OK, after weeks of sitting under constant rain (at one point we were over 600% of “normal” for the month) and with all the “excess rain” events happening around the world, I thought maybe it was time to put some numbers on it.
How much does rain matter?
Folk who have been reading here for a while will know I’ve got the general thesis that the lower atmosphere is dominated by convection. That it completely swamps Infrared. You can see this easily in any given day as the clouds form, water from the seas and trees evaporates, rises into the sky, forms clouds, then falls again as rain.
We’ve also found that the upper atmosphere where heat is dumped via Infrared has that dumping enhanced by CO2. More CO2 means more effective heat loss up high.
That just leaves the middle a bit unexplored (and a minor “dig here”; though if you know one side is getting a load of heat delivered and the other is dumping it faster, the middle is defined by context… but someday I need to look more at heat transport through the stratosphere. For now we’ve mostly just seen how the sun modulates the stratosphere with UV light. As the solar UV has plunged, the stratosphere has cooled, shrunken, and we have gotten colder and wetter (as though the ‘cold end’ of the tropospheric heat engine had gotten colder so more heat was being transported to the stratosphere via evaporation from the ‘hot end’ of the oceans; thus falling as more rain and snow).
Above the Stratosphere, it gets cold. Very cold.
The mesosphere extends from the stratopause to 80–85 km (50–53 mi; 260,000–280,000 ft). It is the layer where most meteors burn up upon entering the atmosphere. Temperature decreases with height in the mesosphere. The mesopause, the temperature minimum that marks the top of the mesosphere, is the coldest place on Earth and has an average temperature around −85 °C (−121 °F; 188.1 K). At the mesopause, temperatures may drop to −100 °C (−148 °F; 173.1 K). Due to the cold temperature of the mesophere, water vapor is frozen, forming ice clouds (or Noctilucent clouds). A type of lightning referred to as either sprites or ELVES, form many miles above thunderclouds in the troposphere.
I note that at those temperatures you could easily have “carbonic gas” instead of CO2 and water vapor, or perhaps it would be “carbonic snow” instead… We really do not know the chemistry at that level of the atmosphere, so any hypothetical CO2 impact is really just that. Hypothetical. What we do know is that it’s cold. Darned cold. Heat is leaving in a tearing hurry and not looking back. And there are clouds, which says water makes it this high and must be considered as part of the heat transport system. And when that water vapor forms those noctilucent clouds? What does THAT mean about IR “trapping” via the “greenhouse gas” of water vapor? When the solar UV dropped low, we had a few more reports of noctilucent clouds. A giant “Dig Here!”, IMHO. What happens to the earth if the sun opens the “Water IR” window above the convecting Troposphere?
For the Stratosphere:
The stratosphere extends from the tropopause to about 51 km (32 mi; 170,000 ft). Temperature increases with height due to increased absorption of ultraviolet radiation by the ozone layer, which restricts turbulence and mixing. While the temperature may be −60 °C (−76 °F; 213.2 K) at the troposphere, the top of the stratosphere is much warmer, and may be near freezing. The stratopause, which is the boundary between the stratosphere and mesosphere, typically is at 50 to 55 km (31 to 34 mi; 160,000 to 180,000 ft). The pressure here is 1/1000 sea level.
OK, so it’s “warm” at “near freezing” at the very top, and due to UV… Still going to be a ‘cold pole’ to the Tropospheric water driven heat engine pumping heat skyward. What happens when the UV plunges (as it just has)? Do we get more rapid trans-stratospheric heat flow? (we ought…) And as that ozone ‘IR window” opens from less UV forming ozone, does the IR transport pick up too? And nearer the bottom of the stratosphere where that Troposphere dump happens? It is -60 C which is “way cold”. Yes, I need to look more closely at stratospheric heat flow to be ‘complete’, but at a first look, it’s a “way cold” end of the Tropospheric Transport and it then has a Solar / UV modulate thickness, temperature, and probably IR heat transport. As there are also stratospheric clouds, there will also be a variety of cloud and water mediated cooling processes as well. The also have a load of interesting gas chemistry going on:
Polar stratospheric clouds (PSCs), also known as nacreous clouds (from nacre, or mother of pearl, due to its iridescence), are clouds in the winter polar stratosphere at altitudes of 15,000–25,000 meters (50,000–80,000 ft). They are implicated in the formation of ozone holes; their effects on ozone depletion arise because they support chemical reactions that produce active chlorine which catalyzes ozone destruction, and also because they remove gaseous nitric acid, perturbing nitrogen and chlorine cycles in a way which increases ozone destruction.
