Question

DESIGN SECONDARY SEDIMEMTATION TANK TO FIND OUT THE DIAMETER THE DEPTH AND NUMBER OF TANKS FOLLOWING INFORMATION:

MLSS = 2600-2900
DIAMETER = 3-50M
WATER DEPTH = 3-5M
40-50% OF WASTERWATER WILL BE RECIRCULATED TO THE AERATION TANK

Answer 1

Secondary sedimentation basin generally follows Zone or Hindered settling, for the design of such a system the column test results must be presented. Hence here the basic sedimentation chamber with recirculation is designed.

let us assume a design inflow, Qi = 1MLD

MLSS = 2600-2900, that is let X = (2600+2900)/2 = 2750 mg/l

Qr = 50% of Q

therefore, total Q = 1.5Q = 1.5MLD

surface overflow rate for secondary clarifiers are in the range of 16-24 m³/m²-d

SOR = 20 m³/m²-d

Total Area required for the tank, A=QSOR

A- SOR

A=1.5*106*10-320=75 m2

1.5 10610-3 = 75 m2 20

A = 75m²

Diameter of one tank be 4m

Therefore, area of one tank, a=π*424=12.56 m2

π*4 12.56m2 а

a = 12.56 m²

number of tanks required, n = A/a = 5.97

approximately provide 6 tanks of 4m diameter

detention time of secondary clarifier is in the range of 90-120minutes

let dt = 90minutes

volume required, V = Q*dt

V = 93.75m³

Depth of each tank D

But volume of each tank is v = D * a

V = 6* D*a

D = V/(6*a)

D = 1.244 m