5. A flocculation basin is to be used to treat a water flow of 0.15 m3/s. If the design detention time is 20 minutes, and the average velocity gradient is 36/s, what power input is required at 17°C?

Flocculation is a process in water treatment that involves the formation of aggregate from small particles in a suspension. A flocculation basin or tank is a place were flocculation is carried out.

Calculations:

The power input that is required in a flocculation tank can be obtained using the expression below;

P = μV[tex]G^{2}[/tex] ................................................1

Where P is the power input required;

μ is the dynamic viscosity of water at a given temperature;

V is the volume of the basin;

G is the average velocity gradient of the basin.

These parameters are given expect the volume which can be gotten with the expression in equation 2;

V = Detention time x flow rate ..............................2

Given

Detention time = 20 minutes = 1200 secs

Flow rate = 0.15 [tex]m^{3}[/tex]/s

V = 1200 secs x 0.15 [tex]m^{3}[/tex]/s

V = 180 [tex]m^{3}[/tex]

Now our parameters are complete;

μ at [tex]17^{0}[/tex]C = 0.001081 kg/ (m.s) (The table attached shows the dynamic viscosity of liquid water at different temperatures)

V = 180 [tex]m^{3}[/tex]

G = 36 /s

Using equation 1 the power input can be expressed as;

P = 0.001081 kg/ (m.s) x 180 [tex]m^{3}[/tex] x [tex](36s)^{2}[/tex]

P = 252.18 [tex]\frac{kg.m^{2} }{s^{3} } =\frac{kg.m }{s^{2} } .(\frac{m}{s} )= N.\frac{m}{s} =\frac{J}{s} = W[/tex]

P = 252.18 W

P = 0.2522 kW