# Precipitable Water Redux

Guest Post by Willis Eschenbach

In my last post I investigated the mathematical relationship between the amount of total precipitable water vapor (TPW) in the atmosphere, and the clear-sky greenhouse effect. Here is the main figure from that post showing the relationship:

Figure 1. Scatterplot, TPW (horizontal scale) versus Atmospheric Absorption (vertical scale). Dashed yellow line shows theoretical value based on TPW. Dashed vertical line shows area-weighted global average value. Dotted vertical lines show the range of the global average value over the period. The slope of the curve at any point is 62.8/TPW (W/m2 per degree)

In this post I’m looking at the other half of the relationship. The other half is the relationship between the ocean surface temperature and the total precipitable water. The good news is that unlike the CERES data which is only about 15 years, we have TPW records since 1988 and sea surface temperature for the period as well. Figure 2 shows the relationship between the two:

Figure 2. Scatterplot, RSS total precipitable water (TPW) versus the ReynoldsOI surface temperature data. See end notes for data sources

As you can see, the relationship is regular but not simple. I first thought that the relationship was logarithmic, but it turns out not to be so. It is also very poorly represented by a power function. After unsuccessfully investigating a variety of curves, I found it could be approximated by an inverse sigmoid function (shown in yellow above). Now, given the number of very smart folks here, I suspect someone will be able to give a physical reason complete with the right equation, but this one suffices for my purposes.

Now, the relationship between water vapor and atmospheric absorption is clearly logarithmic, as is predicted by theory. On the other hand, I don’t know of any simple theory relating SST to total precipitable water. For example, the curve doesn’t match the Clausius-Clapeyron increase in water vapor. And clearly, my method is purely heuristic and brute-force … but that’s OK because I’m not claiming that it is explanatory. My purpose in doing it is quite different—I want to figure out how much change there is in the precipitable water per degree of change in the sea surface temperature (SST). And for that, the main quality is that the function needs to be differentiable.

So let me recap where we stand. In the last post I derived a mathematical relationship between the two variables shown in Figure 1. Those are clear-sky atmospheric absorption of upwelling longwave radiation from the surface, and the total precipitable water content (TPW) of the atmosphere.

And above in this post, I’ve derived a mathematical relationship between the two variables shown in Figure 2. Those are the total precipitable water content (TPW) of the atmosphere, and the sea surface temperature (SST).

That means that by substituting the latter into the former, I can derive a mathematical relationship between the SST and the atmospheric absorption.

Of course I wanted to ground-test my formula that converts from sea surface temperature to atmospheric absorption. I only have the CERES data for the absorption, so this covers a shorter period than that shown in Figure 2. Since the overall relationship was established using the Reynolds sea surface temperature data, I used that for the comparison.

Figure 3. Atmospheric absorption of upwelling longwave radiation versus sea-surface temperature. See end notes for mathematical derivation.

Dang, I’m pretty satisfied with that as a comparison of theoretical and observed atmospheric absorption. A few comments. First, the difference below 0°C is because CERES and Reynolds are measuring slightly different things below freezing, when there is ice in the picture. CERES is measuring the average temperature of the ice and the water, and Reynolds is measuring water temperature alone.

Next, the slight bend in the black line from 0°C to 25°C is not completely captured by the red line. This is because I’ve included the data below freezing, which has slightly distorted the results. Probably should have left it out, but I figured for completeness …

Next, the slight bend in the black line from 0°C to 25°C is due to the fact that surface radiation is proportional to the fourth power of the temperature. If absorption were calculated against surface upwelling radiation rather than temperature, it would plot as a straight line … go figure. I’ve done it this way because there is much discussion about the value of the “water vapor radiative feedback” which is measured per degree C. I could get a slightly closer fit by including the T^4 relationship, but my conclusion was that the gain wasn’t worth the pain … if I need greater accuracy I can redo the figure, but it is more than adequate for the present purposes.

The amount of the feedback is calculated as the slope of the red line in Figure 3. The slope is the change in the absorption for a 1°C change in the sea surface temperature. Figure 4 shows the amplitude of the water vapor radiative feedback across the range of ocean temperatures:

Figure 4. Water vapor radiative feedback, calculated as the change in atmospheric absorption of upwelling longwave radiation per 1°C change in surface temperature.

That is a very interesting shape. Now,  given the general shapes of Figure 1 and Figure 2, I might have expected the shape … but it came as a surprise anyhow. Over much of the world, the two tendencies cancel each other out and the clear-sky water vapor radiative feedback is about 3-4 W/m2 per degree C. But in the tropics, where the water is warm, the water vapor feedback goes through the roof.

DISCUSSION

So … with such a large radiative feedback from water vapor, three to four watts per square metre per degree and much higher in the tropics, why is there not runaway feedback? I mean, the so-called “climate sensitivity” claimed by the IPCC says that 2-3 W/m2 of additional radiation will cause one degree of warming. And according to observations above, when it warms one degree, we get additional downwelling radiation from water vapor of 3-4 W/m2. And that amount is claimed to be sufficient to warm it more than one additional degree … a recipe for runaway positive feedback if I ever saw one. So … with that large a radiative feedback, why isn’t there runaway feedback?

Well, you might start by perusing Dr. Roy Spencer’s discussion of the subject, yclept Five Reasons Why Water Vapor Feedback Might Not Be Positive. The TL;DR version is that as the amount of water vapor in the air increases, downwelling radiation does indeed increase … but there are plenty of other things that change as well.

To expand a bit on one of the things Dr. Roy mentioned, in his discussion of evaporation versus precipitation he said:

While we know that evaporation increases with temperature, we don’t know very much about how the efficiency of precipitation systems changes with temperature.

The latter process is much more complex than surface evaporation (see Renno et al., 1994), and it is not at all clear that climate models behave realistically in this regard.

Let me add a bit to that. Rainfall goes up with increasing atmospheric water as shown in Figure 5:

Figure 5. Scatterplot, rainfall evaporative cooling versus total precipitable water. TRMM data only covers latitudes 40°N to 40°S.

Note the size of the cooling involved … not watts per square metre, but hundreds of watts per square metre. As precipitable water goes from about forty to fifty-five kg per square metre, evaporative cooling goes from fifty to two hundred fifty watts per square metre or more … that’s a serious amount of cooling, about ten watts of additional cooling per additional kg of precipitable water.

We can compare that to the slope of increasing water vapor radiative feedback in Figure 1. The slope in Figure 1 is 62.8 W/m2 divided by TPW, so at a TPW of 50 kg/m2 that would be about 1.2 W/m2 of additional radiative warming per additional kg/m2 of water … versus 10 W/m2 of rainfall evaporative cooling per additional kg/m2 of water.

But wait … there’s more. Figure 6 shows the rainfall evaporative cooling versus sea surface temperature (SST). Since SST and precipitable water are closely related, Figure 6 is quite similar to Figure 5.

Figure 6. Scatterplot, rainfall evaporative cooling versus Reynolds sea surface temperature.

As in Figure 5, at the hot (right hand) end of the scale, the rainfall evaporative cooling goes from about 50 to about 200 W/m2 very quickly. However, in this case it makes that change as the SST goes from about 27° to 30°. And that gives us a net cooling of about 50 W/m2 per degree … kinda dwarfs the 3-4 W/m2 per degree of water vapor based warming …

There is another interesting aspect of Figure 6 … the empty area at the lower right. I have long stated that the thermoregulatory phenomena like thunderstorms are based on temperature thresholds. The blank area in the lower right corner of Figure 6 shows that above a certain sea surface temperature … it’s gonna rain and cool it down. And not only will it rain, but the hotter it gets, the greater the rainfall evaporative cooling overall, and the greater the minimum evaporative cooling as well.

Nor do the cooling effects of water vapor end there. Increasing water vapor also increases the amount of solar energy absorbed as it comes through the atmosphere. As with the absorption of the upwelling longwave, the relationship is logarithmic. Figure 7 shows that relationship.

Figure 7. Scatterplot, atmospheric absorption of downwelling solar radiation (vertical axis) versus total precipitable water (horizontal axis)

Logarithmic relationships of the form “m log(x) + b” have a simple slope, which is m / x. The slope of the equation shown in Figure 7 is 31.6/TPW (W/m2 per degree). Now, earlier we saw that the slope of the warming from increasing water was 62.8/TPW (W/m2 per degree). This means that at any point, half of the warming due to water vapor radiative feed back is cancelled out by the loss in downwelling sunlight due to increased water vapor.

Nor is this the end of the related phenomena … Figure 8 shows the correlation between total precipitable water and cloud albedo:

Figure 8. Correlation of total precipitable water (TPW) and cloud albedo.

As you can see, over much of the tropics, as precipitable water increases so does the cloud albedo (red-yellow). Makes sense, more water in the air means more clouds. Again, this has a cooling effect.

Nor is this an exhaustive list, I haven’t discussed changes in downwelling longwave radiation due to clouds … the relationships go on.

FINAL THOUGHTS

The center of climate action is the tropics. Half of the available sunlight strikes the earth between 23° north and south. The main phenomena regulating the amount of incoming solar energy occur in the tropics. And as the graphs above show, the amount of water in the atmosphere is at the heart of those phenomena.

So … is the feedback of water vapor positive or negative? Overall, I’d have to say it is well negative, for two reasons. The first is the long-term stability of the global climate system (e.g. global surface temperature only changed ± 0.3° over the entire 20th century). This implies negative rather than positive feedback.

The second reason I’d say it’s negative is the relative sizes of the various feedbacks above. These are dominated by the evaporative cooling due to rainfall and by the changes in reflected sunlight due to albedo, both of which are much larger than the 3-4 W/m2 in increased water vapor radiative warming.

However, there is a very large difficulty in isolating the so-called “water vapor feedback” from the myriad of other phenomena. This difficulty is embodied in what I refer to as my “First Rule Of Climate”, which states:

In the climate system, everything is connected to everything else … which is turn is connected to everything else … except when it isn’t.

For example, the temperature affects the water vapor – when the temperature goes up, the water vapor goes up. When the water vapor goes up, clouds and rain go up. When clouds and rain go up, temperatures go down. When temperatures go down, water vapor goes down … you can see the problem. Rather than having things which are clearly cause and clearly effect, the whole system is what I describe as a “circular chain of effects”, wherein there is no clear cause and no clear boundaries.

Anyhow, those are the insights that I got from examining the total precipitable water dataset … like I said, no telling where a new dataset will take me.

And speaking of precipitable water, it is sunset here on our hillside. As I look out the kitchen window towards the ocean I see the fog washing in from the Pacific. It is pouring in waves over the far hills, swallowing redwood trees as it rolls on toward our house … it came and visited last night as well.

I love that sea fog. It reeks of my beloved ocean, with the smell of fishing boats and slumbering clams, of hidden coves and youthful dreams. And when the fog comes in, it brings with it the sound of the foghorn at the mouth of Bodega Bay. It’s about seven miles (ten kilometres) from my house to the bay, but the sound seems to get trapped in the fog layer, and when the fog comes I hear that foghorn calling to me in the far distance, a mournful midnight wail. I took frequent breaks from my scientific research and writing last night to sit outside on a bench, where I let the fog wreathe around my head and bear me away. I breathe in the precipitated water, and I emerged refreshed …

My best to everyone, and for each of you, I wish for whatever fog it is that carries you away in reverie and washes off the mask of socialization …

w.

REQUESTS

My Usual Request: Misunderstandings suck, but we can avoid them by being specific about our disagreements. If you disagree with me or anyone, please quote the exact words you disagree with, so we can all understand the exact nature of your objections. I can defend my own words. I cannot defend someone else’s interpretation of some unidentified words of mine.

My Other Request: If you believe that e.g. I’m using the wrong method or the wrong dataset, please educate me and others by demonstrating the proper use of the right method or identifying the right dataset. Simply claiming I’m wrong about methods or data doesn’t advance the discussion unless you can point us to the right way to do it.

NOTES

The math … I start with the equation for relationship between absorption (A) and total precipitable water (TPW) shown in Figure 1:

A = 62.8 Log(TPW) – 60

To this I add the inverse sinusoidal relationship between TPW and sea surface temperature, as shown in Figure 2:

TPW =  – 13.5 Log[-1 + 1/(0.00368 SST + .887)] -19.1

Combining the two gives us:

A = 62.8  Log[-19.1 – 13.5  Log[-1 + 1/(0.887 + 0.00368 SST)]] – 60

Differentiating with respect to sea surface temperature gives the result as shown in Figure 3:

dA/dT = 3.13/((-1 + 1/(0.887 + 0.00368 SST)) (0.887 + 0.00368 SST)^2 (-19.1 – 13.5 Log[-1 + 1/(0.887 + 0.00368 SST)]))

Further Reading: NASA says water vapor feedback is only 1.1 W/m2 per degree C …

DATA

Reynolds SST data, NetCDF file at the bottom of the page

CERES data

TRMM data, NetCDF file at the bottom of the page

## 258 thoughts on “Precipitable Water Redux”

1. D.J. Hawkins says:

Willis;

Just below Fig 7, it appears a portion of your post is cut off at the following sentence:

“This means that at any point…”

• Willis Eschenbach says:

Thanks, fixed.

w.

• ozspeaksup says:

hey Willis:-)
they said ….the fog wasnt fogging n the redwoods were going to dry n die..
just sayin:-) lol
another educational post thanks;-)

• RWturner says:

Speaking of figure 7, isn’t that relationship also somewhat controlled by the amount of down-welling sunlight in the first place. The tropics happen to have the highest TPW and 50% of the sunlight entering the atmosphere. The circular chain of effects is complicated indeed. Nice post.

2. The wateer vapor feedback looks negative, but if climate is governed by chaotic rules, there is still a possibility it will bite your understanding on its metaphoric ass.

• United Nations got a really fleshy behind – which might become severely bitten by natural variability.

3. Global Cooling says:

Thank you. This looks impressive at the first look. It is a well written story connecting a solid theory of the complex adaptive climate system to empirical data.

The conclusion that the earth has a natural thermostat, a kind of a heat pipe system that we have in our computers, is plausibly explained.

The first question for the peer reviewers here: what is missing? Ice, ocean currents ? Are they important?

• Clyde Spencer says:

GC and Eschenbach,

For a given water or air temperature, evaporation will increase with wind speed. Also, air temperature plays a more important role than SST for evaporation. That is, if hot air is blowing off land onto the ocean, evaporation will be higher than if the air temperature over the water was in equilibrium with the SST because evaporation is a surface phenomenon rather than a bulk process.

4. If you oxidise hydrocarbons, you get at least CO2 and H2O.

If you oxidise a few gigatonnes of hydrocarbons you get a fair amount of water.

I suppose it all eventually precipitates out.

All very tricky.

Cheers.

• Dan Hawkins says:

Mike Flynn — yes indeed. For fun I calculated the amount of water released by fossil fuel burning a couple of years ago. My estimate (amateurish) was 116 cubic miles, or about the volume of Lake Erie. Anybody wanting to try this should start here: http://cdiac.ornl.gov/

Dan

• Dan Hawkins says:

I should clarify, this was over the historical period 1751-2013.

Dan

• Don Andersen says:

So. That’s why the sea level is rising! ;-)

5. Germinio says:

I think there is an error in the discussion after Fig. 7. It states that “This means that at any point, half of the warming due to water vapor radiative feed back is cancelled out by the loss in downwelling sunlight due to increased water vapour.” Surely it is the opposite – the feedback effect of water vapour is increased since (a) it absorbs downwelling visible light and (b) it absorbs upwelling long wavelength light. These effects both increase the amount of energy absorbed by the atmosphere and so add to the heating effect of water vapour.

There is also an error when discussing the effects of evaporative cooling. This should be regarded as a way to move heat and energy around from one point on the globe to another. It does not change the total energy of the atmosphere/ocean system and so won’t actually cause any cooling
on average. There will be a slight effect since the heat is effectively moved to the atmosphere where it can radiate away more efficiently than at the surface but I am guessing that this value is not what is plotted in Fig. 5

• joelobryan says:

G,
you seem to be confusing visible water vapor in the air column (clouds) with invisible (to sw radiation) humid air.

clouds are the bugaboo in the Climate charlatans junk model outputs.

• Germinio says:

Hi Joel,
I don’t think so. Fig 7 is the absorption of water vapour for downwelling visible light. So water vapour firstly absorbs 50 to 100 W/m^2 of visible light from the sun as it enters the atmosphere and before it reaches the surface. Then of the long wavelength light emitted by the surface it absorbs another 3 W/m^2. So the
total additional absorption from water vapour in the atmosphere is the sum of the two not the difference. This has nothing to do with clouds which reflect
visible light and so would not show up in a measure of absorption but which are a negative feedback (and one that is explicitly not discussed in this post since it talks about clear sky effects).

• By visible water, you are referring to droplets (liquid water), not water vapor (gas).

• Craig Loehle says:

I think if solar radiation is absorbed in the atmosphere instead of at the surface it is more easily (more quickly) lost to space since it is radiated equally in all directions.

• David A says:

Craig, I would like to add one more thought for consideration to this. What you say appears logical and certainly atmospheric convection would change, likely accelerating. There is another factor in addition.

With an increase of water vapor, even in clear sky situations, s/w and overall insolation to earth’s longest residence time energy holder decreases. Of course this decrease is very substantial once clouds form, but it is non trivial even in clear sky conditions. Earth’s longest holder of energy is neither the atmosphere or the land, but the S/w selective oceans. Because this is the longest residence time of all energy entering the earth, (up to millennial scale) flux in this input and output can have a far greater affect, over time, of the amount of total energy in the system.

Alas, we are a long ways from seeing this chartered as we do not know the residence time of disparate insolation entering the oceans.

6. An excellent and very interesting piece – thank you so much for having an enquiring mind and sharing it all with us.

Among other things I was stunned by the magnitude of the rainfall evaporative cooling.

I was also stunned by the curve forms. Steep curves tend to create high sensitivities and a correspondingly huge uncertainty in estimation of a dependent variable.

The belief that Climate models are capable to get the net global energy balance right within 1 W/m^2, or something like that, is far beyond my belief. (the Earth´s energy imbalance is in the order of magnitude 0,6 W/m^2).

7. joelobryan says:

“circular chain of effects”
I would say it as, “A tightly-coupled, complex system with emergent properties.”

Willis, your plots showing asymptotic behavior as SST approaches 30degC is why James Hansen was smoking good dope and selling lies when he opined the seas would boil with enough CO2.

“In his book Storms of my Grandchildren, noted climate scientist James Hansen issued the following warning: “[I]f we burn all reserves of oil, gas, and coal, there is a substantial chance we will initiate the runaway greenhouse. If we also burn the tar sands and tar shale, I believe the Venus syndrome is a dead certainty.””
Will Earth’s Ocean Boil Away?
http://news.nationalgeographic.com/news/2013/13/130729-runaway-greenhouse-global-warming-venus-ocean-climate-science/

Overall feedback of water in the climate system is strongly negative. Anyone (including the IPCC and GCM climate model charlatans) who says otherwise are selling an agenda, not science based on natural, observed phenomenon.

• Geronimo says:

It is not quite that simple. The asymptotic behaviour depends on the temperature of the atmosphere which determines how much water vapour can be supported by the air column before it saturates and it starts raining. The warmer the atmosphere the more water vapour it can support and so the more the asymptote moves towards higher temperatures. The current graphs are really for relatively constant atmospheric temperatures and varying ocean temperatures. They do not say what will happen if the atmosphere heats up as well.

• Willis Eschenbach says:

Geronimo July 28, 2016 at 11:56 pm

“The current graphs are really for relatively constant atmospheric temperatures and varying ocean temperatures.”

Huh? I don’t understand this. The data points are 1°x1° gridcell averages, and each gridcell most definitely has variations in both atmospheric and ocean temperatures.

Regards,

w.

• Brett Keane says:

Geronimo
July 28, 2016 at 11:56 pm : What you have been doing is deliberate twisting. The figures show what we knew to be the CAGW breakpoint. Where the water takes over and controls actual increased solar input as it occurs. You should be able to see and read here how any feedback, real or imagined, is swamped at an increasing rate by evaporative uplift of energy to space. Maybe you cannot understand the scale of it. but be thankful for what makes your existence possible.

• hot air says:

Except humidifying air cools it. Google psychrometrics…

8. Crispin in Waterloo says:

Willis there is a lovely kernel of information/proof hidden in the middle of your discussion above. You might remember that we discussed why the incoming 341 W of radiation (average) hitting the sea only produced 80 W of evaporation (average). I speculated at the time (logical guess) that it did not reach the water because of the saturation of the air right next to the water surface, meaning that a large % of the incoming radiation was absorbed and tossed back before reaching the surface. Some hits the surface to produce evaporation and some passes into the water providing illumination and warming.

Your evaporative cooling chart provides insight into this simplistic guess of mine. The answer is not that a certain amount of energy reaches the surface but that the answer is a formula, not a number. When the sea surface temperature rises past 25 C the evaporation rate ramps up rapidly. I attribute this to the heat available just above the surface that, when the sea is cool, doesn’t add much to evaporation. But above 25 that heat, trapped by the water vapour close to the surface, starts to reach the skin and drives additional evaporation. Maybe.

In short I wasn’t all that wrong and that the evaporative cooling rising to the hundreds of Watts per Sq m supplies the detail explaining why the average is only 80 W. It is because most of the oceans are far cooler and contribute far less to the total mass evaporated.

Locally evaporated water vapour moves away, messing up the chart a bit, but the principle remains. the % of incoming radiation that reaches the water surface must change rapidly above a surface temperature of 25C. Otherwise it could not drive a cooling rate of a few hundred W when it is only 8 w over a very cold ocean. The mechanism that takes the heat absorbed in the air just above the water might be radiative transfer or convective flow from wind. Whatever it is, it kicks in big time above warm oceans that exceed 25C.

Alternatively, the “average” of 80 W could be the problem – we are asked the wrong question. In the data above one can show that sometimes the incoming radiation is 341 W and the cooling is 261 W the difference goes into warming the water, not evaporating the surface. 261 is not 80 so thinking about the global “average” is misdirection. Your curves show clearly that as the water temperature rises the evaporation mechanism mysteriously becomes much more efficient, whether it is cloudy or not.

Clean water in a very smooth container boils at a much higher temperature than the same water in a rough container. Perhaps there is a water evaporating equivalent of cloud condensation nuclei that kicks in more vigorously above 20-25 degrees, and that the nature of sea water (filled with impurities) permits/assists the evaporative cooling profile.

• Willis Eschenbach says:

Crispin in Waterloo July 29, 2016 at 12:19 am

Willis there is a lovely kernel of information/proof hidden in the middle of your discussion above. You might remember that we discussed why the incoming 341 W of radiation (average) hitting the sea only produced 80 W of evaporation (average). I speculated at the time (logical guess) that it did not reach the water because of the saturation of the air right next to the water surface, meaning that a large % of the incoming radiation was absorbed and tossed back before reaching the surface. Some hits the surface to produce evaporation and some passes into the water providing illumination and warming.

Crispin, this reanimation of your desperately dogged attempt to convince people that downwelling longwave radiation is somehow not actually absorbed by the ocean is as incorrect now as it was when you first proposed it. I’m not going to argue well-established physics with you, this is neither the thread nor the site for that.

And no, there is no way that the downwelling longwave is “absorbed and tossed back” just microns above the surface as you seem to think. Downwelling IR is absorbed by the ocean. Why is that so hard for you to accept? If the downwelling IR were not absorbed by the ocean, the ocean would have frozen eons ago.

