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, [tex]V_s[/tex] = 6 x 10⁻³ m/s
The overflow rate of the sediments are calculated as follows;
[tex]V_o = \frac{Q_o}{A_s}[/tex]
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
[tex]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[/tex]
The overflow rate;
[tex]V_o = \frac{Q_0}{A_s} = \frac{0.139}{822} = 1.69 \times 10^{-4} \ m/s[/tex]
The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.
