Missing Figures Round - How Do You Make the IPCC's Figures Add Up?

Posted on the 21 May 2020 by Markwadsworth @Mark_Wadsworth

Here's our baseline energy budget, averaged over 24 hours. I have been battling with this for the past 48 hours (hence no post yesterday) and have pinned down its fatal flaw:

We see that the Earth's surface receives, on average, 493 W/m2, consisting of 161 incoming solar radiation (ignore the one-eighth that is reflected straight back) plus 333 back radiation. The average temperature of the surface is 288K. That's our starting point.
If we had no atmosphere, the Earth's surface would receive 298 W/m2 incoming solar radiation, i.e. the 341 minus the one-eighth that would be reflected straight back, and its temperature would be about 250K - 255K (opinions differ). That's another fixed point.
So as a check, to see if we are on the right track, let's see if we can what calculate Earth's average surface temperature would be without an atmosphere.
You convert W/m2 to temperature in K as follows. You need to have a starting position (493 W/m2 and 288K). If W/m2 change, then the change in temperature to the fourth power is proportional to the change in W/m2.
(298/493)^0.25 = 0.605
0.605 x 288^4K (6.88 billion) = 4.16 billion
4.16 billion ^ 0.25 = 254K.
Excellent! It is widely agreed that the average temperature of the Earth, if it had no atmosphere, would be around 254K, so it stacks up so far.
Now, let's do the same exercise for night-time and day-time (working backward from temperature to find W/m2).
Night-time
Typical average Earth night-time temperature is (say) 7C = 280K.
280K^4/288K^4 = 0.893. 0.893 x 493 W/m2= 441 W/m2.
Therefore, Earth's surface must be receiving 441 W/m2 at night.
At night, the earth is clearly getting no incoming solar radiation.
Let's assume the back radiation is the same (333 W/m2).
That gives us a missing figure of 108 W/m2.
Does anybody know where this missing 108 W/m2 comes from?
Day-time
To have an average day-time temperature of (say) 22C = 295K, total incoming radiation must be 541 W/m2 (work it out yourself).
It's seems fair to assume that during the day, the Earth's surface is getting twice the average incoming solar radiation (that is how it is calculated) = 161 x 2 = 322.
The Earth's surface must be getting at least as much back radiation by day as it does on average = 333 W/m2.
That's a total of 655 W/m2 incoming. This would give an average day-time temperature of 309K = 36C, which is clearly nonsense.
To get the 541 W/m2, the amount of back radiation must be 114 W/m2 less than the average, which is also clearly nonsense.
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IMHO, the whole thing is a load of nonsense, it just crumbles under closer inspection, so using it to try and reconcile know and sensible figures is like trying to prove that the square root of the color blue equals a banana.
The correct way to explain/reconcile Earth's average surface temperature with incoming solar radiation - which works by day or by night, with or without an atmosphere - is intuitive, relatively easy to calculate (once you know how), and completely different to their model.
So my next challenge is to re-state all their figures to reconcile W/m2 and temperature for the overall average position; the position if we had no atmosphere; night-time and day-time. I think I know what they missed, possibly deliberately (i.e. what they missed is the latent heat of evaporation/condensation and the temperature/potential energy trade offs when air rises or falls). But maybe I'm wrong :-)