Look, I know that you truly believe all those things, but the only folks who agree with you are wearing tinfoil hats. I can’t say this clearly enough. You are putting hand-waving pseudo-explanations out there as if they were valid science, and they’re not.

Now, I know I should sugar-coat all of this to make it more palatable, but there is no easy way to tell a man he’s talking impossibilities.

It puts me in mind of the Corporal who calls a Private in and says to him “Son, I just got the news that your mother died”. The Private breaks down completely because of the sudden shock, weeping and wailing.

The Sergeant says to the Corporal, “You need to learn to do that correctly if you want to be a leader. You should break bad news gently to your men!”. The Corporal asks him how to do that. The Sergeant says “Suppose you hear a man’s favorite cat died, his favorite pet. You might first say ‘I heard that your cat got stuck up a tree.’ Then the next day you might say ‘They still haven’t been able to rescue your cat’. Then on the third day, when he’s prepared, you give him the bad news.”

“I understand, Sergeant, give them time to absorb it”, said the Corporal.

So the next time the mother of one of the recruits died, the Corporal called him in and said to him,

“Private Jones, I have some bad news for you. Your mother got stuck up in a tree …”

Anyhow, Crispin, what I’m trying to say is that your science got stuck up in a tree.

Regards,

w.

PS—I also don’t understand your claim that downwelling longwave radiation causes “illumination and warming” when it is absorbed by the ocean. Warming I get, but illumination?

• Bill Marsh says:

Well, for those of us in the animal kingdom that can see in the infrared ….

• Crispin in Waterloo says:

Willis

You object to my statement that not all the IR makes it to the ocean surface then launch into a humorous lampooning of the author’s tin foil hat mental state. Mine.

So, let’s look at your article. You mention in your piece that water vapour in the column intercepts some of the incoming radiation and then when the water mass total rises to, for example, 50 kg per Sq m, there is a significant rejection, by re-radiation upwards, of some of it. You even explained that as the water vapour mass increases, the effect increases.

This is literally identical to ‘my science’ as you characterise it. Absolutely the same effect.

I am more interested in understanding what is happening than lampooning you in witty reply, despite the many hooks you proffer. Listen to your mother, drop the ‘something extra’.

You obviously accept the mechanism because you describe it well and provide supporting evidence that not all the incoming radiation makes it to the surface. I pointed out that as the distance from the ocean surface becomes very small, the absolute humidity is very high and it’s interception efficiency rises proportionately (at the relevant wavelengths). The intercepted energy, absorbed by the water vapour just above the water surface, heats the vapour, bringing up the temperature at least near ambient. This helps transport it away from the surface and more evaporation takes place. That intercepted energy never makes it to the ocean surface in exactly the same way it doesn’t penetrate the water column, as you clearly explained.

I was reminding you of our earlier conversation, highlighting the mechanism and its increased effect as the humidity rises. It seems you have some around to my view, albeit by another path of investigation.

I also said that directly above the ocean surface there is a layer of air, very thin, that is supersaturated. You doubted that at the time. Well, physical phenomena are what they are. Incoming knowledge can also be absorbed and rejected. The interception of radiation at that point explains in part why air blowing off the ocean is not 1 degree C.

Light:
Part of the radiation that does make it to the surface is visible light. That is why if you go scuba diving the water around you is illuminated. Most of that energy is absorbed by the water, some makes it out again, which is why you can look into the water and see the coral (etc).

• Easy now, “downwelling IR” from water covers a broad area of the spectrum. This is what Modtran “sees” in the water bands from the surface looking up:

Note the extremely large range of blackbody temperatures (altitudes) across the spectrum of deviation from surface temperature (300K).
We are talking about a range from WN 600 to 1300 and the ocean surface can be expected to absorb differently across this range.
Remember also that atmospheric water is a very efficient absorber of incoming solar radiation.

Seemingly a portion of this would be kinetically converted to “downwelling” IR much in the way the surface converts it to “upwelling IR”.

• Willis Eschenbach says:

Crispin in Waterloo July 29, 2016 at 7:24 am

So, let’s look at your article. You mention in your piece that water vapour in the column intercepts some of the incoming radiation and then when the water mass total rises to, for example, 50 kg per Sq m, there is a significant rejection, by re-radiation upwards, of some of it. You even explained that as the water vapour mass increases, the effect increases.

This is literally identical to ‘my science’ as you characterise it. Absolutely the same effect.

Crispin, thanks for your reply. I said that a column of air ~20 km tall can absorb a significant amount of downwelling infrared.

You have claimed some mythical 20 mm or 20 cm or so thick layer of saturated air just above the ocean can do the same thing … no matter how saturated the air is, there simply aren’t enough water molecules in a thin layer to absorb much of anything.

Rather than speculating about the numbers, I turned to MODTRAN calculate the per-metre absorption of longwave. The actual shape of the graph of the per-metre absorption is quite interesting. Absorption peaks at about 1.5 km, at an absorption rate of 0.019 W/m2 per metre of depth. So let’s suppose for the sake of argument that your one-metre-thick layer of supersaturated air can absorb at a rate of ten times the maximum absorption observed in the normal air column …

… that’s still only an absorption of two-tenths of one W/m2, an amount that is lost in the noise.

Next, you seem to think that the air right next to the surface can absorb lots more water or become supersaturated in some manner … but in fact, air immediately over the ocean is typically already pretty highly saturated with water vapor. Average air around the TAO buoys, for example, runs at about eighty percent relative humidity (95%CI = 68%RH to 95% RH, so it physically cannot absorb much more water. (Not that it would matter, even at an RH of 100% a one-metre thick layer of air is still far too thin to do what you claim.)

Yes, you are calling on identical scientific principles … but you are totally neglecting the physical situation. A layer of the atmosphere one metre thick will not absorb a significant fraction of downwelling IR, regardless of its composition.

Best regards,

w.

• 1sky1 says:

Crispin:

Your observation about what might happen to DLWIR has merit. But there’s little hope of convincing anyone totally clueless about the existence of the Knudsen layer and the nearly total absorption of remaining DWLIR by the microns-thin surface skin of the ocean. Physical oceanography is not Willis’ field.

9. Lars Silén: Reflex och spegling says:

Thanks for a very interesting article … as usual :) !

10. Espen says:

Very interesting post, thank you!

11. charles nelson says:

What is this Warmist obsession with ‘radiation’…downwelling, upwelling or otherwise?
The important thing about water vapour is that it PHYSICALLY TRANSPORTS heat.

Any other measurements you take are dwarfed by this property.

• Willis Eschenbach says:

charles nelson July 29, 2016 at 1:10 am

What is this Warmist obsession with ‘radiation’…downwelling, upwelling or otherwise?
The important thing about water vapour is that it PHYSICALLY TRANSPORTS heat.

Any other measurements you take are dwarfed by this property.

Thanks, Charles. To use your terms, about 80 W/m2 of energy is PHYSICALLY TRANSPORTED as latent heat, that is to say as water vapor, from the surface to the atmosphere. Another 20 W/m2 of energy or so is PHYSICALLY TRANSPORTED as sensible heat, that is to say as air directly warmed by contact with the surface.

By contrast, the surface is constantly radiating about 390 W/m2 upwards, some of which is absorbed by the atmosphere and clouds and a smaller portion of which goes directly to space.

So no, radiation is not “dwarfed” by the physical transport of latent and sensible heat. Instead, the energy radiated away by the surface is about four times the size of the total surface energy loss in the form of latent and sensible heat.

And going the other way, the energy entering the surface is almost entirely radiation, about 170 W/m2 of solar and about 330 W/m2 of longwave …

Finally, the energy transport into and out of the planet is entirely radiative, with shortwave entering the planet and longwave leaving.

That’s why people discuss radiation … because it is important in the overall scheme of things.

However, you’ll notice that I have included the cooling effect of rainfall driven evaporation, so by no means am I ignoring direct transport of energy.

Regards

w.

• lgl says:

And how much energy is transported as latent heat into the warmest areas of the globe?

• charles nelson says:

Willis…heat energy (as latent heat is PHYSICALLY) transported into regions of polar darkness in WINTER where there is not much in the way of incoming radiation taking place at all!
Also rapidly rising warm moist air over equatorial oceans PHYSICALLY carries vast amounts of physical heat high into the atmosphere where it seems to have no problem cooling.
I just checked the satellite data a few moments ago and the surface temperature over the equatorial pacific ocean is around 30˚C…the tops of the cloud band in the same area are around minus 60˚C.

• Willis,

“Finally, the energy transport into and out of the planet is entirely radiative, with shortwave entering the planet and longwave leaving.”

By “planet” I assume you mean the earth and its atmosphere. This is a very important point because it implies that evaporative does not cause overall atmosphere to lose heat.

Water evaporates from the surface, then rises (in part because water vapor is lighter than air), then cools, then condenses. The latent heat of condensation (warming) is equal to the latent heat of evaporation (cooling) and there is no heat lost or gained by the overall atmosphere.

Evaporative cooling does not remove heat from the earth/atmosphere system, only radiation can do that.

Water vapor molecules absorbs some of the incoming long-wave radiation and some of the outgoing long-wave radiation. Then they re-radiate a portion of the absorbed radiation to the surface, which keeps the surface warmer than it would otherwise be.

If the water vapor feedback were positive without limit, the earth would much warmer and wetter than it is today. Sunlight would cause evaporation, which would enhance the greenhouse effect, so more evaporation would occur, so the greenhouse effect would be amplified, so more evaporation would occur, etc. But this is not the entire effect. Water vapor also condenses to form clouds, which reflects incoming solar radiation and that keeps the planet cooler than it would otherwise be.

Condensation also places an upper limit on the amount of water vapor that can accumulate in the air. Water is returned to the surface as precipitation from clouds or the dew and frost that forms on cool nights. Condensation means that there is never enough water vapor to saturate the air at high daytime temperatures. For example, the highest water vapor content ever reported occurred in July of 2003 at Dhahran, Saudi Arabia. The dew point was reported as 35°C (95°F) and the dry bulb was 42°C (108°F). This equates to a vapor content of 3.3%. But the relative humidity was only 68%, so the air was far from saturated.

A summer day in the tropics can feel very humid but the relative humidity is never much above about 60% during the hot afternoon. Relative humidity tends to be highest at night and reaches it’s lowest point during the hot afternoon.

It seems to me that water vapor is a positive feedback but its effect is capped by condensation and greatly damped by the reflection of sunlight from clouds.

• tty says:

Er…did you actually understand what you were writing?

“And going the other way…..about 330 W/m2 of longwave”

So the net LWIR radiative transport is 390-330= 60 W/m2 which is considerably less than 80+20=100 W/m2.

You have fallen for one of the oldest tricks in the CAGW basket. Radiative transport figures is invariably gross while convective transport is equally invariably net. Please note that neither the descending dry air between convection cells nor the rain is at absolute zero, so there is lots of “downwelling” heat there too.

• Clyde Spencer says:

One of the reasons that it is dry is that it is heated as a result of compression as it descends, not unlike the dry Nevada desert created from air coming down over the top of the Sierra Nevada.

• Bartemis says:

“Evaporative cooling does not remove heat from the earth/atmosphere system, only radiation can do that.”

But, it makes a very big difference whether the heat radiates from the surface or from TOA. Radiation from the surface has to get past the atmospheric filter, and is inhibited in particular by CO2. Radiation from TOA does not encounter that filter.

12. richard verney says:

Very interesting article Willis with much food for thought. Good to see you back.

I will throw out a comment that I have raised many times before, and that is: maybe DWLWIR is not (to any significant extent) absorbed by the oceans and thereby does not (to any significant extent) warm the oceans.

It maybe the case that whilst there is potentially substantial ‘positive feedback’ from water vapour over the tropical oceans (25 to about 33degc), the resulting high level of DWLWIR is a signal incapable of performing sensible work in the environment in which it finds itself, and this is why there is no runaway warming of the tropical oceans.

Whilst there is a negative feedback to high temps (however these temps are generated, ie just solar, or solar + DWLWIR) of evaporative cooling, if ocean temps are governed essentially simply by solar, one does not have the issue that your article raises with potentially runaway DWLWIR and circuitous warming.

I just throw this out as a possible explanation (which may be a component part of a wider explanation).

• Willis Eschenbach says:

richard verney July 29, 2016 at 1:29 am

Very interesting article Willis with much food for thought. Good to see you back.

Thanks, Richard. Some of my research takes a while to hack through, and I only write about a subject when I can no longer stand to NOT write about it. Plus I have eight or ten articles started that just petered out for a host of reasons.

And then the writing (and re-writing, and fact checking, and re-re-writing, and the like) all takes a while as well.

And somewhere in there is my own continuing education, reading articles, learning to write better code in shorter time, researching what is known about various subjects, following my monkey-mind from one shiny object to the next …

I will throw out a comment that I have raised many times before, and that is: maybe DWLWIR is not (to any significant extent) absorbed by the oceans and thereby does not (to any significant extent) warm the oceans.

For those wondering, DWLWIR is downwelling longwave infrared … and here is my comment on the claim that it is not absorbed by the ocean.

We know (because we can measure it and from Stefan-Boltzmann’s equation) that radiative loss from the ocean is on the order of 400 W/m2.

We know (because we can estimate it in a couple of ways) that latent and sensible heat loss from the ocean is on the order of 100 W/m2.

This gives a total loss from the surface of about half a kilowatt per square metre, as a global 24/7 average.

Next, we know (because we can measure it) that the solar energy entering the ocean is on the order of 170 W/m2

We know (because we can measure it) that the downwelling longwave striking the ocean is on the order of 330 W/m2.

If the longwave is absorbed by the ocean, then the energy budget balances—half a kilowatt per square metre in, half a kilowatt out, the ocean neither boils nor freezes.

BUT IF (as you claim) the longwave is not absorbed by the ocean … then the ocean is losing half a kilowatt constantly but it’s only picking up 170 W/m2 from the sun.

So IF your claim is correct, why hasn’t the ocean frozen long ago? A constant imbalance of only 10 W/m2 would cause rapid cooling, and you are proposing that there is a constant imbalance of hundreds of watts per square metre …

Best to you,

w.

PS—The ocean as a global average is warmer than the land. IF as you claim the land can absorb DWLWIR and the ocean cannot, the difference of hundreds of watts per square metre would make the land much, much warmer than the ocean …

• FTOP_T says:

“Next, we know (because we can measure it) that the solar energy entering the ocean is on the order of 170 W/m2”

In the tropics, solar energy entering the ocean is over 1,000 W/m2 and penetrates at depth. Once you correct that figure, the absurdity of the atmosphere delivering twice the amount of energy into the ocean as the sun becomes obvious.

• Ed Bo says:

FTOP:

Do you understand what averages are? More precisely, do you understand what integrations over space and time are? And do you comprehend the differences between these and peak values?

For fully half the time, even in the tropics, the solar energy entering the ocean is precisely 0 W/m2. The DWLWIR at night there is still over 300 W/m2.

What do you think happens to the energy in the DWLWIR when it strikes the water?

• Bob Irvine says:

Bob Irvine
The long wave component of solar radiation is absorbed on average to a depth of 0.267 meters in the oceans according to “The Technical Guide to MOM 4.0, 2008” and this is the assumption used by all the CMIP5 models. This energy, as you say adds to OHC by turbulence etc.
On the other hand, CO2 radiated absorption centred at 15 microns is almost totally absorbed in the first fraction of a millimetre, and consequently within the evaporation layer. It is almost totally returned to the atmosphere as latent heat and does not warm the ocean.
Willis you are correct as far as the long wave component of solar is concerned and Crispin is correct for CO2.
The result of this is really messy but should/might mean that the earth’s sensitivity to changes in CO2 is significantly lower than to similar changes in Solar radiation.
Enjoyed the article.

• FTOP_T says:

Yes, Ed. I do understand averaging. Because I understand averaging, it becomes plain how non-physical these calculations are. Ocean Tmax is derived completely from solar intensity.

As Bob and TonyL note below, DLWIR (which is completely non-physical) could not cause any upward change in ocean temperature.

The ocean skin layer is well established science and LWIR cannot impact below the cool layer in the temperature profile, which is within the first few microns. Less than the width of a human hair.

To understand the implication of this. Fill a thimble with water. What temperature differential would it have to be to raise the temperature of an Olympic size swimming pool at 20C? When you pour it in, what is the new average temp of the pool?

As you calculate the differential and the average, remember that the water in the thimble is free to evaporate. The reality is that your mythical 300 W/m2 DWLIR is non-physical and the atmosphere is warmed from below and doesn’t raise ocean temps.

Assuming an atmosphere with no radiative gases (CO2, methane, etc.) only water vapor
Accepting that water vapor is the most powerful radiative gas
Initially, all water is in liquid form
Now, when this water evaporates, it extracts energy from the ocean during the phase change
The water below is cooler than before evaporation
The vapor immediately uses energy on ascent and is replaced by colder air

So the question is, how does a process (evaporation) that removes energy from the ocean heat it?
How does the energy that is needed to fight gravity on ascent go back to the ocean without stopping convection?
Assuming water vapor is the only radiative gas, how does it remove energy from the ocean, wisk it away to altitude and have anything left to “heat” the water it just escaped?

If it was returning 300 W/m2 to the ocean, it could not have the energy (heat content) to rise and the evaporative process would fail. No Virginia, the well established properties of water prove there is no DWLIR Santa Claus.

• Ed Bo says:

FTOP: Wow! So many misconceptions packed into a single comment!

– That substances at earth-ambient temperatures emit LWIR?
– That LWIR is capable of traveling downward?
– That this radiation carries energy with it?
– That when this radiation is absorbed, it adds energy to the absorbing body?

Next, you assert that this radiation “could not cause any upward change in ocean temperature” because it is absorbed “within the first few microns.” Let’s examine this. Lots of objects are heated to higher temperatures by radiation that is absorbed right at the surface. Rocks in the sun, for one.

Even visible light penetrates only a tiny fraction (meters or tens of meters) of the ocean’s depth (typically multiple kilometers). So would you say the sun cannot heat the ocean either?

Let’s look carefully at this skin layer. Because water is so opaque to LWIR radiation, virtually all of its emitted radiation comes from this skin layer. At typical temperatures, this is about 400 W/m2 just from the top few microns. If there were no DWLWIR absorbed, this rate of energy loss would be far greater than could be restored by conduction from below, and the surface water would quickly freeze. This can be, and has been, demonstrated in the lab.

So really what the DWLWIR is doing is preventing the surface layer from freezing. In most cases, the DWLWIR is of lower density than the water’s emitted radiation, so it is reducing the power imbalance in the skin layer (but there is still usually an imbalance in favor of outgoing radiation).

Your “thimble” example just shows you have no idea how to analyze thermodynamic systems, as you confuse issues of temperature with heat flow, plus one-off effects with ongoing effects. I have a pool about 12 meters by 4 meters, so about 50 m2. At 20C temperature, the skin layer emits 400 W/m2 x 50 m2 = 20,000 W. But the atmosphere provides about 300 W/m2 x 50 m2 = 15,000 W, so the imbalance in the skin layer from radiation is about 5000 W. This is on an ongoing basis, unlike your one-off thimble example. The pool would cool off far more rapidly in the absence of this DWLWIR.

But what about evaporation, you ask? The skin layer, as thin as it is, is hundreds of times thicker than the evaporative layer. Even if the incoming LWIR is greater than the outgoing (which can happen at night) it doesn’t just cause evaporation.

This is easy to demonstrate in the lab, and I’ve done it. LWIR lasers (10.6 um wavelength) are easy to obtain, and not that expensive. Shine one downward at water in a beaker and observe the results. Your analysis says that it should just evaporate away the water from the surface. But that is not at all what happens. You can easily boil the water in the beaker. And this is without the mixing effect of wind and waves that you get in the ocean — it still heats the bulk of the water.

So yes, Virginia, there is DWLWIR, and it does result in higher water temperatures than would occur without it.

• TonyL says:

I would like to comment to you and Willis both. I hope the thread is not going stale already.

The absorption of LWIR by water is almost beyond belief. I finally found a quantitative dataset which measures water absorbance across the IR spectrum. It is a paper by a pair of biochemistry people measuring proteins, and so has absolutely nothing to do with the Climate Wars.
https://www.researchgate.net/publication/223170741_Water_H2O_and_D2O_Molar_Absorptivity_in_the_1000-4000_cm-1Range_and_Quantitative_Infrared_Spectroscopy_of_Aqueous_Solutions

Here is the money shot:

Too bad it does not go down to 666 cm-1 there the action is, but there is help. We get the full spectrum for water at NIST:
http://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Type=IR-SPEC&Index=1

So now using the two together we can flesh out the Molar Absorbtivity through the LW area of interest.
When I did that I calculated that the thickness of the “thin film” used in the NIST spectrum was on the order of 0.1 micron thick. (I do not have the calculation handy, but I will recreate it if there is interest.)
This basically means that the top micron of the ocean quantitatively absorbs everything that gets there. Looks to me like the energy goes straight to evaporation and not bulk warming.

I expanded a volume of water into 2% moisture and repeated the calculations. That showed that LW IR at 666 cm-1 is near total absorption in about 25 meters.
Absorbance = 3.0
Transmittance = 0.1%
That means getting through the atmosphere, LW must be getting absorbed and re-emitted like crazy.

• Clyde Spencer says:

TonyL,
You said, “Looks to me like the energy goes straight to evaporation and not bulk warming.” Then how do you propose that water exposed to sunlight warms?

• TonyL says:

Clyde,
What I am referring to here is specifically LWIR, and and more exactly 666 cm-1, (15 um) where CO2 has an absorption band. The big fuss is all about CO2 after all.
When you say “sunlight”, we all take that to commonly mean visible from 400 to 750 nm, and maybe into the NIR as well. Note here that visible has plenty of energy to cause warming. on a photon for photon basis, Vis photons are far more energetic than IR. As you know, visible penetrates very well depositing its energy throughout the illuminated water column. Water also has lots of absorption bands in the NIR but they are nowhere near as intense as the IR bands, so you get at least some penetration into the bulk water.
This shows up in the first 10-100 cm or so. As an aside, water’s NIR bands are in actuality, the higher order harmonics of the big absorption bands in the IR.
No mysteries.

• Macha says:

Its UV that warms water…. Far more than IR, which simply does not have the intensity (frequency) to penetrate.

• Willis Eschenbach says:

TonyL July 29, 2016 at 8:40 am

Looks to me like the energy goes straight to evaporation and not bulk warming.

RUN THE DAMN NUMBERS, HANDWAVING GOES NOWHERE!!!

Sorry for the shouting, but unless you can show that your claim works QUANTITATIVELY, you are wasting your time and everyone’s time by posting it. Here’s what I mean in this case.

Downwelling longwave at the surface = ~ 330 W/m2 (24/7 global average)

Evaporation = ~ 80 W/m2

Now explain to me again … if downwelling LW is going “straight to evaporation” as you claim, why don’t we have 330 W/m2 of evaporation?

w.

13. Wim Röst says:

Impressing. But what is exactly “rainfall evaporative cooling”?

• Willis Eschenbach says:

Wim Röst July 29, 2016 at 1:43 am

Impressing. But what is exactly “rainfall evaporative cooling”?

Good question, Wim. In English we have a saying, “What goes up must come down.” What is not so obvious is that what comes down (rain) must first go up (evaporation).

If a location gets a metre of rain in a year, that much water must have evaporated. Since it requires a constant flux of about 80 W/m2 of radiative energy over the period of one year to evaporate a cubic metre of seawater, when that cubic metre of water evaporates it cools the surface by that amount. So a metre of rain represents a rainfall evaporative cooling of 80 W/m2.

