need to be brief, clear, and interesting. I have nowhere near enough time to answer every random communication.
Sorry,
w.
wobble says:
August 17, 2011 at 8:53 am
tallbloke says:
August 17, 2011 at 8:20 am
we have yet to see any empirical evidence that the DLR has increased relative to the ULR.
Thanks for addressing my comment, and I understand your claim here.
Can you answer this question that I posted?
I don’t know how much of the evaporation is currently caused by DLR and how much of it is caused by solar derived energy, because there seems to be uncertainty about just exactly what the instruments measuring ULR are measuring, the true emission from the ocean surface, or the upward emission of the radiative soup just above it, or a mixture of both, so I can’t answer your question.
This is what Willis affectionately refers to as my “cockamanie ‘dance of the photons’ theory”. 🙂
kadaka (KD Knoebel) says:
5. Add what’s in your wallet and your bank assets together, you have $200.
===============
You are forgetting that in addition to assests, you have liabilities, namely a $100 loan that will need to be repaid. You must subtract your liabilities from your assets to come to your actual amount you have, in otherwords $200assets – $100liabilities = $100
tallbloke says:
What we do know is that however the photons dance, it can’t be driven by more than about a quarter of the DLR. That leaves the rest to warm the top mm of the ocean …
Regarding your “dance of the photons”, you’ve never explained exactly what it is. Somehow your theory (IIRC) had to do with the photons never actually hitting the water surface.
But we know that at least three quarters of the DLR is radiated as thermal radiation. And to be emitted as thermal radiation, first the DLR it has to be converted to thermal energy … making the surface warmer than it would be if there were no DLR>
I don’t see any dancing photons in that. To be radiated as thermal radiation at a different frequency than the incoming energy, the DLR must be first converted to thermal energy, which warms/slows the cooling of the surface.
They can’t just dance above the surface.
w.
PS – Truly, I don’t care what you call it. If a room is cold because a door is open to the frozen outdoors, when you close the door the room gets warmer. You can say closing the door warms the room, you can say closing the door slows the cooling of the room, but my point is simple — both the underlying phenomena and the outcome are the same no matter what we name it, the room ends up warmer.
in the same way we can say that DLR warms the ocean or that it slows the cooling of the ocean, it doesn’t matter. The point is that the ocean is warmer with the DLR than without, much warmer. Our name for it doesn’t change the fact that it’s warmer with DLR.
Nor is it meaningful to say that DLR striking the surface can’t warm the bulk of the ocean. As Tim the Toolman agrees (I think), the bulk of the ocean ends up warmer with DLR than without … so what does “DLR can’t warm the bulk” mean when not just the surface but the bulk ocean undoubtedly ends up warmer with DLR than without DLR?
@Willis Eschenbach says: August 16, 2011 at 9:10 pm
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Willis,
The problem with the point you make is that the very very top surface skin layer is cooler than the layer below and hence conduction is not carrying heat from the very very top surface layer to the ocean below. The significance of this is since DWLWIR can only penetrate only a matter of a few microns into the very very top layer of the ocean, ALL the DWLWIR which found its way into the oceans cannot on your account of the process find its way into deeper parts of the ocean.
The principle that you describe is important as regards the Solar energy since this heats the top 10 meteres of the ocean and heat from this part of the ocean (metre by metre) can be conducted downwards in accordance with the principle you describe.
In my earlier comment, may be I did not sufficiently explain the Peter and Paul point. As I see matters, one should visualise the ocean as follows:
1. There is a very very top layer comprising a matter of microns (lets be generous and say up to 10 microns although we are really interested in the top 3 to 5 microns). You may like to think of it as akin to a blanket that protects the deep ocean from losing too much heat.
2. The very very top layer can absorb to limited extent such DWLWIR that reaches the ocean surface.
3. The ocean below the very very top layer cannot be penetrated by DWLWIR due to the wavelength of the DWLWIR. It is effectively opaque to it. However this part of the ocean can be penetrated by the vast majoirty of the incoming solar radiation and this incoming solar radiation (some ~170 w per sqm) heats the top 10 metres of this part of the ocean. Due to over turning and/or the conduction process that you describe in your comment, heat in this part of the ocean gradually makes its way downwards and goes to heat the deep bulk ocean.
