In order to make this a manageable question, let's add a few simplifications.

  1. The dust particles can be well described as uniform spheres of radius $R$ and density $\rho$.
  2. The space is enclosed and there is no bulk flow, i.e the air is still in a macroscopic sense.
  3. The air is at the standard temperature and pressure (STP); $T=20\ ^\circ\mathrm{C}$ and $P=1\ \mathrm{atm}$.

Under these conditions, what is the settling time for dust particles? At what size/density does Brownian motion of the air become important?


2 Answers 2


Solid particle settling time in air depends mainly on the size of the particle. Different forces become significant depending on what size range you're talking about, so it's hard to give an answer that's both concise and accurate.

I'll do my best to synthesize the important points rather than parrot a reference; that said, where practical applications in the field of air quality are concerned, the text I recommend is Air Pollution Control by Cooper & Alley. In particular, I'm going to pull many of the details for this answer from Section 3.3: Particulate Behavior in Fluids.

Gravitational Settling Overview

Dust doesn't behave like Galileo's bocce balls; small particles of different sizes fall at different rates. For solid particles, variation in settling velocity is due mainly to the influence of drag forces.

You might expect that Brownian motion would "juggle" very small particles around, keeping them from settling. Sufficiently small dust particles can remain entrained indefinitely but, practically speaking, that has more to do with the air never being perfectly still than it does with Brownian motion. In the context of air quality, we care about Brownian motion mainly when considering impaction (e.g., on water droplets in a PM wet scrubber) or deposition (e.g., on foliage near roadways). Neither of these mechanisms are relevant to the case of pure gravitational settling.

In fact, when a solid particle gets small enough to start considering the motion of discrete air molecules, we find that it actually settles a bit more quickly than Stokes' law implies. This is when we apply the experimentally-determined Cunningham slip correction factor to reduce the Stokes drag coefficient. The correction factor in air is related to the particle diameter $d_p$ and the mean free path $\lambda$ by:

$$C = 1 + 2.0 \frac{\lambda}{d_p} \left [ 1.257 + 0.40 \exp(-0.55 \frac{d_p}{\lambda}) \right ]$$

As for what "small enough" actually means, the Cooper & Alley text says:

For particles smaller than 1 micron, the slip correction factor is always significant, but rapidly approaches 1.0 as particle size increases above 5 microns.

That could be justification enough to spare yourself the time or processing cycles required to calculate the correction factor when all you're concerned with are relatively large particles.

Equation of Motion

We can derive an equation of motion in one dimension as follows.

  1. Apply Newton's second law to the particle in terms of its relative velocity in the fluid.* $$m_p v_r' = F_g - F_B - F_D$$
  2. Stokes' law gives the drag force in terms of the viscosity of the fluid and the velocity and diameter of the particle; the buoyant force is equal to the weight of the displaced fluid. $$m_p v_r' = m_p g - m_{air} g - 3 \pi \mu d v_r$$
  3. Divide by the mass of the particle. $$v_r' = g - \frac{m_{air}}{m_p} g - \frac{3 \pi \mu d}{m_p} v_r$$
  4. Express mass as the product of volume and density, where the volume of the particle and the volume of displaced air are the same. $$v_r' = g - \frac{\rho_{air}}{\rho_p} g - \frac{3 \pi \mu d}{\rho_p V} v_r$$
  5. Using $V_{sphere} = \frac{1}{6} \pi d^3$, simplify the drag force term and move it to the left side. $$v_r' + \frac{18 \mu}{\rho_p d^2} v_r = (1 - \frac{\rho_{air}}{\rho_p}) g$$

This is a linear ODE with a known coefficient (at STP) representing the following characteristic time for settling particles: $$\tau = \frac{\rho_p d^2}{18 \mu}$$

The characteristic time is a useful parameter for comparing the behavior of different systems of particles dispersed in fluids, similar to how the Reynolds number can be used to identify when different systems will have similar flow regimes. Applying the Cunningham slip correction factor gives the slip-corrected time $\tau' = C \tau$ and the equation of motion that I'll use in the next section: $$v_r' + \frac{v_r}{\tau'} = (1 - \frac{\rho_{air}}{\rho_p}) g$$

* The coordinate system for this example is defined such that the falling velocity is positive.