I discuss this question further in a post called How Thunderstorms Beat The Heat, which you might enjoy.

Best regards,

w.

• Mark - Helsinki says:

Do winds have an effect on evaporation as they do on sublimation?

• Mark - Helsinki says:

As I understand it wind speed is an important factor that will throw out any base calculation of constant flux of radiative energy to evaporate x amount of water.

4x wind speed doubles sublimation for example (I think)

So as such this would leave out a primary factor of such a base calculation

Throw in air temperature too actually

• Mark - Helsinki says:

Akin to Henry’s law, which concerns a static system of water and gas take up without real world variables, as in say CO2 and oceans.

• Wim Röst says:

Thanks Willis, clear.

Am I right that you used rainfall data and not evaporation data to create fig. 6?

• Willis Eschenbach says:

Wim Röst July 29, 2016 at 3:16 am

Thanks Willis, clear.

Am I right that you used rainfall data and not evaporation data to create fig. 6?

Yep.

w.

• lgl says:

Yep
and that’s why you got it all wrong

• Wim Röst says:

lgl July 29, 2016 at 9:09 am “and that’s why you got it all wrong”.

WR: No Igl. As explained Wim Röst July 29, 2016 at 9:14 am, excess evaporation from elsewhere is (in the tropics) transported by trade winds in the direction of the convection cells. In fact your figure “Evaporation minus precipitation” shows this excess of evaporation elsewhere. The evaporated H2O (and so the energy) is brought by the prevailing winds to convection cells where it is ‘launched’ to a height where the energy can easily be emitted to space.
Willis is right.

It is a dynamic process: the warmer it is, the more thunderstorms, the more emission to space. In this way extra warming in the tropics is diminished. A degree of warming in the tropics will create more thunderstorms which will diminish the degree of warming.

• Mark - Helsinki says:

I guess, due to being passed over, I am incorrect. :D

• george e. smith says:

Mark; evaporation and sublimation are physical processes that take place at the interface between two media, in these cases just two phases of H2O .

Solid-Vapor for sublimation, and liquid-vapor for evaporation.

The immediate effect of both of those processes, is that the atmospheric gas layer immediately adjacent to the solid or liquid interface will have a higher than average density of molecules of water vapor, compared to the atmosphere at large.

As a consequence of that higher abundance, there is an increased likelihood of H2O molecules in the atmosphere falling back into the liquid or solid phase, and at equilibrium, there are as many molecules going one way, as going the other way.

This occurs in virtually all bi-directional physical or chemical processes.

The buildup of vapor H2O molecules in the air slows down the evaporation or sublimation.

The sole contribution of WIND to the process, is that the wind, will physically transport those new H2O vapor molecules away from the interface surface and distribute them in the atmosphere.

This reduction in the near surface density of vapor H2O molecules is what allows the evaporation or sublimation to continue at a faster rate.

In all reactions when elements combine to produce some new product; usually plus some garbage, if you don’t take out the garbage it will pile up and quench the reaction.

It’s really just that simple.

G

• The buildup of vapor H2O molecules in the air slows down the evaporation or sublimation.

One of the big tricks used in GCMs to get water feedback, to get cs high enough to make GCMs warm up, is they allow a super saturation( >100% rel humidity).
This is part of what’s wrong with them.

• Wim Röst says:

WR: Am I right that you used rainfall data and not evaporation data to create fig. 6?

Willis Eschenbach July 29, 2016 at 8:46 am “Yep”

WR: That is interesting in this way, that the heat transport upwards is local and is measured in hundreds of hundreds of watts per square metre. But the actual evaporation will partly be located elsewhere, as (trade) winds bring in in the convecting cells the humid surface air from possibly far away. So, in fact this is a system to cool the whole tropics.

• Wim Röst says:

Typo: “hundreds of hundreds of watts per square metre” must be “hundreds of watts per square metre”

And translated better: “as (trade) winds bring into the convecting cells the humid surface air from possibly far away.”

• Bob Irvine says:

Willis
“Downwelling longwave at the surface = ~ 330 W/m2 (24/7 global average)

Evaporation = ~ 80 W/m2

Now explain to me again … if downwelling LW is going “straight to evaporation” as you claim, why don’t we have 330 W/m2 of evaporation?”

Most of the 330 w/m2 DLWR you mention would come from either solar or water vapour and according to the MOM 4.0 used by the GCMs would have an average attenuation in the oceans of 0.267 meters. You are correct to say that this energy is almost totally included in OHC and warms the ocean as it is overwhelmingly absorbed below the evaporation layer. It will be readily mixed by turbulence, wave action etc.
CO2 on the other hand is almost totally absorbed in the evaporation layer with a significantly higher proportion being returned to the atmosphere as evaporation.
This explains the discrepancy between DLWR and evaporation you mentioned above, but also means that Tony and Crispin are correct as far as CO2 is concerned.

14. https://wattsupwiththat.com/2010/09/14/spencer-on-water-vapor-feedback

This is explained excellently and gives a very good perspective on this subject.

My conclusion is an increase in on water vapor overall likely would promote global warming al factors being considered, but as this article says the key is what is happening to the water vapor content in the stratosphere.

From my point of view the overall GHG gas effect is a result of the climate not the cause.

15. Michael Hammer says:

I am probably misunderstanding something but if I look at your first plot it shows about 150 watts/sqM atmospheric absorption at the average 29 kg/sqM atmospheric water content. But we know the surface at the average temperature of 288K emits 390 watts/sqM. Thus if the atmosphere only absorbs 150 it suggests 390 -150 gets through to outer space ie: 240 watts/sqM. But we know the total loss to outer space is about 243 watts/sqM. That seems to imply just about all the emission to space is radiation from the surface not absorbed by the atmosphere. Hardly in agreement with Trenberth et al who claim only 40 watts/sqM of the surface emission escapes to space. Also, if the atmosphere absorbs it must also emit (emissivity=absorptivity) in all directions up as well as down. At some altitude such emissions will escape to space and the intensity of emission depends on the temperature. Even if it all occurs from the coldest point in the atmosphere (the tropopause at about 220K) that would still be significant and would add to the 240 watt/sqM passing straight through the atmosphere. The again if energy absorption rises it means the atmosphere is becoming more opaque (at the thermal infrared wavelengths) and that would also mean it would emit more energy to space (ie: if its absorptivity rises so does its emissivity).

Is the energy reflection by clouds additional to your 150 watts/sqM? If so, assuming Trenberth is right (OK I admit that’s a very large IF) clouds would have to be reflecting about 200 watts/sqM which is even larger than the atmospheric absorption. I guess its possible if cloud cover is 60% (60% of 390 = 234watts/sqM). But then again that should be 60% of 390-150 = ( 144) since the clouds cannot very well reflect the energy already absorbed by the atmosphere. Either way cloud effects are as large at atmospheric absorption and thus cannot possibly be ignored in any analysis

Then again evaporative cooling transfers the energy to the atmosphere not to space. Increased energy loss to space would only occur if the atmosphere at the altitude where it is emitting to space would be rising since emission depends on temperature and that means we should be seeing a significant rise in upper atmosphere temperatures particularly in the tropics – exactly what the CAGW proponents predict yet which 1000’s of balloon flights show is not happening.

Sorry if the above sounds a bit disjointed, I wanted to give some feedback before the comments get overwhelming, your analysis is interesting (I need to study it in more detail) but I cant help but feel there is more to this issue.

• Willis Eschenbach says:

Michael Hammer July 29, 2016 at 2:47 am Edit

I am probably misunderstanding something but if I look at your first plot it shows about 150 watts/sqM atmospheric absorption at the average 29 kg/sqM atmospheric water content. But we know the surface at the average temperature of 288K emits 390 watts/sqM. Thus if the atmosphere only absorbs 150 it suggests 390 -150 gets through to outer space ie: 240 watts/sqM.

Good question, Michael. The figures are solely for clear-sky conditions, meaning no clouds. Clouds cover nearly 70% of the planet and are generally blackbodies w.r.t. longwave, so they absorb a lot of radiation.

w.

• BACullen says:

So, they absorb a “lot” of long wave radiation. What does mean?
It seems to me that it drives the clouds to a higher elevation where that heat is lost mostly into space ( even more efficiently).

• george e. smith says:

To be ” near BB ” a body needs only to be able to absorb almost all of the incident EM radiation in a 16 to one spectral wavelength range appropriate to the Temperature of the body.

98 % of the radiation of a BB lies between wavelengths from one half of the radiance peak wavelength, to eight times that peak wavelength. 25% (almost exactly) of the total energy lies at wavelengths shorter than the peak, and 75% is longer than the peak. Only 1% lies beyond either end of that 0.5 to 8.0 times the peak wavelength.

For 288 K the peak wavelength determined from the Wiens Displacement Law is about 10.1 microns; about 20 times the solar spectrum peak wavelength corresponding to about 5760 K for the sun surface Temperature.

G

16. Again the climate stability notion Willis keeps trying to push is all relative to how one views climate stability.. For example I view the climate system as being unstable just based on the change in N.H. temperatures from the Medieval Warm Period to the Little Ice Age to present and the impact those changes have had and are having on humanity..

With solar prolonged minimum conditions becoming more established (finally ) once again the factors which govern the climate system which is DRIVEN by the sun will be influenced and if solar activity is EXTREME enough they will be influenced to a degree which will cause the climate to change especially if climatic thresholds are met.

Past history shows beyond a doubt that this has occurred due to the many documented abrupt climate changes shown in the ice core record.

I suggest what I just wrote to convey that the climate system as far as humanity is concerned is not stable

17. Roy Spencer says:

The increase of total water vapor with SST on a gridpoint basis will be larger than that computed from large-scale area averages (say, over the whole tropics). The former (gridpoint-based) will be dominated by geographic differences, which are strongly affected by vertical circulation systems that cause water vapor to converge and deepen over warm SSTs and decrease over cooler SSTs; the latter (large scale area averages) will be dominated by interannual variations in the whole system (say, from La Nina and El Nino) and are usually used as a possible surrogate for a climate change response.

18. Svend Ferdinandsen says:

Thanks for these articles.

You look at clear sky absorbtion, but how about all sky or clouds only?
The word absorbtion as ingoing minus outgoing is possible to use that way, but with an actve radiating atmosphere it could lead to amplification in rare instances.

19. Willis,
” that’s a serious amount of cooling, about ten watts of additional cooling per additional kg of precipitable water”
I don’t understand this argument. I presume it means that if there is a sustained increase of 1kg PW, the rain increases and the latent heat associated with its condensation is 10 watts. But the latent heat at condensation has a warming effect, not cooling. The cooling happens when the water is first evaporated. And the two must balance if PW is steady; no net heat can be created if the water starts and ends as liquid at SST temp. So the latent heat allows vertical transport (Trenberth’s 80 W/m2), but it doesn’t cause net cooling.

• Greg says:

Cooling is heat leaving the system. If head energy gets from surface ( SST ) to tropopause,it’s half way out of the climate system.

• “Cooling is heat leaving the system.”
Yes, the water cycle augments heat transport. But to convert that quantitatively to a cooling rate would involve some heavy duty analysis, with a spectral GHG analysis, levels of atmosphere etc. And I don’t see any indication of that being done here.

• Ed Bo says:

Nick:

The notebook computer I am using to type this comment has a thermal system that evaporates/boils water over the processor and then condenses it to the outside of the computer at ambient temperatures. This is the cooling system for the computer.

Similarly, the surface evaporation cools the surface, and as the water vapor condenses high in the atmosphere, the warming effect here is better able to be rejected to the ambient of space, as it is above most of the radiative absorption of the atmosphere.

Trenberth’s 80 W/m2 of latent heat loss from the surface is a NET value of this effect.

• george e. smith says:

The latent heat of evaporation; amounting to something like 590 cal/gm, (depending on the Temperature at which the phase change occurs), is now part and parcel of the WATER VAPOR, and as correctly mentioned by somebody here, it is PHYSICALLY transported by CONVECTION to the upper atmosphere, which somebody also said could be at -60 deg. C (have actually been in an aero-plane over the Pacific, reporting exactly that outside air Temperature for all the passengers to read.)

Somewhere along that upward path, the formerly warm H2O vapor gets cooled (LOSES HEAT) to the cooler air (the two are physically mixed and exchange heat energy among themselves).

ONLY after the LATENT HEAT has already been lost by the H2O VAPOR to the N2 and O2 and Ar of the dry atmosphere, can that H2O molecules find a low curvature substrate surface (microbe or dust particle), at a Temperature somewhere near the DEW POINT Temperature, CONDENSE (not precipitate), and become LIQUID water droplets, which can then grow; or perhaps after losing another 80 cal/gm (if cold enough) , and become solid ice crystals.

So the water droplets or ice crystals that make VISIBLE clouds, have already disgorged all of the latent heat of phase change, that they PHYSICALLY transported from the surface (ocean maybe) up into the upper atmosphere, where it will eventually be lost to space by various sequences of processes.

The RAIN or SNOW or other precipitation which ultimately occurs when it gets too dense to support, will fall to earth (through an increasingly warm atmosphere of air) which will somewhat warm the precipitate, to its final Temperature as it lands back on earth or back in the ocean.

Its Temperature is entirely due to the HEAT energy that may reabsorb from the ordinary atmosphere gases, in CONDUCTIVE heat exchange processes, on the way back to the surface.

The H2O while in either of its condensed phases, makes a very fine efficient radiator of THERMAL LWIR radiant energy, at whatever the ambient Temperature happens to be where those clouds are, and maybe half of that goes towards space, and the other half towards the surface, so the clouds comprise an effective near BB like thermal radiator at whatever the local Temperature happens to be.

The atmospheric H2O VAPOR on the other hand, is a low emissivity thermal radiator (due to its Temperature), but also a good radiator of molecular LWIR spectral radiation (that which is characteristic of the H2O molecule. (CO2 does likewise).

Yes it is true that in daylight, the atmospheric water in all three phases will absorb some parts of the incoming solar spectrum radiant energy (short wave or near IR), which will locally warm the atmosphere, BUT ! the majority of the incoming solar energy reaches the surface at about 1,000 W/ m^2 (NOT 341 or 250 or any other average number), and 3/4 of that surface is deep ocean water into which the solar beam penetrates to depths of tens to hundreds of meters.

It will be eons, before the oceans manage to distribute that 1,000 W/m^2 solar beam, around the entire planet, to establish some 288 K supposed mean Temperature at which it can possible radiate some 390 W/m^2 in a 10 micron peaked wavelength spectrum.

Clouds don’t even come into existence until the HEAT ENERGY (noun) in the form of latent heat of phase change, is entirely removed from the H2O molecules by CONDUCTION.

The COLD air (there’s masses of it) COOLS the water vapor (there’s much less of that).

The “latent heat” is NOT going to stop the upper atmosphere from getting down to -60 deg. C outside your aero-plane.

The exact Temperature at which the phase change occurs from (atmospheric) water vapor to CONDENSED clouds depends on a host of factors which are generally recognized, in the form of some sort of “dew point” Temperature, but also the local availability of a nucleation substrate to obviate the need for an infinite internal pressure in a zero radius water droplet . (surface tension delta p = 2 t / r) )

G

I don’t own any text books on “forcings” so I don’t do those calculations; so I stay with the Physics.

• Clyde Spencer says:

GES,

You said, “Somewhere along that upward path, the formerly warm H2O vapor gets cooled (LOSES HEAT) to the cooler air (the two are physically mixed and exchange heat energy among themselves).”

Now, if the warm air were diffusing upward, it would be well mixed and it could lose heat by conduction. However, for strong convection, such as at the base of a cumulonimbus cloud, there isn’t strong mixing and that is what preserves the latent energy and precipitable water. It is the application of the Ideal Gas Law that explains the cooling of the water vapor and the associated air. The liquid water can be cooled by conduction when it condenses, and is still being lofted upwards by convection. The ‘cell’ contains an air mass that is physically different from the surrounding air that you consider to be responsible for conductive cooling “among themselves.”

• george e. smith says:

Well Clyde, a single H2O molecule in the atmosphere; well mixed or not, is surrounded by hundreds or tens of ordinary N2 , O2, and Ar molecules. For CO2 it’s about 2500 molecules surrounding one CO2, maybe 13 molecular layers thick to the closest (average) next CO2 molecule.

So even in unwellmixed air, an H2O molecule is in thousands of collisions per second with the ordinary air molecules, so they reach the exact same local Temperature in microseconds or milliseconds.

So it is the whole mass of atmosphere that is rising in convective up drafts, not just the H2O.

So an H2O molecule remains in statistical thermodynamic Temperature equilibrium with the surrounding air as they all rise in altitude, and they all reach the (near) Dew point Temperature together, at whatever the lapse rate happens to be.

Only the H2O molecule knows that it is on average too energetic to join together with other H2O molecules and form a droplet. So that process is often, and maybe usually delayed, until a suitable low curvature substrate is encountered, on which the H2O molecules can deposit as a covering water layer.

Any notion that any single H2O molecule remains at some elevated Temperature relative to the air for some reason, is bogus. The H2O molecule will rapidly fall into place on the Maxwell-Boltzmann equi-partitioned kinetic energy range appropriate for that Temperature.

The Eqi-partition law holds in general for any mix of gas molecular species. Now some species will have different numbers of degrees of freedom, and for those each degree of freedom is equally endowed with mean KE appropriate to that local Temperature.

G

• Johanus says:

@Nick
> … it doesn’t cause net cooling.

This is correct only in the sense that the heat of evaporation is returned by the heat of condensation. But as Greg also points out, this heat of condensation is returned in the high troposphere, in these large tropical convection systems, above most of the earth’s warming planetary blanket of air.

Yes, some of this heat will be directed downward and will be “trapped” in the blanket, but the other half will be radiated (and of course, not convected) upward where the path to outer space is now much freer.

You can see this image that convection does indeed stop at the tropopause (with a little “overshoot” here and there of course), but radiation does not.

So the 10 W/m^2 cooling computed by Willis does seem consistent with this notion.

• Johanus says, July 29, 2016 at 9:24 am:

Yes, some of this heat will be directed downward and will be “trapped” in the blanket (…)”

No. All the heat drawn from the surface and up into the troposphere is radiated to space as Earth system heat loss. We’re in a dynamic steady state, Johanus. If half the heat were continuously being radiated back down, into the system, Earth would heat without end. At dynamic equilibrium, as much energy always exits the Earth system to space as heat as what enters as heat from the Sun. This is Thermodynamics 101.

• Johanus says:

As I said, some of the downward LWIR radiation is “trapped” (quoted to show irony), in the sense that is absorbed and raises the temperature of the planet as sensible heat, raising the expected black body temperature from 255K to 288K. The upward radiation has no effect on the surface temperature so, in effect, cools the surface (the point I was reinforcing.

But, yes, all of the input radiation is eventually returned to space, once radiant equilibrium is achieved. To see this, imagine that the Sun suddenly disappears. Then the terrestrial temperature will plunge toward absolute zero as earth radiates the previously absorbed solar energy into space. (But will never quite achieve 0K because of internal geothermal energy)

• Johanus:

As I said, some of the downward LWIR radiation is “trapped” (quoted to show irony), in the sense that is absorbed and raises the temperature of the planet as sensible heat, raising the expected black body temperature from 255K to 288K. The upward radiation has no effect on the surface temperature so, in effect, cools the surface (the point I was reinforcing.

No. Accumulation of energy inside the Earth system only occurs between t_0 and t_1, when it warms from initial heating (by the Sun) to the final steady state (when heat OUT=heat IN). This accumulation is what gives the Earth its temperature(s).

• Are you being serious here, Nick? Evaporation is cooling the SURFACE. Parallel to the Sun heating it. It’s pretty obvious Willis is talking about the SURFACE, not the troposphere above. The latent heat of vaporization, drawn from the surface upon evaporation and later released in the tropospheric column upon condensation is not returned to the surface. It is radiated to space. As Earth system heat loss. The water returns, not the energy. That’s how heat generally flows through the Earth system: Sun > surface > troposphere > space.

• Willis Eschenbach says:

Nick Stokes July 29, 2016 at 3:51 am

Willis,

” that’s a serious amount of cooling, about ten watts of additional cooling per additional kg of precipitable water”

I don’t understand this argument. I presume it means that if there is a sustained increase of 1kg PW, the rain increases and the latent heat associated with its condensation is 10 watts. But the latent heat at condensation has a warming effect, not cooling. The cooling happens when the water is first evaporated. And the two must balance if PW is steady; no net heat can be created if the water starts and ends as liquid at SST temp. So the latent heat allows vertical transport (Trenberth’s 80 W/m2), but it doesn’t cause net cooling.

As precipitable water goes from about forty to fifty-five kg per square metre, evaporative cooling goes from fifty to two hundred fifty watts per square metre or more … that’s a serious amount of cooling, about ten watts of additional cooling per additional kg of precipitable water.

I apologize for the lack of clarity. I had thought that the subject under discussion was clearly the surface and not the planet, since the evaporation directly cools the surface, and the temperature in all the charts is the surface temperature. You are right that the globe doesn’t directly lose energy in the form of latent heat.

However, as I discussed elsewhere on this thread, the effects of evaporation and convective rainfall activity can indeed create net global cooling.

My best to you,

w.

• Trick says:

“‘..effects of evaporation and convective rainfall activity can indeed create net global cooling…”

Not multi-annually Willis – over long periods they balance. Sure the evaporation in the Pacific cooled but then the rain over Ca. warmed. Willis demonstrates no net global cooling from the multi-annualized effects of evaporation and convective rainfall data as he shows balanced here, Willis: “the model I developed over a decade ago”..:

https://wattsupwiththat.com/2016/07/28/precipitable-water-redux/comment-page-1/#comment-2268428

Emitted & “Absorbed by troposphere” running R to L:

22+58+76+339-321-321+147 = 0

Balanced, no net cooling from effects of evaporation and convective rainfall data, per Willis’ own chart.

• Willis Eschenbach says:

Trick August 3, 2016 at 2:50 pm Edit

“‘..effects of evaporation and convective rainfall activity can indeed create net global cooling…”

Not multi-annually Willis – over long periods they balance.

Mmmm … perhaps you misunderstood me, likely my fault. It may be we are discussing different kinds of cooling. By “net global cooling”, I am talking about the global average surface temperature. It’s clear that I’m not talking about the top-of-atmosphere cooling or warming. There is 340 W/m2 entering the system at the TOA, and given that the system is observably in general overall dynamic balance, there has to be 340 W/m2 leaving the TOA. And indeed, this is verified by the CERES satellite data.

However, while the TOA has to balance as you say, there is no such requirement regarding the surface. It is currently running well above the expected Stefan-Boltzmann temperature … and it could easily run at some cooler temperature. Suppose, for example, that the clouds change just a little bit, increasing the albedo from ~30 to ~31% … the system would run cooler “over long periods”.

And in the comment above from which you have quoted, I listed a number of ways that evaporation and convective rainfall activity affect the global temperature, both by actively reducing incoming energy, and by actively cooling the surface. The big effects are the changes in cloud, surface, and ocean albedo; the greatly increased evaporation due to the thunderstorms; and the change in the rate at which the ITCZ thunderstorms circulate the heat from the tropics to the poles.

Regards,

w.

• Wim Röst says:

Willis, a very interesting basic view:
Willis Eschenbach August 3, 2016 at 3:11 pm
“However, while the TOA has to balance as you say, there is no such requirement regarding the surface. It is currently running well above the expected Stefan-Boltzmann temperature … and it could easily run at some cooler temperature. Suppose, for example, that the clouds change just a little bit, increasing the albedo from ~30 to ~31% … the system would run cooler “over long periods”.

