Question

A circular primary clarifier processes an average flow of 5005 m\(^3\)/day of municipal wastewater. The outflow rate is 35 m\(^3\)/m\(^2\)·d. The diameter of clarifier shall be ........

• A. 10.5 m
• B. 11.5 m
• C. 12.5 m
• D. 13.5 m

Hint:

In clarifier design, surface area is determined by dividing flow rate by surface loading rate. Use \( A = \frac{\pi D^2}{4} \) for circular tanks to get the diameter.

Answer 1

The Correct Option is D

Given:
- Flow rate \( Q = 5005 \, \text{m}^3/\text{day} \)
- Surface loading rate \( SLR = 35 \, \text{m}^3/\text{m}^2/\text{day} \)
The required surface area of the clarifier is: \[ A = \frac{Q}{SLR} = \frac{5005}{35} = 143 \, \text{m}^2 \] Since it's a circular clarifier: \[ A = \frac{\pi D^2}{4} ⇒ D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 143}{\pi}} \approx \sqrt{182.06} \approx 13.49 \, \text{m} \] So, the diameter of the clarifier is approximately 13.5 m.