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,
= 6 x 10⁻³ m/s
The overflow rate of the sediments are calculated as follows;
![V_o = \frac{Q_o}{A_s}](https://tex.z-dn.net/?f=V_o%20%3D%20%5Cfrac%7BQ_o%7D%7BA_s%7D)
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](https://tex.z-dn.net/?f=Q_0%20%28m%5E3%2Fs%29%3D%20%5Cfrac%7B12%2C000%20%5C%20m%5E3%7D%7Bday%7D%20%5Ctimes%20%5Cfrac%7B1%20%5C%20day%7D%7B24%20%5C%20hr%7D%20%5Ctimes%20%5Cfrac%7B1%20%5C%20hour%7D%7B60%20%5C%20%5Cmin%7D%20%5Ctimes%20%5Cfrac%7B1%20%5C%20%5Cmin%7D%7B60%20%5C%20s%7D%20%3D%20%5Cfrac%7B12%2C000%7D%7B24%20%5Ctimes%2060%20%5Ctimes%2060%7D%20%20%28m%5E3%2Fs%29%3D%20%5Cfrac%7B12%2C000%7D%7B86%2C400%7D%20%5C%20m%5E3%2Fs%5C%5C%5C%5CQ_o%20%3D%200.139%20%5C%20m%5E3%2Fs)
The overflow rate;
![V_o = \frac{Q_0}{A_s} = \frac{0.139}{822} = 1.69 \times 10^{-4} \ m/s](https://tex.z-dn.net/?f=V_o%20%3D%20%5Cfrac%7BQ_0%7D%7BA_s%7D%20%3D%20%5Cfrac%7B0.139%7D%7B822%7D%20%3D%201.69%20%5Ctimes%2010%5E%7B-4%7D%20%5C%20m%2Fs)
The settling velocity of the particles (Vs) is greater than the overflow rate (V₀), thus the particles will settle out and will be removed.