Solution :

Data : Daily demand = 6 MLD, Detention time = 4 hours. Flow through velocity = 20 cm/min.

1. Water to be treated per hour =$\frac{6*10^6}{24}$= 0.25 x 106 lit/hr.

2. Capacity of sedimentation tank = Hourly demand x D.T = 0.25 x 1$0^6$ x4=1$0^6$lit=1000$m^3$

3. Length of the tank = Flow velocity x D.T= 0.2 x 4 x 60 = 48 m

4 Assume width of the tank = 10 m

5.Depth of the tank = $\frac{Capacity of the tank}{Length *Width}$=$\frac{1000}{48 *10}$= 2.58m

Assuming free board of 0.42 m, provide overall depth=2.5m

1. Tank dimension 48 m long * 10 m wide *2.5m deep.

2. Over flow rate=$\frac{Discharge rate per hour}{Surface area}$=$\frac{0.25*10^6}{48*10}$=520.83lit/m^2/hr