Sometimes, the model also uses a non-rotating Earth .. but that is going WAY too far.
The formulas below are based on Stefan's law which determines the blackbody temperature of an object at equilibrium.
P * (1-Albedo) = sigma * T^4
In this model,
Ts = [1/sigma (S/4)]^0.25
Ts = [1/sigma (A + S/4)]^0.25
Ta = [1/sigma (2 * A)]^0.25
2 * A = A + S/4 A = S/4
With "real" numbers, the Sun provides
With an albedo of 30%, the surface would be -18.4°C without an atmosphere and 29.7°C with one.
Why the model gives the wrong results
Assuming a 30% albedo and an atmospheric absorption of 80%, the surface temperature would be 16°C (397 W/m2).
S = 1361 Sun TOA S/4 = 340.25 Sun TOA/4 S/4 (1-0.30) = 238.175 Albedo = 30% 2 * A = 80% L The atmosphere absorbs 80% of the surface energy But 'A' is emitted up AND 'A' is emitted down (20% L) + A = S/4 (1-0.30) The amount of heat lost equals the amount from the Sun 20% L + 40% L = S/4 (1-0.30) Just combine to remove 'A' 0.6 * L = S (0.7/4) Simplify L = S (0.7/2.4) = 397 W/m2 => 16.27°C Close to 15°C A = 0.4 * 397 = 158.8 W/m2 158.8 + 238.2 = 397.0 W/m2 The values check A + S = L
At current values, doubling CO2 from 350 ppm will increase the energy absorbed by the atmosphere by 10 W/m2 (based on a line-by-line model that I hope to complete soon). That increases L by 5 W/m2 and produces a new surface temperature of about 17.2°C ... an increase of about 1°C (assuming no cloud or albedo feedbacks).
By the way, if the atmosphere emits 158.8 W/m2, that is associated with an expected temperature of around -30°C (158.8/0.8 = 198.5 W/m2).
Real World Model
2613.9 W/m2 Sun TOA
Just to be clear, the computed values above assumed that 100% of the available energy reached the surface and that only one side of the planet faced the Sun. (Each side actually faces the Sun for 58 days before being dark for 58 days.) Since it is known that about 75% of the solar radiation is reflected by the clouds (and, therefore, never reaches the surface) the actual maximum Greenhouse temperature is much less.
The bottom line is the same - Venus is heated internally and not via some Greenhouse Effect.
Venus - Try 2
On Venus, the atmosphere temperature decreases for about 60 km with a lapse rate of 7.9°K/km. At that point, there are clouds. Above the clouds, the atmosphere cools at about 3.0°K/km for the next 30 km. Unfortunately, I do not know of any way to determine heat loss based on the slope of the lapse rate. My calculator says that CO2 will emit 136 W/m2 at 60 km and 263°K (-10°C). If the clouds are decent blackbodies, they will emit 271 W/m2.
From this rough computation, it is possible that only 136 W/m2 reaching the surface could conceivably produce a 464°C surface temperature because the energy must travel the slow way through the atmosphere before being released to space.
136 W/m2 / 2613.9 W/m2 = 0.052 60 km * 7.9°K/km = 474°K 474°K + 263°K = 737°K (464°C) -> Surface temperature
Implying that only 5% of the available solar radiation could produce a high surface temperature if the atmosphere is considered opaque, thermally conductive, and emits energy at a low temperature.
This also shows how ridiculous the single layer model is. (Climate models only use 4 to 7 constant temperature layers ... and you wonder why I don't trust them.)
Reference: Atmospheric Flight on Venus provides the temperature verses altitude data. The computations are made with simple programs I wrote. These computations should be considered speculative, but they do show how a small amount of energy could create a very high surface temperature.
By the way, because the day to night temperature difference is so small, it is not clear that any heat reaches the surface from the Sun. It is probably more likely that the energy comes from an internal heat source.
The article claims (without a reference) that the atmospheric emissivity is 0.78. (I used 0.80 above.)
Assuming a temperature of 15°C and an emissivity of one, the surface should emit 390.08 W/m2 (using Stefan's equation). Using my line-by-line program, the atmosphere should emit 378.68 W/m2 (assuming 15°C and 1.0% water vapor) producing an effective emissivity of 0.97.
This is a major difference.
If I use their atmospheric temperature of 242.5 °K (-30.5°C), then my program states that 184.25 W/m2 is emitted in each direction (up and down).
184.25 / 390.08 = 0.47 2 * 184.25 / 390.08 = 0.94
The test of these models is to see if they work on Venus. (Hint - theirs doesn't.)
The other main flaw in this model is that only radiative heat is considered. As a result, they claim that both the atmosphere and surface temperatures will increase as more Greenhouse gases are added. (And the model does predict that.) However, when convective heat is added to the model (ie, if the model is more real world), then adding more Greenhouse gases will always cause the temperature of the atmosphere to decrease.