Question

Hello. Please can you help me solve this. It's a wastewater problem.

matenalslu hells Lembo CHAPTER 9 TABLE 9.1 Percent Suspended Solids Removal at Given 7,脚// TIME (min) 2ft 4ft 6ft 8 ft 0ft 20 34 3 70 56 85 66 59 the 20 and 50, 50 and 60, ana atio min (0.27 hr) Data showed an increas 20(6.7/10) + (13/10(60 SOLI nkingWajer treatmen n toSefye Parhces eft The Setting" Velt ty 입..the Somputed for the 30, 4 of the fraction reme Also, a plot of chcles in the wate。 (s linch/min hhhot percentage ef picles willbe emovred w rate is 1420 gal al/day-ft' The r 2,000,000 gal d mep Water. The Plaat tises a water The d day 923 (2167) for sta removed by yfeWdimentaton tank.? 50, and 60% Curves ED Rr(%)

Answer 1

Q = 2 MGD = 2 times 10^6 Gallons /day Q = 268000 ft^3/day settling velocity (v_s) = 1 inch /min surface loading rate (v_0) = Q/surface area = 268000/pi/4 times (40)^2 ft^3/day/ft^2 v_0 = 213.267 ft/day therefore v_0 = 1.777 inch/min therefore percentage of particles removed by sedimentation tank(eta) eta = v_s/v_0 times 100 = 1 inch/min/1.777 inch/min times 100 eta = 56.27%