4. So what is going on in the very very top layer of the ocean. There are three steps/actions to consider.
5. First, water molecules are constantly being evaporated from the very very top layer with the result that the very very top layer is slightly cooler than the layers of ocean immediately below it. This means that there is no ‘additional’ heat to be overturned nor conducted downwards by the process you describe in your post.
6. Second, we come to the Peter Paul point. Some ~320 to 330 w per sq m of DWLWIR make their way to the first few microns of the very very top layer of the ocean. As they do so, in almost instantaneous manner some ~320 to 330 w per sq m are radiated upwards and away from the very very very top layer of the ocean. Accordingly, the DWLWIR is in effect instaneously extinguished and plays no effective role in the heating/warmth of the ocean. If you like this part of the exchange can be seen to cancel itself out so that there is no need to take it into account when considering what keeps the oceans warm
7. Third, the deep ocean gives up some of its heat. It supplies the very very top layer of the ocean (which is very slightly cooler than the ocean immediately below it) with ~170 w per sqm.
8. Fourth, the very very top layer of the ocean having been supplied with this ~ 170 w per sqm of energy from the ocean below now convects ~20 to 30 w per sq m, evaporates some ~70 to 80 w per sqm and radiates some ~70 w per sqm. Lets say 170w per sqm = 20w + 80w + 70w per sqm.
9. The deep ocean is able to provide the very very top layer with some ~170 w per sqm because it receives some ~170 w per sqm of solar energy in accordance with numbered paragraph 3 above.
10. You will note that when looked upon in this manner, the ocean/atmoshere is at equlibrium. In effect, the ocean (ie., that immediately below the top few microns) is receiving some~170 w per sqm of solar energy and is giving up precisely this quantity of energy. Even without the Peter Paul moment, the ocean would never have given up more. We are concerned with net flux, nothing more.
The behaviour of the oceans is extremely complex. We know that the computer models do not model well either the behaviour of oceans nor clouds. I am sure that we do not properly understand what is going on, and I am therefore quite surprised by the bullishness of your stance.
One of the main problems with climate science (and I have commented upon this problem many times and I note some other people have commented on this in the responses to this article) is that there is no attempt to carry out any physical experimental tests employing empirical observation. Many of the fundamental issues could be clarified and arguments laid to rest, if only some proper physical experimentation was conducted.
RE: Myrrh: (August 17, 2011 at 1:09 am)
“Do you know how much thermal infrared is given off by the incandescent bulb it gives as an example in the second parargraph? 95%
“The 5% visible light doesn’t heat anything.”
I have seen statements like this before and let them go by without thinking. This *can’t* be true. The tungsten filament must be heated to a temperature that is close to the same temperature as the surface of the sun. Not quite, I see from one reference that the peak radiation from an incandescent is around one micron so that light does have a deeper tail into the infrared region. Still, I would be curious how the 5% figure came about. I note that the energy-saving fluorescents typically require about 24% of the equivalent incandescent power. Note that a proper energy integral comparison must be done in the frequency or wave number domain. All these spectra seem to have adopted the wavelength domain for their plots, perhaps to avoid squeezing out the detail in the longer wavelength bands.
Again, the higher the frequency or the shorter the wavelength, the more energetic the photon–Planck’s Law.
Willis Eschenbach says:
August 17, 2011 at 10:46 am
What we do know is that however the photons dance, it can’t be driven by more than about a quarter of the DLR. That leaves the rest to warm the top mm of the ocean …
Regarding your “dance of the photons”, you’ve never explained exactly what it is. Somehow your theory (IIRC) had to do with the photons never actually hitting the water surface.
But we know that at least three quarters of the DLR is radiated as thermal radiation. And to be emitted as thermal radiation, first the DLR it has to be converted to thermal energy … making the surface warmer than it would be if there were no DLR
I don’t see any dancing photons in that. To be radiated as thermal radiation at a different frequency than the incoming energy, the DLR must be first converted to thermal energy, which warms/slows the cooling of the surface.
They can’t just dance above the surface.
Hi Willis. I can see a few different possibilies which would account for your observations here. Evaporated molecules form an invisible mist above the ocean surface. The sub visible droplets formed have a much bigger collective surface area than the ocean surface. These droplets become more highly thermalised by radiation emitted both from above and below. The hotter they get, the more buoyant they become, until they rise high enough to cool enough to condense and become visible clouds or a sea fret just above the water. How cloud condensation nuclei grow is a big unknown that Jasper Kirkby and his colleagues are currently working on. We know the gases emitted by decaying plankton which form sulphuric compounds play a part as well as GCR’s.
