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## Q. 12.4

Calculate the fraction of an aerosol of 10 µm diameter particles that would be washed out of the atmosphere in 1 h by rainfall of 1 mm $h^{−1}$, assuming that the raindrops are all of the same radius, namely (i) 100 µm and (ii) 1000 µm. (Hints: Use Figure 12.1 or the method in Problem 1 to calculate the sedimentation velocity v of the raindrops. Assume that the stopping distance of a 10 µm diameter particle impacting on a droplet moving at velocity v is $S_{0}$ = 3 × $10^{4}v$ where $S_{0}$ is measured in m and v is in m $s^{−1}$. The efficiency of impaction $c_{p}$ of the aerosol on the droplet may be assumed similar to that for impaction on cylinders of the same diameter. Then the rainfall rate enables you to calculate how often a raindrop passes through a fixed point in the atmosphere, and if this happens n times per second, the washout coefficient of the particles by the rain drops is $nc_{p}$.)

## Verified Solution

(This problem and solution were written by Dr. A.C. Chamberlain, who was a special professor in the Environmental Physics group at the University of Nottingham)

i. The drops do not obey Stokes’ Law, so their terminal velocities must be calculated by trial and error, seeking to balance the gravitional force by the drag force.

Radius of drop r(µm)                     100                   1000

Projected area A($m^{2}$)                    3.14 × $10^{−8}$     3.14 × $10^{−6}$

Gravitational force $F_{g}$ = mg (N)   4.10 × $10^{−8}$     4.10 × $10^{−5}$

The drag force on a drop with cross-section A, falling at velocity V, is given by $F_{d} = 0.5\rho_{a} Ac_{d} V^{2}$. The dependence of the drag coefficient $c_{d}$ on drop Reynolds’ number is shown in Figure 9.6 or may be calculated from Eq. 9.13. $c_{d} = (24/Re_{p} ) (1+0.17Re^{0.66}_{p} )$ (9.13). Drag forces for a range of fall speeds for the two drop sizes are calculated in the table below:

 Radius of 100 1000 drop r (µm) Velocity 0.5 0.6 0.7 5.0 6.0 7.0 V (m $s^{−1}$) $Re_{p}$ 6.7 8.0 9.3 670 800 930 $c_{d}$ 5.5 5.2 5.0 0.53 0.50 0.48 Drag force (N) 2.65 × $10^{−8}$ 3.6 × $10^{−8}$ 4.7 × $10^{−8}$ 2.6 × $10^{−5}$ 3.4 × $10^{−5}$ 4.5 × $10^{−5}$

By interpolation, the gravitational force and drag force are equal when V = 0.65 m $s^{−1}$ (100 µm drops) and V = 6.7 m $s^{−1}$ (1000 µm drops). These are the terminal velocities of the drops.

ii. Knowing the terminal velocities, the stopping distances $S_{0}$ and Stokes numbers Stk (=$S_{0}$/r) for 10 µm diameter particles impacting on falling drops can be found using Appendix A.6 ($S_{0}$ = 200 µm, Stk = 2.0 (for 100 µm drops); $S_{0}$ = 1700 µm, Stk = 1.7 (for 1000 µm drops)).

iii. The impaction efficiency $c_{p}$ can then be read off Figure 12.3, $c_{p}$ = 0.58 (100 µm drops) and 0.54 (1000 µm drops). Note that, because $c_{p}$ increases with increasing relative velocity between the drop and the particle, it decreases with increasing drop radius, so the net variation in $c_{p}$ with drop size is small.

iv. Now calculate the number of possible impacts on a drop per second. A raindrop has volume 4πr³/3 and projected area πr². Hence each drop, considered as sweeping out a cylinder that just fits it, is equivalent to 4r/3 m of rain. A rainfall rate of 1 mm $h^{−1}$, or $0.28 \times 10^{−6} \; m \; s^{−1}$ is therefore equivalent to $0.28 \times 10^{-6} /(4r/3) = 2.1 \times 10^{-7} r^{-1} \; drops \; s^{-1}$ through each point in the atmosphere. Hence the number of drops per second (n) for 1 mm $h^{−1}$ rainfall is 2 × $10^{−3}$ with 100 µm drops and 2 × $10^{−4}$ with 1000 µm drops. The washout coefficient $\Lambda (s^{-1}) = nc_{p} =1.2\times 10^{-3} (s^{-1})$ for 100 µm drops and $1.1\times 10^{-4} (s^{-1})$ for 1000 µm drops. Hence, for a given rate of rainfall, small drops are more efficient than large drops in removing particles by impaction.

v. The fraction of aerosol remaining after a time t is $f_{w} = \exp -\Lambda t$. Hence, after 1 h (t = 3600 s), $f_{w}$ is 0.013 (1.3%) for the 100 µm rain drops and 0.49 (49%) for the 1000 µm rain drops.