Monday, 23 November 2020

From The Motherlode:

The red area shows radiation emitted from Earth's surface, assuming average temperature 288K (total 390 W/m2). The green area shows actual outgoing radiation measured from space (total 240 W/m2). Their argument is that the red area that is not overlapped by green shows the missing radiation that is 'trapped' by CO2. Game, set and match to the Alarmists!

Here's a simplified version to illustrate the point they are trying to make: the surface emits 390 W/m2 but only 240 W/m2 gets to space. How do we explain the missing radiation? The usual suspect, CO2?

Having given this some thought, it strikes me that this way of looking at things is a massive fudge and ignores the full picture.

Let's go back to the beginning and look at incoming solar radiation. On average it's 685 W/m2 during the day. One-third reaches the land or ocean surface, which reflects 10%. Two-thirds hits clouds, which reflect 40%. Overall, 30% is reflected (i.e. Earth's albedo is 0.3), so 480 W/m2 is absorbed during 12 hours of daylight. Temperatures (and outgoing radiation) don't change much by day and night, so let's assume that a steady 240 W/m2 is radiated back to space by day and by night. So it's all in equilibrium. 480 W/m2 x 12 hours incoming = 240 W/m2 x 24 hours outgoing. (Clouds and the surface beneath them have their own separate equilibrium, which need not concern us further).

Outgoing radiation is the reverse process. We treat clouds as part of the surface when calculating albedo, net incoming radiation and effective temperature, so we also have to treat them as part of the surface when looking at outgoing radiation (or else we get nonsensical answers).

One-third of the surface, land/oceans is 288K (effective temperature as adjusted for gravito-thermal effect and latent heat of evaporation/condensation aka "the lapse rate") and they emit 390 W/m2 (they are close to behaving like a 'black body'). Two-thirds of the effective surface, clouds, emit 165 W/m2*, so the overall average is 240 W/m2. (390 + 165 + 165) ÷ 3 = 240. So there is no missing radiation to explain away in the first place!**

* The calculation for clouds is as follows. Average altitude of clouds/emitting layer = 6 km. At 6 km, the temperature is about 250K, i.e. surface temperature 288K minus 6 x 6.5K for the lapse rate. If they behaved like 'black bodies' with emissivity of 100%, they would be emitting (250^4) ÷ 10^8 x 5.67 = 220 W/m2. But clouds' emissivity is only 75%, so they actually only emit three-quarters of that = 165 W/m2. Emissivity is a bit like 'albedo' but in reverse, look it up.

Clearly, "two-thirds", "6 km", "250K" and "75% emissivity" are arbitrarily chosen reasonable mid-estimates to illustrate the point - you get the same effect if clouds are lower (hence warmer) and with a lower emissivity; or they are higher and emissivity is higher. Nobody really knows what these variables (cloud cover, height etc) are as they change constantly and they largely cancel each other out.

** OK, I accept that the 'missing' radiation appears to be in the wavelengths absorbed and emitted by CO2. They are saying that CO2 absorbs but doesn't re-emit? Or are they saing that CO2 does emit radiation to space, but if there is more CO2, it emits less radiation? You can never pin them down.

Lola said...

So what you are saying is that the alarmists (a) have over-simplified and (b) haven't done their sums. Or is it that they have done their sums but have over-simplified on purpose - i.e. lied?

L, the challenge is to either expose obvious flaws in their 'evidence' and/or to explain that evidence using basic physics without invoking CO2 and radiation at every turn. It's tricky but not impossible, you have to dig around and think a lot.

Bayard said...

Diagram 3 out of 4 contains the following sum, 240/2 = 240, which can't be right.

B, thanks, I wil correct it tomorrow.

Mark said...

Their argument is that the red area that is not overlapped by green shows the missing radiation that is 'trapped' by CO2.

We'd boil in a couple of years, if that were true. That's a LOT of radiation.

The "warming" of CO2 of a degree a century means that the incoming and outgoing balance to down below the 5th significant figure. Since we have no hope of measuring to that level we are just guessing which is bigger.

