Coagulation and Flocculation-II

SOLUTION

 

Design a conventional rectangular horizontal-shaft flocculation tank unit for 10 MLD of settled raw water after coagulant addition and rapid mixing as per design parameters given below:

 

Detention time (t): 10 30 minutes

Velocity gradient (G): 20 75 /s

Gt: 2 x 104 6 x 104

Tank Depth (D): < 5 m

Paddle tip speed (vp): 0.25 0.75 m/s

Velocity of paddle relative to water (v): 0.75 x paddle tip speed

Paddle area (Ap)/Tank section area (AT): 10:100 to 20:100

Coefficient of drag on paddle blade (CD): 1.8

Maximum length of each paddle (l): 5.0 m

Maximum width of each paddle (b): 0.50 m

Kinematic viscosity : 1.003 x 10-6 m2/s

Dynamic viscosity of water : 1.002 x 10-3 N.s/m2

Freeboard: 0.50 m

 

Draw a net sketch of the designed tank (top and front view) clearly showing tank dimensions, paddle shaft position, paddle blade dimensions, water level, etc. Also mention paddle rotation speed and power requirement.

 

Solution:

 

Let the detention time (t) be 25 minutes

Let the velocity gradient (G) be 30 /s

Therefore G.t = 25.(60).(50) = 45000, i.e., within the 20000 60000 limit for Gt.

Volume of the tank (V) = , say 175 m3.

Let the tank depth (D) = 5 m, freeboard = 0.50 m, Total tank depth = 5.5 m

Tank cross sectional area (Acs) =

Let the tank length (L) be 7 m

Therefore, tank width (B) =

Let the paddle be placed length-wise.

Let the diameter of the paddle (Dp) be 4 m.

Let the paddle tip speed (vp) be 0.40 m/s

Then the velocity of paddle relative to water (v) = 0.75.(0.40) = 0.30 m/s

, or,

or, Ap = 6.50 m2

Tank sectional area = (L.D) = 5.(7) = 35 m2

 

, which is between 0.10 and 0.20, hence okay.

 

Let five paddles be provided. Therefore, area of each paddle =

Let the length of each paddle (l) be 5 m

Therefore, breadth of each paddle (b) =

Paddle rotation speed (w, radians/s) = radians/s,

i.e., revolutions per minutes

Power requirement is given by, , or,

or, P = 157 Watts, i.e., provide 0.2 KW motor for driving the impeller at 2 rpm.