Oh, it’s the clouds that make the ozone hole? I thought it was me and my air conditioner… who knew? (No, really. I want to know “WHO, exactly, knew this and when did they know it…”)
But if we’re making clouds, removing O3 (and opening that part of the “IR Window”) removing water vapor into ice clouds, and generally shifting the chemistry and IR character all over the place; you think maybe that might matter as the UV catalyzed effects and UV heating plunge off a cliff? Think that might impact how much water gets sucked up there to dump heat?
But that leads back to the original question of this thread:
How big is rain?
Can rain have a thermal size? If so, how big is it?
Well, yes, it can have a size. What would be ideal would be a total rainfall of the planet historical record. Then we could just measure the impact for the whole planet and be done. But we don’t have that. Huge chunks of the ocean have heavy rains, but no station to measure them.
What we can do is look at just one place. A sample. Then we will have a general idea of the size of things. Is it a whole lot smaller than the 1 W or 2 W of “excess heating” being attributed to CO2 (and thus to people) or is it about the same? Or maybe even a bit larger? Could a 30% increase in precipitation “cover” the added heat budget of the (hypothetical…) CO2 “greenhouse effect”?
That can be done with some fair precision, as we have numbers for some selected places that are pretty good. We can, for example, measure the rainfall over an identified geography and measure the stream flow out of it. Between those two, we have a good idea what fell from the sky and what ran off back to the ocean. (It does NOT capture that which soaked into the ground and was released back to the sky via transpiration of plants, but you can get a rough estimate by comparing the two numbers – what fell and what ran off.)
So is there a well defined geography that is fully isolated from our tendency to dam up rivers and move the water 1000 miles away? Somewhere that is well instrumented and reported in the literature? Sounds like we might be looking for a US Island or Territorial Island. There are many to choose from and it would be interesting to compare a lot of them, but this is the one I found.
If you look at that global precipitation map up top, you will see that Puerto Rico is in that green area just south of Florida and near Cuba. About 1/2 way from the yellow deserts to the dark blue tropical seas. A reasonable “first sample”. (Though please note that the scale is non-linear and that dark blue is 300 mm / month vs about 100 for the dark green).
A hurricane can dump a lot of rain …
Science News, June 11, 2005
The large masses of warm, moist air that fuel hurricanes also prime those windstorms to drop a lot of precipitation in a short time, a phenomenon that residents of Puerto Rico experienced in spades when Hurricane Georges struck their island in 1998. Now, new hydrological analyses indicate just how much storm runoff and sediment washed into the surrounding waters in the wake of that storm.
In the course of a normal year, the 8,700-square-kilometer island of Puerto Rico gets about 1.6 meters of rain, says Matthew C. Larsen, a hydrologist with the U.S. Geological Survey in Reston, Va. That’s about 14 billion cubic meters of precipitation. About 6 billion [m.sup.3] of that water recharges the island’s aquifers, but the other 8 billion [m.sup.3] runs off the island in streams, carrying around 5.9 million metric tons of sediment.
In September 1998, however, Hurricane Georges swept over the island, dumping an islandwide average of 0.3 m of rain–more than 2 months’ worth of precipitation in a mere 2 days. The deluge triggered landslides, flooding, and severe erosion. Data from flow meters in streams indicate that more than 1 billion [m.sup.3] of runoff reached the ocean in those 2 days, along with 2.4 million metric tons of sediment, says Larsen. That’s about 40 percent of the average annual sediment load and amounts to about seven large dump truck loads of sediment from each square kilometer of the island.–S.P.
COPYRIGHT 2005 Science Service, Inc.
