Blowin’ in the wind: dust and fog

The question that I started with was: how does the wind decrease the concentration of dust particles in the atmosphere? I Googled a bit but I was unable to find the answer in terms that made sense to me. I’m sure this question has been done to death by now, but I regret to say I couldn’t find it. My bad, undoubtedly. Anyway, I tried the simplest approach, and that is what I have given below – but it got a bit complicated as it progressed. Still, nowhere as complicated as some computer code written by some group of atmospheric scientists! Nevertheless, the caveat is that the following analysis is not peer-reviewed and should be taken (at best) as heuristic.

The assumption, of course, is that the dust is fairly localized (because the sources are factories or cars in a city), and that the wind blows in clean air from somewhere else. Also, that the wind is steady, and blows in a specific direction at a fixed wind speed. So, the dust particles will only get transported away, from place A to neighbouring place B, but not dissipated as such (the number density of particles remains the same, only the location changes). A similar argument applies to fog droplets. This may also apply to fine rain droplets with a diameter of around 250 microns, as adding a horizontal component to their velocity, but may be less useful for large raindrops.

Q: How does a free particle in the atmosphere of a given diameter – in the range 2.5 to 100 microns -   move (horizontally) when the wind blows horizontally at, say, v = 1 m/s speed? This particle could be a dust particle, or even a water droplet. The density is assumed to be 1 gm/cc, although some reports quote higher values of density for dust particles.

Particles of diameter more than 10 microns tend to settle down to the ground pretty fast, while those whose diameter is less than 10 microns tend to remain suspended in the atmosphere for a significant amount of time.

Assuming the particle is spherical and has a density r_p, its mass is:

                                                          m = r_p4pr³/3

and its effective cross-sectional area for the wind is A = pr²(taking into account the cos(q) factor).

The force due to the wind is: F = (r_av²/2)(A) where the density of air is r_a = 1.225 kg/m³.

Equating this with F = ma, we can find the acceleration of the particle:

                                                         a = [3(v_a)²r_a]/[8rr_p]

For a 2.5 mm diameter particle, and assuming a particle density of 1000 kg/m³, and with wind speed v_a = 1 m/s, the acceleration is a = 180 m/s² or about 18 g’s. If this acceleration were to remain unchanged, the particle would hit sound speed in 2 secs!

Clearly that’s not gonna happen, and to make sure it doesn’t, we have to drag in Stokes’ law.

The force law then is:

                                                        pA - 6phrv = ma

The terminal velocity (horizontal) will be reached when acceleration ‘a’ becomes zero:

                                                        v_T = [r_a(v_a)²r]/[12h]

which is independent of the density of the particle r_p, but depends upon its size r!

The viscosity of air is: h = 1.8x10^-5 kg/m-sec and the density of air is: r_a = 1.225 kg/m³.

This gives the terminal velocity as 0.014 m/s (0.05 kph). Which seems to be pretty low!

Note also that smaller particles have lower terminal velocity, with v_T directly proportional to r.

However, Stokes’ law is only valid if Reynolds number is Re £ 1. It is defined as:

                                                     Re = [Dr_flV_rel]/h_fl

Where r_fl is the density of the fluid (1.2 kg/m³) and h_fl is its viscosity, D is the diameter of the particle, and V_rel is the relative velocity of the particle and the fluid (air). In the above case, even though the terminal velocity of the particle is low (0.014 m/s), the wind-speed (1 m/s) is higher, and Re is calculated on that basis.

When Re>10 or so the flow is turbulent, and the drag force is proportional to the square of the particle velocity v_p:

                                                    F = [r_a(v_p)²C_DA]/2

Where C_D is the dimensionless drag coefficient and the fluid density is the air density. The value of C_D for laminar flow (Re<1 approximately="" by:="" given="" is="" o:p="">

                                                      C_D = 24/Re

For this approximation, the square law force reduces to Stokes’ law. For a spherical particle, and large Re, C_D = 0.47.

