Relation between Magnitude of Greenhouse Effect and the Concentration of GHGs like $CO_2$

Question

The Average Surface Temperature of the Earth is calculated by the following equation:
$$\sigma T_s^4=\frac{(1-A)\Omega}{4}+\Delta E$$ where,
$\sigma$= Stefan-Boltzmann Constant
$T_s$= Average Surface Temperature of Earth
$A$= Global Albido
$\Omega$= Total Solar Irradiance
And, $\Delta E$= Magnitude of Greenhouse Effect

The observed average surface temperature of Earth is about $288K$, Global Albido as seen by satellites is about $0.3$ and the total solar irradiance is about $1370W/m²$. Putting these values in the above equation, $\Delta E$ comes out to be about $150W/m²$.

Now, according to this PDF I found online, the concentration of $CO_2$ is related to $\Delta E$ by the following equation:

$$\Delta E=133.26+0.044[CO_2]$$

When we put $\Delta E=150$, we get the $[CO_2]=380$ which is the actual concentration of $CO_2$ in the atmosphere according to some online sources in ppm. So the above relation seems correct.

How was this relation calculated? And, how to calculate such relations for other Greenhouse gases, such as $H_2O$?

In other words, given the Equation:

$$\Delta E=x+y[H_2O]$$

Find x and y.

What I noticed: The coefficient of $[CO_2]$ in the above mentioned equation is $0.044$. And, the molecular mass of $CO_2$ is $44.01 amu $. So, Molecular mass might be involved in the calculation of $y$.

My approach: I focused on finding $y$ as $x$ can be calculated later from the same equation since $\Delta E$ is known and concentration of $H_2O$ should be available online. According to my understanding of the Greenhouse effect, $y$ represents how much energy per unit area does $1$ $ppm$ of a $GHG$ can trap and transmit down to Earth. I tried searching online for some data on the same, but could find none.

Please throw some light on the topic.

Thank You.

Answer 1

Δ𝐸=133.26+0.044[𝐶𝑂2]

It’s just a numerical coincidence that the 0.044 looks like the molar mass of CO2. That equation gives 0.044 * 380 = 16 W/m2, which is the reduction in surface downwelling IR if you remove all the CO2 from the present day atmosphere (e.g., see Table 1 of Zhong and Haigh, 2013, Weather, https://doi.org/10.1002/wea.2072, pdf).

Looking at water vapour in the same way exposes some of the limits of the model they’ve set up for this tutorial, which was presumably kept simple just to teach some concepts to their students. To do the same thing for water vapour as was done for CO2 would require removing 208 W/m2 of greenhouse effect (again, see Table 1 of that article), more than the total greenhouse effect of 150 W/m2 in their model. The discrepancy is because they have too much solar radiation in the surface energy budget, so the diagnosed greenhouse effect doesn’t have to do as much work to balance a 288 K surface temperature.

A more realistic model would exclude the 80 W/m2 of solar radiation absorbed by the atmosphere, leaving an absorbed solar flux at the surface of 160 W/m2. The balance is then $390 = 160 + \Delta E$, so $\Delta E = 230$ W/m2 and the revised perturbation equation for CO2 is,

$$\Delta E = 214 + 0.044[CO2]$$

The difficulty with H20 is that, unlike CO2, the concentration of water vapour is highly variable in space and time, so the equivalent equation will depend a lot on what you chose as present-day H2O concentration. A ball-park figure for mean H2O concentration is 4000 ppm, which leads to the equation,

$$\Delta E = 22 + 0.052[H2O]$$

I should emphasise that this model isn’t intended for any serious calculation. It’s just a tool to get students used to quantifying parts of the Earth system and how they can trade off against each other.