Problem 7.6.3: A gas is fed to an electrostatic precipitator that has a collection area of 92000 square feet. Gas contains two distinct particles sizes; 30 percent 0.3 microns and 70 percent 0.6 micron. Particulate loading is 5 grains per cubic foot. The gas at 350 °F is flowing at a rate of 3000 cubic feet per second. The particle migration velocity, u, of the smaller particles is 0.1 foot per second. What is the particulate mass collection efficiency? The particulate discharge requirement is 0.08 grain per dry standard cubic foot.

Solution: Migration velocity of the particles is given by C d_pepsilon_o K E² over 3 mu_g.

In this relationship, epsilon_o is particle resistivity, K is dielectric constant, E is electric field strnegth, mu_g is gas viscosity; and C is Cunningham factor. Its value could be considered as 1.

Efficiency of the electrostatic precipitator is given as 1 minus exp ( minus A u over Q), where A is collection area, u is drift velocity and Q, volumetric flow rate.

For 0.3 micron-diameter particles, migration velocity is 0.1 foot per second. Collection efficiency can be found as one minus exp(minus 92000 time 0.1 over 3000) or 0.9534 or 95.34 percent. For 0.6 micron-diameter particles, migration velocity can be found to twice that 0.3 micron-diameter particles. Collection efficiency is calculated to be 0.9978 or 99.78 percent. Overall collection effiicincy is sum of weighted effciencies. That means 0.3 time 0.9534 plus 0.7 time 0.9978 or 0.9845 or 98.45 percent. Exit particulate concentration is 5 time (1 minus 0.9845) or 0.0775 grains per cubic foot.