<p>Air <em>without</em> rain has a mass of about <span class="math-container">$\ce{1200 g/m^3}$</span> at 20°C.</p>

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

<p><a href="https://www.dmcinfo.com/latest-thinking/blog/id/206/geek-challenge-rain-drop-density" rel="nofollow noreferrer">Here's a link</a> 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 <span class="math-container">$\ce{3/4 cm^3}$</span> of water per <span class="math-container">$\ce{m^3}$</span> of rain.</p>

<p>So, given those assumptions, the density of (rain + air) is about .06% higher than the density of air without rain. Interestingly, that's <em>much less</em> 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 <em>still</em> less dense than dry air.</p>

CC BY-SA 4.0
2019-05-12T04:47:16.647Z
2019-05-12T04:47:16.647Z
jeffB
<p>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. </p>

CC BY-SA 4.0
2019-05-10T20:46:46.893Z
2019-05-10T20:46:46.893Z
Basileios
<p>Given a rate of precipitation (in mm<sup>3</sup>/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?</p>
<p>What I mean is, if I were to take a random metric cube of rain + air during rainfall, how much would it weigh?</p>

CC BY-SA 4.0
2019-05-10T20:22:13.920Z
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
<precipitation><rainfall><rain>
Is there a way to approximate the density of air during rain?
bentheiii
# Is there a way to approximate the density of air during rain?

(Asked by: username_0 on 2019-05-10)

Given a rate of precipitation (in mm3/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?

**Tags:** precipitation, rainfall, rain

> **Comments:**
> **username_1**: 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.  

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## 2 answers

### Answer by username_2 on 2019-05-12. Score: 3 | Accepted answer

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](https://www.dmcinfo.com/latest-thinking/blog/id/206/geek-challenge-rain-drop-density) 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.


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### Answer by username_1 on 2019-05-10. Score: 1

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.


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