<blockquote>
<p>Δ=133.26+0.044[2]</p>
</blockquote>
<p>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, <a href="https://doi.org/10.1002/wea.2072" rel="nofollow noreferrer">https://doi.org/10.1002/wea.2072</a>, <a href="https://courses.seas.harvard.edu/climate/eli/Courses/global-change-debates/Sources/CO2-saturation/more/Zhong-Haigh-2013.pdf" rel="nofollow noreferrer">pdf</a>).</p>
<p>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.</p>
<p>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 <span class="math-container">$390 = 160 + \Delta E$</span>, so <span class="math-container">$\Delta E = 230$</span> W/m2 and the revised perturbation equation for CO2 is,</p>
<p><span class="math-container">$$\Delta E = 214 + 0.044[CO2]$$</span></p>
<p>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,</p>
<p><span class="math-container">$$\Delta E = 22 + 0.052[H2O]$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2021-02-27T11:09:43.107Z
2021-02-27T11:09:43.107Z
Deditos
<p>The Average Surface Temperature of the Earth is calculated by the following equation:<br />
<span class="math-container">$$\sigma T_s^4=\frac{(1-A)\Omega}{4}+\Delta E$$</span>
where,<br />
<span class="math-container">$\sigma$</span>= <em>Stefan-Boltzmann Constant</em><br />
<span class="math-container">$T_s$</span>= <em>Average Surface Temperature of Earth</em><br />
<span class="math-container">$A$</span>= <em>Global Albido</em><br />
<span class="math-container">$\Omega$</span>= <em>Total Solar Irradiance</em><br />
And, <span class="math-container">$\Delta E$</span>= <em>Magnitude of Greenhouse Effect</em></p>
<p>The observed average surface temperature of Earth is about <span class="math-container">$288K$</span>, Global Albido as seen by satellites is about <span class="math-container">$0.3$</span> and the total solar irradiance is about <span class="math-container">$1370W/m²$</span>. Putting these values in the above equation, <span class="math-container">$\Delta E$</span> comes out to be about <span class="math-container">$150W/m²$</span>.</p>
<p>Now, according to <a href="https://web.viu.ca/krogh/chem302/CHEM%20302%20Practise%20Questions%208%20scanned%20solns.pdf" rel="nofollow noreferrer">this</a>
PDF I found online, the concentration of <span class="math-container">$CO_2$</span> is related to <span class="math-container">$\Delta E$</span> by the following equation:</p>
<p><span class="math-container">$$\Delta E=133.26+0.044[CO_2]$$</span></p>
<p>When we put <span class="math-container">$\Delta E=150$</span>, we get the <span class="math-container">$[CO_2]=380$</span> which is the actual concentration of <span class="math-container">$CO_2$</span> in the atmosphere according to some online sources in ppm. So the above relation seems correct.</p>
<p>How was this relation calculated? And, how to calculate such relations for other Greenhouse gases, such as <span class="math-container">$H_2O$</span>?</p>
<p>In other words, given the Equation:</p>
<p><span class="math-container">$$\Delta E=x+y[H_2O]$$</span></p>
<p>Find x and y.</p>
<p><em>What I noticed</em>: The coefficient of <span class="math-container">$[CO_2]$</span> in the above mentioned equation is <span class="math-container">$0.044$</span>. And, the molecular mass of <span class="math-container">$CO_2$</span> is <span class="math-container">$44.01 amu $</span>. So, Molecular mass might be involved in the calculation of <span class="math-container">$y$</span>.</p>
<p><em>My approach</em>: I focused on finding <span class="math-container">$y$</span> as <span class="math-container">$x$</span> can be calculated later from the same equation since <span class="math-container">$\Delta E$</span> is known and concentration of <span class="math-container">$H_2O$</span> should be available online. According to my understanding of the Greenhouse effect, <span class="math-container">$y$</span> represents how much energy per unit area does <span class="math-container">$1$</span> <span class="math-container">$ppm$</span> of a <span class="math-container">$GHG$</span> can trap and transmit down to Earth. I tried searching online for some data on the same, but could find none.</p>
<p>Please throw some light on the topic.</p>
<p>Thank You.</p>

<meteorology><atmosphere><temperature><homework><greenhouse-gases>
Relation between Magnitude of Greenhouse Effect and the Concentration of GHGs like $CO_2$
RSR
# Relation between Magnitude of Greenhouse Effect and the Concentration of GHGs like $CO_2$

(Asked by: username_0 on 2021-02-25)

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](https://web.viu.ca/krogh/chem302/CHEM%20302%20Practise%20Questions%208%20scanned%20solns.pdf)
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.

**Tags:** meteorology, atmosphere, temperature, homework, greenhouse-gases


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## 1 answers

### Answer by username_1 on 2021-02-27. Score: 4 | Accepted answer

> Δ=133.26+0.044[2]

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](https://courses.seas.harvard.edu/climate/eli/Courses/global-change-debates/Sources/CO2-saturation/more/Zhong-Haigh-2013.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.


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