<p>I got the solution <a href="https://www.researchgate.net/post/How_to_convert_the_units_of_specific_cloud_liquid_water_from_ERA5_kg_kg_to_kg_m2" rel="nofollow noreferrer">https://www.researchgate.net/post/How_to_convert_the_units_of_specific_cloud_liquid_water_from_ERA5_kg_kg_to_kg_m2</a></p>

<p>and </p>

<p><a href="https://www.nwpsaf.eu/site/download/documentation/rtm/docs_rttov12/rttov_gas_cloud_aerosol_units.pdf" rel="nofollow noreferrer">https://www.nwpsaf.eu/site/download/documentation/rtm/docs_rttov12/rttov_gas_cloud_aerosol_units.pdf</a></p>

<p>In the above link the conversions between <span class="math-container">$\small\mathsf{kg/kg}$</span> , <span class="math-container">$\small\mathsf{g/m^3}$</span> , <span class="math-container">$\small\mathsf{kg/m^2}$</span> can be explicitly derived</p>

<p><strong>EDIT</strong>
The summary of the first hyperlink is:</p>

<blockquote>
  <p><span class="math-container">$\small\mathsf{kg/kg}$</span> stands usually for kg liquid water per kg air.
  Multiplying the term with the calculated air density within the field will result into <span class="math-container">$\small\mathsf{kg/m^3}$</span>, which means kg liquid water per <span class="math-container">$\small\mathsf{m^3}$</span> air. Then you have to multiply with the thickness of the liquid cloud, which will result into <span class="math-container">$\small\mathsf{kg/m^2}$</span>."</p>
</blockquote>

CC BY-SA 4.0
2019-12-05T13:19:42.797Z
The stuff behind a hyperlink has a bad tendency to just vanish. Please put a summary of the method in your answer itself.
Spencer
CC BY-SA 4.0
2019-12-05T13:38:36.590Z
@Spencer I have added the summary
Manmeet Singh
CC BY-SA 4.0
2019-12-05T17:01:26.560Z
OK; I made a little edit to better fit SE formatting conventions.
Spencer
CC BY-SA 4.0
2019-12-05T12:42:36.247Z
2020-01-02T19:37:03.007Z
2020-01-02T19:37:03.007Z
Manmeet Singh
<p>I am trying to initialize my model with the cloud liquid water data from ERA5. My model IITM-ESM which evolved from CFSv2 takes in the cloud liquid water information in <span class="math-container">$\small\mathsf{kg/m^2}$</span>. However the ERA5 specific cloud liquid water data is in kg/kg. I have the following questions</p>

<ol>
<li>Am I right in trying to derive cloud liquid water in <span class="math-container">$\small\mathsf{kg/m^2}$</span> for specific cloud liquid water in kg/kg</li>
<li>How do I go about this ? I found a way to convert kg/kg cloud liquid water content to <span class="math-container">$\small\mathsf{g/m^3}$</span> in the link <a href="https://www.nwpsaf.eu/site/download/documentation/rtm/docs_rttov12/rttov_gas_cloud_aerosol_units.pdf" rel="nofollow noreferrer">https://www.nwpsaf.eu/site/download/documentation/rtm/docs_rttov12/rttov_gas_cloud_aerosol_units.pdf</a>
After getting the cloud liquid water in <span class="math-container">$\small\mathsf{g/m^3}$</span>, will multiplying by layer thickness (data is 3D with lat, lon, levels and the level information is in pressure levels i.e. Pascals) and <span class="math-container">$\small\mathsf{10^{-3}}$</span> be okay to convert it to <span class="math-container">$\small\mathsf{kg/m^2}$</span></li>
</ol>

<meteorology><numerical-modelling><atmospheric-circulation>
How to convert the units of specific cloud liquid water from ERA5 (kg/kg) to kg/m2?
Manmeet Singh
# How to convert the units of specific cloud liquid water from ERA5 (kg/kg) to kg/m2?

(Asked by: username_0 on 2019-12-03)

I am trying to initialize my model with the cloud liquid water data from ERA5. My model IITM-ESM which evolved from CFSv2 takes in the cloud liquid water information in $\small\mathsf{kg/m^2}$. However the ERA5 specific cloud liquid water data is in kg/kg. I have the following questions

1. Am I right in trying to derive cloud liquid water in $\small\mathsf{kg/m^2}$ for specific cloud liquid water in kg/kg
2. How do I go about this ? I found a way to convert kg/kg cloud liquid water content to $\small\mathsf{g/m^3}$ in the link <https://www.nwpsaf.eu/site/download/documentation/rtm/docs_rttov12/rttov_gas_cloud_aerosol_units.pdf>
   After getting the cloud liquid water in $\small\mathsf{g/m^3}$, will multiplying by layer thickness (data is 3D with lat, lon, levels and the level information is in pressure levels i.e. Pascals) and $\small\mathsf{10^{-3}}$ be okay to convert it to $\small\mathsf{kg/m^2}$

**Tags:** meteorology, numerical-modelling, atmospheric-circulation


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## 1 answers

### Answer by username_0 on 2019-12-05. Score: 2

I got the solution <https://www.researchgate.net/post/How_to_convert_the_units_of_specific_cloud_liquid_water_from_ERA5_kg_kg_to_kg_m2>

and

<https://www.nwpsaf.eu/site/download/documentation/rtm/docs_rttov12/rttov_gas_cloud_aerosol_units.pdf>

In the above link the conversions between $\small\mathsf{kg/kg}$ , $\small\mathsf{g/m^3}$ , $\small\mathsf{kg/m^2}$ can be explicitly derived

**EDIT**
The summary of the first hyperlink is:

> $\small\mathsf{kg/kg}$ stands usually for kg liquid water per kg air.
> Multiplying the term with the calculated air density within the field will result into $\small\mathsf{kg/m^3}$, which means kg liquid water per $\small\mathsf{m^3}$ air. Then you have to multiply with the thickness of the liquid cloud, which will result into $\small\mathsf{kg/m^2}$."

> **Comments:**
> **username_1**: The stuff behind a hyperlink has a bad tendency to just vanish. Please put a summary of the method in your answer itself.  
> **username_0**: @username_1 I have added the summary  
> **username_1**: OK; I made a little edit to better fit SE formatting conventions.  

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