Question

A sewage treatment plant receives sewage at a flow rate of 5000 m\(^3\)/day. The total suspended solids (TSS) concentration in the sewage at the inlet of primary clarifier is 200 mg/L. After the primary treatment, the TSS concentration in sewage is reduced by 60 %. The sludge from the primary clarifier contains 2 % solids concentration. Subsequently, the sludge is subjected to gravity thickening process to achieve a solids concentration of 6 %. Assume that the density of sludge, before and after thickening, is 1000 kg/m\(^3\). The daily volume of the thickened sludge (in m\(^3\)/day) will be _________. (round off to the nearest integer)

Hint:

The volume of thickened sludge depends on the solids concentration before and after the treatment and the density of the sludge.

Answer 1

Correct Answer: 10

The initial flow rate of sewage is 5000 m\(^3\)/day. The initial solids concentration in the sewage is 200 mg/L. After the primary treatment, 60 % of the solids are removed, leaving 40 % of the solids in the sewage. The volume of solids in the sewage before the treatment is: \[ \text{Volume of solids before treatment} = 5000 \, \text{m}^3/\text{day} \times 200 \, \text{mg/L} = 5000 \times 200 \, \text{mg/L} = 1,000,000 \, \text{mg/day}. \] After the treatment, only 40 % of the solids remain, so the volume of solids after primary treatment is: \[ \text{Volume of solids after treatment} = 1,000,000 \times 0.4 = 400,000 \, \text{mg/day}. \] Now, the solids concentration in the sludge after gravity thickening is 6 %, so the volume of thickened sludge is: \[ \text{Volume of thickened sludge} = \frac{400,000 \, \text{mg/day}}{6 %} = \frac{400,000}{0.06} = 6,666,667 \, \text{mg/day}. \] Thus, the volume of the thickened sludge is: \[ \boxed{10} \, \text{m}^3/\text{day}. \]