Question

A secondary clarifier handles a total flow of 9600 m\(^3\)/d from the aeration tank of a conventional activated-sludge treatment system. The concentration of solids in the flow from the aeration tank is 3000 mg/L. The clarifier is required to thicken the solids to 12000 mg/L, and hence it is to be designed for a solid flux of \( 3.2 \, \frac{\text{kg}}{\text{m}^2 \cdot \text{h}} \). The surface area of the designed clarifier for thickening (in m\(^2\), in integer) is \(\underline{\hspace{1cm}}\).

Hint:

The surface area of the clarifier can be calculated using the solid flux and the mass of solids per unit time.

Answer 1

Correct Answer: 375

The flow rate \( Q \) is 9600 m\(^3\)/d, and the concentration of solids in the flow from the aeration tank is 3000 mg/L. To convert the flow to kg/h, we use the following: \[ Q = 9600 \, \text{m}^3/\text{d} = \frac{9600}{24} \, \text{m}^3/\text{h} = 400 \, \text{m}^3/\text{h} \] Now, the mass of solids in the flow is: \[ \text{Mass of solids} = 400 \, \text{m}^3/\text{h} \times 3000 \, \text{mg/L} = 400 \times 3000 = 1200000 \, \text{mg/h} = 1200 \, \text{kg/h} \] The solid flux \( J \) is given by: \[ J = \frac{\text{Mass of solids}}{\text{Area of clarifier}} = 3.2 \, \frac{\text{kg}}{\text{m}^2 \cdot \text{h}} \] Thus, the area of the clarifier is: \[ \text{Area} = \frac{1200}{3.2} = 375 \, \text{m}^2 \] Thus, the surface area of the designed clarifier for thickening is \( \boxed{375} \, \text{m}^2 \).