PS – Truly, I don’t care what you call it. If a room is cold because a door is open to the frozen outdoors, when you close the door the room gets warmer. You can say closing the door warms the room, you can say closing the door slows the cooling of the room, but my point is simple — both the underlying phenomena and the outcome are the same no matter what we name it, the room ends up warmer.
in the same way we can say that DLR warms the ocean or that it slows the cooling of the ocean, it doesn’t matter. The point is that the ocean is warmer with the DLR than without, much warmer. Our name for it doesn’t change the fact that it’s warmer with DLR.
As I said before, when common parlance conflates distinct processes, only confusion can result. Better to be more careful with words IMO.
Nor is it meaningful to say that DLR striking the surface can’t warm the bulk of the ocean. As Tim the Toolman agrees (I think), the bulk of the ocean ends up warmer with DLR than without … so what does “DLR can’t warm the bulk” mean when not just the surface but the bulk ocean undoubtedly ends up warmer with DLR than without DLR?
Tim agrees, as do I, that the presence of the radiative flux slows down the cooling of the ocean. Stephen Wylde thinks the magnitude of this effect is fixed by surface pressure – I think. I’ll be discussing that in person with him in a few weeks time.
I don’t think DLR striking the surface can heat the bulk of the ocean, and neither does Tim, for all the reasons we’ve rehearsed several times on this thread and many others.
LW Radiation only penetrates a few nm.
Conduction can’t go downwards because the surface is cooler than the subsurface.
Turbulent convection isn’t significant because if the eddies aren’t strong enough to pull down fingernail sized pieces of saturated toilet paper then they’re not strong enough to survive destructive interference beyond a few inches. Take it from someone who has designed centrifugal pumps.
Cheers
TB.
Willis, you say:
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?”
===============================================
I may be in the wrong discussion here, but I thought the point of it all was that CO2 is adding heat to the system.
My argument against that is – that if, – at the top of the atmosphere, energy in equals’ energy out, – then the “Green House Gas circuit” that has been depicted by our friends Kiehl & Trenberth since the year of our Lord 1997 is totally irrelevant. – It only shows “conservation of energy within the Earth system” – In no way does it show that GHGs add energy to the “Earth System” (ES). – And, -as far as I know, – addition of energy (or use of stored energy) is necessary – anywhere – and on all occasions – if a rise of “Work Done” (WD) which leads to the product we call temperature is to be accomplished.
Therefore, all that DLR can achieve is to help preserve the equilibrium. –And the sucker punch is that the amount of CO2 or any other GHG does not matter. – Only the amount that leaves (the surface) can be returned (to the surface) by GHGs! – According to MIT & the people who measure “Energy in/out” it has not changed (apart from natural fluctuations) since measurements began.
From John Endicott on August 17, 2011 at 10:27 am:
I forgot nothing, and you obviously will never qualify for a career in government finance. The 2010 Social Security Trustees Report shows a “perpetual projected unfunded liability” of $16.1 trillion. The 2009 Social Security and Medicare Trustees Reports give a combined figure for unfunded liabilities of nearly $107 trillion. Yet as seen in the recent budget battle, right about now the US federal debt is only about $15 trillion.
You have $200, maybe $300.
Richard Verney.
An excellent post.You said this;
“PS. I do not understand why so much time in climate science is wasted on peripheral issues. Approximately 70% of the planet is covered by ocean. The ocean contains 99% of the heat capacity of the planet, and it is the oceans that drive the weather.”
i agree. Climate science insists on using models that incorporate not only our extremely flawed ‘global’ land temperature calculations
http://wattsupwiththat.com/2011/05/23/little-ice-age-thermometers-%e2%80%93-history-and-reliability-2/
But insist in believing that the temperatures representing 70% of the planet-SST’s- are anything other than nonsensical as any sort of scientific measure prior to 1960;
http://judithcurry.com/2011/06/27/unknown-and-uncertain-sea-surface-temperatures/
(iI am currently in contact with the Met office concerning SST’s)
Modellers seem to believe that the data they use- in all good faith- is of impeccable quality-it isn’t.
tonyb
@ur momisugly Willis
“The point is that the ocean is warmer with the DLR than without, much warmer.”
not correct. DLW at best very slightly slows the cooling rate
If there are 100 balls in a box. 20 are thrown upwards and 2 return. It does not increase the number of balls in the box to 102
So DLW does not warmer the surface. It might very slightly slow the cooling rate. As the 2 balls return and increase the number to 82 before they almost immediately leave again
l
As an interested layperson following this discussion I must say I am not sure what is being argued. Everyone seems to have a different take. But the thing that stands out to me is this averaging business. Willis uses an average of 170w/m^2 for solar insolation at the earth’s surface and then explains why that is insufficient on its own to keep the oceans liquid. Say again?
