**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

Gt: 2
x 10^{4} – 6 x 10^{4}

Paddle tip speed (v_{p}): 0.25
– 0.75 m/s

Paddle area (A_{p})/Tank
section area (A_{T}): 10:100
to 20:100

Coefficient of drag on paddle blade (C_{D}): 1.8

Maximum width of each paddle (b): 0.50 m

Kinematic viscosity _{}: 1.003
x 10^{-6} m^{2}/s

Dynamic viscosity of water _{}: 1.002
x 10^{-3} N.s/m^{2}

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 m^{3}.

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

Tank
cross sectional area (A_{cs}) = _{}

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 (D_{p}) be 4 m.

Let
the paddle tip speed (v_{p}) be 0.40 m/s

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

_{}, or, _{}

or, A_{p} = 6.50 m^{2}

Tank sectional area = (L.D) = 5.(7)
= 35 m^{2 }

_{}, 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.