Influent BOD = 150 mg/LDesign flow = 50 m3/dayRecirculation ratio = 0.5:1BOD removal efficiency in primary sedimentation = 30%Recirculation factor = F = 1 + R/(1 + R/10)2Primary effluent BOD = (1–0.3) × 150 = 105 mg/LBOD loading = W = 105 × 50 × 1/1,000 = 5.2 kg BOD/dayRemoval efficiency = E = 65%E = 100/1 + 0.4432 By putting all values, we get V = 2.3 m3 A = V/DBOD loading = 5.2/2.3 = 2.2 kg BOD/m3·dHydraulic loading = (1 + 0.5) × 50/1440 × 0.8 = 0.065 m3/m2·minAssume, depth of TF = 2.9 m → Diameter = 1 mSo, to cover the whole flow of 100 m3/d: two TFs have been installedTF1 = dia. 1 m with depth 2.9 m, TF2 = dia. 1 m with depth 2.3 m
 Influent BOD = 150 mg/LDesign flow = 50 m3/dayRecirculation ratio = 0.5:1BOD removal efficiency in primary sedimentation = 30%Recirculation factor = F = 1 + R/(1 + R/10)2Primary effluent BOD = (1–0.3) × 150 = 105 mg/LBOD loading = W = 105 × 50 × 1/1,000 = 5.2 kg BOD/dayRemoval efficiency = E = 65%E = 100/1 + 0.4432 By putting all values, we get V = 2.3 m3 A = V/DBOD loading = 5.2/2.3 = 2.2 kg BOD/m3·dHydraulic loading = (1 + 0.5) × 50/1440 × 0.8 = 0.065 m3/m2·minAssume, depth of TF = 2.9 m → Diameter = 1 mSo, to cover the whole flow of 100 m3/d: two TFs have been installedTF1 = dia. 1 m with depth 2.9 m, TF2 = dia. 1 m with depth 2.3 m