The average figure may be useful for ‘energy budgets’ but why would you consider it a physical reality in this discussion? There would be few places indeed on the earth receiving only that amount every moment of every day.
At the equator, it would vary between zero and 1368 w/m^2 depending on atmospheric conditions, while at the pole far less (I don’t know the figures). Thus the tropical surface cops a lot of heat most of the daylight hours, while the poles get very little most of the time.
Observation shows us that the oceans are liquid in the tropics and ice at the poles. What we see is exactly what Willis asks. If there’s loads of energy at the tropics (probably more than 390 w/m^2 much of the time) the water is liquid, at the poles with little energy it’s frozen. Ocean currents and atmospheric circulation transport heat around which does mean additional heating at the poles and so we see less ice than might otherwise be the case. For example what would happen if there were no currents. More or less sea ice?
So my confusion is, why argue about the effects of a figure that is not real?
@Willis,
“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.”
The air had enough thermal mass to emit it. Were we on fire? The downwelling radiation would be coming from a much greater optical depth of air than the few microns of ocean it is directly interacting with. Since a similar flux of radiation is going upwards and the atmosphere is opague over parts of the infrared range, it can be heating the air. Which is heating which, depends on which was warmer.
Water has about 3 times the heat capacity of a similar mass of air. The mass of air over a meter^2 at sea level is over 10,000 kg. A few hundred watts is going to take awhile to reach high temperatures.
Graeme M
Significant warming due to DLW is based on flawed science that is so obviously flawed that no one should still believe this nonsense
But many do and not just alarmists
@Willis Eschenbach says: August 17, 2011 at 10:25 am
richard verney says:
August 17, 2011 at 8:04 am
Willis,
I note that once again you do not answer my question as to why the ocean at the same latitude (62 degN) freezes at one longitude (19 deg 04 42 E) but not at another longitude (8 deg 45E). Is this because you do not have an answer that fits in with a simple radiative budget?
No, I hadn’t seen the question. Nor do I understand it. Is it your clam that the ocean should freeze everywhere at the same latitude? Why would that be?
And having not understood it, I understand even less what it has to do with the subject under discussion. I don’t have time to read every unrelated post. If you start out with something that doesn’t make sense, or that doesn’t catch my interest, I’m likely to just go “Next!” and move on. I said before, to get someone’s attention you need to be brief, clear, and interesting. I have nowhere near enough time to answer every random communication.
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Willis
I have posted many comments in response to your article; only some of which were addressed to you personally. The comments addressed to you personally are: my comments of 15th August at 6:18 pm & 7:12 pm, of 16th August at 4:36am & 5:58 am & 9:37am, and of 17th August at 8:04 am & 10:56 am.
You have responded to some of these comments. See your comments of 15th August at 10:27pm & 10:33pm, 16th August at 2:25pm, and 17th August at 10:25 am. Generally, I have found your response(s) to be inadequate since these simply side step the issues raised.
We have had this debate before, some months ago, when we exchanged with one another a series of comments, and I seem to recall on one other much earlier occasion (but it may be that that exchange was with someone else), I have now asked you several times to explain why using just your Solar/DLR energy budget the ocean freezes at some areas but not at others. You have persistently refused to detail your explanation. I have repeatedly advised you that the answer to your question as to why the oceans do not freeze lies in the tropics. Notwithstanding the fact that I have told you the answer on sever occasions, you keep on saying/suggesting that I have not answered your question as to why the ocean does not freeze.
You now allege: “NO, I HADN’T SEEN THE QUESTION. Nor do I understand it. Is it your clam that the ocean should freeze everywhere at the same latitude? Why would that be?…” (my emphasis).