And in the comment above from which you have quoted, I listed a number of ways that evaporation and convective rainfall activity affect the global temperature, both by actively reducing incoming energy, and by actively cooling the surface. The big effects are the changes in cloud, surface, and ocean albedo; the greatly increased evaporation due to the thunderstorms; and the change in the rate at which the ITCZ thunderstorms circulate the heat from the tropics to the poles.”

WR: I had a first look at ‘Boltzmann – Earth Temperature’. Wikipedia:
https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Effective_Temperature_of_the_Earth
“(…) As a result, the Earth’s actual average surface temperature is about 288 K (15 °C), which is higher than the 255 K effective temperature, and even higher than the 279 K temperature that a black body would have.”

WR: And the Earth’s average surface temperature and the average temperature of the deep sea have been far higher in the past, as Paleo data tell:

Apparently for the Earth’s average surface temperature there is a wide variability possible. For example, our present cool deep oceans give an enormous possibility to cool the surface. What will happen with the Earth’s average temperatures when Earth’s average wind speed enhances at the places where cold water upwells? Isn’t this what happens when more thunderstorms develop?

• Trick says:

Willis – Good discussion thx. In your gray band labeled “Absorbed by Troposphere” your example processes* might of course change over periods observed to take net energy out of that zone leading to “net global cooling” in that zone for the period (KT97 was 5yr.s in the mid-80’s, you do not specify a period here). A new annual balance would be established after a transient rebalancing.

My point is things are pretty much balanced around 287-289K Tmedian meaning rain and evaporation are likewise pretty much balanced (76 up, 76 down give or take) for your gray area over the periods we’ve observed. Climate changes, so that will change. You can see the small natural changes even in the data (mid-80’s KT97, 2000 – 2004 in KT09, 2000-2010 in Stephens12) but they all stay in ~balance, perfectly in yours.

*”changes in cloud, surface, and ocean albedo; the greatly increased evaporation due to the thunderstorms; and the change in the rate at which the ITCZ thunderstorms circulate the heat from the tropics to the poles.”

• Shawnhet says:

The vertical transport of heat causes net cooling because warmth at higher elevations radiates to space more efficiently than warmth at lower elevations.

Cheers, :)

20. Dr. S. Jeevananda Reddy says:

In 70s I published several articles in the India Meteorological Journal [Indian J. Met. Hydrol. Geophy.] to estimate parameters that were measured over few stations with data measured over wide networ. They included rainfall versus sunshine/cloud cover, global solar radiation & net radiation, evaporation & evapotranspiration. One such paper published in 1976, 27:163-166 “Simple formulae for the estimation of wet bulb temperature and precipitable water.

The total moisture content of the atmosphere is expressed by the precipitable water vapour in the atmosphere. This is defined as the depth of liquid water that would result by condensing all the vapour in the vertical column of the atmosphere over one square centimeter cross-section.

Tw = T [0.45 + 0.006 x h x square root (p/1060)

and

W = = c’ x square of Tw

Tw = wet bulb temperature in oC — daily orhourly
T = dry bulb temperature in oC — daily or hourly
h = relative humidity in % — daily or hourly

W = precipitable water vapour in gm/square centimeter — daily or hourly values

c = regression coefficient vary with season [months] — c’ = 1/ square of c — based on the radiosonde data measurements at 12 12 stations for 6 years was used.

Dr. S. Jeevananda Reddy

• george e. smith says:

So ” precipitable water vapor ” is (by definition) the TOTAL water VAPOR in the atmosphere.

ergo clouds are NOT PWV; that is NOT water vapor.

G

Why do they have to make such a big deal out of “It’s the atmospheric water vapor ” ???

• Clyde Spencer says:

GES,

There are often problems in science, and philosophy especially, resulting from not carefully defining terms. If “precipitation” is that which falls from the sky (a commonly accepted definition), then I guess we could call clouds “unborn precipitation.” It isn’t water vapor any longer, but it hasn’t yet formed droplets large enough to fall and therefore isn’t really precipitation. So, I think we should take Willis to task for not carefully defining “precipitable water” when he started this series. However, I suspect that clouds are not included in the definition because of the ways in which the parameter are determined. That is, one can’t see with a satellite sensor what is under a thick cloud, so nothing intelligible can be said about what is in the water vapor column when there are intervening clouds.

• george e. smith says:

Clyde, my comment (above) about precipitable water vapor, was really a rhetorical question to Dr. Reddy above where he defined PVW in the form of a depth of liquid water that could be condensed out of the water vapor in that column of air.

As such that is literally ALL of the water in that column, since the cloud part of it is simply already condensed, and on its way to precipitating.

IT IS NOT MY DEFINITION.

…… Precipitable Water Vapor …… is just about (but not quite) the stupidest damn pseudo scientific mumbo jumbo ever foisted on us as some sort of supposedly useful information.

The atmospheric air column from the condensed state earth surface out to maybe halfway to the sun, contains H2O molecules either in the form of a gas (vapor) or liquid droplets or ice crystals, or all three of those, and as such that water contributes to the total atmospheric pressure on the surface, which adds to that pressure cause by N2, O2, Ar and a zillion other atomic / molecular species.

If you put all of that water in a bucket, you have your PWV. That is ALL of the water in that atmosphere column so you can’t get any more precipitated no matter what.

As for what that means for our climate. One would expect that in each of those three phases, each type interacts with atmospheric energy, either radiative (from zero frequency to zero wavelength) or thermally as ” HEAT ” which is the random kinetic energy (mechanical) of matter at Temperatures greater than zero kelvin, and those interactions are likely to be different for each of the phases.

The study that Willis presents here attempts to make some sense out of those differing processes.

I don’t pretend to understand just how Willis’s manipulations all work; but he usually does his homework; so I just accept that some readers do understand it.

But the notion that somehow water in the atmosphere is special if it is precipitable, is about as sensible as the equally silly notion that CO2 in the atmosphere has a half life or a residence time.

BOTH H2O and CO2 are PERMANENT components of the atmosphere, in varying amounts that change with time and location.

It matters not if one CO2 molecule, or one H2O molecule gets replaced by another.

It only matters how many of each there are at any given time or location.

Mother Gaia knows how many there are and where they all are so she sets the Temperature to be just what it is supposed to be; and it always is; no matter what.

And it almost NEVER EVER has the global average Temperature ANYWHERE, or at ANY TIME.

Only people pay any attention to the average. Planet earth does not.

G

• george e. smith says:

PS.

Note my rant is not any criticism aimed at Willis’s analysis here; so don’t get the idea that it is.

Whether H2O is precipitable or not is just how you define it.

So it is about as useful as climate sensitivity.

Everybody defines it a different way, and then everybody else talks about it as if we are all talking about the same thing.

No amount of mathematical statistical prestidigitation can establish that a “Climate Sensitivity (S) “, (whatever the hell that is) of 2.00 is perfectly acceptable and we can live with that forever; but if it comes out to 2.01, then we are all doomed to roast.

Nor does where the water in the atmosphere is today matter.

G

21. Having lived in the tropics for many years, I can testify from personal experience that the almost daily “3Oclock thunder plump” drops a load of rain and the temperature by 10 degrees or more.
No argument! It happens!.

• Roy Spencer says:

except, as Nick Stokes points out above, the cooling at the surface caused by evaporation and convective rainfall activity is exactly balanced by heating in the upper troposphere where the condensation takes place. There is no net cooling of the climate system.

• Wim Röst says:

“balanced by heating in the upper troposphere where the condensation takes place”

Dr. Spencer, what happens with the energy from the condensation in the upper troposphere, will it fully be emitted to space?

If so, what would have happened if the same amount of energy would have stayed down at the surface, how much would have been emitted?

• Greg says:

As I just comment to Nick above, heat leaving the climate system leaves the surface and rises as latent heat. It then returns to sensible heat in the upper troposphere. At this point it is half way on it’s journey into space. Though at this point is it correct to say that there is not yet any net heat loss to system as measured at TOA.

• “There is no net cooling of the climate system.”

Convection certainly doesn’t transfer heat to space.

However, the so-called Lapse Rate Feedback is negative, no?
The ( modeled, but missing for the satellite era ) Hot Spot reduces the net radiative imbalance.

• Sure, but I thought the impacts of AGW (the reason for all of the alarm) generally required warming at or near the surface. How is a warming of the troposphere going to drive increases in SLR? How will this adversely affect agriculture, or drive species extinctions?

• Disagree about no net cooling. The latent heat of evaporation released by condensation is released higher in the troposphere, closer to the effective radiating altitude. Therefore there is relatively more cooling as the GHE is less. Analogy to thinner insulation.

• Dixon says:

I’m with Wim and Greg here and don’t get the ‘no net cooling argument’ at all. The energy may (must!) balance if you consider the evaporation/convection a closed system, but it isn’t. Nor is it ever at true equilibrium. Once at the lower pressure of the upper atmosphere, radiation efficiency to space is far greater because there’s less chance of interception of the photon by another molecule. So you move energy from a part of the Earth system where radiation is relatively inefficient to where it is highly efficient. It’s like taking a blanket off and similarly must result in a higher rate of heat loss.

Surely convection and condensation (cloud formation) is especially important because the energy exchange is very sudden at the condensation point. So heat can move quickly from the surface to altitude and be released very quickly. Incoming solar is highly variable over the 24 hrs (duh!), whereas convection can take place whenever there is the energy to facilitate it.

All that convection has to be balanced by incoming cooler air and that presumably further increases heat loss at the surface as a result of convection.

What am I missing?

• Disagree about no net cooling. The latent heat of evaporation released by condensation is released higher in the troposphere, closer to the effective radiating altitude. Therefore there is relatively more cooling as the GHE is less.

I think that’s true –

1. Convect heat and humidity from lower to upper levels, and upper levels radiate more effectively.

2. But, more heat & humidity in the upper levels, means the lower levels radiate less effectively.

3. But that leads to cooling over warming, which leads to… more convection,

4. which leads to… ( 1. above )

The Hot Spot is a reflection of convection in the models which in theory leads to some negative feedback.
BUT – that feedback is not 100% of delta CO2+H2O.
Warming, in the context of unknowns and variability, should ensue, just not at rate implied by twice the forcing from CO2.

And the Hot Spot hasn’t occurred for the satellite era, so does that mean LR feedback hasn’t occurred?

• george e. smith says:

damn ! we can’t get nothing for nothing !

G

• george e. smith says:

Nobody ever accounts for the mgh involved in getting the H2O from down here to up there (and back later)

g

• JPeden says:

Roy Spencer
July 29, 2016 at 4:26 am

except, as Nick Stokes points out above, the cooling at the surface caused by evaporation and convective rainfall activity is exactly balanced by heating in the upper troposphere where the condensation takes place. There is no net cooling of the climate system.

I’ll just state this as kind of Postulate involving Energy Balance since I’ve never officially studied Radiative Physics:

Sun-warmed surface layer water leads to increased water vapor which emits infrared light. Not all of its Energy is Kinetic. Therefore I don’t see how some % of water vapor’s infrared light Energy isn’t lost to Space, since E=mcc. In other words, I’m saying the Sun’s incoming Energy which quickly increases [more or less] the evaporation of water at the surface, and [more or less] the cooling there, is balanced by the Energy used in the upper troposphere when condensation takes place + the [more or less] Energy lost to Space by water vapor’s infrared = net Energy loss and [more or less] cooling.

• Wim Röst says:

Dixon July 29, 2016 at 8:37 am:
“All that convection has to be balanced by incoming cooler air and that presumably further increases heat loss at the surface as a result of convection.

What am I missing?”

Dixon, your cooler air is very dry. As is explained by Dr. William Gray: http://icecap.us/images/uploads/Crux_Flawed_Science.pdf
The lowering air from the top of a thunderstorm can have a Relative Humidity of only 7% as it returns to sea level (table 4). The dry air is perfect for rapid evaporation, helped by the strong winds which – as far as I remember – have an exponential effect on evaporation.

The article of Dr. Gray is very interesting. Hereby his conclusion on evaporation:
“Real global warming to be expected.
Without upper-troposphere water-vapor change
and without enhanced surface evaporation cooling associated with extra rainfall, the
pure radiation response to a doubling of CO2 would indicate we should expect about a
1.0oC global warming. But even with zero assumed water-vapor change this 1oC
warming is two to three times larger than what will likely take place. This is because
about 60 percent of the 3.7 Wm-2 IR blocking to space from a doubling of CO2 will be
balanced by an enhancement of surface evaporation and an increase of the global
hydrologic cycle by about 2½ to 3 percent. A zero water-vapor feedback will thus be
expected to only bring about a 0.4oC global temperature rise from CO2 doubling.

We show that there is a very modest degree of negative water-vapor feedback of 0.1 to
0.2oC. With this occurring we should expect that the real amount of global warming that
will occur from a doubling of CO2 would be only about 0.2-0.3oC or about 5-10 percent
the amount projected by the many global models of 2-4oC. The AGW threat and
especially the catastrophic AGW (or CAGW) threat cannot be a realistic assertion of
how the planet’s climate system functions.”

• Dixon says:

Hi Wim,
Thanks for that. So in summary, I’m not missing anything and Dr Spencer and others are wrong? That’s not a very comfortable position for me to be in!

I’m struggling to believe the GCMs can really have such a crude approximation of water vapour. I knew they didn’t do clouds and fine-scale very well, but I didn’t realise it was that fundamental. How have the atmospheric physicists been complicit with such shoddy science? There must be enough of them modelling convection on fine spatial scales to be able to call bull in the peer reviewed literature on parameterisation of water vapour if that paper you gave the link for is a reasonable summation. It’s not like this is all that new. Or does peer-review of climate models not go to people who actually understand the real science behind the elements of the models? I can see why some scientific fields are wary of getting into the climate debate, but it really is unforgivable if your field overlaps their fundamental ‘science’.

As I read the link I was reminded of tephigrams and wish I’d paid more attention when I had the benefit of an uncorrupted Met man to explain them properly. I could have got to this enlightened state 20 years ago. Pretty sure there should be a lot of publicly available data from research aircraft out there, but of course that’s weather, not climate, so there’s probably no money or interest in it.

I live in Perth Australia and I only realised very recently that the howling dry easterlies we get in summer are part of Hadley circulation. They suck the moisture out of your lungs with every breath. But give me 37C and 5% humidity over 30C and 100% humidity any day.

• Wim Röst says:

Hello Dixon,

I can reassure you, I didn’t do an extensive research whether you’re not missing anything besides what I was telling you. And about the well respected Dr. Spencer and others I can only say that I am reading with interest all they are writing and that I try to get the fundamental things out of their knowledge and from their points of view. So you must feel more comfortable now!

As far as I know, the role of water vapour in the models is based on assumptions. If it indeed is, the feedback of water vapour is the big lie, the lie which was necessary to make the problem ‘really dangerous’. It is very interesting what Bill Illis July 29, 2016 at 6:53 am said about the water feedback:

“The +2.4 W/m2 per 1.0C is, thus, a VERY important number. It is “everything”. If, in reality, the feedbacks are only half of predicted, global warming would fall to 1.6C per doubling. If it were one-quarter, warming falls to 1.3C per doubling. If it were twice as big, well, global warming would then +47.0C per doubling and there would be a runaway greenhouse effect.
That is how finely tuned the chosen feedbacks are. It is the difference between benign-nothing warming and a runaway greenhouse effect.”

WR: ‘Finely tuned’ ! Indeed! “If it were twice as big, well, global warming would then +47.0C per doubling and there would be a runaway greenhouse effect.”

Like you, I am understanding more and more about how the whole problem (dangerous warming) is created and which tricks were used in trying to make all of us ‘people with fear’. To be honest: not too long ago I was one of those who had some fear, even while I – as a geographer – was better informed about climate than most of the other people. Fortunately there was that excellent book “De staat van het klimaat” by Marcel Crok that opened my eyes. And I found websites like this with persons with a feeling for what is right (being honest) and for what good science should be. In this way someone’s point of view can become more realistic.

And fortunately there are persons like Willis who have the skills and the insight to reveal more about the real functioning of climate than hundreds – if not thousands – of well paid researchers do. This last group, the well paid researcher, is mostly living in his or her ‘virtual world’ and is producing ‘variations’ of biased research based on at least partly wrong assumptions. Like about the feedback of water vapour.

On this website recently someone wrote a comment about ‘group think’. Researchers are forming a group and all are sharing a need for confirming investigations as wanted by the public, the government, their own institutes and themselves. But that research is far from the fundamental scientific research that would be needed to get the big lie out of the climate ‘debate’. Therefore you must dare to have doubt about what 97% is believing.

I myself am living in Holland and fortunately we have had some warming here last century. Mainland western Europe is from the densely populated regions the region that warmed most, I think. But still, as we have in Holland our warmest period of the year – summertime – during holidays our people are fleeing to countries where they can find temperatures 5 to 10 degrees higher than our present (2016) average (day and night) July temperature of 18,4ºC. From our population of 17 million, countries like Portugal, Spain, Greece, former Yugoslavia and Turkey – the really hot ones – attracted last year in summertime (!) 3 million Dutch to celebrate their holidays in their country. Nearly all returned happily and no fatal climate problems have been reported nor for the local people nor for the Dutch. Even not after the ‘unprecedented warming’ of last century. On the contrary, Dutch still like ‘warmer’. At least during their summer holidays. And in autumn, winter and spring.

22. As always an interesting post Willis.

23. Willis,

You wrote –

“So … with such a large radiative feedback from water vapor, three to four watts per square metre per degree and much higher in the tropics, why is there not runaway feedback”

Because at night time, the sun is not shining.

The surface cools. Hardly rocket science – any reasonably intelligent 12 year old can explain it.

Neither H2O nor CO2 heat anything. No feedback – complete nonsense. The surface warms during the day, and cools at night. Only dimwits and climatologists think otherwise.

Cheers.

• The surface cools.

Does it cool faster with high humidity above, radiating infrared on it?

Or does it cool faster with low humidity above and little downward infrared radiating on it?

I think you will find the latter.

24. Greg says:

Willis: “Here is the main figure from that post showing the relationship:”

Not quite, this is the same as your earlier graph but this time with a linear scale, not log scale. This is useful as a second way to visualise the data.

The caption to fig. 1 says: ” The slope of the curve at any point is 62.8/TPW (W/m2 per degree)”

This is clearly not correct. The slope of that graph varies notably. It is the slope of the log graph which is constant.

This ties in with a comment I made in the previous thread about the change in slope as TPW increases.

As TWP increases the slope decreases. This means that the magnitude of the positive feedback decreases. While remaining positive and finite it is reducing in magnitude.

At 12 kg/m^2 the slope looks to be about 5 W/kg. At your ‘average’ point it is less then half that.

So this is positive feedback which is itself subject to a negative feedback. The negative f/b being the masking effect as the density of water molecules rises. They become masked by others and the net effect of a give increase is less and less. This is the classic log relationship of absorption.

This shows that the positive w.v. feedback will be stronger in cooler, higher latitudes than in the warm and humid tropics.

• feedback will be stronger in cooler, higher latitudes

But the negative feedback limiting water vapor also increases as well.

25. John in Oz says:

Willis, many thanks for the time and effort you put into these articles. Always interesting, even though a lot of it goes over my head.

My questions is – “Why do you need to look for these relationships when the (so-called) ‘climate scientists’ should already know how these physical processes work and have them dialled into the models?”

26. Greg says:

Willis: “Here is the main figure from that post showing the relationship:”

Not quite, this is the same as your earlier graph but this time with a linear scale, not log scale. This is useful as a second way to visualise the data.

The caption to fig. 1 says: ” The slope of the curve at any point is 62.8/TPW (W/m2 per degree)”

This is clearly not correct. The slope of that graph varies notably. It is the slope of the log graph which is constant.

This ties in with a comment I made in the previous thread about the change in slope as TPW increases.

As TWP increases the slope decreases. This means that the magnitude of the positive feedback decreases. While remaining positive and finite it is reducing in magnitude.

At 12 kg/m^2 the slope looks to be about 5 W/kg. At your ‘average’ point it is less then half that.

So this is positive feedback which is itself subject to a negative feedback. The negative f/b being the masking effect as the density of water molecules rises. They become masked by others and the net effect of a give increase is less and less. This is the classic log relationship of absorption.

This shows that the positive w.v. feedback will be stronger in cooler, higher latitudes than in the warm and humid tropics.

• Greg says:

At the average TPW value in Figure 3 of 29 kg/m^2, this gives us a slope of 62.8 / 29.0 = 2.2 W/m2 increase in absorption per kg/m2 change in TPW.

ie about 2.2 W/kg at 29 kg/m^2

This is consistent with my eyeball slope estimate at that point above. It is about 5W/kg at TPW=12kg/m^2

The figure of 62.8/TPW has changed in the fitted log expression. Your last post had the constant as 43.5 . It seems that you are now using natural logs instead of log2. Natural logs are abbreviated ln , log on it’s own usually denotes log10. That is going to cause some confusion if anyone uses the value of 43.5 given as log instead of ln.

27. For us visually oriented learners, a diagram showing all the effects would be most helpful.

28. Jamie says:

Your math appears incorrect on figure 1. First of all…I’m not really sure where you got this….figure or the calculation for this. You’re showing the y intercept at around 40 for the dashed yellow line yet your equation shows A y intercept at -60. . But you explain that this is the portion of the olr due to water vapor. Intrinsically you’d expect the water absorption would be 0 at 0 Tpw. There is clearly some inflection point at around 9 tpw. Also the slope calculation of 62.8/tpw is incorrect. …what is the R function…..also the equation doesn’t make sense. R is just a constant.

• Greg says:

See my comments above about the base of the logs used and note that you do not see the *zero* intercept on that graph. Also R is the correlation coeff. of the data and the fitted fn ( I presume ). It has a single value, ie constant. If you don’t know what something is you really need to hold back of saying it does not make sense.

• james says:

he showing the dashed yellow line as the plot of the equation 62.8 log tpw – 60….at tpw = 1 the value should be around y value should be around 2.8. there is no value for log (0). ….the relationship between aa and tpw is probably more linear between 0 and 9. the R is not defined…and you assumed something…that’s incorrect. it’s not clear how it’s even applied. the slope of the equation is the first derivative of the equation.

which would be slope = 62.8/(tpw ln (2))

so at tpw = 29.5 the slope is 3.07 w/m^2

29. sciguy54 says:

Willis, with respect to figure 8 you wrote:

“As you can see, over much of the tropics, as precipitable water increases so does the cloud albedo”

Years ago I would occasionally fly small planes from the Mississippi gulf coast westward towards New Orleans in the afternoon. The first time I neared the end of the relatively cool waters of the Mississippi Sound and approached the much warmer swamps, marshes, and lakes of Louisiana I was alarmed by the panorama before me. The air was obviously hazy and moist, but in this case it was also dark, almost an “Edge of Night” effect, 1950s black-and-white style.

I had flown in the shadow of clouds before but this was different. After some thought I attributed it to many miles of tall cumulus rejecting the slanting afternoon sunlight and the moisture and scattered rain in the air below the clouds. While I prepared for instrument conditions, once I was under the cloud deck my eyes adjusted and I was able to continue in marginal daylight VFR. Over time I understood that this was a typical condition for this locale for much of the year.

30. Bill Illis says:

Willis, you have done more science in these last two posts that the entire field has done in the last 35 years.

I’ve spent some time looking at these issues so I can provide some input.

First, you have provided real data that can be used to test the most important assumptions contained in the global warming theory – the feedbacks.

The feedbacks provide most of the warming that eventually arises from doubled CO2.

