Problem 7.6.1: A flue gas flows through a cyclone separator at a rate of 4520 actual cubic feet per minute. Gas has particulate loading of of 5 grains per cubic foot. Gas enters the cyclone separtor at a temperature of 1600 degrees centigrade. The cyclone seprator is 3 ft in diameter. Inlet duct of the separator is rectangular in shape and has a width (B_c) of 0.75 ft and height (H_c) of 1.5 ft. Particulate matter has three distinct sizes 5, 25, and 60 microns. These particles have weight fractions of 0.25, 0.5, and 0.25. Particles have density of 181.05 pouds per cubic foot. The gas viscosity is 3.11 time 10^-5 pound per foot per second. The effective number of turns for this separator (N) is 6. What is the particulate concentration (C₂) at the outlet of the cyclone?

Solution: Cyclone separators efficiency, eta, is given by one over (one plus square of ratio of d_c and d), where d_c is particle diameter collected with 50 percent efficiency. Value of this cut-off diameter can be calculated using the equation: square root of 9 mu B_c over 2 time pi time N time V_i time rho_p minus rho_g.

Flow rate is 4520 cubic feet per minute or 75.33 cubic feet per second. Duct area is 1.5 time 0.75 or 1.125 square feet. Velocity at the inlet duct is 75.33 over 1.125 or 66.96 feet per second.

Now cut-off diameter is found to be 2.14 time 10^-5 foot or 6.5 microns.

Efficiency of the cyclone seprator for 5 micron-diameter particles is 0.369 or 36.9 percent.

Cyclone efficiency for 25-micron diameter particles is 0.936 or 93.6 percent.

Cyclone efficiency for 60-micron diameter particles is 0.988 or 98.8 percent.

Amount of particulates removed is equal to particulate quantity time effciciency. It is equal to sum of the weight fraction time efficiency terms for all fractions.

For 5-micron diameter particles, its value is equal to 0.25 time 0.369 or 0.09225.

For 25-micron diameter particles, its value is equal to 0.5 time 0.936 or 0.468.

For 26-micron diameter particles, its value is equal to 0.25 time 0.988 or 0.247.

Particles removed are 0.0925 plus 0.468 plus 0.247 or 0.807 or 80.7 percent.

So, 19.3 percent of particles are leaving the cyclone, that is equivalent of 0.193 time 5 or 0.963 grains per cubic foot.