Obviously, I cannot comment upon the extent of your understanding, but I can state that your excuse that you had not seen the question is plain wrong! You had seen the question which was set out in my comment of 16th August at 9:37 am since you responded to that comment on 16th August at 2:25 pm. See: “Willis Eschenbach says: August 16, 2011 at 2:25 pm richard verney says: August 16, 2011 at 9:37 am…”
I do not wish to wast time in repeating what I have already said. Please re-read my various comments as detailed above and then please revert with your detailed Solar/DLR budget calculation for:
1. The ocean at 62N 8 deg 45E (say for January)
2 The ocean at 62n 19 deg 04 42E (say for January)
3. The ocean at the tropics (say for January).
Alternatively, if you do not wish to produce actual calculations you can advise whether you:
4. Challenge my assertion that the Solar/DLR energy budget would be broadly similar for the ocean at the same latitude. In this regard, I consider that the incident angle of sunlight, the length of the solar day and the albedo of the ocean will be similar when considering the ocean at the same latitude. Likewise there is no reason why the DLR would be markedly different when considering the ocean at the same latitude given the well mixed nature of greenhouse gases (I am not saying that they would be precisely the same because I can see some reasons which may make modest differences). If the Solar/DLR energy budget is broadly similar at two locations one would IF ONLY considering Solar/DLR energy, expect the oceam to behave in similar manner.
5. Challenge my assertion that the reason why the ocean does not freeze at 62degN 8deg 45 E but does freeze at 62N 19 deg 04 42E is due to the former receiving more of the warm ocean currents eminating from the tropics.
6. Dispute my assertion that the tropical ocean does not freeze when you carry out an energy calculation on a net flux basis.
7. Dispute my assertion that the tropical ocean does not freeze even when you carry out a Solar/DLR energy calculation using your gross figures but ignoring the effects of DWR. If you dispute this, please provide the gross figures which you assert are involved for the tropical ocean (ie., not the average figures that you have used in your article) and the figuiures that you are using for latent heat calculations.
Willis, lets make some progress and actually get to the bottom of this. Please respond constructively and not by side stepping the issues raised.
Ps. I will put my hand up and say that some of my terminology has not been as accurate nor as succint as I would have wished but many of my comments were written at circa 3am to 4am in the early hours of the morning when one is not thinking at their best.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat.
The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
To any here that think thermal infrared does not reach the surface of the Earth, and so does not heat the land and oceans: ALL the heat we feel from the Sun is thermal infrared.
This is what heats the land and oceans to raise the temperature.
I repeat:
Visible light and and the two shortwave energies either side of UV and Nr Infrared are not thermal energies,
That means they do not heat things.
That means they can’t heat land and oceans.
That means the ‘energy’ imput from the Sun from these wavelengths has nothing to do with heating the Earth.
That means they have to be taken out of your calculations.
You are confused between Light and Heat.
Highly energetic does not mean powerful.
Light energies are not powerful enough to create the heat claimed in Your Science Fiction arguments.
It is thermal infrared direct from the Sun which heats the water of the oceans and the land.
Water is a strong absorber of thermal infrared.
In real life applications this is well known.
“The lights use standard HPS and Metal Halide lamps, which are fitted into a water jacket that removes the heat while still allowing all the visible light to reach the plants.”
Read that again.
This is the real world.
You are not in it.
“The project purpose was to develop a mini-farm sufficiently productive to supply food for a manned mission to Mars. It had to be energy efficient, small, light weight, and most importantly, highly productive – all desirable attributes for the home grower as well. The research was done using a high productivity growth chamber (CAAP chamber) fitted with lighting which featured ” ….a recirculating water jacket that absorbs and removes non-photosynthetic energy”. In plain English, they used water cooled lights – the same type of water cooled lights that have been available to home hydroponic growers for several years already.”
http://www.hydroponicsbc.com/watercooledlighting.html
Prove that visible light from the Sun heats water or take it out of your calculations.
You are not scientists if you ignore this challenge.
“”””” Answer to George E Smith. local thermodynamic equilibrium exists everywhere and accepting the exponential transients, most of the time emissivity is approximately equal to absorptivity. “””””
Well Alexander (Duranko), assuming that what you say (or at least said here) is true; and I don’t have any idea what it means; or why I should accept exponential transients (of what ?), it is somewhat irrelevent, in that it isn’t Kirchoff’s Law.
Kirchoff’s Law requires exact equality; and moreover at every possible wavelength (or wave number if you prefer).
People should stop citing Kirchoff’s Law in support of their otherwise unsupportable claims; in all cases where the conditions are NOT those set forth as necessary under Kirchoff’s Law; and that is spectral point by point exact equality for a medium in thermal equilibrium with a radiation field.
