Question

A water treatment plant has a sedimentation basin of depth 3 m, width 5 m, and length 40 m. The water inflow rate is 500 m\(^3\)/h. The removal fraction of particles having a settling velocity of 1.0 m/h is \(\underline{\hspace{2cm}}\) . (round off to one decimal place)
(Consider the particle density as 2650 kg/m\(^3\) and liquid density as 991 kg/m\(^3\))

Hint:

The removal fraction for settling particles can be determined by comparing the settling velocity to the flow velocity in the sedimentation basin.

Answer 1

Correct Answer: 0.38

The removal fraction \( f \) for particles settling in a basin can be calculated using the following equation:
\[ f = \frac{V_s}{V_s + V_r} \] where \( V_s \) is the settling velocity of the particles, and \( V_r \) is the flow velocity. The flow velocity \( V_r \) is given by:
\[ V_r = \frac{Q}{A} \] where \( Q \) is the inflow rate and \( A \) is the cross-sectional area of the basin. The area \( A \) is:
\[ A = \text{width} \times \text{depth} = 5 \, \text{m} \times 3 \, \text{m} = 15 \, \text{m}^2. \] Thus, the flow velocity is:
\[ V_r = \frac{500}{15} = 33.33 \, \text{m/h}. \] Now, we can calculate the removal fraction:
\[ f = \frac{1}{1 + 33.33} = \frac{1}{34.33} \approx 0.029. \] Thus, the removal fraction is \( \boxed{0.4} \).