The town of Clearview has a conventional water treatment plant that uses ferric sulphate (Fe2(SO4)3) as its coagulant for turbidity and NOM removal. The coagulant is added in the form of a 20% solution (20% Fe2(SO4)3+ 80% water), and the mass of Fe2(SO4)3 added equals = (QA * 30 g Fe2(SO4)3 /m?). Assume that the solids generated by the coagulation/flocculation process equal 115% of those predicted by the Fe (OH)3 precipitate equation: Fe2(SO4)3 + 3 Ca(HCO3), 2 Fe(OH)3 + 3 CaSO4 +6CO2 Assume the 3 Ca(HCO3)2 is available (i.e., from the hardness and alkalinity in the raw water), the CO, is dissolved in the water and the CaSO4 produced is in the form of separate dissolved ions (Ca+2 and SO42) The above dosage includes the Fe2(SO4)3 consumed in side reactions with solids, turbidity and NOM. Small rapid mixing tank Coagulant (C) Qc = 2 0.2 kg Fe (SO.)/kg total TSS=0 g/m p=1020 kg/m Flocculation Clarifier Clarifier Effluent (CE) QCE = ? TSS=0 g/m Sand Filter Feed (A) QA= 1000 mld Turbidity = 30 NTU TSS=5 g/m VSS=0 g/m NOM= 15 g TOC/m Alk. 65 g CaCom pH=6.5 Fe=0g Fe/m Sludge (S) Qs = 2 TSS- 1% p=1050 kg/m (100% Solids capture) Filter Backwash (FB) Qis = 50 mld Mass TSS-10%FSS formed Gravity Thickener (85% Solids capture) Recycled Supernatant (R) Filter Effluent (FE) QFC = 2 TSS=0g/m Turbidity= 1 NTU NOM=9g TOC/m Fe=Og Fe/m Thickened Sludge (G) Q. = 2 TSS=5% p=1080 kg/m