1. Application Design sludge storage silo Dr. Hasan Hamouda

2. Design sludge storage silo Design a sludge storage silo for one Year capacity Given are with following WWTP Data: Q In = 50000 m 3, C = 200 mg BOD 5 /l, Q R /Q 0 = 0.58  = 0.9 kg TS / kg BOD 5 SS removal in PS = 60% BOD 5 removal in PS = 30% TS Effluence = 10 mg/l TS Biology = 4 g/l TS max from Thickening tank = 4% TS max from Pressing machine = 18% TS of final dewatering process = 36%

3. Sludge treatment Dewatering Anaerobic Stabilization Landfill agricultural recycling thermal recycling Sludge

4. Thickening Tank Pressing Liming TS~18% TS~4% Silo TS~36% Polymer Dosage Coagulation Flocculation

5. Dewaterd Sludge 36% Excessive Sludge 4% Lime 18% H 2 O 64% Pressed Sludge 18% Sludge 4% H 2 O 96%

7. wastewater treatment plants-treated sludge

9. Solution Load in Influence = 50000 m 3 /d X 200 mg/l = 10000 kg BOD 5, From BOD 5 removal in PS = 30%, Load after PS 10000 kg BOD 5 X 0.7 = 7000 kg BOD 5 From s = 0.9 kg TS / kg BOD 5, calculate the m Excess = 7000 X 0.9 = 6300 kg TS/d From SS removal in PS = 60%, calculate the m Primary sludge = 10000 X 0.6 = 6000 kg TS/d From TS Eff = 10 mg/l, calculate the m Excess, Eff = 50000 X 10 mg/l = 500 kg TS/d (flow in the per flooder) S M total = S m Excess + m Primary sludge + m Excess, Eff = (6300 + 6000 + 500) kg TS/d = 12800 kg TS/d s total = 12800 kg TS/d /10000 kg BOD 5 /d = 1.28 kg TS / kg BOD 5,elimenated S m total, to Dewatering = S m Excess + m Primary sludge = (6300 + 6000 ) kg TS/d = 12300 kg TS/d Assumption: TS primary = 20 g/l and from the Q R /Q 0 Ratio calculate TS R = 10 g/l (for example) So TS to the thickening tank = (20+10)/2 = 15 g/l, Q sludge from the thickening tank = (12300 kg TS/d) / (15 kg/m 3 ) = 820 m 3 /d Q sludge to the Dewatering machine = (12300 kg TS/d) / (40 kg/m 3 ) = 307.5 m 3 /d Q sludge to the liming machine = (12300 kg TS/d) / (180 kg/m 3 ) = 68.3 m 3 /d Q sludge to the sludge Silo = (12300 kg TS/d + 12300 kg TS( lime)/d ) / (360 kg/m 3 ) = 68.3 m 3 /d V silo (for one year) = 68.3 m 3 /d X365 d = 25000 m 3