Table 6

Influent BOD = 150 mg/L Design flow = 50 m ^{3}/dayRecirculation ratio = 0.5:1 BOD removal efficiency in primary sedimentation = 30% Recirculation factor = F = 1 + R/(1 + R/10) ^{2}Primary effluent BOD = (1–0.3) × 150 = 105 mg/L BOD loading = W = 105 × 50 × 1/1,000 = 5.2 kg BOD/day Removal efficiency = E = 65% E = 100/1 + 0.4432 By putting all values, we get V = 2.3 m ^{3} A = V/DBOD loading = 5.2/2.3 = 2.2 kg BOD/m ^{3}·dHydraulic loading = (1 + 0.5) × 50/1440 × 0.8 = 0.065 m ^{3}/m^{2}·minAssume, depth of TF = 2.9 m → Diameter = 1 m So, to cover the whole flow of 100 m ^{3}/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/L Design flow = 50 m ^{3}/dayRecirculation ratio = 0.5:1 BOD removal efficiency in primary sedimentation = 30% Recirculation factor = F = 1 + R/(1 + R/10) ^{2}Primary effluent BOD = (1–0.3) × 150 = 105 mg/L BOD loading = W = 105 × 50 × 1/1,000 = 5.2 kg BOD/day Removal efficiency = E = 65% E = 100/1 + 0.4432 By putting all values, we get V = 2.3 m ^{3} A = V/DBOD loading = 5.2/2.3 = 2.2 kg BOD/m ^{3}·dHydraulic loading = (1 + 0.5) × 50/1440 × 0.8 = 0.065 m ^{3}/m^{2}·minAssume, depth of TF = 2.9 m → Diameter = 1 m So, to cover the whole flow of 100 m ^{3}/d: two TFs have been installedTF1 = dia. 1 m with depth 2.9 m, TF2 = dia. 1 m with depth 2.3 m |

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