# [python #WWH-979089]: Python Kg/m^2/s to Inches Conversion Problem

• Subject: [python #WWH-979089]: Python Kg/m^2/s to Inches Conversion Problem
• Date: Fri, 10 Jul 2020 16:03:11 -0600

```Hello hello! Thanks for reaching out to us.

What sort of trouble are you having? Recreating parts of your provided code I
was able to physically follow your conversion (kg/m2/s * 86400s/24h * 39.37008
inches/m * 1 m3 / 1000 kg (density of water)) and come to inches/hour, and
follow your code to make my own plot. I was able also to recreate the same
using MetPy's units support
(https://unidata.github.io/MetPy/latest/tutorials/unit_tutorial.html) as below
and arrived at the same result:

from metpy.units import units

precip_units = units('kg / m^2 / s')
precip_data = data.variables['Precipitation_rate_surface'][:].squeeze()
precip_with_units = precip_units *
data.variables['Precipitation_rate_surface'][:].squeeze()  # this will create a
Pint Quantity giving our masked array physical units

density_water = units('kg / m^3') * 1000.
precip_converted = (precip_with_units / density_water).to('inches / hour')  #
.to() will perform the calculation to convert our values to inches/hour

where we have to use the density of water (with units!) to convert the given
mass flux (kg/m^2/s) to a volume flux (m^3/m^2/s) which in our 1-m^2
cross-section is our depth flux (m/s) ready to convert to inches/hour. So, from
what I can tell, everything here looks good. If you are having a different

All the best,

Drew

> I'm having trouble converting kg/m^2/s into inches per hour to plot
> precipitation data.
>
> I found the formula here I used here
> https://www.researchgate.net/post/How_do_I_convert_ERA_Interim_precipitation_estimates_from_kg_m2_s_to_mm_day
>
>
> Any help on how I can correctly convert the data would be greatly
> appreciated!

Ticket Details
===================
Ticket ID: WWH-979089
Department: Support Python
Priority: Low
Status: Closed
===================
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```