Design a gravity settler to remove 85% ofthe particulatesfroma5efs airflow with

Question

Design a gravity settler to remove 85% ofthe particulatesfroma5efs airflow with the following characteristics: 1. Temperature 135 Total Particulate mass 400mg/m Particle density 2.2 g/em Viscosity of air 3.3 x 10- lbs s/ft2, 1.7 x 10-5 N"s/m2 I N s/m2-1 Pas 10 poise 1,000 miliPa s 1 ms-1x 10 em/s x10° stokes 1 x 106 centistokes Density of air-1.85 lb/t, 1.24 kg/m Density of water 1 g/cu.cm, 1 g/ml, 1 kg/L. 1000 kg/m, or 62.4 lb/ f Particle Size Distribution Average diameter 100 um 75 um 50 um 30 um 20 um 10 um 5 um 2 um 1 um percent of total mass 3.6 5..3 17.9 18.1 27.5 16 6.9 2.1 1.2 0.5 um Use an aspect ratio of L 3w-3H If four trays are added to the unit, what improvement of particle removal will result? 2, Determine the area of an electrostatic precipitator required to remove 95% of the largest size particulates passing through the gravity settler Assume the dielectric constant of the particle is 6, the distance from wire to plate is 0.075meters, and the average voltage is 24kV. qe -1.6029 x 101 c charge on one electron & 8.85 x 10-12 C/N'm -permittivity of free space K 10 Boltzmann constant 1.3806503 x 10 m2kg sK

Explanation / Answer

For 85% removal, Diameter of particle allowed in air, dp = 50 um

Terminal velocity, ut,dp = 8*dp^2*(density of particle-density of air)/18/viscosity of air = 8*(2*10^-6)^2*(2200-1.24)/(1.7*10^-5)/18=.17 m/s

Settler velocity assumed, U =0.4m/s

H=ut,dp*t

t:residence time

W=Q/H/U

Q, volumetric flowrate of air = 5 cfs = 5*.33^3=.17m3/s

Given W=H

=>H=Q/H/U

=>W=H=(Q/U)^.5 = (.17/.4)^.5=.65 m

L=3H=3*.65=1.95 m

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.