Problem No.1 Alum Coagulation An alum dosage of 125mg/l is required to treat 50MGD of raw water in a water treatment plant. The influent suspended solids concentration of the raw water is 15mg/L. Calculate the following: a) The gallons per day of wet sludge which is concentrated to 2% by weight. (Assuming specific gravity is 2). b) The un-hydrated lime so that no alkalinity is consumed. The chemical reactions involved are: Al2(SO4)314.2H2O+6(HCO3)2Al(OH)3+6CO2+3SO42+14.3H2OCaO+H2OCa(OH)2,1gallon=3.785L,1m3=103L Problem No.2 Ferrous Coagulation: Surface water is coagulated with a dosage of 30mg/L of ferrous sulfate and an equivalent dosage of lime. a) How many Kg of ferrous sulfate are needed per million m3 of water treated?. b) How many Kg of hydrated lime are required assuming a purity of 70%CaO ?, c) How many Kg of Fe(OH)2 sludge are produced per million m3 of water treated [2 Problem No.3 Ferrous Coagulation: Alum reacts with alkalinity as per the following chemical equation. For removing 90% of 300mg/L suspended solids, say 50mg/L Alum (for pH8 solution) is required. If solution also has bicarbonate alkalinity (250mg/L as CaCO3), a) How much Alum would be required for getting 90% removal of suspended solids? b) How much solid waste (in terms of aluminum hydroxide precipitate 2Al(OH)3 and settled suspended solids) would be produced? Al2(SO4)318H2O+3Ca(HCO3)2Al(OH)3+6CO2+3CaSO4+18H2O