Question 6 of 7: Atmospheric Behaviour of SO 2 and Electrostatic Precipitator Terminal Velocity

Question 6: Atmospheric Behaviour of SO₂ and Electrostatic Precipitator Terminal Velocity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) Atmospheric Behaviour of SO₂: Emission to Final Sink

Emission and initial dispersion. SO2 is released primarily from fossil-fuel (especially coal and heavy fuel oil) combustion and metal smelting, entering the atmosphere as a buoyant, hot plume that undergoes the physical dispersion processes of Question 4 — plume rise, then Gaussian/Eddy dilution governed by atmospheric stability and wind speed. As a moderately water-soluble, reactive gas, a fraction of SO2 is removed close to the source by direct dry deposition — uptake onto vegetation, soil and water surfaces at a rate set by the deposition velocity $v_d$.

Atmospheric chemical transformation. The SO2 that is not dry-deposited undergoes gas-phase oxidation by the hydroxyl radical, $\text{SO}_2+\text{OH}\!\cdot\rightarrow\text{HSO}_3\!\cdot$, proceeding on to sulfuric acid and sulfate aerosol over a timescale of days; a parallel and often faster aqueous-phase pathway occurs inside cloud and fog droplets, where dissolved SO2 is oxidized by H2O2 or catalytically by dissolved transition metals (Fe, Mn) to sulfate. This chemical conversion is the pivot of SO2's environmental story: a primary gas is converted into a secondary fine-particulate sulfate aerosol, which itself becomes a major contributor to regional PM2.5 mass and to visibility-reducing haze (Question 5(iii)).

Transport and final sink. Because gas-phase SO2-to-sulfate conversion takes days, SO2 and its sulfate product can be transported hundreds to thousands of kilometres downwind before final removal — this is the physical basis of transboundary acid-rain disputes. Final sinks are: (1) wet deposition — sulfate aerosol and dissolved SO2 are scavenged by precipitation (acid rain/snow, typically the dominant removal pathway for the converted sulfate), acidifying lakes, streams and soils, particularly in areas with thin, poorly buffered (low-carbonate) soils; and (2) dry deposition of the sulfate aerosol itself onto terrestrial and aquatic surfaces. The net result is that a molecule of SO2 emitted at a stack can end its atmospheric life as an acid-rain sulfate ion deposited in a lake or forest far from where it was released, which is why SO2 control (Question 1(ii) FGD) is as much a regional as a local air-quality issue.

(ii) Why ESPs Excel on Small Particles, and the Terminal Electrostatic Velocity

Why ESPs are better on small particles. A mechanical collector (cyclone, gravity settler) relies on inertial or gravitational force, both of which scale with particle mass ($\propto d_p^3$) and therefore weaken sharply as $d_p$ shrinks. An ESP instead relies on electrostatic force: particles are first charged (by ion bombardment in a corona discharge) in proportion to their surface area (roughly $\propto d_p^2$ for field charging in the size range of interest), and the electrostatic driving force in the equation below is directly proportional to that charge, not to particle mass. Because the electrostatic force falls off far more slowly with decreasing particle size than gravitational/inertial force does, an ESP maintains meaningful, controllable collection efficiency for sub-micron particles that a cyclone or gravity settler essentially cannot touch — which is exactly why ESPs (and, similarly, fabric filters relying on diffusional/interception capture) are the technology of choice wherever the fine PM2.5 fraction must be controlled.

Given. A single unit-density particle in an electric field inside an ESP:

Find. The terminal electrostatic (migration) velocity $v_t$.

Approach. Convert the electric field to SI units and substitute directly into the supplied migration-velocity equation.

1. Convert the electric field to SI. $E = 2\ \text{kV/cm} = \dfrac{2000\ \text{V}}{0.01\ \text{m}} = 2.0\times10^5\ \text{V/m}$.
2. Convert the particle diameter to SI. $d = 10\ \mu\text{m} = 1.0\times10^{-5}\ \text{m}$.
3. Substitute into the migration-velocity equation. $$v_t = \frac{n\,e\,E\,C_c}{3\pi\mu d} = \frac{(700)(1.602\times10^{-19})(2.0\times10^5)(1.09)}{3\pi(1.81\times10^{-5})(1.0\times10^{-5})}.$$
4. Evaluate. $$v_t = \boxed{0.0143\ \text{m/s} \approx 1.43\ \text{cm/s}}.$$

This migration velocity is the key sizing parameter in the Deutsch-Anderson equation used in Question 7(i): a higher $w$ (equivalently, a higher $n$, $E$ or lower $d$) directly raises collection efficiency for a fixed plate area and gas flow. Because $v_t$ here does not depend on particle mass at all, doubling the particle diameter changes $v_t$ only through the weak, near-linear $1/d$ term rather than the strong $1/d^2$-type mass penalty a mechanical collector would show, reinforcing why electrostatic collection remains effective well into the sub-micron range.