I want to compare ERA Interim precipitation estimates with ground measurements so i do not know to convert the data from kg/m2/s to mm/day.
Thanks in advance
Nkiaka
You have to consider that:
- 1 kg of rain water spread over 1 square meter of surface is 1 mm in thickness;
- there are 60X60X24=86400 seconds in one day.
Therefore, 1 kg/m2/s = 86400 mm/day.
Considering that this would be an enormous amount of rain in a day by any standard, perhaps 1 kg/m2/s could be in reality 1 kg/m2/year, or per month, or per day. 1 kg/m2 is always 1 mm of rain.
Dear Elias,
You have to divide ERA-I by the density of water so you get m/s and then it's easy to convert into mm/day!
G.C.
You have to consider that:
- 1 kg of rain water spread over 1 square meter of surface is 1 mm in thickness;
- there are 60X60X24=86400 seconds in one day.
Therefore, 1 kg/m2/s = 86400 mm/day.
Considering that this would be an enormous amount of rain in a day by any standard, perhaps 1 kg/m2/s could be in reality 1 kg/m2/year, or per month, or per day. 1 kg/m2 is always 1 mm of rain.
If you are using Netcdf data, there CDO which can be used
cdo mulc,86400 input.nc ouput.nc
From what i later found in the literature, ERA Interim total precipitation is given in metre(s)/day or m/d so i just had to multiply the values i had with 1000 to convert them to mm/day.
thank you all for your answers
Easiest way is to multiply by 86400 following the logic listed above by Roberto.
1 kg/m^2/sec = 1000 gm / (100cm)^2 /sec
= 1000 cm^3/(10000cm^2)/sec (1 cubic cm volume of water is 1 gm of water)
=1 cm/10/sec
=10 mm/10/sec
=1 mm/sec (86400 sec= 1 day, i.e. 1 sec= 1/86400 day)
= 1 mm /(1/86400)day
=86400 mm/day
Hi Elias,
If you have M/L^2/T and you whant L/T, just divide your Kg*m^-2/s by the density of the the water (check the water temperature), and then you have m^2/s what is the same of mm/s. Then you can change mm/s to mm/day.
First, convert days to second using;
60*60*24=86400sec
this is important because 1k.g. of a water spread over 1m^2 of surface 1mm in a thickness.
then if you're linux user, on terminal use the following approach (actually if you do have .nc file)
cdo mulc,86400 ifile ofile
but if you do have different variables first try to select the variable you want to convert using a command
cdo select,name=precip ifile ofile
Use a unit analysis
mm/d=(precip kg/(m^2.s))*(1/ρ_water *m^3/kg)*(86400*s/d)*(1000 mm/m)
precip =precipitation
ρ_water= density of water or rain in kg/m3.
86400: number of seconds in a day
- 1 mm = 1 liter/ m2
- 1 liter = 1 kg
- 1 hr = 3600 s
- 3600 s x 24 hr (1 day) = 86400 s
To convert the precipitation of [kg/(m2 . s)] into (mm/day); we need to multiply 86400.
cdo mulc,86400 input.nc output.nc
The Climate Data Operator (CDO) is the right choice for converting .nc files.
Its correct to multiply the given values by 86400, however I would strongly suggest to adjust also the related "units" attribute, if you deal with NetCDF files. It avoids misunderstands if you share the file or delivers correct plot labels. Hence, I suggest:
cdo -setunit,'mm/day' -mulc,86400. input.nc output.nc
You should also consider a peculiarity of the EraInterim data files. Precipitation is integrated from 0 UTC (Coordinated Universal Time) to 12 UTC and again from 12 UTC to 24 UTC. So if you have files containing data every 6 hours, you can use the first 6 hours (0 UTC to 6 UTC) without any hassle, but the second 6 hour period (6UTC to 12UTC ) contain also the accumulated precipitation of the first period.
Computation of daily precipitation according to ECMWF: https://confluence.ecmwf.int/display/CKB/ERA-Interim%3A+How+to+calculate+daily+total+precipitation
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