Otherwise such citation is the equivalent of asserting that Oliver Cromwell’s axe, in the British Museum, has only had two new heads, and five new handles since Oliver last used it.
Graeme M says:
August 17, 2011 at 1:46 pm
Sensible assessment Graeme. One minor nitpick. The oceanic transport of heat towards the poles is a lot less significant than atmospheric transport north of 65N (top end of Gulf Stream transport)
Another point which is worth mentioning is that the insolation figure doesn’t just average the solar input over latitude, but also between night and day sides of the planet. This is a serious shortcoming of the models because of the non-linearity of the response of water to thermal input in terms of the microclimate it creates above itself.
Yes Tallbloke, that’s another aspect I don’t follow. The energy budgets all show 340 off w/m^2 coming in at TOA and I understand how that is calculated. But the actual physical environment receives something rather different than an average. At the equator at midday the TOA gets the full benefit of 1368 w/m^2, not 342. There has to be a real measurable difference for that extra 1000 or so w/m^2. The climate is not an average. Nor are the effects of the sun on the earth.
This gives the TOA daily insolation figures throughout the year at all latitudes. Some surprises for those who haven’t thought this through carefully
http://curryja.files.wordpress.com/2011/08/thermo-txt1.pdf
“”””” Sorry Myrrh, but the term “thermal radiation” is sometimes confused and misused. It is really a misnomer for EMR, (electromagnetic radiation) which includes visible light which does indeed heat matter when its photons are absorbed. Try here at Wikipedia: “””””
Well “Thermal Radiation” has a much more restrictive meaning than is inferred from what is said here (By Myrrh and later commenter (sorry lost the name)).
Thermal Radiation is a continuum Electromagnetic Radiation, of unrestricted wavelength or wave number range, that arises SOLELY because of the Temperature of an assemblage of molecules of any material that has a Temperature higher than zero Kelvins. The spectrum of such radiation is bounded as a limit case, , and at any wavelength or wave number, by the spectrum of a Black Body Radiator, at the same Temperature. Many dense solid bodies can approach quite closely to the thermal emission of the limit black body; but all materials, whether solid, liquid or gas, at temperatures higher than zero Kelvins can and do emit thermal Radiation. Even a single atom or molecule, as a member of an assemblage of molecules, emits Thermal Radiation.
The mechanism of emission is spelled out in Maxwell’s equations, where any varying electric current in any path of greater than zero length (antenna) must radiate EM waves. A varying Electric Current of course is the same thing as an accelerated Electric Charge (Q); and every particle Physicist knows that accelerated Electric Charges MUST Radiate EM waves. There’s a monument to that phenomenon in Silicon Valley, California. It’s known as the Stanford Linear Accelerator Center; a two mile long pin that holds the San Andreas Fault together. It’s purpose is to accelerate electrons (electric charges) to high kinetic energies without suffering the radiation losses that would occur, if they sent the electrons around a race track. But I digress.
Of course, any isolated (neutral) atom or molecule can be in ballistic flight, subject only to the force of gravity. Gravity is so weak, that it for nearly all practical purposes, has a constant velocity. As such, it of course has no Temperature, which is a consequence of collisions occurring within an assemblage of molecules. In free flight, the molecule has neither acceleration, nor in many cases any electric dipole moment, so it does not radiate EM waves.
However, during the time that a molecule is in collision with another, undergoing an elastic scattering process, due to the electrostatic forces between the two molecules, the molecules or atoms are distorted, because the particle momentum is almost entirely contained in the atomic nuclei. The Proton to electron rest mass ratio, is 1837:1, and then the Neutron has a similar mass, so the nuclear momentum is typically about 3675 times that of the electrons; but the electrostatic forces are of the same order. The result is that one charge species has almost 4,000 times the acceleration of the other, so charge neutrality is moot, since only one sign of charge is undergoing significant acceleration, so the other sign doesn’t matter much.
This results in a distortion of the charge distribution, creating a net electric dipole moment, so during the collision interaction time, the atoms/molecules can and do send and receive EM waves. The wavelength range is unbounded; but it is still Thermal Radiation since it arises as a consequence of the Temperature of the assemblage of atoms/molecules.
Whether EM radiation is capable of warming anything, is irrelevant as to whether it is “Thermal Radiation”.