Question

The Surface Overflow Rate (SOR) in a rectangular sedimentation tank is 45 m\(^3\)/m\(^2\)·day. Minimum diameters of spherical inorganic and organic particles expected to be completely removed in this tank are calculated. Assume that Stoke’s law is applicable. Which of the following options is/are correct:

• A. Minimum diameter of inorganic particles is 24 $\mu m$
• B. Minimum diameter of organic particles is 69 $\mu m$
• C. Minimum diameter of inorganic particles is 15 $\mu m$
• D. Minimum diameter of organic particles is 55 $\mu m$

Hint:

When calculating the minimum diameter of particles to be removed in sedimentation tanks, use Stoke’s law with the correct values for specific gravity, viscosity, and gravity.

Answer 1

The Correct Option is A, B

Given:
- Surface Overflow Rate (SOR) \( V_0 = 45 \, {m}^3/{m}^2{day} = \frac{45}{86400} \, {m/sec} \),
- Kinematic viscosity \( \nu = 1 \times 10^{-6} \, {m}^2/{s} \),
- Gravity \( g = 9.81 \, {m/s}^2 \),
- Specific gravity of inorganic particles \( G = 2.65 \),
- Specific gravity of organic particles \( G = 1.20 \).
By using Stoke's Law for the settling velocity, the diameter of the particles is calculated as:
For inorganic particles:
\[ V_s = \frac{g}{18\nu} (G_s - 1) d^2, \] \[ d_{{inorganic}} = \sqrt{\frac{18 \times 1 \times 10^{-6} \times 45}{86400 \times 9.81 \times (2.65 - 1)}} = 24 \, \mu m. \] For organic particles: \[ V_s = \frac{g}{180 \nu} (G_s - 1) d^2, \] \[ d_{{organic}} = \sqrt{\frac{18 \times 1 \times 10^{-6} \times 45}{86400 \times 9.81 \times (1.20 - 1)}} = 69 \, \mu m. \] Thus, the minimum diameter of inorganic particles is 24 $\mu m$ and the minimum diameter of organic particles is 69 $\mu m$, so the correct answers are (A) and (B).