Question

A 10 million litres per day (MLD) sewage treatment plant (STP) is based on the Activated Sludge Process (ASP). First, the sewage undergoes primary treatment and the resulting treated sewage has BOD$_5$ of 140 mg/L concentration. This is further passed through a 1500 m\(^3\) capacity aeration tank (in ASP), where the mixed liquor volatile suspended solids (MLVSS) concentration is maintained at 3000 mg/L. The concentration of BOD$_5$ of the treated sewage is 5 mg/L. The Food to Microorganisms ratio (F/M) of the ASP is _______ day\(^{-1}\) (rounded off to two decimal places).

Hint:

The F/M ratio is crucial in wastewater treatment as it determines the microbial activity and efficiency of the biological treatment process.

Answer 1

Correct Answer: 0.25

The Food to Microorganisms ratio (F/M) is given by the formula: \[ \frac{F}{M} = \frac{\text{BOD}_5 \times \text{Flow rate}}{\text{MLVSS} \times \text{Volume of aeration tank}}. \] Substitute the given values: \[ \frac{F}{M} = \frac{140 \times 10^3 \times 10}{3000 \times 1500}. \] After calculations, we find: \[ \frac{F}{M} = 0.25 \, \text{day}^{-1}. \] Thus, the value of \( F/M \) is \( 0.25 \).