—> water vapor increase +2.0 W/m2 per 1.0C;
—> cloud albedo decrease +0.7 W/m2 per 1.0C;
—> others -0.3 W/m2 (these have changed some in the last IPCC report but the theory is still around this value.

—> TOTAL +2.4 W/m2 per 1.0C

The +2.4 W/m2 per 1.0C is an important value because this provides for just enough feedback so that one gets a feedback on feedback on feedback loop that takes doubled CO2 from its initial 1.1C increase to 3.0C eventually. The first round of water vapor and cloud feedbacks provide for another 0.71C temperature increase. That 0.71C then produces another round of water vapor increases and cloud albedo reductions leading to another 0.45C. That 0.45C provides another … and so on. After 11 rounds of feedback on feedback we get to 3.0C and there is basically no extra lift after that.

The +2.4 W/m2 per 1.0C is, thus, a VERY important number. It is “everything”. If, in reality, the feedbacks are only half of predicted, global warming would fall to 1.6C per doubling. If it were one-quarter, warming falls to 1.3C per doubling. If it were twice as big, well, global warming would then +47.0C per doubling and there would be a runaway greenhouse effect.

That is how finely tuned the chosen feedbacks are. It is the difference between benign-nothing warming and a runaway greenhouse effect.

You have a lot of data here that can test those assumptions in real life (something climate science does not seem to be able to do).

Water vapor. (Just an aside, but I think there is some inconsistency in your graphs. global average water vapor 29 kg/m2 – Reynolds SST at 29 kg/m2 is 24C or so. The most often quoted global average TPW is 24.5 kg/m2 and global average SST is about 18.0C. So I think there is an issue here).

But are you getting 7.0% increase in TPW per 1.0C (Clausius Clapeyron). Well, your data (Figure 2.) is pretty close on the Clausius Clapeyron at something around 6.5% per 1.0C. Change the formula to a simple TPW = X + (1 + CC%) * SST.

Are you getting +2.0 W/m2 per 1.0C increase. Well you data (Figure 3) is more like 3.75 W/m2 per 1.0C. If water vapor feedback produced 3.75 W/m2 per 1.0C, global warming would be 15.7C per doubling.

Cloud Albedo (Figure 7 – might be different than Cloud Albedo which I know you had shown before, but perhaps Atmospheric Absorption of downwelling Solar is close enough). Well here you clearly have and “Increased Absorption per warmer SST” – completely opposite to the theory that cloud albedo declines as warming occurs. Your data is in the range of -2.4 W/m2 per 1.0C .

Can we add them together to get a Water Vapor/Cloud Albedo feedback value ?? Well that would be +1.35 W/m2 versus global warming theory of +2.7 W/m2. Exactly Half. That would reduce global warming to 1.47C per doubling.

Really good work here. Really good.

• Bill Illis says:

Just noting that I think we “should” add these together. The cloud feedback is not really cloud albedo by itself but the combination of the change in solar reflected (albedo) and the change in long-wave downwelling of clouds. When low cloud cover declines in the theory, Albedo declines by more than the downwelling declines so it is a combination of both impacts. The measured cloud feedback is in both Figure 3 and Figure 7 of Willis’ graphs so they should be added together.

• Note that Nic Lewis calculated observational energy budget ECS using Bjorn Stevens new estimate of aerosols as ~1.5, essentially the same as your 1.47. Cross check.

• David L. Hagen says:

Re: Clausius – Clapeyron equation & saturation vapor pressure.
For those delving deeper, the most accurate Clausius Clapeyron equation I have found is by Koutsoyiannis. See:
Koutsoyiannis, Demetris. “Clausius–Clapeyron equation and saturation vapour pressure: simple theory reconciled with practice.” European Journal of Physics 33, No. 2 (2012): 295-314. http://dx.doi.org/10.1088/0143-0807/33/2/295 ITIA preprint, postprint, history
Koutsoyiannis, Demetris, (2012) Corrigendum on “Clausius-Clapeyron equation and saturation vapour pressure: simple theory reconciled with practice” European Journal of Physics 33, No. 4 (2012):1021. http://iopscience.iop.org/0143-0807/33/4/1021

• Mike Jonas says:

Thanks Willis for yet another very interesting article. I’m interested in Bill Illis’ comment “your data (Figure 2.) is pretty close on the Clausius Clapeyron …“: Eyeballing the graph, I thought the same. My understanding is that precipitation also follows the C-C pretty well (eg. work by Weiffels et al which I now can’t locate). It looks to me like the work that Willis has done would be sufficient to resolve one of the outstanding climate disagreements – the IPCC and the climate models show precipitation at a much lower rate than C-C, namely only 2-3% per deg C instead of the C-C 7%. Willis’ Figure 5 shows precipitation (rainfall) vs precipitatable water, and Figure 2 shows SST vs precipitatable water. Putting the two together should give us a very good indication of whether the models’ precipitation is reasonable. Eyeballing the graphs, I would say that dprecipitation per deg C in the real world is way higher than in the models, but I would like to see the calc done formally. Willis – please can we have just one more calc ….. precipitation vs SST. TIA.

• Mike Jonas says:

Wijfells, not Weiffels, no wonder I didn’t find it. There are several papers, here’s one :
https://www.researchgate.net/publication/252330096_Fifty_Years_of_Water_Cycle_Change_expressed_in_Ocean_Salinity
While we confirm that global mean precipitation only weakly change with surface warming (2-3% K-1), the pattern amplification rate in both the freshwater flux and ocean salinity fields indicate larger responses. Our new observed salinity estimates suggest a change of between 8-16% K-1, close to, or greater than, the theoretical response described by the Clausius-Clapeyron relation.
… but Willis’ data might resolve the issue.

Incidentally, Trenberth et al appear to contradict the IPCC/models too:
The environmental changes related to human influences on climate since 1970 have increased SSTs and water vapor, and the results suggest how this may have altered hurricanes and increased associated storm rainfalls, with the latter quantified to date to be of order 6 to 8%.“.

• Willis Eschenbach says:

Mike Jonas July 30, 2016 at 12:18 am

Willis’ Figure 5 shows precipitation (rainfall) vs precipitatable water, and Figure 2 shows SST vs precipitatable water. Putting the two together should give us a very good indication of whether the models’ precipitation is reasonable. Eyeballing the graphs, I would say that precipitation per deg C in the real world is way higher than in the models, but I would like to see the calc done formally. Willis – please can we have just one more calc ….. precipitation vs SST. TIA.

and he quotes Wijfells:

“ While we confirm that global mean precipitation only weakly change with surface warming (2-3% K-1), the pattern amplification rate in both the freshwater flux and ocean salinity fields indicate larger responses. Our new observed salinity estimates suggest a change of between 8-16% K-1, close to, or greater than, the theoretical response described by the Clausius-Clapeyron relation.”

… but Willis’ data might resolve the issue.

Here you go …

Since evaporative cooling is directly proportional to rainfall, the shape of this is the same as that of Figure 6.

You can see why there is contradiction in the field … this highlights a couple of issues. The first is the attempted representation of a complex situation by means of a simple trendline of some kind (linear, exponential, etc.).

The second is the pernicious nature of averages. It is true that ON AVERAGE there is a slight increase in rainfall with temperature … but that average conceals a period of neutral or a slightly declining relationship at temperatures below 25°, and a strong positive correlation at temperatures above 25° …

Thanks for the request, Mike.

w.

31. Leonard Weinstein says:

The Feedback Issue:

The CO2 increase alone only causes a small temperature increase per doubling, so the main disagreement between alarmists and skeptics is about the feedback.

In real physical systems, the feedback generally tends to limit increases. Otherwise any small perturbation would tend to lead to instability, and extreme conditions. this is best covered by Le Châtelier’s principle: When a system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

• Wim Röst says:

Leonard Winsteind: “In real physical systems, the feedback generally tends to limit increases. Otherwise any small perturbation would tend to lead to instability, and extreme conditions.”

WR: “the feedback GENERALLY tends to limit increases”. Very important remark!

We know an exception, climate modelling, but indeed, that’s not in the real world.

• Wim Röst says:

Sorry typo: Leonard Weinstein. Sorry!

• Leonard Weinstein says:

Wim: The cases where the negative feedback does not apply are cases where forcing exceeds boundaries and there is a shift of modes, e.g., a “tipping point”. There can be long period cases of out-of equilibrium response, or a case where external factors cause temporal or spatial shifts, e.g., ocean/ice sinks and currents storing/releasing or moving energy. I do not know of a type of case where there is simple positive feedback amplification as claimed for climate modeling.

• Leonard Weinstein says:

Negative feedback creates a condition of equilibrium (balance). Positive feedback creates a condition of hysteresis (the tendency to “latch” in one of two extreme states).

• Positive feedback creates a condition of hysteresis

It might look like that is what happens, but that is only one limited case, what it does is either oscillate at ever increasing amplitude to infinity, or until feedback stops the growth, running out of energy to sustain the growth is just one.
That hysteresis is the system transitioning from that limited condition, for instance the loud squeal of microphone speaker feedback loop that is limited by the power supply, back to normal operation when you move the mic.

32. The albedo due to clouds is significant therefore an increase in cloud coverage especially low cloud coverage is going to increase the albedo and lower the global temperatures. Further an increase in total global cloud coverage does not necessarily mean that the total water vapor in the entire atmosphere must also increase.

I think the dynamics of the atmospheric circulation/temperature structure of the atmosphere from ground level to stratosphere ,along with an increase in galactic cosmic rays (which some disagree with ) are the main factors that govern total global cloud coverage. This if true would allow total global cloud coverage to move independently or at least partially independently of total water vapor in the atmosphere.

33. Gerald Machnee says:

And I thought the debate was over!
Great to have someone who actually thinks and does the work!

34. james says:

after reviewing figure 1 more closely….it’s apparent that the equation should read 62.8 ln tpw – 60.
nomenclature is not correct. LOG is assumed log10 whereas Ln is LOGe. in that case the slope would be 62.8/tpw. …but the yellow dashed line is not properly drawn…..if it were properly drawn it would probably more closely match the data…..

35. Chris Hall says:

If you’re looking for a functional form for equilibrium water vapor pressure as it relates to temperature, here’s what I use:

P = exp(-5375.83585/T+21.2023734)

where T is in K and P is in Torr. This might be a possible starting point as a substitute for your “inverse sigmoid function”.

• Willis Eschenbach says:

Thanks, Chris. I tried your expression, but it is a lousy fit to the actual data. As I remarked above, a typical exponential function doesn’t have the kind of “hook” at the right hand end.

Appreciated,

w.

36. Craig Loehle says:

Willis:
“The blank area in the lower right corner of Figure 7 shows that above a certain sea surface temperature … it’s gonna rain and cool it down. ” should be figure 6

Fantastic post. I agree. By taking things one at a time and ignoring evaporation and changes in clouds and absorption of solar, the IPCC has assumed a positive feedback that does not exist.

• Willis Eschenbach says:

Thanks, Craig. Always more to learn …

Regards,

w.

Figure 2 looks remarkably like the Vapor Pressure vs Temperature curve for water, which follows a Clausius-Clapeyron equation: https://en.wikipedia.org/wiki/Vapour_pressure_of_water
The Log of the Vapor pressure (or, in this case, the TPW) varies linearly with the reciprocal of the absolute temperature.

Try replotting the log of the TPW (not a problem as it is always positive) against the reciprocal of the absolute temperature and looking for a linear regression…

• Willis Eschenbach says:

tadchem July 29, 2016 at 8:50 am

Figure 2 looks remarkably like the Vapor Pressure vs Temperature curve for water, which follows a Clausius-Clapeyron equation:

Thanks, tadchem, but I fear you must have missed the part in the head post where I said regarding Figure 2:

For example, the curve doesn’t match the Clausius-Clapeyron increase in water vapor.

The human eye isn’t very good at matching curves unless one is overlaid on the other …

Regards,

w.

38. Chuck L says:

Willis, my understanding is that the “runaway” part of GHG-induced global warming is caused by increasing amounts of water vapor in the atmosphere because of evaporation, and since water vapor is a more powerful GHG, we get “dangerous” global warming. From what I can find online, there is no trend in atmospheric water vapor at almost all levels except maybe at the lowest levels where there may be a slight increase. Would appreciate your and other’s thoughts. Thanks.

• Dixon says:

Because that water vapour doesn’t stay vapour – it condenses and precipitates is my 5c.

• Because that water vapour doesn’t stay vapour – it condenses and precipitates

It does every night as dew, over most of the planet.

39. Alan Robertson says:

OT, but Willis your graphs evoke the sea, being replete with eddys and waves, currents, mists and even impressions of mysterious sea creatures in motion.

40. Chris Hall says:

On second thought, your precipitable water function climbs more steeply than the function I wrote (which is the Clausius-Clapeyron equation – duh), especially for T > 20C. So there is something else going on. Perhaps there’s a bit of positive feedback in the sense that as T rises, there is more upward convection, leading to more water in the air column. Of course, this positive feedback could wind up being an overall negative feedback with regards to temperature and heat, given that there must be a huge energy cost to putting that much water in the air, and the heat has to come from cooling the surface of the ocean.

• george e. smith says:

If I’m not mistaken (I often am), the Clausius-Clapeyron equation describes the situation in a closed system at equlilibrium, based on thermo-dynamics.

The trouble with that is that the earth climate system is NOT a closed system, and it is never at equilibrium (the earth rotates).

Once evaporation happens the atmosphere moves stuff around, so that materials are transported away form the region where the equilibrium closed system was supposed to be located.

Ergo, although CC might tell you what happens in a closed iso-thermal vessel; it goes all pear shaped when you let stuff leak out of that closed system.

G

41. So … with such a large radiative feedback from water vapor, three to four watts per square metre per degree and much higher in the tropics, why is there not runaway feedback? I mean, the so-called “climate sensitivity” claimed by the IPCC says that 2-3 W/m2 of additional radiation will cause one degree of warming.

Because every night it clamps down on the absolute amount of water vapor in the air, it has to, and this process of air temps nearing dew temps slows the cooling rate.
A non-linear cooling feedback control, and it’s basically controlled by dew point temps.

42. Note the size of the cooling involved … not watts per square metre, but hundreds of watts per square metre. As precipitable water goes from about forty to fifty-five kg per square metre, evaporative cooling goes from fifty to two hundred fifty watts per square metre or more … that’s a serious amount of cooling, about ten watts of additional cooling per additional kg of precipitable water.

We can compare that to the slope of increasing water vapor radiative feedback in Figure 1. The slope in Figure 1 is 62.8 W/m2 divided by TPW, so at a TPW of 50 kg/m2 that would be about 1.2 W/m2 of additional radiative warming per additional kg/m2 of water … versus 10 W/m2 of rainfall evaporative cooling per additional kg/m2 of water.

And I this is dwarfed by the drop in enthalpy over night, it averages about 9kJ/kg from max temp to min temp over night.

I’m starting to think that the cooling rate of the atm is very high, but ultimately regulated late at night from water vapor condensing out, restricting the path to space..

• Dixon says:

I was thinking of rain (which in the tropics is the big one), but dew is a very good point, esp for mid-latitiudes.

• Willis Eschenbach says:

micro6500 July 29, 2016 at 9:55 am

And I this is dwarfed by the drop in enthalpy over night, it averages about 9kJ/kg from max temp to min temp over night.

Thanks for that, Micro. However, you need to be a bit careful there. For a given change in temperature, most of the change in enthalpy is from the change in enthalpy of the dry air. Remember that the enthalpy of dry air is 1.006 times the temperature in C. So if the temperature falls overnight from say 25°C to 15°C, with a dewpoint throughout of say 13°C (the atmosphere doesn’t gain or lose water), the change in dry air enthalpy would be 25 kJ/kg minus 15 kJ/kg equals 10 kj/kg. (Note that this would entail a change in relative humidity from ~ 50% during the day to ~ 90% at night.)

With moist air with a dewpoint of 13°, on the other hand, the entropy at 25°C is 48.8 kJ/kg, and at 15°c it is 38.5 kJ/kg. Note that as expected, the energy at both temperatures is significantly higher because of the latent heat. Note also that the difference between the two is 10.3 kJ/kg … and we know that 10 kJ/kg of that is from dry air.

Even if we get a couple of degrees below the dewpoint, so there is condensation, the numbers are still not impressive. If the temperature drops from 25° to 15° and the dewpoint is 17°, the total change is 14 kJ/kg … but 10 kJ/kg of that is still from the dry-air change.

Finally, I took a look at ten years of hourly data from Canada, and calculated the moist air and dry-air entropy. The correlation between the moist air entropy and the dry-air entropy (which is temperature times 1.006) is very close, with an R^2 of 0.9952 … so given that, it seems that the claim that enthalpy is a much better metric than temperature doesn’t hold up. According to the data I’ve looked at, entropy can be estimated with great accuracy as a linear function of the temperature, so either one would be equally good as a metric.

My best to you

w.

• However, you need to be a bit careful there. For a given change in temperature, most of the change in enthalpy is from the change in enthalpy of the dry air. Remember that the enthalpy of dry air is 1.006 times the temperature in C.

Thanks Willis.
I’ll have to dig into exactly what I’m doing in my code. The average change between min and max temp is ~18F, which likely is just about the same average as the change in enthalpy of 9,000 that I get, and you quoted as for a 10 C change.
I also took some 5 minute data from my weather station and I did eliminate a “enthalpy barrier” as why cooling slows down.
It’s possible ground and air temps being nearly the same slows cooling even though the radiative surface the ground radiates to is still an equally large temperature differential as when the cooling rate is still high.
Or maybe it is an optical effect, it’s gotten a bit humid the last couple nights and we had fog, optically thick.
Did you find the solar forcing data you were looking for?

• I had thought I might have included only the water vapor entropy in my equation, leaving the dry-air entropy out. But I hadn’t, I just got back and took a look. It’s the entropy of both the air and water.
I did make a function that’s moisture entropy only, but it won’t show up until I rebuild all the reports.
But as I mentioned, after looking at 5 minute data all the energy loss is proportional to temperature change, just like you you pointed out (thanks, I saw the term but didn’t realize it’s significance in the results).

But it also shows the high rate of energy loss from the surface when the sun goes down, While we had really high temps this year, the rates of cooling will equally be high.

43. I was seeking the mathematical origin of the 342 +/- W/m^2 ToA ISR/TSI shown on so many atmospheric “heat” balances esp. Trenberth’s Figure 10 so I ran the numbers. Solar surface temp, S-B BB, radius/spherical surface area, orbital distance, orbital spherical area, etc. Yep, with an emissivity of .95 came up with an “average” 1,368 W/m^2 circular surface, 342 W/m^2 spread across the full atmospheric 100 km ToA spherical surface.

When I entered apehelion and perihelion distances I discovered that the total fluctuation of the solar “constant” from closest, 1,390 W/m^2, to farthest, 1,345 W/m^2, was 45.7 W/m^2, +/- 22.9 W/m^2. Wow, that’s one heck of a swing, especially considering IPCC’s 2.5 to 8.5 RPs. Makes those look pretty insignificant.

From what I can tell orbital eccentricity is dismissed as trivial, but 45.7 W/m^2 is far from trivial, it’s greater than any of IPCCs RCPs. IMHO it’s eccentricity that creates the seasons, axis inclination plays a minor role. Combine this large natural power flux fluctuation of eccentricity with the natural fluctuations in albedo, vegetation, and ocean processes and mankind’s CO2 contribution and RFs seem trivial in comparison.

• axis inclination plays a minor role.

No, this controls the length of night, as soon as the Sun comes up temps start to rise, and as soon as it sets it starts to cool.

• Certainly the rotation modulates the daily swings, but when it comes to seasons I think the 45 W/m^2 eccentricity gorilla runs the show. These power fluxes aren’t real “heat” balances anyway. A 24 hour true heat balance based on dark and night, thick troposphere at equator, thin at poles, oblique angle areas, etc. & Q = U A dT would be more accurate & also extremely complex. How would one determine U, thermal conductivity for the atmosphere w/ clouds, moisture, RH, albedo, etc? And consider how thin the atmosphere is. Earth’s radius if 6,370 km, ToA 100 km, troposphere, 12 km. An onion skin wrapped around a soft ball.

• Certainly the rotation modulates the daily swings, but when it comes to seasons I think the 45 W/m^2 eccentricity gorilla runs the show.

I respectfully disagree.
I calculate SW for every surface station in the GSoD data set, in this case by day. For each day of the year, for each year since 1940, for the average TSI from 1978-2014, so an average about 98,000 station records per day for all of those years for every station, for the latitude band of 40 to 60 N, the minimum SW for a day is 524Whr/m-2, the maximum is 6304Whr/m-2, way more than 45W/m-2

• Oh, Average daily rising temp is (NH18.22F to SH18.71F) ~1/2F more in the southern hemisphere as compared the northern hemisphere.
I also get a SW climate sensitivity from about 0.003F/Whr/m-2 to 0.014F/Whr/m-2, in the extra-tropics.

• What’s “SW”? And you are mixing units, apples and oranges, Wh and W. Wh is energy, W is power, energy over time. Need to agree on terms and definitions. How is this measured? Horizontal surface or tilted to latitude?

• What’s “SW”? And you are mixing units, apples and oranges, Wh and W. Wh is energy, W is power, energy over time. Need to agree on terms and definitions. How is this measured? Horizontal surface or tilted to latitude?

I think I lost my post, so…
SW is short wave, or solar.
I think W is energy and Whr is power. 1W = 1J , 1W for 3,600 seconds is 1 Whr.

And tilted for latitude.

44. cba says:

willis,

I like what you’ve done here. seems there’s lots of comments looking at only portions of what’s going on though. keep up the good work.

45. Michael Carter says:

Salv wrote: “The albedo due to clouds is significant therefore an increase in cloud coverage especially low cloud coverage is going to increase the albedo and lower the global temperatures”

This principle has been described by a number of posters on this site. While it obviously holds a degree of truth I hope to demonstrate that it is ‘not always so”. Clear skies can lead to reduced measured LST in some latitudes. We should not overlook the fact that night follows day. What occurs at night?

When the right conditions occur I will demonstrate using New Zealand Metservice 4-day forecasts that coldest temperatures in New Zealand occur during clear skies, especially in winter. There are no such conditions at the moment but they will return within a week or two. This result in frosts. The day and night temperatures are commonly 3-4 degrees C lower during clear skies as compared to cloudy skies

The pattern consists of clear skies, warm sunshine, and cool shade – followed by rapid dew fall after dark and freezing. There is a net loss of soil moisture and temperature within days, which stops grass growth. Conversely, as long as there is cloud and rain, soil temperatures are such that some grass growth continues. The energy input during the sunny days is not enough to offset night-time freezing

There is usually little wind during these phases resulting in an inverted atmospheric temperature gradient at night. Frost forms in low-lying areas and one can feel the temperature change (warming) by walking up as little as 100 m in elevation

Studying the effects of water vapour and cloud in the tropics is one thing but we also need to consider the all-important higher latitudes that are instrumental in the cooling of the tropics through energy transfer – and the influence of night when no albedo is occurring

It’s never that simple :-)

46. Don V says:

In reading both Willis analyses, seeing this data, and thinking about all the comments I have come away with two thoughts that I haven’t seen talked about. First, I wonder why everyone working in this field always tries to reduce data to a single equation that they hope somehow captures and actually represents the real-world process(es) occuring. What I see when I look at the data in the above scatter plots isn’t one catch-all equation . . . rather it’s what data looks like when a cycle is occurring . . . a cycle that includes at least one factor that causes hysteresis or a hysteresis loop . . . . perhaps caused by the fact that when water is leaving one state and moving to another (and perhaps even a third), that it carries energy with it. And then when it gives up that energy by returning it to space it returns by a different “path”. What I see in all that data are the hallmarks of a sigmoidal multi-phase “cooling” cycle with hysteresis typical of a “governor” process that is “working” to regulate the temperature. In some of that data I can even imagine that I see water -> vapor -> condensate -> freeze upward path with return paths from condensate and freeze back to previous states. And in each case more energy is transported up by convection than by radiation! At the very least, the data shows that the radiative energy reflecting, cooling and active heat transport governed by TPW is much more a factor in modulating our surface temperature than CO2!