COPYRIGHT 2005 Gale Group
This article has two nice sets of numbers in it. The typical annual precipitation and what a large hurricane looks like. While it is “unusual” to have a hurricane over any one spot on the planet, it is absolutely typical for us to have several ‘cyclonic storms’ happening for months on end ‘somewhere’ on the planet. As those are often ‘out to sea’ where the energy to drive them is found, it wold be good to estimate how much water THEY are pumping to the stratosphere and how much heat that machine is dumping “up high” as it runs for weeks on end…
OK, some folks love numbers, other folks hate them. I try to put things in verbal form most of the time as most folks seem to hate numbers. (I’m rather fond of numbers as they answer so many things so well, but they also trap a lot of folks who think numbers are reason or logic; they are not, they just put sizes on the thinking, so if you have ‘thought well’ you get a very sharp edge, but if you have “thought poorly” the numbers just put a sharp edge on your bed of nails…) But in this case, we are particularly looking for a ‘size of rain’ in heat. We need numbers. So I’m going to “show my work” here (and folks can check if I’ve lost an order of magnitude anywhere or not… it’s easy to do in this kind of thing as we have billions of mega – kilo things…) and then we’ll have a single number at the end that’s easy to grasp. The “Watts per meter” of heat transport represented by that water.
OK, to the slide rule (or spreadsheet for folks not blessed with analog logic and math ;-)
Puerto Rico Area:
8700 square kilometers. As that is 1000 x 1000 meters on a side, the area in square meters is:
8.7 x 10^9 square meters
Puerto Rico Rainfall:
First up, our cross check number is that there is about 1.6 meters of rain per year in Puerto Rico. 1600 mm. As we run the numbers, if at any time we get something like “1.6 mm” of rain, we know we’ve lost a ‘kilo’ somewhere… Yes, it’s an old “sliderule habit”. First you find a very rough ‘guide solution’ then you keep checking against that ‘order of magnitude’ to make sure you have not ‘slipped a digit’…. Yes, we could just use the 1.6 m directly (and I suspect it may be what was used in the first place to calculate the total m^3 of rainfall) but what fun would there be in that? ;-)
14 Billion m^3 of rainfall total / year (13.92 calculated from 1.6 m x 8.7 10^9 m^2)
Rainfall / m^2:
14 10^9 / 8.7 x 10^9 = 1.609 m / m^2
(so we’ve got a ‘cross check’ on that 1.6 m number and it looks like that’s what they used to come up with the total in the first place. OK, at this point we need to pick one and use it, but with the ‘error’ in the 4th digit, I’m not real worried. [ For "young 'uns", in sliderule terms, you just count all the digits without looking at the decimal point to know 'how many digits you are working with'. A sliderule can typically stay precise to 3, but sometimes out to 4. Beyond 3 you are usually 'making things up' so ignore them... the decimal point is 'added back in later' so doesn't matter to 'how many digits'...] but at least we know about how much rain they have.)
This is not all that different from most tropical places and many not-so-tropical. Recent snowfall in the Sierra Nevada has run out to about 50+ feet or 16 meters as measured. That’s about 1.6 meters of rainfall equivalent in snow. So now we have another ‘sanity check’ that says this is a wet place, but that it’s not a particularly uncharacteristic place and even non-tropical places like California can be ‘in the ball park’ of that precipitation level.
OK now lets “find the heat”.
The water evaporates from the ocean or land, so we need the “heat of vaporization”, then it rises to altitude where it typically forms snow or hail in a thunderstorm. That means we need to add the “heat of fusion” too. Now some folks my holler that it’s not always frozen, but the heat of fusion is pretty small in any case. We need to look at it so we can “discover” that, so I’m putting it in. In many cases, like those California mountains, the heat of fusion is not “dealt with” until spring, so for some places, like where I live, it very much does matter.
In theory, we also ought to deal with the heat of the air itself, as it rises, and the expansion of water vapor in the air, and a whole lot of other detail of heat and expansion / compression of air. I’m going to leave those out (if someone want’s to put them back in, be my guest…) We’re looking for how much does the water count, and not for total heat transport, so “how big is the water?” means we look at the water. Leaving out the heat in the “delta T” of about 40 C at the tropical ocean surface to 0 C at altitude is not a large number and it “is an error against our thesis” that water matters, so to some extent leaving it out is “money in the bank” if someone wants to attack the computed number as “too high” later. We can just say “Hang on a moment, and let me add back in the specific heat of the water itself…” Again, if someone else wants to compute that, be my guest (it’s easy …)
So how do we measure this? Our ultimate goal is Watts (instantaneous energy flow) but we have heat over time.
