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ORAC-CC / orac

Optimal Retrieval of Aerosol and Cloud
GNU General Public License v3.0
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Why do we have two equations for the ice water path? #73

Open adamcpovey opened 2 years ago

adamcpovey commented 2 years ago

Within the main processor, we assume ice water path can be calculated from the same formula as liquid water path, but with different values for the density and extinction efficiency.

Within the flux code, we use a formula from Liou (1992).

The second result isn't output, being only used to calculate cloud physical thickness, but does beg the question why we have two equations for the same thing?

And while I'm here, the coefficient we're using for liquid water path (2/3) is derived for a vertically homogeneous cloud. At a quick dig through the literature implies 5/9 would be better. Anyone have an opinion?

andyprata commented 2 years ago
  1. I believe the first point assumes that ice particles and water droplets are spherical and are vertically homogeneous.
  2. I haven't been able to access Liou (1992), so not sure how it was derived. I think it's a 2-parameter fit and probably takes into account the fact that ice particles are not spherical.
  3. Wood and Hartmann (2005) say the 5/9 applies to boundary layer clouds. Question is - does this hold for other types of liquid water clouds? My view is that the 2/3 is the most basic assumption and can be applied generally to all liquid water clouds with the caveat that it will be less accurate for vertically inhomogeneous clouds. However, if we can identify boundary layer cloud then yes why not apply the linear approximation for vertical inhomogeneity (i.e. 5/9) in those cases.