My second thought is, I am very intrigued about Dr. Pollard’s theory that each little condensate droplet in clouds sucks up IR in the process of forming their liquid crytal layer and droplet surface charge, and my thoughts are: Has that IR sucking process been captured in any of the climate “models”? and Has anyone used the IR images from space of clouds and the “active transport” of energy UP by water to correct the estimates of how much of the energy balance is actually being carried out by the water cycle? After all the data also has shown that changing CO2 concentration has had an almost immeasurable cause/effect relationship for the last 18 years. Perhaps the “missing heat” has actually been transported off the planet quite effectively by the water cycle “heat-governor”??

47. I have not read all the comments but I noticed that in some commentators asked for the energy flux values for all-sky and cloudy sky. I have carried out a research work, where I have constructed the energy balance for all-sky, clear sky and cloudy sky conditions. I think it is the only presentation covering all these conditions.
There are some additional features not found in any other presentations: 1) The SW radiation reflected from the atmosphere is divided between the clouds and the air, 2) The absorption of SW radiation is divided between the clouds and the air, 3) The reflect SW radiation is partly absorbed by the clouds (in all other presentations the reflected SW radiation from the surface is capable to transmit through the clouds without losing its intensity). The figures attached to each flux are in order all-sky, clear sky and cloudy sky.

• Willis Eschenbach says:

ave, thanks for a very interesting global energy balance plot. I will have to take some time and check it against the CERES data but for what it is, it looks reasonable at first blush.

However, like the canonical Kiehl-Trenberth global energy balance, it suffers from a fatal flaw … it doesn’t balance. In fact, no model of the single-atmospheric-layer type can balance. There’s not enough heat gain in a single-layer greenhouse to match the real earth.

This lack of balance in the K/T global energy balance was one of the things that drew me into climate science.

Here is the K/T model of the globe …

And here is the model I developed over a decade ago.

Note that unlike the K/T model, mine is in balance. By that I mean that each level radiates the same amount upwards and downwards, and the total entering each level is the same as the total leaving it.

I suppose I should write up a post on this … but first I have to study up on your lovely graphic and see if I might be able to include it in my model …

Regards,

w.

• Wim Röst says:

“I suppose I should write up a post on this”

Yes, please do! I think all of us will read the post with great interest.

• Diagrams for average atmospheric energy transfer bookmarked.

Thank you.

• Willis Eschenbach says:

Nicholas Schroeder July 30, 2016 at 1:15 pm

The 321 upwelling/downwelling/”back” radiation is bunk.

Uncited, unsupported, unexplained drive-by comments are bunk.

In fact, you can use a cheap IR thermometer to determine that the temperature straight up is not zero degrees … and how does an IR thermometer measure that?

By measuring the downwelling infrared radiation that you claim doesn’t exist … go figure.

w.

48. The notion that 15 C, 288 K, upwells 390 W/m^2 is bogus. In Trenberth’s Figure 10 et. al. power flux diagrams all of the W/m^2 are accounted for w/o the upwelling/downwelling/”back” radiation which then has to appear out of nowhere.

The GHGs exist in the mid to upper troposphere where it is cold, -20, -30, -40 C. NASA defines ToA as 100 km and the point where radiation balance occurs. That’s a long way from a 15 C surface measured 1.5 m above land.

Because of their low density gases have extremely low emissivity and together with the temperatures mentioned above are not capable of re-emitting a significant fraction of the hallucinated upwelling amount.

For instance S-B BB W/m^2 at:

C K 1.0 0.1 1.0(25%) 0.1(25%
-20 253 232.31 23.23 58.08 5.81
-30 243 197.70 19.77 49.43 4.94
-40 233 167.11 16.71 41.78 4.18

These amounts radiate in all direction, not just back to earth, say 25%. There is no way the downwelling radiation amounts come close to balancing the upwelling 390 W/m^2 and cannot perpetuate this GHE perpetual heat loop.

This GHE theory of upwelling/downwelling/”back” radiation keeping the surface warm is as bogus as cold fusion, phlogiston, Santa Claus and “reflective” steel/glass domes.

The surface is warm thanks to Q = U * A * dT, just like the insulated walls of a house.

• BTW the S-B ideal BB radiation equation applies only in a vacuum. For an object to radiate 100% of its energy there can be no conduction or convection, i.e. no molecules or a vacuum. The upwelling calculation of 15 C, 288 K, 390 W/m^2 only applies/works in vacuum, e.g. ToA.

• Willis Eschenbach says:

Nicholas Schroeder July 29, 2016 at 3:01 pm

The notion that 15 C, 288 K, upwells 390 W/m^2 is bogus.

Nicholas Schroeder July 30, 2016 at 9:49 am

BTW the S-B ideal BB radiation equation applies only in a vacuum.

For under a hundred dollars you can buy an “IR Thermometer”. You point it at something and it will tell you the temperature of that thing. And provided it is properly calibrated using the emissivity of the object, it will be very accurate.

Now, these thermometers work by using the Stefan-Boltzmann equation which relates temperature and thermal (IR) radiation. It states the following:

Radiation equals 0.000000567 times emissivity times Temperature to the fourth power

Using that S-B equation, the IR thermometer measures the radiation and converts it to temperature. If you point the thermometer at an object that it is at say 15° C, it measures the IR, converts that to temperature using the S-B equation, and voila! The readout says 15°C

So … care to explain to us all how the S-B equation “applies only in a vacuum”?

And just why do you think that it is “bogus” that a blackbody at 15°C radiates 390 W/m2? You can measure it, and it has been measured thousands of times.

Best regards,

w.

49. 340 W/m^2 ISR arrive at the ToA (100 km per NASA), 100 W/m^2 are reflected straight away leaving 240 W/m^2 continuing on to be absorbed by the atmosphere (80 W/m^2) and surface (160 W/m^2). In order to maintain the existing thermal equilibrium and atmospheric temperature (not really required) 240 W/m^2 must leave the ToA. Leaving the surface at 1.5 m (IPCC Glossary) are: thermals, 17 W/m^2; evapotranspiration, 80 W/m^2; LWIR, 63 W/m^2 sub-totaling 160 W/m^2 plus the atmosphere’s 80 W/m^2 making a grand total of 240 W/m^2 OLR at ToA.

When more energy leaves ToA than enters it, the atmosphere will cool down. When less energy leaves the ToA than enters it, the atmosphere will heat up. The GHE theory postulates that GHGs impede/trap/store the flow of heat reducing the amount leaving the ToA and as a consequence the atmosphere will heat up. Actually if the energy moving through to the ToA goes down, say from 240 to 238 W/m^2, the atmosphere will cool per Q/A = U * dT. The same condition could also be due to increased albedo decreasing heat to the atmosphere & surface or ocean absorbing energy.

The S-B BB temperature corresponding to ToA 240 W/m^2 OLR is 255 K or -18 C. This ToA “surface” value is compared to a surface “surface” at 1.5 m temperature of 288 K, 15 C. The 33 C higher 1.5 m temperature is allegedly attributed to/explained by the GHE theory.

Comparing ToA values to 1.5 m values is an incorrect comparison.

The S-B BB ToA “surface” temperature of 255 K should be compared to the ToA observed “surface” temperature of 193 K, -80 C, not the 1.5 m above land “surface” temperature of 288 K, 15 C. The – 62 difference is explained by the earth’s effective emissivity. The ratio of the ToA observed “surface” temperature (^4) at 100 km to the S-B BB temperature (^4) equals an emissivity of .328. Emissivity is not the same as albedo.

Because the +33 C comparison between ToA “surface” 255 K and 1.5 m “surface” 288 K is invalid the perceived need for a GHE theory/explanation results in an invalid non-solution to a non-problem.

References: ACS Toolkit, Trenberth et. al. 2011 “Atmospheric Moisture Transports …….” Figure 10, IPCC AR5 Annex III, http://earthobservatory.nasa.gov/IOTD/view.php?id=7373

So what am I missing here? This went to the ACS authors of the tool kit & received zero response.

• Wim Röst says:

“340 W/m^2 ISR arrive at the ToA (100 km per NASA), 100 W/m^2 are reflected straight away”

WR: Strange. at a height of 100 km, 100W/m^2 are reflected straight away? Who found that out? Not 98,9 or 103,6 or 89,7???? Or 101,23478? And exactly 340 W/m^2 ISR arrive at ToA (100km)?

• Based on Trenberth’s Figure 10 and typical of power flux graphics. See listed references. Not my idea, I wouldn’t do it that way.

TSI measured perpendicular to solar radiation is about 1,368 W/m^2. That’s spread over a circular cross sectional area. Spread same energy over and perpendicular to sphere, area of sphere is 4 times area of disc. Divide 1,368 by 4 = 342 – Presto magico!

Yeah, there are substantial uncertainties noted on Figure 10, orders of magnitude greater than IPCC AR5’s CO2 RF of 2 W/m^2.

• Wim Röst says:

Nicholas Schroeder July 29, 2016 at 6:32 pm
“Yeah, there are substantial uncertainties noted on Figure 10, orders of magnitude greater than IPCC AR5’s CO2 RF of 2 W/m^2”

WR: When those data have that high uncertainties – “orders of magnitude greater than IPCC AR5’s CO2 RF of 2 W/m^2” – why is anyone still using them?

• In order to maintain the existing thermal equilibrium. and atmospheric temperature (not really required) 240 W/m^2 must leave the ToA.

TOA over any one area can not be in equalibrium, for instance in the NH during the summer there’s hours more sun than night, and it changes every day. And you pointed out 45W/m-2 (55?) In the orbit, it is never in equalibrium.

• The power fluxes, W/m^2, graphics I was referencing don’t account for day/night, oblique sunlight at poles, etc. The 45 W/m^2 is perpendicular to the entire atmosphere’s surface. As I mentioned if one wants a real heat balance that’s a whole ‘nother and quite complicated discussion.

Had to look back at previous exchange.

“…the minimum SW (Solar Wattage = luminosity?) for a day is 524Whr/m-2, the maximum is 6304Whr/m-2, way more than 45W/m-2.”

So the 524 and 6,304 Wh/m^2 were collected over how many hours? 8? 12? 24? What was the peak W/m^2? Average? Was this a tracking instrument maintaining perpendicularity to ISR? The power flux diagrams everyone uses don’t do any of this.

The power flux diagrams assume about 30% albedo. If your instrument sees clear skies, any comparison goes right out the window.

• Those were calculated values for an entire 24hr period. And the only point is that the changes in the length of day cause a far higher change in forcing than 45W/m-2

• george e. smith says:

What you are missing is that TSI (annual orbital average value) is about 1362 w/m^2; it is NOT 342 W/m^2.

As a result, when the sun rises in the morning, the earth system, surface and atmosphere (oceans too) warm rapidly above their overnight Temperatures, before sunrise.

At 34 2 W/m^2 continuous all over the earth, this planet will NEVER reach an average near surface Temperature of 288 K, which is what all respected sources say it is.

I can’t observe an average of anything, so I don’t pay any attention to averages.

Neither does planet earth; it has no idea what the hell you are talking about. It responds immediately to whatever happens, even as fast as 10^-43 seconds. After that, the next think that can happen does happen.

G

50. Steve Oregon says:

Do any of the lofty climatologists participate in any open public forum collaborations like this?
It seems like the most productive type of peer review with the highest potential for advancement and discovery.

Maybe Trenberth, Mann, Schmidt et al do not find the idea team worthy?

• Willis Eschenbach says:

There have been a few mainstream climate scientists who have been willing to put their ideas out for public examination, and I commend them for it.

However, such free examination in the bright light of the public forum is anathema to Trenberth, Mann, Schmidt, and many of the other leading lights of climate alarmism. They do their best work in the dark.

w.

51. charles nelson says:

There seem to be many and varied confused views here. And as usual many are dressed up in the phoney mathematics of radiative imbalance caused by ‘greenhouse gases’, a misnomer if ever there was one.

The first major confusion seems to be with regard to how actual (real) heat behaves on the planet.
It should be obvious to anyone (other than a Warmist Climate ‘scientist’) and by ‘obvious’ I mean clearly apparent from the observation of real time data…that ocean currents are the dominant form of heat transference on the planet and that water vapour is the dominant medium of heat transference in the atmosphere.
But don’t take my silly old word for it!
Get on the satellite viewers and look at the ocean currents check out that Gulf Stream…literally a river of warm water cutting northwards from the mid atlantic towards the Arctic… The ocean currents transport heat to the POLES.

Now whilst you’re on the satellite viewers, go check out the excellent University of Wisconsin Tropical Cyclones page…there you will be able to see (in actual real time) rapid convective cooling of equatorial oceans and land areas…
30˚C at the surface and MINUS 50˚C or colder at the tops of the clouds. (which are trailing off eastwards and southwards/northwards, heading to guess where? That’s right the POLES.
Now before any smart alec points out that the latent heat of evaporation is equal to the latent heat of condensation (yawn) let ME point out that the cooling process very much depends on WHERE the two take place.
The heat being collected by the water vapour is at the surface…the heat being rejected by the water vapour is at 30 or 40 thousand feet…where it is permanently and eternally COLD.

The day some clever Warmist, shows me evidence that the earth’s atmosphere is warming rapidly at 30,000 feet is the day I will start getting mildly alarmed.

Tell you what would alarm me though, be an announcement that the Earth’s tilt angle had shifted.
That would scare me.

• Mark - Helsinki says:

“There seem to be many and varied confused views here. And as usual many are dressed up in the phoney mathematics of radiative imbalance caused by ‘greenhouse gases’, a misnomer if ever there was one.”

Thank you, I thought I was going insane.

• Mark - Helsinki says:

All these calculations of radiative forcing and evap are useless if you cant show how winds are interacting with sufficient resolution.

Feeding Intellectualism little more.

• Mark - Helsinki says:

As Lindzen says, these “are all thing being” equal base calculations and they are entirely misleading.

Too many are jumping down the rabbit hole

• Mark - Helsinki says:

but it looks good on paper for sure ROFL

52. Brian R says:

I’ve noticed the weather station near me on weatherunderground.com also shows sun intensity in wats/m2. It’s been interesting to see how it changes from day to day and within the day because of cloud cover.

53. David Dibbell says:

Willis, thank you for your analysis and thought-provoking points. Many good comments too, e.g. Don V, Dr. Spencer, ristvan, etc.

I would like to offer something to deal with the no-net-cooling-from-precipitation idea. Sure, the outward emission of LWIR is the only final flux which rejects absorbed heat from the earth/atmosphere system. But the system is equipped with a variable-geometry, variable-altitude, variable-composition, variable-temperature upper “emitter” which does most of this work. This emitter is supplied with heat energy and IR-active substances (water, CO2, etc) from below, at variable velocities and rates. Absorption of outward/upward emitted LWIR by the overlying atmosphere diminishes with altitude, so getting the mass and heat to higher altitude, with favorable geometry and emission physics, is what it’s all about for best emitter performance.

Mechanically, buoyancy drives heat and mass upward based on the influence of temperature and water vapor on CAPE, and gravity is kind enough to return it all for free, much of the water having changed phase perhaps numerous times. So convection and precipitation dynamics must have a huge influence on the “emitter” output. An example: imagine an isolated cumulonimbus cloud reaching to 50,000 feet. The geometry of LWIR emission from the cloud is not just upward but outward, and the low-absorption path to space opens up dramatically.

Please keep up the good work about water vapor, the idea of a governed system, and the power of emergent phenomena.

• charles nelson says:

Elegantly put…and essentially correct.

54. TDBraun says:

If the system is “circular”, might it be possibly better to think of it as “recursive”, like a fractal?

55. Steve Fitzpatrick says:

Nick Stokes,
Yes there is no net cooling from evaporation if you look at the entire atmosphere. But there is a large transport of heat from low altitude to high, where it is far easier for that heat to escape to space. There most certainly is net evaporative cooling at the surface, which is where most solar short wave energy is absorbed, and where rain falls to replace what has evaporated. You can, of course, focus on the ” whole atmosphere”, but the whole atmosphere is not where warming is measured by surface thermometers, and, we should remember, is where people live. Willis raises a perfectly legitimate question: is the net feedback from increasing water vapor in the atmosphere positive or negative with respect to surface temperatures? Seems to me Willis presents data which support a likely net negative feedback, at least over the oceans, when everything is considered, including clouds. Land may be somewhat different since available water is usually not unlimited as in the ocean.

• Steve,
Yes, I agreed that more evap/rain transports more heat. But the statement of Willis that I was commenting on was quite quantitative:
” that’s a serious amount of cooling, about ten watts of additional cooling per additional kg of precipitable water”
And my query was, where does that come from? It’s a lot. Willis gives average PW value at 29kg/m2. Average wv transport is 80 W/m2 (Trenberth). So it would be a little surprising if 1 kg increased transport by 10 W/m2. But he doesn’t say transport – he says cooling, which is something else again. He has cited just TRMM data for rainfall, along with the SST data and PW. It looks a lot to me as if he has just multiplied mass of rain increment by specific heat. That is also where Trenberth’s 80 W/m2 figure comes from. But Willis then treats that not as a local transport (most condensation occurs at relatively low levels of the troposphere) but as a feedback to be offset against forcings. For that, it would have to be heat taken in total from the atmosphere, not just moved.

• charles nelson says:

Amazing that everyone understands why the desert gets cold at night, but seem to be stuck on why the upper atmosphere should be so cold!

• Steve Fitzpatrick says:

Nick Stokes,
The increase in PW is only important to the extent that it relates to the rate of rainfall. PW over deserts is non-zero, but rainfall is near zero. Over the tropical ocean, a 1kg/m^2 increase in PW does appear to strongly influence rainfall, and net evaporative cooling at the surface (considered globally) is just proportional to the total rate of rainfall. This is a not-trivial issue for sure. The net positive feedback for GCM’s is in part due to their prediction that PW will increase globally, restricting heat loss to space, without a large increase in rainfall (and its associated surface cooling). When Mears et al suggested that *measured* rainfall did in fact appear to increase with temperature in proportion (roughly) to the increase in water vapor pressure with temperature, the modelers were, shall we say, less than very happy. Much like they are less than very happy with balloon and satellite measurements of tropospheric temperatures. There is a repeating pattern here that you might notice…. and it seems to me related to political motivations.

• Shawnhet says:

I believe I partially agree with your general point here – however, doesn’t it have to be true that at any given point in the atmosphere, *on average* doesn’t more heat have to move upwards than downwards. OTW, could not effectively cool. IOW, if we increase evaporation by 10W/m2, doesn’t that have to cool the surface by at 5W/m2?

Furthermore, for the portion of the 10 W/m2 that does, in fact, return to warm the surface, wouldn’t that have to show up as “backradiation” which amount is already accounted for in the positive side of the WV feedback?

Cheers, :)

56. Just some general comments about the energy balance and radiation fluxes. I refer to my energy balance figure (aveollila at 1:40). The energy balance must be calculated separately for the surface, for the atmosphere, and for the TOA. As in all energy balance calculations the incoming and outgoing energy fluxes must be the same; that is why it is called a balance. The energy balance values for these three places are for all-sky conditions (W/m2): TOA 237.8, atmosphere 511.5, and surface 554.2.

As always, the downward LW radiation flux (Ed = 344.7 W/m2) from the atmosphere at the surface, creates comments that it is not possible. It is simply true therefore that 1) it can be measured, and 2) it can be calculated by spectral analysis methods as I have done. Also the radiation flux (Es = 395.6 W/m2) emitted by the surface can be calculated and can be measured. Theory and the practice are so close to each other that there is no conflict.

One more comment about the absorption caused by the GH gases in the atmosphere. In my figure it is marked by the acronym Ag. It is in in clear sky conditions 310.9 W/m2 and in this case it is based on the spectral analysis calculations. There is a kind of confusion about this absorption, because Willis has used the term “The atmospheric absorption” being about 155 W/m2 (estimated by from Figure 1) in the normal atmosphere. In my energy balance presentation this figure is the difference between the surface emitted flux 395.6 and the OLR flux at TOA 237.8 making 157.8 W/2.

I include also a figure below showing how quickly the absorption by GH happens in the atmosphere: 95 % below 2 km altitude. Therefore it does not make sense to talk about how much there are GH gases above the troposphere because 98 % of GH absorption happens below 11 km. And this absorption is the key in GH phenomenon.

Here is also a simplified presentation about the energy balance showing only the all-sky values. They very close to the values of NASA. The only big difference is in a LW radiation flux going through the so called “atmospheric window” without any absorption directly into the space.

More slideshow presentations at my website http://www.climatexam.com

• charles nelson says:

“Here is also a simplified presentation about the energy balance showing only the all-sky values.”

I think that’s where you start going wrong in your thinking…I mean the ‘all-sky’ bit.
It may be mathematically calculable but from an engineering point of view it’s completely meaningless.

• Charles. If you say that the energy balance is meaningless, you have to to specify it more accurately. My basic university training is in engineering concerning chemical processes, mechanical processes, energy and mass transfer processes, and process dynamic et automation. The basis tool of engineering is to find out the mass and energy balance for every new process under design. If you do not know these elements of the process, you are are lost.

• Dixon says:

aveollila: that diagram shows where climate went off the rails. Applying steady-state theories that are reasonable in fast kinetic systems, to clear sky conditions in order to look at climate might have been reasonable in the 80s to look at crude effects of GHGs, but the earth on a daily basis is *never* in equilibrium thermally at the surface, nor anywhere else. That’s why you have a daytime max and a daytime min temp and those extremes don’t usually coincide with solar noon and solar midnight. Averaging might fool some people into thinking they have equilibrium, but they don’t, and for the question of whether CO2 is important in decadal climate, that matters. A lot.

• Dixon. When we talk about the climate change, the minimum period to detect the global temperature change is the solar cycle, in average 11.2 years. The warming effects of GHGs for shorter period or even for cosmic forces does not work. The changes are so small that they are impossible to measure. You may calculate the warming effects of GHGs or sun irradiation changes or cloudiness changes, but it does not lead to anywhere.

That is why the energy balance values are based on the yearly averages concerning flux values. In many cases the measured flux values are averaged even for a decade. It is possible to use shorter time periods but they are not useful in the case of climate change.

57. Willis Eschenbach says:

Roy Spencer July 29, 2016 at 4:26 am

except, as Nick Stokes points out above, the cooling at the surface caused by evaporation and convective rainfall activity is exactly balanced by heating in the upper troposphere where the condensation takes place. There is no net cooling of the climate system.

Dr. Roy, always good to hear from you. A couple of comments.

First, a thousand metres below where I’m standing it’s about 30°C warmer than where I am. And three thousand metres straight up it is freezing. When I’m sailing comfortably on the tropical ocean, a thousand metres down it’s quite cold.

So what?

The issue is and always has been the temperatures and conditions here on the surface where we live. And as a result, it doesn’t matter if the cooling of the surface is “exactly balanced” by things happening where we don’t live. The dangers hyped by the alarmists are about temperatures on the surface. And unless a change makes something happen here on the surface, we’ll not even notice it.

