Question

A water filtration unit is made of uniform-size sand particles of 0.4 mm diameter with a shape factor of 0.84 and specific gravity of 2.55. The depth of the filter bed is 0.70 m and the porosity is 0.35. The filter bed is to be expanded to a porosity of 0.65 by hydraulic backwash. If the terminal settling velocity of sand particles during backwash is 4.5 cm/s, the required backwash velocity is

• A. \( 5.79 \times 10^{-3} \) m/s
• B. \( 6.35 \times 10^{-3} \) m/s
• C. 0.69 cm/s
• D. 0.75 cm/s

Hint:

For backwash operations in filtration systems, the backwash velocity depends on the terminal settling velocity of particles and the change in porosity.

Answer 1

The Correct Option is B

The required backwash velocity can be calculated using the relationship for settling velocity and the porosity change in the filter bed. We use the following relationship for backwash: \[ V_{\text{backwash}} = \frac{V_{\text{settling}} \cdot (\text{initial porosity} - \text{final porosity})}{\text{final porosity}}. \] Given values:
- Terminal settling velocity (\( V_{\text{settling}} \)) = 4.5 cm/s = 0.045 m/s
- Initial porosity = 0.35
- Final porosity = 0.65
Substituting the values, we get: \[ V_{\text{backwash}} = \frac{0.045 \times (0.35 - 0.65)}{0.65} = 6.35 \times 10^{-3} \, \text{m/s}. \] Thus, the required backwash velocity is \( 6.35 \times 10^{-3} \) m/s. Final Answer: \( 6.35 \times 10^{-3} \) m/s