7-5. Calculate the daily volume of sludge produced in Problem 7-1 if the water c
ID: 1885093 • Letter: 7
Question
7-5. Calculate the daily volume of sludge produced in Problem 7-1 if the water contains 12 mg/l of suspended solids and it is assumed that all of the alum and suspended solids are removed during settling. Assume that the sludge settles to 1.9 % solids and that the dry solids have a specific gravity of 2.1.
7-1. Jar test results indicate that an alum [Al2(SO4)3 14H2O] dosage of 20 mg/l will be required to treat a raw water which has an alkalinity of 50 mg/l as CaCO3. Calculate the pounds per day of alum that must be fed to treat 5 MGD of water. ANSWER PROBLEM 7-1: 3800 pounds/day
Explanation / Answer
7-1) Given,
Dose of alum = 20 mg/l
Alkanility = 50 mg/l
Discharge = 5 MGD = 5 * 3.785 = 18.925 MLD (1 MGD = 3.785 MLD)
Total alkalinity = Discharge * Alkalinity = 18.925 * 50 = 946.25 kg/day
Total Alum = Discharge * Alum dose = 18.925 * 20 = 378.5 kg/day
Molecular wt. of alum = 666
Molecular wt. of alkalinity = 100
Alum required = (946.25 * 666) / (3*100) = 2100.675 kg/day
Alum that must be fed = 2100.675 - 378.5 = 1722.175 kg/day = 3800 pounds/day (1kg = 2.204 pounds)
7-5)
Mass of suspended solids = 18.925 * 12 = 227.1 kg/day
Sludge settles to 1.9% solids
dry solids specific gravity of 2.1
density = 20.6 kg/m3
Mass of solids in sludge = 0.019 * 227.1 kg/day = 4.315 kg/day
Volume of sludge = Mass / Density = 4.315 /20.6 = 0.20947 m3/day
Volume of sludge produced is 0.20947 m3/day
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