Terminal Velocity

For a solid particle falling in air, $\dfrac{\rho_{air}}{\rho_p}$ is close to zero. Under that assumption, setting $v_r' = 0$ in the equation of motion gives the terminal settling velocity of the particle: $$v_t = \tau' g$$

Using that terminal velocity, the solution of the equation of motion can be expressed as: $$\frac{v_r}{v_t} = 1 - e^{-t \over \tau'}$$

By the time $t = 4 \tau'$, the particle has already reached 98% of its terminal velocity. If you calculate the characteristic time for dust particles, you'll see that this takes only fractions of a second; dust particles spend most of their settling time falling at terminal velocity. The velocity itself varies significantly with particle diameter, but it can take anywhere from hours to days for fine particulates to settle just a few meters.

Larger Dust

This is all well and good for smaller dust, but what about the bigger stuff that gets in your eyes and makes you cough? Well, bad news from Cooper & Alley:

For a particle larger than 10–20 microns settling at its terminal velocity, the Reynolds number is too high for the Stokes regime analysis to be valid. For these larger particles, empirical means are required to obtain the settling velocity...

"Empirical means" is a nice way of saying figure it out yourself or else get used to reading charts that plot fitted curves with ugly decimal exponents to the results of previous experimentation.


For an individual particle, you could use Stokes' law: $$v_{\text{terminal}}=\frac{2gR^2(\rho_{\text{particle}}-\rho_{\text{air}})}{9\mu}$$ where $\mu$ is the dynamic viscosity of the air, which is easy to calculate. Using Stokes' law assumes you know the initial height of a dust particle, and may not be convenient for large quantities because each particle may start at a completely different position - in other words, it is not a model of a large system of particles.

I found some more accurate data for particles of different radii, given in half-lives; slightly more data is here.

A graph of settling time for coal, iron and cement is given here, further illustrating the non-linear, inverse-exponential relationship between dust radii and settling time.

The theory of settling is applied here to solar nebulae. I'm not sure exactly how many of the formula can be applied here, but some may be useful.

$$t=\frac{\rho_{\text{dust}}}{\rho_{\text{air}}}\frac{R}{v_{\text{thermal}}}$$ where $$v_{\text{thermal}}=\sqrt{\frac{8k_BT}{\pi \mu m_{\text{particle}}}}$$ Set the conditions to STP and you could have a better answer, though the realms of application are vastly different!

  • $\begingroup$ You start with "for an individual particle...". Is the idea also valid for a dense mist of particles? $\endgroup$ May 7, 2015 at 14:01
  • $\begingroup$ @Trilarion It is, but you would have to do different calculations for each one. $\endgroup$
    – HDE 226868
    May 7, 2015 at 14:54
  • 1
    $\begingroup$ @Air Whoops, fixed the math. What I meant about the height was that simply knowing the terminal velocity will not allow you to calculate the settling time; you need to know the initial conditions. $\endgroup$
    – HDE 226868
    May 7, 2015 at 23:37
  • $\begingroup$ True. Those nebula slides are really interesting. They bring up another limitation of the "uniform sphere" approach, which is that sub-micron particles tend to combine with each other to form larger sub-micron and fine particles. Some of it is reactive, too, or forms from precursors in the air. Lots of complexities, and an area of lots of ongoing research. $\endgroup$
    – Air
    May 7, 2015 at 23:43
  • 1
    $\begingroup$ @Air Given how much I love astrophysics, and the specific area - debris disks - being studied, it was quite a surprise to learn something new when researching something quite different, air quality. $\endgroup$
    – HDE 226868
    May 7, 2015 at 23:44

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