The wind-speed for which Stokes’ law is valid is shown in the Figure below. The value goes up, as the particle diameter goes down:

The more general force equation involving the drag coefficient now becomes:

                                                  [r_a(v_a)²A/2] – [r_a(v_p)²C_DA/2] = ma

Since most terms cancel out for a = 0, the terminal particle velocity becomes:

                                                     v_p =  v_a/[C_D]^1/2

According to www.thermopedia.com/content/707 the following expression for the drag coefficient is valid in the range: 0.2 < Re < 2000:

                                                 C_D = [21.12/Re] + [6.3/(Re)^1/2] + 0.25

which reduces approximately to the limit 24/Re for low Re, and to 0.47 for high Re.

                                                                               

Note that the terminal velocity of the particle associated with settling under the action of gravity is given by:

                                                   6phrv = r_p [4pr³/3] g

Which leads to the settling terminal velocity:

                                                   v_sT  = [gd²r_p]/[18h]

The ratio of the above terminal velocity to the settling terminal velocity is:

                                                  [v_T/v_sT] = [3r_a(v_a)²]/[4gdr_p]

For a 25 mm diameter particle of density 1000 kg/m³, the ratio becomes:

                                                  [v_T/v_sT] = 3.67(v_a)²

i.e. even for a wind-speed of 1 m/s, v_T > v_sT. For a particle of smaller diameter, the wind-speed needed to exceed the settling speed increases.

An energy argument can be used, an integral of the force law argument used above, which gives the same result as the force law:

Note the work is done on the particle by drag, so you need to multiply the air pressure term by the distance (v_p dt).

The 1^st term is:

                                                  [r_p(v_p)²A/2] [v_pdt]

The drag term is:

                                                  [6phrv_p] [v_pdt]

The inertial term is:

                                                  (1/2)[d(m(v_p)²/dt](dt)

Removing the common term (v_p dt), you get the equation:

                                                  pA - 6phrv_p  = m[dv_p/dt]

which is the same as the force law case, with Stokes’ law drag force.

Using the above form of drag coefficient as a function of Reynolds number, no assumption need be made that Re < 1; that is, we need not be restricted to the Stokes’ law case, but can do the calculation for the general case, including both the square law and the linear (Stokes’) law cases.

The calculations become a nuisance to do by hand; so I wrote a simple BASIC program to do the calculations. Note that since Re is defined as the relative velocity between the particle and the fluid (the air in the atmosphere), that needs to be incorporated. First the acceleration is calculated, then the particle velocity, and then the relative velocity, and then the drag, based on the C_d that depends on Re.

The end result is as expected: the acceleration starts with a high value and quickly decreases to zero, while, correspondingly, the velocity increases exponentially, and then saturates at the terminal velocity. Note that this is a horizontal terminal velocity, not the vertical terminal velocity that we are accustomed to calculate, which latter is related to the settling rate of large diameter particles.

The velocity vs time graph:

From the above two graphs, it is clear that the terminal velocity is attained within 10 msec (for 250 mm particles), although this time is shorter for smaller particles.

The plot of v_p vs v_a is next. Initially a quadratic plot was attempted for 250 mm particles, and the fit seems ok:

But it isn’t: the extrapolation of the quadratic for zero wind-speed shows a negative value of the particle velocity, which does not make sense.

One alternative is to again fit it to a quadratic, but with the constant term set to zero. This does not give a good fit, except at some points (not shown). The rigorous way is to use the least squares errors method with the constraint that the constant term c = 0. I could not figure that out so I didn’t. Anyway, the expression for the drag coefficient is not valid for Re < 0.2 so v_a = v_p = 0 is not workable. So, instead I have fitted a log-log plot to a quadratic as below:

where the v’ denotes log(v), and the number of points decreases as D decreases, because Re values lower than 0.2 are not considered.

The above graph, and the previous ones, shows something like a break-even velocity, at which the particle velocity equals the wind velocity: v_p = v_a. Below this value of v_abr, v_p < v_a; and above it, v_p > v_a.