From the wiki on heat of fusion and heat of vaporization for water we get:
Water heat of fusion: 334 kJ / kg
Water heat of vaporization: 2257 kJ / kg
Total of the two: 2591 kJ / kg
So we can see right off that the heat of fusion is about 1/10 or “one order of magnitude” smaller than the heat of vaporization. That evaporation at the surface counts for most of the heat, the freezing at altitude is just ‘icing on the cake’…
(The specific heat of water is about 4.18 Joules / cc or 4.18 kJ / kg so you can see why I’m willing to ignore it. It is about 1/500 th the heat of vaporization… at 1/540 it’s the ‘chump change’ in the equation. Though with a 40 C temperature drop it would be 40/500 or about 1/12 of the total. An error term of 0.083 ‘in my favor’ I can live with…)
OK so just what is a Joule? It is one Watt for one Second. A Watt-second. And a kJ is a kiloWatt-second. Run your kiloWatt room heater for one second, you’ve got a kJ of heat energy flowing.
OK, so each of our kg of water now represents 2591 kW-seconds of energy flow.
With 1.6 m^3 of water, that’s 1.6 10^3 kg of water.
(A cubic meter is 100 cm on a side, so 10^6 cc, but a kilo of them is 10^3, after the division you get 10^3 kg / m^3 ) So we’ve got 1600 kg of water.
1600 * 2591 = 4145600 kW-Seconds or 4,145,600,000 W-seconds.
How many seconds are there in a year? About 31,536,000 (using round days and ignoring the fractional bit)
Divide those ‘Watt-Seconds” by “Seconds” and you get:
(A “slide rule guy” would note that there are 4 x 10^9 Watt-seconds and 3 x 10^7 seconds so we’re looking for about 4/3 10^2 Watts, or about 133 as a mental ‘cross check’…)
At this point I’ve done this calculation about 4 times. 2 of them I “lost a kilo” somewhere along the way in the spreadsheet. 2 of them “I found it again” after the cross check sent me back to check my units. So it wouldn’t hurt to have someone else double check that in all the kilo-mega-billion-ton-gram-etc. conversions I didn’t ‘slip a digit’…. but I’m pretty sure I’ve got it right this time ;-)
OK, so we’re looking at an order of magnitude of 130 ish Watts per Square Meter of heat transport by precipitation. Our “size of rain” is about “130 Watts”.
How Big Is the Sunshine?
This, it would seem, is a harder question to answer than to ask. I’ll cut to the chase here. Nobody really knows because we don’t really have a good handle on the albedo impact (how much snow and low angle light on water reflects sunshine back into space) and nobody really knows how much cloud we have. There are some guesses, but that’s all they really are. (Later satellite measures are a bit better. see below). We measure some points in space and some points in time on a fractal surface then try to make up a number that ‘seems right’. It will never be right as the size of the measurement of a fractal surface is directly related to the size of the ruler you use. Change the ruler, change the result…
(See: http://en.wikipedia.org/wiki/Coastline_paradox for an example of the problem).
At this point I could run off into trying to correct for all the clouds and trying to figure the albedo impact of snow and the reflectivity of ocean at low angles of attack of sunshine and a host of other things. But I won’t. You can do that if you wish. I just want a general idea:
Is the rain large or small compared to the sun?
we find (after a lot of figuring in nighttime and the sideways angle of light at the poles):
Over the course of a year the average solar radiation arriving at the top of the Earth’s atmosphere is roughly 1,366 watts per square meter (see solar constant). The radiant power is distributed across the entire electromagnetic spectrum, although most of the power is in the visible light portion of the spectrum. The Sun’s rays are attenuated as they pass though the atmosphere, thus reducing the insolation at the Earth’s surface to approximately 1,000 watts per square meter for a surface perpendicular to the Sun’s rays at sea level on a clear day.
