Table 6
Comprehensive design of trickling filters
| Influent BOD = 150 mg/L Design flow = 50 m3/day Recirculation 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 m3 A = V/D BOD loading = 5.2/2.3 = 2.2 kg BOD/m3·d Hydraulic loading = (1 + 0.5) × 50/1440 × 0.8 = 0.065 m3/m2·min Assume, depth of TF = 2.9 m  → Diameter = 1 mSo, to cover the whole flow of 100 m3/d: two TFs have been installed TF1 = 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 m3/day Recirculation 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 m3 A = V/D BOD loading = 5.2/2.3 = 2.2 kg BOD/m3·d Hydraulic loading = (1 + 0.5) × 50/1440 × 0.8 = 0.065 m3/m2·min Assume, depth of TF = 2.9 m → Diameter = 1 mSo, to cover the whole flow of 100 m3/d: two TFs have been installed TF1 = dia. 1 m with depth 2.9 m, TF2 = dia. 1 m with depth 2.3 m |