Bayard said...

Mark, there are 1.35 x 10e18 m3 of water in the oceans, which is where all the heat is going these days, apparently. (No explanation is given, of course, for why all the heat that has been merrily warming the atmosphere since the end of the last cold period, sorry, the start of the industrial revolution, should now be warming the oceans, but hey ho.) To raise that by 1 degree would take 1.35 x 10e21 x 4.18 J. The Earth has a diameter of 12.7 x 10e6 m, so the area visible to the sun is that squared times pi/4 = 1.27 x 10e14 m2. At 150 W/m2, that equals 1.9 x 10e16 W. Over 12 hrs that equals 8.2 x 10e20 J, or 0.146 C, or 53 C a year. Unless I'm out by an order of magnitude somewhere, yup, boil it is.

Bayard said...

To look at it the other way, to heat the oceans by a degree in a century means a warming rate of 1 degree in 36500 days or 1.58 x 10e9 seconds. 1 C warming takes is 5.64 x 10e21 J from above, or 3.57 x 10e12 W. divide that by the visible area of the Earth, 1.27 x 10e14 m2 and you have a net inflow of 0.0281 W/m2, over five thousand times smaller. E&OE again, of course.

M, I don't think we would boil, radiation can't make something hotter than the body which emitted it (I think). Imagine a sealed mirror box with an LED in it, how hot would the air get in the box? No warmer than the bulb (I expect).

B, the calc's are easier if you just look at 1 m2 of surface area. Say ocean is 10 km deep, that's 10 million kg, times 4 kJ/K/kg = 4 billion Joules required to warm it by 1K.

Incoming solar every day is 12 x 3,600 x 480 = 20 million Joules. So six months of sunshine is in theory enough to warm a 1 m2 vertical column of ocean by 1K, ignoring the fact it always cools down again overnight. And ignoring the fact that trying to explain this sort of thing using radiation is nonsense.

M "The "warming" of CO2 of a degree a century means that the incoming and outgoing balance to down below the 5th significant figure. Since we have no hope of measuring to that level we are just guessing which is bigger."

That is an excellent point.

Dinero said...

In your calculation you have the result of 240 W/m2 radiated from the ground and the figure you are disputing is 390 W/m2. If, *underlined IF*, that second figure is obtained by measurement, incorporating clouds land and sea then , why do climatologist not find your figure when they do the measurement.

M, I don't know how you calculated it. I think that one day's incoming solar/m2 is about 20 million Joules, which is enough to warm atmosphere and top few inches of land or water by about 1 degree.

So - assuming GHG theory is correct - over the past hundred years (36,500 days), one day's worth of solar has been absorbed and not re-emitted = 0.00003. So yes, fifth significant figure and impossible to measure or verify.

Din, 240 is the average outgoing radiation from Earth to space. 390 is radiation from surface level not covered by clouds. It's two quite separate things. I'm not disputing either of those figures. I'm explaining the apparent discrepancy, which as ever is based on a diagonal comparison.

Din, to give an example. Mr Smith's income is £30,000 a year. The average income of everybody in the village is £25,000.

The Alarmists say the £5,000 difference is absorbed by CO2. I say it's because everybody else in the village earns less than Mr Smith.

Bayard said...

"Imagine a sealed mirror box with an LED in it, how hot would the air get in the box? No warmer than the bulb (I expect)."

Well, since LEDs run very cool, compared to other light sources, the corollary to that would be that when the air round the LED reached the same temperature as the LED, the LED would no longer emit heat and only emit light and thus be 100% efficient. Do we know if this is the case?

Robin Smith said...

Does the heat from inside the earth need to be included in any calcs for atmospheric heating. From the radioactive decay of uranium thorium and potassium(what you get in bananas) in the crust and mantle etc?

"Geologists have used temperature measurements from more than 20,000 boreholes around the world to estimate that some 44 terawatts (44 trillion watts) of heat continually flow from Earth’s interior into space. Where does it come from?"

Solar insolation reaching the ground is 15000 TW so that would add a teeny bit to it. About 0.2% extra