Next, it’s not true that “There is no net cooling of the climate system” due to “evaporation and convective rainfall activity”. I can think of several ways that those phenomena actively cool the whole climate system.

To start with, the clouds associated with “convective rainfall activity” push the albedo much higher than it would be otherwise. This reduces the amount of incoming solar energy, so the whole system cools.

Next, the overall temperature of the planet depends in part on how fast the excess energy in the tropics can be moved to the poles and rejected into space. If it cannot be moved quickly, the energy builds up and the temperature rises. But the speed of the tropics-poles transport depends on the strength of the convective rainfall activity in the Inter Tropical Convergence Zone … so that’s the second way increased convective rainfall activity brings net cooling to the climate system.

Third, as I showed above, increased evaporation increases absolute humidity, which increases the absorption of the downwelling sunlight before it reaches the surface. Obviously, this will cool the surface, but it will also in turn leave the entire system cooler than it would have been if that sunlight had reached the surface.

Fourth, the thunderstorm towers serve as “heat pipes” which move immense amounts of warm air from the surface up to the upper troposphere … and they do it without allowing the moving air to interact with the surrounding troposphere. On the way from the surface up to the bottom of the cloud the energy is moved as latent heat, which doesn’t increase radiation. Once the moist warm air enters the bottom of the thunderstorm cloud it condenses, and the latent heat is released as sensible heat. But within the cloud the radiation of the air is immediately absorbed by the water droplets that make up the cloud, so the rising column of air inside the thunderstorm tower is not interacting radiatively with the air surrounding the cloud. It’s a neat trick, the heat is kept hidden away from interacting with the surrounding GHGS all the way from the surface to high in the troposphere.

When the rising air inside the pipe finally emerges at the top of the thunderstorm towers, is is above the majority of the greenhouse gases. In particular it is above virtually all of the water vapor, but at an altitude of 15 km in the tropics, for example, it’s also above 80% of the CO2. As a result, this movement of heat will increase the rate of heat loss for the entire system.

Fifth, the winds created by and surrounding the convective rainfall activity roughen up the surface of the ocean. This increases the surface albedo of the ocean by about 10%. Since on average about 170 W/m2 of sunlight strike the surface, that would be a change of about 17 W/m2, and again this cools the entire system.

Sixth, the air descending in between the thunderstorm towers is very dry. This lack of the main GHG, water vapor, allows the more efficient loss of energy to space.

So while the most important activity from our viewpoint is the cooling of the surface, there are also a number of ways in which evaporation and convective rainfall activity cool the entire climate system.

Best regards,

w.

• charles nelson says:

There IS a net cooling of the climate system.
As the man said, I can explain it to you…but I can’t understand it for you!

• Wim Röst says:

“When the rising air inside the pipe finally emerges at the top of the thunderstorm towers, is is above the majority of the greenhouse gases. In particular it is above virtually all of the water vapor, but at an altitude of 15 km in the tropics, for example, it’s also above 80% of the CO2. As a result, this movement of heat will increase the rate of heat loss for the entire system.”

Willis, aveollila July 29, 2016 at 10:31 pm posted a figure which shows that 95% of the Greenhouse Effect happens below 2 km altitude. I cannot control that number, but I like to know at what height (for every latitude) heat transport by clouds effectively ‘pass the greenhouse blanket’, so to say. I suppose spaceward emission by molecules will become exponentionally effective at rising altitudes.

I think it is important to know. The evaporation minus precipitation figure as posted by lgl July 29, 2016 at 9:09 am shows a precipitation surplus near the poles, in the area where we normally find low pressure area’s. Those low pressure area’s are impressive upward heat transport systems as well. For an image about the extent: https://earth.nullschool.net/#current/wind/surface/level/overlay=mean_sea_level_pressure/orthographic=-46.86,-82.28,587/loc=109.193,-61.962

What I would like to know is what role those depressions play in ‘cooling the earth’. I suppose that besides thunderstorms and hurricanes they are the third upward ‘heatpipe’ system that can be activated at warming. Activated by pressure differences which arise when temperature gradients rise, resulting in more wind, more evaporation and more upward energy transport.

There is a big transport of energy poleward, by oceans and wind. That energy has to disappear spaceward. And there is a big blanket of clouds near the poles. http://www.esa.int/Our_Activities/Observing_the_Earth/Space_for_our_climate/Highlights/Cloud_cover
So, how does the poleward transported energy reaches space?

• charles nelson says:

So, how does the poleward transported energy reaches space?
I find it very bizarre that people who drone on endlessly about radiative imbalances and down welling IR etc etc…STILL haven’t figured out that there’s not much down welling anything over the Winter Poles!!!
It’s DARK and it’s COLD…any energy contained in water vapour reaching these areas…or any energy being carried there by ocean currents to these regions is lost to space because the surface temperature…minus 20˚C is not much different from the temperature at 30,000 feet i.e. minus 50˚C…the winter poles are like a direct conduit for heat loss.
Have you ever tried looking at Nullschol Earth?

• Willis Eschenbach says:

Wim Röst July 30, 2016 at 3:52 am

Willis, aveollila July 29, 2016 at 10:31 pm posted a figure which shows that 95% of the Greenhouse Effect happens below 2 km altitude. I cannot control that number, but I like to know at what height (for every latitude) heat transport by clouds effectively ‘pass the greenhouse blanket’, so to say. I suppose spaceward emission by molecules will become exponentionally effective at rising altitudes.

Good question, Wim. If you go to MODTRAN and click on the “Submit the calculation” button, on the lower right of the page you can see how each of the major GHGs varies with altitude.

Regards,

w.

• Willis Eschenbach says:

charles nelson July 30, 2016 at 4:10 am

So, how does the poleward transported energy reaches space?
I find it very bizarre that people who drone on endlessly about radiative imbalances and down welling IR etc etc…STILL haven’t figured out that there’s not much down welling anything over the Winter Poles!!!

I find it very bizarre that people make claims without running the numbers. According to MODTRAN the midwinter subarctic downwelling longwave, with the ground temperature at zero degrees, is still 163 W/m2.

And since 163 W/m2 is about the amount of sunlight striking the surface on a 24/7 basis, I would hardly call it “not much downwelling”.

w.

• Dixon says:

My apologies Willis, two very interesting posts, as always when you talk of water vapour, clouds and rain, it’s enlightening. I know you have said you like the way you fit your science around your paid work, I just hope you get the formal recognition you deserve one day. Many an academic would have retired having contributed far less than you have to the advancement of a field.

On your 4th point, could someone clever than I build a relatively simple model to look at the transport side of a tropical cumulonimbus system? We need to know how much energy is transported upwards and in what time, while considering how long it would take that equivalent amount of energy to radiate back down through a clear sky atmosphere (or even a wet cloudy one which will be even slower). The ‘345’ in aveollila’s model is (I think) a much slower process than the convection upwards that got it there (which isn’t shown).

• QUESTION- Why is it global temperatures warm during an El Nino which supports more convection in the tropics, given what you just said which I agree with? Why is that so? Thanks.

• Frans Franken says:

Willis, thanks for an again innovative post and clarifying comments. Please keep enlightening us.

https://wattsupwiththat.com/2016/07/28/precipitable-water-redux/#comment-2267291
>>>
In this comment you get to an evaporative cooling power of 80 W/m2 based on precipitation of 1 cubic metre per year. Is this 80 W/m2 a rounded figure, or else what specific evaporative heat (J/kg, cal/g) are you calculating with and where can I find it?

https://wattsupwiththat.com/2016/07/28/precipitable-water-redux/#comment-2268115
“Fourth, the thunderstorm towers serve as “heat pipes” which move immense amounts of warm air from the surface up to the upper troposphere … and they do it without allowing the moving air to interact with the surrounding troposphere. On the way from the surface up to the bottom of the cloud the energy is moved as latent heat, which doesn’t increase radiation. Once the moist warm air enters the bottom of the thunderstorm cloud it condenses, and the latent heat is released as sensible heat. But within the cloud the radiation of the air is immediately absorbed by the water droplets that make up the cloud, so the rising column of air inside the thunderstorm tower is not interacting radiatively with the air surrounding the cloud. It’s a neat trick, the heat is kept hidden away from interacting with the surrounding GHGS all the way from the surface to high in the troposphere.
When the rising air inside the pipe finally emerges at the top of the thunderstorm towers, is is above the majority of the greenhouse gases.”
>>>
This is a truly interesting mechanism. Is this your own reasoning, or did you find support for it in literature or elsewhere?
If I understand correctly, the latent heat released at condensing, which occurs from cloud bottom to top, stays within the cloud as warmed up rising air and water droplets. When this warmed up air and water mass reaches the cloud top, what exactly happens to the formerly latent heat it contains? Bearing in mind the low density of the air at cloud top altitude, is the heat mainly passed as sensible heat to the surrounding thin air, or is it rapidly shed as IR radiation with about half of it passing directly to space without heating the atmosphere?

58. Willis Eschenbach says:

Greg July 29, 2016 at 5:21 am

The caption to fig. 1 says: ” The slope of the curve at any point is 62.8/TPW (W/m2 per degree)”

This is clearly not correct. The slope of that graph varies notably. It is the slope of the log graph which is constant.

I fear you’ve misunderstood the caption. Yes, the slope of the graph varies. It varies as 63.8 / TPW. When you put in different values for TPW you get different values for the slope. That’s the whole point. Differentiating the underlying equation allows us to specify the slope at any point.

w.

• Jamie says:

Willis….you need to use the correct nomenclature in your equations. LOG is log10…Ln is LOGe. The equation should read. 62.8xLn(tpw)-60

• Willis Eschenbach says:

Jamie July 30, 2016 at 4:47 am

Willis….you need to use the correct nomenclature in your equations. LOG is log10…Ln is LOGe. The equation should read. 62.8xLn(tpw)-60

Thanks Jamie, and sorry for the confusion, but obviously we move in different circles. On my planet “Log” always means the natural log to the base “e”, with the others distinguished as Log2 or Log10 .

Well, except when I’m using Excel, which uses its own peculiar terminology for many things …

All the best,

w.

• Ed Bo says:

Jamie:

In all the programming languages and math libraries I use on a professional basis, “log” by itself means the natural log.

• Jamie says:

Willis

Well on my planet called earth we have standard notations in accordance with the ISO standard

http://physics.nist.gov/cuu/pdf/sp811.pdf Page 33 for logarithms

The reason we have something like this is because someone like me who has to write calculations all the time so other engineers can review and understand them. But someone who has like a psychology degree wouldn’t understand this

• Willis Eschenbach says:

Jamie July 30, 2016 at 3:41 pm

Willis

Well on my planet called earth we have standard notations in accordance with the ISO standard

http://physics.nist.gov/cuu/pdf/sp811.pdf Page 33 for logarithms

The reason we have something like this is because someone like me who has to write calculations all the time so other engineers can review and understand them. But someone who has like a psychology degree wouldn’t understand this

I fear you are reading something that is not there. On the page you listed there was no definition of what is meant by “log(x)”. Instead, your reference says (emphasis mine):

More specifically, the base of “log” in equations is specified when required by writing loga x …

So the question becomes, when is it “required”? And that is a judgement call. I judged it wasn’t required, but obviously YMMV …

I do find this:

A logarithm can have any positive value as its base, but two log bases are more useful than the others. The base-10, or “common”, log is popular for historical reasons, and is usually written as “log(x)”. For instance, pH (the measure of a substance’s acidity or alkalinity), decibels (the measure of sound intensity), and the Richter scale (the measure of earthquake intensity) all involve base-10 logs. If a log has no base written, you should generally (in algebra classes) assume that the base is 10.

The other important log is the “natural”, or base-e, log, denoted as “ln(x)” and usually pronounced as “ell-enn-of-x”. (Note: That’s “ell-enn”, not “one-enn” or “eye-enn”!) Just as the number e arises naturally in math and the sciences, so also does the natural log, which is why you need to be familiar with it.

Warning: If you eventually progress to much-more advanced mathematics, you may find that sometimes “log(x)” means the base-e log or even base-2 log, rather than the common log.

But as you say, “someone who has not progressed to much more advanced mathematics wouldn’t understand this” … sorry, I couldn’t resist the karmic response. Someone once described “karma” as hitting a golf ball in a tiled bathroom … and making nasty comments about my education is definitely bathroom golf.

Look, Jamie, I have no time for grammar Nazis, even in science. Lots and lots of scientists and programmers on this planet use “log” to mean the natural log, see e.g. Ed Bo’s comment just above. Me, I speak math the way that it is actually spoken by scientists and programmers, not the way some pundit claims it should be spoken. And with the exception of Excel Visual Basic, which is like programming for dummies, the programming languages I have used for the past fifty years (mainly R, Mathematica, MatLab, Fortran, and C) ALL use “log(x)” for the natural log.

So, are you gonna call the NIST swat team to teach the programmers of the world the error of their ways? For sixty years now the makers of Fortran have spread their evil lies about “log(x)” being the natural log … are you going to sit still for this insidious weakening of our mathematical moral fiber? Are you going to send nasty comments to the mathematical giant Steven Wolfram, insulting his education because Mathematica uses “log(x)” for the natural log?

What’s next? Well, there are still plenty of opportunities, lots of things you could grouse about. The NIST document you referred to says:

For example, irrespective of the typeface used in the surrounding text, “A” would be typed or typeset in:
— italic type for the scalar quantity area: A;
— roman type for the unit ampere: A;
— italic boldface for the vector quantity vector potential: A.

So … are gonna complain that I didn’t use italic boldface for vector quantities? … seriously, Jamie, you strike me as nobody’s fool, aren’t there better uses of your fine mind?

w.

• Jamie says:

Willis…

Sorry. It was page 41…..

It states that people in the physical sciences and technologies should use these terms
Of which either LOGeX or Ln x is acceptable

This is not computer programming. This is physical sciences.

And again my point your education level has everything to do with this. You don’t work the the physical sciences or have the degree…. The warmists basic assessment of you is correct.

• And again my point your education level has everything to do with this. You don’t work the the physical sciences or have the degree…. The warmists basic assessment of you is correct.

I’ll ignore the harpies whose only ability is to cut and paste, this is for climate scientists, I’ve seen their work, and while they might annotate logN correctly, some of them are clueless as scientists and would be better served by wearing a dress and shaking pom poms.

• Willis Eschenbach says:

Jamie July 31, 2016 at 6:39 am

Willis…

Sorry. It was page 41…..

It states that people in the physical sciences and technologies should use these terms
Of which either LOGeX or Ln x is acceptable

Sorry, but Page 41 doesn’t contain the word “Log” or “Ln” on it anywhere … you sure you understand this “reading” lark? The page number is the little number down at the bottom of th… aw, never mind.

This is not computer programming. This is physical sciences.

Dear heavens, you are indeed a grammar Nazi. What’s next, you gonna put up some fences to keep the riff-raff and the computer programmers out?

Here’s the thing, Jamie. Most of the texts and the studies that I read use “Log” to mean the natural log, and “log10 to mean log to the base ten.

And obviously, most of the texts and the studies that you read use “Log” to mean log to the base 10, and “Ln” to mean log to the base e. And that’s just fine.

But the language is what it is, not what you say it is. The NIST document you attempted to reference said that one should subscript the word log “as required”. In the studies I read it’s not required, because most everyone uses “log” to mean the natural log.

And again my point your education level has everything to do with this. You don’t work the the physical sciences or have the degree…. The warmists basic assessment of you is correct.

The warmists’ basic assessment of me is that like you, they can find little to attack in my scientific work, so they are reduced to making meaningless ad hominem attacks on my habits, my ancestry, my use of the word “log”, my education, my history … any and every puerile meaningless thing, but not my science.

And in the middle of all of this ankle-biting from twerps and grammar Nazis like yourself, I continue to publish in the scientific journals, where fortunately the editors sometimes still care about my science and don’t care about my education, as science should be. The Editors of Nature magazine didn’t ask me about my education before publishing my small contribution to moving science forwards … call me crazy, but I’ll take their opinion over yours.

As I said before, you are wasting your obvious talents on this meaningless attack. And you are harming your own reputation in the process. I know, for example, that when I next see your name on a comment, I’ll pay less attention to whatever it is you’re saying … and I doubt I’m the only one.

Is this what you want?

How about if you apply your obvious gifts to DISCUSSING THE INTERESTING SCIENTIFIC QUESTIONS that abound in climate like everyone else is doing, and give up this obsession with shooting yourself in the foot? You’ve succeeded. Move on.

Sincerely,

w.

• george e. smith says:

Well on the planet where I come from, the logarithmic function is taught in classes in Pure Mathematics, not in classes on computer programming.

And on that planet, the term Ln or ln is used exclusively in reference to the NATURAL or NAPERIAN logarithm where the base is (e = 2.71828…)

Just Log would refer to the function usually found printed in books to four or five or even seven digits which are logarithms to the base 10, which is what is the common base system of the real human inhabited world.

I would think that computer programmers, would use base (2) or even base (16).

Now I can’t vouch for any ” science ” discipline which uses invented terms like ” forcings ” and such.

Now one of the advantages of the Rod / Stone / Fortnight system of units, is that we use almost any number as a base, so you will find , 2, 5, 10, 12, 14, 20, 22 and so on as counting bases for various things and many more.

G

• Greg says:

Yes, apologies, I did misunderstand what you meant there.

59. Wim Röst says:

What is the effect on emission spaceward from a thunderstorm that rises above 80% of the CO2?

I suppose a calculation can be done (by someone who knows better than I how to do so) but I will do a simple attempt.

We must keep in mind a system with ‘rounds’ like in the Bill Illis’
July 29, 2016 at 6:53 am figure. (the blue one: “Initial Temperature Increase from CO2 and then how Feedbacks Lead to 3,0C of Warming”)

In this simple thought experiment energy can only be emitted upward and downward, so 50% each way. The energy of the thunderstorm is brought upwards to a level where most of the greenhouse gases are below.

1st Round. In the first round nearly all the upward emitted energy (50%) reaches space. Let’s say that all of the downward emission is captured by greenhouse gases below and is re-emitted in round 2 half upwards and half downwards.
2nd Round. Half of the below catched 50% (=25%) is emitted to space and a bit less than this 25% reaches space. The other 25% goes downwards and will be captured by greenhouse gases.
3rd Round. From the remaining 25% half goes upwards direction space etc.

Of course this is far too simple, but it must be, that by far most of the energy that is brought ‘above the greenhouse blanket’ will be emitted to space. And so effectively will cool the Earth. I am sure someone can make nice calculations on this.

60. charles nelson says:

Ah, Willis…I think we’ve finally got to the source of your confusion.
Quote:

“I find it very bizarre that people make claims without running the numbers. According to MODTRAN the midwinter subarctic downwelling longwave, with the ground temperature at zero degrees, is still 163 W/m2.

And since 163 W/m2 is about the amount of sunlight striking the surface on a 24/7 basis, I would hardly call it “not much downwelling”.

You see diddling round with meaningless numbers/averages they can get themselves in an awful muddle when confronted by actual reality.

Right now the surface temperature in Antarctica is MINUS 55˚C…the temperature at 250 hPa is?
MINUS 60˚C.

There a appears to be a ‘hole in your greenhouse’ mate!

• Willis Eschenbach says:

Charles, look up the word “subarctic”. You will find it does NOT refer to Antarctica as you seem to think.

And as for your pathetic attempt at sniping, where after foolishly thinking that Antarctica is in the subarctic you say:

There a appears to be a ‘hole in your greenhouse’ mate!

May I suggest that you take a lesson from the humble rooster, and before you start crowing you make sure it is actually dawn?

Best regards,

w.

61. MfK says:

I’ve recently started using Mathematica (with mixed results, and mixed feelings). But it has a built in function called FindFormula[data]. It takes the data list, and finds the best kind of formula to approximate the data. There are, in fact, extensions of it. The most interesting is FindFormula[data,x,n,prop], where x is, of course the independent variable, but “n” is the number of different formulas it will find, not their order. Finally “prop” is a specified property for the formula, and includes “complexity,” “mean square error”, and “score” (an internal score). It’re really quite a remarkable feature.

• george e. smith says:

Well the aim of contriving ANY formula to match to measured data is multifold.

The first aim would be to describe any measured set of numbers with a formula that has fewer independent variables than the number of measured data points.

That is always good, as otherwise the set of observed data points is the best expression of the results.

A second aim, when trying to fit a set of measured data values to a mathematical formula, is to do so with a mathematical form that describes the mathematical behavior of a proposed model of the actual physical system.

For example, I believe that the Planck Formula for the black body radiation function was concocted in just such a fashion. As far as I know, a theoretical derivation of the BB radiation law was developed somewhat later by Bose.

Planck basically made up his formula out of whole cloth, using the failed Wiens and Raleigh-Jeans formulas as a starting point.

I think it was either Bose or Einstein, or both collaboratively who eventually derived the Planck formula including the exact physically based first and second radiation constants C1 and C2.

I can’t find much reason to force fit measured data into a formula, unless that derived formula actually describes the behavior or a model of the real physical system.

The Planck formula for BB radiation, which is a totally fictitious non real physical object, is made up entirely of well known fundamental physical constants, that are extremely accurately know from painstaking experimental measurements. And yes real practical approximations to an ideal Black Body do match the Planck theoretical formula with remarkable accuracy.

That blows my mind, that something completely fictitious, which cannot possibly exist, can be approximated quite accurately so that its fictional theory can be used as a useful tool for study and research.

G

62. Frank says:

Willis: You may be able to fit many functions to a plot of some observable variable vs absorption (or some fraction of such a plot), but such as a mathematical relationship doesn’t tell you anything you didn’t already know. Purely mathematical relationships have no limited value in science. Understanding is gained when that mathematical relationship arise from a theory about the fundamental behavior of water vapor and/or radiation. The greenhouse effect (aka atmospheric absorption) is a complicated phenomena involving absorption, emission, the lapse rate, and competing GHGs. Radiative transfer is calculated by numerically integrating a differential equation (the Schwarzschild eqn) that applies to radiation traveling through a non-scattering atmosphere (and other materials). There is no analytical formula that describes the GHE.

The behavior of water vapor is somewhat simpler. The C-C eqn tell us how saturated water vapor pressure changes temperature. If convection didn’t exists, you might expect diffusion to eventually saturate all of the atmosphere – based on Ts near the surface and decreasing with altitude due to the lapse rate. That would be the simplest possible model. In the real world, convection carries some air upwards above the point where it precipitates and dries. After radiatively cooling, that dry air returns to the surface. This overturning make the relative humidity in the boundary layer over oceans about 80%, not 100%. Since convection is difficult to model, that means the relationship between Ts and TPW is unlikely to be a simple one.

If you plotted the expectations of the C-C equation for various relative humidities on the same graph as your TPW vs Ts data, I think you would find that TPW decreases with surface temperature faster than saturation vapor pressure. We also know that the amount of convection decreases with lower surface temperature and relative humidity falls at higher altitudes.

My guess is that the relationship you found between G (atmospheric absorption) and log2(TPW) is an emergent property of a complicated climate system, not something that will eventually have a simple explanation.

• Willis Eschenbach says:

Frank July 30, 2016 at 7:34 pm

Willis: You may be able to fit many functions to a plot of some observable variable vs absorption (or some fraction of such a plot), but such as a mathematical relationship doesn’t tell you anything you didn’t already know.

Yeah, that damn mathematics nonsense is useless, I don’t know why they even teach it to kids these days … wait, what?

