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I need some guidance or clarification regarding the units for precipitation rate or precipitation intensity (esp. for rainfall). I'm used to precipitation intensity being expressed in millimeters per hour (mm h-1), but I noticed that reanalysis datasets generally express precipitation intensity in kilogram per square meter per second (kg m-2 s-1) which somehow confuses me.

I've read somewhere that when converting from kg m-2 s-1 to mm h-1, the values just have to be divided by 3600. And that in terms of precipitation amount/accumulation, 1 mm of rain is equal to 1 kg m-2. I knew that it has something to do with liquid water having a density roughly equal to 1 kg L-1, but the explanations were rushed so I'm still confused on how these two units are related to each other.

So, I would like to ask: how exactly do we convert precipitation rate from kilogram per square meter per second (kg m-2 s-1) to millimeter per hour (mm h-1), and vice versa? How does the dimensional analysis for these units work out, and what is the conversion factor for this? In addition, how exactly do we also convert precipitation amount from mm to kg m-2, and vice versa?

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Thomas' reasoning is approximately correct but relies on the assumption that the density of water is equal to 1000 $\frac{kg}{m^3}$. Since the density of water changes slightly with temperature the above reasoning will fail (slightly) over different temperatures. While these temperature variations are often ignored I think it is still less confusing to express this mathematically. So, if we ignore the time units for now and answer your last question of how to get the depth of water in $mm$ from $\frac{kg}{m^2}$:

Given the density of water is ~ 1000 $\frac{kg}{m^3}$

Then the equivilent depth of water in metres is:

$[m] = \frac{\frac{kg}{m^2}}{1000\frac{kg}{m^3}}$

And so the depth of water in mm is:

$[mm] = [m]*1000[\frac{mm}{m}]$

Back to where we started! Thus we've shown $\frac{kg}{m^2}$ = $mm$ only if the density of water is ~ 1000 $\frac{kg}{m^3}$.

Now if we bring back in the seconds we can do:

$[\frac{m}{s}] = \frac{\frac{kg}{m^2s}}{1000\frac{kg}{m^3}}$

$[\frac{mm}{s}] = [\frac{m}{s}]*1000[\frac{mm}{m}]$

And since there are 3600 seconds in an hour:

$[\frac{mm}{hr}] = [\frac{m}{s}]*3600[\frac{s}{hr}]$

Slightly over complicated but hopefully helps understand the reasoning...

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Yes, this can be confusing, but let's take a look. One kilogram of water is a volume equivalent to 1000 cubic centimeters of water. How do we know that? Well, each cc of water has a mass (or standard equivalent weight) of 1 gram. Consequently, one kilogram of water will cover an area of 1000 square centimeters to a depth of 1 centimeter. But wait (!), a square meter is 10 times the area of 1000 square centimeters. Consequently, if a 1-kg volume of water covering an area of 1000 square centimeters to a 1-centimeter depth were then spread out to cover a square meter of area 10,000 square centimeters (i.e ten times the area of 1000 square centimeters) the resulting depth would be one-tenth as great, or just 1 mm. That means that the volumetric rainfall rate given in liters, or kilograms, per square meter per second, is already given in equivalent linear measure in mm/sec. But we want mm/hr. Consequently, since an hour is 3600 seconds, we would multiply the rainfall rate in kg/sq-m/sec by 3600 to get the rainfall rate in mm/hr. The total rainfall would then be the rainfall rate in mm/hr times the number of hours for a total rainfall in mm.

I think this covers what you were interested in knowing.

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