Question

A water treatment plant with an average flow of Q-1,0 m3/s treats its water with alum at a dose of 30 mg/L. Alum coagulation is used to remove particulate matter, reduce the concentration of organic matter, and reduce the alkalinity of the water according to Equation 10-1. If the organic matter concentration is reduced from 10 mg/L to 2 mg/L. determine the total mass of Al(OH)3 removed per day O 708 kad 652 kg/d 674 O 720 kg

Answer 1

pagetor Answer: According to the given information, ato find this total mass of Alcoh), Eremoved per day. Given, Q: Homols alum dose = 30 mild -- the al kabinety Psi Alla (soulz 't41.3$+ 6H603 = 2AllOH)2 +6CO2 + 14.3 H40 +3505 for kach mole gj alum ackded, six moles of alkalarity (bicarbonate-form) are removed. افرا15 Convert the alum dose into molar unids using of molecular weight of sau. 35 g/mol (30 ng] ) x (6° 31 pm) A sau:35 g/mol - 5.044 x165 mulla of alum. Thus, the 5.0u7x185 molla of alum added, the amount of Hcoz remove is simply 6 times that

page:02 -) 6X 5.0u1x105 molle 2 3.0285x10lu salh of allealinity.. * To find the tola! alkalinity suremoved flay, simply multiply above num by ang flow of plant. = 3:098 5xto4leculk) *1.0 (untles) x 1000(X1647) x 86400(8/day) = 26163. Gug solday. * converting to a mais baric, using the equivalent weight of brcarbonate to be 61 g/mol yiekis, € 26163'Gus Balday x 61 3 /mol : 26163. bus mokbay x61 g/mot § 1595982.528 3 day - 1595-982 kg/day. * for Alloh :- - thus, 5-Quy xeos molla of alum added, the amount of Alloh); remove is simply a times. = 2X5.0u7x16s molla = 1.6094X104 Earl

page:03 for getting oumoval on day basis, multiply it with average -flow day. = 1.0094 xcõleux 10 bome | $x10006198) x Bbwo(Sketay) = 8421.216 mol day - Conventing mest basis, using the molecular weight of Allotz) is a - 8121,216 mmol x789 day mol Geo asursuz glday 2 680x2ity/ckay. total mans of Allohla removed perday & 680:2 kglday. i