# Is there a way to approximate the density of air during rain?

Given a rate of precipitation (in mm^3/hour), and average droplet diameter, is is it possible to calculate (or even ballpark) the density of the air+rain droplets at time of rainfall?

What I mean is, if I were to take a random metric cube of rain+air during rainfall, how much would it weigh?

• The unit you suggest is not correct for rain rate, you would need to divide it by a surface area. That would give you for example mm/hour which equals L/m^2/hour. – Basileios May 10 '19 at 20:22

## 2 Answers

Air without rain has a mass of about $$\ce{1200 g/m^3}$$ at 20°C.

You can figure out the amount of water in falling rain if you know the rate of accumulation (how much rain falls through a given volume in a given time, landing on the ground) and the velocity of the falling rain.

Here's a link to a post elsewhere on this same question. In that discussion, they assume a rainfall rate of 1 inch (2.54cm) per hour, and estimate a raindrop's terminal velocity (from observation) at about 9 m/s. They divide that rate by that velocity (converting units), and come up with about $$\ce{3/4 cm^3}$$ of water per $$\ce{m^3}$$ of rain.

So, given those assumptions, the density of (rain + air) is about .06% higher than the density of air without rain. Interestingly, that's much less than the difference in density between dry and humidity-saturated air (about 1% at 20°C). In other words, air at 100% relative humidity with heavy rain falling through it is still less dense than dry air.

You could calculate the mass of the water per cubic meter of air during rain by dividing the precipitation rate (in kg/m2/s) through the fall velocity (m/s) of the droplets. Or alternatively, from multiplying average droplet mass (volume x water density) with their number concentration.