Design of sedimentation tanks and sample problems

Problem
Maximum daily demand = 12 MLD (Million Litres a Day)
Velocity of flow = 20 cm/min
Detention period = 6 hours
Design a sedimentation tank for the above listed parameters

Solution
Quantity of water in 24 hours = 12000000 l
Detention period = 6 hours
Volume of water to be treated = (Quantity * Detention period) / 24
= 3000000 l = 3000 cu.m (1 cu. m = 1000 l)
Velocity = 20 cm/min = 0.2 m/min
Length of tank = Velocity * Detention period
= 0.2 * 6 *  60 = 72 m
Cross section of tank = Volume of water / length
= 3000 cu. m / 72 m
= 41.6 m2
Assuming depth = 4 m
Width = Cross sectional area / Depth
= 41.6 / 4
= approximately 10.5 m
Apply a free board of 0.5m to the depth
Overall dimensions = 72m (l), 10.5m (w), 4.5m (d)

Problem
Design a sedimentation tank for the following parameters
Daily demand = 9 000 000 l/d
Velocity of flow in sedimentation tank = 22 cm/min
Detention time = 8 h

Problem
Design a sedimentation tank for the following parameters
Water supplied = 1400000
Detention period = 5 h
Velocity of flow = 12 cm/min
Depth of water = 4m
Allowance for sludge = 80 cm


Problem
Design a sedimentation tank for the following parameters
Population = 65,000
Sewage flow = 210 LPCD
Raw sewage BOD = 210 mg/l
Suspended Solids in raw sewage = 300 mg/l
BOD removal in primary treatment = 40%
Overall BOD removal desired = 90%


Problem
Design a activated sludge treatment plant for the following parameters
Population = 50,000
Percapita sewage flow = 150 LPCD
Settled sewage BOD5 = 200 mg/l
Average flow = 22.5 MLD
Effluent BOD5 required = 10 mg/l

Problem
Design a rectangular sedimentation tank.
Water supply =  12 * 10^6 l/d
The tank is mechanically cleaned.
Missing data may be suitably assumed

Problem
Design a circular trickling filter to treat 5 * 10^6 litres of sewage per day
BOD = 150 mg/l