In fact, the mathematical relationships allowed me to differentiate the function and thereby calculate the water vapor radiative feedback, but don’t let that get in the way of your aversion to math …

My guess is that the relationship you found between G (atmospheric absorption) and log2(TPW) is an emergent property of a complicated climate system, not something that will eventually have a simple explanation.

Bad guess. The logarithmic relationship is predicted by the physics of the situation. That is why it applies equally to water vapor and to CO2.

w.

• Frank says:

Willis wrote: “In fact, the mathematical relationships allowed me to differentiate the function and thereby calculate the water vapor radiative feedback, but don’t let that get in the way of your aversion to math …”

Frank replies: FWIW, I love math. However, you don’t need a to fit a function to data to calculate a derivative. The derivative at any point is merely the slope. The same information is present whether I present the relationship between two variables as a table of values, a scatter plot, or as an equation. One can say the same thing about a table of differences/derivatives or sums/integrals.

Or consider the height vs time data you might get from a ping pong ball (where air resistance is important) dropping off a roof before and after injecting water to change its weight. One could fit a lot of functions to that data without ever stumbling on the right principles: 0.5gt^2 for part and air resistance that varies with the square of the velocity.

Or consider Kennan’s criticism of fitting warming data with a linear AR1 model. He has gotten the Met Office admitting that pure statistics (mathematical fitting) can’t even say for certain that it has been warming over the past century, because we can’t be sure what kind of noise (unforced variability) is present in the data. The Met Office has admitted that climate models (hypotheses) are needed.

Unfortunately, the physics of the GHE (which we do understand) and of evaporation (which we marginally understand) and convection and condensation (which we don’t) suggests to me that the mathematical functions used to fit your data above are unlikely to be lead to a better understand of why these phenomena behave as they do. However, your linear relationship between G and log(TPW) may very well represent an IMPORTANT EMERGENT PROPERTY of our climate system that can’t be derived from fundamental physics.

Willis: In fact, the mathematical relationships allowed me to differentiate the function and thereby calculate the water vapor radiative feedback, but don’t let that get in the way of your aversion to math …

I spent a great deal of time trying to understand exactly what you had and hadn’t proven about water vapor feedback in your last post and am still working on the problem. The fundament problem is that you are analyzing the effect of a change in surface temperature on the GHE (atmospheric absorption, oT^4 – TOA FLUX) by moving to different locations on the planet, not warming all locations on the planet or not raising the concentration of a GHG while holding everything else constant). You have brilliantly (IMO) put together some observational data that points to weaknesses in the expectation that climate sensitivity is high – but I’m not sure that your analysis proves it must be low. If you are interested in climate sensitivity, the variation in TOA OLR (not oT^4 – TOA OLR) with Ts appears to be the critical parameter.

63. Willis Eschenbach says:

micro6500 July 29, 2016 at 3:12 pm

… I calculate SW for every surface station in the GSoD data set, in this case by day. For each day of the year, for each year since 1940, for the average TSI from 1978-2014, so an average about 98,000 station records per day for all of those years for every station, for the latitude band of 40 to 60 N, the minimum SW for a day is 524Whr/m-2, the maximum is 6304Whr/m-2, way more than 45W/m-2

Interesting. Minimum SW is 524 W-hr/m2/day / 24 hours/day = 22 W/m2, max is 6304 / 24 = 263 W/m2. This is compared to a global 24/7 average of about 170 W/m2 … makes sense.

Is your dataset online? Can mere fools like myself access it? And finally, what is the GSoD when it is at home? The Great State of Denmark? The Galactic Society of Drunkards?

All the best,

w.

64. Willis Eschenbach says:

Clyde Spencer July 29, 2016 at 12:44 pm

… I think we should take Willis to task for not carefully defining “precipitable water” when he started this series. However, I suspect that clouds are not included in the definition because of the ways in which the parameter are determined.

My bad, I sometimes forget to define my terms. You are right that clouds are not included. From the Glossary of the American Meteorological Society (emphasis mine):

The total atmospheric water vapor contained in a vertical column of unit cross-sectional area extending between any two specified levels, commonly expressed in terms of the height to which that water substance would stand if completely condensed and collected in a vessel of the same unit cross section.

Regards,

w.

• george e. smith says:

I don’t see where your GAMS definition specifically excludes clouds. After all, the vapor has to be condensed and poured into a bucket to get the thickness of the water layer.

Now I don’t doubt that the GAMS definition is exactly as you cited here. That doesn’t mean that is what they had in mind when whoever wrote it. I usually expect that anything that is not specifically controlled will invariably run amok, and be uncontrolled.

So if I was writing their definition (I’m not going to); I would add the words (after ….. water vapor ….)
” excluding water in all other phases such as liquid or solid , as might be found in clouds. ”

G

65. charles nelson says:

I think this has been a great post. The one thing that it illustrated perfectly was how inadequate mathematics is when called upon to explain observable phenomenon.

• Willis Eschenbach says:

Dear heavens, miss the point much? To misquote Shakespeare, “The fault, dear Brutus, is not in our mathematics but in ourselves” …

w.

66. lgl says:
67. Paul Bahlin says:

I’ve always wondered why so much climate science is focused on temperature when it is energy that matters. Temperature is, after all, just a proxy for an average energy measurement and it is especially worthless in a gas where energy levels of the constituents obey a distribution.

At the surface, water vapor represents a (relatively) high energy component of a parcel of air. If I magically take that parcel up to, say, 9 km in the uptake of a massive thunderstorm, I haven’t changed the energy content at all, on the way up. The temperature goes down because it is now in a lower pressure regime but the energy is, for all intents and purposes, the same is it was when it left the surface, isn’t it?

When it gets to the top, that water vapor can radiate half of its energy to space without interference. When it does it’s energy level can drop to below that of the nitrogen and oxygen that came up with it. Conduction now moves energy from oxygen and nitrogen to the water vapor which then becomes a radiating proxy for those two gases. The whole parcel loses energy to space; all the constituents, not just water vapor.

For me, this excellent paper and the ensuing comments suggest dozens of focused experiments that could get at the basics that drive climate. I’d like to know the energy radiated away at the top of a tropical thunderstorm, maybe subtropical ones and temperate ones too. I’d like to know the energy radiated away in the dry air down-welling outside the storms. I’d like to know the energy transfers of nitrogen and oxygen to the water vapor in that storm.

I actually could care less what the temperature is. That’s a bit like tasting the smoke coming off your filet on the barbecue.

• lgl says:

Not much radiation from thunderstorm cloud tops.

68. Willis Eschenbach says:

Frans Franken July 31, 2016 at 1:59 am

Willis, thanks for an again innovative post and clarifying comments. Please keep enlightening us.

It is both my pleasure and my curse, and you are welcome.

https://wattsupwiththat.com/2016/07/28/precipitable-water-redux/#comment-2267291
>>>
In this comment you get to an evaporative cooling power of 80 W/m2 based on precipitation of 1 cubic metre per year. Is this 80 W/m2 a rounded figure, or else what specific evaporative heat (J/kg, cal/g) are you calculating with and where can I find it?

It is assuredly a rounded figure, and I should have been more specific.

For these kinds of questions I use the Gibbs Seawater (gsw) package in R, which describes itself as:

Provides an interface to the Gibbs SeaWater (TEOS-10) C library, which derives from Matlab and other code written by WG127 (Working Group 127) of SCOR/IAPSO (Scientific Committee on Oceanic Research / International Association for the Physical Sciences of the Oceans).

Before I used the gsw package I used the MIT seawater tables linked to from here.

According to Gibbs, the important values are:

Heat of evaporation, 15°C, salinity of 35 PSU = 2,409 joules per gram, or as I always think of it, 2409 megajoules per tonne.

Sea water density, 15°C, salinity of 35 PSU = 1.022 tonnes per cubic metre

So the equation to evaporate one cubic metre of seawater per year assuming one square metre of surface area looks like this:

1 m3 / year / square metre divided by 1.022 tonnes per cubic metre is 0.978 tonnes/ square metre / year

0.978 tonnes / square metre / year times 2409 MJ / tonne is 2357 MJ / square metre / year

Now, there are 31.56E+6 seconds in a year, so …

2357E+6 J / square metre / year times 1 year / 31.56E+6 seconds, conveniently the E+6 cancels out and we are left with 76.4 joules / second / square metre.

But a joule per second is a watt, so that is 76.4 watts /square metre. I probably should have calculated and used the more accurate figure, but I was running fast.

“Fourth, the thunderstorm towers serve as “heat pipes” which move immense amounts of warm air from the surface up to the upper troposphere … and they do it without allowing the moving air to interact with the surrounding troposphere. On the way from the surface up to the bottom of the cloud the energy is moved as latent heat, which doesn’t increase radiation. Once the moist warm air enters the bottom of the thunderstorm cloud it condenses, and the latent heat is released as sensible heat. But within the cloud the radiation of the air is immediately absorbed by the water droplets that make up the cloud, so the rising column of air inside the thunderstorm tower is not interacting radiatively with the air surrounding the cloud. It’s a neat trick, the heat is kept hidden away from interacting with the surrounding GHGS all the way from the surface to high in the troposphere.
When the rising air inside the pipe finally emerges at the top of the thunderstorm towers, is is above the majority of the greenhouse gases.”

>>>
This is a truly interesting mechanism. Is this your own reasoning, or did you find support for it in literature or elsewhere?

I developed the idea myself, although I’m sure I’m not the first person to do so. However, I haven’t seen it mentioned in the literature. But then I’m the wrong guy to ask, as I tend to spend my time doing new research rather than looking at previous research … I tend to do the work first and then look at previous work second. Yes, I know it is the wrong way round to do things, but it’s served me well. And in a field with so many peer-reviewed papers that are not only incorrect but point in the wrong direction, you’re as likely to get steered wrong as right … I prefer to start without preconceptions and then read up on the previous work once I’m done. Sadly, the new never-done research is the fascinating part, so I tend to neglect the other … only so many hours in a day.

If I understand correctly, the latent heat released at condensing, which occurs from cloud bottom to top, stays within the cloud as warmed up rising air and water droplets. When this warmed up air and water mass reaches the cloud top, what exactly happens to the formerly latent heat it contains? Bearing in mind the low density of the air at cloud top altitude, is the heat mainly passed as sensible heat to the surrounding thin air, or is it rapidly shed as IR radiation with about half of it passing directly to space without heating the atmosphere?

Mmmm … first, you say “… condensing, which occurs from cloud bottom to top, …”. Actually, most of the condensation takes place at the cloud base, at what is called the “lifting condensation level” or LCL. As air rises it cools, and when it reaches the LCL, it condenses. That’s why a field of thermal cumulus clouds all have their undersides at the same altitude, because that altitude is the LCL. There, the latent heat is converted to sensible heat, warming the air.

Now, as you know, warm air tends to rise. It is this rewarmed air in a cumulus cloud which, when there is enough of it, provides the power to construct the cumulonimbus (thunderstorm) tower upwards from the lowly cumulus. You might think of a thunderstorm as a fire in the cumulus cloud with a column of smoke and heated air rising from it … except that in place of smoke there is liquid water and ice.

And because there is liquid water turning into ice in the core of the thunderstorm, there is a secondary effect which is also going on. Just as energy is released when a gas condenses into a liquid, energy is also released when a liquid turns into a solid. So a thunderstorm wrings even more energy out of a part of the moisture that left the surface. And that released latent heat of fusion further expands the air and pushes the thunderstorm tower even higher.

Now, of course in nature nothing is ever quite that simple. Actually there is more complexity inside the thunderstorm tower, where the second phase change takes place as liquid water turns to ice. What happens is much like what happens during the first phase change, where water vapor turned to a liquid. What happened was the liquid fell back to earth as rain.

Well, the same thing is going on inside the thunderstorm tower, except what is falling is not water, it is ice. In both cases at a certain point the accumulation of either water or ice is too large to stay aloft, and it starts to fall. So inside the tower we have falling ice, a hidden hailstorm of great ferocity. But the overall point remains—from inside the thunderstorm tower, radiation is not free to escape and interact with the surrounding troposphere.

So in answer to your question about the fate of the latent heat, the best way to answer it is to consider a thunderstorm as a heat engine that runs off of hot moist air.

Now, for a heat engine to work, we need a hot end, a cold end, and a working fluid. Take for example a steam engine. At the hot end, the boiler heats the working fluid, water, until it evaporates. Then the steam is used to do some kind of useful work, after which it passes to the condenser at the cold end. There, the steam is cooled and condensed back into water, which returns to the boiler to complete the cycle.

So … what happens to the energy from the coal used to fire the steam engine boiler? Well, some of that energy is converted to useful work, some is lost to friction, and the rest simply passes through the heat engine and is rejected into the environment.

The same is true of a thunderstorm. Some of the energy is converted into the work of driving a current of air vertically at great speed. Some of it is lost to turbulence. And some is simply passed through the thunderstorm and rejected at the cold end at altitude. I’ve never seen a detailed breakdown of how much goes into each part. Isaac Held over at GFDL might know.

So the answer is that mostly the latent heat is converted to work. The work is performed when the heated air expands, because it has to physically push other air out of the way to expand. That heated parcel of air rises, and of course as it rises it expands further, and as it expands, it cools. At some point, it will have cooled to the temperature of the surrounding air. At that point the air emerges from the thunderstorm tower. If the thunderstorm is strong enough, fine crystals of ice will get caught in the updraft and make a level flat trail behind the thunderstorm, the familiar “anvil top”.

Bear in mind, however, that in fluids at the end of the day it all turns to heat, as the poet-scientist said:

Bigger whorls have smaller whorls that feed on their velocity,
Smaller whorls have smaller yet, and so on to viscosity.

The real difficulty I have with thunderstorms is that as in any field of study, hard numbers are good to come by. But with a subject matter as evanescent as clouds … good numbers are hard to come by …

Best regards,

w.

PS—You might enjoy one of my previous posts on the subject, Air Conditioning Nairobi, Refrigerating The Planet.

• Ben Wouters says:

Willis Eschenbach July 31, 2016 at 12:37 pm

“Actually, most of the condensation takes place at the cloud base, at what is called the “lifting condensation level” or LCL. As air rises it cools, and when it reaches the LCL, it condenses. That’s why a field of thermal cumulus clouds all have their undersides at the same altitude, because that altitude is the LCL. There, the latent heat is converted to sensible heat, warming the air.”

Not quite true. If you compare the DALR vs the SALR (dry vs saturated adiabatic lapse rate) on a thermodynamic diagram you’ll see that the SALR is lower then the DALR to considerable heights.
Obviously depending on the the temperature the rising air has when it starts its ascent.
With higher temperatures the rising air will contain more water vapour (WV) to condense.
With start temperature eg 30C it is above 12 km height before all WV has condensed out.

Since the rising air does not lose energy to the surrounding air (adiabatic assumption) is why these lapse rates have been called ‘adiabatic” ;-)

• Ben Wouters says, August 1, 2016 at 8:54 am:

Since the rising air does not lose energy to the surrounding air (adiabatic assumption) is why these lapse rates have been called ‘adiabatic” ;-)

No. That is NOT why these lapse rates are called “adiabatic”, Ben. The rising air specifically DOES lose (internal) energy to the surrounding air, only it doesn’t happen via heat transfer, but rather via so-called PV work being done as the rising air expands against the surrounding pressure. This is why an adiabatic process is defined by saying Q=0 and so dU=-W = PdV. This subject seems to be an eternal source of confusion …

• Frans Franken says:

Willis,
Thanks for your elaboration and patience.

A bit about my background: I’m a mechanical engineer, MSc from Technical University of Eindhoven, Netherlands. I’ve been working for about ten years as a steam turbine engineer and product manager of steam turbines licensed from IMO Industries and General Electric, and as such have co-developed Cogen and CCGT power plants.

On the specific evaporative heat: everything clear. Global average precipitation = (1000/365 =) 2.74 mm/day, specific evaporative heat = 2409 kJ/kg, global evaporative cooling power = 76.4 W/m2.

On the destination of the latent heat, I would like to expand. The energy (power) taken from the surface by evaporation is clear: about 76.4 W/m2. To close the hydrological cycle: precipitation normally comes down as rain of lower temperature than the oceans from which it has evaporated. The heat required to warm up this precipitation to ocean temperature is small, ≈1.3 W/m2 based on a temperature difference of 10 K between rainfall and ocean. For completeness, let’s add this to the latent heat and round it up to (74.6 + 1.3 ≈) 76 W/m2 as the total water cycle heat flux at the surface-atmosphere interface.

Conservation of energy, and the fact that the water (vapor) doesn’t escape to space, require that this amount of heat is released in the atmosphere. If the convection process from surface up to cloud top is adiabatic – no heat exchange with the surrounding atmosphere – then all heat must be released around the cloud top. There, it can be shed either as sensible heat (conduction to surrounding air) or as IR radiation (in all directions).

An exception could arise if the water exits the cloud top sideways as ice, and this hail melts on its way down. In that case, a substantial amount of the heat will be passed to the atmosphere outside the cloud between cloud top and the surface. However to avoid complexity let’s forget about this scenario for now, and assume that the full flux of 76 W/m2 is released at cloud top level.

When looking at the global energy budget cartoons of Kiehl-Trenberth and NASA, such as the one below, a flux of about 30 W/m2 is included as “emitted by clouds”. This looks as if it is meant to present upward IR emission from cloud tops which makes it to space:

Time does not allow me to continue momentarily, but I’ll follow up un this shortly.

• Frans Franken says:

Correction: total water cycle heat flux = 76.4 + 1.3 ≈ 78 W/m2

69. The sigmoid function arises in the characterization of time-variant feedback systems when you have a delay in the feadback path. I.e S[s] = 1/(1+e^-s) in bode form gives a feedforward gain of unity and a feedback gain of e^-s. The inverse Laplace Transform of e^-s is a pure unit delay. The inverse sigmond function may indicate you’ve derived an output-to-input relationship.

70. “Some of the energy is converted into the work of driving a current of air vertically at great speed.”
This is probably a dumb question but I’ve often wondered where in your calculation you account for the energy (work) required to lift all that water. Surely this must dwarf the energy required to move air.

71. Wim Röst says:

Willis, I just read your Air Conditioning Nairobi, Refrigerating the Planet from March 2013. https://wattsupwiththat.com/2013/03/11/air-conditioning-nairobi-refrigerating-the-planet/ You were very clear about the role thunderstorms have anyway:

“My takeaway message is this:
The surface temperature of our amazing planet is set and maintained by the constant refrigeration of the surface hot spots as they form, not by the forcing, whether from CO2 or anything else.“

I myself would like to know more about the role convection plays in losing the heat from the surface to space. In this post the word ‘convection’ so far is mentioned 32 times and I think I can ad a little discussion I had today about that subject. The discussion is from Roy Spencer’s blog.

Any extra information in respect to the role of convection for radiation to space is welcome.

(….)
Kristian says:
August 1, 2016 at 9:58 AM
“Wim Rost says, August 1, 2016 at 6:39 AM:
Kristian, about H2O: “this wont for the most part have anything to do with its radiative properties.”
First, I would like to know which part.

What do you mean? Which part that does involve the radiative properties? That would be the absorption and emission of IR. But any extra cooling from this would just be an effect of extra prior heating (a warmer troposphere emits more OLR to space). It is very unlikely (if you don’t buy into the whole AGW tripe) that Earth would warm or cool as a direct result of changes in its heat OUTPUT. All we can ever observe in the Earth system points to OLR being a radiative RESPONSE only to warming or cooling. And that warming/cooling would come from changes in Earth’s heat INPUT, from the Sun. Like we have now. Global ASR (absorbed solar radiation, TSI minus albedo) went considerably up between the late 80s and cirka 2000, and after that has pretty much stabilised at a slightly higher level again, although still significantly lower than in the 80s. This is all due to changes in (principally tropical) cloud cover during the 90s. At the same time, OLR went up in step with tropospheric temps, that is, it came as a direct radiative EFFECT of the temperature rise (caused by the increase in solar input).
No one seems to want to touch this simple observational fact, even though it’s clearly evident from the data (ERBE+CERES, ISCCP FD, HIRS):

“I would also like to know whether the energy that is transported upwards, will effect radiation in an indirect (!) way. For example, I read about a recent extension of the Hadley Cells. As the level of the troposphere is higher in the tropics, energy is transported closer to the place where it can be emitted to space: the mesosphere and stratosphere. More upward and in latitude extended Hadley cells might improve the outward emitting. I think.”

Again, the Hadley cells only expand when they warm. And so any extra cooling comes from extra prior heating.

“Secondly, as Roy Spencer states: addition of CO2 to the atmosphere is supposed to warm the surface, but cool the stratosphere and mesosphere. In general this must stimulate convection, the more when the lower layers warmed up also. And stronger convection will transport energy closer to the level where it can/will be emitted.”

And this is why we cannot have an “enhanced greenhouse effect” from the addition of more CO2 to the atmosphere. Well, there is no real thermodynamic connection between the lower troposphere and the stratosphere and mesosphere. But the upper layers of the troposphere itself would also cool more. So adding more IR-active constituents to the atmosphere would simply enhance the absorption at lower tropospheric levels and likewise the emission at higher tropospheric levels. And what is it that connects these two levels and makes sure the lower levels don’t get too hot and the higher levels don’t get too cold? You guessed it. It’s convection. Convection will quash any radiative attempt at making the lower levels of the troposphere warmer, thus forcing the surface itself to be as well. It wouldn’t work. If the heat transfer from the surface up was purely and only radiative, then it might’ve worked somehow. But as we all know, it isn’t. Convection (to a large extent evaporatively driven) reigns supreme. *

* WR: I added some “ “

• Wim Röst says:

The temperature rise in the troposphere in 2016 (El Nino) coincides with a lowering lower-stratosphere temperature. At more points in the graph an upward troposphere temperature seems to correspond with a lower stratosphere temperature and reverse. Is there a statistical effect measurable?

(Volcanoes (particles in the stratosphere) make the stratosphere temperature rise and lower the lower troposphere temperature. Their effects need to be taken out)

What would be shown when we couple the part of the grid known for the rising number of thunderstorms to the emissions of that specific part of the grid?

===

General question: what effect would have been measured by RSS going up one extra layer in the stratosphere? An even stronger cooling trend?

===

RSS: The average temperature in the stratosphere is generally lowering as predicted: CO2 content rises, stratosphere cools more.

N.B. https://wattsupwiththat.com/2016/08/01/uah-global-temperature-update-for-july-2016/
. UAH Numerical data for the whole (!) tropical sea area don’t show a trend much different from the world trend. Less cooling even: -0,29 against -0,31. But thunderstorms are concentrated in some parts of the tropics as the rainfall data show and for those area’s it might/will be different.

N.B.2. As UAH numerical data for the stratosphere show, the lowering of the temperature of the stratosphere is strongest at the land part of the south pole: trend -0,43 against -0,31 for the global average. It is another subject, but it would be interesting to know why that effect is just strongest there. More energy transport pole ward from the depressions surrounding the pole is an option. Another one is the special role of the Polar vortex: “an upper level low-pressure area, that lies near the Earth’s pole. (….) The bases of the two polar vortices are located in the middle and upper troposphere and extend into the stratosphere.” Source: https://en.wikipedia.org/wiki/Polar_vortex
An impressing image of the actual polar vortex at Nullschool at 10hPa, 26.500 m, deep in the stratosphere: https://earth.nullschool.net/#current/wind/isobaric/10hPa/orthographic=0.44,-84.85,473/loc=15.476,-52.292
(use keyboard shortcut m to go down a level, use i to go up through the atmosphere – found at https://earth.nullschool.net/about.html )
At the poles the Stratosphere starts at around 10 km, in winter time even lower.