A concern about the wind blowing the dust or fog away: this assumes that the wind does not change direction randomly. For example, it went from North to South at one moment, to South to North the next, the dust or fog would stay put (on average). But even if the direction of the velocity stays the same, if the Reynolds number exceeds one, the motion of the particle will be turbulent (Re is calculated with the relative velocity). Then all bets are off – the particle could go in any direction, follow a vortex, whatever. Only in laminar flow (Re < 1), will the motion be straight line (or smoothly curved). So the above calculation would only yield sensible results for laminar flow, and that means the wind speed v_a should be < 0.1 m/s  for 250 mm particles, <1 25="" for="" m="" s="" span="" style="font-family: Symbol; mso-ascii-font-family: Calibri; mso-ascii-theme-font: minor-latin; mso-char-type: symbol; mso-hansi-font-family: Calibri; mso-hansi-theme-font: minor-latin; mso-symbol-font-family: Symbol;">m

m particles and <10 2.5="" for="" m="" s="" span="" style="font-family: Symbol; mso-ascii-font-family: Calibri; mso-ascii-theme-font: minor-latin; mso-char-type: symbol; mso-hansi-font-family: Calibri; mso-hansi-theme-font: minor-latin; mso-symbol-font-family: Symbol;">mm particles (as indicated by arrows in the above graph). This is not merely a problem of calculating turbulent particle flow (which it is, at least for me!), but it also means that higher velocity winds will not be especially effective in dispersing dust or fog.

This argument is similar to that used in clean-room design, which all use laminar flow – albeit with different Reynolds number value (Re_tr = 2000) for transition from laminar to turbulent flow. Anyway, there is an optimal speed for the unidirectional flow, and it is around 0.3 m/s (60 ft/min). Increasing the speed does not help, as turbulent flow is not efficient in uniformly removing the dust in the atmosphere.

And this does not even begin to address the other question: how fast does the wind have to blow to kick up a dust devil or a sand-storm! That issue is being brushed under the carpet, buried six feet under and resolutely avoided!

The time taken to move a bunch of particles a distance Dx, would be Dt = Dx/v_p.  As mentioned above, this assumes that the dust (or droplets) are localized within this distance Dx, and clean (droplet-free) air is being brought in by the wind from outside this area.

Suppose we take an area of width 1 km, and v_a = 1 m/s (3.6 kph), with 25 micron diameter particles, v_p = 0.21 m/s, and it would take 4,673 secs (1.3 hrs) to clear the dust away. With a 10 m/s (36 kph) wind, v_p = 4.61 m/s, it would take only 217 secs (3.6 mins). However, as noted above, for 25 micron diameter particles v_a = 10 m/s is probably in the transition zone (1 < Re < 10) between laminar and fully turbulent flow, so this estimate may be questionable.

Now, let us repeat the calculations for 2.5 micron diameter particles: v_a = 1 m/s (3.6 kph), v_p = 0.0805 m/s, and it would take 12,422 secs (3.4 hrs) to clear the dust away. Fine particles are tougher to get rid of! Similarly, with a 10 m/s (36 kph) wind, v_p = 2.14 m/s, it would take only 467 secs (7.8 mins). In this case, 2.5 micron diameter particles: v_a = 10 m/s, is just within the laminar flow regime, and the calculation is correct.

These calculations assume that we can neglect the time involved in accelerating the particles, which seems reasonable, if we look at the ‘a vs t’ and ‘v vs t’ figures plotted earlier (as mentioned above).

Obviously, these calculations are simplistic. You would need to write computer code to take more realistic scenarios into account. But the basic idea is probably correct. A low wind speed is insufficient to blow away fog hanging overhead, but as soon as a brisk wind gets going, the fog quickly lifts.

Another more quantitative question could be: if I have a given number density n of dust particles already in the air, and a dust generating source with a fixed generation rate, s (units of m^-3sec^-1), how does the number density change? The answer is that you would have to solve the continuity equation:

                                                                                [d(nv)/dx]  + [dn/dt]  = s

putting in all the relevant boundary conditions and taking the nonuniform number density into account.

At a naïve level, for s = 0, the continuity equation actually is the same as the rough calculation done above to find the Dt from an assumed Dx, in terms of v_p. For laminar flow, if we assume some n(x), with s = 0, the same ‘shape’ n(x) would only get translated to n(x + v_p Dt). If the distance the dust gets moved is Dx/2, probably the value of n drops by 50%. If there is a dust-generation term, it would have to be added on, at the appropriate value of x.

And it should be in 3D. Easy to talk about: a lot of work to actually do it. It ought to be possible to get an idea of what it looks like by doing a simple 1D case. Some other time!