The actual figure varies with the Sun angle at different times of year, according to the distance the sunlight travels through the air, and depending on the extent of atmospheric haze and cloud cover. Ignoring clouds, the average insolation for the Earth is approximately 250 watts per square meter (6 (kW·h/m2)/day), taking into account the lower radiation intensity in early morning and evening, and its near-absence at night.
OK, we get to ignore clouds. As I’ve noticed they do a lot of that in “climate science” I guess it’s ok /sarcoff>
But there is our one simple number. 250 Watts.
130/250 = 0.52 or roughly 1/2.
Fully one half of all the heat delivered to this part of the planet is taken away by rain alone.
Send In The Clouds, There Must Be Clouds…
From NASA we get this nice picture of the typical cloudiness of the earth. Looks cloudy to me…
One of the biggest sources of uncertainty in computer models that predict future climate is how clouds influence the climate system and how their role might change as the climate warms.
These maps show what fraction of an area was cloudy on average each month. The measurements were collected by the Moderate Resolution Imaging Spectroradiometer (MODIS) on NASA’s Terra satellite. Colors range from blue (no clouds) to white (totally cloudy). Like a digital camera, MODIS collects information in gridded boxes, or pixels. Cloud fraction is the portion of each pixel that is covered by clouds. Colors range from blue (no clouds) to white (totally cloudy).
Uh, yeah, I can agree with that…
As a first approximation, my eye says that the picture is about 1/2 way from dark blue to white. So looks to me like about 1/2 of the sunshine is going to be reflected away. Leaving just about the amount that gets removed by water vapor / rain to be ‘returned’ to altitude via water.
As a first approximation: It’s ALL about the water and CO2 / Infrared radiation is just a fantasy down in the lower atmosphere layers. We also know that “up high” it acts to radiate heat away from the planet, so more CO2 means more effective heat loss. At this point, I’m just not finding anything “left over” for CO2 to do…
OK, at this point some “Warmer Wag” will be setting their tongue to wagging about how some clouds are “warming” and that they “trap heat” via the “back radiation” that moves heat from cold places to warmer places. This, of course, ignores the heat loss that very cloud represents via its formation. But, just to be clear, we have been conservative about how much cloud cover is out there, and how much it will be reflecting heat back to space during the daytime:
has a nice graph of total cloud cover trends here:
Since clouds have a net cooling effect on climate, the above would imply (Svensmark 1998) that the estimated reduction of cosmic ray flux during the 20th century (Marsh and Svensmark 2000) might have been responsible for a significant part of the observed warming. Since 1983, the cooling cover of low clouds have decreased from 29% to about 25% (see below). During the same period the net change of warming high clouds have been small (see below).
which shows two very interesting things. First off, total cloud cover runs about 64% to 70%. There is a whole lot of heat being reflected back out into space by those clouds, so you get to “trap” about 20% of the “heat” just to reach a break even at about 1/2 for water transport. You’ve got a big hill to climb before any “trapping” starts to have a net warming compared to the heat LOST via reflection.
Second, it shows cloud cover dropping from 70% to 63.5% over the period from 1986 to 2000; exactly in sync with the warmer temperatures GISS / CRU / NCDC et.al. found and attribute to CO2.
So, until that impact of clouds is addressed and not just ‘hand waved away’ with a “some warm some cool” it looks to me like we’ve got more Climate Clowns than Clouds in the picture…
So we’ve got things of “order of magnetude” hundreds of Watts driving things (rain and sun) and with variations of 10% range for clouds (so order of 10′s of Watts at least). And how much does all of “human activity” count for as a “forcing function”? Per this chart from the IPCC, less than 2 Watts. We’re ignoring the dollars and dimes and focusing on the pennies when we look at CO2:
OK, I need to polish this a bit more. There are some more bits to add in the way of references to prior articles, etc.. Given a choice of leave this in the hopper, or send it out 3/4 done, I’m going to let folks see it now, and polish later. I also need to do the same exercise for the hurricane water dump to show how mammoth they are. But right now the clouds have parted and the sun is in the garden again. I’m going to “jump on it” while I can and worry about clouds some more when they come back ;-) For now, this is all the “Kitchen Science” that fits in the time available. More later… after a bit ‘o sun and fun…