Question

g A 12,000 m3/day treatment plant has a rectangular sedimentation basin with dimensions 12 meters wide, 3 meters deep, and 25 meters long. Will particles with a settling velocity of 6 x 10-3 m/s be removed in this basin

Answer 1

Answer:

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.

Explanation:

Given;

volumetric flow rate of the treatment, Q₀ = 12,000 m³/day

length of the rectangular tank, L = 25 m

width of the tank, W = 12 m

height of the tank, H = 3 m

settling velocity of the particles, $V_s$ = 6 x 10⁻³ m/s

The overflow rate of the sediments are calculated as follows;

$V_o = \frac{Q_o}{A_s}$

where;

As is the surface area of the tank, m²

Q₀ is the flow rate, m³/s

As = 2LW + 2LH + 2WH

As = (2 x 25 x 12) + (2 x 25 x 3) + (2 x 12 x 3)

As = 822 m²

$Q_0 (m^3/s)= \frac{12,000 \ m^3}{day} \times \frac{1 \ day}{24 \ hr} \times \frac{1 \ hour}{60 \ \min} \times \frac{1 \ \min}{60 \ s} = \frac{12,000}{24 \times 60 \times 60} (m^3/s)= \frac{12,000}{86,400} \ m^3/s\\\\Q_o = 0.139 \ m^3/s$

The overflow rate;

$V_o = \frac{Q_0}{A_s} = \frac{0.139}{822} = 1.69 \times 10^{-4} \ m/s$

The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.